WB JEE Engineering Solved Paper 2022 - Nodrm
WB JEE Engineering Solved Paper 2022 - Nodrm
WB JEE Engineering Solved Paper 2022 - Nodrm
Solved Papers
ENGINEERING
ENTRANCE EXAM 2022
Solved Papers
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FIRST &
FOREMOST
Whenever a student decides to prepare for any examination
his/her first and foremost curiosity is to know about the type of
questions that are expected in the exams. This becomes more
important in the context of competitive examinations where they
have to face neck-to-neck competition.
We feel great pleasure in presenting before you this book
containing Error-Free Solutions of previous years’ (2021-2012)
WB JEE (Engineering) questions and 5 Practice Sets for practice
Going through this book, you will get an exact idea of the
questions generally asked in West Bengal Joint Entrance Exam and
become exam-ready.
We have made maximum efforts to provide correct solutions to
the best of our knowledge and opinion. Detailed explanatory
discussions follow the answers. Discussions are not just sketchy
rather, have been achieved in a manner that the students will
surely be able to solve some other related problems too. In this
way, the students shall be able to judge the extent upto which
they have been able to comprehend the concepts and become
skilled.
We hope, these solved papers along with error-free solutions
would be highly beneficial to the students. We would be grateful if
any discrepancy or mistake in the questions or answers is brought
to our notice so that these could be rectified in subsequent
editions.
PUBLISHER
CONTENTS
SOLVED PAPERS
Solved Paper 2021 1-47
CHEMISTRY
Atoms, Molecules and Chemical Arithmetic mole fraction, molarity, molality and normality.
Dalton's atomic theory; Gay Lussac's law of gaseous Percentage composition, empirical formula and
volume; Avogadro's Hypothesis and its applications. molecular formula; Numerical problems.
Atomic mass; Molecular mass; Equivalent weight; Atomic Structure Concept of Nuclear Atom –
Valency; Gram atomic weight; Gram molecular electron, proton and neutron (charge and mass),
weight; Gram equivalent weight and mole concept; atomic number. Rutherford's model and its
Chemical formulae; Balanced chemical equations; limitations; Extra nuclear structure; Line spectra of
Calculations (based on mole concept) involving hydrogen atom.
common oxidation – reduction, neutralization, and Quantization of energy (Planck's equation E = hJ);
displacement reactions; Concentration in terms of Bohr's model of hydrogen atom and its limitations,
Sommerfeld's modifications (elementary idea); The Directionality of covalent bonds, shapes of poly –
four quantum numbers, ground state electronic atomic molecules (examples); Concept of
configurations of many electron atoms and mono – hybridization of atomic orbitals (qualitative pictorial
atomic ions; The Aufbau Principle; Pauli's Exclusion approach): sp, sp2, sp3 and dsp2.
Principle and Hund's Rule. Molecular orbital energy diagrams for homonuclear
Dual nature of matter and light, de Broglie's diatomic species – bond order and magnetic
relationship, Uncertainty principle; The concept of properties.
atomic orbitals, shapes of s, p and d orbitals (pictorial
Valence Shell Electron Pair Repulsion (VSEPR) concept
approach).
(elementary idea) – shapes of molecules. Concept of
Radioactivity and Nuclear Chemistry Radioactivity resonance (elementary idea), resonance structures
a-, b-, g rays and their properties; Artificial (examples). Elementary idea about electronegativity,
transmutation; Rate of radioactive decay, decay bond polarity and dipole moment, inter- and intra-
constant, half-life and average age life period of molecular hydrogen bonding and its effects on
radio-elements; Units of radioactivity; Numerical physical properties (mp, bp and solubility); Hydrogen
problems. bridge bonds in diborane.
Stability of the atomic nucleus effect of neutron- Coordination Compounds Introduction, Double
proton (n/p) ratio on the modes of decay, group salts and complex salts, coordination compounds
displacement law, radioisotopes and their uses (C, P, (examples only), Werner's theory, coordination
Co and I as examples) isobars and isotones (definition number (examples of coordination number 4 and 6
and examples), elementary idea of nuclear fission and only), colour, magnetic properties and shapes, IUPAC
fusion reactions. nomenclature of mononuclear coordination
The Periodic Table and Chemical Families Modern compounds.
periodic law (based on atomic number); Modern Solid State Classification of solids based on different
periodic table based on electronic configurations, binding forces: molecular, ionic, covalent and metallic
groups (Gr. 1-18) and periods. Types of elements – solids, amorphous and crystalline solids (elementary
representative (s-block and p- block), transition (d- idea). Unit cell in two dimensional and three
block) elements and inner transition (f-block / dimensional lattices, calculation of density of unit cell,
lanthanides and actinides) and their general packing in solids, packing efficiency, voids, number of
characteristics. Periodic trends in physical and atoms per unit cell in a cubic unit cell, point defects,
chemical properties – atomic radii, valency, ionization electrical and magnetic properties. Band theory of
energy, electron affinity, electronegativity, metallic metals, conductors, semiconductors and insulators
character, acidic and basic characters of oxides and and n & p type semiconductors.
hydrides of the representative elements (up to Z =
36). Position of hydrogen and the noble gases in the Liquid State Vapour pressure, viscosity and surface
periodic table; Diagonal relationships. tension (qualitative idea only, no mathematical
derivations).
Chemical Bonding and Molecular Structure
Valence electrons, the Octet rule, electrovalent, Gaseous State Measurable properties of gases.
covalent and coordinate covalent bonds with Boyle's Law and Charles Law, absolute scale of
examples; Properties of electrovalent and covalent temperature, kinetic theory of gases, ideal gas
compounds. Limitations of Octet rule (examples); equation – average, root mean square and most
Fajans Rule. probable velocities and their relationship with
temperature. Daltons Law of partial pressure, Electrolytic Solutions Specific conductance,
Grahams Law of gaseous diffusion. Deviations from equivalent conductance, ionic conductance,
ideal behavior. Liquefaction of gases, real gases, van Kohlrausch's law, Faraday's laws of electrolysis,
der Waals equation; Numerical problems. applications. Numerical problems.
Chemical Energetics and Chemical Dynamics Non-electrolytic Solutions Types of solution, vapour
Chemical Energetics – Conservation of energy pressure of solutions. Raoult's Law; Colligative
principle, energy changes in physical and chemical properties – lowering of vapour pressure, elevation of
transformations. First law of thermodynamics; boiling point, depression of freezing point, osmotic
Internal energy, work and heat, pressure – volume pressure and their relationships with molecular mass
work; Enthalpy. Internal energy change (ΔE) and (without derivations); Numerical problems.
Enthalpy change (ΔH) in a chemical reaction. Hesss
Law and its applications (Numerical problems). Heat Ionic and Redox Equilibria Ionic equilibria –
of reaction, fusion and apourization; Second law of ionization of weak electrolytes, Ostwald's dilution law.
thermodynamics; Entropy; Free energy; Criterion of Ionization constants of weak acids and bases, ionic
spontaneity. Third law of thermodynamics (brief product of water, the pH – scale, pH of aqueous
introduction). solutions of acids and bases; Buffer solutions, buffer
action and Henderson equation.
Chemical Equilibria The Law of mass action,
dynamic nature of chemical equilibria. Equilibrium Acid-base titrations, acid base indicators (structures
constants, Le Chateliers Principle. Equilibrium not required). Hydrolysis of salts (elementary idea),
constants of gaseous reactions (Kp and Kc) and solubility product, common ion effect (no numerical
relation between them (examples). Significance of problems).
ΔG and ΔGº.
Redox Equilibria: Oxidation Reduction reactions as
Chemical Dynamics Factors affecting the rate of electron transfer processes, oxidation numbers,
chemical reactions (concentration, pressure, balancing of redox reactions by oxidation number
temperature, catalyst), Concept of collision theory. and ion-electron methods. Standard electrode
Arrhenius equation and concept of activation energy. potentials (E°), Electrochemical series, feasibility of a
Order and molecularity (determination excluded); redox reaction. Significance of Gibb's equation: ΔG° =
First order reactions, rate constant, half – life – nFΔE° (without derivation), no numerical problems.
(numerical problems), examples of first order and Redox titrations with (examples); Nernst equations
second order reactions. (Numerical problems).
Physical Chemistry of Solutions Colloidal Solutions Hydrogen Position of hydrogen in periodic table,
– Differences from true solutions; Hydrophobic and occurrence, isotopes, preparation, properties and
hydrophilic colloids (examples and uses); Coagulation uses of hydrogen, hydrides-ionic covalent and
and peptization of colloids; Dialysis and its interstitial; physical and chemical properties of water,
applications; Brownian motion; Tyndall effect and its heavy water, hydrogen peroxide – preparation,
applications; Elementary idea of emulsion, surfactant reactions and structure and use; hydrogen as a fuel.
and micelle.
Chemistry of Non-Metallic Elements and their Chemistry in Industry Large scale production
Compounds Carbon – occurrence, isotopes, (including physicochemical principles where
allotropes (graphite, diamond, fullerene); CO and CO2 applicable, omitting technical details) and uses of
production, properties and uses. Sulphuric acid (contact process), Ammonia (Haber's
Nitrogen and Phosphorus occurrence, isotopes, process), Nitric acid (Ostwald's process), sodium bi-
allotopes, isolation from natural sources and carbonate and sodium carbonate (Solvey process).
purification, reactivity of the free elements. Polymers Natural and synthetic polymers, methods
Preparation, properties, reactions of NH3, PH3, NO, of polymerization (addition and condensation),
NO2, HNO2, HNO3, P4O10, H3PO3 and H3PO4. copolymerization, some important polymers –
Oxygen and Sulphur Occurrence, isotopes, allotropic natural and synthetic like polythene, nylon polyesters,
forms, isolation from natural sources and purification, bakelite, rubber. Biodegradable and non-
properties and reactions of the free elements. Water, biodegradable polymers.
unusual properties of water, heavy water (production Surface Chemistry Adsorption – physisorption and
and uses). Hydrogen peroxide and ozone (production, chemisorption, factors affecting adsorption of gases
purification, properties and uses). on solids, catalysis, homogenous and heterogenous
Halogens comparative study, occurrence, physical activity and selectivity; enzyme catalysis colloidal
states and chemical reactivities of the free elements, state distinction between true solutions, colloids and
peculiarities of fluorine and iodine; Hydracids of suspension; lyophilic, lyophobic multimolecular and
halogens (preparation, properties, reactions and macromolecular colloids; properties of colloids;
uses), inter-halogen compounds (examples); Oxyacids Tyndall effect, Brownian movement, electrophoresis,
of chlorine. coagulation, emulsion – types of emulsions.
Chemistry of Metals General principles of metallurgy Environmental Chemistry Common modes of
– occurrence, concentration of ores, production and pollution of air, water and soil. Ozone layer, ozone
purification of metals, mineral wealth of India. Typical hole – important chemical reactions in the
metals (Na, Ca, Al, Fe, Cu and Zn) – occurrence, atmosphere, Smog; major atmospheric pollutants;
extraction, purification (where applicable), properties Green House effect; Global warming – pollution due
and reactions with air, water, acids and non-metals. to industrial wastes, green chemistry as an alternative
Manufacture of steels and alloy steel (Bessemer, tool for reducing pollution, strategies for control of
Open-Hearth and L.D. process). environment pollution.
MATHEMATICS
Algebra
Problems involving both permutations and
A.P., G.P., H.P. Definitions of A. P. and G.P.; General combinations.
term; Summation of first
n-terms of series nnnArithmetic/Geometric Principle of mathematical induction Statement of
series, A.M., G.M. and their relation; Infinite G.P. series the principle, proof by induction for the sum of
and its sum. squares, sum of cubes of first n natural numbers,
divisibility properties like 22n— 1 is divisible by
Logarithms Definition; General properties; Change 3 (n ≥ 1), 7 divides 32n+1+2n+2 (n ≥ 1)
of base.
Binomial theorem (positive integral index)
Complex Numbers Definition and properties of Statement of the theorem, general term, middle term,
complex numbers; Complex conjugate; Triangle equidistant terms, properties of binomial coefficients.
inequality; Square root of complex numbers; Cube
roots of unity; De Moivre's theorem (statement only) Matrices Concepts of m x n (m ≤ 3, n ≤ 3) real
and its elementary applications. Solution of quadratic matrices, operations of addition, scalar multiplication
equation in complex number system. and multiplication of matrices. Transpose of a matrix.
Determinant of a square matrix. Properties of
Quadratic Equations Quadratic equations with real determinants (statement only). Minor, cofactor and
coefficients; Relations between roots and coefficients; adjoint of a matrix. Nonsingular matrix. Inverse of a
Nature of roots; Formation of a quadratic equation, matrix. Finding area of a triangle. Solutions of system
sign and magnitude of the quadratic expression of linear equations. (Not more than 3 variables).
ax2+bx+c
(where a, b, c are rational numbers and a ≠ 0). Sets, Relations and Mappings Idea of sets, subsets,
power set, complement, union, intersection and
Permutation and combination Permutation of n difference of sets, Venn diagram, De Morgan's Laws,
different things taken r at a time (r ≤ n). Permutation Inclusion / Exclusion formula for two or three finite
of n things not all different. Permutation with sets, Cartesian product of sets.
repetitions (circular permutation excluded).
Relation and its properties. Equivalence relation —
Combinations of n different things taken r at a time (r definition and elementary examples, mappings,
≤ n). Combination of n things not all different. Basic range and domain, injective, surjective and bijective
properties. mappings, composition of mappings, inverse of a
mapping.
Statistics and Probability Measure of dispersion, Co-ordinate geometry of three dimensions
mean, variance and standard deviation, frequency Direction cosines and direction ratios, distance
distribution. between two points and section formula, equation of
a straight line, equation of a plane, distance of a point
Addition and multiplication rules of probability,
from a plane.
conditional probability and Bayes' Theorem,
independence of events, repeated independent trails
and Binomial distribution. Calculus
Trigonometry Trigonometric functions, addition and Differential calculus Functions, composition of two
subtraction formulae, formulae involving multiple functions and inverse of a function, limit, continuity,
and sub-multiple angles, general solution of derivative, chain rule, derivative of implicit functions
trigonometric equations. and functions defined parametrically.
Properties of triangles, inverse trigonometric Rolle's Theorem and Lagrange's Mean Value theorem
functions and their properties. (statement only). Their geometric interpretation and
elementary application. L'Hospital's rule (statement
Coordinate geometry of two dimensions Distance
only) and applications. Second order derivative.
formula, section formula, area of a triangle, condition
of collinearity of three points in a plane. Integral calculus Integration as a reverse process of
differentiation, indefinite integral of standard
Polar coordinates, transformation from Cartesian to
functions. Integration by parts. Integration by
polar coordinates and vice versa. Parallel
substitution and partial fraction.
transformation of axes, concept of locus, elementary
locus problems. Definite integral as a limit of a sum with equal
subdivisions. Fundamental theorem of integral
Slope of a line. Equation of lines in different forms, calculus and its applications. Properties of definite
angle between two lines. Condition of integrals.
perpendicularity and parallelism of two lines.
Differential Equations Formation of ordinary
Distance of a point from a line. Distance between two
differential equations, solution of homogeneous
parallel lines. Lines through the point of intersection
differential equations, separation of variables
of two lines.
method, linear first order differential equations.
Equation of a circle with a given center and radius.
Application of Calculus Tangents and normals,
Condition that a general equation of second degree
conditions of tangency. Determination of
in x, y may represent a circle. Equation of a circle in
monotonicity, maxima and minima. Differential
terms of endpoints of a diameter . Equation of
coefficient as a measure of rate.
tangent, normal and chord. Parametric equation of a
circle. Intersection of a line with a circle. Equation of Motion in a straight line with constant acceleration.
common chord of two intersecting circles. Geometric interpretation of definite integral as area,
calculation of area bounded by elementary curves
Definition of conic section, Directrix, Focus and
and Straight lines. Area of the region included
Eccentricity, classification based on eccentricity.
between two elementary curves.
Equation of Parabola, Ellipse and Hyperbola in
Vectors Addition of vectors, scalar multiplication, dot
standard form, their foci, directrices, eccentricities
and cross products, scalar triple product.
and parametric equations.
WB JEE
Engineering Entrance Exam
t t
A block of mass m slides with speed v on a
frictionless table towards another stationary
block of mass m. A massless spring with 13. A uniform thin rod of length L, mass m is
spring constant k is attached to the second lying on a smooth horizontal table. A
block as shown in figure. The maximum horizontal impulse P is suddenly applied
distance, the spring gets compressed through perpendicular to the rod at one end. The
is total energy of the rod after the impulse is
m m k k P2 7 P2 13P 2 2 P2
(a) v (b) v (c) v (d) v (a) (b) (c) (d)
k 2k m 2m m 8m 2m m
17. p q 1 − 1 q 1
(a) (b)
4 π ε0 a b 4 π ε0 a
q 1 − 1 q 1
1 (c) (d)
p0 4 π ε0 R a 4 π ε0 R
2
Z/a
Consider two infinitely long wires parallel to
OM Z-axis carrying same current I in the positive
z-direction. One wire passes through the
point L at coordinates (− 1, + 1) and the other
wire passes through the point M at
The variation of electric field along the Z-axis coordinates (− 1, − 1). The resultant magnetic
due to a uniformly charged circular ring of field at the origin O will be
radius a in XY -plane as shown in the figure. µ 0I $j µ 0I $j
The value of coordinate M will be (a) (b)
2 2π 2π
(a)
1
(b) 2 (c) 1 (d)
1 µ 0I $i µ 0I $i
(c) (d)
2 2 2 2π 4π
21. A metal sphere of radius R carrying charge q 25. A thin charged rod is bent into the shape of a
is surrounded by a thick concentric metal small circle of radius R, the charge per unit
shell of inner and outer radii a and b, length of the rod being λ. The circle is rotated
respectively. The net charge on the shell is about its axis with a time period T and it is
zero.The potential at the centre of the sphere, found the magnetic field at a distance d away
when the outer surface of the shell is (d >> R) from the centre and on the axis,
grounded will be
4 WB JEE (Engineering) Solved Paper 2021
26. 1/χ
A
(c) −
R R
(d)
R R
27. V , ,−
2 2 2 2
Consider a pure inductive AC circuit as 33. A uniform rod of length L pivoted at one end
shown in the figure. If the average power P is freely rotated in a horizontal plane with
consumed is P, then an angular velocity ω about a vertical axis
(a) P > 0 (b) P < 0 passing through P. If the temperature of the
(c) P = 0 (d) P is infinite system is increased by ∆T, angular velocity
WB JEE (Engineering) Solved Paper 2021 5
2p0 C
Category-III (Q. Nos. 36 to 40)
Carry 2 marks each and one or more option(s) p0 A
is/are correct. If all correct answers are not marked B
and also no incorrect answer is marked, then score
V
= 2 × number of correct answers marked ÷ actual V0 2V0
number of correct answers. If any wrong option is
marked or if any combination including a wrong Consider the p - V diagram for 1 mole of an
option is marked, the answer will be considered ideal monatomic gas shown in the figure.
wrong, but there is no negative marking for the Which of the following statement(s) is/are
same and zero marks will be awarded. true?
(a) The change in internal energy for the whole
36. process is zero.
x a (b) Heat is rejected during the process.
O (c) Change in internal energy for process A → B is
v 3
− p0 V0 .
A small bar magnet of dipole moment M is 2
moving with speed v along x-direction (d) Work done by the gas during the entire process
is 2 p0 V0 .
towards a small closed circular conducting
loop of radius a with its centre O at x = 0 (see 39. The potential energy of a particle of mass
figure). Assume x >> a and the coil has a 0.02 kg moving along X -axis is given by
resistance R. Then, which of the following V = Ax (x − 4) J, where x is in metre and A is a
statement(s) is/are true? constant. Which of the following
statement(s) is/are correct?
6 WB JEE (Engineering) Solved Paper 2021
(a) The particle is acted upon by a constant A particle of mass m and charge q moving with
force. velocity v enters region-b from region-a along the
(b) The particle executes simple harmonic normal to the boundary as shown in the figure.
motion.
Region-b has a uniform magnetic field B
(c) The speed of the particle is maximum at
x = 2 m.
perpendicular to the plane of the paper. Also,
π region-b has length L.
(d) The period of oscillation of the particle is s.
5
Choose the correct statement.
b qLB
40. a (a) The particle enters region- c only if v > .
c m
× × ×
qLB
(b) The particle enters region- c only if v < .
× × × m
v
× × × (c) Path of the particle is a circle in region-b.
L (d) Time spent in region-b is independent of velocity v.
Chemistry
Category-I (Q. Nos. 41 to 70) 44. p-nitro–N, N–dimethylaniline cannot be
Carry 1 mark each and only one option is correct. represented by the resonating structures.
In case of incorrect answer or any combination of O O
⊕
more than one answer, 1/4 mark will be deducted. Me2N N Me2N N
O O
41. The exact order of boiling points of the (I) (II)
compounds n-pentane, isopentane, butanone
and 1-butanol is
⊕ ⊕ O ⊕ O
(a) n-pentane < isopentane < butanone <
Me2N N Me2N N
1-butanol O
O
(b) isopentane < n-pentane < butanone <
1-butanol (III) (IV)
(c) butanone < n -pentane < isopentane < (a) I and II (b) II and IV
1-butanol (c) I and III (d) III and IV
(d) 1-butanol < butanone < n-pentane <
isopentane 45. CO2H CH3
The relationship between the pair of The compounds A and B above are
compounds shown above are respectively respectively.
(a) homomer (identical), enantiomer and Ph OMe CO2H
constitutional isomer (a) and Ph
(b) enantiomer, enantiomer and diastereomer
(c) homomer (identical), homomer (identical) and Cl
constitutional isomer Ph OMe
(d) enantiomer, homomer (identical) and (b) and PhCOCH
3
geometrical isomer
Cl
46. The exact order of acidity of the compounds OH
p-nitrophenol, acetic acid, acetylene and Ph OMe CO2H
ethanol is (c) and Ph
(a) p-nitrophenol < acetic acid < acetylene < Cl
ethanol
(b) acetic acid < p-nitrophenol < acetylene < Ph OMe CO2H
ethanol and Ph
(d)
(c) acetylene < p-nitrophenol < ethanol < acetic OMe
acid
(d) acetylene < ethanol < p-nitrophenol < acetic 49. For a spontaneous reaction at all temperatures
acid which of the following is correct?
47. NH2 H (a) Both ∆H and ∆S are positive
(b) ∆H is positive and ∆S is negative
N CO2H
1. Me (c) ∆H is negative and ∆S is positive
(d) Both ∆H and ∆S are negative
O
NH2 H 50. A given amount of Fe2+ is oxidised by x mol of
MnO −4 in acidic medium. The number of moles
N CO2H
2. Me of Cr2O72− required to oxidise the same amount
O
CH3 of Fe2+ in acidic medium is
(a) x (b) 0.83 x
H
(c) 2.0 x (d) 1.2 x
N CO2H
3. H2N 51. An element crystallises in a body centered
cubic lattice. The edge length of the unit cell is
O Me
200 pm and the density of the element is 5.0 g
H cm −3 . Calculate the number of atoms in 100 g
of this element.
N CO2H
4. H2N (a) 2.5 × 1023 (b) 2.5 × 1024
O (c) 5.0 × 10 23
(d) 5.0 × 1024
The dipeptides which may be obtained from 52. Molecular velocities of two gases at the same
the amino acids glycine and alanine are temperature (T) are u1 and u2 . Their masses are
(a) Only 1 (b) Only 2 m1 and m2 respectively. Which of the following
(c) Both 1 and 2 (d) All of them expressions is correct at temperature T?
m1 m2
48. Benzaldehyde + methanol (a) = (b) m1u1 = m2u 2
u12u 22
1. dil. HCl m1 m2
dry
→ A 2 → B (c) = (d) m1u12 = m2u 22
HCl .( CH CO) O, 3 2 u1 u2
CH 3 COONa
8 WB JEE (Engineering) Solved Paper 2021
53. When 20 g of naphthoic acid (C11H 8O 2) is 59. Solubility products (K sp) of the salts of types
dissolved in 50 g of benzene, a freezing point MX , MX 2 and M3 X at temperature T are
depression of 2K is observed. The van’t Hoff 4 .0 × 10 −8 , 3.2 × 10 −14 and 2.7 × 10 −15
respectively. Solubilities (in mol dm −3 ) of the
. K kg mol −1 ]
factor (i) is [K f = 172
salts at temperature T are in the order.
(a) 0.5 (b) 1.0 (c) 2.0 (d) 3.0 (a) MX > MX 2 > M 3 X (b) M 3 X > MX 2 > MX
(c) MX 2 > M 3 X > MX (d) MX > M 3 X > MX 2
54. The equilibrium constant for the reaction
N 2( g )+ O 2( g ) 2NO( g ) is 4 × 10 −4 at
c 60. The reduction potential of hydrogen half-cell
2000 K. In presence of a catalyst the will be negative if
equilibrium is attained 10 times faster. (a) p(H2 ) = 1 atm and [H+ ] = 10
. M
Therefore, the equilibrium constant, in (b) p(H2 ) =1atm and [H+ ] = 2.0 M
presence of the catalyst at 2000 K is (c) p(H2 ) =2 atm and [H+ ] = 10
. M
(d) p(H2 ) =2 atm and [H+ ] = 2.0 M
(a) 4 × 10−4 (b) 4 × 10−3
−5
(c) 4 × 10 (d) 2.5 × 10−4 61. A saturated solution of BaSO 4 at 25°C is
4 × 10 −5 M. The solubility of BaSO 4 in 0.1 M
55. Under the same reaction conditions, initial
concentration of 1.386 mol dm −3 of a Na 2SO 4 at this temperature will be
substance becomes half in 40 s and 20 s . × 10−9 M
(a) 16 . × 10−8 M
(b) 16
−6
through first-order and zero-order kinetics (c) 4 × 10 M (d) 4 × 10−4 M
k
respectively. Ratio 1 of the rate constants 62. A solution is made by a concentrated solution
k0
of Co(NO 3 )2 with a concentrated solution of
for first-order (k1) and zero-order (k0) of the NaNO 2 is 50% acetic acid. A solution of a salt
reactions is containing metal M is added to the mixture,
(a) 0.5 mol −1 dm 3 (b) 0.5 mol dm −3 when a yellow precipitate is formed. Metal ‘
(c) 1.0 mol dm −3 (d) 2.0 mol −1 dm 3 M’ is
56. Which of the following solutions will have (a) magnesium (b) sodium
(c) potassium (d) zinc
highest conductivity?
(a) 0.1 M CH3COOH (b) 0.1 M NaCl 63. Extraction of a metal (M) from its sulphide
(c) 0.1 M KNO 3 (d) 0.1 M HCl ore (M2S) involves the following chemical
57. Indicate the products (X ) and (Y ) in the reactions
following reactions 2 M2S + 3O 2 Heat
→ 2 M2O + 2SO 2 ↑
Na 2S + nS( n = 1 − 8 ) → (X ) M2S + 2 M2O Heat
→ 6 M + SO 2 ↑
Na 2SO 3 + S → (Y ) (a) Zn (b) Cu (c) Fe (d) Ca
(X) (Y) 64. The white precipitate (Y ), obtained on
(a) Na 2S2O 3 Na 2S2 passing colourless and odourless gas (X )
(b) Na 2S( n + 1) Na 2S2O 3 through an ammoniacal solution of NaCl,
(c) Na 2Sn Na 2S2O 3 loses about 37% of its weight on heating and
(d) Na 2S5 Na 2S2O 4 a white residue (Z) of basic nature is left.
Identify (X ), (Y ) and (Z) from following sets.
58. 2.5 mL 0.4 M weak monoacidic base
(K b = 1 × 10 −12 at 25°C) is titrated with 2/15 M (X) (Y) (Z)
HCl in water at 25°C. The concentration of (a) N2 (NH4 )2 CO 3 NH4Cl
H + at equivalence point is (K w = 1 × 10 −14 , at (b) O2 NaNH4CO 3 NaHCO 3
25°C)
(c) CO 2 NH4HCO 3 (NH4 )2 CO 3
. × 10−13 M
(a) 37 . × 10−7 M
(b) 32
(d) CO 2 NaHCO 3 Na 2CO 3
. × 10−2 M
(c) 32 (d) 2.7 × 10−2 M
WB JEE (Engineering) Solved Paper 2021 9
(b) 5 6
(c) 6 5 (c) and
(d) 0 1
NO2 NHCOCH3
70. Which of the following compounds have
magnetic moment identical with OH OH
[Cr(H 2O)6]3+ ?
