IIT-JEE JEE MAIN AND ADVANCED 15 MOCK TEST COMBINED PHYSICS CHEMISTRY MATHEMATICS Prabhat Prakashan 7827007777, 9789353225759

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Physics_Mock Test-1
Physics_Mock Test-2
Physics_Mock Test-3
Physics_Mock Test-4
Physics_Mock Test-5
Chemistry_Mock Test-1
Chemistry_Mock Test-2
Chemistry_Mock Test-3
Chemistry_Mock Test-4
Chemistry_Mock Test-5
Mathematics_Mock Test-1
Mathematics_Mock Test-2
Mathematics_Mock Test-3
Mathematics_Mock Test-4
Mathematics_Mock Test-5

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15 Mock test

cOMBINED pHYSICS, Chemistry AND Mathematics Team Prabhat

Prabhat Prakashan

 Information contained in the work has been obtained by Prabhat Paperbacks, from sources believed to be reliable. However,

neither Prabhat Paperbacks nor its authors guarantee the accuracy or completeness of any information published herein, and neither Prabhat Paperbacks nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that Prabhat Paperbacks and its authors are supplying information but are not attempting to render engineering or other professional services.

 No part of this publication may be reproduced, stored in a

retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, without prior written permission of the publisher.

 All

disputes are subject to Delhi jurisdiction only.

Publisher

Prabhat Prakashan

4/19 Asaf Ali Road, New Delhi–110 002 Ph. 23289555 • 23289666 • 23289777 • Helpline/ 7827007777 e-mail: [email protected] • Website: www.prabhatexam.com © Reserved Edition 2021 Price Two Hundred Eighty Five Rupees ISBN ISBN 978-93-5322-575-9 Printed at Nakshatra Art, Delhi IIT-JEE JEE MAIN AND aDVANCED 15 Mock test cOMBINED pHYSICS, Chemistry AND Mathematics by Team Prabhat Published by Prabhat Prakashan 4/19 Asaf Ali Road, New Delhi–110 002 ISBN ISBN 978-93-5322-575-9 ` 285.00

CONTENTS l

Mock Test – 1 Physics....................................................................................................................................1-18

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Mock Test – 2 Physics..................................................................................................................................19-32

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Mock Test – 3 Physics..................................................................................................................................33-44

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Mock Test – 4 Physics..................................................................................................................................45-58

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Mock Test – 5 Physics..................................................................................................................................59-72

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Mock Test – 1 Chemistry.............................................................................................................................73-91

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Mock Test – 2 Chemistry...........................................................................................................................92-102

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Mock Test – 3 Chemistry.........................................................................................................................103-113

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Mock Test – 4 Chemistry.........................................................................................................................114-124

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Mock Test – 5 Chemistry.........................................................................................................................125-134

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Mock Test – 1 Mathematics.....................................................................................................................135-154

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Mock Test – 2 Mathematics.....................................................................................................................155-168

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Mock Test – 3 Mathematics.....................................................................................................................169-182

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Mock Test – 4 Mathematics.....................................................................................................................183-196

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Mock Test – 5 Mathematics.....................................................................................................................197-212

1

Mock Test-1

Test Booklet Code

A

Mock Test “JEE-Main”

Do not open this Test Booklet until you are asked to do so. Read carefully the Instructions on the Back Cover of this Test Booklet. Important Instructions: 1.

Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited.

2.

The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully.

3.

The test is of 3 hours duration.

4.

The Test Booklet consists of 75 questions. The maximum marks are 300.

5.

There are three parts in the question paper A, B, C consisting of, Physics, Chemistry and Mathematics having 25 questions in each part of equal weightage.

6.

Each question is allotted 4 (four) marks for each correct response. For MCQs - 4 Marks will be awarded for every correct answer and 1 Mark will be deducted for every incorrect answer.

7.

For answer with numeric value - 4 Marks will be awarded for every correct answer and 0 Mark will be deducted for every incorrect answer.

8.

Use Blue/Black Ball Point Pen only for writing particulars/ marking responses on Side-1 and Side-2 of the Answer Sheet. Use of pencil is strictly prohibited.

9.

No candidates is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc., except the Admit Card inside the examination hall/room.

10. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each page and at the end of the booklet. 11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room / Hall. However, the candidates are allowed to take away this Test Booklet with them. 12. The CODE for this Booklet is A. Make sure that the CODE printed on Side-2 of the Answer Sheet and also tally the serial number of the Test Booklet and Answer Sheet are the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the invigilator for replacement of both the Test Booklet and the Answer Sheet. 13. Do not fold or make any stray marks on the Answer Sheet. Name of the Candidate (in Capital letters): Roll Number :

in figures in words

Examination Centre Number: Name of Examination Centre (in Capital letters): Candidate’s Signature:

Invigilator’s Signature:

2

Physics

Read the following instructions carefully: 1.

The candidates should fill in the required particulars on the Test Booklet and Answer Sheet (Side-1) with Blue/Black Ball Point Pen.

2.

For writing/marking particulars on Side-2 of the Answer Sheet, use Blue/Black Ball Point Pen only.

3.

The candidates should not write their Roll Numbers anywhere else (except in the specified space) on the Test Booklet/Answer Sheet.

4.

Out of the four options given for each question, only one option is the correct answer.

5.

For each incorrect response, one-fourth (¼) of the total marks allotted to the question would be deducted for MCQs and No deduction for numeric questions from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Sheet.

6.

Handle the Test Booklet and Answer Sheet with care, as under no circumstances (except for discrepancy in Test Booklet Code and Answer Sheet Code), another set will be provided.

7.

The candidates are not allowed to do any rough work or writing work on the Answer Sheet. All calculations/writing work are to be done in the space provided for this purpose in the Test Booklet itself, marked ‘Space for Rough Work’. This space is given at the bottom of each page and at the end of the booklet.

8.

On completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.

9.

Each candidate must show on demand his/her Admit Card to the Invigilator.

10. No candidate, without special permission of the Superintendent or Invigilator, should leave his/her seat. 11. The candidates should not leave the Examination Hall without handing over their Answer Sheet to the Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the Attendance Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with as an unfair means case. The candidates are also required to put their left hand THUMB impression in the space provided in the Attendance Sheet. 12. Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited 13. The candidates are governed by all Rules and Regulations of the JAB/Board with regard to their conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the JAB/Board. 14. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances. 15. Candidates are not allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, electronic device or any other material except the Admit Card inside the examination hall/room.

Mock Test-1

3

JEE-MAIN: PHYSICS MOCK TEST-1 following shows the relative nature of the liquid columns in the two tubes?

SECTION 1 (Multiple Choice Question)

1.

If a particle moves from point P (2,3,5) to point Q (3,4,5). Its displacement vector be: a. iˆ + ˆj + 10 kˆ b. iˆ + ˆj + 5kˆ c. ˆi + ˆj

2.

A

A

µ α

b.

3.

4.

5.

µα

B

d.

Two rods of different materials having coefficients of Young’s linear expansion α1 , α 2 and

µ α

in the two rods are equal provided Y1 : Y2 is equal to:

a. 2 : 3 c. 3 : 2

d. infinitesimal

A wind-powered generator converts wind energy into electric energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to a. v b. v2 3 c. v d. v4 A spherical hollow is made in a lead sphere of radius R, such that its surface touches the outside surface of lead sphere and passes through the center. What is the shift in the center of mass of lead sphere due to the hollowing? R R a. b. 7 14 R c. d. R 2 The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is t0 in air. Neglect frictional force of water and given that the density of the bob is (4 / 3) × 1000 kg/m3. What relationship is true between t and t0 ?

6.

A

massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If α1 : α 2 = 2 : 3, the thermal stresses developed

1 c.

B

moduli Y1 , Y2 respectively are fixed between two rigid

neglected, then the time after which the bead starts slipping is:

a.

B

c. 7.

A

b.

d. 2iˆ + 4 ˆj + 6kˆ

Long horizontal rod has a bead which can slide along its length and is initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with a constant angular acceleration α. If the coefficient of friction between the rod and bead is µ, and gravity is

B

a.

a. t = t0

b. t = t0 / 2

c. t = 2t0

d. t = 4t0

A capillary tube A is dipped in water. Another identical tube B is dipped in a soap-water solution. Which of the

8.

b. 1 : 1 d. 4 : 9

Three liquids of equal volumes are thoroughly mixed. If their specific heats are s1 , s2 , s3 and their temperature

θ1 , θ 2 ,θ3 and their densities d1 , d 2 , d 3 respectively then the final temperature of the mixture is:

9.

a.

s1 θ1 + s2 θ 2 + s3 θ3 d1 s1 + d 2 s2 + d3 s3

b.

d1 s1 θ1 + d2 s2 θ2 + d3 s3 θ3 d1 s1 + d2 s2 + d3 s3

c.

d1 s1 θ1 + d 2 s2 θ 2 + d3 s3 θ3 d1 θ1 + d 2 θ 2 + d3 θ3

d.

d1 θ1 + d 2 θ 2 + d3 θ3 s1 θ1 + s2 θ 2 + s3 θ3

Two monatomic ideal gases 1 and 2 of molecular masses

m1 and m2 respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to the gas 2 is given by:

a.

m1 m2

c. m1 m2

b.

m2 m1

d. m 2 m1

10. A gas is compressed adiabatically till its temperature is doubled. The ratio of its find volume to initial volume will be: 1 1 b. more than a. 2 2 1 c. less than d. between 1and 2 2

4

Physics

11. A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1 , between 999 nm and 1000 nm is U 2 and between 1499 nm and 1500 nm is U 3 . The Wien constant, b = 2.88 × 106 nm-K. Then,

a. U1 = 0

b. U 3 = 0

c. U1 > U 2

d. U 2 > U1

12. In moving from A to B along an electric field line, the electric field does 6.4 × 10−19J of work on an electron. If

φ1 , φ2 are equipotential surfaces, then the potential difference (VC − VA ) is b. 4V d. 64 V

a. – 4V c. Zero

13. Force acting on a charged particle kept between the plates of a charged condenser is F. If one of the plates of the condenser is removed, the force acting on the same particle becomes: F a. 0 b. 2 c. F d. 2F 14. An electric cable of copper has just one wire of radius 9 mm. Its resistance is 50 Ω. This single copper wire of cable is replaced by 6 different well insulated copper wires, each of radius 3 mm. The total resistance of the cable will now be equal to: a. 7.5 Ω b. 45 Ω c. 90 Ω d. 270 Ω 15. An infinitely long conductor PQR is bent to form a right angle as shown. A current I flows through PQR. The magnetic field due to this current at the point M is H1. Now another infinitely long straight conductor QS is connected at Q so that the current is I/2 in QR as well as in QS, The current in PQ remaining unchanged. The magnetic field at M is now H 2. .The ratio H1 / H 2 is given by M

I –∞

P

90o Q

90o R –∞

S

+∞

1 2 2 c. 3

b. 1

a.

d. 2

16. Double slit interference experiment is carried out with monochromatic light and interference fringes are observed. If now monochromatic light is replaced by white light, what change is expected in interference pattern? a. no change b. pattern disappears c. white and dark fringes are observed throughout the pattern d. a few coloured fringes are observed on either side of central white fringe 17. The electric field of an electromagnetic wave in free space is given by E = 10 cos (107 t + kx ) ˆj V/m, Where, t and x are in seconds and metres respectively. It can be inferred that (a) the wavelength λ is 188.4 m (b) the wave number k is 0.33 rad/m (c) the wave amplitude is 10 V/m (d) the wave is propagating along + x direction Which one of the following pairs of statements is correct? a. (c) and (d) b. (a) and (b) c. (b) and (c) d. (a) and (c)

18. Light of wavelength λ strikes a photosensitive 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 λ ′ where: a. λ ′ = c.

λ 2

λ 2

< λ′< λ

b. λ ′ = 2λ d. λ ′ = λ

19. The ionisation potential of hydrogen atom is 13.6 V. The energy required to remove an electron in the n = 2 state of the hydrogen atom is a. 27.2 eV b. 13.6 eV c. 6.8 eV d. 3.4 eV

20. Hydrogen bomb is based on which of the following phenomenon? a. Nuclear fission b. Nuclear fusion c. Radioactive decay d. None of these

Mock Test-1

5

SECTION 2 (Numeric Value Question) 21. A cube has a side of length 1.2 × 10 −2 m. Calculate its volume: a. 1.7 × 10 −6 m 3

b. 1.73 × 10 −6 m 3

c. 1.70 × 10 −6 m 3

d. 1.732 × 10 −6 m 3

22. A body of mass 2 kg is released at the top of a smooth inclined plane having inclination 30°. It takes 3 seconds to reach the bottom. If the angle of inclination is doubled keeping same height (i.e., made 60° ), what will be the time taken?

a. 3 sec

b.

3 sec

c. 3 3 sec

d. 1.5 sec

23. There are two bodies of masses 100 kg and 10000 kg separated by a distance 1m. At what distance from the Space for rough work

smaller body, the intensity of gravitational field will be zero:

1 1 m b. m 9 10 1 10 m d. m c. 11 11 24. Due to a small magnet intensity at a distance x in the end on position is 9 gauss. What will be the intensity at a a.

x on broad side on position? 2 a. 9 gauss b. 4 gauss c. 36 gauss d. 4.5 gauss 25. When both source and observe approach each other with a velocity equal to the half the velocity of sound the change in frequency of sound as detected by the listener is: a. 0 b. 25 % c. 50% d. 150% distance

6

Physics

JEE ADVANCE PAPER-I Time 3 Hours. Read the Instructions Carefully

Max. Marks 180 (60 for Physics)

Question Paper Format and Marking Scheme: 1. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections. 2. Section 1 contains 6 multiple choice questions with one or more than one correct option. Marking Scheme: +4 if only (all) the correct option(s) is (are) chosen, +3 if all the four options are correct but only three options are chosen, +2 if three or more options are correct but only two options are chosen, both of the options must be correct, +1 if two or more options are correct buy only one option is chosen and it must be correct, -2 (in all other cases). 3. Section 2 contains 8 Numerical value answer type questions. The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive). Marking Scheme: +3 if only correct numerical value is given. 4. Section 3 contains 2 Paragraph based questions (2 paragraphs, each having 2 MCQs with one correct answer only) Marking Scheme: +3 if only the correct option is selected, -1 in all other cases. Note: Possible response Not attempted Partial Correct Correct incorrect

Section 1 0 +1 for each correct option selected +4 -2

Section 2 0 0 +3 0

acting on the conductor, then the correct statement(s) is(are)

SECTION 1 (Maximum Marks: 24)

MCQs with one or more than one correct answer 1.

Section 3 0 0 3 -1

y

A transparent thin film of uniform thickness and refractive

R

R

π/6

index n1 = 1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of

L

refractive index n2 = 1.5, as shown in the figure. Rays of


B is along
B is along
B is along
B is along

a. If

light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f1 from

b. If

the film, while rays of light traversing from glass to air get

c. If

focused at distance f 2 from the film. Then

d. If

R

π/4 R

L

x

zˆ, F ∝ ( L + R)

xˆ, F = 0 yˆ , F ∝ ( L + R)

zˆ, F = 0

n1

3. Air

2.

n2

a. | f1 | = 3R

b. | f1 | = 2.8 R

c. | f 2 | = 2 R

d. | f 2 | = 1.4 R

A conductor (shown in the figure) carrying constant current I is kept in the x-y plane in a uniform magnetic
field B. If F is the magnitude of the total magnetic force

4.

Planck’s constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option(s) is(are) a. M ∝ c

b. M ∝ G

c. L ∝ h

d. L ∝ G

Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies ω1 and ω2 and have total energies E1 and E2 , respectively. The variations of their momenta p with positions x are

Mock Test-1

7

shown in the figures. If

SECTION 2 (Maximum Marks: 24)

a a = n 2 and = n, then the correct b R

Numerical value answer type questions

equation(s) is(are) P

P

Energy = E1

Energy = E2

7.

b x

x

a

5.

R

a. E1ω1 = E2ω2

ω 2 b. 2 = n ω1

c. ω1ω 2 = n 2

d.

E1

ω1

=

B

A

O

ω2 8.

Air

An infinitely long uniform line charge distribution of charge per unit length λ lies parallel to the y-axis in the

3 a (see figure). If the magnitude of the 2 flux of the electric field through the rectangular surface ABCD lying in the x-y plane with its center at the origin is y-z plane at z =

F

α

r

OA = 8 cm OB = 9 cm m = 1.50

Glass

E2

Inner and outer radii of a spool are r and R respectively. A thread is wound over its inner surface and spool is placed over a rough horizontal surface. Thread is pulled by a force F as shown in figure. In case of pure rolling, which of the following statements are false?

R

A small filament is at the centre of a hollow glass sphere of inner and outer radii 8 cm and 9 cm respectively. The refractive index of glass is 1.50. Calculate the position of the image of the filament when viewed from outside the sphere.

λL (ε 0 = permittivity of free space), then the value of n nε 0 is z

a. Thread unwinds, spool rotates anticlockwise and friction acts leftwards b. Thread winds, spool rotates clockwise and friction acts leftwards c. Thread winds, spool moves to the right and friction acts rightwards d. Thread winds, spool moves to the right and friction does not come into existence

L

D

a

C O

A

3 a 2 y

B x

6.

Two ideal batteries of emf V1 and V2 and three resistances R1 , R2 and R3 are connected as shown in the figure. The

current in resistance R2 would be zero if V1

9.

For an atom of an ion having single electron, the wavelength observed λ1 = 2 are units and λ3 = 3 units figure. The value of missing wavelength λ2 is n3 orbit

R1 R2

λ3

λ1

λ2 R3

a. V1 = V2 and R1 = R2 = R3 b. V1 = V2 and R1 = 2 R2 = R3 c. V1 = 2V2 and 2 R1 = 2 R2 = R3 d. 2V1 = V2 and 2 R1 = R2 = R3

V2

n2 orbit n1 orbit

10. Consider an elliptically shaped rail PQ in the vertical plane with OP = 3 m and OQ = 4 m. A block of mass

1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see the given figure). Assuming no frictional losses, the kinetic energy

8

Physics

of the block when it’s reaches Q in (n × 10) joules. The

SECTION 3 (Maximum Marks: 12)

2 Paragraph based questions (2 paragraphs, each having

−2

value of n is (take acceleration due to gravity 10 ms )

2 MCQs with one correct answer only)

Q

Paragraph for Question No. 15 to 16

Light guidance in an optical fiber can be understood by

4m

considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower

90°

3m

O

P

refractive index n2 . The light guidance in the structure takes

11. Two identical uniform discs roll without slipping on two different surfaces AB and CD (see figure) starting at A and C with linear speeds v1 and v2 , respectively, and always

place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with

remain in contact with the surfaces. If they reach B and D

the angle of incidence i less than a particular value im are

with the same linear speed and v1 = 3 m / s, then v2 in m/s

confined in the medium of refractive index n1 . The numerical

is ( g = 10 m / s 2 )

aperture (NA) of the structure is defined as sin im .

A

v1 = 3m / s

C

v2

30 m

n1 > n2 Air

27 m B

D

Cladding θ

n2

Core

n1

i

12. Two spherical stars A and B emit blackbody radiation. The

radius of A is 400 times that of B and A emits 104 times the power emitted from B. The ratio (λ A / λB ) of their wavelengths λA and λB at which the peaks occur in their respective radiation curves is 13. The half life of a freshly prepared radioactive sample is 2 h. If the sample emits radiation of intensity, which is 16 times the permissible safe level, then the minimum time taken after which it would be possible to work safely with source is 14. A Young’s double slit interference arrangement with slits S1 and S 2 is immersed in water (refractive index = 4/3) as

shown in the figure. The positions of maxima on the surface of water are given by x 2 = p 2 m 2 λ 2 − d 2 , where λ is the wavelength of light in air (refractive index = 1), 2d is the separation between the slits and m is an integer. The value of p is

d S2

n2 = 3 / 2, and S 2 with n1 = 8 / 5 and n2 = 7 / 5 and taking the refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is(are) a. NA of S1 immersed in water is the same as that of S 2

immersed in a liquid of refractive index

16 3 15

b. NA of S1 immersed in liquid of refractive index

6 15

is the same as that of S 2 immersed in water c. NA of S 1 placed in air is the same as that of S 2

immersed in liquid of refractive index

4 15

.

d. NA of S1 placed in air is the same as that of S 2 placed

in water

S1

d

15. For two structures namely S1 with n1 = 45 / 4 and

x

Air water

16. If two structures of same cross-sectional area, but different

numerical apertures NA1 and NA2 ( NA2 < NA1 ) are joined longitudinally, the numerical aperture of the combined structure is

Mock Test-1

9

NA1 NA2 NA1 + NA2

a.

Two points K and M are symmetrically located on the

b. NA1 + NA2

c. NA1

opposite faces parallel to the x-y plane (see figure). V1

d. NA2

and V2 are the potential differences between K and M in strips 1 and 2, respectively. Then, for a given current I

Paragraph for Question No. 17 to 18

flowing through them in a given magnetic field strength

In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The

B, the correct statement(s) is(are)

length, width and thickness of the strip are ℓ, w and d,


respectively. A uniform magnetic field B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the zdirection is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The

a. If w1 = w2 and d1 = 2d 2 , then V2 = 2V1 b. If w1 = w2 and d1 = 2d 2 , then V2 = V1 c. If w1 = 2w2 and d1 = d 2 , then V2 = 2V1 d. If w1 = 2w2 and d1 = d 2 , then V2 = V1 18. Consider two different metallic strips (1 and 2) of same dimensions (lengths ℓ, width w and thickness d) with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in

current is assumed to be uniformly distributed on the crosssection of the strip and carried by electrons.

magnetic field B2, both along positive y-directions.

l

between K and M in strips 1 and 2, respectively.

y

•K

I S

I

W d

x

R •M

P

z Q

17. Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are

w1 and w2 and thicknesses are d1 and d2 , respectively. Space for rough work

Then V1 and V2 are the potential differences developed Assuming that the current I is the same for both the strips, the correct option(s) is(are)

a. If B1 = B2 and n1 = 2n2 , then V2 = 2V1 b. If B1 = B2 and n1 = 2n2 , then V2 = V1 c. If B1 = 2B2 and n1 = n2 , then V2 = 0.5V1 d. If B1 = 2 B2 and n1 = n2 , then V2 = V1

10

Physics

JEE ADVANCE PAPER-II Time 3 Hours. Read the Instructions Carefully

Max. Marks 180 (60 for Physics)

Question Paper Format and Marking Scheme: 1. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections. 2. Section 1 contains 6 MCQs with one or more than one correct answer Marking Scheme: +4 if only (all) the correct option(s) is (are) chosen, +3 if all the four options are correct but only three options are chosen, +2 if three or more options are correct but only two options are chosen, both of the options must be correct, +1 if two or more options are correct buy only one option is chosen and it must be correct, -2 (in all other cases) 3. Section 2 contains 8 Numerical value answer type questions. Marking Scheme: +3 if only correct numerical value is given 4. Section 3 contains 4 Matching type questions with 4 options. Marking Scheme: +3 if only the correct option is selected, -1 in all other cases Note: Possible response Not attempted Partial Correct Correct Incorrect

Section 1 0 +1 for each correct option selected +3 -1

Section 2 0 0 +4 -2

Section 3 0 0 +3 0

SECTION 1 (Maximum Marks: 24)

in equilibrium with the sphere P in L1 and sphere Q in L2

MCQs with one or more than one correct answer

and the string being taut (see figure). If sphere P alone in

A fission reaction is given by

L2 has terminal velocity VP and Q alone in L1 has
terminal velocity VQ , then




1.

236 92

U  → 14054 Xe + 94 38Sr + x + y, where x and y are two

particles. Considering

236 92

U to at rest, the kinetic energies

L1

of the products are denoted by K Xe , KSr , K x (2MeV) and L2

K y (2MeV), respectively. Let the binding energies per nucleon of

236 92

U, 140 54 Xe and

94 38

Sr be 7.5 MeV, 8.5 MeV

and 8.5 MeV respectively. Considering conservation laws, the correct option(s) is(are) a. x = n, y = n, KSr = 129 MeV, K Xe = 86 MeV

a.

different

b. x = p, y = e − , K Sr = 129 MeV, K Xe = 86 MeV



c. VP ⋅VQ > 0 3.

c. x = p, y = n, KSr = 129 MeV, K Xe = 86 MeV d. x = n, y = n, KSr = 86 MeV, K Xe = 129 MeV 2.

b.


| VP | η 2
= | VQ | η1



d. VP ⋅VQ < 0

In terms of potential difference V, electric current I, permittivity ε 0 , permeability µ0 and speed of light c, the

c. I = ε 0 cV

ρ 2 , respectively. The spheres are connected by a mass and σ 2 and viscosities η1 and η2 , respectively. They float

Q

dimensionally correct equation(s) is(are) a. µ 0 I 2 = ε 0V 2 b. ε 0 I = µ0V

Two spheres P and Q of equal radii have densities ρ1 and less string and placed in liquids L1 and L2 of densities σ 1


| VP | η1
= | VQ | η 2

P

4.

d. µ0 cI = ε 0V

Consider a uniform spherical charge distribution of radius R1 centred at the origin O. In this distribution, a spherical

Mock Test-1

11

cavity of radius R2, centred at P with distance OP = a = R1 − R2 (see figure) is made. If the electric field


inside the cavity at position r is E (r ), then the correct

8.

statement(s) is(are) R2 P

a

R1

O

A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid mass less rod of length ℓ and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r = 3ℓ from M, the tension in the rod is

 M  zero for m = k   . The value of k is  288 


a. E is uniform, its magnitude is independent of R2 but its




M

direction depends on r

m


b. E is uniform, its magnitude depends on R2 and its

r


direction depends on r


c. E is uniform, its magnitude is independent of a but its
direction depends on a


d. E is uniform and both its magnitude and direction
depend on a 5.

In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement(s) is(are)

9.

m ℓ

The energy of a system as a function of time t is given as E (t ) = A 2 exp( − α t ), where α = 0.2 s −1 . The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E (t ) at t = 5 s is

10. The densities of two solid spheres A and B of the same r  radii R vary with radial distance r as ρ A ( r ) = k   and R 5

r  , respectively, where k is a constant. The R

Strain

ρ B (r ) = k  P

moments of inertia of the individual spheres about axes passing through their centres are I A and I B , respectively.

Q Stress

a. P has more tensile strength than Q b. P is more ductile than Q c. P is more brittle than Q d. The Young’s modulus of P is more than that of Q 6.

A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P(r) is the pressure at r(r < R), then the correct option(s) is(are)

a. P(r = 0) = 0 c.

P(r = 3R / 5) 16 = P (r = 2 R / 5) 21

b.

P(r = 3R / 4) 63 = P(r = 2 R / 3) 80

d.

P (r = R / 2) 20 = P(r = R / 3) 27

SECTION 2 (Maximum Marks: 24)

Numerical value answer type questions 7.

An electron in an excited state of Li2+ ion has angular momentum 3h / 2π . The de Broglie wavelength of the electron in this state is pπ a0 (where a0 is the Bohr radius). The value of p is

If

IB n = , the value of n is I A 10

11. Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0,

π 2π and π. When , 3 3

they are superposed, the intensity of the resulting wave is

nI 0 . The value of n is 12. For a radioactive material, its activity A and rate of change of its activity R are defined as A = −

dN dA and R = − , dt dt

where N(t) is the number of nuclei at time t.

Two

radioactive sources P (mean life τ ) and Q (mean life 2τ ) have the same activity at t = 0. Their rates of change of activities at t = 2τ are R P and RQ , respectively.

RP n = , then the value of n is RQ e

If

12

Physics

13. A monochromatic beam of light is incident at 60° on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle

θ ( n ) with the normal (see the figure). For n = 3 the value of θ is 60° and

dθ = m. The value of m is dn

θ

60°

14. The figure shows a portion of an electric circuit. Resistors are known and are indicated on the diagram and the voltmeters are identical. If the voltmeters V1 and V2 read 7.5 V and 5.0 V respectively, find the reading of the voltmeter V3. V1

R1 = 9

(D) Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is

4. 5 : 27

5. 0 a. A→ 1; B→ 2, 4; C→3; D→ 4 b. A→3; B→1, 2; C→1; D→ 2 c. A→4; B→3, 4; C→2; D→ 1 d. A→2; B→2, 3; C→4; D→ 3 16. You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II. When a current I (steady state for DC or r.m.s. for AC) flows through the circuit, the corresponding voltage V1 and V2. (indicated in circuits) are related as shown in Column I. Column I Column II (A) I ≠ 0, V1 is

1.

proportional to I

V2

V1

V2

6 mH

3 µF

R2 = 14

V3

V

R3 = 24

(B) I ≠ 0, V2 > V1

2.

SECTION 3 (Maximum Marks: 12)

V1

V2

6 mH

2Ω

Matching type questions with 4 options V

15. Match the statement of Column II: Column I (A) In any Bohr orbit of the hydrogen atom, the ratio of kinetic energy to potential energy of the electron is (B) The ratio of the kinetic energy to the total energy of an electron in a Bohr orbit is (C) In the lowest energy level of hydrogen atom, the electron has the angular momentum

Column

with

those

in

(C) V1 = 0, V2 = V

3.

Column II 1 1. − 2

V1

V2

6 mH

2Ω V

(D) I ≠ 0, V2 is

4.

proportional to I

V1 6 mH

5.

h 2π

3µF V

2. 2

3.

V2

V1

V2

1 kΩ

3µF V

a. A→3, 4; b. A→1, 2; c. A→2, 4; d. A→1, 3;

B→2, 3, 4; B→1, 2, 3; B→2, 3, 4; B→1, 3, 4;

C→1, 2; C→3, 4; C→2, 3; C→1, 3;

D→2, 3, 4 D→1, 3 ,4 D→1, 2, 3 D→2, 3, 4

Mock Test-1

13


17. The gravitational field intensity E of earth at any point is defined as the gravitational force per unit mass at that point. If varies from place to place. The variation is shown

in column II with form of position r v s the E in the form
of graphs. The variation of r is given in column I. Choose the correct form of graphs for the corresponding
variations of r .

Column I

Column II


(A) Position r of body measured from surface earth upward
(B) Position r of body

1.

measured from surface of earth along diameter to opposite point on surface of earth
(C) Position r measured

2.

from center of earth to any point

3.

O

R



rE

R



rE


E

O


E


r

O


E


(D) Position r measured from center of hollow sphere to any point

Space for rough work

4.

O


E

R



rE

a. A→1; B→2; C→3; D→4 b. A→4; B→3; C→2; D→1 c. A→4; B→1; C→3; D→2 d. A→3; B→2; C→1; D→4 18. A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the following, state of the lift’s motion is given in Column I and the distance where the water jet hits the floor of the lift is given in Column II. Match the statements from Column I with those in Column II and select the correct using the code given below the lists. Column I Column II (A) Lift is accelerating 1. d = 1.2 m vertically up. (B) Lift is accelerating 2. d > 1.2 m vertically down with an acceleration less than the gravitational acceleration. (C) Lift is moving 3. d < 1.2 m vertically up with constant speed. (D) Lift is falling freely. 4. No water leaks out of the jar a. A→1; B→1; C→1; D→4 b. A→4; B→3; C→2; D→1 c. A→4; B→1; C→3; D→2 d. A→3; B→2; C→1; D→4

14

Physics

ANSWER & SOLUTIONS

M 7M = = mass of remaining sphere 8 8 Choosing the centre of big sphere as the origin, m x + m2 x2 X CM = 1 1 m1 + m2 m2 = M −

JEE-Main 1.

2.

3.

4.

5.

6.

7.

8.

9.

c

a

c

b

c

b

c

b

b

b

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

d

b

b

a

c

d

d

c

d

b

21.

22.

23.

24.

25.

a

b

c

c

c

1.

10.

0=

5.

F cos 60o

y
F

Now t = 2π

60o x

(a) Tangential force (Ft) of the bead will be given by the normal reaction L (N), while centripetal force (Fc) is A Fc provided by friction (Fr). The bead starts sliding when the centripetal force is just equal to the limiting Ft is inwards friction. Therefore, Ft = ma = mα L = N

Ft

∴

Limiting value of friction ( f r ) max = µ N = µ mα L . . . (i)

∴

Angular velocity at time t is ω = α t Centripetal force at time t will be

Fc = mL ω 2 = mL α 2 t 2 Equation (i) and (ii), we get t =

−R 14

(c) In air g eff = g

In water, g eff = g –

F sin 60o

3.

M

Solving, we get x2 =


(c) Displacement vector r = ∆xiˆ + ∆yjˆ + ∆zkˆ

= (3 − 2)iˆ + (4 − 3) ˆj + (5 − 5)kˆ = iˆ + ˆj

2.

( M / 8 ) × ( R / 2 ) + ( 7 M / 8 ) × x2

(b) Soap solution has lower surface tension as compared to pure water so h is less for soap solution.

7.

(c) F1 = F2

⇒

Y1 A α1∆θ = Y2 A α 2 ∆θ ⇒

8.

(b) Let 0°C be the reference temperature for zero heat, then initial heat energy = final heat energy

Y1 r2 3 = = Y2 r1 2

m1s1θ1 + m2 s2θ 2 + m3 s3θ3 = ( m1s1 + m2 s2 + m3 s3 )

. . . (ii)

θθ = =

µ α

m1s1θ1 + m2 s2θ 2 + m3 s3θ3 m1s1 + m2 s2 + m3 s3

Vd1s1θ1 + Vd 2 s2θ 2 + Vd 3 s3θ3 d s θ + d 2 s2θ 2 + d 3 s3θ 3 = 11 1 Vd1s1 + Vd 2 s2 + Vd 3 s3 d1s1 + d 2 s2 + d 3 s3

9.

(b) Speed of sound in a gas is given by v =

∴

v1 = v2

γ RT M

,v ∝

1 M

M2 m2 = M1 m1

Here,γγ =

 d ( ρ × volume )   dm  F = v   = v dt  dt   

 d ( volume )  2 = ρv   = ρ v ( Av ) = ρ Av dt  

10. (b) T1 V1

∴

Power P = ρ Av3 or P ∝ v 3

or

4.

(b) Let ρ be the density of lead. Then M =

Cp CV

γ −1

V1  T1  =  V2  T2 

=

γ −1

= T2 V2

γ −1

5   γ monoatomic =  3 

5 for both the gases 3

V  ⇒ 2   V1  1−γ

 T  = 1   2T2 

=

γ −1

=

1 1−γ

T1 T2

>

1 2

( 2) 4 π R3ρ = 11. (d) Wien’s displacement law is λmT = b (b = Wien’s 3

= mass of total sphere

constant)

3

4 R M  m1 = π   ρ = mass of removed part =   3 2  8 

t , Hence 2t0 = t g eff

6.

⇒

µ For t > , F > ( f r )max i.e. , the bead starts sliding. α c In the figure Ft is perpendicular to the paper inwards. 

(c) Power = F i v = Fv

Buoyant force d g =g− w g= db 4 Mass

∴

λm =

b 2.88 × 106 nm-K = T 2880K

Mock Test-1

∴

15

λ = 1000 nm Energy distribution with wavelength will be as follows:

y-direction. As E is varying with x and t, hence propagation of electromagnetic wave takes place along –x axis. Thus statement (d) is wrong. Comparing the relation, E = 10 cos(107 t + kx) with

standard

equation

of

E = E0 cos

1499 1500

999 1000

499 599

electromagnetic wave

From the fig. it is clear that

U1 (In fact U2 is maximum) U2

−19

W 6.4 × 10 = = 44V V e 1.6 × 10−19 13. (b) In the electric field between plates of parallel plate

⇒

2πυ

∴

F = qE =

qσ

. . . (i)

ε0

or

⇒

F′ 1 = F 2

⇒ F′ =

1 1 or R ∝ 2 ; resistance of 3 mm A r Cable 9 × 5 = 45 Ω In second case 6 wires are connected in parallel, so total 45 resistance of cables = = 7 ⋅ 5Ω 6 15. (c) Magnetic field at any point lying on the current carrying straight conductor is zero. Here H1 = Magnetic field at M due to current in PQ. H2 = Magnetic field at M due to QR + magnetic field at M due to QS + magnetic field at M due to PQ H 2 H 3 = 0 + 1 + H 1 = H1 ⇒ 1 = H2 3 2 2

16. (d) If monochromatic light is replaced by white light, a few coloured fringes are seen on either side of a central white fringe. 17. (d) Electric field of an electromagnetic wave in free space
is given by E = 10cos(107 t + kx) ˆj which is acting along

= 107

2π 1 rad / m m = = 0.033 rad 60π 30 Thus, statement (b) is wrong. k=

2π

λ

=

hc

λ

= −W

. . . (i)

hc = −W λ′

. . . (ii)

λ

⇒

hc 2hc = −W λ′ λ

⇒

λ′ =

14. (a) Resistance of 9 mm cable = 5 Ω As R ∝

λ

hc −W 2hc hc Dividing (ii) by (i) 2 = λ ′ = − 2W = −W hc λ λ′ −W

. . . (ii)

F 2

2π × (3 × 108 )

22 ≈ 188.4 m 7 Thus, statement (a) is correct.

2E =

σ E′ = 2ε 0 ∴

2π   2πυ (υ t + x) = E0 cos  t+ x λ λ  

λ = 60π = 60 ×

18. (c) E =

When one plate is removed, the electric field becomes

σ F ′ = qE′ = 2ε 0

= 107 or

λ

(VC − VA ) =

σ capacitor E = ε0

λ

We have, E0 = 10V / m. Thus statement (c) is correct.

12. (b) Work done by the field W = q (−dV ) = −e(VA − VB )

= e(VB − VA ) = e(VC − VA ) (∵ VB = VC)

2π

1

2

W − λ hc

⇒

1 2 W = − λ ′ λ hc

⇒ λ′ >

λ but less than λ 2

19. (d) Energy required to remove electron in the n = 2 13.6 state = + 2 = +3.4 eV eV (2) 20. (b) Hydrogen bomb is based on nuclear fusion. 21. (a) V = l 3 = (1.2 × 10−2 m)3 = 1.728 × 10 −6 m 3 ∵

Length (l) has two significant figures, the volume (V) will also have two significant figures. Therefore, the correct answer is V = lV = 1.7 × 10 −6 m 3

22. (b) a = g sin θ , l =

h sin θ

⇒

h 1 1 = g sin θt 2 ⇒ t ∝ sin θ 2 sin θ

⇒

t t22 sin θ11 3 1/ 2 1 = 3 s = = = / t2 = 1 = = t1 sin θ22 3 3 3/2 3

16

Physics

GM 1 GM 2 = 2 2 x (1 − x )

23. (c)

⇒ Planet 2

⇒

100 10, 000 = x 2 (1 − x )2

or

1 x2 x2 1 1 = ⇒ = ⇒ x= m 2 100 (1 − x ) 10 1 − x 11

24. (c) In C.G.S. Baxial = 9 =

Bequaterial =

M

x 2  

=

3

r2

1

r1

P

2M

⇒

8M

3  =  − 1 ×100% = 50% 2 

⇒

V2 =

JEE Advance Paper -I 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

6

5

c

a,b,c

a,c,d

b,d

a,c,d

a,b,d

9

6

11.

12.

13.

14.

15.

16.

17.

18.

7

2

8

5

a,c

d

a,d

a,c

1.

2.

1 1.4 − 1.5 1 − 1.4 = + −R −R f2

(a, b, c) F = 2 I ( L + R)[iˆ × B]

3.

(a, c, d) h ≡ [ ML T ], c ≡ [ LT ], G ≡ [ M L T ]

⇒

M∝

4.

(b, d) For first oscillator E1 =

2(L + R)

hc ,L ∝ G

−1

−1

−1 3

⇒ ⇒

1 mω12 a 2 2

8.

a 1 = b mω1

. . .(i)

For second oscillator

1 E2 = mω22 R2 , and mω2 = 1 2

=

E2

ω2

R1 (V1 + V2 ) ⇒ V1 R3 = V2 R1 R1 + R3

R3 (V1 + V2 ) ⇒ V2 R1 = V2 R3 R1 + R3

(9) For refraction at the first surface, u = −8 cm,

µ2 v

−

µ1 u

=

µ 2 − µ1 R1

1.5 1 0.5 ⇒ v ' = −8 cm + = v ' 8 −8 It means due to the first surface the image is formed at the centre. For the second surface u = −9 cm, µ1 = 1.5, R2 = −9 cm m2 m1 m2 − m1 − = v u R2 1 1.9 1 − 1.5 ⇒ v = −9 cm + = v 5 −9 Thus, the final image is formed at the centre of the sphere.

(6) From the figure θ = 60° Line change

and p = mv = mω1a = b ⇒

⇒

−2

hG c3

E1

ω1

R1 = −8 cm, µ1 = 1, µ 2 = 1.5

∴ f2 = 2R

For glass to air.

2

7.

⇒

1.5 1.4 − 1 1.5 − 1.4 (c) For air to glass = + ∴ f1 = 3 R f1 R R

⇒

(a, b, d) V1 =

v/2 vr 3 n =n+ n= n r v 2 % change in frequency

25. (c) n′ = n +

n′ − n  n′  ×100% =  − 1 ×100% n n 

E2

ω22 a 2

6.

From equation (i) and (ii) Bequaterial = 36 gauss.

=

=

(a, c, d) Since, the spool rolls over the horizontal surface, therefore, instantaneous axis of rotation passes through the point of contact of spool with the horizontal surface. About the instantaneous axis of rotation, moment produced by F is clockwise. Therefore, the spool rotates clockwise. In that case acceleration will be rightward and thread will wind. If rotational motion of spool is considered about its own axis then resultant moment on it must be clockwise. But moment produced by the force F is anticlockwise and its magnitude is equal to F.r, Hence, moment produced by the friction (about its own axis) must be clockwise and its magnitude must be greater than F.r. It is possible only when friction acts leftwards. Therefore, option (b) is correct.

. . . (ii)

x3

E1

ω12 a 2

5.

. . . (i)

x3

 a  ω2 = n2  =  b  ω1

θ

3 a 2

a

. . .(ii)

Rectangular surface

So, No. of rectangular surfaces used to form a m =

360° =6 60°

Mock Test-1

9.

(6) As is clear from figure. E1 = E2 + E3

hc

λ1 ∴

17

=

hc

1

λ2

+

1

for the combined structure and hence, correct option is d. 17. (a, d) I1 = I 2

hc

λ3 1

1 1 1 = − = − = = 6 units λ2 λ1 λ3 2 3 6

−mgh + Fd = ∆KE −1×10 × 4 + 18(5) = ∆KE

∆KE = 50 ∴ n = 5

∴

⇒

λ A TB = =2 λB TA

⇒

N 1 1 1 =  = =  N 0  2  16  2 

2.

3.

4.

5.

6.

7.

8.

9.

10.

a

a,d

a,c

d

a,b

b,c

2

7

4

6

11.

12.

13.

14.

15.

16.

17.

18.

3

2

2

3

a

a

d

a

4 d 2 + x 2 − d 2 + x 2 = mλ , m is an 3

15. (a, c) θ ≥ c

90° − r ≥ c

⇒

sin(90° − r ) ≥ c

⇒

cos r ≥ sin c

Using

n2

θ

nm

r

n1

2  ρ −σ2  2  ρ2 − σ1  Vp =  1  g and VQ =  g 9  η2  9  η1 


| VP | η1
= So, and VP .VQ < 0 | VQ | η1 3.

(a, c) BI ℓc ≡ VI

⇒

µ0 I 2 c ≡ VI

⇒

µ0 Ic = V

⇒ µ 02 I 2 c 2 = V 2

⇒

µ0 I 2 = ε 0V 2

⇒ ε 0 cV = I

4.


ρ 
C1C2 (d) E = 3ε 0

i

sin i ni n = and sin = 2 sin r nm n3

We get, sin 2 im =

(a, d) From the given conditions, ρ1 < σ 1 < σ 2 < ρ 2

From equilibrium, σ 1 + < σ 2 = ρ1 + ρ 2

x 2 = 9m 2λ 2 − d 2 ⇒ p = 3

⇒

2.

∴ n=4

integer So,

(a) Q value of reaction = (140 + 94) × 8.5 − 236 × 7.5 = 219 MeV

So, Total kinetic energy of Xe and Sr = 219 − 2 − 2 = 215 MeV So, by conservation of momentum, energy, mass and charge, only option a. is correct.

4

t = n T = 2 × 4 = 8h

14. (3) For maxima,

 B2 n1 =  B1n2

1.

1.

To work safely with the sample, its activity must be 1 reduced to . 16 n

V2 B2v2 w2  B2 w2   n1w1d1 = =  V1 B2v1w1  B1w1   n2 w2 d2

JEE Advance Paper -II

13. (8) Here, T = 2 h, t = ?

From

n1w1d1v1 = n2 w2 d 2 v2 Now,

3 3 m(3)2 + mg (30) = m(v2 ) 2 + mg (27) 4 4 v2 = 7 m / s

2TA = TB and

V1 v1w1 d 2 = = and hence correct choice is a & d. V2 v2 w2 d1

⇒

2

 dQ  4  dQ  2 4 4 2 4 12. (2)   = 10   (400 R) TA = 10 ( R TB )  dt  A  dt  B

So,

d1 w1v1 = d 2 w2 v2

1 2 1  mR  v 3 = mv2 mv +   2 2  2  R2 4 2

So,

⇒

18. (a, c) As I1 = I 2

11. (7) Kinetic energy of a pure rolling disc having velocity of

centre of mas v =

neA1v1 = neA2 v2

Now, potential difference developed across MK, V = Bvw

10. (5) Using work energy theorem Wmg + WF = ∆KE

⇒

⇒

n12 − n22 nm2

C1 ⇒ Centre of sphere and C2 ⇒ centre of cavity. 5.

(a, b) Y =

⇒

1 strain = Y stress

Putting values, we get, correct options as a & c 16. (d) For total internal reflection to take place in both structures, the numerical aperture should be the least one

stress strain

ρ

18

⇒

6. 7.

Physics

1 1 > YP YQ

r1 + r2 = 60°

⇒ YP < YQ

Snell’s Law on 2 surface: n sin r2 = sin θ Using equation (i) and (ii) n sin(60° − r1 ) = sin θ

 r2  (b, c) P (r ) = K 1 − 2   R  nh 3h (2) mvr = = 2π 2π

de-Broglie Wavelength λ = = 8.

h mv

GMm Gm2 − 2 = ma 9ℓ 2 ℓ

Gm2 GMm + = ma ℓ2 16ℓ 2 From both the equations, k = 7

⇒

d  3  dn  4

(

)

 dθ 4n 2 − 3 − 1  = cos θ dn  dθ =2 dn

V1V2 ( R2 − R1 )

. . .(i)

14. (3) V3 =

. . .(ii)

15. (a) A→ 1; B→ 2, 4; C→ 3; D→ 4

V1 ( R3 − R1 ) − V2 ( R3 − R2 )

= 3.0 V

16. (a) A→ 3, 4; B→ 2, 3, 4; C→ 1, 2; D→ 2, 3, 4

2 − at

(4) E (T ) = A e

⇒

2 − at

17. (d) Gravitational field intensity at any point outside due to

GM GM earth is given by E = g = = 2 ( R + h) 2 r
1 where h is height from surface or E ∝ 2 but – ve in sign. r Graph is parabolic. Inside earth g decreases are r increases form surface of earth.
1 x g  −E ∝ 2 E = g 1 +  = g − x r R R 

− at

dE = −α A e dt + 2 AdAe Putting the values for maximum error, dE 4 ⇒ ⇒ % error = 3. = E 100 2 10. (6) I = ∫ ρ 4π r 2 r 2 dr 3

⇒

I A ∝ ∫ ( r )(r 2 )( r 2 ) dr

⇒

I B ∝ ∫ ( r 5 )( r 2 )( r 2 ) dr

∴

 3  1 n  cos r1 − sin r1  = sin θ 2  2 

For θ = 60° and n = 3 ⇒

and for the other m:

9.

⇒

2π r 2π a0 (3) 2 = = 2π a0 3 3 z Li

(7) For m closer to M;

. . .(ii) nd

IB 6 = I A 10

or

11. (3) First and fourth wave interfere destructively. So from the interference of 2nd and 3rd wave only,  2π π  −  = 3I 0 ⇒ n = 3 ⇒ I net = I 0 + I 0 + 2 I 0 I 0 cos   3 3 −λ t 1 1 R ( A λ )e 12. (2) λ P = ; λQ = ⇒ P = 0 P −λ t τ 2τ RQ A0 λ Q e p

E ∝ x is straight line decreasing to 0 at centre of earth and then increases in magnitude but – ve as directed towards the centre always. 1 At any point outside E ∝ 2 (parabolic graph) r Inside earth E ∝ r (linear graph)

In hollow sphere inside E = 0 graph along r-axis; outside

Q

as if whole mass lies at centre E ∝

R 2 At t = 2τ ; P = RQ e 13. (2) Snell’s Law on 1st surface:

3 sin r1 = 2n ⇒

3 4n 2 − 3 cos r1 = 1 − 2 = 4n 2n

1 decreasing r2

parabolic graph after sudden rise at surface of earth.

3 = n sin r1 2

18. (a) A→1; B→1; C→1; D→4

In A, B, C no horizontal velocity is imparted to falling . . .(i)

water, so d remains same. In D, since its free fall, aeff = 0 ∴






Liquid won’t fall with respect to lift.

Mock Test-2

19

JEE-MAIN: PHYSICS MOCK TEST-2 SECTION 1 (Multiple Choice Question) 1.

In the relation p =

α − e β

kθ

−2

b. [ML T]

−1

d. [M 0 L2 T −1 ]

c. [ML T ]

A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in

v

a.

6.

Two satellites A and B go around a planet P in circular orbits having radii 4R and R respectively. If the speed of satellite A is 3v, the speed of satellite B will be: a. 12 v b. 6 v

2

a. [ MLT ]

2.

The total energy of a particle executing simple harmonic motion is: a. ∝ x b. ∝ x2 c. independent x d. ∝ x1/2

p is pressure, Z is distance, k

is Boltzmann constant and θ is the temperature. The dimensional formula of α will be: 0

5.

αZ

c. 7.

4 v 3

d.

3 v 2

The following diagram shows three soap bubbles A, B and C prepared by blowing the capillary tube fitted with stop cocks, S1, S2 and S3. With stop cock S closed and stop cocks S1, S2 and S3 opened.

v

b.

C S

S1

S2

S3 v

c. 3.

A particle, which is constrained to move along 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 + ax 3 . Here, k and a are positive constants. For

a. B will start collapsing with volumes of A and C increasing b. C will start collapsing with volumes of A and B increasing c. C and A both will start collapsing with the volume of B increasing d. Volumes of A, B and C will becomes equal at equilibrium

x ≥ 0, the functional form of the potential energy

U(x) of the particle is U(x)

a.

U(x)

x

b.

U(x)

c.

4.

x

8.

U(x)

x

d.

A carpenter has constructed a toy as shown in the adjoining figure. If the density of the material of the sphere is 12 times that of cone, the position of the center of mass of the toy is given by. a. at a distance of 2R from O b. at a distance of 3R from O c. at a distance of 4R from O d. at a distance of 5R from O

B

A

v

d.

x

steel is 2.0 × 1011 N/m 2 ).

2R

O2

A steel wire of length 20 cm and uniform cross-section 1 mm2 is tied rigidly at both the ends. The temperature of the wire is altered from 40°C to 20°C. What is the magnitude of force developed in the wire? (Coefficient of linear expansion for steel, α = 1.1 × 10–5 /°C and Y for

2R

9.

a. 2.2 × 106 N

b. 16 N

c. 8 N

d. 44 N

The intensity level of two waves of same frequency in a given medium are 20 dB and 60 dB. Then the ratio of their amplitudes is: a. 1 : 4 b. 1 : 16

c. 1 : 104

O1 O 2R

d. 1 : 100

10. The specific heat of a substance varies as (2t2 + t) ×10–3 cal/g °C. What is the amount of heat required to raise the temperature of 100 g of substance through 20°C to 40°C.

20

Physics

a. 37.9 kilocal c. 379 kilocal

b. 3.79 kilocal d. 82 cal

11. A real gas behaves like an ideal gas if its a. pressure and temperature are both high b. pressure and temperature are both low c. pressure is high and temperature is low d. pressure is low and temperature is high 12. The plates of a parallel plate capacitor of capacity of 50µF are charged by a battery to a potential of 100 volt. The battery remains connected the plates are separated from each other so that the distance between them is doubled. How much is the energy spent by battery in doing so? a. 25 × 10–2 J b. –12.5 × 10–2 J –2 c. – 25 × 10 J d. 12.5 × 10–2 J 13. The magnetic moment produced in a substance of 1gm is 6 × 10–7 ampere-metre2. If its density is 5 gm/cm3, then the intensity of magnetisation in A/m will be a. 8.3 × 106 b. 3.0 c. 1.2 × 10–7 d. 3 × 10–6

14. An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the +x direction and a magnetic field along the +z direction, then a. Positive ions deflect towards +y direction and negative ions towards –y direction b. All ions deflect towards +y direction c. All ions deflect towards –y direction d. Positive ions deflect towards –y direction and negative ions towards +y direction 15. A hundred turns of insulated copper wire are wrapped around an iron cylinder of area 1 × 10–3 m2 are connected to a resistor. The total resistance in the circuit is 10 ohms. If the longitudinal magnetic induction in the iron changes from 1 Wb m–2, in one direction to 1 Wb m −2 in the opposite direction, how much charge flows through the circuit? a. 2 × 10–2 C b. 2 × 10–3 C c. 2 × 10–4 C d. 2 × 10–5 C

16. A resistor R, an inductor L and a capacitor C are connected in series to an oscillator of frequency n. If the resonant frequency is nr, then the current lags behind voltage, when a. n = 0 b. n < nr c. n = nr

d. n > nr

17. The intensity ratio at a point of observation due to two coherent waves is 100 : 1. The ratio between their amplitudes is: a. 1 : 1 b. 1 : 10 c. 1 : 100 d. 10 : 1 18. An electromagnetic wave propagating along north has its electric field vector upwards. Its magnetic field vector points towards a. north b. east c. west d. downwards 19. The ionisation energy of 10 times ionised sodium atom is a. 13.6 eV b. 13.6 × 11 eV c.

13.6 eV 11

d. 13.6 × (11)2 eV

20. The example of nuclear fusion is a. Formation of barium and krypton from uranium b. Formation of helium from hydrogen c. Formation of plutonium 235 from uranium 235 d. Formation of water from hydrogen and oxygen SECTION 2 (Numeric Value Question) 21. A force of 5 N acts on a particle along a direction making an angle of 60° with vertical. Its vertical component be: a. 10 N b. 3 N c. 4 N d. 2.5 N 22. A neutral water molecule (H2O) in it's vapor state has an electric dipole moment of magnitude 6.4 ×10–30 C–m. How far apart are the molecules centres of positive and negative charge? a. 4 m b. 4 mm c. 4 µm d. 4 pm 23. A cell of e.m.f. 1.5 V having a finite internal resistance is connected to a load resistance of 2 Ω. . For maximum power transfer, the internal resistance of the cell in ohms should be: a. 4 b. 0.5 c. 2 d. None of these

24. Two plane mirrors are at 45° to each other. If an object is placed between them, then the number of images will be a. 5 b. 9 c. 7 d. 8 25. A photon of energy 8 eV is incident on a metal surface of Threshold frequency 1.6×1015Hz. The kinetic energy of the photoelectrons emitted (in eV) (Take h = 6 × 10 −34 J − s) a. 1.6 c. 2

b. 6 d. 1.2

Mock Test-2

21

JEE ADVANCE PAPER-I a. θ = 45 °

SECTION 1 (Maximum Marks: 24)

b. θ > 45 ° and a frictional force acts on the block towards P.

MCQs with one or more than one correct answer 1.

A piece of wire is bent in the shape of a parabola y = kx

2

(y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-axis is

a. 2.

a gk

b.

a 2 gk

c.

2a gk

d.

c. θ < 45 ° and a frictional force acts on the block towards d. a frictional force acts on the block towards 5.

In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle θ with the horizontal floor. The coefficient of friction between the wall and the ladder is µ1 and that between the floor and the ladder is µ2. The normal reaction of the wall on

a 4 gk

the ladder is N1 and that of the floor is N2. If the ladder is about to slip, then

A block of mass m is placed on a surface with a vertical

µ11

x3 . If the coefficient of friction 6 is 0.5, the maximum height above the ground at which the block can be placed without slipping is: cross-section given by y =

a.

1 m 3

θ

1 m 2

b.

µ2

1 2 d. m m 6 3 A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. Surface BC is frictionless. KA, KB and KC are kinetic energies of the ball at A, B and C, respectively. Then c.

3.

a. µ1 = 0, µ 2 ≠ 0 and N2 tan θ =

mg 2

b. µ1 ≠ 0, µ 2 = 0 and N1 tan θ =

mg 2

c. µ1 ≠ 0, µ 2 ≠ 0 and N1 =

C

A

mg 1 + µ1 µ 2

d. µ1 = 0, µ 2 ≠ 0 and N1 tan θ =

hc hA

mg 2

B

4.

a. hA > hC ; K B > K C

b. hA > hC ; K C > K A

c. hA = hC ; K B = K A

d. hA < hC ; K B > K C

A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with the horizontal. A horizontal force of 1 N acts on the block through its centre of mass as shown in the figure. The block remains stationary if (take g = 10 m/s2).

6.

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are

P

2v

how many elastic collisions, other than that at A, these two particles will again reach the point A? v

A

2v

1N

θ

and

the particles move with constant speeds. After making

Q

O

v

respectively, as shown in the figure. Between collisions,

a. 4

b. 3

c. 2

d. 1

22

Physics

SECTION 2 (Maximum Marks: 24)

Numerical value answer type questions 7.

The densities of two solid spheres A and B of the same radii R vary with radial distance r as ρ A (r ) = k  r  and R 



5

r  , R

ρ B (r ) = k 

respectively, where k is a constant.

The moments of inertia of the individual spheres about axes passing through their centres are IA and IB, respectively. If

8.

IB n = , the value of n is I A 10

Gravitational acceleration on the surface of a planet is

6 g, where g is the gravitational acceleration on the 11 surface of the earth. The average mass density of the 1 planet is times that of the earth. If the escape speed on 3 the surface of the earth is taken to be 11 kms–1, the escape speed on the surface of the planet in kms–1 will be 9.

A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet’s gravity is 1/ 4th of its value at the surface of the planet. If the escape velocity from the planet is vesc = v N , then the value of N is (ignore energy loss due to atmosphere)

10. The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1m at 10°C. Now the end P is maintained at

10°C, while the end S is heated and maintained at 400 °C. The system is thermally insulated from its

surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2 ×10−5 K −1 , the change in length of the wire PQ is:

12. A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice. [Take atmospheric pressure = 1.0 × 105 N / m 2 , density of water = 1000 kg/m3 and g = 10 m/s2. Neglect any effect of surface tension.]

