VIT VITEEE Chapterwise Solutions 2007 to 2014 Engineering Entrance also useful for IITJEE IIT-JEE IIT JEE main advanced MTG

Table of contents :
Cover
......Page 1
Contents......Page 4
PHYSICS......Page 10
Chapter 1 : Electrostatics......Page 12
Chapter 2 :
Current Electricity......Page 22
Chapter 3 : Magnetic Effects of Electric Current and Magnetism......Page 34
Chapter 4 : Electromagnetic Induction and Alternating Current......Page 45
Chapter 5 : Optic
......Page 52
Chapter 6 : Electromagnetic Waves and
Wave Optics......Page 56
Chapter 7 :
Atomic and Nuclear Physics......Page 63
Chapter 8 : Dual Nature of Radiation
and Matter......Page 69
Chapter 9 : Semiconductor Devices and
their Applications......Page 76
CHEMISTRY......Page 82
Chapter 1 :
Atomic Structure......Page 84
Chapter 2 :
p, d and f - Block Elements......Page 87
Chapter 3 : Coordination Chemistry and Solid State Chemistry......Page 93
Chapter 4 :
Thermodynamics, Chemical Equilibrium and Chemical Kinetics......Page 99
Chapter 5 :
Electrochemistry......Page 108
Chapter 6 :
Isomerism in Organic Compounds......Page 113
Chapter 7 :
Alcohols and Ethers......Page 115
Chapter 8 :
Carbonyl Compounds......Page 121
Chapter 9 :
Carboxylic Acids and Their Derivatives......Page 128
Chapter 10 :
Organic Nitrogen Compounds......Page 132
Chapter 11 :
Biomolecules......Page 136
MATHEMATICS......Page 138
Chapter 1 :
Applications of Matrices and Determinants......Page 140
Chapter 2 :
Complex Numbers......Page 146
Chapter 3 :
Analytical Geometry of Two Dimensions......Page 151
Chapter 4 :
Vector Algebra......Page 165
Chapter 5 :
Analytical Geometry of Three Dimensions......Page 170
Chapter 6 :
Differential Calculus......Page 173
Chapter 7 :
Integral Calculus and its Applications......Page 182
Chapter 8 :
Differential Equations......Page 190
Chapter 9 :
Probability Distributions......Page 195
Chapter 10 :
Discrete Mathematics......Page 200
Chapter 11 :
Miscellaneous......Page 202
MODEL TEST PAPERS......Page 218
Model Test Paper 1......Page 220
Model Test Paper 2......Page 244
Model Test Paper 3
......Page 265
Model Test Paper 4
......Page 290
Model Test Paper 5
......Page 312

Citation preview

VITEEE

CHAPTERWISE SOLUTIONS PHYSICS l CHEMISTRY l MATHEMATICS

MTG Learning Media (P) Ltd. New Delhi | Gurgaon

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Contents

About VITEEE Physics 1. 2. 3. 4. 5. 6. 7. 8. 9.

Electrostatics Current Electricity Magnetic Effects of Electric Current and Magnetism Electromagnetic Induction and Alternating Current Optics Electromagnetic Waves and Wave Optics Atomic and Nuclear Physics Dual Nature of Radiation and Matter Semiconductor Devices and their Applications

Chemistry 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Atomic Structure p, d and f-Block Elements Coordination Chemistry and Solid State Chemistry Thermodynamics, Chemical Equilibrium and Chemical Kinetics Electrochemistry Isomerism in Organic Compounds Alcohols and Ethers Carbonyl Compounds Carboxylic Acids and Their Derivatives Organic Nitrogen Compounds Biomolecules

Mathematics 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Applications of Matrices and Determinants Complex Numbers Analytical Geometry of Two Dimensions Vector Algebra Analytical Geometry of Three Dimensions Differential Calculus Integral Calculus and Its Applications Differential Equations Probability Distributions Discrete Mathematics Miscellaneous

Model Test Papers (1-5)

(iv)-(viii) 1-70 .............. 1 ........... 11 ........... 23 ........... 34 ........... 41 ........... 45 ........... 52 ........... 58 ........... 65

1-54 .............. 1 .............. 4 ........... 10 ........... 16 ........... 25 ........... 30 ........... 32 ........... 38 ........... 45 ........... 49 ........... 53

1-78 .............. 1 .............. 7 ........... 12 ........... 26 ........... 31 ........... 34 ........... 43 ........... 51 ........... 56 ........... 61 ........... 63

1-118

iv

VITEEE CHAPTERWISE SOLUTIONS

About VITEEE*  VIT is emerging as a benchmark for continuous journey towards excellence, among the institutions of higher

education in the field of technology in the country. VIT has long since been devoted to providing quality education in various Engineering disciplines of Science and Technology and of late it has proliferated into frontiers of Research as well. The Institute has a rigorous selection procedure to admit students into various programmes. An exclusive entrance examination for all B.Tech. programmes, Vellore Institute of Technology Engineering Entrance Examination (VITEEE), is conducted for all those aspiring for a career in engineering.

Eligibility Nationality:



The applicant should be a Resident / Non Resident Indian National.

Qualifying Examination:

Candidates appearing for the VITEEE-2015 should have either completed or shall be appearing in 2015, in any one of the following qualifying examinations:  The final examination of the 10+2 system of Higher Secondary Examination conducted by the State Board, Central Board of Secondary Education (CBSE, New Delhi), The Council for Indian School Certificate Examination (ISCE, New Delhi).  Intermediate or Two-year Pre-University Examination conducted by a recognized Board/ University.  High School Certificate Examination of the Cambridge University or International Baccalaureate Diploma of the International Baccalaureate Office, Geneva.  General Certificate Education (GCE) examination (London/Cambridge/Srilanka) at the Advanced (A) level. Eligibility Criteria in the Qualifying Examination:

 Candidates appearing for the VITEEE in 2015 should have secured an aggregate of 60% in Physics, Chemistry,

Mathematics/Biology in the qualifying examination (+2/ Intermediate).  The average marks obtained in the subjects Physics, Chemistry and Mathematics or Biology (PCM / PCB) in +2 (or its equivalent) put together should be 50% for the following categories: Age Limit:

Candidates whose date of birth falls on or after 1st July 1993 are eligible to apply for VITEEE-2015. The date of birth as recorded in the High School / SSC / X Certificate will be considered authentic. Candidates should produce this certificate in original as a proof of their age at the time of counselling, failing which they will be disqualified. Admission Procedure  The admission will be purely on the basis of the marks secured in the VITEEE-2015, conducted by VIT University.  Candidates who have not appeared for the VITEEE are not eligible for admission.  The candidates will be short-listed based on their entrance examination marks and will be called for the counselling. *For latest information, please visit www.vit.ac.in

v

VITEEE CHAPTERWISE SOLUTIONS

important date



Issue of Application form Exam Month

: November-December : April



pattern of entrance examination question paper Language of Question Paper- English Type of Questions – Objective Parts : I - Physics, II - Chemistry, III - Mathematics/Biology* * Candidates attempting Physics, Chemistry & Biology are eligible for B.Tech. Bio-Medical Engineering, Biotechnology and Computer Science and Engg. (Spec. in Bioinformatics) programmes only. distribution of marks  All the questions will be mostly from the State Board of Higher Secondary Education and the CBSE syllabus only.  Each part has 40 questions and each question carries one mark.  Each question is followed by 4 alternative answers. The candidate will have to choose the correct answer and click the same.  No negative marks for wrong answers. result Rank List : A merit list will be prepared using Equating Methodology for VITEEE – 2015. Equating is a statistical process that is used to adjust scores on test forms (that are approximately equivalent) so that scores on the forms can be used interchangeably. Thus, equating is used to adjust scores of candidates who have taken different forms of a test, in order to facilitate meaningful and fair comparison for merit list and ranking of these candidates. The statistical procedure of Equipercentile Equating would be used by VIT to identify the percentile ranks for candidates, such that scores on different forms with the same percentile rank are considered to be equivalent. Thus, Percentile Rank is a unique and invariant position of the test taker in that group. A percentile Rank, say 90 Percentile Rank denotes that there are 90 percent of test takers scores below this score level and 10 percent above this test score. Candidates are advised not to make assumptions and predictions of their score or rank based on their own estimates of raw scores.

JJJ

vi

VITEEE CHAPTERWISE SOLUTIONS

SALLABUS

syllabus PART – I - PHYSICS 1.

Electrostatics : Charges and their conservation; Coulomb’s law-forces between two point electric charges - Forces between multiple electric charges superposition principle. Electric field – electric field due to a point charge, electric field lines; electric dipole, electric field intensity due to a dipole - behaviour of a dipole in a uniform electric field. Electric potential - potential difference-electric potential due to a point charge and dipole-equipotential surfaces – electrical potential energy of a system of two point charges. Electric flux-Gauss’s theorem and its applications to find field due to (i) infinitely long straight wire (ii) uniformly charged infinite plane sheet (iii) two parallel sheets and (iv) uniformly charged thin spherical shell (inside and outside) Electrostatic induction-capacitor and capacitance – dielectric and electric polarisation – parallel plate capacitor with and without dielectric medium – applications of capacitor – energy stored in a capacitor - Capacitors in series and in parallel – action of points – Lightning arrester – Van de Graaff generator.

2.

Current Electricity : Electric Current – flow of charges in a metallic conductor – drift velocity and mobility and their relation with electric current. Ohm’s law, electrical resistance - V-I characteristics – electrical resistivity and conductivity-classification of materials in terms of conductivity – Carbon resistors – colour code for carbon resistors - combination of resistors – series and parallel – temperature dependence of resistance – internal resistance of a cell – potential difference and emf of a cell - combinations of cells in series and in parallel. Kirchhoff ’s law – illustration by simple circuits – Wheatstone’s Bridge and its application for temperature coefficient of resistance measurement - Metrebridge - special case of Wheatstone bridge - Potentiometer principle - comparing the emf of two cells.

3.

Magnetic Effects of Electric Current and Magnetism : Magnetic effect of electric current – Concept of magnetic field - Oersted’s experiment – Biot-Savart law-Magnetic field due to an infinitely long current carrying straight wire and circular coil – Tangent galvanometer – construction and working – Bar magnet as an equivalent solenoid – magnetic field lines. Ampere’s circuital law and its application. Force on a moving charge in uniform magnetic field and electric field – cyclotron – Force on current carrying conductor in a uniform magnetic field – Forces between two parallel current carrying conductors - definition of ampere. Torque experienced by a current loop in a uniform magnetic field - moving coil galvanometer – conversion to ammeter and voltmeter – current loop as a magnetic dipole and its magnetic dipole moment - Magnetic dipole moment of a revolving electron. Para, dia and ferro-magnetic substances with examples – Electromagnets – Permanent Magnets

4.

Electromagnetic Induction and Alternating Current : Electromagnetic induction - Faraday’s law - induced emf and current - Lenz’s law. Self induction - Mutual induction - self inductance of a long solenoid - mutual inductance of two long solenoids. Methods of inducing emf - (i) by changing magnetic induction (ii) by changing area enclosed by the coil and (iii) by changing the orientation of the coil (quantitative treatment). AC generator - commercial generator. (Single phase, three phase). Eddy current - applications - transformer - long distance transmission. Alternating current - measurement of AC-AC circuit with resistance - AC circuit with inductor - AC circuit with capacitor - LCR series circuit - Resonance and Q - factor - power in AC circuits.

5.

Optics : Reflection of light, spherical mirrors, mirror formula. Refraction of light, total internal reflection and its applications, optical fibers, refraction at spherical surfaces, lenses, thin lens formula, lens maker’s formula. Magnification, power of a lens, combination of thin lenses in contact, combination of a lens and a mirror. Refraction and dispersion of light through a prism. Scattering of light-blue colour of sky and reddish appearances of the sun at sunrise and sunset.

6.

Electromagnetic Waves and Wave Optics : Electromagnetic waves and their characteristics - Electromagnetic spectrum-radio, microwaves,

infra-red, visible,ultra-violet, X rays, gamma rays.

Wavefront and Huygens’s principle - Reflection, total internal reflection and refraction of plane wave at a plane surface using wavefronts. Interference - Young’s double slit experiment and expression for fringe width - coherent source - interference of light-Formation of colours in thin films - analytical treatment - Newton’s rings. Diffraction - differences between interference and diffraction of light- diffraction grating. Polarisation of light waves - polarisation by reflection - Brewster’s law - double refraction - nicol prism - uses of plane polarised light and Polaroids - rotatory polarisation - polarimeter.

7.

Atomic and Nuclear Physics : Atomic structure – discovery of the electron – specific charge (Thomson’s method) and charge of the electron (Millikan’s oil drop method) – alpha scattering – Rutherford’s atom model. Bohr’s model – energy quantization – energy and wave number expressions – Hydrogen spectrum – energy level diagrams – sodium and mercury spectra – excitation and ionization potentials.

VITEEE CHAPTERWISE SOLUTIONS

vii

Nuclear properties - nuclear radii, masses, binding energy, density, charge - isotopes, isobars and isotones - nuclear mass defect - binding energy - stability of nuclei - Bainbridge mass spectrometer. Nature of nuclear forces - Neutron - discovery - properties - artificial transmutation - particle accelerator. Radioactivity - alpha, beta and gamma radiations and their properties-α -decay, β -decay and γ -decay - Radioactive decay law - half life - mean life - artificial radioactivity - radio isotopes - effects and uses - Geiger - Muller counter. Radio carbon dating - biological radiation hazards. Nuclear fission - chain reaction - atom bomb - nuclear reactor - nuclear fusion - Hydrogen bomb - cosmic rays - elementary particles.

8.

Dual Nature of Radiation and Matter : Photoelectric effect - Light waves and photons - Einstein’s photoelectric equation - laws of photoelectric emission - particle nature of light - photo cells and their applications. Matter waves - wave nature of particles – de Broglie relation – de Broglie wavelength of an electron –Davisson Germer experiment – electron microscope

9.

Semiconductor Devices and their Applications : Semiconductor theory - energy band in solids (Qualitative ideas only) - difference between metals, insulators and semiconductors based on band theory - semiconductor doping - Intrinsic and Extrinsic semiconductors. Formation of P-N Junction - Barrier potential and depletion layer- P-N Junction diode - Forward and reverse bias characteristics - diode as a rectifier - Zener diode-Zener diode as a voltage regulator - LED - seven segment display - LCD. Junction transistors - characteristics - transistor as a switch - transistor as an amplifier - transistor as an oscillator . Logic gates - NOT, OR, AND, EXOR using discrete components - NAND and NOR gates as universal gates -Laws and theorems of Boolean algebra

PART – II - CHEMISTRY 1.

Atomic Structure : Bohr’s atomic model-Sommerfeld’s extension of atomic structure; Electronic configuration and Quantum numbers; Shapes of s,p,d,f-orbitals - Pauli’s exclusion principle - Hund’s Rule of maximum multiplicity- Aufbau principle. Emission spectrum, absorption spectrum, line spectra and band spectra; Hydrogen spectrum – Lyman, Balmer, Paschen, Brakett and Pfund series; de Broglie’s theory; Heisenberg’s uncertainty principle – wave nature of electron – Schrodinger wave equation (No derivation). Eigen values and eigen functions. Hybridization of atomic orbitals involving s,p,d-orbitals.

2.

p,d and f – Block Elements : p-block elements – Phosphorous compounds; PCl3, PCl5 – Oxides. Hydrogen halides, Interhalogen compounds. Xenon fluoride compounds. General Characteristics of d – block elements – Electronic Configuration – Oxidation states of first row transition elements and their colours. Occurrence and principles of extraction: Copper, Silver, Gold and Zinc. Preparation, properties of CuSO4, AgNO3 and K2Cr2O7.

Lanthanides – Introduction, electronic configuration, general characteristics, oxidation state – lanthanide contraction, uses, brief comparison of Lanthanides and Actinides.

3.

Coordination Chemistry and Solid State Chemistry : Introduction – Terminology in coordination chemistry – IUPAC nomenclature of mononuclear coordination compounds. Isomerism, Geometrical isomerism in 4-coordinate, 6-coordinate complexes. Theories on coordination compounds – Werner’s theory (brief), Valence Bond theory. Uses of coordination compounds. Bioinorganic compounds (Haemoglobin and chlorophyll). Lattice – unit cell, systems, types of crystals, packing in solids; Ionic crystals – Imperfections in solids – point defects. X-Ray diffraction – Electrical Property, Amorphous solids (elementary ideas only).

4.

Thermodynamics, Chemical Equilibrium and Chemical Kinetics : I and II law of thermodynamics – spontaneous and non spontaneous processes, entropy, Gibb’s free energy – Free energy change and chemical equilibrium – significance of entropy. Law of mass action – Le Chatlier’s principle, applications of chemical equilibrium. Rate expression, order and molecularity of reactions, zero order, first order and pseudo first order reaction – half life period. Determination of rate constant and order of reaction . Temperature dependence of rate constant, Arrhenius equation, activation energy.

5.

Electrochemistry : Theory of electrical conductance; metallic and electrolytic conductance. Faraday’s laws – theory of strong electrolytes – Specific resistance, specific conductance, equivalent and molar conductance – Variation of conductance with dilution – Kohlrausch’s Law – Ionic product of water, pH and pOH – buffer solutions – use of pH values. Cells – Electrodes and electrode potentials – construction of cell and EMF values, Fuel cells, Corrosion and its prevention.

6.

Isomerism in Organic Compounds : Definition, Classification – structural isomerism, stereo isomerism – geometrical and optical isomerism. Optical activity- chirality – compounds containing chiral centres, R – S notation, D – L notation.

7.

Alcohols and Ethers : Nomenclature of alcohols – Classification of alcohols - distinction between 1°, 2° and 3° alcohols – General methods of

preparation of primary alcohols, properties. Methods of preparation of dihydric alcohols: Glycol – Properties, Uses. Methods of preparation of trihydric alcohols, Properties, Uses. Aromatic alcohols – preparation and properties of phenols and benzyl alcohol. Ethers – Nomenclature of ethers, general methods of preparation of aliphatic ethers, Properties, Uses. Aromatic ethers – Preparation of Anisole, Uses.

8.

Carbonyl Compounds : Nomenclature of carbonyl compounds – Comparison of aldehydes and ketones. General methods of preparation of aldehydes, Properties, Uses. Aromatic aldehydes – Preparation of benzaldehyde, Properties and Uses. Ketones – general methods of preparation of aliphatic ketones (acetone), Properties, Uses. Aromatic ketones – preparation of acetophenone , Properties, Uses, preparation of benzophenone, Properties. Name reactions; Clemmenson reduction, Wolff – Kishner reduction, Cannizzaro reaction, Claisen Schmidt reaction, Benzoin Condensation, aldol Condensation. Preparation and applications of Grignard reagents.

9.

Carboxylic Acids and Their Derivatives : Nomenclature – Preparation of aliphatic monobarboxylic acids – formic acid, Properties, Uses. Monohydroxy mono carboxylic acids; Lactic acid – Synthesis of lactic acid. Aliphatic dicarboxylic acids; Preparation of oxalic and succinic acid. Aromatic

viii

VITEEE CHAPTERWISE SOLUTIONS acids; Benzoic and Salicylic acid – Properties, Uses. Derivatives of carboxylic acids; acetyl chloride (CH3COCl) – Preparation, Properties, Uses. Preparation of acetamide, Properties, acetic anhydride – Preparation, Properties. Preparation of esters, methyl acetate – Properties.

10. Organic Nitrogen Compounds : Aliphatic nitro compounds – Preparation of aliphatic nitroalkanes, Properties, Uses. Aromatic nitro compounds –

Preparation, Properties, Uses. Distinction between aliphatic and aromatic nitro compounds. Amines; aliphatic amines – General methods of preparation, Properties, Distinction between 1°, 2° and 3° amines. Aromatic amines – Synthesis of benzylamine, Properties, Aniline – Preparation, Properties, Uses. Distinction between aliphatic and aromatic amine. Aliphatic nitriles – Preparation, Properties, Uses. Diazonium salts – Preparation of benzene diazoniumchloride, Properties.

11. Biomolecules : Carbohydrates – Distinction between sugars and non sugars, structural formulae of glucose, fructose and sucrose, with their linkages, invert sugar – definition, examples of oligo and polysaccharides. Amino acids – Classification with examples, Peptides-properties of peptide bond. Lipids - Definition, classification with examples, difference between fats, oils and waxes.

PART – III - MATHEMATICS 1.

Applications of Matrices and Determinants : Adjoint, Inverse – properties, Computation of inverses, Solution of system of linear equations by matrix inversion method. Rank of a matrix – elementary transformation on a matrix, consistency of a system of linear equations, Cramer’s rule, non-homogeneous equations, homogeneous linear system and rank method.

2.

Complex Numbers : Complex number system - conjugate, properties, ordered pair representation. Modulus – properties, geometrical representation, polar form, principal value, conjugate, sum, difference, product, quotient, vector interpretation, solutions of polynomial equations, De Moivre’s theorem and its applications. Roots of a complex number - nth roots, cube roots, fourth roots.

3.

Analytical Geometry of two dimensions : Definition of a conic – general equation of a conic, classification with respect to the general equation of a conic, classification of conics with respect to eccentricity. Equations of conic sections (parabola, ellipse and hyperbola) in standard forms and general forms- Directrix, Focus and Latus rectum, parametric form of conics and chords, Tangents and normals, Cartesian form and parametric form- equation of chord of contact of tangents from a point (x1 ,y1 ) to all the above said curves. Asymptotes, Rectangular hyperbola – Standard equation of a rectangular hyperbola.

4.

Vector Algebra : Scalar Product – angle between two vectors, properties of scalar product, applications of dot products. vector product, right handed and left handed systems, properties of vector product, applications of cross product. Product of three vectors – Scalar triple product, properties of scalar triple product, vector triple product, vector product of four vectors, scalar product of four vectors.

5.

Analytical Geometry of Three Dimensions : Direction cosines, direction ratios, equation of a straight line passing through a given point and parallel to a given line, passing through two given points, angle between two lines. Planes – equation of a plane, passing through a given point and perpendicular to a line, given the distance from the origin and unit normal, passing through a given point and parallel to two given lines, passing through two given points and parallel to a given line, passing through three given non-collinear points, passing through the line of intersection of two given planes, the distance between a point and a plane, the plane which contains two given lines (coplanar lines), angle between a line and a plane. Skew lines - shortest distance between two lines, condition for two lines to intersect, point of intersection, collinearity of three points. Sphere – equation of the sphere whose centre and radius are given, equation of a sphere when the extremities of the diameter are given.

6.

Differential Calculus : Derivative as a rate measure - rate of change, velocity, acceleration, related rates, derivative as a measure of slope, tangent, normal and angle between curves, maxima and minima. Mean value theorem - Rolle’s Theorem, Lagrange Mean Value Theorem, Taylor’s and Maclaurin’s series, L’ Hospital’s Rule, stationary points, increasing, decreasing, maxima, minima, concavity, convexity and points of inflexion. Errors and approximations – absolute, relative, percentage errors - curve tracing, partial derivatives, Euler’s theorem.

7.

Integral Calculus and its Applications : Simple definite integrals – fundamental theorems of calculus, properties of definite integrals. Reduction formulae – reduction formulae for ∫ sinn x dx and ∫ cosn x dx , Bernoulli’s formula. Area of bounded regions, length of the curve.

8.

Differential Equations : Differential equations - formation of differential equations, order and degree, solving differential equations (1st order), variables separable, homogeneous and linear equations. Second order linear differential equations - second order linear differential equations with constant co-efficients, finding the particular integral if f (x) = emx, sin mx, cos mx, x, x2.

9.

Probability Distributions : Probability – Axioms, Addition law, Conditional probability, Multiplicative law, Baye’s Theorem Random variable - probability density function, distribution function, mathematical expectation, variance Theoretical distributions - discrete distributions, Binomial, Poisson distributions, Continuous distributions, Normal distribution.

10. Discrete Mathematics : Mathematical logic – logical statements, connectives, truth tables, logical equivalence, tautology, contradiction. Groups-binary operations, semigroups, monoids, groups, order of a group, order of an element, properties of groups.

PHYSICS

1

Electrostatics

1

CHAPTER

1.

2.

If 1000 droplets each of potential 1 V and radius r are combined to form a big drop. Then, the potential of the drop as compared to small droplets will be (a) 1000 V (b) 800 V (c) 100 V (d) 20 V (2014) Four equal charges q each are placed at four corners of a square of side a each. Work done in carrying a charge –q from its centre to infinity is 2q (a) zero (b) pe 0a (c)

3.

Electrostatics

q2 2pe0 a

2q 2 (d) pe0 a

6.

7.

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

Air

d

C 4 C (c) ( K + 1) 4 (a) ( K + 2)

4.

(2013)

5.

(b) ( K + 3)

(d) none of these (2013)

The potential of the electric field produced by point charge at any point (x, y, z) is given by V = 3x2 + 5, where x, y are in metres and V is in volts. The intensity of the electric field at (–2, 1, 0) is (a) +17 V m–1 (b) –17 V m–1 –1 (c) +12 V m (d) –12 V m–1 (2012)

A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown in figure. Let VA, VB, VC be the potentials at points A, B and C respectively. Then

A

C B

(a) VC > VB (c) VA > VB

d

C 4

The potential of a large liquid drop when eight liquid drops are combined is 20 V. Then the potential of each single drop was (a) 10 V (b) 7.5 V (c) 5 V (d) 2.5 V (2012) Two capacitors of capacities 1 mF and C mF are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80 mC, the energy stored in the capacitor of capacity C in mJ is (a) 1800 (b) 1600 (c) 14400 (d) 7200 (2012)

8.

P

(b) VB > VC (d) VA = VC

(2012)

Three charges, each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC, 2a. D and E are the midpoints of BC and CA. The work done in taking a charge Q from D to E is A E

3qQ (a) 8 pe0 a

B

C

D

(b)

qQ 4 pe0 a

2

VITEEE CHAPTERWISE SOLUTIONS

(c) zero 9.

(d)

3qQ  4 pe0 a

(2012)

A parallel plate capacitor has capacitance C. If it is equally filled with parallel layers of materials of dielectric constant K1 and K2 its capacity becomes C1. The ratio of C1 and C is K1K 2 (a) K1 + K2 (b) K1 + K 2 2 K1K 2 K1 + K 2 (c) (d) K1 + K 2 K1K 2 (2012)

10. A glass rod rubbed with silk is used to charge a gold leaf electroscope and the leaves are observed to diverge. The electroscope thus charged is exposed to X-rays for a short period. Then (a) the divergence of leave will not affected (b) the leaves will diverge further (c) the leaves will collapse (d) the leaves will melt (2011) 11. An infinite number of charge, each of charge 1 mC are placed on the x-axis with coordinates x = 1, 2, 4, 8,......∞. If a charge of 1 C is kept at the origin, then what is the net force acting on 1 C charge? (a) 9000 N (b) 12000 N (c) 24000 N (d) 36000 N (2011) 12. A cube of side l is placed in a uniform field E, where E = Ei. The net electric flux through the cube is (a) zero (b) l2E 2 (c) 4l E (d) 6l2E (2011) 13. The capacity of a capacitor is 4 × 10–6 F and its potential is 100 V. The energy released on discharging it fully will be (a) 0.02 J (b) 0.04 J (c) 0.025 J (d) 0.05 J (2011) 14. A thin metallic spherical shell contains a charge Q on it. A point charge q is placed at the centre of the shell and another charge q1 is placed outside it as shown in the figure. All the three charges are positive. Q

q

q1



The force on the charge at the centre is (a) towards left (b) towards right (c) upward (d) zero (2011)

15. As shown in the figure, charges +q and –q are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is A

a B +q

b

b

C –q

1 2a (b) zero ⋅ 4 pe0 a 2 + b 2 q ( −q) 1 1 ⋅ ⋅ (c) (c) 4 pe0 a 2 + b 2 4 pe0 a 2 + b 2 (2011) (a)

16. On moving a charge of 20 C by 2 cm, 2 J of work is done, then the potential difference between the points is (a) 0.1 V (b) 8 V (c) 2 V (d) 0.5 V (2011) 17. The insulation property of air breaks down at 3 × 106 V m–1. The maximum charge that can be given to a sphere of diameter 5 m is nearly (a) 2 × 10–2 C (b) 2 × 10–3 C –4 (c) 2 × 10 C (d) 2 × 10–5 C (2011) 18. Two capacitors of capacities C and 2C are connected in parallel and then connected in series with a third capacitor of capacity 3C. The combination is charged with V volt. The charge on capacitor of capacity C is (a)

1 CV 2

(c) 2CV

(b) CV (d)

3 CV 2

(2011)

19. An electric dipole consists of two opposite charges of magnitude q = 1 × 10–6 C separated by 2.0 cm. The dipole is placed in an external field of 1 × 105 N C–1. What maximum torque does the field exert on the dipole? How much work must an external agent do to turn the dipole end to end, starting from position of alignment (q = 0°)? (a) 4.4 × 106 N-m, 3.2 × 10–4 J (b) –2 × 10–3 N-m, –4 × 103 J

3

Electrostatics

(c) 4 × 103 N-m, 2 × 10–3 J (d) 2 × 10–3 N-m, 4 × 10–3 J

(2010)

20. An arc of radius r carries charge. The linear density of charge is l and the arc subtends p an angle at the centre. What is electric 3 potential at the centre? l l (a) (b) 4e 0 8e 0 (c)

l 12e0

(d)

l 16e0

(2010)

21. A neutral water molecule (H2O) in its vapour 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 charges? (a) 4 m (b) 4 mm (c) 4 µm (d) 4 pm (2010) 22. An infinitely long thin straight wire has 1 C m −1 . uniform linear charge density of 3 Then, the magnitude of the electric intensity at a point 18 cm away is (given e0 = 8.8 × 10–12 C2 N m–2) (a) 0.33 × 1011 N C–1 (b) 3 × 1011 N C–1 (c) 0.66 × 1011 N C–1 (d) 1.32 × 1011 N C–1 (2009) 23. Two points charges –q and +q are located at points (0, 0, –a) and (0, 0, a) respectively. The electric potential at a point (0, 0, z), where z > a is qa q (a) (b) 2 pe 4 4 pe0 z 0a (c)

2qa 2

2

4 pe0 ( z − a )



(d)

2qa

4 pe0 ( z 2 + a 2 )

(2009)

24. The energy stored in the capacitor as shown in the figure (a) is 4.5 × 10–6 J. If the battery is replaced by another capacitor of 900 pF as shown in figure (b), then the total energy of system is

(a) 4.5 × 10–6 J (c) zero

(b) 2.25 × 10–6 J (d) 9 × 10–6 J. (2008)

25. In nature, the electric charge of any system is always equal to (a) half integral multiple of the least amount of charge (b) zero (c) square of the least amount of charge (d) integral multiple of the least amount of charge. (2008) 26. If the force exerted by an electric dipole on a charge q at a distance of 1 m is F, the force at a point 2 m away in the same direction will be (a) F/2 (b) F/4 (c) F/6 (d) F/8 (2008) 27. A solid spherical conductor of radius R has a spherical cavity of radius a (a < R) at its centre. A charge +Q is kept at the center. The charge at the inner surface, outer surface and at a position r(a < r < R) are respectively (a) +Q, –Q, 0 (b) –Q, +Q, 0 (c) 0, –Q, 0 (d) +Q, 0, 0. (2008) 28. A cylindrical capacitor has charge Q and length L. If both the charge and length of the capacitors are doubled, by keeping other parameters fixed, the energy stored in the capacitor (a) remains same (b) increases two times (c) decreases two times (d) increases four times. (2008) 29. A sample of HCl gas is placed in an electric field of 3 × 104 N C–1. The dipole moment of each HCl molecule is 6 × 10–30 C m. The maximum torque that can act on a molecule is (a) 2 × 10–34 C2 N–1 m (b) 2 × 10–34 N m (c) 18 × 10–26 N m (d) 0.5 × 1034 C–2 N m–1. (2008) 30. Two point like charges Q1 and Q2 of whose strengths are equal in absolute value are placed at a certain distance from each other. Assuming the field strength to be positive in the positive direction of x-axis, the signs of the charges Q1 and Q2 for the graphs (field strength versus distance) shown in figure (1), (2), (3) and (4) are:

4

VITEEE CHAPTERWISE SOLUTIONS E QE1 Q1



32. A parallel plate capacitor of capacitance 100 pF is to be constructed by using paper sheets of 1 mm thickness as dielectric. If the dielectric constant of paper is 4, the number of circular metal foils of diameter 2 cm each required for the purpose is (a) 40 (b) 20 (c) 30 (d) 10 (2007)  33. The electric field intensity E, due to an  electric dipole of moment p, at a point on the equatorial line is (a) parallel to the axis of the dipole and opposite to the direction of the dipole moment p (b) perpendicular to the axis of the dipole and is directed away from it (c) parallel to the dipole moment (d) perpendicular to the axis of the dipole and is directed towards it (2007)

E QE1

x Q2 x (1) Q2

Q1

(3)

x Q2 x (2) Q2 (4)

(a) Q1 positive, Q2 negative; both positive; Q1 negative, Q2 positive; both negative (b) Q1 negative Q2 positive; Q1 positive, Q2 negative; both positive; both negative (c) Q1 positive, Q2 negative; both negative; Q1 negative, Q2 positive; both negative (d) both positive; Q1 positive, Q2 negative; Q1 negative, Q2 positive; both negative (2007)

31. ABCD is a rectangle. At corners B, C and D of the rectangle are placed charges +10 × 10–12 C, –20 × 10–12 C and 10 × 10–12 C, respectively. Calculate the potential at the fourth corner. (The side AB = 4 cm and BC = 3 cm) (a) 1.65 V (b) 0.165 V (c) 16.5 V (d) 2.65 V (2007)

Answer Key 1.

(c)

2.

9.

(d)

10. (b)

11. (b)

17. (b)

18. (a)

25. (d)

26. (d)

33. (a)

(d)

3.

(b)

4.

(c)

5.

(c)

6.

(b)

7.

(d)

8.

(c)

12. (a)

13.

(a)

14.

(d)

15.

(b)

16.

(a)

19. (d)

20. (c)

21.

(d)

22.

(a)

23.

(c)

24.

(b)

27. (b)

28. (b)

29.

(c)

30.

(a)

31.

(a)

32.

(d)

5

Electrostatics

e planations

(c) : Let q be charge on each droplet. The potential of each droplet is 1 q V= ...(i) 4 pe0 r If R be radius of the drop formed, then Volume of the drop = Volume of 1000 droplets 1.

4 3 4  pR = 1000  pr 3  3  3 R = 10r The charge on the drop is Q = 1000q The potential of the drop is V′ =



 1 q = 100   = 100V  4 pe0 r 

q

q

O D







(using (i))

q

a

B





a qC

In figure, AC = BD = a 2 + a 2 = a 2 a 2 a = and OA = OB = OC = OD = 2 2 The potential at the centre O due to given charge configuration is q q q  1  q + + + 4 pe0  OA OB OC OD  q q q  1  q = + + +   4 pe0  ( a / 2 ) ( a / 2 ) ( a / 2 ) ( a / 2 ) 

VO =

=

3.



(d) : The situation is shown in figure. A





2q 1  4q   = 4 pe0  ( a / 2 )  pe0 a

Work done in carrying a charge –q from centre O to infinity is WO∞ = – q(V∞ – VO)  2q  = −q  0 − (Q V∞ = 0)  pe0 a  



2q 2 pe0 a

(b) : If A be area of each plate and d is the distance between the plates, then e A C= 0 d When (1/4)th of capacitor is filled with the slab of dielectric constant K as shown in figure, then the arrangement is equivalent to two capacitors in parallel with capacitances

1 Q 1 (1000q) = 4 pe0 R 4 pe0 10r

As V = 1 V (given) \ V′ = 100V = 100(1 V) = 100 V 2.

=

(A/4)

(3A/4)

K

Air

d

K e 0 ( A / 4) K e 0 A = 4d d e0 ( 3 A / 4) 3e0 A and C2 = = 4d d C1 =

As C1 and C2 are in parellel \ The new capacitance C′ is K e A 3e A C ′ = C1 + C2 = 0 + 0 4d 4d e0 A C = ( K + 3) = ( K + 3) 4d 4

 (c) : The intensity of electric field E and potential V are related by the relation   ∂V ^ ∂V ^ ∂V ^  E = − i+ j+ k ∂y ∂z   ∂x Here, V = 3x2 + 5 ∂V \ = 6x, ∂x ∂V ∂V = 0, =0 ∂z ∂y  \ E = −[6 x ^i + 0 + 0] = −6 x ^i At (–2, 1, 0),  ^ ^ E = −6( −2) i = +12 i V m −1 4.



Thus the intensity of electric field at (–2, 1, 0) is + 12 V m–1.

5.

(c) : Let R be radius of the big drop and r be radius of each small drop.

6

VITEEE CHAPTERWISE SOLUTIONS

As volume of the big drop = volume of eight small drops 4 3 4 3 \ pR = 8  pr  3  3 R = 2r …(i) If q be charge on each small drop, then charge on the big drop is Q = 8q …(ii) The potential of each small drop is 1 q V= …(iii) 4 pe0 r and that of the big drop is 1 Q V′ = …(iv) 4 pe0 R

Dividing eqn. (iii) by eqn. (iv), we get 1 q 4 pe q R V 0 r = = 1 Q Qr V′ 4 pe0 R

2  2   mF  (1 mF) +  mF  C = (1 mF)C 3 3 2  2   C 1 mF − mF =  mF  (1 mF) 3  3  

1   2  C  mF =  mF  (1 mF) 3   3 



2   mF  (1 mF) C= 3 = 2 mF 1  m F   3 





The energy stored in the capacitor of capacity C is 1 Q 2 1 (80 × 10 −6 C)2 U= = 2 C 2 ( 2 × 10 −6 F) 7.

= 1600 × 10–6 J = 1600 mJ



(d) : In the given situation, the hollow conducting sphere becomes an equipotential surface. Therefore, the potential at every point on it will be same. Hence, VA = VB = VC

8.

(c) :



A

q 2a

q 2r 1 = = (using eqns. (i) and (ii)) 8q r 4 1 V = V′ 4 As V′ = 20 V (given) 1 \ V = (20 V) = 5 V 4 6. (b) : The situation is shown in figure. 1 F

 (1 mF)C  2 mF =   3  1 mF + C 



E

C F

q B

D

C

q

2a



120 V

As 1 mF and C are connected in series, so their equivalent capacitance is



Ceq =

(1 mF)C (1 mF + C )



The potential difference across combination is V = 120 V The charge on the combination is

Q = CeqV



 (1 mF)C  80 mC =   120 V  1 mF + C 



80 mC  (1 mF)C  = 120 V  1 mF + C 

the

Here, AC = BC = 2a D and E are the midpoints of BC and AC. \ AE = CE = a and BD = CD = a In DADC, (AD)2 = (AC)2 – (CD)2 = (2a)2 – (a)2 = 4a2 – a2 = 3a2 AD = a 3



Similarly, in DBEC, BE = a 3 The potential at D due to the given charge configuration is q q  1  q VD = + +  4 pe0  BD CD AD 







=

q  1 q q + + 4 pe0  a a a 3 

=

q  1  2+  4 pe0 a  3 

…(i)

7

Electrostatics



The potential at E due to the given charge configuration is q q  1  q + + 4 pe0  AE CE BE 



=

q  1 q q + +  4 pe0  a a a 3 



=

q  1  2+  4 pe0 a  3 

VE =



…(ii)

9.

(d) : The capacitance of a parallel plate capacitor is e A C = 0 d where A is the area of each plate and d is the separation between the plates. When it is equally filled with layers of materials of dielectric constant K1 and K2 as shown in figure, then the arrangement is equivalent to two capacitors each of area A and separation d/2 connected in series. A



\

K1

d/2

K2



1  1 × 10 −6 1 × 1 × 10 −6 1 × 1 × 10 −6 + + +  4 pe0  (1)2 ( 2)2 ( 4)2 1 × 1 × 10 −6 ( 8)

=

From eqns. (i) and (ii), it is clear that VD = VE \ The work done in taking a charge Q from D to E is W = Q(VE – VD) = 0 (Q VD = VE)

d/2

F=

−6

 + ......... ∞  

10  1 1  1 1 + + + + ...... ∞   4 pe0  4 16 64  

9

= 9 × 10 × 10

= 9×

 1   1 1−   4

−6 

4 × 10 3 = 12000 N 3

12. (a) : Incoming flux is equal to outgoing flux for a cube placed in a uniform electric field. So net flux through cube will be zero. 13. (a) : Energy released on discharging a capacitor completely = Energy stored in the capacitor.

1 U = CV 2 2 Here, C = 4 × 10–6 F, V = 100 V 1 \ U = × 4 × 10 −6 × (100)2 = 0.02 J 2

14. (d) : Electric field inside the metallic shell is zero. Electric force on the charge at the centre of metallic shell is zero.

d

15. (b) : A

d/2 d/2 1 = + C1 e0 K1 A e0 K2 A

or C1 =

2

a

e0 A  2 K1K2    d  K1 + K2 

C 2 K1K2 \ 1 = C K1 + K2 10. (b) 11. (b) : Given situation is shown in the figure. Y

b

b

B +q



Potential at A, V =



Net force acting on 1C charge is given by

X(m)

+q 2

a +b

2

+

−q 2

a + b2

=0

16. (a) : Potential difference between two points in an electric field is VA − VB =

1C 1C 1C 1C 1C 1C 0 1 2 4 8 16

C –q



W q0

where W is work done by moving charge q0 from point A to B Here, W = 2 J, q0 = 20 C 2 VA − VB = = 0.1 volt 20

8

VITEEE CHAPTERWISE SOLUTIONS

17. (b) : Electric break down of air, E = 3 × 106 V m–1 5 R = m = 2.5 m 2 Qmax E= 4 pe0 R 2 \ Qmax = E × 4pe0 × R2

= 3 × 106 × =

6.25 3 × 10

3

1 9 × 109

× ( 2.5)2

= 2 × 10 −3 C

18. (a) : Given situation is shown in the figure. Q1 C

3C

Q2 2C

Q V Equivalent capacitance of the circuit.



(C + 2C )( 3C ) 3 Ceq = = C C + 2C + 3C 2 3 Q = CeqV = CV 2 1 3 1 C Q1 = Q = × CV = CV 2C + C 3 2 2

(d) : Here, q = 1 × 10–6 C, 2l = 2 cm = 2 × 10–2 m E = 1 × 105 N C–1, tmax = ?, U = ? tmax = pE sin 90° = q2lE = 1 × 10–6 × 2 × 10–2 × 105 = 2 × 10-3 N-m Work done by external agent = change in potential energy of the dipole = Uf – Ui = –pE cos qf + pE cos qi = –q2lE cos 180° + q2lE cos 0°= 2 q2lE = 2 × 1 × 10–6 × 2 × 10–2 × 1 × 105 = 4 × 10–3 J. rp 20. (c) : Length of the arc = rq = l,  3 rp Charge on the arc = ×l r r 3 /3 kq \ Potential at centre = r rp l l 1 = × = 4 pe0 3 r 12e0 19.

21. (d) : There are 10 electrons and 10 protons in a neutral water molecule. So, its dipole moment is p = q(2l) = 10e (2l) Hence length of the dipole i.e., distance between centres of positive and negative charges is 6.4 × 10 −30 p 2l = = 10 e 10 × 1.6 × 10 −19 = 4 × 10–12 m = 4 pm 22. (a) : The magnitude of electric field intensity due to an infinitely long straight wire of uniform linear charge density l is E=

l 1 2l = 2 pe0r 4 pe0 r

where r is the perpendicular distance of the point from the wire. Here, l =

1 = 9 × 109 N m 2 C −2 4 pe0

\

E=



1 C m −1 , r = 18 cm = 18 × 10 −2 m 3



=

1  (9 × 109 N m 2 C −2 )( 2)  C m −1  3  (18 × 10 −2 m)

1 × 1011 N C −1 = 0.33 × 1011 N C −1 3

23. (c) : The situation is shown in figure. z P (0, 0, z) +q (0, 0, a) O

y

–q (0, 0, –a) x



The distance between the points (0, 0, –a) and (0, 0, z) is r1 = z + a and that between the points (0, 0, a) and (0, 0, z) is r2 = z – a The potential at P due to charge –q is

V1 =

( − q) 1 ( − q) 1 = 4 pe0 r1 4 pe0 ( z + a)

and that due to charge +q is

9

Electrostatics

V2 =

q 1 q 1 = 4 pe0 r2 4 pe0 ( z − a)

\ The total potential at P is

 1 1  −    ( z − a) ( z + a) 



q = 4 pe0



q  ( z + a) − ( z − a)  =   4 pe0  ( z − a)( z + a)  =

=

28. (b) : Capacitance of a cylindrical capacitor

q  2a    4 pe0  ( z 2 − a2 )  2qa 2



2

4 pe0 ( z − a ) 2

1Q 24. (b) : Energy stored in the capacitor = 2 C Q = CV = 900 × 10–12 F × 100 V –8 \ Q = 9 × 10 C. Energy of the capacitor when fully charged 2



Q =1 = 4.5 × 10−6 J. 2 C



The total charge is conserved. In figure (b), total capacitance = C ′ = 2 × C = 2 × 900 pF.



\ Final energy =



=

1 Q2 1 Q2 = ⋅ 2 C ′ 2 2C

4.5 × 10 −6 J = 2.25 × 10 −6 J. 2

25. (d) : Electric charge is quantised. It is an integral multiple of e = 1.60 × 10–19 C. 26. (d) : Field of a dipole at a distance d along the axis or perpendicular to the axis is inversely proportional to r3. It means F/8 is the force at the given point, 2 m away if F is the force at 1 m, on the same charge q. 27. (b) : The charge at the inner surface, outer surface and inside the conductor at P = (–Q, +Q, 0) as shown in the figure.

2pe 0 L ln(b / a ) Energy stored in the capacitor, =

Q 2 1 Q 2 ln(b / a ) Q2 = = [const] U=1 . 2 C 2 2pe0 L L If the charge is doubled and length is doubled then Q′2  2 U ′ = [const] = [const] 4  Q  = 2U . ′ L 2 L 

29. (c) : The torque on a dipole moment is   t = p × E.   For the maximum torque p and E are perpendicular to each other so that pE sinq is maximum i.e., sinq = 1. t = (3 × 104 N C–1)(6 × 10–30 C m) = 18 × 10–26 N m. 30. (a) : We know, electric field due to point  ±q ^ charge, E = r 4 pe 0 r 2 Fig. (1) Q1 is positive because E is positive also decreases along positive x-axis. Field due to Q2 is positive and it is along negative of x-axis, so Q2 is negative. Fig. (2) field due to Q1 is explained in fig (1). Field due to Q2 is negative and also directed along negative of x-axis so Q2 is positive. Fig (3) field due to Q1 is negative and direction along positive of x-axis so Q1 is negative. Field due to Q2 is explained in fig (2). Fig. (4) field due to Q1 is explained in fig (3) field due to Q2 is explained in fig. (1). 31. (a) : The situation is summarised in figure. BC = AD = 3 cm, AB = DC = 4 cm, so AC = 5 cm. A

D (10 × 10–12 C)

4 cm

5c

m

4 cm

B

(10 × 10–12 C)

3 cm



V = V1 + V2 ( −q) q 1 1 = + 4 pe0 ( z + a) 4 pe0 ( z − a)

3 cm



C (– 20 × 10–12 C)

10

VITEEE CHAPTERWISE SOLUTIONS



Now potential at A



q q 1 qB 1 1 VA = + . C + . D 4 pe 0 AB 4 pe 0 AC 4 pe 0 AD =



1 4 pe 0



 10 × 10−12 20 × 10−12 10 × 10−12  + −    4 × 10−2 5 × 10−2 3 × 10−2 

 10 20 10  = 9 × 109 × 10−10  − +  5 3 4 =

9 × 10−1 × 11 = 16.5 × 10–1 = 1.65 V 6

2





32. (d) : Suppose there are n metal plates which are separated by dielectric. This combination is equivalent to the (n –1) capacitors connected in the parallel.

(n − 1) k e 0 A d Here, K = 4, C = 100 pF = 100 × 10–12 F \ C=

d = 1 mm = 10–3 m C.d \ (n − 1) = k e0 A

33. (a)

vvv

D A = pr 2 = p   = 3.14 × (1 × 10−2 )2 m 2 2

=

100 × 10−12 × 10−3

4 × 8.85 × 10−12 × 3.14 × 10−4 1000 = = 8.99 4 × 8.85 × 3.14 n = 8.99 + 1 = 9.99 ≈ 10.

11

Current Electricity

2

CHAPTER

1.

2.

Current Electricity

The daniel cell is balanced on 125 cm length of a potentiometer. Now, the cell is short circuited by a resistance of 2 W and the balance point is obtained at 100 cm. The internal resistance of the daniel cell is 4 (a) (b) 1.5 W W 3 (c) 1.25 W (d) 0.5 W (2014) Four resistances of 10 W, 60 W, 100 W and 200 W respectively taken in order are used to form a Wheatstone’s bridge. A 15 V battery is connected to the ends of a 200 W resistance, the current through it will be (a) 7.5 × 10–5 A (b) 7.5 × 10–4 A (c) 7.5 × 10–3 A (d) 7.5 × 10–2 A (2014)

3.

A resistor of 6 kW with tolerance 10% and another resistor of 4 kW with tolerance 10% are connected in series. The tolerance of the combination is about (a) 5% (b) 10% (c) 12% (d) 15% (2014)

4.

If we add impurity to a metal those atoms also deflect electrons. Therefore, (a) the electrical and thermal conductivities both increase. (b) the electrical and thermal conductivities both decrease. (c) the electrical conductivity increases but thermal conductivity decreases. (d) the electrical conductivity decreases but thermal conductivity increases. (2014)

5.

An ammeter and a voltmeter of resistance R are connected in series to an electric cell of negligible internal resistance. Their reading are A and V respectively. If another resistance R is connected in parallel with the voltmeter, then

(a) both A and V will increase. (b) both A and V will decrease. (c) A will decrease and V will increase. (d) A will increase and V will decrease. (2014) 6.

In a hydrogen discharge tube, it is observed that through a given cross-section 3.31 × 1015 electrons are moving from right to left and 3.12 × 105 protons are moving from left to right. The current in the discharge tube and its direction will be (a) 2 mA, towards left (b) 2 mA, towards right (c) 1 mA, towards right (d) 1 mA, towards left (2014, 2012)

7.

When a resistor of 11 W is connected in series with an electric cell, the current flowing in it is 0.5 A. Instead when a resistor of 5 W is connected to the same electric cell in series, the current increases by 0.4 A. The internal resistance of the cell is (a) 1.5 W (b) 2 W (c) 2.5 W (d) 3.5 W (2014)

8.

A battery is charged at a potential of 15 V in 8 h when the current flowing is 10 A. The battery on discharge supplies a current of 5 A for 15 h. The mean terminal voltage during discharge is 14 V. The watt-hour efficiency of the battery is (a) 80% (b) 90% (c) 87.5% (d) 82.5% (2014)

9.

The potential difference across the terminals of a battery is 50 V when 11 A current is drawn and 60 V when 1 A current is drawn. The emf and the internal resistance of the battery are (a) 62 V, 2 W (b) 63 V, 1 W (c) 61 V, 1 W (d) 64 V, 2 W (2014)

12

VITEEE CHAPTERWISE SOLUTIONS

2V A

R

4

1A 2

(a) 4 mC and 10 W (c) 2 mC and 2 W

C 2 F

B

4V

(b) 4 mC and 4 W (d) 8 mC and 4 W (2013)

11. Seven resistances are connected between points A and B as shown in figure. The equivalent resistance between A and B is

15. 2.5 faraday of electricity is passed through a CuSO4 solution. The number of gram equivalents of copper deposited on the cathode is (a) 1 (b) 1.5 (c) 2 (d) 2.5 (2012) 16. Each resistance shown in figure is 2 W. The equivalent resistance between A and B is A

2

10 

3

5

B

8

(a) 5 W (c) 4 W

6

(b) 4.5 W (d) 3 W

6

(2013)

12. In the Wheatstone’s network given, P = 10 W, Q = 20 W, R = 15 W, S = 30 W, the current passing through the battery (of negligible internal resistance) is R

P G

S

Q + – 6V

(a) 0.36 A (c) 0.18 A

(b) zero (d) 0.72 A

(2012)

13. Three resistors 1 W, 2 W and 3 W are connected to form a triangle. Across 3 W resistor a 3 V battery is connected. The current through 3 W resistor is (a) 0.75 A (b) 1 A (c) 2 A (d) 1.5 A (2012) 14. Charge passing through a conductor of cross-section area A = 0.3 m2 is given by q = 3t2 + 5t + 2 in coulomb, where t is in second. What is the value of drift velocity at t = 2 s?

2

2

10 

A

(2012)

2

B 

10 V

(Given, n = 2 × 1025 m–3) (a) 0.77 × 10–5 m s–1 (b) 1.77 × 10–5 m s–1 (c) 2.08 × 10–5 m s–1 (d) 0.57 × 10–5 m s–1

2

10. A network of resistances, cell and capacitor C (= 2 mF) is shown in figure. In steady state condition, the charge on 2 mF capacitor is Q, while R is unknown resistance. Values of Q and R are respectively

(a) 2 W (c) 8 W

(b) 4 W (d) 1 W

(2012)

17. A current of 2 A flows through a 2 W resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 W resistor. The internal resistance of the battery is 1 1 (a) W (b) W 3 4 (c) 1 W

(2012)

(d) 0.5 W

18. Dimensions of a block are 1 cm × 1 cm × 100 cm. If specific resistance of its material is 3 × 10–7 W m, then the resistance between the opposite rectangular faces is (a) 3 × 10–7 W (b) 3 × 10–9 W (c) 3 × 10–5 W (d) 3 × 10–3 W (2011) 19. The magnitude and direction of the current in the circuit shown will be a

1

e

2

b

10 V 4 V

d

3

(a) 7/3 A from a to b through e (b) 7/3 A from b to a through e (c) 1 A from b to a through e (d) 1 A from a to b through e

c

(2011)

13

Current Electricity

20. An electric bulb of 100 W is connected to a supply of electricity of 220 V. Resistance f the filament is (a) 484 W (b) 100 W (c) 22000 W (d) 242 W (2011) 21. Pick out the wrong statement. (a) In a simple battery circuit, the point of lowest potential is the negative terminal of the battery. (b) The resistance of an incandescent lamp is greater when the lamp is switched off. (c) An ordinary 100 W lamp has less resistance than a 60 W lamp. (d) At constant voltage, the heat developed in a uniform wire varies inversely as the length of the wire used. (2011) 22. Five resistances are connected as shown in the figure. The effective resistance between points A and B is 2.5 

B

 7.5  10 A

10 (a) W 3

1

9

(b)

(c) 40 W

10 W 17

(d) 45 W

(2011)

23. A potentiometer is connected across A and B and a balance is obtained at 64.0 cm. When potentiometer lead to B is moved to C, a balance is found at 8.0 cm. If the potentiometer is now connected across B and C, a balance will be found at A

(a) 8.0 cm (c) 64.0 cm

B

C

(b) 56.0 cm (d) 72.0 cm

(2011)

24. A 2 V battery, a 15 W resistor and a potentiometer of 100 cm length, all are connected in series. If the resistance of potentiometer wire is 5 W, then the potential gradient of the potentiometer wire is (a) 0.005 V cm–1 (b) 0.05 V cm–1 –1 (c) 0.02 V cm (b) 0.2 V cm–1 (2011)

25. The ratio of the amounts of heat developed in the four arms of a balance Wheatstone bridge, when the arms have resistances P = 100 W, Q = 10 W, R = 300 W and S = 30 W respectively is (a) 3 : 30 : 1 : 10 (b) 30 : 3 : 10 : 1 (c) 30 : 10 : 1 : 3 (d) 30 : 1 : 3 : 10 (2010) 26. An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20°C? The temperature of boiling water is 100°C. (a) 12.6 min (b) 4.2 min (c) 6.3 min (d) 8.4 min (2010) 27. 50 W and 100 W resistors are connected in series. This connection is connected with a battery of 2.4 V. When a voltmeter of 100 W resistance is connected across 100 W resistor, then the reading of the voltmeter will be (a) 1.6 V (b) 1.0 V (c) 1.2 V (d) 2.0 V (2010) 28. There are a 25 W - 220 V bulb and a 100 W - 220 V line. Which electric bulb will glow more brightly? (a) 25 W bulb (b) 100 W bulb (c) Both will have equal incandescence (d) Neither 25 W nor 100 W bulb will give light (2010) 29. In Meter bridge or Wheatstone bridge for measurement of resistance, the known and the unknown resistances are interchanged. The error so removed is (a) end correction (b) index error (c) due to temperature effect (d) random error (2010) 30. In the adjacent shown circuit, a voltmeter of internal resistance R, when connected across

50 k 100 V

100 V. 3 Neglecting the internal resistance of the battery, the value of R is (a) 100 kW (b) 75 kW (c) 50 kW (d) 25 kW B and C reads



A

B 50 k C

(2009)

14

VITEEE CHAPTERWISE SOLUTIONS

31. A cell in secondary circuit gives null deflection for 2.5 m length of potentiometer having 10 m length of wire. If the length of the potentiometer wire is increased by 1 m without changing the cell in the primary, the position of the null point now is (a) 3.5 m (b) 3 m (c) 2.75 m (d) 2.0 m (2009) 32. Equal amounts of metal are converted into cylindrical wires of different lengths (L) and cross-sectional area (A). The wire with the maximum resistance is the one, which has (a) length = L and area = A (b) length = L/2 and area = 2A (c) length = 2L and area = A/2 (d) all have the same resistance, as the amount of the metal is the same. (2008) 33. A solid sphere of radius R1 and volume r charge density r = 0 is enclosed by a r hollow sphere of radius R2 with negative surface charge density s, such that the total charge in the system is zero. r0 is a positive constant and r is the distance from the centre of the sphere. The ratio R2/R1 is (a)

s r 0

(c)

r0 2s

(b)



2s r0



r (d) 0 s

(2008)

34. Three resistances of 4 W each are connected as shown in figure. If the point D divides the resistance into two equal halves, the resistance between point A and D will be

(a) 12 W (c) 3 W

(b) 6 W (d) 1/3 W.

(2008)

35. The resistance of a metal increases with increasing temperature because (a) the collisions of the conducting electrons with the electrons increase

(b) the collisions of the conducting electrons with the lattice consisting of the ions of the metal increase (c) the number of conduction electrons decreases (d) the number of conduction electrons increases. (2008) 36. In the absence of applied potential, the electric current flowing through a metallic wire is zero because (a) the electrons remain stationary (b) the electrons are drifted in random direction with a speed of the order of 10–2 cm/s (c) the electrons move in random direction with a speed of the order close to that of velocity of light (d) electrons and ions move in opposite direction. (2008) 37. A meter bridge is used to determine the resistance of an unknown wire by measuring the balance point length l. If the wire is replaced by another wire of same material but with double the length and half the thickness, the balancing point becomes 2l, then the value of l is (a) 40 cm (b) 20 cm 300 240 cm cm (c) (d) 7 7 (2008) 38. Identify the incorrect statement regarding a superconducting wire (a) transport current flows through its surface (b) transport current flows through the entire area of cross-section of the wire (c) it exhibits zero electrical resistivity and expels applied magnetic field (d) it is used to produce large magnetic field. (2008) 39. When an electrical appliance is switched on, it responds almost immediately, because (a) the electrons in the connecting wires move with the speed of light (b) the electrical signal is carried by electromagnetic waves moving with the speed of light

15

Current Electricity

(c) the electrons move with speed which is close to but less than speed of light (d) the electron are stagnant (2008) 40. Twelve wires of each of resistance 6 W are connected to form a cube as shown in the figure. The current enters at a corner A and leaves at the diagonally opposite corner G. The joint resistance across the corners A and G is E

F

A

B

G

H

(a) 12 W (c) 3 W

D

C

(b) 6 W (d) 5 W

(2007)

41. A conductor and a semiconductor are connected in parallel as shown in the figure. At a certain voltage both ammeters register the same current. If the voltage of the DC source is increased then the Conductor

A1

Semiconductor

A2

(c) ammeters connected to both semiconductor and conductor will register the same current (d) ammeters connected to both semiconductor and conductor will register no change in the current. (2007) 42. A uniform copper wire of length 1 m and cross-sectional area 5 × 10–7 m2 carries a current of 1 A. Assuming that there are 8 × 1028 free electron/m3 in copper, how long will an electron take to drift from one end of the wire to the other? (a) 0.8 × 103 s (b) 1.6 × 103 s 3 (c) 3.2 ×10 s (d) 6.4 × 103 s (2007) 43. The temperature coefficient of resistance of a wire is 0.00125/K. At 300 K, its resistance is 1 W. The resistance of the wire will be 2 W at (a) 1154 K (b) 1100 K (c) 1400 K (d) 1127 K (2007) 44. If the total emf in a thermocouple is a parabolic 1 function expressed as E = at + bt 2 , which of 2 the following relation does not hold good? (a) neutral temperature tn = −

a b

−2 a b (c) thermoelectric power P = a + bt a (d) tn = b (b) temperature of inversion, ti =

(a) ammeter connected to the semiconductor will register higher current than the ammeter connected to the conductor (b) ammeter connected to the conductor will register higher current than the ammeter connected to the semiconductor

(2007)

Answer Key 1. 9. 17. 25. 33. 41.

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

2. 10. 18. 26. 34. 42.

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

3. 11. 19. 27. 35. 43.

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

4. 12. 20. 28. 36. 44.

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

5. 13. 21. 29. 37.

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

6. 14. 22. 30. 38.

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

7. 15. 23. 31. 39.

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

8. 16. 24. 32. 40.

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

16

1.



VITEEE CHAPTERWISE SOLUTIONS

e planations

(d) : The internal resistance of the cell is l  r = R  1 − 1  l2 

2.



Here, l1 = 125 cm, l2 = 100 cm, R = 2 W \

 125 cm  r = ( 2 W)  − 1 100 cm   = (2 W)(1.25 – 1) = (2 W)(0.25) = 0.5 W

(d) : The Wheatstone’s bridge is shown in figure. 10 

60 





3.

V 15

As 10 W, 60 W and 100 W are in series and their combination is in parallel with 200 W. \ When a 15 V battery is connected across 200 W, then the current through it is 15 V I= = 7.5 × 10 −2 A 200 W (b)

4.

(b) : The electrons increase their number of collisions increasing the thermal and electrical resistance. Hence the electrical and thermal conductivities both decrease.

5.

(d) : The circuit is shown in figure. A

V

R





When a resistance R is connected in parallel with the voltmeter, then reading of ammeter A will increase and voltmeter V will decrease.

6.

(c) : Here, Number of electrons, ne = 3.31 × 1015 Number of protons, np = 3.12 × 1015 The current due to electrons is Ie =



( 3.12 × 1015 )(1.6 × 10 −19 C) t 1s = 4.992 × 10–4 A ≈ 0.5 mA from left to right (i.e. in the direction of motion of protons). \ The total current in the discharge tube is I = Ie + Ip = 0.5 mA + 0.5 mA = 1 mA from left to right. Hence the current in the discharge tube is 1 mA and its direction will be towards right. 7.

0



ne e ( 3.31 × 1015 )(1.6 × 10 −19 C) = 1s t = 5.296 × 10–4 A ≈ 0.5 mA

from left to right (i.e. in the direction of opposite to the motion of electrons). The current due to protons is Ip =

20

100 







np e

=

(c) : Let e and r be emf and internal resistance of the cell. As per question,  r when a resistor of 11 W is connected in series 0.5 A 0.5 A with it, then the current flowing in 11 W is 11  e 0.5 A = 11 W + r e = (0.5 A)(11 W) + r(0.5 A)

= 5.5 V + r(0.5 A) ...(i)  r When a resistor of 5 W is connected in series 0.9 A with it, then the current 0.9 A flowing in 5 W is e 5 0.9 A = 5W+r e = (0.9 A)(5 W) + r(0.9 A) = 4.5 V + r(0.9 A) ...(ii) Equating eqns. (i) and (ii), we get 5.5 V + r(0.5 A) = 4.5 V + r(0.9 A) 5.5 V – 4.5 V = r[0.9 A – 0.5 A] 1V 1 V = r(0.4 A) or r = = 2.5 W 0.4 A 8. (c) : Energy supplied to the battery during charging i.e. input energy = VIt = (15 V)(10 A)(8 h) = 1200 W h Energy supplied by the battery during discharging i.e. output energy = (14 V)(5 A)(15 h) = 1050 Wh \ The efficiency of the battery is

17

Current Electricity

η=

Output energy 1050 W h × 100 = × 100 Input energy 1200 W h

11. (c) :

= 87.5%

10  10 

A

(c) : Let e and r be the emf and the internal resistance of the battery. The potential difference V across the terminals of the battery when a current I is drawn from it is V = e – Ir As per question 50 V = e – (11 A)r ...(i) and 60 V = e – (1 A)r ...(ii) Subtracting eqn. (i) from eqn. (ii), we get 10 V = (10 A)r 9.

10 V =1W 10 A Substituting this value of r in eqn. (i), we get 50 V = e – (11 A)(1 W) 50 V = e – 11 V e = 50 V + 11 V = 61 V r=

10. (a) : In the steady state, capacitor acts as a open circuit, hence no current will flow through 4 W. \ The currents distribution is shown in figure. 10 V

2V A

1A

R

C 4 1A 2

2 F

B

4V

5

B

8

6

6

D





In the given figure, two 10 W resistances and two 6 W resistances are connected in parallel. So their equivalent resistances are 5 W and 3 W respectively. The equivalent circuit is shown in figure. 5

A

3

C

8

5



B 3

D

It is a balanced Wheatstone bridge. Hence the resistance of arm CD becomes ineffective. Now the resistance of arm ACB(= 5 W + 3 W = 8 W) is in parallel with the resistance of arm ADB(= 5 W + 3 W = 8 W). \ The equivalent resistance between A and B is (8 W)(8 W) Req = =4W 8W+8W 12. (a) : A

P

Q

D

The potential difference between C and D is VC – VD = (1 A)(2 W) = 2 V As VD = 0 V (D is earthed) \ VC = 2 V The potential drop across the resistance R = 10 V + 2 V – 2 V = 10 V 10 V \ Resistance, R = = 10 W 1A As there is no current in the 4 W, there is no potential drop across it. \ The potential drop across the capacitor =4V–2V=2V The charge on the capacitor is Q = CV = (2 mF)(2 V) = 4 mC

3

C



=

10



B

R

=

15

 C

G

=

20



D + –

S

=

30



6V

P R i.e., 10 W = 15 W = Q S 20 W 30 W 1 1 or = 2 2 \ The given Wheatstone bridge is balanced. Hence, no current flows through the galvanometer G. Now the resistance of arm ABC (= P + R) is in parallel with the resistance of arm ADC (= Q + S). \ The equivalent resistance of the circuit is ( P + R)(Q + S) Req = ( P + R) + (Q + S)

Q

18

VITEEE CHAPTERWISE SOLUTIONS

2







2

1







3

C

3V

In the given circuit, 1 W and 2 W are in series and their series combination is in parallel with 3 W. \ The equivalent resistance of the circuit is (1 W + 2 W)( 3 W) 3 = W Req = (1 W + 2 W) + ( 3 W) 2 The current drawn from the battery is 3V e I= =2A = Req 3 W 2 As RAC (= 3 W) and RABC (= 1 W + 2 W = 3 W) are equal. \ The current through 3 W resistor I 2A = = =1A 2 2 14. (b) : Here, n = 2 × 1025 m–3, A = 0.3 m2 q = (3t2 + 5t + 2) C The current passing through the conductor is dq d I = = (3t2 + 5t + 2) = (6t + 5) A dt dt At t = 2 s, I = 17 A The drift velocity is I vd = neA 17 A = 25 −3 ( 2 × 10 m )(1.6 × 10 −19 C)(0.3 m 2 )

= 1.77 × 10–5 m s–1

15. (d) : 1 faraday deposits 1 gram equivalent. Hence 2.5 faraday will deposit 2.5 g equivalent of copper.

B

Resistance of upper arm ( = 2 W + 2 W = 4 W) is in parallel with the diagonal resistance (2 W) and the resistance of lower arm ( = 2 W + 2 W = 4 W). \ The equivalent resistance between A and B is

B

A

2



16. (d) : A

2

2

2

(10 W + 15 W)( 20 W + 30 W) = ( 10 W + 15 W) + ( 20 W + 30 W) ( 25 W)( 50 W) 50 = = W 3 ( 25 W) + ( 50 W) The current drawn from the battery is 6V I= = 0.36 A 50 W 3 13. (b) : The circuit is shown in figure.

1 1 1 1 = + + Req 4 W 2 W 4 W 1+ 2 +1 4 = 4W 4W 4W = =1W 4 =



or Req 17. (a) : Let e and r be the emf and the internal resistance of the battery. When a resistor R is connected in series with it, then the current flowing through it is e I= R+r

In first case, I = 2 A, R = 2 W e \ 2A= 2W+r

In second case, I = 0.5 A, R = 9 W e \ 0.5 A = 9W+r Dividing eqn. (i) by eqn. (ii), we get

...(i)

…(ii)

2A 9W+r = 0.5 A 2 W + r 9W+r 4= 2W+r 4(2 W + r) = 9 W + r 8 W + 4r = 9 W + r 4r – r = 9 W – 8 W 3r = 1 W 1 r= W 3 rl 18. (a) : Resistance, R = A Dimensions of the block, 1 cm × 1 cm × 100 cm, r = 3 × 10–7 W m, l = 1 cm = 10–2 m Area of rectangular face of the block , A



= 1 cm × 100 cm = 100 × 10–4 m2 = 1 × 10–2 m2

19

Current Electricity



R = 3 × 10 −7 ×

\

19. (d) :

a

1

1

e

10 −2 1 × 10 −2 2



= 3 × 10 −7 W

2

10 V 4 V I c

3 Here, e1 > e2, so current in the circuit will be clockwise, Using Kirchhoff’s voltage law in the circuit, 10 – 4 – 2I – 3I – 1I = 0 or 6 – 6I = 0 \ I = 1 A

20. (a) : Here, P = 100 W, V = 220 V

We know, P =

I2



1





A

B



10 W 3 1

23. (b) : A B VAB = e1 ∝ 64 VAC = (e1 – e2) ∝ 8 VBC = e2 ∝ l \ 64 – l = 8 or l = 64 – 8 = 56 cm

2

24. (a) : Current in the circuit, I =



⇒ I=

Q

R

S

2

2 2 1 = = A 15 + 5 20 10

C

(say)

e R + Rp

2

2

2

The effective resistance between A and B =

P

 3  I I  × 100 :  I  × 10 :    4  4

I × 300 :   × 30 4

10/3 

B

10 

3 I I= 4 4

Ratio of heat developed per second across P, Q, R and S, HP : HQ : HR : HS 3 = 4

9

10 

330 I 3 = I 330 + 110 4

V

10  A

I 2 = I − I1 = I −

I

10 





In the figure, I1 =

I1

B

7.5  A



( 220)2 = 484 W 100

2.5 

22. (a) :

25. (b) : P = 100 W, Q = 10 W, R = 300 W, S = 30 W

V2 V2 \ R= R P

or R =

21. (b)

5 1 V= V 10 2 So, potential gradient of the potentiometer wire, e 1 1 1 = = × = V m −1 l 2 1 2 = 0.5 V m–1 = 0.005 V cm–1 e = IRp =

b

d

Potential drop across potentiometer wire,

= 90 : 9 : 30 : 3 = 30 : 3 : 10 : 1

26. (c) : Heat gained by water when its temperature changes from 20°C to 100°C is H1 = mc(T2 – T1) = 1000 × 1 × (100 – 20) = 80000 cal = 80000 × 4.2 J Heat developed in time t due to current in resistor is, H2 = V I t = 220 × 4 × t J According to question, H2 = H1 220 × 4 × t = 80000 × 4.2 80000 × 4.2 t= = 381.8 s = 6.3 min 220 × 4 27. (c) : Given situation is shown in the figure. Net resistance of the circuit,

R = 50 +

100 × 100 = 100 W 100 + 100

20

VITEEE CHAPTERWISE SOLUTIONS

V 50 

A

100 

50 k B

100 V

100 

50 k V R 2.4 V

2.4 A 100 Reading of voltmeter = Potential difference across 100 W resistance 2.4 = × 50 = 1.2 V 100 V2 28. (a) : Since power P is given by P = , R 2 V so R = P For the first bulb,





Current through the circuit, I =

 V 2   ( 220)2  R1 =   = 1936 W =  P1   25  For the second bulb,  V 2   ( 220)2  R2 =   = 484 W =  P2   100  Current in series combination is the same in the two bulbs and current I is given by 220 V 220 1 I= = = = A R1 + R2 1936 + 484 2420 11

If the actual powers in the bulbs be P1 and P2 then 2 1 2 P1′ = I R1 =   × 1936 = 16 W  11  2 1 and P2′ = I 2 R2 =   × 484 = 4 W  11  Since P′1 > P′2, 25 W bulb will glow more brightly.

29. (a) : In Meter bridge experiment, it is assumed that the resistance of the L shaped plate is negligible, but actually it is not so. The error created due to this is called end error. To remove this the resistance box and the unknown resistance must be interchanged and then the mean reading must be taken. 30. (c) : When the voltmeter of internal resistance R is connected across B and C, it 100 reads V. 3

C

\ Potential drop across B and C = Voltmeter reading 100 ...(i) i.e., VBC = V 3 As 50 kW and R are in parallel, so their equivalent resistance R′ is ( 50 k W)R R′ = ( 50 k W + R) \ The total resistance of the circuit ( 50 k W)R = 50 k W + R′ = 50 k W + ( 50 k W + R) The current in the circuit is 100 V I=  ( 50 k W)R   50 k W + ( 50 k W + R)  The potential drop across B and C is VBC = IR′

=



100 V  ( 50 k W)R   50 k W + ( 50 k W + R) 

 ( 50 k W)R   ( 50 k W + R)  ...(ii)

Equating eqns. (i) and (ii), we get 100 V  ( 50 k W)R   50 k W + ( 50 k W + R) 

 ( 50 k W)R  100  50 k W + R  = 3 V

( 50 k W)R ( 50 k W)2 + ( 50 k W)R + ( 50 k W)R

=

1 3

(150 kW)R = (50 kW)2 + (100 kW)R (150 kW)R – (100 kW)R = (50 kW)2 (50 kW)R = (50 kW)2 ( 50 k W)2 R= = 50 k W 50 k W 31. (c) : For 10 m long potentiometer wire, the balancing length is 2.5 m. For 11 m long potentiometer wire, the (11 m) ( 2.5 m) balancing length is = 2.75 m (10 m)

21

Current Electricity

l 32. (c) : R = r . To have the maximum A resistance for the same material, l/A should be maximum. Here l′ = 2l, A′ = A/2 gives the maximum resistance.



R1

∫ 4p r

2

dr ⋅

0



R22 r0 = R12 2s

\

R2 = R1



34. (c) : ABD, ACD are in series. They are connected in parallel i.e., 1 + 1 = 1 i.e., R = 3 W. 6 6 3 35. (a) : Resistivity of the conductor, r =

m ne 2 t



In the conductors, average speed of the electrons increases with increase in temperature and collisions between the electrons become more frequently. So, the average time of collision t decreases and resistivity increases. 36. (c) : The free electrons keep moving randomly in all directions throughout the lattice structure of the material due to thermal energy. The thermal speed is very high (~ 106 m s–1). But the direction of motion is random so the average thermal velocity is zero. Therefore there is no flow of current due to the thermal motion of electrons. 37. (c) 38. (b) : In superconductors, the current flows on the surface so that it exactly cancels the magnetic field inside the metal. Therefore statement (b) is not correct. 39. (b) : It is the electric field that is set up which moves with the velocity of light in that medium. 40. (d) : Here R = 6 W

I/3

I/6 D

r0 . 2s

I/6 I/6

I/6

I



B

H

r0 R2 = +4p 1 r 0 r 2

The total negative charge on the outer shell = –4pR22s. But both are equal in magnitude as the sum of the charges is zero.

F

I/3

I/3

33. (c) : The total positive charge on the inner solid sphere

I/6

E

I/3 A

I/6

C

I/3 G

I/3

V

As the resistance of each arm is equal so current is equally distributed in arm AB, AD and AE. Apply KVL in outer loop ADCGA, I I I 5 V =   R +   R +   R = IR ...(i)  3 6  3 6 Assume equivalent resistance of the circuit is Req . So, V = IReq ...(ii) Comparing eqns (i) and (ii), we get



Req =



5 5 R = × 6 = 5W 6 6

41. (c) : Current through semiconductor device is same as that of the current in the conductor. It means both the branches offer  V same resistance  R =  .  I

After increasing DC source voltage, current is increased in each branch by same amount.

42. (d) : Given, l = 1 m, A = 5 × 10–7 m2

I = 1 A, n = 8 × 1028 electrons/m3 t=



or t =



l l = vd ( I / Ane)

(Q I = neA vd )

l Ane I

1 × 5 × 10−7 × 8 × 1028 × 1.6 × 10−19 1 0.00125t = 6.4 × 103 s =

43. (c) : The resistance Rt of a metal conductor at temperature t °C is given by Rt = R0 (1 + at) Here, R300 = 1 W, Rt = 2 W, t = ? a = 0.00125 K–1

22

VITEEE CHAPTERWISE SOLUTIONS

So, 1 = R0 (1 + 0.00125 × 300) ...(i) 2 = R0 (1 + 0.00125 t) ...(ii) From eqns (i) and (ii) 1 + 0.00125 t 2= 1 + 0.00125 × 300 2 + 0.00125 × 300 × 2 = 1 + 0.00125 t 0.00125 t = 1.75, t = 1400 K 44. (d) : Here, E = at +

1 2 bt 2



vvv

\

dE = a + bt dt

At neutral temperature (t = tn),

dE =0 dt

a b The temperature of inversion, (t = ti) 2a ti = 2tn – t0 = 2tn – 0 ; ti = − b dE Thermoelectric power, P = = a + bt dt \ a + btn = 0 ; tn = −

23

Magnetic Effects of Electric Current and Magnetism

3

CHAPTER

Magnetic Effects of Electric Current and Magnetism

1.

A galvanometer has current range of 15 mA and voltage range 750 mV. To convert this galvanometer into an ammeter of range 25 A, the required shunt is (a) 0.8 W (b) 0.93 W (c) 0.03 W (d) 2.0 W (2014)

2.

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

3.

(a) 10 cm

(b) 10 2 cm

(c) 20 cm

(d) 5 2 cm

6.

An electron moves at right angle to a magnetic field of 1.5 × 10–2 T with a speed of 6 × 107 m s–1. If the specific charge of the electron is 1.7 × 1011 C kg–1, the radius of the circular path will be (a) 2.9 cm (b) 3.9 cm (c) 2.35 cm (d) 2 cm (2014)

7.

If a current I is flowing in a loop of radius r as shown in figure, then the magnetic field induction at the centre O will be

(2014)

If a magnet is suspended at an angle 30° to the magnetic meridian, the dip needle makes angle of 45° with the horizontal. The real dip is  3 (a) tan −1  (b) tan −1( 3 )  2   3 (c) tan −1   2 

(a) magnetic intensity is a measure of lines of force passing through unit area held normal to it. (b) magnetic lines of force form a close curve. (c) inside a magnet, its magnetic lines of force move from north pole of a magnet towards its south pole . (d) due to a magnet, magnetic lines of force never cut each other. (2014)

 2  (d) tan −1   3 

 I

(2014) 4.

A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis, where the magnetic induction will

r

(a) zero

th

1 be   to its value at the centre of the coil 8 is (a)

R 3



(c) 2 3R 5.

(b)

(b) R 3 (d)

2 3

R

(2014)

The incorrect statement regarding the lines of force of the magnetic field B is

I

O

8.

m0 I sin q 4 pr

(b)

m0 I q 4 pr

(d)

2m0 I sin q 4 pr 2

(2013)

Two identical magnetic dipoles of magnetic moment 1.0 A m2 each, placed at a separation of 2 m with their axes perpendicular to each other. The resultant magnetic field at a point midway between the dipoles is

24

VITEEE CHAPTERWISE SOLUTIONS

(a) 5 × 10 −7 T (c) 10–7 T 9.

(b) 5 × 10–7 T (d) 2 × 10–7 T

(2013)

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

1 2

(d)

1 4

(2013)

10. In a hydrogen atom, an electron is revolving in the orbit of radius 0.53 Å with 6.6 × 1015 revolutions per second. Magnetic field produced at the centre of the orbit is (a) 0.125 Wb m–2 (b) 1.25 Wb m–2 –2 (c) 12.5 Wb m (d) 125 Wb m–2 (2012) 11. The dipole moment of a short bar magnet is 1.25 A m2. The magnetic field on its axis at a distance of 0.5 m from the centre of the magnet is (a) 1.0 × 10–4 N A–1 m–1 (b) 4 × 10–2 N A–1 m–1 (c) 2 × 10–6 N A–1 m–1 (d) 6.64 × 10–8 N A–1 m–1 (2012) 12. In the figure shown, the magnetic field induction at the point O will be i O

m i (a) 0 2 pr

(c) will turn towards left of direction of motion (d) will turn towards right of direction of motion (2012) 14. Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the centre of the ring is m q m qf (a) 0 (b) 0 2 fR 2R m0q m0qf (c) (d) (2012) 2 pfR 2 pR 15. A galvanometer of resistance, G is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is (a)

S2 (S + G)

(b)

SG (S + G)

(c)

G2 (S + G)

(d)

G (S + G)

(2012)

16. A square loop, carrying a steady current I, is placed in a horizontal plane near a long straight conductor carrying a steady current I1 at a distance d from the conductor as shown in figure. The loop will experience I1 d

I

2r

 m0   i  (b)  4 p   r  ( p + 2)

m  i  m i (c)  0    ( p + 1) (d) 0 ( p − 2)  4p   r  4p r

(2012)

13. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron (a) speed will decrease (b) speed will increase

I

(a) a net repulsive force away from the conductor (b) a net torque acting upward perpendicular to the horizontal plane (c) a net torque acting downward normal to the horizontal plane (d) a net attractive force towards the conductor (2012) 17. The electrochemical equivalent of magnesium is 0.126 mg C–1. A current of 5 A is passed in a suitable solution for 1 h. The mass of magnesium deposited will be (a) 0.0378 g (b) 0.227 g

25

Magnetic Effects of Electric Current and Magnetism

(c) 0.378 g

(d) 2.27 g

(2011)

18. In producing chlorine through electrolysis 100 W power at 125 V is being consumed. How much chlorine per minute is liberated? ECE of chlorine is 0.367 × 10–6 kg C–1. (a) 24.3 mg (b) 16.6 mg (c) 17.6 mg (d) 21.3 mg (2011) 19. A particle carrying a charge equal to 100 times the charge on an electron is rotating per second in a circular path of radius 0.8 m. The value of the magnetic field produced at the centre will be (m0 = permeability for vacuum) (a)

−7

10 m0

(c) 10–16m0

(d) 10–7m0

(2011)

1   1   (d) (iA ⋅ B) (2011) (iA × B) n n 21. In a magnetic field of 0.05 T, area of a coil changes from 101 cm2 to 100 cm2 without changing the resistance which is 2 W. The amount of charge that flow during this period is (a) 2.5 × 10–6 C (b) 2 × 10–6 C –6 (c) 10 C (d) 8 × 10–6 C (2011) (c)

22. An ammeter reads upto 1 A. Its internal resistance is 0.81 W. To increase the range to 10 A, the value of the required shunt is (a) 0.09 W (b) 0.03 W (c) 0.3 W (d) 0.9 W (2011) 23. A straight wire carrying current i is turned into a circular loop. If the magnitude of magnetic moment associated with it in MKS unit is M, the length of wire will be

(c)

4p M 4pi M

(a) (c)

BA 2 m0 p BA 3/ 2 m0 p1/ 2

(b)

(d)

(b) (d)

4pM i Mp i

(2010)

BA 3/ 2 m0 p 2 BA 3/ 2 m0 p1/ 2



(2010)

25. A proton of mass 1.67 × 10–27 kg enters a uniform magnetic field of 1 T at point A as shown in figure, with a speed of 107 m s–1. 45° × A ×

(b) 10–17m0

20. A rectangular loop carrying a current i is placed in a uniform magnetic field B. The area enclosed by the loop is A. If there are n turns in the loop, the torque acting on the loop is given by     (a) ni A × B (b) ni A ⋅ B

(a)

24. Magnetic field at the centre of a circular loop of area A is B. The magnetic moment of the loop will be



× ×

B

×

×

×

×

× C

×

×

×

 ×

×

×

×

The magnetic field is directed normal to the plane of paper downwards. The proton emerges out of the magnetic field at point C, then the distance AC and the value of angle q will respectively be (a) 0.7 m, 45° (b) 0.7, 90° (c) 0.14 m, 90° (d) 0.14 m, 45° (2010)

26. Figure shows a straight wire of length l carrying current i. The magnitude of magnetic field produced by the current at point P is l l

P

i

(a)

2 m0i pl

(b)

(c)

2 m0i 8 pl

(d)

m0i 4 pl m0i 2 2 pl



(2010)

27. A cylindrical conductor of radius R carries a current i. The value of magnetic field at a R point which is distance inside from the 4 surface is 10 T. The value of magnetic field at point which is 4R distance outside from the surface is 4 8 (a) T (b) T 3 3

26

VITEEE CHAPTERWISE SOLUTIONS

40 80 T T (d) (2010) 3 3 28. Find the value of magnetic field between plates of capacitor at a distance 1 m from centre, where electric field varies by 1010 V/m per second. (a) 5.56 × 10–8 T (b) 5.56× 10–3 T (c) 5.56 µT (d) 5.55 T (2010) (c)

29. In the hydrogen atom, the electron is making 6.6 × 1015 rps. If the radius of the orbit is 0.53 × 10–10 m, then magnetic field produced at the centre of the orbit is (a) 140 T (b) 12.5 T (c) 1.4 T (d) 0.14 T (2010) 30. Two bar magnets A and B are placed one over the other and are allowed to vibrate in a vibration magnetometer. They make 20 oscillations per minute when the similar poles of A and B are on the same side, while they make 15 oscillations per minute when their opposite poles lie on the same side. If MA and MB are the magnetic moments of A and B and if MA > MB, the ratio of MA and MB is (a) 4 : 3 (b) 25 : 7 (c) 7 : 5 (d) 25 : 16 (2009) 31. A bar magnet, 10 cm long is kept with its north (N)-pole pointing north. A neutral point is formed at a distance of 15 cm from each pole. Given the horizontal component of earth’s field is 0.4 gauss, the pole strength of the magnet is (a) 9 A m (b) 6.75 A m (c) 27 A m (d) 13.5 A m (2009) 32. A wire of length l is bent into a circular loop of radius R and carries a current I. The magnetic field at the centre of the loop is B. The same wire is now bent into a double loop of equal radii. If both loops carry the same current I and it is in the same direction, the magnetic field at the centre of the double loop will be (a) zero (b) 2B (c) 4B (d) 8B (2009) 33. An infinitely long straight conductor is bent into the shape as shown below. It carries a current of I ampere and the radius of the circular loop is R metre. Then, the magnitude

of magnetic induction at the centre of the circular loop is R O I

m0 I 2 pR m I (c) 0 ( p + 1) 2 pR (a)

m0nI 2R m0 I (d) ( p − 1) 2 pR (2009) (b)

34. When a metallic plate swings between the poles of a magnet (a) no effect on the plate (b) eddy currents are set up inside the plate and the direction of the current is along the motion of the plate (c) eddy currents are set up inside the plate and the direction of the current oppose the motion of the plate (d) eddy currents are set up inside the plate. (2008) 35. A rectangular coil ABCD which is rotated at a constant angular velocity about an horizontal as shown in the figure. The axis of rotation of the coil as well as the magnetic field B are horizontal. Maximum current will flow in the circuit when the plane of the coil is B C

A D

(a) inclined at 30° to the magnetic field (b) perpendicular to the magnetic field (c) inclined at 45° to the magnetic field (d) parallel to the magnetic field (2007) 36. The proton of energy 1 MeV describes a circular path in plane at right angles to uniform magnetic field of 6.28 × 10–4 T. The mass of the proton is 1.7 × 10–27 kg. The cyclotron frequency of the proton is very nearly equal to (a) 107 Hz (b) 105 Hz 6 (c) 10 Hz (d) 104 Hz (2007)

27

Magnetic Effects of Electric Current and Magnetism

37. The magnetic field at the centre of a loop of a circular wire of radius r carrying current I may be taken as B0. If a particle of charge q moving with speed v passes the centre of a semicircular wire, as shown in figure, along the axis of the wire, the force on it due to the current is

I

P O

(a) zero (c)

r

Q

(b)

1 qB v 2 0

1 B qv 4 0

(d) qB0v (2007)

Answer Key 1.

(c)

2.

(b)

3.

(a)

5.

(c)

6.

(c)

7.

(b)

8.

(a)

9.

(b)

10. (c)

11. (c)

12. (b)

13.

(a)

14.

(a)

15.

(c)

16.

(d)

17. (d)

18. (c)

19. (b)

20. (a)

21.

(a)

22.

(a)

23.

(b)

24.

(d)

25. (d) 33. (c)

26. (c)

27. (b)

28. (a)

29.

(b)

30.

(b)

31.

(d)

32.

(c)

34. (c)

35. (d)

36. (d)

37.

(c)



4.

(b)

28

VITEEE CHAPTERWISE SOLUTIONS

e planations

1.

(c) : The current range and voltage range of a galvanometer imply the values of current and voltage which produce full scale deflection in the galvanometer. Thus, Current for full scale deflection, Ig = 15 mA = 15 × 10–3A Voltage for full scale deflection, Vg = 750 mV = 750 × 10–3 V The resistance of the galvanometer is V 750 × 10 −3 V G= g = = 50 W Ig 15 × 10 −3 A Let S be required shunt to convert the given galvanometer into an ammeter of range 25 A. Ig G (15 × 10 −3 A)( 50 W) S= = I − Ig 25 A − 15 × 10 −3 A (0.015 A)( 50 W) = = 0.03 W 25 A − 0.015 A 2.



(b) : If v be velocity acquired by a charged particle on being accelerated through potential difference V, then 2qV 1 2 mv = qV or v = ...(i) m 2 where q is the charge on the particle and m its mass. On entering a region of uniform magnetic field B normally, it follows a circular path of radius r and it is given by



mv m 2qV r= = qB qB m =





(using (i))

1 2mV B q

For the same values of V and B, r∝

\

m q

ra ma qp = = rp mp qa

4m e = 2 m 2e

ra = 2rp = 2 (10 cm ) = 10 2 cm

3.

(a) : If the magnet makes an angle q with the magnetic meridian, then the relation between real dip d and apparent dip d′ (the angle makes by the needle with the horizontal) is tan d tan d′ = cos q tand = tand′cosq Here, d′ = 45°, q = 30°  3 \ tand = tan45°cos30° = (1)    2    3 d = tan −1    2  4.

(b) : The magnetic induction at the centre of a circular current I carrying coil of radius R is

m0 I 2R The magnetic induction at a point on the axis of the coil at a distance x from the centre is m0 IR 2 Baxis = 2( R 2 + x 2 )3 / 2 3/ 2 Bcentre ( R 2 + x 2 )3 / 2  x2  \ = = 1 + 2  Baxis R3  R  1 As Baxis = Bcentre (given) 8 Bcentre =



8  x2  = 1 + 2  1  R 

\



4 = 1+

x2 R

2

3/ 2

or 3 =

x2 R2

x 2 = 3R 2 or x = 3R 5. (c) : Inside a magnet, its magnetic lines of force move from south pole of a magnet towards its north pole. 6.

(c) : Here,



Specific charge of the electron =



e m

= 1.7 × 1011 C kg–1



Magnetic field, B = 1.5 × 10–2 T



Speed, v = 6 × 107 m s–1

29

Magnetic Effects of Electric Current and Magnetism

The radius of the circular path is mv v r= = e eB B m 6 × 107 m s −1 = 11 (1.7 × 10 C kg −1 )(1.5 × 10 −2 T) = 2.35 × 10–2 m = 2.35 cm

As per question W1 = MB(cos 0° – cos90°) = MB(1 – 0) = MB



7.



(b) : 

The magnetic field induction at the centre O is B=

m0 I q 4 pr





8.

(a) : The situation is shown in figure. 1 S1







P B1

O1 N1

1m

B2 2m

N2 2 O2 S2

1m

The point P at which the resultant magnetic field is to be determined is on the axial line of magnetic dipole 1 and on the equatorial line of magnetic dipole 2. \ The magnetic induction at P due to dipole 1 is m 2M (10 −7 T m A −1 )( 2)(1 A m 2 ) B1 = 0 3 1 = 4 p r1 (1 m )3 = 2 × 10–7 T and that due to dipole 2 is B2 =

m0 M2 (10 −7 T m A −1 )(1 A m 2 ) = 4 p r23 (1 m )3

= 1 × 10–7 T As B1 and B2 are perpendicular to each other \ The resultant magnetic field at P is B=

As W1 = nW2 (given)

B12

+ B22

= ( 2 × 10 = 5 × 10

−7

−7

2

T) + (1 × 10

T

n=



I

O r



 1 1 = MB  1 −  = MB  2 2

\ I



and W2 = MB(cos0° – cos60°)

−7

T)

2

9. (b) : The work done in turning a magnet from position q1 to q2 in a magnetic field B is W = MB(cosq1 – cosq2)

W1 MB = =2 1 W2 MB 2

10. (c) : The revolving electron is equivalent to a current loop. If u is the frequency of revolution of the electron, then the current in the loop is I = eu ...(i) The magnetic field produced at the centre of the orbit is m I m eu B = 0 = 0 (using (i)) 2r 2r Here, m0 = 4p × 10–7 Wb A–1 m–1 e = 1.6 × 10–19 C u = 6.6 × 1015 rps r = 0.53 Å = 0.53 × 10–10 m

\ B =



( 4 p × 10 −7 )(1.6 × 10 −19 )(6.6 × 1015 ) 2(0.53 × 10 −10 ) = 12.5 Wb m–2

11. (c) : The magnetic field due to the short bar magnet of magnetic moment M at a point on its axis at a distance r from the centre of the magnet is m 2M B = 0 3 4p r Here, M = 1.25 A m2, r = 0.5 m m0 = 10–7 T m A–1 4p (10 −7 T m A −1 )( 2)(1.25 A m 2 ) \ B = (0.5 m )3

= 2 × 10–6 T = 2 × 10–6 N A–1 m–1

12. (b) : The magnetic field induction at O is E B = BCD + BDEF + BFG

D

i

r O

C 2r



Here,



m i m i BCD = BFG = 0 (sin90° + sin0°) = 0 4p r 4p r

F

i

G

30

VITEEE CHAPTERWISE SOLUTIONS

m0 ip 4p r m i m ip m i B= 0 + 0 + 0 4p r 4p r 4p r m0 i = ( p + 2) 4p r



and BDEF =



\



13. (a) : The total force acting on the electron due to electric and magnetic fields is       F = FE + FB = − eE + ( − e )( v × B)   As v and B are in the same direction, so  v × B = 0.    \ F = − eE + 0 = − eE Since the electric field opposes the motion of the electron, hence its speed will decrease. 14. (a) : The current in the ring is I = qf …(i) The magnitude of magnetic induction at the centre of the ring is m I m qf B = 0 = 0 (using (i)) 2R 2R 15. (c) : Refer to figures (a) and (b).

P 100 So, I = = = 0.8 A V 125 Z = 0.367 × 10–6 kg C–1, t = 1 min = 60 s \ m = ZIt = 0.367 × 10–6 × 0.8 × 60 = 17.6 × 10–6 kg = 17.6 mg 19. (b) : Here, q = 100 e = 1.6 × 10–17 C q = 1.6 × 10 −17 A. t m I m × 1.6 × 10 −17 Bcentre = 0 = 0 [Q r = 0.8 m] 2r 2 × 0.8 = 10–17 m0 I=

20. (a) : Torque acting on the loop,     τ = M × B = ni A × B 21. (a) : Emf induced in the coil, Df DA e=− = −B D t Dt DA or IR = −B Dt BDA |Q| = I Dt = R Here, B = 0.05 T, DA = 1 cm2 = 10–4 m2, R=2W \ Q=

0.05 × 10 −4 = 2.5 × 10 −6 C 2

22. (a) : Here, G = 0.81 W, Ig = 1 A, I = 10 A

To keep the main current I in the circuit unchanged, a resistance R is connected in series with a shunted galvanometer (figure (b)). It will be so if GS +R G +S or R = G – GS G +S





G=

=

G 2 + GS − GS G2 = G +S G +S

16. (d) 17. (d) : Here, Z = 0.126 mg C–1 = 0.126 × 10–6 kg C–1 I = 5 A, t = 1 h = 3600 s, As m = ZIt = 0.126 × 10–6 × 5 × 3600 = 2268 × 10–6 kg = 2.27 g 18. (c) : Here, P = 100 W, V = 125 V





Shunt resistance is given by, S = S=

0.81 × 1 0.81 = = 0.09 W 10 − 1 9

GI g I − Ig

.

