VIT Target VITEEE 2019 Past 13 Years 2018-2006 Solved Papers + 10 Mock Tests 8th Edition Engineering Entrance also useful for IITJEE IIT-JEE IIT JEE main advanced Disha

Table of contents :
Contents......Page 3
VITEEE - 2018......Page 5
VITEEE - 2017......Page 29
VITEEE - 2016......Page 61
VITEEE - 2015......Page 87
VITEEE - 2014......Page 117
VITEEE - 2013......Page 141
VITEEE - 2012......Page 171
VITEEE - 2011......Page 199
VITEEE - 2010......Page 219
VITEEE - 2009......Page 245
VITEEE - 2008......Page 275
VITEEE - 2007......Page 303
VITEEE - 2006......Page 331
Mock Test Paper 1......Page 355
Mock Test Paper 2......Page 367
Mock Test Paper 3......Page 378
Mock Test Paper 4......Page 389
Mock Test Paper 5......Page 400
Mock Test Paper 6......Page 410
Mock Test Paper 7......Page 422
Mock Test Paper 8......Page 434
Mock Test Paper 9......Page 445
Mock Test Paper 10......Page 457
Mock Test Paper Solutions (1-10)......Page 469

Citation preview

EBD_7443

EBD_7443

EBD_7443

EBD_7443

EBD_7443

EBD_7443

EBD_7443

EBD_7443

EBD_7443

EBD_7443

EBD_7443

EBD_7443

EBD_7443

EBD_7443

VITEEE

SOLVED PAPER

2017

(memory based)

GENERAL INSTRUCTIONS

•

• • • •

This question paper contains total 125 questions divided into four parts : Part I : Physics Q. No - 1 to 40 Part II : Chemistry Q. No - 41 to 80 Part III : Mathematics Q. No - 81 to 120 Part IV : English Q. No - 121 to 125 All questions are multiple choice questions with four options, only one of them is correct. For each correct response, the candidate will get 1 mark. There is no negative marking for the wrong answer. The test is of 2½ hours duration.

PART - I (PHYSICS) 1.

2.

A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 m/s. To give an initial upward acceleration of 20 m/s2, the amount of gas ejected per second to supply the needed thrust will be (Take g = 10 m/s2) (a) 127.5 kg/s (b) 137.5 kg/s (c) 155.5 kg/s (d) 187.5 kg/s The power dissipated in the circuit shown in the figure is 30 Watts. The value of R is

4.

5.

6. 5W 10V

3.

(a)

l = h 2mE

(c)

l=

h 2mE

(b) l = (d) l =

2mE h hE

2mE

(c) 1: 3 (d) 3 :1 In which sequence the radioactive radiations are emitted in the following nuclear reaction? A ZX

R

(a) 20 W (b) 15 W (c) 10 W (d) 30 W If the kinetic energy of a moving particle is E, then the de-Broglie wavelength is

Two bodies A and B having masses in the ratio of 3 : 1 possess the same kinetic energy. The ratio of linear momentum of B to A is (a) 1 : 3 (b) 3 : 1

7.

¾¾ ®

A Z + 1Y KA–4

¾¾ ®

A–4 Z–1K

¾¾ ®Z– 1 (a) g, a, b (b) a, b, g (c) b, g, a (d) b, a, g Which of the following does not support the wave nature of light? (a) Interference (b) Diffraction (c) Polarisation (d) Photoelectric effect Six identical conducting rods are joined as shown in figure. Points A and D are maintained at 200°C and 20°C respectively. The temperature of junction B will be

A

B

C

D

EBD_7443 2017-2

8.

9.

10.

Target VITEEE (a) 120°C (b) 100°C (c) 140°C (d) 80°C A hydrogen atom is in ground state. Then to get six lines in emission spectrum, wavelength of incident radiation should be (a) 800 Å (b) 825 Å (c) 975 Å (d) 1025 Å A conducting circular loop of radius r carries a constant current i. It is placed in a uniform r r magnetic field B0 such that B0 is perpendicular to the plane of the loop. The magnetic force acting on the loop is (a) ir B0 (b) 2p ir B0 (c) zero (d) p ir B0 A vessel of depth 2d cm is half filled with a liquid of refractive index m1 and the upper half with a liquid of refractive index m2. The apparent depth of the vessel seen perpendicularly is (a)

æ m1 m 2 ç çm +m 2 è 1

ö ÷d ÷ ø

13.

14.

15.

16.

æ 1 1 ö (b) çç m + m ÷÷ d 2ø è 1

æ 1 æ 1 ö 1 ö ç ç ÷ ÷ ç m + m ÷ 2d (d) ç m m ÷ 2d 2ø è 1 è 1 2ø A smooth sphere of mass M moving with velocity u directly collides elastically with another sphere of mass m at rest. After collision, their final velocities are V and v respectively. The value of v is

(c)

11.

(a)

2uM m

(b)

5000 p The combination of gates shown below yields

(c) 500 p 17.

2um M

2u 2u (d) M m 1+ 1+ m M Two capacitors C1 and C2 in a circuit are joined as shown in figure. The potentials of points A and B are V1 and V2 respectively. Then the potential of point D will be A B D V1 V2 C2 C1

C 2 V1 + C1V2 C1 + C 2

(a)

(V1 + V2 ) 2

(b)

(c)

C1V1 + C 2 V2 C1 + C 2

(d) C 2 V1 + C1V2 C1 + C 2

(d)

A X B

(c)

12.

Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is: (a) < 2.8 × 10-9 m (b) ³ 2.8 × 10-9 m (c) £ 2.8 × 10-12 m (d) < 2.8 × 10-10 m Kerosene oil rises up in a wick of a lantern because of (a) diffusion of the oil through the wick (b) capillary action (c) buoyant force of air (d) the gravitational pull of the wick The current in a coil of L = 40 mH is to be increased uniformly from 1A to 11A in 4 milli sec. The induced e.m.f. will be (a) 100 V (b) 0.4 V (c) 440 V (d) 40 V An alternating voltage of 220 V, 50 Hz frequency is applied across a capacitor of capacitance 2 µF. The impedence of the circuit is 1000 p (a) (b) p 5000

18.

19.

(a) OR gate (b) NOT gate (c) XOR gate (d) NAND gate A hollow insulated conduction sphere is given a positive charge of 10 mC. What will be the electric field at the centre of the sphere if its radius is 2 metres? (a) Zero (b) 5 mCm–2 –2 (c) 20 mCm (d) 8 mCm–2 Two mercury drops (each of radius r) merge to form a bigger drop. The surface energy of the bigger drop, if T is the surface tension, is (a) 25/3 pr2T (b) 4 pr2T 2 (c) 2 pr T (d) 28/3 pr2T

Solved Paper 2017 20.

21.

22.

2017-3

Resistances 1 W, 2 W and 3 W are connected to form a triangle. If a 1.5 V cell of negligible internal resistance is connected across the 3 W resistor, the current flowing through this resistor will be (a) 0.25 A (b) 0.5 A (c) 1.0 A (d) 1.5 A A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes (a) inclined at 45° to the magnetic field (b) inclined at any arbitrary angle to the magnetic field (c) parallel to the magnetic field (d) perpendicular to the magnetic field The value of tan (90° – q) in the graph gives

25.

26.

27.

Strain

28.

29. q

A milli voltmeter of 25 milli volt range is to be converted into an ammeter of 25 ampere range. The value (in ohm) of necessary shunt will be (a) 0.001 (b) 0.01 (c) 1 (d) 0.05 In young’s double-slit experiment, the intensity of light at a point on the screen where the path difference is l is I, l being the wavelength of light used. The intensity at a point where the l path difference is will be 4 I I (a) (b) 2 4 (c) I (d) zero Which of the following is a self adjusting force? (a) Static friction (b) Limiting friction (c) Dynamic friction (d) Sliding friction Which of the following are not electromagnetic waves? (a) Cosmic rays (b) Gamma rays (c) b-rays (d) X-rays Graph of specific heat at constant volume for a monatomic gas is Y

Stress

23.

24.

(a) Young's modulus of elasticity (b) compressibility (c) shear strain (d) tensile strength An electron makes a transition from an excited state to the ground state of a hydrogen - like atom. Then (a) kinetic energy decreases, potential energy increases but total energy remains same (b) kinetic energy and total energy decrease but potential energy increases (c) its kinetic energy increases but potential energy and total energy decrease (d) kinetic energy, potential energy and total energy decrease An A.C. source is connected to a resistive circuit. Which of the following is true? (a) Current leads ahead of voltage in phase (b) Current lags behind voltage in phase (c) Current and voltage are in same phase (d) Any of the above may be true depending upon the value of resistance.

Y 3R

(a)

(b)

cv O

T

Y

(c)

O

T

X

(d) X

cv O

T

X

A charge +q is at a distance L/2 above a square of side L. Then what is the flux linked with the surface? (a)

q 4e 0

2q (b) 3e 0

q 6q (d) e 6e 0 0 The potential energy of a system increases if work is done (a) upon the system by a non conservative force (b) by the system against a conservative force (c) by the system against a non conservative force (d) upon the system by a conservative force

(c) 31.

T

cv

Y

3 —R 2

cv O

30.

X

EBD_7443 2017-4

32.

33.

34.

35.

36.

Target VITEEE Two capacitors when connected in series have a capacitance of 3 mF, and when connected in parallel have a capacitance of 16 mF. Their individual capacities are (a) 1 mF, 2 mF (b) 6 mF, 2 mF (c) 12 mF, 4 mF (d) 3 mF, 16 mF Resonance frequency of LCR series a.c. circuit is f0. Now the capacitance is made 4 times, then the new resonance frequency will become (a) f0/4 (b) 2f0 (c) f0 (d) f0/2. If the light is polarised by reflection, then the angle between reflected and refracted light is (a) 180º (b) 90º (c) 45º (d) 36º The velocity of efflux of a liquid through an orifice in the bottom of the tank does not depend upon (a) size of orifice (b) height of liquid (c) acceleration due to gravity (d) density of liquid On a smooth plane surface (figure) two block A and B are accelerated up by applying a force 15 N on A. If mass of B is twice that of A, the force on B is 15 N

(a) 30 N (c) 10 N 37.

38.

A

B

(b) 15 N (d) 5 N

A potentiometer wire, 10 m long, has a resistance of 40W. It is connected in series with a resistance box and a 2 V storage cell. If the potential gradient along the wire 0.1 m is V/cm, the resistance unplugged in the box is (a) 260 W

(b) 760 W

(c) 960 W

(d) 1060 W

A prism has a refracting angle of 60º. When placed in the position of minimum deviation, it produces a deviation of 30º. The angle of incidence is (a) 30º

(b) 45º

(c) 15º

(d) 60º

39.

Transfer characteristics [output voltage (V0) vs input voltage (Vi)] for a base biased transistor in CE configuration is as shown in the figure. For using transistor as a switch, it is used

V0 I

II

III

Vi

40.

(a) in region (III) (b) both in region (I) and (III) (c) in region (II) (d) in region (I) A bar magnet of magnetic moment M, is placed in a magnetic field of induction B. The torque exerted on it is r r r r (a) M . B (b) - M . B r r r r (c) M ´ B (d) - B. M

PART - II (CHEMISTRY) 41. Schottky defect in crystals is observed when (a) unequal number of cations and anions are missing from the lattice (b) equal number of cations and anions are missing from the lattice (c) an ion leaves its normal site and occupies an interstitial site (d) density of the crystal is increased 42. The cyclobutyl methylamine with nitrous acid gives CH2 OH (a)

(b)

(c)

(d) All of these

43. The exothermic formation of CIF3 is represented by the equation : 2ClF3 (g ) ; Δ H = – 329 kJ CI 2(g ) + 3F2(g ) Which of the following will increase the quantity of CIF3 in an equilibrium mixture of CI 2 , F2 and CIF3 ? (a) (b) (c) (d)

Adding F2 Increasing the volume of the container Removing Cl2 Increasing the temperature

Solved Paper 2017

2017-5

44. For the reaction

47. What is Z in the following sequence of reactions?

2 NO(g) + O2(g) ,

2NO2(g)

( K c = 1.8 ´ 10 -6 at 184°C ) (R = 0.0831 kJ/ (mol. K) When K p and K c are compared at 184°C, it is

found that (a) Whether K p is greater than, less than or equal to K c depends upon the total gas pressure (b) K p = K c (c) K p is less than K c (d) K p is greater than K c CH3 (CH3 CO)2 O

45.

A

Br2 /CH3COOH

B

+

H /H2 O

NH2 What is X? CH3

X

CH3 Br

(a)

Br NH2

CH3

CH3

(c)

COCH3

COCH3 NH2 NH2 46. A compound MpXq has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is

M

(a) MX (c) M2X

(b) MX2 (d) M5 X14

Alkaline

dust

Anhyd. AlCl3

KMnO 4

(a)

1 3 1 a: a: a 2 4 2 2

(b)

1 1 a : 3a : a 2 2

(c)

1 3 3 a: a: a 2 2 2

(d) 1a : 3a : 2a 51. For a first order reaction A®P, the temperature (T) dependent rate constant (k) was found to

(d)

X

CH Cl

(a) Benzene (b) Toluene (c) Benzaldehyde (d) Benzoic acid 48. Which of the following oxy-acids has the maximum number of hydrogens directly attached to phosphorus? (a) H4P2O7 (b) H3PO2 (c) H3PO3 (d) H3PO4 49. The number of geometrical isomers of CH3CH=CH–CH=CH–CH=CHCl is (a) 2 (b) 4 (c) 6 (d) 8 50. If ‘a’ stands for the edge length of the cubic systems : simple cubic, body centred cubic and face centred cubic, then the ratio of radii of the spheres in these systems will be respectively,

(b) NH2

Zn

3¾®Y¾¾ ¾¾® Z Phenol¾¾®X¾¾ ¾

1 + 6.0 . The T pre-exponential factor A and the activation energy Ea, respectively, are (a) 1.0 × 106 s–1 and 9.2 kJ mol–1 (b) 6.0 s–1 and 16.6 kJ mol–1 (c) 1.0 × 106 s–1 and 16.6 kJ mol–1 (d) 1.0 × 106 s–1 and 38.3 kJ mol–1 52. 1-Propanol and 2-propanol can be distinguished by (a) oxidation with alkaline KMnO4 followed by reaction with Fehling solution (b) oxidation with acidic dichromate followed by reaction with Fehling solution (c) oxidation by heating with copper followed by reaction with Fehling solution (d) oxidation with concentrated H2SO4 followed by reaction with Fehling solution

follow the equation log k = – (2000)

EBD_7443 2017-6

Target VITEEE

53. Which group contains coloured ions out of 1. Cu 2+

2. Ti 4+

3. Co 2+ 4. Fe 2 + (a) 1, 2, 3, 4 (b) 1, 3, 4 (c) 2, 3 (d) 1, 2 54. The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2 = 0.301) (a) 23.03 minutes (b) 46.06 minutes (c) 460.6 minutes (d) 230.03 minutes 55. A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives (a) benzyl alcohol and sodium formate (b) sodium benzoate and methyl alcohol (c) sodium benzoate and sodium formate (d) benzyl alcohol and methyl alcohol 56. In the following reaction sequence, the correct structures of E, F and G are

O

O

Ph

OH

* I

Heat

2 ¾¾¾ ®[E] ¾¾¾® [F] + [G] NaOH

[* implies 13C labelled carbon) O

(a)

E=

O F=

* CH3

Ph

* Ph

O

(b)

E=

O * CH3

Ph

F=

* Ph

O

(c)

E=

F=

O

(d)

E= Ph

* – + O Na G = CHI3

Ph O

* CH3

F= Ph

59. The standard reduction potential for Cu 2 + /Cu is + 0.34. Calculate the reduction potential at pH = 14 for the above couple. (Ksp Cu (OH ) 2 = 1 × 10 -19 ) (a) – 0.22 V (b) + 0.22 V (c) – 0.44 V (d) + 0.44 V 60. A substance C4H10O yields on oxidation a compound, C4H8O which gives an oxime and a positive iodoform test. The original substance on treatment with conc. H2SO4 gives C4H8. The structure of the compound is (a) CH3CH2CH2CH2OH (b) CH3CHOHCH2CH3 (c) (CH3)3COH (d) CH3CH2–O–CH2CH3 61. The emf of a particular voltaic cell with the cell reaction Hg 22 + + H 2

2Hg + 2H +

is 0.65 V. The maximum electrical work of this cell when 0.5 g of H 2 is consumed.

– + O Na G = CHI3

O * CH3

Ph

– + O Na G = CHI3

(a) 750 K (b) 1000 K (c) 1250 K (d) 500 K 58. An organic compound (A) on reduction gives compound (B). (B) on treatment with CHCl3 and alcoholic KOH gives (C). (C) on catalytic reduction gives N-methylaniline. The compound A is (a) Methylamine (b) Nitromethane (c) Aniline (d) Nitrobenzene

(a) – 3.12 × 10 4 J

(c) 25.0 ´ 10 6 J (d) None 62. The number of aldol reaction(s) that occurs in the given transformation is : CH3CHO + 4HCHO

– + * O Na G = CH3I

57. Standard entropies of X2 , Y2 and XY3 are 60, 30 and 50 JK–1mol–1 respectively. For the reaction 1 3 X 2 + Y2 ƒ XY3 , DH = – 30 kJ to be at 2 2 equilibrium, the temperature should be:

(b) –1.25´105 J

OH conc. aq. NaOH

OH

¾¾¾¾¾¾¾ ® HO

(a) 1 (c) 3

(b) 2 (d) 4

OH

Solved Paper 2017

2017-7

63. Which of the following is not intermediate in the acid catalysed reaction of benzaldehyde with 2 equivalent of methanol to give acetal ? OCH3 HO H

+ OH 2 HO H (b)

(a)

+

Å

H - C = O- CH 3

H - O- CH 3 HO H (c)

(d)

64. Iron crystallizes in several modifications. At about 911°C, the bcc ' a ' form undergoes a trasition to fcc ' g ' form. If the distance between the two nearest neighbours is the same in the two forms at the transition temperature, the ratio of the density of iron in fcc form (r2 ) to that of iron of bcc form (r1 ) at the transition temperature r1 = 0.918 r2

(b) r1 = 0.718 r2

r (c) 1 = 0.518 r2

r (d) 1 = 0.318 r2

(a)

of a quantum of light with frequency of 8 × 1015 s –1 ? (a) 5 × 10–18 (b) 4 × 101 7 (c) 3 × 10 (d) 2 × 10–25 68. The number of stereoisomers possible for a compound of the molecular formula CH3 – CH = CH – CH(OH) – Me is: (b) 2 (c) 4 (d) 6 (d) 3 69. The optically active tartaric acid is named as D - (+) - tartaric acid because it has a positive (a) optical rotation and is drived from D - glucose (b) pH in organic solvent (c) optical rotation and is derived from D - (+) glyceraldehyde (d) optical r otation when substituted by deuterium 70. Consider the reaction : N 2 + 3H 2 ® 2 NH 3 carried out at constant temperature and pressure. If DH and DU are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ? (a) DH > DU (b) DH < DU (c) DH = DU (d) DH = 0 71. What is D in the following sequence of reactions ? NaBH

4® HBr ( i) Mg, Et 2O O¾¾ ¾¾ ¾® A¾¾¾® B¾¾¾ ¾¾ CH OH

65. The half life of the first order reaction CH 3 .CHO ( g ) ¾ ¾® CH 4 ( g ) + CO (g )

If initial pressure of CH 3CHO (g) is 80 mm Hg and the total pressure at the end of 20 minutes is 120 mm Hg (a) 80 min (b) 120 min (c) 20 min (d) 40 min 66. A compound is soluble in conc. H2SO4. It does not decolourise bromine in carbon tetrachloride but is oxidised by chromic anhydride in aqueous sulphuric acid within two seconds, turning orange solution to blue, green and then opaque. The original compound is (a) a primary alcohol (b) a tertiary alcohol (c) an alkane (d) an ether 67. The values of Planck's constant is 6.63 × 10–34 Js. The velocity of light is 3.0 × 108 m s–1. Which value is closest to the wavelength in nanometres

( ii ) H 2C = O ( iii ) H3O +

3

PCC

C ¾¾¾® D CH 2 Cl 2

(a)

CHO

(b)

COOH

OH CH 3 (d) CHO OH 72. Knowing that the chemistry of lanthanoids(Ln) is dominated by its + 3 oxidation state, which of the following statements is incorrect? (a) The ionic size of Ln (III) decrease in general with increasing atomic number (b) Ln (III) compounds are generally colourless. (c) Ln (III) hydroxide are mainly basic in character. (d) Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic in character. (c)

EBD_7443 2017-8

Target VITEEE

73. What is the R and S configuration for each stereogenic centre in this sugar from top to bottom ? O H

H HO H

OH H OH CH 2 OH (a) R, R, S (b) R, S, S (c) R, S, R (d) S, S, R 74. Saponification of coconut oil yields glycerol and (a) palmitic acid (b) sodium palmitate (c) oleic acid (d) stearic acid 75. A certain reaction is non spontaneous at 298K. The entropy change during the reaction is 121 J K -1 . Is the reaction is endothermic or

exothermic ? The minimum value of DH for the reaction is (a) endothermic, DH = 36.06 kJ (b) exothermic, DH = – 36.06 kJ (c) endothermic, DH = 60.12 kJ DH = – 60.12 kJ (d) exothermic, 76. p -cresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form, the compound B. The latter on acidic hydrolysis gives chiral carboxylic acid. The structure of the carboxylic acid is CH 3

77. Which of the following has maximum number of lone pairs associated with Xe ? (a) XeF4 (b) XeF6 (c) XeF2 (d) XeO3 78. Which one of the following statements is not true regarding (+) Lactose ? (a) On hydrolysis (+) Lactose gives equal amount of D(+) glucose and D(+) galactose. (b) (+) Lactose is a b-glycoside formed by the union of a molecule of D(+) glucose and a molecule of D(+) galactose. (c) (+) Lactose is a reducing sugar and does not exhibit mutarotation. (d) (+) Lactose, C12 H22 O11 contains 8-OH groups. 79. If one strand of DNA has the sequence ATGCTTGA, the sequence in the complimentary strand would be (a) TACGAACT (b) TCCGAACT (c) TACGTACT (d) TACGTAGT 80. The starting reagents needed to make the azo compound shown below CH 3CH 2

N=N

NH 2

+ ethylamine

(a)

(a) OH

CH 2 COOH OH

C2 H5

CH 3 CH 2 COOH

(b)

(b)

OH

+ NH 2

OH

NH 2

CH 3

+

(c)

(c) CH(OH)COOH OH

CH 3 CH(OH)COOH

(d) OH

(d)

NH 2

OH

C 2 H5

NH 2

OH

+ C 2 H5

OH

Solved Paper 2017

PART - III (MATHEMATICS) 81.

82.

83.

87.

sin–1(sin 5) > x2 – 4x holds if (a)

x = 2 – 9 – 2p

(b)

x = 2 + 9 – 2p

(c)

x > 2 + 9 – 2p

(d)

x Î (2 – 9 – 2 p , 2 + 9 – 2 p )

88.

A value of c for which conclusion of Mean Value Theorem holds for the function f (x) = loge x on the interval [1, 3] is (a) log3 e (b) loge3 1 (c) 2 log3e (d) log3e 2 Negation of the proposition : If we control population growth, we prosper (a) If we do not control population growth, we prosper (b) If we control population growth, we do not prosper (c) We control population but we do not prosper (d) We do not control population, but we prosper

84.

The equation z z + (2 - 3i) z + (2 + 3i) z + 4 = 0

85.

represents a circle of radius (a) 2 (b) 3 (c) 4 (d) 6 The function f(x) = sin x – kx – c, where k and c are constants, decreases always when (a) k > 1 (b) k ³ 1 (c) k < 1 (d) k £ 1

86.

2017-9

Equation (a) (b) (c) (d)

1 1 3 = + cos q represents r 8 8

A rectangular hyperbola A hyperbola An ellipse A parabola

The acceleration of a sphere falling through a liquid is (30 – 3v) cm/s2 where v is its speed in cm/ s. The maximum possible velocity of the sphere and the time when it is achieved are (a) 10 cm/s after 10 second (b) 10 cm/s instantly (c) 10 cm/s, will never be achieved (d) 30 cm/s, after 30 second A straight line parallel to the line 2x – y + 5 = 0 is also a tangent to the curve y2 = 4x + 5. Then the point of contact is (a) (2, l)

(b) ( –1, 1)

(c) (1, 3)

(d) (3, 4)

p /2

89.

Value of

ò

0

90.

91.

sin x sin x + cos x

dx is

(a)

p 2

(b)

-p 2

(c)

p 4

(d) None of these

The range of the function f (x) =

1 is 2 - cos3x

(a)

(-2, ¥)

(b) [-2,3]

(c)

é1 ù êë 3 ,1úû

(d)

æ1 ö ç 2 ,1÷ è ø

The area bounded by y –1 = |x|, y = 0 and |x| = will be : (a)

3 4

(b)

3 2

(c)

5 4

(d) None of these

1 2

EBD_7443 2017-10

92.

Target VITEEE

The value of x obtained from the equation x+a

b

g

g

x +b

a

a

b

x+g

=0

96.

Let A be the centre of the circle x2 + y2 – 2x–4y – 20 = 0, and B( 1,7) and D(4,–2) are points on the circle then, if tangents be drawn at B and D, which meet at C, then area of quadrilateral ABCD is (a) 150 (b) 75 (c) 75/2 (d) None of these

97.

ò0 [f ( x)g"( x ) - f " (x ) g (x )] dx is equal to :

will be

(a) 0 and -(a + b + g ) (b) 0 and a + b + g (c) 1 and (a - b - g) 2

2

(d) 0 and a + b + g 93.

The solution of the differential equation log x

94.

95.

2

dy y + = sin 2x is dx x

98.

1

[Given f(0) = g(0) = 0] (a) f(1) g(1) – f(1)g’(1) (b) f(1) g’(1) + f’(1)g(1) (c) f(1) g’(1) – f’(1)g(1) (d) none of these 7-i If z = then z14 = 3 - 4i (a) 27 (b)

(a)

1 y log | x |= C - cos x 2

(b)

1 y log | x |= C + cos 2x 2

(c)

1 y log | x |= C - cos 2x 2

é 3p ù of f (x) = 2 sin x + sin 2x, x Î ê0, ú is – ë 2û

(d)

1 xylog | x |= C - cos 2x 2

(a)

æ x2 xö lim ç - ÷= ç x ®¥ è 3x - 2 3 ÷ø 1 2 (a) (b) 3 3 -2 2 (c) (d) 3 9 r r r r r r If ((a ´ b ) ´ (c ´ d )).( a ´ d ) = 0 , then which of the following is always true ? r r (a) ar, b , cr, d are necessarily coplanar r r (b) either a or d must lie in the plane of r r b and c r r (c) either b or c must lie in the plane of r r a and d r r (d) either a or b must lie in the plane of r r c and d

(c) 99.

214 i

(d)

27 i - 27 i

The difference between greatest and least value

3 3 2

(b)

3 3 -2 2

3 3 +2 (d) None of these 2 100. A and B are two independent witnesses (i.e. there is no collision between them) in a case. The probability that A will speak the truth is x and the probability that B will speak the truth is y. A and B agree in a certain statement. The probability that the statement is true is (c)

(a)

x–y x+y

(b)

xy 1 + x + y + xy

(c)

x–y 1 – x – y + 2 xy

(d)

xy 1 – x – y + 2 xy

101. A and B are events such that P(A È B)=3/4, P(A Ç B)=1/4, P( A ) =2/3 then P ( A Ç B) is (a) 5/12 (b) 3/8 (c) 5/8 (d) 1/4

Solved Paper 2017

2017-11

102. The line which passes through the origin and intersect the two lines x - 1 y + 3 z - 5 x - 4 y + 3 z - 14 = = , = = , is 2 4 3 2 3 4

(a)

x y z = = 1 -3 5

(b)

x y z = = -1 3 5

(c)

x y z = = 1 3 -5

(d)

x y z = = 1 4 -5

p/4

103. If u n = ò 0 (a)

105. The area bounded by f (x) = x2, 0 £ x £ 1,

(b)

(c)

8 3

(d) None of these

normal to the parabola y2 = 4ax is]

(d) 0.87

p = 2aq + ap

p = 2ap + aq

(c)

q 3 = 2ap 2 + aq 2 (d) None of these

7

8

tan -1

1 1 7 + tan -1 + tan -1 is 2 3 8

7 8

(b)

cot -1 15

15 24 109. The parabola having its focus at (3, 2) and directrix along the y-axis has its vertex at tan -1 15

(a) (2, 2)

(c)

æ1 ö ç , 2÷ è2 ø

(d)

tan -1

(b)

æ3 ö ç , 2÷ è2 ø

(d)

æ2 ö ç , 2÷ è3 ø 2 5 ù -4 a - 4ú is ú -2 a + 1ûú

(b) 2 if a = 1 (d) 1 if a = 4

cos x 1 0 2 cos x 1 , then 111. If f (x ) = 1 0 1 2 cos x p/2

2

(a)

(b)

(c) 0.35

(a) 1 if a = 6 (c) 3 if a = 2

x y + = 1 be a p q 3

(b) 0.77

ëê 1

(a)

6

(a) 0.50

é -1

4 3

3 2

2

5

110. The rank of the matrix êê 2

g(x) = - x + 2,1£ x £ 2 and x - axis is

2

4

For the events E = {X is a prime number} and

(c)

1 1 (d) 2n - 1 2n + 1 104. Ten different letters of an alphabet are given, words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated is (a) 69760 (b) 30240 (c) 99784 (d) None of these

3

3

p(X) 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05

(a)

(c)

106. The condition that the line

2

108. The value of tan -1

1 n +1

(b)

1

F = {X < 4}, then P(E È F) is

tan n q dq then un + un–2 is :

1 n -1

X

2

107. A random variable X has the probability distribution

ò f (x)dx is equal to 0

(a)

1 4

(b) –

(c)

1 2

(d) 1

1 3

EBD_7443 2017-12

Target VITEEE

112. The distance of the point (1, –2, 3) from the plane

x - y + z = 5 measured parallel to the line x y z -1 = = is 2 3 -6

(a) 1

(b) 2

(c) 4

(d)

2

2 3

2

x - y cos ec a = 25 is 5

times

the

eccentricity of the ellipse x 2 cos ec 2 a + y 2 = 5 ,

then a is equal to : (a)

(c)

tan

-1

tan

-1

2

2 5

(b)

(d)

sin -1

sin

-1

3 4 2 5

115. The conditional ( p Ù q) Þ p is (a) A tautology (b) A fallacy i.e., contradiction (c) Neither tautology nor fallacy (d) None of these 116. The set of points of discontinuity of the function f(x) = lim

n ®¥ 3 n

(a) R

(c)

(2 sin x ) 2 n - (2 cos x ) 2 n

(b)

2 are connected by the relation V = 5x – x and 6

the volume of water is increasing , at the rate of 5 cm3/sec, when x = 2 cm. The rate at which the depth of water is increasing, is

113. The tangent lines to the curve y2 = 4ax at points where x = a, are (a) parallel (b) perpendicular (c) inclined at 60º (d) inclined at 30º 114. If the eccentricity of the hyperbola 2

117. The volume V and depth x of water in a vessel

(a)

5 cm / sec 18

(b)

1 cm / sec 4

(c)

5 cm / sec 16

(d) None of these

11 8. If vectors aiˆ + ˆj + kˆ, iˆ + bjˆ + kˆ and iˆ + ˆj + ckˆ (a ¹ b ¹ c ¹ 1) are coplanar, then find 1 1 1 + + . 1- a 1 - b 1- c (a) 0

(b) 1

(c) –1

(d) 2

119. If

1 A -=

matrix

é3 - 2 4 ù ú ê A = ê1 2 - 1ú êë0 1 1 úû

and

1 adj (A) , then k is k

(a) 7 (b) – 7 (c) 15 (d) – 11 120. The angle between a pair of tangents drawn from a point T to the circle x 2 + y 2 + 4 x - 6 y + 9 sin 2 a + 13 cos 2 a = 0 is 2a .

is given by p ì ü ínp ± , n Î I ý 3 î þ

p ì ü ínp ± , n Î I ý (d) None of these 6 î þ

The equation of the locus of the point T is (a)

x 2 + y 2 + 4x - 6y + 4 = 0

(b)

x 2 + y 2 + 4x - 6y - 9 = 0

(c)

x 2 + y 2 + 4x - 6y - 4 = 0

(d)

x 2 + y 2 + 4x - 6y + 9 = 0

Solved Paper 2017

PART - IV (ENGLISH) Directions (Qs. 121-123): Study the paragraph and answer the questions that follow: At this stage of civilisation, when many nations are brought into close and vital contact for good and evil, it is essential, as never before, that their gross ignorance of one another should be diminished, that they should begin to understand a little of one another's historical experience and resulting mentality. It is the fault of the English to expect the people of other countries to react as they do, to political and international situations. Our genuine goodwill and good intentions are often brought to nothing, because we expect other people to be like us. This would be corrected if we knew the history, not necessarily in detail but in broad outlines, of the social and political conditions which have given to each nation its present character. 121. The character of a nation is the result of its (a) gross ignorance (b) cultural heritage (c) socio-political conditions (d) mentality 122. According to the author Mentality' of a nation is mainly product of its (a) present character (b) international position (c) politics (d) history 123. The need for a greater understanding between nations (a) is more today than ever before (b) was always there (c) is no longer there (d) will always be there

2017-13

Directions (Q. 124) : In the question below a sentence is given, a part of which is printed in bold and underline. This part may contain a grammatical error. Each sentence is followed by phrases a, b, c and d. Find out which phrase should replace the phrase given in bold/underline to correct the error, if there is any to make the sentence grammatically meaningful and correct. 124. There are any number of skilled writers who can develop content and create marketing materials with a keen eye to using proven methods, but also to developing new and innovative techniques. (a) with a keen eye to using proven methods, but also to developing new and innovative techniques. (b) with a keen eye for using proven methods, and also to developing new and innovative techniques. (c) with a keen eye not only to using proven methods, but also to developing new and innovative techniques. (d) with a keen eye to using proven methods, but to developing new and innovative techniques. 125. Choose the best pronunciation of the word, Sorbet from the following options. (a) Sore-bet (b) Sore-bay (c) Sorb rhymes with orb (d) Shore-bay

EBD_7443 2017-14

Target VITEEE

SOLUTIONS PART - I (PHYSICS) 1.

2.

PB

(d) Given : Mass of rocket (m) = 5000 Kg Exhaust speed (v) = 800 m/s Acceleration of rocket (a) = 20 m/s2 Gravitational acceleration (g) = 10 m/s2 We know that upward force F = m (g + a) = 5000 (10 +20) = 5000 × 30 = 150000 N. We also know that amount of gas ejected æ dm ö F 150000 = 187.5 kg / s ç ÷= = 800 è dt ø v (c) The power dissipated in the circuit. V2 P= Req

PA =

5. 6. 7.

...(i)

P = 30 W Substituting the values in equation (i) (10) 2 æ 5R ö çè ÷ 5 + Rø

8.

15 R = 10 Þ 15R = 50 + 10R 5+ R 5R = 50 Þ R = 10 W

1 E = mv 2 or mv = 2m E 2

so l =

4.

(c) As

h = mv

h 2m E

1 1 2 2 mA v A = mB v B 2 2

vA = vB

mB mA

mA mB

=

mB 1 = mA 3

R

B

C

R

D

2R

æ 5R ö Req = çè ÷ 5 + Rø

(c)

mB mA

(d) (d) Photoelectric effect does not support the wave nature of light. (c) The equivalent electrical circuit of the arrangement is shown in figure.

A

1 1 1 5+ R = + = Req R 5 5R

3.

mA vA

2R

v = 10 volt

30 =

mB vB

=

;

9.

Temperature difference between the end points A and D = 200 – 20 = 180°C As the resistances for the three parts are equal, the temperature difference must be distribuited equally in the three parts (= 180/ 3 = 60°C) \ Temperature of B = 200°C – 60° = 140°C. (c) Number of possible spectral lines emitted when an electron jumps back to ground n(n - 1) state from n th orbit = 2 n(n - 1) =6Þn =4 Here, 2 Wavelength l from transition from n = 1 to n = 4 is given by,

1 16 æ1 1 ö = R ç - 2÷ Þl = = 975 Å è1 4 ø l 15R (c) The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements. By Fleming's Left hand rule on element AB the direction of force will be Leftwards and the magnitude will be dF = Idl B sin 90° = IdlB

Solved Paper 2017

2017-15

x x x x xB x dl dF A x x x x x x 10. 11.

x x x x Cxdl x dF D x Ix x x x x

x x x x

=

| v1 - v 2 | Þ Mu = Mv - MV | u1 - u 2 |

....(ii) From (i) and (ii), 2Mu = (M + m)v 2uM 2u Þ v= Þv = m M+m 1+ M

12.

14.

25 × 10–9 9 = 2.80 × 10–9 nm \ l ³ 2.8 × 10–9 m (b) Kerosene oil rises up in wick of a lantern because of capillary action. If the surface tension of oil is zero, then it will not rise, so oil rises up in a wick of a lantern due to surface tension.

15.

(a)

16.

17. 18.

(c) Consider the potential at D be ‘V’. Potential drop across C1 is (V – V1) and C2 i s (V2 – V)

-3 LdI 40 ´10 (11 - 1) = = 100V dt 4 ´10-3 (d) Impedence of a capacitor is XC = 1/wC 1 1 5000 = = XC = . 6 2 pfC 2 p ´ 50 ´ 2 ´ 10 p (a) The final boolean expression is,

e=

(

++

+ + + + + + + + + +

As q1 = q2 [capacitors are in series] \ C1 (V - V1) = C2 (V2 - V)

13.

C1V1 + C2 V2 C1 + C2

(b) Given : work function f of metal = 2.28 eV Wavelength of light l = 500 nm = 500 × 10– 9m KEmax = KEmax =

hc –f l 6.6 ´ 10-34 ´ 3 ´ 108

5 ´ 10-7 KEmax = 2.48 – 2.28 = 0.2 eV lmin =

h = p

h 2m ( KE )max

– 2.82

)

X = A . B = A + B = A + B Þ OR gate (a) Charge resides on the outer surface of a conducting hollow sphere of radius R. We consider a spherical surface of radius r < R. By Gauss theorem

\ q 1 = C1(V - V1 ), q 2 = C2 (V2 - V)

V=

2 ´ 9 ´10 -31 ´ 0.2 ´1.6 ´10 -19

lmin =

On element CD, the direction of force will be towards right on the plane of the papper and the magnitude will be dF = IdlB. (b) Apparent depth = d/m1 + d/m2 (c) By law of conservation of momentum, Mu = MV + mv ....(i) Also e =

20 ´ 10-34 3

++ +++

+

+ + + S + + E + + + ++ R

O r ++

+ +

rr 1 E.ds = ´ charge s e0

ò

+

enclosed

or

1 ´0 ÞE =0 e0 i.e electric field inside a hollow sphere is zero. 19. (d) Let R be the radius of the bigger drop, then Volume of bigger drop = 2 × volume of small drop 2 E ´ 4pr=

4 3 4 pR = 2 ´ pr 3 Þ R = 21/ 3 r 3 3 Surface energy of bigger drop,

E = 4pR 2T = 4 ´ 2 2 / 3 pr 2T = 28 / 3 pr 2T

EBD_7443 2017-16

20.

Target VITEEE

(b) Equivalent resistance between A and B = series combination of 1 W and 2 W in parallel with 3 W resistor..

26.

(b) For path difference l, phase

2p 2p æ ö x = .l = 2p ÷ difference = 2p ç Q = l l è ø Þ I = I0 + I0 + 2I0 cos 2p Þ I = 4I0 (\ cos 2p = 1) For x =

l p , phase difference = 4 2

\ I' = I1 + I 2 + 2 I1 I 2 cos

1 1 1 2 = + = or R = 1.5 W. R 3 3 3 \ Current in the circuit is I = V/R = 1.5/1.5 = 1A. Since the resistance in arm ACB = resistance in arm AB = 3 W, the current divides equally in the two arms. Hence the current through the 3 W resistor = I/2 = 0.5 A.

21.

(d) stress tan(90° - q) = strain

22.

(a)

23.

(c) U = –K

27. 28.

29.

I I If I1 = I2 = I0 then I ' = 2I0 = 2. = 4 2 (a) Static friction is a self adjusting force in magnitude and direction. (a) Cosmic rays are coming from outer space, having high energy charged particles, like a-particle, proton etc. b-rays are stream of high energy electrons, coming from the nucleus of radioactive atoms. (c) For a monatomic gas 3 R 2 So correct graph is Cv =

ze 2 k ze 2 ; T.E = – r 2 r

k ze 2 . Here r decreases 2 r (c) When resistance is connected to A.C source, then current & voltage are in same phase. (a) Galvanometer is converted into ammeter, by connected a shunt, in parallel with it.

25.

I

3 / 2R

­

K.E =

24.

p 2

Cv T® 30.

q

(c)

L/2

G

q L

L

L S

VG 25 ´ 10-3 GS == G+ S I 25

GS = 0.001W G+S Here S 0 Hence, the solution of differential equation represents family of hyperbolas.

CH 2

[CH3C(Br) = CH2 + CH3CH = CHBr] (A)

5y 1 dy

4 2

n r 1

r3

2

Solved Paper 2015

= P A

=

B C P A A

P A P

83. (b)

2015-23

C A

B

B P C

A

B P

P A

P A

= P A

B

A

84. (d)

n(log 35 – log 36) < log 0.01 n[15441 – 15563] < –2 – 0.0122n < – 2 0.0122n > 2 n > 163.9 So, the least value of n is 164. 86. (d) The given system of equations will be consistent with unique solution, when

B

B

1 1 1

B

1 2 3

C

1 4 k

1 3

1

1

3

2 1

1 = 0 0 1

5

3

9

2

3 0

[Applying R2

R2 – 2R1, R3

1(2k – 12) + 1 (3 – k) + 1(4 – 2) 0 k – 12 + 3 + 2 0 k–7 0 k 7

1

1 4 Let z = – 1 + 8z

87. (d) Given: z

R3 – 3R1]

3

0

5

3

0

0

17 5

1

R3

R3

9 R2 5

rank (A) = 3 85. (b) The probability of getting a double six in one throw of two dice 1 1 = 6 6

ellipse

1 36

1 35 36 36 Now, (p + q)m qn + nC1qn–1p + nC2qn – 2 p2 + .... + nCrqn–r pr +....+ pn The probability of getting atleast one double six in n throws with two dice. = (q + p)n – qn n =1 q

35 36

1| 8

x2

y2

a2

b2

1, then

b2 )2

a 2 b2 m 2 [Here, m = 2, a2 = 9 and b2 = 16]

=

=1

1

|z

|z|

m 2 (a 2

c2

1 , 36 q 1 p

35 36

1 8

1 z 1 z 1 2 4 8 z lies on a circle with centre (–1, 0) and radius 2. 88. (c) If the line y = mx + c is a normal to the

p

=

z

z

1

0

35 36

1

n

=

0.99

n

0.01

2

9 16

9 16

4 49 9 64

2

2 2

4 49 73

196 73

14

c

73 89. (b) Given equation is x2 + x + 1 = 0 2 x = and x Case I : When x = Then 6

n

2

xn

n 1

=( + +(

1 x

2

6

n

2n 2

1

2

n

n 1 2 )2 + ( 2 + 4 )2 + ( 3 4 + 8)2 + ( 5 + 10)2

+ 6 )2 +( 6+

12)2

EBD_7443 2015-24

Target VITEEE

= (–1)2 + (–1)2 + (2)2 + (–1)2 + (–1)2 + (2)2 = 12 Case II: When x = 2 Then 6

2

1

xn

6

xn

n 1

2n

n 2

1

= 12 90. (b) Correct result is as follows: (~ p ~ q) (r s) or ~(p q) r s

91. (a)

A

3

1 2 0

1

= 12

3

1 2 0

1

When m = 1, c = –

7 6

When m = 3, c = –

3 2

y= x–

3 2

1

dx cos x

2

= 12

=

92. (c)

36

12

12 24

36

0

24

12

= 7m2

3(m)

m

1 ,1,3 2

We know that c n (m) + n–1(m) = 0, which in the given case becomes c(2 – 14m + 6m2) + (–14m + 7m2) = 0 c

14m 7m 2 2 14m 6m 2

=

3 sin x

1 dx 2 1 3 cos x sin x 2 2 1 2 cos

2m3

= 3 + 2m – + 2 (m) = –14m + 7 m 2 2 3(m) = 2 – 14m + 6m Now, putting 3(m) = 0, we have 3 + 2m – 7m2 + 2m3 = 0 (1 – m) (1 + 2m) (3 – m) = 0

7 3 and y = 3x – . 6 2

93. (b) The structure (N,.) satisfies the closure property, associativity and commutativity but the identity element 0 does not belong to N. So, N is a semi-group. 94. (a)

3

1 5 x , 2 6

Asymptotes are y =

1

|A| = 1.(4 + 3) – 3 (–2 + 0) + 1(–1 –0) = 7 + 6 – 1 = 12 So, adj (adj A) = |A|n – 2 = A = (12)3 – 2 A = 12A 1

5 6

2

n 1

1

1 ,c 2

So, when m

1 2

dx 3

cos x sin

3

sin x

dx cos x

3

=

1 sec x 2

=

1 x log tan 2 2

3

dx

6

4

C

1 x log tan C 2 2 12 95. (d) Given sphere is x2 + y2 + z2 – 6x – 12y – 2z + 20 = 0 Centre (3, 6, 1) Here, one end of diameter is (2, 7, 3). Let the other end of the diameter be (x, y, z) =

Solved Paper 2015

2015-25

Centre of the sphere will be the mid-point of the ends of diameter.

2 x 7 y 3 z , , 2 2 2 2+x=6 x=4 7 + y = 12 y=5 and 3 + z = 2 z = –1 Therefore, (x, y, z) (4, 5, –1) 96. (d) Given lines are x = my + n, z = py + q and x = m y + n , z = p y + q Above equations can be rewritten as

n1. n 2 n1 n 2

cos

3, 6,1

So

x n m

y 0 1

z q p

x n m

and

y 0 1

z q p

3 3

=

9 9 9

1 3 98. Let , and be the angles made by the line segment OP with X-axis, Y-axis and Z-axis, respectively. and 4 We know that, cos2 cos 2

1 2

A X

C

c os

= OA OB

2 1 iˆ

cos 2

2

1 2

2iˆ ˆj kˆ

2 1 ˆj

1 4 kˆ

= 3iˆ 3jˆ 3kˆ Vector perpendicular to face ABC = n 2 . = AB AC = (–3iˆ 3 ˆj ) ( ˆj kˆ) = 3iˆ 3jˆ 3kˆ Since, angle between faces is equal to angle between their normals.

1 cos 2 4

c os2

Vector perpendicular to face OAB = n1

=

4

2

B

= ˆi 2ˆj kˆ

1

Given:

1

Z

3

1 3

27 27 cos

97. (c)

O

3

9 9 9 9 9 9

=

Lines will be perpendicular, if mm + 1 + pp = 0 mm + pp = –1 Y

3 3

3 + cos2

+ cos2 = 1

cos 2

1

cos2

1

3 2

1

1 4

1 2

4 Hence, direction cosines are cos , cos , cos

1 1 1 , , . 2 2 2 99. (a) ~ p q means F F = F, ~ r means F p means T ~ r] [(~ p q) 100. (b) Let f(x) = x25(1 – x)75, x [0, 1] f (x) = 25x24 (1 – x)75 – 75x25 (1 – x)74 = 25x24 (1 – x)74 {(1 – x) – 3x} = 25 x24 (1 – x)74 (1 – 4x) + – + – 0 1/4 1 i.e.

EBD_7443 2015-26

Target VITEEE We can see that f (x) is positive for x < and f (x) is negative for x >

1 4

104. (a)

1 . 4

O

1 Hence, f (x) attains maximum at x = . 4

101. (b)

z

1 4

= 1 4

= (–z)

1 4

|z|

1 4

1 4

3

11 4

= sin x cos x

on parabola is

–t1x 2at1 at13

2 It also passes through at 2 , 2at 2

So, 2at 2

t1 at 22

2t 2 2t1

t1 t 22 t12

103. (c)

=

1 2

2 t1 a1b1 a2 b2 a12

1

a22

a3b3

a32

b12

2

1

1

1 1 4 4 1 1

2 2 2 6 1 6 6 So, = 0° or = 2 sec 2 = 1 2 = sec–1(1) = sec–1 (1)

=

5 /4

/4 0

sin x cos x dx

/4

cos x sin x dx

cos x sin x

5 /4 /4

+ sin x cos x

3 /4 5x / 4

= 4 2 2 sq units 105. (b) [a + b – c] . [(a – b) × (b – c) = (a + b – c). [a × b – a × c – b × b + b × c] = a. (a × b) – a. (a × c) + a. (b × c) + b. (a × b) – b (a × c) + b. (b × c) – c. (a × b) + c. (a × c) – c.(b × c) = a. (b × c) – b. (a × c) – c. (a × b) = [a b c] – [b a c] – [c a b] =[a b c] + [a b c] – [a b c] = [a b c] = a. (b × c) 106. (a) Surface area A of a cube of side x is given by A = 6x2. On differentiating w.r.t. x, we get dA 12x dx Let the change in x be x = m% of x

t1

–t1

cos

2at1 at13

2

t1 t 2

cos x sin x dx

3 /2

1 11 4 4 102. (b) Equation of the normal at point

at12 , 2at1

/4 0

5 /4

z

t2

3 /2

Required area

= z–

y

/2

b22 b32

mx 100 Change in surface area,

=

A

dA dx

x

12x

x

mx 12x 2 m 100 100 The approximate change in surface area

12x

2m 6x 2 100 = 2m% of original surface area

=

Solved Paper 2015

2015-27

107. (d) Given equation of rectangular hyperbola is x2 – y2 = 82 Length of latusrectum = 2 × (8) = 16 and eccentricity = 2

= hlim0

1 2

cos h 1 tan h

110. (c)

The asymptotes are perpendicular lines. So, x y 0 Now, directrices are

B(0,3) A(3,0)

8

x

O

= 4 2 2 108. (a) Equation of hyperbola is 3x2 – 2y2 = 6

x+y=3 x2+y2=9

x2 y2 1 2 3 So, a2 = 2 and b2 = 3 Given, equation of line is x – 3y = 3.

Area of required region =

1 × Area of circle – Area of OAB 4

1 3 Slope of line perpendicular to given line, m = –3 The equation of tangents are

=

1 × 4

Slope of given line =

y

109. (d)

= –1

a 2 m2

mx

b2

= 3x

2 9 3

= 3x

18 3

= 3x

15

tan 4 tan x 1 lim lim = h 0 x / 4 cos2 x cos 2

x

4

h

4

1 h

1 4 2 111. (c) [a + b, b + c, c + a] = (a + b) . [(b + c) × (c + a)] = (a + b) . [b × c + b × a + c × c + c × a] = (a + b). (b × c + b × a + c × a) [ c × c = 0] = a. (b × c) + a. (b × a) + a.(c × a) + b. (b × c) + b. (b × a) + b . (c × a) = a.(b × c) + b.(c × a) = [a b c] + [a b c] = [a b c] + [a b c] = 2 [a b c] = 9

/2

112. (a) Let I

I

h

0

/2 0

log tan x . sin 2 xdx ...(i)

log tan

2

a

= lim h

0

= hlim0

1 tan h 1 tan h cos

2

0

1 I

2h

1 tan h 1 tan h sin 2h 1 tanh

2 tan h = hlim0 2 sin h cos h 1 tan h

1 ×3×3 2

× (3)2 –

/2 0

x sin 2

f x dx

a 0

log cot x. sin 2x dx

2

x dx

f a x dx

....(ii)

[ sin( – 2x) = sin 2x] On adding eqs (i) and (ii), we get 2I

/2 0

log tan x.sin 2x dx

=

/2 0

/2 0

log cot x sin 2x dx

sin 2 x log (tan x.cot x)dx

[ log m + log n = log (m . n)]

EBD_7443 2015-28

Target VITEEE /2

=

0

I = 0[ /2 0

b1 3iˆ –16jˆ 7kˆ

sin 2x log 1dx

and b2

log 1 = 0]

sin 2x log tan x dx

0

113. (c) Here, mean = 4 and variance = 2 np = 4 and npq = 2 npq np

So,

2 4

Then, p = 1 – q = 1 –

= 1 2

1 2

1 =4 n=8 2 P(X = r) = nCrpr qn–r

n×

8 1 1 p q 2 2 The required probability of atleast 7 successes is P(X 7) = P(X = 7) + P(X = 8)

8 = Cr

1 2

8

C8

=

=

8! 8! 7!1! 8!0!

1 2

8

C7

= 8 1

8

1 2

8

114. (b) Given, lines are

9 256 y 4 16

3iˆ – 16ˆj 7kˆ

and r 10iˆ 30ˆj 4kˆ

3iˆ 8jˆ 5kˆ

On comparing these equations with r = a1 + b1 and r = a2 + µb2, we get a 7iˆ 4jˆ 6kˆ 1

a 2 10iˆ 30jˆ 4kˆ

a in eq. (iii), we get b 2

z 6 7

x 10 y 30 4 z 3 8 5 The vector form of given lines are r

a 2

On putting c

and

7iˆ 4ˆj 6kˆ

72 1224 144 1152 288 units 84 84 21 115. (a) Equation of plane passing through (2, 2, 1) is a(x – 2) + b(y – 2) + c(z – 1) = 0 ....(i) Since, above plane is perpendicular to 3x + 2y + 4z + 1 = 0 and 2x + y + 3z + 2 = 0 3a + 2b + 4c = 0 ....(ii) and 2a + b + 3c = 0 ....(iii) [ for perpendicular, a1a2 + b1b2 + c1c2 = 0] On multiplying eq. (iii) by 2, we get 4a + 2b + 6c = 0 ....(iv) On subtracting eq. (iv) from eq. (ii), we get c

x 7 3

84

=

Mean = np = 4

8

Shortest distance =

a 2 a1 . b1 b2 b1 b2

3iˆ 34ˆj 2kˆ . 24iˆ 36ˆj 72kˆ

1 2

q

3iˆ 8jˆ 5kˆ

= On putting b = we get a(x – 2) –

a 2

a a and c = in eq. (i), 2 2 a a (y – 2) – (z – 1) = 0 2 2

a [2(x – 2) – (y – 2) – (z – 1)] = 0 2 2x – 4 – y + 2 – z + 1 = 0 2x – y – z – 1 = 0 116. (c) Suppose, A : a male is selected B: a smoker is selected Given:

Solved Paper 2015

2015-29

Now, we check the function is maximum or minimum.

7 2 A 2 ,P A B and P 10 5 B 3 The probability of selecting a smoker..

P A

B

P A

P B

B

A B

P

=

3 2 = 5 5 The probability of selecting a male

cos t t

=1–

=

7 10

2 5

P A B

sin t

sin t

t

7 4 6 1 2 10 Probability of selecting a smoker, if a male is first selected, is given by

0

t

1

t

sin t t cos t sin t – 2 lim t t 0 t3

lim

=

tan t t

0

2

Now tlim0 f

P B

3 5

t2

sin t 2 cos t 2 sin t t t2 t3 For maximum or minimum value of f(x), put f (x) = 0

2 3 3 5 2 5 The probability of selecting a non-smoker So, P(B) = 1 – P(B)

P A B

1

1 1 1 2 sin t 2 cos t 2 cos t 3 sin t t t t t

and f t

=

P A

1 cos t t

f t

0 form 0

=

P

B A

P A

=

2 2 5 1

4 5

=

B

P A

sin t t At t = 0, we will check continuity of the function. LHL = f(0 – h)

= hlim0

sin 0 h 0 h

= lim h

0

sin h h

1

RHL = f(0 + h) h

sin 0 h 0 h

0

sin h 1 h and f(0) = 1 LHL = RHL = f(0) So, the function is continuous at t = 0

= lim h

0

cos t

t sin t cos t

3t 2 [using L’ Hospital rule] t

1

=

117. (d) Given: f(t) =

lim

1 2 lim

0

2 sin t lim 3t 0 t

2 1 1 0 3 3 So, function f(t) is maximum at t = 0 118. (d) Consider the function f defined by 1

=

f x

xn 1 xn x2 an ... a n 1 anx n 1 n 2 Since, f(x) is a polynomial, so it is continuous and differentiable for all x. f(x) is continuous in the closed interval [0, 1] and differentiable in the open interval (0, 1). Also, f(0) = 0 and a0

a0 a1 a .... n 1 a n n 1 n 2 i.e. f(0) = f(1) f 1

0 [say]

EBD_7443 2015-30

Target VITEEE Thus, all the three conditions of Rolle’s theorem are satisfied. Hence, there is atleast one value of x in the open interval (0, 1) where f (x) = 0 i.e. a0 xn + a1xn–1 + ....+an = 0 x 1 x

119. (d) Let f(x) = log 1 x f

1 x .1 x.1

1 1 x

x

1 x

1

2

1

= 1 x

1 x

1 x

2

which is positive. [ x > 0] f(x) is monotonic increasing, when x > 0. f(x) > f(0) Now, f(0) = log 1 – 0 = 0 f(x) > 0 log 1 x

x 1 x

0

x log 1 x 1 x Also, for x > 0, x2 > 0 x2 + x > x x(x + 1) > x

x>

....(i)

x

....(ii) x 1 From eqs. (i) and (ii), we get

x < log (1 + x) < x x 1 [ log (1 + x) < x for x > 0]

d dy Its auxiliary equation is m2 – 1 = 0, so that m = 1, –1 Hence, CF = C1ey + C2e–y. where C1, C2 are arbitrary constants

where, D

Now, also PI

x 2

120. (c) We can write given differential equation as, (D2 – 1) x = k ....(i)

1 D

2

1

k

1 e0.y = k. 2 D 1 K.

1

e0.y

K 0 –1 So, solution of eq. (i) is x = C1ey + C2e–y – k ....(ii) Given that x = 0, when y = 0 So, 0 = C1 + C2 – k (From (ii)) C1 + C2 = k ....(iii) Multiplying both sides of eq. (ii) by e –y, we get x. e–y = C1 + C2e–2y –ke–y ....(iv) Given that x m when y , m being a finite quantity. So, eq (iv) becomes x × 0 = C1 + C2 × 0 – (k × 0) C1 = 0 ....(v) From eqs. (iv) and (v), we get C1 = 0 and C2 = k Hence, eq. (ii) becomes x = ke–y – k = k(e–y –1) 2

SOLVED PAPER PART - I (PHYSICS) 1.

2.

3.

The amplification factor of a triode is 50. If the grid potential is decreased by 0.20 V. What increase,in plate potential will keep the plate current unchanged? (a) 5 V (b) 10 V (c) 0.2 V (d) 50 V If the nuclear fission,piece of uranium of mass 5.0 g is lost, the energy obtained in kWh is (a) 1.25 × 107 (b) 2.25 × 107 7 (c) 3.25 × 10 (d) 0.25 × 107 Current in the circuit will be 20 30

20

(a)

5.

(b)

5 A 50

5 5 A (d) A 10 20 An installation consisting of an electric motor driving a water pump left 75 L of water per second to a height of 4.7 m.If the motor consumes a power of 5 kW, then the efficiency of the installation is (a) 39% (b) 69% (c) 93% (d) 96% A potential difference across the terminals of a battery is 50 V when 11 A current is drawn and 60 V,when 1 A current is drawn. The emf and the internal resistance of the battery are (a) 62 V, 2 (b) 63 V, 1 (c) 61 V, 1 (d) 64 V, 2

(c)

4.

5 A 40

5

VITEEE 2014

Beyond which frequency, the ionosphere bands any incident electromagnetic radiation but do not reflect it back towards the earth? (a) 50 MHz (b) 40 MHz (c) 30 MHz (d) 20 MHz 7. A metallic surface ejects electrons. When exposed to green light of intensity I but no photoelectrons are emitted,when exposed to yellow light of intensity I.It is possible to eject electron from the same surface by (a) yellow light of same intensity which is more than I (b) green light of any intensity (c) red light of any intensity (d) None of the above 8. An electron moves at right angle to a magnetic field of 5 × 10–2 T with a speed of 6 × 107 m/s. If the specific charge of the electron is 1.7 × 1011C/kg. The radius of the circular path will be (a) 2.9 cm (b) 3.9 cm (c) 2.35 cm (d) 2 cm 9. A solenoid 30 cm long is made by winding 2000 loops of wire on an iron rod whose cross-section is 1.5 cm2.If the relative permeability of the iron is 6000. What is the self-inductance of the solenoid? (a) 1.5 H (b) 2.5 H (c) 3.5 H (d) 0.5 H 10. A coil of resistance 10 and an inductance 5 H is connected to a 100 V battery. The energy stored in the coil is (a) 325 erg (b) 125 J (c) 250 erg (d) 250 J 11. A galvanometer has current range of 15 mA and voltage range 750 mV. To convert this galvanometer into an ammeter of range 25 A, the required shunt is (a) 0.8 (b) 0.93 (c) 0.03 (d) 2.0 6.

EBD_7443 2014-2

12.

13.

14.

15.

Target VITEEE

The denial cell is balanced on 125 cm length of a potentiometer. Now, the cell is short circuited by a resistance of 2 and the balance is obtained at 100 cm. The internal resistance of the denial cell is 4 (a) (b) 1.5 3 (c) 1.25 (d) 0.5 Four resistance of 10 , 60 , 100 and 200 respectively taken in order are used to form a Wheatstone’s bridge. A 15V battery is connected to the ends of a 200 resistance, the current through it will be (a) 7.5 × 10–5 A (b) 7.5 × 10–4 A –3 (c) 7.5 × 10 A (d) 7.5 × 10–2 A A circuit has a self-inductance of 1 H and carries a current of 2A. To prevent sparking, when the circuit is switched off, a capacitor which can withstand 400 V is used. The least capacitance of capacitor connected across the switch must be equal to (a) 50 µF (b) 25 µF (c) 100 µF (d) 12.5 µF The output Y of the logic circuit shown in figure is best represented as

18.

(a) 10 cm 19.

20.

(a)

A + B.C

(b)

17.

(b)

A E ( A + 1)2

( A -1) E ( A + 1)2 If a magnet is suspended at angle 30° to the magnet meridian, the dip of needle makes angle of 45° with the horizontal, the real dip is æ 3 ö÷ çç ÷ –1 (a) tan èçç 2 ø÷÷ (b) tan –1 ( 3) æ A -1 ö÷ çç ÷ E èç A + 1ø÷ 2

(c)

(c)

16.

æ A ö÷ çç ÷ E èç A + 1ø÷ 2

A + B.C

(d) A + B.C A + B.C A resistor of 6 k with tolerance 10% and another resistance of 4 k with tolerance 10% are connected in series. The tolerance of the combination is about (a) 5 % (b) 10 % (c) 12 % (d) 15 % If we add impurity to a metal those atoms also deflect electrons. Therefore, (a) the electrical and thermal conductivities both increase (b) the electrical and thermal conductivities both decrease (c) the electrical conductivity increases but thermal conductivity decreases (d) the electrical conductivity decrease but thermal conductivity increases

(b) 10 2 cm

(c) 20 cm (d) 5 2 cm An ammeter and a voltmeter of resistance R are connected in series to an electric cell of negligible internal resistance. Their reading are A and V respectively. If another resistance R is connected in parallel with the voltmeter, then (a) both A and V will increase (b) both A and V will decrease (c) A will decrease and V will increase (d) A will increase and V will decrease A neutron is moving with velocity u. It collides head on and elastically with an atom of mass number A. If the initial kinetic energy of the neutron is E, then how much kinetic energy will be retained by the neutron after reflection? (a)

21.

A B A

A proton and an -particle, accelerated through the same potential difference, enter a region of uniform magnetic field normally. If the radius of the proton orbit is 10 cm, then radius of -particle is

22.

23.

(d)

æ 3 ö÷ æ 2 ö ç (c) tan–1 ççç 2 ÷÷÷ (d) tan–1 çç ÷÷÷ è ø èç 3 ø÷ Which has more luminous efficiency ? (a) A 40 W bulb (b) A 40W fluorescent tube (c) Both have same (d) Cannot say The resistance of a germanium junction diode whose V – I is shown in figure is (Vk = 0.3 V) I 10 mA

Vk

2.3V

v

Solved Paper 2014

2014-3

(a) 5 k

(b) 0.2 k

(c) 2.3 k

æ 10 ö (d) ççç ÷÷÷ k è 2.3 ø

24. In hydrogen discharge tube, it is observed that through a given cross-section 3.31 × 10 15 electrons are moving from right to left and 3.12 ×105 protons are moving from left to right. The current in the discharge tube and its direction will be (a) 2 mA towards left (b) 2 mA, towards right (c) 1 mA, towards right (d) 2 mA, towards left 25. In a semiconductor, separation between conduction and valence band is of the order of (a) 0 eV (b) 1 eV (c) 10 eV (d) 50 eV 26. If 1000 droplets each of potential 1V and radius r are mixed to form a big drop. Then, the potential of the drop as compared to small droplets, will be (a) 1000 V (b) 800 V (c) 100 V (d) 20 V 27. A Zener diode, having breakdown voltage equal to 15 V is used in a voltage regulator circuit shown in figure. The current through the diode is

20V

(a)

6 5R

(b)

36 5R

(c)

64 7R

(d) None of these

30. Silver has a work function of 4.7 eV. When ultraviolet light of wavelength 100 nm is incident on it a potential of 7.7 V is required to stop the photoelectrons from reaching the collector plate. How much potential will be required to stop photoelectrons, when light of wavelength 200 nm is incident on it? (a) 15.4 V (b) 2.35 V (c) 3.85 V (d) 1.5 V 31. If the distance of 100 W lamp is increased from a photocell, the saturation current i in the photocell varies with the distance d as (a) i µ d2 (b) i µ d 1 d Following

(c) i µ

32.

(d) i µ process

®e +e hv ¬¾ +

is

(a) Pair production (b) photoelectric effect (c) Compton effect (d) Zeeman effect 33. During charging a capacitor, variations of potential V of the capacitor with time t is shown as V

(a)

V

(b) t

(a) loge

2 5

(b)

5 log e 2

(d) 5 loge 2 (c) 5 log102 29. If the electron in the hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation in term of Rydberg constant is

as

–

15V

(a) 10 mA (b) 15 mA (c) 20 mA (d) 5 mA 28. The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/C counts per minute at t = 5 min. The time, (in minute) at which the activity reduces to half its value, is

1 d2 known

(c)

t

V

(d) t

V t

34. When a resistor of 11 is connected in series with a electric cell. The current following in it is 0.5 A. Instead when a resistor of 5 is connected to the same electric cell in series, the current increases by 0.4A. The internal resistance of the cell is (a) 1.5 (b) 2 (c) 2.5 (d) 3.5

EBD_7443 2014-4

35.

36.

Target VITEEE

A battery is charged at a potential of 15 V in 8 h when the current flowing is 10A. The battery on discharge supplies a current of 5A for 15 h. The mean terminal voltage during discharge is 14V. The watt-hour efficiency of battery is (a) 80% (b) 90% (c) 87.5% (d) 82.5% A circular current carrying coil has a radius R. The distance from the centre of the coil on the 1 axis, where the magnetic induction will be th 8 to its value at the centre of the coil is

(a)

38.

39.

3

(b)

2 3R

(d)

41.

42.

43.

R 3

105 106 Hz (b) Hz 2 2 (c) 2 x 105 Hz (d) 2 x 106 Hz An eye can detect 5 × 104 photons per square meter per sec of green light ( = 5000Å ) while the ear can detect 10–13W/m2. The factor by which the eye is more sensitive as a power detector then ear is close to (a) 5 (b) 10 (c) 106 (d) 15

MnO4– + 8H + + 5e – ¾¾ ® Mn 2+ + 4H 2 O; E° = 1.51 V

E° = 1.23 V E°MnO – | MnO is 4 2

44.

45.

46.

47.

(a)

40.

The sodium extract of an organic compound on acidification with acetic acid and addition of lead acetate solution gives a black precipitate. The organic compound contains (a) nitrogen (b) halogen (c) sulphur (d) phosphorus The volume strength of 1.5 N H2O2 solution is (a) 16.8 L (b) 8.4 L (c) 4.2 L (d) 5.2 L

MnO 2 + 4H + + 2e – ¾¾ ® Mn 2+ + 2H 2 O;

2

R 3 The incorrect statement regarding the lines of force of the magnetic field B is (a) magnetic intensity is a measure of lines of force passing through unit area held normal to it (b) magnetic lines of force forms a close curve (c) inside a magnet, its magnetic lines of force move from north pole of a magnetic towards its south pole (d) due to a magnetic lines of force never cut each other Two coils have a mutual inductance 0.55 H. The current changes in the first coil according to equation I = I0 sin t. where, I0 = 10A and = 100 rad/s. The maximum value of emf in the second coil is (a) 2 (b) 5 (c) (d) 4 An L-C-R circuit contains R = 50 , L = 1 mH and C = 0.1µF. The impedence of the circuit will be minimum for a frequency of (c)

37.

R

PART - II (CHEMISTRY)

(a) 1.70 V (b) 0.91 V (c) 1.37 V (d) 0.548 V A metal has bcc structure and the edge length of its unit cell is 3.04Å. The volume of the unit cell in cm3 will be (a) 1.6 × 1021 cm3 (b) 2.81 × 10-23 cm3 (c) 6.02 × 10-23 cm3 (d) 6.6 × 10-24 cm3 Among [Fe(H2O)6]3+ , [Fe(CN)6]3– , [Fe(Cl)6]3– species, the hybridisation state of the Fe atom are, respectively. (a) d2sp3, d2sp3, sp3d2 (b) sp3 d2 ,d2 sp3 ,d2 sp3 (c) sp3d2,d2sp3,sp3d2 (d) None of the above Which of the following hydrogen bonds are strongest in vapour phase? (a) HF ……….. HF (b) HF ………... HCl (c) HCl ………… HCl (d) HF …………… HI The rate constant for forward reaction and backward reaction of hydrolysis of ester are 1.1 × 10–2 and 1.5 × 10–3 per minute respectively. Equilibrium constant for the reaction is CH3COOC2H5 + H2O

48.

CH3COOH + C2H5OH (a) 33.7 (b) 7.33 (c) 5.33 (d) 33.3 19.85 mL of 0.1 N NaOH reacts with 20 mL of HCl solution for complete neutralisation. The molarity of HCl solution is (a) 9.9 (b) 0.99 (c) 0.099 (d) 0.0099

Solved Paper 2014

2014-5

49. An f-shell containing 6 unpaired electrons can exchange (a) 6 electrons (b) 9 electrons (c) 12 electrons (d) 15 electrons 50. The standard molar heat of formation of ethane, CO2 and water (l) are respectively –21.1, –94.1 and –68.3 kcal. The standard molar heat of combustion of ethane will be (a) –372 kcal (b) 162 kcal (c) –240 kcal (d) 183.5 kcal 51. The solubility product of Ag2CrO4 is 32 × 10– 12. What is the concentration of CrO– ions in 4 that solution? (a) 2 × 10-4 M (b) 16 × 10-4 M -4 (c) 8 × 10 M (d) 8 × 10-8 M 52. The equivalent conductivity of a solution containing 2.54g of CuSO4 per L is 91.0 W–1 cm2 eq–1. Its conductivity would be (a) 2.9 × 10-3 -1 cm-1 (b) 1.8 × 10-2 -1 cm-1 (c) 2.4 × 10-4 -1 cm-1 (d) 3.6 × 10-3 -1 cm-1 53. The half-life of two samples are 0.1 and 0.8 s. Their respective concentration are 400 and 50 respectively. The order of the reaction is (a) 0 (b) 2 (c) 1 (d) 4 54. Which sequence of reactions shows correct chemical relation between sodium and its compounds? HCL(aq) (a) Na + O2 ¾¾ ® Na2O ¾¾ ¾® NaCl ¾¾¾ ® Na ® Na2CO3 ¾¾ CO2

D

O2 CO2 H2O (b) Na ¾¾ ® Na2O ¾¾¾ ® NaOH ¾¾ ¾ ® D Na2CO3 ¾¾ Na ® HCl (c) Na + H2O ¾¾ NaCl ® NaOH ¾¾® CO2 D ® Na ¾¾ ¾ ® Na2CO3 ¾¾

CO2 (d) Na + H2O ¾¾ ® NaOH ¾¾¾ ® Na2CO3 HCl Electrolysis NaCl ¾¾ ¾¾® ¾¾ ® Na

57. In the reaction, 8Al + 3Fe3O4 ¾¾ ® 4Al2O3 + 9Fe the number of electrons transferred from the reductant to the oxidant is (a) 8 (b) 4 (c) 16 (d) 24 58. The bond angles of NH3, NH+4 and NH 2– are in the order (a) NH 2– > NH 3 > NH +4 (b)

NH+4 > NH3 > NH 2–

(c)

NH3 > NH 2– > NH +4

(d) NH > NH +4 > NH 2– 59. A gaseous mixture containing He,CH4 and SO2 was allowed to effuse through a fine hole then find what molar ratio of gases coming out initially? (Given mixture contains He,CH4 and SO2 in 1 : 2 : 3 mole ratio). (a) (b) 2 : 2: 3 2 : 2 :3 (c) 4 : 4 : 3 (d) 1 : 1 : 3 60. According to Bohr ’s theory, the angular momentum for an electron of 3rd orbit is (a) 3 (b) 1.5 (c) 9

2

61. 2.76 g of silver carbonate on being strongly heated yields a residue weighing (a) 3.54 g (b) 3.0 g (c) 1.36 g (d) 2.16 g 62. The final product (IV) in the sequence of reactions PBr

Mg 3 ® I ¾¾¾ CH 3 CHOH ¾¾¾ ® Ether | CH3 CH2—CH2

O III II (a) CH3—CH OCH2CH2OH CH3 (b) CH3 — CHCH2CH2Br

(molten)

55. Purest form of iron is (a) pig iron (b) wrought iron (c) cast iron (d) steel 56. Which has the smallest size? (a) Na+ (b) Mg2+ 3+ (c) Al (d) P5+

(d)

(c)

CH3. CH3—CH —CH2CH2OH

CH3 (d) CH3—CH OCH2CH3 CH3

H2 O

IV is

EBD_7443 2014-6

63.

Target VITEEE 71.

Hg 2+ /H+ Ph — C º C — CH3 ¾¾¾¾ ¾ ® A.

O

Ph—H2C

Ph—C (a)

CH2

(b)

C

O

H3C

H3C

72.

OH

Ph—CH

Ph—C

CH

(c)

(d)

H3C

64. 65.

66.

67.

68.

69.

70.

C—OH H3C

Which of the following has an ester linkage? (a) Nylon-66 (b) Dacron (c) PVC (d) Bakelite Which of the following pairs give positive Tollen’s test? (a) Glucose,sucrose (b) Glucose,fructose (c) Hexanal,acetophenone (d) Fructose,sucrose Peptisation involves (a) precipitation of colloidial particles (b) disintegration of colloidal aggregates (c) evaporation of dispersion medium (d) impact of molecules of the dispersion medium on the colloidal particles Which of the following has the maximum number of unpaired d-electrons? (a) Fe2+ (b) Cu+ (c) Zn (d) Ni3+ Iodine is formed when potassium iodide reacts with a solution of (a) ZnSO4 (b) CuSO4 (c) (NH4)2SO4 (d) Na2SO4 Which of the following does not represent the correct order of the property indicated? (a) Sc3+ > Cr 3+ > Fe3+ > Mn3+ — ionic radii (b) Sc < Ti Ni2+ > Co2+ < Fe2+ —ionic radii (d) FeO < CaO < MnO < CuO —basic nature If the elevation in boiling point of a solution of 10 g of solute (mol. wt. = 100) in 100 g of water is D Tb, the ebullioscopic constant of water is (a) 10 (b) 100 Tb (b) D Tb

(d)

73.

DTb 10

74.

75.

Which of the following compounds cannot be prepared singly by the Wurtz reaction? (a) C2H6 (b) (CH3)2CHCH3 (c) CH3CH2CH2CH3 (d) All of the above can be prepared Which of the following oxides is strongly basic? (a) Tl2O (b) B2O3 (c) Al2O3 (d) Ga2O3 In Langmuir’s model of adsorption of a gas on a solid surface, (a) th e rate of dissociation of adsorbed molecules from the surface does not depend on the surface covered (b) the adsorption at a single site on the surface may involve multiple molecules at the same time (c) the mass of gas striking a given area of surface is proportional to the pressure of the gas (d) the mass of gas striking a given area of surface is independent of the pressure of the gas How many sigma and pi-bonds are there in the molecule of dicyanoethene (CN—CH = CH — CN)? (a) 3 sigma and 3 pi (b) 5 sigma and 2 pi (c) 7 sigma and 5 pi (d) 2 sigma and 3 pi What will be the order of reactivity of the following carbonyl compounds with Grignard’s reagent?

H

H

O

C

C

O

H3C

H I

II

(CH3)3C

H3C

C

C

O

O

(CH3)

H3C III

(a) I > II > III > IV (c) II > I > IV >III

IV (b) IV > III > II > I (d) III > II > I >IV

Solved Paper 2014

2014-7

NH2

PART - III (MATHEMATICS) ¾¾¾ ® A Ac2 O

76.

Br2 ¾¾¾¾ ® CH3COOH¾

H2 O

B ¾¾¾ ®C H+

CH3 The final product ‘C’ in the above reacrtion is

NHCOCH3

NH2

Br (a)

81.

dy x 2 + y 2 + 1 = The solution of , satisfying 2 xy dx

y(1) = 0 is given by (a) hyperbola (c) ellipse 82. If x.

COCH3

(b)

(a)

(b) circle (d) parabola

f ( xy ) dy + y = x. , then f (xy) is equal to f '( xy ) dx x2

(b)

k .e 2

k .e y

xy

2

CH3

CH3

COCH3

NH2

Br (c)

Br

CH3

CH3 77. Which of the following isomerism is shown by ethyl acetoacetate? (a) Geometrical isomerism (b) Keto-enol tautomerism (c) Enantiomerism (d) Diastereoisomerism 78. The final product obtained in the reaction, Br Heavy water Mg/ether ¾¾¾¾ ® A ¾¾¾¾¾ ® is

D

(a)

(b) OH

(c)

(c) k .e x (d) k .e 2 83. The differential equation of the rectangular hyperbola hyperbola, where axes are the asymptotes of the hyperbola, is (a)

(d)

OD

(d)

79. Among the following the strongest nucleophile is (a) C2H5SH (b) CH3COO– (c) CH3NH2 (d) NCCH2 80. Which set has different class of compounds? (a) Tranquillizers-Equanil, heroin,valium (b) Antiseptics-Bithional,dettol,boric acid (c) Analgesics-Naproxen,morphine,asprin (d) Bactericidal-penicillin,aminoglycosides, ofloxacin

2/2

dy dx

y

x

(b)

x

dy dx

y

dy y (d) xdy + ydx = c dx 84. The length of longer diagonal of the parallelogram constructed on 5a + 2b and a – 3b,

(b)

x

if it is given that |a| = 2 2 , |b| = 3 and the angle between a and b is

4

, is

(a) 15

(b)

113

(c)

(d)

369

593

85. If r = b × c + c × a + a × b and [a b c] = 2, then + + is equal to (a) r. [b × c + c × a + a × b] 1 r. (a + b + c) 2 (c) 2r. (a + b + c) (d) 4 86. If a, b, c are three non-coplanar vectors and p, q, r are reciprocal vectors, then (la + mb + nc). (lp + mq + nr) is equal to (a) l + m + n (b) l 3 + m3 + n3 2 2 2 (c) l + m + n (d) None of these 87. If the integers m and n are chosen at random from 1 to 100, then the probability that a number of the form 7n + 7m is divisible by 5, equals to

(b)

(a)

1 4

(b)

1 2

(c)

1 8

(d)

1 3

EBD_7443 2014-8

88.

Let X denote the sum of the numbers obtained when two fair dice are rolled. The variance and standard deviation of X are (a)

89.

Target VITEEE

31 and 6

91.

92.

(b)

35 and 6

35 6

(b)

1 3

theorem for which f (x) = 25 - x 2 in [1,5], is (a) 5 (b) 1 (c) (d) None of these 15 93.

94.

é 3 4ù If A = êê5 7úú , then A . (adj A) is equal to ë û (a) A (b) | A | (c) | A | (d) None of these If there is an error of k% in measuring the edge of a cube, then the percent error in estimating its volume is (a) k (b) 3 k k (d) None of these 3 If the system of equations x + ky – z = 0, 3x – ky – z = 0 and x – 3y + z = 0, has non-zero solution, then k is equal to (a) –1 (b) 0 (c) 1 (d) 2 If the points (1, 2, 3) and (2, –1, 0) lie on the opposite sides of the plane 2x + 3y – 2z = k, then (a) k < 1 (b) k > 2 (c) k < 1 or k > 2 (d) 1 < k < 2

(c)

95.

96.

97.

98.

1 cos x If D (x) = 1 + sin x cos x sin x sin x

1 - cos x 1 + sin x - cos x , 1

D ( x ) dx is equal to

1 1 1 (b) (c) 0 (d) 4 2 4 Let f (x), be differentiable " x. If f (1) = –2 and f (x) ³ 2 " x [1, 6], then (b) f (6) ³ 8 (a) f (6) < 8 (d) f (6) £ 5 (c) f (6) ³ 5 m

2r 1

99.

1 (c) (d) None of these 4 If the vertices of a triangle are A(0, 4, 1), B(2, 3, –1) and C(4, 5, 0), then the orthocentre of ABC, is (a) (4, 5, 0) (b) (2, 3, –1) (c) (–2, 3, –1) (d) (2, 0, 2) The equation of normal to the curve y = (1 + x)y + sin–1(sin2 x) at x = 0 is (a) x + y = 1 (b) x – y = 1 (c) x + y = –1 (d) x–y = –1 The value of c from the Lagrange’s mean value

p/ 4

ò0

(a)

17 31 17 35 (c) and (d) and 6 6 6 6 A four digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd, is

(a) zero

90.

31 6

then

If

r

= m

2

1

2

Cr

1

m

m 1

, then

sin 2 (m 2 ) sin 2 (m) sin 2 (m 1)

å D , is m

the value of (a) 1 (c) 2 100. Two lines

r =0

r

(b) 0 (d) None of these

x -1 y + 1 z -1 = = and 2 3 4

x -3 y - k = = z intersect at a point, if k is 1 2 equal to

(a)

2 9

(b)

1 2

(c)

9 2

(d)

1 6

x 101. The minimum value of log x is 1 e (c) e 2 (d) e 3 102. The triangle formed by the tangent to the curve f (x) = x2 + bx –b at the point (1,1) and the coordinate axes lies in the first quadrant. If its area is 2, then the value of b is (a) –1 (b) 3 (c) –3 (d) 1 103. The statement (p Þ q) Û (~ p Ù q) is a (a) tautology (b) contradiction (c) Neither (a) nor (b) (d) None of these

(a) e

104. If x + i y = equal to (a) 3x – 4 (c) 4x + 3

(b)

3 , then x2 + y2 is 2 + cos q + i sin q

(b) 4x –3 (d) None of these

Solved Paper 2014

2014-9

105. The negation of (~ p Ù q) Ú (p Ù ~q) is (a) (p Ú ~q) Ú (~p Ú q) (b) (p Ú ~q) Ù (~p Ú q) (c) (p Ù ~q) Ù (~p Ú q) (d) (p Ù ~q) Ù (p Ú ~q) 106. The normals at three points P, Q and R of the parabola y2 = 4ax meet at (h, k). The centroid of the D PQR lies on (a) x = 0 (b) y = 0 (c) x = –a (d) y = a 107. The minimum area of the triangle formed by any tangent to the ellipse coordinate axes is

x2 y2 + = 1 with the a2 b2

( a + b) 2 2 ( a - b) 2 (c) ab (d) 2 108. If the line lx + my – n = 0 will be a normal to the

(a) a2 + b2

hyperbola, then

(b)

a2 b2 (a2 + b2 )2 , where - = l 2 m2 k

k is equal to (a) n (b) n 2 3 (c) n (d) None of these 109. If cos + i sin , b = cos + i sin , c = cos cos ( to

+ i sin ) + cos (

b c a and + + = 1 , then c a b ) + cos ( ) is equal

3 3 (b) 2 2 (c) 0 (d) 1 110. If | z + 4 | £ 3, then the greatest and the least value of | z + 1| are (a) –1, 6 (b) 6, 0 (c) 6, 3 (d) None of these 111. The angle between lines joining the origin to the

(a)

point of intersection of the line 3 x + y = 2 and the curve y2 – x2 = 4 is 2 p (a) tan –1 3 (b) 6 æ 3 ö÷ p (c) tan–1 ççç ÷÷ (d) çè 2 ø÷ 2 112. If the area of the triangle on the complex plane formed by the points z, z + i z and iz is 200, then the value of 3 | z | must be equal to

(a) 20 (b) 40 (c) 60 (d) 80 113. Equation of the chord of the hyperbola 25x2 –16y2 = 400 which is bisected at the point (6, 2) is (a) 6x – 7y = 418 (b) 75x – 16y = 418 (c) 25x – 4y = 400 (d) None of these 114. If a plane meets the coordinate axes at A,B and C such that the centroid of the triangle is (1, 2, 4), then the equation of the plane is (a) x + 2y + 4z = 12 (b) 4x + 2y + z = 12 (c) x + 2y + 4z = 3 (d) 4x + 2y + z = 3 115. The volume of the tetrahedron included between the plane 3x + 4y – 5z – 60 = 0 and the coordinate planes is (a) 60 (b) 600 (c) 720 (d) 400 116.

ò

2x 0

(sin x+ | sin x |) dx is equal to

(a) 0 (c) 8

(b) 4 (d) 1

117. The value of

ò

2 0

[ x 2 ] dx, where [ . ] is the

greatest integer function, is (a) 2 – 2 (c)

(b) 2 +

2 –1

118. If l(m, n) =

ò

(d) 1 0

2

2 –2

t m (1 + t)n dt, then the expression

for l(m, n) in terms of l(m + 1, n + 1) is

n 2n . l (m +1, n -1) m +1 m +1 n . l (m + 1, n -1) (b) m +1 n 2n l. ( m +1, n -1) + (c) m + 1 m +1 m . l ( m + 1, n -1) (d) n +1 119. The area in the first quadrant between x2 + y2 = 2 and y = sin x is (a)

3

(a)

-8 4

3

(b)

4

3 -16 -8 (b) 4 2 120. The area bounded by y = xe |x| and lines | x | = 1, y = 0 is (a) 4 sq units (b) 6 sq units (c) 1 sq unit (d) 2 sq units 3

(c)

EBD_7443 2014-10

Target VITEEE

SOLUTIONS PART - I (PHYSICS) 1.

(b) Amplification factor of a triode,

=

3.

6

8.

As wavelength less than that of yellow colour and hence can initiate photoelectric effect irrespective of intensity. (c) Radius of circular path

Vg

Vs = –50 (–20 ) = 10V (a) As we know, energy E = mc2 = 0.5 × 10 –3 × (3×108)2 = 4.5 × 1013J 4.5 1013

(a)

Vp

Vp

2.

7.

r=

1.25 107 kWh

3.6 10 (b) Here, diode in lower branch is forward and in upper branch is reversed biased

9.

5 A i= 20 30 50 (b) Power consumed by motor = 5 kW = 5 × 10 3 W = 5000W

Power used in lifting water =

mgh t

Power used 100% Power consumed

=

10.

11.

6.

6 107

2 r. 0N A

I

600 4

10 7

2000

2

0.3

(d) Current I =

V 100 = = 10A R 10

1 Energy, E= LI2 2

3454.5 100 69% 5000 (c) For a closed circuit cell supplies a constant current in the circuit.

For cell E = V + Ir For V = 50 V E = 50 + 11r Similarly, for V = 60 V E = 60 + r From eqs. (i) and (ii), we get E = 61V (b) The ionosphere can reflect electromagnetic waves of frequency less than 40 MHz but not of frequency more than 40 MHz.

V e B m

= 1.5 H

=

5.

mV B

1.7 1011 15 10 2 = 2.35 × 10–2 m = 2.35 m (a) As we know, Self-inductance of the solenoid

L=

= 7.5 × 9.8 × 4.7 = 3454.5 kW Efficiency =

violet

r=

5

4.

red

1 = × 5 × (10)2 = 250J 2 (c) Given : V=750×10 –3 V; Ig=15×10–3A and I = 25A Using, a y Ig

=

750 10 3

=50 15 10 3

Ig =

S ×I S a

15 × 10–3 = S = 0.03

S × 25 S 50

1.5 10 4

Solved Paper 2014

12.

2014-11

I1 125 1 (d) Here, r = R I 1 =2 100 2

5 1 1 = 2 × = 0.05 4 4 (d) Resistances 10 , 60 and 100 are in series and they together are in parallel to 200 resistance. When a potential difference of 15 V is applied across 200 then current through it

18.

(b) Radius of path r time = r rp

=2

13.

I= 14.

C= 15.

19.

V2

2 400

2 2 = 25

20.

17.

2

2 m 2 m1 k A 1 k retained m1 m 2 A 1 After collision kinetic energy retained by

neutron

R2

10 10 6 4 = 100 100 = 0.6 + 0.4 = 1 R 1 = 10 % R 10 (b) If the number of electrons increase, their number of collision, increasing the thermal and electrical resistance. So, electrical and thermal conductivities both decrease.

Nucleus at rest

Neutron

A + B. C = Y (b) In series combination equivalent resistance, R = R1 + R2 =6 + 4 = 10 k Error in combination, R1 +

V I Here, as R decreases, so V decrease and I should increase. (c) Fraction retained by nucleus m1 = 1 m2 = A

According to Ohm’s law, V = IR or, R =

F

(d) Boolean expression for Logic gate-1 B. C = Y’ Boolean expression for Logic gate-II

R=

V

(R)

A+ B.C Y" Boolean expression for Logic gate-III 16.

r 4 r 10 2cm . 2 10 (d) The effective resistance will decrease when resistance R is connected in parallel with the voltmeter.

1 2 LI 2

1

=

q

A

(b) Energy stored in capacitor = energy stored in inductance

LI2

qp

m mp

or,

15 = 7.5×10–2A 200

1 2 i.e., CV 2

2mv q

21.

(d) Here, tan tan or

22.

=

A 1 A 1

2

E

tan ' = cos

1 3/2

= tan–1

=

=

tan 45 cos 30

2 3

2 3

(b) Luminous efficiency for the same power supply, 40 W fluorescent tube gives more light. Hence, 40 W fluorescent tube has greater

EBD_7443 2014-12

23.

Target VITEEE

V I

(b) Resistance, R =

2.3 0.3 10 10

5 log 2 7 T = 5 loge 2 Now, let t’ be the time after which activity reduces to half

–1 =

3

2 103 = 0.2 × 103 = 0.2k 10 (c) Here, number of electrons n e = 3.13 × 1015 and number of protons n p = 3.12 × 1015 Current I = n eqe + np qp = 3.13 × 10 15 × 1.6 × 10–17 + 3.12 × 10 15 × 1.6 × 10 –19 = 1 × 10 –3 = 1 mA Now, due to excess charge on electrons, the direction of the current will be towards right. (b) In conductor separation between conduction and valence bands is zero and in insulator, it is greater than 1eV. Hence, in semiconductor the separation between conduction and valence band is 1 eV. (c) Potential of big drop = q × n 2/3 = 100 V

R=

24.

25.

26. 27.

(d)

250V

l'/ 5log e 2

1 1 2 2 t' = 5 ologe2

29.

1

30.

15V

31.

20 15 5 = 20 mA 250 250 iZener = 20 – 15 = 5mA (d) After n half-lives

i250 =

N

1 2

N0 e

n

cN0 cN0

1 2

t/7

1 2

1

1

2

3

2

1 1 4 9

2 =R

5R 36

=

hc

hc '

0

ev'0 eV0

'

2ev’0 + 2

= ev0 +

=

100 200

1 2

ev 0 7.7 4.7 = 1.5V 2 2 (d) As we know, Photoelectric current depends on the intensity of incident radiation i.e., i I

ev’0 =

15 i1 = mA = 15 mA 1 R = 250

N N0

9 4 36

and ev0

For R = 1k

28.

R

36 5R (d) From Einstein’s Photoelectric equation,

ev0 +

i2

20V

=R

1

=

i1

i

(b) As we know

5/7

1 so, i d2 d2 (a) The creation of an elementary particle and its antiparticle usually from a photon (or another neutral boson ) is called Pair production. This is allowed, provided there is enough energy available to create the pair.

But, intensity of radiation I

32.

–

5/7

1 1 e 2 Taking log on both sides, we get 5 1 log 1 – log e = log 7 2

1

e 0

Neutron Photon )

po sitr on

0 +

e

Solved Paper 2014 33.

2014-13

(a) For charging the capacitor, q = q 0 t CR 1 e

37. 38.

And, Potential difference V = V0 (1 – e–t/CR) 39.

V V0 0.632 V0

(c) Inside a magnet, magnetic lines of force move from south pole to north pole. di d (i sin t) = 0.005 × dt dt 0 = 0.005 × i0 cos t emax = 0.005 × 10 × 100 = 5 (a) Impedance of L-C-R circuit will be minimum for a resonant frequency so,

(b) E.M.F. e = M

v0 =

Growth of potential

34.

40.

R r

E 11 r

E = 5.5 + 0.5 r

E

or, E = 4.5 + 0.9r 5 r On solving we get r = 2.5 (c) Power of battery, when charged is given by P1 = V1I1 Electrical energy dissipated id given by E1 = P1t1 E1 = V1I1 t1 = 15 × 10 × 8 = 1200 Wh Similarly, the electrical energy dissipated during the discharge of battery is given by E2 = V2 I2t2 = 14 × 5 × 15 = 1050 Wh Hence, watt-hour efficiency of the battery E2 100 0.875 × 100 = 87.5 % E1 0.9 =

35.

E

(c) Here, i = 0.5 =

2

t

(b) Here the ratio,

BCentre x = 1 Baxis R2

1 Also, Baxis = Bcentre 8

8 1

3=

1

x2

x2

X2 = 3R2

R2

or, x =

4=1+

R2

3R

=

105 Hz 2

12375 = 2.475 eV = 4 × 10–19 J 5000 Minimum intensity to which the eye can respond. leye = (photon flux) × energy of a photon leye = (5 × 104) × 4 × 10–19 = 2 × 10–14 W/m2 Now, lesser the intensity required by a detector for detection more sensitive it will

(a) Energy =

10 13 2 10 14

5

PART - II (CHEMISTRY) 41.

42. 43.

3/ 2

LC

1 10 3 0.1 10 6

lear be = l eye

2 3/ 2

36.

2

1

= CR

1

(c) The organic compounds containing sulphur when react with sodium metal give Na 2S. The Na 2 S when react with lead acetate forms black ppt. of PbS. Na2S + (CH3COO)2Pb PbS + 2CH3COONa Black ppt (b) Volume strength = 5.6 × normality = 5.6 × 1.5 = 8.4 L (a) On subtracting eqn. (ii) from (i) we get

MnO 4 + 4H+ + 3e–

x2 R2

1.51 5 2 1.23 3 E3 = 1.70 V (b) Volume of unit cell (V) = a3 = ( 3.04 × 10–8 cm )3 = 2.81 × 10–23 cm3

–E3 =

44.

MnO2+2H2O

EBD_7443 2014-14

45.

Target VITEEE

(c)

Now, eqs. 2 × (b) + 3 × (c) – (a) 3d

3 O 2CO2 + 3H2O 2 2 Heat of combustion of Ethane x = 2(–94.1) + 3 (–68.3) – (–21.1) = (–188.2) + (–204.9) – 21.1 = – 372 kcal

4d

4p

4s

C2H6+

Fe= 3+

'' '' '' '' '' ''

[Fe(H2O)6 ] =

3 2

sp d hybridisation 3–

51.

'' '' '' '' '' ''

[Fe(CN)6] = 2

S

S=

3 2

47.

(a) A compound havin g maximum electronegative element will form strong hydrogen bond. F is the most negative element among halogens hence form strongest hydrogen bond. (b) Given kf = 1.1 × 10–2 , kb = 1.5 × 10–3 kc=

48.

49.

kf 1.1 10 2 = kb 1.5 10 3

(d)

52.

K sp 4

2 10 4 M (a) We know that,

= 91 0.1 1

1/3 32 10 12 4

1/3

= eq .C Given, eq = 91.0

7.33

normality (c) Molarity of base = Acidity

S

KSP = (2s)2 s = 4s .s = 4s3

sp d hybridisation

46.

2S 2

'' '' '' '' '' ''

[Fe(Cl)6] =

2Ag+ + Cr O 24

Ag2CrO4

3

d sp hybridisation 3–

(a)

1

cm 2 eq 1

2.54 eq.cm 3 159 / 2 1000

0.1

M1V1 = M2V2 0.1 × 19.85 = M2 × 20 M2 = 0.09925 0.099

–1 cm2 eq–1

53.

1cm 1

= 2.9 × 10–3 (b) It is known, t1/2 1 t1/2 2

a2 a1

(n 1)

Here, n = order of the reaction Given, (t1/2)1 = 0.1 s, a1= 400 (t1/2)2 = 0.8 s, a2= 50 On putting the values, 5 + 4 + 3 + 2 + 1 = 15 Equation of normal of x = 0 and y = 1 is y – 1 = –1 (x – 0) y – 1 = –x x + y = 1 50.

(a) Given, (a) 2C + 3H2 (b) C + O2 (c) H2 + 1 O2 2

0.1 0.8

CO2 ; H2O ;

H = –21.1 kcal H = – 94.1 kcal H = –68.3 kcal

(n 1)

Taking log on both sides log

C2H6 ;

50 400 0.1 0.8

n 1 log

1 1 n 1 log 8 8 n–1=1 n=2 log

50 400

Solved Paper 2014 54.

2014-15

2NaOH + H2 Na2CO3 + H2O 2NaCl + H2O + CO2

(d)

2Na + 2H2O 2NaOH + CO2 Na2CO3 + 2HCl

NaCl Electrolysis (molten)

+

+e

55. 56. 57. 58.

+

Na

–

Cl

62.

CH3–CH–OH

–e

n He n 'CH

1 16 2 4

CH3 CH2–CH2 O CH3–CH–CH2CH2OMgBr CH3 H2O CH3–CH–CH2CH2OH CH3 3-methyl butanol

63.

n'

2+

Ph–C C–CH3+H2O

2

2+

O Ph–C–CH2CH3

4 3

64.

n 'He : n 'CH : n 'SO 4:4:3. 4 2 (a) Angular momentum, mvr nh 2

1.5h

2Ag

CO2

1 O2 2

276g 216g As 276g of Ag2CO3 gives = 216g of Ag Hence, 2.76g of Ag2CO3 will give =

OH

2.76 216 276

2.16g

–H2O

Ph–C–CH2CH3 OH

(b) Condensation of diacid with dialcohol leads to ester linkage,

h

Ag2CO3

+

–COOH nHOCH2CH2OH+nHOOC– Ethylene glycol Terephthalic acid (dialcohol) (diacid)

h 2 (d) Silver carbonate on being strongly heated decomposes as = 3h

61.

3 h 2

OH – Ph–C–CH–CH 3 Hg /H

So, molar ratio will be,

=

+

Hg /H

1 1

1 64 3 4

He

n 'SO

60.

(a)

–

4

CH3 CH3–CH–MgBr

(no/p) (one/p) (two/p) (c)

Mg CH3–CH–Br ether

–

NH 4 > NH3 > NH 2

59.

PBr3

CH3

–

Na Cl (b) The purest form of iron is wrought or malleable. (d) P5+ having maximum nuclear charge per electron. Therefore, its size is smallest. (d) 8Al 8 Al3+ + 24e– 8/3+ 9 Fe + 24e– 9Fe Total 24 electrons are transferred. (b) On increases the number of lone pairs of electrons, bond angle decreases. Therefore, order of bond angle is

'

(c)

–H2O 65.

66.

–O–CH2CH2–OOC–

–CO–

n Ester linkage dacron (b) Aldehydes and -hydroxy ketones give positive Tollen’s test. Glucose being a polyhydroxy aldehyde and fructose an -hydroxy ketone give positive Tollen's test. (b) Peptisation is the process in which freshly prepared precipitate disintegrates into colloidal solution.

EBD_7443 2014-16

67.

68.

Target VITEEE

(a) Fe2+(24) = [Ar] 3d6 4s0 It has 4 unpaired electrons Cu+(28) = [Ar] 3d104s0 It has 0 unpaired electron Zn(30) = [Ar] 3d10 4s2 It has 0 unpaired electron Ni3+(25) = [Ar] 3d7 4s0 It has 3 unpaired electrons (b) CuSO4 + 2Kl

76.

NH2

70.

2Cul + l2 Cuprous iodide (a) The correct order of ionic radii is Cr 3+ > Mn3+ > Fe3+ > Sc3+. 1000 K b w (c) We know that, Tb W M 1000 K b w M= W Tb

CH3 (A)

72.

73.

74.

75.

(CH3)3C (CH3)3C

C–O

NH2

NHCOCH3 Br

Br

HOH

CH3 (B)

CH3 (C)

Tb

71.

Br2/CH3COOH

CH3

Cul2 + K2SO4

1000 K b 10 100 100 Tb K b (b) If two different alkyl halides (R1–X and R2–X) are used, a mixture of three alkanes is obtained which is difficult to separate. (b) On moving down the group, the nature of the oxides of group 13 elements changes from weakly acidic to amphoteric and amphoteric to basic. Tl 2 O in aqueous solution gives TlOH which soluble and a strong base. (c) According to Langmuir’s adsorption isotherm, the mass of gas striking a given area of surface is proportional to the x k 'p pressure of the gas as m 1 kp (c) 2 N C–C=C–C N 2 H H 7 sigma and 5 pi (a) As the number and the size of the alkyl groups increases, reactivity decreases. Hence, the correct order of reactivity is H H H3C – – C O> C O> C– O > H H3C H3C

NHCOCH3

Ac2O

2Cul2

69.

(d)

O

O

77.

(b) CH3 –C–CH2–C–OC2H5 (Keto)

OH

O

(enol) MgBr

Br 78.

(b)

Mg ether (A)

D Heavy water

79.

80.

(B) (a) Nucleophiles are the species which have tendency to donate a pair of electrons. They can be neutral or negatively charged. The nucleophilic power depends on the tendency of species to donate the electrons. Due to the presence of +I effect, it increases. Hence, higher the +I effect, higher the nucleophilic power. (a) Heroin is a narcotic analgesic and not used as tranquilizer.

Solved Paper 2014

2014-17

PART - III (MATHEMATICS) 81.

(a)

x2

dy dx

y2 1 2xy

84.

2xy dy

x 2 1 dx y 2dx

xd y2

y2dx

x2 1

x2

x2

d

y2 x

1

y2 x

x

1 C x

y

2

x

2

1 x2

82.

i.e.,

d xy dx

x

f ' xy f xy

.dx

dx

x2 e 2 .eC

83.

85.

a.b c

abc

593

a.c a

a.a b

0 0

Similarly, r.b =

abc and r.c =

[abc]

1 1 r. a b c r.a r.b r.c 2 2 1 abc = 2

C

x2 / 2 C

0

(b) r.a = =

xdx

x2 k.e 2

dy dx

16 8 25 9 40 6

1 2 2 (c) p, q and r are reciprocal vectors of a, b and c respectively. So, p.r = 1, p.b = 0 = p.c q.a = 0, q.b = 1, q.c = 0 r.a = 0, r.b = 0, r.c = 1 (la + mb + nc) . (lp + mq + nr) = l2 + m2 + n2 (a) Let I = 7n + 7m , then we observe that 7i, 72, 73 and 74 ends in 7, 9, 3 and 1, respectively. Thus, 7i ends in 7, 9, 3 or 1 according as i is of the form 4k + 1, 4k+2, 4k–1, respectively. If S is the sample space, then n(S) = (100)2 7m + 7n is divisible by 5, if (i) m is of the form 4k + 1 and n is of the form 4k – 1 or (ii) m is of the form 4k + 2 and n is of the form 4k or

=

(b) The differential equation of the rectangular hyperbola xy = c2 is y+ x

(4a)2 (5 b 2 ) 2 | 4a || 5b | cos 45

86.

x2 2

1 2

288 9 12 6 225 15 other diagonal is (5a + 2b) – (a – 3b) = 4a + 5b Its length is

x dx

d xy

e

36 8 9 12 2 2 3

f (x, y) f ' x, y

log f xy f xy

(c) Given : a 2 2, b 3 One diagonal is 5a + 2b + a – 3b = 6a – b Length of one diagonal

1 Cx

f ' xy d xy f xy

y

= 6a b

f (xy) x. f ' xy

y

dy dx

36a2 b 2 2 6 a . b .cos 45

When x = 1, y = 0 0 = 1–1 + C C 0 The solution is x2–y2 = 1 i.e., hyperbola. dy (a) x. dx

x

87.

EBD_7443 2014-18

Target VITEEE (iii) m is of the form 4k–1 and n is of the form 4k+1 or (iv) m is of the form 4k and n is of the form 4k + 1 or So, number of favourable ordered pairs (m, n) = 4 × 25 × 25 4 25 25

Required probability = 88.

100

2

2 2 1 32 2 42 3 52 4 6 2 5 7 2 6 82 5 9 2 4 102 3 112 2 122 1

=

1 4

1974 49 36 1974 1764 = 36

=

(b) Let x denote the sum of the numbers obtained when two fair dice are rolled. So, X may have values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

=

1 P (X = 2) = P (1,1) = 36

P (X = 3) = P{(1, 2), (2, 1)} =

2 36

=

89.

5 6 ; P(X=7) = ; 36 36 5 ; 36

4 3 ; P (X=10) = ; P(X=11) 36 36

2 ; 36

1 36 Probability distribution table is given below

2

3

4

5

6

7

8

9

10 11 12

1 2 3 4 5 6 5 4 3 2 1 P(X) 36 36 36 36 36 36 36 36 36 36 36

Mean X

=

XP X

36 7

Variance =

35 6 (d) Given digits are 1, 2, 3, 4. Possibilities for units place digit (either 1 or 3) = 2 Possibilities for ten’s digit = 3 Possibilities for hundred’s place digit = 2 Possibilities for thousand place’s digit = 1 Number of favourable outcomes = 2 × 3 × 2 × 1 = 12 Number of numbers formed by 1, 2, 3, 4 (without repetitions) = 4!

90.

12 4 3 2

X 2P X

X

2

1 2

(b) Vertices of ABC are A(0, 4, 1) , B (2, 3, –1) and C (4, 5, 0). AB =

(2 0) 2

=

4 1 4

BC = =

2 1 3 2 4 3 5 4 6 5 7 6 8 5 9 4 10 3 11 2 12 1 252 36

35 6

Required probability =

P (X = 12) =

X

35 6

And, SD =

P(X=8) = P (X = 9) =

210 36

Variance =

3 4 P (X = 4) = , P(X=5) = : 36 36

P (X = 6) =

72

36

4 2

2

4 4 1

and CA =

4 0

3 4

2

1 1

2

3 5 3

2

0 1

2

2

0 1

3 2

5 4

2

= 16 1 1 3 2 AB2 + BC2 = AC 2 ABC is a right angled triangle. We know that, the orthocentre of a right angled triangle is the vertex containing the right angle. Orthocentre is point B (2, 3, –1).

Solved Paper 2014 91.

2014-19

(a) Given curve is y = (1 + x)y + sin –1 (sin2 x) On differentiating w.r.t x, we get

dy dx

1 x

y dy log 1 x 1 x dx

y

2sin x cos x

+ dy dx at 0,1

92.

at x

1

1 sin 4 x 0, y 1

f' (c) =

f 5

f 1

5 1

0 2 6 4

=

6 2 4 c2 = 6 (25 – c2) 4c2 = 150 – 6c2

10 c2 = 150

c2

15

25 c2

= 15

A=

A

96.

=–

c = 15 (c)

95.

2

25 c

c= 1,5

21 20 1

x 3x3

x

3kx 3 3k 3 .x 100 100 = 3K% of original volume (c) The system has non- zero solution, if

1

k

3

k

1

1

3

1 =0 1

1(–k –3) –k (3 + 1) –1(–9 + k) = 0 – 6k + 6 = 0 k=1 (d) The points (1, 2, 3) and (2, –1, 0) lie on the opposite sides of the plane 2x + 3y – 2z – k = 0 So, (2 + 6 –6 – k) (4 – 3 –k) < 0 .... (i) (k – 1) (k – 2) < 0 1< k p1 If a gas mixture contains 2 moles of O2 and 4 moles of Ar at temperature T, then what will be the total energy of the system (neglecting all vibrational modes) (a) 11 RT (b) 15 RT (c) 8 RT (d) RT In the adjoining figure, two pulses in a stretched string are shown. If initially their centres are 8 cm apart and they are moving towards each other, with speed of 2cm/s, then total energy of the pulses after 2 s will be

8 cm

35.

(a) Zero (b) Purely kinetic (c) Purely potential (d) Partly kinetic and partly potential When two waves of almost equal frequency n 1 and n2 are produced simultaneously, then the time interval between succesive maxima is (a)

1 n1 + n2

(c)

1 1 n1 n2

38.

(a) 220 V , 2.2 A (b) 100 V, 2.0 A (c) 220 V , 2.0 A (d) 100 V, 2.2 A If the work done in turning a magnet of magnetic moment M by an angle of 90º from the magnetic meridian is n times the corresponding work done to turn it through an angle of 60º, then the value of n is (a) 1 (b) 2

1 1 (d) 2 4 The capacitance of a parallel plate capacitor with air as dielectric is C. If a slab of dielectric constant K and of the same thickness as the separation between the plates is introduced so as to fill 1/4th of the capacitor (shown in figure), then the new capacitance is

(b) 39.

Air

K

t

t

(b)

1 1 + n1 n2

(a) (K + 2)

C 4

(b) (K + 3)

(d)

1 n1 - n2

(c) (K + 1)

C 4

(d) None of these

C 4

Solved Paper 2013

2013-5

40. Seven resistance are connected between points A and B as shown in adjoining figure. The equivalent resistance between A and B is 10

10

A

3

5

B

CH2COOH

6

8

43. Benzene diazonium chloride on treatment with hypophosphorous acid and water yield benzene. Which of the following is used as a catalyst in this reaction? (a) LiAlH4 (b) Red p (c) Zn (d) Cu+ 44. Consider the following reaction sequence,

CH2COOH (a) 5 W (c) 4 W

(b) 4.5 W (d) 3 W

PART - II (CHEMISTRY) 41. Which of the following does not undergo benzoin condensation?

CHO

CH2CHO (a)

(b)

CHO

CHO

(c)

(d)

COOH + NaHCO3 ®

42.

CO2 +

COONa +

*

C is with the product (a) CO2

(b)

COONa

(c) Both (a) and (b) (d) None of the above

A

D

Alcoholic KOH

B

CF3CO 3H

O3, H2O

C

Alkaline KMno4

E

Isomers are (a) C and E (b) C and D (c) D and E (d) C,D and E 45. When a monosaccharide forms a cyclic hemiacetal, the carbon atom that contained the carbonyl group is identified as the …. Carbon atom,because (a) D,the carbonyl group is drawn to the right (b) L, the carbonyl group is drawn to the left (c) acetal,it forms bond to an –OR and an – OR’ (d) anomeric, its substituents can assume an b or position 46. Which of the following is/ are - amino acid? (a)

CH3

OCH3

P/Br 2

NH3 Q

CO2

(b)

Q

N

CO2

H H (c) Both (a) and (b) (d) None of these 47. Calculate pH of a buffer prepared by adding 10 mL of 0.10 M acetic acid to 20 mL of 0.1 M sodium acetate [pKa (CH3COOH) = 4.74] (a) 3.00 (b) 4.44 (c) 4.74 (d) 5.04 48. The equivalent conductance of silver nitrate solution at 250°C for an infinite dilution was found to be 133.3 –1 cm2 equiv-1. The transport number of Ag+ ions in very dilute solution of AgNO3 is 0.464. Equivalent conductances of Ag+ and NO-3 (in –1 cm2 equiv–1) at infinite dilution are respectively (a) 195.2, 133.3 (b) 61.9, 71.4 (c) 71.4, 61.9 (d) 133.3, 195.2

EBD_7443 2013-6

49.

Target VITEEE

Treating anisole with the following reagents, the major product obtained is I. (CH3)3 CCl, AlCl3 II. Cl2, FeCl3 III. HBr, Heat OH

(c) Both (a) and (b) (d) None is correct

Br

Br

Br

(a)

53.

(b) C(CH3)3

(b)

(a)

KHSO

HCIO

4 A Glycerol A –A and B respectively are

O

O

(a)

C(CH3)3

CH2

CH CH Cl

OCH3

OH Br

(b) CH2

(d)

O

CHCH OH

C(CH3)3

C(CH3)3

50.

51.

(c)

Ketones [R—C—R’] where, R = R’ = alkyl

CH3CH2CHO

Cl

CH3CH2CH

OH Cl

O

O group can be obtained in one step by (a) Hydrolysis of esters (b) Oxidation of primary alcohols (c) Oxidation of secondary alcohols (d) Reaction of acid halide with alcohols An optically active compound ‘X’ has molecular formula C4H8O3. It evolves CO2 with aqueous NaHCO3. ‘X’ reacts with LiAlH4 to give an achiral compound.’X’ is

54.

Phenol is heated with phthalic anhydride in the presence of conc. H2SO4. The product gives pink colour with alkali. The product is (a) phenolphthalein (b) bakelite (c) salicylic acid (d) flurorescein

(a)

55.

C6H5NH2

CH 3CH 2CHCOOH

(d) CH2

CH 3CHCOOH

56. (d) CH 3CHCH2C OOH

Y 2 Z, Z is identified as (a) C6H5 — NH—CH3 (b) C6H5 — CH2—NH2 (c) C6H5 — CH2—COOH (d) C6H5—COOH B can be obtained from halide by van-Arkel method. This involves reaction

(a)

OH

Product is/are

CuCN

+

OMe

OH

X

0ºC

H O/H

(b) CH 3CHCOOH (c)

CHCH ClO

OH

52.

OH

O

Cl

(c)

B,

(b) conc.H2SO4

products.

2Bl3 Red hot W or Ta Red hot W or Ta

(c) Both (a) and (b) (d) None of the above

Solved Paper 2013

2013-7

57. NH4Cl(s) is heated in a test tube. Vapours are brought in contact with red litmus paper, which changes it to blue and then to red. It is because of (a) formation of NH4OH and HCl (b) formation of NH3 and HCl (c) greater diffusion of NH3 than HCl (d) greater diffusion of HCl than NH3 58. Out of H2S2O3, H2S2O4, H2SO5 and H2S2O8 peroxy acids are (a) H2S2O3, H2S2O8 (b) H2SO5, H2S2O8 (c) H2S2O4, H2SO5 (d) H2S2O3, H2S2O4 59. The density of solid argon is 1.65 g per cc at 233°C. If the argon atom is assumed to be a sphere of radius 1.54 × 10-8 cm, what per cent of solid argon is apparently empty space? (Ar = 40) (a) 16.5% (b) 38% (b) 50% (d) 62% 60. When 1 mole of CO2(g) occupying volume 10L at 27°C is expanded under adiabatic condition, temperature falls to 150 K. Hence,final volume is (a) 5 L (b) 20 L (c) 40 L (d) 80 L 61. Acid hydrolysis of ester is first order reaction and rate constant is given by k=

64. Select the correct statements(s). (a) LiAlH4 reduces methyl cyanide to methyl amine (b) Alkane nitrile has electrophilic as well as nucleophilic centres (c) saponification is a reversible reaction (d) Alkaline hydrolysis of methane nitrile forms methanoic acids conc.HNO3 + conc.H2SO4

65.

(a)

V¥ = Vt

(b) V¥ = (Vt –V0)

(c)

V¥ = 2Vt –V0

(d) V¥ = 2Vt + V0

62. A near UV photon of 300 nm is absorbed by a gas and then re-emitted as two photons. One photon is red with wavelength of the second photon is (a) 1060 nm (b) 496 nm (c) 300 nm (d) 215 nm 63. Which of these ions is expected to be coloured in aqueous solution? I. Fe3 + II. Ni2+ III. Al3+ (a) I and II (b) II and III (c) I and III (d) I, II and III

Cl2/FeCl3

The product Y is (a) p-chloro nitrobenzene (b) o-chloro nitrobenzene (c) m-chloro nitrobenzene (d) o, p-dichloro nitrobenzene 66. End product of the following reaction is O + HBr

O

(a)

O

O Br

V -V0 2.303 where, V0, Vt and V¥ log ¥ t V¥ - Vt

are the volumle of standard NaOH required to neutralise acid present at a given time, if ester is 50% neutralised then

X

OH (b) HO

Br

(c)

Br

OH HO

(d) HO

OH Br

Y

EBD_7443 2013-8

67.

Target VITEEE

Following compounds are respectively … geometrical isomers Cl

Cl

P

68.

Cl

Cl

Q

Cl

(c)

O

69.

70.

(a)

CH—CH 2 2

71.

CH 3CH

72.

73.

74.

(i) CH3MgBr(one mole)

(ii) H 2O

reaction is

A formed in this

O

H3C CH3

CH3

OH

OH

For the cell reaction 2Ce4+ + Co ® 2Ce3+ + Co3+ ; Eºcell cell is 1.89 V. If ECo2 + /Co is – 0.28 V,,

(a) 0.28 V (b) 1.61 V (c) 2.17 V (d) 5.29 V A constant current of 30 A is passed through an aqueous solution of NaCl for a time of 1.00 h. What is the volume of Cl2 gas at STP produced? (a) 30.00 L (b) 25.08 L (c) 12.54 L (d) 1.12 L Consider the following reaction, kA kB

Chair Boat The reaction is of first order in each diagram, with an equilibrium constant of 104. For the conversion of chair form to boat form e–Ea/RT = 4.35 × 10-8 m at 298 K with pre-exponential factor of 1012 s-1. Apparent rate constant (= kA / kB) at 298 K is (a) 4.35 × 104 s-1 (b) 4.35 × 108 s-1 -8 -1 (c) 4.35 × 10 s (d) 4.35 × 1012 s-1

O

O O || || CH3 C CH 2CH 2 COCH 2 CH3

O

H3C

º what is the value of E Ce4 + /Ceo3+ ?

(b) CH3CH2CCH3 (d) H2C

O

(d)

O

(c)

O

(b)

P Q R (a) cis cis trans (b) cis trans trans (c) trans cis cis (d) cis trans cis Which is more basic oxygen in an ester?

R— C—O—R (a) Carbonyl oxygen, (b) Carboxyl oxygen, b (c) Equally basic (d) Both are acidic oxygen In a Claisen condensation reaction (when an ester is treated with a strong base) (a) a proton is removed from the -carbon to form a resonance stabilised carbanion of the ester (b) carbanion acts as a nucleophile in a nucleophilic acyl substitution reaction with another ester molecule (c) a new C—C bond is formed (d) All of the above statements are correct An organic compound B is formed by the reaction of ethyl magnesium iodide with a substance A, followed by treatment with dilute aqueous acid, Compound B does not react with PCC or PDC in dichloromethane. Which of the following is a possible compound for A? O O

O

(a)

Cl

R

OH

75.

Cu+ Zn2+ If for the cell reaction, Zn + Cu2+ Entropy change Sº is 96.5 J mol-1K-1, then temperature coefficient of the emf of a cell is (a) 5 × 10-4 VK-1 (b) 1 ×10-3 VK-1 (c) 2 × 10-3 VK-1 (d) 9.65 × 10-4 VK-1

Solved Paper 2013

2013-9

76. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, n = 4 to n = 2 of He+ spectrum? (a) n = 4 to n = 2 (b) n = 3 to n = 2 (c) n = 2 to n = 1 (d) n = 4 to n = 3 77. What is the degeneracy of the level of H-atom

(d)

æ R ö that has energy çç- H ÷÷÷ ? èç 9 ø

(b) 9 (a) 16 (c) 4 (d) 1 78. Match the following and choose the correct option given below. Compound/Type Use A. Dry ice I. Anti-knocking compound B. Semiconductor II. Electronic diode or triode C. Solder III. Joining circuits D. TEL IV. Referigerant for preserving food A B C D (a) I II IV III (b) II III I IV (c) IV III II I (d) IV II III I 79. Which of the following ligands is tetradentate? N H

(a)

(b) N N H2 NH

81. The relation R defined on set A = {x : | x | < 3, x e I } by R = {(x, y) : y = | x |} is (a) {–2, 2), (–1, 1), (0, 0), (1, 1), (2, 2) } (b) {(–2, –2), (–2, 2), (–1, 1), (0, 0), (1, –2), (1, 2), (2, –1), (2, –2)} (c) {0, 0), (1, 1), (2, 2)} (d) None of the above 82. The solution of the differential equation dy yf '( x) - y 2 = is dx f ( x)

x+ 2 x+3 x+4

N

N

PART - III (MATHEMATICS)

(a) f(x) = y+C (b) f(x) = y(x+C) (c) f(x) = x+C (d) None of the above 83. If a,b and c are in AP, then determinant

NH2

NH2

80. What is the EAN of [Al(C4O4)3]3–? (a) 28 (b) 22 (c) 16 (d) 10

NH2

x + 2a x + 2b is x + 2c

(b) 1 (a) 0 (c) x (d) 2x 84. If two events A and B. If odds against A are as 2:1 and those in favour of A È B are as 3:1, then (a)

1 3 £ P( B) £ 2 4

(b)

(c)

1 3 £ P(B) £ 4 5

(d) None of these

(c) NH

x+3 x+4 x+5

5 3 £ P( B) £ 12 4

85. The value of 2 tan–1 (cosec tan–1 x – tan cot–1 x) is (a) tan–1 x (b) tan x (c) cot x (d) cosec–1 x

EBD_7443 2013-10

86.

87.

88.

Target VITEEE

The proposition ~ ( p Û q) is equivalent to (a) (p ~ q) Ù (q Ù ~ p) (b) (p Ù ~ q) (q Ù ~ p) (c) (p Ù ~q) Ù (q Ù ~ p) (d) None of the above If truth values of P be F and q be T. Then, truth value of ~(~ p q) is (a) T (b) F (c) Either T or F (d) Neither T not F The rate of change of the surface area of a sphere of radius r, when the radius is increasing at the rate of 2 cm/s is proportional to 1 1 (b) r r2 (c) r (d) r2 If N denote the set of all natural numbers and R be the relation on N × N defined by (a, b) R (c, d), if ad (b + c) = bc (a + d ), then R is (a) symmetric only (b) reflexive only (c) transitive only (d) an equivalence relation A complex number z is such that arg

(a) 4 £ x2 + y2 £ 64 (b) x2+ y2 £ 25 (c) x2+ y2 ³ 25

94.

(a)

89.

90.

(c) 3 cot b – cot

95.

96.

æ 1 3r 7r 15r ö r n çç + + + +...m terms÷÷is C (–1) å rç ÷ r 2r 3r 4r 2 2 è2 2 ø÷ r =0

complex number will lie on (a) an ellipse (b) a parabola (c) a circle (d) a straight line If a1, a2 and a3 be any positive real numbers, then which of the following statement is true? (a)

3a1a2 a3 £ a13 + a23 + a33

(b)

a1 a2 a3 + + ³3 a2 a3 a1

(c)

æ1 1 1ö (a1 + a2 + a3 )ççç + + ÷÷÷ ³ 9 çè a1 a2 a3 ÷ø

(d)

æ1 1 a ö (a1. a2 . a3 )ççç + + 3 ÷÷÷ ³ 27 è a1 a2 a3 ÷ø

92. 93.

(a)

If | x2 – x – 6 | = x + 2, then the values of x are (a) –2, 2, –4 (b) –2, 2, 4 (c) 3, 2, –2 (d) 4, 4, 3 The centres of a set of circles, each of radius 3, lie on the circle x2 + y2 = 25. The locus of any point in the set is

2 mn -1 2 mn (2 n -1)

(b)

2 mn -1 2 n -1

2mn + 1 (d) None of these 2n + 1 The angle of intersection of the circles x2+ y2 – x + y – 8 = 0 and x2 + y2 + 2x + 2y – 11 = 0 is

(c) 97.

3

(d) cot b – 3 cot

and b are the roots of x2 – ax + b = 0 and if n + b n = V , then n (a) Vn+1 = aVn + bVn-1 (b) Vn+1 = aVn + aVn–1 (c) Vn+1 = aVn – bVn–1 (d) Vn+1 = aVn-1 – bVn The sum of the series If

n

æ z - 2 ö÷ z çç ÷ = . The points representing this èç z + 2 ø÷ 3

91.

(d) 3 £ x2 + y2 £ 9 A tower AB leans towards west making an angle with the vertical.The angular elevation of B, the top most point of the tower is b as observed from a point C due east of A at a distance ‘d’ from A. If the angular elevation of B from a point D due east of C at a distance 2d from C is r, then 2 tan can be given as (a) 3 cot b – 2 cot (b) 3 cot – 2 cot b

98.

æ19 ö (a) tan–1 ççç ÷÷÷ è9ø

(b) tan –1(19)

æ9ö (c) tan–1 ççç ÷÷÷ è19 ø

(d) tan–1 (9)

The vector b = 3j + 4k is to be written as the sum of a vector b1 parallel to a = i + j and a vector b2 perpendicular to a. Then b1 is equal to (a)

3 (i + j) 2

(b)

2 (i + j) 3

(c)

1 (i + j) 2

(d)

1 (i + j) 3

Solved Paper 2013

2013-11

99. If the points (x1, y1), (x2, y2) and (x3, y3) are collinear, then the rank of the matrix é x1 y1 1ù ê x y 1ú will always be less than ê 2 2 ú êë x3 y3 1úû (a) 3 (b) 2 (c) 1 (d) None of these 100. The value of the determinant

1 cos ( cos

cos ( 1 cos

)

) cos cos 1

(b) x – 3 y – 2 = 0 3 x+y–2 3 =0

(d) x +

3y – 2 = 0

103. The line 2x + x2–2y2 (a)

6y = 2 is a tangent to the curve = 4. The point of contact is (b)

(4, – 6)

(7, – 2 6)

(c) (2, 3) (d) ( 6,1) 104. The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices (0, 0), (0, 21) and (21, 0) is (a) 133 (b) 190 (c) 233 (d) 105 105.

ò (1 + x - x

—1

) e x+ x

x x 1

(a)

( x 1) e

(b)

( x 1)e x

x 1

—1

C

C

C

1

x x x C (d) xe 106. If f (x) = x – [x], for every real number x, where [x] 1

is the integral part of x. Then, ò f ( x ) dx is equal -1

to (a) 1

(b) 2

(c) 0

(d)

1 2

107. The value of the integral 2 2 é ù êçæ x + 1÷÷ö + çæ x -1 ÷÷ö - 2ú dx is ç ç ò–1/2 êêçè x -1÷ø çè x +1÷ø úú ë û 1/ 2

1/2

æ4ö (a) log ççèç ø÷÷÷ 3

æ3ö (b) 4 log çççè ø÷÷÷ 4

æ4ö (c) 4 log ççç ÷÷÷ è3ø

æ3ö (d) log ççç ÷÷÷ è4ø

108. If a tangent having slope of -

3x–y–2 3 =0

(c)

x 1

is

2 2 2 2 (a) (b) (d) 0 (c) 1 101. The number of integral values of K, for which the equation 7 cos x + 5 sin x = 2K + 1 has a solution, is (a) 4 (b) 8 (c) 10 (d) 12 102. The line joining two points A(2,0), B(3,1) is rotated about A in anti-clockwise direction through an angle of 15°. The equation of the line in the now position,is

(a)

xe x

(c)

dx is equal to

4 to the ellipse 3

x2 y 2 + = 1 intersects the major and minor axes 18 32 in points A and B respectively, then the area of OAB is equal to (O is the centre of the ellipse) (a) 12 sq units (b) 48 sq units (c) 64 sq units (d) 24 sq units 109. The locus of mid points of tangents intercepted between the axes of ellipse

x2 y2 + = 1 will be a2 b2

(a)

a2 b2 + =1 x2 y2

(b)

a 2 b2 + =2 x2 y2

(c)

a 2 b2 + =3 x2 y2

(d)

a 2 b2 + =4 x2 y2

110. If PQ is a double ordinate of hyperbola x2 y2 - = 1 . Such that OPQ is an equilateral a2 b2 triangle, O being the centre of the hyperbola, then the eccentricity ‘e’ of the hyperbola satisfies

EBD_7443 2013-12

Target VITEEE

(a) 1 < e
3 2 111. The sides AB, BC and CA of a ABC have respectively 3, 4 and 5 points lying on them.The number of triangles that can be constructed using these points as vertices is (a) 205 (b) 220 (c) 210 (d) None of these (c)

e=

a + bx 112. In the expansion of , the coefficient of ex x r is

(a)

a-b r!

(b)

a - br r!

(c)

a - br (-1) r r!

(d) None of these

113. If n = (1999) !, then å log n x is equal to 1999 x=1

(a) 1

(b) 0

(c) (d) –1 1999 114. P is a fixed point (a, a, a) on a line through the origin equally inclined to the axes, then any plane through P perpendicular to OP, makes intercepts on the axes, the sum of whose reciprocals is equal to 1999

(a) a

(b)

3 2a

3a (d) None of these 2 115. For which of the following values of m, the area of the region bounded by the curve y = x – x2

(c)

9 2 (a) – 4 (b) – 2 (c) 2 (d) 4 116. If f : R R be such that f (1) =3 and f (1) = 6. and the line y = mx equals

ïìï f (1 + x ) ïüï ý equals to Then, lim íï x®0 ï î f (1) ïïþ (a) 1 (b) e 1/2 2 (c) e (d) e 3

ìï p ïï(1+ | sin x |)a /|sin x| , - < x < 0 ïï 6 b , x=0 , then 117. If f (x) = í ïï p tan 2 x /tan 3 x ïe , 0< x (slope)1 p2 < p1 (a) Total energy U= 2

u f

A12

nR T P

V–T graph is a straight line with slope

1 f

1 1 u 1 or m f this is the equation of a straight line whose slope

R=

(b) In an isobaric process, p = constant Hence, V T i.e., V =

u f

1 1 m

30.

32.

2

37. 0

1 cos 2 or = 60° (a) Given, V1 = V2 = 300V; V3 = ?, i = ?

As, V =

V32 (V1 V2 ) 2

220 = V32 I=

V3

V3 = 220V

V3 220 = = 2.2A R 100

Solved Paper 2013 38.

2013-17

(b) We have, W = – MB(cos 2 – cos 1) So, W1 = –MB (cos 90° – cos0°) = MB and W2 = – MB (cos 60° – cos 0°) =

42.

*

COOH NaHCO3

(a)

1 MB 2

As W1 = nW2 n=

39.

W1 W2

*

COONa CO2 H 2O

MB =2 1 MB 2

N=NCl 43.

0A

(b) Capacitance, C =

H 3 PO 2

(d)

H 2O

d As one–fourth of capacitor is filled with dielectric of constant K, then,

C1 =

K 0A / 4 d

+ N2 + HCl 44.

3A / 4 and C2 = d Both C1 and C2 are in parallel. K 0A CP = C1 + C2 = 4d

Cu

(c) CH2COOCH

P/Br2

BrCHCOOCH

CH2COOCH

HVZ-reaction

CH2COOCH A

CHCOOCH

Alc. KOH

3 0A 4d

CHCOOH

C 0A (K 3) 4d 4 (c) The equivalent circuit of the given circuit is

B (maleic acid)

= (K + 3)

5

A

3

C

5

8

COOH

HO—C—H H—C—OH

B

COOH

D d(+) and l(–) tartaric acid (racemic mixture)

3

CHCOOH

O3 H2O

B

2

COOH COOH

C (oxalic acid)

COOH H—C—OH

This is a balanced Wheatstone bridge. Therefore, the arm CD becomes in effective. Hence 5 and 3 are in series and they together are in parallel with (5 + 3)

(5 3) (5 3) (5 3) (5 3)

4

PART - II (CHEMISTRY) 41.

CHCOOH

H—C—OH

D

Net resistance =

CF3CO3H

Alkaline KMnO4

40.

(a) Benzoin condensation is performed by aromatic aldehydes (i.e., compounds in which –CHO group is directly attached with benzene ring).

COOH

E meso-tartaric acid

45.

(d) When two cyclic forms of a carbohydrate differ in configuration only at hemiacetal carbons, they are said to be anomers. Thus, anomers are cyclic forms of carbohydrates that are epimeric at hemiacetal carbon and this carbon (C–1 of aldose) is called anomeric carbon, e.g.,

EBD_7443 2013-18

Target VITEEE 50. Anomeric carbon

(c) Oxidation of Ketones, yield secondary alcohol R

HO—C—H

H—C—OH H—C—OH HO—C—H H–C

47.

R

HO—C—H R

O

H—C

R

CH2OH -D-glucose

– amino acids are bifunctional organic compounds. They contain a basic amino group (–NH2) on the –carbon and one acidic carboxyl group (–COOH). (d) [CH3COOH] = millimoles of CH3COOH = 0.1 × 10 = 1.0 [CH3COONa] = millimoles of CH3COONa = 0.1 × 20 = 2.0 From, Henderson Hasselbalch equation,

51.

CH2Cl2

CH3

CH3

R R

C= O

C=O R

[(CH 3)3CO]3Al

Achiral carbon

CH2OH H—C—COOH

52.

(d)

= 133.3 – 61.9 = 71.4 OCH3

OH

(3° carbocation (Y))

53.

(b)

Y is less soluble than (X) due to lack of symmetry Chiral carbon. This is reduced to —CH2OH

CH2OH

Friedel-Crafts

CHOH

C(CH 3) 3

CH2OH

OCH3

Cl2/FeCl3

H can be lost only from this carbon (3° carbocation (X) 1, 2-methyl shift

(Ag ) –1 cm2 equiv–1

Activating group

Cl

HBr,

Cl

OH2

conc. H2SO4

(NO3 )

OCH 3

(CH3)3CCl/AlCl3

CH2OH

(a)

(AgNO3 )

(NO3 )

H—C—CH2OH

CH3

–1 cm2 equiv–1

(Ag )

LiAlH 4

CH3

–H2O

(AgNO3 )

C=O

(c) Since, X NaHCO3 CO 2 Hence, it must contain —COOH group.

2 = 4.74 + 0.30 = 5.04 1 (Ag + ) = tr an sport number of

Ag+ × A(AgNO3 ) = 0.464 × 133.3 = 61.9 By Kohlrausch’s law

49.

PCC

R

= 4.74 + log (b)

C=O

K2Cr2O 7/H2SO4 R

(c)

[conjugate base] pH = pKa + log [acid]

48.

CHOH +

R

[O]

CHOH

R

H—C—OH

CH2OH -D-glucose

46.

R

H—C—OH O

CHOH

KHSO4

CH2 CH

CH2 – CH

CHO

CHO –

OH–Cl

CH2OH CHCl CHO

C(CH3) 3

C(CH3)3

Solved Paper 2013 54.

2013-19

(a)

O O

H2S2O4

O

C

HO—S—S—OH

O

O

O

HO—S—O—O—H

H2SO5

C

O

O

H

O +

C

H

O

conc. H2 SO4

OH

OH

OH

O (d) Volume of one molecule

OH

4 3 4 r = (1.54 × 10–8)3 cm3 3 3 = 1.53 × 10–23 cm3 Volume of molecules in 1.65 g Ar

(Phenolphthalein test)

Pink colour

1.65 N0 1.53 10 23 = 0.380 cm3 40 Volume of solid containing 1.65 g Ar = 1 cm3 Empty space = 1 – 0.380 = 0.620 Per cent of empty space = 62% (d) In adiabatic expansion

=

(d) NH2

NaNO2 /HCl

N

0°C

CuCN

NCl

60.

T2 T1

CN Y H2O/H

57.

58.

Red hot w or Ta filament

2B + 3I2

1 2

NH 4 Cl(s) NH 3 HCl Graham’s law of diffusion says, lighter gas will diffuse most rapidly. Therefore, NH3 will be (mol. wt. = 17) diffused rapidly than HCl. (mol. wt. = 36.5). (b) Peroxy acids contain —O—O— linkage.