13. Consider two solid spheres P and Q each of density 8 gm

cm−3 and diameters 1 cm and 0.5 cm, respectively. Sphere P is dropped into a liquid of density 0.8 gm cm−3 and viscosity η = 3 poiseulles. Sphere Q is dropped into a liquid of density 1.6 gm cm −3 and viscosity η = 2 poiseulles. The ratio of the terminal velocities of P and Q is:

14. A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross-sectional area is 4.9 × 10−7 m 2 . If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s −1. If the Young’s modulus of the material of the wire is

n × 109 Nm −1 , the value of n is SECTION 3 (Maximum Marks: 12)

Paragraph based questions (2 paragraphs, each having 2 MCQs with one correct answer only) Paragraph for Question No. 15 to 16 A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction upto the point Q, as shown in the figure, below, is 150 J. (Take the acceleration due to gravity, g = 10 m / s −2 ).

11. Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m2. The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is n 0.04 N/m. Find the ratio B , where nA and nB are the nA number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity.]

R

P

30°

Q

R

Mock Test-2

23

15. The speed of the block when it reaches the point Q is a. 5 ms −1 b. 10 ms − 1 c. 10 3 m s − 1

d. 20 ms −1

16. The magnitude of the normal reaction that acts on the block at the point Q is a. 7.5 N b. 8.6 N c. 11.5 N d. 22.5 N Space for rough work

Paragraph for Question No. 17 to 18 Two waves y1 = A cos(0.5 π x − 100π t ) y2 = A cos

(0.4 π x − 92π t ) are travelling in a pipe placed along x-axis. 17. Find the number of times intensity is maximum in time interval of 1 sec a. 4 b. 6 c. 8 d. 10 18. Find wave velocity of louder sound a. 100 m/s b. 192 m/s c. 200 m/s d. 96 m/s

24

Physics

JEE ADVANCE PAPER-II SECTION 1 (Maximum Marks: 24)

horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in rad/s) is

MCQs with one or more than one correct answer 1.

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8×10 7 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate (Take

L

m

g = 9.8 ms−2 ) :

2.

3.

a. 2.45×10 –3 kg

b. 6.45×10 –3 kg

c. 9.89×10 –3 kg

d. 12.89 ×10 –3 kg

6.

If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that a. linear momentum of the system does not change in time b. kinetic energy of the system does not change in time c. angular momentum of the system does not change in time d. potential energy of the system does not change in time A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2




z P




5.

a.

2R 15

b. R

c.

4R 15

d.




a. LO and LP do not vary with time







b. LO varies with time while LP remains constant.





A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m. The ball is rotated on a




d. LO and LP both vary with time. SECTION 2 (Maximum Marks: 24)

Numerical value answer type questions 7.

8.

2 15

R 4




c. LO remains constant while LP varies with time

−1

b. Momentum of 5 kg mass after collision is 4 kg ms −1 c. Kinetic energy of the centre of mass is 0.75 J d. Total kinetic energy of the system is 4 J A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be

m

O

for the system of these two masses?

a. Total momentum of the system is 3 kg ms




denoted by LO and LP respectively, then.

ms −1 . Which of the following statement(s) is (are) correct

4.

a. 9 b. 18 c. 27 d. 36 A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x-y plane with centre at O and constant angular speed ω . If the angular momentum of the system, calculated about O and P are

9.

A body sends waves 500 mm long through medium A and 0.25 m long in medium B. If velocity of waves in medium A is 16 m/s, what is the velocity (in m/s) of waves in medium B? A light pointer fixed to one prong of a tuning fork touches gently a smoked vertical plate. The fork is set vibrating and the plate is allowed to fall freely. Two complete oscillations are traced when the plate falls through 40 cm. What is the frequency (in Hz) of the tuning fork? A water tank is 20 m deep. If the water barometer reads 10 m at that place, then what is the pressure at the bottom of the tank in atmosphere?

Mock Test-2

25 −4

10. There is a soap bubble of radius 2.4 × 10 m in air cylinder

SECTION 3 (Maximum Marks: 12)

which is originally at the pressure of 105 N/m 2 . The air in

Matching type questions with 4 options

the cylinder is now compressed isothermally until the radius of the bubble is halved. The pressure of air in the

15. Work is defined as dot product of force and displacement

W = ∫ dW = ∫ F ⋅ dS . It is a scalar quantity. The total work

cylinder now becomes n ×105 N/m 2 . The surface tension of soap film is 0.08Nm –1 . Find the integer value of n.

11. A length of wire carries a steady current I. It is bent first to form a circular plane coil of one turn. The same length is now best more sharply to give double loop of smaller radius. If the same current I is passed, the ratio of the magnitude of magnetic field at the centre with its first value is.

12. A current 1 amp is flowing in the sides of an equilateral triangle of side 4⋅5 ×10−2 m. the magnetic field at the centroid of the triangle in the unit of (10 −5 T) is.

13. There are two infinite long parallel straight current carrying wires, A and B separated by a distance r (Fig.) B

A I

Q

P

r

done will depend on the displacement and the force, which may be constant or variable. Thus in different situations of variable force applied in column I, the final expression for work done can be expressed as in column II. Column I Column II (A) Force constant in magnitude acts 1 1. kx 2 cos θ at constant angle θ with direction 2 of motion (B) Force constant in magnitude acts F 2. sin θθ K at angle θ which varies as θ = kx

(C) Force varies with distance x as

3. Fx sin θ

F = k x but angle θ is constant

(D) Force is constant in magnitude 4. zero but changes in direction with changing angle and always acts along radius of circular path a. A→1, B→2, C→3, D→4 b. A→3, B→2, C→1, D→4 c. A→4, B→3, C→2, D→1 d. A→2, B→1, C→4, D→3 16. Four charges Q1 , Q2 , Q3 and Q4 of same magnitude are fixed

I

along the x-axis at x = −2a, − a, + a and +2a respectively.

r

A positive charge q is placed on the positive y-axis at a

The current in each wire is I. The ratio of magnitude of

distance b > 0 . Four options of the signs of these charges

magnetic field at points P and Q when points P and Q lie

are given in column I. the direction of the forces on the

in the plane of wires is

charge q is given in column II. Match column I with column II and select the correct answer using the code

14. A length of wire carries a steady current I. It is bent first to form a circular plane coil of one turn. The same length

given below the lists. +q

(0, b)

is now bent more sharply to give double loop of smaller radius. If the same current I is passed, the ratio of the magnitude of magnetic field at the centre with its first

Q1

value is

(−2a,0) (−a,0)

a

P O

2a

Q3

Q4

(+ a,0)

(+2a,0)

Q2

Column I

Column II

(A) Q1 , Q2 , Q3 , Q4 all positive

1. + x

(B) Q1 , Q2 positive; Q3 , Q4 negative

2. − x

(C) Q1 , Q4 positive; Q2 , Q3 negative

3. + y

(D) Q1 , Q3 positive; Q2 , Q4 negative

4. − y

26

Physics

a. A→3, B→1, b. A→4, B→2, c. A→3, B→1, d. A→4, B→2,

C→4, C→3, C→2, C→1,

D→2 D→1 D→4 D→3

17. The vibration of body can be under various types of forces. The vibration are classified mentioned in column I under the conditions mentioned in column II. Match the type of vibrations in column I with conditions in column II. Column I

Column II

(A) Free vibrations

1. A body vibrating in viscous medium

(B) Forced vibrations (C) Resonant vibrations

(D) Damped vibrations

a. A→1, B→2, C→3, D→4 b. A→4, B→3, C→2, D→1 Space for rough work

2. A body vibrating under its natural restoring force 3. A body vibrating under the influence of another vibrating body 4. A body vibrating with its natural frequency under the influence of another vibrating body of same frequency

c. A→2, B→3, C→4, D→1 d. A→3, B→4, C→1, D→2 18. To determine specific heat of different substances we use different types of calorimeters. Can you match the type of colorimeter used named in column I to the substance whose specific heat is determined by corresponding calorimeter in column II.

Column I

Column II

(A) Regnault’s calorimeter

1. To determine specific heat of solids at very low temperatures

(B) Joly’s differential calorimeter

2. To determine specific heat of solids or gases at constant pressures

(C) Calender and Barmer’s calorimeter

(D) Nernst’s vacuum calorimeter

3. Used to measure specific heat at constant volume

4. Specific heat of liquids and gases at constant pressure

a. A→1, B→2, C→3, D→4 b. A→2, B→3, C→4, D→1 c. A→3, B→4, C→1, D→2 d. A→4, B→3, C→2, D→1

Mock Test-2

27

ANSWER & SOLUTIONS

2k a From the given function we can see that F = 0 at x = 0 i.e., slope of U – x graph is zero at x = 0. U ( x ) = Negative for x >

JEE-Main 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

a

a

d

c

c

b

c

d

d

a

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

d

a

b

c

a

d

d

b

d

c

21.

22.

23.

24.

25.

d

d

c

c

c

1.

∴

4.

1 16 2 m1 = π ( 2 R ) ( 4 R ) ρ = π R 3 ρ 3 3

h 4R = =R 4 4 From O on the line of symmetry, i.e., y1 = R

And its centre of mass will be at a height

(a) [V] =

[GM] [M −1L3 T −2 M]  kθ  = = [L2 T −2 ] ⇒ [ α] =   [r ] [L] Z

Further

[ p] =  β 

Similarly the mass of the sphere 4 m2 = π R3 (12 ρ ) = 16π R 3 ρ = 3m1 3 And its centre of mass will be at its centre O 2 , i.e.,

α   

α 

 kθ 

 



(c) If the density of cone be ρ, then its mass will be

[β] =  p  =  Zp 

y2 = 4 R + R = 5R (from O).



y

Dimensions of k θ are that to energy. 2R

[Kθ] [ML2 T −2 ] Hence, [ α ] = = = [MLT −2 ] [ z] [L]

2.

C

(a) Since, the block rises to the same heights in all the four cases, from conservation of energy, speed of the block at highest point will be same in all four cases. Say it is v0 Equation of motion will be N + mg = or N =

m2 O2

O2

4R m1 O1

O1

2 0

mv R

O

O 2R

mv02 − mg R

Now treating the sphere and cone as point masses with their masses concentrated at their centres of mass respectively and taking the line of symmetry as y-axis with origin at O, the centre of mass of the toy is given by m y + m2 y2 m1 × R + 3m1 × 5 R YCM = 1 1 = = 4R m1 + m2 m1 + 3m1

R (the radius of curvature) in first case is minimum. Therefore, normal reaction N will be maximum in first case. v0

i.e., centre of mass of the toy is at a distance 4R from O on the line of symmetry, i.e., at the apex of the cone. N + mg

Note: In the question it should be mentioned that all the four tracks are frictionless. Otherwise, v0 will be different in different tracks. dU dx

3.

(d) F = −

∴

dU = − F . dx U ( x) =

∫

x

3 or U ( x) = − (−kx + ax ) dx 0

kx 2 ax 4 − 2 4

U ( x ) = 0 at x = 0 and x =

2k a

5.

(c) The total energy of a particle in simple harmonic motion is constant.

6.

(b) Orbital speed of a satellite v0 =

⇒

vB r 4R = A = =2 vA rB R

⇒

vB = 2v A = 2 × ( 3v ) = 6v

7.

(c) Excess pressure inside soap bubble is inversely proportional to the radius of bubble, i.e., ∆P ∝ 1/r′. This means that bubbles A and C posses greater pressure inside it than B. So the air will move from A and C, towards B.

GM r

28

8.

Physics

(d) F = Y A α∆θ ∆

= ( 2.0 × 10

11

−3

) × (1×10 ) × (1.1×10 ) ( 40 − 20) = 44 N −6

9.

I (d) L2 − L1 = 10log10 2 I1

⇒

60 − 20 = 10log10

15. (a) dQ =

−5

16. (d) The current will lag behind the voltage when reactance of inductance is more than the reactance of condenser. 1 1 or ω > Thus, ω L > ωC LC

I2 I1

or

I2 a2 = 104 ⇒ 22 = 104 I1 a1 ⇒

n>

1 2π LC

or n > nr

where, nr = resonant frequency.

a2 = 100 ⇒ a1 : a2 = 1 : 100 a1

17. (d) I ∝ a 2 ⇒

a1 = a2

I1 100 10 = = I2 1 1

18. (b) The direction of propagation of electromagnetic wave

is given by the direction of ( E × B ). Here, the em wave is

10. (a) dQ = mc dt t2

Q = ∫ mc dt

propagating along north. The electric field vector is acting upwards, so the magnetic field vector will point towards east.

t1

t2 = 40

=

dφ nAdB 100 × 1× 10 × 2 = = = 2 × 10−2C R R 10

40

3 t2  2 −3 −1  2t 100 × 2 + × 10 t t dt = 10 + ( )   ∫ 2  20  3 t1 = 20

1 3 3 2 2 = 10 −1  ( 40 ) − ( 20 )  + ( 40 ) − ( 20 )   2 3  

= 3.79 ×103 cal = 3.79 kilocal

19. (d) (Eion)Na = Z 2 ( Eion ) H = (11) 213.6 eV 20. (c) Fast neutrons can escape from the reaction. So as to proceed the chain reaction. Slow neutrons are best. 21. (d) The component of force in vertical direction

11. (d) A real gas behaves like an ideal gas at low pressure and high temperature.

F cos 60o

y

12. (a) When separation between the plates is doubled the

capacitance becomes one half i.e., C ′ = 25 µFF


F

60

o

F sin 60o

Energy spent by battery = qV

1 = 2.5 N 2 22. (d) There are 10 electrons and 10 protons in a neutral water molecule.

= ( C ′V )V = C ′V 2

= F cos θ = F cos 60° = 5 ×

= 25 ×10−6 × (100 ) = 25 ×10−2 J 2

13. (b) I =

M M = , V mass/density

So it's dipole moment is p = q ( 2l ) = 10e(2l ) Hence length of the dipole i.e,. distance between centres of positive and negative charges is

Given mass = 1gm = 10−3 kg And density = 5gm / cm3 = Hence I =

6 × 10

x

−7

× 5 × 10

10

−3

5 ×10−3 kg −2 3

3

(10 ) m

= 5 ×103 kg / m3

3

=3

14. (c) As the electric field is switched on, positive ion will start to move along positive x-direction and negative ion along negative x-direction. Current associated with motion of both types of ions is along positive x-direction. According to Fleming's left hand rule force on both types of ions will be along negative y-direction.

2l =

p 6.4 × 10 −20 = = 4 × 10 −12 m = 44 pm 10e 10 × 1.6 × 10 −19

23. (c) For maximum power Rext = Rint = 2 Ω 24. (c) n =

360° −1 = 7 45°

25. (c) W = hv0 = 6 ×10−34 ×1⋅ 6 ×1015 joule =

∴

6 × 10−34 × 1 ⋅ 6 × 1015

eV = = 6 eV 1 ⋅ 6 × 10−19 eV−–6 eV==22eV Ek = hv − W = 88eV 6 eV eV

Mock Test-2

29

If θ > 45° ; mg sin θ > F cos θ

JEE Advance Paper -I 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

b

c

a,b,c

a,c

c,d

c

6

3

2

7

11.

12.

13.

14.

15.

16.

17.

18.

6

6

3

4

b

a

a

c

1.

(b) tan θ =

a g

y N cos θ

dy = 2kx tan θ = dx ⇒

x=

∴

Friction acts towards Q. If θ < 45 ° F cos θ > mg sin θ

∴

Friction acts towards P.

5.

(c, d) Condition of translational equilibrium N1 = µ2 N 2

⇒

N 2 = µ1 N1 = Mg , Solving N2 =

⇒

N1 =

N

θ

N

ma

a 2 gk

sin θ

θ

(c) mg sin θ = µ s mg cos θ [when partied is just balanced]

⇒

tan θ = us ⇒ tan θ =

µ2 mg Applying torque equation about corner (left) 1 + µ1 µ2

point on the floor

x

ma

2.

mg 1 + µ1µ2

⇒

dy  x 2  = dx  2 

ℓ mg cos θ = N1ℓ sin θ + µ1 N1ℓ cos θ 2

Solving tan θ = 6.

1 − µ1µ 2 2µ 2

(c) Velocity will exchange after each collision A

V

2V

fs

N

Mg cos θθ

Mg sin θ

2π / 3

2π / 3

2V

V 2π / 3

st

1 collision

∴

x2 1 = 0.5 ⇒ x = 1 ⇒ y = 2 6

3.

(a, b, c) EA = mghA + K A

7.

2 (6) I = ∫ ρ 4π r 2 r 2 dr 3

⇒

EB = K B

⇒

I A ∝ ∫ ( r )( r 2 )( r 2 ) dr

⇒

EC = mghC + K C

⇒

I B ∝ ∫ ( r 5 )( r 2 )( r 2 ) dr

∴

IB 6 = I A 10

V

Using conservation of energy E A = EB = EC ⇒

K B > KC

⇒

KB > K A

⇒

K − KA Mg (hA − hC ) + ( K A − K C ) = 0 ⇒ hA − hC = C Mg

4.

(a, c) f = 0 (θ = 45°)

8.

(3)

So, for block to be at rest F cos θ = mg sin θ LHS F cos θ = 1 × cos 45° =

1

=

RHS mg sin θ = 0.1 × 10 ×

2

2V

R' 3 6 = R 22

⇒

V 'esc R '2 ρ ' 3 ∝ = ⇒ v 'esc = 3km/s. R2 ρ Vsec 11

9.

(2) At height R from the surface of planet acceleration due

2 1

2 collision

g' 6 ρ' 2 ; = = g 11 ρ 3

Hence,

1

nd

2

to planet’s gravity is

F cos θ

1 th in comparison to the value at the 4

surface F

F

θ

mg

{LHS = RHS)

GMm 1 2 GMm GMn 1 2 + mv = − and − + mvesc = 0 R 2 R+R R 2

So,

−

∴

vesc = v 2

30

Physics

10. (1)

2

Q, R

1m

P 10°C

140°C 2 k , α1

∵

1m

S

400°C

k ,α 2

200 − 150 =

⇒

v = 10 m/s

∆ℓ = ∫ d ℓ

16. (a) N − mg cos 60° =

 θ − 10  130 ⇒  = x  1

θ = 10 + 130x

⇒

1

∆ℓ = 130α1

0

∆ℓ = 130 × 1.2 × 10−5 × 11. (6) PA = P0 +

JEE Advance Paper -II

R

RB

8 N / m2

4T = 12N / m2 ⇒ RB

1.

2.

4.

5.

6.

7.

8.

9.

d

a

a,c

a

d

c

8

7

3

11.

12.

13.

14.

15.

16.

17.

18.

4

4

8

4

b

a

c

3.

a

1. 3

nB PB  RB  =   =6 n A PA  R A 

PV 0 0 = PV

2. 3.

500 mm 500 mm

⇒

105 [ A(500 − H )] = 98 × 103 [ A(500 − 200)]

⇒

H = 206 mm

(a, c) By conservation of linear momentum

. . .(i)

By Newton’s experimental law of collision u =v+2

. . .(ii) 5 kg

1 kg

u

Level fall = 206 – 200 = 6 mm

before collision

13. (3) 6πη rv + ρ LVg = ρ 0Vg

1 kg

after collision

Using (i) and (ii) we have v = 1m/s and u = 3 m/s

rQ .ηQ r 3 (8 − 0.8) = P × 3 ηP .rP (8 − 1.6) rQ

Kinetic energy of the centre of mass 1 = msystem v 2cm = 0.75J 2

2

 r   ηQ   7.2  7.2 2 × =3 =  P  ×  ×  = 4× 6.4 3  rQ   ηP   6.4 

15. (b) Using work energy theorem

2 3 2R MR 2 = Mr 2 ⇒ r = 5 2 15

4.

(a)

5.

2 (d) mωmax r = Tmax

⇒

ωmax =

14. (4) ω = YA

mL

5 kg

V

6πηQ rQ VP (ρ PVP − ρ LVP ) g = × VQ 6πηP rP ( ρQVQ − ρ LVQ )

1 2 mv 2

8

(d) Let m mass of fat is used. 1 m(3.8 ×107 ) = 10(9.8)(1)(1000) 5 9.8 × 5 m= = 12.89 × 10 −3 kg 3.8 ×103 (a) Linear momentum remains constant if net external force on the system of particles is zero. u = 5v − 2

200 mm

mg R sin 30° + W f =

10.

4 cm

12. (6) P = P0 − ρ gh = 98 × 10 3 N / m 2

H

5 = 7.5 N 2

N = 5+

m/s 18. (c) v1 = v2 = 200 m /s

B

2 cm

mv 2 R

17. (a) f1 − f 2 = 4 s −1

4T = 16N / m2 RA

RA

PB = P0 +

x2 2

1 = 0.78 mm ≃ 1 2

A

⇒

v 2

d ℓ = dxα1 (θ − 10)

∆ℓ = ∫ (130 x)α1dx

⇒

⇒

Tmax 324 = = 1296 = 36 rad/s 0.5 × 0.5 mr

Mock Test-2

6.

31


(c) LO = (mω r 2 )kˆ

P2 = P′ + θ


LO

Using Boyle’s law, we have, P1 V1 = PV 2 2


LP

or

7.


Direction of LP varies with time as its direction will be

or

perpendicular to string i.e. changing with time.

LP = Variable LO = Constant ,

or

(8) Here, λ1 = 500 mm = 0.5m

λ2 = 0.25 m;υ1 = 16 m/s,υ2 = ?

∴

υ 2 λ2 0.25 1 = = = υ1 λ1 0.5 2

⇒

υ2 =

8.

l , n = 1 Magnetic field 2π induction at the centre of circular coil due to current I is B=

υ1 16 = = 8m/s υ2 2

9.

µ0 2π nI µ0 2π × 1× I µ0π I = = 4π (l / 2π ) 4π r l

When wire is taken in the form of double loop, then

Time taken by the plate to fall down,

m 2 = = 7Hz. t 2/7

(3) The pressur at the bottom, P = P0 + h ρ g

I and n = 2 4π Magnetic field induction at the centre of the circular coil. l = 2 × 2π r1 or r1 =

∴

2h 2 × 40 2 = = sec g 980 7

Frequency of fork, n =

4S 1  8S  =  P′ +  R 9 R 24 S 24 + 0.08 P′ = 8P + = 8 ×105 + = 8.08 ×105 = n ×105 R 2.4 ×10−4 n = 0.08 ≈ 8.

P+

coil, then length, l = 2π r or r =

(7) Here, .m = 2, h = 40cm, n = ?

t=

4S  8S  V1   P+ V1 =  P′ +  R  R 8  

11. (4) When wires is taken in the form of one turn circular

As frequency of body is fixed, say n, therefore, υ1 = λ1 n;υ2 = λ2n ⇒

4S 8S   =  P′ +  ( R / 2)  R 

B1 = ∴

µ0 2π × n × I µ0 2π × 2 × I πI = = 4 × µ0 4π r1 4π (l / 4π ) l

B1 =4 B

12. (4) Refer Fig. A

Where, p0 = atmospheric pressure, therefore,

I

hρ g atmosphere P =1+ P0

O

20 ρ g = 1+ = 3 atmosphere. 10 ρ g

I

60° 60° r B

10. (8) Here, p = 105 N/m 2 , R = 2.4 × 10 −4 m; S = 0.08 N/m Initial pressure inside the soap bubble in air cylinder, 4S P1 = P + . R Let V1 be the initial volume of the soap bubble;

4 V1 = π R3 3 After compression, volume of the soap bubble; 3

4 R V V2 = π   = 1 3 2 8 If is the air pressure after compression in the cylinder, then pressure inside the bubble is

I

D a

C

The magnetic field induction at the centriod O due to current I through one side BC of the triangle will be

µ0 I (sin θ1 + sin θ 2 ) 4π r It will be acting perpendicular to the plane of triangle upwards. Total magnetic field induction at O due to current through all the three sides of the triangle will be B1 =

B = 3B1 =

3µ 0 I [sin θ1 + sin θ 2 ] 4π π

Here, I = 1 A, θ1 = 60° = θ 2 and r = OD =

BD a/2 = tan 60° 3

Physics

32 −2

= ∴

a 4 ⋅ 5 × 10 = m. 3 2 3

r r r r ∴ Displacements S = x while angle θ between F and x is

2

constant

1 × [sin 60° + sin 60°] B = 3 × 10 × (4 ⋅ 5 × 10 −2 / 2 3) −7

On solving, B = 4 × 10 −5 T.

∴

13. (8) Magnetic field at P due to currents in two wires will be acting perpendicular to the plane of wires, upwards and is given by.

BP =

µ0 2 I µ 0 2 I µ 0 I + = 4π 2r 4π r 4π r BP ( 2µ0 I / π r ) = = 8. BQ ( µ0 I / 4π r )

by BQ =

∴

14. (4) When wire is taken in the form of one turn circular coil, then length l = 2π r or

l , n =1 2π Magnetic field induction at the centre of circular coil due to current I is r=

B=

µ0 2π nI µ0 2π × 1× I µ0π I = = l 4π r 4π ( l / 2π )

When wire is taken in the form of double loop, then

l and n = 2 4π Magnetic field induction at the centre of the circular coil, l = 2 × 2π r1 or r1 =

∴

B1 = ⇒

r uur dθ F = ∫ F ⋅ dx = ∫ F ⋅ cos θ ⋅ == sin θ K K uur uur r uur W = ∫ dW = ∫ F ⋅ dx = ∫ Kx ⋅ dx

µ0 2 I µ 2I 2µ I + 0 = 0 πr 4π ( r / 2 ) 4π ( r / 2 )

Magnetic field at Q due to current in A is perpendicular to the plane of wire upwards and due to current in B is perpendicular to the plane of wire downwards and is given

µ0 2π × n × I µ0 2π × 2 × I πI = = 4 × µ0 r1 l 4π 4π ( l / 4π )

B1 = 4. B

15. (b) A→3, B→2, C→1, D→4 r r W = F ⋅ S = Fx cos θ

dθ dθ = K or dx = dx k r uur ∫ dW = ∫ F ⋅ dS

θ = kx, hence

∴

1 = K ∫ x dx cos θ == Kx 2 cos θ 2 r r F ⊥ x at every point r uur dW = F ⋅ dx = 0

Q θ = 90° = Fdx cos90° = 0

16. (a) A→3; B→1; C→4; D→2 17. (c) A→2, B→3, C→4, D→1 When a body is disturbed from its mean position and left to vibrate under restoring force, e.g., A tunning force struck with rubber pad, vibrations are called free or natural vibrations. When a body is mode to vibrate with another vibrating body placed nearby, e.g., stem of a vibrating tunning fork when pressed on top of sonometer makes the sonometer wire to vibrate having forced vibrations. When a body is forced to vibrate by another vibrating body the vibrations are forced on other body. But if frequency of forced vibrations is equal to the natural frequency of forced body then vibrations of forced body are called resonant vibrations. When a body is vibrating under the action of viscous force, gradually the amplitude of vibration decreases because energy of vibrating body is dissipated as work done against viscous force in the form of heat etc. and body gradually stops vibrating. Such vibratious are called damped vibrations. 18. (a) A→1, B→2, C→3, D→4 The particular names of calorimeters are after the names of scientists who designed them for specific heat measurements in different conditions.

33

Mock Test-3

JEE-MAIN: PHYSICS MOCK TEST-3 SECTION 1 (Multiple Choice Question) 1.

7.

A solid sphere of radius R made of material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid when a mass m is placed on the piston to compress the liquid, the fractional change in the δR radius of the sphere, is: R mg mg a. b. AK 3 AK mg mg c. d. A 3 AK

8.

The wavelength of light observed on the earth, from a moving star is found to decreases by 0.05%. Relative to the earth, the star is:

ur If A = 2iˆ + 4 ˆj − 5 kˆ the direction of cosines of the vector

ur A are: a. c. 2.

3.

4.

2

,

45 4

4

and

45 , 0 and

45

−5 45

4 45

b. d.

1

,

2

45

45

3

2

,

45

and

3 45

and

45

5 45

A particle is projected vertically upwards and it is at a height h after 2 seconds and again after 10 seconds. The height h is: a. 196 m b. 98 m c. 9.8 m d. 19.8 m An insect crawls up a hemispherical surface very slowly (see the figure). α The coefficient of friction between the surface and the insect is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle α is given b. tan α = 3 a. cot α = 3 c. sec α = 3 d. cosecα = 3 An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unscratched. Then the maximum extension in the spring is

4Mg a. k

2Mg b. k

Mg Mg d. k 2k A particle executes simple harmonic motion between x = − A and x = + A. The time taken for it to go from c.

5.

0 to A/2 is T1 and to go from A/2 to A is T2 . Then:

6.

a. T1 < T2

b. T1 > T2

c. T1 = T2

d. T1 = 2T2

A large number of liquid drops each of radius r coalesce to form a single drop of radius R. The energy released in the process is converted into the kinetic energy of the big drop so formed. The speed of the big is (Given surface tension of liquid is T, density of liquid is ρ) a.

T 1 1  − ρ  r R 

b.

2T  1 1  − ρ  r R 

c.

4T  1 1  − ρ  r R 

d.

6T  1 1  − ρ  r R 

a. moving away with a velocity of 1.5 ×105 m/s b. coming closer with a velocity of 1.5 ×105 m/s c. moving away with a velocity of 1.5 ×104 m/s d. coming closer with a velocity of 1.5 ×104 m/s 9.

An ideal gas expands isothermally from a volume V1 to

V2 and

then

compressed

to

original

volume

V1

adiabatically. Initial pressure is p1 and final pressure is

p3 . The total work done is W. Then, a. p3 > p1 ,W > 0

b. p3 < p1 ,W < 0

c. p3 > p1 ,W < 0

d. p3 = p1 ,W = 0

10. At NTP one mole of diatomic gas is compressed adiabatically to half of its volume ( γ = 1.40). The work done on the gas will be ( 2 ) a. 1280 J c. 1792 J

0.4

= 1.40 : b. 1610 J d. 2025 J

11. Two identical conducting rods are first connected independently to two vessels, one containing water at 100°C and the other containing ice at 0ºC. In the second case, the rods are joined end to end connected to the same vessels. Let q1 and q2 gram per second be the rate of melting of ice in the two cases respectively. The ratio q1 / q2 is

1 2 4 c. 1

a.

2 1 1 d. 4 b.

Physics

34

12. The energy density (energy per unit volume) in an electric field caused by a point charge falls off with the distance from the point charge as: a. 1/ r

b. 1/ r 2

c. 1/ r 3

d. 1/ r 4

 3 a. sin −1    8 

 3 b. cos −1    8 

 3 c. tan −1    8 

 3 d. cot −1    8 

1 If power factor is in a series RL circuit R = 100 Ω . 2 ac mains is used then L is a.

3 henry π

b. π henry

π

henry d. None of these 2 15. The amplitude ratio of two superposing waves is 2 : 1. The ratio of maximum and minimum intensities is: a. 1 : 1 b. 2 : 1 c. 4 : 1 d. 9 : 1 c.

16. The electric and the magnetic field, associated with an e.m. wave, propagating along the + z-axis, can be represented by r r r r a.  E = E0iˆ, B = B0 ˆj  b.  E = E0 kˆ, B = B0iˆ    r r r r c.  E = E0 ˆj , B = B0iˆ  d.  E = E0 ˆj , B = B0 kˆ    17. If the wavelength of the first line of the Balmer series of hydrogen is 6561 Å, the wavelength of the second line of the series should be a. 13122 Å

b. 3280 Å

c. 4860 Å

d. 2187 Å 11

11

SECTION 2 (Numeric Value Question) 20. For television broadcasting, the frequency employed is normally a. 30–300 MHz b. 30–300 GHz c. 30–300 kHz d. 30–300 Hz 21. The potential difference V and current i flowing through an appliance in an ac circuit are given by V = 5cos ωt volt, i = 5sin ωt amp and power dissipated in the appliance is a. 0 W c. 5 W

b. 10 W d. 2.5 W

22. A wire has a mass (03. ± 0.003) g, radius (0.5 ± 0.005) mm and length (6 ± 0.06) cm. The maximum percentage error in the measurement of its density is: a. 1 b. 2 c. 3 d. 4 23. An aircraft with a wing-span of 40 m flies with a speed of 1080 km h–1 in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of earth's magnetic field is 8.3 × 10−4 T. Then the e.m.f. that develops between the tips of the wings is: a. 0.5 V b. 0.35 V c. 1 V d. 2.1 V

24. If an observer is walking away from the plane mirror with 3 m/sec. Then the velocity of the image with respect to observer will be a. 6 m/sec b. −6 m / sec d. 3 m / sec

+

X stand for

Space for Rough Work

C

a. NPN transistor b. PNP transistor c. Forward biased PN junction diode d. Reverse biased NP junction diode

c. 12 m / sec

18. In the nuclear reaction 6 C →5 B + β + X, what does a. An electron c. A neutron

E

B

13. The needle of a deflection galvanometer shows a deflection of 60° due to a short bar magnet at a certain distance in tan A position. If the distance is doubled, the deflection is

14

19. The symbol given in figure represents

b. A proton d. A neutrino

25. The eccentricity of earth’s orbit is 0.0167. The ratio of its maximum speed in its orbit to its minimum speed is: a. 2.507 b. 1.033 c. 8.324 d. 1.000

Mock Test-3

35

JEE ADVANCE PAPER-I length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K? a. 4 if wires are in parallel b. 2 if wires are in series c. 1 if wires are in series d. 0.5 if wires are in parallel

SECTION 1 (Maximum Marks: 24)

MCQs with one or more than one correct answer 1.

In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be: a. 12 A b. 14 A c. 8 A d. 10 A

2.

When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10 −4 ms −1. If the electron density in the wire is

SECTION 2 (Maximum Marks: 24)

Numerical value answer type questions 7.

3.

a. 1.6 × 10 −8 Ω m

b. 1.6 × 10 −7 Ω m

c. 1.6 × 10 −6 Ω m

d. 1.6 ×10−5 Ω m

4.

5.

centroid of the triangle in the units of (10−5 T ) is

d. 3 Ω

The temperature dependence of resistances of Cu and undoped Si in the temperature range 300–400 K, is best described by: a. Linear increase for Cu, linear increase for Si b. Linear increase for Cu, exponential increase for Si c. Linear increase for Cu, exponential decrease for Si d. Linear decrease for Cu, linear decrease for Si Two ideal batteries of emf V1 and V2 and three resistances

R1, R2 and are R3 connected as shown in the figure. The current in resistance R2 would be zero if

a. V1 = V2 and R1 = R2 = R3

R1 R2

c. V1 = 2V2 and 2R1 = 2R2 = R3

6.

12. A microwave telephone link operating at the central frequency of 10 GHz has been established. If 2% of this is available for microwave communication channel and each telephone is allotted a bandwidth of 8 kHz, the number of telephone channels which can be simultaneously granted is 2.5 × 10 a. What is the integer value of a ?

13. A uniform circular disc of mass 1.5 kg and radius is V1

b. V1 = V2 and R1 = 2R2 = R3 d. 2V1 = V2 and 2R1 = R2 = R3

If the maximum values of signal and carrier waves are 4 V and 5V respectively, the percentage of amplitude modulation is a × 10%. What is the value of a ? 9. A signal wave of frequency 4.5 kHz is modulated with a carrier wave of frequency 3.45 MHz. The bandwidth of FM wave is kHz is 10. What is the maximum usuable frequency (in MHz) for E-layer of atmosphere having critical frequency 4 MHz, when the angle of incidence is 60°? 11. A T.V. Tower has a height 100 m. In order to triple its coverage range, the height of tower to be increased is a × 10 2 m. What is the integer value of a?

8.

A galvanometer having a coil resistance of 100 Ω gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A, is: a. 0.01 Ω b. 2 Ω

c. 0.1 Ω

A current 1 amp is flowing in the sides of an equilateral triangle of side 4.5 ×10−2 m . The magnetic field at the

−3

8 ×10 m , the resistivity of the material is close to: 28

R3

V2

initially at rest on a horizontal frictionless surface. Three forces of equal magnitude F = 0.5 N are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in rad s –1 is F

Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of

X O

Y Z

F

F

36

Physics

14. A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two mass less spring toyguns, each carrying a steel ball of mass 0.05 kg are attached to the platform at a distance 0.25 m from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of

9 ms −1 with

respect to the ground. The rotational speed of the platform in

16. Consider two different metallic strips (1 and 2) of same dimensions (lengths ℓ , width w and thickness d) with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in magnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developed between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips, the correct option(s) is(are) a. If B1 = B2 and n1 = 2n2 , then V2 = 2V1

b. If B1 = B2 and n1 = 2n2 , then V2 = V1 c. If B1 = 2 B2 and n1 = n2 , then V2 = 0.5V1

rad s –1 after the balls leave the platform is

d. If B1 = 2 B2 and n1 = n2 , then V2 = V1

SECTION 3 (Maximum Marks: 12)

Paragraph based questions (2 paragraphs, each having 2 MCQs with one correct answer only) Paragraph for Question No. 15 to 16 In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are l, w and d,  respectively. A uniform magnetic field B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross-section of the strip and carried by electrons. l

y

•K

I

W d

P

I

x

R

S

•M

z Q

15. Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are w1 and w2 and thicknesses are d1 and d2, respectively. Two points K and M are symmetrically located on the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the correct statement(s) is(are) a. If w1 = w2 and d1 = 2d2, then V2 = 2V1 b. If w1 = w2 and d1 = 2d2, then V2 = V1 c. If w1 = 2w2 and d1 = d2, then V2 = 2V1 d. If w1 = 2 w2 and d1 = d 2 , then V2 = V1

Paragraph for Question No. 17 to 18 The capacitor of capacitance C can be charged (with the help of a resistance R) by a voltage V source V, by closing switch S1 R

while keeping switch S 2 open. The capacitor can be connected in series with an inductor ‘L’ by closing switch S 2 and opening S1

C

L

S1

S2

17. Initially, the capacitor was uncharged. Now, switch S1 is closed and S2 is kept open. If time constant of this circuit is τ, then a. after time interval τ, charge on the capacitor is CV/2 b. after time interval 2τ , charge on the capacitor is

CV (1 − e −2 ) c. the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged d. after time interval 2τ , charge on the capacitor is

CV (1 − e −1 ) 18. After the capacitor gets fully charged, S1 is opened and S2 is closed so that the inductor is connected in series with the capacitor. Then a. at t = 0, energy stored in the circuit is purely in the form of magnetic energy b. at any time t > 0, current in the circuit is in the same direction c. at t > 0, there is no exchange of energy between the inductor and capacitor d. at any time t > 0, instantaneous current in the circuit may V

C L

Mock Test-3

37

JEE ADVANCE PAPER-II SECTION 1 (Maximum Marks: 24)

MCQs with one or more than one correct answer 1.

The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field

4.

increases with time. I 1 and I 2 are the currents in the segments ab and cd. Then,

a.

b µ 0 NI In   2(b − a )  a 

b.

 b + a  µ 0 NI In   2(b − a )  b – a 

c.

µ 0 NI  b  In    a  2b

d.

µ 0 NI  b+ a  In    b – a  2b

A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector kˆ is coming out of the plane of the paper. The magnetic moment of the current loop is y

d

c a

b I

x

a a

a. I1 > I 2 b. I1 < I 2 c. I 1 is in the direction ba and I 2 is in the direction cd d. I1 is in the direction ab and I 2 is in the direction dc 2.

A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

5.

a. a 2 Ikˆ

π  b.  + 1 a 2 Ikˆ 2 

π  c. −  + 1 a 2 Ikˆ 2 

d. (2π + 1) a 2 Ikˆ

An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform current density along its length. The magnitude of the magnetic



field, | B | as a function of the radial distance r from the axis is best represented by.  B

 B

a.

b. r

R/2

a. IBL 3.

IBL

b.

c.

π

IBL 2π

d.

IBL 4π

Along insulated copper wire is closely wound as a spiral of 'N' turns. The spiral has inner radius 'a' and outer radius 'b'. The spiral lies in the X-Y plane and a steady current 'I' flows through the wire. The Z component of the magnetic field at the center of the spiral is Y

r

R

R/2

R

R/2

R

 B

 B

c.

d. r

r

R/2

6.

R

A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows

I

a

b

X

through the smaller loop, then the flux linked with bigger loop is

a. 9.1 × 10 −11 weber

b. 6 × 10 −11 weber

c. 3.3 × 10 −11 weber

d. 6.6 × 10 −9 weber

38

Physics

SECTION 2 (Maximum Marks: 24) Numerical value answer type questions 7.

12 5

The isotope

β - decay to

12 6

B having a mass 12.014 u undergoes

C. 126 C has an excited state of the nucleus

( 126 C* ) at 4.041 MeV above its grounds state. If 12 5

B decays to

12 6

C* , the maximum kinetic energy of the

β - particle in units of MeV is: ( 1u = 931.5MeV / c2,, where c is the speed of light in vacuum)

8.

9.

A binary star consists of two stars A (mass 2.2Ms) and B (mass 11Ms), where Ms is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is Two homogeneous spheres A and B of masses m and 2m having radii 2a and a respectively are placed in contact. The ratio of distance of c.m. from first sphere to the distance of c.m. from second sphere is:

10. A non-uniform thin rod of length L is placed along X-axis so that one of its ends is at the origin. The linear mass density of rod is λ = λ0 x ⋅ The centre of mass of rod divides the length of the rod in the ratio:

11. A sphere of mass 5 kg and diameter 2 m rotates about a tangent. What is its moment of inertia? 12. A uniform rod of length 1 m and mass 0.5 kg rotates at angular speed of 6 rad/sec about one of its ends. What is the KE of the rod? 13. A particle performing uniform circular motion has angular momentum L. When its angular velocity is doubled and KE is also doubled, the new angular momentum becomes x times. What is x? 14. The fundamental frequency of a closed organ pipe is equal to first overtone frequency of an open organ pipe. If the length of the open pipe is 36 cm, what is the length (in cm) of the closed pipe? SECTION 3 (Maximum Marks: 12) Matching type questions with 4 options 15. Match Column I with Column II and select the correct answer using the codes given below the Columns:

Column I (A) Boltzmann Constant

Column II

(B) Coefficient of viscosity

2. [ML−1T −1 ]

(C) Plank Constant

3. [MLT −3 K −1 ]

(D) Thermal conductivity

4. [ML2 T −2 K −1 ]

a. A-4; b. A-1; c. A-4; d. A-1;

1. [ML2 T −1 ]

B-1, 4; C- 4; D-2 B- 1,4; C-4; D-4 B-1,4; C-4; D-2 B- 1,4; C-2; D-4

16. Column I give a list of possible set of parameters measured in some experiments. The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graph given in Column II. Column I

Column II

(A) Potential energy of a simple pendulum (y axis) as a function of displacement (x axis)

1. y

(B) Displacement(y axis) as a function of time (x axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive x-direction

2. y

(C) Range

3. y

of

a

projectile

O

O

x

x

(y-axis) as a function of its velocity

(x-axis)

when

projected at a fixed angle

(D) The square of the time period (y-axis) of a simple pendulum as a function of its length (x -axis)

O

x

5. y

O

x

a. A-4; B- 1,4; C- 4; D- 2 b. A-1; B- 1,4; C- 4; D- 4 c. A-2; B- 1,4; C- 4; D- 2 d. A-1; B- 1,4; C- 2; D- 4 17. Motion is defined as rate of change of position. In the motions described in column I, match the facts about that motion in column II for the changes in position vector.

Mock Test-3

Column I (A) Magnitude only (B) Direction only (C) Both in magnitude and direction (D) Remains invariant

39

Column II 1. A coin dropped from roof of house 2. A coin thrown at any angle with horizontal 3. A coin held in your hand 4. A coin in rotated in circular path with variable speed

a. A-1; B-2; C-3; D-4 b. A-4; B-2; C-3; D-1 c. A-1; B-4; C-2; D-3 d. A-3; B-2; C-4; D-1 20.

Match the statement of Column with those in Column II: Column I Column II (A) In any Bohr orbit of the 1 1. − hydrogen atom, the ratio 2

Space for Rough Work

of kinetic energy to potential energy of the electron is (B) The ratio of the kinetic energy to the total energy of an electron in a Bohr orbit is (C) In the lowest energy level of hydrogen atom, the electron has the angular momentum (D) Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is a. A-1; B-3,4; C-4; D-2,3 b. A-1; B-3,4; C-4; D-3 c. A-2; B-3,4; C-1; D-3 d. A-1; B-2,4; C-3; D-4

2. 2

3.

h 2π

4. 5 : 27

40

Physics

ANSWER & SOLUTIONS

6.

(d) Work done, W = T ∆A

JEE-Main

or

W = T [(4π R 2 ) n − 4π R 2 ] = T .4π [nr 2 − R 2 ]

∴

4 4 1 Where n × π r 3 = π R 3 and W = mυ 2 3 3 2 1 2 mυ = T .4π (nr 2 − R 2 ) 2 1 4 3 2 × π R ρυ = T .4π (nr 2 − R 2 ) 2 3

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

a

b

a

b

b

d

b

b

c

c

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

d

c

a

d

a

c

d

a

a

21.

22.

23.

24.

25.

a

d

c

a

b

or



1.

(a) A = 2ˆi + 4 ˆj − 5 kˆ

∴

| A | = (2) 2 + (4 ) 2 + (−5 ) 2 = 45

2

∴

cos α =

2.

(b) h = ut −

45

or

, cos β =

4 45

, cos γ =

7.

−5 45

⇒

1 2 gt 2 1 1 2 2 h = u × 2 − g × (10 ) = u × 10 − g × (10 ) 2 2 2u − 2 g = 10u − 50 g ⇒ u = 6g

∴

h = 2u − 2 g = 2 × 6 g − 2 g = 10 g = 98 m

3.

(a) Equilibrium of insect gives

⇒

µN α mg cos α

∴

α N

8.

mg sin α

⇒

mg

N = mg cos α

. . . (i)

µ N = mg sin α

. . . (ii)

From equation (i) and (ii) we get cot α =

4.

∴

5. ∴

⇒

(b) Let x be the maximum extension of the spring. From conservation of mechanical energy decrease in gravitational potential energy = increase in elastic potential energy

k M

v=0

M

x

1 2 2 Mg kx or x = 2 k (b) For block P, friction will provide the necessary restoring force. fmax = mω2 A with ω 2 = k = k m + m 2m Mg x =

Hence, f max

kA  k  = m A= 2  2m 

6T  nr 2 R 2  6T  nr 2 1  6T  1 1  − = −  3 − 3 =  ρ  nr 3 R  ρ  r R  R  ρ R

∆P mg / A ∆V mg = = ∆V / V ∆V / V V AK 4 But for a sphere V = πR 2 3 4 Differentiating δV = π3 R 2 (δR ) 3 4 π(3R 2 )δR δV 3 3δR = = 4 3 V R πR 3 3δV mg δR mg From (i) = ⇒ = 3 AK R AK R (b)

. . .(i)

0.05 v λ , As ∆λ = λ 100 c 0.05 v λ = λ or v = 5 × 10 −4 c 100 c

(b) ∆λ =

v = 5 × 10 −4 × 3 × 108 = 1.5 × 105 m/s

As λ decreases, the star is approaching the observer.

9.

1 =3 µ

v=0

or

υ=

(c) Slope of adiabatic process p C at a given state (p, V, T) is more than the slope of A B isothermal process. The corresponding p-V graph for V the two processes is as shown in figure. In the figure, AB is isothermal and BC is adiabatic. WAB = positive (as volume is increasing) and

WBC = negative (as volume is decreasing) plus,

WBC > WAB , as area under p-V graph gives the work done. Hence, WAB + WBC = W < 0 From the graph itself, it is clear that p3 > p1 . Hence, the correct option is Note: At point B, slope of adiabatic (process BC) is greater than the slope of isothermal (process AB).

Mock Test-3

41 γ

V  10. (c) P1 V1γ = P2 V2γ ⇒ P2 = P1  1   V2  work =

γ −1  R 1   V1   P1   − 1 ( P2 V2 − P1 V1 ) = 1− γ 1 − γ   V2   

1 0.40 × 105 × 22.4 ×10−3 × ( 2 ) − 1   1 − 1.40 1 = × 22.4 ×102 [1.32 − 1] = 1792 J 0.40

17. (c) The wavelength of spectral line in Balmer series is

1 1 = R 2 − 2  n  2 For first line of Balmer series, n = 3 given by

⇒

=

11. (c) or

dQ  dm  Temperature difference  dm  = L = L  or  Thermal resistance dt  dt   dt 

dm 1 1 ⇒q∝ ∝ dt Thermal resistance R In the first case rods are in parallel and thermal resistance R while in second case rods are in series and thermal 2 resistance is 2R. is

⇒

q1 2R 4 = = q2 R / 2 1

12. (d) Energy u ∝ E2 and u ∝

1 1 for a point charge; so u ∝ 4 r2 r

13. (c) For short bar magnet in tan A-position

µ0 2M = H tan θ . . . (i) 4π d 3 When distance is doubled, then new deflection θ ′ is given by µ0 2 M . . . (ii) = H tan θ ′ 4π (2 d )3 ∴

tan θ ′′ 1 = tan θ 8

⇒

tan θ ′ =

14. (a) cos φ =

1 ⇒ φ = 60° 2

1 1  3R 1 = R 2 − 2  = λ2 2 4   16 20 λ2 20 = ⇒ λ1 = × 6561 = 4860 Å 27 λ1 27

18. (d) 6 C 11 → 5 B 11 + β+ + γ because β+ = 1 e 0 19. (a) The base is always thin 20. (a) VHF (Very High Frequency) band having frequency range 30 MHz to 300 MHz is typically used for TV and radar transmission.

π  21. (a) V = 5cos ωt = 5sin  ωt +  ⇒ I = 2sin ωt 2  ∴

φ=

π 2

⇒ P = Erms I rsm cos

22. (d) Density ρ = ∴

π 2

=0 W

m

π r2L

∆ρ

∆r ∆L   ∆m × 100 =  +2 +  ×100 m r L   After substituting the values, we get the maximum percentage error in density = 4%

ρ

23. (c) L = 40 m, v = 1080 km, ⇒

I max ( a1 + a2 )  2 + 1  9 = =  = I min ( a1 − a2 )2  2 − 1  1

e = Blv = 8.3 ×10−4 × 40 × 300 = 1 V

24. (a) Relative velocity of image w.r.t. object = 3 – (–3) = 6 m/sec

ωL 3 ⇒L= H R π 2

15. (d)

∴

1  1 1  5R = R 2 − 2  = ; For second line n = 4. λ1  2 3  36

h–1 = 300 m sec−1 and B = 8.3 × 10−4 T

 3  tan θ tan 60° 3 = = ⇒ θ ′ = tan –1   8 8 8  8 

tan 60° =

⇒

1

λ

–3m/sec

3m/sec

O

I

2

16. (a) In the given question, the electromagnetic wave is propagating along + z axis. In e.m. wave, the electric field   ( E) and magnetic field ( B ) are perpendicular to each other and also perpendicular to the direction of propagation of wave,   So, E = E0 iˆ and B = B0 ˆj.

25. (b) Angular momentum of satellite about center of earth remains constant, i.e., mvr = constant 1 v r ⇒ v∝ ⇒ 2 = 1 2 v1 r2 Speed is maximum at position 2,

∴

vmax r1 (1 + ε ) a 1 + ε 1 + 0.0167 = = = 1.033 = = vmin r2 (1 − ε ) a 1 − ε 1 − 0.0167

42

Physics A

JEE Advance Paper -I 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

a

d

a

c

a,b,d

b,d

4

8

9

8

11.

12.

13.

14.

15.

16.

17.

18.

8

4

2

4

a,d

a,c

b

d

I O

B

1.

1 1 1 (a) = + R e q R1 R2

⇒

V2 V2 V2 = + + ..... P= R e q R1 R2 P = VI

⇒

P = 2500 = 220 I ⇒ I =

2.

(d) υd =

i Ane

V V = R ρ ℓ A

⇒

i=

⇒

V ρ= υd ⋅ ne ℓ

⇒

ρ=

∴

B = 3B1 =

i = υd ⋅ ne A i ℓ A

∴

BD a/2 a 4.5 × 10 −2 = = = m tan 60° 3 2 3 2 3

1

( 4.5 ×10

−2

/2 3

)

× [sin 60° + sin 60°]

On solving, B = 4 × 10–5 T.

8.

(8) Here, Am = 4 V ; Ac = 5 V Percentage of modulation,

9.

Am 4 × 100 = × 100 = 80% = a × 10%. ⇒ a = 8 Ac 5

(9) Here, vs = 4.5kHz; vc = 3.45MHz = 3450 kHz

Bandwidth = 2 vs = 2 × 4.5 = 9.0 kHz. 10. (8) Here, vc = 4 MHz, i = 90°

Here ig = 10 −3 A

(a) S =

⇒

G = 102 Ω, I = 10 A ⇒ S = 10−2 Ω

4.

(c) For conductor (Cu) resistance increases linearly and for semiconductor resistance decreases Exponentially in given temperature range.

Maximum usuable frequency = vc sec i

= 4 × sec60° = 4 × 2 = 8MHz 11. (8) d ' = 2h ' R = 3d = 3 2hR

or

h ' = 9 h = 9 ×100 = 900 m Increase in height of tower = 900 − 100 = 800 = a × 10 2

5.

R (V + V2 ) (a, b, d) V1 = 1 1 R1 + R3

⇒

V2 =

6.

(b, d) H =

⇒

t1 = 2 min. ⇒ t2 = 0.5 min.

7.

(4) Refer Fig., The magnetic field induction at the centroid O due to current I through one side BC of the

⇒ V1 R3 = V2 R1 ⇒ V2 R1 = V2 R3

V2 V2 V2 4= t1 t2 R R / 2 R /8

triangle will be B1 =

B = 3 × 10−7 ×

µ=

3.

R3 (V1 + V2 ) R1 + R3

3µ0 I [sin sin θ2 2] sin θ11++sin 4π r

And r = OD =

∴ρ= V

2.5 ×10−4 × 8 ×1028 ×1.6 ×10−19 × 0.1 2 1 = = × 10−4 = 1.6 ×10−5 Ω m 8 × 1.6 × 10−4 6.4 1 − ig

C

Here, I = 1 A, θ1 = 60° = θ2

250 125 = 22 11

5

ig G

I

It will be acting perpendicular to the plane of triangle upwards. Total magnetic field induction at O due to current through all the three sides of the triangle will be

W = 15 × 40 + 5 × 100 + 80 × 5 + 1× 1000 = 2500 w ⇒

I

60° 60° r D a

µ0 I . (sin 1θ+1 + sin2 θ2) 4π r

⇒

a =8

12. (4) Microwave frequency used in telephone link = 10 GHz = 10 × 109 Hz = 1010 Hz Frequency available for microwave communication 2 ×1010 = 2 × 108 Hz = 2% of 10 GHz = 100 Bandwidth of each telephone channel = 8 kHz = 8 × 103 Hz Number of microwave telephone channels

=

2 × 108 = 2.5 × 10 4 = 2.5 × 10 a ∴ a = 4 8 × 103

Mock Test-3

43

13. (2) τ = Iα

F

⇒

3FR sin 30° = Iα

⇒

MR 2 I = 2

R 30° R sin 30°

F

⇒

α=2

⇒

ω = ω0 + αt ⇒ ω = 2 rad / s

(c) 2T sin

⇒

Tdθ = BiRdθ

⇒

T = BiR =

3.

(a) ∫

0 = Iω + 2mv (r / 2); comparing magnitude

4.

⇒

neAv 1 1 = neA2v2 ⇒ d1w1v1 = d2 w2v2

I 

J=

R2   4 

R/2

r R

I 4I = 3R 2 3π R 2 π 4

Amperian loop

R  R Field for  r ≤  B = 0 For < r ≤ R 2 2  Applying Ampere’s Law B.2π r

18. (d) q = Q0 cos ω t

dq = Q0ω sin ωt dt

∴

C L

= µ0

 4I R 2  4µ I  R2  .π  r 2 −  = 02  r 2 −  2 3π R  4  3R  4 

B=

2 µ0 I  R2  R r −   ⇒ At r = , B = 0 3π R 2  4r  2

R filed is continuous. 2 From the above expression as r increases B increases.

At r =

For r ≥ R B.2π r = µ 0 I

JEE Advance Paper -II

1.

 a2  2  π  2  + a  + 1 a 2   2



⇒

a

a

π  R2 −

Q = CV (1 − e−t /τ ) after time interval 2τ .

ikax = CωV = V

4

a a

+ a2 = π 

(d) Let current be I.

17. (b) Q = Q0 (1 − e − t /τ )

⇒

π a2

5.

Correct options are a & c.

i=−

µ IN dr b = 0 ln   2(b − a)  a  a r b

a A = 4 ×   + a2 2

π  M =  + 1 a 2 Ikˆ 2  

V2 B2v2 w2  B2 w2   n1w1d1  B2 n1 = =  = V1 B2v1w1  B1w1   n2 w2 d2  B1n2

⇒

sin(dθ / 2)

2(b − a) ∫

Current density J =

16. (a, c) As I1 = I2 n1w1d1v1 = n2 w2d2v2

∴

µ0 IN

dr =

π  2

V1 v1w1 d2 = = and hence, correct choice is a & d. V2 v2 w2 d1

Now,

2T

∴

Now, potential difference developed across MK V = Bvw ⇒

T

(b) M = IA

= 2×

15. (a, d) I1 = I2

T

2

ω=4

⇒

2r (b − a)

a

0.5  0.45 × 0.5 × 0.5  ×2   ω = 0.05 × 9 × 2 2  

∴

µ0 IN

b

T cos(dθ / 2)

T cos (dθ / 2)

BiL 2π

F

14. (4) Since net torque about centre of rotation is zero, so we can apply conservation of angular momentum of the system about centre of disc Li = L f

∴

dθ = BiRdθ (for θ small) 2

2.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

d

c

a

b

d

a

9

6

2

2/3

11.

12.

13.

14.

15.

16.

17.

18.

7

3

1

9

c

a

c

d

(d) According to Lenz’s Law, current will be in anticlockwise sense as magnetic field is increasing into the plane of paper.

µ0 I 2π r

⇒

B=

⇒

At r = R,

∴B∝

B=

1 r

µ0 I 2π R

From inside expression at r = R ⇒

B=

2µ0 I  R 2  µ0 I . R −  = 3π R2  4R  2π R

This proves the continuity in the graph at r = R. From the above only correct option is (d).

44

6.

Physics

(a) First we find the mutual inductance of the assembly.

R I

Let current I flow through the larger loop. The strength of induction at the smaller loop

1 2 1  ml 2  2 1 0 ⋅ 5 × 12 Iω =  × 62 = 3 J ⋅ ω = × 2 2 3  2 3 13. (1) Here, L1 = I1ω1 , When ω2 = 2ω1 ,

µ0 IR2 = 2( R 2 + d 2 )3 / d

KE =

∴

Flux through smaller loop =

∴

Mutual inductance =

∴

Flux linked through coil =

7.