(a) [Cu(H2O)6 ]2+ (b) [Mn(H2O)6 ]3+
(d) and
(c) [Fe(H2O)6 ]3+ (d) [Mn(H2O)6 ]3 +
COCH3
Category-II (Q. Nos. 71 to 75)
NO NH2
Carry 2 marks each and only one option is
correct. In case of incorrect answer or any 73. The atomic masses of helium and neon are
combination of more than one answer, 1/2 mark 4.0 and 20.0 amu respectively. The value of
will be deducted. the de-Broglie wavelength of helium gas at
− 73 °C is M times the de-Broglie wavelength
71. Among the following chlorides the of neon at 727°C. The value of M is
compounds which will be hydrolysed most (a) 5 (b) 25
easily and most slowly in aqueous NaOH 1 1
(c) (d)
solution are respectively: 5 25
10 WB JEE (Engineering) Solved Paper 2021
74. The mole fraction of a solute in a binary 77. The compounds X and Y are respectively
solution is 0.1 at 298 K, molarity of this 1. Mg, ether
solution is same as its molality. Density Br
2. Acetaldehyde
CH3 X
of this solution at 298 K is 2.0 g cm −3 . 3. Br2 / NaOH
The ratio of molecular weights of the 4. H2O+
solute and the solvent (Msolute / Msolvent) is 1. SOCl2
1 1 2. NH3
(a) 9 (b) (c) 4.5 (d) Y
9 4.5 3. Br2 / NaOH
Mathematics
Category-I (Q. Nos. 1 to 30) 6. The value of the integral
1
x + 1 2 x − 1 2 2
1
Carry 1 marks each and only one option is 2
5
18. Consider the real valued function h : {0, 1, 2,
12. The value of ∫ max{ x 2 , 6 x − 8} dx is
0
..... 100} → R such that h(0) = 5, h(100) = 20
1
(a) 72 (b) 125 and satisfying h(p) = { h(p +1) + h(p – 1)} for
(c) 43 (d) 69 2
every p = 1, 2 ……99. Then the value of h() 1 is
13. A bulb is placed at the centre of a circular (a) 5.15 (b) 5.5
track of radius 10 m. A vertical wall is erected (c) 6 (d) 6.15
touching the track at a point P. A man is
running along the track with a speed of 10 19. If|z| = 1 and z ≠ ± 1, then all the points
m/sec. Starting from P the speed with which z
representing lie on
his shadow is running along the wall when 1 − z2
he is at an angular distance of 60° from P is
(a) a line not passing through the origin
(a) 30 m/sec (b) 40 m/sec
(b) the line y = x
(c) 60 m/sec (d) 80 m/sec
(c) the X -axis
14. Two particles A and B move from rest along a (d) the Y-axis
straight line with constant accelerations f
and f ′ respectively. If A takes m sec. more 20. Let C denote the set of all complex numbers.
than that of B and describes n units more Define A = {(z , w)| z , w ∈ C and|z| = |w|},
than that of B in acquiring the same velocity,
B = { z , w)|z , w ∈ C and z 2 = w 2 }. Then
then
(a) (f + f ′ ) m2 = ff ′ n (b) (f − ff ′ ) m2 = ff ′ n (a) A = B (b) A ⊂ B
1 1 (c) B ⊂ A (d) A ∩ B = ϕ
(c) (f ′ − f ) n = ff ′ m2 (d) (f + f ′ ) m = ff ′ n2
2 2 21. Let α , β be the roots of the equation
15. Let α , β , γ be three non-zero vectors which x 2 − 6 x − 2 = 0 with α > β. Ifa n = α n − β n for
are pairwise non-collinear. If α + 3β is a − 2a 8
n ≥ 1, then the value of 10 is
collinear with γ and β + 2 γ is collinear with α, 2a 0
then α + 3β + 6 γ is (a) 1 (b) 2
(a) γ (b) 0 (c) 3 (d) 4
(c) α + γ (d) α
22. For x ∈ R , x ≠ − 1, if
16. Let f :R → R be given by f (x) = | x 2 − 1|, x ∈ R.
(1 + x)2016 + x (1 + x)2015 + x 2 (1 + x)2014
Then 2016
(a) f has a local minimum at x = ± 1but no local + .... + x 2016 = Σ a i . x i , then a17 is equal to
i =0
maximum.
(b) f has a local minimum at x = 0 but no local 2016! 2016!
(a) (b)
minimum. 17 !1999! 16!
(c) f has a local minima at x = ± 1and a local 2017 ! 2017 !
(c) (d)
maxima at x = 0. 2000! 17 !2000!
(d) f has neither a local maxima nor a local minima
at any point. 23. Five letter words, having distinct letters, are
17. Let a , b, c be real numbers, each greater than to be constructed using the letters of the
2 3 5 word ‘EQUATION’ so that each word
1, such that log b a + log c b + log a c = 3. contains exactly three vowels and two
3 5 2
consonants. How many of them have all the
If the value of b is 9, then the value of ‘a’
vowels together?
must be
27 (a) 3600 (b) 1800
(a) 3 81 (b) (c) 18 (d) 27
2 (c) 1080 (d) 900
WB JEE (Engineering) Solved Paper 2021 13
24. What is the number of ways in which an 31. Let T and U be the set of all orthogonal
examiner can assign 10 marks to 4 questions, matrices of order 3 over R and the set of all
giving not less than 2 marks to any question? non-singular matrices of order 3 over R
(a) 4 (b) 6 respectively. Let A = { − 1, 0 , 1}, then
(c) 10 (d) 16 (a) there exists bijective mapping between A and T,
U.
25. The digit in the unit’s place of the number (b) there does not exist bijective mapping between
1! + 2 ! + 3 ! + .... + 99 ! is A and T, U.
(a) 3 (b) 0 (c) there exists bijective mapping between A and T
(c) 1 (d) 7 but not between A and U.
(d) there exists bijective mapping between A and U
26. If M is a 3 × 3 matrix such that (0, 1, 2) but not between A and T.
M = (1 0 0), (3, 4 5) M = (0, 1, 0), then
(6 7 8) M is equal to 32. Four persons A , B, C and D throw and
(a) (2 1 −2 ) (b) (0 0 1) unbiased die, turn by turn, in succession till
(c) (−1 2 0) (d) (9 10 8) one gets an even number and win the game.
What is the probability that A wins the game
1 0 0 if A begins?
27. Let A = 0 cos t sin t (a)
1
(b)
1
(c)
7
(d)
8
0 − sin t cos t
4 2 15 15
Let λ 1 , λ 2 , λ 3 be the roots of det(A − λI 3) = 0, 33. The mean and variance of a binomial
where I 3 denotes the identity matrix. If distribution are 4 and 2 respectively. Then
the probability of exactly two successes is
λ 1 + λ 2 + λ 3 = 2 + 1, then the set of possible
7 21 7 9
values of t, − π ≤ t < π is (a)
64
(b)
128
(c)
32
(d)
32
π
(a) a void set (b)
4
π π
(c) − ,
π π
(d) − ,
34. Let Sn = cot −1 2 + cot −1 8 + cot −118
4 4 3 3
+ cot −1 32 + ... to nth term. Then lim Sn is
n→ ∞
28. Let A and B two non singular skew
π π π π
symmetric matrices such that AB = BA, then (a) (b) (c) (d)
A 2 B2 (A T B)−1 (AB−1)T is equal to 3 4 6 8
(a) A 2 (b) − B2 35. If a > 0 , b > 0 then the maximum area of the
(c) − A 2
(d) AB parallelogram whose three vertices are O(0 , 0),
A(a cos θ , bsin θ) and B(a cos θ , − bsin θ) is
29. If a n(> 0) be the nth term of a G.P. then (a) ab when θ =
π
(b) 3ab when θ =
π
4 4
log a n log a n + 1 log a n + 2 π
(c) ab when θ = − (d) 2ab
log a n + 3 log a n + 4 log a n + 5 is equal to 2
log a n + 6 log a n + 7 log a n + 8 36. Let A be the fixed point (0, 4) and B be a
(a) 1 (b) 2 (c) −2 (d) 0 moving point on X -axis. Let M be the
midpoint of AB and let the perpendicular
30. Let A , B, C be three non-void subsets of set S. bisector of AB meets the Y -axis at R. The
Let (A ∩ C) ∪ (B ∩ C ′) = φ where C ′ denote the locus of the midpoint P of MR is
complement of set C in S. Then 1
(a) y + x2 = 2 (b) x2 + ( y − 2 )2 =
(a) A ∩ B = φ (b) A ∩ B ≠ φ 4
1
(c) A ∩C = A (d) A ∪C = A (c) ( y − 2 )2 − x2 = (d) x2 + y2 = 16
4
14 WB JEE (Engineering) Solved Paper 2021
37. A moving line intersects the lines x + y = 0 44. The locus of the centre of a variable circle
and x − y = 0 at the points A , B respectively which always touches two given circles
such that the area of the triangle with externally is
vertices (0,0 ), A and B has a constant area C. (a) an ellipse (b) a hyperbola
The locus of the mid-point AB is given by the (c) a parabola (d) a circle
equation
45. A line with positive direction cosines passes
(a) ( x2 + y2 )2 = C 2 (b) ( x2 − y2 )2 = C 2
through the point P(2 , −1, 2) and makes equal
(c) ( x + y)2 = C 2 (d) ( x − y)2 = C 2 angle with co-ordinate axes. The line meets
the plane 2 x + y + z = 9 at point Q. The length
38. The locus of the vertices of the family of of the line segment PQ equals.
parabolas 6 y = 2 a 3 x 2 + 3 a 2 x − 12 a is (a) 1 unit (b) 2 unit
105 64 (c) 3 unit (d) 2 unit
(a) xy = (b) xy =
64 105
35 16 5 x + 12 1 − x 2
(c) xy = (d) xy = 46. For y = sin −1
16 35 ;| x| ≤ 1, if
13
39. A ray of light along x + 3 y = 3 gets a(1 − x ) y 2 + bxy1 = 0 then (a , b) =
2
7
(a) lx + my ± tanα l 2 + m2 = 0 7K
68. The remainder when 7 7 (22 time 7) is
(b) lx + my ± z tanα l 2 + m2 + 1 = 0 divided by 48 is
(c) lx + my ± z tanα l 2 + 1 = 0 (a) 21 (b) 7
(c) 47 (d) 1
(d) lx + my ± z tanα l + m = 02 2
Answers
Physics
1. (d) 2. (c) 3. (c) 4. (a) 5. (a) 6. (c) 7. (*) 8. (a) 9. (a) 10. (c)
11. (a) 12. (b) 13. (d) 14. (a) 15. (b) 16. (b) 17. (d) 18. (b) 19. (b) 20. (d)
21. (a) 22. (b) 23. (c) 24. (b) 25. (d) 26. (d) 27. (d) 28. (c) 29. (c) 30. (c)
31. (b) 32. (c) 33. (b) 34. (b) 35. (b) 36. (b,c) 37. (a,b) 38. (b,c) 39. (b,c) 40. (a,d)
Chemistry
41. (b) 42. (d) 43. (a) 44. (b) 45. (c) 46. (d) 47. (d) 48. (d) 49. (c) 50. (b)
51. (d) 52. (d) 53. (a) 54. (a) 55. (a) 56. (d) 57. (b) 58. (d) 59. (d) 60. (c)
61. (d) 62. (c) 63. (b) 64. (d) 65. (a) 66. (b) 67. (b) 68. (a) 69. (a) 70. (d)
71. (a) 72. (c) 73. (a) 74. (a) 75. (d) 76. (b,c) 77. (d) 78. (c, d) 79. (b, c) 80. (b,c,d)
Mathematics
1. (c) 2. (b) 3. (c) 4. (c) 5. (b) 6. (c) 7. (c) 8. (c) 9. (b) 10. (a)
11. (d) 12. (c) 13. (b) 14. (c) 15. (b) 16. (c) 17. (d) 18. (a) 19. (d) 20. (c)
21. (c) 22. (d) 23. (c) 24. (c) 25. (a) 26. (c) 27. (c) 28. (c) 29. (d) 30. (a)
31. (b) 32. (d) 33. (a) 34. (b) 35. (a) 36. (a) 37. (b) 38. (a) 39. (b) 40. (a)
41. (b) 42. (b) 43. (b) 44. (b) 45. (c) 46. (b) 47. (c) 48. (c) 49. (c) 50. (b)
51. (b) 52. (c) 53. (a) 54. (b) 55. (c) 56. (a) 57. (c) 58. (b) 59. (d) 60. (c)
61. (b) 62. (b) 63. (d) 64. (c) 65. (a) 66. (b, c) 67. (d) 68. (b) 69. (c,d) 70. (a,c)
71. (a) 72. (a) 73. (a, d) 74. (b, c) 75. (d)
The molar mass of O2 is 32g/mol. Substituting in Eq. 21. (a) The given situation is shown below
(iii), we get
1 Mass +
⇒ = +
2 32g / mol b
–
⇒ Mass of O2 = 16 g –
+q +
19. (b) We know that, latent heat of fusion at ice is – r a –
79.7 cal per gram. – –
+ –
Let final temperature be T. +
P
Then, +
m1 s∆T = m2L
⇒ 300 × 1 × (25 − T) = 100 × 75 A radial electric field E exists in the region
⇒ 25 − T = 25 between the two shells due to charge on inner
shell only.
⇒ T = 0° C
Electric field at point P is calculated according to
After that total energy left, Q = 4.7 × 100 = 470 cal
Gauss’s law,
Total mass of water = 400 g q
Amount of water again converted into ice, i.e φE = E ⋅ 4 πr 2 =
ε0
Q 470
m= = q
Lice 79.7 ⇒ E= …(i)
4 π ε0 r 2
⇒ m = 59
. g
The potential at the centre of sphere,
Thus, whole mass is converted into water at 0°C and b b b q
about 5.9 g water is again converted into ice whose V = − ∫ E ⋅ dr = ∫ E ⋅ dr = ∫ dr
a a a 4 πε r 2
temperature is also 0°C. 0
After achieving the temperature of 0°C, latent heat [from Eq. (i)]
of fusion is required firstly for conversion of water q 1 1
⇒ V= −
into ice, then further lowering of temperature as 4 π ε0 a b
possible.
So, the final temperature will be 0°C. 22. (b) The given situation is shown in the figure
= p −
∆L = αL∆T 4T 4 3
πc
Now, new length of the rod, c 3
L ′ = ∆L + L = αL∆T + L = L(α∆T + 1) where, T is surface tension of soap bubble.
⇒ L ′ = L(1 + α∆T) …(i) ⇒ p[c 3 − (a 3 + b 3)] − 4T(a 2 + b 2 − c 2) = 0
Now, new angular momentum, p(c 3 − a 3 − b 3)
⇒ T=
ML ′2 ω 4(a 2 + b 2 − c 2)
L f = I 2ω2 = .
3 2
By the law of conservation of angular momentum, 36. (b,c)
M
L f = Li O
a
ML2 x
I 2ω2 = ⋅ω
3
ML ′2 ω ML2
⋅ = ⋅ω Magnetic field at the centre O of the circular coil
3 2 3
due to bar magnet (axial position)
L ′2 µ 2M
⇒ = L2 B= 0. 3
2 4π x
L2(1 + α∆T)2
⇒ = L2 [from Eq. (i)] Hence, option (a) is not true.
2 dφ d
Induced emf, e = − = − ⋅ BA
⇒ (1 + α∆T)2 = 2 dt dt
⇒ 1 + α 2∆T 2 + 2α∆T = 2 d µ 2M
= − ⋅ 0 ⋅ 3 πa 2
dt 4 π x
Since, α is very small, hence the term α 2∆T 2 may be
µ Ma 2 d 1
neglected. =− 0 ⋅
∴ 1 + 2α∆T = 2 ⇒ 2α∆T = 1 2 dt x 3
−3 dx
⇒ ∆T =
1 = − µ 0 Ma 2. 4
2α x dt
3µ 0 Ma 2 dx
34. (b) As the gravitational field is uniform, therefore e= ⋅
centre of gravity and the centre of mass are at x4 dt
same location. 1
⇒ e∝ 4
∴The location of centre of mass is x
∞ Hence, option (b) is true.
∫ hdm
h= 0∞ e 3µ 0 Ma 2 dx
Induced current, I = = …(i)
∫0dm R x 4 R dt
∞ 3µ 0 Ma 2 dx 2
∴ Magnetic moment, µ = IA = πa
∫ hρdh
h = 0∞ …(i) x 4 R dt
3 π µ 0 M dx 4
∫0 ρdh ⇒ µ= (a )
x 4 R dt
But from barometric formula and gas law,
µ ∝ a4
ρ = ρ0 e − Mgh/ RT
⇒ Hence, option (c) is true.
26 WB JEE (Engineering) Solved Paper 2021
. × 10
527 −19 Potential energy,
= eV U = Ax(x − 4) J
1.6 × 10−19
⇒ U = Ax 2 − 4 Ax
= 329
. eV
dV d
∴Maximum kinetic energy of photoelectron, Force, F=− = − (Ax 2 − 4 Ax)
dx dx
K max = − (eV0) = − [e(−2)] = 2eV
⇒ F = − 2Ax + 4 A …(i)
38. (b,c) Since, given p-V diagram is not a cyclic This force is dependent on x, hence particle is not
process, hence the change in internal energy for acted upon by a constant force.
the whole process is not zero.
From Eq. (i), it is clear that f ∝ − x
WB JEE (Engineering) Solved Paper 2021 27
Hence, particle executes simple hormonic motion. Since, charge particle enters into magnetic field B
Speed of particle is maximum at equilibrium perpendicular to it, hence it performs a circular
position, i.e. when F = 0 path in magnetic field. Radius of circular path,
−2Ax + 4 A = 0 mv
i.e. r=
⇒ x = 2m Bq
Period of oscillation, Charge particle will enter in region c, when
m L < 2r
T = 2π …(i) mv
k ⇒ L< 2
From Eq. (i), F = −2Ax + 4 A Bq
LBq qLB
Since, value of k will be obtained in terms of A. ⇒ < v ⇒ v>
Therefore, the value of time period will be also 2m 2m
obtained in terms of A, hence option (d) is not ⇒
v>
1 qLB
correct. 2 m
qLB
40. (a,d) The given situation is shown below i.e. When v > , then particle surely will inter
b m
a qLB qLB
× × × c into region c, because >
m 2m
× × ×
Hence, option (a) is correct.
× × ×
Since, charge particle enters into region c, hence
× × × path of the particle is a circle not in region b.
× × × Hence, option (c) is not correct.
× × × Time spent in region b is given as
v 2πm
× × × T= which is independent of v.
Bq
L Hence, option (d) is correct.
Chemistry
41. (b) The exact order of the boiling points of the Therefore, the maximum number of atom in one
given compound is isopentane < n-pentane plane in this case are 15.
< butanone < 1-butanol. 43. (a) The total number of triple bonds present in
The highest boiling will be of butanol because in cyclo (18) carbon are 9. Its structure is shown
butanol, molecules are associated due to extensive below:
intermolecular hydrogen bonding. C ≡≡ C — C≡≡
Now butanone has higher boiling point from —
C C—
≡≡
All the correct resonating structures of p-nitro-N, The general formula of amino acid is
N-dimethyl aniline are as follows H 2 N CH(R) COOH, R is CH 3 for alanine and H
+ + + for glycine. The structure are as follows
NMe2 NMe2 NMe2 NMe2 O O
– – H 2N CH C OH + H 2N C H C OH →
CH 3 H
–
NH2 NH2
+ +
N N N+ N+
NH COOH + NH COOH +
O O– O O– O O– O O– CH3 CH3
CH3
45. (c) COOH CH3 O O
H H
76. (b,c) Butyne (Me C ≡≡ C Me) on reaction with Na / NH 3(liq.) undergoes reduction to give but-2-ene which
on reaction with dil. alkaline KMnO 4 gives diol.
H H OH H H OH
Na/NH3(liq.) Dil. alk. KMnO 4
Me—C== C—Me Me—C== C—Me Me—C—C—Me + Me—C—C—Me
Ethanol, –33ºC
But-2-ene
H OH HO H
Butan-2, 3-diol
Br2/NaOH,
H3O+ SOCl2
HOOC CH3 ClOC CH3
Br2/NaOH O
H2N CH3 C CH3
Hoffmann H2N
bromamide
(Y) degradation 4-methyl benzamide
78. (c, d)Buffer solutions are made by mixing weak acid with its salt of strong base or mixing weak base with its
salt of strong acid. It is never formed by mixing strong acid and strong base. The pair(s) of solutions which
− + − +
form a buffer upon mixing are (c) HNO3 and CH 3 COONa, (d) i.e. CH 3 COOH and CH 3 COONa.
79. (b, c) Reaction of silver nitrate solution with phosphorus acid produces phosphoric acid and metallic silver.
Complete reaction is as follows:
H 3 PO3 + AgNO3 + H 2 O −→ H 3 PO4 + Ag ↓ + 2HNO3
Phosphorus Phosphoric
acid acid
80. (b, c, d) N 2 H 4 and H 2 O2 show similarity in reducing and oxidising nature. They have same hybridisation of
central atoms i.e. sp 3. The structure of hydrazine and hydrogen peroxide as follows.
H
H H
N N O O
H
H H
Hydrazine Hydrogen peroxide
(N2H4 ) (H2O2 )
34 WB JEE (Engineering) Solved Paper 2021
Mathematics
x x2 2 1
dt − ∫ 2 dt
a
b 2 ∫ t
e − x −1 − =−
t
1. (c) Given, I = lim 2
x→ 0 2
x
2 t− 2 + 1
= − 2 ∫ log() t −a + C
b − 2 + 1
x x2
= − 2 log | t | + + C
e − x −1 − 2 a
sin 2 b t
x 2 x2
ex − x − 1 − 2 a
2 ∴ I=− log | a + b cos x | + a + b cos x + C
= lim × b2
x→ 0 x2 x2
e − x −1 −
x
2
2 ⇒ α=−
x2 b2
x2 4. (c) Given, g(x) = ∫ f ()
2x t
dt, x > 0
ex − x − 1 −
= lim 2 0 form x t
x→ 0 x2 0 f (2x) d f (x) d
⇒ g′(x) = (2x) − (x)
Q lim sin x
= 1
2x dx x dx
x→0 x (Using Leibnitz Rule for Differentiation)
ex − 1 − 0 − x f (2x) f (x)
= lim 0 form = (2) −
2x x
x→ 0 2x 0
f (2x) − f (x)
ex − 1 =
= lim x
x→ 0 2 f (x) − f (x)
= (Q f (2x) = f (x))
1 ex − 1
= lim × x x
2x→0 x ⇒ g′(x) = 0
1 ex − 1 ⇒ g(x) is constant function
= lim × lim x
2 x → 0 x x→ 0
5. (b) We have,
=0
3 | x + 1|
=∫ dx
2. (b) Given, | f ′(x) |≤ 5 for all x 1| x − 2| + | x − 3|
1 1 1 2 x −1 3 x −1
⇒ ∫0 − 5dx ≤ ∫0 f ′(x)dx ≤ ∫0 5 dx = ∫1 − x + 2 − x + 3 dx + ∫2 x − 2 − x + 3 dx
⇒ − 5 ≤ f()
1 − 0≤ 5 (Q| x | = x, x ≥ 0, − x, x < 0)
⇒ − 5 ≤ f()1 ≤5 2x −1 3
=∫ dx + ∫2(x − 1)dx
⇒ 1 ∈ [− 5, 5]
f() 1 − 2x + 5
sin 2x 3
3. (c) Let I = ∫ dx 1 2− 2x + 2 x2
(a + b cos x)2 =− ∫
2 − 2x + 5
1
dx +
2
− x
2
2 sin x cos x
⇒ I=∫ dx 1 2− 2x + 5 − 3
dx + − 3 − (2 − 2)
9
2 ∫1 − 2x + 5
(a + b cos x)2 =−
2
Put a + b cos x = t
t−a 1 2− 2x + 5 2 3 3
2 ∫1 − 2x + 5 ∫1 − 2x + 5 dx +
⇒ cos x = ⇒ − b sin x dx = dt =− dx −
b 2
t − a 1
I = − ∫ 1 2
dx + log [− 2x + 5]1 +
2 3 3
2 ∫1
2
dt =−
b b t2 2 2
2 t−a 3
= − 2 ∫ 2 dt = log 3 + 1
b t 4
WB JEE (Engineering) Solved Paper 2021 35
1 1
= [log e t − log e (t + 1)]1t
2
2
2
2
6. (c) ∫ x + 1 + x − 1 − 2 dx t
t
1 x − 1 x + 1
= log e
−
2 t + 1 1
1 1
t 1
2 x + 1 x − 1 2 2 = log e − log e
= ∫ − dx t + 1 2
1 x − 1 x + 1
− 2t
2 = log e
1
t +1
2
4x 2t 3
∴ =
= ∫ 2 dx
− 1 2t + 1 2
1 x
−
2 ⇒ 4t = 3t + 3
1 ⇒t = 3
0
4x 4x 2
Now, ex − 1 = t
= ∫ 2
− 1
dx − ∫ 2 dx
− 1 ⇒ ex − 1 = 3
1 x 0 x
−
2
1
⇒ ex = 4
0
x 2
x ⇒ x = log 4
=−4 ∫ 1 − x 2 dx + 4∫ 1 − x 2 dx 8. (c) Equation of normal
1 0
−
2 −1
1
Y − y= (X − x)
dy / dx
= 2[log(1 − x 2)]0 1 − 2 [log(1 − x 2)]02
−
2
When y = 0
G = x + y , 0
dy
= − 2 log 1 − − 2 log 1 −
1 1 Then,
dx
4 4
3 4 According to the question,
= − 4 log = 4 log dy dy
4 3 x+ y = 2x ⇒ y = x
x dx dx
1
7. (c) Let I = ∫ e x
−1
dx ⇒ ∫ y dy = ∫ x dx
loge 2
y2 x 2 y2 − x 2
Put ex − 1 = t ⇒ ex = t + 1 ⇒ = + C ⇒ =C
2 2 2
e x dx = dt
⇒ which is hyperbola.
dt dt
⇒ dx = x = 9. (b) We know that the equation of an ellipse
e t +1
x2 y2
When x = x, t = e − 1 x whose centre is at origin is + =1
a2 b2
and when x = log e 2, t = eloge 2
−1= 2−1 =1 Differentiate both sides w.r.t x, we get
t 2x 2y dy
1 + 2 =0
I= ∫ t(t + 1) dt a2 b dx
1
2y dy 2x
1 A B ⇒ =− 2
= + b 2 dx a
t(t + 1) t t +1
y dy b2
1 = A(t + 1) + B()
t ⇒ =− 2
x dx a
t = 0, A = 1
Again differentiate both sides w.r.t x, we get
t + 1 = 0 ⇒ t = − 1, B = − 1
y d dy d y
∴
1 1
= −
1 + y′ =0
x dx dx dx x
t(t + 1) t t + 1
xy′ − y × 1
y ′′ + y′
y
t
1 1 ⇒ =0
⇒ I = ∫ − dt x x2
1
t t + 1
36 WB JEE (Engineering) Solved Paper 2021
xy′− y
y ′′ + y′
y
⇒ =0 13. (b) Given, radius of circular track is 10 m.
x x2
⇒ xyy ′′ + x(y′)2 − yy′ = 0
y
Which is the required differential equation.
= ke 2 dt
= sec2 60° ⋅ (10)
11. (d) − y = (x − 1) − 1 2
= (2)2 × 10
y = 1 − (x − 1)2 = 40 m/sec.
Hence the required area is
1 14. (c) Let B travels x units,
1
(x − 1)3 v = u + at
∫ [1 − (x − 1) ] dx = x − 3
2
⇒ β = λ 2α − 2γ ...(ii) 1
⇒ a = (36) 2 = 33 = 27
From Eqs. (i) and (ii), we get
λ1 α
γ − = λ 2α − 2γ 18. (a) Given, h(p) = 1 {h(p + 1) + h(p − 1)}
3 3 2
λ for every p = 1, 2, ..., 99
α λ 2 + = γ 1 +
2
1
⇒
3 3 2h(p) = h(p + 1) + h(p − 1)
1 λ ⇒ h(p − 1), h(p), h(p + 1) are in AP.
⇒ λ 2 + = 0 and 1 + 2 = 0
3 3 Now, h(100) = 20
1 λ1 ⇒ h(0) + 99d = 20
⇒ λ 2 = − and =−2
3 3 ⇒ 5 + 99d = 20 (Qh(0) = 5)
1 15 5
⇒ λ1 = − 6 and λ 2 = − ⇒ d= =
3 99 33
α ∴ 1 = h(0) + d
From Eqs. (i) and (ii), β = − 2γ − h()
3 5
= 5+ = 5 + 0.15
α 33
∴ α + 3β + 6γ = α + 3 − 2γ − + 6γ
3 =5.15
=0
19. (d) Let z = e iθ
16. (c) Given, f (x) = | x − 1 |, x ∈ R.
2
e iθ 1
Also, let w = =
The graph of f (x) is clearly, from graph. 1 − e 2iθ e − iθ − e iθ
f(x)=x2–1 1 i
= =
− 2 ⋅ i sin θ 2 sin θ
⇒ w is purely imaginary
⇒ Locus of point is Y-axis.
x=–1 x=1
20. (c) We have, z 2 = w2
⇒ z 2 − w2 = 0
f (x) has a local minima at x = ± 1 and has local
maxima at x = 0 ⇒ (z − w)(z + w) = 0
17. (d) Given, ⇒ z = w, z = − w
2 3 5 ⇒ |z|= |w|
log b a + log c b + log a c = 3
3 5 2 ∴B⊂ A
2 log a 3 log b 5 log c
⇒ + + =3 21. (c) Given, x 2 − 6 x − 2 = 0
3 log b 5 log c 2 log a
2 3 5 ⇒ x n − 2(x 2 − 6 x − 2) = 0
log b a + log c b + log a c
Since, A.M = 3 5 2 ⇒ x n − 6 x n − 1 − 2x n − 2 = 0
3
3 When, n = 10,
= =1
3 Then, x10 − 6 x 9 − 2x 8 = 0
1
⇒ x10 − 2x 8 = 6 x 9
G.M = log b a ⋅ log c b ⋅ log a c
2 3 5 3
and
3 5 2 Since, α and β are roots of (i).