O H—O—S—O—H O

10 V2

1 2

(c)

H2S2O3

1

10 V2

150 300

(a) According to Van–Arkel method, pyrolysis of BI3 is carried out in the presence of red hot W or Ta filament. 2BI3

V1 V2

for CO2 (triatomic gas) is, = 1.33

+

COOH

56.

O

=

(phenolphthalein)

NaOH

55.

HO—S—O—O—S—OH

H2S2O8 59.

O

1 8

3

0.33

0.33

10 V2

10 V2

V2 = 80 L 61.

(c)

H

RCOOR H2O RCOOH R OH At t = 0, a 0 0 At time t, a – x x x At time , a – a a a At t = 0, V0 = volume of NaOH due to H+ (catalyst) Vt = x + V0

EBD_7443 2013-20

Target VITEEE 66.

V = a + V0

If ester is 50% hydrolysed then, x =

a 2

(d)

a V0 2 or a = 2Vt – 2V0 V = 2Vt – 2V0 + V0 = 2Vt – V0 (b) Energy values are additives. E = E1 + E2

or Vt =

62.

hc

hc

hc

1

2

1 300

1 760

O

O

O 67.

cis p.

2

68.

Fe2+ = Ar 4 unpaired electrons, Coloured ion,

3d

69.

Ni2+ = Ar 8

3d 2 unpaired electrons, Coloured ion III. Al3+ = [Ne] No unpaired electron in 3d, colourless ion.

64.

(b) CH3C

N:

65.

OH

Cl

Cl'

cis Q.

Cl'

Cl Cl trans R

O

(a)

R—C—O—R –O oxygen atom can donate lone pair of electron more easily, therefore, it is more basic than –oxygen. (d) When two molecules of an ethylacetate undergo condensation reaction, in presence of sodium ethoxide involving the reaction is called as Claisen condensation and product is a –keto ester.

O

2CH3CH2COCH3

CH3C = N :

electrophilic

HO

(a)

1

6

II.

Enolisation

OH

Cl

1 1 300 760 2 = 495.6 nm = 496 nm

(a) I.

Br

Br

1

63.

H Br –

O– 1,4-addition

O

O

nucleophilic

(i) CH3O–

(ii) H

+

O

(c)

NO2

CH3

conc. HNO3 + conc. H 2SO 4

Mechanism Step I

X

O

NO2

O

CH3CHCOCH3

Cl2/FeCl 3

H Y

Cl

–

CH3O

CH3CHCOCH3 + CH3OH

Solved Paper 2013

2013-21

71.

Step II

O

(c) Keto group is more reactive for addition of Grignard reagent.

O

CH3CHCOCH 3 + CH3CH2COCH3 –

O O || || CH3 C CH3 CH 2 C OCH 2 CH3

–

O

(i)CH3MgBr(one mole)

CH3CH2COCH3

(ii) H3O

CH3CHCOCH3 O Step III

CH3 CH3—C—CH2CH2 OH

–

O

O

CH3CH2C—OCH3

O

O

72.

1 alocohol

C

(b)

E cell

E ox

E red E

CH3CH2 Mgl OH

CH3CH2CHCH2CH3

CH3

E

73.

H 3O

CH3 CH CH2 CH3

= – 1.879 – 0.28 = 1.61 V

Ce 4 / Ce3

2Cl (aq)

(d) HCHO CH 3 CH 2 Mgl

H3 O

CH 3 CH 2 CH 2 OH 1 alcohol

H 2 ( g ) 2OH ( aq) Cl2 ( g ) 2e

2H2 O(l ) 2Cl ( aq)

H2 ( g ) Cl2 ( g ) + 2OH– (aq) OH– formed = NaOH formed = Z i t

=

E i t 96500

=

40 30 1 60 60 = 44.77 g 96500

=

44.77 = 1.12 mol 40

-Carbon atom OH 2° alcohol

/ Ce3

(c) Due to electrolysis

3°alcohol

(c) CH 3CHO CH 3 CH 2 Mgl

Ce4 / Ce3

1.89 = – (–0.28) + E Ce 4

2H 2 O(l ) 2e

O

H3O

O CH2

= E Co / Co 2

H3O

CH 3CH 2CH 2 CH 2OH

(b) CH3CH2CCH3

C=O

O

(b) B is a tertiary alcohol based on given properties. (a) CH2 CH2 CH3CH2MgI

CH3—C—CH2

C2H5O

CH3CH2CCHCOCH3 – + CH3O CH3

CH3CHCOCH3

70.

CH3

Cl 2 formed =

1 mol of NaOH 2

1.12 = 0.56 mol 2 = 0.56 × 22.4 L at STP = 12.54 L

=

EBD_7443 2013-22

74.

Target VITEEE

(b) KB = Ae E a / RT = 1012 × 4.35 × 10–8 = 4.35 × 104 s–1

n2 where, Z = atomic number, n2 = degeneracy

kA Also equilibrium constant, k = k = 104 B

75.

dE G = H – nFT dT

(a)

and G = S nF

or

76.

For H–atom, En =

× 104 = 4.35 × 108 s–1

kA = kB

P

78.

dE cell dT

1 10 2

P

= 5 × 10–4 VK–1

He

2 R H ZH

1

1

n12

n 22

1

1

n12

n 22

1

1

n12

n 22

1

1

n12

n 22

79.

1

4

4

R H Z2

He

1

1

2

42

2

77.

Semicon ductor

C. D.

Solder TEL

Referigerant for preserving food Ge Electronic diode and triode in computer Sn/Pb Joining circuits (C2H5)4 Pb Antiknocking compound for petroleum products

N

NH2;

NH2

4 1 16

NH

(c) tetradentate

12 22 Hence, trnasition n2 = 2 to n1 = 1 will give spectrum of the same wavelength as that of Balmer transition, n2 = 4 to n1 = 2 in He+. (b) Energy of the electron in the nth orbit in terms of RH is

80.

N

CH3 CH2

NH2

3 4

N (b) tridentate

C=O

NH

1 1 1 4

Uses

(c)

NH2 (a) tridentate

1 1 4 16

3 4

1

B.

N H

If n1 = 1, then n2 = 2, 3, .... For first line n2 = 2, n1 = 1 1

(d)

3

(c) We have to compare wavelength of transition in the H–spectrum with the Balmer transition n = 4 to n = 2 of He+ spectrum. H=

n2

Compound Symbol/ formla CO2 A. Dry ice

dE dT P dE dT P

R H (1)2

RH RH 9 n2 n2 = 9

T–T S

96.5 2 96500

R H Z2

En =

C=O CH3 (d) bidentate

(b) Effective atomic number EAN = Atomic number – oxidation number + 2 × coordination number For [Al (C2O4)3]3– Z = 13 ON = 3 CN = 6 EAN = 13 – 3 + 2 × 6 = 22

Solved Paper 2013

2013-23

PART - III (MATHEMATICS) 81.

(a) A = { x : |x| < 3, x I} A = {x : – 3 < x < 3, x I} = {–2, –1, 0, 1} Also, R = {(x, y) : y = |x|} R = {(–2, 2), (–1, 1), (1, 1), (0, 0), (2, 2)} 2

82.

(b)

P(B) P (A

85.

5 3 P(B) 4 12 (a) 2 tan –1 (cosec tan–1 x – tan cot–1 x)

dy yf (x) y f (x) dx yf (x) dx – f(x) dy = y2 dx yf (x)dx f (x)dy

y2

d

f (x) y

–1 = 2 tan–1 cosec cosec

= dx

= dx

f (x) =x+C y f (x) = y (x + C) x 2 x 3 x 2a

= x 3 x 4 x 2b x 4 x 5 x 2c

(a) Let

1 = 2

x 2 x 3

x 2a

0 2(2b a c) 0 x 4 x 5 x 2c

= 2 tan–1

1 x2 x

= 2 tan–1

1 x2 1 x

0

0

84.

x 4 x 5 x 2c

1 3 , P (A B) = 3 4 P (A B) = P(A) + P (B) – P (A B) P(A) + P (B)

1 P(B) 3

5 12 Also, B A B

1 cos sin

2sin 2

=0 = 2. 86.

2

~ (p

2 tan

2

(b) ~ (p

(b) P (A) =

3 4

= 2 tan–1

= 2 tan–1 tan

Since, all elements of R2 are zero.

1 x

2sin .cos 2 2

x 2 x 3 x 2a 0

sec 1 (put x = tan ) tan

tan–1

1

1 x

= 2 tan–1

=2

(using R2 2R2 – R1 – R3) But a, b and c are in AP using 2b = a + c, we get 1 = 2

1 x2 x tan tan

On integration, we get

83.

3 4

B) =

q)

1

x

~ [( p

q) (~ (q

q)

(q

(p ~ q) (q ~ p)

87.

(b)

p)]

p)) ( De–Morgan’s law)

p q ~ p ~ p q ~ (~ p q ) F T T T F

P(B)

Truth value of ~ (~ p q) is F..

EBD_7443 2013-24

88.

Target VITEEE

(c) Surface area of sphere, S = 4 r2

91.

89.

90.

4

2r

dr dt

z 2 z 2

1 a1

(a1.a2.a3) 92.

93.

1 a2

1 a3

27

(a1.a2.a3)

1 a1

1 a2 1 a2

3

1 a3 1 a3

3

27

(b) |x2 – x – 6| = x + 2, then Case I : x2 – x – 6 < 0 (x – 3) (x + 2) < 0 –2 0 1>x x nr A parallel plate capacitor has capacitance C. If it is equally filled the parallel layers of materials of dielectric constant K 1 and K2 its capacity becomes C1. The ratio of C1 and C is

EBD_7443 2012-2

Target VITEEE

(a) K1 + K2 (c) 8

9.

10

11

K1 K 2 K1K 2

(b)

K1K 2 K1 K 2

(d)

2K1K 2 K1 K 2

The potential of the electric field produced by point charge at any point (x, y, z) is given by V = 3x2 + 5, where x, y are in metres and V is in volts. The intensity of the electric field at (–2, 1, 0) is (a) +17 Vm–1 (b) –17 Vm–1 –1 (c) +12 Vm (d) –12 Vm–1 The potential of a large liquid drop when eight liquid drops are combined is 20 V. Then the potential of each single drop was (a) 10 V (b) 7.5 V (c) 5 V (d) 2.5 V A an d B are two metals with threshold frequencies 1.8 × 1014 Hz and 2.2 × 104 Hz. Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted by (Take h = 6.6 × 10–34 J-s) (a) B alone (b) A alone (c) Neither A nor B (d) Both A and B In the Wheatstone's network given, P = 10 , Q = 20 , R = 15 , S = 30 , the current passing through the battery (of negligible internal resistance) is

P

(a)

15

16.

17.

18.

S

+ –

12

13.

6V (a) 0.36 A (b) Zero (c) 0.18 A (d) 0.72 A Three resistors 1 , 2 and 3 are connected to form a triangle. Across 3 resistor a 3V battery is connected. The current through 3 resistor is (a) 0.75 A (b) 1 A (c) 2 A (d) 1.5 A In a common emitter amplifier the input signal is applied across

1 3

(b)

3

1 (d) 3 3 A radioactive substance contains 10000 nuclei and its half-life period is 20 days. The number of nuclei present at the end of 10 days is (a) 7070 (b) 9000 (c) 8000 (d) 7500 A direct X-ray photograph of the intenstines is not generally taken by radiologists because (a) intenstines would burst an exposure to X-rays (b) the X-rays would be not pass through the intenstines (c) the X-rays will pass through the intenstines without causing a good shadow for any useful diagnosis (d) a very small exposure of X-rays causes cancer in the intenstines Charge passing through a conductor of crosssection area A = 0.3 m 2 is given by q = 3t2 +5t + 2 in coulomb, where t is in second. What is the value of drift velocity at t = 2s? (Given, m = 2 × 1025/m3) (a) 0.77 × 10–5 m/s (b) 1.77 × 10–5 m/s (c) 2.08 × 10–5 m/s (d) 0.57 × 10–5 m/s Two capacitors of capacities 1 µF and C µF are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80 µC, the energy stored in the capacitor of capacity C in µJ is (a) 1800 (b) 1600 (c) 14400 (d) 7200 A hollow conducting sphere is placed in an electric field produced by a point charg placed at P as shown in figure. Let VA, VB, VC be the potentials at points A, B and C respectively. Then (c)

R G

Q

14

(a) anywhere (b) emitter-collector (c) collector-base (d) base-emitter The kinetic energy of an electron get tripled then the de-Broglie wavelength associated with it changes by a factor

19.

A

C B

P

Solved Paper 2012

20.

21.

22.

23.

24.

25.

26.

27.

(a) VC > VB (b) VB > VC (c) VA > VB (d) VA = VC In a hydrogen discharged tube it is observed that through a given cross-section 3.13 × 1015 electrons are moving from right to left and 3.12 × 1015 protons are moving from left to right. What is the electric current in the discharge tube and what is its direction? (a) 1 mA towards right (b) 1 mA towards left (c) 2 mA towards left (d) 2 mA towards right In CuSO4 solution when electric current equal to 2.5 faraday is passed, the gm equivalent deposited on the cathode is (a) 1 (b) 1.5 (c) 2 (d) 2.5 In hydrogen a atom, an electron is revolving in the orbit of radius 0.53 Å with 6.6 × 10 15 radiations/s. Magnetic field produced at the centre of the orbit is (a) 0.125 Wb/m2 (b) 1.25 Wb/m2 2 (c) 12.5 Wb/m (d) 125 Wb/m2 The dipole moment of the short bar magnet is 12.5 A-m2. The magnetic field on its axis at a distance of 0.5 m from the centre of the magnet is (a) 1.0 × 10–4 N/A-m (b) 4 × 10–2 N/A-m (c) 2 × 10–6 N/A-m (d) 6.64 × 10–8 N/A-m The turn ratio of transformers is given as 2:3. If the current through the primary coil is 3 A, thus calculate the current through load resistance (a) 1A (b) 4.5 A (c) 2 A (d) 1.5 A In an AC circuit, the potential across an inductance and resistance joined in series are respectively 16 V and 20 V. The total potential difference across the circuit is (a) 20.0 V (b) 25.6 V (c) 31.9 V (d) 33.6 V If hydrogen atom is its ground state absorbs 10.2 eV of energy. The orbital angular momentum is increase by (a) 1.05 × 10–34 J/s (b) 3.16 × 10–34 J/s (c) 2.11 × 10–34 J/s (d) 4.22 × 10–34 J/s Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of Helium nucleus is (14)1/3. The atomic number of nucleus will be

2012-3

28.

(a) 25 (b) 26 (c) 56 (d) 30 Each resistance shown in figure is 2 . The eqivalent resistance between A and B is 2

2

A 2

29.

30.

31.

32.

B

2 2

(a) 2 (b) 4 (c) 8 (d) 1 If in a triode value amplification factor is 20 and plate resistance is 10 k , then its mutual conductance is (a) 2 milli mho (b) 20 milli mho (c) (1/2) milli mho (d) 200 milli mho The output wave form of full-wave rectifier is (a)

(b)

(c)

(d)

Calculate the energy released when three -particles combined to form a 12C nucleus, the mass defect is (Atomic mass of 2He4 is 4.002603 u) (a) 0.007809 u (b) 0.002603 u (c) 4.002603 u (d) 0.5 u In the figure shown, the magnetic field induction as the point O will be

2r

O

(a)

(c)

0i

(b)

2 r 0

4

i ( r

1) (d)

0

4 i ( 4 r 0

i ( r 2)

2)

EBD_7443 2012-4

33.

34.

35.

Target VITEEE

In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. Th e corresponding stopping potential is (a) 1.3 V (b) 0.5 V (c) 2.3 V (d) 1.8 V A current of 2 A flows through a 2 resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 resistor. The internal resistance of the battery is (a) 1/3 (b) 1/4 (c) 1 (d) 0.5 The current i in a coil varies with time as shown in the figure. The variation of induced emf with time would be i

37.

38.

a change in the collector current from 10 mA to 20 mA. The current gain is (a) 75 (b) 100 (c) 25 (d) 50 A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron (a) speed will decrease (b) speed will increase (c) will turn towards left of direction of motion (d) will turn towards right of direction a motion Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the centre of the ring is (a)

0 T/4 T/2 3T/4 T

t

0 qf

2R 0q

(c)

emf

(a)

39.

0 T/4 T/2 3T/2 T t

(b)

0

T/4 T/2 3T/2 T

(c)

0

40.

T/4

T/2 3T/4 T

t

2fR 0 qf

(d)

S2 (S G)

(b)

SG (S G)

G G2 (d) (S G) (S G) Three charges, each + q, are placed at the corners of an isosceles triagle ABC of sides BC and AC, 2a. D and E are the mid-points of BC and CA. The work done in taking a charge Q from D to E is A (c)

t

emf

0q

2 fR 2 R A galvanometer of resistance, G is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is

(a)

emf

(b)

E

emf

(d) 36.

0

B

T/4 T/2 3T/4 T

t

A transistor is operated in common emitter configuration at VC = 2 V such that a change in the base current from 100 µA to 300 µA produces

(a)

eqQ 8 0a

(c) Zero

D

C qQ

(b)

4

(d)

3qQ 4 0a

0a

Solved Paper 2012

2012-5

PART - II (CHEMISTRY) 41.

42.

43.

A bubble of air is underwater at temperature 15°C and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25°C and the pressure is 1.0 bar, what will happen to the volume of the bubble ? (a) Volume will become greater by a factor of 1.6 (b) Volume will become greater by a factor of 1.1 (c) Volume will become smaller by a factor of 0.70 (d) Volume will become greater by a factor of 2.9 Match List-I with List-II for the compositions of substances and select the correct answer using the codes given below the lists. List-I List-II (Substances) (Composition) A. Plaster of Paris 1. CaSO4.2H2O B.

Epsomite

2.

C. D.

Kieserite Gypsum

3. 4. 5.

Codes : A B C D (a) 3 4 1 2 (b) 2 3 4 1 (c) 1 2 3 5 (d) 4 3 2 1 The pairs of species of oxygen and their magnetic behaviours are noted below. Which of the following presents the correct description? (a)

O 2 , O 22

–

Both diamagnetic

(b)

O , O 22

–

Both paramagnetic

(c)

O2 , O2

–

Both paramagnetic

– O, O 22 Consider the reactions

Both paramagnetic

(d)

44.

1 CaSO4 . H 2 O 2 MgSO4.7H2O MgSO4.H2O CaSO4

(i)

(CH3 ) 2 CH CH 2 Br

C 2 H5OH

(CH3 ) 2 CH CH 2OC2 H5 HBr

(ii)

(CH3 ) 2 CH CH 2 Br

C 2 H5 O

45.

46.

47.

(CH3 )2 CH CH 2OC2 H5 Br The mechanisms of reactions (i) and (ii) are respectively (a) SN1 and SN2 (b) SN1 and SN1 (c) SN2 and SN2 (d) SN2 and SN1 Which of the following complex compounds will exhibit highest paramagnetic behaviour? (At. no. Ti = 22, Cr = 24, Co = 27, Zn = 30) (a) [Ti(NH3)6]3+ (b) [Cr(NH3)6]3+ (c) [Co(NH3)6]3+ (d) [Zn(NH3)6]2+ Which of the following oxide is amphoteric? (a) SnO2 (b) CaO (c) SiO2 (d) CO2 The following reactions take place in the blast furnace in the preparation of impure iron. Identify the reaction pertaining to the formation of the slag. (a) Fe 2 O3 (s) 3CO(g)

2Fe(l) 3CO 2 (g)

(b)

CaCO 3 (s)

(c)

CaO(s) SiO 2 (s)

CaO(s) CO 2 (g)

CaSiO 3 (s)

(d)

48.

49.

50.

2CO(g) 2C(s) O 2 (g) Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is (a) Mg < Ca < Cl < P (b) Cl < P < Mg < Ca (c) P < Cl < Ca < Mg (d) Ca < Mg < P < Cl The reaction, 2A(g) B(g) 3C(g) D(g)

is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression (a) [(0.75)3 (0.25)] ÷ [(1.00)2 (1.00)] (b) [(0.75)3 (0.25)] ÷ [(0.50)2 (0.75)] (c) [(0.75)3 (0.25)] ÷ [(0.50)2 (0.25)] (d) [(0.75)3 (0.25)] ÷ [(0.75)2 (0.25)] Which of the following expressions correctly represents the equivalent conductance at infinite dilution of Al2(SO4)3 ? Given that SO 24

Al3

and

are the equivalent conductances at

infinite dilution of the respective ions?

EBD_7443 2012-6

(a)

Target VITEEE 2

(b)

Al3 Al3

(c)

3

SO 24

SO 24

3

Al3

6

SO 24

1 1 3 2 Al 3 2 SO4 The pressure exerted by 6.0g of methane gas in a0.03 m3 vessel at 129°C is (Atomic masses : C =12.01, H = 1.01 and R = 8.314 JK–1 mol–1) (a) 215216 Pa (b) 13409 Pa (c) 41648 Pa (d) 31684 Pa Match List I (Equations) with List II (Types of process) and select the correct option.

(d)

51.

52.

List-I (Equations) A. KP > Q

53.

54.

55.

57.

List-II (Types of process) 1. Non-spontaneous

B. C.

G RT ln Q 2. KP = Q 3.

D.

T

H S

56.

4.

Equilibrium Spontaneous and endothermic Spontaneous

Codes : A B C D (a) 1 2 3 4 (b) 3 4 2 1 (c) 4 1 2 3 (d) 2 1 4 3 Among the following which one has the highest cation of anion size ratio? (a) CsI (b) CsF (c) LiF (d) NaF Which of the following species is not electrophilic in nature? (a)

Cl

(b) BH3

(c)

H 3O

(d)

58.

List-I List-II (Substances) (Processes) A. Sulphuric acid 1. Haber's process B. Steel 2. Bessemer's process C. Sodium hydroride 3. Leblanc process D. Ammonia 4. Contact process Codes : A B C D (a) 1 4 2 3 (b) 1 2 3 4 (c) 4 3 2 1 (d) 4 2 3 1 When glycerol is treated with excess of HI, it produces (a) 2-iodopropane (b) allyl iodide (c) propene (d) glycerol triiodide Some statements about heavy water are given below. (i) Heavy water is used as moderator in nuclear reactors (ii) Heavy water is more associated than ordinary water (iii) Heavy water is more effective solvent than ordinary water Which of the above statements are correct? (a) (i) and (ii) (b) (i), (ii) and (iii) (c) (ii) and (iii) (d) (i) and (iii) Which one of the following compounds will be most readily dehydrated ? O

(a)

H3C OH O

(b) H3C OH

O

(c)

H3C

OH

NO2

Match List I (Substances) with List II (Processes employed in the manufacture of the substances) and select the correct option.

OH

(d) H3C O

Solved Paper 2012 59.

60.

2012-7

Which one of the following complexes is not expected to exhibit isomerism? (a) [Ni(NH3)4(H2O)2]2+ (b) [Pt(NH3)2Cl2] (c) [Ni(NH3)2Cl2] (d) [Ni(en)3]2+ Which of the following conformers for ethylene glycol is most stable ?

64.

(iii) CH 3 CHOH | CH3

OH

H

H

(a) H

H OH

OH OH

65.

(b) H H

HH

OH H

(b) Isostructural with same hybridisation for the central atom (c) Isostructural with different hybridisation for the central atom (d) Similar in hybridisation for the central atom with different structures Following compounds are given (i) CH3CH2OH (ii) CH3COCH3

66.

Which of the above compound(s) on being warmed with iodine solution and NaOH, will give iodoform? (a) (i), (iii) and (iv) (b) Only (ii) (c) (i), (ii) and (iii) (d) (i) and (ii) Fructose reduces Tollen's reagent due to (a) asymmetric carbons (b) primary alcoholic group (c) secondary alcoholic group (d) en olisation of fructose followed by conversion to aldehyde by base In the following reaction, C6 H5CH 2Br

(c)

HO H

HH OH OH

H

67.

(d) H 61.

62.

63.

H H

The IUPAC name of the compound CH3CH = CHC CH is (a) pent-4-yn-2-ene (b) pent-3-en-1-yne (c) pent-2-en-4-yne (d) pent-1-yn-3-ene Which of the following oxidation states is the most common among the lanthanoids? (a) 4 (b) 2 (c) 5 (d) 3 Some of the properties of the two species, NO3– and H3O+ are described below. Which one of them is correct? (a) Dissimilar in hybridisation for the central atom with different structures

68.

(iv) CH3OH

(i) Mg, Ether (ii) H3O

X,

the product 'X' is (a) C6H5CH2OCH2C6H5 (b) C6H5CH2OH (c) C6H5CH3 (d) C6H5CH2CH2C6H5 Which of the following is not a fat soluble vitamin? (a) Vitamin-B complex (b) Vitamin-D (c) Vitamin-E (d) Vitamin-A Which of the statements about 'Denaturation' given below are correct? Statements : (i) denaturation of proteins causes loss of secondary and tertiary structures of the protein. (ii) Denaturation leads to the conversion of double strand of DNA into single strand'. (iii) Denaturation affects primary structure which gets destroyed. (a) (ii) and (iii) (b) (i) and (iii) (c) (i) and (ii) (d) (i), (ii) and (iii)

EBD_7443 2012-8

69.

70.

Target VITEEE

Which has the maximum number of molecules among the following ? (a) 44 g CO2 (b) 48 g O3 (c) 8 g H2 (d) 64 g SO2 Which of the following compounds undergoes nucleophilic substitution reaction most easily ?

75.

Consider the following processes H (kJ/mol) 1 A 2

3B

Cl

E A

(a) NO2

For B D

Cl

76.

(b) CH3 Cl Cl

(c)

(d) OCH3

71.

72.

73.

A 0.1 molal aqueous solution of a weak acid is 30% ionised. If Kf for water is 1.86° C/m, the freezing point of the solution will be (a) –0.18°C (b) –0.54°C (c) –0.36°C (d) –0.24°C Which of the following carbonyls will have the strongest C – O bond? (a) Mn(CO)+6 (b) Cr(CO)6 (c) V(CO)–6 (d) Fe(CO)5 The order of reactivity of phenyl magnesium bromide (PhMgBr) with the following compounds H 3C

H

C (I)

74.

O

H 3C

H 3C

C

O and

(II)

Ph

Ph

C

O

77.

B 150

2C D 125 2D 350 E 2C, H will be

(a) 525 kJ/mol (b) –175 kJ/mol (c) –325 kJ/mol (d) 325 kJ/mol Match the compounds given in List-I with List-II and select the suitable option using the codes given below List-I List-II A. Benzaldehyde 1. Phenolphthalein B. Phthalic 2. Benzoin anhydride condensation C. Phenyl benzoate 3. Oil of wintergreen D. Methyl 4. Fries rearrangement salicylate Codes : A B C D (a) 4 1 3 2 (b) 4 2 3 1 (c) 2 3 4 1 (d) 2 1 4 3 Which of the following compound is the most basic ? (a)

(b)

O2N

NH2

CH2NH2

(III)

(a) III > II > I (b) II > I > III (c) I > III > II (d) I > II > III A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y–) will be (a) 275.1 pm (b) 322.5 pm (c) 241.5 pm (d) 165.7 pm

(c)

N H

(d)

NH2

COCH3

Solved Paper 2012 78.

2012-9

Which of the following structures is the most preferred and hence of lowest energy for SO3?

84.

O

S O

O

O

S

S

(d)

O

O O

80.

85.

What is the value of electron gain enthalpy of Na+ if IE1 of Na = 5.1 eV ? (a) –5.1 eV (b) –10.2 eV (c) +2.55 eV (d) +10.2 eV The unit of rate constant for a zero order reaction is (a) mol L–1s–1 (b) L mol–1s–1 2 –2 –1 (c) L mol s (d) s –1

PART - III (MATHEMATICS) 81.

1 (1 x 2 ) 2

87.

88.

is

(a) y (1 + x2) = C + tan–1x

89.

y

= C + tan–1x 1 x2 (c) y log (1 + x2) = C + tan–1x (d) y (1 + x2) = C + sin–1x If x, y and z are all distinct and (b)

82.

86.

The solution of the differential equation 2yx dy dx 1 x 2

90.

x x 2 1 x3 y y 2 1 y3

z

83.

z

2

1 z

0, then the value of xyz is

91.

3

(a) –2 (b) –1 (c) –3 (d) None of these The probability that atleast one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then P(A) P(B) is (a) 0.4 (c) 1.2

(b) 0.8 (d) 1.4

4 5 (c) 90° (d) None of these If 2 tan –1(cos x) = tan–1(2 cosec x), then the value of x is (a)

sin

(a)

3 4

1

3 5

(b)

(b)

sin

1

4

(c)

O

79.

O

O O

(c)

(b)

O

S

(a)

If 3p and 4p are resultant of a force 5p, then the angle between 3p and 5p is

92.

(d) None of these 3 Let a be any element in a boolean algebra B. If a + x = 1 and ax = 0, then (a) x = 1 (b) x = 0 (c) x = a (d) x = a' Dual of (x + y) . (x + 1) = x + x . y + y is (a) (x . y) + (x . 0) = x . (x + y) . y (b) (x + y) + (x . 1) = x . (x + y) . y (c) (x . y) (x . 0) = x . (x + y) . y (d) None of the above The function f : R R defined by f (x) = (x – 1)(x – 2)(x – 3) is (a) one-one but not onto (b) onto but not one-one (c) both one-one and onto (d) neither one-one nor onto If the complex numbers z 1, z2 and z3 are in AP, then they lie on a (a) a circle (b) a parabola (c) line (d) ellipse Let a, b and c be in AP and |a| < 1, |b| < 1, |c| < 1. If x = 1 + a + a2 + ... to , y = 1 + b + b2 + ... to , z = 1 + c + c2 + ... to , then x, y and z are in (a) AP (b) GP (c) HP (d) None of these The number of real solutions of the equation

9 3 x x 2 is 10 (a) 0 (b) 1 (c) 2 (d) None of these The lines 2x – 3y – 5 = 0 and 3x – 4y = 7 are diameters of a circle of area 154 sq units, then the equation of the circle is (a) x2 + y2 + 2x – 2y – 62 = 0 (b) x2 + y2 + 2x – 2y – 47 = 0 (c) x2 + y2 – 2x + 2y – 47 = 0 (d) x2 + y2 – 2x + 2y – 62 = 0

EBD_7443 2012-10

93.

Target VITEEE

The angle of depressions of the top and the foot of a chimney as seen from the top of a second chimney, which is 150 m high and standing on the same level as the first are and respectively, then the distance between their tops when 4 and tan 3

tan

5 is 2

150 m (b) 100 3 m 3 (c) 150 m (d) 100 m If one root is square of the other root of the equation x2 + px + q = 0, then the relations between p and q is (a) p3 – (3p – 1) q + q2 = 0 (b) p3 – q (3p + 1) + q2 = 0 (c) p3 + q (3p – 1) + q2 = 0 (d) p3 + q (3p + 1) + q2 = 0 The coefficient of x53 in the following expansions (a)

94.

95.

100

100

Cm (x 3)100 m .2 m is

m 0 100C

96.

97.

100C

(a) (b) 47 53 (d) –100C 100 (c) –100C 53 If (–3, 2) lies on the circle x2 + y2 + 2gx + 2fy + c = 0, which is concentric with the circle x2 + y2 + 6x + 8y – 5 = 0, then c is equal to (a) 11 (b) –11 (c) 24 (d) 100 If a = i + j + k, b = i + 3j + 5k and c = 7i + 9j + 11k, then the area of Parallelogram having diagonals a + b and b + c is (a)

4 6 sq. units

(b)

1 21 sq. units 2

(c)

6 sq. units 2

(d)

6 sq. units

1

98.

99.

If A

independent of independent of independent of and None of the above

100. Themaximum valueof 4 sin2 x – 12 sin x + 7 is (a) 25 (b) 4 (c) does not exist (d) None of these 101. A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A, its equation is (a) 3x – 4y + 7 = 0 (b) 4x + 3y = 24 (c) 3x + 4y = 25 (d) x + y = 7 102. The tangent at (1, 7) to the curve x2 = y – 6 touches the circle x2 + y2 + 16x + 12y + c = 0 at (a) (6, 7) (b) (–6, 7) (c) (6, –7) (d) (–6, –7) 103. The equation of straight line through the intersection of the lines x – 2y = 1 and x + 3y = 2 and parallel 3x + 4y = 0 is (a) 3x + 4y + 5 = 0 (b) 3x + 4y – 10 = 0 (c) 3x + 4y – 5 = 0 (d) 3x + 4y + 6 = 0 104.

dx sin x cos x 1 x tan 2 2

(a) (b) (c)

1 x tan 2 2

1 2

cot

C

8 C

8

1 x cot 2 2

(d)

equals to

2

C

8 x 2

C

8 1

105. The value of integral 0

1 x dx is 1 x

5 7

0

7

9 , then trace of matrix A is

11

8

9

(a) 17 (b) 25 (c) 3 (d) 12 The value of the determinant cos

sin

sin

cos

cos(

(a) (b) (c) (d)

)

sin(

1 1 is ) 1

(a)

2 (c) –1

1

(b)

2 (d) 1 1

106. The value of I

xx 0

1

1 dx is 2

(a)

1 3

(b)

1 4

(c)

1 8

(d) None of these

Solved Paper 2012

2012-11

107. The eccentricity of the ellipse, which meets the x 7

straight line

y 2

1 on the axis of x and the

x y 1 on the axis of y and whose 3 5 axes lie along the axes of coordinates, is

straight line

3 2 7 3 7

(a) (c) x2

(b)

2 6 7

(d) None of there

y2

2 is (a) x2 – y2 = 1 (b) xy = 1 (c) 2xy – 4x + 4y + 1 = 0 (d) 2xy + 4x – 4y – 1 = 0 110. There are 5 letters and 5 different envelopes. The number of ways in which all the letters can be put in wrong envelope, is (a) 119 (b) 44 (c) 59 (d) 40 111. The sum of the series 1

is

12 22 2!

12 22 32 12 22 32 42 ... 3! 4!

17 13 19 e (c) e e (d) 6 6 6 n 112. The coefficient of x in the expansion of loga(1 + x) is

(a) 3e

(a)

(b)

( 1)n 1 n

(b)

( 1)n n

n 1

1

log a e

n

( 1) ( 1) log e a (d) log a e n n 113. If a plane meets the coordinate axes at A, B and C in such a way that the centroid of ABC is at the point (1, 2, 3), then equation of the plane is

(c)

(a)

x 1

y 2

z 1 3

(b)

(c)

x 1

y 2

z 3

(d) None of these

1 3

x 3

y 6

the circle x2 + y2 = 4, the line x is (a) (c)

sq units 3

z 9

1

(b)

sq units

115. The value of lim

1 (a b) and x2 – y2 = c2 cut at a 2 b2 right angles, then (a) a2 + b2 = 2c 2 (b) b2 – a2 = 2c 2 2 2 2 (c) a – b = 2c (d) a2b2 = 2c 2 109. The equation of the conic with focus at (1, –1) directrix along x – y + 1 = 0 and with eccentricity

108. If

114. Area lying in the first quadrant and bounded by

x

2

mx 1,

x

sin x n, x

1

1x

x

is

2

is continuous at

2

, then

(a) m = 1, n = 0 (c)

tan

(b) 1 (d) e

116. If f (x)

2

sq units

2

(d) None of these

(a) 0 (c) –1

x

3y and x-axis

n

m

2

(b)

m

n 2

(d)

m

n

117. The domain of the function f (x)

1

2

4 x2 sin 1 (2 x)

is (b) [0, 2) (a) [0, 2] (c) [1, 2) (d) [1, 2] 118. The general solution of the differential equation (1 + y2) dx + (1 + x2) dy = 0 is (a) x – y = C (1 – xy) (b) x – y = C (1 + xy) (c) x + y = C (1 – xy) (d) x + y = C (1 + xy) 119. The order and degree of the differential equation 1

dy dx

2 32

are, respectively

2

d y dx 2

(a) 2, 2 (b) 2, 3 (c) 2, 1 (d) None of these 120. The relation R defined on the set of natural numbers as {(a, b) : a differs from b by 3} is given (a) {(1, 4), (2, 5), (3, 6),... } (b) {(4, 1), (5, 2), (6, 3),... } (c) {(1, 3), (2, 6), (3, 9),... } (d) None of the above

EBD_7443 2012-12

Target VITEEE

SOLUTIONS PART - I (PHYSICS) 1.

6.

I1

(d)

F1

d

(d) When reactance of inductance is mroe than the reactance of cndenser, the current will lag behind the voltage. Thus

I F4

or n

1

7.

(c) Capacitance, CA =

F3 C1 Ceq = C 2

F2 = – F4 F1 =

0 I1Il

2 dl

=

0 I1I F2 = 2 (d l) F1 > F3 Fnet = F1 – F3 So, wire attract loop.

2.

(a) Here, V0 = = =

6.62 10

E V e 34

34

4.

5.

3.3 1014 )

19

4.9 1033

6.62 4.9 10 1 1.6 V0 = 2 volt (a) In forward biasing, resistance of p–n junctin diode is zero, so whole voltage appears across the resistance. (d) BE of Li7 = 39.20 MeV and He4 = 28.24 MeV Hence binding energy of 2He4 = 56.84 MeV Energy of reaction = 56.84 – 39.20 = 17.28 MeV

=

3.

(8.2 1014

1.6 10 6.62 10 1.6

8.

h(v v0 ) e

(b)

v

(Z b)

LC

or n > nr 2 LC nr = resonant frequency

F2

I

1

1 or c

L

9.

CACB CA CB

K1 0 A , CB d 2

K2

0A

d 2

2K1K 2 K1 K 2

2K1K2 K1 K2

0A

d

0A

e

(d) Intensity of the electric field, E =

d dV = 6x dx

Potential (v) = 3x2 + 5 E at x = –2 = 6 (–2) = – 12V/m (c) Volume of 8 small drops = Volume of big drop

4 3 4 3 r 8 R 3 3 2r = R ...(i) According to charge conservation 8q = Q ...(ii) Potential of one small drop (V ) = 4 Similarly, potential of big drop (V) = 4 V q R V Q r V = 5V

Now,

V 20

9 2r 8q r

q 0r

Q 0R

Solved Paper 2012 10.

(b) Threshold energy of A EA = hvA = 6.6 × 10–34 × 1.8 × 1014 = 11.88 × 10–20 J =

11.

2012-13

11.88 10

I2

20

eV = 0.74 eV

19

1.6 10 Similarly, EB = 0.91eV As the incident photons have energy greater than EA but less than EB So, photoelectrons will be emitted from metal A only. (a) Balanced wheatstone bridge condition P R Q S No, current flows through galvanometer Now, P and R are in series, so Resistance R1 = P + R = 10 + 15 = 25 Similarly, Q and S are in series, so Resistance R2 = R + S = 20 + 30 = 50 Net resistance of the network as R1 and R2 are in parallel 1 R

1 R1

1 R2

25 50 R = 25 50

50 3

3 = 2A (3/ 2)

I=

I = IA 2 (b) In CE amplifier, the input signal is applied across base–emitter junction.

Current in 3 resistor =

13.

C

C1

RC

RB

V1

O V0 VCC

RBB

(c) de–Broglie wavelength of an electron =

h mv

h 2mK 1 3K

1

or K 1

K 1 3

2

or

3

=

3 i.e. de–Broglie wavelength will change by

C

factor

3V

Here, two reisistance of 1 and 2 are in series, which form 3 which is in parallel with 3 resistance. Therefore, the effective resistance

(1 2) 3 (1 2) 3

O Output

E

Input

14.

C2

B

B

A

I1 + I2 = 2A

Current in the circuit,

V 6 = 0.36A R 50 / 3 (b) The arrangement is shown in figure.

1

I2 = 1A

3

I1 = 1A

Hence, current, I =

12.

2

1

3 2

15.

1

3 (a) We know, N N0

N=

1 2

.

t/T

10000

2

N 10000

1 2

10000 = 7070 1.414

10 20

EBD_7443 2012-14

16. 17.

Target VITEEE

(c) As X-rays pass through the intestine without casting a clear shadow. (b) Given : A = 0.3 m2 n = 2 × 1025/m3 q = 3t2 + 5t + 2

23.

= 10–7 ×

dq i= = 6t + 5 = 17 dt

Drift velocity, vd = =

24.

17 2 10 25 1.6 10 19 0.3

17

=

18.

i neA

25.

= 1.77 × 10–5 m/s

0.96 106 (b) Capacitance 1 F and C F are connected in series,

26.

C 1 C Given, V = 120 V and q = 80 C q = CeqV

C 20 C 1 or C = 2 F Energy stored in the capacitor of capicity C

27.

6

1 80 10 6 80 10 6 2 2 10 6 U = 1600 J Conducting surface behaves as equipotential surface. I = neqe + npqE = 1 mA (towards right) 1 faraday deposited 1 g equivalent The magnetic field

d3

= 2 × 10–6 N/A–m nS nS

3 3 i.e. I 2 or, IS = 2A S (b) Voltage VR2

VC2 =

(20)2

(16)2

= 25.6V (a) Electron goes to its first excited state (n = 2) from ground state (n = 1) after absorbing 10.2 eV energy h 2

6.6 10 34 6.28 = 1.05 × 10–34 J–s (b) Using R = R0 A1/3 R1 R2

(80 10 6 )2

2 10

2N

.

=

1q 2 C

1 2

(0.5)3

4

Increase in momentum =

80 =

=

2 1.25

0

IP (c) Transformation ratio, I S

V=

Ceq =

U=

(c) The magnetic field, B =

A1 A2

R R He

A 4

1/3

1 3

=

(d)

20. 21. 22.

(a) (a) (c)

B=

0

4

.

= 10–7 ×

28.

2 (qv) r

2 3.14 (1.6 10

= 12.5 Wb/m3

2

2 19

0.53 10

1.6 1015 )

A

B

2

10

2 2

19.

1

A 3 A = 56 4 So, = 56 – 30 = 26 (a) Given circuit is a balanced Wheatstone bridge.

(14)1/3 =

Equivalent resistance of upper arms

Solved Paper 2012

2012-15

From Eqs. (i) and (ii), we have

=2+2=4 Equivalent resistance of lowre arms =2+2=4 RAB = 29.

4 4 =2 4 4

(a) Mutual conductance gm = R P

35.

20

= 2 × 10–3 = 2 milli mho 10 103 (c) Full-wave rectifier output wave form =

30.

2 0.5

9 r 2 r

4=

9 r 2 r

1 3 (d) We know, induced emf

3r = 1

r=

e=–L

di dt

During 0 to

T di , = constant 4 dt

So, e = –ve T T di =0 to , 4 2 dt i.e., e = 0

For

31.

32.

(c) Mass defect m = Total mass of –particles – mass of 12C nucleus = 3 × 4.002603 – 12 = 12.007809 – 12 = 0.007809 unit (a) Field due to a straight wire of infinite length 0i

is

if the point is on a line perpendicular 4 r to its length while at the centre of a i semicurcular coil is 4 r 0

T 3T di , = constant to 4 4 dt i.e., e = +ve

For

36.

a 38.

r b

=

O c

=

i 0 i 0 i . . 4 r 4 r 4 r

=

0

0 i ( 2) out of the phase 4 r (b) Stopping potential = Maximum KE eV = KEmax

=

33. 34.

(d) Current i = 2=

E 2 r

0.5 =

E 9 r

39.

10 10

3

0i 2R

q = it

B = Ba + Bb + Bc

(20 100)mA (300 100)mA

= 50 200 10 6 (a) Field B not applied only force. Field E will apply a force opposite to velocity of the electron hence, speed will decreases. (a) We know magnetic field =

37.

IC IB

(d) Current gain, =

i=

q = qf t

0 qf

2R (c) If resistance remains same so current will be unchanged. R G G

E

S

R r

... (i) ...(ii)

G=

GS R G S

or, R =

G2 G S

R=G–

GS G S

EBD_7443 2012-16

40.

Target VITEEE

(c) Here, AC = BC

q

O2

C

*

O+ = 1s 2 ,2s 2 , 2 p1x , 2 p1y , 2 p1z

44.

PART - II (CHEMISTRY) p1V1 T1

(a)

p2V2 (By ideal gas equation) T2

1.5 V1 1 V2 288 298 V2 = 1.55 V1 i.e, volume of bubble will be almost 1.6 times to initial volume of bubble.

or

1 HO 2 2 Epsomite = MgSO4 · 7H2O Kieserite = MgSO4 · H2O Gypsum= CaSO4 · 2H2O molecular orbital configurations of

42.

(b) (A) Plaster or Paris = CaSO4

43.

(B) (C) (D) (c) The

O2 , O2 , O 22

O2

*

*

1s 2 , 1s 2 , 2s 2 , 2s 2 , 2 p 2z , 2 px2 *

*

2 p 0y *

*

O22

*

1s 2 , 1s 2 , 2s 2 , 2s 2 , 2 pz2 , 2 px2 2 p 2y , 2 p x2

*

2 p1y

*

*

1s 2 , 1s 2 , 2s 2 , 2s 2 , 2 pz2 , 2 px2 *

2 p 2y , 2 p 2x

45.

As O2 , O2 ,O 2 ,O and O+ have unpaired electrons, hence are paramagnetic. (a) C2H5OH being a weaker nucleopbile, when used as a solvent in case of hindered 1° halide, favours SN1 mechanism while C2H5O–being a strong nucleophile in this reaction favours SN2 mechanism. (b) (a) Electronic configuration of Ti 3+ in [Ti(NH3)6]3+ Ti3+ = 3d1;

3d

4s

4p

2

d sp3 hybridisation (b) Electronic configuration of Cr 3+ in [Cr(NH3)6]3+ Cr3+ = 3d3; 3d

4s

4p

and O2 are

2 p 2y , 2 p1x

O2

2 p1y

O = 1s 2 ,2s 2 , 2 p2x , 2 p1y ,2 p1z

q B

D

*

and the electronic configuration of O and O+ are

D

VD = VE = V W = Q[VE – VD] W = Q [V – V] W= 0

41.

*

2 p 2y , 2 p1x

E q A

*

1s 2 , 1s 2 , 2s 2 , 2s 2 , 2 pz2 , 2 px2

*

2 p 2y

2

d sp3 hybridisation

(c) Electronic configuration of Co3+ in [Co(NH3 )6]3+; Co3+ = 3d6. 3d 4s 4p

In the presence of strong field ligand NH3, pairing of electrons takes place and hence, octahedral complex, [Co(NH3)6]3+ is diamagnetic. inner orbital or [Co(NH3)6]3+ low spin complex (6NH3 molecles)

Solved Paper 2012

2012-17

3d ×× ××

4s

4p

××

×× ×× ××

d 2sp3 hybridisation (d) Electronic configuration of Zn2+ in [Zn (NH3)6]2+

3d

49.

Zn2+ = 3d10;

On the other hand, on increasing the number of electron in the same shell, the atomic radii decreases because effective nuclear charge is increases. In Mg, P and C1, the number of electrons are increasing in the same shell, thus the order of their atomic radii is C1 < P < Mg In case of Ca, the electron is entering in higher shell. So, its atomic radii is highest. Thus, the order of radii is Cl < P < Mg < Ca (b) The reaction2 A(g) B (g)

4s

4d

4p

Initial

1

0

At equil 1 0.50 1 0.25

3

sp d 2 hybridisation

46.

[Zn(NH3 ) 6 ] 2+ is an outer orbital complex and is diamagnetic. (a) SnO2 reacts with acids as well as bases to form corresponding salts. So it is an amphoteric oxide. SnCl4 + 2H2O SnO2 + 4HCl SnO2 + 2NaOH

47.

48.

Na 2SnO3 sod.stannate

+ H2O

(c) A slag is an easily fusible material which is formed when gangue still present in the roasted or the calcined ore combines with the flux. For example, in the metallurgy of iron, CaO (flux) combines with silica gangue to form easily fusible calcium silicate (CaSiO3) slag. CaO + SiO2 CaSiO3 (slag) (b) Atomic radii increases, as the number of shells increases. Thus, on moving down a group atomic radii increases. The electronic configuration of the given element is Mg12 = [Ne] 3s2 Ca20 =[Ar] 4s2 p15 = [Ne] 3s23p3 Cl17 = [Ne]3s23p5

K= 50.

(b)

0.75

0 0.25

(0.75)3 (0.25) (0.50)2 (0.75)

Al2 (SO4 )3 2Al3 3SO24 We can calculate th e equivalent conductance only for ions, so the equivalent conductance at infinite dilution, eq

51.

3C (g) D(g)

1

Al3

SO24

(c) w(given mass of methane) = 6g temperature, T = 129 + 273 = 402 K mol mass of methane, M = 12. 01 + 4 1.01 =16.05

From, ideal gas equation, pV = nRT

nRT v

6 8.314 402 = 41648 Pa 0.03 16.05 (c) (A) If kp > Q and goes in forward direction than reaction is spontaneous (B) Given, G°< RT ln Q, thus, G° = + ve and hence, the reaction is non– spontaneous.

p=

52.

P

EBD_7443 2012-18

Target VITEEE 57.

(C) At equilibrium, Kp = Q H S or T S = H This is valid condition for spontaneous endothermic reactions (as G H – T S) (b) The size of cation is in order ofLi+ < Na+ < Cs+ and the size of anions in the order ofI– > F – Thus, when the cation is largest and anion is smallest, the ratio of their sizes is maximum. Hence, cation to anion size ratio is maximum for CsF. (c) Electron deficient species are known as electrophiles.

(D) T >

53.

54.

55.

56.

Among the given, H3 O has lone pair of electrons for donation, so it is not electron deficient and hence, not an electrophile. (d) Contact process is used for sulphuric acid, steel is manufactured by Bessemer's process, Leblanc process is used for the production of NaOH while Haber's process is used for NH3 production. (a)

CH2OH

CHOH CH2OH

3HI –3H 2O

CH2I

CH2

CHI

CH

–I2

CH2I (highly unstable)

O

O

< H3C

H3C

(a)

(b) and (d) O

+ < H3C (c)

59.

60.

Hence, compound given in option (c) readily undergoes dehydration. (c) Compounds having tetrahedral geometry does not exhibit isomerism due to presence of symmetry elements. Here, [Ni(NH3)2Cl2] has tetrated ral geometry. (d)

H

OH

H

H

CH3

CHI

–I2

CH2I (unstable)

H

CH CH2 CH3

HI

H-bonding

OH

CH2I allyl iodide

CH3 (HI (excess)

58.

(a) In nuclear reactors heavy water is used as a moderator. It has higher boiling point as compared to the ordinary water. Thus, it is more associated as compared to ordinary water. The dielectric constant is however higher for H2O, thus, H2O is a more effective solvent as compare to heavy water (D2O). (c) Dehydration of alcohols involve formation of carbocation intermediate. Higher the stability of carbocation, higher is the ease of dehydration. The order of stability of carbocation, is

This conformation is most stable due to intramolecular H-bonding. 4

3

2

1

(b)

62.

(d) The most common oxidation state exhibited by lanthanoids is +3.

63.

(a) In NO3 ,

CHI CH3 2-iodopropane

5

61.

C H3 — C H

CH — C

pent-3-en-1-yne

H =

CH

1 [5 + 0 – 0 + 1] = 3. So, sp3 2

Solved Paper 2012

2012-19

hybridization. Thus, it has trigonal planar geometry. O

67. 68.

N –

O

O

In H3O+, 1 [6 + 3 – 1 + 0] = 4; So, sp3 2 hybridization and it has pyramidal geometry due to the presence of one lone pair of electrons.

H =

69.

70.

(a) Fat soluble vitamins are A, D and E. Whereas vitamin-B complex is soluble in water. (c) In the process denaturation secondary and tertiary structures of protein destroyed but primary structure remains undisturbed. Heat, acid and alkali denature DNA molecule and double strand of DNA converts into single strand. (c) 44 g CO2 = 1 mol CO2 = NA molecules of CO2 48g O3 = 1 mol O3 = NA molecules of O3 8 g H2 = 4 mol H2 = 4 × NA molecules of H2 64 g SO2 = 1 mol SO2 = NA molecules of SO2 NA = 6.023 × 1023 (a) Cl

O+ H

64.

H NO2 It has electron withdrawing group — NO2 which reduces the double bond character between carbon of benzene ring and chlorine. Hence, the correct order of nucleophilic substitution reactions are,

H

O || (c) Compoun ds having eith er CH3 C—

group or CH3CHOH— group, give iodoform when warmed with I2 and NaOH. Thus, compounds

Cl

65.

66.

(c)

C6H5CH2Br

Mg,Ether

C6H5CH2MgBr Grignard reagent

Br H3O /H

C6 H5CH3 Mg toluene

OH

Cl

Cl

O || CH3 C HOH,CH3 — C— CH 3 , CH 3 — CHOH | | H CH3

give iodoform when heated with I2 and NaOH. (Note : NaOI oxidises CH3CH2OH to CH3CHO, and gives positive iodoform test.) (d) In aqueous medium, fructose is enolised and converted into aldehyde in basic medium. Generally all aldehydes reduce Tollen's reagent, thus fructose can also reduces Tollen's reagent.

Cl

>

NO2

71.

72.

>

>

CH3

OCH3

(d) Freezing point depression ( Tf) = iKf m HA H+ + A– 1– 1 – 0.3 0.3 0.3 i = 1 – 0.3 + 0.3 + 0.3 i = 1.3 Tf = 1.3 × 1.86 × 0.1 = 0.2418°C Tf = 0 – 0.2418°C = – 0.2418°C (a) As positive charge on the central metal atom increases, the less readily the metal can donate electron density into the anti– bonding –orbitals of C–O ligand to weaken the C—O bond. Thus, the C–O bond would be strongest in Mn(CO)6+

EBD_7443 2012-20

73.

H3C

Target VITEEE

(d) Since alkyl group has +I–effect and aryl group has + R–effect, Hence greater the number of alkyl and aryl groups attached to the carbonyl group, its reactivity towards nucleophilic addition reaction. Secondly, as the steric crowding on carbonyl group increases, the r eactivity decreases accordingly. Correct reactivity order for reaction with PhMgBr is C

H

H3C

76.

H

H

conc. H2SO4/

O

–H2O

C

O

O >

C

O

(II)

(III)

(c) Radius ratio of NaCl like crystal =

O phthalic anhydride C

r

O

r C

100 = 241.5 pm 0.414

1 A B ; H = 150 kJ/mol (b) ...(i) 2 3B 2C + D; H = – 125 kJ/mol ...(ii) E + A 2D; H = + 350 kJ/mol ...(iii) By [2 × (i) + (ii)] – (iii), we have B + D E + 2C H = 150 × 2 + (–125) – 350 = –175 kJ/mol (d) (a) Benzoin condensation : Heating ethanolic solution with strong alkali like KCN or NaCN, benzoin is obtained. O 2C6H5 — C — H

CN

OH

OH

C

Ph

= 0.414 or r – = 75.

OH

Ph

C

H3C (I)

74.

O >

OH

O phenolphthalein

(c) Fries rearrangement Phenyl benzoate heated with anhydrous AlCI 3 in the presence of inert solvent gives ortho–and Para–hydroxybenzophenone. In this rearrangement, there is only a benzoyl group migration from the phenolic oxygen to an ortho–and para–position. O

OH

O—C—C6H5

C—C6H5

AlCl3/

–

O

phenyl benzoate

O p-hydroxybenzophenone

C6H5 — C — CH — C6H5 OH benzoin

(b) Formation of phenolphthalein phenol is treated with phthalic anhydride in the presence of conc. H 2 SO 4 , it gives phenolphthalein, an indicator.

OH

+

O

C — C6H5 p-hydroxybenzophenone

Solved Paper 2012

2012-21

(d) Methylsalicylate

PART - III (MATHEMATICS)

OH COOCH3

81.

(a)

dy dx

2yx 1 x

1

2

(1 x 2 ) 2

which is a linear differential equation. Here, P =

(A chief constituent of oil of wintergreen) 77.

78.

(b)

1 × bonding electrons. 2 For Lewis structure of SO3

electrons –

.. ..

O

.O...

S

=

1 (1 x 2 ) 2

P dx

2 dx 2 = (1 + x ) 2 e1 x elog(1 x ) Solution of differential equation is

1

y. (1 + x2) =

(1 x 2 ) 2

.(1 x 2 )dx C

1

dx C 1 x2 y (1 + x2) = tan–1 x + C y (1 + x2) =

x x 2 1 x3

82.

(b)

y y2 1 y3 = 0 z

z2

1 z3

x x2 1

x

x2

x3

y y2 1

y

y2

y3 = 0

z2 1

z

z2

z3

1 x x2

Formal charge on S atom

y y2 1

xyz 1 y y 2

1 × 12 = 0 2 Formal charge on three O atoms

z

1 4 0 2 (a) IE1 of Na = – Electron gain enthalpy of Na+ ion = –5.1eV. (a) For zero order reaction, Rate = k [Reactants]° Rate = k and unit of k = mol L–1 s –1

z2 1

1 z

z2

=0

x x2 1

6 4

80.

,Q

x x2 1

=6–0–

79.

2

2x

z

..O..

1 x

Now, IF = e

CH2NH2

Compound is most basic due to localised lone pair of electrons on nitrogen atom While in other compounds, because of resonance, the lone pair of electrons on nitrogen atom gets delocalised over benzene ring and thus is less easily available for donation. (d) Formal charges help in selection of the lowest energy structure from a number of possible Lewis structures for a given species. Generally the lowest energy structure is the one with the smallest formal charges on the atoms. Formal charge on an atom = total no. of valence electrons – non -bonding

2x

(1 + xyz)

y y2 1 z

z2 1

=0

(1 + xyz) [x(y2 – z2) –y (x2 – z2) + z (x2 – y2)] = 0 (1 + xyz) (x – y) (y – z) (z – x) = 0 1 + xyz = 0 xyz = –1

EBD_7443 2012-22

83.

Target VITEEE

(c) P (A B) = 0.6 and P (A B) = 0.2 we know that P (A B) = P (A) + P(B) – P(A B) 0.6 = P(A) + P(B) – 0.2 P(A) + P(B) = 0.8

87. 88.

1 – P( A ) + 1 – P ( B ) = 0.8 – [P( A ) + P( B )] = 0.8 – 2 P( A ) + P( B ) = 1.2 84.

R sin sin( ) 2 Also, (5P) = (4P)2 + (3P)2 + 2 (4P) (3P) cos ( + ) 25P2 = 16p2 + 9P2 + 24P2 cos ( + ) 24P2 cos ( + ) = 0 cos ( + ) = 0 = cos90° + = 90°

(b) Q =

89.

z1 z 3 2 So, B is the mid–point of the line AC. A, B and C are collinear. z1, z2 and z3 lie on a line.

z2 =

4P R

5P

90.

sin =

5P sin sin 90

85.

4 5

2 cos x 1 cos 2 x

2cos x 1 cos 2 x 2 cos x sin 2 x

86.

= tan–1 (2 cosec x)

= 2 cosec x

= 2 cosec x

sinx = cos x

1 1 a

=

1 1 b

=

1

1 c Since, a, b and c are in AP. 1 – a, 1 – b and 1 – c are also in AP.

(b) 2 tan–1 (cos x) = tan –1 (2 cosec x) tan–1

=

and z = 1 + c + c2 + ....

4 5

= sin–1

(c) x = 1 + a+ a2 + ... y = 1 + b + b2 + ....

3P Now, 4P =

(a) (x + y) . (x + 1) = x + x.y + y Replace ‘.’ by ‘+’, ‘+’ by ‘.’, ‘1’ by ‘0’, we get (x . y) + (x . 0) = x.(x + y) . y (b) f (x) = (x – 1) (x – 2) (x – 3) f(1) = f(2) = f (3) = 0 f(x) is not one–one. For each y R, there exists x R such that f (x) = y. f is onto. Note that if a continuous function has more than one roots, then the function is always many–one. (c) Let z1, z2 and z3 be affixes of points A, B and C, respectively. Since, z1, z2 and z3 are in AP, therefore 2z2 = z1 + z3

x=

4 (d) Given conditions are a + x = 1 and ax = 0. These two conditions will be true, if x = a .

91.

1 1 1 , and are in HP.. 1 a 1 b 1 c x, y and z are in HP. Note that if the common ratio of a GP is not less than 1, then we do not determined the sum of an infinite GP that series. (a) Let f(x) = – 3 + x – x2 Then, f(x) < 0 for all x because coefficient of x2 < 0 and disc < 0. Thus, LHS of the given equation is always positive whereas the RHS is always less than zero. Hence, the given equation has no solution. Alternate Solution : Given, equation is 9 = –3 + x – x2 10

Solved Paper 2012

2012-23

In ABE,

Y

150 d d = 150 cot

tan =

y= 9 10

2 = 60 m 5 In DCE,

= 150 ×

X

X

y = – 3 + x – x2

tan =

Y 9 , therefore 10 y = –3 + x – x2

Let y =

y=

x2

x

1 4

1 4

3

94.

2

92.

1 11 x y+ 4 2 It is clear from the graph that two curves do not intersect. Hence, no solution exists. (c) The centre of the required circle lies at the intersection of 2x – 3y – 5 = 0 and 3x – 4y – 7 = 0. Thus, the coordinates of the centre are (1, –1). Let r be the radius of the circle. r2 = 154 22 2 r = 154 r = 7 7 Hence, the equation of required circle is (x –1)2 + (y + 1)2 = 72 x2 + y2 – 2x + 2y – 47 = 0

93.

(d) Given : tan =

96.

4 5 and tan = 3 2

E 97.

x D

A

h 150 m

d

C

d

B

4 4 h h 60 h = 80 m 3 d 3 Now in DCE, DE2 = DC2 + CE2 x2 = 602 + 802 = 10000 x = 100 m (a) Given equation x2 + px + q = 0 has roots and 2. Sum = + 2 = –p and Product = 3 = q ( + 1) = –p 3 [ 3 + 1 + 3 ( + 1)] = – p3 q (q + 1 – 3p) = –p3 p3 – (3p – 1) q + q2 = 0 100

95.

h d

(c)

100

C m (x 3)100

m

.2m

m 0

Above expansion can be rewritten as [(x – 3) + 2]100 = (x – 1)100 = (1 – x)100 x53 will occur in T54. T54 = 100C53 (–x)53 Required coefficient is – 100C53. (b) Equation of family of concentric circles to the circle x2 + y2 + 6x + 8y – 5 = 0 is x2 + y2 + 6x + 8y + = 0 which is similar to x2 + y2 + 2gx + 2fy + c = 0 Thus, the point (–3, 2) lies on the cirlce x2 + y2 + 6x + 8y + c = 0 (–3)2 + (2)2 + 6(–3) + 8 (2) + c = 0 9 + 4 – 18 + 16 + c = 0 c = –11 (a) a = i + j + k, b = i + 3j + 5k and c = 7i + 9j + 11k Let A = a + b = (i + j + k) + (i + 3j + 5k) = 2i + 4j + 6k and B = b + c = (i + 3j +5k) + (7i + 9j + 11k) = 8i + 12j + 16k Area of parallelogram

EBD_7443 2012-24

Target VITEEE We know that, –1 sin x 1

1 |A B| 2 ( A and B are diagonals)

=

i j k 1 = 2 4 6 2 8 12 16

1 |–8i + 16j – 8k| 2

=

( 4)2

(8)2

=

96 = 4 6 sq units

1 4 sin x

3 2

2

25 4

2

25 2

3= and 4 =

a ii

i 1

a 0 2 0 b 2

23

a=6 b=8

Y

7

9 , then 9

8

P(0, b)

tr (A) = 1 + 7 + 9 = 17

A(3, 4)

cos

sin

1

sin

cos

1

cos(

sin(

)

[Applying R3

) 1

R3 – R1(cos ) + R2 (sin )]

cos

sin

1

= sin 0

cos

1

0

1 sin

) (cos2

cos

= (1 + sin – cos + sin2 ) = 1 + sin – cos , which is independent of . 100. (d) 4 sin 2 x – 12sin x + 7 = 4 (sin2x – 3 sin x) + 7 =4

3 2

5 7

If A = 0 11

(a) Given,

sin x

1 2

3 2 2 101. (b) A is mid point of line PQ.

( 4)2

(a) We know that, tr (A) = 1

99.

3 2

1 4 sin x

n

98.

sin x

1 4

1 = |i (64 – 72) – j (32 – 48) + k (24 – 32)| 2

=

5 2

–

2

3 2

sin x

= 4 sin x

3 2

= 4 sin x

3 2

9 4

2

9 7 2

2

7

Q(a, 0)

O

X

Thus, equation of line is x 6

y =1 8 4x + 3y = 24 102. (d) The tangent at (1, 7) to the curve x2 = y – 6 is 1 (y + 7) – 6 2 2x = y + 7 – 12 y = 2x + 5 which is also tangent to the circle x2 + y2 + 16x + 12y + c = 0 i.e., x2 + (2x +5)2 + 16x + 12 (2x + 5) + c = 0 5x2 + 60x + 85 + c = 0, which must have equal roots. Let and are the roots of the equation. Then + = – 12 = –6 ( = ) x = – 6, y = 2x + 5 = – 7 Point of contact is (–6, –7).

x=

Solved Paper 2012

2012-25

103. (c) The intersection point of lines x – 2y = 1

Since, required line is parallel to 3x + 4y = 0. 3 . 4 Equation of required line which passes

1 x

1

= 0

1 x2

Therefore, the slope of required line is

through

7 1 , is given by 5 5 3 x 4

1 y 5

7 5

=

=

=

1 x2

1 cos x

dx x 2 2 sin 2 2

=

2

cos x cos

4

4

1

[t]10

2

1/ 2

=– 4

0

2

1 2 2

cosec

cot .

x cot 2 2

x 2 1 2

1

8

1

x

0

1 x2

dx

dx

2t dt = – 2x dx

8

8

x 2

x2 = 4

x3 3

=

1 16

1 24

=

6 4 96

dx

C

C

2

0t 1

t

dt

1

=

12 96

1 dx 2

x x

= 0

1/ 2

8

x 2

2

1 x2

1 1 dx 106. (c) Let I = 0 x x 2

1

1

x

I = (sin–1 1 – sin–1 0) +

dx

2

1 0

t dt = –x dx

sin x cos x

1

dx

Put t2 = 1 – x2

dx

2 sin x sin

=

0

dx

= [sin 1 x]10

dx

=

1

1

=

3x 21 1 y 4 20 5 3x + 4y – 5 = 0

104. (c) Let I =

1 x dx 0 1 x 1

105. (b) Let I =

7 1 , and x + 3y = 2 is 5 5

1 8

0

1/ 2 1

x 2 dx 1/2

1

1/ 2

x3 3

1 1 3 4

x x

x x2

x2 4

1 1/2

1 1 24 16

32 24 4 6 96

1 dx 2 x dx 2

EBD_7443 2012-26

Target VITEEE

107. (b) Let the equation of the ellipse be

x2

y2

a2

b2

= 1. It is given that it passes through (7, 0) and (0, –5). Therefore, a2 = 49 and b2 = 25 The eccentricity of the ellipse is given by

108. (c)

a2 25 49

= 1

24 2 6 = 49 7

x2

y2

2x

2y dy . =0 b 2 dx

=1 a 2 b2 ...(i) On differentiating w.r.t. x, we get

a

2

xb 2

dy dx

a2y

dy =0 dx

dy dx

c1

c2

b2 x x . = –1 a2y y

x2

y2

a2

b2

x2

y2

2

2

a

b

1 2

x

2

y 1

1 1 1 1 1 = 44 1! 2! 3! 4! 5! Note that if r (0 r n) objects occupy the original places and none of the remaining (n – r) objects occupies its original places then the number of such arrangements = nCr. (n – r)!

= 5! 1

1 1 1 .... ( 1) n 1! 2! 3!

111. (b) Tn = =

= –1

(y 1) 2

2 (x – 1)2 + (y + 1)2 = (x – y + 1)2 2xy – 4x + 4y + 1 = 0 110. (b) Required numbers

and

x dy = y dx The two curves will cut at right angles, if

dy dx

b2 = c2 2 a2 – b2 = 2c2 109. (c) Let P (x, y) be any point on the conic. Then,

1

x2 – y2 = c2 On differentiating w.r.t. x, we get 2x – 2y

a2 2

(x 1) 2

b2

e= 1

[using eq. (i)] On substituting these values in x2 –y2 = c2, we get

12 n2 n!

22

=

2

1 (n r)!

32 ... n 2 n!

n(n 1)(2n 1) 6n!

=

1 2n 3 3n 2 6 n!

=

1 n3 2. 6 n!

3n 2 n!

n

n n!

Sum of the series =

1 n3 n2 2 3 6 n 1 n! n 1 n!

=

1 (2 5e 3 2e e) 6

=

1 (10e 6e e) 6

17 e 6

n n! n 1

Solved Paper 2012

2012-27

112. (b) loga (1 + x) = loge (1 + x) logae = logae

( 1) n 1

n 1

n

x n

So, the coefficient of xn in loga (1 + x) is ( 1) n 1 log a e. n 113. (b) Let the equation of the required plane be

x y z = 1. a b c This meets the coordinate axes at A, B and C, the coordinates of the centroid of ABC

3 2

2 sin

=

3 2

2

1 2

1

3 2

1

2 sin

0

3 = sq units 2 3

6

115. (b) Let y = lim

1

tan

2

x

x

Taking log on both sides, we get

1 log x 2

log y = lim x

tan

1

x

form

form

a b c , , 3 3 3

are

=

1

a b c 1, 2, 3 3 3 3 a = 3, b = 6, c = 9 Hence, the equation of the plane is

1 x2

= lim x

2

x y z =1 3 6 9 114. (c) Required area

tan

1

x

(using L ‘Hospitals’ rule) 2x

Y

= lim x

x = 3y

(1 x 2 )2 1 1 x2

(using L Hospital’s rule) X

X

O 2

= lim x

2

x +y =4

2x 1 x2

=0

y = e° = 1

116. (c) f(x) is continuous at x = Y

So, lim f (x) = =

1 0 1 0

(x 2

x

x1 )dy

( 4 y2

1 2 = 2y 4 y

m

3y) dy

1 (4)sin 2

1

y 2

3y 2 2

1 0

m

2

2

1 sin

.

lim f (x) x

2

2

2

1 1 n

2

n

m =n 2

EBD_7443 2012-28

Target VITEEE

117. (c) f(x) =

4 x2 sin 1 (2 x)

1

2 4 x 2 is defined for 4 – x 0. x2 4 –2 x 2 and sin–1 (2 – x) is defined for –1 2 – x 1 –3 – x – 1 1 x 3 Also, sin–1 (2 – x) = 0 for x = 2 Domain of f (x) = [–2 , 2] [1, 3] – {2} = [1, 2) 118. (c) (1 + y2) dx + (1 + x2) dy = 0

dx

dy

dy dx

=0

1 x 2 1 y2 On integrating, we get tan–1 x + tan–1 y = tan –1 C

x y =C 1 xy x + y = C (1 – xy)

119. (a)

=

3 2 2

d2 y dx 2 d2 y dx 2

1

dy dx

2 3/ 2

On squaring both sides, we get 2

d2y dx 2

2

1

dy dx

2 3

Clearly, it is a second order differential equation of degree 2. Note that the higher order derivative is in the transcendental, then we do not determined the degree of that equation. 120. (b) Let R = {(a, b) : a, b N, a – b = 3} = [{(n + 3), n} : n N] = {(4, 1), (5, 2), (6, 3), ...}

SOLVED PAPER PART - I (PHYSICS) 1.

2.

3.

4.

5.

6.

A glass rod rubbed with silk is used to change a gold leaf electroscope and the leaves are observed to diverge. The electroscope thus charged is exposed to X-rays for a short period. Then (a) the divergence of leave will not affected (b) the leaves will diverge further (c) the leaves will collapse (d) the leaves will melt An infinite number of charge, each of charge 1 C are placed on the x-axis with coordinates x=1, 2, 4, 8,...... If a charge of 1C is kept at the origin, then what is the net force acting on 1C charge? (a) 9000 N (b) 12000 N (c) 24000 N (d) 36000 N A cube of side is placed in a uniform field E, where E = Ei . The net electric flux through the cube is (a) zero (b) l2E 2 (c) 4l E (d) 6l2E The capacity of a capacitor is 4×10-6 F and its potential is 100 V. The energy released on discharging it fully will be (a) 0.02 J (b) 0.04 J (c) 0.025 J (d) 0.05 J Dimensions of a block are 1cm×1cm×100cm. If specific resistance of its material is 3×10-7 m, then the resistance between the opposite rectangular faces is (a) 3×10-7 (b) 3×10-9 (c) 3×10-5 (d) 3×10-3 The magnitude and direction of the current in the circuit shown will be a

1

2

e

10V

b

4V

3

c

2011

(a) 7/3 A from a to b through e (b) 7/3 A from b to a through e (c) 1A from b to a through e (d) 1A from a to b through e 7. An electric bulb of 100 W is connected to a supply of electricity of 220 V. Resistance of the filament is (a) 484 (b) 100 (c) 22000 (d) 242 8. Pick out the wrong statement. (a) In a simple battery circuit, the point of lowest potential is the negative terminal of the battery. (b) The resistance of an incandescent lamp is greater when the lamp is switched off. (c) An ordinary 100W lamp has less resistance than a 60 W lamp. (d) At constant voltage, the heat developed in a uniform wire varies inversely as the length of the wire used. 9. The electrochemical equivalent of magnesium is 0.126 mg/C. A current of 5A is passed in a suitable solution for 1h. The mass of magnesium deposited will be (a) 0.0378 g (b) 0.227 g (c) 0.378 g (d) 2.27 g 10. In producing chlorine through electrolysis 100 W power at 125 V is being consumed. How much chlorine per minute is leberated? (ECE of chlorine is 0.367 × 10–6 kg/C) (a) 24.3 mg (b) 16.6 mg (c) 17.6 mg (d) 21.3 mg 11. A particle carrying a charge to 100 times the charge on an electron is rotating per second in a circular path of radius 0.8 m. The value of the magnetic field produced at the centre will be ( 0= permeability for vacuum) (a)

d

VITEEE

(c)

7

10

(b) 10-17

0

10-6

0

(d) 10-7

0 0

EBD_7443 2011-2

Target VITEEE

12. A rectangular loop carrying a current i is placed in a uniform magnetic field B. The area enclosed by the loop is A. If there are n turns in the loop, the torque acting on the loop is given by (a) ni A × B (b) ni A · B l l (iA×B) (d) (iA · B) n n In a magnetic field of 0.05 T, area of a coil changes from 101 cm2 to 100 cm2 without changing the resistance which is 2 . The amount of charge that flow during this period is (a) 2.5 × 10–6 C (b) 2 × 10–6 C –6 (c) 10 C (d) 8 × 10–6 C A solenoid has 2000 turns wound over a length of 0.30 m . The area of its cross-section is 1.2 × 10-3 m2. Around its central section, a coil of 300 turn is wound. If an initial current of 2A in the solenoid is reversed in 0.25 s, then the emf induced in the coil is (a) 6× 10–4 V (b) 4.8× 10–3V –2 (c) 6× 10 V (d) 48mV An inductive circuit contains a resistance of 10 and an inductance of 2.0 H. If an AC voltage of 120 V and frequency of 60 Hz is applied to this circuit, the current in the circuit would be nearly (a) 0.32 A (b) 0.16 A (c) 0.43 A (d) 0. 80 A In a Millikan's oil drop experiment the charge on an oil drop is calculated to be 6.35 × 10–19 C. The number of excess electrons on the drop is (a) 3.2 (b) 4 (c) 4.2 (d) 6

(c)

13.

14.

15.

16.

17. The values

1 and 2

1 of spin quantum 2

number show (a) rotation of electron clockwise and anticlockwise directions respectively (b) rotation of electron anti-clockwise and clockwise directions respectively (c) rotation in any direction according to convention (d) None of the above 18. The frequency of incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is (a) double the earlier value (b) unchanged (c) more than doubled (d) less than doubled

19. Light of two different frequencies whose photons have energies 1 eV and 2.5 eV, respectively, successively illuminate a metal whose work function is 0.5 eV. The ratio of the maximum speed of the emitted electrons will be (a) 1:5 (b) 1:4 (c) 1:2 (d) 1:1 20. An electron accelerated under a potential difference V volt has a certain wavelength . Mass of proton is some 2000 times of the mass of the electron. If the proton has to have the same wavelength , then it will have to be accelerated under a potential difference of (a) V volt (b) 2000 V volt (c)

V volt 2000

(d)

2000 V volt

21. The ratio of momentum of an electron and -particle which are accelerated from rest by a potential difference of 100 V is (a) (c)

1 me / m

(b)

2m e / m a

(d)

me / 2m

22. Sky wave propagation is used in (a) radio communication (b) satellite communication (c) T V communication (d) Both T V and satellite communication 23. The frequency of an FM transmitter without signal input is called (a) the centre frequency (b) modulation (c) the frequency deviation (d) the carrier sweing 24. What is the age of an ancient wooden piece if it is known that the specific activity of C14 nuclide in its amounts is 3/5 of that in freshly grown trees? Given the half of C nuclide is 5570 yr. (a) 1000 yr (b) 2000 yr (c) 3000 yr (d) 4000 yr 25. A thin metallic spherical shell contains a charge Q on it. A point charge q is placed with the centre of the shell and another charge q1s placed outside it as shown in the figure. All the three charges are positive.

Solved Paper 2011

2011-3

Q

30. Five resistances are connected as shown in the figure. The effective resistance between points A and B is 2.5 B

q

q1

7.5

10

1 The force on the charge at the centre is (a) towards left (b) towards right (c) upward (d) zero 26. As shown in the figure, charges + q and – q are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is

A

a b

B

(a)

(c)

1 4 1 4

b

+q

.

.

2a a2 b2 q 2

2

-q C

(b) zero

(d) 4

1

.

( q)

a b a2 b2 27. On moving a charge of 20 C by 2 cm, 2 J of work is done, then the potential difference between the points is (a) 0.1 V (b) 8 V (c) 2 V (d) 0.5 V 28. The insulation property of air breaks down at 3 × 106V/m. The maximum charge that can be given to a sphere of diameter 5 m is nearly (a) 2×10–2C (b) 2×10–3C –4 (c) 2×10 C (d) 2×10–5C 29. Two capacitors of capacities C and 2C are connected in parallel and then connected in series with a third capacitor of capacity 3C. The combination is charged with V volt. The charge on capacitor of capacity C is

(a)

1 CV 2

(b) CV

(c)

2CV

(d)

3 CV 2

A (a)

9

10 3

(b)

10 17

(c) 40 (d) 45 31. A potentiometer is connected across A and B and a balance is obtained at 64.0 cm. When potentiometer lead to B is moved to C, a balance is found at 8.0 cm. If the potentometer is now connected across B and C, a balance will be found at D B C (b) 56.0 cm (a) 8.0 cm (c) 64.0 cm (d) 72.0 cm 32. In an electromagnetic wave, the average energy density associated with magnetic field is

(a)

L i 20 /2

(c)

2

0B

(b) B2 / 2 /2

(d)

0

0

/ 2B2

33. An electromagnetic wave going through vacuum is described by E=E0 sin (kx – t) Which of the following is/are independent of the wavelength? 2 (a) k (b) (c) k/ (d) k 34. An ammeter reads upto 1A. Its internal resistance is 0.81 . To increase the range to 10 A, the value of the required shunt is (a)

0.09

(b) 0.03

(c) 0.3 (d) 0.9 35. A coil of resistance 10 and inductance 5 H is connected to a 100 V battery. Then the energy stored in the coil is (a) 250 J (b) 250 erg (c) 125 J (d) 125 erg

EBD_7443 2011-4

Target VITEEE

36. A nucleus

A Z

X emits an

- particle. The

resultant nucleus emits a + – particle. The respective atomic and mass number of final nucleus will be (a) Z–3, A–4 (b) Z–1, A–4 (c) Z–2, A–4 (d) Z, A–2 37. In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is = I. The intensity of light at a point where the path difference becomes /3 is (a) I / 4 (b) I / 3 I (d) I 2 38. Polarising angle for water is 53°4'. If light is incident at this angle on the surface of water and reflected the angle of refraction is (a) 53°4' (b) 126°56' (c) 36°56' (d) 30°4' 39. A 2V battery, a 15 resistor and a potentiometer of 100 cm length, all are connected in series. If the resistance of potentiometer wire is 5 , then the potential gradient of the potentiometer wire is (a) 0.005 V/cm (b) 0.05 V/cm (c) 0.02 V/cm (d) 0.2 V/cm 40. The output voltage of a transformer connected to 220 V line is 1100 V at 2 A current. Its efficiency is 100%. The current coming from the line is (a) 20A (b) 10A (c) 11A (d) 22A

(c)

2 42. Amongst Ni(CO)4, [Ni(CN)4]2– and NiCl4

Ni(CO)4 and NiCl 24 are diamagnetic but [Ni(CN)4]2– is paramagnetic (b) Ni(CO)4 and [Ni(CN)4]2– are diamagnetic

(a)

but NiCl 24 is paramagnetic (c)

NiCl24 and [Ni(CN)4]2– are diamagnetic but Ni(CO)4 is paramagnetic

(d) Ni(CO)4 is diamagnetic but NiCl 24 and [Ni(CN)4]2– is paramagnetic 43. The equivalent conductances of two ions at infinite dilution in water at 25ºC are given below o Ba 2 o Cl

= 127.00 Scm2/ equiv.. = 76.00 Scm2/equiv..

The equivalent conductance (in S cm2/equiv) of BaCl2 at infinite dilution will be (a) 203 (b) 279 (c) 205.5 (d) 139.5 44. The product formed when phthalimide is treated with a mixture of Br2 and strong NaOH solution is (a) aniline (b) phthalamide (c) phthalic acid (d) anthranilic acid 45. In a set of reactions acetic acid yielded a product D. CH3 COOH

SOCl2

A

Benzene anhy. AlCl3

PART - II (CHEMISTRY)

C

41. An alkene having molecular formula C8H12 on ozonolysis yields glyoxal and 2, 2-dimethyl butane-1, 4-dial. The structure of alkene is CH3

(a)

H 3C

(b)

CH3 CH3

The structure of D would be

OH CH2 — C — CH3 (a)

CN OH

CH3

(c)

H3C (d) H3C

B

C — COOH (b)

CH3

HCN

H 2O

D

Solved Paper 2011

2011-5

CN

This reaction is called (a) Reimer-Tiemann reaction (b) Lederer-Manasse reaction (c) Sandmeyer reaction (d) Kolbe's reaction

C — CH3

(c)

OH COOH CH2 — C — CH3 (d)

O || 52. CH 3 C CH 3

OH

46. The alcohol having molecular formula C4H9OH, when shaken with a mixture of anhydrous ZnCl2 and conc. HCl gives an oily layer product after five minutes. The alcohol is (a) H3C — (CH2)3 — OH (b) (CH3)2CH — CH2OH (c) (CH3)3C — OH (d) H3C — CH(OH) CH2 — CH3 47. p-toluidine and benzyl amine can be distinguished by (a) Sandmeyer's reaction (b) Dye test (c) Molisch test (d) Gattermann reaction 48. CH3CH2Br undergoes Wurtz reaction. We may expect some of the following product A : CH3CH2CH2CH3 B : CH2 = CH2 C : CH3 — CH3 Select correct product. (a) Only A (b) A and B (c) A, B and C (d) A and C 49. Sometimes explosion occurs while distilling ethers. It is due to the presence of (a) peroxides (b) oxides (c) ketones (d) aldehydes 50. Glycerine is used as a preservative for fruits and eatables because (a) it makes them sweet (b) it acts as an insecticide (c) it keeps the food moist (d) all of the above OH

51.

OH HCHO

OH or H

OH CH2OH

+

SeO 2

X Se H 2 O : X

(a)

O O O || || || CH 3 C C H (b) CH3 C OCH3

(c)

O || C CH2OH (d) None of the above

CH3

53. Which of the following will give Cannizzaro reaction? (a) CH3CHO (b) CH3COCH3 (c) (CH3)3C – CHO (d) CH3CH2CHO 54. The secondary structure of a protein refers to (a) -helical backbone (b) hydrophobic interactions (c) sequence of -amino acids (d) fixed configuration of the polypeptide backbone 55. Self condensation of two moles of ethyl acetate in the presence of sodiumethoxide after acidification yields (a) acetic acid (b) acetoacetic ester (c) ethyl propionate (d) ethyl butyrate 56. Which one of the following will be most basic? (a) Aniline (b) p-methoxyaniline (c) p-methyl aniline (d) Benzylamine 57. Mn2O7 dissolves in water to give an acid. The colour of the acid is (a) green (b) blue (c) purple (d) red 58. "925 fine silver" means an alloy of (a) 7.5% Ag and 92.5% Cu (b) 92.5% Ag and 7.5% Cu (c) 80% Ag and 20% Cu (d) 90% Ag and 10% Cu

EBD_7443 2011-6

59. In which of the following octahedral complexes of Co (At. no. 27), will the magnitude of 0 be the highest? (a) [Co(CN)6]3– (b) [Co(C2O4)3]3– 3+ (c) [Co(H2O)6] (d) [Co(NH3)6]3+ 2+ 60. Assertion (A) Cu and Cd2+ are separated by first adding KCN solution and then passing H2S gas. Reason (R) KCN reduces Cu2+ to Cu+ and forms a complex with it. The correct answer is (a) Both (A) and (R) are true and (R) is the correct explanation of (A) (c) Both (A) and (R) are true but (R) is not the correct explanation of (A) (c) (A) is true but (R) is not true (d) (A) is not true but (R) is true 61. The effective atomic number of cobalt in the complex [Co(NH3)6]3+ is (a) 36 (b) 24 (c) 33 (d) 30 62. The IUPAC name for th e complex [Co(NO2)(NH3)5]Cl2 is (a) nitrito-N-pentammine cobalt (III) chloride (b) nitrito-N-pentammine cobalt (II) chloride (c) pentaminenitrito-N-cobalt (II) chloride (d) pentaminenitrito-N-cobalt (III) chloride 63. The radio-isotope used for treatment of thyroid disorders is (a) Na-24 (b) P-32 (c) Co-60 (d) I-131 64. Tetragonal crystal system has the following unit cell dimensions (a) a = b = c, = = = 90º (b) a = b c, = = = 90º (c) a b c, = = = 120º (d) a = b c, = = 90º, = 120º 65. A crystalline solid (a) changes rapidly from solid to liquid (b) has no definite melting point (c) undergoes deformation of its geometry easily (d) soften easily 66. Two glass bulbs A and B are connected by a very small tube having a stop-cock. Bulb A has a volume of 100 cm3 and contained the gas while bulb B was empty. On opening stop-clock, the pressure fell down to 40%. The volume of the bulb B must be

Target VITEEE 75 cm3

(a) (b) 125 cm3 3 (c) 150 cm (d) 250 cm3 67. 20 mL of 0.2 M NaOH is added to 50 mL of 0.2 M acetic acid. The pH of this solution after mixing is (Ka = 1.8 × 10–5) (a) 4.5 (b) 2.3 (c) 3.8 (d) 4 68. Consider the followin g equation, which represents a reaction in the extraction of chromium from its ore 2Fe2O3 . Cr2O3 + 4Na2CO3 + 3O2 2Fe 2 O 3

4Na 2 CrO 4

4CO 2

Which one of the following statements about the oxidation states of the substances is correct? (a) The iron has been reduced from +3 to +2 state. (b) The chromium has been oxidised from +3 to +6 state. (c) The carbon has been oxidised from +2 to +4 state. (d) There is no change in the oxidation state of the substances in the reaction. 69. The freezing point of a solution composed of 10.0 g of KCl in 100 g of water is 4.5ºC. Calculate the van't Hoff factor, i for this solution. (a) 2.50 (b) 1.8 (c) 1.2 (d) 1.3 70. In the reversible reaction, 2NO 2

k1 k2

N 2O 4

the rate of disappearance of NO2 is equal to 2k1 2 NO 2 k2 (b) 2k1[NO2]2 – 2k2[N2O4] (c) 2k1[NO2]2 – k2[N2O4] (d) (2k1 – k2) [NO2] 71. A chemical reaction was carried out at 300 K and 280 K. The rate constants were found to be k l and k2 respectively. Then (a) k2 = 4kl (b) k2 = 2kl (c) k2 = 0.25 kl (d) k2 = 0.5 kl 72. The rate constant of a reaction at temperature 200 K is 10 times less than the rate constant at 400 K. What is the activation energy of the reaction? (a) 1842.4 R (b) 460.6 R (c) 230.3 R (d) 921.2 R 73. A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. The value of K if the total pressure at equilibrium is 0.8 atm, is

(a)

Solved Paper 2011

2011-7

(b) 3 atm (a) 1.8 atm (c) 0.3 atm (d) 0.18 atm 74. For the reaction 2A + B C, H = x cal, which one of the following conditions-would favour the yield of C on the basis of Le-Chatelier principle? (a) High pressure, high temperature (b) Only low temperature (c) High pressure, low temperature (d) Only low pressure 75. The EMF of the cell, Mg | Mg2+ (0.0IM) || Sn2+ (0.1 M) | Sn at 298K is

Eo

Mg 2 / Mg

2.34V, E o

(a) 2.17 V (c) 2.51 V 76. Heat of formation,

Sn 2 / Sn

0.14V

(b) 2.23 V (d) 2.45 V H of of an explosive

compound like NCl3 is (a) positive (b) negative (c) zero (d) positive or negative 77. For the reaction, C3 H8 (g) 5O 2 (g)

3CO 2 (g) 4H 2 O(l)

at constant temperature, H – E is (a) RT (b) – 3RT (c) 3RT (d) – RT 78. The favourable conditions for a spontaneous reaction are (a) T S > H, H = + ve, S = +ve (b) T S > H, H = + ve, S = –ve (c) T S = H, H = –ve, S = –ve (d) T H = H, H = + ve, S = +ve 79. Compound A and B are treated with dil. HCl separately. The gases liberated are Y and Z respectively. Y turns acidified dichromate paper green while Z turns lead acetate paper black. The compound A and B are respectively. (a) Na2CO3 and NaCl (b) Na2SO3 and Na2S (c) Na2S and Na2SO3 (d) Na2SO3 and Na2SO4 80. Which of the following is correct comparison of the stability of the molecules?

(a)

CN < O2

(b) CN = N2

(c)

N2 < O2

(d) H 2

He 2

PART - III (MATHEMATICS) 81. To the lines ax 2 a 2 x2

2hxy by2

2h(a b)xy b2 y2

0 , the line

0 are

(a) equally inclined (b) perpendicular (c) bisector of the angle (d) None of the above 82. If R be a relation from A={1, 2, 3, 4} to B={1, 3, 5} such that (a, b) R a < b, then ROR–1 is (a) {(1,3) , (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} (b) {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} (c) {(3, 3), (3, 5), (5, 3), (5, 5) (d) {(3, 3), (3, 4), (4, 5)} 83. If x + iy =(1 – i 3 ) 100 , then find (x, y). 299 , 299 3

(a)

299 , 299 3

(c)

84. For a GP, an common ratio. (a) 2 (c) 3

(d) None of these

3 2n ,

86. If

a

,

(a)

x 11 or x

3 2

(b)

x 11 or x

1

3 2

(d)

1

b

,

c

will be

(b) G P (d) None of these

2x 7 0 (c) m = 0 (d) m = 3 101. A line making angles 45 and 60 with the positive directions of the axes of x and y makes with the positive direction of z-axis, an angle of (a) 60 (b) 120 (c) 60 or 120 (d) None of these

102. If I =

3 /4

p2 of q

2c

(c)

2x 2 x2

95. If the roots of the equation x 2 ax b 0 are c and d, then one of the roots of the equation

q2 p

(b) 2A (d) 4A

B=

1 0 0 1

, J=

cos sin

sin cos

0

1 1 0

and

, then B is equal to

(b) I sin (a) I cos J sin J cos (c) I cos (d) J sin I cos J sin 103. Which of the following is correct? (a) Determinant is a square matrix (b) Determinant is a number associated to a matrix (c) Determinant is a number associated to a square matrix (d) All of the above

Solved Paper 2011 104. If , and

the value of

2011-9

are the roots of x3 + ax2 +b = 0,

then

is

(a) –a3 (b) a3–3b (c) a3 (d) a2–3b 105. If the axes are shifted to the point (1, –2) without solution, then the equation 2x2 + y2 – 4x + 4y = 0 becomes (a) 2X2+3Y2 = 6 (b) 2X2 + Y2=6 (c) X2 + 2Y2 = 6 (d) None of these 106. If f(x)=

x2 , x 0 , then x = 0 is 2sin x, x 0

(a) point of minima (b) point of maxima (c) point of discontinuity (d) None of the above 107. In a group (G,*), then equation x * a = b has a (a)

unique solution b * a

1

(b) unique solution a 1 * b (c) unique solution a 1 * b 1 (d) many solutions 108. A die is rolled twice and the sum of the numbers appearing on them is observed to be 7. What is the conditional probability that the number 2 has appeared at least once? 1 2 2 1 (b) (c) (d) 3 3 5 2 109. The locus of the mid-points of the focal chord of

(a)

the parabola y 2 4ax is (b) y2 =2a(x – a) (a) y2 = a(x – a) 2 (c) y = 4a(x – a) (d) None of these 110. Find the value of sin 12° sin 48° sin 54° . (a)

1 2

1 (b) 4

1 1 (d) 6 8 111. In an equilateral triangle, the inradius, circumradius and one of the exradii are in the ratio (a) 2 : 3 : 5 (b) 1 : 2 : 3 (c) 1 : 3 : 7 (d) 3 : 7 : 9 112. Let p and q be two statements. Then, p q is false, if (a) p is false and q is true (b) both p and q are false

(c)

(c) both p and q are true (d) None of the above 113. In how many ways 6 letters be posted in 5 different letter boxes? (a) 56 (b) 65 (c) 5! (d) 6! 114. If A and B be two sets such that A×B consists of 6 elements. If three elements A × B are (1, 4), (2, 6) and (3, 6), find B×A. (a) {(1, 4), (1, 6), (2, 4), (2, 6),(3, 4), (3, 6)} (b) {(4, 1), (4, 2), (4, 3), (6, 1),(6, 2), (6, 3)} (c) {(4, 4), (6, 6)} (d) {(4, 1), (6, 2), (6, 3)} 115. Let f : R R be defined as f (x) = x2+1, find f –1(–5). (a) (b) (d) {–5, 5} (c) {5} 116. If X is a poisson variate such that P(X = 1) = P(X = 2), then P(X=4) is equal to (a)

1 2e

(b)

2

2

(c)

(d)

2

1 3e 2 1

3e e2 117. The area enclosed by y = 3x – 5, y = 0, x = 3 and x = 5 is (a) 12 sq units (b) 13 sq units 1 sq units (d) 14 sq units 2 118. The order and degree of the differential equation

(c)

13

1 4 (a)

dy dx

1,

2 3

2/3

4

d2y dx 2

are respectively (b) 3, 2

2 3 119. The solution of the differential equation

(c)

2, 3

(d) 2,

dy 2 4 x y 1 , is dx (a) (4x + y + 1) = tan (2x+ C) (b) (4x + y +1)2 = 2tan (2x + C) (c) (4x + y +1)3 = 3tan (2x + C) (d) (4x + y + 1) = 2tan (2x + C) 120. The system of equations 2x + y –5=0, x – 2y + 1 = 9, 2x –14y – a = 0, is consistent. Then, a is equal to (a) 1 (b) 2 (c) 5 (d) None of these

EBD_7443 2011-10

Target VITEEE

SOLUTIONS PART - I (PHYSICS) 1.

2.

5.

(b) As charge on glass rod is +(ve) so charge on gold leaves will also be +(ve). Due to X-rays more electrons from leaves will be emitted, so leaves becomes more positive and diverge further. (b) In figure there is the schematic diagram of distribution of charges on x-axis

1C

1 C 1 C x=1

1 C

1 C

x=2 x=3

x=8

(a) Here length l = 1 cm = 10 –2 m Area of cross-section, A = 1 cm × 100 cm = 10 –2 m2 1 cm 1 cm

100 cm

Resistance, R = 6.

A

= 3 × 10 –7 ×

10 10

2 2

= 3 × 10 –7 (d) The current in the circuit will be clockwise Since E1 (10 V) > E2 (4V) a

E1

1C

e

10 V

E2

2

b

4V

Using principle of superposition total force acting on 1 C charge F

1 4

0

1 1 10 (1)2

1 1 10 (4)2 10 4

6

0

6

6

1 1 10 (2) 2

1 1 10 (8)2

1 1 1 1 4 16

6

3

Applying Kirchhoff’s junction rule –1 × i – 10 – 4 × –2 × i –3i = 0 i = 1A (a to b via e)

6

...

Current I =

1 ... 64

7. 9 10

6

1

4.

1 4

4 12000 N 3 (a) Here, net flux is zero, as there is no charge residing inside the cube. (a) Energy released on discharge of capacitor 9 109 10

3.

8.

1

109

U

9.

6

1 CV 2 2

1 = × 4 × 10 – 6 × (100)2 = 0.02J 2

10.

V 10 4 = =1A R 6

V 2 220 220 = = 484 W 100 R (b) When switched off, resistance of an incandescent lamp increases. (d) Using Faraday’s law of electrolysis, Mass m = Zit = 0.216 × 10 – 3 × 5 × 3600 = 2.27 g (c) P = 100 W and V = 125 V

(a) Power P =

P 100 A = V 125 Mass of chlorine liberated = zit

P = VI

I=

= 0.367 × 10– 6 ×

100 × 60 = 17.6 mg 125

Solved Paper 2011 11.

2011-11

q = 100 × e t

(b) Current, i =

0

Magnetic field, Bcentre 0

4

.

4

.

2 i r

12.

200 1.6 10 19 10 4 0.8 (a) Torque, T = ni (A × B)

13.

(a) Magnetic Flux,

0

B dA

v1 = v2

20.

0

or

2.5 10 6 C

4

R2

(b) Here, impedence z = (10)

2

(2

21.

v

60 2) = 753.7

(c)

me mp

Qe V Qp

Now, pe =

1 2 mv 2

eV

2eV m

p = mv =

2meV

2me e 100

and p = 4m

Q Number of electron n = e

17.

h 2m pQ pVp

m p QpV p

(d) As we know,

X L2

2

3.9

P

V volt 2000

V 120 = = 0.159 A Z 753.7 (b) As we know, Q = ne

6.35 10 19 1.6 10 19

e

1 (1)V 2000

Current i =

16.

1 2

h 2meQeV

3

2000 300 1.2 10 0.30 2 ( 2) 0.25 –3 = 48.2 × 10 V = 48 mV

15.

1 0.5 2.5 0.5

Vp

N1 N 2 A di . l dt

10

1 2 mv 2

me QeV

7

g

(c) According to question

di (d) Induced emf e = M dt

=

y

2( E W0 ) m

v=

d Charge dQ = R 6

r

E – W0 =

17

= 0.05(101 – 100) × 10 – 4 = 5× 10 – 6 Wb

14.

19.

the wevelength of red light is greater than threshold wavelength and hence no electrons are emitted. (c) If E is the energy of incident photon and W0 the work function, then Available energy = E – W0

B A

Change in flux, d

5 10 2

(b) Since,

100e r

2

0

=

18.

e 100

pe me = 2m p

4

22. 23.

(a) (c) Frequency deviation is the frequency of an FM transmitter without signal input.

EBD_7443 2011-12

24.

Target VITEEE

(d) Here, N0 = e

t

3 N0e 5

t

Charge on capacitor, C =

5 3

=

25. 26.

or

5 t = loge 3

T 0.5 0.693

T

0.693

5570 0.5 yr 4018.7 yrs 0.693 (d) Field inside the metallic shell is zero. (b) Potential at A

31.

q

0 a b a b2 (a) Potential difference between two points in an electric field 2

2

W VA – VB = q0

where W is work done by moving charge q0 from point A to B Here, W = 2J, q0 = 20 c VA – VB =

1 R

2 = 0.1 volt 20

(b) Maximum electric field, Emax

10 3 (b) Here, E1 64 E1 – E2 8 E2 l 64 – l = 8 or l = 64 –8 = 56 cm

or

32. 33.

(c)

34.

GI g (a) Shunt S = I I

35.

= 2×10 – 3 C 29.

(a) In parallel combination, the potential difference across each individual capacitor is the same as the potenital difference across the combination. So, potential difference across capacitor of capacity C is

V . 2

is independent of

0.81 1 10 1

g

0.81 9

0.09

E =10 A R Energy stored in the magnetic field

(a) Final current, I =

1 Qmax 4 0 R2

3 10 6 2.5 2.5 9 109

k

=

U=

9 10 Qmax 2.5 2.5

Qmax

R

B2 (b) Emag = 2 0

9

3 × 106 =

1 1 1 10 10 10

or

q

28.

30.

t = 1 loge 5 3

2

27.

CV 2 (a) It is a parallel combination of three resistances, each of 10 . The effective resistance between points A and B.

Charge on capacitor of capacity C =

log e t = loge 5 3 or

Q V

1 Li 2

1 × 5 × (10)2 = 250 J 2 36.

(a)

A Z

X

A 4 Z 2

Y

4 2

He e

A 2 Z 2 A 4 Z 3

Y

Y'

+ emission.

During 1 1 + 1p 0n + The proton changes into neutron. So, charge number decreases by 1 but mass number remains unchanged.

Solved Paper 2011 37.

2011-13

4a 2 cos2

(a) Intensity, I

4a

( BaCl2 )

2

4a 2 cos 2

38.

2 or I ' 3

44.

C

(d)

C

e ( R Rh

40.

2 5 0.5 V / m (15 5 0) 1 = 0.005 V/cm (b) For 100% efficiency Vsis =Vpip 1100 × 2 = 220 × ip i.e., Primary current, ip = 10A

O

R r) L .

C OH NH2

O

45.

(b) CH3COOH

C |

CH 3

CHO

CHO | CHO

2, 2 - dimethyl butane glyoxal 1, 4 - dial 42.

SOCl2

CH3COC1

Benzene Anhy.AlCl3

O

C CH3

(b) The products formed as a result of ozonolysis suggests CH CH 2

+ 2NaBr + Na2CO3+H2O

anthranilic acid

PART - II (CHEMISTRY)

|

NH + Br2 + 4NaOH

phthalimide

(a) Potential gradient

CH 3

(Cl )

O

39.

41.

2

O

a2

I' 1 1 or I ' I 4 4 (c) Here, p + r = 90° r = 90° – p or r = 90° – ' = 36°56 dv dx

( Ba 2 )

= 127 + 2 × 76 = 127 + 152 = 279 Scm2/equiv.

2 3

In the second case, I'

(b) The equivalent conductance of BaCl2 can be calculated as

2

In the first case,

I'

43.

2

(B) OH

CH3 CH3

C CH3 CN

OH +

C CH3

H

COOH

(C)

(C8H12)

(D)

(b) The electronic configuration of Ni is Ni(28) = [Ar] 3d8, 4s2 Ni2+ = [Ar] 3d 8 Both Ni and Ni 2+ have two unpaired electrons. CO and CN– are strong field ligands and thus unpaired electrons get paired. Hence, Ni(CO)4 and [Ni(CN)4]2– are diamagnetic. Cl– is a weak field ligand hence, no pairing of e– will take place. hence NiCl 24 paramagnetic.

HCN

is

46.

(d) Secondary alcohol, when shaken with a mixture of anthydrous ZnCl2 and conc. HCl (Lucas regent) gives an oily layer product after five minutes. H3C CH CH 2 CH3 |

Lucas reagent

OH oily layer product after 5 min. The amines in which amino group is directly attached to benzene ring undergo diazotisation reaction.

EBD_7443 2011-14

47.

Target VITEEE

(b) p-toluidine contains amino group attached directly to benzene ring, thus undergo diazotisation reaction and gives red dye. In benzyl amine, the amino group is not directly attached to benzene ring. Hence, it will not undergo diazotisation reaction.

O CH3 C CH3

CH2NH2

2CH3

(c) C2H5Br + Na•

|

CH3

CH3CH 2 + CH3CH 2

CH3

54.

CH3CH2 – CH2CH3

CH2

CH2 + CH3 – CH3

C |

55.

2CH3 COOC2 H5

(a) Ethers, In the presence of air and light form peroxides (c) Glycerine is used as a preservative because it keeps the fruit moist. (b) OH

OH –

+ HCHO

CH2OH

+

OH or H

acetoacetic ester

56.

57.

CH2OH (a) In presence of SeO2 compounds containing active methylene (i.e., CH2 next to the carbonyl group) oxidises to another CO group.

CH3COCHNaCOOC2H 5

CH3COCH2COOC2H5

OH

52.

NaOC2 H5

+HCl

(Lederer - Menasse reaction)

+

|

OH

(a) Secondary structure involves -helical and -pleated sheet like structure. Where as Primary strucure involves sequence of -amino acids polypeptide chain. (b)

(disproportion reaction)

51.

|

CH2OH + CH3— C — COOK

CH3

n-butane

CH3CH 2 + CH3CH 2

CH3

CH3

CH3CH 2 + NaBr

Intermediate free radical CH 3CH 2 combines to form CH3CH2 – CH2CH3 (as a main product) and also CH2 == CH2 and CH3CH3 by disproportion

CHO + KOH

C

Thus, these two can be distinguished by dye test.

50.

||

(c) Aldehydes which have no -H atom give Cannizzaro’s reaction, (CH3)3C – CHO does not contain -H atom, hence it will give Cannizzaro reaction.

benzyl amine

p-toluidine

49.

||

CH3 C C H

CH3

CH3

48.

SeO 2

Se H 2O

53.

NH2

O O

||

58.

(d) In aniline, p-methoxyaniline and p-methyl aniline, the lone pair of electrons on the Natom is delocalised on the benzene ring while in benzylamine it is delocalised, and more available for donation. Hence benzylamine is most basic among the given. (c) Mn 2 O 7 dissolves in water to give permanganic acid which is purple in colour. Mn2O7 + H2O 2HMnO4 Purple (b) “925 fine silver” means 925 parts of pure Ag in 1000 parts of an alloy. Therefore, in terms of percentage, it will be 92.5% Ag and 7.5% Cu.

Solved Paper 2011 59.

2011-15

(a) Spectrochemical series. According to the strength of splitting is as: CO

CN

H2O

O

NO2 2

ox

en 2–

NH3

OH

F

py

68.

61.

62. 63. 64.

65.

66.

NCS

67.

40 p 100 V2 = 250 cc (Total volume) Volume of bulb B = 250 – 100 = 150 cc (a) NaOH + CH3COOH CH3COONa + H2O m mol 20 × 0.2 50 ×0.2 0 0 added = 4 = 10 m mol after reaction 0 6 4 4

100 × p = V2 ×

6 4 : [CH3COONa] = 70 70 Now since for a basic buffer

[CH3COOH =

pH = –logka+ log

[salt] [base]

pH = – log (1.8 ×10–5) + log = 4.56

4 / 70 6 / 70

3

2 Fe2 O3

C1

SCN S Br I Crystal field stabilisation energy (CFSE) for octahedral complex, depends on the strength of negative ligand. [Higher the strength more will be the CFSE] (b) KCN forms complex with Cu+ and Cd2+ as K3[Cu(CN)4] and K2[Cd(CN)4] respectively. On passing H2S, only Cd2+ complex gets decomposed to yellow CDS. (a) EAN =at. no. of central atom – oxidation state +2 (number of ligands) = 27 –3 + 2(6) = 27 –3 +12 = 36 (d) [Co(NO2)(NH3)5]Cl2 Pentamminenitrito-N-cobalt (III) chloride. (d) I-131 radio-isotope is used for thyroid disorders. (b) Tetragonal crystal system has the unit cell dimensions a = b c, = = = 90° (c) A crystalline solid undergoes deformation of its geometry by application of pressure or heat.

(c)

3

2 Fe 2 O3 Cr 2 O3 3

2

60.

(b)

69.

4Na 2 CO3 3O2 6

4Na 2 Cr O 4

4CO 2

Hence, the oxidation state of chromium increase from + 3 to + 6. T f = iK f m (b) 10 1000 1.34m 74.55 100 [Molecular Mass of KCl = 74.55] m=

i 70.

Tf Kfm

(c) 2NO2

k1 k2

4.5 C (1.86 C / m)(1.34m)

1.8

N2O4

1 d[NO 2 ] = k1[NO2]2– k2[N2O4] 2 dt Rate of disappearance of NO2 i.e.,

Rate =

d[NO 2 ] =2k1[NO2]2 – k2[N2O4] dt (c) For every 10° rise in temperture, Rate gets double. Hence, in this case rate constant will become four time. (d) i.e., k1 = 4k2 k2 = 0.25k1 as the Rise in temperature given is 20°C At T1 = 200 K, if k1 = k then at T2 = 400 K, k2 = 10k

–

71.

72.

Ea 400 200 10k = 2.303R 400 200 k Ea = 921.2R

log 73.

(a)

CO2(g) + C(s)

2CO(g)

Initial 0.5 atm pressure At equi (0.5 – x) Total no. of moles at equilibrium = (0.5 – x) + 2x = 0.8 x = 0.3 atm. pCO = 0.5 – 0.3 = 0.2 2 PCO = 2x = 2 0.3 = 0.6 atm 2 pCO K= p CO 2

0.6 0.2

2

= 1.8 atm

EBD_7443 2011-16

74.

75.

Target VITEEE

(a) 2A + B

C, H = x cals.

1 0 =0.5 2 Molecular orbital configuration of He2+ is

BO =

Since the reaction is endothermic and np < nr. Hence, high pressure and high temperature will favour forward reaction. (b) The cell reaction is Mg + Sn2+ Mg2+ + Sn

He+2(3) = 1s2, * 1s1 2 1 = 0.5 2 Hence, stability of H+2 > stability of He+2.

BO =

2

Ecell = E°cell –

0.0591 [Mg ] log 2 [Sn 2 ]

0.0591 10 log = (2.34 – 0.14) – 2 10

76.

78.

79.

= 2.23 V

81.

(a)

(b) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) ng = np – nr = 3 – 6 = –3 H = E + ngRT H = E – 3RT H – E = –3RT (a) G= H–T S For a spontaneous reaction, G should be negative. Hence, H = +ve, S = +ve and T S> H (b) Gas (Y) SO2 which can be obtained from Na2SO3(A) and gas (Z) is H2S which can be obtained from Na2S(B). This can be easily understood by the following reactions. Na2SO3 + 2HCl 2NaCl + SO2 + H2O (A) (Y) Na2S + 2HCl 2NaCl + H2S (B) (Z) K2Cr2O7 + H2SO4 + 3SO2 K2SO4 + Cr2(SO4)3 + H2O PbS + 2CH3COOH

black

(d) Higher the bond order higher will be the stability. When the bond order of two molecules are same, the molecule with least number of antibonding electrons is more stable. Molecular orbital configuration of H2+ is H+2 (1) = 1s1

The equation of the bisectors of the angle between the lines given by ax2 + 2hxy + by2 = 0 is x 2 y2 xy ...(i) a b h And the equation of the bisectors of the angle between the lines given by a2x2 + 2h (a + b) xy + b2y2 = 0 is

H f is positive.

Pb(CH3COO)2 + H2S 80.

1

(a) Explosive compound has high heat content i.e., it is formed by absorption of heat. Hence,

77.

PART - III (MATHEMATICS)

2

x2

y2

2

2

a

82.

83.

(c)

(c)

b

xy h(a b)

x2 y2 x ...(ii) a b h From eqs. (i) and (ii), it is clear that both the pair of straight lines have the same bisector, hence, the given two pairs of straight lines are equally inclined. A = {1, 2, 3, 4} and B = {1, 3, 5} R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} and R–1 = {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} ROR–1 = {(3, 3), (3, 5), (5, 3), (5, 5)} (1 i 3)100

= 2100 100 =2

100

2100

1 2

i 3 2

100

= 2100 1 2

3i 99 =–2 2

Now, x + iy = (1 i 3)100 = 299 299 3i x = –299, y = 299 3 (x, y) = (–299, –299

3)

299 3i

Solved Paper 2011 84.

(a)

Given, an = 3 (2 ) an+1 = 3 (2n+1)

(c)

3(2n 1 )

an 1 an

r=

85.

2011-17 n

89.

(b)

The centres of the given circles x + y2 = 4 and x2 + y2 – 6x – 8y = 24 are C1 (0, 0) and C2 (3, 4) respectively. Their radii are r 1 = 2 and r2 = 7 respectively. C1C2 = 5 < sum of radii But C1C2 = difference of radii Thus, the given circles touch each other internally. Hence, number of common tangent is only one.

90.

(d)

1 ex

=2

3(2 n ) a, b, c are in HP.

1 1 1 , , are in AP.. a b c a b c a b c a b c , , are in AP.. b c a

a c a b b c ,1 ,1 1 are in AP.. b c a

2

1

,

b

,

c

86.

(d)

x

2x 7 2x 3 x2

6

91.

0

x 2 10x 11 2x 3

0

(x 11)(x 1) 2x 3

0

(d)

88.

(d)

=

2 lim

(1, ) 2 sin

lim

x

x 2

x

0

x 2

x

0

1

x= –

0

and RHL at x = 0 is

2

1 2

.

1 cos x does not exist. x x 0

Hence, lim

,

4 3

1

It can be easily seen that LHL at x = 0 is

3 ( 1,11) 2 The number of faces is equal to the number of colours. Therefore the number of ways of painting them is 1. Equation of line perpendicular to 3x + y – 3 = 0 is x – 3y + c = 0 Since, it passes through (2, 2). c=4 Equation of line perpendicular to 3x + y – 3 = 0 is x – 3y + 4 = 0 For y - intercept, put x = 0 y=

0

sin

(2x 3) 2 (x – 11) (x + 1) (2x + 3) < 0

(a)

, 0)

1 cos x x

lim

x

x

(x 11)(x 1)(2x 3)

87.

(

x

1 x

6

2x 7 2x 3

x

1 1 log1 x

1

1 1 0 x

are in HP..

b c c a a b 2

0

1 1 ex

b c a c a b , , are in AP.. b c a a

1

92.

(a)

Let I =

1

= x

2

x2 4 x 4 16

dx

4 x dx = 16 x2

Putting x So that 1

x

4 = t, x 4 x2

4

1

2

dx = dt

x2 4 x

dx

2

8

EBD_7443 2011-18

Target VITEEE

=

1 2 2

=

t2

(b)

(2 2 )2

2 2

1

(a)

x 4 tan 1 2 2 2x 2

/4

=

94.

(b)

1 dx 1 cos x

3 /4

98.

(b)

x2 1 ax b = 0 x 1

lim

x

C

x 2 (1 a) (a b)x b 1 =0 x 1

lim

3 /4

1 cos x

/4

sin 2 x

dx 99.

(b)

(cosec2 x – cot x cosec x ) dx

/4

=

cot x cosec x

3 /4 /4

100. (a)

= (1 2) ( 1 2) = 2 Let a and b are two numbers. Then, A =

a b 2

...(i)

q p

b q

1 – a = 0 and a + b = 0 a = 1, b = –1 Let a = i – 2j + k and b = 4i – 4j + 7k, then projection of a on b is equal to

a.b 4 8 7 19 = |b| 16 16 49 9 Here, a + b + c = 0 |a + b + c|2 = 0 |a|2 + |b|2 + |c|2 + 2 {a.b + b.c + c.a} = 0 a.b + b.c + c.a 1 {|a|2 + |b|2 + |c|2} < 0 2 m 0 as well as f (0+) > 0 and f (0) = 0 Hence, it is a point of minima. x * a = b (x * a) * a–1 = b * a–1 x *(a * a–1) = b * a–1 x *e = b * a–1 x = b * a–1 Let A and B be two events such that A= getting number 2 at least once B = getting 7 as the sum of the numbers on two dice Here, A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)} and B = {(2, 5), (5, 2), (6, 1), (1, 6), (3, 4), (4, 3)}

12 ) sin 72 cos 36 sin 72

[sin12 sin(60

=

–

107. (a)

sin12 sin(60

sin12

non differentiable, x 2 cos x,

Any chord PQ which bisected point R (h, k) is T = S i.e., ky – 2a (x + h) = k2 – 4ah Since, it is a focal chord, so it must pass through focus (a, 0). k (0) – 2a (a + h) = k2 – 4ah k2 = 2ah – 4a2 Hence, locus is y2 = 2a (x – a) sin 12° sin 48° sin 54° = sin 12° sin (60° – 12°) sin (90° – 36°)

a3

abc 4

s a

3a

a 2

3

3 / 4a 2 a/2

3 a 2 Required ratio = r : R : r 1 a

:

a

:

3 a = 1 : 2 : 3. 2

2 3 3 p q is false only when both p and q are false.

EBD_7443 2011-20

113. (a)

114. (b)

115. (b)

116. (c)

Target VITEEE Since, each letter can be posted in any one of the five different letter boxes. So, a letter can be posted in 5 ways. Since, there are six letters and each can be posted in 5 ways. So, total number of ways = 5 × 5 × 5 × 5 × 5 × 5 = 56 Since, (1, 4), (2, 6) and (3, 6) are the elements of A × B, therefore 1, 2, 3 are the elements of A and 4, 6 are the elements of B. Also, A × B has 6 elements A = {1, 2, 3} and B = {4, 6} B × A = {(4, 1), (4, 2), (4, 3), (6, 1), (6, 2), (6, 3)} Let f –1 (–5) = x. Then, f (x) = – 5 x2 + 1 = – 5 x2 = –6 x = ± 6 which does not belong to R. f –1 (–5) = Given : P (X = 1) = P (X = 2) e 1!

119. (d)

dy dx dv dx

P (X = 4) = 117. (d)

5

Required area = 3x 2 = 2

5

5x 3

3

3e2

(3x 5)dx

75 25 = 2

48 10 = 14 sq units 2 Given differential equation is

=

118. (c)

2

dy 3 1 4 dx

2

4

d2 y dx 2

27 15 2

4

1 tan 2

120. (d)

2

dx 2

…(i)

4

4 = v2

v

2!

2 4!

dv dx

dv

=2 2 4

3

d2 y

dy = (4x + y + 1)2 dx Put 4x + y + 1 = v

1

(

From eq. (i))

dx

v 2

x C

4x y 1 2x C 2 4x + y + 1 = 2 tan (2x + C) Given system of equations are 2x + y – 5 = 0 x – 2y + 1 = 0 and 2x – 14y – a = 0 This system is consistent tan

e

4

3

Here, order is 2 and degree is 3.

2

e

2

dy 1 4 dx

1

2 1

1 2

2

14

...(i) ...(ii) ...(iii)

5 1 =0 a

2 (2a + 14) – 1 (–a –2) – 5 (–14 + 4) = 0 4a + 28 + a + 2 + 50 = 0 5a = –80 a = –16

SOLVED PAPER PART - I (PHYSICS) 1.

6.

A straight wire carrying current i is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is M, the length of wire will be (a)

4 M

(b)

4 M i

M 4 i (d) i M The ratio of the amounts of heat developed in the four arms of a balance Wheatstone bridge, when the arms have resistances P=100 ,

3.

4.

Q=10 , R=300 and S=30 respectively is (a) 3 : 30 : 1 : 10 (b) 30 : 3 : 10 : 1 (c) 30 : 10 : 1 : 3 (d) 30 : 1 : 3 : 10 An electric kettle takes 4 A current at 220V. How much time will it take to boil 1 kg of water from temperature 20 °C ? The temperature of boiling water is 100°C. (a) 12.6 min (b) 4.2 min (c) 6.3 min (d) 8.4 min Magnetic field at the centre of a circular loop of area A is B. The magnetic moment of the loop will be (a)

(c) 5.

BA 2 0

BA3 / 2 0

1/ 2

(b)

(d)

4.4 × 106N-m, 3.2 × 10–4J

(b) –2 × 10–3N-m, – 4×103J (c)

4 × 103N-m, 2 × 10– 3J

(d) 2 × 10–3N-m, 4 × 10–3J 7.

The electron of hydrogen atom is considered to be revolving round a proton in circular orbit of radius h2/me2 with velocity e2/h, where h= /2 . The current i is (a)

(c) 8.

BA 3/ 2 2BA3 / 2

4

2

me5 2

4

2

m 2e 2 3

(b)

(d)

4

2

me5 3

4

2

m 2 e5 3

In a double slit experiment, 5th dark fringe is formed opposite to one of the slits, the wavelength of light is (a)

d2 6D

(b)

d2 5D

(c)

d2 15D

(d)

d2 9D

0

0

2010

An electric dipole consists of two opposite charges of magnitude q =1×10–6 C separated by 2.0 cm. The dipole is placed in an external field of 1×105 NC–1. What maximum torque does the field exert on the dipole? How much work must an external agent do to turn the dipole end to end, starting from position of alignment ( = 0°) ? (a)

(c)

2.

VITEEE

1/ 2

In Young's double slit experiment, the spacing between the slits is d and wavelength of light used is 6000 Å. If the angular width of a fringe formed on a distance screen is 1°, then value of d is (a) 1 mm (b) 0.05 mm (c) 0.03 mm (d) 0.01 mm

9.

Which of the following rays is emitted by a human body? (a)

X-rays

(b) UV rays

(c)

Visible rays

(d) IR rays

EBD_7443 2010-2

10.

Target VITEEE

A proton of mass 1.67 × 10–27kg

enters a uniform magnetic field 1T of at point A shown in figure with a speed of 107 ms–1. 45

B

A

C

15. A small coil is introduced between the poles of an electromagnet so that its axis coincides with the magnetic field direction. The number of turns is n and the cross-sectional area of the coil is A. When the coil turns through 180° about its diameter, the charge flowing through the coil is Q. The total resistance of the circuit is R. What is the magnitude of the magnetic induction? (a)

The magnetic field is directed normal to the plane of paper downwards. The proton emerges out of the magnetic field at point C, then the distance AC and the value of angle will respectively be (b) 0.7 m, 90 (a) 0.7 m , 45 (c) 0.14 m, 90 (d) 0.14 m, 45 11. A neutral water molecule (H2O) in its vapour state has an electric dipole moment of magnitude 6.4×10–30C-m. How far apart are the molecules centres of positive and negative charges? (a) 4 m (b) 4mm (c) 4 m (d) 4 pm 12. Figure shows a straight wire length l carrying current i. The magnitude of magnetic field produced by the current at point P is P

i

(a)

2 0i l

(b)

0i

4 l

2 0i 0i (d) 2 2 l 8 l 13. Zener diode is used for (a) producing oscillations in an oscillator (b) amplification (c) stabilisation (d) rectification 14. Two light sources are said to be coherent if they are obtained from (a) two independent point sources emitting light of the same wavelength (b) a single point source (c) a wide source (d) two ordinary bulbs emitting light of different wavelengths

(c)

QR nA

(b)

2QR nA

QR Qn (d) 2nA 2RA 16. The attenuation in optical fibre is mainly due to (a) absorption (b) scattering (c) neither absorption nor scattering (d) Both (a) and (b) 17. An arc of radius r carries charge. The linear density of charge is and the arc subtends an

(c)

angle

at the centre. What is electric potential 3 at the centre? (a)

4 0

(b)

8 0

(c)

(d) 12 0 16 0 18. Sinusoidal carrier voltage of frequency 1.5 MHz and amplitude 50 V is amplitude modulated by sinusoidal voltage of frequency 10 kHz producing 50% modulation. The lower and upper side-band frequencies in kHz are (a) 1490, 1510 (b) 1510, 1490 1 1 1 1 , , (d) 1490 1510 1510 1490 19. 50 and 100 resistors are connected in series. This connection is connected with a battery of 2.4 V. When a voltmeter of 100 resistance is connected across 100 resistor, then the reading of the voltmeter will be (a) 1.6V (b) 1.0V (c) 1.2V (d) 2.0V 20. In space charge limited region, the plate current in a diode is 10 mA for plate voltage 150V. If the plate voltagte is increased to 600V, then the plate current will be (a) 10 mA (b) 40 mA (c) 80 mA (d) 160 mA

(c)

Solved Paper 2010

2010-3

21. Light of wavelength strikes a photo-sensitive surface and electrons are ejected with kinetic energy E. If the kinetic energy is to be increased to 2E, the wavelength must be changed to wher (a)

2

(b)

2

(c)

(d) 2 22. The maximum velocity of electrons emitted from a metal surface is v, when frequency of light falling on it is f. The maximum velocity when frequency becomes 4f is (a) 2v (b) > 2v (c) < 2v (d) between 2v and 4v 23. The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. Light source is put on and a saturation photo-current is recorded. An electric field is switched on which has a vertically downward direction, then (a) the photo-current will increase (b) the kinetic energy of the electrons will increase (c) the stopping potential will decrease (d) the threshold wavelength will increase 24. A cylindrical conductor of radius R carries a current i. The value of magnetic field at a point R distance inside from the surface is 4 10 T. The value of magnetic field at point which is 4R distance outside from the surface

which is

(a)

4 T 3

(b)

8 T 3

40 80 T T (d) 3 3 25. The power of a thin convex lens (ang= 1.5) is 5.0 D. When it is placed in a liquid of refractive index an , then it behaves as a concave lens of focal length 100cm. The refractive index of the liquid anl will be (a) 5/3 (b) 4/3 (c) (d) 5/4 3 26. Find the value of magnetic field between plates of capacitor at a distance 1m from centre, where electric field varies by 1010 V/m per second. (a) 5.56×10-8T (b) 5.56×10-3T (c) 5.56 T (d) 5.55T

(c)

27. Using an AC voltmeter the potential difference in the electrical line in a house is read to be 234V. If line frequency is known to be 50 cycles/s, the equation for the line voltage is (a) V=165sin (100 t) (b) V=331sin (100 t) (c) V=220sin (100 t) (d) V=440sin (100 t) 28. There are a 25W – 220 V bulb and a 100W–220V line.Which eletric bulb will glow more brightly? (a) 25W bulb (b) 100W bulb (c) Both will have equal incadescene (d) Neither 25 W nor 100 W bulb will give light 29. Silver has a work function of 4.7 eV. When ultraviolet light of wavelength 100 nm is incident upon it , potential of 7.7 V is required to stop photoelectrons from reaching the collector plate. The potential required to stop electrons when light of wavelength 200 nm is incident upon silver is (a) 1.5V (b) 1.85V (c) 1.95V (d) 2.37V 30. Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2, Respectively. The ratio of masses of X and Y is (a) (R1/R2)–2 (b) (R2/R1) 2 (c) (R1/R2) (d) (R1/R2) 31. According to the Bohr's theory of hydrogen atom, the speed of the electron, energy and the radius of its orbit vary with the principal quantum number n, respectively, as (a)

1 1 2 , ,n n n2

(b)

1 2 1 ,n , 2 n n

1 1 (d) n, 2 , 2 n n n 32. In the hydrogen atom, the electron is making 6.6 × 1015 rps. If the radius of orbit is 0.53 × 10–10 m, then magnetic field produced at the centre of the orbit is (a) 140T (b) 12.5T (c) 1.4T (d) 0.14T (c)

n2 ,

1

2

, n2

EBD_7443 2010-4

Target VITEEE

33. Two identical light sources S1 and S2 emit light of same wavelength . These light rays will exhibit interference if (a) their phase differences remain constant (b) their phases are distributed randomly (c) their light intensities remain constant (d) their light intensities change randomly 34. In Meter bridge or Wheatstone bridge for measurement of resistance, the known and the unknown resistances are interchanged. The error so removed is (a) end correction (b) index error (c) due to temperature effect (d) random error 35. A fish, looking up through the water, sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12cm below the surface of water, the radius of the circle in centimetre is (a) (c)

12 3 5

(b) 12×3× 5

12 3

(d) 12 3 7 7 36. Radio waves diffract around building althrough light waves do not. The reason is that radio waves (a) travel with speed larger than c (b) have much larger wavelength then light (c) carry news (d) are not electromagnetic waves 37. In the Bohr model of a hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If a0 is the radius of the ground state orbit, m is the mass and e is charge on the electron and 0 is the vacuum permittivity, the speed of the electron is (a)

0

e

(b)

e

0a 0 m

4

0a0m

(d) 4 0a0m e 38. A potential difference of 2V is applied between the opposite faces of a Ge crystal plate of area 1 cm2 and thickness 0.5 mm. If the concentration of electrons in Ge is 2×1019/m2 and mobilities of electrons and holes are 0.36 m2V-1s-1 and 0.14 m2V-1s-1 respectively, then the current flowing through the plate will be (a) 0.25 A (b) 0.45 A (c) 0.56 A (d) 0.64 A (c)

39. An AM wave has 1800 W of total power content. For 100% modulation the carrier should have power content equal to (a) 1000 W (b) 1200 W (c) 1500 W (d) 1600 W 40. Two light rays having the same wavelength in vacuum are in phase initially. Then the first ray travels a path l1 through a medium of refractive index n1 while the second ray travels a path of length l2 through a medium of refractive index n2. The two waves are then combined to observe interference. The phase difference between the two waves is (a) (c)

2 2

l2 l1 n2 l2

(b) n1l1

(d)

2 2

n1l2 l1 n1

n2 l1 l2 n2

PART - II (CHEMISTRY) 41. The correct formula of the complex tetraammineaquachlorocobalt (III) chloride is (a) [Cl(H2O) (NH3)4 Co] Cl (b) [CoCl(H2O) (NH3)4] Cl (c) [Co (NH3)4(H2O)Cl] Cl (d) [CoCl (H2O) (NH3)4] Cl2 42. The equivalent conductance at infinite dilution of a weak acid such as HF (a) can be determined by extrapolation of measurements on dilute solutions of HCl, HBr and HI (b) can be determined by measurement on very dilute HF solutions (c) can best be determined from measurements on dilute solutions of NaF, NaCl and HCl (d) is an undefined quantity 43.

C2 H5 I

Alcoholic KOH

X

Br2 CCl4

KCN

Y A

H3O

Z +

The product 'A' is (a) succinic acid (b) melonic acid (c) oxalic acid (d) maleic acid 44. For a reaction of type A + B products, it is observed that doubling concentration of A causes the reaction rate to be four times as great, but doubling amount of B does not affect the rate. The unit of rate constant is (a) s – 1 (b) s–1 mol L–1 –1 –1 (c) s mol L (d) s s–1 mol–2 L2

Solved Paper 2010

2010-5

45. A chemical reaction was carried out at 320 K and 300 K. The rate constants were found to be k1 and k2 respectively. Then (a) k2 = 4k1 (b) k2 = 2k1 (c) k2 = 0.25 k1 (d) k2 = 0.5 k1 46. The formula of ethyl carbinol is (a) CH3OH (b) CH3CH2OH (c) CH3CH2CH2OH (d) (CH3)3COH 47. Which of the following gives red colour in Victor Meyer's test? (a) n-propyl alcohol (b) Isopropyl alcohol (c) tert-butyl alcohol (d) sec-butyl alcohol 48. Enthalpy of a compound is equal to its (a) heat of combustion(b) heat of formation (c) heat of reaction (d) heat of solution 49. For which one of the following reactions will there be a positive S? H 2 O (l ) (a) H 2 O (g) (b)

H2

(c)

CaCO 3 (s)

I2

51.

52.

53.

(a) (b)

COCl

H 2 / Pd BaSO 4

AlCl3

CaO(s) CO 2 (g)

N 2 (g) 3H 2 (g) 2NH 3 (g) Across the lanthanide series, the basicity of the lanthanide hydroxides (a) increases (b) decreases (c) first increases and then decreases (d) first decreases and then increases When p-nitrobromobenzene reacts with sodium ethoxide, the product obtained is (a) p-nitroanisole (b) ethyl phenyl ether (c) p-nitrophenetole (d) no reaction occurs A radioactive element X emits 3 , 1 and 1 particles and forms 76Y235. Element X is (a) 81 X 247 (b) 80 X 247 246 (c) 81 X (d) 80 X 246 For the reaction,

2AB 2 (g) 2A(g) B2 (g) the equilibrium constant, Kp at 300 K is 16.0. The

value of Kp for AB2 (g)

A(g) + 1/2 B2 (g)

is (a) 8 (b) 0.25 (c) 0.125 (d) 32 54. Frenkel defect is generally observed in (a) AgBr (b) AgI (c) ZnS (d) All of the above 55. Most crystals show good cleavage because their atoms, ions or molecules are (a) weakly bonded together (b) strongly bonded together

(c)

CHO

+ HCl + HCN (i) Anhy. AlCl3 (ii) H 3O

(d)

CHO

H + CO + HCl

2HI

(d)

50.

(c) spherically symmetrical (d) arranged in planes 56. [Co (NH3)4Cl2]NO2 and [Co (NH3)4ClNO2]Cl exhibit which type of isomerism? (a) Geometrical (b) Optical (c) Linkage (d) Ionisation 57. Which of the following compounds is not coloured? (a) Na2[Cu(Cl4] (b) Na[Cd(Cl)4] (c) K4[Fe(CN)6] (d) K3[Fe(CN)6] 58. Which of the following is a Gattermann aldehyde synthesis?

CH3

CrO 2Cl 2

CHO CHO

59. Aldol is (a) -hydroxybutyraldehyde (b) -hydroxybutanal (c) -hydroxypropanal (d) None of the above 60. Nitrobenzene can be converted into azobenzene by reduction with (a) Zn, NH4Cl, (b) Zn/NaOH, CH3OH (c) Zn/NaOH (d) LiAlH4, ether 61. The one which is least basic is (a) NH3 (b) C6H5NH2 (c) (C6H5)3N (d) (C6H5)2NH 62. Coordination number of Ni in [Ni(C2O4)3]4– is (a) 3 (b) 6 (c) 4 (d) 5 63. Mg is an important component of which biomolecule occurring extensively in living world? (a) Haemoglobin (b) Chlorophyll (c) Florigen (d) ATP

EBD_7443 2010-6

Target VITEEE

64. Sterling silver is (a) AgNO3 (b) Ag2 S (c) Alloy of 80% Ag + 20% Cu (d) AgCl 65. Identify the statement which is not correct regarding CuSO4 (a) It reacts with KI to give iodine (b) It reacts with KCl to give Cu2Cl2 (c) It reacts with NaOH and glucose to give Cu2O (d) It gives CuO on strong heating in air 66. Transition metals usually exhibit highest oxidation states in their (a) chlorides (b) fluorides (c) bromides (d) iodides 67. The number of Faradays needed to reduce 4 g equivalents of Cu2+ to Cu metal will be (a) 1 (b) 2 1 (d) 4 2 Which one of the following cells can convert chemical energy of H2 and O2 directly into electrical energy? (a) Mercury cell (b) Daniel cell (c) Fuel cell (d) Lead storage cell On treatment of propanone with dilute Ba(OH)2, the product formed is (a) aldol (b) phorone (c) propionaldehyde (d) 4-hydroxy-4-methyl-2-pentanone Which of the following converts CH3CONH2 to CH3NH2? (a) NaBr (b) NaOBr (c) Br2 (d) None of the above Which metal aprons are worn by radiographer to protect him from radiation? (a) Mercury coated apron (b) Lead apron (c) Copper apron (d) Aluminimised apron The standard Gibb's free energy change, Gº is related to equilibrium constant, Kp as

(c)

68.

69.

70.

71.

72.