µ0πIR 2 r 2 2( R 2 + d 2 )3 / 2

µ 0πR 2 r 2 .i 2( R 2 + d 2 ) 3 / 2

∴

14. (9) Let l1 be the length of closed pipe l2 = 36cm= length

4π × 10−7 × π × 10−2 × 9 × 10−6 × 2 2(4 ×10 −2 + 2.25 ×10−2 )3 / 2

=

16 10 × 18 × × (10 +3−15 ) = 9.1× 10 −11 weber 2 15.625 12 5

of open pipe Fundamental frequency of closed pipe n1 = υ

4l1

Frequency of first overtone of open pipe,

12 6

B  → C + β− +γ −

n2 = 2 ×

Q = (m5 B12 − m612C * )c 2 = [m5 B12 − (m6C 12 + ∆m)]C 2

= (m5 B12 − m6C12 )C 2 − ∆mC 2 = 0.014 × 931 − 4.041 = 9

∴

2 1 1 2 2 2

Ltotal m r = +1 LB mr

8.

(6)

9.

(2) Assuming that masses of the two spheres are concentrated at their centres A and B and taking x = 0 at A, Fig., we get m 2m C m × 0 + 2m × 3a = 2a = AC xcm = A ( m + 2m ) 2a a B

AC 2a = = 2⋅ CB a 10. (2/3) As shown in Fig., mass of small element of length dx is dm = λ dx = λ0 x dx dm x

L

∫ xdm ∫ x(λ xdx )

dx

0

⇒

xcm =

0

∫ dm

=

0

L

L

∫ λ xdx 0

0

xcm =

[ x 3 / 3]0L L3 / 3 2 L = = [ x 2 / 2]0L L2 / 2 3

If x1 =

υ 4l1

=

υ 2l1

υ 36

=

υ l2

=

;4l1 = 36

υ 36

As n1 = n2 ⇒ l1 =

36 = 9cm. 4

15. (c) A-4; B-1, 4; C- 4; D-2

3 KT ⇒ [ ML2T −2 ] = K ′[ K ] ⇒ K ′ = [ ML2T −2 K −1 ] 2 B. F = 6πη rv ⇒ [ MLT −2 ] = η [ L ] [ LT −1 ] ⇒ η [ ML−1T −1 ] A. KE =

C. E = hf ⇒ [ ML2T −2 ] =

CB = a ∴

L

1 1  I 2ω22 = 2  I1ω12  2 2   1 1 1 I 2 (2ω1 ) 2 = 2 × I1ω12 ; I 2 = I1 2 2 2 L 1 L2 = I 2ω2 = I1 × 2ω1 = I1ω1 = L1 ⇒ x = 2 = 1 ⋅ L1 2

and KE =

µ0 R 2 r 2 2( R 2 + d 2 )3 / 2

=

(9)

2 = 1m ; I = ? 2 Moment of inertia of the sphere about a tangent to the 7 7 sphere, I = mr 2 = × 5 × 12 = 7kgm2 ⋅ 5 5 12. (3) Here, l = 1 m, m = 0.5 kg, ω = 6 rad/sec 11. (7) Here, m = 5 kg; r =

r

2L 2L L x 2L / 3 1 , x2 = L − = ⇒ 1 = = ⋅ 3 3 3 x2 L/3 2

D.

⇒

h ⇒ h = [ ML2T −1 ] [T ]

dQ K ′A(∆T ) [ ML2T −2 ] k[ L2 ] [ K ′] = ⇒ = dt ∆x [T ] [ L]

K ′ = [ MLT −3 K −1 ]

16. (a) A-4; B- 1,4; C- 4; D- 2 17. (c) c. A-1; B-4; C-2; D-3 When coin is dropped if falls is straight line where direction is same but magnitude changes with change in position. When a coin is rotated in circular path direction goes on changing but radius remains constant. In projectile motion both direction and magnitude of r changes. A coin held in your hand has constant position vector and hence it is invariant. 18. (d) A-1, B-2,4, C-3, D-4






Mock Test-4

45

JEE-MAIN: PHYSICS MOCK TEST-4 SECTION 1 (Multiple Choice Question) 1.

a.

Which of the following sets have different dimensions? a. Pressure, Young’s modulus, Stress b. Emf, Potential difference, Electric potential c. Heat, Work done, Energy d. Dipole moment, Electric flux, Electric field

2.

The 100 coplanar forces each equal to 10 N act on a body. π with the preceding force. Each force makes angle 50 What is the resultant of the forces? a. 1000 N b. 500 N c. 250 N d. Zero

3

A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by

π 12

 8 10π c.  −  3 16

The radius of earth is 6400 km and g = 10 m/s2. In order that a body of 5 kg weighs zero at the equator, the angular speed of earth in radian/sec is: 1 1 b. a. 80 400 1 1 c. d. 800 1600

7.

The lower end of a capillary tube of radius r is placed vertically in water. Then with the rise of water in the capillary, heat evolved is:

m

π 2 r 2h 2 J

c. − 8.

4.

2 Mg

b.

2 mg

c.

( M + m) 2 + m 2 g

d. ( ( M + m) 2 + M 2 ) g

9.


a
a

c.

2J

π r 2 h 2dg J

A wire of length L and cross-sectional area A is made of a material of Young’s modulus Y. If the wire is stretched by the amount x, the work done is: b.

YAx 2 L

d. Yax 2 L

Which of the following graphs correctly represent the dV / dp variation of β = =− with p for an ideal gas at V constant temperature?

d. a.


a

5.

2J d. −

YAx 2 2L YAx c. 2L




π 2 r 2 h2 dg

2

a.

A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector a is correctly shown in a. b.

dg b. +

π r h dg 2

a.

4 π  d.  −  MR 2 3 6

 2  MR 

6.

a. +

M

4 π  b.  −  MR 2 3 4

MR 2


a

Four holes of radius R are cut form of a thin square plate of side 4R and mass M. The moment of inertia of the remaining portion about z-axis is: y

x

b. p

c.

p

d. p

p

10. Two rods of different materials and identical cross-sectional area are joined face to face at one end and their free ends are fixed to the rigid walls. If the temperature of the surroundings is increased by 30°C, the magnitude of the displacement of the joint of rods is: (length of the

Physics

46

rods l1 = l2 = 1 unit,

Young’s

a. tan −1 ( 3 / 2)

b. tan −1 ( 3)

modulii, Y1 / Y2 = 2 ; coefficients of linear expansion are α1

c. tan −1 ( 3 / 2)

d. tan −1 (2 / 3)

ratio

of

their

and α2)

11

a. 5(α 2 − a1 )

b. 10(α1 − a2 )

c. 10(α 2 − 2a1 )

d. 5(2α1 − a2 )

A steel ball of mass m1 = 1 kg moving with velocity 50 ms–1 collides with another stationary steel ball of mass m2 = 0.200 kg. During the collision, their internal energies change equally. If T1 and T2 are rise in temperature of balls m1 and m2 respectively, specific heat of steel = 0.105 and mechanical equivalent of heat J = 4.2 J/cal, then: 3

3

a. T1 = 7.0×10 °C, T2 = 1.4×10 °C, b. T1 = 1.4 ×103 °C, T2 = 7.0×103 °C, c. T1 = 1.4 °C, T2 = 7.0 °C,

15. Two straight long conductors AOB and COD are perpendicular to each other and carry currents i1 and i2 .

The magnitude of the magnetic induction at a point P at a distance a from the point O in a direction perpendicular to the plane ACBD is µ0 µ0 a. b. (i1 + i2 ) (i1 − i2 ) 2π a 2π a µ I1 I 2 µ0 2 c. d. 0 ( I1 + I 22 )1/ 2 2π a ( I1 + I 2 ) 2π a 16. A circular coil and a bar magnet placed nearby are made to move in the same direction. The coil covers a distance of 1 m in 0.5 sec and the magnet a distance of 2 m in 1 sec. The induced e.m.f. produced in the coil: y A

d. T1 = 7.0 °C, T2 = 1.4 °C,

B

12. Three identical dipoles are arranged as shown below.

x

 1  What will be the net electric field at P  k = ? πε 4 0   →

–Q

p

+Q

x +Q

–Q P x

→

p

x

–Q

→

p

+Q

k. p x3

b.

c. Zero

d.

a.

2kp x3

C

a. Zero b. 1 V c. 0.5 V d. Cannot be determined from the given information 17. Following figure shows an ac generator connected to a "box" through a pair of terminals. The box contains possible R, L, C or their combination, whose elements and arrangements are not known to us. Measurements outside the box reveals that e = 75 sin(sin ωt ) volt, i = 1.5 sin

(ωt + 45°) amp, then, the wrong statement is

2 kp x3

13. The area of the plates of a parallel plate capacitor is A and the distance between the plates is 10 mm. There are two dielectric sheets in it, one of dielectric constant 10 and thickness 6 mm and the other of dielectric constant 5 and thickness 4 mm. The capacity of the capacitor is: 2 12 a. b. ε 0 A ε 0A 3 35 5000 c. d. 1500 ε 0 A ε0A 7 14. If a magnet is suspended at an angle 30o to the magnetic meridian, it makes an angle of 45o with the horizontal. The real dip is

D

~

?

a. There must be a capacitor in the box b. There must be an inductor in the box c. There must be a resistance in the box d. The power factor is 0.707 18. The focal length of a concave mirror is f and the distance from the object to the principle focus is x. The ratio of the size of the image to the size of the object is f f +x a. b. x f c.

f x

d.

f2 x2

Mock Test-4

47

19. The energy of incident photon is 12.375 eV while the energy of scattered photon is 9.375 eV. Then the kinetic energy of recoil electron is a. 3 eV b. less than 3 eV c. more than 3 eV d. 21.75 eV 20. In a transistor circuit shown here the base current is 35 µA. The value of the resistor Rb is E

C

23. A sound wave of wavelength 32 cm enters the tube at S as shown in the figure. Then the smallest radius r so that a minimum of sound is heard at detector D is.

r D

S

a. 7cm c.21cm

B Rb

RL

b. 14 cm d. 28 cm

24. Three rods made of the same material and having the same cross-section have been joined as shown in the

9V

figure. Each rod is of the same length. The left and right

a. 123.5 kΩ c. 380.05 k Ω

b. 257 k Ω d. None of these

ends are kept at 0°C and 90ºC respectively. The temperature of junction of the three rods will be 90°C

SECTION 2 (Numeric Value Question) 21. A wooden block of mass 10 g is dropped from the top of a cliff 100 m high. Simultaneously a bullet of mass 10 g is fired from the foot of the cliff upwards with a velocity of 100 m/s. The bullet and the wooden block will meet, each other, after a time: a. 10 s b. 0.5 s c. 1 s d. 7 s 22. The displacement of a particle varies according to the relation x = 4(cos πt + sin πt ). The amplitude of the

0°C

90°C

a. 45°C c. 30° C

b. 60°C d. 20° C

25. Six resistance of 6 ohm each are connected to form a hexagon. The resistance between any two adjacent terminals is:

particle is: a. 8

b. – 4

a. 6 Ω

b. 36 Ω

c. 4

d. 4 5

c. 5 Ω

d. 1 ohm

Space for Rough Work

48

Physics

JEE ADVANCE PAPER-I SECTION 1 (Maximum Marks: 24)

MCQs with one or more than one correct answer 1.

2

α

d

a.

1 + tan α 1 − tan α

b.

1 + sin α 1 − cos α

c.

1 + sin α 1 − sin α

d.

1 + cos α 1 − cos α

1

a. air from end 1 flows towards end 2. No change in the volume of the soap bubbles b. air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases c. no changes occurs d. air from end 2 flows towards end 1. volume of the soap bubble at end 1 increases 2.

d

A glass tube of uniform internal radius r has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius r. End 2 has sub-hemispherical soap bubble as shown in figure. Just after opening the valve,

5.

On heating water, bubbles being formed at the bottom of the vessel detaches and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r d F

c. d A > d F

d. d A + d B = 2d F

Mock Test-4

7.

49

SECTION 2 (Maximum Marks: 24)

SECTION 3 (Maximum Marks: 12)

Numerical value answer type questions

Paragraph based questions (2 paragraphs, each having 2 MCQs with one correct answer only)

A 25 watt bulb, which is producing monochromatic light of wavelength 6600 Å is used to illuminated a metal surface. If the surface has 3% efficiency for photoelectric effect, then the photoelectric current produced in deci ampere is (use h = 6.6 × 10 −34 Js) :

8.

de-Broglie wavelength associated with an electron accelerated through a potential difference 4 V is λ1 . When accelerating voltage is decreased by 3 V, its de-Broglie wavelength is λ2 . The ratio

Paragraph for Question No. 15 to 16 The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure.

λ1 is λ2

ω Q

9.

P

What is the energy in eV that should be added to an electron of energy 2 eV to reduce its de-Broglie wavelength from 1 nm to 0.5 nm?

10. The angular momentum of electron in 2nd orbit of hydrogen atom is L. The angular momentum of electron in 4th orbit of hydrogen atom is n L where n = ? 11. For an atom of an ion having single electron, the wavelength observed λ1 = 2 are units and λ3 = 3 units figure. The value of missing wavelength λ2 is n3 orbit λ3

λ1

n2 orbit λ2

n1 orbit

60º. What is the resultant intensity at the point?

14. A pn-junction diode can withstand currents upto 10 mA. When it is forward biased, the potential drop across it is 1.0 V. Assuming that this potential drop is independent of the current, find the maximum voltage of the battery used to forward bias the diode when a resistance of is connected in series with the diode.

y

When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed ω , the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotating of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed ω in this case. Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical and parallel to x-z plane; Case (b) the disc with its face making an angle of 45° with x-y plane and its horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed ω about the z-axis. z

I and 2I cross each other at a point with a phase diff. of

Q

x

ω

12. How many different wavelength may be observed in the spectrum from a hydrogen sample if the atoms are excited to 3rd excited state? 13. Light waves from two coherent sources having intensities

P

z

Q

ω

y

Q

P

P x

45° y

x

15. Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct? a. It is

2ω for both the cases

b. It is ω for case a.; and

ω 2

for case b.

c. It is ω for case a.; and ω for case b. d. It is ω for both the cases

50

Physics

16. Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct? a. It is vertical for both the cases a. and b. b. It is vertical for case a.; and is at 45°to the x-z plane and lies in the plane of the disc for case b. c. It is horizontal for case a.; and is at 45° to the x-z plane and is normal to the plane of the disc for case b. d. It is vertical for case a.; and is at 45° to the x-z plane and is normal to the plane of the disc for case b. Paragraph for Question No. 17 to 18 A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω is an example of a non-inertial frame of reference.


Where vrot is the velocity of the particle in the rotating frame of
reference and r is the position vector of the particle with

respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed ω about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along
the rotation axis (ω = ω kˆ ). A small block of mass m is gently
placed in the slot at r = ( R / 2)iˆ at t = 0 and is constrained to move only along the slot.

17. The distance r of the block at time t is: R R a. cos 2ωt b. ( e 2ωt + e −2ωt ) 2 4

ω

c.

R cos ωt 2

d.

R ωt (e + e − ω t ) 4

R

18. The net reaction of the disc on the block is : m

R/2


The relationship between the force Frot experienced by a particle of mass m moving on the rotating disc and the force
Fin experienced by the particle in an inertial frame of reference






is, Frot = Fin + 2m (vrot × ω ) + m (ω × r ) × ω , Space for Rough Work

a. mω 2 R sin ω tjˆ − mgkˆ b. − mω 2 R cos ω tjˆ − mgk 6ˆ c.

1 mω 2 R(eωt − e −ωt ) ˆj + mgkˆ 2

d.

1 mω 2 R (e 2ω t − e −2ωt ) ˆj + mgkˆ 2

Mock Test-4

51

JEE ADVANCE PAPER-II

a. F1 = F2 = 0

b. F1 is radially inwards and F2 is radially outwards

c. F1 is radially inwards and F2 = 0

d. F1 is radially outwards and F2 = 0

SECTION 1 (Maximum Marks: 24)

MCQs with one or more than one correct answer 1.

2.

3.

One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is a. 0.25 b. 0.50 c. 2.00 d. 4.00

6.

U 1U  ∝ T 4 and pressure P =   . If the shell now V 3 V  undergoes an adiabatic expansion the relation between T and R is: u=

The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is: (For steel Young’s modulus 2 × 1011 Nm–2 and coefficient of thermal expansion is 1.1 × 10–5 K–1 )

a. 2.2×107 Pa

b. 2.2×106 Pa

c. 2.2×108 Pa

d. 2.2×109 Pa

A pendulum made of a uniform wire of cross-sectional area A has time period T. When an additional mass M is

c. T ∝

4.

  T 2  A c. 1 −  M     T   Mg

  T 2  A d. 1 −      TM   Mg

Function x = A sin 2 ωt + B cos 2 ωt + C sin ωt cos ωt represents SHM. θ

1 R

d. T ∝

Numerical value answer type questions 7.

An electron in an excited state of Li2+ ion has angular momentum 3 h / 2π . The de-Broglie wavelength of the electron in this state is pπa0 (where a0 is the Bohr radius). The value of p is

8.

Consider a hydrogen atom with its electron in the nth orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is (hc = 1242 eV nm) A hydrogen atom in its ground state is irradiated by light of wavelength 970 Å. Taking hc/ e = 1.23 7 × 1 0 − 6 eV m and the ground state energy of hydrogen atom as –13.6 eV, the number of lines present in the emission spectrum is:

9.

10. To determine the half life of a radioactive element, a

L

student plots a graph of ℓ n I

5.

1 R3

SECTION 2 (Maximum Marks: 24)

Young's modulus of the material of the wire is Y then 1/Y is equal to: (g = gravitational acceleration)

 T 2  Mg b.  M  − 1  A  T 

b. T ∝ e −3R

a. T ∝ e − R

added to its bob, the time period changes to TM . If the

 T  2  A a.  M  − 1  T   Mg

Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume

I 6

b. If A = – B; C = 2B, amplitude = | B 2 |

5

c. If A = B; C = 0 d. If A = B; C = 2B amplitude = |B| Two coaxial solenoids of different radii carry current I in
the same direction. Let F1 be the magnetic force on the
inner solenoid due to the outer one and F2 be the magnetic force on the outer solenoid due to the inner one. Then:

ℓn|dN(t)/dt|

a. For any value of A, B and C (except C = 0)

dN (t ) versus t. dt

4 3 2 1

2

3

4

5 Years

6

7

8

Here dN(t)/dt is the rate of radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 years, the value of p is

52

Physics

11. The activity of a freshly prepared radioactive sample is 10

10

disintegrations per second, whose mean life is

10 s. The mass of an atom of this radioisotope is 10−25 kg. 9

kg. The mass (in mg) of the radioactive sample is

12. A freshly prepared sample of a radioisotope of half-life 1386s has activity 103 disintegrations per second. Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80s after preparation of the sample is 13. A nuclear power plant supplying electrical power to a village uses a radioactive material of half life T years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is 12.5 % of the electrical power available from the plant at that time. If

(D) If different masses have m1 same momentum, the ratio 4. m 2 of their kinetic energy is a. A-1, B-2, C-3, D-4 b. A-2, B-1, C-4, D-3 c. A-4, B-3, C-2, D-1 d. A-3, B-2, C-1, D-4 16. Waves are classified on the basis of frequency range. Column I mentions the type of waves while column II gives the corresponding frequency range. Can you match the proper options? Column I Column II (A) Infrasonic 1. Objects having velocity more than velocity of sound in air at 0°C (B) Audible 2. Frequency < 20 Hz (C) Ultrasonics

3. Frequency > 20,000 Hz

(D) Supersonics

4. Frequency lies between 20 Hz to 20, 000 Hz b. A-2, B-4, C-3, D-1 d. A-3, B-2, C-1, D-4

the plant is able to meet the total power needs of the village for a maximum period of nT years, then the value of n is

14. For a radioactive material, its activity A and rate of change

dA dN of its activity R are defined as A = − and R = − , dt dt where N(t) is the number of nuclei at time t. Two

a. A-4, B-3, C-2, D-1 c. A-4, B-4, C-1, D-2

17. Various electromagnetic waves are given in column I and various frequency ranges in column II Column I Column II (A) Radiowaves 1. 1×1016 to 3×10 21 Hz

radioactive sources P (mean life τ ) and Q (mean life 2τ )

(B) γ-rays

2. 1×109 to 3×1011 Hz

have the same activity at t = 0. Their rates of change of

(C) Microwaves

activities at t = 2τ are RP and RQ , respectively. If

3. 3×1018 to 5×1022 Hz

(D) X-rays

4. 5×105 to 109 Hz

RP n = , then the value of n is RQ e

a. A-1, B-2, C-3, D-4 c. A-3, B-4, C-1, D-2

SECTION 3 (Maximum Marks: 12) Matching type questions with 4 options 15. There is a definite relation between velocity, mass and acceleration of body to know about the work done by the forces applied on body. Column I has some statements showing some relations which are related with some statements in Column II. Match the correct options.

Column I (A) If kinetic energy is K the momentum P is 0 (B) If momentum p is zero the kinetic energy is 0 (C) If different masses have same kinetic energy the ratio of their momenta is 0

Column II 1. zero 2. 3.

2 mk m2 m1

b. A-2, B-3, C-4, D-1 d. A-4, B-3, C-2, D-1

18. In each of the following questions, match column I and column II, and select the correct match out of four given choices. Column I Column II (A) Transducer 1. Range of frequencies over which communication system works (B) Attenuation 2. A device that has input in electrical form or provides output in electrical form (C) Range 3. Loss of strength of a signal during propagation. (D) Bandwidth 4. Argest distance between transmitter and receiver. a. A-,2 B-4, C-1, D-3 b. A-4, B-3, C-1, D-2 c. A-1, B-2, C-3, D-4 d. A-2, B-3, C-4, D-1

Mock Test-4

53

∴

ANSWERS & SOLUTIONS

I = I square − 4I hole

JEE-Main 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

d

d

d

c

c

c

b

a

a

c

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

c

c

a

c

a

b

b

a

b

21.

22.

23.

24.

25.

c

d

b

b

c

1.

2.

3.

π

= 2π 50 So, all the force will pass through one point and all forces will be balanced. i.e. their resultant will be zero.

8  8 10π = MR 2 − 10mR 2 =  − 3  3 16

(b) When the tube is placed vertically in water, water rises

through height h given by h =

2T cos θ rdg

Upward force = 2πr ×T cos θ Work done by this force in raising water column through height h is given by ∆W = (2πrT cos θ)h

 rhdg  2 2 = (2πrh cos θ)T = (2πrh cos θ)   = πr h dg  2 cos θ 

R

However, the increase in potential energy ∆E p of the raised

Mg + mg

h water column = mg , where m is the mass of the raised 2 column of water.

Pulley is in equilibrium under four forces. Three forces as shown in figure and the fourth, which is equal and opposite to the resultant of these three forces, is the force

∴ m = π r 2 hd

applied by the clamp on the pulley (say F).

2 2  gh  π r h dg So, ∆E p = (π r 2 hd )   = 2  2 

Resultant R of these three forces is R = ( (M + m)2 + M 2 )g Therefore, the force F is equal and opposite to R as shown

Further, ∆W − ∆E p =

in figure.

π r 2 h 2 dg

2 The part (∆W − ∆E p ) is used in doing work against viscous

∴

F = ( ( M + m) + M ) g .

4.


(c) Net acceleration a of the bob in position

2

forces and frictional forces between water and glass surface and appears as heat.

A


a

g 10 1 rad/s = = R 6400 × 103 800

7.

Mg

2

(c) g ′ = g − Rω 2 (at equator)

ω=

F

T=Mg

 2  MR 

⇒

(d) Free body diagram of pulley is shown in figure.

mg

 mR 2  M 16 R 2 + 16 R 2 ) − 4  + m ( 2 R 2 ) ( 12 2  

Given g ′ = 0 ⇒ g = Rω 2

(d) Dipole moment = (charge) × (distance) Electric flux = (electric field) × (area) (d) Total angle = 100 ×

=

6.

T = Mg

So, Heat released


an B 
at

8.

B has two components.
(a) an = radial acceleration (towards BA)
(b) at = tangential acceleration (perpendicular to BA)
Therefore, direction of a is correctly shown in option (c).

5.

Moment of inertia of the remaining portion,

(c) M = Mass of the square plate before cutting the holes  M  2 πM Mass of one hole m =  πR = 2  16  16 R 

J

=

π r 2h 2 dg

2J 1 1 (a) Energy stored, U = × (force)×(extension) = Fx 2 2 FL YAx As Y = ⇒ F= Ax L

1  YA ∴ U=  x 2 L 9.

∆W − ∆E p

1 YAx 2  x = 2 L 

dV / dp = compressibility of gas V 1 1 = under isothermal = and β = p Bulk modulus of elasticity

(a) β == −

Physics

54

conditions. Thus, β versus p graph will be a rectangular hyperbola.

14. (a) Let the real dip be φ, then tan φ =

KA (θ1 − θ 2 ) l temperature difference across the rods are 20°C and 10°C respectively. ∴ Displacement of the joint = α 2t2 − α1t1

For apparent dip,

10. (c) Applying, Q =

tan φ ′ = =

BVV BV 2 BV = = BH cos ββ BH cos 30° 3 BH −1  3  .tan φ or φ = tan   3  2 

2

or tan 45° =

= 10 α 2 − 20α1 = 10 (α 2 − 2α1 )

15. (c) At P : Bnet = B12 + B22

1 1 ∆ 11 11. (c) ⋅ m1v 2 = JJ mm11ss ∆θ 2 2

B2 P a

( 50 ) v2 == = 1.4 °C 44Js 4 × 4.2 × 103 × 0.105 2

⇒

∆ 1= = ∆θ

2

 µ 2i   µ 2i  =  0 1  + 0 2   4π a   4π a 

12. (c) Point P lies at equatorial positions of dipole 1 and 2 and axial position of dipole 3. Hence field at P 1

+Q

–Q E1

–Q

2

+Q

k. p (towards left) x3 k. p Due to dipole 2 E2 = 2 (towards left) x k .(2 p ) Due to dipole 3 (towards right) E3 = x3 So, Net field at P will be zero.

13. (c) The two capacitor C1 and C2 are connected in series 6 mm 4 mm Kε A K ε A C1 = 1 0 , C2 = 2 0 d1 d2

1 1 1 d1 d2 = + = , C C1 C2 K1 ε 0 A K 2 ε 0 A

1 14 ×10−3 ε 0 A 10

This gives C =

2 m/s = 2m /s 1

5000 ε0 A 7

N

S v1

v2

17. (b) Since voltage is lagging behind the current, so there must be no inductor in the box. 18. (b)

I f ; where u = f + x = O f −u

∴

I f =− O x

19. (a) Kinetic energy of recoil electron, Ek = hv − hv′

= 12.375 − 9.375 = 3eV 20. (b) Vb = ib Rb ⇒ Rb =

1  d1 d 2  + ε 0 A  K1 K 2 

1  6 ×10−3 4 × 10−3  = +  ε 0 A  10 5  =

(i12 + i22 )1/ 2

E1 =

Due to dipole 1

=

µ0 2π a

1 = 2 m/s m/s 0.5 Relative speed between coil and magnet is zero, so there is no induced emf in the coil.

3

E3

=

Speed of the coil v2 =

P E2

2

16. (a) Speed of the magnet v1 =

+Q

D

O i2

B

m 1 ∆θ 2 = 1 ∆θ1 = ×1.4 = 7°C m2 0.2

–Q

B1 A i1

C

m1 s ∆θ1 = m2 s ∆θ 2 ⇒

BV BH

9 = 257 kkΩ Ω 35 ×10−6

21. (c) s1 + s2 = 100 10 mm

⇒

1 2 1 gt + ut − gt 2 = 100 m 2 2

⇒

ut = 100 m ⇒ t =

100 100 = == 1ss 100 u

55

Mock Test-4

22. (d) x = 4(cos πt + sin πt )

1  1  4 2 cos π t + sin π t  2 2  

π π  π   = 4 2 sin cos π t + cos sin π t  = 4 2 sin  π t +  4 2  2  

⇒

M shell =

⇒

σ shell =

2

2l

θ

2Y

θ1

(a) P′(54 − h) = P0 8 (Boyel’s Law)

⇒

P′ =

l

⇒

8 P0 + ρ gh = P0 54 − h

Y

⇒

ρ gh = 

32 = 16 2

θ2

16 = 14cm π−2 24. (b) Let θ be the temperature of the junction (say B). Thermal resistance of all the three rods is equal. Rate of heat flow through AB + Rate of heat flow through CB = Rate of heat flow through BD

⇒

P0 8 54 − h

A 90°C

(54-h) P0

⇒

⇒

0°C 90°C C

P’

4.

 46 − h  h=  (76)  54 − h 

⇒ 54h − h2 = 76 × 46 − 76h

h2 − 130h + 76 × 46 = 0 ⇒ ( h − 38) ( h − 92) = 0 h = 38,92 ⇒ h = 38cm ↓ Not possible (a) Pressure at 0 level is same from both sides.

90°− θ 90°− θ θ − 0 + = , Here, R =Thermal resistance R R R 3θ = 180°, or θ = 60°C

∴ ∴

25. (c) Resistance

( 6 × 5) = 30 Ω

2.

3.

4.

5.

6.

7.

d1 (1 − sin α ) = d1 (1 − cos α ) + d 2 [cos α + sin α ]

8.

9.

10.

6

2

b

a

a

a

c

a,b,d

4

2

12.

13.

14.

15.

16.

17.

18.

6

6

4

2

d

a

d

c

2. ⇒

(b) P1 = pressure just inside the bubble at the end 4T 2 = P0 + R (a) If water is half filled M shell g = ρ ( ∆V ) displaced g

V  ( ∆V ) =  shell   2 

d

α

0

11.

1.

α α

and 6 Ω are in parallel; so

30 × 6 =5Ω equivalent resistance R = 30 + 6 JEE Advance Paper -I 1.

54 cm

h

⇒

B

⇒ P′ + ρ gh = P0

 46 − h   P0 ( P0 = ρ g (76))  54 − h 

r=

D

1 ⇒ R.d = . 2

3.

a=4 2 =

ρliquid 2

 M  ρliquid ⇒  shell  = 2  Vshell 

1 , Then the cylinder should be more than half 2 filled so that it is half submerged and floating.

Comparing the given equation with standard equation λ

2

If R.d
P1 > P3

5.

(d) Power ∝1/ R

Using equation (i) and (ii) n sin(60° − r1 ) = sin θ

6.

(c) G1 is acting as voltmeter and G 2 is acting as ammeter.

7.

(3) For v1 =

 3  1 n  cos r1 − sin r1  = sin θ 2  2 

4.

(c) P =

⇒

Snell’s Law on 2

50 m, u1 = −25m 7

25 18 × = 3 kmph. 30 5

7 −1 7 1 (2) − = 4 4V1 −24 6

⇒ V1 = 21 cm

4/3 7/4 ⇒ − =0 V2 21 ⇒ V2 = 16 cm ⇒ x = 18 − 16 = 2 cm

(7) Image by mirror is formed at 30 cm from mirror at its right and finally by the combination it is formed at 20 cm on right of the lens. So in air medium, magnification by

7 lens is unity. In second medium, µ = , focal length of 6 the lens is given by,  1 1  1 (1.5 − 1)  −   R1 R2  10 = 1  1.5 1   1 − 1  −   f  7 / 6   R1 R2  35 cm 2 So, in second medium, final image is formed at 140 cm to the right of the lens. Second medium does not change the magnification by mirror. ⇒

. . .(ii) nd

Speed of object =

9.

3 4n 2 − 3 = 4n 2 2n

r1 + r2 = 60°

d  3  dn  4

25 ⇒ v2 = m, u2 = −50m 3

8.

cos r1 = 1 −

(

surface: n sin r2 = sin θ

)

 dθ 4n2 − 3 − 1  = cos θ dn 

For θ = 60 ° and n = 3 ⇒

dθ =2 dn

1 2 mv = qV 2 h λ= mv

11. (3) ⇒ ⇒

λ = 8 ≃ 3.

12. (7) Stopping potential =

hc

λ

−W

= 6.2 eV − 4.7 eV = 1.5 eV

⇒

V =

Kq = 1.5 r

⇒

n=

1.5 × 10 −2 = 1.05 × 107 9 × 109 × 1.6 × 10 −19

⇒

Z =7

13. (7)

∴

f =

1 2 1 (120)e2 [By Conservation of energy] mv∞ = 2 4πε 0 (10 fm)

Assuming the nucleus to be considerably massive, we can disregard its motion. Let momentum of proton be p = mv∞

∴

1 (120)e2 p2 = 2m p 4πε 0 (10 × 10−15 )

∴

5 120 × e 2 p = 9 × 109 × 2 × × 10 −27 × 3 10 −14

72

Physics

⇒

p = 30 × 120 × 10

∴

p = 3600 × 10 −4 × e 2

∴

p = 60 × 10 −2 × e

∴

λ=

∴

λ = 7fm

9 − 27 +14

×e

Primers contain materials which make obtuse angle of contact due to their surface tension and do not allow capillary rise of water in pours of bricks. Interstellar dust particles attract each other by gravitational pull to build a star. Oil has smaller surface tension than water and is lighter than water so it spreads move over the surface of water to decrease height of tides.

2

h 42 × 10−15 42 h = = × 10−15 m = p e × 60 × 105 60 × 10−2 6

14. (1) Slope of graph is h/e = constant ⇒ 1

17. (c) A-1; B-2; C-3; D-4

15. (a) A→1, B→3, C→3, D→1,2,3 A, C and D in case of concave mirror convex lens image can be real, virtual, diminished magnified or of same size. B is case of convex mirror image is always virtual (for real object) 16. (d) A-3; B-4; C-2; D-1 According to Bernoulli’s theorem the spinning of ball along with horizontal throw will change the velocity of streamlines of air above and below the ball which will change the pressure on ball above and below it which cause change in its direction of motion. This is called magnus effect.

18. (b) A-4; B-3; C-2; D-1 F l Longitudinal stress Y= × = = Young’s modulus A ∆l Longitudinal strain

F V Normal stress × = = Bulk modulus A ∆V Volumetric strain = Volume elasticity Lateral strain Poisson’s ratio = Longitudinal strain K=

⇒






∆D ∆D l σ= D = × ∆l D ∆l l

Mock Test-1 Test Booklet code

73

A

Mock Test 1 “JEE-Main”

Do not open this Test Booklet until you are asked to do so. Read carefully the Instructions on the Back Cover of this Test Booklet. Important Instructions: 1.

Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited.

2.

The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully.

3.

The test is of 3 hours duration.

4.

The Test Booklet consists of 90 questions. The maximum marks are 360.

5.

There are three parts in the question paper A, B, C consisting of, Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.

6.

Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each question. 1/4 (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.

7.

There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.

8.

Use Blue/Black Ball Point Pen only for writing particulars/ marking responses on Side-1 and Side-2 of the Answer Sheet. Use of pencil is strictly prohibited.

9.

No candidates is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc., except the Admit Card inside the examination hall/room.

10. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each page and at the end of the booklet. 11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room / Hall. However, the candidates are allowed to take away this Test Booklet with them. 12. The CODE for this Booklet is A. Make sure that the CODE printed on Side-2 of the Answer Sheet and also tally the serial number of the Test Booklet and Answer Sheet are the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the invigilator for replacement of both the Test Booklet and the Answer Sheet. 13. Do not fold or make any stray marks on the Answer Sheet. Name of the Candidate (in Capital letters): Roll Number :

in figures  in words

Examination Centre Number:  Name of Examination Centre (in Capital letters): Candidate’s Signature:

Invigilator’s Signature:

74

Chemistry

Read the following instructions carefully: 1.

The candidates should fill in the required particulars on the Test Booklet and Answer Sheet (Side-1) with Blue/Black Ball Point Pen.

2.

For writing/marking particulars on Side-2 of the Answer Sheet, use Blue/Black Ball Point Pen only.

3.

The candidates should not write their Roll Numbers anywhere else (except in the specified space) on the Test Booklet/Answer Sheet.

4.

Out of the four options given for each question, only one option is the correct answer.

5.

For each incorrect response, one-fourth (¼) of the total marks allotted to the question would be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Sheet.

6.

Handle the Test Booklet and Answer Sheet with care, as under no circumstances (except for discrepancy in Test Booklet Code and Answer Sheet Code), another set will be provided.

7.

The candidates are not allowed to do any rough work or writing work on the Answer Sheet. All calculations/writing work are to be done in the space provided for this purpose in the Test Booklet itself, marked ‘Space for Rough Work’. This space is given at the bottom of each page and at the end of the booklet.

8.

On completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.

9.

Each candidate must show on demand his/her Admit Card to the Invigilator.

10. No candidate, without special permission of the Superintendent or Invigilator, should leave his/her seat. 11. The candidates should not leave the Examination Hall without handing over their Answer Sheet to the Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the Attendance Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with as an unfair means case. The candidates are also required to put their left hand THUMB impression in the space provided in the Attendance Sheet. 12. Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited 13. The candidates are governed by all Rules and Regulations of the JAB/Board with regard to their conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the JAB/Board. 14. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances. 15. Candidates are not allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, electronic device or any other material except the Admit Card inside the examination hall/room.

Mock Test-1

75

JEE-MAIN: CHEMISTRY MOCK TEST-1 1.

2.

3.

Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately a. 1.86 Å b. 3.22 Å c. 5.72 Å d. 0.93 Å Which of the following is the energy of a possible excited state of hydrogen? a. 13.6 eV b.  6.8 eV c.  3.4 eV

4.

d.  6.8 eV

The number of atoms in 100 g of an fcc crystal with –3 density =10.0 g cm and cell edge equal to 200 pm is

equal to a. 5  10 24 5.

b. 5  10 25

c. 6  10 23

d. 2  10 25

The following reaction is performed at 298 K.

2NO(g)  O2 (g)

2NO2 (g)

The standard free energy of formation of NO (g) is 86.6 kJ/mol at 298 K. What is the standard free energy of 12 formation of NO 2 (g) at 298 K? (Kp  1.6 10 )

a. R(298) n (1.6  10 )  86600 12

b. 86600  R(298) n (1.6  1012 ) c. 86600 

n (1.6  1012 ) R(298)

The vapour pressure of acetone at 20C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20  C, its vapour pressure was 183 torr. The molar mass (g mol–1) of the substance is: a. 32 c. 128

7.

b. 64 d. 488

The standard Gibbs energy change at 300 K for the reaction 2A

b. reverse direction because Q  K c c. forward direction because Q  K c d. reverse direction because Q  K c 8.

Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is: a. 0 g b. 63.5 g c. 2 g d. 127 g

9.

What is the half life period of a radioactive substance if 87.5% of any given amount of the substance disintegrates in 40 minutes? a. 160 min b. 10 min c. 20 min d. 13 min 20 sec

10. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is : a. 18 mg b. 36 mg c. 42 mg d. 54 mg 11. The ionic radii (in Å) of N 3 , O 2 and F are respectively: a. 1.36, 1.40 and 1.71 c. 1.71, 1.40 and 1.36

b. 1.36, 1.71 and 1.40 d. 1.71, 1.36 and 1.40

12. In the context of the Hall-Heroult process for the extraction of A , which of the following statements is

false? a. CO and CO 2 are produced in this process b. A 2 O3 is mixed with CaF2 which lowers the melting

d. 0.5[2  86600  R(298) n (1.6 1012 )] 6.

a. forward direction because Q  K c

In a galvanic cell, the salt bridge a. does not participate chemically in the cell reaction. b. stops the diffusion of ions from one electrode to another. c. is necessary for the occurrence of the cell reaction d. ensures mixing of the two electrolytic solutions

B  C is 2494.2 J. At a given time, the

composition of the reaction mixture is [A]  12 , [B]  2 and [C]  12 . The reaction proceeds in the: [ R  8.314 J / K / mol, e  2718 ]

point of the mixture and brings conductivity c. A 3 is reduced at the cathode to form A d. Na 3A F6 serves as the electrolyte 13. From the following statements regarding H 2 O 2 , choose

the incorrect statement a. It can act only as an oxidising agent b. It decomposes on exposure to light c. It has to be stored in plastic or wax lined glass bottles in dark d. It has to be kept away from dust 14. Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy? a. CaSO 4 b. BeSO 4 c. BaSO 4

d. SrSO 4

76

Chemistry

15. Which among the following is the most reactive? a. Cl 2

b. Br2

c. I 2

16. Which one has the highest boiling point? a. He b. Ne c. Kr

d. ICl

21. Which compound would give 5-keto-2-methyl hexanal upon ozonolysis? CH 3

d. Xe

CH 3

17. The number of geometric isomers that can exist for square

CH 3

a.

b.

CH 3

planar [Pt(Cl)(py) (NH 3 )(NH 2 OH)] is (py = pyridine): a. 2

b. 3

c. 4

CH 3

d. 6

CH 3

18. The color of KMnO 4 is due to a. M  L charge transfer transition

c.

b. d  d transition c. L  M charge transfer transition d.    * transition 19. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is: (Atomic mass Ag  108, Br  80) a. 24

b. 36

c. 48

d. 60

23. In the following sequence of reactions: KMnO 2 SOCl2 H 2 / Pd Toluene  A   B   C, the BaSO 4

product C is:

CH3

a. C6 H5COOH

b. C6 H5CH3

c. C6 H5CH 2OH

d. C6 H 5CHO

24. In the reaction NH3

O

a.

CH 3

22. The synthesis of alkyl fluorides is best accomplished by : a. Free radical fluorination b. Sandmeyer’s reaction c. Finkelstein reaction d. Swarts reaction

20. In the reaction shown below, the major product(s) formed is/are H | N

NH 2

NaNO 2 / HCl CuCN / KCN   D    E  N2 The product E 0  5 C 

O

NH 2

CH3 is

H | N

b. O

CH 3

+ CH 3 COOH

COOH

O

H | N

a.

CH 3

O c.

CH 3

c.

O



H | N

CH3

CH 3

O

CN

+ H 2O

N H 3CH 3COO 

O

CH3

b.

CH3

H | N

O

d.

d. H3C

+ H 2O

d. H

3

CH

25. Which polymer is used in the manufacture of paints and lacquers? a. Bakelite b. Glyptal c. Polypropene d. Poly vinyl chloride

Mock Test-1

77

26. Which of the vitamins given below is water soluble? a. Vitamin C b. Vitamin D c. Vitamin E d. Vitamin K 27. Which of the following compounds is not colored yellow? a. Zn 2 [Fe(CN)6 ] b. K 3 [Co(NO 2 )6 ] c. (NH 4 )3 [As(Mo3O10 ) 4 ]

d. BaCrO 4

28. Match the statement of Column with those in Column II: Column I Column II H 1. Pseudo first (A) C12 H22 O11  H 2 O  order C 6 H12 O 6  C 6 H12 O 6 HOH (B) CH 3 COOC 2 H 5   H  or OH 

2. Zero order

CH 3 COOH  C 2 H 5 OH hv (C) H 2  Cl 2   2HCl

3. Second order

(D) CH3Cl  OH   CH3OH  Cl

4. First order

a. A1; B3, 4; C2; D3 b. A2; B4; C3; D1 c. A1; B3, 2; C2; D4 d. A4; B1; C3; D2 Space for rough work

Assertion and Reason Note: Read the Assertion (A) and Reason (R) carefully to mark the correct option out of the options given below: a. If both assertion and reason are true and the reason is the correct explanation of the assertion. b. If both assertion and reason are true but reason is not the correct explanation of the assertion. c. If assertion is true but reason is false. d. If the assertion and reason both are false. e. If assertion is false but reason is true 29. Assertion: Enthalpy and entropy of any elementary substance in the standard state are taken as zero. Reason: At zero degree absolute, the constituent particles become completely motionless. 30. Assertion: Molecularity has no meaning for a complex reaction. Reason: The overall molecularity of a complex reaction is equal to the molecularity of the slowest step.

78

Chemistry

JEE ADVANCE PAPER-I Time 3 Hours. Read The Instructions Carefully

Max. Marks 264 (88 for Chemistry)

Question Paper Format and Marking Scheme: 1. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections. 2. Section 1 contains 8 questions. The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive). Marking Scheme: +4 for correct answer and 0 in all other cases. 3. Section 2 contains 10 multiple choice questions with one or more than one correct option. Marking Scheme: +4 for correct answer, 0 if not attempted and –2 in all other cases. 4. Section 3 contains 2 “match the following” type questions and you will have to match entries in Column I with the entries in Column II. Marking Scheme: for each entry in Column I, +2 for correct answer, 0 if not attempted and – 1 in all other cases.

NOTE: It’s the mock test as per previous year’s papers but sometimes IIT changes the test paper pattern and marking scheme too.

SECTION 1 (Maximum Marks: 32) 

This section contains EIGHT questions.



The answer to each question is a SINGLE DIGIT INTEGER

4.

The total number of lone pairs of electrons in N 2 O 3 is

5.

For the octahedral complexes of

difference between the spin-only magnetic moments in Bohr magnetons (When approximated to the nearest integer) is [Atomic number of Fe = 26]

For each question, darken the bubble corresponding to the correct integer in the ORS.



1.

Marking scheme: +4

If the bubble corresponding to the answer is darkened.

0

In all other cases.

6.

The total number of stereoisomers that can exist for M is CH 3 H 3C

H 3C

3.

O M The number of resonance structures for N is OH NaOH  N

Among the triatomic molecules/ions, BeCl 2 , N 3 , N 2 O, NO 2 , O 3 ,SCl 2 , ICl 2 , I3 and XeF2 , the total number of

If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is 0.0558C, the number of

linear molecule(s)/ion(s) where the hybridization of the central atom does not have contribution from the dorbital(s) is [Atomic number: S = 16, Cl = 17, I = 53 and Xe = 54]

chloride(s) in the coordination sphere of the complex is [ K f of water = 1.86 K kg mol–1] 2.

SCN 

(thiocyanato-S) and in CN  ligand environments, the

ranging from 0 to 9, both inclusive. 

Fe3 in

7.

Not considering the electronic spin, the degeneracy of the second excited state (n = 3) of H atom is 9, while the degeneracy of the second excited state of H  is

8.

All the energy released from the reaction X  Y.  r G 0  193 kJ mol 1 is used for oxidizing M  as M   M3

2e , E 0  0.25V. Under

standard

conditions,

the

number of moles of M  oxidized when one mole of X is converted to Y is

[F  96500 C mol –1 ]

Mock Test-1

79

SECTION 2 (Maximum Marks: 40)    

9.

CH 3

This section contains TEN questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. Marking scheme: +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened. 0 If none of the bubbles is darkened –2 In all other cases 

 3

 4

For the reaction: I  ClO  H 2SO 4   HSO  I 2

CH 3

c.

14. The major product of the reaction is H 3C CO 2 H NaNO2 , aqueous HCl  0 C

CH 3

 a. Stoichiometric coefficient of HSO4 is 6

H 3C

c.

H

a.

Br

CH 3

CH 3

d. H 3C

H

Br

c.

2

CH 3

CH 3

O a.

O b.

O

O CH 3

c.

d.

12. In the following reaction, the major product is CH 3 1 equivalent HBr CH 2   H 3C CH 3

a. H 3C

Br

CH 3

CH 3

b.

H 3C

Br

OH

CO2 H CH3

OH NH 2

H 3C

CO 2 H CH 3

d.

CH 3

OH

15. The reactivity of compound Z with different halogens under appropriate conditions is given below:

Z



H3C

b.

OH

H 3C

CH 3

KOH , H O    C H 3 i. ii. H , heat

NH 2 CH 3

OH

11. The major product of the following reaction is O

O

NH 2

H 3C

10. Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is (are) Br H H Br CH3 CH 3 a. H3C b. H 2 C

Br

H 3C

13. The correct statement(s) for orthoboric acid is/are a. It behaves as a weak acid in water due to self ionisation b. Acidity of its aqueous solution increases upon addition of ethylene glycol c. It has a three dimensional structure due to hydrogen bonding. d. It is a weak electrolyte in water

The correct statement(s) in the balanced equation is/are b. Iodide is oxidised c. Sulphur is reduced d. H 2 O is one of the products.

d.

Br

H 3C

mono halo substituted derivative when X 2  I 2

X2

di halo substituted derivative when X 2  Br2 tri halo substituted derivative when X 2  Cl 2

C(CH 3 )3

The observed pattern of electrophilic substitution can be explained by a. the steric effect of the halogen b. the steric effect of the tert-butyl group c. the electronic effect of the phenolic group d. the electronic effect of the tert-butyl group 16. An ideal gas in thermally insulated vessel at internal pressure  P1 , volume  V1 and absolute temperature

 T1 expands irreversibly against zero external presssure, CH 3

as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are P2 , V2 and T2 , respectively. For this expansion.

Pext  0

Pext  0 P1 , V2 ,T2

P1 , V2 ,T2 Thermal insulation

80

17.

Chemistry

a. q  0

b. T2  T1

c. P2 V2  P1V1

d. P2 V2  P1V1



Fe3 is reduced to Fe2  by using a. H 2 O2 in presence of NaOH



b. Na 2 O 2 in water c. H 2 O2 in presence of H 2SO 4 d. Na 2 O 2 in presence of H 2SO 4 18. The % yield of ammonia as a function of time in the

reaction N 2 (g)  3H 2 (g)

2NH 3 (g), H  0 at (P, T1 )

is given below:

For each entry in Column I, darken the bubbles of all the matching entries. For example, if entry (A) in Column I, matches with entries (2), (3) and (4), then darken these three bubbles in the ORS. Similarly, for entries (B), (C) and (D). Marking scheme: For each entry in Column I +2 If only the bubble(s) corresponding to all the correct match(es) is(are) darkened 0 If none of the bubbles is darkened –1 In all other cases

19. Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from Column -I with an appropriate structure from Column –II is: P   R  R O  CO 

T1

1

O R

time

If this reaction is conducted at (P, T2 ), with T2  T1 , the

O R O (Peroxyester)

% Yield

% Yield

1

Column I (A) Pathway A

time

T2

% Yield

T1

d. time

(B) Pathway B



This section contains TWO questions.



Each question contains two columns, Column I and Column II



Column I has four entries (A), (B), (C) and (D)



Column II has five entries (1), (2), (3), (4) and (5)



Match the entries in Column I with the entries in Column II



One or more entries in Column I may match with one or more entries in Column II The ORS contains a 4  5 matrix whose layout will be similar to (A) (1) (2) (3) (4) (5) (B) (1) (2) (3) (4) (5) (C) (1) (2) (3) (4) (5) (D) (1) (2) (3) (4) (5)

2.

(C) Pathway C

3.

O O

4.

CH 3

O O CH 3 O CH 2 C6 H5 CH 3

C6 H5 CH 2

(D) Pathway D

O O

C6 H 5

CH 3

O

C6 H 5

SECTION 3 (Maximum Marks: 16)

the one shown below:

O O

C6 H5 CH 2

T2

time



Column II O 1.

time

T1 % Yield

1

T1

b.

c.

R   RCO2  R O    CO   CO  1

T2

a.

1

R  X  carbonyl compound  R  X  carbonyl compound  S    RCO 2  R O    R  R O  CO 

% yield of ammonia as a function of time is represented by T1

Q   R  R O    CO 

O

CH 3

C6 H5 CH 3

a. →1; B→3; C→4; D→2 b. →2; B→4; C→4; D→1 c. →1; B→1; C→2; D→4 d. →3.; B→2; C→1; D→3 20. Match the orbital overlap figures shown in Column -I with the description given in Column -II and select the correct

Mock Test-1

81

answer using the code given below the lists. (en H 2 NCH 2 CH 2 NH 2 ; atomic numbers :

3.

p-d  bonding

4.

d-d  antibonding

(C)

Ti  22, Cr  24, Co  27; Pt  78) Column I

Column II 1.

p-d  antibonding

2.

d-d  bonding

(A)

(D) a. →2; B→3; C→1; D→4

(B)

b. →2; B→4; C→4; D→1 c. →1; B→1; C→2; D→4 d. →3.; B→2; C→1; D→3

Space for rough work

82

Chemistry

JEE ADVANCE PAPER-II Time 3 Hours. Read The Instructions Carefully

Max. Marks 240 (80 for Chemistry)

Question Paper Format and Marking Scheme: 1. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections. 2. Section 1 contains 8 questions. The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive). Marking Scheme: +4 for correct answer and 0 in all other cases. 3. Section 2 contains 8 multiple choice questions with one or more than one correct option. Marking Scheme: +4 for correct answer, 0 if not attempted and –2 in all other cases. 4. Section 3 contains 2 “paragraph” type questions. Each paragraph describes an experiment, a situation or a problem. Two multiple choice questions will be asked based on this paragraph. One or more than one option can be correct. Marking Scheme: +4 for correct answer, 0 if not attempted and – 2 in all other cases.

NOTE: It’s the mock test as per previous year’s papers but sometimes IIT changes the test paper pattern and marking scheme too. SECTION 1 (Maximum Marks: 32)

II

This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS. Marking scheme: +4 If the bubble corresponding to the answer is darkened. 0 In all other cases.

1.

In dilute aqueous H 2SO 4 , the complex diaquodioxalato-

III

IV

4.

H2O   → 100 ° C

COCl H2  → Pd − BaSO 4

CO2 Me DIBAL-H  → Toluene, −78° C H 2 O

The crystal of a solid is square packing of identical

ratio of the rate of change of [H + ] to the rate of change of

spheres in each layer and spheres of one layer are placed just above the voids made by spheres in previous layer. The packing efficiency of such crystal (in %) is

[MnO −4 ] is

[π = 3.15, 2 = 1.4]

ferrate (II) is oxidized by MnO −4 . For this reaction, the

2.

CHCl2

The number of hydroxyl group(s) in Q is +

aqueous dilute KMnO4 (excess) H   →P  →Q heat 0°C

H HO

H3C 3.

CH3

Among the following, the number of reaction(s) that produce(s) benzaldehyde is I

CO, HCl  → Anhydrous AlCl3 / CuCl

5.

Among the complex ions, [Co(NH 2 - CH 2 - CH 2 -NH2)2 Cl 2 ]+ , [CrCl 2 (C 2 O 4 ) 2 ]3– , [Fe(H 2 O) 4 (OH) 2 ]+ , [Fe(NH3)2] (CN) 4 ]− , [Co(NH 2 − CH 2 − CH 2 − NH 2 ) 2 (NH 3 )Cl]2 + and

[Co(NH 3 ) 4 (H 2 O)Cl]2+ , the number of complex ion(s) that

show(s) cis-trans isomerism is

Mock Test-1

6.

83

methanol. The number of moles of boron containing product formed is 7.

The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity

O a.

U

O

O

H

H | O

O

is (consider degree of ionisation of both acids to be O2− > F−

Zeff

(10.0 g cm –3 × (200 pm)3 × 10−30 cm 3 × 6 × 10 23 atoms) 4 −1

=

Thus, 12gmol contains = N A atoms = 6 × 10 atoms 23

6 × 10 23 × 100 = 5 × 10 24 atoms 12

12. (d) 13. (a) It can act as an oxidising as well as reducing agent. 14. (b) BaSO 4 is least soluble, BeSO 4 is most soluble. 15. (d) The interhalogen compounds are generally more reactive than halogens (except F2).

∴

100 g contains =

5.

(d) ∆GRxn = 2∆GNO2 − 2∆G NO

16. (d) Xe has the highest boiling point.

∆G NO2 = ∆G NO + 12 ∆G Rx

17. (b) No. of Geometrical isomers of

= ∆G NO + 12 (− RT ln e K p )

6.

[Pt(Cl)(py)(NH 3 )(NH 2 OH)]+ = 3

= 0.5[2 × 86600 − R(298) ln (1.6 × 1012 )]

18. (c)

n1 P0 − P = Xsolute = (b) P0 n1 + n 2

19. (a) % of Bromine =

⇒

185 − 183 2 1.2 / M = = 185 185 1.2 + 100 M 58 M = 64

7.

(b) ∆G = −RT ln e K c

20. (a) Only amines undergo acetylation and not acid amides. O NH

1 = 0.36 2.718

CH 3

O

O H3C

C

C

C

O

+ CH 3 COOH

OH

C

NH 2

O

21. (b) CH3

CH3

(B)(C) 2 × 12 Q= = =4 [A]2 [1/ 2]2 Q > K c , i.e. backward reaction.

C

NH 2

2494.2 = 8.314 × 300 ln e K c ⇒ K c = e −1

K c = e −1 =

80 141 mg × ×100 = 24% 188 250 mg

O3  → Zn / H 2 O

CH3

O CHO CH3

5-keto-2-methyl hexanal

NH 2

86

Chemistry

Total no. of stereoisomers of M = 2 CH 3 H 3C

22. (d) R  I  AgF   R  F  AgI (Swarts Reaction) CH 3

23. (d)

COOH SOCl2  

KMnO 2 

Toluene

H 3C

Benzoic acid

COCl

CHO

3.

O

(9)

HO H 2 / Pd   BaSO4

Benzoyl chloride

NaOH  N

Benzaldehyde

O–

24. (c) 

N Cl

NH3

N is

CN

O–

 



NaNO2 / HCl 05 C

CH3

CH3

O

O

 N2

CuCN / KCN 

1

CH3

2 O

O

25. (b) Glyptal is used in the manufacture of paints and lacquers.

4

26. (a) Vitamin ‘B’ and ‘C’ are water soluble.

28. (a) A1; B3, 4; C2; D3 (A) Inversion of cane sugar is pseudo first order reaction. (B) Hydrolysis of ester in the presence of acid is first order while in the presence of base is second order reaction. (C) Photochemical reactions are of zero order. (D) SN2 reactions are of second order.

7

4.

5.

29. (c) Enthalpy is zero but entropy is not zero. Vibrational motion exists even at absolute zero.

O

5 O

27. (a) Zn 2 [Fe(CN)6 ] is bluish white ppt.

3 –

6 O–

O 8

9

O:

+

:O  N –– N

(8) :O  N

N  O O Total no. of lone pairs = 8

:O:

(4) [Fe(SCN) 6 ]3 and [Fe(CN) 6 ]3

In both the cases the electronic configuration of Fe3 will be 1s 2 , 2s 2 , 2p 6 ,3s 2 ,3p 6 ,3d 5 .

30. (b) Molecularity of a reaction can be defined only for an elementary reaction because complex reaction does not take place in one single step and it is almost impossible for all the total molecules of the reactants to be in a state of encounter simultaneously.

Since SCN is a weak field ligand and CN is a strong field ligand, the pairing will occur only in case of [Fe(CN)6 ]3 . Case-1 3d 5

(high spin) (no pairing)

JEE Advance Paper -I 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

1

2

9

8

4

4

3

4

a,b,d

b,d

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

a

d

b,d

c

a,b,c

a,b,c

a,b

b

a

a

1.

Case-2

Case – 1   n(n  2)  5(5  2)  35  5.91 BM Case – 2   n(n  2)  1(1  2)  3  1.73 BM

(1) T f  iK f m

Difference in spin only magnetic moment  5.91  1.73  4.18  4

0.0558  i  1.86  0.01  i  3



Complex is [Co(NH 3 )5 Cl]Cl2 .

2.

(2) Bridging does not allow the other 2 variants to exist.

(low spin) (pairing)

6.