2 log a 3 log b 5 log c
= × × =1 ∴ α10 − 2α 8 = 6α 9
3 log b 5 log c 2 log a
and β10 − 2β 8 = 6β 9
∴ A.M. = G.M
∴
2 log a
= 1 ⇒ a2 = b3 ∴ (α10 − β10) − 2(α 8 − β 8) = 6(α 9 − β 9)
3 log b
38 WB JEE (Engineering) Solved Paper 2021
Y B′
37. (b) x–y=0 On putting y = 0, we get x = 3
B
∴Coordinates of A are ( 3, 0)
Now, let B(0, 1) then B′ = (0, − 1)
O
X y−0 x− 3
∴Equation of AB′ is =
A 0+1 3−0
⇒ 3y = x − 3
x+y=0
Let coordinates of A are (α, − α) and coordinates of B 40. (a)
are (β, β) A
∴β + α = 2h and β − α = 2k
1
Now, Area of ∆AOB = × OA × OB M O
x2+y2=4
2 (–4,0)
1 B
= 2α 2
2β 2 = |αβ | = c
2
⇒ α 2β 2 = c 2 Given, equation of circle x 2 + y 2 = 4
⇒ 16α 2β 2 = 16c 2 Centre = (0, 0) and radius = 2
WB JEE (Engineering) Solved Paper 2021 41
e− t
a
48. (c) Given, f1 (x) = x, f2(x) = 2 + log e x Let I= ∫t − a −1
dt
Now, Let h(x) = f2(x) − f1 (x) a −1
= (2 + log e x) − x a
e − ( a + a − 1 − t)
= 2 + log e x − x
I= ∫a + a −1 − t − a −1
dt
a −1
Here, h(0+ ) < 0, h()
1 > 0, h(e) > 0 and h(e 2) < 0 and
b b
value of h(x) for all x ≥ e 2 is negative. Q ∫ f (x) dx =
a
∫ f (a + b − x) dx
∴h(x) = 0 has two roots is (0, 1) and (e , e 2) a
a t − 2a + 1
e
49. (c) Given equation, 6x + 8x = 10x I= ∫a − 2− t
dt
x x a −1
⇒ 6 + 8 =1
put t − (a − 1) = x ⇒ dt = dx
10 10
x x when t = a, x = 1 and t = a − 1 ⇒ x = 0
⇒ 3 + 4 = 1 1 + a − 1 − 2a + 1
ex
5 5 ∴ I= ∫ a − 2 − x − a + 1 dx
3 4 0
Let = sinθ ⇒ = cosθ
ex ⋅ e− a
1
5 5
∴ (sinθ)x + (cosθ)x = 1
⇒ I= ∫ − (x + 1) dx
0
it is possible only x = 2 1
ex
∴given equation has exactly one real root. ⇒ I = − e− a∫ dx
0
x +1
50. (b) We have, f : D → R where D = [0, 1] ∪ [2, 4] 1 et
if x ∈ [0, 1] ⇒ I = − e − a(b) Q ∫ dt = b
x, +
f (x) = 0 t 1
4 − x , if x ∈ [2, 4]
⇒ I = − be − a
Y
0,2 53. (a) We have,
2
f (x) = log e (1 + e10 x ) − tan− 1 (e 5x )
1 0,1
10e10 x 5e 5x
X′ X ⇒ f ′(x) = −
O (1,0) (2,0) (3,0) (4,0) 1+ e 10 x
1 + e10 x
10e10 x − 5e 5x
⇒ f ′(x) =
1 + e10 x
Y′
WB JEE (Engineering) Solved Paper 2021 43
10 − 5 5 2 3/ 2 2
⇒ f ′(0) = = = 5× [y ]0
1+1 2 3
10 3/ 2
We know that, = [2 − 0]
f (x + ∆x) − f (x) = f ′(x) dx 3
10
5
⇒ f (x + ∆x) − f (x) = × 0.2 [Qdx = 0.2] = ×2 2
2 3
20
⇒ ∆f (x) = 0.5 = 2sq unit.
3
54. (b) We have,
57. (c) We have,
f (x) = 1 + x
a(α × β) + b(β × γ) + c(γ + α) = 0
f (c) = 1 + c
Clearly a(α × β), (β × γ ) and (γ + α) are coplanar
⇒ c =1 + c [Q f (c) = c] vector∴α , β and γ are also coplanar
⇒ (c − 1)2 = ( c)2
58. (b) Given equation
⇒ c 2 − 2c + 1 = c y 2 = 2x 3
⇒ c 2 − 3c + 1 = 0 dy
⇒ 2y = 6x2
3± 5 dx
⇒ c=
2 dy 3x 2
⇒ =
∴ f has unique fixed point in [1, ∞] dx 4
4x
dy 3h2
55. (c) Let L = lim 3x − 1 ⇒
dx ( h , k)
=
k
x → ∞ 3x + 1
3x − 1 Slope of tangent is perpendicular to line
lim − 1 4x
L=e
x → ∞ 3x + 1 4 x = 3y
3x − 1 − 3x − 1
3h2 4
lim 4x ∴ = −1
L=e
x→∞ 3x + 1 k 3
lim
− 2× 4 ⇒ 4h2 = − k …(i)
x→∞ 1
3+
L=e x and k = 2h
2 3
…(ii)
From Eqs. (i) and (ii)
L = e− 8/ 3
(4h2)2 = 2h3
2
56. (a) Given curve, y = 4 x 2, y = x and y = 2 ⇒ 16h4 = 2h3
9
1
Graph of given curve ⇒ h=
y=4x2 8
2
k = − 4
–1 1 1
,2 2 ∴
,2
√ 2 y=2 √ 2 y=x /9 8
1 −1
(–3/√2,2)
0,2 (3/√2,2) ⇒ k=−4× =
64 64
1 − 1
∴ (h, k) = , only
8 16
= 2(p + q + r) = 12 1 b2 c2
=p+ q+ r=6 = (a + b + c + 10) 1 b + 10
2 2 2 2
c2
∴p = 3, q = 2, r = 1 1 b 2
c + 10
2
= 3
6! 6!
∴Coefficient of a 3b 4 c 5 is
3! 21
! ! 3! 3! R2 → R2 − R1 and R3 → R3 − R1
1 b2 c 2
60. (c) Given a , b , c are in G.P and log 5c , log 7b . = (a + b + c + 10) 0 10
2 2 2
0
2a 5c
log and A.P
2a 0 0 10
7b
= (a 2 + b 2 + c 2 + 10) (100)
∴b 2 = ac
QIf is divisible by 100.
2log = log + log
7b 5c 2a
and
5c 2a 7b 62. (b) We have,
2 S = {(x , y) : y = x + 1, 0 < x < 2}
49b 5c
⇒ = For reflexive (x , x) ∈ S
25c 2 7b
x = x + 1 ∉S
⇒ (7b)3 = (5c)3 ∴S is not reflexive
⇒ 7b = 5c ⇒ c =
7
b ∴S is not equivalence relation
5 T = {(x , y) : x − y is an integer}
and b 2 = ac For reflexive (x , x) ∈ T
x = x − x = 0∈R
b 2 = a b
7
⇒ T is reflexive
5
For symmetric
7
⇒ b= a (x , y) = x − y is an integer
5
(y, x) = y − x is also integer
5b 7
Sides are , b, b ∴T is symmetric
7 5
For Transitive
∴ a , b , c are the length of the sides of scalene
triangle. (x , y) ∈ T, (y, z) ∈ T a⇒(x , z) ∈ T
(x , y) = x − y is an integer
61. (b) We have, (y, z) = y − z is also integer
a 2 + 10 ab ac ∴(x − y) + (y − z) = x − z is an integer
ab b 2 + 10 bc ∴ (x , z) ∈ T
ac bc c 2 + 10 Hence, T is an equivalence relation.
∴ P3 = lx + my + nz = 0
Taking common a , b , c from R1 , R2 and R3 angle between P1 and P3 = α
respectively
l 2 + m2
a 2 + 10 b2 c2 ∴ cosα =
abc l + m + n2 l 2 + m2
2 2
= a2 b 2 + 10 c2
abc l 2 + m2
a2 b2 c 2 + 10 ⇒ cosα =
l + m2 + n2
2
Applying C1 → C1 + C2 + C3, we get
a 2 + b 2 + c 2 + 10 b2 c2 l 2 + m2
⇒ cos2 α =
l + m2 + n2
2
= a 2 + b 2 + c 2 + 10 b 2 + 10 c2
a + b + c + 0
2 2 2 2
c + 10
2 l 2 + m2 + n2 n2
b ⇒ sec2 α = =1 + 2
l + m
2 2
l + m2
WB JEE (Engineering) Solved Paper 2021 45
⇒
3
<I<
2 68. (b) We have
8 6 7... 7
77 (22 time 7)
7
66. (b, c) have | z + i | − | z − 1 | = | z | − 2 ⇒ 7 77
is an odd number
| z | − 2⇒| z | = 2represent a circle. ∴ 7( 2m + 1) = 72m × 7
= (72)m × 7
= (49)m × 7
|z|=2
= (48 + 1)m × 7
= 71
( + 48k)
= 7 + (48 × 7) k
∴ remainder = 7
46 WB JEE (Engineering) Solved Paper 2021
2 2 2
α β r2
⇒ Locus of R is x − + y − =
dx
69. (c, d) (a) I1 = ∫4+ x 2 2 2 4
−2
0
n
It is possible to put x = tan t + …+
∴ Option (b) is also incorrect [n + 4 (n − 1)]3
n −1
Hence, option (c) and (d) are correct. n
⇒ I = lim
n→ ∞
∑
70. (a, c) Let equation of plane be r=0 (n + 4r)3
x y z n −1
+ + =1 1 n n
a b c ⇒ I = lim
n→ ∞ n
∑
r = 0 (n + 4r)
3
Centroid = , ,
a b c
n −1
3 3 3 1 1
a b c
⇒ I = lim
n→ ∞ n
∑ 3/ 2
∴ = 1, = r, = r 2 r=0 1 + 4r
3 3 3 n
⇒ a = 3, b = 3r c = 3r 2 1 dx
Equation of plane be
⇒ I= ∫0 (1 + 4 x )3/ 2
x y z
+ + =1 (1 + 4 x)− 3/ 2 + 1
1
3 3r 3r 2 ⇒ I=
Now plane is passes through (5, 5, − 12) 4(− 3 / 2 + 1) 0
5 5 12 −1
∴ + − =1 ⇒ I= [(1 + 4 x) −1 / 2]10
3 3r 3r 2 2
⇒ 5r 2 + 5r − 12 = 3r 2 − 1 − 1/ 2
⇒ I= [5 − 1]
2
⇒ 2r 2 + 5r − 12 = 0
−1 1
⇒ 2r 2 + 8r − 3r − 12 = 0 ⇒ I= − 1
2 5
= (2r − 3) (r + 4) = 0 − 1 1 − 5
⇒ I=
⇒ r = 3 / 2, − 4 2 5
71. (a) Let the equation of circle is 5 −1 5( 5 − 1) 5 − 5
⇒ I= = =
(x − α)2 + (γ − β)2 = r 2 2 5 10 10
∴ x = α + r cosθ 0, if −1≤ x < 0
y = β + r sinθ 73. (a, d) We have, f (x) = 1, if x=0
∴ P ≡ (α + r cosθ, β + r sinθ) 2, if 0< x ≤1
θ = (a , b) x
Let R = (h, k) F(x) = ∫ f ()t dt
α + r cosθ β + r sinθ −1
∴ h= ,k =
2 2 0 1
The left compartment is filled with a given 20. Four identical point masses, each of mass m
mass of an ideal gas of molar mass 32 while and carrying charge + q are placed at the
the right compartment is filled with an equal corners of a square of sides a on a frictionless
mass of another ideal gas of molar mass 18 at plane surface. If the particles are released
same temperature. What will be the distance simultaneously, the kinetic energy of the
of P from the left wall A when equilibrium is system when they are infinitely far apart is
established? q2 q2
(a) (2 2 + 1) (b) ( 2 + 2 )
(a) 2.5 m (b) 1.8 m (c) 3.2 m (d) 2.1 m a a
q2 q2
18. When 100 g of boiling water at 100°C is (c) ( 2 + 4) (d) ( 2 + 1)
a a
added into a calorimeter containing 300 g of
cold water at 10°C, temperature of the 21. A very long charged solid cylinder of radius a
mixture becomes 20°C. Then, a metallic block contains a uniform charge density ρ.
of mass 1 kg at 10°C is dipped into the Dielectric constant of the material of the
mixture in the calorimeter. After reaching cylinder is K. What will be the magnitude of
thermal equilibrium, the final temperature electric field at a radial distance x (x < a) from
becomes 19°C. What is the specific heat of the axis of the cylinder?
the metal in CGS unit? (a) ρ
x
(b) ρ
x
(a) 0.01 (b) 0.3 (c) 0.09 (d) 0.1 ε0 2 kε0
x2 x
19. As shown in the figure, a point charge (c) ρ (d) ρ
−6
q1 = + 1 × 10 C is placed at the origin in 2 aε0 2k
G O r a
t
As shown in the figure, a wire is bent to form If all the cells have negligible internal
a D-shaped closed loop, carrying current I, resistance, what will be the current through
where the curved part is a semi-circle of the 2 Ω resistor when steady state is reached?
radius R. The loop is placed in a uniform (a) 0.66 A (b) 0.29 A (c) 0 A (d) 0.14 A
magnetic field B, which is directed into the
28. Consider a conducting wire of length L bent
plane of the paper. The magnetic force felt by
in the form of a circle of radius R and another
the closed loop is
conductor of length a (a < < R) is bent in the
1
(a) zero (b) IRB (c) 2IRB (d) IRB form of a square. The two loops are then
2
placed in same plane such that the square
25. 2R 2R 2R
loop is exactly at the centre of the circular
A loop. What will be the mutual inductance
between the two loops?
R R R πa2 πa2
(a) µ 0 (b) µ 0
L 16 L
B πa2 a2
(c) µ 0 (d) µ 0
4L 4 πL
What will be the equivalent resistance
between the terminals A and B of the infinite 29. An object, is placed 60 cm in front of a
resistive network shown in the figure? convex mirror of focal length 30 cm. A plane
( 3 + 1)R ( 3 − 1)R mirror is now placed facing the object in
(a) (b)
2 2 between the object and the convex mirror
R
(c) 3 (d) ( 3 + 1) R such that it covers lower half of the convex
2 mirror. What should be the distance of the
WB JEE (Engineering) Solved Paper 2020 5
plane mirror from the object, so that there block, if the specific heat of the block is 0.1
will be no parallax between the images CGS unit? Assume g =10 ms−2 and uniform
formed by the two mirrors? rise in temperature throughout the whole
(a) 40 cm (b) 30 cm (c) 20 cm (d) 15 cm block. [Ignore absorption of heat by the
table]
30. A thin convex lens is placed just above an
empty vessel of depth 80 cm. The image of a (a) 0.0025°C (b) 0.025°C
(c) 0.001°C (d) 0.05°C
coin kept at the bottom of the vessel is thus
formed 20 cm above the lens. If now water is 34. Consider an engine that absorbs 130 cal of
poured in the vessel upto a height of 64 cm, heat from a hot reservoir and delivers 30 cal
what will be the approximate new position of heat to a cold reservoir in each cycle. The
the image? Assume that refractive index of engine also consumes 2 J energy in each
water is 4/3. cycle to overcome friction. If the engine
(a) 21.33 cm, above the lens works at 90 cycles per minute, what will be
(b) 6.67 cm, below the lens the maximum power delivered to the load?
(c) 33.67 cm, above the lens [Assume the thermal equivalent of heat is
(d) 24 cm, above the lens 4.2 J/cal]
(a) 816 W (b) 819 W
Category II (Q. Nos. 31 to 35) (c) 627 W (d) 630 W
Carry 2 marks each and only one option is correct.
In case of incorrect answer or any combination of 35. Two pith balls, each carrying charge + q are
more than one answer, 1/2 mark will be deducted. hung from a hook by two strings. It is found
that when each charge is tripled, angle
31. A conducting circular loop of resistance 20 Ω between the strings double. What was the
and cross-sectional area 20 × 10 −2 m 2 is initial angle between the strings?
placed perpendicular to a spatially uniform (a) 30° (b) 60° (c) 45° (d) 90°
magnetic field B, which varies with time t as
B = 2 sin(50 πt) T. Find the net charge flowing Category III (Q. Nos. 36 to 40)
through the loop in 20 ms starting from t = 0. Carry 2 marks each and one or more option(s)
(a) 0.5 C (b) 0.2 C (c) 0 C is/are correct. If all correct answers are not marked
(d) 0.14 C
and also no incorrect answer is marked then score
32. A pair of parallel metal plates are kept with a = 2 × number of correct answers marked ÷ actual
separation d. One plate is at a potential + V number of correct answers. If any wrong option is
and the other is at ground potential. A marked or if any combination including a wrong
narrow beam of electrons enters the space option is marked, the answer will be considered
between the plates with a velocity v0 and in
wrong, but there is no negative marking for the
a direction parallel to the plates. What will be
same and zero marks will be awarded.
the angle of the beam with the plates after it
travels an axial distance L? 36. A point source of light is used in an
−1
eVL −1 eVL experiment of photoelectric effects. If the
(a) tan (b) tan 2
mdv 0 mdv 0 distance between the source and the
eVL eVL photoelectric surface is doubled, which of the
(c) sin−1 (d) cos −1 2
following may result?
mdv 0 mdv 0
(a) Stopping potential will be halved.
(b) Photoelectric current will decrease.
33. A metallic block of mass 20 kg is dragged
(c) Maximum kinetic energy of photoelectrons will
with a uniform velocity of 0.5 ms−1 on a decrease.
horizontal table for 2.1 s. The coefficient of (d) Stopping potential will increase slightly.
static friction between the block and the
table is 0.10. What will be the maximum 37. Two metallic spheres of equal outer radii are
possible rise in temperature of the metal found to have same moment of inertia about
6 WB JEE (Engineering) Solved Paper 2020
their respective diameters. Then, which of 39. A 400 Ω resistor, a 250 mH inductor and a
the following statement(s) is/are true? 2.5 µF capacitor are connected in series with
(a) The two spheres have equal masses. an AC source of peak voltage 5 V and angular
(b) The ratio of their masses is nearly 1.67 : 1. frequency 2kHz. What is the peak value of
(c) The spheres are made of different materials. the electrostatic energy of the capacitor?
(d) Their rotational kinetic energies will be equal
(a) 2 µJ (b) 2.5 µJ
when rotated with equal uniform angular speed
(c) 3.33 µJ (d) 5 µJ
about their respective diameters.
38. A simple pendulum of length l is displaced, 40. A charged particle moves with constant
so that its taught string is horizontal and velocity in a region, where no effect of gravity
then released. A uniform bar pivoted at one is felt but an electrostatic field E together
end is simultaneously released from its with a magnetic field B may be present. Then,
horizontal position. If their motions are which of the following cases are possible?
synchronous, what is the length of the bar? (a) E ≠ 0, B ≠ 0
(a)
3l
(b) l (c) 2l (d)
2l (b) E ≠ 0, B = 0
2 3 (c) E = 0, B = 0
(d) E = 0, B ≠ 0
Chemistry
Category-I (Q. Nos. 41 to 70) OH OH OH COOH
Carry 1 mark each and only one option is correct.
In case of incorrect answer or any combination of
more than one answer, 1/4 mark will be
NO2
deducted.
NO2 CH3 OCH3
41. (I) O2N CO2CH3 I II III IV
(a) II < IV < III < I (b) II < III < I < IV
(c) II < III < IV < I (d) III < II < I < IV
(II) MeO CO2CH3
44. For the following carbocations, the correct
order of stability is
(III) Me CO2CH3 I. ⊕ CH 2 COCH 3 II. ⊕ CH 2 OCH 3
III. ⊕ CH 2 CH 3
For the above three esters, the order of rates
(a) III < II < I (b) II < I < III
of alkaline hydrolysis is
(c) I < II < III (d) I < III < II
(a) I > II > III (b) II > III > I
(c) I > III > II (d) III > I > II 45. The reduction product of ethyl
È 3-oxobutanoate by NaBH 4 in methanol is
50% aq.NaOH
42. Ph CDO → Ph COOH + an OH OH O
Warm
alcohol.
(a) OEt (b) OH
This alcohol is
OH O O
(a) Ph CHD OH (b) Ph CHD OD
(c) Ph CD2 OH (d) Ph CD2 OD
(c) OEt (d) OH
43. The correct order of acidity for the following
compounds is :
WB JEE (Engineering) Solved Paper 2020 7
46. What is the major product of the following The equilibrium constant for the reaction
reaction ? 1
P - A + B at 25°C is
O O 2
CHO 1. NaOCOEt 1 1
(a) (b) 20 (c) (d) 21
+ C C 2. H3O+ 20 42
Et O Et
O2N
50. Among the following, the ion which will be
H more effective for flocculation of Fe(OH)3
solution is
C COOH
(a) PO 34 − (b) SO 24 −
(a) C
(c) SO 23 − (d) NO −3
O2N CH3
51. The mole fraction of ethanol in water is 0.08.
H
Its molality is
C COOH (a) 6.32 mol kg − 1 (b) 4.83 mol kg − 1
(b) C (c) 3.82 mol kg − 1 (d) 2.84 mol kg − 1
O2N
H 52. 5 mL of 0.1 M Pb(NO 3 )2 is mixed with 10 mL
CH3 of 0.02 M KI. The amount of PbI2 precipitated
will be about
(c)
C COOH (a) 10− 2 mol (b) 10− 4 mol
C (c) 2 × 10− 4 mol (d) 10− 3 mol
O2N H 53. At 273 K temperature and 76 cm Hg pressure,
CH3 the density of a gas is 1964
. g L− 1 . The gas is
C (a) CH4 (b) CO (c) He (d) CO 2
(d) H
C
54. Equal masses of ethane and hydrogen are
O2N COOH
mixed in an empty container at 298 K. The
fraction of total pressure exerted by hydrogen is
47. The maximum number of electrons in an (a) 15:16 (b) 1:1 (c) 1:4 (d) 1:6
atom in which the last electron filled has the
quantum numbers n = 3, l = 2 and m = − 1 is
55. An ideal gas expands adiabatically against
vacuum. Which of the following is correct for
(a) 17 (b) 27
the given process?
(c) 28 (d) 30
(a) ∆S = 0 (b) ∆T = − ve
48. In the face centred cubic lattice structure of (c) ∆U = 0 (d) ∆P = 0
gold the closest distance between gold atoms
. K kg mol − 1 . The temperature
56. K f (water) = 186
is (‘a’ being the edge length of the cubic unit
cell) at which ice begins to separate from a
a a mixture of 10 mass % ethylene glycol is
(a) a 2 (b) (c) (d) 2 2a
2 2 2 (a) − 186
. °C (b) − 372
. °C
(c) − 3.3° C (d) − 3° C
49. The equilibrium constant for the following
reactions are given at 25°C 57. The radius of the first Bohr orbit of a
hydrogen atom is 0 .53 × 10 − 8 cm. The velocity
2A - B + C , K 1 = 10
.
of the electron in the first Bohr orbit is
2B - C + D, K 2 = 16 (a) 2.188 × 108 cm s − 1 (b) 4.376 × 108 cm s − 1
2C + D - 2 P, K 3 = 25 (c) 1.094 × 108 cm s − 1 (d) 2.188 × 109 cm s − 1
8 WB JEE (Engineering) Solved Paper 2020
58. Which of the following statements is not true 66. To a solution of a colourless efflorescent
for the reaction, 2F2 + 2 H 2O → 4 HF + O 2 ? sodium salt, when dilute acid is added, a
(a) F2 is more strongly oxidising than O 2 colourless gas is evolved along with
(b) F F bond is weaker than O == O bond formation of a white precipitate. Acidified
(c) H F bond is stronger than H O bond dichromate solution turns green, when the
(d) F is less electronegative than O colourless gas is passed through it. The
sodium salt is
59. The number of unpaired electrons in the
(a) Na 2SO 3 (b) Na 2S
uranium (92 U) atom is
(c) Na 2S2O 3 (d) Na 2S4O 6
(a) 4 (b) 6 (c) 3 (d) 1
67. The reaction for obtaining the metal (M)
60. How and why does the density of liquid from its oxide ( M2O 3) ore is given by
water change on prolonged electrolysis?
Heat
(a) Decreases, as the proportion of H2O increases M2O 3(s) + 2 Al()l → Al 2O 3 () l + 2 M(s),
(b) Remains unchanged (s = solid, l = liquid) in that case, M is
(c) Increases, as the proportion of D2O increases (a) copper (b) calcium
(d) Increases, as the volume decreases (c) iron (d) zinc
61. The difference between orbital angular 68. In the extraction of Ca by electro reduction of
momentum of an electron in a 4 f -orbital and molten CaCl 2 some CaF2 is added to the
another electron in a 4 s-orbital is electrolyte for the following reason :
(a) 2 3 (b) 3 2 (a) To keep the electrolyte in liquid state at
(c) 3 (d) 2 temperature lower than the m.p. of CaCl 2
(b) To effect precipitation of Ca
62. Which of the following has the largest
(c) To effect the electrolysis at lower voltage
number of atoms?
(d) To increase the current efficiency
(a) 1 g of Ag (b) 1 g of Fe
(c) 1 g of Cl 2 (d) 1 g of Mg 69. The total number of alkyl bromides
(including stereoisomers) formed in the
63. Indicate the correct IUPAC name of the reaction Me3C CH == CH 2 + HBr → will be
coordination compound shown in the figrue. (a) 1 (b) 2
NH3 (c) 3 (d) No bromide forms
NH3
Cl Cr NH3 Cl 70. 1. Mg/diethyl ether
Cl Br Product
Cl 2. CH2O
NH3 3. H2O
+
dil.alkaline KMnO 4
Me Me
→ Product(s)
(c) HO H (d) HO H
H OH HO H
Me Me
Mathematics
Category-I (Q. Nos. 1 to 50) (b) cos − 1 (f( x))2 − (φ( x))2 + c
Carry 1 mark each and only one option is correct. f( x) φ( x) − 1
(c) 2 tan− 1 +c
In case of incorrect answer or any combination of 2
more than one answer, 1/4 mark will be deducted. f ( x) φ( x) + 1
(d) 2 tan− 1 +c
n 2
y x
1. Let cos − 1 = log . Then
b n 4. The value of
10 − 2n 10 2 n + 1
(a) x y2 + xy1 + n y = 0
2 2
[ a > 0 ] attains its maximum and minimum at 20. Let I (n) = nn, J (n) = 13
. .5 ........ (2 n − 1) for all
p and q respectively such that p 2 = q, then a is (n > 1), n ∈ N , then
equal to (a) I(n) > J(n) (b) I(n) < J(n)
1
(a) 2 (b)
1
(c)
1
(d) 3 (c) I(n) = J(n) (d) I(n) = J(n)
2 4 2
13. If a and b are arbitrary positive real numbers, 21. If c0 , c1 , c2, ....., c15 are the binomial
then the least possible value of
6 a 10 b
+ is coefficients in the expansion of (1 + x)15, then
5b 3a c c c c
6 the value of 1 + 2 2 + 3 3 + … + 15 15 is
(a) 4 (b) c0 c1 c2 c14
5
10 68 (a) 1240 (b) 120
(c) (d) (c) 124 (d) 140
3 15
12 WB JEE (Engineering) Solved Paper 2020
3 − t 1 0 29. Four persons A , B, C and D throw an unbiased
die, turn by turn, in succession till one gets
22. Let A = − 1 3 − t 1 and det A = 5, then
0 − 1 0
an even number and win the game. What is
the probability that A wins if A begins?