23. (b) : Magnetic moment of a circular loop carrying current i is given by M pi



M = iA = ipr2; r =



Length of wire, l = 2 pr = 2 p

M = pi

4p M i

24. (d) : Magnetic field at the centre of the circular loop is given by B=

m0 I 2 Br ;I= 2r m0

Also, A = p r2; r =

A p

31

Magnetic Effects of Electric Current and Magnetism



Magnetic moment of the loop, M = IA or, M =

M=

2 Br 2B A= m0 m0

A A p



m0 p1/ 2



25. (d) : From the symmetry of figure, the angle q = 45°. The path of moving proton in a normal magnetic field is circular. If r is the radius of the circular path, then from the figure

AC = 2r cos 45° = 2r × As Bqv = AC =



mv 2 r

1 2

or r =

= 2r

...(i)

mv Bq

2 mv 2 × 1.67 × 10 −27 × 107 = Bq 1 × 1.6 × 10 −19 = 0.14 m

26. (c) : The given situation can be redrawn as follows. As we know the general formula for finding the magnetic due to a finite length of wire l

45° l



B=

P

i

m0 i ⋅ (sin f1 + sin f 2 ) 4p r

Here, f1 = 0°, f = 45° \

m i B = 0 ⋅ (sin 0° + sin 45°) 4p r 2 m0 i m i = 0⋅ = 4p 2 l 8 pl

27. (b) : Magnetic field inside the long cylindrical conductor, m 2 Ir Bin = 0 2 4p R Magnetic field outside the cylinder at a distance r′ from the axis m 2I B0 = 0 4p r′

\

R 3R = 4 4

B0 = 10 ×

2 B A 3/ 2



r=R−

Bin r r ′ R2 = 2 ; B0 = Bin B0 R r r′

Here, Bin = 10 T, r′ = R + 4R = 5R

8 R2 = T  3R  3  4  ( 5R)

28. (a) : Magnetic field between the plates of a capacitor at a distance r from the centre of plate, m I m df B = 0 D = 0 e0 e 2 pr 2 pr dt m0 d = (E × p r2 ) e 2 p r 0 dt m e dE m0e0r dE = 0 0 p r2 = 2 pr 2 dt dt dE Here, r = 1 m, = 1010 V m −1 s −1 dt 1 1 100 \ B= × × 1010 = × 10 −8 T 16 2 9 × 10 18 = 5.56 × 10–8 T 29. (b) : Here, u = 6.6 × 1015 rps = 6.6 × 1015 Hz q = e = 1.6 × 10–19 C, r = 0.53 × 10–10 m I = qu m I m qu Magnetic field, B = 0 = 0 (Q I = qu) 2r 2r =

4 p × 10 −7 × 1.6 × 10 −19 × 6.6 × 1015 2 × 0.53 × 10 −10

= 12.5 T

30. (b) : The frequency of vibration of a bar magnet in a vibration magnetometer is u=

1 2p

MBH I

where M and I be the magnetic moment and moment of inertia of the magnet respectively and BH is the horizontal component of earth’s magnetic field. When the similar poles of magnets A and B are on the same side, then Net magnetic moment = MA + MB Net moment of inertia = IA + IB \ The frequency of vibration is



u1 =

1 2p

( MA + MB )BH ( IA + IB )

...(i)

32

VITEEE CHAPTERWISE SOLUTIONS

When the opposite poles of A and B lie on the same side, then Net magnetic moment = MA – MB (if MA > MB) Net moment of inertia = IA + IB \ The frequency of vibration is u2 =

1 ( MA − MB )BH 2p IA + IB

...(ii)

Dividing eqn. (i) by eqn. (ii), we get u1 = u2

MA + M B MA − M B

20 = 15

MA + MB MA − MB

Here, u1 = 20 oscillations per minute u2 = 15 oscillations per minute \

MA + MB MA − MB

4 = 3

Squaring both sides, we get 16 MA + MB = 9 MA − MB 16MA – 16MB = 9MA + 9MB 7MA = 25MB MA 25 = MB 7 31. (d) : The situation is shown in figure. N

BH N

15 cm 5 cm

O

r 5 cm

S

15 cm

BH P B

S When the magnet is placed with its north pole pointing north, the neutral point (P) is obtained on the equatorial line of the magnet. \ The magnetic field due to the magnet is m m( 2l) B = 0 2 2 3/ 2 4 p (r + l )



where m is the pole strength of the magnet, r is the distance of the point on the equatorial line from its centre, 2l is the length of the magnet. At the neutral point P, B = BH m0 m( 2l) = BH \ 2 4 p r ( + l 2 )3 / 2



or m =





BH (r 2 + l 2 )3 / 2 m0 ( 2l) 4p

Here, BH = 0.4 G = 0.4 × 10–4 T (Q 1 G = 10–4 T) 2l = 10 cm = 10 × 10–2 m 10 \ l= cm = 5 cm 2 From figure, r2 = (15 cm)2– (5 cm)2 = 225 cm2 – 25 cm2 = 200 cm2 \ r2 + l2 = 200 cm2 + 25 cm2 = 225 cm2 = 225 cm2 = 225 × 10–4 m2 (0.4 × 10 −4 T) ( 225 × 10 −4 m 2 )3 / 2 \ m= (10 −7 T m A −1 ) (10 × 10 −2 m) (0.4 × 10 −4 T)((15)3 × 10 −6 m 3 ) = (10 −7 T m A −1 )(10 × 10 −2 m) = 13.5 A m 32. (c) : When the wire of length l is bent into a circular loop of radius R, then l = 2pR ...(i) The magnetic field at the centre of the loop is m I B = 0 ...(ii) 2R Now, the same wire is bent into a double loop of equal radii. If r be radius of each loop of the double loop, then l = 2 pr 2 2 pR R l or r = = = (using (i)) 4 p 4p 2

The magnetic field due to the first loop at the centre of the double loop is m I m0 I B1 = 0 = (Q r = R/2) 2r 2( R / 2)



m I  = 2  0  = 2B  2R 

(using (ii))

33

Magnetic Effects of Electric Current and Magnetism



Similarly for the second loop of the double loop, B2 = 2B Since the current in both loops is in the same direction, \ The net magnetic field at the centre of the double loop = B1 + B2 = 2B + 2B = 4B 33. (c) :

m = 1.7 × 10–27 kg, u = ?

\ u=

I

34. (c) : Eddy currents are set up when a plate swings in a magnetic field. This opposes the motion. 35. (d) : Emf induced in the rectangular coil is given by, e = NAB w sin wt

qB 2 pm

Here, q = 1.6 × 10–19 C, B = 6.28 × 10–14 T



The magnetic field at O due to straight portion is m I B1 = 0 2 pR The magnetic field at O due to circular loop is m I B2 = 0 2R Both these fields act in the same direction. \ The resultant magnetic field at O is B = B1 + B2 m I m I m I = 0 + 0 = 0 (1 + p) 2 pR 2 R 2 pR

e will be maximum if sin wt = 1. or wt = 90°. Hence current in the coil will be maximum if normal to the plane is perpendicular to the magnetic field or plane of the coil is parallel to the magnetic field.

36. (d) : We know, cyclotron frequency, u =

R O







1.6 × 10−19 × 6.28 × 10−4 2 × 3.14 × 1.7 × 10−27

= 0.94 × 104 Hz ≈ 104 Hz.

37. (c) : Magnetic field at the centre of a loop of a circular wire of radius r carrying current I, m I B0 = 0 2r

Net magnetic field at point O, is given by BN =

B0

+ Bstraight =

B0

2 2 [Q Bstraight = 0, as straight wires pass through the point O] Required force on the charge q at point O,    F = q v × BN  1 1 | F | = q vBN sin q = q vB0 sin 90° = q vB0 2 2

vvv

34

VITEEE CHAPTERWISE SOLUTIONS

4

CHAPTER

1.

2.

3.

Electromagnetic Induction and Alternating Current

A coil of resistance 10 W and an inductance 5 H is connected to a 100 V battery. The energy stored in the coil is (a) 325 erg (b) 125 J (c) 250 erg (d) 250 J (2014, 2011) A circuit has a self inductance of 1 H and carries a current of 2 A. To prevent sparking, when the circuit is switched off, a capacitor which can withstand 400 V is used. The least capacitance of capacitor connected across the switch must be equal to (a) 50 mF (b) 25 mF (c) 100 mF (d) 12.5 mF (2014)



7.

The natural frequency of the circuit shown in figure is

L

V



t

(c) 8.

(d) t

(2014)

Two coils have a mutual inductance 0.005 H. The current changes in the first coil according to equation I = I0 sinwt. where, I0 = 10 A and w = 100p rad s–1. The maximum value of emf in the second coil is (a) 2p V (b) 5p V (c) p V (d) 4p V (2014) An LCR circuit contains R = 50 W, L = 1 mH and C = 0.1 mF. The impedance of the circuit will be minimum for a frequency of 10 5 Hz (a) 2p

C

C

(a)

t

5.

A solenoid 30 cm long is made by winding 2000 loops of wire on an iron rod whose crosssection is 1.5 cm2. If the relative permeability of the iron is 600, what is the self inductance of the solenoid? (a) 1.5 H (b) 2.5 H (c) 3.5 H (d) 0.5 H (2014)

(b)

t

(c)



(2014)

L

V

(a) V

(d) 2p × 106 Hz

6.

During charging a capacitor, the variation of potential V of the capacitor with time t is shown as V

4.

(c) 2p × 105 Hz

106 Hz (b) 2p

1 2p LC 2 2p LC

1



(b)



(d) zero

2 p 2 LC (2013)

In the given circuit, if the reading of voltmeter V1 and V2 are 300 volts each, then the reading of the voltmeter V3 and ammeter A are respectively L

C R = 100 

V1

V2

V3

A

(a) 220 V, 2.2 A (c) 220 V, 2.0 A 9.

~ 220 V, 50 Hz

(b) 100 V, 2.0 A (d) 100 V, 2.2 A

(2013)

The turns ratio of a transformer is 3 : 2. If the current through the primary coil is 3 A, then the current through load resistance is

35

Electromagnetic Induction and Alternating Current

(a) 1 A (c) 2 A

(b) 4.5 A (d) 1.5 A

(2012)

10. In an AC circuit, the potential difference across an inductance and resistance joined in series are respectively 16 V and 20 V. The total potential difference across the circuit is (a) 20.0 V (b) 25.6 V (c) 31.9 V (d) 33.6 V (2012) 11. The current i in a coil varies with time as shown in the figure. The variation of induced emf with time would be i 0



T/4 T/2 3T/4 T

t

emf

(a) 0

T/4 T/2 3T/4 T

T/4 T/2 3T/4 T

t

t

emf

(c) 0

T/4

T/2 3T/4 T

t

emf

(d) 0

T/4

T T/2 3T/4

(b) 4.8 × 10–3 V (b) 48 mV

(2011)

14. An inductive circuit contains a resistance of 10 W and an inductance of 2.0 H. If an AC voltage of 120 V and frequency of 60 Hz is applied to this circuit, the current in circuit would be nearly (a) 0.32 A (b) 0.16 A (c) 0.43 A (d) 0.80 A (2011) 15. The output voltage of a transformer connected to 220 V line is 1100 V at 2 A current. Its efficiency is 100%. The current coming from the line is (a) 20 A (b) 10 A (c) 11 A (d) 22 A (2011) 16. A small coil is introduced between the poles of an electromagnet so that its axis coincides with the magnetic field direction. The number of turns is n and the cross-sectional area of the coil is A. When the coil turns through 180° about its diameter, the charge flowing through the coil is Q. The total resistance of the circuit is R. What is the magnitude of the magnetic induction? QR 2QR (a) (b) nA nA Qn QR (c) (d) (2010) 2 RA 2nA

emf

(b) 0

(a) 6 × 10–4 V (c) 6 × 10–2 V

t



(2012)

12. A resistor R, an inductor L and 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 (2012) 13. A solenoid has 2000 turns wound over a length of 0.30 m. The area of its cross-section is 1.2 × 10–3 m2. Around its central section, a coil of 300 turn is wound. If an initial current of 2 A in the solenoid is reversed in 0.25 s, then the emf induced in the coil is

17. Using an AC voltmeter the potential difference in the electrical line in a house is read to be 234 V. If line frequency is known to be 50 cycles/s, the equation for the line voltage is (a) V = 165 sin (100 pt) (b) V = 331 sin (100 pt) (c) V = 220 sin (100 pt) (d) V = 440 sin (100 pt) (2010) 18. The following series L-C-R circuit, when driven by an emf source of angular frequency 70 kilo-radians per second, the circuit effectively behaves like 100 H 1 F

10 

36

VITEEE CHAPTERWISE SOLUTIONS

(a) purely resistive circuit (b) series R-L circuit (c) series R-C circuit (d) series L-C circuit with R = 0

(2009)

19. Two identical incandescent light bulbs are connected as shown in figure. When the circuit is an AC voltage source of frequency f, which of the following observations will be correct?

(a) both bulbs will glow alternatively (b) both bulbs will glow with same brightness 1 provided f = (1 / LC ) 2p (c) bulb b1 will light up initially and goes off, bulb b2 will be ON constantly (d) bulb b1 will blink and bulb b2 will be on constantly. (2008) 20. Atransformer rated at 10 kW is used to connect a 5 kV transmission line to a 240 V circuit. The ratio of turns in the windings of a transformer (a) 5 (b) 20.8 (c) 104 (d) 40. (2008) 21. Three solenoid coils of same dimension, same number of turns and same number of layers of windings are taken. Coil 1 with inductance L1 was wound using a Mn wire of resistance 11 W/m, coil 2 with inductance L2 was wound using the similar wire but the direction of winding was reversed in each layer; coil 3 with inductance L3 was wound using a superconducting wire. The self inductance of the coils L1, L2, L3are (a) L1 = L2 = L3 (b) L1 = L2 ; L3 = 0 (c) L1 = L3 ; L2 = 0 (d) L1 > L2 > L3 (2008) 22. The transmission of high frequencies in a coaxial cable is determined by 1 , where L and C are inductance (a) ( LC )1/ 2 and capacitance

(b) (LC)2 (c) the impedance L alone (d) the dielectric and skin effect

(2007)

23. There are two solenoids of same length and inductance L0 but their diameters differ to the extent that one can just fit into the other. They are connected in three different ways in series. (1) They are connected in series but separated by large distance, (2) they are connected in series with one inside the other and senses of the turns coinciding, (3) both are connected in series with one inside the other with senses of the turns opposite as depicted in figures 1, 2 and 3, respectively. The total inductance of the solenoids in each of the case 1, 2 and 3 are respectively

1

3

2

(a) 0, 4L0, 2L0 (c) 2L0, 0, 4L0

(b) 4L0, 2L0, 0 (d) 2L0, 4L0, 0

(2007)

24. From figure shown below a series L-C-R circuit connected to a variable frequency 200 V source. L = 5 H, C = 80 µF and R = 40 W. Then the source frequency which drive the circuit at resonance is L=5H

C = 80 µF

R = 40 

V = 200 volt

(a) 25 Hz (c) 50 Hz

25 Hz p 50 (d) Hz p (b)

(2007)

25. If the coefficient of mutual induction of the primary and secondary coils of an induction coil is 5 H and a current of 10 A is cut-off in 5 × 10–4 s, the emf induced (in volt) in the secondary coil is

37

Electromagnetic Induction and Alternating Current

(a) 5 × 104 (c) 25 × 105

(b) 1 × 105 (d) 5 × 106

C = 400 mF. The frequency (in Hz) of the source at which maximum power is dissipated in the above is (a) 51.5 (b) 50.7 (c) 51.1 (d) 50.3 (2007)

(2007)

26. A voltage of peak value 283 V and varying frequency is applied to a series L-C-R combination in which R = 3 W, L = 25 mH and

Answer Key 1.

(d)

2.

9.

(b)

3.

(a)

4.

(b)

5.

(a)

6.

(a)

7.

(a)

8.

(a)

(c)

10. (b)

11. (d)

12. (d)

13.

(d)

14.

(b)

15.

(b)

16.

(d)

17. (b)

18. (c)

19. (a)

20. (b)

21.

(b)

22.

(d)

23.

(d)

24.

(b)

25. (b)

26. (d)

38

VITEEE CHAPTERWISE SOLUTIONS

e planations

(d) : Here, R = 10 W, L = 5 H, V = 100 V The current in the coil is V 100 V I= = = 10 A R 10 W The energy stored in the coil is 1 1 U = LI 2 = ( 5 H)(10 A)2 = 250 J 2 2 2. (b) : The least capacitance is such that the energy stored in capacitor is equal to that stored in inductance. If C be required capacitance, then LI 2 1 1 CV2 = LI 2 or C = 2 2 2 V Here, L = 1 H, I = 2 A, V = 400 V 1.

\

C=

(1 H) ( 2 A)

2

( 400 V)2 = 25 × 10–6 F = 25 mF

(a) : During charging a capacitor, voltage develop across capacitor at time t is V = V0(1 – e–t/RC) where V0 is the source voltage and RC is the capacitative time constant of the circuit. Hence the variation of V with t will be shown as in (a).



6.

(a) : The self inductance of the solenoid is m m N2A L= r 0 l Here, mr = 600, m0 = 4p × 10–7 H m–1 N = 2000, A = 1.5 cm2 = 1.5 × 10–4 m2 l = 30 cm = 30 × 10–2 m

L=

(600)( 4 p × 10 −7 H m −1 )( 2000)2 (1.5 × 10 −4 m 2 ) ( 30 × 10 −2 m )



= 1.5 H

7.

(a) :

(b) : The emf induced in the second coil is dI e=M dt where M is the mutual inductance between dI two coils and is the rate of change of dt current in the first coil.

As I = I0 sinwt (given) d \ e = M ( I0 sin wt) = MI0 wcoswt dt The maximum value of emf is emax = MI0w Here, M = 0.005 H, I0 = 10 A, w = 100p rad s–1 \ emax = (0.005 H)(10 A)(100p rad s–1) = 5p V

L



(a) : The impedance of the LCR circuit will be minimum for a resonant frequency ur. 1 ur = 2 p LC

L

given circuit, two capacitors are in In the series and so are the two inductors. So, their equivalent capacitance and inductance is 1 1 1 2 C = + = or Cs = Cs C C C 2 and Ls = L + L = 2L \ The natural frequency of the circuit is

u=

1 2 p Ls Cs

=

8.

1 C  2 p ( 2 L)   2

L

C R = 100 

V1

V2

=

1 2 p LC

(a) :



5.

C

C

= 0.25 × 10 −4 F

3.

4.

Here, L = 1 mH = 1 × 10–3 H C = 0.1 mF = 0.1 × 10–6 F 1 10 5 \ ur = = Hz 2 p (1 × 10 −3 H)(0.1 × 10 −6 F) 2 p

V3

A ~ 220 V, 50 Hz



As V = VR2 + (VL − VC )2

Here, V = 220 V VL = VC = 300 V

\

220 V = VR2 + ( 300 V − 300 V)2

or VR = 220 V Hence the reading of voltmeter V3 is 220 V.

39

Electromagnetic Induction and Alternating Current

VR 220 V = = 2.2 A R 100 W Hence the reading of ammeter A is 2.2 A. I=

9.

(c) : For a transformer, Ip N  Np = s or I s =  Is Np  Ns



where Ip, Np be the current and number of turns in the primary coil and Is, Ns be corresponding quantities in the secondary coil. N Ns 3 = (Q Turns ratio = s ) Here, Ip = 3 A, Np Np 2



  Ip 

2 \ I s =   ( 3 A) = 2 A 3 Hence the current through load resistance is 2 A.

10. (b) : Here, VL = 16 V, VR = 20 V The potential difference across the series LR circuit is V =

=

VL2

+ VR2

(16 V)2 + ( 20 V)2 = 25.6 V

dI dt where L is the self inductance of the coil.

11. (d) : The induced emf is e = – L

i

0









T/4 T/2 3T/4 T

t

T di , = a positive constant as 4 dt i–t graph is a straight line with positive constant slope. So, e is negative and constant. T di T For to , = 0 as i –t graph is a straight 2 dt 4 line parallel to t-axis. So, e = 0. 3T di T For to , = a negative constant as 4 dt 2 i–t graph is a straight line with negative constant slope. So, e is positive and constant. During 0 to

3T di to T, = 0 as i = 0. 4 dt So, e is zero. From the above analysis, the variation of induced emf with time will be as shown in (d). For

12. (d) : In a series LCR circuit, the current lags behind voltage if, XL > XC 1 1 1 i.e., wL > ;w> or n > wC LC 2p LC 1 As nr = 2p LC \ n > nr 13. (d) : Here, A = 1.2 × 10 –3 m2, N1 = 2000, N2 = 300 dI | 2 − ( −2)| l = 0.3 m, = = 16 A s −1 dt 0.25 dI m N N A dI Induced emf, |e| = M = 0 1 2 ⋅ dt l dt

4 p × 10 −7 × 2000 × 300 × 1.2 × 10 −3 × 16 0.30 = 48.2 × 10–3 V = 48.2 mV =

14. (b) : Impedence, Z = R 2 + XL2

Current, I =

V = Z

V 2

R + XL2 Here, R = 10 W, L = 2.0 H, V = 120 V u = 60 Hz, w = 2pu = 120 p rad s–1 XL = wL = 240 p = 754 W V = 120 V 120 \ I= = 0.16 A 2 10 + 754 2 15. (b) : Here, h = 100% = 1 P h = S ⇒ PS = PP or VSIS = VP I P P P

or 1100 × 2 = 220 × IP \

16. (d) : Induced emf, e = −

IP = 10 A

Dφ Dt

eDt Dφ =− R R nBA =− (cos θ2 − cos θ1 ) R nBA 2nBA =− (cos 180° − cos 0°) = R R QR \ B= 2 nA Q = I Dt =

17. (b) : Here, Vrms = 234 V, V0 = Vrms 2 = 234 2 = 331 V w = 50 cycles/s = 100 p rad s–1 \ Line voltage eqn., V = V0 sin wt V = 331 sin (100 pt)

40

VITEEE CHAPTERWISE SOLUTIONS 100 H 1 F



10 

18. (c) :

19. 20.

21.

22.

Here, w = 70 kilo rad s–1 = 70 × 103 rad s–1 L = 100 µH = 100 × 10–6 H, C = 1 µF = 1 × 10–6 F, R = 10 W The inductive reactance is XL = wL = (70 × 103 rad s–1)(100 × 10–6 H) = 7 W The capacitive reactance is 1 1 = wC (70 × 10 3 rad s −1 )(1 × 10 −6 F) 1 100 = W= W −2 7 7 × 10

XC =

23. (d) : Case 1 : there is no mutual inductance, Leq = L1 + L2 = L0 + L0 = 2L0 Case 2 and 3 : Due to mutual inductance, equivalent inductance is given by

As XC > XL, hence the circuit effectively behaves as series R-C circuit. (a) : This is a parallel circuit. For oscillation, the energy in L and C will be alternately maximum. (b) : A 5 kV has to be connected to a 240 V circuit. This is a step down transformer. The ratio of primary to secondary np 5000 V/ 240 V = . ns Primary has 20.8 times the turns in the secondary. (b) : The self-inductance of a coil L = m0n2Al where m0 = permeability of air, n = number of turns per unit length, A = area of cross-section and l = length of the solenoid. This depends on the geometry of the inductor such as cross-sectional area, length and number of turns and not on the material, even if it is made of a super conducting material. If the superconductor is below the critical temperature, the current will continuously flow and the inductance may not have the property of inductance any more. \ L1 = L2 ; L3 = 0. (d) : In the transmission of power through coaxial cable, the dielectric medium separating the inner conductor from outer one plays a vital role. These dielectric materials are good insulators only at low frequencies. As the frequency increases, the energy loss becomes significant.

A steady signal flowing in a wire, uniformly distributes itself throughout the cross-section of the wire. A high frequency signal, on the other hand distributes itself uniformly, there being a concentration of current on the outer surface of the conductor. If the frequency of the current is very high, the current is almost wholly confined to the surface layers. This is called Skin effect.



Leq = L1 + L2 ± 2 L1 L2 where (+) ve sign is for positive coupling and (–) ve sign is for negative coupling. L1 = L2 = L0. For positive coupling, Leq = 4L0. For negative coupling, Leq = 0

24. (b) : Resonance frequency is given by, 1 ur = 2 p LC Here, L = 5 H, C = 80 mF = 80 × 10–6 F 1 \ ur = 2 p 5 × 80 × 10−6 1 25 Hz = = p 2 p × 20 × 10−3 25. (b) : Induced emf e in the secondary is given by Dφ DI DI e=− =−M ; or | e | = M dt Dt Dt Here, M = 5 H, DI = 10 A, Dt = 5 × 10–4 s 10 \ |e| = 5 × = 1 × 10 5 V 5 × 10−4 26. (d) : Power dissipated in a series L-C-R circuit is maximum at resonance. Resonance frequency is given by 1 ur = 2 p LC Here, L = 25 mH = 25 × 10–3 H C = 400 mF = 400 × 10–6 F 1 ur = 2 × 3.14 × 25 × 10−3 × 400 × 10−6 = 50.3 Hz

vvv

41

Optics

5

CHAPTER

Optics

1.

Which has more luminous efficiency? (a) A 40 W bulb (b) A 40 W fluorescent tube (c) Both have same (d) Cannot say (2014)

2.

If e0 and m0 represent the permittivity and permeability of vacuum and e and m represent the permittivity and permeability of medium, then refractive index of the medium is given by

3.

(a)

e0m0 em

(b)

em e0m0

(c)

m0e0 e

(d)

m0e0 m

b ca c (c) b 4.

6.

A fish, looking up through the water, sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface of water, the radius of the circle in centimeter is 12 × 3 (a) (b) 12 × 3 × 5 5 12 × 3 (c) (d) 12 × 3 × 7 7 (2010)

7.

The two lenses of an achromatic doublet should have (a) equal powers (b) equal dispersive powers (c) equal ratio of their power and dispersive power (d) sum of the product of their power and dispersive power equal to zero (2009)

8.

The communication using optical fibres is based on the principle of (a) total internal reflection (b) Brewster angle (c) polarisation (d) resonance. (2008)

9.

Light travels with a speed of 2 × 108 m/s in crown glass of refractive index 1.5. What is the speed of light in dense flint glass of refractive index 1.8? (a) 1.33 × 108 m/s (b) 1.67 × 108 m/s (c) 2.0 × 108 m/s (d) 3.0 × 108 m/s. (2008)

b a

(a)

The power of a thin convex lens (ang = 1.5) is 5.0 D. When it is placed in a liquid of refractive index anl, then it behaves as a concave lens of focal length 100 cm. The refractive index of the liquid anl will be (a) 5/3 (b) 4/3 (c) 3 (d) 5/4 (2010)

(2013)

A student plots a graph between inverse of magnification 1/m produced by a convex thin lens and the object distance u as shown in figure. What was the focal length of the lens used?

1/m

5.

c

u

bc a b (d) c

(b)

(2013)

The attenuation in optical fibre is mainly due to (a) absorption (b) scattering (c) neither absorption nor scattering (d) both (a) and (b) (2010)

42

VITEEE CHAPTERWISE SOLUTIONS

10. If the speed of a wave doubles as it passes from shallow water into deeper water, its wavelength will (a) unchanged (b) halved (c) doubled (d) quadrupled. (2008)

11. Rising and setting sun appears to be reddish because (a) diffraction sends red rays to earth at these times (b) scattering due to dust particles and air molecules are responsible (c) refraction is responsible (d) polarization is responsible (2007)

Answer Key 1.

(b)

2.

(b)

9.

(b)

10. (c)

3.

(c)

11. (b)

4.

(d)

5.

(a)

6.

(c)

7.

(d)

8.

(a)

43

Optics

1.

e planations

(b) : A 40 W fluorescent tube has more luminous efficiency than 40 W bulb.

2.

(b) : The velocity of light in vacuum is 1 c= e0m0 and in medium is 1 v= em The refractive index of the medium is







em c m= = v e0m0

1 1 = u+1 m  f  This is the equation of a straight line whose 1 slope is . f From graph,

4.

1/m

b a

c

6.

\

 1 1  − 5 = (1.5 − 1)    R1 R2 

...(ii)

Also, f = –100 cm = –1 m, anl = ?  1 1  1.5 1  = − 1  −  R R −1  nl  1 2

...(iii)

From eqn. (ii) and (iii), 0.5 −1 1.5 −1= ;  1.5  nl 10  n − 1  l 

1.5 1 9 15 5 =1− = ; nl = = nl 10 10 9 3



(c) : Given situation is shown in the figure. Here h = 12 cm 4 m= 3 r tan qc = r h  r = h tan qc h  sin qc =h cos qc sin qc =h 1 − sin 2 qc



Here, sin qc =



c

c



\

5.

(a) : According to lens maker formula,

7.



1  ng − nm   1 1  = −  f  nm   R1 R2  ng = 1.5 Here, a ng = na



1 m

1 m

u

(d) : A very small part of light energy is lost from an optical fibre due to absorption or due to light leaving the fibre area resulting scattering of light sideways by impurities in the glass fibre.



1 =5D f

−5 =

(c) : According to thin lens formula 1 1 1 − = v u f Multiplying by u on both sides, we get u u −1= v f u 1 v −1= (as m = ) m f u

1 b = f c c f = b

P=

\

3.

\



r=h

1−



...(i)

=

12 × 3 7

1 m

2

=

h 2

m −1

=

12 16 −1 9



cm

(d) : The condition for an achromatic doublet is w1 w 2 + =0 f1 f2   1 P1w1 + P2w2 = 0  P = f (focal length)  

44

VITEEE CHAPTERWISE SOLUTIONS



where P1 and P2 be the powers of the lenses and w1 and w2 be their dispersive powers.



8.

(a) : The principle of transmission of information through optical fibres is the total internal reflection.



9.

(b) : Refractive index µ Velocity of light in air = Velocity of light in medium c mcrown glass = 1.5 = 2 × 108 ⇒ c = 1.5 × 2 × 108 = 3 × 108 m/s. 3 × 108 m = 1.8 = For dense flint glass cflint glass

\

cflint glass =



3 × 108 = 1.67 × 108 m/ s 1.8

10. (c) : The speed of a wave doubles as it passes from a shallow to a deeper water.

vdeep = 2 vshallow,v = ul. Assuming that frequency remains the same (by analogy with light). vdeep 2 v \ l1 deep = = shallow . u u v \ l 2 shallow = shallow . u \ ldeep = 2lshallow.

11. (b) : At the time of sunrise and sunset, the sun is near the horizon. The rays from the sun have to travel a larger part of the atmosphere. As lb < lr, and intensity of scattered light 1 ∝ , therefore, most of the blue light is l4 scattered away, only red colour, which is least scattered enters our eyes and appears to come from the sun. Hence, the sun looks red both at the time of sunrise and sunset.

vvv

45

Electromagnetic Waves and Wave Optics

6

CHAPTER

1.

2.

3.

4.

Electromagnetic Waves and Wave Optics

Beyond which frequency, the ionosphere bends any incident electromagnetic radiation but do not reflect it back towards the earth? (a) 50 MHz (b) 40 MHz (c) 30 MHz (d) 20 MHz (2014) The amplitude of an electromagnetic wave in vacuum is doubled with no other changes made to the wave. As a result of this doubling of the amplitude, which of the following statement is correct? (a) The frequency of the wave changes only. (b) The wavelength of the wave changes only. (c) The speed of wave propagation changes only. (d) None of the above is correct. (2013) If a radio receiver amplifies all the signal frequencies equally well, it is said to have high (a) sensitivity (b) selectivity (c) distortion (d) fidelity (2013) Sky wave propagation is used in (a) radio communication (b) satellite communication (c) TV communication (d) Both TV and satellite communication (2011)

5.

The frequency of a FM transmitter without signal input is called (a) the centre frequency (b) modulation (c) the frequency deviation (d) the carrier swing (2011)

6.

In an electromagnetic wave, the average energy density associated with magnetic field is B2 Li 2 (a) 0 (b) 2m 0 2

(c) 7.

8.

m0B2 2

(d)

m0

2B2



(2011)

An electromagnetic wave going through vacuum is described by E = E0 sin(kx – wt) Which of the following is/are independent of the wavelength? (a) k (b) w2 (c) k/w (d) kw2 (2011) In Young’s double slit experiment, the intensity of light at a point on the screen where the path different is l = I. The intensity of light at point where the path difference l becomes is 3 I I (a) (b) 4 3 I (d) I (2011) 2 Polarising angle for water is 53°4′. If light is incident at this angle on the surface of water and reflected the angle of refraction is (a) 53°4′ (b) 126°56′ (c) 36°56′ (d) 30°4′ (2011) (c)

9.

10. In Young’s double slit experiment, the spacing between the slits is d and wavelength of light used is 6000 Å. If the angular width of a fringe formed on a distant screen is 1°, then value of d is (a) 1 mm (b) 0.05 mm (c) 0.03 mm (d) 0.01 mm (2010) 11. In a double slit experiment, 5th dark fringe is formed opposite to one of the slits, the wavelength of light is (a)

d2 6D

(b)

d2 5D

46

VITEEE CHAPTERWISE SOLUTIONS

(c)

d2 15D

(d)

d2 9D

(2010)

12. Which of the following rays is emitted by a human body? (a) X-rays (b) UV rays (c) Visible rays (d) IR rays (2010) 13. Two light sources are said to be coherent if they are obtained from (a) two independent point sources emitting light of the same wavelength (b) a single point source (c) a wide source (d) two ordinary bulbs emitting light of different wavelengths (2010) 14. Sinusoidal carrier voltage of frequency 1.5 MHz and amplitude 50 V is amplitude modulated by sinusoidal voltage of frequency 10 kHz producing 50% modulation. The lower and upper side-band frequencies in kHz are (a) 1490, 1510 (b) 1510, 1490 1 1 , (c) 1490 1510

1 1 , (d) 1510 1490 (2010)

15. Two identical light sources S1 and S2 emit light of same wavelength l. These light rays will exhibit interference if (a) their phase differences remain constant (b) their phases are distributed randomly (c) their light intensities remain constant (d) their light intensities change randomly (2010) 16. Radio waves diffract around building although light waves do not. The reason is that radio waves (a) travel with speed larger than c (b) have much larger wavelength than light (c) carry news (d) are not electromagnetic waves (2010) 17. An AM wave has 1800 W of total power content. For 100% modulation the carrier should have power content equal to (a) 1000 W (b) 1200 W (c) 1500 W (d) 1600 W (2010)

18. Two light rays having the same wavelength l in vacuum are in phase initially. Then the first ray travels a path l1 through a medium of refractive index n1 while the second ray travels a path of length l2 through a medium of refractive index n2. The two waves are then combined to observe interference. The phase difference between the two waves is 2p 2p (l2 − l1 ) (a) (b) (n l − n2l2 ) l l 11 (c)

2p (n l − n1l1 ) l 22

(d)

2p l

 l1 l2  n − n   1 2 (2010)

19. In the Young’s double slit experiment, the intensities at two points P1 and P2 on the screen are respectively I1 and I2. If P1 is located at the centre of a bright fringe and P2 is located at a distance equal to a quarter of I fringe width from P1, then 1 is I2 1 2

(a) 2

(b)

(c) 4

(d) 16

(2009)

20. In Young’s double slit experiment, the 10th maximum of wavelength l1 is at a distance of y1 from the central maximum. When the wavelength of the source is changed to l2, 5th maximum is at a distance of y2 from its y  central maximum. The ratio  1  is  y2  2 l1 l2 l1 (c) 2l2 (a)

2l2 l1 l (d) 2 2 l1

(b)

(2009)

21. Four light sources produce the following four waves : (i) y1 = a sin (wt + f1) (ii) y2 = a sin 2wt (iii) y3 = a′ sin (wt + f2) (iv) y4 = a′ sin (3wt + f) Superposition of which two waves give rise to interference? (a) (i) and (ii) (b) (ii) and (iii) (c) (i) and (iii) (d) (iii) and (iv) (2009, 2007)

47

Electromagnetic Waves and Wave Optics

22. A radio station has two channels. One is AM at 1020 kHz and the other FM at 89.5 MHz. For good results you will use (a) longer antenna for the AM channel and shorter for the FM (b) shorter antenna for the AM channel and longer for the FM (c) same length antenna will work for both (d) information given is not enough to say which one to use for which. (2008) 23. A parallel beam of fast moving electrons is incident normally on a narrow slit. A screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statement is correct? (a) Diffraction pattern is not observed on the screen in the case of electrons (b) The angular width of the central maximum of the diffraction pattern will increase (c) The angular width of the central maximum will decrease (d) The angular width of the central maximum will remain the same. (2008) 24. Two beams of light will not give rise to an interference pattern, if (a) they are coherent (b) they have the same wavelength (c) they are linearly polarized perpendicular to each other (d) they are not monochromatic. (2008) 25. A slit of width a is illuminated with a monochromatic light of wavelength l from a distant source and the diffraction pattern is observed on a screen placed at a distance D from the slit. To increase the width of the central maximum one should (a) decrease D (b) decrease a (c) decrease l (d) the width cannot be changed. (2008) 26. A thin film of soap solution (n = 1.4) lies on the top of a glass plate (n = 1.5). When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths 420 and 630 nm. The

minimum thickness of the soap solution is (a) 420 nm (b) 450 nm (c) 630 nm (d) 1260 nm. (2008) 27. Radar waves are sent towards a moving aeroplane and the reflected waves are received. When the aeroplane is moving towards the radar, the wavelength of the wave (a) decreases (b) increases (c) remains the same (d) sometimes increases or decreases (2007) 28. The antenna current of an AM transmitter is 8 A when only the carrier is sent, but it increases to 8.93 A when the carrier is modulated by a single sine wave. Find the percentage modulation. (a) 60.1% (b) 70.1% (c) 80.1% (d) 50.1% (2007) 29. Following diffraction pattern was obtained using a diffraction grating using two different wavelengths l1 and l2. With the help of the figure identify which is the longer wavelength and their ratios. 2 1 12 1

I

12

1 2 1 

(a) l2 is longer than l1 and the ratio of the longer to the shorter wavelength is 1.5 (b) l1 is longer than l2 and the ratio of the longer to the shorter wavelength is 1.5 (c) l1 and l2 are equal and their ratio is 1.0 (d) l2 is longer than l1 and the ratio of the longer to the shorter wavelength is 2.5 (2007) 30. In Young’s double slit experiment, the interference pattern is found to have an intensity ratio between bright and dark fringes is 9. This implies that (a) the intensities at the screen due to two slits are 5 units and 4 units respectively (b) the intensities at the screen due to the two slits are 4 units and 1 units, respectively (c) the amplitude ratio is 7 (d) the amplitude ratio is 6 (2007)

48

VITEEE CHAPTERWISE SOLUTIONS

31. Indicate which one of the following statements is not correct? (a) Intensities of reflections from different crystallographic planes are equal (b) According to Bragg’s law higher order of reflections have high q values for a given wavelength of radiation

(c) For a given wavelength of radiation, there is smallest distance between the crystallographic planes which can be determined (d) Bragg’s law may predict a reflection from a crystallographic plane to be present but it may be absent due to the crystal symmetry (2007)

Answer Key 1.

(b)

2.

9.

(d)

3.

(d)

4.

(a)

5.

(a)

6.

(b)

7.

(c)

8.

(a)

(c)

10. (c)

11. (d)

12. (d)

13.

(b)

14.

(a)

15.

(a)

16.

(b)

17. (b)

18. (b)

19. (a)

20. (a)

21.

(c)

22.

(b)

23.

(c)

24.

(c)

25. (b)

26. (b)

27. (a)

28. (b)

29.

(c)

30.

(b)

31.

(a)

49

Electromagnetic Waves and Wave Optics

1.

2.



e planations

(b) : The ionosphere can reflect electromagnetic waves of frequency less than 40 MHz but do not reflect electromagnetic waves of frequency more than 40 MHz. (d) : The speed of an electromagnetic wave in vacuum is 1 c= = 3 × 108 m s −1 m0ε0

(d) : If a radio receiver amplifies all the signal frequencies equally well, it is said to have high fidelity.

4.

(a)

5.

(a)

6.

B2 (b) : UB = 2m0

7.

(c) : Here, E = E0sin(kx – wt)



2p c , w = 2 pu = 2 p l l k 1 \ = which is independent of l. w c k=



Case I : ∆x = l , f =



IR = I

2p 2p ∆x = × l = 2p l l

So, I R = 4 I0 cos 2 ( p) = 4 I0 ⇒ I0 = l 2p l 2p Case II : ∆x = , f = ⋅ = . 3 l 3 3

f 2

I 4



 2p  I So, I R′ = 4 I0 cos 2   = I0 ⇒ I R′ = .  3  4

9.

(c) : For polarisation due to reflection, qp + r = 90° \ r = 90° – qp = 90° – 53°4′ = 36°56′

10. (c) : Angular fringe width, q =

or, d =

= 3.44 × 10–5 m = 0.03 mm

11. (d) : nth

l q

l d

dark

fringe

is

formed

at

Dl 2d 9Dl \ y5 = (Q n = 5) 2d d Also, y5 = 2 yn = ( 2n − 1)



(a) : Resultant intensity, I R = 4 I0 cos 2





Here, l = 6000 Å = 6000 × 10–10 m p q = 1° = rad 180 6000 × 10 −10 \ d= p / 180

So,

8.





It is independent of amplitude, frequency and wavelength of the electromagnetic wave.

3.





d 9 Dl d2 = ;l= 2 2d 9D

12. (d) : Generally, temperature of human body is 37°C corresponding to which IR and microwave radiations are emitted from the human body. 13. (b) : Two light sources are said to be coherent if they have same wavelength, same frequency and a constant phase difference. 14. (a) : Here fc = 1.5 MHz = 1500 kHz,

fm = 10 kHz \ Lower side-band frequency



= fc – fm = 1500 kHz – 10 kHz = 1490 kHz



Upper side-band frequency



= fc + fm = 1500 kHz + 10 kHz = 1510 kHz

15. (a) 16. (b) : Diffraction takes place when the wavelength of waves is comparable with the size of the obstacle in path. The wavelength of radio waves is greater than the wavelength of light waves. Therefore, radio waves are diffracted around building. 17. (b) : Total power in AM wave

 m2  PT = PC  1 +  2  



Here, m = 1,



PT = 1800 W, PC = ?

(Q 100% modulation)

50



VITEEE CHAPTERWISE SOLUTIONS

\

frequency and having zero or a constant phase difference. Hence, superposition of y1 = a sin (wt + f1) and y3 = a′ sin (wt + f2) will give rise to interference as these two waves have same frequency w and constant phase difference (f1 – f2).

 1 3 1800 = PC  1 +  = PC  2 2

\ PC = 1200 W

18. (b) : Optical path for 1st ray = n1l1 Optical path for 2nd ray = n2l2

Phase difference, ∆f = =

2p (n l − n2 l2 ) l 1 1

2p ∆x l

19. (a) : The intensity at any point P on the screen is f I = 4 I 0 cos 2 2 where I0 is the intensity of either wave and f0 is the phase difference between the two waves. At P1, f = 0 \ I1 = 4I0 ...(i) p At P2 , f = 2 p \ I 2 = 4 I0 cos 2 = 2 I0 ...(ii) 4 Dividing eqn. (i) by eqn. (ii), we get I1 4 I 0 = =2 I 2 2 I0





20. (a) : The distance of nth maximum (nth bright fringe) from the central maximum (central bright fringe) is nlD d where l is the wavelength, D is the distance of the screen from the slits and d is the distance between the two slits. As per question, 10 l1D For 10th maximum, y1 = ...(i) d y=



5l 2 D d Dividing eqn. (i) by eqn. (ii), we get For 5th maximum, y2 =



22. (b) : The frequency of AM channel is 1020 kHz whereas for the FM it is 89.5 MHz (given). For higher frequencies(MHz), space wave communication is needed. Very tall towers are used as antennas. For 1020 kHz, ground wave propagation is used. For this, antennas need not be very tall. 23. (c) : When the speed of the electron is increased, the wavelength decreases because, l = h/p. The central maximum extends from – l/a to + l/a. As l decreases, the width decreases. The angular width of the central maxima decreases. 24. (c) : Two beams having the same wavelength, monochromatic or white radiation, having the same initial phase (coherent sources), can give interference pattern by superposition. But when their vibrations are perpendicular to each other, interference will not be possible. 25. (b) : The width of the central maximum for a given D, the distance from the screen for a given wavelength l and a, the slit width, is 2Dl/a. By decreasing a, the width can be increased. 26. (b) : For reflection at the air-soap solution interface, the phase difference is p.

...(ii)

10 l1D 10 l1 2 l1 y1 d = = = 5 l2D l2 5 l2 y2 d

21. (c) : The phenomenon of interference takes place between two waves which have same



For reflection at the interface of soap solution to glass also there will be a phase difference of p. \ The condition for maximum intensity 2mt = nl. m = 1.4, 2 × 1.4 × t = n 420 nm = (n – 1) 630 nm.

51

Electromagnetic Waves and Wave Optics



n(630 – 420) = 630, \ n(210) = 630 \ n = 630 = 3 210



2

This is the maximum order where they coincide 2 × 1.4 × t = 3 × 420 ⇒ t = 3 × 420 = 450 nm. 2 × 1.40 27. (a) : Assume radar actual frequency be be u. Apparent frequency received by the aeroplane u′. According to Doppler’s effect, u′ =



u v Here, vp is velocity of aeroplane. Also, vp > 0, So, u′ > u or, l′ < l Hence, the wave received by the aeroplane or reflected by the aeroplane has decreased wavelength. 2

PT  IT  m2 =   =1+ PC  IC  2

 8.93  m2 m2  8  = 1 + 2 ; 2 = 0.246 m2 = 0.492, m = 0.701 = 70.1%

29. (c) 30. (b) : We know,

v + vp

28. (b) : We know,

Given that, IT = 8.93 A, IC = 8 A, m = ?

Here, \



I max =9 I min

a1 + a2 =3 a1 − a2

a1 =2 a2 Intensity ratio at the screen due to slits, or 2 a1 = 4 a2 or

31. (a)

vvv

I max ( a1 + a2 )2 = I min ( a − a )2 1 2

I1 a12 = =4 I 2 a2 2

52

VITEEE CHAPTERWISE SOLUTIONS

7

CHAPTER

1.

The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minute at t = 5 min. The time (in minute) at which the activity reduces to half its value is 2 (a) log e 5 (c) 5log102

2.

Atomic and Nuclear Physics

(b)

5 log e 2

(d) 5loge2

A radioactive substance contains 10000 nuclei and its half-life period is 20 days. The number of nuclei present at the end of 10 days is (a) 7070 (b) 9000 (c) 8000 (d) 7500 (2012)

7.

A direct X-ray photograph of the intestines is not generally taken by radiologists because (a) intestines would burst on exposure to X-rays. (b) the X-rays would not pass through the intestines. (c) the X-rays will pass through the intestines without causing a good shadow for any useful diagnosis. (d) a very small exposure of X-rays causes cancer in the intestines. (2012)

8.

If a hydrogen atom in its ground state absorbs 10.2 eV of energy, the orbital angular momentum is increased by (a) 1.05 × 10–34 J s (b) 3.16 × 10–34 J s –34 (c) 2.11 × 10 J s (d) 4.22 × 10–34 J s (2012)

9.

Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of Helium nucleus is (14)1/3. The atomic number of nucleus will be (a) 25 (b) 26 (c) 56 (d) 30 (2012)

(2014)

If the electron in the hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation in term of Rydberg constant is (a)

6 5R

(b)

36 5R

(c)

64 7R

(d) none of these (2014)

3.

If in a nuclear fission, piece of uranium of mass 0.5 g is lost, the energy obtained in kW h is (a) 1.25 × 107 (b) 2.25 × 107 7 (c) 3.25 × 10 (d) 0.25 × 107 (2014)

4.

As the electron in Bohr’s orbit of hydrogen atom passes from state n = 2 to n = 1, the kinetic energy (K) and the potential energy (U) changes as (a) K four fold, U also four fold (b) K two fold, U also two fold (c) K four fold, U two fold (d) K two fold, U four fold (2013) An element with atomic number Z = 11 emits Ka X-ray of wavelength l. The atomic number of element which emits Ka X-ray of wavelength 4l is (a) 4 (b) 6 (c) 11 (d) 44 (2013)

5.

6.

10. When three a-particles are combined to form a C12 nucleus, the mass defect is (Atomic mass of 2He4 is 4.002603 u) (a) 0.007809 u (b) 0.002603 u (c) 4.002603 u (d) 0.5 u (2012) 11. If the binding energies per nucleon of Li 7 and He4 nuclei are 5.60 MeV and 7.06 MeV

53

Atomic and Nuclear Physics

respectively, then the energy of the reaction is 7

4

Li + p → 2 2He is (a) 19.6 MeV (c) 8.4 MeV

(b) 2.4 MeV (d) 17.3 MeV

(2012)

12. The graph between the square root of the frequency of a specific line of characteristic spectrum of X-ray and the atomic number of the target will be (a)





(b)



Z

Z

(c)





(d)



Z

Z

(2012)

13. In a Millikan’s oil drop experiment the charge on an oil drop is calculated to be 6.35 × 10–19 C. The number of excess electrons on the drop is (a) 3.2 (b) 4 (c) 4.2 (d) 6 (2011) 1 1 14. The values + and − of spin quantum 2 2 number shows (a) rotation of electron clockwise and anti-clockwise directions respectively (b) rotation of electron anti-clockwise and clockwise directions respectively (c) rotation in any direction according to convention (d) none of the above (2011) 15. What is the age of an ancient wooden piece if it is known that the specific activity of C14 nuclide in its amounts is 3/5 of that in freshly grown trees? Given the half life of C nuclide is 5570 yr. (a) 1000 yr (b) 2000 yr (c) 3000 yr (d) 4000 yr (2011) 16. A nucleus emits an a-particle. The resultant nucleus emits a b+-particle. The respective atomic and mass number of final nucleus will be (a) Z – 3, A – 4 (b) Z – 1, A – 4 (c) Z – 2, A – 4 (d) Z, A – 2 (2011) A ZX

17. The electron of hydrogen atom is considered to be revolving round a proton in circular orbit of radius  2/me2 with velocity e2/ , where  = h/2p. The current i is (a) (c)

4 p 2 me 5 h2



4 p 2 m2 e 2 h3

(b)

(d)

4 p 2 me 5 h3 4 p 2 m2 e 5 h3

(2010)

18. According to the Bohr’s theory of hydrogen atom, the speed of the electron, energy and the radius of its orbit vary with the principal quantum number n, respectively, as 1 1 2 1 2 1 (a) , 2 , n (b) , n , 2 n n n n (c) n2 ,

1 n2

, n2

(d) n,

1 1 , n2 n2

(2010)

19. In the Bohr model of hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If a0 is the radius of the ground state orbit, m is the mass and e is charge on the electron and e0 is the vacuum permittivity, the speed of the electron is e (a) 0 (b) e0 a0 m (c)

e 4 pe0 a0 m



(d)

4 pe0 a0 m e

(2010)

20. Two radioactive materials X1 and X2 have decay constants 10l and l respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be 1/e after a time 1 1 (a) (b) 10l 11l (c)

11 10l

(d)

1 9l

(2009)

21. Ka and Kb X-rays are emitted when there is a transition of electron between the levels (a) n = 2 to n = 1 and n = 3 to n = 1 respectively (b) n = 2 to n = 1 and n = 3 to n = 2 respectively

54

VITEEE CHAPTERWISE SOLUTIONS

(c) n = 3 to n = 2 and n = 4 to n = 2 respectively (d) n = 3 to n = 2 and n = 4 to n = 3 respectively. (2008) 22. A certain radioactive material ZXA starts emitting a and b particles successively such that the end product is Z – 3YA – 8. The number of a and b particles emitted are (a) 4 and 3 respectively (b) 2 and 1 respectively (c) 3 and 4 respectively (d) 3 and 8 respectively. (2008)

27. The half-life of a radioactive element is 3.8 days. The fraction left after 19 days will be (a) 0.124 (b) 0.062 (c) 0.093 (d) 0.031 (2007) 28. The magnetic moment of the ground state of an atom whose open sub-shell is half-filled with five electrons is (a)

25.

U has 92 protons and total 234 nucleons in its nucleus. It decays by emitting an alpha particle. After the decay it becomes (a) 232U (b) 232Pa 230 (c) Th (d) 230Ra. (2008) 234

26. Ionisation power and penetration range of radioactive radiation increases in the order (a) g, b, a and g, b, a respectively (b) g, b, a and a, b, g respectively (c) a, b, g and a, b, g respectively (d) a, b, g and g, b, a respectively (2007)

(b) 35 µB (d) µ B 35

(c) 35 µ B

(2007)

29. Identify the graph which correctly represents the Moseley’s law?

23. The radioactivity of a certain material drops to 1/16 of the initial value in 2 hours. The half-life of this radio nuclide is (a) 10 min (b) 20 min (c) 30 min (d) 40 min. (2008) 24. An observer A sees an asteroid with a radioactive element moving by at a speed = 0.3c and measures the radioactivity decay time to be TA. Another observer B is moving with the asteroid and measures its decay time as TB. Then TA and TB are related as below. (a) TB < TA (b) TB = TA (c) TB > TA (d) either (a) or (c) depending on whether the asteroid is approaching or moving away from A. (2008)

35 µ B

(a)

f

O

(c)

Z

O

f

O

f

(b)

(d)

Z

f O

Z

Z

(2007)

30. Assuming f to be frequency of first line in Balmer series, the frequency of the immediate next (i.e., second) line is (a) 0.50 f (b) 1.35 f (c) 2.05 f (d) 2.70 f (2007) 31. The radius of nucleus is (a) proportional to its mass number (b) inversely proportional to its mass number (c) proportional to the cube root of its mass number (d) not related to its mass number (2007) 32. Radio carbon dating is done by estimating in specimen the (a) amount of ordinary carbon still present (b) amount of radio carbon still present (c) ratio of amount of 14C6 to 12C6 still present (d) ratio of amount of 12C6 to 14C6 still present (2007)

Answer Key 1. 9. 17. 25.

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

2. 10. 18. 26.

(b) (a) (a) (b)

3. 11. 19. 27.

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

4. 12. 20. 28.

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

5. 13. 21. 29.

(b) (b) (a) (b)

6. 14. 22. 30.

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

7. 15. 23. 31.

(c) (d) (c) (c)

8. 16. 24. 32.

(a) (a) (a) (c)

55

Atomic and Nuclear Physics

e planations

1. (d) : The activity of the sample at time t is R = R0e–lt where l is the decay constant and R0 is the activity at time t = 0. Here, R0 = N0 counts per minute N R = 0 counts per minute e N0 1 \ = N0 e − lt or e − lt = = e −1 e e lt = 1 1 1 1 l= = = min −1 t 5 min 5

The half life of the sample is log e 2 log e 2 = = 5 log e 2 min T1/ 2 = 1 l −1 min 5 Thus the time at which the activity reduces to half its value is 5 loge2 min. 2.

(b) : The wavelength of the emitted radiation for the electron transition from ni to nf (nf < ni) is

 1 1 1 = R 2 − 2  l  n f ni    Here, ni = 3, nf = 2

3.

or

36 5R

=

4.5 × 1013 3.6 × 10

6

kW h (  1 kW h = 3.6 × 106 J)

= 1.25 × 107 kWh (a) : The kinetic energy of an electron in n orbit is



Kn =



me 4 8n2e02 h 2

or Kn ∝

1 n2

th

K 2 (1)2 1 = = or K1 = 4 K 2 K1 ( 2 ) 2 4 U 2 (1)2 1 = = or U1 = 4U 2 U1 ( 2)2 4



and



Thus, both K and U change as four fold.

5.

(b) : For Ka X-ray,



where R is the Rydberg constant and Z is the atomic number of the element. \

l1 ( Z2 − 1)2 = l 2 (Z1 − 1)2

\

l ( Z2 − 1)2 = 4 l (11 − 1)2

1 3 R(Z − 1)2 = l 4

Here, l1 = l, Z1 = 11 l2 = 4l, Z2 = ?

6.

(a) : According to Einstein’s mass-energy equivalence relation E = mc2 = (0.5 × 10–3 kg)(3 × 108 m s–1)2 = 4.5 × 1013 J

4.

l=

and potential energy of the electron in nth orbit is 1 me 4 Un = − 2 2 2 or Un ∝ 2 4n e 0 h n \ When the electron passes from n = 2 to n = 1, then



 9 − 4  5R 1 1 = R −  = R =  36  36 4 9







1 1 1 = R 2 − 2  l 3  2

\



1 ( Z2 − 1)2 = 4 (10)2 (Z2 – 1)2 = 25 Z2 – 1 = 5 Z2 = 5 + 1 = 6

(a) : The number of nuclei of a radioactive substance present after time t is t /T



 1  1/ 2 N = N0   2 where N0 is the initial number of nuclei and T1/2 is the half-life of the substance. Here, N0 = 10000, t = 10 days T1/2 = 20 days



\



1 = 10000   2

7.

1 N = 10000   2

(c)

10 days / 20 days

1/ 2

=

10000 2

= 7070

56

VITEEE CHAPTERWISE SOLUTIONS

8.

(a) : According to Bohr’s quantisation condition, nh Ln = 2p When an electron in hydrogen atom in ground state (n = 1) absorbs 10.2 eV energy, it goes to the first excited state (n = 2). \ The increase in orbital angular momentum is DL = L2 – L1 h h = − p 2p h 6.6 × 10 −34 J s = =1.05 × 10–34 J s = 2p 2( 3.14) 9. (b) : The radius of a nucleus is R = R0A1/3 where R0 is a constant and A be its mass number. 1/ 3 R R  A \ As = (14)1/3 (given) =    R RHe 4 He  A \ (14)1/3 =   4

1/ 3

A or A = 56 4 The atomic number of nucleus is Z = A – N = 56 – 30 = 26

; 14 =

4

10. (a) : The given reaction is 3 2He → 6C The mass defect for this reaction is DM = [3MHe – MC] = [3(4.002603 u) – 12 u] = 12.007809 u – 12 u = 0.007809 u

12. (b) : According to Moseley’s law υ = a(Z – b) where a and b are constants. Hence the graph between represented by (b).

15. (d) : Here, T1/2 = 5570 yr. 3 A = A0 , t = ? 5 3 5 As A = A0 e − lt ⇒ = e − lt ⇒ e lt = 5 3 5 or lt = ln   3 T 0.693  5  \ t = 1/ 2 ln   T =   0.693  3  l  5570 or t = × 0.5 = 4018.7 yr ≈ 4000 yr. 0.693 16. (a) : ZA X → 42 He + ZA−−24Y A− 4 + A− 4 Z− 2 Y → e + Z− 3 Y ′ During b+ emission. 1 1 + 1p → 0 n + b The proton changes into neutron. So, charge number decreases by 1 but mass number remains unchanged.

17. (b) : Current, i =

12

11. (d) : The binding energy of Li7 = 7(5.60 MeV) = 39.2 MeV and that of He4 = 4(7.06 MeV) = 28.24 MeV \ The energy of the given reaction Li7 + p → 22He4 is Q = 2(BE of 2He4) – BE of Li7 = 2(28.24 MeV) – 39.2 MeV = 56.48 MeV – 39.2 MeV = 17.28 MeV ≈ 17.3 MeV

14. (c)

υ and Z is

13. (b) : Excess charge on the oil drop, q = 6.35 × 10–19 C Excess number of electron on the drop, Q 6.35 × 10 −19 n= = = 3.97 = 4 e 1.6 × 10 −19



=

ev 2pr

Q e e = = t T 2pr / v ...(i)

Here, r =



2 2

,v=

2

e h ,= 2p 

me Eqn. (i) becomes, e2 e× 5 2 5  = me = 4 p me i= 2 2 p 3 h3 2p me 2 18. (a)

19. (c) : Centripetal force = Force of attraction of proton on electron e mv 2 1 e2 v = ; = 2 4 pe0 ma0 a0 4 pe0 a0 20. (d) : Let N0 be initial number of nuclei in X1 and X2. The number of nuclei of X1 after a time t is ...(i) NX = N0 e −10 lt 1

and that of X2 is NX = N0 e − lt 2 Dividing eqn. (i) by eqn. (ii), we get NX

NX

1

2

=

N0 e −10 lt N0 e − lt

1 = e −9 lt =   e

9 lt

...(ii)

57

Atomic and Nuclear Physics



As \

by an Aluminium sheet, only 0.02 mm thick. b-particles have very small mass, so their penerating power is large. g-rays have very large penetrating power.

NX

1 = (given) NX e 2 1 1 = 9 lt or t = 9l 1

21. (a) : When colliding electrons remove an electron from inner most K-shell of atom and electron from some higher shell jumps to K-shell to fill up this vacancy and characteristic X-ray of K-series are obtained.

27. (d) : Here, T1/2 = 3.8 days, t = 19 days t Number of half lives, n = =5 T1/ 2 After n half lives, remaining nuclei is given n





1 by, N = N0   2 The fraction of remaining nuclei left after 19 days is 5

22. (b) : Z X A → Z – 3Y A – 8 2a particles have been emitted decreasing A by 8 and Z by 4. One b is emitted increasing Z by 1. N N 23. (c) : In 4 × T1/2, N = 0 1 1 1 = 0 16 2 2 2 2 According to question, 4 × T1/2 = 2 hrs ⇒ T1/2 = 30 min.

( )( )( )( )

24. (a) : T =





T0 2

1 − v2 c



28. (d) : The magnetic moment of the ground state of an atom is µ = n (n + 1) µ B Here, n = 5 (number of electrons) \ µ = 5 ( 5 + 1) µ B = 35 µ B 29. (b) : According to Moseley’s law

T is the time observed by the person on earth in relative motion with respect to the asteroid. T0 is measured by the person at rest \ TA > TB. (TB = T0).

234 a 230 25. (c) : 92 U → 90 Th + a The mass number of thorium is 230 and its atomic number, Z is 90.

26. (b) : Because of large mass and large velocity a-particles have large ionising power. Each a-particle produces thousands of ions before being absorbed. The b-particles ionise the gas through which they pass, but their ionising 1 th that of a-particles. power is only 100 g-rays have got small ionising power. Because of large mass, the penetrating power of a-particles is very small, it being 1/10 times that of b-rays and 1/10000 times that of g-rays. a-particles can be easily stopped

N 1 1 = = = 0.031 N0  2  32

f = ( z − b) This is represented by graph given in option (b).

30. (b) : In the Balmer series, lower energy state is nf = 2. Frequency of spectral lines of Balmer series  1 1  υ = R Z2c  −  n2 n2  For first spectral line, n = 3. For second spectral line, n = 4.  1 1 5  4 − 9  υ1 20 36 \ = = = 3 27 υ2  1 1   4 − 16  16 f 20 = ; f ′ = 1.35 f . f ′ 27 31. (c) : Radius of a nucleus with mass number A is given by, R = R0 A1/3 ; R ∝ A1/3 Thus, the radius of nucleus is proportional to the cube root of its mass number. 32. (c)

vvv

58

VITEEE CHAPTERWISE SOLUTIONS

8

CHAPTER

Dual Nature of Radiation and Matter

1.

Silver has a work function of 4.7 eV. When ultraviolet light of wavelength 100 nm is incident on it a potential of 7.7 V is required to stop the photoelectrons from reaching the collector plate. How much potential will be required to stop photoelectrons, when light of wavelength 200 nm is incident on it? (a) 15.4 V (b) 2.35 V (c) 3.85 V (d) 1.5 V (2014, 2010)

2.

If the distance of 100 W lamp is increased from a photocell, the saturation current i in the photocell varies with the distance d as (a) i ∝ d2 (b) i ∝ d 1 1 (c) i ∝ (d) i ∝ 2 (2014) d d

3.

4.

5.

The following process is known as hu → e + + e − (a) pair production (b) photoelectric effect (c) Compton effect (d) Zeeman effect

6.

When a certain metallic surface is illuminated with monochromatic light of wavelength l, the stopping potential for photoelectric current is 3V0. When the same surface is illuminated with a light of wavelength 2l, the stopping potential is V0. The threshold wavelength for this surface for photoelectric effect is (a) 4l (b) 6l 4 (c) 8l (d) l (2013) 3

7.

A and B are two metals with threshold frequencies 1.8 × 1014 Hz and 2.2 × 1014 Hz. Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted by (Take h = 6.6 × 10–34 J s) (a) B alone (b) A alone (c) neither A nor B (d) both A and B (2012)



8.

The kinetic energy of an electron get tripled, then the de Broglie wavelength associated with it changes by a factor 1 (a) (b) 3 3 1 (c) (d) 3 (2012) 3

9.

In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is (a) 1.3 V (b) 0.5 V (c) 2.3 V (d) 1.8 V (2012)

(2014)

An eye can detect 5 × 104 photons per square metre per second of green light (l = 5000 Å) while the ear can detect 10–13 W m–2. The factor by which the eye is more sensitive as a power detector than the ear is close to (a) 5 (b) 10 (c) 106 (d) 15 (2014) A metallic surface ejects electrons when exposed to green light of intensity I but no photoelectrons are emitted when exposed to yellow light of same intensity I. It is possible to eject electrons from the same surface by (a) violet light of intensity more than I (b) yellow light of any intensity (c) red light of any intensity (d) none of these (2014)

10. The threshold frequency for a photosensitive metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly

59

Dual Nature of Radiation and Matter

(a) 2 V (c) 5 V

(b) 3 V (d) 1 V

(2012)

11. The frequency of incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is (a) double the earlier value (b) unchanged (c) more than doubled (d) less than doubled (2011) 12. Light of two different frequencies whose photons have energies 1 eV and 2.5 eV, respectively, successively illuminate a metal whose work function is 0.5 eV. The ratio of the maximum speed of the emitted electrons will be (a) 1 : 5 (b) 1 : 4 (c) 1 : 2 (d) 1 : 1 (2011) 13. An electron accelerated under a potential difference V volt has a certain wavelength l. Mass of proton is some 2000 times of the mass of the electron. If the proton has to have the same wavelength l, then it will have to be accelerated under a potential difference of (a) V volt (b) 2000 V volt (c)

V volt 2000

(d)

2000 V volt (2011)

14. The ratio of momentum of an electron and a-particle which are accelerated from rest by a potential difference of 100 V is (a) 1 (c)

 me  m   a

(b)

 2me  m   a 

(d)

 me   2m   a

(2011)

15. Light of wavelength l 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 l′ where l (a) l′ = (b) l′ = 2l 2 l < l′ < l (c) (d) l′ > l 2 (2010) 16. The maximum velocity of electrons emitted from a metal surface is v, when frequency of

light falling on it is f. The maximum velocity when frequency becomes 4f is (a) 2v (b) > 2v (c) < 2v (d) between 2v and 4v (2010) 17. The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. Light source is put on and a saturation photo-current is recorded. An electric field is switched on which has a vertically downward direction, then (a) the photo-current will increase (b) the kinetic energy of the electrons will increase (c) the stopping potential will decrease (d) the threshold wavelength will increase (2010) 18. Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2, respectively. The ratio of masses of X and Y is (a) (R1/R2) (b) (R2/R1) (c) (R1/R2)2 (d) (R2/R1)2 (2010) 19. The work function of a certain metal is 3.31 × 10–19 J. Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength 5000 Å is (Given, h = 6.62 × 10–34 J s, c = 3 × 108 m s–1, e = 1.6 × 10–19 C) (a) 2.48 eV (b) 0.41 eV (c) 2.07 eV (d) 0.82 eV (2009) 20. A photon of energy E ejects a photoelectron from a metal surface whose work function is W0. If this electron enters into a uniform magnetic field of induction B in a direction perpendicular to the field and describes a circular path of radius r, then the radius r is given by, (in the usual notation) (a)

2 m( E − W0 ) eB

(b)

(c)

2 e( E − W0 ) mB

(d)

2 m( E − W0 ) eB 2 m( E − W0 ) eB

(2009)

60

VITEEE CHAPTERWISE SOLUTIONS

21. A light whose frequency is equal to 6 × 1014 Hz is incident on a metal whose work function is 2 eV. [h = 6.63 × 10–34 Js, 1 eV = 1.6 × 10–19 J] The maximum energy of the electrons emitted will be (a) 2.49 eV (b) 4.49 eV (c) 0.49 eV (d) 5.49 eV. (2008) 22. An electron microscope is used to probe the atomic arrangements to a resolution of 5 Å. What should be the electric potential to which the electrons need to be accelerated? (a) 2.5 V (b) 6 V (c) 2.5 kV (d) 6 kV. (2008) 23. Which phenomenon best supports the theory that matter has a wave nature? (a) Electron momentum (b) Electron diffraction (c) Photon momentum (d) Photon diffraction. (2008) 24. The velocity of a particle at which the kinetic energy is equal to its rest energy is  3c  (a)   2

(b) 3

c 2

(c)

( 3c )1/ 2 2

(d)

c 3 2

(2007)

25. One electron and one proton is accelerated by equal potential. Ratio in their de-Broglie wavelength is m (a) 1 (b) e mp (c)

mp me



(d)

me mp

(2007)

26. Two electrons are moving in opposite direction with speeds 0.8 c and 0.4 c, where c is the speed of light in vacuum. Then the relative speed is about (a) 0.4 c (b) 0.8 c (c) 0.9 c (d) 1.2 c (2007) 27. A photo-sensitive material would emit electrons, if excited by photons beyond a threshold. To overcome the threshold, one would increase the (a) voltage applied to the light source (b) intensity of light (c) wavelength of light (d) frequency of light (2007)

Answer Key 1.

(d)

2.

(d)

3.

(a)

9.

(b)

10. (a)

11. (c)

17. (b)

18. (c)

19. (b)

25. (c)

26. (c)

27. (d)

4.

(a)

5.

(a)

6

(a)

7.

(b)

8.

(c)

12. (c)

13.

(c)

14.

(d)

15.

(c)

16.

(b)

20. (d)

21.

(c)

22.

(b)

23.

(b)

24.

(d)

61

Dual Nature of Radiation and Matter

e planations

1.

(d) : According to Einstein’s photoelectric equation hc eV0 = − f0 l where V0 is the stopping potential, l is the incident wavelength and f0 is the work function of the metal. hc or = eV0 + f 0 l For wavelengths l1 and l2, hc = eV01 + f 0 ...(i) l1 hc = eV02 + f 0 and ...(ii) l2 Dividing eqn. (i) by eqn. (ii), we get l 2 eV01 + f 0 = l1 eV02 + f 0 Here, l1 = 100 nm, l2 = 200 nm \

200 nm eV01 + f 0 = 100 nm eV02 + f 0 2=



eV01 + f 0 eV02 + f 0

2eV02 + 2f0 = eV01 + f0 2eV02 = eV01 – f0

eV01 − f 0 2 As V = 7.7 V and f0 = 4.7 eV (given) 01 e(7.7 V) − 4.7 eV \ eV02 = 2 7.7 eV − 4.7 eV 3 eV = = = 1.5 eV 2 2 V02 = 1.5 V eV02 =

2.

3.

(d) : Saturation current depends upon the intensity of light which is inversely proportional to the square of the distance of the lamp from the photosensitive surface. 1 Thus i ∝ 2 d (a) : The given process is known as pair production. In this process a photon of sufficient energy materializes into an electron (e –) and a positron (e+). It is illustrated in figure.

Atom

4.

h

e+ e–

(a)

5. 6

(a) (a) : According to Einstein’s photoelectric equation hc eV0 = − f0 l where V0 is the stopping potential, l is the wavelength of incident light and f0 is the work function of the surface. As per question

hc − f0 l hc and eV0 = − f0 2l Multiplying eqn. (ii) by 3, we get e( 3V0 ) =

e( 3V0 ) =

3hc − 3f 0 2l

Equating eqns. (i) and (iii), we get hc 3hc − f0 = − 3f 0 l 2l 3hc hc 3f 0 − f 0 = − 2l l hc hc 2f 0 = or f 0 = 2l 4l \ The threshold wavelength is hc hc l0 = = = 4l f 0 ( hc / 4 l ) 7.

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

...(iii)

(b) : Here, Threshold frequency of A, u0A = 1.8 × 1014 Hz Threshold frequency of B, u0B = 2.2 × 1014 Hz The work function of A is f0A = hu0A = (6.6 × 10–34 J s)(1.8 × 1014 Hz) = 11.88 × 10–20 J =

11.88 × 10 −20

eV (Q 1 eV = 1.6 × 10–19 J) 1.6 × 10 −19 = 0.742 eV and that of B is f0B = hu0B = (6.6 × 10–34 J s)(2.2 × 1014 Hz) = 14.52 × 10–20 J

62

VITEEE CHAPTERWISE SOLUTIONS



=

14.52 × 10 −20 −19

eV = 0.908 eV

1.6 × 10 Since the incident energy 0.825 eV is greater than 0.742 eV but less than 0.908 eV, so photoelectrons are emitted by A alone.

8.

(c) : de Broglie wavelength associated with an electron is h l = 2 mK

where m be mass of the electron and K its kinetic energy.



or l ∝



\



l′ = l l′ =

1 K K = K′

K 1 = 3K 3

If frequency of incident light is doubled then, KE′ = hu′ – f = 2hu – f = 2(hu – f) + f KE′ = 2 KE + f.

12. (c) : As KE = hu – f = E – f KE1 v12 E1 − f Also, = = KE2 v22 E2 − f Here, E1 = 1 eV, E2 = 2.5 eV f = 0.5 eV v12 (1 − 0.5) eV 0.5 1 \ = = = 2 2 4 v2 ( 2.5 − 0.5) eV v1 1 = v2 2

13. (c) : Wavelength of a particle,

l

3 Hence, de Broglie wavelength changes by 1 .

l=

10. (a) : According to Einstein’s photoelectric equation eV0 = hu – hu0 where, u = incident frequency u0 = threshold frequency V0 = cut-off or stopping potential h( u − u0 ) or V0 = e Here, u0 = 3.3 × 1014 Hz u = 8.2 × 1014 Hz h = 6.63 × 10–34 J s e = 1.6 × 10–19 C

V0 =

(6.63 × 10 −34 J s)(8.2 × 1014 Hz − 3.3 × 1014 Hz) 1.6 × 10 −19 C

≈2V

11. (c) : According to Einstein’s photoelectric eqn, the kinetic energy of photoelectron, KE = hu – f

h 2mqV

Here, lp = le h

3

9. (b) : Here, Kmax = 0.5 eV If V0 is the stopping potential, then Kmax = eV0 \ 0.5 eV = eV0 or V0 = 0.5 V

h = p

2mpqpVp





=

h 2me qeVe

m q  me qe Ve = mp qp Vp or Vp =  e   e  Ve  mp   qp  Here, Ve = V, mp = 2000 me, qe = qp = e 1 V \ Vp = ×1×V = volt. 2000 2000

14. (d) : P = 2mKE = 2mqV \

Here, Ve = Va = 100, qa = 2qe So,



2me qeVe Pe = Pa 2ma qaVa Pe me 1 me = ×1 = Pa ma 2 2ma

15. (c) : According to Einstein’s photoelectric equation, kinetic energy of a photoelectron is given by, hc K.E. = −f l According to question, hc hc E= − f and 2 E = −f l l′

63

Dual Nature of Radiation and Matter



\



l′ E + f 1 + f / E 1 = = > l 2E + f 2 + f / E 2 l′ 1 l > ; l′ > l 2 2

16. (b) : Maximum velocity of photoelectron,



vmax = v =

2( E − f) = m

2( hf − f) m

If the frequency becomes 4f then maximum velocity of photoelectron, 2( hf − f / 4) 2( h × 4 f − f) =2 v′ = m m \ v′ > 2v

17. (b) : In electric field, photoelectron will experience force and accelerate opposite to the field so its KE increases (i.e., stopping potential will increase), no change in photoelectric current and threshold wavelength.



According to Einstein’s photoelectric equation



hc − f0 l (6.62 × 10 −34 J s)( 3 × 108 m s −1 )

Kmax = =

( 5000 × 10 −10 m)



= 3.972 × 10–19 J – 3.31 × 10–19 J



= 0.662 × 10–19 J =



0.662 × 10 −19





2qV 1 mv 2 = qV ; v = 2 m

Now particle enters a region of uniform magnetic field and describes a circular path of radius mv m R= = qB qB

19.

1

2qV  2 mV  2 1 = × m B  q 

Here, V, q and B are same for both the charged particles So, R ∝ (m)1/2 \



...(i)

mX  R1  = mY  R2 

2

(b) : Here, Work function, f0 = 3.31 × 10–19 J Wavelength of incident radiation, l = 5000 Å = 5000 × 10–10 m

(Q 1 eV = 1.6 × 10 −19 J)

20. (d) : According to Einstein’s photoelectric equation



1 mv 2 = E − W0 2

2( E − W0 ) ...(i) m The radius of the circular path of the electron is



v=

r=

mv m = eB eB

=

18. (c) : Kinetic energy gained by charge after acceleration,

eV

= 0.41 eV



E

1.6 × 10 −19

− 3.31 × 10 −19 J

2( E − W0 ) m

(using (i))

2 m( E − W0 ) eB

21. (c) : Frequency of light = 6 × 1014 Hz. 6 1014 × 6.63 × 10−34 Energy in electron volt = × 1.6 × 10−19

= 2.49 eV. Work function of the metal = 2 eV. \ Kinetic energy maximum of photo electrons = 0.49 eV. 22. (b) : Minimum wavelength of electron = 5 Å.

l=

° h ° = 12.2 A = 5 A. 2 meV V

Accelerating potential = 6.0 V.

23. (b) : Wave nature of matter of de Broglie was proved when accelerated electrons showed diffraction by metal foil in the same manner as X-ray diffraction. 24. (d) : Rest energy = m0c2

...(i)



Relativistic kinetic energy of a particle,



K = (m – m0)c2

64



VITEEE CHAPTERWISE SOLUTIONS

    m0 − m0  c 2 m0 c 2 =  2   1− v   c2 1=



1 1 − v2 / c2 1−

v2

=

−1;2=

So, l ∝

[Using eqn (i)]

1 1 − v2 / c2

c 3 1 v2 3 ; = ;v= 4 c2 4 2

c2 25. (c) : de-Broglie wavelength is given by, l=

h = p

h

2mqV Here, q, V are same for electron and proton.

1

\

le = lp

mp

me m 26. (c) : Relative velocity is given by, u′ − v u= u′ v 1− c2 Here, u′ = 0.8 c, v = – 0.4c 0.8 c − ( −0.4 c ) 1.2 c \ u= = = 0.9c (0.8 c ) ( − 0.4 c ) 1.32 1− c2 27. (d) : The emission of photoelectrons takes place only when the frequency of the incident light is above the threshold frequency. To overcome the threshold, one would increase the frequency of light.

vvv

65

Semiconductor Devices and their Applications

9

CHAPTER

1.

Semiconductor Devices and their Applications

The output Y of the logic circuit shown in figure is best represented as

what increase in plate potential will keep the plate current unchanged? (a) 5 V (b) 10 V (c) 0.2 V (d) 50 V (2014)

A B C

2.

6.

(a) A + B ⋅ C

(b) A + B ⋅ C

(c) A + B ⋅ C

(d) A + B ⋅ C

20 

(2014)

30 

The resistance of a germanium junction diode whose V – I is shown in figure is (Vk = 0.3 V)

20 

I

5 (a) A 40

10 mA Vk 2.3 V

3.

4.

(c)

V

(a) 5 kW

(b) 0.2 kW

(c) 2.3 kW

 10  (d)   kW 2.3 

1 k

–

(a) 10 mA (c) 20 mA

(b) 15 mA (d) 5 mA

(2014)

The amplification factor of a triode is 50. If the grid potential is decreased by 0.20 V,

5 A 50

(d)

5 A 20

(2014)

8.

Mobilities of electrons and holes in a sample of intrinsic germanium at room temperature are 0.36 m2 V–1 s–1 and 0.17 m2 V–1 s–1. The electron and hole densities are each equal to 2.5 × 1019 m–3. The electrical conductivity of germanium is (a) 4.24 S m–1 (b) 2.12 S m–1 –1 (c) 1.09 S m (d) 0.47 S m–1 (2013)

9.

In a common emitter amplifier, the input signal is applied across (a) anywhere (b) emitter-collector (c) collector-base (d) base-emitter (2012)

250  15 V

(b)

To get an OR gate from a NAND gate, we need (a) only two NAND gates. (b) two NOT gates obtained from NAND gates and one NAND gate. (c) four NAND gates and two AND gates obtained from NAND gates. (d) none of these (2013)

A Zener diode, having breakdown voltage equal to 15 V is used in a voltage regulator circuit shown in figure. The current through the diode is

20 V

5 A 10

5V

7. (2014)

In a semiconductor, separation between conduction and valence band is of the order of (a) 0 eV (b) 1 eV (c) 10 eV (d) 50 eV (2014)

+

5.

The current in the circuit will be

10. If in a triode valve amplification factor is 20 and plate resistance is 10 kW, then its mutual conductance is

66

VITEEE CHAPTERWISE SOLUTIONS

(a) 2 milli mho (c) (1/2) milli mho

(b) 20 milli mho (d) 200 milli mho (2012)

11. The output waveform of a full wave rectifier is (a)





(c)



(b)



(d) (2012)

16. A potential difference of 2 V is applied between the opposite faces of a Ge crystal plate of area 1 cm2 and thickness 0.5 mm. If the concentration of electrons in Ge is 2 × 1019/m3 and mobilities of electrons and holes are 0.36 m2 V–1 s–1 and 0.14 m2 V–1 s–1 respectively, then the current flowing through the plate will be (a) 0.25 A (b) 0.45 A (c) 0.56 A (d) 0.64 A (2010) 17. Currents flowing in each of the following circuits A and B respectively are

12. A transistor is operated in common emitter configuration at VC = 2 V such that a change in the base current from 100 mA to 300 mA produces a change in the collector current from 10 mA to 20 mA. The current gain is (a) 75 (b) 100 (c) 25 (d) 50 (2012)

4

4

4

4

8V (Circuit A)

(a) 1 A, 2 A (c) 4 A, 2 A

13. For the given circuit of p-n junction diode, which of the following statements is correct ?

8V (Circuit B)

(b) 2 A, 1 A (d) 2 A, 4 A

(2009)

R

18.

(a) In forward R is V. (b) In forward R is 2V. (c) In reverse R is V. (d) In reverse R is 2V.

V

biasing the voltage across biasing the voltage across biasing the voltage across biasing the voltage across (2012)

14. Zener diode is used for (a) producing oscillations in an oscillator (b) amplification (c) stabilisation (d) rectification (2010) 15. In space charge limited region, the plate current in a diode is 10 mA for plate voltage 150 V. If the plate voltage is increased to 600 V, then the plate current will be (a) 10 mA (b) 40 mA (c) 80 mA (d) 160 mA (2010)





In the circuit shown above, an input of 1 V is fed into the inverting input of an ideal Op-amp A. The output signal Vout will be (a) +10 V (b) – 10 V (c) 0 V (d) infinity. (2008)

19. When a solid with a band gap has a donor level just below its empty energy band, the solid is (a) an insulator (b) a conductor (c) p-type semiconductor (d) n-type semiconductor. (2008) 20. A p-n junction has acceptor impurity concentration of 1017 cm–3 in the P side and donor impurity concentration of 1016 cm–3 in the N side. What is the contact potential at the junction (KT = thermal energy, intrinsic carrier concentration ni = 1.4 × 1010 cm–3)? (a) (KT/e) ln(4 × 1012) (b) (KT/e) ln(2.5 × 1023) (c) (KT/e) ln(1023) (d) (KT/e) ln(5 × 1012) (2008)

67

Semiconductor Devices and their Applications

21. A Zener diode has a contact potential of 1 V in the absence of biasing. It undergoes Zener breakdown for an electric field of 106 V/m at the depletion region of p-n junction. If the width of the depletion region is 2.5 mm,what should be the reverse biased potential for the Zener breakdown to occur? (a) 3.5 V (b) 2.5 V (c) 1.5 V (d) 0.5 V (2008) 22. In Colpitt oscillator the feedback network consists of (a) two inductors and a capacitor (b) two capacitors and an inductor (c) three pairs of RC circuit (d) three pairs of RL circuit. (2008) 23. The reverse saturation of p-n diode (a) depends on doping concentrations (b) depends on diffusion lengths of carriers (c) depends on the doping concentrations and diffusion lengths (d) depends on the doping concentrations, diffusion length and device temperature. (2008) 24. Two identical p-n junctions are connected in series in three different ways as shown below to a battery. The potential drop across the p-n junctions are equal in p n

n p 1 p n

n p

n p

p n 3

2

(a) circuits 2 and 3 (b) circuits 1 and 2 (c) circuits 1 and 3 (d) none of the circuit

(2007)

25. The temperature coefficient of a Zener mechanism is (a) negative (b) positive (c) infinity (d) zero (2007) 26. Identify the logic gate from the following truth table: Input Output A B Y 0 0 1 0 1 0 1 0 0 1 1 0 (a) NOR gate (b) NOT gate (c) AND gate (d) NAND gate (2007) 27. In Boolean algebra, A ⋅ A is equal to (a) A ⋅ B (c) A ⋅ B

(b) A + B (d) A + B

(2007)

28. The output stage of a television transmitter is most likely to be a (a) plate-modulated class C amplifier (b) grid-modulated class C amplifier (c) screen-modulated class C amplifier (d) grid-modulated class A amplifier (2007)

Answer Key 1.

(d)

2.

9.

(b)

3.

(b)

4.

(d)

5.

(b)

6.

(b)

7.

(b)

8.

(b)

(d)

10. (a)

11. (c)

12. (d)

13.

(a)

14.

(c)

15.

(c)

16.

(d)

17. (c)

18. (b)

19. (d)

20. (d)

21.

(b)

22.

(b)

23.

(d)

24.

(a)

25. (a)

26. (a)

27. (d)

28. (b)

68

1.

VITEEE CHAPTERWISE SOLUTIONS

e planations

(d) :



A B

B

A+B·C

B·C

C



The output Y of the given logic circuit is



Y = A + B ⋅C

2.

Y



10 mA



4.

From graph, Resistance of the germanium junction diode is 2V DV 2.3 V − 0.3 V R= = = DI 10 mA − 0 10 × 10 −3 A

(b) : To obtain an OR gate from NAND gates, we need two NOT gates obtained from NAND gates and one NAND gate as shown in figure. A

A Y

(b) : In a semiconductor, separation between conduction and valence band (i.e. energy gap) is the order of 1 eV. (d) : The current distribution in the circuit is shown in figure. I 250  20 V –

Iz 15 V

I 1 k

B

B



The Boolean expression is



Y = A ⋅ B = A + B (by De Morgan’s theorem)



= A + B ( A = A and B = B) It is same as of OR gate.

8. (b) : The electrical conductivity of germanium is s = e[neme + nhmh]

I

Here, Zener diode is used as a voltage regulator. \ The voltage across 1 kW = 15 V. The current through 1 kW is 15 V I′ = = 15 × 10 −3 A = 15 mA 3 1 10 × W The voltage across 250 W = 20 V – 15 V = 5 V The current through 250 W is 5V 1 I= = A = 0.02 A = 20 mA 250 W 50 The current through the zener diode is IZ = I – I′ = 20 mA – 15 mA = 5 mA 5.

(b) : The amplification factor of a triode is DVp m= DVg

(b) : In the given circuit, diode in the lower branch is forward biased so it conducts and diode in the upper branch is reverse biased so it does not conduct. \ The total resistance in the circuit = 20 W + 30 W = 50 W 5V 5 The current in the circuit is I = = A 50 W 50

= 0.2 × 103 W = 0.2 kW

+



7.

V

Vk 2.3 V

3.

6.



I

(b) :



Here, m = 50, DVg = 0.2 V \ DVp = (50)(0.2 V) = 10 V

or DVp = mDVg

Here, me = 0.36 m2 V–1 s–1 mh = 0.17 m2 V–1 s–1 ne = nh = 2.5 × 1019 m–3 \ s = (1.6 × 10 –19 C)[(0.36 m2 V–1 s–1) (2.5 × 1019 m–3) 2 –1 –1 + (0.17 m V s )(2.5 × 1019 m–3 )] = 2.12 S m–1

9.

(d) : In a common emitter amplifier, the input signal is applied across base-emitter junction.

10. (a) : The mutual conductance gm, amplification factor m and plate resistance rp of a triode are related by the relation m = rpgm m or gm = rp

Here, m = 20, rp = 10 kW = 10 × 103 W

69

Semiconductor Devices and their Applications



gm =

\

20

= 2 × 10–3 mho

10 × 10 3 W = 2 milli mho



11. (c) : The output waveform of a full wave rectifier is represented by (c). 12.

(d) : Here, Change in base current, DIB = 300 mA – 100 mA = 200 mA Change in collector current, DIC = 20 mA – 10 mA = 10 mA The current gain is DI 10 mA 10 × 10 −3 A = b= C = = 50 DI B 200 mA 200 × 10 −6 A

13. (a) : In forward biasing, the resistance of p-n junction diode is very low. So practically all the voltage will be dropped across the resistance R i.e. voltage across R is V. In reverse biasing, the resistance of p-n junction diode is very high. So the voltage across R is zero. 14. (c) : Zener diode has an unusual reverse current characteristic which is particularly suitable for voltage regulating purposes. Due to this characteristic, it is used as voltage stabilizer in electronics. 15. (c) : In space charge limited region, the plate current is given by

3/2

IP = K VP

3/ 2

3/ 2  VP   600  \ = 2 = = ( 4)3 / 2 = 8  150  I P  VP  1 1 IP = 8 IP = 8 × 10 mA = 80 mA

IP

2





2

1

16. (d) : Conductivity, s = ne (µe + µh) = 2 × 1019 × 1.6 × 10–19 (0.36 + 0.14) = 1.6 (W m)–1

0.5 × 10 −3 25 l l = = W = 8 A sA 1.6 × 10 −4 V 16 2 i= = = A = 0.64 A R 25 / 8 25



( 4 W)( 4 W) = = 2W 4W+4W (As both 4 W resistors are in parallel) The current in the circuit A is 8V I= =4A 2W 4

4 8V (Circuit B)

In circuit B, upper diode is forward biased and lower diode is reverse biased. So upper diode will conduct and lower diode will not conduct. \ The total resistance in the circuit B =4W The current in the circuit B is 8V I= =2A 4W 18. (b) : This is an operational or inverting amplifier. R V Voltage gain, A = 0 = − f V Ri i Given Vi = 1 V, Rf = 10 k W, Ri = 1 k W Rf \ Vo = −Vi = −1 × 10 1 Ri

R=ρ



4 8V (Circuit A)

⇒ Vo = –10 V. Vo is negative because Vinput is + 1 V (positive).

19. (d) : Donor level

4

17. (c) :

In circuit A, both diodes are forward biased. So both conduct. The total resistance in the circuit A

(T > 0 K)

20. (d) : Contact potential, V0 =

KT  N A N D  ln  . 2 e  ni 

Given ND = 1016 cm–3, NA = 1017cm–3,

70

VITEEE CHAPTERWISE SOLUTIONS

ni = 1.4 × 1010 cm–3, \ V0 =

17

16

KT  10 × 10  ln   e  (1.4 × 1010 ) 2 

KT ln(5 × 1012 ) . e 21. (b) : The break down field of the Zener diode = 106 V/m. The width of the depletion region = 2.5 × 10–6 m. \ Vbreakdown= E × d = 106 × 2.5 × 10–6 ⇒ Vbreakdown= 2.5 V.

26. (a) : NOR gate is a combination of OR gate and NOT gate. Output of NOR gate is inverted form of OR gate. This is illustrated in the truth table for NOR gate.

V0 =

24. (a) : In circuit 2, both p-n junctions are in reverse bias hence potential drop across them will be same. In circuit 3, both p-n junctions are in forward bias and same current flows hence they have same potential drop across them. 25. (a) : Zener breakdown occurs in heavily doped p-n junction. As the temperature increases the Zener breakdown voltage decreases. Hence, the temperature coefficient of a Zener mechanism is negative.

Output

Y=A+B

B

Y =A+B A

Y

B

Input

22. (b) : In the Colpitt oscillator, the tank circuit has a single inductor, tapped in the middle with two capacitors instead of two inductors and a single capacitor as in the case of the Hartley oscillator. 23. (d) : The width of the potential barrier and its magnitude is the most important factor in the functioning of a p-n diode. These factors depend on the nature of the semiconductor, the difusion length and the concentration of the charge carriers. The rise in temperature will affect all semiconductors.

Input

A



Output

A

B

OR(Y ′)

NOR(Y)

0

0

0

1

0

1

1

0

1

0

1

0

1

1

1

0

The Boolean expression for NOR function is

Y=A+B 27. (d) : A . B = A + B =A+B

[Using X . Y = X + Y ] [Using X = X ]

28. (b) : Because of their inherent distortion class C amplifiers are never used as audio amplifiers, but they are primarily employed in the final stage of TV transmitters. Also, owing to low efficiency of class A operations, these amplifiers are not employed where large RF (radio frequency) power is involved e.g., to excite transmitting antenna. In such cases class C amplifiers are used, since a class C amplifier has a very high efficiency, it can deliver more load power. In output stage of a TV transmitter grid modulated class C amplifier is used.

vvv

CHEMISTRY

1

Atomic Structure

1

CHAPTER

1.

Atomic Structure

According to Bohr’s theory, the angular momentum for an electron of 3rd orbit is (a) 3h (b) 1.5h h (c) 9h (d) 2 π (2014)

5.

2.

An f-shell containing 6 unpaired electrons can exchange (a) 6 electrons (b) 9 electrons (c) 12 electrons (d) 15 electrons. (2014)

6.

The wavelengths of electron waves in two orbits is 3 : 5. The ratio of kinetic energies of electrons will be (a) 25 : 9 (b) 5 : 3 (c) 9 : 25 (d) 3 : 5 (2009)

3.

A near UV photon of 300 nm is absorbed by a gas and then re-emitted as two photons. One photon is red with wavelength 760 nm. Hence, wavelength of the second photon is (a) 1060 nm (b) 496 nm (c) 300 nm (d) 215 nm (2013) What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, n = 4 to n = 2 of He+ spectrum? (a) n = 4 to n = 2 (b) n = 3 to n = 2 (c) n = 2 to n = 1 (d) n = 4 to n = 3 (2013)

7.

Electrons with a kinetic energy of 6.023 × 104 J/mol are evolved from the surface of a metal, when it is exposed to radiation of wavelength of 600 nm. The minimum amount of energy required to remove an electron from the metal atom is (a) 2.3125 × 10–19 J (b) 3 × 10–19 J (c) 6.02 × 10–19 J (d) 6.62 × 10–34 J (2009)

8.

On an X-ray diffraction photograph the intensity of the spots depends on (a) neutron density of the atoms/ions (b) electron density of the atoms/ions (c) proton density of the atoms/ions (d) photon density of the atom/ions. (2007)

4.

What is the degeneracy of the level of H-atom  RH  that has energy  − ?  9  (a) 16 (c) 4

(b) 9 (d) 1

(2013)

Answer Key

1.

(a)

2.

(d)

3.

(b)

4.

(c)

5.

(b)

6.

(a)

7.

(a)

8.

(b)

2

VITEEE CHAPTERWISE SOLUTIONS

1.

(a) : Angular momentum,



nh 3 × h mvr = = = 3h 2π 2π

2.



e planations h    h = 2π 

(d) : The number of exchanges that can take place in f-shell containing 6 unpaired electrons are as follows : 2

1 5-exchanges by electron 1



3

4

3-exchanges by electron 3

2-exchanges by electron 4

5



4-exchanges by electron 2

1-exchange by electron 5



1  1  4 − 1  2 − 2  = 4 ×  16   n1 n2  1 3  1  2 − 2= 4  n1 n2  If n1 = 1, then n2 = 2, 3, ..... For first line n2 = 2, n1 = 1 1 1 1 3 1  2 − 2= 1− 4 = 4 1 2 



Hence, transition n2 = 2 to n1 = 1 will give spectrum of the same wavelength as that of Balmer transition, n2 = 4 to n1 = 2 in He+ spectrum.

5.

(b) : Energy of the electron in the nth orbit in terms of RH is



En = −

RHZ 2



Total number of exchanges possible = 5 + 4 + 3 +2 + 1 = 15 electrons



n2 where, Z = atomic number, n2 = degeneracy

3.

(b) : Energy values are additives. E = E1 + E2 hc hc hc = + l l1 l 2 1 1 1 or = + l l1 l 2



For H-atom, En = −



−



n2 = 9

6.

(a) : According to de-Broglie’s equation,



1 1 1 = + 300 760 l 2



or



or



On solving, l2 = 495.6 nm ≈ 496 nm

4.

(c) : We have to compare wavelength of transition in the H-spectrum with the Balmer transition n = 4 to n = 2 of He+ spectrum.  lH = lHe+ 1 1 = l H l He +



1 1 1 = − l 2 300 760

1 1 1 2  1 2 \ RHZH  2 − 2  = RH ZHe +  2 − 2    2 4 n n  1 2   1 1 1 1  1×  −  = 4 ×  −  2 2  4 16   n1 n2 



RH (1) 2 n2

RH R =− H 9 n2

l=

h h2 h2 ⇒ l 2 = 2 2 ⇒ mv 2 = mv m v ml 2

\

K.E. =

1 mv 2 2

h2 1 × 2 ml12



K.E.1 = K.E.2 1 h2 × 2 ml 22 2

2

K.E.1 l 22  l 2  25 5 = = =  = 3 K.E.2 l12  l1  9



K.E.1 : K.E.2 = 25 : 9

7.

(a) : 1 mol = 6.023 × 1023 atoms K.E. of 1 mol = 6.023 × 104 J or K.E. of 6.023 × 1023 atoms = 6.023 × 104 J

3

Atomic Structure

\

K.E. of 1 atom =

hnenergy

hc = = l

6.023 × 10 4

6.023 × 10 23 = 1.0 × 10–19 J 6.626 × 10 −34 × 3 × 108 600 × 10 −9

= 3.313 × 10–19 J





Minimum amount of energy required to remove an electron from the metal ion (i.e., Threshold energy) = hn – K.E. = 3.313 × 10–19 – 1.0 × 10–19 = 2.313 × 10–19 J

8.

(b)

vvv

4

VITEEE CHAPTERWISE SOLUTIONS

2

CHAPTER

+

Which of the following has the maximum number of unpaired d-electrons? (a) Fe2+ (b) Cu+ (c) Zn (d) Ni3+ (2014)

6.

Out of H2S2O3, H2S2O4, H2SO5 and H2S2O8 peroxy acids are (a) H2S2O3, H2S2O8 (b) H2SO5, H2S2O8 (c) H2S2O4, H2SO5 (d) H2S2O3, H2S2O4 (2013)

(2012)

Which of the following oxidation states is most common among the lanthanoids? (a) 4 (b) 2 (c) 5 (d) 3 (2012)

10. Which of the following structures is the most preferred and hence of lowest energy for SO3? O S (a) (b) S O

(c)

S O

O



(d)

S

O

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

D 3 4 1 1

O

8.

Which of these ions is expected to be coloured in aqueous solution? I. Fe3+ II. Ni2+ III. Al3+ (a) I and II (b) II and III (c) I and III (d) I, II and III (2013)

C 2 3 2 3

Haber’s process Bessemer’s process Leblanc process Contact process

O

Iodine is formed when potassium iodide reacts with a solution of (a) ZnSO4 (b) CuSO4 (c) (NH4)2SO4 (d) Na2SO4 (2014)

9.

B 4 2 3 2

1. 2. 3. 4.

O

5.

7.

Codes A (a) 1 (b) 1 (c) 4 (d) 4

(2014)

The bond angles of NH3, NH4 and NH2– are in the order + (a) NH2– > NH3 > NH4 + (b) NH4 > NH3 > NH2– + (c) NH3 > NH2– > NH4 + (d) NH3 > NH4 > NH2– (2014)

Sulphuric acid Steel Sodium hydroxide Ammonia

O

4.

A. B. C. D.

List II (Processes)

O

3.

Which has the smallest size? (a) Na+ (b) Mg2+ 3+ (c) Al (d) P5+

List I (Substances)

O

2.

Purest form of iron is (a) pig iron (b) wrought iron (c) cast iron (d) steel. (2014)

O

1.

p, d and f - Block Elements

(2012)

11. “925 fine silver” means an alloy of (a) 7.5% Ag and 92.5% Cu (b) 92.5% Ag and 7.5% Cu (c) 80% Ag and 20% Cu (d) 90% Ag and 10% Cu

(2011)

12. Assertion (A) : Cu and Cd are separated by first adding KCN solution and then passing H2S gas. 2+

2+

5

p, d and f - Block Elements



Reason (R) : KCN reduces Cu2+ to Cu+ and forms a complex with it. The correct answer is (a) both (A) and (R) are true and (R) is the correct explanation of (A) (b) both (A) and (R) are true but (R) is not the correct explanation of (A) (c) (A) is true but (R) is not true (d) (A) is not true but (R) is true. (2011)

13. Consider the following equation, which represents a reaction in the extraction of chromium from its ore 2Fe2O3·Cr2O3 + 4Na2CO3 + 3O2 → 2Fe2O3 + 4Na2CrO4 + 4CO2 Which one of the following statements about the oxidation states of the substances is correct? (a) The iron has been reduced from +3 to +2 state. (b) The chromium has been oxidised from +3 to +6 state. (c) The carbon has been oxidised from +2 to +4 state. (d) There is no change in the oxidation state of the substances in the reaction. (2011) 14. Across the lanthanide series, the basicity of the lanthanide hydroxides (a) increases (b) decreases (c) first increases and then decreases (d) first decreases and then increases. (2010) 15. Sterling silver is (a) AgNO3 (b) Ag2S (c) Alloy of Ag and Cu (d) AgCl

(2010)

16. Identify the statement which is not correct regarding CuSO4. (a) It reacts with KI to give iodine. (b) It reacts with KCl to give Cu2Cl2. (c) It reacts with NaOH and glucose to give Cu2O. (d) It gives CuO on strong heating in air. (2010) 17. Transition metals usually exhibit highest oxidation states in their

(a) chlorides (c) bromides

(b) fluorides (d) iodides.