(4) BeCl 2 , N 3 , N 2 O, NO 2 , O 3 ,SCl 2 , ICl 2 , I3 , XeF2

Mock Test-1

87

BeC 2   sp   linear

 (2) H 3 C  CH — C — CH 2 — CH 3  H 2 /Pt

 3

N   sp   linear

H

Br H 3 C — CH — C — CH 2 — CH 3

N 2 O   sp   linear

H Br Optically inactive



NO 2   sp   linear O 3   sp 2   linear

H 2 /Pt  (3) H 2 C  C — C — CH 3  H2C H H Br H 3 C — C — C — CH3

SCl 2   sp 3   linear I 3   sp 3 d   linear

ICl 2   sp 3 d   linear

H2C

H Br Optically inactive

XeF2   sp 3d   linear

So, among the following only four (4) has linear shape and no d-orbital is involved in hybridization. 7.

H /Pt  (4) H2 C  CH —C — CH2 —CH3  2

(3) Single electron species don’t follow the (n  ) rule but

H

Br

H3C — CH2 — C—CH2 —CH3

multi electron species do.

H Br Optically inactive

Ground state of H  1s2 First excited state of H   1s1 , 2s1

O

11. (a)

Second excited state of H   1s1 , 2s 0 , 2p1



CH 2 O

Px Py Pz

:OH 

CH 3

O

H2O  

(3 degenerate orbitals)

8.

(4) X   Y  r G  193KJ / mol 0

O

0 M    M 3   2e  E  0.25V

CH 3

G for the this reaction is 0

O

G   nFE  2  ( 0.25)  96500  48250 J / mol 0

0

48.25 KJ/mol So the number of moles of M  oxidized using X   Y will be 

9.

12. (d)

CH 3 OH H+ ,H2O/Δ  

O

CH 3 | H   Br  H 2 C  C — CH  CH 2  

193  4 moles 48.25 

CH CH 3

CH 3 |

 3

 (a, b, d) 6I  ClO  6H 2SO4 

H 3 C — C — CH  CH 2  

Cl   6HSO 4  3I 2  3H 2 O

10. (b, d)

H3C

CH 3

(1) H 3C — CH  CH — C H

H3C



:



Br C  CH — CH2  

H3C H3C

C  CH — CH2 — Br

13. (b, d) (a) H 3 BO 3 is a weak monobasic Lewis acid. Br

CH 3

  H 3C — CH 2 — CH 2 — C H 2 /Pt

H Br Optically active

H3 BO3  H  OH

B(OH)4  H 

. . .(i)

(b) Equilibrium (i) is shifted in forward direction by the addition of syn-diols like ethylene glycol which forms a

stable complex with B(OH) 4 .

88

Chemistry

O

H

H

O

H O

forward reaction would be greater than at lower temperature that is why the percentage yield of NH3 too would be more initially.

O

H

19. (a) →1; B→3; C→4; D→2 O O O CH3 (A) H5C6CH2

B H

O

O

H

O H –

O

O

H O

 4H 2 O

B O

O

(stable complex)

(c) It has a planar sheet like structure due to hydrogen bonding. (d) H3BO3 is a weak electrolyte in water.

C6 H5  CH2  CO2  CH2O O CH 3 O CH 3 O (B) H 5 C6 CH 2 CH 2 C6 H 5

O | C6 H5 — C H 2  CO2  Ph — CH 2 — C — CH3 | CH3

14. (c) H3C H3C

H3C H3C

OH | C  O CH —CH2 —CH

NaNO2 / HCl   

NH2

H3C H3C

:OH | C  O 

N

CH —CH2 —CH —COOH   H2O OH

H3C H3C

  Ph — C H 2  CH3 — CO — CH3

CH —CH2 —HC



CH —CH2 —HC

O

N

O

O H

O | C6 H5 — CO 2  CH3  C — CH3 | CH3

I I2   Rxn (i)

CMe 3

OH

OH

O O CMe 3 Br OH

Cl Cl2  Rxn (iii)

Cl

2OH

C6 H 5

O

CH 3

20. (a)→2; B→3; C→1; D→4

CMe 3

(A)

16. (a, b, c) Since container is thermally insulated so, q = 0, and it is a case of free expansion therefore W = 0 and E 0. So, T1 T2 . Also, P1V1 P2 V2 . H 2 O2

(D)

C6 H 5  CO 2  CH 3O   C6 H 5  CO2

Cl

3

  Ph  CH 3 — CO — Ph  CH 3

Br Br2   Rxn (ii)

CMe 3

17. (a, b) 2Fe

C6 H 5

C== O

OH

15. (a, b, c)

O

(C) C6 H 5

CH 3 CH 3

2Fe

2

  d-d  bonding

(B)

  p-d  bonding

(C)

  p-d  antibonding

2H2 O O2

Na 2 O2  H 2 O   H 2 O 2  NaOH 18. (b) N 2 (g)  2H 2 (g) N 2 (g)  2H 2 (g)

2NH 3 (g); H  0

Increasing the temperature lowers equilibrium yield of ammonia. However, at higher temperature the initial rate of

(D)

  d-d  antibonding

Mock Test-1

89

JEE Advance Paper-II 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

8

4

4

75

5

6

3

9

a

a

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

1 is shared in second layer 2 1 Z   4  0.5  1 8

b

a

b,c

c,d

b

b,c,d

a

b

c

d

(2r 2 )  h2  x 2 fi h = 2r

1.



Area = 4r 2 [ a 2 = (2r ) 2 fi a = 2r ]

(8) [Fe(C 2 O 4 )(H 2 O)]2   MnO 24   8H   

Vol  4r 2  2r  4 2r 3 4 1  r 3  3 D   0.75  75% 3 4 2r 3 2

Mn 2   Fe3  4CO 2  6H 2 O

So,

the ratio of rate of change of [H  ] to that of rate of change of [MnO4 ] is 8.

2.

(4)

5.

(5) [Co(en) 2 Cl 2 ]   will show cis-trans isomerism [CrCl 2 (C 2 O 4 ) 2 ]3   will show cis-trans isomerism



H

H    +

HO

[Fe(H 2 O) 4 (OH) 2 ]   will show cis-trans isomerism [Fe(CN) 4 (NH 3 ) 2 ]   will show cis-trans isomerism [Co(en) 2 (NH 3 )Cl]2   will show cis-trans isomerism

 

[Co(NH 3 ) 4 (H 2 O)Cl]2  

(P) aqueous dilute KMnO4 (excess) 0C

6.

will not show cis-trans

isomerism (although it will show geometrical isomerism) (6) B 2 H 6  6MeOH   2B(OMe)3  6H 2 1 mole of B2 H 6 reacts with 6 moles of MeOH to give 2

OH

moles of B(OMe)3 HO

3.

give 6 moles of B(OMe)3 .

(Q) 7.

(4) CO, HCl   Anhydrous AlCl3 / CuCl

I

CHCl2

II

4.

3 moles of B2 H 6 will react with 18 moles of MeOH to

OH

HO

IV

CO2 Me

H2   Pd  BaSO 4

BIBAL  H   Toluene, 78 C H 2 O

H  Y

HY

 m for HX   m1

CHO

  

COCl

Ka 

[H  ] [X  ] [HX]

Ka 

[H  ] [Y  ] [HY]

CHO

H2O 100  C

III

H  X

(3) HX

 m1  CHO

1 m 10 2

Ka  C 2

 m  Ka1  C1   0 1    m1 

CHO

 m  Ka 2  C2   0 2    m2 

(75)

Ka1 C1  Ka 2 C2

2r h

 m for HY   m 2

2

2

 m1     m 2 

2

2



x

0.01  1    0.001 0.1  10 

pKa1  pKa 2  3 1 Z  1 [ th is shared in first layer and 8

8.

(9) In conversion of

238 92

U to

206 82

Pb, 8 -

particles and 6β particles are ejected.

90

Chemistry

The number of gaseous moles initially  1 mol

12. (a)

NH 2

The number of gaseous moles finally  1  8 mol; (1 mol (a)

Ph  CH 2

C  C

H Ph  CH 2 H

cis

–

N2 Cl Napthol/NaOH β  

 NaNO3, HCl 0C

from air and 8 mol of 2 He 4 ) So the ratio  9 /1  9 9.



(V)

CH 3

N  N  Ph

H

OH

H

C  C

CH 3

trans

(W)

10. (a)

O || C—H

H 3C

H 3C

(i) O3   (ii) Zn.H 2 O

..

(I) H — O — Cl . .  O

CH 2 — C — H || O

R



CH 2 — CH — NH3 || O

O

OH



H 3C

NH

NH2

14. (c, d) Cu 2 , Pb 2 , Hg 2 , Bi 3 give ppt. with H 2S in presence

of dilute HCl.

CH3 CH3 | | H O 15. (b) Cl — Si — Cl  2  HO — Si — OH   | | CH3 CH3

CH CH 2 OH

OH

CH 3 CH3 | | H — O — Si — O — Si ——O — H | | CH 3 CH 3 n

2H 2 O

H 3C

..

(II) H — O — Cl  O || O

O || (III) H — O — Cl  O (IV) || O

O || C—H

H 3C NH3  

H 3C

13. (b, c) H — O — Cl

N Me3SiCl, H 2 O

(S)

Me Me Me

O || C—H

H 3C

CH 3 CH 3 | | Si — O — Si — O — Si ——O — Si | | CH 3 CH 3 n

Me Me Me

..

CH 2 — CH — NH 2 || OH 11. (b)



 H3C — CH — CH3  

O2  

H 3C

CH T

U CH3

16. (b, c, d) * Adsorption of O 2 on metal surface is

exothermic. * During electron transfer from metal to O 2 electron * occupies  2p orbital of O2 .

CH3 | C — CH3 | O O

* Due to electron transfer to O 2 the bond order of O 2 decreases hence bond length increases. For Question Nos. 17 to 18

H

17. (c) E S1 

E1H  32 9   2.25  E1H . 4 22

Mock Test-1

91

18. (b) S2 is 3p  l = 1

O || C — CH3

For Question Nos. 19 to 20

CH2  CH2  OH

C8 H 6    double bond equivalent

 8 1

6 6 2 C  CH

HgSO 4 , H 2SO 4 , H 2O

(i) EtMgBr (ii) H 2 O

CH  CH2

(1) B2 H8 (2) H 2O 2 , NaOH, H 2O

(X)

OH CH 3 | | H  / heat Ph — C — CH3   Ph — C  CH — CH 3 | (Y) Et 19. (c) 20. (d)

  

92

Chemistry

JEE-MAIN: CHEMISTRY MOCK TEST-2 1.

2 g of oxygen contain same number of atoms as contained by a. 0.5 g hydrogen b. 4.0 g sulphur c. 7.0 g nitrogen d. 2.3 g sodium

2.

The axial angles in triclinic crystal system are a.       90 b.     90,   90 c.       90

3.

d.       90

A sample of drinking water was found to be severely contaminated with chloroform CHCl3 , supposed to be a carcinogen. The level of contamination was 15 ppm (by mass). Determine the molality of chloroform in the water sample.

4.

a. 2.12  104 mol kg 1

b. 1.26  10 4 mol kg 1

c. 0.12  10 4 mol kg 1

d. 5.36  10 4 mol kg 1

The temperature at which hydrogen molecules will have the same root mean square velocity as oxygen molecules have at 27C is a. 25C b. 7.93C c. 248C

5.

16 3R

b.

2R 16

c.

3R 16

d.

4R 16

Select the compound in which chlorine is assigned the oxidation number +5 a. HClO 4 b. HClO 2 c. HClO3

10. The dissociation constant of CH 3 COOH is 1.8 105.

The hydrolysis constant for 0.1 M sodium acetate is a. 5.56  10 4

b. 5.56 1010

c. 1.8  105

d. 1.8  109

11. The half-life of a first order reaction having rate constant

K  1.7 105 s 1 is a. 12.1 h c. 11.3 h

b. 9.7 h d. 1.8 h

12. Which characteristic is true in respect of colloidal particle a. They always have two phases b. They are only in liquid state c. They can't be electrolysed d. They are only hydrophilic 13. The heat of transition (H t ) of graphite into diamond

would be, where

An electron makes a transition from orbit n = 4 to the orbit n = 2 of a hydrogen atom. The wave number of the emitted radiations (R = Rydberg's constant) will be a.

6.

d. 127C

a. More cis-2-pentene is formed b. Equilibrium is shifted in the forward direction c. Equilibrium remains unaffected d. Additional trans-2-pentene is formed

d. HCl

C(graphite)  O 2 (g)   CO2 (g); H  x kJ C(diamond)  C2 (g)   CO 2 (g) ; H  y kJ

a. (x  y) kJ mol 1

b. (x  y) kJ mol 1

c. (y  x) kJ mol 1

d. None of these

14. All the nuclei from the initial element to the final element constitute a series which is called a. g-series b. b-series c. b-g series d. Disintegration series

7.

The molar conductivity is maximum for the solution of concentration a. 0.001 M b. 0.005 M c. 0.002 M d. 0.004 M

15. The following compound will undergo electrophilic substitution more readily than benzene a. Nitrobenzene b. Benzoic acid c. Benzaldehyde d. Phenol

8.

The number and type of bonds between two carbon atoms in calcium carbide are a. One sigma, one pi b. One sigma, two pi c.Two sigma, one pi d. Two singma, two pi

16. Cis and trans 2-butene are a. Conformational isomers b. Optical isomers c. Position isomers d. Geometrical isomers

9.

The standard state gibbs free energy change for the given isomerization reaction cis-2-pentene

trans-2-pentene

is –3.67 kJ / mol at 400K. If more trans-2-pentene is added to the reaction vessel, then

17. Electrolysis of a concentrated solution of sodium fumarate gives a. Ethylene b. Ethane c. Acetylene d. Vinyl alcohol

Mock Test-2

18. Identify

93

X

and

Y

in the following sequence

X Y C 2 H 5 Br   product   C3 H 7 NH 2

a. X  KCN, Y  LiAlH 4 b. X  KCN, Y  H 3O  c. X  CH 3Cl, Y  AlCl3 / HCl d. X  CH 3 NH 2 , Y  HNO 2 19. Consider the following alcohols (i) 1-Phenyl-1-propanol (ii) 3-Phenyl-1-propanol (iii) 1-Phenyl-2-propanol The correct sequence of increasing order of reactivity of these alcohol in their reaction with HBr is a. (i), (ii), (iii) b. (ii), (i), (iii) c. (i), (iii), (ii) d. (ii), (iii), (i) 20. Reaction between diethyl cadmium and acetyl chloride leads to the formation of a. dimethyl ketone b. ethylmethyl ketone c. diethyl ketone d. acetaldehyde 21. Which of the following is basic? a. CH 3CH 2 OH b. H 2O2 c. HOCH 2 CH 2 OH 22.

d. CH3COOH

Na, C2 H5OH CH3CN   X. The compound X is:

a. CH 3CH 2 NO 2

b. CH3CH 2COOH

c. C6 H 5 N(CH 3 ) 2

d. C6 H 5CONH 2

23. Which of the following is a synthetic polymer a. Rubber b. Perspex c. Protein d. Cellulose Space for rough work

24. The correct statement in respect of protein haemoglobin is that it a. Acts as an oxygen carrier in the blood b. Forms antibodies and offers resistance to diseases c. Functions as a catalyst for biological reactions d. Maintains blood sugar level 25. Which one of the following is known as broad spectrum antibiotics a. Streptomycine b. Ampicillin c. Chloramphenicol d. Penicillin G 26. The element or elements whose position is anomalous in the periodic table is a. Halogens b. Fe, Co and Ni c. Inert gases

d. Hydrogen

27. Flux added in the extraction of iron is a. Silica b. Felspar c. Limestone d. Flint 28. The correct order of the increasing ionic character is a. BeCl 2  MgCl 2  CaCl 2  BaCl 2 b. BeCl 2  MgCl 2  BaCl 2  CaCl 2 c. BeCl 2  BaCl 2  MgCl 2  CaCl 2 d. BaCl 2  CaCl 2  MgCl 2  BeCl 2 29. Nesseler's reagent is a. K 2 HgI 4 c. K 2 HgI 2  KOH

b. K 2 HgI 4  KOH d. K 2 HgI 4  Hg

30. The primary valence of the metal ion in the co-ordination

compound K 2  Ni  CN 4  is a. Four b. Zero c. Two

d. Six

94

Chemistry

JEE ADVANCE PAPER-I SECTION 1 Contains 8 Questions

SECTION 2 Contains 10 Multiple Choice Questions

The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

With one or more than one correct option

1.

At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probable speed of gas Y at 60 K. The molecular weight of the gas Y is

2.

Based on VSEPR theory, the number of 90 degree F−Br−F angles in BrF5 is

3.

20% surface sites have adsorbed N 2 . On heating N 2 gas evolved from sites and were collected at 0.001 atm and 298 K in a container of volume is 2.46 cm3. Density of

9.

Sodium metal crystallises in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately a. 1.86 Å b. 3.22 Å c. 5.72 Å d. 0.93 Å

10. 18 g glucose (C6 H I2 P6 ) is added to 178.2 g water. The vapor

pressure of water (in torr) for this aqueous solution is : a. 7.6 b. 76.0 c. 752.4 d. 759.0 11. For one mole of a van der Waals gas when b = 0 and T = 300 K, the PV vs. 1/V plot is shown below. The value of the van der Waals constant a (atm. liter2 mol2) is. PV (liter-atm mol–1)

surface sites is 6.023 1014 / cm2 and surface area is 1000 cm2, find out the no. of surface sites occupied per molecule of N 2 . 4.

The total number of contributing structure showing hyperconjugation (involving C–H bonds) for the following carbocation is H3C

5.

CH 2 CH 3



complete precipitation of chloride ions present in 30 mL

6.

7.

2.0 3.0 1 (mol liter 1 ) V

b. 4.5

c. 1.5

d. 3.0

12. A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0C. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be (R  8.314 J / mol K) (ln 7.5  2.01)

of 0.01 M solution of [Cr(H 2 O)5 Cl]Cl2 , as silver chloride

a. q  208 J, w  208 J

b. q  208 J, w  208 J

is close to

c. q  208 J, w  208 J

d. q  208 J, w  208 J

Among PbS, CuS, HgS, MnS, Ag 2S, NiS, CoS, Bi 2S3 and

13. The first ionisation potential of Na is 5.1 eV. The value of

SnS2 , the total number of black coloured sulphides is

electron gain enthalpy of Na  will be a. –2.55 eV b. 5.1eV

29.2% (w/w) HCl stock solution has a density of 1.25 g

c. 10.2 eV

1

volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is The maximum number of isomers (including stereoisomers) that are possible on monochlorination of the following compound is

CH 3 CH 2

d. 2.55 eV

1

mL . The molecular weight of HCl is 36.5 g mol . The

8.

(Graph not to scale)

0

a. 1.0

The volume (in mL) of 0.1 M AgNO3 required for

24.6 23.1 21.6 20.1

14. In a galvanic cell, the salt bridge a. does not participate chemically in the cell reaction. b. stops the diffusion of ions from one electrode to another. c. is necessary for the occurrence of the cell reaction d. ensures mixing of the two electrolytic solutions   15. Stability of the species of Li2 , Li2 and Li2 increase in

CH 3

the order of

C

a. Li 2  Li 2  Li 2

b. Li 2  Li 2  Li 2

c. Li 2  Li 2  Li 2

d. Li 2  Li 2  Li 2

H

CH 2 CH 3

Mock Test-2

95

16. How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2? a. 0.1 L b. 0.9 L c. 2.0 L d. 9.0 L 17. NiCl2 {P(C2 H 5 ) 2 (C6 H 5 )}2 exhibits temperature dependent

magnetic behavior (paramagnetic / diamagnetic). The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively. a. tetrahedral and tetrahedral b. square planar and square planar c. tetrahedral and square planar d. square planar and tetrahedral 18. Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is

(C) CO 2 (P  1 atm, T  273K)

3. P  nRT

(D) Real gas with very large molar volume

4. P(V  nb)  nRT

a. A  2; B  3; C  2; D  1, 4 b. A  1, 3; B  1; C  2; D  4 c. A  3; B  4; C  3, 4; D  1, 2 d. A  1, 2; B  3; C  1, 2; D  1, 4 20. The standard reduction potential data at 25ºC is given below: E (Fe 3  , Fe 2  )   0.77 V;

E (Fe 2  , Fe)   0.44V

E (Cu 2  , Cu)   0.34V;

E (Cu  , Cu)   0.52V

E[O 2 (g)  4H   4e    2H 2 O]   1.23V;

expected to have the highest E 0M3 / M 2 value?

E[O 2 (g)  2H 2 O  4e    4OH  ]   0.40V

a. Cr(Z  24)

b. Mn(Z  25)

E (Cr 3  , Cr)   0.74V;

c. Fe(Z  26)

d. Co(Z  27)

SECTION 3 Contains 2 Match The Following Type Questions You will have to match entries in Column I with the entries in Column II.

19. Match gases under specified conditions listed in Column-I with their properties/laws in Column-II. Column I Column II (A) Hydrogen gas (P  200 atm, 1. Compressibility factor  1 T  273K) (B) Hydrogen gas (P 0, T  273K) Space for rough work

2. Attractive forces are dominant

E (Cr 2  , Cr)   0.91 V

Match E of the redox pair in Column I with the values given in Column II. Column I

Column II 1. – 0.18 V

3

(A) E (Fe , Fe) (B) E(4H 2 O

4   4OH  )

2. – 0.4 V

(C) E(Cu 2   Cu   2Cu  )

3. – 0.04 V

(D) E (Cr 3  , Cr 2  )

4. – 0.83 V

a. A  2; B  3; C  1; D  4 b. A  1; B  2; C  3; D  4 c. A  3; B  4 ; C  1; D  2 d. A  2; B  3; C  1; D  4

96

Chemistry

JEE ADVANCE PAPER-II

The maximum number of electrons that can have principal quantum number, n = 3, and spin quantum number, 1 ms   , is 2

2.

The dissociation constant of a substituted benzoic acid at 25ºC is 1.0 × 10–4. The pH of a 0.01M solution of its sodium salt is

3.

The total number of cyclic structural as well as stereo isomers possible for a compound with the molecular formula C5 H10 is

4.

5. 6.

To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mol of He and 1.0 mol. of an unknown compound (vapour pressure 0.68 atm. At 0ºC) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at 0ºC is close to The coordination number of Al in the crystalline state of AlCl3 is Among the following, the number of compounds than can react with PCl5 to give POCl3 is O 2 , CO 2 , SO 2 , H 2 O, H 2SO 4 , P4 O10

7.

8.

If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is –0.0558ºC, the number of chloride(s) in the coordination sphere of the complex is [Kf of water = 1.86 K kg mol–1] A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titer value is

SECTION 2 Contains 8 Multiple Choice Questions

11. For a gaseous state, if most probable speed is denoted by

C*, average speed by C and mean square speed by C, then for a large number of molecules the ratio of these speeds are a. C* : C : C  1.225 :1.128 :1 b. C* : C : C  1.128 :1.1225 :1 c. C* : C : C  1:1.228 :1.225 d. C* : C : C  1:1.225 :1.128 12. The species having bond order different from that in CO is a. NO– b. NO+ c. CN– d. N2 13. Which of the following is the wrong statement? a. ONCF and ONO  are not isoelectronic b. O3 molecules is bent c. Ozone is violet–black in solid state d. Ozone is diamagnetic gas 14. The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of KCl, CH 3 OH and CH 3 (CH 2 )11 OSO 3 Na  at room temperature. The correct

assignment of the sketches is

I

Concentration

II

Concentration

Surface tension

1.

10. Mixture(s) showing positive deviation from Raoult’s law at 35ºC is(are) a. carbon, tetrahedral + methanol b. carbon disulphide + acetone c. benzene + toluene d. phenol + aniline

Surface tension

The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

Surface tension

SECTION 1 Contains 8 Questions.

III

Concentration

With one or more than one correct option

9.

The correct statement(s) regarding defects in solids is(are) a. Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion b. Frenkel defect is a dislocation defect c. Trapping of an electron in the lattice leads to the formation of F-centre d. Schottky defects have no effect on the physical properties of solids

a. I : KCl II : CH 3 OH III : CH 3 (CH 2 )11 OSO 3 Na  b. I : CH 3 (CH 2 )11 OSO 3 Na  II : CH 3 OH III : KCl c. I : KCl II : CH 3 (CH 2 )11 OSO 3 Na  III : CH 3 OH d. I : CH 3 OH II : KCl III: CH 3 (CH 2 )11 OSO 3 Na  15. Using the data provided, calculate the multiple bond energy (kJ mol–1) of CC bond in C2H2. That energy is (take the bond energy of a C—H bond as 350 kJ mol–1)

Mock Test-2

97

2C(s) + H 2 (g)   C 2 H 2 (g);  H  225 kJ mol

1

3

18. The solubility product (K sp ; mol dm ) of MX2 at 298 K

2C(s)   2C(g)

based on the information available for the given concentration cell is (take 2.303 × R × 298 / F = 0.059 V)

H  1410 kJ mol 1 H 2 (g)   2H(g)

 H  330 kJ mol 1

a. 1165 c. 865

b. 837 d. 815

a. carbanion c. carbocation

b. carbene d. free radical

SECTION 3 Contains 2 Paragraph Type Questions Each paragraph describes an experiment, a situation or a problem. Two multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.

Paragraph for Question No. 17 to 18

The electrochemical cell shown below is a concentration cell. 2

solution

of

sparingly

soluble

salt,

on the difference in concentration of M 2  ions at the two electrodes. The emf of the cell at 298 K is 0.059 V. 17. The value of  G (kJ mol 1 ) for the given cell is (take 1 F = 96500 C mol 1 )

Space for rough work

b. 4  1015

c. 11012

d. 4  1012

Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023  10 23 ) are present in a few grams of any chemical

compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na  23, Hg  200, 1 Faraday  96500 coulombs)

3

MX 2 ) || M (0.001 mol dm ) | M. The emf of the cell depends

a. –5.7 c. 11.4

a. 11015

Paragraph for Question Nos. 19 to 20

16. A solution of (–)-1-chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of SbCl5 , due to the formation of

M|M 2  (saturated

9

b. 5.7 d. –11.4

19. The total number of moles of chlorine gas evolved is a. 0.5 b. 1.0 c. 2.0 d. 3.0 20. If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is a. 200 b. 225 c. 400 d. 446

Chemistry

98

ANSWER & SOLUTIONS

⇒

JEE-Main 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

b

c

b

c

c

c

a

b

a

b

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

a

b

d

d

d

c

a

d

b

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

a

a

b

a

c

d

c

a

b

c

1.

(b) Number of atoms in 2 g of oxygen =

2 × 6 × 1023 16

∴

T 400 = 2 32 400 × 2 T= = 25 K or T = 25 − 273 = −248°C 32

5.

(c) Wave number

6.

(c) HClO3

1 1  1 1  3R =R 2 − 2  =R −  = λ  4 16  16  n1 n 2 

1

∗

⇒ 1 + x − 2 × 3 = 0 ; x = 6 − 1 = +5 (a) Since molar conductance ∝

8.

(b) 1σ and 2π

9.

(a) Equilibrium shifts backward by Le-chatelier’s principle.

23 23 (1) 6 ×10 × 0.5 = 3×10

6 ×1023 × 4 = 7.5 ×1022 (2) 32 (3)

6 ×1023 × 7 = 1.5 × 1023 28

6 ×10 × 2.3 = 6 ×1022 23 The correct answer is (b). 23

10. (b) K b =

(4)

∴ 2.

3.

(c) The axial angles in triclinic crystal system are different and none is perpendicular to any of the others i.e.,

K w 1× 10−14 = = 5.56 × 10−10 K a 1.8 × 10−5

11. (c) K = 1.7 ×10−5 s−1 t1/ 2 =

0.693 0.693 = × 105 = 11.32h K 1.7

α ≠ β ≠ γ ≠ 90°.

12. (a) Dispersion medium and dispersed phase are phases of colloid.

(b) Level of contamination = 15 ppm

13. (b) Graphite → diamond ∆H t = ( x − y ) kJ mol −1 .

= 15 parts in 106 parts Mass % of CHCl3

15 = 6 ×100 = 1.5 ×10−3 10 Thus, the water sample contains 1.5 ×10−3% (by mass) of CHCl3 Amount of CHCl3 =

15g = 0.126mol 119.5g mol−1

Mass of water (solvent) = 10 6 g = 10 3 kg So, molality of CHCl3 in solution =

4.

1 Molarity

7.

0.126 mol = 1.26 × 10 −4 mol kg −1 103 kg

(c) r.m.s. (H 2 ) =

3RT 2

3R × 400 32 T = 273 + 127 = 400 K) (

r.m.s. (O 2 ) =

3 RT 3R × 300 = 2 32

14. (d) Definition of disintegration series. 15. (d) Phenol will undergo electrophilic substitution more readily than benzene. 16. (d) Cis and trans 2-butene are geometrical isomers.

CHCOONa Electrolysis 17. (c) || + 2H 2 O  → CHCOONa

CH ||| CH Acetylene

+ 2CO 2 + 2KOH + H 2 LiAlH4 (Y) KCN(X) → C2 H5CN  → 18. (a) C2 H5 Br 

C2 H5CH 2 NH 2 (C3H 7 NH 2 ) X = KCN,Y = LiAlH 4 19. (d) HBr reacts with alcohols through the formation of carbocation. The stable is the carbocation formed, more is the reactivity of alcohol with HBr. The carbocation formed are CH+CH 2CH3 CH2CH2CH+2

(1)

(2)

(3)

CH2 − CH+ − CH3

Mock Test-2

99

CH 2CH 3

20. (b) Cd

CH 2CH 3

2.

 2CH3COCl  

2CH3COCH 2CH3  CaCl2

F

21. (a) Ethanol is the weakest acid among these, hence it is most basic.

So among the following only four (4) has linear shape and no d-orbital is involved in hybridization.

2

3.

PV 0.001 2.46  103   1.0  107 RT 0.0821 298 Now molecules of n N2 

25. (c) Chloramphenicol is broad spectrum antibiotic used in the treatment of typhoid, dysentry, acute fever. 26. (d) It shows similarities with both alkali metals as well as halogens.

N 2  6.023  1023 1107  6.023 1016 Now total surface sites available

27. (c) Impurities of SiO 2 is present in iron ore so basic flux

CaCO3 is added.



CaO  SiO2   CaSiO3 Slag



28. (a) BeCl 2  MgCl 2  CaCl 2  BaCl 2

As we go down the group I.E. decreases. Hence ionic character increases. 29. (b) 2KI  HgI 2   K 2 Hgl 4  KOH Nessler's reagent

30. (c) Primary valencies are also known as oxidation state. K 2 [ Ni (CN) 4 ], 2  x  4  0  x  2 JEE Advance Paper -I 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

4

0

2

6

6

6

8

8

a

c

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

a

b

a

b

d

c

d

d

c

1.



400  3 2  60  40 x 120 , x4 30  x

12.04  1016 2 6.02  1016

4.

(6) 6  H  atoms are there

5.

(6) Number of ionisable Cl in [Cr(H 2 O)5 Cl]Cl 2 is 2

 Millimoles of Cl  30  0.01 2  0.6

 Millimoles of Ag  required  0.6  0.6  0.1 V 

6.

V  6 ml

(6) Black coloured sulphides {PbS, CuS, HgS, Ag 2S, NiS, CoS}

* Bi2S3 in its crystalline form is dark brown but Bi 2S3 precipitate obtained is black in colour.

M.W.(X gas)  40;M.W.(Ygas)  x 2RT1 M2

 6.023 1014 1000  6.023 1017 Surface site used to adsorb 20 N2   6.023  1017  12.04  1016 100 Sites occupied per molecule of N2 

(4) Vrms ( Xgas )( 400 K )  Vmp ( Ygas )( 60K )

3RT1  M1

(2) PN2  0.001 atm, T  298 K, V  2.46 cm2

By ideal gas, PV  nRT

Hb 4  4O2   Hb 4 O8

Impurity

F

F F All four planar bonds (F−Br−F) will reduce from 90° to 84.8° after p  bp repulsion.

23. (b) Perspex is a synthesised polymer. 24. (a) Four Fe ions of each haemoglobin can bind with 4 molecules of O 2 and it is carried as oxyhaemoglobin.

Br

F

Na, C2 H5OH  CH3CH 2 NH 2 (X) 22. (a) CH3CN  4H 

Flux

(0)

7.

(8) Molarity 

W  d sol n W W 100  d so ln      1000 V M w  v M w  Wsol Wsol M w  100

W %  d  10 W Mw

Chemistry

100

⇒

29.2 ×1.25 ×10 = 10 M M1V1 = M 2 V2 36.5 10 × V1 = 0.4 × 200

⇒

V=

1 PV = −a ×   + RT V y = mx + C fi Slope = – a

=

0.4 × 200 = 8ml 10

Slope

12. (a) q = + 208 J, (as it absorb heat)

Cl 8.

(8)

Br

CH 3

Br

Cl

v  w Re w = −2.303 nRT log10  2   v1 

CH 3 CH 3 CH 3 CH 2 C*− CHCH 3 Two Enantiomeric pairs = 4 * H CH 3 CH 3 CH 2 C − CH 2 CH 3

1

H CH 2 Cl 1

10. (c) Moles of glucose =

n Total = 10

⇒

∆P 0.1 = P° 10

⇒

18 = 0.1 180

178.2 = 9.9 18

PS = 760 − 7.6 = 752.4 torr 11. (c) b = 0, T = 300 K, n = 1   (V − nb) = RT 

a    P + 2  (V) = RT V   ⇒

PV +

Hence stability order = Li −2 < Li +2 < Li 2 0.1 0.01

⇒ 1 + v = 10 ⇒ v = 10 − 1 = 9L

In strong ligand 3d8

In weak ligand 3d8 Hybridisation is dsp2 (diamagnetic) Square planar

∆P = 0.01P° = 0.01× 760 = 7.6 torr

 an 2 P + 2 V 

15. (b) B.O. of Li +2 = 0.5 , B.O. of Li −2 = 0.5

Ni +2 (28) : [Ar]4s°3d 8

3 3 a= × 4.29 Å = 1.85 Å 4 4

⇒

14. (a)

3a = 4r

(a) For BCC 6 unit cell,

Moles of water =

∴ EA of Na + = −5.1eV

17. (c) [NiCl2 P(C 2 H 5 ) 2 (C6 H 5 ) 2 ]+2

H Total = 2 + 4 + 1 + 1 = 8

r=

 375  = −2.303 × (0.04 × 8.314 × (310) log10   = −208 J  50  13. (b) E.A = Ionisation potential

16. (d) 0.1× 1 = (1 + v) × 0.01 ⇒ 1 + v =

CH 3 CH 2 C*− CH 2 CH 3

9.

y2 − y1 20.1 − 21.6 = = 1.5 x 2 − x1 3−2

a = RT V

Hybridisation is sp3 (paramagnetic) tetrahedral

18. (d) Factual 19. (d) A → 1, 2 ; B → 3 ; C → 1, 2 ; D → 1, 4 (A) Z =

PVm at high pressure and low temperature. RT

 an 2  Equation  P + 2  (V − nb) = nRT reduces to V   P(V − nb) = nRT.

Mock Test-2

101

(B) For hydrogen gas value of Z = 1 at P = 0 and it increases continuously on increasing pressure. (C) CO2 molecules have larger attractive forces, under normal conditions. PVm , at very large molar volume Z  1. (D) Z  RT

3.

20. (c) A  3; B  4 ; C  1 ; D  2

4.

(A)

G

 G



o 3  FEo(Fe3 / Fe)   1 FE (Fe  (2  FEoFe2 / Fe ) 3 / Fe2 )



E oFe3 / Fe   0.04 V

(B)

O 2 (g)  2H 2 O  4e    4OH E   0.40 V . . .(i)

o Fe3 / Fe

o Fe3 / Fe2

 G

4H   4OH

4H 2 O

For 3rd structure 2 cis-trans and 1 optical isomer are possible. Total 7 isomers.

o Fe2 / Fe

(7) Let unknown is X. p He  p total  Px  (1  0.68) atm  0.32 atm

Now p He  n He

2H 2 O   O 2 (g)  4H   4e  E   1.23 V . . .(ii)

So

(7) Cyclic C5 H10

. . .(iii)

rd

E For III reduction  0.40  1.23   0.83 V. o o  (1 F  ECu ) (C) G o(Cu2 / Cu)  1 FECu 2  / Cu  / Cu

RT 0.10  0.082  273 =7  p He 0.32



v

5.

(6) Coordination number of Al is 6. It exists in ccp lattice with 6 coordinate layer structure.

6.

(5)

7.

(1) Tf = iK f m 0.0558  i  1.86  0.01

o 2  FE oCu 2 / Cu  1 FE oCu 2 / Cu   (1 F  E Cu )  / Cu

i3



  0.18 V.

Complex is [Co(NH 3 )5 Cl]Cl2



E

(D)

G oCr3 / Cr 2  G oCr3 / Cr  G oCr / Cr 2

8.

(3)

1 F  E oCr3 / Cr 2  3  F  E oCr 3 / Cr  (2  F  E oCr / Cr2 )

9.

(b, c)



o Cu 2 / Cu

10. (a, b) (a) CCl 4  CH 3 OH  Positive deviation from Raoult’s

E oCr 3 / Cr 2   0.4 V.

law

JEE Advance Paper -II 1.

1.

2.

3.

4.

O 5.

6.

7.

8.

9.

10.

9

8

7

7

6

5

1

3

b, c

a, b

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

a

a

d

d

c

d

b

b

d

(9) Number of orbital for n  3 is  n  9 2

Number of electrons for n  3 and m s   2.

(b) CS2  

1 9 2

Kh 

C6 H5COOH OH 1 0.01h

K w 0.01h 2  Ka 1 h

1014 10 2 h 2 (1–h 1)  104 1 h [OH  ]  0.01h  0.01 104  106 [H  ]  10 8 pH  8



CH 3

Positive deviation from Raoult’s law (c) C 6 H 6  C 7 H 8  Ideal solution

(d)

pH of 0.01M C6 H 5COONa 0.01(1 h)

C

CH 3

NH 2

OH

(8) K a (C6 H 5COOH) = 1 × 10–4

C6 H5COO  H 2

RT V

+

Negative deviation from Raoult’s law.

0.01h

11. (c) C*: C : C 

2RT 8RT 3RT   M M M

8  3 3.14 1 : 1.128 : 1.225 2:

12. (a) NO  (16 electron system) Bond order =2.

102

Chemistry 



NO , CN and N2 are isoelectronic with CO therefore all

have same bond order (=3). 13. (a) ONCF and ONO– are isoelectronic in nature. 14. (d) Impurities affect surface tension appreciably. It is observed that impurities which tend to concentrate on surface of liquids, compared to its bulk lower the surface tension.

Substances like detergents, soaps [CH 3 (CH 2 )11 SO 3 Na ] decreases the surface tension sharply. Those like alcohol (e.g., –CH 3 OH, C 2 H 5 OH) lower the

16. (c) Racemises slowly due to formation of intermediate carbocation. For Question Nos. 17 to 18 17. (d) G  nFE cell  2  96500  0.059  11.387 kJ mol1  11.4 kJ

18. (b) E 

0.059 

surface tension slightly. This can also be related to the fact that CH 3 OH has smaller dielectric constant. Dielectric

10 2 

constant is directly proportional to surface tension. So, on adding CH 3 OH in water, overall dielectric constant decreases and surface tension decreases. Inorganic impurities present in bulk of a liquid such as KCl tend to increase the surface tension of water.

0.0591 C log10 3 2 10

0.0591  C  log10  3  2  10 

C C  [M 2  ]  10 5 M. 10 3

K sp  [M 2 ] [X  ]2  4s3 = 4(10 -5 )3 = 4 ¥ 10-15 For Question Nos. 19 to 20 19. (b) NaCl   Na   Cl 

At anode: 2Cl    Cl 2

15. (d) 2C(s)  H 2 (g)   C 2 H 2 (g) (H  C  C  H) 

BE(H2) + BE(H2) + Hsub (C) – BE (C – H) × 2 +



BE(C  C ) Hrxn 330  1410  [350  2  x]  225  x= 815

Moles of Cl  2 in 500 ml. Therefore 1 mole of Cl2 evolves. 20. (d) Na—Hg (amalgam) formed = 2 moles at cathode.

  

Mock Test-3

103

JEE-MAIN: CHEMISTRY MOCK TEST-3 1.

The volume of 1.0 g of hydrogen in litres at N.T.P. is a. 2.24 b. 22.4 c. 1.12 d. 11.2

2.

Which set of characteristics of ZnS crystal is correct? a. Coordination number (4: 4) : ccp; Zn2+ ion in the alternate tetrahedral voids b. Coordination number (6: 6); hcp; Zn2+ ion in all tetrahedral voids c. Coordination number (6: 4) : hcp; Zn2+ ion in all octahedral voids d. Coordination number (4: 4); ccp; Zn2+ ion in all tetrahedral voids

3.

PtCl4.6H2O can exist as a hydrated complex. 1 molal aqueous solution has depression in freezing point of 3.72º Assume 100% ionisation and K f (H 2O)  1.86 mol1 kg,

4.

5.

6.

8.

9.

0.1 M NH 4 CN at 25ºC is a. 1.4

b. 7.2  10 15

c. 7.2  101

d. 1.4  106

11. For the reaction A + B  C it is found that doubling the concentration of A increases the rate by 4 times, and doubling the concentration of B doubles the reaction rate. What is the overall order of the reaction? a. 4 b. 3/2 c. 3 d. 1 12. The coagulation power of an electrolyte for arsenious sulphide decreases in the order

then complex is a. [Pt(H2O)6 ]Cl4

b. [Pt(H 2 O) 4 Cl2 ]Cl 2 2H 2O

a. Na + , Al+3 , Ba +2

b. PO 4-3 , SO 4-2 , Cl -

c. [Pt(H 2O)3 Cl3 ],Cl.3H 2O

d. [Pt(H 2O 4 )Cl 4 ]4H 2 O

c. Al+3 , Ba +2 , Na +

d. None of these

A weather balloon filled with hydrogen gas at 1 atm and 27ºC has volume equal to 12000 litres. On ascending, it reaches a place where temperature is –23ºC and pressure 0.5 atm. The volume of the balloon is a. 24000 L b. 12000 L c. 10000 L d. 20000 L In Bohr model of the hydrogen atom, the lowest orbit corresponds to a. Infinite energy b. The maximum energy c. The minimum energy d. Zero energy When KMnO 4 is reduced with oxalic acid in acidic solution, the oxidation number of M n changes from a. 7 to 4 b. 6 to 4 c. 7 to 2 d. 4 to 2

7.

10. Ammonium cyanide is salt of NH4OH(Kb = 1.8 × 10–5) and HCN (Kb = 4.0 × 10–10). The hydrolysis constant of

The unit of molar conductivity is a. –1cm–2 mol–1 b.  cm–2 mol–1 c. –1cm2 mol–1 d.  cm2 mol–1 In a double bond connecting two atoms, there is a sharing of a. 2 electrons b. 1 electron c. 4 electrons d. All electrons In a reversible reaction, the catalyst a. Increases the activation energy of the backward reaction b. Increases the activation energy of the forward reaction c. Decreases the activation energy of both, forward and backward reaction d. Decreases the activation energy of forward reaction

13. Correct relationship between heat of fusion (H fus ), heat

of vaporisation (H vap ) and heat of sublimation (Hsub ) is a. H fus  H vap  H sub b. H vap  H fus  H sub c. H sub  H vap  H fus d. H sub  H vap  H fus 14. The number of neutrons in the parent nucleus which gives

N14 on - emission is a. 7 b. 14

c. 6

d. 8

15. Which represents nucleophilic aromatic substitution reaction? a. Reaction of benzene with Cl 2 in sunlight b. Benzyl bromide hydrolysis c. Reaction of NaOH with dinitrofluorobenzene d. Sulphonation of benzene 16. Which one of the following is the chiral molecule? a. CH3Cl

b. CH 2 Cl2

c. CHBr3

d. CHClBrI

17. n-Propyl chloride and benzene react in the presence of

anhydrous AlCl3 to form a. ethyl benzene c. n-propyl benzene

b. methyl benzene d. iso-propyl benzene

Chemistry

104

18. 1-chlorobutane reacts with alcoholic KOH to form a. 1-butene b. 2-butane c. 1-butanol d. 2-butanol 19. An alcohol having molecular formula C5H11OH on dehydration gives an alkene, which on oxidation yield a mixture of ketone and an acid. The alcohol is a. CH3CH2CH(OH)CH2CH3 b. CH 3CHCH 2 CH 2 CH 3 | OH c. (CH3 )2 CHCH(OH)CH3

d. (CH3 )2 CCH2OH

20. m-chlorobenzaldehyde on reaction with conc. KOH at room temperature gives a. potassium m-chlorobenzate and m-chlorobenzyl alcohol b. m-hydroxy benzaldehyde and m-chlorobenzyle alcohol c. m-chlorobenzyl alcohol and m-hydroxy benzyle alcohol d. potassium m-chlorobenzoate and m-hydroxy bezaldehyde 21. Acetamide is treated separately with the following reagents. Which one of these would give methyl amine? a. PCl5

b. NaOH  Br2

c. Sodalime

d. Hot conc. H 2SO 4

22. On treating aniline with nitrous acid and HCl at 0–5ºC gives a. An alcohol b. Diazonium salt c. Nitro aniline d. Aniline hydrogen chloride 23. ‘Rayon’ is a. Natural silk c. Natural plastic or rubber Space for rough work

b. Artificial silk d. Synthetic plastic

24. The waxes are long chain compounds of fatty acids, which belong to the class of a. Esters b. Ethers c. Alcohols d. Acetic acid 25. Which of the following is a local anaesthetic? a. Diazepam b. Procaine c. Mescaline d. None of these 26. The electronic configuration of the element which is just above the element with atomic number 43 in the same periodic group is a. 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 2 b. 1s 2 2s 2 2p 6 3s 2 3p 6 3d10 4s 2 4p5 c. 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s1 d. 1s 2 2s 2 2p 6 3s 2 3p 6 3d10 4s1 4p 6 27. Complex is formed in the extraction of a. Na b. Cu c. Ag 28.

d. Fe

MgCl 2 .6H 2O when heated gives

a. Magnesium oxychloride c. Magnesium oxide

b. Magnesium dichloride d. Magnesium chloride

29. Acidified potassium dichromate on reacting with a sulphite is reduced to a. CrO 2Cl2

b. CrO 24

c. Cr 3

d. Cr 2

30. In the extraction of which of the following, complex ion forms? a. Cu

b. Ag

c. Fe

d. Na

Mock Test-3

105

JEE ADVANCE PAPER-I SECTION 1 Contains 8 Questions The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

1.

The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.10 M). If  X0   Y0 , 



the difference in their pK a values, pK a (HX) pK a (HY), is (consider degree of ionization of both acids to be 0 and ∆Ssurrounding > 0

(B) KI(0.1M)+ AgNO3 (0.01 M) X

Y

b. ∆Ssystem > 0 and ∆Ssurrounding < 0 c. ∆Ssystem < 0 and ∆Ssurrounding > 0 d. ∆Ssystem < 0 and ∆Ssurrounding < 0 18. Which of the following exists as covalent crystals in the solid state? a. Iodine b. Silicon c. Sulphur d. Phosphorus

(C) CH 3 COOH + KOH X

Y

(D) NaOH + HI X

Y

SECTION 3 Contains 2 Match The Following Type Questions You will have to match entries in Column I with the entries in Column II.

19. According to Bohr’s theory, E n = Total energy Vn = Potential energy Space for rough work

K n = Kinetic energy

rn = Radius of nth orbit

a. A → 1 ; B → 2 ; C → 3 ; D → 4 b. A → 2 ; B → 4 ; C → 3 ; D → 1 c. A → 3 ; B → 4 ; C → 2 ; D → 1 d. A → 3 ; B → 4 ; D → 1 ; D → 2

Column II 1. Conductivity decreases and then increases 2. Conductivity decreases and then does not change much 3. Conductivity increases and then does not change much 4. Conductivity does not change much and then increases

Mock Test-3

107

JEE ADVANCE PAPER-II SECTION 1 Contains 8 Questions The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

1.

2.

Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is KCN K 2SO 4 (NH 4 ) 2 C 2O 4 NaCl

Zn(NO3 ) 2

FeCl3

d. The unit cell edge length is 2 2 times the radius of the atom

K 2CO3

NH 4 NO3

LiCN

10. The standard Gibbs energy change at 300 K for the

When the following aldohexose exists in its D– configuration, the total number of stereoisomers in its pyranose form is CHO | CH 2 | CHOH | CHOH | CHOH | CH 2 OH

3.

The number of resonance structures for N is

4.

The concentration of R in the measured as a function of time and obtained: [R] (molar) 1.0 0.75 t (min). 0.0 0.05 The order of the reaction is

reaction RP was the following data is 0.40 0.12

The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is

6.

The total number of diprotic acids among the following is H 3 PO 4 H 2SO 4 H 3 PO 3 H 2CO3 H 2S2 O 7 H3BO3

H 3 PO 2

H 2 CrO 4

controlled nuclear fission to 8.

H 2SO 3

The number of neutrons emitted when 142 54

Xe and

90 38

235 92

U undergoes

Sr is

In dilute aqueous H 2SO 4 , the complex diaquodioxalatoferrate (II) is oxidised by MnO 4 . For this reaction, the ratio of the rate of change of [H  ] to the rate of change of [MnO 4 ] is

SECTION 2 Contains 8 Multiple Choice Questions With one or more than one correct option

9.

reaction 2A

The correct statement(s) for cubic close packed (cep) three dimensional structure is (are)

B  C is 2494.2 J. At a given time, the

composition of the reaction mixture is [A]  12 , [B]  2 and [C]  12 . The reaction proceeds in the:

[ R  8.314 J / K / mol, e  2718 ] a. forward direction because Q  K c b. reverse direction because Q  K c c. forward direction because Q  K c d. reverse direction because Q  K c 11. 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= X=

0.10 0.18

5.

7.

a. The number of the nearest neighbours of an atom present in the topmost layer is 12 b. The efficiency of atom packing is 74% c. The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively

a. MX

b. MX 2

c. M 2 X

d. M 5 X14

12. Which of the following is the energy of a possible excited state of hydrogen? a. 13.6 eV b.  6.8 eV c.  3.4 eV

d.  6.8 eV

13. Assuming 2s-2p mixing is not operative, the paramagnetic species among the following is a. Be2 b. B2 c. C2 d. N2+ 14. In allene(C3H4), the type(s) of hybridization of the carbon atoms is (are) a. sp and sp3

b. sp and sp 2

c. only sp 2

d. sp 2 and sp3

Chemistry

108

15. Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure I.

Cotton wool Soaked in X

and

II.

The correct order of their boiling point is a. I > II > III b. III > II > I c. II > III > I d. III > I > II CH3 Cl 2 , h v   N (isomeric products)

16. H3C

d

Initial formation of The product

Cotton wool Soaked in Y

17. The value of d in cm (shown in the figure), as estimated from Graham’s law, is a. 8 b. 12 c. 16 d. 20

and

II.

L = 24 cm

Paragraph for Question Nos. 19 to 20

CH3 fractional distillation C5H11Cl   M(isomeric products)

What are N and M ? a. 6, 6 c. 4, 4

18. The experimental value of d is found to be smaller than the estimate obtained using Graham’s law. This is due to a. larger mean free path for X as compared to that of Y b. larger mean free path for Y as compared to that of X c. increased collision frequency of Y with the inert gas as compared to that of X with the inert gas d. increased collision frequency of X with the inert gas as compared to that of Y with the inert gas

b. 6, 4 d. 3, 3

SECTION 3 Contains 2 Paragraph Type Questions

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7ºC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol 1 ),

Each paragraph describes an experiment, a situation or a problem. Two multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.

this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M

Paragraph for Question Nos. 17 to 18 X and Y are two volatile liquids with molar weights of 10g mol–1 and 40g mol–1 respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.

NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6C was measured. (Consider heat

Space for rough work

acetic acid (K a  2.0  10 5 ) was mixed with 100 mL of 1.0 M

capacity of all solutions as 4.2 J g 1 K 1 and density of all solutions as 1.0 g mL1 ) 19. Enthalpy of dissociation (in kJ mol 1 ) of acetic acid

obtained from the Expt. 2 is a. 1.0 b. 10.0 c. 24.5 20. The pH of the solution after Expt. 2 is a. 2.8 b. 4.7 c. 5.0

d. 51.4 d. 7.0

109

Mock Test-3

ANSWER & SOLUTIONS JEE-Main 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

d

a

c

d

c

c

c

c

c

b

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

c

c

d

b,c

d

c

a

c

a

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

b

d

b

a

b

a

c

c

c

b

1.

3.

(a) ZnS has zinc blende type structure (i.e., ccp structure). The S2– ions are present at the corners of the cube and at the centre of each face. Zinc ions occupy half of the tetrahedral sites. Each zinc ion is surrounded by four sulphide ions which are disposed towards the corner of regular tetrahedral. Similarly, S2– ions surrounded by four Zn2+ ions.

i=

4.

9.

(c) Decreases the activation energy of both forward and backward reaction.

10. (b) K h = =

KW Ka × Kb

1 × 10 −4 = 1.4 4.0 × 10 −4 × 1.8 ×10 −5

12. (c) According to Hardy-Schulze rule. 13. (c) Heats of combustion are always exothermic except oxidation of N as, N 2 + O 2  → 2NO ; ∆H = + ve

(∆Tf )obs 3.72 = = 2 = 1 + (n − 1)α . (∆Tf )cal 1.86

Hence,α = 1,∴ n = 2 Two species will be produced from single species which is only possible for [Pt(H 2 O)3 Cl3 ],Cl3H Cl3H22OO (d) P1 = 1 atm,

11. (c) A + B → C On doubling the concentration of A rate of reaction increases by four times. Rate ∝[A]2. However on doubling the concentration of B, rate of reaction increases two times. Rate ∝[B] Thus, overall order of reaction = 2 + 1 = 3

N 2 + 12 O 2  → N 2 O ; ∆H = + ve

(c) (∆Tf )cal = K f × m

(∆Tf )cal = 1.86 ×1 = 1.86

∴

(c) In a double bond connecting two atoms sharing of 4 electrons take place as in H2C = CH2.

(d) 2 g of hydrogen occupy volume = 22.4 L

22.4 ×1 = 11.2 L 1 g of hydrogen occupies volume = 2 2.

8.

β 14. (d) 6 X14  → 6+1 N14

in 6 X14 no. of neutrons 14 – 6 = 8.

15. (b, c)

NO

NO

Di nitro fluoro benzene Dinitrofluorobenzene

V2 = ?

T2 = −23 + 273 = 250 K

P1 V1 P2 V2 = T1 T2 P1 V1 T2 1 × 12000 × 250 = = 20, 000 L P2 T1 0.5 × 300

or

V2 =

5.

(c) In hydrogen atom, the lowest orbit (n = 1) corresponds to minimum energy (– 13.6 eV).

OH − Na +

OH +

O2

 →

V1 = 12000 L,

T1 = 27 + 273 = 300 K P2 = 0.5 atm,

F

NO

Di nitro phenol phenol Dinitro

H | 16. (d) I — C* — Br | Cl A carbon atom which is attached to four different atoms or groups is called a chiral or asymmetric carbon atom. such a carbon atom is often marked by an asterisk.

CH 2 CH 2CH 3

AlCl3 +2 +7 17. (c) COOH + CH 3 CH 3 CH 2 Cl  → + 2KMnO4 + 3H2SO4  → K2SO4 + 2MnSO4 +10CO2 + 8H2O 5 | COOH 18. (a) CH 3CH 2 CH 2CH 2 − Cl + KOH(alc.)  → K 2SO 4 + 2MnSO 4 + 10CO 2 + 8H 2 O CH 3CH 2 − CH = CH 2 + KCl + H 2 O In this reaction oxidation state of Mn change from +7 to +2. 1-butene

6.

(c)

7.

(c) Molar conductivity =

1

ρM

So, its unit will be Ω–1cm2 mol–1

19. (c)

CH 3 — CH — CH — CH 3 Dehydration  → | | CH 3 OH

Chemistry

110

CH 3 — C  CH — CH 3 | CH 3

CH 3 — C  O | CH 3

Oxidation

O

C — CH 3

heat

28. (c) MgCl 2 .6H 2O   MgO  5H 2O  2HCl

 2Cr 3  3SO24  4H 2O 29. (c) Cr2 O72  8H   2SO32  | OH

Sodium dicyno argentate

20. (a)

2Na [Ag(CN) 2 ]  Zn   Na 2 [Zn (CN) 4 ] 2Ag

Cl

Cl

Cl

Sodium tetracynozincate (ppt)



 KOH  

JEE Advance Paper-I

COOK

CHO

CH2OH

1.

Cannizzaro's reaction 

O O || | Ph — C — H  HO    Ph — C — H | OH I

1.

O ||  CH 3 NH 2 21. (b) CH 3 — C — NH 2  Br2  NaOH 

7.

26. (a)

25

9.

4

2

7

8

5

6

a

a

14.

15.

16.

17.

18.

19.

20.

b

a, b, d

a

a

d

b

b

b

a

c

H  X

(3) HX

H  Y

Ka 

[H  ][X  ] [HX]

Ka 

[H  ][Y  ] [HY]

 m for HY   m2

1 m 10 2

Ka  C 2 2

 m  Ka 2  C2   0 2    m2 

2

2

2

2.

0.01  1    0.001  pKa1  pKa 2  3 0.1  10 

(8)

92

6 2  U 238   80 X 214   82 Pb 214

(6 , 2  ), total 8 particles.

3.

(4)

H HO



H   

+

 

(P) aqueous dilute KMnO4 (excess) 0C

Mn  3d 4s . 2

OH

27. (c) Hydrometallurgy

Ag 2S  4NaCN   2Na[Ag(CN)2 ]  Na 2S 2Na[Ag(CN)2 ]  Zn   Na 2 [Zn(CN)4 ]  2Ag

10.

13.

(b) Local anasthetic – affect only the part of body e.g. Xylocaine, Procain etc. 5

8.

8



24. (a) Waxes are esters of higher fatty acids. 25. (b) The anaesthetics produce temporary insensitibility to the vital function of all type of cells, specially of nervous system and are used during surgical operations. These are classified as (a) General anasthetic – produces unconsciousness all over the body e.g. N 2 O, Cyclopropane, chloroform

6.

12.

Ka1 C1  m1     Ka 2 C2  m 2 

O

23. (b) ‘Rayon’ is man-made fibre which consists of purified cellulose in the form of long threads. Rayon resembles silk in appearance. Hence called as artificial silk.

5.

3

 m  Ka1  C1   0 1    m1 

NH2 O

4.

11.

 m1 

HCl  HONO   

ææ Æ

3.

 m for HX   m1

N 2 Cl

NH2

2.

HY

“Hofmann’s bromamide reaction” 22. (d)

2Na  Ag  CN 2   Na 2S

30. (b) Ag 2S  4NaCN

OH

HO HO

(Q)

Mock Test-3

111

4.

O ||  (2) —C — O and — NH 2 are basic groups in lysine.

5.

(7) Let the solubility of AgCl is x

11. (b)

 12. (a, b, d) ClO3  6I  6H 2SO4  3I2  Cl  6HSO4  3H 2O

Ag   Cl and that of CuCl is y mol litre 1

AgCl

x

x

Cu   Cl

CuCl 

mol litre 1

10. (a)

y

13. (a) 0.3010 

y

K sp of AgCl  [Ag  ] [Cl1 ] 1.6  1010  x(x  y)

. . .(i)



Similarly K sp of CuCl  [Cu  ] [Cl ] 1.6  106  y(x  y)

6.