(a) (c)

Kp

RT ln Gº (b) K p

Kp

G RT

(d) K p

e RT

e

Gº

Gº/ RT

73. The yield of the product in the reaction A 2 (g) 2B (g) C (g) Q kJ would be higher at (a) high temperature and high pressure (b) high temperature and low pressure (c) low temperature and high pressure . (d) low temperature and low pressure 74. In which of the following case, does the reaction go farthest to completion? (a) K = 102 (b) K = 10 (c) K = 10–2 (d) K = 1 75. Formation of cyanohydrin from a ketone is an example of (a) electrophilic addition (b) nucleophilic addition (c) nucleophilic substitution (d) electrophilic substitution 76. Glycerol on treatment with oxalic acid at 110ºC forms (a) formic acid (b) allyl alcohol (c) CO2 and CO (d) acrolein 77. The activity of an old piece of wood is just 25% of the fresh piece of wood. If t 1/2 of C-14 is 6000 yr, the age of piece of wood is (a) 6000 yr (b) 3000 yr (c) 9000 yr (d) 12000 yr 78. The radius of Na+ is 95 pm and that of Cl– ion is 181 pm. Hence, the coordination number of Na + will be (a) 4 (b) 6 (c) 8 (d) unpredictable 79. The reaction, ROH + H2CN2 in the presence of HBF4, gives the following product (a) ROCH3 (b) RCH2OH (c) ROHCN2N2 (b) RCH2CH3 80. The fatty acid which shows reducing property is (a) acetic acid (b) ethanoic acid (c) oxalic acid (d) formic acid

PART - III (MATHEMATICS) 81. If F is function such that F (0) = 2, F (1) = 3, F (x+2)= 2F (x) - F (x+1) for x 0, then F (5) is equal to (a) –7 (b) –3 (c) 17 (d) 13 82. Let S be a set containing n elements. Then, number of binary operations on S is (a)

nn

(c)

nn

(b) 2n 2

(d) n2

2

Solved Paper 2010

2010-7

83. The numerically greatest term in the expansion 1 , is 5 (b) 55 × 36 (a) 55 × 39 9 (c) 45 × 3 (d) 45 × 36 84. The number of solutions of the equation sin (ex) = 5x + 5–x, is (a) 0 (b) 1 (c) 2 (d) infinitely many 85. If ax = by = cz = du and a, b, c, d are in GP, then x, y, z, u are in (a) AP (b) G P (c) HP (d) None of these 86. If z satisfies the equation |z|–z = 1+2i, then z is equal to

of (3–5x)11 when x =

(a)

3 + 2i 2

(c)

2–

87. If z = (a) (c)

1 i 3 1 i 3

16 29

(b)

1 15

(c)

27 59

(d)

42 107

93. The value of tan

3 (d) 2 + i 2

(c)

tan

(b) 120 (d) 300

log10 x 2 . The set of all values of x for which f (x) is real, is (a) [–1, 1] (b) [1, ] (c) (– , –1] (d) (– , –1] [1, ) 89. For what values of m can the expression 2x2 + mxy + 3y2 – 5y – 2 be expressed as the product of two linear factors? (a) 0 (b) 1 (c) 7 (d) 49 90. If B is a non-singular matrix and A is a square matrix, then det (B–1AB) is equal to (a) det (A–1) (b) det (B–1) (c) det (A) (d) det (B) 91. If f (x), g(x) and h (x) are three polynomials of degree 2 and 88. If f (x) =

(x)

(a)

(a)

, then arg (z) is

60 240

1 1 1 , , respectively. A box is selected 5 6 7 at random and a screw drawn from it at random is found to be defective. Then, the probability that it came from box A, is

A, B, C are

3 –2i 2

(b)

3 i 2

92. The chances of defective screws in three boxes

f (x)

g(x)

h(x)

f' (x)

g' (x)

h' (x)

f'' (x) g'' (x) h'' (x)

then (x) is a polynomial of degree (a) 2 (b) 3 (c) 0 (d) atmost 3

cos 1 sin

2

4

4

2

is equal to (b) tan (d) tan

4 4

2 2

94. If 3sin 5cos 5 , then the value of 5 sin – 3 cos is equal to (a) 5 (b) 3 (c) 4 (d) None of these 95. The principal value of sin (a)

(b)

6

sin

1

5 6

is

5 6

7 (d) None of these 6 96. A rod of length l slides with its ends on two perpendicular lines. Then, the locus of its mid point is

(c)

(a)

x2

y2

l2 4

(b) x 2

y2

l2 2

l2 (d) None of these 4 97. The equation of straight line through the intersection of line 2x + y=1 and 3x + 2y=5 and passing through the origin is (a) 7x + 3y = 0 (b) 7x – y = 0 (c) 3x + 2y = 0 (d) x + y = 0 98. The line joining (5,0) to (10 cos , 10 sin ) is divided internally in the ratio 2:3 at P. If varies, then the locus of P is (c)

x2

y2

EBD_7443 2010-8

Target VITEEE

(a) a straight line (b) a pair of straight lines (c) a circle (d) None of the above 99. If 2x + y + k = 0 is a normal to the parabola y2 = – 8x, then the value of k, is (a) 8 (b) 16 (c) 24 (d) 32 1 1.2

100. nlim

1 2.3

1 1 ..... n( n 1) 3.4

is equal

to (a) 1 (b) –1 (c) 0 (d) None of these 101. The condition that the line lx + my =1 may be normal to the curve y2 = 4ax, is (a)

al 3 2alm2

m 2 (b) al 2

2alm3

m2

(c)

al 3

m3 (d) al 3

2alm2

m2

2alm 2

f ( x)dx

102. If

f ( x), then

1 f (x) 2

(a)

3

103. sin

4 x 2 8 x 13

(a) (x+1) tan (b)

1

(c) ( x 1) tan

2x 2 3

3 x 1 tan 2

(d) 104. If

the 2

1

2x 2 3

equation 2

3x 2 y 6 x 8 y 5 following are true ?

(a)

e

x2

y2

a2

b2 b2

x

y=

b2

(c)

a2

b2

y= x

(a)

, 2 2

(c)

, 2 2

{1}

(d)

f (x)

2

(c)

0,

3 log 4 x 2 8 x 13 4

of

an

elipse

3 (b) centre is (–1, 2) (c) foci are (–1, 1) and (–1, 3) (d) All of the above

b2

1 , are

, 2 2

{1}

(d) None of these

2

2

x2

1

(b)

0,

(d)

0,

1 x2

, is

2 2

sec n

cos , y

cos n

is equal to

(a)

n2 ( y 2

(c)

n2 ( y 2 4)

(b) n2 (4 y 2 )

4)

(d) None of these

c

109. If y= x c c

, then

2

dy dx

(x 2 4)

y

x

, then

y

dy is dx

equal to y

(a)

y

is

0 , then which of the

1

a2

(b)

107. Range of the function y = sin 0,

3 log 4 x 2 8 x 13 2

x2

a 2 b2 (d) y= x 106. Domain of the function f (x) log x cos x, is

(a)

c

y2

a2

a2

3

dx is equal to

1 and

(b) y= x

f (x)

3 4 x 2 8 x 13 log 4 9

2x 2 3

1

(a)

(b)

3 4 x 2 8 x 13 log 4 9

2x 2 3

1

dx is equal to

hyperbolas

108. If x = sec

(2 x 2)

1

3 tan 2

2

3

f (x)

(c)

2

f ( x)

105. The equation of the common tangents to the two

2

2y x

110. If 1

2

x

t t2 1

2 3 2

y3 2y

2

x 2 xy 1

(d) None of these

x

dt

(a)

(b)

2x

y3

(c)

(c)

x

=

6

, then x can be equal to (b)

3

(d) None of these

Solved Paper 2010

2010-9

111. The area bounded by the curve y= sin x , x-axis and the lines x = , is (a) 2 sq unit (b) 1 sq unit (c) 4 sq unit (d) None of these 112. The degree of the differential equation of all curves having normal of constant length c is (a) 1 (b) 3 (c) 4 (d) None of these ˆ ˆ ˆ ˆ 113. If a 2i 2j 3k, b i 2jˆ kˆ and c 3iˆ ˆj , then a tb is perpendicular to c , if t is equal to (a) 2 (b) 4 (c) 6 (d) 8 114. The distance between the line r

2iˆ 2ˆj 3kˆ

r ˆi 5jˆ kˆ

iˆ ˆj 4kˆ and the plane 5 , is

10 3

(a)

(b)

10

10 (c) (d) 9 3 3 115. The equation of sphere concentric with the

sphere x 2 y2 z 2 4x 6y 8z 5 which passes through the origin, is (a)

x2

y2

z 2 4x 6y 8z

(b)

x2

y2

z 2 6y 8z

(c)

x2

y2

z2

0

(d)

x2

y2

z2

4x 6y 8z 6

the

x 3 1 is

lines

y k 2

x 1 2

0 and

0

0

y 1 3

3 2

(b)

0

z 1 4

and

z intersect, then the value of k, 1

9 2

2 3 (d) 9 2 x 117. The two curves y = 3 and y = 5x intersect at an angle

(c)

(a)

tan

1

log 3 log 5 1 log 3log 5

(b)

tan

1

log 3 log 5 1 log 3log 5

(c)

tan

1

log 3 log 5 1 log 3log 5

(d)

tan

1

log 3 log 5 1 log 3log 5

118. The equation

3

10

116. If

(a)

x2

4xy y 2

x 3y 2

0

represents a parabola, if is (a) 0 (b) 1 (c) 2 (d) 4 119. If two circles 2 x 2

2 y2 3x 6 y k

0 and

x 2 y 2 4x 10y 16 0 cut orthogonally,, then the value of k is (a) 41 (b) 14 (c) 4 (d) 1 120. If A (–2, 1), B (2, 3) and C (–2, –4) are three points. Then, the angle between BA and BC is

(a)

tan

(c)

tan

1

2 3

(b) tan

1

1 3

(d) tan

1

3 2

1

1 2

EBD_7443 2010-10

Target VITEEE

SOLUTIONS PART - I (PHYSICS) 1.

(b) Here, length l = 2 r or r

4.

1 2

(d) Magnetic field, B

0i

2 i 4 r 0

2r

0

2

Area of circular loop A r Magnetic moment M = iA = i r2 M

2.

4

i 330 3 i 330 110 4 Here i = total current Similarly, current through arms of resistances R and S in series i1

M

2 BA

5.

1 i 4

3.

0

t

D

1

2 BA3/ 2

10 :

0

D d 180

180

d

1/ 2

Y D

1 D

0=

(width Y = )

d

rad and = 6 × 10–7 m 6 10

7

0.03 mm

0

6. 1 i 4

2

30

= 30 : 3 : 10 : 1 (c) Heat taken by water when its temperature changes from 20°C to 100°C. H1 mc( 2 1 ) 1000 1 (100 20) cal = 1000 × 80 × 4.2 J Heat produced in time t due to current in resistor H2 = Vit = 220 × 4 × t J According to question, 220 × 4 × t = 1000 × 80 × 4.2 1000 80 4.2 220 4

1/ 2

Y D Angular fringe width

0

2

300 :

A

(c) Here, sin

H P : HQ : H R : H S

2

A

0

i 110 1 i 330 110 4 Heat developed per second = i2R Ratio of heat developed per sec

3 100 : i 4

2 Br

iA

0

i2

2

1/ 2

Magnetic moment,

2

4 M l i (b) Current through arms of resistances P and Q in series

3 i 4

A

Also, A = r2 or r

l2

i

2Br

or i

381.8 s = 6.3 min

(d) Here, charge q = ± 1 × 10–6 C 2a = 2.0 cm = 2.0 × 10–2 m E = 1 × 105 NC–1, max = ? W = ?, 1 = 0°, 2 = 180° max = pE = q(2a)E = 1 × 10–6 × 2.0 × 10–2 × 1 × 105 = 2 × 10–3 Nm W

7.

pE (cos 1 cos 2 ) = (10–6 × 2 × 10–2)(105) (cos 0° – cos 180°) = 4 × 10–3 J e e ev (b) Current, i t 2 r /v 2 r Here, v

e2

2

and r

me 2

Solved Paper 2010

2010-11

e (e 2 / )

i

2 (

2

e3 me 2

/ me2 )

3

2

11.

me5

2

3

h (given) 2

me5

i

4 2 me5 3

h

h 2 (d) For dark fringe, 2

8.

xd D

(2m 1)

12.

B

d 2

Here, D

B

(2 5 1)

2

9

13.

d2 9D (d) Generally, temperature of human body is 37°C (= 98.4°F) corresponding to which IR and microwave radiations are emitted from the human body. (d) The path of moving proton in a normal magnetic field is circular. If r is the radius of the circular path, then from the figure, From the symmetry of figure, the angle = 45°.

AC

2r cos 45

As Bqv

mv 2 or r r

2mv Bq = 0.14 m

AC

2r

1 = 0°, 0

4

= 45°

i (sin 0 r

sin 45 )

0

4

i 2l

2 0i 8 l (c) Zener diode is suitable for voltage regulating purpose. It is used as voltage stabilizer in many applications in electronics. (a) If two independent sources emitting light of the same wavelength are said to be coherent. (d) Induced charge B

Wavelength,

10.

2)

P

45°

i

d 2

d d 2 D

sin

1

l

P

S2

i (sin r

l

x=

9.

0

4

d 2

S1

d2 D

6.4 10 30 p 4 10 12 m 10e 10 1.6 10 19 (c) Magnetic field due to finite length of a wire, 2l

3

2

Here, m = 5, x

or

(d) In a neutral water molecule, there are 10 electrons and 10 protons. So, its dipole moment p = q (2l) = 10 e (2l) Hence length of the dipole = distance between centres of positive and negative charges

1 2

2r

1 1.6 10

15.

19

107

2

nBA (cos180 R

cos 1 ) cos 0 )

QR 2nA (d) From an optical fibre due to absorption or light leaving the fibre area resulting scattering of light sideways by impurities in the glass fibre. And due to this reason a very small part of light energy is lost.

B

16. 27

nBA (cos R

Q

...(1)

mv Bq

2 1.67 10

14.

EBD_7443 2010-12

17.

Target VITEEE r 3

(c) Length of the arc = r

21.

Potential at centre v 1 4

0

r 3 r

hc W0 '

hc W0 and 2E 2

1 W0 / E 2 W0 / E

E W 2 E W0

r 3

Charge on the arc =

kq r

(c) Here, E

(1 W0 / E ) Since (2 W / E ) 0

12

22.

0

l,

1 So 2

2 (b) According to Einstein's photoelectric equation, E

W0

1 2 mvmax 2

2(hf

vmax

W0 )

m

If frequency becomes 4f then

r

r 3

18.

19.

(a) Herefc = 1.5 MHz = 1500 kHz, fm = 10 kHz Lower side-band frequency = fc – fm = 1500 kHz – 10 kHz = 1490 kHz Upper side-band frequency = fc + fm = 1500 kHz + 10 kHz = 1510 kHz (c) Equivalent resistance of the circuit Req = 100 Current through the circuit, i

V R

23.

24.

v ' 2v (b) In electric field photoelectron will experience force and accelerate opposite to the field so its KE increases (i.e., stopping potential will increase), no change in photoelectric current, and threshold wavelength. (b) Magnetic field inside the cyclindrical

current is given by Child's law i p

V p1

600 150

3/ 2

i p1 8 10 8 mA

(4)3/ 2

80 mA

8

rr ' R2

10 Bout

R

R ( R 4 R) 4 R2

8 T 3 (a) By using lens maker's formula, Bout

25.

i p1

or, i p 2

Bin Bout

KV p3/ 2

Thus, Vp 2

0 2i 4 r'

distance r' from its axis, Bout

Since the voltmeter and 100 resistance are in parallel, the voltmeter reads the same value i.e., 1.2 V. (c) In space charge limited region, the plate

ip2

m

conductor Bin

2.4 50 1.2 V 100

3/ 2

2

0 2ir 4 R2 (R = radius of cylinder and r = distance of observation point from axis of cylinder) Magnetic field outside the cylinder at a

2.4 A 100

Potential difference across combination of voltmeter and 100 resistance

20.

2(h 4 f W0 ) m

v'

W0 4

2 hf

1 f

(

1)

1 R1

5 (1.5 1)

1 R2

2 R

...(i)

Solved Paper 2010

2010-13

If a lens of refractive index g is immersed in a liquid of refractive index 1, then its focal length in liquid 1 fl

(

1 1) R1

g

1 R2

1.5 1 n

1

2 R

12375 12.375 eV 1000 Now, 7.7 = 12.375 – 0 or 0 = 12.375 – 7.7 = 4.675 eV In the second case, Energy corresponding to 2000 Å

...(ii) 0.5n 1.5 n 7.5 4.5n

Dividing, (i) by (ii) 5 7.5 5n

29.

0.5n

12375 eV 6.1875 eV 2000 Now, 4.7 = 6.1875 – '0 or '0 = 6.1875 – 4.7 = 1.4875

75 5 45 3 (a) Magnetic field

n

26.

27.

0 0r

dE 1 1010 2 dt 9 1016 2 = 5.56 × 10–8 T (b) E E0 sin t Voltmeter read rms value B

30.

V2 , R For the first bulb,

(a) Power, P

R1

V2 P1

R

(220)2 25

V2 P2

V

i 2 R1

and P '2

1 11 i 2 R2

qBR m 2 qV m

Hence,

qBR or R m

2mV q

1936 16 W

Vn

1 n4

4W

1 B

2

e2 or vn 0 (h / 2 )

1 n

(ii) The energy of the electron in the nth orbit En

me4

13.6

8n2 20 h2

2

n

eV or En

1 n2

(iii) The radius of the electron in the nth orbit rn

n2 h2 me

2

484

R1 R2

1/ 2

(a) According to Bohr's theory of hydrogen atom, (i) The speed of the electron in the nth orbit

2

1 11

q B v

1936

220 1 220 A R1 R2 1936 484 2420 11 If the actual powers in the two bulbs be P1 and P2 then

P '1

mv 2 R

m1 And, m2

484

2qV m

qV or v

Here, V, q and B are constants. R m

Current in series combination is the same in the two bulbs, i

v

31.

(220)2 100

1 2 mv 2

1.5 V

V2 P

For the second bulb, R2

(c) As we know,

Centripetal force

E0 2 234 V 331 V and t 2 nt 2 50 t 100 t Thus, the equation of the line voltage E = 331 sin (100 t) 28.

Since P '1 P '2 , so, 25 W bulb will glow more brightly. (a) Given : = 100 nm = 1000 Å Energy corresponding to 1000 Å

where a0

0 2

n 2 a0 or rn

h2 0 me

5.29 10

n2 11

m

EBD_7443 2010-14

32.

Target VITEEE

(b) Current, i = qv 0i 2r

B

4

7

10

19

1.6 10

10

2

34.

35.

1.6 6.6 12.513 T 5.3 (a) For interference phase difference must be constant. (a) To remove the error, resistance box and the unknown resistance must be interchanged and then the mean reading must be taken. r or r h

(c) Here, tan ic or r

h

i

6.6 1015

2 0.53 10

33.

h

1

1

2

2

sin ic 1 sin ic

41.

12

16 1 9

12 3

36.

37.

cm 7 (b) Diffraction takes places when the wavelength of waves is comparable with the size of the obstacle in path. Pradio > Plight Hence, radio waves are diffracted around building. (c) Centripetal force = force of attraction of nucleus on electron. mv 2 a0

38.

e2

1 4

0

a02

4

e

h)

2 1019 1.6 10 = 1.6 ( m)–1

19

43.

0.64A

Pc 1

ma 2 2

ma 2

1

Pc 1

2

x

( n1l1 n2 l2 )

(0.36 + 0.14)

(d) The correct formula of the given complex is tetraammine aqua chlorocobalt (III) chloride is [CoCl(H2O)(NH3)4]Cl2, because in it the oxidation number of Co is +3. While in rest other options O. No. of Co is +2 [CoCl(H2O)(NH3)4]Cl2 x + (–1) + 0 + (0 × 4) + (–1) 2 = 0 x – 3 = 0 x = +3 (c) According to Kohlrausch's law, equivalent conductance at infinite dilution of HF, HF = NaF + HCl – NaCl (a)

C2 H5 I

Alc. KOH C2 H 4 (dehydrohalogenation) ethylene 'X'

Br | CH 2 CH 2 | Br

1, 2-dibromoethane 'Y'

H 3O

0 ma0

(d) As we know, conductivity ne(

42.

e

v

25 8

PART - II (CHEMISTRY) 2

1

1.6 10

4

40.

2

h

3

1 Pc 1200 W 2 (b) Optical path for ray 1 = n1l1 Optical path for ray 2 = n2l2 Phase difference, 1800

1

h

2 25 / 8

V R

0.5 10

(b) Total power Pt

1 r

l A

39.

h tan ic

sin ic or r cos ic

But sin ic

l A

R

0 qv 2r

KCN

Br2 CCl 4

CN | CH 2 — CH 2 | CN 'Z'

COOH | CH 2 — CH 2 | COOH

succinic acid 'A'

44.

(c) Let the initial rate be R and order with respect to A be x and B be y. Thus, rate law can be written as, rate, R = [A]x [B]y ...(i) After doubling the concentration of A, rate becomes 4R,

Solved Paper 2010

2010-15

4R = [2A]x [B]y

...(ii) After doubling the concentration of B, rate remains R, R = [A]x [2B]y ...(iii) From Eq. (i) and (ii), we get 2

x

R 1 1 1 4R 2 2 2 So, x = 2 From Eq. (i) and (iii), we get o

y

45.

46.

ethyl

47.

50.

y

1 k1 0.25 k1 4 (c) The other name of methanol is carbinol. So, the formula of ethyl carbinol is CH3CH2 CH2OH 4k 2

49.

x

R 1 1 1 R 2 1 2 So, Y = 0 Hence, the rate law is, rate R = [A]2 [B]0 This clearly shows that the order of this reaction is 2 and for second order reaction units of rate constant are mol–1 Ls–1. (c) As we know that for every 10° rise in temperature, rate constant, k becomes doubled. Hence, on rising the temperature 20°, the rate constant will be four times, i.e., k1

48.

51.

OC2H5

Br + C2H5ONa

k2

carbinol

NO2

52.

(a) In Victor Meyer's test, Red colour is given by primary alcohols (1°) (alcohols having – CH2 OH). The structres of the given Alcohols are

XA

76 Y

(b) CH3 C H — CH3 | OH

iso-propyl alcohol (2°)

sec-butyl alcohol (2°)

Hence, n-propyl alcohol is a 1° alcohol and gives red colour in Victor-Meyer's test.

32 He 4

particle

1e

0

particle

53.

On observing the reaction, mass number of 'X' is A = 235 + 12 + 0 = 247 On observing the reaction, atomic number of 'X' is Z = 76 + 6 – 1 = 81 Hence, element 'X' is 81X247. (b) For the given reaction, 2 A( g )

B2 ( g )

2 AB2 ( g )

the equilibrium constant,

tert -butyl alcohol (3°)

(d) CH 3 C HCH 2 CH3 | OH

235

particle

n propyl alcohol (1°)

CH3 | C CH3 | OH

NO2

p-nitrophenetole

(a) The complete nuclear reaction is z

(a) CH3 CH 2 CH 2 OH

(c) CH3

(b) Heat of formation is equal to enthalpy of a compound. (c) S (entropy change) is the measure of randomness and thus in solid, liquid and gas, the order of entropy is gas > liquid > solid Thus, S is positive for the reaction given in option (c) because solid CaCO3 is forming gaseous CO2. (b) We know that basicity is depend on ionic character. So, as the ionic size of lanthanide decreases, the covalent character of their hydroxide increases. Hence, their basicity decreases. (c) Commonly aryl halides do not take part in Williamson's synthesis, due to their high stability but due to the presence of strong electron withdrawing group like –NO2 makes the C–X bond weaker and substitution of –Br takes place by –OR.

Kp

2 p AB 2

p 2A p B2

16

...(i)

For the other given reaction, AB2 ( g )

A( g )

1 B2 ( g ) 2

EBD_7443 2010-16

Target VITEEE The equilibrium constant,

59.

2 p A p1/ B2

K 'p

(a) Aldol is -hydroxybutyraldehyde (or 3-hydroxybutanal) i.e.

...(ii)

p AB2

CH 3

On squaring Eq. (ii), we obtain, (K ' p )

2

60.

p A2 p B2

...(iii)

2 p AB 2

Now, from Eq. (i) and (iii), we obtain, K p .( K ' p ) 2

1

16.( K ' p ) 2

(b) Nitrobenzene can be converted into azobenzene, on reduction in the presence of Zn/NaOH in CH3OH. NO2

1

(

OH | CH CH 2 CHO

Zn/NaOH in CH 3OH

Kp = 16.0)

–N = N–

8[H]

1/2

1 1 K 'p 0.25 16 4 (d) When the size of cation is much smaller than the anion then frenkel defect is observed. Hence, AgBr, AgI and ZnS all exhibit Frenkel defect. (d) Crystals show good cleavage when, the constitutents are arranged in orderly pattern, i.e., in planes, (d) On ionisation they give different ions (K 'p )

54.

55. 56.

[Co(NH 3 ) 4 Cl 2 ]NO2

[Co(NH 3 ) 4 Cl2 ]

61.

62.

(c) Due to the presence of electron withdrawing group like Ph group decreases the electron density of nitrogen and hence, the lone pair of nitrogen are not available for donation. So, (C6 H5 )3 N is least basic due to the presence of three electron withdrawing Ph(C6H5) groups. (b) Coordination number of Ni in [Ni(C2O4)3]4– is 6 because C2 O42 (oxalate) is a bidentate

NO 2

[Co(NH3 ) 4 ClNO 2 ]Cl

57.

azobenzene

[Co(NH3 ) 4 ClNO 2 ] Cl So, they show ionisation isomerism. (c) We know that complex compound having no unpaired electron is colourless. Among the given complexes, K4[Fe(CN)6] has no unpaired electron as CN– is a strong field ligand and causes pairing of electrons. So, it is colourless.

63.

(b)

64.

(c)

65.

(b)

66.

(b)

67.

(d)

CH=NH

58.

(c)

+ HCN + HCl

Anhy. AlCl3

CHO H3 O

+

–NH3 (Benzaldehyde)

This is Gattermann aldehyde synthesis.

68.

ligand and each have two sites to coordinated with the central atom . Chlorophyll is rich source of Mg, the green pigment of plants. An alloy of 80% Ag and 20% other metals, usually copper is sterling silver. As CuSO4 reacts with KI to give white precipitate of Cu2I2 and to evolve I2.CuSO4 does not react with KCl. Due to highest reduction potential of fluorine. Transition metals exhibit highest oxidation states in their fluoride. Cu 2

1 mol 1 mol 2 1 gequi.

2e

2F

1F 1F

Cu

1 mol 1 mol 2 1g equi.

Thus, to reduce 4 g equivalent of Cu2+ into Cu 4F are required. (c) Fuel cell, which convert chemical energy of fuels like H2, O2, CH4, etc, is converted into electric energy, e.g., H2–O2 fuel cell.

Solved Paper 2010 69.

2010-17

(d) In Aldol condensation propanone gives diacetone alcohol in presence of Ba(OH)2

CH2 – OOC 110°C

O CH3–C–CH3 + HCH2COCH3

–H2O

Ba(OH) 2

OH

CH3

–CO 2

4-hydroxy-4-methyl pentanone-2 (diacetone alcohol)

72.

73.

74. 75.

RT ln K p

G K p e G / RT ln K p RT (c) A2 ( g ) 2 B ( g ) C ( g ) Q kJ Since, the reaction is exothermic, So, it is favoured by low temperature. In addition, the number of moles of products is lesser than the number of moles of reactants, thus high pressure favours the forward reaction. (a) Larger the value of K more the reaction moves towards completion. (b) As CN is a nucleophile. So it is an example of nucleophilic addition

CN

R2C = O +

–

R2 – C – O

CN

–

CHOH + HCOOH formic acid

CH2OH

77.

76.

(a)

t

78.

(b)

79.

(a)

80.

0.693 t1/ 2

COOH

0.693 1.155 10 6000

4

N 2.303 100 2.303 log 0 log 4 k N 1.155 10 25 = 12000 yr (age of piece of wood)

Radius of cation, r

95 181

0.525 Radius of anion, r As this value lies in between 0.414 – 0.732, thus, the coordination number of Na + ion will be 6. HBF

4 ROH H 2 CN 2 ROCH3 N 2 (d) The compounds having –CHO group reduces Tollen's reagent, Fehling solution etc. Thus, formic acid (HCO/ OH) has reducing property.

PART - III (MATHEMATICS)

R2 – C – OH

(d)

F ( x 2) 2 F ( x) F ( x 1) Putting x = 0, we get F (2)

CHOH + CH2OH

+

81. HOOC

(d) As we know that, k

CN

CH2OH

CHOH

CH2OH

nucleophile

H

CH2OH

CH2OH

(b) CH3CONH 2 NaOBr CH3 NH 2 Na 2 CO3 (b) Radiographer to protect themself from radiation worn lead apron. (d) G° and Kp are related as

G

COOH

OH H CH2 OOCH

H3C – C – CH2COCH3

70. 71.

CHOH

...(i)

2 F (0) F (1)

F (2)

2(2) 3 { F (0)

2, F (1) 3}

EBD_7443 2010-18

Target VITEEE F (2) 4 3 F (2) 1 Putting x = 1, in eq. (i), we get F (3)

2 F (1) F (2)

84.

F (3) 5 Putting x = 2, in eq. (i), we get F (4)

(a) We have, sin(e x ) 5x 5 x Let 5x = t, then eq. (i), reduces to

2F (2) F (3)

t

1 t

sin(e x )

t

sin(e x )

2(1) 5 { F (2) 1, F (3) 5}

F (4) 3 Putting x = 3, in eq. (i), we get F (5)

2F (3) F (4)

x

2(5) 3 { F (3) 5, F (4)

82.

sin(e )

3}

F (5) 13 (c) The number of binary operations on a set S

83.

(a)

(3 5x)

3

1 3

311 1

Now, r

85.

5 1 . 3 5

311 1

11

1 5

x

3

C2 (1)

1 3

1 (11 1) 3 1 1 3

2

311

4 4 3

C3 (1)8

ax

by

5x

0,

t exists}

2

cz

du cz

Let, a x

by

a

k1/ x , b

du

k

k1/ y , c

k1/ z , d

k1/ u ...(i)

3

d c

k 1/ y

k 1/ z

k 1/ u

k 1/ x

k 1/ y

k 1/ z

1 1 x

1 y

1 x

1 1 y

1 1 z

kz

1 z

{using Eq. (i)}

ku

1 y

1 u

1 z

1 1 1 1 , , , are in A.P.. x y z u

11 10 3 1.2.9

1 3

c b

ky

86. 11

(c)

b a

11

= 55 × 39 and T4

2

t

a, b, c, d are in GP.

| x | (n 1) |x| 1

9

2

11

r 3 Therefore, 3rd (T3) and (3 + 1) = 4th (T4) terms are numerically greatest in the expansion of (3 – 5x)11. So, greatest term = T3 11 11

...(i)

1. which is not possible as also sin Thus, given equation has no solution.

11

5x 1 3

1

t

sin(e x )

2

11

55 39

1 2 2 t

{ 5x

having n elements in n n . 11

1 27

Greatest term (numerically) = T3 = T4 = 55 × 39

{ F (1) 3, F (2) 1}

2(3) 1

11 10 9 . 1.2.3

311

x, y, z, u are in H.P. (b) Given : | z | – z = 1 + 2i If z = x + iy, then this equation reduces to | x iy | ( x iy ) 1 2i ( x2

y2

x) ( iy) 1 2i

Solved Paper 2010

2010-19

On comparing real and imaginary parts of both sides of this equation, we get x2

y2

y2

x

2

2

y

2

y

x2

1 x

1 x

2

y2

(1 x )2

x iy

z

87.

(c)

x

3 2

z

91.

1 3 2 3i 1 3

49 2

(c) det (B–1AB)

2

2 2 3i 4

Now,

( x)

x

1 and x 1

m2 4

m

7

0

49

det( B 1 ) det A det B

'( x)

ax2 2hxy by 2 2 gx 2 fy c, we get

h '''( x )

f ( x)

g ( x)

h( x )

f '( x)

g '( x)

h '( x)

0

f '( x)

g '( x)

h '( x )

f '( x)

g '( x)

h '( x )

f ''( x) g ''( x) h ''( x ) f ( x)

g ( x)

h( x)

f ''( x) g ''( x) h ''( x) f ''( x) g ''( x) h ''( x)

x ( , 1] [1, ) (c) The given expression is

2 x 2 mxy 3 y 2 5 y 2 Comparing the given expression with

g '''( x )

f ''( x ) g ''( x) h ''( x)

0 1

0 ( 2)

0

m2

therefore f '''( x )

3 i 2

x2

m2 2

0

det( B 1B) det A det I .det A = 1 . det A = det A (c) Since f(x), g(x) and h(x) are the polynomials of degree 2,

log10 x 2 is real, if

f ( x)

log10 x

89.

m2 2

1 i 3 1– i 3

1 2

5 2

0, f

det( B 1 ) det B det A

z cos 240 i sin 240 Thus, arg (z) = 240°

(d)

25 2

12

3 2i 2

1 i 3 1 i 3

z

88.

...(i)

90.

1 i 3 z

2, g

25 4

(2)(3)( 2) 2(0) 2

1 i 3

z

3, c

abc 2 fgh af 2 bg 2 ch 2

1 2x 3

m ,b 2

2x

1 2x and –y = 2 y 2 Putting this value in eq. (i), we get

2x

2, h

The given expression is resolvable into linear factors, if

x 1

x2

( 2)2

a

'( x )

f ( x)

g ( x)

h ( x)

f '( x )

g '( x )

h '( x)

f '''( x)

g '''( x ) h '''( x)

0 0 0

0

( x) constant. Thus, (x) is the polynomial of degree zero.

EBD_7443 2010-20

92.

Target VITEEE

(d) Let E 1, E2 and E3 denote the events of selecting boxes A, B, C respectively and A be the event that a screw selected at random is defective. Then, P ( E1 )

1 , P ( E3 ) 3

2 Now, 5 sin – 3 cos

1 3

2 tan 5.

1 tan 2

By Baye's rule, the required probability

3 5 5. 9 1 25

A E1

P( E1 ) P

A 1 ,P E2 5

A E1

A 1 ,P E3 6

P( E1 ) P

A E1

P( E2 ) P

A E2

P( E3 ) P

1 5 1 6

A E3

1 5

1 7

95.

(a)

sin

1

93.

(c)

cos 1 sin

2 sin

cos

94.

(b) 3sin

5cos

2

4

2

4

2

3.2 sin cos 2 2

150 48 34

sin

6

1

102 34

sin

1

sin

3.

6

) sin ]

6

2

4

5

9 25 3. 9 1 25

[ Principal value [0, / 2]]

2 cos

sin

2

[ sin(

2

1 cos

sin

1 tan 2

2

1

16 25 9 1 25

sin sin

2

6 3.

42 107

3.

2.

1 1 . 3 5 1 1 1 1 1 1 . . . 3 5 3 6 3 7

E P 1 A

1 tan 2

2

1 7

P

E P 1 A

1 , P ( E2 ) 3

3 5

tan

cos

4

2

96.

2

4

2

tan

6 which is the required principal value. (a) Let both of the ends of the rod are on x-axis and y-axis. Let AB be rod of length l and coordinates of A and B be (a, 0) and (0, b), respectively.

Y 4

2

B (0, b) 3sin

5(1 cos )

5.2 sin 2 sin

P (h, k) 2

A (a, 0)

2 sin cos 2 2

and 1 cos

2 sin 2

O 2

X

Solved Paper 2010

2010-21

Let P (h, k) be the mid point of the rod AB. Then,

0 a 2 b 0 2

h k

a 2 b 2

[Using the internal section formula] h 3 k cos and sin 4 4 Squar ing and adding both of these equations,

...(i)

(h 3)2 16

In OAB, OA

2

OB

a

2

b

2

2

AB l

2

l2

[using eq. (i)]

2

l 4 The equation of locus is

h2

x2 97.

k2

(a) Let L1

2x

99.

L1

L2

0

(2 x y 1) (3 x 2 y 5) 0 Since, this line passes through the origin also (0 0 1) 5

(0 0 5) 1 5

1

y 1)

2

98.

3 x 5

1 1 2 1

lim

n

1

1 2 3

1 1 ..... 3 4 n( n 1)

1 2

1 2

1 3

1 1 3 4 1 n

1 n 1

lim 1

1 (3 x 2 y 5) 5

0

n

4 cos

2(10 sin ) 3(0) 2 3

4 sin

n

lim

2 y 1 1 0 5

2(10 cos ) 3(5) 2 3

and k

lim

n

n

7 3 x y 0 7x 3y 0 5 5 (c) Let coordinates of P be (h, k), then h

100. (a)

0

Required line is (2 x

...(i) y2 8 x is y = mx + 4m + 2m3 (using equation of normal of parabola in slope form y = mx – 2am – am3 and a = –2) The given normal is 2x y k 0 2x k ...(ii) y Comparing eqs. (i) and (ii), we get m = – 2 and –4m –2m3 = k k = 8 + 16 = 24

y 1 0

L2 3x 2 y 5 0 The equation of straight line passing through the intersection point of the lines L1 and L2 is given by

sin 2

( x 3) 2 y 2 16 which is a circle. (c) The equation of any normal to the parabola

l2 4

y2

cos 2

(h 3)2 k 2 16 Therefore, locus of point P is

2

(2h)2 (2k )2

k2 16

3

n 1

1 n

lim

n

1 n 1

n n 1 1

lim

n

......

1

1 n

1

101. (d) Let P(x1, y1) be a point on the curve

y2

...(i)

4ax

On differentiating y 2 2y

dy dx

dy dx

4a

( x1 , y1 )

2a y1

4ax w.r.t. 'x', we get

EBD_7443 2010-22

Target VITEEE Thus, the equation of normal at (x1, y1) is y

103. (a) Let I

y – 1 ( x x1 ) 2a

y1

...(ii)

y1 ( x1 2a)

y1 x 2ay

...(iii) But lx + my = 1 is also a normal. Therefore, coefficients of eqs. (ii) and (iii), must be proportional. i.e.,

y1 l

y1 ( x1 2a ) 1

2a m

2al and x1 m

y1

2

4a

4a 2l 2 m al

102. (a)

3

m

2

1 2a l

f ( x ) dx

d f ( x) dx

2am l

sin

1

I

sin

1

2x 2

dx

(2 x 2)2 32

1

3tan 3sec

3 2 sec d 2

I

3 2

sec 2 d

I

3 2

tan

al

2alm

2

m

2

3 [ tan 2

I

3 tan 2

tan d

f ( x) f ( x )dx

2 { f ( x)}2 dx { f ( x)}2

1 { f ( x )}2 2

1

( x 1) tan

d f ( x) f ( x )dx dx dx f ( x ) f ( x ) dx

log | sec |] c

d f ( x)dx dx

2x 2 3

2x 2 3

log 1

2x 2 3

1

2

c

2x 2 3

3 4 x 2 8 x 13 log 4 9

(integrating by parts)

{ f ( x)}2 dx

dx

8x 4 9

(integrating by parts) 3

f ( x)

f ( x) f ( x)

4x

3 sin 1 (sin ) sec 2 d 2

f ( x)

f ( x ) f ( x)dx

2x 2 2

I

f ( x)

Now, { f ( x )}2 dx

dx

3sec 2 d , we get

sin

I

1 2a l

2

I

2dx

4 a 8a 2 l l

2

4 x 2 8 x 13

Substituting 2x + 2 = 3 tan ,

Putting these values of x1 and y1 in eq. (i), we get

2al m

2x 2

1

sin

c

104. (c) The equation of ellipse is

3 x2

2 y2 6x 8 y 5 0

3( x 2 2 x) 2( y 2 4 y) 5 0 3( x 2

2 x 1) 2( y 2 4 y 4) 5 3 8 0

Solved Paper 2010

2010-23

3( x 1)2

2( y 2) 2

( x 1)2 ( y 2)2 2 3 Comparing with ( x h) 2

( y k )2

106. (d) We have

6

f ( x) log x cos x f (x) is defined for cos x > 0. x > 0, x 1

1

2 Also, x > 0, x 1

1 , we get

a2 b2 h = –1, k = 2, a2 = 2, b2 = 3 Here, centre (h, k) = (–1, 2) And using a2 = b2 (1 – e2) 2

2 3(1 e )

107. (b)

3 And foci are (h, k + be) and (h, k – be) =(–1, 2 + 1) and (–1, 2 – 1) = (–1, 3) and (–1, 1) 105. (b) We have the two hyperbolas as x2

y2

a2

b2

and

1

y2

x2

2

2

1

a2 2 2

2

b (m x 2

(b m

2

x2

...(i)

Now, y

...(ii)

x2

c

2

a )x

2

2

2 2

2mcx ) – a x 2

2

2mcb x b (c

2

a b 2

a )

0

4m 2 c 2 b 4

c

2

4(b2 m 2

a 2 ).b 2 (c 2

4b2 (b2 m 2c 2 b 2 m2 a2

a

2

a 2 m2

2

b m b2

108. (c)

x

0

a2 ) a4 )

2

a2

(a 2 b 2 )m 2

a 2 c2

b 2 m 2 [ using Eq. (iii)]

a2 b2

m 1 m2 1 Hence, the equation of common tangent are

y

x

a 2 b2

sin

1

1 x2 R.

x2

1

1 x2

sin y

x

sin y 1

0

sin y 1 sin y

y

2 Thus, range of the given function is

2

. cos

sec dx d

For the tangency, it should have equal roots (2mcb2 )2

x2

1 x For the existance of x sin y 0 and 1 – siny > 0

0, 2 2

1 x2

2

1

b2

x2

1

sin

{1}

2

which is true for all x

where c a 2 m 2 – b2 … (iii) But this tangent touches the parabola eq. (ii) also –

y

For y to be defined

a b Any tangent to the hyperbola eq(i) y = mx + c

( mx c ) 2

2

Domain of f is 0,

1

e

x

cos x > 0

y

sec tan

secn dy d

dy dx dy dx

sin ,

cosn

n sec n

n

n

1

sec tan

n cos n

1

sin

(sec n tan cos n 1 sin ) (sec tan sin )

(sec n (sec

cos n ) tan cos ) tan

EBD_7443 2010-24

Target VITEEE

n(sec n (sec

dy dx dy dx

2

dy dx

2

y

Y

n 2 {(sec n – cos n ) 2 (sec – cos ) 2

n (y x

( x2 109. (d)

111. (c) Required area = Shaded area

cos n ) cos )

2

2 2

y

x

y2

x

y

y2

x

y

4}

4

4) x=–

x) 2 y

0

y ......

x

2[cos x]0 2(cos = 4 sq units 112. (d) Length of normal = c

y ......

y2

y

2 xy 1)

y2

1

2y

2

dy ( y2 dx

(2 y 3

x

110. (a)

dy 1 dx

xy )

3

2y

x

dy dx

x) y2

[sec 1 t ]1x sec

1

sec

1

dy dx x

2 xy 1

x

2 3

cos 0)

2

dy dx

y2 1

x

c 2

c2

Clearly, this is the differential equation of degree 2. 113. (d) a 2iˆ 2 ˆj 3kˆ

b

–iˆ 2 ˆj kˆ

c 3iˆ Now,

ˆj

a tb

(2iˆ 2 ˆj 3kˆ) ( tiˆ 2tjˆ tkˆ) (2 t )iˆ (2 2t ) ˆj (3 t )kˆ

6

Since, a tb is perpendicular to c

(a tb) c

6

x sec 1 1 x 0

dy dx

y 1

dy dx

dt | t | t2 1

2 | sin x | dx

| sin x | dx

2( y 3

dy dx

x=

n2 ( y 2 4)

( y 2 x)2 2 y On differentiating both sides w.r.t. x, we get 2( y 2

X

O

4

dy 4) dx

x

2

6

0

{(2 t )iˆ (2 2t ) ˆj (3 t ) kˆ} (3iˆ

6 x

3(2 t ) 2 2t sec

6

6 3t 2 2t 114. (c) The given line is r 2iˆ 2 ˆj 3kˆ

ˆj ) 0

0 0

t

(iˆ

ˆj 4kˆ)

On comparing it with r

8

a tb , we get

Solved Paper 2010

2010-25

a 2iˆ 2 ˆj 3kˆ, b Also, the plane is r (iˆ 5 ˆj kˆ) 5

iˆ

ˆj 4kˆ

On comparing it with r n

3 2 Taking second and third part of eq. (iii), we get 3r – 1 – k = 8r +2 3r 1 k 8r 2 0 k 5r 3

r

d , we get

n iˆ 5 ˆj kˆ and d = 5 Since, b n (iˆ ˆj 4kˆ) (iˆ 5 ˆj kˆ) = 1 – 5 + 4 =0 Given line is parallel to the given plane. Now, distance between the line and the plane is given by required distance a n d | (2iˆ 2 ˆj 3kˆ) (iˆ 5 ˆj kˆ) 5 |

1 25 1

10

3 3 27 115. (a) The equation of the sphere concentric with the sphere

x2

y2

2

2

z 2 4 x 6 y 8z 5 0 is 2

...(i) x y z 4 x 6 y 8z c 0 Since, this sphere eq. (i) passes through origin, therefore 0+0+0–0–0–0+c=0 c 0 Hence, the required equation of sphere is

x 2 y 2 z 2 4 x 6 y 8z 116. (b) We have the lines x 1 2

and

y 1 3

x 3 1

z 1 4

y k 2

3

15 3 2

k

r

3 2

9 2

k

3x

117. (a) Given curves are y

dy dx

3x log 3 and

dy dx

...(i)

(0,1)

Let a point (2r 1, 3r 1, 4r 1) be on the line Eq. (i). If this is an intersection point of both the lines, then it will lie on Eq. (ii), also 2r 1 3 3r 1 k 4r 1 ...(iii) 1 2 1 Taking first and third part of eq. (iii), we get 2r – 2 = 4r + 1

5 x log 5

log 3 and dy dx

log 5 (0,1)

m1 m2 1 m1m2

tan

log 3 log 5 1 log 3 log 5

tan

tan

...(ii)

dy dx

m1 log 3 and m2 log 5 Angle between these curves is given by

0

...(i) z 1

3 2

5

and y 5 x ...(ii) intersect at the point (0, 1). Now, differentiating eqs. (i) and (ii) w.r.t. x, we get

|n|

| 2 10 3 5 |

k

1

log 3 log 5 1 log 3log 5

118. (d) We have the given equation as

x 2 4 xy y 2 x 3y 2 0 On comparing this equation with ax2 2hxy by 2 2 gx 2 fy c get

0 , we

3 ,c 2 2 2 Since, given equation represents a parabola ,h

a

h2

ab

2, b 1, g

4

1

, f

4

EBD_7443 2010-26

Target VITEEE

119. (c) The given two circles are

2x

2

x2

2y

2

y2

3x 6 y k

0 k 2

3 x 3y 2

120. (a) A (–2, 1), B (2, 3) and C (–2, –4) are three given points. Slope of the line BA 0

...(i)

and x2 y 2 4 x 10 y 16 0 Since, general equation of circle is

3 , f1 4

3 , c1 2

k 2

and g2 2, f 2 5, c2 16 Both the circles cut orthogonally, 2( g1 g2 f1 f 2 ) c1 c2

2 18

3 15 2 2 k 16 2

k 16 2 k 2

2

Using slope formula, m

4

y2 x2

y1 x1

Slope of the line BC 4 3 7 2 2 4 Now, angle between AB and BC is given by

m2

tan

m1 m2 1 m1m2

10 15

tan k

1 2

...(ii)

...(iii) x2 y 2 2 gx 2 fy c 0 Therefore, comparing eqs. (i) and (ii) with eq. (iii), we get g1

1 3 2 2

m1

tan

1

2 3

1 2

7 4 1 7 1 2 4

2 3

tan tan

1

2 3 [ | x | x]

VITEEE

SOLVED PAPER 4.

PART - I (PHYSICS) 1

When a wave traverses a medium the displacement of a particle located at x at a time t is given by y = a sin (bt – cx). Where a, b and c are constants of the wave. Which of the following is a quantity with dimensions? (a)

y a

(b) bt

b c A body is projected vertically upwards at time t = 0 and it is seen at a height H at time t1 and t2 second during its flight. The maximum height attained is (g is acceleration due to gravity)

(c) cx

2.

(a)

5.

t1 )2

(b)

8

6.

g (t1 t2 ) 2 4

g (t1 t2 ) 2 g (t2 t1 )2 (d) 8 4 A particle is projected up from a point at an angle with the horizontal direction. At any time t, if p is the linear momentum, y is the vertical displacement, x is horizontal displacement, the graph among the following which does not represent the variation of kinetic energy KE of the particle is (c)

3.

y

(C) KE

t

(a) graph (A) (c) graph (C)

p2 (b) graph (B) (d) graph (D)

13 –1 ms 2 A body is projected vertically upwards from the surface of the earth with a velocity equal to half the escape velocity. If R is the radius of the earth, maximum height attained by the body from the surface of the earth is R R (a) (b) 6 3 2R (c) (d) R 3 The displacement of a particle executing SHM is given by

7.

8.

13 ms–1

y

(D) KE

x

1 th of its original velocity.. 10 Then the mass of the second body is (a) 4.09 kg (b) 0.5 kg (c) 5 kg (d) 5.09 kg A particle of mass 4 m explodes into three pieces of masses m, m and 2m. The equal masses move along X-axis and Y-axis with velocities 4 ms–1and 6 ms–1 respectively. The magnitude of the velocity of the heavier mass is (a) (b) 2 13 ms–1 17 ms–1

(c)

(B) KE

(A) KE

A motor of power P0 is used to deliver water at a certain rate through a given horizontal pipe. To increase the rate of flow of water through the same pipe n times, the power of the motor is increased to P1. The ratio of P1 to P0 is (a) n : 1 (b) n2 :1 3 (c) n : 1 (d) n4 :1 A body of mass 5 kg makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to

(d)

g (t2

2009

(d)

5sin 4t

3 If T is the time period and the mass of the particle is 2g, the kinetic energy of the particle when t = T/4 is given by (a) 0.4 J (b) 0.5 J (c) 3 J (d) 0.3 J

EBD_7443 2009-2

9.

Target VITEEE

If the ratio of lengths, radii and Young's modulus of steel and brass wires shown in the figure are a, b and c respectively, the ratio between the increase in lengths of brass and steel wires would be (a)

b2a 2c

(b)

bc 2a2

Which of the following curves represents the equivalent cyclic process? A

(a)

P

Brass 2kg

11.

12.

13.

14.

A

B

P

(c)

D

D

T

B

C

C

(d)

A

B

D

C

V

T T 15. An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states are Q1 = 6000 J, Q2 = –5500 J; Q3 = –3000 J; Q4 = 3500 J W1= 2500 J; W2 = –1000 J; W3 = –1200 J; W4 = x J. The ratio of the net work done by the gas to the total heat absorbed by the gas is ). The values of x and respectively are (a) 500 ; 7.5 % (b) 700; 10.5 % (c) 1000 ; 21 % (d) 1500; 15 % 16. Two cylinders A and B fitted with pistons contain equal number of moles of an ideal monoatomic gas at 400 K. The piston of A is free to move while that of B is held fixed. Same amount of heat energy is given to the gas in each cylinder. If the rise in temperature of the gas in A is 42 K, the rise in temperature of the gas in B is (a) 21 K (b) 35 K (c) 42 K (d) 70 K 17. Three rods of same dimensional have thermal conductivity 3 K, 2 K and K. They are arranged as shown in the figure below 50°C T 2K 3K 100°C K 0°C Then, the temperature of the junction in steady state is

(a)

200 C 3

(c) 75°C

V

C

P

B

V

T

A

(c)

10.

A

(b) D

2

a ba Steel (d) 2b2c 2c 2kg A soap bubble of radius r is blown up to form a bubble of radius 2r under isothermal conditions. If T is the surface tension of soap solution, the energy spent in the blowing (a) 3 Tr2 (b) 6 Tr2 2 (c) 12 Tr (d) 24 Tr2 Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of 6 cm-s–1. If they coalesce to form one big drop, what will be the terminal speed of bigger drop? (Neglect the buoyancy of the air) (a) 1.5 cm-s–1 (b) 6 cm-s–1 –1 (c) 24 cm-s (d) 32 cm-s–1 A clock pendulum made of invar has a period of 0.5 s, at 20°C. If the clock is used in a climate where the temperature averages to 30°C, how much time does the clock lose in each oscillation? (For invar, = 9 × 10–7/°C, g = constan t) (a) 2.25 × 10–6 s (b) 2.5 × 10–7 s –7 (c) 5 × 10 s (d) 1.125 × 10–6s A piece of metal weighs 45 g in air and 25 g in a liquid of density 1.5 × 103 kg-m–3 kept at 30°C. When the temperature of the liquid is raised to 40° C, the metal piece weights 27 g. the density of liquid of 40°C is 1.25 × 103 kg–m–3. the coefficient of linear expansion of metal is (a) 1.3 × 10–3/°C (b) 5.2 × 10–3/°C –3 (c) 2.6 × 10 /°C (d) 0.26 × 10–3/°C An ideal gas is subjected to a cyclic process ABCD as depicted in the p-V diagram given below:

B

(b)

100 C 3

(d)

50 C 3

Solved Paper 2009

2009-3

18. Two sources A and B are sending notes of frequency 680 Hz. A listener moves from A and B with a constant velocity u. If the speed of sound in air is 340 ms–1, what must be the value of u so that he hears 10 beats per second? (a) 2.0 ms–1 (b) 2.5 ms–1 –1 (c) 3.0 ms (d) 3.5 ms–1 19. Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously? (a) 0.01 (b) 0.02 (c) 0.03 (d) 0.04 20. In the Young's double slit experiment, the intensities at two points P1 and P2 on the screen are respectively I1 and I2. If P1 is located at the centre of a bright fringe and P2 is located at a distance equal to a quarter of fringe width fromP1,

23.

24.

I1 then I is 2 1 2 (c) 4 (d) 16 21. In Young's double slit experiment, the 10th maximum of wavelength 1 is at a distance of y1, from the central maximum. When the wavelength of the source is changed to 2, 5th maximum is at a distance of y2 from its central maximum. The

(a) 2

(b)

y1 ratio y 2

(a)

(b)

2 1

2 2 1 2

(ii)

y1

a sin( t

y2

a sin 2 t

(iii) y3 a sin( t (iv) y4

a sin(3 t

1 cm –1 . Then, the 3 magnitude of the electric intensity at a point 18 cm away is (given 0 = 8.8 × 10–12 C2Nm–2) (a) 0.33 × 1011 NC–1 (b) 3 × 1011 NC–1 (c) 0.66 × 1011 NC–1 (d) 1.32 × 1011 NC–1 Two point charges –q and +q are located at points (0, 0, – a) and (0, 0, a) respectively. The electric potential at a point (0, 0, z), where z > a is

linear charge density of

(d) 2 2 2 1 22. Four light sources produce the following four waves: (i)

26.

is

2 1

(c)

25.

Superposition of which two waves give rise to interference? (a) (i) and (ii) (b) (ii) and (iii) (c) (i) and (iii) (d) (iii) and (iv) The two lenses of an achromatic doublet should have (a) equal powers (b) equal dispersive powers (c) equal ratio of their power and dispersive power (d) sum of the product of their powers and dispersive power equal to zero Two bar magnets A and B are placed one over the other and are allowed to vibrate in a vibration magnetometer. They make 20 oscillations per minute when the similar poles of A and B are on the same side, while they make 15 oscillations per minute when their opposite poles lie on the same side. If MA and MB are the magnetic moments of A and B and if MA > MB, the ratio of MA and MB is (a) 4 : 3 (b) 25 : 7 (c) 7 : 5 (d) 25 : 16 A bar magnet is 10 cm long is kept with its north (N)-pole pointing north. A neutral point is formed at a distance of 15 cm from each pole. Given the horizontal component of earth's field is 0.4 Gauss, the pole strength of the magnet is (a) 9 A-m (b) 6.75 A-m (c) 27 A-m (d) 1.35 A-m An infinitely long thin straight wire has uniform

1)

2)

)

27.

(a)

(c)

qa 4

0z

(b)

2

2qa 4

0 (z

2

2

a )

(d)

q 4

0a

2qa 4

0 (z

2

a2 )

EBD_7443 2009-4

28.

Target VITEEE

In the adjacent shown circuit, a voltmeter of internal resistance R, when connected across B

I ampere and the radius of the circular loop is R metre. Then, the magnitude of magnetic induction at the centre of the circular loop is

100 V. Neglecting the internal 3 resistance of the battery, the value of R is

and C reads

I

A (a)

50k B

100V

50k C

29.

30.

(a) 100 k (b) 75k (c) 50k (d) 25k A cell in secondary circuit gives null deflection for 2.5 m length of potentiometer having 10 m length of wire. If the length of the potentiometer wire is increased by 1 m without changing the cell in the primary, the position of the null point now is (a) 3.5 m (b) 3 m (c) 2.75 m (d) 2.0 m The following series L-C-R circuit, when driven by an emf source of an gular frequency 70 kilo-radians per second, the circuit effectively behaves like 100 µH

32.

(b)

0 nI 2R

0I 0I ( 1) ( – 1) (d) 2 R 2 R The work function of a certain metal is 3.31 × 10–19 J. Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength 5000 Å is (Given, h = 6.62 × 10–34 Js, c = 3 × 108 ms–1, e = 1.6 × 10–19 C) (a) 2.48 eV (b) 0.41 eV (c) 2.07 eV (d) 0.82 eV A photon of energy E ejects a photoelectron from a metal surface whose work function is W0. If this electron enters into a uniform magnetic field of induction B in a direction perpendicular to the field and describes a circular path of radius r, then the radius r is given by, (in the usual notation)

(c)

33.

34.

(a)

(c)

31.

0I 2 R

2m(E W0 )

eB

1µH 10

(a) purely resistive circuit (b) series R-L circuit (c) series R-C circuit (d) series L-C circuit with R = 0 A wire of length l is bent into a circular loop of radius R and carries a current I. The magnetic field at the centre of the loop is B. The same wire is now bent into a double loop of equal radii. If both loops carry the same current I and it is in the same direction, the magnetic field at the centre of the double loop will be (a) Zero (b) 2 B (c) 4 B (d) 8 B An infinitely long straight conductor is bent into the shape as shown below. It carries a current of

R O

2e(E W0 )

(b)

(d)

2m(E W0 )eB 2m(E W0 )

eB mB 35. Two radioactive materials x1 and x2 have decay constants 10 and respectively. if initially they have the same number of nuclei, then the ratio of the number of nuclei of x1 to that of x2 will be 1/e after a time (a) (1/10 ) (b) (1/11 ) (c) 11/(10 ) (d) 1/(9 ) 36. Current flowing in each of the following circuit A and B respectively are 4

4

4

4

8V (Circuit A)

8V (Circuit B)

Solved Paper 2009

2009-5

(a) 1A, 2A (b) 2 A, 1 A (c) 4 A, 2 A (d) 2 A, 4 A 37. A bullet of mass 0.02 kg travelling horizontally with velocity 250 ms–1 strikes a block of wood of mass 0.23 kg which rests on a rough horizontal surface. After the impact, the block and bullet move together and come to rest after travelling a distance of 40 m. The coefficient of sliding friction of the rough surface is (g = 9.8 ms–2) (a) 0.75 (b) 0.61 (c) 0.51 (d) 0.30 38. Two persons A and B are located in X-Y plane at the points (0, 0) and (0,10) respectively. (The distances are measured in MKS unit). At a time t = 0, they start moving simultaneously with ˆ 1 va 2 ˆjms 1 and vb 2ims respectively. The time after which A and B are at their closest distance is (a) 2.5s (b) 4s

velocities

10

s 2 39. A rod of length l is held vertically stationary with its lower end located at a point P, on the horizontal plane. When the rod is released to topple about P, the velocity of the upper end of the rod with which it hits the ground is (c) 1s

g l

(a)

(c)

(d)

3

g l

(b)

3gl

(d)

3g l

40. A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string is wrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad-s–2 is produced in it due to the torque. Then, moment of inertia of the wheel is ( g = 10 ms–2) (a) 2 kg-m2 (b) 1 kg-m2 2 (c) 4 kg-m (d) 8 kg-m2

PART - II (CHEMISTRY) 41. Given that Hf(H) = 218 kJ/mol, express the H—H bond energy in kcal/mol. (a) 52.15 (b) 911 (c) 104 (d) 52153

42.

Identify the alkyne in the following sequence of reactions,

Alkyne

H2 Lindlar 's catalyst

A

Ozonolysis

Wacker Process

B

only

CH 2

CH 2

(a) H3C — C — C — CH3 (b) H3C — CH2 — C CH (c) H2C = CH— C CH (d) HC C — CH2 — C CH 43. Fluorine reacts with dilute NaOH and forms a gaseous product A. The bond angle in the molecule of A is (a) 104°40' (b) 103° (c) 107° (d) 109°28' 44. One mole of alkene X on ozonolysis gave one mole of acetaldehyde and one mole of acetone. The IUPAC name of X is (a) 2-methyl-2-butene (b) 2-methyl-1-butene (c) 2-butene (d) 1-butene 45. The number of p – d ‘pi’ bonds present in XeO3 and XeO4 molecules, respectively are (a) 3, 4 (b) 4, 2 (c) 2, 3 (d) 3, 2 46. The wavelengths of electron waves in two orbits is 3 : 5. The ratio of kinetic energy of electrons will be (a) 25 : 9 (b) 5 : 3 (c) 9 : 25 (d) 3 : 5 47. Which one of the following sets correctly represents the increase in the paramagnetic property of the ions? (a) Cu2+ > V2+ > Cr2+ > Mn2 + (b) Cu2+ < Cr 2+ < V2+ < Mn2+ (c) Cu2+ < V 2+ < Cr2+< Mn2+ (d) V2+< Cu2+ < Cr2+ < Mn2+ 48 . Electrons with a kinetic energy of 6.023 × 104 J/ mol are evolved from the surface of a metal, when it is exposed to radiation of wavelength of 600 nm. The minimum amount of energy required to remove an electron from the metal atom is (a) 2.3125 × 10–19 J (b) 3 × 10–19 J (c) 6.02 × 10–19 J (d) 6.62 × 10–34 J 49. The chemical entities present in thermosphere of the atmosphere are (a)

O2 , O , NO

(b) O3

(c) N2, O2, CO2, H2O (d) O3, O2 , O2

EBD_7443 2009-6

50.

51.

52.

53.

54.

55.

56.

57.

58.

Target VITEEE

The type of bonds present in sulphuric anhydride are (a) 3 and three p – d (b) 3 , one p – p and two p – d (c) 2 and three p – d (d) 2 and two p – d In Gattermann reaction, a diazonium group is replaced by X using Y. X and Y are X Y (a)

Cl –

Cu/HCl

(b)

Cl

CuCl2/HCl

(c) Cl – CuCl2/HCl (d) Cl2 Cu2O/HCl Which pair of oxyacids of phosphorus contains ' P—H' bonds? (a) H3PO4, H3PO3 (b) H3PO5, H4 P2O7 (c) H3PO3, H3PO2 (d) H3PO2, HPO3 Dipole moment of HCl = 1.03 D, HI = 0.38 D. Bond length of HC1 = 1.3 Å and HI =1.6 Å. The ratio of fraction of electric charge, , existing on each atom in HCl and HI is (a) 12 : 1 (b) 2.7 : 1 (c) 3.3 : 1 (d) 1 : 3.3 SiC14 on hydrolysis forms ‘X’ and HC1. Compound ‘X’ loses water at 1000°C and gives ‘Y’. Compounds ‘X’ and ‘Y’respectively are (a) H2SiCl6 , SiO2 (b) H4SiO4, Si (c) SiO2, Si (d) H4 SiO4, SiO2 1.5 g of CdCl 2 was found to contain 0.9 g of Cd. Calculate the atomic weight of Cd. (a) 118 (b) 112 (c) 106.5 (d) 53.25 Aluminium reacts with NaOH and forms compound ‘X’. If the coordination number of aluminium in ‘X’ is 6, the correct formula of X is (a) [Al(H2O)4(OH)2]+ (b) [Al (H2O)3(OH)3] (c) [Al(H2O)2 (OH)4 ]– (d) [Al (H2O)6 ](OH)3 The average kinetic energy of one molecule of an ideal gas at 27°C and 1 atm pressure is (a) 900 cal K –1mo1–1 (b) 6.21 × 10– 21 JK–1 molecule–1 (c) 336.7 JK–1 molecules–1 (d) 3741.3 JK–1 mol–1 Assertion (A) : K. Rb and Cs form superoxides. Reason (R) : The stability of the superoxides increases from ‘K’ to ‘Cs’ due to decrease in lattice energy.

59.

60.

61.

62.

63.

The correct answer is (a) Both (A) and (R) are true and (R) is the correct explanation of (A) (b) Both (A) and (R) are true but (R) is not the correct explanation of (A) (c) (A) is true but (R) is not true (d) (A) is not true but (R) is true How many mL of perhydrol is required to produce sufficient oxygen which can be used to completely convert 2 L of SO2 gas to SO3 gas? (a) 10 mL (b) 5 mL (c) 20 mL (d) 30 mL pH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27°C. The buffer capacity of the solution is (a) 0.1 (b) 10 (c) 1 (d) 0.4 Match the following List I List II (A) Flespar (I) [Ag3Sb3] (B) Asbestos (II) Al2O3. H2O (C) Pyrargyrite (III) MgSO4. H2O (D) Diaspore (IV) KAlSi3O8 (V) CaMg3(SiO3)4 The correct answer is (A) (B) (C) (D) (a) IV V II I (b) IV V I II (c) IV I III II (d) II V IV I Which one of the following order is correct for the first ionisation energies of the elements? (a) B < Be < N < O (b) Be < B < N < O (c) B < Be < O < N (d) B < O < Be < N What are X and Y in the following reaction sequence? Cl

64.

Cl

2 2 X Y C 2 H 5 OH (a) C2H5C1, CH3CHO (b) CH3CHO, CH3CO2H (c) CH3CHO, CC13CHO (d) C2H5C1, CCl3CHO What are A, B, C in the following reactions?

(i)

(CH 3CO 2 ) 2 Ca

(ii)

CH3 CO2 H

(iii) 2CH3CO 2 H

A

HI Red P P4 O10

B C

Solved Paper 2009

2009-7

A B C (a) C2H6 CH3COCH3 (CH3CO)2O (b) (CH3CO)2O C2H6 CH3COCH3 (c) CH3COCH3 C2H6 (CH3CO)2O (d) CH3COCH3 (CH3CO)2O C2H6 65. One per cent composition of an organic compound A is, carbon : 85.71% and hydrogen 14.29%. Its vapour density is 14. Consider the following reaction sequence A

Cl2 /H2O

B

(i) KCN/EtOH

69.

70.

C2H5Cl + AgCN

C

(ii)H3O

Identify C. (a)

CH3 — CH — CO2 H |

OH

(b)

HO — CH 2 — CH 2 — CO 2 H

(c)

HO — CH 2 — CO 2 H

(d) CH3 — CH 2 — CO 2 H 66. How many tripeptides can be prepared by linking the amino acids glycine, alanine and phenyl alanine? (a) One (b) Three (c) Six (d) Twelve 67. A codon has a sequence of A and specifies a particular B that is to be incorporated into a C. What are A, B, C? A B C (a) 3 bases amino acid carbohydrate (b) 3 acids carbohydrate protein (c) 3 bases protein amino acid (d) 3 bases amino acid protein 68. Parkinson's disease is linked to abnormalities in the levels of dopamine in the body. The structure of dopamine is CH2CH2CH2NH2

CH2NH2 (a)

71.

72.

73.

(c)

CH2–CHCO2H NH2

(d)

OH OH

OH OH

Which one of the following statements is true for X? (I) It gives propionic acid on hydrolysis (II) It has an ester functional group (III) It has a nitrogen linked to ethyl carbon (IV) It has a cyanide group (a) IV (b) III (c) II (d) I For the following cell reaction,

G f (AgCl)

–109kJ / mol

G f (Cl – )

129 kJ/ mol

E° of the cell is (a) –0.60 V (b) 0.60 V (c) 6.0 V (d) None of these The synthesis of crotonaldeh yde from acetaldehyde is an example of ....... reaction. (a) nucleophilic addition (b) elimination (c) electrophilic addition (d) nucleophilic addition-elimination At 25°C, the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl 2 are 248 × 10–4 , 126 × 10–4 and 280 × 10–4 Sm2 mol–1 respectively,

OH

CH2CH2NH2

X (major)

Gf (Ag ) 78kJ / mol

OH

OH

EtOH/H2O

Ag | Ag | AgCl | Cl – | Cl2,Pt

(b)

OH

During the depression in freezing point experiment, an equilibrium is established between the molecules of (a) liquid solvent and solid solvent (b) liquid solute and solid solvent (c) liquid solute and solid solute (d) liquid solvent and solid solute Consider the following reaction,

74.

m Ba(OH)2

in Sm2 mol–1 is (a) 52.4 × 10–4 (b) 524 × 10–4 –4 (c) 402 × 10 (d) 262 × 10–4 The cubic unit cell of a metal (molar mass = 63.55 g mol–1) has and edge length of 362 pm. Its density is 8.92g cm -3 . The type of unit cell is (a) primitive (b) face centred (c) body centred (d) end centred

EBD_7443 2009-8

75.

Target VITEEE

The equilibrium constant for the given reaction is 100. N2 (g) + 2O2(g)

2NO2 (g)

What is the equilibrium constant for the reaction given below? 1 N 2 (g) O 2 (g) 2 (a) 10 (b) 1 (c) 0.1 (d) 0.01 For a first order reaction at 27°C, the ratio of time required for 75% completion to 25% completion of reaction is (a) 3.0 (b) 2.303 (c) 4.8 (d) 0.477 The concentration of an organic compound in chloroform is 6.15 g per 100 mL of solution. A portion of this solution in a 5cm polarimeter tube causes an observed rotation of –1.2°. What is the specific rotation of the compound? (a) +12° (b) –3.9° (c) –39° (d) +61.5° 20 ml of 0.1 M acetic acid is mixed with 50 mL of potassium acetate. Ka of acetic acid = 1.8 × 10–5 at 27°C. Calculate concentration of potassium acetate if pH of the mixture is 4.8. (a) 0.1 M (b) 0.04 M (c) 0.4 M (d) 0.02 M Calculate H° for the reaction, Na2SO4 (g) Na2O(s) + SO3(g) given the following : NO2 (g)

76.

77.

78.

79.

(A) Na(s) + H2O(l) (B) Na2SO4(s) + H2O(l)

80.

1 H (g) 2 2 H°= –146 kJ

NaOH(s) +

2NaOH(s) + SO3(g) H° = + 418 kJ 4Na(s) + 2H2O(l) (C) 2Na2O(s) + 2H2 (g) H° = + 259 kJ (a) +823 kJ (b) –581 kJ (c) –435 kJ (d) +531 kJ Which one of the following is the most effective in causing the coagulation of an As2S3 sol? (a) KCl (b) AlCl3 (c) MgSO4 (d) K3Fe(CN)6

PART - III (MATHEMATICS) 81. If f : [2, 3] R is defined by f(x) = x3 + 3x – 2, then the range f(x) is contained in the interval (a) [1, 12] (b) [12, 34] (c) [35, 50] (d) [–12, 12] 82. The number of subsets of {1, 2, 3, ..., 9} containing at least one odd number is (a) 324 (b) 396 (c) 496 (d) 512 83. A binary sequence is an array of 0's and l's. The number of n-digit binary sequences which contain even number of 0's is (a) 2n –1 (b) 2n –1 n –1 (c) 2 – 1 (d) 2n 84. If x is numerically so small so that x2 and higher powers of x can be neglected, then 3/2

2x .(32 5 x ) –1/5 3 is approximately equal to 1

(a)

32 31x 64

31 32 x 64 The roots of

(c)

85.

(b)

31 32 x 64

(d)

1 2x 64

( x a )( x a 1) ( x a 1)( x a 2) ( x a)( x a 2) 0 a R are always (a) equal (b) imaginary (c) real and distinct (d) rational and equal 86. Let f (x)= x2 + ax + b, where a, b R. If f(x) = 0 has all its roots imaginary, then the roots of f (x)+ f (x) + f (x) = 0 are (a) real and distinct (b) imaginary (c) equal (d) rational and equal 87. If f(x)= 2x4 – 13x2 + ax + b is divisible by x2 – 3x + 2, then (a, b) is equal to (a) (–9, – 2) (b) (6, 4) (c) (9, 2) (d) (2, 9) 88. If x, y, z are all positive and are the pth, qth and rth terms of a geometric progression respectively, then the value of the determinant

log x p 1 log y q 1 equals log z r 1

(a) log xyz (c) pqr

(b) (p – 1) (q – 1)(r – 1) (d) 0

Solved Paper 2009

2009-9

89. The locus of z satisfying the inequality z 2i 2z i

(a)

96.

1 , where z = x + iy, is

x2

y2

(b)

1

x2

y2

1

(c) x 2 y 2 1 (d) 2 x 2 3 y 2 1 90. If n is an integer which leaves remainder one when divided by three, then (1 3i ) n (1 – 3i ) n equals (a) –2n + 1 (b) 2n + 1 n (c) –(–2) (d) –2n

97.

91. The period of sin 4 x cos 4 x is 4

(b)

2 92. If 3cos x

(c)

(a)

93.

98.

(d)

4

2

2 2sin x, then the general solution of

sin 2 x – cos 2 x

99.

2 sin 2 x is

( 1) n

n

2

(b)

n ,n 2

(c)

(4 n 1)

(d)

(2n 1) , n

cos

1

,n

Z

1 2

,n

Z

Z

2sin

1

1 2

3cos

1

1

2

4 tan 1 ( 1) equals 19 35 (b) (c) 12 12 In a ABC

(a)

94.

47 12

43 12

(d)

(a b c )(b c a )(c a b)(a b c )

4b2 c 2 equals (a) cos2A (b) cos2B 2 (c) sin A (d) sin2 B 95. The angle between the lines whose direction cosines satisfy the equations l + m + n = 0, l2

(a)

m2

6

n2

(b)

0 is

4

1 2 1 3 (b) (c) (d) 3 3 2 4 The area (in square unit) of the circle which touches the lines 4x + 3y = 15and 4x + 3y = 5is (a) 4 (b) 3 (c) 2 (d) The area (in square unit) of the triangle formed by x + y + 1 = 0 and the pair of straight lines x2 – 3xy + 2y2 = 0 is

1 7 5 1 (b) (c) (d) 6 12 12 12 100. The pairs of straight lines x2 – 3xy + 2y2 = 0 and x2 – 3xy + 2y2 + x – 2 = 0 form a (a) square but not rhombus (b) rhombus (c) parallelogram (d) rectangle but not a square 101. The equations of the circle which pass through the origin and makes intercepts of lengths 4 and 8 on the x and y-axes respectively are (a) x2 + y2 ± 4x ± 8y = 0 (b) x2 + y2 ± 2x ± 4y = 0 (c) x2 + y2 ± 8x ± 16y = 0 (d) x2 + y2 ± x ± y = 0 102. The point (3, –4) lies on both the circles x2 + y2 – 2x + 8y + 13 = 0 and x2 + y2 – 4x + 6y + 11 = 0 Then, the angle between the circles is

(a)

Z

2

a3 iˆ ˆj kˆ, and a4 –iˆ 3 ˆj kˆ , then the correct order of m1, m2, m3 and m4 is (a) m3 < m1 < m4 < m2 (b) m3 < m1 < m2 < m4 (c) m3 < m4 < m1 < m2 (d) m3 < m4 < m2 < m1 If X is a binomial variate with the range {0, 1, 2, 3, 4, 5, 6} and P(X = 2) = 4P(X = 4), then the parameter p of X is

(a)

2

(a)

If m 1, m 2, m 3 and m 4 are respectively the magnitudes of the vectors a1 2iˆ ˆj kˆ, a2 3iˆ 4 ˆj 4kˆ,

(c)

3

(d)

(a) 60°

tan

1

1 2

3 (d) 135° 5 103. The equation of the circle which passes through the origin and cuts orthogonally each of the

(c)

tan

circles x 2 2

(b)

x2

1

y2 6x 8

y2 2x – 2 y

7 is

0 and

EBD_7443 2009-10

Target VITEEE 3x2 + 3y2– 8x – 13y = 0 3x2 + 3y2 – 8x + 29y = 0 3x2 + 3y2 + 8x + 29y = 0 3x2 + 3y2 – 8x – 29y = 0

(a) (b) (c) (d) 104. The number of normals drawn to the parabola y2 = 4x from the point (1, 0) is (a) 0 (b) 1 (c) 2 (d) 3 105. If the circle x2 + y2 = a2 intersects the hyperbola xy = c2 in four points (x1, y1) for i = 1, 2, 3 and 4, then y1 + y2 + y3 + y4 equals (a) 0 (b) c (c) a (d) c 4 106. The mid point of the chord 4x – 3y = 5 of the hyperbola 2x2 – 3y2 = 12 is 5 3

0,

(a)

(b) (2, 1)

5 11 ,0 ,2 (d) 4 4 107. The perimeter of the triangle with vertices at (1, 0, 0), (0, 1, 0) and (0, 0, 1) is (a) 3 (b) 2 (c) 2 2 (d) 3 2 108. If a line in the space makes angle , and with the coordinate axes, then (c)

cos 2

cos 2

sin 2

cos 2

sin 2 2

(b) 0 (c) 1 (a) –1 109. The radius of the sphere x2 + y2 + z2 = 12x + 4y + 3z is (a) (b) (c) 110. lim x

x 5 x 2

(d) 52

x 3

equals (d) e5

2sin x sin 2 x , if x 0 2 x cos x , a, if x 0

then the value of a so that f is continuous at 0 is (a) 2 (b) 1 (c) –1 (d) 0 112. x

cos

1

1 1 t2

y

sin

1

, t

1 t

2

113.

dy dx is equal to

(b) tan t (d) sin t cos t

d x 1 a tan 1 x b log dx x 1 a – 2b is equal to (a) 1 (b) –1 (c) 0 a sin 1 x

(1 x 2 ) yn

114. y e equal to

(n2 a 2 ) yn

(a) (c)

(n2

a 2 ) yn

2

1

x

4

1 (d) 2

(2n 1) xyn

(b)

(n2 – a 2 ) yn

(d)

(n2 a2 ) yn

1

is

115. The function f ( x) x3 ax 2 bx c, a 2 3b has (a) one maximum value (b) one minimum value (c) no extreme value (d) one maximum and one minimum value 116.

2 sin 2 x x e dx is equal to 1 cos 2 x (b) ex cot x + c (a) – ex cot x + c x (c) 2e cot x + c (d) – 2ex cot x + c

117. If I n

sin n x dx , then

nI n (n 1) I n 2

equals (a)

sin n 1 x cos x

sin n

(c)

(b) e2 (c) e3 R is defined by

(a) e 111. If f : R f ( x)

equals sin (d) 2

(a) 0 (c) 1

1

x cos x

(b) (d)

cos n

1

cosn

x sin x 1

x sin x

118. The line x

divides the area of the region 4 bounded by y = sin x, y = cos x and x-axis

0

x

into two regions of areas A1 and 2 A2. Then A1 : A2 equals (a) 4 : 1 (b) 3 : 1 (c) 2 : 1 (d) 1 : 1 119. The solution of the differential equation dy sin ( x y ) tan ( x y ) – 1 is dx (a) cosec (x + y) + tan (x + y) = x + c (b) x + cosec (x + y) = c (c) x + tan (x + y) = c (d) x + sec (x + y) = c 120. If p (~ p q) is false, the truth value of p and q are respectively (a) F, T (b) F, F (c) T, F (d) T, T

Solved Paper 2009

2009-11

SOLUTIONS Maximum height attained

PART - I (PHYSICS) 1.

(d) Here, y a sin (bt cx) Comparing this equation with geneal wave equation

y

2 t a sin T

H max

(a) Dimensions of

2

[ L] = Dimensionless [ L]

3.

(a) Momentum, p y

[T ] 2 (b) Dimensions of bt t T [T ] = Dimensionless

2 T 2

T

t'

t2 t1 2

vx

...(i)

1 p2 m 2 2 m

1 2 mv 2

[ LT 1 ]

Hmax

B H

1 constant 2m Hence, the graph between KE and p2 will be linear. 1 2 mv 2 The velocity component at point P,

(u sin

gt ) and vx u cos Resultant velocity at point P, v v y ˆj vx iˆ gt ) ˆj u cos iˆ

(u sin A Total time taken to reach point C t1 t '

t1

2t1 t2 2

t1

1 2 p 2m

p2

or, KE

vy

t'

T

x

Now, kinetic energy KE =

C

t1

(x, y)

O Kinetic energy, KE

(b) Let t' be the time taken by the body to fall from point C to B. Then t1 2t ' t2

p m

v

v

P

[ L] x (c) Dimensions of cx [ L] = Dimensionless b (d) Dimensions of c

m v

vy

2

2.

1 g (t1 t 2 ) 2 m 8

Or, H max

y a

2

1 (t1 t2 )2 g 2 4

2 x

2 ,c T

we get b

t t 1 g 1 2 2 2

1 g (T ) 2 2

t2

t1 2 t1 t2 2

(u cos )2

|v|

u 2 cos 2

u 2 sin 2

u 2 (cos 2

(u sin

gt ) 2

g 2t 2 2ugt sin

sin 2 ) g 2 t 2 2ugt sin

EBD_7443 2009-12

Target VITEEE 1 m (u 2 2

KE

g 2t 2

If rate of flow of water is increased by n times, i.e., (nx). Increased power,

2ugt sin )

i.e., KE t 2 Hence, graph will be parabolic intercept on y-axis. Hence, the graph between KE and t. Now, in case of height 1 m(v 2 ) and v 2 2

KE

(u

2

2 gy )

mgy ' y' mg t t The ratio of power, P1

P1 P0

5.

1 m(u 2 2 gy ) 2

KE

KE

1 mu 2 2

mgy

Intercept on y-axis =

1 mu 2 2

Now, KE

(a) Mass of the first body m1 = 5 kg and for elastic collision coefficient of restitution, e = 1. u =u m1 1 M

u . 10 Let after collision velocity of M block becomes (v2). By conservation of momentum m1v1 m2 v2

m1u1 m2 u2

A

or

5u M 0

1 2 mv 2

or

5u

u 2

5

or

KE

4.

u v2 10

Mv2

...(i) e(u1 u2 )

1(u ) or

u u 10

v2

11u v2 ...(ii) 10 Substituting value of v2 in Eq. (i) from Eq. (ii)

or

x

1 x m t 2

u 10

Mv2

Since, v1 v2

KE

n :1

after collision velocity becomes

–mg = tan

O

nmgx ; P1 : P0 mgx

...(ii)

u2 = 0 Let initially body m1 moves with velocity v

B 1 2 mu 2

nmgx

2

5u

i.e., KE x 2 . Thus graph between KE and x will be parabolic. (a) Power of motor initially = P0 Let, rate of flow of motor = (x) Since, power,

P0

work time

y t

x = rate of flow of water = mgx ...(i)

mgy t

mg

y t

or 5

6.

u 11u M 2 10 1 2

M

11 10

M

9 10 2 11

45 11

4.09 kg (c) Let third mass particle (2m) moves making angle with X-axis. The horizontal component of velocity of 2m mass particle = u cos And vertical component = u sin

Solved Paper 2009

2009-13

Velocity of particle

Y

dy dt

u=?

6ms

–1

2m1

5d sin 4t dt

5cos 4t

3

m 4m

m

–1

4ms

dy dt

From conservation of linear momentum in X-direction m1u1 m2 u2 m1v1 m2 v2 or 0

or

or 4 2u cos or 2 u cos ...(i) Again, applying law of conservation of linear momentum in Y-direction 0 m 6 2m(u sin ) 6 u sin or 3 u sin ...(ii) 2 Squaring Eqs. (i) and (ii) and adding,

(4) (9) u cos

2

2

u sin

u 2 (cos2

sin 2 )

u

gR gR 2 R 2 2 Height, h gR 3 gR 2 gR 2 2 (d) Particle executing SHM.

5sin 4t

3

2

1 2 mu 2

10 9.

2 gR 2 gR gR v2 v2 4 2 2 Now, putting value of v2 in Eq. (i), we get

Displacement y

20 cos

KE

ve 2

2

T

3

20sin

3

1 2 10 2

3

( 10 3) 2

100 3 0.3J

l1 r Y a, 1 b, 1 c l2 r2 Y2 Free body diagram of the two blocks brass and steel are

(d) Given,

T'

T

Brass

Steel

R 3

2g

...(i)

3

10 3 The kinetic energy of particle,

v2 R

or v

8.

2 T

2 4 2 Now, putting value of T in Eq. (ii), we get

...(i) 2 gR v 2 Velocity of body = half the escape velocity i.e., v

3

or T =

2

u 13 ms 1 (b) Here, maximum height attained by a projectile h

T 4

20cos T

As

or 13 u 2 7.

u

3

...(ii) 3 Comparin g the given equation with standard equation of SHM. y a sin ( t ) We get, = 4

m 4 2m(u cos )

2

20 cos 4

T t 4

20cos 4t

T 4

Velocity at t

X

4

3

T

2g

EBD_7443 2009-14

Target VITEEE Let Young's modulus of steel is Y1 and of brass is Y2. F1 l1 A1 l1

Y1

F2 l2 A2 l2 Dividing Eq. (i) by (ii), F1 A1 F2 A2

Y1 Y2

...(ii)

2 r2 ( )g ...(i) 9 Let terminal velocity becomes v' after coalesce, then 6

l1 l1 l2 l2

2 R2 ( )g 9 Dividing Eq. (i) by (ii), we get v'

Y1 F1 A2 l1 l2 ...(iii) Y2 F2 A1 l2 l1 Force on steel wire from free body diagram T = F1 = (2g) newton Force on brass wire from free body diagram F2 T ' T 2 g (4 g ) newton Now, putting the value of F1, F2, in Eq. (iii), we get

or

Y1 Y2

r22

2g 4g

or c

l1 l2

r12

1 1 2 b2

a

6 v'

12.

l2 l1

l2 l1

l1 a l2 2b 2 c (d) Initially area of soap bubble, A1 = 4 r2 Under isothermal condition radius becomes 2r.

A 2( A2

A1 )

2(16 r 2 4 r 2 )

24 r 2

Energy spent, 11.

W T A T 24 r 2 24 Tr 2 J (c) Let radius of big drop = R. 4 3 R 3

4 3

r3 8

R 2r Here r = radius of small drops. Now, terminal velocity of drop in liquid

)g )g

6 v'

r2 (2r ) 2

or v' = 24 cm s–1 (a) Time period of oscillation, l g

T

2

As,

dl l

dT T

1 dl 2 l

dt

13.

1 1 9 10 7 (30 20) dt 2 2 = 4.5 × 10–6 Loss in time = 4.5 × 10–6 × 0.5 = 2.25 × 10–6 s (c) Volume of the metal at 30°C

V30

Area A2 4 (2 r )2 16 r 2 Increase in surface area

2 r2 ( 9 2 R2 ( 9

...(ii)

dT T

or

10.

)g

where is coefficient of viscosity and is density of drop is density of liquid. Terminal speed drop is 6 cm s–1

...(i)

and Y2

2 r2 ( 9

vT

loss of weight specific gravity × g

(45 25) g 13.33 cm3 1.5 g Similarly, volume of metal at 40°C V40

(45 27) g 1.25 g

Now, V40 or

V30 [1

14.40 cm 3

(t2 t1 )]

V40 V30 14.40 13.33 V30 (t 2 t1 ) 13.33(40 30) = 8.03 × 10–3/°C

Solved Paper 2009

2009-15

Coefficient of linear expansion of the metal 8.03 10 3

14.

3 (a) Process A clockwise. During A

B

3

3

2.6 10 C

D

/ C

B , pressure is constant and

1 ie., T is V constant. During process C D , both p an d V changes and process D A

follows p

A

(2500 1000 1200 700) 100 6000 3500

A is

C , process follows p

B

W1 W2 W3 W4 100 Q1 Q4

16.

R (T1 T2 ) 1

1 i.e., T is constant. V

W

B

For monoatomic gases,

p

W

V Hence, equivalent cyclic process is has follows. A

5 3

1;

1 R 3 R 42 (442 400) 5 2 1 3 or W = 63R But U 0 , for cylinder A Q 0 63R 63R For cylinder B volume is constant,

C D

B

W = 0 and Q CV T For monoatomic gas

C

3 3 R Q 1 R T 2 2 As heat given on both cylinder is same CV

p D

17.

(b) From first law of thermodynamics U W or U Q W Q U1

Q1 W1

3 KA(100 T ) l

6000 2500 3500 J

U2

Q2 W2

5500 1000

4500 J

U3

Q3 W3

3000 1200

1800 J

U 4 Q4 W4 3500 x For cyclic process U 0 3500 4500 1800 3500 x or x 700 J

Efficiency,

3 R T T 42 K 2 (a) From figure, H = H1 + H2 63R

T

15.

1000 100 10.5% 9500 (c) From first law of thermodynamics U W Q For cylinder A pressure remains constant. Work done by a system

output 100 input

300 3T

2T 100 T

KA(T l

6T

0)

400

200 C 3 (b) Let listener go from A B with velocity ( ). As Bs T

18. 0

2 KA(T 50) l

680 Hz 680 Hz And the apparent frequency of sound from source A by listener using Doppler's effect,

EBD_7443 2009-16

Target VITEEE v vo n' n v vs

constructive inference takes place. The position of fringe at p2.

340 u n ' 680 340 0

n '' n

v vo v vs

680

n D d

x

The apparent frequency of sound from source B by listener

Given, '

340 u 340 0

4

P2

Listener hear 10 beats per second. Hence, n'' – n' = 10

680

340 u 340

680

340 u 340

S1

10

d

2(340 u 340 u) 10

19.

u 2.5 ms –1 (b) When both the wires vibrate simultaneously, beats per second, n1 n2 6 or

1 2l

T m

1 T' 2l m

1 T' 2l m

6

T m

6

1 2l

21.

1 T' 1 T' 600 6 606 2l m 2l m Given that fundamental frequency 1 T 600 2l m Dividing Eq. (i) by (ii), 1 T' 2l m 1 T 2l m

T' T

...(i)

n D or n d

I1 I2

a2

1 4

16 :1 2 a 4 (a) Position fringe from central maxima

10 1 D d For second source y1

...(ii)

5

y2

2D

d

...(i)

...(ii)

10 1 D 2 1 d 5 2D 2 d (c) Inteference takes place between two waves having equal frequency and propagate in same direction. Hence, y1 a sin( t 1 ) y1 y2

T' T

(1.02)%

D d Let be the amplitude of the place where

(d) Fringe width

D 4d

n 1D d Given, n = 10

606 600

(1.01)

D

y1

22.

T ' T (1.02) Increase in tension T ' T 1.02 T (0.02T ) T = 0.02 20.

P1 S2

23.

y3 a 'sin( t 2 ) will give interference as the two waves have same frequency . (d) The two lenses of an achromatic doublet should have, sum of the product of their powers and dispersive power = zero.

Solved Paper 2009 24.

(b) Ratio of magnetic moments Ts2 Ts2

Td2 Td2

MA MB

25.

2009-17

1 20

2

1 15

2

vs2 vs2

According to Gauss theorem

vd2 vd2

1 15

2

1 20

2

0

400 225 400 225

SA

(OP 2

0.4 10

4

27.

10

0.4 10 10

4

7

26.

1

25 10 4 )3 / 2

M

0

q

2a

1

M

4

q ( z a)

0

q 4

and r = 18 × 10–2 m

l

q ( z a)

P

Z

A B (0,0,z) (0,0,–a) (0,0,a) x Total potential at P due to electric dipole V V1 V2

or 28.

l

4

y

M 1.35 A-m (a) Charge density or charge per unit length of long wire 1 Cm 3

2 0r

q

1 4

–q

(225 10 4 )3/ 2

0.4 103 10 6 (225)3/ 2

2

2 0r 2

1

V2

M 4

0 rl

4 0 (z a) Potential at P due to (–q) charge of dipole

AO 2 )3/ 2 (200 10

2

1 1 2 3 18 10 2 = 0.33 × 1011 NC–1 (c) Potential at P due to (+q) charge of dipole

V1

5 cm

7

q /l 2 0r

9 109

M

0

4

0

q

E

OP 200 cm 225 25 Since, at the neutral point, magnetic field due to the magnet is equal to BH, BH

q

or E 2 rl

0

BN O

q

E dS

M A : M B 25 : 7 (d) Here, length of magnet = 10 cm = 10 × 10–2 m, r = 15 × 10–2 m P 15 cm cm 15

5 cm

q

E dS

V

1 4

0

q ( z a)

( z a z a) ( z a)( z a)

0

2 qa 4

0 (z

2

a2 )

(c) Internal resistance of voltmeter = R. Effective resistance across B and C 1 R'

or R '

1 R

1 50

50 R 50 R

50 R 50 R

EBD_7443 2009-18

Target VITEEE

A 50 k

100V

or

l I 11 A

2.5 11 10

l

VL 30.

50 k

(c) ~

VC

Impedance, Z

( X L ~ X C )2

C According to Ohm's law, V '

100 3

or

IR '

50R 50 R

I

100 50 R I 3 50R Now, total resistance of circuit 50

Now, V '' 100

29.

...(i)

50 R 50 R

1 C

XC

102 7

2

100 50 R 2500 100 R 3 50 R (50 R )

150 R 2500 100R or R 50k (c) Here length of potentiometer wire, l = 10 m Resistance of potentiometer wire

31.

2.5 A 10

R'

2.5 A 10 Again the length of potentiometer wire is increased by 1 m. Resistance of null position Potential, V '

R '' V ''

I R'

I

l 11 A IR '' and V

V'

6

7

6

100 7

I 2 R

0

B

...(i)

R2

4

l

l 10 or R A A The value of 2.5 m length wire 10 2.5 A 10

R2

7 10 As XC > XL So, circuit behaves like R-C circuit. (c) At the centre of the loop, magnetic field

R

R'

I

1 70 103 1 10

1

IR ''

VR

R2

XL L 70 103 100 10 Capacitance reactance

(2500 100 R ) (50 R )

or R ''

2

1 C Inductive reactance L~

or Z

or

R ''

2.75 m

100 H 1 F 10

B

R V

I 2.5 A 10

B R I For the wire which is looped double let radius becomes r Then, B'

or B '

l 2

2 r or

I 2 r 2

0

r2

4 0

4

l 4

I

l 2 2 2 1 4

(r )

Solved Paper 2009 or B '

2009-19

4

I l

4

1 2

R

2

l 2

= 0.662 × 10–19 J

...(ii)

2

l 0

Now, B

2

Il 16

0

= (3.972 – 3.31) × 10–19

...(iii)

34.

I l 16 2 B' 4 l2 4 2 B 0 Il 4 4 l2 or B' = 4B (c) Magnetic field due to long wire at O point 0

0.41 eV 1.6 10 (d) From Einstein's photoelectric equation E

W0

F

evB

or r

R

l (upward) 2 R Magnetic field due to loop at O point 0

B2

0

R2

I B2 (in upward direction) 2 R Resultant magnetic field at centre O B = B1 + B2 0I

33.

2

R

36.

(

1)T

5000 10 10 m According to Einstein's photoelectric equation E W0 KE

KE

3.31 10

3.31 10

19

KE

6.62 10

19

37. 34

5000 10

3 108

e

1

e(

and N 2 10

)t

N0 e e

t

9 t

1 9 (c) Here current flows in circuit A as both (p-n) junction diode act as forward biasing. Total resistnace R.

19

or 8 I B 4 or I B 2 A (c) After collision the bullet and block move together and comes to rest after covering a distance of 40 m.

10

6.62 3 10 5 = (–3.31 × 1.324 × 3) × 10–19 3.31 10

1 e

eB

1 1 1 2 or R = 2 R 4 4 4 According to Ohm's law V I AR or 8 = IA × 2 or IA = 4A In circuit B, lower p-n-junction diode is reverse biased. Hence, no current will flow but upper diode is forward biased so current cna flow through it V IB R

(b) Work function W0 3.31 10 19 J Wavelength of incident radiation

hc

10 t

N0 e

2m( E W0 )

r

t

0

B

2( E W0 ) m eB

(d) Here, N1 N1 N2

I 2 R

4

mv eB

or r 35.

B1

mv 2 r

m

O

I

19

19

2( E W0 ) 1 2 v mv or 2 m A charged particle placed in uniform magnetic field experience a force

Dividing Eq. (ii) by (iii),

32.

0.662 10

or E

19

m

250 ms

m = 0.2 kg

–1

0.23 kg

u2 = 0

EBD_7443 2009-20

Target VITEEE By conservation of momentum m1u1 m2 u2 m1v1 m2 v2 0.02 250 0.23 0

ml 2 . 3 By law of conservation of energy

I about point A

0.02v 0.23v

5 0 v (0.25) or v 20 ms 1 Now, by conservation of energy 1 Mv 2 2

or

R d

40.

1 0.25 400 or 2

38.

mg

0.25 9.8 40

200 0.51 9.8 40 (a) Let after the time (t) the position of A is (0, vAt) and position of B = (vBt, 10). Distance between them

(0 vB t ) 2

y

(v At 10) 2

or y 2

(2t )2 (2t 10)2

or y 2

l

Now,

dl dt

4 10 0.4

41.

0

or

39.

B

CG

l/2

v Because centre of gravity of stick lies at the middle of the rod, PE of metre stick =

H;

42.

2H;

mgl 2

Rotational kinetic energy E

CH 2

CH 2

H = 218 kJ/mol H = 436 kJ/mol

1 O2 2

PdCl2 CuCl2 H 2O

CH3CHO (B)

O by double bond

H3C

Therefore, alkyne must be CH3–C

C– CH3

CH3

C =C

H

H

'A'

H2

Lindlar's catalyst

H3C 2

2 kg.m 2

Since only alkenes produce aldehydes, on ozonolysis hence 'A' must be an alkene. Now to find the structure of alkene we should add two molecules of aldehyde and replace O by double bond H3C CH3 C=O+O=C H H ‘B’ ‘B’

(alkyne)

1 I 2

16 8

436 104.3 kcal/mol 4.18 Hence, H–H bond energy is 104.3 kcal/mol. (a) In Wacker process,alkene is oxidised into aldehyde.

Replacement of

A

3gl

I 8 or I

1 H2 2

or H 2

40 d 2l 2.5 s 16 (+ve) 16 dt 2 So, l will be minimum. (b) Here, potential energy of the metre stick will be converted into rotational kinetic energy.

2

(c) Given : H f (H) = 218 kJ/mol i.e.,

t

1 ml 2 vB l 2 3

2

PART - II (CHEMISTRY)

4t 2 4t 2 100 40t

(16t 40)

1 I 2

By solving, we get vB (a) Given, r = 0.4 m, = 8 rad s–1, m = 4 kg, I = ? Torque, I mgr I

8t 2 100 40t

l

l 2

H

C =C 'A'

CH3 H

Solved Paper 2009 43.

(b)

2F2

2009-21

2NaOH

2NaF OF2

dilute

H2O

( A)

K1 K2

The structure of 'A' (OF2) is as O

44.

F 103.2 F bonds made by O = 2 Due to repulsion between two lone pairs of electrons, its shape gets distorted. Therefore, the bond angle in the molecule is 103°. (a) To decide the structure of alkene that undergoes ozonolysis, add the products and replace O by double (=) bond. Thus,

H3C

H

CH3

C=O+O=C

48.

CH3

acetone 4

acetaldehyde

of O by double bond

3

2

C= C

H

hvenergy

(a) Structure of XeO3

O O 3p -d pi bonds. Structure of XeO4 O

49.

O

O O 4p -d bonds. (a) According to de-Broglie's equation.

2 or mv

h2

2

m 2v 2 h2 m

1 2 mv 2

KE ( K )

1 h2 2m 2

6.626 10

34

50.

3 108 9

3.313 10 19 1.0 10 19 = 2.313 × 10–19 J Hence minimum amount of energy required to remove an electron from the metal ion will be 2.313 × 10–19J. (a) The earth's thermosphere also includes the region of the atmosphere, called the ionosphere. The ionosphere is the region of the atmosphere that is filled with charged

particles such as O 2 , O+, NO+. The high temperature in the thermosphere can cause molecules to ionize. (b) The formula of sulphuric anhydride is SO3 and its structure is as follows : O

2

KE( K )

hc

600 10 = 3.313 × 10–19 J Now since Threshold energy hv KE

O

h mv

6.023 104

6.023 1023 = 1.0 × 10–19 J

CH3

Xe

2

Cu 2 V2 Cr 2 Mn 2 23 (a) 1 mol = 6.023 × 10 atoms KE of 1 mol = 6.023 × 104 J or KE of 6.023 × 1023 atoms = 6.023 × 104 J

CH3

Xe

46.

1

KE of 1 atom =

2- methyl-2-butene

45.

5 3

K1 : K 2 25 : 9 (c) Paramagnetic nature depends upon the number of unpaired electrons. Higher the number of unpaired electrons, higher the paramagnetic property will be. Cu2+ = [Ar] 3d9, no. of unpaired electrons = 1 V2+ = [Ar] 3d3, no. of unpaired electrons = 3 Cr2+ = [Ar] 3d4, no. of unpaired electrons = 4 Mn2+ = [Ar] 3d5, no. of unpaired electrons = 5 Hence, correct order is

1

H3C

Replacement

47.

2 2

S O

O 3 , 1p -p , 2 p -d bonds are present.

EBD_7443 2009-22

Target VITEEE CH3

+

N2Cl

51.

57.

CH3

–

–

Cl Cu/HCl

(a)

+ N2

Gattermann reaction

H 52.

(c) H3PO3

P–H bonds

58.

H – O – P – O – H}

1 are super oxides 2 and is represented as O2–. Usually these are formed by active metals such as KO2, RbO2 and CsO2. For the salts of larger

O

an oxidation state of

O

H3PO2

P – OH

H

H

two P–H bonds

53.

(c) Dipole moment, ( ) d where, = magnitude of electric charge d = distance between particles (or bond length)

59.

HCl HI

HCl

d HI

d HCl

HI

1.03 1.6 1.3 0.38

54.

(d) SiCl4

H 4SiO 4 55.

3.3 :1

4H 2 O 1000 C

SO2

H 4SiO4

SiO2

2H 2 O

0.9 100 60% 1.5 Mass percent of Cl2 in CdCl2 = 100 – 60 = 40% 40% part (Cl2) has atomic weight = 2 × 35.5 = 71.0 60% part (Cd) has atomic weight 71.0 60 40

56.

4HCl

(c)

2Al 2NaOH 2H 2 O

106.5

2NaAlO 2 3H 2 sodium meta aluminate

Sodiummetaaluminate, thus formed, is soluble in water and changes into the complex [Al(H2 O) 2 (OH) 4 ] – , in which coordination number of Al is 6.

1 O2 2

SO3

1L

1 L 2

1L

2L

1L

2L

Since, 100 mL of oxygen is obtained by = 1 mL of H2O2 1000 mL of oxygen will be obtained by

(c) Mass percent of Cd in CdCl 2 =

anions (like O2 ), lattice energy increases in a group. Since, lattice energy is the driving force for the formation of an ionic compound and its stability, the stability of the superoxides from 'K' to 'Cs' also increases. (a) 30% solution of H 2 O 2 is known as perhydrol . H2O2 decomposes as 2H 2 O2 2H 2 O O 2 Volume strength of 30% H2O2 solution is 100 i.e., 1 mL of th is solution on decomposition gives 100 mL oxygen.

d

or

(b) Average kinetic energy per molecule 3 kT 2 3 R 3 8.314 T 300 or 2 N0 2 6.023 10 23 = 6.21 × 10–21 JK–1 molecule–1 (c) The species having an O–O bond and O in

1 1000 mL of H O = 10 mL of H O 2 2 2 2 100

60.

dCHA , d pH where, dCHA = no. of moles of acid added per litre dpH = change in pH

(d) Buffer capacity,

dCHA = =

moles of acetic acid volume 0.12/60 250/1000

1/125 0.02

1 2.5

1 125 0.4

Solved Paper 2009 61.

62.

63.

2009-23

(b) (A) Felspar (orthoclase) (KAlSi3O8) (B) Asbestos {CaMg3(SiO3)4} (C) Pyrargyrite (Ruby silver) (Ag3SbS3) (D) Diaspore (Al2O3 . H2O) (c) Along a period first ionisation energy increases. Thus, the first IE of the elements of the second period should follow the order Be < B < N < O But in practice. The first IE of these elements follows the order B < Be < O < N The lower IE of B than that of Be is because in B (1s2, 2s2 2p1), electron is to be removed from 2p which is easy while in Be (1s2, 2s2), electron is to be removed from 2s which is difficult. The low IE of O than that of N is because of the half-filled 2p orbitals in N (1s2, 2s2 2p3). (c) CH 3 CH 2 OH Cl2 CH 3CHO 2HCl

KCN

EtOH

H 3O

CH 2 CH 2 COOH | OH

66.

(c) Tripeptides are amino acids polymers in which three individual amino acid units, called residues, are linked together by amide bonds. glycine (NH2 –CH2 –COOH), alanine (CH3

CH COOH) and phenyl alanine | NH2

C6 H5 CH 2

X

C Cl3 CHO

3HCl

64.

(c) I.

CH3 COO

Ca

CH3CO O

II. CH3 COOH Red P

III. 65.

–CaCO3

P4O 10

(b) C = 85.71%

CH3CO CH3CO

CH3COO H

85.71 12

68.

7.14;

CH2CH2NH2 2H 2O O + H2O

7.14 7.14

1

14.29 14.29 2 14.29; 7.14 1 Empirical formula = CH2 and, empirical formula weight = 12 + 2 = 14 Now since molecular formula weight = 2 × vapour density = 2 × 14 = 28

69.

OH OH (a) Freezing point of a substance is the temperature at which the solid and the liquid forms of the substance are in equilibrium.

70.

(b)

H = 14.29% =

28 2 14 Molecular formula = (CH2)2 = C2H4

n

CH 2

( A)

CH 2 HOCl

CHCOOH can be linked | NH2

in six different ways. (d) A codon is a specific sequence of three adjacent bases on a strand of DNA or RNA that provides genetic code information for a particular amino acid. (c) The IUPAC name of dopamine is 2-(3,4-dihydroxyphenyl) ethylamine and its structure is as follows :

6HI

CH3 CH3 3I 2

CH3CO OH

67.

Y chloral

CH3COCH3

CN

(C )

Acetaldehyde 3Cl 2

CH 2 CH 2 | OH

CH 2 CH 2 Cl | OH (B)

EtOH / H2 O

C 2 H 5 Cl AgCN

C2H 5

71.

N C AgCl

(N-linked to ethyl carbon)

(a) For the given cell, Ag | Ag+ | AgCl | Cl– | Cl2, Pt The cell reactions are as follows At anode :

Ag

At cathode : AgCl e Net cell reaction : AgCl

Ag

e

Ag( s ) Cl Ag

Cl

EBD_7443 2009-24

Target VITEEE G oreaction

G op

G Ro

= (78 – 129) – (–109) = + 58 kJ/mol G

o

nFE

58 10 J

72.

58 1000 0.6 V 96500 (d) Crotonaldehyde is produced by the aldol condensation of acetaldehyde.

H

H

CH3 – C = O + H – C – CHO acetaldehyde

H

1 1 or K 2 0.1 100 10 (c) For a first order reaction,

(nucloeophilic addition)

t

OH H

CH3CH = CHCHO

t1 t2

crotonaldehyde

aldol

73.

(b)

BaCl2

2NaOH

m Ba(OH) 2

Ba(OH) 2

m BaCl2

2

m NaOH

m NaCl

10 3

8.92 6.023 10 (362 10 ) 63.55 = 4.0 Thus, metal has face centred unit cell.

75.

(c)

N2

K1

0.6020 0.1249

dN 0 a3 M 23

2O 2

2NO 2

77. 78.

[ ]

[ ] observed l C

[NO2 ]2

or 100

[NO2 ]2 [N 2 ][O2 ]2

4.8 ..(i)

4.81

1.2 39 6.15 5 1000 (a) Let the concentration of potassium acetate is x. According to Henderson's equation,

(c)

pH = pK a

[N 2 ][O2 ]2

log 4 log 4 / 3

2 0.3010 2 0.3010 0.4771

N0 a3

where, Z = number of atoms in unit cell

Z

remains constant]

log 4 log 4 log 3

MZ

(b) Density, d

100 100 75 [ 100 log 100 25

log

log

= 280 × 10–4 + 2 × 248 × 10–4 – 2 × 126 × 10–4 = (280 + 496 – 252) × 10–4 = 524 × 10–4 Sm2 mol–1 74.

a

log10

100 25 100 log 75

2NaCl 2

2.303

a x Let initial amount of reactant is 100.

H H –H2O (elimination)

1

K 22

acetaldehyde

CH3 – C – C – CHO

...(ii)

[NO 2 ]2

Eqs. (i) × (ii), we get 100 K 22

76.

Dil. NaOH

[N 2 ][O 2 ]2

or K 22

o Ecell

O2

[N 2 ]1/2 [O2 ] [NO2 ]

K2

o o 1 96500 Ecell

3

1 N2 2

Again, [NO2 ]

log

[salt] [acid]

log (1.8 10 5 ) log

4.8 4.74 log 25x or log 25 x

0.06

x 50 20 0.1M

Solved Paper 2009

2009-25

25x 1.148

x 79.

84.

0.045M C 2

(b) By ' 2 A

80.

259 418 2

[1 x ]2

H 580.5 581 kJ (a) According to Hardy Schulze rule, greater the valency of the coagulating ion, greater is its coagulating power. Thus, out of the

PART - III (MATHEMATICS)

85.

(c)

82.

83.

x2 1 f (x) is either increasing or decreasing. At x = 2, f (2) = 23 + 3(2) – 2 = 12 At x = 3, f (3) = 33 + 3(3) – 2 = 34 f ( x ) [12, 34] . (c) The total number of subsets of given set is 29 = 512 Even numbers are {2, 4, 6, 8}. Case I : When selecting only one even number = 4C1 = 4 Case II : When selecting only two even numbers = 4C2 = 6 Case III : When selecting only three even numbers = 4C3 = 4 Case IV : When selecting only four even numbers = 4C4 = 1 Required number of ways = 512 – (4 + 6 + 4 + 1) – 1 = 496 [Here, we subtract 1 for due to the null set] (a) The required number of ways = The even number of 0's i.e., {0, 2, 4, 6, …} n! n! n! n ! 2!(n 2)! 4!( n 4)! n

C0

n

C2

n

C4 ... 2

n 1

32 31x 64 (neglecting x2 term) ( x a)( x a 1) ( x a 1)( x a 2)

t 2 t t 2 3t 2 t 3 2t

3x 2 3 0

0

1

( x a )( x a 2) Let x – a = t, then t (t 1) (t 1)(t 2) t (t 2) 0

f '( x) 3 x2 3 Put f '( x)

1 5 x 5 32 (neglecting higher powers of x)

1

(1 x)(32 x) 64

x3 3 x 2

(b) Given, f ( x)

1/5

1 x (1 x) 1 32 2

given, AlCl3 (Al3 ) is most effective for causing coagulation of As2S3 sol.

81.

(32 5x)

3 2x 5 x (32) 1/ 5 1 2 3 32 (neglecting higher powers of x)

1

Na 2SO4 ;

2 146

3/ 2

2x 3

1

1/5

B ' , we get

Na 2 O SO3 H

(a)

3t

2

6t 2

6

t

x

0

0

36 24 2(3) 3

x a

0

6 2 3 2(3)

3 3

a

3

3

3 Hence, x is real and distinct.

86.

(b)

f ( x)

x2 ax b has imaginary roots.

Discriminant, D f '( x ) 2 x a f ''( x ) Also, f ( x ) x2

x x

2

a 2 4b

0

0

2

f '( x)

f ''( x)

0

...(i)

ax b 2 x a 2 0

(a 2) x b a 2 0 ( a 2)

( a 2) 2 2

( a 2)

a2 2

4( a b 2)

4b 4

EBD_7443 2009-26

Target VITEEE Since, a 2

4b

3x 2 3 y 2

0

a 2 4b 4 0 Hence, Eq. (i) has imaginary roots. 87.

(c)

90.

(c)

2 x4 13x 2 ax b is divisible by

f ( x)

3i) n (1

(1

1

2

2a b

2(1)4 13(1)2 a b

and f (1)

88.

0

...(ii) a b 11 On solving Eqs. (i) and (ii), we get a = 9, b = 2 (d) Let a and R be the first term and common ratio of a GP respectively. So, Tp

aR

aR q

Tq

1

log x log y

aRr

1

92.

q 1

log a (q 1) log R

q 1

log z

r 1

log a (r 1) log R

r 1

log a

p 1

( p 1) log R

p 1

log a

q 1

(q 1) log R

q 1

log a

r 1

(r 1)log R

r 1

z 2i 2z i

(2 x )

x2

2

1

1 2(sin 2 x )2 4

1

1 (1 cos 4 x ) 4

sin x cos 2 x

2 4

r 1 1

3 4

2 sin 2 x

1 cos x (2cos 2 1)

2 2sin x cos x

2

3cos x 2sin x cos x 0 cos x (2sin x 3cos x) 0 cos x

93.

(d)

( 2sin x 3cos x

0,

x

2n

x

(4n 1) , n 2

cos

1 2

1

2sin

Z 1 2

1

3cos

1

1

2

4 tan 1 ( 1) cos

1 2

1

2

cos

3

6

1

1 2

4 tan 1(1)

( y 2) 2

y2 4 4 y

0)

2

1

(2 y 1)

cos 4 x 4

2

2

(C2 C2 C3 ) two columns are identical)

=0+0= 0 ( (c) Let z = x + iy

( x)2

r 1

sin 4 x cos4 x

2

p 1 p 1 1

log a 1 q 1 log R q 1 q 1 1

Given :

(c)

p 1

log y

1 r 1

( 2)n

)

Period of f ( x)

log a ( p 1) log R

2

(sin 2 x cos2 x)2 2sin 2 x cos2 x

z

log a (r 1) log R

p 1

2

(d) Let f ( x)

log a (q 1)log R

1 p 1

89.

91.

x

y and Tr

n

3i

( 2 )n

)

( 2)n (

log a ( p 1) log R

and log z log x

p 1

1

( 2)n [( 2 )3r 1 ( )3r 1 ] ( n = 3r + 1, where r is an integer)

...(i)

20

2 n

1

2

2

( 2

y2

3i) n n

3i

( x 2)( x 1) .

f (2) 2(2)4 13(2) 2 a (2) b 0

x2

3

2

1

4x2 4 y2 1 4 y

3

3

3

3

3 4

3

4 43 12

4

4

Solved Paper 2009 94.

2009-27

(c) We know that, 2s = a + b + c

3 p2 2 p 1 0 ( p 1)(3 p 1) 0

(a b c)(b c a)(c a b)(a b c) 4b2c2

2 s (2 s 2a )(2 s 2b)(2 s 2c ) 4b c

98.

s( s a) (s b)( s c) 4 bc bc

A A 4cos sin 2 2 2 2

95.

(c)

0,

l m n

and l 2

n2

( m n) 2

2m2

d

2

sin A

l

m2

m n

99.

0

m2 n 2

10 5

42 32 d 2 diameter of the circle Radius of circle = 1 Area of circle = r2 = sq unit

x2 2 xy xy 2 y 2

0 0

x 2 y, x y Also, x + y + 1 = 0 On solving Eqs. (i) and (ii), we get

m 0 or m n 0 If m = 0, then l = –n

m2 1

15 5

( x 2 y )( x y )

2m(m n) 0

l2 0

(c)

0

2mn 0

l1 m1 n 1 0 1 and if m + n = 0

1 ( p cannot be negative) 3 (d) Given lines are parallel. p

2 2

A

2 1 , ,B 3 3

1 1 , , C (0, 0) 2 2

2 3 1 1 2 2 0

n , then l = 0

m

n2 1

Area of ABC

ie., (l1 , m1 , n1 ) ( 1, 0, 1) and (l2 , m2 , n2 )

(0, 1, 1)

0 0 1

cos

1 0 1 0 1 1

(a)

x2 3xy 2 y 2

m1

| a1 |

m2

| a2 |

m3

| a3 |

and m4 97.

22

( 1)2

(1)2

32 ( 4)2 12 12

| a4 |

m3 m1 m4 m2 (a) Here, n = 6 According to the question 6

C2 p 2 q 4 q2

(1 p)2

0

and ( x 2 y 2)( x

x 2y 11

y)

x 2 0

0, x y

and x 2 y 2

y 1)

0

0

0, x y 1 0

The lines x 2 y

0, x 2 y 2 0 and

x y 0, x y 1 0 are parallel. Also, angle between x – 2y = 0 and x – y = 0 is not 90°. It is a parallelogram.

4 6C4 p 4 q 2

4 p2

( x 2 y )( x 41

3

( 1)2 (3)2 (1)2

0

and x2 3 xy 2 y 2

6

( 4) 2

( 1) 2

1 1 3 1 1 2 0 1

1 1 1 1 1 1 2 6 12 2 3 6 100. (c) Given pair of lines are

1 2

3

96.

...(i) ...(ii)

4 p2

EBD_7443 2009-28

Target VITEEE OAC, OC2 = 22 + 42 = 20

101. (a) In

y 2

y = 4x

x'

C(±2, ±4)

4 4

A

2 32

O

y'

Also, point (1, 0) is the focus of the parabola. It is clear from the graph that only one normal is possible.

Required equation of circle is

( x 2)2 ( y 4)2

20

x2 y 2 4 x 8 y 102. (d) Given circles are x

2

y

and x

2

r1

y

2

4), C2

(1,

C1C2 d2

cos

3)

2

2

( 3 4) 2 4 2

2 2

2

2 1

2

2

135 103. (b) Let the required equation of circle be 2

x y 2 gx 2 fy 0 The above circle cuts the given circles orthogonally. 2( 3 g ) 2 f (0)

and 2 g 2 f

2g

8

7

x2

y2

7

8 x 3

29 y 3

5 and 2 x2 3 y 2

8 3

5 3y 4

2

3y2

0

12

12 12

15 y 2 30 y 71 0 y

30

900 4260 30

10 x 2 40

3360 30

12

40 x 61 0 1600 4 10 61 2 10

40

840 20

and B 2

1

2

4x 5 3

2 Also, 2 x 3

Points are A 2

0

or 3 x2 3 y 2 8 x 29 y 104. (b) Given curve is y2 = 4x.

c4

(25 9 y 2 30 y ) 3y2 8

x

8 29 3 3 Required equation of circle is

2f

y2 )

106. (b) Solving 4 x 3 y 2

r12 r22 2r1r2

c4

y 4 a 2 y 2 c4 0 Let y1, y2, y3 and y4 are the roots. y1 y2 y3 y4 0

2

(2 1)

x2 y 2

y 2 (a2

(2,

4 9 11

So, d

105. (a)

4 x 6 y 11 0

1 16 13

and r2

0

2 x 8 y 13 0

2

Here C1

2

x

(1, 0)

840 ,1 20

840 20

2

840 ,1 20

3360 . 30

Mid point of AB is (2, 1).

3360 30

Solved Paper 2009

2009-29

107. (d) Let A = (1, 0, 0), B = (0, 1, 0) and C = (0, 0, 1) So, AB

2

(0 1)

(1 0)

02 (0 1)2

BC

2

0

2

(1 0)2

2 2

and CA (1 0)2 02 (0 1) 2 Perimeter of triangle = AB + BC + CA 2 2 cos 2 cos 2 cos 108. (c) + sin

2

2

sin

sin 2 ) (cos2

(cos2

sin 2 ) sin 2

2

2

x

y

2

z

2

and y

2

sin 2 ) sin 2

sin 2

y

36 4

110. (c)

x 5 x 2

lim

x

lim

x

1

9 4

lim 1 x 2 3

3

3 2

169 4

3

111. (d)

lim f ( x)

x

0

lim x

0

lim

0

1 x

x b log

1 ,b 2

a

2

1 1 x

2

1

1 tan 2

1 4

1

1

y e a sin x On differentiating w.r.t. x, we get y1

e a sin

1

x

a

y1 1 x 2

1 1 x2

ay

(1 x a2 y2 Again differentiating w.r.t. x, we get 2

) y12

(1 x 2 )2 y1 y2 2 xy12 (1 x 2 ) y2

dx

x 1 x 1

1 1 2 2 4

a 2b

0 form 0

x 1 x 1

x 1 1 log 4 x 1

3( x 3) x 2

2sin x sin 2 x 2 x cos x

2 cos x 2 cos 2 x 2(cos x x sin x )

1

x 2

e3

tan 1 t

1

x b log

a tan

114. (c)

x

1

x 3

2 sin x sin 2 x , if x 0 2 x cos x if x 0 a,

f ( x)

dy dx

x

1 2

13 2

1

e

tan 1 t and y

2

x 2

3 x lim 3 2 x 1 x

1 t2

d x 1 1 a tan 1 x b log 4 dx x 1 x 1 On integrating both sides, we get

a tan

(2)2

x 3

x

0

3 . 2

(6)2

Radius of sphere

113. (b)

t

1

sin

x

2

Centre of sphere is 6, 2,

1 t2

3 2

sin

12 x 4 y 3z

1

1

cos

112. (c) We have, x

cos cos 1 cos 109. (a) Given equation of sphere is 2

0

x

2

2

2

(cos2

2 2 0 x 0 2(1 0) Since, f (x) is contineous at x = 0 f (0) = lim f ( x) a 0 lim

a 2 2 yy1

xy1 a 2 y

0

1

x

EBD_7443 2009-30

Target VITEEE Using Leibnitz's rule,

(1 x 2 ) yn

n

2

n

xyn

1

(1 x 2 ) yn

2

n

C1 yn 1 ( 2 x)

C2 yn ( 2)

a 2 yn

C1 yn

Y

118. (d)

0

y = sin x

xyn 1 ( 2n 1)

X'

yn [ n(n 1) n a ] 0 (2n 1) xyn

2

1

3b

Y'

f '( x) 3x 2 2ax b Put f '( x) 3x 2 x

/4

Area, A1

0 0 2

4a 2 3

[cos x]0 / 4

12b

and area, A2

cosec 2 xe x dx cot xe x

cos x dx /4

[sin x]

2 sin 2 x x e dx 1 cos 2 x

2 1

A1 : A2

119. (b)

cot x e x dx

( cot x ) e x dx

2

/2 /4

1 2 1

:

2

1

2 1

2

2

1:1

dy sin ( x y) tan ( x y ) 1 dx Put x + y = z dy dx

1

cot x e x dx c

dz dx

dz 1 sin z tan z 1 dx

c

cos z

dz dx sin 2 z Putting sin z t cos z dz

n

sin x dx, then

sin n 1 x cos x n 1 In n n where n is a positive integer. In

nI n (n 1) I n

2

/2

2 2sin x cos x x e dx 2sin 2 x

In

1

2

Since, a 2 3b , x has an imaginary value. Hence, no extreme value of x exists.

cot xe x 117. (c) We know that, if

1

2 1

2a 2 a 2 3b 3

116. (a) Let I

sin x dx 0

2ax b 2a

X

(n2 a 2 ) yn

x 3 ax 2 bx c, a 2

115. (c) Given, f ( x )

/2

/4

O

2

(1 x 2 ) yn

y = cos x

2

1 dt t2

2

sin n 1 x cos x

x c

cosec z

120. (c)

x c

1 t

dt , we have

x c x cosec( x

y) c

p (~ p q) is false means p is true and ~ p q is false. p is true and both ~p and q are false. p is true and q is false.

SOLVED PAPER 6.

PART - I (PHYSICS) 1.

2.

3.

4.

5.

Two beams of light will not give rise to an interference patern, if (a) they are coherent (b) they have the same wavelength (c) they are linearly polarized perpendicular to each other (d) they are not monochromatic A slit of width 'a' is illuminated with a monochromatic light of wavelength from a distant source and the diffraction pattern is observed on a screen placed at a distance 'D' from the slit. To increase the width of the central maximum one should (a) decrease D (b) decrease a (c) decrease (d) the width cannot be changed A thin film of soap solution (n = 1.4) lies on the top of a glass plate (n = 1.5). When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths 420 and 630 nm. The minimum thickness of the soap solution is (a) 420 nm (b) 450 nm (c) 630 nm (d) 1260 nm If the speed of a wave doubles as it passes from shallow water into deeper water, its wavelength will be (a) unchanged (b) halved (c) doubled (d) quadrupled A light whose frequency is equal to 6 × 1014 Hz is incident on a metal whose work function is 2eV. h 6.63 10

34

Js

1eV 1.6 10

19

J

The maximum energy of the electrons emitted will be (a) 2.49 eV (b) 4.49 eV (c) 0.49 eV (d) 5.49 eV

7.

8.

9.

10.

11.

VITEEE 2008

An electron microscope is used to probe the atomic arrangements to a resolution of 5Å. What should be the electric potential to which the electrons need to be accelerated ? (a) 2.5 V (b) 5 V (c) 2.5 kV (d) 5 kV Which phenomenon best supports the theory that matter has a wave nature ? (a) Electron momentum (b) Electron diffraction (c) Photon momentum (d) Photon diffraction The radioactivity of a certain material drops to 1 of the initial value in 2 hours. The half life of 16 this radionuclide is (a) 10 min (b) 20 min (c) 30 min (d) 40 min An observer 'A' sees an asteroid with a radioactive element moving by at a speed = 0.3 c and measures the radioactivity decay time to be TA. Another observer 'B' is moving with the asteroid and measures its decay time as TB. Then TA and TB are related as below (a) TB < TA (b) TA = TB (c) TB > TA (d) Either (A) or (C) depending on whether the asteroid is approaching or moving away from A 234U has 92 protons and 234 nucleons total in its nucleus. It decays by emitting an alpha particle. After the decay it becomes (a) 232U (b) 232Pa 230 (c) Th (d) 230Ra K and K x-rays are emitted when there is a transition of electron between the levels (a) n = 2 to n = 1 and n = 3 to n = 1 respectively (b) n = 2 to n = 1 and n = 3 to n = 2 respectively (c) n = 3 to n = 2 and n = 4 to n = 2 respectively (d) n = 3 to n = 2 and n = 4 to n = 3 respectively

EBD_7443 2008-2

12.

Target VITEEE radioactive material ZXA starts emitting

A certain and particles successively such that the end product is Z 3 Y A 8 . The number of particles emitted are (a) 4 and 3 respectively (b) 2 and 1 respectively (c) 3 and 4 respectively (d) 3 and 8 respectively

1k +1 V

15.

16.

17.

and

19.

10k

13.

14.

18.

– +

A

Vout

In the circuit shown above, an input of 1V is fed into the inverting input of an ideal Op-amp A. The output signal Vout will be (a) +10 V (b) –10 V (c) 0 V (d) infinity When a solid with a band gap has a donor level just below its empty energy band, the solid is (a) an insulator (b) a conductor (c) a p-type semiconductor (d) an n-type semiconductor A p-n junction has acceptor impurity concentration of 1017 cm–3 in the p-side and donor impurity concentration of 1016 cm–3 in the n-side. What is the contact potential at the junction (kT = thermal energy, intrinsic carrier concentration ni = 1.4×1010 cm–3) ? (a) (kT/e) ln (4×1012) (b) (kT/e) ln (2.5×1023) (c) (kT/e) ln (1023) (d) (kT/e) ln (109) A Zener diode has a contact potential of 1V in the absence of biasing. It undergoes Zener breakdown for an electric field of 106 V/m at the depletion region of p-n junction. If the width of the depletion region is 2.5 m, what should be the reverse biased potential for the Zener breakdown to occur ? (a) 3.5 V (b) 2.5 V (c) 1.5 V (d) 0.5 V In Colpitt oscillator the feedback network consists of (a) two inductors and a capacitor (b) two capacitors and an inductor (c) three pairs of RC circuit (d) three pairs of RL circuit

20.

21.

22.

The reverse saturation of p-n diode (a) depends on doping concentrations (b) depends on diffusion lengths of carriers (c) depends on the doping concentrations and diffusion lengths (d) depends on the doping concentrations, diffusion length and device temperature A radio station has two channels. One is AM at 1020 kHz and the other FM at 89.5 MHz. For good results you will use (a) longer antenna for the AM channel and shorter for the FM (b) shorter antenna for the AM channel and longer for the FM (c) same length antenna will work for both (d) information given is not enough to say which one to use for which The communication using optical fibers is based on the principle of (a) total internal reflection (b) Brewster angle (c) polarization (d) resonance In nature, the electric charge of any system is always equal to (a) half integral multiple of the least amount of charge (b) zero (c) square of the least amount of charge (d) integral multiple of the least amount of charge The energy stored in the capacitor as shown in Fig. (a) is 4.5×10–6 J. If the battery is replaced by another capacitor of 900 pF as shown in Fig. (b), then the total energy of system is

+ + + + +

– – – – – 900pF

+– 100V Fig. (a) (a) 4.5 × 10–6 J (c) zero

– – – – 900pF + –– + – + – + 900pF Fig. (b) + + + +

(b) 2.25 × 10–6 J (d) 9×10–6 J

Solved Paper 2008 23.

2008-3

Equal amounts of a metal are converted into cylindrical wires of different lengths (L) and cross-sectional area (A). The wire with the maximum resistance is the one, which has (a) length = L and area = A (b) length =

28.

Three resistances of 4 each are connected as shown in figure. If the point D divides the resistance into two equal halves, the resistance between point A and D will be A

L and area = 2A 2

A 2 (d) all have the same resistance, as the amount of the metal is the same If the force exerted by an electric dipole on a charge q at a distance of 1m is F, the force at a point 2m away in the same direction will be

4

(c) length = 2L and area =

24.

(a)

F 2

(b)

25.

D

B (a)

F 4

F F (d) 6 8 A solid sphere of radius R1 and volume charge

29.

0

density

is enclosed by a hollow sphere r of radius R2 with negative surface charge density , such that the total charge in the system is zero. 0 is a positive constant and r is the distance from the centre of the sphere. The ratio R2 R1 is

(a) (c) 26.

27.

30.

0 0

/ 2

(b)

2 / 0

(d)

0

A solid spherical conductor of radius R has a spherical cavity of radius a (a < R) at its centre. A charge + Q is kept at the centre. The charge at the inner surface, outer surface and at a position r (a < r < R) are respectively (a) + Q, – Q, 0 (b) – Q, + Q, 0 (c) 0, – Q, 0 (d) + Q, 0, 0 A cylindrical capacitor has charge Q and length L. If both the charge and length of the capacitor are doubled, by keeping other parameters fixed, the energy stored in the capacitor (a) remains same (b) increases two times (c) decreases two times (d) increases four times

31.

12

C

4

(b)

6

1 3 The resistance of a metal increases with increasing temperature because (a) the collisions of the conducting electrons with the electrons increase (b) the collisions of the conducting electrons with the lattice consisting of the ions of the metal increase (c) the number of conduction electrons decreases (d) the number of conduction electrons increases In the absence of applied potential, the electric current flowing through a metallic wire is zero because (a) the electrons remain stationary (b) the electrons are drifted in random direction with a speed of the order of 10–2 cm/s

(c)

(c)

4

3

(d)

(c) the electrons move in random direction with a speed of the order close to that of velocity of light (d) electrons and ions move in opposite direction A meter bridge is used to determine the resistance of an unknown wire by measuring the balance point length l. If the wire is replaced by another wire of same material but with double the length and half the thickness, the balancing point is expected to be 1 8 (c) 8

(a)

1 4 (d) 16

(b)

EBD_7443 2008-4

32.

33.

34.

35.

36.

Target VITEEE

Identify the INCORRECT statement regarding a superconducting wire (a) transport current flows through its surface (b) transport current flows through the entire area of cross-section of the wire (c) it exhibits zero electrical resistivity and expels applied magnetic field (d) it is used to produce large magnetic field A sample of HCl gas is placed in an electric field 3×104 NC–1. The dipole moment of each HCl molecule is 6×10–30cm. The maximum torque that can act on a molecule is (a) 2 × 10–34 C2Nm–1 (b) 2 × 10–34 Nm (c) 18 × 10–26 Nm (d) 0.5×1034 C–2 Nm–1 When a metallic plate swings between the poles of a magnet (a) no effect on the plate (b) eddy currents are set up inside the plate and the direction of the current is along the motion of the plate (c) eddy currents are set up inside the plate and the direction of the current oppose the motion of the plate (d) eddy currents are set up inside the plate When an electrical appliance is switched on, it responds almost immediately, because (a) the electrons in the connecting wires move with the speed of light (b) the electrical signal is carried by electromagnetic waves moving with the speed of light (c) the electrons move with the speed which is close to but less than speed of light (d) the electrons are stagnant. Two identical incandescent light bulbs are connected as shown in the Figure. When the circuit is an AC voltage source of frequency f, which of the following observations will be correct ? R C

Bulb b1

R L Bulb b2

(a) both bulbs will glow alternatively (b) both bulbs will glow with same brightness provided frequency f

37.

38.

39.

40.

1 2

1/ LC

(c) bulb b1 will light up initially and goes off, bulb b2 will be ON constantly (d) bulb b1 will blink and bulb b2 will be ON constantly A transformer rated at 10 kW is used to connect a 5kV transmission line to a 240V circuit. The ratio of turns in the windings of the transformer is (a) 5 (b) 20.8 (c) 104 (d) 40 Three solenoid coils of same dimension, same number of turns and same number of layers of winding are taken. Coil 1 with inductance L1 was wound using a Mn wire of resistance 11 /m; Coil 2 with inductance L2 was wound using the similar wire but the direction of winding was reversed in each layer; Coil 3 with inductance L3 was wound using a superconducting wire. The self inductance of the coils L1, L2, L3 are (a) L1 = L2 = L3 (b) L1 = L2; L3 = 0 (c) L1 = L3; L2 = 0 (d) L1 > L2 > L3 Light travels with a speed of 2 × 108 m/s in crown glass of refractive index 1.5. What is the speed of light in dense flint glass of refractive index 1.8 ? (a) 1.33 × 108 m/s (b) 1.67 × 108 m/s 8 (c) 2.0 × 10 m/s (d) 3.0 ×108 m/s A parallel beam of fast moving electrons is incident normally on a narrow slit. A screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statement is correct ? (a) diffraction pattern is not observed on the screen in the case of electrons (b) the angular width of the central maximum of the diffraction pattern will increase (c) the angular width of the central maximum will decrease (d) the angular width of the central maximum will remain the same

Solved Paper 2008

2008-5

PART - II (CHEMISTRY) 41.

42.

43.

44.

45.

46.

47.

48. 49.

CH 3CH 3

HNO 3

675 K

(a) CH3CH2NO2 (b) CH3CH2NO2 + CH3NO2 (c) 2CH3NO2 (d) CH2 = CH2 When acetamide is hydrolysed by boiling with acid, the product obtained is : (a) acetic acid (b) ethyl amine (c) ethanol (d) acetamide Which will not go for diazotization ? (a) C6H5NH2 (b) C6H5CH2NH2 (c) H 2N (d) H2N C 6H 4 C6H4 H 3C O2N Secondary nitroalkanes can be converted into ketones by using Y. Identify the Y from the following R R CHNO2 + Y C=O R R (b) Aqueous NaOH (a) Aqueous HCl (c) KMnO4 (d) CO Alkyl cyanides undergo Stephen reduction to produce (a) aldehyde (b) secondary amine (c) primary amine (d) amide The continuous phase contains the dispersed phase throughout, Example is (a) Water in milk (b) Fat in milk (c) Water droplets in mist (d) Oil in water The number of hydrogen atoms present in 25.6 g of sucrose (C12H22O11) which has a molar mass of 342.3 g is (a) 22 × 1023 (b) 9.91 × 1023 23 (c) 11 × 10 (d) 44 × 1023 Milk changes after digestion into : (a) cellulose (b) fructose (c) glucose (d) lactose Which of the following sets consists only of essential amino acids ? (a) Alanine, tyrosine, cystine (b) Leucine, lysine, tryptophane (c) Alanine, glutamine, lycine (d) Leucine, proline, glycine

50. 51. 52.

53.

54. 55.

56.

Which of the following is ketohexose ? (a) Glucose (b) Sucrose (c) Fructose (d) Ribose The oxidation number of oxygen in KO3, Na2O2 is (a) 3, 2 (b) 1, 0 (c) 0, 1 (d) –0.33, –1 Reaction of PCl3 and PhMgBr would give (a) bromobenzene (b) chlorobenzene (c) triphenylphosphine (d) dichlorobenzene Which of the following is not a characteristic of transition elements ? (a) Variable oxidation states (b) Formation of colored compounds (c) Formation of interstitial compounds (d) Natural radioactivity Cl - P - Cl bond angles in PCl5 molecule are (a) 120° and 90° (b) 60° and 90° (c) 60°and 120° (d) 120° and 30° The magnetic moment of a salt containing Zn 2+ ion is (a) 0 (b) 1.87 (c) 5.92 (d) 2 The number of formula units of calcium flouride CaF2 present in 146.4 g of CaF2 are (molar mass of CaF2 is 78.08 g/mol) (a)

1.129 1024 CaF2 (b) 1.146 1024 CaF2

7.808 1024 CaF2 (d) 1.877 1024 CaF2 The IUPAC name of the given compound

(c)

57.