(a) t = 1 (b) t = 2 (c) t = − 1 (d) t = − 2 (a)
1
(b)
1
(c)
7
(d)
8
125 24 4 2 12 15
23. Let A = x 6 2 . The value of x for 30. A rifleman is firing at a distant target and
− 1 − 2 3 has only 10% chance of hitting it. The least
number of rounds he must fire to have more
which the matrix A is not invertible is
than 50% chance of hitting it at least once, is
(a) 6 (b) 12 (c) 3 (d) 2
(a) 5 (b) 7 (c) 9 (d) 11
a b
24. Let A = be a 2 × 2 real matrix with 31. cos(2 x + 7) = a(2 − sin x) can have a real
c d
solution for
det A = 1. If the equation det(A − λI 2) = 0 has
imaginary roots (I 2 be the identity matrix of (a) all real values of a (b) a ∈[2, 6]
(c) a ∈ (− ∞, 2 ) \ {0} (d) a ∈ (0, ∞ )
order 2), then
(a) (a + d )2 < 4 (b) (a + d )2 = 4 32. The differential equation of the family of
(c) (a + d ) > 4
2
(d) (a + d )2 = 16 curves y = e x (A cos x + Bsin x) where A , B are
arbitrary constants is
a2 c 2 + ac
bc
d2y d2y dy
25. If a 2 + ab b2 ca = ka 2 b2 c 2 , (a) 2
− 9 x = 13 (b) −2 + 2y = 0
dx dx2 dx
ab b + bc
2
c 2
d2y
2
(d) +
dy dy
(c) + 3y = 4 − xy = 0
then k = dx2 dx dx
(a) 2 (b) − 2 (c) − 4 (d) 4 π
33. The equation rcos θ − = 2 represents
26. If f : S → R, where S is the set of all 3
non-singular matrices of order 2 over R and (a) a circle (b) a parabola
a b (c) an ellipse (d) a straight line
f = ad − bc, then
c d 34. The locus of the centre of the circles which
(a) f is bijective mapping touch both the circles x 2 + y 2 = a 2 and
(b) f is one-one but not onto x 2 + y 2 = 4 ax externally is
(c) f is onto but not one-one
(d) f is neither one-one nor onto (a) a circle (b) a parabola
(c) an ellipse (d) a hyperbola
27. Let the relation ρ be defined on R by a ρ b holds
35. Let each of the equations x 2 + 2 xy + ay 2 = 0
if and only if a − b is zero or irrational, then
(a) ρ is equivalence relation and ax 2 + 2 xy + y 2 = 0 represent two straight
(b) ρ is reflexive and symmetric but is not transitive lines passing through the origin. If they have
(c) ρ is reflexive and transitive but is not symmetric a common line, then the other two lines are
(d) ρ is reflexive only given by
28. The unit vector in ZOX plane, making angles (a) x − y = 0, x − 3 y = 0 (b) x + 3 y = 0, 3 x + y = 0
(c) 3 x + y = 0, 3 x − y = 0 (d) (3 x − 2 y) = 0, x + y = 0
45° and 60° respectively with α = 2 $i + 2 $j − k$
and β = $j − k$ is 36. A straight line through the origin O meets the
1
parallel lines 4 x + 2 y = 9 and 2 x + y + 6 = 0
(a) $i + 1 $j (b)
1 $i − 1 k$ at P and Q respectively. The point O divides
2 2 2 2
1 the segment PQ in the ratio
(c) $i − 1 $j (d)
1 $i + 1 k$
2 2 2 2 (a) 1 : 2 (b) 3 : 4 (c) 2 : 1 (d) 4 : 3
WB JEE (Engineering) Solved Paper 2020 13
37. Area in the first quadrant between the (a) 8 x + 14 y + 13 z + 37 = 0
(b) 8 x − 14 y − 13 z − 37 = 0
ellipses x + 2 y = a and 2 x + y = a is
2 2 2 2 2 2
(c) 8 x − 14 y − 13 z + 37 = 0
a2 1 3a2 1 (d) 8 x − 14 y + 13 z + 37 = 0
(a) tan− 1 (b) tan− 1
2 2 4 2
5a2 − 1 1 9 πa2 44. The sine of the angle between the straight
(c) sin (d) x −2 y−3 z− 4
2 2 2 line = = and the plane
3 4 5
38. The equation of circle of radius 17 unit, 2 x − 2 y + z = 5 is
with centre on the positive side of X -axis and 2 3 2
through the point (0, 1) is (a) (b)
5 10
(a) x2 + y2 − 8 x − 1 = 0 4 5
(c) (d)
(b) x2 + y2 + 8 x − 1 = 0 5 2 6
(c) x2 + y2 − 9 y + 1 = 0
45. Let f (x) = sin x + cos ax be periodic function.
(d) 2 x2 + 2 y2 − 3 x + 2 y = 4
Then,
39. The length of the chord of the parabola (a) a is any real number
y 2 = 4 ax (a > 0) which passes through the (b) a is any irrational number
vertex and makes an acute angle α with the (c) a is rational number
axis of the parabola is (d) a = 0
(a) ± 4acotα cosec α (b) 4acotα cosec α 1
(c) − 4acotα cosec α (d) 4a cosec 2 α 46. The domain of f (x) = − (x + 1) is
x
40. A double ordinate PQ of the hyperbola (a) x > − 1 (b) (− 1, ∞ ) \ {0}
x2 y2 5 − 1 1 − 5
− = 1 is such that ∆OPQ is equilateral, (c) 0, (d) , 0
a 2 b2 2 2
O being the centre of the hyperbola. Then the
eccentricity e satisfies the relation 47. Let y = f (x) = 2 x 2 − 3 x + 2. The differential of
2 2 y when x changes from 2 to 1.99 is
(a) 1 < e < (b) e =
3 3 (a) 0.01 (b) 0.18 (c) − 0.05 (d) 0.07
3 2
(c) e = (d) e > 1/ x 1/ x
2 3 1 + cx 1 + 2 cx
48. If lim = 4, then lim is
x → 0 1 − cx x → 0 1 − 2 cx
41. If B and B′ are the ends of minor axis and
x 2 y2 (a) 2 (b) 4 (c) 16 (d) 64
S and S′ are the foci of the ellipse + = 1,
25 9 49. Let f : R → R be twice continuously
then the area of the rhombus SBS′ B′ will be
differentiable (or f ′ ′ exists and is continuous)
(a) 12 sq units (b) 48 sq units
such that f (0) = f ()
1 = f ′ (0) = 0 . Then
(c) 24 sq units (d) 36 sq units
(a) f ′ ′(c ) = 0 for some c ∈ R
42. The equation of the latusrectum of a parabola (b) there is no point for which f ′ ′( x) = 0
is x + y = 8 and the equation of the tangent (c) at all points f ′ ′( x) > 0
at the vertex is x + y = 12. Then, the length of (d) at all points f ′ ′( x) < 0
the latusrectum is
50. Let
(a) 4 2 units (b) 2 2 units
(c) 8 units (d) 8 2 units f (x) = x 13 + x 11 + x 9 + x 7 + x 5 + x 3 + x + 12.
Then
43. The equation of the plane through the point (a) f( x) has 13 non-zero real roots
(2 , − 1, − 3) and parallel to the lines (b) f( x) has exactly one real root
x −1 y + 2 z x y −1 z − 2
= = and = = is (c) f( x) has exactly one pair of imaginary roots
2 3 −4 2 −3 2 (d) f( x) has no real root
14 WB JEE (Engineering) Solved Paper 2020
Category-II (Q. Nos. 51-65) 57. Let A = { x ∈ R : − 1 ≤ x ≤ 1} and f : A → A be a
Carry 2 marks each and only one option is mapping defined by f (x) = x | x |. Then f is
correct. In case of incorrect answer or any (a) injective but not surjective
combination of more than one answer, 1/2 mark (b) surjective but not injective
will be deducted. (c) neither injective nor surjective
(d) bijective
51. The area of the region
{(x , y) : x 2 + y 2 ≤ 1 ≤ x + y} is 58. Let f (x) = x 2 − 3 x + 2 and g(x) = x be two
π2 π π 1 π2 given functions. If S be the domain of fog and
(a) (b) (c) − (d)
2 4 4 2 3 T be the domain of gof , then
π (a) S = T (b) S ∩ T = φ
52. In open interval 0 , , (c) S ∩ T is a singleton (d) S ∩ T is an interval
2
(a) cos x + xsin x < 1 59. Let ρ1 and ρ2 be two equivalence relations
(b) cos x + xsin x > 1 defined on a non-void set S. Then
(c) no specific order relation can be ascertained (a) both ρ1 ∩ ρ2 and ρ1 ∪ ρ2 are equivalence
between cos x + xsin x and 1 relations
1 (b) ρ1 ∩ ρ2 is equivalence relation but ρ1 ∪ ρ2 is not
(d) cos x + xsin x <
2 so
(c) ρ1 ∪ ρ2 is equivalence relation but ρ1 ∩ ρ2 is not
53. If the line y = x is a tangent to the parabola so
y = ax 2 + bx + c at the point (1, 1) and the (d) neither ρ1 ∩ ρ2 nor ρ1 ∪ ρ2 is equivalence relation.
curve passes through (− 1, 0), then
x2 y2
(a) a = b = − 1, c = 3
1
(b) a = b = , c = 0 60. Consider the curve +
= 1. The portion
2 a 2 b2
1 1 1 of the tangent at any point of the curve
(c) a = c = , b = (d) a = 0, b = c =
4 2 2 intercepted between the point of contact and
the directrix subtends at the corresponding
54. If the vectors α = $i + a$j + a 2k$ , β = $i + b$j + b2k$ focus an angle of
and γ = $i + c$j + c 2 k$ are three non-coplanar π π π π
(a) (b) (c) (d)
4 3 2 6
a a2 1 + a3
vectors and b b2 1 + b3 = 0 , then the 61. A line cuts the X -axis at A(7 , 0) and the
c c 2
1+ c 3 Y -axis at B(0 , − 5). A variable line PQ is drawn
perpendicular to AB cutting the X -axis at
value of abc is P(a , 0) and the Y -axis at Q(0 , b). If AQ and BP
(a) 1 (b) 0 (c) − 1 (d) 2 intersect at R, the locus of R is
(a) x2 + y2 + 7 x + 5 y = 0
55. Let z1 and z 2 be two imaginary roots of
(b) x2 + y2 + 7 x − 5 y = 0
z 2 + pz + q = 0 , where p and q are real. The
(c) x2 + y2 − 7 x + 5 y = 0
points z1 , z 2 and origin form an equilateral
(d) x2 + y2 − 7 x − 5 y = 0
triangle if
(a) p2 > 3q (b) p2 < 3q 1 /( k + α )
dx
(c) p = 3q
2
(d) p = q
2 62. Let 0 < α < β < 1. Then, lim
n→ ∞ ∫ 1 +x
is
1 /( k + β)
56. If P(x) = ax 2 + bx + c and Q(x) = − ax 2 + dx + c, (a) loge
β
(b) loge
1+ β
where ac ≠ 0 [a , b, c , d are all real], then α 1+ α
P(x) ⋅ Q(x) = 0 has 1+ α
(c) loge (d) ∞
(a) atleast two real roots (b) two real roots 1+ β
(c) four real roots (d) no real root
WB JEE (Engineering) Solved Paper 2020 15
1 1 (log 3 x ) 2 −
9
log 3 x + 5
63. lim − 68. The equation x 2 = 3 3 has
x → 1 lnx (x − 1)
(a) at least one real root
(a) Does not exist (b) 1
1 (b) exactly one real root
(c) (d) 0 (c) exactly one irrational root
2
(d) complex roots
1
64. Let y = , then 69. In a certain test, there are n questions. In this
1 + x + ln x test 2 n − i students gave wrong answers to at
dy
(a) x + y = x
dy
(b) x = y( yln x − 1) least i questions, where i = 1, 2 , …, n. If the
dx dx total number of wrong answers given is 2047,
2
then n is equal to
(d) x = y − x
dy dy
(c) x2 = y2 + 1 − x 2
dx dx (a) 10 (b) 11 (c) 12 (d) 13
65. Consider the curve y = be − x / a , where a and b 70. A and B are independent events. The
1
are non-zero real numbers. Then probability that both A and B occur is and
x y 20
(a) + = 1is tangent to the curve at (0, 0) the probability that neither of them occurs is
a b
x y 3
(b) + = 1is tangent to the curve, where the . The probability of occurrence of A is
a b 5
curve crosses the axis of y 1 1 1 1
(a) (b) (c) (d)
x y
(c) + = 1is tangent to the curve at (a, 0) 2 10 4 5
a b
x y 71. The equation of the straight line passing
(d) + = 1is tangent to the curve at (2 a, 0)
a b through the point (4, 3) and making intercepts
on the coordinate axes whose sum is − 1 is
x y x y
Category-III (Q. Nos. 66 to 75) (a) − =1 (b) + =1
2 3 −2 1
Carry 2 marks each and one or more option(s) x y x y
is/are correct. If all correct answers are not (c) − + = 1 (d) − = 1
3 2 1 2
marked and no incorrect answer is marked, then
score = 2 × number of correct answers marked ÷ x 2 y2
72. Consider a tangent to the ellipse + =1
actual number of correct answers. If any wrong 2 1
option is marked or if any combination including at any point. The locus of the mid-point of
a wrong option is marked, the answer will be the portion intercepted between the axes is
considered wrong, but there is no negative x2 y2 x2 y2
(a) + =1 (b) + =1
marking for the same and zero marks will be 2 4 4 2
1 1 1 1
awarded. (c) 2 + =1 (d) 2 + =1
3x 4 y2 2x 4 y2
66. The area of the figure bounded by the x2 d2y
parabola x = − 2 y 2 , x = 1 − 3 y 2 is 73. Let y = . Then, 2 is
(x + 1) (x + 2)
2
dx
1 4
(a) sq unit (b) sq unit 3 3 4
3 3 (a) 2 − +
(c) 1 sq unit (d) 2 sq units ( x + 1)4
( x + 1 )3
( x + 2 )3
2 4 5
67. A particle is projected vertically upwards. If it (b) 3 + −
has to stay above the ground for 12 sec, then ( x + 1)3
( x + 1) 2
( x + 2 )3
6 4 3
(a) velocity of projection is 192 ft/sec (c) − +
(b) greatest height attained is 600 ft ( x + 1)3 ( x + 1)2 ( x + 1)3
(c) velocity of projection is 196 ft/sec 7 3 2
(d) − +
(d) greatest height attained is 576 ft ( x + 1)3 ( x + 1)2 ( x + 1)3
16 WB JEE (Engineering) Solved Paper 2020
1 (a) the differential equation of the curve is
74. Let f (x) = x sin x − (1 − cos x). The smallest dy
3 3x + y = 0
f (x) dx
positive integer k such that lim k ≠ 0 is (b) the differential equation of the curve is
x→ 0 x dy
3x − y = 0
(a) 4 (b) 3 (c) 2 (d) 1 dx
(c) the curve passes through , 2
1
75. Tangent is drawn at any point P(x , y) on a 8
curve, which passes through (1, 1). The (d) the normal at (1, 1) is x + 3 y = 4
tangent cuts X -axis and Y -axis at A and B
respectively. If AP : BP = 3 :1, then
Answers
Physics
1. (b) 2. (d) 3. (b) 4. (a) 5. (a) 6. (d) 7. (a) 8. (d) 9. (b) 10. (a)
11. (a) 12. (a) 13. (c) 14. (b) 15. (c) 16. (a) 17. (b) 18. (d) 19. (a) 20. (c)
21. (b) 22. (a) 23. (d) 24. (a) 25. (d) 26. (c) 27. (c) 28. (b) 29. (a) 30. (a)
31. (c) 32. (b) 33. (a) 34. (c) 35. (b) 36. (b) 37. (d) 38. (a) 39. (d) 40. (a,c,d)
Chemistry
41. (c) 42. (c) 43. (b) 44. (d) 45. (c) 46. (a) 47. (d) 48. (b) 49. (a) 50. (a)
51. (b) 52. (b) 53. (d) 54. (a) 55. (c) 56. (c) 57. (a) 58. (d) 59. (a) 60. (c)
61. (a) 62. (d) 63. (d) 64. (b) 65. (b) 66. (c) 67. (c) 68. (a) 69. (c) 70. (d)
71. (d) 72. (b) 73. (d) 74. (c) 75. (d) 76. (a, b) 77. (a, b,d) 78. (a, b, d) 79. (a, c, d) 80. (a, c)
Mathematics
1. (a) 2. (a) 3. (c) 4. (d) 5. (b) 6. (d) 7. (c) 8. (c) 9. (c) 10. (b)
11. (c) 12. (a) 13. (a) 14. (a) 15. (c) 16. (b) 17. (c) 18. (d) 19. (b) 20. (a)
21. (b) 22. (d) 23. (c) 24. (a) 25. (d) 26. (d) 27. (b) 28. (b) 29. (d) 30. (b)
31. (c) 32. (b) 33. (d) 34. (d) 35. (b) 36. (b) 37. (a) 38. (a) 39. (b) 40. (d)
41. (c) 42. (d) 43. (*) 44. (b) 45. (c) 46. (c) 47. (c) 48. (c) 49. (a) 50. (b)
51. (c) 52. (b) 53. (c) 54. (c) 55. (c) 56. (a) 57. (d) 58. (d) 59. (b) 60. (c)
61. (c) 62. (b) 63. (c) 64. (b) 65. (b) 66. (b) 67. (a, d) 68. (a, c) 69. (b) 70. (c, d)
71. (a, b) 72. (d) 73. (a) 74. (c) 75. (a, c)
Z X A
→ Z + 1 Y A
+ −1 e
0
+ ν+ Q RC=500 W
IC
∴ Mass defect, mx = M x − Zme Y
and m y = M y − (Z + 1)me
where, M x is atomic mass of X, M y is atomic mass lB=200 mA
VCE
of Y and me is mass of an electron.
Total mass defect, ∆m = mx − m y = (M x − M y − me)
∴Binding energy, E = ∆mc 2
IC
So, maximum energy of β-particle emitted We know that, current gain, β =
IB
= (M x − M y − me) c 2
IC
⇒ 48 =
5. (a) Given, binding energy per nucleon of nuclei 200 × 10−6
with mass number 119 = 7.6 MeV
⇒ I C = 48 × 200 × 10−6
Binding energy per nucleon of nuclei with mass
number 238 = 8.6 MeV = 96 × 10−4 A
According to question, Now, VCC = I C RC + VCE
fission of nucleus 238 =119 + 119 ⇒ VCE = VCC − I C RC
Hence, total energy of nucleus before fission, From figure, VCC = 5 V and RC = 500 Ω
E i = 238 × 8.6 MeV ⇒ VCE = 5 − 96 × 10−4 × 500
Total energy of nucleus after fission,
= 5 − 4.8 = 0.2 V
E f = 238 × 7.6 MeV
∴Energy released in the process of fission 8. (d) Given, frequency, ν = kδE
= Ei − E f where, k is constant and δE is change in energy.
= 238 × 8.6 − 238 × 7.6 ν ν
⇒ k= = [QδE = hν]
= 238 MeV δE hν
Hence, released energy is closest to 214 MeV. 1
=
h
6. (d) Given, load resistance, RL = 6 kΩ = 6000 Ω
QWe know that, energy, E = hν
Base voltage, VB = 15 mV = 15 × 10−3 V E ML2 T −2
⇒ h= =
Base current, I B = 20µA = 20 × 10−6 A ν T −1
2 −1
Collector current, I C = 1.8 mA = 1.8 × 10−3 A h = [ML T ]
∴Voltage gain = β (Current gain) × Resistance 1 1
∴Dimension of k = = = [M−1 L−2 T]
gain h [ML2 T −1 ]
I R
= C × L 9. (b) Given, A = $i + $j − k$ , B = 2i$ − $j + k$ ,
I B RB
1 $ $ $)
I R I R C = (i − 2 j + 2k
= C L = C L [QVB = I B RB ] 5
I B RB VB
$i $j k $
1.8 × 10−3 × 6000
= Q A×B=1 1 $
−1 = − 3$j − 3k
15 × 10−3
2 −1 1
= 720
1 $ $ ) (− 3$j − 3k
$)
7. (a) Given, β of transistor = 48 ∴ C ⋅ (A × B) = (i − 2$j + 2k
5
Base current, I B = 200 µA = 200 × 10−6 A 6 6
= − =0
5 5
WB JEE (Engineering) Solved Paper 2020 19
µ mg µ mg
10. (a) Given, speed of fighter plane, v = 360 km/h = =
1 + µ2 1 + µ2
5
= 360 × m/s = 100 m/s
18 1 + u2
Altitude, h = 500 m µ ⋅ mg
⇒ Fmin =
v=100m/s 1 + µ2
12. (a) Given, a tennis ball hits the floor with a speed
h=500 m v making an angle θ with the normal as shown in
the figure below.
ev cos q
ev
Now, from equation of motion,
1 v q f
h = ut + gt 2
2
v sin q O e
1 Q u = 0 and
⇒ 500 = 0()t + × 10 × t 2 g =10 m/s2 f
2
⇒ t 2 =100 ⇒ t = 10 s ev cos q
A
∴The bomb should be dropped at the distance, v sin q B
x = vt = 100 × 10 = 1000 m The coefficient of restitution = ε
11. (a) The block diagram is as shown below, Now, from the above diagram,
vsinθ tanθ
F sin q In ∆AOB, tan φ = =
εv cosθ ε
tanθ
φ = tan−1
N F
∴Angle of reflection,
f= mN ε
q
F cos q
13. (c) Time period of second pendulum,T = 2 s
Displacement equation of simple pendulum,
mg x = Asinωt … (i)
Given, t = 2. 25 s = 2 + 0.25
According to above diagram, 1
= 2+
mg = N + F sinθ 4
⇒ N = mg − F sinθ … (i) ∴ t=T+
T
and F cosθ = µ N 8
⇒ F cosθ = µ (mg − F sinθ) [using Eq. (i)] Velocity of second pendulum,
µmg v=
dx
⇒ F=
cosθ + µ sinθ dt
d
For Fmin , = ⋅ (Asinωt) [from Eq. (i)]
dt
d
(cosθ + µ sinθ) = 0 v = Aω cosωt
dθ
Velocity at t = T,
⇒ − sinθ + µ cosθ = 0
2π
⇒ tanθ = µ v = A ω cos ⋅ T = Aω = v0 … (ii)
T
µ 1
Q sinθ = and cosθ = T
1 + µ2 1 + µ2 Velocity at t = T + ,
8
µmg
v1 = Aω cosω T +
∴ Fmin = T
1 µ ⋅µ 8
+
1+µ 2
1 + µ2 2π T
= Aω cos T +
T 8
20 WB JEE (Engineering) Solved Paper 2020
9 From given figure,
= Aω cos 2π ⋅
8 FS × x = FB × (15 − x)
π
= Aω cos 2π + ⇒ 2 × 107 × x = 107 × (15 − x)
4
⇒ 2x = 15 − x
π
= Aω cos ⇒ 3x = 15 ⇒ x = 5 cm
4
Aω 15. (c) We know that,
=
2 2r 2
terminal velocity, vT = (ρS − ρL) g
v1 =
v0
[from Eq. (ii)] 9η
2 1
⇒ vT ∝ , where η is viscosity of the liquid.
By equation of motion, η
v12 = v02 − 2gh Hence, graph between terminal velocity vT and
2 2 viscosity of liquid η is as shown below.
⇒ v0 = v2 − 2gh ⇒ h = v0
0
2 4g vT
a
a a
2
x 5− x
⇒ =
M / 32 M / 18
⇒ 32x = 18 (5 − x)
q a q
⇒ 32x = 90 − 18 x
1 q1 q 2 k q1 q 2
⇒ 50 x = 90 QPotential energy, PE = =
90 4 πε0 r r
⇒ x= = 1.8 m
50 Here, q1 = q 2 = q
and r=a
18. (d) Given, mass of boiling water = 100 g k q q kq 2
⇒ PE = =
Mass of cold water = 300 g a a
Fall in temperature = 100 − 20 = 80°C Due to four identical point masses,
Let heat capacity of calorimeter = ms 4kq 2 2kq 2
PE i = +
Specific heat of water = 1 cal g −1 °C −1 a 2a
∴Heat lost by boiling water = ms∆T = 100 × 1 × 80 According to law of conservation of energy,
and heat gained by cold water KE i + PE i = KE f + PE f
= 300 × 1 × 10 + ms × 10 Initially, KE i = 0
Now, according to the principle of calorimetry, and at infinity, PE f = 0
heat lost = heat gained 4kq 2 2kq 2
∴ KE f = PE i = +
100 × 1 × 80 = 300 × 1 × 10 + ms × 10 a 2a
⇒ ms = 500 cal/°C kq 2
= (4 + 2)
Let specific heat of metal is s b. a
Mass of block = 1 kg =1000 g QIn CGS unit, k =1
Rise in temperature of metallic block, q2
∆t = 19 − 10 = 9°C ∴ KE = (4 + 2)
a
Then, again from principle of calorimetry,
mmixture × s∆T + ms∆T = mblock × s b × ∆t 21. (b) Given, radius of solid cylinder = a
(100 + 300) × 1 × (20 − 19) + 500 × (20 − 19) Charge density of cylinder = ρ
= 1000 × s b × 9 Dielectric constant of material of cylinder = k
⇒ s b = 01
. cal/g°C The electric field at a radial distance x(x < a) as
shown is calculated as
19. (a) Given, q1 = 1 × 10−6 C, q 2 = 3 × 10−6 C
∴ q 2 = 3q1 ; q 2 > q1
1 a
As we know that, electric field, E ∝
x2
So, the graph showing variation of E x in
x
x-direction is as given below. l
Ex
x
O (10,0)
22 WB JEE (Engineering) Solved Paper 2020
From Gauss’ law, 23. (d) According to given figure,
Q
∫ E ⋅ dA = k εin0 , where k is dielectric constant of B
material.
I
ρ(πx 2l)
⇒ E(2πxl) = [Qcharge, Qin = ρ(πx l )]
2
O r
k ε0 G I
a
ρx
⇒ E=
2k ε0
a
22. (a) A galvanometer can be converted to a
voltmeter of full scale deflection V0 by connecting Given, a : r = 8 : π
a 8 πa
a series resistance R1 , then = ⇒r = … (i)
V r π 8
R1 = 0 − G µ 0I
Ig Magnetic field due to square, B1 = × 2,
a
π
⇒ V0 = I g (G + R1) … (i) 2
where, I g = current through the galvanometer outward
µ 0I
and G = resistance of galvanometer. Magnetic field due to circular loop, B2 = ,
2r
Similarly, a galvanometer can be converted to an
ammeter by connecting a shunt resistance R2, inward
then Strength of magnetic field at the common centre
Ig O,
R2 = G µ I 2 2µ 0 I
I − I B = B1 + B2 = 0 −
0 g
2r πa
R2 Substituting value of r from Eq (i), we get
⇒ I g = I0 … (ii)
G + R2 µ I × 8 2 2 µ 0I
B= 0 −
where, I 0 = total current. 2πa πa
Now, from Eq. (i), we get µ 0I
= (4 − 2 2)
V0 πa
− R1 = G … (iii) µ I
Ig = 0 2 2 ( 2 − 1)
πa
From Eq. (ii), we get
I0 24. (a) A wire is bent to form a D-shaped loop
G + R2 = × R2 carrying current I, where the curved part is
Ig
semi-circle of radius R. The loop is placed in a
I0 uniform magnetic field B which is directed into
⇒ G= × R2 − R2 … (iv)
Ig the plane of the paper. As single current by flowing
in the loop, so net magnetic force on a closed
From Eqs. (iii) and (iv), we get current loop in a uniform magnetic field B is zero.
V0 I
− R1 = 0 × R2 − R2 25. (d) For equivalent resistance, between the
Ig Ig terminals A and B of the infinite resistance
V0 I 0 network, we redraw the given circuit as
⇒ − × R2 = R1 − R2
Ig Ig 2R In parallel combination
1 1 1 = X+R
⇒ (V0 − I 0 R2) = R1 − R2 +
R X RX
Ig X R X
V − I 0 R2
⇒ Ig = 0
R1 − R2
RX
Now the 2R and is in series, so
X+ R
WB JEE (Engineering) Solved Paper 2020 23
RX QIn steady state, the combination of resistance
X = 2R +
R+ X and capacitance in loop CEFD will not work. So,
we can eliminate this combination and hence the
X 2 − 2RX − 2R2 = 0
equivalent circuit is as shown below.
2R ± 4R2 − 41
() (− 2R2) 2W i
⇒ X= i
2 B C
2R ± 12R2 2R ± 2 3R
⇒ X= = = (1 ± 3)R
2 2 4W 8W
⇒ X = (1 + 3) R
2V
26. (c) When a DC voltage is applied at the two ends A D
of a circuit kept in a closed box, it is observed that 2V
the current gradually increases from zero to a QIn the above circuit, the current i will flow from
certain value and then remains constant. As B to C and same current i will flow from C to B.
capacitor blocks DC. So, the circuit must contains Hence, total current through the 2Ω resistor is zero.
a resistor and an inductor in series as shown
below. 28. (b) A conducting wire of
length L bent in the form
of a circle of radius R and
S
another conductor of R
a\4
length a is bent in the form
of a square. The two loops
R L are then placed in same
From Graph plane such that the square
loop is exactly at the centre
Growth of current, I = I 0 (1 − e − t / τ L )
of circular loop as shown in the figure
At t = 0, I = 0 For circular loop,
At t = τ L, 2πR = L … (i)
I = I 0 1 − = 0.693 I 0
1 For square of side S,
e 4S = a … (ii)
I µ 0 IS 2
Flux linked, φ =
2R
I0 φ µ S2
Mutual inductance, M = = 0
I 2R
0.693 I0
Substituting R from Eq. (i) and S from Eq. (ii), we get
t µ a 2 2π µ 0 a 2 π
M= 0 =
L 216
( ) L 16 L
where, τ L = = time constant and
R
E 29. (a) The given situation is as shown in figure below.