. . .(ii)

14. (a) K 3 [Fe(CN)6 ]   3K   [Fe(CN)6 ]3 , i  4 Tf  K f  i 

x7

(8) Total no. of N – Co – O bond angles is 8. O

O || O — C — CH 3

O

N

O

COOH

O

Co

N

(Aspirin)

O O

CH 3

16. (b)

(5) 3Br2  3CO32    5Br   BrO3  3CO 2

8.

(6) Let number of glycine units  n Mass of decaeptide = 796 Mass of H 2 O needed = 162 g, Total mass = 958 g 958 



9.

n

47  75  n 100

958  47 6 100  75

O CHO

CH 3

At

17. (b)

H 2 O( )

CH 3

5-keto-2-methyl hexanal

100C

and

1



Ssystem  0 and Ssurrounding  0

18. (b) Silicon exists as covalent crystal in the solid state. 19. (a) A  3; B  2; C 1; D 4

Number of O atoms  4

(A)

Number of Al3  4  m Number of Mg 2   8  n Due to charge neutrality 4(2)  4m(3)  8n( 2)  0 m

1 1 and n  2 8

pressure

Stotal  0 and Ssystem  Ssurrounding  0

20. (c) A  3; B 4; C  2; D 1

Number of tetrahedral voids  8

atmosphere

H 2 O(g) is at equilibrium. For equilibrium

(a) In ccp lattice:

Number of Octahedral voids  4



CH 3 O3   Zn / H 2 O

O 7.

m 1000 0.1 1000   1.86  4   M W 329 100

 2.3  102 fi Tf  2.3 102

15. (d)

O

Ea  310  300  2.303  8.314  10 3  310  300 

E a  53.6 kJ mol

On solving (i) and (ii) [Ag  ]  1.6  107



0.3010 

Ea  T2  T1    2.303R  T1T2 

(C 2 H 5 )3  N  CH 3 COOH   (C 2 H 5 )3 NH  CH 3 COO  X

Y

Initially conductivity increases due to ion formation after that it becomes practically constant because X alone cannot form ions. Hence (3) is the correct match. (B) KI(0.1 M)  A gNO3 (0.01 M)   AgI   KNO3 X

Y

Number of ions in the solution remains constant until all the AgNO3 precipitated as AgI. Thereafter conductance increases due to increase in number of ions. Hence (4) is the correct match.

Chemistry

112

(C) Initially conductance decreases due to the decrease in the 5. number of OH ions thereafter it slowly increases due to 2MnO 4KOH + O2 (2) → is2K 4 + 2H 2O the increase in number of H2 ++ions. Hence the2 MnO correct match. 6. (D) Initially it decreases due to decrease in H+ ions and then increases due to the increase in OH ions. Hence (1) is the 7. correct match. 8.

JEE Advance Paper -II 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

3

8

9

5

6

6

3

9

b,c,d

b

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

b

c

c

b

b

b

c

d

a

b

1.

(3) K CN , K 2CO3 ,- LiCN are basic in nature and their

2.

aqueous solution turns red litmus paper blue. (8)

9.

O.S. of Mn = +6in K 2 MnO4 H 3 PO3 , H H22SO SO33 (6) H 2SO 4 , H 2 CO3 , H 2S2 O7 , H 2 CrO 4 , H3PO3, (3)

92

U 235 → 54 Xe142 + 38Sr 90 + 30 n1

  a Ê a0 ˆ 2.303 2.303 (9) K = log Á , K= log  0 ˜ t t1/8 Ë a0 - x ¯  1 a0 8

    

(b, c, d) (a) For any atom in topmost layer, coordination number is not 12 since there is no layer above topmost layer, (b) Fact, (c) Fact, (D)

2 a = 4R

10. (b) ∆G = − RT ln e K c

2494.2 = 8.314 × 300 ln e K c ⇒ Kc = e

K c = e −1 =

Total No. of stereoisomers = 24 = 16 which contains 8D – Configuration and 8 – L Configuration. NaOH →

−1

1 = 0.36 2.718

2 × 12 (B)(C) = =4 2 [A] [1/ 2]2

Q=

OH

(9)

(Potassiummanganate)

So, a = 2 2 R

CH 2 OH O OH H HO H H OH H

3.

(6) 2MnO2 + 4KOH + O2 → 2K 2 MnO4 + 2H 2O O.S. of Mn = +6i

Q > K c , i.e. backward reaction.

N

1 8

11. (b) X : 8 × + 6 ×

HO

 →N NaOH

M=

N is

O–

1 =4 2

1 × 4 + 1 = 2; M 2 X 4 = MX 2 4

12. (c) Energy in 1st excited state = −3.4 eV O–

O

13. (c) Assuming that no 2s-2p mixing takes place

O

(a) Be 2 → σ 1s 2 , σ *1s 2 , σ 2s 2 , σ * 2s 2 (diamagnetic) x (diamagnetic) (b) B2 → σ 1s 2 , σ *1s 2 , σ 2s 2 , σ * 2s 2 , σ 2p z2 , ππ 2p 2p 0 0

1

2

3 O–

O

4

5 O

7

(diamagnetic)

2 2 2 2 2 π 2p π *2p 0 (c) C2 → σ 1s , σ *1s , σ 2s , σ * 2s , σ 2p z , π 2p1x , π *2p0x , σ * 2p z (param

6

C2 8

y

O

O σ 1s 2 , σ *1s 2 , σ 2s 2 , σO* 2s 2 , σ 2p z2 , –

9

π 2p1x π 2p1y

,

π *2p 0x π *2p 0y

0

y

y

, σ * 2p 0z (paramagnetic)

(d) N 2 → σ 1s 2 , σ *1s 2 , σ 2s 2 , σ * 2s 2 , σ 2p z2 , π

2 2 (5) From two data, (forNzero order σ 1s kinetics) , σ *1s 2 , σ 2s 2 , σ * 2s 2 , σ 2p z2 , ππ 22 pp x2 , 2 y x 0.25 KI = = =5 t 0.05 H 14. (b) x 0.60 ⇒ K II = = =5 H t 0.12

4.

1

π

π *2 p 0x , π *2 p 0y

σ * 2p 0z (diamagnetic)

C == C == C sp 2 sp sp 2

H (allene) H

,π π

, σ * 2p 0z (diam

Mock Test-3

113

Energy evolved due to neutralization of HCl and NaOH  0.1 57  5.7 kJ  5700 Joule

15. (b) III > II > I More the branching in an alkane, lesser will be the surface area, lesser will be the boiling point. CH3 CH3 16. (b) Cl * * H3C H3C CH3 CH3

CH3

CH3 CH3

H3C

ms. t  912 m.s  5.7  912 ms  160 Joule / C [Calorimeter constant]

CH2Cl

Md, 1 cannot be separated by fractional distillation.

Energy evolved by neutralization of CH 3COOH and

For Question Nos. 17 to 18

NaOH  200  4.2  5.6  160  5.6  5600 Joule

17. (c)



Energy used to increase temperature of calorimeter  5700  4788  912 Joule

1,d

1,d

H3C Cl

Energy used to increase temperature of solution  200  4.2  5.7  4788 Joule

So

rX d 40   2 rY 24  d 10

energy used in dissociation of 0.1 mole CH 3COOH  5700  5600  100 Joule Enthalpy of dissociation  1 kJ / mole

d  48  2d 3d  48 d  16 cm

1100 1  200 2 1100 1 CH3CONa   200 2 [salt] pH  pK a  log [acid]

20. (b) CH3COOH 

18. (d) As the collision frequency increases then molecular speed decreases than that expected. For Question Nos. 19 to 20

 NaCl  H 2O 19. (a) HCl  NaOH 

pH  5  log 2  log

n  100  1  100 m mole  0.1 mole

  

1/ 2  pH= 4.7 1/ 2

Chemistry

114

JEE-MAIN: CHEMISTRY MOCK TEST-4 1.

a. 1.0 104

The number of moles of SO 2 Cl2 in 13.5 g is a. 0.1 c. 0.3

b. 0.2 d. 0.4

2.

Arrangement of sulphide ions in zinc blende is a. simple cubic b. hcp c. bcc d. fcc

3.

A pressure cooker reduces cooking is increased a. Heat is more evenly distributed b. Boiling point of water inside the cooker is increased c. The high pressure tenderizes the food d. All of the above

4.

Two flasks of equal volume contains SO 2 and CO 2 respectively at 25 C and 2 atm pressure. Which of the

5.

6.

9.

12. Which of the following is property of colloid? a. Scattering of light b. They show attraction c. Dialysis d. Emulsion 13. Which of the following is an example of endothermic reaction?

Oxygen has oxidation states of +2 in the a. H2O2 b. CO 2

14. The nuclear binding energy for Ar (39.962384 amu) is: (given mass of proton and neutron are 1.007825 amu and 1.008665 amu respectively) a. 343.81 MeV b. 0.369096 MeV

d. OF2

Given l / a  0.5 cm 1 , R  50 ohm, N  1.0. The equivalent conductance of the electrolytic cell is

8.

c. 1.0 1010 d. 1.0 1014 11. According to Arrhenius theory, the activation energy is a. The energy it should possess so that it can enter into an effective collision b. The energy which the molecule should possess in order to undergo reaction c. The energy it has to acquire further so that it can enter into a effective collison d. The energy gained by the molecules on colliding with another molecule

following is equal in them ? a. masses of the two gas b. number of molecules c. rates of effusion d. molecular structure The ratio of the kinetic energy to the total energy of an electron in a Bohr orbit is a. –1 b. 2 c. 1 : 2 d. None of these

c. H 2 O 7.

b. 1.0 1010

a. 10 ohm 1 cm 2 gm eq 1

b. 20 ohm 1 cm 2 gm eq 1

c. 300 ohm 1 cm 2 gm eq 1

d. 100 ohm 1 cm 2 gm eq 1

In N 2 molecule, the atoms are bonded by a. One , Two 

b. One , One 

c. Two , One 

d. Three  bonds

For

the

reaction

H 2 (g)  I 2 (g)

2HI(g), the

equilibrium constant changes with a. Total pressure b. Catalyst c. The amounts of H 2 and I 2 taken d. Temperature 10. A certain weak acid has a dissociation constant of 4

1.0  10 . The equilibrium constant for its reaction with a strong base is.

a. C 2 H 2  2H 2   C 2 H 6 ; E  314.0 kJ b. C  O 2   CO 2 ; E  393.5kJ c. N 2  O 2   2NO; E  180.5kJ d. 2H 2  O 2   2H 2 O; E  571.8 kJ

c. 931 MeV

d. None of these

15. Which is an electrophile? a. AlCl3 b. CN 

c. NH 3

d. CH 3OH

16. Cyanide and isocyanide are isomers of type a. Positional b. Functional c. Tautomer d. Structural 17. Acetone will be formed by the ozonolysis of a. Butene-1 b. Butene-2 c. Isobutene d. Butyne-2 18. Which of the H 2 C  C  C  CH 2 ?

following

Zn / CH3 OH a. CH 2 Br  CBr  CH 2 ææææ Æ Aq.K 2 CO 3 b. HC  C  CH 2  COOH   o 40 C

c. CH 2 Br  C  C  CH 2 Br  Zn

Heat

 d. 2CH 2  CH  CH 2 I 

reactions

gives

Mock Test-4

115

19. Which of the following does not form phenol or phenoxide? a. C6 H 5 Cl b. C 6 H 5 COOH c. C6 H5 N 2 Cl

d. C 6 H 5SO 3 Na

20. Identify the final product (Z) in the following sequence of reactions: 

H 3O H 2SO4 Me 2 C  O  HCN   X   Y   Z; Heat

a. (CH 3 ) 2 C(OH)COOH

b. CH 2  C(CH 3 )COOH

c. HOCH 2 CH(CH 3 )COOH d. CH 3CH  CHCOOH 21. When propionic acid is treated with aqueous sodium bicarbonate, CO 2 is liberated. The C of CO 2 comes from? a. methyl group c. methylene group

b. carboxylic acid group d. bicarbonate group

22. Aniline undergoes condensation to form Schiff base on reacting with a. Acetyl chloride b. Ammonia c. Acetone d. Benzaldehyde 23. The mass average molecular mass and number average molecular mass of a polymer are respectively 40,000 and 30,000. The polydispersity index of polymer will be a. < 1 b. > 1 c. 1 d. 0 24. Hardening of oils is caused by a. H 2 b. N 2 c. O 2 Space for rough work

d. CO 2

25. Which of the following is molecular disease? a. Allergy b. Cancer c. German measeles d. Sickel-cell-anaemia 26. Hydrogen can be put in halogen group because a. It has deuterium and tritium as isotopes b. It forms hydrides like chlorides c. It contains one electron only d. It is light 27. Which of the following metal is amalgamation process? a. Tin b. Silver c. Copper d. Zinc

extracted

by

28. Which of the following hydroxide is insoluble in water? a. Be(OH)2

b. Mg(OH) 2

c. Ca(OH)2

d. Ba(OH)2

29. The product of oxidation of I ion by MnO4 in alkaline

medium is a. I2

b. IO3

c. IO4

d. I3

30. Potassium ferrocyanide is a a. Normal salt c. Double salt

b. Mixed salt d. Complex salt

Chemistry

116

JEE ADVANCE PAPER-I SECTION 1 Contains 8 Questions

6.

The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

1.

CH3 — CH 2 — CH — CH 3 | OH

HCl gas is passed into water, yielding a solution of

enantiomeric (  )

OH | CH3 — C — CH3 ; | CH3

1

density 1.095g mL and containing 30% HCl by weight. Calculate the molarity of the solution. 2.

A sample contains a mixture of NaHCO3 and Na 2 CO3 . HCl is added to 15.0 g of the sample, yielding 11.0 g of NaCl. What percent of the sample is Na 2 CO3 ?

CH 3 — CH 2 — CH 2 — CH 2 OH

7.

Reactions are:

CH 3 — CH — CH 2 OH | CH 3

Total number of isomers (including stereoisomers) is Total number of isomers CH 3 CH 2 CH3 CH 3 CH 3

 Na 2 CO 3  2HCl   2NaCl  CO 2  H 2 O     NaCl  CO 2  H 2 O   NaHCO 3  HCl 

Mw of NaCl  58.5, Mw of NaHCO3  84, Mw of

Na 2 CO3  106g mol1 3.

CH 3

CH 3 H

CH 3

H

H CH 3

H

Amongst the following, the total number of compounds soluble in aqueous NaOH is H 3C

N

CH 3

COOH

OH

NO 2

H 3C OCH 2CH 3 CH 2OH

N

meso

8.

Mirror

CH 3

H

H

CH 3

Pair of enantiomers

In the scheme given below, the total number of intramolecular aldol condensation products formed from ‘Y’ is 1. O3 1. NaOH(aq)   Y    2. Zn, H2 O 2. heat

CH 3

OH SECTION 2 Contains 10 Multiple Choice Questions With one or more than one correct option

9. CH 2CH3 CH 2CH3

4.

5.

COOH

The total number of contributing structures showing hyperconjugation (involving C—H bonds) for the following carbocation is. The total number(s) of stable conformers with non-zero dipole moment for the following compound is (are) Cl Br

CH 3

Br

Cl CH 3

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is Assuming concentration of solute is 2C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take K b  0.76 K kg mol1 )

a. 724 b. 740 c. 736 d. 718 10. Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T1 are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then I raised to T2. The final pressure pf is: T1

T1 Pi , V

T1 Pi , V



T2 Pf ,V

Pf ,V

Mock Test-4

117

 TT  a. p i  1 2   T1  T2 

 T1  b. 2p i    T1  T2 

 T2  c. 2p i    T1  T2 

 TT  d. 2p i  1 2   T1  T2 

11. A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/ (where  is wavelength associated with electron wave) is given by: a. meV b. 2 meV c.

d.

meV

2 meV

13. For the following electrochemical cell at 298 K, 4

2

Pt(s) | H 2 (g, 1 bar) H (aq, 1 M) || M (aq), M (aq) | Pt(s)

E cell  0.092 V when

2

[M (aq)]  10 x. [M 4  (aq)]

Given: E 0M4 / M2  0.151 V; 2.303 The value of x is a. –2 c. 1

RT  0.059 V F

b. –1 d. 2

15. Aqueous solutions of HNO3, KOH, CH3COOH and CH3COONa of identical concentrations are provided. The pair (s) of solutions which form a buffer upon mixing is (are) a. HNO3 and CH 3COOH b. KOH and CH 3COONa c. HNO3 and CH 3COONa d. CH 3COOH and CH 3COONa

B is

(Given  r H o298K  54.07 kJ mol1 , rSo298K  10 JK1mol1 1

1

and R  8.314 JK mol ; 2.303 × 8.314 × 298 = 5705 ) a. 5

b. 10

c. 95

c. H 2 C  C  O

d. H 2 C  C  CH 2

You will have to match entries in Column I with the entries in Column II.

19. According to Bohr’s theory, E n  Total energy

K n  Kinetic energy

Vn  Potential energy

rn  Radius of nth orbit

Match the following: Column I

Column II

(A) Vn / K n  ?

1. 0

(B) If radius on nth orbit

2. –1

E ,x ? x n

(C) Angular momentum in lowest orbital (D)

14. Which one of the following molecules is expected diamagnetic behaviour? a. C2 b. N2 c. O2 d. S2

16. The value of log10K for a reaction A

18. Amongst the given options, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any), is (are) H H H CC a. b. H  C  C — C H2C CH 2 CH 2

SECTION 3 Contains 2 Matches The Following Type Questions

12. Reduction of the metal centre in aqueous permanganate ion involves a. 3 electrons in neutral medium b. 5 electrons in neutral medium c. 3 electrons in alkaline medium d. 5 electrons in acidic medium 

17. The number of structural isomers for C6H14 is a. 3 b. 4 c. 5 d. 6

d. 100

3. –2 4. 1

1  Zy , y  ? n r

a. A  3; B  2; C  4; D  1 b. A  1; B  2; C  3; D  4 c. A  3; B  1; C  2; D  4 d. A  3; B  2; C 1; D 4 20. Match the thermodynamic processes given under Column I with the expression given under Column II: Column I

Column II

(A) Freezing of water at 273 K and 1 atm

1. q = 0

(B) Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions

2. w = 0

(C) Mixing of equal volumes of two ideal gases at constant

3. Ssys  0

Chemistry

118

temperature and pressure in an isolated container (D) Reversible heating of

cooling to 300 K at 1 atm 5. 4. U  0

G  0

a. A  3,5; B  1,2,4 ; C  1,2,4; D  1,2,4,5

H 2 (g) at 1 atm from

b. A  1,2,4; B  1,2,4,5; C  3,4; D  1,2,4

300 K to 600 K, followed by reversible

c. A  3,5; B  1,2,4; C  1,2,4,5; D  1,2,4

Space for rough work

d. A  1,2,4; B  3,5; C  1, 4,5; D  1,2,4

Mock Test-4

119

JEE ADVANCE PAPER-II SECTION 1 Contains 8 Questions The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

1.

0.45 g of an acid (mol wt. = 90) required 20 ml of 0.5 N KOH for complete neutralization. Basicity of acid is

2.

The co-ordination number of copper in cuprammonium sulphate is

3.

The co-ordination number of cobalt in the complex [Co(en) 2 Br2 ]Cl2 is

4.

The primary valence of the metal ion in the co-ordination compoun K 2 [Ni  CN 4 ] is

a.

h2 4 2 ma 02

b.

h2 16 2 ma 02

c.

h2 32 2 ma 02

d.

h2 64 2 ma 02

12. The equilibrium

Cu  Cu II

2Cu I

in aqueous

medium at 25ºC shifts towards the left in the presence of a. NO 3

b. Cl

c. SCN

d. CN 

13. The bond energy (in kcal mol–1) of C – C single bond is approximately a. 1 b. 10 c. 100 d. 1000 14.

N 2  3H 2

2NH 3 Which is correct statement if N 2

5.

The oxidation number of Cr in [Cr(NH 3 )6 ]Cl3 is

6.

The number of equivalent Cr  O bonds in CrO 24 is.

7.

The number of the following reagents that produce ppt.

is added at equilibrium condition? a. The equilibrium will shift to forward direction because according to IInd law of thermodynamics the entropy must increases in the direction of spontaneous reaction

with ZnSO 4 solution is.

b. The condition for equilibrium is G N2  3G H2  2G NH3

NaOH, N 2 CO3 , NaCl, Na 2 HPO 4 , Na 2S, CH 3CO3 Na

where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent c. The catalyst will increase the rate of forward reaction

8.

The change in the magnetic moment value when 2

Cu  H2O 4  is converted to Cu  NH3 4 

2

is.

SECTION 2 Contains 8 Multiple Choice Questions With one or more than one correct option

9.

A gas described by van der Waal’s equation a. behaves similar to an ideal gas in the limit of large molar volumes b. behaves similar to an ideal gas in the limit of large pressures c. is characterized by van der Waal’s coefficients that are dependent on the identity of the gas but are independent of the temperature d. has the pressure that is lower than the pressure exerted by the same gas behaving ideally

10. Assuming that Hund’s rule is violated, the bond order and

magnetic nature of the diatomic molecule B 2 is a. 1 and diamagnetic c. 1 and paramagnetic

b. 0 and diamagnetic d. 0 and paramagnetic

11. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius].

by α and that of backward reaction by . d. Catalyst will not alter the rate of either of the reaction 15. For a first order reaction AP, the temperature (T) dependent rate constant (k) was found to follow the 1 equation log k  (2000)  6.0. The pre-exponential T factor A and the activation energy Ea, respectively, are 6 1 a. 1.0 10 s and 9.2 kJmol 1 1 b. 6.0s and 16.6 kJmol 1 6 1 c. 1.0 10 s and 16.6 kJmol 1 6 1 d. 1.0 10 s and 38.3 kJmol 1

16. The initial rate of hydrolysis of methyl acetate (1 M) by a weak acid (HA, 1M) is 1/100th of that of a strong acid (HX, 1M), at 25ºC. The Ka of HA is a. 1104

b. 1105

c. 1106

d. 1103

Chemistry

120 SECTION 3 Contains 2 Paragraph Type Questions

b. At the start of the reaction, dissociation of gaseous X 2

Each paragraph describes an experiment, a situation or a problem. Two multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.

takes place spontaneously c. βequlibrium = 0.7 d. K C < 1

Paragraph for Question No. 17 to 18 Thermal decomposition of gaseous X 2 to gaseous X at 298 K

Paragraph for Question No. 19 to 20

takes place according to the following equation: X 2 (g) 2X(g)

Tollen’s reagent is used for the detection of aldehyde when a solution of AgNO 3 is added to glucose with NH 4 OH then

The standard reaction Gibbs energy, ∆r G° of this reaction is

gluconic acid is formed

positive. At the start of the reaction, there is one mole of X 2

Ag + + e −  → Ag ; E ored = 0.8 V

and no X. As the reaction proceeds, the number of moles of X formed is given by β . Thus β equilibrium is the number of moles of

→ Gluconic acid C6 H12 O 6 + H 2 O 

X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: R = 0.083 L bar K −1 mol −1 )

Ag(NH3 ) 2+ + e −  → Ag (s) + 2NH3 ; E ooxd = 0.337 V [Use 2.303 ×

17. The equilibrium constant K p for this reaction at 298 K, in

19.

terms of β equilibrium , is a.

c.

2 8βequilibrium

2 − βequilibrium 2 4βequilibrium

2 − βequilibrium

b.

d.

2 8βequilibrium 2 4 − βequilibrium 2 4β equilibrium 2 4 − βequilibrium

(C6 H12 O7 ) + 2H + + 2e− ; E ooxd = − 0.05 V RT F = 0.0592 and = 38.92 at 298 K F RT

2Ag − + C6 H12 O6 + H 2O  → 2Ag (s) + C6 H12 O7 + 2H + Find ln K of this reaction. a. 66.13 c. 28.30

b. 58.38 d. 46.29

20. When ammonia is added to the solution, pH is raised to 11. Which half-cell reaction is affected by pH and by how much? a. E oxd will increase by a factor of 0.65 from Eooxd

18. The incorrect statement among the following, for this reaction, is a. Decrease in the total pressure will result in formation of more moles of gaseous X Space for rough work

b. E oxd will decrease by a factor of 0.65 from Eooxd c. E red will increase by a factor of 0.65 from E ored d. E red will decrease by a factor of 0.65 from E ored

121

Mock Test-4

ANSWER & SOLUTIONS

13. (a) For exothermic reactions H p < H R .

JEE-Main

For endothermic reactions H p > H R .

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

a

d

b

b

a

d

a

a

d

c

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

a

a

a

a

b

c

c

b

c

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

d

d

b

a

b

b

b

a

b

d

1.

2.

Mass defect = [m × p + m × n] − 39.962384

= [1.007825 ×18 + 1.008665 × 22] − 39.962384

15. (a) AlCl3 is lewis acid i.e., electron deficient compound. 2−

(d) Arrangement of sulphide ions (S ) in zinc blende 2+

ions occupy alternate tetrahedral

So it is electrophile. r 16. (b) R — C ≡≡ N and R — N = C are functional isomers. Cyanide

(b) The temperature at which a liquid boils increases with increase in pressure.

4.

(b) Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Isocyanide

CH 3C == O | CH 3

17. (c) CH 3 — C == CH 2 | O3  → CH 3 H2O, Zn

Acetone

Zn 18. (c) CH2Br − C ≡ C − CH2Br → CH 2 = C = C = CH 2 ∆

19. (b) Benzoic acid.

5.

(a) K.E. = – (T.E.)

6.

(d) Oxygen have + 2 oxidation state in OF2 .

20. (b)

−1

(a) l / a = 0.5 cm , R = 50 ohm

Ra 50 = = 100 l 0.5 1000 1 1000 1 1000 Λ = k× = × = × = 10 ohm−1 cm2 gm eq−1 N p N 100 1

p=

π

Total no of neutrons = 22

= 0.369 × 931 = 343.62MeV

3.

7.

Ar 40 Total no of protons = 18

Binding energy = mass defect × 931

13.5 = 0.1 135

(ZnS) is fcc while Zn voids.

18

= [18.14085 + 22.19063] − 39.962384 = 0.369

(a) Molecular mass of SO 2 Cl 2 = 32 + 2 × 16 + 35.5 × 2 = 135 Moles =

14. (a)

OH | H3O+ → CH 3 — C — CN  Me2C = O + HCN  → | CH 3 (X )

OH | H2SO4 → CH 2 == CCOOH CH 3 — C — COOH  | | CH3 CH 3 (Z)

(Y )

σ

*

8.

(a) N

→ CH 3CH 2 COONa + H 2 O 21. (d) CH 3 — CH 2 — COOH + Na H CO3 

N *

π 9.

(d) Equilibrium constant changes with temperature, pressure and the concentration of either reactant or product.

10. (c) HA : K a = 10−4

HA + NaOH HA + NaOH

NaA +H NaA + 2HO O

Clearly, the reverse reaction is the hydrolysis reaction.

⇒

K Required =

*

CH 3 — CH 2 — COOH + Na H CO 3  → CH 3CH 2 COONa + H 2 O + C O 2

K 1 10−4 = a = −14 = 1010 K h K w 10

11. (c) The definition of activation energy. 12. (a) Scattering of light is a property of colloid.

22. (d) C6H5NH2 + O = CHC6H5  →C6H5 − N = CHC6H5 + H2O Schiff 's base

23. (b) Average number molecular weight Mn = 30,000 Average mass molecular weight M w = 40,000 Polydispersity index (PDI) =

M w 40,000 = = 1.33 Mn 30,000

Ni 24. (a) Oil (unsaturated) + H 2  → Fat (saturated)

25. (b) “Cancer” is known as molecular disease.

Chemistry

122

26. (b) Hydrogen, forms hydrides like halides, e.g. HCl.

4.

27. (b) Cu 2 Cl 2  Ag 2S   Cu 2S  2AgCl

(6) These are total 6 — H to sp carbon and they all can 2

participate in hyperconjugation.

2AgCl  Hg   Hg 2 Cl 2  2Ag



H

AgCl  Hg   Ag  HgCl

 4



 3

29. (b) KI  MnO   K IO  Mn

2

30. (d) In K 4 Fe(CN) 6 , the species retains its identity in solid

as well as in solution state. JEE Advance Paper -I 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

9

9

4

6

3

5

7

1

a

c

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

d

a, d

d

c

c. d

b

c

b, c

d

a

Three structures

5.

(3)

6.

(5)

7.

(7)

8.

(1)

9.

(a) TL  k b  m  k b 

2  0.76 

(9) M 

2.

(9) Let x be the percentage of Na 2 CO3 . Then, weight of NaHCO3  (15  x)g

11.0 g Moles of NaCl produced   0.18 mol 58.5g

 x  The NaCl is produced by the reaction of   mol of  106  Na 2 CO3 and

3.

1.35 100  9.0% Na 2 CO3 15

11. (d) K.E.  eV



COOH

OH

h 2meV



N CH 3

 2meV

MnO 4  8H   5e    Mn 2   4H 2 O

In neutral medium, MnO 4  2H 2 O  3e    MnO 2  4OH  Hence, number of electrons loose in acidic and neutral medium are 5 and 3 respectively. 13. (d) E cell  E ocell 

0.092  0.151  H 3C

h



12. (a, d) In acidic medium

H 3C  CH 2 CH 3

OH

1 1     T2 T1 

 T  T2   T2  2Pi  Pf  1   Pf  2Pi   T1  T1T2   T1  T2 

(4) Aromatic alcohols and carboxylic acids form salt with NaOH, will dissolve in aqueous NaOH.

COOH

Ws Ws 2.5  18    36.0 Wsolv 9.5  100 M solv

Pi Pi Pf Pf 2Pi     Pf T1 T1 T2 T1 T1

NaHCO3  (15  1.35)  13.6 g

% Na 2 CO3 

Ws M s  Wsolution

10. (c) Initial moles = final moles Pi  V Pi  V Pf  V Pf  V    RT1 RT1 RT2 RT1

(15  x) mol of NaHCO 3 . Each mol of 84

2 x 15  x   0.188 Solve x :  13.5g Na 2 CO3 , 106 84

Two structures

x  760  36  724

Na 2 CO3 produces 2 mol of NaCl. 



2.5  1000 0.76  2.5  1000 Ms   9.5 Ms 100 100  2

ns 760  x  Xs  760 n solution

% by weight 10  d 30  10 1.095   9M Mw2 36.5

1.



 

28. (a) The solubility of hydroxides of alkaline earth metals in water increases on moving down the group.

H

H



x2

0.059 [M 2  ][H  ]2 log10 2 [M 4  ]pH 2

0.059 log10 10x 2

123

Mock Test-4

14. (c) O2 is expected to be diamagnetic in nature but actually it is paramagnetic. 15. (c, d) In option (c), if HNO3 is present in limiting amount then this mixture will be a buffer and the mixture given in option (d), contains a weak acid (CH3COOH) and its salt with strong base NaOH, i.e. CH3COONa. 16. (b) ∆G ° = ∆H° − T∆S° = −54.07 × 1000 − 298 × 10

= −57050 J mol−1 − 57050 = −5705log10 K

4.

(2) Primary valencies are also known as oxidation state. K 2 [ Ni (CN) 4 ], 2 + x − 4 = 0 ⇒ x = +2

5.

(3) x + 6 × (0) + 3 × (−1) = 0

6. 7.

x – 3 = 0, x + 3, Oxidation number of Cr is = +3 (4) (4)

8.

(0)

log10 K = 10 9.

Hence (b) is correct. 17. (c) C6 H14 H 3C — CH 2 — CH 2 — CH 2 — CH 2 — CH 3 H 3 C — CH — CH 2 — CH 2 — CH 3 | CH 3 H 3C — CH — CH — CH 2 — CH 3 | CH 3

 n2a  temperature. The term  P + 2  represents the pressure V   exerted by an ideal gas while P represents the pressure exerted by a real gas.

H 3C — CH — CH — CH 3 | | CH 3 CH 3

CH3 | H 3C — C — CH 2 — CH 3 , Hence (c) is correct | CH 3

* * 10. (a) B2 (10) = σ1s2 σ 1s2σ 2s2σ 2s2 π 2p2

x

6−4 = 1 (nature is diamagnetic as no 2 unpaired electron)

Bond order

18. (b, c) 19. (d) A → 3; B → 2; C →1; D →4

11. (c) mvr =

20. (a) A → 3,5; B → 1,2,4; C → 1,2,4; D → 1,2,4,4 JEE Advance Paper -II 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

2

4

6

2

3

4

4

0

a, c, d

a

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

b,c,d

c

b

d

a

b

c

b

a

1.

(2) Normality = N =

∴

Molec. Wt 90 Basicity = = =2 Eq. Wt 45

2.

(4) In

Cuprammonium

sulphate

[Cu(NH3 )4 ]SO4 co-

ordination no. of Cu is 4. 3.

nh mv2 e2 = 2 ⇒ mv2 r = e2 and 2π r r

e2 × 2π nh ∴ (mvr = ) nh 2π

v=

⇒

1 4 me × 4π 2 1 2 K.E. = mv = 2 2 2 2 nh

Expression for a 0 =

0.45 × 1000 = 45 0.5 × 20

Eq. Wt =

=

⇒

WB × 1000 Eq.wt × V

∴

 n 2a  (a, c, d)  P + 2  (V − nb) = nRT V   At low pressure, when the sample occupies a large volume, the molecules are so far apart for most of the time that the intermolecular forces play no significant role, and the gas behaves virtually perfectly. a and b are characteristic of a gas and are independent of

(6) [Co(en)2 Br2 ]Cl2

C.N.of Co = 2 × number of bidentate ligand

+1× number of monodentate ligand = 2 × 2 + 1× 2 = 6.

. . .(i)

h2 4π 2 me2

⇒

me2 =

h2 4π 2a 0

. . .(ii)

⇒

K.E. =

h2 1 × 2 2 2 8ma 0π n

. . .(iii)

For n = 2 K.E. =

h2 32π 2 ma 02

12. (c, d) Cu2+ ions will react with CN– and SCN– forming

[Cu(CN) 4 ]3 − and [Cu(SCN) 4 ]3− leading the reaction in the backward direction.

Chemistry

124

Cu

2



 2CN   Cu(CN) 2

2Cu(CN) 2   2CuCN  (CN) 2

Px(g)

CuCN  3CN    [Cu(CN) 4 ]3 Cu 2   4SCN    [Cu(SCN) 4 ]3

2

Cu 2  also combines with CuCl2 which reacts with Cu to produce CuCl pushing the reaction in the backward direction.

So

CuCl2  Cu   2CuCl 

 2 eq.   Ptotal   (Px) 2  2   eq.  Kp   (Px 2 )  2   eq.   Ptotal   (2  )  eq  

 Kp 

13. (c) 14. (b)

Since, log K  log A 

Ea 2.303RT

6 1 So, A  10 sec

[H  ]weak acid 



[H ]weak acid



C  102



K a  104



1 (rate in strong acid) 100

Kp 

1 [H  ]strong acid 100

8 eq2 4  eq2



8  (0.7)2  1, but 4  (0.7)2

(d) Correct statement.

As G  0 & G  RT n K p

1  M  10 2 M 100

G  1, so K p should be less than 1. So

K 1 K p  K c (RT) ng. (RT  1)

Kc 

For Question Nos. 17 to 18 17. (b) Paragraph-1 X 2 (g)

Initial mole t  t eq.

 8eq2    P  total 2 2   4  eq.  4  eq 

place spontaneously. (c) Incorrect statement as

and Ea  38.3kJ / mole 16. (a) Rate in weak acid 

2 4eq.

18. (c) (a) Correct statement. As one decrease in pressure reaction will move in the direction where no. of gaseous molecules increases. (b) Correct statement At the start of reaction Qp  K p so dissociation of X 2 take

2000 15. (d) Given, log K  6  T



   eq   2eq  Ptotal    Ptotal   1  eq   2  eq    2

2X(g)

1 (1   )

Given 2  equilibrium So,  

0 2

equilibrium 2

  eq  Total mole at equilibrium  (1   )  (1     2 

  eq.  1   2 P    2   eq P    2   eq P  Px 2     total total  total  1   eq.   2   eq   2   eq  2  

So

Kp

RT Kc  1

 Kc  Kp

For Question Nos. 19 to 20 RT 19. (b) E ocell  ln K nF 1 0.0592 (0.8  0.05)   ln K 2 2.303 (0.8  0.05)  22.303 ln k   58.38 0.0592 20. (a) On increasing concentration of NH3 the concentration of H+ ion decreases. Therefore, Ered increases.

  

Mock Test-5

125

JEE-MAIN: CHEMISTRY MOCK TEST-5 1.

2.

3.

One mole of potassium dichromate completely oxidises the following number of moles of ferrous sulphate in acidic medium a. 1 b. 3 c. 5 d. 6 In a solid lattice, the cation has left a lattice site and is located at an interstitial position, the lattice defect is a. Frenkel defect b. Schottky defect c. F-centre defect d. Valency defect At 300 K, the vapour pressure of an ideal solution containing 3 mole of A and 2 mole of B is 600 torr. At the same temperature, if 1.5 mole of A and 0.5 mole of C (nonvolatile) are added to this solution the vapour pressure of solution increases by 30 torr. What is the value of PBo ? a. 940 c. 90

4.

5.

6.

7.

b. 405 d. None of these

If the root mean square speed of helium is 4.75 m s–1 at 25C, then its speed will become 9.50 m s–1 at a. 100C

b. 323C

c. 919C

d. 1192C

Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon? a.

3s

b. 2p

c.

2s

d.

1s

The formation of nitric oxide by contact process 2NO. H  43.200 kcal is favoured by

N 2  O2

a. Low temperature and low pressure b. Low temperature and high pressure c. High temperature and high pressure d. High temperature and excess reactants concentration 10. The pH of an aqueous solution Mg  OH 2 is 9.0. If the

solubility product of Mg  OH 2 is 1 10 11 , what is [ Mg 2  ] ? a. 1  10 5

b. 1.0  104

c. 1 102

d. 0.1

11. If ‘I’ is the intensity of absorbed light and C is the concentration of AB for the photochemical process AB  hv  AB*, the rate of formation of AB* is directly

proportional to a. C c. I 2

b. I d. C.I

12. Gold number is maximum for the lyophilic sol is a. Gelatin b. Haemoglobin c. Sodium oleate d. Potato starch 13. For the allotropic change represented by equation C(diamond)  C(graphite); the enthalpy change is H  1.89 kJ. If 6 g of diamond and 6 g of graphite are

separately burnt to yield carbon dioxide, the heat liberated in the first case is a. Less than in the second case by 1.89 kJ

The product of oxidation of I  with M nO 4 in alkaline

b. More than in the second case by 1.89 kJ

medium is a. IO 3

c. Less than in the second case by 11.34 kJ b. I 2

c. IO 

d. IO 4

The standard reduction electrode potentials of four elements are A   0.250 V, B   0.136 V, C   0.126 V and D   0.402 V. The element that displaces A from its compounds is a. B c. D

8.

9.

b. C d. None of these

The bond angle in carbon tetrachloride is approximately a. 90 b. 109 c. 120 d. 180

d. More than in the second case by 0.945 kJ 14. The half-life of 6 C14 if its K or  is 2.31 104 is a. 2  10 2 yrs

b. 3  10 3 yrs

c. 3.5  10 4 yrs

d. 4  103 yrs

15. Most stable carbonium ion is 



a. C 2 H 5

b. (CH 3 )3 C 

c. (C 6 H 5 )3C



d. C 6 H 5CH 2

16. Which of the following compounds will exhibit cis-trans isomerism a. 2-butene b. 2-butyne c. 2-butanol d. Butanone

Chemistry

126

17. Acetylene gas is obtained by the electrolysis of a. Sodium fumarate b. Sodium succinate c. Sodium maleate d. Both (a) and (c) 18. The compound added to prevent chloroform to form phosgene gas is a. C2 H5OH b. CH3COOH c. CH 3COCH 3 19. Epoxides are a. Cyclic ethers b. Not ethers c. Aryl-alkyl ethers d. Ethers with another functional group NaOH / H  reacts with

a. C6 H 5OCH 3

b. CH 3 OH

O || c. CH 3 — C — CH 3

d. C2 H5OH 

21. In the reaction C8 H 6O 4   X   NH3

The compound X is a. Phthalic anhydride b. Phthalic acid c. o-xylene d. Benzoic acid Zn , HCl 22. In the reaction CH 3 (CH 2 ) 4 CN   A  HONO O A   B   C ; C is

a. Pentanal c. 2-Hexanone

b. Pentanone d. Hexanal

23. A polymer containing nitrogen is a. Bakelite b. Dacron c. Rubber d. Nylon-66 Space for rough work

25. Which of the following is an antidiabatic drug a. Insulin b. Penicillin c. Chloroquine d. Aspirin 26. The correct sequence of elements in decreasing order of first ionisation energy is a. Na  Mg  Al b. Mg  Na  Al c. Al  Mg  Na

d. CH 3 OH

20.

24. The base present in DNA, but not in RNA is a. Guanine b. Adenine c. Uracil d. Thymine

d. Mg  Al  Na

27. In the metallurgical extraction of zinc from ZnO the reducing agent used is a. Carbon monoxide b. Sulphur dioxide c. Carbon dioxide d. Nitric oxide 28. In the lime (kiln), the reaction CaCO3 (s)   CO2 (g)

goes to completion because a. Of high temperature b. CaO is more stable than CaCO3 c. CO2 escapes simultaneously d. CaO is not dissociated 29. Amalgams are a. Highly coloured alloys b. Always solid c. Alloys which contain mercury as one of the contents d. Alloys which have great resistance to abrasion 30. Generally, a group of atoms can function as a ligand if a. They are positively charged ions b. They are free radicals c. They are either neutral molecules or negatively charged ions d. None of these

Mock Test-5

127

JEE ADVANCE PAPER-I SECTION 1 Contains 8 Questions

8.

The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

1.

3.

4.

takes place simultaneously by SN 2 and SN1 pathways. A plot of 

The change in the number of unpaired electrons when

the slope equal to 2  103 mol 1 L h 1 and intercept equal to

A compound of mol. wt. 180 is acetylated to give a compound of mol. wt. 390. The number of amino groups in the initial compound is?

consumption of RX when the reaction is carried out taking

4

1102 h 1. Calculate the initial rate (mol L1 min 1 ) of 1mol L1 of RX and 0.1 mol L1 of [OH  ] ions.

Specific rotations of -anomer of glucose is 112  and

SECTION 2 Contains 10 Multiple Choice Questions

for -anomer is + 19  . Specific rotation of equilibrium

With one or more than one correct option

mixture is 52.6 . Calculate % composition of -and anomers in the equilibrium mixture.

9.

N3 , O2 and

F are

a. 1.36, 1.40 and 1.71 b. 1.36, 1.71 and 1.40 c. 1.71, 1.40 and 1.36

are also reacted separately). The total number of ketones that give a racemic product(s) is/are Amongst the following, the total number of compounds soluble in aqueous NaOH is

The ionic radii (in Å) of respectively:

Consider all possible isomeric ketones including stereoisomers of MW = 100. All these isomers are independently reacted with NaBH 4 (Note: Stereoisomers

5.

1 d[R  X] vs. [OH  ] is a straight line of [RX] dt

 Fe  H 2 O 6  is changed into  Fe  CN 6  is

2

2.

Hydrolysis of an alkyl halide (RX) by dilute alkali [OH  ]

d. 1.71, 1.36 and 1.40 10. From the following statements regarding H 2O 2 , choose

the incorrect statement a. It can act only as an oxidizing agent

H3C

CH3

b. It decomposes on exposure to light

COOH

N

OCH2CH3 CH2OH3

OH

c. It has to be stored in plastic or wax lined glass bottles in dark d. It has to be kept away from dust

OH

NO2

11.

CH2CH3 CH2OH3

gradually and the conductivity of the solution was

COOH

measured. The plot of conductance (^) versus the volume of AgNO3 is

N H 3C 6.

7.

+

CH3

The half-life period of a radioactive substance is 2 min. The time taken for 1 g of the substance to reduce to 0.25 g will be  f H  of hypothetical MgCl is 125 kJ mol 1 and for

MgCl2

is

AgNO3 (aq.) was added to an aqueous KCl solution

642 kJ mol 1 .

The

enthalpy

disproportionation of MgCl is –49x. Find the value of x.

of

a. 

+

+ ++

+

volume (a)

c.  ++ + + + + + + volume (c)

b. 

+ + +

+ + + volume (b)

+ + d.  + + + + volume (d)

+

Chemistry

128

12. The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is: a. ion-ion interaction b. ion-dipole interaction c. London force d. hydrogen bond 13. According to Molecular Orbital Theory a. C 22  is expected to be diamagnetic

–1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25C is a. + 2900 kJ b. – 2900 kJ c. – 16.11 kJ d. +16.11 kJ 18. Which of the following compounds is not colored yellow? a. Zn 2 [Fe(CN) 6 ]

b. O 22  is expected to have a longer bond length than O2

b. K 3[Co(NO2 )6 ]

c. N and N have the same bond order

c. (NH 4 )3 [As(Mo3O10 ) 4 ]

d. He 2 has the same energy as two isolated He atoms

d. BaCrO 4

 2

14. For

 2

the

first

order

reaction

2N 2 O5 (g)   4NO 2 (g)  O 2 (g)

a. the concentration of the reactant decreases exponenttially with time b. the half-life of the reaction decreases with increasing temperature c. the half-life of the reaction depends on the initial concentration of the reactant d. the reaction proceeds to 99.6% completion in eight half-life duration

 R  S the time taken for 75% 15. In the reaction, P  Q  reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is [Q]0

[Q]

Time

a. 2

b. 3

c. 0

d. 1

16. Choose the correct reason(s) for the stability of the lyophobic colloidal particles. a. Preferential adsorption of ions on their surface from the solution b. Preferential adsorption of solvent on their surface from the solution c. Attraction between different particles having opposite charges on their surface d. Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles 17. The standard enthalpies of formation of CO 2 (g), H2O(l )

and glucose(s) at 25C are –400 kJ/mol, –300 kJ/mol and

SECTION 3 Contains 2 Match The Following Type Questions You will have to match entries in Column I with the entries in Column II.

19. Match each of the diatomic molecules in Column I with its property/properties in Column II. Column - I Column – II 1. Paramagnetic (A) B2 (B) N 2

2. Undergoes oxidation

(C) O 2

3. Undergoes reduction

(D) O2

4. Bond order ≥ 2 5. Mixing of ‘s’ and ‘p’ orbitals

a. A  5, 3, 1, B  5, 4, C  2, 1, D  4, 1, 2 b. A  1, 3, 5, B  1, 2, 4 C  3, 2, D  1, 2, c. A  1, 3, B  1, 2, 4 C  1, 2, 3 D  4, 5, d. A1, 3, 5, B4, 5, C1, 2, D1, 2, 4 [According to MOT] 20. Match the compounds/ions in Column I with their properties/reactions in Column II. Column I Column II (A) C6H5CHO

(B) CH3C  CH

1. gives precipitate with 2, 4-dinitrophenylhydrazine 2. gives precipitate with AgNO3

(C) CN 

3. is a nucleophile

(D) I 

4. is involved in cyanohydrin formation

a. A  1, 2, 3, B  2, C  1, 3, 4, D  2, 3 b. A  1, 3, 4, B  2, 3 C  1, 2, 3, D  2 c. A  1, 3, B  1, 4, C  1, 3, D  2, 3 d. A  2, 3, B  1, 2, C  1, 3, D  1, 2

Mock Test-5

129

JEE ADVANCE PAPER-II SECTION 1 Contains 8 Questions

10. Among the electrolytes Na 2SO4 , CaCl2 , Al2 (SO 4 )3 and

The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

1.

The number of rings formed in

 Ca  EDTA  

2

is

…………… 2.

The number of equivalent Cl  O bonds in Cl2O7 is

NH4Cl, the most effective coagulating agent for Sb2S3 sol is a. Na 2SO4

b. CaCl2

c. Al2 (SO4 )3 d. NH4Cl

11. The compound(s) with TWO lone pairs of electrons on the central atom is(are)

……………

a. BrF5

b. ClF3

3.

In fructose, the possible optical isomers are……………

c. XeF4

d. SF4

4.

Oxidation of glucose is one of the most important reactions in a living cell. What is the number of ATP molecules generated in cells from one molecule of glucose?

5.

10 gm of a mixture of hexane and ethanol reacts with sodium to give 200 ml. of H 2 at 27  C and 760 mm pressure. The percentage of ethanol is……………

6.

The polymerisation of propene to linear polypene is represented by the reaction

 CH 3   CH 3  |  |   n  CH  CH 2    — CH — CH 2— n Where n has large integral value, the average enthalpies of bond dissociation for (C  C) and (C  C) at 298 K are +590 and 331kJ mol 1 , respectively. The enthalpy of polymerization is 360 kJ mol1 . Find the value of n. 7.

In the case of a first order reaction, the time required for 93.75% of reaction to take place is x times that required for half of the reaction. Find the value of x.

8.

In 1 L saturated solution of

12. Solubility product constant MX, MX 2 and

M 3X

4.0  108 , 3.2  1014

and

(K sp ) of salts of types

at

temperature 2.7  1015 ,

are

respectively.

3

Solubilities (mole dm ) of the salts at temperature ‘T’ are in the order a. MX  MX 2  M 3 X

b. M 3 X  MX 2  MX

c. MX 2  M 3 X  MX

d. MX  M 3 X  MX 2

13. According to the Arrhenius equation a. a high activation energy usually implies a fast reaction b. rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy c. higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant d. the pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy

added. The resultant concentration of Ag  in the solution

14. For a linear plot of log (x/m) versus log p in a Freundlich adsorption isotherm, which of the following statements is correct? a. Both k and 1/n appear in the slope term b. 1/n appears as the intercept c. Only 1/n appears as the slope d. log (1/n) appears as the intercept.

is 1.6 10 x. Calculate the value of x.

15. When O2 is adsorbed on a metallic surface, electron

AgCl (K sp of AgCl

 1.6  1010 ), 0.1mol of CuCl (Ksp CuCl  1.0 106 ) is

transfer occurs from the metal to SECTION 2 Contains 8 Multiple Choice Questions With one or more than one correct option

9.

‘T’

Which of the following atoms has the highest first ionization energy? a. Rb b. Na c. K d. Sc

O2 . The TRUE

statement(s) regarding this adsorption is(are) a. O2 is physisorbed b. heat is released * c. occupancy of  2p of O2 is increased

d. bond length of O2 is increased

Chemistry

130

16. Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25 °C. For this process, the correct statement is a. The adsorption requires activation at 25 ° C b. The adsorption is accompanied by a decrease in enthalpy c. The adsorption increases with increase of temperature d. The adsorption is irreversible SECTION 3 Contains 2 Paragraph Type Questions Each paragraph describes an experiment, a situation or a problem. Two multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.

Paragraph for Question Nos. 17 to 18 Carbon–14 is used to determine the age of organic material. The procedure is based on the formation of 14 C by neutron capture in the upper atmosphere 14 7

1 N + 0 n1  → 14 6 C + 1n

14

C is absorbed by living organisms during photosynthesis.

The 14 C content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of 14 C in the dead being, falls due to the decay which C 14 undergoes 14 6

− → 14 C  7 N + β

The half life period of 14 C is 5770 years. The decay constant (λ ) can be calculated by using the following formula

λ=

0.693 . The comparison of the β - activity of the dead t1/ 2

matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of 14 C to

12

12

C in living matter is 1 : 10 .

17. Which of the following option is correct? a. In living organisms, circulation of 14 C from atmosphere is high so the carbon content is constant in organism Space for rough work

b. Carbon dating can be used to find out the age of earth crust and rocks c. Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content remains constant in living organism d. Carbon dating cannot be used to determine concentration of 14 C in dead beings 18. What should be the age of fossil for meaningful determination of its age? a. 6 years b. 6000 years c. 60,000 years d. It can be used to calculate any age Paragraph for Question Nos. 19 to 20 Rocket propellants consist of rocket engines powered by propellants. These are used both in space vehicles as well as in offensive weapons such as missiles. The propellants are chemical substances which on ignition provide thrust for the rocket to move forward. These substances are called rocket propellants. A propellant is a combination of an oxidiser and a fuel which when ignited undergoes combustion to release large quantities of hot gases. The passage of hot gases through the nozzle of the rocket motor provides the necessary thrust for the rocket to move forward according to Newton's third law of motion. 19. A biliquid propellant contains a. Liquid hydrazine b. A mixture of liquid fuel and a liquid oxidiser c. A solid rocket fuel d. A liquid fuel which can also act as an oxidiser 20. A hybrid rocket propellant uses a. A liquid oxidiser and a solid fuel b. A composite solid propellant c. A biliquid propellant d. A solid, liquid and gas as a propellant

131

Mock Test-5

11. (b) In photochemical reaction the rate of formation of product is directly proportional to the intensity of absorbed light.

ANSWER & SOLUTIONS JEE-Main 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

a

c

b

d

d

d

a

c

c

d

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

b

d

d

b

c

a

d

a

a

c

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

a

d

d

d

a

d

a

d

c

c

1.

2.

3.

13. (d) C(graphite)  → C(diamond ) , ∆H = 1.9 kJ

(d) Cr2 O 7−− → Cr 3+ ; Fe ++ → Fe+++ n=1 n=6 eq. of K 2 Cr2 O7 = eq. of FeSO 4 1× 6 = x × 1

C(graphite) + O 2  → CO 2 , ∆H = −∆H1 C(diamond) + O 2  → CO 2 , ∆H = −∆H 2

(a) When an ion (generally cation due to its small size) is missing from its normal position and occpy an interstitial site between the lattice points, the lattice defect obtained is known as Frenkel defect. (c) P = PAo x A + PBo x B

 3  o 2  o o 600 = PAo   + PB   ;3PA + 2PB = 3000  3+ 2  2+3 4.5 2   o  o o 630 = PAo   + PB  ;4.5PA + 2PB = 4410 4.5 + 2 + 0.5 4.5 + 2 + 0.5     1.5PAo = 1410; PAo = 940 and PBo = 90

4.

(c) r.m.s. speed ∝ T

4.75 298 = 9.50 T or T = 1192 K

5.

2

298  4.75  or   = T  9.50  or T = 1192 − 273 = 919°C

(d) When electron jumps to lower orbit photons are emitted while photons are absorbed when electron jumps to higher orbit, 1s orbital is the lower most, electron in this orbital can absorb photons but cannot emit.

6.

(a) 6MnO 4− + I − + 6OH −  → 6MnO 24− + IO 3− + 3H 2 O

7.

(c) A is displace from D because D have a E° = −0.402 V

8.

(b) CCl4 is sp3 hybridised so bond angle will be

9.

approximately 109 ° . (d) High temperature and excess concentration of the reactant concentration.

10. (d) pH = 9 ∴ [H + ] = 10 −9 [OH − ] =

Mg(OH) Mg(OH) 2

−14

1× 10 10−9

= 10−5

Mg 2 + Mg + 2OH −2OH

K sp = [Mg 2 + ][OH − ]2 ⇒ 1 ×10 −11 = [Mg 2 + ][10 −5 ]2

⇒

[Mg 2 + ] =

1× 10−11 = 10−1 = 0.1 (10−5 )2

12. (d) Gold number shows the protective power of a lyophilic solution. Lesser the gold number, greater will be the protecting power of that colloid. Gelatin is one of the best protective colloid. Among the given colloids, potato starch has maximum gold number.

( − ∆H1 ) − ( − ∆H 2 ) = 1.9 kJ or ∆H 2 = ∆H1 + 1.9 For combustion of 6g, ∆H 2 > ∆H1 by 1.9 / 2 = 0.95 kJ.

0.693 0.693 = = 0.3 ×104 yrs = 3.0 × 10 3 yrs. k 2.31× 10−4 15. (c) In the triphenyl methyl carbonium ion the π electrons of all the three benzene rings are delocalised with the vacant p-orbital of central carbon atom. So, it is resonance stabilised. It is the most stable of all the carbonium ions given 14. (b) t1/ 2 =

CH3 | The ion CH 3 — C+ is stabilised by hyperconjugation, a | CH3 second order resonance.

16. (a) CH 3

CH 3

C == C

H

H

CH 3

C == C

CH 3

H

Cis 2 butene

H

Trans 2 butene

Cis-trans isomerism shown by compound which have double or triple bond by which they restrict their rotation, since 2 butyne have no hydrogen on triple bonded carbon. CH 3 — C ≡≡ C — CH 3 [It does not show cis-trans] 2 butyne

CH — COONa CH COO− 17. (d) || + 2Na +  → || − CH — COONa CH COO Sodium salt of maleic acid or fumaric acid

CH COO− ||  → CH COO−

CH ||| + 2CO 2 + 2e − CH Acetylene

18. (a) Because it float over chloroform and prevent its oxidation. 19. (a) Cyclic ethers are called epoxies. CH 2 — CH 2

O

Chemistry

132

3. The -anomer of D-glucose has a specific rotation of +112 degrees in water. The β-anomer of D-glucose has a specific rotation of + 19 degrees (18.7 actually, but rounding up to 19). If the fraction of glucose present as the -anomer is xand the fraction present as the β-anomer is y, and the rotation of the mixture is +52.6 degree, we have X (+112.2 degree) + y(18.7) = 52.6 degree ……..(i) There is a very little of the open chain form present, so the fraction present as the α-anomer (x) plus the fraction present as the β-anomer (y) should account for all the glucose, i.e., x + y = 1 or y = 1 – x. Putting value of y in equation (i), we get x (+112.2 degree) + (1 – x)(18.7) = 52.6 degree ……..(ii) Solving equation (ii) for (x), we have, x = 0.36 or 36%. Thus (y) must be (1 – 0.36) = 0.64 or 64%. So, percentage composition of - and -anomers in the equilibrium mixture is 36% and 64% respectively. 4. (5)

dil NaOH 20. (c) 2CH 3  CO  CH 3   

OH O | || CH 3 — C — CH 2 — C — CH 3 | (Diacetone alchol) CH 3

21.

(a) COOH

O || C

 

COOH

C || O CONH 2

 NH 3  

COOH

Zn , HCl 22. (d) CH 3 (CH 2 ) 4 CN  

HONO CH 3 (CH 2 ) 4 CH 2 NH 2  

O CH 3 (CH 2 ) 4 CH 2OH   CH 3 (CH 2 ) 4 CHO Hexanal

23. (d) Nylon-66- It is a polymer containing nitrogen

5.

H O   H ||  |  |  — N — (CH 2 ) 6 — N — C — (CH 2 ) 4 — C —  ||   O n 

(4) H3C

CH3 N

OH

OH

COOH

Nylon-66

24. (d) Thymine is present in DNA while in RNA there is Uracil. 25. (a) Insulin is an antidiabatic drug. 26. (d) Mg  Al  Na. This is due to the presence of fully

filled s-orbital in Mg. 27. (a) ZnO  CO    CO 2  Zn 28. (d) Lime stone – CaCO3, Clay – silica and alumina Gypsum – CaSO44 .2H 2O

N H3C 6.

t1/ 2 t1/ 2 0  5g  0  25g t 25%  4min (2) 1g 

7.