(2010)

18. Fluorine reacts with dilute NaOH and forms a gaseous product (A). The bond angle in the molecule of (A) is (a) 104°40′ (b) 103° (c) 107° (d) 109°28′ (2009) 19. The number of pp-dp bonds present in XeO3 and XeO4 molecules, respectively are (a) 3, 4 (b) 4, 2 (c) 2, 3 (d) 3, 2 (2009) 20. Which one of the following sets correctly represents the increase in the paramagnetic property of the ions? (a) Cu2+ > V2+ > Cr2+ > Mn2+ (b) Cu2+ < Cr2+ < V2+ < Mn2+ (c) Cu2+ < V2+ < Cr2+ < Mn2+ (d) V2+ < Cu2+ < Cr2+ < Mn2+ (2009) 21. The type of bonds present in sulphuric anhydride are (a) 3s and three pp-dp (b) 3s, one pp-pp and two pp-dp (c) 2s and three pp-dp (d) 2s and two pp-dp. (2009) 22. Which pair of oxyacids of phosphorus contains P– H bonds? (a) H3PO4, H3PO3 (b) H3PO5, H4P2O7 (c) H3PO3, H3PO2 (d) H3PO2,HPO3 (2009) 23. Dipping iron article into a strongly alkaline solution of sodium phosphate (a) does not affect the article (b) forms Fe2O3·xH2O on the surface (c) forms iron phosphate film (d) forms ferric hydroxide. (2008) 24. Reaction of PCl3 and PhMgBr would give (a) bromobenzene (b) chlorobenzene (c) triphenylphosphine (d) dichlorobenzene. (2008) 25. Which of the following is not a characteristic of transition elements? (a) Variable oxidation states (b) Formation of coloured compounds (c) Formation of interstitial compounds (d) Natural radioactivity (2008)

6

VITEEE CHAPTERWISE SOLUTIONS

26. Cl – P – Cl bond angles in PCl5 molecule are (a) 120° and 90° (b) 60° and 90° (c) 60° and 120° (d) 120° and 30° (2008) 27. The magnetic moment of a salt containing Zn2+ ion is (a) 0 (b) 1.87 (c) 5.92 (d) 2 (2008) 28. Hair dyes contain (a) copper nitrate (b) gold chloride (c) silver nitrate (d) copper sulphate.

(2008)

29. Which of the following compounds volatilises on heating? (a) MgCl2 (b) HgCl2 (c) CaCl2 (d) FeCl3 (2007) 30. AgCl dissolves in a solution of NH3 but not in water because (a) NH3 is a better solvent than H2O (b) Ag+ forms a complex ion with NH3 (c) NH3 is a stronger base than H2O (d) the dipole moment of water is higher than NH3. (2007) 31. Among the following the compound that is both paramagnetic and coloured is (a) K2Cr2O7 (b) (NH4)2[TiCl6] (c) VOSO4 (d) K3[Cu(CN)4] (2007)

Answer Key 1.

(b)

2.

(d)

3.

(b)

9.

(d)

10. (d)

11. (b)

17. (b)

18. (b)

25. (d)

26. (a)

4.

(a)

5.

(b)

6.

(b)

7.

(a)

8.

(d)

12. (b)

13.

(b)

14.

(b)

15.

(c)

16.

(b)

19. (a)

20. (c)

21.

(b)

22.

(c)

23.

(a)

24.

(c)

27. (a)

28. (c)

29.

(b)

30.

(b)

31.

(c)

7

p, d and f - Block Elements

e planations

1.

(b) : Wrought or malleable iron is the purest form of iron.

2.

(d) : For cationic species, atomic decreases with increase in charge.

3.

(b) : As the number of lone pairs of electrons increases, bond angle decreases. Thus, the order of bond angle is



NH4 > NH3 > NH 2– . (1 l⋅p) (2 l⋅p)

(b) : Peroxy acids contain S

O O



H2S2O3 ⇒ H O S O H

linkage.

O

O

O

H2S2O8 ⇒ HO

S

H

O



O O

O

S

OH

II. Ni2+ (3d8) :

3d

5 unpaired electrons, coloured ion.

17. (b) : Transition metals usually exhibit highest oxidation states in their fluorides. This is due to the highest reduction potential of fluorine.

5

3d

18. (b) : 2F2 + 2NaOH(dil.)  OF2  + 2NaF + H2O

2 unpaired electrons, coloured ion.

III. Al = [Ne] No unpaired electron thus, colourless ion. 3+

+4

16. (b) : CuSO4 reacts with Kl to give white precipitate of Cu2I2 and to evolve I2. But it does not react with KCl in the same way.

O

(a) : I. Fe (3d ) : 3+

+6

14. (b) : As the ionic size of lanthanide decreases, the covalent character of their hydroxide increases due to which, their basicity decreases. 15. (c) : Sterling silver is an alloy of 92.5% Ag and 7.5% other metals, usually copper.

(A)

The structure of A(OF2) is pm

O

+4

2 Fe 2O 3 + 4 Na 2Cr O 4 + 4 CO 2

141

S

+3

+3

H2S2O4 ⇒ HO S S OH

O



12. (b) : KCN forms complex with Cu+ and Cd2+ as K3[Cu(CN)4] and K2[Cd(CN)4] respectively. K2[Cd(CN)4] is less stable than K3[Cu(CN)4] thus, the solution contains more free Cd2+ than Cu2+ and so, only for CdS, the Ki > Ksp. Hence, on passing H2S only Cd2+ complex gets precipitated as yellow ppt. of CdS. +3

O

7.

11. (b) : “925 fine silver” means 925 parts of pure Ag in 1000 parts of an alloy. Therefore, in terms of percentage, it will be 92.5% Ag and 7.5% Cu.

13. (b) : 2 Fe 2O 3 ⋅ Cr2O 3 + 4 Na 2CO 3 + 3O 2

O O

H2SO5 ⇒ HO

:

6.

Iodine

:

Cuprous iodide



S 10. (d) : O O has maximum number of covalent bonds and hence is most stable.

:

(no l⋅p)

5. (b) : [CuSO4 + 2KI → CuI2 + K2SO4] × 2 2CuI2 → 2CuI + I2 2CuSO4 + 4KI → 2CuI + 2K2SO4 + I2



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

+

4. (a) : Fe = [Ar] 3d6 4s0 ⇒ 4 unpaired electrons Cu+ = [Ar] 3d10 4s0 ⇒ 0 unpaired electron Zn = [Ar] 3d10 4s2 ⇒ 0 unpaired electron Ni3+ = [Ar] 3d7 4s0 ⇒ 3 unpaired electrons



9.

size

2+



(d) : Sulphuric acid is manufactured by Contact process, steel is manufactured by Bessemer’s process, Leblanc process is used for the production of NaOH while NH3 is obtained by Haber’s process.

:



8.

F

O 103° F

8

VITEEE CHAPTERWISE SOLUTIONS

Due to repulsion between two lone pairs of electrons, its shape gets distorted. Thus bond angle of molecule (OF2) is 103°.

Xe



H4P2O7 



19. (a) : Structure of XeO3 :

O O O

O



Structure of XeO4 : O

Xe



O O

O



 Four p - d bonds

20. (c) : Ion Electronic configuration No. of unpaired electrons



Cu2+

V2+

Cr2+

Mn2+

3d9

3d3

3d4

3d5

1

3

4

5

Greater the number of unpaired electrons, more is the paramagnetism. Thus, the correct order is Cu2+ < V2+ < Cr2+ < Mn2+.

21. (b) : Sulphuric anhydride is SO3. It is sp2 hybridized. The three sp2 orbitals of sulphur overlap with p-orbitals of oxygen to form three S – O, s-bonds. The sulphur atom is now left with one p-and two d-orbitals which overlap with p-orbitals of oxygen to form three p-bonds. 3s

3p

O

H3PO3  H



OH

P

HO HO

P

O

OH OH

OH

H O HPO3  HO

P

O

23. (a) : As Fe is less reactive than Na it cannot displace Na from sodium phosphate, so, no reaction takes place. 24. (c) : Reaction between PCl3 and PhMgBr gives triphenyl phosphine. 3C6H5MgBr + PCl3 → (C6H5)3P + 3MgClBr 25. (d) : The variable oxidation states of a transition metals are due to the involvement of (n — 1)d and ns electrons. The colour of transition metal ions arises from the excitation of electrons from the d-orbitals of lower energy to the d-orbitals of higher energy. Small non-metallic atoms such as H, B, C, N, etc. are able to occupy interstitial spaces of the lattices of the d-block elements to form combinations which are termed interstitial compounds. But natural radioactivity is the characteristic of f-block elements. 26. (a) : Shape of PCl5 is



axial angles are 90° and equatorial angles are 120°.

27.

(a) : Magnetic moment, µ = n (n + 2) B.M where n = number of unpaired electrons Zn2+ : 3d10 4s0 i.e number of unpaired electrons is zero hence, magnetic moment will also be zero.

28. (c) : Hair dye contains silver nitrate. 29. (b) : HgCl2 compound is easily volatile. They are insoluble in water and soluble in acids.

OH

OH O H3PO5 



P OH

O

P

P

3d

S in the second excited sp2-hybridisation, Forms one Forms two state p - p double p - d double hybrid orbitals form three S – O, bond with O bonds with O  bonds

22. (c) : H3PO4  HO

P O HO OH O

H3PO2  H

 Three p - d bonds

O

OH

30. (b) : AgCl dissolves in the solution of NH3 but not in water because Ag+ forms a complex ion with NH3. [Ag(NH3)2]Cl AgCl + 2NH3 Diammine silver (I) chloride

9

p, d and f - Block Elements

31. (c) : In K2Cr2O7, O.S. of Cr is +6 i.e. Cr6+, the electronic configuration of Cr6+ is 3d0 4s0 i.e. it has no unpaired electron. Thus it is diamagnetic and colourless (absence of d-electrons) In (NH4)2[TiCl6], the O.S. of Ti is +4, i.e., Ti4+. The configuration of Ti4+ is 3d0 4s0, i.e., it has no unpaired electron, hence it is diamagnetic and colourless (because of absence of d-electrons).





vvv

In VOSO4, the O.S. of vanadium in VOSO4 is +4. V[Z = 23] : [Ar]3d34s2 V4+ : [Ar]3d14s0 It has one unpaired electron hence, it is paramagnetic and coloured (because of presence of d-electrons). In K3[Cu(CN)4], the O.S. of Cu is +1, i.e., Cu+. Its configuration is 3d10 4s0. It has no unpaired electron so it is diamagnetic and colourless.

10

VITEEE CHAPTERWISE SOLUTIONS

3

CHAPTER

Coordination Chemistry and Solid State Chemistry

1.

A metal has bcc structure and the edge length of its unit cell is 3.04 Å. The volume of the unit cell in cm3 will be (a) 1.6 × 1021 (b) 2.81 × 10–23 –23 (c) 6.02 × 10 (d) 6.6 × 10–24 (2014)

6.

2.

Among [Fe(H2O)6]3+, [Fe(CN)6]3–, [FeCl6]3– species, the hybridisation state of the Fe atom are, respectively (a) d2sp3,d2sp3, sp3d2 (b) sp3d2, d2sp3, d2sp3 (c) sp3d2, d2sp3, sp3d2 (d) none of the above. (2014)

7.

3.

4.

The density of solid argon is 1.65 g per cc at –233°C. If the argon atom is assumed to be a sphere of radius 1.54 × 10–8 cm, what percent of solid argon is apparently empty space? (Ar = 40) (a) 16.5% (b) 38% (c) 50% (d) 62% (2013) Which of the tetradentate ? (a) NH2 (b)

(c)

N H

ligands

is

N N

NH NH

N CH3

N H2

C



O

(d) CH2 C

O

(2013)

CH3

What is the EAN of [Al(C2O4)3]3– ? (a) 28 (b) 22 (c) 16 (d) 10

8.

A solid compound XY has NaCl structure. If the radius of the cation (X+)is 100 pm, the radius of the anion (Y–) will be (a) 275.1 pm (b) 322.5 pm (c) 241.5 pm (d) 165.7 pm (2012)

9.

Which of the following complex compounds will exhibit highest paramagnetic behaviour?



(At. no. Ti = 22, Cr = 24, Co = 27, Zn = 30) (a) [Ti(NH3)6]3+ (b) [Cr(NH3)6]3+ 3+ (c) [Co(NH3)6] (d) [Zn(NH3)6]2+ (2012)

10. Amongst Ni(CO)4, [Ni(CN)4]2– and [NiCl4]2– (a) Ni(CO)4 and [NiCl4]2– are diamagnetic but [Ni(CN)4]2– is paramagnetic (b) Ni(CO)4 and [Ni(CN)4]2– are diamagnetic but [NiCl4]2– is paramagnetic (c) [NiCl4]2– and [Ni(CN)4]2– are diamagnetic but Ni(CO)4 is paramagnetic (d) Ni(CO)4 is diamagnetic but [NiCl4]2– and [Ni(CN)4]2– are paramagnetic. (2011)

NH2

NH2

5.

following

Which one of the following complexes is not expected to exhibit isomerism? (a) [Ni(NH3)4(H2O)2]2+ (b) [Pt(NH3)2Cl2] (c) [Ni(NH3)2Cl2] (d) [Ni(en)3]2+ (2012) Which of the following carbonyls will have the strongest C—O bond? (a) [Mn(CO)6]+ (b) Cr(CO)6 (c) [V(CO)6]– (d) Fe(CO)5 (2012)

(2013)

11. In which of the following octahedral complexes of Co (At. no. 27), will the magnitude of Do be the highest ? (a) [Co(CN)6] 3– (b) [Co(C2O4)3]3– 3+ (c) [Co(H2O)6] (d) [Co(NH3)6]3+ (2011)

11

Coordination Chemistry and Solid State Chemistry

12. The effective atomic number of cobalt in the complex [Co(NH3)6]3+ is (a) 36 (b) 24 (c) 33 (d) 30 (2011) 13. The IUPAC name for the complex [Co(NO2)(NH3)5]Cl2 is (a) nitrito-N-pentaammine cobalt (III) chloride (b) nitrito-N-pentaammine cobalt (II) chloride (c) pentaamminenitrito-N-cobalt (II) chloride (d) pentaamminenitrito-N-cobalt (III) chloride. (2011) 14. Tetragonal crystal system has the following unit cell dimensions (a) a = b = c, a = b = g = 90° (b) a = b ≠ c, a = b = g = 90° (c) a ≠ b ≠ c, a = b = g = 120° (d) a = b ≠ c, a = b = 90°, g = 120° (2011) 15. A crystalline solid (a) changes rapidly from solid to liquid (b) has no definite melting point (c) undergoes deformation of its geometry easily (d) soften easily. (2011) 16. The correct formula of the complex tetraammineaquachloridocobalt(III) chloride is (a) [Cl(H2O)(NH3)4Co]Cl (b) [CoCl(H2O)(NH3)4]Cl (c) [Co(NH3)4(H2O)Cl]Cl (d) [CoCl(H2O)(NH3)4]Cl2 (2010) 17. Frenkel defect is generally observed in (a) AgBr (b) AgI (c) ZnS (d) all of the above. (2010) 18. Most crystals show good cleavage because their atoms, ions or molecules are (a) weakly bonded together (b) strongly bonded together (c) spherically symmetrical (d) arranged in planes. (2010) 19. [Co(NH3)4Cl2]NO2 and [Co(NH3)4ClNO2]Cl exhibit which type of isomerism? (a) Geometrical (b) Optical (c) Linkage (d) Ionisation (2010) 20. Which of the following compounds is not coloured?

(a) Na2[CuCl4] (c) K4[Fe(CN)6]

(b) Na[CdCl4] (d) K3[Fe(CN)6]

(2010)

21. Coordination number of Ni in [Ni(C2O4)3]4– is (a) 3 (b) 6 (c) 4 (d) 5 (2010) 22. Mg is an important component of which biomolecule occurring extensively in living world? (a) Haemoglobin (b) Chlorophyll (c) Florigen (d) ATP (2010) 23. The radius of Na+ is 95 pm and that of Cl– ion is 181 pm. Hence, the coordination number of Na+ will be (a) 4 (b) 6 (c) 8 (d) unpredictable. (2010) 24. The cubic unit cell of a metal (molar mass = 63.55 g mol–1) has an edge length of 362 pm. Its density is 8.92 g cm–3. The type of unit cell is (a) primitive (b) face centred (c) body centred (d) end centred. (2009) 25. The number of unpaired electrons calculated in [Co(NH3)6]3+ and [CoF6]3– are (a) 4 and 4 (b) 0 and 2 (c) 2 and 4 (d) 0 and 4 (2008) 26. The IUPAC name of the given compound [Co(NH3)5Cl]Cl2 is (a) pentaaminocobaltchloride chlorate (b) cobaltpentaaminenchloro chloride (c) pentaamminechlorocobalt(III) chloride (d) pentaaminocobalt(III) chlorate (2008) 27. When SCN – is added to an aqueous solution containing Fe(NO3)3, the complex ion produced is (a) [Fe(OH2)2(SCN)]2+ (b) [Fe(OH2)5(SCN)]2+ (c) [Fe(OH2)8(SCN)]2+ (d) [Fe(OH2)(SCN)]6+ (2008) 28. Schottky defects occur mainly in electrovalent compounds where (a) positive ions and negative ions are of different size (b) positive ions and negative ions are of same size

12

VITEEE CHAPTERWISE SOLUTIONS

(c) positive ions are small and negative ions are big (d) positive ions are big and negative ions are small. (2008) 29. When an ion leaves its regular site and occupies a position in the space between the lattice sites, it is called (a) Frenkel defect (b) Schottky defect (c) impurity defect (d) vacancy defect. (2007) 30. The 8 : 8 type of packing is present in (a) MgF2 (b) CsCl (c) KCl (d) NaCl

(2007)

31. Which of the following is hexadentate ligand? (a) Ethylenediamine (b) Ethylenediaminetetraacetic acid (c) 1, 10-Phenanthroline (d) Acetylacetonate (2007) 32. Which of the following complexes has zero magnetic moment (spin only)? (a) [Ni(NH3)6]Cl2 (b) Na3[FeF6] (c) [Cr(H2O)6]SO4 (d) K4[Fe(CN)6] (2007) 33. The IUPAC name of [Ni(PPh3)2Cl2] is (a) bis dichloro (triphenylphosphine) nickel(II) (b) dichloro bis (triphenylphosphine) nickel (II) (c) dichloro triphenylphosphine nickel(II) (d) triphenyl phosphine nickel(II) dichloride. (2007)

Answer Key 1.

(b)

2.

9.

(b)

10. (b)

11. (a)

17. (d)

18. (d)

25. (d)

26. (c)

33. (b)

(c)

3.

(d)

4.

(c)

5.

(b)

6.

(c)

7.

(a)

8.

(c)

12. (a)

13.

(d)

14.

(b)

15.

(a)

16.

(d)

19. (d)

20. (c)

21.

(b)

22.

(b)

23.

(b)

24.

(b)

27. (b)

28. (b)

29.

(a)

30.

(b)

31.

(b)

32.

(d)

13

Coordination Chemistry and Solid State Chemistry

e planations

1. (b) : Volume of unit cell (V) = a3 = (3.04 × 10–8 cm)3 = 2.81 × 10–23 cm3 4s

4p

4d

[Fe(H2O)6] : As H2O is a weak field ligand, pairing of electrons does not take place. 4s

4p

4d

sp3d 2 hybridisation



[Fe(CN6)] : As CN – is a strong field ligand, pairing of electrons takes place. 4s 4p 3d 3–







[FeCl6]3– : As Cl– is a weak field ligand, pairing of electrons does not take place. 4p 3d 4s 4d

4 3 pr 3

3.

(d) : Volume of one molecule =



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



1.65 × N 0 × 1.53 × 10 −23 = 0.380 cm3 40 Volume of solid containing 1.65 g of Ar = 1 cm3 \ Empty space = 1 – 0.380 = 0.620 \ Percent of empty space = 62%

4.

(c) :

 : NH2;

N

(b) Tridentate

N

:

(a) Tridentate

N

:

H2N: 

N H

:



=

:



:

CH2

C O 

NH2

CH3 (d) Bidentate

5.

(b) : Effective atomic number (EAN) = Atomic number – Oxidation number + 2 × Coordination number For [Al(C2O4)3]3 –,



Z = 13, Oxidation no. = + 3, Coordination no. = 6 \ EAN = 13 – 3 + 2 × 6 = 22

6.

(c) : [Ni(NH3)2Cl2] has tetrahedral geometry and thus, does not exhibit isomerism due to presence of symmetry elements.

7. d2sp3 hybridisation

sp3d2 hybridisation



C O 

(c) Tetradentate

3+

3d

NH

CH3

:

(c) : Fe(26) : 3d64s2 3d



NH

: :

2.

NH2

(a) : As positive charge on the central metal atom increases, the less readily the metal can donate electron density into the anti-bonding p-orbitals of CO ligand to weaken the C O bond. Hence, the C O bond would be strongest in [Mn(CO)6 ]+. r 8. (c) : Radius ratio of NaCl like crystal = + r− = 0.414 100 or r− = = 241.5 pm 0.414 9.

(b) : Ti : [Ar] 3d24s2, Ti3+ : [Ar] 3d14s0 (1 unpaired electron)



Cr : [Ar] 3d44s2, Cr3+ : [Ar] 3d34s0 (3 unpaired electrons)



Co : [Ar] 3d74s2, Co3+ : [Ar] 3d64s0 (0 unpaired electrons because of pairing)



Zn : [Ar] 3d104s2, Zn2+ : [Ar] 3d10 (no unpaired electrons) [Cr(NH3)6]3+ exhibits highest paramagnetic behaviour as it contains 3 unpaired electrons.



10. (b) : Ni(28) : [Ar] 3d84s2 Ni2+ : [Ar] 3d8 Ni and Ni2+ both have two unpaired electrons.

14

VITEEE CHAPTERWISE SOLUTIONS

CO and CN– are strong field ligands and thus, unpaired electrons get paired. Hence, Ni(CO)4 and [Ni(CN)4]2– are diamagnetic. Cl– is a weak field ligand hence, [NiCl4]2– is paramagnetic. 11. (a) : Higher the strength of ligand more will be the CFSE. Spectrochemically it has been found that the strength of splitting varies as: CO > CN– > NO2– > en > NH3 > py > NCS– > H2O > O2– > C2O42– > OH– > F– > Cl– > SCN– > S2– > Br– > I– 12. (a) : EAN = At. no. of central atom – oxidation state + 2 (number of ligands) = 27 – 3 + 2(6) = 27 – 3 + 12 = 36 13. (d) : [Co(NO2)(NH3)5]Cl2 Pentaamminenitrito-N-cobalt (III) chloride.

21. (b) : C2O42– (oxalate) is a bidentate ligand and thus, each coordinates with the central atom at two sites. 22. (b) : Mg is present in chlorophyll, the green pigment of plants. 23. (b) : Radius of cation, r+ = 95 = 0.525 Radius of anion, r− 181

This value lies in between 0.414 – 0.732, thus, the coordination number of Na+ ion will be 6. Z×M 24. (b) : d = N A × a3

where d = 8.92 g cm–3 , M = 63.55 g mol–1 , a = 362 pm = 362 × 10–10 cm



Z=? d × N A × a3 Z= M 8.92 × 6.022 × 10 23 × ( 362 × 10 −10 )3 Z= =4 63.55

14. (b) 15. (a) : A crystalline solid has a sharp melting point i.e. it changes abruptly into liquid state. 16. (d) : The correct formula of the complex tetraammineaquachloridocobalt(III)chloride is [CoCl(H2O)(NH3)4]Cl2, because in it the oxidation number of Co is +3. [CoCl(H2O)(NH3)4]Cl2 x + (–1) + 0 + (0 × 4) + (–1) 2 = 0 x–3=0 x = +3 17. (d) : Frenkel defect is generally observed when the size of cation is much smaller than the anion. Hence, AgBr, AgI and ZnS all exhibit Frenkel defect. 18. (d) : In crystals, the constituents are arranged in orderly pattern i.e. in planes, thus they show good cleavage. 19. (d) : [Co(NH3)4Cl2]NO2 and [Co(NH3)4ClNO2]Cl give different ions on ionisation as [Co(NH3)4Cl2]NO2 [Co(NH3)4Cl2]+ + NO–2 [Co(NH3)4ClNO2]Cl [Co(NH3)4ClNO2]+ + Cl– Thus, they show ionisation isomerism. 20. (c) : Complex compound with no unpaired electron is colourless. Among the given complexes, K4[Fe(CN)6] has no unpaired electron as CN– being strong field ligand causes pairing of electrons.



Thus, metal has face centred unit cell.

25. (d) : [Co(NH3)6]3+ : Inner orbital complex, d2sp3 hybridisation



Co3+ → 3d6 4s0

[Co(NH3)6]3+ : NH3 is a strong field ligand, it causes pairing of electrons.

No. of unpaired electrons = zero (diamagnetic) [CoF6]3– : Outer orbital complex, sp3d2 hybridisation



No. of unpaired electrons = four (paramagnetic)

26.

(c) : Let O.S of Co in the given complex, [Co(NH3)5Cl]Cl2 be x \ x + 5 × 0 – 1 = + 2 ⇒ x = 3 Now the IUPAC name of the complex will be pentaamminechlorocobalt(III) chloride.

15

Coordination Chemistry and Solid State Chemistry

27. (b) : When SCN – is added to an aqueous solution of Fe(NO3)3, blood red colour pentahydrated complex is formed. Fe3+ + SCN– + H2O → [Fe(OH2)5SCN]2+ 28. (b) : Schottky defect is caused if some of the lattice points are unoccupied. The points which are unoccupied are called vacancies or holes. The number of missing positive and negative ions is the same keeping the crystal neutral. Cations and anions are of similar size.

OOCCH2 N OOCCH2

CH2

CH2

N

Ethylenediaminetetraacetate ion

CH2COO CH2COO

32. (d) : [Fe(CN)6]4– : All orbitals are doubly occupied, hence [Fe(CN)6]4– is diamagnetic in nature hence, K4[Fe(CN)6] complex has zero magnetic moment. 3d

4s

4p

××

×× ×× ××

Fe atom

29. (a) Fe (II)

30. (b) : The 8 : 8 type of packing is present in caesium chloride (CsCl). In this structure each Cs+ ion is surrounded by 8Cl– ions and each Cl– ion is also surrounded by 8Cs+ ions. 31. (b) : Ethylenediaminetetraacetic acid is a hexadentate ligand because it has six donor centres.

[Fe(CN)6]4–

×× ××

CN – CN – CN – CN – CN – CN – d2sp3 hybridisation

33. (b)

vvv

16

VITEEE CHAPTERWISE SOLUTIONS

4

CHAPTER

1.

2.

Thermodynamics, Chemical Equilibrium and Chemical Kinetics

The standard molar heat of formation of ethane, CO2 and water are respectively –21.1, – 94.1 and –68.3 kcal. The standard molar heat of combustion of ethane will be (a) –372 kcal (b) 162 kcal (c) –240 kcal (d) 183.5 kcal (2014)



The solubility product of Ag2CrO4 is 32 × 10–12. What is the concentration of CrO42– ions in that solution? (a) 2 × 10–4 M (b) 16 × 10–4 M (c) 8 × 10–4 M (d) 8 × 10–8 M (2014)

where, V0, Vt and V∞ are the volumes of standard NaOH required to neutralise acid present at a given time, if ester is 50% neutralised then (a) V∞ = Vt (b) V∞ = (Vt – V0) (c) V∞ = 2Vt – V0 (d) V∞ = 2Vt + V0 (2013)

7.

Consider the following reaction,

3.

The half-lives of two samples are 0.1 s and 0.8 s. and their respective concentrations are 400 mol L–1 and 50 mol L–1. The order of the reaction is (a) 0 (b) 2 (c) 1 (d) 4 (2014)

4.

The rate constants for forward reaction and backward reaction of hydrolysis of ester are 1.1 × 10–2 min–1 and 1.5 × 10–3 min–1 respectively. Equilibrium constant for the reaction



5.

6.

CH3COOC2H5 + H2O is (a) 33.7 (c) 5.33

CH3COOH + C2H5OH (b) 7.33 (d) 33.3



V − V0 2.303 log ∞ t V∞ − Vt

kA boat

kB

chair



The reaction is of first order in each diagram, with an equilibrium constant of 104. For the conversion of chair form to boat form e–Ea /RT = 4.35 × 10 –8 at 298 K with preexponential factor of 1012 s–1. Apparent rate constant (= kA/kB) at 298 K is (a) 4.35 × 104 s–1 (b) 4.35 × 108 s–1 (c) 4.35 × 10–8 s–1 (b) 4.35 × 1012 s–1 (2013)

8.

The reaction, 2A(g) + B(g)



is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression

(2014)

When 1 mole of CO2(g) occupying volume 10 L at 27°C is expanded under adiabatic condition, temperature falls to 150 K. Hence, final volume is (a) 5 L (b) 20 L (c) 40 L (d) 80 L (2013) Acid hydrolysis of ester is first order reaction and rate constant is given by

k=

3C(g) + D(g)

(a) [(0.75)3(0.25)] ÷ [(1.00)2(1.00)] (b) [(0.75)3(0.25)] ÷ [(0.50)2(0.75)] (c) [(0.75)3(0.25)] ÷ [(0.50)2(0.25)] (d) [(0.75)3(0.25)] ÷ [(0.75)2(0.25)] 9.

(2012)

Match List I (Equations) with List II (Types of process) and select the correct option.

Thermodynamics, Chemical Equilibrium and Chemical Kinetics

List I (Equations)

List II (Types of process)

A. B. C.

1. Kp > Q DG° < RT ln 2. 3. Q Kp = Q

D.

T>

Codes A (a) 1 (b) 3 (c) 4 (d) 2

DH DS B 2 4 1 1

4.

C 3 2 2 4

Non-spontaneous Equilibrium Spontaneous and endothermic Spontaneous

D 4 1 3 3

(2012)

10. Consider the following processes : DH (kJ/mol) 1 A B; + 150 2 3B 2C + D; – 125 E + A

2D;

+ 350

For B + D E + 2C , DH will be (a) + 525 kJ/mol (b) – 175 kJ/mol (c) – 325 kJ/mol (d) + 325 kJ/mol

(2012)

11. The unit of rate constant for a zero order reaction is (a) mol L–1 s–1 (b) L mol–1 s–1 2 –2 –1 (c) L mol s (d) s–1 (2012) 12. In the reversible reaction, 2NO2

k1 k2

N2O4

the rate of disappearance of NO2 is equal to 2 k1 [NO 2 ]2 k2 (b) 2k1[NO2]2 – 2k2[N2O4] (c) 2k1[NO2]2 – k2[N2O4] (d) (2k1 – k2)[NO2] (a)

(2011)

13. A chemical reaction was carried out at 300 K and 280 K. The rate constants were found to be k1 and k2 respectively. Then (a) k2 = 4k1 (b) k2 = 2k1 (c) k2 = 0.25 k1 (d) k2 = 0.5 k1 (2011, 2010)

17

14. The rate constant of a reaction at temperature 200 K is 10 times less than the rate constant at 400 K. What is the activation energy of the reaction? (a) 1842.4 R (b) 460.6 R (c) 230.3 R (d) 921.2 R (2011) 15. A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. The value of K if the total pressure at equilibrium is 0.8 atm, is (a) 1.8 atm (b) 3 atm (c) 0.3 atm (d) 0.18 atm. (2011) 16. For the reaction 2A + B C, DH = x cal, which one of the following conditions would favour the yield of C on the basis of Le-Chatelier’s principle? (a) High pressure, high temperature (b) Only low temperature (c) High pressure, low temperature (d) Only low pressure (2011) 17. Heat of formation, DHf° of an explosive compound like NCl3 is (a) positive (b) negative (c) zero (d) positive or negative. (2011) 18. For the reaction, C3H8(g) + 5O2(g) → 3CO2(g) + 4H2(O)(l) at constant temperature, DH – DE is (a) RT (b) –3RT (c) 3RT (d) –RT (2011) 19. The favourable conditions for a spontaneous reaction are (a) TDS > DH, DH = + ve, DS = + ve (b) TDS > DH, DH = + ve, DS = – ve (c) TDS = DH, DH = – ve, DS = – ve (d) TDS = DH, DH = + ve, DS = + ve (2011) 20. For a reaction of type A + B → products, it is observed that doubling concentration of ‘A’ causes the reaction rate to be four times but doubling amount of ‘B’ does not affect the rate. The unit of rate constant is (a) s–1 (b) s–1 mol L–1 (c) s–1 mol–1 L (d) s–1 mol–2 L–2 (2010)

18

VITEEE CHAPTERWISE SOLUTIONS

21. Enthalpy of a compound is equal to its (a) heat of combustion (b) heat of formation (c) heat of reaction (d) heat of solution. (2010) 22. For which one of the following reactions will there be a positive DS? (a) H2O(g) → H2O(l) (b) H2(g) + I2(g) → 2HI(g) (c) CaCO3(s) → CaO(s) + CO2(g) (d) N2(g) + 3H2(g) → 2NH3(g) (2010) 23. For the reaction, 2 A( g ) + B2( g )  2 AB2( g )

the equilibrium constant, Kp at 300 K is 16.0. The value of Kp for AB2( g )  A( g ) + 1 / 2 B2( g ) is (a) 8 (c) 0.125

(b) 0.25 (d) 32

(2010)

24. The standard Gibb’s free energy change, DG° is related to equilibrium constant, Kp as DG °

 e  (a) Kp = – RT ln DG° (b) K p =    RT  DG (c) K p = − (d) K p = e − DG °/ RT RT (2010) 25. The yield of the product in the reaction, A2( g ) + 2 B( g )  C( g ) + Q kJ would be higher at (a) high temperature and high pressure (b) high temperature and low pressure (c) low temperature and high pressure (d) low temperature and low pressure. (2010) 26. In which of the following cases, does the reaction go farthest to completion? (a) K = 102 (b) K = 10 (c) K = 10–2 (d) K = 1 (2010) 27. The activity of an old piece of wood is just 25% of the fresh piece of wood. If t1/2 of C-14 is 6000 yr, the age of piece of wood is (a) 6000 yr (b) 3000 yr (c) 9000 yr (d) 12000 yr (2010) 28. Given that DHf (H) = 218 kJ/mol, express the H H bond energy in kcal/mol. (a) 52.15 (b) 911 (c) 104 (d) 52153 (2009)

29. The equilibrium constant for the given reaction is 100. N 2( g ) + 2O 2( g )  2 NO 2( g )

What is the equilibrium constant for the reaction given below? 1 NO 2( g )  N 2( g ) + O 2( g ) 2 (a) 10 (b) 1 (c) 0.1 (d) 0.01 (2009)

30. For a first order reaction at 27°C, the ratio of time required for 75% completion to 25% completion of reaction is (a) 3.0 (b) 2.303 (c) 4.8 (d) 0.477 (2009) 31. Calculate DH° for the reaction, Na2O(s) + SO3(g) → Na2SO4(g) Given the following: 1 (A) Na( s) + H 2O( l ) → NaOH( s) + H 2( g ) ; 2 DH° = –146 kJ (B) Na2SO4(s) + H2O(l) → 2NaOH(s) + SO3(g) ; DH° = +418 kJ (C) 2Na2O(s) + 2H2(g) → 4Na(s) + 2H2O(l) ; DH° = +259 kJ (a) +823 kJ (b) –581 kJ (c) –435 kJ (d) +531 kJ (2009) 32. The standard free energy change of a reaction is DG° = – 115 kJ at 298 K. Calculate the equilibrium constant Kp in log Kp (R = 8.314 JK–1 mol–1). (a) 20.16 (b) 2.303 (c) 2.016 (d) 13.83 (2008) 33. If an endothermic reaction occurs spontaneously at constant temperature T and pressure P, then which of the following is true? (a) DG > 0 (b) DH < 0 (c) DS > 0 (d) DS < 0 (2008) 34. If a plot of log10 C versus t gives a straight line for a given reaction, then the reaction is (a) zero order (b) first order (c) second order (d) third order. (2008) 35. A spontaneous process is one in which the system suffers (a) no energy change (b) a lowering of free energy (c) a lowering of entropy (d) an increase in internal energy. (2008)

19

Thermodynamics, Chemical Equilibrium and Chemical Kinetics

36. The half life period of a first order reaction is 1 min 40 seconds. Calculate its rate constant. (a) 6.93 × 10–3 min–1 (b) 6.93 × 10–3 sec–1 (c) 6.93 × 10–3 sec (d) 6.93 × 103 sec (2008) 37. When a solid melts reversibly (a) H decreases (b) G increases (c) E decreases (d) S increases.



(2007)

42. According to Arrhenius equation, the rate constant (k) is related to temperature (T) as

38. Enthalpy is equal to  ∂ (G / T )  (a) T 2    ∂T  P  ∂ (G / T )  (c) T 2    ∂T V

What change will shift the equilibrium to the right? (a) Decreasing the temperature (b) Increasing the temperature (c) Dissolving NaCl crystals in the equilibrium mixture (d) Dissolving NH4NO3 crystals in the equilibrium mixture (2007)

 ∂ (G / T )  (b) −T 2    ∂T  P  ∂ (G / T )  (d) −T 2    ∂T V

(a) ln

k2 Ea  1 1 =  −  k1 R  T1 T2 

(b) ln

E k2 =− a k1 R

(c) ln

k2 Ea  1 1 =  +  k1 R  T1 T2 

(d) ln

E 1 k2 1 =− a  +  k1 R  T1 T2 

(2007) 39. Condition for spontaneity in isothermal process is (A is work function.) (a) DA + W < 0 (b) DG + U < 0 (c) DA + U > 0 (d) DG – U < 0 (2007) 40. Given : 2C(s) + 2O2(g) → 2CO2(g); DH = –787 kJ

1 1  −   T1 T2 

(2007)

43. Equivalent amounts of H2 and I2 are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to HI, the Kc at this temperature is (a) 64 (b) 16 (c) 0.25 (d) 4 (2007)

1 H 2( g ) + O 2( g ) → H 2O( l ) ; DH = − 286 kJ 2 1 C 2H 2( g ) + 2 O 2( g ) → 2 CO 2( g ) + H 2O( l ) ; 2 DH = –1310 kJ The heat of formation of acetylene is (a) –1802 kJ (b) +1802 kJ (c) +237 kJ (d) –800 kJ (2007)

44. For the reaction H2(g) + I2(g) → 2HI(g), the equilibrium constant Kp changes with (a) total pressure (b) catalyst (c) the amount H2 and I2 (d) temperature. (2007)



41. Given that the equilibrium system, + NH4Cl(s) → NH4(aq) + Cl–(aq) (DH = +3.5 kcal/mol)

Answer Key 1.

(a)

2.

9.

(c)

(a)

3.

(b)

5.

(d)

6.

(c)

7.

(b)

8.

(b)

12. (b)

13.

(c)

14.

(d)

15.

(a)

16.

(a)

19. (a)

20. (c)

21.

(b)

22.

(c)

23.

(b)

24.

(d)

26. (a)

27. (d)

28. (c)

29.

(c)

30.

(c)

31.

(b)

32.

(a)

33. (c)

34. (b)

35. (b)

36. (b)

37.

(d)

38.

(b)

39.

(a)

40.

(c)

41. (b)

42. (a)

43. (a)

44. (d)

10. (b)

11. (a)



17. (a)

18. (b)

25. (c)

4.

(b)

20

VITEEE CHAPTERWISE SOLUTIONS

e planations

1.

(a) : Given, (i) 2C + 3H2 → C2H6; DH = – 21.1 kcal (ii) C + O2 → CO2; DH = – 94.1 kcal



(iii) H2 +



1 O2 → H2O; DH = – 68.3 kcal 2 Eqs. 2 (ii) + 3 (iii) – (i)



C 2H 6 +



7 O2 → 2CO2 + 3H2O 2 DH = 2DHf (CO2) + 3 DHf (H2O) – DHf (C2H6)

DH = 2(– 94.1) + 3(– 68.3) – (– 21.1) = – 372 kcal 2.

(a) : Ag2CrO4 → 2Ag+ + CrO42– s





2s

2

Ksp = (2s) s = 4s3 K sp

32 × 10 −12 = 2 × 10 −4 M 4



s=3

3.

(b) : We know that,



(t1/ 2 )1



s

4

=3

n −1

a  = 2 (t1/ 2 ) 2  a1  where, n = order of the reaction Given, (t1/2)1 = 0.1 s, a1 = 400 mol L–1 (t1/2)2 = 0.8 s, a2 = 50 mol L–1 On substituting the values, n −1



0.1  50  =  0.8  400 



On taking log both sides



Here, for CO2 (triatomic gas), g = 1.33



 150   10  \  =  300   V 

0.33

2

0.33



 1   10  or   =   2 V  2



1 10 or   = 2 V2



or



or V2 = 80 L

6.

(c) : RCOOR′ + H2O

3

1 10 = 8 V2

H+

RCOOH + R′OH



At t = 0,

a

0

0



At time t,

a – x

x

x



At time ∞,

a – a

a

a

At t = 0, V0 = volume of NaOH used due to H+ (catalyst) Vt = x + V0 V∞ = a + V0 a If ester is 50% hydrolysed then, x = 2 a or Vt = + V0 2 or a = 2Vt – 2V0 \ V∞ = 2Vt – 2V0 + V0 = 2Vt – V0 7. (b) : kB = Ae – Ea / RT = 1012 × 4.35 × 10 –8 = 4.35 × 104 s–1



0.1 50 = (n − 1)log 0.8 400 1 1 log = (n − 1)log 8 8 – 0.90 = (n – 1) (– 0.90) n–1=1 ⇒ n=2

4.

(b) : kf = 1.1 × 10–2, kb = 1.5 × 10–3



Initial conc.



Equilibrium conc. 1 – 0.50 1 – 0.25



Kc =



log

kf kb

=

1.1 × 10 −2 1.5 × 10 −3

= 7.33

5.

(d) : For adiabatic expansion,



T2  V1  = T1  V2 

g −1



Also equilibrium constant, K =



\ kA = kB × 104 = 4.35 × 108 s–1 2A(g) + B(g)

8. (b) :

1



K=

kA = 10 4 kB

3

[C ] [ D] 2

[ A ] [ B]

=

1

3C(g) + D(g) 0 0 0.75 0.25

= 0.50 = 0.75 3

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

9. (c) : (A) When Kp > Q, the reaction goes in forward direction, i.e., the reaction is spontaneous.

21

Thermodynamics, Chemical Equilibrium and Chemical Kinetics

(B) DG = DG° + RT ln Q, DG° < RT ln Q, thus, DG = +ve and hence, the reaction is non-spontaneous. (C) At equilibrium, Kp = Q DH (D) T > or TDS > DH DS This condition is true for spontaneous, endothermic reactions (as DG = DH – TDS). 1 10. (b) : A B; DH = +150 kJ/mol ...(i) 2 3B 2C + D; DH = –125 kJ/mol ...(ii) E+A 2D; DH = + 350 kJ/mol ...(iii) By [2 × (i) + (ii)] – (iii), we have B+D E + 2C \ DH = 150 × 2 + (–125) – 350 = – 175 kJ/mol 11. (a) : For zero order reaction, Rate = k[Reactants]0 \ Rate = k and unit of k = mol L–1 s–1 12. (b) : 2NO2

k1 k2

N2O4

1 d[NO 2 ] Rate of reaction = − 2 dt = k1[NO2]2 – k2[N2O4]



\ Rate of disappearance of NO2 i.e., d[NO 2 ] − = 2k1[NO2]2 – 2k2[N2O4] dt 13. (c) : Rate constant becomes double for every 10° rise in temperature. Hence, for 20° rise in temperature, rate constant will become four times. i.e., k1 = 4k2 1 k2 = k1 4 k2 = 0.25k1

14. (d) : At T1 = 200 K, if k1 = k then at T2 = 400 K, k2 = 10 k

Ea  400 − 200  10 k =   k 2.303R  400 × 200  Ea = 921.2 R

K=

2 16. (a) : 2A + B C, DH = x cal Here np < nr and reaction is endothermic. Hence, high pressure and high temperature will favour forward reaction.

17. (a) : Explosive compound has high heat content i.e., it is formed by absorption of heat. Hence, DHf° is positive. 18.

(b) : C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Dng = np – nr = 3 – 6 = – 3 DH = DE + DngRT DH = DE – 3RT DH – DE = – 3RT 19. (a) : Q DG = DH – TDS For a spontaneous process, DG should be negative. Hence, DH = + ve, DS = + ve and TDS > DH. 20. (c) : Let the initial rate is R and order with respect to A is x and order with respect to B is y. Thus, rate law is, R = [A]x [B]y ...(i) On doubling the concentration of A, rate becomes 4R, 4R = [2A]x [B]y ...(ii) On doubling the concentration of B, rate remains R, R = [A]x [2B]y ...(iii) From eqns (i) and (ii), we get

CO2(g) + C(s)

2CO(g)



Initial pressure

0.5 atm

0



At equi.

(0.5 – x)

2x

\ \

(0.5 – x) + 2x = 0.8 x = 0.3 atm pCO2 = 0.5 – 0.3 = 0.2 atm pCO = 2x = 2 × 0.3 = 0.6 atm

R 1 = 4 R  2  2

x

x



1 1  2  =  2  \ x = 2 From eqns (i) and (iii), we get



R 1 = R  2 

log

15. (a) :

( pCO )2 (0.6)2 = = 1.8 atm pCO 0.2



y

y

1 1=    2 \ y = 0 Thus, actual rate law is, rate, R = [A]2 [B]0. The order of this reaction is 2 and for second order reaction unit of rate constant are s–1 mol–1 L.

22

VITEEE CHAPTERWISE SOLUTIONS

21. (b)



22. (c) : DS (entropy change) is the measure of randomness and thus in solid, liquid and gas, the order of entropy is gas > liquid > solid. Thus, DS is positive for the reaction given in option (c). 23.

(b) : 2A(g) + B2(g) 2AB2(g) Kp = 16.0 On reversing the equation, 2AB2(g) 2A(g) + B2(g) K p′ =



1 1 = K p 16

Dividing the equation by 2 1 A( g ) + B2( g ) AB2(g) 2



K p′′ = K p′ =



1 1 = = 0.25 16 4

24. (d) : Standard free energy change, DG° and equilibrium constant Kp are related as DG° = –RT ln Kp DG° ln K p = − RT



K p = e − DG°/ RT



25. (c) : A2(g) + 2B(g) C(g) + Q kJ np < nr and the reaction is exothermic, hence it is favoured by low temperature and high pressure. 26. (a) : Higher value of K indicates that the reaction goes farthest to completion. 27. (d) : k =

0.693 0.693 = = 1.155 × 10−4 6000 t1/ 2

N 2.303 log 0 k N 2.303 100 log = = 12000 yr −4 25 1.155 × 10

t=



28. (c) : Given : DHf (H) = 218 kJ/mol

i.e.,

1 H → H; DH = 218 kJ/ mol 2 2

or H2 → 2H; DH = 436 kJ/mol 436 = = 104.3 kcal/ mol 4.18

Thus, 104.3 kcal/mol energy is absorbed for breaking one mole of H H bonds. Hence, H H bond energy is 104.3 kcal/mol.

29. (c) : N 2( g ) + 2O 2( g )  2 NO 2( g )



K1 = or

[NO 2 ]2

[N 2 ][O 2 ]2 100 =

[NO 2 ]2

Again, NO 2( g )  K2 =

[N 2 ]1/2 [O 2 ] [NO 2 ]

or

K22 =

1 N + O 2( g ) 2 2( g )

[N 2 ][O 2 ]2

...(ii)

[NO 2 ]2



Equations (i) × (ii), we get,



100 × K2 = 1



...(i)

[N 2 ][O 2 ]2

2

K22 =

or

1 100

or K2 =

1 = 0.1 100

30. (c) : For a first order reaction, 2.303 a log t= ( a − x) k For 75% completion,

2.303 100 log (100 − 75) k

t1 = =

2.303 log 4 k

For 25% completion, 2.303 100 log t2 = (100 − 25) k 2.303 4 = log k 3 t1 log 4 = 4 t2 log 3 2 × 0.3010 log 4 = = = 4.81 log 4 − log 3 2 × 0.3010 − 0.4771 (C ) 31. (b) : By 2(A) + − ( B), we get 2 Na2O + SO3 → Na2SO4; 259 − 418 2 or DH = – 580.5 ≈ – 581 kJ



DH = − 2 × 146 +

23

Thermodynamics, Chemical Equilibrium and Chemical Kinetics

32. (a) : The standard free energy change of a reaction, DG° = – 2.303 RT log Kp –115 × 103 = – 2.303 RT log Kp

115 × 10 3 log K p = 2.303 × 8.314 × 298 log Kp = 20.1547 ≈ 20.16

33. (c) : DG = DH – TDS For an endothermic reaction, DH = +ve For reaction to be spontaneous, DG should be negative which is possible only when DS > 0 at constant temperature and pressure. 34. (b) : Integrated rate equation for a first order reaction is given as :

[C ] 2.303 log 0 k= [C ] t



⇒ log[C ] =



Thus if log [C] values are plotted against time ‘t’ the graph obtained is a straight line −k with slope = . 2.303

−k t + log[C ]0 2.303



DG = − S DT From eqns (i) and (ii)



 DG  G= H +T  DT 





 ∂G  or G = H + T   ∂T  P −



...(ii)

H T

2

=−

G T

2

 ∂ (G / T )  1  ∂G  =  T  ∂T  P  ∂ T  P

+

 ∂ (G / T )  H = − T2    ∂T P

39. (a) : The condition for spontaneity in an isothermal process is that DG should be –ve. Q DG = DA + PDV DG = DA + W (Q W = PDV) \ For isothermal process DA + W < 0 40. (c) : 2C(s) + 2O2(g) 1 H 2( g ) + O 2( g ) 2

2CO2(g) ; DH = –787 kJ

log [C ]



t

35. (b) : Negative value of DG indicates that the process is spontaneous. 36. (b) : The half life of a first order reaction is given by t1/ 2 = 0.693 . k k = 0.693 = 6.93 × 10−3 sec−1 . (Q t1/ 2 = 100 sec) 100



...(i)

H2O(l); DH = –286 kJ

...(ii)

1 C 2 H 2( g ) + 2 O 2( g ) 2CO2(g) + H2O(l) ; 2 DH = – 1310 kJ ...(iii) Add the eqns (i) and (ii) and subtract eqn. (iii) we get, 2C + H2 C2H2; – 787 – 286 + 1310 = +237 Hence, the heat of formation of acetylene is +237 kJ.

41. (b) : NH4Cl(s)

NH+4(aq) + Cl–(aq);

37. (d) : When a solid melts, its entropy (S) increases because on changing from solid to liquid disorder or randomness of molecules increases.

DH = +3.5 kcal/mol This is an endothermic reaction hence, increasing the temperature will shift the equilibrium to the right.

38.

42. (a) : k1 = Ae

(b) : We know that, G = H – TS ...(i) G = E + PV – TS [Q H = E + PV] DG = DE + PDV + VDP – TDS – SDT From 1st and 2nd laws, TDS = DE + PDV DG = VDP – SDT At constant pressure, DP = 0



k2 = Ae

− Ea / RT1

− Ea / RT2

ln k1 = ln A −

Ea RT1



ln k2 = ln A −

Ea RT2



From eqns (i) and (ii), we get

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

24



VITEEE CHAPTERWISE SOLUTIONS

ln k2 − ln k1 = ln A −



E 1 k 1 ln 2 = a  −  k1 R  T1 T2 

or ln

Ea E − ln A + a RT2 RT1

E  1 k2 1 =− a  −  k1 R  T2 T1 

43. (a) : Initially



H2 + I2

2HI

1 1 0 At equilibrium (1 – 0.8) (1 – 0.8) (2 × 0.8) = 0.2 = 0.2 = 1.6

Kc =

2.56 [HI]2 (1.6)2 = = = 64 [H 2 ] [I 2 ] 0.2 × 0.2 0.04

44. (d) : Equilibrium constant of a particular reaction is constant and depends only on temperature.

vvv

25

Electrochemistry

5

CHAPTER

Electrochemistry

1.

The equivalent conductivity of a solution containing 2.54 grams of CuSO4 per litre is 91.0 W–1 cm2 eq–1. Its conductivity would be (a) 2.9 × 10–3 W–1 cm–1 (b) 1.8 × 10–2 W–1 cm–1 (c) 2.4 × 10–4 W–1 cm–1 (d) 3.6 × 10–3 W–1 cm–1 (2014)

2.

For the cell reaction, 2Ce4+ + Co → 2Ce3+ + Co2+; E°cell is 1.89 V. If ECo2+/Co is – 0.28 V, what is the value of E°Ce4+/Ce3+ ? (a) 0.28 V (b) 1.61 V (c) 2.17 V (d) 5.29 V (2013)

3.

A constant current of 30 A is passed through an aqueous solution of NaCl for a time of 1.00 h. What is the volume of Cl2 gas produced at STP? (a) 30.00 L (b) 25.08 L (c) 12.54 L (d) 1.12 L (2013)

4.

If for the cell reaction, Zn + Cu2+ Cu + Zn2+; entropy change DS is 96.5 J mol–1 K–1, then temperature coefficient of the emf of a cell is (a) 5 × 10–4 V K–1 (b) 1 × 10–3 V K–1 –3 –1 (c) 2 × 10 V K (d) 9.65 × 10–4 V K–1 (2013)

5.

6.

Calculate pH of a buffer prepared by adding 10 mL of 0.10 M acetic acid to 20 mL of 0.1 M sodium acetate [pKa(CH3COOH) = 4.74] (a) 3.00 (b) 4.44 (c) 4.74 (d) 5.04 (2013) The equivalent conductance of silver nitrate solution at 250°C for an infinite dilution was found to be 133.3 W–1cm2 equiv–1. The transport number of Ag+ ions in very dilute solution of AgNO3 is 0.464. Equivalent conductances of Ag+ and NO–3 (in W–1 cm2 equiv–1) at infinite dilution are respectively (a) 195.2, 133.3 (b) 61.9, 71.4 (c) 71.4, 61.9 (d) 133.3, 195.2 (2013)

7.

Which of the following expressions correctly represents the equivalent conductance at infinite dilution of Al2(SO4)3? Given that ° 3+ and L°SO2– are the equivalent conductances LAl 4 at infinite dilution of the respective ions? (a) 2L°Al3+ + 3L°SO2–4 (b) L°Al3+ + L°SO2–4 1 1 (c) (L°Al3+ + 3L°SO2–4 ) × 6 (d) L° 3+ + L° 2− Al 3 2 SO4 (2012)

8.

The equivalent conductances of two ions at infinite dilution in water at 25°C are given below : l∞Ba2+ = 127.00 S cm2/equiv. l∞Cl– = 76.00 S cm2/equiv. The equivalent conductance (in S cm2/ equiv.) of BaCl2 at infinite dilution will be (a) 203.0 (b) 279.0 (c) 205.5 (d) 139.5 (2011)



9.

20 mL of 0.2 M NaOH is added to 50 mL of 0.2 M acetic acid. The pH of this solution after mixing is (Ka = 1.8 × 10–5) (a) 4.5 (b) 2.3 (c) 3.8 (d) 4 (2011)

10. The EMF of the cell, Mg|Mg2+ (0.01 M)||Sn2+(0.1 M)|Sn at 298 K is (E°Mg2+/Mg = –2.34 V, E°Sn2+/Sn = – 0.14 V) (a) 2.17 V (c) 2.51 V

(b) 2.23 V (d) 2.45 V

(2011)

11. The equivalent conductance at infinite dilution of a weak acid such as HF (a) can be determined by extrapolation of measurements on dilute solutions of HCl, HBr and HI (b) can be determined by measurement on very dilute HF solutions (c) can best be determined from measurements on dilute solutions of NaF, NaCl and HCl (d) is an undefined quantity. (2010)

26

VITEEE CHAPTERWISE SOLUTIONS

12. The number of Faradays needed to reduce 4 g equivalents of Cu2+ to Cu metal will be (a) 1 (b) 2 (c) 1/2 (d) 4 (2010)

18. The molar conductivities of KCl, NaCl and KNO3 are 152, 128 and 111 S cm2 mol–1 respectively. What is the molar conductivity of NaNO3? (a) 101 S cm2 mol–1 (b) 87 S cm2 mol–1 (c) –101 S cm2 mol–1 (d) – 391 S cm2 mol–1 (2008)

13. Which one of the following cells can convert chemical energy of H2 and O2 directly into electrical energy? (a) Mercury cell (b) Daniell cell (c) Fuel cell (d) Lead storage cell (2010)

19. The electrochemical cell stops working after sometime because (a) electrode potential of both the electrodes becomes zero (b) electrode potential of both the electrodes becomes equal (c) one of the electrodes is eaten away (d) the cell reaction gets reversed. (2008)

14. pH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27°C. The buffer capacity of the solution is (a) 0.1 (b) 10 (c) 1 (d) 0.4 (2009)

20. The amount of electricity required to produce one mole of copper from copper sulphate solution will be (a) 1 Faraday (b) 2.33 Faraday (c) 2 Faraday (d) 1.33 Faraday. (2008)

15. For the following cell reaction, Ag |Ag+| AgCl | Cl–| Cl2,Pt DGf°(AgCl) = –109 kJ/mol

– DG°f (Cl ) = – 129 kJ/mol DG°f (Ag+) = 78 kJ/mol E° of the cell is (a) – 0.60 V (b) 0.60 V (c) 6.0 V (d) none of these. (2009)

21. How long (in hours) must a current of 5.0 A be maintained to electroplate 60 g of calcium from molten CaCl2? (a) 27 h (b) 8.3 h (c) 11 h (d) 16 h (2007)

16. At 25°C, the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl2 are 248 × 10–4, 126 × 10–4 and 280 × 10–4 S m2 mol–1 respectively. l°m Ba(OH)2 in S m2 mol–1 is (a) 52.4 × 10–4 (b) 524 × 10–4 –4 (c) 402 × 10 (d) 262 × 10–4 (2009)

22. For strong electrolytes the plot of molar conductance vs (a) parabolic (c) sinusoidal

C is

(b) linear (d) circular.

(2007)

23. If the molar conductance values of Ca2+ and Cl– at infinite dilution are respectively 118.88 × 10–4 m2 mho mol–1 and 77.33 × 10–4 m2 mho mol–1 then that of CaCl2 is (in m2 mho mol–1) (a) 118.88 × 10–4 (b) 154.66 × 10–4 –4 (c) 273.54 × 10 (d) 196.21 × 10–4 (2007)

17. 20 mL of 0.1 M acetic acid is mixed with 50 mL of potassium acetate. Ka of acetic acid = 1.8 × 10–5 at 27°C. Calculate concentration of potassium acetate if pH of the mixture is 4.8. (a) 0.1 M (b) 0.04 M (c) 0.4 M (d) 0.02 M (2009)

Answer Key 1.

(a)

2.

(b)

3.

(c)

9.

(a)

10. (b)

11. (c)

17. (b)

18. (b)

19. (b)

4.

(a)

5.

(d)

6.

(b)

7.

(b)

8.

(a)

12. (d)

13.

(c)

14.

(d)

15.

(a)

16.

(b)

20. (c)

21.

(d)

22.

(b)

23.

(c)

27

Electrochemistry

e planations

2.54   = (91 W–1 cm2 eq–1)  eq ⋅ cm −3   (159 / 2) × 1000 



Total volume after mixing = 30 mL \ [CH3COOH] = 1/30 M and [CH3COONa] = 2/30 M



= 2.9 × 10–3 W–1 cm–1



2.

(b) : E°cell = E°Ce4+/Ce3+ – E°Co2+/Co

pH = pK a + log



⇒ 1.89 = E°Ce4+/Ce3+ – (– 0.28)

1.

(a) : k = Leq ⋅ C





⇒ E°Ce4+/Ce3+ = 1.89 – 0.28 = 1.61 V

3. (c) : Due to electrolysis, – 2H2O(l) + 2e – → H2(g) + 2OH (aq) 2Cl–(aq)

→ Cl2(g) + 2e

–

– 2H2O(l) + 2Cl–(aq) → H2(g) + Cl2(g) + 2OH (aq)

OH – formed = NaOH formed = Z i t E = ×i×t 96500 40 = × 30 × 1 × 60 × 60 96500 = 44.77 g

44.77 = 1.12 mol 40 1 Cl2 formed = mol of NaOH 2



4.

=

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

 ∂E  (a) : DG = DH – nFT    ∂T  P

6.

[Salt ] [Acid]

 2 / 30  = 4.74 + 0.3010 = 5.04 = 4.74 + log   1 / 30 

(b) : l∞eq(Ag+) = transport number of Ag+ × L∞eq(AgNO3)

= 0.464 × 133.3 = 61.9 W–1cm2 equiv–1 By Kohlrausch’s law, L∞eq(AgNO3) = l∞eq(Ag+) + l∞eq(NO–3 )

\ l∞eq(NO–3 ) = L∞eq(AgNO3) – l∞eq(Ag+)

7.



= 133.3 – 61.9 = 71.4 W–1 cm2 equiv–1

(b) : Al2(SO4)3 2Al3+ + 3SO42– Since equivalent conductance is given only for ions, the equivalent conductance at infinite dilution, ° ° L∞ eq = L 3+ + L 2 Al

SO4

8. (a) : L∞eq(BaCl2) = l∞eq(Ba2+) + l∞eq(Cl–)

= 127 + 76 = 203 S cm2/equiv.

9.

(a) : NaOH + CH3COOH → CH3COONa + H2O

Millimoles 20×0.2 added

50×0.2

= 4

Millimoles after reaction 0





and DG = DH – TDS





\

DS  ∂E  =  nF  ∂T  P





or

96.5  ∂E  =  2 × 96500  ∂T  P





 ∂E  \  cell  = 5 × 10 −4 V K −1  ∂T  P

5.

(d) : 10 mL of 0.10 M acetic acid = 10 × 0.10 = 1.0 millimole 20 mL of 0.1 M sodium acetate = 20 × 0.1 = 2.0 millimole



0

6

4

4

As total volume after mixing is 70 mL so, 6 [CH3COOH] = 70 4 and [CH3COONa] = 70 [Salt] pH = pKa + log [Acid]  4/70  −5 pH = − log(1.8 × 10 ) + log   6/70  = 4.74 + (– 0.17) = 4.56

10. (b) : Mg + Sn2+

0

= 10

Ecell = E°cell −

Mg2+ + Sn

[Mg 2+ ] 0.0591 log 2 [Sn 2+ ]

28



VITEEE CHAPTERWISE SOLUTIONS

= ( 2.34 − 0.14) −

0.0591 10 −2 log = 2.23 V 2 10 −1

17. (b) : pH = pKa + log

11. (c) : According to Kohlrausch’s law equivalent conductance at infinite dilution of HF l°HF = l°NaF + l°HCl – l°NaCl 12. (d) : Cu2+ + 1 mol

2e– 2F

1 mol 2 1 g equiv.



1F 1F

Cu 1 mol 1 mol 2 1 g equiv.



Thus, to reduce 4 g equivalents of Cu2+ into Cu four Faradays are required. 13. (c) : In fuel cell, the energy produced by the combustion of fuels like H2, O2, CH4, etc, is converted into electric energy e.g., H2 O2 fuel cell. ∂b 14. (d) : Buffer capacity (φ) = ∂( pH) where ∂b = no. of moles of acid or base added to 1 L solution ∂(pH) = change in pH 0.12 / 60 ∂b = = 8 × 10 −3 250 / 1000 8 × 10 −3 \ φ = = 0.4 0.02 15.

(a) : For the given cell, – Ag | Ag+ | AgCl | Cl | Cl2,Pt the cell reactions are as follows: At anode : Ag → Ag+ + e– – At cathode : AgCl + e– → Ag + Cl



Net cell reaction : AgCl → Ag+ + Cl





[Salt] [Acid]

4.8 = − log(1.8 × 10 −5 ) + log 4.8 = 4.74 + log 25 [Salt] log 25 [Salt] = 0.06 25 [Salt] = 1.148 1.148 [Salt] = = 0.045 M 25

[Salt ] × 50 20 × 0. 1 M

18. (b) : l°K + + l°Cl – = 152 S cm2 mol–1

...(i)



l°Na+ + l°Cl – = 128 S cm mol

...(ii)



l°K + + l°NO 3– = 111 S cm mol

...(iii)



2

2

–1

–1

Now, L°NaNO3 = L°NaCl + L°KNO3 – L°KCl



L°NaNO 3 = 128 + 111 – 152



L°NaNO 3 = 87 S cm2 mol–1

19. (b) : In electrochemical cell, potential difference is created. Cathode is at higher potential and anode is at lower potential. Current flows due to potential difference, from higher potential to lower potential. The electrochemical cell stops working when potential difference is zero i.e both the electrode potentials become equal. 20. (c) : The electrode reaction is Cu 2+ + 2e− → Cu 1 mol



2 × 96500

63.5 g of Cu or 1 mol of Cu deposited by passing = 2 × 96500 coulombs = 2 F of electricity.

21. (d) : Equivalent weight of calcium –



° \ DGreaction = ΣDGP° − ΣDGR° = (78 –129) – (–109) = + 58 kJ/mol DG° = –nFE° 58 × 103 J = –1 × 96500 × E°cell −58 × 1000 ° Ecell = = − 0.6 V 96500



16. (b) : BaCl2 + 2NaOH → Ba(OH)2 + 2NaCl l°m Ba(OH) = l°m BaCl + 2 l°m NaOH − 2 l°m NaCl



=

Atomic weight 40 = = 20 Valency 2

According to Faraday’s first law of electrolysis : equivalent weight ×i×t 96500 20 \ 60 = ×5×t 96500 96500 × 60 t= s 20 × 5 96500 × 60 = h 20 × 5 × 60 × 60 W = Zit =



= 280 × 10 + 2 × 248 × 10 – 2 × 126 × 10





= (280 + 496 – 252) × 10

22. (b) : For strong electrolytes the plot of molar



= 524 × 10–4 S m2 mol–1

2 –4

2

–4

–4

–4

= 16 h conductance ( L m ) vs C is linear.

29

Electrochemistry

Str

on

m

ge

lec

tro

23. (c) : According to Kohlrausch’s law, ∞ Lm = l +∞ + l −∞ For CaCl2, ∞ Lm (CaCl 2 ) = l ∞ 2+ + 2 l ∞ − Ca Cl = 118.88 × 10–4 + 2 × 77.33 × 10–4

lyt

e



C



Variation of molar conductance (Lm) with C for strong electrolyte

vvv

= 273.54 × 10–4 m2 mho mol–1

30

VITEEE CHAPTERWISE SOLUTIONS

6

CHAPTER

Isomerism in Organic Compounds

1.

Which of the following isomerism is shown by ethyl acetoacetate ? (a) Geometrical isomerism (b) Keto-enol tautomerism (c) Enantiomerism (d) Diastereoisomerism (2014)

2.

Following compounds are respectively_____ geometrical isomers. Cl Cl

(P)

P (a) cis (b) cis (c) trans (d) cis

Cl Cl

(Q)

Q cis trans cis trans

3.

The concentration of an organic compound in chloroform is 6.15 g per 100 mL of solution. A portion of this solution in a 5 cm polarimeter tube causes an observed rotation of –1.2°. What is the specific rotation of the compound? (a) +12° (b) –3.9° (c) –39° (d) +61.5° (2009)

4.

Maleic acid and fumaric acid are (a) position isomers (b) geometric isomers (c) enantiomers (d) functional isomers.

Cl

(R)

Cl

5.

R trans trans cis cis

Identify, which of the below does not possess any element of symmetry. (a) (+)-Tartaric acid (b) Carbon tetrachloride (c) Methane (d) meso-Tartaric acid (2007)

(2013)

Answer Key 1.

(b)

2.

(a)

3.

(b)

4.

(b)

5.

(2008)

(a)

31

Isomerism in Organic Compounds

O

1.

(b) : CH3 C

O CH2 (Keto)

C

OC2H5

OH



2.

e planations

CH3 C Cl

(a) : cis (P)

Cl

O CH

(Enol)

C

Cl

cis (Q)

Cl

OC2H5

3.

(b) : [α] =

4.

(b) : H

H

Cl

trans (R)

Cl

5.

vvv

C C

[α]observed = l×C

−1.2 = − 3.9° 6.15 5× 100

COOH

H

COOH

HOOC

Maleic acid (Cis-form)

C C

COOH

H

Fumaric acid (Trans-form)

Both are geometrical isomers of each other. (a) : (+) or (–)-Tartaric acid does not possess any element of symmetry.

32

VITEEE CHAPTERWISE SOLUTIONS

7

CHAPTER

1.

Alcohols and Ethers

The final product (IV) in the sequence of reactions

3.

CH3



CH

OH

CH3



PBr3

CH2

(II)

(a) CH3

CH

Mg (I) Ether

O

(b) CH3 (c) CH3

4.

(a)

(c)



(b) CH2

(2014)

(b)

Br

C(CH3)3

OCH3 Br

OH

C(CH3)3

(d)

B

CH

O

C

H , CH2 Cl

CH

(2013)

C

O H,

(c) CH3CH2CHO ,

CH2

CH

OH

Cl

CH3CH2CH

C

H

OH Cl

O

(d) CH2

CH

C



H,

CH2CH2CHO ClO

(2013) 5.

Phenol is heated with phthalic anhydride in the presence of conc. H2SO4. The product gives pink colour with alkali. The product is (a) phenolphthalein (b) bakelite (c) salicylic acid (d) fluorescein. (2013)

6.

End product of the following reaction is



O

Cl

C(CH3)3

CH

C H

OH

O



C(CH3)3



HClO



Br

Br

(b)

KHSO

CH2CH2OH

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



4 Glycerol A A and B respectively are

(a) CH2

CH3

CH3



Product(s) is/are

O

CH2CH2Br

CH

products.

(c) both (a) and (b) (d) none of these. (2013)

(IV) is



CH3 (d) CH3 CH OCH2CH3

2.

H2O

(III)

OCH2CH2OH

CH

conc.H2SO4

(a)

CH2

CH3



OH

O + HBr

33

Alcohols and Ethers

OH

(a) O

O

(b) HO Br

Br

(c) Br

OH (d) HO Br

HO

7.

OH



8.

(d) H2C

O

CH3



(2013)

Which one of the following compounds will be most readily dehydrated? O (a) H3C

OH O

(b) H3C OH O

OH

(c) H3C

13.

9.

O

(2012)

When glycerol is treated with excess of HI, it produces (a) 2-iodopropane (b) allyl iodide (c) propene (d) glycerol triiodide. (2012)

OH

OH

(d)

H H



H

H

H

C2H5OH

(CH3)2CH — CH2OC2H5 + HBr C2H5O –

(CH3)2CH — CH2OC2H5 + Br–

The mechanisms of reactions (i) and (ii) are respectively (a) SN1 and SN2 (b) SN1 and SN1 (c) SN2 and SN2 (d) SN2 and SN1 (2012)

OH – H+

+ HCHO or

OH



HH

12. The alcohol having molecular formula C4H9OH, when shaken with a mixture of anhydrous ZnCl2 and conc. HCl gives an oily layer product after five minutes. The alcohol is (a) H3C(CH2)3OH (b) (CH3)2CH—CH2—OH (c) (CH3)3C—OH (d) H3C—CH(OH)CH2—CH3 (2011) OH

OH

(d) H3C

HO H

H

(i) (CH3)2CH — CH2Br



C H

OH H

(ii) (CH3)2CH — CH2Br

O

(c) CH3

OH

H H



11. Consider the reactions,

C

H

(2012)

CH2 CH2

H

(c)

OH OH

H

(b)

(2013)

O

(b) CH3

OH

H

(a)

An organic compound B is formed by the reaction of ethyl magnesium iodide with a substance A, followed by treatment with dilute aqueous acid. Compound B does not react with PCC or PDC in dichloromethane. Which of the following is a possible compound for A? O (a) H2C

10. Which of the following conformers for ethylene glycol is most stable?



CH2OH +

This reaction is called (a) Reimer-Tiemann reaction (b) Lederer-Manasse reaction

OH

CH2OH

34

VITEEE CHAPTERWISE SOLUTIONS

(c) Sandmeyer reaction (d) Kolbe’s reaction.

20. The reaction, ROH + H2CN2 in the presence of HBF4, gives the following product (a) ROCH3 (b) RCH2OH (c) ROH⋅CH2N2 (d) RCH2CH3 (2010)

(2011)

14. Sometimes explosion occurs while distilling ethers. It is due to the presence of (a) peroxides (b) oxides (c) ketones (d) aldehydes. (2011)

21. Hydroboration oxidation of 4-methyloctene would give (a) 4-methyloctanol (b) 2-methyldecane (c) 4-methylheptanol (d) 4-methyl-2-octanone. (2008)

15. Glycerine is used as a preservative for fruits and eatables because (a) it makes them sweet (b) it acts as an insecticide (c) it keeps the food moist (d) all of the above. (2011) 16. The formula of ethyl carbinol is (a) CH3OH (b) CH3CH2OH (c) CH3CH2CH2OH (d) (CH3)3COH

22. When ethyl alcohol is heated with conc. H2SO4, the product obtained is (a) CH3COOC2H5 (b) C2H2 (c) C2H6 (d) C2H4 (2008)

(2010)

23. Anisole is the product obtained from phenol by the reaction known as (a) coupling (b) etherification (c) oxidation (d) esterification. (2008)

17. Which of the following gives red colour in Victor Meyer’s test? (a) n-Propyl alcohol (b) iso-Propyl alcohol (c) tert-Butyl alcohol (d) sec-Butyl alcohol (2010)

24. Ethylene glycol gives oxalic acid on oxidation with (a) acidified K2Cr2O7 (b) acidified KMnO4 (c) alkaline KMnO4 (d) periodic acid. (2008)

18. Glycerol on treatment with oxalic acid at 110°C forms (a) formic acid (b) allyl alcohol (c) CO2 and CO (d) acrolein. (2010) 19. When p-nitrobromobenzene reacts with sodium ethoxide, the product obtained is (a) p-nitroanisole (b) ethyl phenyl ether (c) p-nitrophenetole (d) no reaction occurs. (2010)

25. Phenol is more acidic than (a) p-chlorophenol (b) p-nitrophenol (c) o-nitrophenol (d) ethanol. (2007)

Answer Key 1.

(c)

2.

9.

(a)

10. (d)

11. (a)

17. (a)

18. (a)

19. (c)

25. (d)

(d)

3.

(a)

4.

(b)

5.

(a)

6.

(d)

7.

(b)

8.

(c)

12. (d)

13.

(b)

14.

(a)

15.

(c)

16.

(c)

20. (a)

21.

(a)

22.

(d)

23.

(b)

24.

(c)

35

Alcohols and Ethers

1.

e planations

(c) : CH3 CH

PBr3

OH

CH3 CH

CH3

Br

5.

Mg Ether

(a) :

O

(I)

CH2CH2OMgBr

CH2 CH2 O

O

C

CH3 CH3

O

CH

H

CH3

CH3 CH

C

O

MgBr

O +

C H conc. H2SO4  (–H2O)

(II)

CH3

(III)

OH

H2O



CH3

CH

CH2CH2OH

CH3 3-Methylbutanol (IV)

6.

+

OCH3

Cl

Cl2/FeCl3

(II)

(a) :

O

C(CH3)3

7.

OH

Cl

HBr,  (III)

C(CH3)3

3.

O

C(CH3)3 +

OH

OH2

conc. H2SO4

–H +

4.

(b) : CHOH

– 2H2O

CH2 CH

–

CH2OH

CHO

Glycerol

Acrolein (A) – +



HO–Cl

Br OH

Enolisation

OH

HO

(b) : ‘B’ is a tertiary alcohol based on given properties. Tertiary alcohols are formed by the reaction of ketones with Grignard reagent.

CH3 CH2

C ‘A’

CH3 + CH3CH2MgI

CH3CH2 C

+

+

CH2 CH

CHCl CHO (B )

H3O +

CH2CH3

CH3 ‘B ’ (3° Alcohol)



Tertiary alcohols do not have oxidisable hydrogen linked to carbon atom bearing hydroxyl group, thus, are stable to oxidation under ordinary conditions.

8.

(c) : Dehydration of alcohols involve formation of carbocation intermediate. More the stability of carbocation, higher is the ease of dehydration. The order of stability of carbocations is

CHO CH2OH

H + Br – 1,4-addition

OH

(More stable)

KHSO4

O–

O

O



–H2O

CH2OH

O Br

Friedel-Crafts reaction (I)

OCH3



OH

Pink colour

(d) :

(CH3)3CCl/AlCl3

(d) :

OH

Phenolphthalein NaOH (Phenolphthalein test)

OCH3

2.

OH

Phenol (2 molecules)

36

VITEEE CHAPTERWISE SOLUTIONS

O

O


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

4.

5.

C

II (CH3)3C (CH3)3C

In a Claisen condensation reaction (when an ester is treated with a strong base) (a) a proton is removed from the a-carbon to form a resonance stabilised carbanion of the ester (b) carbanion acts as a nucleophile in a nucleophilic acyl substitution reaction with another ester molecule (c) a new C—C bond is formed (d) all of the above statements are correct. (2013) O

6.



CH3

C

O CH2

CH2

(a) CH3

OH

O

C CH 2 CH2

C

Ketones [R — C — R] where, R = R′ = alkyl O group can be obtained in one step by

CH3

O

CH2CH3

CH3

O

(b) CH3 C CH2

IV (b) IV > III > II > I (d) III > II > I > IV (2014)

CH2

‘A’ formed in this reaction is

O

O

O

(i) CH3MgBr(one mole) A (ii) H3O+

O

C

C

O

(c) H3C H3C

CH2

C OH

CH3

O

CH3

(d) CH3

C

CH2

CH3 CH2

C

CH3

OH

(2013)

39

Carbonyl Compounds

7.

Which of the following does not undergo benzoin condensation? CH2CHO

(a)

CHO



(c)

(d)

OCH3

8.



9.



(2013)

(2012)

List II

Codes A (a) 4 (b) 4 (c) 2 (d) 2

B 1 2 3 1

C 3 3 4 4

1. 2. 3. 4.

D 2 1 1 3

O

C

C

H

C

OCH3

O C

CH2OH

(2011)

12. Which of the following will give Cannizzaro reaction? (a) CH3CHO (b) CH3COCH3 (c) (CH3)3CCHO (d) CH3CH2CHO (2011) 13. Which of the following is a Gattermann aldehyde synthesis? H2/Pd

(a)

COCl BaSO 4

(b)

H + CO + HCl

CHO AlCl3

(c)

Fries rearrangement

(2012)

+ HCl + HCN (i)Anhy. AlCl3 (ii)H3O+

(d)

Phenolphthalein Benzoin condensation Oil of wintergreen

CHO

10. Match the compounds given in List I with List II and select the suitable option using the codes given below. A. Benzaldehyde B. Phthalic anhydride C. Phenyl benzoate D. Methyl salicylate

O

X + Se + H2O : X is

(III)

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

List I

SeO 2

(d) none of the above.

CH3 Which of the above compounds on being warmed with iodine solution and NaOH, will give iodoform? (a) (i), (iii) and (iv) (b) Only (ii) (c) (i), (ii) and (iii) (d) (i) and (ii) (2012) The order of reactivity of phenyl magnesium bromide (PhMgBr) with the following compounds H3C H3C Ph C O C O C O H H3C Ph

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

(b) CH3 (c) CH3

Following compounds are given (i) CH3CH2OH (ii) CH3COCH3 (iii) CH3CHOH (iv) CH3OH

(II)

X is

CH3

O

CH3

(I)

C

(a) CH3

CHO



11. CH3

(b)

CHO

O

CH3

CrO2Cl2

CHO CHO

(2010) 14. Aldol is (a) b-hydroxybutyraldehyde (b) a-hydroxybutanal (c) b-hydroxypropanal (d) none of the above.

(2010)

15. On treatment of propanone with dilute Ba(OH)2, the product formed is (a) aldol (b) phorone (c) propionaldehyde (d) 4-hydroxy-4-methyl-2-pentanone. (2010)

40

VITEEE CHAPTERWISE SOLUTIONS

16. Formation of cyanohydrin from a ketone is an example of (a) electrophilic addition (b) nucleophilic addition (c) nucleophilic substitution (d) electrophilic substitution. (2010)

(a) H2O (c) CO

(b) HCHO (d) CH3CHO (2008)

23. The gas evolved on heating alkali formate with soda-lime is (a) CO (b) CO2 (c) hydrogen (d) water vapour. (2008)

17. The fatty acid which shows reducing property is (a) acetic acid (b) ethanoic acid (c) oxalic acid (d) formic acid. (2010)

24. Alkyl cyanides undergo Stephen reduction to produce (a) aldehyde (b) secondary amine (c) primary amine (d) amide. (2008)

18. What are X and Y in the following reactions sequence? Cl 2 Cl 2 C 2H 5OH  → X  →Y (a) C2H5Cl, CH3CHO (b) CH3CHO, CH3CO2H (c) CH3CHO,CCl3CHO (d) C2H5Cl, CCl3CHO (2009)

25. In the Grignard reaction, which metal forms an organometallic bond? (a) Sodium (b) Titanium (c) Magnesium (d) Palladium (2007) 26. Aldol condensation is given by (a) trimethylacetaldehyde (b) acetaldehyde (c) benzaldehyde (d) formaldehyde.

19. The synthesis of crotonaldehyde from acetaldehyde is an example of ....... reaction. (a) nucleophilic addition (b) elimination (c) electrophilic addition (d) nucleophilic addition-elimination (2009)

(2007)

20. A Wittig reaction with an aldehyde gives (a) ketone compound (b) a long chain fatty acid (c) olefin compound (d) epoxide. (2008)

27. In which of the below reactions do we find a, b-unsaturated carbonyl compounds undergoing a ring closure reaction with conjugated dienes? (a) Perkin reaction (b) Diels-Alder reaction (c) Claisen rearrangement (d) Hofmann reaction (2007)

21. Cannizzaro reaction is given by (a) HCHO (b) CH3COCH3 (c) CH3CHO (d) CH3CH2OH

28. The catalyst used in Rosenmund reaction is (a) Zn/Hg (b) Pd/BaSO4 (c) Raney Ni (d) Na in ethanol. (2008)

NH

3 →? 29. C6H5CHO  (a) (C6H5CHN)2CHC6H5 (b) C6H5NHCH3 (c) C6H5CH2N2 (d) C6H5NHC6H5

22.

Identify the reactant.

(2007)

Answer Key 1.

(a)

2.

5.

(d)

6.

(c)

7.

(a)

8.

(c)

9.

(d)

10. (d)

(b)

3.

11. (a)

(a)

4.

12. (c)

(c)

13.

(c)

14.

(a)

15.

(d)

16.

(b)

17. (d)

18. (c)

19. (d)

20. (c)

21.

(a)

22.

(c)

23.

(c)

24.

(a)

25. (c)

26. (b)

27. (b)

28. (b)

29.

(a)

41

Carbonyl Compounds

(a) : Ph

C

C

Hg2+

CH3

as Claisen condensation and product is a b-keto ester. O

(i) +H2O:

:

1.

e planations Ph

C



C

CH3

(ii) –H +



Hg2+ OH Ph

C

2CH3

CH2

C

CH3

Ph

(ii) Hg 2+

C

Hg +

C

CH3

CH3

CH2



O

(a) : As the number and the size of the alkyl groups increase, reactivity decreases. Hence, the reactivity order is H

C

I

O >

H H3C

C

II

O >







H3C

C

H3C

III

(CH3)3C

>

4.

(CH3)3C

R R R R R R

CHOH

[O] K2Cr2O7/H2SO4

CHOH

PCC CH2Cl2

CHOH +

R R

CH3 CH3

C

R R

C

CH3 + CH3OH

O

CH3 + CH3OH

CH3

C

O

CH3

H CH3O –

O CH3



CH –

C

Step II O

O

C

CH

CH3CH – IV

C

O OCH3 + CH3CH2

O

C

OCH3

O

CH3

O CH3

(c) : Secondary alcohols, on oxidation yield ketones.

CH2

C

CH3

CH

C

O

CH3

O

O



Step III O

C

O



CH3

CH2

C O

CH3

CH

CH3 C

O

CH3

O

O

[(CH3)3CO]3Al

R R

5.

CH3



(b) : Aldehydes and a-hydroxy ketones give positive Tollens’ test. Glucose is a polyhydroxy aldehyde and fructose is an a-hydroxy ketone.

H

O

O

CH3



C



Mechanism : Step I

CH2

3.

CH



Ph C

(i) CH3O – (ii) H +

CH3 O

C

H Tautomerises

2.

O

O

OH (i) +H+

C

C

CH3 CH2

O

(d) : When two molecules of an ester undergo a condensation reaction, the reaction is called

6.

O

O

C

CH C CH3

O

CH3 + CH3O

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

42

VITEEE CHAPTERWISE SOLUTIONS

O

O

CH3 C

CH2CH2

C

O

CH2CH3

CH3 CH3

C



8.



– C2H5OH

C O

CH2

(a) : Phenylethanal (C6H5CH2CHO) does not have CHO group directly attached to phenyl group i.e., it is not an aromatic aldehyde. Only aromatic aldehydes undergo benzoin condensation. O (c) : Compounds having either CH3 C group or CH3CHOH group, give iodoform when warmed with I2 and NaOH. Thus, O H

(i)

C

CH3, CH3

CHOH CH3

(ii)

(iii)

(d) : Since alkyl group has +I effect and aryl group has +R effect, thus greater the number of alkyl and aryl groups attached to the carbonyl group, its reactivity towards nucleophilic addition reaction decreases. Secondly, as the steric crowding on carbonyl group increases, the reactivity decreases accordingly. \ Correct reactivity order for reaction with PhMgBr is H3 C H3 C Ph C O > C O > C O H H3 C Ph

10. (d)

(I)

(II)

(III)

11. (a) : SeO2 oxidises the reactive methylene (CH2) group adjacent to carbonyl group to another CO group. O SeO2 CH C CH 3







Acetone

3

CH3

Methyl glyoxal

CH3

CH3 C CH2OH + CH3 C COONa CH3



CH3

13. (c) : Gattermann aldehyde synthesis reaction is + HCN + HCl



Anhy. AlCl3

Benzene

CH

NH·HCl

CHO

H3O+ –NH4Cl



Benzaldehyde

14. (a) : Aldol is b-hydroxybutyraldehyde or 3-hydroxybutanal i.e. H



CH3 C O + H CH2CHO Acetaldehyde Acetaldehyde

dil. NaOH or K2CO3

H CH3 C CH2CHO OH

3-Hydroxybutanal (Aldol)



15. (d) : In the presence of Ba(OH)2, propanone (acetone) undergoes aldol condensation as O



CH3

C

Acetone

CH3 + HCH2COCH3

Ba(OH)2

Acetone

OH H3C

O O CH3 C C H + Se + H2O



CHO + NaOH

CH3



O





O

2CH3 C

C

CH3CHOH, CH3

9.

CH3

CH3 C CH2

C2H5O

7.

(ii) H3O +

12. (c) : (CH3)3CCHO does not contain a-H atom, hence it will undergo Cannizzaro reaction.

CH3

CH2CH2

OH

(i) CH3MgBr (one mole)



C

CH2COCH3

CH3 4-Hydroxy-4-methylpentan-2-one (Diacetone alcohol)

43

Carbonyl Compounds

22. (c) :

CN

16. (b) : R2C

CN –

O+

Nucleophile

H+





C6H6 + CO + HCl

CN



It is a direct method for introducing an O

R2 C

OH

17. (d) : Formic acid (HCO OH), because of the presence of — CHO group also reduces Tollens’ reagent, Fehling’s solution etc. Thus, it has reducing property. Cl2, [O] –HCl

Ethyl alcohol

3Cl2

CH3CHO Acetaldehyde



CCl3CHO

–3HCl

Chloral

(X)

(Y)

H

19. (d) : CH3

C

H

H

Acetaldehyde Acetaldehyde



O + H CH2CHO

CH3

C

C

OH H

Dil. NaOH

(Nucleophilic addition)

Heat CHO –H2O (Elimination)

CH3 CH



CHCHO

20. (c) : Wittig reaction affords an important and useful method for the synthesis of alkenes (olefins) by the treatment of aldehydes or ketones with alkylidene triphenyl phosphorane (Ph3P CR2) or simply known as phosphorane. –



+

C O + :CH2 P(C6H5)3

23. (c) : The gas evolved on heating alkali formate with soda-lime is hydrogen. HCOONa + NaOH CaO  → H 2 + Na 2CO3 ” D 24. (a) : When a solution of alkyl cyanides in ether is reduced with stannous chloride and conc. hydrochloric acid at room temperature, imine hydrochloride is obtained which on subsequent hydrolysis with boiling water, forms an aldehyde (Stephen’s reduction.)

RCHO + NH4Cl

25. (c) : In the Grignard reaction, magnesium metal forms an organometallic bond. RX + Mg

Dry ether

R

C CH2 + (C6H5)3P O

21. (a) : In the presence of a strong base, aldehydes having no a-hydrogens undergo self oxidation reduction i.e. disproportionation reaction. This is known as Cannizzaro reaction. As formaldehyde does not have a-hydrogen hence, it undergoes Cannizzaro reaction.

Mg

X

Grignard reagent

26. (b) : Aldol condensation is given by acetaldehyde due to the presence of a-hydrogen atom. H



CH3

C

O+H

Acetaldehyde

CH2CHO

Dil. NaOH

Acetaldehyde

CH3

CH

CH2

CHO

OH

3-Hydroxybutanal (Aldol)

Dry ether or THF

H2O/H +

RCH

R





Crotonaldehyde

C6H5CHO + HCl Benzaldehyde

aldehydic group ( C H) in benzene. It is called Gattermann-Koch synthesis when a mixture of dry HCl gas and carbon monoxide is used in presence of anhydrous AlCl3.

Thus, it is an example of nucleophilic addition because the attacking species is a nucleophile.

18. (c) : CH3CH2OH

anhyd. AlCl3

O–

R2 C

27. (b) : The addition of a, b-unsaturated carbonyl compound, with conjugated diene is called Diels-Alder reaction. Z Z + Diene

(Dienophile)

Substituted cyclohexene

28. (b) : Acid chlorides can be reduced into aldehydes with hydrogen in boiling xylene

44

VITEEE CHAPTERWISE SOLUTIONS

using palladium or platinum as catalyst supported over barium sulphate. This reaction is called Rosenmund reaction.

29. (a) : C6H5CH C6H5CH

O C

Cl + H2



Benzoyl chloride

Pd/BaSO4

Boiling xylene

CHO



+ HCl Benzaldehyde



vvv

O H2 N H + + O O H2 N H

CH

C6H5

Benzaldehyde (Three molecules)

C6H5CH

N

C6H5CH

N

CH

Hydrobenzamide

C6H5 + 3H2O

45

Carboxylic Acids and Their Derivatives

9

CHAPTER

1.

Carboxylic Acids and Their Derivatives

An optically active compound ‘X’ has molecular formula C4H8O3. It evolves CO2 with aqueous NaHCO3. ‘X’ reacts with LiAlH4 to give an achiral compound. ‘X’ is (a) CH3

CH2

(b) CH3 CH

CH

COOH

OH COOH

OMe

(c) CH3



Isomers are (a) C and E (c) D and E

5.

In a set of reactions acetic acid yielded a product D.



CH3COOH



The structure of D would be OH

CH COOH

(a)

CH2OH

(d) CH3

CH

CH2

COOH

OH

2.

(b)





CO2 +

*

(c) (2013)

COONa

C

C

D

CH3

CN C

COOH

CH3

(2013)

4. Consider the following reactions sequence, CH2COOH P/Br KOH (alc.) O ,H O 2 A B 3 2 C CH2COOH CF3 CO3H Alkaline D E KMnO4

C

CH3

OH CH2

C OH

CH3

(2011)

6.

Self condensation of two moles of ethyl acetate in the presence of sodium ethoxide after acidification yields (a) acetic acid (b) acetoacetic ester (c) ethyl propionate (d) ethyl butyrate. (2011)

7.

C2H5I

COONa

(c) both (a) and (b) (d) none of the above.

Benzene HCN B anhy.AlCl3 H 2O

COOH

(d)

C is with the product (a) CO2

(b)

CH2

A

CN

*  COOH + NaHCO 3

3.

SOCl 2

OH

(2013)

Which is more basic oxygen in an ester? O R C O R (a) Carbonyl oxygen, a (b) Carboxyl oxygen, b (c) Equally basic (d) Both are acidic oxygen.

(b) C and D (d) C, D and E. (2013)

Br2 H3O+ Alcoholic KCN X Y Z A CCl4 KOH

The product ‘A’ is (a) succinic acid (c) oxalic acid

(b) melonic acid (d) maleic acid. (2010)

46

VITEEE CHAPTERWISE SOLUTIONS

8.

What are A, B, C in the following reactions? ∆ →A (I) (CH 3CO 2 )2Ca 



 →B (II) CH 3CO 2H Red P

11. The weakest acid amongst the following is (a) ClCH2COOH (b) HCOOH (c) FCH2CH2COOH (d) CH2(I)COOH (2007)

HI

PO

4 10 →C (III) 2CH 3CO 2H  A B (a) C2H6 CH3COCH3 (b) (CH3CO)2O C2H6 (c) CH3COCH3 C2H6 (d) CH3COCH3 (CH3CO)2O

9.

C (CH3CO)2O CH3COCH3 (CH3CO)2O C2H6 (2009)

12. Trans esterification is the process of (a) conversion of an aliphatic acid to ester (b) conversion of an aromatic acid to ester (c) conversion of one ester to another ester (d) conversion of an ester into its components namely acid and alcohol. (2007)

Consider the following reaction, EtOH/H O

2 C 2H 5Cl + AgCN  → X (major)



13. The IUPAC name for O

Which one of the following statements is true for X? (I) It gives propionic acid on hydrolysis. (II) It has an ester functional group. (III) It has a nitrogen linked to ethyl carbon. (IV) It has a cyanide group. (a) IV (b) III (c) II (d) I (2009)

O

H3C CH2 C C H2 CH2 C OCH3 is

(a) ethyl-4-oxoheptanoate (b) methyl-4-oxoheptanoate (c) ethyl-4-oxohexanoate (d) methyl-4-oxohexanoate.

(2007)

H O

2 →? 14. (CH2CO)2O + RMgX  (a) ROOC(CH2)COOR (b) RCOCH2CH2COOH (c) RCOOR (d) RCOOH

10. When acetamide is hydrolysed by boiling with acid, the product obtained is (a) acetic acid (b) ethyl amine (c) ethanol (d) acetamide. (2008)

(2007)

Answer Key 1.

(c)

2.

(a)

9.

(b)

10. (a)

3.

(a)

11. (b)

4.

(c)

12. (c)

5.

(b)

6.

(b)

13.

(d)

14.

(b)

7.

(a)

8.

(c)

47

Carboxylic Acids and Their Derivatives

NaHCO3(aq.)

1.

(c) :  X



\ It must contains CH2OH

H *C COOH



e planations 5.

CO2

COOH group.

LiAlH4

2.

to it. Hence, a-oxygen atom can donate lone pair of electrons more easily, and is more basic than b-oxygen. *

COOH + NaHCO3 *

4.

(c) : CH2COOH CH2COOH

6.

H

C

OH

COOH

CHCOOH

CF3CO3H (anti addition)

D d(+) and l(–) Tartaric acid (Racemic mixture)

CHCOOH

O3 H2O

B

alk. KMnO4

H

(syn addition)

C

CH3COCH2COOC2H5 + C2H5OH Acetoacetic ester



This reaction is called Claisen condensation.

7.

(a) : C2H5I

Br

CH2

COOH H C OH H C OH COOH E meso -Tartaric acid

COOH COOH

C (Oxalic acid)

KCN

CH2

C2H4 Ethylene (X) Br2

CCl4

1, 2-Dibromoethane (Y)

(Z)

H3O+

CH2CH2 Br

CN

CHCOOH

2

Alc. KOH (Dehydrohalogenation)

CN

B (Maleic acid)

COOH

NaOC2H5

Ethyl acetate



CHCOOH

HO





A

Alc. KOH

CH3

(b) : 2CH3COOC2H5

CH2COOH

Succinic acid

CH3

(B)

(D)

BrCHCOOH

P/Br2 HVZ-reaction

O

COOH

COONa + CO2 + H2O



C

Benzene Anhy. AlCl3

C

HCN

OH

H +/H2O



(a) : R C O R b-oxygen has lesser electron density because

(a) :

CH3

(C)

(Achiral)

CO group attached

CH3COCl (A)

CN

CH3

of electron withdrawing

3.

C

H C CH2OH

(X, C4H8O3)

SOCl2

OH

CH2OH

CH3

O

(b) : CH3COOH

COOH CH2

CH2 COOH

Succinic acid (A)

8.

(c) :



(I) (CH3CO2)2Ca





Calcium acetate

(II) CH3CO2H + 6HI Acetic acid

(III) 2CH3CO2H Acetic acid

CH3COCH3 + CaCO3

Acetone (A) Red P

CH3 CH3 + 2H2O + 3I2 Ethane (B)

P4 O10, 

CH3CO CH3CO

O + H2O

Acetic anhydride (C)

48 9.

VITEEE CHAPTERWISE SOLUTIONS

13. (d) :

EtOH

(b) : C2H5



Cl + AgCN H O 2 C2H5

NC + AgCl

N-linked to ethyl carbon

10. (a) : Acid amides when boiled with mineral acids alkalies are hydrolysed to corresponding acids and ammonia.

CH3CONH2 + H2O

11. (b)

O

12. (c) : CH3

C

H + or OH – 

CH3COOH + NH3

6



H3C

O

5

CH2

C

4

3

O

2

CH2

CH2

C

1

Methyl-4-oxohexanoate

OCH3

O

14. (b) :

CH2

C

CH2

C

O + RMgX

O

Succinic anhydrideOMgX

OC2H5 + CH 3 OH

H+

Ethyl acetate (One ester)

CH2

C

CH2

C

C

O

H2O –Mg(OH)X

O

O CH3

R

OCH3 + C2H5OH

Methyl acetate (Another ester)

vvv

O R

C

CH2

CH2

COOH

49

Organic Nitrogen Compounds

10 CHAPTER

Organic Nitrogen Compounds (a) LiAlH4 (c) Zn

NH2 Ac2O

1.

Br2

NHCOCH 3 Br

(c)

H+

CH3

NH2

(b)

COCH3 Br CH3

C6H5NH2



C

5.

(d)

NaNO2/HCl 0°C

X

COCH3

CH3 NH2

Y

(2013)

Which of the following compound is the most basic? (a) O2N NH2 (b)

CH2NH2

(c)

N

COCH3

H

Br

(d) (2014)

H2O/H+

7.

p-Toluidine and benzyl amine can be distinguished by (a) Sandmeyer’s reaction (b) Dye test (c) Molisch’s test (d) Gattermann reaction. (2011)

8.

Which one of the following will be most basic? (a) Aniline (b) p-Methoxyaniline (c) p-Methylaniline (d) Benzyl amine (2011)

9.

Nitrobenzene can be converted into azobenzene by reduction with (a) Zn, NH4Cl, D (b) Zn/NaOH,CH3OH (c) Zn/NaOH (d) LiAlH4, ether (2010)

(2013)

Benzenediazonium chloride on treatment with hypophosphorous acid and water yields benzene. Which of the following is used as a catalyst in this reaction?

(2012)

The product formed when phthalimide is treated with a mixture of Br2 and strong NaOH solution is (a) aniline (b) phthalamide (c) phthalic acid (d) anthranilic acid. (2011)

Z,

Select the correct statement. (a) LiAlH4 reduces methyl cyanide to methyl amine. (b) Alkane nitrile has electrophilic as well as nucleophilic centres. (c) Saponification is a reversible reaction. (d) Alkaline hydrolysis of methane nitrile forms methanoic acid. (2013)

NH2

6.