I0 = = maximum current.
R
16 cm
y-component of velocity, v y = t = ⋅ ×
qE e V L
80 cm m m d v0
64 cm Object
eVL
⇒ vy =
mdv0
Given, object distance, u = − 80 cm v eLV eVL
∴ tanθ = y = =
Image distance, v = + 20 cm vz mdv0 . v0 mdv02
Using lens formula, we get eVL
1 1 1 ⇒ θ = tan−1
= − 2
mdv0
f v u
1 1 1 5 33. (a) Due to dragging movement of the block, work
⇒ = + =
f 20 80 80 (W f ) will be done against force of friction ( f).
⇒ f = 16 cm Now, force of friction, f = µmg
Now, water is poured in the vessel upto height of where, µ = coefficient of static friction = 0.1,
64 cm, then object distance, mass, m = 20 kg and g = 10 ms−2.
64 64 × 3 ⇒ f = 0.1 × 20 × 10 = 20 N
u = 16 + = 16 + = 64 cm
µ 4 This work done (W f ) will produce heat energy,
(given, refractive index, µ = 4 / 3) H = mc∆T
Again, using lens formula, Hence, W f = H
1 1 1
= − ⇒ Force × Displacement = mc∆T
f v u ⇒ f × vt = mc∆T [Qs = v ⋅ t]
1 1 1 Here, v = 0.5 ms−1 , t = 2.1 s
⇒ = + (Qu = −64 cm)
16 v 64 . CGS unit = 0.1 × 4.2 × 103 SI unit
and c = 01
1 3
⇒ = ⇒ 20 × 0.5 × 2.1 = 20 × 0.1 × 4.2 × 103 × ∆T
v 64
20 × 0.5 × 2.1
∴ v = 21.33 cm, above the lens. ⇒ ∆T = = 0.0025° C
20 × 0.1 × 4.2 × 103
WB JEE (Engineering) Solved Paper 2020 25
2 tan θ
34. (c) We have, power = work done As, tan 2θ =
time 1 + tan2 θ
Given, time = 1 min = 60 s and sin 2θ = 2 sin θ cos θ
Now, work done per cycle = (130 − 30) 2 9sin2 θ 9
= 100 cal = 100 × 4.2 J ⇒ = =
1 − tan θ 4sin θ cos θ 4 cos2 θ
2 2 2
q T sin q O 3
1
⇒ tanθ = = tan 30°
L sin q L sin q 3
mg ⇒ θ = 30°
Now, T sinθ = Fe …(i) Hence, the angle between the strings
T cosθ = mg …(ii) = 2θ = 2 × 30° = 60°
On dividing Eq. (i) by Eq. (ii), we get 36. (b) The photoelectric current depends on the
T sinθ F q ×q intensity of incident radiation by relation,
= e =k 1 2 2
T cosθ mg r i ∝I
Here, q1 = q 2 = q But the intensity of radiation is inversely
Distance, r = 2L sinθ proportional to the square of distance between the
kq 2 source and the photoelectric surface,
⇒ tanθ = … (iii) 1
(2L sinθ)2 ⋅ mg i.e. I∝ 2
d
Now, in the second case, when the charges are 1
tripled, new charge q ′ = 3q ∴ i∝ 2
d
As the angle between the strings are doubled, θ So, when the distance(d) is doubled, the
becomes 2θ, hence from Eq. (iii), we get photoelectric current will decrease. The stopping
k(3q)(3q)
tan 2θ = potential and maximum kinetic energy are
(2L sin 2θ)2 ⋅ mg independent of distance, so they remain same.
9kq 2 37. (d) As, inner radii are not given, so we cannot
tan 2θ = … (iv)
(2L sin 2θ)2 mg calculate their masses. Also, it is not given about
On dividing Eq. (iv) by Eq. (iii), we get the nature of material of spheres.
tan 2θ 9kq 2 4L2 sin2 θ ⋅ mg The rotational kinetic energy of sphere,
= 2 2 × 1
tanθ 4L sin 2θ ⋅ mg kq 2 (KE)rot = Iω2
2
tan 2θ 9sin2 θ where, I = moment of inertia
=
tanθ sin2 2θ and ω = angular speed.
26 WB JEE (Engineering) Solved Paper 2020
It is given that, the two metallic spheres have As the motions of pendulum and bar are
same moment of inertia (I) about their respective synchronous, so their angular velocities will be
diameters. So, their rotational kinetic energies will equal, i.e.
be equal when rotated with equal uniform ω1 = ω2
angular speed (ω) about their respective diameters. 2gsinθ 3gsinθ
⇒ =
38. (a) The velocity of the particle at extreme l l′
position, 2 3 3
⇒ = ⇒ l′ = l
v = 2gh [from v 2 − u 2 = 2gh] l l′ 2
3
Let when released from extreme position, the ∴The length of the bar is l.
pendulum traverse an angular distance of θ as 2
shown below. 39. (d) Given, R = 400 Ω, L = 250 mH = 250 × 10−3 H,
String C = 2.5µF = 2.5 × 10−6 F, VS = 5 V
I
and ω = 2 kHz = 2000 rad/s = 2 × 103 rad/s
q
∴ Inductive reactance,
w1 I sin q X L = ωL = 2 × 103 × 250 × 10−3 = 500 Ω
Capacitive reactance,
1 1
XC = = = 200 Ω
v ωC 2 × 103 × 2.5 × 10−6
Here, h = l sin θ Impedance, Z = R2 + (X L − X C)2
Then, linear velocity, v = 2gl sinθ
= (400)2 + (500 − 200)2
v 2gl sinθ
and angular velocity, ω1 = = = 250000 = 500 Ω
l l
2gsinθ The peak voltage across capacitor,
= V 5
l (VP)C = I P X C = S X C = × 200 = 2 V
For a bar moved through the same angular Z 500
distance θ as shown below. ∴Peak value of electrostatic energy of capacitor,
1
Bar (U P) C = C(VP)C2
I¢ 2
1
q = × 2.5 × 10−6 × (2)2
I¢/2 sin q 2
w2 = 5 × 10−6 J
or (U P)C = 5µJ
l′ 40. (a, c, d)
Here, distance travelled, h = sinθ, where l′ is the
2 It is given that the charged particle is moving with
length of bar. constant velocity in a region of no gravity, so
Potential energy = Rotational kinetic energy following cases are possible
1
mgh = Iω22 (i) The particle may move in a straight line in any
2 direction, when E = 0 and B = 0.
l′ 1 m(l ′)2 ml 2 (ii) The particle may move in a circle with constant
⇒ mg sinθ = × ω22 Q for rod, I =
2 2 3 3 velocity. The centripetal force for circular
motion is provided by magnetic field.
(l ′)2 2
⇒ gl ′sinθ = ω2 So, E = 0 and B ≠ 0.
3
(iii) The particle may move in a helical path with
3gl ′sinθ 3gsinθ
⇒ ω2 = = constant velocity. For helical motion, the
(l ′)2 l′ condition is that both fields must be present,
i.e. E ≠ 0 and B ≠ 0.
WB JEE (Engineering) Solved Paper 2020 27
Chemistry
41. (c) Alkaline hydrolysis of an ester (carboxylic acid acids are more acidic than phenol due to more
derivative) follows acyl SN 2 mechanism. It is as stability of carboxylate ion than phenoxide ion.
follows: So, compound IV is the most acidic among the
given compounds.
O O
Further, acidicity of phenols increases by the
–
OH presence of electron withdrawing groups (EWG)
C C + RO–
OR OH such as NO 2. The effect of EWG is more
pronounced at ortho- and para-positions than at
E
COOH OH OH OH
Electron withdrawing group (− R > − I) increases
the rate of SN 2 as it promotes the nucleophilic
attack whereas electron donating group (+ R > + I) > > >
decreases the rate of SN 2 reaction as it demotes NO2
the nucleophilic attack.
OCH3 NO2 CH3
Here, the nature and order of functional groups
attached para to benzene ring are: (IV) (I) (III) (II)
NO 2(I) > Me(III) > OMe(II)
( − R) ( + I) ( + R) 44. (d) Carbocations are stabilised by the presence of
electron donating group (+ R, + I , + H) and
So, the order of rate of alkaline hydrolysis is destabilised by presence of electron withdrawing
I > III > II group(− R, − H , − I).
O
42. (c) Given reaction proceed via shift of D − .
+
It is shown as follows : I. CH 2 C CH 3
( − I)
O
O O– O
OH – Ph C D
Ph C q Ph C OH − I-effect of C CH 3 destabilises the
r.d.s.
carbocation. So, it is least stable carbocation.
D D
II. + +
–
O O [CH2—O CH3 CH2==OCH 3]
43. (b) Among the given compounds, compound IV is 45. (c) NaBH 4 reduces aldehydes, ketones and acid
a carboxylic acid whereas rest are phenols. Carboxylic chlorides into alcohols. It cannot reduce ester and
28 WB JEE (Engineering) Solved Paper 2020
amides. Therefore, in the given reaction, NaBH 4 Since, five orientations of d-orbitals are
reduces only the keto group without affecting the degenerated, therefore m = − 1 can be assigned to
ester group. any one of them which means all the 3d-orbitals
The reaction takes place as: are filled. Therefore, the expected maximum
number of electrons in an atom of the given
O O O
OH quantum numbers are 30.
NaBH4
C So, the correct option is (d).
C C CH
H 3C OC2H5 Methanol
H 3C OC2H5
48. (b) The given crystal lattice is face centred cubic
Ethyl-3- oxobutanoate Ethyl-3- hydroxybutanoate
lattice, which has Zeff = 4.
a
Thus, correct option is (c). Also, for fcc, atomic radius (r) =.
2 2
46. (a) The given reaction proceeds as follows: The arrangement of Au-atom in a fcc lattice is
O O – shown below:
O O
C C C C
CH3 –
OOCEt CH3
Et O CH Et O CH
r
Nucleophile
H
– r
O O O
O2N C C C
2r
CH O Et r
H
CH3
86 3 1
: [Rn] 5 f 6d 7s . 2
Cl NH3
92 U –
Cr Cl
It consist of your electrons, i.e. three from f-orbital Cl NH3
and one from d-orbital.
NH3
So, the correct answer is option (a).
Let the oxidation state of central atom (Cr) be x.
60. (c) On prolonged electrolysis, the density of liquid
Then,
water increases as the proportion of D 2O increases.
(x × 1) + (4 × 0) + (2 × − 1) = + 1
This is because on prolonged electrolysis, the
concentration of heavy water (D 2O) in liquid water x − 2= + 1
increases which has higher density than ordinary x=+ 3
water (H 2O). If electrolysis continued, then almost Thus, the correct IUPAC name of complex is
pure D 2O is obtained. cis-tetraamminedichlorochromium (III) chloride.
61. (a) Orbital angular momentum = l(l + 1) h . So, the correct answer is (d).
2π
64. (b) We know,
For 4 f-orbital, l = 3,
mass of N A atoms = 12 g of C
h
∴ Orbital angular momentum = 3(3 + 1) or mass of 6.023 × 1023 atoms = 12 g
2π
12
=2 3
h ∴Mass of 1 atom = g = 1.9923 × 10− 23g
2π 6.023 × 1023
For 4s-orbital, l=0 So, the correct option is (b).
∴ Orbital angular momentum = 0.
WB JEE (Engineering) Solved Paper 2020 31
65. (b) We know that, 69. (c) The given reaction proceeds as follows:
Bond order CH3
¾
1
= [number of electrons in bonding orbitals – H + +
2 Me3C—CH==CH2 CH3—C—CH—CH3
¾
number of electrons in antibonding orbitals] (2°carbocation)
HBr CH3
The electronic configuration and bond order less stable
(B.O.) of given options is as follows: CH3 Methyl shift
¾
* CH3 H
H3C—C—–CH—CH3
2− 2
¾
¾
¾
B.O = =0 CH3—C — C—CH3
2 CH3 Br +
¾
(ii) He+2 (3 e −) = σ(1s)2 σ * (1s)1 (Minor product) CH3
( Exists in two (3°carbcation)
2−1 1
B.O = = stereoisomers R and S) more stable
2 2
+Br
(iii) He22 + (2 e −) = σ(1s)2
CH3 H
2− 0
B.O = =1
¾
2 CH3—C — C—CH3
Therefore, the bond order of He2, He+2 and He22 + are
¾
1 Br CH3
respectively 0, , 1.
2 (Major product)
66. (c) Colourless, efflorescent sodium salt is Na2S2O 3. Therefore, the total number of alkyl bromides
The colourless gas evolved in SO 2. Complete (including stereoisomers) formed in the reaction
reaction is as follows : are 3. Thus, the correct option is (c).
Na 2S2O 3(s) + 2H + (dil.) → 70. (d) The given reaction can be completed as
(Colourless, efflorescent) follows:
S(s) + 2Na+(aq) + SO 2 ↑ + H 2O()
l Mg
White ppt. Cl Br ClMg MgBr
ether
Now, when the colourless gas is passed through it
acidified dichromate solution turns green.
The reaction taking place is (2eq.) HCHO
73. (d) Given, 75. (d) For the given reaction statement (d) is correct
K sp (SrCO 3) = 7.0 × 10− 10 and applicable. For this reaction, a mixed SN1 and
SN 2 pathway is followed. The given reaction can
− 10
K sp (SrF2) = 7.9 × 10 proceed via SN1 pathway, So as,
[CO 23 − ] = 1.2 × 10− 3 M CH 3 O CH 2 Cl →
+
Now, for the reaction, [CH3—O —CH2 CH3—O ==CH2]
2+
SrCO 3 - Sr + CO 23 −
Here, the carbocation is stabilised by resonance.
K sp = [Sr 2 + ][CO 23 − ] The reaction can also be proceeded via SN 2 path,
since the transition state is resonance stabilised.
7 × 10− 10 = [Sr 2 + ] × 1.2 × 10− 3
d+ –
7 × 10− 10 OH
∴ [Sr 2 + ] = CH3 — O— CH2—Cl CH3 —O—CH2 —OH
1.2 × 10− 3
Also, SrF2 - Sr 2 + + 2F − So, the correct option is (d).
2+ − 2
K sp = [Sr ] [F ] 76. (a, b) Both reactions (a) and (b) give a meso-
K sp compound as the main product the reactions are as
[F − ]2 = 2+
follows :
[Sr ] Br
K sp H
[F − ] = 2+ (a)
Br2
[Sr ] CH2Cl2
H
− 10 Br
7.9 × 10
= Meso-compound
7 × 10− 10
1.2 × 10− 3 H H
− −2 H2
∴ [F ] = 3.7 × 10 M (b) H3C CH3
Pd-C
So, correct option is (d).
Meso-compound
WB JEE (Engineering) Solved Paper 2020 33
H H 79. (a, c, d) SiO 2 is attacked by HF, hot NaOH and
(c) CººC C==C fluorine. Reactions involved are as follows :
H2
Lindlar’s catalyst (a) SiO 2(s) + 6HF(aq) → H 2SiF6(aq) + 2H 2O()
l
Br (c) SiO 2(s) + (hot) 2NaOH(aq) →
H Na 2SiO 3(s) + H 2O()
l
(d) Br2
H (d) SiO 2(s) + F2(g) → SiF4 (aq) + O 2(g)
CCl4
H Br So, the correct options are (a) (c) and (d).
H
80. (a, c) Complete reaction is as follows :
So, options (a) and (b) are correct. Na / NH (liq.)
Me C ≡≡ C Me
3
→
77. (a, b, d) For spontaneous polymerisation, EtOH, − 33° C
∆G = negative Me H
Polymerisation process involves the following C==C
reaction: H (x) Me
nA → — ( A− A− A− A— )n
So, ∆S = negative due to association of atoms. Dill. KMnO4
Mathematics
1. (a) We have, d2y dy
⇒ x2 + x = − n2 y
n dx 2 dx
cos− 1 = log
y x
b n d2y dy
⇒ x2 2 + x + n2 y = 0
dx dx
⇒ y = b cos(n log x − n log n)
On differentiating both sides w.r.t.x, we get 2. (a) We have,
dy n φ(x) = f (x) + f (1 − x) and f ′′(x) < 0 in [0, 1]
= − b sin(n log x − n log n) ×
dx x φ′(x) = f ′(x) − f ′(1 − x)
dy For monotonic increasing, f ′(x) − f ′(1 − x) ≥ 0
⇒ x = − nb sin(n log x − n log n)
dx ⇒ f ′(x) ≥ f ′(1 − x)
Again differentiating both sides w.r.t. x, we get Since, f ′(x) is decreasing.
d 2 y dy n2b cos(n log x − n log n) 1
x 2 + =− ∴x ≤ 1 − x ⇒ x ≤
dx dx x 2
34 WB JEE (Engineering) Solved Paper 2020
⇒ φ(x) is monotonic increasing in 0, and ⇒ I=0+ 2 −1 + 2 3 − 2 2 + 6 − 3 3
1
2
⇒ I = 5− 2− 3
monotonic decreasing in , 1
1
2 6. (d) Given, curve
y2 = x 3
3. (c) Let
f (x)φ′(x) + φ(x) f ′(x) On differentiating both sides w.r.t. x, we get
I= ∫ ( f (x)φ(x) + 1) f (x)φ(x) − 1
dx
2y
dy
= 3x 2
dx
Put f (x)φ(x) − 1 = t 2 dy 3x 2
⇒ =
⇒( f (x)φ′(x) + φ(x) f ′(x))dx = 2t dt dx 2y
2tdt
∴ I=∫ 2 dy 3(m4) 3
(t + 2)t ⇒ = = m
dx ( m 2 , m 3) 2m3 2
dt
⇒ I = 2∫ 2 3
t + 2 ∴Slope of tangent at (m2 , m3) is m.
2
tan− 1 + c
2 t Now slope of normal at (M 2 , M 3) to the curve
=
2 2
y 2 = x 3 is
f (x)φ(x) − 1
⇒ I= 2 tan− 1 + c −
dx − 2y
=
2 dy 3x 2
4. (d) We have, dx − 2M 3 − 2
− 2n
⇒ − = =
10 10 2n + 1 dy ( M 2 , M 3) 3M 4 3M
I= ∑ ∫ sin x dx + ∑ ∫ sin
27 27
x dx
n =1 − 2n − 1 n =1 2n Given, slope of tangent = slope of normal
3m − 2
10 − 2n 2n + 1 ∴ =
I = ∑ ∫ sin27 x dx + ∫ sin 27
x dx 2 3M
n =1
− 2n − 1
−4
2n ⇒ mM =
− 2n 9
I1 = ∫ sin
27
Let x dx 7. (c) Given, x 2 + y2 = a 2
− 2n − 1
On differentiating both sides w.r.t. x, we get
put x = − t ⇒ dx = − dt dy
When x = − 2n − 1 2x + 2y =0
dx
⇒ t = 2n + 1 and x = − 2n dy − x
⇒ =
⇒ t = 2n dx y
2n
2 2
∴ I1 = ∫ sin (− t) (− dt) dy = x
27
⇒
2n + 1 dx y2
2n a 2
= ∫ sin 1 + dx
27 dy
∫
t dt Let I=
2n + 1 dx
0
2n10 2n + 1 a
x2
Hence, I = ∑ ∫ sin27 x dx + ∫ sin27 x dx I= ∫ 1+ dx
n =1 y2
2n + 1 2n 0
a
10 2n + 1 2n + 1 x 2 + y2
I = ∑ − ∫ sin27 x dx + ∫ sin27 x dx = 0 ⇒ I= ∫ y2
dx
n =1
2n 2n 0
a
a
⇒ I= ∫ [Q x 2 + y 2 = a 2]
2 dx
5. (b) Let I = ∫ [x 2] dx 0 a − x
2 2
0 a
I = a sin− 1
x
1 2 3 2 ⇒
I = ∫ 0. dx + a 0
⇒ ∫ dx + ∫ 2dx + ∫ 3 dx
0 1 2 3
WB JEE (Engineering) Solved Paper 2020 35
I = a sin− 1 − sin− 1 0
a dv dx
⇒ ⇒ − =
a cosec v x
⇒ I = a[sin− 1 1 − sin− 1 0] dx
π π
⇒ ∫ − sin v dv = ∫ x
= a − 0 = a
2 2 ⇒ cos v = log xc
y
⇒ cos = log cx …(i)
8. (c) We have, x
n
1 j π
n→ ∞
lim ∑ f
n
Put x = 1, y = in Eq. (i), we get
2
j =0n
π
By standard formula, we write cos = log c
n 1 2
1 j
lim
n→ ∞
∑ f =
n ∫ f (x) dx ⇒ log c = 0
j =0n 0 ⇒ c =1
lim (as j = 0) = 0
j y
Where, cos = log x
n→ ∞ n x
lim (as j = n) = 1
j
and 11. (c) We have,
n → ∞ n
f (x) = 1 − x2
9. (c) We have, f (x) = 1 − | x |
y′ + yf ′(x) − f (x) f ′(x) = 0 1 + x , x < 0
f (x) =
⇒ y′ + yf ′(x) = f (x) f ′(x) 1 − x , x ≥ 0
Which is a linear differentiable equation of the
dy Graph of f (x)
form + Py = Q
dx Y
∫ f ′( x ) dx
∴ I.F = e =e f ( x)
(0,1)
f(x
x
Solution of given differential equation is
1+
y ⋅ e f ( x) =∫ f (x) f ′(x) ⋅ e dx )=
)=
f ( x)
1–
f (x
X¢ x X
y⋅ e f ( x)
= ∫ te t dt [Qput f (x) = t ⇒ f ′(x) dx = dt] (–1,0) O (1,0)
⇒ y ⋅ e f ( x ) = te t − e t + c
⇒ y ⋅ e f ( x ) = f (x)e f ( x ) − e f ( x ) + c
Y¢
⇒ y = f (x) − 1 + ce − f ( x)
12 24 5 Apply C2 → C2 − C3
23. (c) Given, A = x 6 2 0 0 1
A = 2abc 2 a + b b − a a
− 1 − 2 3
b b c
Since, A is not invertible.
∴ | A| = 0 A = 2abc 2 |(a + b)b − (b − a) b |
| A | = 12 (18 + 4) − 24 (3x + 2) + 5(− 2x + 6) = 0 = 2abc 2 | ab + b 2 − b 2 + ab|
⇒ 264 − 72x − 48 − 10 x + 30 = 0 A = 4a 2b 2c 2
⇒ 82x = 246 ⇒ x = 3 ∴ k= 4
a b 26. (d) We have,
24. (a) Given, A = , | A| = 1 and | A − λI | = 0 has
c d f :S→ R
imaginary roots. a b
f = ad − bc
|A | = ad − bc = 1
c d
a − λ b
A − λI = 4 3
c d − λ f = 12 − 6 = 6
2 3
| A − λI | = (a − λ) (d − λ) − bc
2 6
|A − λI | = ad − (a + d) λ + λ2 − bc f = 12 − 6 = 6
1 6
= λ2 − (a + d)λ + 1 (Q ad − bc = 1)
∴Hence, f is not one-one.
| A − λI |has imaginary roots.
Since, O ∈ R but S does not contain any singular
∴ (a + d)2 − 4 < 0 ⇒ (a + d)2 < 4 matrix.
a2 bc c 2 + ab ∴ f is not onto.
25. (d) Let A = a + ab2
b2 ca Hence, f is neither one-one nor onto.
ab b 2 + bc c2 27. (b) Given,
Taking common a, b, c from C1 , C2 and C3 aρb = a − b is zero or irrational.
respectively. For reflexive,
a c c+ a (a , a) = a − a = 0. Hence, ρ is reflexive.
∴ A = abc a + b b a For symmetric,
b b+ c c (a , b) ∈ρ ⇒(b , a) ∈ρ
⇒ a − b is zero or irrational then b − a is zero or
Apply R1 → R1 + R2 + R3 irrational.
2(a + b) 2(b + c) 2(c + a)
∴ ρ is symmetric.
A = abc a+ b b a For transitive,
b b+ c c If a = 3, b = 3 and c = 5
a+ b b+ c c+ a (a , b) ∈ρ, (b , c) ∈ρ ⇒(a , c) ∉ρ
A = 2abc a + b b a ∴ ρ is not transitive.
b b+ c c
28. (b) Let the unit vector in ZOX plane be
Apply R1 → R1 − R2 a = x$i + zk $ ,|a | = 1
0 c c
a ⋅ α = |a ||α | cos 45°
A = 2abc a + b b a
$ ) ⋅(2$i + 2$j − k
⇒ ( x i + zk
$ $ ) =1 × 3 × 1
b b+ c c 2
$]
[Qα = 2$i + 2$j − k
38 WB JEE (Engineering) Solved Paper 2020
3 dy
2x − z = ⇒ = y + e x (− A sin x + B cos x)
2 dx
and a ⋅ β = |a ||β | cos 60° d 2 y dy
1 ⇒ = + e x ( − A sin x + B cos x)
⇒ (x$i + zk$) ⋅($j − k$) = 1 × 2 × [Qβ = $j − k$ ] dx 2 dx
2 + e x (− A cos x − B sin x)
1
−z=
d 2 y dy dy
2 ⇒ = + − y − y
z=−
1
and x =
1 dx dx dx
2 2 d 2 y 2dy
1 $ 1 $ ⇒ = − 2y
∴ a = i− k dx 2 dx
2 2
d 2 y 2dy
⇒ − + 2y = 0
29. (d) Four person A, B, C, D throw an unbiased die, dx 2 dx
turn by turn, in succession till one gets an even
number and win the game. 33. (d) We have,
π
Probability of an even number =
1 r cos θ − = 2
2 3
π π
r cos θ cos + sin θ sin = 2
1
∴ P(A) = P(B) = P(C) = P(D) =
2 3 3
1 r cos θ r sin θ ⋅ 3
P(A) = P(B) = P(C) = P(D) = + = 2,
2 2 2
Required probability
x+ 3y = 4 [Q x = r cos θ, y = r sin θ]
= P(A) + P(A) P(B) P(C) P(D) P(A)
+ P(A) P(B) P(C) P(D) P(A) P(B) P(C) P(D) P(A) + K ∴It represents a straight line.
5 9
1 1
+ + + K
1 34. (d) Let (h, k) be the centre and r be the radius of
=
2 2 2 the variable circle.
1/ 2 1/ 2 8 Since, variable circle touches the circle x 2 + y 2 = a 2
= = =
1 − (1 / 2)4 15/16 15
∴ h2 + k2 = a ...(i)
30. (b) Given, P = 10 = 1 Also variable circle touches the circle x 2 + y 2 = 4ax
100 10
q =1 − P =1 −
1
=
9 ∴ (h − 2a)2 + k2 = 2a + r ...(ii)
10 10
From Eqs. (i) and (ii), we get
50 1
P(X ≥ 1) ≥ = (h − 2a)2 + k2 − h2 + k2 = a
100 2
P(X ≥ 1) = 1 − P(X = 0) ∴Locus of centre of variable circle is
0 n
(x − 2a)2 + y 2 − x 2 + y 2 = a
P(X ≥ 1) = 1 − C0
n 1 9
10 10 which represents the hyperbola.
n
1 − ≥
9 1
35. (b) We have, x 2 + 2xy + ay2 = 0 and
10 2
ax 2 + 2xy + y 2 = 0 have common line.
n
9 ≤1
y = x satisfies both lines.
10 2
∴ x 2 + 2x 2 + ax 2 = 0
For n = 7 the least number of round to must so fire.
3x 2 + ax 2 = 0 ⇒ a = − 3
31. (c) By using Sandwich theorem. Equation of lines are
32. (b) We have, x 2 + 2xy − 3y 2 = 0 and − 3x 2 + 2xy + y 2 = 0
y = e x (A cos x + B sin x) (x + 3y) (x − y) = 0 and (3x + y) (x − y)= 0
dy The other lines are x + 3y = 0 and 3x + y = 0.
= e x (A cos x + B sin x) + e x (− A sin x + B cos x)
dx
WB JEE (Engineering) Solved Paper 2020 39
36. (b) Given lines, In ∆OAB
AB
4 x + 2y = 9 and 2x + y + 6 = 0 sin α =
6 9 OA
Lines having distance from origin are and
2 5 5 ⇒ OA = AB cosec α = 2at cosec α
respectively. ∴ OA = 2a(2 cot α) cosec α
The point O divide the segment PQ in the ratio OA = 4a cot α cosec α
=
9
÷
6
= 3: 4 ∴Length of chord = 4a cot α cosec α
2 5 5 2 2
40. (d) A double PQ of the hyperbola x 2 − y2 = 1
37. (a) Given curve a b
x + 2y = a and 2x + y = a
2 2 2 2 2 2 Such that ∆OPQ is an equilateral where O is origin.