Co NH3

58.

5

Cl Cl 2 is

(a) pentaamino cobalt chloride chlorate (b) cobalt pentaamine chloro chloride (c) pentaamine chloro cobalt (III) chloride (d) pentaamino cobalt (III) chlorate When SCN– is added to an aqueous solution containing Fe(NO3)3,the complex ion produced is 2

SCN –

(a)

Fe OH 2

2

(b)

Fe OH 2

5

(c)

Fe OH 2

8

(d)

Fe OH 2 SCN

2

SCN –

2

SCN – 6

EBD_7443 2008-6

59. 60.

61.

Target VITEEE

Hair dyes contain (a) copper nitrate (b) gold chloride (c) silver nitrate (d) copper sulphate Schottky defects occurs mainly in electrovalent compounds where (a) positive ions and negative ions are of different size (b) positive ions and negative ions are of same size (c) positive ions are small and negative ions are big (d) positive ions are big and negative ions are small The number of unpaired electrons calculated in Co NH3

62.

3

6

and Co F6

3

64.

65.

66.

67.

69.

70.

are

(b) 0 and 2 (a) 4 and 4 (c) 2 and 4 (d) 0 and 4 The standard free energy change of a reaction is G 115kJ at 298 K. Calculate the equilibrium constant k p in log k p

63.

68.

R 8.314 Jk 1 mol 1 . (a) 20.16 (b) 2.303 (c) 2.016 (d) 13.83 If an endothermic reaction occurs spontaneously at constant temperature T and P, then which of the following is true ? (a) G>0 (b) H0 (d) S AB) is (a) circle (b) ellipse (c) parabola (d) hyperbola The directrix of the parabola y2 + 4x + 3 = 0 is (d)

2

2

The equation r 2r.c h 0, c h, represents (a) circle (b) ellipse (c) cone (d) sphere The simplified expression of sin (tan–1 x), for any real number x is given by 1

(a) 1098 (b) 1096 (c) 1097 (d) 1095 The shortest distance between the straight lines through the points A1 = (6, 2, 2) and A2 = (–4, 0, –1), in the directions of (1, –2, 2) and (3, –2, –2) is (a) 6 (b) 8 (c) 12 (d) 9 The center and radius of the sphere x2 + y2 + z2 + 3x – 4z + 1 = 0 are

(b)

89.

2

(a)

84.

1

(a) 45° (b) 135° (c) 225° (d) 240° In a triangle ABC, the sides b and c are the roots of the equation x2 – 61x + 820 = 0 and 4 A = tan–1 , then a2 is equal to 3

(a)

If a, b, c be three unit vectors such that

1 3i is 2 i

Argument of the complex number

90.

(a)

x 1 x2

x 1 x2

4 3

0

(b)

x

1 4

3 1 0 (d) x 4 4 If g (x) is a polynomial satisfying g (x) g(y) = g(x) + g(y) + g(xy) – 2

(c)

91.

x x

0 0

for all real x and y and g (2) = 5 then Lt g(x) is x

(a) 9 (c) 25

(b) 10 (d) 20

3

EBD_7443 2008-8

92.

Target VITEEE

The value of f(0) so that

( ex

2x ) x

97. may be

4 x

continuous at x = 0 is (a) 93.

95.

0

6

2 log(1

2)

(b) 3

2 log(1

2)

6

2 log(1

2)

(d) 3

2 log(1

2)

(c)

x5

If I

1 x3 5

(a) (b)

2 (1 x 3 ) 2 9 log

y2

8 is (b) 2 (d) 4

98.

The value of

a

x

x

0

dx is

a 2

(b)

a 4

a a (d) 2 4 99. Let y be the number of people in a village at time t. Assume that the rate of change of the population is proportional to the number of people in the village at any time and further assume that the population never increases in time. Then the population of the village at any fixed time t is given by (a) y = ekt + c, for some constants c < 0 and k>0 (b) y = cekt, for some constants c > 0 and k < 0 (c) y = ect + k, for some constants c < 0 and k>0 (d) y = kect, for some constants c > 0 and k < 0 100. The differential equation of all straight lines touching the circle x2 + y2 = a2 is

(c)

dy dx

2

(a)

y

(b)

y x

dy dx

(c)

y x

dy dx

(d)

y

dy dx

a2 1 2

a2 1

3

2 (1 x3 ) 2 3

1 x3

log

x

(d)

2 (1 x 3 ) 2 9

3

2

dy dx

C

dy dx

dy dx

a2 1

a2 1

dy dx

C

101. The differential equation (c)

2

dx , then I is equal to

1 x3

x

2

a

(a)

(b) f (x) is continuous at x = 0 (c) f (x) is not differentiable at x = 0 (d) f (x) = 1 A spherical balloon is expanding. If the radius is increasing at the rate of 2 centimeters per minute, the rate at which the volume increases (in cubic centimeters per minute) when the radius is 5 centimetres is (a) 10 (b) 100 (c) 200 (d) 50 The length of the parabola y2 = 12x cut off by the latus-rectum is (a)

96.

(b) 0

lim f (x) does not exist

x

2

(a) (c) 3

(c) 4 (d) – 1 + log 2 Let [ ] denote the greatest integer function and f (x) = [tan2 x]. Then (a)

94.

1 log 2

Area enclosed by the curve

C 1

2 (1 x3 ) 2 3

C

dy dx

admits (a) infinite number of solutions (b) no solution (c) a unique solution (d) many solutions

y

3

0

Solved Paper 2008

2008-9

102. Solution of the differential equation x2

xdy ydx

y2 dx

(a)

y

x2

y2

C x2

(b)

y

x2

y2

C x2

(c)

x

x2

y2

C y2

0 is

x x 2 y2 C y 2 Let P, Q, R and S be statements and suppose that P Q R P. if ~ S R , then (b) ~ Q S (a) S ~ Q (c) ~ S ~ Q (d) Q ~ S In how many number of ways can 10 students be divided into three teams, one containing four students and the other three? (a) 400 (b) 700 (c) 1050 (d) 2100 If R be a relation defined as a R b iff |a – b| > 0, then the relation is (a) reflexive (b) symmetric (c) transitive (d) symmetric and transitive Let S be a finite set containing n elements. Then the total number of commutative binary operation on S is

(d)

103.

104.

105.

106.

(a)

n

n(n 1 2

(b)

n

n(n 1 2

2

(c) n (n ) (d) 2(n 2 ) 107. A manufacturer of cotter pins knows that 5% of his product is defective. He sells pins in boxes of 100 and guarantees that not more than one pin will be defective in a box. In order to find the probability that a box will fail to meet the guaranteed quality, the probability distribution one has to employ is (a) Binomial (b) Poisson (c) Normal (d) Exponential 108. The probability that a certain kind of component 3 . The 4 probability that exactly 2 of the next 4 components tested survive is

will survive a given shock test is

(a)

9 41

(c)

1 5

109. Mean and standard deviation of marks obtained in some particular subject by four classes are given below. Report the class with best performance (a) 80, 18 (b) 75, 5 (c) 80, 21 (d) 76, 7 110. A random variable X follows binomial distribution with mean and variance . Then (a) 0 < < (b) 0 < < (c) 0. Hence to have a negative G. H < T S. As T & P are constant. T S must be positive to give the total value a negative sign. Hence S> 0. 64. (b) Any first order reaction follows the equation

k t + log [A]o 2.303 it resembles equation of straight line y = mx + C y = log [A] i.e. log10 C

log [A] =

k if x = t & C = log [A]0 2.303 hence the plot is for a 1st order reaction.

m=

65. (b) The driving forces which are responsible for a process to be spontaneous are : i) Tendency for minimizing energy ii)Tendency for maximum randomness. i.e. maximum entropy 66. (b) For 1st order reaction. t½ (half life time) = Hence K =

0.693 K

0.693 0.693 = = 0.693 × t½ (60 40)sec

10–2 sec–1 = 6.93 × 10–3 sec–1 67. (b) Since NaNO3 is formed by the reaction Nacl KNO3 NaNO3 KCl hence, using Kohlrausch's law o

m NaNO 3

o

o NaCl

o KNO 3

KCl

= 128 + 111 – 152 = 87 S cm2 mol–1 68. (c) As we know that Ecell = Ecathode – Eanode when Ecathode = Eanode Ecell = 0 If Ecell = 0 no net reaction occurs. The reactants and products are at equilibrium and no current will flow. Note that it is only possible to obtain electrical work from a system that is not at equilibrium. In order for current to flow, there must be a net reaction occurring. As the oxidation- reduction reaction proceeds toward equilibrium, and the concentrations of the reacting species approach their equilibrium values, the EMF of the cell decreases to zero. When the system is at equilibrium, the cell potential is zero and we have a dead battery.

69. (c) In CuSO4 solution, oxidation state of Cu is +2. Hence one mole of copper sulphate will require charge equal to two moles of electrons to form metallic Cu. Mole charge = IF. Hence 2 Faraday is required. 70. (c) Rusting of iron is generally promoted in an acidic aqueous medium. Alkaline medium prevents availability of H+ ions. Sodium phosphate will cause formation of a protective film of iron phosphate on the iron preventing rusting. These solutions are used in car radiators to prevent rusting of iron parts.

Solved Paper 2008

2008-19

71. (a) Hydroboration-oxidation of alkenes give alcohols containing same number of carbon atoms. The addition follow anti-Markowi Koff's rule. Boron atom act as an electrophile. Main two steps re involved. Reagent used BH3 & NaOH/H2O2. H H H | | | 3H3C — C — C— C — | | | H H H

Me H | | C — C— C | | | H H H

76. (c) The witting reaction is a chemical reaction of an aldehyde or ketone with triphenyl phosphonium ylide to give an alkene and triphenyl phosphine oxide. R C = O + (C6 H5)3 P = C

OR

CH2 BH 3

R

where R=4-methyl octyl

C=C

3CH 3 (CH 2 ) 3 CH (CH 2 ) 3 |

R´

OH

4 Methyloctanol

72. (d) Ethyl alcohol on treating with conc. H2SO4 undergoes dehydration to form alkene i.e. ethylene Conc H 2SO4

C2 H 4

H 2O

OCH3

+ O = P(C6 H5)3 Triphenyl phosphine oxide

Me

73. (b) Anisole is O

R´

Phosphorus ylide

R 3B

C2 H5 OH

R

– (C6H5)3P – C +

4 - methyl octene

H 2O 2 NaOH

R´

77. (a) Cannizaro's reaction is for those aldehydes which does not contain - hydrogen atom. This is also called self oxidation - reduction reaction. Among the given carbonyl compounds only HCHO does not have hydrogen.

CHO , Phenyl methyl ether. It

78. (c)

AlCl3 HCl

O + CO

O

can be prepared by treating phenol first with a base like NaOH to form phenoxide ion. The phenoxide ion will then substitute the halide of an R-X molecule, to form methyl phenyl ether.

When a mixture of CO and HCl gas is passed through benzene in presence of catalyst consisting anhydrous AlCl3, benzaldehyde is formed.

C6 H5ONa CH3 Cl

CO HCl

CH3 O C6 H5 Anisole

74. (c) Alkaline Potassium permanganate is a strong oxidising agent. It oxidises ethylene glycol to oxalic acid.

CH2OH

CH2OH Ethylene glycol

alk. KMnO 4

COOH COOH Oxalic acid

75. (a) In the structure of diamond each carbon atom in sp3 hybridised & is covalently bonded with four other carbon atom held at the corners of a regular tetrahedron by covalent bonds. This results in a very big three dimensional polymeric structure in which C – C distance is 154 pm and bond angle is 109.5°. Owing to very strong covalent bonds by which atoms are held to gether diamond is the hardest substance known.

C6 H 6

HOCl formyl chloride

HOCl

AlCl3

C6 H 5 CHO HCl

79. (b) Maleic acid & fumaric acid are both the isomers of butene dioic acid. Maleic acid is the cis isomer & fumaric is the trans-isomer. O

O

O

OH OH

OH

OH

O

Maleic acid

Fumaric acid

80. (c) Alkali formate i.e. HCOONa with soda-lime i.e. NaOH + CaO will react to give Na2CO3 and hydrogen gas is liberated. HCOONa

NaOH

CaO

Na 2 CO3 H 2

EBD_7443 2008-20

Target VITEEE

PART - III (MATHS)

sin

1 b 2

81. (c) a (b c)

1 b [Vector triple product] 2

(a.c)b (a.b)c

1 b 2

( a . c cos 2 )b ( a . b cos 1 )c

a b

[

c

1 and cos 2

cos

2

cos and cos 3

1

2

(r

r

2

c)

R2

r.c c.r

c.c

c).(r

r.r

c

R2

2

2(r.c)

c

2

R2

r

2

2(r.c)

c

2

R2

If h

c

2

R 2 i.e. c

0

h , then

2

r 2(r.c) h 0 Hence, the given equation represents a sphere. 83. (b) Let tan–1 x =

x

tan

sin cos

sin 1 sin 2

1 sin 2

sin

x 2 (1 sin 2 )

sin 2

x

x2

sin 2

1 x2

1 x2

x 1 x2

5Z 1

5 x iy 1

(x 25) iy 5 (x 1 iy Squaring both sides, we get (x – 25)2 + y2 = 25 {(x – 1)2 + y2} x 2 50x 625 y 2

2 5i 3i 2 1 3i 1 3i 2 i 2 i 2 i 2 i 4 i2 = –1 – i Now, let us put –1 = r cos , –1 = r sin Squaring and adding, r2 = 2 i.e., r = 1 1 So that cos = , sin = 2 2 Therefore, = 225° Thus, argument is 225°. 86. (c) Since b and C are the roots of x2 – 61x + 820 = 0, so b + c = 61, bc = 820

4 tan A 3 Now, using the formula,

sin

x

1

84. (c) Given that

A

sin 2 (1 x 2 )

x2

1 x2

25x 2 50x 25 25y 2 24x2 + 24 y2 – 600 = 0 x2 + y2 – 25 = 0 |x + iy|2= 25 |Z|2 = 52 |Z| = 5 85. (c) We have,

R2

r

x

1

=

x iy 25

and 1 3 2 82. (d) Vector equation of a sphere in central form with centre having position vector c and radius R is

R

sin

z 25 Z 25 5 z 1 Let Z = x + iy, then

2

r c

x

1]

0

1

1

1 x2

Now, sin (tan–1x) = sin sin

Equating the coefficients of b and c on both sides, we get cos 2

tan

x

1

x 1 x2

tan

cos A =

1

b2

c2 a 2 2bc

4 3

2

cos A

3 5

Solved Paper 2008 a2

b2

2008-21

c2 –

x2 + y2 + z2 + 3x – 4z + 1 = 0,

= + 2bc cos A = (b + c)2 – 2bc – 2bc cos A = (b + c)2 – 2bc (1 + cos A) = (61)2 – 2 × 820 1

3 5

x 4 y Z 1 (say) l 3 2 2 x = 3l – 4, y = –2l, z = –2l –1 Hence, general point on the second line, Q (3l – 4, – 2l, –2l – 1) Direction ratios of PQ are 3l – 4 – k –6, –2l + 2k –2, – 2l –1 –2k – 2 i.e. 3l – k – 10, –2l + 2k– 2, – 2l –2k – 3 Now |PQ| will be the shortest distance between the two lines if PQ is perpendicular to both the lines. Hence, 1(3l k 10) ( 2)

( 2l 2k 2) 2( 2l 2k 3) 0 and 3 (3l – k – 10) + (–2) (–2l + 2k – 2) + (–2) (–2l – 2k – 3) = 0 i.e. 3l – 9k = 12 or l – 3k = 4 ...(i) and 17l – 3k = 20 ... (ii) Subtracting equation (i) from (ii), we get 16l = 16 l=1 Putting this value of l in equation (i), we get – 3k = 3, k = – 1 P (–1 + 6, – 2 (– 1) + 2, 2 (– 1) + 2) (5, 4, 0) Similarly, Q = (–1, –2, –3) Hence, shortest distance, PQ, ( 2 4)2

( 3 0)2

( 6)2 ( 6)2 ( 3)2 = 36 36 9 = 9 units 88. (c) Since the centre and radius of the sphere x2 + y2 + z2 + 2 ux + 2vy + 2wz + d = 0 are

=

(–u, –v, – w) and u 2 v 2 respectively. So, for the sphere

w2

2

02

(2)2 1

9 21 4 1 4 2 89. (b) Let two fixed points be A (ae, 0) and B (–ae, 0). Let C (x, y) be a moving point such that AC + CB = constant = 2a (say)

x 6 y 2 z 2 k(say) 1 2 2 x = k + 6, y = –2k + 2, z = 2k + 2 Hence, general point on the first line, P (k + 6, – 2k + 2, 2k + 2) Equation of second line,

( 1 5)2

3 2

Radius =

= 3721 – 2624 = 1097 87. (d) Equation of first line,

=

3 , 0, 2 , and 2

Centre

d

i.e. ,

(x ae) 2 (y 0)2 (x ae) 2

Or

x2

y2

a 2e2

(y 0)2

2a

2aex

x 2 y 2 a 2e 2 2aex 2a ...(1) Or l + m = 2a ...(2) Where, l2 = x2 + y2 + a2e2 – 2aex ...(3) and m2 = x2 + y2 + a2e2 + 2aex ...(4) From, (3) and (4) m2 – l2 = 4aex or (m – l) (l + m) = 4 aex 2a (m – l) = 4aex [From (2)] m – l = 2ex ...(5) Adding (2) and (5), we get m = a + ex ...(6) From (4) and (6), a2 +e2 x2 + 2aex = x2 +y2 + a2e2 + 2aex x2 (1 – e2) + y2 = a2 (1 – e2) Dividing both sides by a2 (1 – e2), we get x2

y2

a2

a 2 (1 e 2 )

x2

1

y2

1 , where b2 = a2 (1 – e2) a2 b2 This is the equation of ellipse. 90. (d) The equation of the parabola is

Or

y2

4x 3

0

3 4 The directrix of the parabola Y2 = – 4aX is X = a.

or

y2

4 x

...(1) ... (2)

EBD_7443 2008-22

Target VITEEE On comparing the equation (1) and (2), we get

and X

4a = 4

3 4

x

3 or a= 1 and X x 4 Hence the directrix of the parabola (1) is

3 1 1or x 0. 4 4 91. (b) g (x). g(y) = g(x) + g (y) + g (x y) – 2 ...(1) Put x = 1, y = 2, then g (1). g(2) = g (1) + g (2) + g (2) – 2 5g (1) = g (1) + 5 + 5 – 2 4g (1) = 8 g(1) = 2 1 in equation (1) , we get x

g(x).g

1 x

g(x) g

1 x

g(1) 2

g(x).g

1 x

g(x) g

1 x

2 2

[ g(1) This is valid only for the polynomial g (x) = 1 + xn ... (2) Now g (2) = 5 (Given) 1 + 2n = 5 [Using equation (2)] + 2n = 4, 2n = 4, –4 Since the value of 2n cannot be –Ve. So, 2n = 4, n = 2 Now, put n = 2 in equation (2), we get g (x) = 1 + x2

1 =x

ex

h

0

0

lim [tan 2 (0 h)]

h

0

2

= lim [tan h] [tan 0] [0] h

0

0

Now, determine the value of f(x) at x = 0. f (0) = [tan 2 0] = [0] = 0 Hence, f (x) is continuous at x = 0. 94. (c) Let r and V be the respectively radius and volume of the balloon. Let t represents the time. The rate of increament in radius is dr 2 cm/minute. The volume of the dt balloon is given by

2]

3

2x

4 3 r 3 Differentiating w.r. to t, we get V

dV dt

4 dr (3r 2 ) 3 dt

Substituting the values of and

dr , we get dt

dV 4 (3 52 2) 200 cm3 / minute dt 3 95. (a) On comparing the equation of the parabola y2 = 12 x with the standard equation, y2 = 4 ax, we get 4 a = 12 or a = 3. y

x

A(3, 6)

x x2 (1 1! 2!

log 2 x3 ..) 1 x 3! 1! (bg 2) 2 2 x 2!

f(x) = log 2 1

0

0

2

= 1 + 9 = 10, – 8 92. (d) f (x)

0

lim [tan 2 (0 h)]

h

2 2 = lim [tan h] [tan 0] [0]

x

Lt g(x) = Lt (1 x ) = 1 + (3)2 3 x

x

2 = lim [tan x]

2

x

2 = lim [tan x]

R.H.L. (at x = 0)

x

Put y

93. (b) Check the continuity of the function f (x) = [tan2 x] at x = 0. L.H.L. (at x = 0)

(log 2)3 3 x .. 3!

x {(log 2) 2 1} 2!

(0, 0) O

C (3, 0)

x2 {(log 2 )3 1} .... 3!

Putting x = 0, we get f(0) = log 2 – 1 + 0 + 0 + .... = – 1 + log 2.

B (3, –6)

x

Solved Paper 2008

2008-23

Hence, the focus, point C will be at (3, 0) and the extremities of the latus-rectum AB will be at (a, 2a) and (a, –2a). So the coordinates of A and B are (3, 6) and (3, –6) respectively. Now we need to find the length ( curve AOB) of the parabola. As it is not a straight line so we cannot directly find the length of this curve as we cannot directly apply Pythagorous theorem. Let us consider a small length ds on the parabola. Using pythagorous theorem for this length, ds =

(dx)

2

(dy)

6

2y 12 6 6

2 = 6

6

y

2

1dy

= 6 2 6log

x5

=

6

2

1 dy

2 0

2 t3 3 3

Using

y2 36 dy 36

t

a2 log x 2

a a

2

x

We get s

1 y 2 6 3 2

log y

2

6

y

2

2

2

x

C

y2

2 2 t 1)dt 3 (1 x 3 )3 / 2 3

2 3

C

2)2

[4(x

6 dy

x

x 3 .x 2

dx

2) 2

(x

2)2

(x

2)2 2

6 2

C

8

y2

y 2

2

2

y2 2

2

2

1

...(1)

This is the equation of the ellipse having centre ( 2, 0 ). Observe the figure of ellipse (1). The centre

C 0

C 0 18log 0

C

y2 ] 8

2

2

6

1 6 2 6 62 18log 6 = 3 2

1

(1 x 3 ) 2

2 2 (1 x 3 )3 / 2 (1 x 3 )1/ 2 9 3 97. (d) The given curve is

4(x

x 2 a 2

2)

2 td t 3

0

2

6

=

2

2

2)

2 (t 2 1) t dt 3 t

...(1)

Putting in (1),

18 log 6

dx 1 x3 1 x3 Let 1 + x3 = t2, so that 3x2 dx = 2t dt

96. (d) I

x 2 dx

y2 12

6(1

= 6 ( 2 log(1

I

x=

y 6 y 6

s

=

1 .dy

From y2 = 12 x

dx dy

2

dx dy

1 3.6 2 18log 6 6 2 3

2

dx dy

6

s

2

=

2

6

6

62

0

P is ( 2, 0 ). A and B are

2

C

2

2

2,

, 0 respectively..

2

, 0 and

EBD_7443 2008-24

Target VITEEE Q

y

1

=8

1 (0) + sin –1 (1)–0–0

1

8

4 square units. 2 98. (c) Let x = cos2 , so that dx = – 2a sin cos d 2 ,0

2

2 ,0

2

Now

( 2, 0)

O

A

B

P

a

a

x

x

0

dx

0

a a cos2

2

a cos 2

( a sin cos )d

[ =a 0

2

1 cos 2

/2

The required area = 4 × area of figure PQB 2

=4

8

2

2) 2 dx

4(x

=a

2

2

4 2

=8

=a

2

2

=4

2

2

2

2

(x

2

2

2

=

1 x a2 2

x 2 dx

8

2

2

2

1

2

sin

1

( 2

2

x

1

2

2 2/

a sin 2 2 x

( 2

x a

1

2)

C

2sin 2

2

1

sin 2 2

=a

sin 2

2

2

2

2 2/

f (x) dx

sin .sin cos cos

=a

=a

/2 0

(1 cos 2 ) d

1 2sin 2 ]

/2

0

(0 0)

a 0 2 2 99. (b) According to the question, dy dt

2

sin

0

y

dy dt

ky

Separating the variables, we get

2

1

0

a]

=a

2

2

2) 2

2

2

2

x2

2

/2

2

t

f (x) dx

[ cos 2

2)2

(2 / ) sin 2

/2 0

2) 2 dx

(x

2

x

a2

0 t

ydx

2

0 at x

0;

2sin cos d

cos 2

2

at x

dy dt

kdt

dy k dt y log y = k t + M (as y cannot be –ve) y = ekt+M y = eM . ekt y = C ekt, where C = eM Constant k cannot be positive because the population never increases in time. And Integrating both sides, we get

Solved Paper 2008

2008-25

another constant C cannot be negative because of eM > 0 always. Hence y = Cekt, for some constants C > 0 and k < 0. 100. (b) The given circle is, x2 + y2 = a2 Differentiating with respect to x, we get dy dx

2x 2y

x y x2

0

0

dy dx

dy 2xy dx

y

x x

dy dx

2

y

2

y

2

dy 2xy dx

dy dx

x

a2 1

dv

dy dx

2

dy dx

2

dy dx

2

dy dx

2

y

both sides)

dy dx

y2

0, y

a2 0

3 3

y

dy y 3 0 is not possible. Hence dx Therefore, the given differential equation has no solution. 102. (b) The given differential equation is x2

xdy ydx xdy

dy dx

y2 dx

dv . Then dx

1 v2

v

dx x

dx x

2

1 v2 ) log x log C

log(v

y

x2

3 0 Since

y

dv dx

1 v2

1 v

2

dy dx

2

v x

x 2 v2 x 2 x

vx

dv

2

dy dx

2

dy dx

Integrating both sides, we get

x2 101. (b)

dv dx

v x

0

2

2

dy dx

0

2

dy dx

2 (Adding y

y x

v x

(Squaring both sides)

y2

x

2

dy dx

Put y = vx, so that

2

dy dx

2xy

2

x y

This is the linear differential equation.

0

y

x2

y2 dx

(y

x2 x

y2

v

1 v2

y x

1

x2

y2

y2 x2

Cx Cx

[ y

vx]

Cx 2

103. (b) ~ Q S ~S Q But ~ S R Q R, True [As P Q R P] Hence ~ Q S is true 104. (d) We know that the number of ways of dividing (m+n+p) things into three groups containing m, n and p things respectively (m n p)! m!n!p! Further if any two groups out of the three have same number of things then number of ways (m n p)! m!n!p! 2 Hence number of ways to divide 10 students into three teams one containing four students and each remaining two teams contain three

10! 10 9 8 7 6 5 = 4! 3! 3! 2 = = 2100 3 2 3 2 2 105. (d) We observe the following properties : Reflexivity - Let a be an arbitrary element. Then,

a a 0| 0 a R/ a This, R is not reflexive on R.

EBD_7443 2008-26

Target VITEEE Symmetry – Let a and b be two distinct elements, then (a, b) R |b – a| > 0 |a – b| > 0 a b

b a

(b, a) R Thus, (a, b) R (b, a) R. So, R is symmetric. Transitivity – Let (a, b) R and (b, c) R. Then |a – b| > 0 and |b – c| > 0 |a – c| > 0 (a, c) R So, R is transitive. 106. (a) Let S = {ai } where i = 1.2.....n From commutative operations, a i *a j a j *ai … (i) i, j 1, 2,3....n

where * represents a binary operation Number of distinct elements in S × S i.e., {a i } {a j} subject to the condition (i) i 1,2...n

n

n(n 1) 2 No. of commutative binary operations = No. of functions f : S × S S subject to (i)

n (n 1) (n 2) .... 2 1

n(n 1)

n(n 1) times n 2 2 107. (b) Poisson distribution is a probability distribution which is obtained when the probability (p) of the happening of an event is same in all the trials and there are only two events in each trials generally says successes and failures probability (p) of the happening of the event in trial is very less but number of trials (n) is very large. = n.n.n....

5 1 is very less and 100 20 n = 100, is very large. Hence, one has to employ the Poisson distribution in the given question. 108. (d) The probability that a component survives

Here, p = 5% =

1 4 4 [ p q 1] n takes the value 4 and r = 2. Hence the required probability is

is p =

4

. Then q = 1 – p = 1

4

2

3 4

C2

1 4

2

3 3 1 27 4 4 4 4 128 109. (b) Performance of the class will be best if mean of the marks obtained is maximum but standard deviation of the marks obtained is minimum. Hence the class which has mean and standard deviation of the marks obtained as 75 and 5 respectively performs best. 110. (b) Mean, np = ; and variance, npq = where n = number of trials and p + q = 1.

So,

= n{(a1, a1 ), (a1, a 2 )......(a1, a n ), ...(a n 1, a n 1 ), (a n 1 , a n ), (a n ,a n )

r

=6

j 1,2...n

(a 2 , a 2 ), (a 2 , a 3 ),....(a 2 , a n ),

Cr pr q n

npq np

q

(1 p)

0 p 1 –1< – p < 0

0(CH3) 3N (b) (CH3)3N > (CH3)2 NH>CH3NH2 (c) (CH3)3N > CH3 NH2= (CH3 )2 NH (d) (CH3)2N H > (CH3 )3 N> CH3 NH2 When aqueous solution of benzene diazoniumchloride is boiled, the product formed is (a) C6H5CH2OH (b) C6H6+N2 (c) C6H5COOH (d) C6H5OH Carbylamine reaction is given by aliphatic (a) primary amine (b) secondary amine (c) tertiary amine (d) quaternary ammonium salt NH3 H 2 ,Ni

51.

54.

55.

56.

57.

(a)

137 52Te

(b)

144 55 Cs

(c)

137 56 Ba

(d)

144 56 Ba

AgCl dissolves in a solution of NH3 but not in water because (a) NH3 is a better solvent than H2O (b) Ag+ forms a complex ion with NH3 (c) NH3 is a stronger base than H2O (d) the dipole moment of water is higher than NH3 Which of the following is hexadenate ligand? (a) Ethylene diamine (b) Ethylene diamine tetra acetic acid (c) 1, 10- phenanthroline (d) Acetyl acetonato A coordinate bond is a dative covalent bond. Which of the below is true? (a) Three atoms form bond by sharing their electrons (b) Two atoms form bond by sharing their electrons (c) Two atoms form bond and one of them provides both electrons (d) Two atoms form bond by sharing electrons obtained from third atom Which of the following complex has zero magnetic moment (spin only)? (a)

58.

[Ni(NH3 )6 ]Cl2 (b)

Na 3[FeF6 ]

(c) [Cr(H 2O)6 ]SO4 (d) K 4 [Fe(CN)6 ] The IUPAC name of [Ni(PPh3)2Cl2]2+ is (a) bis dichloro (triphenylphosphine) nickel (II) (b) dichloro bis (triphenylphosphine) nickel (II) (c) dichloro triphenylphosphine nickel (II) (d) triphenyl phosphine nickel (II) dichloride

Solved Paper 2007 59.

60.

61.

62.

63.

64.

Among the following the compound that is both paramagnetic and coloured is (a) K2Cr2O7 (b) (NH4)2 [TiCl6] (c) VOSO4 (d) K3Cu (CN)4 On an X-ray diffraction photograph the intensity of the spots depends on (a) neutron density of the atoms/ions (b) electron density of the atoms/ions (c) proton density of the atoms/ions (d) photon density of the atoms/ions An ion leaves its regular site occupy a position in the space between the lattice sites is called (a) Frenkel defect (b) Schottky defect (c) Impurity defect (d) Vacancy defect The 8:8 type of packing is present in (a) MgF2 (b) CsCl (c) KCl (d) NaCl When a solid melts reversibly (a) H decreases (b) G increases (c) E decreases (d) S increases Enthalpy is equal to (a)

(c) 65.

66.

2007-7

G

–T 2

T

(b)

T

(d)

T

T

P

69.

G

2

T

V

P

Condition for spontaneity in an isothermal process is (b) (a) A W 0 G U 0 (c) (d) A U 0 G–U 0 Given: 2C s

2O 2 g

C2 H 2 g

1 O2 g 2

1 2 O2 g 2

70.

2CO2 g ; H

H2 g

68.

H 2O( );

H

2CO 2 g

; H

Given the equilibrium system: NH4Cl (s) NH4+(aq) + Cl–(aq) ( H = +3.5kcal/mol). What change will shift the equilibrium to the right? (a) Decreasing the temperature (b) Increasing the temperature (c) Dissolving NaCl crystals in the equilibrium mixture (d) Dissolving NH 4NO 3 crystals in the equilibrium mixture According to Arrhenius equation, the rate constant (k) is related to temperature (T) as (a)

In

k2 k1

Ea 1 1 – R T1 T2

(b)

In

k2 k1

–

(c)

In

k2 k1

(d)

In

k2 k1

G/T

–T 2

V

G/T

T2

67.

–787kJ

–286kJ

H 2O( )

–1310kJ

The heat of formation of acetylene is (a) –1802 kJ (b) +1802 kJ (c) +237 kJ (d) –800 kJ

71.

72.

Ea 1 1 – R T1 T2

Ea 1 R T1

1 T2

Ea 1 R T1

1 T2

–

Equivalent amounts of H2 and I2 are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to HI, the Kc at this temperature is (a) 64 (b) 16 (c) 0.25 (d) 4 For the reaction H2(g)+I2(g) 2HI(g), the equilibrium constant Kp changes with (a) total pressure (b) catalyst (c) the amount H2 and I2 (d) temperature How long (in hours) must a current of 5.0 amperes be maintained to electroplate 60g of calcium from molten CaCl2? (a) 27 hours (b) 8.3 hours (c) 11 hours (d) 16 hours For strong electrolytes the plot of molar conductance vs (a) parabolic (c) sinusoidal

C is (b) linear (d) circular

EBD_7443 2007-8

73.

74.

75.

76.

77.

78.

79.

Target VITEEE conductance values of Ca2+ and Cl–

If the molar at infinite dilution are respectively 118.88×10–4 m2 mho mol–1 and 77.33×10–4 m2 mho mol–1 then that of CaCl2 is (in m2 mho mol–1) (a) 118.88 × 10–4 (b) 154.66 × 10–4 (c) 273.54 × 10–4 (d) 196.21× 10–4 The standard reduction potentials at 298K for the following half reactions are given against each Zn2+ (aq) + 2e– Zn(s) E0 = – 0.762 V Cr3+ (aq) + 3e– Cr(s) E0 = – 0.740 V 2H+ (aq) + 2e– H2(g) E0 = 0.00 V Fe3+ (aq) + 3e– Fe2+(aq) E0 = + 0.762 V The strongest reducing agent is (a) Zn(s) (b) Cr(s) (c) H2(g) (d) Fe2+(aq) The epoxide ring consists of which of the following ? (a) Three membered ring with two carbon and one oxygen (b) Four membered ring with three carbon and one oxygen (c) Five membered ring with four carbon and one oxygen (d) Six membered ring with five carbon and one oxygen In the Grignard reaction, which metal forms an organometallic bond? (a) Sodium (b) Titanium (c) Magnesium (d) Palladium Phenol is less acidic than (a) p-chlorophenol (b) p-nitrophenol (c) p-methoxyphenol (d) ethanol Aldol condensation is given by (a) trimethylacetaldehyde (b) acetaldehyde (c) benzaldehyde (d) formaldehyde Give the IUPAC name for H3C CH 2

(a) (b) (c) (d)

O || C H 2 C CH 2

Ethyl-4-oxoheptonate Methyl-4-oxoheptonate Ethyl-4-oxohexonate Methyl-4-oxohexonate

O || C OCH3

80.

In which of the below reaction do we find , unsaturated carbonyl compounds undergoing a ring closure reaction with conjugated dienes? (a) Perkin reaction (b) Diels-Alder reaction (c) Claisen rearrangement (d) Hoffman reaction

PART - III (MATHEMATICS) 81.

82.

Let the pairs a , b and c , d each determine a plane. Then the planes are parallel if (a)

a c

b d

0

(b)

a c

b d

0

(c)

a b

c d

0

(d)

a b

c d

0

The area of a parallelogram with 3iˆ ˆj – 2kˆ and

ˆi – 3jˆ 4kˆ as diagonals is

83.

84.

85.

(a)

72

(b)

73

(c)

74

(d)

75

cos2 x

If cosx + =1then the value of sin12 x+ 3sin10 x + 3sin8x + sin 6 x–1 is equal to (a) 2 (b) 1 (c) –1 (d) 0 The product of all values of (cos +i sin )3/5 is equal to (a) 1 (b) cos +i sin (c) cos3 +i sin3 (d) cos5 +i sin5 The imaginary part of

1 i

2

is

i 2i –1

(a)

4 5

(b) 0

(c)

2 5

(d)

–

4 5

Solved Paper 2007 86.

–1 If sin –1 x sin y

2007-9

2

, then cos–1x + cos–1y is

93.

equal to (a)

87.

(b)

2

4

3 (c) (d) 4 The equation of a directrix of the x 2 y2 1 is 16 25 (a) 3y = 5 (b) y = 5 (c) 3y = 25 (d) y = 3 If the normal at (ap2, 2ap) on the parabola y2 = 4ax, meets the parabola again at (aq2, 2aq), then (a) p2 + pq + 2 = 0 (b) p2 – pq + 2 = 0 (c) q2 + pq + 2 = 0 (d) p2 + pq + 1 = 0 The length of the straight line x – 3y = 1 intercepted by the hyperbola x2 – 4y2 = 1is

94.

ellipse

88.

89.

(a)

10

1

91.

(d)

(c) 2 tan u

(d) tan u

92.

3 4

3 (c) 1 (d) – 4 The function f(x) = x2 e–2x, x > 0. Then the maximum value of f(x) is

(a) (c)

1 e

1 e2

(b) (d)

1 2e

4 e4

u y

The angle between the tangents at those points on the curve x = t2 + 1 and y = t2 – t – 6 where it meets x-axis is (a)

tan –1

4 29

(b)

tan –1

5 49

(c)

tan –1

10 49

(d)

tan –1

8 29

4

95.

x – 3 dx is equal to

The value of 1

(a) 2

(b)

5 2

1 2

(d)

3 2

(c) 96.

3 with the positive x-axis, then f '(3) 4 is equal to

(b)

y

(b) cosec u

The area of the region bounded by the straight lines x = 0 and x = 2 and the curves y = 2x and y = 2x –x2 is equal to (a)

2 4 – log 2 3

(b)

3 4 – log 2 3

(c)

1 4 – log 2 3

(d)

4 3 – log 2 2

an angle

(a) – 1

u x

(a) sin u

6 5

6 10 5 10 The curve described parametrically by x = t2 + 2t–1, y = 3t + 5 represents (a) an ellipse (b) a hyperbola (c) a parabola (d) a circle If the normal to the curve y = f(x) at (3,4) makes

(c)

90.

(b)

If (x +y )sin u = x2y2, then x

97.

dx

The value of 0

a2

x2

7

is equal to

(a)

231 1 2047 a13

(b)

235 1 2048 a13

(c)

232 1 2047 a13

(d)

231 1 2048 a13

EBD_7443 2007-10

98.

Target VITEEE

x The value of the integral e

(a)

(b)

(c)

1– x

ex

1 x

1 x

d2 y

dx is

dx

If x sin

y dy x

and y 1

y sin

y – x dx x

y , then the value of cos x 2

103. is 104.

1 x

(a) x

(b)

(c) log x

(d) ex

105.

100. The differential equation of the system of all circles of radius r in the XY plane is

(a)

(b)

(c)

(d)

1

1

dy dx

1

dy dx

1

dy dx

3

2e3x is given by

e3x 8

y

c1 c 2 x e x

(b)

y

c1 c 2 x e – x

e –3x 8

(c)

y

c1 c2 x e – x

e3x 8

2

r2

3 2

r

2

r

2

2 3

2 3

d2 y

2

dx 2

d2 y

3

r2

2

dx 2

d2 y

dx 2

3

c1 c2 x e x

xy

(a) 10

(b)

1 10

1 5 107. A box contains 9 tickets numbered 1 to 9 inclusive. If 3 tickets are drawn from the box one at a time, the probability that they are alternatively either {odd, even, odd} or {even, odd, even} is

(c) 5

dx 2

d2 y

106.

y

1 3 y C (b) xy = y4 + C 3 (c) y4 = 4xy + C (d) 4y = y3 + C The number of positive integral solutions of the equation x1 x2x3x4x5=1050 is (a) 1870 (b) 1875 (c) 1865 (d) 1880 Let A= {1,2,3,....., n}and B ={a,b,c}, then the number of functions from A to B that are onto is (a) 3n – 2n (b) 3n – 2n –1 n (c) 3 (2 – 1) (d) 3n – 3(2n – 1) Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is (a) 9 (b) 12 (c) 10 (d) 14 If(G,*) is a group and the order of an element a G is 10, then the order of the inverse of a* a is

(a)

equal to

dy dx

y

(a)

(d)

(d) ex(1–x) + C 99.

dy dx

e –3x 8 102. The solution of the differential equation ydx+(x–y3)dy = 0 is

C

2

2

2

C

1 x2

ex

1 x

2

101. The general solution of the differential equation

2

C

1 x2

ex

1– x

(d)

(a)

5 17

(b)

4 17

(c)

5 16

(d)

5 18

Solved Paper 2007 108. If P(A) =

p A

2007-11

1 1 5 and P(B/A)= , P(B) = then 15 12 12

89 180

(b)

90 180

(c)

91 180

(d)

92 180

109. If the probability density function of a random x in 0 x 2, then 2 P(X > 1.5 X > 1) is equal to

variable X is f(x) =

7 16

(b)

21 7 (d) 64 12 110. If X is a poisson variate such that 2P(X = 0)+ P(X = 2 ) = 2P(X = 1) then E(X) is equal to (a) 1 (b) 2 (c) 1.5 (d) 1.75 1

(cos2

– tan

tan 1

and AB = I, then

(c) b>0, a>0, c>0

2

–4 a – 4 is 1, –2 a 1

(a) A( )

(b)

A

(c) A(– )

(d)

A

2x 1 4 2x 112. If x = –5 is a root of 2 6 7

(d) b>0, a>0, c 0 and log3x + log3( x ) + log3 ( 4 x ) + 8

x + log3

16

x +....= 4, then x equals

(a) 9 (b) 81 (c) 1 (d) 27 117. The number of real roots of the equation

x

1 x

3

x

1 x

0 is

(a) 0 (c) 4

(b) 2 (d) 6

118. If H is the harmonic mean between P and Q, then the value of

)B is equal to

H P

H is Q

2

(a) 2

(b)

PQ P Q

– 2

(c)

1 2

(d)

P Q PQ

8 2 = 0, then 2x

the other roots are (a) 3, 3.5 (b) 1, 3.5 (c) 1, 7 (d) 2, 7 113. The simultaneous equations Kx + 2y–z =1, (K –1)y–2z = 2 and (K + 2)z = 3 have only one solution when (a) K = –2 (b) K = –1 (c) K = 0 (d) K = 1

5

then the value of a is (a) –1 (b) 2 (c) –6 (d) 4 115. If b2 4ac for the equation ax4 + bx2 + c = 0, then all the roots of the equation will be real if (a) b>0, a0 (b) b0, c>0

log3

3 4

(c)

111. If A

2

1

B is equal to

(a)

(a)

114. If the rank of the matrix

–1

119. If a and b are two unit vectors, then the vector ( a + b ) × ( a × b ) is parallel to the vector

(a)

a +b

(b) 2 a + b

(c) a – b (d) 2 a – b 120. If is the angle between the lines AB and AC where A, B and C are the three points with coordinates (1,2,–1), (2,0,3), (3,–1,2) respectively, then 462 cos is equal to (a) 20 (b) 10 (c) 30 (d) 40

EBD_7443 2007-12

Target VITEEE

2007 SOLUTIONS PART - I (PHYSICS) 1.

(d) The magnetic moment of the ground state of an atom whose open sub-shell is half filled with n electrons is given by

n(n 2). B where B is the gyromagnetic moment of the atom. Here, n = 5. 2.

3.

5(5 2). B 35 B (a) Bragg's law gives 2dsin = n , n= order of reflection, d= distance between planes. For same and d, n for a given , smallest d for least n, can be found. If crystal is symmetric reflections from different planes may cancel out. (b) According to Moseley's law, square root of frequency of X-ray is plotted against atomic number it gives straight line, the relation is

5.

m0

m

(for

f 0, Z c , for Z = 0, f –c ) Option (b) is correct. as Z can not be negative. (b) Balmer series is given for n 1 = 2 and n2 = 3, 4, ... Rc

1

.

c2

(m – m0)c2 = m0c2 provides

m0 v

1

m0 c 2

2

m 0c2

c2

1

–

22

1

Rc

1

1 2

2

1

or,

1

1 n2

2

Rc

1 1 – 4 9

f

1

2

2

–

1

4

2

Rc

Rc

1 1 – 4 16

4 –1 16

1–

v2 c

v2 c

1

v2

c

2

3 4

3 c. 2

f

Rc

1 4

1

2

v

5 36 Rc Rc f 36 5 For second line n 2 = 4

or,

c2

v2

c

2

v2

1

n 22

–

c2

or,

or,

1 1

v2

1

For Ist line in spectrum n2 = 3

2

v2

c Z –1 where c = constant

f

4.

3 3 36 Rc f 1.35f . 16 16 5 (d) Here, Kinetic energy = Rest energy we know that Kinetic energy (Relativistic) = (m – m0)c2, and Rest energy = m0c2, where m0 = rest mass, c = velocity of light Also, mass (m) of a particle moving with velocity v is given by 2

2

v=

2

=

1 4

=1 3 c 2

1 3 = 4 4

3 2 v2 = c 4

1 4

Solved Paper 2007 6.

2007-13

(d) de-Broglie wavelength ( ) of a particle of mass m and moving with a velocity v is

10.

h , where h is Planck's mv

given by,

constant. When a particle having charge q is accelerated through a potential V then

11.

1 mv 2 2

qV

m 2 v2 2m

or, qV

mv

2mqV 12.

(c) Radio carbon dating is done by measuring ratio of 14C present in the sample, since proportion of 14C and 12 C in a body is same; but after death 14C decays. Hence knowing the present ratio of 14C to 12 C, sample can be dated. (b) particles are positively charged He nucleus, it can accept 2e– , rays are negatively charged which are similar to e–, can donate 1e–, are radiations. Hence ionisation power of is maximum. are most energetic and is least energetic Penetration power of is maximum (d) Given half life T= 3.8 days; t = 19 days

h

N N0

2mqV

? now

Hence, de-Broglie wavelength of electron, e

2me eV

and de-Broglie wavelength of proton, h p

2m p eV

, where e represents the

7.

mp

p

me

13.

(c) Relative speed is given by u

u' v u 'v 1 c2

14.

Here, u' = 0.8 c and v = 0.4 c u

1

0.8c 0.4c (0.8c)(0.4c)

1.2 c 1.32

15.

(d) By Einstein's equation, photoemission occurs when h > 0 or h > h 0 i.e., frequency of incident photon is greater than threshold frequency.

9.

(c) Radius of nucleus is given by R R

R 0 A1/ 3 , where A = mass number

A1/ 3

5

1 32

A B hence it is NOR gate. 16.

8.

1 2

5

N0 0.03 N 0 32 (a) If the p-n junctions are identical, then their resistances would be same. In circuit 1, the first p-n function is forward biased and second is reverse biased. Hence the voltages across them would be different. In circuit 2 both are reverse biased. So potential drop would be same. In circuit 3 both are forward biased; again voltages would be same. (a) Zener diode is a semiconductor device and for semiconductors, temperature coefficient is negative (a) The truth table corresponds to the logic

0.9c.

c2

19 3.8

N

charge of electron (or proton) e

t

1 T 2

N N0

h

t T

17.

(d)

A.B in Boolean algebra, can be written as

A.B A B A B (a) By Dopplers effect, if source moves towards the observer, frequency received will increase. As

1

wavelength will decrease.

EBD_7443 2007-14

18.

19.

20.

Target VITEEE

(d) A coaxial cable consists of a conducting wire surrounded by a dielectric space, over which there is a sleeve of coppermesh covered with a shield of PVC insulation. The power transmission is regulated by dielectric. At high frequencies energy loss due to mesh is significant (called skin effect). (b) Due to their inherent distortion, platemodulated class C amplifiers are not used as audio amplifiers. Also, class A amplifiers are not used owing to low efficiency. A gridmodulated amplifier has very high frequency. Therefore, in output stage of a TV transmitter, grid-modulated class C amplifier is used. (b) The rms value of carrier current is Ic = 8A. The rms value of modulated current = Ic + Im = 8.93 A = It Percentage modulation = ma × 100 The current relation in AM wave is It 2

1

Ic2

ma 2 2

It2

ma

Ic2

8.93

Modulation index, m a

79.7 –1 2 64

–1 2

82

2

–1 2

(d) For a point charge, E

22.

VA

(a)

4 cm

3 cm D(q 3 ) 4 cm

0.9

23.

q1 AB

10 10 –12 4 10

q2 AC

q3 AD

–20 10 –12

–2

5 10

10 4

10 10 –12

–2

3 10 –2

–20 10 5 3

150 – 240 200 60

0.9 110 60

1.65V

(d) The capacitance (C) of a parallel plate capacitor with dielectric is given by 0

C

A

d – t 1–

1 K

C = 100 pF = 100 × 10–12 F, = 8.85 × 10–12, A = r2 = 3.14×(10–2)2 (r = lcm) t = 10–3m, d = t, K = 4 8.85 10 –12 3.14 10 –4 1 10 –3 –10 –3 (1 – ) 4

C1

27.79 10 –16 10 –3 – 0.75 10 –3

27.79 10 –13 0.25 27.79 10 –13 0.25

111.16 10 –13 F

(for 1 set) Required C = 100 10

3 cm C(q 2)

0

9 109 10 –10

1

B(q 1)

4

9 109

. For positive x2 charges, electric field will decrease in positive direction as distance increases, (case 1), for negative charge, as distance increases field will increase (case 4). As we move from positive charge to negative charge, field will keep on decreasing (case 2), As we move from negative to positive charge, field will keep on increasing (case 3) A

1

VA

1.24 –1 2

= 0.701 Percentage modulation = ma × 100 =70.1% 21.

AC 42 32 25 5cm. Potential at A = VB + VC + VD.

n

C C1

–12

F

100 10 –12 111.16 10 –13

10

Solved Paper 2007 24.

2007-15

(a) The resultant intensity at a point on

26.

equatorial line is E . E is parallel and opposite to direction of p .

E E1 –q 25.

V are I same. If voltage is increased by small value then following the same relation V I for constant R, both conductor and semiconductor show same current. (d) Time taken by free electrons to cross the conductor

reading hence their resistances R

E2 3 cm

+p

O

(c) Since conductor and semiconductor are connected in parallel hence voltage across them is same. If ammeters show same

27.

(d) Let a total current of 6I enter at A. It divides into three equal parts, each of 2I, along AE, AB and AD. At E, B and D each the current 2I divides into two equal parts, each of I, along EF, EH, BF, BC, DH and DC. At F, H and C, the two currents each of I, combine together to give a current of 2I at each corner. Thus, at G we get the same current 6I as shown in the figure.

6I

E 2I I A 2I H 2I I D I

I

t

B 2I I 2I G 6I

1 vd

t

28.

1.6 10 –19 5 10 –7

64 10–2

64 102 s

(b) Temperature coefficient, t 2 – t1

2I

C

Let r be the value of resistance of each arm of the cube and R be the joint resistance across the conners A and G. Applying Kirchhoff's law along the loop AEFGA, we get 2Ir + Ir + 2Ir = E or, 5Ir = E ........... (1) Also, by Ohm's law, 6I × R = E 6IR = 5Ir, using (1) 5 r. 6

8 10

28

10 –2 m/s . 64

t 2 – t1

E

or, R

1

vd

F I

I neA

where drift velocity v d

vd

6.4 103 sec

R 2 – R1 R1 t 2 – t1

R 2 – R1 R1 2 –1 1 0.00125

1 t1 800 300 1100K 0.00125 (d) Torque on the coil is = nIB A cos . If coil is set with its plane parallel to direction of magnetic field B, then =0°, cos = 1 = nIBA.1 = nIBA = maximum. Hence, I = maximum (as n,B, A are constant) t2

29.

30.

(d)

1 2 bt is the parabolic equation for 2 thermo emf. The thermoelectric power is E

at

S

dE dt

Here, r = 6 R

5 6 5 6

S = a + bt. The graph between S and straight line.

dE dt

EBD_7443 2007-16

Target VITEEE When t = 0, S = a (intercept). dE At neutral temperature dt

t = tn

–2a b (d) Energy of proton = K.E. = 1MeV = 106 eV.

junction t i

31.

0 and

2

5 80 10

35.

Bq 2 m

–M

6.28 10 –4 1.6 10 –19 2 3.14 1.7 10 –27

32.

e

36.

magnetic field of strength B is given by

F

I( F

33.

B)

34.

2

L

2n

2

hence inductance L = 4L0 In third case, the sense of turns is opposite. So net inductance cancel each other L= 0 (b) At resonance, impedance of circuit is minimum.When impedance of capacitor and inductor is same, XL = XC Resonance frequency

10

5

1 10 5 V

–4

5 10 (d) Average power of an LCR circuit is P = EvIvcos . Maximum power is dissipated at resonance when XL =XC. The resonance frequency is given by 2

Here, = 90° F I B I R B RIB Therefore, a force of magnitude ( RIB) will act on each wire. The direction of the forces on each wire will be same. Thus, total force on the wire AB = RIB + RIB = 2 RIB (d) In first case, the two inductances are in series hence total inductance = L0 + L0 = 2L0 In second case, the current is same, but no. of turns has doubled since the sense of turning is same. n

dI , (–ve sign shows direction) dt

1

I B sin

Thus L

400 10 –6

103 1000 25 40 40 (b) M = 5 H, I = 10 A, t = 5×10–4 s. Now, emf induced in secondary is e

0.9 10 4 Hz 10 4 Hz (d) Force on a conductor of length due to a

2

20 10 –3

2

2t n

Frequency

1 –6

1

–a and at cold b

tn

0 = a + btn

1

1 2

LC

LC

1 25 10 –3 400 10 –6

2 3.14

1 2 3.14 5 20 10 –5 10 105 3.14 200 10

103 6.28

37.

38.

10

159.2 3.16

50.31 Hz

(d) Coherent sources sh ould have same frequency and wavelength and constant or zero phase difference. Frequency is same only for wave X1 and X4 and phase difference = (c) Since position of central maximum is same, hence asin = is same for both wavelengths. i.e, a = constant, = same, is same for both waves and

1 2

1

Solved Paper 2007 39.

2007-17

(b) For bright fringes, I max = (a + b)2

43.

For dark fringes, Imin = (a – b)2 Im ax Now Imin

9

a + b = 3(a – b) a b

a2

2

b

a b

2

a–b

2

a b a–b

a + b = 3a–3b

COOH | H C OH | HO C H | COOH

3

2a = 4b

Ratio of intensities of the two slits, a2

I1 I2

40.

b

2

and

b
0 f '(x) = x2.e–2x(–2) + e–2x.2x put f '(x) = 0 2e–2x. x (–x + 1) = 0 x = 1 or x = 0 f"(x) = (–4x2 – 6x + 1)e–2x f"(1) = –9e–2x < 0 f"(0) = e–2x > 0 value of f(x) is maximum at x = 1

1 f(1) = e–2 = 2 e (d) Given : (x + y) sin U = x2y2 f(x) = x2.e–2x

93.

x 2 y2 sinU = = v (let) x y

Euler's theorem x. sin U x

x.cos U

y

U x

sin U cos U

v x

y.

sin U y y.cos U.

v y

94.

t 2 3t 2t 6

(t 3)(t 2)

0

U y

0

dy dt 2t 1 dx 2t dt Slope of the tangent at point (10, 0)

dy Now, dx

dy dy 5 dx x 10 dx t 3 6 Slope of the tangent at point (5, 0),

m1

dy dy 5 5 dx x 5 dx t 2 4 4 If be the angle beween two tangents, then

m2

m 2 m1 1 m1m 2

tan

15 10 12 24 25 24

tan

5 4

5 6 5 5 1 6 4

5 12 49 24

10 49

10 49

tan

sin U

0

t 3 or 2 when t = 3, then x = 10 when t = –2, then x = 5 Hence, the points where the curve meets the X-axis are (10, 0) and (5, 0).

nv

sin U

tan U

U U y tan U x y (c) Equation of the given curve in parametric form, x = t2 + 1 and y = t2 – t – 6 Y-coordinate of the point, where the given curve meets X-axis is 0. When y = 0, then t2 – t – 6 = 0

Here n = 2 – 1 = 1

x

U y

t(t 3) 2(t 3)

1 at x = 3 or f '(x)

92.

y

x

y= 3 x 2 2

91.

U x

x

1

10 49

tan

1

10 49

EBD_7443 2007-24

Target VITEEE 4

95.

3

4

| x 3 | dx

(b) 1

(x 3)dx 1

97.

dx

(d) Let I

3

3

x2 2

(x 3)dx

x2 2

3x 1

2

x 2 )7

a tan

dx

0

4

Put x

3

limit at x

3x

(a

0

a sec2 d

0&x

2

1 2 2 1 2 2 [3 1 ] 3[3 1] .[4 3 ] 3[4 3] 2 2

4 6

7 3 2

I

5 2

(b) Required area

1 (y 2

a13

y1 )dx

0

2 0

1 sec12

2

x

y=2

(2,4)

(x2y2) Q

m n

& B(m, n)

x' (2,0)

x

1

I

a13

2x –x

2

o

m n

y=

2

1 B(m, n) 2

2x x )dx

0

x3 3

2

1 2a13

0

4 8 1 4 log 2 3 log 2

3 log 2

4 3

.

.

1 2

13 2 1 . 2 1 2

2a13

2

&

0

1

I

2x x2 log 2

0

sin 0 . cos12 d

1 , n 2

m

(2

a13

2

. cos12 .d

sin 2m 1 .cos 2n 1 . d

P

x

1

0

(1,1)

2

d

d

But

y

(0,1)

a14 (1 tan 2 )7

0

2

96.

a sec 2

2

1 1 .(7) 3(1) (8) 3(2) 2 2

13 2 13 2

11 9 7 5 3 1 . . . . . . 2 2 2 2 2 2 6.5.4.3.2.1

231 1 . 2048 a13

Solved Paper 2007 98.

2007-25

(c) Let I

ex

I

x

e

(1 x)2 (1 x 2 )2

1 x 2 2x

(1 x 2 )2

1 (1 x 2 )

1

I

(1 x 2 )

dx 2 e x . ex .

.e x

I

2

1 x2

ex x

e

99.

(c) Given : x sin

y Put x x.

ex x 2 2

(1 x )

y sin

y x

dz dx

z

dx

1 (y b)

d2 y

x dx

cos z

log x c

x 1, y

z cosec z

dy dx

2

dy dx

0

...(3)

2

...(4)

dy dx

...(5)

d2 y

dx x

dx Substituting the values of (x – a) & (x – b) in (1), we get dy dx

1

2 2

d2 y dx

1

2

2

2

log x c

2

...(2)

dx 2 On putting the value of (y – b) in equation (2), we get

x a

sin z dz

0

d2 y

1

dz z.1 x. dx

cosec z

y cos x

dy dx

dx 2

x

zx sin z x x sin z

dy dx

dy 0 dx Differentiating (2) w.r.t. x, we get

y x sin x dy dx

0

(x a) (y b)

(y b)

y x

y sin

z

But y(1)

dx

(1 x 2 )2

+C

dz dx

x

2(x a) 2(y b)

ex x

1

y dy x

c

y log x x 100. (c) The equation of the family of circles of radius r is (x – a)2 + (y – b)2 = r2 ...(1) Where a & b are arbitrary constants. Since equation (1) contains two arbitrary constants, we differentiate it two times w.r.t x & the differential equation will be of second order. Differentiating (1) w.r.t. x, we get

x

1 x2

dy dx

dx

dx

(1 x 2 )2 2

I

x (1 x 2 )2

(1 x 2 ) 2 2xdx

2

ex

dx

(1 x 2 )2

(1 x )

ex

I

2x

2 2

log(1) c

2

cos

dx

1 x2

ex

cos

dx

dy dx

2

1

2

d2 y

2

dy dx

dy dx

dx 2 2 3

r

2

d2 y

dx 2

2

2 2

2

r2

EBD_7443 2007-26

Target VITEEE d2 y

101. (c) Given :

dx 2

dy 2 dx

y

2e

104. (d) Number of onto functions: If A & B are two sets having m & n elements respectively such that 1 n m then number of onto functions from A to B is

3x

The auxiliary equation is D2 + 2D + 1 = 0 or m2 + 2m + 1 = 0 (m 1)(m 1)

0

m

n

1, 1

i.e., repeated roots Complementary function = (c1 + c2x)e–x Now Particular Integral (P.I.)

1 D2

2D 1

. 2e3x

1

P.I.

[D = 3] 2e3x 16

3x

. 2.e 32 2.3 1 Solution y = C. F. + P. I.

dx dy

x y3

I.f .

x × I.f.

Px

e

Q

Pdy

e 8

1 x y

1 ,Q y e

y 2 .y dy C1

4xy

y4

y4

y

2

log y

x1 x 2 x 3 x 4 x 5

3n

C1 1

n

–1

C2 2n

3

xy

y4

C1

4xy C ,

1050

2 3 52

3 2 3

7

Each of 2, 3 or 7 can take 5 places and can be disposed in 15 ways. Hence, number of positive integral solution = 53 × 15 = 1875

C3 3

n

n

–1

3 3

3! n 3 3! 0!

3n

105. (b) Let there are n persons in the room. The total number of hand shakes is same as the number of ways of selecting 2 out of n. C2

n n 1

66

66

2!

n 132

0

0

n 12

106. (a) We know that, let (G, 0) be a group & e be the identity then (a * a)–1 = a–1 o a–1 = (a–1)–1 = a. 107. (d) No. of tickets = 9 No. of odd numbered tickets = 5 No. of even numbered tickets = 4 Required probability = P {odd, even, odd} + P (even, odd, even) 5

52

C2 2

C3 3n

n 12 n 11 y4 4

n

3 2n 1

n2

where C = –4C1 103. (b) Given, x1x 2 x3 x 4 x 5

3

C1

y

4C1

4xy 4C1

3 13

3 3. 2n

y2

Q I.f .dy C1

x.y

. Cr r

r 1

n

1 dy y

3–r 3

3! 3! n 2 2!1! 2!1!

dx dy

P

e

–1

3

compare this equation to general equation dx i.e. dy

3

3

0

. Cr r n

Given A = {1, 2, 3, ---- n} & B = {a, b, c} Number of onto fun ction s

1

102. (c) Given ydx + (x – y3) dy = 0

y

n r n

r 1

e3x 8

3x

c1 c2 x e – x

y

–1

9

5 9

C1

4

C1

8

4 8

C1

4

C1

7

4 7

4 9

C1

4

C1

9

5 8

3 7

C1

5

C1

8

5 18

C1

3

C1

C1

7

C1

Solved Paper 2007

2007-27

108. (a) Given 1 ,P B 12

P A

5 ,P B/ A 12 P A

We know that P B / A P A

1 15

B

B

P A

1 15 12

B

sec

1 180

sec

B

P A

1 180

2 1.5

f x dx , where f x

x 2

x dx 2

2 1.5

1 12

B

1 . 2

2

1 x 2 2

xdx

89 180

cos 2

1.5

2

m2 2

m 2

m2 .e 2!

0

m

m.e 1!

4m 4

1 tan

.B 1

.B

tan , A 22

1 tan tan 1

tan 1

A

m

0

4 4x 14

8 12 14x 8x 3

0

24x 4x 2 12 16x

56 96 112 x

0

8x 3 4x 2 152x 140 0 (x + 5) is a factor of above equation

8x 3

40x 2 36x 2 180x 28x 140

8x

0

2

x 5

36x x 5

28 x 5

36x 28

0 0

0

9x 7

0

I 4 x 5 2x 2

1

tan 1

2 & AB

1

1 tan

A

x 5 4 2x 2

tan

I A , A11 1, A12

|A|

2

.

x 5 8x 2

2

tan

A 21

m

m2

2m

1

A

B

where r = 0, 1, 2, .....

m0 e m 0!

2

111. (c)

m

m .e r!

P(X = r)

10 01

2x 1 4x 2 12

1 1 175 7 4 2.25 1.75 4 4 400 16 110. (b) If 2 P(X = 0 ) + P (X = 2) = 2 P (X = 1) Let probability distribution of X be given by r

.

tan 1

1 tan

112. (b) Given : 2x 1 4 8 2 2x 2 7 6 2x

2

2

1.5

sec 2

.

2

.A

sec2

B

P A

5 12

P A

109. (a)

P B

2

1

B

But,

P A

2

1

B

1/12

sec 1

B

P A

1

1

A

1 15

tan ,

4 x 5

1

7x 2x 7

x 2x 7

1 2x 7

4 x 5 2x 7 x 1 1 tan

2

sec

2

x

5, 3.5,1

0

0

0

EBD_7443 2007-28

Target VITEEE 118. (a) Given : H is the harmonic mean between P&Q

113. (b) For only one solution | A | 0 k 2 0 k 1 0

1 2

0

k 2

k k 1 k 2

0

k

2.

0, k 1, k

2 H k

1

114. (c) Let A

1 2

2 5 4 a 4

1 2 5 0 0 a 6

1

2

0

R2

2PQ P Q

H

0

R2

a 1

2R1 , R 3

0 a 6

1 Q

P Q 2PQ

1 H

1 P

H P

H Q

119. (c) We have a b

a b

a

a b

a b

b

a .b a

a .a b

b.b a

a b b.b

115. (c) Given : ax 4 bx 2 c 0 Equation will be real if D 0 4ac

b2

0

116. (a) Given : log 3 x log 3 log 3 8 x

log3

1 1 1 1 1 x 2 4 8 16

log3 x

1

1 1 2

log3 x 2

117. (a)

x

1 x x

log 3 16

log 3

3

x 1 x

1 x x

x

x a b where x

x 4

a 1 r

S

x2

4

4

4

4 34

x

9

2

1

2

1, a .a

a

2

1

0

1 x 0 x2 1 0 x i x Thus, the given equation has no real roots.

a.b 1 is a scalar

The given vector is parallel to a b . 120. (a) Given : A, B & C are three points with coordinates (1, 2, –1), (2, 0, 3) & (3, –1, 2) respectively. Now, direction ratio's of AB = 2 – 1, 0 – 2, 3 + 1 = 1, – 2, 4 & direction ratio's of AC = 3 –1, –1 –2, 2 + 1 = 2, – 3, 3 we know that a1a 2

cos

0 1 x

b

a .b 1 a b

4ac x

b.a b

a .b a b

R 3 R1

clearly rank of A is 1 if a = –6

b2

2

a12

cos

b12

b1b 2 c1c 2 2 c1 . a 2 2 b 2 2

1 2

2

3

4 3

1 4 16 . 4 9 9 2 6 12 21 . 22

462 cos

20

20 462

c22

SOLVED PAPER 4.

PART - I (PHYSICS) 1

2.

3.

A potential difference of 300 V is applied to a combination of 2.0µF and 8.0 µF capacitors connected in series. The charge on the 2.0µF capacitor is (a) 2.4 × 10–4C (b) 4.8 × 10–4C –4 (c) 7.2 × 10 C (d) 9.6 × 10–4 C Two point charges 4 C and – 2 C are separated by a distance of 1 m in air. Then the distance of the point on the line joining the charges, where the resultant electric field is zero, is (in metre) (a) 0.58 (b) 0.75 (c) 0.67 (d) 0.81 Figure shows a triangular array of three point charges. The electric potential V of these source charges at the midpoint P of the base of the triangle is 1 4

9 109 Nm 2 C

5.

+

6.

2

–6

q3 = 3 × 10 C

7. 0.300 m –6

(a) 55kV (c) 63kV

–6

0.200 m

P

0.200 m

(b) 45kV (d) 49kV

q2 = –2 × 10 C

2006

A current of 5A is passing through a metallic wire of cross-sectional area 4 × 10–6m2. If the density of the charge carriers in the wire is 5 × 1026m–3, the drift speed of the electrons will be [e = 1.602 × 10–19C] (a) 1.56×10–2ms–1 (b) 1.98×10–2ms–1 (c) 2.42×10–2ms–1 (d) 2.84×10–2ms–1 The series combination of two capacitors shown in figure is connected across 1000V. The magnitude of the charges on the capacitors will be + – + – – 3pF 6pF

1000V

0

q1 = 1 × 10 C

VITEEE

8.

(b) 2 × 10–9 C (a) 3 × 10–9 C (c) 2.5 × 10–9 C (d) 3.5 × 10–9 C Three resistances of values 2 , 3 and 6 are to be connected to produce an effective resistance of 4 . This can be done by connecting (a) 6 resistance in series with the parallel combination of 2 and 3 (b) 3 resistance in series with the parallel combination of 2 and 6 (c) 2 resistance in series with the parallel combination of 3 and 6 (d) 2 resistance in parallel with the parallel combination of 3 and 6 The resistance of a field coil measures 50 at 20ºC and 65 at 70ºC. The temperature coefficient of resistance is (a) 0.0086/ºC (b) 0.0068/ºC (c) 0.0096/ºC (d) 0.0999/ºC The electrolyte used in Lechlanche cell is (a) copper sulphate solution (b) ammonium chloride solution (c) dilute sulphuric acid (d) zinc sulphate

EBD_7443 2006-2

9.

10.

11.

Target VITEEE

A galvanometer has a resistance of 50 . If a resistance of 1 is connected across its terminals, the total current flow through the galvanometer is [Ig represents the maximum current that can be passed through the galvanometer] (a) 42 Ig (b) 53 Ig (c) 46 Ig (d) 51 Ig In a tangent galvanometer, a current of 1A produces a deflection of 30º. The current required to produce a deflection of 60º is (a) 3A (b) 2A (c) 4A (d) 1A In the presence of magnetic field ‘B’ and electric field ‘E’, the total force on a moving charged particle is (a)

F

(b)

F q[(v E) B]

(c)

F q[(v B) E]

17.

18.

19.

(a)

v[(q B) E]

13. 14.

15.

16.

F B[(q E) v] A circular coil of radius 40 mm consists of 250 turns of wire in which the current is 20mA. The magnetic field in the center of the coil is [ = 4 × 10–7 Hm–1] (a) 0.785 G (b) 0.525 G (c) 0.629 G (d) 0.900 G RMS value of AC is _______ of the peak value. (a) 7% (b) 7.7% (c) 70% (d) 70.7% Q-factor can be increased by having a coil of (a) large inductance, small ohmic resistance (b) large inductance, large ohmic resistance (c) small inductance, large ohmic resistance (d) small inductance, small ohmic resistance A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in 0.5 second. The magnetic flux between the pole pieces is known to be 8 × 10–4 Wb. The emf induced in the wire is (a) 16 m V (b) 1.6 V (c) 1.6 m V (d) 16V Current in the LCR circuit becomes extremely large when (a) frequency of AC supply is increased (b) frequency of AC supply is decreased (c) inductive reactance becomes equal to capacitive reactance (d) inductance becomes equal to capacitance

2 3

(b) 3

1 3 (d) 2 2 In an experiment on Newton’s rings, the diameter of the 20th dark ring was found to be 5.82mm and that of the 10th ring 3.36 mm. If the radius of the plano-convex lens is 1 m, the wavelength of light used is (a) 5646 Aº (b) 5896 Aº (c) 5406 Aº (d) 5900 Aº What is the angular momentum of an electron in the fourth orbit of Bohr’s model of hydrogen atom?

(c)

20.

(d)

12.

Our eyes respond to wavelengths ranging from (a) 400 nm to 700 nm (b) 700 nm to 800 nm (c) 0 to (d) – to + A new system of units is evolved in which the values of 0 and 0are 2 and 8 respectively. Then the speed of light in this system will be (a) 0.25 (b) 0.5 (c) 0.75 (d) 1 A ray of light strikes a piece of glass at an angle of incidence of 60º and the reflected beam is completely plane polarised. The refractive index of glass is

21.

(a)

h 2

23.

24.

2h

h 4 The transition of an electrom from n2 = 5,6, .......... to n1 = 4 gives rise to (a) Pfund series (b) Lyman series (c) Paschen series (d) Brackett series The ground state energy of hydrogen atom is – 13.6 eV. What is the potential energy of the electron in this state? (a) –27.2 eV (b) –13.6 eV (c) +13.6 eV (d) 0 eV The longest wavelength that can be analysed by a sodium chloride crystal of spacing d = 2.82 Aº in the second order is (a) 2.82 Aº (b) 5.64 Aº (c) 8.46 Aº (d) 11.28 Aº

(c) h

22.

(b) (d)

Solved Paper 2006 25.

26.

27.

Which is the incorrect statement of the following? (a) Photon is a particle with zero rest mass (b) Photon is a particle with zero momentum (c) Photons travel with velocity of light in vacuum (d) Photons even feel the pull of gravity The deBroglie wavelength associated with a steel ball of mass 1000 gm moving at a speed of 1 ms–1 is [h = 6.626 × 10–34 Js] (a) 6.626 × 10–31m (b) 6.626 × 10–37m (c) 6.626 × 10–34m (d) 6.626 × 1034m The velocity v, at which the mass of a particle is double its rest mass is (a) v = c

(c) 28.

29.

30.

v

(b) 3 c 2

32.

v

(d) v = 2c

(b)

33.

34.

35.

3 c 4

How much energy is produced, if 2 kg of a substance is fully converted into energy? [c = 3 × 108 ms–1] (a) 9 × 1016 J (b) 11 × 1016 J (c) 15 × 1016 J (d) 18 × 1016 J The difference between the rest mass of the nucleus and the sum of the masses of the nucleons composing a nucleus is known as (a) packing fraction (b) mass defect (c) binding energy (d) isotopic mass The half life period of Radium is 3 minute. Its mean life time is (a) 1.5 minute

31.

2006-3

36.

37.

38.

3 minute 0.6931

(c) 6 minute (d) (3 × 0.6931) minute ‘Pair production’ involves conversion of a photon into (a) a neutron-electron pair (b) a positron-neutron pair (c) an electron-proton pair (d) an electron-positron pair The sub atomic particles proton and neutron fall under the group of (a) mesons (b) photons (c) leptons (d) baryons

When the conductivity of a semiconductor is only due to the breaking up of the covalent bonds, the semiconductor is known as (a) donor (b) extrinsic (c) intrinsic (d) acceptor In a P-type semiconductor, the acceptor impurity produces an energy level (a) just below the valence band (b) just above the conduction band (c) just below the conduction band (d) just above the valence band An oscillator is essentially (a) an amplifier with proper negative feedback network circuits (b) converts alternating current into direct current (c) an amplifier with no feedback network (d) an amplifier with proper positive feedback network circuits Which of the following gates can perform perfect binary addition? (a) AND gate (b) OR gate (c) EXOR gate (d) NAND gate The frequency of an FM transmitter without signal input is called (a) the centre frequency (b) modulation factor (c) the frequency deviation (d) the carrier swing The fundamental radio antenna is a metal rod which has a length equal to (a) in free space at the frequency of operation (b)

in free space at the frequency of 2 operation

(c)

in free space at the frequency of 4 operation

3 in free space at the frequency of 4 operation Vidicon works on the principle of (a) electrical conductivity (b) photoconductivity (c) thermal conductivity (d) SONAR

(d)

39.

EBD_7443 2006-4

40.

Target VITEEE

The maximum range, dmax, of radar is (a) proportional to the cube root of the peak transmitted power (b) proportional to the fourth root of the peak transmitted power (c) proportional to the square root of the peak transmitted power (d) not related to the peak transmitted power at all

PART - II (CHEMISTRY) 41.

42.

43.

44.

45.

46.

47.

48.

The equivalent weight of potassium permanganate when it acts as oxidising agent in ferrous ion estimation is (a) 158 (b) 31.6 (c) 79 (d) 39.5 The magnetic moment of lanthanide ions is determined from which one of the following relation? (a)

n(n 2)

(b)

g J(J 1)

(c)

g n(n 1)

(d)

2 n(n 1)

Which one of the following has maximum number of unpaired electrons? (a) Mg2+ (b) Ti3+ 3+ (c) V (d) Fe2+ Excess of NaOH reacts with Zn to form (a) ZnH2 (b) Na2ZnO2 (c) ZnO (d) Zn(OH)2 How many isomers does Co(en)2Cl2+ have? (a) 1 (b) 3 (c) 2 (d) 4 NH3 group in a coordination compound is named as (a) ammonium (b) ammine (c) amine (d) ammonia Name the complex Ni(PF3)4 (a) tetrakis (phosphorus (III) fluoride) nickel (0) (b) tetra (phosphorus (III) fluoride) nickel (c) Nickel tetrakis phosphorus (III) fluoride (d) (phosphorus (III) tetrakis fluoride) nickel (0) The purple colour of KMnO4 is due to (a) charge transfer (b) d-d transition (c) f-f transition (d) d-f transition

49.

50.

51. 52.

53.

54.

55.

56.

57.

58.

59.

How many lattice points belong to a face centered cubic unit cell? (a) 1 (b) 2 (c) 4 (d) 3 Schottky defect in solids is due to (a) a pair of cation and anion vacancies (b) occupation of interstitial site by a pair of cation and anion (c) occupation of interstitial site by a cation (d) occupation of interstitial site by an anion Which one of the following is amorphous? (a) Polystyrene (b) Table salt (c) Silica (d) Diamond The metal that crystallises in simple cubic system is (a) Po (b) Na (c) Cu (d) Ag When ideal gas expands in vacuum, the work done by the gas is equal to (a) PV (b) RT (c) 0 (d) nRT For a closed system consisting of a reaction N2O4(g) 2NO2(g), the pressure (a) remains constant (b) decreases (c) increases (d) becomes zero 6 moles of an ideal gas expand isothermally and reversibly from a volume of 1 litre to a volume of 10 litres at 27ºC. What is the maximum work done? (a) 47 kJ (b) 100 kJ (c) 0 (d) 34.465 kJ The reaction, Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s) is an example of a (a) spontaneous process (b) isobaric process (c) non-spontaneous process (d) reversible process For the reaction, H2 (g) + I2 (g) 2HI (g) (b) Kc = 0 (a) Kp = –Kc (c) Kp = Kc (d) Kp = 0 The increase of pressure on ice water at a constant temperature will cause (a) water to vaporize (b) water to freeze (c) no change (d) ice to melt The order of the reaction N2O5 (a) 3 (c) 1

N2O4(g) +

1 O (g) is 2 2 (b) 2 (d) 0

Solved Paper 2006 60.

61.

62.

63.

64.

65.

66.

67.

2006-5

The reactions with low activation energy are always (a) adiabatic (b) slow (c) non-spontaneous (d) fast For a cell reaction to be spontaneous, the standard free energy change of the reaction must be (a) zero (b) positive (c) infinite (d) negative Equivalent conductance of an electrolyte containing NaF at infinite dilution is 90.1 Ohm–1cm2. If NaF is replaced by KF what is the value of equivalent conductance? (a) 90.1 Ohm–1cm2 (b) 111.2 Ohm–1cm2 (c) 0 (d) 222.4 Ohm–1cm2 The tendencies of the electrodes made up of Cu, Zn and Ag to release electrons when dipped in their respective salt solutions decrease in the order (a) Zn > Ag > Cu (b) Cu > Zn > Ag (c) Zn > Cu > Ag (d) Ag > Cu > Zn The electrode reaction that takes place at the anode of CH4 – O2 fuel cell is (a) 2O2 + 8H+ + 8e– 4H2O (b) CH4 + 2H2O CO2 + 8H+ + 8e– CO2 + 2H2O (c) CH4 + 2O2 (d) 2H+ + 2e– H2 What is the hybridization of oxygen atom in an alcohol molecule? (a) sp 3 (b) sp (c) sp 2 (d) p 2 O LiAlH 4 ? R–C–OH (b) RCHO (a) RCH2CH2OH (c) RCOR (d) RCH2OH Which one of the following is correct? (a) RCH2OH

KMnO4

(b) CH3CH2OH

Na 2Cr2O7 ,H 2SO4

68.

69.

O

CH3–C–CH3 +

(c) CH3CHO Na 2Cr2O7 ,H 2SO4

alkaline KMnO 4

No reaction

O–CH2

(c) 70.

71.

72.

73.

74.

CH3

H3C

C

?

COOH

+

H3C

HCl

CH2OH

(a) CH3–CHOH (b)

No reaction

CH3 (d) CH3–C–OH

CH2OH

CH 3

No reaction No reaction

Which one of the following products obtained when diethyl ether is boiled with water in presence of dilute acid? (a) Glycol (b) Ethy1 alcohol (c) Ethylene oxide (d) Peroxide Identify the product for the following reaction

COOH

O–CH2

H3C C–OH H3C CH3–CHOH

(d) No reaction What is the reaction of acetaldehyde with concentrated sulphuric acid? (a) No reaction (b) Decomposition (c) Charred to black residue (d) Polymerisation Calcium Acetate on heating under distillation gives (a) Acetaldehyde and Calcium Oxide (b) Calcium Carbonate and Acetic acid (c) Acetone and Calcium Carbonate (d) Calcium Oxide and CO2 Identify the correct statement (a) Aldehydes on reduction give secondary alcohols (b) Ketones on reduction give primary alcohols (c) Ketones reduce Fehling’s solution and give red cuprous oxide (d) Ketones do not react with alcohols The O – H stretching vibration of alcohols absorbs in the region 3700 – 3500 cm–1. The O – H stretching of carboxylic acids absorb in the region (a) 3900 – 3700 cm–1 (b) 3000 – 2500 cm–1 (c) 3700 – 3500 cm–1 (d) 1700 – 2000 cm–1 Which among the following reduces Fehling’s solution? (a) Acetic acid (b) Formic acid (c) Benzoic acid (d) Salicylic acid

EBD_7443 2006-6

75.

Target VITEEE

Determine the experimental condition for the following reaction COOH

82.

76.

77.

78.

79.

80.

OH

(a) in presence of KOH (b) on heating (c) in presence of NaOH (d) in presence of HCl Which one of the following is an ingredient of Pthalic acid manufacture by catalytic oxidation (a) Benzene (b) Salicylic acid (c) Anthranilic acid (d) naphthalene On comparison with H C H bond angle of methane, the C N C bond angle of trimethylamine is (a) higher (b) no change (c) not comparable (d) lower The treatment of acylazide (RCON3) with acidic or alkaline medium gives (a) RCONH2 (b) R – NH2 (c) RCH2 NH2 (d) RCOCHNH The sequence of basic strength of alky1 amines follows the order (a) RNH2 < R2NH > R3N (b) R2NH2 < R2NH < R3N (c) R2NH < RNH2 < R3N (d) RNH2 < R2NH < R3N Activation of benzene ring in aniline can be decreased by treating with (a) dil. HCl (b) ethyl alcohol (c) acetic acid (d) acetyl chloride

83.

84.

85.

A

1

x

1

1 x

(a) ±1 (c) ±3

2 2x 2 is singular, is 2

(b) ±2 (d) ±4

1

3

2

2

5

t

86.

, then the values of t

6

for which inverse of A does not exist (a) –2, 1 (b) 3, 2 (c) 2, –3 (d) 3, –1 The non integer roots of 3x 3

2x 2

1 (a) (3 2

13),

3x 1 0

1 (3 2

13)

(b)

1 (3 2

13),

1 (3 2

(c)

1 (3 2

17 ),

1 (3 2

(d)

1 (3 2

17),

1 (3 2

If e x (a)

y 1 x (e 2 x

87. 1

Let A

x4

The value of x, for which the matrix 2 x

two roots are (a) 3, 7 (b) 2, 7 (c) 3, 6 (d) 2, 6 The values of for which the system of equation x + y + z = 1, x + 2y + 4z = , x + 4y + 10z = 2 is consistent are given by (a) 1, –2 (b) –1, 2 (c) 1, 2 (d) 1, 1

4 7 t

PART - III (MATHEMATICS) 81.

0 , then other

7 6 x

+ CO2 OH

x 3 7 If x = –9 is a root of 2 x 2

13) 17 )

17)

1 y2 , then the value of y is e

x

)

(b)

x 2

1 x (e 2 x

e

x

x 2

)

(c) e e (d) e e Consider an infinite geometric series with the first term a and common ratio r. If its sum is 4 and the second term is

3 , then 4

(a) a =

4 3 ,r= 7 7

(b) a = 2, r =

3 8

(c) a =

3 1 ,r= 2 2

(d) a = 3, r =

1 4

Solved Paper 2006 88.

If and are the roots of the equation ax2 + bx + c = 0, then the value of 3 + (a)

(c)

89.

3abc b3

3abc b

a 3 b3 3abc

(b)

a3

3 is

95.

96. 3

3

(3abc b )

(d)

a3 a The volume of the tetrahedron with vertices P (–1, 2, 0), Q ( 2, 1, –3), R (1, 0, 1) and S (3, –2, 3) is

(a) (c) 90.

2006-7

3

1 3

(b)

1 4

(d)

ˆ b iˆ 2jˆ 3k,

If a

then tan (a)

2 3

92.

3 4

97.

ˆi 2jˆ kˆ and

to c will be equal to (a) 5 (b) 4 (c) 6 (d) 2 An equation of the plane passing through the line of intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0 and passing through (1, 1, 1) is (a) 2x + 3y + 4z = 9 (b) x + y + z = 3 (c) x + 2y + 3z = 6 (d) 20x + 23y + 26z = 69 The length of the shortest distance between the ˆ and lines r 3iˆ 5jˆ 7kˆ (iˆ 2ˆj k)

r

ˆi ˆj kˆ

ˆ is (7iˆ 6jˆ k)

(a) 83 units 93.

(b)

6 units

(d) 2 29 units (c) 3 units The region of the argand plane defined by

z i (a) (b) (c) (d)

z i

4 is

interior of an ellipse exterior of a circle interior and boundary of an ellipse interior of a parabola 13

94.

The value of the sum

(i n

n 1

i 1 equals (a) i (c) – i

A B tan is equal to 2 2

2 3

(b)

3 2

1 (d) 5 5 In a model, it is shown that an arc of a bridge is semielliptical with major axis horizontal. If the length of the base is 9m and the highest part of the bridge is 3m from horizontal; the best approximation of the height of the arch, 2m from the centre of the base is

(c)

c 3iˆ ˆj then t such that a tb is at right angle

91.

If sin , cos , tan are in G.P. then cos9 + cos6 + 3cos5 – 1 is equal to (a) –1 (b) 0 (c) 1 (d) 2 If in a triangle ABC, 5cosC + 6cosB = 4 and 6cosA + 4cosC = 5,

(b) i –1 (d) 0

i n 1 ) where

(a)

11 m 4

(b)

8 m 3

7 m (d) 2 m 2 The number of real tangents through (3,5) that can be drawn to the ellipses 3x2 + 5y2 = 32 and 25x2 + 9y2 = 450 is (a) 0 (b) 2 (c) 3 (d) 4 If the normal to the rectangular hyperbola xy =

(c)

98.

99.

c2 at the point ct,

c t

meets the curve again at

c , then t' (a) t3t' = 1 (b) t3t' = –1 (c) tt' = 1 (d) tt' = –1 100. An equilateral triangle is inscribed in the parabola y2 = 4x one of whose vertex is at the vertex of the parabola, the length of each side of the triangle is ct ',

(a) (c)

3 2

8

3 2

3 2

(b)

4

(d)

8 3

EBD_7443 2006-8

Target VITEEE

101. If f(2) = 4 and f '(2) = 1, 2 0

(a) 0 1 (b) 2

(a) 0

(c)

105. If f ' x

(a)

2 (1 3

u y y

(b) cos 2u (d) tan 2u x 1 x

3 x) 2

and f(0) = 0, then f(x) =

6(1

(c)

a2 12

(d)

1 x) 2

1

(a) (b)

d2 y dx

2

2

d2 y dx 2 d2 y

(c)

a

(d)

2a

dx 2 d2 y dx 2

dy dx

3

(b)

3

(c)

2 (1 x) 2 3 3

(d)

2 (1 x) 2 3

3(1

1 x) 2

2

1

4(1 x) 2

a2 24

0 3

dy dx

0 3

dy dx

0

dy dx

3

0

110. The solution of xcosec 2 (1 x) 2 3

a2 8

1

x 3 y3 u x y , then x x

(a) sin 2u (c) sec2 2u

(b)

(a) e 3 (b) 4e3 (c) 4(e3 – e) (d) 4e3 – 2e 109. The differential equation that represents all parabolas each of which has a latus rectum 4a and whose axes are parallel to the x – axis is

1 e e

(d)

104. If u = tan –1

a2 6

108. The value of the integral e t dt

1

1 ee

(a)

9

is

(b) ee

(a) e

(d) 2 4 107. What is the area of a loop of the curve r = asin3 ?

x

1 x

(b) 1

(c)

(c) 1 (d) 2 102. What is the least value of k such that the function x2 + kx + 1 is strictly increasing on (1,2) (a) 1 (b) –1 (c) 2 (d) –2 103. The maximum value of

log(tan x)dx

106. The value of the integral

xf (2) 2f (x) is equal to then lim x 2 x 2

2

is (a)

log x

cos

x y

c

(b)

log x

cos

y x

c

(c)

log x

sin

x y

c

(d)

log x

sin

y x

c

1

3(1 x) 2 1

y x

y dx + xdy = 0

Solved Paper 2006

2006-9

d2 y

111. The particular integral of (a) x2 –1

dx 2

2y

2

x is

(b) x2 + 1

1 2 1 2 (x 1) (x 1) (c) (d) 2 2 112. The solution of (D2 + 16) y = cos4x is

(a) Acos4x + Bsin 4x +

x sin 4x 8

(b) Acos4x + Bsin 4x

x sin 4x 8

(c) Acos4x + Bsin 4x

x sin 4x 4

(d) Acos4x + Bsin 4x

x sin 4x 4

113. Determine which one of the following relations on X = {1,2,3,4} is not transitive. (a) R1= , the empty relation (b) R2 = X x X, the universal relation (c) R3 = {(1,3), (2,1)} (d) R4 = {(1,1), (1,2), (2,3), (1,3), (4,4)} 114. Find the number of ways in which five large books, four medium-size books, and three small books can be placed on a shelf so that all books of the same size are together. (a) 5 × 4 × 3 (b) 5! × 4! × 3! (c) 3 × 5! × 4! × 3! (d) 3! × 5! × 4! × 3! 115. Consider the set Q of rational numbers. Let be the operation on Q defined by a b = a + b – ab. The identity element under is (b) 1 (a) 0 (c) 2 (d) not exist 116. The statement ~ p q is equivalent to (a) p q (b) ~ p q (c)

~p

117. In rolling two fair dice, what is the probability of obtaining a sum greater than 3 but not exceeding 6?

~q

(d)

p

~q

(a)

1 2

(b)

1 3

(c)

1 4

(d)

1 6

118. Team A has probability

2 of winning whenever 3

it plays. Suppose A plays four games. What is the probability that A wins more than half of its games? (a)

16 27

(b)

19 27

(c)

19 81

(d)

32 81

119. An unprepared student takes five-questions of true-false type quiz and guesses every answer. What is the probability that the student will pass the quiz if at least four correct answers is the passing grade? (a)

1 16

(b)

3 16

(c)

1 32

(d)

3 32

120. The probability density f(x) of a continuous random variable is given by f(x) = Ke

x

,

x

. Then the value of K is

(a)

1 2

(b) 2

(c)

1 4

(d) 4

EBD_7443 2006-10

Target VITEEE

2006 SOLUTIONS PART - I (PHYSICS) 1.

3.

(b) V= 300V, C1= 2.0 F, C2= 8.0 F,

1 Cs

Net capacitance,

1 C1

1 C2

4

4

E=

4.

0

q1

1 0

4 C x2 2 1– x

2

2

2

x

2 1– x

x

1 1.414 x

0.58m

0

2 10 0.2

6

3 10 0.3

6

1– x

nAev

v

1

x2

1– x

5 10

1.414 2.414

x

26

4 10

6

1.60 2 10 –19

4 1.602 101 10 –1 =1.56 ×10 –2 m/s 6.408 (b) In series combination of capacitors, charges on both capacitors will be same. Q C1

Q C2

1000

Q

1 C1

1000

Q

C1 C2 C1C2

Q

1.414 –1.414x x

5

2

x

I nAe

1

Vs

2

45 103 V = 45 kV

drift velocity, v

5.

r22

–2 C

2

1– x

1.414 1– x

x

x2

2 C

Taking root

1.414

4

4 C

4

q2

1

2 0 r1

4

6

q3 r3

(a) In a metal, conduction current is due to electrons given by I

0

1

1 10 0.2

9 103 5

C C P x Let the point P where resultant field is zero be x m from 4 C charge and (1– x) m distance apart from –2 C charge. Since field is zero at this point then,

E1 E 2

0

q2 r2

9 109 5 –10 10 10 –6

2 8 16 = 1.6 F.. 2 8 10 Now total charge, Q = Vs ×Cs= 300 × 1.6 × 10–6 = 4.8 ×10– 4 C. In series charge is same on capacitors Charge on 2 F capacitor is 4.8 × 10– 4C

E

0

1

V

Cs

(a)

q1 r1

1

V

C1C2 C1 C2

Cs

2.

(b) The net electric potential is algebraic sum of potential due to individual point charges.

1 C2

1000 C1C 2 C1 C2

Q 1000

3 10 –12 6 10–12

2 10 –9 C

3 6

10

–12

18 10 –9 9

Solved Paper 2006 6.

(c) Parallel combination of 3 effective resistance, 3 6 3 6

Rp

7.

2006-11

18 9

2

and 6

( n = no. of turns, I = current through coil, r = radius of coil)

gives

. This in series with

2 gives net ressistance as 4 . (b) The value of temperature coefficient of resistance is given by R 2 – R1 R1 t 2 – t1

65 – 50 50 70 – 20

8. 9.

Ig(galvanometer) = I

13.

(d)

G

I1 I2

tan tan

2

1 I2

(a)

15.

(c)

1.7321 0.5774

Fe F

(a) In a circular coil of n turns, magnetic field is B

0 nI

2r

16.

4

10 –7 250 20 10 –3

2 40 10 –3

–d . Assuming, small dt

8 10– 4 80 10 – 4 5 0.5 – 4 =16 × 10 =1.6×10–3 V = 1.6 mV (c) Current through an LCR circuit is maximum when impedance is minimum. Now impedance

R2

Z

L–

1 C

2

is minimum at

1 and C Z= R = minium i.e., inductive reactance ( L) is equal to capacitive reactance (1/ C) (a) Our eyes respond to visible range from 400 nm to 700 nm (a) Velocity of electromagnatic wave in space

resonance frequency when

17. 18.

qE q v B

Induced emf e

|e| =

2.999Å

Fm

1 L R C If resistance R is decreased, Q increases and inductance L is increased, Q increases.

Q-factor is given by Q

change in flux d =8×10 – 4 Wb change in time dt = 0.5s

tan 30 tan 60

I 2 3A (c) Lorentz force on a charged particle in presence of magnetic and electic field is

F

12.

1

tan 60 tan 30

I2

11.

14.

G S 50 1 Ig 51Ig S 1 (a) Current in tangent galvanometer H tan G Where G = galvanometer constant H= earth's horizontal field = constant

I0

0.707I 0 2 I0 = peak value it is 70.7% of peak value.

S

I

785 10 –7 0.785 10 –4 tesla = 0.785 gauss RMS value of A.C is

Iv

Ig

I

10.

S

3

250 3.14 10 –7

(t1 and t2 are in °C) 15 0.006 / C 50 50 (b) In Leclanche cell a strong solution of ammonium chloride acts as an electrolyte. (d) In the galvanometer, Ig = max. current th rough galvan ometer, S = shunt resistance, G = galvanometer resistance then

250 20 10 –7 –3 2 40

4

B

1

is c

0

1 16

1 4

c 0

0.25

1 2 8

L

EBD_7443 2006-12

19.

20.

Target VITEEE

(b) According to Brewster's law, reflected light is plane polarised if unpolarised light falls at the interface of air and medium at an angle ip called polarising angle then = tan ip (glass) = tan 60 3 (a) Newton's ring arrangement is used for determin ing the wavelength of monochromatic light. For this the diameter of nth dark ring (Dn) and (n + p)th dark ring (Dn + p) are measured then D2(n p)

4(n p) R and D2n

D2n

4n R

25.

m0 m 1 – v2 / c2 hence m0 (photon)= 0 photon has zero rest mass.

Momentum of photon = 26.

27.

4pR

Here, n = 10, n + p = 20; p = 10; R = 1 m, D10 = 3.96 × 10–3 m, D20 = 5.82 × 10–3 m D 220

2 D10

4pR 3 2

21.

is mvr mvr

22.

23.

nh for 4th orbit, n = 4 2 4h 2

E

KZe 2 2r

28.

K.E. P.E.

–KZe2 r P.E. = 2 × E P.E.=2 × ( –13.6)= –27.2 eV (a) Bragg's condition is 2dsin = n for second order n = 2 , sin =1. For longest d=2 =d

30.

1 2

1–

v2 c2

v2

2m 0 1 –

v2 c2

1 v2 =1 – 2 4 c

c2

1–

m0

1 4

3 4

3 c 4 (d) By Einstein's equation E= mc2 where m = 2kg E = 2×(3 × 10 8 )2 = 2 × 3 × 3× 10 16 = 18 × 1016 J (b) By definition, the difference between the sum of the masses of neutrons and protons forming a nucleus and mass of nucleus is called mass defect (b) Mean life time = 1.44 T where T is half life period of an atom T 3 minute 0.6931 0.6931 (d) (by conservation of charge) (d) Baryons are proton, neutron, lamda,

= 1.44 T =

31. 32.

sigma (

Potential energy in the orbit P.E.

24.

m 1–

v

29.

(d) According to Bohr's, Brackett series is obtained when an electron jumps to 4th orbit from any other outer orbit (a) Total energy of electron

6.626 10 –34 m

–3

m0

c2

(5.82 10 ) (3.36 10 ) 4 10 1

= 5646 Å (b) Angular momentum in any stationary orbit

6.626 10 –34

1000 10 1 (b) Let the velocity of a particle be v where mass m is double the rest mass i.e., m = 2m0 then

v2 3 2

h

(c) de Broglie wavelength is given by h mv

D2n

p

(b) Photon moves with speed of light ie, v = c and rest mass of a particle is

33.

–), X i

0

,

–

(c) As donor and acceptor impurities are added to semiconductor to make an extrinsic semiconductor, intrinsic semiconductor is formed by internal generation of e– by breaking up of covalent bonds.

Solved Paper 2006 34.

35.

36.

2006-13

(d) In p-type semiconductor, valency = 3, thus Y there is one unformed bond or hole created. This hole is in valence band and is able to cause hole current. The energy levels of acceptor are in forbidden gap just above valence band (d) In an oscillator, L-C circuit is coupled with transistor amplifier in such a way that there is a positive feed back to the LC circuit i.e., proper energy supply to LC at proper timings. So that total energy of LC circuit remains same. (c) The gates AND, OR, NAND do not give binary addition, however in EXOR gate truth table is

37.

A 0 0 1 1

B 0 1 0 1

max

fc

40.

K2SO4 + 2MnSO4 + 8H2O + 5Fe2(SO4)3 Eq. mass of KMnO4 Molecular mass change in oxidation number

= 42.

l/

2

l

p and d

p

1/ 4

158 = 31.6 5

(b) In case of lanthanoids, 4f orbitals lie too deep and hence the magnetic effect of the motion of the electron in its orbital is not quenched out. Here spin contribution S and orbital contribution L couple together to give a new quantum number J. given by,

g J(J 1)

where J = L – S when the shell is less than half fill J = L + S when the shell is more than half fill

held vertically

(b) Maximum range of radar d max l and power transmitted by antenna of length l is

Molecular mass 5

Thus magnetic moment of lanthanoids is

)

4 with its lower end touching the ground. (b) The vidicon is a storage-type camera tube in which a charge-density pattern is formed by the imaged scene radiation on a photoconductive surface which is then scanned by a beam of low-velocity electrons. The fluctuating voltage coupled out to a video amplifier can be used to reproduce the scene being imaged. The electrical charge produced by an image will remain in the face plate until it is scanned or until the charge dissipates.

p

(b) The oxidation of ferrous ion by KMnO4 takes place in acidic medium as per following reaction 2KMnO4 + 8H2SO4 + 10FeSO4

(c) The common antenna is a straight conductor of length l

39.

41.

This shows it gives perfect binary addition (d) In FM, carrier frequency is the constant frequency which is modulated by signal amplitude. It is also called carrier swing. (Centre frequency is fc in AM wave, frequency deviation f max – fc , mod ulation factor

38.

Y 0 1 1 0

PART - II (CHEMISTRY)

and g 1 43.

1 2

S(S 1) L(L 1) 2J(J 1)

(d) Mg2+ = 1s2, 2s2, 2p6 (No unpaired electrons) Ti3+ = 1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d1 (One unpaired electrons) V3+ =

1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d2 (Two unpaired electrons)

Fe2+ = 1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d6 (Four unpaired electrons) Fe2+

has highest number of unpaired electrons. 44.

(d) Zn + 2NaOH

Na2ZnO2 + H2 Sod. zincate

EBD_7443 2006-14

45.

Target VITEEE +

(c)

+

en

en

Cl

54.

(b) As the system is closed, hence the reaction will be reversible, hence according to Le-chatelier principle pressure decreases since the volume is increasing.

55.

(d) W = – 2.303 nRT log

Cl

Co

Co Cl

Cl

Given n = 6, T = 27°C = 273 + 27 = 300 K V1 = 1 L, V2 = 10 L

en

en

V2 V1

l - cis form

d - cis form

W = – 2.303 × 6 × 8.314 × 300 log +

Cl

56. en

Co

en

57. Cl trans form (meso)

46.

47. 48.

49.

(b) Neutral ligands are given the same names as the neutral molecule. However, two very important exceptions to this rule are: H2O Aquo (Aqua) NH3 Ammine. (a) Ni(PF3)4 – tetrakis phosphours (III) fluoride nickel (0). (a) The colour of KMnO4 is due to charge transfer. The configuration of manganese in permagnate ion is d0 but it is coloured because its electrons are photo-exited. (c) In face centred cubic lattice, the atoms are present at eight corners of faces and one each at 6 faces. Lattice points belonging to face centred 1 1 6 4 8 2 (a) Schottky defect is caused when equal number of cations and anions are missing from their lattice sites. (a) Polystyrene is thermoplastic substance. (a) Po – Simple cubic lattice Na – bcc Cu – fcc

58.

59. 60.

61.

62.

cubic unit cell = 8

50. 51. 52.

53.

(c)

Wirr

Pext

R P1T2

P2 T1

P1P2

During expansion in vaccum Pext = 0 work done = 0.

63.

64. 65. 66.

10 1

= 34.465 kJ (a) It is spontaneous process because zinc is more reactive than copper, hence can easily repace Cu from CuSO4. (c) Kp = Kc (RT) n n = np(g) – nr(g) = 2 – 2 = 0 Kp = Kc (d) Ice Water The volume of ice is more than water. Therefore when pressure is increased the equilibrium shifts in forward direction. It favoures melting of ice. (c) It is a first order reaction because rate of reaction [N2O5] (d) The reactions with low activation energies are always fast whereas the reactions with high activation energy are always slow. (d) For spontaneous reaction free energy change is negative. G = – nFE (a) Because at infinite dilution the equivalent conductance of strong electrolytes furnishing same number of ions is same. (c) Reducing character i.e tendency to loose electron decreases down the series, hence the correct order is Zn > Cu > Ag. (b) At anode the following reaction takes place CH4 + 2H2O CO2 + 8H + + 8e – (a) Oxygen atom in alcohol molecule is sp3 hybridised. (d) In this reaction LiAlH4 acts as reducing

O || agent. R C OH

LiAlH4

RCH2OH

Solved Paper 2006 67.

68.

2006-15

(d) 3° alcohols are resistant to oxidation and are oxidised only by strong oxidising agents like conc. HNO3. They are resistant to oxidation in neutral or alkaline KMnO4. (b) C2H5 – O – C2H5

H 2O boil, dil.acid

2C2H5OH

73.

74.

(d) The O – H strecting of carboxylic acid absorb in region of 1700-2000 cm–1

(b) Formic acid

69.

(b)

CH3

75.

(b) Conc. H 2SO4 drops Room temp.

O

3 CH3CHO Dil. H2SO4

CH3 CH

+ CO2 This is decarboxylation reaction. 76.

COOH COOH Phthalic acid 77.

Conc. H 2SO4 or HCl gas

H3C – CH – O – CH – CH3

dil H 2SO 4

H3C – CH – O – CH – CH3

O

O

distillation

CH3COCH3 + 2CaCO3 Acetone Calcium carbonate (d) Ketones do not react with alcohol. reduction

Aldehydes e.g., CH3CHO

(ii) Ketones CH3

reduction

C=O

1° alcohol

CH 3CH 2OH

2° alcohol CH3

CHOH CH3 CH3 (iii) Ketones do not reduce Fehling solution but aldehydes do so.

(b) In both the cases carbon is sp3 hybridised and bond angle is 109°28'. (b) This reaction is known as curtius rearrangement. RCON3

N2

RNCO

2NaOH

Na2CO3 + RNH2

Metaldehyde (used as solid fuel in spirit lamps)

(i)

Catalytic oxidation

(d)

O

78.

72.

OH

O

H3C

(c) 2(CH3COO)2Ca

OH

Napthalene

CH3 Para aldehyde (pleasant smelling liquid, used as hypnotic and soporofic) (sleep producing)

71.

NaOH

CH

CH

4 CH3CHO

(c)

O – CH2

H3C O – CH2 cyclic ketal 70.

COOH

HCl

H3C

has – CHO

group and therefore it reduces Fehling solution.

ethyl alcohol O CH 2OH || C CH3 + | CH 2OH

O || H C OH

79.

1° amine is formed. (a) It is expected that the basic nature of amines should be in order tertiary > secondary > primary but the observed order in the case of lower members is found to be as secondary > Primary > tetriary. This anomalous behaviour of tetriary amines is due to steric factors i.e crowding of alkyl groups cover nitrogen atom from all sides thus makes the approach and bonding by a proton relatively difficult which results the maximum steric strain in tetiary amines. The electrons are there but the path is blocked, resulting the reduction in basicity. Thus the correct order is R2 NH > R NH2 > R3N.

EBD_7443 2006-16

80.

Target VITEEE

(d) On acetylation aniline is converted into acetamide which is resonance stablised and therefore less reactive. NH2

(x 9)(x 2 9x 14)

(x 9) (x 2 7x 2x 14)

NHCOCH3

+ CH3COCl

0

(x 9) (x 7) (x 2)

83.

0

0

x 9, 7, 2 (c) We have

Acetanilide

PART - III (MATHEMATICS) 81.

|A|

|A|

1 x

2x

1

1 x

2

2 2x

2x 2x x3

x2

2

0

2 x

2 2

2

2x

& R3

1 2 x x

1 1 1 : ~ 0 1 3 :

0

1 2

R2

1 R1

R 3 R1

1 1 2

0 0 0 :

3

2

0

applying R 3

R 3 3R 2

But the system is consistent 0

2

0

84.

1 3 7 0

x 3 67x 126 0 But given (x = 9) is a root of given determinant. (x + 9) is a factor x 3 9x 2 9x 2 81x 14x 126 x (x 9) 9x(x 9) 14(x 9)

3

0 0

2

0

2 or 1 ( 2) ( 1) 0 (c) We know that inverse of A does not exist only when |A| = 0 1

3

2

2

5

t

4 7 t

x[x 2 12] 3[2x 14] 7[12 7x] 0

2

1

0

x 1 0

(b) Given 2 x 2 7 6 x

2

0 3 9 :

1 2 2x

2 2x

(x 1) (x 2 1)

1

applying R 2

x (x 1) 1(x 1)

x

:

1 1 1 :

2

x

:

4

~ 0 1 3 :

2 2x 2x x

3

1

2

1

2 [0] x

1 1

1 2

1 4 10 :

(a) We know that, A is singular if |A| = 0 2 x

82.

A:B

0

6

( 30 7t t 2 ) 3( 12 4t)

2(14 2t 20) 30 7t t 2 36 12t 12 4t t2

t 6

t 2 3t 2t 6

0

t(t 3) 2(t 3)

0

(t 3)(t 2)

t

0

2, 3

0 0

0

Solved Paper 2006 85.

(a)

2006-17

Given x4 – 3x3 – 2x2 + 3x + 1 = 0

(x 1) [x

3

x

2

3x

2

16r 2 16r 3

3x x 1] 0

2

(x 1) x (x 1) 3x(x 1) 1(x 1)

(x 1) (x 1) (x 2 3x 1)

0

0

x = 1, – 1 or Now x2 – 3x – 1 = 0

3

3

b2 2a

4ac

88.

86.

1 13), (3 2

e

x

y

13)

1 y

e 2x 1

87.

2e

y

x

1 2

ex

e

3

3

4

3 4 3 4r

1 r

b a

b3

3bc

3

a2

b 3 3abc

1 PQ PR PS 6

Now, PQ

& an

c 3 . a

)

a3 (b) Given : The vertices of tetrahedron are P(–1, 2, 0), Q(2, 1, –3), R(1, 0, 1) & S(3, –2, 3)

(2 1)iˆ (1 2)ˆj ( 3)kˆ

& PS

a

(

Volume of tetrahedron

...(1) S

3

b a

3

Similarly, PR

1 r

a

3

)3 3

x

3 4

4 & a2

a

& ar

3

3

89.

3

c a

(

a

(d) First term = a & common ratio = r Given S

3

2

Squaring both side, we have e2x + y2 – 2exy = 1 + y2 2ex y = e2x – 1 y

3

Now,

1 y2

y

1 4

4

3 & r

13

(b) Given e x

3

b & a

2 non-integer roots of given equation are 1 (3 2

1 4

1 4 (c) Given : & are roots of equation ax2 + bx + c = 0

a

b

0

1 then a 4

when r

9 4 2 x

x

3 or r 4

r

4

0

(4r 3) (4r 1)

x2 – 3x – 1 = 0

x

3 4r(1 r)

Equation (1) becomes

By using Hit & trial method, we have (x – 1) is a factor of given equation (x – 1) (x3 – 2x2 – 4x – 1) = 0

ar n

1

2iˆ 2ˆj kˆ

4iˆ 4jˆ 3kˆ

Volume of tetrahedron 3 1 2 6 4

1

3

2

1

4

3

2 3

3iˆ ˆj 3kˆ

EBD_7443 2006-18

90.

Target VITEEE

ˆ t( ˆi 2ˆj k) ˆ (a) We have, a tb (iˆ 2ˆj 3k)

b1 b2

ˆi( 2 6) ˆj(1 7) k( ˆ 6 14)

b1 b2

4iˆ 6jˆ 8kˆ

(1 t)iˆ (2 2t)ˆj (3 t)kˆ It is

3iˆ ˆj

to c

Shortest distance

If 3(1 t) (2 2t) (3 t) (0) 91.

0

3 3t 2 2t 0 t 5 (d) The equation of the plane through the line of intersection of the given planes is (x + y + z – 6) + (2x + 3y + 4z + 5) = 0 ... (1) If equation (1) passes through (1, 1, 1), we have 3 14

3 in (1), we obtain the 14 equation of the required plane as 3 (x y z 6) (2x 3y 4z 5) 14

PQ 93.

0

b1 b 2 . a 2 a1

13

94.

(b)

in

13

1

n 1

i

i

>

>

>

in

in

n 1

1 i13 1 i (1 i) (1 i)

i2 (1 i) (1 i)

13

in

1

n 1

i13 1 i i 1

>

>

>

>

>

>

(b) Given : sin , cos , tan cos 2

sin tan

cos3

1 cos 2

(cos3

A (a1)

P

r = (3i + 5j + 7k) + (i – 2j + k)

And b1 b2

95.

are in G.P.. cos3

sin 2

cos 2 ) 1

...(1)

Cubic both sides, we have

>

>

>

L1

2 29 units

This equation represent the interior and boundary of ellipse with foci at (0, 1) & (0, –1), whose major axis is along the y-axis.

b1 b 2

B (a 2) 6j + k) – i 7 ( µ j–k+ r = –i – Q L2

2 29

| z (0 i) | | z (0 i) | 0

4iˆ 6jˆ 8kˆ

a 2 a1

116

(c) Given, | z i | | z i | 4

ˆi ˆj kˆ 3iˆ 5jˆ 7kˆ

Now, a 2 a1

116

116

20x 23y 26z 69 0 (d) Shortest distance PQ

16 36 64

116

Putting

92.

16 36 64 PQ

3 14

0

ˆ . ( 4iˆ 6ˆj 8k) ˆ (4iˆ 6ˆj 8k)

PQ

cos

cos6

3cos5 . (cos3

cos2 ) 1

cos 6

1

ˆi 1

ˆj kˆ 2 1

cos

7

6 1

cos

3cos5

[Using equation (1)] cos6

3cos5

1 0

Solved Paper 2006 96.

2006-19

(c) Given : 5cosC + 6cosB = 4 6cosA + 4cosC = 5 Adding eq. (1) & (2), we have 9cosC + 6(cosA + cosB) = 9 A B A B . cos 2 2

9cos C 6 2cos 9 cos C 9 12cos

2

9(cos C 1) 12sin 9 1 2sin 2

18sin 2 3sin

C 2

C 1 2

B

3cos

2

3 cos

C A B . cos 2 2

C A B . cos 2 2

12sin

4x 2 81

y1

0

0

97.

A B . tan 2 2

0

x2

y2

1 a2 b2 Here, 2a = 9

a

x2 81 4

9 , 2

y2 9

....... (1)

b= 3 1

X

P(2, y1)

0

Since point P lies on the ellipse (2) y12 9

4 4 81

y12 9

1

y12

65 9

16 81

1

81 16 81

y1

65 81

65 3

8 m 3

Hence, best approximation of the height of

A B cos . cos 2 2

B 1 A tan .tan 2 2 5 (b) Equation of the semielliptical bridge

2m Q

Here, OQ = 2 m, let PQ = y1

A B 2

1

P(2, y)1

Y'

A B A B sin .sin 2 cos .cos 2 2 2 2

5 tan

O

X'

A B A B .cos sin .sin 2 2 2 2

A B 5sin . sin 2 2

1 ....... (2)

Y

A B 2

2cos

y2 9

9

C A B .cos 2 2

C A B C 12sin .cos 2 2 2 2 cos

...(1) ...(2)

the arch 98.

8 m. 3

(c) Given : Equations of ellipses 3x2 + 5y2 = 32

...(1)

& 25x2 + 9y2 = 450

...(2)

Tangents to the ellipse (1) & (2) are passing through the point (3, 5) 3(3)2 + 5(5)2 – 32 = 27 + 75 – 32 > 0 So the given point lies outsides the ellipse. Hence, two real tangents can be drawn from the point to the ellipse, & 25(3)2 + 9(5)2 – 450 = 225 + 225 – 450 = 0 The point lie on the ellipse. Hence one real tangent can be drawn. No. of real tangents = 3

EBD_7443 2006-20

99.

Target VITEEE

(b) The equation of tangent at ct,

& f '(x) = a f '(2) = a = 1 a = 1 2 × 1 + b = 4 b = 2 [using equation (1)] f(x) = x + 2

c is t

ty = t3x – ct4 + c

Now, lim

c then If it passes through ct ', t' tc t'

3

t ct ' ct

4

x

c

& BM

sin 30º

4x 2(x 2) x 2

lim

2(x 2) (x 2)

2

x

t . t ' t 3 t '(t '.t) t 3 t ' 1 Note : If we take the co-ordinate axes along the asymptotes of a rectangular hyperbola, then the general equation x2 – y2 = a 2 becomes xy = c2, where c is a constant.

100. (d) Let AB = , then AM

lim

x

t 3 t '2 t 4 t ' t '

t

3

cos 30º

2

2

2

xf (2) 2f (x) x 2

2

lim

x

2

2x 4 x 2

2

102. (d) Let f(x) = x2 + kx + 1 f '(x) = 2x + k f(x) is strictly increasing on (1, 2) if f '(x) > 0 for x (1, 2) 2x + k > 0 for x (1, 2) k > –2x for x (1, 2) Now, 1 < x < 2 2 < 2x < 4 –2 > –2x > –4 – 4 < –2x < –2 k 2 Hence least value of k = –2.

Y

1 x

103. (c) Let y

B

x

y

x

x

Then log y = –x log x 30º 30º

A

X´

M

1 dy y dx

X

C

or

Y´

&

So, the coordinates of B are

3 2

,

2

4

4

d y

dx

dx

3

Now,

2

16 . 3 8 3 2 101. (d) Let f(x) = ax + b Given f(2) = 4 & f '(2) = 1 f(2) = a . 2 + b = 4 2a + b = 4

y(1 log x)

y.

2

d y

Since, B lies on y2 = 4x 2

dy dx

(1 log x)

2

1 x

2

dy dx

l og x

...(1)

x

1 e

1 dy (1 log x). x dx x 1

(1 log x).

0

1 log x

1

log e

0

log

1 e

dy dx

Solved Paper 2006

2006-21

d2 y

Also,

dx 2

1 1 is e e

at x

1 e

Let 1 + x = t2 dx = 2t . dt

0

dy dx

0

1 So, x is a point of local maxima. e Maximum value = value of y when x z x

104. (a) Euler's theorem x

y

x3 y3 x y

t2 1 . 2t dt t

f (x)

f (x)

2

t3 3

f (x)

2

(1 x)3/ 2 3

z y

nz

But f (0)

z x

x

z y

y

x

z (let) f (x)

2z

tan U y

2

U x . sec U . x

sec 2 U . x

x.

U x

0

x

106. (a) Let I

1 x x

1 x x 1 x

dx

c

c

0

(1 x)1/ 2

2

log(tan x)dx ...(1)

2

Then, I

log tan 0

2

x

a

dx

a

f (x)dx 0

0

4 3

0

2 tan U

2 tan U

2

sin U . cos2 U cos U

, f (0)

1 1 3

2

(1 x)3/ 2 3

sin 2U

f '(x)dx f (x)

U y

y

2.

U y y

105. (b) Given : f '(x)

2 tan U

2

U U y x y

U x x

.tan U

U y.sec U . y

2

(1 x)1/ 2

2 (1 x)3 / 2 3(1 x)1/ 2 3

f (x)

y

c

4 4 c 0 c 3 3 Equation (1) becomes

n= 3–1=2 x

t

2 (t 2 1)dt

.......... (1)

x3 y3 x y

Given : U = tan–1

tan U

1 ee

1 e

x = t2 – 1

2

I

log(cot x)dx 0

dx 2

log

I 0

1 dx tan x

f (a x)dx 0

EBD_7443 2006-22

Target VITEEE 2

I

log tan x

1

2

dx

0

I

a2 6

log(tan x)dx

0

I

2I 0

0

0 r

0

Thus 0&

3 2

2 6 a

there 3

3 0

is

sin 2 d

0

I 0

[Using eq. (1)] 107. (d) If curve r = a sin 3 To trace the curve, we consider the following table : 3

2

2

2 a

a

2 3 0

loop

5 2 5 6 a

a 2 2 1 cos 2 . d 6 2 0

cos 2

1 2sin

3

0

between

as r varies from r = 0 to r = 0.

a2 . 12

sin 2 2

a2 . 12 2

sin

9

108. (b) Let I

2 0

a2 24

e t dt

1

Put t = x2 dt = 2x . dx For limit : x = 1 & x = 3 3

I

3

e x . 2x dx

2 e x . x dx

1

1

X 2 x . ex

I

xe x

2

Hence, the area of the loop lying in the

positive quadrant

I

3 1

3

3 1

ex

e x dx

1 3 1

2 3e3 e1 e3 e1

2e3 .2

4e3

109. (d) Equation of the family of such parabolas is (y – k)2 = 4a(x – h) ...(1) where h & k are arbitrary constants Differentiating w.r.t. x, we get

13 2 r d 2 0

dy 2a dx Differentiating again (y k)

3

1 1 sin 2 . d 2 3

...(2)

0

[On putting, 3

d

1 d ] 3

(y k)

d2 y dx 2

dy dx

2

0

...(3)

Solved Paper 2006

2006-23

Putting value of (y – k) from (2) in (3), we get

2a

d2 y

dx 2 equation.

0 , which is required

1

y x

y dx x dy

0

D 2

1 . 1 2

y dx x dy

0

y x

x y . sin

ysin

y x

y x

y dy x

x

y

0

zx

dz z.1 x. dx

D

dz z x dx

But

Equation (1) becomes dz dx

zx.sin z x x sin z

x

dz dx

cos ecz

z cosec z

sin z dz

y log | x | cos x 111. (c) If

d2 y dx 2

(D2

2y

2)y

c

x2

x2

D

d dx

. (x 2 )

D3 ......

D2 2

2

..... (x 2 )

D2 2 (x ) 2

dx x

2

16

.cos 4x

1 D

2

a

2

cos ax

x sin ax 2a

x x .sin 4x sin 4x 8 2 4 Solution y = Complementary function + Particular Integral P.I.

y = A cos 4x + B sin 4x +

log | x | cos z c

1

1 . x2 1 2 112. (a) If (D2 + 16)y = cos 4x Here the auxiliary equation is m2 + 16 = 0 m =±4 Complementary function = (A cos 4x + B sin 4x) & Particular Integral (P.I.)

1

z x

2

.x 2

P.I.

...(1)

y x sin x z

dy dx

dx x sin

D2 2

1 . x2 2

P.I.

2

1 D2 . 1 2 2

2

(1 D) 1 1 D D2 P.I.

y sin x

.x

2

2 1

x

Put

D

3

dy dx

110. (b) Given : x cos ec

dy dx

1

Particular integral (P.I.)

x sin 4x 8

113. (c) 114. (d) Let us make one packet for each of the books on the same size. Now, 3 packets can be arranged in P(3, 3) = 3! ways 5 large books can be arranged in 5! ways 4 medium size books can be arranged in 4! ways 3 small books can be arranged in 3! ways Required number of ways = 3! × 5! × 4! × 3! ways

EBD_7443 2006-24

Target VITEEE

115. (a) An identity relation is one in which every element of a set is related to itself only. a * b = a + b – ab As in identity relation 'a' is related to 'a', so the correct option will be the one which gives the value of the relation = 'a'. So, equating a + b – ab = a, we get b(1 – a) = 0. Now putting the values of a, we find b and the option in which a = b, will be the answer. For a = 0, b = 0, so the correct option. For a = 1, b(1 – 1) = 0 b can have multiple values. For a = 2, b(1 – 2) = 0 b = 0 but a = 2. 116. (a) p q ~p ~ pvq p q T T F T T T F F F F F T T T T F F T T T 117. (b) Let S be the sample space n(S) 36 Events [sum greater than 3 but not exceeding 6] = {(2, 2), (3, 1), (1, 3), (4, 1), (1, 4), (5, 1) (1, 5), (3, 2), (2, 3), (4, 2), (2,4), (3, 3)}

4

C3

3

1 3

.

4! 8 1 . . 3!1! 27 3

4 3

4

C4

2 3

4

.

1 3

4! 16 . 4! 0! 81

5! 5! 5 1 6 4!1! 5! 0!

m

6

Since to pass the quiz, student must give 4 or 5 true answers. m 6 3 p p n 32 16 120. (a) Since f(x) is the probability density function of random variable X.

Hence, p

f (x) 1 n(E) n(S)

12 36

1 3

118. (a) Let 'p' denote the probability of winning of team A whenever it plays 2 2 1 p &q 1 3 3 3 Let X denotes the number of winning games out of 4 games i.e. n = 4 The probability of r success P(X = r) = ncr pr qn – r, r = 0, 1, 2, 3, 4 Probability of winning more than half games = P(X > 2) = P (X = 3) + P(X = 4)

4 4

32 16 48 16 81 81 81 27 119. (b) n = total number of ways = 25 = 32 Since each answer can be true or false & m = favourable number of ways = 5C4 + 5C5

n(E) 12

Required probability =

2 3

Now we have Ke | x | dx 1

2 K . e | x | dx 1 0

2 K . e x dx 1 0

2K . e

K

1 2

x 0

1

2K 1

MOCK

VITEEE Mock Test Paper

1

Max. Marks : 125

Time : 2½ hrs 6.

PART - I : PHYSICS 1.

2.

If s, k and r are coefficients of static friction, sliding friction and rolling friction, then (a) (b) k < r < s s< k< f (c) < < (d) r k s r= k= s In a region of space having a uniform electric field E, a hemispherical bowl of radius r is placed. The electric flux through the bowl is (a)

3.

(c) 2 r 2 E (d) r2 E Which of the following relation is true ? 9 Y (a) 3Y K (1 ) (b) K Y (c)

4.

5.

7.

(c) 4 r 2 E

2 rE

A capacitor C 1 is charged to a potential difference V. The charging battery is then removed and the capacitor is connected to an unchar ged capacitor C 2 . The potential difference across the combination is (a)

VC1 (C1 C 2 )

(c)

V 1

(b) V 1

C1 C2

(d)

(a) 2R

05.Y

(d)

)Y

(6K

Kirchhoff’s first law of electricity follows (a) law of conservation of energy only (b) law of conservation of charge only (c) law of conservation of both energy and charge (d) sometimes law of conservation of energy and some other times law of conservation of charge In a Wheatstone’s bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is (b)

R 4

R (d) R 2 A particle is acted by a force F = kx, where k is a +ve constant. Its potential energy at x = 0 is zero. Which curve correctly represents the variation of potential energy of the block with respect to x

(c)

8.

U

C2 C1

(a)

VC 2 (C1 C 2 )

In the network shown in the Fig, each resistance is 1 . The effective resistance between A and B is

U

x (b)

U

x

U

1 1

A

1 1

x (d)

x

B

1

4 3

(b)

(c) 7

(d)

(a)

(c)

1

1

3 2 8 7

9.

A straight wire of diameter 0.5 mm carrying a current of 1 A is replaced by another wire of 1 mm diameter carrying same current. The strength of magnetic field far away is

EBD_7443 MT-2

10.

11.

Target VITEEE (a) twice the earlier value (b) same as the earlier value (c) one-half of the earlier value (d) one-quarter of the earlier value A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will (a) continue to move due east (b) move in a circular orbit with its speed unchanged (c) move in a circular orbit with its speed increased (d) gets deflected vertically upwards. Power dissipated across the 8 resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the 3 resistor is 1

3

16.

17.

i1

(a)

i 8

12.

13.

14.

15.

(b)

(c) 18.

0 0

(d)

0 0

Th e Young’s double slit experiment is performed with blue and with green light of wavelengths 4360Å and 5460Å respectively. If x is the distance of 4th maxima from the central one, then (a) x (blue) = x (green) (b) x (blue) > x (green) (c) x (blue) < x (green) (d)

19.

0 0

0

i2

(a) 1.0 (b) 0.5 (c) 3.0 (d) 2.0 A battery is charged at a potential of 15V for 8 hours when the current flowing is 10A. The battery on discharge supplies a current of 5A for 15 hour. The mean terminal voltage during discharge is 14V. The “Watt-hour” efficiency of the battery is (a) 87.5% (b) 82.5% (c) 80% (d) 90% A coil of 40 henry inductance is connected in series with a resistance of 8 ohm and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is (a) 20 seconds (b) 5 seconds (c) 1/5 seconds (d) 40 seconds In an inductor of self-inductance L = 2 mH, current changes with time according to relation i = t2e–t. At what time emf is zero? (a) 4s (b) 3s (c) 2s (d) 1s A varying current in a coil changes from 10A to zero in 0.5 sec. If the average e.m.f induced

in the coil is 220V, the self-inductance of the coil is (a) 5 H (b) 6 H (c) 11 H (d) 12 H In a region of uniform magnetic induction B = 10–2 tesla, a circular coil of radius 30 cm and resistance 2 ohm is rotated about an axis which is perpendicular to the direction of B and which forms a diameter of the coil. If the coil rotates at 200 rpm the amplitude of the alternating current induced in the coil is (a) 4 2 mA (b) 30 mA (c) 6 mA (d) 200 mA If 0 and 0 are the electric permittivity and magnetic permeability in vacuum, and are corresponding quantities in medium, then refractive index of the medium is

x (blue) x (green)

5460 4360

A cylindrical metal rod is shaped into a ring with a small gap as shown. On heating the system :

(a) (b) (c) (d)

x decreases, r and d increase x and r increase, d decreases x, r and d all increase x and r decreased, d remains constant

MT-3

Mock Test 1 20.

21.

The angular separation d between two wavelength and + d in a diffraction grating is directly proportional to (a) frequency of light (b) grating element (c) spatial frequency of grating (d) wavelength of light In the Bragg scattering of a beam of electrons each of mass m and velocity v by a nickel crystal, the first maximum is observed at = 30° ( being the angle the beam makes with the crystal plane). What is the inter-planar distance d for the crystal? (a)

h mv

(b)

26.

27.

I

2h mv

C

23.

h mv (d) 2mv h What is the ground state energy of positronium? (The ground state energy of hydrogen is –13.6 eV) (a) – 3.4 eV (b) – 6.8 eV (c) – 13.6 eV (d) – 27.2 eV The minimum wavelength in Lymann series of hydrogen spectra is 91.2 nm, the longest wavelength in this series must be

(a)

eh Vc

(b)

ch eV

V

25.

28.

(a) A and B will have different intensities while B and C will have different frequencies (b) B and C will have different intensities while A and C will have different frequencies (c) A and B will have different intensities while A and C will have equal frequencies (d) A and B will have equal intensities while B and C will have different frequencies Which figure shows the correct force acting on the body sliding down an inclined plane? (m mass, fs force of friction) N

eV cV (d) ch eh The glancing angle in a X-rays diffraction experiment is 30° and the wavelength of the Xrays used is 20 mm. The interplanar spacing of the crystal diffracting these X-rays will be (a) 40 nm (b) 20 nm (c) 15 nm (d) 10 nm

(c)

24.

(a)

(b)

The velocity of a body of rest mass m o is

3 c (Where c is the velocity of light in 2 vacuum). The mass of this body is (a)

(c)

3 mo 2 2mo

(b)

(d)

3

mo

fs

A

N

B

sin mg C

(c)

fs

mg mg cos

fs

sin mg

mg mg cos

C

N

(d)

B

sin mg mg C mg cos

N

1 mo 2

2

B A

(c)

22.

Einstein’s photoelectric equation is . In this equation Ek refers to Ek h (a) kinetic energy of all the emitted electrons. (b) mean kinetic energy of emitted electrons (c) maximum kinetic energy of emitted electrons. (d) minimum kinetic energy of emitted electrons. In a photoelectric experiment anode potential is plotted against plate current

sin mg C

fs

mg mg cos

A B

A

B

A

EBD_7443 MT-4

29.

30.

31. 32.

33.

34.

35.

36.

Target VITEEE U235

The energy released per fission of a 92 nucleus is nearly (a) 200 eV (b) 20 eV (c) 200 MeV (d) 2000 eV Half-lives of two radioactive substances A and B are respectively 20 minutes and 40 minutes. Initially, the sample of A and B have equal number of nuclei. After 80 minutes, the ratio of remaining number of A and B nuclei is (a) 1 : 16 (b) 4 : 1 (c) 1 : 4 (d) 1 : 1 The binding energy per nucleon is largest for (a) 56Fe (b) 16O (c) 4He (d) 208Pb The radius of Ge nucleus is measured to be twice the radius of 9Be4 nucleus. How many nucleons are there in the Ge nucleus ? (a) 72 (b) 96 (c) 120 (d) 144 A Zener diode has a breakdown voltage of 9.1 V with a maximum power dissipation of 364 mW. What is the maximum current that the diode can withstand ? (a) 0.04 A (b) 0.4 A (c) 4.0 A (d) 40 A The voltage across a diode in a full wave rectifier having input voltage of peak value Vm during its non conducting period is (a) 0 (b) – Vm (c) – 2Vm (d) – 4Vm A n-p-n transistor conducts when (a) both collector and emitter are negative with respect to the base (b) both collector and emitter are positive with respect to the base (c) collector is positive and emitter is negative with respect to the base (d) collector is positive and emitter is at same potential as the base The diagram of a logic circuit is given below. The output F of the circuit is represented by

37.

38.

(a)

39.

W Y (a) W . (X + Y) (c) W + (X . Y)

(b) W . (X . Y) (d) W + (X + Y)

(b)

t1 t2

d1t 1 d2 t2 (d) d 2t 2 d1 t 1 Two masses ma and mb moving with velocities va and v b in opposite dir ection collide elastically and after the collision ma and mb move with velocities Vb and Va respectively. Then the ratio ma/mb is

(a)

Va Va

40.

41.

Vb Vb

(b)

ma mb ma

1 2 The most suitable device at present for solid state picture tube is (a) LED (b) LCD (c) Silicon (d) Quartz crystal PART - II : CHEMISTRY

(c) 1

42.

F

d1 t 1 d2t2

(c)

W X

In frequency modulation (a) the amplitude of modulated wave varies as frequency of carrier wave (b) the frequency of modulated wave varies as amplitude of modulating wave (c) the amplitude of modulated wave varies as frequency of carrier wave (d) the frequency of modulated wave varies as frequency of modulating wave Two liquids of densities d1 and d2 are flowing in identical capillary tubes uder the same pressure difference. If t1 and t2 are time taken for the flow of equal quantities (mass) of liquids, then the ratio of coefficient of viscosity of liquids must be

(d)

Some Gem stones used to show colour. Ruby shows colour due to __________. (a) d-d transition of Al3+ and Cr3+ (b) d -d transition of Al3+ (c) d -d transition of Cr 3+ (d) d-d transition of Cr 3+ The lanthanide contraction is responsible for the fact that (a) Zr and Y have about the same radius (b) Zr and Nb have similar oxidation state (c) Zr and Hf have about the same radius (d) Zr and Zn have the same oxidation states (Atomic numbers : Zr = 40, Y = 39, Nb = 41, Hf = 72, Zn = 30)

MT-5

Mock Test 1 43.

At equilibrium, if Kp = 1, then (a) (b) G 0

50.

G 1 (d) None of these

44.

45.

(c) G 1 Carbon - 14 dating method is based on the fact that: (a) C-14 fraction is same in all objects (b) C-14 is highly insoluble (c) Ratio of carbon-14 and carbon-12 is constant (d) all the above 235 92 U

1 0n

236 92

U

46.

47.

48.

49.

235 92 U

53.

K 3[Al(C 2O 4 )3 ] is called (a) Potassium alumino oxalate (b) Potassium trioxalateoaluminate (III) (c) Potassium aluminium (III) oxalate (d) Potassium trioxalato aluminate (VI) A co-ordination complex compound of cobalt has the molecular formula containing five ammonia molecules, one nitro group and two chlorine atoms for one cobalt atom. One mole of this compound produces three mole ions in an aqueous solution. On reacting this solution with excess of AgNO3 solution, we get two moles of AgCl precipitate. The ionic formula for this complex would be (a) [Co(NH3)4 (NO2) Cl] [(NH3) Cl] (b) [Co (NH3)5 Cl] [Cl (NO2)] (c) [Co (NH3)5 (NO2)] Cl2 (d) [Co (NH3)5] [(NO2)2Cl2] Which of thefollowing does not have a metalcarbon bond? (a)

Al(OC 2 H 5 )3

(b) C 2 H 5MgBr

(c)

K[Pt (C 2 H 4 )Cl 3 ]

(d) Ni(CO) 4

Which one of the following octahedral complexes will not show geometric isomerism? ( A and B are monodentate ligands) (a) [MA5B] (b) [MA2B4] (c) [MA3B3] (d) [MA4B2]

(c) Zn2+

(d) Cr3+

When electrons are trapped into the crystal in anion vacancy, the defect is known as :

(c) Stoichiometric defect (d) F-centres

finally

(b) 16.40 × 107 kJ (d) 6.50 × 106 kJ

(b) Cu+

(b) Frenkel defect

52.

The energy released when 1 g of

(a) Ti4+

(a) Schottky defect

fission products +

neutrons + 3.20 × 10–11 J undergoes fission is (a) 12.75 × 108 kJ (c) 8.20 × 107 kJ

51.

Which one of the following ionic species will impart colour to an aqueous solution?