CH3

CuCN

Z is identified as (a) C6H5—NH—CH3 (b) C6H5—CH2—NH2 (c) C6H5—CH2—COOH (d) C6H5—COOH

4.

H2O

The final product ‘C’ in the above reaction is



3.

B

CH3COOH

CH3

(a)

2.

A

(b) Red P (d) Cu+

50

VITEEE CHAPTERWISE SOLUTIONS

10. The one which is least basic is (a) NH3 (b) C6H5NH2 (c) (C6H5)3N (d) (C6H5)2NH

15. Secondary nitroalkanes can be converted into ketones by using Y. Identify Y from the following. R R

(2010)

11. Which of the following converts CH3CONH2 to CH3NH2? (a) NaBr (b) NaOBr (c) Br2 (d) None of the above. (2010)

R (a) Aqueous HCl (c) KMnO4

16. Urea on slow heating gives (a) NH2CONHNO2 (b) NH2CONHCONH2 (c) HCNO (d) NH2CONH2⋅HNO3

12. In Gattermann reaction, a diazonium group is replaced by X using Y. X and Y are X Y – (a) Cl Cu/HCl + (b) Cl CuCl2/HCl – (c) Cl CuCl2/HCl (d) Cl2 Cu2O/HCl (2009)

18. When aqueous solution of benzene diazonium chloride is boiled, the product formed is (a) C6H5CH2OH (b) C6H6 + N2 (c) C6H5COOH (d) C6H5OH (2007)

(2008)

14. Which will not go for diazotisation? (a) C6H5NH2 (b) C6H5CH2NH2 (c)



(2007)

17. The correct sequence of base strengths in aqueous solution is (a) (CH3)2NH > CH3NH2 > (CH3)3N (b) (CH3)3N >(CH3)2NH > CH3NH2 (c) (CH3)3N > CH3NH2 = (CH3)2NH (d) (CH3)2NH > (CH3)3N > CH3NH2 (2007)

13. (a) CH3CH2NO2 (b) CH3CH2NO2 + CH3NO2 (c) 2CH3NO2 (d) CH2 CH2

R (b) Aqueous NaOH (d) CO (2008)

19. Carbylamine reaction is given by aliphatic (a) primary amine (b) secondary amine (c) tertiary amine (d) quaternary ammonium salt. (2007)

(d) (2008)

Answer Key 1.

(d)

9.

(b,d) 10. (c)

17. (a)

2.

(d)

18. (d)

3.

(b)

11. (b) 19. (a)

4.

(d)

12. (a)

5.

(b)

6.

(d)

7.

(b)

8.

(d)

13.

(b)

14.

(b)

15.

(a)

16.

(b)

51

Organic Nitrogen Compounds

e planations

NH2

1.

NHCOCH3 Ac2O

(d) :

6.

(d) :

CH3 (A)

NHCOCH3 Br

NH2 HOH

CH3

Br

O C C

CH3

(B)

Phthalimide

(d) : +

2NaBr + Na2CO3 + H2O +

Benzenediazonium chloride

(X )



CuCN

COOH

H2O/H +

Benzoic acid

3.

(b) :



(a)

(Y )

LiAlH4

CH3 CN

CH3

Methyl cyanide +

(b) CH3C N:

CH3C

8.

(d) : In aniline, p-methoxyaniline and p-methylaniline, the lone pair of electrons on the N-atom is delocalised on the benzene ring while in benzyl amine it is localised and more available for donation. Hence, benzyl amine is most basic among the given compounds.

9.

(b, d) :

CH2 NH2 Ethylamine

–

N:

NO2



(c) Saponification is an irreversible reaction. (d) CH3

CN

OH –

Nitrobenzene



N2+Cl–

4.

(d) :

Cu+

Benzenediazonium chloride

+ N2 + HCl + H3PO3

Benzene

N

N

Azobenzene

+ 4H2O

NO2 2

+ H3PO2 + H2O

Zn/NaOH in CH3OH 8[H]

CH3COOH Ethanoic acid

NH2 Anthranilic acid

(b) : p-Toluidine undergoes diazotisation reaction and gives red dye while benzyl amine does not undergo diazotisation reaction.

Electrophilic Nucleophilic



C OH

7. CN

Phenyl cyanide

(Z )

O

–

N NCl

Aniline



NH + Br2 + 4NaOH

O

(C)

NaNO2/HCl NH2 0°C



(b) : Benzylamine is most basic due to localised lone pair of electrons on nitrogen atom while in other compounds, as a result of resonance, the lone pair of electrons on nitrogen atom gets delocalised over benzene ring and thus, is less easily available for donation.

Br2/CH3COOH

CH3

2.

5.

+ 8[H]

Nitrobenzene (2 molecules)

LiAlH4 Ether

N

N

Azobenzene

+ 4H2O

10. (c) : Presence of electron withdrawing group like phenyl group decreases the electron density of nitrogen and hence, the lone pair of nitrogen is not available for donation.

52

VITEEE CHAPTERWISE SOLUTIONS

Thus, (C6H5)3N is least basic because of the presence of three electron withdrawing phenyl groups.

11. (b) : NaOBr converts amide group (—CONH2) to amine group (—NH2).

CH3CONH2

NaOBr

Acetamide

12. (a) :

Benzenediazonium chloride

H2NCONH H + H2N CONH2

Cl

(X) Gattermann reaction Chlorobenzene

HNO 3 C 2H 6  → C 2H 5NO 2 425 − 675 K

H2NCONHCONH2 + NH3



Methyl amine

Cu/HCl(Y)



Urea (2 molecules)

+ N2

13. (b) : Ethane undergoes nitration to give nitroethane and nitromethane.

16. (b) : Urea on slow heating gives biuret.

CH3NH2

+

N NCl–

HCl(aq.)



18. (d) :

15. (a) : Secondary nitroalkanes on hydrolysis form ketones.

N2+ Cl – + H2O

+ CH 3NO 2

14. (b) : Diazotisation reaction is given by an aromatic primary amine (having NH2 group directly attached to benzene ring).

Ammonia

Biuret

17. (a) : Tertiary amine is less basic than primary and secondary amine due to steric effect of three methyl groups present on nitrogen atom. Thus the correct order is (CH3)2NH > CH3NH2 > (CH3)3N.

Benzenediazonium chloride

OH Warm

+ N2 + HCl Phenol

19. (a) : Carbylamine reaction is given by aliphatic and aromatic primary amines. Secondary and tertiary amines do not show this reaction.

vvv

53

Biomolecules

11 CHAPTER

1.

2.

Biomolecules

When a monosaccharide forms a cyclic hemiacetal, the carbon atom that contained the carbonyl group is identified as the _____ carbon atom, because (a) D, the carbonyl group is drawn to the right (b) L, the carbonyl group is drawn to the left (c) acetal, it forms bond to an —OR and an —OR′ (d) anomeric, its substituents can assume an a or b-position. (2013)

(ii) Denaturation leads to the conversion of double strand of DNA into single strand. (iii) Denaturation affects primary structure which gets destroyed. (a) (ii) and (iii) (b) (i) and (iii) (c) (i) and (ii) (d) (i), (ii) and (iii) (2012) 5.

The secondary structure of a protein refers to (a) a-helical backbone (b) hydrophobic interactions (c) sequence of a-amino acids (d) fixed configuration of the polypeptide backbone. (2011)

6.

How many tripeptides can be prepared by linking the amino acids glycine, alanine and phenyl alanine? (a) One (b) Three (c) Six (d) Twelve (2009)

7.

Milk changes after digestion into (a) cellulose (b) fructose (c) glucose (d) lactose.

Which of the following is/are a-amino acid? NH3 CO2

(a)

(b)

N H

H

CO2

(c) Both (a) and (b) (d) None of these. (2013) 3.

4.

Fructose reduces Tollens’ reagent due to (a) asymmetric carbons (b) primary alcoholic group (c) secondary alcoholic group (d) enolisation of fructose followed by conversion to aldehyde by base. (2012) Which of the statements about ‘Denaturation’ given below are correct? (i) Denaturation of proteins causes loss of secondary and tertiary structures of the protein.

(2008)

8.

Which of the following sets consists only of essential amino acids? (a) Alanine, tyrosine, cystine (b) Leucine, lysine, tryptophan (c) Alanine, glutamine, lysine (d) Leucine, proline, glycine (2008)

9.

Which of the following is a ketohexose? (a) Glucose (b) Sucrose (c) Fructose (d) Ribose (2008)

Answer Key 1.

(d)

9.

(c)

2.

(c)

3.

(d)

4.

(c)

5.

(a)

6.

(c)

7.

(c)

8.

(b)

54

1.

VITEEE CHAPTERWISE SOLUTIONS

e planations

(d) : A pair of stereoisomers which differ in configuration only around C–1 are called anomers and the C–1 carbon is called the anomeric carbon. Anomeric carbon

H

C

OH

H

C

OH

HO

C

H

H

C

OH

H

C CH2OH -D-glucose

5.

6. 7.

HO C H



H C OH O

HO C H

O

3.

(d) : In aqueous solution, fructose is enolised and then converted into aldehyde in basic medium. All aldehydes generally reduce Tollens’ reagent, thus fructose also reduces Tollens’ reagent.

4.

(c) : During denaturation, secondary and tertiary structures of protein are destroyed but primary structure remains intact.

H

9.

(c) : Fructose is a ketohexose. It is a pentahydroxy ketone and its open chain and closed chain structures can be represented as :

CH2OH -D-glucose

(c) : a-amino acids contain a basic amino group (–NH2) on the a-carbon to the carboxyl group.

(Milk)

(b) : Essential amino acids are those amino acids which are supplied to our bodies through diet because they cannot be synthesized in the body. Essential amino acids are listed as Valine, Leucine, Isoleucine, Phenyl alanine, Tryptophan, Threonine, Methionine, Lysine, Arginine and Histidine.

C

2.

H O

2 Lactose  → Glucose + Galactose +

8.

H C OH H

(a) : Primary structure involves sequence of a-amino acids in a polypeptide chain. Secondary structure involves a-helical and b-pleated sheet like structure. (c) (c) : Milk contains lactose which is a disaccharide. Disaccharide on hydrolysis yields two monosaccharides.

vvv

MATHEMATICS

1

Applications of Matrices and Determinants

1

CHAPTER

1.

2.

Applications of Matrices and Determinants

3 4 If A =   , then A·(adj A) is equal to 5 7  (a) A (b) |A| (c) 2 |A| (b) None of these (2014) If the system of equations x + ky – z = 0, 3x – ky – z = 0 and x – 3y + z = 0, has non-zero solution, then k is equal to (a) –1 (b) 0 (c) 1 (d) 2 (2014) 2r - 1

3.

2

If D r = m - 1 2

2

m

Cr

1

m

m+1

2

2

2

, then

sin (m ) sin (m) sin (m + 1) the value of (a) 1 (c) 2 4.



(a) 0 (c) x 7.

8.

(b) 2 (d) None of these (2013)

(b) a2 – b2 (d) 0

(b) 1 (d) 2x

 1 -5 7    If A =  0 7 9  , then trace of matrix A is 11 8 9  (a) 17 (b) 25 (c) 3 (d) 12 (2012) The value of the determinant cos a - sin a 1 sin a cos a 1 is cos(a + b) - sin(a + b) 1 (a) independent of a (b) independent of b (c) independent of a and b (d) None of the above

9.

(2013)

(2012)

If x, y and z are all distinct and x x2 1 + x3 y y 2 1 + y 3 = 0 , then the value of xyz is

The value of the determinant 1 cos ( a - b) cos a cos (a - b) 1 cos b is cos a cos b 1 (a) a2 + b2 (c) 1

If a, b and c are in A.P., then determinant x + 2 x + 3 x + 2a x + 3 x + 4 x + 2b is x + 4 x + 5 x + 2c

(b) 0 (d) None of these (2014)

y1 1  y2 1 will always be less than y3 1

(a) 3 (c) 1 5.

∑ D r is

If the points (x1, y1), (x2, y2) and (x3, y3) are collinear, then the rank of the matrix  x1   x2  x3





m

r =0

6.



z2

1 + z3

(b) –1 (d) None of these (2012) 1 0  0 1 10. If I =  , J =   and 0 1  -1 0 

(2013)

z

(a) –2 (c) –3

 cos q sin q  B=  , then B is equal to  - sin q cos q (a) I cos q + J sin q (b) I sinq + J cos q (c) I cos q – J sin q (d) –I cos q + J sin q (2011)

2

VITEEE CHAPTERWISE SOLUTIONS

(c) a = b2

11. Which of the following is correct? (a) Determinant is a square matrix. (b) Determinant is a number associated to a matrix. (c) Determinant is a number associated to a square matrix. (d) All of the above. (2011) 12. If a, b and g are the a then the value of b g (a) – a3 (c) a3

19. If x, y, z are different from zero and

roots of x3 + ax2 + b = 0, b g g a is a b (b) a3 – 3b (d) a2 – 3b (2011)



then the value of the expression (b) – 1 (d) 2

a b c + + is x y z (2008)

 1 tan q 20. If A(q) =   and AB = I, then 1   - tan q (sec2 q)B is equal to q (a) A(q) (b) A   2  -q  (d) A    2 

(c) A (–q)

15. If f(x), g(x) and h(x) are three polynomials of degree 2 and

(2007)

2x + 1 4 8 21. If x = –5 is a root of 2 2 x 2 = 0 , then 7 6 2x the other roots are (a) 3, 3.5 (b) 1, 3.5 (c) 1, 7 (d) 2, 7 (2007)

f ( x) g( x) h( x) D( x) = f ′( x) g ′( x) h′( x) , f ′′( x) g ′′( x) h′′( x)

22. The simultaneous equations Kx + 2y – z = 1, (K – 1)y – 2z = 2 and (K + 2)z = 3 have only one solution when (a) K = –2 (b) K = –1 (c) K = 0 (d) K = 1 (2007)

(2010)

16. The system of equations x + y + z = 0, 2x + 3y + z = 0, x + 2y = 0 has (a) a unique solution; x = 0, y = 0, z = 0 (b) infinite solutions (c) no solution (d) finite number of non-zero solutions (2008) 4 0 a  17. If   = I , then b 0 (a) a = 1 = 2b



a b-y c-z D = a-x b c - z = 0, a-x b-y c

(a) 0 (c) 1

14. If B is a non-singular matrix and A is a square matrix, then (B–1 AB) be equal to (a) det (A–1) (b) det (B–1) (c) det (A) (d) det (B) (2010)

then D(x) is a polynomial of degree (a) 2 (b) 3 (c) 0 (d) at most 3

(2008)

18. If D = diag(d1, d2, ..., dn) where di ≠ 0, for i = 1, 2, ..., n, then D–1 is equal to (a) DT (b) D (c) Adj(D) (d) diag(d1–1, d2–1, ..., dn–1) (2008)

13. The system of equations 2x + y – 5 = 0, x – 2y + 1 = 0, 2x – 14y – a = 0, is consistent. Then, a is equal to (a) 1 (b) 2 (c) 5 (d) None of these (2011)



(d) ab = 1

 -1 2 5  23. If the rank of the matrix  2 -4 a - 4  is 1,    1 -2 a + 1  then the value of a is (a) –1 (b) 2 (c) –6 (d) 4 (2007)

(b) a = b

Answer Key 1.

(b)

2.

(c)

9.

(b)

10. (a)

17. (d)

18. (d)

3.

(b)

4.

(b)

5.

(d)

6.

(a)

7.

(a)

8.

(a)

11. (c)



12. (c)

13.

(d)

14.

(c)

15.

(c)

16.

(b)

19. (d)

20. (c)

21.

(b)

22.

(b)

23.

(c)

3

Applications of Matrices and Determinants

e planations 1.

 3 4   7 -4  \ A (adj A) =      5 7   -5 3  1 0 = = A. 0 1

2.

5.

3 4 (b) : Given, A =   5 7  |A| = 21 – 20 = 1

(c) : The system has non-zero solution, if 1 k -1 3 - k -1 = 0 1 -3 1

⇒ 1(–k –3) – k(3 + 1) –1(–9 + k) = 0 ⇒ –6k + 6 = 0 ⇒ k = 1 2r - 1 3.

m

2 m+1 (b) : D r = m - 1 2 2 2 sin (m ) sin (m) sin 2 (m + 1) m

m

∑ ( 2r - 1)

m

r =0

2

m -1

\ ∑ Dr = r =0

2

2

m2 - 1 2

= m -1





m

m ∑ Cr

r =0

2

sin (m )



4.

1

Cr

2

m

2

m

6.

x + 2 x + 3 x + 2a (a) : Let A = x + 3 x + 4 x + 2b x + 4 x + 5 x + 2c



1 = 2

∑1

r =0

m+1 2

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

m+1

2

m+1

m

(d) : On solving the determinant, 1(1 – cos2b) – cos (a – b) [cos (a – b) – cos a ⋅ cos b] + cos a [cos b ⋅ cos (a – b) – cos a] = 1 – cos2b – cos2a – cos2(a – b) + 2 cos a ⋅ cos b ⋅ cos (a – b) = 1 – cos2b – cos2a + cos(a – b) [2 cos a ⋅ cos b – cos (a – b)] 2 = 1 – cos b – cos2a + cos (a – b) [cos(a + b) + cos (a – b) – cos(a – b)] = 1 – cos2b – cos2a + cos(a – b) cos(a + b) = 1 – cos2b – cos2a + cos2a ⋅ cos2b – sin2a ⋅ sin2b 2 2 = 1 – cos b – cos a (1 – cos2b) – sin2a ⋅ sin2b = 1 – cos2b – cos2a ⋅ sin2b – sin2a ⋅ sin2b = (1 – cos2b) – sin2b (sin2a + cos2a) = sin 2b – sin2b ⋅ 1 =0



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



A=



Since, all elements of R2 are zero.

7.

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

sin 2 (m2 ) sin 2 (m) sin 2 (m + 1) = 0

(Q two rows are identical)

 x1  (b) : The given matrix is  x2  x3

y1 1  y 2 1 , y 3 1 Using R2 → R2 – R1, R3 → R3 – R1  x1 y1 1   D =  x2 - x1 y 2 - y1 0  = 0  x3 - x1 y 3 - y1 0  (Q points are collinear i.e., area of triangle = 0) x - x1 y 2 - y1 ⇒ 2 =0 x3 - x1 y 3 - y1 So, the rank of matrix is always less than 2.

x+2 x+3 x + 2a 0 0 2( 2b - a - c ) x+4 x+5 x + 2c

1 2

x + 2 x + 3 x + 2a 0 0 0 =0 x + 4 x + 5 x + 2c n

∑ aii

i=1

8.

 1 -5 7    \ If A =  0 7 9  , then 11 8 9  tr(A) = 1 + 7 + 9 = 17 cos a - sin a 1 (a) : Given, sin a cos a 1 cos(a + b) - sin(a + b) 1

4



VITEEE CHAPTERWISE SOLUTIONS

Applying R3 → R3 – R1(cos b) + R2 (sin b) cos a - sin a 1 = sin a cos a 1 0 0 1 + sin b - cos b = (1 + sin b – cos b)(cos2 a + sin2 a) = 1 + sin b – cos b, which is independent of a. x x2

9.

1 + x3

(b) : Given, y y 2 1 + y 3 = 0 z x x2

1

z2

1 + z3

x x2

⇒ y y2 1 + y y2

2

1

x x2

1

z

z

z

z

x3 y3 = 0

2

z

1 x x2

z

z2

1 z

1 x x2

⇒ (1 + xyz) y y

2

z2

1 1 =0

2

z z 1 ⇒ (1 + xyz)[x(y2 – z2) – y(x2 – z2) + z(x2 – y2)] = 0 ⇒ (1 + xyz)(x – y)(y – z)(z – x) = 0 ⇒ 1 + xyz = 0 ⇒ xyz = –1  cos q sin q  10. (a) : B =    - sin q cos q

cos q 0   0 sin q = +  0 cos q sin q 0     1 0  0 1 = cos q   + sin q   0 1  -1 0  = I cos q + J sin q

11. (c) : According to the definition of determinant, determinant is a number associated to a square matrix. 12. (c) : Since, a, b, g are the roots of given equation, therefore a + b + g = – a ab + bg + ga = 0 and abg = – b

Now,

a b g b g a = - (a + b + g ) g a b

⇒ (a2 + b2 + g2 – ab – bg – ga) = – (a + b + g) [(a + b + g)2 – 3(ab + bg + ga)] = – (–a) (a2 – 0) = a3 13.

3

⇒ y y 2 1 + xyz 1 y y 2 = 0



(d) : Given, system of equations are 2x + y – 5 = 0 x – 2y + 1 = 0 and 2x – 14y – a = 0 Since, this system is consistent 2 1 -5 \ 1 - 2 1 =0 2 - 14 - a

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



⇒ 2(2a + 14) – 1 (– a – 2) – 5 (– 14 + 4) = 0 ⇒ 4a + 28 + a + 2 + 50 = 0 ⇒ 5a = – 80 ⇒ a = – 16

14.

(c) : det (B–1 AB) = det (B–1) det A det B = det (B–1)⋅ det B⋅det A = det (B–1B) det A = det I⋅det A = 1⋅det A = det A

15. (c) : Since, f(x), g(x) and h(x) are the polynomials of degree 2, therefore f ′′′(x) = g′′′(x) = h′′′(x) = 0 f ( x)



g( x)

h( x)

Now, D( x) = f ′( x) g ′( x) h′( x) f ′′( x) g ′′( x) h′′( x) f ′( x ) g ′( x ) h ′( x ) ⇒ D ′( x ) = f ′( x ) g ′( x ) h ′ ( x ) f ′′( x) g′′( x) h′′( x) f ( x) g( x) h( x) + f ′′( x) g ′′( x) h′′( x) f ′′( x) g ′′( x) h′′( x)







⇒ D′(x) = 0 + 0 + 0 = 0 ⇒ D(x) = constant Thus, D(x) is the polynomial of degree zero.

f ( x) g( x) h( x) + f ′( x ) g ′( x ) h ′( x ) f ′′′( x) g ′′′( x) h′′′( x)

5

Applications of Matrices and Determinants

16.

(b) : The given system of equations are x + y + z = 0, 2x + 3y + z = 0, and x + 2y = 0



1 1 1 Here, 2 3 1 = 1(0 – 2) – 1(0 – 1) + 1 (4 – 3) 1 2 0



=–2+1+1=0 \ This system has infinite solutions.



 1 tan q 20. (c) : Q A(q) =   1   - tan q

4

0 a  17. (d) :   =I b 0



2



0 a  0 a  0 a  Now,   =   b 0   b 0 b 0







 a2b 2 ⇒   0

Also, AB = I. ⇒ B = A–1  1 - tan q 1 =  2  tan q 1  1 + tan q  =

 ab 0  =   0 ab  4 2 2 0 a  0 a  0 a  = and       b 0 b 0 b 0  ab a   ab 0  =    0 ab   0 ab   a2b 2 0  =  2 2 a b   0 4 0 a  1 0 But   =  b 0 0 1

⇒ abz + acy - ayz - abz + bxz + ayz – xyz + cxy – acy – xyz + ayz = 0 ⇒ ayz + bxz + cxy - 2xyz = 0 ⇒ ayz + bxz + cxy = 2xyz ayz bxz cxy ⇒ + + =2 xyz xyz xyz a b c ⇒ + + =2 x y z



 1 - tan q   1  tan q sec q  1

2

 1 - tan q ⇒ (sec 2 q) B =   = A(–q). q 1  tan 

21. (b) : (given)

a  1 0 =  a b  0 1 2 2

⇒ a2b2 = 1 ⇒ ab = 1

18. (d) : If D = diag (d1, d2, ..., dn)

\ D–1 = diag (d1-1 , d2-1 ,..., dn-1 ) a b-y c-z 19. (d) : a - x b c-z =0 a-c b-y c b c-z a-x c-z ⇒ a - (b - y) b y c a-x c a-x b +(c - z) =0 a-x b-y  ⇒ a(bc – bc + bz + cy – yz) – (b – y) (ac – cx – ac + az + cx – xz) + (c – z) (ab – ay – bx + xy – ab + bx) = 0 ⇒ a(bz + cy - yz) - (b – y)(az – xz) + (c – z)  (xy – ay) = 0

Applying R1 → R1 + R2 + R3 ⇒







2x + 1 4 8 2 2x 2 = 0 7 6 2x

2 x + 10 2 x + 10 2 x + 10 2 2x 2 =0 7 6 2x

1 1 1 ⇒ ( 2 x + 10) 2 2 x 2 = 0 7 6 2x Applying C3 → C3 – C1, C2 → C2 – C1 1 0 0 ⇒ ( 2 x + 10) 2 2 x - 2 0 =0 7 -1 2x - 7 ⇒ (2x + 10)(2x – 2)(2x – 7) = 0 ⇒ x = - 5 , 1,

7 2

7 Hence, other roots are 1 and . 2 22. (b) : The system of given equations are Kx + 2y – z = 1 ...(i) (K – 1)y – 2z = 2 ...(ii) and (K + 2)z = 3 ...(iii)

6



VITEEE CHAPTERWISE SOLUTIONS

This system of equations has a unique solution, if K 2 -1 0 K - 1 -2 ≠ 0 0 0 K+2 ⇒



K 2 ( K + 2) ≠0 0 K -1

⇒ (K + 2)(K)(K – 1) ≠ 0 ⇒ K ≠ –2, 0, 1 i.e., K = –1, is a required answer.

 -1 2 5    23. (c) : Since, the rank of matrix  2 -4 a - 4   1 -2 a + 1  is 1, then

vvv

2 5 =0 -4 a - 4 ⇒ 2a – 8 + 20 = 0 ⇒ 2a + 12 = 0 ⇒ a = – 6

7

Complex Numbers

2

CHAPTER

1.

2.

Complex Numbers

3 If x + i y = , then x2 + y2 is 2 + cos q + i sin q equal to (a) 3x – 4 (b) 4x – 3 (c) 4x + 3 (d) None of these (2014) If a = cos a + i sin a, b = cos b + i sin b b c a c = cos g + ising and + + = 1, then c a b cos (b – g) + cos(g – a) + cos(a – b) is equal to 3 2 (c) 0 (a)

3 (b) − 2 (d) 1

(2014)

3.

If |z + 4|≤ 3, then the greatest and the least value of |z + 1| are (a) –1, 6 (b) 6, 0 (c) 6, 3 (d) None of these (2014)

4.

If the area of the triangle on the complex plane formed by the points z, z + iz and iz is 200, then the value of 3|z| must be equal to (a) 20 (b) 40 (c) 60 (d) 80 (2014)

5.

6.

7.

If z satisfies the equation |z| – z = 1 + 2i, then z is equal to 3 3 (a) + 2i (b) − 2i 2 2 3 3 (c) 2 − i (d) 2 + i (2010) 2 2

8.

If z =

9.

1− i 3

1+ i 3 (a) 60° (c) 240°

, then arg(z) is (b) 120° (d) 300°

(2010)

The roots of (x – a)(x – a – 1) + (x – a – 1)(x – a – 2) + (x – a) (x – a – 2) = 0, a ∈ R are always (a) equal (b) imaginary (c) real and distinct (d) rational and equal (2009)

10. The locus of z satisfying the inequality z + 2i < 1, where z = x + iy, is 2 z+i (a) x2 + y2 < 1 (c) x2 + y2 > 1

(b) x2 – y2 < 1 (d) 2x2 + 3y2 < 1

A complex number z is such that

(2009)

z−2 p arg  = . The points representing this  z + 2  3 complex number will lie on (a) an ellipse (b) a parabola (c) a circle (d) a straight line (2013)

11. If n is an integer which leaves remainder one when divided by three, then

If x + iy = (1 − i 3 )100 , then find (x, y).

12. If z − 25 = 5, the value of |z| is z−1 (a) 3 (b) 4 (c) 5 (d) 6

(a) ( 299 , 299 3 )

(b) ( 299 , −299 3 )

(c) ( −299 , 299 3 )

(d) None of these (2011)



(1 + 3i )n + (1 − 3i )n equals (a) –2n + 1 (b) 2n + 1 n (c) – (–2) (d) –2n

(2009)

(2008)

8

VITEEE CHAPTERWISE SOLUTIONS

(b) cos 3a + i sin 3a (d) cos 5a + i sin 5a

 −1 − 3i  13. Argument of the complex number   2 + i  is (a) 45° (b) 135° (c) 225° (d) 240° (2008)

(2007)

(1 + i )2 15. The imaginary part of is i( 2i − 1) 4 5 2 (c) 5

14. The product of all values of (cos a + i sin a)3/5 is (a) 1 (b) cos a + i sin a

(a)

(b) 0 (d) −

4 5

(2007)

Answer Key 1.

(b)

2.

(d)

9.

(c)

10. (c)

3.

(b)

11. (c)

4.

(c)

12. (c)

5.

(c)

6.

(c)

7.

(b)

13.

(c)

14.

(c)

15.

(d)

8.

(c)

9

Complex Numbers

1.



 2. 

3 2 + cos q + i sin q 3( 2 + cos q − i sin q) = ( 2 + cos q)2 + sin 2 q

e planations

(b) : x + iy =

6 + 3 cos q − 3i sin q 4 + cos 2 q + 4 cos q + sin 2 q 6 + 3 cos q − 3i sin q = 5 + 4 cos q  6 + 3 cos q   −3 sin q  = +i  5 + 4 cos q   5 + 4 cos q  On equating real and imaginary parts, we get −3 sin q 3( 2 + cos q) x= and y = 5 + 4 cos q 5 + 4 cos q 9 \ x2 + y2 = ( 5 + 4 cos q)2 [4 + cos2 q + 4 cos q + sin2q] =

=

 6 + 3 cos q  9 = 4 − 3 = 4x − 3  5 + 4 cos q  5 + 4 cos q

(d) : We have, a = cos a + i sin a b = cos b + i sin b and c = cos g + i sin g Now, b = cos b + i sin b × cos g − i sin g c cos g + i sin g cos g − i sin g = cos b ⋅ cos g + sin b ⋅ sin g + i [sin b ⋅ cos g – sin g⋅ cos b]



b ⇒ = cos (b – g) + i sin (b – g) ...(i) c c Similarly = cos (g – a) + i sin (g – a) ...(ii) a a and = cos (a – b) + i sin (a – b) ...(iii) b On adding Eqs. (i), (ii), and (iii), we get cos(b – g) + cos(g – a) + cos(a – b) + i [sin(b – g) + sin(g – a) + sin(a – b)] = 1



  b c a Q c + a + b = 1,given    On equating real parts, we get cos(b – g) + cos(g – a) + cos(a – b) = 1

3.

(b) : We have, z + 4≤ 3 ⇒ –3 ≤ z + 4 ≤ 3 ⇒ – 6 ≤ z + 1 ≤ 0





⇒ 0 ≤ – (z + 1) ≤ 6 ⇒ 0 ≤ z + 1 ≤ 6 Hence, the greatest and least values are 6 and 0.

4.

(c) : Let z = x + iy, then z + iz = x + iy + i(x + iy) = (x – y) + i(x + y) and iz = i(x + iy) = –y + ix, Then, the area of the triangle formed by these lines is ∆=





x y 1 1 ( x − y) ( x + y) 1 2 −y x 1

Applying R2 → R2 – (R1 + R3), x y 1 1 1 ∆= 0 0 −1 = ( x 2 + y 2 ) 2 2 −y x 1 1 2 | z| = 200 2

(given) ⇒ |z|2 = 400 ⇒ |z| = 20 \ 3|z| = 3 × 20 = 60 5.

z−2 p (c) : arg  =  z + 2  3



 x − 2 + iy  p = ⇒ arg   x + 2 + iy  3



⇒ arg (x – 2 + iy) – arg (x + 2 + iy) =



 y   y  p ⇒ tan −1  − tan −1  =  x − 2   x + 2  3 4y = 3 ⇒ 2 x + y2 − 4



3(x2 + y 2 ) − 4y − 4 3 = 0 which is equation of a circle.

6.

(c) : We have,



(1 − i 3 )100 = 2100  − 1 + i



⇒

2

3  2 

100



=2



=2



 1 3i = 2100  − +   2 2 



= – 299 + 299 3 i

100 100

w

100

⋅w

p 3

10

VITEEE CHAPTERWISE SOLUTIONS

10. (c) : Let z = x + iy

100

x + iy = (1 − i 3 )

\

2

= – 299 + 299 3 i x = – 299, y = 299 3

\

(x, y) = (– 2 , 2 99

99

3)



\

7.

(b) : We have, |z| – z = 1 + 2i If z = x + iy, then this equation reduces to |x + iy| – (x + iy) = 1 + 2i



⇒

( x + y − x) + (−iy) = 1 + 2i 2

x2 + y 2 − x = 1



...(i)

1 − 3 − 2 3i −2 − 2 3i = 1+ 3 4

1 3 − i 2 2 ⇒ z = cos 240° – i sin 240­° Thus, arg(z) = 240° ⇒ z=−

9. (c) : Given, (x – a)(x – a – 1) + (x – a – 1)(x – a –2) + (x – a)(x – a – 2) = 0 Let x – a = t, then t(t – 1) + (t – 1)(t – 2) + t(t – 2) = 0 ⇒ t2 – t + t2 – 3t + 2 + t2 – 2t = 0 ⇒ 3t2 – 6t + 2 = 0

⇒ t=

2( 3)

6±2 3 = 2( 3)

3± 3 3± 3 ⇒ x=a+ 3 3 Hence, x is real and distinct. ⇒

x−a=

⇒ x2 + y2 + 4 + 4y < 4x2 + 4y2 + 1 + 4y ⇒ 3x2 + 3y2 > 3 ⇒ x2 + y2 > 1

11. (c) : Now, (1 + 3i )n + (1 − 3i )n



  1 − 3i     1 + 3i   = 2     + 2    2     2   = (–2w2)n + (–2w)n

n

= (–2)n [(w2)3r + 1 + (w)3r + 1] (Q n = 3r + 1, where r is an integer) = (–2)n(w2 + w) = – (–2)n

1 − i 3  1 − i 3 1 − i 3 × (c) : z =   = 1 + i 3  1 + i 3 1 − i 3

6 ± 36 −24

2

( x) + ( y + 2) z + 2i b) and x2 – y2 = c2 cut at right a2 b 2 angles, then (a) a2 + b2 = 2c2 (b) b2 – a2 = 2c2 2 2 2 (c) a – b = 2c (d) a2b2 = 2c2 (2012)

22. If

+

23. The equation of the conic with focus at (1, –1), directrix along x – y + 1 = 0 and with eccentricity 2 , is (a) x2 – y2 = 1 (b) xy = 1 (c) 2xy – 4x + 4y + 1 = 0 (d) 2xy + 4x – 4y – 1 = 0

(2012)

24. To the lines ax + 2hxy + by = 0, the lines a2x2 + 2h(a + b)xy + b2y2 = 0 are (a) equally inclined (b) perpendicular (c) bisector of the angle (d) None of the above (2011) 2

2

14

VITEEE CHAPTERWISE SOLUTIONS

25. A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. What is its y-intercept? 1 2 (a) (b) 3 3 4 (2011) 3 26. The number of common tangents to the circles x2 + y2 = 4 and x2 + y2 – 6x – 8y = 24 is (a) 0 (b) 1 (c) 3 (d) 4 (2011) (c) 1

(d)

27. If the axes are shifted to the point (1, –2) without solution, then the equation 2x2 + y2 – 4x + 4y = 0 becomes (a) 2X2 + 3Y2 = 6 (b) 2X2 + Y2 = 6 2 2 (c) X + 2Y = 6 (d) None of these (2011) 28. The locus of the mid-points of the focal chord of the parabola y2 = 4ax is (a) y2 = a(x – a) (b) y2 = 2a(x – a) 2 (c) y = 4a(x – a) (d) None of these (2011) 29. A rod of length l slides with its ends on two perpendicular lines, then the locus of its mid point is l2 (a) x + y = 4 l2 (c) x 2 - y 2 = 4 2

2

l2 (b) x + y = 2 2

2

(d) None of these (2010)

30. The equation of straight line through the intersection of line 2x + y = 1 and 3x + 2y = 5 passing through the origin is (a) 7x + 3y = 0 (b) 7x – y = 0 (c) 3x + 2y = 0 (d) x + y = 0 (2010) 31. The line joining (5, 0) to (10 cos q, 10 sin q) is divided internally in the ratio 2 : 3 at P. If q varies, then the locus of P is (a) a straight line (b) a pair of straight lines (c) a circle (d) None of the above (2010) 32. If 2x + y + k = 0 is a normal to the parabola y2 = – 8x, then the value of k, is (a) 8 (b) 16 (c) 24 (d) 32 (2010)

33. If the equation of an ellipse is 3x2 + 2y2 + 6x – 8y + 5 = 0, then which of the following are true? 1 (a) e = 3 (b) centre is (–1, 2) (c) foci are (–1, 1) and (–1, 3) (d) All of the above (2010) 34. The equation of the common tangents to the two hyperbolas are

x2 a2

-

y2 b2

= 1 and

y2 a2

-

x2 b2

= 1,

2 2 (a) y = ± x ± b2 - a2 (b) y = ± x ± a - b

(c) y = ± x ± a2 + b2 (d) y = ± x ± (a2 – b2) (2010) 35. The two curves y = 3x and y = 5x intersect at an angle  log 3 - log 5  (a) tan -1   1 + log 3 log 5   log 3 + log 5  (b) tan -1   1 - log 3 log 5   log 3 + log 5  (c) tan -1   1 + log 3 log 5   log 3 - log 5  (d) tan -1   1 - log 3 log 5 

(2010)

36. The equation lx2 + 4xy + y2 + lx + 3y + 2 = 0 represents a parabola, if l is (a) 0 (b) 1 (c) 2 (d) 4 (2010) 2 2 37 . If two circles 2x + 2y – 3x + 6y + k = 0 and 2 2 x + y – 4x + 10y + 16 = 0 cut orthogonally, then the value of k is (a) 41 (b) 14 (c) 4 (d) 1 (2010)

38. If A(–2, 1), B(2, 3) and C(– 2, – 4) are three points. Then, the angle between BA and BC is -1  2  (a) tan   3

-1  3  (b) tan   2

1 (c) tan -1   3

1 (d) tan -1   2

(2010)

15

Analytical Geometry of Two Dimensions

39. The area (in square unit) of the circle which touches the lines 4x + 3y = 15 and 4x + 3y = 5 is (a) 4p (b) 3p (c) 2p (d) p (2009) 40. The area (in square unit) of the triangle formed by x + y + 1 = 0 and the pair of straight lines x2 – 3xy + 2y2 = 0 is 7 12 1 (c) 12 (a)

5 12 1 (d) 6 (b)

(2009)

41. The pairs of straight lines x2 – 3xy + 2y2 = 0 and x2 – 3xy + 2y2 + x – 2 = 0 form a (a) square but not rhombus (b) rhombus (c) parallelogram (d) rectangle but not a square (2009) 42. The equations of the circle which pass through the origin and makes intercepts of lengths 4 and 8 on the x and y-axes respectively are (a) x2 + y2 ± 4x ± 8y = 0 (b) x2 + y2 ± 2x ± 4y = 0 (c) x2 + y2 ± 8x ± 16y = 0 (d) x2 + y2 ± x ± y = 0 (2009) 43. The point (3, – 4) lies on both circles x2 + y2 – 2x + 8y + 13 = 0 and x2 + y2 – 4x + 6y + 11 = 0 Then, the angle between the circles is 1 (a) 60° (b) tan -1   2 3 (c) tan -1   5

(d) 135°

46. If the circle x2 + y2 = a2 intersects the hyperbola xy = c2 in four points (xi, yi),for i = 1, 2, 3 and 4, then y1 + y2+ y3 + y4 equals (a) 0 (b) c (c) a (d) c4 (2009) 47. The mid point of the chord 4x – 3y = 5 of the hyperbola 2x2 – 3y2 = 12 is (a)  0 , - 5   3 

(b) (2, 1)

5  (c)  , 0  4 

 11  (d)  , 2  4 

(2009)

48. The perimeter of the triangle with vertices at (1, 0, 0), (0, 1, 0) and (0, 0, 1) is (a) 3 (b) 2 (c) 2 2

(d) 3 2

(2009)

49. The length of the parabola y = 12x cut off by the latus-rectum is (a) 6( 2 + log(1 + 2 )) (b) 3( 2 + log(1 + 2 )) 2

(c) 6 ( 2 - log(1 + 2 )) (d) 3( 2 - log(1 + 2 ))

(2008)

50. Let A and B are two fixed points in a plane then locus of another point C on the same plane such that CA + CB = constant, (> AB) is (a) circle (b) ellipse (c) parabola (d) hyperbola (2008) 51. The directrix of the parabola y2 + 4x + 3 = 0 is 4 = 0 3 3 (c) x - = 0 4 (a) x -

(2009)

44. The equation of the circle which passes through the origin and cuts orthogonally each of the circles x 2 + y 2 – 6x + 8 = 0 and x2 + y2 – 2x – 2y = 7 is (a) 3x2+ 3y2 – 8x –13y = 0 (b) 3x2 + 3y2 – 8x + 29y = 0 (c) 3x2 + 3y2 + 8x + 29y = 0 (d) 3x2 + 3y2 – 8x – 29y = 0 (2009) 45. The number of normals drawn to the parabola y2 = 4x from the point (1, 0) is (a) 0 (b) 1 (c) 2 (d) 3 (2009)

1 =0 4 1 (d) x - = 0 4 (b) x +

(2008)

52. The equation of a directrix of the ellipse x2 y 2 + = 1 is 16 25 (a) 3y = 5 (c) 3y = 25

(b) y = 5 (d) y = 3

(2007)

53. If the normal at (ap2, 2ap) on the parabola y2 = 4ax, meets the parabola again at (aq2, 2aq), then (a) p2 + pq + 2 = 0 (b) p2 – pq + 2 = 0 (c) q2 + pq + 2 = 0 (d) p2 + pq + 1 = 0 (2007)

16

VITEEE CHAPTERWISE SOLUTIONS

54. The length of the straight line x – 3y = 1 intercepted by the hyperbola x2 – 4y2 = 1 is 6 (a) 10 (b) 5 1 6 (c) (d) (2007) 10 5 10

55. The curve described parametrically by x = t2 + 2t – 1, y = 3t + 5 represents (a) an ellipse (b) a hyperbola (c) a parabola (d) a circle

(2007)

Answer Key 1.

(b)

2.

(b)

3.

(c)

9.

(a)

10. (a)

11. (b)

17. (b)

18. (b)

25. (d)

4.

(b)

5.

(c)

6.

(b)

7.

(a)

8.

(c)

12. (d)

13.

(d)

14.

(d)

15.

(a)

16.

(c)

19. (d)

20. (c)

21.

(b)

22.

(c)

23.

(c)

24.

(a)

26. (b)

27. (b)

28. (b)

29.

(a)

30.

(a)

31.

(c)

32.

(c)

33. (d)

34. (b)

35. (a)

36. (d)

37.

(c)

38.

(a)

39.

(d)

40.

(c)

41. (c)

42. (a)

43. (d)

44. (b)

45.

(b)

46.

(a)

47.

(b)

48.

(d)

49. (a)

50. (b)

51. (d)

52. (c)

53.

(a)

54.

(d)

55.

(c)

17

Analytical Geometry of Two Dimensions

1.

e planations

(b) : The equation of any normal to 2

x y - 2 = 1 is 2 a b ax cos f + by cot f = a2 + b2 ⇒ ax cos f + by cot f – (a2 + b2) = 0 ...(i) The straight line lx + my – n = 0 will be normal



2

The equation of tangent at point (1, 1) is



 dy  y - 1 =   ( x - 1)  dx (1, 1)



⇒ y – 1 = (b + 2) (x – 1)



⇒ (2 + b) x – y = 1 + b

2

x y - 2 == 11, then Eq. (i) and 2 a b lx + my – n = 0 represent the same line. to the hyperbola





2

a cos f b cot f a2 + b 2 ∴ = = l m n nb na ⇒ sec f = 2 and tan f = 2 2 m( a + b 2 ) l( a + b )

⇒

x y =1  1 + b  (1 + b)  2 + b  Y B



Q sec2 f – tan2 f = 1



⇒





a 2 b 2 ( a 2 + b 2 )2 ⇒ 2 - 2 = l m n2 But given equation of normal is



1+ b 2+b and OB = – (1 + b)



Now, area of DAOB =



a 2 b 2 ( a 2 + b 2 )2 = k l 2 m2 2 ∴ k = n



⇒ 4 (2 + b) + (1 + b)2 = 0 ⇒ 8 + 4b + 1 + b2 + 2b = 0 ⇒ b2 + 6b + 9 = 0 ⇒ (b + 3)2 ⇒ b = – 3

4.



(b) : We know that, the sum of ordinates of feet normals drawn from a point to the parabola, y2 = 4ax is always zero. Now, as normals at three points P, Q and R of parabola y2 = 4ax meet at (h, k). ⇒ The normals from (h, k) to y2 = 4ax meet the parabola at P, Q and R. ⇒ y – coordinates y1, y2, y3, of three points P, Q and R will be zero. ⇒ y – coordinate of the centroid of DPQR



y1 + y2 + y3 0 = =0 3 3 Hence, centroid lies on y = 0.

5.

(c) : On homogenising y2 – x2 = 4 with the

n2 a2 n2b 2 =1 l 2 ( a 2 + b 2 )2 m 2 ( a 2 + b 2 )2

2.

(b) : Given, vertices of DABC are A(0, 4, 1) B(2, 3, –1) and C(4, 5, 0).



Now, AB = ( 2 - 0)2 + ( 3 - 4)2 + ( -1 - 1)2



BC =



=

4 + 1+ 4 = 3 ( 4 - 2)2 + ( 5 - 3)2 + (0 + 1)2

= 4+4+1 = 3

and CA = ( 4 - 0)2 + ( 5 - 4)2 + (0 - 1)2

= 16 + 1 + 1 = 3 2 \ AB2 + BC2 = AC2 \ DABC is right angled triangle. We know that, the orthocentre of a right angled triangle is the vertex containing 90° angle. \ Orthocentre is point B(2, 3, –1).

3.

(c) : Given curve is y = f(x) = x2 + bx – b On differentiating w.r.t. x, we get f ′(x) = 2x + b

(1, 1)



A

X

So, OA =

1 (1 + b)[ -(1 + b)] × =2 2 ( 2 + b)

i.e.,

3x + y = 2, we get

help of the line 2



2

O





y = x + bx – b

y -x

2

( =4

3x+y 4

)

2

18

⇒ 4 x 2 + 2 3xy = 0 On comparing with ax2 + 2hxy + by2 = 0, we get a = 4 , h = 3 and b = 0



\ The angle between the lines is tan q = 2 ⋅

6.

h 2 - ab

a+b  3 ⇒ q = tan -1    2 

=

2⋅ 3-0 4+0



 19  9 ⇒ cosq =  or tan q =    19   442 



9 ⇒ q = tan -1    19 

9.

(a) : Here, slope of AB =



(b) : Given equation of hyperbola is 25x2 – 16y2 = 400. If (6, 2) is the mid-point of the chord, then equation of chord is T = S1 ⇒ 25(6x) – 16(2y) = 25(36) – 16(4) ⇒ 75x – 16y = 450 – 32 (divide by 2) ⇒ 75x – 16y = 418

1 1 ⇒ tan q = m1 = 1 or q = 45° Thus, slope of new line is tan (45° + 15°) = tan 60° = 3 (Q it is rotated anti-clockwise, so the angle will be 45° + 15° = 60°) Y C B (3, 1) °



⇒ y 2 - x 2 = 3 x 2 + y 2 + 2 3 xy

15



VITEEE CHAPTERWISE SOLUTIONS

(a) : Let (h, k) be any point in the set, then equation of circle is (x – h)2 + (y – k)2 = 9 7.

8 2

But (h , k) lies on x2 + y2 = 25, then h2 + k2 = 25 \ 2 ≤ Distance between the two circles ≤ 8

⇒ 2 ≤ h 2 + k 2 ≤ 8 ⇒ 4 ≤ h2 + k2 ≤ 64



So, locus of (h, k) is 4 ≤ x2 + y2 ≤ 64

8.

(c) : We know that, angle of intersection between two circles is given by





10 17 + 13 r12 + r22 - d 2 4 cosq = = 2 2r1r2 17 2 ⋅ 13 2  2 2 here, r =  1  +  - 1  + 8 1 2  2   17  = ,  2  r2 = 13   and d = c1c 2 = 10  2

45°

X

A (2, 0)



Hence, the equation is y = 3 x + c , but it still passes through (2, 0), hence c = - 2 3.



Thus, required equation is y = 3x - 2 3

10. (a) : On solving the equation of line and curve, we get, 2



 2 - 2x  x - 2  =4  6  1 ⇒ x 2 - × 4(1 + x 2 - 2 x) = 4 3 ⇒ 3x2 – 4 – 4x2 + 8x = 12 ⇒ – x2 + 8x – 16 = 0 ⇒ x2 – 8x + 16 = 0 ⇒ (x – 4)2 = 0 ⇒ x = 4



and



So, point of contact is 4 , - 6 .



2

6 ⋅ y = 2 - 2( 4) = - 6 ⇒ y = - 6

(

)

11. (b) : x + y = 21 B(0, 21)

C (0, 0)

A (21, 0)

19

Analytical Geometry of Two Dimensions



The number of integral solutions to the equation x + y < 21, i.e., x < 21 – y \ Number of integral coordinates = 19 + 18 + ... + 1 19 (19 + 1) 19 × 20 = = = 190 2 2 12. (d) : Let P(x1, y1) be a point on the ellipse.

x2 y2 x2 y 2 ...(i) + =1 ⇒ 1 + 1 =1 18 32 18 32 The equation of the tangent at (x1, y1) is



xx1 yy1  18  + = 1. This meets the axes at A  , 0  18 32  x1   32  and B 0 , It is given that slope of the  y  . 1 4 tangent at (x1, y1) is - . 3 x1 32 4 Hence, - ⋅ =18 y1 3 x1 3 x1 y1 ⇒ (say) = = K = ⇒ y1 4 3 4 \ x1 = 3K, y1 = 4K Putting x1, y1 in (i), we get

K2 = 1



=

1 18 32 1 (18) ( 32) 1 (18) ( 32) ⋅ ⋅ = = 2 x1 y1 2 ( x1y1 ) 2 ( 3K ) ( 4 K ) 24

K2 13. (d) : Let mid-point of part PQ which is in between the axes is R(x1, y1), then coordinates of P and Q will be (2x1, 0) and (0, 2y1), respectively. y x \ Equation of line PQ is + =1 2 x1 2 y1 y  or y = -  1  x + 2 y1  x1 



If this line touches the ellipse

x2



a2 then it will satisfy the condition,



c2 = a2m2+ b2



2

-y  i.e., ( 2 y1 ) 2 = a 2  1  + b 2  x1 



 a2   b2   a2   b2  ⇒ 4 =   +   ⇒   +   = 4 2 2 2 2  x1   y1   x1   y1 

\ Required locus of (x1, y1) is

+

y2 b2

=1

a2 x2

+

b2 y2

=4

14. (d) : Let P (a sec q, b tan q); Q (a sec q, – b tan q) be end points of double ordinate and (0, 0) is the centre of the hyperbola. Now, PQ = 2b tan q

OQ = OP = a 2 sec 2 q + b 2 tan 2 q Since, OQ = OP = PQ \ 4b2 tan2 q = a2sec2 q + b2 tan2 q ⇒ 3b2 tan2 q = a2 sec2 q (0, 0) O



P

(a sec , b tan )

Q (a sec , – b tan )





(Q K 2 = 1)

= 24 sq units



 a 2 y 2  1 + b2 ⇒ 4 y12 =  2   x1 



1 Now, area of DOAB = OA ⋅ OB 2

=



⇒ 3b2 sin2 q = a2 ⇒ 3a2 (e2 – 1) sin2 q = a2 ⇒ 3(e2 – 1) sin2 q = 1 1 ⇒ = sin 2 q < 1, (Q sin 2 q < 1) 2 3( e - 1) ⇒

1 2

< 3 ⇒ e2 - 1 >

e -1 4 2 ⇒ e 2 > \ e > 3 3

1 3

15. (a) : In all there are 3 + 4 + 5 = 12 points in a plane. The number of required triangles = (The number of triangles formed by these 12 points) – (The number of triangles formed by the collinear points)

= 12C3 – (3C3 + 4C3 + 5C3) = 220 – (1 + 4 + 10) = 205

16. (c) : The centre of the required circle lies at the intersection of 2x – 3y – 5 = 0 and 3x – 4y – 7 = 0. Thus, the coordinates of the centre are (1,–1).

20

VITEEE CHAPTERWISE SOLUTIONS

Let r be the radius of the circle. Then, pr2 = 154 22 2 ⇒ r = 154 ⇒ r = 7 7 Hence, the equation of required circle is (x – 1)2 + (y + 1)2 = 72 ⇒ x2 + y2 – 2x + 2y – 47 = 0



17. (b) : Equation of family of concentric circles to the circle x2 + y2 + 6x + 8y – 5 = 0 is x2 + y2 + 6x + 8y + l = 0 which is similar to x2 + y2 + 2gx + 2fy + c = 0 Thus, the point (–3, 2) lies on the circle x2 + y2 + 6x + 8y + c = 0 \ (–3)2 + (2)2 + 6(–3) + 8(2) + c = 0 ⇒ 9 + 4 – 18 + 16 + c = 0 ⇒ c = –11 18. (b) : Since, A is mid-point of line PQ. y a+0 ∴ 3= 2

⇒ b = 8



Thus, equation of line is x y + =1 6 8 ⇒ 4x + 3y = 24





A(3, 4)



O

Q(a, 0)

x

19. (d) : The tangent at (1 ,7) to the curve x2 = y – 6 is



1 x = ( y + 7) - 6 2 ⇒ 2x = y + 7 – 12 ⇒ y = 2x + 5 which is also tangent to the circle x2 + y2+ 16x + 12y + c = 0 i.e., x2 + (2x + 5)2 + 16x + 12(2x + 5) + c = 0 ⇒ 5x2 + 60x + 85 + c = 0, which must have equal roots. Let a and b be the roots of the equation. Then a + b = –12 ⇒ a = – 6 (Q a = b) \ x = – 6, y = 2x + 5 = –7 Hence, point of contact is (–6, –7).

20. (c) : The intersection point of lines x – 2y = 1 7 1 and x + 3y = 2 is  ,  5 5



1 3 7 7 1 through  ,  is y - = -  x -  5 4 5 5 5 3 x + 4 y 21 + 4 3x 21 1 ⇒ +y= + ⇒ = 4 20 4 20 5 ⇒ 3x + 4y – 5 = 0

21. (b) : Let the equation of the ellipse be x2 y 2 + = 1. It is given that it passes through a2 b2 (7, 0) and (0, –5). Therefore, a2 = 49 and b2 = 25



P(0, b)

⇒ a = 6 0+b and 4 = 2



Since, required line is parallel to 3x + 4y = 0. 3 Therefore, the slope of required line is - . 4 \ Equation of required line which passes

The eccentricity of the ellipse is e = 1 = 1-

25 = 49

x2

y2



=1 a2 b2 and x2 – y2 = c2



On differentiating w.r.t. x, we get



a2 ⇒







+

a2

24 2 6 = 49 7

22. (c) : Given,

2x

b2

+

...(i)

2 y dy dy ⋅ = 0 and 2 x - 2 y =0 dx b2 dx

dy dy x xb2 = - 2 and = dx dx y a y

The two curves will cut at right angles, if  dy   dy   dx  ×  dx  = - 1 c c 1

2

x2 y 2 b2 x x ⇒ - 2 ⋅ = -1 ⇒ 2 = 2 a b a y y ⇒

x2 a2

=

y2 b2

=

1 x2 y 2 [using eq. 2 + 2 = 1] 2 a b

On substituting these values in x2 – y2 = c2, we get a2 b2 = c2 2 2 ⇒ a2 – b2 = 2c2

23. (c) : Let P(x, y) be any point on the conic.

 x - y + 1 2 2 Then, ( x - 1) + ( y + 1) = 2   2 

21

Analytical Geometry of Two Dimensions



⇒ (x – 1)2 + (y + 1)2 = (x – y + 1)2



⇒ 2xy – 4x + 4y + 1 = 0

24. (a) : The equation of the bisectors of the angle between the lines given by ax2 + 2hxy + by2 = 0 is

x 2 - y 2 xy ...(i) = a-b h And the equation of the bisectors of the angle between the lines given by a2x2 + 2h(a + b)xy + b2y2 = 0 is x2 - y2 2

a -b

2

=

xy h( a + b)

x 2 - y 2 xy = ...(ii) a-b h From Eqs. (i) and (ii), it is clear that both the pair of straight lines have the same bisector, hence, the given two pairs of straight lines are equally inclined. ⇒

25. (d) : Equation of line perpendicular to 3x + y – 3 = 0 is x – 3y + c = 0 Since, it passes through (2, 2). \ c = 4 \ Equation of line perpendicular to 3x + y – 3 = 0 is x – 3y + 4 = 0 For y-intercept, put x = 0 4 \ y = 3 26. (b) : The centres of the given circles x2 + y2 = 4 and x2 + y2 – 6x – 8y = 24 are C1 (0, 0) and C2 (3, 4) respectively. Their radii are r1 = 2 and r2 = 7 respectively. We have, C1C2 = 5 < sum of radii But C1C2 = difference of radii Thus, the given circles touch each other internally. Hence, number of common tangent is only one. 27. (b) : Substituting x = X + 1 and y = Y – 2 in given equation, we get 2(X + 1)2 + (Y – 2)2 – 4 (X + 1) + 4 (Y – 2) = 0 ⇒ 2X2 + Y2 = 6 28. (b) : Any chord PQ which is bisected at point R(h, k) is T = S or i.e., ky – 2a(x + h) = k2 – 4ah



Since, it is a focal chord, so it must pass through focus (a, 0). \ k(0) – 2a(a + h) = k2 – 4ah ⇒ k2 = 2ah – 2a2 Hence, locus is y2 = 2a(x – a) 29. (a) : Let both the ends of the rod be on x-axis and y-axis respectively. Let AB be rod of length l and coordinates of A and B be (a, 0) and (0, b) respectively. Let P(h, k) be the mid point of the rod AB. Then, 0 + a a =  h= 2 2 ...(i)  b + 0 b and k = = 2 2  2 Now, in DOAB, OA + OB2 = AB2 ⇒ a2 + b2 = t2 Y

B(0, b)

P(h, k) A(a, 0)

O

X



⇒ (2 h)2 + (2 k)2 = l2 [using eq. (i)]



⇒ h2 + k 2 =



\ The equation of locus is x 2 + y 2 =

l2 4

l2 4

30. (a) : Let L1 ≡ 2x + y – 1 = 0 L2 ≡ 3x + 2y – 5 = 0 We know that the equation of straight line passing through the intersetion point of the lines L1 and L2 is given by L1 + l L2 = 0 ⇒ (2x + y – 1) + l(3x + 2y – 5) = 0 Since, this line passes through the origin, also \ (0 + 0 – 1) + l(0 + 0 – 5) = 0

⇒ –5l = 1 ⇒ l = -

1 5

\ Required line is 1 ( 2 x + y - 1) - ( 3x + 2 y - 5) = 0 5   3 2 ⇒  2 -  x + 1 -  y - 1 + 1 = 0   5 5 7 3 ⇒ x + y = 0 ⇒ 7x + 3y = 0 5 5

22

VITEEE CHAPTERWISE SOLUTIONS

31. (c) : Let coordinate of P be (h, k), then 2(10 cos q) + 3( 5) h= = 4 cos q + 3 2+3





34. (b) : We have the hyperbolas

2(10 sin q) + 3(0) = 4 sin q 2+3  Using the internal section formula,    h = m1x2 + m2 x1 , k = m1y2 + m2 y1    m1 + m2 m1 + m2

and k =

h-3 k = cos q and = sin q 4 4 Squaring and adding both equations ⇒

of

32. (c) : The equation of any normal to the parabola y2 = –8x is y = mx + 4m + 2m3 ...(i) (using equation of normal of parabola in slope form y = mx – 2am – am3 and a = –2) But, the given normal is 2x + y + k = 0 ⇒ y = –2x – k ...(ii) Comparing eqs. (i) and (ii), we get m = –2 and – 4m – 2m3 = k ⇒ k = 8 + 16 = 24 33.

(d) : We have equation of ellipse is 3x2 + 2y2 + 6x – 8y + 5 = 0 ⇒ 3(x2 + 2x) + 2(y2 – 4y) + 5 = 0 ⇒ 3(x2 + 2x + 1) + 2(y2 – 4y + 4) + 5 – 3 – 8 = 0 ⇒ 3(x + 1)2 + 2(y – 2)2 = 6



( x + 1)2 ( y - 2)2 + =1 2 3 Comparing with ( y - k )2



+ = 1, we get a2 b2 h = –1, k = 2, a2 = 2, b2 = 3



Now, centre (h, k) = (–1, 2)



And using a2 = b2 (1 – e2) ⇒ 2 = 3(1 – e2)





35.

⇒

( x - h)2

⇒ e=

1

3 And foci are (h, k + be) and (h, k – be) = (–1, 2 + 1) and (–1, 2 – 1) = (–1, 3) and (–1, 1)

a

2

-



and

y2 b2 y2

= 1 x2

= 1 a2 b2 Any tangent to the hyperbola (i) is, y = mx + c where c = ± a2 m2 - b2

these

( h - 3) 2 k 2 + = cos 2 q + sin 2 q 16 16 ⇒ (h – 3)2 + k2 = 16 Therefore, locus of point P is (x – 3)2 + y2 = 16 which is a circle.

x2

-



...(ii)

... (iii)

But this tangent touches the hyperbola (ii), also ∴

( mx + c )2 a

2

-

x2 b2

=1

⇒ b2(m2x2 + c2 + 2mcx) – a2x2 = a2b2 ⇒ (b2m2 – a2)x2 + 2mcxb2 + b2c2 – a2b2 = 0 ⇒ (b2m2 – a2)x2 + 2mcb2x + b2 (c2 – a2) = 0 For the tangency, it should have equal roots ⇒ (2mcb2)2 = 4(b2m2 – a2)⋅b2(c2 – a2) ⇒ 4m2c2b4 = 4b2(b2m2c2 – b2m2a2 – a2c2 + a4) ⇒ m2c2b2 = b2m2c2 – b2m2a2 – a2c2 + a4 ⇒ a2c2 = a4 – b2m2a2 ⇒ c2 = a2 – b2m2 ⇒ a2m2 – b2 = a2 – b2m2 [using eq. (iii)] ⇒ (a2 + b2)m2 = a2 + b2 ⇒ m2 = 1 ⇒ m = ± 1 Hence, the equation of common tangents are y = ± x ± a2 - b2 (a) : Given curves y = 3x ...(i) and y = 5x ...(ii) Intersect at the point (0, 1). Now, differentiating eqs. (i) and (ii) w.r.t. x, we get dy dy = 3x log 3 and = 5x log 5 dx dx  dy   dy  ⇒   = log 3 and   = log 5  dx ( 0 , 1)  dx ( 0 , 1) ⇒ m1 = log 3 and m2 = log 5 Angle between these curves is given by m - m2 tan q = 1 1 + m1m2 log 3 - log 5 1 + log 3 ⋅ log 5  log 3 - log 5  ⇒ q = tan -1   1 + log 3 log 5  ⇒ tan q =



...(i)

23

Analytical Geometry of Two Dimensions

36.

(d) : We have the given equation lx2 + 4xy + y2 + lx + 3y + 2 = 0 On comparing this equation with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get l 3 a = l, h = 2, b = 1, g = , f = , c = 2 2 2 Since, given equation represents a parabola \ h2 = ab ⇒ 4 = l ⋅ 1 ⇒ l = 4

37. (c) : We have given two circles are 2x2 + 2y2 – 3x + 6y + k = 0



3 k x + 3 y + = 0 ...(i) 2 2 and x2 + y2 – 4x + 10y + 16 = 0 ...(ii) Since, general equation of circle is x2 + y2 + 2gx + 2fy + c = 0 ...(iii) Therefore, comparing eqs. (i) and (ii) with eq. (iii), we get 3 3 k g1 = - , f1 = , c1 = 4 2 2 and g2 = –2, f2 = 5, c2 = 16 Since, both the circles cut orthogonally \ 2(g1g2 + f1 f2) = c1 + c2



 3 15  k ⇒ 2  +  = + 16 2 2  2





⇒ x2 + y 2 -

⇒ 18 =

k + 16 2

38. (a) : A(–2, 1), B(2, 3) and C(–2,–4) are three given points. Slope of the line BA 1 -3 1 m1 = = -2 - 2 2  y2 - y1   using slope formula m = x - x   2 1 Slope of the line BC -4 - 3 7 m2 = = -2 - 2 4 Now, angle between AB and BC is given by



1 7 2 4 1 7 1+ ⋅ 2 4

10 2 ⇒ tan q = ⇒ tan q = 15 3

2 3

2 ⇒ q = tan -1   3

[Q| - x| = x] 39. (d) : Since, given lines are parallel 15 - 5 10 ∴ d= = 2 2 5 4 +3 ⇒ d = 2 = diameter of the circle \ Radius of circle = 1 \ Area of circle = pr2 = p sq. unit 40.

(c) : Given, x2 – 2xy – xy + 2y2 = 0 ⇒ (x – 2y)(x – y) = 0 ⇒ x = 2y, x = y Also, x + y + 1 = 0 On solving eqs. (i) and (ii), we get



 2 1  1 1 A  - , -  , B  - , -  , C( 0 , 0)  3 3  2 2 2 3 1 1 Area of DABC = 2 2 0 -

∴



k ⇒ =2 ⇒ k=4 2

m - m2 tan q = 1 = 1 + m1m2



⇒ q = tan -1 -

41.

=

1 2

1 3 

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

1 1 3 1 1 2 0 1 -

1 1 1 1 = = 6  2  6  12

(c) : Given pair of lines are x2 – 3xy + 2y2 = 0 and x2 – 3xy + 2y2 + x – 2 = 0 \ (x – 2y)(x – y) = 0 and (x – 2y + 2)(x – y– 1) = 0 ⇒ x – 2y = 0, x – y = 0 and x – 2y + 2 = 0, x – y – 1 = 0 Since, the lines x – 2y = 0, x – 2y + 2 = 0 and x – y = 0, x – y – 1 = 0 are parallel. Also, angle between x – 2y = 0 and x – y = 0 is not 90°. \ It is a parallelogram.

42. (a) : In DOAC, OC2 = 22 + 42 = 20 y

4

C(±2, ±4)

4

A

O

2

2

x

24

VITEEE CHAPTERWISE SOLUTIONS

\ Required equation of circle is (x ± 2)2 + (y ± 4)2 = 20 ⇒ x2 + y2 ± 4x ± 8y = 0

43. (d) : Given circles are x2 + y2 – 2x + 8y + 13 = 0 and x2 + y2 – 4x + 6y + 11 = 0. Here, C1 = (1, – 4), C2 = (2, –3) ⇒



and r2 = 4 + 9 - 11 = 2



Now, d = C1C2 = ( 2 - 1)2 + ( -3 + 4)2 = 2 d

- r12

2r1r2

r22



r1 = 1 + 16 - 13 = 2



2



2-4-2

1



⇒ 15y2 – 30y + 71 = 0 y=

⇒

30 ± 900 - 4260 30

=1±

2

-3360 30

 4x - 5  Also, 2 x 2 - 3  = 12  3  ⇒ 10x2 – 40x + 61 = 0 x=

⇒

40 ± 1600 - ( 4 × 10 × 61) 2 × 10

44. (b) : Let the required equation of circle be x2 + y2 +2gx + 2fy = 0. Since, the above circle cuts the given circles orthogonally. \ 2(–3g) + 2f(0) = 8 8 ⇒ 2 g = - and –2g – 2f = –7 3 8 29 ⇒ 2f = 7 + = 3 3 \ Required equation of circle is 8 29 x2 + y 2 - x + y=0 3 3 or 3x2 + 3y2 – 8x + 29y = 0



40 ± -840 -840 =2± 20 20  -840 ∴ Points are A  2 + ,1+ 20 



\ Mid point of AB is (2, 1).

45. (b) : Given curve is y2 = 4x. Also, point (1, 0) is the focus of the parabola. It is clear from the graph that only one normal is possible. y



∴

cosq =

=

2×2× 2

=-

2



⇒ q = 135°

y2 = 4x

x

(1, 0)

x

y



and CA = (1 - 0)2 + 0 2 + (0 - 1)2 = 2 \ Perimeter of triangle = AB + BC + CA = 2+ 2+ 2

2



3  dy  \ Required length = 2 ∫ 1 +   dx  dx  0 2

2



BC = 0 2 + (0 - 1)2 + (1 - 0)2 = 2

=3 2 49. (a) : Given equation of parabola is y2 = 12x ...(i) and equation of latus rectum is x = 3 ...(ii) From Eqs. (i) and (ii), we get y2 = 36 ⇒ y = ± 6 \ Coordinates of end points of a latus rectum are (3, 6) and (3, – 6).

47. (b) : Given, 4x – 3y = 5 and 2x2 – 3y2 = 12

-840 -3360  ,1 20 30 

Now, AB = (0 - 1)2 + (1 - 0)2 + 0 2 = 2





4

 and B  2 

-3360   30 

48. (d) : Let A = (1, 0, 0), B = (0, 1, 0) and C = (0, 0, 1)

(a) : Given, x y = c ⇒ y2(a2 – y2) = c4 ⇒ y4 – a2y2 + c4 = 0 Let y1, y2, y3 and y4 are the roots. \ y1 + y2 + y3 + y4 = 0

46.

2 2

=

∴

 5 + 3y  2 - 3 y 2 = 12  4 



⇒

( 25 + 9 y 2 + 30 y ) - 3 y 2 = 12 8



3 3 6 12 x + 36 = 2 ∫ 1 +   dx = 2 ∫ dx 12 x y 0 0 3

= 2∫

0

x+3 x 2 + 3x

dx

 2 3 = 2  x + 3x + log 2 

3

 3  2  x + 2  + x + 3 x  0

25

Analytical Geometry of Two Dimensions



 3 9  3  3  = 2  3 2 + log  + 3 2  - log    2  2  2  2 



= 2[3 2 + 3 log( 2 + 1)] = 6[ 2 + log(1 + 2 )]

50. (b) : If A and B are two fixed points in a plane, then the locus of another point C on the same plane such that CA + CB = constant, (> AB) is an ellipse. 51. (d) : The equation of parabola is y2 + 4x + 3 = 0

 3 ⇒ y2 = - 4  x +   4



x2 y 2 + =1 16 25

16 3 = 25 5  5  \ Equation of directrix is y = ±   3 / 5 

25 3 ⇒ 3y = 25 is one of equation of directrix. 53. (a) : We know that the normal drawn at 2 a point P(at1, 2at1) to the parabola y2 = 4ax 2 meets again the parabola at Q(at2, 2at2), then 2 t2 = - t1 t1 Here, t1 = p and t2 = q

2 p

⇒ p2 + pq + 2 = 0.

 13 6  A(1, 0) and B  - , -   5 5 which are the points of intersection of straight line and hyperbola. \ Length of straight line intercepted by the hyperbola 2



 13   6 =  - - 1 +  -   5   5 =

324 + 36 = 25

2

2

 18   6 = -  + -   5  5

2

360 6 = 10 25 5

55. (c) : Given that x = t2 + 2t – 1 ...(i) and y = 3t + 5 ...(ii) y-5 ⇒ t= 3 On putting the value of t in eq. (i), we get 2

where a = 4, b = 5 and e = 1 -

=±





Let X = x +

52. (c) : The equation of the ellipse is

q = -p -

54. (d) : The equations of straight line and hyperbola are respectively x – 3y = 1 ...(i) and, x2 – 4y2 = 1 ...(ii) On solving eqs. (i) and (ii), we get



3 and Y = y 4 \ Equation of parabola becomes Y 2 = – 4x The equation of directrix of parabola is X = 1 (Q a = 1) 3 1 ⇒ x+ =1 ⇒ x- =0 4 4

∴

 y - 5  y - 5 x= + 2 -1  3   3  1 2 ⇒ x = { y 2 - 10 y + 25} + ( y - 5) - 1 9 3 1 2 ⇒ x = { y - 10 y + 25 + 6 y - 30 - 9} 9 1 2 = { y - 4 y - 14} 9 ⇒ 9x + 14 = y2 – 4y + 4 – 4 ⇒ 9x + 18 = (y – 2)2 ⇒ (y – 2)2 = 9(x + 2) This is an equation of a parabola. Hence, given parametric equations represent a parabola.

vvv

26

VITEEE CHAPTERWISE SOLUTIONS

4

CHAPTER

1.

The length of longer diagonal of the   parallelogram constructed on 5a + 2b and     a − 3b , if it is given that | a | = 2 2 ,|b |= 3and   π the angle between a and b is is 4 (a) 15

2.

4.

5.

(b)

6.

113

(c) 593 (d) 369 (2014)        If r = a(b × c ) + b(c × a ) + g( a × b ) and   [ a b c ] = 2, then a + b + g is equal to        (a) r ⋅ [b × c + c × a + a × b ] 1        (b) (c) 2r ⋅ ( a + b + c ) r ⋅(a + b + c ) 2 (d) 4

3.

Vector Algebra

7.

(a)

5 6 10

(b)

19 9

9 6 (d) (2011) 19 19    If a ,b , c are three non-zero vectors such that         a + b + c = 0 and m = a ⋅ b + b ⋅ c + c ⋅ a , then (a) m < 0 (b) m > 0 (c) m = 0 (d) m = 3 (2011) (c)

(2014)

If a, b, c are three non-coplanar vectors and p, q,  r are reciprocal vectors, then      (la + mb + nc ). (lp + mq + nr ) is equal to (a) l + m + n (b) l3 + m3 + n3 2 2 2 (c) l + m + n (d) None of these (2014)  ^ ^ The vector b = 3 i + 4 k is to be written as the  ^ ^ sum ofa vector b1 parallel to a = i + j and a  vector b2 perpendicular to a. Then, b1 is equal to 3 ^ ^ 2 ^ ^ (a) ( i + j ) (b) ( i + j ) 2 3 1 ^ ^ 1 ^ ^ (c) ( i + j ) (d) ( i + j ) (2013) 2 3 P is a fixed point (a, a, a) on a line through the origin equally inclined to the axes, then any plane through P perpendicular to OP, makes intercepts on the axes, the sum of whose reciprocals is equal to 3 (a) a (b) 2a 3a (c) (d) None of these 2 (2013)

 ^ ^ ^  ^ ^ ^ If a = i + j + k , b = i + 3 j + 5 k and  ^ ^ ^ c = 7 i + 9 j + 11 k , then the area of parallelogram     having diagonals a + b and b + c is 1 21 sq units (a) 4 6 sq units (a) 2 6 (c) sq units (d) 6 sq units 2 (2012) The projection of the vector iˆ − 2 ˆj + kˆ on the vector 4iˆ − 4 ˆj + 7 kˆ is

8.

9.

A line making angles 45° and 60° with the positive directions of the axes of x and y makes with the positive direction of z-axis, an angle of (a) 60° (b) 120° (c) 60° or 120° (d) None of these (2011) ^ ^ ^  ^ ^ ^  10. If a = 2 i + 2 j + 3 k , b = − i + 2 j + k and  ^ ^    c = 3 i + j , then a + t b is perpendicular to c , if t is equal to (a) 2 (b) 4 (c) 6 (d) 8 (2010) 11. The distance between the line  ^ ^ ^ ^ ^ ^ r = 2 i − 2 j + 3 k + λ ( i − j + 4 k ) and the plane  ^ ^ ^ r ⋅ ( i + 5 j + k ) = 5, is (a)

10 3

(b)

10 3

27

Vector Algebra

(c)

10 3 3



(d)

10 9

π π ,q = 6 2 3 π π (c) q1 = , q2 = 2 3 (a) q1 =

(2010)

12. The angle between the lines whose direction cosines satisfy the equations l + m + n = 0, l2 + m2 – n2 = 0 is π π (a) (b) 6 4 π π (c) (d) (2009) 3 2

16. The shortest distance between the straight lines through the points A1 = (6, 2, 2) and A2 = (– 4, 0, – 1), in the directions of (1, – 2, 2) and (3, – 2, – 2) is (a) 6 (b) 8 (c) 12 (d) 9 (2008)

13. If m1, m2, m3 and m4 are respectively, the magnitudes of the vectors  ^ ^ ^  ^ ^ ^ a1 = 2 i − j + k , a2 = 3 i − 4 j − 4 k   ^ ^ ^ ^ ^ ^ a3 = i + j − k and a4 = − i + 3 j + k then the correct order of m1, m2, m3 and m4 is (a) m3 < m1 < m4 < m2 (b) m3 < m1 < m2 < m4 (c) m3 < m4 < m1 < m2 (d) m3 < m4 < m2 < m1 (2009)  ^ ^ ^  ^ ^ ^ 14. a = i − j + k and b = 2 i + 4 j + 3 k are one of the sides and medians respectively, of a triangle through the same vertex, then area of the triangle is (a)

1 83 2

(b)

π π ,q = 3 2 6 π π (d) q1 = , q2 = 3 2 (2008) (b) q1 =

       17. If b is a unit vector, then ( a ⋅ b ) b + b × ( a × b ) is      (a) | a |2 b (b) | a ⋅ b |a   (c) a (d) b (2007) 18. If q is the angle between the lines AB and AC where A, B and C are the three points with coordinates (1, 2, –1), (2, 0, 3), (3, –1, 2) respectively, then 462 cos q is equal to (a) 20 (b) 10 (c) 30 (d) 40 (2007)     19. Let the pairs a , b and c , d each determine a plane. Then the planes are parallel, if      (a) ( a × c ) × (b × d ) = 0     (b) ( a × c) ⋅ (b × d) = 0    (c) ( a × b ) × (c × d ) = 0     (2007) (d) ( a × b ) ⋅ (c × d ) = 0

83

1 (d) 86 (2008) 85 2    15. If a , b , c be three unit vectors such that     1  a × (b × c ) = b , b and c being non-parallel. 2   If q1 is the angle between a and b and q2 is   the angle between a and c , then (c)

^

^

^

^

^

and i − 3 j + 4 k as diagonals is (a)

72

(b)

73

(c)

74

(d)

75

(2007)

Answer Key 1.

(c)

2.

(b)

3.

(c)

4.

(a)

9.

(a)

10. (d)

11. (c)

12. (c)

17. (c)

18. (a)

19. (c)

20. (d)

^

20. The area of a parallelogram with 3 i + j − 2 k

5.

(d)

6.

(a)

7.

(b)

8.

(a)

13.

(a)

14.

(d)

15.

(c)

16.

(d)

28

VITEEE CHAPTERWISE SOLUTIONS

e planations

1.

  (c) : Given that, | a | = 2 2 ,|b | = 3       The longer vector is 5a + 2b + a − 3b = 6 a − b   Length of one diagonal = |6 a − b |     = 36 a 2 + b 2 − 2 × 6| a |⋅|b |⋅ cos 45° = ( 36 × 8) + 9 − 12 × 2 2 × 3 ×



1 2

= 288 + 9 − (12 × 6) = 225 = 15   Other diagonal is 4 a + 5b . Its length is     = ( 4 a )2 + ( 5b )2 + 2 × | 4 a || 5b |cos 45° = (16 × 8) + ( 25 × 9) + ( 40 × 6) = 593

2.

(b) : Now,            r ⋅ a = a( a ⋅ (b × c )) + b( a ⋅ (c × a )) + g( a ⋅ ( a × b ))    = a [ a b c] + 0 + 0       Similarly, r ⋅ b = b[ a b c ] and r ⋅ c = g [ a b c ] 1    1       ∴ r ⋅ ( a + b + c ) = (r ⋅ a + r ⋅ b + r ⋅ c ) 2 2  1 = (a + b + g ) [ a b c ] 2 1 = (a + b + g ) × 2 = a + b + g 2    3. (c) : Q p , q and r are reciprocal vectors of    a , b and c respectively.       ∴ p . a = 1, p . b = 0 = p . c      q . a = 0, q . b = 1, q . c = 0       r . a = 0, r . b = 0, r . c = 1       ∴ (la + mb + nc ) . (lp + mq + nr ) = l2 + m2 + n2   (a) : Since, b1 || a , therefore    ˆ  b1 = a ( iˆ + j ) b2 = b – b1 = (3 – a) iˆ – ajˆ + 4 kˆ   Also, b2 . a = 0 3 ⇒ ( 3 − a) − a = 0 ⇒ a = 2  3 ˆ ˆ Hence, b1 = i + j 2 (d) : Since, the line is equally inclined to the axes and passes through the origin, its

4. 5.

( )



6.

direction ratios are < 1, 1, 1 >. So, its equation x y z is = = . 1 1 1 A point P on it is given by (a, a, a). So, equation of the plane through P(a, a, a) and perpendicular to OP is 1(x – a) + 1 (y – a) + 1 (z – a) = 0 (Q OP is normal to the plane) i.e., x + y + z = 3a y x z ⇒ + + =1 3a 3a 3a Intercepts on axes are 3a, 3a and 3a, therefore sum or reciprocal of these intercepts 1 1 1 1 = + + = 3a 3a 3a a   (a) : Given, a = i^ + ^j + ^k , b = i^ + 3 ^j + 5 k^  ^ ^ ^ and c = 7 i + 9 j + 11 k    ^ ^ ^ ^ ^ ^ Let A = a + b = ( i + j + k ) + ( i + 3 j + 5 k )    ^ ^ ^ = ( 2 i + 4 j + 6 k ) and B = b + c ^

^

^

^

^

^

= ( i + 3 j + 5 k ) + (7 i + 9 j + 11 k ) ^

^

^



= 8 i + 12 j + 16 k   Since A and B are diagonals.



1   \ Area of parallelogram = | A × B| 2 ^

^

^

i j k 1 = 2 4 6 2 8 12 16 7.



1 ^ ^ ^ = | i (64 − 72) − j( 32 −48) + k( 24 − 32)| 2 1 ^ ^ ^ = | −8 i + 16 j − 8 k | 2 = ( −4)2 + (8)2 + ( −4)2 = 96 = 4 6 sq units   b = 4iˆ − 4 ˆj + 7 kˆ , (b) : Let a = iˆ − 2 ˆj + kˆ and   then projection of a on b is equal to   4+8+7 19 a ⋅b =  = b 16 + 16 + 49 9

29

Vector Algebra

   (a) : We have, a + b + c = 0   ⇒ | a + b + c |2 = 0 2 2        ⇒ |a| + |b| + |c|2 + 2{a . b + b . c + c . a } = 0     ⇒ a . b + b . c + c . a  1   = − {| a |2 + |b |2 + |c |2|< 0 2 ⇒ m < 0 8.

9.

(a) : Let g be the required angle, then



cos 45° + cos 60° + cos g = 1 2

2

2

1 1 + + cos 2 g = 1 2 4 3 1 ⇒ cos 2 g = 1 − = 4 4 1 ⇒ cos g = ⇒ g = 60° 2  ^ ^ ^ 10. (d) : We have, a = 2 i + 2 j + 3 k  ^ ^ ^  b = − i + 2 j + k and c = 3 i^ + ^j   ^ ^ ^ ^ ^ ^ Now, a + t b = ( 2 i + 2 j + 3 k ) + ( −t i + 2t j + t k )





⇒

^

^

^



= ( 2 − t ) i + ( 2 + 2t ) j + ( 3 + t ) k    Since a + t b is perpendicular to c    \ ( a + t b ) ⋅ c = 0



⇒ {( 2 − t ) i + ( 2 + 2t ) j + ( 3 + t ) k} ⋅ ( 3 i + j ) = 0



⇒ 3(2 – t) + 2 + 2t = 0 ⇒ 6 – 3t + 2 + 2t = 0 ⇒ t = 8





^

^

^



⇒ 2m2 + 2mn = 0 ⇒ 2m(m + n) = 0 ⇒ m = 0 or m + n = 0 If m = 0, then l = – n l1 m1 n1 ∴ = = −1 0 1 and if m + n = 0 ⇒ m = –n, then l = 0 l2 m2 n2 ∴ = = 0 −1 1 i.e., (l1, m1, n1) = (–1, 0, 1) and (l2, m2, n2) = (0, –1, 1) ∴

cosq =

⇒

q=

^

11. (c) : We have the given line  ^ ^ ^ ^ ^ ^ r = 2 i − 2 j + 3 k + λ ( i − j + 4 k)    On comparing it with r = a + t b , we get  ^ ^ ^  ^ ^ ^ a = 2 i − 2 j + 3 k, b = i − j + 4 k  ^ ^ ^ Also, the plane is r ⋅ ( i + 5 j + k ) = 5   On comparing it with r ⋅ n = d , we get  ^ ^ ^ n = i + 5 j + k and d = 5   ^ ^ ^ ^ ^ ^ Since, b ⋅ n = ( i − j + 4 k ) ⋅ ( i + 5 j + k ) = 0 \ Given line is parallel to the given plane. Now, distance between the line and the plane is given by required distance   ^ ^ ^ ^ ^ ^ a ⋅ n − d |( 2 i − 2 j + 3 k ) ⋅ ( i + 5 j + k ) − 5| = =  | n| 1 + 25 + 1 | 2 − 10 + 3 − 5| 10 = = 27 3 3 12. (c) : Given, l + m + n = 0, ⇒ l = – m – n and l2 + m2 – n2 = 0 \ (–m –n)2 + m2 – n2 = 0

=

1 2

π 3

13. (a) : Given,

^

0+0+1 1+ 0 +1 0 +1+1

 m1 = | a1 | = 2 2 + ( −1)2 + (1)2 = 6  m2 = | a2 | = 32 + ( −4)2 + ( −4)2 = 41  m3 = | a3 | = 12 + 12 + ( −1)2 = 3



 2 2 2 and m4 = | a4 | = ( −1) + ( 3) + (1) = 11



\

m3 < m1 < m4 < m2

14. (d) : Area of triangle iˆ jˆ kˆ

= 1 −1 1 2 4 3 = −7 iˆ − jˆ + 6 kˆ = 49 + 1 + 36 = 86

   1 15. (c) : Since, a × (b × c ) = b 2       1 ⇒ ( a ⋅ c ) b − ( a ⋅ b ) c = b 2 On comparing both sides, we get   1   a ⋅ c = and a ⋅ b = 0 2   1 Now, a ⋅ c = 2   1 ⇒ a ⋅ c cosq2 = 2   π ⇒ cos q2 = (Q a and b are unit vectors) 2 π π ⇒ cos q2 = cos ⇒ q2 = 3 3

30

VITEEE CHAPTERWISE SOLUTIONS

    and a ⋅ b = 0 ⇒ a ⋅ b cosq1 = 0   π ⇒ cos q1 = cos (Q a and b are unit vectors) 2 π π π ⇒ q1 = . Hence, q1 = and q2 = . 2 2 3

  16. (d) : Here, a1 = 6iˆ + 2 jˆ + 2 kˆ , a2 = −4iˆ + 0 jˆ − kˆ ,   b1 = iˆ − 2 jˆ + 2 kˆ , and b2 = 3iˆ − 2 jˆ − 2 kˆ ,  ( a − a ) ⋅ (b × b )  \ Shortest distance =  2 1  1 2    b1 × b2   =

( −10iˆ − 2 jˆ − 3kˆ ) ⋅ ( 8iˆ + 8 jˆ + 4 kˆ ) 64 + 64 + 16



^



^

^

^

( i − 2 j + 4 k) ⋅ (2 i − 3 j + 3 k) 1 + 4 + 16 4 + 9 + 9

⇒

^

^

^

∴

^

Area of parellelogram = ^



^

^

462

462 cos q = 20.   19. (c) : Since, a and b are coplanar, therefore   a × b is a vector perpendicular to the plane     containing a and b . Similarly, c × d is a vector perpendicular to the plane containing   c and d. Thus, the two planes will be parallel    if their normals, i.e., ( a × b ) and (c × d ) are parallel.      ⇒ ( a × b ) × (c × d ) = 0

^

= 2 i − 3 j + 3k ^

cos q =

^

21 122

20

=

^

are 3 i + j − 2 k and i − 3 j + 4 k .

18. (a) : Since, the coordinates of A, B and C are (1, 2, –1), (2, 0, 3) and (3, –1, 2), then  ^ ^ ^ AB = ( 2 − 1) i + (0 − 2) j + ( 3 + 1) k ^ ^ ^ = i − 2 j + 4k  ^ ^ ^ and AC = ( 3 − 1) i + ( −1 − 2) j + ( 2 + 1) k ^



2 + 6 + 12

20. (d) : Since, the diagonals of a parallelogram

−108 = =9 12       17. (c) : ( a ⋅ b )b + b × ( a × b )           = (a ⋅ b ) b + ( b ⋅ b ) a − (a ⋅ b ) b = a



=



vvv

^

^

i j k 1 = × Mod of 3 1 −2 2 1 −3 4 1 ^ ^ ^ = | −2 i − 14 j − 10 k | 2 =

1 1 × 4 + 196 + 100 = 300 2 2

=

1 × 2 75 = 75 2

1   d × d2 2 1

31

Analytical Geometry of Three Dimensions

5

CHAPTER

1.

2.

Analytical Geometry of Three Dimensions

If the points (1, 2, 3) and (2, –1, 0) lie on the opposite sides of the plane 2x + 3y – 2z = k, then (a) k < 1 (b) k > 2 (c) k < 1 or k > 2 (d) 1 < k < 2 (2014)

6.

If a line in the space makes angles a,b and g with the coordinate axes, then cos 2a + cos 2b + cos 2g + sin2 a + sin2 b + sin2 g equals (a) –1 (b) 0 (c) 1 (d) 2 (2009)



x -1 y +1 z-1 = = and 2 3 4 x-3 y-k = = z intersect at a point, if k is 1 2 equal to 2 1 9 1 (a) (b) (c) (d) 9 2 2 6 Two lines

7.

The radius of the sphere x2 + y2 + z2 = 12x + 4y + 3z is (a) 13/2 (b) 13 (c) 26 (d) 52

8.

The plane through the point (– 1, – 1, – 1) and containing the line of intersection of the ^ ^  ^ ^  ^ planes r ⋅ ( i + 3 j - k ) = 0 and r ⋅ ( j + 2 k ) = 0 ^ ^ ^  ^ ^  ^ (a) r ⋅ ( i + 2 j - 3 k ) = 0 (b) r ⋅ ( i + 4 j + k ) = 0 ^ ^  ^ ^ ^  ^ (c) r ⋅ ( i + 5 j - 5 k ) = 0 (d) r ⋅ ( i + j + 3 k ) = 0 (2008)

9.

The equation     r 2 - 2r ⋅ c + h = 0 ⋅|c | > h , represents (a) circle (b) ellipse (c) cone (d) sphere

(2014, 2010) 3.

4.

5.

If a plane meets the coordinate axes at A, B and C such that the centroid of the triangle is (1, 2, 4), then the equation of the plane is (a) x + 2y + 4z = 12 (b) 4x + 2y + z = 12 (c) x + 2y + 4z = 3 (d) 4x + 2y + z = 3 (2014) The volume of the tetrahedron included between the plane 3x + 4y – 5z – 60 = 0 and the coordinate planes is (a) 60 (b) 600 (c) 720 (d) 400 (2014)

(2009)

(2008)

10. The center and radius of the sphere x2 + y2 + z2 + 3x – 4z + 1 = 0 are

The equation of sphere concentric with the sphere x2 + y2 + z2 – 4x – 6y – 8z – 5 = 0 and which passes through the origin, is (a) x2 + y2 + z2 – 4x – 6y – 8z = 0 (b) x2 + y2 + z2 – 6y – 8z = 0 (c) x2 + y2 + z2 = 0 (d) x2 + y2 + z2 – 4x – 6y – 8z – 6 = 0 (2010)

(a)  - 3 , 0 , - 2  ; 21 (b)  3 , 0 , 2  ; 21 2   2  2  3  21 21 (c)  - 3 , 0 , 2  ; (d)  - , 2 , 0  ;  2  2  2  2 (2008)

Answer Key 1. 9.

(d) (d)

2. (c) 10. (c)

3.

(b)

4.

(b)

5.

(a)

6.

(c)

7.

(a)

8.

(a)

32

1. 2.

3.

VITEEE CHAPTERWISE SOLUTIONS

e planations

(d) : Since, the points (1, 2, 3) and (2, –1, 0) lie on the opposite sides of the plane 2x + 3y – 2z – k = 0 ⇒ (2 + 6 – 6 – k) (4 – 3 – k) < 0 ⇒ (k – 1) (k – 2) < 0 ∴ 1 < k < 2

x -1 y +1 z-1 = = = r(say ) 2 3 4 ⇒ x = 2 r + 1, y = 3r – 1, z = 4 r + 1 Since, the two lines intersect. So, putting above values in second line, we get 2r + 1 - 3 3r - 1 - k 4r + 1 = = 1 2 1 st rd Taking 1 and 3 terms, we get 2r – 2 = 4r + 1 ⇒ r = –3/2 Also taking 2nd and 3rd terms, we get 3r – 1 – k = 8r + 2 15 9 -3= ⇒ k = –5r – 3 = 2 2 (b) : Let the equation of the plane is x y z + + =1 a b g Then, A(a, 0, 0), B(0, b, 0) and C(0, 0, g) are the points on the coordinate axes. Since, the centroid of the triangle is (1,2, 4) \

a =1 ⇒a=3 3 b =2 ⇒b=6 3 g = 4 ⇒ g = 12 3



and



\ The equation of the plane is x y z + + =1 3 6 12

⇒ 4x + 2y + z = 12





(c) :



4.



(b) : The given equation of the plane is 3x + 4y – 5z – 60 = 0, it can be written in the form x y z + + =1 20 15 -12

5.

which meets the coordinates axes at the points A(20, 0, 0), B(0, 15, 0) and C(0, 0, –12). The coordinates of the origin are (0, 0, 0). Therefore, volume of the tetrahedron OABC is 20 0 0 1 1 = 0 15 0 = | 20 × 15 × ( -12)| = 600 6 6 0 0 -12 (a) : The equation of the sphere concentric with the sphere x2 + y2 + z2 – 4x – 6y – 8z – 5 = 0 is x2 + y2 + z2 – 4x – 6y – 8z + c = 0 ...(i) Since, the sphere (i) passes through origin, therefore, 0+0+0–0–0–0+c=0 ⇒ c = 0 Hence, the required equation of sphere is x2 + y2 + z2 – 4x – 6y – 8z = 0

6. (c) : cos 2a + cos 2b + cos 2g + sin2 a + sin2 b + sin2g 2 2 2 2 = (cos a – sin a) + (cos b – sin b) + (cos2 g – sin2 g) + sin2 a + sin2 b + sin2 g = cos2 a + cos2 b + cos2 g = 1 7.

(a) : Given equation of sphere is x2 + y2 + z2 – 12x – 4y – 3z = 0



3  \ Centre of sphere is  6 , 2 ,  .  2



3 Radius of sphere = (6)2 + ( 2)2 +   2

8.

= 36 + 4 + =

2

9 169 = 4 4

13 2

(a) : The cartesian form of an equation of plane is x + 3y – z = 0 and y + 2z = 0 The line of intersection of two planes is (x + 3y – z) + l(y + 2z) = 0 Since, it is passing through (– 1, –1, – 1)

33

Analytical Geometry of Three Dimensions

9.

\ (– 1 – 3 + 1) + l (– 1 – 2) = 0 ⇒ – 3 – 3l = 0 ⇒ l = – 1 \ (x + 3y - z) - 1 (y + 2z) = 0 ⇒ x + 2y – 3z = 0 Hence, equation of plane is  r ⋅ ( iˆ + 2 jˆ - 3kˆ ) = 0



(d) : The given equation is     r 2 - 2r ⋅ c + h = 0 , c > h





This is an equation of sphere in diameter     form i.e., (r - a ) ⋅ (r - b ) = 0

10. (c) : The given equation of sphere is x2 + y2 + z2 + 3x – 4z + 1 = 0



vvv

On comparing this equation with general equation of sphere x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0, 3 we get u = , v = 0, w = – 2 and d = 1 2 \ Coordinates of centre of sphere = (–u, –v, –w)  3  =  - ,0,2   2  and radius of sphere = u2 + v 2 + w 2 - d =

9 9 + 12 + 4-1 = 4 4

=

21 2

34

VITEEE CHAPTERWISE SOLUTIONS

6

CHAPTER

1.

2.

Differential Calculus

The equation of normal to the curve y = (1 + x)y + sin–1(sin2x) at x = 0 is (a) x + y = 1 (b) x – y = 1 (c) x + y = –1 (d) x – y = –1 (2014)

4.

5.

6.

If there is an error of k% in measuring the edge of a cube, then the percent error in estimating its volume is (a) k (b) 3k k (c) (d) None of these 3 (2014) Let f ′(x) be differentiable ∀ x. If f(1) = –2 and f ′(x) ≥ 2 ∀ x ∈[1, 6], then (a) f(6) < 8 (b) f(6) ≥ 8 (c) f(6) ≥ 5 (d) f(6) ≤ 5 (2014) The minimum value of

x is log x

1/ x

(a) 1 (c) e2

(b)

(c) e2

(d) e3

(2014)

The minimum area of the triangle formed by any tangent to the ellipse the coordinate axes is

x2 a2

8.



2 3 , 3 2 3 3/ 2 (c) ,e 2 (a)

9.