Let P(a sec θ, b tan θ) and Q(a sec θ, − b tan θ)
Y (0, a)
∆OPQ is an equilateral
(0, a/√2) a/√3, a/√3
∴ OP = PQ = OQ
a sec θ + b tan2 θ = 4b 2 tan2 θ
2 2 2
X′ X
O (a, 0) a 2 sec2 θ = 3b 2 tan2 θ
b2 sec2 θ 1
(a/√2, 0) = = cosec2 θ
a 2
3 tan2 θ 3
Y′
b2 cosec2 θ 1
a2 a/ 2 e = 1+ = 1+ = 3 + cosec2 θ
∴ Required area = + ∫a/ a − 2x dx × 2
2 2 2
a 3 3
6 3
2
e> [Qcosec2 θ ≥ 1]
a2
tan−1
1 3
=
2 2
41. (c) Equation of ellipse
38. (a) Given, the radius of circle = 17 x2 y2
+ =1
With centre on positive sides of X-axis 25 9
∴ Equation of circle is (x − a)2 + y 2 = 17
Since, it passes through (0, 1) B (0,3)
∴ a 2 + 1 = 17
⇒ a=4
∴ Equation of circle is S¢ S(4,0)
(–4,0)
x 2 + y2 − 8 x − 1 = 0
B¢ (0,–3)
39. (b) Given, parabola y2 = 4ax
A (at2,2at)
Here, B = (0, 3), B′ = (0, − 3)
S = (4, 0), S ′ = (0, − 4)
1
a Area of rhombus SBS ′ B′ = SS ′ × BB′
2
O B (at2,0) 1
= × 8 × 6 = 24 sq units
2
lim 1
44. (b) Given line, 1 + cx
⇒ ex → 0 − 1 × = 4
x−2 y−3 z−4 1 − cx x
= =
3 4 5 lim
2 cx 1
and plane 2x − 2y + z = 5 ⇒ ex → 0 × =4
1 − cx x
Angle between line and plane is
(3)(2) + 4(− 2) + 51
() ⇒ e 2c = 4
sin θ =
32 + 42 + 52 22 + (− 2)2 + ()
12 1
1 + 2cx x
6−8+ 5 Now, lim = e 4 c = (e 2c)2 = (4)2 = 16
sinθ = x→ 0 1 − 2cx
50 × 3
1 2 49. (a) Given,
sin θ = =
5 2 10 f is twice continuously and differentiable and
1 = f ′(0) = 0
f (0) = f ()
45. (c) Given, f (x) = sin x + cos ax ∴ f (x) = Kx 2(x − 1)
sin x is periodic with 2π. f (x) = K (x 3 − x 2)
f (x) is periodic only on sin x and cos x both are
periodic. f ′(x) = K (3x 2 − 2x)
2π 2
∴ cos ax is periodic at . f ′(x) = 0 ⇒ x = 0and
a 3
2π Hence, Rolle’s theorem is applicable in f ′(x).
LCM of 2π and is possible only a is rational
a
∴ f ′ ′(c) = 0 for some c ∈ R
number.
50. (b) We have,
46. (c) Given, f (x) = 1 − x + 1 f (x) = x13 + x11 + x 9 + x 7 + x 5 + x 3 + x + 12
x
1 f ′(x) = 13x12 + 11 x10 + 9 x 8 + 7 x 6 + 5x 4 + 3x 2 + 1
f (x) is defined − x + 1 ≥ 0 and x > 0
x f ′(x) > 0 for all value of x ∈R.
x 2 + x − 1 ≤ 0, x > 0 Hence, f (x) has exactly are real roots.
−1 ± 5 51. (c) Area of region
x= , x>0
2 {(x , y) = x 2 + y 2 ≤ 1 < x + y}
WB JEE (Engineering) Solved Paper 2020 41
∴ x 2 + y 2 = 1 and x + y = 1 54. (c) Given α = $i + a$j + a 2k$ , β = $i + b$j + b 2k$ and
Y γ = $i + c$j + c 2k$
1 a a2
(0, 1) ∴ 1 b b2 ≠ 0
B 1 c c2
X¢
O
X a a2 1 + a3
A
(1, 0) Now, b b 2 1 + b 3 = 0
c c2 1 + c3
a a2 1 a a2 a3
Y¢ b b 1 + b b2 b3 = 0
2
c c2 1 c c2 c3
Area of shaded region = Area of quadrant − Area of
triangle
1 a a2
π 2 1
= () 1 − ×1 ×1 1 b b 2 (1 + abc) = 0
4 2
π 1 1 c c2
= −
4 2
1 a a2
52. (b) Let f (x) = cos x + x sin x − 1
⇒ abc + 1 = 0 Q 1 b b 2 ≠ 0
f ′(x) = − sin x + x cos x + sin x 1 c c 2
π
f ′(x) = x cos x > 0; x ∈ 0, ⇒ abc = −1
2
π
f (x) is increasing on x ∈ 0, .
55. (c) Given, z1 and z2 be two imaginary roots of
2 z 2 + pz + q = 0
∴ cos x + x sin x − 1> 0 ∴ z1 + z2 = − p, z1 z2 = q
⇒ cos x + x sin x > 1
z1 , z2 and origin form an equilateral triangle.
∴ z12 + z22 = z1 z2
53. (c) Given parabola
⇒ (z1 + z2)2 − 2z1 z2 = z1 z2
y = ax 2 + bx + C
⇒ p 2 − 2q = q ⇒ p 2 = 3q
put x = 1, y = 1, we get
1=a+ b+ c ...(i) 56. (a) Given,
y = x is tangent of the parabola at (1, 1). P(x) = ax 2 + bx + c, Q(x) = − ax 2 + dx + c
∴ dy = 2ax + b
Discriminant for P(x) = b 2 − 4ac
dx (1 , 1)
and Discriminant for Q(x) = d 2 + 4ac
⇒ 1 = 2a + b ...(ii)
If ac> 0, Q(x) has real roots and ac < 0, P(x) has
Also curve passes through (− 1, 0) real roots.
∴ 0=a−b+ c ...(iii) ∴P(x) ⋅ Q(x) = 0 has atleast two real roots.
From Eqs. (i), (ii) and (iii), we get
1 57. (d) We have,
a=c= f (x) = x| x |, x ∈ [− 1, 1]
4
1 − x 2 , − 1 ≤ x < 0
and b= f (x) = 2
2 0≤ x <1
x ,
42 WB JEE (Engineering) Solved Paper 2020
Graph of f (x) is Here, P is the orthocentre of ∆ABQ.
Y ∴Slope of BR × Slope of AR = − 1
⇒ k + 5 × k = − 1
h h − 7
f(x)=x2
⇒ x 2 + y 2 − 7 x + 5y = 0
X¢ X
–1 O 1
62. (b) Let
f(x)=x2 n 1 / ( k + α)
dx
I = lim
n→ ∞
∑ ∫ + x
k = 1 (1 / k + β) 1
Y¢
1
n
Clearly f (x) is bijective function.
∑ [log(1 + x)] 1
k+ α
I = lim
n→ ∞
k =1 k+ β
58. (d) We have,
f (x) = x 2 − 3x + 2
n 1 1
I = lim
n→ ∞
∑ log 1 + − log 1 +
k + α
k + β
g(x) = x k =1
n k + α + 1 k + β
fog = f (g(x)) = x − 3 x + 2 I = lim
n→ ∞
∑ log
k + α k + β + 1
k =1
fog is defined x −3 x + 2 ≥ 0 and x ≥ 0
α + 2 β + 1 α + 3 β + 2
( x − 1) ( x − 2) ≥ 0 I = lim log
n→ ∞ α + 1 β + 2 α + 2 β + 3
x ∈ (− ∞ , 1] ∪ [4, ∞)
α + n + 1 β+ n
gof = g ( f (x)) = x 2 − 3x + 2 ....
α+ n β + n + 1
gof is defined x 2 − 3x + 2 ≥ 0 β + 1 α + n + 1
I = lim log
(x − 1) (x − 2) ≥ 0 n→ ∞ α + 1 β + n + 1
x ∈ (− ∞ , 1]∪ [2,∞ )
β + 1
∴Domain of fog is (−∞ , 1] ∪ [4, ∞) I = log e
α + 1
Domain of gof is (− ∞ , 1] ∪ [2,∞)
∴ S ∩ T is an interval. 1
63. (c) Let P = lim 1 −
x→1 log x x − 1
59. (b) Given, ρ1 and ρ2 be two equivalent relation by
theorem, x − 1 − log x
P = lim
x → 1 (x − 1) log x
ρ1 ∩ ρ2 is equivalence relation but ρ1 ∪ ρ2 is not so.
60. (c) Since, the length of tangent between the point Apply L-Hospital rules,
of contact and the point where it meets the 1−
1
directrix subtends right angle at the P = lim , x
x →1 x −1
corresponding focus. + log x
x
61. (c) Y
x −1
=
(0,b) (x − 1) + x log x
Q
Again apply L-Hospital rule,
R (x,y) 1
P = lim
x→1 x
1 + + log x
X¢ X x
P (a,0) A (7, 0)
1
= lim
x → 1 2 + log x
B (0,–5) 1 1
P= =
Y¢ 2+ 0 2
WB JEE (Engineering) Solved Paper 2020 43
1
64. (b) We have, = 2∫ (1 − y 2) dy
0
1
y= y3
1
1 + x + log x = 2 y −
3 0
(1 + x + log x) y = 1
= 21 − = sq unit
On differentiating both sides, we get 1 4
3 3
dy
+ 1 + y = 0
1
(1 + x + log x)
dx x 67. (a, d) We have, v = u − gt
dy − (x + 1) y at t = 6, v = 0
⇒ =
dx x(1 + x + log x) = u − gt = 0
dy 1 ⇒ u = 6 g = 192ft/sec (Q g = 32 ft / sec2)
⇒ x = − (x + 1) (y 2) Q y = 1 + x + log x 1
dx Now, h = ut − gt 2
2
dy
⇒ x = − y(x + 1) y 1
dx = 192.6 − × 32 × 6 × 6
2
dy
⇒ x = y(y log x − 1) [Q(1 + x) y = 1 − y log x] = 576 ft
dx
−
x 68. (a, c) Given, equation
65. (b) Given, curve y = be a
x(log3 x) 29
−
log 3 (x + 5) = 3 3
x 2
dy b −
⇒ (log 3 x)2 − log 3 x + 5 log 3 x = log 3 3
⇒ =− e a 9 3
dx a 2 2
∴ dy (0, b) = − b put log 3 x = t
dx a t 2 − 9 t + 5 t = 3
∴
∴Equation of tangent 2 2
y − b= −
b
(x − 0) ⇒ 2t 3 − 9t 2 + 10t − 3 = 0
a
(t − 3) (t − 1) t − = 0
1
⇒
⇒ y−b=−
bx 2
a 1
⇒ t = 3, 1,
y − x 2
⇒ −1 =
b a 1
⇒ log 3 x = 3, 1,
x y 2
⇒ + =1 1
a b
⇒ x = 33, 31 , 32
66. (b) Intersecting points of both curves are P(− 2, 1)
and Q(− 2, − 1). 69. (b) Given, total number of questions are n and
total wrong answers are 2047
x =–2y2 P (–2,1)
Also, total wrong answer given
Y = Σk × (2n − k − 2n − ( k + 1))
P (–2,1)
(1,0) ⇒ 2n − 1 + K + 1 = 2047
X¢ X
O ⇒ 2n = 2048 = 211
x =1– 3y2 ⇒ n = 11
Q
Y¢ 70. (c, d) Given, P(A′ ∩ B′) = 3
(–2,–1) 5
3
⇒ 1 − P(A ∪ B) =
∴Area of the bounded region 5
1
= ∫− 1 ((1 − 3y ) − (− 2y )) dy 2
2 2
⇒ P(A ∪ B) =
5
44 WB JEE (Engineering) Solved Paper 2020
2 x2
⇒ P(A) + P(B) − P(A ∩ B) = 73. (a) Given, y =
5 (x + 1)2 (x − 2)
2
⇒ P(A) + P(B) − P(A) P(B) = ...(i) By using partial fraction,
5
x2 4 3 1
(since A and B are independent events) = − +
(x + 1)2 (x + 2) (x + 2) x + 1 (x + 1)2
1
Also, given P(A ∩ B) = 4 3 1
20 ∴ y= − +
1 x + 2 x + 1 (x + 1)2
⇒ P(A) P(B) = ...(ii)
20 On differentiating both sides w.r.t. x, we get
dy −4 3 2
From Eqs. (i) and (ii), we get = + −
1 1 dx (x + 2)2 (x + 1)2 (x + 1)3
P(A) = or
4 5 Again differentiating w.r.t. x
d2y 8 6 6
71. (a, b) Equation of straight line having intercepts = − +
on the coordinate axes are a and b, are dx 2 (x + 2)3 (x + 1)3 (x + 1)4
x y
+ =1 4 3 3
...(i) = 2 − +
a b
( x + 2) 3
( x + 1) 3
( x + 1)4
Given, a + b = − 1 ⇒ b = − (a + 1)
Line (i) passing through the point (4, 3). 74. (c) Given, f (x) = 1 x sin x − (1 − cos x)
3
4 3
∴ + =1 f (x) x sin x − 31
( − cos x)
a b ∴ lim = lim
4 3
x→ 0 xk x→ 0 3x k
⇒ − =1 x x x
a a +1 2x sin cos − 6 sin2
= lim 2 2 2
⇒ 4(a + 1) − 3a = a(a + 1) x→ 0 3x k
⇒ a2 + a − a = 4 sin x 2x cos x − 6 sin x
1
⇒ a=± 2 = lim 2 ⋅ lim 2 2
→ x → k −1
When a = 2, then b = − 3 3 x 0 x 0 2x
2
and when a = − 2, then b = 1
x y x y x cos x − 3 sin x
∴ Equation of straight line are − = 1, + =1 1
2 3 −2 1 = lim 2 2
k − 1
2 2
3x→0 x
72. (d) Equation of tangent to the ellipse x + y = 1
2 1 f (x)
at point (x1 , y1) is For, lim ≠ 0⇒k −1 =1⇒k = 2
x → 0 xk
xx1 yy
+ 1 =1 ...(i)
2 1 75. (a, c) Given, AP : BP = 3 : 1
Let mid-point of intercept be P(h, k). Y
1
∴ h=
x1
B
1
⇒ x1 =
h 1
1
and k= P (x,y)
2y1 3
X¢ X
1 O A
⇒ y1 =
2k
∴Required locus of the mid-point
1 1
= 2 + =1 Y¢
2x 4 y2
WB JEE (Engineering) Solved Paper 2020 45
Equation of tangent AB is Y − y =
dy
(X − x) ⇒ 3 log y + log x = c
dx ⇒ xy 3 = c
x dy − y at x = 1, y = 1, we get c = 1
∴Coordinates of A are dx , 0 . ∴ xy 3 = 1
y′
⇒ x(3y 2)
dy
+ y3 = 0
dx
and coordinates of B are 0, y − x .
dy
dx dy
⇒ 3x + y=0
dx
1 × 0 + 3 × y − x
dy
dy y
dx ⇒ =−
Now, y = dx 3x
4
∴ dy =−
1
⇒ 4 y = 3y − 3x
dy
dx (1 , 1) 3
dx
dy ∴Slope of normal = 3
⇒ 3x + y=0
dx ∴Equation of normal = y − 1 = 3(x − 1)
⇒
3dy dx
+ =0 ⇒ y − 3x + 2 = 0
y x
WB JEE (Engineering) Solved Paper 2019 1
WB JEE
Engineering Entrance Exam
3. A proton and an electron initially at rest are 7. The correct dimensional formula for impulse
accelerated by the same potential difference. is given by
2 WB JEE (Engineering) Solved Paper 2019
(a) ML2 T− 2 (b) MLT− 1 (c) ML2 T− 1 (d) MLT− 2 13. A compressive force is applied to a uniform
8. The density of the material of a cube can be rod of rectangular cross-section so that its
estimated by measuring its mass and the length decreases by 1%. If the Poisson’s ratio
length of one of its sides. If the maximum for the material of the rod be 0.2, which of
error in the measurement of mass and length the following statements is correct?
are 0.3% and 0.2% respectively, the “The volume approximately ...........”
maximum error in the estimation of the (a) decreases by 1% (b) decreases by 0.8%
density of the cube is approximately. (c) decreases by 0.6% (d) increases by 0.2%
(a) 1.1% (b) 0.5% (c) 0.9% (d) 0.7% 14. A small spherical body of radius r and density
ρ moves with the terminal velocity v in a fluid
9. Two weights of the mass m1 and m2(> m1) are
of coefficient of viscosity η and density σ.
joined by an inextensible string of negligible What will be the net force on the body?
mass passing over a fixed frictionless pulley. 4π 3
The magnitude of the acceleration of the (a) r (ρ − σ ) g (b) 6πηrv
3
loads is (c) Zero (d) Infinity
m2 − m1
(a) g (b) g
m2 15. Two black bodies A and B have equal surface
m1 m − m1 areas and are maintained at temperatures
(c) g (d) 2 g
m2 + m1 m2 + m1 27ºC and 177ºC respectively. What will be the
ratio of the thermal energy radiated per
10. A body starts from rest, under the action of second by A to that by B?
an engine working at a constant power and (a) 4 : 9 (b) 2 : 3 (c) 16 : 81 (d) 27 : 177
moves along a straight line. The displacement
s is given as a function of time ()
t as 16. What will be the molar specific heat at
(a) s = at + bt 2 , a and b are constants constant volume of an ideal gas consisting of
rigid diatomic molecules?
(b) s = bt , b is a constant
2
3 5
(c) s = at 3 / 2 , a is a constant (a) R (b) R
2 2
(d) s = at , a is a constant (c) R (d) 3R
11. Two particles are simultaneously projected in 17. Consider the given diagram. An ideal gas is
the horizontal direction from a point P at a contained in a chamber (left) of volume V
certain height. The initial velocities of the and is at an absolute temperature T. It is
particles are oppositely directed to each other allowed to rush freely into the right chamber
and have magnitude v each. The separation of volume V which is initially vacuum. The
between the particles at a time when their whole system is thermally isolated. What will
position vectors (drawn from the point P) are be the final temperature of the equilibrium
mutually perpendicular, is has been attained?
v2 v2
(a) (b)
2g g
4v 2 2v2
(c) (d) V V
g g Ideal gas Vacuum
18. Five identical capacitors, of capacitance 20µF 22. An electric current ‘I’ enters and leaves a
each, are connected to a battery of 150 V, in a uniform circular wire of radius r through
combination as shown in the diagram. What diametrically opposite points. A particle
is the total amount of charge stored? carrying a charge q moves along the axis of
the circular wire with speed v. What is the
magnetic force experienced by the particle
when it passes through the centre of the circle?
µ 0i µ 0i
(a) qv (b) qv
a 2a
µ i
(c) qv 0 (d) Zero
2 πa
26. When the value of R in the balanced 30. When the frequency of the light used is
Wheatstone bridge, shown in the figure, is changed from 4 × 10 14 s− 1 to 5 × 10 14 s− 1 , the
increased from 5Ω to 7 Ω, the value of S has to angular width of the principal (central)
be increased by 3Ω in order to maintain the maximum in a single slit Fraunhoffer
balance. What is the initial values of S? diffraction pattern changes by 0.6 radian.
What is the width of the slit (assume that
P Q the experiment is performed in vacuum)?
. × 10− 7 m
(a) 15 (b) 3 × 10− 7 m
G −7
(c) 5 × 10 m (d) 6 × 10− 7 m
R S
Category II (Q. Nos. 31 to 35)
Carry 2 marks each and only one option is
correct. In case of incorrect answer or any
(a) 2.5 Ω (b) 3 Ω combination of more than one answer, 1/2 mark
(c) 5 Ω (d) 7.5 Ω will be deducted.
27. When a 60 mH inductor and a resistor are 31. A capacitor of capacitance C is connected in
connected in series with an AC voltage source, series with a resistance R and DC source of
the voltage leads the current by 60º. If the emf E through a key. The capacitor starts
inductor is replaced by a 0 .5 µF capacitor, the charging when the key is closed. By the
voltage lags behind the current by 30º.What is time the capacitor has been fully charged,
the frequency of the AC supply? what amount of energy is dissipated in the
1 1
(a) × 104 Hz (b) × 104 Hz resistance R?
2π π
3 1
(c) × 104 Hz (d) × 108 Hz C
2π 2π
R
28. A point object is placed on the axis of a thin E
convex lens of focal length 0.05 m at a distance
of 0.2 m from the lens and its image is formed 1 E2
(a) CE 2 (b) 0 (c) CE 2 (d)
on the axis. If the object is now made to 2 R
oscillate along the axis with a small amplitude
of A cm, then what is the amplitude of 32. A horizontal fire hose with a nozzle of
5
oscillation of the image? cross-sectional area × 10 − 3 m 2 delivers
1 21
you may assume, 1 + x ≈ 1 − x , where x < < 1 a cubic metre of water in 10s. What will be
the maximum possible increase in the
4A 5A temperature of water while it hits a rigid
(a) × 10− 2 m (b) × 10− 2 m
9 9 wall (neglecting the effect of gravity)?
A A
(c) × 10− 2 m (d) × 10− 2 m (a) 1ºC (b) 0.1ºC (c) 10ºC (d) 0.01ºC
3 9
33. Two identical blocks of ice move in opposite
29. In Young’s experiment for the interference of directions with equal speed and collide with
light, the separation between the slits is d and each other. What will be the minimum
the distance of the screen from the slits is D. If speed required to make both the blocks
D is increased by 0.5% and d is decreased by melt completely, if the initial temperatures
0.3% then for the light of a given wavelength, of the blocks were − 8 ºC each?
which one of the following is true? (Specific heat of ice is 2100 Jkg − 1 K − 1 and
“The fringe width ............. latent heat of fusion of ice is 3.36 × 105 Jkg − 1 )
(a) increases by 0.8% (b) decreases by 0.8% (a) 840 ms − 1 (b) 420 ms − 1
−1
(c) increases by 0.2% (d) decreases by 0.2% (c) 8.4 ms (d) 84 ms − 1
WB JEE (Engineering) Solved Paper 2019 5
34. A particle with charge q moves with a velocity 37. The initial pressure and volume of a given
v in a direction perpendicular to the Cp
directions of uniform electric and magnetic mass of an ideal gas with = γ , taken in
fields, E and B respectively, which are CV
mutually perpendicular to each other. Which a cylinder fitted with a piston, are p 0 and V0
one of the following gives the condition for respectively. At this stage the gas has the
which the particle moves undeflected in its same temperature as that of the surrounding
original trajectory? medium which is T0 . It is adiabatically
V
y compressed to a volume equal to 0 .
2
E
Subsequently the gas is allowed to come to
thermal equilibrium with the surroundings.
v
x What is the heat released to the
B surrounding?
p0 V0
(a) 0 (b) (2 γ − 1 − 1)
z γ −1
E B E B p0 V0
(a) v = (b) v = (c) v = (d) v = q (c) γp0 V0 ln2 (d)
B E B E 2(γ − 1)
35. A parallel plate capacitor in series with a 38. A projectile thrown with an initial velocity of
resistance of 100 Ω, an inductor of 20 mH 10 ms − 1 at an angle α with the horizontal,
and an AC voltage source of variable has a range of 5 m. Taking g = 10 ms − 2 and
frequency shows resonance at a frequency of neglecting air resistance, what will be the
1250 estimated value of α?
Hz.
π (a) 15º (b) 30º (c) 45º (d) 75º
If this capacitor is charged by a DC voltage 39. In the circuit shown in the figure all the
source to a voltage 25 V, what amount of resistance are identical and each has the
charge will be stored in each plate of the value r Ω. The equivalent resistance of the
capacitor? combination between the points A and B will
(a) 0.2 µC (b) 2 mC (c) 0.2 mC (d) 0.2 C remain unchanged even when the following
pairs of points marked in the figure are
Category III (Q. Nos. 36 to 40) connected through a resistance R.
Carry 2 marks each and one or more option(s) 1 2 3 4 5
is/are correct. If all correct answers are not marked r r r r r r
and also no incorrect answer is marked then score A B
= 2 × number of correct answers marked ÷ actual r 6 r 7 r
number of correct answers. If any wrong option is
marked or if any combination including a wrong (a) 2 and 6 (b) 3 and 6
option is marked, the answer will be considered (c) 4 and 7 (d) 4 and 6
wrong, but there is no negative marking for the 40. A metallic loop is placed in a uniform
same and zero marks will be awarded. magnetic field B with the plane of the loop
perpendicular to B. Under which condition(s)
36. Electrons are emitted with kinetic energy T given an emf will be induced in the loop?
from a metal plate by an irradiation of light
of intensity J and frequency ν. Then, which “If the loop is ............”
of the following will be true? (a) moved along the direction of B
(b) squeezed to a smaller area
(a) T ∝ J
(c) rotated about its axis
(b) T linearly increasing with ν
(d) rotated about one of its diameters
(c) T ∝ time of irradiation
(d) Number of electrons emitted ∝ J
Chemistry
Category I (Q. Nos. 41 to 70) 44. The compound, which evolves carbon dioxide
Carry 1 mark each and only one option is correct. on treatment with aqueous solution of
In case of incorrect answer or any combination of sodium bicarbonate at 25ºC, is
more than one answer, 1/4 mark will be (a) C 6H5OH (b) CH3COCI
deducted. (c) CH3CONH2 (d) CH3COOC 2H5
OCOCH3 OH
49. For the equilibrium, H 2O( l ) - H 2O(v),
which of the following is correct?
CO2H CO2H (a) ∆G = 0, ∆H < 0, ∆S < 0
COCH3 (b) ∆G < 0, ∆H > 0, ∆S > 0
(c) (d) (c) ∆G > 0, ∆H = 0, ∆S > 0
COCH3 (d) ∆G = 0, ∆H > 0, ∆S > 0
ab
OH OH 50. For a van der Waals’ gas, the term
represents some V2
43. Cyclopentanol on reaction with NaH followed
(a) pressure (b) energy
by CS2 and CH 3I produces a/an
(c) critical density (d) molar mass
(a) ketone (b) alkene
(c) ether (d) xanthate 51. In the equilibrium, H 2 + I 2 -
2HI, if at a
given temperature the concentrations of the
WB JEE (Engineering) Solved Paper 2019 7
reactants are increased, the value of the 60. The melting points of (i) BeCl 2 (ii) CaCl 2 and
equilibrium constant, K C , will (iii) HgCl 2 follows the order
(a) increase (a) i < ii < iii (b) iii < i < ii
(b) decrease (c) i < iii < ii (d) ii < i < iii
(c) remain the same
61. Which of these species will have non-zero
(d) cannot be predicted with certainty
magnetic moment?
52. If electrolysis of aqueous CuSO 4 solution is (a) Na + (b) Mg (c) F − (d) Ar +
carried out using Cu-electrodes, the reaction
taking place at the anode is 62. The first electron affinity of C, N and O will
− be of the order
(a) H + e → H
+
69. The conformations of n-butane, commonly 73. At constant pressure, the heat of formation
known as eclipsed, gauche and of a compound is not dependent on
anti-conformations can be interconverted by temperature, when
(a) rotation around C H bond of a methyl group (a) ∆C p = 0 (b) ∆C v = 0
(b) rotation around C H bond of a methylene (c) ∆C p > 0 (d) ∆C p < 0
group
(c) rotation around C1-C2 linkage 74. A copper coin was electroplated with Zn and
(d) rotation around C2-C3 linkage then heated at high temperature until there
is a change in colour. What will be the
70. The correct order of the addition reaction resulting colour?
rates of halogen acids with ethylene is (a) white (b) black
(a) hydrogen chloride > hydrogen bromide (c) silver (d) golden
> hydrogen iodide
(b) hydrogen iodide > hydrogen bromide
75. Oxidation of allyl alcohol with a peracid gives
> hydrogen chloride
a compound of molecular formula C3H 6O 2 ,
which contains an asymmetric carbon atom.
(c) hydrogen bromide > hydrogen chloride
The structure of the compound is
> hydrogen iodide
(d) hydrogen iodide > hydrogen chloride OH
> hydrogen bromide O
(a) O OH (b) CH3
H
Category II (Q. Nos. 71 to 75)
H H
Carry 2 marks each and only one option is O
correct. In case of incorrect answer or any OH
(c) H3C (d) H3C
combination of more than one answer, 1/2 mark O OH
will be deducted.
71. The total number of isomeric linear Category III (Q.Nos. 76 to 80)
dipeptides which can be synthesised from Carry 2 marks each and one or more option(s)
racemic alanine is is/are correct. If all correct answers are not
(a) 1 (b) 2 (c) 3 (d) 4 marked and also no incorrect answer is marked
72. The kinetic study of a reaction like vA → P at then score = 2 × number of correct answers
marked ÷ actual number of correct answers. If
300 K provides the following curve, where
−3 any wrong option is marked or if any combination
concentration is taken in mol dm and time
including a wrong option is marked, the answer
in min.
will considered wrong, but there is no negative
r0 : Initial rate
[A0] : Initial
marking for the same and zero mark will be
concentration of A awarded.