The number of atoms contained in a fcc unit cell of a monoatomic substance is (a) 1

(b) 2

(c) 4

(d) 6

The second order Bragg diffraction of X-rays with = 1.00 Å from a set of parallel planes in a metal occurs at an angle 60º. The distance between the scattering planes in the crystal is (a) 0.575 Å (c) 2.00 Å

54.

(b) 1.00 Å (d) 1.15 Å

Identify the correct statement regarding entropy: (a) At absolute zero of temperature, entropy of a perfectly crystalline substance is taken to be zero (b) At absolute zero of temperature, the entropy of a perfectly crystalline substance is +ve (c) At absolute zero of temperature, the entropy of all crystalline substances is taken to be zero (d) At 0ºC, the entropy of a perfectly crystalline substance is taken to be zero

55.

Standard Gibb’s free energy change for isomerization reaction cis-2 pentene trans-2-pentene is – 3.67 kJ/mol at 400 K. If more trans-2 pentene is added to the reaction vessel, then (a) more cis-2 pentene is formed (b) equilibrium remains unaffected (c) additional trans-2 pentene is formed (d) equilibrium is shifted in forward direction

EBD_7443 MT-6

56.

Target VITEEE In the following reaction, how is the rate of appearance of the underlined product related to the rate of disappearance of the underlined reactant ? BrO 3( aq )

5Br

6 H (aq )

( aq )

3Br2( l) 3H 2 O ( l)

57.

58.

59.

60.

(a) K(510.1 – 457.6) ln (3.07 × 10

(a)

d[Br2 ] dt

5 d[Br ] 3 dt

(b)

d[Br2 ] dt

d[ Br ] dt

(c)

d[Br2 ] dt

3 d[ Br ] 5 dt

(d)

d[Br2 ] dt

61.

62.

3 d[Br ] 5 dt

The rate of reaction between two reactants A and B decreases by a factor of 4 if th e concentration of reactant B is doubled. The order of this reaction with respect to reactant B is: (a) 2 (b) 2 (c) 1 (d) 1 According to Le-chatelier’s principle, adding

63.

heat to a solid

64.

liquid equilibrium will cause

the (a) temperature to increase (b) temperature to decrease (c) amount of liquid to decrease (d) amount of solid to decrease. Which of the following is true at chemical equilibrium? (a) ( G)T,p is minimum and ( S)U,V is also minimum (b) ( G)T,V is minimum and ( S)U,V is maximum (c) ( G)T,V is maximum and ( S)U,V is zero (d) ( G)T,p is zero and ( S)U,V is also zero The racemisation of - pinene is first order reaction. In the gas the specific reaction rate was found to be 2.2 10 3.07 10 is

3

5

mm

1

at 457.6 K and

at 510.1 K. The energy of activation

(b) 3 . 048 10

3

(c)

5

/ 2.2 10

5

)

457.6 K cal.

(d) (510 .1 457 .6 / 52 .5) R ln (307)/2.2) cal. The ionic conductance of Ba2+ and Cl– are respectively 127 and 76 ohm–1 cm2 at infinite dilution. The equivalent conductance (in ohm– 1 cm2) of BaCl at infinite dilution will be : 2 (a) 139.5 (b) 203 (c) 279 (d) 101.5 On passing a current of 1.0 ampere for 16 min and 5 sec through one litre solution of CuCl2, all copper of the solution was deposited at cathode. The strength of CuCl2 solution was (Molar mass of Cu= 63.5; Faraday constant = 96,500 Cmol–1) (a) 0.01 N (b) 0.01 M (c) 0.02 M (d) 0.2 N Standard electrode potentials are : Fe+2/Fe Eº = –0.44; Fe+3/Fe+2 Eº = + 0.77 ; If Fe+2, Fe+3 and Fe blocks are kept together, then (a) Fe+2 increases (b) Fe+3 decreases (c)

65.

ln 2. 2 10

3

Fe

2

remains unchanged Fe 3 (d) Fe+2 decreases The most durable metal plating on iron to protect against corrosion is (a) nickel plating (b) copper plating (c) tin plating (d) zinc plating. Reaction of CH — CH with RMgX leads to 2 2 O formation of (a) RCHOHR (b) RCHOHCH3 R

(c) RCH2CH2OH 66.

(d)

R

CHCH 2OH

Increasing order of acid strength among p-methoxyphenol, p-methylphenol and p-nitrophenol is (a) p-Nitrophenol, p-Methoxyphenol, p-Methylphenol (b) p-Methylphenol, p-Methoxyphenol, p-Nitrophenol (c) p-Nitrophenol, p-Meth ylphenol, p-Methoxyphenol. (d) p-Methoxyphenol, p-Methylphenol, p-Nitrophenol

MT-7

Mock Test 1 67.

68.

When phenol is treated with excess bromine water. It gives (a) m-Bromophenol (b) o-and p-Bromophenols (c) 2,4-Dibromophenol (d) 2,4, 6-Tribomophenol. In the reaction: CH3 | CH3 CH CH 2 O CH 2 CH3 HI

73.

74.

Heated

Which of the following compounds will be formed? (a)

CH3 CH CH3 CH3CH 2 OH | CH3

(b)

CH3 CH CH 2 OH CH3CH3 | CH3

(c)

69.

70.

71.

72.

CH3 | CH3 CH CH 2 OH CH3 CH 2

75.

I

CH3 | (d) CH3 CH CH 2 I CH3CH 2OH n-Propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent? (a) PCl5 (b) Reduction (c) Oxidation with potassium dichromate (d) Ozonolysis The thermodynamic efficiency of cell is given by (a) H/ G (b) nFE/ G (c) nFE/ H (d) nFE Pinacolone is (a) 2, 3-Dimethyl-2 3-butanediol (b) 3, 3-Dimethyl-2 butanone (c) I-Phenyl-2Propanone (d) 1,1-Diphenyl-2-ethandiol. An ester is boiled with KOH. The product is cooled and acidified with concentrated HCl. A white crystalline acid separates. The ester is (a) Methyl acetate (b) Ethyl acetate (c) Ethyl formate (d) Ethyl benzoate

76.

Aspirin is an acetylation product of (a) o-hydroxybenzoic acid (b) o-dihydroxybenzene (c) m-hydroxybenzoic acid (d) p-dihydroxybenzene Which of the following is incorrect? (a) NaHSO3 is used in detection of carbonyl compound (b) FeCl3 is used in detection of phenolic group (c) Tollen reagent is used in detection of unsaturation (d) Fehling solution is used in detection of glucose Formic acid is obtained when (a) Calcium acetate is heated with conc. H2SO4 (b) Calcium formate is heated with calcium acetate (c) Glycerol is heated with oxalic acid at 373 K (d) Acetaldehyde is oxidised with K2Cr2O7 and H2SO4. When aniline reacts with oil of bitter almonds (C 6 H 5CHO ) condensation takes place and

77.

78.

benzal derivative is formed. This is known as (a) Million's base (b) Schiff's reagent (c) Schiff's base (d) Benedict's reagent The consituent of the powerful explosive RDX is formed during the nitration of (a) toluene (b) phenol (c) glycerol (d) urotropine Aniline in a set of reactions yielded a product D.

The structure of the product D would be: (a) C6H5NHOH (b) C6H5NHCH2CH3 (c) C6H5CH2NH2 (d) C6H5CH2OH

EBD_7443 MT-8

Target VITEEE

83.

(a) 4 (b) 3 (c) 2 (d) 1 If tan x + tan (x + /3) + tan (x + /3) = 3, then (a) tan x = 1 (b) tan 2x = 1 (c) tan 3x =1 (d) none of these.

84.

If b and c are any two non-collinear mutually

C N

79.

+ CH3MgBr

H 3O

P

OCH3

Product 'P' in the above reaction is OH

O

CH – CH3

C – CH3

(a)

(b)

then ( a b) b ( a . c) c

OCH3

OCH3

(c)

(a)

OCH3

85.

What is the decreasing order of basicity of primary, secondary and tertiary ethylamines and NH3 ? (a)

NH 3

C 2 H 5 NH 2

| b c |2

(b

c)

(b) 2 a

a

(c)

(d) OCH3

a . ( b c)

is equal to :

COOH

CHO

80.

perpendicular unit vectors and a is any vector,,

(d) None 3a Which one of the following graphs represents the function y = 1+ |x| for all x R? Y

(a)

(C 2 H 5 ) 2 NH

(–1,0)

(1,0) O

X

(C 2 H 5 )3 N (b) (C2 H 5 )3 N

(C2 H 5 )2 NH C2 H 5NH 2

(c)

(C2 H 5 )2 NH

(C 2 H 5 ) 2 NH

NH 3

(b)

C2 H 5 NH 2 (C 2 H 5 )3 N

(d)

Y

Y

(0,1)

NH 3 .

If R be a relation “less than” from A = {1, 2, 3, 4} B = {1, 3, 5}, i.e. (a, b) R iff a < b, then ROR–1 is (a) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} (b) {(3,1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} (c) {(3, 3), (3, 5), (5, 3), (5, 5)} (d) {(3, 3), (3, 4), (4, 5)} The smallest positive integer n for which 1 i 1– i

n

1 is

X

NH 3

(c)

PART - III : MATHEMATICS

82.

(1,0)

O

(C 2 H 5 ) 3 N C 2 H 5 NH 2

81.

(–1,0)

X

O

Y (0,1)

(d)

86.

(1,0) (–1,0) O

X

The point (2a, a) lies inside the region bounded by the parabola x2 = 4y and its latus rectum. Then, (a) 0 a 1 (b) 0 < a < 1 (c) a > 1 (d) a < 0

MT-9

Mock Test 1 87. The linear inequations for which the shaded area in the following figure is the solution set, are

91.

92.

x–y=1

Three concurrent lines with direction cosines l1, m1, n1; l2, m2, n2 and l3, m3, n3 are coplanar l1 m1 then l 2 m 2 l3 m 3 (a) 0 (c) –1 The resultant ˆi 2ˆj 3kˆ , 2ˆi

n1 n2 n3

(b) 1 (d) 2 moment of three forces 3ˆj 4kˆ and ˆi ˆj kˆ acting

on a particle at a point P (0, 1, 2) about the point A (1, –2, 0) is X

X 2x + y = 2

88.

89.

x + 2y = 8

93.

(a) x + y 1, 2x + y 2, x – 2y 8, x 0, y 0 (b) x – y 1, 2x + y 2, x + 2y 8, x 0, y 0 (c) x – y 1, 2x + y 2, x + 2y 8, x 0, y 0 (d) x + y 1, 2x + y 2, x + 2y 8, x 0, y 0 The roots of the equation (3 – x)4 + (2 – x)4 = (5 – 2x)4 are (a) two real and two imaginary (b) all imaginary (c) all real (d) none of these Coloured balls are distributed in four boxes as shown in the following table Box I II III IV

Colo ur Black W hite Red 3 4 5 2 2 2 1 2 3 4 3 1

Blue 6 2 1 5

A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black. Probability that the ball drawn from Box III, is (a) 0.161 (b) 0.162 (c) 0.165 (d) 0.104 p

90.

If

p x

q y r z q

p x q y p x (a) (c)

q y 0 2

r z

(a)

0 , then the value of

r

94.

6 2

(b)

140

(c) (d) None 21 th th th If the p , q , r , terms of a G.P. are respectively x, y, z, then the value of x q r y r p z p q is (a) xyz (b) pqr (c) 1 (d) none of these A signal which can be green or red with probability 4/5 and 1/5 respectively, is received by station A and then trasmitted to station B. The probability of each station receiving the signal correctly is 3/4. If the signal received at station B is given, then the probability that the original signal is green, is (a)

3 5

(b)

6 7

20 9 (d) 23 20 If A and B are two events, the probability that at most one of these events occurs is : (a) P(A' ) P(B' ) P(A ' B' )

(c)

95.

(b)

96.

P(A' ) P(B' ) P(A

B) 1

(c) P(A B' ) P(A' B) P(A' B' ) (d) All above are correct. If x, y, z are distinct number in A.P. with common differences d and the rank of the matrix 4 5

x

5 6

y is 2 then the values of d and k are : z

6 k

x ; arbitrary number 4 (b) arbitrary number, 7

(a)

r is z

(b) 1 (d) 4pqr

(c) x, 5

(d)

x ,6 2

EBD_7443 MT-10

97.

Target VITEEE

The solution of sin–1 x – sin –12x = 1 3

(a)

3 2

(c)

(b)

1 4

(d) ±

1 2

2

98.

If y

tan

and x

99.

1 1 u cos 1 2 1 u2 2u

1 u

2

, then

3

is

1 2u sin 1 2 1 u2

dy dx

3c 4b 2c

(b)

3c 4b 2b

4c 3b (d) none of these. (c) 2c 100. The octal numeral 23450 is equal to the binary number (a) 10011100101000 (b) 10011100011000 (c) 10111000101000 (d) 10011100111000 101. nCr + 2 nCr–1 + nCr–2 is equal to: (a) n+2Cr (b) nCr+1 n–1 (c) Cr+1 (d) None of these 102. The domain of the function defined by f x sin 1 x 1 is (a) [1, 2] (b) [–1, 1] (c) [0, 1] (d) None of these 103. Let f(x) be a continuous function such that the area bounded by the curve y = f(x), x-axis and

the lines x = 0 and x = a is then f (a) 1 (c)

1 3

2

length

a2 2

a sin a + cos a, 2 2

= (b)

1 2

(d) None of these

40 on the line y – 2x = 1, then a is

equal to (a) 1 (c) –1 106. If sin

(a) –1 (b) 0 (c) 1 (d) None of these If the sides of a triangle ABC are in A.P., and a is the smallest side, then cos A equals to : (a)

104. The number of all five digit numbers which are divisible by 4 that can be formed from the digits 0, 1, 2, 3, 4 (without repetition) is (a) 36 (b) 30 (c) 34 (d) None of these 2 105. If the parabola x = ay makes an intercept of

(b) 3 (d) 2 1

x dx

f (x ) sin

(a) f(x) = x, g(x)

1

x g (x ) c . Then

1 x2

(b) f(x) = 1 x 2 , g(x) = 1 + x2 x

(c) f(x) = x, g(x) =

1 x2

(d) none of these. 107. The line x cos the ellipse

y sin

x2

y2

a2

b2

p is a tangent to

1 if

(a)

p

(b)

p2

a 2 cos 2

b 2 sin 2

(c)

p2

a 2 sec 2

b 2 tan 2

(d)

p

a cos

a sec

b sin

b tan

3x 4 and g 5x 7 7x 4 : A B is defined by g x , where A 5x 3 3 7 =R– and B = R – and IA is an 5 5 identity function on A and IB is identity function on B, then (a) fog = IA and gof = IA (b) fog = IA and gof = IB (c) fog = IB and gof = IB (d) fog = IB and gof = IA

108. If f : B

A is defined by f x

MT-11

Mock Test 1 109. In [0,1] Lagranges Mean Value theorem is NOT applicable to

(a)

1 x 2 2 1 x 2

f (x)

x x

1 2 1 2

114. The locus of a point, such that the sum of the squares of its distances from the planes x + y + z = 0, x – z =0 and x – 2y + z = 0 is 9, is (a)

f ( x)

sin x , x x 1, x

(c)

f ( x)

xx

(d)

f ( x)

x

0 0

x x

(a)

(b)

(c)

2 1

r.

4 1

4 1

dx is equal to

x

x

x

2log

2log

2log

(d) None of these.

z2

y2 y2

z2 z2

3 6 9 12

115. The distance of the point (–5, –5, –10) from the point of intersection of the line

(1 y 2 ) x dx (1 x 2 ) y dy 0 represents a family of : (a) ellipses of constant eccentricity (b) ellipses of variable eccentricity (c) hyperbolas of constant eccentricity (d) hyperbolas of variable eccentricity

1

x2

(d) x 2

110. With respect to multiplication modulo 5, the set G = {1, 2, 3, 4} is (a) a finite abelian group of order 4 (b) not a finite abelian group of order 4 (c) both (a) and (b) (d) none 111. Let f be function such that f (x + y) = f(x) + f(y) for all x and y and f(x) = (2x2 + 3x) g (x) for all x, where g (x) is continuous and g(0) = 3. Then f’(x) is equal to (a) 9 (b) 3 (c) 6 (d) none of these 112. The differential equation

113.

y2

(b) x 2 y 2 z 2 (c)

(b)

x2

1

x 1

1

x 1

1

x 1

1

x 1

1

x 1

1

x 1

c

c

c

2iˆ ˆj 2kˆ

plane r · ˆi ˆj kˆ (a) 13

3iˆ 4ˆj 2kˆ

and the

5 is

(b) 12

(c) 4 15 (d) 10 2 116. The asymptotes of the hyperbola xy – 3x + 4y + 2 = 0 are (a) x = – 4 and y = 3 (b) x = 4 and y = –3 (c) x = –4 and y = –4 (d) x = –3 and y = 3 117. The integer n for which (cos x 1)(cos x e x ) is a finite non-zero x 0 xn number is (a) 1 (b) 2 (c) 3 (d) 4 118. The order and degree of the differential equation whose solution is lim

y cx c 2 3c 3 2 2 , where c is a parameter,, are respectively (a) 1 and 4 (b) 1 and 3 (c) 1 and 2 (d) none of these 119. A football is inflated by pumping air in it. When it acquires spherical shape its radius increases at the rate of 0.02 cm/s. The rate of increase of its volume when the radius is 10 cm is ___________ cm/s (a) 0 (b) 2 (c) 8 (d) 9 120. Let B is a set containing the elements of boolean algebra and a, b B , then a. (a + b) =

(a) b (c) a + b

(b) a . b (d) a

EBD_7443 MT-12

Target VITEEE PART - IV : (ENGLISH)

Direction (Qs. 121 - 123) : Read the passage carefully and answer the questions given below. At this stage of civilization, when many nations are brought in to close and vital contact for good and evil, it is essential, as never before, that their gross ignorance of one another should be diminished, that they should begin to understand a little of one another's historical experience and resulting mentality. It is the fault of the English to expect the people of other countries to react as they do, to political and international situations. Our genuine goodwill and good intentions are often brought to nothing, because we expect other people to be like us. This would be corrected if we knew the history, not necessarily in detail but in broad outlines, of the social and political conditions which have given to each nation its present character. 121. According to the author the 'Mentality' of a nation is mainly product of its(a) present character (b) international position (c) politics (d) history

122. The character of a nation is the result of its(a) gross ignorance (b) cultural heritage (c) socio-political conditions (d) mentality 123. The need for a greater understanding between nations(a) is more today than ever before (b) was always there (c) is no longer there (d) will always be there 124. Direction: Choose the correct meaning of the idiom from the options given below. At one's wit's end (a) Perplexed (b) Clear Up (c) Explain (d) Enlighten 125. Choose the best pronunciation of the word, Poem, from the following options. (a) poy-em (b) poe-um (d) poh-om (d) poi-um

MOCK

VITEEE Mock Test Paper

2

Max. Marks : 125

Time : 2½ hrs

PART - I : PHYSICS 1.

2.

In an inelastic collision, which of the following does not remain conserved? (a) Momentum (b) kinetic energy (c) Total energy (d) Neither momentum nor kinetic energy Three capacitors C1, C2 and C3 are connected to a battery as shown. With symbols having their usual meanings, the correct conditions are

V1 Q1 C1

6.

3.

(a)

5.

2C

8.

(b) 2C

C

C (d) 2 2 At the centre of a cubical box + Q charge is placed. The value of total flux that is coming out a wall is

(c)

4.

7.

(a) Q / o (b) Q / 3 o (c) Q / 4 o (d) Q / 6 o What will be the equivalent resistance of circuit shown in figure between two points A and D

9.

B

10 D

10

V3

(a) Q1 = Q2 = Q3 and V1 = V2 = V (b) V1 = V2 = V3 = V (c) Q1 = Q2 + Q3 and V = V1 + V2 (d) Q2 = Q3 and V2 = V3 A parallel plate condenser with oil between the plates (dielectric constant of oil K = 2) has a capacitance C. If the oil is removed, then capacitance of the capacitor becomes

10

C

C3

V

10

10

V Q2 2 C 2

Q3

10

A

10

10

(b) 20 (a) 10 (c) 30 (d) 40 According to stokes law, the relation between terminal velocity (vt) and viscosity of the medium (n) is (a) vt = n (b) vt n 1 n (d) vt is independent of n. Kirchhoff’s first law, i.e. i = 0 at a junction, deals with the conservation of (a) charge (b) energy (c) momentum (d) angular momentum The potential difference between the terminals of a cell in an open circuit is 2.2 V. When a resistor of 5 is connected across the terminals of the cell, the potential difference between the terminals of the cell is found to be 1.8 V. The internal resistance of the cell is

(c)

vt

(a)

7 12

(b)

10 9

(c)

9 10

(d)

12 7

Direct current is passed through a copper sulphate solution using platinum electrodes. The elements liberated at the electrodes are (a) copper at anode and sulphur at cathode (b) sulphur at anode and copper at cathode (c) oxygen at anode and copper at cathode (d) copper at anode and oxygen at cathode

EBD_7443 MT-14

10.

11.

12.

Target VITEEE

Two electric bulbs, one of 200 V, 40W and other of 200 V, 100W are connected in a domestic circuit. Then (a) they have equal resistance (b) the resistance of 40W bulb is more than 100W bulb (c) the resistance of 100W bulb is more than 40 W bulb (d) they have equal current through them A beam of metal supported at the two edges is loaded at the centre. The depression at the centre is proportional to

(a) Y 2 (b) Y (c) 1/Y (d) 1/Y 2 A charged particle of charge q and mass m enters

16.

17.

18.

19.

perpendicularly in a magnetic field B . Kinetic energy of the particle is E; then frequency of rotation is

13.

14.

(a)

qB m

(b)

qB 2 m

(c)

qBE 2 m

(d)

qB 2 E

t

If N is the number of turns in a coil, the value of self inductance varies as (a) N0 (b) N (c) N2 (d) N–2 In an A.C. circuit with voltage V and current I the power dissipated is (a) dependent on the phase between V and I (b)

15.

20.

(a) 10 s (b) 0.1 s (c) 0.01 s (d) 1 s Two coils of self inductances 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is (a) 6 mH (b) 4 mH (c) 16 mH (d) 10 mH Which one of the following electromagnetic radiations has the smallest wavelength? (a) Ultraviolet waves (b) X-rays (c) -rays (d) Microwaves Which one of the following phenomena is not explained by Huygen’s constr uction of wavefront? (a) Refraction (b) Reflection (c) Diffraction (d) Orgin of spectra Two light waves having their intensities in the ratio 16 : 9 interfere to produce interference pattern. What is the ratio of maximum intensity to minimum intensity in this pattern ? (a) 4 : 3 (c) 25 : 7 (b) 625 : 49 (d) 49 : 1 A slab of thickness t and refractive index 1.5 is placed in between point A and B as shown in the figure given below. The optical path length between A and B is

1 2

y

B

n = 1.5

VI

(c) 1 VI 2 (d) VI A magnetic field of 2 × 10–2 T acts at right angles to a coil of area 100 cm2, with 50 turns. The average e.m.f. induced in the coil is 0.1 V, when it is removed from the field in t sec. The value of t is

x

A

21.

(a)

x

t 2

(c)

x

3 t y 2

y

(b) (x + t + y)

(d)

x

5 t y 2

If according to the Bohr model of the hydrogen atom, the ionization energy of the atom in its ground state is 13.6 eV, then the energy required to ionize the atom from its first excited state will be (a) 6.8 eV (b) 3.4 eV (c) 1.7 eV (d) 0.85 eV

MT-15

Mock Test 2 22.

Match List-I with List-II and select the correct answer using the codes given belowthe Lists List-I List-II A.

Balmer series

1. v

RH

B.

Brakett series

2. v

RH

C.

Lyman series

3. v

RH

D.

25.

RH

1

1

2

n2

1

1

2

2

1

2

n

1

1

2

2

4

n

27.

5

x(m)

15

28.

13

3.2 10

(b) 1.6 10

J

13

J

(c) 4.8 10 13 J (d) 6.4 10 13 J The ratio of molar specific heat at constant pressure CP, to molar specific heat at constant volume Cv for a monoatomic gas is (a)

CP CV

3 5

(b)

CP CV

5 3

(c)

CP CV

7 9

(d)

CP CV

9 7

29.

Which of the following are true regarding forces between nucleons inside the nucleus : 1. attractive in nature 2. electrical in nature 3. extremely short range 4. strongest forces in nature Which of the above are correct ? (a) 1, 2 and 4 (b) 2 and 3 (c) 1, 3 and 4 (d) 3 and 4

30.

2 1H

20

The work done by the particle as it moves from x = 0 to 20 m is (a) 37.5 J (b) 10 J (c) 45 J (d) 22.5 J The population inversion necessary for laser action used in solid state lasers is (a) electrical discharge (b) inelastic atom-atom collision (c) direct conversion (d) optical pumping The photo electric threshold of Tungsten is 2300Å. The energy of the electrons ejected from the surface by ultraviolet light of wavelength 1800Å is (a) 0.15 eV (b) 1.5 eV (c) 15 eV (d) 150 eV

Radiations of two photon’s energy, twice and ten times the work function of metal are incident on the metal surface successsively. The ratio of maximum velocities of photoelectrons emitted in two cases is (a) 1 : 2 (b) 1 : 3 (c) 1 : 4 (d) 1 : 1 If an electron and positron annihilate, then the energy released is (a)

3

0

24.

4. v

1 n2

Codes : A B C D (a) 3 2 1 4 (b) 3 4 2 1 (c) 4 3 2 1 (d) 2 3 1 4 A force Fx acts on a particle such that its position x changes as shown in the figure.

Fx (N)

23.

Paschen series

1 32

26.

2 1H

3 2 He

1 0n

13 MeV

In the above nuclear reaction, the binding energy of 21 H is 2.2 MeV. The binding energy of 23 He is

31.

(b) 8.6 MeV (a) 4.4 MeV (c) 13 MeV (d) 17.4 MeV A radioactive nucleus has a half-life of 100 years. If the number of nuclei at t = 0 is N0, what will be the number of nuclei which have decayed in 300 years ? (a)

7N0 8

(b)

N0 2

(c)

3N 0 4

(d)

N0 8

EBD_7443 MT-16

(a) f3 – f2

The diagram of a logic circuit is given below. W X

(c) 36.

W Y

33.

34.

The output F of the circuit is represented by (a) W + (X + Y) (b) W + (X.Y) (c) W, (X + Y) (d) W, (X.Y) NAND and NOR gates are called universal gates primarily because they (a) are available universally (b) can be combined to produce OR, AND and NOT gates (c) are widely used in Integrated circuit packages (d) are easiest to manufacture Of the diodes shown in the following diagrams, which one is reverse biased ? +10 V

(a) (b)

37.

The current gain will be (a) 4.9 (b) 7.8 (c) 49 (d) 78 A student takes 50gm wax (specific heat = 0.6 kcal/kg°C) and heats it till it boils. The graph between temperature and time is as follows. Heat supplied to the wax per minute and boiling point are respectively 250

38.

R

39.

+5 V

R

(d)

0.707 A max

A max

(c) 40.

(c) f3 f 4

sin 1 (n 2 n1 )

tan

1

n2 n1

(b) sin (d)

1

tan

n12 1

n 22

n1 n2

The length of a metal is 1 when the tension in it is T1 and is 2 when the tension is T2. The original length of the wire is (a)

f1 f 2

1 2 3 4 567 8 Time (Minute)

Consider telecommunication through optical fibres. Which of the following statements is not true ? (a) Optical fibres may have homogeneous core with a suitable cladding (b) Optical fibres can be of graded refractive index (c) Optical fibres are subject to electromagnetic interference from outside (d) Optical fibres have extremely low transmission loss What should be the maximum acceptance angle at the aircore interface of an optical fibre if n 1 and n2 are the refractive indices of the core and the cladding, respectively (a)

The frequency response curve of RC coupled amplifier is shown in figure. The band with of the amplifier will be

200 150 100 50 0

(d) 1000 cal, 200°C

–5 V

35.

Ic is 0.98. Ie

In common emitter amplifier the

(c) 1500 cal, 200°C

R

–10 V

(d) f3 – f1

(b) 1000 cal, 100°C

+5 V

(c)

f 4 f2 2

(a) 500 cal, 50°C

R

–12 V

(b) f4 – f2

Temperature (°C)

32.

Target VITEEE

1

2

(b)

2 1T2

2T1

T2 T1

(d)

1T2

2T1

T1 T2 T1T2

1 2

MT-17

Mock Test 2 41.

42. 43.

44.

45.

46.

47.

PART - II : CHEMISTRY Identify the correct statement for change of Gibbs energy for a system ( Gsystem) at constant temperature and pressure : (a) If Gsystem = 0, the system has attained equilibrium (b) If Gsystem = 0, the system is still moving in a particular direction (c) If G system < 0, the process is not spontaneous (d) If G system > 0, the process is not spontaneous Which is a planar molecule ? (a) XeO4 (b) XeF4 (c) XeOF4 (d) XeO2F2 Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is (a)

Ssystem

Ssurroundings

0

(b)

Ssystem

Ssurroundings

0

(c)

Ssystem

48.

a b c and the axial angles are

49.

50.

(a)

90

(b)

90

(c)

90 ,

90

90 (d) Which one of th e following will show paramagnetism corresponding to 2 unpaired electrons?(Atomic numbers : Ni = 28, Fe = 26) (a) [Fe F6]3– (b) [Ni Cl4]2– (c) [Fe (CN)6]3– (d) [Ni (CN)4]2– In gaseous equilibrium the correct relation between Kc and Kp is (a)

Kc

K p ( RT) n (b) K p

K c ( RT) n

Kp Kc (K p ) n (d) (K c ) n RT RT Which one of the following reaction occurs at the cathode? (a) 2OH H 2 O O 2e

(c) 51.

0 only

Ssurroundings 0 only (d) In the solid state, MgO has the same structure as that of sodium chloride. The number of oxygens surrounding each magnesium in MgO is (a) 6 (b) 1 (c) 2 (d) 4 A solid with high electrical and thermal conductivity from the following is (a) Si (b) Li (c) NaCl (d) Ice Four successive members of the first row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionization enthalpy? (a) Vanadium (Z = 23) (b) Chromium (Z = 24) (c) Manganese (Z = 25) (d) Iron (Z = 26) Which one of the following is expected to exhibit optical isomerism? (en = ethylenediamine) (a) cis-[Pt(NH3)2 Cl2] (b) trans-[Pt(NH3)2Cl 2] (c) cis-[Co(en)2Cl 2] (d) trans-[Co(en)2Cl 2]

For orthorhombic system axial ratios are

52.

53.

54.

(b)

Ag

(c)

Fe 2

Ag

e

Fe 3

e (d) Cu 2 2e Cu Which of the following coordination compounds would exhibit optical isomerism? (a) pentamminenitrocobalt(III) iodide (b) diamminedichloroplatinum(II) (c) trans-dicyanobis (ethylenediamine) chromium (III) chloride (d) tris-(ethylendiamine) cobalt (III) bromide Lanthanoids are (a) 14 elements in the sixth period (atomic no. = 90 to 103) that are filling 4f sublevel (b) 14 elements in the seventh period (atomic no. = 90 to 103) that are filling 5f sublevel (c) 14 elements in the sixth period (atomic no. = 58 to 71) that are filling 4f sublevel (d) 14 elements in the seventh period (atomic no. = 58 to 71) that are filling 4f sublevel

The radioactive isotope

60 27 Co

which is used in

the treatment of cancer can be made by (n, p) reaction. For this reaction the target nucleus is (a)

59 28 Ni

(b)

59 27 Co

(c)

60 28 Ni

(d)

60 27 Co

EBD_7443 MT-18

55.

56.

57.

58.

59.

60.

61.

The age of most ancient geological formations is estimated by (a) Potassium–argon method (b) Carbon-14 dating method (c) Radium-silicon method (d) Uranium-lead method. The activation energy for a simple chemical reaction A B is Ea in forward direction. The activation energy for reverse reaction (a) Is always double of Ea (b) Is negative of Ea (c) Is always less than Ea (d) Can be less than or more than E a Select the rate law that corresponds to data shown for the following reaction A+B products. [A] [B] Initial rate Exp. 1 0.012 0.035 0.1 2 0.024 0.070 0.8 3 0.024 0.035 0.1 4 0.012 0.070 0.8 (a) rate = k [B]3 (b) rate = k [B]4 (c) rate = k [A] [B]3 (d) rate = k [A]2 [B]2 The plot of concentration of the reactant vs. time for a reaction is a straight line with a negative slope. The reaction follows a (a) zero order rate equation (b) first order rate equation (c) second order rate equation (d) third order rate equation A substance 'A' decomposes by a first order reaction starting initially with [A] = 2.00 m and after 200 min, [A] becomes 0.15 m. For this reaction t1/2 is (a) 53.72 min (b) 50.49 min (c) 48.45 min (d) 46.45 min Which one of the following information can be obtained on the basis of Le Chatelier principle? (a) Dissociation constant of a weak acid (b) Entropy change in a reaction (c) Equilibrium constant of a chemical reaction (d) Shift in equilibrium position on changing value of a constraint. Equivalent conductances of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 ohm–1 cm2 respectively. The equivalent conductance of CH3COOH at infinite dilution would be (a) 101.38 ohm–1 cm2 (b) 253.62 ohm–1 cm2 (c) 390.71 ohm–1 cm2 (d) 678.90 ohm–1 cm2

Target VITEEE 62.

The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25°C. The equilibrium constant of the reaction would be (Given F = 96500 C mol –1 ; R = 8.314JK–1mol–1) (a)

63.

64.

65.

2.0 1011

(b) 4.0 1012

(c) 1.0 10 2 (d) 1.0 1010 In the silver plating of copper, K[Ag(CN)2] is used instead of AgNO3. The reason is (a) A thin layer of Ag is formed on Cu (b) More voltage is required (c) Ag+ ions are completely removed from solution (d) Less availability of Ag+ ions, as Cu cannot displace Ag from [Ag(CN)2]– ion If 0.5 amp current is passed through acidified silver nitrate solution for 100 minutes. The mass of silver deposited on cathode, is (eq.wt.of silver nitrate = 108) (a) 2.3523 g (b) 3.3575 g (c) 5.3578 g (d) 6.3575 g For the reaction A + B C + D. The variation of the concentration of the products is given by the curve Y Z

Conc W Time

66.

67.

X

(a) Z (b) Y (c) W (d) X Acetamide and ethylamine can be distinguished by reacting with (a) Aqueous HCl and heat (b) Aqueous NaOH and heat (c) Acidified KMnO4 (d) Bromine water. Aniline when diazotized in cold and when treated with dimethyl aniline gives a coloured product. Its structure would be (a) CH3NH (b) CH3

N=N N=N

(c) (CH3)2N

N=N

(d) (CH3)2N

NH

NHCH3 NH2

MT-19

Mock Test 2 68.

69.

70.

Electrolytic reduction of nitrobenzene in weakly acidic medium gives (a) N-Phenylhydroxylamine (b) Nitrosobenzene (c) Aniline (d) p-Hydroxyaniline If an organic compound shows a streching ( C = O) frequency of 1720 cm–1. The organic compound is __________. (a) An aliphatic aldehyde (b) An , unsaturated aldehyde (c) A phenolic aldehyde (d) An aromatic aldehyde In the given reaction, NO 2

(b) SnCl2

(c) – Cl (d) – NH 3 Cl In a set of reactions acetic acid yielded a product D. SOCl2

A

Benzene Anhy.AlCl3

B

HCN

H.OH

D

C The structure of D would be: COOH CH2 – C – CH3

(a)

OH OH CH2 – C – CH3

(c) 72.

CN

77.

CN

(b)

C – CH3

OH

78.

OH

(d)

C – COOH

CH3

Ketones [ R — C — R1 , where R = R1 = alkyl groups]\

79.

||

73.

Ethanol and dimethyl ether form a pair of functional isomers. The boiling point of ethanol is higher than that of dimethyl ether, due to the presence of (a) H-bonding in ethanol (b) H-bonding in dimethyl ether (c) CH3 group in ethanol (d) CH3 group in dimethyl ether Propan - 1- ol may be prepared by the reaction of propene with (a) H3BO3 (b) H2SO4/H2O (c) B2H6,NaOH–H2O2 (d)

76.

HCl

X- stands for (a) – NH 2

CH3COOH

75.

X

Sn

71.

74.

O can be obtained in one step by (a) oxidation of primary alcohols (b) hydrolysis of esters (c) oxidation of tertiary alcohols (d) reaction of acid halides with alcohols Phenylmethyl ketone can be converted into ethylbenzene in one step by which of the following reagents? (a) LiAlH4 (b) Zn-Hg/HCl (c) NaBH4 (d) CH3MgI

80.

O || CH 3 C O O H

Sodium formate on heating yields (a) Oxalic acid and H2 (b) Sodium oxalate and H2 (c) CO2 and NaOH (d) Sodium oxalate. An ester (A) with molecular fomula, C9H10O2 was treated with excess of CH3MgBr and the complex so formed was treated with H2SO4 to give an olefin (B). Ozonolysis of (B) gave a ketone with molecular formula C8H8O which shows +ve iodoform test. The structure of (A) is (a) C6H5COOC2H5 (b) C2H5COOC6H5 (c) H3COCH2COC6H5 (d) p — H 3CO — C 6 H 4 — COCH 3 Self condensation of two moles of ethyl acetate in presence of sodium ethoxide yields (a) acetoacetic ester (b) methyl acetoacetate (c) ethyl propionate (d) ethyl butyrate Methanol is industrially prepared by (a) Oxidation of CH4 by steam at 900°C (b) Reduction of HCHO using LiAIH4 (c) Reaction HCHO with a solution of NaOH (d) Reduction of CO using H2 and ZnO – Cr2O3. Indicate which of the nitrogen compounds amongst th e following would undergo Hoffmann’s reaction, i,e., reaction with Br 2 and strong KOH to furnish the primary amine (R – NH2) ? (a) R CO NHCH 3 (b) R – CO – ONH4 (c) R – CO – NH2 (d) R – CO – NHOH.

EBD_7443 MT-20

Target VITEEE PART - III : MATHEMATICS

81.

82.

If sin (y + z – x), sin (z + x – y) and sin (x + y – z) are in A.P. (with x, y, z /2), then tan x, tan y and tan z are in (a) A.P. (b) G.P. (c) H.P. (d) none of these. The number of roots of equation tan

83.

1

2x

tan

1

is 4 (a) 0 (b) 1 (d) infinite (c) 2 In order that the equation

III

0 may have real

roots, n belongs to interval (a)

[ 2, 2]

(b) (

(b)

4

89.

(a) A

1 , B 4

(b) A

1 , B 8

(c) A

0, B

87.

Cr

1

(a)

n 1

(c)

n 2

Cr

Cr Cr

1

2

n

90.

(b)

n 2

1

(d)

n 1

Cr

A printing company prints two types of magazines A and B. The company earns `10 and

6

60

A sin 4 x B sin 6 x C sin 8 x D .

1 , C 3 1 , C 4

1 , D 8 1 , D 3

1 , C 6

R R

1 , D 8

1 4

(b)

8 15

(c)

1 2

(d)

15 8

If

( x)

(c) Cr

2

(a)

R

log 5 log 3 x , then ' (e) =

(a) 1

Cr equals

1

36 50

(d) none of these. In a ABC, the area of the triangle is

things taken r at a time, then the expression n

3 2

b 2 (c a ) 2 , then tan B is equal to

If nCr denotes the number of combination of n n

2 5

Then

3

2 There are 25 trees at equal distances of 5 meters in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. The total distance the gardener will cover in order to water all the trees is (a) 3550 m (b) 3434 m (c) 3370 m (d) 3200 m

A(x) B(y) Time available

The number of constraints is (a) 3 (b) 4 (c) 5 (d) 6 If sin 3 x cos 5 dx

2)

2 (d) 3

(c)

86.

,

88.

(c) (2, ) (d) ( , ) The portion of a tangent to a parabola y2 = 4ax cut off between the directrix and the curve subtends an angle at the focus, where = (a)

85.

Magzine Machine I II

3x

2ax(ax nc) (n 2 2)c 2

84.

`15 on each magazine A and B respectively. These are processed on three machines I, II & III and total time in hours available per week on each machine is as follows:

91.

1 log5e e

(b) e loge5 (d) loge5

If f(x) = x log x and f(0) = 0, then the value of for which Rolle’s theorem can be applied in [0, 1] is (a) –2 (b) –1 (c) 0 (d) 1/2

MT-21

Mock Test 2 92.

The spheres x 2

y2

z2

x y z 1 0

If sin 1 (1 x )

98.

2 sin

1

x

2

2 2 2 and x y z x y z 5 0

(a) (b) (c) (d) 93.

94.

95.

96.

intersect in a plane intersect in five points do not intersect none of these.

(a)

f :R

99.

defined

S,

dy y) dx

(x

(a) x y

y

(b)

x 4

y a

c

(d) none of these m

97.

If

(k 2 1)k! 1999

2000! , then m is

k 1

(a) 1999 (c) 2001

(d) none of these

(3a 5)iˆ (a 5)ˆj (a 2)kˆ are (a) coplanar if a < 0 (b) coplanar if a > 0 by

(c) always coplanar (d) never coplanar 100. If B is a non-singular matrix and A is a square matrix, then det (B–1 AB) is equal to

101.

(a) det (A–1)

(b) det (B–1)

(c) det (A)

(d) det (B)

3

7

x 1

3

x 4 dx is equal to

(a)

8/7 21 3 4 1 x C 32

(b)

8/7 32 3 4 1 x C 21

(c)

8/7 7 3 4 1 x C 32

(d)

None of these.

102. For a binary operation * on the set {1, 2, 3, 4, 5}, consider the following multiplication table.

y c a

tan

1 2

ˆ (2a 1)iˆ (2a 3)ˆj (a 1)k,

a is

a tan 1 ( x y) c

(c) y 1

(b) x

If a is a non-zero real number, then the vectors

2

a tan

1 2

ˆ aiˆ 2ajˆ 3ak,

f ( x) sin x 3 cos x 1, is onto, then the interval of S is (a) [ –1, 3] (b) [–1, 1] (c) [ 0, 1] (d) [0, 3] The locus of the mid points of the focal chords of the parabola y2 = 4ax is another parabola whose vertex is given by (a) (a, 0) (b) (0, a) (c) (–a, 0) (d) None of these The solution to the differential equation 2

0,

(c) x = 0 only

The function f : R R defined by f (x) = (x – 1) (x – 2) (x – 3) is (a) one-one but not onto (b) onto but not one-one (c) both one-one and onto (d) neither one-one nor onto If

x

, then :

(b) 2000 (d) None of these

*

1

2

3

4

5

1

1

1

1

1

1

2

1

2

1

2

1

3

1

1

3

1

1

4

1

2

1

4

1

5

1

1

1

1

5

Which of the following is correct? (a) (2 * 3) * 4 = 1 (b) 2 * (3 * 4) = 2 (c) * is not commutative (d) (2 * 3) * (4 * 5) = 2

EBD_7443 MT-22

Target VITEEE

103. If x cos

y sin x2

p , a variable chord of the y2

1 subtends a right angle a 2a 2 at the centre of the hyperbola, then the chords touch a fixed circle whose radius is equal to

hyperbola

(a)

2

2a

(c) 2a 104. If x 2 cos x dx

(b)

3a

(d)

5a

f (x ) sin x g(x ) cos x c .

Then (a) f(x) = x2, g(x) = x (b) f(x) = x2 + 2, g(x) = x (c) f(x) = x2, g(x) = 2x – 1 (d) f(x) = x2 – 2, g(x) = 2x 105. Let P be a variable point on the ellipse x2

y2

1 with foci F and F . If A is the area 1 2 a 2 b2 of the triangle PF1F2, then the maximum value of A is (a) 2abe (b) abe

1 abe (d) None of these 2 106. In each of a set of games it is 2 to 1 in favour of the winner of the previous game. The chance that the player who wins the first game shall win three at least of the next four is

(c)

(a)

8 27

(b) a tumbler is half empty if and only if it is half full (c) Both (a) and (b) (d) None of the above 109. A car will hold 2 in the front seat and 1 in the rear seat. If among 6 persons 2 can drive, then the number of ways in which the car can be filled is (a) 10 (b) 20 (c) 30 (d) None of these 110. The value of

(b)

4 81

4 (d) None (c) 9 107. The normal to the curve represented parametrically by x = a (cos + sin ) and y = a (sin – cos ) at any point , is such that it (a) makes a constant angle with x-axis (b) is at a contant distance from the origin (c) touches a fixed circle (d) satisfies (b) and (c) 108. Consider the following statements p : A tumbler is half empty. q : A tumbler is half full. Then, the combination form of “p if and only if q” is (a) a tumbler is half empty and half full

2

2

log e sin x dx 0

(a)

log e cos x dx is 0

log 2

2

(b)

2

log 2

1 (d) none of these. 2 111. If A, B, C are the angles of a triangle then the

(c)

log

sin 2A

sin C

sin B

value of determinant sin C sin B

sin 2B

sin A

sin A

sin 2C

is (b) 0 (a) (c) 2 (d) None of these 112. The length of projection of the segment joining P(–1, 2, 0) and Q (1, –1, 2) on the plane 2x y 2z 4 is (a) 3 (b) 4 (c) 5

(d) 3 3 x sin1 / x , x 0

113. f(x) =

0 , x 0 at x = 0 is (a) continuous as well as differentiable (b) differentiable but not continuous (c) continuous but not differentiable (d) neither continuous nor differentiable 114. Area enclosed between the curves y = sin 2x, y = cos2x and y = 0 in the interval [0, /2] is

(a)

1 2 3

(b)

1 2

1 sq. units

3 sq. units

1 2 sq. units 4 (d) (2 + 3) sq. units

(c)

MT-23

Mock Test 2 PART - IV : ENGLISH

115. If ABCDEF is a regular hexagon and

AB AC AD AE AF n AD . Then n is (a) 1

(b) 2

(c) 3

5 (d) 2

116. The altitude through vertex C of a triangle ABC, with position vectors of vertices a , b, c respectively is : b c c a a b

(a)

(c) 117. If

a b c

(b)

b a b c c a a b b a

b a

(d) None of these

the

solution of the differential dy ax 3 equation represents a circle, then dx 2y f the value of ‘a’ is (a) 2 (b) – 2 (c) 3 (d) – 4 118. Let A = {1,2,3,4,5,6,7,8} which of the following relations from A to R is NOT a function? (a)

R1 {( x, f ( x ) ) : x

A, f ( x ) 6 x 7}

(b)

R2

{( x, f ( x)) : x

A, f ( x ) | x | 9}

(c)

R3

(x , f ( x )) : x

A, f ( x )

(d)

R4

{( x, f ( x )) : x

A, f ( x )

119. Let f : R

1 x

2

7

4x}

R be such that f(1) = 3 and

f ' (1) = 6. Then xlim0

f (1 x ) f (1)

1/ x

(a) 1 (b) e½ 2 (c) e (d) e 3 120. The value of (2 – ) (2 – 2) (2 – (a) 49 (b) 16 (c) 16 (d) 49 2

equals

10) (2

–

11) is

Direction (Qs. 121 - 125) : Read the passage carefully and answer the questions given below. I felt the wall of the tunnel shiver. The master alarm squealed through my earphones. Almost simultaneously, Jack yelled down to me that there was a warning light on. Fleeting but spectacular sights snapped into ans out of view, the snow, the shower of debris, the moon, looming close and big, the dazzling sunshine for once unfiltered by layers of air. The last twelve hours before re-entry were particular bonechilling. During this period, I had to go up in to command module. Even after the fiery re-entry splashing down in 81o water in south pacific, we could still see our frosty breath inside the command module. 121. The word 'Command Module' used twice in the given passage indicates perhaps that it deals with (a) an alarming journey (b) a commanding situation (c) a journey into outer space (d) a frightful battle. 122. Which one of the following reasons would one consider as more as possible for the warning lights to be on? (a) There was a shower of debris. (b) Jack was yelling. (c) A catastrophe was imminent. (d) The moon was looming close and big. 123. The statement that the dazzling sunshine was "for once unfiltered by layers of air" means (a) that the sun was very hot (b) that there was no strong wind (c) that the air was unpolluted (d) none of above 124. His musical tastes are certainly ____ ; he has recordings ranging from classical piano performances to rock concerts, jazz and even Chinese opera. (a) antediluvian (b) eclectic (c) harmonious (d) sonorous 125. Choose the best pronunciation of the word, Poignant, from the following options. (a) poig-nant (b) pohing-nant (c) poing-nat (d) poi-nyant

EBD_7443

MOCK

VITEEE Mock Test Paper

3

Max. Marks : 125

Time : 2½ hrs

PART - I : PHYSICS 1.

2.

3.

5.

At temperature T, the emissive power and absorption power of a body for certain wavelength are e and a respectively, then (a) e = a (b) e > a (c) e < a (d) there will not be any difinite relation between e and a The work done in placing a charge of 8 × 10–18 coulomb on a condenser of capacity 100 microfarad is (a) 3.1 × 10–26 joule (b) 4 × 10–10 joule (c) 32 × 10–32 joule (d) 16 × 10–232 joule ABC is an equilateral triangle. Charges +q are placed at each corner as shown as fig. The electric intensity at centre O will be

6.

(a) 7.

+q A

r r +q B

(a)

1 4

o

q r

O

r

8.

+q C

(b)

1 4

q o

r2

1 3q (d) zero 4 o r2 A metallic sphere is placed in a uniform electric field. The line of force follow the path (s) shown in the figure as

(c)

4.

1 2 3 4

(a) 1 (c) 3

1 2 3 4

(b) 2 (d) 4

In an explosion, a body breaks up into two pieces of unequal masses. In this (a) both parts will have numerically equal momentum (b) lighter part will have more momentum (c) heavier part will have more momentum (d) both parts will have equal kinetic energy When a wire of uniform cross–section a, length l and resistance R is bent into a complete circle, resistance between any two of diametrically opposite points will be

9.

R 4

(b) 4R

(c)

R 8

(d)

R 2

Three resistances P, Q, R each of 2 and an unknown resistance S form the four arms of a Wheatstone bridge circuit. When a resistance of 6 is connected in parallel to S the bridge gets balanced. What is the value of S? (a) 3 (b) 6 (c) 1 (d) 2 A jar is filled with two non-mixing liquids 1 and 2 having densities 1 and, 2 respectively. A solid ball, made of a material of density 3 , is dropped in the jar. It comes to equilibrium in the position shown in the figure.Which of the following is true for 1, 1and 3? (a) 3 < 1 2 (b)

1>

3

2

(c)

1
r2 ) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is

EBD_7443 MT-36

9.

10.

Target VITEEE

(a)

r1 r2 2

(b)

r1 r2 2

(c)

r1 r2

(d)

r1 r2

If 25W, 220 V and 100 W, 220 V bulbs are connected in series across a 440 V line, then (a) only 25W bulb will fuse (b) only 100W bulb will fuse (c) both bulbs will fuse (d) none of these A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars. A mass m is suspended from the mid point of the wire. Strain in the wire is

14.

15.

16.

2L

x m

(a)

x 2L 2

(b)

(a) x L 2

x x (d) L 2L An electron enters a region where magnetic field (B) and electric field (E) are mutually perpendicular, then (a) it will always move in the direction of B (b) it will always move in the direction of E (c) it always possesses circular motion (d) it can go undeflected also. A wire carries a current. Maintaining the same current it is bent first to form a circular plane coil of one turn which produces a magnetic field B at the centre of the coil. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre of the double loop, caused by the same current is (a) 4B (b) B/4 (c) B/2 (d) 2B A coil of inductive reactance 31 has a resistance of 8 . It is placed in series with a condenser of capacitative reactance 25 . The combination is connected to an a.c. source of 110 volt. The power factor of the circuit is (a) 0.64 (b) 0.80 (c) 0.33 (d) 0.56

(c)

(c)

11.

12.

13.

The core of a transformer is laminated because (a) the weight of the transformer may be reduced (b) rusting of the core may be prevented (c) ratio of voltage in primary and secondary may be increased (d) energy losses due to eddy currents may be minimised A step-up transformer operates on a 230 V line and supplies a load of 2 ampere. The ratio of the primary and secondary windings is 1 : 25. The current in the primary is (a) 25 A (b) 50 A (c) 15 A (d) 12.5 A An inductance L having a resistance R is connected to an alternating source of angular frequency . The quality factor Q of the inductance is

17.

18.

19.

20.

R L

R L

(b) 1/ 2

(d)

L R

2

L R

Assuming no heat losses, the heat released by the condensation of x g of steam at 100°C can be used to convert y g of ice at 0°C into water at 100°C, the ratio x : y is : (a) 1 : 1 (b) 1 : 2 (c) 1 : 3 (d) 3 : 1 Interference is possible in (a) light waves only (b) sound waves only (c) both light and sound waves (d) neither light nor sound waves Two sources of light are said to be coherent if they emit light (a) of equal amplitudes (b) having the same wavelength (c) having a constant phase relationship (d) having the same intensity The figure gives the potential energy function U(x) for a system in which a particle is in onedimensional motion. In which region the magnitude of the force on the particle is greatest

MT-37

Mock Test 4 25.

26.

23.

24.

(b) K.E.max.

K.E.max.

(a)

(c)

(d)

27.

If n bullets each of mass m are fired with a velocity v per second from a machine gun, the force required to hold the gun in position is (a) (n + 1) mv

(b)

mv n2

mv n

(d)

mnv

(c) 28.

The ratio of de-Broglie wavelengths of proton and -particle having same kinetic energy is (a)

29.

K.E.max.

22.

photoelectrons with applied frequency ( ) is

K.E.max.

21.

(b) AB (a) OA (c) BC (d) CD The frequency of the radiation emitted by a hydrogen atom for the transition between n = 2 and n = 1 states is v0. What is the frequency of the radiation emitted by the hydrogen atom for transition between n = 4 and n = 1 states ? (a) 3v0/2 (b) 2v0 (c) 4v0 (d) 5v0/4 Match List-I (Series Spectra of Hydrogen) with List-II (Region in which the series lies) and select the correct answer using the codes given below the Lists : List-I List-II (Series spectra (Region in which the of Hydrogen) series lies) A. Lyman 1. Visible B. Balmer 2. Infra red C. Paschen 3. Ultra violet D. Brackett 4. X-ray 5. -ray Codes : A B C D (a) 3 1 2 2 (b) 1 3 2 4 (c) 3 1 4 5 (d) 1 2 3 5 In Millikan oil drop experiment a drop of charge Q and radius r is kept constant between two plates of potential difference of 800 volt. Then charge on other drop of radius 2 r which is kept constant with a potential difference of 3200 V is (a) Q/2 (b) 2 Q (c) 4 Q (d) Q/4 White X-rays are called white due to the fact that (a) they are electromagnetic radiations having nature same as that of white light. (b) they are produced most abundantly in X ray tubes. (c) they have a continuous wavelength range. (d) they can be converted to visible light using coated screens and photographic plates are affected by them just like light.

The photo electric work function for a metal surface is 4.125 eV. The cut-off wavelength for this surface is (a) 4125 Å (b) 3000 Å (c) 6000 Å (d) 2062 Å The variation of maximum kinetic energy

2 :1

(b)

2 2 :1

(c) 2 : 1 (d) 4 : 1 Match List-I (Classification) with List-II (Elementary Particles) and select the correct answer using the codes given below the Lists : List-I List-II (Classification) (Elementary particles) A. Baryons 1. Nucleon B. Mesons 2. Neutrino C. Leptons 3. Pion D. Bosons 4. Photon Codes : A B C D (a) 1 3 2 4 (b) 2 3 1 4 (c) 1 4 2 3 (d) 2 4 1 3

EBD_7443 MT-38

30.

31.

32.

33.

34.

35.

36.

Target VITEEE

Mp is the mass of proton and mn is the mass of neutron. If the mass of nucleus of an atom ZXA is measured and found to be M, then the nuclear binding energy would be (c is the velocity of light) : (a) {Zmp + (A – Z)mn – M}c2 (b) {Amn + Zmp + M}c2 (c) {(Z – A)mp + Amn – M}c2 (d) {M – Zmp– (A – Z)mn}c2 When a radioactive element decays by gamma radiation (a) its mass number will decrease by one unit with no change in atomic number (b) its mass number will not change but the atomic number will increase by one unit (c) both mass number and atomic number of the element change (d) there will be no change in either mass number or atomic number of the element In the uranium radioactive series, the initial nucleus is 92U238 and that the final nucleus is 206 82Pb . When uranium nucleus decays to lead, the number of particles and particles emitted are (a) 8 , 6 (b) 6 , 7 (c) 6 , 8 (d) 4 , 3 In semiconductors at a room temperature (a) the conduction band is completely empty (b) the valence band is partially empty and the conduction band is partially filled (c) the valence band is completely filled and the conduction band is partially filled (d) the valence band is completely filled The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10V. The d.c. component of the output voltage is (a) 20/ V (b) 10/ 2 V (c) 10/ V (d) 10V An oscillator is nothing but an amplifier with (a) positive feedback (b) large gain (c) no feedback (d) negative feedback The following configuration of gate is equivalent to A B

OR

Y AND NAND

37.

38.

(b) XOR (a) NAND (c) OR (d) NOR A ball is thrown up at an angle with the horizontal. Then the total change of momentum by the instant it returns to ground is (a) acceleration due to gravity × total time of flight (b) weight of the ball × half the time of flight (c) weight of the ball × total time of flight (d) weight of the ball × horizontal range A particle of mass m has momentum p. Its kinetic energy will be (a) mp (b) p2m p2 p2 (d) m 2m The reading of Centigrade thermometer coincides with that of Fahrenheit thermometer in a liquid. The temperature of the liquid is (a) – 40ºC (b) 313°C (c) 0°C (d) 100°C If a rubber ball is taken at the depth of 200 m in a pool, its volume decreases by 0.1%. If the density of the water is 1 × 103 kg/m3 and g = 10m/s2, then the volume elasticity in N/m2 will be (a) 108 (b) 2 × 108 9 (c) 10 (d) 2 × 109

(c) 39.

40.

PART - II : CHEMISTRY 41.

42.

43.

The alcohol which does not give a stable compound on dehydration is (a) Ethyl alcohol (b) Methyl alcohol (c) n-Propyl alcohol (d) n-Butyl alcohol IR spectrum of an organic compound is found about 1715 cm–1. The organic compound is (a) An aliphatic ketone (b) An aliphatic aldehyde (c) , - unsaturated ketones (d) Phenolic ketone The most suitable method of separation of a mixture ortho and para-nitrophenols mixed in the ratio of 1 : 1 is (a) Steam distillation (b) Crystallisation (c) Vapourization (d) Colour spectrum

MT-39

Mock Test 4 44.

45.

46.

When diethyl ether is treated with excess of Cl 2 in the presence of sunlight, then the product formed is (a)

CH 3CHCl O CH 2 CH 3

(b)

CH 3CHCl O CHClCH 3

(c)

CCl 3CCl 2OCCl 2 CCl3

(d) CH 3CCl 2 O CHClCH 3 The ether that undergoes electrophilic substitution reactions is (a) CH3OC2H5 (b) C6H5OCH3 (c) CH3OCH3 (d) C2H5OC2H5 What is the Eºcell for the reaction 2+ 2+ + Cu (aq) + Sn (aq)

47.

48.

49.

50.

4+ Cu (S) + Sn (aq)

at 25ºC if the equilibrium constant for the reaction is 1 × 106? (a) 0.5328 V (b) 0.3552 V (c) 0.1773 V (d) 0.7104 V The ionic conductance of Ba 2+ and Cl – are respectively 127 and 76 ohm–1 cm2 at infinite dilution. The equivalent conductance (in ohm–1 cm2) of BaCl2 at infinite dilution will be : (a) 139.5 (b) 203 (c) 279 (d) 101.5 A solution of [Ni(H2O)2+ 6 ] is green due to (a) d-d transition from t2g eg (b) d-d transition from eg t2g (c) d-d electronic transition from t 2g eg state associated with an amount of energy which comes under visible green region. (d) d-d electronic transition from t2g* eg* Prevention of corrosion of iron by Zn coating is called (a) Galvanization (b) Cathodic protection (c) Electrolysis (d) Photoelectrolysis Indicate which of the nitrogen compounds amongst th e following would undergo Hoffmann’s reaction, i,e., reaction with Br 2 and strong KOH to furnish the primary amine (R – NH2) ? (a) R CO NHCH 3 (b) R – CO – ONH4 (c) R – CO – NH2 (d) R – CO – NHOH.

51.

52.

53.

54.

Indicate which of the nitrogen compounds amongst th e following would undergo Hoffmann’s reaction, i,e., reaction with Br 2 and strong KOH to furnish the primary amine (R – NH2) ? (a) R CO NHCH 3 (b) R – CO – ONH4 (c) R – CO – NH2 (d) R – CO – NHOH. Benzaldehyde reacts with ethanoic KCN to give (a) C6H5CHOHCN (b) C6H5CHOHCOC6H5 (c) C6H5CHOHCOOH (d) C6H5CHOHCHOHC6H5 Which of the following compound will undergo self aldol condensation in the presence of cold dilute alkali ? (a)

CH 2

CH CHO

(b)

CH

(c)

C 6 H 5CHO

(d)

CH 3CH 2CHO.

C CHO

3CH 3COCH 3

HCl 3H 2O

(A) (CH 3 )2 C CH CO CH

55.

56.

C(CH 3 ) 2

(B) This polymer (B) is obtained when acetone is saturated with hydrogen chloride gas, B can be (a) phorone (b) formose (c) diacetone alcohol (d) mesityl oxide. Formic acid is obtained when (a) Calcium acetate is heated with conc. H2SO4 (b) Calcium formate is heated with calcium acetate (c) Glycerol is heated with oxalic acid at 373 K (d) Acetaldehyde is oxidised with K2Cr2O7 and H2SO4. In the Freidel Craft's acylation reaction, the effective electrophile is (a) RCOCl (b) AlCl3 (c) RCOCl

(d)

RCO

EBD_7443 MT-40

57.

58.

59.

60.

61.

62.

Target VITEEE

The total number of possible isomers for the complex compound [CuII (NH3)4] [PtII Cl4] (a) 3 (b) 6 (c) 5 (d) 4 Identify the incorrect statement among the following: (a) Lanthanoid contraction is the accumulation of successive shrinkages. (b) As a result of lanthanoid contraction, the properties of 4d series of the transition elements have no similarities with the 5d series of elements. (c) Shielding power of 4f electrons is quite weak. (d) There is a decrease in the radii of the atoms or ions as one proceeds from La to Lu. Wooden artifact and a freshly cut down tree give 7.6 and 15.2 counts min –1 g –1 of carbon (t1/2 = 5760 years) respectively. The age of the artifact is (a) 5760 years (b) 5760 × (7.6/15.2) years (c) 5760 × (15.2/7.6) years (d) 5760 × (15.2 – 7.6) years Half-life of a radioactive particle is 1 second. The rate of dissociation of A is 1000 per second. Then after 3 sec, (A) will be (a) 500 (b) 250 (c) 125 (d) 333 Half life of a first order reaction is 4 s and the initial concentration of the reactants is 0.12 M. The concentration of the reactant left after 16 s is (a) 0.0075 M (b) 0.06 M (c) 0.03 M (d) 0.015 M When a biochemical reaction is carried out in laboratory in the absence of enzyme then rate of reaction obtained is 10–6 times, then activation energy of reaction in the presence of enzyme is 6 RT (b) Different from Ea obtained in laboratory (c) P is required (d) Can't say anything The temperature dependence of rate constant (k) of a chemical reaction is written in terms of

(a)

63.

(a)

65.

66.

(b)

k vs T

k vs

NH2

NH2

(a)

(b) COCH3

COCH3

(c)

NH2

(d)

NHCOCH3

COCH3

67.

Which of the following reagents will convert pmethylbenzenediazonium chloride into pcresol? (a) Cu powder (b) H2O (c) H3PO2 (d) C6H5OH

68.

[A ]

69.

[ C] N Methylanil ine , A is (a) Formaldehyde (b) Trichloromethane (c) Nitrobenzene (d) Toluene A nitro alkane reacts with HONO to give insoluble product in alkali which turns blue on treatment with an alkali. The nitric alkane is (a) CH 3 CH 2 NO 2

reduction

[ B]

CHCl 3 KOH

reduction

(b)

CH 3 CH CH 2 NO 2 | CH 3

(c)

(CH 3 ) 2 CHNO 2

*

-E / RT Arrhenius equation, k = Ae a . Activation

energy (E*a ) of the reaction can be calculated by plotting

1 log T

1 1 (d) log k vs log T T The activation energy for a simple chemical reaction A B is Ea in forward direction. The activation energy for reverse reaction (a) Is always double of Ea (b) Is negative of Ea (c) Is always less than Ea (d) Can be less than or more than E a If is the fraction of HI dissociated at equilibrium in the reaction, 2 HI (g) H2 (g) + I2 (g), starting with 2 moles of HI, the total number of moles of reactants and products at equilibrium are (a) 2 + 2 (b) 2 (c) 1 + (d) 2 – Aniline is an activated system for electrophilic substitution. The compound formed on heating aniline with acetic anhydride is

(c)

64.

log k vs

(d) (CH 3 )3 CNO 2

MT-41

Mock Test 4 73.

70.

CH 3 COONH 4

( i ) CO 2

P

( ii ) H 3O

In the above reaction product 'P' is CHO

74.

COOH

(a)

(b) OH O

(c) 71.

Indenfity Z in the sequence

MgBr

(d)

C 6H5

||

C C6H 5

75.

In a set of the given reactions, acetic acid yielded a product C. CH3COOH PCl5

C6 H6

A

Anh.AlCl3

B

C2H 5MgBr Ether

C

76.

Product C would be C2 H 5 | C (OH)C 6 H 5

(a)

CH 3

(b)

CH 3CH(OH)C 2 H 5

(c)

CH 3COC 6 H 5

(d)

CH 3CH(OH)C6 H 5

77.

78.

CH2CH3 (i ) KMO 4 / OH

72.

X

( ii ) H / H 2O

Predict ‘X’ in the above reaction CH2COOH

(a)

79. CH2 CHO

(b)

X

P2 O5

Y

Proteins

(c)

CHO

(d)

Z

(a) CH3CH2CONH2 (b) CH3CN (c) CH3COOH (d) (CH3CO)2O Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction? (a) Exothermic and increasing disorder (b) Exothermic and decreasing disorder (c) Endothermic and increasing disorder (d) Endothermic and decreasing disorder What is the entropy change (in JK–1 mol–1) when one mole of ice is converted into water at 0º C? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol–1 at 0ºC) (a) 21.98 (b) 20.13 (c) 2.013 (d) 2.198 The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm–3, respectively. If the standard free energy difference ( Gº) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is (a) 9.92 × 105 Pa (b) 9.92 × 108 Pa 7 (c) 9.92 × 10 Pa (d) 9.92 × 106 Pa In the fluorite structure, the coordination number of Ca2+ ion is : (a) 4 (b) 6 (c) 8 (d) 3 On doping Ge metal with a little of In or Ga, one gets (a) p-type semi conductor (b) n-tpe semi conductor (c) insulator (d) rectifier During the process of digestion, the proteins present in food materials are hydrolysed to amino acids. The two enzymes involved in the process Enzyme( A)

Polypeptides Enzyme(B)

COOH

H 2O / H

are respectively (a) Diastase and Lipase (b) Pepsin and Trypsin (c) Invertase and Zymase (d) Amylase and Maltase

Amino acids

EBD_7443 MT-42

80.

Target VITEEE (a) 42 (c) 50

Which one of the following chemical units is certainly to be found in an enzyme? (a)

O HO

OH O

H

N—C

(b)

HO

86.

O

O O

N

(d)

(c) N

O O O

R

O

R

R

O

1

1

cos 2 tan

3 4

(a) not real

sin 2 cot

(b)

1

1 2

87.

4

88.

(c) less than 4 4 The general value of that satisfies the equation 2cot 2

(a)

83.

84.

4cosec

(b)

6

8 n

0 is : 6

2n (d) (c) 2n 6 6 Which of the following functions is NOT one-one ? (a) f : R R defined by f (x) 6x 1

(b)

f:R

R defined by f (x) x 2

(c)

f:R

R defined by f (x) x 3

(d)

f : R {7}

sin 2 cos 1

(a)

6 25

3 5

R defined by f (x)

(b)

6

3 4

(a)

1 , 2

1 9 , 2 2

(b)

(1, –2, 3)

(c)

3 , 2

3 5 , 2 2

(d)

None of these

log

2 x dx is 2 x

The value of

90.

(a) 0 (b) 2 (c) loge2 (d) log e 2 The values of p, for which both the roots of equation 4x2 – 20px + 25p2 + 15p – 66 = 0 are less than 2, belong to :

1

2x 1 x 7

24 25

(d) 3 2 If A, B, C are points (1, 0, 4), (0, –1, 5) and (2, –3, 1) respectively, then the coordinates of foot of the perpendicular drawn from A to the line BC are

89.

4 ,2 5 (c) (2, )

(a)

is equal to (b)

1 (d) –1 2 2 If the eccentricity of the hyperbola x – y2 sec2 = 4 is 3 times the eccentricity of the ellipse x2sec2 + y2 = 16, then the value of equals

1

7

4 24 (d) – 5 25 Suppose that the number of elements in set A is p, the number of elements in set B is q and the number of elements in A × B is 7. Then p2 + q 2 =

(c)

85.

n

2 3 cot

for x 0 is to be sin x 3 continuous at x = 0, then f(0) must be defined as: (a) 0 (b) 1

(c)

(c) greater than

82.

log e (1 x 2 tan x )

If f ( x )

(a)

is

equal to

49 51

(c)

PART - III : MATHEMATICS 81. tan

(b) (d)

91.

Let

(b) (d) 2iˆ kˆ ,

A

B

4 5 (– , –1) 1,

ˆi ˆj kˆ

C 4ˆi 3ˆj 7 kˆ . The vector satisfies the equations R

B

C

B and R . A

R

and which

0 is given by

(a)

2ˆi kˆ

(b)

ˆi 8ˆj 2kˆ

(c)

1 ˆ ˆ (i j 2kˆ ) 6

(d)

None of these

MT-43

Mock Test 4 92.

93.

94.

95.

A and Let A = {1, 2, 3, 4}. The function f : A g:A A are defined in the table given below.. x 1 2 3 4 f (x) 3 2 4 1 g (x) 4 3 2 2 The value of x, for which (fog) (x) = (gof) (x), is : (a) 1 (b) 2 (c) 3 (d) 4 The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009? (a) 3125 (b) 31250 (c) 21350 (d) 12350 If a, b, c are sides of a triangle and a2

b2

c2

(a 1) 2 (a 1) 2

(b 1) 2 (b 1) 2

(c 1) 2 (c 1) 2

0 , then

(a) ABC cannot be equilateral triangle (b) ABC is a right angled isosceles triangle (c) ABC is an isosceles triangle (d) None of these Let A, B, C be three events. If the probability of occurring exactly one event out of A and B is 1– a, out of B and C is 1– 2a and out of C and A is 1 – a, and that of occurring three events simultaneously is a2, then the probability that at least one out of A, B, C will occur is (a)

1 2

(b)

1 2

1 (d) none of these 2 The number of values of x ( x ) which satisfy the equation

(c)

96.

2

97.

81 |cos x| |cos x| ..... to 43 is (a) 2 (b) 4 (c) 6 (d) 8 If a point (h, k) satisfies an inequation ax + by 4, then the half plane represented by the inequation is (a) The half plane containing the point (h, k) but excluding the points on ax + by = 4 (b) The half plane containing the point (h, k) and the points on ax + by = 4 (c) Whole xy-plane (d) None of these

98.

The minimum value of |z| + | z – i| is (a) 0 (b) 1 (c) 2 (d) none

99.

If x m y n

(a)

dy ( x y ) m n then dx

my mx

(b)

ny nx

y (d) none of these x 100. If the system of equations x + 2y – 2z = 1, 4x + 2 y – z = 2, 6x + 6y + z = 3 has a unique solution, then (a) 1 (b) 2 (c) 3 (d) None of these 101. A plane passes through a fixed point (a, b, c). The locus of the foot of the perpendicular to it from the origin is the sphere

(c)

(a)

x2

y2

z 2 ax by cz

(b)

x2

y2

z2

0

2ax 2by 2cz

0

(c) x 2 y 2 z 2 4ax 4by 4cz 0 (d) None of these 102. The family of curves for which the area of the triangle formed by the x-axis, the tangent drawn at any point on the curve and radius vector of the point of tangency is constant equal to 2a2, is given by (a)

x

cy

a2 y

(b)

y

cx

a2 x

(c) x 2 ay 2 cy (d) a 2 x 2 y 2 cy 103. Out of 7 consonants and 4 vowels, how many words can be made each containing 3 consonants and 2 vowels? (a) 120 (b) 25200 (c) 4200 (d) None of these 104. The equation of normal to the curve x y where it cuts x-axis, is : (a) y = x (b) y = x + 1 (c) y = x – 1 (d) x + y = 1

xy,

105. If a is any vector, then ˆi

a ˆi

ˆj

a ˆj

kˆ

a kˆ is equal to

(a)

a

(b)

2a

(c)

3a

(d)

0

EBD_7443 MT-44

Target VITEEE

106. Sum of the series

1

2

1 12 14

1 22

111.

3 24

1 32 34

....

up to n terms is equal to n2

(a)

2(n n2

(c)

n

2

n 1

2

n2

(b)

n 1) n 1

2(n

n n 1)

n2 n

(d)

n 1

2

2(n

2

n 1)

107. There is 30% chance that it rains on any particular day. Given that there is at least one rainy day, then the probability that there are at least two rainy days is 14 5

(a)

1

7 10 7 10

7

7 10

6

13 5

(c)

6

6

14 17

14 15

7 10

7 10

(b)

1

(d)

7 10

1

b 0 1 1 0

X 0 1 0 0

6

7

is

(b) X = a + b (c) (a) X = a . b X = a'. b (d) X = a . b' 109. < N, + > where N is set of natural numbers is (a) a semi group (b) a monoid group (c) both (a) and (b) (d) none 110. The angle between the lines whose direction cosines are given by the equations 3l m 5n 0 , 6 nm 2 nl 5lm 0 is (a)

cos

1

(c)

cos

1

1 6

(b)

cos

1

2 3

(d)

cos

1

1 6 5 6

x x 4 3x 2 1

dx is equal to

(a)

log x

1 x

x2

(b)

log x

1 x

x2

(c)

log x

x2 3

1 x2 1 x2

3

C

3

C

C

(d) None of these 112. Let f (z) = sinz and g(z) = cosz. If * denotes a composition of functions, then the value of (f + ig) * (f – ig) is : (a)

i e– e

–iz

(b)

ie

eiz

– e– i z

108. The logical expression X, in its simplest form for the truth table a 1 1 0 0

( x 2 1)

(d) None of these (c) – i e 113. Suppose that the probability that an item produced by a particular machine is defective equals 0.2. If 10 items produced from this machine are selected at random, the probability that not more than one defective is found is (a)

1 e

2

(b)

2 e2

3 (d) none of these e2 114. Four distinct points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on a circle for (a) only one value of k (b) 0 < k < 1 (c) k < 0 (d) all integral values of k

(c)

|x| 1 , x 0 , x 0 then lim f (x) exists for 115. If f (x) = 0 x a |x| 1 , x 0

all (a) a 1 (b) a 0 (c) a –1 (d) a 2 116. Area bounded by the parabola y = x2 – 2x + 3 and tangents drawn to it from the point P(1, 0) is equal to (a)

4 2 sq. units

(b)

4 2 sq. units 3

(c)

8 2 sq. units 3

(d)

16 2 sq. units 3

MT-45

Mock Test 4 117. The length of the line segment joining the vertex of the parabola y2 = 4ax and a point on the parabola where the line segment makes an angle to the x-axis is

4am . Here, m and n n

respectively are (a) sin , cos (b) cos , sin 2 (c) cos , sin (d) sin2 , cos 118. A balloon ascends with uniform acceleration of 5 122 cm/sec units, at the end of half a minute a 8

do? The administration set up remains week mainly because the employees do not have the right example to follow and they are more concerned about being in the good books of the bosses than doing work. 121. The employees in our country (a) are quite punctual but not duty conscious (b) are not punctual, but somehow manage to complete their work (c) are somewhat lazy but good natured (d) are not very highly qualified

body is released from it. The time that elapses before the body reaches the ground in sec is : (a) 10 (b) 15 (c) 20 (d) none 1 - 1- x 2 is: 119. The domain of f (x) = 2x -1 ù1 é (a) ú ,1ê (b) [ – 1, [ úû 2 êë (c) [1, [ (d) None of these 120. The number of ways of selecting 8 books from a library which has 9 books each of Mathematics, Physics, Chemistry and English is : (a) 165 (b) 27C8 8 (c) 3 (d) None of these

122. According to the writer, the administration in India

PART - IV : ENGLISH

124. Choose the word which is most dissimilar to the word given in bold.

(a) is by and large effective (b) is very strict and firm (c) is affected by red tape (d) is more or less ineffective 123. The word 'assessment' means (a) enquiry (b) report (c) evaluation (d) summary

Direction (Qs. 121 - 123) Read the passage carefully and answer the questions given below What needs to be set right is our approach to work. It is a common sight in our country of employees reporting for duty on time and at the same time doing little work. If an assessment is made of time they spent in gossiping, drinking tea, eating "pan" and smoking cigarettes, it will be shocking to know that the time devoted to actual work is negligible. The problem is the standard which the leadership in administration sets for the staff. Forgot the ministers because they mix politics and administration. What do top bureaucrats do? What do the below down officials

Abate (a) augment (b) free (c) provoke (d) wane 125. Choose the best pronunciation of the word, Genre, from the following options. (a) zhon-ruh

(b)

jen-ner

(c)

(d)

zon-ra

zon-er

EBD_7443

MOCK

VITEEE Mock Test Paper

5

Max. Marks : 125

Time : 2½ hrs

PART - I : PHYSICS 1.

Six charges of equal magnitude, 3 positive and 3 negative are to be placed on PQRSTU corners of a regular hexagon, such that field at the centre is double that of what it would have been if only one +ve charge is placed at R. P

U T

2.

3.

R S

(b) –, + , +, +, –, –

7.

(c) –, + , +, –, +, – (d) + , –, +, –, +, – Steam is passed into 22 g of water at 20°C . The mass of water that will be present when the water acquires a temperature of 90°C is (Latent heat of steam is 540 cal/gm) (a) 24.8 gm (b) 24 gm (c) 36.6 gm (d) 30 gm Two capacitors of capacitances C1 and C2 are connected in parallel across a battery. If Q1 and Q2 respectively be the charges on the capacitors, then

Q1 will be equal to Q2

(a)

C2 C1

(c) 4.

6.

Q O

(a) + , +, +, –, –, –

5.

C12 C 22

(b)

(d)

8.

C1 C2

C 22 C12

Streamline flow is more likely for liquids with (a) high density and low viscosity (b) low density and high viscosity (c) high density and high viscosity (d) low density and low viscosity

9.

In a metre-bridge, the balancing length from the left end when standard resistance of 1 is in right gap is found to be 20 cm. The value of unknown resistance is (a) 0.25 (b) 0.4 (c) 0.5 (d) 4 A wire has a resistance of 3.1 at 30ºC and a resistance 4.5 at 100ºC. The temperature coefficient of resistance of the wire (a) 0.0064 ºC–1 (b) 0.0034 ºC–1 –1 (c) 0.0025 ºC (d) 0.0012 ºC–1 Two batteries, one of emf 18 volt and internal

resistance 2 and the other of emf 12 volt and internal resistance 1 , are connected as shown. The voltmeter V will record a reading of (a) 30 volt (b) 18 volt (c) 15 volt (d) 14 volt A galvanometer can be converted into a voltmeter by connecting (a) A high resistance in parallel (b) A low resistance in series (c) A high resistance in series (d) A low resistance in parallel An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the circle. The radius of the circle is proportional to (a)

B v

(b)

B v

(c)

v B

(d)

v B

MT-47

Mock Test 5 10.

11.

12.

13.

14.

15. 16.

17.

The Young’s modulus of a perfectly rigid body is (a) unity (b) zero (c) infinity (d) some finite non-zero constant A beam of electron passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off, and the same magnetic field is maintained, the electrons move (a) in a circular orbit (b) along a parabolic path (c) along a straight line (d) in an elliptical orbit. A heating coil is labelled 100 W, 220 V. The coil is cut in half and the two pieces are joined in parallel to the same source. The energy now liberated per second is (a) 25 J (b) 50 J (c) 200 J (d) 400 J Eddy currents are produced when (a) a metal is kept in varying magnetic field (b) a metal is kept in steady magnetic field (c) a circular coil is placed in a magnetic field (d) through a circular coil, current is passed If the number of turns per unit length of a coil of solenoid is doubled, the self-inductance of the solenoid will (a) remain unchanged (b) be halved (c) be doubled (d) become four times The time constant of C-R circuit is (a) 1/CR (b) C/R (c) CR (d) R/C In a circuit, L, C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45°. The value of C is (a)

1 f (2 fL R )

(b)

1 2 f (2 fL R )

(c)

1 f (2 fL R )

(d)

1 2 f (2 fL R )

The frequency of electromagnetic wave, which is best suited to observe a particle of radius 3 × 10–4 cm is of the order of (a) 1015 (b) 1014 13 (c) 10 (d) 1012

18.

A parallel beam of monochromatic light of wavelength 5000Å is incident normally on a single narrow slit of width 0.001 mm. The light is focussed by a convex lens on a screen placed in focal plane. The first minimum will be formed for the angle of diffraction equal to (a) 0° (b) 15° (c) 30° (d) 50° 19. The angular separation d between two wavelength and + d in a diffraction grating is directly proportional to (a) frequency of light (b) grating element (c) spatial frequency of grating (d) wavelength of light 20. A stationary body of mass 3 kg explodes into three equal pieces. Two of the pieces fly off in two mutually perpendicular directions, one with a velocity of 3iˆ ms

1

and the other with a

velocity of 4jˆ ms 1. If the explosion occurs in 10–4 s, the average force acting on the third piece in newton is y (a)

ˆ 10 4 (3iˆ 4j)

(b)

ˆ 10 4 (3iˆ 4j)

(c) (d) 21.

22.

23.

1 4ˆj

(3iˆ 4ˆj) 10 4

ˆ 104 (3iˆ 4j)

1×

(–

(3

i+

) 4j

1 3iˆ

x

Which one of the following is accompanied by the characteristic X-rays emission ? (a) -particle emission (b) Electron emission (c) Positron emission (d) k-electron capture A spectral line results from the transition n = 2 to n = 1 in the atoms/species given below. Which one of these will produce th e shortest wavelength emission ? (a) Hydrogen atom (b) Singly ionised helium (c) Doubly ionised lithium (d) Deuterium atom K wavelength emitted by an atom of atomic number Z = 11 is . Find the atomic number for an atom that emits K radiation with wavelength 4 (a) Z = 6 (b) Z = 4 (c) Z = 11 (d) Z = 44

EBD_7443 MT-48

24.

25.

26.

27.

28.

The glancing angle in a X-ray diffraction is 30º and the wavelength of X-rays used is 20 nm. the interplanar spacing of the crystal dffracting these X-rays will be (a) 40 nm (b) 20 nm (c) 15 nm (d) 10 nm If the energy of a photon is 10 eV, then its momentum is (a) 5.33 × 10–23 kg m/s (b) 5.33 × 10–25 kg m/s (c) 5.33 × 10–29 kg m/s (d) 5.33 × 10–27 kg m/s In a photoelectric experiment the stopping potential for the incident light of wavelength 4000Å is 2 volt. If the wavelength be changed to 3000 Å, the stopping potential will be (a) 2 V (b) Zero (c) Less than 2 V (d) More than 2 V A particle with rest mass m0 is moving with speed of light c. The de-Broglie wavelength associated with it will be (a) (b) zero (c) m0 c/h (d) h /m0c Figure shows three forces applied to a trunk that moves leftward by 3 m over a smooth floor. The force magnitudes are F1 = 5N, F2 = 9N, and F3 = 3N. The net work done on the trunk by the three forces

Target VITEEE He4)

32.

33.

34.

35.

M(2 =4.00387 amu, then in the nuclear reaction 6 2 2 2He4, the energy released will be 3Li + 1H (a) 20 MeV (b) 22 MeV (c) 50 MeV (d) 44 MeV What is the product aXb of the following nuclear reaction ? 14 4 16 b 7N + 2He 8 O + aX (a) Electron (b) Neutron (c) Proton (d) Deutron The following truth table belongs to which of the following four gates? A B Y 1 1 0 1 0 0 0 1 0 0 0 1 (a) NOR (b) XOR (c) NAND (d) OR The transfer ratio of a transistor is 50. The input resistance of the transistor when used in the common emitter configuration is 1 k . The peak value of the collector A.C. current for an A.C. input voltage of 0.01 V peak is (a) 100 A (b) 0.01 mA (c) 0.25 mA (d) 500 A Which of the following gates will have an output of 1? 1 1

0 1 (A)

0 1

(D)

(C)

29.

30.

31.

(a) 1.50 J (b) 2.40 J (c) 3.00 J (d) 6.00 J The activity of a radioactive sample is measured as 9750 counts per minute at t = 0, as 975 counts per minute at t = 5 min. The decay constant is approximately, (a) 0.230 per minute (b) 0.461 per minute (c) 0.691 per minute (d) 0.922 per minute A radioactive substance has a half life of four months. Three fourth of the substance will decay in (a) three months (b) four months (c) eight months (d) twelve months If the masses of the nuclei are M(1H2) = 2.0147 amu, M(3Li6) =6.017 amu and

36.

(B)

0 0

(a) D (b) A (d) C (c) B The following figure shows a logic gate circuit with two inputs A and B and the output C. The voltage waveforms of A, B and C are as shown below A B

Logic gate circuit

C

1 A 1

t

B 1

t

C

t

The logic circuit gate is (a) NAND gate (b) NOR gate (c) OR gate (d) AND gate

MT-49

Mock Test 5 37.

38.

39.

40.

A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. Then the coefficient of friction is (Take g = 10 ms–2) (a) 0.06 (b) 0.03 (c) 0.04 (d) 0.01 Advantage of optical fibre is (a) high bandwidth and EM interference (b) low bandwidth and EM interference (c) high and width, low transmission capacity and no EM interference (d) high bandwidth, high data transmission capacity and no EM interference If 1 and 2 are the refractive indices of the materials of core and cladding of an optical fibre, then the loss of light due to its leakage can be minimised by having (a) (b) 1 < 2 1> 2 (c) = (d) none of these 1 2

NH 2

(b) m2 >> m2

(c) m1 = m2

(d) m1 = 2m2

(a) Br

42.

(c) 44.

Sn / HCl

B

(1) HONO ( 2 ) H 3 PO2

C

Cl

Br

(b) Br

Br

(c) None of these

Br

Br

Which of the following gives the best yield of metabromo aniline ? (a)

HNO3

(1) Sn , HCl

H 2SO 4

(2 ) NaOH

Br2 / FeCl3

(b)

HNO3

Br2

H 2SO 4

FeBr3

(1) Sn , HCl (2 ) NaOH

(c)

One organic alcohol is irradiated with IR source and it produces three streching frequency such as 3391 cm–1, 2981 cm–1 and 1055 cm–1. Name of that organic alcohol is (a) Methanol (b) Ethanol (c) Isobutanol (d) Isopropanol The following reaction is

Br2

HNO3

FeBr3

H 2SO 4

(1) Sn , HCl

HNO3

(2 ) NaOH

H 2SO 4

(1) Sn , HCl (2 ) NaOH

(d) Br2 FeBr3

NO2

NO2

OH

45.

heat

+ KOH (solid)

NO2

+

OH

43.

A

NO 2 Cl

PART - II : CHEMISTRY 41.

( 2 ) CuCl

NO 2

A particle of mass m1 moving with velocity v strikes with a mass m2 at rest, then the condition for maximum transfer of kinetic energy is (a) m1 >> m2

(1) HONO

Br

(a) nucleophilic substituton (b) electrophilic substitution (c) fee radical substitution (d) electrophilic addition The product – (C) obtained in the following sequence of reactions is

46.

Which reaction sequence would be best to prepare 3-chloroanilne from benzene ? (a) Chlorination, nitration, reduction (b) Nitration, chlorination, reduction (c) Nitration, reduction, chlorination (d) Nitration, reduction, acetylation, chlorination, hydrolysis General electronic configuration of lanthanides is (a) (n – 2) f1 –14 (n –1) s2p6d0 – 1 ns2 (b) (n – 2) f10 –14 (n –1) d0 – 1 ns2 (c) (n – 2) f0 –14 (n –1) d10 ns2 (d) (n – 2) d0 –1 (n –1) f1 – 14 ns2

EBD_7443 MT-50

47.

48.

49.

Target VITEEE

According to IUPAC nomenclature sodium nitroprusside is named as (a) Sodium pentacyanonitrosyl ferrate (III) (b) Sodium nitroferrocyanide (c) Sodium nitroferrocyanide (d) Sodium pentacyanonitrosyl ferrate (II) The number of unpaired electrons in the complex ion [CoF6]3– is (Atomic no.: Co = 27) (a) Zero (b) 2 (c) 3 (d) 4 Which of the following is considered to be an anticancer species ? Cl

(a)

CH2 CH2

Pt Cl

Cl

(c) H3N

50.

51.

52.

(d)

Pt Cl

55.

Pt

Cl

Cl

H3N

Cl

56.

Cl

H3N

54.

Cl

Cl

(b)

53.

NH3

The main reason for larger number of oxidation states exhibited by the actinoids than the corresponding lanthanoids, is (a) more energy difference between 5f and 6d orbitals than between 4f an d 5d orbitals. (b) lesser energy difference between 5f and 6d orbitals than between 4f an d 5d orbitals. (c) larger atomic size of actinoids than the lanthanoids. (d) greater reactive nature of the actinoids than the lanthanoids. Colour of cupric chloride is blue is due to __________. (a) p -d electronic transition causes emission of energy (b) d-d electronic transition causes emission of energy (c) d-d electronic transition causes absorption of energy at red visible wave length (d) p-d electronic transition of Cu2+ state. Which metal Aprons are worn by radiographer to protect him from radiation? (a) Mercury Coated Apron (b) Lead Apron (c) Copper Apron (d) Aluminimised Apron

(b) 2.3 K (a) 3 K 2 (c) 2 K (d) K In acidic medium the rate of reaction between BrO3 and Br ion is given by the expression

Pt

Cl

Which one of the following radioisotopes is used in the treatment of blood cancer ? (a) P32 (b) Co60 131 (c) I (d) Na24 For a first order reaction A B the reaction rate at reactant concentration of 0.01 M is found to be 2.0 10 5 mol L 1 S 1. The half life period of the reaction is (a) 30 s (b) 220 s (c) 300 s (d) 347 s The rate of certain reaction increases by 2-3 times when the temperature is raised from 300 K to 310 K. If K is the rate constant at 300 K then rate constant at 310 K will be

d(BrO3 ) 2 = K [BrO3 ][Br ][H ] dt Which of the following is correct ?

57.

58.

(a) Doubling the concentration of H ions will increase the reaction rate by 4 times. (b) Rate constant of overall reaction is 4 sec–1. (c) Rate of reaction is independent of the conc. of acid (d) The change in pH of the solution will not affect the rate. The reaction 2N2O5 2N2O4 + O2 is (a) Bimolecular and second order (b) Unimolecular and first order (c) Bimolecular and first order (d) Bimolecular and zero order In a two-step exothermic reaction 3C(g) D(g) A2(g) + B2(g) Step 1

Step 2

Steps 1 and 2 are favoured respectively by (a) high pressure, high temperature and low pressure, low temperature (b) high pressure, low temperature and low pressure, high temperature (c) low pressure, high temperature and high pressure, high temperature (d) low pressure, low temperature and high pressure, low temperature

MT-51

Mock Test 5 59.

60.

The OH group of an alcohol or the –COOH group of a carboxylic acid can be replaced by –Cl using (a) phosphorus pentachloride (b) hypochlorous acid (c) chlorine (d) hydrochloric acid A and B in the following reactions are OH B R– C R–C–R' HCN/ A KCN CH2NH2 R' O CN (a) A = RR'C , B = LiAlH4 OH OH (b) A = RR'C , B = NH3 COOH CN (c) A = RR'C , B H 3O OH

61.

65.

66.

The appearance of colour in solid alkali metal halides is generally due to (a) Schottky defect (b) Frenkel defect (c) Interstitial positions (d) F-centres CsBr crystallises in a body centered cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being 6.02 × 1023 mol–1, the density of CsBr is (a) 0.425 g/cm3 (b) 8.25 g/cm3 3 (c) 4.25 g/cm (d) 42.5 g/cm3

67.

Cu aq is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction :

CaCO 3

CH3COOH

A

heat

B

I2 NaOH

C

The molecular formula of C is

68.

OH

(a)

|

CH 3 C CH 3 |

(b) ICH2 — COCH3

I

62.

63.

64.

Cu 2 (aq) Cu (s) choose correct Eº for above reaction if Eº Cu2+/Cu = 0.34 V and Eº Cu2+/Cu+ = 0.15 V (a) –0.38 V (b) +0.49 V (c) +0.38 V (d) –0.19 V Specific conductance of 0.1 M sodium chloride solution is 1.06 × 10–2 ohm–1 cm–1. Its molar conductance in ohm–1 cm2 mol–1 is (a) 1.06 × 102 (b) 1.06 × 103 4 (c) 1.06 × 10 (d) 5.3 × 102 Hydrogen-Oxygen fuel cells are used in space craft to supply (a) Power for heat and light (b) Power for pressure (c) Oxygen (d) Water Corrosion is basically a (a) altered reaction is presence of H2O (b) electrochemical phenomenon (c) interaction (d) union between two light metals and a heavy metal When during electrolysis of a solution of AgNO3 9650 coulombs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be (a) 1.08 g (b) 10.8 g (c) 21.6 g (d) 108 g How many isomers of C5H11 OH will be primary alcohols ? (a) 5 (b) 4 (c) 2 (d) 3. HBr reacts fastest with (a) 2-Mehtylpropan-1-ol 2Cu

(d) A = RR'CH2CN, B = NaOH Consider the following transformations :

(c) CHI3 (d) CH3I Caboxylic acid group does not give the usual addition and elimination reactions of aldehydes and ketones because (a) O–H bond is more polar than C = O group (b) Carboxylate ion gets ionised (c) Carboxylate ion gets stabilised by resonance (d) it exists as – COOH and there is no carbonyl group The enthalpy and entropy change for the reaction Br2(l) + Cl2 (g) 2 BrCl (g) are 30kJ mol–1 and 105 JK–1 mol–1 respectively. The temperature at which the reaction will be in equilibrium is (a) 273 K (b) 450 K (c) 300 K (d) 285.7 K In graphite electrons are : (a) Localised on each carbon atom (b) spread out between the sheets (c) localised on every third carbon atom (d) present in antibonding orbital.

69.

70.

71.

72.

73.

( aq )

EBD_7443 MT-52

74.

75.

76.

77.

78.

79.

80.

Target VITEEE

(b) 2-Methylpropene-2-ol (c) propan-2-ol (d) propan-1-ol. Which one of the following on oxidation gives a ketone ? (a) Primary alcohol (b) secondary alcohol (c) tertiary alcohol (d) All of these An aromatic ether is not cleaved by HI even at 525 K. The compound is (a) C6H5OCH3 (b) C6H5OC6H5 (c) C6H5OC3H7 (d) Tetrahydrofuran Acetic anhydride reacts with diethyl ether in the presence of anhydrous AlCl3 to give (a) CH3COOCH3 (b) CH3CH2COOCH3 (c) CH3COOCH2CH3 (d) CH3CH2OH Phenol is (a) a base stronger than ammonia (b) an acid stronger than carbonic acid (c) an acid weaker than carbonic acid (d) a neutral compound CH 2

O

CH 2

O

CH 2

O

The above shown polymer is obtained when a carbonyl compound is allowed to stand. It is a white solid. The polymer is (a) Trioxane (b) Formose (c) Paraformaldehyde (d) Metaldehyde. The compound formed when malonic ester is heated with urea is (a) Cinnamic acid (b) Butyric acid (c) Barbituric acid (d) Crotonic acid. Schotten-Baumann reaction is a reaction of phenols with (a) Benzoyl chloride and sodium hydroxide (b) Acetyl chloride and sodium hydroxide (c) Salicylic acid and conc. H2SO4 (d) Acetyl chloride and conc H2SO4

82.

The equation k sin + cos 2 = 2k – 7 possesses a solution if : (a) 2 k 6 (b) k > 2 (c) k > 6 (d) k < 2 If the tangents are drawn from the point (x1, y1) to the parabola y2 = 4ax, then the length of their chord of contact is

1 ( y12 |a|

2ax1 )( y12

4a 2 )

(b)

1 ( y12 |a|

4ax1 )( y12

4a 2 )

1 ( y12 4ax1 )( y12 |a| (d) None of these

2a 2 )

(c)

83. In a ABC, a = 5, b = 4 and cos( A B)

31 . 32

Then c is (a) 6 (c) 4 84.

(b) 5 (d) 3 ˆ ˆ ˆ ˆ ˆ Let a 2i j k, b i 2 j kˆ and a unit vector cˆ be coplanar. If cˆ is perpendicular to aˆ , then cˆ = (a)

1 2

( ˆj kˆ)

(b)

1 3

( ˆi ˆj kˆ )

1 ˆ ˆ 1 ˆ ˆ ˆ (i 2 j) (i j k ) (d) 5 3 85. Let X = {1, 2, 3, 4} A function is defined from X

(c)

86.

to N as R ( x, f (x )) : x X, f (x ) xPx 1 . The range of f is (a) {1, 2, 3, 4} (b) {1, 4, 9, 16} (c) {1, 2, 6, 24} (d) {1} If f(x) be a quadratic polynomial such that f(0) = 2, f’(0) = – 3 and f’’ (0) = 4 then 1

f (x ) dx is equal of 1

16 3 (c) 0 (d) none of these. The tangent at a point P on the hyperbola

(a) –3

87.

x2

(b)

y2

1 meets one of its directrices in F. If a 2 b2 PF subtends an angle at the corresponding focus, then equals

PART - III : MATHEMATICS 81.

(a)

3 (b) (c) (d) 4 2 4 12 In a right angled triangle one side is thrice the other side in length. The is suspended by a string attached at the right angle. The angle that the hypotenuse of the will make with the vertical is

(a)

88.

MT-53

Mock Test 5 sin–1 (3/5)

89.

90.

sin–1 (4/5)

(b) (a) (c) 60º (d) None of these | (a × b).c| = |a| |b||c| , if (a) a.b = b. c = 0 (b) b.c = c. a = 0 (c) c.a = a.b = 0 (d) a.b = b.c = c.a = 0 3 1 If, P( B) , P(A B C ) 4 3 1 and P( A B C ) , then P (B C) is 3 1 1 1 1 (c) (d) (b) 6 15 9 12 If m sin = n sin ( + 2 ), then tan ( + ) . cot is equal to m n m–n (a) (b) m–n m n

(a)

91.

m n mn 2 92. If P A , P B 5

(c)

(d)

m–n mn

3 and P A 10

then P A | B . P B | A

94.

95.

x

x2

x (1

3

6

x)

x

0 |

|

dx is equal to

cos 2

sin cos sin cos

cos 2 cos 2

3

|

3 2/3 x 6 tan 1 x1 / 6 C 2 3 2/3 x 6 tan 1 x1 / 6 C (b) 2 3 2/3 x 6 tan 1 x1 / 6 C (c) 2 (d) None of these. If the system of equations x + 2y – 3z = 1, (p + 2) z = 3(2p + 1) y + z = 2 is inconsistent, then the value of p is

(a)

96.

3

(d)

sin cos

2

then

cannot exceed (a) 1 (b) 0 1 (c) (d) None of these 2 101. Area of the region between th e curves x2 + y2 = 2, y = sin x and y-axis in first quadrant is (a)

|

0 and

sin 2 sin sin 2

is equal to

x 2 + bx + c = 0 , where, c < 0 < b, then (a) 0 < < (b) 0

(c)

(b) (c) (d) 0 3 2 4 98. For binary operation * defined on R – {1} such a that a * b is b 1 (a) not associative (b) not commutative (c) commutative (d) both (a) and (b) 99. The value of the boolean expression [a.b' c].a ' if a = 0, b = 0, c = 1 (a) 1 (b) 0 (c) 2 (d) none 100. Suppose , , R are such that (a)

1 , 5

B

5 5 25 (b) (c) (d) 1 6 7 42 Solution of differential equation xdy – ydx = 0 represents: (a) rectangular hyperbola. (b) parabola whose vertex is at origin. (c) circle whose centre is at origin. (d) straight line passing through origin. If and ( < ) are the roots of the equation

0

97.

(b)

sin , sin , sin

(a) 93.

1 (c) 0 (d) 2 2 3 The equation of two curves are x – 3xy2 = a2 and 3x2y – y3 = b2. The angle of intersection of these curves is

(a) –2

8

4 2

(c)

4

8

3

sq. units

(b)

4 8

2

sq. units (d)

8

4

sq. units sq. units

3 2 2i 0 represent the 102. If the roots of z iz vertices of a ABC in the Argand plane, then the area of the triangle is

3 7 3 7 (b) 2 4 (c) 2 (d) None 103. If n + 2C8 : n – 2P4 = 57 : 16, then the value of n is: (a) 20 (b) 19 (c) 18 (d) 17 104. The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. The length of the greatest side is (a) 4 (b) 5 (c) 6 (d) none of these (a)

EBD_7443 MT-54

Target VITEEE

105. A family of curves is given by the equation

x2

= 1. The differential equation a 2 b2 representing this family of curves is given by

xy

d2 y

dy Ax dx

dx 2

2

– y

dy = 0. The value of A dx

is (a) 0 (b) 1 (c) 3 (d) 5 106. A brick manufacture has two depots A and B, with stocks of 30000 and 20000 bricks respectively. He receive orders from three builders P, Q and R for 15000, 20,000 and 15000 bricks respectively. The cost (in `) of transporting 1000 bricks to the builders from the depots as given in the table. To

From

Transportation cost per 1000 bricks (in `) P

Q

40

20

20

B

20

60

40

sin(a 1)x sin x , x 0 x c ,x 0 x bx 2 bx

x

3/ 2

continuous at x = 0 are 3 1 (a) a ,c , b= 0 2 2 3 1 (b) a ,c ,b 0 2 2

,x

0

0

1 5 cos 1 2 3

3

(b)

(a)

5

is

3

5

2 2 5 (c) (d) none of these 6 110. If 3f (cos x) + 2 f (sin x) = 5x, then f’(cos x) = (a)

R

A

a

109. The value of tan

(c)

The manufacturer wishes to find how to fulfill the order so that transportation cost is minimum. Formulation of the L.P.P., is given as (a) Minimize Z = 40x – 20y Subject to, x + y 15, x + y 30, x 15, y 20, x 0, y 0 (b) Minimize Z = 40x – 20y Subject to, x + y 15, x + y 30, x 15, y 20, x 0, y 0 (c) Minimize Z = 40x – 20y Subject to, x + y 15, x + y 30, x 15, y 20, x 0, y 0 (d) Minimize Z = 40x – 20y Subject to, x + y 15, x + y 30, x 15, y 20, x 0, y 0 107. Total number of ways of selecting five letters from letters of the word INDEPENDENT is (a) 70 (b) 72 (c) 75 (d) 80 108. The values of a, b and c which make the function

f (x)

3 1 ,c ,b 2 2 (d) none of these

(c)

y2

5 cos x

(b)

5 sin x

5 cos x

(d) none of these

111. Let A {x : 1 x 1} = B and S be the subset of A×B defined by S = {(x,y) : x2 + y2 = 1}. This defines (a) an one-one function from A into B (b) a many - one function from A into B (c) a bijective function from A into B (d) Not a function 112. A variable plane moves so that the sum of reciprocals of its intercepts on the three coordinate axes is constant . It passes through a fixed point, which has coordinates (a) ( , , ) (c)

,

(b) ,

(d)

1 1 1 , , 1

,

1

,

1

113. The probability of a man hitting a target is 1/4. The number of times he must shoot so that the probability he hits the target, at least once is more than 0.9, is [use log 4 = 0.602 and log 3 = 0.477] (a) 7 (b) 8 (c) 6 (d) 5 114. The coordinates of a point on the line 6 x 2 y 1 z 3 at a distance of from 2 3 2 2 the point (1, 2, 3) is (a) (56,,43, 111) (c) (2, 1, 3)

56 43 111 , , 17 17 17 (d) (–2, –1, –3)

(b)

MT-55

Mock Test 5 115. Suppose that X has a Poisson distribution. If P(X 1) 0.3 , and P(x 2) 0.2 then P(X = 0) = (a) e 1/3 (b) e –3/2 (c) e –4/3 (d) 0.5 116. The value of x which satisfy the equation 1 + sin2ax = cos x, where a is irrational is/are , (d) 0, 2 2 117. A curve passes through the point (5, 3) and at any point (x, y) on it, the product of its slope and the ordinate is equal to its abscissa. The curve is (a) parabola (b) ellipse (c) hyperbola (d) circle 118. The lines whose vector equations are ˆ (2iˆ pjˆ 5k) r 2iˆ 3jˆ 7kˆ

(a) 0

(b) 0,

(c)

ˆ and r iˆ 2jˆ 3kˆ (3iˆ pjˆ pk) are perpendicular for all values of and if p = (a) 1 (b) –1 (c) – 6 (d) 6 119.

lim

x

sin( cos 2 x ) x2

0

equals

(a)

(b)

(c)

(a) aR1b

|a|

(c) aR 3b

a divides b (d) aR 4 b

(d) 1

2 120. Which one of the following relations on the set of real numbers R is an equivalence relation ?

|b|

(b) aR 2 b

a

b

a b

PART - IV : ENGLISH Direction (Qs. 121 -123) : Read the passage carefully and answer the questions given below. Naval architects never claim that a ship is unsinkable, but the sinking of the passenger-and-car ferry Estonia in the Baltic surely should have never have happened. It was well designed and carefully maintained. It carried the proper number of lifeboats. It had been thoroughly inspected the day of its fatal voyage. Yet hours later, the Estonia rolled over and sank in a cold, stormy night. It went down so quickly that most of those on board, caught in their dark, flooding cabins, had no chance to save themselves: Of those who managed to scramble overboard, only 139 survived. The rest died of hypothermia before the rescuers could pluck them from the cold sea. The final death toll amounted to 912

souls. However, there were an unpleasant number of questions about why the Estonia sank and why so many survivors were men in the prime of life, while most of the dead were women, children and the elderly. 121. One can understand from the reading that ----. (a) the lifesaving equipment did not work well and lifeboats could not be lowered (b) design faults an d incompetent crew contributed to the sinking of the Estonia ferry (c) 139 people managed to leave the vessel but died in freezing water (d) most victims were trapped inside the boat as they were in their cabins 122. It is clear from the passage that the survivors of the accident ----. (a) helped one another to overcome the tragedy that had affected them all (b) were mostly young men but women, children and the elderly stood little chance (c) helped save hundreds of lives (d) are still suffering from severe post-traumatic stress disorder 123. According to the passage, when the Estonia sank, ----. (a) there were only 139 passengers on board (b) few of the passengers were asleep (c) there were enough lifeboats for the number of people on board (d) faster reaction by the crew could have increased the Estonia's chances of survival 124. Fill in the blanks with the appropriate words from the option to complete the sentence. A rising China and the anti-India _________of Kathmandu's hill elite, however, have the potential to neutralize, over the longer term, some of Delhi's natural _________advantages in Nepal. (a) happiness, unnecessary (b) good nature, unimportant (c) resentments, strategic (d) enjoyment, common 125. Choose the best pronunciation of the word, Debris, from the following options. (a) deb-rees (b) de-bree (c) deb-ris (d) deb-re

EBD_7443

MOCK

VITEEE Mock Test Paper

6

Max. Marks : 125

Time : 2½ hrs

PART - I : PHYSICS 1.

(a)

Two particles A and B, move with constant velocities v1 and v2 . At the initial moment their condition for particles A and B for their collision is

2.

r1.v1

(b)

r1 v1

r2

(c)

r1

v1 v 2

(d)

r1 r2 | r1 r2 |

0

r2 .v2

r2

v2

d2

(c) Less than

4.

v2 v1 | v2 v1 |

4

4

d2 q2

0

d2

C1= 4µF 14 volt C2

C3 4F

2µF

E

d (a) Negative, increase (b) Positive, decrease (c) Negative, increase (d) Positive, increase Here, distance increases so, potential energy increases. Two charged spherical conductors (each of radius R) are at a distance d part (d > 2R). The charge on one is +q and that on the other is – q. The magnitude of the force between them will be :

0

1

In the figure, a proton moves a distance d in a

+ p

q2

1

(d) zero There capacitors C1, C2 and C3 are connected as shown in the figure. A potential difference of 14 volts is applied to the input terminals. What is the charge on C3 (in µC) ?

uniform electric field E as shown in the figure. Does the electric field do a positive or negative work on the proton? Does the electric potential energy of the proton increase or decrease?

3.

4

(b) more than

position vectors are r1 and r2 respectively. The

(a)

q2

1

5.

(a) 8 (b) 4 (d) 100 (c) 42 The radii of two concentric spherical conducting shells are r 1 and r2 (> r1). The charge on the outer shell is q. What will be the charge on the inner shell which is connected to the earth ? (a)

6.

r1 r2

q

(b)

r2 r1

q

(c) – q (d) zero Figure shows three points A, B and C in a region of uniform electric field E . The line AB is perpendicular and BC is parallel to the field lines. Then which of the following holds good?

MT-57

Mock Test 6 E

A B

3

C

7.

(a) VA = VB = VC (b) VA = VB > VC (c) VA = VB < VC (d) VA > VB = VC Where VA, VB and VC represent the electric potentials at the points A, B and C respectively. The work done in carrying a charge q once around a circle or radius r with a charge Q placed at the centre will be (a)

Qq(4

0r

2

)

(c) Zero 8.

9.

10.

(b)

Qq /(4

(d)

Qq 2 /(4

12.

0r) 0r)

FAB

(c)

3 FAB

3FBA

(b)

FAB

FBA

FBA

(d)

FAB

4FBA

An aluminium (Al) rod with area of cross-section 4 × 10–6 m2 has a current of 5 ampere, flowing through it. Find the drift velocity of electron in the rod. Density of Al = 2.7 × 103 kg/m3 and Atomic wt. = 27. Assume that each Al atom provides one electron (a) 8.6 × 10–4 m/s (b) 1.29 × 10–4 m/s (c) 2.8 × 10–2 m/s (d) 3.8 × 10–3 m/s Taking the internal resistance of the battery as negligible, the steady state current in the 2 resistor shown in the figure will be 2

A 0.2 µF

6

13.

6

A6

B

11.

Y

(a) 18 (b) 6 (c) 3 (d) 2 The figure shows three situations when an electron with velocity v travels through a uniform magnetic field B . In each case, what is the direction of magnetic force on the electron? y y y v

B

x v

B z

14.

x

B

x

v

z z 1 2 3 (a) + ve z-axis, – ve x-axis, + ve y-axis (b) – ve z-axis, – ve x-axis and zero (c) + ve z-axis, + ve y-axis and zero (d) – ve z-axis, + ve x-axis and zero The magnetic field at P on the axis of a solenoid having 100 turns/metre and carrying a current of 5 A is:

6 2.8 6V (a) 1.8 A (b) 2.9 A (c) 0.9 A (d) 2.8 A Six resistors, each of value 3 area connected as shown in the fig.

E

B 3 A cell of emf 3V is connected across AB. The effective resistance across AB and the current through the arm AB will be (a) 0.6 , 1 A (b) 1.5 , 2 A (c) 0.6 , 2 A (d) 1.5 , 1 A In a given network, each resistance has value of 6 . The point X is connected to point A by a copper wire of negligible resistance and point Y is connected to point B by the same wire. The effective resistance between X and Y will be

X

B

3

3

F

A

An object A has a charge of +2 µC and the object B has charge of +6 µC. Which statement is true? (a)

3

3

C

3

D

45°

45° P

EBD_7443 MT-58

Target VITEEE

(a) 250 15.

(b) 500 2

0

0

20.

(c) 500 0 (d) 250 2 0 In an A.C. circuit resistance R, inductance L and capacitance C are in parallel. The value of R is 50 , inductive reactance X L = 40 and capacitive reactance XC = 25 . Then the reading of ammeter will be: C L

In a series LCR circuit, if the applied voltage V and the current in the circuit I at any instant t are given as : V= V0 sin t and I = I0 sin ( t – ) then which of the following holds good : 1 1 (a) L (b) L C C 1 (d) none C A particle tied to a string describes a vertical circular motion of radius r continually. If it has a

(c)

21.

L

velocity A

R 200V

16.

17

18.

19.

(a) 10.2 A (b) 5 A (c) 17 A (d) 7 A The current i passed in any instrument in an A.C. circuit is i = 2 sin t amp and potential difference applied is given by V = 5 cos t volt. Power loss in the instrument is: (a) 10 watt (b) 5 watt (c) zero watt (d) 20 watt In A.C. circuit in which inductance and capacitance are joined in series. Current is found to be maximum when the value of inductance is 0.5 H and the value of capacitance is 8 F. The angular frequency of applied alternating voltage will be: (a) 400 Hz (b) 5000 Hz (c) 2 × 105 Hz (d) 500 Hz In a series LCR circuit the frequency of a 10 V a.c. voltage source is adjusted in such a fashion that the reactance of the inductor measures 15 and that of the capacitor 11 . If R = 3 , the potential difference across the series combination of L and C will be: (a) 8 V (b) 10 V (c) 22 V (d) 52 V What direct current will produce the same amount of thermal energy in a resistance R = 2 as an alternating current that a peak value of 4.24 A and frequency 50 Hz? (a) 3 A (b) 2 A (c) 5 A (d) 4 A

22.

23.

3 gr at the highest point, then the

ratio of the respective tensions in the string holding it at the highest and lowest points is (a) 4 : 3 (b) 5 : 4 (c) 1 : 4 (d) 3 : 2 A generator at a utility company produces 100 A of current at 4000 V. The voltage is stepped up to 2,40,000V by a transformer before it is sent on a high voltage transmission line. The current in transmission line is (a) 3.67 A (b) 2.67 A (c) 1.67 A (d) 2.40 A A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is (a) 0 (b) (c)

24.

(d) 2 2 The electric field of a plane electromagnetic wave in free space travelling along x-direction is given by : Ez = 60 sin (kx + t) V/m. The magnetic field component will be given as : (a)

Bz

60 sin (kz c

t)T

(b)

Bz

60 sin (kx c

t)T

(c)

By

60 sin (ky c

t)T

60 t)T sin (kx c Here c represents the speed of light in free space.

(d)

By

MT-59

Mock Test 6 25.

2

The binding energy of deuteron ( 1 H ) is 1.15

32.

4

MeV per nucleon and an alpha particle ( 2He )

For the given case which figure is correctly showing the after inelastic collision situation? m1 m2 u 1

has a binding energy of 7.1 MeV per nucleon. Then in the reaction 2 1H

26.

27.

28.

29.

2 H 1

2 2 He

v1

Q

the energy released Q is : (a) 5.95 MeV (b) 26.1 MeV (c) 23.8 MeV (d) 289.4 MeV When a slow neutron is captured by U235 nucleus. A fission results, which release 200 MeV energy. If the output power of a nuclear reactor is 1.6 Mwatt. The rate of nuclei undergoing fission is: (a) 4 × 1016 (b) 5 × 1016 (c) 3 × 1016 (d) none of these Light of frequency 5 v0 falls on a photosensitive surface of the threshold frequency v 0 . The velocity of emitted electrons is 4 × 105 m/s. If now light of frequency 9 v0 is incident on the surface, then velocity of electron will become: (a) 5.64 × 105 m/s (b) 7.2 × 105 m/s (c) 8 × 105 m/s (d) 2 × 105 m/s What is the maximum wavelength of photon that would excite an electron in the valence band of diamond to the conduction band ? The energy gap for diamond is 5.5 eV : (a) 169 nm (b) 205 nm (c) 226 nm (d) 350 nm If the electron in hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation is given by: (a)

36 5R

(b)

(a)

(b) (c)

33.

34.

35.

5R 36

R 5 (d) 6 R For Balmer series that lies in the visible region, the shortest wavelength corresponds to quantum number: (a) n = 1 (b) n = 2 (c) n = 3 (d) n = A p type semiconductor has acceptor levels 57 meV above the valance band. The maximum wavelength of light required to create hole is (Planck constant h = 6.6 × 10–34Js) (a) 11.61 × 10–33 Å (b) 217105Å (c) 57 × 10–3 Å (d) 57 Å

31.

36.

v2 m2

m1 m1

m2 v m1

m2

–v1

+v1

m1 m2 –v Intrinsic semiconductor is electrically neutral, extrinsic semiconductor having large number of current carriers would be: (a) electrically neutral (b) positively charged or negatively charged depending upon the type of impurity that has been added (c) positively charged (d) negatively charged A transistor has an = 0.95. It has change in emitter current of 100 mA then the change in collector current is: (a) 95 mA (b) 99.05 mA (c) 100.96 mA (d) 100 mA In a p-n junction diode acting as a half-wave rectifier, which of the following statements is not true? (a) The average output voltage over a cycle is non-zero (b) The drift current depends on biasing (c) The depletion zone decreases in forward biasing (d) The diffusion current increases due to forward biasing A transistor with = 0.98 is operated as an amplifier in a common base circuit with a load resistance of 5 k . If the dynamic resistance of the emitter base junction is 70, the voltage gain and power gain of the circuit will be: (a) 70, 68.6 (b) 72, 70.6 (c) 74, 72.6 (d) none of these (d)

(c)

30.

Before collision u 2= 0

EBD_7443 MT-60

37.

A transmitter radiates 10 kW of power with the carrier unmodulated and 11.8 kW with the carrier sinusoidally modulated. The modulation factor is (a) 56% (b) 60 % (c) 72 % (d) 84% The graph given is a stress-strain curve for

Stress (N/ m2)

38.

Target VITEEE

1.0

43. 0.5

44.

0

39.

40.

0.5

1.0 Strain

(a) elastic objects (b) plastics (c) elastomers (d) None of these In Bernoulli’s theorem which of the following is conserved? (a) Mass (b) Linear momentum (c) Energy (d) Angular momentum A bubble of n mole of helium is submerged at a certain depth in water. The temperature of water increases by t°C. How much heat is added approximately to helium during expansion? nc p (a) ncv t (b) t (c)

n 2c v t

(d) ncp t

(c) There are three unpaired electrons in the central atom which is due to transfer of odd

45.

to a

2

strong field in [Ni(CN)4 ] , which orbitals get lowered in energy ? Assume tetragonal distortion along the Z-axis. (a)

d

(c)

d xy , d

x 2 y2 z2

(b)

d

(d)

d xz , d yz , d

z2 z2

The complex [ Fe ( H 2 O) 5 NO ] 2 ion has a magnetic moment of 3.87 B.M. This is indicative of the fact that (a) Fe in this complex exists in + 1 oxidation

attained by sp 3d 2 hybridization.

When d8 arrangement in nickel ion changes from a weak octahedral field in [Ni (H2O)6 ]2

Pick out the correct statement of the following (a) The stability of either of HgCl2 and SnCl2 is not affected when present simultaneously in aqueous solution (b) Both Cu(OH)2 and Fe(OH)2 are soluble in aqueous NH3 (c) Copper (I) salts are not known in aqueous solution (d) White precipitate of Zn(OH)2 is obtained on adding excess of NaOH to aqueous ZnSO4. The E.A.N. of iron in [Fe(CN)6]3– is : (a) 32 (b) 35 (c) 38 (d) 41

state and nitrosyl as NO ( nitrosonium ion) (b) The complex is octahedral in geometry as

46.

PART - II : CHEMISTRY 41.

42.

47.

electron of NO to Fe 2 (d) The complex is an octahedral low spin complex with Fe in + 2 oxidation state. Ethylenediaminetetraacetic acid (EDTA) is the antidote for lead poisoning. It is administered in the form of (a) free acid (b) sodium dihydrogen salt (c) Calcium dihydrogen salt (d) none of these Which of the following coordination compounds would exhibit optical isomerism? (a) pentamminenitrocobalt(III) iodide (b) diamminedichloroplatinum(II) (c) trans-dicyanobis (ethylenediamine) chromium (III) chloride (d) tris-(ethylendiamine) cobalt (III) bromide When electrons are trapped into the crystal in anion vacancy, the defect is known as : (a) Schottky defect (b) Frenkel defect (c) Stoichiometric defect (d) F-centres

MT-61

Mock Test 6 48.

49.

50.

51.

52.

53.

Copper metal has a face-centred cubic structure with the unit cell length equal to 0.361 nm. Picturing copper ions in contact along the face diagonal. The apparent radius of a copper ion is(a) 0.128 (b) 1.42 (c) 3.22 (d) 4.22 For a reaction to be spontaneous at all temperatures (a) G and H should be negative (b) G and H should be positive (c) G= S= 0 (d) H < G and The standard entropies of CO2 (g), C(s) and O2(g) are 213.5, 5.740 and 205 JK–1 respectively. The standard entropy of formation of CO2 is: (a) 2.76 JK–1 (b) 2.12 JK–1 –1 (c) 1.12 JK (d) 1.40 JK–1 In a closed insulated container, a liquid is stirred with a paddle to increase the temperature, which of the following is true? E W 0, q 0 (a) (b)

E

W

q

(c)

E 0, W q

54.

H

55.

56.

0

0

(d) W 0, E q 0 Le-Chatelier’s principle is applicable only to a : (a) system in equilibrium (b) irreversible reaction (c) homogeneous reaction (d) heterogeneous reaction The plot shown below represents the variation of molar conductivity of NaOH and CH3OH

57.

NaOH m

CH3COOH

58.

c 0 m

for CH3COOH is obtained from the :

1000 K c (b) Intersection of NaOH and CH3 COOH curves

(a) Relation

m

(c) Intercept of CH3COOH plot (d) Kohlrausch’s law For the reaction,

59.

CH3COCH3 + I2 products The rate is governed by expression dx k[acetone][H ] dt The order w.r.t. I2 is: (a) 1 (b) 0 (c) 3 (d) 2 1.08 g of pure silver was converted into silver nitrate and its solution was taken in a beaker. It was electrolysed using platinum cathode and silver anode. 0.01 Faraday of electricity was passed using 0.15 volt above the decomposition potential of silver. The silver content of the beaker after the above shall be (a) 0 g (b) 0.108 g (c) 0.108 g (d) 1.08 g In a fuel cell, CH3OH is used as fuel and oxygen gas is used as an oxidiser. The reaction is 3 CH3OH( ) O 2 (g) CO2 (g) H 2O( ) 2 If standard enthalpy of combustion for methanol is – 726 kJ/mol, what will be the cell efficiency. Given that G f values (kJ/mol) : CH3OH(l) = – 166.3 CO2(g) = – 394.4 H2O (l) = – 237.1 (a) 70.23% (b) 96.7% (c) 80% (d) none of these Specific conductance of a N/10 KCl solution at 23°C is 0.0112 ohm–1 cm–1.The resistance of cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be : (a) 0.316 cm–1 (b) 0.616 cm–1 (c) 6.16 cm–1 (d) 616 cm–1 1-Phenylethanol can be prepared by the reaction of benzaldehyde with (a) ethyl iodide and magnesium (b) methyl iodide and magnesium (c) methyl bromide and aluminium bromide (d) methyl bromide When phenol is treated with excess bromine water. It gives (a) m-Bromophenol (b) o-and p-Bromophenols (c) 2,4-Dibromophenol (d) 2,4, 6-Tribomophenol.

EBD_7443 MT-62

60. 61.

Target VITEEE

Ether in air forms : (a) Oxide (b) Peroxide (c) Alkenes (d) Alkanes Which one of the following will most readily be dehydrated in acidic conditions ? O

(a)

62. 63.

64.

65.

66.

67. 68.

O

O

( i) Mg, Et 2O

PCC

( ii ) H 2C O (iii ) H 3O

CH 2Cl 2

CHO

70.

71.

COOH

(b)

OH (d) CHO

CH 3 OH With Tollen’s reagent a silver mirror will be formed by: (a) acetic acid (b) acetaldehyde (c) acetyl chloride (d) acetone Identify the respective final products in the following two reactions. (c)

(i)

CH3CH2COOH

Br2 / PCl 3

[A]

Br2

[B]

H2O

(ii) CH3CH2COOH

Cl 2 / PBr3

[D]

Cl 2

[C] [E]

H2O

(a) (b) (c) (d)

[F] CH3CHOHCOOH and CH3CHOHCOOH CH3CHOHCOCl and CH3CHOHCOBr CH3CHBrCOOH and CH3CHClCOOH CH3CHClCOOH and CH3CHBrCOOH.

O Product .

72. Here the product is O

(a)

O

(b) O

O

(b) OC2H2

(c)

CH 3OH

(a)

(d)

HBr

NaBH 4

O

(b)

OH OH Prestone is a mixture of: (a) Glycol + H2O (b) Glycerol + H2O (c) Acetone + H2O (d) propanal + H2O Benzyl ethyl ether can be prepared by which combination of components ? (a) C6H5CH2Br + C2H5ONa (b) C6H5CH2ONa + C2H5Br (c) both (a) and (b) (d) C6H5CH2OH + C2H5OH Which is not true about acetophenone ? (a) It reacts with 2,4-dinitrophenylhydrazine to form 2, 4-dinitrophenylhydrazone (b) It reacts with Tollen’s reagent to form silver mirror (c) It reacts with I2/NaOH to form iodoform (d) On oxidation with alkaline KMnO4 followed by hydrolysis it gives benzoic acid Which of the following does not turn Schiff’s reagent to pink ? (a) Formaldehyde (b) Benzaldehyde (c) Acetone (d) Acetaldehyde When acetic acid is heated with P 2 O5 the compound formed will be : (a) (CH3CO)2O (b) CH3COCH3 (c) CH3CHO (d) CH4 Acetone on distillation with conc. H2SO4 forms : (a) phorone (b) acrolein (c) mesitylene (d) mesityl oxide The major product in the reaction of -bromo cyclohexanone with potassium ethoxide is

(a)

What is product of the following sequence of reactions

OH

OH

O

(c)

69.

(d)

CO2C2H5

COOH

(c)

O

COOH

(d)

COOH

MT-63

Mock Test 6 73.

What is the main product when

77.

COOH is heated ? COOH

HOOC

Identify (C) and (D) in the following series of reactions CH3 NH 2

excess of

[A]

CH 3I

AgOH

heat

COOH COOH

(a)

[B] [C] [D]

(a) (CH3)3COH, CH3NH2 (b) (CH3)2C = CH2, CH3NH2

O

(b)

(c) (CH3)3N, CH3OH (d) (CH3)2C=CH2, CH3OH

(c)

HOOC

O

78.

Consider the following sequence of reactions : [A]

(d) 74.

HNO 2

[B]

the compound [A] is

C C O

(a) CH3CN

(b) CH3NC

(c) CH3CH2CN

(d) CH3NO2

Acetyl chloride does not react with

79.

(b) Sodium acetate

The bad smelling substance formed by the action of alcoholic caustic potash on chloroform and aniline is (a) nitrobenzene

(c) 2-methylpropene

(b) phenyl isocyanide

(d) It reacts with all the three

(c) phenyl cyanide

Formic acid is obtained when (a) Calcium acetate is heated with conc. H2SO4 (b) Calcium formate is heated with calcium acetate

(d) phenyl isocyanate 80.

When aniline is treated with benzene diazonium chloride at low temperature in weakly acidic medium, the final product obtained is

(c) Glycerol is heated with oxalic acid at 373 K (d) Acetaldehyde is oxidised with K2Cr2O7 and H2SO4. 76.

CH 3 CH 2 OH

O

(a) Water

75.

Reduction

NH2

(a)

N

(b)

N

N

Which statement is true regarding the following structure?

N

CH3

C3H7 C2H5

(c)

N

N

N

NH2

NH

(a) It is a chiral molecule (b) It exists in two resolvable optically active forms (c) Both (a) and (b) (d) Neither a) nor b)

NH2 (d)

N

N

EBD_7443 MT-64

Target VITEEE PART - III : MATHEMATICS

81. For the set A = {1, 2, 3}, define a relation R in the set A as follows R = {(1, 1), (2, 2), (3, 3), (1, 3)} Then, the ordered pair to be added to R to make it the smallest equivalence relation is (a) (1, 3) (b) (3, 1) (c) (2, 1) (d) (1, 2) 82.

86. The region represented by the inequalities x 6, y 2, 2x + y 10, x 0, y 0 is (a) unbounded (b) a polygon (c) exterior of a triangle (d) None of these 87.

The matrices

P=

u1

v1

u2

v2

u3

v3

2

w1

1 w2 ; Q = 9 w3

2

1

13

–5 m

–8

1

88.

5

u1

v1

w1

u2

v2

w2

x y

u3

v3

w3

z

1 =

(c) 84.

85.

b2 q

2

a2 p2

ac pr

(b)

bc qr

Let function

c2 r

2

89.

ab pq

(d) None of these f :R

90.

R be defined by

f (x) = 2x + sin x for x R , then f is (a) one-one and onto (b) one-one but NOT onto (c) onto but NOT one-one (d) neither one-one nor onto Eighteen guests are to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side of the table. The number of ways in which the seating arrangement can be done equals (a) 11C4 (9!)2 (b) 11C6 (9!)2 6 5 (c) P0 × P0 (d) None of these

(c) 2 log e 3 (d) None A variable plane moves so that the sum of reciprocals of its intercepts on the three coordinate axes is constant . It passes through a fixed point, which has coordinates

(c)

Let 1 , 2 and 1 , 2 be the roots of ax2 + bx + c = 0 and px2 + qx + r = 0 respectively. If the system of equations 1y + 2z = 0 and 1y + 2z = 0 has a non-trivial solution, then (a)

(b) 2 log e 2

(a) ( , , )

1 , the 5

value of y comes out to be –3. Then the value of m is equal to (a) 27 (b) 7 (c) – 27 (d) – 7

83.

2 2 2 + + + ............= 1.2.3 3.4.5 5.6.7

(a) 2 log e 4

are such that PQ = I, an identity matrix. Solving the

equation

1+

91.

,

1 1 1 , ,

(b)

,

1

(d)

,

1

,

1

The perpendicular distance of A(1, 4, –2) from BC, wher e coordinates of B and C ar e respectively (2, 1, –2) and (0, –5, 1) is (a)

3 7

(b)

26 7

(c)

3 26 7

(d)

26

2 tan 1 is 3 3 5 6 3 4 (a) (b) (c) (d) 17 17 17 17 If cos–1x + cos–1y + cos–1z = 3 then xy + yz + zx is equal to : (a) 1 (b) 0 (c) –3 (d) 3

The value of cot cosec

92. Let f : R f

x

y 3

15

R be a function such that

f ( x) f ( y) , f (0) 3

0 and

f ' (0) 3 . Then (a) f(x) is a quadratic function (b) f(x) is continuous but not differentiable (c) f(x) is differentiable in R (d) f(x) is bounded in R

MT-65

Mock Test 6 93.

94.

95.

3 i sin

If x + iy =

then 4x – x2 – y2

cos 2 reduces to (a) 2 (b) 3 (c) 4 (d) 5 The locus of the centre of a circle which touches the circle | z – z1 | = a and | z – z2 | = b externally (z, z1 & z2 are complex numbers) will be (a) an ellipse (b) a hyperbola (c) a circle (d) none of these

In a parallelogram ABCD, | AB | a,| AD | b and | AC | c , the value of DB . AB is (a)

a

2

b2 c 2 2 2

b 3c 2 Solve for x : (c)

96.

3a 2

a 2 3b 2 c 2 2

(b)

2

a

(d)

{ x cos(cot–1x) + sin(cot–1 x)}2 = (a) 97.

98. 99.

1

(b)

2

2

3b 2

2

c

2

51 , 50

1 5 2

(c) 2 2 (d) 5 2 The number of solution of the equation 1 | cot x | cot x (0 x 2 ) is : sin x (b) 1 (a) 0 (c) 2 (d) 3 If a group contains only four elements, then it is (a) non-abelian (b) abelian (c) cyclic (d) none of these If S1, S2, S3, ...., Sn are the sum of infinite geometric series whose first terms are 1, 2, 3, ..., n and whose 1 1 1 1 , , ,..., 2 3 4 (n 1) respectively, then the value of

comon

ratios

are

S12 S 22 S32 ... S22 n

1

is equal to

(a)

1 [ n ( 2 n 1)( 4 n 1) 3] 3

(b)

1 [n (2n 1)(4n 1) 3] 3

1 [ n ( 2 n 1)( 4n 1) 3] 3 (d) None of these

(c)

100. The point (2a, a) lies inside the region bounded by the parabola x2 = 4y and its latus rectum. Then, (a) 0 a 1 (b) 0 < a < 1 (c) a > 1 (d) a < 0 101. If the normal at the end of a latus rectum of an ellipse passes through one extermity of the minor axis, then (a)

e4

e2 1 0

(b)

e4

e2 1 0

(c) e 4 e 2 1 0 (d) None of these 102. The asymptotes of a hyperbola, having centre at the point (1, –1), are parallel to the lines x + 2y – 3 = 0 and 2x + y + 5 = 0. If the hyperbola passes through the point (2, 1), then its equation is (a)

2x 2

2 y 2 5xy x y 21 0

(b)

2x 2

2 y 2 5xy x y 21 0

(c) 2x 2 2 y 2 5xy x y 21 0 (d) None of these 103. If a , b , c are vectors such that [ a b c ] = 4, then [ a b b c c a ] (a) 16 (b) 64 (c) 4 (d) 8 104. If f (x) is differentiable and strictly increasing f (x 2 ) f ( x) is x 0 f ( x ) f ( 0) (b) 0 (a) 1 (c) –1 (d) 2 105. The line y = mx bisects the area enclosed by lines x =0, y = 0 and x = 3/2 and the curve y = 1 + 4x – x2. Then the value of m is

function, then the value of lim

(a)

13 6

(b)

13 2

13 13 (d) 5 7 106. If angle is divided into two parts such that the tangent of one part is K times the tangent to other and is their difference, then sin is equal to

(c)

(a)

K 1 sin 2 K –1

(b)

K 1 sin K –1 2

(c)

K 1 sin K –1

(d)

K –1 sin K 1

EBD_7443 MT-66

Target VITEEE

107. The expansion of cos x by Maclaurin’s series is 2

(a)

cos x 1

4

x 2!

x 4!

( 1)n / 2 .

(b)

(c)

6

x 6!

x6 6!

1

x3 3!

x cos x 1 1!

.....

[Use Bernoulli's formula]

xn ..... where n is even n!

x 2 x4 2! 4! where n is even cos x

xn ..... n!

.....

x5 5!

.....