+

( a + b) 2

y2 b2

= 1 with

(a) a2 + b2

(b)

(c) ab

( a − b )2 (d) 2

(2014)

(b) e1/2 (d) e3

(2013)

(b)

2 2/ 3 ,e 3

(d) None of these (2013)

The solution of the differential equation dy yf ′( x ) − y 2 is = dx f ( x) (b) f(x) = y(x + C) (d) None of the above (2013)

10. The rate of change of the surface area of a sphere of radius r, when the radius is increasing at the rate of 2 cm s–1 is proportional to 1 r (c) r (a)

2

equals to

 p (1+ |sin x |)a /|sin x|, − 6 < x < 0  b , x=0 , If f ( x ) =   tan 2 x /tan 3 x p  e , 0 p  2 p x = , then 2 np (a) m = 1, n = 0 (b) m = +1 2 p p (c) n = m (d) m = n = 2 2 (2012) 1 − cos x 14. Find the value of the lim . x→0 x (a) 0 (b) 1 (c)

2

(d) does not exist (2011)

15. The values of constants a and b so that

 x2 + 1  lim  − ax − b  = 0 are x →∞  x + 1  (a) a = 0, b = 0 (b) a = 1, b = –1 (c) a = –1, b = 1 (d) a = 2, b = –1

(2011)

x≤0  x 2 , 16. If f ( x ) =  , then x = 0 is  2 sin x , x > 0 (a) point of minima (b) point of maxima (c) point of discontinuity (d) None of the above

 dy  ( x 2 + 4)   is equal to  dx  (a) n2(y2 – 4) (b) n2(4 – y2) (c) n2(y2 + 4)

(d) None of these (2010)

dy 20. If y = x + y + x + y + ... ∞ , then is dx equal to y+x y3 − x (a) 2 (b) 2 y 2 − 2 xy − 1 y − 2x (c)

y3 + x 2y2 − x

x + 5 21. lim   x→∞  x + 2 



(d) None of these (2010)

x+3

equals

(a) e (c) e3

 1 1 1 1  + + + ... + 17. lim   is equal n→∞  1 ⋅ 2 2⋅3 3⋅4 n(n + 1)  to (a) 1 (b) –1 (c) 0 (d) None of these (2010) 18. The condition that the line lx + my = 1 may be normal to the curve y2 = 4ax, is (a) al3 – 2alm2 = m2 (b) al2 + 2alm3 = m2 (c) al3 + 2alm2 = m3 (d) al3 + 2alm2 = m2 (2010)

(2009)

22. If f : R → R is defined by  2 sin x − sin 2 x , if x ≠ 0  f ( x) =  2 x cos x  a, if x = 0  then the value of a so that f is continuous at 0 is (a) 2 (b) 1 (c) –1 (d) 0 (2009)  1 −1 23. If x = cos   1 + t 2

(2011)

(b) e2 (d) e5



dy is equal to dx (a) 0 (c) 1

  t  , y = sin −1   1 + t 2 

  

then

(b) tan t (d) sin t cos t

(2009)

 x −1  1 d  −1  a tan x + b log  = 4  dx   x + 1   x − 1 then a – 2b is equal to (a) 1 (b) –1 (c) 0 (d) 2 (2009)

24. If

25. If y = ea sin–1 x then (1 – x2) yn + 2 – (2n + 1)x yn + 1 is equal to (a) –(n2 + a2)yn (b) (n2 – a2)yn 2 2 (c) (n + a )yn (d) –(n2 – a2)yn (2009)

36

VITEEE CHAPTERWISE SOLUTIONS

26. The function f(x) = x3 + ax2 + bx + c, a2 ≤ 3b has (a) one maximum value (b) one minimum value (c) no extreme value (d) one maximum and one minimum value (2009) 27. If g(x) is a polynomial satisfying g(x) g(y) = g(x) + g(y) + g(xy) – 2 for all real x and y and g(2) = 5, then lim g( x ) is x→3 (a) 9 (b) 10 (c) 25 (d) 20 (2008) 28. The value of f (0) so that

( − e x + 2x ) may be x

31. If the normal to the curve y = f(x) at (3, 4) 3p makes an angle with the positive x-axis, 4 then f ′(3) is equal to

(d) – 1 + log 2 (2008)

29. Let [·] denote the greatest integer function and f (x) = [tan2x]. Then f ( x ) does not exist (a) lim x →0 (b) f (x) is continuous at x = 0 (c) f (x) is not differentiable at x = 0 (d) f (x) = 1

(2008)

30. A spherical balloon is expanding. If the radius is increasing at the rate of 2 centimetres per minute, the rate at which the volume increases (in cubic centimetres per minute) when the radius is 5 centimetres is (a) 10p (b) 100p (c) 200p (d) 50p (2008)

(b)

(c) –1

(d) −

3 4

(2007)

32. The function f(x) = x2 e–2x, x > 0. Then the maximum value of f(x) is 1 1 (a) (b) e 2e

continuous at x = 0 is (a) log  1  (b) 0 2 (c) 4

3 4

(a) 1

(c)

1 e

2



(d)

4 e4

33. If (x + y) sin u = x2y2, then x to (a) sin u (c) 2 tan u



(2007)

∂u ∂u is equal +y ∂x ∂y

(b) cosec u (d) 3 tan u

(2007)

34. The angle between the tangents at those points on the curve x = t2 + 1 and y = t2 – t – 6 where it meets x-axis is  4  (a) ± tan −1    29 

−1  5  (b) ± tan    49 

 10  (c) ± tan −1    49 

 8  (d) ± tan −1    29  (2007)

Answer Key 1. 9. 17. 25. 33.

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

2. 10. 18. 26. 34.

(c) (c) (d) (c) (c)

3. 11. 19. 27.

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

4. 12. 20. 28.

(b) (b) (d) (d)

5. 13. 21. 29.

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

6. 14. 22. 30.

(c) (d) (d) (c)

7. 15. 23. 31.

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

8. 16. 24. 32.

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

37

Differential Calculus

e planations

1.

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



dy  y dy  = (1 + x )y  + log(1 + x )  dx dx  1 + x +



 dy  = 1 ⇒    dx  at( 0 ,1)



2 sin x cos x 1 − sin 4 x

[Q at x = 0, y = 1]

Slope of normal at (x = 0) = –1 \ Equation of normal at x = 0 and y =1 is y – 1= –1 (x – 0) ⇒ y – 1= –x ⇒ x + y = 1



(c) : It is clear that f(x) has a definite and unique value for each x ∈ [1, 5]. Thus, for every point in the interval [1, 5], the value of f(x) exists. So, f(x) is continuous in the interval [1, 5]. −x Also, f ′(x) = which clearly exists for 25 − x 2 all x in an open interval (1, 5). Hence, f ′(x) is differentiable in (1, 5) So, there must be a value c ∈ [1, 5] such that



f ′(c) =

2.



f ( 5) − f (1) 0 − 24 0 − 2 6 − 6 = = = 4 2 5−1 4 −c But f ′(c) = 25 − c 2 −c

6 2



\



⇒ 4c2 = 6(25 – c2)



⇒ 4c2 = 150 – 6c2 ⇒ 10c2 =150



⇒ c2 = 15 ⇒ c = ± 15



\ c =

25 − c

2

=−

Now the change in volume,  dV  Dx = 3x 2 ( Dx ) DV =   dx 

15 ∈ [1, 5] 3. (b) : We know that, the volume V of side x is given by V = x3 On differentiating w.r.t. x, we get dV = 3x2 dx kx Let the change in x be Dx = k% of x = 100



 k x  3x 3k = 3x 2  =  100  100



∴ The approximate change in volume



=

4.

(b) : Since, f ′(x) is differentiable ∀ x ∈ [1, 6]. ∴ By Lagrange’s mean value theorem,



f (6) − f (1) 6−1 Q f ′(x) ≥ 2 ∀ x ∈ [1, 6]



3kx 3 3k 3 = ⋅ x = 3k% of original volume 100 100

f ′( x ) =

f ( 6) + 2 ≥ 2 5 ⇒ f (6) ≥ 10 – 2 ⇒ f (6) ≥ 8 ⇒

(given) [Q f(1) = –2]

x log x On differentiating w.r.t. x, we get log x − 1 f ′( x ) = (log x )2 For maxima and minima, put f ′(x) = 0 Now, log x – 1 = 0 ⇒ x=e 2 log x 1 (log x )2 ⋅ − (log x − 1) ⋅ x x f ′′( x ) = (log x )4 1 −0 1 = >0 ⇒ f ′′( e ) = e e 1 ∴ f(x) is minimum at x = e Hence, minimum value of f(x) at x = e is e f (e) = =e log e 6. (c) : Equation of tangent at (a cos q, b sin q) to the ellipse is x y cos q + sin q = 1 a b Coordinates of P and Q are 5.

(a) : Let f(x) =



b   a    cos q , 0  and  0 , sin q  , respectively.

38

VITEEE CHAPTERWISE SOLUTIONS Y



Q

O

X�

7.

8.



(a cosθ, b sinθ) X P

Now, area of DOPQ

X�

1 a b ab × = 2 cos q sin q sin 2q

=



f ( x) = x + C ⇒ f(x) = y (x + C) y

10. (c) : Surface area of sphere, S = 4pr2 dr and =2 dt

\

dS dr = 4 p × 2r = 8 pr × 2 = 16 pr dt dt



⇒

dS ∝r dt

∴ Minimum area = ab (c) : Limit = lim

= e x→0

1 lim [log f (1+ x ) − log f (1)] e x→0 x

f ′(1+ x )/ f (1+ x ) 1

= e f ′(1)/f(1) = e6/3 = e2

 p a / sin x , − 0 and f(0) = 0 Hence, it is a point of minima. ⇒

1 1  + −  + 1   n n 

18. (d) : Let P(x1 , y1) be a point on the curve y2 = 4ax ...(i) On differentiating y2 = 4ax w.r.t. ‘x’, we get dy 2y = 4a dx



x sin 2 = 2 lim x→ 0 x 1



 1 1 1 1 1 lim   1 −  +  −  +  −  + ... 2 2 3 3 4

n→∞  

 n 1  lim  1 − = lim n→∞  n + 1  n→∞ n + 1 n 1 lim = lim =1 n→∞  1 1  n→∞  + 1 + 1   n  n 

p . 2

So, lim f ( x) = lim f ( x) x→

 1 1 1 1  17. (a) : lim  + + + .... +  n→∞  1 ⋅ 2 2⋅3 3⋅4 n(n + 1) 





 dy  ⇒    dx ( x

1 , y1 )

=

2a y1

Thus, the equation of normal at (x1, y1) is y y − y1 = − 1 ( x − x1 ) 2a ⇒ y1x + 2ay = y1(x1 + 2a) ...(ii) But lx + my = 1 ...(iii) is also a normal. Therefore, coefficients of eqs. (ii) and (iii), must be proportional y 2 a y1 ( x1 + 2 a) \ 1= = 1 l m 2 al 1 ⇒ y1 = and x1 = − 2 a l m Putting these values of x1 and y1 in eq. (i), we get 2



 2 al  1  4a2l 2 4a − 8a2l =  m  = 4 a  l − 2 a  ⇒ l m2 ⇒ al3 = m2 – 2am2l ⇒ al3 + 2alm2 = m2

19. (c) : x = sec q – cos q dx ⇒ = sec q tan q + sin q, dq y = secn q – cosn q

dy = n sec n−1 q sec q tan q + n cosn−1 q sin q dq (sec n q tan q + cosn−1 q sin q) dy \ =n (sec q tan q + sin q) dx ⇒

40

VITEEE CHAPTERWISE SOLUTIONS

⇒

(sec n q + cosn q) tan q dy =n dx (sec q + cos q) tan q

⇒

dy n(sec n q + cosn q) = dx (sec q + cos q)



2 sin x − sin 2 x 2 x cos x 0   0 form  2 cos x − 2 cos 2 x = lim x→ 0 2 (cos x − x sin x) 2−2 = lim =0 x→ 0 2(1 − 0) Since, f(x) is continuous at x = 0 Now, lim f ( x) = lim x→ 0

2

n2 {(sec n q + cosn q)2 + 4}  dy  ⇒   =  dx  (sec q − cos q)2 + 4 2

n2 ( y 2 + 4 )  dy  ⇒   =  dx  x2 + 4



2

 dy  ⇒ ( x 2 + 4 )   = n2 ( y 2 + 4 )  dx 



⇒ y 2 = x + y + x + y + ... ∞ ⇒ y =x+ y+y ⇒ y2 = x + 2y ⇒ (y – x) = 2y On differentiating both sides w.r.t., x, we get



2

dy  dy  2( y 2 − x)  2 y − 1 = 2  dx  dx



⇒

=e

 1 + lim 3  x →∞  1+ 

3 x 2  x

22. (d) : Given,



  

⇒ x = tan–1 t and y = tan–1 t \ y = x dy =1 dx

 x − 1  1 d  −1  a tan x + b log  = 4  dx   x + 1   x − 1 On integrating both sides, we get  x − 1 a tan −1 x + b log   x + 1  1  1 1  − 2  2  dx ∫ 2  x − 1 x + 1   x −1  ⇒ a tan −1 x + b log   x + 1  =



 3  = lim  1 + x→∞  x + 2 

3( x + 3) x+2  x + 2  3 

  3 = lim   1 + x→∞  + x 2   

 t and y = sin −1   1 + t 2

  

24. (b) : Given,



y2 − x dy = 3 dx 2 y −2 xy − 1 x+3

x→ 0

⇒



dy dy − ( y 2 − x) = dx dx dy 3 2 ⇒ ( 2 y − 2 xy − 1) =y −x dx ⇒ 2( y 3 − xy )

 x+5  21. (c) : lim   x→∞  x + 2 





2

2

f (0) = lim f ( x) ⇒ a = 0

\

 1 23. (c) : Given, x = cos −1   1 + t 2

20. (d) : y = x + y + x + y + ... ∞

x→ 0

  

x+3



=

= e3

 2 sin x − sin 2 x , if x ≠ 0  2 x cos x f ( x) =   a, if x = 0 

 x − 1 1 1 log  − tan −1 x 4  x + 1  2

⇒

1 1 a= − ,b= 2 4

\

a − 2b = −

1 1 − 2  = −1 4 2 −1

25. (c) : Given, y = e a sin x On differentiating w.r.t. x, we get −1 1 y1 = e a sin x a ⋅ ⇒ y1 1 − x 2 = ay 2 1− x 2 2 ⇒ (1 – x )y1 = a2y2

41

Differential Calculus

Again, differentiating w.r.t. x, we get (1 – x2)2y1y2 – 2xy12 = a22yy1 ⇒ (1 – x2)y2 – xy1 – a2y = 0 Using Leibnitz’s rule, (1 – x2)yn + 2 + nC1yn + 1(–2x) + nC2yn(–2) – xyn + 1 – nC1yn – a2yn = 0 ⇒ (1 – x2)yn + 2 + xyn + 1(–2n – 1) + yn[–n(n – 1) – n – a2] = 0 ⇒ (1 –x2)yn + 2 – (2n + 1)xyn + 1 = (n2 + a2)yn 26.



x=

−2 a ± 4 a2 − 12b 2×3

=

\ lim f ( x ) = lim[tan 2 x ] = 0 x→ 0

and f (0) = [tan 0] = 0 Thus, f (x) is continuous at x = 0.



\ Volume, V =



dV 4 dr = p3r 2 dt 3 dt = 4 pr 2

= 4pr2 . (2)

−2 a ± 2 a2 − 3b



⇒

3



Since, a2 ≤ 3b, \ x has an imaginary value. Hence, no extreme value of x exist.



...(i)

4 3 pr 3

dr dt



27. (b) : Since, g(x) g(y) = g(x) + g(y) + g(xy) – 2

x→ 0

2

30. (c) : Let r be the radius of spherical balloon.

(c) : Given, f(x) = x3 + ax2 + bx + c, a2 ≤ 3b. On differentiating w.r.t. x, we get f ′(x) = 3x2 + 2ax + b Put f ′(x) = 0 ⇒ 3x2 + 2ax + b = 0 ⇒



29. (b) : Given, f (x) = [tan2 x]

dV = 8 pr2 cm3 / min dt Now, when r = 5 cm dV 2 \ = 8 p(5) = 200 p cm3 / min dt

 dr  Q dt = 2 

31. (a) : Let ‘m’ be the slope of the tangent to the curve at point (3, 4) such that  −1 3p  m ×  tan  = − 1 or m = =1  3p  4   tan    4 



Now, at x = 0, y = 2, we get



g(0) g(2) = g(0) + g(2) + g(0) – 2



⇒ 5g(0) = 5 + 2g(0) –2



⇒ f ′(3) = 1



⇒ 3g(0) = 3 ⇒ g(0) = 1 g(x) is given in a polynomial and by the relation given g(x) can not be linear. Let g(x) = x2 + k



⇒ g(x) = x2 + 1 [Qg(0) = 1] \ From Eq. (i),

32.

(c) : Q f(x) = x2e–2x \ f ′(x) = 2x e–2x – 2x2e–2x = 2x(1 – x)e–2x Put f ′(x) = 0 for maxima or minima, we get 2x(1 – x)e–2x = 0 ⇒ x = 0, 1



Now, f ′′(x) = 2x(–1)e–2x + 2(1 – x)e–2x



(x2 + 1) (y2 + 1) = x2 + 1 + y2 + 1 + x2 y2 + 1 – 2



–2⋅2x(1 – x)e–2x



\ lim g(x) = g(3) = 3 + 1 = 10



f ′′(0) = 0 + 2ec = 2

[Qg(2) = 5]

2

x→3

−e + 2 28. (d) : Q f ( x ) = x and f (x) is continuous at x = 0 x



x

\ lim f (x) = f (0) x→ 0

−e x + 2x = f ( 0) x→ 0 x

⇒ lim

− e x + 2 x log 2 = f ( 0) x→ 0 1 (using L-Hospital’s rule)

⇒ lim

⇒ – e + 2 log 2 = f (0) ⇒ f (0) = – 1 + log 2 0



and f ′′(1) = − 2 e −2 + 0 − 0 = −

2 e2

1.5|X > 1) is equal to 7 3 (a) (b) 16 4 7 21 (c) (d) (2007) 12 64 17. If X follows a binomial distribution with 1 parameters n = 100 and p = , then P(X = r) 3 is maximum when r is equal to (a) 16 (b) 32 (c) 33 (d) none of these (2007)

13. A random variable X follows binomial distribution with mean a and variance b. Then (a) 0 < a < b (b) 0 < b < a (c) a < 0 < b (d) b < 0 < a (2008) 14. A box contains 9 tickets numbered 1 to 9 inclusive. If 3 tickets are drawn from the box

Answer Key 1.

(a)

2.

9.

(a)

10. (b)

17. (c)

(b)

3.

(d)

11. (d)

4.

(b)

12. (a)

5.

(c)

6.

(b)

7.

(c)

8.

(d)

13.

(b)

14.

(d)

15.

(a)

16.

(c)

58

VITEEE CHAPTERWISE SOLUTIONS

e planations

(a) : Let l = 7n + 7m, then we observe that 71, 72, 73 and 74 ends in 7, 9, 3 or 1 respectively. Thus, 7i ends in 7, 9, 3 or 1 according as i is of the form 4k + 1, 4k + 2, 4k – 1 or 4k, respectively. If S is sample space, then n(S) = (100)2 7m + 7n is divisible by 5, if (i) m is of the form 4k + 1 and n is of the form 4k – 1 or (ii) m is of the from 4k + 2 and n is of the from 4k or (iii) m is of the from 4k – 1 and n is of the from 4k + 1 or (iv) m is of the from 4k and n is of the from 4k + 1. Thus, number of favorable ordered pairs (m, n) = 4 × 25 × 25. 4 × 25 × 25 1 = Hence, required probability = 4 (100)2 (b) : Let X denote the sum of the numbers obtained when two fair dice are rolled. So, X may have values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. 1 p (x = 2) = p(1, 1) = 36 2 p (x = 3) = p{(1, 2), (2, 1)} = 36 4 3 p (x = 4) = ; p(x = 5) = ; 36 36 5 6 5 p (x = 6) = ; p(x = 7) = ; p(x = 8) = ; 36 36 36

1.



2.



4 2 3 ; p(x = 10) = ; p(x = 11) = ; 36 36 36



p (x = 9) =



p (x = 12) = x p(x)

2

3

4

1 36 5

6

7

8

9

10 11 12

1 2 3 4 5 6 5 4 3 2 1 36 36 36 36 36 36 36 36 36 36 36

( )



Mean X = Sxp(x)



( 2 × 1) + ( 3 × 2) + ( 4 × 3) + ( 5 × 4) + ( 6 × 5) +     (7 × 6) + (8 × 5) + (9 × 4) + (10 × 3) +    (111 × 2) + (12 × 1) = 36





252 =7 36 Variance = Sx2p(x) – (Sxp(x))2 =

=

 ( 2 2 × 1) + ( 32 × 2) + ( 4 2 × 3) + ( 52 × 4)   2  2 2 2  +(6 × 5) + (7 × 6) + (8 × 5) + (9 × 4)   2 2 2  + (10 × 3) + (11 × 2 ) + (12 × 1) 



36 1974 1974 − 1764 210 35 = − 49 = = = 36 36 36 6



∴ Variance =

3. 4.

35 35 . Hence, S.D. = 6 6 (d) : Given numbers are 1, 2, 3, 4. Possibilities for unit’s place digit (either 1 or 3) =2 Possibilities for ten’s digit = 3 Possibilities for hundred place digit = 2 Possibilities for thousand’s places digit = 1

∴ Number of favourable outcomes = 2 × 3 × 2 × 1 = 12 Number of number formed by 1, 2, 3, 4 (without repetitions) = 4! 12 1 = ∴ Required probability = 4×3×2 2 (b) : Q P( A) =

1 3 , P( A ∪ B) = 3 4

\ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

≤ P(A)+ P(B)

3 1 5 ⇒ ≤ + P( B) ⇒ P( B) ≥ 4 3 12 Also, B ≤ A ∪ B 3 ⇒ P( B) ≤ P( A ∪ B) = 4 5 3 \ ≤ P( B) ≤ 12 4 5.

− 72

(c) : Given, P(A ∪ B) = 0.6 and P(A ∩ B) = 0.2 We know that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ⇒ 0.6 = P(A) + P(B) – 0.2 ⇒ P(A) + P(B) = 0.8

59

Probability Distributions

⇒ 1 − P( A) + 1 − P( B) = 0.8 ⇒ − [ P( A) + P( B)] = 0.8 − 2 ⇒ P( A) + P( B) = 1.2



6. (b) : Let A and B be two events such that A = getting number 2 at least once B = getting 7 as the sum of the numbers on two dice We have, A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)} and B = {(2, 5), (5, 2), (6, 1), (1, 6), (3, 4), (4, 3)} 11 6 , P( B) = 36 36 2 P( A ∩ B) = 36 Now, required probability \ P( A) =

P( A / B) =



9.

(a) : Here, n = 6 According to the question 6 C2 p2q4 = 4⋅6C4 p4q2 ⇒ q2 = 4p2 ⇒ (1 – p)2 = 4p2 ⇒ 3p2 + 2p – 1 = 0 ⇒ (p + 1)(3p – 1) = 0 ⇒

2

=

−l 2





P( A / E1 ) =

1 1 1 , P( A / E2 ) = , P( A / E3 ) = 5 6 7

Now, by Baye’s theorem, the required probability P( E1 ) P( A / E1 ) P( E1 / A) = P( E1 )P( A / E1 ) + P( E2 )P( A / E2 ) + P( E3 ) P( A / E3 )

1 1 ⋅ 3 5 ⇒ P( E1 / A) = 1 1 1 1 1 1 ⋅ + ⋅ + ⋅ 3 5 3 6 3 7 1 42 5 = = 1 1 1 107 + + 5 6 7



2

4! 9 1 24 9 1 27 × × = × × = 2 ! 2 ! 16 16 4 16 16 128

12. (a) : x =

P( A ∩ B) 2 / 36 2 1 = = = P( B) 6 / 36 6 3

(d) : Let E1, E2 and E3 denote the events of selecting boxes A, B and C respectively and A be the event that a screw selected at random is defective. Then, 1 1 1 P( E1 ) = , P( E2 ) = , P( E3 ) = 3 3 3

(Q p cannot be negative)

3 1 11. (d) : Required probability = 4 C2     4 4

e−l l e l ⇒ l=2 = 1! 2! e −2 2 4 2 \ P( X = 4) = = 4! 3e 2 8.

1 3

10. (b) : Required probability distribution is poisson distribution.

7. (c) : Given, P (X = 1) = P (X = 2) ⇒

p=

8 + 12 + 13 + 15 + 22 70 = = 14 5 5

x

(x − x )

( x − x )2

8

–6

36

12

–2

4

13

–1

1

15

+1

1

22

+8

64

S x = 70

S (x – x ) = 0

S (x – x )2 = 106

∴ σ=

∑(x − x ) N

2

=

106 = 21.2 = 4.604 5

13. (b) : For binomial distribution 0 < variance < mean ⇒ 0 < b < a 14. (d) : In out of 9 tickets, 5 tickets are odd number and 4 tickets are even number. Required probability



4 4 4 5 3  5 C C C C C C  = 9 1 × 8 1 × 7 1 + 9 1 × 8 1 × 7 1 C1 C1 C1 C1 C1   C1

=

5 4 4 4 5 3 × × + × × 9 8 7 9 8 7

80 + 60 140 5 = = 504 504 18 1 5 15. (a) : Q P( A) = , P( B) = 12 12 =



B 1 and P   =  A  15

60



VITEEE CHAPTERWISE SOLUTIONS

We know that,  B  P( A ∩ B) 1 P( A ∩ B) P  = ⇒ = 1  A P( A) 15 12 1 ⇒ P( A ∩ B) = 180 Also, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 15 + 75 − 1 89 1 5 1 = + − = = 180 180 12 12 180

x 16. (c) : Q f ( x) = 2 Now, P( X > 1.5) =

2

( 0 ≤ x ≤ 2)



 X > 1.5  P( X > 1.5) Hence, P  = P( X > 1)  X > 1 



=

0.4375 7 = 0.75 12



101 is not an iteger. 3 Therefore, P(X = r) is maximum when



 101  r=  = 33.  3 

17. (c) : Since, (n + 1) p =

2

x ∫ 2 dx = 0.4375 1.5

x and P( X > 1) = ∫ dx = 0.75 2 1

vvv

61

Discrete Mathematics

10 CHAPTER

1.

2.

Discrete Mathematics

The proposition ~ (p ⇔ q) is equivalent to (a) (p ∨ ~q) ∧ (q ∧ ~p) (b ) (p∧ ~q) ∨ (q ∧ ~p) (c) (p ∧ ~q) ∧ (q ∧ ~p) (d ) None of the above (2013)

4.

If truth values of p be F and q be T. Then, truth value of ~(~p ∨ q) is (a) T (b) F (c) Either T or F (d) Neither T nor F (2013)

(b) x = 0 (d) x = a′

Dual of (x + y)·(x + 1) = x + x · y + y is (a) (x · y) + (x · 0) = x · (x + y) · y (b) (x + y) + (x · 1) = x · (x + y) · y (c) (x · y)(x · 0) = x · (x + y) · y (d) None of the above

6.

The negation of (~p∧q) ∨ (p∧~q) is (a) (p∨~q) ∨ (~p∨q) (b) (p∨~q) ∧ (~p∨q) (c) (p∧~q) ∧ (~p∨q) (d) (p∧~q) ∧ (p∨~q) (2014)

3.

5.

(a) x = 1 (c) x = a

The statement (p ⇒ q) ⇔ (~p ∧ q) is a (a) tautology (b) contradiction (c) Neither (a) nor (b) (d) Both (a) and (b) (2014)

(2012)

(2012)

7.

Let p and q be two statements. Then p ∨ q is false , if (a) p is false and q is true (b) both p and q are false (c) both p and q are true (d) None of the above (2011)

8.

If p ⇒ (~ p ∨ q) is false, the truth value of p and q are respectively (a) F, T (b) F, F (c) T, F (d) T, T (2009)

9.

Let p, q, r and s be statements and suppose that p → q → r → p. If ~ s → r, then (a) s → ~ q (b) ~ q → s (c) ~ s → ~ q (d) q → ~ s (2008)

Let a be any element in a boolean algebra B. If a + x = 1 and ax = 0, then

Answer Key 1.

(c)

9.

(b)

2.

(b)

3.

(b)

4.

(b)

5.

(d)

6.

(a)

7.

(b)

8.

(c)

62

1.

VITEEE CHAPTERWISE SOLUTIONS

e planations

(c) : p

q

~p

p⇒q

T T F F

T F T F

F F T T

T F T T

~ p ^ q (p ⇒ q) ⇔ (~p ^ q) F F F T T T F F



Hence, given statement is neither tautology nor contradiction.

2.

(b) : Let S : (~ p ∧ q) ∨ ( p∧ ~ q) ⇒ ~ S : ~ [(~ p ∧ q) ∨ (p∧ ~ q)] ⇒ ~ S :~ (~ p ∧ q) ∧ ~ (p∧ ~ q) ⇒ ~ S : ( p ∨ ~ q) ∧ (~p ∨ q)

6.

(a) : Given, (x + y)⋅(x + 1) = x + x⋅y + y Replace ‘⋅’ by ‘+’ and ‘1’ by ‘0’, we get (x⋅y) + (x⋅0) = x⋅(x + y)⋅y

7.

(b) : p ∨ q is false only when both p and q are false.

8.

(c) : p ⇒ (~ p ∨ q) is false means p is true and ~ p ∨ q is false. ⇒ p is true and both ~ p and q are false. ⇒ p is true and q is false.

9.

(b) : Q p → q → r → p and ~ s → r

3. (b) : ~(p ⇔ q) ≡ ~((p ⇒ q) ∧ (q ⇒ p)) ≡ ~(p ⇒ q) ∨ (~(q ⇒ p)) (Qby De-Morgan’s law) ≡ (p∧ ~ q) ∨ (q∧ ~ p) 4.

(b) : p

q

~p

~p∨q

~(~ p ∨ q)

F

T

T

T

F



\ Truth value of ~ (~ p ∨ q) is F.

5.

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



vvv

s

q

r

~s

~q

r

~s → r

~s → s

T

T

T

T

F

T

T

F

T

T

T

T

T

T

T

T

F

T

T

F

F

T

F

F

F

F

T

F

T

T

T

T

T

T

F

T

F

F

T

T

T

F

F

T

T

F

T

T

F

T

F

F

F

F

F

F

F

F

F

F

T

F

T

T

⇒ Options (c) and (d) are not true. Also ~ s → r \ Option (a) is not true. Hence, option (b) is correct.

63

Miscellaneous

11 CHAPTER

1.

2.

Miscellaneous

If N denote the set of all natural numbers and R be the relation on N × N defined by (a, b)R(c, d), if ad(b + c) = bc(a + d), then R is (a) symmetric only (b) reflexive only (c) transitive only (d) an equivalence relation (2013) If a1, a2 and a3 be any positive real numbers, then which of the following statement is not true ? (a) 3a1a2a3 ≤ a13 + a23 + a33 (b)

a1 a2 a3 + + ≥3 a2 a3 a1

1 1 1 (c) ( a1 + a2 + a3 )  + +  ≥ 9  a1 a2 a3  1 1 1 (d) ( a1 ⋅ a2 ⋅ a3 )  + +  ≥ 27  a1 a2 a3  3.

4.

5.



6.

A tower AB leans towards West making an angle a with the vertical. The angular elevation of B, the top most point of the tower is b as observed from a point C due East of A at a distance ‘d’ from A. If the angular elevation of B from a point D due East of C at a distance 2d from C is r, then 2 tan a can be given as (a) 3 cot b – 2 cot g (b) 3 cot g – 2 cot b (c) 3 cot b – cot g (d) cot b – 3 cot g (2013) If a and b are the roots of x2 – ax + b = 0 and if an + bn = Vn, then (a) Vn + 1 = aVn + bVn – 1 (b) Vn + 1 = aVn + aVn – 1 (c) Vn + 1 = aVn – bVn – 1 (d) Vn + 1 = aVn – 1 – bVn (2013)

1

n

3r

7r



15r

∑ ( -1)r n Cr  2r + 22r + 23r + 24r + ... m terms 

r =0



is (a) (c)

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



(b)



2mn - 1 2n - 1

(d) None of these (2013)

7.

The number of integral values of K, for which the equation 7cos x + 5 sin x = 2K + 1 has a solution, is (a) 4 (b) 8 (c) 10 (d) 12 (2013)

8.

In the expansion of

(2013)

If |x2 – x – 6|= x + 2, then the values of x are (a) –2, 2, – 4 (b) –2, 2, 4 (c) 3, 2, – 2 (d) 4, 4, 3 (2013)

The sum of the series

a + bx

xr is

a-b r! a - br (c) ( -1)r r! (a)

9.

If n = (1999)!, then

ex (b)

, the coefficient of a - br r!

(d) None of these (2013) 1999

∑ log n x

is equal to

x =1

(a) 1 (c)

1999

(b) 0 1999

(d) –1

(2013)

10. The domain of the function 1 f ( x) = + x + 2 is log 10 (1 - x ) (a) ]–3, – 2.5[∩] – 2.5, –2[ (b) [–2, 0[∪]0, 1[ (c) ]0, 1[ (d) None of the above

(2013)

64

VITEEE CHAPTERWISE SOLUTIONS

11. The relation R defined on set A = {x :|x|< 3, x ∈I} by R = {(x, y) : y = |x|} is (a) {(–2, 2), (–1, 1), (0, 0), (1, 1), (2, 2)} (b) {(–2, –2), (–2, 2), (–1, 1), (0, 0), (1, –1), (1, 2), (2, –1), (2, –2)} (c) {(0, 0), (1, 1), (2, 2)} (d) None of the above (2013) 12. The value of 2 tan (cosec (tan x)–tan (cot x)) is (a) tan–1x (b) tanx (c) cotx (d) cosec–1x (2013) –1

–1

–1

13. The function f : R → R defined by f(x) = (x – 1) (x – 2)(x – 3) is (a) one-one but not onto (b) onto but not one-one (c) both one-one and onto (d) neither one-one nor onto (2012) 14. If the complex numbers z1, z2 and z3 are in A.P., then they lie on a (a) circle (b) parabola (c) line (d) ellipse (2012) 15. Let a, b and c be in A.P. and |a| < 1, |b| < 1, |c| < 1. If x = 1 + a + a2 + .... ∞, y = 1 + b + b2 + .... ∞, z = 1 + c + c2 + .... ∞, then x, y and z are in (a) A.P. (b) G.P. (c) H.P. (d) None of these (2012) 16. The number of real solutions of the equation 9 2  10  = –3 + x – x is (a) 0 (b) 1 (c) 2 (d) None of these (2012) 17. The angle of depressions of the top and the foot of a chimney as seen from the top of a second chimney, which is 150 m high and standing on the same level as the first are q and f respectively, then the distance between their tops when tan q = (a)

150

m 3 (c) 150 m

5 4 and tan f = is 2 3

(b) 100 3 m (d) 100 m

(2012)

18. If one root is square of the other root of the equation x2 + px + q = 0, then the relations between p and q is (a) p3 – (3p – 1)q + q2 = 0 (b) p3 – q(3p + 1) + q2 = 0 (c) p3 + q(3p – 1) + q2 = 0 (d) p3 + q(3p + 1) + q2 = 0 (2012) 19. The coefficient of x53 in the following expansion (a)

100

(c) –

C47

100

100

∑ 100Cm ( x - 3)100 -m ⋅ 2m is

m=0

C53

(b)

100

(d) –

C53

100

C100

(2012)

20. There are 5 letters and 5 different envelopes. The number of ways in which all the letters can be put in wrong envelope, is (a) 119 (b) 44 (c) 59 (d) 40 (2012) 21. The sum of the series

12 + 2 2 12 + 2 2 + 32 + 2! 3! 12 + 2 2 + 32 + 4 2 + + ...is 4! 17 e (a) 3e (b) 6 13 19 e e (c) (d) (2012) 6 6 1+

22. The coefficient of xn in the expansion of loga(1 + x) is (a)

( -1)n -1 n

(b)

( -1)n -1 log a e n

(c)

( -1)n -1 log e a n

(d)

( -1)n log a e n

(2012)

23. If a plane meets the coordinate axes at A, B and C in such a way that the centroid of DABC is at the point (1, 2, 3), then equation of the plane is x y z x y z (a) + + = 1 (b) + + = 1 1 2 3 3 6 9 x y z 1 (c) + + = (d) None of these 1 2 3 3 (2012) 4 - x2 24. The domain of the function f ( x ) = sin -1( 2 - x ) is

65

Miscellaneous

(a) [0, 2] (c) [1, 2)

(b) [0, 2) (d) [1, 2]

(2012)

25. The relation R defined on the set of natural numbers as {(a, b) : a differs from b by 3} is given by (a) {(1, 4), (2, 5), (3, 6),...} (b) {(4, 1), (5, 2), (6, 3),...} (c) {(1, 3), (2, 6), (3, 9),...} (d) None of the above (2012) 26. If 3p and 4p are resultant of a force 5p, then angle between 3p and 5p is -1  4  3 (a) sin -1   (b) sin   5 5 (c) 90° (d) None of these (2012) 27. If 2 tan–1(cos x) = tan–1(2 cosec x), then the value of x is 3p p (a) (b) 4 4 p (c) (d) None of these 3 (2012) 28. If R be a relation from A = {1, 2, 3, 4} to B = {1, 3, 5} such that (a, b) ∈R ⇔ a < b, then ROR–1 is (a) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} (b) {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} (c) {(3, 3), (3, 5), (5, 3), (5, 5)} (d) {(3, 3), (3, 4), (4, 5)} (2011) 29. For a G.P. an = 3(2n)," n ∈ N. Find the common ratio. 1 (a) 2 (b) 2 1 (c) 3 (d) (2011) 3 a b c 30. If a, b, c are in H.P., then will , , b+c c+a a+b be in (a) A.P. (b) G.P. (c) H.P. (d) None of these (2011) x2 + 2x + 7 31. If < 6, x ∈R, then 2x + 3 3 (a) x > 11 or x < – (b) x > 11 or x < –1 2 3 (c) – < x < – 1 2 3 (d) –1 < x < 11 or x < – (2011) 2

32. The number of ways of painting the faces of a cube of six different colours is (a) 1 (b) 6 (c) 6! (d) 36 (2011) 33. If D is the set of all real x such that 1 – e(1/x) – 1 is positive, then D is equal to (a) (– ∞, 1] (b) (– ∞, 0) (c) (1, ∞) (d) (– ∞, 0) ∪ (1, ∞) (2011) 34. If one A.M. ‘A’ and two G.M. p and q are inserted between two given numbers, then find the value of (a) A (c) 3A

p2 q2 + . q p (b) 2A (d) 4A

(2011)

35. If the roots of the equation x + ax + b = 0 are c and d, then one of the roots of the equation x2 + (2c + a)x + c2 + ac + b = 0 is (a) c (b) d – c (c) 2d (d) 2c (2011) 2

36. The sum of the coefficients of (6a – 5b)n, where n is a positive integer, is (a) 1 (b) –1 (c) 2n (d) 2n –1 (2011) 37. Find the value of (7.995)1/3 correct to four decimal places. (a) 1.9995 (b) 1.9996 (c) 1.9990 (d) 1.9991 (2011) 38. In a group (G, *), the equation x* a = b has a (a) unique solution b* a–1 (b) unique solution a–1 * b (c) unique solution a–1 * b –1 (d) many solutions (2011) 39. Find the value of sin 12° sin 48° sin 54° 1 2 1 (c) 6 (a)

1 4 1 (d) 8 (b)

(2011)

40. In an equilateral triangle, the inradius, circumradius and one of the exradii are in the ratio (a) 2 : 3 : 5 (b) 1 : 2 : 3 (c) 1 : 3 : 7 (d) 3 : 7 : 9 (2011) 41. In how mny ways 6 letters be posted in 5 different letter boxes? (a) 56 (b) 65 (c) 5! (d) 6! (2011)

66 42. If A and B be two sets such that A × B consists of 6 elements. If three elements A × B are (1, 4), (2, 6) and (3, 6), find B × A. (a) {(1, 4), (1, 6), (2, 4), (2, 6), (3, 4), (3, 6)} (b) {(4, 1), (4, 2), (4, 3), (6, 1), (6, 2), (6, 3)} (c) {(4, 4), (6, 6)} (d) {(4, 1), (6, 2), (6, 3)} (2011) 43. Let f : R → R be defined as f(x) = x2 + 1, find f –1(–5). (a) {f} (b) f (c) {5} (d) {–5, 5} (2011) 44. If F is function such that F(0) = 2, F(1) = 3, F(x + 2) = 2F(x) – F(x + 1) for x ≥ 0, then F(5) is equal to (a) –7 (b) –3 (c) 17 (d) 13 (2010) 45. Let S be a set containing n elements. Then, number of binary operations on S is 2 (a) nn (b) 2n n2 2 (c) n (d) n (2010) 46. The numerically greatest term in the 1 expansion of (3 – 5x)11 when x = , is 5 (a) 55 × 39 (b) 55 × 36 (c) 45 × 39 (d) 45 × 36 (2010) 47. The number of solutions of the equation sin(ex) = 5x + 5–x, is (a) 0 (b) 1 (c) 2 (d) infinitely many (2010) 48. If a x = b y = c z = d u and a, b, c, d are in GP, then x, y, z, u are in (a) AP (b) GP (c) HP (d) None of these (2010) 2 49. If f ( x) = log10 x . The set of all values of x for which f(x) is real, is (a) [–1, 1] (b) [1, ∞) (c) (–∞, –1] (d) (–∞, –1] ∪ [1, ∞) (2010)

50. For what value of m can the expression 2x2 + mxy + 3y2 – 5y – 2 be expressed as the product of two linear factors? (a) 0 (b) ±1 (c) ±7 (d) 49 (2010)

VITEEE CHAPTERWISE SOLUTIONS

cos q is equal to 1 + sin q  p q q p (a) tan  -  (b) tan  - -   4 2 2 4

51. The value of

p q (c) tan  -  4 2

p q (d) tan  +  4 2

(2010)

52. If 3 sin q + 5 cos q = 5, then the value of 5 sin q – 3 cos q is equal to (a) 5 (b) 3 (c) 4 (d) None of these (2010)  5p  53. The principal value of sin -1 sin  is 6  p 5p (a) (b) 6 6 7p (c) (d) None of these 6 (2010) 54. Domain of the function f(x) = logx cos x, is  p p  p p (a)  - ,  - {1} (b)  - ,  - {1}  2 2  2 2  p p (c)  - ,   2 2

(d) None of these (2010)

 x2  , is 55. Range of the function y = sin -1  2 1 + x   p (a)  0 ,   2  p (c)  0 ,   2

 p (b) 0 ,   2  p (d) 0 ,   2

(2010)

56. If f : [2, 3] → R is defined by f(x) = x3 + 3x – 2, then the range f(x) is contained in the interval (a) [1, 12] (b) [12, 34] (c) [35, 50] (d) [–12, 12] (2009) 57. The number of subsets of {1, 2, 3, .....,9} containing at least one odd number is (a) 324 (b) 396 (c) 496 (d) 512 (2009) 58. A binary sequence is an array of 0’s and 1’s. The number of n-digit binary sequences which contain even number of 0’s is (a) 2n–1 (b) 2n–1 n–1 (c) 2 – 1 (d) 2n (2009)

67

Miscellaneous

59. If x is numerically so small so that x2 and higher powers of x can be neglected, then 3/ 2



 2x  -1/ 5  1 + 3  ⋅ ( 32 + 5x) is approximately equal to 32 + 31x (a) 64 31 - 32 x (c) 64

66. In D ABC ( a + b + c )(b + c - a)(c + a - b)( a + b - c)

31 + 32 x (b) 64 1 - 2x (d) 64

(2009)

60. Let f(x) = x2 + ax + b, where a, b ∈ R. If f(x) = 0 has all its roots imaginary, then the roots of f(x) + f′(x) + f ′′(x) = 0 are (a) real and distinct (b) imaginary (c) equal (d) rational and equal (2009)

equals (a) cos2 A (c) sin2 A

4b 2 c 2 (b) cos2 B (d) sin2 B

(2009)

67. In how many number of ways can 10 students be divided into three teams, one containing four students and the other three? (a) 400 (b) 700 (c) 1050 (d) 2100 (2008) 68. If R be a relation defined as aRb iff |a – b| > 0, then the relation is (a) reflexive (b) symmetric (c) transitive (d) symmetric and transitive (2008)

61. If f(x) = 2x4 – 13x2 + ax + b is divisible by x2 – 3x + 2, then (a, b) is equal to (a) (–9, –2) (b) (6, 4) (c) (9, 2) (d) (2, 9) (2009)

69. Let S be a finite set containing n elements. Then the total number of commutative binary operation on S is

62. If x, y, z are all positive and are the pth, qth and rth terms of a geometric progression respectively, then the value of the determinant

(2008)



log x p 1 log y q 1 equals log z r 1 (a) log xyz (b) (p – 1)(q – 1)(r – 1) (c) pqr (d) 0 (2009)

63. The period of sin x + cos x is 4

4

p 2 p (c) 4 (a)

4

(2009)

64. If 3 cos x ≠ 2 sin x, then the general solution of sin2 x – cos 2 x = 2 – sin 2x is p (a) np + ( -1)n , n ∈ Z 2 np p (b) (c) ( 4n ± 1) , n ∈ Z , n ∈Z 2 2 (d) (2n – 1)p, n ∈ Z (2009)  -1   -1  1 65. cos -1   - 2 sin -1   + 3 cos -1   2  2  2  –1 – 4 tan (–1) equals (a)

19 p 12

(b)

 n ( n - 1)   

(b) n  2  ( n2 ) (d) 2

70. Probability of getting positive integral roots of the equation x2 – n = 0 for the integer n, 1 ≤ n ≤ 40 is 1 1 3 1 (a) (b) (c) (d) 5 10 20 20 (2008) 71. The number of real roots of the equation x 4 + x 4 + 20 = 22 is (a) 4 (b) 2 (c) 0 (d) 1

2

p 2 p (d) 2 (b)

 n ( n + 1)   

(a) n  2  ( n2 ) (c) n

47 p 35p 43p (c) (d) 12 12 12 (2009)

(2008)

72. Let a, b be the roots of the equation x – ax + b = 0 and An = an + bn. Then An+1 – aAn + bAn–1 is equal to (a) – a (b) b (c) 0 (d) a – b (2008) 2

73. If the sides of a right-angle triangle form an A.P., the ‘sin’ of the acute angles are b c (a) , (b) a , b a a b c a c (c) , (d) none of these c b (2008)

74. The simplified expression of sin(tan–1x), for any real number x is given by 1 x (a) (b) 2 1+ x 1 + x2

68

VITEEE CHAPTERWISE SOLUTIONS

(c) -

1 1+ x

2



(d) -

x 1+ x

2



80. If x > 0 and log3x + log3 ( x ) + log 3 ( 4 x )

(2008)



75. In a triangle ABC, the sides b and c are the roots of the equation x2 – 61x + 820 = 0 and 4 A = tan -1   , then a2 is equal to 3 (a) 1098 (b) 1096 (c) 1097 (d) 1095 (2008)

equals (a) 9 (c) 1

+ log 3 ( 8 x ) + log 3 (16 x ) + ... = 4 , then x (b) 81 (d) 27

(2007)

81. The number of real roots of the equation 3

 1 1  x + x  + x + x = 0 is (a) 0 (b) 2 (c) 4 (d) 6

76. Let A = {1, 2, 3,..., n} and B = {a, b, c}, then the number of functions from A to B that are onto is (a) 3n – 2n (b) 3n – 2n – 1 n (c) 3(2 – 1) (d) 3n – 3(2n – 1) (2007)

(2007)

82. If H is the harmonic mean between P and H H Q,then the value of + is P Q PQ (a) 2 (b) P+Q

77. Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is (a) 9 (b) 12 (c) 10 (d) 14 (2007)

(c)

1 2

(d)

P+Q PQ

(2007)

83. If cos x + cos2 x = 1, then the value of sin12 x + 3 sin10 x + 3 sin8 x + sin6 x – 1, is equal to (a) 2 (b) 1 (c) –1 (d) 0 (2007) p 84. If sin -1 x + sin -1 y = , then cos–1 x + cos–1 y is 2 equal to p p (a) (b) 2 4 3p (c) p (d) (2007) 4

78. In a group G = {1, 3, 7, 9} under multiplication modulo 10, the inverse of 7 is (a) 7 (b) 3 (c) 9 (d) 1 (2007) 79. If b2 ≥ 4ac for the equation ax4 + bx2 + c = 0, then all the roots of the equation will be real if (a) b > 0, a < 0, c > 0 (b) b < 0, a > 0, c > 0 (c) b > 0, a > 0, c > 0 (d) b > 0, a > 0, c < 0 (2007)

Answer Key 1.

(d)

2.

(d)

3.

(b)

9.

(a)

10. (b)

11. (a)

17. (d)

18. (a)

25. (b)

4.

(c)

5.

(c)

6.

(a)

7.

(b)

8.

(c)

12. (a)

13.

(b)

14.

(c)

15.

(c)

16.

(a)

19. (c)

20. (b)

21.

(b)

22.

(b)

23.

(b)

24.

(c)

26. (b)

27. (b)

28. (c)

29.

(a)

30.

(c)

31.

(d)

32.

(a)

33. (d)

34. (b)

35. (b)

36. (a)

37.

(b)

38.

(a)

39.

(d)

40.

(b)

41. (a)

42. (b)

43. (b)

44. (d)

45.

(c)

46.

(a)

47.

(a)

48.

(c)

49. (d)

50. (c)

51. (c)

52. (b)

53.

(a)

54.

(d)

55.

(b)

56.

(b)

57. (c)

58. (a)

59. (a)

60. (b)

61.

(c)

62.

(d)

63.

(d)

64.

(c)

65. (d)

66. (c)

67. (d)

68. (d)

69.

(a)

70.

(c)

71.

(b)

72.

(c)

73. (a)

74. (b)

75. (c)

76. (d)

77.

(b)

78.

(b)

79.

(b)

80.

(b)

81. (a)

82. (a)

83. (d)

84. (a)

69

Miscellaneous

e planations

1. (d) : For (a, b), (c, d) ∈ N × N (a, b) R (c, d) ⇒ ad (b + c) = bc (a + d) Reflexive : Since, ab (b + a) = ba (a + b), " a,b ∈ N \ (a, b) R (a, b) So, R is reflexive. Symmetric : For (a, b), (c, d) ∈ N × N, Let (a, b) R (c, d) \ ad (b + c) = bc (a + d) ⇒ bc (a + d) = ad (b + c) ⇒ cb (d + a) = da (c + b) ⇒ (c, d) R (a, b) So, R is symmetric. Transitive : For (a, b), (c, d), (e, f ) ∈ N × N Let (a, b) R (c, d), (c, d) R (e, f ) \ ad (b + c) = bc (a + d), cf (d + e) = de (c + f ) ⇒ adb + adc = bca + bcd ...(i) and cfd + cfe = dec + def ...(ii) On multiplying Eq. (i) by ef and Eq. (ii) by ab and adding, then, we get adbef + adcef + cfdab + cfeab = bcaef + bcdef + decab + defab ⇒ adcf (b + e) = bcde (a + f ) ⇒ af (b + e) = be (a + f ) ⇒ (a, b) R (e, f ) So, R is transitive. Hence, R is an equivalence relation. 2.

(d) : QG.M. ≥ H.M.



⇒ ( a1 ⋅ a2 ⋅ a3 )1/ 3 ≥



⇒ ( a1 ⋅ a2 ⋅ a3 ) ≥

3 1 1 1  a + a + a  1 2 3 27

1 1 1  a + a + a  1 2 3

3

3



1 1 1 ⇒ ( a1 ⋅ a2 ⋅ a3 )  + +  ≥ 27  a1 a2 a3 

3. (b) : Given, |x2 – x – 6| = x + 2 Case I : x2 – x – 6 < 0 ⇒ (x – 3) (x + 2) < 0 ⇒ – 2 < x < 3 In this case, the equation becomes x2 – x – 6 = – x – 2 or x2 – 4 = 0 \ x = ± 2



Clearly, x = 2 satisfies the domain of the equation in this case. So, x = 2 is a solution. Case II : x2 – x – 6 ≥ 0 So, x ≤ – 2 or x ≥ 3 Then, equation reduces to x2 – x – 6 = x + 2 i.e., x2 – 2x – 8 = 0 or x = – 2, 4 Both these values lie in the domain of the equation, so x = – 2, 4 are the roots. Hence, roots are x = – 2, 2, 4.

4.

(c) : By m – n theorem at C. (d + 2d) cot b = d cot g – 2d cot (90° + a)



Y(N) B α γ

β

(W) X ′

A

C

d

2d

D

X(E)

Y ′(S)



⇒ 3d cot b = d cot g + 2d tan a ⇒ 3 cot b = cot g + 2 tan a \ 2 tan a = 3 cot b – cot g

5.

(c) : Multiplying x2 – ax + b = 0, by xn –1 xn+1 – axn + bxn – 1 = 0 ...(i) a, b are roots of x2 – ax + b = 0, therefore they will satisfy (i). Also, an+1 – aan + ban–1 = 0 ...(ii) n+1 n n–1 and b – ab + bb = 0 ...(iii) On adding Eqs. (ii) and (iii), we get



(an+1 + bn +1) – a (an + bn) + b (an–1 + bn – 1) = 0 or Vn + 1 – aVn + bVn– 1 = 0 ⇒ Vn + 1 = aVn – bVn– 1 (given, an + bn = Vn)

6.

(a) :

n

∑ (-1)r ⋅n Cr

r= 0

 1  3r 7r  r + 2r + 3r + ...upto m terms  2  2 2



=

n

∑

r=0

( -1)r ⋅nC r

1 2r

+

n

∑

r=0

+

( -1)r ⋅nC r ⋅

n

3r 2 2r 7r

∑ ( -1)r ⋅ nCr ⋅ 2 3r

r=0

+ ...

70

VITEEE CHAPTERWISE SOLUTIONS n

n

n

 1  3  7 =  1 -  +  1 -  +  1 -  + ...upto m terms  2  4  8 1

1

1

+ ...upto m terms



=



1  1  1 - mn  n   2 mn - 1 2 2 = =  1  2 mn ( 2 n - 1)  1 - n  2

7.

(b) : - 7 2 + 5 2 ≤ (7 cos x + 5 sin x) ≤ 7 2 + 5 2



So, for solution,

2

n

+

4

n

+

8n



- 74 ≤ ( 2 K + 1) ≤ 74 ⇒ – 8.6 ≤ (2K + 1) ≤ 8.6



⇒ – 9.6 ≤ 2K ≤ 7.6 ⇒ – 4.8 ≤ K ≤ 3.8 So, integral values of K are – 4, – 3, – 2, – 1, 0, 1, 2, 3 (eight values)

8.

(c) : (a + bx) e   x x2 x3 xn = ( a + bx) 1 - + + ... + ( -1)n + ...  1! 2 ! 3 !  n! \ The coefficient of xr



9.

1999

14. (c) : Let z1, z2 and z3 be affixes of points A, B and C, respectively. Since, z1, z2 and z3 are in A.P, therefore 2z2 = z1 + z3 z + z3 ⇒ z2 = 1 2 ⇒ B is the mid-point of the line AC. ⇒ A, B and C are collinear. ⇒ z1, z2 and z3 lie on a line.

x =1

15. (c) : Given, x = 1 + a + a2 + .... ∞ =

–x



= a⋅

(a) :

( -1)r ( -1)r -1 ( -1)r +b r! (r - 1)! = r ! ( a - br )

∑ log n x



= log(1999)! 1 + log(1999)! 2 + ... + log(1999)! 1999 = log(1999)! (1⋅2⋅ 3 ... 1999) = log(1999)! (1999)! = 1

10.

(b) : x + 2 ≥ 0, i.e., x ≥ – 2 or – 2 ≤ x Q log10 (1 – x) ≠ 0 ⇒ 1 – x ≠ 1 ⇒ x ≠ 0 Again, 1 – x > 0 ⇒ 1 > x ⇒ x < 1 All these can be combined as – 2 ≤ x < 0 and 0 < x < 1

11.

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

12. (a) : 2 tan–1 (cosec (tan–1 x) – tan (cot–1 x))

  = 2 tan -1 cosec cosec-1  



 1 + x2 1  = 2 tan -1  -   x x  1 + x 2 - 1    = 2 tan -1   x    sec q 1 = 2 tan -1   (put x = tan q)  tan q   q  2 sin 2    1 cos q -1 2  = 2 tan -1   = 2 tan   q  sin q   2 sin ⋅ cos q   2 2  q q = 2 tan -1  tan  = 2 ⋅ = q = tan -1 x  2 2 13. (b) : Given, f(x) = (x – 1)(x – 2)(x – 3 ⇒ f(1) = f(2) = f(3) = 0 ⇒ f(x) is not one-one. For each y ∈ R, there exists x ∈ R such that f(x) = y. Therefore, f is onto.

1 + x 2  x 

1   - tan tan -1   x 

1 1- b 1 2 and z = 1 + c + c + ... ∞ = 1- c Since, a, b and c are in A.P. ⇒ 1 – a, 1 – b and 1 – c are in A.P. 1 1 1 ⇒ , and are in H.P. 1- a 1- b 1- c ⇒ x, y and z are in H.P. Note. If the common ratio of a G.P. is not less than 1, then we can not determine the sum of an infinite G.P. series.



1 1- a

y = 1 + b + b2 + .... ∞ =

16. (a) : Let f(x) = –3 + x – x2 Then, f(x) < 0 for all x because coefficient of x2 < 0 and disc < 0. Thus, L.H.S of the given equation is always positive whereas the R.H.S is always less than zero. Hence, the given equation has no solution. 4 5 17. (d) : Given that, tan q = and tan f = 3 2

71

Miscellaneous

22. (b) : We have, loga (1 + x) = loge (1 + x) logae

150 d



In DABE, tan f =



2 ⇒ d = 150 cot f = 150 × = 60 m 5



In DDCE, tan q =



h d

x



E h

4 h  150 m D = C d 3 d 4  ⇒ h = × 60 3 d B A ⇒ h = 80 m 2 2 2 Now in DDCE, DE = DC + CE 2 2 2 ⇒ x = 60 + 80 = 10000 ⇒ x = 100 m ⇒

18. (a) : Given, equation x2 + px + q = 0 has roots a and a2. ⇒ a + a2 = –p and a3 = q ⇒ a(a + 1) = –p ⇒ a3[a3 + 1 + 3a(a + 1)] = –p3 ⇒ q(q + 1 – 3p) = –p3 ⇒ p3 – (3p – 1)q + q2 = 0 19. (c) : The given sigma expansion 100



∑

100

m=0

Cm ( x - 3)100 - m ⋅ 2 m

can be rewritten as [(x – 3)+ 2]100 = (x – 1)100 = (1 – x)100 \ x53 will occur in T54. ⇒ T54 = 100C53(–x)53 \ Required coefficient is – 100C53.

21. (b) : We have, Tn =



=

∑ n2

=

2

2

1 + 2 + 3 + ... + n n!

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

2

n! 1  2n3 + 3n2 + n  1  n3 3n2 n  =  + +   = 2 ⋅ 6 n! n! n!   6  n! \ Sum of the series



=

1 6

∞ 2 ∞  ∞ n3 n n + 3∑ +∑  2∑  n =1 n!  n =1 n ! n =1 n ! 

1 ( 2 × 5e ) + ( 3 × 2 e ) + e  6 1 17 = (10 e + 6 e + e ) = e 6 6 =

( -1)n - 1 log a e. n

23. (b) : Let the equation of the required plane be x y z + + = 1. a b c

This meets the coordinate axes at A, B and C, the coordinates of the centroid of DABC are a b c  3 , 3 , 3  .

a b c = 1, = 2 , = 3 ⇒ a = 3, b = 6, c = 9 3 3 3 Hence, the equation of the plane is x y z + + =1 3 6 9 4 - x2 24. (c) : Given, f ( x) = sin -1( 2 - x) \

4 - x 2 is defined for 4 – x2 ≥ 0. ⇒ x2 ≤ 4 ⇒ –2 ≤ x ≤ 2 and sin–1(2 – x) is defined for –1 ≤ 2 – x ≤ 1 ⇒ –3 ≤ – x ≤ – 1 ⇒ 1 ≤ x ≤ 3 Also, sin–1(2 – x) = 0 for x = 2 \ Domain of f(x) = [–2, 2] ∩ [1, 3] – {2} = [1, 2)

20. (b) : Required numbers  1 1 1 1 1 = 5 ! 1 - + - + -  = 44  1! 2 ! 3 ! 4 ! 5 !  2



 ∞ xn  = log a e  ∑ ( -1)n - 1  n  n = 1  So, the coefficient of xn in loga (1 + x) is

25. (b) : Let R = {(a, b) : a, b ∈ N, a – b = 3} = {(n + 3), n) : n ∈ N} = {(4, 1), (5, 2), (6,3),...} R sin a sin (a + b) Also, (5P)2 = (4P)2 + (3P)2 + 2(4P)(3P) cos(a + b) 2 2 2 ⇒ 25P = 16P + 9P + 24P2 cos (a + b) ⇒ 24P2 cos (a + b) = 0 ⇒ cos (a + b) = 0 = cos 90° 4P ⇒ a + b = 90° 5P sin a R Now, 4 P = sin 90° 5P  4 ⇒ sin a =  5 3P 4 ⇒ a = sin -1   5 26. (b) : Since, Q =

72

VITEEE CHAPTERWISE SOLUTIONS

27. (b) : Given, 2 tan–1 (cos x) = tan–1 (2 cosec x)



 2 cos x  ⇒ tan -1  = tan -1 ( 2 cosec x) 2   1 - cos x  2 cos x ⇒ = 2 cosec x 1 - cos 2 x 2 cos x ⇒ = 2 cosec x sin 2 x ⇒ sin x = cos x p ⇒ x= 4

28. (c) : Given, A = {1, 2, 3, 4} and B = {1, 3, 5} \ R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} and R–1 = {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} Hence, ROR–1 = {(3, 3), (3, 5), (5, 3), (5, 5)} 29. (a) : Given, an = 3(2n)

an + 1 = 3(2n +1) a 3 ( 2 n +1 ) Now, r = n+1 = =2 an 3(2n )

30. (c) : Since, a, b, c are in H.P.

1 1 1 ⇒ , , are in A.P. a b c



a+b+c a+b+c a+b+c , , ⇒ are in A.P. a b c



⇒ 1 +



⇒

b+c a+c a+b , , are in A.P. a b c



⇒

a b c are in H.P. , , b+c c+a a+b

b+c a+c a+b are in A.P. ,1 + ,1 + a b c

31. (d) : Given,

x2 + 2x + 7 0. For reflexive : aRa iff |a – a| > 0 Which is not true. So R is not reflexive. For symmetric : aRb iff |a – b| > 0 |a – b| = |b – a| ⇒ |b – a| > 0 ⇒ bRa Thus, R is symmetric. For transitive : aRb iff |a – b| > 0 bRc iff |b – c| > 0 ⇒ |a – c| = |a – b + b – c| > 0 ⇒ |a – c| > 0 ⇒ aRc Thus, R is also transitive.

77

Miscellaneous

69. (a) : Q S be a finite set containing n elements. Then total number of commutative binary  n ( n + 1)   2  

operations on S is n 

.

70. (c) : Given, x – n = 0



⇒ x2 – 41x –20x + 820 = 0 ⇒ (x – 41) (x – 20) = 0 ⇒ x = 41, 20 Let b = 41 and c = 20



3 4 Also, A = tan -1   ⇒ cos A = 5 3

2



⇒ x=± n \ n = 1, 4, 9, 16, 25, 36 6 3 \ Required probability = = 40 20

71. (b) : x 4 + x 4 + 20 = 22

⇒ x 4 + 20 + x 4 + 20 = 22 + 20

(

)

4

4

⇒ x + 20 + x + 20 = 44 4



Let x + 20 = y \ y2 + y – 44 = 0 Hence, the number of real roots of the equation is 2.

72.

(c) : Q a and b be the roots of x2 – ax + b = 0. \ a2 – aa + b = 0 and b2 – ab + b = 0 Now, An+1 – aAn + bAn–1 = an+1 + bn+1 – a(an + bn) + b(an–1 + bn–1) = an–1 (a2 – aa + b) + bn – 1 (b2 - ab + b) =0

73. (a) : Q b, c and a are in A.P. a b c \ = = sin A sin B sin C b c ⇒a= =  (Q ∠A = 90°) sin B sin C b c ⇒ sin B = ,sin C = a a Hence, option (a) satisfies this equation. 74. (b) : Let tan–1 x = q ⇒ tan q = x sin (tan–1 x) = sin q 1 + x2

A

=



1

C x

B

BC x = AC 1 + x2

75. (c) : The given equation is x2 – 61x + 820 = 0

\ By cosine formula, a2 = b2 + c2 – 2bc cos A

= 412 + 202 – 2 × 41 × 20 ×



= 2081 – 984 = 1097

3 5

76. (d) : Q A = {1, 2, 3,..., n} and B = (a, b, c} \ n(A) = n and n(B) = 3 Number of onto function from A to B =

3

∑ (-1)3 - r

r =1

Cr r n

3

= (–1)2 3C1(1)n + (–1)1 3C2(2)n + (–1)0 3C3 3n = 3n – 3⋅2n + 3 = 3n – 3(2n – 1)

77. (b) : Let the total number of persons in the room = n. \ Total number of handshakes = nC2 But number of handshakes = 66 n! \ = 66 - 2)! 2 !( n

⇒ n2 – n – 132 = 0 ⇒ (n – 12)(n + 11) = 0 ⇒ n = 12 (Q n ≠ –11)

78. (b) : The identity element for multiplication modulo 10, is 1 and 3X10 7 = 1. So, the inverse of 7 is 3. 79. (b) : If b2 – 4ac ≥ 0, then the equation ax 4 + bx2 + c = 0 has all roots real, if b < 0, a > 0, c > 0 80. (b) : log 3 x + log 3 x + log 3 4 x + log 3 8 x + ... =4 1 1 1 ⇒ log 3 x + log 3 x + log 3 x + log 3 x + ... 2 4 8 =4  1 1 1  ⇒ log 3 x 1 + + + + ... = 4 2 4 8  

 1  log 3 x  =8  1- 1 / 2  ⇒ log3 x = 4 ⇒ x = 34 = 81 ⇒

78

VITEEE CHAPTERWISE SOLUTIONS 3

  1 1 81. (a) :  x +  +  x +  = 0   x x



⇒

H 2Q H 2P and = = P P+Q Q P+Q

\

2( P + Q) H H 2Q 2P + = + = = 2. P Q P+Q P+Q P+Q



2   1   1  =0 + + 1 x x +  x    x   

83.

1 1 1 Since x + ≥ 2 x ⋅ ⇒ x + ≥ 2 " x ∈ R x x x 2



⇒

 1 and  x +  + 1 is always + ve  x \ number of real roots of given equation is 0.

82. (a) : Q H is the harmonic mean between P and Q. 2 PQ \ H= P+Q

(d) : Q cos x + cos2 x = 1 ⇒ cos x = 1 – cos2 x = sin2 x sin12 x + 3 sin10 x + 3 sin8 x + sin6 x – 1 = cos6 x + 3 cos5 x + 3 cos4 x + cos3 x – 1 = (cos2 x + cos x)3 – 1 = 1 – 1 = 0

84. (a) : sin -1 x + sin -1 y =

vvv

\ ⇒

p 2

p p p - cos -1 x + - cos -1 y = 2 2 2 p cos -1 x + cos -1 y = . 2

MODEL TEST PAPERS PHYSICS • CHEMISTRY • MATHEMATICS

Model Test Paper-1

1

Model Test Paper Time : 2 Hours 30 Minutes

physics 1.

2.

3.

4.

5.

Two parallel large thin metal sheets have equal surface charge densities (s = 26.4 × 10–12 C m–2) of opposite signs. The electric field between these sheets is (a) 1.5 N C–1 (b) 1.5 × 10–10 N C–1 –1 (c) 3 N C (d) 3 × 10–10 N C–1 A gang capacitor is formed by inter locking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 cm and the overlapping area of the plates is 5 cm2. The capacity of the unit is (a) 1.06 pF (b) 4 pF (c) 6.36 pF (d) 12.72 pF The saturation current of a p-n junction diode made from germanium at 27°C is 10–5 A. What will be the required potential in order to obtain a current of 250 mA in forward bias? (Boltzman constant k = 1.38 × 10–23 J K–1) (a) 0.26 V (b) 2.6 V (c) 0.52 V (d) 1.04 V When a p-n junction diode made from germanium or silicion is forward-biased, energy is released at the junction due to the recombination of electrons and holes. This energy is in (a) Visible region (b) Infrared region (c) UV region (d) X-ray region The Newton’s rings are seen in reflected light of wavelength 5896Å. The radius of curvature of plano-convex lens is 1.0 meter. An air film is replaced by a liquid whose refractive index is to be calculated under the conditions if 16th ring and its diameter is 5.1 mm. (a) 1.45 (b) 2.90 (c) 1.73 (d) 2.50

1

Max. Marks : 120

6.

Two flash light electric incandescent lamps, each requiring 3 A at 1.5 V are placed in series and connected to a 6 V cell. What resistance must be connected in series to operate them? (a) 22 W (b) 2 W (c) 6 W (d) 1 W

7.

In the figure shown below the e.m.f. of the cell is 2 V and internal resistance is negligible. The resistance of the voltmeter is 80 W. The reading of voltmeter will be (a) 2.00 V (b) 1.33 V (c) 1.60 V (d) 0.80 V

V

8. Four cells of identical emf E and internal resistance r are connected in series to a variable resistor. The following graph shows the variation of terminal voltage of the combination with current. The emf of each cell used is V(volt) 5.6 4.2 2.8 1.4 0

(a) 1.4 V (c) 2 V

9.

0.5

1

1.5 2 I(amp)

(b) 5.6 V (d) 1 V

In the given circuit, the voltmeter records 5 V. The resistance of the voltmeter in ohm is

(a) 200 (c) 10

(b) 100 (d) 50

2

VITEEE CHAPTERWISE SOLUTIONS

10. ABCD is a rectangle whose side AB = 10 cm and side BC = 24 cm. A charge of 0.104 mC is lying at the center O of rectangle. If the mid-point of side BC is E, then the work done in carrying 100 mC charge from B to E will be (a) 1.152 J (c) 4.082 J

(b) 2.304 J (d) 23.4 J

11. Name the gate represented by the following circuit



(a) OR gate (c) AND gate

(b) XOR gate (d) NAND gate

12. A 220 V input is supplied to a transformer. The output circuit draws a current of 2.0 A at 440 V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is (a) 3.6 A (c) 2.5 A

(b) 2.8 A (d) 5.0 A

13. Three similar cells, each of emf 2 V and internal resistance r W send the same current through an external resistance of 2 W, when connected in series or in parallel. The strength of the current flowing through the external resistance is (a) 1 A (c) 2 A

(b) 1.5 A (d) 0.75 A

of solenoid is 25 cm, its magnetic moment is equal to (a) 1 A m2 (b) 10 A m2 2 (c) 12.5 A m (d) 125 A m2 18. A circular coil of radius 0.1 m has 80 turns of wire. If the magnetic field through the coil increases from 0 to 2 T in 0.4 s and the coil is connected to a 11 W resistor, what is the current through the resistor during the 0.4 s? 8 (a)   A 7  (c) 8 A

7  (b)   A 8 (d) 7 A

19. A power of transformer with an 8 : 1 turn ratio has 60 Hz, 120 V across the primary, the load in the secondary is 104 W. The current in the secondary is (a) 96 A (b) 0.96 A (c) 9.6 A (d) 96 mA 20. Each hadron consists of a proper combination of a few elementary components called (a) photons (b) vector bosons (c) quarks (d) meson-baryon pairs 21. A calcite crystal is placed over a dot on a piece of paper and rotated. On viewing through calcite, one will see (a) a single dot (b) two stationary dots (c) two rotating dots (d) one dot rotating about the other.

14. In the Bohr’s model of hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in nth quantum state is

22. The hydrogen in a hydrogen bomb is converted into (a) helium (b) tritium (c) plutonium (d) uranium

15. Half-life of a radioactive substance is 20 minute. The time between 20% and 80% decay will be

23. Two copper balls, each weighing 10 g, are kept in air 10 cm apart. If one electron from every 106 atoms is transferred from one ball to the other, the coulomb force between them is (atomic weight of copper is 63.5) (a) 2.0 × 1010 N (b) 2.0 × 104 N (c) 2.0 × 108 N (d) 2.0 × 106 N

(a) –1 (c) –2

(a) 20 min (c) 40 min

(b) +1 (d) +2

(b) 30 min (d) 25 min

16. A transformer is used to light 140 W, 24 V lamp from 240 V ac mains. The current in the mains is 0.7 A. The efficiency of transformer is nearest to (a) 90% (b) 80% (c) 70% (d) 60% 17. The magnetic flux through a coreless solenoid carrying current I is 5 × 10–6 Wb. If the length

24. A torque of 10–5 N m is required to hold a magnet at 90° with the horizontal component of the earth’s magnetic field. The torque required to hold it at 30° will be 1 (a) 5 × 10–6 N m (b) × 10–5 N m 2 (c) 5 3 × 10–6 N m (d) 2 × 10–6 N m

3

Model Test Paper-1

25. The potential difference in volt across the resistance R3 in the circuit shown in figure, is (Given, R1 = 15 W, R2 = 15 W, R3 = 30 W, R4 = 35 W) (a) 5 (b) 7.5 (c) 15 (d) 12.5 26. An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closet approach is of the order of (a) 10 Å (b) 10–10 cm (c) 10–12 cm (d) 10–15 cm 27. An ac voltage source of variable angular frequency w and fixed amplitude V0 is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When w is increased. (a) the bulb glows dimmer (b) the bulb glows brighter (c) total impedance of the circuit is unchanged (d) total impedance of the circuit increases. 28. An electron of mass m and charge q is travelling with a speed v along a circular path of radius r at right angles to a uniform magnetic field B. If speed of the electron is doubled and the magnetic field is halved, the resulting path would have a radius of r r (a) (b) 2 4 (c) 2r (d) 4r 29. Gieger-Muller tube is a (a) gas ionization detector (b) cloud chamber (c) Spectro Photometer (d) fluorescent detector 30. The radiocarbon dating method is a realiable method for dating remains up to (a) 50 years (b) 500 years (c) 5000 years (d) 50000 years 31. Light of the wavelength 1500 Å fall on aluminium surface. Work function of aluminium is 4.2 eV. What is the kinetic energy of the fastest emitted photoelectrons? (a) 2 eV (b) 1 eV (c) 4 eV (d) 0.2 eV

32. If the series limit wavelength of Lyman series for the hydrogen atom is 912 Å, then the series limit wavelength for Balmer series of hydrogen atom is (a) 912 Å (b) 912 × 2 Å 912 (c) 912 × 4 Å (d) Å 2 33. One of the refracting surfaces of a prism of angle 30° is silvered. A ray of light incident at an angle of 60° retraces its path. The refractive index of the material of prism is 3 (a) 3 (b) 2 (c) 2

(d)

2

34. The focal length of objective and eye-piece of a microscope are 1 cm and 5 cm respectively. If the magnifying power for relaxed eye is 45, then length of the tube is (a) 9 cm (b) 15 cm (c) 12 cm (d) 6 cm 35. Which one of the following is not correct about Lorentz Force?  (a) In presence of electric field E(r ) and magnetic field B(r) the force on a moving   electric charge is F = q[ E(r ) + v × B(r )] . (b) The force, due to magnetic field on a negative charge is opposite to that on a positive charge. (c) The force due to magnetic field become zero if velocity and magnetic field are parallel or anti-parallel. (d) For a static charge the magnetic force is maximum. 36. A 4 A current carrying loop consists of three identical quarter circles of radius 5 cm lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin  joined together, value of B at the origin is m m (a) 0 (i + j − k ) T (b) 0 ( −i + j + k ) T 10 10 m0    (i + j + k ) T (c) (d) 10m0 (i + j + k ) T 5 37. Two radioactive nuclei A and B have disintegration constants lA and lB and initially NA and NB number of nuclei of them are there. The time after which their undistintegrated nuclei are same is (a)

 NB  l Al B  NB  1 ln (b) ln ( l A − l B )  N A  ( l A + l B )  N A 

(c)

 NB   NB  1 1 (d)  ln ln ( l B − l A )  N A  ( l A − l B )  N A 

4

VITEEE CHAPTERWISE SOLUTIONS

38. The magnitude of angular momentum, orbit radius and frequency of revolution of electron in hydrogen atom corresponding to quantum number n are L, r, and u respectively. Then, according to Bohr’s theory of hydrogen atom, which one is constant for all orbits. (a) ur2L (b) urL (c) u2rL (d) urL2 39. If angle of incidence is twice the angle of refraction in a medium of refractive index m, then angle of incidence is −1 

m (a) 2 cos   2 

(c) 2 cos–1(m)

−1 

m (b) 2 sin    2

(d) 2 sin–1(m)

40. A telescope uses light having wavelength 5000 Å and lenses of focal lengths 2.5 cm and 30 cm. If the diameter of the aperture of the objective is 10 cm, then the resolving limit of telescope is (a) 6.1 × 10–6 rad (b) 5.0 × 10–6 rad –4 (c) 8.3 × 10 rad (d) 7.3 × 10–3 rad

Chemistry 41. Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+ and M3+ in its oxide. Fraction of the metal which exists as M3+ would be (a) 5.08% (b) 7.01% (c) 4.08% (d) 6.05% 42. In the synthesis of glycerol from propene, the steps involved are (a) glycerol-b-chlorohydrin and allyl chloride (b) glyceryl trichloride and glycerol-a-chlorohydrin (c) allyl alcohol and glycerol-b-chlorohydrin (d) allyl alcohol and monosodium glycerolate. 43. A solution containing one mole per litre of each Cu(NO3)2; AgNO3; Hg2(NO3)2; is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are Ag+/Ag = +0.80, Hg2+ 2 /2Hg = +0.79 Cu2+/Cu = +0.34, Mg2+/Mg = –2.37 With increasing voltage, the sequence of deposition of metals on the cathode will be (a) Ag, Hg, Cu, Mg (b) Mg, Cu, Hg, Ag (c) Ag, Hg, Cu (d) Cu, Hg, Ag 44. A greenish yellow gas reacts with an alkali metal hydroxide to form a halate which can be used in fire works and safety matches. The gas and halate respectively are

(a) Br2, KBrO3 (c) I2, NaIO3

(b) Cl2, KClO3 (d) Cl2, NaClO3

45. The formula of the complex diamminechloroido (ethylenediamine)nitro-platinum (IV) chloride is (a) [Pt(NH3)2Cl(en)NO2]Cl2 (b) Pt[Pt(NH3)2(en)Cl2NO2] (c) Pt[(NH3)2(en)NO2]Cl2 (d) Pt[(NH3)2(en)NO2Cl2] 1 46. For the reaction, SO 2( g ) + O 2( g )  SO 3( g ) , 2 if Kp = Kc(RT)x where the symbols have usual meaning then the value of x is (assuming ideality) (a) 1 (b) – 1 (c) −

1 2

(d)

1 2

47. In which reaction, aromatic aldehyde is treated with acid anhydride in the presence of corresponding salt of the acid to give unsaturated aromatic acid? (a) Friedel–Crafts reaction (b) Perkin’s reaction (c) Wurtz reaction (d) None of these. 48. Which of the following is not a function of proteins? (a) Nail formation (b) Skin formation (c) Muscle formation (d) Providing energy for metabolism 49. The yield of NH3 in the reaction, N2 + 3H2 2NH3; DH = –22.08 kcal is affected by (a) change in pressure and temperature (b) change in temperature and concentration of N2 (c) change in pressure and concentration of N2 (d) change in pressure, temperature and concentration of N2. 50. Lipids are (a) nucleic acids occurring in plants (b) proteins occurring in animals (c) carbohydrates occurring in plants (d) fats of natural origin.

5

Model Test Paper-1

51. The rate law for a reaction between the substances A and B is given by rate = k [A]n [B]m. On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as 1 (a) (b) (m + n) m+n 2 (c) (n – m) (d) 2(n – m) 52. In SN1 (substitution nucleophilic unimolecular) reaction, racemization takes place, it is due to (a) inversion of configuration (b) retention of configuration (c) conversion of configuration (d) both (a) and (b). 53. Though the five d-orbitals are degenerate, the first four d-orbitals are similar to each other in shape whereas the fifth d-orbital is different from others. The fifth orbital is (a) dx2 – y2 (b) dz2 (c) dxz (d) dxy 54. Consider the following diazonium ions : +

Me2N

N2 ,

(II)

(I) +

N2 ,

CH3O (III)



+

N2 ,

O2N

+

N2

CH3 (IV)

The order of reactivity towards diazo-coupling with phenol in the presence of dil. NaOH is (a) I < IV < II < III (b) I < III < IV < II (c) III < I < II < IV (d) III < I < IV < II

55. Which of the following ions are colourless? Ti3+, Sc3+, Ag+, Cd2+, Cu2+ I II III IV V (a) I and V only (b) II, III and IV only (c) I, III and V only (d) III and IV only 56. Which of the following compounds does not react with ethanolic KCN? (a) Ethyl chloride (b) Acetyl chloride (c) Chlorobenzene (d) Benzaldehyde 57. Which of the following processes is a nonspontaneous process? (a) Dissolution of salt or sugar in water. (b) Mixing of different gases through diffusion.

(c) Precipitation of copper when zinc rod is dipped in aqueous solution of copper sulphate. (d) Flow of heat from a cold body to a hot body in contact. 58. Which of the following compounds is the most basic in aqueous medium? (a)



(b)

(c)



(d)

59. The relationship between standard reduction potential of a cell and equilibrium constant is shown by n (a) E°cell = log K c 0.059 (b) E°cell =

0.059 log K c n

(c) E°cell = 0.059 n log Kc log K c n 60. In the Liebermann’s nitroso reaction, sequential changes in the colour of phenol occurs as (a) Brown or red → green → deep blue (b) Red → deep blue → green (c) Red → green → white (d) White → red → green. (d) E°cell =

61. When benzenediazonium chloride in hydrochloric acid reacts with cuprous chloride, then chlorobenzene is formed. The reaction is called (a) Gattermann reaction (b) Perkin reaction (c) Etard reaction (d) Sandmeyer reaction. 62. Which of the following elements does not show allotropy? (a) Nitrogen (b) Bismuth (c) Antimony (d) Arsenic 63. The electronic configuration of Cu (II) is 3d 9 whereas that of Cu (I) is 3d10. Which of the following is correct? (a) Cu (II) is more stable. (b) Cu (II) is less stable. (c) Cu (I) and Cu (II) are equally stable. (d) Stability of Cu (I) and Cu (II) depends on nature of copper salts.

6

VITEEE CHAPTERWISE SOLUTIONS

64. Primary nitro compounds react with nitrous acid to form nitrolic acids which dissolve in NaOH giving (a) yellow solution (b) blue solution (c) colourless solution (d) red solution. 65. In an aqueous solution of 1 L, when the reaction 2Ag+(aq) + Cu(s) Cu2+(aq) + 2Ag(s) reaches equilibrium, [Cu2+] = x M and [Ag+] = y M. If volume of solution is doubled by adding water, then at equilibrium x 2 x 2 + (b) [Cu ] > 2 (c) [Cu 2 + ] < x 2 x (d) [Cu 2 + ] < 2

(a) [Cu 2 + ] =

y 2 y + M , [Ag ] > 2 y M , [Ag + ] > 2 y M , [Ag + ] < 2 M , [Ag + ] =

69. The radial wave equation for hydrogen atom is Y=

1

1 16 4  a0 

3/ 2

[( x − 1)( x 2 − 8 x + 12)] e − x / 2

where, x = 2r / a0; a0 = radius of first Bohr orbit. The minimum and maximum position of radial nodes from nucleus are a (a) a0 , 3a0 (b) 0 , 3a0 2 a0 a0 (c) (d) ,a , 4 a0 2 0 2 70. D(+)-glucose reacts with hydroxyl amine and yields an oxime. The structure of the oxime would be

M M

(a)

(b)

(c)

(d)

M M

66. In which of the following coordination entities the magnitude of Do (CFSE in octahedral field) will be maximum? (a) [Co(C2O4)3]3– (b) [Co(H2O)6]3+ 3+ (c) [Co(NH3)6] (d) [Co(CN)6]3– 67. Which one/ones of the following reactions will yield propan-2-ol? (i) (ii) CH3CHO (iii) HCHO

(i) CH3MgI

71. The solubility of AgCl at 20°C is 1.435 × 10–3 g/L. The solubility product is

(ii) H2O/H +

(i) C2H5MgI

(a) 1.0 × 10–10 (c) 1.035 × 10–5

(ii) H2O/H +

(iv) Choose the right answer. (a) (i) and (ii) (b) (ii) and (iii) (c) (i) and (iii) (d) (ii) and (iv)  d( DG)    dT  P

68. If DG = DH – TDS and DG = DH + T 

then variation of EMF of a cell, with temperature T, is given by DS nF DH (c) nF

(a)

DS nF DG (d) nF

(b) −

(b) 2 × 10–10 (d) 108 × 10–3

72. On heating with conc. HNO3, proteins give yellow colour. This test is called (a) oxidising test (b) xanthoproteic test (c) Molisch’s test (d) acid-base test. 73. The correct statement about the compounds A, B and C is



7

Model Test Paper-1

(a) A and B are identical. (b) A and B are diastereomers. (c) A and C are enantiomers. (d) A and B are enantiomers. 74. The wave function (Y) of 2s is given by



1/ 2

r  − r / 2 a0  2 −  e a  0 At r = r0, radial node is formed. Thus for 2s, r0 in terms of a0 is (a) r0 = a0 (b) r0 = 2a0 (c) r0 = a0 / 2 (d) r0 = 4a0

Y 2s =

1

1 2 2 p  a0 

75. Which of the following arrangements does not represent the correct order of the property stated against it? (a) Sc < Ti < Cr < Mn : number of oxidation states (b) V2+ < Cr2+ < Mn2+ < Fe2+ : paramagnetic behaviour (c) Ni2+ < Co2+ < Fe2+ < Mn2+ : ionic size (d) Co3+ < Fe3+ < Cr3+ < Sc3+ : stability in aqueous solution. 76. The pair of compounds in which both the compounds give positive test with Tollens’ reagent is (a) glucose and sucrose (b) fructose and sucrose (c) acetophenone and hexanal (d) glucose and fructose. 77. Which one of the following nitrophenols is the strongest acid? CH3

(a)

(b) NO2

(c)

(d)

78. When vapours, at atmospheric pressure were gradually heated from 25°C their colour was found to deepen at first and then fade as the temperature was raised above 160°C. At 600°C, the vapours were almost colourless, but their colour deepened when the pressure was raised at this temperature. The vapours were of (a) bromine (b) a mixture of nitrogen dioxide and dinitrogen tetraoxide (c) pure nitrogen dioxide (d) pure dinitrogen tetraoxide. 79. Acetal formation is a reversible reaction +



Under what conditions, the reaction can be forced to proceed only in right (forward) direction? (a) Using excess of alcohol. (b) Using high temperature. (c) Using dilute acid and excess of alcohol. (d) Using dry acid and excess of alcohol.

80. How many coulombs are required for the oxidation of one mole of H2O2 to O2? (a) 93000 C (b) 1.93 × 105 C –5 (c) 1.93 × 10 C (d) 4.53 × 10–4 C

Mathematics  2x   1 + 6x  + tan −1  81. If y = tan −1  . 2  3 − 2 x   1 + 8x 

Find

OH

(a)

CH3

(c)

dy ? dx 4

1 + 16 x 2

2

1 − 16 x 2



(b)



(d)

2 1 + 16 x 2 4 1 − 16 x 2



82. Find the greatest lateral surface area of a cylinder that can be inscribed inside a given sphere. (a) 4pa (b) 2pa (c) 6pa (d) 8pa

8

VITEEE CHAPTERWISE SOLUTIONS

83. Evaluate :

∫ sin

3

(a) coplanar (c) both (a) & (b)

x cos 3xdx

(a)

3  − cos 4 x  2  cos 2 x  1  cos 6 x   + 9  2  + 10  6  + C 4  4

(b)

3  − cos 4 x  3  cos 2 x  1  cos 6 x   + 8  2  + 8  6  + C 8  4

(c) –cos4x + 3cos6x + C (d) 2(cos4x) + 12(cos2x) + C 84. Evaluate the definite integral 2

2



∫0 [x



where [x] represents integral part of x.

− 1]dx

(a) 1 − 2

(b) 1 + 2

(c) 2 + 4

(d) 2 + 6

85. Find the value of x, if



 1 3 2  1    [1 x 1]  2 5 1  2  = 0 15 3 2   x  (a) 14, 2 (b) 10, 1 (c) –14, –2 (d) 5, 13

86. If A, B and C are angles of a triangle, then −1 cos C cos B find the determinant cos C −1 cos A cos B cos A −1 (a) –10 (b) 0 (c) 2 (d) 6  1 −1 1   2 A = 87. If  2 −1 0  , then find A =  1 0 0  (a) 0 (b) A–1 (c) I (d) A3 88. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 unit and after 3 s, it is 6 unit. Find the radius of balloon after t s. (a) 60t (b) 63t (c) (63t + 27)1/3 (d) 27t 89. The four points A, B, C and D with position ^ ^ ^ vectors 4 i^ + 5 j^ + k^ , − j^ + k^ , −3 i + 9 j + 4 k and ^ ^ ^ 4( − i + j + k ) respectively, are

90. Find

parallelopiped  ^ ^ ^ determined by the vectors a = i + j + k ,  ^ ^ ^  b = 2 i + 4 j − k and c = i^ + j^ + 3 k^ , if the base is taken to the parallelogram determined by   a and b. (a) 4 units (b) 16 units 4 (c) 38 units (d) units 38 1 1 91. If 2 f ( x) + 3 f   = − 2 , x ≠ 0 , then x x

2

∫1

the

altitude

(b) collinear (d) none of these of

a

f ( x) dx is equal to

2 1 (a) − log 2 + 5 2 2 1 (c) log 2 + 5 2

2 1 (b) − log 2 − 5 2 (d) None of these

92. If (–1, 2) and (2, 4) are two points on the curve y = f(x) and if g(x) is the gradient of the curve at the point (x, y), then (a) 2 (c) 0

2

∫−1 g( x) dx is

(b) –2 (d) 1

93. A curve y = f(x) passes through the origin and lies entirely in the first quadrant. Through any point p(x, y) on the curve, lines are drawn parallel to the coordinate axes. If the curve divides the area formed by these lines and coordinates axes in m : n, then the value of f(x) is equal to (a) cxm + n (b) cxm – n m/n (c) x (d) cxm/n 94. If the solution of the differential equation dy ax + 3 represents a circle, then the = dx 2 y + f value of ‘a’ is (a) 2 (c) 3

(b) –2 (d) –4     95. If the three planes r ⋅ n1 = p1 , r ⋅ n2 = p2 and   r ⋅ n3 = p3 have a common line of intersection,       then p1(n2 × n3 ) + p2 (n3 × n1 ) + p3 (n1 × n2 ) is equal to (a) 1 (b) 2 (c) 0 (d) –1

9

Model Test Paper-1

96. If

3 + i = ( a + ib)(c + id), then

b d tan −1   + tan −1   has the value a c

p p + 2np , n ∉ I (b) np + , n ∈ I 3 6 p p (c) np − , n ∈ I (d) 2np − , n ∈ I 3 3 97. The complex number z which satisfies the i+z condition = 1 lies on i−z (a)

(a) circle x2 + y2 = 1 (c) y-axis

(b) x-axis (d) line x + y = 1

98. A line through (0, 0) cuts the circle x2 + y2 – 2ax = 0 at A and B, then locus of the centre of the circle drawn AB as diameter is (a) x2 + y2 – 2ay = 0 (b) x2 + y2 + ay = 0 (c) x2 + y2 + ax = 0 (d) x2 + y2 – ax = 0 99. The parametric equation of a parabola is x = t2 + 1, y = 2t + 1. The cartesian equation of its directrix is (a) x = 0 (b) x + 1 = 0 (c) y = 0 (d) None of these 100. The line lx + my + n = 0 cuts the ellipse x2 y 2 + = 1 in points whose eccentric angles a2 b2 p differ by , if a2l2 + b2m2 is equal to 2 2 (a) n (b) 2n2 (c) 2n (d) n 101. Minimum area of the triangle by any tangent to the ellipse axes is

x2 a

2

+

y2 b2

= 1 with the coordinate

( a + b) 2 2 ( a − b) 2 (c) ab (d) 2 102. The equation of the hyperbola whose (a) a2 + b2

(b)

eccentricity is 2 and the distance between the foci is 16, taking transverse and conjugate axes of the hyperbola as x and y axes respectively, is (a) x2 – y2 = 0 (b) x2 – y2 = 32 2 2 (c) x – y = 2 (d) None of these

103. The function xx is increasing, when 1 e (c) x < 0

1 e (d) for all real x

(a) x >

(b) x
Hg 22+ > Cu2+ > Mg2+ +0.80 +0.79 +0.34 –2.37 (in V) Mg2+ will not be reduced as its reduction potential value is much lower than that of water (–0.834 V). On the basis of above reduction potential values, the deposition of metals will be in the order Ag, Hg, Cu.

(Greenish yellow gas)

KClO 3 + 5KCl + 3H2O

KClO3 is used in fire works, safety matches and as an oxidizing agent. 45. (a) : [Pt(NH3)2Cl(en)NO2]Cl2 Diamminechlorido (ethylenediamine nitroplatinum (IV) chloride 46. (c) : For the reaction, 1   SO 2( g) + O 2( g )    SO 3( g ) 2 Using formula, Kp = Kc(RT)Dng where, Dng = no. of moles of gaseous products – no. of moles of gaseous reactants

 1 1 = 1 − 1 +  = − = x  2 2

O

H2O/H+

Allyl chloride

NaOH(aq.)



–H2O

CH

CH – CO CH3 – CO

HOCH2 – CHOH – CH2OH

44. (b) : 3Cl2 + 6KOH

CH3COONa

Hydrolysis

ClCH2 – CH

HOCl

Glycerol--monochlorohydrin

O

Acetic anhydride (Aliphatic acid anhydride)

CH – COOH

O

+ CH3COOH

Cinnamic acid (Unsaturated aromatic acid)

aq. Na2CO3 150°C, 12 atm

HOCH2 – CHCl – CH2OH

CH3 – C

Benzaldehyde (Aromatic aldehyde)

0.04 = × 100 = 4.08% 0.98

Propene

H2CH – C

48. (d) : Proteins are known as the building blocks of body. They do not provide energy for metabolism. 49. (d) : N2 + 3H2 2NH3, DH = – 22.08 kcal The reaction is exothermic and accompanied with decrease in number of moles. So, the equilibrium is affected by temperature, pressure and concentration. 50. (d) 51. (d) : Rate1 = k [A]n [B]m On doubling the concentration of A and halving the concentration of B, we get Rate2 = k [2A]n [B/2]m Ratio of new rate to earlier rate =

k[2 A]n [ B / 2]m n

m

k[ A] [ B]

1 = 2n ×   2

m

= 2 n− m

52. (d) : In SN1 reaction, the intermediate carbocation which is planar can be attacked from either side resulting in formation of products of both types i.e. with retention as well as inversion of configuration. 53. (b) : The shape of four orbitals dxy, dyz, dxz, dx2 – y2 is clover leaf like while dz2 has a doughnut shaped electron cloud in the centre. 54. (b) : Diazonium ion acts as an electrophile in coupling reaction; greater the electron withdrawing power of the group attached at p-position higher is the electrophilicity. Thus, the correct order is : I < III < IV < II 55. (b) : Ti3+ and Cu2+ are coloured while Sc3+ is colourless due to empty d-orbitals. Ag+ and Cd2+ are colourless due to completely filled d-orbitals.

16

VITEEE CHAPTERWISE SOLUTIONS

56. (c) : Ethyl chloride (C2H5Cl) and acetyl chloride (CH3COCl) react with alc. KCN by nucleophilic substitution reaction. Benzaldehyde (C6H5CHO) undergoes benzoin condensation. C2H5Cl

alc. KCN

CH3COCl

C2H5CN + KCl

alc. KCN

alc. KCN

CH3COCN + KCl

C6H5CH(OH)COC6H5 2C6H5CHO Thus, only chlorobenzene does not react.

57. (d) : Heat cannot flow from a cold body to a hot body. 58. (c) :

+

–

N2Cl

Cl HCl/CuCl

+ N2



62. (b) : Bismuth does not show allotropy. 63. (a) : Though Cu (I) has 3d10 stable configuration while Cu (II) has 3d9 configuration, yet Cu (II) is more stable due to greater effective nuclear charge of Cu (II) (i.e., to hold 17 electrons instead of 18 in Cu (I)). 64. (d) :

NOH

R – C – NO2

RCH2NO2 + HNO2

Nitrolic acid

1° Nitro compound

(a)

Sodium nitrolate (red)

65. (c) : 2Ag+(aq) + Cu(s) (little tendency to accept a proton)

(c)

NONa R – C – NO2

(little tendency to accept a proton)

(b)

NaOH



K=

[Cu 2 + ] + 2

[Ag ]

=

Cu2+ + 2Ag(s) (aq) x

y2

After dilution, reaction quotient, Q

=

( x / 2) ( y / 2)2

=

2x y2

Here, Q > K, so the equilibrium will shift in backward direction. Carbocation (I) is more stable since it has two equivalent resonance structures. (d)

Hence, [Cu2+]
. 2 2

66. (d) : Greater the charge on central metal ion and stronger the field of ligand, greater is the Do value (CFSE). According to the spectrochemical series, the increasing order of Do is 2– – C2O4 < H2O < NH3 < CN . 67. (a) : (i)

Carbocation II is less stable than I since it has only one resonating structure. Thus, compound (c) is most basic. (Propan-2-ol)

59. (b) 60. (a) : Phenol on reaction with sodium nitrite (NaNO2) and conc. H2SO4 gives a sequential change in colour as Brown or red → green → deep blue. 61. (d) : The described reaction is Sandmeyer reaction.

(ii)

(i) CH3MgI

(ii) H3O +

(Propan-2-ol)

(iii) Formaldehyde gives primary alcohol with Grignard reagent. (iv) Propene reacts with alkaline KMnO4 to give diols.

17

Model Test Paper-1

68. (a) : By comparing given two equations, we get  d( DG)  d( − nFE)  dE  DS = −  = nF    ; DS = − dT  dT  dT  P  dE  DS   = dT nF

\

69. (b) : At radial node, Y = 0 \ From given equation,

x – 1 = 0 and x2 – 8x + 12 = 0



x–1=0 ⇒ x=1

i.e.,

Thus, (A) and (B) are enantiomers.

a 2r = 1; r = 0 (Minimum) 2 a0

74. (b) : When r = r0, y2s = 0, then from the given equation, r 2 − = 0 ⇒ r = 2a0 a0

x2 – 8x + 12 = 0 ⇒ (x – 6) (x – 2) = 0 when x – 2 = 0 ⇒ x = 2 i.e.,

2r = 2 , r = a0 (Middle value) a0

when x – 6 = 0 ⇒ x = 6 i.e.,

2r = 6 ; r = 3a0 (Maximum) a0

70. (d) :

O

C H

CH

H C OH HO C H

H C OH + NH2OH

H C OH

NOH

H C OH CH2OH D(+)-Glucose

HO C H H C OH H C OH CH2OH Glucoxime

71. (a) : Solubility of AgCl = 1.435 × 10–3 g/L

=

AgCl

1.435 × 10 −3 = 10 −5 mol L−1 143.5 (mol. wt. of AgCl = 143.5) Ag+ + Cl– –

Ksp = [Ag+][Cl ] = (10–5)(10–5) = 10–10 72. (b) : Xanthoproteic test involves heating of proteins with conc. HNO3 which produces yellow colour. 73. (d) : Rotation of (B) through 180° within the plane of the paper gives (D) which is an enantiomer of (A).

75. (b) : Number of unpaired electrons in Fe2+ is less than Mn2+, so Fe2+ is less paramagnetic than Mn2+. 76. (d) : Glucose is an aldose containing an aldehydic group (–CHO) so it responds to Tollens’ test. Fructose is a ketose containing a ketonic group C O and it undergoes rearrangement in presence of basic medium (provided by Tollens’ reagent) to form glucose (containing –CHO group), therefore fructose also gives Tollens’ test in basic medium. 77. (a) : o- and p-nitrophenols are stronger acids than m-nitrophenol. As a result, nitrophenols (a) and (c) are stronger acids than (b) and (d). In (c), the NO2 group is flanked by two CH3 groups which push the NO2 group out of the plane of the benzene ring. As a result of this steric hindrance the electron withdrawing resonance effect of the nitro group will be reduced (also called steric inhibition resonance) and hence, the acidic character of the nitrophenol will decrease. Therefore, nitrophenol (a) is the strongest acid. 78. (d) : N2 O4

(Colourless)

160°C

2NO2

(Brown)

600°C

2NO + O2 (Colourless)

On heating the vapours, colour deepens due to conversion of colourless N2O4 to brown coloured NO2. But beyond 160°C NO2 gets converted to colourless mixture of NO and O2 thus, now colour fades away. But if pressure is increased at 600°C then, according to

18

VITEEE CHAPTERWISE SOLUTIONS

Le-Chatelier’s principle equilibrium will again shift towards brown coloured NO2 due to lesser number of moles and hence, colour again deepens. Thus, originally vapours contain pure N2O4. 79. (d) : Being reversible reaction, the backward reaction i.e., acetal-hemiacetal step can be restricted by minimizing water content, i.e. by using dry HCl. The backward step hemiacetalaldehyde can be restricted by using excess of alcohol. 80. (b) : H2O2 → O2 + 2H+ + 2e– 1 mole of H2O2 ≡ 2 moles of e– ≡ 2 × 96500 C ≡ 193000 C = 1.93 × 105 C



S(x) at x = 0, a and a / 2 . we have, S(0) = S(a) = 0.



 a  a a2 2 and S  4 a = p − = 2 pa  2  2 2 Hence, the required greatest surface area is 2pa.