√r0
Slope = 4.0 76. Haloform reaction with I2 and KOH will be
responded by
O
O
[A]0 Me
(a) I Ph (b) Ph
Identify the correct order(n) and rate constant(k) OH
(a) n = 0, k = 4.0 mol dm− 3 min− 1 OH O
(b) n = 1 / 2, k = 2.0 mol1/ 2 dm− 3 / 2 min− 1
(c) n = 1, k = 8.0 min− 1 (c) Me
Me
Ph (d) Ph N Me
H
(d) n = 2, k = 16.0 dm3 mol − 1 min− 1
WB JEE (Engineering) Solved Paper 2019 9
Mathematics
Category I (Q. Nos. 1 to 50) 4. The value of the integration
π /4
Carry 1 mark each and only one option is µ sin x
correct. In case of incorrect answer or any ∫ λ |sin x | +
1 + cos x
+ γ dx
− π /4
combination of more than one answer, 1/4 mark
will be deducted. (a) is independent of λ only
(b) is independent of µ only
1. lim (x n ln x), n > 0 (c) is independent of γ only
x → 0+ (d) depends on λ, µ and γ
1 sin 2 t
(a) does not exist (b) exists and is zero a a
sin 2 t
(c) exists and is 1 (d) exists and is e − 1 5. The value of lim ∫ e dt − ∫e dt is
x→0x
y x + y
x
2. If ∫ cos x log tan dx equal to
2 2 2
y
x (a) esin (b) e 2sin y (c) e| sin y | (d) ecosec y
3 n n n 1 1
8. lim 1 + + + 14. If log 62 + = log 2 2 x + 8 , then the values
n→ ∞ n
n+3 n+6 n+9 2x
n of x are
+ ... +
n + 3(n − 1) 1 1
(a) ,
1 1
(b) , (c) −
1 1
,
1
(d) , −
1
4 3 4 2 4 2 3 2
(a) does not exist (b) is 1
(c) is 2 (d) is 3 15. Let z be a complex number such that the
principal value of argument, arg z > 0.
9. The general solution of the differential
x Then, arg z − arg(− z) is
x π
equation 1 + e y dx + 1 − e x / y dy = 0 is (a) (b) ± π (c) π (d) − π
y 2
(C is an arbitrary constant) 16. The general value of the real angle θ, which
x x satisfies the equation,
(a) x − ye = C y
(b) y − xe = C y (cos θ + i sin θ) (cos 2θ + i sin 2θ) ...
x x (cos nθ + i sin nθ) = 1 is given by, (assuming k
(c) x + ye y = C (d) y + xe y = C is an integer)
2 kπ 4kπ
dy (a) (b)
10. General solution of (x + y)2 = a 2 , a ≠ 0 is n+2 n(n + 1)
dx 4kπ 6kπ
(c) (d)
(C is an arbitrary constant) n+1 n(n + 1)
x y
(a) = tan + C (b) tan xy = C
a a 17. Let a , b, c be real numbers such that
y+C x+ y a + b + c < 0 and the quadratic equation
(c) tan( x + y) = C (d) tan =
a a ax 2 + bx + c = 0 has imaginary roots. Then,
11. Let P(4 , 3) be a point on the hyperbola (a) a > 0, c > 0 (b) a > 0, c < 0
x 2
y 2 (c) a < 0, c > 0 (d) a < 0, c < 0
2
−= 1. If the normal at P intersects the
a b2 18. A candidate is required to answer 6 out of
X -axis at (16, 0), then the eccentricity of the 12 questions which are divided into two parts
hyperbola is A and B, each containing 6 questions and
5 he/she is not permitted to attempt more than
(a) (b) 2 (c) 2 (d) 3 4 questions from any part. In how many
2
different ways can he/she make up his/her
12. If the radius of a spherical balloon increases choice of 6 questions?
by 0.1%, then its volume increases (a) 850 (b) 800 (c) 750 (d) 700
approximately by
(a) 0.2% (b) 0.3% (c) 0.4% (d) 0.05%
19. There are 7 greeting cards, each of a different
colour and 7 envelopes of same
13. The three sides of a right angled triangle are 7 colours as that of the cards. The number
in GP (geometric progression). If the two of ways in which the cards can be put in
acute angles be α and β, then tanα and tanβ envelopes, so that exactly 4 of the cards go
are into envelopes of respective colour is,
5+1 5 −1 5+1 5 −1 (a) 7C 3 (b) 2.7 C 3 (c) 3!4 C 4 (d) 3!7 C 3 4C 3
(a) and (b) and
2 2 2 2
1 5 2 20. 7 2n + 16 n − 1 (n ∈ N) is divisible by
(c) 5 and (d) and
5 2 5 (a) 65 (b) 63 (c) 61 (d) 64
WB JEE (Engineering) Solved Paper 2019 11
21. The number of irrational terms in the 28. Let the relation ρ be defined on R as aρb if
84
1 1 1 + ab > 0 . Then,
expansion of 3 8 + 5 4 is (a) ρ is reflexive only.
(b) ρ is equivalence relation.
(a) 73 (b) 74 (c) 75 (d) 76 (c) ρ is reflexive and transitive but not symmetric
(d) ρ is reflexive and symmetric but not transitive.
22. Let A be a square matrix of order 3 whose all
entries are 1 and let I 3 be the identity matrix 29. A problem in mathematics is given to
of order 3. Then, the matrix A − 3 I 3 is 4 students whose chances of solving
1 1 1 1
(a) invertible (b) orthogonal individually are , , and . The probability
(c) non-invertible 2 3 4 5
(d) real Skew Symmetric matrix that the problem will be solved at least by
one student is
23. If M is any square matrix of order 3 over ú 2 3
(a) (b)
and if M′ be the transpose of M, then 3 5
adj(M ′) − (adj M)′ is equal to (c)
4
(d)
3
(a) M (b) M′ 5 4
(c) null matrix (d) identity matrix
30. If X is a random variable such that σ(X ) = 2.6,
5 5 x x then σ(1 − 4 X ) is equal to
24. If A = 0 x 5 x and| A2 | = 25, then| x | is (a) 7.8 (b) − 10.4 (c) 13 (d) 10.4
0 0 5
− sin x
31. If e −e
sin x
− 4 = 0, then the number of
equal to real values of x is
1
(a) (b) 5 (c) 52 (d) 1 (a) 0 (b) 1
5
(c) 2 (d) 3
25. Let A and B be two square matrices of order 3
32. The angles of a triangle are in the ratio
and AB = O3 , where O3 denotes the null
2 : 3 : 7 and the radius of the circumscribed
matrix of order 3. Then,
circle is 10 cm. The length of the smallest
(a) must be A = O3 , B = O3 side is
(b) if A ≠ O3 , must be B ≠ O3
(a) 2 cm (b) 5 cm
(c) if A = O3 , must be B ≠ O3
(c) 7 cm (d) 10 cm
(d) may be A ≠ O3 , B ≠ O3
26. Let P and T be the subsets of k , y-plane 33. A variable line passes through a fixed point
(x1 , y1) and meets the axes at A and B. If the
defined by rectangle OAPB be completed, the locus of P
P = {(x , y) : x > 0, y > 0 and x 2 + y 2 = 1} is, (O being the origin of the system of axes).
x1 y
T = {(x , y) : x > 0, y > 0 and x 8 + y 8 < 1} (a) ( y − y1 )2 = 4( x − x1 ) (b) + 1 =1
x y
Then, P ∩ T is x2 y2
(a) the void set φ (b) P (c) x2 + y2 = x12 + y12 (d) 2 + 2 = 1
(c) T (d) P − T C 2 x1 y1
35. A variable line passes through the fixed point 42. P is the extremity of the latusrectum of
(α , β). The locus of the foot of the ellipse 3 x 2 + 4 y 2 = 48 in the first quadrant.
perpendicular from the origin on the line is The eccentric angle of P is
π 3π π 2π
(a) x2 + y2 − αx − βy = 0 (a) (b) (c) (d)
8 4 3 3
(b) x2 − y2 + 2αx + 2βy = 0
(c) αx + βy ± (α 2 + β 2 ) = 0 43. The direction ratios of the normal to the
x 2
y 2 plane passing through the points (1, 2 , − 3),
(d) + =1 x − 2 y +1 z
α 2
β 2 (− 1, − 2 , 1) and parallel to = = is
2 3 4
36. If the point of intersection of the lines (a) (2, 3, 4) (b) (14, − 8, − 1)
2 ax + 4 ay + c = 0 and 7 bx + 3 by − d = 0 lies in (c) (− 2, 0, − 3) (d) (1, − 2, − 3)
the 4 th quadrant and is equidistant from the 44. The equation of the plane, which bisects the
two axes, where a , b, c and d are non-zero line joining the points (1, 2, 3) and (3, 4, 5)
numbers, then ad : bc equals to at right angles is
(a) 2 : 3 (b) 2 : 1 (c) 1 : 1 (d) 3 : 2 (a) x + y + z = 0 (b) x + y − z = 9
(c) x + y + z = 9 (d) x + y − z + 9 = 0
37. A variable circle passes through the fixed
point A(p , q) and touches X -axis. The locus of 45. The limit of the interior angle of a regular
the other end of the diameter through A is polygon of n sides as n → ∞ is
(a) ( x − p)2 = 4qy (b) ( x − q )2 = 4 py π 3π 2π
(a) π (b) (c) (d)
(c) ( y − p) = 4qx2
(d) ( y − q ) = 4 px
2 3 2 3
46. Let f (x) > 0 for all x and f ′(x) exists for all x.
1 3
38. If P(0 , 0), Q(1, 0) and R , are three given If f is the inverse function of h and
2 2 1
h′ (x) = . Then, f ′(x) will be
points, then the centre of the circle for which 1 + log x
the lines PQ, QR and RP are the tangents is
(a) 1 + log(f( x)) (b) 1 + f( x)
1 3 1 1 (d) 1 , − 1
(a) ,
1 1 (c) 1 − log(f( x)) (d) log f( x)
(b) , (c) ,
2 4 2 4 2 2 3 2 3
47. Consider the function f (x) = cos x 2. Then,
x 2
y 2 (a) f is of period 2 π (b) f is of period 2 π
39. For the hyperbola − = 1, which (c) f is not periodic (d) f is of period π
cos α sin α 2 2
Category II (Q.Nos. 51 to 65) 56. For any non-zero complex number z, the
Carry 2 marks each and only one option is minimum value of| z | + | z − 1| is
correct. In case of incorrect answer or any 1 3
(a) 1 (b) (c) 0 (d)
combination of more than one answer, 1/2 mark 2 2
will be deducted.
57. The system of equations
51. Let a = min{ x 2 + 2 x + 3 : x ∈ R} and λ x + y + 3z = 0
1 − cos θ n
2 x + µy − z = 0
b = lim
θ→ 0 θ2
. Then ∑ a r bn − r is 5x + 7 y + z = 0
r=0
has infinitely many solutions in R. Then,
2n + 1 − 1 2n + 1 + 1
(a) (b) (a) λ = 2, µ = 3 (b) λ = 1, µ = 2
3⋅2n 3⋅2n
n+1 (c) λ = 1, µ = 3 (d) λ = 3, µ = 1
4 −1 1
(c) (d) (2 n − 1)
3⋅2n 2 58. Let f : X → Y and A , B are non-void subsets
52. Let a > b > 0 and I (n) = a1/ n − b1/ n, of Y , then (where the symbols have their
usual interpretation)
J (n) = (a − b) 1/ n
for all n ≥ 2, Then
(a) f − 1( A) − f − 1(B) ⊃ f − 1( A − B) but the opposite
(a) I(n) < J(n) (b) I(n) > J(n) does not hold.
(c) I(n) = J(n) (d) I(n) + J(n) = 0
(b) f − 1( A) − f − 1(B) ⊂ f − 1( A − B) but the opposite
$ γ$ be three unit vectors such that
53. Let α$ , β, does not hold.
1 (c) f − 1( A − B) = f − 1( A) − f − 1(B)
α$ × (β$ × γ$) = (β$ + γ$) where
2 (d) f − 1( A − B) = f − 1( A) ∪ f − 1(B)
α$ × (β$ × γ$) = (α$ ⋅ γ$)β$ − (α$ ⋅ β$)γ$ . If β$ is not parallel
59. Let S, T , U be three non-void sets and
to γ$ , then the angle between α$ and β$ is f : S → T , g : T → U be so that gof : s → U is
5π π π 2π surjective. Then,
(a) (b) (c) (d)
6 6 3 3 (a) g and f are both surjective
(b) g is surjective, f may not be so
54. The position vectors of the points A , B, C and
(c) f is surjective, g may not be so
D are 3 $i − 2 $j − k$ , 2 $i − 3 $j + 2 k$ , 5 $i − $j + 2 k$ and (d) f and g both may not be surjective
4 $i − $j − λk$ , respectively. If the points A , B, C π
60. The polar coordinate of a point P is 2, − .
and D lie on a plane, the value of λ is 4
(a) 0 (b) 1 (c) 2 (d) − 4 The polar coordinate of the point Q which is
55. A particle starts at the origin and moves such that line joining PQ is bisected
1 unit horizontally to the right and reaches perpendicularly by the initial line, is
1 π π π π
P1 , then it moves unit vertically up and (a) 2, (b) 2, (c) − 2, (d) − 2,
2 4 6 4 6
1
reaches P2 , then it moves unit horizontally
4 61. The length of conjugate axis of a hyperbola is
1 greater than the length of transverse axis.
to right and reaches P3 , then it moves unit Then, the eccentricity e is
8
vertically down and reaches P4 , then it moves 1
(a) = 2 (b) > 2 (c) < 2 (d) <
1 2
unit horizontally to right and reaches P5
16 x q
and so on. Let Pn = (x n , y n) and lim x n = α 62. The value of lim is
and lim y n = β. Then, (α , β) is n → ∞ x → 0+ p x
n→ ∞
[q ]
(b) , (c) , 1 (d) , 3
4 2 2 4 (a) (b) 0 (c) 1 (d) ∞
(a) (2, 3)
3 5 5 3 p
14 WB JEE (Engineering) Solved Paper 2019
63. Let f (x) = x 4 − 4 x 3 + 4 x 2 + c, c ∈ R. Then, 67. Two particles A and B move from rest along a
straight line with constant accelerations f
(a) f( x) has infinitely many zeros in (1, 2) for all c
and h, respectively. If A takes m seconds more
(b) f( x) has exactly one zero in (1, 2) if −1 < c < 0
than B and describes n units more than that
(c) f( x) has double zeros in (1, 2) if − 1 < c < 0
of B acquiring the same speed, then
(d) whatever be the value of c, f( x) has no zero in
(1, 2) (a) (f + h) m2 = fhn
(b) (f − fh)m2 = fhn
64. The graphs of the polynomial x 2 − 1 and cos x
1
intersect (c) (h − f )n = fhm2
2
(a) at exactly two points 1
(b) at exactly 3 points (d) (f + h) n = fhm2
2
(c) at least 4 but at finitely many points.
(d) at infinitely many points. 68. The area bounded by y = x + 1 and y = cos x
10 and the X -axis, is
65. A point is in motion along a hyperbola y = 3
x (a) 1 sq unit (b) sq unit
so that its abscissa x increases uniformly at a 2
rate of 1 unit per second. Then, the rate of 1 1
(c) sq unit (d) sq unit
change of its ordinate when the point passes 4 8
through (5, 2) 69. Let x1, x 2 be the roots of x 2 − 3 x + a = 0 and
1
(a) increases at the rate of unit per second x 3 , x 4 be the roots of x 2 − 12 x + b = 0 .
2
1 If x1 < x 2 < x 3 < x 4 and x1 , x 2 , x 3 , x 4 are in
(b) decreases at the rate of unit per second
2 GP, then ab equals
2
(c) decreases at the rate of unit per second 24
5 (a) (b) 64
5
2
(d) increases at the rate of unit per second (c) 16 (d) 8
5
1 − i cos θ
70. If θ ∈ ú and is real number, then
Category III (Q. Nos. 66 to 75) 1 + 2 i cos θ
Carry 2 marks each and one or more option(s) θ will be (when I : Set of integers)
π 3nπ
is/are correct. If all correct answers are not (a) (2 n + 1) , n ∈ I (b) , n∈I
marked and also no incorrect answer is marked 2 2
then score = 2 × number of correct answers (c) nπ, n ∈ I (d) 2nπ, n ∈ I
marked ÷ actual number of correct answers. If 3 0 3
any wrong option is marked or if any combination
71. Let A = 0 3 0 . Then, the roots of the
including a wrong option is marked, the answer 3 0 3
will considered wrong, but there is no negative
marking for the same and zero marks will be equation det(A − λI 3) = 0 (where I 3 is the
awarded. identity matrix of order 3) are
(a) 3, 0, 3 (b) 0, 3, 6
1
(c) 1, 0, − 6 (d) 3, 3, 6
66. Let I n = ∫ x tann −1
x dx . If a nI n + 2 + bnI n = cn
0 72. Straight lines x − y = 7 and x + 4 y = 2
for all n ≥ 1, then intersect at B. Points A and C are so chosen
(a) a1, a2 , a3 are in GP on these two lines such that AB = AC . The
(b) b1, b2 , b3 are in AP equation of line AC passing through (2 ,− 7) is
(c) c1, c 2 , c 3 are in HP
(a) x − y − 9 = 0 (b) 23 x + 7 y + 3 = 0
(d) a1, a2 , a3 are in AP
(c) 2 x − y − 11 = 0 (d) 7 x − 6 y − 56 = 0
WB JEE (Engineering) Solved Paper 2019 15
Answers
Physics
1. (b) 2. (b) 3. (d) 4. (d) 5. (d) 6. (b) 7. (b) 8. (c) 9. (d) 10. (c)
11. (c) 12. (a) 13. (c) 14. (c) 15. (c) 16. (b) 17. (a) 18. (d) 19. (a) 20. (a)
21. (d) 22. (d) 23. (*) 24. (c) 25. (c) 26. (d) 27. (a) 28. (d) 29. (c) 30. (c)
31. (a) 32. (a) 33. (a) 34. (a) 35. (c) 36. (b, d) 37. (b) 38. (a, d) 39. (a, c) 40. (b, d)
Chemistry
41. (c) 42. (a) 43. (d) 44. (b) 45. (a) 46. (b) 47. (c) 48. (b) 49. (d) 50. (b)
51. (c) 52. (d) 53. (c) 54. (c) 55. (c) 56. (a) 57. (b) 58. (d) 59. (a) 60. (b)
61. (d) 62. (b) 63. (a) 64. (b) 65. (d) 66. (b) 67. (d) 68. (d) 69. (d) 70. (b)
71. (d) 72. (d) 73. (a) 74. (d) 75. (a) 76. (a, b) 77. (a, b) 78. (b, c, d) 79. (a, c) 80. (d)
Mathematics
1. (b) 2. (b) 3. (d) 4. (b) 5. (a) 6. (d) 7. (d) 8. (c) 9. (c) 10. (d)
11. (b) 12. (b) 13. (b) 14. (b) 15. (c) 16. (b) 17. (d) 18. (a) 19. (b) 20. (d)
21. (b) 22. (c) 23. (c) 24. (a) 25. (d) 26. (a) 27. (d) 28. (d) 29. (c) 30. (d)
31. (a) 32. (d) 33. (b) 34. (b) 35. (a) 36. (b) 37. (a) 38. (c) 39. (c) 40. (c)
41. (a) 42. (c) 43. (b) 44. (c) 45. (a) 46. (a) 47. (c) 48. (c) 49. (a) 50. (b, c)
51. (c) 52. (a) 53. (d) 54. (d) 55. (b) 56. (a) 57. (c) 58. (c) 59. (b) 60. (a)
61. (b) 62. (a) 63. (b) 64. (a) 65. (c) 66. (b, d) 67. (c) 68. (b) 69. (b) 70. (a)
71. (b) 72. (a, b) 73. (a, c) 74. (a, c) 75. (d)
Answer with Explanations
Physics
1. (b) According to the question, 3. (d) As we know that de-Broglie wavelength is
h h
given as λ = = …(i)
p mv
n
e0 e where, h = Planck constant
θ θ p = momentum of particle
v = velocity of particle
and m = mass of the particle.
We know that incident ray, reflected ray and
normal lie in the same plane. Eq. (i) can be written as,
and angle of incidence = angle of reflection h h
λ= = [QKE = qv]
$ will be along the angle bisector of e$ and
Therefore n 2m(KE) 2mqv
− e$ 0 , where, KE = Kinetic energy of particle
$ = e + (− e 0)
$ $ 1
i.e. n …(i) Hence, λ ∝
|e$ − e$ 0 | m
[QBisector will along a vector dividing in same λ proton melectron
ratio as the ratio of sides forming that angle] Now, =
λ electron mproton
$ is a unit vector where|e$ − e$ 0 | = OC
But n
= 2OP = 2|e$ | Given, mass of proton, mproton = 2000 melectron
cosθ = 2cosθ λ proton 1
=
Substituting this value in Eq. (i), we get λ electron 2000
$ = e − e0
$ $
n λ proton 1 λe
2cosθ = ⇒ λp =
λ electron 20 5 20 5
e$ = e$ 0 + (2cosθ) n $
e$ = e$ 0 − 2(n$ ⋅ e$ 0)n $ ⋅ e$ 0 = − cosθ]
$ [Qn 4. (d) According to the Bohr’s atomic model,
nh
2. (b) According to the question, Angular momentum, L = mvr = …(i)
α-particle β-particle
2π
X → Y → Z(say) Since, angular velocity ω =
v
According to Rutherford Soddy law of radioactive r
decay, the rate of decay of radioactive along at any ⇒ v = rω
instant is proportional to the number of atoms From Eq. (i), we get
present at that instant nh
dN m(rω) r =
= − λN 2π
dt nh
Since, λ = constant (decay constant) and mωr 2 =
2π
N = number of atoms nh
Rate of decay of Y particle is given as, ω= …(ii)
2πmr 2
dN Y
λ X N X − λ Y NY = =0
dt Since, the radius of the electron in nth orbit of
[QDecay rate for β-particle become n2h2ε0
Bohr’s atomic model is given as, r = …(iii)
constant after some time] πmze 2
Given, rate of emission of β-particle = 107/h Squaring the Eq. (iii) and substituting its value in
Eq. (ii), we get
∴ λ X N X = λ Y N Y = 10 /h
7
nh(πmze)2 1
ω= ⇒ω ∝ 3
2πm(n4 h4 ε20) n
WB JEE (Engineering) Solved Paper 2019 17
5. (d) According to the question, 9. (d) According to the question, we can draw the
following diagram
1 kΩ
A
10 V 1 kΩ 6V
T T
B m1
a
m2 a
Input voltage VS = 10V m1g
Source resistance RS = 1kΩ
m 2g
Zener diode voltage VZ = 6V
Q Breakdown voltage of zener diode is = 6V, and the Here, T is the tension in the string.
potential difference across the zener diode 5V. In equilibrium condition,
∴Current flow in zener diode I Z = 0 For mass m1 , T − m1 g = m1 a …(i)
For mass m2, m2 g − T = m2a …(ii)
6. (b) By using the de-Morgan’s law
After adding Eqs. (i) and (ii), we get
A + B = A⋅ B
m2 g − m1 g = m1 a + m2a
Hence, option (b) is the correct. (m2 − m1) g = (m1 + m2)a
(m − m1)
7. (b) As we know that, a= 2 g
Impulse, I = force (F) × small time interval (m1 + m2)
I = ma × t [Q F = ma] 10. (c) Given, Power (P) = constant
I = [M] [LT− 2] × [T] 1
Kinetic Energy (KE) = mv2
I = [M L T− 1 ] 2
Hence, the correct dimensional formula for impulse KE mv2
We know that, P = ⇒P=
is given by [M L T− 1 ] . ∆t 2
Q P = constant,
8. (c) Given,
Hence, velocity of the body v ∝ t …(i)
Maximum error in the measurement of mass = 0.3%
ds
Maximum error in the measurement of length As, Velocity v = …(ii)
= 0.2% dt
We know that, From Eqs. (i) and (ii), we get
ds
Error in density is given as, So, ∝ t
mass (m) m dt
Density, ρ = = Integrating the above equation w.r.t. time ()
t,
volume (V) L3
ds
where, L = side of cube ∫ dt ∝ ∫ t
Error in density is given as,
we get, displacement of the body s ∝ t 3/ 2
∆ρ ∆m 3∆L
= + Q Displacement s = at 3/ 2, where a is constant.
ρ m L
∆ρ ∆m 3∆L 11. (c) According to the question,
or × 100 = + × 100
ρ m L (r2) v v(r1)
P
Substituting the given values, we get
∆ρ
= (0.3% + 3(0.2)%) = 0.3% + 0.6%
ρ max
O
∴ Maximum percentage error measurement of Representation of position vectors
of two particles (drawn from the point P)
∆ρ
density = 0.9%.
ρ max
18 WB JEE (Engineering) Solved Paper 2019
In two dimension, the position vectors r1 and r2 13. (c) Given, Decrement in the length = 1%.
represented as
Poisson’s ratio for material of the rod, σ = 0.2
1
r1 = vt $i − gt 2$j …(i) As we know that,
2
1 Volume, v = πr 2l ,[where, l is the length of the rod]
r2 = vt(− $i) − gt 2($j) …(ii)
∆V 2∆r ∆l
2 = + …(i)
Q We know that, when the two vectors are V r l
mutually perpendicular, i.e. − ∆D / D
Since, Poisson’s ratio, σ =
θ = 90º ∆L/ L
So, r1 ⋅ r2 = r1 r2 cos 90º − ∆r / r
=
r1 ⋅ r2 = 0 ∆L/ L
Substituting the values r1 and r2 in the above So, Eq. (i) can be written as,
relation, we get ∆V ∆l
1 1 = (1 − 2σ)
((vt)$i − gt 2$j) ⋅ (vt(− $i) − gt 2($j)) = 0 V l
2 2 ∆V ∆l
1 24 × 100 = × 100 × (1 − 2σ)
− v t + 4 g t = 0 (where, i ⋅ $i = $j ⋅ $j = k$ ⋅ k$ = 1)
2 2 $ V l
4 = − 1 × [1 − 2 × (0.2)]
1
v2t 2 = g2t 4 = − 1 × [1 − 0.4]
4
1 22 = − 0.6%
⇒ v = gt
2
Here, negative sign shows the decrement in the
4
1 volume.
∴Magnitude of velocity of the particles, v = gt
2 14. (c) When the spherical body falls with constant
We know that, separation distance between velocity, i.e. terminal velocity then the net force
particles at a time t becomes zero, i.e the weight of body is equal to
∆x = 2vt the buoyancy force.
2v 4v2 Hence, Fnet = 0
∆x = 2 × v × ⇒ ∆x =
g g 15. (c) According to the question,
Area of both bodies A and B = A
12. (a) According to the Kepler’s third law
Temperature of body A = 27ºC = 27 + 273 K
T2 ∝ r3
Temperature of body B = 177ºC = 177 + 273 K
where, T = time period of revolution
Now, by Stefan-Boltzmann law, thermal energy
r = radius
2 3 2/ 3
radiated per second by a body
T r r T Q = σAT 4 ,
Now, E = E = E = E
TP rP rP TP where, A = Area
2/ 3
rE 2π /ωE Q T = 2π T = temperature
=
rP 2π /ωP ω and σ = Stefan-Boltzmam’s constant
2/ 3 So, the ratio of thermal energy radiated per second
rE ωP by A to that by B is
=
rP ωE 4
Q1 σ A T1
According to the question, ωP = 2ωE and rE = R =
Q2 σA T2
2/ 3
R 2ωE 4
⇒ = Q1 T1
rP ωE Now, =
Q2 T2
R
= (2)2/ 3 4
rP Q1 273 + 27
=
R Q2 273 + 177
rP = = R(2)− 2/ 3
(2)2/ 3 Q1 300
4
=
Q2 450
WB JEE (Engineering) Solved Paper 2019 19
4
Q1 2
= =
16 Similarly, in branch CD,
Q2 3 81 CCD = 10 µF
Ratio of thermal energy radiated per second Now, C AB and CCD are connected in parallel
combination. Hence, the equivalent capacitance of
Q1 : Q2 = 16 : 81 the circuit,
16. (b) For a gas at temperature T, the internal energy, Ceq = 10 + 10 = 20 µF
f We know that, charge Q = CeqV
U= µRT
2 Q = 20 × 10− 6 × 150 V
where, f = degree of freedom Amount of charge stored Q = 3 × 10− 3C
f
⇒ Change in energy, ∆U = µR ∆T …(i)
2 19. (a) The clock diagram is as shown below
Also, as we know for any gas heat supplied at +Q
+Q +Q
constant volume 12
11 1
(∆QV) = µCV ∆T = ∆U …(ii) +Q
10 2
where, CV = molar specific heat at constant volume
r
From Eqs. (i) and (ii), we get +Q 9 3 +Q
1 C
CV = fR
2 8 4 +Q
+Q
For diatomic gas, degree of freedom, f = 5 7 6 5
5 +Q +Q
CV = R +Q
2
In the above diagram, charge + Q is not placed at
17. (a) Since, the whole system is isolated, this 10 h position.
means, there is no transfer of heat between the
So, net electric field strength at centre C,
system and the surrounding. Also, the right side
container is initially vacuum, so the gas could E net = E1 + E 2 + E 3 + E 4 + E 5 + E 6
easily rush there without any resistive force. As + E 7 + E 8 + E 9 + E11 + E12
the right side container has no temperature. Thus, ⇒ E net = E1 + E 2 + E 3 + E 4
the temperature of the ideal gas would remain + E 5 + E 6 + (− E1 − E 2 − E 3 − E 5 − E 6)
same even if it enters right side chamber. Q
⇒ E net = E 4 = , from centre towards the
Therefore, final temperature attained at the 4 πε0 r 2
equilibrium will be T. mark 10.