( 1)2n

1

x 2n 1 ..... (2n 1)!

x x 2 x3 xn ..... (d) cos x 1 n! 1! 2! 3! 108. A particle is moving along a straight line path according to the relation s2 = at2 + 2bt + c s represents the distance travelled in t seconds and a, b, c are constants. Then the acceleration of the particle varies as (a) s – 3 (b) s 3/2 –2/3 (c) s (d) s 2 2

x

109. With the usual notation

112. Evaluate : x 3e 2x dx

2

x

2

dx is

1

(a)

1 2x e x 2

(b)

e 2x x

(c)

1 2x 3 e x 2

(d)

e 2x x 4

4 3

3 2 x 4

3 2 x 2 3 2 x 2

3

2 –

(b) 4 –

2 +

3

(c) 4 – 2 – 3 (d) None of these 110. Let and be the distinct roots of ax2 + bx = c = 0, then lim

1 cos(ax 2

(a)

)2

(x

x

a2 ( 2

bx c)

)2

is equal to

113. The solution of differential equation dy = cos (x + y) is – dx

x y 2

(a)

tan

(b)

cosec

(c)

cot

x C

x y 2

x y 2

x C

x C

(d) None of these 114. The solution of x 3

1 a2 ( )2 (c) (d) ( )2 2 2 111. If f (x) = (x + 1)cot x be continuous at x = 0 then f (0) is equal to: (a) 0 (b) – e (c) e (d) None

dy dx

4 x 2 tan y

e x sec y

satisfying y (1) = 0 is : (a)

(x 2) e x log x

tan y

e x ( x 1) x

4

tan y ( x 1)e x x

3

x (d) sin y e (x 1) x

3

(b) sin y (c)

(b) 0

3 4

2 2 x 3

x3

equal to(a) 4 +

3x 2

115. G is a set of all rational numbers except –1 and * is defined by a* b = a + b + ab for all a, b G . In the group (G, *) 2 1 * x *3 (a) 71 (c) 63/5

1

the solution of

5 is

(b) 68 (d) 72/5

MT-67

Mock Test 6 116. Which of the following is a contradiction? (a)

(p q) ~ (p q)

(b) p ( p q) (c) (p q) p (d) None of these 117. If P = n(n 2 – 12) (n2 – 22) (n2 – 32) ...... (n2 – r2), n > r, n N then P is necessarily divisible by (a) (2r + 2) ! (b) (2r + 4) ! (c) (2r + 1) ! (d) None of these 118. For a biased dice, the probability for the different faces to turn up are Face

1

2

3

4

5

6

P

0.10

0.32

0.21

0.15

0.05

0.17

The dice is tossed and it is told that either the face 1 or face 2 has shown up, then the probability that it is face 1, is 1 16 (b) (c) 10 21 119. The domain of the function

(a)

3

+ log10 (x3 – x), is 4 – x2 (– 1, 0) (1, 2) (1, 2) (2, ) (– 1, 0) (1, 2) (2, ) (1, 2)

f(x) = (a) (b) (c) (d)

5 5 (d) 16 21

120. If E and F are the complementary events of events E and F respectively and if 0 < P(F) < 1, then (a) P(E \ F) + P( E \ F) = 1 (b) P(E \ F) + P(E \ F ) = 1 (c) P( E \ F) + P(E \ F ) = 1 1 2 PART - IV : ENGLISH

(d) P(E \ F ) + P( E \ F ) =

Direction (Qs. 121 -123) : Read the passage carefully and answer the questions given below Erosion of America's farmland by wind and water has been a problem since settlers first put the prairies and grasslands under the plow in the nineteenth century. By the 1930s, more than 282 million acres of farmland were damaged by erosion. After 40 years of

conservation efforts, soil erosion has accelerated due to new demands placed on the land by heavy crop production. In the years ahead, soil erosion and the pollution problems it causes are likely to replace petroleum scarcity as the nation's most critical natural resource problem. 121. As we understand from the reading, today, soil erosion in America(a) causes humans to place new demands on the land (b) is worse than it was in the nineteenth century (c) happens so slowly that it is hardly noticed (d) is the most critical problem that the nation faces 122. The author points out in the passage that erosion in America(a) has damaged 282 million acres ever since settlers first put the prairies and grasslands under the plow (b) has been so severe that it has forced people to abandon their settlements (c) occurs only in areas with no vegetation (d) can become a more serious problem in the future 123. It is pointed out in the reading that in America(b) petroleum is causing heavy soil erosion and pollution problems (b) heavy crop production is necessary to meet the demands and to prevent a disaster (c) soil erosion has been hastened due to the overuse of farming lands (d) water is undoubtedly the largest cause of erosion 124. Realists in India can't object to a good neighborly relationship between Nepal and China _________in Beijing know the dangers of moving too far and too fast and __________an Indian reaction. (a) Pragmatists, provoking (b) escapist, delighting (c) evader, make happy (d) idealist, rude 125. Choose the best pronunciation of the word, Bowl, from the following options. (a) bohl (b) ba-owl (c) bo-ul (d) ba-oul

EBD_7443

MOCK

VITEEE Mock Test Paper

7

Max. Marks : 125

Time : 2½ hrs

The force ‘F’ acting on a particle of mass ‘m’ is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8 s is : 6.

F(N)

6 3 0 –3

4

2

6

8

t(s)

2.

3.

4.

(a) 24 Ns (b) 20 Ns (c) 12 Ns (d) 6 Ns An air filled parallel plate condenser has a capacity of 2 pF. The separation of the plates is doubled and the interspace between the plates is filled with wax. If the capacity is increased to 6pF, the dielectric constant of wax is : (a) 2 (b) 3 (c) 4 (d) 6 Four metal conductors having different shapes : 1. a sphere 2. cylinder 3. pear 4. lightning conductor are mounted on insulating stands and charged. The one which is best suited to retain the charges for a longer time is : (a) 1 (b) 2 (c) 3 (d) 4 After terminal velocity is reached, the acceleration of a body falling through a fluid is (a) equal to g (b) zero (c) less than g (d) greater than g

7.

An electric dipole consists of two opposite charges each 0.05 C separated by 30 mm. The dipole is placed in an uniform external electric field of 106 NC–l. The maximum torque exerted by the field on the dipole is (a) 6 × l0–3 N-m (b) 3 × 10–3 N- m (c) 15 × l0–3 N- m (d) 1.5 × l0–3 N- m A particle A has a charge q and particle B has charge + 4q with each of them having the mass m. When they are allowed to fall from rest through same potential difference, the ratio of their speeds VA : VB will be : (a) 4 : 1 (b) 1 : 4 (c) 1 : 2 (d) 2 : 1 The inward and outward electric flux from a closed surface are respectively 8 × 103 and 4 × 103 units. Then the net charge inside the closed surface is : (a) – 4 × 103 coulomb (b) 4 × 103 coulomb 4 103 (c) coulomb 0

8.

(d) – 4 × 103 0 coulomb The adjacent graph shows the extension ( l) of a wire of length 1m suspended from the top of a roof at one end with a load W connected to the other end. if the corss-sectional area of the wire is 10–6m2, calculate the Young’s modulus of the material of the wire

–4

1.

5.

(l ×10 )m

PART - I : PHYSICS

4 3 2 1 20 40 60 80 W(N)

MT-69

Mock Test 7 (a) 2 × 1011 N/m2 (b) 2 × 10–11 N/m2 –12 2 (c) 2 × 10 N/m (d) 2 × 10–13 N/m2 9. In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2 , the balancing length becomes 120 cm. The internal resistance of the cell is (a) 0.5 (b) 1 (c) 2 (d) 4 10. A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of a wire made of 'A'. then for the two wires to have the same resistance, the ratio lB/lA of their respective lengths must be (a) 1

1 (d) 2 4 A current of 6 A enters one corner P of an equilateral triangle PQR having 3 wires of resistances 2 each and leaves by the comer R. Then the current I1 and I2 are 6A

P I2

I1 2 Q

2 2

15.

1 2

(b)

(c)

11.

14.

R

(a) 2 A, 4 A (b) 4A, 2A (c) 1 A, 2A (d) 2A, 3A 12. The electron drift speed is small and the charge of the electron is also small but still, we obtain large current in a conductor. This is due to the (a) conducting property of the conductor (b) resistance of the conductor is small (c) electron number density of the conductor is small (d) electron number density of the conductor is enormous 13. A wire in the form of a circular loop of one turn carrying a current produces a magnetic field B at the centre. If the same wire is looped into a coil of

16.

17.

two turns and carries the same current, the new value of magnetic induction at the centre is : (a) 3 B (b) 5 B (c) 4 B (d) 2 B How many calories of heat will be produced approximately in a 210 watt electric bulb in 5 minutes? (a) 80000 cal (b) 63000 cal (c) 1050 cal (d) 15000 cal As the temperature of a liquid is raised, the coefficient of viscosity (a) decreases (b) increases (c) remains the same (d) may increase or decrease depending on the nature of liquid On a linear temperature scale Y, water freezes at – 160° Y and boils at – 50° Y. On this Y scale, a temperature of 340 K would be read as : (water freezes at 273 K and boils at 373 K) (a) – 73.7° Y (b) – 233.7° Y (c) – 86.3° Y (d) – 106.3° Y The upper half of an inclined plane of inclination is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by (a)

=

2 tan

(b)

= 2 tan

(c)

= tan

(d)

=

1 tan

18. The tank circuit used in a radio transmitter should have (a) high effective Q (b) low effective Q (c) loosely coupled load (d) Both (a) and (c) 19. A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be

EBD_7443 MT-70

Target VITEEE

(a) halved (b) the same (c) doubled (d) quadrupled 20. A coil of wire having inductance and resistance has a conducting ring placed coaxially within it. Thecoil is connected to abattery at timet = 0, so that a time-dependent current l1(t) starts flowing through the coil. If I2(t) is the current induced in the ring, and B(t) is the magnetic field at the axis of the coil due to I1(t), then as a function of time (t > 0), the product I2(t) B(t) (a) increases with time (b) decreases with time (c) does not vary with time (d) passes through a maximum 21. In a LCR circuit capacitance is changed from C to 2 C. For the resonant frequency to remain unchanged, the inductance should be changed from L to (a) L/2 (b) 2 L (c) 4 L (d) L/4 22. One conducting U tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure . If each tube moves towards the other at a constant speed v, then the emf induced in the circuit in terms of B, l and v where l is the width of each tube, will be

X X X X

X

X

A v X

X X

X

X

I

× × × × × × × × × × I ×

I

D

(1)

(2) × × × × × × D I × × × × × × × I D × × × × × (3) (4) (a) i1 > i2 > i3 > i4 (b) i1 = > i2 > i3 > i4 (c) i1 > i2 > i3 = i4 (d) i1 = i2 > i3 = i4 25. A conducting circular loop is placed in a uniform magnetic field of 0.02 tesla with its plane perpendicular to the magnetic field. If the radius of the loop starts striking at a constant rate of 1 mm/s. Then the emf induced in the loop at the instant when the radius is 4 cm will be: (a) 8 V (b) 5.0 V (c) 2.0 V (d) 2.5 V 26. Two inductors 0.4 H and 0.6 H are connected in parallel. If this combination is connected in series with an inductor of inductance 0.76 H. The equivalent inductance of the circuit will be: (a) 0.1 H (b) 0.2 H (c) 1 H (d) 2 H 27. Which one of the following is not true ?

× × × × × × ×

× × × × × × ×

(a) Ampere's law is :

B.d

µ0iin

BX X

v X C

(b) Faraday's law is : C

0

X

(a) – Blv (b) Blv (c) 2 Blv (d) zero 23. If the number of turns in a coil becomes doubled then its self inductances will be: (a) unchanged (b) four times (c) halved (d) doubled 24. Four identical circular conducting loops are placed in uniform magnetic fields that are either increasing (I) or decreasing (D) in magnitude at identical rates. Arrange the magnitude of the currents induced (i) in the loops.

× × ×

(c) Biot-Sawart law is dB (d) Gauss's law is

0

d dt

µ0 i r . 3 4 r

E .d A

qin

28. Two slits separated by a distance of 1 mm are illuminated with red light of wavelength 6.5 × 10–7 m. The interference fringes are observed on a screen placed 1 m from the silts. The distance of the third dark fringe from the central fringe will be equal to : (a) 0.65 mm (b) 1.30 mm (c) 1.62 mm (d) 1.95 mm

MT-71

Mock Test 7 29. In an interference pattern produced by the identical coherent sources of monochromatic light, the intensity at the site of central maxima is I. The intensity at the same spot when either of the two slits is closed is I0 then we must have (a) I and I0 are not related (b) I = 4I0 (c) I = I0 (d) I = 2I0 30. Energy E and frequency v of a photon is correctly represented by: E

(a)

E

(b)

E

E

If the average power radiated by the star is 10 16 watt, the deuteron supply of the star is exhausted in the time of the order of the mass of the nuclei follows: M(1H2) = 2.014 amu, M(p) = 1.007 amu, M(n) = 1.008 amu, M(He4) = 4.001 amu (a) 1016 sec (b) 1012 sec (c) 106 sec (d) 108 sec 35. Radiation from a hydrogen discharge tube is incident on the cathode of a photocell. The work function of the cathode surface is 3.2 eV. To reduce the photocurrent to zero, the voltage (in volts) of the anode relative to the cathode must be made : (a) – 3.2 (b) + 3.2 (c) – 10.4 (d) – 13.6 36. The circuit has two oppositively connected ideal diodes in parallel. What is the current flowing in the circuit?

4 (c)

D1

(d)

12V 31. A freshly prepared radioactive source of half life 2 hours emits radiation of intensity of which is 64 times the permissible safe level. The minimum time after which is would be possible to work safely with this source is: (a) 64 hrs (b) 24 hrs (c) 6 hrs (d) 12 hrs 32. An -particle of 10 MeV collides head on with copper nucleus (Z = 29) and is deflected back. What is the maximum limit of radius of copper nucleus? (a) 10.4 × 10–15 m (b) 6.4 × 10–15 m (c) 4.2 × 10–15 m (d) 8.4 × 10–15 m 33. Ultra violet light of wavelength 300 nm and intensity 1.0 W/m2 falls on the surface of a photoelectric material. If one percent of incident photons produce photoelectrons, then the number of photoelectron emitted from an area of 1.0 cm2 of the surface is nearly: (a) 2.13 × 1011/sec (b) 1.52 × 1012/sec 12 (c) 3.02 × 10 /sec (d) None of these 34. A star initially has 1040 deuterons. It produces energy via the process 2 2 3 1H + 1H 1H + p 4 2 3 2He + n 1H + 1H

3

D2 2

(a) 1.71 A (b) 2.00 A (c) 2.31 A (d) 1.33 A 37. If the lattice constant of this semiconductor is decreased, then which of the following is correct? conduction band width

band gap

Ec Eg

valence band width

Ev

(a) All Ec, Eg, Ev increase (b) Ec and Ev increase, but Eg decreases (c) Ec and Ev decrease, but Eg increases (d) All Ec, Eg, Ev decrease 38. The logic circuit shown below has the input waveforms ‘A’ and ‘B’ as shown. Pick out the correct output waveform. A B

Y

EBD_7443 MT-72

Target VITEEE PART - II : CHEMISTRY

Input A

41. Which of the following is diamagnetic ? (a) d 5 arrangement in Fe3 octahedral field.

Input B

Output is

is strong

(b) d7 arrangement in Co 2 is weak octahedral field.

(a) (b) (c)

42.

(d) 39. In the circuit below, A and B represent two inputs and C represents the output. 43.

A C

44.

B The circuit represents (a) NOR gate (b) AND gate (c) NAND gate (d) OR gate 40. A conductor and a semi-conductor are connected in parallel as shown in the figure. At a certain voltage both ammeters registers the same current. If the voltage of the DC source is increased then Conductor A1 Semiconductor A2

(a) the ammeter connected to the semiconductor will register higher current than the ammeter connected to the conductor (b) the ammeter connected to the conductor will register higher current than the ammeter connected to the semiconductor (c) the ammeters connected to both semiconductor and conductor will register the same current (d) the ammeter connected to both semiconductor and conductor will register no change in the current

45.

(c) d 5 arrangement in Mn 2 is weak octahedral field. (d) d10 arrangement of Ni is tetrahedral Ni(CO)4. The covalent character of the following chlorides follow the order (a) HgCl2 < CdCl2 < ZnCl2 (b) ZnCl2 < CdCl2 < HgCl2 (c) CdCl2 < ZnCl2 < HgCl2 (d) HgCl2 < ZnCl2 < CdCl2 Which particle contains 2 neutrons and 1 proton ? (a) 1H2 (b) 2He4 3 (c) 1T (d) 1D2 An aqueous solution of [Ti(H2O)6]3+ ion has a mild violet colour of low intensity. Which of the following statements is incorrect ? (a) The ion absorbs visible light in the region of ~ 5000 Å. (b) The colour results from an electronic transition of one electron from the t2g to an eg orbital ? (c) The low colour-intensity is because of a low probability of transition. (d) The transition is the result of metal-ligand back bonding. The IUPAC name of the red coloured complex [Fe(C 4 H 7 O 2 N 2 ) 2 ] obtained from the reaction

of Fe 2 and dimethyl glyoxime (a) bis (dimethyl oxime) ferrate (II) (b) bis (dimethyl oxime) iron (II) (c) bis (2, 3- butanediol dioximato) iron (II) (d) bis (2, 3- butanedione dioximato) iron (II) 46. [Co(NH3)4 (NO2)2] Cl exhibits (a) linkage isomerism, ionization isomerism and geometrical isomerism (b) ionization isomerism, geometrical isomerism and optical isomerism (c) linkage isomerism, geometrical isomerism and optical isomerism (d) linkage isomerism, ionization isomerism and optical isomerism

MT-73

Mock Test 7 47. The appearance of colour in solid alkali metal halides is generally due to (a) Schottky defect (b) Frenkel defect (c) Interstitial positions (d) F-centres 48. The second order Bragg’s diffraction of X-rays of wavelength 100 pm from a set of parallel lattice planes in a metal occurs at a grazing angle of 30º. The spacing between the successive scattering planes in the cystal is (a) 50 pm (b) 100 pm (c) 150 pm (d) 200 pm 49. The decomposition of limestone CaCO3 (s)

CaO(s) + CO 2 (g)

is non-spontaneous at 298 K. The H and S value for the reaction are 176.0 kJ and 160 JK–1 respectively. At what temperature the decomposition becomes spontaneous? (a) At 1000 K (b) below 500 C (c) Between 500 C and 600 C (d) above 827 C 50. Calculate the entropy change for 3H2(g) + CO(g), using CH4(g) + H2O(g) the following data : Substance CH4(g) H2O(g) H2(g) CO(g) S°/JK–1 186.2 188.7 130.6 197.6 mole–1 The entropy change is: (a) – 46 JK–1 mole–1 (b) + 46 JK–1 mole–1 (c) – 214.5 JK–1 mole–1 (d) + 214.5 JK–1 mole–1 51. According to the third law of thermodynamics which one of the following quantities for a perfectly crystalline solid is zero at absolute zero? (a) Free energy (b) Entropy (c) Enthalpy (d) Internal energy 52. Which reaction is not affected by change in pressure? (a)

H 2 (g) + I 2 (g)

(b)

N 2 (g) + 3H 2 (g)

(c)

2C(s) + O 2 (g)

(d)

SO 2 (g) + 2O 2 (g)

2HI(g)

2NH3 (g) 2CO(g) 2SO3 (g)

53.

54.

55.

56.

57.

58.

59.

The order of a reaction, with respect to one of the reacting component Y, is zero. It implies that: (a) the reaction is going on at a constant rate (b) the rate of reaction does not vary with temperature (c) the reaction rate is independent of the concentration of Y (d) the rate of formation of the activated complex is zero The specific rate constant of a first order depends on the : (a) concentration of the product (b) concentration of the reactant (c) temperature (d) all of these The resistance of 1 N solution of acetic acid is 250 ohm, when measured in a cell of cell constant 1.15 cm –1 . The equivalent conductance (in ohm–1 cm2 equiv–1) of 1 N acetic acid will be (a) 4.6 (b) 9.2 (c) 18.4 (d) 0.023 Zn amalgam is prepared by electrolysis of aqueous ZnCl2 using Hg cathode (9gm). How much current is to be passed through ZnCl2 solution for 1000 seconds to prepare a Zn Amalgam with 25% Zn by wt. (Zn = 65.4) (a) 5.6 amp (b) 7.2 amp (c) 8.85 amp (d) 11.2 amp Rusting of iron is prevented by zinc coating over its surface. This process is called : (a) Electrolysis (b) Galvanization (c) Photoelectrolysis (d) Cathodic protection Ethanol and dimethyl ether form a pair of functional isomers. The boiling point of ethanol is higher than that of dimethyl ether, due to the presence of (a) H-bonding in ethanol (b) H-bonding in dimethyl ether (c) CH3 group in ethanol (d) CH3 group in dimethyl ether Reaction of CH — CH with RMgX leads to 2

O

formation of (a) RCHOHR

2

(b) RCHOHCH3 R

(c) RCH2CH2OH

(d)

R

CHCH2OH

EBD_7443 MT-74

60.

61.

Target VITEEE

Acetylene, on reacting with formaldehyde under high pressure, gives (a) propanol (b) propyl alcohol (c) allyl aldehyde (d) butyndiol. In the following reaction : C2H5OH

62.

63.

64.

Conc. H 2SO 4 443 K

Z.

Identify Z : (a) CH2 = CH2 (b) CH3 – CH2 – O – CH2 | CH3 (c) CH3 – CH2 – C | SO4 (d) (CH3CH2)2 – SO4 The most suitable reagent for the conversion of RCH2OH RCHO is : (a) KMnO4 (b) K2Cr2O7 (c) CrO3 (d) PCC (pyridine chloro chromate) Arrange the following in decreasing order of solubility in water

O

O

O

I

II

III

(a) I > III > II (b) III > II > I (c) II > III > I (d) All are equally soluble Which of the following reaction can produce R – CO – Ar ? (a)

ArCOCl H Ar

(b)

RCOCl ArMgX

(c)

ArCOCl RMgX

67. Which of the following is formed, when calcium acetate is heated ? (a) CH3CHO (b) CH3COCH3 (c) CH3CH2OH (d) CH3CHO 68. The reaction of an organic compound with ammonia followed by nitration of the product gives a powerful explosive, called RDX. The organic compound is: (a) phenol (b) toluene (c) glycerine (d) formaldehyde 69. Which carbonyl compound will be purified with sodium bisulphite(a) 2-Pentanone (b) 3-Hexanone (c) 4-Heptanone (d) All of these 70. Which of the following will not give HVZ reaction? (a) Trichloroethanoic acid (b) 2-methyl butanoic acid (c) Ethanoic acid (d) Propanoic acid 71. When propionic acid is treated with aqueous NaHCO3, CO2 is liberated. The C of CO2 comes from (a) NaHCO3 (b) – CH3 group (c) – COOH group (d) – CH2 group 72. Both propanoic acid, CH3CH2COOH as well as 2-chloropropanoic acid give precipitate with aq. AgNO3. How you distinguish the precepitates in the two ? (a) By aq. NH4OH (b) By aq. HNO3 (c) By either of the two (d) By none of the two.

AlCl3

O

73.

NCl

CH3CH2CH2COOH + O

AlCl3

65.

66.

(d) RCOCl H Ar The compound which reacts with Fehling solution is (a) C6H5COOH (b) HCOOH (c) C6H5CHO (d) CH2ClCH3 Among the following which one does not act as an intermediate in hofmann rearrangement? (a) RNCO (b) RCON (c)

RCONHBr

(d) RNC

Cl

(a)

|

CH 3CH 2 C HCOOH

(b) ClCH2CH2CH2COOH Cl

(c)

|

CH 3 C HCH 2 COOH

(d) All the three

HCl

Product is

MT-75

Mock Test 7 74. Which of the following is correct ? (a)

(COOH) 2

heat

(b)

(COOH) 2

heat

(c)

HOOCCH

CHCOOH

HCOOH CO 2 CO 2

CH 2

(d)

COOH O

O 77.

CO H 2 O

O

COOH HOOC COOH HOOC

(d)

COOH COOH

(b) RCONH2

(c) CH3CH2NH2

(d)

In the reaction sequence NH2 NaNO 2 ,HCl 0 C

CuCN

B

79.

C, the product ‘C’ is: (a) benzonitrile (b) benzaldehyde (c) benzoic acid (d) benzylamine The end product (Z) of the following reaction is N 2Cl

H /H2 O

[Z]

A

LiAlH 4

80.

(Y)

Cu/KCN (X)

NaOH CaO,

(Z)

(b) A carboxylic acid (a) A cyanide (c) An amine (d) An arene Consider the following ions (I)

+

Me2N

(III) H3C (IV) H3C

HOOC (c)

(a)

N +

(II) O 2N

COOH

HOOC (b)

78.

[Y] oxidation

Compound Z should be

CH2

O

heat

(ii ) Heat with Cu powder

CH 3

(A) , product (A) is :

CHCOOH CO2

(i ) CHN 2COOC 2 H 5

(a)

In the reaction

CH3

heat

75. Introduction of a methyl group in ammonia markedly increases the basic strength of ammonia in aq. solution, introduction of the second methyl group increases only marginally the basic strength of methyl amine in water. This is due to (a) different type of hybridisation in the two amines (b) protonated dimethyl amine is more solvated than methyl amine (c) protonated dimethyl amine is more solvated than the protonated methyl amine (d) protonated dimethyl amine is less stable than the protonated methyl amine 76.

CH 2 N

N

N N +

N

O +

N

N

N

The reactivities of these ions in azo-coupling reactions (under similar conditions) will be such that (a) (I) < (IV) < (II) < (III) (b) (I) < (III) < (IV) < (II) (c) (III) < (I) < (II) < (IV) (d) (III) < (I) < (IV) < (II)

EBD_7443 MT-76

Target VITEEE PART - III : MATHEMATICS 1

81.

2

The rank of the matrix 2

x x , if 0 x 1 x, if 2

a

b

x

2

(c) equals If x, a tan – 1

87.

(c) 1

1

R, where

3a 2 x – x 3 a 3 – 3ax 2

is equal to 2

(d) does not exist

2

a 3

x

a 3

, then

is equal to x a

x a

(b) –3 tan – 1

3 tan – 1

a x

(d) None of these

x 5 If sin 1 , then the value cosec 1 5 4 2 of x is (a) 4 (b) 5 (c) 1 (d) 3 The sum of the series log 4 2 – log 8 2 + log16 2 + ..................to is (a) 1 – log e 2 (c)

log e 2 – 1

10 , then a b c is

equal to (a) 5 2

(b) 50

(c)

(d) 10

10 2

89. The acute angle that the vector 2iˆ 2ˆj kˆ makes with the plane contained by the two vectors 2ˆi 3ˆj kˆ and ˆi ˆj 2kˆ is given by

(d) 2

(b) equals –

2

8 and c a

6,

(a)

cos

1

(c)

tan

1

1 3

2

(b)

sin

1

(d)

cot

1

1 5

( 2)

x 1 y 1 z 3 and the 3 2 1 plane : x 2y z . Of the following assertions, the only one that is always true is (a) L is to (b) L lies in (c) L is paralel to (d) None of these If f (x) = sin x + cos x, g (x) = x2 – 1, then g (f (x)) is invertible in the domain

90. Given the line L :

(a) 3 tan – 1 (c) 86.

2 is 3

c r is

1 cos{2( x 2)} x 2

lim

(a) equals

85.

x x

0 has a non-trivial solution,

then value of p q (a) –1 (b) 0 84.

b c

(a) differentiable at x = 2 (b) not differentiable at x = 2 (c) continuous at x = 2 (d) None of these Suppose p, q, r 0 and system of equation (p + a)x + by + cz = 0, ax (q b) y cz 0 , ax by (r c)z

a , b and c are perpendicular to b c , c a

an d a b respectively an d if a b

(b) 2 if a = 1 (d) none of these

82. The function f x

83.

5

4 a 4 is 2 a 1

1

(a) 1 if a = 6 (c) 3 if a = 2

88.

(b) 1 + log e 2 (d) log 2 e

91.

(a)

0,

2

(b)

, 4 4

, (d) [0, ] 2 2 92. Shamli wants to invest `50,000 in saving certificates and PPE. She wants to invest atleast `15,000 in saving certificates and at least `20,000 in PPF. The rate of interest on saving certificates is 8% p.a. and that on PPF is 9% p.a. Formulation of the above problem as LPP to determine maximum yearly income, is (a) Maximize Z = 0.08x + 0.09y Subject to, x + y 50,000, x 15000, y 20,000 (b) Maximize Z = 0.08x + 0.09y Subject to, x + y 50,000, x 15000, y 20,000 (c) Maximize Z = 0.08x + 0.09y Subject to, x + y 50,000, x 15000, y 20,000 (d) Maximize Z = 0.08x + 0.09y Subject to, x + y 50,000, x 15000, y 20,000

(c)

MT-77

Mock Test 7 are the roots of x2 – x + 1 = 0, then the

93. If and equation whose roots are 100 and 100 are (a) x2 – x + 1 = 0 (b) x2 + x – 1 = 0 2 (c) x – x – 1 = 0 (d) x2 + x + 1 = 0 94. If ( 1) is a cube root of unity, then 1

1 2

1 1

1

(b)

2

1

r

2y 5 3

2iˆ

1

passing through c 2 ,

2iˆ

passing through c 2 , 3ˆ j pkˆ 2

then p is

equal to (a) 0 (b) 1 (c) 2 (d) 3 96. If A = sin8 + cos14 , then for all values of : (a) A 1 (b) 0 < A 1 (c) 1< 2A 3 (d) none of these 97. A circular wire of radius 7 cm is cut and bent again into an arc of a circle of radius 12 cm. The angle subtended by the arc at the centre is (a) 50° (b) 210° (c) 100° (d) 60° 98. The locus of the mid points of chords of the parabola y

2

c 2

and

c 2

and

4ax, which pass through a fixed

point (h, k), is given by (a)

y 2 2ax ky 2ah

0

(b)

y 2 2ax ky 2ah

0

(c)

y 2 2ax ky 2ah

0

c 2

(c) a parabola with its vertex at 0,

z 1, is

5ˆ ˆ j k 2

a circle having its centre at 0,

1 2

1

(b) 1 (a) 0 (c) – 4 (d) 2 95. If vector equation of the line x 2 2

the axis of x is always a constant c. The locus of P is (a) a straight line having equal intercepts c on the axes

(d) None of these 99. If the tangent to the parabola y2 = 4ax meets the axis in T and tangent at the vertex A in Y and the rectangle TAYG is completed, then the locus of G is (a) y2 + 2ax = 0 (b) y2 + ax = 0 2 (c) x + ay = 0 (d) x2 + 4ay = 0 100. A point P moves such that the difference between its distances from the origin and from

c 2

(d) None of these 101. The projections of a line segment on the coordinate axes are 12, 4, 3. The direction cosine of the line are: (a)

-

4 3 12 ,- , 13 13 13

(b)

4 3 12 ,- , 13 13 13

12 4 3 , , (d) None of these 13 13 13 102. A parabola is drawn with its vertex at (0, – 3), the axis of symmetry along the conjugate axis of the

(c)

x 2 y2 1 and passing through 49 9 the two foci of the hyperbola. The co-ordinates of the focus of the parabola are

hyperbola

(a)

0,

11 6

(b)

0,

11 6

(c)

0,

11 12

(d)

0,

11 12

103. A value of c for which conclusion of Mean Value Theorem holds for the function f (x) = loge x on the interval [1, 3] is (a) log3 e (b) loge3 1 log3 e 2 104. Which of the following function is an odd function ?

(c) 2 log3e

(d)

EBD_7443 MT-78

Target VITEEE 1 x x2

(a)

f (x)

(b)

f (x)

x

(c)

f ( x)

log

1 x x2

ax 1 ax 1 1 x2 2

1 x (d) f(x) = k, k is a constant 105. Let p : I am brave, q : I will climb the Mount Everest. The symbolic form of a statement, 'I am neither brave nor I will climb the mount Everest' is (a) p q (b) ~ (p q) (c) ~ p ~ q (d) ~ p q 106. The expansion of log cosh x in powers of x by Maclaurin’s theorem is x3 5

x5 25

x6 ...... 125

(a)

x 1

(b)

x2 2

x4 12

x6 ........... 45

(c)

x2 2

x4 12

x6 ........... 45

x x3 x 5 x 6 (d) ........... 1 5 25 125 107. The current flowing through a tangent galvanomenter is obtained from the relation i = k tan where is the deflection and k is a constant. The relative error in the value of i due to error in the observation of is

(a)

d sin

(b)

2d sin

d 2d (d) sin 2 sin 2 108. If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number (a) 601 (b) 600 (c) 603 (d) 602

(c)

/2

109. Evaluate 0

sin x sin x

cos x

dx

(a) /3 (b) /2 (c) /4 (d) /6 110. The area of the plane region bounded by the curves x + 2y2 = 0 and x + 3y2 = 1is equal to (a)

5 3

(b)

1 3

4 2 (d) 3 3 111. The area between the curve y = x (x –1) (x – 2) and x-axis is (a) 1/4 (b) 1/2 (c) 1 (d) 0

(c)

/6

112. Evaluate :

cos7 3x dx

0

(a)

15 150

16 105

(b)

105 16 (d) 150 150 113. The differential equations of all conics whose axes coincide with the co-ordinate axis is

(c)

(a)

xy

(b)

xy

(c)

xy

d2 y dx

2

d2 y dx

2

d2 y dx

2

d2 y

dy dx

2

x

dy dx

2

x

dy dx

2

x

2

y

dy dx

0

x

dy dx

0

y

dy dx

0

dy dy y 0 2 dx dx dx 114. The degree and order of the differential equation of the family of all parabolas whose axis is X - axis, are respectively. (a) 2, 3 (b) 2, 1 (c) 1, 2 (d) 3, 2 115. In the group of non-zero rational numbers under the binary operation * given by a* b = ab/5 the identity element and ht inverse of 8 are respectively. (a) 5 and 5/8 (b) 5 and 25/8 (c) 5 and 8/25 (d) none of these

(d)

xy

x

MT-79

Mock Test 7 ( x 1) ( x 3) is a real valued function in ( x 2)

116. f ( x)

the domain (a)

(

, 1] [3, ) (b) (

, 1]

(2, 3]

(c) [ 1, 2) [3, ) (d) none of these 117. In a chess tournament, where the participants were to play one game with another, two chess players fell ill, having played 3 games each. If the total number of games played is 84, the number of participants at the beginning was (a) 15 (b) 16 (c) 20 (d) 21 118. The equation sin –1x – cos–1x = cos–1

3 2

has

(a) unique solution (b) no solution (c) infinitely many solutions (d) None of these 119. There is a five-volume dictionary among 50 books arranged on a shelf in random order. If the volumes are not necessarily kept side by side, the probability that they occur in increasing order from left to right is : (a) (c)

1 5

(b)

1 550

1

(d) None 505 120. If A and B play a series of games in each of which probability that A wins is p and that B wins is q = 1 – p. Therefore the chance that A wins two games before B wins three is p2

(a)

p2 1 3q

(c)

p 2 (1+2q+3 q 2 ) (d) None of these

(b)

1 3q 2

PART - IV : ENGLISH Direction (Qs. 121 - 123) : Read the passage carefully and answer the questions given below Dolphins are regarded as the friendliest creatures in the sea and stories of them helping drowning sailors have been common since Roman times. The more we learn about dolphins, the more we realize that their society is more complex than people previously

imagined. They look after other dolphins when they are ill, care for pregnant mothers and protect the weakest in the community, as we do. Some scientists have suggested that dolphins have a language but it is much more probable that they communicate with each other without needing words. Could any of these mammals be more intelligent than man? Certainly the most common argument in favor of man's superiority over them that we can kill them more easily than they can kill us is the least satisfactory. On the contrary, the more we discover about these remarkable creatures, the less we appear superior when we destroy them. 121. It is clear from the passage that dolphins(a) don't want to be with us as much as we want to be with them (b) are proven to be less intelligent than once thought (c) have a reputation for being friendly to humans (d) are the most powerful creatures that live in the oceans 122. The fact that the writer of the passage thinks that we can kill dolphins more easily than they can kill us(a) means that they are better adapted to their environment than we are (b) shows that dolphins have a very sophisticated form of communication (c) proves that dolphins are not the most intelligent species at sea (d) does not mean that we are superior to them 123. One can infer from the reading that(a) dolphins are quite abundant in some areas of the world (b) communication is the most fascinating aspect of the dolphins (c) dolphins have skills that no other living creatures have such as the ability to think (d) dolphins have some social traits that are similar to those of humans 124. Choose the word opposite in meaning to the word given in bold. Tremulous (a) Steady (b) Obese (c) Young (d) Healthy 125. Choose the best pronunciation of the word, Asthma, from the following options. (a) ash-thama (b) as-tma (c) as-tha-maa (d) azma

EBD_7443

MOCK

8

VITEEE Mock Test Paper

Max. Marks : 125

Time : 2½ hrs

PART - I (PHYSICS) 1.

2.

3.

Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectively. The ratio of the energy radiated per second by the first sphere to that by the second is (a) 1 : 1 (b) 16 : 1 (c) 4 : 1 (d) 1 : 9. A current is flowing in a circular conductor of radius r. It is lying in a uniform magnetic field B such that its plane is normal to it. The magnetic force acting on the loop will be (a) 0 (b) /rB (c) 2 /rB (d) irB A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = -kx + ax3. Here k and a are positive constants. For x 0 , the functional form of the potential energy U(x) of the particle is

4.

5.

6.

U(x)

(a)

X

U(x)

(b)

X

7. U(x)

(c)

In a half wave rectifier, the r.m.s. value of the A.C. component of the wave is (a) equal to d.c. value (b) more than d.c. value (c) less than d.c. value (d) zero In a Wheatstone's bridge, three resistances P, Q and R connected in the three arms and the fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for the bridge to be balanced will be (a)

P Q

2R S1 S2

(b)

P Q

R (S1 S2 ) S1S2

(c)

P Q

R (S1 S2 ) 2S1S2

(d)

P Q

R

S1 S2 A 10 resistance, 5 mH coil and 10 µF capacitor are joined in series. When a variable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequency (a) is halved (b) is doubled (c) remains unchanged (d) is quadrupled An object at rest in space suddenly explodes into three parts of same mass. The momentum of the two parts are 2pˆi and pˆj . The momentum of

the third part

X

(a) will have a magnitude p 3 U(x)

(d)

X

(b) will have a magnitude p 5 (c) will have a magnitude p (d) will have a magnitude 2p.

MT-81

Mock Test 8 If vs, vx and vm are the speed of soft gamma rays, X-rays and microwaves respectively in vacuum, then (a) vs > vx > vm (b) vs < vx < vm (c) vs > vx < vm (d) vs = vx = vm 9. The resistance of a wire is R. If the length of the wire is doubled by stretching, then the new resistance will be (a) 2R (b) 4R (c) R (d) R/4 10. Which of the following curves correctly represents the variation of capacitive reactance (Xc) with frequency n?

of photoelectrons would be about (a) 1.49 eV (b) 2.2 eV (c) 3.0 eV (d) 5.0 eV The load versus elongation graph for four wires has been shown in the figure. The thinest wire is Load a b (a) a c

8.

XC

15.

(b) b

Elongation (d) d A charged particle shows an acceleration of 4.2 × 1010 ms–2 under an electric field at low speed. The acceleration of the particle under the same field when its speed becomes 2.88 × 108 ms–1 will be (a) 2.88 × 108 ms–2 (b) 1.176 × 1010 ms–2 (c) 4.2 × 1010 ms–2 (d) None of these The real time variation of input signals A and B are as shown below. If the inputs are fed into NAND gate, then select the output signal from the following.

16.

XC

(a)

(b) n

n

17. (c)

XC

XC

(d) n

n

The half-life of radioactive Radon is 3.8 days. The time at the end of which (1/20)th of the Radon sample will remain undecayed is (given log10e = 0.4343) (a) 13.8 days (b) 16.5 days (c) 33 days (d) 76 days 12. Find equivalent resistance between the points A and B. 11.

A

1

2R

2R 2

d

(c) c

3

A

A

B t (s)

(a)

Y

0 2 4 6 8

R 4

Y

B

t (s)

B

(a) R/2 (b) R/4 (c) R (d) R/8 13. Water boils in the electric kettle in 15 minutes after switching on. If the length of heating wire is decreased to 2/3 of its initial value, then the same amount of water will boil with the same supply voltage in (a) 8 minutes (b) 10 minutes (c) 12 minutes (d) 15 minutes 14. The threshold wavelength of the tungsten is 2300 Å. If ultraviolet light of wavelength 1800 Å is incident on it, then the maximum kinetic energy

Y (b)

0 2 4 6 8

t (s)

(c) Y

0 2 4 6 8 (d)

t (s)

Y 0 2 4 6 8

t (s)

EBD_7443 MT-82

18.

19.

20.

Target VITEEE

When a plastic thin film of refractive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes. The thickness of the film will be, if the wavelength of light is 5890Å. (a) 6.544 × 10–4 cm (b) 6.544 × 10–4 m (c) 6.54 × 10–4 cm (d) 6.5 × 10–4 cm A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be (a) 0, 1 (b) 1, 1 (c) 1, 0.5 (d) 0, 2 In the circuit given below, the charge in C, on the capacitor having 5 F is 2 F 3 F 5 F

f 4 F

a

+ 6V

A

A

27.

0

, represents (a) fission (b) -decay (c) -decay (d) fusion A small current element of length d and carrying current is placed at (1, 1, 0) and is carrying current in ‘+z’ direction. If magnetic field at origin be

(a)

Z 1Y

1e

B1

B2

(b) | B1 | | 2B2 |

(c) B1 (d) B1 B2 2B2 According to Maxwell’s equation the velocity of light in any medium is expressed as (a) (c)

24.

26.

b

B1 and at point (2, 2, 0) be B2 then

23.

25.

(a) 4.5 (b) 9 (c) 7 (d) 15 A nuclear reaction is given by ZX

22.

c

1 0 o

/

(b) (d)

1

G 2V

12V

d

e

21.

500

28.

R

B

A

(a) 100 (b) 200 (c) 1000 (d) 500 When light is incident on a metal surface the maximum kinetic energy of emitted electrons (a) vary with intensity of light (b) vary with frequency of light (c) vary with speed of light (d) vary irregularly If the electric flux entering and leaving an enclosed surface respectively is 1 and 2, the electric charge inside the surface will be (a) ( 2 + 2) × o (b) ( 2 – 2) × o (c) ( 1 + 2) × o (d) ( 2 – 1) × o Total internal reflection can take place only if (a) light goes from optically rarer medium (smaller refractive index) to optically denser medium (b) light goes from optically denser medium to rarer medium (c) the refractive indices of the two media are close to different (d) the refractive indices of the two media are widely different Doubly ionised helium atom and hydrogen ions are accelerated, from rest, through the same potential difference. The ratio of final velocities of helium and hydrogen is (a) 1 : 2

(b)

2 :1

(c) 1 : 2 (d) 2 : 1 29. Figure here shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over s = 0 to 20 m will be f(N)

0

In the circuit , the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be

15 10 5 0

0

5

10

20

x(m)

MT-83

Mock Test 8 (a) 250 J (b) 200 J (c) 150 J (d) 10 J 30. In cyclotron the resonance condition is (a) the frequency of revolution of charged particle is equal to the frequency of A.C. voltage sources (b) the frequency of revolution of charged particle is equal to the frequency of applied magnetic field (c) the frequency of revolution of charged particle is equal to the frequency of rotation of earth (d) the frequency of revolution of charged part icle, frequency of A.C. source an d frequency of magnetic field are equal 31. The length of a potentiometer wire is . A cell of emf E is balanced at a length /3 from the positive end of the wire. If the length of the wire is increased by /2. At what distance will be the same cell give a balance point? (a) 2 /3 (b) /2 (c) /6 (d) 4 /3 32. Two drops of the same radius are falling through air with a steady velocity of 5 cm per sec. If the two drops coalesce, the terminal velocity would be (a) 10 cm per sec (b) 2.5 cm per sec (c) 5 × (4)1/3 cm per sec (d) 5

35.

36.

i

–e

(a)

–2V

+2V

(b)

–4V

–3V

(c)

3V

5V

(d)

0V

–2V

V R

x

force on electron at this instant is: (a)

(c) 37.

0

eviR 2 x

2 (x 2 0

(b)

R 2 )3 / 2

eviR 2 x

4 (x 2

eviR 2 x 0

(x 2

R 2 )3 / 2

(d) zero

R 2 )3 / 2

In a photoelectric experiment anode potential is plotted against plate current I

C

B A V

3 cm per sec

33. In Young’s double slit expt. the distance between two sources is 0.1 mm. The distance of the screen from the source is 20 cm. Wavelength of light used is 5460 Å. The angular position of the first dark fringe is (a) 0.08º (b) 0.16º (c) 0.20º (d) 0.32º 34. The forward biasd diode is

A nucleus ruptures into two nuclear parts which have their velocity ratio equal to 2 : 1. What will be the ratio of their nuclear radius ? (a) 21/3 : 1 (b) 1 : 21/3 1/2 (c) 3 : 1 (d) 1 : 31/2 An electron with velocity V along the axis approaches a circular current carrying loop as shown in the figure. The magnitude of magnetic

38.

(a) A and B will have different intensities while B and C will have different frequencies (b) B and C will have different intensities while A and C will have different frequencies (c) A and B will have different intensities while A and C will have equal frequencies (d) A and B will have equal intensities while B and C will have different frequencies A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be

q1

q2

–q1 (a) due to q1 and q2 only (b) due to q2 only (c) zero (d) due to all

EBD_7443 MT-84

39.

Target VITEEE

Two identical batteries, each of e.m.f. 2 volt and internal resistance 1.0 ohm are available to produce heat in a resistance R = 0.5 , by passing a current through it. The maximum power that can be developed across R using these batteries is 1

2V

2V

1

0.5

40.

(a) 1.28 W (b) 2.0 W (d) 3.2 W (c) 8/9 W To use a transistor as an amplifier (a) emitter-base junction is forward biased and collector base junction is reverse biased (b) both junctions are forward biased (c) both junctions are reverse biased (d) it does not matter how the transistor is biased, it always works as an amplifier

PART - II (CHEMISTRY) 41.

42.

Paramagnetism of Cr (Z = 24), Mn2+ (Z = 25) and Fe3+ (Z = 26) are x, y and z respectively. They are in the order (a) x = y = z (b) x > y > z (c) x = y > z (d) x > y = z Nitrobenzene can be reduced to aniline by H 2 / Ni

Sn / HCl

Zn / NaOH

LiAlH4

I

III

43.

44.

II

IV

(a) I, II and III (b) I and II (c) I, II and IV (d) only II Which reaction is used for converting a lower carboxylic acid into its next higher homologue? (a) Curtius reaction (b) Baeyer – Villiger reaction (c) Darzen glycidic ester synthesis (d) Arndt – Eistert synthesis Which of the following reagent reacts in different ways with CH3 CHO, HCHO and C6H5CHO ?

(a) Fehling solution (b) C6H5NHNH2 (c) Ammonia (d) HCl 45. The IUPAC name for th e coordination compound Ba[BrF4 ]2 is (a) Barium tetrafluorobromate (V) (b) Barium tetrafluorobromate (III) (c) Barium bis (tetrafluorobromate) (III) (d) none of these 46. The rate constant of the reaction : A 2 B, is 1.0 × 10–3 mol L–1 min–1. If the initial concentration of A is 1.0 mol L –1, what would be the concentration of B after 100 minutes: (a) 0.1 mol L–1 (b) 0.2 mol L–1 –1 (c) 0.9 mol L (d) 1.8 mol L–1 47. Which of the following statements is correct for a strong electrolyte : ½ (a) m increases linearly with C (b)

m

increases linearly with C2

(c)

m

decreases linearly with C2

½ (d) m decreases linearly with C 48. If all the reactants and the products in a reaction are in their standard states of unit activity, then which is true of the following ? (a) Gº = 0 (b) G = 0 (c) G = Gº (d) Hº – T Sº = 0 49. EMF of an H2 – O2 fuel cell (a) is independent of partial pressures of H2 and O2

(b) decreases on increasing PH 2 and PO2 (c) increases on increasing PH 2 and PO2

(d) varies with the concentration of OH– ions in the cathodic and anodic compartments. 50. When K2 Cr 2O7 is heated with concentrated H2SO4 and soluble chloride such as KCl (a) Cl– ion is oxidised to Cl2 gas (b) (c) (d) 51. The

C 2 O 72 ion is reduced to green Cr3+ ion red vapour of CrO2Cl2 is evolved CrCl3 is formed energy of a photon is given as E/atom

3.03 10 19 J atom 1 . Then the wavelength ( ) of the photon is (a) 65.6 nm (b) 656 nm (c) 0.656 nm (d) 6.56 nm 52. The unit cell of diamond is made up of (a) 6 carbon atoms, 4 atoms constitute ccp and two atoms occupy half of octahedral voids

MT-85

Mock Test 8 (b) 8 carbon atoms, 4 atoms constitute ccp and 4 atoms occupy all the octahedral voids (c) 8 carbon atoms, 4 atoms form fcc lattice and 4 atoms occupy half of the tetrahedral voids alternately (d) 12 carbon atoms, 4 atoms form fcc lattice and 8 atoms occupy all the tetrahedral holes.

57.

58.

OH NaOH

53.

[X]

CH 2=CHCH 2Cl

[Y] .

Here [Y] is a (a) single compound (b) mixture of two compounds (c) mixture of three compounds (d) no reaction is possible 54. In the reaction A + B Products, initial concentration of both A and B equal to 0.1 M,each is reduced to 1.0 × 10–2 M. The half-life is increased to ten fold. The rate of the reaction is : (a) proportional to first power of concentration (b) proportional to second power of concentration (c) independent of concentration (d) insufficient information 55. Aomic size of Zr and Hf is almost the same because of (a) both have (n – 1)d2ns2 configurations (b) both have the same number of d-electrons in penultimate energy level (c) lanthanide contraction (d) large screening effect of 4f-electrons over the valence shell. 56. Ethanol can be prepared more easily by which reaction? (i)

CH 3 CH 2 Br H 2 O

60.

61.

62.

CH 3 CH 2 OH

(ii) CH3CH 2 Br Ag2 O (in boiling water) CH 3CH 2 OH

(a) (b) (c) (d)

59.

by (i) reaction by (ii) reaction both reactions proceed at same rate by none

63.

In some fission process, the mass number of fission fragments are 144 and 90 respectively. If the K.E. of heavy fragment is 70 MeV, the total fission energy is (a) 200 MeV (b) 182 MeV (c) 190 MeV (d) 170 MeV At constant pressure, addition of helium to the reaction system : 2 NH 3 ( g ) N 2 ( g ) 3H 2 ( g ) (a) favours the formation of ammonia (b) reduces the formation of ammonia (c) reduces the dissociation of ammonia (d) does not affect the position of equilibrium Fehling solution can be used for distinguishing between (a) CH3CHO and C6H5CHO (b) CH3CHO and CH3COCH2OH (c) both (a) & (b) (d) none Which one does not belong to the same compound ? (a) Paraformaldehyde (b) Paraldehyde (c) Trioxane (d) Formalin An iron piece treated with acidified KMnO4 (a) displaces copper from a dilute H2SO4 (b) liberates hydrogen from dilute H2SO4 (c) displaces copper from CuSO4 solution but does not liberate H2 from dilute H2SO4 (d) neither displaces copper from CuSO4 solution nor liberates H2 from dilute H2SO4 .

A

B, k1

1011 e

X

Y , k2

1010 e

3000 / T 2000 / T

If the two reactions have all the reactants at unit molarity each, at what temperature their initial rates will be equal? (a) 2000 K (b) 3000 K (c) 1000 K (d) 1000/2.303 K Molar conductances of BaCl2, H2SO4 and HCl at infinite dilutions are x1, x2 and x3 respectively. Equivalent conductance of BaSO4 at infinite dilution will be : (a) (x1 + x2 – x3) /2 (b) x1 + x2 – 2x3 (c) (x1 – x2 – x3) /2 (d) (x1 + x2 – 2x3) /2

EBD_7443 MT-86

64.

Target VITEEE

Geometrical isomerism can be shown by (a) [Ag( NH 3 )(CN)] (b)

Na 2 [Cd( NO 2 ) 4 ]

(c) [Pt Cl 4 I 2 ] 65.

(d) [Pt ( NH 3 )3 Cl][Au(CN) 4 ] Which of the following statement is not true ? (a) At room temperature, formyl chloride is present in the form of CO and HCl (b) Acetamide behaves as a weak base as well as a weak acid. LiAlH

(c)

66.

67.

68.

4 CH 3 CONH 2 CH 3CH 2 NH 2 (d) None of the three A carboxylic acid can best be converted into acid chloride by using (a) PCl5 (b) SOCl2 (c) HCl (d) ClCOCOCl In the Mac-Arther Cyanide process for extraction of silver, a small amount of KNO3 is added. The function of KNO3 is (a) to oxidise the sulphur present in argentite ore to SO2 (b) to form anions with Ag+ which is then reduced to metallic silver by zinc (c) to oxidise Ag in natural form to Ag+ (d) to oxidise the impurities of lead and zinc present in it Dehydration of an alcohol in presence of sulphuric acid gives alkene

H SO

69.

70.

71.

2 4 CH 3CH 2 OH CH 2 CH 2 H 2 O Here sulphuric acid acts as (a) an acid (b) a base (c) a catalyst (d) all the three Dopping of AgCl crystals with CdCl2 results in (a) Schottky defect (b) Frenkel defect (c) Substitutional cation vacancy (d) Formation of F- centres An organic compound A on heating with ethanol gives compounds B and C, of which compound C is again a derivative of the compound B. The compound A is (a) CH3COOH (b) (CH3CO)2O (c) CH3COOC2H5 (d) CH3 CH2OH When aniline is treated with acetyl chloride in presence of anhydrous aluminium chloride, the main product is

(a) o - aminoacetophenone (b) p-aminoacetophenone (c) both (a) and (b) (d) m-aminoacetophenone 72. A coordination complex having which one of the following descr iptions would be paramagnetic to the maximum extent? (a)

d 6 (octahedral, low-spin)

(b)

d 8 (octahedral)

(c)

d 6 (octahedral, outer orbital)

(d) d 4 (octahedral,low-spin) 73. Which of the following processes takes place with decrease of entropy? (a) Solid gas (b) sugar + water solution (c) NH3(g) + HCl(g) NH4Cl(s) (d) A(g) + B(g) mixture 74. Identify (C) and (D) in the following series of reactions CH 3 NH 2

excess of CH3I

[A ]

AgOH

heat

[B] [ C] [ D ]

(a) (CH3)3COH, CH3NH2 (b) (CH3)2C = CH2, CH3NH2 (c) (CH3)3N, CH3OH (d) (CH3)2C=CH2, CH3OH 75. Which one of the following is not an oxidation product of a primary amine? (a) A hydroxylamine (b) A nitroso compound (c) A nitro compound (d) None of these 76. Which method can be used to distinguish [Co(NH3 )6 ][Cr(NO 2 ) 6 ] and

(a) (b) (c) (d)

[Cr(NH3)6][Co(NO2)6] by measurement of their conductivity by titration method by precipitation method with AgNO3 by electrolysis of their aqueous solutions

77. Which electrphile is likely to be formed as an intermediate in the following electrophilic substitution reaction ?

MT-87

Mock Test 8 OH

82.

OH CHO

CHCl3 NaOH

(a)

C HCl 2

(b) C HO

(c)

: C Cl 2

(d) : CCl 2

83. 78. In p-type semiconductor, the added impurity to silicon is ----- and conduction of electric current is due to the movement of ----(a) As, electrons (b) P, holes (c) Ga, holes (d) Ga, electrons and holes 79. In the radioactive decay zA

A

z 2B

A 4

z 1C

A 4

zA

conc . H 2SO 4

85.

A 4

the sequence of radiation emitted is (a) (b) (c) (d) 80. In the following reaction, the reagent X should be RCOOH [ X ]

84.

86.

87.

RNH 2

(a) NH3 (b) HN3 (c) Either of the two (d) None of the two

88.

PART - III (MATHEMATICS) 2 3 3

81. Let f :

,

[0, 4] be a function

defined as f (x) = by

3 sin x cos x 2 . Then, f–1(x) is given

sin

1

(b) sin

1

(a)

x 2 2 x 2 2

6 6

x 2 2 cos 1 (c) 3 2 (d) None of these

89.

Which of the following statement is correct? (a) Every L.P.P. admits an optimal solution (b) A L.P.P. admits a unique optimal solution (c) If a L.P.P. admits two optimal solutions, it has an infinite number of optimal solutions (d) The set of all feasible solutions of a L.P.P. is not a convex set. The period of tan 3 is (a) (b) 3 /4 (c) /2 (d) None of these The function f (x)

x x

x 1 is – 1 2

e (a) an odd function (b) an even function (c) neither an odd nor an even function (d) a periodic function The equation of the right bisector plane of the segment joining (2, 3, 4) and (6, 7, 8) is (a) x + y + z + 15 = 0 (b) x + y + z – 15 = 0 (c) x – y + z – 15 = 0 (d) None of these What is the angle between the line 6x = 4y = 3z and the plane 3x + 2y – 3z = 4 ? (a) 0 (b) /6(c) /3 (d) /2 An infinite G.P. has first term ‘x’ and sum ‘5’, then x belongs to (a) x < – 10 (b) – 10 < x < 0 (c) 0 < x < 10 (d) None of these By examining the chest X-ray, the probability that TB is detected when a person is actually suffering is 0.99. The probability of an healthy person diagnosed to have TB is 0.001. In a certain city, 1 in 1000 people suffers from TB, A person is selected at random and is diagnosed to have TB. Then, the probability that the person actually has TB is (a)

110 221

(b)

2 223

(c)

110 223

(d)

1 221

If f(z) =

7

z

1 z2

, where z = 1+ 2i, then |f(z)| is

equal to : |z| 2 (c) 2 | z |

(a)

(b) | z | (d) None

EBD_7443 MT-88

90.

Target VITEEE

| z1 |

(c) | z2 | = 91.

92.

93.

(b) sin + cos (a) sin – cos (c) tan + cot (d) tan – cot 98. If the eccentricity of the hyperbola x2 – y2 sec2 = 4 is 3 times the eccentricity of the ellipse x2sec2 + y2 = 16, then the value of equals

If arg(z1 ) arg(z 2 ) , then (a) z2 = kz1–1 (k > 0) (b) z2 = kz1(k > 0) (d) None of these If nPr = nPr + 1 and nCr = nCr – 1, then the values of n and r are (a) 4, 3 (b) 3, 2 (c) 4, 2 (d) None of these If (G, *) is a group such that (a * b)2 = (a * a) * ( b * b) for all a, b, G , then G is (a) abelian (b) finite (c) infinite (d) None Let f : R

R be a function defined by

f (x) = min {x 1, x

1} ,Then which of the

following is true ? (a) f (x) is differentiable everywhere (b) f (x) is not differentiable at x = 0 (c) f (x) 1 for all x R (d) f (x) is not differentiable at x = 1 94.

95.

96.

The value of

C1 2

C3 4

C5 6

....... is equal to

(a)

2n 1 n 1

(b)

(c)

2n 1 n 1

2n 1 (d) n 1

32 k 2 If 42 k 5 k of k is (a) 2 (c) –1 If f(x) =

2n n 1

42 32 3 k 52 42 4 k = 0, then the value 62 52 5 k

(b) 1 (d) 0 x 2 – 4x – 5, then f(A), where

2 1 2 2 2 1

equals

(a) O (c) –I 97.

cos 1 tan

(c)

(b)

6

(d)

3

3 4 2

99. In a triangle ABC, a cos A b cos B c cos C is a b c equal to (a)

r R

(b)

R r

2r R (d) R 2r 100. The value of sin (cot–1 (cos (tan –1x)) is

(c)

(a)

x2

2

(b)

x2 1

x2 1 x2

x

(c)

x

2

2

1

(d)

2

101. The equation sin

1

5 x

sin

x

2

2

1 12

x

2

has

(a) no solution (b) only one solution, x = 13 (c) only one solution, x = – 13 (d) two solutions, x = ± 13 102. A point P moves such that the sum of twice its distance from the origin and its distance from the y-axis is a constant equal to 3. P describes (a) A circle with its centre at (–1, 0) and radius 2 3 (b) An ellipse centred at (–1, 0) and of 1 2 (c) A hyperbola centred at (1, 0) and of eccentricity 2 (d) None of these 103. The equation of the tangent to the parabola

eccentricity

1 2 2

A=

(a)

(b) I (d) 2I sin 1 cot

is equal to

y 2 6x at the point whose ordinate is 6, is (a) x + 2y + 6 = 0 (b) 2x – y + 6 = 0 (c) x – 2y + 6 = 0 (d) x – y + 6 = 0

MT-89

Mock Test 8 x2 y 2 1 16 25 and e2 is the eccentricity of the hyperbola passing through the foci of the ellipse and e1e2 = 1, then equation of the hyperbola is :

104. If e1 is the eccentricity of the ellipse

(a)

x2 9

y2 16

x2 (b) 16

1

y2 9

1

x2 y 2 x2 y 2 (c) (d) 1 1 9 25 9 36 3 2 105. The two curves x – 3xy + 2 = 0 and 3x2y – y3 – 2 = 0 (a) Cut at right angles (b) Touch each other (c) Cut at an angle /3 (d) Cut at an angle /4 106. For the curve yn = an – 1 x if the subnormal at any point is a constant then n is equal to (a) 1 (b) 2 (c) –2 (d) –1 107. The coordinates of a moving particle at time t are given by x = c t2 and y = b t2. The speed of the particle is given by

(a) 2 t (c + b) (c) 108.

t (c 2

b2 )

1

lim

h 0 h3 8 h

(a)

1 8

1 2h

(b) 2 t (c 2

b2)

(d) 2 t (c 2

b2 )

equals to

(b)

1 8

5 . Then, the probability that he will knock 6 down fewer than 2 hurdles is

(a) (c)

2 69

59 2 610

equal to (a)

1 6

1 2

f (1)

(b)

1 1 4f (0) f 2 6

f (1)

(c)

1 1 f (0) f 2 6

4(1)

(d)

1 1 f (0) f 2 6

f (1)

f (0) 4f

1

(b) (d)

510 2 610

510 2 69

f (x ) dx and given f (0) = 1.

112. If f’(x) =f (x) + 0

Then f (x) = (a)

2e x 2 e

(c)

2e x 2 e

113. If p

1 e 1 e

(u

v)

(a) 5 (c) 15

1 1 (d) 48 48 109. Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many ways can we place the balls so that no box remains empty? (a) 50 (b) 100 (c) 150 (d) 200 110. In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is

f (x) dx is 0

and [ u v w ]

(c)

5

1

111. If f(x) is a quadratic in x, then

(b)

2e x 3 e

(d)

2e x 3 e

(v

w)

1 e 3 e

(w

u)

1 , then is equal to 5 (b) 10 (d) None of these

114. Solve : (D2 + 6D + 8)y = e–2x. Be 2x

(a)

y

Ae 4x

(b)

y

Ae

4x

Be

(c)

y

Ae

4x

Be 2x

2x

1 xe 2x 2 1 xe 2x 2 1 xe 2

2x

1 xe 2x 2 115. The area of the triangle formed by the tangent

(d)

y

Ae 4x

Be

2x

and normal at the point 1, 3 on the circle x2 + y2 = 4 and the x-axis is (a) 3 sq. units (b) 2 3 sq. units (c) 3 2 sq. units (d) 4 sq. units

EBD_7443 MT-90

Target VITEEE

116. The temperature T of a cooling object drops at a rate proportional to the difference T – S, where S is constant temperature of surrounding medium. If initially T = 150°C, find the temperature of the cooling object at any time t. (a) S – (S + 150) ekt (b) S + (150 – S)ekt (c) S + (150 + S)et (d) S – (150 – S)ekt 117. The acute angle between the medians drawn through the acute angle of an isosceles right angled triangle is (a)

cos

1

2 3

(b) cos

1

3 4

4 1 5 (d) cos 5 6 118. A girls walks 4 km towards West. Then, she walks 3 km in a direction 30° East to North and stops. The girls displacement from her initial point of departure is

(c)

cos

1

3ˆ 3 3ˆ i j 2 2

(a)

5ˆ 3ˆ i j 2 2

(b)

5ˆ 3 3ˆ i j (d) None of these 2 2 119. For the probability density function (c)

2x

2e , x 0 ,x F(2) is equal to

f (x)

(a) (c)

e

4

e

e

121. Modern medicine is primarily concerned with (a) promotion of good health (b) people suffering from imaginary illness (c) people suffering from real illness (d) increased efficiency in work 122. The passage suggests that (a) health is an end in itself (b) health is blessing

0 0

(c) health is only means to an end

1

(b)

4

5/ 2

In the world today we make health and end in itself. We have forgotten that health is really means to enable a person to do his work and do it well. a lot of modern medicine and this includes many patients as well as many physicians pays very little attention to health but very much attention to those who imagine that they are ill. Our great concern with health is shown by the medical columns in newspapers. the health articles in popular magazines and the popularity of television programmes and all those books on medicine. We talk about health all the time. Yet for the most part the only result is more people with imaginary illness. The healthy man should not be wasting time talking about health: he should be using health for work. The work does the work that good health possible.

1

(d)

e

3

e e

1 3

5

e e 120. Consider the switching circuit

1 5

(d) we should not talk about health 123. Talking about the health all time makes people (a) always suffer from imaginary illness (b) sometimes suffer from imaginary illness (c) rarely suffer from imaginary illness (d) often suffer from imaginary illness 124. Select the word or group of words that is most similar in meaning to the word in capital letters.

a' b'

RUMINATE c

Logical expression corresponding to the complimentary to the above circuit is (a) a . b . c ' (b) a b c ' (c)

a '. b '.c

(d) none

PART - IV (ENGLISH) Direction (Qs. 121 - 123) : Read the passage carefully and answer the questions given below.

(a) to run fast

(b) to reprimand

(c) to think deeply (d) to spend lavishly 125. Choose the best pronunciation of the word, Video, from the following options. (a) wee-diyo (b) vid-i-yoh (c) vehi-dyoh

(d) vee-dio

MOCK

VITEEE Mock Test Paper

9

Max. Marks : 125

Time : 2½ hrs

PART - I (PHYSICS) 1.

2.

4.

If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio of the radiant energy received on earth to what it was previously will be (a) 32 (b) 16 (c) 4 (d) 64 Two thin flat metal plates having large surface area are charged separately to acquire charge densities + and – . The plates are then brought near to each other and held parallel to each other (Fig.):

A

C

B

V

1

(b) (c) (d) 3.

EA

EC

EA

EB

EC

EA

EC

0, E B

EA

EC

6.

E. (c) Electric field at the surface of a charged conductor is always normal to the surface. (d) The electric potential decreases along a line of force in an electric field. In figure, a carriage P is pulled up from A to B. The relevant coefficient of friction is 0.40. The B work done will be kg 0 5 P (a) 10 kJ 30 m m 50 (b) 23 kJ

0

0, E B

0

0

(c) 25 kJ

2 0

An application of Bernoulli’s equation for fluid flow is found in (a) dynamic lift of an aeroplane (b) viscosity meter (c) capillary rise (d) hydraulic press

3

5.

+vely charged

(a)

2

4 x (b) E2 > E4 > E1 = E3 (a) E4 > E2 > E3 > E1 (c) E1 > E2 > E3 > E4 (d) E1 > E3 > E2 > E4 Identify the WRONG statement : (a) In an electric field two equipotential surfaces can never intersect. (b) A charged particle free to move in an electric field shall always move in the direction of

–vely charged If EA, EB and EC denote the electric fields at the points A, B and C respectively, then which of the following will be true :

The figure shows electric potential V as a function x. Rank the four regions according to the magnitude of x-component of the electric field E within them, greatest first:

7.

A

C

(d) 28 kJ A metallic spherical shell of radius R has a charge –Q on it. A point charge +Q is placed at the centre of the shell. Which of the graphs shown below may correctly represent the variation of the electric field E with distance r from the centre of the shell ?

EBD_7443 MT-92

Target VITEEE E

E

(a)

(b) R

0

r

R

0

E

2

1 r

E

3 (c)

(d) R

0

8.

r

R

0

r

11.

A solid sphere of radius R1 and volume charge density

0

is enclosed by a hollow r sphere of radius R2 with negative surface charge density , such that the total charge in the system is zero. 0 is a positive constant and r is the distance from the centre of the sphere. The

(a) i2 > i3 > i1 (b) i2 > i1 > i3 (c) i1 > i2 > i3 (d) i1 > i3 > i2 The variable point B of a 80 rheostat AC has been set exactly in the midway such that the resistance of the part AB is equal to the resistance of the part BC. The rheostat is connected with a resistance of 20 and a battery of 8.0 V as shown in the figure. The current supplied by the battery is : A B

R2 ratio R is 1

(a)

(c) 9.

10.

C

(b)

0 0/ 2

(d)

2 / 0

8.0V

(a)

0

A battery has an emf of 15 V and internal resistance of 1 . Is the terminal to terminal potential difference less than, equal to or greater than 15 V if the current in the battery is (1) from negative to positive terminal, (2) from positive to negative terminal (3) zero current? (a) Less, greater, equal (b) Less, less, equal (c) Greater, greater, equal (d) Greater, less, equal The figure shows three circuits with identical batteries, inductors and resistances. Rank the circuits according to the currents through the battery just after the switch is closed, greatest first:

1 A 2

(b)

1 A 5

2 1 (d) A A 15 3 Two cells, having the same e.m.f., are connected in series through an external resistance R. Cells have internal resistances r1 and r 2 (r1 > r 2 ) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is

(c)

12.

(a)

r1 r2 2

(b)

r1 r2 2

(c)

r1 r2

(d) r1 r2

13. An electron is moving in an orbit of radius R with a time-period T as shown in the figure. The magnetic moment produced may be given by :

MT-93

Mock Test 9 (a)

15.

16.

17.

18.

2 e A T

R 2 e A T e e A (c) m T e A (d) m A R2 T | e | represents the magnitude of the electron charge. A galvanometer has a resistance of 50 . If a resistan ce of 1 is connected across its terminals, the total current flow through the galvanometer is [Ig represents the maximum current that can be passed through the galvanometer] (a) 42 Ig (b) 53 Ig (c) 46 Ig (d) 51 Ig An LC circuit contains a 20 mH inductor and a 50 F capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0. At what time is the energy stored completely magnetic ? (a) t = 0 (b) t = 1.54 ms (d) t = 3.14 ms (d) t = 6.28 ms A uniformly wound solenoidal coil of selfinductance 1.8 × 10–4 H and resistance 6 is broken up into two identical coils. These identical coils are then connected in parallel across a 12 V battery of negligible resistance. The time constant for the current in the circuit is: (a) 0.1 × 10–4 s (b) 0.2 × 104 s –4 (c) 0.3 × 10 s (d) 0.4 × 10–4 s A 3 kg ball strikes a heavy rigid wall with a speed of 10 m/s at an angle of 60º. It gets reflected with the same speed and angle as shown here. If the ball is in contact with the wall for 0.20s, what is the average force exerted on the ball by the wall? (b)

14.

m

m

19.

A conducting square loop is placed in a magnetic field B with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate . The induced emf in the loop at an instant when its side is ‘a’ is (a) 2a B (b) a2 B 2 (c) 2a B (d) a B A rectangular loop of wire with dimensions shown is coplanar with a long wire carrying current I. The distance between the wire and the left side of the loop is r. The loop is pulled to the right as indicated. What are the directions of the induced current in the loop and the magnetic forces on the left and right sides of the loop as the loop is pulled? r

I

20.

a Induced current Force on Force on left side right side (a) Counterclockwise To the left To the left (b) Counterclockwise To the right To the left (c) Clockwise To the right To the left (d) Clockwise To the left To the right An equilateral triangular loop ABC made of uniform thin wires is being pulled out of a region with a uniform speed v, where a uniform magnetic

field B perpendicular to the plane of the loop exists. At time t = 0, the point A is at the edge of the magnetic field. The induced current (I) vs time (t) graph will be as A v ××××××××××××× ××××××××××××× ××××××××××××× ××××××××××××× ××××××××××××× × × × ×B× × × × ×C× × × × ×××××××××××××

60º I

60º

(a) 150N

(b) Zero

(c)

(d) 300N

150 3N

b

(a) t

EBD_7443 MT-94

Target VITEEE (a) + ve y-axis, x-axis (b) – ve z-axis, y-axis (c) + ve z-axis, x-axis (d) + ve x-axis, x-axis 25. Fpe represents electrical force on proton due to

I

(b) t I

I

(d) t

The magnitude of flux linked with the circuit of resistance 2 k varies with t second according to equation: 5t 2 4t 12 The induced current in the circuit at t = 0.2 second is (a) 2 A (b) 1 A (c) 2 mA (d) 1 mA Two bulbs mar ked 200 V–100 W an d 200 V–200 W are joined in series and connected to a power supply of 200 V. The total power consumed by the two will be near to : (a) 33 watt (b) 66 watt (c) 100 watt

23.

24.

(d) 300 watt 200 V An unpolarised beam of light is incident on a place surface separating air and glass at an angle equal to the Brewster angle. Then : (a) the reflected light has electric component only perpendicular to the incident plane. (b) the reflected light has electric component only in the plane of incidence. (c) the electric component parallel to the plane of incidence in refracted ray completely disappear. (d) the magnetic component of the refracted light completely disappear. If the magnetic field B of a polarised electromagnetic wave oscillates parallel to y-axis and is given by : By = Bm sin (kz – t). What is the direction of propagation of the em-wave and parallel to which axis does the associated electric field oscillates ?

(a)

Fpe

Fep

(b)

F'pe

' 0 Fep

(c)

Fpe

Fpe ' Fep

(d)

Fpe

Fep and Fpe '

0

Fep '

0 Fep '

26. The photo electric work function for a metal surface is 4.125 eV. The cut-off wavelength for this surface is (a) 4125 Å (b) 3000 Å (c) 6000 Å (d) 2062 Å 27. Let

,

e

and

e

p represent the wavelengths

of electrons, positrons and proton, respectively, of the same momentum then : (a) (c)

e

e

p

(b)

e

e

p

(d)

e

e

p

e

e

p

28. Figure represents a graph of kinetic energy of most energetic photoelectrons, Kmax (in eV), and frequency ( ) for a metal used as cathode in photoelectric experiment. The threshold frequency of light for the photoelectric emission from the metal is :

Kmax

t

22.

hydrogen atom. Similarly, F'pe represents the gravitational force on proton due to electron and

F'ep the corresponding force on electron due to proton. Which of the following is NOT true ?

(c)

21.

electron and Fep on electron due to proton in a

3 2 1 0

15

10 Hz

1.5 × 1014 Hz (a) 1 × 1014 Hz (b) 14 (c) 2.1 × 10 Hz(d) 2.7 × 1014 Hz 29. Using the following data,

MT-95

Mock Test 9 mass of hydrogen atom = 1.00783 u mass of neutron = 1.00867 u mass of nitrogen atom 7 N 4 =14.00307 u the calculated value of the binding energy of the nucleus of the nitrogen atom 7 N 4 is close to (a) 56 MeV (b) 98 MeV (c) 104 MeV (d) 112 MeV 30. The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10V. The d.c. component of the output voltage is (a) 20/ V (b) 10/ 2 V (c) 10/ V (d) 10V 31. The diagram of a logic circuit is given below.

34.

35.

A B

F

W X

Reverse bias Vz d

a c

b

36.

37.

I (mA)

Forward bias

V (V)

38.

e I ( A)

(a) a b (b) b c (c) c d (d) d e 33. Carbon, silicon and germanium have four valence electrons each. These are characterized by valence and conduction bands separated by energy band gap respectively equal to (Eg)C , (Eg) Si and (Eg) Ge. Which of the following statements is true? (a) (Eg)C = (Eg)Si = (Eg)Ge (b) (Eg)C > (Eg)Si > (Eg)Ge

Logic gate circuit

C

1

W X

The output F of the circuit is given by : (a) W. (X + Y) (b) W. (X. Y) (c) W + (X. Y) (d) W + (X + Y) 32. The graph given below represents the I-V characteristic of a Zener diode. Which part of the characteristic curve is most relevant for its operation as a voltage regulator ?

(c) (Eg)C < (Eg)Ge > (Eg)Si (d) (Eg)Si < (Eg)Ge > (Eg)C In semiconductors at a room temperature (a) the conduction band is completely empty (b) the valence band is partially empty and the conduction band is partially filled (c) the valence band is completely filled and the conduction band is partially filled (d) the valence band is completely filled The following figure shows a logic gate circuit with two inputs A and B and the output C. The voltage waveforms of A, B and C are as shown below

39.

A 1

t

B 1

t

C

t

The logic circuit gate is (a) NAND gate (b) NOR gate (c) OR gate (d) AND gate Two bodies A and B having masses in the ratio of 3 : 1 possess the same kinetic energy. The ratio of linear momentum of B to A is (a) 1 : 3 (b) 3 : 1 (c) 1: 3 (d) 3 :1 Main function of the RF amplifiers in a superheterodyne receiver is to (a) amplify signal (b) reject unwanted signal (c) discriminate against image frequency signal and IF-signal (d) all the above Optical fibre are used for long distance communication because (a) it amplifies signals to be transmitted (b) it transfer signals faster than electrical cables (c) it pre-emphasise weak signals (d) it provide little attenuation as compared to electrical cable for light propagation A super heterodyne receiver is designed to receive transmitted signals between 5 and 10 MHz. The tuning range of the local oscillate for IF frequency 600 kHz for high-side tuning would be (a) 4.6 to 9.6 MHz (b) 5.6 to 10.6 MHz (c) 4.6 to 10.6 MHz (d) 5.6 to 9.6 MHz

EBD_7443 MT-96

40.

Target VITEEE

Two straight metallic strips each of thickness t and length are rivetted together. Their coefficients of linear expansions are 1 and 2 . If they are heated through temperature T, the bimetallic strip will bend to form an arc of radius (a) t /{ 1 T} 2 (b) (c) (d)

t /{( 2 1 T} t( 1 T 2 T t( 2 1

PART - II (CHEMISTRY) 41.

42.

43.

44.

Colour of cupric chloride is blue due to _____. (a) p -d electronic transition causes emission of energy (b) d-d electronic transition causes emission of energy (c) d-d electronic transition causes absorption of energy at red visible wave length (d) p-d electronic transition of Cu2+ state. Stainless steel does not rust because: (a) chromium and nickel combine with iron (b) chromium forms an oxide layer and protects iron from rusting (c) nickel present in it, does not rust (d) iron forms a hard chemical compound with chromium present in it. Excited state configuration of Mn2+ is (a)

t 42g

(b) t 32g e2g

(d)

t 42g e2g

(d) t 52g e0g

[Ti(H2O)6]3+ shows purple colour due to (a) d yz d 2 2 electronic transition x

(b)

d xz

d 2 2 electronic transition x y

(c)

d xy

eg [d

(d) d xz 45.

y

x2 y2

,d

z2

] electronic transition

eg [d 2 2 , d 2 ] electronic transition x y z

Amongst the following which are not true ? (a) EAN of iron in Fe(C5 H 5 ) 2 is 36. (b) [Fe(H 2 O) 6 ]2 has paramagnetism due to 4 unpaired electrons. (c)

[Cr ( NH 3 ) 6 ]3 is paramagnetic.

(d) [Co I 4 ]2 has square planar geometry..

46. A transition metal ion exists in its highest oxidation state. It is expected to behave as (a) a chelating agent (b) a central metal in a coordination compound (c) an oxidising agent (d) a reducing agent 47. Non-stoichiometr ic compound have the properties of: (a) electrical conductance (b) isolation (c) insulation (d) none of these 48. The lattice parameters of a crystal are a = 5.62 Å, b = 7.41 Å C = 9.48 Å. The three co-ordinates are mutually perpendicular to each other. The crystal is – (a) tetragonal (b) orthorhombic (c) monoclinic (d) trigonal 49. Enthalpy is equal to –T 2

(a)

T2

(c)

50. Given that SoCl

2

G T

(b) –T 2 V

G/T T

S oH 2

(d)

T2

1

1

V

131 JK mol

G/T T

P

G T

P

,

223 JK 1mol 1 and

SoHCl 187 JK 1mol 1 . The standard entropy change in the formation of 1 mole of HCl(g) from H 2 (g ) and Cl 2 (g ) will be

(a) 20 JK–1 (b) 10 JK–1 –1 (c) 187 JK (d) 374 JK–1 51. What is the entropy change (in JK–1 mol–1) when one mole of ice is converted into water at 0º C? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol–1 at 0ºC) (a) 21.98 (b) 20.13 (c) 2.013 (d) 2.198 52. Two substances R and S decompose in solution independently, both following first order kinetics. The rate constant of R is twice that of S. In an experiment, the solution initially contained 0.5 millimoles of R and 0.25 of S. The molarities of R and S will be equal just at the end of time equal to (a) twice the half life of R (b) twice the half life of S (c) the half life of S (d) the half life of R

MT-97

Mock Test 9 53. The first order rate constant for a certain reaction

54.

55.

56.

57.

increases from 1.667 × 10 6 s 1 at 727ºC to 1.667 × 10–4 s–1 at 1571ºC. The rate constant at 1150ºC, assuming constancy of activation energy over the given temperature range is (a) 3.911 × 10–5 s–1 (b) 1.139 × 10–5 s–1 (c) 3.318 × 10–5 s–1 (d) 1.193 × 10–5 s–1 Law of mass action can be applied to aqueous solutions of weak electrolyte since(a) Reaction is always carried out in closed vessel. (b) Pressure, temperature of surroundings remain constant. (c) State of equilibrium exist between the ionised and unionised molecules (d) Nothing can be predicted A 0.5 M NaOH solution offers a resistance of 31.6 ohm in a conductivity cell at room temperature. What shall be the approximate molar conductance of this NaOH solution if cell constant of the cell is 0.367 cm–1 . (a) 234 S cm2 mole–1 (b) 232 S cm2 mole–1 (c) 4645 S cm2 mole–1 (d) 5464 S cm2 mole–1 An electrochemical cell is set up as: Pt; H2 (1atm)|HCl(0.1 M) || CH3COOH (0.1 M)| H2 (1atm); Pt. The e.m.f of this cell will not be zero, because (a) the temperature is constant (b) e.m.f depends on molarities of acids used (c) acids used in two compartments are different (d) pH of 0.1 M HCl and 0.1 M CH 3COOH is not same The half cell potential for the quinhydrone electrode O

(a)

CH3

C6H5 C — O — OH respectively CH3 (b) CH3CH2– O – O– CH2CH3 and CH3 C6H5 CH

+

OOH

(c)

C6H5 C — O — OH respectively CH3 CH3 |

(d) No reaction and C 6 H 5 C – OOH |

CH3

59.

60.

61. OH

set up at pH = 4 will be (quinhydrone = 1 : 1 molecular compound of quinone (Q) and hydroquinone (QH2), Eº = 0.699 V ) (a) 0.699 V (b) 0.463 V (c) 0.935 V (d) 0.817 V 58. Predict the compounds A and B in the following reactions CH 3CH 2 – O CH 2 CH 3

C6H5CH

CH3

CH3

O2

O2, 95 – 135º

h

B

A ;

|

CH 3 C H – O – CH 2 CH 3 and

CH3

–

O

respectively

CH2 – O – OH

OH

+ 2H + 2e

CH 3CH 2 – O O CH 2 CH 3 and

respectively Which of the following diols would cleave into two fragments with HIO4 (a) 1, 3-hexanediol (b) 2, 4-hexanediol (c) 1, 6-hexanediol (d) 3, 4-hexanediol One organic alcohol is irradiated with IR source and it produces three streching frequency such as 3391 cm–1, 2981 cm–1 and 1055 cm–1. Name of that organic alcohol is (a) Methanol (b) Ethanol (c) Isobutanol (d) Isopropanol Which of the following reactions will not result in the formation of anisole? (a) Phenol + dimethyl sulphate in presence of a base (b) Sodium phenoxide is treated with methyl iodide (c) Reaction of diazomethane with phenol (d) Reaction of methylmagnesium iodide with phenol

OH 62.

CHCl3 KOH

X

50% KOH

The final product of this reaction is/are:

EBD_7443 MT-98

Target VITEEE

OH

OH

(b) X is

CH

(a)

and Y is

OH

OH

CH2OH

(b)

COOH +

(c) OH

(c) Both are

COOK

+ (d) Both are

COOK

67. The condensation product of benzaldehyde and acetone is O || (a) C 6 H 5 CH CH C CH 3

O COOK

C

(d) 63.

(b)

In the reaction, C6 H5 OH

64.

65.

66.

C 6 H 5CH C(CH 3 ) 2

NaOH

(A)

CO 2 140 C,(4–7 atm)

(B)

HCl

The compound(C) is: (a) benzoic acid (b) salicylaldehyde (c) chlorobenzene (d) salicylic acid An ester (A) with molecular fomula, C9H10O2 was treated with excess of CH3MgBr and the complex so formed was treated with H2SO4 to give an olefin (B). Ozonolysis of (B) gave a ketone with molecular formula C8H8O which shows +ve iodoform test. The structure of (A) is (a) C6H5COOC2H5 (b) C2H5COOC6H5 (c) H3COCH2COC6H5 (d) p — H 3CO — C 6 H 4 — COCH 3 The reagent (s) which can be used to distinguish acetophenone from benzophenone is (are) (a) 2,4- Dinitrophenylhydrazine (b) Aqueous solution of NaHSO3 (c) Benedict reagent (d) I2and Na2CO3. In the following reactions, X and Y are Y (a) X is and Y is

(c)

(C) :

68.

C 6 H 5 CH 2

O || C

CH

CH 2

O || (d) C H C CH CH.CH 3 6 5 During reduction of aldehydes with hydrazine and potassium hydroxide, the first is the formation of : — N — NH 2 (a) R — CH —

(b) (c)

R —C N R — C — NH 2

(d)

O R — CH— — NH

||

69. Which one of the following has maximum acid strength? (a) p-nitrophenol (b) p-nitrobenzoic acid (c) m-nitrobenzoic acid (d) o-nitrobenzoic acid 70. Which of the following does not give benzoic acid on hydrolysis? (a) Phenyl cyanide (b) Benzoyl chloride (c) Benzyl chloride (d) Methyl benzoate 71. When succinic acid is heated, product formed is: (a) succinic anhydride (b) acetic acid (c) CO2 and methane (d) propionic acid

MT-99

Mock Test 9 72. Which of the following carboxylic acids undergoes decarboxylation easily ? (a) C6H5CHOHCOOH (b) C6H5COCOOH (c) C6H5COCH2COOH (d) C6H5COCH2CH2COOH 73. The reaction of HCOOH with conc. H2SO4 gives (a) CO2 (b) CO (c) Oxalic acid (d) Acetic acid 74. If acetylchloride is reduced in the presence of BaSO 4 and Pd, then (a)

CH 3CHO is formed

(b)

CH 3COOH is formed

(c)

CH 3CH 2OH is formed

(d) CH 3COCH 3 is formed 75. The treatment of acylazide (RCON3) with acidic or alkaline medium gives (a) RCONH2 (b) R – NH2 (c) RCH2 NH2 (d) RCOCHNH 76. Which of the following compounds will form alcohol on treatment with NaNO2, HCl/H2O at 0°C? (a) (CH3)2CHNH2 (b) C6H5NH2 CH3

NH 2

(d) H2N

NH2

(c)

79.

80.

During the preparation of arenediazonium salts, the excess of nitrous acid, if any is destroyed by adding. (a) Aq NaOH (b) Aq Na2CO3 (c) Aq NH2CONH2 (d) Aq KI A nitrogen containing organic compound gave an oily liquid on heating with bromine and potassium hydroxide solution. On shaking the product with acetic anhydride, an antipyretic drug was obtained. The reactions indicate that the starting compound is : (a) aniline (b) benzamide (c) acetamide (d) nitrobenzene

PART - III (MATHEMATICS) 81.

82.

If A is a non-zero column matrix of order m × 1 and B is a non-zero row matrix of order 1× n, then rank of AB is equal to (a) n (b) m (c) 1 (d) None of these The number of solutions of equation x 2 – x 3 = 1, – x1 + 2 x 3 = 2, x1 – 2x 2 = 3 is (a) zero (b) one (b) two (d) infinite

83.

Shaded region in the following figure is represented by Y

(0,20) x + y = 20 20 , 40 B C(10,16) 3 3

77. The refluxing of (CH3)2NCOCH3 with acid gives (a) 2 CH3NH2 + CH3COOH (b) 2 CH3OH + CH3CONH2 (c) (CH3)2NH + CH3COOH (d) (CH3)2NCOOH + CH4 78. Which of the following reactions will not give N, N- dimethyl benzamide ? (a)

CO.O.C 2 H5 (CH3 ) 2 NH

(b)

CONH 2 CH 3Mg I

(c)

COCl (CH 3 ) 2 NH

(d)

CO.O.CO.

+ (CH 3 ) 2 NH

2x+5y = 80

A(20,0)

(a) (b) (c) (d)

X (40,0)

2x 5y 80, x y 20, x

0, y 0

2x 5y 80, x y 20, x

0, y 0

2x 5y 80, x y 20, x

0, y 0

2x 5y 80, x y 20, x

0, y 0

84.

If | a

b | |a

85.

and b are adjacent sides of (a) a rectangle (b) a square (c) a rhombus (d) None of these If three positive numbers a, b, c, are in H.P. then (a) b 2 ac (c) Both correct

b | then the vectors a

(b) a n c n 2 b n (d) none correct

EBD_7443 MT-100

86.

Target VITEEE 2 sin sin 1 cos

If

1 cos

then

87.

=

1 (a) y (c) 1 – y If

(b) y (d) 1 + y

1 sin 1 5

88.

sin

1 sin

1

sin

(c) a = 3 , b = 3 (d) None of these 92. The length intercepted by a line with direction ratios 2, 7, –5 between the lines

= y,

sec 1(2)

1 2

2tan 1

1 3

sec 1 (5)

2 tan 1( 3) = k , then k =

(a) 1 (c) 4 If the vectors

(b) 2 (d) 5

aˆi ˆj kˆ ,

ˆi bˆj kˆ ,

ˆi ˆj ckˆ

89.

1 is 1 c (a) 1 (b) –1 (d) None (c) 2 A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is ( , , ) . The equation of the plane is

(a)

1 1 a

x

(b) (c) 90.

91.

1 1 b

y

x x

z

y y

3

1 2

(d) None of these

A sphere is inscribed in a cube of side 2a. If the cube is bounded by the coordinate planes as its three faces, then the equation of the sphere is (a)

x2

y2

z 2 2ax 2ay 2az a 2

(b)

x2

y2

z 2 2ax 2ay 2az 2a 2

(c)

x2

y2

z 2 ax ay az a 2

(d)

x2

y2

z2

0 0

0

4a 2

Let a , b R , such that 0 0, (dG)T, P < 0 (b) (dS)V, E = 0, (dG)T, P = 0 (c) (dS)V, E = 0, (dG)T, P > 0 (d) (dS)V, E < 0, (dG)T, P < 0 Bordeaux used as fungicide is a mixture of (a)

CuSO 4

Ca OH

2

(b)

CaSO 4

Cu OH

2

(c) CuCO3 Cu OH 2 (d) CuO CaO The conductivity of a saturated solution of BaSO4 is 3.06 × 10–6 ohm –1 cm –1 and its equivalent conductance is 1.53 ohm –1 cm 2 equiv–1. The Ksp for BaSO4 will be (a) 4 × 10–12 (b) 2.5 × 10–9 –13 (c) 2.5 × 10 (d) 4 × 10–6 In a uranium mineral, the atomic ratio NU–238/NPb–206 is nearly equal to one. The age (in years) of the mineral is nearly (given that the half-life of U-238 is 4.5 × 109 years) (a) 3.0 × 108 (b) 4.5 × 108 9 (c) 3.0 × 10 (d) 4.5 × 109 M is started with 10.0 g of The reaction L L. After 30 and 90 minutes 5.0 g and 1.25 g of L respectively are left. The order of the reaction is (a) 0 (b) 1 (c) 2 (d) 3

MT-109

Mock Test 10 56.

57.

58.

59.

60.

61.

62.

Metallothermic processes involving Lanthanides are called as (a) Aluminothermic process (b) Lanthanido-thermic process (c) Reduction process (d) Oxidation process Ethanol when reacted with PCl5 gives A, POCl3 and HCl. A reacts with silver nitrite to form B (major product) and AgCl. A and B respectively are (a) C2H5Cl and C2H5OC2H5 (b) C2H6 and C2H5OC2H5 (c) C2H5Cl and C2H5NO2 (d) C2H6 and C2H5NO2 A compound of the formula C4H10O reacts with sodium and undergoes oxidation to give a carbonyl compound which does not reduce Tollen’s reagent, the original compound is (a) Diethyl ether (b) n-Butyl alcohol (c) Isobutyl alcohol (d) sec-Butyl alcohol If Germanium crystallises in the same way as diamond, then which of the following statement is not correct (a) Every atom in the structure is tetrahedrally bonded to 4 atoms. (b) Unit cell consists of 8 Ge atoms and coordination number is 4. (c) All the octahedral voids are occupied. (d) All the octahedral voids and 50% tetrahedral voids remain unoccupied. Which of the following cannot be made by using Williamson’s synthesis? (a) Methoxybenzene (b) Benzyl p-nitrophenyl ether (c) Methyl tertiary butyl ether (d) Di-tert-butyl ether The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23, 24, 25, and 26. Which one of these may be expected to have the highest second ionization enthalpy ? (a) Cr (b) Mn (c) Fe (d) V The following reaction is known as : OH

HCl HCN

(a) (b) (c) (d)

anhy. ZnCl 2

Perkin reaction Gattermann reaction Kolbe reaction Gattermann-aldehyde reaction

OH CHO

63.

For the reaction CH 4 (g) 2O 2 (g) CO2 (g) + 2H2O(l), –1 rH = –170.8 k J mol Which of the following statements is not true ? (a) The equilibrium constant for the reaction is given by K p

[CO 2 ] [CH 4 ][O 2 ]

(b) Addition of CH4 (g) or O2 (g) at equilibrium will cause a shift to the right (c) The reaction is exothermic (d) At equilibrium, the concentrations of CO2 (g) and H2O(l) are not equal CH 3

64.

CH 3

CH 3

65.

The above compound describes a condensation polymer which can be obtained in two ways : either treating 3 molecules of acetone (CH3 COCH3) with conc. H2SO4 or passing propyne (CH3 C CH) through a red hot tube. The polymer is (a) Phorone (b) Mesityl oxide (c) Deacetonyl alcohol (d) Mesitylene. The pair in which both species have same magnetic moment (spin only value) is : (a) [Cr(H2O)6]2+ , [CoCI4]2– 2 3 (b) [Cr(H 2O)6 )] , [Fe(H 2O)6 ]

(c)

66.

[Mn (H 2 O) 6 ) 2 , [Cr(H 2 O) 6 ]2

(d) [CoCl 4 ) 2 , [Fe(H 2 O) 6 ]2 Which one of the following pairs is not correctly matched ? (a) C O CH 2 (Clemmensen 's reduction) (b)

C

(c)

COCl

(d)

C

O

N

CHOH (Wolf -Kishner 's reduction)

CHO (Rosenmund 's reduction) HO (Stephen 's reduction)

EBD_7443 MT-110

67.

68.

69.

Target VITEEE

A drop of a solution (volume = 0.05 ml) contains 6 10 7 mol of H+. If rate of disapperanace of H+ is 6.0 × 105 mol lit–1 s–1, how long will it take for the H+ in the drop to disappear ? 2

(a)

2.0 10

s

(b)

(c)

8.0 10 8 s

2 .0 10

8

73.

74.

Which of the following has Frenkel defects? (a) Sodium chloride (b) Graphite (c) Silver bromide (d) Diamond Which of the following will give maximum number of isomers? (a) [Co(NH3)4Cl2] (b) [Ni(en)(NH3)4]2+ (c) [Ni(C2O4)(en)2] (d [Cr(SCN)2(NH3)4]2+

75.

232 90 Th

s

6

(d)

6.0 10 s The weakest acid among the following is (a) CHCl2COOH (b) BrCH2COOH (c) ClCH2COOH (d) FCH2COOH A compound of a metal ion M x Z

24 has a

spin only magnetic moment of 15 Bohr Magnetons. The number of unpaired electrons in the compound are (a) 2 (b) 4 (c) 5 (d) 3 C N

76.

+

+ CH3MgBr

70.

Q

H3 O

P

77.

OCH3

The product ‘P’ in the above reaction is CH

OH

CO

CH3

CH3

(a)

(b) OCH3

CHO

(c) 71.

COOH

78.

(d)

OCH3 OCH3 The reaction of acetyl chloride on benzene in the presence of AlCl3 gives

(a)

C6 H5Cl

(b)

(b) Propanone and mesityl oxide (c) Propan one and 2,6-dimeth yl-2, 5-heptadien-4-one (d) Propanone and mesitylene oxide

(c)

72.

(a)

(c)

C

(b)

NHCOCl

Cl

208 terminates at 82 Pb . The number of and particles emitted in this process is (a) 6 and 4 (b) 4 and 6 (c) 10 and 6 (d) 6 and 10 Which of the following shows oxidation state of +8 ? (a) Ir (b) Pt (c) Ru (d) Os An electrochemical cell is set up as follows : Pt (H2, 1 atm)/0.1 M HCl/0.1 M acetic acid/(H2, 1 atm) Pt EMF of this cell will not be zero because (a) the temperature is constant (b) the pH of 0.1 M HCl and 0.1 M acetic acid is not the same (c) acids used in the two compartments are different (d) EMF of a cell depends on molarities of the acids used An organic compound ‘A’ has the molecular formula C3H6O. It undergoes iodoform test. When staturated with HCl it gives ‘B’ of molecular formula C 9 H 14 O. ‘A’ an d ‘B’ respectively are

(a) Propanal and mesitylene

C6 H5COCH3

CH3 CO CH3 (d) C 6 H 4 Cl2 Aniline reacts with phosgene and KOH to form OH O

is the starting material of decay of chain series. This series of successive decays

79.

Which of the following is not formed when glycerol reacts with HI? (a) CH2 = CH – CH2I

(d)

NCO

(b) CH2OH–CHI–CH2OH (c) CH3–CH = CH2 (d) CH3–CHI–CH3

MT-111

Mock Test 10 80.

The factor of G values is important in metallurgy. The G values for the following reactions at 800ºC are given as : 2O 2 ( g )

2SO 2 ( g ) ; G =–544 kJ

2 Zn (s )

S 2(s)

2 ZnS (s ) ; G = –293 kJ

2 Zn (s )

O 2( g )

2 ZnO (s ) ; G = – 480 kJ

S 2 (s )

85.

86.

Then G for the reaction : 2 ZnS (s )

3O 2 ( g )

2 ZnO (s )

will be : (a) –357 kJ (c) –773 kJ

2SO 2( g )

87.

(b) –731 kJ (d) –229 kJ 88.

PART - III (MATHEMATICS) If z = rei , then |eiz| equals (a) er sin (b) e–r sin –r cos (c) e (d) er cos 82. The equation of two lines through the origin, x 3 y 3 z at which intersect the line 1 1 2

(c) (AB) B A (d) AB = 0 A = O or B = O 89.

cos (90 sin (360

90.

equals (a) 2 (b) 1 (c) –1 (d) 0 Let the angles A, B, C of ABC be in A.P. and let

angles of (a) (b) (c) (d) 83.

84.

1 i = A + iB, then A2 + B2 equals to 1 i 2 (a) 1 (b) (c) –1 (d) – 2 Which of the function defined below is one to one?

f : (0,

91.

(b)

f : [0,

92.

)

x R , f (x) = e

(d) f : R

R, f (x)

1

93.

ex

n(x 2

x 1)

If

1

(a)

R , f (x) = x2 + 4x – 5

(c) f : R

Solution of y is 0 y (a) 5 /4 (c) /2

2

R , f(x) = x2 – 4x + 3

)

) sec ( ) tan (180 ) )sec (180 ) cot (90 )

b : c = 3 : 2 . Then angle A is (a) 75° (b) 45° (c) 60° (d) none of these

If

(a)

value of a is: (a) a < 12 (b) a 12 (d) None of these (c) a 12 Number of different 3-letter words can be formed with the letter of word 'JAIPUR' when A and I are always to be excluded is (a) 24 (b) 4 (c) 20 (d) 10 If A and B are two matrices and (A + B) (A – B) = A2 – B2, then (a) AB = BA (b) A2 + B2 = A2 – B2 (c) A'B' = AB (d) none of these If A and B are any two square matrices of the same order, then (a) (AB) A B (b) adj(AB) = adj(A) adj(B)

81.

each, are 3 x y z x y z ; 1 2 1 1 1 2 z x y z x y ; 1 1 1 2 1 2 x y z x y z ; 1 1 1 2 1 2 None of the above

a 2x , 5 + 5–2x are in A.P., then the 2

51+x + 51–x,

sin

cos 1

1

1

cos 1 ( ( 3 / 2)) for y, where

4 sin 5

4 cos 5 2

(b) /6 (d) 5 /6 1

1 and 3

11

3

then (b)

1

2

(c) (d) none of these 1 2 Value of for which the line y = x + touches the ellipse 9x2 + 16y2 =144, is (a) ± 5 (b) ± 2 (c) ± 8 (d) ± 7

EBD_7443 MT-112

94.

Target VITEEE

(at2 ,

The point x

2

y

a

2

b2

2bt) lies on the hyperbola,

2

1 for

(a) all values of t 95.

(b)

t2

a2

y2

+

= 1 to the tangent and normal at a

b2

point whose eccentric angle is

(a)

(c)

97.

5

(c) t 2 2 (d) no real value of t. 5 The area of the rectangle formed by the perpendiculars from the centre of the ellipse x2

96.

2

a2

b 2 ab

a2

b2

a2

b2

ab a 2

b2

4

a2

(b)

(d)

b2 ab

a2

b2

a2

b2 b2

For all values of x, function f (x) = 2x3 + 6x2 + 7x – 19 is (a) monotonic increasing (b) monotonic decreasing (c) not monotonic (d) none of these x sin ,0 x 1 f (x) = , then 2 3 2x, x 1

(a)

99.

3 7

f (x)

ex

e

x

ex

e

x

4 3 (c) (d) 3 4 Let f '(x) < 0 and g' (x) > 0 for all real x, then (a) f (g(x + 1)) > f (g (x + 5)) (b) f (g(x)) < g (f(x +2)) (c) g (f(x)) < g (f (x +2)) (d) g (f (x)) > g(f (x – 2))

(b) loge

x 1 3 x

1/ 2

(a)

1/2

1/ 2

x 3 z 2 (z x) 2

0 3

8xz 4x

2

3

8xz

z

(b)

11 / 3

2

(c) 2

21 / 3

4

is equal to

1 23 / 3

2 .z (d) None of these

z

105. The intergral

0 , then the

7 (b) 3

1/ 2

x 3 x 1

x 2 x 1 (d) loge x 3 x 2 102. Area bounded by the circle x2 + y2 = 1 and the curve | x | + | y | = 1 is (a) 2 (b) – 2 (c) (d) + 3 cosax cos bx m 103. lim is equal to , where m and x 0 cos cx 1 n n are respectively (a) a2 + b2, c2 (b) c2, a2 + b2 2 2 2 (c) a – b , c (d) c2, a2 – b2

x

13

2 is

(c) loge

104. lim

(a) f (x) has a local minimum at x = 1 (b) f (x) has a local maximum at x = 1 (c) f (x) does not have any local maximum or minimum at x = 1 (d) f (x) has a global minimum at x = 1 98. If the points (1, 1, p) and (–3, 0, 1) be equidistant from the plane r. 3iˆ 4j 12kˆ value of p is

10! (b) 3! 7! 6 10 (d) None of these (c) P3 . 7! 101. The inverse of the function

(a)

(a) loge

, is

ab a 2

100. Ten persons, amongst whom are A, B and C to speak at a function. The number of ways in which it can be done if A wants to speak before B and B wants to speak before C is

dx x 1 x3

(a)

1 1 x3 1 log 3 1 x3 1

C

(b)

1 1 x3 1 log 3 1 x3 1

C

(c)

2 1 log 3 1 x3

(d)

1 log | 1 x 3 | C 3

C

is equal to

MT-113

Mock Test 10 106. A function f (x) which satisfies the relation f (x)

e

x

1

x

e f (x) dt , then f (x) is 0

2a

ex (a) 2 e (c) 2ex

2) ex

(b) (e – (d)

ex/2

107. The circle x2 + y2 = a2 is revolved about x-axis. Then the volume of the sphere so formed is (a)

4 2 a cub. units (b) 3

4 a cub. units 3

4 3 a cub. units (d) none of these 3 108. The solution of differential equation

(c)

dy – y = 3 represents a family of dx (a) circles (b) straight lines (c) ellipses (d) parabola 109. The general solution of the differential equation

2x

dy = (ex – e–x) is (ex +e–x) dx (a) y = log |ex + e–x| + c (b) y = log | ex – e–x | – c (c) y = – log | ex –e–x| +c (d) none of these

110. If vectors a c

3iˆ

113. If a , b and c are unit coplanar vectors, then the scalar triple product

iˆ

b, 2 b

(a) 0

(b) 1

(d) 3 3 114. If p, q, r are statement with truth values F, T, F respectively then the truth value of (~ p ~ q) r is (a) true (b) false (c) false if r is true (d) false if q is false 115. The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is (a)

28 256

(b)

219 256

(c)

128 256

(d)

37 256

116. A random variable X has Poisson distribution with mean 2. Then P (X > 1.5) equals

2

pjˆ 5kˆ are coplanar, then the value of

p is (a) 2 (b) 6 (c) – 2 (d) – 6 111. The angle between the lines 2x = 3y = –z and 6x = –y = –4z, is (a) 0° (b) 30° (c) 45° (d) 90° 112. Six dice are thrown. The probability that different numbers will turn up is (a)

129 1296

(b)

1 54

(c)

5 324

(d)

5 54

(b) 0

e2

(c) 1

iˆ 2 ˆj kˆ and

a

(c)

(a)

ˆj kˆ , b

c, 2 c

3

(d)

2

3 2

e e 117. The normal at the point (at 12, 2at 1) on the parabola, cuts the parabola again at the point whose parameter is (a)

t2

(c)

t2

2 t1

t1 t1

2 t1

(b)

t2

t1

2 t1

(d) None of these

118. For real numbers x and y, wewrite x R y x – y + 2 is an irrational number. Then, the relation R is (a) Reflexive (b) Symmetric (c) Transitive (d) None of these 119. The set Z' of all non-negative integers under addition is (a) a monoid (b) not a monoid (c) semi-group (d) none

EBD_7443 MT-114

Target VITEEE

120. Consider Max. z = – 2x – 3y subject to x y x y 1, 1 , x, y 0 3 2 2 3 The max value of z is : (a) 0 (b) 4 (c) 9

(d) 6

PART - III (ENGLISH) Direction (Qs 121 - 123) : Read the passage carefully and answer the questions given below Sometimes we went off the road and on a path through the pine forest. The floor of the forest was soft to walk on; the frost did not happen it as it did the road. But we did not mind the hardness of the road because we had nails in the soles and heels nails bit on the frozen ruts and with nailed boots it was good walking on the road and invigorating. It was lovely walking in the woods. 121. 'Frozen ruts' means (a) very cold roads (b) wheel marks in which frost had become hard (c) the road covered with frost (d) hard roads covered with snow

122. We did not mind the hardness of road because (a) we had nailed boots on (b) it was good walking on the road (c) the walk was refreshing (d) the nails bit on the frozen roads 123. Sometimes we walked through the pine forest as (a) the path was unaffected by the frost (b) it was good walking with nails in the boots (c) the walks was invigorating (d) it was sheer joy to walk in the forest 124. To drive home (a) To find one's roots (b) To return to place of rest (c) Back to original position (d) To emphasise 125. Choose the best pronunciation of the word, Extempore, from the following options. (a) ex-tem-pour (b) ex-tem-purree (c) ex-tem-poray (d) ex-tem-pure

MT-115

Solutions

SOLUTIONS MOCK TEST 1 1.

(c)

2.

(d)

E (ds) cos

3.

(d) Y

2 (1

4.

(a) Charge Q=C1V Total capacity of combination (parallel) C = C1+C2

s>

k> r

E ( r 2 ) cos 0º 0.5Y

)

A

Q C1V C C1 C 2 (d) At A current is distributed and at B currents are collected. Between A and B, the distribution is symmetrical. It has been shown in the figure. It appears that current in AO and OB remains same. At O, current i4 returns back without any change. If we detach O from AB there will not be any change in distribution. Now, CO & OD will be in series hence its total resistance = 2 It is in parallel with CD, so, equivalent

6. 7.

2 1 2 2 1 3 This equivalent resistance is in series with AC & DB, so, total resistance 2 1 1 3

8 3

i4

i2

R

D

i1

O

0

16 14

U ( x)

B

i1

8 is parallel to AB, that is, 2 3 total resistance

16 / 3 14 / 3

2R

8 7

x

Fdx or

U=

Now

8/3 2 8/3 2

R

(b) We know that U = – W Therefore

i2

A

so

B

2R

R

x

i4

r7

R

8.

i3

O

r1

r6

8 1 3 1 7 or R R 8 2 8 7 (b) The first law concerns the conservation of charge. (d) Since, wheatstone's bridge is balanced, then resistance of galvanometer will be uneffective.

R

C

r5

Between C & D, the equivalent resistance is given by 1 1 3 1 1 1/ r 2 2 r3 (r4 r5 ) Equivalent resistance along 2 8 1 ACDB 1 3 3 Effective resistance between A and B is

resistance =

=

D

r4

r2

[Alt :

P.D.

5.

r3

C

r2 E .

U

kxdx 0

U ( 0)

kx 2 given U(0) = 0, 2

kx 2 2 9. (b) The current is same in both the wire, hence magnetic field induced will be same. 10. (b) In a perpendicular magnetic field, the path of a charged particle is a circle, and the magnetic field does not cause any change in energy. U(x) =

EBD_7443 MT-116

11. (c)

Target VITEEE Power = V . I = I2R i2

Power R

2 8

1 4

Therefore refractive index of the medium

1 Potential over 8 = Ri 2 8 4V 2 This is the potential over parallel branch. So, 4 1 i1 4 Power of 3 = i12R = 1 × 1 × 3 = 3W output input

12.

(a) Efficiency is given by

13.

5 15 14 0.875 or 87.5 % 10 8 15 (b) Time constant is L/R Given, L = 40H & R = 8 = 40/8 = 5 sec. (c) L = 2mH, i = t2e–t

14.

(I L 2

I1 ) t

L

=

19.

17.

(0 10) 0.5

20. 21.

wave in vacuum (v0 )

b

d

velocity of electromagnetic wave in

0 0

n

D d

x green

1 ; n 2

n 2 sin

1

1 2

1 2

h mv (According to de-Broglie, wavelength

or d

associated with an electron 22.

(b) For positronium reduced mass

h ) mv mm m m

m2 m 2m 2 Hence Rydberg’s constant for positronium is half of that for hydrogen atom. Hence, ground state energy for positronium 13.6 2

and

0 0

=

g

sin 30

20L

E 0 nBA R R Given, n = 1, B = 10–2 T, A = (0.3)2m2, R = 2 f= (200/60) and = 2 (200/60) Substituting these values and solving, we get I0 = 6 × 10–3 A = 6mA (d) We know that velocity of electromagnetic I0

1

1/

(b) Material expands outward and so x, r increases. Due to linear expansion diameter of rod will increase. (c) (a) For Bragg’s scattering of a beam of electron, the inter-planar distance ‘d’ is given by 2d sin = n Here = 30°

220 11H 20 [where L = Self inductance of coil]

(c)

0 0

x blue

or, L

16.

1/

(c) Distance of n th maxima, x As

di L[ t 2 e t 2 te t ] E= dt when E = 0, –e–t t2 + 2te–t = 0 or, 2t e–t = e–t t2 t = 2 sec. (c) Initial current (I1) = 10 A; Final current (I2)= 0; Time (t) = 0.5 sec and induced e.m.f. ( ) = 220 V. Induced e.m.f. ( )

= L dI dt

Vel. of E.M.wave in vacuum (v 0 ) Vel. of E.M. wave in medium (v)

( )

18.

L

15.

1 .

medium is (v)

1 2

23.

(b)

eV

hc min

6.8 eV

or

min

hc eV

MT-117

Solutions 24.

(b) Using formula n 2d sin

d d

25.

(c)

20 nm

mo

mo

m

1/ 2

v2

1 34

1/ 2

2m o

29. 30.

(c) No. n2

31.

32.

34.

(c)

35.

(c)

36. 37.

(c) (b)

38.

(a)

39.

(c) As velocities are exchanged on perfectly elastic collision, therefore masses of two objects must be equal.

c2

(c) (a) From the graph it is clear that A and B have the same stopping potential and therefore the same frequency. Also B and C have the same intensity. (b) If a body slides down, then the force of friction acts upwards along the plane weight(mg) act vertically downwards. (c) Energy released per fission is 200 MeV

28.

of

half

80 40

n1

lives

80 20

4;

2 n1

N1 N 0A

1 2

N2 N 0B

1 2

N1 N2

1 4 16 1

1 2

n2

1 2

1 4

4

1 16

for A

1 4

for B

( N 0A

N 0B )

2

364 10 3 0.04 A 9.1 Peak value Vm = P.D. across R – (– Vm) = – (Vm + Vm) = –2Vm When the collector is positive and emitter is negative w.r.t. base it causes the forward biasing for each junction, which causes conduction of current. (W + X) . (W + Y) = W + (X . Y) The process of changing the frequency of a carrier wave (modulated wave) in accordance with the audio frequency signal (modulating wave) is known as frequency modulation (FM). Maximum current

3

20 10 1 2 2

(a) Applying formula

3

1 20 10 2sin 30

n 2sin

1

26. 27.

33.

(a) From the graph of binding energies of the nuclei, the curve indicates that binding energy increases to a maximum of about 8.8 MeV per nucleon around mass number 56 and then decreases slowly to about 7.6 MeV per nucleon at A = 238. Thus among the options, the binding energy per nucleon is largest for 56Fe. (a) 9 × (2)3 = 9 × 8 = 72 Hence number of nucleons in the Ge nucleus = 72.

40. 41.

ma 1 or m a m b . mb (a) LED is generally and most suitably used in picture tubes of TVs. (d) d-d transition of Cr3+.

Ruby is Alumina (Al2O3), contains about 0.5 to 1.5% Cr3+ ions which are randomly distributes in place of Al3+ ions. As Cr3+ has d 3 configuration, it behaves like octahedral complex. At the points wherever they are present within the lattice, they undergo d-d transitions and thus produces colour. Ground state configuration of Cr 3+ 3 e0 ) Cr3+ 3d3 ( t2g g d xz , d xy , d yz , d

eg t2g

x 2 y2

,d

z2

d x2–y2 d z2

dxy dxz dyz

e–

Any one of the electron from t2g moves into eg state which is associated with some specific amount of energy, gives rise to colour.

EBD_7443 MT-118

42.

43. 44.

Target VITEEE

(c) We know that regular decrease in the size of the atoms and ions is called lanthanide contraction. In vertical column of transition elements there is a very small change in size and some times size is found same from second member to third member.The similarity in size of the atoms of Zr and Hf is evident due to the object of lanthanide contraction. Therefore Zr and Hf both have same radius 160 pm. (a) G° = –2.303 RT logKpwhen Kp = 1, G° = 0since log 1 = 0 (c) By carbon dating method

colour is due to the hydrated chromium (III) in [Cr(H 2 O) 6 ]3 . 51.

(d) When electrons are trapped in anion vacancies, these are called F-centres. +ve –ve ion ion

52.

F- centre in crystal (c) The no. of atoms is a unit cell may be calculated by the the formula n n n n Z= c + b + f + e 8 1 2 4 Where nc = no. of atom at the corner n b = no. of atoms at body centre nf = no. of atoms at face centre ne = no. of atoms at edge centre. An Fcc crystal contains

N0 2.303 t½ log 0.693 N

Age of wood

ratio of C14 / C12 in living wood ratio of C14 / C12 in dead wood

Hence it is based upon the ratio of C14 and C12. 45.

(c) 1 atom of

235 92 U on

fission gives energy

53.

= 3.2 × 10–11 J

6.023 × 1023 atom (1 mole) on fission gives energy = 3.2 × 10–11 × 6.023 × 1023 J 235 gm of = 46. 47.

235 92 U

on fission gives energy

[Co( NH3 ) 5 NO 2 ]Cl 2 2Cl

48.

50.

2AgNO 3

2Cl

54.

55.

2AgCl 2 NO 3

(a) Triethoxyaluminium has no Al – C linkage

Al 49.

[Co( NH 3 ) 5 NO 2 ]

O CH 2 CH 3 O CH 2 CH 3 O CH 2 CH 3

(a) MA3 B3 – 2 geometrical isomers MA2 B4 – 2 geometrical isomers MA4 B2 – 2 geometrical isomers The complexes of general formula Ma6 and Ma5b octahedral geometry do not show geometrical isomerism. (d) We know that chromium (III) salts dissolve in water to give violet solution. The violet

2 3 d 1.15Å 2 3 (where d = Difference between the scattering planes) (a) We know from the third law of thermodynamics, the entropy of a perfectly crystalline substance at absolute zero temperature is taken to be zero. (a) If more trans-2-pentene is added, then its concentration in right hand side will increase. But in order to maintain the K constant, concentration of cis-2-pentene will also increase. Therefore more cis-2pentene will be formed. (d) Rate of reaction

2 1 2.d.

6.023 3.2 1012 J = 8.2 × 107 kJ 235

(b) IUPAC name is potassium trioxalatoaluminate (III). (c) As it forms two moles of silver chloride thus it has two moles of ionisable Cl.

8 6 = + = 4 atoms in a unit cell. 8 2 (d) Given : Order of Bragg diffraction (n) = 2 ; Wavelength ( ) = 1 Å and angle ( ) = 60º. We know from the Bragg’s equation n = 2d sin or 2 × 1 = 2d sin 60º

56.

–

57.

(b)

1 d[Br ] 5 dt

1 d[Br2 ] 3 dt

d[Br2 ] 3 d[Br – ] – dt 5 dt RateI = k [A]x[B]y

Rate I 4

= k [A]x [2B]y

... (1) ... (2)

MT-119

Solutions = 4k[A]x[2B]y

or Rate1 From (1) and (2) we get

58.

59.

2B B

1 y 4 =2

y

or 2–2 = 2y or y = –2. (d) Solid Liquid It is an endothermic process. So when temperature is raised, more liquid is formed. Hence adding heat will shift the equilbrium in the forward direction. (d) 3Fe(s) + 4H2O (steam) Fe3O4 (s) + 4 H2(g) Kp =

64.

Fe 3 / Fe 2 E º 0.77 The metals having higher negative electrode potential can displaced metals having lower values of negative electrode potential from their salt solutions. (d) This is because zinc has higher oxidation potential than Ni, Cu and Sn. The process of coating of iron surface with zinc is known as galvanization. Galvanized iron sheets maintain their lustrue due to the formation of protective layer of basic zinc carbonate. (c) We know that

65.

Fe

(d)

K

(pH 2 ) 4

66.

0.693 45 2.303 100 45 log 0.693 100 99.9

2.303 45 3 1 7 hours . 0.693 2 (a) The equivalent conductance of BaCl2 at infinite dilution of BaCl2

1 2

of Cl –

of Ba

127 76 2

62.

(a) By Faraday's Ist Law,

67.

139.5 ohm –1 cm 2 W E

no. of equivalent Volume (in litre)

(d) Electron-donating groups (– OCH3, – CH3 etc.) tend to decrease an d electron withdrawing groups (– NO2, – OCH3 etc.) tend to increase the acidic character of phenols. Since – OCH3 is a more powerful electron-donating group than – CH3 group, therefore, p-methylphenol is slightly more acidic that p-methoxyphenol while pnitrophenol is the strongest acid. Thus, option (d), i .e. p-methoxyphenol, pmethylphenol, p-nitrophenol is correct. (d) With Br 2 water, phenol gives 2, 4, 6tribromophenol.

H 2O

1 100 1

= 0.01 N

R

+ 3Br2 (excess)

q 96500

W it 1 965 1 E 96500 96500 100 (where i= 1 A, t = 16×60+5 = 965 sec.) Since, we know that

OH

OH

(where q it = charge of ion ) we know that no of equivalent

Normality

OMgX R CH2 – CH2

–Mg(OH)X

=

61.

CH2 – CH2

H 2O

( pH 2 O) 4

further t

/ Fe

H2C – CH2+RMgX O

only gaseous products and reactants. 60.

Eº = –0.44

(b)

k[A]x [B]y = k[A]x[2B]y 4

1 [ B]y = [2B]y or 4 4

2

63.

Br

OH

Br +3HBr

Br 2, 4, 6 Tribromphenol

68.

(c) In the cleavage of mixed ethers with two different alkyl groups, the alcohol and alkyl iodide that form depend on the nature of alkyl group. When primary or secondary alkyl groups are present, it is the lower alkyl group that forms alkyl iodide therefore

EBD_7443 MT-120

Target VITEEE CH 3

CH3

69.

CH2OH HOOC

CH CH 2 O CH 2 CH 3 HI | CH3

|

CHOH |

CH3 | CH CH2 OH CH3CH 2 I

H3C

CH – OH

[O]

[O] H3C

|

CH 2OH

Benzaldehyde

HCOOH

Formic acid

Aniline

C 6 H 5C H

NC 6 H 5 Benzyliden e aniline

Ketone

71.

nFE H (b) Pinacolone is 3,3-dimethyl-2 butanone. CH3 | CH3 C C CH 3 | || CH 3 O

77.

78.

H 2O

This is know as Schiff’s base reaction. (d) RDX is prepared by treating urotropine with fuming nitric acid. When the inner bridge system is destroyed by oxidation and the peripheral nitrogen atom are nitrated, it forms cyclonitrite (or RDX). (d)

(d) Methyl acetate and ethyl acetate on hydrolysis give CH3COOH which is a liquid. Similarly ethyl formate on hydrolysis will give formic acid which is also a liquid. Only ethyl benzoate on hydrolysis will give benzoic acid which is a solid. (a) OH

OCOCH3

COOH + ClCOCH3 O-hydroxy benzoic acid (Salicylic acid)

acetyl chloride

COOH

Pyridine

Aspirin

(c) Tollen's reagent is used to detect of aldehydes. Tollen's reagent is an ammonical solution of silver nitrate. When aldehyde is added to Tollen's reagent, silver oxide is reduced to metallic silver which deposits as mirror. RCHO + Ag2O

75.

|

CHOH

C6 H 5CH O C 6 H 5 NH 2

C=O

(c) Thermodynamic efficiency is given by

74.

CHOH

(c) Benzaldehyde reacts with primary aromatic

CH 3 CH 2 COOH

70.

73.

CH 2OH H 2O

amine to form schiff's base

isopropyl alcohal

72.

|

CH 2OH

|

n propyl alcohal

CH 3 CH 2 CHO

|

CHOH

CH 2OH

76.

[O]

oxalic acid

CH2OOCH CO2

|

100 110 C H 2O

HOOC

CH2OH

(c) Primary alcohol on oxidation give aldehyde which on further oxidation give carboxylic acid whereas secondary alcohols give ketone.

CH 3CH 2 CH 2 OH

CH 2OOC.COOH

|

RCOOH + 2Ag

(c) When glycerol is heated with oxalic acid following reaction occurs.

Benzyl alcohol

C N

79.

(b)

+ CH3MgBr

OCH3 H3C –C = NH

H3C –C = NMgBr H 3O

OCH3

+

OCH3

MT-121

Solutions COCH3

84. (a) Since, b, c and b c are mutually perpendicular vectors, therefore any vector

OCH3

80.

81.

82.

1 i 1– i

1 i2

1 i 1 i 1– i 1 i

1 i 1– i

83.

b, c and b

Let a x b

n

1

2i

1– i in

2

2i 2

tan x

tan x tan / 3 tan x tan 2 / 3 1 tan x tan / 3 1 tan x tan 2 / 3

tan x

tan x

tan x

z

a. b

tan x

3

1

3 tan x

(tan x

3)(1 (1

tan x 1

3

3 tan x

3 tan x)

3 tan x) (tan x 1 3tan 2 x

8 tan x 1 3 tan 2 x

3

tan x (1 3 tan 2 x ) 8 tan x 1 3 tan 2 x

3(3 tan x tan 3 x )

3

3 1 3 tan 2 x 3 tan 3x 3 tan 3x 1

3)

0 0 z | b c |2

c)

a . ( b c) | b c |2

x b . b y c . b z .0

x

Taking dot product with c in eq. (i), we get a. c

0 y 0

y

a. c

a ( a . b) b ( a . c ) c

86.

3

c in eq. (i),

Taking dot product with b in eq. (i), we get

1

Clearly the smallest value of n is 4. (c) The given equation can be written as

....(i)

y c z ( b c)

we get, a. ( b

85.

i

c.

Taking dot product with b

C 2 H 5 NH 2 NH 3 . This anomolous behaviour of tertiary ethyl amine is due to steric factors i.e., crowding of alkyl groups cover nitrogen atom from all sides and thus makes the approach and bonding by a lewis acid relatively difficult which results the maximum steric strain in tirtiary amines. The electrons are there but the path is blocked resulting the reduction in its basicity. (c) R A × B under given condition a < b is given by R = {(1, 3), (1,5), (2,3) (2,5), (3,5) (4,5)} R–1 = {(3,1), (5, 1), (3,2), (5,2), (5, 3), (5,4)} RoR–1 = {(3, 3), (3, 5), (5,3), (5, 5)}. (a)

can be expressed in ter ms of

a

(d) All aliphatic amines are stronger bases than NH3 and among different ethylamines order of basictity is 2° > 3° > 1°. Thus, the correct order is (d) i.e., (C 2 H 5 ) 2 NH (C 2 H 5 )3 N

a . ( b c) | b c |2

( b c)

(c) Since y 1 | x | 1 x R, therefore, graph of the given function must lie on or above the line y = 1. hence only (c) option is correct. (b) Let S x2 – 4y Since the point (2a, a) lies inside the parabola, S (2a, a) = 4a2 – 4a < 0 or a (a – 1) < 0 ...(1) Also, the vertex A (0, 0) and the point (2a, a) are on the same side of the line y = 1 (the equation of latus rectum) So, a – 1 < 0 i.e. a < 1 ...(2)

Y 3

S (0, 1)

H3 O

+

(2a, a) 3

A

y=1

X

From (1) and (2), we have a (a – 1) < 0 or 0 < a < 1

EBD_7443 MT-122

Target VITEEE

87. (c)

Let L1 : x + 2y = 8; L2 2x + y = 2; L3 : x – y = 1 Since the shaded area is below the line L1, we have x + 2y 8. Since the shaded area is above the line L2, we have 2x + y 2. Since the common region is to the left of the line L3, we have x – y 1 88. (a) We have, (3 – x)4 + (2 – x)4 = (5 – 2x)4 = (3 – x + 2 – x)4 (3 – x)4 + (2 – x)4 = (3 – x)4 + (2 – x)4 + 4 (3 – x) (2 – x)3 + 4 (3 – x)3 (2 – x) + 6 (3 – x)2 (2 – x)2 2 (3 – x) (2 – x) [2 (3 – x)2 + 3 (3 – x) (2 – x) + 2 (2 – x2)] = 0 (3 – x) (2 – x) (18 + 2x2 – 12x + 18 – 9x – 6x + 3x2 + 8 + 2x2 – 8x) = 0 (3 – x) (2 – x) (7x2 – 35x + 44) = 0 x = 2, 3 and 7x2 – 35x + 44 = 0 Since discriminant of 7x2 – 35x + 44 = 0 is negative, it has no real roots. Hence the gievn equation has two real roots and two imaginary roots. 89. (c) Let A, E1, E2, E3 and E4 be the events as defined below A : a black ball is selected E1 : box I is selected E2 : box II is selected E3 : box III is selected E4 : box IV is selected Since, the boxes are chosen at random, Therefore, P(E1) = P(E2) = P(E3) = P(E4) = Also, P(A/E1) =

P(E3/A) = P E3 P A / E3 P E1 P A / E1

+ P(E4)P(A/E4) =

90.

(c).

1 1 4 7 1 1 1 1 4 4 4 7

1 3 4 18

p

1 4 4 13

= 0.165

q y r z

p x

q

r z

p x q y

we get

0

r

R1 – R3 and R2

Apply R1 x

0

z

0

y

z

p x q y

R2 – R3,

0

r

x[ yr z(q y)] z[0 y(p x)] 0 [Expansion along first row]

xyr xzq xzy yzp zyx 0 xyr zxq yzp

91.

1 4

3 2 , P(A/E2) = , P(A/ 18 8

1 4 E3) = and P(A/E4) = 7 13

P(box III is selected, given that the drawn ball is black) = P(E3/A) By Bayes’ theorem,

P E2 P A / E2 +P E3 P A / E3

92.

2 xyz

p x

q y

r z

2

(a) Let l, m, n be the d.c.’s of the normal to the plane which contains the two concurrent lines with d.c’s l1, m1, n1 and l2, m2, n2. The line whose d.c.,’s are l, m, n is perpendicular to the line whosed d.c.,’s are l1, m1, n1 and l2, m2, n2. l1 l + m1 m + n 1 n = 0 ...(1) and l2 l + m2 m + n2 n = 0 ...(2) If the third line whose d.c’s are l3, m3, n3 also lies in this plane, then it is at right angles to the normal to this plane. l3 l + m3 m + n 3 n = 0 ...(3) Eliminating l, m, n from (1), (2) and (3), we get l1 l2

m1 m2

n1 n2

l3

m3

n3

0,

(b) If F be the resultant of the three given forces then

MT-123

Solutions ˆ (2iˆ 3jˆ 4k) ˆ ( ˆi ˆj k) ˆ (iˆ 2jˆ 3k)

F

OP OA

AP

P BG

ˆ (iˆ 2j) ˆ (ˆj 2k)

F

P BG

ˆ (2iˆ 4jˆ 2k) ˆ ( ˆi 3jˆ 2k)

93.

G

1 2

95.

80 40

1 2

P BG

2iˆ 6jˆ 10kˆ The magnitude of the moment

4 5

5 8

G

P BG G B P G P G G Now, P(G/BG) =

ˆi 3jˆ 2kˆ Vector moment of the given forces about A = vector moment of F about A = AP

5 8

10 16

P BG / G

OA = p.v. of A = ˆi 2ˆj

4 5

46 ,P G 80

P BG

2iˆ 4jˆ 2kˆ If O be the origin, then OP = p.v. of P = ˆj 2kˆ

20 23

(d)

U

= 4 36 100 140 (c) Let the G.P. be A, AR, AR2,............ Then x

p th term AR p 1 ,

y

q th term AR q 1 ,

z

r th term AR r

Now, x q AR

A

r

R

1

AR

(p 1)(q r)

q 1 r p

A

r p

A A

AR

R

P(A' B' )

(q 1)(r p)

p q

R

AG 3 4 BG

4 5

1 5

P(A' B' )

1 4

3 4

1 4

AR

AR

AG

13 44

1 1 4 4

BR

BG

3 1 4 4 BR

B) 1.

B' ) P (A ' B) P (A' B' )

B) P(A' B) P( A

B' ) P(A

B)

[See the Venn diagram]. (b) rank = 2 4 5 x

R

B)]

P[( A' )' B' ] P[ A' ( B' )' ] P( A' B' )

= P (A P( A

B),

P (A' ) P(B' ) P(A' B' )

P(A ' ) P(B' ) P(A Finally, since

96.

G

B A' B

P (A ' ) P(B' ) [1 P(A

(r 1)(p q)

R pq pr q r qr qp r p rp rq p q A R 1 94. (c) From the tree diagram, it follows that

3 4

A

this is P(A' B' ) 1 P(A Now,

r 1 p q

q r r p p q

S

B'

At most one of two events occurs if the event A' B' occurs. The probability of

yr p zp q

p 1 q r

q r

A

5 6 y 6 k z

4 5

4

5

x

0

5 6 y 0 k 7 0

0

rank

3 4

Also

BG

independent of the value of d

5 6

0

2, if k

k

7

7

EBD_7443 MT-124

Target VITEEE 101. (a) Let A = Cr + 2 Cr–1 + Cr–2 n

97. (d)

Given, sin–1 x – sin–1 2x = ± sin–1 x – sin–1 2x = sin–1 3 2

sin–1 x – sin–1 3 4

sin–1 x 1

3 3 2

= sin–1 2x 3 2

3 1 x2 2

n![(n 2

1 x2

98.

(c)

y

tan

1 (2 tan 2

1

tan(tan 1 u tan 1 u ) tan tan

99.

1

1 2

u)

tan(2 tan 1 u )

2u

2u

1 u2

1 u2

1

x=±

1 (2 tan 2

u)

x

b2

=

b

2

c

4r r r] r!(n r 2)!

( 2 b c) 2bc

102. (a)

2

sin 1 x 1

a2 2

b 2 c 2 4b 2 c 2 2bc

0

4bc

4bc 3b 2 4c 3b 2bc 2c 100. (a) Replacing each octal digit by the corresponding 3-digit binary number, we have (23450)8 = (010 011 100 101 000)2 = (10011100101000)2.

a sin a 2

2

cos a

Differentiating w.r.t. a, we get f(a) = a +

Put a =

2

2

1 x 1 1 0 x–1 R2 (for 100 W) 11. (c) For a beam, the depression at the centre is

flux

4Ybd3 [f, L, b, d are constants for a particular beam]

so total flux = Q/ o Since cube has six face so flux coming out through one wall or one face is Q/6 o. 5.

(c)

10

A

O 10

F 10

10 C

G

12.

10

E 10

1 Y (b) For circular path in magnetic field, mr 2 = qvB

i.e.

B

10 D

10

qvB mr

2

Equivalent Circuit F

10 A

10 10

O

qB q(r )B mr m If is frequency of roatation, then 2

E 10

D

G

6. 7.

As v r

10

10

Equivalent

fL

given by,

o

Resistance of

circuit

20 20 10 10 10 10 30 = 10 20 20 (c) According to Stoke's law , F= 6 vTR (a) We know from the Kirchhoff's first law that the algebraic sum of the current meeting at any junction in the circuit is zero (i.e. i = 0) or the total charge remains constant. Therefore, Kirchhoff's first law at a junction deals with the conservation of charge.

qB 2 m

2

13.

(c)

N ; i

L L

14.

N i

BA; B 0 Ni

2R

A

0 Ni

2R 0N

2R

2

A

L

N2

(a) Power dissipated = Erms. Irms = (Erms) (Irms) cos Hence, power dissipated depends upon phase difference.

MT-129

Solutions 15.

(b)

( 2

e

1)

( 0 NBA ) t

t

NBA 50 2 10 –2 10 –2 e 0. 1 (b) Mutual Inductance of two coils t

16.

M1 M 2

M

17.

18. 19.

2mH 8mH

NBA t

21.

E1

0.1 s

22. 4mH

series v

a1 a2

4 3

....(1)

In interference, the maximum and minimum resultant amplitudes are (a1 + a 2) and (a1 – a2) respectively Maximum intensity I max Minimum intensity I min

23.

24.

Imax I min

I max Imin

2

a2

4a 2 3

a2

49 1

RH

v

RH

v

RH

25.

26.

x + (t × ) + y = x + (1.5t) + y x

3 t 2

y

1

1

2

1

1

2

5 3 5 3 10 3 = 45 J 2 2 (c) A Febry Perot interferometer consists of two plane parallel glass plates. With this interferometer fringes of constant inclination are obtained by transmitted light after multiple reflection between the glass plate.

(b)

Ek

=

(c) The optical path length between A and B is

1

2

3 n2 (c) W = area of F – x graph = area of + area of rectangle + area of

2

49 :1

1

=

4a 2 3 4a 2 3

v

1 n2 For Paschen series, electron jumps from higher orbits to third stationary orbit (nf = 3)

But from equation (1), a1

1

2

4 n2 For Lyman series, electron jumps from higher orbit to the first stationary orbit i.e., nf = 1

(a1 a 2 )2 (a1 a 2 ) 2

1

RH

2 n2 For Brackett series, the electron jumps from higher orbits to fourth stationary orbit (nf = 4)

16 9

a 22

3.4 eV

Hence, energy required to ionize atom from its first excited state will be 3.4 eV. (b) For Balmer series, electron jumps from higher orbit to secondary orbit, thus for

Wavelength [Range in m] X-rays 1 × 10–11 to 3 × 10–8 -rays 6 × 10–14 to 1 × 10–11 Microwaves 10–3 to 0.3 Radiowaves 10 to 104 Wavelength of U.V. rays ranges from 6 × 10–8 to 4 × 10–7. (d) Huygen's construction of wavefront does not apply to origin of spectra which is explained by quantum theory. (d) Let the amplitude of the waves be a1, a2 and the intensities I1 and I2. The ratio of this intensities is 16 : 9 a12

13.6 4

22

(c) Rays

I1 Then I 2

20.