83. (b) : I = ∫ sin 3 x cos 3xdx

Mathematics

81. (a) : We have,

 3 sin x − sin 3 x  = ∫  cos 3x dx  4 [sin 3x = 3 sin x – 4 sin3 x] 3 1 sin x cos 3 xdx − ∫ sin 3x cos 3x dx ∫ 4 4 3 1 = ∫ (sin 4 x − sin 2 x)dx − ∫ sin 6 xdx 8 8 3  − cos 4 x  3  cos 2 x  1  cos 6 x  =   + 8  2  + 8  6  + C. 8 4 =

 2x   1 + 6x  y = tan −1  + tan −1  2  3 − 2 x   1 + 8x   4x − 2x   1 / 3 + 2x  = tan −1  + tan −1   + ⋅ 1 4 x 2 x    1 − 1 / 3 ⋅ 2 x 

84. (a) : We have, [x2 – 1] = 1, 0 ≤ x < 1



= tan–1(4x) – tan–1(2x) + tan–1(1/3) + tan–1(2x) \ y = tan–1(4x) + tan–1(1/3)





Hence,



dy 4 = dx 1 + 16 x 2

= 2, 3 ≤ x < 2 Hence, we have 2

82. (b) : Let x be the radius and 2y be the height of the cylinder. From the figure, y

a 2y



x



we can see that x2 + y2 = a2 ...(i) where a denotes the radius of the given sphere. The lateral surface area of the cylinder is given by

2 2 S = 2px⋅2y = 4 px ⋅ a − x , 0 ≤ x ≤ a Now, we have



 dS = 4 p  a2 − x2 −  dx 

= 0, 1 ≤ x < 2 = 1, 2 ≤ x < 3



 2 2  = 4p (a − 2x ) a2 − x 2  a2 − x2 x2

which exists everywhere in (0, a) and vanishes at x = a / 2 only. To find the greatest and least values, we need to check the values of

∫0 [x

2

− 1]dx

1



= ∫ − 1dx + ∫ 0

0

1

2

0dx + ∫

=  − x  0 + 0 +  x 

3 2

3 2

+  x 

2

1 dx + ∫ 1 dx 3

2 3

= −1+ 3 − 2 + 2 − 3 = 1 − 2.

 1 3 2 85. (c) : Now, [1 x 1]  2 5 1   15 3 2 

= [1 + 2 x + 15 3 + 5 x + 3 2 + x + 2] (using row by column multiplication) = [16 + 2 x 6 + 5x 4 + x] \



⇒ ⇒ ⇒ ⇒

 1   [16 + 2 x 6 + 5x 4 + x]  2  = 0  x  [16 + 2x + 12 + 10x + 4x + x2] = 0 (using row by column multiplication) [x2 + 16x + 28] = 0 ⇒ x2 + 16x + 28 = 0 x2 + 14x + 2x + 28 = 0 x(x + 14) + 2(x + 14) = 0 ⇒ x = –14, –2

19

Model Test Paper-1

86. (b) : Given A, B and C are the angles of a triangle. \ A + B + C = p ⇒ A + B = p – C Now, cos(A + B) = cos (p – C) = – cos C ⇒ cos A cos B – sin A sin B = – cos C ⇒ cos A cos B + cos C = sin A sin C ...(i) Similarly, cos A cos C + cos B = sin A sin C ...(ii) and sin (B + C) = sin (p –A) = sin A ...(iii)



−1 cos C cos B Now, let D = cos C −1 cos A cos B cos A −1 Expanding along R1, we get D = –1(1 – cos2 A) + cos C (cos C + cos A cos B) + cos B (cos A cos C + cos B) = – sin2 A + cos C (sin A sin B) + cos B (sin A sin C) [using eqs. (i) and (ii) and cos2 A + sin2 A = 1] = – sin2 A + sin A (sin B cos C + cos B sin C) = – sin2 A + sin A sin (B + C) [Q sin (x + y) = sin x cos y + cos x sin y] = – sin2 A + sin2 A = 0 [using eq. (iii)]

 1 −1 1   87. (b) : We have, A =  2 −1 0   1 0 0 









 1 −1 1  1 −1 1     Then, A 2 = A × A =  2 −1 0  ×  2 −1 0   1 0 0   1 0 0   1 − 2 + 1 −1 + 1 + 0 1 + 0 + 0    =  2 − 2 + 0 −2 + 1 + 0 2 − 0 + 0   1 + 0 + 0 −1 + 0 + 0 1 + 0 + 0 

(multiplying row by column)

0 0 1   A 2 = 0 −1 2   1 −1 1 3 A = A2 × A 0 0 1  1 −1 1     = 0 −1 2  ×  2 −1 0   1 −1 1  1 0 0   0 + 0 + 1 0 + 0 + 0 0 + 0 + 0   = 0 − 2 + 2 0 + 1 + 0 0 + 0 + 0   1 − 2 + 1 −1 + 1 + 0 1 + 0 + 0 

1 0 0   A3 = 0 1 0  = I 3 0 0 1 On multiplying both sides by A–1, we get A–1A3 = A–1I \ IA2 = A–1 (Q A–1 A = I) \ A2 = A–1. 88. (c) : Let the rate of change of the volume of the balloon be k (where, k is constant) d (Volume) = Constant dt d 4 3 ⇒ pr  = k  dt  3  4 3 Q volume of sphere = 3 pr  4  dr  ⇒  p   3r 2  = k 3  dt  On separating the variables, we get 4pr2 dr = k dt ...(i) On integrating both sides, we get 4 p ∫ r 2 dr = k ∫ dt r3 ⇒ 4 p = kt + C 3 ⇒ 4pr3 = 3(kt + C) ...(ii) Now, initially t = 0 and r = 3, \ 4p(3)3 = 3(k × 0 + C) ⇒ 108p = 3C ⇒ C = 36p Also, when t = 3 and r = 6, then from eq. (ii), we get 4p(6)3 = 3 (k × 3 + C) ⇒ 864p = 3(3k + 36p) ⇒ 3k = 288p – 36p = 252p ⇒ k = 84p On substituting the values of k and C in eq. (ii), we get 4pr3 = 3(84 pt + 36p) ⇒ 4pr3 = 4p(63t + 27) ⇒ r3 = 63t + 27 ⇒ r = (63t + 27)1/3 which is the required radius of the balloon at time t. 89. (a) : We know that, the four points A, B, C and D are coplanar, if the three vectors AB,   AC and AD are coplanar i.e.,     AB AC AD  = 0    ^ ^ ^ ^ ^ Now, AB = − ( j + k ) − ( 4 i + 5 j + k ) ^ ^ ^ = − 4i − 6 j − 2k  ^ ^ ^ ^ ^ ^ AC = ( 3 i + 9 j + 4 k ) − ( 4 i + 5 j + k ) ^ ^ ^ = − i + 4 i + 3k

20

VITEEE CHAPTERWISE SOLUTIONS

 ^ ^ ^ ^ ^ ^ and AD = 4( − i + j + k ) − ( 4 i + 5 j + k ) ^ ^ ^ = − 8 i − j + 3k −4 −6 −2    [ AB AC AD] = −1 4 3 −8 −1 3

\ 90.





92. (a) : Since, g(x) = f′(x)

2

∫−1 g( x) dx

\

2

^

... (i)

−1 = 4 – 2 [Q f(–1) = 2 and f(2) = 4] =2 93. (d) : Area (OAPB) = xy x



^

j



^

Area (OAPO) = ∫ f (t ) dt y

k

^

^

^

^

^

Volume of parallelopiped = 38 × Altitude



[From eq. (ii)]

⇒ 4 = 38 × Altitude [From eq. (i)] 4 ⇒ Altitude = units 38

1 1 91. (a) : We have, 2 f ( x) + 3 f   = − 2 ...(i) x x

1 ⇒ 2 f   + 3 f ( x) = x − 2 x

x

^

= i ( −5) − j ( −3) + k ( 2) = − 5 i + 3 j + 2 k   \ | a × b | = ( −5)2 + ( 3)2 + ( 2)2

⇒

y = f(x) P(x, y)

^

= 25 + 9 + 4 = 38 ...(ii) Also, Volume of parallelopiped = Area of base × Altitude

0

B

= i ( −1 − 4) − j ( −1 − 2) + k ( 4 − 2) ^

2

 2 3x2 2  =  − log x + − x 10 5   5 1

  a×b= 1 1 1 2 4 −1

3x 2  dx − 5 5 

= ∫ f ′( x) dx = [ f ( x)]2−1 = f ( 2) − f ( −1)

= (12 + 1) – (6 + 1) + (2 – 4) = 4 cu units   Now, area of the base = | a × b |



2

2

∫1  − 5x +

2 1 = − log 2 + 5 2

1 1 1 = 2 4 −1 1 1 3

^

f ( x) dx =

 2 6 4  3 2 =  − log 2 + −  −  −   5 5 5   10 5 

= – 4(12 + 3) + 6 (–3 + 24) + (–2)(1 + 32) = – 4 × 15 + 6 × 21 – 2 × 33 = – 60 + 126 – 66 = –126 + 126 = 0 Hence, A, B, C and D are coplanar. (d) : We know that,    Volume of parallelopiped [a b c ]

i

2

∫1

\

...(ii)

On solving the eqs. (i) and (ii), we get 2 3x 2 + − f ( x) = − 5x 5 5

O

x

A y

x



Therefore, area (OBPO) = xy − ∫ f (t ) dt



According to the given condition

0

x

xy − ∫ f (t ) dt x



∫0

0

f (t ) dt

=

m n x

⇒ nxy = ( m + n) ∫ f (t ) dt 0

On differentiating w.r.t. x, we get



 dy  n  x + y  = ( m + n) f ( x) = ( m + n) y , as y = f(x)  dx 



⇒



c is a constant ⇒ y = cxm/n ⇒ f(x)= cxm/n

m dx dy m ⋅ = ⇒ ⋅ ln x = ln y − ln c , where n x y n

dy ax + 3 = dx 2 y + f ⇒ (ax + 3) dx = (2y + f) dy

94. (b) : We have,

21

Model Test Paper-1



On integrating, we obtain

x2 a + 3x = y 2 + f y + c 2 a ⇒ − x2 + y 2 − 3x + f y + c = 0 2 This will represent a circle, if a − =1 [Q Coeff . of x 2 = Coeff . of y 2 ] 2 9 and, + f 2 − c > 0 [Using g 2 + f 2 − c > 0] 4 2 ⇒ a = –2 and 9 + 4f – 4c > 0 95. (c) : Equation of the plane passing through the line of intersection of the first two    planes is r ⋅ (n1 + l n2 ) = p1 + lp2 , where l is a parameter. Since three planes have a common line of intersection, the above equation should be   identical with r ⋅ n3 = p3 for some l. That is for some l,    n1 + l n2 = k n3 ...(i) and p1 + lp2 = kp3 ...(ii) where K is a scalar From eq. (i)     n1 × n3 + l n2 × n3 = 0     and n1 × n2 = k n3 × n2 ...(iii)

Froms eq. (ii)     ( p1 + lp2 )(n2 × n3 ) = kp3 (n2 × n3 )   = p3 (n2 × n1 ) ... (iv)       p1(n2 × n3 ) + p2 l (n2 × n3 ) + p3 (n1 × n2 ) = 0 ⇒       ⇒ p1(n2 × n3 ) + p2 (n3 × n1 ) + p3 (n1 × n2 ) = 0 [Using eq. (iii)] Hence, (c) is the correct answer. 96. (b) : Q 3 + i = ( a + ib)(c + id) = (ac – bd) + i (ad + bc)

Here, ac − bd = 3 and ad + bc = 1

b d Now, tan −1   + tan −1   a c  bc + ad   1  = tan −1  = tan −1   3   ac − bd  p = n p + , n ∈I 6 97. (b) : Given,

Let z = x + iy

i+z =1 i−z

\

⇒

i + x + iy | x + i(1 + y )| =1 ⇒ =1 i − ( x + iy ) | − x + i(1 − y )| x 2 + (1 + y )2 = ( − x)2 + (1 − y )2

⇒ x2 + 12 + y2 + 2y = x2 + 12 + y2 – 2y ⇒ 4y = 0 ⇒ y = 0 Hence, the given condition lies on x-axis.

98. (d) : Let AB is a chord and its equation is y = mx ...(i) y

B x2 + y2 – 2ax = 0

M (0, 0) A

C(a, 0)

x

Equation of CM which is perpendicular to AB, is x + my = l It passes through the centre (a, 0) ⇒ x + my = a ...(ii) On eliminating m from eqs. (i) and (ii), we get x2 + y2 = ax ⇒ x2 + y2 – ax = 0 is the locus of the centre of the required circle.

99. (a) : Given parametric equation of parabola is x = t2 + 1, y = 2t + 1 2



 y − 1 ⇒ x= +1  2 



⇒ Y 2 = 4X where Y = y – 1, X = x – 1 Vertex is (1, 1), length of latus rectum = 4 Clearly, equation of directrix is X = –1 ⇒ x – 1 = –1 ⇒ x = 0.

⇒ (y – 1)2 = 4(x – 1)

100. (b) : Suppose the line lx + my + n = 0 cuts the ellipse at P(a cos q, b sin q)  p  p  and Q a cos  + q  , b sin  + q   .     2 2  Then, these two points lie on the line la cos q + mb sin q + n = 0 and –la sin q + mb cos q + n = 0 ⇒ la cos q + mb sin q = –n and –la sin q + mb cos q = –n ⇒ (la cos q + mb sin q)2 + (–la sin q + mb cos q)2 = n2 + n2 2 2 ⇒ l a + m2b2 = 2n2.

22

VITEEE CHAPTERWISE SOLUTIONS

101. (c) : Equation of tangent at (a cos q, b sin q) is y

(a cos , b sin )

Q x

O

P

x

E P   = P (scooter driver met with an accident)  E1  1 = 0.01 = 100

 E and P   = P (motorcycle driver met with  E2  an accident) 2 = 0.02 = 100 \ The probability that insured motorcycle met with an accident is given by

y



y x cos q + sin q = 1 a b \ Coordinates of P and Q are  a   b   cos q , 0  and  0 , sin q  respectively.



Area of DOPQ =

\

1 a b ab × = 2 cos q sin q |sin 2q|

\ (Area)minimum = ab. 102. (b) : Let the equation of the hyperbola be x2



y2

=1 a2 b2 The coordinates of the foci are (ae, 0) and (–ae, 0). \ 2 ae = 16 ⇒ 2 a 2 = 16 ⇒ a = 4 2 Also, b2 = a2(e2 –1) = 32(2 – 1) = 32 Thus, a2 = 32 and b2 = 32 Hence, the required equation is −

x2 y 2 − = 1 ⇒ x 2 − y 2 = 32 32 32

103. (a) : Let y = xx On differentiating w.r.t. x, we get dy = x x (1 + log x) dx dy For increasing function, >0 dx ⇒ xx(1 + log x) > 0 ⇒ 1 + log x > 0 1 1 ⇒ log e x > log e ⇒ x > e e 1 Function is increasing when x > . e 104. (b) : Given, number of scooters = 2000 and number of motorcycles = 3000 \ Total number of vehicles = 2000 + 3000 = 5000 Let E1 and E2 be the events of insured scooter and motorcycle, respectively.

2000 2 3000 3 = and P( E2 ) = = 3000 5 5000 5 Let E be the event that insured vehicle met with an accident. Then, \



P( E1 ) =



E  P( E2 ) ⋅ P( E / E2 ) P 2  =  E  E  E   P( E1 ) ⋅ P   + P( E2 ) ⋅ P    E  1  E2    (by Baye’s theorem) 6 3 2 × 5 5 100 = = 2 1  3 2  2 6  5 × 100  +  5 × 100   5 + 5  6 5 6 3 = × = = 5 8 8 4 Hence, the probability that the insured 3 motorcycle met with an accident is . 4

105. (a) : Let man tosses the coin n times. Now, given that P(having atleast one head) > 80% 80 i.e., P ( X ≥ 1) > 100 where, X is the number of heads.

\

1 − P ( X = 0) >

⇒ 1 − nC0 p0q n > 0

80 100

80 100 [using P(X) = nCr pr qn – r] n

1 1 8 ⇒ 1 − nC0     > 2 2 10 Q p = probability of getting a head    1 1 1  once = and q = 1 − p = 1 − =    2 2 2 1 4 1 1 1 4 ⇒ 1− n > ⇒ < 1− ⇒ < n n 5 5 5 2 2 2 ⇒ 2n > 5 ...(i) Inequality (i) is satisfied for n ≥ 3. Hence, coin must be tossed 3 or more times. 106. (a)

23

Model Test Paper-1

107. (d) : We know that ~ (p ∨ q) = ~ p ∧ ~ q 108. (a) : We have, f(x) = x3 + 2x2 + x + 5, x ∈ R and f ′(x) = 3x2 + 4x + 1 = (x + 1)(3x + 1), x ∈R Draw the number line for f ′(x), we have +ve

–ve

+ve

–1



–1/3

f(x) strictly increases in (–∞, –1) f(x) strictly decreases in (–1, –1/3) and f(x) strictly increases in (–1/3, ∞) Also, we have f(–1) = –1 + 2 – 1 + 5 = 5

 −1  −1 2 1 4 and f   = + − +5=5− ≈ 4.85   3 27 9 3 27 The graph of f(x) shows that f(x) cuts the X-axis only once. Now, we have f(–3) = –27 + 12 – 3 + 5 = –13 and f(–2) = –8 + 8 – 2 + 5 = 3



�1

�1/3

which are of opposite signs. This proves that the curve cuts the X-axis somewhere between –2 and –3. ⇒ f(x) = 0 has a root a lying between –2 and –3 Hence, [a] = –3. 109. (d) : The given equation can be written as

dx 1 1 +x = dy y ln y 2 y whose I.F. = e

∫

dy y ln y 2 1

= e2

1 1/ y ∫ ln y dy

= e2

ln (ln y )

= ln y The equation becomes exact, and its primitive is



x ln y = ∫ i.e., x =

ln y y

dy =



f ′( x ) =

1 x

2

[ x 2 + f ′( x)] +

−2

∫

x

x3 2

[t 2 + f ′(t )] dt

+ f ′(t )] dt



P ′′ − y′ 2 2 1 i.e., y 3 y ′′ = y 2 P ′′ − y 2 y ′ 2 2 1 1 = PP ′′ − ( P ′ )2 [using results (i) and (ii)] 2 4 d 3 Hence, ( y y ′′ ) dx 1 1 1 1 = P ′P ′′ + P ′P ′′′ − P ′P ′′ = P ′P ′′′ 2 2 2 2 i.e., yy ′′ =

112. (d) : we have f(x) = 3x2 + 12x – 1, –1 ≤ x ≤ 2 = 37 – x, 2 < x ≤ 3 and f ′(x) = 6(x + 2), –1 < x ≤ 2 = –1, 2 < x ≤ 3 (a) f(2–) = 3⋅22 + 12⋅2 – 1 = 35 = f(2) f(2) = 37 – 2 = 35 ⇒ continuity of f(x) on [–1, 3] (b) f′(x) = 6(x + 2) > 0 ∀ x ∈ (–1, 2) f(x) is increasing in [–1, 2] (c) f′(2–) = 6(2 + 2) = 24 f′(2+) = –1 ⇒ f′(2) does not exist. (as f′(2–) ≠ f′(2 +) 113. (d) : We have x

lim

x→∞





x)2 dx  ∞  (tan −1 x)2 = lim   x  ∞  x→∞ 1 + x2

∫0 (tan



110. (d) : We have, 1 x f ( x) = 2 ∫ [t 2 + f ′(t )] dt x 2 Differentiating w.r.t. x, we have

2

1 f ′( 2 ) + 0 4 i.e., f ′(2) = 4/3. 111. (d) : We have, y2 = P(x) ...(i) Differentiating w.r.t. x, we have 2yy′ = P′ ...(ii) Differentiating again w.r.t. x,we have 2y′2 + 2yy′′ = P′′

(ln y )3/ 2 +C 3/2

2 C ln y + 3 ln y

2

∫2 [t

=1+



Hence, we have 1 ( −2) f ′( 2) = [4 + f ′( 2)] + 4 8

−1

= lim (tan −1 x)2 ⋅ lim x→∞

x→∞

1+ x

2

1 + x2

x

1 | x| 1 + 2 p2 x = ⋅ lim x 4 x→∞ p2 − p2 = as x → + ∞ and as x → − ∞ 4 4 Hence, limit does not exist.

24

VITEEE CHAPTERWISE SOLUTIONS

114. (a) : We have, f ( x) =

We can see that f ( a) + f ( − a) =



ea

1 + ea

Thus, we have

^ ^  ^  ^ +[( k⋅ k )a − ( k⋅ a ) k ]  ^ ^  ^ ^  ^  ^ = a − ( i ⋅ a ) i + a − ( j⋅ a ) j + a − ( k ⋅ a ) k ]

ex 1 + ex

+

e−a

1 + e−a a 1 e = + =1 1 + ea 1 + ea

^ ^

...(i)

f ( a)

I1 =



x g{ x(1 − x)} dx

f (f −( aa))

∫

=





∫

(1 − x) g {(1 − x)x} dx

f ( − a)

  using = I2 – I1

[using (i)]



b



b

∫a f ( x) dx = ∫a f (a + b − x) dx

i.e., 2I1 = I2 gives

I2 =2 I1

115. (c) : adj(Q–1 BP–1) = adj Q–1⋅adj B⋅adj P–1 P Q = A⋅ = PAQ P | | | Q| 116. (d) : Given, M = [auv]n×n = [sin (qu – qv) + i cos (qu – qv)]

⇒ M = [sin(qu − qv ) − i cos(qu − qv )]



T

M = [sin(qu − qu ) − i cos(qv − qu )] = [– sin (qu – qv) – i cos (qu – qv)] = –[sin (qu – qv)+ i cos (qu – qv)] = –M.

1 log x y log x z 117. (a) : Let D = log y x 1 log y z log z x log z y

1

= 1(1 – logzy logyz) –logxy(logyx – logyz logzx) + logxz (logyx logzy – logzx) = (1 – 1) – logxy (logyx – logyz logzx) + logxz (logyx logzy – logzx) = (1 – 1) – (1 – logxy logyx) + (logxz logzx–1) = 0 (since, logxy logy x = 1) = 0 – (1 – 1) + (1 – 1) = 0  ^ ^  ^ ^  ^ ^ 118. (b) : [ i × ( a × i ) + j × ( a × j ) + k × ( a × k )] ^ ^  ^  ^ ^ ^  ^  ^ = [( i ⋅ i )a − ( i ⋅ a ) i ] × [( j⋅ j )a − ( j⋅ a ) j ]

^ ^

^ ^

[Q i ⋅ i = j⋅ j = k⋅ k = 1]  ^  ^ ^  ^ ^  ^ = 3a − [( i ⋅ a ) i + ( j⋅ a ) j + ( k⋅ a ) k ]  ^ ^ ^ Let a = a1 i + a2 j + a3 k . Then, ^  ^ ^ ^ ^ i ⋅ a = i ⋅ ( a1 i + a2 j + a3 k ) ^ ^

^ ^

^ ^

= a1 i ⋅ i + a2 ( i ⋅ j ) + a3 ( i ⋅ k )



= a1(1) + a2(0) + a3(0) = a1 ^  ^  Similarly, j⋅ a = a2 , k⋅ a = a3  ^ ^ ^ \ L.H.S = 3a − ( a1 i + a2 j + a3 k )    = 3 a − a = 2 a = R.H.S.    119. (a) : As a , b and c are linearly dependent vectors.  [a b c ] = 0 ⇒

⇒

1 1 1 4 3 4 =0 1 a b



⇒ 1(3b – 4a) – 1(4b – 4) + 1(4a – 3) = 0 ⇒ –b + 1 = 0 ⇒ b = 1



Now,|c | = 3



⇒



⇒ a2 = 1 ⇒ a = ± 1

2 1 + a 2 + b2 = 3 ⇒ 1 + 1 + a = 3

120. (b) : Let an equation of the required plane be x y z + + =1 a b c



vvv

This meets the coordinates axes in A(a, 0, 0), B(0, b, 0) and C(0, 0, c) So that the coordinates of the centroid of the a b c DABC are  , ,  = (1, r , r 2 ) (given) 3 3 3 ⇒ a = 3, b = 3r, c = 3r2 Hence, the required equation of the plane is x y z + + 2 = 1 or r2x + ry + z = 3r2. 3 3r 3r

Model Test Paper-2

2

Model Test Paper Time : 2 Hours 30 Minutes

Max. Marks : 120

physics 1.

Current versus time and voltage versus time graph of a circuit element is shown below



The type of the circuit element is (a) capacitor of 2 F (b) rsistor of 2 W (c) capacitor of 1 F (d) a voltage source of emf 1 V.

2.

In the circuit shown in figure, charge stored in 4 mF capacitor is (a) 20 mC (c) 30 mC

3.

4.

5.

25

6.

Pick out the correct statement from the following (1) Electron emission during b-decay is always accompanied by neutrino. (2) Nuclear force is charge independent. (3) Fusion is the chief source of stellar energy. (a) (1), (2) are correct (b) (1), (3) are correct (c) Only (1) is correct (d) (2), (3) are correct

7.

A battery of emf 10 V and internal resistance 3 W is connected to a resistor. The current in the circuit is 0.5 A. The terminal voltage of the battery when the circuit is closed, is (a) 10 V (b) 0 V (c) 1.5 V (d) 8.5 V

8.

The current in the conductor varies with time t as I = 2t + 3t2 where I is in ampere and t in seconds. Electric charge flowing through a section of conductor during t = 2 s to t = 3 s is (a) 10 C (b) 24 C (c) 33 C (d) 44 C

9.

Which phenomenon best supports the theory that matter has a wave nature? (a) Electron momentum (b) Electron diffraction (c) Photon momentum (d) Photon diffraction

(b) 40 mC (d) zero

Pick out the incorrect statement from the following : (a) Mercury vapour lamp produces line emission spectrum. (b) Oil flame produces line emission spectrum. (c) Band spectrum helps us to study molecular structure. (d) Sunlight spectrum is an example for line absorption spectrum. A cell develops the same power across two resistances R1 and R2 separately. The internal resistance of the cell is R + R2 (a) R1 + R2 (b) 1 2 R1R2 (c) R1R2 (d) 2 A voltmeter with resistance 500 W is used to measure the emf of a cell of internal resistance 4 W. The percentage error in the reading of the voltmeter will be (a) 0.2% (b) 0.8% (c) 1.4% (d) 2.2%

10. Logic gates X and Y have the truth tables shown below

A

B

C

P

R

0

0

0

0

1

1

0

0

1

0

0

1

0

–

–

1

1

1

–

–

When the output of X is connected to the input of Y, the resulting combination is equivalent to a single (a) NOT gate (b) OR gate (c) NOR gate (d) NAND gate

26

VITEEE CHAPTERWISE SOLUTIONS

11. A transistor is operated in common-emitter configuration at constant collector voltage, Vc = 1.5 V such that a change in the base current from 100 mA to 150 mA produces a change in the collector current from 5 mA to 10 mA. The current gain is (a) 67 (b) 75 (c) 100 (d) 50 12. In the given figure, the electron enters into the magnetic field as shown. It deflects in (a) +ve X direction (b) –ve X direction (c) +ve Y direction (d) –ve Y direction 13. In Young’s double slit experiment, intensity at a point is (1/4)th of the maximum intensity. Angular position of this point will be (a) sin–1 (l/d) (b) sin–1 (l/2d) –1 (c) sin (l/3d) (d) sin–1 (l/4d) 14. Magnetic field induction at the centre O of a square loop of side ‘a’ carrying current I as shown in figure is

m0 I 2 pa m I (c) 0 2pa

(a)

(b)

2m0 I pa

(d) zero

15. Two wires of same length are shaped into a square and a circle. If they carry same current, ratio of their magnetic moments is (a) 2 : p (b) p : 2 (c) p : 4 (d) 4 : p 16. If a resistance of 100 W, an inductance of 0.5 H and a capacitance of 10 × 10–6 F are connected in series through 50 Hz ac supply, the impedance will be (a) 1.87 W (b) 101.3 W (c) 18.7 W (d) 189.7 W 17. Which gas is present in Gieger Muller Counter? (a) Argon (b) Neon (c) Hydrogen (d) Both (a) and (b)

18. Radiocarbon dating is based on the principle that all living matter possesses a certain amount of a radioactive form of (a) Carbon (b) Sodium (c) Hydrogen (d) Potassium 19. One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in (a) visible region (b) infrared region (c) ultraviolet region (d) microwave region 20. A solenoid of 1.5 m length and 4.0 cm diameter possesses 10 turns per cm. A current of 5 A is flowing through it. The magnetic induction at axis inside the solenoid is (a) 2p × 10–3 T (b) 2p × 10–5 T –2 (c) 2p × 10 G (d) 2p × 10–5 G 21. The first antiparticle found was the (a) positron (b) hyperon (c) quark (d) baryon 22. A wire of length l is bent in the form of a circular coil of some turns. A current I flows through the coil. The coil is placed in a uniform magnetic field B. The maximum torque on the coil can be (a)

IBl 2 2p

(b)

IBl 2 4p

(c)

IBl 2 p

(d)

2 IBl 2 p

23. A loop of area 0.5 m2 is placed in a magnetic field of strength 2 T in direction making an angle of 30° with the field. The magnetic flux linked with the loop will be (a)

1 Wb 2

(c) 2 Wb

(b)

3 Wb 2

(d)

3 Wb 2

24. A long straight wire carrying current of 30 A rests on a table. Another wire AB of length 1 m, mass 3 g carries the same current but in the opposite direction, the wire AB is free to slide up and down. The height upto which AB will rise is (a) 0.6 cm (b) 0.7 cm (c) 0.4 cm (d) 0.5 cm

27

Model Test Paper-2

25. A circular coil of radius 10 cm having 100 turns carries a current of 3.2 A. The magnetic field at the centre of the coil is (a) 2.01 × 10–3 T (b) 5.64 × 10–3 T –4 (c) 2.64 × 10 T (d) 5.01 × 10–4 T

32. In the given circuit with steady current, the potential drop across the capacitor must be

26. A 2 mC charge moving around a circle with a frequency of 6.25 × 1012 Hz produces a magnetic field 6.28 T at the centre of the circle. The radius of the circle is (a) 2.25 m (b) 0.25 m (c) 13.0 m (d) 1.25 m

V 2 V 2V (c) (d) 3 3 33. In the circuit shown below, switch S is closed at a time t = 0. The charge which passes through the battery in one time constant is (a) EL

27. Four capacitors and a battery are connected as shown in the figure. If the potential difference across the 7 mF capacitor is 6 V, then which of the following statements is incorrect? (a) The potential drop across the 12 mF capacitor is 10 V. (b) The charge in the 3 mF capacitor is 42 mC. (c) The potential drop across the 3 mF capacitor is 10 V. (d) The emf of the battery is 30 V. 28. Light transmitted by Nicol prism is (a) unpolarised (b) plane polarised (c) circularly polarised (d) elliptically polarized. 29. In Newton’s rings experiment the diameter of 5th dark ring is reduced to half of its value after placing a liquid between plane glass plate and convex surface. Calculate the refractive index of liquid. (a) 2 (b) 4 (c)

2

(d) 3

30. In a free space electron is placed in the path of a plane electromagnetic wave, it will start moving along (a) centre of earth (b) equator of earth (c) magnetic field (d) electric field 31. The radii of curvature of the two surfaces of a lens are 20 cm and 30 cm and the refractive index of the material of the lens is 1.5. If the lens is concavo-convex, then the focal length of the lens is (a) 24 cm (b) 10 cm (c) 15 cm (d) 120 cm

(a) V

(b)

eR2 (b) eL ER

L eR2 E (d) E   R L 34. A sinusoidal voltage of peak value 300 V and an angular frequency w = 400 rad s–1 is applied to series L-C-R circuit, in which R = 3 W, L = 20 mH and C = 625 mF. The peak current in the circuit is (c)

(a) 30 2 A

(b) 60 A

(c) 100 A

(d) 60 2 A

35. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is (a) 1215 Å (b) 1640 Å (c) 2430 Å (d) 4687 Å 36. When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a distance of 10 cm. If the object is moved with a speed of 9 m s–1, the speed with which image moves is (a) 10 m s–1 (b) 1 m s–1 (c) 9 m s–1 (d) 0.9 m s–1 37. When the electromagnetic radiations of frequencies 4 × 1015 Hz and 6 × 1015 Hz fall on the same metal, in different experiments, the ratio of maximum kinetic energy of electrons liberated is 1 : 3. The threshold frequency for the metal is (a) 2 × 1015 Hz (b) 1 × 1015 Hz 15 (c) 3 × 10 Hz (d) 4 × 1015 Hz

28 38.

VITEEE CHAPTERWISE SOLUTIONS 7 3

Li nucleus has three protons and four neutrons. Mass of 73 Li nucleus is 7.016005 amu. Mass of proton is 1.007277 amu and mass of neutron is 1.008665 amu. Mass defect of lithium nucleus in amu is (a) 0.04048 (b) 0.04050 (c) 0.04052 (d) 0.04055

39. In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used have light nuclei. Heavy nuclei will not serve the purpose because (a) they will break up (b) elastic collision of neutrons with heavy nuclei will not slow them down (c) the net weight of the rector would be unbearably high (d) substances with heavy nuclei do not occur in liquid or gaseous state at room temperature 40. A plastic sheet (refractive index = 1.6) covers one slit of a double slit arrangement meant for the Young’s experiment. When the double slit is illuminated by monochromatic light (wavelength in air = 6600 Å), the centre of the screen appears dark rather than bright. The minimum thickness of the plastic sheet to be used for this to happen is (a) 3300 Å (b) 6600 Å (c) 2062 Å (d) 5500 Å

Chemistry 41. 0.6 mole of PCl5, 0.3 mole of PCl3 and 0.5 mole of Cl2 are taken in a 1 L flask to obtain the following equilibrium: PCl5(g) PCl3(g) + Cl2(g)

If the equilibrium constant Kc for the reaction is 0.2 predict the direction of the reaction. (a) Forward direction (b) Backward direction (c) Direction of the reaction cannot be predicted. (d) Reaction does not move in any direction.

42. Which of the following orders about the basic strengths is correct? (a) NH3 < NH2OH < HN3 < NH2NH2 (b) NH2OH < HN3 < NH2NH2 < NH3 (c) HN3 < NH3 < NH2OH < NH2NH2 (d) HN3 < NH2OH < NH2NH2 < NH3

43. In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen? (a) 3 → 2 (b) 5 → 2 (c) 4 → 1 (d) 2 → 5 44. Which of the following enzymes is not useful in the digestion of proteins? (a) Chymotrypsin (b) Pepsin (c) Trypsin (d) Lipase 45. Which of the following conformations is optically inactive? (a)

(c)





(b)

(d)

46. Which of the following statements is not true? (a) XeO3 is an explosive compound. (b) Noble gases neither burn nor help in burning. (c) Two S – O bonds in SO2 molecule are not equal. (d) The electron affinity of noble gases is zero. 47. Which of the following compounds will exhibit cis-trans (geometrical) isomerism? (a) 2-Butene (b) 2-Butyne (c) 2-Butanol (d) Butanal 48. The correct order of decreasing thermal stability of the hydrides of group 16 elements is (a) H2S > H2Te > H2Po > H2O > H2Se (b) H2Se > H2S > H2Po > H2O > H2Te (c) H2O > H2Se > H2Te > H2S > H2Po (d) H2O > H2S > H2Se > H2Te > H2Po 49. The time for half life period of a certain reaction A Products is 1 hour. When the initial concentration of the reactant A is 2.0 mol L–1, how much time does it take for its concentration to come from 0.50 to 0.25 mol L–1 if it is a zero order reaction? (a) 1 h (b) 4 h (c) 0.5 h (d) 0.25 h

29

Model Test Paper-2

50. An ether is more volatile than an alcohol having the same molecular formula. This is due to (a) dipolar character of ethers (b) alcohols having resonance structures (c) intermolecular hydrogen bonding in ethers (d) intermolecular hydrogen bonding in alcohols. 51. Which of the following does not show stereoisomerism?

CH3 (a) C

H CH3 (b) C

H

CO – NH NH – CO CO – NH HN – CO

CH3 C H H C CH3

H (c) CH3 – C – COOH

H COOH (d)



CF3CH2OH I II (a) I < IV < III < II (c) IV < I < III < II

53. The probability of finding out an electron at a point within an atom is proportional to the (a) square of the orbital wave function i.e., y2 (b) orbital wave function i.e., y (c) Hamiltonian operator i.e., H (d) principal quantum number i.e., n 54. What is the correct order of acidity from weakest to strongest acid for these compounds?

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

56. Which of the following will produce a buffer solution when mixed in equal volumes? (a) 0.1 mol dm–3 NH4OH and 0.1 mol dm–3 HCl (b) 0.05 mol dm–3 NH4OH and 0.1 mol dm–3 HCl (c) 0.1 mol dm–3 NH4OH and 0.05 mol dm–3 HCl (d) 0.1 mol dm–3 CH3COONa and 0.1 mol dm–3 NaOH

57.

Identify A and B.

COOH 52. An organic compound of molecular formula C3H6O does not produce any precipitate with 2,3-dinitrophenylhydrazine and does not react with sodium metal. This compound is (a) CH3COCH3 (b) CH2 CH – OCH3 (c) CH3CH2CHO (d) CH2 CHCH2OH



55. Which of the following does not give oxygen on heating? (a) K2Cr2O7 (b) (NH4)2Cr2O7 (c) KClO3 (d) Zn(ClO3)2

H – C – OH H – C – OH



(a)

(b)

(c)

(d) A and B =

30

VITEEE CHAPTERWISE SOLUTIONS

58. The emf of the three galvanic cells are represented by E1, E2 and E3. I. Zn | Zn2+ (1 M) || Cu2+ (1 M) | Cu II. Zn | Zn2+ (0.1 M) || Cu2+ (1 M) | Cu III. Zn | Zn2+ (1 M) || Cu2+ (0.1 M) | Cu Which of the following is true? (a) E1 > E2 > E3 (b) E3 > E2 > E1 (c) E3 > E1 > E2 (d) E2 > E1 > E3 59. Non-stoichiometric metal deficiency is shown in the salts of (a) all metals (b) alkali metals only (c) alkaline earth metals only (d) transition metals only. 60. The product of acid hydrolysis of P and Q can be distinguished by

(a) Lucas reagent (b) 2,4-DNP (c) Fehling’s solution (d) NaHSO3 61. Interstitial compounds are (a) non-stoichiometric and are ionic in nature (b) non-stoichiometric and are covalent in nature (c) non-stoichiometric and are neither typically ionic nor covalent (d) stoichiometric and are neither ionic nor covalent. 62. Calorific value is in the order (a) fats > carbohydrates > proteins (b) carbohydrates > fats > proteins (c) proteins > carbohydrates > fats (d) fats > proteins > carbohydrates.

3a 2

(d) rCs + + rCl − =

66. Which of the following changes will increase the emf of the cell? Co(s) | CoCl2(M1) || HCl(M2) | H2(g) | Pt(s) (a) Increase the volume of the CoCl2 solution from 100 to 200 mL (b) Increase M2 from 0.010 to 0.500 M (c) Increase the pressure of H2(g) from 1.00 to 2.00 atm (d) Increase the mass of the Co-electrode from 10.0 to 20.0 g 67. The magnitude of CFSE (Crystal Field Splitting Energy, Do) can be related to the configuration of d-orbitals in a coordination entity as (a) if Do < P, the configuration is t2g3 eg1 = weak field ligand and high spin complex (b) if Do > P, the configuration is t2g3 e1g = strong field ligand and low spin complex (c) if Do > P, the configuration is t2g4 eg0 = strong field ligand and high spin complex (d) if Do = P, the configuration is t2g4 eg0 = strong field ligand and high spin complex. 68. In an antifluorite structure, cations occupy (a) octahedral voids (b) centre of cube (c) tetrahedral voids (d) corners of cube.

63. CsCl crystallises in body-centred cubic lattice. If ‘a’ is its edge length then which of the following expressions is correct? (a) rCs + + rCl − = 3a (b) rCs+ + rCl– = 3a (c) rCs + + rCl− =

65. Which of the following orders of relative strengths of acids is correct? (a) FCH2COOH > ClCH2COOH > BrCH2COOH (b) ClCH2COOH > BrCH2COOH > FCH2COOH (c) BrCH2COOH > ClCH2COOH > FCH2COOH (d) ClCH2COOH > FCH2COOH > BrCH2COOH

3 a 2

64. In which of the following reactions, Kp > Kc ? (I) PCl5(g) PCl3(g) + Cl2(g) (II) N2O4(g) 2NO2(g) (III) N2(g) + 3H2(g) 2NH3(g) (IV) 2SO2(g) + O2(g) 2SO3(g) (a) (III) and (IV) (b) (II) and (III) (c) (I) and (II) (d) (I), (II) and (IV)

69. Which one of the following is an example of Hell-Volhard-Zelinsky reaction? (a) RCOOH

(i) B2H6

(ii) H3O+

(b) R2CHCOOH

(c)

(i) Br2/Red P (ii) H2O

COOH COOH

(d) RCOOH

PCl5

RCH2OH R2 C Br

COOH O

NH3 

O O

RCOCl

31

Model Test Paper-2

70. A radiation of wavelength l illuminates a metal and ejects photoelectrons of maximum kinetic energy of 1 eV. Another radiation l of wavelength , ejects photoelectrons of 3 maximum kinetic energy of 4 eV. What will be the work function of metal? (a) 4 eV (b) 2.5 eV (c) 0.5 eV (d) 1.5 eV 71. Ketones do not reduce Tollens’ reagent, but fructose with a keto group reduces it. It is attributed to (a) enolisation of keto group of fructose and then, its transformation into aldehyde – group in presence of OH which is present in Tollens’ reagent (b) CHOH group which is also oxidised to keto group (c) both statements are correct (d) none of these. 72. Assuming that water vapour is an ideal gas, the internal energy change (DU) when 1 mol of water is vapourised at 1 bar pressure and 100°C, (given : molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol–1 and R = 8.3 J mol–1 K–1) will be (a) 41.00 kJ mol–1 (b) 4.100 kJ mol–1 (c) 3.7904 kJ mol–1 (d) 37.904 kJ mol–1 73. White phosphorus on reaction with NaOH gives PH3 as one of the products. This is a (a) dimerization reaction (b) disproportionation reaction (c) condensation reaction (d) precipitation reaction. 74. The solubility product of AgI at 25°C is 1.0 × 10–16 mol2 L–2. The solubility of AgI in 10–4 N solution of KI at 25°C is approximately (in mol L–1) (a) 1.0 × 10–8 (b) 1.0 × 10–16 (c) 1.0 × 10–12 (d) 1.0 × 10–10 75. What is the decreasing order of basicity of primary, secondary and tertiary ethylamines and NH3? (a) NH3 > C2H5NH2 > (C2H5)2NH > (C2H5)3N (b) (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3 (c) (C2H5)2NH > C2H5NH2 > (C2H5)3N > NH3 (d) (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 76. The unit cell of aluminium is a cube with an edge length of 405 pm. The density of aluminium is 2.70 g cm–3. What is the structure of unit cell of aluminium?

(a) Body-centred cubic cell (b) Face-centred cubic cell (c) End-centred cubic cell (d) Simple cubic cell 77. The most appropriate set of reagents for carrying out the following conversion is O

(a) EtMgBr ; HCl (c) C2H5Li ; HCl

Cl

OH

(b) (C2H5)2CuLi ; HCl (d) HCl ; EtMgBr

78. In the nitration of benzene with a mixture of concentrated nitric acid and concentrated sulphuric acid, the active species involved is (a) nitrite ion (b) nitrate ion (c) nitronium ion (d) nitric oxide. 79. Activation energy (Ea) and rate constants (k1 and k2) of a chemical reaction at two different temperatures (T1 and T2) are related by E 1 k 1 (a) ln 2 = − a −   k1 R  T1 T2  (b) ln

E  1 k2 1 =− a −  k1 R  T2 T1 

(c) ln

E  1 k2 1 =− a +  k1 R  T2 T1 

(d) ln

k 2 Ea  1 1 = + k1 R  T1 T2 

80. Low spin complex of d6 cation in an octahedral field will have the following energy −12 −12 (a) (b) D +P D + 3P 5 o 5 o −2 −2 (d) D + 2P D +P 5 o 5 o (Do = Crystal field splitting energy in an octahedral field, P = Electron pairing energy) (c)

Mathematics 81. If |z – i| ≤ 2 and z0 = 5 + 3i, then the maximum value of |iz + z0| is (a) 2 + 31

(b) 7

(c)

(d) None of these

31 − 2

82. The continued product of four roots of p p   cos + i sin  3 3 (a) 0 (c) i

3/ 4

is equal to (b) 1 (d) –1

32

VITEEE CHAPTERWISE SOLUTIONS

83. If capital letters denote the cofactors of the corresponding small letters in a1 b1 c1 A1 B1 C1 D = a2 b2 c2 , then D ′ = A B C is 2 2 2 a3 b3 c3 A B C (a) D2 (c) 0 1 1 1 84. If + + = 0 , then a b c is equal to (a) – abc (c) abc

(b) 2D (d) D

3

3

3

1+ a 1 1 1 1+ b 1 1 1 1+ c (b) 0 (d) None of these

 1 2 2  6 −2 −6      85. If A =  2 3 0  and adj. A =  −4 2 x .  0 1 2   y −1 −1 Then x + y is equal to (a) –1 (b) 1 (c) 6 (d) – 6 86. The position vectors of three points are         2 a − b + 3c , a − 2b + mc and na − 5b where    a , b , c are non-coplanar vectors and m, n are scalars. The three points are collinear if and only if 9 9 (a) m = −2 , n = (b) m = , n = −2 4 4 9 9 (c) m = 2 , n = (d) m = − , n = 2 4 4 87. If the vectors   a = iˆ + ˆj + kˆ , b = 4iˆ + 3 ˆj + 4 kˆ  and c = iˆ + aˆj + bkˆ  are linearly dependent and c = 3 , then (a) a = ±1, b = 1 (b) a = 1, b = –1 (c) a = 1, b = ±1 (d) a = –1, b = ±1 88. On each evening a boy either watches Doordarshan channel or Ten Sports. The probability that he watches Ten Sports is 4/5. If he watches Doordarshan, there is a chance of 3/4 that he will fall asleep, while it is 1/4 when he watches Ten Sports. On one day, the boy is found to be asleep. The probability that the boy watched Doordarshan is 5 2 (a) (b) 7 7 3 4 (c) (d) 7 7

89. Three groups A, B and C are competing for the positions on the Board of Directors of a company. The probabilities of their winning are 0.5, 0.3 and 0.2, respectively. If the group A wins, the probability of introducing a new product is 0.7 and other corresponding probabilites for groups B and C are, respectively, 0.6 and 0.5. The probability that new product will be introduced is (a) 0.43 (b) 0.53 (c) 0.63 (d) 0.73 90. The range of random variable X is {1, 2, 3} and P(X = 1) = 3l 3, P(X = 2) = 4l – 10l 2, P(X = 3) = 5l – 1 where l is constant. Then P(2 ≤ X ≤ 3) is equal to 8 2 (a) (b) 9 3 4 1 (c) (d) 9 3 91. In a book of 750 pages, there are 500 typographical errors. Assuming Poisson law for the number of errors per page, find the probability that a random sample of 5 pages will contain no error. (a) e–10 (b) e–2/3 10 (c) e (d) e–10/3 92. The area enclosed within the curve |x| + |y| = 1 is (a) 2 (b) 4 (c) 6 (d) none of these 93. The radius of the circle having centre at (2, 1), whose one of the chord is a diameter of the circle x2 + y2 – 2x – 6y + 6 = 0 is (a) 1 (b) 2 (c) 3

(d)

3

94. The equation of the parobola whose focus is the point (0, 0) and the tangent at the vertex is x – y + 1 = 0 is (a) x2 + y2 + 2xy – 4x – 4y + 4 = 0 (b) x2 + y2 – 2xy – 4x + 4y – 4 = 0 (c) x2 + y2 – 2xy + 4x – 4y – 4 = 0 (d) x2 + y2 + 2xy – 4x + 4y – 4 = 0 95. The angle between pair of tangents drawn to the ellipse 3x2 + 2y2 = 5 from the point (1, 2) is 6 12 (a) tan −1 (b) tan −1 5 5 (c) tan −1

12 5



(d) tan −1

12 5

33

Model Test Paper-2

96. For hyperbola

x2 2

−

y2 2

= 1, which of

cos a sin a the following remains constant with change in ‘a’? (a) abscissa of vertices (b) abscissa of foci (c) eccentricity (d) directrix

97. Find the maximum value of f(x) = |x loge x| in (0, 1). (a) e (b) –e (c) –1/e (d) none of these 98. Semi-vertical angle of a cone is 45°. The height of cone is 30.05 cm. Find the approximate volume. (a) 1325p c.c (b) 9045p c.c (c) 2020p c.c (d) 330p c.c xn 99. If y = n!

sin x cos x sin(np / 2) cos(np / 2) , then value p2

p dn y

at x = 0, is dxn (a) 0 (c) p2(p + 1)

of

100. If

p3

x

y

a

a

(b) p(p + 1) (d) none of these

∫ f (t)dt − ∫ g(t)dt = b, then the value of

dy dx

at (x0 , y0), is

f ( x0 ) g( y0 ) (c) g(x0) – f(y0) (a)

101. The function f(x) = has

g( x0 ) f ( y0 ) (d) f(x0) – g(y0)

103. Let f(x) =

x

∫0 g(t)dt ,

where g is such that

–1/2 ≤ g(t) ≤ 0 for t ∈ [0, 1] –1/2 ≤ g(t) ≤ 1 for t ∈ [1, 3] g(t) ≤ 1 for t ∈ [3, 4] Then f(4) satisfies the inequality (a) 1/2 ≤ f(4) ≤ 3 (b) 0 ≤ f(4) ≤ 2 (c) f(4) ≤ 3 (d) none of these 104. ∫

[ x]

0

[ x] dx , is equal to

[ x]2 [ x]([ x] + 1) (b) 2 2 [ x]([ x] − 1) ([ x] − 1)([ x] + 1) (c) (d) 2 2 105. Find the shortest distance between the curves y2 = x3 and 9x2 + 9y2 – 30y + 16 = 0. (a) 0.5 (b) 0.2 (c) 1 (d) 3 (a)

106. The false statement in the following is (a) (p ⇒ q) ⇔ (~ q ⇒ ~ p) is a contradiction (b) ~ (~ p) ⇒ p is a tautotogy (c) p ∨ (~ p) is a tautology (d) p ∧ (~ p) is a contradiction 107. The inverse of the proposition (p ∧ ~ q) ⇒ r is (a) ~ r ⇒ ~ p ∨ q (b) r ⇒ p ∧ ~ q (c) ~ p ∨ q ⇒ ~ r (d) None of these 108. Which venn diagram represents the truth of the statement ‘No policeman is a thief’?

(b)

x  −p p  , x ∈ ,  2 2  1 + x tan x

 p (a) one minima in  0 ,   2  −p  (b) one maxima in  , 0  2   −p p  (c) no extrema in  ,  2 2  (d) none of these

102. The function f(x) = (x – 1)m (x – 2)m+1 has (a) inflection at x = 1, 2 (b) inflection at x = 1 or x = 2 (c) no point of inflection (d) two points of extrema

(a)

P

T



U

P

T

U P

(c)

(b)

T



(d) None of these

U

109. A line is uniquely determined if (i) it passes through a given point and has given direction (ii) it passes through two given points (iii) it passes through a point and is parallel to a line. (iv) it passes through two given points and is perpendicular to the plane (a) only (i) (b) (i) and (ii) (c) (i), (ii) and (iii) (d) all of these

34

VITEEE CHAPTERWISE SOLUTIONS

110. If a, b, c are real numbers and planes ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 meet in a line then (a) a2 + b2 + c2 = 0 (b) a2 + b2 + c2 = ab + bc + ca (c) 3abc = a2 + b2 + c2 (d) 3(a2 + b2 + c2) = 2(ab + bc + ca) 111. The equation of the plane containing the y z line + = 1, x = 0 and parallel to the line b c x z + = 1, y = 0 must be a c x y z x y z (a) + + = 1 (b) − − = 1 a b c a b c x y z x y z (c) − − + 1 = 0 (d) + − = 1 a b c a b c p 3p 112. If Kc reaction will proceed in backward direction. 42. (b) : Basicity in nitrogen compounds is attributed to the availability of lone pair of electrons. Putting more electronegative groups on N will decrease its basicity. Therefore, the order of basicity of these compounds is NH3 > NH2NH2 > HN3 > NH2OH. 43. (b) : The lines at the red end belong to Balmer series. These are obtained for jumps to n = 2 from outer orbits. The third line from the red end (yellow region) corresponds to jump from n = 5 to n = 2. 44. (d) : Lipase functions in digestion of lipids. 45. (c) : Other given conformations do not have any plane of symmetry or centre of symmetry. 46. (c) : One S—O bond is pp-pp double bond and other is pp-dp double bond with oxygen atom. They appear to be different but due to resonance they are equal. 47. (a) : Cis-trans (geometrical) isomerism is exhibited by those compounds which have same structural formula but different spatial arrangement of groups around C C (carbon-carbon) double bond. H3C – C – H H – C – CH3

and

Trans –

H3C – C – H H3C – C – H Cis –

2-Butene

48. (d) : As the size of the element increases down the group, the E–H bond dissociation energy decreases and hence E–H bond breaks more easily.

39

Model Test Paper-2

49. (d) : For a zero order reaction, t1/2 is given as

[ A] t1/2 = 0 2k

[ A]0 or k = 2t1/2

Given, t1/2 = 1 h, [A]0 = 2 M

\

k=

2 = 1 mol L−1 h −1 2 ×1

Integrated rate law for zero order reaction is [A] = –kt + [A]0 Here, [A]0 = 0.5 M and [A] = 0.25 M ⇒ 0.25 = –t + 0.5 ⇒ t = 0.25 h 50. (d) : The reason for the lesser volatility of alcohols than ethers is the intermolecular association of a large number of molecules due to hydrogen bonding as – OH group is highly polarised. R

R

R

Zn(ClO3)2 ZnCl2 + 3O2 56. (c) : In 0.1 mol dm–3 NH4OH and 0.05 mol dm–3 HCl, total amount of HCl reacts with NH4OH to form NH4Cl and some NH4OH will be left unreacted. Thus, the resultant solution contains NH4Cl and NH4OH which will produce a buffer solution. 57. (d) : LiAlH4 and NaBH4 both can reduce to . 58. (d) : Ecell = E°cell − In first case,

R



51. (c) : CH3 – C – COOH

H

(NH4)2Cr2O7 KClO3

N2 + Cr2O3 + 4H2O KCl + 3/2O2

[Cu 2 + ]

= 1 thus Ecell = E°cell

1

0.0591 log 10 −1 n 0.0591 = E°cell + n

\ Ecell = E°cell –

H

2K2CrO4 + Cr2O3 + 3/2O2

[Zn 2+ ]

[Cu 2 + ]

No such hydrogen bonding is present in ethers.

55. (b) : 2K2Cr2O7

RT [Zn 2 + ] ln nF [Cu 2 + ]

2+ In second case, [Zn ] = 0.1

- - - - O – H - - - - O – H - - - - O – H - - - - O – H- - - -

No chiral carbon and thus, does not show stereoisomerism. 52. (b) : Since the compound does not react with 2,3-dinitrophenylhydrazine, it cannot be carbonyl compound. No reaction with sodium metal indicates that compound cannot be an alcohol. Hence it could only be CH2 CH – O – CH3. 53. (a) : y2 is known as probability density and is always positive. From the value of y2 at different points within an atom it is possible to predict the region around the nucleus where electron will most probably be found. 54. (c) : Carboxylic acids are stronger acids than phenols and phenols are stronger acids than alcohols. Among alcohols, due to strong –I effect of F atoms, CF3CH2OH is a stronger acid than benzyl alcohol. Thus, the overall acidic character increases in the order : IV < I < III < II.

but cannot affect

In third case,

[Zn 2+ ] [Cu

2+

]

=

1 = 10 0.1

\ Ecell = E°cell − 0.0591 log 10 = E°cell − 0.0591 n

n

Therefore E2 > E1 > E3 Thus greater the value of [Zn2+]/[Cu2+], less is the emf. 59. (d) : Metal deficiency defect is shown by transition metals because they possess variable valency. 60. (c) :

H3C Propan-2-one

O CH3

40

VITEEE CHAPTERWISE SOLUTIONS

Propan-2-one and propanal can be distinguished by Fehling’s solution as aldehydes give red precipitate with Fehling’s solution while ketones do not react. 61. (c)

62. (a)

63. (d) : In a body-centred cubic (bcc) lattice, oppositely charged ions touch each other along the cross-diagonal of the cube. In case of CsCl, 3 a 2 rCs + + 2rCl − = 3a or, rCs+ + rCl − = 2 64. (c) : Kp = Kc(RT)Dn for Kp > Kc , Dn = +ve (I) PCl5(g) PCl3(g) + Cl2(g) ; Dn = 2 – 1 = 1 (II) N2O4(g) 2NO2(g) ; Dn = 2 – 1 = 1 (III) N2(g) + 3H2(g) 2NH3(g) ; Dn = 2 – 4 = –2 (IV) 2SO2(g) + O2(g) 2SO3(g) ; Dn = 2 – 3 = –1 65. (a) : –I effect of the substituent follows the order : F > Cl > Br Hence, order of relative acidic strength is : FCH2COOH > ClCH2COOH > BrCH2COOH + 66. (b) : Co(s) + 2H(aq) → Co2+ (aq) + H2(g)



K=

[Co 2 + ] [H + ]2

E = E° −

( pH ) 2

2+

0.0591 [Co ] log ( pH ) 2 2 [H + ]2

If [H+]2 increases or [Co2+] decreases or pH 2 decreases then E will increase. 67. (a) : If CFSE (Do) < P (Energy required for pairing), the electrons do not pair up and fourth electron goes to eg of higher energy. Hence, high spin complex is formed. Pairing of electrons does not take place in case of weak field ligands. 68. (c) : In an antifluorite structure, anions are in ccp arrangement while the cations occupy half of the tetrahedral voids. 69. (b) : HVZ halogenation reaction halogenates carboxylic acids at the a-carbon. 70. (c) : Absorbed energy = Threshold energy + Kinetic energy of photoelectrons c h = E0 + 1 eV ...(i) l c 3h = E0 + 4 eV ...(ii) l From equations (i) and (ii), we get



3(E0 + 1 eV) = E0 + 4 eV E0 = 0.5 eV

71. (a) : Unlike simple ketones, fructose can reduce Tollens’ reagent. This is probably due to formation of an equilibrium mixture of glucose, mannose and fructose in alkaline solution. 72. (d) : DU = DH – DnRT = 41000 – 1 × 8.3 × 373 = 41000 – 3095.9 = 37904.1 J mol–1 = 37.904 kJ mol–1 73. (b) : In disproportionation reaction, the same element of compound is oxidized and reduced. 0

+1

P4 + 3NaOH + 3H2O

–3

3NaH2PO2 + PH3

Disproportionation reaction

74. (c) : AgI  Ag+ + I – – – Ksp = [Ag+] [I ], [I ] = 10–4 M (from KI) [10–4 N KI = 10–4 M KI]

[Ag + ] =

1 × 10 −16 10 − 4

= 1 × 10 −12 M

75. (d) : In case of ethylamines, the combined effect of inductive effect, steric effect and solvation effect gives the order of basic strength as : (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3

(2°)

(3°)

76. (b) : d =

Z×M NA ×V

⇒ 2.70 =

(1°)

Z × 27 23

6.023 × 10 × ( 405 × 10 −10 ) 3 Z=4 Hence, structure of aluminium unit cell is fcc or face-centred. 77. (d) : 1, 4-addition of HCl furnishes 4-chlorobutanone, which reacts with Grignard reagent to get the desired product. H+ O

HCl

Cl

78. (c) :

HNO3 + 2H2SO4

O

(i) EtMgBr (ii) H3O+

Cl

OH

NO2+ + H3O+ + 2HSO4¯

Nitronium ion

41

Model Test Paper-2

79. (b) : k1 = Ae–Ea/RT1 k2 = Ae–Ea/RT2 ln k1 = ln A – Ea/RT1 ln k2 = ln A – Ea/RT2 From eq. (i) and (ii), we have E E ln k 2 − ln k1 = ln A − a − ln A + a RT2 RT1 ⇒ ln

k 2 Ea  1 1 = − k1 R  T1 T2 

⇒ ln

E  1 k2 1 =− a −  k1 R  T2 T1 

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

80. (b) : C.F.S.E. = (– 0.4x + 0.6y)Do + zP where x = number of electrons occupying t2g orbital y = number of electrons occupying eg orbital z = number of pairs of electrons For low spin d6 complex electronic configuration 6 0 2, 2, 2 0 = t2g eg or t2g eg

\ x = 6, y = 0, z = 3 C.F.S.E. = (– 0.4 × 6 + 0 × 0.6)Do + 3P

=

−12 D + 3P 5 o

Mathematics 81. (b) : |iz + z0| = |i(z – i) – 1 + 5 + 3i| = |i(z – i) + 4 + 3i| ≤ |i| × |z – i| + |4 + 3i| ≤1×2+5 ≤7 3/ 4

 ip  3

1

( ip)   p p  82. (b) :  cos + i sin  = e 3  4 = e 4   3 3 = (– 1)1/4 Hence continued product of four roots of (– 1)1/4 = (– 1)4–1 (– 1) = 1 (– 1)1/4 = (cos p + i sin p)1/4 The continued product of (cos p + i sin p)1/4 = – (cos p + i sin p) {(z)1/4 = – z if n is even} = (– 1) = 1

D 0 0 83. (a) : D ⋅ D ′ = 0 D 0 = D 3 0 0 D ⇒ D′ = D2

1+ a 1 1 1 1 + b 1 84. (c) : 1 1 1+ c By expanding determinant, we get  1 1 1 abc  1 + + +  = abc(1 + 0) = abc  a b c 85. (c) : |A| = 6 – 8 + 4 = 2 Now, A(adj. A) = |A|I  1 2 2   6 −2 −6   2 0 0       x  = 0 2 0  2 3 0   −4 2  0 1 2   y −1 −1  0 0 2  2 y − 2 0 2x − 8  2 0 0     2 3x − 12  =  0 2 0   0  2 y − 4 0 x − 2   0 0 2  Comparing L.H.S. and R.H.S. 2y – 2 = 2, 2y – 4 = 0 2x – 8 = 0, 3x – 12 = 0, x – 2 = 2 ⇒ x = 4  x+y=6 ⇒ y=2 86. (b) : Let the points be P, Q and R, respectively. Then P,Q,R are collinear if and only if   PQ = xQR for some scalar x. This implies and is implied by       − a − b + ( m − 3)c = x[(n − 1)a − 3b − mc ] ⇔ x(n – 1) = –1, –3x = –1 and m – 3 = –xm 1 9 ⇔ x = , m = and n = –2 3 4    87. (a) : Since the vectors a , b , c are linearly dependent, we have 1 1 1 4 3 4 =0 1 a b

⇒ (3b – 4a) – (4b – 4) + (4a – 3) = 0 ⇒ – b + 1 = 0 ⇒ b = 1  Now c = 3 implies 12 + a2 + b2 = 3 ⇒ a2 = 1 ⇒ a = ±1 88. (c) : Let E1 and E2 be the events of the boy watching Doordarshan and Ten Sports, respectively. It is given that

42

VITEEE CHAPTERWISE SOLUTIONS

1 4 and P( E2 ) = 5 5 Let E be the event of the boy falling asleep. Again by hypothesis 3 1 P( E/E1 ) = and P( E/E2 ) = 4 4 Now,

P( E1 ) =

E = E ∩ (E1 ∪ E2) = (E1 ∩ E) ∪ (E2 ∩ E) So that P(E) = P(E1)P(E/E1) + P(E2)P(E/E2) By Bayes’ theorem P( E1 )P( E /E1 ) P( E1 /E) = P( E1 )P( E /E1 ) + P( E2 )P( E /E2 ) (1/ 5) × ( 3 / 4) 3 = = (1/ 5) × ( 3/ 4) + ( 4 / 5) × (1/ 4) 7

89. (c) : Let P, Q and R be the winning events of A, B and C, respectively. E is the introduction of new product. Then E = E ∩ (P ∪ Q ∪ R) = (E ∩ P) ∪ (E ∩ Q) ∪ (E ∩ R) Therefore, P(E) = P(E ∩ P) + P(E ∩ Q) + P(E ∩ R) = P(P)P(E/P) + P(Q)P(E/Q) + P(R)P(E/R) 5 7 3 6 2 5 = × + × + × 10 10 10 10 10 10 63 = = 0.63 100 90. (a) : We have 3

∑ P( X = K ) = 1

K =1 ⇒ 3l3 – 10l2 + 9l – 1 = 1 ⇒ 3l3 – 10l2 + 9l – 2 = 0 ⇒ (l – 1)(3l – 1)(l – 2) = 0 l can not be 1 and 2. Therefore l = 1/3. Now P(2 ≤ X ≤ 3) = P(X = 2) + P(X = 3) 1 1 5 = 4   − 10   + − 1 3 9 3 10 8 =− +2= 9 9 91. (d) : The average number of errors per page in the book is 500 2 l= = 750 3 The probability of r errors per page is P( X = r ) =

r

l r − l  2  1 −2 / 3 e =  e  3  r! r!

Therefore P(X = 0) = e–2/3

The required probability that a random sample of 5 pages will contain no error is (P(X = 0))5 = (e–2/3)5 = e–10/3 92. (a) : The given lines are ± x ± y = 1 i.e., x + y = 1, x – y = 1, x + y = –1, x – y = –1 These lines form a quadrilateral whose vertices are A(– 1, 0), B(0, – 1), C(1, 0) and D(0, 1). Clearly ABCD is a square. Length of the side of this square ABCD = AB = (0 + 1)2 + ( −1 − 0)2 = 2 \ Required area i.e., area of square ABCD

= ( 2 )2 = 2. ( 5 )2 + 2 2 = 3

93. (c) : Radius AD =

D (2, 1) 3 A

2

5

C (1, 3)

B

94. (d) : Distance of (0, 0) from the tangent 1 x – y + 1 = 0 is . 2 The directrix is parallel to the tangent 1 x – y + 1 = 0 and is at a distance from it. 2 \ Equation of directrix is x – y + k = 0 k 1 where ⇒k=2 = 2⋅ 2 2 \ Equation of parabola is x−y+2 = ( x − 0)2 + ( y − 0)2 2 ⇒ x2 + y2 + 2xy – 4x + 4y – 4 = 0 95. (c) : c2 = a2m2 + b2 = 8 × 4 + 4 = 36 ⇒ c = ± 6. 96. (b) : ae = a2 + b2 = cos 2 a + sin 2 a = 1 So, (b) is the choice. 97. (c) : f(x) = |x loge x|, x ∈ (0, 1) = – x loge x 1 \ f ′(x) = − x ⋅ − log e x = – (1 + loge x) x 1 \ f ′(x) = 0 ⇒ x = e 1 f ″(x) = − < 0 x 1 \ f(x) will be maximum at x = and e 1 1 1 maximum value of f(x) is − ⋅ log e   = . e e e

43

Model Test Paper-2

98. (b) : r = h, h = 30, dh = 0.05 1 1 dV V = pr 2 h = ph 3 ⇒ dV  dh 3 3 dh ⇒ dV = ph2 ⋅ dh = p(30)2(0.05) = 45p \ Approximate vol. = V + dV 1 = p( 30)3 + 45 p = 9045p c.c. 3 99. (a) : We have xn y = n!

 −p p  , . and x1 in the interval  2 2

Now, we have f ′(–x1–) < 0, f ′(–x1+) > 0 ⇒ minima at –x1 and f ′(x1–) > 0, f ′(x1+) < 0 ⇒ maxima at x1.

sin x cos x sin(np / 2) cos(np / 2) p2

p

dn and

Thus, f ′(x) vanishes at two points, say –x1

dn y dxn

p3 ( xn )

dxn n! =

dn

dn

dxn sin(np / 2)

(cos x) dxn cos(np / 2)

p2

p3

p

(sin x)

102. (b) : We have f(x) = (x – 1)m(x – 2)m+1 and f ′(x) = m(x – 1)m–1(x – 2)m+1 + (m + 1)(x – 1)m (x – 2)m+1 m–1 m = (x – 1) (x – 2) [m(x – 2) + (m + 1)(x – 1)] = (x – 1)m–1(x – 2)m[(2m + 1)x – (3m + 1)] It m – 1 is odd, then m is even. In that case f ′(x) does not change sign at x = 2 ⇒ inflection at x = 2 only.

  np  np  n! sin  x +  cos  x +   2 2  = n! sin(np / 2) cos(np / 2)

p2

p

p3

d y Hence,  = 0 .   dxn  x =0 n

x

It m – 1 is even, then m is odd. In that case f ′(x) does not change sign at x = 1 ⇒ inflection at x = 1 only. 103. (c) : We have

[R1 = R2]

0

4

3

3 −1 ⋅ 2 ≤ ∫ g(t) dt ≤ 1 ⋅ 2 1 2

and

∫

4

3

g(t) dt ≤ 1 ⋅ 1

Adding the above inequalities, we have f(4) < 3. 104. (c) : We have

x  −p p  f ( x) = , x ∈ ,  2 2  1 + x tan x (1 + x tan x) − x(tan x + x sec 2 x) and f ′( x) = (1 + x tan x)2 2 2 cos 2 x − x 2 1 − x sec x = = 2 2 (1 + x tan x) cos x(1 + x tan x)2

� 2

3

1

1 −1 ⋅ 1 ≤ ∫ g(t) dt ≤ 0 ⋅ 1 0 2

101. (a, b) : We have

Y

1

0

Now, we have

y

100. (a) : We have ∫a f (t )dt − ∫a g(t )dt = b Differentiating w.r.t. x, we have dy f ( x) − g( y ) = 0 dx dy f ( x) = gives dx g( y ) f ( x0 )  dy  Hence,   =  dx ( x , y ) g( y ) 0 0 0

From the graph shown alongside, we can see that y = x2 and y = cos2 x intersect at two points in the interval  − p , p  .  2 2

4

0

f( 4) = ∫ g(t) dt = ∫ g(t) dt + ∫ g(t ) dt + ∫ g(t ) dt

[ x]

∫0

1

2

3

0

1

2

[ x] dx = ∫ 0 dx + ∫ 1 dx + ∫ 2 dx + ... + [ x]

∫[ x]−1([x] − 1)dx

= 0 + 1 + 2 + ... + ([x] – 1) [ x]([ x] − 1) = 2

105. (b) : We have 9x2 + 9y2 – 30y + 16 = 0 which is a circle having

1  2

X

 5 centre =  0 ,  and radius =  3

2

5 16 =1   − 3 9

44

VITEEE CHAPTERWISE SOLUTIONS

Y Let us choose any point on the curve y2 = x3 as A(t2, t3). If (0,1) B is the point on the C circle and nearest to B A then AB = AC – radius O

y = (x3/2)

A(t2, r3)

X

2

 5 = (t 2 − 0 ) 2 +  t 3 −  − 1  3 2

5  Let f (t ) = t 4 +  t 3 −   3  5 and f ′(t ) = 4t 3 + 2  t 3 −  ⋅ 3t 2  3     5 = 2t 2  2t + 3  t 3 −     3   2 3 = 2t (3t + 2t – 5) = 2t2(t – 1)(3t2 + 3t + 5) The value of t at which AB attains minima is given by the equation f ′(t) = 0 gives, t = 0, 1 But t = 0 is not a point of extrema, since f ′(t) does not change sign in the neighbourhood of t = 0. However, t = 1 is a point of minima, since f ′(1–) < 0 and f ′(1+) > 0. Hence the minimum value of AB is 2

 5 1 +  1 −  − 1  0.2  3 106. (a) 107. (c) 108. (b) 109. (d) 110. (b) : Three planes meet in a line the system of equation must have inifinite solutions. a b c b c a =0 c a b ⇒ (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = 0 x z 111. (a) : Let D.R’s of + = 1 , y = 0 are a1, b1, c1. a c a1 c Hence + 0 × b1 + 1 = 1 and a c 0 × a1 + b1 + 0 × c1 = 0 a1 b c ⇒ = 1= 1 1 0 1/a 0− c  −1 1 So D.R’s  , 0 ,   c a y z + − 1 + lx = 0 is equation of the Now b c plane through the line, which is parallel to

the line. Direction ratios are

−1 1 , 0, . c a

 −1  1 1 1 then l   + × 0 + × = 0  c  b c a ⇒ l = 1 ay z x therefore, + − 1 + = 0 b c a 112. (b) : E = 2 cos q (cos q + i sin q) p 3p since, < q < 2 2 \ r = 2 cos q = –ve \ E = (– 2 cos q)(– cos q – i sin q) E = – 2 cos q [cos(p + q) + i sin(p + q)] 3p p where p + < p + q < p + 2 2 3p 5p < p+q< \ 2 2  3p 5p  i.e. arg(p + q) lies in  ,  2 2   p p 5p  3p  − 2p, − 2 p  i.e., in  − ,  or in   2 2  2 2 which is a subset of (–p, p). 113. (b) : Three relations are necessary to eliminate two constants. Thus, besides the given relation, we require two more and they will be obtained by differentiating the given relation twice successively. Thus, we have dy ..(i) ( x − h) + ( y − k ) = 0 dx 2 d 2 y  dy  ...(ii) 1 + ( y − k) +  =0 dx 2  dx  From eqn. (i) and (ii), we obtain y−k=−

and x − h =

 dy  1+    dx 

2

d2 y

dx 2   dy  2  dy 1 +      dx   dx d2 y

dx 2 On substituting these value in the given relation, we obtian 3

2   dy  2   d2 y  2 1 +    = a     dx    dx 2  which is the required differential equation.

45

Model Test Paper-2

114. (d) : The given differential equation is (1 + y2)dx = (tan–1 y – x)dy This equation can be rewritten as

1

117. (a) : We have, I ( m , n) = ∫ t m (1 + t )n dt 1

 t m +1  n 1 ⇒ I ( m , n) = (1 + t )n ⋅ (1 + t )n−1 t m +1dt  −  m + 1  0 m + 1 ∫0 n 2n ⇒ I ( m , n) = − I ( m + 1, n − 1) m+1 m+1 1 1 2 3n  118. (d) : lim  + + ... +   4n  n→∞ n n + 1 n + 2 2 3n   1  1 n n = lim  + + ... + n  1 2 3n n→∞ n 1+  1+ 1+  n n n   r  3 x 1 3n  n  = lim ∑  = dx r  ∫0 1 + x n→∞ n r =1 1+   n 3 x +1−1 3 3 1 =∫ dx = ∫ dx − ∫ dx 0 (1 + x) 0 0 1+ x

−1

tan y dx x + = 2 dy 1 + y 1 + y2 1 tan −1 y Here, p = and Q = 2 1+ y 1 + y2 1

dy ∫ 2 −1 \ IF = e 1+ y = e tan y Therefore, required solution is

xe tan

−1

y

= ∫ e tan

−1

y

tan −1 y

dy + C 1 + y2 –1 –1 ⇒ xetan y = (tan–1 y – 1)etan y + C –1 ⇒ x = tan–1 y – 1 + Ce–tan y ⋅

dy y y = − cos 2 dx x x xdy − ydx  y  = −  cos 2  dx  x x y  xdy − ydx  dx sec 2  =−  2   x x x y  y  dx sec 2 d   = − x x x y tan = − log x + C x p x = 1, y = , C = 1 4 y tan = 1 − log x x y log x = 1 − tan   x

115. (a) : Given, ⇒ ⇒ ⇒ ⇒ At \ ⇒

3

= [x – log(x + 1)]0 = 3 – log 4 119. (c) : By definition of Gamma function, \

y 1− tan   x

\

x=e 116. (d) : Required area

y

2

y = 2x (0, 1)

= ∫ [2 x − ( 2 x − x 2 )]dx 0 2

= ∫ ( 2 x − 2 x + x 2 )dx 0

2

x

0

x

O (2, 0)  2x x3  = − x2 +  y y = 2x – x2 3  0  log 2 4 3 4 8 1 = −4+ − = − sq unit log 2 3 log 2 log 2 3

∞ −x 3

∞ − x 4 −1

∫0 e x dx = ∫0 ∞ −x 3 ∫0 e x dx = 6

e x

dx = 4 = 3!

120. (c) : It is clear that f(x) has a definite and unique value of each x ∈ [1, 5]. Thus, every point in the interval [1, 5], the value of f(x) is equal to the limit of f(x). So, f(x) is continuous in the interval [1, 5]. −x Also, f ′( x) = , which clearly exists 25 − x 2 for all x in an open interval (1, 5). Hence, f(x) is differentiable in (1, 5). So, there must be a value c ∈ (1, 5) such that f ( 5) − f (1) 0 − 24 − 6 f ′( c ) = = = 5−1 4 2 −c But f ′( c) = 25 − c 2 −c − 6 \ = 2 2 25 − c

vvv

⇒ 4c2 = 6(25 – c2) ⇒

c = ± 15

Clearly, c = 15 ∈(1, 5) such that Lagrange’s theorem is satisfied.

46

VITEEE CHAPTERWISE SOLUTIONS

3

Model Test Paper Time : 2 Hours 30 Minutes

Max. Marks : 120

(a) 2 mC (c) 2 mC

physics 1.

Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius but uncharged is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is F 4 F (c) 8

(a)

2.

3.

4.

5.

6.

3F 4 3F (d) 8

7.

A 10 mF capacitor is charged to a potential difference of 1000 V. The terminals of the charged capacitor are disconnected from the power supply and connected to the terminals of an uncharged 6 mF capacitor. What is the final potential difference across each capacitor? (a) 167 V (b) 100 V (c) 625 V (d) 250 V

8.

The exchange particle of the electromagnetic force is the (a) gluon (b) muon (c) proton (d) photon

9.

Radiation can be a hazard to living organisms because it (a) produces ionization along its path of travel (b) disrupts chemical bond (c) generates free polyatmic ions (d) all of the above

(b)

Two point charges A and B of values +5 ×10–9C and +3 × 10–9 C are kept 6 cm apart in air. When the charge B is moved by 1 cm towards charge A, then work done is equal to (a) 4.5 × 10–6 J (b) 4.5 × 10–7 J –7 (c) 3.5 × 10 J (d) 4.5 × 10–8 J The maximum wavelength of radiation that can produce photoelectric effect in certain metal is 200 nm. The maximum kinetic energy acquired by electron due to radiation of wavelength 100 nm will be (a) 12.4 eV (b) 6.2 eV (c) 100 eV (d) 200 eV When a hydrogen atom emits a photon in going from n = 5 to n = 1, its recoil speed is almost (a) 10–4 m s–1 (b) 8 × 102 m s–1 (c) 2 × 10–2 m s–1 (d) 4 m s–1 In Newton’s rings experiment, the diameter of the 15th ring was found to be 0.59 cm and that of the 5th ring was 0.336 cm. If the radius of the plano convex lens is 100 cm, compute the wavelength of light used. (a) 5880 Å (b) 4880 Å (c) 588 Å (d) 4980 Å An electric dipole is placed at an angle of 30° with an electric field of intensity 2 × 105 N C–1. It experiences a torque equal to 4 N m. Calculate the charge on the dipole if the dipole length is 2 cm.

(b) 4 mC (d) 4 mC

10. In the circuit shown in figure, potential difference between points A and B is 16 V. The current passing through 2 W resistance will be 4

9V 1

3V

4 B

A



(a) 2.5 A (c) 4.0 A

2

(b) 3.5 A (d) zero

11. n identical cells are joined in series with two cells A and B with reversed polarities. EMF of each cell is e and internal resistance is r. Potential difference across cell A or B is (n > 4) 

1



2

(a)

2e n

(b) 2 e  1 −  n 

(c)

4e n

(d) 2 e  1 −  n 

47

Model Test Paper-3

12. A straight wire carrying a current of 13 A is bent into a semi-circular arc of radius 2 cm as shown in figure. The magnetic field is 1.5 × 10 –4 T at the centre of arc, then the magnetic field due to straight segment is m

2c

(a) 1.5 × (c) zero

10– 4

10– 4

T

(b) 2.5 × T (d) 3.0 × 10– 4 T ^

13. An element of 0.05 i m is P placed at the origin as shown in figure which 1 m carries a large current of 10 A. The magnetic field ∆x = 0.05 i^ m at a distance of 1 m in perpendicular direction is (a) 4.5 × 10–8 T (b) 5.5 × 10–8 T (c) 5.0 × 10–8 T (d) 7.5 × 10–8 T 14. Two Nicols are oriented with their principal planes making an angle of 60°. Then the percentage of incident unpolarised light which passes through the system is (a) 100 (b) 50 (c) 12.5 (d) 37.5 15. A helium nucleus makes a full rotation in a circle of radius 0.8 m in 2 s. The value of the magnetic field induction B in tesla at the centre of circle will be (a) 2 × 10–19 m0

(b)

10 −19 m0

(c) 10–19 m0

(d)

2 × 10 −19 m0

16. A torque required to hold a small circular coil of 10 turns, 2 × 10–4 m2 area and carrying 0.5 A current in the middle of a long solenoid of 103 turns/m carrying 3 A current, with its axis perpendicular to the axis of the solenoid, is (a) 12p × 10–7 N m (b) 6p × 10–7 N m (c) 4p × 10–7 N m (d) 2p × 10–7 N m 17. If a charged particle of charge 5 mC and mass 5 g is moving with constant speed 5 m s–1 in a uniform magnetic field B on a curve x2 + y2 = 25, where x and y are in metre. The value of magnetic field will be (a) 1 kT along x-axis (b) 1 kT along z-axis (c) 5 kT along the x-axis (d) 1 kT along any line in the x-y plane

18. Light of certain colour has 2000 waves to the millimetre in air. What will be the wavelength of this light in a medium of refractive index 1.25? (a) 1000 Å (b) 2000 Å (c) 3000 Å (d) 4000 Å 19. A conducting circular loop of radius a and resistance R is kept on a horizontal plane. A vertical time varying magnetic field B = 2t is switched on at time t = 0. Then (a) power generated in the coil at any time t is constant (b) flow of charge per unit time from any section of the coil is constant (c) total charge passed through any section  4 pa 2    R 

between time t = 0 to t = 2 s is  (a) all of the above.

20. A coil in the shape of an equilateral triangle of side l is suspended between the pole  pieces of a permanent magnet, such that B is in plane of the coil. If due to a current I in the  triangle, a torque t act on it, the side l of the triangle is (a)

2  t   3  BI 

(c)

2  t   3  BI 



(b) 2  

1/ 2



(d)

t   3BI 

1/ 2

1  t   3  BI 

21. The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are (a) 10 cm, 10 cm (b) 15 cm, 5 cm (c) 18 cm, 2 cm (d) 11 cm, 9 cm 22. At a point on the right bisector of a magnetic dipole, the magnetic (a) potential varies as

1 r2

(b) potential is zero at all points on the right bisector (c) field varies as r3 (d) field is perpendicular to the axis of dipole. 23. In an LR circuit u = 50 Hz, L = 2H, V = 5 V, R = 1 W, energy stored in inductor is (a) 50 J (b) 25 J (c) 3.66 × 10–4 J (d) 6.33 × 10–5 J

48

VITEEE CHAPTERWISE SOLUTIONS

24. Three unequal resistors in parallel are equivalent to a resistance 1 ohm. If two of them are in the ratio 1 : 2 and if no resistance value is fractional, the largest of the three resistances in ohm is (a) 4 (b) 6 (c) 5 (d) 12 25. In the Wheatstone bridge shown below, in order to balance the bridge, we must have

(a) R1 = 3 W; R2 = 3 W (b) R1 = 6 W; R2 = 15 W (c) R1 = 1.5 W; R2 = any finite value (d) R1 = 3 W; R2 = any finite value 26. In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is l. The magnification of the telescope is L (a) l L −1 (c) l

L (b) + 1 l L+1 (d) L −1

27. A galvanometer of resistance 50 W is connected to a battery of 3 V along with a resistance of 2950 W in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be (a) 6050 W (b) 4450 W (c) 5050 W (d) 5550 W 28. In the part of a circuit shown in figure, the potential difference between points G and H will be

(a) 0 V (b) 12 V (c) 7 V (d) 3 V 29. A cell of emf 1.5 V and internal resistance 2 W is connected to two resistors of 5 W and 8 W in series. The potential difference across the 5 W resistor will be (a) 3.3 V (b) 1 V (c) 0.5 V (d) 0.33 V 30. In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be (a) 100 Hz (b) 70.7 Hz (c) 50 Hz (d) 25 Hz 31. An electric charge of 8.85 × 10–13 C is placed at the centre of a sphere of radius 1 m. The electric flux through the sphere is (a) 0.2 N C–1 m2 (b) 0.1 N C–1 m2 (c) 0.3 N C–1 m2 (d) 0.01 N C–1 m2 32. A deflection mangnetometer is adjusted in the usual way. When a magnet is introduced, the deflection observed is q, and the period of oscillation of the needle in the magnetometer is T. When the magnet is removed, the period of oscillation is T0. The relation between T and T0 is 2 (a) T2 = T0cosq (b) T = T0 cosq (c) T =

T0 cosq

(d) T 2 =

T02 cosq

33. The most stable particle in the Baryon group is (a) neutron (b) proton (c) lamda particle (d) sigma particle 34. The phenomenon of polarization shows that the nature of light is (a) particle (b) transverse (c) longitudinal (d) dual 35. A convex lens forms an image of an object placed 20 cm away from it at a distance of 20 cm on the other side of the lens. If the object is moved 5 cm towards the lens, the image will move (a) 5 cm towards the lens (b) 5 cm away from the lens (c) 10 cm towards the lens (d) 10 cm away from the lens 36. The Fraunhofer diffraction pattern of a single slit is formed in the focal plane of a lens of focal length 1 m. The width of slit is 0.3 mm. If third minimum is formed at a distance of 5 mm from central maximum, then wavelength of light will be (a) 5000 Å (b) 2500 Å (c) 7500 Å (d) 8500 Å

49

Model Test Paper-3

37. Energy required to remove an electron from an aluminium surface is 4.2 eV. If light of wavelength 2000 Å falls on the surface, the velocity of faster electrons ejected from the surface is (a) 2.5 × 1018 m s–1 (b) 2.5 × 1013 m s–1 (c) 6.7 × 1018 m s–1 (d) 8.4 × 105 m s–1 38.

The half-life for the a-decay of Uranium 238 92 U

is 4.47 × 109 yr. If a rock contains sixty percent of its original



238 atoms, 92 U

its age is

[Given, log 6 = 0.778; log 2 = 0.3] (a) 3.3 × 109 yr (b) 6.6 × 109 yr 8 (c) 1.2 × 10 yr (d) 5.4 × 107 yr

39. A ray of light incident at an angle q on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is made of a material of refractive index 1.5, the angle of incidence is (a) 7.5° (b) 5° (c) 15° (d) 2.5° 40. The current voltage relation of diode is given by I = (e1000 V/T – 1) mA, where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring ± 0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA? (a) 0.05 mA (b) 0.2 mA (c) 0.02 mA (d) 0.5 mA

Chemistry 41. For a zero order reaction, the plot of concentration vs time is linear with (a) positive slope and zero intercept (b) negative slope and zero intercept (c) positive slope and non-zero intercept (d) negative slope and non-zero intercept. 42. The most probable structural formula for the compound whose empirical formula is C3H6O, and which can react with Benedict’s reagent is O (a) CH3CH CH2 (b) CH3CH2CHO (c) CH3OCH CH2 (d) CH2 CHCH2OH

43. The structural feature which distinguishes proline from other natural a-amino acids? (a) It is optically inactive. (b) It contains aromatic group. (c) It contains two amino groups. (d) It is a secondary amine. 44. Which of the following rules is not correct regarding IUPAC nomenclature of complex ions? (a) Cation is named first and then anion. (b) In coordination sphere, the ligands are named alphabetically. (c) Positively charged ligands have suffix ‘ate’. (d) More than one ligand of a particular type are indicated by using di, tri, tetra, etc. 45. Which of the following leads to the formation of methyl t-butyl ether? (a) (C2H5)3CONa + CH3Cl (b) CH3ONa + (CH3)3CCl (c) (CH3)3CONa + C2H5Cl (d) (CH3)3CONa + CH3Cl 1 A → 2 B rate of disappearance 2 of A is related to the rate of appearance of B by the expression d[ A] 1 d[ B] d[ A] d[ B] (a) − (b) − = =4 dt 2 dt dt dt

46. For a reaction,

d[ A] 1 d[ B] d[ A] d[ B] (d) − = = dt 4 dt dt dt 47. Which element from group 15 gives most basic compound with hydrogen? (a) Nitrogen (b) Bismuth (c) Arsenic (d) Phosphorus (c) −

48. Which one of the following compounds does not decolourise an acidified aqueous solution of KMnO4? (a) Sulphur dioxide (b) Ferric chloride (c) Hydrogen peroxide (d) Ferrous sulphate 49. The correct order of the packing efficiency in different types of unit cells is (a) fcc < bcc < simple cubic (b) fcc > bcc > simple cubic (c) fcc < bcc > simple cubic (d) bcc < fcc > simple cubic.

50

VITEEE CHAPTERWISE SOLUTIONS

50. Which of the following will exhibit highest boiling point? (a) CH3CH2OCH2CH2CH3 (b) CH3CH2CH2CH2CH2OH (c) CH3CH2CH2CH(CH3)OH (d) CH3CH2C(CH3)2OH 51. The equivalent conductance of NaCl at concentration C and at infinite dilution are lC and l∞, respectively. The correct relationship between lC and l∞ is given as (where, the constant B is positive) (a) lC = l∞ + (B) C (b) lC = l∞ + (B)C (c) lC = l∞ – (B)C

(d) lC = l∞ – B C

52. Which of the following reagents cannot be used to oxidise primary alcohols to aldehydes? (a) CrO3 in anhydrous medium (b) KMnO4 in acidic medium (c) Pyridinium chlorochromate (d) Heat in the presence of Cu at 573 K 53. The correct order of acidity of oxoacids of halogens is (a) HClO < HClO2 < HClO3 < HClO4 (b) HClO4 < HClO3 < HClO2 < HClO (c) HClO < HClO4 < HClO3 < HClO2 (d) HClO4 < HClO2 < HClO3 < HClO 54. The data for the reaction A + B → C, is Exp. 1 2 3 4

[A]0 0.012 0.024 0.024 0.012

[B]0 0.035 0.070 0.035 0.070

Initial rate 0.10 0.80 0.10 0.80

The rate law corresponds to the above data is (a) Rate = k[A][B]3 (b) Rate = k[A]2[B]2 (c) Rate = k[B]3 (d) Rate = k[B]4

55. Reaction of cyclohexanone with dimethylamine in the presence of catalytic amount of an acid forms a compound if water during the reaction is continuously removed. The compound formed is generally known as (a) a Schiff’s base (b) an enamine (c) an imine (d) an amine. 56. An organic compound P has 76.6% C and 6.38% H. Its vapour density is 47. It gives a characteristic colour with aq. FeCl3. P when treated with CO2 and NaOH at 140°C under pressure gives Q which on acidification gives R. R reacts with acetyl chloride to give S, which is

(a)

(b)

(c)



(d)

57. Oils and fats are esters of higher fatty acids with (a) ethanol (b) glycol (c) glycerol (d) methanol. 58. At 20°C, the Ag + ion concentration in a saturated solution of Ag2CrO4 is 1.5 × 10–4 mol L–1. At 20°C, the solubility product of Ag2CrO4 would be (a) 3.3750 × 10–12 (b) 1.6875 × 10–10 –12 (c) 1.6875 × 10 (d) 1.6875 × 10–11 59. In corrosion of iron, (a) an electrochemical (galvanic) cell is formed in which Fe acts as anode and the site where O2 is reduced acts as cathode. (b) electrons flow from anode to cathode through the metal while ions flow through the water droplets (c) dissolved O2 oxidises Fe2+ to Fe3+ before it is deposited as rust (Fe2O3.H2O) (d) all of the above takes place. 60. Phenyl isocyanide is prepared by which of the following reactions? (a) Rosenmund’s reduction (b) Carbylamine reaction (c) Reimer-Tiemann reaction (d) Wurtz reaction 61. Zr and Hf have almost equal atomic and ionic radii because of (a) diagonal relationship (b) lanthanoid contraction (c) actinoid contraction (d) belonging to the same group. 62.

The product X in the above series of reactions is (a)

(c)

OCH3 Br

Br



(b)

Cl

C(CH3)3

C(CH3)3

Br

OH

Br



C(CH3)3

(d)

Cl

C(CH3)3

51

Model Test Paper-3

63. Mark the correct increasing order of reactivity of the following compounds with HBr/HCl. CH2OH

CH2OH

NO2 (II)

(I)

(a) I < II < III (c) II < III < I

CH2OH

Cl (III) (b) II < I < III (d) III < II < I

64. Which of the following is most likely structure of CrCl3.6H2O if 1/3 of total chlorine of the compound is precipitated by adding AgNO3 to its aqueous solution? (a) CrCl3.6H2O (b) [Cr(H2O)3Cl3](H2O)3 (c) [CrCl2(H2O)4]Cl.2H2O (d) [CrCl(H2O)5]Cl2.H2O 65. Which of the following statements is not correct? (a) Methylamine is more basic than NH3. (b) Amines form hydrogen bonds. (c) Ethylamine has higher boiling point than propane. (d) Dimethylamine is less basic than methylamine. 66. At temperature T, a compound AB2(g) dissociates according to the reaction,   2 AB2( g )    2 AB( g ) + B2( g )



with a degree of dissociation x, which is small compared to unity. The expression for Kp, in terms of x and the total pressure, P is 3 (a) Px

2 (b) Px

Px 3 (c) 3

Px 2 (d) 2

2



I2 + H2S → 2HI + S



DG°f (HI) = 1.8 kJ mol–1, DG°f (H2S) = 33.8 kJ mol–1 (a) Non-spontaneous in forward direction (b) Spontaneous in forward direction (c) Spontaneous in backward direction (d) Non-spontaneous in both forward and backward directions

69. In which of the following reactions, the product shown is incorrect? O

CHO

(a)

Conc. NaOH

H+

C O

CHO

(b) O OH

(c) Ph—C

O

KCN alcohol,

Ph

H

(d)

Ph—C—C—H

O

NH2NH2

KOH(alc.),

CH3 Cl

Cl CH3 70. For H3PO3 and H3PO4, the correct choice is (a) H3PO3 is dibasic and reducing (b) H3PO4 is dibasic and non-reducing (c) H3PO4 is tribasic and reducing (d) H3PO3 is tribasic and non-reducing.

71. The correct order of decreasing acid strengths of different groups in the given amino acid is

3

67. Which of the following is correct method to convert p-toluidine to p-toluic acid? (a) Diazotisation, CuCN, H2/Pd (b) CHCl3/NaOH, KCN, Sn/HCl (c) Diazotisation, CuCN/KCN, H2O/H+ (d) Diazotisation, NaCN, NaOH 68. For the reaction given below the values of standard Gibbs free energy of formation at 298 K are given. What is the nature of the reaction?

(a) X > Z > Y (b) Z > X > Y (c) X > Y > Z (d) Y > X > Z 72. The equilibrium constants Kp and Kp for 1 2 the reactions X 2Y and Z Q + R, respectively are in the ratio of 1 : 9. If degree of dissociation of X and Z be equal then the ratio of total pressures at these equilibria is (a) 1 : 9 (b) 1 : 36 (c) 1 : 1 (d) 1 : 3

52

VITEEE CHAPTERWISE SOLUTIONS

73. Propan-1-ol can be prepared from propene by (a) H2O/H2SO4 (b) Hg(OAc)2/H2O followed by NaBH4 (c) B2H6 followed by H2O2 (d) CH3CO2H/H2SO4 74. What products are expected from the disproportionation reaction of hypochlorous acid? (a) HClO3 and Cl2O (b) HClO2 and HClO4 (c) HCl and Cl2O (d) HCl and HClO3 75. What is the colour corresponding to the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from n = 4 to n = 2? (a) Blue (b) Red (c) Yellow (d) Green 76. An optically active amine (A) of molecular formula C4H11N is subjected to Hofmann’s exhaustive methylation process and is followed by hydrolysis, an alkene (B) is produced which upon ozonolysis and subsequent hydrolysis yields formaldehyde and propanal. The amine (A) is (a) CH3 CH CH2CH3

(c) acetaldehyde, iso-propyl alcohol (d) acetone, iso-propyl alcohol. 79. Which of the following statements is incorrect about peptide bond? (a) C—N bond length in proteins is longer than usual bond length of C—N bond. (b) Spectroscopic analysis shows planar C NH Bbond. structure of

O

(c) C—N bond length in proteins is smaller than usual bond length of C—N bond. (d) None of these. 80. What is the effect of Frenkel defect on the density of ionic solids? (a) The density of the crystal increases. (b) The density of the crystal decreases. (c) The density of the crystal remains unchanged. (d) There is no relationship between density of a crystal and defect present in it.

mathematics 81. The equation of the tangent to the curve y = 9 − 2 x 2 at the point where the ordinate and the abscissa are equal, is

NH2

(b) CH3 NH CH CH3 (c)

(a) 2x + y – 3 3 = 0 (b) 2x + y + 3 = 0 (c) 2x + y – 3 = 0 (d) None of these

C2H5

(d) CH3CH2CH2CH2 NH2 77. Which of the following statements is not correct? (a) La(OH)3 is less basic than Lu(OH)3. (b) In lanthanoid series, ionic radius of Ln3+ ions decreases. (c) La is actually an element of transition series rather than lanthanoid series. (d) Atomic radii of Zr and Hf are same because of lanthanoid contraction. 78. An organic compound A reacts with methyl magnesium iodide to form an addition product which on hydrolysis forms the compound B. Compound B gives blue colour salt in Victor Meyer’s test. The compounds A and B respectively are (a) acetaldehyde, t-butyl alcohol (b) acetaldehyde, ethyl alcohol

82. If a0, a1, a2, ..., an are real numbers such that a a a a a0 + 1 + 2 + ... + n −1 + = 0 , (n ≠ 1), 2 3 n n+1 then the equation a0 + a1x + a2x2 + ... + anxn = 0 has a real root lying between (a) 0, 1 (b) 1, 2 (c) 1, 3 (d) None of these 83. The approximate value of f(5.001), where

f(x) = x3 – 7x2 + 15, is (a) –34.995 (b) –33.995 (c) –33.335 (d) –35.993 x

84. If f(x) = (ab – b2 – 2)x + ∫ (cos 4 q + sin 4 q)dq is 0

decreasing function of x for all x ∈R and b ∈R, b being independent of x, then (a) a ∈(0 , 6 )

(b) a ∈( − 6 , 6 )

(c) a ∈( − 6 , 0)

(d) None of these

53

Model Test Paper-3

85. The minimum intercepts made by the axes x2 y2 on the tangent to the ellipse + = 1 is (a) 25 (b) 7 16 9 (c) 1 (d) None of these 86. A foot of the normal from the point (4, 3) to a circle is (2, 1) and a diameter of the circle has equation 2x – y = 2. Then the equation of the circle is (a) x2 + y2 + 2x – 1 = 0 (b) x2 + y2 – 2x – 1 = 0 (c) x2 + y2 – 2y – 1 = 0 (d) None of these 87. If the chord of contact of tangents from a point P to the parabola y2 = 4ax, touches the parabola x2 = 4by, then the locus of P is a/an (a) circle (b) parabola (c) ellipse (d) hyperbola 88.

x2

y2

= 1 will represent the r 2 − r − 6 r 2 − 6r + 5 ellipse, if r lies in the interval (a) (– ∞, 2) (b) (3, ∞) (c) (5, ∞) (d) (1, ∞) +

89. The equation of the asymptotes of the hyperbola 2x2 + 5xy + 2y2 – 11x – 7y – 4 = 0, are

(a) 2x2 + 5xy + 2y2 – 11x – 7y – 5 = 0



(b) 2x2 + 4xy + 2y2 – 7x – 11y + 5 = 0



(c) 2x2 + 5xy + 2y2 – 11x – 7y + 5 = 0



(d) None of the above

90. The derivative of tan −1 cos −1

1+ 1+ x 2 1 + x2

2

1 + x2 − 1 x

w.r.t.

is

(a) 1 (b) –1 1 (c) (d) None of these 2 91. The probability of getting qualified in IIT/JEE and EAM/CET by a student are respectivley 1 3 and . The probability that the student 5 5 gets qualified for at least one of these test, is 3 8 (a) (b) 25 25 17 22 (c) (d) 25 25

1 92. If the mean of a poisson distribution is , 2 then one ratio of P(X = 3) to P(X = 2) is (a) 1 : 2 (b) 1 : 4 (c) 1 : 6 (d) 1 : 8 93. In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability 1 that he makes a guess is . The probability 3 1 that he copies is and the probability that 6 his answer is correct given that he copied it is 1 . The probability that he knew the answer 8 to the question given that he correctly answered it, is 24 1 (a) (b) 29 4 3 1 (c) (d) 4 2 94. Let S be a non-empty subset of R. Consider the following statement: P : There is a rational number x ∈ S such that x > 0. Which of the following statements is the negation of the statement P ? (a) There is a rational number x ∈ S such that x ≤ 0. (b) There is no rational number x ∈ S such that x ≤ 0. (c) Every rational number x ∈ S satisfies x ≤ 0. (d) x ∈ S and x ≤ 0 ⇒ x is not rational. 95.