(8) 2MgCl   Mg  MgCl 2

30. (c) Ligand must have capacity to donate lone pair of electrons to form coordinate bond.

Mg(s)  Cl 2 (g)   MgCl 2 ; H 2  642 kJ mol 1 H  H 2  2H1  642  (2  125)  392 kJ mol 1 

8.

JEE Advance Paper -I 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

4

5

2

5

4

2

8

5

c

a

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

d

b

a, c

c

b,c,d

b

c

b

b

a

(4)

2.

(5) Difference in mass of compound = 390 – 180 = 210 wt. of CH 3 CO – group is = 43 210 Therefore no. of  NH 2 group =  4.88  5 . 43

H  ?

Mg(s)  1/ 2Cl2 (g)   MgCl ; H1  125 kJ mol 1

29. (c) Amalgams are alloys which contain mercury as one of the contents.

1.

CH3

49 x  392 

(5)

x8

d[RX]  k 2 [RX][OH  ] (by SN 2 pathway) dt

k 2  rate constant of SN2 reaction d[RX]  k 2 [RX][OH  ]  k1[RX] dt



1 d[RX]  k 2 [OH  ]  k1 [RX] dt

This is the equation of a straight line for 

1 d[RX] [RX] dt

vs [OH  ] plot with slope equal to k 2 and intercept equal to k1 .

Mock Test-5

133 1

1

From question: k 2  2  10 mol L hr , k1  1 10 hr 3

2

H combustion  (6  H f CO 2  6  H f H 2 O)  H f C6 H12 O 6

1

[RX]  1.0 M and [OH ]  0.1M

 (6  400  6  300)  (1300)

d[RX]  2  103  1 0.1  1 102  1 dt  300 mol L1 hr 1  5 mol L1 min 1

 2900 kJ / mol



Hence,

9.

3

2

 2900 /180 kJ / g  16.11 kJ / g



(c) Ionic Radii order: N  O  F

18. (a) Zn 2 [Fe(CN)6 ] is bluish white ppt.

10. (a) It can act as an oxidising as well as reducing agent. + +  + +++ 11. (d)

+

19. (d) A  1, 3, 5, B  4, 5, C  1, 2, D  1, 2, 4 [According to MOT] 20. (a) A  1, 2, 3, B  2, C  1, 3, 4, D  2, 3

(Note: Assuming AgNO3 is ammonical)

NO2

Volume

12. (b) Ion-dipole interaction

NH  NH 2  

(a) PhCHO  O 2 N

13. (a,c) (a) C 22  Total no. of electrons = 14 so it is

O2N

diamagnetic

PhHC  N  NH

(b) O 22  Bond order = 3; O2 Bond order = 2 

Bond length in O 22  is less than bond length in O2 .  2

NH3 PhCHO  Ag 2 O   PhCOO   Ag 

 2

( whiteppt )

(c) Bond order of N  2.5 ; Bond order of He  1/ 2 

CN | KCN PhCHO   Ph — C — O  | H

Some energy is released during the formation of He 2 from two isolated He atoms.

14. (a, b, d) For first order reaction [A]  [A]0 e  kt

ammonical AgNO3  CH3  C  C Ag   (b) CH3C  CH 

Hence concentration of [NO2 ] decreases exponentially,

(White ppt)

0.693 Also, t1/ 2  , which is independent of concentration K and t1/ 2 decreases with the increase of temperature. t 99.6 

NO2

(ppt.)

CN |

KCN (c) PhCHO   Ph — C — O  ;

| H

2.303  100  log   K  0.4 

2.303 0.693 (2.4)  8   8 t1/ 2 K K 15. (d) Overall order of reaction can be decided by the data given t 75%  2t 50%

AgNO3  CN    AgCN 

(d) AgNO3  I    AgI 

t 99.6 



It is a first order reaction with respect to P. From graph [Q] is linearly decreasing with time, i.e., order of reaction with respect to Q is zero and the rate expression is r  k[P]1[Q]0 .

16. (a, d) (a) Preferential adsorption of ions on surface from the solution (c) Attraction between particles having same charges on their surface accounts for the Brownian motion. (d) Definition of Zeta Potential. 17. (c) Combustion of glucose

C6 H12 O6  6O2   6CO2  6H 2 O

JEE Advance Paper –II 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

5

6

8

38

10

5

4

7

d

c

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

b, c

d

b,c,d

b,c,d

b

b

c

b

2

1

1. 2.

(5) (6)

3. 4.

(8) Fructose has three chiral centres and hence 23  8 optical isomers are possible. (38) C 6 H 12 O 6  6O 2  6CO 2  6H 2 O  38ATP

5.

(10) 22400 ml of H 2 is produced by 46 gms of ethanol.

200 ml of H 2 is produced by

46  200 23  22400 56

Chemistry

134

 Square planar

23  100  9.6% 10% 56  10 (5) Energy released = Energy due to formation of two

Percentage of C 2 H 5OH 

6.

 Two lone pair on central atom (Xe)  One lone pair

on central atom (S) Only, ClF3 & XeF4 contains two lone pair of electrons on

single bonds  2  331  662 kJ mol1 of propene

central atom.

 H polymerisation/mol  590  662  72 kJ mol 1

H polymerisation  72  n  360  n  5

7.

(4) t1/ 2

Also, t 93.75  8.

(MX)  4 108  2 104 Solubility of (MX 2 )  8 105

2.303 100 2.303 100 log log   k1 100  93.75 k1 6.25

Solubility of (M 3X)  1 104



2.303 4  2.303 log 2 4  0.693 log 24    4t1/ 2 k1 k1 k1 Ag  Cl (K sp  1.6  1010 )

dk Ea K dT RT 2

(S x )

S



Cu  Cl  (K sp  1  10 6 ) x

(x S )

S(S  x)  1.6 10

10

; x(x  S)  10

dk  Ea dT A  Frequency factor

6

S  1.6  10 4 x

= No of collisions per unit time per unit volume.



S  (1.6 104 )x.  x(x  1.6 104 x)  106



x  103



S  1.6 107  1.6 10 x The value of x = 7

9.

14. (c) According to the Freundlich adsorption isotherm

x x 1  kP1/ n log  log K  log P m m n

(d) I. P1 = Sc > Na > K > Rb

15. (b, c, d) Adsorption of O2 on metal surface is exothermic.

During electron transfer from metal to O2 electron

10. (c) As Sb2S3 is a negative sol, so, Al2 (SO4 )3 will be the

* occupies  2p orbital of O2 .

most effective coagulant due to higher charge density on

Due to electron transfer to O2 the bond order of O2

Al3 in accordance with Hardy-Schulze rule. Order of effectiveness of cations:

decreases hence bond length increases.

Al3  Ca   Na   NH 4

16. (b) Adsorption of methylene blue on activated charcoal is physical adsorption hence it is characterised by decrease in enthalpy.

11. (b, c) F

F

17. (c) Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content remains constant in living organism

F Cl

Br F

F

F

(Square Shape)

F

18. (b) 6000 years

F (T-Shape)

 One lone pair on central atom (Br)  Two lone pair

on central atom (Cl) F F

F Xe

F

F

MX  M 3 X  MX 2 .

13. (b, c, d) K  A e  Ea / RT



(7) AgCl CuCl



12. (d) Solubility of

0.0693  k1

F

S F F (See-Saw Shape)

19. (b) Biliquid Propellant – A double base propellant is a high strength, high modulus gel of cellulose nitrate (gun cotton) in glyceryl trinitrate or a similar solvent. 20. (a) Hybrid Propellant – A hybrid propellant consists of a solid fuel and liquid oxidiser to provide propulsion energy and working substance e.g., Solid acrylic rubber and liquid N 2 O4 .

  

135

Mock Test-1 Test Booklet code

A

Mock Test “JEE-Main”

Do not open this Test Booklet until you are asked to do so. Read carefully the Instructions on the Back Cover of this Test Booklet. Important Instructions: 1.

Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited.

2.

The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully.

3.

The test is of 3 hours duration.

4.

The Test Booklet consists of 90 questions. The maximum marks are 360.

5.

There are three parts in the question paper A, B, C consisting of, Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.

6.

Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each question. 1/4 (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.

7.

There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.

8.

Use Blue/Black Ball Point Pen only for writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet. Use of pencil is strictly prohibited.

9.

No candidates is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc., except the Admit Card inside the examination hall/room.

10. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each page and at the end of the booklet. 11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them. 12. The CODE for this Booklet is A. Make sure that the CODE printed on Side-2 of the Answer Sheet and also tally the serial number of the Test Booklet and Answer Sheet are the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the invigilator for replacement of both the Test Booklet and the Answer Sheet. 13. Do not fold or make any stray marks on the Answer Sheet. Name of the Candidate (in Capital letters): Roll Number:

in figures in words

Examination Centre Number: Name of Examination Centre (in Capital letters): Candidate’s Signature:

Invigilator’s Signature:

Mathematics

136

Read the Following Instructions Carefully: 1.

The candidates should fill in the required particulars on the Test Booklet and Answer Sheet (Side-1) with Blue/Black Ball Point Pen.

2.

For writing/marking particulars on Side-2 of the Answer Sheet, use Blue/Black Ball Point Pen only.

3.

The candidates should not write their Roll Numbers anywhere else (except in the specified space) on the Test Booklet/Answer Sheet.

4.

Out of the four options given for each question, only one option is the correct answer.

5.

For each incorrect response, one-fourth (¼) of the total marks allotted to the question would be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Sheet.

6.

Handle the Test Booklet and Answer Sheet with care, as under no circumstances (except for discrepancy in Test Booklet Code and Answer Sheet Code), another set will be provided.

7.

The candidates are not allowed to do any rough work or writing work on the Answer Sheet. All calculations/writing work are to be done in the space provided for this purpose in the Test Booklet itself, marked ‘Space for Rough Work’. This space is given at the bottom of each page and at the end of the booklet.

8.

On completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.

9.

Each candidate must show on demand his/her Admit Card to the Invigilator.

10. No candidate, without special permission of the Superintendent or Invigilator, should leave his/her seat. 11. The candidates should not leave the Examination Hall without handing over their Answer Sheet to the Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the Attendance Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with as an unfair means case. The candidates are also required to put their left hand THUMB impression in the space provided in the Attendance Sheet. 12. Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited. 13. The candidates are governed by all Rules and Regulations of the JAB/Board with regard to their conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the JAB/Board. 14. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances. 15. Candidates are not allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, electronic device or any other material except the Admit Card inside the examination hall/room.

137

Mock Test-1

JEE-MAIN: MATHEMATICS MOCK TEST-1 1.

2.

The domain of definition of f ( x) = a. R − {−1, −2}

b. (−2, ∞)

c. R − {−1, −2, −3}

d. (−3, ∞) − {−1, −2} 2 2 have = 1− x −1 x −1 b. Two d. None

10.

12. From a 60 meter high tower angles of depression of the top and bottom of a house are α and β respectively. If

a. sin α sin β

60sin ( β − α ) , then x = x b. cos α cos β

c. sin α cos β

d. cos α sin β

the height of the house is

+(3 + 3ω + 5ω ) =

4.

c. A−1 = 5.

b. 0 d. None of these

From the following find the correct relation a. ( AB)′ = A′B′ b. ( AB )′ = B′A′

adj A A

6.

7.

d. ( AB) −1 = A−1B −1

3 3 3 3 3 3 b. ca = bd c. a b = c d d. ab = cd

a. 2 n

b. n. 2n

c. n. 2n−1

d. n. 2n+1

1+

b. a + b

c. log a + log b

d. log a − log b

a. –1

dy , at θ = 3π , is dx 4 b. 1

c. − a 2

d. a 2

15. The minimum value of [(5 + x )(2 + x)]/[1 + x ] for non-

(log e n) 2 (log e n) 4 + +K = 2! 4!

negative real x is a. 12 b. 1

b. 1/ n

1 1 d. (e n + e − n ) ( n + n −1 ) 2 2 How many words can be made from the letters of the word INSURANCE, if all vowels come together. a. 18270 b. 17280 c. 12780 d. None of these c.

9.

a. a − b

14. If x = a cos 4 θ , y = a sin 4 θ , then

C1 + 2C2 + 3C3 + 4C4 + .... + nCn =

a. n

8.

log(1 + ax) − log(1 − bx) is not defined x at x = 0. The value which should be assigned to f at x = 0 so that it is continuous at x = 0, is

13. The function f ( x) =

3 2 If the roots of the cubic equation ax + bx + cx + d = 0 are in G.P., then

3 3 a. c a = b d

There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then, the probability that only two tests are needed, is a.

1 3

b.

1 6

c.

1 2

d.

1 4

d. 1

b. 1 d. None of these

2 2

a. 4 c. – 4

c. 1/4

a. – 1 c. 0

ω is the cube root of unity, then (3 + 5ω + 3ω 2 ) 2 +

If

 π   3π  sin   sin  =  10   10  a. 1/2 b. – ½

11. Find real part of cosh −1 (1)

How many roots the equation x − a. One c. Infinite

3.

log 2 ( x + 3) is: x 2 + 3x + 2

c. 9

d. 8

−1

16.

∫

e tan x dx = 1 + x2

a. log(1 + x 2 ) + c c. e tan

−1

x

b. log etan

−1

d. tan−1 etan

+c

+c

x

−1

x

+c

17. The measurement of the area bounded by the co-ordinate axes and the curve y = loge x is

a. 1 c. 3

b. 2 d. ∞

18. The solution of the equation a. e y = e x +

x3 +c 3

c. e y = e x + x 3 + c

dy = e x − y + x 2 e − y is dx

b. e y = e x + 2 x + c d. y = e x + c

Mathematics

138

19. The distance between 4 x + 3 y = 11 and 8 x + 6 y = 15, is

7 2 7 c. 10

b. 4

a.

26. The sum to infinity 1 1 1 1 − + − + K is n 2 n 2 3n 3 4 n 4

d. None of these

20. The area of a circle whose centre is (h, k) and radius a is a. π (h2 + k 2 − a2 )

b. π a 2 hk

c. π a 2

d. None of these

another parabola with directrix

c. x = 0

d. x =

a 2

a 2

r r 22. If a = (2, 5) and b = (1, 4), then the vector parallel to r r ( a + b ) is

a. (3, 5) c. (1, 3)

b. (1, 1) d. (8, 5)

 n +1 a. loge    n 

 n  b. loge    n +1

 n −1  c. loge    n 

 n  d. loge    n −1 

c. A = −1, B = 1 28.

∫

23. The acute angle between the line joining the points (2, 1,

 7  a. cos −1    5 10 

 1  b. cos −1    10 

 3  c. cos −1    5 10 

 1  d. cos −1    5 10 

b. p ∧ ~ q

c. ~ p ∧ q

d. ~ p ∧ ~ q

29. The

a. 3 Space for rough work

b. 2

c. 1

equations

1 (3 + 5 x 4 )3/ 2 + c 5

d. None of these of

tangents

2

line 5 x + 12 y + 8 = 0 are

a. 12 x − 5 y + 8 = 0, 12 x − 5 y = 252 b. 12 x − 5 y = 0, 12 x − 5 y = 252 c. 12 x − 5 y − 8 = 0,12 x − 5 y + 252 = 0 d. None of these

~ p ∧ q is logically equivalent to a. p → q

d. 0

b.

to

the

circle

x + y − 22 x − 4 y + 25 = 0 which are perpendicular to the 2

30.

25. The number of solutions of the system of equations 2x + y − z = 7, x − 3 y + 2z = 1, x + 4 y − 3z = 5 is

continuous

x 3 3 + 5 x 4 dx =

1 (3 + 5 x 4 )3/ 2 + c 30

c.

24. Which of the following is logically equivalent to ~ (~ p ⇒ q) a. p ∧ q

series

d. A = −1, B = 0

a. (3 + 5 x 4 )3/ 2 + c

x −1 y z + 3 = = –3), (–3,1,7) and a line parallel to 3 4 5 through the point (–1, 0, 4) is

given

π  x≤−  −2sin x, 2  π π  is f ( x) =  A sin x + B, − < x < , 2 2  π  x≥  cos x, 2  everywhere are a. A = 0, B = 1 b. A = 1, B = 1

focus to a moving point on the parabola y 2 = 4 ax is

b. x = −

the

27. The values of A and B such that the function

21. The locus of the mid-point of the line segment joining the

a. x = − a

of

b. q → p c. ~ ( p → q ) d. ~ (q → p )

139

Mock Test-1

JEE ADVANCE PAPER-I Time 3 Hours. Read The Instructions Carefully

Max. Marks 264 (88 for Mathematics)

Question Paper Format and Marking Scheme: 1. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections. 2. Section 1 contains 8 questions. The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive). Marking Scheme: +4 for correct answer and 0 in all other cases. 3. Section 2 contains 10 multiple choice questions with one or more than one correct option. Marking Scheme: +4 for correct answer, 0 if not attempted and –2 in all other cases. 4. Section 3 contains 2 “match the following” type questions and you will have to match entries in Column I with the entries in Column II. Marking Scheme: For each entry in Column I, +2 for correct answer, 0 if not attempted and –1 in all other cases.

NOTE: It’s the mock test as per previous year’s papers but sometimes IIT changes the test paper pattern and marking scheme too.

5.

SECTION 1 (Maximum Marks: 32) This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. For each question, darken the bubble corresponding to the correct integer in the ORS.

Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then the value of

Marking scheme: +4

If the bubble corresponding to the answer is darkened.

0

In all other cases.

1.

If area enclosed between the curves y = ln( x + e) and

2.

1 x = ln   and the axis of x is λ sq unit, then the value of  y 22 6 must be The number of distinct solutions of the equation

5 cos 2 2 x + cos 4 x + sin 4 x + cos 6 x + sin 6 x = 2 4 interval [0, 2π ] is 3.

4.

m is n

6.

TP and TQ are any two tangents to a parabola and the tangent at a third point R cuts then in P' and Q', then the TP ′ TQ ′ value of must be = TP TQ

7.

Let

f ( x) =

f :R→R

be

a

function

defined

by

{[0,x], xx ≤> 22, where [x] is the greatest integer less 2

in

the

Let the curve C be the mirror image of the parabola y 2 = 4 x with respect to the line x + y + 4 = 0. If A

than or equal to x. If I =

xf ( x 2 ) ∫ 2 + f ( x + 1) dx, then the value −1

of (4 I − 1) is

8.

A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed

and B are the points of intersection of C with the line y = −5, then the distance between A and B is

inner volume of V mm3 , has a 2 mm thick solid wall and

The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96 is

is open at the top. The bottom of the container is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the container.

Mathematics

140

If the volume of the material used to make the container is minimum when the inner radius of the container is 10 mm, then the value of V / 250π is SECTION 2 (Maximum Marks: 40) This section contains TEN questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is (are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. Marking scheme: +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened. 0 If none of the bubbles is darkened –2 In all other cases

9.

r uuur r uuur Let ∆PQR be a triangle. Let a = QR, b = RP, and r rr r uuur r c = PQ. If | a | = 12, | b | = 4 3 and b.c = 24, then which of the following is (are) true? r r | c |2 r | c |2 r − | a | = 12 − | a | = 30 a. b. 2 2 r r r r rr c. | a × b + c × a | = 48 3 d. a.b = −72

3

13. In R , let L be a straight line passing through the origin. Suppose that all the points on L are at a constant distance from

the

two

P1 : x + 2 y − z + 1 = 0

planes

and

P2 : 2 x − y + z − 1 = 0. Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane

P1 . Which of the following points lie(s) on M?

5 2  a.  0, − , −  6 3   1 1 1 b.  − , − ,   6 3 6 1  5 c.  − , 0,  6  6 2  1 d.  − , 0,  3  3 14. Let P and Q be distinct points on the parabola y 2 = 2 x such that a circle with PQ as diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the area of the triangle ∆OPQ is 3 2, then which of the following is (are) the coordinates of P?

10. Let X and Y be two arbitrary, 3×3, non-zero, skewsymmetric matrices and Z be an arbitrary 3×3, non-zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric?

a. Y 3 Z 4 − Z 4Y 3

b. X 44 + Y 44

4 3 3 4 c. X Z − Z X

d. X 23 + Y 23

11. Which of the following values of α satisfy the equation

(1 + α)

2

(1 + 2α)

2

(1 + 3α)

2

(2 + α) 2 (2 + 2α) 2 (2 + 3α) 2 = −648α ?? (3 + α)2 (3 + 2α)2 (3 + 3α)2

a. (4, 2 2)

b. (9,3 2)

1 1  c.  ,  4 2

d. (1, 2)

15. Let y( x) be a solution of the differential equation (1 + e x ) y ′ + ye x = 1.

If

y(0) = 2, then which of the

following statements is (are) true ?

a. y(−4) = 0 b. y(−2) = 0 c. y( x) has a critical point in the interval ( −1, 0) d. y( x) has no critical point in the interval ( −1, 0)

a. −4

b. 9

c. −9

d. 4

12. In R 3 , consider the planes P1 : y = 0 and P2 : x + z = 1. Let P3 be a plane, different from P1 and P2 , which passes through the intersection of P1 and P2 . If the distance of the point (0, 1, 0) from P3 is 1 and the distance of a point

(α, β, γ) from P3 is 2, then which of the following relations is (are) true?

16. Consider the family of all circles whose centers lie on the straight line y = x. If this family of circles is represented by the differential equation Py ′′ + Qy ′ + 1 = 0, where P, Q

dy d2 y , y ′′ = 2 ), then dx dx which of the following statements is (are) true? a. P = y + x are functions of x, y and y ′(here y ′ =

b. P = y − x

a. 2α + β + 2γ + 2 = 0

b. 2α – β + 2γ + 4 = 0

c. P + Q = 1 − x + y + y ′ + ( y ′) 2

c. 2α + β – 2γ – 10 = 0

d. 2α – β + 2γ – 8 = 0

d. P − Q = x + y − y ′ − ( y ′) 2

141

Mock Test-1

g:R→R

17. Let

be

a

differential

function

with

matching entries. For example, if entry (A) in Column I, matches

g (0) = 0, g ′(0) = 0 and g ′(1) ≠ 0. Let

 x g ( x ), f ( x) =  | x |  0,

with entries (2), (3) and (5), then darken these three bubbles in the

x≠0

and h ( x ) = e| x| for all

x=0

x ∈ R. Let ( f o h) (x) denote

For each entry in Column I, darken the bubbles of all the

ORS. Similarly, for entries (B), (C) and (D). Marking scheme: For each entry in Column I

f ( h ( x )) and (h o f )( x)

denote h( f ( x)). Then which of the following is (are)

+2 If only the bubble(s) corresponding to all the correct match(es) is(are) darkened 0

true?

If none of the bubbles is darkened

–1 In all other cases

a. f is differentiable at x = 0 b. h is differentiable at x = 0

19. Match the Column: Column I (A) In R 2 , if the magnitude of

c. f o h is differentiable at x = 0 d. h o f is differentiable at x = 0

Column II 1. 1

the projection vector of the

18. Let

π π  f ( x) = sin  sin  sin x   2  6

g ( x) =

π 2

for all

x∈R

and

is

sin x for all x ∈ R. Let (f o g)(x) denote f (g(x))

and (g o f )(x) denote g(f (x)). Then which of the following is (are) true?

 1 1 a. Range of f is  − ,   2 2  1 1 b. Range of f o g is  − ,   2 2 c. lim x→0

vector αiˆ + β ˆj on

f ( x) π = g ( x) 6

d. There is an x ∈ R such that ( g o f )( x) = 1

3iˆ + ˆj

3 and if α | = 2 + 3β,

then possible value(s) of | α | is (are)

(B) Let a and b be real numbers such

that

f (x) =

the

2. 2

function

{

−3ax2 −2, x 1 and

is(are) darkened.

9.

0

If none of the bubbles is darkened

–2

In all other cases

f ′( x) =

Let m≤

∫

1

1/ 2

y1 > 0. The common

tangent to H and S at P intersects the x-axis at point M. If

(l , m) is the centroix of the triangle ∆PMN, then the

3

192x 1 for all x ∈ R with f   = 0. If 4 2 + sin π x 2

f ( x ) dx ≤ M , then the possible values of m and M

correct expression(s) is (are) 1 dl = 1 − 2 for x > 1 a. dx1 3x1

dm = dx1 3

c.

dl 1 = 1 + 2 for x1 > 1 dx1 3x1

d.

dm 1 = for y1 > 0 dy1 3

are 1 1 ,M = 4 2

a. m = 13, M = 24

b. m =

c. M = −11, M = 0

d. M = 1, M = 12

10. Let S be the set of all non-zero real numbers α such that the quadratic equation α x 2 − x + α = 0 has two distinct real roots x1 and x2 satisfying the inequality | x1 − x2 | < 1. Which of the following intervals is(are) a subset(s) of S ?

 1 1  a.  − , −  5  2

 1  , 0 b.  −  5 

 1  c.  0,  5 

 1 1 ,  d.   5 2

6 4 11. If α = 3 sin −1   and β = 3 cos −1   , where the inverse  11  9 trigonometric functions take only the principal values, then the correct option(s) is (are) a. cos β > 0 b. sin β < 0

x1

b.

(

x12 − 1

)

for x1 > 1

14. The option(s) with the values of a and L that satisfy the following equation is(are) 4π

∫ (sin π ∫ e (sin

6

0

1

0

6

at + cos 4 at ) dt at + cos 4 at ) dt

= L?

a. a = 2, L =

e4π − 1 eπ − 1

b. a = 2, L =

e4π + 1 eπ + 1

c. a = 4, L =

e4π − 1 eπ − 1

d. a = 4, L =

e4π + 1 eπ + 1

15. Let f , g :[−1, 2] → R be continuous functions which are

origin. The major axes of E1 and E 2 lie along the x-axis

twice differentiable on the interval (–1, 2). Let the values of f and g at points –1, 0 and 2 be as given in the following table: x = −1 x=0 x=2 3 6 0 f ( x) 0 1 –1 g ( x)

and the y-axis, respectively. Let S be the circle x 2 + ( y − 1) 2 = 2. The straight line x + y = 3 touches the

In each of the intervals (–1, 0) and (0, 2) the function ( f − 3g )′′ never vanishes. Then the correct statements(s) is

curves S, E1 and E 2 at P, Q and R, respectively Suppose

(are) a. f ′( x) − 3 g ′( x) = 0 has exactly three

c. cos(α + β ) > 0

d. cos α < 0

12. Let E1 and E 2 be two ellipse whose centers are at the

2 2 . If e1 and e2 are the eccentricities of that PQ = PR = 3 E1 and E2 , respectively, then the correct expression(s) is(are)

solutions in

(−1, 0) ∪ (0, 2)

b. f ′( x) − 3 g ′( x) = 0 has exactly one solution in (–1, 0)

145

Mock Test-1

c. f ′( x) − 3 g ′( x) = 0 has exactly one solution in (0, 2) d. f ′( x) − 3 g ′( x) = 0 has exactly two solutions in (–1, 0) f ( x ) = 7 tan 8 x + 7 tan 6 x − 3 tan 4 x − 3 tan 2 x

for all

b.

 π π x ∈  − ,  Then the correct expression(s) is (are)?  2 2 1 a. ∫ xf ( x ) dx = 0 12 π /4 1 c. ∫ xf ( x ) dx = 0 6 π /4

b.

∫

π /4

d.

∫

π /4

0

0

∫

x F ′( x ) dx = − 12 and

3

1

F ′′( x ) dx = 40, then

the

∫

3

1

f ( x ) dx = 12

c. 9 f ′(3) − f ′(1) + 32 = 0 d.

f ( x ) dx = 0

f ( x ) dx = 1

1

2

correct expression(s) is(are) a. 9 f ′(3) + f ′(1) − 32 = 0

and exactly two solutions in (0, 2)

16. Let

∫

18. If

3

∫

3

1

f ( x ) dx = − 12

Paragraph-II Let

n1 and

n2 be the number of red and black balls,

respectively, in box I. Let n3 and n4 be the number of red and

SECTION 3 (Maximum Marks: 16)

black balls, respectively, in box II.

This section contains FOUR questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR

MORE THAN ONE of these four option(s) is(are) correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. Marking scheme: +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened.

19. One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball 1 was drawn from box II is , then the correct option(s) with 3 the possible values of n1 , n2 , n3 and n4 is (are)

0

If none of the bubbles is darkened

a. n1 = 3, n2 = 3, n3 = 5, n4 = 15

–2

In all other cases

b. n1 = 3, n2 = 6, n3 = 10, n4 = 50 c. n1 = 8, n2 = 6, n3 = 5, n4 = 20

Paragraph-I

d. n1 = 6, n2 = 12, n3 = 5, n4 = 20

Let F : R → R be a thrice differentiable function. Suppose that F (1) = 0, F (3) = −4 and

F ′( x) < 0 for all

f ( x ) = xF ( x ) for all x ∈ R.

17. The correct statement(s) is (are) a. f ′(1) < 0 b. f (2) < 0 c. f ′( x) ≠ 0 for any x ∈ (1, 3) d. f ′( x) = 0 for some x ∈ (1, 3) Space for rough work

1  x ∈  , 3  . Let 2 

20. A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, 1 after this transfer, is , then the correct option(s) with 3 the possible values of n1 and n2 is(are) a. n1 = 4, n2 = 6

b. n1 = 2, n2 = 3

c. n1 = 10, n2 = 20

d. n1 = 3, n2 = 6

Mathematics

146

8.

ANSWER & SOLUTIONS JEE-Main

are

1.

2.

3.

4.

5.

6.

7.

8.

9.

d

d

c

a

a

c

c

d

a

c

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

d

b

a

c

c

d

a

c

c

22.

23.

24.

25.

26.

27.

28.

29.

30.

c

c

a

d

d

a

c

c

a

d

1.

(d) x + 3 > 0 and x 2 + 3 x + 2 ≠ 0

2.

(d) If x ≠ 1, multiplying each term by ( x − 1), the given

9.

(b) The probability that only two tests are needed = probability that the first machine tested is faulty × probability that the second machine tested is faulty

=

2 1 1 × = 4 3 6

10. (c) sin

equation reduces to x( x − 1) = ( x − 1) or given equation has no roots.

= (3 + 3ω + 3ω + 2ω ) + (3 + 3ω + 3ω + 2ω ) 2

2

10

(

12. (d) α

then A 3 = Product of the roots = −

d a

1/ 3

d ⇒ A = −  a Since A is a root of the equation.

aA3 + bA2 + cA + d = 0

⇒

 d  d a −  + b −   a  a

⇒ 6.

d  b  a

2/3

2/3

1/ 3

d  = c  a

⇒ b3

⇒

h=

60sin( β − α ) sin β cos α cos β cos β

x = cos α sin β .

13. (b) Since limit of a function is a + b as x → 0, therefore to be continuous at a function, its value must be a + b at

x=0

+d =0

2

d d ⇒ b3d = c3a. = c3 a2 a

(c) Trick: Put n = 1, 2, 3,....

Now by alternate (c), put n = 1, 2 S1 = 1.20 = 1, S 2 = 2.21 = 4

(c) 1 +

60 tan β = 60 − h tan α

⇒ − h = 60 tan α − 60 tan β ⇒ tan β

1/ 3

 d + c −   a

S1 = 1, S 2 = 2 + 2 = 4

7.

β

H = d tan β and H − h = d tan α

⇒

ax3 + bx2 + cx + d = 0

∴

H=60m

d

A , A, AR be the roots of the equation R

Let

)

3

(b) It is understandable. (a)

)

2 2

h

5.

5 −1 5 +1 1 = . . 4 4 4

cosh −1 (1) = log 1 + 12 − 1 = log1 = 0.

= (2ω )2 + (2ω 2 )2 = 4ω 2 + 4ω 4 = 4(−1) = −4. 4.

3π = sin18°.sin 54° 10

(

(1 + ω + ω = 0, ω = 1) 2

sin

11. (c) We know that cosh −1 x = log x + x 2 − 1

∴

(c) (3 + 5ω + 3ω 2 )2 + (3 + 3ω + 5ω 2 )2

π

= sin18°.cos 36° =

( x − 1) 2 = 0 or

x = 1, which is not possible as considering x ≠ 1. Thus

2

6! × 4 ! = 8640 2!

10.

21.

3.

(d) IUAENSRNC Obviously required number of words

n + n −1 (log e n) 2 (log e n) 4 eloge n + e − loge n . = + +K = 2 2! 4! 2

⇒ f (0) = a + b. 14. (a) y = a sin 4 θ ⇒

dy = 4a sin 3 θ cosθ dθ

4 and x = a cos θ

⇒

dx = −4a cos3 θ sin θ dθ

∴

dy dy / dθ − sin 2 θ = = = − tan 2 θ dx dx / dθ cos 2 θ

∴

 dy  2  3π   3π = − tan   dx θ =  4 4

  = −1. 

147

Mock Test-1

15. (c) Given f ( x ) =

[(5 + x )(2 + x )] [1 + x ]

21. (c) Let P(h, k ) be the mid-point of the line segment joining the focus (a,0) and a general point Q( x, y) on the

4 4 f ( x) = 1 + + (5 + x) = (6 + x) + 1+ x (1 + x ) ⇒

parabola.

x+a y , k = ⇒ x = 2 h − a, y = 2 k . 2 2 Put these values of x and y in y 2 = 4 ax, we get Then h =

4 = 0; f '( x ) = 1 − (1 + x) 2

x2 + 2 x − 3 = 0 ⇒

4 k 2 = 4 a (2 h − a )

x = −3, 1 Now f ′′ ( x ) =

⇒

8 , (1 + x)3

So, locus of P(h, k ) is y 2 = 2 ax − a 2 ⇒

f ′′ (−3) = −ve, f ′′ (1) = +ve Hence minimum value at x = 1

16. (c) Putting t = tan −1 x ⇒ dt =

a  y 2 = 2a  x −  2  Its directrix is

a a = − ⇒ x = 0. 2 2 r r 22. (c) a + b = 3iˆ + 9 ˆj = 3(iˆ + 3 ˆj ). Hence it is parallel to (1, 3). x−

(5 + 1)(2 + 1) 6 × 3 = = 9. f (1) = (1 + 1) 2

1 dx, we get 1 + x2

23. (a) Direction ratio of the line joining the point

(2, 1, − 3), (− 3, 1, 7) are (a1 , b1 , c1 )

−1

∫

4 k 2 = 8ah − 4 a 2 ⇒ k 2 = 2 ah − a 2

−1 e tan x dx = ∫ et dt = et + c = e tan x + c. 2 1+ x

⇒

(− 3 − 2, 1 − 1, 7 − (−3))

⇒ (− 5, 0, 10)

17. (d) Area

Direction ratio of the line parallel to line x −1 y z + 3 are = = 3 4 5 (a2 , b2 , c2 )

∞

A = ∫ log x.dx 0

= ( x log x − x)∞0 = ∞ Y

⇒ (3, 4, 5) Angle between two lines, a1a2 + b1b2 + c1c2 cos θ = 2 a1 + b12 + c12 a22 + b22 + c22

y =loge x (1,0)

X

cosθ = 18. (a)

⇒

dy = e x − y + x 2 e − y = e − y (e x + x 2 ) dx

⇒ cos θ =

e y dy = ( x 2 + e x ) dx

x3 Now integrating both sides, we get e = + e x + c. 3 15 19. (c) 4x + 3 y = 11 and 4 x + 3 y = 2 y

15 11 − Therefore, 2 = 7 . D= 5 10 20. (c) Since area = π r 2 , where

r =a ⇒ Area = π a 2 .

(− 5 × 3) + (0 × 4) + (10 × 5) 25 + 0 + 100 9 + 16 + 25 35 25 10 

7    5 10 

⇒ θ = cos −1  24.

(d) Since ~ ( p ⇒ q) ≡ p ∧ ~ q ~ (~ p ⇒ q) = ~ p ∧ ~ q

25.

2 1 −1 (d) ∆ = 1 −3 2 1 4 −3

= 2(9 − 8) − 1(−3 − 2) − 1(4 + 3) = 7 − 7 = 0 Hence, number of solutions is zero.

Mathematics

148

26. (a)

2

=

JEE Advance Paper-I

1 1 1 1 − 2 + 3 − 4 +K n 2n 3n 4n 3

1. 4 11. b,c

4

1 (1/ n) (1/ n) (1/ n) − + − +K n 2 3 4

 1  n +1 = log e  1 +  = log e  . n    n 

1.

2. 8 12. b,d

3. 4 13. a,b

(4)

4. 8 14. a,d

5. 5 15. a,c

6. 1 16. b,c

7. 0 17. a,d

8. 4 18. a,b,c

9. a,c,d 19. c

10. c, d 20. c

7 6

27. (c) For continuity at all x ∈ R , we must have

⇒

5

 π f  −  = lim − (−2sin x)  2  x→( −π / 2)

4

=

2

3

lim ( A sin x + B)

x →( −π / 2)+

1

2 = −A + B

–2 –1

. . .(i)

O

π  and f   = lim − ( A sin x + B )  2  x→(π / 2)

1

2

3

4

5

–1 –2

= lim + (cos x)

–3

x →(π / 2)

⇒ 0 = A+ B

. . .(ii)

From (i) and (ii),

Required area = 4 × (1× 1) = 4 sq unit

∴

1 1 1 1 + + + +K∞ 4 8 16

A = −1

and

λ λ λ λ K∞ = λ 2

B = 1.

x3

1 3 + 5 x 4 dx = ∫ t 1/ 2 dt 20

2 1 1 = × .t 3/ 2 + c = (3 + 5 x 4 )3/ 2 + c. 3 20 30

1/ 2

= λ 1−1/ 2 = λ = 4

2.

5 (8) cos 2 2 x + cos 4 x + sin 4 x + cos 6 x + sin 6 x = 2 4

⇒

5 cos 2 2 x − 5cos 2 x sin 2 x = 0 4

⇒

tan 2 2 x = 1, where 2 x ∈ [0, 4π ]

3.

Number of solutions = 8 (4) Image of y = −5 about the line x + y + 4 = 0

⇒

Distance AB = 4

4.

(8) Let coin was tossed ‘n’ times

28. (c) Put 3 + 5 x 4 = t ⇒ 20 x3 dx = dt , then

∫

λ =4

is

x =1

29.

(a) Equation of line perpendicular to

5x + 12 y + 8 = 0 is 12 x − 5 y + k = 0.

1 n Probability of getting at least two heads = 1 −  n + n  2 2 

Now it is a tangent to the circle, if Radius of circle

= Distance of line from centre of circle 121 + 4 − 25 =

12(11) − 5(2) + k 144 + 25

⇒ k = 8 or −252. Hence equations of tangents are

12 x − 5 y + 8 = 0 and 12x − 5 y = 252 30. (d) ~ p ∧ q =~ (q → p) .

⇒

 n + 1 1 −  n  ≥ 0.96  2 

⇒

2n ≥ 25 ⇒ n ≥ 8 n +1

5.

(5) n = 6!.5!

(5 girls together arranged along with 5

5

boys) m = C4 .(7! − 2.6!).4! (4 out of 5 girls together arranged with others – number of cases all 5 girls are together) m 5 ⋅ 5 ⋅ 6!⋅4! = 6! ⋅ 5! n

149

Mock Test-1

6.

(1) Let Parabola be y = 4ax and coordinates of P and Q 2

10. (c,d) (Y Z − Z Y ) = ( Z ) (Y ) − (Y ) ( Z ) 3

is the point of intersection of tangents at t1 and t2. ∴

or

TP′ t3 − t2 = TP t1 − t 2

(0) I =

⇒

2

TQ′ t1 − t3 = TQ t1 − t2

x ⋅0

⇒ 1

x ⋅0

2

x ⋅1

0

1

5 + 4α

5 + 6α

2

2

(4) Let inner radius be r and inner length be l

⇒

2

| λ − 1|

λ +2 2

=1 ⇒ λ = −

1 2

P3 ≡ 2 x − y + 2 z − 2 = 0 Distance of P3 from (α , β , γ ) is 2

Volume of material be M, M = π ( r + 2) 2 (l + 2) − π r 2 l

4V 8V dM = − 2 − 3 + 8π + 0 + 4π r dr r r

0

12. (b,d) Let the required plane be x + z + λ y − 1 = 0

4I = 1 = 0

πr l =V

⇒

4

−2α 2 −5α 2 −9α 2 − 3 −2α −2α 3 + 6α = − 648α 0

1

2

⇒

T

= − 648α ( R1 → R1 − R2 ; R3 → R3 − R2 )

⇒ 8.

3

α 2 − 2 4α 2 − 2 9α 2 − 2 3 + 2α 3 + 4α 3 + 6α

∫ 2 + 0 dx + ∫ 2 + 1 dx + ∫ 2 + 0 dx + 0 = 4

−1

T

= − 648α ( R3 → R3 − R2 ; R2 → R2 − R1 )

TP ′ TQ ′ = =1 TP TQ 0

7.

3

symmetric X 44 + Y 44 is symmetric

5 + 2α

λ=

T

(1 + α) 2 (1 + 2α) 2 (1 + 3α) 2 11. (b,c) We get 3 + 2α 3 + 4α 3 + 6α

t3 − t2 t1 − t2

∴

4

X 23 + Y 23 skew symmetric.

P′ ≡ {at3 , t1 , a (t3 + t1 )}

Let TP′ : TP = λ : 1

Similarly,

T

X 4 Z 3 − Z 3 X 4 skew symmetric

Q′ ≡ {at2 , t3 , a (t2 + t3 )}

or

⇒

Coordinates of T ≡ {at1 , t2 , a (t1 + t2 )} Similarly ,

3 T

4

= − Z 4Y 3 + Y 3 Z 4

on this parabola are P ≡ ( at , 2at1 ) and Q ≡ ( at , 2 at 2 ); T 2 2

2 1

4

| 2α − β + 2γ − 2 |

⇒

=2 4 ×1 + 4 2α − β + 2λ + 4 = 0 and 2α − β + 2λ − 8 = 0

dM = 0 when r = 10 dr V = 1000π

13. (a,b) Line L will be parallel to the line of intersection of P1 and P2

V =4 250π

⇒

Let a, b and c be the direction ratios of line L a + 2b − c = 0 and 2a − b + c = 0

⇒

a : b : c ::1 : − 3 : − 5

r r r 9. (a,c,d) | b + c | = | a | r2 r2 r r r2 ⇒ | b | + | c | = 2b ⋅ c = | a | r ⇒ 48 + | c |2 + 48 = 144 r r | c |2 r ⇒ |c|=4 3∴ − | a | = 12 2 r r r Also, | a + b | = | c |

r2 r2 r r r2 ⇒ | a | + | b | = 2a ⋅ b = | c | r r r r r ⇒ a ⋅ b = − 72 a + b + c = 0 r r r r r r r r r r ⇒ a × b = c × a ⇒ | a × b + c × a | = 2| a × b | = 48 3

Equation of line L is

x−0 y−0 z −0 = = 1 −3 −5

Again foot of perpendicular from origin to plane P1 is

∴

 1 1 1 − ,− ,   6 3 6 Equation of project of line L on plane

P1

is

1 2 1 y+ z− 6= 6 = 6 =k 1 −3 −5

x+

5 2   1 1 1 Clearly points  0, − , −  and  − , − ,  satisfy the 6 3   6 3 6 line of projection i.e. M

Mathematics

150

 16a 8a  14. (a,d) P( at 2 , 2at ) ⇒ Q  2 , −  t   t ∆OPQ =

⇒

17. (a,d) Differentiability of f ( x ) at x = 0

0 + g (−δ )  f (0) − f (0 − δ )  LHD f ′(0− ) = lim  =0  = lim → δ →0 δ 0 δ δ  

1 OP ⋅ OQ 2

1 a (4) 16 at t 2 + 4 ⋅ +4 =3 2 2 t t2

⇒

t − 3 2t + 4 = 0

Differentiabiligy of h( x) at x = 0

t = 2, 2 2

h′(0+ ) = 1, h( x) is an even function

 t2  Hence, P ( at 2 , 2at ) = P  , t  2 

Hence non diff. at x = 0

2

⇒

Differentiability of f ( h ( x )) at x = 0 f ( h( x )) = g (e| x| ) ∀ x ∈ R

t = 2 ⇒ P(1, 2)

LHD f ′( h(0 − )) = lim

t = 2 2 ⇒ P(4,2 2)

δ

= lim

x ex dx = eIn(1+ e ) = 1 + e x x e 1+ e

δ

c =1

⇒

x+4 y= 1 + ex

⇒

y (−4) = 0

⇒

(1 + e x ) − ( x + 4)e x y′ = =0 (1 + e x ) 2

= g ′(1) f ( h(0 + δ )) − f ( h(0))

δ

δ →0

δ

= lim

g (e ) − g (1)

δ →0

⇒

⇒

δ

= g ′(1) Since g ′(1) ≠ 0

f ( h ( x )) is non diff. at x = 0 Differentiability of h ( f ( x )) at x = 0

{

( f ( x )| , h( f ( x )) = e 1,

(1 + e ) − ( x + 4)e (1 + e x ) 2 x

x≠0 x=0

LHD. h′( f (0 − δ )) = lim

h( f (0)) − h( f (0 − δ ))

δ →0

δ

| g ( − δ )|

= lim δ →0

x

π 2

sin x ∀ x ∈ R

1  f ( x) = sin  g ( g ( x))  3 

 1 3 2 1− 1 +  − e e  e >0

2 2   3 3

  is  

5 cos 2 2 x + cos 4 x + sin 4 x + cos6 x + sin 6 x = 2 in 4 interval [0, 2π ] is 5.

x

[0, 1]. If

1 1 1 1 a. f   < and f   > 2 2 3 3

The value of  1 1 1 1 6 + log 3  4− 4− 4−  3 2 3 2 3 2 2 3 2 

4.

Let f be a non-negative function defined on the interval

0

10 a −5 , a −4 , 3a −3 , 1, a 8 and a with a > 0 is

3.

9.

Consider

the

set

of

eight

vectors

V = {aiˆ + bjˆ + ckˆ; a, b, c ∈{−1,1}}. Three non-coplanar vectors

can be chosen from V in 2 p ways. Then p is.

1 2 2  12. If A =  2 1 −2  is a matrix satisfying the equation  a 2 b  AAT = 9 I , where I is 3× 3 identity matrix, then the ordered pair (a, b) is equal to: a. (2, −1) b. (−2,1)

d. ( −2, −1)

c. (2,1) 13. The sum of first 0.7, 0.77, 0.777, ..... is 7 (179 − 10 −20 ) 81 7 c. (179 + 10 −20 ) 81

a.

20

terms

of

the

7 (99 − 10 −20 ) 9 7 d. (99 + 10 −20 ) 9

b.

sequence

158

Mathematics

14. Coefficient

11

x in

of

(1 + x ) (1 + ) (1 + x ) 2 4

3 7

4 12

the

expansion

of

is

a. 1051 c. 1113

b. 1106 d. 1120

(B) Let a and b be real numbers such that the function

f (x) =

15. The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is a. 55 b. 66 c. 77 d. 88 16. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations values 3, 4 and 5 are added to the data, then the mean of the resultant data, is a. 16.8 b. 16.0 c. 15.8 d. 14.0 17. Let ABC be a triangle such that ∠ACB =

π

and let a, b 6 and c denote the lengths of the sides opposite to A, B and C respectively. The value (s) of x for which a = x 2 + x + 1, b = x 2 − 1 and c = 2 x + 1 is (are)

{

−3ax2 −2, x 0 and

x∈R

for

which

and

Y = {9( n − 1) : n ∈ N },

where N is the set of natural numbers, then X ∪ Y is equal to: a. N b. Y – X c. X d. Y

10. A value of b for which the equations x 2 + bx − 1 = 0 x + x + b = 0, Have one root in common is

a. − 2 b. −i 3 c. i 5 d. 2 11. The variance of first 50 even natural numbers is: 833 437 a. b. 833 c. 437 d. 4 4 tan A cot A 12. The expression can be written as + 1 − cot A 1 − tan A a. sin A cos A + 1

b. sec A cosec A + 1

c. tan A + cot A

d. sec A + cosec A

n  23   The value of cot  ∑ cot −1  1 + ∑ 2k   is  k =1    n =1

13.

and z, respectively, then the value of 2 x + y + z is

6.

distinct

2

are at a distance of 6 units from each other. Let P be the

5.

of

3

With one or more than one correct option

x 2 sin( β x ) lim = 1. Then x → 0 α x − sin x

6 (α + β ) equals

4.

number

SECTION 2 Contains 8 Multiple Choice Questions

9.

( x0 , y0 , z0 ) with y0 z0 ≠ 0, is Let

total

1+ x x x 2 2x 4x 1 + 8 x 3 = 0 is 3x 9 x 2 1 + 27 x 3

equations

2cos3θ 2sin3θ + y z

( xyz )sin 3θ = ( y + 2 z ) cos3θ + y sin 3θ have

3.

The

2

The number of all possible values of θ , where 0 < θ < π , for

Let f :IR → IR be defined as f(x) = | x | + | x 2 − 1| . The total

2

x y + = 1 are 9 5

f 2 < 0. Let

P1 and

P2 be two parabolas with a common vertex at (0, 0) and

a.

23 25

b.

25 23

24 23 d. 23 24 14. A man is walking towards a vertical pillar in a straight

c.

with foci at ( f1 , 0) and (2 f 2 , 0), respectively. Let T1 be

path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the

a tangent to P1 which passes through (2 f 2 , 0) and T2 be

pillar is 30°. After walking for 10 minutes from A in the

a tangent to P2 which passes through ( f1 , 0). The m1 is

same direction, at a point R, he observes that the angle of

the slope of T1 and m2 is the slope of T2 , then the value  1  of  2 + m22  is m 

elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, from B to reach the pillar, is:

a. 6

b. 10

c. 20

d. 5

160

15.

Mathematics

lim x →0

(1 − cos 2 x) (3 + cos x ) is equal to x tan 4 x

a. −

1 4

b.

1 2

c. 1

d. 2

18. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2 is

 π π 16. Let the function g : (−∞, ∞) →  − ,  be given by  2 2 g (u ) = 2 tan −1 (eu ) −

π 2

a.

116 181

b

formula ∫ f ( x )dx =

a. even and is strictly increasing in (0, ∞ )

a

b. odd and is strictly decreasing in ( −∞ , ∞ ) d. neither even nor odd, but is strictly increasing in ( −∞ , ∞ )

a

Each paragraph describes an experiment, a situation or a problem. Two multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.

55 181

Paragraph for Question No. 17 to 18 A box B1 contains 1 white ball, 3 red balls and 2 black balls.

∫ sin x dx

balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls. 17. If 1 ball is drawn from each of the boxes B1 , B2 and B3 , the probability that all 3 drawn balls are of the same colour is 558 648

( f (a ) + f (b)), for more accurate

d.

566 648

is equal to

0

c.

Another box B2 contains 2 white balls, 3 red balls and 4 black

c.

d −a 2

π /2

19.

a.

Space for rough work

d.

b−a ( f (a) + f (b) + 2 f (c)). 4

∫ f ( x)dx =

SECTION 3 Contains 2 Paragraph Type Questions

90 648

65 181

b−c c−a ( f ( a ) + f (c)) + 2 2 a+b ( f (b) + f (c)). When c = , 2 b

b.

c.

result for c ∈ ( a, b) F (c) =

c. odd and is strictly increasing in ( −∞ , ∞ )

82 648

126 181

Paragraph for Question No. 19 to 20 Suppose we define the definite integral using the following

.

Then, g is

a.

b.

20. If

π 8

(1 + 2)

π 8 2

b. d.

π 4

(1 + 2)

π 4 2

f ′′( x ) < 0 ∀x ∈ (a, b) and c is a point such that

a < c < b , and (c, f (c )) is the point lying on the curve for

which F ( c ) is maximum, then f ′( c ) is equal to

f (b ) − f ( a ) b−a 2 f (b ) − f ( a ) c. 2b − a a.

b.

2( f (b) − f (a )) b−a

d. 0

161

Mock Test-2

ANSWER & SOLUTIONS JEE-Main 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

a

d

c

d

b

a

d

d

b

b

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

a

c

a

b

c

a

a

a

b

b

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

b

d

b

a

d

c

a

d

a

b

1.

5.

(d) It is understandable.

6.

9 27 1 3 (a) Given series is 3 + 4 + 6 + ........ = 3 + + + ..... 2 4 2 4

32 33 34 35 + + + + ..... (in G.P.) 2 4 8 16 3 Here a = 3, r = , then sum of the five terms 2 = 3+

 3  5   5  3   − 1 1  3 − 1 a(r n − 1)  2   =  32  S5 = =  3 1 r −1 −1 2 2

(a) Use the identity f ( f − 1( x)) = x replace x by f − 1( x), in the given function we get

1 f ( x)

f ( f − 1( x )) = f − 1( x) +

1 , solve to find f − 1x. f ( x)

⇒

x = f − 1( x) +

2.

(d) Given x2 + x + 1 = 0

∴

1 1 1 x = [−1 ± i 3] = (−1 + i 3), (−1 − i 3) = ω, ω 2 2 2 2

−1

Hence the equation will be same.

⇒

7.

64 = 2(1 + a2 + a4 + ...)

∴ 8.

2π   2π  (c) ω n = cos   + i sin   n    n 

ω3 = cos

(d) (1 + x − 2 x 2 ) 6 = 1 + a1 x + a2 x 2 + .... + a12 x12 . Putting x = 1 and x = −1 and adding the results

But α 19 = ω19 = ω and β 7 = ω 14 = ω 2 .

3.

9  243 − 32  211× 3 633 = 6 = = = 39 .  32 16 16 16  

−1

2π 2π 1 i 3 + i sin =− + =ω 3 3 2 2

a2 + a4 + a6 + .... + a12 = 31. (d) Tn =

Σn n(n + 1) = n! 2 (n)!

=

1  (n + 1)  1  n − 1 2  +  =   2  (n − 1)! 2  (n − 1)! (n − 1)!

=

1 1 2  (e + 2e) 3e + = .  = 2  (n − 2)! (n − 1)! 2 2

2

2π 2π  4π 4π  and ω32 =  cos + i sin + i sin  = cos 3 3  3 3  1 i 3 =− − = ω2. 2 2

∴

( x + yω3 + zω ) ( x + yω + zω3 ) 2 3

2 3

= ( x + yω + zω 2 ) ( x + yω 2 + zω )

= x + y + z − xy − yz − zx. 2

2

6

4.



∑  sin

(d)

k =1

9.

(b) The given series reduces to 3 4  n  log e 2 + log e   + log e   + K + log e   2 3  n −1 

= log e 2 + log e 3 − log e 2 + log e 4 − log e 3 + K + log e (n) − log e (n − 1) = log e n. 10. (b) 210 − 1 = 1023, corresponds to none of the lamps is

2

2kπ 2kπ  − i cos  7 7 

 6 i 27kπ    2 kπ 2kπ   = ∑ −i  cos + i sin  = −i ∑ e  7 7   k =1  k =1 

being switched on. 11. (a) The event that the fifth toss results in a head is independent the event that the first four tosses result in tails. ∴ Probability of the required event = ½

6

= −i{ei 2π / 7 + ei 4π / 7 + ei 6π / 7 + ei 8π / 7 + ei10π / 7 + ei12π / 7 }

  ei 2π / 7 − ei14π / 7 )  (1 − ei12π / 7 )  = −i  e i 2 π / 7 = − i    i 2π / 7 1 − ei 2π / 7    1− e  i 2π / 7

e − 1 (Q ei14π / 7 = 1) = −i  =i i 2π / 7  1 − e 

12. (c) x.

1 1 3.2 x . = ⇒ = 2 ⇒ x=8. 2 4 2.3 4 2

13. (a) sin θ + sin 3θ + sin 2θ = sin α

⇒ 2sin 2θ cos θ + sin 2θ = sin α ⇒ sin 2θ (2 cos θ + 1) = sin α Now, cos θ + cos 3θ + cos 2θ = cos α 2 cos 2θ cos θ + cos 2θ = cos α

. . .(i)

162

⇒

Mathematics

cos 2θ (2cos θ + 1) = cos α

. . .(ii)

19. (b) Put log x = t ⇒

From (i) and (ii),

⇒

tan 2θ = tan α ⇒ 2θ = α ⇒ θ = α / 2 .

  5 7 9 i  14. (b) Imaginary part of sin −1  −  16 16      

1 dx = ∫ x(log x)2

∫

Y

20. (b)

x=4

x =1 y =

 9 9 + 1 +  = − log(2). = − log  16   16

2x

C

D O

1 dx = dt , then x 1 1 1 dt = − + c = − + c. 2 t t log x

A

X

B

15. (c)

D′

C′

h

Required area = CDD ' C ' = 2 × ABCD 60°

30°

= 2∫

d

d = h cot 30 ° − h cot 60 ° and time = 3 min. ∴

h (cot 30 o − cot 60 o ) per minute 3

Speed =

21. (b)

It will travel distance h cot 60° in h cot 60 × 3 = 1.5 minute. h(cot 30o − cot 60 o )

16. (a) For continuous lim f ( x ) = f (2) = k x→2

k = lim x→2

28 2 sq. unit. 3

dy 1 + x 2 1  + = 0 ⇒ dy +  + x  dx = 0 dx x x 

⇒

x 3 + x 2 − 16 x + 20 ( x − 2) 2

∫

sec y log(sec y + tan y ) dy

= ∫ sec x log(sec x + tan x) dx Put log(sec x + tan x) = t and log(sec y + tan y ) = z

[log(sec x + tan x)]2 [log(sec y + tan y )]2 = + c. 2 2

17. (a) y = e x + y ⇒ log y = ( x + y ) log e

dy 1 dy dy y = 1+ ⇒ . = y dx dx dx 1 − y

A (2, –1)

23. (b)

18. (a) Let y = sin p x.cos q x

dy = p sin p −1 x.cos x.cos q x + q cos q −1 x.(− sin x)sin p x dx dy = p sin p −1 x.cos q +1 x − q cos q −1 x.sin p +1 x dx dy = 0, Put dx p ∴ tan 2 x = q ⇒

∴

tan x = ±

p q

Point of maxima x = tan −1

x2 + c = 0. 2

22. (d) cos y log(sec x + tan x)dx = cos x log(sec y + tan y)dy

( x 2 − 4 x + 4) ( x + 5) = lim = 7. x→2 ( x − 2) 2

⇒

2 x 1/ 2 dx =

On integrating, we get y + log x +

o

⇒

4

1

60o D x + 2y –1 = 0

| AD | = ⇒

⇒ ⇒

p . q

⇒

2 − 2 −1

tan 60° =

12 + 2 2

=

1 5

AD BD

1/ 5 BD 1 BD = 15 3=

BC = 2 BD = 2 / 15 .

C

163

Mock Test-2

24.

(a) It represents a circle, if a = b

⇒

3 k 1 = ⇒ k= . 3 4 4

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

8

8

4

8

4

4

7

5

c

c

25.