(b) Energy in first excited state

(b)

hc 1 e

1

(in eV)

0

3 108 1010 1800 1.6 10

6.6 10

34

19

1 m 2

2 1

2 W0

1 m 2

2 2

10 W0

W0 W0

1010 2300 = 1.5 eV

W0 and 9W0

EBD_7443 MT-130

Target VITEEE W0 9 W0

1 2

27.

1 3

32.

(b) Energy of electron = mec2 Energy of positron= mpc2

me

W Y

mp c

speed of light. Thus according to conservation of energy released 2m e c 2

28.

31

2 9.1 10

3 108

2

33. 34. 35. 36.

1.6 10 13 Joules . (b) For a monoatomic gas, the average energy

of a molecule at temperature T is

3 k BT . 2

dU 3 R dT 2 For an ideal gas, CP – CV = R

CV =

5 CP 5 R and 2 CV 3 (c) Forces between nucleons inside the nucleus are (i) attractive in nature (ii) extremely short range (iii) strongest forces in nature

30.

(d)

2 1H

2 1H

3 2 He

1 0n

13 MeV

31.

300 3 100 Now the number of atoms left after n halflives is

37.

38. 39.

N0

N0

N0 8

40.

7N0 8

W + (X.Y)

0.98

collector current base current

1

sin

n12

n 22

is the original length of wire, then (c) If change in length of first wire, ( 1 ) Change in length of second wire, 1

( 2

2

or

T1

or

T2 A

1

T2

1

or T1

)

T1 A

Now, Y

3

n

1 1 N0 N0 2 2 8 Hence the number of nuclei which have decayed in 300 years is N

L1

L3 (W + Y)

0.98 0.98 49. 1 1 0.98 0.02 (c) Since specific heat = 0.6 kcal/g × °C = 0.6 cal/g × °C From graph it is clear that in a minute, the temperature is raised from 0°C to 50°C. Heat required for a minute = 50 × 0.6 × 50 = 1500 cal. Also from graph, Boiling point of wax is 200°C. (c) Optical fibres are not subjected to electromagnetic interference from outside. (b) Core of acceptance angle

2 1H

is 2.2 The binding energy of (2.2) × 2 = 13 MeV + Binding energy (4.4) = – 13 MeV + Binding energy Binding energy of He = 13 + 4.4 = 17.4 (a) Half live of given radioactive nucleus = 100 years Number of half lives is 300 years

(W + X)

(b) (d) Positive terminal is at lower potential (–10V) and negative terminal is at higher potential 0V. (b) (c) We know that for common base

ic ib

CP =

29.

L1

i c collector current i e emmiter current & for common emitter

3 RT 2

Internal energy U

(b) W X

T1

or

1

2 2

– T1

= T2

T2 1 T1 T2 T1

2

2

T2 2

1

– T2

MT-131

Solutions 41.

42. 43. 44. 45. 46.

(a) If Gsystem = 0 the system has attained equilibrium is right choice. In it alternative (d) is most confusing as when G > 0, the process may be spontaneous when it is coupled with a reaction which has G < 0 and total G is negative, so right answer is (a). (b) XeF4 is planar (a) For a spontaneous process, S total is always positive (a) Since MgO has a rock salt structure. In this structure each cation is surrounded by six anions and vice versa. (b) Out of the given substances, only Li has high electrical and thermal conductivity. (c) For third ionization enthalpy last configuration of V – 4s0 3d3 4s 3d Cr

N

47.

48. 49. 50. 51. 52.

(b) (b) (b) (d) (d)

Enantiomers

en

3+

N

N N en

l form

53.

The two optically active isomers are collectivity called enantiomers. (c) Lanthanides are 4 f-series elements starting from cerium (Z= 58) to lutetium ( Z = 71). These are placed in the sixth period and in third group.

54.

(c)

[X ] 10 n

60 Co 1 H 1 27

Balancing the mass and atomic numbers on both sides

60 1 28 X 0

Thus X should be 55.

56.

(c)

Transform of [M(AA) 2 a 2 ]n± does not shows optical isomerism. 90 For orthorhombic system –2 As in [NiCl4] Chloride ion being a weak ligand is not able to paired the electron in d orbital. Relation is Kp = Kc (RT) n Reduction occurs at cathode. The optical isomers are pair of molecules which are non super imposable mirror images of each other

Co

N

N

N d form

Fe – 4s0 3d6 For third Ionization enthalpy Mn has stable configuration due to half filled d-orbital.

en

en

en

– 4s0 3d4

Mn – 4s0 3d5

N

N Co

N

N

3+

en – N

57.

n

60 27

Co 11 H

60 28 Ni

(d) Age of geological formations (i.e. predicting the age of the earth and rocks) is estimated by U– Pb method, also known as helium dating. Note : C-14 dating method is used to predict the age of fossils or dead animals or a fallen tree. (d) The activation energy of reverse reaction will depend upon whether the forward reaction is exothermic or endothermic. As H = Ea (forward reaction) – Ea(backward reaction) For exothermic reaction H = –ve – H = Ea(f) – Ea(b) or Ea(f) = Ea(b) – H Ea(f) < Ea(b) for endothermic reaction H = + ve H = Ea(f) + Ea(b) or Ea(f) = H + Ea(b) Ea(f) > Ea(b). (a) From data 1 and 3, it is clear that keeping (B) const, [A] is doubled, rate remains unaffected. Hence rate is independent of [A]. from 1 and 4, keeping [A] constant, [B] is doubled, rate become 8 times. Hence rate [ B]3 .

EBD_7443 MT-132

58.

Target VITEEE

on intergration ,

ò

Copper being more electropositive readily precipitate silver from their salt solution

dx = k(a - x) dt

(b) For a first reaction

dx = (a - x)

ò

k dt

i.e – n (a –x) = kt + c or kt = n a – n (a–x) or kt = 2.303 [log a – log (a – x)] Thus if we plot a graph between log a & t we get

64.

st

1 order

log[a]

Cu + 2AgNO3 ¾¾ ® Cu(NO 3 ) 2 + Ag whereas in K[Ag (CN)2] solution a complex anion [Ag(CN)2]– is formed and hence Ag+ are less available in the solution and therefore copper cannot displace Ag from its complex ion. (b) Given current (i) = 0.5 amp; Time (t) = 100 minutes × 60 = 6000 sec Equivalent weight of silver nitrate (E) = 108. Accordin g to Far aday's fir st law of electrolysis Eit 108 0.5 6000 3.3575 g. 96500 96500 (b) Order of reaction is independent of given factors. (b) Acetamide and ethylamine can be distinguished by heating with NaOH solution.Acetamide evolves NH3 but ethylamine does not. W

time t

59.

60.

61.

65.

(a) Given initial concentration (a) = 2.00 m; Time taken (t) = 200 min and final concentration (a – x) = 0.15 m. For a first order reation rate constant, (d) According to Le-chatelier's principle" whenever a constraint is applied to a system in equilibrium, the system tends to readjust so as to nullify the effect of the constraint. (c) By Kohlraush's law, 0 Na

0 Cl

0 H

0 Cl

0 CH 3OO –

0 eq NaCl

126.45

CH 3CONH 2

CH 3CH 2 NH 2

....(1)

Ù0(CH

3COOH)

62.

(d)

0 Na

67.

(c)

NH 2

NaNO 2 ,H Cl 0 5 C

91 ....(3) NMe2 N 2 Cl

+

H

517.16 126.45

390.71ohm

1

N=N

cm 2

0.0591 log 10 K n Here, n 2, E 0.295 E

2 0.295 = 10 or K 1010 0.0591 (d) In th e silver platin g of copper, log 10 K

63.

NaOH

No reaction.

on adding (2) and (3) then subtract (1) from it 0 CH 3COO

NaOH CH 3COONa NH 3

126.45

426.16 ...(2) 0 Na

66.

K[Ag(CN ) 2 ] is used instead of AgNO3.

68.

NMe 2

(c) Electrolytic reduction of Nitroalkane in weakly acidic medium give aniline NO2

NH2 Electrolytic reduction (weakly acidic medium) Aniline

Whereas in strongly acidic medium it gives p-hydroxyaniline

MT-133

Solutions NO2

NHOH

4[O]

electrolytic

(CH3)3 COH

reduction in presence of conc. H2SO4 (Strongly acidic medium)

rearrangement

73.

(a) An aliphatic aldehyde. Generally H – C = O streching frequency is about 2830 – 2695 cm–1 but for aliphatic aldehyde is 1740 – 1720 cm–1. (a) Nitrobenzene is reduced to aniline in the presence of Sn/HCl.

(b)

Zn-Hg/HCl

74.

75.

(c)

CH3–CH=CH2

NH 2

SOCl 2

B 2H 6

NaOH / H 2 O 2

B 2H 6

CH3–CH2–CH2–OH

aOH / H 2 O 2

Aniline Nitrobenzene Hence, option (a) is correct. (d) CH 3COOH

ethyl benzene

This reaction is known as clammensen's reduction. (a) Due to H-bonding, the boiling point of ethanol is much higher than that of the isomeric diethyl ether.

Sn / HCl

71.

CH2–CH3

Phenyl methyl ketone

p-Hydroxyaniline

NO 2

CH3COOH+2CO2+3H2O

C–CH3

OH

70.

8[O]

O

NH2

69.

CH3COCH3+CO2+2H2O

Propanol

COONa

76.

(b)

2HCOONa

+ H2

|

360

COONa

CH 3 COCl

sod.formate

77.

(a) OMgBr CH3MgBr

C6 H5 COOC2 H5

|

C6 H5

C

OC 2 H 5

|

CH3

O Mg(OC2 H5 )Br

C6 H5

||

C CH3

Excess CH3MgBr

OMgBr

C6 H5

C |

CH3

CH3

H2 O

|

C6 H5 — C — CH3 |

CH3

OH

CH3 Con. H 2SO 4

|

C 6 H 5 — C CH 2

Ozonolysis

'B' C 6 H 5COCH 3 HCHO

72.

(c) By oxidation of tertiary alcohol with stronger oxidising agents ketones may be formed along with carboxylic acid.

3I

78.

4 NaOH

2 C 6 H 5 COCH 3 CHI 3 (a) It is an example of Claisen condensation. The product is acetoacetic ester.

EBD_7443 MT-134

Target VITEEE

O || CH 3 C

OC 2 H 5

H

It has real roots, So

O || C OC 2 H 5

CH 2

D

O O || || CH 3 C CH 2 C OC 2 H 5

0

(2anc) 2

n2

2(n 2

4 2a 2 (n 2 2)

79.

CO 2H 2

ZnO Cr2O3 300 C

R CONH 2

Br2 / KOH

Compressed gas

80.

81.

n2

(d) Water gas is mixed with half its volume of hydrogen.The mixture is compressed to approximately 200 – 300 atmospheres. It is then passed over a catalyst |ZnO + Cr 2O3| at 300°C. Methyl alcohol vapours are formed which are condensed

(c)

2 n + –2

Methyl alcohol

84.

RNH 2

This is known as Hoffmann’s Bromamide reaction. (a) For the given A.P., we have sin (z + x – y) – sin (y + z – x) = sin (x + y – z)– sin (z + x – y) 2 cos z sin (x – y) = 2 cos x sin (y – z)

1

2x

tan

tan 1

1

5x 1 6x 2

x+a=0

tan

1)( x

5x 1

1)

0

0

x

1 . 6 (a) The given equation is 2 2

2a x

2ancx (n

y

2

a ( t 2 1) t

O

X

S

Thus, (1) meets the directrix at

1

negative, so x

83.

a at 2

a,

a ( t 2 1) t

Now, slope of PS is m1 =

1 or – 1 6 Now for x = – 1, LHS of equation becomes

(6 x

+ 2

2at 0 at

2

a

2t t

2

1

and slope of QS is m2

6x 2

1

1

0

n [ 2, 2]

4

2 x 3x 1 2 x.3x

2)

Q

Q 3x

2

2)(n

2 at) P (a t1 , 2

tan x – tan y = tan y – tan z, showing that tan x, tan y and tan z are in A.P. tan

(n

n 2

ty

sin y cos z cos y sin z cos y cos z

(b)

4a 2 c 2 is positive]

(c) The equation of the tangent at P (at2, 2at) to y2 = 4ax is ty = x + at2 ...(1) It meets the directrix x = –a

sin x cos y cos x sin y cos x cos y

82.

0

–

–2

CH3OH

4

0

0

[

(ethyl acetoacetate )

2)c 2

85.

a ( t 2 1) 0 ( t 2 1) t = a a 2t Since m1m2 = –1, therefore PQ subtends a right angle at the focus. (c) Obviously the well (W) must be on one side of the trees T1, T2,......., T25, W

5m T1

2 )c

2

0

5m T2

T3...................T24

T25

MT-135

Solutions The total distance covered by the gardener tan B

B 2 2 B 1 tan 2

1 2 15 16

8 15

(x)

log 5 log 3 x

log 5

loge x log e 3

2 tan

WT1 (2WT1 T1T2 ) (2WT2 T2T3) ..... (2WT24 T24T25]

10 ( 2 10 5) (2 15 5) ............ to 25 terms

10 (25 35 45 ..........to 24 terms) 24 [ 2 25 ( 24 1) 10] 10 12 [50 230 ] 3370 2

10

86.

(c)

n

n

Cr 1 n

Cr

n

Cr 1 2 Cr n

1

n 1

87. (c)

n

Cr

n

Cr

n 1

Cr

90.

(c)

If

log5 log e x

1

n 2

Cr Cr 1 Cr 1 Constraints are 2x + 3y 36; 5x + 2y 50; 2x + 6y 60, x 0, y 0

log e (log e x ) log 5 (log e 3) log e 5

(a) I =

sin 3 x. cos 5 x dx

Put sin x = t

cos x dx

t 3 2t 5

t 7 dt

dt

1 4 t 4

sin 3 x . cos5 x dx

Put cos x = t I=

2

1 t t

5

91.

2

t 3 1 t 2 dt 2 6 t 6

1 1 4 sin 6 x = sin x 4 3 ALTERNATE :

I=

1 8 t c 8

1 sin 8 x 8

lim

x

dt

[b 2 (c a ) 2

t

t

5

dt

t8 8

t6 6

4(s a )(s c)

(b c a ) (b c a )

s(s b) 16(s a )(s c) (s a )(s c) s(s b) 1 4

1 16

tan 2

lim x log x

x

B 2

1 16

0

c

lim

x

x

0

0

0

x

x

2(s a )2(s c)] Squaring and simplifying we, get

B 2

log x

– sin x dx = dt

s(s a )(s b)(s c)

tan

0

0

0

1/ x

Applyin g L.H. Rule lim

1 1 cos 8 x cos 6 x c 8 6 [Not that this result is in terms of powers of cos x]

(b)

lim f ( x ) x

c

=

89.

1 1 1 1 . . log e 5 log e e e e log e 5 (d) For Rolle’s theorem in [a, b], f(a) = f(b), In [0, 1] f(0) = f(1) = 0 the function has to be continuous in [0, 1] f (0)

sin 2 x . cos5 x . sin x dx

7

0

' (e )

I = sin 3 x . cos 4 x . cos x dx =

1 1 1 . . log e 5 log e x x

' ( x)

The number of constraints are 5. 88.

log5 log e 3

0

1

x

0

0

0

92. (c) 93. (b) In the function f (x) = (x – 1) (x – 2) (x – 3) for more than one value of x, i.e. x = 1, x = 2 and x = 3, value of the function is zero. So, the function is not one-one. Range of the function is the set of all real number i.e. R. Since Range = Co-domain = R, the function is onto. Thus the given function f(x) is onto but not one-one. 94. (a) f ( x) is onto S = range of f (x) Now f (x) = sin x

3 cos x 1

1 sin x

3

2sin x 1

3

1

EBD_7443 MT-136

Target VITEEE

1 2sin x

95.

(k 1)(k 1)k! 2k!

1 3

3

Tk

f ( x) [ 1, 3] S (a) The equation of the parabola is y2 = 4ax...(1) Let (x1, y1) be the mid point of a focal chord of the parabola. Equation of chord having (x1, y1) as its mid point is

= (k 2 3)(k 1)! 2(k)! (k 2)! 3(k 1)! 2(k)! T1

T2

2 or yy1 – 2ax = y1 2ax 1 Since it passes through the focus (a, 0)

2

Tk

z2

dz 1 dx

= =

dz dx

1

98.

a2 z

z2

2

z

z

a

z

2

a

a2 tan a

z

a tan

z a x y

97.

dx

a2

[1

1

1

z a

tan a tan

z a

z

z

D

] dz

2

x x

x

x a y c , a

2 sin 1 x

sin 1 (1 x )

sin

1

sin

1

99.

D

x

x

2

sin

(1 x )

sin

1 x

sin

2

1 2x 2

sin

1 x

1 2 Only x = 0 satisfies the given equation. (d) We have

0

a [

D

m 1999

2

2

cos

x (2x 1)

]

x

0 or x

2a

3a

2a 1 2a 3 a 1 3a 5

a 5

a 2

a{(2a 3)(a 2) (a 5)(a 1)} 2a{(2a 1)(a 2) (3a 5)(a 1)}

D

3a{(2a 1)(a 5) (3a 5)(2a 3)}

D

a (15a 2 31a 37) , on simplification where c

(a) The general term Tk

(c)

1 2 sin 2

2

dz

T3 ........... ..... Tm

(m 2)! 2(m 1)! (m 1)!(m 2 2)

Now, cos 2

dz dx z a2 Integrating we get 2

2(m)!

3(2)! 2(1)! 2(2)! (m 1)! (m 2)! 3(m 1)!

and

z2

z

T1 T2

where,

a2 2

2

2

3(4)! 2(3)! ......... ........ ......... ........ 2)! 3(m 1)!

m(m 1)!

1

a2

2(2)! ;

Now, m(m 1)! 1999 2000!

dy dz dx dx The equation then becomes z

3(3)!

k 1

2a ( x 1 a ) = 0 Thus, the locus of the mid point (x1, y1) is y2 = 2a (x – a), which is a parabola whose vertex is (a, 0).

(a) Put x y

(4)!

2(1)! ;

3(2)!

m

2ax1 or

2 y1

96.

(3)!

T3 (5)! ..... ........ ..... ........ Tm (m

2 T = S1 i.e. yy1 – 2a (x + x1) = y1 4ax 1

0 – 2a2 = y1

(k 1)(k 1)! 2(k )!

(k 2 1)k! (k 2 1 2)(k )!

D

15a

a

31 30

2

[for all non-zero a]

1259 900

0

MT-137

Solutions Hence, the given vectors are non-coplanar, for all a 0 . 100. (c)

1

det(B ) det A det B

1

det( B 1B). det A

det( B ). det B. det A

det(1). det A 1. det A

3

7

x 1

3

a

1

det(B AB)

101. (a) Put 1 x

1

4 /3

z

7

x 4 dx

z

1/3

x

det A. 21 6 z dz 4

dx

21 6 z dz 4

8

21 z 4 8

2

1 2a

p2

p 2 2a 2 p 2a The equation of the chord is x cos y sin 2a Which is the equation of straight line always

touching the circle x 2 fixed circle of radius

C

8/7 21 3 4 1 x C 32 102. (a) We have, (2 * 3) * 4 = 1 * 4 = 1 and 2 * (3 * 4) = 2 * 1 = 1 Since, each row is same as their corresponding column. Therefore, the operation * is commutative. We have (2 * 3) = 1 and (4 * 5) = 1 Therefore, (2 * 3) * (4 * 5) = 1 * 1 = 1 103. (a) Equation of the pair of straight lines joining the centre (origin) and the intersection points of the variable chord and the hyperbola is

x 2 sin x

2 x sin xdx d x sin x dx dx dx

x 2 sin x 2 x sin x dx

[Integrating by parts again taking x as first function] x 2 sin x 2 x ( cos x )

105. (b) Let P

S (ae, 0) X

2a .

[Integrating by parts taking x2 as first function]

2 x cos x

2) sin x x2

f (x)

F

( 2a) 2 , a

d 2 x cos x dx dx dx

= x 2 cos x dx

(x 2

P

y2

x 2 cos x dx

104. (d) I =

x 2 sin x

Y

C

1 2

cos x dx

2 sin x

c

2x cos x c 2, g( x )

2x

(a cos , b sin ) P(acos , bsin )

x

x2

y2

2

2

a e

F2 (-ae, 0) F1(ae, 0)

x cos

2

y sin

0 ....(i)

p a 2a [By homogenising the equation of hyperbola with the help of the equation of the straight lines. See the pair of straight lines]. The equation (i) represents a pair of perpendicular lines, if coefficient of x2 + coefficient of y2 = 0 i.e.,

1

cos 2

a2

p2

1 2a 2

sin 2 p2

0

Then, A = area of PF1F2 1 2

a cos ae ae

b sin 0 0

1 1 1

1 2ae b sin 2

abe | sin | Clearly, A is maximum when | sin | 1 Maximum value of A = abe 106. (c) Let W stand for the winning of a game and L for losing it. Then there are 4 mutually exclusive possibilities (i) W, W, W (ii) W, W, L, W (iii) W, L, W, W (iv) L, W, W, W.

EBD_7443 MT-138

Target VITEEE [Note that case (i) includes both the cases whether he losses or wins the fourth game.] By the given conditions of the question, the probabilities for (i), (ii), (iii) and (iv) respectively are 2 2 2 2 2 1 1 2 1 1 2 . . ; . . . ; . . . 3 3 3 3 3 3 3 3 3 3 3 1 1 2 2 . . . . 3 3 3 3 Hence the required probability and

8 4 4 4 36 4 . 27 81 81 81 81 9 [ The probability of winning the game if

=

2 2 1 2 3 and the probability of winning the game if

previous game was also won is

previous game was a loss is

1 1 2

1 ]. 3

happens. We can express this fact as If a tumbler is half empty, then it is half full. If a tumbler is half full, then it is half empty. We combine these two statements and get the following. A tumbler is half empty, if and only if it is half full. 109. (d) Since 2 persons can drive the car, therefore we have to select 1 from these two. This can be done in 2C1 ways. Now from the remaining 5 persons we have to select 2 which can be done in 5C2 ways. But the front seat and the rear seat person can interchange among themselves. Therefore, the required number of ways in which the car can be filled is 5C2 × 2C1 × 2! = 20 × 2 = 40. 2

110. (b)

107. (d) We have dx a ( sin d and dy d

a (cos

...(i)

log sin x dx 0

sin

cos )

2

a cos

2

log sin 0

log cos x dx ...(ii)

x dx

2

0

From (i) and (ii) cos

sin )

a sin ,

dy dy / d tan dx dx / d So, the slope of the normal is

2 0 /2

log cos x dx 0

log

1 cot , which varies as . dy / dx The equation of the normal at any point is y a(sin cos )

cot [x a(cos

2

log sin x dx

2

0

sin 2 x dx 2

2

2

log sin 2x dx 0

log 2 dx ; 0

sin )]

x cos y sin a. Clearly, it is a line at a constant distance | a | from the origin. Further, the equation x cos + y sin = a represents the equation of a tangent to the circle x2 + y2 = a2 for any value of , which is a fixed circle. 108. (b) The given statements are p : A tumbler is half empty. q : A tumbler is half full. We know that, if the first statement happens, then the second happens and also if the second happens, then the first

2

log sin 2x dx

2

0

Put 2x t 2

2dx

1 log sin t dt 2 0

log 2

dt

2

log 2

2 1 log 2 . 2 log sin t dt 2 2 0 [Since sin ( – t) = sin t, see property No. 8]

2

2

log 2

2

log e 2

MT-139

Solutions sin 2A

sin C

sin B

sin C

sin 2B

sin A

sin B

sin A

sin 2C

111. (b) Let

function is continuous at x = 0 For function to be differentiable : f (0 + h)= f (0 – h) f (0 + h) =

2 sin A cos A

sin C

sin B

sin C

2 sin B cos B

sin A

sin B

sin A

2 sin C cos C

1 0 1 h lim sin lim h h 0 h h 0 which does not exist. h sin

The above determinant is the product of two determinants, sin A cos A 0

cos A sin A 0

sin B cos B 0 sin C cos C 0

cos B sin B 0 cos C sin C 0

112. (b) Given plane is 2x y 2z 4

0.

R

N

M

f

(0

| NQ |

22

( 1) 2

( 2) 2

| 2.1 ( 1) 2.2 4 | 2 2 ( 1) 2

( 2) 2

| PQ |

(1 1)

| NM | | QR |

( 1 2) PQ

2

RP

2

4

5 3

2

17 1

=

17 4.

= 0

= =

1 2 4

/2

1 cos 2x dx 2

1 2 4

1 2

cos 2 x dx

/4

1 sin 2x x = 2 2

0

=0

/2

sin 2 x dx

/4

0

0

X

/2

0

= 0 × (a finite quantity) = 0 Also, lim x sin 1/x = 0 × (a finite quantity) x

,y

Required area = area of shaded region

= 0 × (a finite quantity) = 0 f (0 – h) = lim – h sin (1/–h) h

4

/4

O

113. (c) For function to be continuous : f(0 + h) = f(0 – h) = f(0) f(0 + h) = lim h sin 1/h h

x

1

8, 3

( 2 0)

h

0

Y

/4 2

h

1 h h 0 which does not exist. So function is not differentiable at x = 0 114. (c) y = sin2x ...(i) and y = cos2x ...(ii) Solving (i) and (ii), we get sin2x = cos2x tan2x = 1

Also 2

= lim

h)

0

= lim sin

From points P and Q draw PM and QN perpendiculars on the given plane and QR MP . | MP |

–

tan x

| 2( 1) 2 0 4 |

1 h

( h)sin

0

P

Q

f (0 h) f (0) h

4

/4

/4 0

1 cos 2x dx 2

1 x 2

1 2

1 0 2 2 4

1 2

1 2

4

1 4

sin 2x 2

/2 /4

1 2 2 sq. units.

EBD_7443 MT-140

Target VITEEE

115. (c)

E

D

F

a 2 C

and, A

B

AB AC AD AE AF

118. (d)

|b c

c

|b

a

119. (c)

ax 3 2y f

120. (a)

a

a b|

x2 + 3x = y2 + fy + C (Integrating) 2 a 2 2 x + y – 3x + fy + C = 0 2

This will represent a circle, if

6x 7

f (1 x) lim f (1) x 0

1/ x

1 f (1 x) f (1) lim x f (1) x 0 e

10 =

9.

= , 11 = (2 – ) (2 – 2) (2 – = (2 – )2 (2 – 2)2

f '(1) f (1)

e2

9. 2 = 10) (2 –

2 11)

Now x3 – 1 = (x – 1) (x – ) (x – 2), dividing by x – 1 x2 + x + 1 = (x – ) (x – 2) or (x2 + x + 1)2 = (x – )2 (x – 2)2. Put x = 2, we get (2 – )2 (2 – 2)2 = (7)2 = 49

a|

(ax + 3) dx = (2y + f) dy

In case of R1 , f ( x)

e

117. (b) We have,

dy dx

f2 – C > 0[Using : g2 + f 2 – c > 0] 4

clearly every element of A has a unique image hence, R1 represents a function. Similarly, R2 and R3 also represent functions. In case of R4, f (x) = ± 4x every element of A has two unequal images for example f(1) = ± 4, f(2) = ± 8, etc. R4 is not a function

AD AD AD 3AD n 3. 116. (a) Let CM be the altitude through C. C(c)

CM

Coeff. of x2 = Coeff. of y2]

a = – 2 and 9 + f 2 – 4C > 0

ED AC AD AE CD = (AC CD) AD (AE ED)

B(b) M A(a) Then, area of triangle ABC 1 1 (AB)(CM) | b a | CM ....(i) 2 2 Again, area of triangle ABC 1 1 | AB AC | | (b a ) ( c a ) | 2 2 1 |b c c a a b | ...(ii) 2 From (i) and (ii),

9 4

1 [

121. (c) 122 (c) 123 (d) 124 (b) The second part of the sentence explains or amplifies the part before the semicolon. Since his recordings range widely, his tastes could be described as wide-ranging, or eclectic. 125 (d)

MT-141

Solutions

MOCK TEST 3 1.

(a)

2.

(c) Work done =

3.

(8 10 18 ) 2 1 q2 = 2 c 2 100 10 8 = 32 × 10–32 J (d) Unit positive charge at O will be repelled equally by three charges at the three corners of triangle.

9.

1 mv 2 qV ..........(i) 2 As the magnetic field curves the path of the ions in a semicircular orbit

By symmetry, resultant E at O would be zero. 4. 5.

(d) (a) If m1, m2 are masses and u1, u2 are velocity then by conservation of momentum m1u1

mv 2 BqR v R m Substituting (ii) in (i)

Bqv

+ m2u2 = 0 or | m1u1 | | m2 u 2 | R/2

6.

1 BqR m 2 m

(a)

R R . R1 R 2 2 2 R Re q R eq R R R1 R 2 4 2 2 (a) A balanced wheatstone bridge simply requires

P R 2 2 Q S 2 S Therefore, S should be 2 . A resistance of 6 is connected in parallel. In parallel combination, 1 R

8.

1 R1

2

qV or

.......... (ii)

q m

2V B2 R 2

Since V and B are constants,

R/2

7.

Also 3 > 1 otherwise the ball would have floated in liquid 1. From the above discussion we conclude that 1 < 3 < 2. (a) In mass spectrometer, when ions are accelerated through potential V

q m

10.

that 1 2 Also 3 < 2 otherwise the ball would have sink to the bottom of the jar.

R2 (d) Magnetic moment, m = IA qv ( R2 ) 2 R

I 11.

1 R2

1 1 1 S 3 2 6 S (d) From the figure it is clear that liquid 1 floats on liquid 2. The lighter liquid floats over heavier liquid. Therefore we can conclude

1

q and T T

2 R v

(a) Electron moves undeflected if force exerted due to electric field is equal to force due to magnetic field. q | v || B | q | E |

12.

qvR 2

|v|

|E| | B|

(a) Conservative force is negative gradient of potential

–dV(x) dx (b) Efficiency of the transformer F(x) =

13.

Poutput Pinput

14.

100

100 100 220 0.5

90.9%

(d) Condition for which the current is maximum in a series LCR circuit is,

EBD_7443 MT-142

Target VITEEE

1 LC 1

1000

L(10 10 6 ) = L = 100 mH

15.

1 (d) At resonance L = , C

=

1 LC

E R Power dissipated at Resonance = i2R

22. 23.

(a) (c) The stimulating radiation is an electromagnetic wave of suitable frequency.

24.

(d)

16.

17.

In electromagnetic waves, E

25.

5 10 11 1.6 10 19

(d)

B B0 sin( t kx) The electromagnetic waves travel in the

h

18. 19.

20. 21.

(a) (a)

(b) (c)

ze2 1 ze 2 4 0 2rn 4 0 2rn or En = 2Un Here En = –21.76 × 10–19 J 1

27.

(d)

hc

c f

and

so

h mv

eV0

2 4

W0 and eV

0

1

i.e.

2q V m

hc

pc

2q Vm ;

h

q m q pm p

p

pc

hf c

p

1 m v 2 or m v 2

qm

2 2 hc 2 0

Subtracting them we have e ( V0

V)

or V

V0

tan(90

hc

1

0

1 2

hc 2 0

hc 2e 0 )

stress strain

28.

(a)

29.

(b) Bason group photon, pion, graviton, kaon (c) The half life of the sample is

30.

0.693

N0 N

J

mc

so, E

qV

so

10

h

hc

direction of (E B). 0. 4 ' 0.3 mm 4 3 As the floor exerts a force on the ball along the normal, & no force parallel to the surface, therefore the velocity component along the parallel to the floor remains constant. Hence V sin = V1 sin 1. From Brewster’s law = tan i i = tan–1 = tan–1(1.5) = 57° The total energy of the electron in n th orbit is given by En = Tn + Un

3 108

eV = 24.82 keV

p

But, p

mc 2 mc.c

E p

34

0.5 10

6.6 10 34 3 108

E E

26.

6.6 10

(a) Moment of photon

B. Both

E and B are in the same phase. In electromagnetic waves E E0 sin( t kx)

hc

Energy

=

Current through circuit i =

d(i 0 sin t ) di M = 0.005 (b) e dt dt = –0.005 × i0 ×( cos t) e max . 0.005 i 0 (when cos t = –1) = 0.005 × 10 × 100 = 5 V (c) Variation in magnetic field causes electric field and vice versa.

21.76 10 19 2 = – 10.88 × 10–19 J

En 2

Thus U n

W0

;

MT-143

Solutions 31.

(a)

7 3 Li

1 1H

8 4 Be

4 2 He

39. 40.

4 2 He

(b) (c) here, m = 4 , g = 4 × 10–3 kg

32.

33.

34.

(c)

ZX

A

Z 2X

A 4

Z 2 1X

dx dt

A 4

no. of neutron = A – 4 – (Z – 2 + 1) = A – 4 – (Z – 1) =A– Z– 4 + 1 =A– Z– 3 (d) During the formation of a junction diode. Holes from p-region diffuse into n-region and electrons from n-region diffuse into pregion. In both cases, when an electrons meets a hole, they cancel the effect to each other and as a result, a thin layer at the junction becomes free from any of charges carriers. This is called depletion layer there is a potential gradient in the depletion layer, negative on the p-side, and positive on the n-side. The potential difference thus developed across the junction is called potential barrier.

2

(c) As we know that I

Ic

36. 37.

Ib Ic

1

10 106 9

E

d 2 x dx dt dt 2 dt

2

8t 2 32 10 t 2 0 = 32 × 10–3 [4(2)2 + 2 – 0] = 576 mJ 3

41.

42.

(b) For reaction 3A B C If it is zero order reaction r = k [A]0, i.e the rate remains same at any concentration of 'A'. i.e independent upon concentration of A. (b) For a first order reaction, A products

r

k[A] or k

k

43.

r [ A]

1.5 10 2 = 3 × 10–2 0.5

Further, t1/ 2

0.693 k

0.693

23.1 3 10 2 (c) We know that the activation energy of chemical reaction is given by formula E a é T2 - T1 ù k2 = k = 2.303 êê T T úú , where k 1 is the 1 ë 1 2 û rate constant at temperature T1 and k2 is the rate constant at temperature T2 and Ea is the activation energy. Therefore activation energy of chemical reaction is determined by evaluating rate constant at two different temperatures.

2

1.2 1012 m 3

44.

(b)

r r

S

m

0

(c) Frictional force is always opposite to the direction of motion N W

fdx

2

1

1

8

dt 2

32 10 3 (8t 1)dt

Ib

1 (a) In reverse bias the current through a p-n junction is almost zero. (a) The critical frequency of a sky wave for reflection from a layer of atmosphere is given by fc = 9(Nmax)1/2 10 × 106 = 9(Nmax)1/2

Nmax 38.

1

d2 x

(4 10 3 )(8)(8 t 1) dt

Divide by I c both side Ie Ic

8t 1

0

(b) Output of upper AND gate = AB Output of OR gate, Y A B BA This is boolean expression for XOR gate.

t

Work done, W

Output of lower AND gate = AB

35.

4t 2

x

This nuclear reaction is not possible

k[A]n k[A]0

if n = 0

EBD_7443 MT-144

Target VITEEE or r = k thus for zero order reactions rate is equal to the rate constant.

45.

(b)

H

27 =9 3 No. of gm eq. of H+ = no. of mole of H+ Hence Mass of H+ = 0.5 × 1g = 0.5 g We know that, 2g H2 at STP = 22.4 L

Eq. wt of Al3+ =

Ea(f ) Ea(b)

Thus energy of activation for reverse reaction depend upon whether reaction is exothermic or endothermic If reaction is exothermic, H ve

Ea(b)

49.

Ea(f )

If reaction is endothermic

Ea(b) 46.

H

ve

Ea(f )

(d) Given molarity = 0.01 M Resistance = 40 ohm; Cell constant

l A

0.4 cm

1

.

50.

Specific conductivity ( ) cell constant 0.4 = 40 resistance

= 0.01 ohm

1

51.

cm 1

Molar conductance (

47.

1000 .01 10 3 ohm .01 (d) Given for the reaction

Cu (2aq )

Zn (s)

1000 Molarity

m) 1

2

cm mol

Cu (s )

(iii) R — C — N —Br

2.20 log10 k eq = 37.22 0.059 or Keq = 1.66 × 1037 (d) No. of gram equivalent of H+ = No. of gm eq. of

+

K

R — C — N + KBr O

R

N••

••

(iv) O = C

R—N=C=O (rearrangement)

(v) R N C O 2KOH 52.

(d)

NH2

0.0591 log10 K eq 2

4.5 0.5 9

R — CO N Br H 2O

O

Zn 2(aq ) ,

0.0591 log10 K eq n here (n number of exchan ge of electrons)

Al3+ =

OH

RCONHBr HBr

1

E cell

48.

Supplied heat to the system, Q = w + E E = Q –w (c) (CH3)2NH is most basic because two electron releasing groups are present on Nitrogen. Also aromatic amines are less basic then aliphatic amines. The basic character of amines follow the order R2NH > RNH2 > C6H5NHCH3 > NH3 (c) The mecahnism of Hoffmann bromide reaction is (i) RCONH 2 Br2 RCONHBr HBr (ii)

Eº = +1.10 V. At equilibrium

or 1.10

22.4 ´ 0.5 = 5.6 L 2 (a) Thermodynamic is based on conservation of energy. According to Ist law of thermodynamics Supplied heat to any system is used as work and change in internal energy ( E )

0.5 g H2 at STP = =

K 2 CO 3

RNH 2

NHCOCH3 (CH 3CO)2O

Br2 CH3COOH

CH3

CH3

(A)

NHCOCH3 Br

CH3 (B)

NH2

+

Br

H /H2O

CH3 (C)

MT-145

Solutions 53.

54.

55.

(c) Due to greater electron-withdrawing effect of NO2, and also due to additional resonance due to –NO2 group in the phenoxide ion, nitrophenols are stronger acids than chlorophenol. Further since resonance in p-nitrophenoxide ion is more important than that in m-nitrophenoxide ion, former is more stable, hence p-nitrophenol is more acidic than m-nitrophenol. o-Nitrophenol is least acidic among the three isomers due to orthoeffect. (b) Only amides (but not acids and esters) undergo hydrolysis in presence of sodalime to form sodium salt of a carboxylic acid and ammonia gas. Further, since the given compound is a liquid, it must be formamide, because propanamide is a solid (b) RCONH 2 NaOH RCOOH NH 3 59 g (1 mole )

59.

x-ray density–pyknometric density x-ray density

60.

61.

RT ln K nF Again Gº = nFEº put in (i)

57.

58.

c

0.13´103

2.178 ´103

(c) For a spontaneous reaction G(–ve), which is possible if S = +ve, H = +ve and T S > H [As G = H – T S] (c) [Co(NH3)6]3+ Co3+(27 –3 = 24) 2

3

d sp

(inner octahedral complex & diamagnetic)

[Cr(NH3)6]3+ Cr3+(24 –3 = 21) 2

3

d sp

(inner octahedral complex & paramagnetic)

[Ni(NH3)6]2+ Ni2+ (28 – 2 = 26) 3 2

sp d

(outer octahedral complex & paramagnetic)

[Zn(NH3)6]2+ Zn 2+(30 – 2 = 28)

= 0 and Q = KC

3 2

sp d

…(i)

…(ii)

G º RT RT lnK lnK ; G º nF nF (c) A body-centred cubic system consists of all eight corners plus one atom at the centre of cube. (d) If in an ionic crystal of the type A+, B–, equal number of cations and anions are missing from their lattice sites so that the electrical neutrality is maintained. The defect is called Schottky defect.

=

130 = 5.96 × 10–3 2178

17 g (1 mole )

RT ln Q E Eº nF

Eº

2.178´103

=

R = 59 – 44 = 15 Molecular mass of R as 15 corresponds to CH3 group, hence RCOOH should be CH3COOH (a) The relation between free energy change and equilibrium constant is given by Nernst equation

At equilibrium, E

2.178´103 - 2.165´103

=

Thus wt. of 1 mole of RCONH2 = 59 In other words, R + 12 + 16 + 14 + 2 = 59

56.

(a) Fraction of unoccupied sites in NaCl crystal

62.

(outer octahedral complex &diamagnetic)

(a) The given compound may have linkage isomerism due to presence of NO2 group which may be in the form –NO2 or –ONO. It may have ionisation isomerism due to presence of two ionisable group –NO2 & – Cl. It may have geometrical isomerism in the form of cis-trans form as follows : [Co(NH3 ) 4 Cl(NO 2 )]NO 2 & [Co(NH3 ) (NO2)2]Cl ––– ionisation isomers. [Co(NH3)5(NO2)2]Cl & [Co(NH3)5(ONO)2Cl ––– Linkage isomers

EBD_7443 MT-146

Target VITEEE NO2

H3N CO NH3

63.

66.

NO2

NH3

H3N

NH3

NH3

67.

NO2

CO

NH3

NO2

NH3

Trans-form

cis-form

Geometrical isomers (b) Lesser is the number of unpaired electrons smaller will be the paramagnetic behaviour. As Cr++, Mn++, Fe++ & Ni++ contains. Cr++ (3d4) =

68. 69.

70.

= 4 unpaired

e–.

Mn++ (3d5) = 71.

= 5 unpaired e–. Fe++ (3d6)= = 4 unpaired e–. Ni++ (3d8) = = 2 unpaired e–. As has minimum no. of unpaired e– thus this is least paramagnetic. (c) Given t1/2 = 12.3 years Initial amount (N0) = 32 mg Total time = 49.2 years No. of half lives ( n ) now N t 65.

N0

1 2

T t1 / 2

n

= 32

1 2

49.2 12.3 4

=

32 16

72.

73.

Hence 32 mg becomes 2 mg in 49.2 years (d) Remaining activity = 0.01M after 24 hrs Remaining activity = Initial activity Used half life time (n) = So,

Total time T1 / 2

0.01 = Initial activity

1 2

4

Initial activity = 0.01 × 16 = 0.16M

1 2

24 6

n

74. 4

2

5

C 6 H 5 CH 2 CH 2 OH ;

4

2 mg

1

(d) 2-Phenylethanol, C H 2OH C H 2C 6 H 5 , is a 1º alcohol which can be prepared from C6H5MgBr by treating with ethylene oxide (note that HCHO will introduce only one carbon atom, i.e. it will give C6H5CH2OH and not C6H5CH2CH2OH). O C H MgBr 6

Ni++

64.

(a) Chloro diaquatriammine cobalt (III) chloride is [CoCl( NH 3 )3 (H 2O) 2 ]Cl 2 (a) The basic character of the transition metal monoxide is TiO > VO > CrO > FeO because basic character of oxides decrease with increase in atomic number. Oxides of transitional metals in low oxidation state i.e., + 2 and + 3 are generally basic except Cr2O3. (c) According to VSEPR theory, a molecule with six bond pairs must be octahedral. (b) Due to –I effect of the –COOH group, Hbonds in acids are much stronger than in alcohols; while aldehydes do not exhibit H-bonding. (d) Since acetic acid freezes at 16.6ºC while water freezes at 0ºC, glacial acetic acid is obtained by crystallising, separating and melting acetic acid.

C 6 H 5 MgBr HCHO C 6 H 5 CH 2 OH (a) Among acetic acid, phenol and n-hexanol only CH 3 COOH reacts with NaHCO3 to evolve CO2 gas. CH3COOH + NaHCO3

CH3COONa + CO2 + H2O (d) Iodoform test is exhibited by ethyl alcohol acetaldehyde, acetone methyl ketones and those alcohols which possess CH3CH(OH)group. As 3-pentanone does not contain CH3CO-group as therefore it does not give iodoform test. (b) Acetaldehyde reacts only with nucleophiles. Since the mobile p electrons of carbon– oxygen double bond are strongly pulled towards oxygen, carbonyl carbon is electron-deficient and carbonyl oxygen is electron-rich. the electron deficient (acidic) carbonyl carbon is most susceptible to attack by electron rich nucleophilic

MT-147

Solutions

75.

reagents, that is, by base. Hence the typical reaction of aidehydes and ketones is nucleophilic addition. (d) Dihydrogen sodium phosphate (NaH2PO4) does not have a lone pair of electrons on the P atom. As such it can not act as a nucleophile and hence does not react with aldehydes and ketones.

725 - 720 cm–1.

80.

(a)

81.

Generally in case of long chain alkanes C – H rocking frequency classically characterized in the region of 725 - 720 cm–1. (b) We have 2y cos = x sin or

OH

C

76.

CH

HO

C

or 2 x cos

O C–CH3

–H2O

77.

or

2x kx

or

2 k

Acetophenone

(d) Out of given compound only acetaldehyde can form optical active hydroxy acid as it contains one asymmetric carbon atom as marked below : O CH3

|

HCN

C H

CH3 C CN

y sin

3

y 2ky

3

1 2k

3 , giving k =

[from (1)]

|

O CH3

|

(c) Chlorine is electron withdrawing group. Further Inductive effect is stronger at position than -position. i.e., CH 2 ClCOOH

79.

(c)

CH 2 Cl CH 2 OOH

eg [d 2 2 , d 2 ] x y z electronic transition If we see the d elctronic configuration of Ti, it has d 1 configuration. Since [Ti(H2O)6]3+ is an octahedral compound hence the d-orbital splitting will be d xy

Energy

d x 2–y2 , d z2

eg

electron transition

Here an electron which is present in t2g excites and transition occuring from d xy eg[d 2 2 , d 2 ] . y

z

4 y2

f (x)

y2

x2 x

1 , or x2 + 4y2 =4

x

2

( x2

2

x 1

x

x 1) 1 2

x 1

1

1 x

1 2

2

3 4

We can see here that as x , f (x) 1 which is the min value of f (x). i.e. fmin = 1. 1 2

Also f (x) is max when x

t2g e dxy dxz dyz

x

x2 4 82. (c) We have

=y

x2

cos2 + sin2 =

|

C* H COOH

78.

1 2

x and sin 2 Squaring and adding we get

CH hydrolysis

k (say)

We now get cos

OH

||

sin 2y

Then cos = kx and sin = 2ky ...(1) Again 2x sec – y cosec = 3

CH 3

H 2SO 4 /HgSO 4

(a)

cos x

2

3 is min 4

which is so when x = – 1/2 i.e. when x fmax 1

1 2

1 3/4

2

3 4 7/3

3 . 4

Rf = (1, 7/3]

EBD_7443 MT-148

83.

Target VITEEE

(d) 2 + 2x + x

86.

2

1+x

(b) We have PQ PQ

x

1 1+x

cos(tan–1x) =

1 2 2x x 2

(d) 1

sin 1

2

1 x2 1 x= 2 2

2 tan 1 x

2

2

tan–1 x

=

sin–1

tan–1 + tan–1 = 2 tan –1 x 2 tan 1 x

tan–1 tan 1

+ tan–1 = tan–1 x 1

C

Q

1 2tan–1 + 2tan–1 = 2tan–1 x

[

P

B

1 2

1 + 2x = 0 2

sin

1

87. 2x 1 x2

AB CQ and AB || CQ ABQC is a parallelogram. Q is a fixed point. (c) Since, A is obtuse angle 90

]

2x 1 x2

tan 1 x

90

180

90

B C 0

B 88.

90

A 180

(B C) 180

C

B C 90 tan B tan(90

C)

tan B cot C tan B tan C 1 (c) Chord PQ is normal to the parabola at Q and is equation is y = mx – 2am – am3 ...(1) P

x y tan–1 x + tan–1 y = tan–1 1 xy ]

[

PQ CP

A

1 x2

2 2x x

sin 1

AB PC

1

1

So,

AP PB PC

AB PQ PC = CP PQ CQ .

1

sin cot 1 1 x

84.

2

O

X

90º

x

85.

1 (b) We have (gof) (1) = g (f (1)) = g (3) = 1 (fog) (1) = f (g (1)) = f (4) = 1 (gof) (2) = g (f (2)) = g (5) = 3 (fog) (2) = f (g (2)) = f (1) = 3 (gof) (3) = g (f (3)) = g (3) = 1 (fog) (3) = f (g (3)) = f (1) = 3 (gof) (4) = g (f (4)) = g (1) = 4 (fog) (4) = f (g (4)) = f (2) = 5 (gof) (5) = g (f (5)) = g (2) = 1 (fog) (5) = f (g (5)) = f (3) = 3 gof = {(1, 1), (2, 3), (3, 1), (4, 4), (5, 1)} and fog = {(1, 1), (2, 3), (3, 3), (4, 5), (5, 3)}.

Q

The slope of normal PQ is m = tan . The joint equation of OP and OQ is obtained by making the equation of parabola y2 = 4ax homogeneous with the help of (1). Thus joint equation of lines OP, OQ is y2 = 4ax or

mx

y

2am am3 m (2 + m2) y2 + 4xy – 4mx2 = 0 ...(2)

Since

POQ =

, th e sum of the 2 coefficients of x2 and y2 in (2) must be zero. m (2 + m2) – 4m = 0 or m (m2 – 2) = 0

MT-149

Solutions 0, we get m2 = 2 or m = ± 2 .

As m 89.

Thus, m = 2 . (b) Let (x1, y1)be the mid point of the variable chord AP (where A is fixed while P varies) of the hyperbola

x2

y2

2

b2

a

xx1 a

yy1

x12

y12

2

2

b2

2

b

a

91.

(T = S1)

As it passes through the fixed point A (a, 0) ax1

x12

y12

2

2

2

a

a

4x12

b

4ax1 a

4x12

or

a

2

4y12 2

a2 ( 2x 1 a )

2

4y12 b2

0

Hence locus of (x1,y1) is

(2x a) 2

4y 2

a2

b2

92. 1

y

1 6

5 . 6

P (A wins )

1 6

5 6

2

1 6

5 6

4

probabilities 1 1 6

5 6

1 6

1 1

5 6

1 5 , 6 6

2

2

1 5 , 6 6

4

z 3 iz 2

ABC

1 ...... ] 6

94. 1 36 . 6 36 25

6 11

1

1

a

1

1

1 x 2 x 2i

(z i )(z 2

2iz 2)

0

0 1 1 2 1

1

1

1 1 1 1

1 | 2 2| 2 2

.........

6 5 . 11 11 i.e. the probability of winning of the person P(B wins)

(c)

2

5 6

x sin

cos sin )

z i, 1 i, – 1 – i Let A = (0,1), B = (1,–1), C=(–1, –1),

1 .....to 6

[Since, A can win at the 1st, 3rd, 5th, ..... mutually exclusive trials with respective 2

1

dy dx

93.

sin 2 . then

and a

sin 1 (sin cos sin

1 and that of ‘not getting a 6

three’ = 1

m!(m 1)! (m n 1)!

Pn

sin 2

(c) Put x

(a) Let the persons be A and B. Suppose that A has the first throw. We know that the probability of getting ‘a three’ in a single throw =

m 1

1

b

Pn ways.

Hence the required no. of ways = m!

4y12 2

m 1

these (m + 1) places in

1

b

a2

90.

4 x1 a

2

5 . 11 Hence the ratio of chances of their winning = 6:5. (c) The m men can be seated in m! ways When they are seated, there are (m + 1) places, shown by × where n women can sit. Then no two women would be together as shown below × M × M × M × ...... × M × Then n women can arrange themselves in

other =

1

Equation of the chord having (x1, y1) as its mid points is

6 and that of the 11

having the first throw =

(b) Given mean = np =

35 6

...(i)

35 36 Solving (i) and (ii), we get

and variance = npq =

1 1 , p= 1 6 6 we get n = 7

q=

5 6

...(ii)

EBD_7443 MT-150

Target VITEEE nC r

pr qn – r

P (x = r) = distribution where r = 7

5 6

P (x > 6) = 7C7

7

.

for Binomial 7 7

1 6

5 6

=

7

Thus R is equivalence relation. 98.

(a) It is given that a (b c) Note: a (b c)

95. (b) From the graph of f(x) = |sinx|, it is clear that f(x) is continuous everywhere but not differentiable at x = n , n Z.

1 b 2

X

– /2 O

/2

3 /2 2 5 /2 3

.

(c) We have y2

x 4a x a sin a

........( i)

dy x 4a 1 cos dx a

Since, the tangent is parallel to dy x axis , dx This gives x 4a 1 cos a

97.

1 2

a.b

1 2

99.

(a)

|A|

3 2 0 1

0

0

x cos a

1

x sin a

0

(a , b) R (a , b) (a , b) refiexive (ii) Let ad

N N . Thus R is

bc (a, b) and (c, d)

100. (a)

3 0

1

3

0

0

1

2 3

0

1 1 3 0

A

1

exists

2 3

2 3

(A 1 ) 3

1 1 27 1

1 1 27 0

8 1 9

1 = ) 2

. ( cos–1

3

2 1 3 0

2

0

2 1 3 0

1 1 27 0

3

Given curves are y and 2y – x + 3 = 0

x

2 3

26 27

.... (i) .... (ii)

Y

y= x

N N

bc ad cb da (c, d) R (a , b) . Thus R is symmetric (iii) Again let (a,b) R (c,d) and (c,d) R (e,f) ad bc and cf de adcf bcde af de Thus R is transitive.

A

0 i.e., a . c

1

Adj. A

From (i) y 2 4a x 0 y 2 4ax, which is the required locus. (b) The relation is defined on N N . since the elements of N N are ordered pairs hence the relation on N N relates two ordered pairs. Now, (i) a , b N ab ba

(a, b) R (c, d)

a .c

1 and a . b = 0 2 Since a , b , c are the un it vectors therefore

Differentiating w.r. t x, 2y

(a .c)b (a .b)c

1 ) b (a .b)c 0 ...(i) 2 Since b and c are non-parallel, therefore for the existence of relation (i), the coeff. of b and c should vanish separately.. Therefore, we get

f(x) = y = |sin x|

96.

(a.c)b (a .b)c

(a .c

Y

– 3 – 3 /2 – 2 – 3 /2 –

1 b 2

X

2y – x + 3 = 0 X

O

3 –3/2

(a , b) R (e, f ) Y

MT-151

Solutions On solving eqs. (i) and (ii), we get 2 x

x x

2

2

3

0

2 x

x 3

x 2 y 2 z 2 ax by cz 0 Radius of this sphere is equal to 2k

3

0

x 1

0

a 2 b 2 c 2 4(2k) 2 16k 2 Also let (z, y, z) be the co-ordinates of the centroid of the tetrahedron OABC, then

x 3 ( x = –1 is not possible) y=3 Required area

=

3 x 0 2

3 0

x1 dy

y 2 dy

2y 3

9 9 9

2

Since cos x sin x

x+

4

1

2

2

cos x

1, 1

4

3 7 , 4 4 102. (c) Let the roots of the equation be 2 , then m l

and

l and 3(m l ) Eliminating , we get l2 9( m l ) 2

1 2(l m)

2 2

0, t 1, When

I

2

, t

0

and

1 2(l m)

t1 t 3

2 2 = 3t

.

2l 2 9l 9m

2

1

1 t2

.

dt

0

0

5 t2

dt

0

1

9 8 103. (c) Let co-ordinate of the points A, B, C be (a, 0, 0), (0, b, 0) and (0, 0, c) respectively. Then equation of the sphere OABC is

m

cos .sin 2 .sin d

0

1 l m

Since l is real, so (–9)2 – 4 . 2 . 9m

sin3 d

0

x=

2

k2 .

Put cos t sin d dt [Note that power of sin is odd] also, when

= 0, 2 , 4 6 , ..... or , 3 , 5 , .....

l

z2

0

2,

1 1

cos

104. (a) I =

2

We must have cos x sin x 4

y2

16k 2

9

101. (c) Given equations is cos x sin x

cos x

16x 2 16y 2 16z 2

or x 2

3

y3 3 0

= y 2 3y

1 1 1 a, y b, z c 4 4 4 so that a = 4x, b = 4y, c = 4z Eliminating a, b, c from (1) and (2), the required locus is x

7

2 2 t 7

1

0

2 3

2 7

8 21

105. (a) Since each number consisting of 6 digits starts with 35, 3 and 5 are fixed in the first and second places. The other four places can be filled up with remaining 8 digits in 8P4 ways = 8 . 7 . 6 . 5 = 1680 ways. Hence the required number of telephone numbers is 1680. 106. (b) The given equation can be converted to linear form by dividing both the sides by cos2y. We get dy 1 sec 2 y 2 tan y x 3 ; dx x

EBD_7443 MT-152

Target VITEEE Put tan y

sec 2 y

z

dy dx

dz dx dz dx

The equation becomes

2 z x

x3 ,

=

which is linear in z The integrating factor is 2 dx x

2 log x

e e I.F. e Hence, the solution is z( x 2 )

log x 2

x 3 ( x 2 ) dx a

x

zx 2

110

2

1

1

0

0

0

0

2 3

1

2 3

have 1

dv dt

differential v

dv v

dt

dv . Hence, dt equation is

t c

109. (a)

Given that a * b

(3) 2 , i.e.,

v V

t

v

ab a, b 4

Q

e

2 3 , , . 29 29 29 Projection of PQ on RS 4 29 29

2

( 4 2) ( 8 2 6)

Ve

t

3

( 4 3)

29

29

(6 4 )

0

which implies that PQ RS 111. (c) As given, Mean = np = 3 and Variance = npq = 2 npq np

...(ii)

When t 0, v V C log V The equation (ii) becomes log v = – t + logV t

( 2) 2

4

are

2 3

q=

and p = ( 1– q) = 1 –

...(i)

Integrating, we get log v

v log V

Dividing each by ( 4) 2

1

1.

0 or

108. (d) The retardation at time t the

The four points are P (–2, 3, 4), Q (–4, 4, 6), R (4, 3, 5), S (0, 1, 2). direction ratios of RS are 0 – 4, 1 – 3, 2 – 5, i.e., 4, 2, 3.

=

= 3( 1)(1 ) 3(1 2 ) For the system to be consistent, we must 2

S

29 , the actual direction cosines' of RS

we use R1 R 1 R 2 followed by C 2 C 2 C1 to obtain 1

P

(a)

0

To evaluate 1

3 160

R

2 2 3

1 40 4

a,

(6tan y)x 2 x 6 6a 6x 2 tan y x 6 c, [c = 6a] 107. (a) The system of equations will be consistent if 1

3

1 3* 40

Q

x6 6

a is constant of integration.

1

1 1 3* 5 2 4

1 1 3* * 5 2

2 3

2 3

1 3

1 n=9 3 3 P(X > 8) = P(X = 8) + P(X = 9) n.

9 = C8

=9

1 3

1 3 8

8

2 3

2 1 1 3 3

9

C9

9

=

1 3

9

18 1

19

9

39

3

MT-153

Solutions 112. (c)

Tn

n2 (2n 1) . 2n . (2n 1) . (2n 2) n 2(2 n 1)(2 n 1)(2 n 2)

1 1 = 2 12 (2n 1)

1 4 (2n 1)

1 3 (2n 2)

1 1 24 2n 1

3 2n 1

4 2n 2

1 1 = 24 2n 1

4 1 2n 1

4 2n 2

=

1

1

1

1

4

1

1 1 1 3 3 5

1 3

1 4

1 5

1

1 .... 7

1 1 ... 5 6

1 = 24 1 4 log e 2

1

1 2

1 1 1 (4 log e 2 1) log e 2 24 6 24 Note that sum of n terms of the first series

1 . which tends to 1 as n 2n 1 Thus the sum of the first infinite series is 1. 113. (b) At time t, let OA = x and OB = y. dx 1 2 ft./sec. Then dt 2

is 1

Y

O

Putting the values of y and

2a ...(i) m If (i) is a tangent to the circle, x2 + y2 = 2a2

X

dy . dt From right angled AOB, we have AB2 = OA2+OB2 132 = x2 + y2 ....(1) Differentiating with respect to t, we get

2a

2a

m m2 1 m2(1 + m2) = 2 ( m2 + 2)(m2 – 1) = 0; m = 1. So from (i), y = (x + 2a). 115. (c) Let the no. of pencils in a bag be x Let the no. of pens in a bag be y. There should be at least 5 items in a bag we have x + y 5 cost of pencils in a bag = `5x cost of pens in bag `10y Total cost of a bag = 5x + 10y, The total cost has to minimized Objective function is minimize C = 5x + 10y subject to x + y 5, x 0, y 0 sin 2 x 4

cos x

13ft

and we have to find

0

y = mx+

116. (c) Let I 2, 4 A

5 dy 5 2 dt

dx in (2), dt

dy dy 0; 6ft / sec . dt dt Hence the end B is moving at the rate of 6ft/sec. The negative sign shows the downward tendency of B i.e.y decreases when t increases) 114. (b) Any tangent to the parabola y2 = 8ax is

then,

=

B

5.

or 30 5

Putting n = 1, 2, 3, 4 and adding, we get 1 24

13 2 12 2

or y

we get 12

= 2 2n 1 2n 1 4 2 n 1 2 n 2

S

dx dy y 0 ....(2) dt dt When OA = 12, i.e. x = 12 then from (1), 132 = 122 + y2 x

sin x 3

3 cos x

dx

1 tan x 3

1 tan x 1 tan 2 x 3 1 tan 3 x 3

c

sin x

1 I 0, 2 3

3

3 . cos x c

1 tan x 3

c

EBD_7443 MT-154

Target VITEEE ALTERNATE : sin 2 x

tan 2 x . sec 2 x dx

dx

4

cos x

put tan x

of eq. (ii) and the given line are perpendicular. ˆi ˆj 2kˆ .

t

b1 b 2 2b3

1 3 1 t c tan 3 x c 3 3 117. (b) We have ae = 5 [Since focus is (±ae, 0)] t 2 . dt

a and e

36 5

b1 b2 Similarly,

is 118. (d)

x2 36

y2 11

5 6

1

2

n((a n)n 1)

119. (c)

ˆi 2ˆj 3kˆ

...(iv)

b2 2 3 1 1

b3 1 1 3

1

b 1ˆi b2ˆj b3kˆ

3iˆ 5jˆ 4kˆ

Substituting the value of b in eq. (i), we obtain

0

0

r

1 (a n)n 1 a n . n The equation of line passing through (1, 2, 3) and parallel to b is given by r

b1 1 1 2

b tan x x

0

b1 b 2 b3 3 5 4 Therefore, the direction ratios of b are –3, 5 and 4.

1.

sin nx lim n . lim (a n )n x 0 nx x 0

0

b1ˆi b 2 ˆj b3 kˆ 0 (3b1 + b2 + b3) = 0 ...(v) From eqs. (iv) and (v), we obtain

25 b a (1 e ) 36 1 11 36 Thus, the required equation of the ellipse 2

2b3

0

3iˆ ˆj kˆ .

a e

since directrix is x

On solving we get a = 6 and e = 2

b1ˆi b 2 ˆj b3 kˆ

b1iˆ b 2 ˆj b3 kˆ ...(i)

ˆi 2jˆ 3kˆ

3iˆ 5jˆ 4kˆ

This is the equation of the required line. 120. (b) 121 (d) 122 (b)

The equations of the given planes are

123 (b)

r. ˆi ˆj 2kˆ

124 (d) Foment means to instigate or stir up (an undesirable or violent sentiment or course of action). Therefore, Incite is the synonym of Foment.

5

...(ii)

6 and r. 3iˆ ˆj kˆ ...(iii) The line in eq. (i) and plane in eq. (ii) are parallel. Therefore, the normal to the plane

125 (c)

MT-155

Solutions

MOCK TEST 4 1.

(b) In shell , q charge is uniformly distributed over its surface it behaves as a conductor.

7.

X q

X R

X X

X X

8.

q

and inside V 4 0R 4 0R Because of this it behaves as an equipotential surface. (b) Net field along, AB

kQ

a2

a

2

=0 E

2 –Q

–Q

Q 2 1

Q 2

(c) W= U =

4.

(a) In air Fair =

4

6.

r

4

100 25

i.e., V25 and V2

q1q 2 0

Kr 2

i.e., V100

Fair Fm 1 Fm Fair K K (decreases K-times) (c) By using m = ZIt Where, Z is the electrochemical equivalent of copper. m 30 10 5 1.5 10 60 = 0.27 gm. (a) Kirchoff's first law deals with conservation of electrical charge & the second law deals with conservation of electrical energy.

r1 r2

R

R = r 1 – r2 (a) As for an electric appliance R (Vs2 / W ) , so for same specified voltage Vs

So, V1

2

1

2r1

0

4

i.e, R 25 4R with R100 R Now in series potential divides in proportion to resistance.

Q (2 2 1) 4

q1q 2 0

2Er1 (r1 r2 ) R

2Er1 r1 r2 R

R 25 R 100

1 2 q /C 2

1

In medium Fm =

5.

q

2q

1 CV 2 2

3.

9.

–Q

–Q

2E r1 r2 R

P.D. across first cell = E – ir 1 E r1 E (r1 r2 ) R Now, E

kq

( 2a ) 2

V 5V R eq 18 (d) Current in the circuit E E r1 r2 R

q

2

18 5

i

X

kQ

6 9 6 9

R eq

V = potential at surface

2.

(d) It is a balanced Wheatstone bridge. Hence resistance 4 can be eliminated.

R1 V (R 1 R 2 )

4 440 352V 5 R2 V ( R1 R 2 ) 1 440 5

88V

From this, it is clear that voltage across 100 W bulb (= 88 V) is lesser than specified (220 V) while across 25 W bulb (= 352 V) is greater than specified (220 V), so, 25 W bulb will fuse. 10. 11.

Change in length Original length (d) When the deflection produced by electric field is equal to the deflection produced by

(a) strain =

EBD_7443 MT-156

Target VITEEE magnetic field, then the electron can go undeflected.

0i and so it is independent of 2 r thickness. (a) Let I be current and l be the length of the wire.

B

12.

0 In

For Ist case : B 2 r

13.

2r

0 nI

0 2I

2r '

2

15.

16.

r'

2(2 r' ) 4

0I

l 4

l 4

21.

CD. (d) For hydrogen atom

1 C

2

8

8

(31 25)2 82

6 2 82

Ep

dU

1

R2

1 25

Ns Es Es = 25 Ep But EsIs = EpIp 25Ep × 2 = Ep × Ip (d) Quality factor

1

1

n12

n 22

y 1 (100 – 0)

2, n 2

v0 ; n1

v0

R

1

(2)

1

1 2

(1)

R

2

1 1 4 1

3 4 R or R v0 4 3 For second condition v0

4 (constant) n1 = 4, n2 = 1 3

R 1

4 1 v0 3 (4)2

4 1 1 v0 3 16 1 22.

(a)

23.

(b)

QV

24. 25.

1 (1)2

4 v0 3

15 16

5 v0 4

4 3 r g and 3

QV

QV 4 (2 r )3 g so QV 3

or Q

8 QV V

Ip = 50 A

I L L = IR R (c) The heat lost in condensation = x × 540 cal. x 540 = y 80

R

1

Potential drop across capacitor or inductor = Potential drop across R

17.

(d) | F | = , which is greatest in the reagion dx

For first condition

0.8 10 (d) When there is change of flux in the core of a transformer due to change in current around it, eddy current is produced. The direction of this current is opposite to the current which produces it, so it will reduce the main current. We laminate the core so that flux is reduced resulting in the reduced production of eddy current.

(b)

20.

R

(b) Power factor,

Np

19.

4B

l

L

14.

where

l

1 . 3

=

(c) Interference is a wave phenomenon shown by both the light waves and sound waves. (c) Two sources of light are said to be coherent if they emit light having a constant phase relationship.

18.

l and n = 1

For IInd Case : l B'

0I

x y

or

8Q

800 3200

8 2Q

(c) (b) Since work function for a metal surface is W

hc 0

where

0

is threshold wavelength or cut-

MT-157

Solutions off wavelength for a metal surface. here W = 4.125, eV = 4.125 × 1.6 × 10–19 Joule

26.

27. (d)

28.

6.6 10

0

(c) de Broglie wavelength,

[

m mp

E K.E ( p

31.

32.

33.

3 108

E= h wF KE max i.e. till a certain valve of , KE remains 0, it only starts increasing once the Work function (WF) of the metal surface is achieved. By Newton’s second law of motion F = n(mv) = nmv

p

29. 30.

34

3000A º 4.125 1.6 10 19 (b) As per Einstein’s photoelectric equation :

so

=

)

can cross over to conduction band at 0 K. But at room temperature some electrons in the valence band jump over to the conduction band due to the small forbidden gap, i.e., 1 eV.

=

34.

(c)

35.

(a)

36.

(b) A

10

V

Y1 Y Y2

Y1

2mEK.E

4m p

37.

E K.E ( p) ]

A B, Y2

= mg 38.

A.B

Y (A B).AB AB BA This expression is for XOR (c) Change in momentum of the ball = mv sin – (– mv sin ) = 2 mv sin

2 1

(a) (a) Binding energy E = [Zmp + (A – Z)mn – M]c2 It can be shown as Mass defect M = Mass of nucleons – Mass of nucleus. From Einstein mass energy relation E = Mc2 E = {Zmp + Nmn) – M}c2 If A is the mass number, then N = A – Z Binding energy E = [Zmp + (A – Z)mn – M]c2 (d) -rays are electrically neutral. The -rays are electromagnetic waves like X-rays. They are called -photons. Thus, there will be no change in either mass number or atomic number of the element. (a) Let no. of -particles emitted be x and no. of particles emitted be y. Diff. in mass no. 4x = 238 – 206 = 32 x = 8 Diff. in charge no. 2x – 1y = 92 – 82 = 10 16 – y = 10, y = 6 (c) In semiconductors, the conduction is empty and the valence band is completely filled at 0 K. No electron from valence band

V0

B

h

mp

V

2vsin g

= weight of the ball × total time of flight (d) Let the velocity of the particle be v m/s. Momentum of the particle (p) = mv Kinetic energy of the particle (E) =

1 mv 2 2

1 (mv )2 . 2 m

p2 2m (a) Let the readings of two thermometers coincide at C = F = x

E

39.

As

C 5

F 32 9

x x 32 5 9 or 9x = 5x – 160 4x = – 160, x = – 40°C

40. (d) K

P V /V

h g V /V

200 103 10 0.1/100

2 109

EBD_7443 MT-158

41.

Target VITEEE

(b) Dehydration of CH3 OH gives carbene (methylene), an unstable intermediate. H 2SO 4

CH 3OH

42. 43.

44

45.

(a) Galvanization

50.

(c)

Carbene

O

O

C CH 3

C OCH 3

||

R CONH 2

Br2 / KOH

R CONH 2

Br2 / KOH

51.

52.

(c)

H O —C +C—

Benzoin

53.

(d) Aldehydes which contain a -hydrogen on a saturated carbon, i.e., CH3CH2CHO undergo aldol condensation.

46.

(c)

CH3CH 2 – C

2.303RT log K eq nF

127 2

48.

76

1 2

of Cl –

of Ba

139.5 ohm

CH3 CH3

.0591 log K eq n

(a) The equivalent conductance of BaCl2 at infinite dilution of BaCl2

|

O H H C CHO

propanol

||

C OH

.0591 log 10 6 = .0591 × 3 = 0.1773 = 2

47.

CH3

H

O

E cell

KCN (alc)

O H H O —C +C— OH

||

N

RNH 2

This is known as Hoffmann’s Bromamide reaction. (b) When benzaldehyde is refluxed with aqueous alcoholic potassium cyanide, two molecules of benzaldehyde condense together to form benzain

|

C

RNH 2

This is known as Hoffmann’s Bromamide reaction.

[: CH 2 ] H 2O

(a) The streching frequency for C = O in case of aliphatic ketones is 1715 cm –1. (a) Due of chelation (intramolecular Hbonding), o-nitrophenol has less b.p. than the p-nitrophenol in which association of molecules is possible due to intermolecular H-bonding. Hence the two isomers have considerably different boiling points. (c) Ethers react with Cl2 in presence of light to form fully chlorimated product (all hydrogen atoms of the alkyl groups are replaced by chlorine). (b) Only –OCOCH3 is electron-releasing due to presence of lone pair of electrons on the key atom. .. O ..

49.

–1

cm

OH

|

|

CH3CH2 C |

CHCHO

OH

3-hydroxy-2-methyl pentanal

54.

(a)

When 3 molecules are combined in presence of dry HCl gas they condense to form phorone. CH3

CH3 C = O + H2 CHCOCH H2 + O = C

CH3

2

(c) d-d electronic transition from t2g eg state associated with an amount of energy which comes under visible green region. Here in the given compound Ni2+ is d8 system so it is an octahedral compound hence splitting pattern will be

HCl –2H 2O

CH3 CH 3

CH 3 C CH.CO.CH

C CH 3

CH 3

Note : Two molecules of acetone condense to form mesityl oxide. CH3

C = O + H2 CHCOCH3 CH3

eg electronic transition

t2g

CH 3 C = CH.CO.CH3

CH 3

MT-159

Solutions 55.

(c) When glycerol is heated with oxalic acid following reaction occurs. CH 2OH HOOC |

CHOH |

|

HOOC oxalic acid

CH 2OH |

CHOH

CHOH CH 2OH |

CHOH |

CH2OH

61.

CH2OH

HCOOH

56.

(d) R COCl + AlCl3

57.

(d) The total number of isomers for the complex compound

R C

Electrophile

62.

[Cu ( NH 3 )Cl 3 ] [Pt ( NH 3 ) 3 Cl] ,

2.303

log

Activity of fresh wood Activity of dead wood

0.693 t1 / 2 t

15.2 2.303 5760 log 7.2 0.693

( 2) 3

125 .

T = 16

T t1 / 2

16 4

4

( T = n × t½)

Where Å = initial concentration & A = concentration left after time t (b) The presence of enzyme (catalyst) increases the speed of reaction by lowering the energy barrier, i.e. a new path is followed with lower activation energy. ET E'T Energy

[CuCl 4 ][Pt ( NH 3 ) 4 ] and Cu( NH 3 ) 4 PtCl 4 . The isomer [Cu (NH3)2 Cl2][Pt (NH3)2 Cl2] does not exist due to both parts being neutral. (b) There is a steady decrease in the radii as the atomic number of the lanthanide elements increases. For every additional proton added in nucleus the corresponding electron goes to 4f subshell. The shape of f -orbitals is very much diffused and they have poor shielding effect. The effective nuclear charge increases which causes the contraction in the size of electron charge cloud. This contraction in size is quite regular and known as Lanthanoid contraction. Since the change in the ionic radii in the lanthanide series is very small, thus their chemical properties are similar. (a) Age of artifact t =

1000

æ 1 ön æ 1 ö4 0.12 = 0.0075 M A = A o çç ÷÷÷ = 0.12 ´çç ÷÷÷ = çè 2 ø èç 2 ø 16

–

+ AlCl4

[Cu II ( NH 3 ) 4 ][Pt II Cl 4 ] is four.. These four isomers are [Cu ( NH 3 )3 Cl] [Pt ( NH 3 )Cl 3 ],

(a) t1/2 = 4 s n

Formic acid

O

59.

3 . The rate of

dissociation after 3t1/2 =

|

CH 2OH H 2O

3 1

|

100 110 C H 2O

|

58.

(c) Number of t 1 / 2

CH2OOC.COOH

CH 2OOCH CO 2

60.

Ea

Ea

Products 1

Reactants + catalyst Progress of reaction

63.

Here ET is the threshold energy. Ea and Ea is energy of activation of reaction 1 in absence and presence of catalyst respectively. (d) k = Ae–Ea/RT or log k = log A -

Ea 2.303 RT

Comparing the above equation with y = mx + c 1 T Thus A plot of log10k vs 1/T should be a straight line, with slope equal to – Ea/2.303 RT and intercept equal to log A

y = log k, x

Slope =

log k

5760 years 1/T

Ea 2.303RT

EBD_7443 MT-160

Target VITEEE

64.

65.

66.

O

Ea 2.303R

Slope

or E a –2.303R Slope (d) The activation energy of reverse reaction will depend upon whether the forward reaction is exothermic or endothermic. As H = Ea (forward reaction) – Ea(backward reaction) For exothermic reaction H = –ve – H = Ea(f) – Ea(b) or Ea(f) = Ea(b) – H Ea(f) < Ea(b) for endothermic reaction H = + ve H = Ea(f) + Ea(b) or Ea(f) = H + Ea(b) Ea(f) > Ea(b). (b) According to equation 2HI

H2

At t = 0 (2 moles) At equilibrium (2 – 2 ) moles

0 mole

+

I2 0 mole

C O

'P' Benzoic acid

71. (a) COCH3

+ N2

67.

Cl

–

OH H

CH3

73.

(c)

CH 3COONH 4

heat H 2O

CH 3 CONH 2 X

p- cresol

P2O5 H 2O

(c)

CH 3CN

H3O

CH 3COOH Z

Y

Sn HCl

(B) N C

Nitrobenzene

(A)

74.

(a) Measure of disorder of a system is nothing but Entropy. For a spontaneous reaction, G < 0. As per Gibbs Helmnoltz equation, G= H– T S Thus G is –ve only When H = –ve (exothermic) and S = +ve (increasing disorder)

75.

(a)

C HC l 3 KOH

NH–CH3

(C)

Re duction Na/C2 H5OH

N-methylaniline

70.

MgBrC2H5

(c) Remember that alkyl group present on benzene ring when oxidised by strong oxidising agent is always oxidised to – COOH group, ignoring its length.

CH3

reduction

69.

OMgBr

+

72.

NO2

68.

Friedle Craft reaction

(C)

OH + N2+HCl

C6H6 AnhydAlCl3

C2H5 – C – CH3 ether C2H5 – C – CH3 hydrolysis

+ CH3–C–OH

H 2O

(b)

CH3COCl [A]

CH3COOH +PCl5

O

CH 3–C–O –C –CH 3

H + Mg(OH)Br

NH–C–CH3

O

+

O

O O

H3O

(i) CO2

Total moles at equilibrium = 2 – 2 + + = 2 mole (d) Aniline when treated with acetic anhydride forms acetanilide (nucleophilic substitution) NH2

C OMgBr

MgBr

(c) See action of HONO on P, S and T nitro alkanes. (b) Grignard reagent forms addition product with bubbled carbondioxide which on hydrolysis with HCl yields benzoic acid.

S

H T

S( per mole )

H per mole T

21.98 JK 1mol 1

6000 273

MT-161

Solutions 76.

(b)

G = – P V = Work done V is the change in molar volume in the conversion of graphite to diamond. V

12 12 3.31 2.25

10 3 L

sin 2 cot

2 2

1.91 10 3 L

1 (2)

Work done = –(–1.91 × 10–3) × P × 101.3 J P

1895J mol –1

1.91 10 3 101.3 1 atm = 105 × 1.013 Pa

77.

78.

79.

Pr oteins

Pepsin/ HCl Stomach

82.

Peptidases (Intistine)

1

4 cos ec cot

2

3 ) 2 (cos ec cosec 2

4

8 0

4 cos ec

5 0

2) 2

0

1

3 and cosec = – 2

2n

6

6

,n

I

84.

(d)

Peptides

sin 2 cos 1

Amino acids 2 sin cos 1

2sin

2

85. 7 25

2 25

4 5

(b) We have f (x) = x2 + 7, x R. f(–2) = (–2)2+ 7 =11, f (2) = (2)2 + 7 = 11 The images of distinct elements –2 and 2 of R are equal. ‘f’ cannot be one-one.

1 tan 2

9 16 9 1 16

2

(cot

3 5 3 5

3 cos 1 5

6 3 sin cos 1 5 5

1

7 25

1

83.

1 tan 2

3 cos 2 tan 1 4

2 3 cot

General solution is

2 tan

1 tan

1

tan

lies is the 4th quadrant and

O

and cos 2

1

tan

cot

N C

sin 2

27 25

3)

cot 2

(b) Peptide bonds are present in enzyme.

(c)

4 5

2

(cot

H

81.

2

(c) Given equation is

2 cot 2

Proteases and Peptones

Trypsin Chemotrypsin (Pancreatic juice Intestine)

80.

tan

1

sin 2 tan

Hence, given expression

9794 atm

P 9.92 108 Pa (c) In fluorite structure each F – ion is surrounded by four Ca++ ions whereas each Ca++ is surrounded by eight F– ions, giving a body centred cubic arrangement. Thus the co-ordination number of Ca++ = F– are 8 and 4. (a) p-type of semiconductors are produced (a) due to metal deficiency defects (b)by adding impurity containing less electrons (i.e. atoms of group 13) Ge belongs to Group 14 and In to Group 13. Hence on doping, p-type semicondutor is obtained. (b) Pepsin and Trypsin are two enzymes involved in the process (hydrolysis of proteins)

1 2

1

cos cos 1

3 5

3 5

6 9 sin sin 1 1 5 25

24 6 4 5 5 25 (c) n(A) = p, n(B) = q n(A × B) = pq = 7 So, possible values of p, q are 7, 1 p2 + q2 = (7)2 + (1)2 = 50.

EBD_7443 MT-162

86.

(b)

Target VITEEE log e (1 x 2 tan x )

f ( x)

sin x

3

1 1

1

;

(

f(0) must be equal to Lt f ( x ) x

3

( 1)

1) (3

0

1 1

1 3 1

( 2)

1) 2(1 3 )

0

1 . 5

0

A (1, 0, 4)

1

Lt

x

2 x 2 tan x log(1 x 2 tan x ) x tan x

sin x 3

0

1

tan x x 3 2 log(1 x 2 tan x ) x tan x x 0 x sin x 3 Lt

= (1) (1) loge e = 1

Given: x2 – y2 sec2 = 4 and x2 – sec2 + y2 = 16 x2 4

1 5 , 1 1 5

y2

4 4cos 2 4

88.

89.

1 + cos2 = 3(1 – cos2 ) 4 cos2 = 2 1 3 cos , 4 4 2 (d) Let M be the foot of the perpendicular drawn from the point A to the line BC. Since the point M lies on the line BC, it must divide BC in some ratio, say : 1, ( 1) . Then coordinates of M are given by

0

.( 3) 5 , 1

2 1 , 1 2

2 1

i.e.,

3

,

1,

3

1

1 1

0,

4

1 3 1 1 3 . , , 1 1 1

Now AM is to BC. Using the condition of perpendicularity, we get

log

2 x 2 x

log

2 x 2 x

1

2 x f (x) 2 x f(x) is an odd function 1

log 1

90.

2 x dx 2 x

0

(d) Since both the roots are less than 2, we have 22 5 (ii) 4f (2) > 0 4 (25p2 – 25p – 50) > 0 p2 – p – 2 > 0 p < –1 or p > 2

(i)

D>0

15p – 66 < 0

p
is not a monoid. 110. (b) The given equations are ....(i) 3l m 5n 0 and 6mn 2nl 5lm 0 ...(ii) From (i), we have m = –3l – 5n. Putting m = –3l – 5n in (ii), we get 6( 3l 5n )n 2nl 5l ( 3l 5n) 0

1

3

7

7 10

1

1

ALTERNATE : The equations are 3l m n and 6nm – 2nl + 5lm = 0

P (B A) P(A)

P (B / A)

i(cos

i( ie iz )

i sin ) iz

e ie = ie = ie 113. (c) We suppose the distribution to be Binomial with n = 10, p = 0.2, q = 1 – p =0.8

MT-167

Solutions The probability that not more than one defective is found. P( k

n pqn 1

qn

0) P (k 1)

x

1

(0.8)10 10(0.2)(0.8)9 . (0.8)9 [0.8 2]

exists and for a > 0, lim f (x) = lim x – 1 = a – 1 exists.

0) P(k 1)

e m

k = 0 or k =

X

P(1, 0)

Thus equation of AB is 1 . (y 0) = x.1 – (2 + 1) + 3 i.e., y = 4 2 x coordinates of points A and B will be given by, x2 – 2x + 3 = 4 i.e., x2 – 2x –1 = 0 x = 1

Thus AB = 2 2 units. x 1, x 0 0 , x 0 x 1 , x 0

Let us first check the existence of limit of f (x) at x = 0. At x = 0, RHL = lim f (x)

1 (2 2).4 4 2 sq. units 2 Now area bounded by line AB and parabola is equal to

Hence

1

2

1

2

PAB

(4 2 (x 2

2x 3)) dx

0

f (0 h) lim(0 h) 1 = lim h 0 h 0

= lim h – 1 = 0 – 1 = – 1 h

0

LHL = lim f (x) = limf (0 h) x

h

0

= lim h

0

0

(0 h) 1

1 0 1 1 = limh h 0 RHL LHL At x = 0, limit does not exist. Note that for any a < 0 or a > 0, limf (x) exists, x

B

2

| x| 1 , x 0 f (x) = 0 , x 0 | x| 1 , x 0

x

y=4

A

me m

5 13 all the four points are distinct)

k(13k – 5) = 0

a

y = x2 – 2x + 3

e 2 2e 2 3e 2 114. (a) The equation of the circle through (1, 0), (0, 1) and (0, 0) is x2 + y2 – x – y = 0 It passes through (2k, 3k) So, 4k2 + 9k2 – 2k – 3k = 0 or 13k2 – 5k = 0

But k 0 [ 5 k . 13 115. (b) Given,

x

2.8(0.8)9

This value is very small so the Binomial probabilities are approximated by Poisson probabilities then m np 10 0.2 2 The probability that not more than one defective is found. P(k

a

Hence, lim f(x) exists for all a 0. x a 116. (c) Let the drawn tangents be PA and PB. AB is clearly the chord of contact of point P.

a

as for a < 0, limf (x) = lim – x + 1 = –a + 1 x

a

x

a

4 2 sq. units. 3 Thus required area =

4 2 8 2 sq. units. 3 3 117. (c) Let any point P(h, k) will satisfy y2 = 4ax i.e, k2 = 4ah ...(i) Let a line OP makes an angle from the xaxis. PA In OAP, sin OP k sin l = 4 2

EBD_7443 MT-168

Target VITEEE k = l sin and cos cos

Y

where p(x) =

OA OP h h = cos l

1 2x 1

Now, Domain of p(x) exist when 2x – 1 0 x=

1 and 2x – 1 > 0 2

x=

1 and x 2

P(h, k)

X

l

k

h

A

O

and q(x) = 1 x 2

X

1 2

1 , 2

x

and domain of q(x) exists when 1 x2

Y Hence, from eq.(i), we get l2sin2 = 4a × l cos (put k = l sin , h = l cos ) l

4a cos

sin 2 118. (b) For the balloon, u 0, f

5 122 cm / sec 2 8

981 8

g cm / sec 2 8

and t = 30 seconds. Let h be the height and v the velocity of the balloon after 30 seconds. Then v

0

g .30 8

and 0

1 ,1 2 120. (a) In the absence of any condition, no book need to be selected on a particular subject and all the selected books may be on a single subject. If xn stands for the selection of n (0 n 8) books on one subject, the total number of ways of selecting 8 books out of the lot.

= the coeff. of x8 in (x 0 x1 x 2 .... x 8 )4 , 4 being the number of subjects = the coeff.

1 2 gt 2 or 2 t 2 15t 225 0 or ( t 15) ( 2t 15) 15 t 15 sec . sin ce t 2

119. (a) Given, f(x) =

15 gt 4

1 2x 1

= p(x) – q(x)

1 x2

|x| 1

1

Common domain is

1 g 2 . .30 2 8

225 g 4

x2

1 x 1

15 g cm / sec 4

225 g 4 Now the initial velocity of the released body will be 15/4 g cm/sec in the upward direction. Suppose the body reaches the ground in t sec. h

0

of x8 in

1 x9 1 x

4

= the coeff. of x8 in (1 x) 4 (1 x 9 ) 4 = the coeff of x8 in (1 x )

4

11

C8

165

121. (a) 122 (d) 123 (c)

0

124. (a) Abate means to reduce in degree or intensity; Augment means to increase; Provoke means to stimulate or give rise to (a reaction or emotion, typically a strong or unwelcome one) in someone; Wane means to decrease gradually in size, number, strength, or intensity 125 (a)

MT-169

Solutions

MOCK TEST 5 1.

(c) Let us place the charges as shown on the regular hexagon

5.

(a) Let unknown resistance be X. Then condition of Wheatstone's bridge gives X 20r , where r is resistance of wire per R 80r cm.

Q +

P

U

+R +T

R=1

X

S

E due to each charge at O is same in magnitude, the arrow shows the direction of E due to each charge

2.

100°C Steam

(H1= m × 540)

100°C Water

20 1 R 1 0.25 80 4 (a) R1 = 3.1 at T = 30°C R2 = 4.5 at T = 100°C We have, R = R0 (1 + t) R1 = R0 [1 + (30)] R2 = R0 [1 + (100)]

6.

90°C Water

3.

Q1 Q2

4.

(b)

C1V C2V

C1 C2

(100 – 20) cm

X

[H2= m ×1× (100× 90)]

Heat gained by water (20°C) to raise it's temperature upto 90°C = 22 × 1× (90 – 20) Hence , in equilibrium, heat lost = Heat gain m × 540 + m × 1 × (100 – 90) = 22 × 1 × (90 – 20) m = 2.8 gm The net mass of the water present in the mixture = 22 +2.8 =24.8 gm. (b) In parallel, potential is same, say V

Q = 80r

P = 20r 20 cm

E R E u E 0 where E 0 is electric field due to +ve charge placed at R (a) Let m g of steam get condensed into water (By heat loss). This happens in following two steps.

R1 R2

1 30 1 100

3.1 4.5

1 30 1 100

E1

7.

8.

9.

r (d) V = 11 r1

0.0064 C

1

E2

18 12 r2 2 1 1 = 1 1 2 1 r2

14 V

(Since the cells are in parallel). (c) A galvanometer can be converted into a voltmeter by connecting the high ressistance in series with the galvanometer so that only a small amount of current passes through it. (d)

r

mv qB

r

v B

EBD_7443 MT-170

10.

11.

12.

13.

14.

Target VITEEE

(c) For a perfectly rigid body strain produced is zero for the given force applied, so Y = stress/strain = (a) If the electric field is switched off, and the same magnetic field is maintained, the electrons move in a circular orbit and electron will travel a magnetic field to its velocity. (d) Power of heating coil = 100 W and voltage (V) = 220 volts. When the heating coil is cut into two equal parts and these parts are joined in parallel, then the resistance of the coil is reduced to one-fourth of the previous value. Therefore energy liberated per second becomes 4 times i.e., 4 × 100 = 400 J. (a) Eddy currents are produced when a metal is kept in a varying magnetic field. (d) Self inductance of a solenoid = n2

So, self induction

15.

n2 A

1 C (R

So, inductance becomes 4 times when n is doubled. (c) The time constant for resonance circuit,

c

= 1014 Hertz However, when frequency is higher than this, wavelength is still smaller. Resolution becomes better. (c) For first minimum, a sin = n = 1 5000 10 –10

sin

19. 20.

1 = L) 2 f (R 2 f L)

3 108

3 10 6

18.

L

(b) Size of particle

a

0.001 10 – 3

0.5

= 30° (c) (d) According to law of conservation of momentum the third piece has momentum

ˆ kg ms–1 1 –(3iˆ 4j) Impulse = Average force × time

T CR Growth of charge in a circuit containing capacitance and resistance is given by the

16.

1

C= 17.

R

Impulse time

Average force

formula, q q 0 (1 e / CR ) CR is known as time constant in this formula. (d) From figure,

Change in momentum time –(3iˆ 4ˆj)kg ms –1

10 –4 s

21. 22.

1 tan 45º = C

1 C

L R

23.

(d) K-electron capture is accompanied by the characterstic X-rays emission. (c) Among given atoms/species, doubly ionised lithium will produce shortest wavelength emission from the transition n = 2 to n = 1. (a) According the Moseley’s law f

a(Z b)

f

C

a 2 (Z b) 2

for k

line, b = 1

L =R

a 2 (Z b) 2

MT-171

Solutions From (i) 2

(Z1 1)2

1

(Z 2 1)2

24.

6

n 2 sin

1 20 2 sin 30º

27.

(b)

28.

(a)

32.

19

10 1.6 10

33.

3 10 8

5.33 10

(d)

20 nm E c

(d) Momentum of a photon =

26.

Z2

(b) 2 d sin = n or d

25.

31.

(Z2 1)2

10 2

Z2 1

(11 1)2

4

eVs

hc

27

n = 2 i.e. in two half lives t = nT = 2 × 4 = 8 months (b) Mass defect, M = 8.03170 – 8.00774 = 0.02396 (3Li6 + 1H2 2 2He4) 6.017 + 2.0147 2(4.00387) 8.0317 8.00774 Mass defect = 8.0317 – 8.00774 = 0.02396 and 1 a.m.u. = 931 MeV Thus is this case the mass defect 0.02396 a.m.u. is equivalent to (0.02396 × 931 MeV) 22 MeV (d) 7N14 + 2He4 8O16 + aXb Balancing the eqation, we have, a = 1, b = 2 2 1X whichis deutron (a) The given truth table is of (OR gate + NOT gate) NOR gate

1

kg ms

h 1 v2 / c2 m0 v

0 (

9ˆ i 2

iˆ 2

c)

(c) is a NAND gate so output is 0 1 0 1 (d) is a XOR gate so output is 0 0 0 0 1

1

Following is NAND Gate

s = –3iˆ .

W = F .s =

9 3 2

36.

3 ˆj .( 3iˆ)

Y A.B (d) On the basis of given graph, following table is possible.

= 1.5 J. 29.

(b)

R R0

e

A B C

975 9750

t

5

30.

e

log e 1 log e 10

N N0

3 N0 4 1 4

0 1 0

5

log e 10

2.3026 log 10 10 0.461 min 5 (c) Substance left undecayed -

N0

A

(b) is a NOR gate so output is 0 1 1 0

3 ˆj

iˆ 2

6

(d) (a) is a NAND gate so output is 1 1 1 0

9 3ˆ j 3 ˆj 2

9 3 2

500 10

35.

F = –5iˆ 9cos 60 iˆ 9sin 60 ˆj 3 ˆj

= –5iˆ

Vs 0.01 = 50 R in 1000 = 500 A

(d)

W0 . If decreases, V increases s

h mv

=

ic

34.

0 1 0

1 0 0 It is the truth table of AND gate.

1

37.

1 N0 4 1 2

0 1 1

n

38.

(a)

a

g

6 10

[using v = u + at]

6 6 = 0.06 10 g 10 10 (d) Few advantage of optical fibres are that the number of signals carried by optical fibres

EBD_7443 MT-172

39. 40.

(a) (c)

Target VITEEE presence of electron withdrawing – NO2 group which makes o - and p - carbon electron deficient and hence liable to be attacked by OH– (a nucleophile)

is much more than that carried by the Cu wire or radio waves. Optical fibres are practically free from electromagnetic interference and problem of cross talks whereas ordinary cables and microwave links suffer a lot from it. 43.

1 m1u12 , 2

Ki

Ki

Br

HONO CuCl Sandomeye

NO 2

m1 m 2 u1 m1 m 2

Cl

Br

Br

Sn / HCl

1 1 m1u12 m1 v12 2 2 1 m1u12 2

Kf

Cl

Br

Br

(c)

Fractional loss Ki

NH 2

NO 2

1 m1v12 , v1 2

Kf

Br

NH 2 Cl

Br

Br

HONO H 3PO 2 elimenatio n of NH 2

v12

1

1

u12

m2

m ; m1 Kf Ki

1

m1 m 2 2 m1 m 2

2

4m1m 2 m1 m 2

1 n

2

n 12

42.

HNO3 , H 2SO 4

NO2

2

Br2

1

n 2 1 2n

41.

nitration

(b)

4n

4n

m1

44.

nm

Energy transfer is maximum when K f

1 n

NO2

2

0

4n

1 n2

0

FeBr3

Br

NH 2

2n Sn / HCl

0

n 1 ie.

NaOH

m2

m,

NO2

m

Transfer will be maximum when both masses are equal and one is at rest. (b) As far as characterstic of IR spectrum for CH3CH2OH is O–H streching frequency 3391 cm–1 O–H streching frequency 2981 cm–1 O–H streching frequency 1055 cm–1 (a) Although benzene does not undergo nucleophilic substitution, nitrobenzene undergoes such reaction due to the

Br

45.

HNO 3, H2SO 4

(b)

Cl2/Fe

333 K

NO2 Cl /Fe

NH2 reduction

Cl

Cl

(a) In case chlorination is done earlier than nitration, chlorobenzene formed

MT-173

Solutions at first step would introduce –NO2 group in ortho-position, not in mposition

53. 54.

(b) Again if –NO2 group is reduced earlier than the chlorination step, –NH2 group formed on reduction will again introduce –Cl in o-position 46.

(b) Cobalt - 60 is used for the treatment of blood cancer. (d) Given [A] = 0.01 M Rate = 2.0 × 10–5 mol L–1 S–1 For a first order reaction Rate = k[A]

2.0 10 5 k= = 2 × 10–3 [0.01]

(a) The Lanthanides are transition metals from Atomic numbers 58 (Ce) to 71(Lu). Hence the electron configulation becomes:

47.

48.

(n –2) f 1– 14 (n – 1) s2p6 d0 – 1 ns2. (a) IUPAC name of sodium nitroprusside Na 2 [Fe(CN) 5 NO] is sodium pentacyanonitrosoyl ferrate (III) because in it NO is neutral ligand. Hence 2×O.N. of Na + O.N. of Fe + 5×O.N. of CN 1×O.N. of NO = 0 2×(+1) + O.N. of Fe + 5 ×(–1) +1×0 = 0 O.N. of Fe = 5 – 2 = +3, Hence ferrate (III) (d) Co here is in +3 oxidation state Co

t1/2 = 55.

49.

(c)

50.

(b)

51.

52.

(c)

(b)

K 310 = 2.3 K 300

K 310 = 2.3 K 300

(a) The rate law equation contains [H ]2 , term. The rate will charge with change in pH and new rate will be

57.

[2]2 = 4 times the old rate (c) It is bimolecular first order reaction since Rate [N2O5]

58.

(d)

Co3+ Unpaired electrons = 4 and sp 3 d 2 hybridisation and octahedral shape. Diaminodichloroplatinum (II) commonly known as cis - platin is found to have anticancer property. The cause of showing different oxidation is due to the fact that there is only a small difference between the energies of electron in the ns orbitals and (n – 1)d orbitals with the result both ns as well (n – 1)d electrons can be used for compound formation. Lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals result in larger no. of oxidation state. d-d electronic transition causes absorption of energy at red visible wave length. In cupric chloride CuCl 2, cupper is in +2 state and hence it is d9 system. In five dorbitals there is one unpaired electron present which absorbs energy due to d-d transition. Lead Apron since it absorbs harmful radiations.

= 347 sec. 2 10 3

(b) Temperature coefficient = or

56.

0.693

59.

A 2 (g) B 2(g )

D(g) 3C(g) step-1 step-2 since the steps 1 and 2 are exothermic hence low temprature will favour both the reactions. In step - 1 moles are increasing hence low pressure will favour it. In step 2 moles are decreasing, hence high pressure will favour it. (a) Among the given options, only PCl5 can convert an alcoholic group as well as a carboxyl group to chloride.

RCH 2OH

RCOOH

PCl5

PCl5

60.

(a)

RCOCl OH

O ||

RCH 2Cl

R – C– R

HCN KCN

|

R– C

– CN

|

R (A ) OH Re duction by LiAlH 4 ( B)

|

R – C – CH 2 NH 2 |

R

EBD_7443 MT-174

61.

(c)

Target VITEEE CH 3COOH CaCO3

(CH 3COO) 2 Ca Heat

I 2 NaOH

62.

CH 3 CH 3

C

O

O C

O

(d) We know that G= H– T S When the reaction is in equilibrium, G = 0 0

H T S

H S

T

2 30 1000 285.7 K 2 105 (b) In graphite, the electrons are spread out between the sheets. (d) The appearance of colour in solid alkali metal halide is due to presence of F-centres found as defect in the crystal structure. (d) In body centred cubic lattice one molecule of CsBr is within one unit cell. Atomic mass of unit cell = 133 + 80 = 213 a.m.u Volume of cell = (436.6 × 10–12)3 T

64. 65. 66.

Density =

Density =

E3º E ºcell

( 436.6 10

67.

(c)

2 Cu

2e

Cu

Cu

2

2

º Cu ; E1

0.34V; ...(i)

0.15F E3º F

0.68 0.15

0.53V

º º E cathode (Cu / Cu ) E anode (Cu 2 / Cu )

1000 M

1.06 10 2 1000 0.1

69. 70. 71.

(b) Mass deposited =

72.

m

Eq.wt i t 96500

108 9650 10.8 g 96500 (b) Four primary alcohols of C5 H11 OH are possible. These are:

(i)

CH 3CH 2 CH 2 CH 2 CH 2 OH

(ii)

CH 3CH 2 CH CH 2 OH | CH 3

(iii) CH 3 CH CH 2CH 2 OH | CH 3

)

Cu

G 3º

1.06 10 2 (b) H2 – O2 fuel cell supply power for pressure. (b) Corrosion is an electrochemical phenomenon.

12 3

42.5 gm / cm 3

G º2

1 E 3º F

(a)

(iv) CH 3

a 3 NA 6.02 (436.6) 3

2 0.34F

68.

ZM

213 107

... (iii)

= 0.53 – 0.15 = 0.38 V.

213 6.02 10

?

0.15V; ...(ii)

G 3º

1 0.15F ,

0.68F

n ´ at.wt. Av.no.´ vol.of unit all cell 23

nFE1º

Again, G1º

O O

Cu; E3º

e

G º2

º Cu ; E 2

2

Now, G1º

NaI CH 3COONa 3H 2 O

O

Cu

Cu

CO

(c) Due to resonance in carboxylate ion, the double bond character of C = O bond in carboxylic acids is greatly reduced as compared to that in aldehydes and ketones. C

63.

CHI3

e

73.

CH 3 | C CH 2 OH | CH 3

(b) Greater the stability of the intermediate carbocation, more reactive is the alcohol. Since 2-methylpropan-2-ol generates 3° carbocation, therefore, it reacts fastest with HBr.

MT-175

Solutions 74.

(b) Secondary alcohols on oxidation give ketones. Note : – Primary alcohols from aldehydes.

R

R

[O]

CHOH

C=O

R

R

Isopropyl alcohol

75.

76.

Ketone

(c)

82.

Ester

(c) Conjugate base of phenol, i.e. phenoxide ion is a weaker base than the HCO3– , conjugate base of H2CO3. This is due to resonance, in HCO3– in which all resonating structures are equivalent. On keeping

79.

(a)

(c)

3 HCHO aq. solution

CH 2

CH2

CH2 O Trioxane (meta formaldehyde)

COOH

+

H2N

CO

H2 N

80.

CH 2

(a)

CO NH CO

CO NH Barbituric acid O.COC6H5

+ C6H5COCl

6

4ax

( x1, y1 )

... (1) , at 1 Q( 2

R (a

) 2at 1

The function of NaOH is (i) To convert phenol to more stronger nucleophile PhO– (ii) To neutralize the acid formed

(at12

at 22 ) 2

( 2at 1 2at 2 ) 2

a 2 ( t1 t 2 ) 2 [( t1 t 2 ) 2 | a || t1 t 2 | ( t1

| a | ( t1

aq. NaOH

phenyl benzoate

t2 2, 2 at2 )

Let the tangents from P touch the parabola at Q(at12, 2at1) and R (at22, 2at2), then P is the point of intersection of tangents. x1 = at1 t2 and y1 = a (t1 + t2) or t1t2 = x1/a and t1 + t2 = y1 / a ...(2) Now QR

OH

2 k

P (x1,y 1)

O

Malonic ester 2 H 2O

Let P

1

1

(b) Given parabola is y 2

CH2

COOH

1 sin

2 k 4 2

CH 3COOCH 2 CH 3

O

k 4 2

1

Acid anhydride

2.

k 4 but 2

sin

(CH 3CH 2 ) 2 O (CH 3 CO) 2 O anhy. AlCl3

78.

(a) Given, k sin + cos 2 = 2k – 7 k sin + 1 – 2 sin2 = 2k – 7 2sin2 – k sin + (2k – 8) = 0 For the existence of real roots, discriminant 0. k2 – 4 × 2 (2k – 8) 0 (k – 8)2 0, which is always true. Roots of the quadr atic equation are k 4 , 2, but sin 2

(b) Due to greater electronegativity of sp2 hybridized carbon atoms of the benzene ring, diaryl ethers are not attacked by nucleophiles like I–. Ether

77.

81.

|a|

y12 a2

1 ( y12 |a|

t 2 )2

t 2 )2

4] 4

4 t1t 2 . ( t1

t 2 )2

4

4x1 y12 . a a2

4 [using (2)]

4ax1 ).(y12

4a 2 )

EBD_7443 MT-176

83.

Target VITEEE

(a) We have,

f (1) 1! 1, f (2) 2! 2,

A B 2

tan

31 1 32 31 1 32

1 cos( A B) 1 cos( A B)

1 63

Using Napier’s analogy,

86.

A B a b C cot 2 a b 2 From (i) and (ii) we get, tan

a b C cot a b 2

cot

1

9

tan

63

C 2

a2

52 84.

b2

2x 2

4

x

2

1 8

1 dx

2

(1 1) 18

1 3 2

Thus, we may take c (c)

f (x) x Px 1 n

Pr

1 2

y tan b

1

( j k ).

( j k ).

x! x! (x x 1)!

n! , see Permutatio ns (n r )!

P F X

S

C

)

1 18 2

x sec a

Y

2

1

0

1 16 1 3 3 (b) The equation of tangent at any pont P (a sec , b tan ) is

3(2

c

x

4

x

2 . Thus, c = (a – 2b) = (–3j + 3k) = 3 (– j + k) | c |2 9

2 dx

1

x3 4 3

0

87.

2 2 1)

2 2x 2

2 dx

0

1

2ab cos C

6)

1

3x

1

c=6 (a) As c is coplanar with a and b, we take, c= a+ b ...(1) where , are scalars. As c is pependicular to a, c . a = 0 from (1) we get, 0= a. a+ b.a 0

85.

1

[Separating even & odd parts]

1 2 5 4 = 36 8

42

(b) Let f (x) = ax 2 bx c . f (0 ) 2 c 2 Also f’(x) = 2ax + b and f”(x) = 2a f”(0) = 4 2a = 4 a = 2 and f’(0) = – 3 b = –3 f(x) = 2x2 – 3x + 2

7 3

7 9 7 9

C 1 2 cos C C 1 1 tan 2 2 Now, using cosine rule, 1 tan 2

c2

...(ii)

5 4 C 1 cot 5 4 2 63 [given a = 5, b = 4]

63

C 2

f (3) 3! 3 2 1 6 , f (4) = 4! = 4 × 3 × 2 × 1 = 24 Range of f = {1, 2, 6, 24}

...(i)

a e

Its point of intersection with the directrix 2

x=

a b(sec e) a is F , e e tan e

Slope of SF, m1

b(sec

a tan (e where S is the focus (ae, 0).

Also the slope of PS, m 2 Clearly m1m 2 2

.

e) 2

1)

,

b tan a (sec e)

1 . That is SF

r

PF .

MT-177

Solutions 88.

(a) In a right angled triangle the mid-point of the hypotenuse is equidistant from the vertices A

–G

C

x x

/2

91.

b

/2 x

3b

3 1 1 1 . 4 3 3 12 (a) We have, m sin = n sin( + 2 ) 2 sin m sin n Using componendo and dividendo, we get sin 2 sin m n sin 2 – sin m–n

D

B W

2 – 2 2 – sin 2 cos m n sin m–n m n tan( + ) . cot = m–n 2 2 2 2 2 sin 2 cos

2 sin

DB = DC = DA Hence, DAB is isosceles

2 cos

A

B

2 where is the angle which BC makes with the vertical. When the triangle is suspended from right angle A by a string then line of action of W, i.e., the vertical through G must pass through A, so, that median AD is vertical. Now sin = 2 sin ( /2) cos ( /2) 2.

b

3b

. b 10 b 10 sin

89.

1

3 5

3 5

P A

(d) We have | (a × b).c | = | a || b || c | | a || b || sin n.c.| = | a || b || c | | a || b | | c || sin cos | = | a || b || c | |sin | |cos

90.

6 10

|

and

A

B

C

B

P(B

B

C

B

2 , P B 5

P A

1 P A

P B

1 P B

B

P A

B

P B

P A

P A |B

B

P B |A

3 , P A 10

2 5

3 5

3 10

7 10

B

2 3 5 10

1 1

P A

1 P A

P A

P B

B

B

.

B

1 5

1 5

1 2

1

1 2

P A

1 2

B

P A

1/ 2 1/ 2 1 50 25 . 7 /10 3 / 5 4 21 42 (d) Given: xdy – ydx = 0 Dividing by xy on both sides, we get:

=

93.

C) P(A

y = log c y = cx or y – cx = 0 x which represents a straight line passing through origin.

log

C)

P(B) P(A

P A

m n m–n

dy dx dy dx =0 y x y x By integrating on both sides, we get, log y = log x + log c

C

A

P A

=0

2 a b and c | | n a b and c is perpendicular to both a and b a, b, c are mutually perpendicular Hence, a.b. = b.c = c.a. = 0 (a) From venn diagram, we can see that

A

(c)

92.

cos

B

C)

EBD_7443 MT-178

94.

Target VITEEE ,c

(b) Given b Clearly,
0] n2 + n – 1 – 379 = 0 n2 + n – 380 = 0 (n + 20) (n – 19) = 0 n = – 20, n = 19 n is not negative. n = 19 104. (c) Let the sides of the ABC be a = n, b = n + 1, c=n +2 Where n is a natural number Clearly, C is the greatest and A is the least angle As given, C = 2 A sin C = sin 2 A i.e. sin C = 2in A cos A

x2

a

or 1

2

xy dy =0 b 2 dx y2

xy dy =0 b2 dx

x2

1

y2

a2 b2 b2 Differentiating again w.r.t. x, we get

EBD_7443 MT-180

Target VITEEE

2y dy b2 dx or

106. (c)

xy

y dy b2 dx

x dy dy . b 2 dx dx

d2 y

dy dx

2

x

2

y

xy d 2 y b 2 dx 2

=0

dy =0 dx

dx comparing with the given differential equation, we get A = 1. The given information can be expressed as given in the diagram: In order to simply, we assume that 1 unit = 1000 bricks Suppose that depot A supplies x units to P and y units to Q, so that depot A supplies (30 – x – y) bricks to builder R. Now, as P requires a total of 15000 bricks, it requires (15 – x) units from depot B. Similarly, Q requires (20 – y) units from B and R requires 15 – (30 – x – y) = x + y – 15 units from B. Using the transportation cost given in table, total transportation cost. Z = 40x + 20y + 20(30 – x – y) + 20(15 – x) + 60(20 – y) + 40(x + y – 15) = 40x – 20y + 1500 Obviously the constraints are that all quantities of bricks supplied from A and B to P, Q, R are non-negative.

x 0, y 0 107. (b) There are 11 letters in the given word which are as follows (NNN) (EEE) (DD)IPT Five letters can be selected in the following manners : (i) All letters different : 6C5 = 6 (ii) Two similar and three different : 3C1. 5C = 30 3 (iii) Three similar and two different : 2C1. 5C = 20 2 (iv) Three similar and two similar : 2C1. 2C1 =4 (v) Two similar, two similar and one different : 3C . 4C = 12 2 1 Total selections = 6 + 30 + 20 + 4 + 12 = 72 108. (c) In the definition of the function, b 0, for then f(x) will be undefined in x > 0. f(x) is continuous at x = 0, LHL = RHL = f(0) sin(a 1)x sin x 0 x

lim

x x 0

` 20

0 `4

y

P 15 units

`2 0

x

y

B 20 units

) ` 40 –y +6 5 x 1 (7 – y – 8– =x+

x 0, y 0, 30 – x – y 0, 15 – x 0, 20 – y 0, x + y – 15 0. Since, 1500 is a constant, hence instead of minimizing Z = 40x – 20y + 1500, we can minimize Z = 40x – 20y. Hence, mathematical formulation of the given LPP is Minimize Z = 40x – 20y, subject to the constraints: x + y 15, x + y 30, x 15, y 20,

0

x

1 2

a+2

109. (b) Let cos

3 ,c 2 1

5 3

0

c (1 bx ) 1

bx ( 1 bx 1)

1

lim

a+2

a

lim

x

1 bx 1

0

sin

c

c

c

1 ,b 2

0

, then 0

5 3

cos

c

sin x x

1 bx 1 bx

(a 1) 1

R 15 units

y –

`

20

15 – `2 x 0

60

Q 20 units

0

lim

x–

x

3/ 2

sin(a 1)x x

lim

x

30 –

bx

x 0 x 0

A 30 units

x

x bx 2

lim

1 cos 2

2 3

2

and

MT-181

Solutions 1 So, tan cos 2 1 cos sin

Hence, the variable plane (1) always passes

5 3

1

3

tan

2

through the fixed point

5 2

110. (c) Given 3f (cos x) + 2f (sin x) = 5x Replace x by 3f cos

5

2

2

.....(i)

2 x

2f sin

2

x

4 3

x

( sin x)f '(cos x)

5

2

x ...(ii)

5

5 sin x 111. (d) The relation S is defined by,

f '(cos x)

S {( x, y) : x 2

y2

Clearly to each x

1}

3 4

C0

1 P( X

0)

0.9

3 4

n

x y z 1 ...(1) a b c The intercepts on the coordinate axes are a, b, c. The sum of reciprocals of intercepts is constant , therefore 1 c

1 1 1 , ,

(1 / ) b

n

1 10

10

n(log 4 log 3) 1

1 8 0.125 Therefore the least number of trials required is 8. 114. (b) Given equation of line is n

x 2 y 1 z 3 (let) 2 2 3 x = 3 – 2y = 2 – 1, z = 2 + 3 ...(i) Coordinates of any point on the line are (3 – 2, 2 – 1, 2 + 3) The distance between this point and

6 2

A , there exists two

1 x values of y given by y S cannot represent a function. 112. (b) Let equation of the variable plane be

(1 / ) a

0.9

n

(1, 2, 3) is

A B

2

or

0.9

n(0.602 0.477) 1

Solving (i) and (ii) simultaneously, we get f(cos x) = 5x – Differentiating we get

1 b

n

1

or 3f (sin x ) 2f (cos x )

1 a

113. (b) Let n denote the required number of shots and X the number of shots that hit the target. Then X ~ B(n, p), with p = 1/4. Now, P(X 1)

x we get

1 1 1 , , .

(1 / ) c

1

lies on the plane (1)

3

2 1

2

2

1 2

2

2

3 3

(3 – 3)2 + (2 – 3)2 + (2 )2 =

2

6 2

36 2

Squaring on both sides 9 2 + 9 – 18 + 4 2 + 9 – 12 + 4 (17 – 30) = 0

2 = 18

30 17 Substituting the values of in eq. (i) we get the required point (–2, –1, 3) and 0,

56 43 111 , , 17 17 17

EBD_7443 MT-182

Target VITEEE

115. (c) We are given that e–m . m = 0.3 and

e

m

m2 2!

118. (d)

4 3

0.4 0.3

119. (b)

P (X = 0) = e–m = e–4/3 116. (a) The given equation can be written as 1 cos x sin 2 ax

or sin 2

x 2

x sin 2

0

0

m a;

2n

x

ydy

dy .y dx

x2 sin( sin 2 x)

sin 2 x

0

sin 2 x

x2

sin 2 x

x2 120. (a) The relation R1 is an equivalence relation x

0

R, | a | | a |, i.e. aR1a a

a

R

R1 is reflexive. m

2an

...(1)

Again aR1b Also,

R, | a | | b |

|b| |a|

bR 1a . Therefore R is symmetric. a, b,c

R, | a | | b | and | b | | c |

aR1b and bR 1c

x (Given)

y2 x2 C i.e y 2 x 2 C 2 2 2 Since the curve passes through the point (5, 3), 9 = 25 + C C = – 16 So the curve is y2 = x2 – 16 i.e x2 – y2 = 16 which is clearly a rectangular hyperbola.

a, b

|a| |c|

aR1c

R1 is transitive R2 and R3 are not symmetric. R4 is neither reflexive nor symmetric.

xdx

Integrating.

sin( sin 2 x)

lim

Now m and n are integers but a is irrational hence the result (1) will hold good if m = 0 and n = 0 and in that case x = 0 is the only solution. 117. (c) We have,

x2

0

lim

m a

sin 2 x}

sin{

0

x

2n and x

are

x2

0

lim

x n and 2 ax m where m, n are integers

x

x

x

0 and sin ax

3iˆ pjˆ pkˆ

sin( cos 2 x)

lim

lim

0

sin 2 ax

and

perpendicular 2 × 3 + p (–p) + 5 (p) = 0 p = –1 or p = 6 Hence for p = 6, the lines are perpendicular.

= 0.2 Dividing, m

2iˆ pjˆ 5kˆ

121 122 123 124

(d) (b) (c) (c)

125 (b)

MT-183

Solutions

MOCK TEST 6 1.

(d)

V2 = 3V1 ...(ii) From (i) and (ii), V1 = 3.5 V and V2 = 10.5 V Thus charge of C3, q3= C3V2 = 4 × 10.5 = 42 V.

For collision V B/A should be along

B

A rA/B

So,

V 2 V1 V2 V1

r1 r2 r1 r2

5.

(a)

r2

V2

V1

q

r1 q1

B

A A

2.

3.

B

(a) Force = eE Workdone = force × distance Force and distance are in opposite direction, so work is negative. W=–eE×d Here, distance increases so, potential energy increases. q –q – + – + – + + (b) – – + – +

d Because of induction, the effective force between them becomes equal to greater than F= 4.

q2

1

4

0

d2

Let q1 be the charge on the inner shell. On being earth its potential will be zero. or

V=0 =

q' = 6.

7.

q1 r1

1 4

r1 r2

0

q2 r2

q

(b) Here the electric field at A and B is same but at C it will decreases, s o VA = VB ( V = – Edr) and VC is less than VA or VB VA = VB > VC (c) In a round trip, displacement is zero. Hence, work done is zero. +q r

.

+Q

(c) C1 = 4µF

8.

14V 2µF

4µF

+2 C

The capacitance of parallel capacitors 2 4 4 µF 2 4 3 If V1 and V2 are the p.d. s across C1 and parallel capacitors, then V1 + V2 = 14 ...(i) 4 V2 3

9.

+6 C

A

d

FAB =

FBA

F AB

=

and 4 V1 =

(b) The magnitude of force on A due to B and on B due to A is same but both will act in opposite directions as the electrostatic force between them is repulsive.

B

N M where d = density, N = Avogadro number, M = atomic weight

(b) Electron density, n

d

FBA

EBD_7443 MT-184

Target VITEEE When a cell of emf 3V is connected across A & B, 3 C D

2.7 103 6.02 10 26 so, n (atom/kg) 27 28 = 6.02 × 10 electrons/m3

6.02 10

28

1.6 10

19

4 10

6

= 1.29 × 10–4 m/s (c) No current flows through 6 resistor as a capacitor offers infinite resistance to a D.C. The equivalent circuit is thus as shown below in which 2 and 3 are in parallel 2

3V 2A 1.5 This current is equally distributed along arms AB and CD. Hence, current along AB is 1A. (d) One end of all the three resistors are connected at the point X and the other at the point Y. Therefore, the equivalent circuit is given below : i

12.

A

B

3

3V The current through the circuit is given by

5

or vd 10.

A

I neA

Drift velocity, vd

B

3

6

6

2.8

6V

6 1 1 1 or R ' 5 2 3 R' Current drawn from the battery

X

1.2

6

The equivalent resistance is given by

E 6 1.5 Ampere R 4 Potential difference between A and B V IR ' 1.5 1.2 1.8V Current through 2 resistor I

11.

E

3 3

3 F

A

1 R

V 1.8 0.9 A 2 2 (d) Rearranging the resistors CDEF, we get a balanced wheatstone bridge as shown in the figure. D 3 3

C

3 A

B 3

B 3 effective resistance between C and E 6 6 =3 6 6 Now, effective resistance between A and B

3 3 3 3

1.5

Y

13.

14.

1 1 1 6 6 6 Req 2

3 6

1 2

(b) According to Fleming's Right hand Rule In (1) force is in – z axis In (2) force is in – x-axis In (3) force is zero because F = q (v B) = qvB sin 180 i.e. F = 0 (d) Given: n = 100 turns/m I = 5A 0 nI (sin 1 sin 2 ) B= 2 B=

0

=

0

=

0

100 5 (sin 45 2 500 2

1

500 2

2

= 250 2 0

sin 45 )

1

2

2

2

=

0 250

2

MT-185

Solutions 15.

I2R

(IC

IL )2

IR

E R

200 50

4A

IC

E XC

200 25

8A

IL

E XL

200 40

5A

(4)2

(8 5)2

(b) I =

I

16.

17

æ 4.24 ÷ö2 4.24 = 2.99A Id = çç ÷ = çè 2 ÷ø 2

20.

21.

=

1

=

mv2 – mg 2mg ( vtop = 3gr ) r Tension at the lowest point Tbottom = 2mg + 6mg = 8mg

Ttop Tbottom 22.

1

LC

0.5 8 10

23. 6

1 1000 2 = 500 Hz (a) XL = 15 ; XC = 11 ; R = 3

=

18.

Impedance, Z = =

19.

R2

(X L

9 16

I0

24.

5

E rms 10 2A Irms = = 5 Z VL = Irms × XL = 2 × 15 = 30 V VC = Irms × XC = 2 × 11 = 22 V VL – VC = 8V (a) Peak current, I0 = 4.24 A I rms

X C )2

26.

2

Id

2 I rms

Direct current,

2

I rms RT

4000 100 A = 1.67 A 240000 Vs (d) In a single slit diffraction pattern, the path difference ( x) for nth minima is given by 1, 2, 3,... x = n ,, where n

or phase difference, n(2 ) For 1st minimum, n = 1. 2 (d) Given Ez= 60sin (kx + t) V/m. The magnetic field component must be perpendicular to Ez and x. E0 c

60 . Thus c

60 sin (kx + t). c 4 (c) Given, 1H2 + 1H2 2He + Q The total binding energy of the deutrons = 4 × 1.15 = 4.60 MeV The total binding energy of alpha particle = 4 × 7.1 = 28.4 MeV The energy released in the process = 28.4 – 4.60 = 23.8 MeV. (b) Number of fissions per second

By =

Thermal energy produced, 2 H = Id RT

1 . 4

Vp I p

Also B0 =

4.24

2

2mg 8mg

(c) For a transformer, Vp Ip = Vs Is

Is

6

1 4 10

(b) Given V = V0 sin t; i = i0sin( t – ) Here current lays the potential. This can be 1 possible when XL > XC or L > . C (c) Tension at the highest point

Ttop

I 16 9 25 I = 5A (a) i = 2 sin t V = 5 cos t I (max) = 2 amp, V = 5V P = VI = 5 × 2 = 10 watt. (d) Current is max, when at resonance condition

So,

3A

26.

=

Total energy produced / sec. Energy released per fission

EBD_7443 MT-186

Target VITEEE =

1.6 106 J / s

200 1.6 10

13

1019 200

J

31.

(b) Given:

E = 57 meV = 57× 10–3 × 1.6 × 10–19 J

E=

1000 1016 200 = 5 ×1016 fissions per second.

=

27.

(a) We have 1 mv 2 2 1 mv12 2

28.

0

=

0

0)

....(1)

h(9

0

0)

....(2)

8 0 4 0

v12

h

h(5

1 mv2 2 2 (2)/(1)

v22

h

=

32. 33.

2

v2 = 2v1 2 4 105 = 5.64 × 105m/sec h (a) = 2mK

6.63 10

34.

34

2 9.1 10 31 5.5 1.6 10 16 = 169 nm 29.

1

1 1 1

30.

1

(a) We have, =R

= R

1

1

22

32

1 =R 4

1

1

n12

n 22

36 5R (d) For shortest wavelength n1 = 2, n2 =

1 1

ic

35. 36.

R

1 2

22

26

ic 100 0.95 100mA

RL Ri

(b)

Pc

Pt 1

5 103 70

0.98 2

.

RL Ri

70

0.98 70

68.6

m2 2

11.8 10 1

38.

3 108

57 1.6 10

Power gain = 37.

34

i c 95mA (d) In forward biasing, the value of diffusion current decreases. (a) = 0.95; RL = 5k Ri = 70

AV =

1

R(0.5) , shows visible region.

6.6 10

6.6 3 10

0.95 =

1 9

5 36

hc E

57 1.6 10 22 = 0.2171052 × 10–4 m = 217105.2 × 10–10 m = 217105 Å (Approx.) (b) When m1 > m2 & m2 at rest then the bodies collide in elastically and move together as one body without changing the direction. (b) Extrinsic semiconductor is due to the type of impurity atom's valency. If impurity atoms are trivalent, then carriers are hole. And if impurity atoms are pentavalent then carriers are electron. (a) = 0.95 ie = 100 mA ic = ? ic = i e

9 4 =R 36 =R

hc

m2 2

% modulation = 60%. m 0.6 (c) The given graph does not obey Hooke's law and there is no well defined plastic region. So the graph represents elastomers.

MT-187

Solutions 39. 40. 41.

42. 43.

44.

45.

46.

(c) In Bernoulli’s theorem only law of conservation of energy is obeyed. (d) H = ncp t (d) In a tetragonal distortion, the ligands along the z-axis move away. Their repulsion significantly decreases for the orbitals involving z-direction. These are lowered in energy. (c) Cu + ion in aqueous solution disproportionates to Cu(s) and Cu 2(aq) . (b) In [Fe(CN)6]3– Fe is in +3 state. Fe (Z = 26) in Fe3+ we have 23 electrons. Thus, EAN of Fe = 26 – 3 + 12 = 35 [12 electrons gained from six ligands] (d) Magnetic moment n ( n 2) 3.87 B.M n 3 (unpaired electrons). The comlex may be regarded as a high spin d7 complex of Fe+ and No+ (c) EDTA forms a more stable coordination complex with lead than with calcium or sodium and in this form passes out from the body without harmful effects. The calcium salt is used so that any excess of EDTA will not remove Ca 2 ions from the body. (d) The optical isomers are pair of molecules which are non super imposable mirror images of each other

N

N

N Co

N

N

3+

en – N en

en

Co

N

N

47.

Enantiomers

49.

50.

51.

52. 53.

3+

2a 2 0.361 nm = 0.128 = 4 4 (a) For a reaction to be spontaneous G should be negative. We know G = H – T S Thus, for a reaction to be spontaneous at all temperatures G and H should be negative. (a)

C(s) O 2 (g) CO 2 (g) Thus entropy of formation of CO2 SCO2 (SC SO2 )

= [213.5 – (5.740 + 205)] JK–1 = 2.76 JK–1. (a) Internal energy is dependent upon temperature and according to first law of thermodynamics total energy of an isolated system remains same, i.e., in a system of constant mass, energy can neither be created nor destroyed by any physical or chemical change but can be transformed from one form to another E q w For closed insulated container, q = 0, so, E w , as work is done by the system (a) Le-Chatelier’s principle is applicable only to a system in equilibrium. (d) We can obtain for NaOH at infinite dilution by extrapolating the line of NaOH but we cannot obtain for CH3COOH because on extrapolating the curve of weak electrolyte it does not approach a limiting value. But molar concluctivity at infinite dilution for a weak electrolyte can be calculated by using Kohlrausch's law. According to this law Ù°m for AxBy = x

A

y

B

The relation of molar conductivity and 1000K , C is not useful because when C 0, Ùm (b) The order w.r.t. I2 is zero because the rate is not depends on the concentration of I2.

specific conductivity i.e., Ùm =

l form

54. 55.

+ve –ve ion ion

For a face-centred cube, we have. radius =

N

The two optically active isomers are collectivity called enantiomers. (d) When electrons are trapped in anion vacancies, these are called F-centres.

F- centre in crystal

(a)

N N en

en N d form

en

48.

(a)

Ag

e

1F

Ag 108 g

1 F = 1 mole of electrons = 96500 C 0.01F = 1.08 g Ag; Ag left = 1.08 – 1.08 = 0

EBD_7443 MT-188

56.

(b)

Target VITEEE G reaction

Products

G (CO2 )

Reactants

G (H 2O)

62.

G (CH3OH)

3 G (O2 ) 2

= [– 394.4 + (– 237.1)] – [(–166.3) + (0)] = – 702.3 kJ mol–1 702.3 G 100 100 = % Efficiency = 726 n = 96.7%

57.

(b) Cell constant =

61.

Specific conductance Conductance

63. 64. 65.

0.0112 or 0.0112 × 55 1/55 = 0.616 cm–1.

=

58.

(b)

CH 3 I Mg

Dry ether

C O

CHOH

water molecule from two CH 3COOH and thus, acetic anhydride is formed. CH 3CO OH

CH O MgI

CH3

CH3

66.

heated with CH 3COOH, it eliminates a

C6H5 CH3MgI + H C6H5

C6H5

CH 3 MgI

(c) Option (c) is the answer. Dehydration will be most easier when –OH group is trans to C = O group. (a) Prestone is a mixtrue of glycol & H2O. It has freezing point much below 0°C, hence presence of glycol (CH2OHCH2OH) in H2O reduce its freezing point. Therefore, it is used as an antifreeze for automobiles radiators. (c) C6H5CH2Br + NaOC2H5 or C6H5CH2ONa + C2H5Br C6H5CH2OC2H5 (b) Acetophenone (C6 H5 COCH3 ) being a ketone, does not reduce Tollen’s reagent. (c) Aldehydes restore the pink colour of Schiff's reagent (Schiff's reagent is a dilute solution of rosaniline hydrochloride in water whose red colour has been discharged by passing sulphur dioxide). Ketones do not restore Schiff's reagent colour. (a) P2O5 is a dehydrating agent. When it is

+ CH 3COOH

H 2O

P2O 5 Heat

CH 3CO

2 mole of acetic acid

Mg ( OH ) I

67.

Acetic anhydride

(c) Mesitylene is formed.

(1-Phenyl ethanol)

59.

O || 3CH3 C CH3

(d) With Br 2 water, ph enol gives 2, 4, 6- tribromophenol.

OH

H2O

OH

CH3 H 2SO 4 3H 2O

68.

Br +3HBr

(b) Ether exists as peroxide in air.

CH3

(d) It is Favorskii rearrangement – ring contraction. The transformation of -haloketones to esters with rearranged carbon skeleton by the treatment with alkoxide ions is called Favorskii rearrangement

Br 2, 4, 6 Tribromophenol

60.

CH3

Trimethyl benzene or Mesitylene

+ 3Br2 (excess)

Br

O + H2O

CH3CO

RO

MT-189

Solutions 69.

NaBH 4

O

(a)

CH 3OH

HBr

CH2OOC.COOH

OH

|

CHOH

Br

(ii ) H 2C

CO2

|

(i) Mg. Et 2O

CH2OH

O

CH2OOCH

CH 2OH

|

CH 2 OH

|

H 2O

CHOH

CHOH |

|

CH2OH

PCC

CHO

CH 2Cl 2

70.

76.

(a)

CH3CHO + 2[Ag(NH3)2]+ + 3OH– Tollen's reagent

CH3COO– + 2H2O + 2Ag

+ 4NH3 Silver mirror

71.

Br2 PCl3

(c) (i) CH3CH2COOH Br2

77.

CH3CH2COCl

Cl2 PBr3

(ii) CH3CH2COOH

73.

Cl2

CH 3

|

C

CH3COCl

CH 2

ZnCl2 or AlCl 3

CH3

|

C CH 2COCH3 |

Cl (c) When glycerol is heated with oxalic acid following reaction occurs.

CH2OH HOOC |

CHOH |

CH 2OH

|

HOOC oxalic acid

(A)

heat

(CH3 )4 N OH

(C) and (D)

78.

(a) Consider the given sequence [A]

Reduction

[B]

HNO 2

100 110 C H 2O

CH3 CH 2 OH

Since, [B] on reaction with HNO2 form a primary alcohol (CH3CH2OH) so [B] is a primary aliphatic amine. i.e., [B] is CH3CH2NH2. Since, [A] on reduction forms CH3CH2NH2 so [A] is CH3CN. Reduction

CH3

75.

(CH3 )4 N I

(CH3 )N CH 3OH

CH3CH(Cl)COOH. (a) This is an example of Baeyer-Villiger oxidation (c) 1, 6- and 1, 7-dicarboxylic acids on heating form cyclic ketones (Blanc rule) (d)

CH3I

excess

(B)

CH2CHClCOBr

CH3

74.

CH3 NH 2 AgOH

H2O

72.

(c)

CH3CH(Br) COCl CH3CH(Br) COOH.

H2 O

CH3CH2COBr

CH2OH 3 In amines, N is sp hybridised and thus has

pyramidal shape. In the given structure, since the three alkyl groups are different, and the fourth corner of the pyramid is occupied by lone pair of electrons, the molecule is chiral. However, the two enantiomers of the amine are not resolvable because of their rapid interconversion through a transition state having planar structure (sp2 hybridised nitrogen)

(b) Acetaldehyde is the only compound among the given compounds which can reduce Tollen’s reagent to silver mirror. Acetaldehyde

HCOOH

Formic acid

79.

CH 3 CN CH 3CH 2 NH 2 (b) When alcoholic KOH reacts with chloroform and aniline the following reaction take place:

NH2 + CHCl3 + 3KOH

N=C + 3KCl + 3H2O Phenyl isocyanide

EBD_7443 MT-190

80.

Target VITEEE

This is also called Carbylarmine reaction. It is used for detection of 1° amines. The isocyanide formed has a very offensive smell which is used to detect its presence. (c) Diazonium cation reacts with aniline in weakly acidic medium resulting in N, Ncoupling rather than C-coupling. N

N

N

N

NH

81. (b) The given relation is R = {(1, 1), (2, 2),(3, 3), (1, 3)} on the set A = {1, 2, 3}. Clearly, R is reflexive and transitive. To make R symmetric, we need (3, 1) as (1, 3) R. If (3, 1) R, then R will be an equivalence relation. Hence, (3, 1) is the single ordered pair which needs to be added to R to make it the smallest equivalence relation. 82.

(d)

PQ = I P–1 = Q Now the system in matrix notation is PX = B X = P–1 B = QB

2 1 13 9 8

x y z y

83.

1 1

2

5 m 1 1

5

5

1 (13 5 5m) 9

– 27 = 8 + 5m m=–7

(Given y = – 3)

= 0 i.e.

1

2

1

1

2

1

pb qa

pc ra

b

2

q2

2–

2 1=0

2

1 2

2

1 2

ac pr

9=x

2

B(0, 10)

10

1

Y

y=

or

2

one Also as x , f (x) and x – f (x) – Range of f (x) = R = domain of f (x) f (x) is onto. Thus, f (x) is one-one and onto. 85. (b) After sending 4 to one side and 3 to other side. We have to select 5 for one side and 6 for other side from remaining. This can be done in 11C5 × 6C6 ways = 11C5 Now, there are 9 on each side of the long table and each can be arranged in 9! ways. Required number of ways = 11C5 × 9! × 9! = 11C6 × (9!)2 [ nCr = nCn – r] 86. (d) The graph of the inequalities 2x + y 10, x 0, y 0 is the region bounded by AOB. This region has no point common with the region {(x, y) : x 6, y 2} as is clear from the figure . Hence, the region of the given inequalities is the empty set.

2x +

(a) Since 1, 2 and 1, 2 are the roots of ax2 + bx + c = 0 and px2 + qx + r = 0 respectively, therefore c b ...(1) 1+ 2= a , 1 2= a q r ...(2) and 1 + 2 = p , 1 2 = p Since the given system of equation has a non-trivial solution 1

R

f ' (x) > 0, x R f (x) is strictly increasing and hence one-

HN H

–H

84. (a) Given that f (x) = 2x + sin x , x f ' (x) = 2 + cos x But – 1 cos x 1 1 2 + cos x 3 1 2 + cos x 3

y=2

A (5, 0)

(6, 0)

X

MT-191

Solutions 87.

(b) The general term of the series 2 1.2.3

tn =

2 3.4.5

2 , n = 1, 2, 3,...... (2n 1) 2n (2n 1)

1 2 = 2n 1 2 n 1 = 2n 1

1+

2 1.2.3

1 2n 1

1 2n

1 2n

2 3.4.5

2 5.6.7

1 1 1 2

=1

88.

[ | BA | BA

2 ....... 5.6.7

ˆi 3ˆj

(1 / ) a

2

6 3

1

3

90.

0

(b)

5

3

2 , tan 1 3 5 , tan 3

cosec

4 , cot 3

2 3 3 2

13

2

1 1 1 , , .

3

=

91.