Consider the following statements P : Suman is brilliant Q : Suman is rich R : Suman is honest The negation of the statement “Suman is brilliant and dishonest if and only if Suman is rich” can be expressed as (a) ~ Q ↔ ~ P ∧ R (b) ~ (P ∧ ~ R) ↔ Q (c) ~ P ∧ (Q ↔ ~ R) (d) ~ (Q ↔ (P ∧ ~ R))    96. If a , b and c are unit vectors, then       | a − b |2 + |b − c |2 + |c − a |2 does not exceed (a) 4 (b) 9 (c) 8 (d) 6

54

VITEEE CHAPTERWISE SOLUTIONS

97. The diagonals of a parallelogram are given   ^ ^ ^ ^ ^ ^ by d1 = 2 i + 3 j − 6 k and d2 = 3 i + 4 j − k then area is 1 (a) 1 50 sq. units (b) 1005 sq. units 2 2 1 (c) 1105 sq. units (d) None of these 2 98. Equation of the sphere for which the circle x2 + y2 + z2 + 7y – 2z + 2 = 0, on the plane 2x + 3y + 4z – 8 = 0 be great circle, must be (a) x2 + y2 + z2 + 4x + 2y – 6z + 10 = 0 (b) x2 + y2 + z2 – 4x + 2y – 6z + 10 = 0 (c) x2 + y2 + z2 – 6x – 4y – 2z – 10 = 0 (d) x2 + y2 + z2 – 2x + 4y – 6z + 10 = 0 99. The equations of the perpendicular from the origin to the 2x + 3y + 4z + 5 = 0 and x + 2y + 3z + 4 = 0 must be (a) x + 2y – z = 0 = 3x – 2y – z (b) 2x + y + z = 0 = x – 2y – z (c) x + 2y – z = 0 = 3x + 2y + z (d) x – 2y + z = 0 = 3x + 2y + z 100. Find the value of l if the following equations are consistent

x + y – 3 = 0, (1 + l)x + (2 + l)y – 8 = 0,



x – (1 + l)y + (2 + l) = 0.

−1 2 9 (c) 2 , 5 101. Evaluate (a) 0 ,

(b) 1,

−5 3

(d) 4, –3



−1/

 x4   2x  dx cos −1   4  1 + x 2  x 1 −   3

 −2  3 + 1 p (a) p  + log  +  3  3 − 1  6    (b)

 3 − 1 p p  −2 log   +  3  3  3 + 1  4 

 3 + 1 p p  −2 + log  (c)  +  2  3  3 − 1  6   (d) p  −2 log 

(c) k e–1

(d) – k e

103. If f(x) be a continuous function such that f(a – x) + f(x) = 0 for all x ∈ [0, a], then evaluate a

dx

∫ 1 + e f ( x) . 0

1 1 (a) − a (b) a 2 2 (c) 3a (d) 2a 104. If A1 is the area of the parabola y2 = 4ax lying between vertex and the latus rectum and A2 is the area between the latus rectum and A the double ordinate x = 2a, then 1 = A2 1 (a) 2 2 − 1 (b) ( 2 2 + 1) 7 1 (c) ( 2 2 − 1) (d) none of these 7 105. Solution of the differential equation tan y sec2x dx + tan x sec2y dy = 0 is tan x (a) (b) tan x tan y = K =K tan y (c) tanx + tany = K (d) tanx – tany = K 106. Which of the following functions is a solution of the differential equation? 2

1/ 3

∫

1

 n!  n 102. Evaluate lim   , where k (≠ 0) is a n →∞  ( kn)n    function and n ∈ N. 1 −1 1 −1 (a) e (b) − e k k

(

)

p 3 −1 +  6



 dy   dy   dx  − x  dx  + y = 0 ? (a) y = 2x2 – 4 (c) y = 2x

(b) y = 2x – 4 (d) y = 2

107. For two independent events A and B, 3 8 P( A ∩ B) = , P( A′ ∩ B) = , then P(B) = 25 25 3 7 (a) (b) 11 25 11 (c) (d) none of these 25 108. If the line ax + by + c = 0 is a normal to the curve xy = 1, then (a) a > 0, b > 0 (b) a > 0, b < 0 (c) a < 0, b < 0 (d) none of these

55

Model Test Paper-3

 3 1    2  , A =  1 1 and Q = PAPT 109. If P =  2    1 3 0 1 −   2 2  and x = PTQ2005P, then x is equal to  1 2005 (a)   1  0  4 + 2005 3 (b)   2005

  4 − 2005 3  6015

1 2 + 3 (c) 4   −1

  2 − 3 

1  2005  4  2 + 3

2 − 3  2005 

(d)

111. If Aij is the cofactor 2 −3 determinant 6 0 1 5 value of a32·A32. (a) 200 (c) 110

of the element aij of the 5 4 ,then write the −7 (b) 150 (d) 90

112. A balloon, which always remains spherical on inflation is being inflated by pumping in 900 cm3/s of gas. Find the rate at which the radius of the balloon increases when the radius is 15 cm. (a) 11 cm/s (b) 2p cm/s (c) 1/p cm/s (d) p2 cm/s 113. Find the point on the curve y = x3 – 11x + 5, at which the tangent is y = x – 11. (a) (4, –7) (b) (0, 3) (c) (–2, –13) (d) (2, –9) 114. Evaluate (a) 0 (c) –1

∫ 0



 4 + 3 sin x  log  dx.  4 + 3 cos x  (b) –2 (d) 2

 dy  log   = 3 x + 4 y  dx  e −4 y = e3x + C − 4 (c) e –4y = e –3x + C

1

x a b    110. The factors of  a x b  are  a b x  (a) x – a, x – b and x + a + b. (b) x + a, x + b and x + a + b. (c) x + a, x + b and x – a – b. (d) x – a, x – b and x – a – b.

p/ 2

115. Sketch the region lying in the first quadrant and bounded by y = 9x2, x = 0, y = 1 and y = 4. Find the area of region using integration. 5 (a) sq. units (b) 10 sq. units 3 14 (c) sq. units (d) 9 sq. units 9 116. Solve the following differential equation

(a)

e −4 y e 3 x = +C 3 −4 (d) e 4y = e –3x + C

(b)

117. Let p be real and |p|≥ 2. If A, B and C are variable angles such that

p 2 − 4 tan A + p tan B + p 2 + 4 tan C = 6 p

then the minimum value of tan2A + tan2B + tan2C is (a) 8 (b) 12 (c) 18 (d) 6   118. Let a and b be two non-collinear unit vectors.          If a = a − ( a ⋅ b )b and β = a × b , then |β|is     (a) |a (b) |a | + |a ⋅ a | |        (c) |a | + |a ⋅ b | (d) |a | + a ⋅ ( a + b ) 119. The odds in favour of a book reviewed by three independent critics are, respectively, 5 : 2, 4 : 3 and 3 : 4. The probability that majority of the critics give favourable remark is 210 209 (a) (b) 343 343 211 205 (c) (d) 343 343 120. Bag A contains 5 white and 3 black balls. Bag B is empty. Four balls are taken at random from A and transferred to empty bag B. From B, a ball is drawn at random and is found to be black. Then, the probability that among the transferred balls three are black and one is white is 1 7 (a) (b) 8 8 6 1 (c) (d) 7 7

56

VITEEE CHAPTERWISE SOLUTIONS

Answer Key 1.

(d)

2.

(b)

3.

(b)

4.

9.

(d)

10. (b)

11. (d)

17. (b)

18. (d)

25. (d)

26. (a)

33 (b)

(d)

5.

(a)

6.

(a)

7.

(c)

8.

(d)

12. (c)

13.

(c)

14.

(c)

15.

(c)

16.

(a)

19. (d)

20. (b)

21.

(c)

22.

(b)

23.

(d)

24.

(b)

27. (b)

28. (c)

29.

(c)

30.

(a)

31.

(b)

32.

(a)

34. (b)

35. (d)

36. (a)

37.

(d)

38.

(a)

39.

(a)

40.

(b)

41. (d)

42. (b)

43. (d)

44. (c)

45.

(d)

46.

(c)

47.

(a)

48.

(b)

49. (b)

50. (b)

51. (d)

52. (b)

53.

(a)

54.

(c)

55.

(b)

56.

(d)

57. (c)

58. (c)

59. (d)

60. (b)

61.

(b)

62.

(d)

63.

(c)

64.

(c)

65. (d)

66. (a)

67. (c)

68. (b)

69.

(d)

70.

(a)

71.

(a)

72.

(b)

73. (c)

74. (d)

75. (a)

76. (a)

77.

(a)

78.

(c)

79.

(c)

80.

(c)

81. (a)

82. (a)

83. (a)

84. (b)

85.

(b)

86.

(b)

87.

(d)

88.

(c)

89. (c)

90. (a)

91. (c)

92. (c)

93.

(a)

94.

(c)

95.

(d)

96.

(b)

97. (d)

98. (d)

99. (d)

100. (b)

101. (c)

102. (a)

103. (b)

104. (b)

105. (b)

106. (b)

107. (c)

108. (b)

109. (a)

110. (a)

111. (c)

112. (c)

113. (d)

114. (a)

115. (c)

116. (b)

117. (b)

118. (a,c) 119. (c)

120. (d)

e planations

Physics 1 q2 4 pe 0 d 2

1.

(d) : Initially, F =

...(i)



when the third equal conductor touches B, the charge of B is shared equally between them.



q \ Charge on B = = charge on third 2 conductor

Now this third conductor with charge    2



 q touches C, their total charge  q +  is equally  2

shared between them,



3q on C = = charge 4

\ Charge on third conductor \ New force between B and C

2.



 1 1 −   r2 r1 

= kq1q2 

= 9 × 109 × 5 × 10–9 × 3 × 10–9

  1 1 × −  5 × 10 −2 6 × 10 −2 

 q







Initial distance between charges (r1) = 6 × 10–2 m Final distance between charges (r2) = 5 × 10–2 m In moving the charge, Work done = Final P.E. – Initial P.E.

F′ =

1  q 3q  3  ×  = F (using (i)) 4 pe 0 d 2  2 4  8

(b) : qA = +5 ×

10–19

C, qB = 3 ×

10–9C



= 135 × 10 −9 ×

1 30 × 10 −2

= 4.5 × 10–7 J

3.

(b) : Here, l0 = 200 nm, l = 100 nm, hc = 1240 eV nm



Maximum kinetic energy =



=

hc hc (in eV) − l l0

 1 1  hc  1 1  − = 1240  = 6.2 eV −  100 200  e  l l 0 

57

Model Test Paper-3

4.

(d) : Energy of photon emitted,



 1 24 1 E = 13.6  2 − 2  eV = 13.6 × eV = 13.06 eV 1  25 5 E



Momentum of photon =



The momentum of hydrogen atom is equal and opposite to the momentum of photon. If m is the mass of hydrogen atom ( = 1.67 × 10–27 kg) and v is recoil speed of hydrogen atom, then



mv =

5.

v = 4.17 m s–1 ≈ 4 m s–1 (a) : Given D15 = 5.9 × 10–3 m, D5 = 3.36 × 10–3 m p = 10 and R = 1.0 m Wavelength of light, l=

4 pR

[( 5.9)2 − ( 3.36)2 ] × 10 −6 = 4 × 10 × 1.0

6.

l = 5880Å    (a) : Torque, t = p × E



\ or



= 4 × 10 or p = 2 × 10 5 × sin 30° Dipole moment, p = ql

7.

\

t = pEsinq 4 = p × 2 × 105 sin30° 4

q=

−5

Cm

p 4 × 10 −5 = = 2 × 10 −3 C = 2 mC l 0.02

(c ) : After charging, total charge on the capacitor q = CV (where C = 10 mF) \ q = 10 × 10–6 × 1000 = 10–2 C When this charged capacitor is connected to uncharged capacitor then total charge remains same. \ q = q1 + q2 10–2 = (C1 + C2)V 8. 9.

\

V=

10 −2

16 × 10 −6

= 625 V

(d) (d) : Radiation can be a hazard to living organisms because it produces ionization along its path of travel. This ionization can disrupt chemical bonds in essential macromolecules such as DNA and produce molecular fragments, which are free polyatomic ions that can interfere with enzyme action and other essential cell

3V

4Ω

I2 I1 + I 2

A

c

E c E 13.06 × 1.6 × 10 −19 v= = mc 1.67 × 10 −27 × 3 × 108

D(2n + p ) − Dn2

functions. 4 Ω I1 9 V 1 Ω 10. (b ) :

I1 B

2Ω

VA – VB = 16 V \ 4I1 + 2(I1 + I2) – 3 + 4I1 = 16 ...(i) Using Kirchhoff’s second law in the closed loop, we have 9 – I2 – 2(I1 + I2) = 0 ...(ii) Solving equations (i) and (ii), we get I1 = 1.5 A and I2 = 2 A \ Current through 2 W resistor = 2 + 1.5 = 3.5 A 11. (d ) : Current in the circuit will be I =

( n − 4)e . nr

Hence, potential difference across A or B is V = e + Ir = e +

 ( n − 4)e 2 r = 2e  1 −   nr n

12. (c ) : Since dl and r for each element of the straight segments are either parallel or antiarallel. Therefore

   dl × r = 0

Hence, B due to straight segment is also zero.

13. (c ) : dB =

m 0 Idl sin q 4p r 2

Here, dl = Dx = 0.05 m, I = 10 A, r = 1 m and sinq = sin90° = 1, \

dB = 10 −7 ×

10 × 0.05 × 1 2

(1) = 0.50 × 10 –7 = 5.0 × 10 –8 T 14. (c ) : Suppose intensity of unpolarised light = 100. \ Intensity of polarised light from first nicol prism I 1 = 0 = × 100 = 50 2 2 According to law of Malus, 2

1 I = I0cos2 q = 50(cos60°)2 = 50 ×   = 12.5 2 m 2 pI 15. (c ) : B = 0



4p r 2 e 2 × 1.6 × 10 −19 Here, I = = = 1.6 × 10 −19 A t 2 −19 m I m × 1.6 × 10 \ B= 0 = 0 2r 2 × 0.8



= m0 × 10–19 T

58

VITEEE CHAPTERWISE SOLUTIONS

16. (a ) : Magnetic dipole moment of circular loop is m = NIA = 10 × 0.5 × 2 × 10–4 = 10–3 A m2 Magnetic field inside the solenoid carrying current B = m0nI = 4p × 10–7 × 103 × 3 = 12p × 10–4 T Torque, t = m B sin q = 10–3 × 12p × 10–4 × sin 90° = 12p × 10–7 N m 17. (b) : x2 + y2 = 25 \ r = 5 m

 r=

mv qB

or 5 =

5 × 10 −3 × 5

5 × 10 −3 × 5 5 × 10 −6 × B



\ B=



The magnetic field will be 1 kT along z-axis.

5 × 10 −6 × 5

18. (d) : l air =





= 103 T = 1 kT

1 mm 2000

= 5 × 10–4 mm = 5 × 10–7 m = 5000 Å l medium =

l air 5000 Å = = 4000 Å 1.25 m

19. (d) : Here, B = 2t

\

dB =2 dt dφ dB =A = 2 pa 2 dt dt

Induced e.m.f., | e | =



Flow of charge per unit time through any section of the coil = induced current,





e 2 pa 2 = = constant. R R



Also, power generated, P = I2R = constant Total charge passed through any section between t = 0 to t = 2 s is  2 pa 2  4 pa 2 q = It =  ( 2 − 0) =  R  R 

 t  3l 2 t = IB × or l = 2  4  3BI 

fo =9 fe

…(i)

where fo and fe are the focal lengths of the objective and eyepiece respectively Also, fo + fe = 20 cm …(ii) On solving (i) and (ii), we get fo = 1 8 cm, fe = 2 cm

22. (b) : Magnetic potential at any point is the amount of work done in bringing a unit north pole from infinity to that point. At any point on the right bisector, the potentials due to the two poles are equal and opposite. 23. (d) : Impedance of the LR circuit

Z = R 2 + 4p 2 u2 L2 = 12 + 4 p 2 ( 50)2 2 2 = 394385 ≈ 628



I=

V 5 = A Z 628

Energy stored in the inductor, 2



U=

 5  1 2 1 LI = × 2 ×  J = 6.33 × 10 −5 J 2 2  628 







R1 1 or R2 = 2 R1 = R2 2

The equivalent combination

resistance

of

parallel

1 1 1 3 1 1 1 1 1 = + + = + = + + R R1 R2 R3 R1 2 R1 R3 2 R1 R3 1 1 3 3 or = − =1− 2 R1 R3 1 2 R1 or 1 = R3 −

3 R3 3 R3 or R3 = 1 + 2 R1 2 R1

Since no resistance is in fraction, therefore minimum value of R3 2 = R1 3 3 2 \ R3 = 1 + × = 2 W and R1 = 3 W 2 3



1 1 3 2 A = × Base × Height = × l × l sin 60° = l 2 2 4 \





20. (b) : Normal to the plane of the coil will be  perpendicular to the field B. \ t = IBAsin90° = IBA Area of equilateral triangle,



24. (b) : Here ,



I=

21. (c) : Magnifying power, m =

1/ 2

The maximum resistance value is R2 = 2R1 = 2 × 3 = 6 W

25. (d) : The bridge ABCD is balanced if

10 30 = or R1 = 3 W R1 9

When this bridge is balanced, no current flows in arm BD. Therefore, R2 can have any finite value.

59

Model Test Paper-3

26. (a) : Let fo and fe be the focal lengths of the objective and eyepiece respectively. For normal adjustment, distance between the objective and the eyepiece (tube length) = fo + fe. Treating the line on the objective as the object, and the eyepiece as the lens, u = – (f0 + fe) and f = fe.

1 1 1 − = v −( f o + f e ) f e

fo 1 1 1 = − = v fe fo + fe ( fo + fe ) fe (f + f )f or v = o e e . fo

or

Magnification =

\

f image size l . v = e = = u f o object size L

fo L = = magnification of telescope in fe l

VG − 8 V + 3 V − 4 V + 2 V = VH VG − VH = 7 V

29. (c) : Here, e = 1.5 V, r = 2 W, Req = 5 W + 8 W, R =5W The potential difference across 5W resistance,



30. (a) : Frequency of full wave rectifier = 2 × input frequency = 2 × 50 = 100 Hz. 31. (b) : According to Gauss’s law, the electric flux through the sphere is φ=

normal adjustment.

27. (b) : Total initial resistance = G + R = 50 W + 2950 W = 3000 W Current , I =

3V = 1 × 10 −3 A = 1 mA 3000 W

If the deflection has to be reduced to 20 divisions, then current 1 mA

2

I ′ = 30 × 20 = 3 mA Let x be the effective resistance of the circuit, then 3 V = 3000 W × 1 mA = x W × or x = 3000 × 1 ×

2 mA 3

3 = 4500 W 2

\ Resistance to be added = (4500 W – 50 W) = 4450 W 28. (c) : The current distribution in a circuit is as shown in the figure.

 e    1.5 V V = R=  5 W = 0.5 V  Req + r   ( 5 W + 8 W) + 2 W   

qin 8.85 × 10 −13 C = = 0.1 N C −1 m 2 e 0 8.85 × 10 −12 C 2 N −1 m −2

32. (a) : In the usual setting of deflection magnetometer, field due to magnet (F) and horizontal component (H) of earth’s field are perpendicular to each other. Therefore, the net field on the magnetic needle is

F2 + H 2

\



I

T = 2p

2

M F +H

When the magnet is removed, I MH

T0 = 2 p

…(ii)

F = tanq H



Also,



Dividing (i) by (ii), we get



T = T0 =

⇒



…(i)

2

H 2

F + H2



H 2

2

H tan q + H T2 T02

= cosq

2

=

H H sec 2 q

= cosq

\ T2 = T20 cosq

Let VG and VH be the potentials at points G and H respectively.

33. (b) : The most stable particle in the baryon group is proton.

\ VG − ( 2 A)( 4 W) + 3 V − ( 2 A)( 2 W)

34. (b) : The phenomenon of polarization shows that light has transverse nature.

+ ( 2 A)(1 W) = VH

60

VITEEE CHAPTERWISE SOLUTIONS

35. (d) : Clearly, 2f = 20 cm or f = 10 cm Now, u = –15 cm, f = 10 cm Using lens formula, 1

1

1

1

1 1 1 − = v −15 10 1

1

or v + 15 = 10 or v = 10 − 15 or

1 3−2 1 = = or v = 30 cm v 30 30

The change in image distance is (30 – 20) cm i.e., 10 cm away from the lens.

40. (b) : As, I = (e1000 V/T – 1) mA ...(i) Here, I = 5 mA at T = 300 K dV = 0.01 V \ 5 = (e1000 V/T – 1) ⇒ e(1000 V/T) = 6 mA. Differentiating eqn. (i), we get  1000  (1000 V /T ) dI =  e dV  T  =

1000 (6)(0.01) = 0.2 mA. 300

36. (a) : As for minima, ax ax  x or l = nl = asinq =  sin q =  f nf  f

Here, a = 0.3 mm = 0.3 × 10–3 m, x = 5 mm = 5 × 10–3 m, n = 3, f = 1 m. 0.3 × 10 −3 × 5 × 10 −3 = 5 × 10 −7 m = 5000 Å. 3×1 hc 1 37. (d) : From mv 2 = − φ0 (in eV) el 2 −34 6.6 × 10 × 3 × 108 1 \ mv 2 = − 4.2 = 6.2 − 4.2 2 2000 × 10 −10 × 1.6 × 10 −19 \ l=

= 2 eV = 2 × 1.6 × 10–19 J

⇒

v= =



2 × 2 × 1.6 × 10 −19 9.1 × 10 −31 6.4 × 106 m s −1 = 0.84 × 106 m s −1. 9.1

= 8.4 × 105 m s–1

38. (a) : Here, T1/2 = 4.47 × 109 yr, N = 60 N0

100 N  1 60  1  10 n = ⇒ = or 2 = N0  2  100  2  6 n



n

⇒ n log 2 = log 10 – log 6 = 1 – 0.778 = 0.222 \

n=

Chemistry 41. (d) : For a zero order reaction, A P r= −

d[ A] = k  or –d[A] = kdt dt

When t = 0, [A] = [A]0 At t = t, [A] = [A] Hence,

[ A] [ A]0

t

− d[ A] = ∫ kdt 0

or [A]0 – [A] = kt or [A] = [A]0 – kt Thus, plot of [A] vs t is linear with negative slope, k and intercept [A]0. 42. (b) : Aldehydes reduce Benedict’s solution. 43. (d) : Proline is a secondary amine. 44. (c) : Positively charged ligands have suffix ‘ium’. 45. (d) :

CH3

CH3 C ONa + CH3Cl CH3

CH3 CH3 C O CH3 + NaCl

0.222 0.222 = = 0.74 log 2 0.3

Now, t = nT1/2 = 0.74 × 4.47 × 109 yr = 3.3 × 109 yr. 39. (a) : Here, A = 5°, m = 1.5, i = q, e = 0° As the emergent ray is normal to the refracting surface of the prism Hence, for a small angled prism, d = (m – 1)A, d = (1.5 – 1)5° = 2.5° Since, A + d = i + e, ⇒ 5° + 2.5° = q + 0° or q = 7.5°

∫

CH3

whereas,

Methyl t-butyl ether

CH3 CH3 C Cl + CH3ONa

CH3

CH3 CH3 C CH2 + Alkene

CH3OH + NaCl Secondary and tertiary alkyl halides readily undergo elimination reaction rather than substitution.

61

Model Test Paper-3

46. (c) : For the reaction, 1 A → 2 B 2 Rate = − ⇒

−

1 d[ A] 1 d[ B] = 1 / 2 dt 2 dt

\ Keeping [A] constant, [B] is doubled, rate becomes 8 times. Dividing eq. (iii) by eq. (i), we get x

0.10  0.024  =  ⇒ 2x = 1 ⇒ x = 0 0.10  0.012 



d[ A] 1 d[ B] = dt 4 dt

47. (a) : Nitrogen forms most basic compound with hydrogen among group 15 elements. 48. (b) : Ferric chloride, FeCl3(i.e., Fe3+) cannot be further oxidised.

\ Keeping [B] constant, [A] is doubled, rate remains unaffected. Hence, rate is independent of [A]. Rate ∝ [B]3 55. (b) :

49. (b) : Packing efficiency of ccp or fcc = 74% Packing efficiency of bcc = 68% Packing efficiency of simple cubic = 52.4% 50. (b) : Boiling point of alcohols is higher than ethers due to H-bonding. In alcohols, the boiling point decreases with branching due to decrease in surface area. Hence, n-pentanol will have highest boiling point. 51. (d) : According to Debye Hückel-Onsager equation, lC = l∞ – (B) C B is a constant depending upon the type of the electrolyte, the nature of the solvent and the temperature. 52. (b) : KMnO 4 in acidic medium oxidises 1° alcohol to acid. + RCH2OH KMnO4/H RCOOH 53. (a) : Acidic strength of oxoacids of a particular halogen atom increases with increase in oxidation number thus, the order of acidic strength is

+1

+3

+5

+ (CH3)2NH

(i) H+ (ii) dehydration Enamine

56. (d) : Element

%

Relative no. Simple of atoms ratio

C

76.6

76.6 = 6.38 12

6

H

6.38

6.38 = 6.38 1

6

O

17.02

17.02 = 1.06 16

1

\ Empirical formula (P) = C6H6O Empirical formula weight = 94 Molecular weight = 2 × VD = 2 × 47 = 94 \ Molecular formula of P is C6H6O. Since P gives colour with aq. FeCl3 it has a phenolic group. Compound P should be C6H5OH (phenol).

+7

HClO < HClO 2 < HClO3 < HClO 4 .

54. (c) : A + B → C Let rate = k[A]x [B]y where order of reaction is (x + y) Putting the values of exp. 1, 2, and 3, we get following equations: 0.10 = k [0.012]x [0.035]y ...(i) 0.80 = k [0.024]x [0.070]y ...(ii) 0.10 = k [0.024]x [0.035]y ...(iii) Dividing eq. (ii) by eq. (iii), we get

N(CH3)2

O

y

0.80  0.070  y =  ⇒2 =8 ⇒ y=3 0.10  0.035 

Aspirin (analgesic)

Salicylic acid

57. (c) 58. (c) : Ag2CrO4  2Ag+ + CrO42– It is given that [2Ag+] = 1.5 × 10–4 mol L–1 \ [Ag+] = 0.75 × 10 – 4 mol L–1 [Ag+] = [CrO42–] = 0.75 × 10–4 mol L–1 Ksp = [2Ag+]2 [CrO 24– ] = (1.5 × 10–4)2 (0.75 × 10–4) = 1.6875 × 10–12

62

VITEEE CHAPTERWISE SOLUTIONS

59. (d) : Corrosion is an electrochemical process in which Fe2+ are oxidised to Fe3+. Steps involved in the formation of rust are : Fe → Fe2+ + 2e– (Anode) O2 + 4H+ + 4e– → 2H2O (Cathode) 4Fe2+ + O2 + 4H2O → 2Fe2O3 + 8H+ Fe2O3 + xH2O → Fe2O3 ⋅ xH2O

(Rust)

65. (d) : Dimethylamine is more basic than methylamine.   2AB2(g)    2AB(g) + B2(g)

66. (a) : Initially

1

At equilibrium (1 – x)

60. (b) :

0

0

x

x/2

Total no. of moles at equilibrium

NH2 + CHCl3 + 3KOH (alc.) Aniline

N





C + 3KCl + 3H 2O

61. (b) : Due to lanthanoid contraction, atomic and ionic radii of Zr and Hf are almost equal. 62. (d) : OCH3

OCH3 (CH3)3CCl

Cl2/FeCl3

Anhyd. AlCl3

= (1 − x ) + x +

OCH3 group is a stronger o, pC(CH3)3 directing group than C(CH3)3

2 ×p pAB B 2 pAB

2

2

  x 2   2   x  2 + x × P ×  2 + x × P        2  2 Px 3 = = 2 ( 2 + x )(1 − x )2  1− x  ×P 2+x    2

Since x Z > Y. 72. (b) :

Initial moles.

    X   2Y ; Z   Q+R 1

At equilibrium 1 – a

1 − a2 = 1 ⇒ 9 a 2 P2



KOH (alc.), 

0

1

0

0

2a

1–a

a

a

4 P1 1 = P2 9

⇒

P1 1 = P2 36

73. (c) : Hydroboration-oxidation gives antiMarkownikoff’s product.

74. (d) : 3HClO(aq) → HClO3(aq) + 2HCl(aq) It is a disproportionation reaction of hypochlorous acid where the oxidation number of Cl changes from +1 (in ClO–) to +5 – (in ClO3–) and –1 (in Cl ). 75. (a) :

1 = R  1 − 1  = 109677  1 − 1  cm −1  2 2  22 42  l  n1 n2 

l=

= 20564.4 cm–1

1 = 486 × 10−7 cm 20564.4 cm −1

or 486 × 10–9 m = 486 nm Colour corresponding to this wavelength is blue.

64

VITEEE CHAPTERWISE SOLUTIONS

76. (a) : The only optically active molecule is

Mathematics 81 (a) : Given, x1 = y1

CH2

CH—CH2—CH 3 (B)

\

x1 = 9 − 2 x12

⇒

x12 = 9 − 2 x12

Since, y1 > 0, therefore, the point is ( 3 , 3 ).

Now, y = 9 − 2 x 2 On differentiating w.r.t.x, we get dy 2x dy 2y = −4 x ⇒ =− dx dx y

\

79. (c) : Due to resonance, C—N bond acquires some double bond character.

O–

O C

.. NH

+

C

NH

As a result C—N bond length in proteins becomes smaller than usual C—N bond length. 80. (c) : In Frenkel defect, ions get displaced from their original position and move to interstitial sites. Hence, there is no change in the density of the crystal.

 dy   dx  (

= −2 3 , 3)

So, the required equation of tangent is



77. (a) : La(OH)3 is more basic than Lu(OH)3. 78. (c) : In Victor Meyer’s test, blue colour salt is given by secondary alcohols. It means compound B must be a secondary alcohol. Only aldehydes with Grignard reagent yield secondary alcohols. Aldehyde should have one carbon less than the secondary alcohol since Grignard reagent contains methyl group. These conditions are satisfied by acetaldehyde and iso-propyl alcohol.

⇒ x1 = ± 3.

( y − 3 ) = −2( x − 3 )

⇒ 2x + y − 3 3 = 0 82. (a) : Consider the polynomial

f ( x ) = a0 x +

a1x 2 a2 x 3 x n +1 + + ... + an 2 3 n+1

...(i)

We have, f(0) = 0 an a1 a2 =0 and f (1) = a0 + + + ... + 2 3 n+1

Therefore, 0 and 1 are roots of f(x). Hence, by algebraic interpretation of Rolle’s theorem, f ′(x) = 0 i.e., a0 + a1x + a2x2 + .... + anxn = 0 must have a root lying between 0 and 1. 83. (a) : Firstly, break the number 5.001 as x = 5 and Dx = 0.001 and use the relation f(x + Dx)  f(x) + Dx f ′(x). Consider f(x) = x3 – 7x2 + 15 ⇒ f ′(x) = 3x2 – 14x Let x = 5 and Dx = 0.001 Also, f(x + Dx)  f(x) + Dx f′(x) Therefore, f(x + Dx)  (x3 – 7x2 + 15) + Dx(3x2 – 14x) ⇒ f(5.001)  (53 – 7 × 52 + 15) + (3 × 52 – 14 × 5)(0.001) (as x = 5, Dx = 0.001) = 125 – 175 + 15 + (75 – 70)(0.001) = –35 + (5)(0.001) = –35 + 0.005 = –34.995 84. (b) : f′(x) = (ab – b2 – 2) + cos4x + sin4x < 0 = ab – b2 – 2 + (sin2x + cos2x)2 – 2sin2xcos2x < 0

65

Model Test Paper-3



⇒

1 1 ab − b 2 − 1 <   sin 2 2 x < 2 2

⇒ 2ab – 2b2 – 2 < 1 ⇒ 2b2 – 2ab + 3 > 0 \ (–2a)2 – 4.2.3 < 0

[ b2 – 4ac < 0]

⇒ a2 < 6 ⇒ − 6 < a < 6 ⇒ a ∈( − 6 , 6 )

85. (b) : Any tangent to the ellipse is x y cos t + sin t = 1, where the point of contact 4 3 is (4cost, 3sint) x y or + =1 4 sec t 3 cos ec t It means the axes Q (4 sect, 0) and R(0, 3cosect). \ The distance of the line segment QR is QR2 = D = 16 sec 2t + 9cosec 2t So, the minimum value of D is (4 + 3)2 or QR = 7.

86. (b) : The line joining (4, 3) and (2, 1) is also along a diameter. So, the centre of the circle is the intersection of the diameters 2x – y = 2 3−1 (x – 4). 4−2 On solving these two equations, the coordinates of centre of the circle are (1, 0). Also, the radius of circle = the distance and y – 3 =

between (1, 0) and ( 2 , 1) = 2 . Equation of circle is (x – 1)2 + y2 = 2 ⇒ x2 + y2 – 2x – 1 = 0 87. (d) : Let P(h, k) be a point. Then, the chord of contact of tangents from P to y2 = 4ax is ky = 2a(x + h) ...(i) 2 This touches the parabola x = 4by. So, it should be of the form b x = my + m Eq. (i) can be rewritten as

...(ii)

 k  x=  y−h ...(iii)  2a  Since, Eqs. (ii) and (iii) represent the same line.

\

k m= 2a

b and = −h m Eliminating m from these two equations, we get 2ab = – hk Hence, the locus of P(h, k) is xy = –2ab, which is a hyperbola.

88.

(c) : r2 – r – 6 > 0 and r2 – 6r + 5 > 0 ⇒ (r – 3)(r + 2) > 0 and (r – 1)(r – 5) > 0 ⇒ {r < –2 or r > 3} ∩ {r < 1 or r > 5} ⇒ r < –2 or r > 5

89. (c) : Equation of hyperbola is 2x2 + 5xy + 2y2 – 11x – 7y – 4 = 0 ...(i) Let asymptotes of the hyperbola (i) be 2x2 + 5xy + 2y2 – 11x – 7y + c = 0 ...(ii) Since, Eq (ii) represents a pair of lines 2 5 / 2 −11 / 2 \ 5/2 2 −7 / 2 = 0 −11 / 2 −7 / 2 c ⇒

⇒

\

−

11  35  7 55   25  − + 11 +  −7 +  + c  4 −  = 0  2 2  4 4   4 

9 99 189 90 = ⇒c=5 c=− + 4 8 8 8 Equation of the required asymptotes is 2x2 + 5xy + 2y2 – 11x – 7y + 5 = 0

 1 + x2 − 1  90. (a) : Let y = tan −1     x and t = cos −1



Put x = tanq



\





\

2 1 + x2

 sec q − 1   1 − cos q  y = tan −1  = tan −1   tan q   sin q  q q  = tan −1  tan  =  2 2

−1 and t = cos



1 + 1 + x2

1 + sec q 1 + cos q = cos −1 2 sec q 2

q q  = cos −1  cos  =  2 2 y=

q =t ⇒ 2

dy =1 dt

66

VITEEE CHAPTERWISE SOLUTIONS

91. (c) : We have P( A) =

Required probability = P( A)P(B ) + P( A)P(B) + P( A)P(B) =

1 3   1  3 1 3 17 1 −  + 1 −  ⋅ + ⋅ =  5 5  5  5 5 5 25

92. (c) : We have, m =



94 (c) : The given statement is P : at least one rational x ∈ S such that x > 0. The negation would be : There is no rational number x ∈ S such that x > 0 which is equivalent to all rational numbers x ∈ S satisfy x ≤ 0.

1 3 , P(B) = 5 5

As P(X = r ) =

95. (d) : The statement can be written as P ∧ ~R ⇔ Q Thus the negation is ~ (Q ↔ P ∧ ~ R)

1 2

e − mmr r!

3

1 1 e −1/ 2   e −1/ 2     2 P(X = 3) 2 = ÷ P(X = 2) 3! 2!

\

e

−

=

1 2

2

3

1  2  × 3!

2! e

−

1 2

1  2 

2

=1:6 Answer is correct

93. (a) :

96. (b) : Given expression       6 − 2( a ⋅ b + b ⋅ c + c ⋅ a )    As |a + b + c | ≥ 0       ⇒ 3 + 2( a ⋅ b + b ⋅ c + c ⋅ a ) ≥ 0       ⇒ − 3 − 2( a ⋅ b + b ⋅ c + c ⋅ a ) ≤ 0       ⇒ 6 − 2( a ⋅ b + b ⋅ c + c ⋅ a ) ≤ 9 1   1 97. (d) : | d1 × d2 | = 698 2 2 ⇒ (d) is correct. 98.

Guesses 1 3

P(G) 





Knows Copies 1 1 P(C )  2 6

P( K ) 

  Knows P  Answer correct   Answer correct  P(Knows) × P   Knows  =  Answer correcct  P(Guesses) × P   +  Guesses





 Answer correct  P(Knows) × P   +  Knows  Answer correct  P(Copies) × P   Copies  1 ×1 24 2 = = 1 1 29 1 1 1 × + ×1+ × 6 8 3 4 2



(d) : Let the required sphere x2 + y2 + z2 + 7y – 2z + 2 + l(2x + 3y + 4z – 8) = 0 whose centre must lie on the plane 2x + 3y + 4z – 8 = 0   1 Hence centre :  − l , ( −7 − 3l), (1 − 2 l)    2 ⇒

3 2( − l) + ( −7 − 3 l) + 4(1 − 2 l) = 8 2

⇒ l = –1 Hence, x2 + y2 + z2 – 2x + 4y – 6z + 10 = 0

99. (d) : The plane passing through the origin and containing the given planes is (2x + 3y + 4z + 5) + k (x + 2y + 3z + 4) = 0 and The value of k can be obtained by putting x=y=z=0 5 We get k = − ⇒ 3x + 2 y + z = 0 4

Let P is any point on the line of intersection of given plane. Let D.R’s of the line of intersection l, m, n which are related to the D.R’s of normals of the planes. ⇒ 2l + 3m + 4n = 0 and l + 2m + 3n = 0 l m n ⇒ = = 1 −2 1

67

Model Test Paper-3



Hence, 1(x – 0) – 2(y – 0) + 1(z – 0) = 0 i.e., x – 2y + z = 0 So, x – 2y + z = 0 = 3x + 2y + z is correct.

100. (b) : Here, there are 3 equations in 2 unknowns. The equations are consistent if D = 0 1 1 −3 i.e., if 1 + l 2 + l −8 = 0 1 −1 − l 2 + l (Operating C2 – C1, C3 + 3C1) 1 0 0 if 1 + l 1 −5 + 3l = 0 1 −2 − l 5+l i.e., if (5 + l) + (2 + l)(–5 + 3l) = 0 −5 i.e., if 3l2 + 2l – 5 = 0 i.e., if l = 1, 3 1/ 3  4  x p −1  101. (c) : I = ∫  1 − x 4   2 − 2 tan x  dx −1/ 3



p = 2

1/ 3

= 2⋅





∫

−1/ 3

p 2

x4 1− x

1/ 3

∫

dx − 2 4

x4



1

r

lim log   n 1 n→∞ n r∑ =1 = e k

=

1

1  1 2 3 n n = lim  ⋅ ⋅ ...  n →∞ k  n n n n 

1 0 ⋅e k

log x dx

=

1 1 1  x log x  − ∫ x⋅ dx 0 x 1 0 e

k

I=∫

− f ( x) 0 1+ e a f ( x) 0

\

dx e

e

f ( x)

+1

a

2I = ∫

dx

dx = ∫ 1dx = a f ( x) 0

1 I = a. 2

104. (b) : A1 = 2 ∫ 4 axdx = 0



a

1 + e f ( x)

0 1+ e

⇒

(by given condition)

Adding, we get

a

1

∫

⇒



 3  1 1+ x 1 = p  − x + log e + tan −1 x  2 1− x 2 0   3 + 1 p p 2 = − + log  +  2  3  3 − 1  6 

1



=∫

−1/

∫

a

a





 x4  tan −1 x dx  4 x − 1   3

0 1− x [ first integrand is even function and second integrand is an odd function of x ] 1 1 1/ 3 = p ∫ −1 + 2 2 + 2 2 dx 1− x 1+ x 0

1 n

a

dx dx 103. (b) : I = ∫ =∫ f ( x) f ( a− x) + e + e 1 1 0 0

1/ 3

dx − 0 4

n 102. (a) : lim  n !  n n n →∞  k n 



      log x   ∞    lim( x log x ) = lim   1   ∞   x→0 x→0     x  1   = e −1   1 k     = lim x = lim( − x ) = 0   1 x→0 x→0 − 2   x  

2a

a

0

0

8 a2 . 3

A2 = 2 ∫ 4 ax − ∫ 4 ax dx = \

16 8 a2 2 a2 − 3 3

A1 1 2 2 +1 = = . 7 A2 2 2 − 1

105. (b) : Given differential equation is

sec 2 y dy sec 2 x dx + =0 tan x tan y

Integrating, log|tanx| + log|tany| = C ⇒ tan x tany = ± eC = K. dy 106 (b) : Clearly, y = 2x – 4 gives = 2 and these dx two equations together satisfy the given differential equation. ( 22 – x(2) + 2x – 4 = 0 is true for all x) 107. (c) : Since A and B are independent, P(A ∩ B) = P(A)P(B) and P(A′ ∩ B) = P(A′)P(B) 3 8 ⇒ P( A)P(B) = and (1 − P( A))P(B) = 25 25

⇒

P(B) =

11 . 25

68

VITEEE CHAPTERWISE SOLUTIONS

dy 1 1 108. (b) : y = x ⇒ dx = − 2 x \ Slope of normal is x2.

Slope of line ax + by + c = 0 is – \

a a a x2 = − ⇒ − > 0 ⇒ < 0 b b b



a b

dV dr = 4 pr 2 dt dt On substituting the given values, we get dr 900 = 4 p × (152 ) × dt ⇒

109. (a) : PTP = I Q = PAPT so that



x = PTQ2005P = PT(PAPT)2005P = PTPAPT(PAPT)2004P



=A

2005



 1 2005 = . 1  0

110. (a) : Operate R2 – R1 and R3 – R1, we get x a b a−x x−a 0 a−x b−a x−b



x a b = ( x − a) −1 1 0 a−x b−a x−b

Operate C1 + C2 + C3



x+a+b a b = ( x − a) 0 1 0 0 b−a x−b = (x – a)(x + a + b)(x – b).

2 −3 5 a11 4 = a21 111. (c) : Let A = 6 0 1 5 −7 a31



a12 a22 a32

a13 a23 a33

Here, a32 = 5 Given that, Aij is the cofactor of the element aij of A. Then, A32 = ( − 1)3+ 2 \

4 3 pr 3

On differentiating both sides w.r.t. t, we get dV 4 dr = p × 3r 2 × dt 3 dt

\ a and b have opposite signs.



Volume of the balloon (sphere) is V =

2 5 = ( −1)5 (8 − 30) = − ( −22) = 22 6 4

a32·A32 = 5 × 22 = 110

112. (c) : Let r be the radius and V be volume of the balloon. dV Given, r = 15 cm, = 900 cm3/s dt

dr 900 1 = = cm/s dt 4 p × 225 p Hence, the radius of the balloon is increasing 1 at the rate of cm/s, when the radius is p 15 cm. \

113. (d) : Given curve is y = x3 – 11x + 5 ...(i) and given line is y = x – 11 ⇒ x – y – 11 = 0 ...(ii) On differentiating Eq. (i) both sides w.r.t. x, we get dy = 3x 2 − 11 dx \ Slope of tangent, m1 = 3x2 – 11 From Eq. (ii), Slope of tangent, m2 = 1  coefficient of x   m = − coefficient of y  Since, line of Eq. (ii) is the tangent to the curve. \ m1 = m2 ⇒ 3x2 – 11 = 1 ( slope of the curve = slope of the tangent) ⇒ 3x2 = 12 ⇒ x2 = 4 \ x = ± 2 When x = 2 Then, y = (2)3 – 11(2) + 5 = 8 – 22 + 5 = –9 When x = –2 Then, y = (–2)3 – 11(–2) + 5 = –8 + 22 + 5 = 19 \ Points are (2, –9) and (–2, 19). The point (2, –9) lies on the line, y = x – 11 but the point (–2, 19) does not lie on the line y = x – 11. Hence, the required point is (2, –9).

69

Model Test Paper-3

114. (a) : Let I =

p/ 2

∫ 0

=

p/ 2

∫



 4 + 3 sin x  log  dx  4 + 3 cos x 

log( 4 + 3 sin x )dx −

0

∫

log( 4 + 3 cos x )dx

0

m    log n = log m − log n  

=

p/ 2

∫ 0

p/ 2   p  log  4 + 3 sin  − x   dx − ∫ log( 4 + 3 cos x )dx 2   0

  

=

p/ 2

p/ 2

∫

a

∫ 0

 f ( x )dx = ∫ f ( a − x )dx   0

log( 4 + 3 cos x )dx −

0

\

a

p/ 2

∫

log( 4 + 3 cos x )dx

0

I = 0

115. (c) : Given curve is y = 9x2 1 1 ⇒ x2 = y ⇒ x = y 9 3

...(i)

Y D y=4 4 C 3 y = 9x2 2 1 Ay=1 B

X





O Y

It is a parabolic curve, which opens upwards. And it is symmetrical about Y-axis and passes through the origin. \ Required area of bounded region ABCDA is 4



X

= ∫ xdy = 1

4

1 y dy 3 ∫1 4

[from Eq. (i)]



1  y3/ 2  1 2 3/ 2 3/ 2 =   = × [( 4) − (1) ] 3  3 / 2  3 3



2 14 sq units = [( 2)3 − 1] = 9 9



Therefore, the required area is

1

14 sq units. 9

116. (b) : Given differential equation is  dy  log   = 3 x + 4 y  dx  dy \ = e3x + 4 y dx dy ⇒ = e3x ⋅ e4 y dx On separating the variables, we get dy = e 3 x dx 4y e On integrating both sides, we get

−4 y

dy = ∫ e 3 x dx



 e ax  ax  ∫ e dx =  a   which is the required solution. ⇒



∫e

−4 y

3x

e e = +C −4 3

117. (b) : Consider the vectors  a = p 2 − 4 iˆ + pjˆ + p 2 + 4 kˆ and  b = (tan A)iˆ + (tan B) ˆj + (tan C )kˆ So that   a ⋅ b = p 2 − 4 tan A + p tan B + p 2 + 4 tan C   Let q be the angle between a and b .   36 p 2 = ( a ⋅ b )2     =| a |2|b |2 cos 2 q ≤ | a |2|b |2 = (p2 – 4 + p2 + p2 + 4)(tan2 A + tan2B + tan2C) = 3p2(tan2 A + tan2 B + tan2 C) Hence, tan2 A + tan2 B + tan2 C ≥ 12 and is equal to 12 when q is either 0 or p.   118. (a,c) : Let q be the angle between a and b . Then     a ⋅ b = | a || b |cos q = cos q      Now |a |2 = | a − ( a ⋅ b )b |2       = | a |2 + | a ⋅ b )2 b |2 −2( a ⋅ b )2 = 1 + cos2q – 2cos2q = sin2q     = (| a || b |sin q)2 ( | a | = |b | = 1)    = | a × b |2 = |β|2 Therefore   |β| = |a |

70

VITEEE CHAPTERWISE SOLUTIONS



Hence, (a) is correct. Now        a ⋅ b = [ a − ( a ⋅ b )b ] ⋅ b       = a ⋅ b − ( a ⋅ b )(b ⋅ b )      = a ⋅b − a ⋅b ( |b | = 1) =0 Since       |a | + |a ⋅ b | = |a | + 0 = |a | = |β| Therefore (c) is correct.





5 4 3 , P(E2 ) = and P(E3 ) = 7 7 7

Let E be the event that majority reviewed favourably. Therefore E = (E1 ∩ E2 ∩ E3 ) ∪ (E1 ∩ E2 ∩ E3 ) ∪ (E1 ∩ E2 ∩ E3 ) ∪ (E1 ∩ E2 ∩ E3 ) Hence P(E) = P(E1 )P(E2 )P(E3 ) + P(E1 )P(E2 )P(E3 )

+ P(E1 )P(E2 )P(E3 + P(E1 )P(E2 )P(E3 ) 5 4  3   5  4 3 =  × × 1 −   +  1 −  × ×  +   7 7   7 7 7 7    5  4  3 5 4 3  × 1 −  ×  +  × ×  7  7  7  7 7 7 



80 + 24 + 45 + 60 209 = 7 ×7 ×7 343

120. (d) : Let Bj be the number of black balls transferred ( j = 0, 1, 2, 3). B is the event of drawing a black ball. Therefore

119. (c) : Let E1, E2 and E3 be the events of the critics giving favourable remarks. Then P(E1 ) =

=

5 4 4 2 4 3 5 3 3 = × × + × × + × × + 7 7 7  7 7 7  7 7 7  5 4 3  7 × 7 × 7 





P(B0 ) = P(B1 ) = P(B2 ) = P(B3 ) =

5

C4

=

8

C4

3

5 70

C1 × 5 C3 8

C4

3

C2 × 5 C2 8

C4

3

C3 × 5 C1 8

=

30 70

=

30 70

=

5 70

C4 Also P(B/B0) = 0 ( no black ball is transferred) 1 2 P(B / B1 ) = P(B / B2 ) = 4, 4 P(B / B3 ) =

3 4

Therefore by Bayes’ theorem, P(B )P(B / B3 ) P(B3 / B) = 3 3

∑ P(Bi )P(B / Bi )

i =0





vvv

5 3 × 70 4 = 30 1 30 2 5 3 5 ×0+ × + × + × 70 4 70 4 70 4 70 =

15 15 1 = = 30 + 60 + 15 105 7

Model Test Paper-4

Model Test Paper Time : 2 Hours 30 Minutes

physics 1.

A resistor has a colour code of green, blue, brown and silver. What is its resistance? (a) 56 W ± 5% (b) 560 W ± 10% (c) 560 W ± 5% (d) 5600 W ± 10%

2.

Consider the following statements regarding the network shown in the figure (1) The equivalent resistance of the network between points A and B is independent of value of G. (2) The equivalent resistance of the network 4 between points A and B is R. 3 (3) The current through G is zero. Which of the above statements is/are true? (a) (1) only (b) (2) only (c) (2) and (3) (d) (1), (2) and (3)

3.

4.

Particles that interact by the strong force are called (a) leptons (b) hadrons (c) muons (d) electrons The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. How much electrical energy is stored by the capacitor? (a) 2.55 × 10–6 J (b) –2.55 × 10–6 J –6 (c) –2.05 × 10 J (d) 2.05 × 10–6 J

4

71

Max. Marks : 120

7.

Rising and setting sun appears to be reddish because of (a) diffraction (b) scattering due to dust particles and air molecules (c) refraction (d) polarization.

8.

A conducting sphere of radius 10 cm is charged with 10 mC. Another uncharged sphere of radius 20 cm is allowed to touch it. After that if the spheres are separated, the surface density of charges on the spheres will be in the ratio of (a) 1 : 4 (b) 1 : 2 (c) 1 : 3 (d) 2 : 1

9.

A 4 m long wire of resistance 8 W is connected in series with a battery of emf 2 V and a resistor of 7 W. The internal resistance of the battery is 1 W. What is the potential gradient along the wire? (a) 1.00 V m–1 (b) 0.75 V m–1 –1 (c) 0.50 V m (d) 0.25 V m–1

10. Excitation energy of a hydrogen like ion in its first excitation state is 40.8 eV. Energy needed to remove the electron from the ion in ground state is (a) 54.4 eV (b) 13.6 eV (c) 40.8 eV (d) 27.2 eV

5.

The flux entering and leaving a closed surface are 5 × 105 and 4 × 105 MKS units respectively, then the charge inside the surface will be (a) –8.85 × 10–7 C (b) 8.85 × 10–7 C (c) 8.85 × 107 C (d) 6.85 × 10–7 C

11. Two coils have a mutual inductance 0.005 H. The current changes in the first coil according to equation I = I0sinwt, where I0 = 10 A and w = 100p rad s–1. The maximum value of emf in the second coil is (in V) (a) 2p (b) 5p (c) p (d) 4p

6.

Which of the following is not made by quarks? (a) Neutron (b) Positron (c) Proton (d) p-meson

12. Which of the energy band diagrams shown below corresponds to that of a semiconductor?

72

VITEEE CHAPTERWISE SOLUTIONS

(a)



(b)

(c)



(d)

13. In the circuit shown in figure, the current gain, b = 100 for the transistor. What would be the base resistance RB so that VCE = 5V? (Neglect VBE). (a) 2 × 103 W (c) 1 × 106 W

(a) nicol prism (c) biprism

18. In an electromagnetic wave, the electric and magnetic fields are 100 V m–1 and 0.265 A m–1. The maximum energy flow will be (a) 79 W m–2 (b) 13.2 W m–2 –2 (c) 53.0 W m (d) 26.5 W m–2

RB

1k Ω

IB

IC C

B VBE

RL

IE

E

10 V VCC VCE

(b) 2 × 105 W (d) 500 W

14. In the propagation of electromagnetic waves, the angle between the direction of propagation and plane of polarisation is (a) 0° (b) 45° (c) 90° (d) 180° 15. A long straight wire carrying a current of 30 A is placed in an external uniform magnetic field of induction 4 × 10–4 T. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point 2 cm away from the wire is (m0 = 4p × 10–7 H m–1) (a) 10–4 (b) 3 × 10–4 –4 (c) 5 × 10 (d) 6 × 10–4 16. The rate of change of current is 500 A s–1 at the instant the key is pressed in the circuit shown in figure. The current through the circuit is

(a) 2 A (c) 0.5 A

(b) mirror (d) half wave plate

(b) 1 A (d) 1.5 A

17. One of the devices to produce plane polarised light is

19. There are three wavelengths : 10–8 m, 10–2 m, 108 m. Their respective names are (a) visible rays, g-rays, ultraviolet rays (b) ultraviolet, microwaves, radiowaves (c) X-rays, visible rays, radiowaves (d) radiowaves, X-rays, microwaves. 20. A Geiger counter is able to provide an indirect measure of radioactivity because radiation has a property of (a) ionization (b) making matter glow in the dark (c) fogging photographic film (d) altracting electrons 21. In the circuit shown in the figure, the switch S is closed at time t = 0.  L  Given,R =  C The current through the capacitor and inductor will be equal at time t equals (a) RC (b) RC ln 2 1 (c) (d) LR RC ln 2 22. What is the net force on the rectangular coil shown in figure?

(a) 25 × 10–7 N towards wire (b) 25 × 10–7 N away from wire (c) 35 × 10–7 N towards wire (d) 35 × 10–7 N away from wire 23. An electron having momentum 2.4 × 10–23 kg m s–1 enters a region of uniform magnetic field of 0.15 T. The field vector makes an angle of 30° with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be

73

Model Test Paper-4

(a) 2 mm (c)

3 mm 2

(b) 1 mm (d) 0.5 mm

24. An electron is moving in a cyclotron at a speed of 3.2 × 107 m s–1 in a magnetic field of 5 × 10–4 T perpendicular to it. What is the frequency of this electron? (e = 1.6 × 10–19 C, me = 9.1 × 10–31 kg) (a) 1.4 × 105 Hz (b) 1.4 × 107 Hz 6 (c) 1.4 × 10 Hz (d) 1.4 × 109 Hz 25. A ray of light moves from denser to rarer medium. Which of the following is correct? (a) Energy increases (b) Frequency increases (c) Phase change by 90° (d) Velocity increases 26. A 150 ohm resistor and an inductance L are connected in series to a 200 V, 50 Hz source of negligible impedance. The current comes to 1.0 A. When the source is changed to 400 V, 100 Hz, the current will be (a) less than 1.0 A (b) 1.0 A (c) between 1 A and 2 A (d) between 4 A and 2 A 27. The activity of a radioactive sample is measured as N0 counts per minute at t = 0 and N0/e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is (a) log e

2 5

(b)

5 log e 2

(a) zero (c) 0.02

(b) 0.01 (d) 0.03

30. The distance between plates of a parallel plate capacitor is 5d. The positively charged plate is at x = 0 and negatively charged plate is at x = 5d. Two slabs one of conductor and the other of a dielectric of same thickness d are inserted between the plates as shown in figure. Potential (V) versus distance (x) graph will be

(a)

(b)

(c)

(d)

28. The output Y of the logic circuit as shown in figure is

31. Let K1 be the maximum kinetic energy of photoelectrons emitted by light of wavelength l1 and K2 corresponding to wavelength l2. If l1 = 2l2 then (a) 2K1 = K2 (b) K1 = 2K2 (c) K1 < K2/2 (d) K1 > 2K2



32. A rigid circular loop of radius r and mass m lies in the x-y plane on a flat table and has a current I flowing in it. At this particular place,  the earth’s magnetic field is B = Bx i + Bz k . What is the value I so that one edge of the loop lifts from the table?

(c) 5log10 2

(a) ( A + B) C (c) (B + C) A

(d) 5loge 2

(b) ( A + C) B (d) A + B + C

29. The circuit shown in the figure contains two diodes each with a forward resistance of 30 W and with infinite backward resistance. If the battery is 3 V, the current through the 50 W resistance (in ampere) is

(a)

mg pr

Bx2

mg (c) prBx

+

Bz2



(b) (d)

mg prBz mg pr Bx Bz

74

VITEEE CHAPTERWISE SOLUTIONS

33. In the Young’s double slit experiment apparatus shown in figure, the ratio of maximum to minimum intensity on the screen is 9. The wavelength of light used is l, then the value of y is

39. A convex lens of focal length 0.15 m is made of a material of refractive index 3/2. When it is placed in a liquid, its focal length is increased by 0.225 m. The refractive index of the liquid is (a) 7 4 (c) 9 4

(a)

lD d

(b)

lD 2d

(c)

lD 3d

(d)

lD 4d

34. Find the phase velocity of electromangetic wave having electron density and frequency for D layer, N = 400 electron per c.c., u = 200 kHz. (a) 3 × 108 m s–1 (b) 3.4 × 108 m s–1 (c) 6.9 × 108 m s–1 (d) 1.1 × 109 m s–1 35. The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye-piece is found to be 20 cm. The focal lengths of the lenses are (a) 18 cm, 2 cm (b) 11 cm, 9 cm (c) 10 cm, 10 cm (d) 15 cm, 5 cm 36. An object is placed 30 cm to the left of a diverging lens whose focal length is of magnitude 20 cm. Which one of the following correctly states the nature and position of the virtual image formed? Nature of image Distance from lens (a) inverted, enlarged 60 cm to the right (b) erect, diminished 12 cm to the left (c) inverted, enlarged 60 cm to the left (d) erect, diminished 12 cm to the right 37. The angular momentum of electron in 3d orbital of an atom is  h   h  (a) 2   (b) 3    2p   2p  (c)

 h  6    2p 

(d)

(b) 5 4

 h  12    2p 

38. The binding energy of an electron in the ground state of He is equal to 24.6 eV. The energy required to remove both the electrons is (a) 49.2 eV (b) 24.6 eV (c) 38.2 eV (d) 79.0 eV

(d)

3 2

40. A transparent thin plate of a polaroid is placed on another similar plate such that the angle between their axes is 30°. The intensities of the emergent and the unpolarized incident light will be in the ratio of (a) 1 : 4 (b) 1 : 3 (c) 3 : 4 (d) 3 : 8

Chemistry 41. In cyclotrimetaphosphoric acid, number of P—O—P bonds, P O bonds and P—OH bonds are respectively (a) 6, 3, 3 (b) 5, 0, 3 (c) 4, 3, 0 (d) 3, 3, 3 42. Which of the following carbonyl compounds is most polar? (a)

(c)

(b)



(d)

43. In an irreversible process taking place at constant T and P and in which only pressurevolume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria (a) (dS)V, E < 0, (dG)T, P < 0 (b) (dS)V, E > 0, (dG)T, P < 0 (c) (dS)V, E = 0, (dG)T, P = 0 (d) (dS)V, E = 0, (dG)T, P > 0 44. The reagent with which both acetaldehyde and acetone react easily is (a) Fehling’s reagent (b) Grignard reagent (c) Schiff’s reagent (d) Tollens’ reagent. 45. If the solubility of PbCl2 at 25°C is 6.3× 10–3 mol L–1 its solubility product at that temperature is (a) (6.3 × 10–3) × (6.3 × 10–3)2 (b) (6.3 × 10–3) × (12.6 × 10–3)2 (c) (6.3 × 10–3) × (12.6 × 10–3) (d) (12.6 × 10–3) × (12.6 × 10–3)

75

Model Test Paper-4

46. Which of the following organic compounds gives positive Fehling’s test as well as iodoform test? (a) Methanal (b) Ethanol (c) Propanone (d) Ethanal 47. What is the maximum number of emission lines when the excited electron of a hydrogen atom in n = 6 drops to ground state? (a) 6 (b) 15 (c) 30 (d) 10 48. Which one of the following species is not a pseudohalide? (a) CNO– (b) RCOO– – (c) OCN (d) NNN– 49. The number of radial nodes of 3s and 2p orbitals are respectively (a) 2, 0 (b) 0, 2 (c) 1, 2 (d) 2, 1 50. The number of structural and configurational isomers of a bromo compound C5H9Br, formed by the addition of HBr to pent-2-yne respectively are (a) 1 and 2 (b) 2 and 4 (c) 4 and 2 (d) 2 and 1 51. Alkali halides do not show Frenkel defect because (a) cations and anions have almost equal size (b) there is a large difference in size of cations and anions (c) cations and anions have low coordination number (d) anions cannot be accommodated in voids. 52. The structure of 2R, 3S-dibromocinnamic acid is (a) H

CO2H

Br Br

H

(c) H

(b) Br

CO2H H Br

H

Ph

Ph

CO2H

CO2H

Br H

Br

Ph

(d) Br

Ph

0.32

(b) 10 0.0295

0.26

0.32

(c) 10 0.0295

(d) 10 0.0591

54. Interhalogen compounds are more reactive than the individual halogens because (a) they are prepared by direct combination of halogens (b) X X′ bond is weaker than X X or X′ X′ bond (c) they are thermally more stable than halogens (d) there is a large difference in their electronegativities. 55. Ground state energy of H-atom is (– E1), the velocity of photoelectrons emitted when photon of energy E2 strikes stationary Li2+ ion in ground state will be (a) v =

2( E2 − E1 ) (b) v = m

2( E2 + 9 E1 ) m

(c) v =

2( E2 − 9 E1 ) (d) v = m

2( E2 − 3E1 ) m

56. Oils are converted into fats by (a) hydration (b) decarboxylation (c) hydrogenation (d) dehydrogenation. 57. Glycol is added to aviation petrol because (a) it prevents freezing of petrol (b) it minimises the loss of petrol (c) it increases the efficiency of fuel (d) it prevents the engine from heating up. 58. Which of the following expressions correctly represents the equivalent conductance at ° 3+ infinite dilution of Al2(SO4)3? Given that L Al

° and L are the equivalent conductances SO 2− 4

at infinite dilution of the respective ions. ° ° ° ° (a) 2 L (b) L +L + 3L Al 3+ SO 24− Al 3+ SO 2− 4

H H

Br

0.32

(a) e 0.0295

53. The EMF of the cell, Zn | Zn2+ (0.01 M) | | Fe2+ (0.001M) | Fe at 298 K is 0.2905 V then the value of equilibrium constant for the cell reaction is

1° 1° ° ° (c) ( L +L ) × 6 (d) LAl3+ + L SO2− Al 3+ SO 2− 4 4

3

2

59. The correct order of ionic radii of Ce, La, Pm and Yb in +3 oxidation state is (a) La3+ < Pm3+ < Ce3+ < Yb3+ (b) Yb3+ < Pm3+ < Ce3+ < La3+ (c) La3+ < Ce3+ < Pm3+ < Yb3+ (d) Yb3+ < Ce3+ < Pm3+ < La3+

76

VITEEE CHAPTERWISE SOLUTIONS

60. The IUPAC name of the following compound O O is O (a) propionic anhydride (b) dipropanoic anhydride (c) ethoxy propanoic acid (d) propanoic anhydride. 61. Consider the reactions,





NO 2 

1 N + O 2 , K1 2 2



N2O4

2NO2,

K2

Give the equilibrium constant for the formation of N2O4 from N2 and O2. (a)

1 K12

+

1 K2

1 K1K2

(c)

(b)

1 1 + 2K1 K2

(d)

K2 K1

OH

SO3H

(a)

Br

Br



Br

SO 3H OH

OH

Br

Br

(b)

Br

(c)

OH

Br

(d)

Br Br Br Br SO 3 H 63. Which of the following arrangements shows schematic alignment of magnetic moments of antiferromagnetic substances?

(a) (b)

Which of the following sets of quantum numbers is not possible? (a) (i), (ii), (iii) and (iv) (b) (ii), (iv) and (v) (c) (i) and (iii) (d) (ii), (iii) and (iv)

66. Which of the following is not formed when glycerol reacts with HI? (a) CH2 CH – CH2I (b) CH2OH – CHI – CH2OH (c) CH3 – CH CH2 (d) CH3 – CHI – CH3

aqueous Br2 (3.0 equivalents)

OH

65. Consider the following sets of quantum numbers: n l m s (i) 3 0 0 +1/2 (ii) 2 2 1 +1/2 (iii) 4 3 –2 –1/2 (iv) 1 0 –1 –1/2 (v) 3 2 3 +1/2

62. The major product of the following reaction is

Br

64. The ease of liquefaction of noble gases decreases in the order (a) He > Ne > Ar > Kr > Xe (b) Xe > Kr > Ar > Ne > He (c) Kr > Xe > He > Ar > Ne (d) Ar > Kr > Xe > He > Ne

67. For preparing a buffer solution of pH = 6 by mixing sodium acetate and acetic acid, the ratio of the concentration of salt and acid should be (Ka = 10–5) (a) 1 : 10 (b) 10 : 1 (c) 100 : 1 (d) 1 : 100 68. Which one of the following is hydride transfer reaction? (a) (b) 2

(c)

(c) (d)

(d)

O

O

(i) Mg, Benzene (ii) H2O

Base

OH O

HO OH

77

Model Test Paper-4

69. Which of the following compounds can be detected by Molisch’s test? (a) Sugars (b) Amines (c) Primary alcohols (d) Nitro compounds 70. Hess’s law is applicable for the determination of heat of (a) transition (b) formation (c) reaction (d) all of these. 71. Which of the following statements is true regarding main cause of lanthanide contraction? (a) Poor shielding of 4f-electron by another in the subshell (b) Poor shielding of 5d-electrons (c) Effective shielding of 4f-electrons (d) Effective shielding of 5d-electrons by 4f-electrons 72. Which of the following will be most readily dehydrated in acidic conditions? O

OH

(a)

OH



(b) O

O

(c)



(d) OH

OH

73. An organic compound with the formula C6H12O6 forms a yellow crystalline solid with phenylhydrazine and gives a mixture of sorbitol and mannitol when reduced with sodium. Which among the following could be the compound? (a) Fructose (b) Glucose (c) Mannose (d) Sucrose 74. Which of the following statements is incorrect? (a) t1/2 ∝ a, for zero order reaction. (b) t1/2 is independent of a, for first order reaction. (c) t1/2 ∝ (d) t1/2 ∝

1 a2

, for third order reaction.

1 a1−n

, for nth order reaction

75. An ester (A) with molecular formula C9H10O2 was treated with excess of CH3MgBr and the compound so formed was treated with conc.

H2SO4 to form olefin (B). Ozonolysis of (B) gave ketone with formula C8H8O which shows positive iodoform test. The structure of (A) is (a) C6H5COOC2H5 (b) CH3OCH2COC6H5 (c) CH3COC6H4COCH3 (d) C6H5COOC6H5 76. In the reduction of nitrobenzene, which of the following is the intermediate? (a) C6H5 – N O (b) C6H5NH – NHC6H5 (c) C6H5 – N N – C6H5 O N – C6H5

(d) C6H5 – N

77. Which of the following ligands will not show chelation? (a) EDTA (b) DMG (c) Ethene-1, 2-diamine (d) SCN – 78. pKa of a weak acid is 5.76 and pKb of a weak base is 5.25. What will be the pH of the salt formed by the two? (a) 7.255 (b) 7.005 (c) 10.225 (d) 4.255 79. Which of the following compounds does not contain chiral carbon? (a) CH3 CH CH2 (b) O

CH3

CH3

(c)

H

—CH C— 2 CH3

(d) H

—CH C— 2 CH3

80. Increasing order of reactivity of the following alkyl halides in the Williamson’s synthesis is I. CH2 CHCH2Cl II. CH3CH2CH2Br III. (CH3)3CCH2Br IV. CH3CH2CH2Cl (a) II < III < IV < I (b) III < II < IV < I (c) IV < III < I < II (d) III < IV < I < II

78

VITEEE CHAPTERWISE SOLUTIONS

Mathematics 81. If the point of intersection of the lines x−2 y−6 z−3 x y−2 z+3 and = = = = 2 2 4 1 2 3 is (x, y, z), then y + x is. (a) 9 (b) –8 (c) 8 (d) 1       82. If a = 6 and b = 5 , then [( a × b ) × b ] × b is equal to     (a) 5(b × a ) (b) 5( a × b )    (c) 6( a × b ) (d) 6(b × a )

83. A bag contains 4 balls of unknown colours. A ball is drawn at random from it and is found to be white. The probability that all the balls in the bag are white is 4 1 (a) (b) 5 5 3 2 (c) (d) 5 5 84. A natural number is selected at random from the first 100 natural numbers. Let A, B and C denote the events of selection of even number, a multiple of 3 and a multiple of 5, respectively. Then 3 4 (a) P(A ∩ B) = (b) P(B ∩ C) = 50 25 1 (c) P(C ∩ A) = (d) All of these 10 85. Find the distance between the following parallel planes 2x – y + 2z + 3 = 0 and 4x – 2y + 4z + 5 = 0 (a) 2/3 (b) 4 (c) 1/6 (d) 2/5 86. Find the angle between the pair of lines given by  r = 3i + 2 j − 4 k + l(i + 2 j + 2 k )  and r = 5i − 2 j + m( 3i + 2 j + 6 k )  19   21  (a) cos −1   (b) cos −1    21   19   19  (c) cos    20 

 20  (d) cos −1    21 

87. Solve the following differential equation dy 1 + cos 2 y + = 0. dx 1 − cos 2 x (a) tan x = cot y + C (b) cot x = tan y + C (c) cot y = tan x + C (d) tan y = cot x + C

88. Solve the following differential equation (1 + x2)dy + 2xy dx = cot x dx, (x ≠ 0). log sin x C (a) sin x + C (b) + 2 1+ x 1 + x2 (c) 1 + x2 + C (d) log|x| + C 89. Find the area bounded by curves {(x, y) : y ≥ x2 and y ≤ |x|}. (a) 3 sq. units (b) 5 sq. units (c) 2 sq. units (d) 1/3 sq. units 90. Find the area bounded by the curves y = sin x between x = 0 and x = 2p. (a) 4 sq. units (b) 6 sq. units (c) 2 sq. units (d) 8 sq. units  −1  91. If f(x) = cos–1  x − x  , then f ′(x) is  x −1 + x  (a) odd (b) even (c) periodic (d) None of these  ln( ex 4 )   ln( e / x 2 )  –1 92. If y = cot–1  + cot  ln( e 2 / x 2 )  ,  ln( ex 2 )  d2 y then equals dx 2 (a) –1 (b) 0 (c) 1 (d) 2 93. If fn(x) = efn–1(x) ∀ n ∈ N and f0(x) = x, then f n′ (x) equals (a) (c)

n

n

∑ f ( x)

(b)

∏ f ′( x)

(d) None of these

r =1 n−1 r =1

r

r

∏ f ( x) r =1

r

p 94. The function f(x) = sin   is strictly decx reasing in the interval (a) (2n + 3, 2n + 5), n ∈ I  3 3  (b)  ,n∈I ,  6n + 2 6n + 1  (c)  2 , 2  , n ∈ I   4n + 1 4n − 1  (d) None of these

95. For x > 1, y = ln x satisfies the inequality 1 (a) y < x – 1 (b) y < 1 – x (c) y < x2 – 1 (d) None of these 96. For any function f(x), if f ′(a) = f ′′(a) = …… = f (n –1)(a) = 0 but f (n)(a) ≠ 0, then f(x) has a minima at x = a if (a) n is even and f (n)(a) > 0 (b) n is even and f (n)(a) < 0 (c) n is odd and f (n)(a) > 0 (d) n is odd and f (n)(a) < 0

79

Model Test Paper-4

n2 97. The largest term in the sequence an = 3 , n + 200 is (a) a1 (b) a7 (c) a8 (d) None of these p

98. The value of ∫0 [tan x]dx ([⋅] denotes integral part), is equal to (a) –p/2 (b) 0 (c) –1 (d) None of these 99. If w is an imaginary cube root of unity, then 1 1 1 the value of is + − 1 + 2w 2 + w 1 + w (a) –2 (c) 1

(b) –1 (d) 0

100. If (2 + i)(2 + 2i) (2 + 3i) …… (2 + ni) = x + iy, then 5. 8. 13. …… (4 + n2) is equal to (a) x2 – y2 (b) x2 + y2 4 4 (c) x – y (d) x4 + y4 101. The value of the expression

 1  1   1  1   1 +   1 + 2  +  2 +   2 + 2  w w w w



  1  1  1  1  +  3 +   3 + 2  + ...... +  n +   n + 2  ,   w w w  w 



where w is an imaginary cube root of unity is n(n2 + 3) 3 2 (c) n(n + 1) 3 (a)

(b)

n(n2 + 2) 3

(d) None of these

102. If a straight line through the point P(3, 4) p makes an angle with x-axis and meets the 6 line 12x + 5y + 10 = 0 at Q, find the length of PQ. (a) 132 (b) 12 3 + 5 −132 (c) (d) 5 12 3 + 5 103. Find the orthocentre of the triangle whose sides have equations x – 2 = 0, y – 5 = 0 and 5x + 2y – 10 = 0. (a) (5, 2) (b) (0, 2) (c) (5, 0) (d) (2, 5) 104. The lengths of the tangents from any point on the circle 15x2 + 15y2 – 48x + 64y = 0 to the two circles 5x2 + 5y2 – 24x + 32y + 75 = 0 and 5x2 + 5y2 – 48x + 64y + 300 = 0 are in the ratio (a) 1 : 2 (b) 2 : 3 (c) 3 : 4 (d) None of these

105. If the point P(4, –2) is the one end of the focal chord PQ of the parabola y2 = x, then the slope of the tangent at Q is 1 1 (a) – (b) 4 4 (c) 4 (d) – 4 106. The focus of the parabola y2 – x – 2y + 2 = 0 is 1  (a)  , 0  (b) (1, 2) 4  3  (c)  , 1 4 

(d)  5 , 1 4 

107. If P is a variable point on the ellipse x2 y 2 = 1 with AA′ as the major axis. Find + a2 b2 the maximum value of the area of the triangle APA′. (a) b (b) a2 (c) ab (d) a 108. If e and e′ be the eccentricities of a hyperbola 1 1 and its conjugate, then 2 + 2 is equal to e e′ (a) 0 (b) 1 (c) 2 (d) None of these 109. The number of solutions of the system of equations 3x – 2y + z = 5, 6x – 4y + 2z = 10 and 9x – 6y + 3z = 15 is (a) 0 (b) 1 (c) 2 (d) infinite. 110. If A is a square matrix satisfying the equation A2 – 4A– 5I = 0, then A–1 = 1 (a) A – 4I (b) (A – 4I) 3 1 1 (c) (A – 4I) (d) (A – 4I) 4 5 111. If p : It rains today, q : I go to the school, r : I shall meet my friends and s : I shall go for a movie, then which of the following is the proposition : If does not rain or if I do not go to school, then I shall meet my friend and go for a movie. (a) ~ (p ∧ q) ⇒ (r ∧ s) (b) (~ p ∧ ~ q) ⇒ (r ∧ s) (c) ~ (p ∨ q) ⇒ (r ∨ s) (d) None of these 112. If p : Ajay is tall q : Ajay is intelligent then the symbolic statement ~ p ∨ q means (a) Ajay is not tall or he is intelligent. (b) Ajay is tall or he is intelligent. (c) Ajay is not tall and he is intelligent. (d) Ajay is not tall then he is intelligent.

80

VITEEE CHAPTERWISE SOLUTIONS

113. (~(~p)) ∧ q is equal to (a) ~ p ∧ q (b) p ∧ q (c) ~ p ∧ ~ q (d) p ∧ ~ q 114. Negation of the compound proposition If the Examination is difficult, then I shall pass if I study hard. (a) The Examination is difficult and I study hard but I shall not pass. (b) The Examination is not difficult and I study hard and I shall pass. (c) The Examination is difficult and I study hard and I shall pass. (d) None of these. 115. If x2 + y2 = 1, then (a) yy′′ – (2y′)2 + 1 = 0 (b) yy′′ + (y′)2 + 1 = 0 (c) yy′′ – (y′)2 – 1 = 0 (d) yy′′ + 2(y′)2 + 1 = 0 116. The differential equation of all parabolas whose axes are parallel to y-axis, is 3

(a)

dy = 0 dx 3

(b)

d y =0 dx 2

(c)

d 2 y dy = 0 + dx 2 dx

(d)

d 2 y dy +y=0 + dx 2 dx

2

117. The integrating factor of the differential equation (y log y) dx = (log y – x) dy is 1 (a) (b) log(log y) log y (c) 1 + log y (d) log y   118. If the vectors a = (2, log3 x, a) and b = (–3, a log3 x, log3 x) are inclined at an acute angle, then (a) a = 0 (b) a < 0 (c) a > 0 (d) None of these    119. If a , b , c are linearly independent vectors and    a b c       D = a ⋅ a a ⋅ b a ⋅ c , then       a ⋅c b ⋅c c ⋅c (a) D = 0 (b) D = 1 (c) D = any non-zero value (d) None of these 120. The p.c of the centre of the sphere 2  r + r ⋅ (i + j − k ) − 9 = 0 is 1 (a) i + j − k (b) (i + j − k ) 2 1 (c) − (i + j − k ) (d) −i + j + k 2

Answer Key 1.

(b)

2.

(d)

3.

(b)

9.

(d)

10. (a)

11. (b)

17. (a)

18. (d)

25. (d)

26. (c)

33. (c)

4.

(a)

5.

(a)

6.

(b)

7.

(b)

8.

(d)

12. (d)

13.

(b)

14.

(a)

15.

(c)

16.

(b)

19. (b)

20. (a)

21.

(b)

22.

(a)

23.

(d)

24.

(b)

27. (d)

28. (a)

29.

(c)

30.

(b)

31.

(c)

32.

(c)

34. (c)

35. (a)

36. (b)

37.

(c)

38.

(d)

39.

(b)

40.

(d)

41. (d)

42. (d)

43. (b)

44. (b)

45.

(b)

46.

(d)

47.

(b)

48.

(b)

49. (a)

50. (b)

51. (a)

52. (d)

53.

(b)

54.

(b)

55.

(c)

56.

(c)

57. (a)

58. (b)

59. (b)

60. (d)

61.

(a)

62.

(b)

63.

(d)

64.

(b)

65. (b)

66. (b)

67. (b)

68. (a)

69.

(a)

70.

(d)

71.

(a)

72.

(a)

73. (a)

74. (d)

75. (a)

76. (a)

77.

(d)

78.

(a)

79.

(c)

80.

(d)

81. (c)

82. (a)

83. (d)

84. (d)

85.

(c)

86.

(a)

87.

(d)

88.

(b)

89. (d)

90. (a)

91. (a)

92. (b)

93.

(b)

94.

(c)

95.

(a,c) 96.

(a)

97. (b)

98. (a)

99. (d)

100. (b)

101. (b)

102. (c)

103. (d)

104. (a)

105. (c)

106. (d)

107. (c)

108. (b)

109. (d)

110. (d)

111. (a)

112. (a)

113. (b)

114. (a)

115. (b)

116. (a)

117. (d)

118. (d)

119. (c)

120. (c)

81

Model Test Paper-4

Physics 1.

2.

e planations

(b) : Numbers corresponding to green, blue, brown and silver are 5, 6, 1 and 10% respectively. Therefore, the resistance of given resistor = 56 × 101 W ± 10% = 560 W ± 10%

7. 8.

R 2R , hence the given network = R 2R is a balanced Wheatstone bridge. Therefore, no current flows through G, so equilvalent circuit diagram is shown in the figure. (d) : As

R

Neutron and proton are baryons. Positron is a lepton. p-meson is a meson. (b) (d) : Common potential, q +q 10 + 0 10 V= 1 2 = = C1 + C2 4 pe0 (0.1 + 0.2) 4 pe0 (0.3) Charges after contact q′1 = C1V = 4pe0(0.1) ×

R

A

\

q′ σ1 4 pr22 q1′  r2  = 12 × =   q2′ q2′  r1  σ 2 4 pr1

2R E



2R A

B 4R

E

Hence, the equivalent resistance of the network between points A and B is ( 2 R)( 4 R) 4 Req = = R 2R + 4R 3 3. 4.

5.

6.

10 20 = mC 4 pe0 (0.3) 3 The ratio of surface density of charges

and q′2 = C2V = 4pe0(0.2) ×

B 2R

10 10 = mC 4 pe0 (0.3) 3

(b) (a) : Here, A = 90 cm2 = 90 × 10–4 m2 d = 2.5 mm = 2.5 × 10–3 m, V = 400 V 1 1 e0 A 2 Electrical energy stored = CV 2 = V 2 2 d 1 8.85 × 10 −12 × 90 × 10 −4 ( 400)2 = × −3 2 2.5 × 10 = 2.55 × 10–6 J (a) : Net flux leaving the surface, Df = 4 × 105 – 5 × 105 = –105 MKS \ Charge must be negative q = fe0 = –105 × 8.85 × 10–12 = – 8.85 × 10–7 C (b) : Baryons are made of three quarks and mesons are made of one quark and one antiquark. On the other hand, leptons are not made of quarks.

9.

2

2

=

10 3  0.2  1 ×   = ×4=2:1 3 20  0.1  2

(d) : Current in the potentiometer, 2 1 I= = A 8+7 +1 8 Voltage drop across potentiometer wire 1 V= ×8=1V 8 \ Potential gradient of potentiometer wire V 1 = = 0.25 V m–1 l 4

10. (a) : Excitation energy

1 1 DE = E2 − E1 = 13.6Z 2  2 − 2  2  1 3 40.8 = 13.6Z 2 × 4

or Z = 2 So, energy required to remove the electron from ground state =+

13.6Z 2 (1)2

= 13.6( 2)2 = 54.4 eV

11. (b) : Induced emf in second coil,

dI1 d = 0.005 × ( I 0 sin wt ) dt dt



e2 = M

\

= 0.005 × I0wcoswt e2max = 0.005 × 10 × 100p = 5p V

82

VITEEE CHAPTERWISE SOLUTIONS

12. (d) : In a semiconductor, the energy gap (Eg) between valence band and conduction band corresponds to thermal energy (kBT) at room temperature (T). Hence option (d) represent the correct diagram. 13. (b) : Here, b = 100, VCE = 5 V, VCC = 10 V I I I As b = C or IB = C = C …(i) IB b 100 Also, VCE = VCC – ICRL or 5V = 10 V – IC × 1000 5V \ IC = = 5 × 10–3 A 1000W 5 × 10 −3 A = 5 × 10–5 A 100 V − VBE Thus, RB = CC IB and IB =



=

10 V 5 × 10 −5 A

(Using(i))

= 2 × 105 W

(neglecting VBE) 14. (a) : Plane of vibration is perpendicular to direction of propagation and also perpendicular to plane of polarisation. Therefore, angle between plane of polarisation and direction of propagation is 0°.

m0 I ( 4 p × 10 −7 ) × 30 = = 3 × 10–4 2 pr 2 p × 0.02 The direction of B1 will be perpendicular to B2 (= 4 × 10–4 T) Hence, resultant magnetic field at given point is

21. (b) : Growth of current in RC circuit IC = I0e–t/RC) Growth of current in LR circuit  IL = I0(1 – e–Rt/L) L R 1  or =  R =  = I0(1 – e–t/RC) C L RC   But IC = IL \ I0e–t/RC = I0(1 – e–t/RC) or 2e–t/RC = 1 or et/RC = 2 t or = ln 2 RC or t = RC ln 2.  2I I  22. (a) : FAB = km 1 2  l = 30 × 10–7 N (attractive)  r  (as l = 15 cm = 15 × 10–2 m, r = 2 cm = 2 × 10–2 m)  2I I  FCD = km  1 2  l = 5 × 10–7 N (repulsive)  r′  (as r′ = (10 + 2) cm = 12 × 10–2 m) Fnet = FAB – FCD = 25 × 10–7 N towards wire (as FBC = FAD = 0) 23. (d) : The radius of the helical path of the electron in the uniform magnetic field is

15. (c) : B1 =

2 2 B = B1 + B2 = [(3 × 10–4)2 + (4 × 10–4)2]1/2 = 5 × 10–4 T. 16. (b) : Induced e.m.f. across L is dI e=L = 10–2 × 500 = 5 V dt \ Potential difference across R is, V = 60 – 5 = 55 V 55 V \ I = = =1A 55 R

mv⊥ mv sin q = eB eB ( 2.4 × 10 −23 kg m s −1 ) × sin 30°

r= =

(1.6 × 10 −19 C) × (0.15 T)

= 5 × 10–4 m = 0.5 × 10–3 m = 0.5 mm 24. (b) : Here, v = 3.2 × 107 m s–1, B = 5 × 10–4 T The frequency of electron is

u=

1.6 × 10 −19 × 5 × 10 −4 eB = 2 pme 2 × 3.14 × 9.1 × 10 −31

= 1.4 × 107 Hz = 14 MHz



25. (d) : When a ray of light moves from one medium to other, its velocity changes. This change depends on refractive index of the medium. Here, light travels from denser to rarer medium, i.e., from medium of higher refractive index to lower refractive index. So, in rarer medium its velocity increases.

17. (a) : A nicol prism produces plane polarised light. 18. (d) : Maximum energy flow in an 26. electromagnetic wave, S = E0 × B0 = 100 × 0.265 = 26.5 W m–2 19. (b) : l = 10–8 m → ultraviolet rays, l = 10–2 m → microwaves, or R2 + X L2 = 40000 l = 108 m → radiowaves 20. (a)

(c) : As I = or

R

2

+ X L2

200 R

2

+ X L2

= 1 or

R2 + X L2 = 200

= 40000 X L2 ⇒ 40000 − (150)2 or X L2 = 17500

X L2 ⇒ 40000 − (150)2 or X L2 = 17500

Since u′ = 100 Hz and u = 50 Hz, \ u′ = 2u

83

Model Test Paper-4 As X L = 2 puL , or

X′L 2 puL = = 2 or X′L = 2 X L X L 2 puL

X L′ 2 = 4 X L2 = 4 × 17500 = 70000

⇒

Z′ = R2 + X′L 2 = (150)2 + 70000 W = 304 W I′ =

\

V ′ 400 V = = 1.3 A Z′ 304 W

i.e., between 1 A and 2 A 27. (d) 28. (a) :

32. (c) : The torque on the loop must be equal to the gravitational torque exerted about an axis tangent to the loop. The gravitational torque t1 = mgr …(i) Only Bx causes a torque. Therefore torque to the magnetic  field  t2 = M × B = MB sin90° = pr2IBx …(ii) Hence from equation (i) and (ii), we get t1 = t2, ⇒ mgr = pr2IBx \ 33. (c) :

(

)

Y = ( A ⋅ B) ⋅ C = A + B C = ( A + B) C

29. (c) : In the circuit the upper diode D1 is reverse biased and the lower diode D2 is forward biased. Thus there will be no current across upper diode junction. The effective circuit will be as shown in figure.

I=

2

I max  I1 / I 2 + 1 9 =  = 1 I min  I1 / I 2 − 1

Current in circuit , I =

30. (b) : Electric field E = – (slope of V-x graph) and E inside a conductor = 0 \ slope of V-x graph between x = d to x = 2d should be zero. also E in air > E in dielectric \ |slope in air|>|slope in dielectric| hc 31. (c) : K1 = −W l1 hc and K 2 = −W l2

…(i) …(ii)

Here, W is work function of given metal surface. Substituting l1 = 2l2 in equation (i), we get hc − W ⇒ = 1  hc  − W = 1 ( K + W ) − W K1 = 2l 2 2  l 2  2 2

)

I1 = x 2 = 22 = 4 I2

and

I1 = 4I2 I2 = I0 ⇒ I1 = 4I0 1

δ

δ δ ⇒ I0 = 4I0 cos2 2 2

cos2 2 = 4

δ 1 p = = cos \ δ = 2p 2 2 3 3 yd 2p δ= Dx and Dx = and D l lD \  2 p   yd  = 2 p ⇒ y =  l  D 3d 3

or

3V V = = 0.02 A R 150 W

x = I1 / I 2

x=2

\

⇒

(

x+1 =3 x −1

or

Let

mg prBx

Since I2 = I1 cos2

Total resistance of circuit R = 50 + 70 + 30 = 150 W

 hc   l = K 2 + W  2

K2 W − 2 2 K K1 < 2 2 K1 =

cos

34. (c) : Since refractive index of D layer, m = 1−

81.45 N u2

Here, N = 4000 electrons per c.c. = 400 × 106 electrons per m3 u = 200 kHz = 200 × 103 Hz \

m = 1−

As m = \

81.45 × 400 × 106 ( 200 × 10 3 )2

c v

Phase velocity, v =



= 0.43

c 3 × 108 = m 0.43

= 6.9 × 108 m s–1

84

VITEEE CHAPTERWISE SOLUTIONS

35. (a) : As for a telescope magnifying power, f M= 0 fe

\

f 9 = 0 or f0 = 9 fe fe

Also, L = f0 + fe or 20 = f0 + fe or 20 = 9fe + fe or 20 = 10 fe ⇒ fe = 2 cm \ f0 = 9 × 2 cm = 18 cm 36. (b) : When an object is placed between 2f and f (focal length) of the diverging lens, the image is virtual, erect and diminished as shown in the ray diagram.

index of the material of the lens. \ For a convex lens in air

= – 12 cm (to the left of the diverging lens.) 37. (c) : The angular momentum is given by  h  L = l(l + 1)    2p  For 3d electron, l = 2.  h   h  \ L = 2( 3)   = 6    2p   2p  38. (d) : Helium atom has 2 electrons. When one electron is removed, the remaining atom is hydrogen like atom, whose energy in first orbit is E1 = –(2)2(13.6 eV) = – 54.4 eV Therefore, to remove the second electron from the atom, the additional energy of 54.4 eV is required. Hence, total energy required to remove both the electrons = 24.6 + 54.4 = 79.0 eV. 39. (b) : According to lens maker’s formula  1 1 1  m2 − − 1  = f  m1   R1 R2 

where m1 is the refractive index of the medium in which lens is placed and m2 is the refractive

 1 1 1  mg − = − 1   R R fl  m l   1 2

…(ii)

Dividing (i) by (ii), we get

fl m g − m a ml = × fa ma m g − ml

Here, fl = 0.150 + 0.225 = 0.375 m Substituting the given values, we get 3  −1 ml 0.375  2  = × 0.15 1 3   − m l  2 ml 5 or 5(3 – 2ml) = 2ml = 2 3 − 2m l

⇒

1 1 1 1 1 1 = − ⇒ = − f v u −20 v 30 ( 20)( 30) ⇒ v=− 20 + 30

…(i)

For the same convex lens in liquid



To calculate the distance of the image from the lens, we apply

 1 1 1  mg − = − 1   R R fa  m a   1 2 



12ml = 15

\

ml =

15 5 = 12 4

40. (d) : Let I0 be the intensity of unpolarized light, then intensity of light from first transparent thin plate of a polaroid is I=

I0 2

Now this light will pass through the second similar plate whose axis is inclined at an angle of 30° to that of first plate. According to Malus law, the intensity of emerging light is I ′ = I cos 2 30° =

\

2

I0  3  3   = I0 2 2  8

I′ 3 = I0 8

Chemistry 41. (d) :

Cyclotrimetaphosphoric acid, (HPO3)3

42. (d) : HCHO will be most polar due to lowest electron density on carbon of carbonyl group.

85

Model Test Paper-4

43. (b) : For spontaneity, change in entropy (dS) must be positive, means it should be greater than zero. Change in Gibbs free energy (dG) must be negative means that it should be lesser than zero. (dS)V, E > 0, (dG)T, P < 0. 44. (b) : Fehling’s solution, Schiff’s reagent and Tollens’ reagent react only with aldehydes, but Grignard reagent reacts with both aldehydes and ketones. O OMgBr OH R C H + RMgBr

R

C

C

R + RMgBr

R

C H

R

R

OMgBr

OH

O R

H

R C

R

R

R

C

R

R

–

45. (b) : PbCl2  Pb2+ + 2Cl – (Ksp) = [Pb2+][Cl ]2 = (6.3 × 10–3) × (2 × 6.3 × 10–3)2 = (6.3 × 10–3) × (12.6 × 10–3)2 46. (d) : Ethanal CH3CHO, being aldehyde gives O positive Fehling’s test and it has CH3 C group hence gives positive iodoform test. 47. (b) : Maximum number of emission lines

n(n − 1) 2 6(6 − 1) 30 = = = 15 lines 2 2 =

48. (b) : Pseudohalides are monovalent ions made two or more electronegative atoms of which at least one is nitrogen and have properties similar to those of halide ions. The corresponding dimers of pseudohalides are known as pseudohalogens. RCOO– is not a pseudohalide. 49. (a) : Number of radial nodes = (n – l – 1) For 3s: n = 3, l = 0 (Number of radial node = 2) For 2p: n = 2, l = 1 (Number of radial node = 0) 50. (b) :

CH3 – C

C – CH2 – CH3 + HBr CH3 – CH

C(Br) – CH2 – CH3

CH3 – C(Br)

+

Thus, two structural isomers are formed and each of them exists in a pair of geometrical isomer. Thus, in total 2 structural and 4 configurational (geometrical) isomers are formed. 51. (a) : Frenkel defect is not shown by alkali metals because cations and anions have almost same size and cations cannot be accommodated in interstitial sites. 52. (d) :

CH – CH2CH3

53. (b) : Cell reaction is, Zn + Fe2+ → Zn2+ + Fe 0.059 10 −2 log 2 10 −3 2+ –2 2+ (n = 2, [Zn ] = 10 M, [Fe ] = 10–3 M) Since E = 0.2905 V \ 0.2905 = E° – 0.0295 or E° = 0.2905 + 0.0295 = 0.32 V 0.059 Again E° = log K eq 2

Using, E = E° −

\

0.32 =

0.32

0.059 log K eq or K eq = 10 0.0295 2

54. (b) : X – X′ bond is weaker than X – X or X′ – X′ bonds 55. (c) : En = Z2 × E1 for hydrogen - like species. Threshold energy of Li2+ = 9E1 Absorbed energy = Threshold energy + Kinetic energy of photoelectrons 1 E2 = 9 E1 + mv 2 ⇒ mv2 = 2(E2 – 9E1) 2 2( E2 − 9 E1 ) v= m 56. (c) 57. (a) : Glycol serves as an antifreeze for petrol. 58. (b) : Equivalent conductivity of any electrolyte at infinite dilution is the sum of the equivalent conductivities of the cation and the anion. l°eq = l°c + l°a

59. (b) : The overall decrease in atomic and ionic radii from La3+ to Lu3+ is called lanthanoid contraction. Hence, the correct order is Yb3+ < Pm3+ < Ce3+ < La3+.

86

VITEEE CHAPTERWISE SOLUTIONS

60. (d)

CH2 – I

CH2OH

61. (a) : The required equation is N2 + 2O2 N2O4; K = ? The given equations are 1 N + O 2 ; K1 2 2

...(i)

N2O4 2NO2; K2 Multiply eqn. (i) by 2 2NO2 N2 + 2O2, K′ = K21 Invert eqn. (ii) 2NO2

N2O4; K′′ =

1

1 K2

OH

p-Hydroxybenzene sulphonic acid

O

O

Br H

CH3

CH3 HI

CH – I

–I2

CH CH2

CH3

CHI CH2I

(Unstable)

Propylene

[Salt] [Acid]

[Salt] = 6 − 5 = 1 , Taking antilog, [Acid] [Salt] = 10 , i.e., [Salt] : [Acid] = 10 : 1 [Acid]

log

OH Br

S

CH3

pKa = – log Ka = – log 10–5 = 5

Br

Br

– SO3

O

HI

pH = pKa + log

K12

62. (b) :

SO3H

Allyl iodide

Iso-propyl iodide

1 N2 + 2O2 N2O4; 2 + K1 K2 ______________________________

Br

CH2I

(Unstable)

67. (b) : We know for acidic buffers,

1

3Br2(aq)

CH2 – I

Glycerol

...(iv)

2NO2 N2O4; K2 ______________________________

OH

CH

CH2OH

...(iii)

1

2NO2;

CH – I

...(ii)

Invert eqn. (iii) and add to eqn. (iv), N2 + 2O2

CH2

–I2

68. (a) : Aldehydes like benzaldehyde, having no a-hydrogen atoms on treatment with NaOH or KOH, are converted to equal amounts of the corresponding carboxylate anion and alcohol via Cannizzaro reaction which is proton (H+)-hydride (H–) transfer reaction.