(d) Tangent to the curve y 2 = 8 x is y = mx + 2 / m.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

d

d

c

c

c

d

b

d

c

a

JEE Advance Paper-I

So, it must satisfy xy = −1. ⇒

∴

2 2  x  mx +  = −1⇒ mx 2 + x + 1 = 0 m m  Since, it has equal roots. D=0 z

1.

i.e., x 9 , x1+8 , x 2+ 7 , x 3+ 6 , x 4+5 , x1+ 2+ 6 , x1+3+5 , x 2+3+ 4 and coefficient in each case is 1

4 − 4m = 0 ⇒ m3 = 1 ⇒ m2 ⇒ m = 1 Hence, equation of common tangent is y = x + 2. r r r r r r 26. (c) Since c = ( x − 2) a + b and d = (2 x + 1) a − b are r r collinear, therefore c = λ d r r r r ⇒ ( x − 2) a + b = λ (2 x + 1) a − λ b r r or [( x − 2) − λ (2 x + 1)] a + (λ + 1)b = 0

⇒ Coefficient of x9 = 1 + 1 + 1 + .........+ 1 = 8 8 times

2.

a −5 + a −4 + 3a −3 + 1 + a8 + a10 ≥ 8

3.

(4) 6 + log3/ 2

put x = 4 −

x − 2 + 2x +1 = 0 1 ⇒ x= . 3

x2 = 4 −

r r r r and | a − c | = | b − c | r r r r ⇒ | a − c |2 = | b − c |2 r r r ∴ a + b = 2c

r r r  r ar + b  Therefore, (b − a ) ⋅  c − =0 2  

(2 + 10 − 12) = cos −1 (0) 4 + 25 + 16 1 + 4 + 9

⇒ θ = 90° x y z + + = 1 , where p = a b c

or

1 1 1 1 + 2+ 2 = 2 2 a b c p

1

3 2

1

3 2

4−

4−

1

3 2

1

3 2

4−

1

3 2

4−

1

3 2

...

...

x 3 2

3 2 x 2 = 12 2 − x

r r r r r r r r   r  27. (a) (b − ar ) ⋅  cr − a + b  = b ⋅ cr − b ⋅  a + b  − ar ⋅ cr + a (ar + b ) 2 2 2    

(a)Plane is

1/8 a −5 + a −4 + 3( a −3 ) + 1 + a 8 + a10 ≥ ( a −5 a −4 ( a −3 ) 3 (1) ( a 8 )( a10 ) ) = 1 1+1+ 3 +1+1+1

∴

⇒

29.

(8)

(using AM ≥ GM)

( x − 2) − λ (2 x + 1) = 0, λ + 1 = 0 r r (Q a , b are linearly independent)

28. (d) θ = cos −1

(8) x9 can be formed in 8 ways

1 Σ (1/ a )

3 2 x 2 + x − 12 2 = 0

4 2 ; x = − 3/ 2 not possible 3

⇒

x=

⇒

 1 4 2 6 + log 3/ 2  ×  3  3 2

⇒

 4 6 + log3/ 2   ⇒ 6 − 2 = 4 9

4.

5 (8) cos 2 2 x + cos 4 x + sin 4 x + cos 6 x + sin 6 x = 2 4

⇒

5 2 cos 2 x − 5cos2 x sin 2 x = 0 4

⇒

tan 2 2 x = 1, where 2 x ∈ [0, 4π ]

2

a b c Now according to equation, x = , y = , z = 4 4 4 Put the values of x, y, z in (i), we get the locus of the centroid of the tetrahedron. 30. (a) ~ ( p ∨ q) ≡ ~ p ∧ ~ q .

Number of solutions = 8 . . .(i)

5.

(4) Let inner radius be r and inner length be l

π r 2l = V Volume of material be M

M = π (r + 2)2 (l + 2) − π r 2 l dM 4V 8V = − 2 − 3 + 8π + 0 + 4π r dr r r

164

Mathematics

dM = 0 when r = 10 dr

⇒

⇒

V = 1000π

⇒

V =4 250π

6.

(4) Image of y = −5 about the line x + y + 4 = 0 is x = 1

⇒

7.

⇒

Distance AB = 4 1

(7) G (1) = ∫ t | f ( f (t )) | dt = 0

⇒

⇒

F ( x) − F (1) 1 F ( x) f (1) x −1 lim = lim = = x →1 G ( x ) x →1 G ( x ) − G (1) | f ( f (1)) | 14 x −1

⇒

1/ 2 1 = | f (1/ 2) | 14

8.

find the number of ways of selecting co-planar vectors. Observe that out of any 3 coplanar vectors two will be collinear (anti parallel) Number of ways of selecting the anti parallel pair = 4 Number of ways of selecting the third vector = 6 Total = 24 Number of non co-planar selections

= C3 − 24 = 32 = 2 , p = 5 8

5

=

α 8 (α 2 − 2) − β 8 ( β 2 − 2) 2(α 9 − β 9 )

=

α 8 .(6α ) − β 8 (6 β ) 2(α 9 − β 9 )

−

6 (α 9 − β 9 ) =3 2 2(α 9 − β 9 )

8

X =

1 − cos 30θ 1 = 2sin θ 4 sin 2°

12. (d) AAT = 9I

1 2 2  1 2 a      2 1 −2   2 1 2  = 9 I  a 2 b   2 −2 b 

1 ⇒ f   = 7. 2

(5) Let (1,1,1), ( −1,1,1), (1, −1,1), ( −1, −1, −1) be vectors r r r r r r r r a , b , c , d rest of the vectors are −a, −b , −c , −d and let us

8

2(sin θ ) X = 1 − cos 2θ + cos 2θ − cos 4θ + ... + cos 28θ − cos30θ

f (− x ) = − f ( x ) 1 2

10

11. (d) X = sin θ + sin 3θ + ... + sin 29θ

−1

Given f (1) =

a10 − 2a8 (α − β ) − 2(α − β ) = 2a9 2(α 9 − β 9 ) 10

⇒

9 0 a + 4 + 2b  9 0 0    0 9 2a + 2 − 2b  ⇒  0 9 0    a + 4 + 2b 2a + 2 − 2b a 2 + 4 + b 2   0 0 9  Equation

a + 4 + 2b = 0

⇒ a + 2b = −4

. . .(i)

and 2a + 2 − 2b = 0

⇒ 2a − 2b = −2

. . .(ii)

a2 + 4 + b2 = 0

⇒ a2 + b2 = 5

. . .(iii)

Solving a = −2, b = −1

Alternate Required value =

∴ 9.

8× 6× 4 3!

p=5

(c) f ′ = ± 1 − f

2

⇒

f ( x) = sin x or f ( x) = − sin x (not possible)

⇒

f ( x) = sin x

Also,

x > sin x ∀ x > 0

10. (c) α is a roots of equation

α 2 − 6α − 2 = 0; β 2 − 6 B − 2 = 0

α 2 − 6α − 2 = 0 ⇒

α 2 − 2 = 6α

13. (c) 0.7 + 0.77 + 0.777 + ..... + 0.777...7 =

7 [0.9 + 0.99 + 0.999 + ... + 0.999...9] 9

=

7 [(1 − 0.1) + (1 − 0.01) + (1 − 0.001...1) + ... + (1 − 0.000...1)] 9

=

7 1 1 1  1  20 −  + 2 + 3 + ... + 20 9 10  10 10 10

1  1 − 20 7 1 10 =  20 − . 9 10 1 − 1 10 

  

  7 1  1020 − 1    =  20 − .   9  1020    9  

7  1  7  −20 180 −  1 − 20   = [179 + 10 ] 81   10   81

165

Mock Test-2

14. (c) 2 x1 + 3 x2 + 4 x3 = 11 Possibilities are (0, 1, 2); (1, 3, 0); (2, 1, 1); (4, 1, 0). ∴ Required coefficients = ( 4 C0 ×7 C1 ×12 C2 ) + ( 4 C1 ×7 C3 ×12 C0 ) + ( 4 C2 ×7 C1 ×12 C1 ) + ( 4 C4 ×7 C1 × 1)

= (1× 7 × 66) + (4 × 35 × 1) + (6 × 7 × 12) + (1× 7)

= 462 + 140 + 504 + 4 = 113. 15. (c) Coefficient of x10 in ( x + x 2 + x 3 )7 Coefficient of x in (1 + x + x )

2 7

3

Coefficient of x3 in

(1 − x ) (1 − x) 3 7

=

−7

=

7 + 3+ 7

x 0).

1 1 18. (d) f k = (sin k x + cos k x) f 6 ( x) = (sin 6 x + cos6 x) 4 6 1 f 4 ( x) = (sin 4 x + cos 4 x) 4

1 3  f 6 K = 1 − sin 2 2 x  6 4 

2ab =4 a+b

⇒

ab =2 a+b

⇒

(5 − d )(5 + 2 d ) = 2(5 − d + 5 + 2d ) = 2(10 + d )

⇒

25 + 10d − 5d − 2d 2 = 20 + 2d

⇒

2d 2 − 3d − 5 = 0

⇒

d = −1, d =

⇒

| 2 d | = 2,5

20.

a b c ∆ = b c a = − 1 (a + b + c)[( a − b) 2 + (b − c) 2 + (c − a )2 ] 2 c a b

5 2

(A) If a + b + c ≠ 0

1  sin 2 2 x  f 4 ( x ) = 1 =  4 2 

and a 2 + b 2 + c 2 = ab + bc + ca

 1 sin 2 2 x   1 sin 2 2 x  1 1 1 f 4 ( x) − f 6 ( x) =  − − − = − = 8  6 8  4 6 12 4 19. (A)

. . .(i)

3α + β = 3 2

⇒

∆ = 0 and a = b = c ≠ 0

⇒ the equations represent identical planes. (B) a + b + c = 0 and a 2 + b 2 + c 2 ≠ ab + bc + ca ⇒ ∆=0

⇒ the equations have infinitely many solutions.

3α + β = ± 2 3

. . .(i)

ax + by + ( a + b) z

Given α = 2 + 3β

. . .(ii)

bx + cy = (b + c) z

From equation (i) and (ii), we get α = 2 or − 1

⇒

(b 2 − ac ) y = (b 2 − ac ) z ⇒

So |α| = 1 or 2

⇒

ax + by + cy = 0 ⇒ ax = ay ⇒ x = y = z

y=z

166

Mathematics

(C) a + b + c ≠ 0 and a + b + c ≠ ab + bc + ca ⇒ ∆≠0 ⇒ the equation represent planes meeting at only one point. 2

2

2

4.

(8) cos α =

2 2 1 sin α = 3 3 C1

A

(D) a + b + c = 0

α

C2

2 2 P α 3

1

B

and a + b + c = ab + bc + ca 2

2

2

R α

⇒ a =b=c =0 ⇒ the equation represent whole of the three dimensional space.

C

JEE Advance Paper -II 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

2

3

7

8

9

4

5

2

d

b

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

b

b

b

d

d

c

a

d

a

a

1.

(2) Let f ( x ) = x 4 − 4 x 3 + 12 x 2 + x − 1 = 0

tan α =

5.

2 2 = 8 units. tan α r r r r (9) s = 4 p + 3q + 5r r r r r r r r r r r s = x(− p + q + r ) + y ( p − q + r ) + z (− p − q + r ) r r r r s = (− x + y − z ) p + ( x − y − z )q + ( x + y + z )r

⇒

−x + y − z = 4 ⇒ x − y − z = 3 ⇒ x + y + z = 5

⇒

f ′( x ) = 4 x 3 − 12 x 2 + 24 x + 1 = 4( x 3 − 3 x 2 + 6 x ) + 1 f ′′( x ) = 12 x 2 − 24 x + 24 = 12( x 2 − 2 x + 2)

f ′′( x ) has 0 rl roots f(x)has maximum 2 distinct real roots

∴

R=

On solving we get x = 4, y =

as f (0) = −1.

2.

2 2 R

(3) ( y + z ) cos 3θ − ( xyz )sin 3θ = 0

. . .(i)

xyz sin 3θ = (2cos3θ ) z + (2sin 3θ ) y

. . .(ii)

⇒ 6.

( y + z ) cos 3θ = (2 cos 3θ ) z + (2sin 3θ ) y = ( y + 2 z ) cos 3θ + y sin 3θ y

2x + y + z = 9 (4) The equation of P1 is y 2 − 8x = 0 and P2 is

y 2 + 16x = 0

(cos 3θ − 2sin 3θ ) = z cos 3θ and y (sin θ − cos3θ ) = 0

Tangent to y 2 − 5x = 0 passes through (–4, 0)

⇒ 0 ⇒ sin 3θ − cos 3θ = 0 ⇒ ⇒

sin 3θ = cos 3θ

∴

3θ = nπ + π / 4

3.

  β x β x x2  β x − + − ......  3! 5! x sin β x  = lim  (7) lim 3 5 x →0 α x − sin x x →0   x x α x −  x − − − .......  3! 5!   3 3

5 5

⇒

0 = m1 (−4) +

⇒

1 =2 m12

2

⇒

  β 3 x2 + .....  x3  β − 3!   =1 = lim x→0 x3 x5 (α − 1) x + − + ....... 3! 5! α −1 = 0

⇒

α = 1, Limit = 6 β = 1

1 6

⇒

β=

⇒

1 7  6(α + β ) = 6  1 +  = 6 × = 7 6 6 

9 7 ,z=− 2 2

2 m1

Also tangent to y 2 + 16x = 0 passes through (2, 0)

4 m2

⇒

0 = m2 × 2 −

⇒

m22 = 2

⇒

1 + m 22 = 4 m12

7.

(5) fx = x + x 2 − 1

x ≤ −1  x 2 − x − 1 if  2 − x − x + 1 if –1 ≤ x 0

⇒ increasing. 17. (a) P (required) = P (all are white) + P (all are red) + P (all are black)

1 2 3 3 3 4 2 4 5 5 = × × + × × + × × × 6 9 12 6 9 12 6 9 12 12 =

6 36 40 82 . + + = 648 648 648 648

18. (d) Let A : one ball is white and other is red

E1 : Both balls are from box B1 E2 : Both balls are from box B2

168

Mathematics

E3 : Both balls are from box B3 E  Here, P (required) = P  2   A  A P   .P ( E 2 )  E2  = A  A  A P   .P ( E1 ) + P   .P ( E2 ) +   .P ( E.3 ) E E  1  2  E3  C1 × C1 1 1 × 9 C2 3 55 6 = 1 = = C1 × 3C1 1 2 C1 × 3C1 1 3 C1 × 4 C1 1 1 1 2 181. + + × + 9 × + 12 × 6 C2 3 C2 3 C2 3 5 6 11 2

3

π /2

∫

19. (a)

π sin x dx = 2

0

=

π 8

 π  +0 0+ 2 π   sin(0) + sin   + 2sin  4  2  2   

(1 + 2)

20. (a) F ′(c ) = (b − a ) f ′(c ) + f ( a ) − f (b) F ′′(c ) = f ′′(c )(b − a ) < 0 ⇒

F ′(c) = 0

⇒

f ′(c) =

f (b ) − f ( a ) b−a

     

169

Mock Test-3

JEE-MAIN: MATHEMATICS MOCK TEST -3 1.

 −1 x < 0  Let g ( x) = 1 + x − [ x] and f ( x) =  0 x = 0. Then for all  1 x >1 

a.

1 2

b.

7 15

c.

2 15

d.

1 3

x, f {g ( x)} is equal to: a. x 2.

b. 1

5.

b. – I

b. 2 < x < 3 d. None of these

c. 1

a.

−1  2 3  8  4 2 

b.

−1  3 2  8  2 4 

c.

1  2 3 8  4 2 

d.

1  3 2 8  2 4 

The sum of 100 terms of the series .9 + .09 + .009......... ill be 100

106

 1 c. 1 −    10 

3π , then 2

100

 1 b. 1 +    10 

a.

2 sin α

b. −

2 sin α

c.

1 sin α

d. −

1 sin α

a. π / 3

b. π / 4

 3 −1  c. log   2   

d. None of these

12. A house of height 100 metres subtends a right angle at the window of an opposite house. If the height of the window be 64 metres, then the distance between the two houses is a. 48 m b. 36 m c. 54 m d. 72 m  sin x + cos x, when  13. If f ( x) =  x  2, when

100

 1 d. 1 +    10 

a. lim f ( x) ≠ 2

If n is an integer greater than 1, then a − C1 ( a − 1) + n

7.

b. 0

c. a 2

d. 2 n

The sum of

3e 2 c. 2e a.

8.

9.

d. 3e

If the letters of the word KRISNA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word KRISNA is a. 324 b. 341 c. 359 d. None of these Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently, equals

then

x=0 x →0 −

c. f ( x ) is continuous at x = 0 d. None of these 14. If x y = e x − y , then

2 6 12 20 + + + +K is 1! 2! 3! 4! b. e

x≠0

b. lim f ( x) = 0

x →0 +

n

C2 (a − 2) +.... + ( −1) n (a − n) = a. a

1 − cos α 1 + cos α = + 1 + cos α 1 − cos α

 3 i 11. Find real part of cos −1   2 + 2   

d. – 1

 2 −3 The inverse of   is  −4 2 

 1 a. 1 −    10 

6.

10. If π < α
0.

the

Statements /Expressions in Column I with the Statements/ Expressions in Column II Column I Column II

(A) L1 , L2 , L3 are

1.

k = −9

2.

k=−

concurrent, if

(B) One of L1 , L2 , L3 is

2

1 4x2 + 3x 3 1 2

π

L2 : 3 x − ky − 1 = 0, L3 : 5 x + 2 y − 12 = 0 Match

d. 20( 3 − 1)

b. −

4. π

5.

c. 20 2

1 2x + 3x 3

π 3

5/6

b. 40( 3 − 2)

a.

3.

π  then the value of f   is 6

a. 40( 2 − 1)

Then f ( x) is

2π 3

a

(C) The

17. A bird is sitting on the top of a vertical pole 20 m high and is elevation from a point O on the ground is 45° . It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (in m/s) of the bird is:

f (1) = 1, and

2.

b

(B) If

parallel to at least of the other two, if

(C) L1 , L2 , L3

form

a

3. k =

triangle, if

(D) L1 , L2 , L3 do not form a triangle, if

4.

5 6

k =5

6 5

173

Mock Test-3

JEE ADVANCE PAPER-II SECTION 1 Contains 8 Questions.

8.

The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

1.

( x − 3) 2 + ( y + 2) 2 = r 2 , then the value of r 2 is

distances of the point P from the lines x − y = 0 and

SECTION 2 Contains 8 Multiple Choice Questions

x+ y =0

With one or more than one correct option

respectively. The area of the region R

plane and satisfying 2 ≤ d1 ( P ) + d 2 ( P) ≤ 4, is _______ The

slope

of

the

tangent

to

the

1

9.

the

number

of

such

points

for

 π π   , where − < θ < . 4 4 

1/ 2

∫

c.

−1/ 2

d Then the value of (f (θ )) is d(tan θ )

(−∞, ∞) such that

1/ 2

∫

d.

f ( t ) esin π t dt =

3 + 1 apart. If the chords subtend at the centre, angles of

f (1 − 1)esin π t dt π /2

and

Let a, b, c be positive integers such that

∫ (2 cos ec x) π

17

dx is equal to

/4

log(1+ 2 )

∫

a.

2(eu + e − u )16 du

0

log(1+ 2 )

b is an integer. a

∫

b.

(eu + e −u )17 du

0

If a, b, c are in geometric progression and the arithmetic

7.

∫

1/ 2

11. The following integral

2π , where k > 0, then the value of [k] is [Note: [k] k k denotes the largest integer less than or equal to k]. 6.

f ( x) = f (1 − x) and

1

Two parallel chords of a circle of radius 2 are at a distance

π

1 4

1  f  x +  sin x dx = 0 2 

0

5.

d.

1 b. f ′   = 0 2

2

 −1  sin θ Let f (θ ) = sin  tan   cos 2θ 

1 2

a. f ′′( x) vanishes at least twice on [0, 1]

which

x + y + z ≤ 100 is 4.

c.

1 f ′   = 0. Then 4

−3 x + z = 0 −3x + 2 y + z = 0. 2

b. 1

defined on

the system of homogeneous equations: 3x − y − z = 0 ,

2

x→0+

10. Let f ( x) be a non-constant twice differentiable function

Let (x, y, z) be points with integer coordinates satisfying

Then

Let p = lim (1 + tan 2 x ) 2 x then log p is equal to:

a. 2

curve

( y − x5 )2 = x(1 + x2 )2 at the point (1, 3) is 3.

points of its latus rectum are tangents to the circle

For a point P in the plane, let d1 ( P ) and d 2 ( P) be the

consisting of all points P lying in the first quadrant of the

2.

If the normals of the parabola y 2 = 4 x drawn at the end

log(1+ 2 )

a 2 + a − 14 mean of a, b, c is b + 2, then the value of is a +1

c.

Let ω = eiπ / 3 , and a, b, c, x, y, z be non-zero complex

d.

∫

(eu − e −u )17 du

0

log(1+ 2 )

∫

2(eu − e −u )16 du

0

numbers such that a+b+c = x

12.

a + bω + cω 2 = y

a + bω 2 + cω = z. Then the value of

| x |2 + | y |2 + | z |2 is | a |2 + | b |2 + | c |2

∫x

x2 − 1 3

2 x4 − 2x2 + 1

dx is equal to

a.

2x4 − 2x2 + 1 +c x2

b.

2 x4 − 2 x2 + 1 +c x3

c.

2 x4 − 2 x2 + 1 +c x

d.

2 x4 − 2 x2 + 1 +c 2 x2

Mathematics

174

13. If

y ( x) satisfies the differential y '− y tan x = 2 x sec x and y(0) = 0, then.

equation

2 π  π a. y   = 4 8 2 2 π  π b. y '   =  4  18

π  π c. y   = 3 9

2

2  π  4π 2π d. y '   = + 3 3 3 3

SECTION 3 Contains 2 Paragraph Type Questions Each paragraph describes an experiment, a situation or a problem. Two multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.

Tangents are drawn from the point P(3, 4) to the ellipse

x2 y 2 + = 1 touching the ellipse at points A and B. 9 4 The coordinates of A and B are a. (3, 0) and (0, 2)

17.

 8 2 161   9 8 b.  − ,  and  − ,   5 5  5 15 

14. Given an isosceles triangle, whose one angle is 120° and

 8 2 161  c.  − ,  and (0, 2)  5 15 

radius of its incircle = 3. Then the area of the triangle in sq. units is

 9 8 d. (3, 0) and  − ,   5 5

a. 7 + 12 3 b. 12 − 7 3 c. 12 + 7 3

The equation of the locus of the point whose distances from the point P and the line AB are equal, is a. 9 x 2 + y 2 − 6 xy –54 x –62 y + 241 = 0

18.

d. 4π

b. x 2 + 9 y 2 + 6 xy –54 x + 62 y – 241 = 0 15. The circle passing through the point (–1, 0) and touching the y-axis at (0, 2) also passes through the point

 3  a.  − , 0   2   5  b.  − , 2   2   3 5 c.  − ,   2 2 d. (–4, 0)

c. 9 x 2 + 9 y 2 –6 xy –54 x –62 y – 241 = 0 d. x 2 + y 2 – 2 xy + 27 x + 31 y –120 = 0 Paragraph for Question No. 19 to 20 Let S = S1 ∩ S 2 ∩ S3 , where S1 = { z ∈ C :| z | < 4},    z − 1 + 3i  S2 =  z ∈ C : Im   > 0  1 − 3i   

and S3 = { z ∈ C : Re Z > 0}.

19. Area of S = 16

The normal at a point P on the ellipse x 2 + 4y 2 = 16 meets

a.

b.

the x-axis at Q. If M is the mid-point of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points

10π 3

20π 3

c.

16π 3

d.

32π 3

 3 5 2 a.  ± , ±   2 7   3 5 19  b.  ± ,±   2 4  

1  c.  ±2 3, ±  7   4 3 d.  ±2 3, ±   7  

20.

min |1 − 3i − z |= z∈S

a.

2− 3 2

b.

2+ 3 2

c.

3− 3 2

d.

3+ 3 2

175

Mock Test-3

ANSWER & SOLUTIONS

7.

(d) Let S =

JEE-Main 1.

2.

3.

4.

5.

6.

7.

8.

9.

S1 = 2 + 6 + 12 + 20 + K + Tn

10.

b

c

d

a

a

b

d

a

b

b

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

b

a

c

a

d

c

d

c

b

c

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

a

c

a

a

b

a

c

a

a

d

S1 =

1.

g ( x) < 0 g ( x) = 0

Tn = 2 + 4 + 6 + 8 + .......upto n terms

∴

g ( x) > 0

n Tn = [2 × 2 + (n − 1) 2] = n(2 + n − 1) = n(n + 1) 2 nth term of given series Tn =

Since g ( x) ≥ 1 > 0

⇒

∞

∞ 1 1 + 2∑ = e + 2e = 3e. n =1 ( n − 2)! n =1 ( n − 1)!

(c) x = 6 + x , x > 0 ⇒ x 2 = 6 + x, x > 0

8.

2

x − x − 6 = 0, x > 0

⇒

−1 (d) z + z = 1 ⇒ z 2 − z + 1 = 0 2 z = −ω or −ω

For z = − ω , z100 + z −100 = (−ω )100 + (−ω ) −100

=ω+

1

ω

= ω + ω 2 = −1

For z = −ω 2 , z100 + z −100 = (−ω 2 )100 + (−ω 2 ) −100

= ω 200 +

1

ω 200

= ω2 +

1

ω2

= ω2 + ω

9.

∴

putting 10 balls in a row. Now the number of ways in which no two black balls put together is equal to the number of ways of choosing 3 places marked '− ' out of eight places. −W − W − W − W − W − W − W −

(a) Let The matrix of cofactors of the elements of A viz.  c11  c21

c12   2 −( −4)   2 4  = = c22   −( −3) 2   3 2 

adjA = transpose of the matrix of cofactors of elements 2 3 of A =   4 2

∴

5.

∴

6.

This can be done in 8 C3 ways. 8

∴

A(adj A) =| A | I . (a) Series is a G.P. with a = 0.9 =

S100

(b) The number of ways of placing 3 black balls without any restriction is 10 C3 . Since, we have total 10 places of

= −1. 4.

(a) Words starting from A are 5 ! = 120 Words starting from I are 5 ! = 120 Words starting from KA are 4 ! = 24 Words starting from KI are 4 ! = 24 Words starting from KN are 4 ! = 24 Words starting from KRA are 3 ! = 6 Words starting from KRIA are 2 ! = 2 Words starting from KRIN are 2 ! = 2 Words starting from KRISA are 1 ! = 1 Words starting from KRISNA are 1 ! = 1 Hence rank of the word KRISNA is 324.

⇒ x = 3, x > 0. 3.

1 2 n(n + 1) n(n + 1) or Tn = or Tn = + (n − 2)! (n − 1)! n! n(n − 1)!

Now, sum = ∑

Hence g ( g ( x)) = 1 2.

2 + 6 + 12 + KTn −1 + Tn

0 = 2 + 4 + 6 + 8 + K upto n terms − Tn

⇒  −1  (b) f ( g ( x)) = 0 1 

2 6 12 20 + + + + K and let 1! 2! 3! 4!

1   1 − r100  9  1 − 10100 = a =   1 − r  10  1 − 1 10 

9 1 and r = = 0.1 10 10

  1  = 1 − 100 . 10  

(b) L.H.S. = a[C0 − C1 + C2 − C3 + ...( −1) .Cn ]

Required probability =

10. (b) =

C3 8× 7 × 6 7 = = 10 9 8 15 × × C3

10

1 − cos α 1 + cos α 1 − cos α + 1 + cos α + = 1 + cos α 1 − cos α 1 − cos 2 α 2 2 3π   ,  since π < α < = . ± sin α − sin α  2 

11. (b) Q Expression cos −1 (cos θ + i sin θ )

n

+[C1 − 2C2 + 3C3 − .... + ( −1) n −1 n.Cn ] = a.0 + 0 = 0

= sin −1 sin θ − i log( sin θ + 1 + sin θ ), where θ =

π 6

Mathematics

176

 3 i 1 π ∴ Real part of cos −1  = . +  = sin −1  2 4  2 2

 b a 2 − b 2 cos θ + 0 − ab  −b a 2 − b 2 cos θ − ab  19. (b)     b 2 cos 2 θ + a 2 sin 2 θ  b 2 cos 2 θ + a 2 sin 2 θ    

12. (a) 64 cot θ = d Also (100 − 64) tan θ = d or (64)(36) = d , 2

∴

d = 8 × 6 = 48 m.

13.

(c) f (0+) = f (0−) = 2 and f (0) = 2 Hence f ( x) is continuous at x = 0.

=

−[b 2 ( a 2 − b 2 ) cos 2 θ − a 2b 2 ] (b 2 cos 2 θ + a 2 sin 2 θ )

=

b 2 [a 2 − a 2 cos 2 θ + b 2 cos 2 θ ] b 2 cos 2 θ + a 2 sin 2 θ

=

b 2 [a 2 sin 2 θ + b 2 cos2 θ ] = b2 b 2 cos 2 θ + a 2 sin 2 θ

14. (a) x y = e x− y ⇒

y log x = x − y

⇒

y=

⇒

20. (c) O

x 1 + log x

45° A

dy = log x(1 + log x) −2 = log x[log ex ]−2 . dx

AB =

AC = BC =

z = x 3 (20 − x) 2

z = 400 x 3 + x5 − 40 x 4

 d 2z  d 2z 3 2 Now = −ive 2400 x 16 x 480 x ; = + −  2 dx 2  dx  x =12

21.

then it reduces to 2 2 2 ∫ t 4 dt = (tan x )5 + c = tan 5 x + c. 5 5

∫

17. (d) Required area =

2

0

[2 x − (2 x − x 2 )] dx

2

 2x x3  4 8 1 3 4 = − x2 +  = = − . −4+ − log 2 3 log 2 3 log 2 log 2 3  0

18. (c) Given equation

⇒

dy = e x (sin x + cos x ) dx

dy = e x (sin x + cos x) dx

On integrating, we get y = e x sin x + c.

. 2 =1

(a)

Tangent as focal chord

A

2 P θ θ (4,0)

x'

C(6,0)

x

2 B

y'

Here, the focal chord of y 2 = 16 x is tangent to circle

2

sec x dx = dt , 2 x

2

y

1 tan 4 x .sec2 x dx x

Put tan x = t ⇒

1

∴ Area of the circle = π (OB ) 2 = π

Hence x = 12 is the point of maxima x = 12, y = 8.

∫

2 1 = 2 2

In ∆ OBC , OB = BC cosec 45° =

dz = 1200 x 2 + 5 x 4 − 160 x 3 dx dz = 0, then x = 12, 20 Now dx

16. (c)

2 , O be centre of the circle

and let OC be the perpendicular from O on AB. Then

and x 3 . y 2 = z ⇒ z = x 3 . y 2

∴

2

Let AB be the chord of length

15. (d) Let x + y = 20 ⇒ y = 20 − x

⇒

B

C

( x − 6) 2 + y 2 = 2.

⇒

focus of parabola as (a,0) ie, (4,0) Now, tangents are drawn from (4,0) to ( x − 6) 2 + y 2 = 2. Since, PA is tangent to circle.

BC = −1 = 1, or BP 2 ∴ Slope of focal chord as tangent to circle = ±1. r r r r r r r r r r r 22. (c) a ⋅ b = a ⋅ c ⇒ a ⋅ b − a ⋅ c = 0 ⇒ a ⋅ (b − c ) = 0 r r r r r r r r ⇒ Either b − c = 0 or a = 0 ⇒ b = c or a ⊥ (b − c ). ∴

tan θ = slope of tangent =

AC = AP

23. (a) Given lines are, x−5 y−7 z+ 2 = = = r1 , (say) 3 1 −1

2

177

Mock Test-3

x+3 y−3 z−6 = = = r2 , (say) 2 4 −36 x = 3r1 + 5 = −36r2 − 3 ,

Let AE is a vertical lamp-post. Given, AE = 12m

y = −r1 + 7 = 3 + 2r2 and z = r1 − 2 = 4r2 + 6

AC = AE = 12m m

and ∴

AB =

24. (a) ~ ( p ∧ q) ≡ ~ p ∨ ~ q . BC =

(b) From the given identity

2bx + b + c = 8 x + 3

⇒

b = 4, c = −1

26. (a) As given H =

∴ 27.

⇒

2 pq p+q

H H 2q 2p 2( p + q ) + = + = =2. p q p+q p+q p+q

O

y= x

x−3 2

x

A(3,0) (9,0)

y'

To find the area between the curves, y = x , 2 y + 3 = x

(c) Plan: A square matrix M is invertible if f det(M) or | M |≠ 0.

and x-axis in the Ist quadrant (we can plot the above condition as); Area of shaded portion OABO 9

=∫

α α  M =  ⇒ | M |= 0 α α  M is non-invertible.

2  1  81  9  =  ⋅ 27  −  − 27  −  − 9   3 2 2 2      

0

30. (d) The equation of tangent at ( x1 , y1 ) is xx1 − 2 yy1 = 4, which is same as 2 x + 6 y = 2

M is non-invertible.

⇒

 a 0 (c) As given M =   ⇒ | M |= ac ≠ 0 0 c  ( Q a and c are non-zero) M is invertible. a b (s) M =   ⇒ | M |= ac − b 2 ≠ 0 b c  ac is not equal to square of an integer. M is invertible. D

∴

x1 2y 4 =− 1 = 2 6 2

⇒

x1 = 4 and y1 = − 6

JEE Advance Paper-I 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

5

9

9

5

3

2

4

5

a

d

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

b

d

d

b

c

d

a

c

a

C

E

1.

450

600 A

9

1 = 18 − (18) = 9 sq unit 2

a = b = c = α (let)

28. (a)

9

9

 x3 / 2  1  x 2   x −3 x dx − ∫  dx =   −  − 3x   3  2   3 / 2 0 2  2 3

 a  b  (a) Given that   =   ⇒ a = b = c = α (let) b   c 

⇒

∴

AC 2 − AB 2 = 144 − 48 = 96 = 4 6

y=

x'

Again | M |= 0

Q

=4 3

B

(b) Given that [b c] = [a b] ⇒

3

y

29. (a)

a b Let, M =   b c 

⇒

AE

Area = = AB × BC = 4 3 × 4 6 = 48 2 sq sq.cm. . cm.

b( x + 1) 2 + c ( x + 1) + d − (bx 2 + cx + d ) = 8 x + 3 ⇒

AE AB

tan 60° =

5 10 On solving, we get x = 21, y = , z = 3 3 Trick: Check through options.

25.

AE AC

tan 45° =

B

 5 (5) | 2z − 6 + 5i |= 2 z −  3 −   2  5  5 ≥ 2 3 −  3 − i  (corresponding Pt A) = 2   = 5  2  2

Mathematics

178

2.

a b c (9) Let M = d e f  g h i 

7.

0   −1 M 1  =  2  ⇒ b = −1, e = 2, h = 3 0   3 

∴

8.

⇒

Sum of diagonal elements = 9. (9) Let seventh term be ‘a’ and common difference be‘d’

6 S7 = S11 11

⇒ a = 15d Hence, 130 < 15 d < 140 ⇒ d = 9 4. (5) Clearly, 1 + 2 + 3 + K + n − 2 ≤ 1224 ≤ 3 + 4 + K n ⇒

(n − 2) (n − 1) (n − 2) ≤ 1224 ≤ (3 + n) 2 2

⇒ ⇒ ⇒

n 2 − 3n − 2446 ≤ 0 and n 2 + n − 2454 ≥ 0 49 < n < 51 n = 50

∴ ⇒

5.

⇒

c= =

∴

6.

r=

(

e

n

−e

( cos(α

−1) n

least

3

elements

1 2 Not possible. As condition for two distinct real root is

f ′( x) = 0 ⇒ x 2 = − f (α ) f ( β ) = 0

(where are α , β roots of f ′( x) = 0)

11. (c) Let z = ω 1+ z 1+ ω −ω 2 = = =ω 1+ z 1+ ω2 −ω 1+ z arg = arg ω = θ (put z = cosθ + i sin θ ) 1+ z

Now

∴

12. (b) | A | I = AAT

αm

e e (cos(α )

Number of subsets having 2 elements each 8! 8× 7 = 8C2 = = = 28 2!6! 2 Number of subsets having at = 256 − 1 − 8 − 28 = 256 − 37 = 219

⇒

225 × 3 ∆ 2 = 3. ⇒ r = 6 + s  10 + 14    2  

α →0

α →0

Number of subsets having 1 element each = 8C1 = 8

6 + 10 − 2 × 6 × 10 × cos120 ° = 14

(2) lim

lim

Number of subsets having 0 elements = 8C0 = 1

2

( )

C2 = 2n

f ′( x) = 6 x 2 + 3

a 2 + b 2 − 2 ab cos C

cos α n

C2 − n = n

∴ Total number of subsets of ( A × B) = 28 = 256

2∆ 2 ×15 3 3 = = ⇒ C = 120° ab 6 ×10 2

2

n

n

10. (d) f ( x) = 2 x3 + 3x + k

1 (3) ∆ = ab sin C 2

sin C =

(5) Number of red lines = n C2 − n

n(n − 1) = 2n 2 ⇒ n −1 = 4 ⇒ n = 5. 9. (a) Set A has 4 elements Set B has 2 elements ∴ Number of elements in set ( A × B) = 4 × 2 = 8

∴

n(n + 1) − (2k + 1) = 1224 2 k = 25 ⇒ k − 20 = 5

⇒

2

Simultaneously, we get (2, 4) and (0, 0) Focus is (2, 0) 1 Are = × 2 × 4 = 4sq. units. 2

Hence,

1  0  M 1 =  1  ⇒ g + h + i = 12 ⇒ i = 7 1 12 

Given

2

Number of blue lines = n

1 1 M  −1 =  1  ⇒ a = 0, d = 3, g = 2  0   −1

∴ 3.

(4) Solving y = 8x and x + y − 2 x − 4 y = 0 2

=−

e 2

) (cos α )

n

) −1 α α m

− 1) 2n

if and only if 2n − m = 0

α 2n = −

e 2

⇒

1 0  5a −b   5a 3  (10a + 3b)  =    0 1   3 2   −b 2 

⇒

25a 2 + b2 = 10a + 3b & 15a − 2b = 0 & 10a + 3b = 13

⇒

10a +

3.15a = 13 ⇒ 65a = 2 ×13 2

179

Mock Test-3

⇒

2 ⇒ 5a = 2 5 2a = 6 ⇒ b = 3

∴

5a + b = 5

⇒

⇒ OB − OA = 20( 3 − 1)

a=

Hence distance covered in one second by the bird is

AB = 20( 3 − 1) Thus speed of bird

13. (d) Let the numbers be a, ar, ar2 is G.P. a + ar 2 Given a, 2ar, ar are in A.P. the 2 ar = ( a ≠ 0) 2

= 20( 3 –1) − 1) m/s /

2

which gives r = 2 + 3, as the G.P. is an increasing G.P.

14. (d) Theoretically the number of terms are 2N + 1 (i.e. odd) But As the number of terms being odd hence considering that number clubbing of terms is done hence the solutions follows: Number of terms = ∴

n+2

C2 = 28

⇒

⇒ f ( x ) = cx 2 + 1

3x Also f (1) = 1 3

Sum of coefficient = 3n = 36 = 729 Put x = 1

15. (b)

⇒ x 2 f ′( x ) − 2 x f ( x ) + 1 = 0

⇒ c=2

n=6

n +1

t 2 f ( x) − x 2 f (t ) =1 t →x t−x

18. (a) lim

C3 − C3 = 10 n

On solving n = 5

Hence f ( x ) =

r r 19. (A) a − b = 1 + 3 = 2 r r | a |= b, | b |= 2 cos θ =

16. (c) Let x = P (computer turns out to be defective given that it is produced in plate T2 ), ⇒ ⇒

θ=

7 1 4 = (10 x) + x 100 5 5 7 = 200 x + 80 x

⇒

⇒

,

(B)

but its

2π as its opposite to side of maximum 3

∫ (f (x) − 3x) dx = a b

∫ f (x)dx = ⇒

4 × 273 2 × 273 546 78 = = = 210 + 4 × 273 105 + 2 × 273 651 93 P

2

− b2

a

a

4 4  273  5(1 − x) 5  280  = 1 4 1  280 − 70  4  273  (1 − 10 x) + (1 − x) + 5 5 5  280  5  280 

Q

π 2π

b

P( A ∩ B) P(B)

17. (d)

2 1 = 2× 2 2

3 3 length.

7 ⇒ x= 280 P (produced in T2 / not defective) =

2 2 1 x + . 3 3x

(C)

3 2 −a 2 + b 2 (b − a 2 ) + a 2 − b 2 = 2 2

f (x) = x. 5/ 6 π 2  ln (sec π x + tan π x) 7 / 6     ln 3  π



=



π 

5π 5π 7π 7π  + tan − ln sec + tan  ln sec  = π. ln 3  6 6 6 6 

(D) Let u =

1 1 ⇒ z = 1− 1− z u

| z |= 1 ⇒ 1 −

1 =1 2

⇒ | u − 1|=| u | B

A

Here, AP = QB = 20m

∠POA = 45°, ∠QOB = 30°

⇒ OA = 20; OB = 20 3

O

∴ locus of u is perpendicular bisector of line segment joining 0 and 1

⇒ maximum arg u approaches

π 2

but will not attain.

Mathematics

180

20.

x + 3 y − 5 = 0 and 5 x + 2 y − 12 = 0 intersect at (2, 1)

 −1 sin θ  (1) f (θ ) = sin  tan  2cos θ  

4.

Hence 6 − k − 1 = 0 k = 5

1 3 = ⇒ k = −9 3 −k 3 −k −6 . = ⇒k= for L2 , L3 to be parallel 5 2 5 −6 for k ≠ 5, − 9, they will form triangle 5 −6 for k = 5, k = −9, they will not form triangle 5

 sin θ = sin  sin −1 2 sin θ + cos 2θ 

for L1 , L2 to be parallel

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

6

8

7

1

5

4

3

2

3

All

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

a

d

a,d

c

d

a

d

a

b

c

1.

(6)

x= 2

d d(tan θ ) f (θ ) = =1 d(tan θ ) d(tan θ )

5

(2) 2cos

⇒

cos

Let

θ 2

θ 2

k

Now put x = 1, y = 3

6.

= 3 +1

θ 2

θ 2

= =

= t ⇒ 2t 2 + t −

3 +1 2 3 +1 2 3 +3 =0 2

−1 ± 1 + 4(3 + 3) −1 ± (2 3 + 1) −2 − 2 3 3 , = = 4 4 4 2

=

(4) ⇒ c=

π 6

θ 2

=

3 2

⇒ k = 3.

12 4

b2 a +b+c ⇒ =b+2 a 3

a − 2b + c = 6

a − 2b + ⇒

dy = m = 8. dx

(7) 3x − y − z = 0 −3 x + z = 0

b c = = (integer) b 2 = ac a b

a + b + c = 3b + 6 ⇒

dy = m. 2(3 − 1)( m − 5) = 1(4) + (1)(4)(2) dx

1−

b2 =6 a

2b b 2 6 + = a a2 a 2

6 b   − 1 = a = 6 only a a  

7.

(3) The expression may not attain integral value for all a, b, c If we consider a = b = c, then x = 3a

⇒

y = 0 and z = 3 x

⇒

x 2 + y 2 + z 2 = x 2 + z 2 = x 2 + 9 x 2 = 10 x 2 ≤ 100

y = a(1 + ω + ω 2 ) = a(1 + i 3)

⇒

x2 ≤ 10

z = a(1 + ω 2 + ω ) = a(1 + i 3)

⇒

x = 0, ± 1, ± 2, ± 3

There are such seven points.

sin θ sin θ = = tan θ | cos θ | cos θ

3 +1 2

=

− 1 + cos

2

π

=

2

= 1(1 + x 2 )2 + ( x)(2(1 + x 2 )(2 x))

3.

θ

t ∈ [ −1,1],cos

 dy  (8) 2( y − x 5 )  − 5 x 4   dx 

m = 5+3 = 8 ⇒

k

Q ⇒

k

= 0, cos θ + cos

t=

= 8 − 2 = 6 sq.. units

⇒

π

⇒

Area of region = ((2 2) − ( 2) )

m−5 =

+ cos

cos

2

⇒

+ 2cos

⇒

2 2 ≤ 2α ≤ 4 2

and

π

π 2k

π 2k

P(α , β ) x=2 2

2

⇒

2 cos 2

2 ≤α ≤ 2 2

2.

2

⇒

For P (α , β ), α > β

⇒

sin θ + (cos θ − sin θ ) 2

y=x

2 ≤ d1 ( p ) + d 2 ( p ) ≤ 4

⇒

sin θ

=

JEE Advance Paper-II

  

∴

| x |2 + | y |2 + | z |2 = 9 | a |2 +4 | a |2 +4 | a |2 = 17 | a |2

181

Mock Test-3

8.

(2) Equation of normals are x + y = 3 and x − y = 3.

⇒

Distance from (3, −2) on both normals is ‘r’

⇒ ⇒

9

|3−2−3| 2

− ∫ (eu + e − u )17

−u

(e − e ) du 2cosec x cot x ln(1+ 2 )

0

= −2

=r

∫

(eu + e − u )16 du =

ln(1+ 2 )

r = 2.

∫

2(eu + e − u )16 du

0

2

(3) P = lim+ (1 + tan 2 x − 1)

1 2x

x →∞

P=e ⇒

⇒

u

lim

(1+ tan 2 x −1 )

x→0+

1 2x

1 2

=e

lim

x →0+

(tan x )2 2( x ) 2

then log p = 1

= e2

1 2

log P = log e =

 1 1   3 − 5  dx x x  12. (d) ∫  2 1 2− 2 + 4 x x

Let 2 −

⇒

10. (a,b,c,d)

⇒ 1/4

3/4 1/2

⇒

2 1 + =z x2 x4

1 dz 4∫ z 1 × z +c 2 1 2 1 2− 2 + 4 +c 2 x x

13. (a,d) y '− y tan x = 2 x sec x

f ( x) = f (1 − x)

I.F. = e∫

Put x = 1/ 2 + x ⇒

= elog cos x = cos x

∴

y cos x = ∫ 2 x sec x.cos x dx

⇒

y cos x = x 2 + c

Hence f ( x + 1/ 2) is an even function or f ( x + 1/ 2)sin x

⇒

y ⋅ cos x = x 2 (Q y (0) = 0)

an odd function.

⇒

y = x 2 sec x

∴

2 2 π  π  π  4π 2π y  = and y'   = + 4 8 2 3 3 3 3

1  1  f  + x = f  − x 2  2 

Also, f '( x) = − f '(1 − x) and for x = 1/ 2, We have f '(1/ 2) = 0. 1

∫

Also,

1/ 2

0

f (1 − t )esin π t dt = − ∫ f ( y)esinπ t dy 1/ 2

(obtained by putting, 1 − t = y ). Since f ′(1/ 4) = 0, f ′(3/ 4) = 0. Also f ′(1/ 2) = 0

⇒

tan x dx

f ′( x) = 0 atleast twice in [0, 1] (Rolle’s Theorem)

3 2 b 4 sin120° sin 30° Also = a b

14. (c) ∆ =

⇒

a = 3b and ∆ = 3s and s =

π 2

11. (a)

∫ (2cosec x) π

17

⇒ ∆=

dx

4

Let eu + e− u = 2cosec x, x =

⇒ u = ln(1 + 2), x =

π 4

1 (a + 2b) 2

3 (a + 2b) 2

From equation (i) and (ii), we get ∆ = (12 + 7 3)

15. (d) Circle touching y-axis at (0, 2) is (x − 0) 2 + (y − 2) 2 + λ x = 0

π 2

⇒

u=0

∴

⇒

cosec x + cot x = eu and cosec x − cot x = e − u

∴

⇒

cot x

eu − e −u (eu − e − u ) dx = −2cosec x cot xdx 2

. . .(i)

passes through (–1, 0) 1+ 4 − λ = 0 ⇒ λ = 5 x 2 + y 2 + 5x − 4y + 4 = 0

Put y = 0 ⇒ x = −1, − 4

∴

Circle passes through (–4, 0)

. . .(ii)

Mathematics

182

16. (a) Any point on the line can be taken as Q ≡ {(1 − 3µ ), ( µ − 1), (5µ + 2)} PQ = {−3µ − 2, µ − 3, 5 µ − 4}

Now 1(−3µ − 2) − 4( µ − 3) + 3(5µ − 4) = 0 ⇒

− 3 µ − 2 − 4 µ + 12 + 15 µ − 12 = 0

18. (a) Locus is parabola 3x 4 y Equation of AB Is + =1 9 4 x ⇒ + y = 1 ⇒ x + 3y − 3 = 0 3

( x + 3 y − 3) 2 10 2 2 10 x + 90 – 60 x + 10 y + 160 – 80 y = x 2 + 9 y 2 + 9 + 6 xy – 6 x – 18

( x − 3)2 + ( y − 4) 2 =

8 µ = 2 ⇒ µ = 1/ 4

17. (d) y = mx + 9 m 2 + 4 4 − 3m =

9m 2 + 4

10 x 2 + 90 – 60 x + 10 y 2 + 160 – 80 y = x 2 + 9 y 2 + 9 + 6 xy – 6 x – 18 y

16 + 9m 2 − 24m = 9m 2 + 4 ⇒ m =

Equation is y − 4 =

12 1 = 24 2

⇒

9 x 2 + y 2 – 6 xy – 54 x – 62 y + 241 = 0.

19. (b)

y + 3x = 0

1 ( x − 3) 2

60°

2y − 8 = x − 3 ⇒

⇒

⇒ ⇒

x 2 + y 2 < 16

x − 2y + 5 = 0 Let B = (α , β )

⇒

60°

xα y β + −1 = 0 9 4 α / 9 β / 4 −1 = = ⇒ 1 5 −2 9 8 α =− , β = 5 5

 9 8 B ≡  − , .  5 5

Area of region S1 ∩ S 2 ∩ S3 = shaded area

=

π × 42 4

+

42 × π 6

 1 1  20π = 42 π  +  = 3 4 6 20. (c) Distance of (1, –3) from y + 3 x = 0

>

−3 + 3 × 1 3 − 3 > 2 2

183

Mock Test-4

JEE-MAIN: MATHEMATICS MOCK TEST-4 1.

a. 3 2.

b. 1

4.

5.

d. 0

b. 1, 4 d. None of these

1+ 3 i is a root of equation x 4 − x 3 + x − 1 = 0 then 2 its real roots are a. 1, 1 b. – 1, – 1 c. 1, – 1 d. 1, 2

10.

If

 1 0 0 Let A =  5 2 0  , then the adjoint of A is  −1 6 1   2 −5 32  a.  0 1 −6   0 0 2 

 −1 0 0  b.  −5 −2 0   1 −6 1 

 −1 0 0  c.  −5 −2 0   1 −6 −1

d. None of these

232 990

b.

232 c. 990

11.

π  π  tan  + θ  − tan  − θ  = 4  4  a. 2 tan 2θ b. 2 cot 2θ c. tan 2θ

232 9990

232 d. 9909

The sum of the coefficients of even power of x in the

a.

tanh x + tanh y 1 − tanh x tanh y

b.

tanh x + tanh y 1 + tanh x tanh y

c.

tanh x − tanh y 1 − tanh x tanh y

d.

tanh x − tanh y 1 + tanh x tanh y

12. The length of the shadow of a pole inclined at 10° to the vertical towards the sun is 2.05 metres, when the elevation of the sun is 38°. The length of the pole is 2.05 sin 38° 2.05 sin 42° a. b. sin 42 ° sin 38° c.

7.

b. 128

In the expansion of

c. 512

d. 64

1 − 2 x + 3x 2 , the coefficient of x5 ex

will be

8.

9.

a.

71 120

b. −

71 120

c.

31 40

d. −

31 40

The total number of seven digit numbers the sum of whose digits is even is a. 9000000 b. 4500000 c. 8100000 d. None of these If E and F are events with P ( E ) ≤ P ( F ) and P ( E ∩ F ) > 0, then

2.05 cos 38° cos 42 °

d. None of these

x≠0

, then

x=0

a. f (0 + 0) = 1

b. f (0 − 0) = 1

c. f is continuous at x = 0

d. None of these

14. If y =

expansion of (1 + x + x 2 + x 3 )5 is a. 256

d. cot 2θ

tanh( x + y ) equals

1  2  x sin , when 13. If f ( x) =  x  0, when

The value of 0.234 is a.

6.

c. 2

The real roots of the equation x 2 + 5 | x | + 4 = 0 are a. – 1, 4 c. – 4, 4

3.

a. occurrence of E ⇒ occurrence of F b. occurrence of F ⇒ occurrence of E c. non-occurrence of E ⇒ non-occurrence of F d. None of the above implication holds

The number of solutions of log 4( x − 1) = log 2( x − 3) is:

a.

( x − a)( x − b) dy , then = ( x − c)( x − d ) dx

1 1 1  y 1 + − − 2  x − a x − b x − c x − d 

1 1 1   1 + − − b. y  x a x b x c x d  − − − −  c.

1 1 1 1 1  + − − 2  x − a x − b x − c x − d 

d. None of these 15. If

f ( x) =

x2 − 1 , for every real number x, then the x2 + 1

minimum value of f a. Does not exist because f is unbounded b. Is not attained even though f is bounded c. Is equal to 1 d. Is equal to –1

Mathematics

184

16.

a

∫

x

1 − a2x

23. A line makes the same angle θ , with each of the x and

dx =

z-axis. If the angle β , which it makes with y-axis is such

1 a. sin −1 a x + c log a

b. sin a + c

that sin 2 β = 3sin 2 θ , then cos 2 θ equals

1 cos −1 a x + c log a

d. cos −1 a x + c

a.

c.

−1

x

3 5 1 c. 5

17. The area bounded by the circle x 2 + y 2 = 4, line x = 3 y and x-axis lying in the first quadrant, is

a.

π

b.

2

π 4

c.

π 3

dy = 2 is dx 2 b. y = c − c. y = 2 cx x

24. d. π

2 x

d. y = c − the

3 x2

point

y = x tan α + c, c > 0 is

c. c sec 2 α

d. c cos 2 α

20. The circle passing through point of intersection of the circle S = 0 and the line P = 0 is a. S + λ P = 0 b. S − λ P = 0 and λ S + P = 0 c. P − λ S = 0 d. All of these 21. Axis of a parabola is y = x and vertex and focus are at a distance 2 and 2 2 respectively from the origin. Then equation of the parabola is-

a. ( x − y ) 2 = 8( x + y − 2)

b. ( x + y ) 2 = 2 ( x + y − 2)

c. ( x − y ) 2 = 4 ( x + y − 2) d. ( x + y ) 2 = 2 ( x − y + 2) r r r r 22. If a and b be unlike vectors, then a ⋅ b = r r r r a. | a | | b | b. − | a | | b | c. 0 Space for rough work

b. p ∧ q

c. p ∧ ~ q

d. ~ p ∧ ~ q

x2 + 6 x + λ = 0 and 3α + 2β = −20, then λ =

length of perpendicular from ( a cos α , a sin α ) upon the straight line

b. c sin 2 α

d. None of these

25. If α and β are the roots of the equation

19. The

a. c cosα

2 3

(~ (~ p)) ∧ q is equal to a. ~ p ∧ q

18. The general solution of x 2 a. y = c +

b.

d. None of these

a. –8

b. –16

c. 16

d. 8

26. If a, b, c, d are in H.P., then ab + bc + cd is equal to a. 3ad

b. (a + b)(c + d )

c. 3ac

d. None of these

27. If y = 3 x + 6 x 2 + 10 x 3 + ...., then the value of x in terms of y is

a. 1 − (1 − y ) −1/ 3

b. 1 − (1 + y )1/ 3

c. 1 + (1 + y ) −1/ 3

d. 1 − (1 + y ) −1/ 3

28. The number of arrangements of the letters of the word BANANA in which the two N’s do not appear adjacently, is a. 40 b. 60 c. 80 d. 100 29. Differential coefficient of sin −1 x w.r.t cos −1 1 − x 2 is 1 a. 1 b. 1 + x2 c. 2 d. None of these 30. The interval in which the function x 2 e − x is non decreasing, is a. (−∞, 2]

b. [0, 2]

c. [2, ∞)

d. None of these

185

Mock Test-4

JEE ADVANCE PAPER-I SECTION 1 Contains 8 Questions.

SECTION 2 Contains 10 Multiple Choice Questions

The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

With one or more than one correct option

1.

 kπ   kπ  For any integer k, let α k = cos  + i sin    , where  7   7 

9.

If y = sec(tan −1 x), then

a.

12

i = −1. The value of the expression

Σ | α k +1 − α k |

k =1 3

is

Σ | α 4 k −1 − α 4 k − 2 |

k =1

2.

Let m be the smallest positive integer such that the

1 2

(1 + x) 2 + (1 + x)3 + ... + (1 + x)49 + (1 + mx)50 is (3n + 1) 51C3

c. for all x in the interval [1, ∞),f (x + 2) − f (x) > 2 d. f’ (x) is strcily decreasing in the interval [1, ∞)

The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads

a.

1 1− 1 (1 + nx n ) n + K n(n − 1)

c.

1+ 1 (1 + nx n ) n + K n(n + 1)

than or equal to x. Let f be a real valued function defined

Ï x - [ x] on the interval [−10, 10] by f (x) = Ì Ó1 + [ x] - x 2 Then the value of π

if [ x] is odd if [ x] is even

10

10 −∫10

f ( x)cosπ x dx is

5.

 d2  The value of ∫ 4 x3  2 (1 − x 2 )5  dx is ________  dx  0

6.

Let f be a real-valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If the y-intercept of the tangent at any point P(x, y) on the curve y = f(x) is equal to the cube of the abscissa of P, then the value of

g ( x) dx equals

2x + y = 1 is

tangent

b.

1 1− 1 (1 + nx n ) n + K n −1

d.

1+ 1 (1 + nx n ) n + K n +1

1

1

12. Let f :[0, 2] → R be a function which is continuous on

x2

F ( x) =

to

the

hyperbola

intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is r r r If a , b and c are unit vectors satisfying r r r r r2 r r2 r r2 a − b + b − c + c − a = 9, then 2a + 5b + 5c is.

f ( t ) dt for x ∈ [0,2]. If F ′( x ) = f ′( x ) for all

x ∈ (0, 2), then F (2) equals

a. e2 − 1

b. e4 − 1

c. e − 1

d. e4

13. The function y = f ( x ) is the solution of the differential equation

dy dy xy xy xx 4 + 2xx in (–1,1) satisfying ++ 2 == dx x − 11 1 − x2

f (0) = 0. Then

2

x y − = 1. If this line passes through the point of a 2 b2

∫ 0

f (−3) is equal to

8.

n− 2

[0, 2] and is differentiable on (0, 2) with f (0) = 1. Let

1

2

∫x

f occurs n times

For any real number x, let |x| denote the largest integer less

line

x for n ≥ 2 and (1 + x n )1/ n

g ( x) = ( f o f oKo f )( x). Then 144244 3

is at least 0.96 is

The

2

d.

b. lim f '(x) = 1

11. Let f ( x ) =

7.

c. 1

1 10. For function f(x) = x cos , x ≥ 1, x a. for atleast one x in interval [1, ∞),f (x + 2) − f (x) < 2

for some positive integer n. Then the value of n is

4.