4 3

3 2

1

=

6 2 1 9 8

AD =AB sin AB.

| BC BA |

| BC BA |

| BC | . | BA |

| BC |

6 17

4 3 3 2 (d) As given: cos–1x + cos–1y + cos–1 z = 3

and we know that 0 C (0, –5, 1)

cot cot 1 cot cot

cot

Now,

A (1, 4, –2)

D

7

3 26 7

5 Let cosec 1 3

1

(c)

3 26

4

Hence, the variable plane (1) always passes

B (2, 1, –2)

9iˆ 3jˆ 12kˆ

4 36 9

lies on the plane (1)

through the fixed point

kˆ

and | BC |

AD

(1 / ) c

ˆj

........

cot

(1 / ) b

ˆi

92 32 (12) 2

1 c

1 1 1 , ,

89.

and BA

| BC BA |

x y z 1 ...(1) a b c The intercepts on the coordinate axes are a, b, c. The sum of reciprocals of intercepts is constant , therefore

or

2ˆi 6ˆj 3kˆ

1 2n 1

1 1 .... 3 4

1 b

Now BC

BC BA

1 1 1 1 .... 2 3 4 5 = 1 + loge2 + loge2 – 1= 2 loge2 (b) Let equation of the variable plane be

1 a

AB]

cos –1 x

cos 1 x , cos 1 y , cos 1 z x = y = z = cos = –1. xy + yz + zx = (–1) (–1) + (–1) (–1) + (–1) (–1) = 1+ 1 + 1 = 3

EBD_7443 MT-192

92.

Target VITEEE x y 3

(c) We have f

f ( x ) f ( y) , 3

94.

(b) Let the the variable circle be | z z 0 | r .

f(0) =0 and f (0) = 3 f (x )

3x 3h 3 h

f lim

h

0

0

f (3h ) f (0) lim 3h h 0

f(x) = 3x + c, f(x) = 3x

93.

(b) x + iy =

a z1

z2 b Then | z 0 z1 | a r and | z0 z2 | b r Eliminating r, we get, | z 0 z1 | | z 0 z 2 | a b Hence locus of z0 is a hyperbola

f (3x ) f (0) 3 h 3

f(0) = 0

3 i sin

cos

r

r

f (x)

f (3x ) f (3h ) 3 lim

h

z0

f (x h) f (x) lim h 0 h

c=0

95.

(a) Let AB

a , AD

cos

2 i sin 3

x iy

2

x iy (x iy) (x iy) 1 [(cos 3 x x

2

y x

x

2

y

x

2

y

x

2

y

2

1 x

2

2 3

y2

2

2

(3 4 x ) 1

0

x2 y2

y2 3

c2

b2

a2 2

b .a

c2

a 2 b2 c 2 2ab C

c –

A

y x

a .a

b2 2

)

b

1 sin 3

2

y

4x x 2

a2

ab cos

b ).( a )

D

1 cos 3

2

3 4x

a2

2)

Squaring and adding, we get x

(a

In ABC , cos(

y

and

Now, D B. A B

2) i sin ]

2 3

2

a b

3a 2

1 (cos 3

2

c

when a , b and c are non-collinear coplanar vectors. DB AB AC

1

b and AC

2

96. 1 9

B

a

(b) {x cos(cot–1x) + sin(cot–1x)}2 = 1 x cos tan 1 x

1

2

1 sin tan 1 x

1

xcos cos 1

51 50

1 x2

2

1 x

sin sin 1 1

51 50

1 x2

MT-193

Solutions =

Note that ac is either b or e. If ac = c, then a = e, which is not so. So, ac = b. Similarly bc = a and ca = b. Next, complete the table keeping in mind that each element in a row (or a column) must be distinct. Then the composition table shows that G is abelian.

51 50

x2

1 x

2

x2 1

2

x

2

2

1

1 x

2

51 50

51 50

1

99.

(a) We have Sn

1 51 x 5 2 50 0, then |cot x| = cot x, the equation

x2 1

(c) If cot x

1 0 , which is not posible sin x Now, if cot x < 0, then | cot x| = – cot x, the equation becomes 1 cot x cot x sin x cos x cos x 1 sin x sin x sin x 2 cos x 1 1 0 cos x sin x 2

becomes

2n

2 ,n 3

Hen ce x

2 4 , 3 3

x

98.

I and 0 x 2 .

i.e., number of

solution is 2. (b) A group containing exactly four elements is always abelian. Let G = {e, a, b, c} be a finite group of order 4 with identity element e, then e–1 = e. Case I : If a–1 = a, b–1 = b, c–1 = c, then each element is its own inverse and hence G is abelian. Case II : If a–1 a, say a–1 = b then b–1 = a and hence c–1 = c. So, in this case ab = e, ba = e. The composition table becomes : e a b c e e a b c a a c c b b b e c a c

c

b a

e

S2n

1

1

n 1

n 1

(n 1) 2

2 So, S12 S22 ... S2n

1

22 32 ... (2n)2 1 (2n )(2n 1)(4n 1) 1 6 1 [n (2n 1)(4 n 1) 3] 3 100. (b) Let S x2 – 4y Since the point (2a, a) lies inside the parabola, S (2a, a) = 4a2 – 4a < 0 or a (a – 1) < 0 ...(1) Also, the vertex A (0, 0) and the point (2a, a) are on the same side of the line y = 1 (the equation of latus rectum) So, a – 1 < 0 i.e. a < 1 ...(2)

Y

S (0, 1)

97.

n (n 1) n 1 1

n

(2a, a)

A

y=1

X

From (1) and (2), we have a (a – 1) < 0 or 0 < a < 1 101. (a) Let the equation of the ellipse be x2

y2

a2

b2

1.

EBD_7443 MT-194

Target VITEEE y B

S

(b

=( a

b ).

(m a ) c (m c ) a )

A

x

(where m

L

B

ae,

Coordinates of L are

b2 a

and

[ a b c ]2

104. (c)

ay

a

b

But b 2

(a 2

42

16

f ( x 2 ) f (x ) x 0 f (x ) f (0) Using L.H. Rule f (x 2 ).2 x f (x ) f ( x) 0

lim

x

2

2 xf ( x 2 ) 1 0 1 0 f ( x)

lim

(see previous illustration) If it passes through B’ (0, – b), then 0 + ab = a2 – b2 a 2b 2

c )

lim

coordinates of B are (0, –b). Equation of the normal at L is 2

b

{( a b ). c }.{( a ( b c )}

Let the normal at the extremity L of the latus rectum passes through the extremity B of the normal axis

ax e

x

105. (a) y 1 4x x 2

Y

b2 ) 2

y = mx

a 2 (1 e 2 ).

1 e 2 e 4 or e 4 e 2 1 0 102. (b) Let the asymptotes be x + 2y + k1 = 0 …(1) and 2x + y + k2 = 0 …(2) Since (1) and (2) pass through the centre of the hyperbola i.e (1, –1), k1 = 1 and k2 = –1 So, the asymptotes are x + 2y + 1 = 0 and 2x + y – 1 = 0 Since the equation of a hyperbola and its asymptotes differ in constant terms only, therefore equation of hyperbola is ( x 2 y 1)(2x y 1) 0 Since it passes through the point (2, 1), (2 2 1)(4 1 1) 0 20 Thus the equation of hyperbola is ( x 2 y 1)(2x y 1) 20 0

2x 2

2 y 2 5xy x y 21 0

103. (a) We have, [ a

b b

1

5 ( x 2) 2

y = 1+ 4x – x2

a 2 a 2 (1 e 2 ) [a 2 a 2 (1 e 2 )]2

i.e

a)

b ).

L

C

A

c) (c

= (a

c

c

a]

3 2

X 2 3/ 2

3/ 2 2

(1 4x x )dx

We have

2

0

mx dx 0

3 9 2 2 4

1 27 3 8

m

13 6 + . Then, tan

9 4

On solving, we get m 106. (c) Let

=

= K tan

tan K tan 1 Applying componendo and dividendo, we have

or

tan tan

tan – tan

K 1 K –1

MT-195

Solutions cos cos

sin sin

or

cos – cos

sin – sin Given that, – Therefore,

or sin

y2

sin

then

yn

sin x

x

2

cos

3 2

cos

3

1 cos a ( x

)( x

(x

x

(x

)( x 2

)

)

2

)

)2

(x

=

1

x

sin 2 a

2 )2

(x

a2 (

=

x2 f ''(0) ... 2!

)2 2

)2 ( x 4

)2

.

lim f (x)

x

f (0)

0

cot x

Now, lim (1 x) x

s

)2

111. (c) Given f (x) = (x + 1) cot x is continuous at x =0

x6 xn ... ( 1) n / 2 ... 6! n!

b

)

)2 ( x 4

a2 ( x

if n is even.

at

) (x 2

(x

a2 ( x

1

n = 0, if n is odd 2

,

lim (1

0

x

x cot x = lim e

x

a.s (at b) ds . dt s2 at b as (at b) s 2 s

1 dx 1

3

x

ds ds 2 at 2 b or dt dt again differentiating

dt

2 –

2

3 dx –

1

= lim

2s

2

2 dx +

110. (a) Given limit = lim

lim

n/2

2

2

1 dx +

2sin 2 a

f (0) xf '(0)

x2 x4 cos x 1 2! 4! when n is even. 108. (a) s2 = at2 + 2bt + c

d 2s

0

0

= ( 1) Then f (x)

0

=4–

n In general (y n )0 cos 2 Now, put n = 4; then (y4)0 = cos 2

Hence (y n )0

2

2

n 2

cos x

cos

2

2 109 (c) I = [x 2 ] dx – [x ] dx

x

2

b) 2

s 3.

dt 2

1

2 2

cos x

Now put x = 0; then y0 1, (y1 )0 0, (y 2 )0

(y3 )0

= .

K 1 sin . K –1

=

( at s3

d 2s

a +

as 2

dt 2

=

cos x, y1

y

d 2s

K 1 K –1

K 1 K –1 = and

i.e.,

K 1 sin sin K –1 107. (a) We have

sin sin

= e

x

x

x 0 tan x

f (0) = e 112. (c)

lim 1 x

0

lim

0

= e

1 x) x

0

x 0 tan x

lim

x

1 x

1

x 3e 2x dx

u = x3

dv = e2x dx v = 1/2 e2x

x cot x

e

EBD_7443 MT-196

Target VITEEE u' = 3x2

v1 = 1/4 e2x v2 = 1/8 e2x v3 = 1/16 e2x

116. (a)

u'' = 6x u''' = 6 Using Bernoulli's formula u dv

p q p q p q ~ (p q) (p q) ~ (p q) F F T T T T

T F F T F F

uv u ' v1 u "v 2 ....

We get x 3 e 2x dx

1 2x e 2

(x 3 )

1 2x e 8

(6x)

1 2x 3 e x 2

Put x + y = v,

1 2x e 16

(6)

3 2 x 2

3x 2

dy cos (x dx

113. (a) We are given that

1 2x e 4

(3x 2 )

3 4

y)

dy dv dy dy 1 dx dx dx dx So, the given equation becomes

1 v sec 2 dv 2 2

dy x cos y dx

3

4 x sin y

d 4 (x sin y ) dx

xe x dx c

Since, y (1)

0 so, c = 0.

Hence, sin y

115. (a)

1

x

4

2*(23)

( x 1) e x

3

5 21

±2

+

–1 0 1 x (–1, 0) (1, ) Domain of f(x) is (–1, 0) (1, 2)

=

c

P(E F) P(F) P( E

( x 1) e x

2*(5 3 15) 2 23 46

P F

0.10 0.42

(2, ).

120. (a) P (E / F) + P ( E / F)

1

2 * x *3 5 x 2*(5*3)

P E

x 4 – x2 D (g(x)) = R – {–2, 2} h (x) = log10 (x3 – x) x3 – x > 0 x (x + 1) (x – 1) > 0 – + –

x

xe x

x 4 sin y

F

P F

119. (c) Let g(x) =

1.dx

xe

P E

= P E/F

dx

v x y tan x C x C 2 2 which is the required solution. Rewriting the given equation in the form 4

F F F

E F E F E P(E) = 0.10 and P(F) = P(1) + P(2) = 0.10 + 0.32 = 0.42 Hence, required probability

tan

114. (b)

F F T

(p q) (~ (p q)) is a contradiction. 117. (c) P = n (n + 1) (n – 1) (n + 2) (n – 2)......... (n + r) (n – r) = {n (n + 1) (n 2).......(n + r)} {(n – 1) (n – 2).......(n – r)} = (n + r) (n + r – 1)......(n + 1) (n) (n – 1)...... (n – r) Clearly P is product of (2r + 1) consecutive integers, so divisible by (2r + 1) !

dv 1 cos v dx

1 2v 1 sec dv dv dx 2 2 1 cos v Integrating both sides, we get

T T F

118. (d) Let E : ‘face 1 comes up’ and F: ‘face 1 or 2 comes up’

so that 1+

dv 1 cos v dx

F F F

71

121 122 123. 124. 125

(b) (d) (c) (a) (a)

P(E F) P(F)

F) P(E P(F)

F)

P(F) P(F)

1

MT-197

Solutions

MOCK TEST 7 1.

(c) Change in momentum, p = Fdt = Area of F-t graph = ar of – ar of + ar of 1 2 6 3 2 4 3 2 = 12 N-s (d) When air is filled in between the plates, then capacitance is given by A C= 0 d When interspace between the plates is filled with wax, then capacitance is K 0A C = 2d A K 0 or C = d 2

or

3.

4.

(a)

(b)

5.

(d)

6.

(c)

C =C

K 2

K 6=2 K=6 2 In case of a spherical metal conductor, the charge spreads uniformly over the entire surface. Therefore, the sphere will retain the charge for longer time. When terminal velocity is reached then body moves with constant velocity hence, accelesation is zero. The maximum torque experienced by a dipole of dipole moment p placed in an uniform external electric field E is given by max = pE Here, p = qa = 0.05 × 10–6 × 30 × 10–3 = 1.5 × 10–9 cm E = 106 NC–1 –9 6 max = (1.5 × 10 × 10 ) Nm = 1.5 × 10–3 Nm. 1 As, mv2 = vq 2 2vq or, v = m

2vq m

and,

2v 4q m

vB =

7.

vA 1 1 Hence, v 4 2 B or, vA : vB = 1 : 2. (d) Net electric flux = (4 × 103 – 8 × 103) = – 4 × 103 Nm2c–1 q =

8.

q = 0 = – 4 × 103 0 coulomb. (a) From the graph l = 10–4m, F = 20 N A = 10–6m2, L = 1m

=

2.

Thus, vA =

0

FL Al

Y

9.

20 1 10

6

10

20 1010 2 1011 N/m 2 (c) The internal resistance of the cell, 1

r=

2

R

2

10.

(d)

12.

240 120 2=2 120

B=2 A dB = 2dA

RB = RA

11.

4

B B

A A

AB

AA

B

A

d B2

A

A

B

d A2

2 A

4d A2 d A2

2

(a) From Kirchhoff’s junction law at junction P, I1 + I 2 = 6 ...(1) From Kirchhoff’s voltage law to the closed ciruit PQRP, –2I1 – 2I1 + 2I2 = 0 –4I1 + 2I2 =0 2I1 – I2 = 0 ...(2) Adding (1) and (2), we get 3I1 = 6 I1 = 2A From (1), I2 = 6 – 2 = 4A. (d) The number of electrons per unit volume, i.e., the electron number density in a conductor is very large ( 10 28 m –3).

EBD_7443 MT-198

13.

14.

15.

16.

Target VITEEE

Therefore, a large current is obtained in a conductor irrespective of small drift speed of electron and small charge. (c) Magnetic field at the centre of a current carrying loop is given by ni B= 0 2r For n = 1 turn i B= 0 ... (1) 2r When n = 2 turns and radius r r2 = , i2 = i 2 2 i B= 0 r 2 2 2 0i 2 or B = ...(2) 2r Now, from eqn. (1) and (2) B =4 B Hence, B = 4B (d) Heat developed by the electric bulb, W P 210 5 60 H= = 4.2 J J = 15000 Cal. (a) As the temperature rises the atoms of the liquid become more mobile and the coefficient of viscosity falls. (c)

v2 = u2 + 2a S/2 = 2 (g sin ) S/2 = gS sin For lower half of inclined plane 0 = u2 + 2 g (sin – cos ) S/2 – gS sin = gS ( sin – cos ) 2 sin = cos =

(d) Harmonics and fundamental frequency voltages cross-modulate with each other producing fundamental frequency currents whose relation with the normal fundamental current vary with amount of harmonics in the oscillator circuit. The phase angle of the resultant fundamental frequency current thus varies resulting in frequency variation. Such frequency variation can be minimized by using a tank circuit having high effective Q having loosely coupled load.

19.

(b) P

(b)

S/2 ugh Ro

r2

r2 d dt N 2 A2

P

Case 1 : P1

y ( 160) 50 ( 160)

S/2 h oot Sm

E2 R =

dB dt

2

r2 d ( NBA)2 dt

2

N 2r 2 N 2r 2

(4 N )2 (r / 2) 2 4 When we decrease the radius of the wire, its length increases but volume remains the same]

Case 2 : P2

67 y 160 100 110 y = – 86.3° y

17.

= 2 tan

18.

Reading on any scale – LFP UFP LFP = constant for all scales 340 273 373 273

2 sin cos

P1 P2

S/2 sin

S/2 sin

For upper half of inclined plane

20.

1 1

Power remains the same. (b) The magnetic field at the centre of the coil B(t) = µ0nI1. As the current increases, B will also increase with time till it reaches a maximum value

MT-199

Solutions (when the current becomes steady). The induced emf in the ring

where is magnetic flux = B.2r

d d d A (µ0 nI1 ) ( B. A) dt dt dt The induced current in the ring

e

dI1 decreases with time and hence I2 also dt decreases with time.] Where I1 = Imax (1 – e –t/ ) The relevant graphs are

4 1 22 0.02 2 100 1000 7 = 5.02 × 10–6V = 5 V (approx.) (c) The equivalent inductance can calculated as the combination of resistances L1 = 0.4 H; L2 = 0.6 H both are in parallel

=

µ0 nA dI1 R dt

|e| R

I 2 (t )

dr dt

26.

L1 L2 ie, L' = L L 1 2

0.24 H

Now, L' and L3 = 0.76 H are connected in series ie, L = L' + L3 = 0.24 + 0.76 = 1H. 27.

(b) Ampere's law is :

B.d

µ0i in

Faraday law is : 21.

(a) For resonant frequency to remain same LC should be const. LC = const L 2 (c) Relative velocity = v + v = 2v emf. = B.l (2v) (b) Self inductance of a small circular coil of

23.

radius r, L =

24. 25.

0N

2

A

2r

0N

d =– dt

d (B r 2 ) dt

28.

( r2 ) 2r

1 m / sec 1000 (Since radius is decreasing) = ? ; when r = 4 cm = 0.04 m 1mm / sec

µ0 id r . 4 r3

Gauss's law is :

2

i.e. L N 2 If no. of turns are doubled then self inductance will be four times. (d) The change in flux in both 1 and 2 are same while in (3) and (4) is zero. (b) B = 0.02T ; = 90° dr dt

dB

L'

LC = L' × 2C

22.

d d not e 0 dt dt Biot-Savart's law is :

e=

(c) Given d = 1mm, D = 1m.

0

E.dA

q.

= 6.5 × 10–7m.

For n th dark fringe yn=

2n 1 D 2 d

For third fringe, 2 3 1 2 = 1.625 mm.

y3 =

29. 30.

31.

1 6.5 10

1 10

3

(b) I = I0 + I0 + 2 I0 I0 4I0 I/I0= 4 (c) E = h E It is a straight line graph (c) (d) Given N0 = 64 N N = N0

1 2

n

7

EBD_7443 MT-200

Target VITEEE N

1 2

64N

n

6

32.

1 64

n

1 2

total number of reactions, n =

r=

1

2Ze 0 E

4 9

=

Etotal = n × E

n

1 1 i.e., 2 2 n=6 but n = T/T1/2; T1/2 = 2 hrs T = T1/2 × n T = 2 × 6 = 12 hrs. (d) The distance of closest approach of particle

9 10

1040 3.875 10 3 = 1.29 × 1028 J We have Etotal = Pt

=

-

t=

2

35.

2 29 1.6 10

19

10 1.6 10

13

1.6 10

19

36.

hc

(b) Energy of a photon = =

6.62 10

34

3 108

IA (hc / )

(b)

1 10

4 19

6.62 10

=

1H

2

1H

3

31 H 2

3 P 1H 2 He

4

4 2 He

n P n

M = M(2He4) + M(p) + M(n) – 3M (1H2) = (4.001 + 1.007 + 1.008) – (3 × 2.014) = (6.016 – 6.042) = – 0.026 amu | E | = | M | c2 = 0.026 × 931.5 MeV = 0.026 × 931.5 × 1.6 × 10–13J = 3.875 × 10–12J As each reaction involves 3 deutrons, the

12 6

2A

38.

(d) Here y ( A B ) A.B A B . Thus it is an AND gate for which truth table is

1019 10 4 6.62

1 1015 = 1.52 × 1012/sec 100 6.62

2 2 1H 1H

1016

(c) A crystal structure is composed of a unit cell, a set of atoms arranged in a particular way; which is periodically repeated in three dimensions on a lattice. The spacing between unit cells in various directions is called its lattice parameters or constants. Increasing these lattice constants will increase or widen the band-gap (E g), which means more energy would be required by electrons to reach the conduction band from the valence band. Automatically Ec and Ev decreases.

1015 6.62 If one percent of incident photons produce photo electrons, th en the no. of photoelectrons emitted

34.

1.29 10 28

E total P

37.

=

=

J

(a) Given W = 3.2 eV Vs = – 3.2 V. (b) D2 is forward biased whereas D1 is reverse biased. So effective resistance of the circuit R= 4+2 =6 i

300 10 9 = 6.62 × 10–19 J No. of photons emitted/sec/m2

=

12

= 1.29 × 1012 s.

= 8.35 ×10–15 m = 8.4 × 10–15 m. 33.

1040 3

A B

y

0

0

0

0

1

0

1

0

0

1

1

1

MT-201

Solutions 39.

(d)

A

NO2

C B The truth table for the above logic gate is :

A 1 1 0 0

40.

B 1 0 1 0

C 1 1 1 0

This truth table follows the boolean algebra C = A + B which is for OR gate. (c) Since conductor and semiconductor are connected in parallel hence voltage across them is same. If ammeters show same

H3N CO NH3

47. 48.

42. 43.

49.

V are I same. If voltage is increased by small value then following the same relation V I for constant R, both conductor and semiconductor show same current. (d) In d10 arrangement of Ni in Ni(CO)4 all electrons are paired. (b) ZnCl2 and CdCl2 are ionic, but HgCl2 is covalent. (c) For a particle that contains 2 neutrons and 1 proton. Z = 1, A = 2 + 1 = 3. Thus it is 1T3.

(d)

45. 46.

(d) (a) The given compound may have linkage isomerism due to presence of NO2 group which may be in the form –NO2 or –ONO. It may have ionisation isomerism due to presence of two ionisable group –NO2 & – Cl. It may have geometrical isomerism in the form of cis-trans form as follows : [Co(NH3 ) 4 Cl(NO 2 )]NO 2 & [Co(NH3 ) (NO2)2]Cl ––– ionisation isomers. [Co(NH3)5(NO2)2]Cl & [Co(NH3)5(ONO)2Cl ––– Linkage isomers

H3N

NH3

NH3

CO

NO2 NH3

NO2

NH3

Trans-form

cis-form

Geometrical isomers (d) The appearance of colour in solid alkali metal halide is due to presence of F-centres found as defect in the crystal structure. (d) d-spacing n 2 100 200pm 2 sin 2 sin 30 º (d) Using the relation, G = H – T S, we get For a reaction to be spontaneous the value of G should be negative. When G = 0, the equilibrium exists. Substituting the given values, we get G = 176.0 × 103 – T × 160 or 0 = 176.0 × 103 – T × 160 (at equilibrium) or 160 T = 176.0 × 103 176.0 103 = 1100 K. 160 The reaction will be spontaneous above 1100 K at which it is in equilibrium. Since, 1100 K = (1100 – 273)°C = 827°C. So, the reaction will be spontaneous above 827°C.

or

Ti(H2O )36 has one unpaired electron in its d-subshell which gives rise to d-d transition to impart colour.

44.

NH3

=

reading hence their resistances R

41.

NO2

50.

(d)

T

S° = S (products)

S (reactants)

= [3 S (H 2 ) S (CO)] [S (CH 4 ) S (H 2O)] = [3 130.6 197.6] [186.2 188.7]

51.

= [391.8 197.6] [374.9] = 214.5 JK–1 mol–1 (b) Entropy states the randomness or disorderness of the system. At absolute zero, the movement of molecules of the system or randomenss of the system is zero, hence entropy is also zero.

EBD_7443 MT-202

52.

53.

54.

Target VITEEE

(a) A reaction in which ng is zero is not affected by change in pressure, considering various given reactions (a) ng = 2 – 2 = 0 (b) ng = 2 – (3 + 1) = –2 (c) ng = 2 – 1 = + 1 [Consider only gaseous substances] (d) ng = 2 – (2 + 1) = –1 Thus, (a) is the correct option. (c) Let us consider a reaction, x X + yY aA+bB rate = [X]x [Y]y It is given that order of reaction w.r.t. component Y is zero. Hence, rate = [X]x i.e., rate becomes independent of the concentration of Y. (c) When concentration of each of the reactant is taken as unity, the rate of a reaction is called specific rate constant. Specific rate constant is depend on time as illustrated in equ. (i) and also on temperature as illustrated in Arrhenius equation (ii)

[A ] 2.303 log 0 t [A] k = Ae–Ea/RT

k=

Cell constant R

k

–Mg(OH)X

60.

(d)

HC

56.

1.15 1000 4. 6 250 1 (c) Let x gm of Zn deposit on 9 gm of Hg

% of Zn in Amalgam =

x 9 x

HC

|

C CH 2 OH H C

O

CCH 2 OH But 2 yn, 4 diol

61.

Therefore, acetylene with formaldehyde under high pressure forms by butyndiol. (a) If ethyl alcohol is dehydrolysed by conc. H 2SO 4 at 443°K, then ethylene is formed. C 2 H 5OH

63.

3 2 Eq. of Zn = 65.4

59.

C CH 2 OH

HOCH 2

conc . H 2SO 4 H 2O

.. .O. I electron pair localised

64.

65.

C2H 4

(d) The most suitable reagent for converting alcohol to acetaldehyde is PCC. Other reagent will converted into acid. (a) Higher the electron density on O, stronger is the H-bond with water and thus more is the solubility. Thus solubility of the three ethers follow the order

x = 3gm

58.

High pressure

HCHO

H

× 100 = 25

6 96500 Current = × = 8.85 amp. 65.4 1000 (b) The process of prevention of rusting of iron by coating zinc over its surface is called Galvanization. (a) Due to H-bonding, the boiling point of ethanol is much higher than that of the isomeric diethyl ether. (c) We know that

R

OH

formaldehyde

HC

62.

1.15 250

eq

57.

CH

acetylene

........ (ii)

(a)

OMgX R CH2 – CH2

H 2O

........ (i)

55.

CH2 – CH2

H2C – CH2+RMgX O

>

.. .O. III e pair delocalised

>

.. .O. II e pair more delocalised

(a) ArCOR’ can be prepared by the combination of ArH + R’COCl and not by ArCOCl + RMgX because here the ArCOR formed will further react with RMgX to form 3º alcohol, ArC(OH) R2 as the final products. (b) Like aldehydes but unlike other acids, formic acid reduces Tollen's reagent, Fehling

MT-203

Solutions solution, mercuric chloride and potassium permanganate HCOOH

66.

2CuO

CO 2 H 2 O

Fehling sol.

Cu 2 O Cuprous oxide (reddish brown)

(d) Hoffman Rearrangement O || R C NH 2

Br2

-

72.

From propanoic acid

AgCl (Insoluble in HNO 3 ) From chloropropanoic acid

Br

Br

74. 75.

Alkyl siocyanate O || C

67.

N R

Ca

CH 3COO Calcium acetate

68.

(a) N-Chloro- or N-bromo-succinimide is the latest reagent used for -halogenation. (a) (d) The basic character of an amine in water is determined by (i) electron availability on the N atom and (ii) the extent of stabilization of the cation (protonated amine) due to solvation by hydrogen bonding

OH H 2O

RNH 2 CO32 .

CH3

CH 3 CH 3

H

H

70.

+

H

H

OH2

CH3

N

H

+

CH3

CH3

OH2

N

+

H

OH2

CH3

Dimethyl amine

76.

;

OH

H H

Protonated dimethyl amine + (Lesser stabilized than CH 3NH 3)

COOC 2 H 5 | C N | NH

(urotropine)

[HNO3 is a nitrating agent]. (a) The reaction is used in purification of all types of aldehydes and methyl ketones only because the ketones having bulky alkyl form unstable adduct. (a) In H.V.Z reaction, in presence of small amount of phosphorus, aliphatic carboxylic acids containing -hydrogen atom reacts with Cl2, Br2 to yield an acid halide in which -hydrogen has been replaced by halogen. So, trichloroethanoic acid, among the given options which do not contains -hydrogen atom, will not give HVZ reaction.

N

2 + Protonated methyl amine, CH 3NH 3 (Highly stabilized)

(CH 2 )6 N 4 6H 2 O

RDX (or cyclonite)

CH3

H

CO + CaCO 3

C3H 6O6 N6 3HCHO NH3

+

Methyl amine

Acetone

(d) 6HCHO 4NH 3

N

[Urotropine is a drug for treatment of urinary infection.] (CH2)6N4 or C6H12N4 + 3HNO3

69.

OH2

H

So (d) is the answer as this intermediate is not present in the reaction. (b) When calcium acetate is heated, then acetone is formed. CH 3COO

So, CO2 comes from HCO3 of NaHCO3 (b) Precipitates in two cases is due to different compounds. CH 3CH 2 CH 2 COO Ag + (Soluble in HNO3)

73.

O || C=N–R

: :

O || R–C–N

(a) Carboxylic acids decompose bicarbonates into CO2 gas and H2O to form sodium salts of carboxylic acids.

O || R C NHBr

O || R C N

KOH

71.

(c)

Cu powder heat ( N 2 )

[Y] Pyrazoline derivative

CHCOOC 2 H 5

oxidation

CHCOOH

HOOC COOH [Z]

EBD_7443 MT-204

77.

Target VITEEE

(a) Completing the reaction, we get CH3 CH3

CH2 – N +

LHD = Lf 2 CH3 CH3

+ N 'A'

(d)

h

h

O°C Aniline (diazotisation)

2 3

2 1 2

2

2 h

h

2 h 2 h 0

h

h

0

2 h 1

= lim

CuCN

h

0

h

h 1 2 h

2

h

0

h

2 1 2

h

= lim = lim

f 2

h

0

0

h

1

1

f 2 h

2 h 1 2 h

= lim h

h h

lim

RHD = Rf (2) = lim

(A) benzene diazonium chloride

f 2

h

0

N NCl NaNO2 ,HCl

h

0

= lim

+

f 2 h

lim

2 h 2 h

= lim

HO

NH2

x x

at x = 2

CH3

78.

x x , if 0 x 1 x, if 2

82. (b) Here, f x

O–

h

2

0

3h 2 2 h

h h 3 h 2 3h lim =3 h h h 0 h 0 LHD RHD f(x) is not differentiable at x = 2. (a) Since the system is homogeneous and has a non-trivial solution

= lim

79.

(d)

N 2Cl H /H2O

80.

(b)

81.

(b) Let

COOH

CN

83.

Sodalime

2

5

1 2

b

c

p

a

q b

c

0

q

a

b

r c

a

b

5

2

4 a 4 ~ 0

0 a 6

1

2

0 a 6

a 1

0

using R1 R 1 R 2 and R 2 Expanding along C1 we get p[q(r c) br ] aqr 0

[R 2 R 2 2R1, R 3 R 3 R1 ] Clearly rank of A is 1 if a = –6 1 2 5 2 4 3 0 Also, for a 1, | A | 1 2 2 and

p a 0 1 A

Cu/KCN

2

5

6 20 14

4 3 rank of A is 2 if a = 1

0

pqr pqc prb qra a p

84.(d) lim x

2

b q

c r

(at x = 2)

lim

x

0 r r c

R 2 R 3.

0

1.

1 cos{2( x 2)} x 2 L.H.L =

q

2

lim

x

2 sin( x 2)

2

2 sin( x 2) ( x 2)

x 2 1

MT-205

Solutions

2

x

(at x = 2)

Thus L.H.L

1

(a) log42 – log82 + log162........ = log 2 2 2 log 23 2 log 2 4 2 .........

R.H.L

(at x = 2)

(at x = 2)

2

85.

(a)

y

Let If y

tan

3a x x

1

x a

a

3

3ax

tan

2

3

2

3

2

3 tan

3

sin

1

sin

x3

x a

1

tan

tan 3

6

1

y 1

3 tan

5 4

Let cos

1

4 5

sin

2

cos A

1

x a

1

cos A

a

2 1

2

b

2

b

2

2

2

2a.b

2

a

c

2

a b c

5 4

c

2

c

2

2

...(i)

36 2b.c

64

2c.a 100

a.b b.c c.a

2

...(ii) ...(iii)

100 ...(iv)

a.b b.c c.a

100

1

2

a

a b c

/ 2]

4 5

0

a

4 5

b

1 C 5

3 4

x=3

B

b 2

2

c

2

2 a.b b.c c.a

[using (iv)]

100 10

2ˆi 3ˆj kˆ and b ˆi ˆj 2 kˆ , then ˆi ˆj kˆ 2 3 1 5ˆi 5ˆj 5kˆ

89. (d) Let a

… (i)

3 A 5 cos–1(4/5) = sin –1 (3/5) equation (i) become, x 3 x 3 sin 1 sin 1 5 5 5 5

2

a b c

4 5

3 sin A = 5

A = sin–1

b

Now,

cosec

2

x 5

2

and c

a

cosec

x 5

6

Similarly, b

, then

1 = 1 – loge2

On adding eqs.(i), (ii) and (iii), we get

3

x 5

a b

a

1 ...... 4

1 1 1 ........ 2 3 4

1

a2

[ sin 1 x cos 1 x

sin

88. (d)

a3 x2

x a

1

x (d) Let sin 1 5 1

1

6

a

x

3

1 3

tan , then y

a

86.

3

1 3

=

exist. 2

1 2

=

1 cos{2( x 2)} does not x 2

Hence, lim x

87.

2 sin( x 2) ( x 2)

R.H.L = lim

1

2

unit vector perpendicular to the plane of a and b is 1 ˆ ˆ ˆ (i j k ) . If is the required angle, then 3 2ˆi 2ˆj kˆ 1 ˆ ˆ ˆ cos . (i j k ) = 1 2 3 3 3

sin

1

1 3

tan

1

1 2

cot 1 ( 2 )

EBD_7443 MT-206

Target VITEEE

90.

(b) Since 3 (1) + 2 (–2) + (–1)(–1) = 3 – 4 + 1 = 0 given line is to the normal tothe plane i.e., given line is parallel to the given plane. Also (1, –1, 3) lies on the plane x 2 y z 0 if 1 – 2 (–1) –3 = 0 i.e. 1 + 2 – 3 = 0 which is true L lies in plane . 91. (b) f (x) = sin x + cos x, g (x) = x2 – 1 g (f (x)) = (sin x + cos x)2 – 1 = sin 2x Clearly g (f (x)) is invertible in – [

2 sin is invertible when – /2

2x

95. (a)

x 2 2

100

94.

1

b

2 1 2 1 2

1

96.

2

2( 2 1

2

97.

2

3

)

3

4

sin 2 cos14

14 and cos

sin 2

cos 2

cos 2

1

A 1

2

0

2( 3

sin 8

2

2 0

3ˆ j 0kˆ . So, its vector equation 2

2iˆ

sin 8

1 1

2iˆ

5ˆ ˆ 3ˆ 2iˆ j 0kˆ . j k 2 2 Hence, p = 0 (b) We have, sin 8 0 and cos14 0 8 A = sin + cos14 0 But sin8 + cos14 = 0 is possible only if sin = 0 and cos = 0 simultaneously which is not true for any value of A 0 or A > 0 Also, 0 sin2 1 r

1 2

5ˆ ˆ j k and is 2

is

2

1

2iˆ

parallel to the vector

2 ( 2 )100 so that the required equation is x2 + x + 1 = 0. (c) Apply c1 c1 c 2 c 3 , then determinant is

2

and has

2,

position vector a

100

and

5 2, , 1 2

3 , 0 . Thus, given 2 line passes through the point having

direction ratios

)100

(

y

through the point

x 4 4 92. (a) Let Shamali invest ` x in saving certificate and ` y in PPF. x + y 50000, x 15000 and y 20000 8 9 x y Total income = 100 100 Given problem can be formulated as Maximize Z = 0.08x + 0.09y Subject to, x + y 50000, x 15000, y 20000.

Thus

z 1,

5 2 z 1 3 0 2 This shows that the given line passes

2 /2]

(d) We have x 3 1 ( x 1) ( x 2 x 1) Therefore, and are the complex cube roots of –1 so that we may take =– and =– 2, where 1 is a cube root of unity..

2y 5 3

x 2 2

–

93.

The given line is

4

)

Hence, we get 0 A 1 (b) Circumference of a circular wire of radius 7 cm is = 2 × 7 = 14 l As we know, = r 14 7 180 = = 210°. 12 6

MT-207

Solutions 98.

parabola is y2 = 4ax.

(c) The given Let (x1, y1) be the midpoint of the chord passing through the point (h, k). Equation of the chord having (x1, y1) as its mid point is T = S1 i.e yy1 – 2a(x + x1) = y12 – 4ax1 or yy1 – 2ax = y12 – 2ax1. Since it passes through (h, k), ky1 – 2ah = y12 – 2ax1 The locus of the mid point (x1, y1) is ky – 2ah = y2 – 2ax, y 2 2ax ky 2ah

or 99.

(b)

0

Let P (at2, 2at) be any point on the parabola

y2 = 4ax. The equation of the tangent at P is ty = x + at2 Since the tangent meets the axis of parabola in T and tangent at the vertex A in Y, coordinates of T and Y are (–at2, 0) and (0, at) respectively.

2at) P (at , 2

T

X

Let direction cosine of the line be l, m, n where given, cos =

DC's = r cos = rl rl = 12 … (i) similarly r m = 4 … (ii) and r n = 3 … (iii) Squaring and adding equations (i), (ii) and (iii), we get r2(l2 + m2 + n2) = 122 + 42 + 32 r2 = 169 ( l2 + m2 + n2 = 1) r = 13 Now, l =

projection on x axis length of line segment 4 ,n 13

y1 a

2

y12

ax 1

Also e

y2

1

b2 a

2

x

c

2

2cy

0

3 13

12 4 3 , , . 13 13 13

1

9 49

1

58 7

i.e. ( 58 ,0) . Now equation of parabola with vertex at (0,–3) and axis along y-axis is x 2 It passes through ( 58

y| c

2c y

12 13

Foci of hyperbola ( ae,0),

(0 3)

Taking only positive sign and squaring 2

DC 's r

x 2 y2 49 9 Its conjugate axis is y-axis.

The locus of G (x1, y1) is y2 + ax = 0 100. (c) Let (x, y) be the co-ordinates of point P. Given, | x2

c . 2

102. (a) Eqn. of hyperbola is

Let the coordinates of G be (x1, y1) Since TAYG is a rectangle. So, x1 = –at2 and y1 = at Eliminating t, we get a

101. (c)

and passing th rough

Hence, direction cosines are

A

x1

c 2,

c 2

Similarly, m =

Y

G

0,

at

c , 2

which represents a parabola with its vertex

Parabola is x 2 Its focus is 0, 3

( y 3)

58 ,0) . 58 3 58 ( y 3) 3 58 4.3

or 0,

11 6

EBD_7443 MT-208

Target VITEEE

103. (c) Using Lagrange's Mean Value Theorem Let f(x) be a function defined on [a, b] f (b) f (a) b a c [a, b] Given f(x) = logex 1 f '(x) = x equation (i) become

then, f '(c)

c

1 c

f (3) f (1) 3 1

1 c

log e 3 log e 1 2

2 log e 3

f ( x)

x

a

1

a

x

1

f (–x) = ( x)

( x)

1 ax 1 a

x

log

f ( x)

a

1

a

x

1

ax 1 x

1

y2

sech 2 x

(y) 0

y

1 x2

1 tanh 2 x

2y 2 y3 )

0,(y2 )0

1, (y3 )0

0,

x2 (y ) .... 2! 2 0

x(y1 )0

f ( x)

log cosh x

107. (d) Given i

f ( x)

1 y12

x2 2

x4 12 di d

k tan

x6 – .... 45 k.sec 2

Let i be the error in i, corresponding to error in di d

f (x)

i i

k sec 2

2

. The relative

i i

k sec2 . k tan

cos 2

105. (c) 106. (b) We have y log cosh x tanh x

0, (y1 )0

1

It is an even function (d) f(x) = k f (–x) = k = f(x) It is an even function

y1

y 2 y3

error in i is

1 x2 1 x

2(y1y4

i

1 x2

log

y5

We obtained

x

a It is an even function (c) f ( x)

y 22 )

y

1 x x2

x

2(y1 y3

(y 4 )0 2, (y5 )0 0 and (y6 ) 0 16 etc. Substitute the values in maclaurin’s series

1 x x2 1 x x2 f(x) is an odd function

(b) f ( x )

y4

(y)0

c = 2 log3e

x

2y1y2

y6 2(y1 y5 4y2 y4 3y32 ) Now putting x = 0, we obtain

log e3 2

1 x x2

104. (a) (a) f ( x )

....(i)

y3

.

1 cos 2 sin cos

cos . sin

i 2 2 i sin .cos 2sin .cos sin 2 108. (a) Alphabetical order is A, C, H, I, N, S No. of words starting with A = 5! No. of words starting with C = 5! No. of words starting with H = 5! No. of words starting with I = 5! No. of words starting with N = 5! SACHIN – 1 Sachin appears at serial no. 601

MT-209

Solutions /2

The required area is ACBDA, given by

sin x

109. (c) Let I =

sin x

0

dx

cos x

... (i)

1

/2

sin( / 2 x)

I=

sin( / 2 x)

0

/2

=

cos( / 2 x)

5

= 1 – 3 – –1 3

=2

2 3

1 –1

4 sq. units. 3

111. (b) Given curve meets x-axis at x = 0,1, 2

cos x cos x

0

–1

5

dx

5 y3 y– 3

2

(1 – 3 y – 2 y )dy

=

Then,

2

sin x

dx ... (ii)

Y

Adding (i) and (ii), we get /2

2I = 0

/2 0 /2

sin x sin x

= 0 /2

x=2

O

cos x dx cos x sin x

The required area is symmetrical about the point x = 1 as shown in the diagram. So Reqd. area

cos x dx cos x

1

1

2 (x 3 3x 2

= 2 y dx 1.dx [x]0 / 2 /2

= 4

0

0

0

2

0

sin x sin x

cos x

dx

4

y2

0

–

1 – ( x – 1) 3 [Left handed parabola with vertex at (1, 0)] Solving the two equations we get the points of intersection as (–2, 1), (–2, –1) x 3y2

y2

1

Y

A (–2, 1)

D (1, 0)

C

X B (–2, –1)

Y

X

/6

112. (b)

1

x

3

x

2

cos7 3x dx

x 0 t 0

cos 3x dx 0

1 3

/2

1 1 1 = 1/2 4

3x

t

Put 3x = t 3dx = dt dx = 1/3 dt 7

=2 0

0

/6

2x) dx

0

x4 2 = 4

x 110. (d) 2 [Left handed parabola with vertex at (0, 0)]

x 2 y2

X

x=1

sin x dx cos x sin x

/6 /2

cos7 t dt.

0

1 6 4 2 16 . . 3 7 5 3 105 113. (c) Any conic whose axes coincide with coordinate axis is ax2 + by2 = 1 .....(i) Diff. both sides w.r.t. 'x', we get

2ax + 2by

dy =0 dx

EBD_7443 MT-210

Target VITEEE dy =0 dx Differential again,

i.e. ax + by

a+b y

d2 y

C 2 6 84 = n – 2C2 (Two players played three games each)

2

=0

a From (2), b

ydy / dx x

a From (3), b

y

y

xy 114. (c)

n 2

dy dx

dx 2

y2

dy dx x

d2 y

dx

y

x

2

dx

2

dy dx

dx 2 2

dy dx

2

dy =0 dx

y

4a

ae a e 5 Let x be the inverse of 8. Then

116. (c)

x *8

8x 5

5

or or

(x 1)(x 3) (x 2) 2) 2

2

+ —

2 1 x

25 8

0

+

–1 2

or

3 x

C5 45! 50!

x

or [ x ( 1)](x 2)(x 3) 0, x

–

50

0

(x 1) (x 2) (x 3) (x

5

5

0

{g ( x )} is real if g(x)

(n 2)(n 3) 156 0

n 15 .

…(i)

6

…(ii) 2 Adding equations (i) and (ii), we get 2 sin 1 x 2sin–1 x = 3 3 3 x sin 3 2 Given equation has unique solution. 119. (d) The number of ways of arranging 50 books = 50 P 50 = 50!. The number of ways of choosing places for the five volume dictionary is 50C5 and the number of ways of arranging the remaining 45 books = 45P = (45)! Thus the number of favourable 45 ways is ( 50C5) (45 !). Hence the probability of the required event

2

a for all a

e

78

sin–1x + cos–1 x=

yy1 2a y12 yy1 0 Degree = 1, order = 2. 115. (b) Let e be the identity element. Then

x *8

C2

118. (a) Given, sin–1x – cos–1x = dy dx

d2 y

n 2

.....(iii)

n 2 5n 150

d2 y

4a(x h), 2yy1

a *e

Then the number of games played by (n – 2) players

.....(ii)

3

Domain = [–1, 2) [3 ) 117. (a) Let there be n participants in the beginning.

50! 5! 45!

45! 50!

1 1 5! 120 120. (c) With respect to A the following results are in favour of the required event . WW; WLW; WLLW; LWW; LWLW; LLWW So, the desired probability = p.p + pqp + pqqp + qpp + qpqp + qqpp = p2(1+ 2q + 3q2)

121 122. 123 124

(c) (d) (d) (a) Tremulous means shaking or quivering slightly. Therefore, Steady is the opposite of Tremulous. 125: (d)

MT-211

Solutions

MOCK TEST 8 1. 2. 3.

R2T4.

(a) Energy radiated (a) (d) dU(x) = – Fdx

13.

kx 2 2

x

Ux

Fdx

0

ax 4 4

5. 6. 7.

S1S 2 P R where S S1 S 2 Q S (c) Resonance frequency is independent on resistance. (b) Total momentum = 2pˆi pˆj Magnitude of total momentum ( 2 p)

p

2

5p

2

11.

1 2

(b)

1 C n

N N0

2 3

15 60

600 s

h

h

t

10 minutes.

1

hc

0

6

hc 1.24 10

1 0

108 18

108 23

1.24 100 (23 18) 18 23

15.

(b)

16.

(b) m*

17.

m0

m0

v

1

2

1 (.96)

2

1.49 eV

m0 0.28

2

c anew = 0.28 (4.2 × 1010) =1.176 × 1010 ms–2 (b) From input signals, we have,

(d) (b) Rnew = n2R XC

V2 (2/ 3) R

Use hc 1.24 10 6 (eV) m

5p

This must be equal to the momentum of the third part.

10. (c)

15 60

(a) K max

(b)

=

8. 9.

14.

x = 0, F is zero. (b) The r.m.s. value of a.c. component of wave is more than d.c. value due to barrier voltage of p-n junction used as rectifier.

2

V2 R

or t 2k ; also at a

U = 0 at x = 0 and at x 4.

(b) H

A B Output NAND gate 1 or X C 2 nC

1 n

1 gives n = 4.32 20

t = nT = 4.3 × 3.8 = 16.5 days 12. (a) Point 1 and 3 are at same potential. Similarly point 2 and 4 are at same potential. Joining resistance between 1 and 2, 2 and 3, 3 and 4 we find that they all are in parallel.

2,4

0 0

1 1

0

0

1

1 0

1 0

0 1

The output signal is shown at B. 18.

(a)

X0 t

1,3

0 1

19.

(

1)t

5 5890 10 0.45

(0.45)t

5 10

5890 10

10

= 6.544 × 10–4 cm

(a) Clearly v1 = 2 ms –1, v2 = 0 m1 = m (say), m2 = 2m v1' = ?, v'2 = ?

So equivalent resistance 1 Rp

1 2R

1 2R

1 R

2 R

or Rp = R/2

v1 ' v2 ' e= v v 2 1

....(i)

EBD_7443 MT-212

Target VITEEE By conservation of momentum, 2m = mv1' + 2mv2' From (i), 0.5 =

20.

... (ii)

v2 ' v1 ' 2

v2' = 1 + v1' From (ii), 2 = v1'+ 2 + 2 v1' v1 = 0 and v2 = 1 ms–1 (b) Potential difference across the branch de is 6 V. Net capacitance of de branch is 2.1 µF So, q = CV q = 2.1 × 6 µC q = 12.6 µ C Potential across 3 µF capacitance is 12.6 4.2 volt 3 Potential across 2 and 5 combination in parallel is 6 – 4.2 = 1.8 V So, q' = (1.8) (5) = 9 µC

(b)

sin i 2 sin r 1 where r = 90º for particular incidence angle called critical angle. When the incidence angle is equal to or greater than i c, then total internal reflection occurs. It takes place when ray of light travels from optically denser medium ( 1 > 2) to optically rarer medium.

28. (a) q V

0

is emitted with 1 e particle or electron), so it is known as -decay. r 0 id (c) B for B1 , r ( ˆi ˆj) 4 r3 i ˆ 0 B1 k ( ˆi ˆj) ..... (1) 4 2 2 for B , r ˆi ˆj 2

B2

0

ikˆ (iˆ ˆj)

......... (2)

4 2 2 From (1) and (2)

23.

B1 B2 and | B1 | | B2 | (b) Velocity of light in a medium,

1

c

0 o

v He vH

A

i

12 – 2 = (500 ) i

q He qH

2q V i.e. v m

2e m e 4m

=

2V

R

1

2 29. (a) Loss in K.E = Area under the curve 30. (a) 31. (b) Let x be the desired length E0 E0

/3

x

E

E

i

1 50

E0

E

.

E0 3 /2

2E 0 3

2E 0 3 From equations (i) and (ii),

E

10 500

q m

mH m He

E0 3 3 Potential gradient in second case

r r

12V

1 m v 2 or v 2

Potential gradient in the first case

1

500

24. (a)

R = 100

27. (b) According to Snell’s Law,

0( –

22.

1 50

25. (b) Max. K.E. = h – W0 ; so Max. K.E. 26. (d)

is known as – particle & is known as antineutrino. Since in this reaction 1e

12 500 R

500 + R = 600

V

21.

Again, i =

E0 3

(x)

2E 0 x 3

x

2

E0

…(i)

…(ii)

MT-213

Solutions 32. (c) If R is radius of bigger drop formed, then 4 R3 3

As

2

4 3 r or R = 21/3 r 3

r2

v0

v 01 v0

42. 43.

R r

41.

2

1/ 3

(2

2

r

r)

2

22 / 3

2

or v 01 v 0 2 2 / 3 5 (4)1 / 3 33. (b) The position of nth dark fringe. So position of first dark fringe in x1 D / 2d . d = 20 cm, D = 0.1mm, = 5460 Å, x1 = 0.16 –2V It can be sketched as 34. (d) 0V

0V

45. 46.

–2V

35. (b) As momentum is conserved, therefore, m1 m2 R1 R2

A1 A2

v2 v1 A1 A2

1/ 3

1 2 1 2

1/ 3

0.5

Power 40. (a)

0.5 0.5

(2) 2 0.5 2 W.

RCH 2 COOH

48. 49.

(c) A plot of m vs C is linear with negative slope. (c) (c) In H2 – O2 fuel cell, anode :

1 : 21 / 3

coil is equal to external resistance (= 0.5 ), hence power developed is maximum by cells in circuit. 2

H 2O

(c) With ammonia, HCHO forms hexamethylenetetramine, CH3CHO gives acetaldehydeammonia addition product, while C6H5CHO gives hydrobenzamide. (b) (c) Units of k show zero order; rate = k = constant [A] reacted after 100 min = 100 × k = 100 ×1.0 × 10–3 = 0.1 mol L–1. [B] after 100 min = 2 × 0.1 = 0.2 mol L–1

2H 2 O( ) 2e

1 O 2 (g) 2e 2 Hence, net reaction : H 2O

H 2 (g)

50.

1 O2 (g) 2

H 2 O( )

(c) K 2 Cr2 O 7

6H 2 SO 4

4KCl

4NaHSO4

51. (b)

E

h hc E

2A

52.

2OH (aq)

0.0592 1/ 2 log P PO H2 2 2

º Ecell

Ecell

. Since internal resistance of

Current through R

Ag 2O

RCOCHN 2

H 2 (g) 2OH (aq) Cathode :

ˆ ˆ 0 36. (d) F q [v ( i)] B (i) 37. (a) From the graph it is clear that A and B have the same stopping potential and therefore the same frequency. Also B and C have the same intensity. 38. (d) The electric field deduced by Gauss’ law is due to all the charges. 39. (b) Total internal resistance of two cells 1 1 1 1

RCOCl

Diazoketone

44.

47.

depletion layer

SOCl 2

(d) R COOH CH 2 N 2

Immobile ions

– + P – + N – +

(d) Number of unpaired electrons in Cr, Mn2+ and Fe3+ are 6, 5 and 5 respectively. (b) LiAlH4 does not reduce –NO2 group to –NH2 group.

(c)

hc

2KHSO 4

2CrO 2Cl 2

3H 2O

;

6.63 10 34 (3 108 ) 3.03 10 19

656 nm .

EBD_7443 MT-214

Target VITEEE

53. (c) –

OH

O

O

O :– .–. CH 2 = CHCH2Cl

OH

OCH2CH = CH2

CH2CH = CH2

+

+

OH

59. (a) Aromatic aldehydes are weak reducing agents, hence these do not reduce Fehling solution. On the other hand, aliphatic aldehydes and –hydroxyketones are easily oxidised by Fehling solution, i.e., they reduce Fehling solution. 60. (b) Paraldehyde is a solid trimer of acetaldehyde, while others are different forms of formaldehyde. 61. (d) Iron becomes passive on treatment with conc. HNO3 or acidified KMnO4, H2O2 or H2CrO4. 62. (d) Under the given condition, k1 = k2 1011 e–3000/T = 1010 e–2000/T ; 10 = e1000/T ln 10 = 1000/T, 2.303 log 10 = 1000/T; T= 1000/2.303 K 63. (d)

54.

(b)

(t1/ 2 )2 (t1/ 2 )1

a1 a2

n

0.1 0.01

; 10

n 1

N1

orbital. CH3CH 2 X Ag (–AgX)

57.

[CH3CH 2 .... X .... Ag] + CH3CH 2 OH

–

from H 2O

CH 3CH 2OH

(b) The momenta of fission fragments are equal. K .E.

1 2 mv 2

m2 v 2 2m

p2 2m

K.E. of lighter fragment K.E. of heavier fragment

2

( HCl )

º m ( BaSO 4 )

CO HCl

(b) CH 3CONH 2 HCl CH 3CO N H 3Cl (Acetamide as a weak base) 2 CH3CONH2 HgO (CH3CONH) 2 Hg (Acetamide as a weak acid) (c) CH 3CONH 2

LiAlH 4 or Na / C2H 5OH

E 70

144 90

Thus, total fission energy = 70 + 112 = 182 MeV

CH 3CH 2 NH 2

H 2O

66. (d) Use of SOCl 2 and ClCOCOCl forms gaseous by-products which can be easily removed, giving better yield of RCOCl. Further, oxalyl chaloride is particularly easy to use becasue any excess of it can be easily evaporated due to its low b.p. (62ºC) O O

||

E 112 MeV

(b)

1 2

º e

64. (c) 65. (d)

O

(p = momentum)

58.

m ( H 2 SO4 )

2 x3 ;

||

reaction because of presence of empty

slow

x2

(a) H C Cl

(c) (b) Heavy metal ions, particularly Ag+, catalyse S

x1

SO42

O

2

Rate = k [A] [B ] 55. 56.

Ba 2

m ( BaCl2 )

CH2CH = CH2 n 1

m ( BaSO4 )

||

||

R C OH Cl C C Cl

R

O ||

R C Cl HCl CO CO 2 67. (d) KNO3 is added to oxidise the lead and zinc impurities.

68. (d) CH 3CH 2 OH

H (from H 2SO 4 )

CH 3CH 2 O H 2

MT-215

Solutions CH3 C H 2

HSO4 (from H 2SO 4 )

CH 2

CH2

On electrolysis of aqueous solution of the complex ion of chromium [i.e., {Cr (NH 3 ) 6 } 3+ ] of the complex [Cr(NH3 ) 6 ][Co(NO 2 ) 6 ] moves towards cathode (i.e., negative electrode) and on this electrode chromium would finally be deposited.

H 2SO4

Thus note that H2SO4 is acting as an acid in step (i), as a base in step (iii). Since it is regenerated back as such it also acts as a catalyst. 69. (c) Since two Ag+ ions are replaced by one Cd2+ ion to maintain electrical neutrality, there are cation vacancies.

77.

–

(d) HCCl3 OH

H 2 O : CCl3

: CCl 2 Cl Dichlorocarbene (a neutral electrophile)

C2 H5OH

70. (b) (CH 3CO) 2 O ( A)

O

CH 3COOH CH 3COOC 2 H 5 ( B)

:

( C)

71. (d) In presence of AlCl3 (a Lewis acid), aniline is converted into anilinium cation, which being m-dir ecting gives aminoacetophenone. 6

2

O –

H CCl2

:CCl2

–

–

O

O

CHCl2

OH

3

72. (c) (a) d : d sp (n = 0) (b) d 8 : sp3 d 2 (n = 2)

78. 79. 80.

3 2

6

(c) d : sp d (n = 4) (d) d 4 : (d 2 sp3 ) (n = 2) 73. (c) 74. (c)

CH 3 NH 2

CH3I excess

( A)

heat

(CH 3 ) 4 N OH ( B)

81. (C) and (D)

[O]

1º Amine

H

[O]

R – .N. – H

[O]

.. R–N=O Nitroso

R – .N. – H

[O]

2 3 3

(b) f :

,

f (x)

+ R–N Nitro

O O–

84.

76. (d) The given compounds are isomeric. They can be distinguished by electrolysis of their aqueous solution. On electrolysis of its aqueous solution the complex ions of cobalt [i.e., {Co(NH3)6}3+] of the complex [Co(NH3 )6 ][Cr(NO 2 )6 ] moves towards cathode (i.e., negative electrode) and on this electrode finally cobalt would be deposited.

CO 2

,

3 sin x cos x 2

(f 1 (x))

82. 83.

RNH 2

[0, 4],

f(x) = 2sin x

[O]

Hydroxylamine

conc . H 2SO 4

HN 3

(CH 3 ) N CH 3OH

75. (d) All of the three are oxidation products of a 1º amine. H OH

R – .N. – H

(c) (b) (b) Reaction is an example of Schmidt reaction in which carboxylic acids are heated with hydrazoic acid in presence of a mineral acid to form primary amines. RCOOH

AgOH

(CH 3 ) 4 N I

CHO

–

2

6

sin

x 2 2

1

6 (c) (d) tan is of period so that tan 3 is of period /3. x x 1 (b) f (x) x e 1 2 = f ( x)

2x xe x

x

2 (e x 1)

x xe x x

1

1

x xe x 2 (e x 1)

x xe x

2 (e x 1) 2 (e 1) f (–x) = f (x) for all x f (x) is an even function.

1

1

EBD_7443 MT-216

Target VITEEE

85. (b) If the given points be A (2, 3, 4) and B (6, 7, 8), then their mid-point N(4, 5, 6) must lie on the plane. The direction ratios of AB are 4, 4, 4, i.e. 1, 1, 1.

Let E denote the event that the person is diagnosed to have TB. P(A) =

A (2, 3, 4)

1 ,P B 1000

999 1000

E E 0.001 0.99, P A B The required probability is given by P

N

A P E B (6, 7, 8)

The required plane passes through N (4, 5, 6) and is normal to AB. Thus its equation is 1(x 4) 1( y 5) 1(z 6) 0 x y z 15 86. (a) The equation of the given line is 6x = 4y = 3z which is written in symmetric form as x 0 y 0 z 0 1/ 6 1/ 4 1/ 3 1 1 1 , , Direction ratios of this line are 6 4 3 and equation of the plane is 3x + 2y –3z – 4 =0 If be the angle between line and plane, then direction ratios of the normal to this plane is (3, 2, – 3) sin

a12

a1a 2

b1b2

b12

c12 a 22

E A

P B

P

=

0.99 0.99 0.001 0.999

=

990 990 999

89. (a) z

1 2i

f ( z)

990 1989

|z|

b 22 c 22

x x 5 r 1 1 r 5 Since, G.P. contains infinite terms –1 < r < 1

x 1 0 x 10, x 5 5 88. (a) Let A denote the event that the person has TB Let B denote the event that the person has not TB.

E B

110 221

1 z

2

3 i 2 2i

1 (1 2i ) 2

6 2i 4 4i

z1 z1 z1

| z1 |2 z1

3 i 2 2i

|3 i| | 2 2i |

9 1

90. (a) z1

5

7 1 2i

7 z

5 2

4 4

0

P

0.99 0.99 0.999

1 4

6 2i 1 (1 4 4i )

| f ( z) |

= 0° 87. (c)

E A

1 0.99 1000 = 1 999 0.99 0.001 1000 1000

c1c 2

1 1 1 3 2 ( 3) 6 4 3 1 1 1 9 4 9 36 16 9

P A

P A

|z| 2

1

arg(z1–1) = arg (z1 ) = arg (z2) z2 = kz1–1 (k > 0) 91. (b) We have, nPr = nPr + 1 n! n! 1 1 (n r )! (n r 1)! (n r) ...(1) or n–r=1 Also, nCr = nCr – 1 r + r – 1 = n 2r – n = 1 ...(2) Solving (1) and (2), we get r = 2 and n = 3

1 1

92. (a)

(a * b)2

(a * a)* (b * b) for all a, b

(a *b)*(a *b)

G

(a *a)*(b * b) for all

MT-217

Solutions

[Applying R3 – R2 and R2 – R1 in second det.]

a, b G a *(b *a) *b

9 k 16 3 7 9 1 = 0 [Applying R3 – R2]

a *(a * b) *b for all

a, b G

2

b * a a * b for all a, b G (by cancellation laws) G is abelian f (x) 93. (a) f (x) = min {x + 1, | x | + 1} =x+1 x R Y

y=x+1

y=–x+1

96.

2

1 =0

2

0

0

[Applying C2 – C1] 2 (7 – k – 6) = 0 k=1 (a) Given f(x) = x2 – 4x – 5

Now, A

X

Y' Hence, f (x) is differentiable everywhere for all x R. C1 C 3 C 5 ..... 94. (d) 2 4 6 n 2

7

2

(n 1)n n 1 2!

A

1 n 1

[ n 1C2

5

2

2n

n 1

8 8 9

4A 5I

32

97.

(b)

C6 .....]

k 5

2

4

2

6

2

5

2

32

k + 4

2

k

2

5

9 k 16 3 0+ 7 9 1 =0 9

11 1

k

42

k 5

4 =0

k 6

2

5

1 0 0

4 2 1 2

5 0 1 0

2 2 1

0 0 1

O

sin 1 cot

cos 1 tan

=

=

3

2

1 2 2

0 0 0

C0

k

8 9 8

0 0 0

n 1

C4

1 1 n 1

42

k

2 2 1 2 2 1

0 0 0

2n 1 n 1 n 1 95. (b) Breaking the given determinant into two determinants, we get 1

4

9 8 8 8 9 8

8 8 9

(n 1)n (n 1)(n 2) 4!

(n 1)n (n 1)(n 2)(n 3)(n 4) ..... 6!

2

1 2 2 1 2 2 2 1 2 2 1 2

n (n 1)(n 2) 4!

1

k

4A 5I

9 8 8 2

n (n 1)(n 2)(n 3)(n 4) ...... + 6!

32

A2

f (A )

(–1, 0)

0

9 k 7 k 3

(0, 1) X'

2

cos sin 1 cos

sin cos 1 sin

cos 2 cos sin

sin 2 cos sin

cos2 sin 2 = cos + sin cos sin Given: x2 – y2 sec2 = 4 and x2 – sec2 + y2 = 16

= 98. (b)

x2 4

y2 4 cos 2

1

EBD_7443 MT-218

Target VITEEE This represents an ellipse centred at (–1, 0)

x2

y2 1 and 16 16 cos2 According to problem

4 4cos 2 4

16 16 cos 2 16

3

1 + cos2 = 3(1 – cos2 )

1

cos 99.

1 4 3 = . 2 4 103. (c) Let x1 be the abscissa of the point whose ordinate is 6. Since the point (x1, 6) lies on y2 = 6x, 36 = 6x1 or x1 = 6 The point is (6, 6) Equation of tangent at (6, 6) is

and of eccentricity =

4 cos2 = 2

3 4 4 ,

2

3 ( x 6) 2 or x 2 y 6 0 y 6

(a) We have, a cos A b cos B c cos C a b c R (sin 2A sin 2B sin 2C) 2R (sin A sin B sin C)

A B C sin sin 2 2 2

100. (b) We have, sin [cot = sin cot

1

sin cot

1

1

e1

1

x ))]

1 1 tan tan 1 x

dy dx

1 x2

dy dx

1

1

1 1 x2

1 x2

1 1

2 x2

1 x2 101. (b) 5 and 12 are two sides of a right angled triangle and

x = hypotenuse = 102. (b)

2 x2

52 122

y2 + x = 3

y2 (x 1) 2 + = 1. 3 4

16 25

y2 25

1 is

3 5

x2 y 2 1 16 9 ......... (1) 105. (a) We have x3 – 3xy2 + 2 = 0 and 3x2y – y3 – 2 = 0 ......... (2) Diff. (1) and (2) w.r.t. x, we obtain

1

1

x2 16

5 ( e1e2 = 1) 3 foci of ellipse = (0, 3) Equation of hyperbola is

2

1 cot 2 cot

1

yy1 = 2a (x + x1))

e2

r R

(cos (tan

(

104. (b) The eccentricity of

4 sin A sin B sin C A B C 2 4 cos cos cos 2 2 2

4 sin

2

C1

= =

x2 y2 and 2xy 2xy

x y2 Since m1 × m2 = – 1, therefore the two curves cut at right angles. 106. (b) For the curve yn = an – 1 x On putting n = 2, we get y2 = ax Which is curve of the form of parabola, where the subnormal at any point is a constant. n = 2.

107. (d)

dx dt

C2

2

2 c t and

dy dt

2bt

13 v

dx 2 dt { 2ct 2

dy dt

2

2bt 2 }

2t c 2

b2

MT-219

Solutions 108. (d)

lim

h

lim

h

0

2

3

9

8 h

2h.3 8 h

0 2h.3 8

8 (8 h)

111. (a)

1

h{82/3 81/3. (8 h)1/3 (8 h)2/3}

1 48 109. (c) Let the boxes be marked as A, B and C. We have to ensure that no box remains empty and all five balls have to put in. There will be two possibilities : (i) Any two box containing one ball each and 3rd box containing 3 balls. Number of ways = A(1) B(1) C(3) = 5C1 . 4C1 . 3C3 = 5 . 4 . 1 = 20 (ii) Any two box containing 2 balls each and third containing 1 ball, the number of ways = A(2) B(2) C(1) = 5C2 . 3C2 . 1C1 = 10 × 3 × 1 = 30 Since, the box containing 1 ball could be any of the three boxes A, B, C. Hence, the required number of ways = 30 × 3 = 90. Hence, total number of ways = 60 + 90 = 150. 110. (d) It is a case of Bernoulli trials, where success is not crossing a hurdle successfully. Here, n = 10.

0

1

112. (b) Given f’(x) = f(x) + f (x ) dx

=

1 1 5 9 C1 6 6

...(1)

0

Differentiating we get, f " (x )

f " (x ) 1 f ' (x )

f ' (x ) 0

f " (x) dx f ' (x)

On integrating ln f ' ( x )

x

c

dx ex

f ' (x )

c

ke x

where k = ec, a constant Again integrate kex D

f ' (x) dx

kex D ...(2)

f ( x)

k+ D= 1 ...(3)

Put x = 0 in (1), f(0) = ke°+D (given f(0) = 1) 1

f ( x ) dx

Also from (1), f’(x) = f(x) + 0

ke x

1

ke x

D

(ke x

D) dx

0

ke x

Dx

1 0

D 0

ke D k

D 0

k (e 1) 2D 0 Solving (3) and (4) , we get 2 and D 3 e

k

= 10Cr

10

1 + f (1) = 2a + 3b + 6c 2

f (0) + 4f

1 n = 10 and p = 6 P(X = r) = nCrqn–r. pr

1 0 5 10 0 C0 6 6

1 a b = + + c, f (1) = a + b + c 2 4 2

f (0) = c, f

5 1 5 q= 6 6 6 let X be the random variable that represents the number of times the player will knock down the hurdle. Clearly, X has a binomial distribution with

1 r 5 10 r 6 6 P (player knocking down less than 2 hurdles) = P(X < 2) = P(X = 0) + P(X = 1)

1 (2a + 3b + 6c) 6

f (x) dx =

p = P (success) = 1

10

5 5 10 510 6 6 6 2 69 2 If f (x) = ax + bx + c,

=

2e x 3 e

f (x )

113. (d)

p

(u p.w

1 e 3 e

1 e 3 e

v)

(v

(u

v ). w

w ) v( w (v

u)

w ). w

EBD_7443 MT-220

Target VITEEE v( w

u ). w

=

[u v w] 0 0

v 5( p . w ) 5( p . u ) 5( p . v ) v

w)

1 D

=

3 2

6D 8

e

2x

1 e 2x (D 4)(D 2) Since, f(D) = (D + 2) q (D) 1 1 xe 2x xe ( 2) 2 Hence the general solution is y

Ae

4x

1 3

0

1 2

1 3 1 3

12

4x

x2 2

1

1 16 1 2

4 4 1 15 2

4

3 2

9 2 3

3 3 3 2 3 sq. units 2 2 116. (b) Let T be the temperature of the cooling object at any time t. =

depends on the vectors Hence, 114. (b) The characteristic equation is p2 + 6p + 8 = 0 (p + 4) (p + 2) = 0 p = – 4 and – 2 The C.F. is Ae–4x + Be–2x Particular integral, 2

3

5 ( p . u ) and v 5 ( p . v )

Similarly,

P.I. =

=

1

x2 2

5

5( p . w )

5 p.(u

3

Be

2x

1 xe 2

2x

2x

dT dt

dT k(T S) dt where k is negative. (T S)

T S

ce kt ,

T S ce kt When t = 0, T = 150 150 = S + c c = 150 – S The temperature of the cooling object at any time is T = S + (150 – S)ekt 117. (c) Without loss of generality, let the right angled OAB be such that OA = OB = a units. Along OA take unit vector as B

115. (b) The tangent on x2 + y2 = 4 at 1, 3 is x

3y

4 and equation of normal at

1, 3 is y

D

x 3 O

Y

iˆ and along OB taken unit vector as ˆj , so that

(1, 3 )

X

(4, 0)

(0, 0)

1

Required area = 0

OA

ˆ OE ai;

AD

OD OA

BE

OE OB

4

x 3 dx

A

E

4 x dx 3 1

AD . BE

aˆ i; OB 2

ˆ OD aj;

aˆ ˆ j ai ; 2

aˆ ˆ i aj ; 2

aˆ ˆ aˆ ˆ j a i . i aj 2 2

aˆ j; 2

MT-221

Solutions 5a 2 4

5a 2 cos 4

a2

4 4 ; cos 1 5 5 Let O and B be the initial and final positions of the girl respectively. Then, the girl’s position can be shown as in the figure. Now, we have OA = 4iˆ

5ˆ 3 3ˆ 8 3 ˆ 3 3ˆ i j i j 2 2 2 2 Hence the girl’s displacement from her

=

cos

118. (c)

AB

ˆi AB cos 60

B 3 km

W

30° 60° A 4 km

3 ˆ 3 3ˆ i j 2 2

4

=

initial point of departure is 2

119. (a) F(2)

F(X

2)

f (x)dx

ˆj AB sin60 2

N

2e

2x

dx

2.

e

0

[e

E

O

1 ˆ 3 3ˆ 3 3 ˆ ˆ j 3 i j = i 3 2 2 2 2 By the triangle law of vector addition, we have 4iˆ

3ˆ 3 3ˆ i j 2 2

4

1] 1 e

2x 2

2

0

e4 1

4

e4

120. (a) Logical expression corresponding to above circuit is = a ' b' c Logical expression corresponding to the complimentary to the above circuit is

S (AB cos 60° is component of AB along Xaxis and AB sin 60° is component of AB along Y-axis)

OB = OA + AB =

5ˆ 3 3 ˆ i j 2 2

(a ' b' c)' (a ' b' )' . c' = (a ' )' . (b' )'. c' = a . b . c' 121. 122. 123. 124.

[(a ' )' a ; (b ') '

b]

(b) (c) (d) (c) Ruminate means to think deeply about something. 125. (b)

EBD_7443 MT-222

Target VITEEE

MOCK TEST 9 1.

(d)

R2

A

E R 2T 4

R 22 T24

E2 E1

7.

R12 T14

put R 2

2R, R1

2T, T1

T2

64 8.

.B

EA =

2

EB =

2

3.

(a)

4.

(b) E =

0,

0

2

0

0

2

0

dV dx

dV dx (2)

.C

EC = 0,

5. 6.

r2

1

qin

–Q

Q

or E × 4 r2 = 0 E=0 (c) Let us first calculate the total charge on the soild sphere. Let us consider a concentric sphere of radius r and thickness dr. Then volume of the sphere, dV = 4 r2dr Given, the volume charge density of the R1

0

r

r dr

Charge on this sphere, dQ = .dV

dV dx (4)

R1

Qs

4 50 5

8 kJ

dQ

4

R1 0 0 rdr

R12 2 0 R12 2 ...(1) QS = 2 0R12 Now, the total charge on the hollow sphere, ... (2) Qh = – (4 R22) By question, Qs + Qh = 0 2 0R12 = 4 R22

dV dx (3) (for region 1 and 3, potential

= 0.4 50 10

0

= 4

s

2

0

.4 r dr = 4 .rdr.. r Total charge on the whole solid sphere,

slope of V x curve

mg cos

0

. 0

remains constant so E = 0) So, E2 > E4 > E1 = E3 (b) The negative charge will move opposite to the direction of field. (b) Work done against gravity Wg = 50 × 10 × 30 = 15 kJ Work done against friction Wf

q

R

sphere =

Since for region–2, V = +ve E2 = –Ve and for region-4, V = –ve E4 = +ve

dV dx (1)

1

23 kJ

R: 0

–

A.

R: E = 4

E.dA

T

R 2T 4

(c)

(b) For r For r

R

(2R) 2 (2T) 4

E2 E1 2.

Total work done Wg Wf 15 kJ 8 kJ

AT 4

E

R2 R1

9.

0

2 0

2

R2 R1

0

2

(a) Terminal potential difference, V = E – Ir (i) This is the case when cell is discharged so, V < E. (ii) This is the case when cell is charged so, V > E (iii) When I = 0 so, V = E

MT-223

Solutions 10.

(a) In circuit 1 , on closing the switch, the current through the inductor is zero. (i1 = 0) i1

R1

2Er1 r1 r2 R

E

In circuit 2 , on closing the switch, the current through the inductor is zero and i 2

13.

2r1

(d)

e T Magnetic moment is given by

The equivalent current, i

14.

R1 R 2

e A T (d) In the galvanometer, Ig = max. current th rough galvan ometer, S = shunt resistance, G = galvanometer resistance then iA

M

E

Ig(galvanometer) = I

40

20

A

40 C

60

15.

B

T 8V

R=

60 40 60 40

24

V 8 1 A R 24 3 (d) Current in the circuit

Current, i =

12.

E E r1 r2 R

r1

2E r2 R

P.D. across first cell = E – ir 1 E

E r1 (r1 r2 ) R

S

Ig

=2

16.

51Ig

LC

= 2 20 10 3 50 10 = 6.28 × 10–3 s. The required time will be

40

The equivalent resistance of the circuit

S G

G S 50 1 Ig S 1 (b) The time period of oscillation I

Thus i2 > i3 > i1 (d)

R

i

In circuit 3, on closing the switch , the current through the inductor is zero and

11.

r1 r2

R

E R2

i3 =

0

R = r 1 – r2

R2

=

2Er1 (r1 r2 ) R

Now, E

6

T 1.54 ms 4 (c) The self induction and resistance of each part become 0.9 × 10–4 H and 3 respectively. The effective value of self inductance

t

=

L

0.9 10 2

4

0.45 10 4 H

3 1.5 2 Thus time constant

and R

L R

0.45 10 4 = 0.3 × 10–4 1.5

EBD_7443 MT-224

17.

Target VITEEE

(c) Change in momentum along the wall = mv cos60º – mv cos 60º = 0 Change in momentum perpendicular to the wall = mv sin60º – (– mv sin60º) = 2mv sin60º Applied force = Change in momentum Time

=

18.

19.

(b)

oscillations of E will be x.

××× ××× ××× ××× ××× ××× ×××

I

R

h

× × ×b× × × × ××××××× ××××××× ×B× × × × ×C× ××××××× ××××××× ×××××××

| d / dt | R

Thus (d) option is not correct. (b) Since work function for a metal surface is 0

where 0 is threshold wavelength or cut-off wavelength for a metal surface. here W = 4.125, eV = 4.125 × 1.6 × 10–19 Joule so 27.

29.

e R

2 2 103

= 10–3 A= 1 mA

34

3 108

4.125 1.6 10

19

3000Å

h = . As each of them has p same momentum, so

(b) Wavelength,

e p. e (d) We have h = h 0 + Kmax h K max 0= h

6.63 10

34

1015 3 1.6 10

6.63 10

19

34

2.7 × 1014 Hz

= (c) Mass of nucleons

7 1.00783 7 1.00867 14.1155 u The mass defect, m = 14.1155 – 14.00307 = 0.11243 u Thus binding energy = 0.11243 × 931 = 104 MeV.

d 1 dh h.b Bb dt 2 dt Bbv R R b t I t (d) R = 2k = 5t2 – 4t + 12, t = 0.2 sec.

I=

6.6 10

0

=

B.

d Induced emf e = dt e = – 10t + 4 = –10 × 0.2 + 4 = + 2 V

hc

W

B

21.

F ep ' .

26.

××× ××× ××× ××× ××× ××× ×××

dA dt R

Fep and F pe '

(d)

28.

A v

Fpe

25.

v

D C As flux decreases the flux current in loop is clockwise. Force on DA due to long wire and BC is towards left while on wire BC is towards right.

1 P2

P1 P2 100 200 66 W P1 P2 100 200 (a) The related light becomes plane polarised. (c) The direction of propagation of em-wave is z. As electric field and magnetic field must be perpendicular and so direction of

d emf induced E = – dt E = – B.2 (a0 – t) (0 – ) = +2 aB r B A

(d)

1 P1

P

23. 24.

= 50 3 3 = 150 3 newton (a) At any time t, the side of the square a = (a0 – t), where a0 = side at t = 0. At this instant, flux through the square : = BA cos 0° = B (a0 – t)2

(b) The power consumed in series is given by 1 P

2 mv sin 60º 2 3 10 3 = 0.20 2 0.20

I

20.

22.

V0

10

30.

(c)

31. 32.

(a) F (W X ). (W Y ) W .( X Y ) (d) Voltage regulator needed constant voltage and so de part is most relevant for its operation.

V

V

MT-225

Solutions 33.

34.

35.

(b) Because atomic mass M of carbon, MC < MSi < MGe So energies are inversely proportional to that of atomic mass. (c) In semiconductors, the conduction is empty and the valence band is completely filled at 0 K. No electron from valence band can cross over to conduction band at 0 K. But at room temperature some electrons in the valence band jump over to the conduction band due to the small forbidden gap, i.e., 1 eV. (d) On the basis of given graph, following table is possible.

41.

42.

A B C 0 1 0

0 1 1

0 1 0

1

0

0

It is the truth table of AND gate. 36.

(c) As

1 2 mA v A 2

1 2 mB v B 2 PB ; P A

mB

vA vB

mA

44.

mA vA mB mA

mB

5 0.6

1 3

Energy

5.6 MHz

r

2

r ( 2

r1 r2 1)

T

or

1)

2

r

So, r

t

t (

2

1)

z

electron transition

t2g e dxy dxz dyz

Here an electron which is present in t2g excites and transition occuring from d xy eg[d 2 2 , d 2 ] . x

45.

y

z

( d) (a) In [Fe(C5 H 5 ) 2 , EAN of Fe 26 electron of Fe 10 electron from two C5 H 5 ions

t

(

1

y

d x 2–y2 , d z2

eg

f LOU 10 0.6 10.6 MHz (b) Let the angle subtended by the arc formed be . Then or

x

electronic transition. If we see the d elctronic configuration of Ti, it has d1 configuration. Since [Ti(H2O)6]3+ is an octahedral compound hence the dorbital splitting will be

(d) All options (a), (b) and (c), are correct. (d) (b) For high-side tunning fLO = fm + fIF , PIF 600 kHz 0.6 MHz f LOL

40.

mB vB

mA

mB mA

37. 38. 39.

43.

(c) d-d electronic transition causes absorption of energy at red visible wave length. In cupric chloride CuCl2, cupper is in +2 state and hence it is d9 system. In five dorbitals there is one unpaired electron present which absorbs energy due to d-d transition. (b) Stainless steel is an iron-carbon alloy with a minimum of 10.5% chromium content. Stainless steel has higher resistance to oxidation (rust) and corrosion in many natural and man made environments because the chromium forms a passivation layer of chromium (III) oxide (Cr2O3) when exposed to oxygen. The layer is too thin to be visible, which means that the metal remains lustrous. It is, however, impervious to water and air, protecting the metal beneath. Also, this layer quickly reforms when the surface is scratched. (b) After crystal field spliting, the five dorbitals. get separated as three t2g and two orbitals. For Mn 2 +, last shell has 5 e–s i.e., 3d5 4s0 . So according to Hund's rule of maximum multiplicity the excited state configuration will be t2g3 eg2. (c) Colour is due d xy eg [d 2 2 , d 2 ]

T

T

(b) [Fe(H 2 O) 6 ]2 : 4p

.. .. .. .. ..

3d

6

4d

( H 2O a weak field ligand)

..

4s

sp3 d 2

EBD_7443 MT-226

Target VITEEE 3

(c) [Cr( NH 3 ) 6 ]

3d

6

..

4s

:

47.

48.

sp3 (tetrahedral) (c) A transition metal ion that exists in its highest oxidation state can be expected to behave as a oxidising agent as it can not be further oxidised but can be reduced. (a) Non-stoichiometr ic compounds are chemical compounds with an elemental composition that cannot be represented by a ratio of well-defined natural numbers, and are therefore in violation of the law of definite proportions. They are most often solids that contain random defects, resulting in the deficiency of one element. So, none of given option is correct. (b) In orthorhombic geometry, y

50.

P

T

Gx T

– P

P

51.

H T

G T

S

1 S 2

HCl

1 S 2

H2

Cl2

1 (131 223) = 10 JK-1 2

H T H per mole T

6000 273

21.98 JK 1mol 1 (a) Substance R Substance S 2k k rate constant t½ 2 t½ Half life period T = n × t1/2 where n = number of half life period

Amount of R left =

0. 5 ; ( 2) T / t ½

Amount of S left

0 .25 ( 2)T / 2 t ½

Equating both or 2

(2)

T /t

0.5 0.25

( 2) ( 2)

T/ t1 / 2

T/ 2 t1 / 2

1/ 2

T 2 t 1 / 2 . 2t1/2 is half life of S and twice the half -life of R

....(4) P

Equation (4) is on alternative form of Gibbs Helmholtz equation Dividing equ. (4) by T2, we get

P

1 1 H (g ) Cl 2( g ) HCl (g ) 2 2 S° = S° (Products) – S° (reactants)

....(2)

G T

(a)

T2

S(per mole )

= – Sy – ( – Sx) S = – ( Sy – Sx) = ......(3) where S change in entropy on combining equation (1) & (3) we get G

P

H

–T 2

187

a

Gy

(b)

S

x z Here, a b c; = = = 90º (d) Gibbs Helmholtz Equation– .......(1) G H–T S differentiate this equation w.r.t. temperature at constant pressure G T

P

H

52.

49.

T

–

T

c b

G

1 T

G

4s

.. .. ..

46.

T

..

3d

(d) [CoI 4 ]2 : 4p

2

on rearrangement, we get

d 2 sp3 (3 unpaired electrons) 7

H

2

T

4p

.. .. ..

G

53.

(c) 2.303 log =–

1.667 10

4

1.667 10

6

Ea R

1 1 1844 1000

MT-227

Solutions 844 E 2.303 × 2 = a × 1844 1000 R

.......(1)

4.606 1844 1000 Ea = 844 R

2.303 log =

0.1M CH3 COOH is (weak acid) not same, therefore the cell voltage is not zero. 57.

6

58.

1423 1000 Ea × 1423 1000 R

423 Ea × .......(2) 1423 1000 R Dividing equation (2) by equation (1)

=

log

Eº

[QH 2 ] 0.0592 log 2 [Q][H ]2

59.

0.699 0.0592 4 0.463V (c) Ethers, when exposed to air in presence of light, form ether peroxides. Cumene (isopropylbenzene), when heated in presence of oxygen also forms peroxide, here oxidation occurs at benzylic carbon. (d) CH 3 CH 2 CH CH CH 2 CH 3 | | OH OH HIO4

K 1.667 10 2

60.

6

423 1844 1000 = 1423 1000 × 844 log

E

E º 0.0592pH

K 1.667 10

(b)

61.

K 1.667 10

6

423 1844 = 1.299 1423 844 K = 19.9 × 1.667 × 10–6 = 3.318 × 10–5 s–1

2 CH 3 CH 2 CHO (b) As far as characterstic of IR spectrum for CH3CH2OH is O–H streching frequency 3391 cm–1 O–H streching frequency 2981 cm–1 O–H streching frequency 1055 cm–1 (d) Phenol has active (acidic) hydrogen so it reacts with CH3MgI to give CH4, and not anisole C 6 H 5 OH CH 3 MgI

CH 4

C 6 H 5 OMgI

= 2×

54. 55.

56.

(c) (b) Here, R = 31.6 ohm 1 1 ohm 1 = 0.0316 ohm–1 C= R 31.6 Specific conductance = conductance × cell constant. = 0.0316 ohm–1 × 0.367 cm–1 = 0.0116 ohm–1 cm–1 Now, molar concentration = 0.5M (given) = 0.5 × 10–3 mole cm–3 k 0.0116 Molar conductance = C 0.5 10 3 = 23.2 s cm2 mol–1 (d) For a concentration cell having different concentrations of ions. c 0.0591 log 1 n c2 If all the concentrations are identical then obviously the cell voltage is zero. But as the pH of 0.1 M HCl (strong acid) & pH of

62.

(b)

OH 50%KOH

OH

+

Cannizzaro reaction

63.

COOK

CH2OH

(d) Completing the given equation C6 H5OH NaOH C6H5ONa H2 O

'A '

OCOONa CO 2 140 C 4 7 atm

OH

E=

COONa

Rearrangement

'B'

EBD_7443 MT-228

Target VITEEE

OH

(a)

(d) Carboxylic acids are more stronger acids than phenols due to greater resonance stablization of carboxylate ion than phenoxide ion. So we have to choose strongest acid from ortho, para or meta – nitrobenzoic acid. And nitrobenzoic acid is strongest among three due to ortho–effect. CH2Cl

70.

(c)

COOH

+HCl NaCl

64.

69.

'C' ‘C’ is 2-hydroxybenzoic acid (or salicylic acid) Note: If the reaction is carried out at a higher temperature p-hydroxybenzoic acid is produced.

acid.

OMgBr CH3MgBr

C6 H5 COOC2 H5

C6 H 5

|

C |

OC2 H5

71.

(a)

Mg(OC2 H5 )Br

C6 H 5

||

C CH3

OMgBr

C6 H5

C |

Excess CH3MgBr

CH3 H2 O

CH3

-Keto carboxylic acids undergo decarboxylation very easily.

73.

(b)

HCOOH

H C

3I

65.

66.

H C

Ozonolysis

4 NaOH

2 C 6 H 5 COCH 3 CHI 3 (d) I2 and Na2CO3 react with acetophenone (C6H5COCH3) to give yellow ppt. of CHI3 but benzophenone (C 6H5COC6H5 ) does not and hence can be used to distinguish between them. (a)

74.

75.

76.

OCH3

CH 3OH

68.

(a) (a) is th e product of cross aldol Condensation O (a) R C H + NH2 NH2 R C N NH2 H Aldehyde

Hydrazine

Aldehyde hydrazone

H+

(a)

77.

H 2O + CO

H C

O

O

H2O

OH 2

CO H

CH 3 COCl 2 H

Pd / BaSO 4

CH 3CHO HCl (Rosenmund's reduction) (b) This reaction is known as curtius rearrangement. RCON3

–

67.

O

H 2SO4

OH

CH3

'B' C 6 H 5COCH 3 HCHO

Succinic anhydride

(c)

|

|

O

72.

|

C 6 H 5 — C CH 2

CH2CO

Succinic acid

OH

Con. H 2SO 4

CH2CO

heat (235°C) H2 O

CH 2 COOH

C6 H5 — C — CH3

CH3

CH 2 COOH |

CH3

O

on hydrolysis will not yield an

N2

2NaOH

RNCO Na2CO3 + RNH2 1° amine is formed. (a) NaNO2 with HCl at 0 °C gives nitrous acid (HNO2) which on reaction with primary aliphatic amines gives alcohol with quantitative evolution of N2 gas. Since (CH3)2CHNH2 is the only primary aliphatic amine among the given options, hence it gives alcohol on reaction with NaNO2, HCl/ H2O at 0°C. The reactions are as follows: NaNO2 + HCl HNO2 + NaCl (CH3)2CHNH2 + HNO2 (CH3)2CHOH + N2 + H2O (c) (CH3 )2 NCOCH3 is dimethylacetamide, hence on refluxing with acid, it undergoes hydrolysis in the following way (CH 3 ) 2 NCOCH 3

H ,reflux

( CH 3 ) 2 NH HOOCCH 3

MT-229

Solutions 78.

(b) (a)

C 6 H 5COOC 2 H 5 + (CH 3 ) 2 NH

a11

C 6 H 5CON (CH 3 ) 2 + C 2 H 5OH 81.

(b) C 6 H 5CONH 2 + CH 3 Mg I

a 21

(c) Let A

.... a m1

C 6 H 5CO NH MgI + CH 4 (c)

and B [b11 b12 b13 .....b1n ] B two non-zero column and row matrices respectively.

C 6 H 5COCl + (CH 3 ) 2 NH C 6 H 5CON (CH 3 ) 2 + HCl

(d) 79.

80.

C 6 H 5CO.O.COC 6 H 5 + (CH 3 ) 2 NH

a 11b11

C 6 H 5CON (CH 3 ) 2 + C 6 H 5COOH (c) Urea decompose HNO2 to N2, CO2 and H2O NH2CONH2 + 2HONO 2N2 + CO2+ 3H2O (b) Let X be the nitrogen containing compound. As we get an oily liquid on treating X with Br2 and KOH, it indicates toward Hoffman's Bromamide degradation in which a carbon atom is degraded from acid amide and amine is formed as product. All this information indicates that the starting compound must be one from benzamide and acetamide.

CONH2

...

a11b1n

a 21b11 a 21b12 ... a 21b1n ... ... ... ... a m1b11 a m1b12 ... a m1b1n Since A, B are non-zero matrices, matrix AB will be a non zero matrix. The matrix AB will have atleast one non-zero element obtained by multiplying corresponding non-zero elements of A and B . All the tworowed minors of AB clearly vanish. rank of AB = 1 AB

82.

(a)

The system is 0x1 + x2 – x3 – x1 + 0x2 + 2x3 = x1 – 2x2 + 0x3 =

0 1

NH2

1

....(i)

+ Br2 + KOH

a 11b12

1 x1 2 x2 0 x3

1 0 2

= 2 3

1

1 2 or AX

B

3

Clearly | A | = 0

Aniline

Now Adj A = 4

CH3CONH2 + Br2 + KOH ¾¾ ® CH 3 NH 2 Methyl amine

2 2

2 1 1 2 1 1

.....(ii)

When benzamide is treated with (CH3CO)2O an antipyretic, paracetamol is formed which confirms that starting compound is benzamide. NH2 NH2

83. 84.

(Adj A) B

system is inconsistent

a +b

b

(CH3CO)2O

a –b O

COCH3 Paracetamol As an oily product is formed when benzamide is treated with Br2 and KOH, it clearly indicated that final product is aniline which is oily in nature. Hence starting compound is benzamide.

0

(c) In all the given equations, the origin is present in shaded area, answer (c) satisfies this condition. (a) B C

A

a

Let OA a and OB parallelogram OACB. a b Again a

b

b . Complete the

OA

OB

OC

|a

b | OC

OA

OB

BA

|a

b | BA

EBD_7443 MT-230

Target VITEEE Given | a b | |a b | OC BA Diagonals of the parallelogram OACB are equal. OACB is a rectangle.

85.

sin

or or

G.M. H.M ac b b ac .......(i ) Again, for the positive numbers an, cn

an

cn

ac

2 an

86.

cn

(b) We have,

cos

87.

2

cn

b

n

88.

2

cos

2 sin sin

2 sin

2

sin

2

2 2

sin

1

1

1 3

sec 1 (2) cosec

1 3

1

2

1

2

1 3

1

1 cot 1 3

k k

k

2 + =k 2 =k k=2

(a) Since , therefore

are coplanar vectors,

,

a 1 1 ] 0

1 b 1

0

1 1 c

a (bc 1) 1(c 1) 1(1 b) 0 abc a c 1 1 b 0 ...(1) = y

abc

Now, 2 2

2cot

1 2 tan 1 3

(Using (i))

=y cos

(a b c) 2

1 1 a

1 1 b

1 1 c

2

(1 b)(1 c) (1 a)(1 c) (1 a)(1 b) (1 a)(1 b)(1 c)

2

3 2(a b c) (ab bc ca ) 1 (a b c) (ab bc ca ) (abc)

cos cos

sec 1(5) sec 1(2) sin 2 tan

2

[

=y

3 2(a b c) (ab bc ca ) 1 (a b c) (ab bc ca ) (a b c) 2 [From (1)]

2

1 – cos sin = y.. 1 sin Trick: Put value of = 30° and check. (b) The given question can be written as

1 5

or or or

a n cn

2

2 sin 1 cos sin

2 cos 2

2 sin

cn 2

an

n

2

2b n

4 sin

Then,

an

a n c n and A.M.

Since A.M. > G.M.

1 5

1

cos

ac 2

G.M.

1 5

2tan

a and b are adjacent sides of a rectangle. (b) Clearly b is H.M. of a and c

Also the G.M. of a and c is Since

1

1

x a

1 2

2 tan 1 ( 3)

89.

3 2(a b c) (ab bc ca ) 1. 3 2(a b c) (ab bc ca ) (a) Let the equation of the required plane be y b

z c

1 ...(i)

It meets co-ordinate axes in points A (a, 0, 0), B(0, b, 0), C(0, 0, c). k

The centroid of

ABC is

or a 3

,

b 3

,

c 3

a b c , , 3 3 3

MT-231

Solutions Equating real and imaginary parts,

a 3 ,b 3 ,c 3 Hence the required plane is

90.

x y z x y z 1 i.e., 3. 3 3 3 (b) Since the sphere is inscribed in the cube whose faces are x = 0, x = 2a, y = 0, y = 2a, z = 0 and z = 2a, therefore, the centre of the sphere is the centre of the cube i.e., midpoint of any of its diagonals and radius is half the distance between the parallel faces. Hence the radius of the sphere is 1 (2a ) a and the centre is mid-point of 2 the segment joining (0, 0, 0) and (2a, 2a, 2a) which are the corners of a diagonal. So the centre is 0 2a 0 2a 0 2a , , (a , a , a ) . 2 2 2 Hence, the equation of the sphere is

91.

(x a)2

(y a)2

x2

z2

y2

(z a ) 2

a2

2ax 2ay 2az 2a 2

0

(a)

z2 0 Clearly from the figure z 0 1

= cos z2

z1

1 3 i 2 2

or, 1 bi

=

a 2

i

i sin

3 2

1

2

3

Q( 3 3s, 3 2s, 6 4s) . Direction ratios of PQ are 3 3s 5 3t, 3 2s 7 t, 6 4s 2 t

8 3s 3t, 4 2s t, 8 4s t i.e., If PQ is the desired line then direction ratios of PQ should be proportional to < 2, 7, –5>, therefore, 8 3s 3t 4 2s t 8 4s t 2 7 5 Taking first and second numbers, we get 8 4s 2t 56 21s 21t … (i) 25s 23t 48 Taking second and third members, we get 20 10s 5t 56 28s 7 t … (ii) 38s 2t 36 Solving (i) and (ii) for t and s, we get s = –1 and t = –1. The coordinates of P and Q are respectively (5 3( 1), 7 ( 1), 2 1) (2, 8, 3) and ( 3 3( 1), 3 2( 1), 6 4( 1)) (0, 1, 2) The required line intersects the given lines in the points (2, 8, –3) and (0, 1, 2) respectively. Length of the line intercepted between the given lines

= | PQ | = ( 0 2) 2 93. (d) We have 3

a

1 3 a b b 2 3 2 2 92. (b) The general points on the given lines are respectively P(5 3t , 7 t , 2 t ) and

lim x

4

(1 8) 2

1 2

1 i 3 2 3 a 2

( 2 3) 2

4 2 (cos x sin x)5 1 sin 2x 5

( a i) 3 2

3

a 2

5

2 2 [(cos x sin x) 2 ]2 = lim 2 (1 sin 2x) x 4

5

5

(1 sin 2x) 2 2 2 = lim (1 sin 2x) 2 x 4

78 .

EBD_7443 MT-232

Target VITEEE 5

97.

5

t 2 22 = lim , where t = 1 + sin 2x t 2 t 2 5

5 (2) 2 = 2

94.

(c)

1

| z | | z || Arg(z )

cos

95.

| 1

arg(z) arg( )

2

(b) Let the point A represents the complex number z, B represents z and C represents z. & are complex cube roots of unity clearly z means rotation of z by z (

y

C (w, z)

2 also 3 OA = OB = OC = |z|. That is the ABC is equilateral.

COA =

Now AC = 2AD = 2 (OA cos 30°) = 2 |z| =

96.

3 2

3 |z|

3 3 3 2 (side)2 = |z| 2 2 (a) From sin x + sin2 x = 1, we get sin x = cos2 x ...(1) Now the given expression is equal to cos6 x (cos6 x + 3cos4 x + 3 cos2 x + 1) – 1 = cos6 x (cos2 x + 1)3 –1 = sin 3 x (sin x + 1)3 –1 [From (1)] = (sin 2 x + sin x)3 – 1 = 1 – 1 = 0 Area of ABC =

cos

1

z

1 x 2 1 y2 ]

i.e

xy

1 x2 1 y2

i.e

xy z

x 2 y2

( z)

cos

1

z] z

1 x 2 1 y2

z2

2 xyz

(1 x 2 )(1 y 2 ) 1 x 2

98.

y2

x 2 y2

x 2 y 2 z 2 2 xyz 1 (b) The number of choices available to him = 5C4 × 8C6 + 5C5 × 8C5 5! 4! 1!

8! 6! 2!

5! 5! 0!

8! 5! 3!

8 7 6 7 +1× 3 2 2 =5×4×7+8×7 = 140 + 56 = 196 99. (a) a = 0, b = 1 a + b = Max (0, 1) = 1 100. (a) The equation of axis of the parabola is x – 4 = 0 which is parallel to y-axis. So the ray of light is parallel to the axis of the parabola. We know that any ray parallel to the axis of a parabola passes through the focus after reflection. The ray must pass through the point (4, –1). 101. (a) The equation of a tangent to the given circle is

=5×

x

BOC =

z

y

1

A (z)

D

AOB =

x cos

1

1

[Since cos

=

O

y cos

Squaring both sides, we get

2 and 3

2 z) means rotation of z by . 3

B (w, z)

1

1

cos 1 ( z)

1

2

x cos

cos 1[xy

i.e

arg(z) arg z

1

cos

5 2

| | z ||

(b) It is given that

y

8

mx

a 2 m2

b2

...(i) y mx a 2 m 2 b2 Equation of a line perpendicular to the above tangent and passing through any of the foci (±ae, 0) is my + x = ± ae ...(ii) The locus of point of intersection of the lines represented by equations (i) and (ii)

MT-233

Solutions can be found by eliminating m from the two equations. Squaring and adding (i) and (ii), we have (1 m 2 )y2 a 2 m2

(1 m 2 )x 2

b2

a 2 e2

a 2 m 2 a 2 (1 e 2 ) a 2 e 2

a 2 (1 m 2 )

x 2 y 2 a 2 , which is the equation of the desired locus.

102. (a) Let eq. of hyperbola be

x2

y2

1 centre

a 2 b2 is (0, 0), vertex (a, 0), focus (ae, 0).

1 2 a a

e = eccentricity = 2ae 3

that a

4 2 2 a e 9

a2 9a

b 2 . It is given

2

4a

2

4 2 (a 9

b2 )

2

2

4b

b

Eq. of hyperbola is

(400 2x)

D

1

x2

y2

a2

5a 2 4

(440 x – 2x2)

(440 – 4x)

x = 110 ;

dx 2

= – ve

A is max. when x = 110. 2r =

440 2x

=

a 16 / 3

x 4

y 4

16 / 3

x4

y4

a 16 / 3 (a 4 ) a 4 / 3 106. (b) Since x2 – 5x + 4 = (x – 4) (x – 1), the function f is defined for all real numbers except x = 4 and x = 1. Hence, the domain of f is R – {1, 4}.

107. (b) After dividing by cos2 x to numerator and denominator of integration sec 2 x dx 4 tan 2 x 4 tan x 5 sec2 x dx

B

d 2A

a 16 / 3

a

I

C

x

q 4/3

1

r

1

[ cos x 1, 100x 99 cos x 99] f’(x) > 0 f(x) is increasing on ( /2, ). 105. (a) 4x3 + 4y3 (dy/dx) = 0 dy/dx = –x3/y3 Equation of tangent, Y – y = – x3/y3 (X–x) y3 Y + x 3 X = x 4 + y4 = a 4 X Y 1 4 3 4 a /x a / y3

p 4/3

r

dA dx

x100 sin x 1.

f '(x) 100x 99 cos x If 0 < x < /2, then f’(x) > 0, therefore f(x) is increasing on (0, /2). If 0 < x < 1, then 100x99 > 0 and cos x > 0 [ x lies between 0 and 1 radian] f’(x) = 100x99 + cos x > 0 f(x) is increasing on (0, 1). If /2 < x < , then 100 x99 > 100 [ x > 1, x99 > 1] 99 100x + cos x > 0

5 2 a 4

x

A

f (x)

Here, p = a4/x3, q = a4/y3

5x 2 4y 2 5a 2 103. (b) Perimeter = 440 ft. 2x + r+ r = 440 or 2x + 2 r = 440 ..........(1) A = Area of rectangular portion = x. 2r

A= x

104. (d)

440 220 = 70 ft 22 / 7

(2 tan x 1)2

4

1 2 tan x 1 tan 1 2 22

C

108. (a) The given function is f(x) = x3 + 3x – 5 The first derivative is f '(x) = 3x2 + 3 The second derivative is f"(x) = 6x . The function is concave upward when f"(x) > 0 , that is, when x > 0. The function is concave downward when f"(x) < 0 , that is, when x < 0.

EBD_7443 MT-234

Target VITEEE At point of inflection, f"(x) = 0 , that is, x = 0. At x = 0, f(x) = – 5 Hence, f is concave upward in the interval (0, ), concave downward in the interval (– , 0) a nd (0, –5) is the point of inflection. /2

109. (a) Let I

(x) (x)

0

/2

2

x

2

I 0

dx then

x

dx

x

2

(x)

1;

f ' (x) =

(x)

0

2

1.dx

x

d2y dx

(x)

/2

= /2

0

2 , BC

0

–1

1

y = |x| –2

2

y = 1 – |x –1|

D –2 C

Thus, bounded area = ( 2) (2 2) = 4 sq. units. 111. (a) Given, A = {x : |x| < 3, x I} A = {x : –3 < x < 3, x I} = {–2, –1, 0, 1, 2} Also, R = {(x, y) : y = |x|} R = {(–2, 2), (–1, 1), (0, 0), (1, 1), (2, 2)} 112. (d) f (x) = max. {x, x3}

=

x ; x ; 3

x ;

1

x

3

x

1 0

dy dx

dy dx

0

y

x 1

x 1

0

0,

x cos x c ....(ii) c 1

x2 2

x sin x

115. (b) 116. (a)

x ;

Integrating we get

y

A(1, 1)

–2

....(i)

x2 sin x x D ...(iii) 2 D 0 When x 0, y 0 The particular solution is

4 4 2 2

B

x 1

Equation (ii) is

I = /4 110. (a) Bounded figure ABCD is a rectangle.

1 1

x 1

dy x cos x 1 dx Integrating again we get

0

AB

0

1 sin x

When x

/2

1;

0

Clearly f is not differentiable at – 1, 0 and 1. 113. (c) The given equation can be rewritten as y = A cos (x + B) – Cex, where A = c1 + c2, B = c3, C = c4e c5 As the minimum number of parameters is 3, order of the differential equation = 3. 114. (b) The differential equation is

dx

x

x

1

3x ;

x

2

2I

3x ; 2

Adding /2

1

x

2

8n 7n 1 (1 7) n 7 n 1 Which is a multiple of 49 of n 2

For n 1, 8n For

n

7n 1 8 7 1 0

2, 8 n

7n 1 64 14 1 49,

a multiple of 49. So, 8n 7n 1 is a multiple of 49 for all n N . Hence X contains all elements which are multiples of 49 for n N . Also Y contains all elements which are multiples of 49. X Y Note that Y contains all multiples of 49 but X does not contain all multiples of 49 but some, i.e. Y / X .

MT-235

Solutions 117. (a) Let us first consider 2 letters and 2 envelopes, then there is only one way to place both the letters in wrong envelope. Next, we consider 3 letters and 3 directed envelopes. The number of ways of putting all letters in wrong envelopes = Total number of possible arrangements – Number of ways in which all letters are in correct envelopes – Number of ways in which 1 letter in correct envelope = 3 ! – 1 – 3C1 × 1 = 2 [ The case of two letters in correct envelope and one in wrong envelope is not possible] Further, we consider 4 letters and 4 directed envelopes. The number of ways of putting all letters in wrong envelopes = Total number of possible arrangements – number of ways in which all letters are in correct envelope – Number of ways in which 1 letter is in correct envelopes (3 in wrong envelope) – Number of ways in which 2 letters are in correct envelope (2 in wrong envelope) = 4 ! – 1 – 4C1 × 1 = 9. NOTE : Such problems are called problems of deragement. Hence, using the formula of deragement. The required number of ways of placing all letters in wrong envelope 1 = 4! 1 1!

=

1 1 2! 2!

m

m2 = 0.2 2! 0.4 0.3

P (X = 0) = e–m = e–4/3

7

1 2

C7 n

C7

4 3

n 7

1 2 1 2

n

n

n

C9

C9 1 2

1 2

9

1 2

n 9

n

That is, nC7 = nC9 = nCn–9, yielding 7 = n – 9 or n = 16. Hence 1 2) 16 C 2 2

P( X

16

16 15 2

1 2

16

15 213

120. (a) Let E1 be the event that exactly two players scored more than 50 runs then P(E1) =

1 1 3 9 2 3 4 10 +

1 2

2 3

1 9 4 10

1 2

2 3

3 1 4 10

1 2

1 3

1 9 4 10

1 2

1 3

3 1 4 10

1 2 1 1 65 2 3 4 10 240 Let E2 be the event that A and B scored more than 50 runs, then 1 1 3 9 2 3 4 10 Desired probability

P(E1

E2 )

= P ( E 2 / E1 )

4! 4! 4! 12 4 1 9 2 ! 3! 4 !

Dividing, m

n

1 4!

118. (c) We are given that e–m . m = 0.3 and

e

119. (c) Let n be the number of tosses and X the number of times heads occurs. Then X ~ B (n, p), with p = 1/2. Since P(X = 7) = P(X = 9), we have

121 122 123. 124.

P ( E1 E 2 ) P ( E1 )

27 240

27 65

(c) (d) (c) (b) Pertinacious means holding firmly to an opinion or a course of action. Insipid means lacking vigour or interest. 125 (d)

EBD_7443 MT-236

Target VITEEE

MOCK TEST 10 1.

(b)

V

t k1

4

4 Q t A k1

V

2.

d t k2

, where is the linear charge 2 0r density on the inner cylinder.

(c)

E

d t k2

b

and V

v2

v1

v2 v2

u1

v which gives 1 v2

(d)

V

q 4

0

1 4

I

J.dA

q

V E E

4

1 v1 / v 2 , 1 v1 / v 2

From (1) : I

1 ...... 8

1 4

6.

1 1 ...... 4 16

Er (R r) V = 0 at R = 0 R= , V=E

r A

r

(1–f)R0

1 4

1 4q 4 0 3 (c) Let K be the dielectric constant of solid capacitor and if C0 is capacitance of air capacitor then solid capacitor will have capacitance KC0. After charge sharing the common potential becomes V.

V

C1V1 C2 V2 C1 C2

V

CV0 KC(0) C KC

E r f R 0 (1 f ) i max if f R0 (1 – f) = 0 f = 0 or 1 (a) Power maximum when r = R. So, power consumed by it i

E

4.

(c)

A

1

1

2 v ln (b / a)

f R0(1–f)

0

0

2 0 ln (b / a) 0

f R0

1

q

.2 dr

(b) V = E – ir

0

4 4

7.

(2q)

q

....... (1)

....... (2)

0

1 2

1

E.dA

I

1 0

0r

2 0r Current per unit length will be

V V

2

b a

ln

Now,

1 e 1 e

1 2

1

v1 v1

E.d a

(a) As u2 = 0 and m1 = m2, therefore from m1 u1 + m2 u2 = m1 v1 + m2 v2 we get u1 = v1 + v2 Also, e

3.

5.

K

V0

V

V

8.

P

r=R

will decrease for R > r.

R

MT-237

Solutions

02

1

So the emf induced between its tips, e = BV v But as by definition of angle of dip, BV tan i.e., BV BH tan BH So,

01

2

e

9.

(a) In parallel combination, total power P = P1 + P2.

10.

(b)

0

v v2 v1

4

10 7.8 –1.2 8.5 10 13.2 7.8 –1 = 6.25 ×10–4 cms–1 0 I B [sin 90 sin( )] 4 a I = 0 (1 sin ) 4 a

i.e., e

=

11.

(b)

(BH tan )v

16. 17. 18.

(c) (d) (a)

2 nt

I

(b)

(a,0)

0 I 1 4 a 0i (sin 4 d

B

=

b a

2

1

10 –7 100

3 1

13.

(d)

14.

(a) E = – NA i=

40

R = 30

X L2

302

40 2

50

Vrms 200 4A Z 50 (a) VL and VC will be in opposite phase so they will cancel each other being equal in magnitude. Resultant potential difference = applied p.d. = 100 volts Hence, Z = R Vrms 100 I rms 2 amp. 50 Z

b2

sin

2

)

3 2

1 = 5 × 10–6 T 2

21.

(d)

NA [75 × (–200)e–200t] × 10–4 R 50 7 10 8 =+ [15000e–1] 30 175 10 5 175 10 5 = = e 2.73 = 0.64 × 10–3 A = 0.64 mA (b) As the plane is flying horizontally it will cut the vertical component of earth’s field BV .

2 0E 0

is electric energy density..

B2 is magnetic energy density.. 2 0

FG dB IJ H dt K

NA d E = (75e –200t) × 10–4 R R dt

1 2

2 0E 0

B20 2 0

(d) Shift of fringe pattern = (

1)

So, total energy = 22.

1 2

30D (4800 10 d

=

15.

R2

0.4

I rms

(0,0)

12.

0.173V

n = 200

400 t

50

Z

20.

=

( 3) 10 V

3 250 20

r.m.s. current = I0 / 2 = 50 amp. (d) Here, X L L 2 fL 2

A (0, b)

1

50 2 amp.

I0 19.

5

2 10

30 4800 10 t

10

30 4800 10 0.6

10

)

tD d

(0.6)t

D d

0.6t 10

1.44 10 0.6

5

24 10

23.

(b)

24. (d)

6

EBD_7443 MT-238

25.

Target VITEEE

(a) Energy of photon corresponding to first line 2 1 1 of Balmer series = (13.6) (2) 4 9 Energy need to eject electron from n = 2 level in H atom

1 = (13.6) = (13.6) 4

4 1 9

1 4

33.

(c)

dN n dt dN (n N

n 1

11 4.155 eV 36 (a) According to relation, Bqv

Mv2 r

r

r

m2v B'q

B

27.

(a) 28. (d)

29.

(d)

hc

eV0

Mv Bq

t

2B'

2x

hc 2 0

t

e

W0

N

n

n

Subtracting them, we have e ( V0

or 30. 31.

V)

hc 0

V

V0

1 2

hc 2 0

hc 2e 0

34. 35. 36.

(b) (d) Balancing vertical forces, we have N Fsin 60 mg For the block not to move, we must have Fcos 60 N i.e., Fcos 60 1 F 2 or F

32.

1

1 2 3

1 2

1 4

(Fsin 60 F

3 2

or F 4 10N 40N (d) According to principle of continuity, for a streamline flow of fluid through a tube of non-uniform cross-section the rate of flow of fluid (Q) is same at every point in the tube. i.e., Av = constant A1v1 = A2v2 Therefore, the rate of flow of fluid is same at M and N.

37.

N N0

N)

n n

N N0

log e

n n

N0 N

n

dN N

t

t

t

N0 N

(n N0 e

t

N0 ) e t

N

(b) Change in momentum = F × t = 10 × 10 = 100 Ns or 100 kg m/s (c) (a) Voltage across the collector load resistance (RC) = 6V, = 0.97; RC = 3 k . The voltage across the collector resistance is, RC = ICRC = 6V 6 6 2mA Hence, IC = R 3 103 C 0.97 1 0.97

1

32.33

2 10 3 61.86 A 32.33 (a) Rf = 740 k , R'f = 650 k , R1 = 10 k , R'1 = 100 k . Let V1 be the output voltage of the first amplifier IC

IB

(2 3)(10)N

N0

log e

Current gain

mg)

10N

0

n n

n

N

1

dt

log e (n

1

W0 and eV

0

t

N

= 13.6

26.

N)dt

dN

N0

eV

N

V1

1

Rf V R1 in

1

740 103 10 103

Vin

75Vin V0

Rf V1 R in 487.5 Vin

650 103 100 103

75Vin

MT-239

Solutions 38.

(a)

Na

4.4 10 28 8

4 10

ni2

6.25 1038 ,

pp

Na

np

ni2 Na

CH3CH 2 – C

OH

6.25 1038

1.1 10

|

3.hydroxy,2 methyl pentanal

44.

3

(d) In body centred cubic lattice one molecule of CsBr is within one unit cell. Atomic mass of unit cell = 133 + 80 = 213 a.m.u Volume of cell = (436.6 × 10–10)3 n ´ at.wt. Density = Av.no.´ vol.of unit cell 6.02 10 23 (436.6 10 10 )3 For body centered cubic crystal n = 2

I 4.8 10 3 A 10 6 3 = 4.8 × 10 A/m2. Suppose the drift velocity is vd then I = neAvd. 3

3 m / sec 10 22 1.6 10 19 10 6 Now the time taken by the electron to travel distance full length of the sample t speed vd

6 10 3

2

2 10 2 sec

40. (d) 41. (a) The equivalent conductance of strong electrolyte little increases with dilution. 42. (a) Normally NaBH4 as well as LiBH4 reduce only–CHO group without effecting carboncarbon double bond, however when it is present in conjugation with benzene ring and aldehydic group it is also reduced along with the reduction of – CHO group. LiAlH 4

213

Density =

ZM

So, current density J

4.8 10

|

CHCHO

OH

5.68 1018

=

|

CH3CH 2 C

20

0.22 ni 2.5 1020 (a) Current I = 4.8 mA = 4.8 × 10–3 A Width b = 4 × 10–3 m, Thickness t = 25 × 10–5 m Length = 6 × 10–2 m Free electron density n = 1022/m3. The current density J = I/A Area = b × t = 4 × 10–3 × 25 × 10–5 = 10–6 m2

vd

|

O H H C CHO

CH3 CH3

1.1 1020 atom / m3

5.68 10 atom / m

39.

|

propanol

18

np

CH3

H

1.1 10 20 atom / m3 ,

a3NA 2 213 107

8.50 gm / cm3

6.02 (436.6)3

45. 46.

(c) (d)

H3 C C = O + CH 3 MgI H3 C H3 C C

O

H OH MgI

H2 O

H3 C CH3 CH3 | C C OH Mg(OH)I | CH3 Tert butyl alcohol

C6 H5CH 2CH 2CH 2 OH C6 H5CH CH CHO NA C6 H5CH 2 CH 2CH 2OH 47. (c) At equilibrium C6 H5CH CH CHO t1 / 2 of A 43. (d) Aldehydes which contain a -hydrogen on a saturated carbon, i.e., CH3CH2CHO N0 N0 undergo aldol condensation. 1 x 228 232 7 1.4 1010

NB t1 / 2 of B

EBD_7443 MT-240

Target VITEEE where N0

Avogadro no.

55.

(b) After every 30 minutes the amount is

and x g is wt. of radium

48. 49.

on solving x = 5 × (b) As volume is constant hence work done in this proces is zero hence heat supplied is equal to change in internal energy. (a) Nitroethane on reduction in acidic medium, like Sn + HCl, Fe + CH3COOH or H2 – Ni gives. Primary amine.

1 and t1/2 is independent of 2 initial concentration for 1st order, hence reaction is 1 order.

reduced to

10–10g

56.

(b)

57.

(c) C 2 H 5 NO 2

AgNO 2

B

C 2 H 5 OH

PCl 5

C 2 H5C l POCl3 HCl [ A]

Sn HCl

(d)2 O Since the compound (C4H10O) react with CH 3CH 2 NH58. 2 2H Sn HCl sodium, it must be alcohol (option b, c, or d). CH3CH2 NO2 6H CH3CH2 NH2 2H2O As it is oxidised to carbonyl compound whereas alkyl nitrite on reduction in neutral which does not reduce Tollen’s reagent, the medium gives alcohol. carbonyl compound should be a ketone and Area Concentration thus C 4 H10 O should be a secondary 50. (a) Conductivity length alcohol, i.e., sec-butyl alcohol; other two given alcohols are 1º. k Area Concentration 59. (c) The statement (c) is incorrect. length 60. (d) The two components should be (CH3)3CONa + (CH3)3CBr. However, tertConductivity length alkyl halides tend to undergo elimination k Area concentration reaction rather than substitution leading to 1 the formation of an alkene, Me2C = CH2 Sm m –1 61. (a) Electronic configuration of Cr is S m mol = m 2 mol m 3 3d 4s 51. (a) For spontaneous reaction, dS > 0 and G and dG should be negative, i.e. < 0. So, due to half filled orbital I.P. is high of Cr. 52. (a) CuSO 4 Ca(OH) 2 is Bordeaux mixture. 62. (d) The reaction is known as Gatterman 6 1000 K 1000 3.06 10 aldehyde reaction. 53. (d) Solubility 1.53 63. (a) First option is incorrect as the value of KP eq given is wrong. It should have been = 2 × 10–3 Ksp = S2 = 4 × 10–6 PCO 2 KP N 2.303 0 54. (d) t log PCH 4 [PO2 ]2 Nt 64. (d) Acetone forms mesitylene (1,3,5-trimethyl N0 0.693 benzene) on distillation with conc. H2SO4. where log 65. (b) [Cr(H2O)6]2+ Cr is in Cr2+ form t1 2 Nt CH 3CH 2 NO 2

6H

where N0 = original amount Nt = amount left after time t t

or

t

2.303

2.303

log

Cr 2 24

U 238 Pb206

log 1

U 238

Pb206

U 238 Given Pb206 : U238 = 1 2.303 4.5 109 log 2 0.693

66. 4.5 10 9 years

3d

4s

Fe 3 3d 4s 26 In [Fe(H2O)]2+ Fe2+ form. Both will have 4 unpaired electrons. (b) Like Clemmensen reduction, Wolf-Kishner reduction involves reduction of >C=O to >CH2 , of course by different reagent.

MT-241

Solutions 67.

6 10

(b) [H+] =

7

0.05 10

3

= 1.2 10

Rate of disappearance of H+ r =

2

exhibit geometrical and linkage isomerism. M [H ] ; t

2 [H ] 1.2 10 M = r 6 105 M s 1 T = 2 × 10–8 s (b) Acidity increases with increases with increasing electronegativity of elements. Fluorine is highly electronegative so produces weaker + I effect and stabilize the carboxylate ion. So, (d) is the strongest acid. Order of electronegativity is F Cl Br . Bromoacetic acid is the weakest acid.

t =

68.

69.

70.

(d) Magetic moment n n 2 where n = number of unpaired electrons

(b)

15 n=3 C

nn 2

H3C – C = NH

N CH3 MgX

OCH3

Q

OCH3

H3C – C = O H 3O

P

O 71.

||

+ CH3 – C– Cl

(b) Benzene

Anhyd.AlCl3

Acetyl chloride

COCH3 + HCl

75. 76. 77. 78.

6 4 208 208 (a) 90 Th 232 78 X 82 Pb (d) Os shows + 8 oxidation state. (b) It is an example of concentration cell, Ecell cannot be zero since [H+] are different (HCl is strong and CH3COOH weak acid). (c) Since, compound A (C3H6O) undergoes iodoform test, it must be CH3 COCH3 (propanone). Further the compound B obtained from A has three times the number of carbon atoms in A (propanone), B must be phorone, i.e., 2, 6-dimethyl 2, 5- heptadien -4- one. (CH 3 ) 2 C O H 3 CCOCH 3 O C(CH 3 ) 2

A , propanone ( 3 molecules )

HCl

(CH3 )2 C CHCOCH C(CH3 )2 2, 6 dimethyl 2,5 heptadiene 4 one

79.

(b) Since, 1º alcohol is more reactive than 2º alcohol toward HI, latter can react only when both 1º alcoholic groups have reacted. 80. (b) For the reaction 2 ZnS 2 Zn + S2 ; G1º = 293 kJ ..........(1) 2 Zn + O2 ZnO ; G2º = – 480 kJ ..........(2) S2 + 2 O2 2 SO2 ; G3º = –544 kJ .........(3) Gº for the reaction 2 ZnS + 3 O2 2 ZnO + 2 SO2 can be obtained by adding eqn. (1), (2) and (3) Gº = 293 – 480 – 544 = – 731 kJ 81. (b) If z = rei = r (cos + i sin ) Then iz = ir (cos + i sin ) = – r sin + ir cos or eiz = e(–r sin +ir cos )) = e–r sin ei r cos or |ei z| = |e– r sin | | er i cos | = e–r sin [cos2 (r cos ) + sin2 (r cos )]1/2 = e–r sin 82. (b) Given equation of line is x 3 y 3 z 2 1 1 DR’s of the given line are 2, 1, 1

DC’s of the given line are

2

,

1

6

6

,

1

6 This reaction is known as Friedel Craft’s acetycation. Since, required lines make an angle with 3 72. (d) C6 H 5 NH 2 COCl 2 KOH C 6 H 5 NH.COCl HCl the given line C6 H5 NH 2 COCl2 KOH C6 H5 NH.COCl HCl The DC’s of the required lines are C 6 H 5 NCO HCl 1 2 1 1 1 2 , , and , , 73. (c) AgBr exhibit Frenkel defect. 6 6 6 6 6 6 74. (d) The complex ion [Cr(SCN)2(NH3)4]2+ can respectively. Acetophenone

EBD_7443 MT-242

Target VITEEE Also, both the required lines pass through the origin. Equation of required lines are x 1

83.

y 2

z x y z and 1 1 1 2 1 i 1 i (a) A + iB = A – iB = 1 i 1 i

84.

(a)

1

90. (a)

A2 + B2 =1 (b) By definition only f (x) = x2 + 4x – 5 with domain [0, ) is one to one.

85. (d)

1 x

5

5

x

x

5(5 )

a = 5t

5 2 t t

a = 5.5 Let 5x = t

a= t

2

1

5

5

x 2

1

2

(5 )

We know that y

t2 92.

(c)

tan

1

1

4 tan 3

tan 2

2

1

5 /6.

1

1

2 2

8 2 3

6 2 4 2 4 sin 1 sin 5 2

1

1 3 1

b [ add & subtract 2a 5 2

2

5 2

2

]

93.

(a)

94.

(b) We have t 4

2

2

5 17 17 a= A a . 2 4 4 (a) After leaving A and I, we are remained with 4 different letters which are to be used for forming 3-letter words. Hence the required number = 4P3 = 4.3.2 = 24

If the line y = x + touches the ellipse 9x2 + 16y2 = 144, then c2 = a2m2 + b2 2 = 16 × 12 + 9 2 = 25, =±5

t2

2

86.

3 1 3 2 2 A = 120° – 45° = 75°

.

( 3 / 2) .

1

1 2 5 t t

5A

3

2

sin 60

(d) The expression cos 1 ( ( 3 / 2)) can be interpreted as the angle whose cosine is

1 A Put t t a = A2 + 5A – 2

a = A2

2

sin C

91.

2

3: 2

3: 2]

C = 45°

x

(5 )

b: c=

[

2x

1 5 t t

t2

1 a= t t

2x

1 x

BA

( sin ) (sec ) ( tan ) 1 (sin )( sec ) tan Since A, B, C are in A. P. A + C = 2B A + B + C = 3B 180° = 3B B = 60° A + C = 120° Now sin B : sin C =

a Given : 51 x 51 x , ,52x 5 2x are in A.P.. 2 We know that if a, b, c are in A.P. then 2b = a+c

a 2. 2

By matrix distribution law, A2 – AB + BA – B2 = A2 – B2 BA – AB = 0 BA = AB

88. (c) It is a known fact that (AB) 89. (c) Given expression

(1 i ) (1 i ) (1 i ) (1 i )

(A iB) (A iB)

87.

(A + B) (A – B) = A2 – B2

95.

(a)

4

4t 2

16 4 2

2

1

2

5

t2 > 0 t2 2 5 Equation of the tangent at /4 is x y + – 2 =0 a b Equation of the normal at /4 is

.......(1)

MT-243

Solutions 8 20 12p 13 13 Taking +ve sign, we get

a b x y ........(2) – = – b 2 a 2 b a p1 = length of the perpendicular from the centre to the tangent

2

=

1

1

2 ab

=

a 2 b2 a 2 b2 p2 = length of the perpendicular from the centre to the normal

=

a b 2

b a 2

1

1

a2

b2

a2

=

b2

2 a2

Area of the rectangle = p1 p2 =

96.

97.

b2

ab a 2

b2

a 2 b2 (a) f ' (x) = 6x2 + 12x + 7 = 6 (x2 +2x) + 7 = 6 (x +1)2 + 1 which is positive for all value of x. Hence f(x) is monotonic increasing function.

(b) f (x) =

98. (b)

20 12p 8 p=1 13 13 Taking –ve sign, we get

cos x, 0 x < 1 2 2 = – 2, x 1 f (x) changes sign from positive to negative from the left side of x = 1 to the right side of x = 1. f (x) changes from an increasing function to a decreasing function. f (x) has a local maximum at x = 1. The distance of point (1, 1, p) from the plane r . 3iˆ 4ˆj 12kˆ 13 0

20 12p 8 28 7 p 13 13 12 3 99. (a) Given, f (x) < 0 and g (x) > 0 therefore g (x) is an increasing function and f (x) is a decreasing function x + 1 < x + 5 g (x + 1) < g (x + 5) f (g (x + 1)) > f(g (x + 5)) Again x < x + 1 g (x + 1) f (g (x)) > f (g (x + 1)) x < x + 2 f (x) > f (x + 2) g (f (x)) > g (f (x + 1)) x > x – 2 f (x) < f (x – 2) g (f (x)) < g (f (x – 2)) 100. (a) For A, B, C to speak in order of alphabets, 3 places out of 10 may be chosen first in 10 C3 ways. The remaining 7 persons can speak in 7! ways. Hence, the number of ways in which all the 10 persons can speak is 10C3 .

7! = 101. (b)

y 2 ex e x 1 ex e x Applying comp. and dividendo.

y 1 3 y

or (in cartesian form) 3x + 4y – 12z + 13 = 0, is d1

3 1 4 1 12 p 13 32

42

12

3 4 12p 13

2

3

3

20 12p 13

4 0 12 1 13 2

32 42 12 According to the given condition, d1 = d2 20 12p 13

8 13

2e x 2e

e2x

x

1 y 1 y 1 log = log 2 3 y 3 y Hence, the inverse of the function

1/2

x

(i) 169 The distance of the point (–3, 0, 1) from the plane 3x + 4y – 12z + 13 = 0 is d2

10! 10! . 6 3!

8 13

ex

e

1/2

x

x 1 x x 3 x e e 102. (b) By changing x as –x and y as–y, both the given equation remains unchanged so required area will be symmetric with respect to both the axis, which is shown in the figure. So, f (x)

2 is loge

1

Required area = 4 0

1 x2

(1 x) dx

EBD_7443 MT-244

Target VITEEE x 1 x2 =4 2

1 sin 1 x x 2

x2 2

1

0 x 4/3 [ 3 8z

x

0

3

1 1 = 4 0 . 1 = –2 2 2 2 Y

2

23/3

.z

x 2 dx

x 1 x3 x3 1 x3 3 2 Put 1 – x = t –3x2dx = 2t dt 2 tdt 3 (1 t 2 ) t

X

2 dt 3 1 t2

1 1 x3 1 log 3 1 x3 1 1

ex

f (x)

1 1 t log 3 1 t

C

ex f (x) dt = ex + kex, where

0

cos ax cos bx 0 cos cx 1

103. (c) We have, lim x

8z ]4

1 4

dx

integral =

106. (a)

3

4x

2z

2 3 8z

105. (b)

O

x 4/3 3 2z x

lim

1

f (t) dt

k 0

(a b) (a b)x 2 sin x sin 2 2 = lim x 0 cx 2 sin 2 2

1

k

(e t

ke t ) dt

e ke 1 k ,

0

(a b)x (a b)x sin .sin 2 2 2 . x = lim 2 x 0 cx x sin 2 2 2

cx 4 (a b)x (a b)x sin sin 2 2 c 2 2 . . = lim x 0 (a cx (a b)x 2 b)x 2 . sin 2 . 2 a b 2 2 a b

a b a b 4 a 2 b2 . Hence m 2 2 2 c c2 and n are a2 – b2 and c2 respectively.

k

e 1 2 e

e 1 ex 2 e 2 e 107. (c) The sphere is the solid of revolution generated by the revolution of a semi circular area about its diameter. y

Thus f (x)

ex 1

B(–a,0)

O

=

104. (b)

x

lim

x

0 3

lim

x

3 2

z

(z x) 2

8xz 4x 2

0 (3

x

3

3

8xz

3

x

4

2xz x 2

x 3 8z 4x

A(a,0)

8z 3 x ) 4

Let the equation of the circle be x2 + y2 = a2 for the semi circle about the x-axis the variable varies form x = –a to x = a. Volume of the sphere

MT-245

Solutions a

Angle between two lines is given by

a

2

y .dy

=

(a

a

2

2

x ) dx

a

(a 2

2

a1a 2

cos

a

a12

x 2 )dx

a2 – x2 is an even function)

( x3 3

2

2

a x

a

2

a3

0

a3 3

108. (c)

(by + k) dy = (ax + h) dx

a 2 x + hx + C 2 For this to represent a parabola, one of the two terms x2 or y2 is zero. Therefore, either a = 0, b 0 or a 0 , b= 0

109. (a) We have (ex +e–x) ex

e

x

ex

e

x

e

dy = (ex – e–x) dx

2

3

p

x

y 1/ 3

0

6

113. (a) a , b , c

are

z x and 1 1/ 6

p

vectors,

b

q

2b

c

2c

a

0

4 and variance = npq = 2

1 and n 2

8

8

C2

6

1 2

1 2

2

28 28

k

P(x

1 e

y 1

coplanar

2 a b , 2 b c , and 2 c a are also coplanar vectors.

116. (c) P(x

1 =0 5

1 1 , 1, 6 4

6

5 324

6

P(2 success)

1

Direction ratios of two lines are and

( 1)

1 1 1 1 1 9 36 16

1 2 3 4 5 6

6

x

(10 + p + 3) – (6 – 5 – p) = 0 111. (d) The given lines may be written as x 1/ 2

6!

114. (b) 115. (a) mean = np

dt y= + c = log | t | + c t Hence y = log | ex + e–x | + c which is the reqd. general solution. 110. (d) a, b, c are coplanar [a b c] = 0

1

=

Thus 2 a

e dx + c ex e x x –x x Put e + e = t so that (e –e–x) dx = dt

Integrating y =

1

1 4

1 4

Total number of ways = 66

by2 + ky =

1

1 ( 1) 3

c 22

Hence, required probability

ax h by k

dy dx

cos

b 22

90 112. (c) The number of ways of getting the different numbers 1, 2, ....., 6 in six dice = 6 !.

4 3 a cub. units. 3

dy dx

b12 c12 a 22

1 1 2 6

0

b1b 2 c1c 2

p=–6 z 1/ 4

1 1 , , 1 2 3

k)

e

k!

2) 1 P ( x

e

1!

0) P ( x 1)

1

3 e2

.

117. (c) Let the normal at ‘t1’ cuts the parabola again at the point ‘t2’. the equation of the normal at (at12, 2at1) is y + t1x = 2at1 + at13 Since it passes through the point ‘t2’ i.e (at22, 2at2) 2at2 + at1t22 = 2at1 + at13 2a(t1 – t2) + at1(t12 – t22) = 0 2 + t1(t1 + t2) = 0 ( t1 t 2 0) 2 + t12 + t1t2 = 0 t1t2 = –(t12 + 2)

t2

t1

2 t1

EBD_7443 MT-246

Target VITEEE

118. (a) Reflexive: For any x

R, we have

Y

x x 2 2 an irrational number.. x R x for all x. So, R is reflexive. Symmetric: R is not symmetric, because 2 R 1 but 1R 2 , Transitive: R is not transitive also because

(0, 3) (0, 2) X'

2 R 1 and 1 R 2 2 but 2 R 2 2 . 119. (a) For a, b Z' a + b Z' Z' is closed under addition Also, a + b = b + a Addition on Z' is commutative

+

2y

X 2x + 3y = 6

= 6 Y' Shaded portion shows the feasible region. Now, the corner points are (0, 2), (2,0),

6 6 , , (0, 0). 5 5

At (0, 2), value of z = – 2(0) – 3(2) = – 6 At (2, 0), value of z = – 2(2) – 3(0) = – 4

x y x y 1 , x, y 0 1, 3 2 2 3 First convert these inequations into equations we get 3x + 2y = 6 ...(i) 2x + 3y = 6 ...(ii) on solving these two equation, we get point

Subject to

6 6 , . 5 5 Now, we draw the graph of these lines.

(3, 0)

(2, 0) 3x

a b c a b c , a, b, c Z '. Addition is associative on Z'. a 0 0 a a, a Z ' 0 Z'is an additive identity < Z', + > is a monoid. 120. (a) Given problem is max z = – 2x – 3y

of intersection is

(6/5, 6/5)

At

30 =–6 5 At (0, 0), value of z = – 2(0) – 3(0) = 0 The max value of z is 0.

=

121 122 123. 124. 125

(c) (a) (a) (d) (b)

6 6 6 6 , –3 , Value of z = – 2 5 5 5 5