Br 2, 4, 6-Tribromophenol

63. (d) : Alignment of magnetic moments of antiferromagnetic solids are such that the resultant magnetic moment is zero. 64. (b) : Ease of liquefaction of noble gases increases down the group since van der Waals forces of attraction increase down the group with increasing atomic size. Thus, order of ease of liquefaction of noble gases is Xe > Kr > Ar > Ne > He. 65. (b) : (i) represents an electron in 3s orbital. (ii) is not possible as value of l must vary from 0, 1, ... (n – 1). (iii) represents an electron in 4f orbital. (iv) is not possible as value of m varies from – l ... +l. (v) is not possible as value of m varies from –l ... +l, it can never be greater than l. 66. (b) : During the reaction of glycerol with HI, all are formed except CH2OH – CHI – CH2OH.

– :O:

:

NO2

CHOH + 3HI

–3H 2O

O C

C

H

2

OH

–

O H H

C

OH O

O–

C

C

OH



–H+

H

H

(H+ exchange)

+H+

O



C

OH

Carboxylate ion

C

–

O

+

H

H

Alcohol

69. (a) : It is a general test for carbohydrates. A drop or two of alcoholic solution of a-naphthol is added to 2 mL of glucose solution. 1 mL of conc. H2SO4 is added along the sides of the tube and a violet ring is formed at the junction.

87

Model Test Paper-4

70. (d) : According to this law, the total enthalpy change is independent of intermediate steps involved in the change. It depends only on initial and final values of enthalpy change. So it can be used for the calculation of heat of formation, reaction or transition. 71. (a) 72. (a) : b-Hydroxy ketones or aldols readily undergo dehydration due to the presence of active methylene group. OH

O

O + 2 [H]

CH2OH

Na-Hg

H

H2O

C

(CHOH)3

CH2OH

CH2OH

Fructose (C6H12O6)

NHOH

NH2 2[H]

Aniline Phenylhydroxylamine – 77. (d) : SCN is a monodentate ligand hence cannot show chelation.

1 [pKw + pKa – pKb] 2 pH = 7 + 1 (5.76 − 5.25) = 7.255 2

78. (a) : pH =

79. (c) : It does not contain chiral carbon.

CH2OH C

H

(CHOH)3 CH2OH Mannitol

74. (d) : For nth order reaction, t1/ 2 ∝ 75. (a) : COOC2H5

1 a

n−1

OH

2CH3MgBr

(A)

C – CH3

H2 O

CH3

(C9H10O2) (i) O3

C

(ii) Zn/H2O

CH3

Nitrosobenzene

Sorbitol

+ HO

(C8H8O)

Nitrobenzene

OH

(CHOH)3

C



2[H]

+ H2O

H H 73. (a) : Fructose undergoes osazone formation with phenylhydrazine. However, on partial reduction with sodium amalgam and water, a mixture of two epimeric alcohols, sorbitol and mannitol is obtained. This is due to the creation of a new asymmetric carbon atom at C-2. C

2[H]

O

H+

CH2OH

76. (a) : In the reduction of nitrobenzene, nitrosobenzene (C6H5 – N O) and phenylhydroxylamines (C6H5 – NHOH) are obtained as intermediate depending upon the pH of reaction medium. NO2 NO

CH3

(B) Olefin

O + HCHO

Formaldehyde

CH2

H2SO4 (conc.)

80. (d) : C Br bond is weaker than C Cl bond therefore II reacts faster than I and IV. CH2 CH is electron withdrawing while CH3CH2 is electron donating group, hence nucleophilic attack occurs faster in I than IV i.e. I is more reactive than IV. Further, as the reaction follows SN2 mechanism III is least reactive due to steric hinderance.

Mathematics 81. (c) : Let L1 and L2 be the given lines, respectively. P is a point on L1 and Q is a point on L2. Therefore, P = (t, 2 + 2t, –3 + 3t) and Q = (2 + 2s, 6 + 2s, 3 + 4s) where t, s ∈ R. Now P = Q implies t = 2 + 2s or t – 2s = 2 …(i) 2 + 2t = 6 + 2s or 2t – 2s = 4 …(ii) and – 3 + 3t = 3 + 4s or 3t – 4s = 6 …(iii) From Eqs. (i) and (ii) t = 2, s = 0 which also satisfy Eq. (iii). Therefore the point of intersection is (2, 6, 3) = (x, y, z) so \ y + x = 8

88

VITEEE CHAPTERWISE SOLUTIONS

82. (a) : We have             [( a × b ) × b ] × b = [( a × b ) ⋅ b ]b − (b ⋅ b )( a × b )      = 0(b ) − 5( a × b ) ⇒ = 5(b × a ) 83. (d) : Let Wj (j = 1, 2, 3, 4) denote 1, 2, 3 and 4 white balls are in the bag. Let W be the ball drawn is white. Then 1 P(W1) = P(W2) = P(W3) = P(W4) = 4 1 2 P(W/W1) = ⇒ P(W/W2) = 4 4 3 P(W/W3) = ⇒ P(W/W4) = 1 4 Therefore by Bayes’ theorem P(W4 )P(W / W4 ) P(W4/W) = 4 ∑ P(Wj )P(W / Wj ) j =1



1 ×1 4 2 4 = ⇒= = 11 2 3 4 10 5  + + +  4 4 4 4 4

84.

(d) : We have Number of even numbers ≤ 100 is equal to 50. Number of multiples of 3 ≤ 100 is 33. Number of multiples of 5 ≤ 100 is 20. Number of common multiples of 2 and 3 is 16. Number of common multiples of 3 an 5 is 6. Number of common multiples of 2 and 5 is 10. Number of common multiples of 2, 3 and 5 is 3. Now, 20 50 33 P(A) = , P(B) = , P(C) = 100 100 100 6 16 P(A ∩ B) = , P(B ∩ C) = , 100 100 10 P(C ∩ A) = 100 85. (c) : Given equations of planes are 2x – y + 2z + 3 = 0 …(i) and 4x – 2y + 4z + 5 = 0 …(ii)



The given planes are parallel as their DR’s are proportional. Let P(x1, y1, z1) be any point lie on the plane (i). \ 2x1 – y1 + 2z1 + 3 = 0 ⇒ 2x1 – y1 + 2z1 = – 3 …(iii) \ Perpendicular distance d from the point P(x1, y1, z1) to plane (ii) is



=

4 x1 − 2 y1 + 4 z1 + 5



=

2( 2 x1 − y1 + 2 z1 ) + 5



=



=



d



4x – 2y + 4z + 5 = 0

16 + 4 + 16 2( −3) + 5 36

−6 + 5 6 1 = 6

[from Eq. (iii)]

86. (a) : Given lines are  …(i) r = 3i + 2 j − 4 k + l(i + 2 j + 2 k )       and r = 5i − 2 j − m( 3i + 2 j + 6 k ) …(ii)    On comparing with r = a + lb , we get   b1 = i + 2 j + 2 k and b2 = 3i + 2 j + 6 k The angle q between the two lines is given by   b ⋅b cos q = 1 2 b1 b2 (i + 2 j + 2 k ) ⋅ ( 3i + 2 j + 6 k )



=



=



=



 19  q = cos–1    21 

2x – y + 2z + 3 = 0 P(x1, y1, z1)

( 4)2 + ( −2)2 + ( 4)2

\

(1)2 + ( 2)2 + ( 2)2 ( 3)2 + ( 2)2 + ( 6)2 (1)( 3) + ( 2)( 2) + ( 2)(6) 1 + 4 + 4 9 + 4 + 36 3 + 4 + 12 9 49

=

19 19 = 3 × 7 21

87. (d) : Given differential equation is dy 1 + cos 2 y + =0 dx 1 − cos 2 x

⇒

dy 1 + cos 2 y =− dx 1 − cos 2 x



⇒

dy 2 cos 2 y =− dx 2 sin 2 x

89

Model Test Paper-4



(\ 1 + cos2q = 2cos2q and 1 – cos2q = 2sin2q) dy cos 2 y ⇒ =− dx sin 2 x On separating the variables, we get



The points of intersection of the parabola and line are (–1, 1) and (1, 1). As the area to be found is symmetrical about Y-axis.



\ Required area = 2(Area of shaded region in the first quadrant) 1 = 2∫0 [y2 (line) – y1(parabola)]dx



= 2∫0 (x – x2)dx



= 2[∫0 x dx – ∫0 x2 dx]

dy dx =− 2 cos y sin 2 x

sec2 y dy = – cosec2x dx On integrating both sides, we get ∫sec2 y dy = – ∫ cosec2 x dx ⇒ tan y = cot x + C which is the required solution. 88. (b) : Given differential equation is (1 + x2)dy + 2xy dx = cot x dx (Q x ≠ 0) Above equation can be rewritten as; (1 + x2)dy + (2xy – cotx)dx = 0 ⇒ (1 + x2)dy = (cotx – 2xy)dx On dividing both sides by 1 + x2, we get cot x − 2 xy 2 xy dy cot x dy = dx ⇒ = − 2 1 + x2 dx 1 + x 1 + x 2 dy cot x 2x ⇒ …(i) y= + dx 1 + x 2 1 + x2 This is a linear differential equation of 1st order and is of the form dy + Py = Q …(ii) dx On comparing Eqs. (i) and (ii), we get 2x cot x P= and Q = 2 1+ x 1 + x2 2x dx 2 Pdx ∫ Now, I.F = e ∫ = e 1+ x2 = e log 1+ x = 1 + x 2 2x 2 [Q I1 = ∫ 1 + x2 dx, put 1 + x = t ⇒ 2x dx = dt] dt ⇒ ∫ = log|t| + C = log|1 + x2| + C] t Now, solution of linear differential equation is given by, y × I.F = ∫(Q × I.F)dx cot x y(1 + x2) = ∫ × (1 + x2)dx 1 + x2 ⇒ y(1 + x2) = ∫cot x dx ⇒ y(1 + x2) = log|sin x| + C (Q ∫ cot x dx = log |sin x|) log sin x C + ⇒ y = 1 + x2 1 + x2 which is the required solution. 89. (d) : Given, curve y = x2 …(i) and equation of line y = |x| …(ii) In first quadrant, the equation of the line is y = x and in second quadrant, the equation of the line is y = – x

1

1

(Q y = x and y = x2)

1



  x2 1  x3 1   1  1  = 2   −    = 2  − 0  −  − 0    3    2  0  3  0   2



1 1 1 1 = 2  −  = 2   = sq. unit 2 3 6 3

1 sq. unit. 3 90. (a) : The graph y = sin x is shown below.

Therefore, the required area is

Required area = Area of region OABO + Area of region BCDB

=

∫

p

0

sin xdx +

∫

2p

0

sin xdx

(since, in region BCDB the graph is below the X-axis, which area come out to be negative, therefore we take the absolute value)

= [ − cos x ]0 + [ − cos x ]p = (– cosp + cos0) + |– cos2p + cos p| = (1 + 1) + |– 1 – 1| = 2 + |– 2| = 2 + 2 [Q |–2| = – (–2) = 2] = 4 sq. units Therefore, the required area is 4 sq. units. p

2p

90

VITEEE CHAPTERWISE SOLUTIONS

91. (a) : We have

and f ′(x) =



=

 1 − x2   −1  f(x) = cos–1  x − x  = cos–1   1 + x 2   x −1 + x  −1  1 − x2  1−   1 + x 2  −1 2

⋅

2

⋅

(1 + x 2 )( −2 x) − (1 − x 2 )( 2 x) (1 + x 2 )2

−4 x 2x = (1 + x 2 ) x (1 + x 2 )

4x which is an odd function, since f ′(–x) = f ′(x). 92. (b) : We have  ln( e / x 2 )   ln( ex 4 )  y = cot–1  + cot–1  2   ln( ex )   ln( e 2 / x 2 )  2  1 + ln x 2  –1  2 − ln x  = tan–1  + tan    1 − ln x 2   1 + 2 ln x 2 

= tan–1 1 + tan–1(ln x2) + tan–1 2 – tan–1(ln x2)  −1 −1 −1  A + B     tan A + tan B = tan  A − B   y = tan–1 1 + tan–1 2. 2 Hence, d y = 0. dx 2 93. (b) : We have, f0(x) = x and fn(x) = efn–1(x) …(i) Thus, we have x f1(x) = ex, f2(x) = ee , …… Diferentiating equation (i) w.r.t. x, we have f n′(x) = f n–1 ′ (x)efn–1(x) = f n–1 ′ (x)fn(x) Thus, we have f n′ (x) = f n–1 ′ (x)fn(x) = [f n–2 ′ (x)fn–1(x)]fn(x) = [f n–3 ′ (x)fn–2(x)]fn–1(x)fn(x) ……… ……… ……… n

= f 0′ (x) f1(x)f2(x) ……fn(x) = ∏ fr ( x) . r =1 94. (c) : We have p f(x) = sin   x −p p and f ′(x) = 2 cos   x x which is negative for value of x, given by p cos   > 0 x p p p i.e., 2np – < < + 2np, n ∈ I 2 x 2 2 2 i.e., 1 (a) Let f(x) = y – (x – 1) = ln x – x + 1 1 Now, f ′(x) = –11 x ⇒ f is decreasing in (1, ∞) Thus, f(x) < f(1) = 0 ⇒ y < x – 1.  1 1 (b) Let f(x) = y –  1 −  = ln x – 1 +  x x 1 1 x−1 Now, f ′(x) = − 2 = 2 > 0 ∀ x > 1 x x x ⇒ f is increasing in (1, ∞) 1 Thus, f(x) > f(1) = 0 ⇒ y > 1 – . x (c) Let f(x) = y – (x2 – 1) = ln x – x2 + 1  1 −2  x 2 −  1   2 Now, f ′(x) = – 2x = 1 x x ⇒ f is decreasing in (1, ∞) Thus, f(x) < f(1) = 0 ⇒ y < x2 – 1. 96. (a) 97. (b) : We have

x2 x 3 + 200 ( x 3 + 200)2 x − x 2 ⋅ 3x 2 Then, f ′(x) = ( x 3 + 200)2 3 − x( x − 400) = ( x 3 + 200)2 From the sign scheme for f ′(x), we can see that f attains maxima at x = (400)1/3.

f(x) =

Now, we have 7 < 4001/3 < 8 72 49 a7 = 3 = 7 + 200 543 82 64 and a8 = 3 = 8 + 200 712 49 64 Since , therefore a7 is the largest > 543 712 term of the given sequence. p

98. (a) : Let I = ∫0 [tan x]dx p

p

Also, I = ∫0 [tan(p – x)]dx = ∫0 [– tan x]dx

p

= ∫0 (–[tanx] – 1)dx ( [–(I + f )] = – I – 1) =–I–p gives I = – p/2.

91

Model Test Paper-4

99. (d) :

1 1 1 + − 1 + 2w 2 + w 1 + w 1  1+ w − 2 − w  = + 1 + 2w  ( 2 + w )(1 + w )  1 −1 = + 1 + 2w 2 + 3w + w 2 1 1 = =0 − 1 + 2w 1 + 2w (Q 2 + 3w + w2 = (1 + 2w) + (1 + w + w2))

100. (b) : We have (2 + i)(2 + 2i)(2 + 3i) …… (2 + ni) = x + iy …(i)

103. (d) : Clearly, the triangle is right angled as x – 2 = 0 and y – 5 = 0 are perpendicular. \ The point of intersection (2, 5) is the orthocentre of the triangle. 104. (a) : Let P(h, k) be a point on the circle 15x2 + 15y2 – 48x + 64y = 0. Then the lengths PT1 and PT2 of the tangents from P(h, k) to 5x2 + 5y2 – 24x + 32y + 75 = 0 and 5x2 + 5y2 – 48x + 64y + 300 = 0 respectively are

PT1 =

h2 + k 2 −

24 32 h + k + 15 5 5

h2 + k 2 −

48 64 h + k + 60 5 5

⇒ (2 + i)(2 + 2i)(2 + 3i) …… (2 + ni) = x + iy ⇒ (2 – i)(2 – 2i) (2 – 3i) …… (2 – ni) = x – iy …(ii) Multiplying (i) and (ii),we get (4 – i2)(4 – 4i2)(4 – 9i2) …… (4 – n2i2) = x2 – i2y2 ⇒ 5 . 8 . 13. …… (4 + n2) = x2 + y2.



and PT2 =



Since, (h, k) lies on 15x2 + 15y2 – 48x + 64y = 0



\

h2 + k2 =

101. (b) : Given exp. = (1 + w2)(1 + w) + (2 + w2)(2 + w) + (3 + w2)(3 + w) + …… + (n + w2)(n + w) = (1 + (w + w2) + w3) + (4 + 2(w + w2) + w3) + (9 + 3(w + w2) + w3) + …… + (n2 + n(w + w2) + w3) = (1 – 1 + 1) + (4 – 2 + 1) + (9 – 3 + 1) + …… + (n2 – n + 1) 2 = (1 + 4 + 9 + …… + n ) – (1 + 2 + 3 + …… + n) + n



\

PT1 =





=

n(n + 1)( 2n + 1) n(n + 1) n(n2 + 2) − +n= 6 2 3

line through P(3, 4) p and making angle with x-axis, is 6 x−3 y−4 = r, where r is the distance of = p p cos sin 6 6 any point on the line from P(3, 4). Coordinates of any point Q on it are p p    r cos + 3, r sin + 4  6 6

102. (c) : Equation



of

If Q lies on 12x + 5y + 10 = 0, then  3  r 12  3 + r  + 5  4 +  + 10 = 0  2  2



⇒ r =



\

−132 12 3 + 5

Length PQ =

− 132 12 3 + 5

.



and



\

48 64 h− k 15 15

= PT2 = = 2

48 64 24 32 h − k − h + k + 15 15 15 5 5 32 24 k − h + 15 15 15 48 64 48 64 h − k − h + k + 60 15 15 5 5 32 −24 h + k + 15 = 2(PT1) 15 15

PT1 : PT2 = 1 : 2

105. (c) : The equation of tangent at (4, –2) to y2 = x is 1 –2y = (x + 4) or x + 4y + 4 = 0 2

1 . Therefore, the slope of the 4 perpendicular line is 4. Since, the tangents at the end of the focal chord of a parabola are at right angle, therefore, the slope of the tangent at Q is 4. Its slope is –

106. (d) : We have y2 – 2y = x – 2 or (y – 1)2 = x – 1. Shifting the origin at (1, 1), we have x = X + 1, y = Y + 1. The equation (y – 1)2 = (x – 1) reduces to Y2 = X. This represents a parabola with L.R. = 1. Coordinates of the focus w.r.t. new 1 axes are ( , 0). So, the coordinates of the 4 5 focus w.r.t. old axes are ( , 1). 4

92

VITEEE CHAPTERWISE SOLUTIONS

107. (c) : Area of DAPA′ is maximum when its height is maximum because its base is fixed.



\

Maximum area =

1 × 2a × b = ab. 2

x2 y 2 − = 1, b2 = a2(e2 – 1) a2 b2 b2 a2 + b2 or e2 = 1 + 2 = a a2 x2 y 2 Conjugate hyperbola 2 − 2 = – 1 a b 2 2 y x a2 a2 + b2 or − 2 = 1 \ e′2 = 1 + 2 = 2 b a b b2 2 2 1 1 a +b \ = 1. + 2 = 2 2 e e′ a + b2

108. (b) :

 3 −2 1 5 109. (d) : Augmented Matrix = 6 −4 2 10  9 −6 3 15  3 −1 1 5 R2 − 2 R1   0 −2 0 0  R3 − 3 R1  0 −3 0 0   3 −1 1 5 3   R3 − R2 0 −2 0 0  2 0 0 0 0  Therefore the given equations have infinitely many solutions. 110. (d) : A2 – 4A – 5I = 0 ⇒ A(A – 4I) = 5I 1 ⇒ A ⋅ (A – 4I) = I 5 1 ⇒ A–1 = (A – 4I) (Q AA–1 = 1) 5 111. (a) 112. (a) 113. (b) 114. (a) 115. (b) : Given, x2 + y2 = 1 On differentiating w.r.t. x, we get 2x + 2yy′ = 0 ⇒ x + yy′ = 0 Again, differentiating, we get 1 + yy′′ + (y′)2 = 0 116. (a) : The equation of a member of the family of parabolas having axis parallel to y-axis is

y = Ax2 + Bx + C, where A, B and C are arbitrary constant. d3 y d2 y dy ⇒ = 2Ax + B ⇒ 2 = 2A ⇒ =0 dx dx 3 dx 117. (d) : Given, differential equation can be rewritten as dx (log y − x ) = dy y log y dx x 1 ⇒ + = dy y log y y 1

∫ y log y dy

log log y = e ( ) = logy  118. (d) : Since, the vectors a = 2i + log 3 x j + ak  and b = −3i + a log 3 x j + log 3 xk are inclined at acute angle. Therefore,   a ⋅b > 0 ⇒ – 6 + a(log3x)2 + alog3x > 0 for all x > 0 ⇒ – 6 + ay2 + ay > 0, where y = log3x ⇒ ay2 + ay – 6 > 0 for all y ⇒ a > 0 and a2 + 24a < 0 ⇒ a > 0 and a ∈ (–24, 0)  But this  is not possible. Hence, the vectors a and b are not inclined at acute angle for any real value of a. 119. (c) : If D = 0, then    a b c       =0 a ⋅a a ⋅b a ⋅c       a ⋅c b ⋅c c ⋅c  ⇒ la + mb + nc = 0 (Where l, m, n are consistant)    ⇒ a , b , c are L.D., which is a contradiction. Hence, D can take any non-zero real value.  120. (c) : Let r = xi + y j + zk be the position vector.



\



The given equation is 2  r + r ⋅ (i + j − k ) − 9 = 0 ⇔ 2 xi + y j + zk + ( xi + y j + zk ) ⋅ (i + j − k ) − 9 = 0



⇔ x2 + y2 + z2 + x + y – z – 9 = 0, which is the equation of sphere, whose centre is  1 1 1  − , − ,  2 2 2 \ p.v. of the centre is



vvv

I.F = e

1 1 1 1 − i − j + k = − (i + j − k ) . 2 2 2 2

Model Test Paper-5

Model Test Paper Time : 2 Hours 30 Minutes

Max. Marks : 120

spheres are connected by a conducting wire, then, in equilibrium position, the ratio of the magnitude of electric fields at the surface of the spheres A and B is (a) 1 : 4 (b) 4 : 1 (c) 1 : 2 (d) 2 : 1

physics 1.

2.

3.

4.

5.

A rigid container with thermally insulated walls contains a coil of resistance 100 W, carrying current 1 A. Change in internal energy after 5 min will be (a) zero (b) 10 kJ (c) 20 kJ (d) 30 kJ Four charges equal to –Q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is Q Q (a) − (1 + 2 2 ) (b) (1 + 2 2 ) 4 4 Q Q (c) − (1 + 2 2 ) (d) (1 + 2 2 ) 2 2 Out of the following statements, which is not correct ? (a) When unpolarised light passes through a Nicol’s prism, the emergent light is elliptically polarised. (b) Nicol’s prism works on the principle of double refraction and total internal reflection. (c) Nicol’s prism can be used both, to produce and analyse polarised light. (d) Calcite and Quartz are both, doubly refracting crystals. A coil, a capacitor and an AC source of voltage 24 V (rms) are connected in series. By varying the frequency of the source, a maximum rms current of 6 A is observed. If this coil is connected to a battery of emf 12 V and internal resistance 4 W, the current through it will be (a) 2.0 A (b) 1.5 A (c) 3.0 A (d) 2.5 A Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the

5

93

6.

The sides of a rectangular block are 2 cm, 3 cm and 4 cm. The ratio of maximum to minimum resistance between its parallel faces is (a) 4 (b) 3 (c) 2 (d) 1

7.

What is the value of inductance L for which the current is maximum in a series LCR circuit with C = 10 mF and w = 1000 s–1? (a) 1 mH (b) 10 mH (c) 100 mH (d) cannot be calculated unless R is known

8.

The refractive index of the material of a prism is 2 , and its refracting angle is 30°. One of the refracting surfaces of the prism is made a mirror inwards. A beam of monochromatic light entering the prism from the other face retraces its path, after reflection from mirrored surface, if its angle of incidence on prism is (a) 0° (b) 30° (c) 45° (d) 60°

9.

Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr’s theory, the spectral lines emitted by hydrogen will be (a) one (b) two (c) three (d) four

10. Potentiometer wire of length 1 m is connected in series with 490 W resistance and 2 V battery. If 0.2 mV cm–1 is the potential gradient, then resistance of the potentiometer wire is (a) 4.9 W (b) 7.9 W (c) 5.9 W (d) 6.9 W

94 11. A conducting wire having 1029 free electrons per m3 carries a current of 20 A. If the crosssection of the wire is 1 mm2, then the drift velocity of electrons will be (Take e = 1.6 × 10–19 C) (a) 1.25 × 10–4 m s–1 (b) 1.25 × 10–3 m s–1 (c) 1.25 × 10–5 m s–1 (d) 6.25 × 10–3 m s–1 12. A proton is accelerating on a cyclotron having oscillating frequency of 11 MHz in external magnetic field of 1 T. If the radius of its dees is 55 cm, then its kinetic energy (in MeV) is (mp = 1.67 × 10–27 kg, e = 1.6 × 10–19 C) (a) 13.36 (b) 12.52 (c) 15.89 (d) 14.49 13. A straight wire having mass of 1.2 kg and length of 1 m carries a current of 5 A. If the wire is suspended in mid-air by a uniform horizontal magnetic field, then magnitude of field is (a) 0.65 T (b) 1.53 T (c) 2.4 T (d) 3.2 T 14. In Young’s double slit experiment, the distance between two sources is 0.1 mm. The distance of the screen from the source is 20 cm. Wavelength of light used is 5460 Å. The angular position of the first dark fringe is (a) 0.08° (b) 0.16° (c) 0.20° (d) 0.32° 15. The length of germanium rod is 0.928 cm and its area of cross-section is 1 mm2. If for germanium ni = 2.5 × 1019 m–3, mh = 0.19 m2 V–1s–1 and me = 0.39 m2 V–1s–1, then the resistance of rod is (a) 2.5 kW (b) 4 kW (c) 5 kW (d) 10 kW 16. A common emitter amplifier has a voltage gain of 50, an input impedance of 100 W and an output impedance of 200 W. The power gain of the amplifier is (a) 500 (b) 1000 (c) 1250 (d) 100 17. Light rays of wavelength 6000 Å and of photon intensity 39.6 W m–2 is incident on a metal surface. If only 1% of photons incident on surface emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be (Given, h = 6.64 × 10–34 J s, c = 3 × 108 m s–1) (a) 12 × 1018 (b) 10 × 1018 18 (c) 1.2 × 10 (d) 12 × 1016

VITEEE CHAPTERWISE SOLUTIONS 257 18. If the atom 100 Fm follows the Bohr model 257 and the radius of fifth orbit of 100 Fm is N times the Bohr radius, then the value of N is (a) 100 (b) 200 (c) 4 (d) 1/4

19. What is the refractive index of material of a planoconvex lens, if the radius of curvature of the convex surface is 10 cm and focal length of the lens is 30 cm? 7 6 (a) (b) 4 5 4 2 (d) 3 3 20. Spectrum of sunlight is an example for (a) band emission spectrum (b) line absorption spectrum (c) continuous emission spectrum (d) continuous absorption spectrum (c)

21. Two concentric and coplanar circular coils have radii a and b as shown in figure. Resistance of the inner coil is R. Current in the other coil is increased from 0 to I, then the total charge circulating the inner coil is (a)

m0 Iab 2R

(b)

m0 Iapb2 2 ab

(c)

m0 Ib 2 pR

(d)

m0 Ia2 2 Rb

22. The dielectric strength of air is 3 × 106 V m–1. A parallel plate capacitor has area 20 cm2 and plate separation 0.1 mm. Find the maximum rms voltage of an ac source which can be connected. (a) 210 V (b) 300 V (c) 435 V (d) 490 V 23. A galvanometer of resistance 20 W gives a full scale deflection when a current of 0.04 A is passed through it. It is desired to convert it into an ammeter of range 20 A. The only shunt available is 0.05 W. The resistance that must be connected in series with the coil of the galvanometer is (a) 4.95 W (b) 5.94 W (c) 9.45 W (d) 12.62 W

95

Model Test Paper-5

24. At a given plane on the earth’s surface, the horizontal component of earth’s magnetic field is 3 × 10–5 T and resultant magnetic field is 6 × 10–5 T. Angle of dip at this place is (a) 30° (b) 40° (c) 50° (d) 60° 25. In a Young’s double slit experiment, the fringes are displaced by a distance x when a glass plate of refractive index 1.5 is introduced in the path of one of the beams. When this plate is replaced by another plate of same thickness, the shift of fringes is (3/2)x. The refractive index of second plate is (a) 1.75 (b) 1.50 (c) 1.25 (d) 1.00 2 26. The binding energy of deuteron (1 H) is 1.15 MeV per nucleon and an alpha 4 particle ( 2 He) has binding energy of 7.1 MeV per nucleon. Then in the reaction 2 H +12 H → 42 He + Q, the energy released Q is 1 (a) 5.95 MeV (b) 26.1 MeV (c) 23.8 MeV (d) 289.4 MeV

27. In Young’s experiment the wavelength of red light is 7.5 × 10–5 cm and that of blue light 5.0 × 10–5 cm. The value of n for which (n + 1)th blue bright band coincides with nth red band is (a) 8 (b) 4 (c) 2 (d) 1 28. Two batteries of emf 4 V and 8 V with internal resistance 1 W and 2 W are connected in a circuit with a resistance of 9 W as shown in figure. The current and potential difference between the points P and Q are

1 (a) A and 3 V 3 1 (c) A and 9 V 9

1 (b) A and 4 V 6 1 (d) A and 12 V 2

29. In a common emitter amplifier, using output resistance of 5000 W and input resistance of 2000 W, if the peak value of input signal voltage is 10 mV and b = 50, then peak value of output voltage is (a) 5 × 10–6 V (b) 12.5 × 10–4 V (c) 1.25 V (d) 125 V

30. Three identical coils, A, B and C are placed with their planes parallel to one another. Coils A and C carry currents as shown in figure. Coils B and C are fixed in the position and coil A is moved towards B. Then, current induced in B is in (a) clockwise direction (b) anticlockwise direction (c) no current is induced in B (d) current is induced only when both coils move. 31. A coil has an inductance of 0.7 H and is joined in series with a resistance of 220 W. When an alternating e.m.f. of 220 V at 50 cycles per second, is applied to it, then wattless component of current in the circuit is (a) 7 A (b) 5 A (c) 0.7 A (d) 0.5 A 32. At the present time, the elementary particles are considered to be the (a) photons and baryons (b) leptons and quarks (c) baryons and quarks (d) baryons and leptons. 33. Polaroid glass is used in sun glasses because (a) it reduces the light intensity to half on account of polarisation (b) it is fashionable (c) it has good colour (d) it is cheaper. 34. A cylindrical rod is reformed to half of its original length keeping volume constant. If its resistance before this change were R, then the resistance after reformation of rod will be R (a) R (b) 4 3R R (c) (d) 4 2 35. A current carrying wire in the shape of a circle as the current progresses along the wire the direction of current density changes in an exact manner while the current I remains unaffected. The responsible factor for it is (a) the charges ahead (b) electric field produced by charges accumulated on the surface of wire. (c) the charges just behind a given segment of wire which push them, right away by repulsion. (d) None of these

96

VITEEE CHAPTERWISE SOLUTIONS

36. A and B are two points on a uniform ring of resistance 15 W. The ∠AOB = 45°. The equivalent resistance between A and B is (a) 1.64 W (b) 2.84 W (c) 4.57 W (d) 2.64 W

Chemistry

45°

37. Two wires of same dimension but resistivities r1 and r2 are connected in series. The equivalent resistivity of the combination is (a) r1 + r2

(b)

r1r2

(c)

1 (r1 + r2) 2

(d) 2(r1 + r2)

38. A small current element of length dl is placed at (1, 1, 0) and is carrying current in +z direction.  If magnetic field at origin be B1 and at point  (2, 2, 0) be B2 , then 



(a) B1 = B2 









(b) B1 = 2B2



(c) B1 = −B2

(d) B1 = −2B2

39. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is p/2 at point A and p at point B. Then the difference between the resultant intensities at A and B is (a) 2I (b) 4I (c) 5I (d) 7I 40. A radioactive nucleus X decays to a stable nucleus Y. Then the graph of rate of formation of Y against time t will be

(a)

(c)





(b)

(d)

41. What will be the standard internal energy change for the reaction at 298 K? OF2(g) + H2O(g) → O2(g) + 2HF(g); DH° = – 310 kJ (a) –312.5 kJ (c) –310 kJ

(b) –125.03 kJ (d) –156 kJ

42. If electron, hydrogen, helium and neon nuclei are all moving with the velocity of light, then the wavelengths associated with these particles are in the order : (a) electron > hydrogen > helium > neon (b) electron > helium > hydrogen > neon (c) electron < hydrogen < helium < neon (d) neon < hydrogen < helium < electron 43. Dye test can be used to distinguish between (a) ethylamine and acetamide (b) ethylamine and aniline (c) urea and acetamide (d) methylamine and ethylamine. 44. Which of the following is a specific test for proteins? (a) Beilstein test (b) Biuret test (c) Benedict’s test (d) Molisch’s test 45. When one mole of each of the following complexes is treated with excess of AgNO3, which will give maximum amount of AgCl? (a) [Co(NH3)6]Cl3 (b) [Co(NH3)5Cl]Cl2 (c) [Co(NH3)4Cl2]Cl (d) [Co(NH3)3Cl3] 46. Match the compounds given in Column I with the hybridisation, shape given in Column II and mark the correct option. Column I Column II (A) XeF6 1. sp3d3 - distorted octahedral (B) XeO3 2. sp3d2 - square planar (C) XeOF4 3. sp3 - pyramidal (D) XeF4 4. sp3d2 - square pyramidal A B C D (a) 1 4 2 3 (b) 2 3 4 1 (c) 1 3 4 2 (d) 4 2 3 1 47. Arrange the following carbonyl compounds in increasing order of their reactivity in nucleophilic addition reactions. I. Ethanal II. Propanal III. Propanone IV. Butanone (a) I < IV < III < II (b) II < III < IV < I (c) IV < III < II < I (d) I < III < IV < II

97

Model Test Paper-5

48. How many six-fold axis of symmetry does a hexagonal close packing have? (a) 2 (b) 3 (c) 1 (d) 4 49. Nitrogen forms stable N2 molecule but phosphorus is converted to P4 from P2 because (a) pp - pp bonding is strong in phosphorus (b) pp - pp bonding is weak in phosphorus (c) triple bond is present in phosphorus (d) single P — P bond is weaker than N — N bond. 50. Which of the following is not metabolized in the body? (a) Sucrose (b) D-Glucose (c) L-Glucose (d) Aspartame 51. The number of nodal planes in a px orbital is (a) one (b) two (c) three (d) zero. 52. The optically active tartaric acid is named as D-(+)-tartaric acid because it has a positive (a) optical rotation and is derived from D-glucose (b) pH in organic solvent (c) optical rotation and is derived from D-(+)-glyceraldehyde (d) optical rotation only when substituted by deuterium. 53. The pH of neutral water at 25 °C is 7.0. As the temperature increases, ionisation of water increases, however, the concentration of H+ ions and OH – ions are equal. What will be the pH of pure water at 60 °C? (a) Equal to 7.0 (b) Greater than 7.0 (c) Less than 7.0 (d) Equal to zero 54. The carboxyl functional group (–COOH) is present in (a) picric acid (b) barbituric acid (c) ascorbic acid (d) aspirin. 55. The reaction is SO 2 + Cl 2  SO 2Cl 2 exothermic and reversible. A mixture of SO2(g), Cl2(g) and SO2Cl2(g) is at equilibrium in a closed container. Now a certain quantity of extra SO2 is introduced into the container, the volume remains the same. Which of the following is true? (a) The temperature will decrease. (b) The temperature will increase. (c) The temperature will not change.

(d) The pressure inside the container will not change. 56. Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? (a) 863.73 s (b) 500 s (c) 275 s (d) 925 s 57. What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid? (a) Cr3+ and Cr2O72– are formed. (b) Cr2O72– and H2O are formed. (c) CrO42– is reduced to +3 state of Cr. (d) CrO42– is oxidised to +7 state of Cr. 58. Which compound is not a lipid? (a) Lecithin (b) Lysine (c) Cerebroside (d) Cephalin 59. In a cubic packed structure of mixed oxides, the lattice is made up of oxide ions, one-fifth of tetrahedral voids are occupied by divalent (X2+) ions while one-half of the octahedral voids are occupied by trivalent ions (Y3+), then the formula of the oxide is (a) XY2O4 (b) X2YO4 (c) X4Y5O10 (d) X5Y4O10 O¯

O

CH3 CH3 attacks H 60. Ph to form a hydroxy ketone. How many chiral carbons would be there? (a) 1 (b) 2 (c) 3 (d) 0

61. The correct order of the thermal stability of hydrogen halides (H – X) is (a) HI > HBr > HCl > HF (b) HF > HCl > HBr > HI (c) HCl < HF > HBr < HI (d) HI > HCl < HF > HBr 62. Which of the following configurations does not follow Hund’s rule of maximum multiplicity? (a) 1s2 2s2 2p6 3s2 3p2 (b) 1s2 2s2 2p6 3s2 3p6 4s2 3d6 (c) 1s2 2s2 2p6 3s2 3p6 4s1 3d5 (d) 1s2 2s2 2p6 3s2 3p4 4s2

98

VITEEE CHAPTERWISE SOLUTIONS

63. Among the following sets of reactants which one produces anisole? (a) CH3CHO ; RMgX (b) C6H5OH ; NaOH ; CH3I (c) C6H5OH ; neutral FeCl3 (d) C6H5CH3 ; CH3COCl ; AlCl3 64. The Ksp of CuS, Ag2S and HgS are 10–31, 10–44 and 10–54 respectively. The solubility of these sulphides are in the order : (a) Ag2S > CuS > HgS (b) Ag2S > HgS > CuS (c) HgS > Ag2S > CuS (d) CuS > Ag2S > HgS. 65. In the reaction 2N2O5 → 4NO2 + O2, the rate is expressed as (i) −



d  N 2O 5  dt

(ii)

d  NO 2 

(iii)

d [O2 ]

dt

dt

= k1  N 2O 5 

= k 2  N 2O 5 

= k3 [ N 2 O5 ]

Relation between k1, k2 and k3 is (a) 2k1 = 4k2 = k3 (b) 2k1 = k2 = 4k3 (c) 2k1 = k2 = 2k3 (d) k1 = 4k2 = 2k3.

66. Identify X, Y and Z in the following reactions. NaN3

X

N2O

Y + HNO2

H2O

Z (a) NaOH, NH4OH, HNO2 (b) NaNH2, NH2OH, H2N2O2 (c) NaOH, NH2OH, HNO2 (d) NaNH2, NH4OH, H2N2O2.

67. Which of the following is a method of converting an unsaturated ketone into unsaturated hydrocarbon? (a) Aldol condensation (b) Reimer-Tiemann reaction (c) Cannizzaro’s reaction (d) Wolff-Kishner reduction. 68. A sample of air consisting of N2 and O2 was heated at 2500 K until the equilibrium N2(g) + O2(g) 2NO(g) was established with an equilibrium constant KC = 2.1 × 10–3. At equilibrium, the mole % of

NO was 1.8. Estimate the initial composition of air in mole fraction of N2 and O2. (a) 50%, 50% (b) 75%, 25% (c) 79%, 21% (d) 85%, 15% 69. Which of the following compounds does not react with NaNO2 and HCl? (a) C6H5OH (b) C6H5NH2 (c) (CH3)3CNO2 (d) (CH3)2CHNO2 70. If E1, E2 and E3 represent respectively the kinetic energies of electron, an alpha particle and a proton each having same de-Broglie wavelength then (a) E1 > E3 > E2 (b) E2 > E3 > E1 (c) E1 > E2 > E3 (d) E1 = E2 = E3. 71. 1 M NaCl and 1 M HCl are present in an aqueous solution. The solution is (a) not a buffer solution with pH < 7 (b) not a buffer solution with pH > 7 (c) a buffer solution with pH < 7 (d) a buffer solution with pH > 7. 72. Propan-1-ol and propan-2-ol can be best distinguished by (a) oxidation with KMnO4 followed by reaction with Fehling’s solution (b) oxidation with acidic dichromate followed by reaction with Fehling’s solution (c) oxidation by heating with copper followed by reaction with Fehling’s solution (d) oxidation with concentrated H2SO4 followed by reaction with Fehling’s solution. 73. The colour of light absorbed by an aqueous solution of CuSO4 is (a) orange-red (b) blue-green (c) yellow (d) violet. 74. If the radius of an octahedral void is r and radius of atoms in close packing is R, the relation between r and R is (a) r = 0.414R (b) R = 0.414r (c) r = 2R (d) r = 2R 75. The product (R) in the following reaction sequence would be

99

Model Test Paper-5

Mathematics (a)

(c)

(b)



(d)

76. Elements of which of the following groups will form anions most readily? (a) Oxygen family (b) Nitrogen family (c) Halogens (d) Alkali metals. 77. Dehydrogenation of 1-bromo-1,2-diphenylpropane proceeds through E2 mechanism to form 1,2-diphenylpropene, what do you expect about this reaction? (a) It is a stereoselective reaction. (b) It is a stereospecific reaction. (c) Both (a) and (b). (d) None of these. 78. When enthalpy and entropy change for a chemical reaction are –2.5 × 103 cal mol–1 and 7.4 cal deg–1 mol–1, respectively predict the reaction at 298 K is (a) irreversible (b) reversible (c) spontaneous (d) non-spontaneous. 79. Which of the following ions is paramagnetic? (a) [Ni(H2O)6]2+ (b) [Fe(CN)6]4– (c) [Ni(CO)4] (d) [Ni(CN)4]2– 80. Five-membered ring structures of fructose are given below. Mark the incorrect statement.

81. The equation of the sphere with centre (3, 6, –4) and touching the plane 2x – 2y – z – 10 = 0 is (a) x2 + y2 + z2 – 6x + 12y – 8z + 45 = 0 (b) x2 + y2 + z2 + 6x – 9y + 8z + 35 = 0 (c) x2 + y2 + z2 – 6x + 9y + 8z + 35 = 0 (d) x2 + y2 + z2 – 6x – 12y + 8z + 45 = 0 82. Consider the planes 3x – 6y + 2z + 5 = 0 and 4x – 12y + 3z = 3. The plane 67x + 162y + 47z + 44 = 0 bisects the angle between the planes which (a) contains origin (b) is acute (c) Both (a) and (b) (d) None of these 83. The position vector of a point lying on the line joining the points whose position vectors are i + j − k and i − j + k is (a) j (c) k

(b) i  (d) 0

  84. If the volume of the parallelopiped with a , b  and c as coterminous edges is 40 cu unit, then the volume of the parallelopiped having       b + c , c + a and a + b as coterminous edges in cubic unit is (a) 80 (b) 120 (c) 160 (d) 40 85. Solution

of

x dx + y dy x dx − y dy (a)

the =

y3 x3

differential is given by

y x3 2 + y 3 2 3 log   + log x 2 x3 2 y + tan −1   x

(b)

(a) The five-membered ring structures are named as furanose structures. (b) The cyclic structures represent two anomers of fructose. (c) Five-membered ring structures are named as pyranose structures. (d) These are also called Haworth structures.

equation

32

x3 2 + y 3 2 y 2 log   + log 3 x x3 2 + tan − 1

(c)

y +c =0 x

 3 y x+ y 2 −1 y log   + log   + tan  3 3 x x x

(d) None of the above

+c=0

2



2

+c=0

100

VITEEE CHAPTERWISE SOLUTIONS

86. Solution of the differential equation 2

3

dy (xy) 2  dy  (xy) 3  dy  x = 1 + xy +   +   + ... is dx 2 !  dx  3 !  dx  (a) y = loge(x) + c

(b) y = (loge x)2 + c

(c) y = ± (log e x) 2 + 2c (d) xy = xy + k

1 , 2 x = 2, y = log x and y = 2x, then the area of this region, is 4 5 (a) sq unit (b) sq unit 3 3

87. The region bounded by the curves x =

3 (d) None of these sq unit 2 88. If a = cos a + i sin a, b = cos b + i sin b, b c a c = cos g + i sin g and + + = 1, then c a b cos(b – g) + cos (g – a) + cos (a – b) is equal to 3 3 (a) (b) − 2 2 (c)

(c) 0

(d) 1

 2p   2p  89. Let w n = cos   + i sin   , i 2 = −1 , then n n (x + yw3 + zw23) (x + yw23 + zw3) is equal to (a) 0 (b) x2 + y2 + z2 (c) x2 + y2 + z2 – yz – zx – xy (d) x2 + y2 + z2 + yz + zx + xy 90. Locus of the image of the point (2, 3) in the line (x – 2y + 3) + l(2x – 3y + 4) = 0 is (l ∈ R) (a) x2 + y2 – 3x – 4y – 4 = 0 (b) 2x2 + 3y2 + 2x + 4y – 7 = 0 (c) x2 + y2 – 2x – 4y + 4 = 0 (d) None of the above 91. Let A(2, –3) and B(–2, 1) be vertices of a DABC. If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex C is the line (a) 2x + 3y = 9 (b) 2x – 3y = 7 (c) 3x + 2y = 5 (d) 3x – 2y = 3 92. Two circles with radii a and b touch each other externally such that q is the angle between the direct common tangents (a > b ≥ 2), then, a−b (a) q = 2 cos −1   a+b a+b (b) q = 2 cos −1   a−b

a+b (c) q = 2 sin −1   a−b a−b (d) q = 2 sin −1   a+b 93. The equation of the parabola, the extremities of whose latus rectum are (1, 2) and (1, –4), is (a) (x + 1)2 = 3(2y – 5) (b) (y + 1)2 = 3(2x – 5) (c) (y + 1)2 = –3(2x + 5) (d) None of these 94. If a point (x, y) = (tan q + sin q, tan q – sin q), then locus of (x, y) is (a) (x2y)2/3 + (xy2)2/3 = 1 (b) x2 – y2 = 4xy (c) (x2 – y2)2 = 16 xy (d) x2 – y2 = 6xy 95. The value of m, for which the line y = mx +

25 3 3

is a normal to the conic

x2 y2 − = 1, is 16 9 (a) −

2 3



(b)

3

3 (d) None of these 2 96. Using the method of integration, find the area bounded by the curve |x| + |y| = 1 (a) 3 sq. units (b) 2 sq. units (c) 5 sq. units (d) 10 sq. units (c) −

97. Form the differential equation of the family of circles having centre on Y-axis and radius 3 units. (a) (x2 – 5) (y′)2 + x2 = 0 (b) (x + 2) (y′)2 + x2 = 0 (c) (x2 + 9) (y′)2 + x2 = 0 (d) xy′ = 0 98. For the differential equation dy xy = (x + 2)( y + 2), find the solution curve dx passing through the point (1, –1). (a) y + x – 2 = log [x(y + 2)] (b) –y – x + 2 = log [x2(y + 2)2] (c) y – x + 2 = log [x(y + 2)] (d) y – x + 2 = log [x2(y + 2)2]

101

Model Test Paper-5

   99. Let a = i + j + k , b = 4i − 2j + 3k and c = i − 2j + k and find a vector of magnitude 6 units which    is parallel to the vector 2a − b + 3c. (a) 6 i + 8j − 10 k

(b) 2i − 4j + 4 k

(c) 2 i + 4j − 4 k (d) none of these             100. (b × c ) ⋅ ( a × d ) + (c × a ) ⋅ (b × d ) + ( a × b ) ⋅ (c × d ) is equal to (a) 0

(b) 1

(c) 2

(d) 3

(b) OY (d) all of these

102. If 10 ten rupee coins, 5 five rupee coins are to be placed in a line, then the probability that the extreme coins are five rupee coins is (a)

1 15!

(b)

1 10!

(c)

16

C2



40

C5

16

C 2 × 23C 2 40

C5

(b)



120 504

(d)

167 1024

106. A plane whose equation is 2x – y + 3z + 5 = 0 is rotated through 90° about its line of intersection with the plane 5x – 4y – 2z + 1 = 0. Find the equation of the plane in the new position.

23

C2

40

C5



(d) None of these

104. Mean and variance of a binomial variable X are 2 and 1 respectively, then P(X ≥ 1) is (a)

2 3

(b)

7 8

(c)

4 5

(d)

15 16

(c) 27x – 26y = 130 (d) 9x + 12y – 13z = 10 107. Find the plane formed by the lines AB and AC, where A is (–1, 0, 2) and d⋅r’s of AB and AC are 4, 1, 3 and 1, 7, 3. (a) 2x + y – z + 6 = 0 (b) 2x + y + 3z – 8 = 0 (c) x + 2y + 3z = 8 (d) 2x – y – 3z + 8 = 0

5 !10 ! (c) (d) None of these 15 ! 103. From a set of 40 cards numbered 1 to 40, 5 cards are drawn at random and arranged in ascending order of magnitude x1 < x2 < x3 < x4 < x5. The probability x3 = 24, is (a)

(b)

(a) 9x – 12y + 13z = 13 (b) 27x – 24y – 26z = 13

101. Show that the vector i + j + k is equally inclined to axes OX, OY and OZ. (a) OX (c) OZ

150 523 176 (c) 1024 (a)

105. Ten coins are thrown simultaneously. Then, the probability of getting atleast seven heads, is

108. OABC is a tetrahedron such that OA = OB = OC = k and ∠AOB = ∠BOC = ∠COA = q. The range of q is  p (a)  0,   2

p p (b)  ,  4 2

 2p   p 2p  (c)  ,  (d) 0,  . 3 3  3   109. The vertices of a triangle are A(m , n), B(12, 19) and C(23, 20), where m and n are integers. If its area is 70 and the slope of the median through A is –5, then m + n is (a) 47 (b) 50 (c) 20 (d) – 27 110. Two vertices of a triangle are (3, –2) and (–1, 5). If its orthocentre is the origin, its circumradius is (a)

65

(b)

65 2

(c)

65 2

(d)

65 . 3

102

VITEEE CHAPTERWISE SOLUTIONS

111. The locus of the midpoints of the chords of the circle 4x2 + 4y2 – 12x + 4y + 1 = 0 that

positive then (a) abc > – 11 (c) abc > – 8

2p subtend an angle of at its centre is 3 (a)

x2

+

y2

x x2

31 + 3x – y − =0 16

(b) x2 + y2 – 3x + y +

(d)

+

y2

31 =0 16

31 – 3x + y − =0 16

described on AB as diameter is cut by the directrix, then the length of the intercept is a (a) a (b) 2 (c) 2a (d) none of these 113. The point on the ellipse x2 + 2y2 = 6 whose distance from the line x + y = 7 is minimum is (a) (– 2, – 1) (b) (2, – 1) (c) (2, 1) (d) (– 1, 2) 114. If a, b, c are roots of y3 – 3y2 + 3y + 26 = 0 and w is cube roots of unity, then the value of a−1 b−1 c −1 equals + + b−1 c −1 a−1 (b) 3w2 (d) 2w2

 rp   rp  115. If zr = cos   + i sin   then  10   10  equals (a) 0 (c) – 1

(a) 16

(b) – 16

(c) 24

(d) None of these

2p

112. Let AB be a focal chord of y2 = 4ax. If the circle

(a) –3w2 (c) w2

x3

117. If f (x) = 1 2x 3x 2 , then f ′(2) is equal to 0 2 6x

(c) x2 + y2 + 4x – 30 = 0 x2

(b) abc < – 8 (d) None of these

4

∏ zr r =1

118.

 a + b sec x 

∫ log  a − b sec x  dx equals 0

(a)

p(a + b) a−b

(b)

(c)

p 2 2 (a – b ) 2

(d) 0 y

x

119. If

∫

p 2

(3 − sin 2 z ) dz + cos z dz = 0 then

∫

p/3

0

dy equals dx (a)

sin 2 x − 3 cos y

(b)

(c)

3 − sin 2 x cos y

(d) −

cos y 3 − sin 2 x



3 − sin 2 x cos y

3

120. x (3 − x)999 dx equals

∫ 0

(b) 1 (d) None of these

(a)

31001 (1000)(1001)

(b)

a 1 1 1 b 1 1 1 c

(c)

31000 (1000) × (1001)

(d) None of these

116. If the value of determinant

is

31000 999 × 1000

103

Model Test Paper-5

Answer Key 1. (d) 9. (c) 17. (c) 25. (a) 33. (a) 41. (a) 49. (b) 57. (b) 65. (b) 73. (a) 81. (d) 89. (c) 97. (c) 105. (c) 113. (c)

2. (b) 10. (a) 18. (d) 26. (c) 34. (b) 42. (a) 50. (c) 58. (b) 66. (b) 74. (a) 82. (c) 90. (c) 98. (c) 106. (b) 114. (b)

3. 11. 19. 27. 35. 43. 51. 59. 67. 75. 83. 91. 99. 107. 115.

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

4. (b) 12. (d) 20. (b) 28. (a) 36. (a) 44. (b) 52. (c) 60. (b) 68. (c) 76. (c) 84. (a) 92. (d) 100. (a) 108. (d) 116. (c)

5. 13. 21. 29. 37. 45. 53. 61. 69. 77. 85. 93. 101. 109. 117.

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

6. 14. 22. 30. 38. 46. 54. 62. 70. 78. 86. 94. 102. 110. 118.

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

7. 15. 23. 31. 39. 47. 55. 63. 71. 79. 87. 95. 103. 111. 119.

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

8. 16. 24. 32. 40. 48. 56. 64. 72. 80. 88. 96. 104. 112. 120.

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

e planations Physics 1.

(d) : As container is having thermally insulated walls. Hence, W = 0. Therefore, from first law of thermodynamics, DU = DQ = I2Rt = (1)2(100) (5 × 60)J = 30 kJ

2.

(b)

3.

(a) : The emergent light from a nicol prism is not elliptically polarised.

4.

(b) : With an ac source, current in the circuit is maximum when, Z = Zmin = R (the resistance of coil) ∴

I=

12 r+R

(where r is the internal resistance of source) 5.

6.

12 = = 1. 5 A 4+4

(d) : When joined by a wire, the two spheres attain common potential V. V 1 qA \ Electric field, EA = = 2 4 pe0 RA RA

(a) : R1 =

V RB

EA RB 2 = = EB RA 1

\

rl rl1 ; R2 = 2 A2 A1 rl1 l1 rl12 × = A1 l1 V



R1 =

R2 =

rl2 l2 rl22 × = A2 l2 V

\

R1 l12 = R2 l22

For maximum resistance, l1 = 4 cm For minimum resistance, l2 = 2 cm R1 16 \ =4 = R2 4

24 R= =4W 6

When connected with an dc source of emf 12 V then current through the coil

Similarly, EB =

7.

(c) : In series LCR circuit, maximum at resonance. At resonance, XL = XC or wL = or w2 = 1 LC

or L = 12 wC

is

1 wC

Given w = 1000 s–1 and C = 10 mF ∴ L=

current

1 = 0.1 H = 100 mH 1000 × 1000 × 10 × 10 −6

104 8.

VITEEE CHAPTERWISE SOLUTIONS

(c) : Here m = 2 , A = 30°. On reflection from mirrored surface, the ray will retrace its path, if it falls normally on the surface. In DAED





∴ sin i = m sin r = 2 sin 30° = 2 ×

1 1 = 2 2

or i = 45° (c) : Ionisation potential of hydrogen atom is 13.6 eV. Energy required for exciting the hydrogen atom in the ground state to orbit n is given by E = En – E1

or

13.6  −13.6  13.6 −  = − 2 + 13.6 n2  12  n

−1.5 =

−13.6 13.6 or n2 = = 9 or n = 3 n2 1.5

Number of spectral lines emitted n(n − 1) 3 × 2 = = =3 2 2

10. (a) : Potential across potentiometer wire Q \

V=

−3

(0.2 × 10 )V × 1m 10 −2 m

= 0.02 V

Re V= r+R R 0.02 = ×2 r+R

(where R is resistance of potentiometer wire and r is resistance connected in series.) or 0.02(490 + R) = 2R ⇒ R = 4.9 W 11. (b) : Here, I = 20 A, n = 1029 m–3 A = 1 mm2 = 10–6 m2, e = 1.6 × 10–19 C

I Drift velocity, vd = nAe 20A

=

10 29 m −3 × 10 −6 m 2 × 1.6 × 10 −19 C

= 1.25 × 10–3 m s–1 12. (d) : Here, uc = 11 MHz = 11 × 106 Hz B = 1 T, R = 55 cm = 55 × 10–2 m,

e = 1.6 × 10–19C and mp = 1.67 × 10–27 kg, q 2 B2 R 2 K.E. = 2m =

(1.6 × 10 −19 )2 × (1)2 × ( 55 × 10 −2 )2

2 × 1.67 × 10 −27 = 23.19 × 10–13 J 23.19 × 10 −13

eV = 14.49 × 106 eV 1.6 × 10 −19 = 14.49 MeV 13. (c) : For mid-air suspension, the upward force F on wire due to magnetic field B must be balanced by the force due to gravity, ⇒ IlB = mg mg B= Il Here, m = 1.2 kg, g = 10 m s–2, I = 5 A, l = 1 m 1.2 × 10 \ B = = 2.4 T 5×1

sin i As m = sin r

12.1 = −

\

30° + 90° + ∠D = 180° or ∠D = 60° Also ∠ D + ∠r = 90° or ∠r = 90° – 60° = 30°

9.



=

14. (b) : Position of dark fringes/minima on the screen is given by x =

( 2n − 1) lD , 2 d

where n = 1, 2, 3… for first, second, third… dark fringes. For first dark fringe (n = 1) lD lD = 2 d 2 d x l Angular position, q = = D 2d

\

x = ( 2n − 1)



5460 × 10 −10 radian 2 × 10 −4 180 = 2730 × 10–6 × degree = 0.16°. p





=

15. (b) : We know, rl l R= = A ni e(m e + m h ) A

1 1    r = σ = n e(m + m )   i e h 

∴R =

0.928 × 10 −2 19

2.5 × 10

× 1.6 × 10 −19 (0.39 + 0.19) × 10 −6

= 4000 W = 4 kW 16. (c) : ac power gain is the ratio of change in output power to the change in input power. ac power gain Changein output power DVC DIC = = Changein input power DVi DI B  DV   DI  =  C  ×  C  = AV × bAC  DVi   DI B 

105

Model Test Paper-5

where AV is voltage gain and bAC is ac current gain. R  Also, AV = bAC ×  0  .  Ri  Given, AV = 50, R0 = 200 W, Ri = 100 W 200 Hence, 50 = bAC × 100 \ bAC = 25 Now, ac power gain = AV × bAC = 50 × 25 = 1250 17. (c) : Useful intensity for the emission of electron is 1 I′ = 1% of I = × 39.6 = 0.396 W m–2 100 hc Energy of each photon = l 8

) × ( 3 × 10 ) = 3.32 × 10–19 J 6000 × 10 −10 Number of photoelectrons emitted per second per unit area 0.396 = ≈ 1.2 × 1018 3.32 × 10 −19 =

(6.64 × 10

−34

18. (d) : As rn = number. For \ or

n2 a0, where n is the orbit Z

257 100 Fm

, Z = 100

25 1 a0 = a0 100 4 r5 1 1 ⇒ N= = 4 a0 4

r5 =

19. (d) 20. (b) : Spectrum of sunlight is an example for line absorption spectrum. 21. (d) : Initial flux linked with inner coil when I = 0 is zero. Final flux linked with inner coil m I when I = I is  0  pa2 .  2 pb  m I \ Change in flux, df =  0  pa2  2 pb  df As dq = R \ Total charge circulating the inner coil  m I  pa2 m0 Ia2 = =  0   2 pb  R 2 Rb V 22. (a) : Electric field, E = d

\

V = Ed = 3 × 106 × (10–4) = 300 V V Vrms = 0 = 300(0.707) ≈ 210 V 2

23. (a) : Let R be resistance connected in series with the galvanometer.

From figure, Ig S = I − Ig G + R R=S

 I  –G  I − 1  g 

Substituting the given values, we get  20  R = 0.05  − 1 – 20  0.04  R = 4.95 W 24. (d) : Horizontal component magnetic field, BH = Be cosq

of

earth’s

BH 3 × 10 −5 1 = = Be 6 × 10 −5 2 q = 60°

cosq = \

D (m – 1)t d For same D, d and t, we have, 2 1 x (1.5 − 1) = ⇒ = 3 3 2(m − 1) (m − 1) x 2 1 4 3 = ⇒ m−1= ⇒ m−1 3 4 7 \ m = = 1.75 4

25. (a) : As, y0 =

2

2

4

26. (c) : Given, 1H + 1H → 2He + Q The total binding energy of the deutrons = 4 × 1.15 = 4.60 MeV The total binding energy of alpha particle = 4 × 7.1 = 28.4 MeV \ The energy released in the process = 28.4 – 4.60 = 23.8 MeV 27. (c) : For bright fringe, n1l1 = n2l2 \ n(7.5 × 10–5) = (n + 1)(5 × 10–5) ⇒ 2.5 × 10–5n = 5 × 10–5 or

n=

5.0 × 10 −5 2.5 × 10 −5

=2

106

VITEEE CHAPTERWISE SOLUTIONS

(Given : volume remains constant)

28. (a) :

r12

or

r22

=

Now, R2 = e1 = 4 V and e2 = 8 V As, e2 > e1 current flows from Q to P. Effectice emf 8−4 4 I= = = Net Resistance 1 + 2 + 9 12 1 ∴ I= A 3

\

R So, V0 = Vi × b 0 Ri

1 ×9= 3 V 3

V0 R =b 0 ; Vi Ri

Here, Vi = 10 mV, b = 50, R0 = 5000 W, Ri = 2000 W \

V0 = 10 × 50 ×

5000 = 1250 mV = 1.25 V 2000

30. (b) : As coil A is moved closer to B, field due to A intercepting B is increasing. Induced current in B must oppose this increase. Hence the current in B must be anticlockwise. 31. (c) : Here, XL = wL = 2pnL = 2p × 50 × 0.7 = 220 W R = 220 W Z=

rl2

pr22 l1 r22 × l2 r12

R1 rl1 = R2 pr12

or

R1 l1 l1  l1  = × = = (2)2 R2 l2 l2  l2 

rl2

pr22

=

2

=

29. (c) : As voltage gain, Av =

…(i)

\

R1 =4 R2

Potential difference across PQ



l2 l1

R2 + X L2 = 220 2 + 220 2 = 220 2 W.

ev 220 = Z 220 2 1 = = 0.707 A 2 Through L, the current is wattless. 32. (b) 33. (a) : Polaroid glass polarises light reducing the light intensity to half its original value. 34. (b) : The resistance of rod before reformation rl rl rl   R1 = R = 12  R = A = 2  pr1 pr  

R2 =

\

(using(i))

R1 R = 4 4

35. (b) : The direction of current density is the direction of flow of positive charge in the circuit which is possible due to electric field produced by charges accumulated on the surface of wire. 36. (a) : Resistance per unit length of ring, R r= 2pr Length of sections ADB and ACB are rq and r(2p – q)

45°

Resistance of section ADB, R1 = rrq R Rq rq = = 2 pr 2p \

Iv =

and resistance of section ACB, R2 = rr(2p – q) R R( 2 p − q) = r( 2 p − q) = 2 pr 2p Now, R1 and R2 are connected in parallel between A and B then Rq R( 2 p − q) × R1R2 2p Req = = 2p R1 + R2 Rq R( 2 p − q) + 2p 2p Rq × R( 2 p − q)

Now the rod is reformed such that l l2 = 1 2 2 2 \ pr1 l1 = pr2 l2



=



=

( 2 p)2 Rq + R( 2 p − q) 2p Rq( 2 p − q) 4 p2

107

Model Test Paper-5

Putting q = 45° =

Req =

=

15 ×

41. (a) : DH = DU + DngRT

p  p  15p  7 p  ×  2p −    4  4 = 4  4  4 p2 4 p2

DH = –310 × 103 J, Dng = 3 – 2 = 1, R = 8.314 J K–1mol–1, T = 298 K DU = –310 × 103 – (1 × 8.314 × 298)

105 W = 1.64 W 64

37. (b) : R = R1 + R2 = r1

Chemistry

p rad and R = 15 W 4

R = (r1 + r2)

l pr 2

l pr 2

+ r2

42. (a) : l ∝ 1 ,

l

m

pr 2



…(i)

For equivalent resistance

R=r

2l

pr 2



…(ii)

From equations (i) and (ii), we have 2r = r1 + r2 or

r=

(



\

lNe < lHe < lH < le

43. (b) : Dye test is used to distinguish between primary aromatic amines and primary aliphatic amines. 45. (a) : [Co(NH3)6]Cl3 gives 3 moles of AgCl.

  m0 Idl × r 38. (c) : From Biot-Savart Law, B = 4p r 3  For B1, r = −i − j  m Idl  …(i) \ B1 = 0 k × − i − j 4p 2 2  For B , r = i + j

\

since, me < mH < mHe < mNe

44. (b) : Biuret test is used for proteins.

r1 + r2 2

2

= –312.477 × 103 J = –312.5 kJ

) ( )

(

m Idl    B2 = 0 k× i+ j 4p 2 2

)

[Co(NH3)5Cl]Cl2 gives 2 moles of AgCl. [Co(NH3)4Cl2]Cl gives 1 mole of AgCl. [Co(NH3)3Cl3] will not give AgCl. 46. (c) : XeF6 : sp3d3-distorted octahedral

…(ii)



From (i) and (ii)

    B1 = − B2 or B1 = B2

39. (b) : Since resultant intensity …(i)

I = I1 + I 2 + 2 I1 I 2 cosf

Applying equation (i) when phase difference is p/2 Ip/2 = I + 4I ⇒ Ip/2 = 5I Again applying equation (i) when the phase difference is p Ip = I + 4I + 2 I 4 I cosp

\ Ip = I From equations (i) and (ii) we get, Ip/2 – Ip = 5I – I = 4I

Hence the graph representation.

(c)

sp3-pyramidal O

XeOF4 :

the

correct

O

F

O

pyramidal F

Xe F

XeF4 :

F

O sp3d2-square planar F

dN = + lN0e–lt dt

is

Xe

sp3d2-square

…(ii)

40. (c) : Since N = N0e–lt ⇒ NY = N0(1 – e–lt) Rate of formation of Y , R =

XeO3 :

F Xe

F

F

108

VITEEE CHAPTERWISE SOLUTIONS



CH3 CH3

Propanal

C O

Propanone

48.

49.

50. 51.

C O

CH3 CH3 CH2

C O

Butanone

As we move from ethanal → propanal → propanone → butanone, the +I effect of the alkyl group increases. As a result, electron density on the carbon atom of the carbonyl group progressively increases and hence attack by the nucleophile becomes slower and slower. Thus, the reactivity increases in the reverse order, i.e., butanone < propanone < propanal < ethanal IV III II I (c) : A six-fold axis is that axis around which when a crystal is rotated by 60°, we get the same arrangement. Hence the imaginary line joining the end centres of a hexagonal close packing is called the six-fold axis of symmetry. A hcp can have only one six‑fold axis of symmetry. (b) : pp – pp bonding in nitrogen is strong hence it can form triple bond with another N. Single N — N bond is weaker than P — P bond due to high interelectronic repulsions of non-bonding electrons. Hence, N ≡≡ N is stable and P2 is not. (c) : L-Sugars are not metabolized in the body. (a) : Number of nodal planes in any orbital = l. px orbital is dumb-bell shaped consisting of two portions known as lobes. This is a plane in which the electron density is almost nil. This is known as nodal plane. Y

px Nodal plane Z

For 2px orbital, nodal plane is yz.

NO2

OH

NO2

NO2 (2,4,6-Tinitrophenol)



HN

NH

C

OH CH3

O O

O

(Pyrimidine-2,4,6-trione)

OH

Ascorbic acid O

O

O

(Vitamin-C)

Picric acid O

O

OH

Ethanal

H

OH

C O

Similarly for 2py and 2pz orbitals, the nodal planes are zx and xy respectively. 52. (c) : Any compound that can be prepared from, or converted into, D(+)-glyceraldehyde will belong to D-series and any compound that can be prepared from, or converted into, L(–) glyceraldehyde will belong to L-series. 53. (c) : The pH of neutral water at 25 °C is 7.0. \ [H+] = [OH–] = 10–7 ( pH = – log [H+]) Now, Kw = [H+][OH–] = (1 × 10–7)2 = 1 × 10–14 As the temperature increases, ionisation of water increases, thus [H+] and [OH–] increases equally. Now Kw = [H+] [OH–] > 1 × 10–14 ( [H+]=[OH–]) or [H+]2 > 1 × 10–14 \ [H+] > 1 × 10–7 and pH < 7 54. (d) : OH

H

CH3CH2

H

47. (c) : CH3

(2-Acetoxybenzoic acid)

Barbituric acid Aspirin  55. (b) : SO2 + Cl2 SO2Cl2 + heat Introduction of extra SO2 will shift the equilibrium in forward direction which is in accordance with Le-Chatelier’s principle. The forward reaction is accompanied with the liberation of heat, hence the temperature will increase. 56. (a) : I = 1.5 A, W = 1.45 g Ag, t = ?, E = 108, n = 1 Using Faraday’s 1st law of electrolysis, E W = ZIt or, W = It F 108 × 1.5 t or, 1.45 g = 96500 or, t =

1.45 × 96500 = 863.73 s 1.5 × 108

57. (b) : Dilute nitric acid converts chromate into dichromate and H2O. Yellow

Orange

109

Model Test Paper-5

58. (b) : Lecithin, cerebroside and cephalin are phospholipids while lysine is an amino acid. 59. (c) : In ccp, anions occupy primitives of the cube while cations occupy voids. In ccp there are two tetrahedral voids and one octahedral hole per anion. For one oxygen atom there are two tetrahedral holes and one octahedral hole. Since one-fifth of the tetrahedral voids are occupied by divalent cations (X2+). \ Number of divalent cations in 1 tetrahedral voids = 2 ×

(b) C6H5OH + NaOH

–H2O

C6H5OCH3 + NaI Methyl phenyl ether (Anisole)

(c)

C6H5ONa Sodium phenoxide

CH3I  (Williamson’s synthesis)

(Neutral)

5

Since one-half of the octahedral voids are occupied by trivalent cations (Y3+). \

Number of trivalent cations = 1 ×

1 2

So, the formula of the compound is (X )

2×

1 (Y ) 1 ( O)1 1× 5 2

(d)

or X 2 Y1 O1 or X4Y5O10. 5 2

60. (b) : The chiral carbons are : O¯ Ph



CH3

O H

CH3 O Ph

O¯ * *

CH3

CH3 61. (b) : As the size of the halogen atom increases from F to I, H – X bond length in HX molecules also increases from H – F to H – I (H – F < H – Cl < H – Br < H – I). The increase in H – X bond length decreases the strength of H – X bond from H – F to H – I (H – F > H – Cl > H – Br > H – I). The decrease in the strength of H – X bond is evident from the fact that H – X bond dissociation energies decrease from H – F to H – I. Due to successive decrease in the strength of H – X bond from H – F to H – I, thermal stability of HX molecules also decreases from HF to HI (HF > HCl > HBr > HI).

62. (d) : The configuration does not follow Hund’s rule of maximum multiplicity because 3p will be fully filled before the electrons go to 4s. 63. (b) : (a)

64. (a) : Let solubility be s Cu2+ + S2–

CuS

Ksp = [Cu2+] [S2–] 10–31 = (s) (s) ⇒ s = 10 −31 = 3.1 × 10 −16 2Ag+ + S2–

Ag2S

Ksp = [Ag+]2 [S2–] 10–44 = (2s)2 (s) ⇒ s = Hg2+ + S2–

HgS Ksp =

[Hg2+]

3

10 −44 ≈ 1.35 × 10 −15 4

[S2–]

10–54 = (s) (s) ⇒ s = 10 −54 = 10 −27 Hence solubility order of given salts is Ag2S > CuS > HgS. 65. (b) : 1 d [N 2O 5 ] 1 d [NO 2 ] d [O 2 ] = = dt dt dt 2 4 1 1 Rate = − k1 [N 2O 5 ] = k 2 [N 2O 5 ] = k 3 [N 2O 5 ] 2 4 On dividing all the terms by [N2O5] and multiplying by 4 we get, 2k1 = k2 = 4k3. Rate = −

110

VITEEE CHAPTERWISE SOLUTIONS

66. (b) : NaN3

X NaNH2

N2O

NH2OH + HNO2 Y

H2 O

H2N2O2 Z

67. (d) : In Wolff-Kishner reduction, only ketone group is reduced to alkyl group whereas double bond remains intact. R

C

R

= O + NH

Carbonyl compound

2

NH 2

Hydrazine

KOH Wolff-Kishner reduction N 2 + H 2 O + R – CH 2 – R Hydrocarbon

68. (c) : Let the total number of moles of N2 and O2 initially = 100 and number of moles of N2 initially = a N2(g) + O2(g) 2NO(g) Initially a 100 – a 0 At equit. (a – x) (100 – a – x) 2x Given :

2x 1.8 = ⇒ x = 0.9 100 100

Substituting the values in the relation

KC =

[NO]2 [N 2 ][O 2 ]

2.1 × 10 −3 = 2.1 × 10

−3

( 2 x )2 ( a − x )(100 − a − x )

( 2 × 0.9)2 = ( a − 0.9)(100 − a − 0.9)

On usual calculations, a = 79 % of N2 in air = 79% % of O2 in air = 100 – 79 = 21% 69. (c) : Tertiary nitroalkanes do not react with NaNO2 and HCl i.e HNO2 due to lack of a-hydrogen atom. 70. (a) : Since K .E. = ∴ K .E. =

⇒

1 h mv 2 and l = 2 mv

1 h2 h2 m⋅ = 2 2 2 m l 2ml 2

K .E. ∝

1 [As l is same.] m

Increasing order of mass is me < mp < ma-particle me = mass of electron = 9.1 × 10–28 g mp = mass of proton = 1.67 × 10–24 g ma-particle = 2(mp + mn),

where mn = mass of neutron (1.675 × 10–24 g) Thus order of kinetic energy K.E. of electron (E1) > K.E. of proton (E3) > K.E. of a-particle (E2) 71. (a) : HCl is a strong acid and its salt do not form buffer solution. As the resultant solution is acidic, hence pH is less than 7. 72. (c) : CH3CH2CH2OH Cu, 

CH3CH2CHO

CH3CH2CHO gives Fehling’s solution.

test

Propan-1-ol

CH3 – CHOH – CH3

positive

Cu, 

Propan-2-ol

Propanal (Aldehyde)

with

CH3 – CO – CH3 Propanone (Ketone)

CH3COCH3 does not give positive test with Fehling’s solution. With KMnO4 or acidic dichromate 1-propanol is oxidised to propanoic acid which cannot be tested with Fehling’s solution, whereas with conc. H2SO4 propanol undergoes dehydration to form propene which again cannot be tested with Fehling’s solution. 73. (a) : The colour of aqueous solution of CuSO4 is blue green. Thus it absorbs orangered colour and exhibits the complementary colour. 74. (a) : (2R)2 = (R + r)2 + (R + r)2

2 R = R+r ( 2 − 1)R = r 0.414R = r 75. (c) :

111

Model Test Paper-5

76. (c) : Halogens are just one electron short of noble gas configuration. Therefore, they have great tendency to accept electrons and form anions readily.

77. (c) : Erythro

In [Fe(CN)6]4–,

Strong field ligands like CO, CN– generally result in inner orbital complexes. In [Ni(H2O)6]2+, H2O is a weak field ligand, pairing of electrons will not take place. Hence [Ni(H2O)6]2+ is paramagnetic in nature. 80. (c) : Five-membered rings are named as furanose while six-membered rings are named as pyranose rings.

Mathematics

Threo

81. (d) : Radius of the sphere = Distance of (3, 6, – 4) from the given plane = 4. Hence, the required equation of the sphere is (x – 3)2 + (y – 6)2 + (z + 4)2 = (4)2. ⇒ x2 + y2 + z2 – 6x – 12y + 8z + 45 = 0 82. (c) : 3x – 6y + 2z + 5 = 0 ...(i) – 4x + 12y – 3z + 3 = 0 ...(ii) Equation of plane bisects the angle between the planes that contains the origin is

Thus the reaction is stereoselective as well as stereospecific. 78. (c) : Enthalpy change, DH = –2.5 × 103 cal mol–1 Entropy change, DS = 7.4 cal deg–1 mol–1 T = 298 K As, DG = DH – TDS DG = –2.5 × 103 – 298 × 7.4 = –4.70 × 103 cal mol–1 For spontaneity of reaction, negative value of DG is required, so the reaction is spontaneous. 79. (a) : In [Ni(CO)4],

3x − 6y + 2z + 5 9 + 36 + 4

16 + 144 + 9

83. (b) : The position vector of mid-point of line joining the points whose position vectors are i + j − k and i − j + k i + j − k + i − j + k 2i = = i 2 2

84. (a) : Given, In [Ni(CN)4]2–,

−4x + 12y − 3z + 3

or 13(3x – 6y + 2z + 5) = 7(–4x + 12y – 3z + 3) or 39x – 78y + 26z + 65 = – 28x + 84y – 21z + 21 or 67x – 162y + 47z + 44 = 0 ...(iii) Let q be the angle between (i) and (iii), then find cos q and then we obtain |tan q| < 1.

=



=

 volume of parallelopiped [a b c ] = 40 \ Volume of parallopiped     = [b + c c + a a + b ]  = 2[a b c ] = 2 × 40 = 80 cu unit

112

VITEEE CHAPTERWISE SOLUTIONS y

85. (d ) : We have, x dx + y dy

⇒

⇒

x dx − y dy d(x

3/ 2

d(x

3/ 2

y3

=

(x, y1) P

x3

) + d( y

3/ 2

)

) − d( y

3/ 2

)

=

y

x3/ 2

⇒ u du + v dv = v du – u dv

⇒

u2 + v 2

=

d (u2 + v 2 ) 2

2

log ⇒

+

v2)

u2 + v 2

= –2

v   + c u

y 1 log(x 3 + y 3 ) + tan −1   2 x

86. (c) : We have, x = e

xy

⇒

3/ 2

log x = xy

2

y = ± (log e x) + 2c

87. (d) : Since, the inverse of a logarithmic function is an exponential function and viceversa and these two curves are on the opposite sides of the line y = x. Thus, y = 2x and y = log x do not intersect. Their graphs are shown in figure. The shaded region in figure shows the area bounded by the given curves. Let us slice this region into vertical strips as shown in figure. For the approximating rectangle shown in figure, we have Area = (y1 – y2) dx

y

\

x=2

=

2

∫1/2 (2

x

1 and x = 2. 2

2

∫1/2 (y1 − y2 ) dx

Required area =

− log x) dx

 P(x, y1) and Q(x, y 2 ) lie on     y = 2x and y = log x respectively. ∴ y = 2 x and y = log x.   1  2

c = 2

dy dx

Length = (y1 – y2), Width = Dx

x = 1/2



dy dx log x dx ⇒ y dy = log x d(log x) ⇒ y dy = x y 2 (log x) 2 = +c On integrating, we get 2 2 ⇒ y2 = (loge x)2 + 2c ⇒

x

(1/2, 0)

horizontally between x =

  v  = −2d  tan −1    + c  u 

tan–1

(1, 0) (2, 0)

As the approximating rectangle can move

vdu − udv

u +v On integrating, we get (u2

Q (x, y2)

O

x

du + dv v = , where u = x 3 / 2 and v = y 3 / 2 du − dv u ⇒ u du + u dv = v du – v dv udu + vdv

y = log x

3/ 2

⇒

⇒

y = 2x

2

 2x  = − x log x + x   log 2  1/2   2  4 1 1  + log 2 +  = − 2 log 2 + 2 −  2    log 2 2  log 2 =

(4 − 2 ) 5 3 − log 2 + sq unit log 2 2 2

Hence, (d) is the correct answer. 88. (d) : We have, a = cos a + i sin a b = cos b + i sin b and c = cos g + i sin g Now,

b cos b + i sin b cos g − i sin g = × c cos g + i sin g cos g − i sin g

= cos b cos g + sin b sin g + i[sin b cos g – sin g cos b] b = cos(b − g ) + i sin (b − g ) c c Similarly, = cos( g − a) + i sin ( g − a) a ⇒

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

113

Model Test Paper-5

a = cos(a − b) + i sin (a − b) ...(iii) b On adding Eqs. (i), (ii) and (iii), we get cos(b – g) + cos(g – a) + cos (a – b) + i [sin (b – g) + sin (g – a) + sin(a – b)] = 1 On equating real parts to on both sides, we get cos(b – g) + cos (g – a) + cos(a – b) = 1 and

 2p   2p  89. (c) : We have, w n = cos   + i sin   n n w 3 = cos

⇒

2p 2p + i sin 3 3

2p 2p   and w 23 =  cos + i sin  3 3 = cos ⇒ \

2

1 i 3 =− + =w 2 2

4p 4p 1 i 3 + i sin =− − = w2 3 3 2 2

w23 = w2 and w3 = w (x + yw3 + zw23) (x + yw23 + zw3) = (x + yw + zw2) (x + yw2 + zw) = x2 + y2 + z2 – xy – yz – zx

90. (c) : The family of lines (x – 2y + 3) + l(2x – 3y + 4) = 0 are concurrent at point P(1, 2). If image of point A(2, 3) in the above variable line is B(h, k), then AP = BP ⇒ (h – 1)2 + (k – 2)2 = (2 – 1)2 + (3 – 2)2 Hence, locus of point P is x2 + y2 – 2x – 4y + 4 = 0 91. (a) : Let the coordinates of vertex C is (x, y) and the coordinates of centroid of the triangle is (x1, y1). y − 3+1 x+2−2 \ x1 = and y1 = 3 3 y−2 x and y1 = 3 3 Since, the centroid lies on the line 2x + 3y = 1. ( y − 2) x \ 2x1 + 3y1 = 1 and 2   + 3 =1 3 3 ⇒ 2x + 3y – 6 = 3 ⇒ 2x + 3y = 9 ⇒

x1 =

92. (d) :

C N a M



B

b L

D E



A

From DMLN, a−b sin a = a+b \ \

a−b a = sin −1   a+b Angle between AB and AD a−b = 2a = 2 sin − 1  . a+b

93. (d) : Let the extremities be L = (1, 2) and L′ = (1, – 4) 2+4 6 = =∞ \ Slope of L′ L = 1−1 0 Hence, latus rectum L′L is perpendicular to x-axis. Therefore, axis of the parabola will be parallel to x-axis and tangent at the vertex will be parallel to y-axis. Let the equation of the parabola be (y – b)2 = 4a(x – a) Now, L′L = |4a| \ 6 = |4a| ⇒ 4a = ± 6 ...(i) Hence, from Eq. (i), equation of parabola becomes (y – b)2 = ± 6(x – a) ...(ii) Since, L and L′ lie on Eq. (ii), therefore (2 – b)2 = ± 6(1 – a) ...(iii) and (4 + b)2 = ± 6(1 – a) ...(iv) From Eqs. (iii) and (iv), we get (2 – b)2 = (4 + b)2 ⇒ 12b = –12 ⇒ b = –1 \ From Eq. (iii), −1 5 9 = ± 6 (1 − a) ⇒ a = , 2 2 On putting the values of a and b in Eq. (ii), we get  1 ( y + 1) 2 = 6  x +  ⇒ ( y + 1) 2 = 3 (2x + 1) 2  5 and ( y + 1) 2 = − 6  x −  ⇒ ( y + 1) 2 = − 3 (2x − 5) 2 94. (c) : Put the value of (x, y) = (tan q + sin q, tan q – sin q) in the given option, we get the required result. On putting the value of x and y in option (c), we get [(tan q + sin q)2 – (tan q – sin q)2]2 = 16 (tan q + sin q) × (tan q – sin q)

114

VITEEE CHAPTERWISE SOLUTIONS

⇒ [(tan2 q + sin2 q – tan2q – sin2q + 4 tan q sin q]2 = 16(tan2q – sin2q) ⇒ (4 tan q ⋅ sin q)2 = 16(tan2q – sin2q) ⇒ 16 tan2q sin2q = 16 tan2q (1 – cos2q) ⇒ 16 tan2q sin2q = 16 tan2q sin2q Hence, the option (c) satisfies. 25 3 95. (d) : Given that, y = mx + 3 and

x2 y 2 − =1 16 9

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

1

  x2  1  = 4  x −  = 4   1 −  − 0   2 0 2   =4×

1 = 2 sq. units 2

Therefore, the required area is 2 sq. units. 97. (c) : The equation of the family of circle having centre on Y-axis and radius 3 units is x2 + (y – b)2 = 9 ...(i) Y

Here, Eq. (i) is normal to Eq. (ii), then (a 2 + b 2 ) 2 c2



=

a2 m2

−

b2 1

(0, 3)

(16 + 9) 2 × 9 16 9 = − 625 × 3 m2 1

⇒

16

⇒

m

2

= 12 ⇒ m = ±

X

2 3

96. (b) : The given curve is |x| + |y| = 1. In first quadrant, (x > 0, y > 0) Then, the line AB is x + y = 1. In second quadrant, (x < 0, y > 0) Then, the line BC is – x + y = 1. In third quadrant, (x < 0, y < 0) Then, the line CD is – x – y = 1. In fourth quadrant, (x > 0, y < 0) Then, the line DA is x – y = 1. Y B (0, 1) y=x+1 (–1, 0) X C

x+y=1 X A (1, 0)

O

y = –x – 1

y=x–1 D (0, –1) Y

When we plot above lines they form a square ABCD, \ Required area = 4 (Area of shaded region in the first quadrant) = 4

1

X

1

∫0 y (line AB) dx = 4 ∫0 (1 − x) dx

(Q x + y = 1 ⇒ y = 1 – x)

Y

On differentiating Eq. (i) w.r.t. x, we get 2x + 2(y – b) y′ = 0 x ⇒ y − b = − ...(ii) y′ On substituting this value of (y – b) in Eq. (i), we get 2

 x x2 +  −  = 9  y′  ⇒ x2[(y′)2 + 1] = 9(y′)2 ⇒ (x2 – 9) (y′)2 + x2 = 0 which is the required differential equation. 98. (c) : The given curve is dy xy = (x + 2)( y + 2) dx On separating the variables, we get y x+2 dy = dx y+2 x

...(i)

On integrating both sides, we get y x+2 dy = dx y+2 x

∫

∫

y + 2− 2 dy = y + 2 

∫ 

⇒

∫ y + 2 dy − ∫ y + 2 dy = ∫ x dx + ∫ x dx

y+2

2

∫

x+2 dx x

⇒

x

2

115

Model Test Paper-5

⇒

∫

dy − 2

∫

1 1 dy = dx + 2 dx y+2 x

∫

∫

⇒ y – 2 log |y + 2| = x + 2log |x| + C ...(ii) Now, the curve passes through the point (1, –1). So, put x = 1 and y = – 1 Then, – 1 – 2 log |– 1 + 2 | = x + 2 log |x| + C ⇒ – 1 – 2 log 1 = 1 + 2 log 1 + C ⇒ C = – 2 (Q log 1 = 0) From Eq. (ii), we get y – 2 log |y + 2| = x + 2 log |x| – 2 ⇒ y – x + 2 = 2 log |x(y + 2)] (Q log m + log n = log mn) ⇒ y – x + 2 = log [x2(y + 2)2] (Q log mn = n log m) which is the required equation of the curve.   99. (b) : Given that, a = i + j + k , b = 4i − 2j + 3k  and c = i − 2j + k    Firstly, we find the vector 2a − b + 3c    \ 2a − b + 3c = 2(i + j + k ) − (4i − 2j + 3k ) + 3(i − 2j + k )

= 2i + 2j + 2k − 4i + 2j − 3k + 3i − 6j + 3k = i − 2j + 2k

Now, we find a unit vector in the direction    of vector 2a − b + 3c which is equal to    2a − b + 3c    | 2a − b + 3c | = =

i − 2j + 2k

            =[(b ⋅ c )× a ]⋅ d +[(c × a )× b ]⋅ d +[b ×(c × d )]⋅ a               =[(b ⋅ a )c − (c ⋅ a )b ]⋅ d +[(c ⋅ b )a − (a ⋅ b)c]⋅ d        +[b ⋅ d )c − (b ⋅ c )d ]⋅ a             [(a ⋅ b )(c ⋅ d )]–[(c ⋅ a )(b ⋅ d )]+[(c ⋅ b )(a ⋅ d )]–             [(a⋅ b )(c ⋅ d )]+[(a ⋅ c )(b ⋅ d )]–[(a ⋅ d )(b ⋅ c )] =0  101. (d) : Let a = i + j + k  \ Direction ratios of a are 1, 1, 1.  Now, |a | = (1) 2 + (1) 2 + (1) 2 = 1 + 1 + 1 = 3  So, direction cosines of a are


 Let the vector a makes angles a, b and g respectively, with positive axes OX, OY and OZ. 1 1 1 Then, cos a = , cos b = and cos g = 3 3 3 1   \ a = b = g = cos −1   3   Hence, the vector a is equally inclined to the axes OX, OY and OZ. 102. (d) : Total number of 10 ten and 5 five rupee coins = 10 + 5 \ Total number of arrangement in a row 15 ! = 10 ! 5 ! Now

(1) 2 + (−2) 2 + (2) 2 i − 2j + 2k

=

i − 2j + 2k 1 2 2 = i − j + k 3 3 3 3

9 So, vector of magnitude 6 units parallel to the vector. 1 2 2     2a − b + 3c = 6  i − j + k  3 3 3    = 2i − 4 j + 4k          100. (a) : Let b × c = u, c × a = v and c × d = w          ∴ u ⋅ (a × d ) + v ⋅ (b × d ) + (a × b ) ⋅ w          = (u × a ) ⋅ d + (v × b ) ⋅ d + a ⋅ (b × w)

five rupee coin

five rupee coin

As two coins of rupee 5 are fixed at extreme position so the remaining coins can be (10 + 3)! arranged in a row are = which are 10 ! 3 ! the favourable number of cases. 13 ! 10 ! × 3 ! \ Required probability = 15 ! 10 ! × 5 !

=

13 ! 10 ! 5 ! 5×4 2 × = = 10 ! 3 ! 15 ! 15 × 14 21

116

VITEEE CHAPTERWISE SOLUTIONS

103. (c) : Sample points of S = 40C5 We need third card valued x3 = 24 which means x1, x2 < 24 and x3 = 24, x4, x5 > 24. \ We draw the card from 23 cards. Required ways = 23C2, x3 = 24 can be drawn by one way only. The remaining 2 cards are drawn from remaining 16 cards which can be done by 16C2. 23 C 2 × 16C 2 \ Required probability = 40 C5 1 104. (d) : np = 2, npq = 1, \ q = 2 and we know that 1 1 p+q=1⇒p+ = 1 ⇒ p = 2 2 1 1 So, n = 4, p = , q = 2 2 Then, P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0) = 1 – nC0 p0 qn–0 0

1 1 = 1 − 4C 0     2 2 15 \ P(X ≥ 1) = 16

4

= 1−1×

1 16 − 1 = 16 16

105. (c) : p = Probability of getting a head = 1 2

1 2



q = Probability of getting a tail =

\

Probability of getting atleast seven

heads is P (X ≥ 7) = P(7) + P(8) + P(9) + P(10) 7

3

8



1 1 1 1 = 10C7     + 10C 8     2 2 2 2



1 1 1 + 10C 9     + 10C10   2 2 2



 1  10 10 10 10 =   { C7 + C8 + C9 + C10} 2



=

9

10

1   2

2

0

Now, (i) 7 (5x − 4y − 2z + 1) = 0 4 ⇒ 27x – 24y – 26z = 13. ⇒

2x − y + 3z + 5 −

107. (d) : The normal   AB and AC.



is

perpendicular

i j k = 4 1 3 = − 9 (2i − j − 3k ) 1 7 3 C

A (–1, 0, 2) B

\ The plane is 2x – y – 3z = d It passes through A(–1, 0, 2). \ 2x – y – 3z + 8 = 0.       108. (d) : Let OA = a , OB = b , OC = c    a + b + c  The centroid of DABC is G   3     1       OG 2 = OG ⋅ OG = 9 (a + b + c ) ⋅ (a + b + c ) 1 = (3 + 6 cos q) k 2 > 0 9 1  2p  ⇒ cosq ≥ − \ q ∈0,  . 2  3 109. (a) : D is the midpoint of BC.  35 39  Coordinate of D is  ,   2 2  Slope of AD is –5. A(m ,n)

10

C(23, 20)

120 + 45 + 10 + 1 176 = 1024 1024

106. (b) : The new equation is 2x – y + 3z + 5 + l (5x – 4y – 2z + 1) = 0 ....(i) (2 + 5l)x – (1 + 4l)y + (3 – 2l)z + 5 + l = 0 It is perpendicular to 2x – y + 3z + 5 = 0. \ 2(2 + 5l) + (1 + 4l) + 3(3 – 2l) = 0 7 ⇒ 14 + 8l = 0 ⇒ l = − 4

B(12, 19)

D

35 , 39 2 2

39 2 = −5 35 m− 2 or 2n – 39 + 5 (2m – 35) = 0 ⇒ 5m + n = 107 \

n−

......(i)

to

117

Model Test Paper-5

Area of DABC = 70 m n 1 \ 12 19 1 = 140 ⇒ –m + 11n = 337 ....(ii) 23 20 1 Solving (i) and (ii), m = 15, n = 32, m + n = 47. 110. (c) : Let A= (3, –2), B = (–1, 5), C = (a, b) be the vertices. The orthocentre O = (0, 0).  b   −7  OC ⊥ AB ⇒     = − 1 ⇒ 4a − 7b = 0 a 4 

...(i)

 2  b−5  OA ⊥ BC ⇒  −   = − 1 ⇒ 3a − 2b = 13 ...(ii)  3   a + 1  (i), (ii) ⇒ a = – 7, b = – 4 ⇒ C = (– 7, – 4)  5 1 The centroid, G =  − , −  3  3 \

But OS =

3  5 1 OG \ S =  − , −  2  2 2 2

 5  1 Circumradius = SA =  3 +  +  2 −  2 2   =



3 1 Centre C =  , −  , radius = 2 2

1 =0 4

9 1 1 3 + − = 4 4 4 2

AB is a chord with midpoint P(x, y) 3 p CB = , ∠PCB = 2 3 \

CP = CB cos

p 3 1 3 = ⋅ = 3 2 2 4 C

A

B

P traces a circle with centre C and radius 2

  1 1 ⇒ y 2 − 2ay  t −  − 4a 2 + a 2 (1 + t 2 )  1 +  = 0  t  t2 

2

2

⇒

 1  1 y 2 − 2a  t −  y + a 2  t −  = 0 t t  

or

  1  y − a  t − t   = 0

2

 1  1 \ y = a t −  , a t −  t  t  \ The length of the intercept is zero. The directrix touches the circle. 113. (c) : The distance of a point from the line if tangent at that point is parallel to the line. Let the point be (x1, y1). Tangent at (x1, y1) is x1x + 2y1y = 6 x Slope = − 1 = − 1, slope of x + y = 7 2 y1 \ x1 = 2y1 As x12 + 2y12 = 6 ⇒ 4y21 + 2y12 = 6, ⇒ y1 = 1, x1 = 2 The point is (2, 1) 114. (b) : From given equation (y – 1)3 = – 27 3

 y − 1  −3  = 1

y −1 = (1)1/3 −3 y −1 \ = 1, w, w2 −3 As a, b, c are roots \ a – 1 = – 3, b – 1 = – 3w, c – 1 = – 3w2 \

3/2

P(x, y)

The directrix is x = – a  a  2a  \ (−a − at 2 )  −a −  + ( y − 2at)  y +  = 0 2 t    t 



3 1 ,– 2 2  3

The circle on AB as diameter is  a 2a   (x − at 2 )  x −  + ( y − 2at)  y +  = 0 2 t    t 

2

65 . 2

111. (b) : The circle x2 + y2 – 3x + y +

 a −2a  112. (d) : A(at 2 , 2at), B =  ,  t 2 t 

2

\

 3  1 3  x − 2  +  y + 2  =  4 

or

31 x 2 + y 2 − 3x + y + = 0. 16

3 4

\

a − 1 b − 1 c − 1 1 1 w2 + + = + + b − 1 c − 1 a − 1 w w w3



= w 2 + w2 + w2 = 3w2

118

VITEEE CHAPTERWISE SOLUTIONS 4

115. (c) : z1 z2 z3 z4 =

2p

∏ zr r =1

\

z1z2z3z4



p p  2p 2p   =  cos + i sin   cos + i sin  10 10   10 10  

0

p

 a + b sec x  f(x) = 2 log  dx  a − b sec x 

∫ 0

3p 3p   4p 4p   ×  cos + i sin   cos + i sin  10 10   10 10  

f (x) = 2

 p 2p 3p 4p  + + = cos  +  10 10 10 10 

0

=

 a + b sec(p − x) 

∫ log  a − b sec(p − x)  dx p



f (x)  a + b sec x  = − log  dx = −  2  a − b sec x 

∫ 0

2p

\ f(x) = 0

= abc – (a + b + c) + 2

 a + b sec x 

∫ log  a − b sec x  dx = 0

i.e.

0

As D > 0 \ abc + 2 > a + b + c ...(A) Using A.M. > G.M. a+b+c ⇒ > (abc)1/3 3 or a + b + c > 3(abc)1/3 ...(B) By (A) and (B), we have \ abc + 2 > a + b + c > 3(abc)1/3 ⇒ abc + 2 > 3(abc)1/3 ⇒ 2 + x3 > 3x (x = (abc)1/3) ⇒ x3 – 3x + 2 > 0 ⇒ (x – 1)2 (x + 2) > 0 \ x>–2 \ (abc)1/3 > – 2 ⇒ abc > – 8 x x2

 a + b sec x 

∫ log  a − b sec x  dx 0

= cos p + i sin p =–1

a 1 1 116. (c) : D = 1 b 1 1 1 c

p

p



 p 2p 3p 4p  + i sin  + + +  10 10 10 10 

 a + b sec(2p − x) 

∫ log  a − b sec(2p − x)  dx

118. (d) : f(x) =

119. (d) : Given x

∫p / 3

(3 − sin 2 z) dz +

y

∫0 cos z dx = 0

Differentiating w.r.t. x, we get  dy  3 − sin 2 x + cos y   = 0 dx ⇒

dy − 3 − sin 2 x = cos y dx 3

120. (a) : x (3 − x)999 dx

∫ 0

Let 3 – x = y 3

⇒

x3

117. (c) : f(x) = 1 2x 3x 2 0 2 6x f(x) = x(12x2 – 6x2) – x2 (6x – 0) + x3 (2 – 0) = 6x3 – 6x3 + 2x3 \ f(x) = 2x3 Now f ′(x) = 6x2 f ′(2) = 6(2)2 = 24

vvv

∫ ( 3 − y) ( y)

999

0

dy

3

3  y1000  y1000 ⇒ − (3 − y) dy  + (−1)  1000  0 1000

∫ 0

1000

3

  y y1001  ⇒ − (3 − y) − 1000 (1000) (1001)  0 

1001   (3)1001 ⇒ − 0 − (3)  ⇒ (1000)(1001)  (1000) 1001 