1 2

b.

x →∞

coefficient of x 2 in the expansion of

3.

dy at x = 1 is equal to dx

3 2

∫

f ( x ) dx is

3 2

3 3 3 3 π π π b. − c. − d. − 3 2 3 4 6 4 6 2 14. Let O (0, 0), P (3, 4), Q (6, 0) be the vertices of the triangle a.

π

−

OPQ. The point R inside the triangle OPQ is such that the triangles OPR, PQR, OQR are of equal area. The coordinates of R are

Mathematics

186

4  a.  , 3  3 

 2 b.  3,   3

 4 c.  3,   3

4 2 d.  ,  3 3

19.

x2 y 2 − = 1, parallel 9 4 to the straight lien 2 x − y = 1. The points of contact of the

15. Tangents are drawn to the hyperbola

(B) The

tangents on the hyperbola are.

1   9 , a.   2 2 2

1   9 , b.  −   2 2 2

c. 3 3, −2 2

d. −3 3, 2 2

72

33

d.

45

f (x) = sin

3. [2, ∞) tan θ 1 − tan θ

1 f (θ ) = − tan θ −1

then

the

(D) If

set

f (x) = x 3 / 2 (3x − 10),

x ≥ 0,

through the intersection of P1 and P2 . If the distance of the

increasing in

then

f(x)

b. 2α – β + 2γ + 4 = 0 c. 2α + β – 2γ – 10 = 0

20.

x2 − 6 x + 5 Match the conditions/expressions x2 − 5 x + 6 in Column I with statements in Column II Column I Column II (A) If −1 < x < 1, then 1. 0 < f ( x) < 1

Let f ( x) =

(B) If 1 < x < 2, then

18. The negation of ~ s ∨ (~ r ∧ s ) is equivalent to b. s ∧ ( r ∧ ~ s ) d. s ∧ r

SECTION 3 Contains 2 Match The Following Type Questions You will have to match entries in Column I with the entries in Column II.

Space for rough work

is

f ( x) satisfies

d. 2α – β + 2γ – 8 = 0

a. s ∧ ~ r c. s ∨ ( r ∨ ~ s )

4. (−∞, − 1] ∪ [1, ∞)

5. (−∞, 0] ∪ [2, ∞)

point (0, 1, 0) from P3 is 1 and the distance of a point

a. 2α + β + 2γ + 2 = 0

1 tan θ , 1

π  f (θ ) : 0 ≤ θ <  is 2 

P3 be a plane, different from P1 and P2 , which passes

relations is (are) true?

2. (−∞, 0) ∪ (0, ∞)

−1

(C) If

17. In R3 , consider the planes P1 : y = 0 and P2 : x + z = 1. Let

(α, β, γ) from P3 is 2, then which of the following

the

 8(3)  is  2( x −1)  1− 3 

through A is

b.

of

x−2

sides of a triangles ABC , then the length of the median

a. 18

domain

function

16. If the vectors AB = 3iˆ + 4kˆ and AC = 5iˆ − 2 ˆj + 4kˆ are the

c.

Match the statements given in Column I with the interval/union of intervals given in Column II Column I Column II 1. (−∞, − 1) ∪ (1, ∞)   2iz   : z (A) The set Re  is   2   1 − z  a complex number, is | z |= 1, z ≠ ±1

2. f ( x) < 0

f ( x) satisfies (C) If 3 < x < 5, then

3. f ( x) > 0

f ( x) satisfies (D) If x > 5, then f ( x) satisfies

4. f ( x ) < 1

187

Mock Test-4

JEE ADVANCE PAPER-II SECTION 1 Contains 8 Questions.

SECTION 2 Contains 8 Multiple Choice Questions

The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

With one or more than one correct option

1.

The

least

period

of

the

9.

Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B having 3 or more elements is a. 256 b. 220 c. 219 d. 211

10.

The

function

 π [ x]  πx  π [ x]  sin   + cos   + tan   is λ, then the value  12   4   3  of 201λ must be (where [⋅] denotes the greatest integer

function)

2.

3.

4.

dy  ax − b  If y = tan −1   then the value of (2008) dx  bx + a 

∫ sin

−1

11.

k

for

which

z1 − 2 z 2 is unimodular and z2 is not unimodular. Then 2 − z1 z 2

  2x + 2  2x + 2    dx = ( x + 1) tan −1   2    3   (4 x + 8 x + 13) 

A particle moves in a straight line with a velocity given dx by = ( x + 1) (x is the distance travelled). If the time dt taken by a particle to traverse a distance of 99 m is λ.

the point z1 lies on a

a. straight line parallel to x-axis b. straight line parallel to y-axis c. circle of radius 2 d. circle of radius

2 1 4 4

12.

If the adjoint of a 3 × 3 matrix P is 2 1 7  , then the 1 1 3

possible value(s) of the determinant of P is (are) a. –2 b. –1 c. 1 d. 2

If a triangle has its orthocenter at (1, 1) and circum 3 3 centre at  ,  and if centroid and nine point centre are 2 4

(α , β ) and

(γ ,δ ) respectively, then the value of

13.

m is equal to: a. 102 c. 100

intersection of the lines 3 x + 5 y = 1 and

λ 25

, then the value of

λ must be 8.

A, B, C, D are any four points in the space. If uuur uuur uuur uuur uuur uuur | AB × CD + BC × AD + CA × BD | = λ (area of ∆ABC ) then the value of 125λ must be

2

2

 3  2  1 2 1  +  2  +  3  + 4  5  5  5

If the radius of a circle which passes through the point (2, 0) and whose centre is the limit of the point of

(2 + c) x + 5c 2 y = 1 as c → 1 is

If the sum of the first ten term of the series 2

6α + 12β + 4γ + 8δ must be 7.

equation,

A complex number z is said to be unimodular if | z |= 1.

then the value of 20 λ log10 e must be

6.

the

Suppose z1 and z2 are complex numbers such that

+ λ ln(4 x 2 + 8 x + 13) + c, then the value of − 4λ must be

5.

number

2 x + 3 x + k = 0 has two distinct real roots in [0, 1] a. lies between 1 and 2 b. lies between 2 and 3 c. lies between –1 and 0 d. does not exist

x =−1

must be The indicated horse power I of an engines is calculated PLAN π from the formula where A = d 2. I= 33000 4 Assuming that error of 10% may have been made in measuring P, L, N and d. If the greatest possible in I is λ % then λ must be If

real

3

14.

2

+  4 4  + ..., is 16 m, 5 



5

then

b. 101 d. 99

Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is a. 264 b. 265 c. 53 d. 67

Mathematics

188

15.

Let θ ,ϕ ∈ 0,2π be

such

that

2cos θ 1 − sin ϕ

θ θ  = sin 2 θ  tan + cot  cos ϕ − 1 tan 2π − θ > 0 and 2 2 

−1 < sin θ < − a. 0 < ϕ
0 Length of the perpendicular drawn on line from point (a cos α , a sin α )

p=

− a sin α + a cos α tan α + c 1 + tan α 2

; p=

c = c cos α sec α

21.

y

(a)

y=x

Alternate: f ′( x) =

N (1,1) V

( x + 1)2 x − ( x − 1)2 x 4x = 2 ( x 2 + 1) 2 ( x + 1) 2 2

F (2,2)

P M

2

x'

f ′( x) = 0 ⇒ x = 0

∴

Y

dy 2 2 = 2 ⇒ dy = 2 dx, Now integrate it. dx x x

maximum value.

∴

=

arc

Hence f ( x) has minimum value –1 and also there is no

f ′′ ( x) =

3

20. (d) It is a fundamental concept.

2 ≤2 x +1 2

O

x+y–2=0 x

y'

( x + 1) 4 − 4 x.2( x + 1)2 x ( x 2 + 1) 4 2

2

π 4 π = . = . 6 2 3

2 x2 −1 x2 + 1 − 2 = = 1− 2 x2 + 1 x2 + 1 x +1

15. (d) f ( x) =

x 1  x2  4 x 4 − x 2 + sin −1    + 2 2 3  2  0  2

by

dy y  1 1 1 1  =  + − − . dx 2  ( x − a) ( x − b) ( x − c) ( x − d ) 

4 − x 2 dx

3  3 2π  π + π − − = . 2  2 3  3 Trick: Area of sector made

1 log y = [log( x − a ) + log( x − b) − log( x − c) − log( x − d )] 2 Differentiating w.r.t. x we get

Thus

3

=

 ( x − a )( x − b)  14. (a) y =    ( x − c )( x − d ) 

1 dy 1  1 1 1 1  =  + − −  y dx 2  ( x − a) ( x − b) ( x − c) ( x − d ) 

x dx + ∫ 3

2

3

=

Hence f ( x) is continuous at x = 0.

⇒

3

2

2

Since, distance of vertex from origin is

( x 2 + 1)4 − 16 x( x) −12 x 2 + 4 = = ( x 2 + 1)3 ( x 2 + 1)3

∴

f ′′(0) > 0

∴

2 and focus is

2 2. V(1, 1) and F(2, 2) (ie, lying on y = x) where, length of latusrectum = 4a = 4 2 By definition of parabola

(Q a =

2)

PM 2 = (4 a )( PN )

There is only one critical point having minima.

Where, PN is length of perpendicular upon x + y − 2 = 0

Hence f ( x) has least value at x = 0.

(ie, tangent at vertex).

f min = f (0) =

−1 = −1. 1

16. (a) Put a x = t ⇒ a x log e a dx = dt , then

∫ =

ax 1 − a2x

dx =

1 log e a ∫

dt 1− t2

1 sin −1 (a x ) sin −1 (t ) + c = + c. log e a log e a

⇒

( x − y )2  x+ y−2 = 4 2  2 2  

⇒

( x − y ) 2 = 8( x + y − 2)

r r r r 22. (b) a ⋅ b = − | a | | b | , (Q cos θ = −1)

23. (a) Here, l = cosθ , m = cos β , n = cos θ , (Q l = n) Now, l 2 + m 2 + n 2 = 1

191

Mock Test-4

⇒

2cos θ + cos β = 1

⇒

2 cos θ = sin β Given, sin β = 3sin θ

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

⇒

2 cos 2 θ = 3sin 2 θ ⇒ 5cos 2 θ = 3,

4

5

3

4

2

9

2

3

a

b,c,d

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

a

b

b

c

a,b

c

b,d

d

b

a

2

2

JEE Advance Paper-I

2

2

2

2

3 cos 2 θ = . 5 24. (b) (~ (~ p)) ∧ q = p ∧ q .

∴

12

Σ e

k =1

25. (b) α + β = −6

. . .(i)

1.

i

kπ 7

. . .(ii)

and given 3α + 2β = −20

. . .(iii)

π

(4) 3

αβ = λ

i

e 7 −1 i

12 =4 3

=

π

Σ ei (4 k − 2) e 7 − 1

k =1

(5) Coeff. x 2

Solving (i) and (iii), we get β = 2, α = −8

2.

Substituting these values in (ii), we get λ = −16 .

⇒

2

C2 + 3C2 + 4C2 + ....... + 49C2 + 50C2 m2 = (3n + 1) 51C3

(a) Since a , b, c, d are in H.P., therefore b is the H.M. of

⇒

3

26.

C3 + 3C2 + 4C2 + ....... + 49C2 + 50C2 m2 = (3n + 1) 51C3

⇒

n

Cr + nCr −1 = n+1Cr ⇒

⇒

50

C3 + 50C2 + (m2 − 1) 50C2 = 3n

⇒

51

⇒

2ac and c is the H.M. of b and d a+c 2ac 2bd 2bd i.e. c = , ∴ ( a + c)(b + d ) = . b+d b c ab + ad + bc + cd = 4ad ⇒ ab + bc + cd = 3ad .

C3 + (m2 − 1) 50C2 = 51n ⋅ 50C2 + 51C3

⇒

m2 − 1 = 51n m2 = 51n + 1

a and c i.e. b =

1 1 1 Trick : Check for a = 1, b = , c = , d = . 2 3 4

3.

1 + y = 1 + 3 x + 6 x 2 + 10 x 3 + ...

⇒ 1 + y = (1 − x )

−3

⇒ 1 − x = (1 + y )

−1/ 3

⇒ x = 1 − (1 + y )

6! = 60 3!2! The number of arrangements of words BANANA in 5! = 20 3! Required number of arrangements = 60 – 20 = 40 which two N’s appear adjacently =

(8) Let coin was tossed ‘n’ times

⇒

 n + 1 1 −  n  ≥ 0.96  2 

⇒

2n ≥ 25 ⇒ n ≥ 8 n +1

=

4.

(4) 1

29. (a) Let y1 = sin −1 x and y 2 = cos −1 1 − x 2

–2 –1

1 dy1 = dx 1 − x2

dy2 1 1(−2 x) 1 dy =− = ⇒ 2 = 1. 2 2 dx dy1 1 − (1 − x ) 2 1 − x 1− x 30. (b) Let y = f ( x ) = x 2e − x

10

∴

∫

I=

f ( x ) cos π x dx

−10

10

2

= 2 ∫ f ( x ) cos π x dx = 2 × 5∫ f ( x ) cos π x dx 0

0

  10 =  ∫ (1 − x ) cos π xdx + ∫ ( x − 1) cos π x dx  = 10( I1 + I 2 ) 0 1  1

dy = 2 xe− x − x 2e− x = e− x (2 x − x 2 ) dx Hence f '( x) ≥ 0 for every x ∈ [0, 2], therefore it is non-

3

2

1

 x − 1, 1 ≤ x < 2 f (x) is periodic with period 2 f ( x) =  1 − x , 0 ≤ x < 1

Differentiating w.r.t. x of y1 and y 2 , we get

decreasing in [0, 2].

51 50 ⋅ C2 + 51C3 3

n 1 Probability of getting at least two heads = 1 −  n + n  2 2  

−1/ 3

28. (a) Total number of arrangements of word BANANA

⇒

C3 + 50C2 ⋅ m2 = (3n + 1) 51C3

Min value of m2 for 51n + 1 is integer for n = 5.

27. (d) We have y = 3 x + 6 x 2 + 10 x 3 + ....

∴

50

2

2

I 2 = ∫ ( x − 1) cos π x dx put x − 1 = t 1

Mathematics

192

1= a m −b

0

1 = 4a2 − b2

2

I 2 = − ∫ t cos π t dt 1

1

I1 = − ∫ (1 − x ) cos π x dx = − ∫ x cos(π x ) dx 0

1=

0

1

∴

1  40 π  1 = −20  − 2 − 2  = 2 ∴ I = 4 π  π 10  π 2

1

5.

(2) ∫ 4 x 3 0

I

8. ⇒

d2 (1 − x 2 ) 5 dx dx 2 II 1

⇒

1

d d   =  4 x 3 (1 − x 2 )5  − ∫12 x 2 (1 − x 2 )5 dx dx dx  0 0

1

⇒

 (1 − x 2 )6  = 0 − 0 − 12[0 − 0] + 12∫ 2 x(1 − x 2 )5 dx = 12 ×  −  6 0  0  1 = 12 0 +  = 2  6 6.

(9) y − y1 = m ( x − x1 )

e

1 f dx x

= e− ln x =

1 1 y × = ∫ − x 2 × dx x x

7.

Also, 1 = a2 m2 − b2

3 r r r r r r a ⋅b + b ⋅c + c ⋅ a ≥ − 2 Equation (i) and (ii) are simultaneously true 1 r r r r r r If a ⋅ b = b ⋅ c = c ⋅ a ≥ − 2 r 2 r r Now, 2a + 5b + 5c

. . .(i)

. . .(ii)

(a) y = sec(tan −1 x) 1 dy = sec(tan −1 x ).tan (tan −1 x ). dx 1 + x2

⇒

2 1  dy   dx  = 1 + 1 = 2   x =1

1 10. (b,c,d) For f (x) = x cos   , x ≥ 1 x

1 1 1 f '(x) = cos   + sin   → 1forx → ∞ x x x also f '(x) =

x3 3 f ( x) = − + x ∴ f (−3) = 9. 2 2 a  (2) Substituting  , 0  in y = −2x + 1 e  2a 0=− +1 e e 2a =1 a = e 2

3 r r r r r r − = a ⋅b + b ⋅c + c ⋅a 2

⇒

y y x2 = −∫ xdx ⇒ = − + c x x 2

⇒

r r2 r r2 r r2 (3) a − b + b − c + c − a = 9 r r r r r r 2 − 2a ⋅b + 2 − 2 b ⋅c + 2 − 2 c ⋅ a = 9

r r r = 54 − 10 − 25 − 10 = 9 ⇒ 2a + 5b + 5c = 3 9

1 x

a = 1, e = 2

r r r r r r = 4 + 25 + 25 + 20 a ⋅ b + 50b ⋅ c + 20( a ⋅ c )

3 Put x = 0, to get y intercept y1 = mx1 = x1

dy y1 − x1 = x13 dx dy x − y = − x3 dx dy y − = − x2 dx x

4e 2 − b2 4

r r r Now, | a + b + c |2 ≥ 0 r r r r r r ⇒ 1+1+1+ 2 a ⋅b + b ⋅c + c ⋅ a ≥ 0

1   1 1 =  4 x 3 × 5(1 − x 2 )4 (−2 x )  − 12   x 2 (1 − x 2 )5  − ∫ 2 x (1 − x 2 )5 dx  0 0   0

1

2

b2 = e2 = 1. Also, b 2 = a 2 ( e1 − 1)

   sin π x cos π x  + I = 10  −2∫ x cos π x dx  = −20  x π π 2  0   0  1

∴

2

1

=

1 1 1 1 1 1 sin   − 2 sin   − 3 cos   2 x x x x x x

1 1 cos   < 0 forx ≥ 1 3 x x

⇒

f '(x) is decreasing for [1, ∞)

⇒

f '(x + 2) < f '(x). Also, 1 1  lim f (x + 2) − f (x) = lim (x + 2)cos − x cos  = 2 x →∞ x →∞ x + 2 x  

∴

f (x + 2) − f (x) > 2∀x ≥ 1

193

Mock Test-4

11. (a) Here ff ( x ) =

f ( x) x = [1 + f ( x ) n ]1/ n (1 + 2 x n )1/ x

x (1 + 3 x n )1/ n

⇒

fff ( x ) =

⇒

g ( x ) = ( f o f oK o f )( x ) =

(3, 4)

x (1 + nx n )1/ n

n times

Hence I = ∫ x n−2 g ( x) dx = ∫ 1 n 2 x n −1dx 1 = 2∫ = 2 n 1/ n n (1 + nx ) n

∴

I=

14. (c) Since, ∆ is isosceles, hence centroid is the desired point.

∫

1 1− n n

1 (1 + nx ) n(n − 1)

xn−1dx (1 + nx n )1/ n

(0, 0)

15. (a,b) Slope of tangent = m = 2 Equation of tangent in slope form is

d (1 + nx n ) dx dx (1 + nx n )1/ n

y = mx ± a 2 m 2 − b 2 , y = 2 x ± 4 2

+K

 ma 2 −b 2  , and point of contact is  −  c   c

12. (b) F (0) = 0

4   9 1   2×9 ≡ − ,− ,±  ≡ ±  2  ±4 2 ±4 2   2 2

F ′( x ) = 2 xf ( x ) = f ′( x )

f ( x) = e x

2

(6, 0)

+c

R

2

f ( x ) = e x (Q f (0) = 1)

Q

16. (c) T

x2

(α , β , γ )

F ( x ) = ∫ e dx x

S

0

iˆ ˆj kˆ 1 uuur uuur 1 Area of base (PQRS) = | PR × SQ |= 3 1 −2 2 2 1 −3 −4

2

F ( x) = e x − 1 (Q F (0) = 0) ⇒ F (2) = e 4 − 1 dy x x4 + 2x + 2 y= dx x − 1 1 − x2 This is a linear differential equation

13. (b)

x

I.F. = e

∫ x2 −1dx

1

= e2

ln| x 2 −1|

⇒ Solution is y 1 − x = ∫ or

⇒

=

= 1 − x2

2

x ( x 3 + 2) 1 − x2

f ( x) 1 − x 2 =

⋅ 1 − x dx

x5 + x2 5

3/2

Now,

∫

f ( x )dx =

− 3/2 3/2

= 2

∫ 0

x

1− x

2

∫

1 − x2

π /3

2

dx = 2

17. (b, d) Let the required plane be x + z + λ y − 1 = 0

⇒

x2

− 3/2 2

 2  Volume = (5 3)   = 10 cu. units  3

⇒ 3/2

∫ 0

dx (Using property)

sin θ cos θ dθ cos θ

(Taking x = sin θ ) π /3

1 | −10iˆ + 10 ˆj − 10kˆ |= 5 | iˆ − ˆj + kˆ |= 5 3 2

1− 2 + 3 2 Height = proj. of PT on iˆ − ˆj + kˆ = = 3 3

2

x5 y 1 − x 2 = ∫ ( x 4 + 2 x )dx = + x 2 + c 5 f (0) = 0 ⇒ c = 0

P

| λ −1 |

λ +2 2

=1 ⇒ λ = −

P3 ≡ 2 x − y + 2 z − 2 = 0 Distance of P3 from (α , β , γ ) is 2

⇒

| 2α − β + 2γ − 2 | 4 ×1 + 4

2α − β + 2λ + 4 = 0 and 2α − β + 2λ − 8 = 0

18. (d) ~ [~ s ∨ (~ r ∧ s )] π /3

1 2

θ sin 2θ  = 2 ∫ sin 2 θ dθ =  − 4  0 2 0

= ~ (~ s )∧ ~ (~ r ∧ s )

 3 π 3 π  . = 2   − 2   = − 4 6  8  3

= (s ∧ r ) ∨ (s ∧ ~ s)

= s ∧ (r ∨ ~ s)

= s∧r∨ F = s∧r

=2

Mathematics

194

19. (A) z =

2i ( x + iy ) 2i ( x + iy ) = 1 − ( x + iy ) 2 1 − ( x 2 − y 2 + 2ixy )

Using 1 − x 2 = y 2 Z= Q

1 1 ≤ −1 or − ≥ 1. y y

2.

3.

4.

5.

6.

7.

8.

9.

10.

4824

1004

50

3

40

30

1601

500

c

d

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

a,d

b

c

a,c,d

c

a

b

b

c

1.

8.3x − 2 ≤1 1 − 32( x −1)

(B) For domain −1 ≤

π [ x]    π [ x]  = sin  2π +  = sin   12    12 

3 x − 3x − 2 −1 ≤ 0 1 − 32 x − 2

 π [ x]  Similarly the period of tan   is 3 and the period of  3 

(3x − 1)(3x − 2 − 1) ≥0 (32 x − 2 − 1)

π x  2π cos  =8  is  4  π /4 Hence, the period of the given function λ = LCM of 24, 8, 3 = 24

⇒ x ∈ ( −∞, 0] ∪ (1, ∞ ).

Case (ii):

3 x − 3x − 2 +1 ≥ 0 1 − 32 x − 2

∴

(3x − 2 − 1)(3x + 1) ⇒ ≥0 (3x.3x − 2 − 1)

2.

⇒ x ∈ ( −∞, 1) ∪ [2, ∞ ).

So, x ∈ (−∞, 0] ∪ [2, ∞)

0 1 − tan θ

2 tan θ = 2(tan 2 θ + 1) = 2sec 2 θ . 1

3 1/ 2 15 1/ 2 3/ 2 (D) f ′( x) = ( x) (3x − 10) + ( x) × 3 = ( x) ( x − 2) 2 2 Increasing, when x ≥ 2. 20.

dy 1 = −0 dx 1 + x 2

∴

dy dx

⇒

3. y=

1

2

3

5

x

b    x− a   ax − b  −1 (1004) Q y = tan    = tan   bx + a  1+ b ⋅ x  a   −1

∴

( x − 1)( x − 5) . The graph of f ( x) is shown f ( x) = ( x − 2)( x − 3) y

201 λ = 201× 24 = 4824

b = tan −1 x − tan −1   a

(C) R1 → R1 + R3 0 f (θ ) = − tan θ −1

 π [ x]  2π (4824) The period of sin  = 24 as  is  12  π /12  π [ x + 24]   π ([ x] + 24)  sin   = sin   12 12    

3x − 3x − 2 ≤ 1. 1 − 32 x − 2

Case (i): ⇒

1.

2ix − 2 y 1 =− . y 2 y 2 − 2ixy

−1 ≤ y ≤ 1 ⇒ −

⇒ −1 ≤

JEE Advance Paper-II

∴

⇒

= x = −1

(2008)

1 1 1 = = 2 1 + (−1) 1 + 1 2

dy dx

= 2008 × x = −1

1 = 1004 2

π  PL  d 2  N PLAN 4  (50) I = = 33000 33000 ∆I ∆P ∆L 2∆d ∆N = + + + I P L d N  ∆I   ∆P   ∆L   ∆d   ∆N  ×100  +  ×100  + 2  ×100  +  ×100   ×100  =  I P L d N          

= 10% + 10% + 2 ×10% + 10% = 50% (A) If −1 < x < 1 ⇒ 0 < f ( x) < 1 (B) If 1 < x < 2 ⇒ f ( x) < 0 (C) If 3 < x < 5 ⇒ f ( x) < 0 (D) If x > 5

⇒ f ( x) < 1

∴

4.

λ = 50   2x + 2  dx (3) Let I = ∫ sin −1   (4 x 2 + 8 x + 13)   

195

Mock Test-4

 2x + 2 I = ∫ sin −1   (2 x + 2)2 + 32  Put 2 x + 2 = 3tan θ

∴

  dx  

7.

3x + 5 y = 1,  1 − 3x  2 (2 + c) x + 5c 2   = 1, then, (2 + c ) x + c (1 − 3 x ) = 1 5  

2 dx = 3sec θ dθ

3sec2 θ dθ 3 Then I = ∫ θ ⋅ = {θ ⋅ tan θ − ln secθ } + c 2 2 2 3  2 x + 2   2 x + 2   −1  2 x + 2  =  ⋅ − + ln 1 tan      + c 2  3  3  3      

 2x + 2  3 2 = ( x + 1) tan −1   − ln(4 x + 8 x + 13) + c 3 4  

∴

⇒ ⇒

(40)

dx = x +1 dt

2

∴

dx = dt x +1 ln( x + 1) = t + c

∴

⇒

∴

8.

∴

20λ log10 e = 20 × 2log e 10 × log10 e = 40

2

G ( x, y )

1

3 2 C ,  2 4

4 5 Centroid is  ,  and nine point centre is the mid point 3 6 of orthocenter and circumcentre.

 1 + ( 3/ 2 ) 1 + ( 3/ 4 )  , Nine point centre is  , 2 2   5 7 i.e.,  ,  4 8 4 5 5 7 and γ = , δ = a= ,β = 3 6 4 8 6α + 12β + 4γ + 8δ 4 5 5 7 + 12 × + 4 × + 8 × = 8 + 10 + 5 + 7 = 30 3 6 4 8

r r r r (500) Let PV of A, B, C and D be a , b , c and 0 uuur uuur r r r r r r r AB × CD = (b − a ) × − c = − b × c + a × c uuur uuur r r r r r r r BC × AD = ( c − b ) × − a = − c × a + b × a uuur uuur r r r r r r r and CA × BD = ( a − c ) × −b = − a × b + c × b

(30) Since, centroid divides the orthocenter and circum center in the ratio 2 : 1 (internally) and if centroid

= 6×

2

λ 64 1 1601 1601 + = = = 25 625 625 25 25 λ = 1601

∴

∴

3 3 2 × + 1× 1 2 × + 1× 1 4 5 2 4 = and y = = x= 2 +1 3 2 +1 6

∴

2   1  − 2 +  − − 0 5 25    

Radius of the required circle = 

=

For x = 99, t = ln100 = 2log e 10

G ( x, y ), then O (1, 1)

c →1

1  2 Therefore the centre of the required circle is  , −   5 25  but circle passes through (2,0)

⇒ t = ln( x + 1)

6.

2 1+ c ⇒ x= 3c + 2 5 1 − 3 x 1 − (6 / 5) 1 = =− y= 5 5 25

x = lim

∴

3 4

Putting t = 0, x = 0 we get c = 0

∴

1 − c2 2 + c − 3c 2 (1 + c )(1 − c ) 1+ c or x = = (3c + 2)(1 − c ) 3c + 2

x=

∴

Then, −4λ = 3

5.

(1601) Solving the equation

2

Hence λ = −

(2 + c ) x + 5c y = 1 and 2

∴

Adding all we get uuur uuur uuur uuur uuur uuur r r r r r r AB × CD + BC × AD + CA × BD = − 2 ( a × b + b × c + c × a ) uuur uuur uuur uuur uuur uuur r r r r r r | AB × CD + BC × AD + CA × BD | = 2 | a × b + b × c + c × a | uuur uuur r r r r = 2 | ( c − a ) × (b − a ) | = 2 | AC × AB | 1 uuur uuur = 4 ⋅ | AC × AB | = 4 (Area of ∆ABC ) 2 λ=4

Then,

9.

125 λ = 500

(c) A = [ x, y ] B = {a, b, c, d}

A × B Having 2 × 4 = 8 elements

∴

Total subsets of A × B is 28 = 256 Total no. of subsets of A × B having 3 or more elements

 = 256 −  1 + 8 + 8 C2  null set single ton set subset having 2 elements  = 256 − 1 − 5 − 28 = 219

   

Mathematics

196

10. (d) f ( x) = 2 x + 3x + k

  1 Also, 2cosθ (1 − sin φ ) = sin θ  θ  sin cos θ  2 2

3

2

f ′( x) = 6 x 2 + 3 f ′( x) = 0 ⇒

1 2 Not possible. As condition for two distinct real root is x2 = −

f (α ) f ( β ) = 0 (where are α , β roots of f ′( x) = 0) 11. (c)

z1 − 2 z 2 2 2 = 1 ⇒ | z1 − 2 z2 | = | 2 − z1 z2 | 2 − z1 z 2

⇒

( z1 − 2 z2 )( z1 − 2 z2 ) = (2 − z1 z2 )(2 − z1 z2 )

⇒

z1 z1 + 4 z2 z2 = 4 + z1 z1 z2 z2

⇒

4 + | z1 |2 | z2 |2 − 4 | z2 |2 − | z1 |2 = 0

⇒

⇒

2cos θ − 2 cos θ sin φ = 2sin θ cos φ − 1

⇒

1 + 2 cos θ = 2sin(θ + φ )

⇒ ⇒

1 sin(θ + φ ) = + cosθ 2 1 π 4π < sin(θ + φ ) < 1 ⇒ < φ < 2 2 3

16. (c) 1 + x 2  ( x cos cot −1 x + sin cot −1 x ) 2 − 1

1/ 2

2   1  x = 1 + x  x cos cos −1 + sin sin −1 − 1     1 + x2 1 + x2   

(| z1 |2 −1) ⋅ ( | z2 |2 − 4) = 0

 x 2 1 = 1 + x 2  + 2  1 + x 1 + x2 

Hence, z lies on a circle of radius 2 centered at origin.

1 4 4  12. (a,d) adj P =  2 1 7  ⇒ | adj P |= 4 1 1 3 

17. (a) a1 = 1

a2 = 2

a3 = 3

18. (b) 1 2

2

2

 8   12   16   20   24  13. (b)   +   +   +   +   + ... 5  5   5   5   5 

=

82 122 162 202 242 (4n + 4) 2 + 2 + 2 + 2 + 2 + .... Tn = 2 5 5 5 5 5 52

Sn = =

1 52

10

∑16(n + 1) n =1

2

=

16 10 2 ∑ (n + 2n + 1) 25 n =1

16 10 × 11× 21 2 × 10 × 11 16  16 + + 10  = × 505 = m  25  6 2 5  25

⇒ m = 101 14. (c) Number of required ways = 5!− {4 ⋅ 4}!− 4 C2 ⋅ 3!+ 4 C3 ⋅ 2!− 1} = 53.

15. (a,c,d,) Conditions: − tan(θ ) > 0 ⇒ tan θ < 0 and ∴

1  3π 5π  θ ∈  ,  ⇒ 0 < cosθ < 2  2 3 

1/ 2

−1 < sin θ < −

3 2

a4 = 5

an is following Fibonacci series. Hence a17 = a16 + a15 .

| adj P | = | P |2 ⇒ 4 = | P |2 ⇒ | P | = ± 2 2

2    − 1   

= 1 + x2 ( x2 + 1 − 1)1/ 2 = x 1 + x 2 .

We know, | adj P | = | P |n −1 (where n is order of matrix)

2

1/ 2

2

But | z2 | ≠ 1, ∴ | z2 | = 2

⇒

   cos φ − 1  

1

Total no of ways exactly two consecutive two is = 5 ways Total no of ways exactly three consecutive two is = 2 ways Total no of ways exactly four consecutive two is = 1 ways So, b6 = 5 + 2 + 1 = 8

19. (b) Case (i): One odd, 2 even Total number of ways = 2×2×3+1×3×3+1×2×4 = 29. Case (ii): AII 3 odd Number of ways = 2×3×4 = 24 Favourable ways = 53 53 53 Required probability = . = 3 × 5 × 7 105 20. (c) Here 2 x2 = x1 + x3 ⇒

x1 + x3 = even Hence number of favourable ways = 2C1 ⋅4 C2 +1 C 1 ⋅3 C1 = 11.

197

Mock Test-5

JEE-MAIN: MATHEMATICS MOCK TEST-5 1.

The domain of definition of the function y ( x ) given by

10.

the equation 2 x + 2 y = 2 is: a. 0 < x ≤ 1 c. −∞ < x ≤ 0 2.

3.

4.

b. 2

c. 3

c. 2i sinh α sin β

d. 2 cosh α cos β

c. 2(1 − 320 )

c.

an +1 1 = . Then an 3

20

∑a

r

is

7.

( x + 2) + ... + ( x + 2)

d. None of these

1+

2 3 4 + + + K∞ = 3! 5! 7!

a. e 8.

9.

b. 2e

15. c. e/2

d. e/3

The number of ways in which the following prizes be given to a class of 20 boys, first and second Mathematics, first and second Physics, first Chemistry and first English is. a. 204 ×192

b. 203 ×193

c. 202 × 194

d. None of these

If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form

1 b. 7

1 c. 8

b. 2 d. None of these

dy = dx

x + log(1 + x) 1+ x

x is maxima at 1 + x tan x a. x = sin x c. x =

16.

π

b. x = cos x d. x = tan x

3

∫

1 dx = x 1 + log x

a.

2 (1 + log x )3/ 2 + c 3

c. 2 1 + log x + c

b. (1 + log x)3/ 2 + c d.

1 + log x + c

17. The area of the triangle formed by the tangent to the

7 m + 7 n is divisible by 5, equals 1 a. 4

d. l cot α cot β

 x  + log(1 + x)  c. (1 + x) x  1 + x  d. None of these

Cr (3n− r − 2 n − r )

c. n Cr (3r + 2 n− r )

n

c. l tan α tan β

b.

b.

r

b. l tan α cot β

 x  + log ex  a. (1 + x) x  1 + x 

n −1

a. Cr (3 − 2 n ) n

1 (cosh 2 x + 1) 2

the tower is a. l tan β cot α

14. If y = (1 + x) x , then

( x + 3) n −1 + ( x + 3) n − 2 ( x + 2)

+ ( x + 3)

d.

a. 1 c. 4

r =1

d. None of these

2

1 (cosh 2 x − 1) 2

k (2 x − x 2 ), when x < 0 is continuous at x = 0, is f ( x) =  cos x, when x ≥ 0 

Coefficients of x r [0 ≤ r ≤ ( n − 1)] in the expansion of n −3

b. cosh 2 x + 1

13. The value of k so that the function

b. Idempotent d. Symmetric

1   b. 3 1 − 20   3 

20 [4 + 19 × 3] 2

d. − sin θ

12. A tower subtends an angle α at a point A in the plane of its base and the angle of depression of the foot of the tower at a point l meters just above A is β . The height of

1 −5  is 0 

For a sequence < an >, a1 = 2 and

c. sin θ

sinh 2 x equals

d. 4

b. 2 cosh α cosh β

 0 −4 Matrix  4 0   −1 5

b. –1

a. cosh 2 x − 1

a. 2 sinh α sinh β

a.

6.

11.

cosh(α + i β ) − cosh(α − i β ) is equal to

a. Orthogonal c. Skew-symmetric 5.

a. 1 b. x ≤ x ≤ 1 d. −∞ < x < 1.

A real root of the equation log 4 {log 2 ( x + 8 − x )} = 0 is a. 1

sin(π + θ ) sin(π − θ ) cosec2θ =

1 d. 49

hyperbola xy = a 2 and co-ordinate axes is

a. a 2

b. 2a 2

c. 3a 2

d. 4a 2

Mathematics

198

18. The solution of

dy = x log x is dx

24.

~ ( p ∨ (~ q )) is equal to a. ~ p ∨ q

x2 a. y = x 2 log x − + c 2

b. (~ p ) ∧ q c. ~ p ∨ ~ p

x2 b. y = log x − x 2 + c 2 1 1 c. y = x 2 + x2 log x + c 2 2 d. None of these

d. ~ p ∧ ~ q 25. If the sum of the roots of the equation λ x + 2x + 3λ = 0 be equal to their product, then λ = a. 4 b. –4 c. 6 d. None of these 26. If a , b, c and u , v , w are the complex numbers representing 2

19. The distance of point (–2, 3) from the line x − y = 5 is a. 5 2 b. 2 5 c. 3 5

the vertices of two triangles such that c = (1 − r )a + rb and

d. 5 3

w = (1 − r )u + rv, where r is a complex number, than the

20. If the coordinates of one end of the diameter of the circle

x + y − 8x − 4 y + c = 0 are (–3, 2), then the coordinates 2

2

of other end are a. (5, 3) c. (1, –8)

b. (6, 2) d. (11, 2)

two triangles a. have the same area b. are similar c. are congruent d. None of these

a. x 2 = y

1 4 2 3 27. The rank of the matrix, A =  0 1 2 − 1 is   0 −2 −4 2  a. 2 b. 3 c. 1 d. Indeterminate

b. y 2 = 2 x

28. The

21. Let ( x, y) by any point on the parabola y 2 = 4 x. Let P be the point that divides the line segment from (0, 0) to

( x, y) in the ratio 1: 3. Then, the locus of P is

c. y 2 = x

of

xn

in

the

expansion

of

−n

(1 + x + x + ....) is 2

d. x = 2 y 2

r r r r r r 22. If a , b , c are unit vectors such that a + b + c = 0, then r r r r r r a ⋅b + b ⋅c + c ⋅ a = a. 1 b. 3 c. – 3/2 d. 3/2 23. The direction cosines of three lines passing through the origin are l1 , m1 , n1 ; l2 , m2 , n2 and l3 , m3 , n3 . The lines will be coplanar, if

l1 a. l2 l3

coefficient

n1 n2

m1 m2 = 0

n3

m3

c. l1l2l3 + m1m2 m3 + n1n2 n3 = 0 d. None of these

l1 b. l2 l3

m2 m3

n3 n1 = 0

m1

n2

a. 1 b. (–1)n c. n d. n + 1 29. Three boys and two girls stand in a queue. The probability that the number of boys ahead of every girl is atleast one more than the number of girls ahead of her, is a. 1/2 b. 1/3 c. 2/3 d. 3 4 30. The imaginary part of tan −1 (cos θ + i sin θ ) is a. tanh −1 (sin θ ) b. tanh −1 (∞) c.

1 tanh −1 (sin θ ) 2

d. None of these

199

Mock Test-5

JEE ADVANCE PAPER-I SECTION 1 Contains 8 Questions. The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

1.

A variable plane is at a constant distance p form the origin and meets the axes in A, B and C. If the locus of the centroid of the tetrahedron OABC is

7.

f (60, 48) + f (80, 48) + f (13, 5) must be 8.

If x = a (cos t + t sin t ); y = a (sin t − t cos t ), then the value of 120)

x −2 + y −2 + z −2 = λ p −2 then the value of 160 λ must be 2.

 y2 y2  If f  2 x 2 + , 2 x 2 −  = xy , then the value of 8 8  

d2y must be dx 2 t =π / 3

The normal to the parabola y 2 = 8 x at the point (2, 4)

SECTION 2 Contains 10 Multiple Choice Questions

meets it again at (18, –12). If length of normal chord is λ ,

With one or more than one correct option

then the value of λ must be 2

3.

An electric component manufactured by ‘RASU electronics’ is tested for its defectiveness by a sophisticated testing device. Let A denote the event “the device is defective” and B the event “the testing device reveals the component to be defective”. Suppose 1. P( A) = α and P( B / A) = P( B '/ A ') = 1 − α , where 0 < α < < 1. If the probability that the component is not defective, given that the testing device reveals it to be defective, is λ, then the value of 2008 λ must be

4.

The coefficient of x 50 in the polynomials after parenthesis have been removed and like terms have been collected in

9.

Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A × B , each having at least three elements is: a. 219 b. 256 c. 275 d. 510

10. The quadratic equation p( x) = 0 with real coefficients has

purely

imaginary

roots.

Then

the

equation

p( p( x)) − 0 has a. only purely imaginary roots b. all real roots c. two real and two purely imaginary roots d. neither real nor purely imaginary roots

the expansion (1 + x )1000 + x (1 + x )999 + x2 (1 + x )998 + ... + x100011. A value of θ for which 2 + 3i sin θ is purely imaginary, is: 1 − 2i sin θ λ! (1 + x ) 998 + ... + x1000 is , then the value of λ + 2µ +3v must be π π µ ! v! a. b. 3 6 (v > µ )  3  1  c. sin −1  d. sin −1    4  5. The value of  3   1 7π  2π  5π  −π    −1  −1  −1  200Let cot −1ω cos 12. be a complex cube root of unity with ω ≠ 1 and  216 sin  sin  + 27 cos  cos  + 28 tan  tan +  6  3  4  4  π     i+ j P[ Pij ] be a n × n matrix with pij = ω . Then when 2π  5π  −π    −1  −1  P 2 ≠ 0, n = 7 cos −1  cos  + 28 tan  tan  + 200 cot  cos  3  4  4     a. 57 b. 55 must be c. 58 d. 56 r r r 6. Let a, b and c be three non-coplanar unit vectors such 13. If m is the A.M. of two distinct real numbers l and n ( l, n > 1) and G1, G2 and G3 are three geometric means π that the angle between every pair of them is . If 3 between l and n, then G14 + 2G24 + G34 equals, r r r r r r r a × b + b × c = pa + qb + rc , where p, q and r are a. 4l 2 mn b. 4l m2 n scalars, then the value of

p 2 + 2q 2 + r 2 is _______ q2

c. 4l mn2

d. 4l 2 m2 n 2

Mathematics

200 4n

14. Let Sn = ∑ (−1)

k ( k +1) 2

SECTION 3 Contains 2 Match The Following Type Questions

k 2 . Then S n can take value(s)

k =1

a. 1056

b. 1088

c. 1120

d. 1332

15. If 3x = 4 x −1 , then x = a.

2 log 3 2 2 log 3 2 − 1

b.

2 2 − log 2 3

c.

1 1 − log 4 3

d.

2 log 2 3 2 log 2 3 − 1

16. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is: a. 216 b. 192 c. 120 d. 72

You will have to match entries in Column I with the entries in Column II.

19. Match the statements given in Column I with the interval/union of intervals given in Column II Column I Column II 1. (−∞, − 1) ∪ (1, ∞)   2iz   : z is a (A)The set Re  2    1 − z  complex number, | z |= 1, z ≠ ±1 is (B) The domain of the function x −2

 8(3) 2( x −1)  1− 3

f (x) = sin −1 

17. A ship is fitted with three engines E1 , E2 and E3 . The

(C)If f (θ ) = − tan θ

X 1 , X 2 and X 3 denote respectively the events that the

(D)If

engines E1 , E2 And E3 are functioning. Which of the following is (are) true? 3 a. P[ X 1c | X ] = 16 b. P [Exactly two engines of the ship are functioning 7 [X ] = 8 5 c. P[ X | X 2 ] = 16

d. P[ X | X 1 ] =

7 16

triangle opposite to the angles at P, Q and R respectively.

2 sin P − sin 2 P Then equals 2 sin P + sin 2 P a.

3 4∆

Space for rough work

b.

−1

then

45 4∆

  is 

1 tan θ , 1

the

3. [2, ∞ )

set

π  f (θ ) : 0 ≤ θ <  is 2  f (x) = x 3 / 2 (3x − 10),

x ≥ 0, then increasing in

f(x)

4. (−∞, − 1] ∪ [1, ∞)

is

5.

 π interval  0,  2 

(B) Value (s) of k for which the plane kx + 4 y + z = 0, 4 x + ky + 2 z = 0 and 2 x + 2 y + z = 0 intersect

in

 3    4∆ 

c. 

 45    4∆ 

d. 

2

2. 2

a

straight line (C) Value (s) of k for which

3. 3

| x − 1 | + | x − 2 | + | x + 1 | x + 2 |= 4 k

has integer solution (s) (D) If y′y + 1 and y (0) = 1 then value 2

(−∞, 0] ∪ [2, ∞ )

20. Match the statements/expressions in Column I with the values given in Column II. Column I Column II (A) The number of solutions of the 1. 1 equation xesin x − cos x = 0 in the 

7 18. Let PQR be a triangle of area ∆ with a = 2, b = and 2 5 c = , where a, b and c are the lengths of the sides of the 2

tan θ 1 − tan θ

1

engines function independently of each other with 1 1 1 respective probabilities , and . For the ship to be 2 4 4 operational at least two of its engines must function. Let X denote the event that the ship is operational and let

2. (−∞, 0) ∪ (0, ∞)

4. 4

(s) of y (ln2)

5. 5

201

Mock Test-5

JEE ADVANCE PAPER-II

SECTION 1 Contains 8 Questions. The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).

1.

If the approximate value of log10 (4.04) 0 abcdef, It is

9.

 2x tan −1 y = tan −1 x + tan −1  2  1− x Then a value of y is

Let

a.

3 x − x3 1 − 3x 2

b.

3x + x 3 1 − 3x 2

c.

3 x − x3 1 + 3x 2

d.

3x + x 3 1 + 3x 2

given that log 4 4 = 0.6021 and log10 e = 0.4343, then the value of abcd must be

2.

 tan x − 1  If ∫ ( tan x + cot x )dx = a tan −1   + c, then the  b tan x  value of a 4 + b 5 must be

3.

The solution of the differential equation

d3y d2y −8 2 = 0 3 dx dx

1 satisfying y (0) = , y1 (0) = 0and y2 (0) = 1is 8

y1 =

4.

If

e8 x − 8 x + 7

λ m1 and

m2 are

the

roots

of

the

equation

7.

d. 2 : 3

11. Let g ( x) = integers,

3: 2

b.

( x − 1) n ; log cos m ( x − 1)

0 < x < 2,

m ≠ 0, n > 0, and let

p

m and n are

be the left hand

a. n = 1, m = 1

b. n = 1, m = −1

c. n = 2, m = 2

d. n > 2, m = n

2

2

x →1

Tangents are drawn form P(6, 8) to the circle x 2 + y 2 = r 2 ,

12. Let g ( x) = log( f ( x)) where f ( x ) is a twice differentiable

then the radius of the circle such that the area of the ∆ formed

positive function on (0, ∞) such that f ( x + 1) = x f ( x).

by tangents and chore of contact is maximum must be

Then, for N = 1, 2, 3, ...,

‘P’ is any arbitrary point on the circum circle of the equilateral triangle of side length 26 unit, then the value uuur uuur uuur of | PA |2 + | PB |2 + | PC |2 must be

 1 1  1 + ... + a. −4 1 + + 2 (2 N − 1)   9 25

The lines

x + 4 y + 6 z −1 and 3 x − 2 y + z + 5 = 0 = = 3 5 −2

= 2 x + 3 y + 4 z − k are coplanar for k is equal to 8.

3 :1

formed by the lines y = m1 x, y = m2 x2 , and y = c is

(a + b)c , then the value of 2008 (a + b ) must be

6.

a.

derivative of | x − 1| at x = 1. If lim+ g ( x) = p, then

2

1 . 3

ratio, AB : BC is

x + ( 3 + 2) x + ( 3 − 1) = 0 and if area of the triangle 2

5.

| x |
0, ∀x ∈ (0, π / 2)

(sin x + cos x ) dx (sin x cos x )

Put sin x − cos x = t

f (0) < 0 and f (π / 2) > 0 So, one solution.

(B) Let (a, b, c) is direction ratio of the intersected line, then

⇒

1 − sin 2 x = t 2

∴

(cos x + sin x) dx = dt

ak + 4b + c = 0

Then, I = ∫

4a + kb + 2c = 0 a b c = = 2 8 − k 4 − 2 k k − 16

 t = 2 sin −1 t = 2 tan −1  2 1− t   1− t    2 

2(8 − k ) + 2(4 − 2 k ) + ( k 2 − 16) = 0

k = 2, 4.

⇒

3.

k can take value 2, 3, 4, 5. −4x

4x

4 − 2x

−2

dy

(D)

∫ y + 1 = ∫ dx

⇒

f ( x ) = 2e x − 1

⇒

f (ln 2) = 3

2x − 4

6

−1

2

We, get a = 2, b = 2 Then, a 4 + b 5 = 4 + 32 = 36

(C) Let f ( x) =| x + 2 | + | x + 1| + | x − 1| + | x − 2 | ⇒

dt

 +c 

 sin x − cos x  −1  tan x − 1  = 2 tan −1   + c = 2 tan  +c sin 2 x    2 tan x 

We must have ⇒

f ′( x) =

1

⇒ Or

(64) We have,

d 3 y / dx 3 =8 d 2 y / dx 2

d2y = 8x + c dx 2 ln y2 = 8 x + c ln

Putting x = 0 we have,

2

c = log y2 (0) = ln1 = 0

∴

ln y2 = 8 x

⇒

y2 = e8 x i.e.,

y1 =

e8 x +D 8

Mathematics

210

1 Again putting x = 0, then y1 (0) = + D 8 1 ⇒ 0= +D 8 1 ∴ D=− 8

 r4  Then QR = 2 ⋅ QM = 2 {(OQ)2 − (OM ) 2 } = 2  r 2 −  100   P(6, 8)

Q

e8 x 1 e8 x x − ⇒ y= − +E ⇒ y1 = 8 8 64 8 1 1 Putting x = 0, we have y (0) = −0+ E = 64 8 1 1 7 ∴ E= − = 8 64 64 Hence, y =

4.

e8 x x 7 (e8 x − 8 x + 7) − + = ⇒ λ = 64 64 8 64 64

(5522) Since m1 and m2 are the roots of the equation x 2 + ( 3 + 2) x + ( 3 − 1) = 0

Then m1 + m2 = − ( 3 + 2), m1 m2 = ( 3 − 1)

∴

Area of ∆QPR =

∴

∆2 =

∴

dz 1 = {r 2 ⋅ 3(100 − r 2 ) 2 ⋅ (−2r ) + (100 − r 2 )3 ⋅ 2r} dr 1000

=

And

6.

11 33 ,b= 4 4 33 11 44 11 + = = a 2 + b2 = 16 16 16 4 11 = 502 ×11 = 5522 2008( a 2 + b 2 ) = 2008 × 4

or a =

(5) Equation of chord of contact (QR ) is 6 x + 8 y − r 2 = 0

OM =

6 ⋅ 6 + 8⋅8 − r2

0 + 0 − r2 (6 + 8 ) 2

2

=

dz = 0, then we get dr

=

r2 10

100 − r 2 10

and

d 2z dr 2

= −ve r =5

∴ Area of triangle is also maximum at r = 5.

33 11 ,b= 4 4

On comparing, a =

(62 + 82 )

2r (100 − r 2 )2 {100 − r 2 − 3r 2 } 1000

r = 5, ( r ≠ 10 as P is outside the circle)

 33 + 11  2 1 11( 3 + 1) = c2 =   c 2 ( 3 − 1)( 3 + 1)  4 

PM =

2

r 2 (100 − r 2 )3 = z (say) 1000

For maximum or minimum

1 c c ×c − ×c 2 m1 m2

1  1 1  1 2 m2 − m1 1 11 = c2  − = c2 = c 2  m1 m2  2 m1m2 2 ( 3 − 1)

5.

1 ⋅ QR ⋅ PM 2

 1 r 4  100 − r ∆ (say) = ⋅ 2  r 2 − ⋅ 2 100  10 

Hence, the required area of triangle

⇒

r

∴

and coordinates of the vertices of the given triangle are (0, 0) (c / m1 , c ) and (c / m2 , c)

∴

R O (0, 0)

m1 − m2 = (m1 + m2 )2 − 4m1m2 = (3 + 4 + 4 3 − 4 3 + 4) = 11

=

M r

∴

r r r r (1352) Let PV of P, A, B and C are p , a , b and c r respectively and O (0) be the circumcircle of the

equilateral triangle ABC. r r r a+b +c Then, =0 3 r r r r 26 and | p | = | a | = | b | = | c |= unit 3 uuur r r r r Now, | PA |2 = | a − p |2 = a 2 + p 2 − 2a ⋅ p uuur r r Similarly | PB |2 = b 2 + p 2 − 2b ⋅ p and uuur r r | PC |2 = c 2 + p 2 − 2c ⋅ p uuur r r r r Σ | PA |2 = a 2 + b 2 + c 2 + 3 p 2 − 2 p ( a + b + c ) = 6 p 2 − 0 [From equation (i) and (ii)]

=6

(26) 2 = 2(26) 2 = 1352 3

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

211

Mock Test-5

7.

(4) Any point on the first line in symmetrical form is (3r − 4, 5r − 6, − 2r + 1). If the lines are coplanar, this

11. (c) From graph, p = −1

point must lie on both the planes which determine the second line. . . .(i) ⇒ 3(3r − 4) − 2(5r − 6) − 2r + 1 + 5 = 0 and 2(3r − 4) + 3(5r − 6) + 4(−2r + 1) − k = 0

. . .(ii)

From equation (i), we get r = 2 Now substituting r = 2 equation (ii), then k = 4

8.

(16) Let the equation of circle x 2 + y 2 + 2 gx + 2 fy + k = 0 xy = c 2

⇒

⇒

 n ⋅ h n−1   h n−1  n lim  = −   lim    = −1, which holds if h →0 m ⋅ ( − tan h)  m  h→0  tan h   

n = m = 2.

t1 , t2 , t3 , t4 then t1t2t3t4 = 1

9.

2x   x + 1 − x2 (a) tan ( y ) = tan   1 − x.2 x2 1− x 

⇒

−1

1 4   1 g ′′  2 +  − g ′′ 1 +  = − .......... 2 9   2

    

1 1 4   g ′′  N +  − g ′′  N −  = − 2 2 (2 N − 1)2   Summing up all terms

 3x − x3  3x − x3 tan −1 ( y) = tan −1  y = ⇒ 2  1 − 3x 2  1 − 3x 

10. (a)

1 h = AQ 3

 1  1 1  1 . Hence, g ′′  N +  − g ′′   = −4 1 + + ... + 2  2 (2 N − 1)   2  9 13. (c) J − I = ∫

P

A

⇒

45° B

∴

60° C

AQ = 3h

Similarly, BQ = h

Q

e x (e 2 x − 1) ( Z 2 − 1) dx = ∫ 4 dz 4x 2x e + e +1 z + z2 +1

1  1 − 2  dz 1  e x + e − x − 1  z  where z = e x = ∫  = ln  x − x +c 2 2  e + e +1  1   z +  −1 z 

h

30°

1 x2

 1 1 g ′′ 1 +  − g ′′   = −4  2 2

This equation being fourth degree in t. Let the roots be

−1

g ( x + 1) − g ( x) = log x

⇒ g ′′( x + 1) − g ′′( x) = −

c 2t 4 + 2 gct 3 + kt 2 + 2 fct + c 2 = 0

16 t1t2t3t4 = 16 ×1 = 16

h→0

  hn lim   = −1 h →0 log cos m h  

⇒

t

∴

lim g ( x) = −1 ⇒ lim g (1 + h ) = −1

x →1+

12. (a) g ( x + 1) = log( f ( x + 1) = log x + log( f ( x)) = log x + g ( x)

2 Then c 2 t 2 + c2 + 2 gct + 2 fc + k = 0

t

1

⇒

. . .(ii)

c Put x = ct and y = in equation (i) t

⇒

0

. . .(i)

and the equation of the rectangular hyperbola is

x–1

–x+1

1  e2 x − e x + 1  J − I = ln  2 x x  + c. 2  e + e +1 π

14. (d) I = ∫ 1 + 4sin 2 0

π x x x − 4sin I = ∫ 1 − 2sin dx 2 2 2 0

π

h 3

⇒

CQ =

∴

3 AB AQ − BQ ( 3 − 1)h = = = 1 BC BQ − CQ  h  h −  3 

π

3 π x   I = ∫  1 − 2sin dx + ∫ −  1 − 2sin dx 2 2    π 0 3

π

π

x3  x  1 =  x + 4 cos  +  − x − 4 cos  2 0  2 π 

3

Mathematics

212

1=

18. (c) Plane is given by −( x + 1) − 7( y + 2) + 5( z + 1) = 0

 3   3 π  + 4  − 1 −  π −  − 4  0 −  3 3  2   2  

π

I=

π 3

+2 3 −4−

⇒

π 2π +2 3 = 4 3−4− 3 3

15. (a) Rearranging the equation we get, 1 dp (t ) = dt p (t ) − 400 2 Integrating (1) on both sides we get p ( t ) = 400 + k e t / 2 ,

⇒ distance =

2, − 2, 2i

. . .(i)

and − 2i respectively.

⇒

3x + y = 1 is 150°

Inclination of line L = 150° ± 60° = 210°, 90°

| z1 − 2 |2 + | z1 − 2 |2 + | z1 − 2i |2 + | z1 − 2i |2 | z1 |2 +2 3 = = 2 | z2 + 2 |2 + | z2 − 2 |2 + | z2 − 2i |2 + | z2 − 2i |2 | z1 | +2 4

20. (c) AG = 2 ∴

Slope of line L = tan 210° = tan 30° =

PA2 + PB 2 + PC 2 + PD 2 QA2 + QB 2 + QC 2 + QD 2 =

the relation is P (t ) = 400 − 300 e1/ 2

16. (b) Inclination of line

1 + 7 − 5 + 10 13 . = 75 75

19. (a) Let A, B, C and D be the complex numbers,

where k is a constant of integration. Using p (0) = 100, we get k = −300

∴

x + 7 y − 5z + 10 = 0

AT1 = T1G =

T1 is the vertex and BD is the directrix of parabola].

1

Also T2T3 is latus return

3

Equation of = Line L

⇒

⇒

1

y+2=

3

1 [as A is the foucs, 2

D

M

( x − 3)

i

j

T1

T2 A

3y − x + 3 + 2 3 = 0

C G B

k

17. (b) 3 1 2 = −iˆ − 7 ˆj + 5kˆ 1 2 3

−iˆ − 7 ˆj + 5kˆ Hence unit vector will be . 5 3

1 2

∴

T2T3 = 4 ×

∴

Area of ∆ T1 T2T3 = ×

1 2

1 4 × =1 2 2