18. (d) The given circuit shows a balanced 20. (a) According to the question,
Wheatstone bridge. Now, the circuit becomes Q
20µF 20µF
C D r
d
θ θ E
x –q
A B
20µF 20µF d
Q
150 V Force experienced by the charge − q due to charge Q
In the branch AB, both capacitor are arranged in 2kQq 1
F = − 2 cosθ … (i) where, k =
the series combination. Hence, its equivalent r 4 πε 0
capacitance is given by x
1 1 1 2 1 From diagram, cosθ =
= + = = r
C AB 20 20 20 10
By substituting the value of cosθ in Eq. (i)
C AB = 10 µF
20 WB JEE (Engineering) Solved Paper 2019
F=−
2kQq x
⋅ Q Net magnetic field at the centre = 0
r2 r So, magnetic force Fmagnetic = q(v × B) = 0
2kQx
or F=− So, the magnetic force experienced by the particle
r3 when it passes through the centre is F = 0
2kQx Q r 2 = x 2 + d 2
or F=− 23. (c) According to the question,
(x + d 2)3/ 2
2
r = x 2 + d 2 P
For, x < < d, so x can be neglected
2
2kQx
F=− 3 3m 4m
d
So, the force developed by negative charge (− q) due
to the system of the charges as shown in the figure A B
is, 5m
− 4kQqx
F= µ0 × I
d3 Magnetic field due to first wire (A) is B1 =
2π × 3
⇒ F∝x
µ ×I
So, the forced developed by negative charge is Magnetic field due to second wire (B) is B2 = 0
directly proportional to the distance x. 2π × 4
21. (d) If velocity of particle, v is perpendicular to Net magnitude of magnetic field B = B12 + B22
magnetic field, B i.e. θ = 90º, then particle will µ 0I 1 1 µ 0I × 5
experience maximum magnetic force, i.e. B= + =
2π 9 16 2π × 3 × 4
Fmax = qvB. This force acts in a direction
5 µ 0I
perpendicular to the motion of charged particle. Magnetic field B = ×
Therefore the trajectory of the particle is a circle. 24 π
In this case path of charged particle is circular and 24. (c) As we know that, magnetic field due to a long
magnetic force provides the necessary centripetal wire at a distance x from it is given by
force,
µ I
mv2 B = 0 , where, I is the current flowing through
i.e. qvB = 2πx
r the wire β ∝ I
mv ∴Magnetic flux associated with the square loop,
⇒ Radius of path, r =
qB φ ∝β ∝ I
r=
p
[Q p = mv] Now, if the current increase, then φ also increases.
qB Direction of long wire will be q.
where, p = momentum of the particle This means, magnetic field due to induced
∴ r ∝ momentum current will be opposite to the existing magnetic
field, i.e. according to Lenz’s law,
Hence, option (d) is correct.
The induced current in the loop will be in the
22. (d) Given, electric current in circular wire = I anti-clockwise direction. Now,
Radius of wire = r
Charge on particle = q FAD
Speed of the particle = v
The given loop can be as shown in the figure below. D A
I/2 FAB
FCD
r1
I I
C B
FBC
r2
I/2
WB JEE (Engineering) Solved Paper 2019 21
Since, wires attract each other, if current flowing 5 7
=
through them is in same direction and repel each S S+ 3
other, if currents are in opposite direction.
5 (S + 3) = 7S
∴ Part CD of the loop will experience a force of
5 S + 15 = 7S
repulsion, whereas part AB will experience
attraction. Parts BC and AD will not experience any 2S = 15
force. Thus, the overall force will be a force of S = 7.5 Ω.
repulsion because AB is closer to straight. The force
between two current carrying conductors is 27. (a) Given, inductance of inductor, L = 60 mH
inversely proportional to the distance between = 60 × 10− 3 H
them
Phase difference between voltage and current in
1
F∝ L-R circuit, θ1 = 60º
r
Capacitance of capacitor, C = 0.5 µF = 0.5 × 10− 6F
Q r1 < r2
Phase difference between voltage and current in
So, FCD > FAB
R-C circuit, θ2 = 30º
Fnet = FCD − FAB
For L-R circuit,
Hence, the loop will moves away from the wire. X
tanθ1 = L
25. (c) According to the question, R
ωL
I1 R R I2 R R tanθ1 = …(i) [Q X L = ωL]
I=I1/8 R
I0 I1/2 I 1 /4 I 1 /8 I Similarly, for R-C circuit,
2R V 2R 2R 2R 2R
X
tanθ2 = C
R
Total resistance of the given circuit 1
…(ii) Q X C =
1
Req = 2R tanθ2 = ωC
R ωC
Now, circuit
From Eqs. (i) and (ii), we get
V
2R=I tanθ1 ωL
= = ω2LC
tanθ2 R × 1
I0
ωCR
2R V 2R tan 60º
= ω LC
2
tan 30º
B 3
= ω2LC
V / 2R 1/ 3
∴ I= (Q I1 = V / 2R)
8 ω2LC = 3
So, current I in the circuit I =
V Q I = I1 3
ω2 =
16R 8 LC
3
26. (d) According to the balanced condition of ω2 =
Wheatstone bridge, 60 × 10− 3 × 0.5 × 10− 6
In the first case, 3
ω2 =
P
=
R 30 × 10− 9
Q S
ω = 108
P 5
= …(i)
Q S ω = 104
In the second case As we know that,
P
=
7
…(ii) ω = 2πf
Q S+ 3 ω 104 1
f = = Hz = × 104 Hz
From Eqs. (i) and (ii), we get 2π 2π 2π
22 WB JEE (Engineering) Solved Paper 2019
28. (d) According to the question, we can draw the 30. (c) Given, initial frequency of light,
following diagram f1 = 4 × 1014 s− 1
f=0.05 m Final frequency of light, f2 = 5 × 1014 s− 1
Change in wavelength , ∆λ = λ1 − λ 2
c c
axis or ∆λ = −
0.2 m
f1 f2
3 × 108 3 × 108
= −
4 × 1014
5 × 1014
Convex lens
3 × 108 1 1
Given, u = − 0.2 m and f = 0.05 m = −
1014 4 5
As we know that,
3 × 108 1
1 1 1
= − …(i) = ×
f v u 1014 20
1 1 1 = 1.5 × 10− 7 m
= +
v f u Now, we know that,
1 1 1 Angular width of central maxima,
= −
v 0.05 0.2 2λ 2∆λ
θ= or ∆θ =
1 100 10 d d
= −
5 2 2∆λ
v d=
1 1 ∆θ
= 20 − 5 ⇒ v = m
v 15 Here, ∆θ = 0.6 radian
Now, differentiating eq. (i), we get 2 × 1.5 × 10− 7
d= = 5 × 10−7 m
dv du v2 0.6
− 2 = − 2 ∴ dv = du 2
v u u 31. (a) We know that, energy stored in the capacitor
2
2
1
A max = A × ×
1 1 2
=CE
15 (− 0.2) 2
1 and energy supplied by the source of emf
A max = A × × 25
225 = CE 2
A ∴ Energy dissipated in resistance R
A max =
9 = Energy supplied by the source of emf E
Here A is in cm. − Energy stored in the capacitor
A
Hence, A max = × 10− 2 m 1
= CE − CE 2
2
9 2
29. (c) As we know that, 1 1
= 1 − CE 2 = CE 2
λD 2 2
fringe width, β =
d
32. (a) Given, cross sectional area of nozzle
where, λ = wavelength of light 5
D = Distance of the screen from the slits A= × 10− 3 m2
21
d = separation between the slits 1
∆β ∆D ∆d and rate of heat transfer Q = = 10− 1 m3 / s
Now, × 100 = × 100 + × 100 10
β D d
Q Heat transfer Q = AV
According to the question, Q 10− 1 m3 / s
∆β ∴ V= = = 2 21 × 10 m/s
× 100 = 0.5 + (− 0.3) = 0.5 − 0.3 = 0.2% A 5
× 10− 3 m2
β 21
Hence, the fringe width increases by 0.2%.
WB JEE (Engineering) Solved Paper 2019 23
When it hits a rigid wall then maximum possible E
So, v = is the condition for which the particle
increase in temperature of water can be expressed B
as moves undeflected in its original trajectory.
1 2
mv = ms∆T …(i) 35. (c) In given, Series R - L - C circuit
2
Resistance R = 100 Ω
where, m = mass, s = specific heat of water
Inductance of Inductor, L = 20 mH = 20 × 10− 3H
= 1 cal / g = 4.2 × 103 J / kg
1250
and ∆T = increase in temperature Resonance frequency f = Hz
π
1 v2
From Eq. (i), ∆T = Source voltage VDC = 25V
2s
According to resonant frequency,
(2 21 × 101)2
= ω0 = 2πf0 =
1
2 × 4.2 × 103 LC
84 × 102 1
= or (2πf0) =
2
8.4 × 103 LC
= 1° C 2 1250 × 1250 1
or 4π =
π× π LC
33. (a) Maximum loss in K . E = Total K . E energy
1000
before collision of two ice blocks or C=
1250 × 1250 × 4 × 20
1 1
K . Emax loss = mv2 + mv2 = mv2
2 2 (QBy substituting L = 20 × 10− 3)
This maximum loss in kinetic energy will be equal or C = 8 × 10− 6F
to loss in heat to melt two ice block. We know that, Charge QS = CV
⇒ mv2 = (ms ∆ θ + mL)2 …(i) QS = 8 × 10− 6 × 25 = 0.2 mC
According to question, So, the amount of charge stored in each plate of
. × 105 J kg − 1
L = 336 capacitor is 0.2 mC.
S = 2100 J kg − 1 K − 1 36. (b,d) In photoelectric effect, if the incident light
∆θ = 8º C had a frequency less than a minimum frequency
From Eq. (i) v = 2(s ∆ θ + L) ν0 , then no electrons are ejected regardless of the
light’s amplitude. This minimum frequency is also
= 2(2100 × 8 + 336
. × 105) called the threshold frequency, and the value of ν0
depends on the metal. For frequency greater than
v = 840 ms −1
ν0 , electrons would be ejected from the metal.
Furthermore, the kinetic energy of the
34. (a) According to the question,
photoelectrons is proportional to the light
Charge on particle is = q frequency (ν).
Velocity of particle = v The relationship between photoelectron kinetic
Due to uniform electric field, energy T and light frequency ν is shown in graph
electric force on particle below
Felectric = qE …(i)
Due to a uniform magnetic field,
magnetic force on particle is given by, Light
kinetic
Fmagnetic = q(v × B) … (ii) energy
When v is perpendicular to E and B, which are (T)
mutually perpendicular to each other,
Felectric = Fmagnetic
v0
From Eqs. (i) and (ii) Light frequency (n) Hz
qE = qvB
Graph shows that kinetic energy T is linearly
E
∴ v= increasing with light frequency (ν).
B
24 WB JEE (Engineering) Solved Paper 2019
However, for a given photosensitive material and 39. (a, c) The circuit diagram is as shown below,
frequency of incident light, number of
photoelectrons emitted per second is directly r r r r r r
proportional to the intensity of incident light
i.e., Number of electrons emitted ∝ J. A B
Also, photoelectric emission is an instantaneous r r r
process.
According to question, when given pair of point in
37. (b) The equation of state for an ideal gas options are connected through a resistance R, then
undergoing adiabatic process is given as, equivalent resistance between point A and B (R AB)
TV γ −1 = constant remains unchanged. It is only possible when current
does not flows through resistance R and circuit become
Let the temperature after adiabatic compression Wheatstone bridge as shown in the figure below,
as given in the question be T then,
γ −1 C
T0 V γ − 1 = T 0
V
2 P Q
γ −1
⇒ T = T0 2
A G B
V0
Now, heat released at volume to achieve
2 R S
temperature T0 .
D
The net heat released can be determined by the
equation. P R
At the balanced condition, =
∆Q = µCV ∆T Q S
R
=µ × (T0 2 γ − 1 − T0) The current flow in CD branch will be zero.
γ −1 Now by checking each option,
µRT0 γ − 1 from option (a), circuit is,
= (2 − 1) (QµT0 R = p0 V0)
γ −1 2
p V
∴ Heat released = 0 0 (2 γ − 1 − 1) 2r 4r
γ −1
A R B
38. (a, d) Given, initial velocity of particle,
−1
u = 10 ms
r 2r
Range, R = 5 m
6
Gravitational acceleration,
g = 10 ms− 2 P R 2r r 1 1
Now,Q = ⇒ = ⇒ =
Q S 4r 2 r 2 2
As we known that,
Hence, option (a) is correct.
u2 sin 2θ
Range of projectile, R = From option (b), circuit is
g
3
(10)2 sin 2α
5=
10 3r 2r
5
sin 2α =
10 A R B
1
sin 2α = r 2r
2
α = 15º 6
or (90° − 15° = 75º) P R 3r r
Now, = ⇒ ≠
Q S 3r 2r
WB JEE (Engineering) Solved Paper 2019 25
So, option (b) is also incorrect. P R 4r r
Now, = ⇒ ≠
From option (c) circuit is, Q S 2r 2r
4 So, option (d) is also incorrect.
Hence, option (a) and (c) are correct.
4r 2r
40. (b, d) According to the question,
A R B B
r 2r
7 Metallic loop
P R 4r 2r 2 2
Now, = ⇒ = ⇒ =
Q S 2r r 1 1
So, option (c) is also correct. Given, A metallic loop is placed in uniform
From option (d), circuit is magnetic field B. Magnetic field B and metallic loop
4 are perpendicular to each other. Then, if metallic
loop is moved along B or rotated about its own axis,
4r 2r the net flux associated with it remains constant.
Thus, no emf will be induced in these cases.
A R B However, when the loop is squeezed to a smaller
area or rotated about one of its diameters. Then its
r 2r flux changes. Thus, emf is induced. So, option (b)
7 and (d) both are correct.
Chemistry
41. (c) 2, 2, 2-trichloroacetophenone reacts with 42. (a) When salicylic acid reacts with acetic anhydride
aqueous KOH to form potassium Salt of benzoic in the presence of conc.H 2SO 4 and heat, the product
acid, which undergoes hydrolysis to form benzoic will be (Q) acetyl salicylic acid (C9 H 8O 4).
acid as the final product. CO2H CO2H
acetic anhydride
O O
Conc.H2SO4 (cat) Heat
C C HO H3COCO
(i) aq KOH 'Q'
CCl3 OH
(C9 H8 O4)
(ii) H3O+
Mechanism In presence of acid anhydride,
2, 2, 2-trichloro Benzoic acid nucleophilic acyl substitution reaction takes place.
acetophenone P O O O
Mechanism C C C
OH H 3C O CH3
+
O O
Acetic anhydride
HO (C4 H6 O3)
C C
CCl3 OK Salicylic acid
+ aq KOH + CHCl3 (C7 H6 O3) Conc. H2SO4 ∆
H3O+ O
2, 2, 2-trichloro
C
acetophenone O OH O
C O +
OH C H
H3C O
C
'P'
Benzoic acid O CH3
Acetyl salicylic acid Acetic acid
The product of the above reaction P is benzoic acid. (C9 H8 O4) (C2 H4 O2)
Thus, option (c) is correct answer. Aspirin
So, the option (a) is correct answer.
26 WB JEE (Engineering) Solved Paper 2019
0.5 / 15
43. (d) When cyclopentanol on reaction with NaH ∴ pH = pK a + log ⇒ pH = pK a
followed by CS2 and CH 3I produces a xanthate. 0.5 / 15
OH The value of pK a is the lowest for the mixture of
NaH CH 3COOH and CH 3COONa (acidic buffer).
Xanthate
CS2/CH3I Therefore, it has lowest pH.
Cyclopentanol
Hence, pH value for option (c) is the lowest.
Mechanism 48. (b) Given, for two first order reactions.
–+ A → B, k = 0.693 min− 1 …(1)
OH ONa
NaH S=C=S and A → C, t1 / 2 = 0.693 min …(2)
For first order reaction
Cyclopentanol S S 0.693
–+
t1 / 2 = ,
O—C—SNa O—C—S— CH3 k
CH3-I
+ NaI Thus, for the reaction A → B
0.693 0.693
Xanthate t1 / 2 = = = 1.0 min
k 0.693
So, the option (d) is correct answer.
Also, for first order reaction, lower the t1 / 2 value,
44. (b) CH 3 C Cl hydrolyses to form CH 3COOH faster is the reaction.
1
Q t1 / 2 ∝
O k
even at 25ºC, which subsequently reacts with 0.693
NaHCO 3 (sodium bicarbonate) present in the and k for A → C is = 1 min− 1
0.693
same medium to form carbon dioxide (CO 2).
Thus, reaction (2) is faster than reaction (1) or
The reaction involved is given below. reaction (1) is slower than reaction (2).
CH 3COCl + NaHCO 3 → CH 3COONa + HCl + CO 2 Hence, option (b) is the correct answer.
Thus, the option (b) is correct answer.
51. (c) In the equilibrium H 2 + I 2 - 2HI, if at a In the crystalline solid MSO 4 ⋅ nH 2O, the value of
n is 5.
given temperature, the concentrations of the
reactants are increased, the value of the Hence, option (c) is correct answer.
equilibrium constant K C will remain the same 56. (a) Given, mass of gas (W) = 7.5 g
because equilibrium constant does not depend on
the molar concentration of reactants. Volume of gas at STP (V) = 56. L
V(L) W
The option (c) is correct answer. Q Moles of gas at STP = =
22 . 4(L) M
52. (d) Electrolysis of copper sulphate solution using
Cu-electrodes, Where M = molar mass of gas.
Ions present : Cu 2 + , H+ , SO 24 − , OH − W × 22 . 4
∴ M=
V
At cathode Cu 2 + (aq) + 2e − → Cu (s)
7.5 × 22.4
At anode Cu(s) − 2e − → Cu 2 + (aq) = = 30.00 g mol −1
56
.
So, the option (d) is correct answer.
Among the given options
53. (c) The three quantum numbers n, l and m that Molar mass(M) of
describe an orbital has integer values of 0, 1, 2, 3 (a) NO = 14 + 16 = 30.00 g mol −1
and so on.
(b) N 2O = 28 + 16 = 44.00 g mol −1
Name Symbol Range of Values (c) CO = 12 + 16 = 28.00 g mol −1
Principal quantum n 1≤ n (d) CO 2 = 12 + 32 = 44.00 g mol −1
number
Q Molar mass of NO = 30 g mol −1 .
Azimuthal quantum l 0≤ l ≤ n − 1
number Thus, the given gas is NO.
Magnetic quantum m − l ≤ m≤ l Hence, (a) is the correct option.
number 57. (b) Let, initial concentration (a) = 100
The electronic arrangement of option (c) is absurd Given, half-life (t1 / 2) = 60 days
because To find, radioactivity,
n= 2 i.e. (a − x) after time T (180 days)
l=0 Q T = n × t1 / 2
m= 0 where, n = no. of half-lives
To sum up, where n = 2, l = 0, m = 0 180 = n × 60
Hence, for l = 0, value of m should be zero. 180
n= =3
60
54. (c) The quantity hv / K B corresponds to a 100 100
and (a − x) = n = 3 =
temperature. hv = 3 / 2 K B T 2 2 8
where, T = temperature (a − x) = 12 .5%
K B = Boltzmann constant Hence, (b) is the correct answer.
The percentage of H 2O in the solid is
58. (d) Given, 82 A
210
→ B → C → 82D
206
= (100 − 64) = 36%
hv 3 i.e. difference in mass no. between A and D is of 4
= T
KB 2 units i.e., mass no. is decreased by 4 units while
atomic no. remains the same, i.e. 82 for A and D.
3 hv
Since, is a constant, the value of corresponds Also,
2 KB
(i) On emission of each β-particle, mass no.
to temperature. Therefore option (c) is correct.
remains the same but atomic no. is increased
55. (c) Mass of H 2O = 36 × 250 = 90 g by one unit.
100 (ii) On emission of each α-particle, mass no. is
90
Moles of H 2O = = 5mol decreased by 4 units, while atomic no. is
18 decreased by 2 units.
28 WB JEE (Engineering) Solved Paper 2019
Thus, on emission of 2-β and 1-α particle, (ii) Smaller the size, more is value of electron affinity, i.e.
we get 82 D 206, more (−) ve will be the electron gain enthalpy.
−β −β −α On moving across a period, the size of atoms
i.e. 82 A
210
→ 83 A210 → 84 A210 → 82 A206 decreases due to increase in nuclear charge.
Thus, option (d) is the correct answer. (i) Electron configuration of the elements are
shown below.
59. (a) Key Point For similar electronic configuration of
C(Z = 6) = 1s 2 2s 2 2p 2
outer most shell, size will decide the value of ionisation
energy. More the size, lesser is the value of ionisation N(Z = 7) = 1s 2 2s 2 2p 3
energy. O(Z = 8) = 1s 2 2s 2 2p 4
Electronic configuration IInd I.E (ii) Due to half-filled electronic configuration of
Zn+ ion (Z = 30) = [Ar] 3d 10 4s1, (1734 kJ/mol). nitrogen (i.e. 3 electrons in 3p orbitals), it is
highly stable and has (+) ve value of electron
Cd + ion (Z = 48) = [Kr] 4d 10 5 s1, (1631 kJ/mol) gain enthalpy (i.e. about 30.9 kJ/mol).
+
Hg ion(Z = 80) = [Xe] 4f ⋅ 5d 14 10 1
6s (1809 kJ/mol) (iii) As size of ‘O’, is smaller than that of ‘C’, ‘O’
atom has higher (−) ve electron gain enthalpy
Q Size of Cd+ > Zn+ , thus it has lower IInd (about − 141.1 kJ / mol) than that of carbon
ionisation energy while due to lanthanoid effect (C-atom) [about − 122.3 kJ / mol].
(i.e., poor screening by 4 f and 5d electrons, Hg + has Hence, correct order is O > C > N and (b) is
higher IInd ionisation-energy. the correct option.
Hence, correct order is,
Zn > Cd < Hg and option (a) is the correct answer.
63. (a) Key point As the percentage of s character in a
bond increases, the bond angle also increases. Therefore,
60. (b) Key Point A compound having more ionic as the bond angle increases, the percentage of
character, has more melting point. p-character decreases.
Be and Ca belong to the same group and ionic In the given options,
character increases down the group. Ammonia (NH 3) has bond angle = 107.6º
Thus, BeCl 2 is less ionic than CaCl 2. This means and Phosphine (PH 3) has bond angle = 93. 5º
that melting point of CaCl 2 is higher than that of 1
BeCl 2. Q Bond angle ∝ and ∝ s-character
p - character
Also, Hg is a transition metal, its compound is less
ionic than BeCl 2. Therefore, order of melting point Bond angle of NH 3 > PH 3 due to lone-pair lone-pair
will be repulsion.
CaCl 2 > BeCl 2 > HgCl 2 Thus, the lone pair on NH 3 has less p-character.
or (ii) > (i) > (iii) Hence, option (a) is the correct answer.
Hence, option (b) is the correct answer. 64. (b) Chlorine bleach is NaOCl. The hypochlorite
ion in chlorine bleach dissociates to give nascent
61. (d) Key point The species which has one or more oxygen as shown below.
unpaired electrons have non-zero magnetic moment.
OCl − → [O] + Cl −
Electronic configuration of
Therefore, the reactive species in chlorine bleach is
Na+(Z = 11) = 1s 2 2s 2 2p 6
OCl − .
Mg (Z = 12) = 1s 2 2s 2 2p 6 3s 2 So, the option (b) is correct.
F − (Z = 9) = 1s 2 2s 2 2p 6 65. (d) Co-ordinate compound of Co (III) dissociates
Ar + (Z = 18) = 1s 2 2s 2 2p 6 3s 2 3p 5 into 3 ions in the solution means it has two
ionisable-ions out side the coordination sphere
As, Ar + has an unpaired electron, it has non-zero with oxidation state of cobalt (Co) = + 3
magnetic moment. Hence, (d) is the correct option.
In option (a)
62. (b) Key Point Name : Hexaammine cobalt (III) chloride
(i) Half-filled/fully-filled configuration have (−) ve or Formula : [Co(NH 3)6]Cl 3
low electron affinity i.e., (+) ve or high electron gain Q Oxidation state of Co = (+) 3
enthalpy.
WB JEE (Engineering) Solved Paper 2019 29
In option (b) anti-conformations can be interconverted by
Name : Pentaammine sulphatocobalt (III) chloride rotation around C2-C3 linkage.
Formula : [Co(NH 3)5(SO 4)]Cl CH3 CH3 CH3
CH3
Q Oxidation state of Co = + 3 H H H CH3
In option (c)
Name : Pentaamminechloridocobalt (III) sulphate H H H H H H
Formula : [Co(NH 3)5Cl]SO 4 H H
CH3 H
Q Oxidation state of Co = + 3
Anti- Gauche- Eclipsed-
In option (d) conformer conformer conformer
Name : Pentaamminechloridocobalt (III) chloride
Hence, the option (d) is correct answer.
Formula : [Co(NH 3)5Cl]Cl 2
QOxidation state of Co = + 3 70. (b) In halogen acids, as the size of halogen atom
increases, the bond between halogen and
Dissociation of this compound is shown below:
2+
hydrogen atom weakens. Therefore, the case with
[CO(NH 3)5Cl]Cl 2 → CO(NH 3)5Cl] + 2Cl − which the bond can be broken increases.
144424443
3 ions The correct order of the addition reaction rates of
This compound can give total 3-ions, as follows. halogen acids with ethylene is hydrogen iodide
(HI) > Hydrogen bromide (HBr)
one [Co(NH 3)5Cl]2 + and two Cl − ions.
> Hydrogen chloride (HCl)
Hence, option (d) is the correct answer. So, the option (b) is correct answer.
66. (b) In the Bayer’s process, the leaching of
alumina is done by using NaOH. 71. (d) Total number of isomeric linear dipeptides
which can be synthesised from racemic mixture of
Al 2O 3(s) + 2NaOH(aq)
→
150º C, 35 atm
alanine are four (4), by (R) type and (S) type.
Bauxite ore
2Na[Al(OH)4 ] +3H 2O (Q Has two chiral centres)
Sodium i.e.
aluminate
(soluble in water) O
CH3 H
2Na[Al(OH)4 ] + CO 2 → Al 2O 3 ⋅ xH 2O + 2NaHCO 3 H 2N COOH C
Al 2O 3 ⋅ xH 2O(s) 1470
K
→ Al 2O 3(s) + xH 2O(g) H 2N COOH
Pure alumina H 3C H NH
So, the option (b) is correct answer. CH3 H
74. (d) A copper coin was electroplated with zinc Thus, options (a, b) are correct.
(Zn) and then heated at high temperature until
there is a change in colour. The resulting colour 77. (a, b)
will be golden. (a) In CrO 5 oxidation no. of Cr is + 6.
The golden colour is due to the zinc migrating
–2
through the copper to form the alpha-form of brass –1O O O –1
alloy (percentage of Cu > 65% and that of Zn <
35%). Cr
∆
Zn + Cu → Brass –1O O –1
Thus, the option (d) is correct answer. (b) For the reaction (in ideal case)
75. (a) N 2 O4 (g) → 2NO2 (g)
OH Peracid
O
OH ∆H = ∆U + p. ∆V
Q ∆H = ∆U + ∆n g RT
RCO3H
Allyl
[O]
(C3H6O2) (Q p∆V = ∆n g RT)
alcohol
Where, ∆n g = (gaseous moles of product)
Oxidation of allyl alcohol with a peracid gives a
compound of molecular formula C3H 6O 2, which − (gaseous moles of reactant)
contains an asymmetric carbon atom. ∆n g = 2 − 1 = 1
The structure of the compound is Thus, ∆H > ∆U
O (c) pH of 0.1 N H 2SO 4 is less than of 0.1 HCl at
OH
25º is a wrong statement.
pH of H 2SO 4
So, the option (a) is correct answer.
Q pH = − log[H + ] = − log[10− 2]
76. (a, b) Haloform reaction with I 2 and KOH
Q (N = C × Z) and (Z = 2for H 2SO 4 )
(a)
O O pH = 2and pH of HCl
I Ph I /KOH
2 I
Ph pH = − log[10− 1 ] = 1
α CHI3
I Thus, pH of H 2SO 4 > pH of HCl.
I
O RT
(d) = 0.0591 at 25ºC is a wrong statement.
+ Ph—CH2—C—ONa F
Where, R = Gas constant = 8.314
T = Temperature = 273 + 25 = 298 K
F = Charge over one mole of electrons
= 96500C
WB JEE (Engineering) Solved Paper 2019 31
RT 8.314 × 298 • No reaction with water
Thus, = = 0.02567
F 96500 Thus, NH 4NO 3 can be used to label all three
2.303 RT beakers.
The correct value is = 0.0591
F (c) (NH 4)2CO 3 will evolve pungent smelling gas
Hence, only option (a) and (b) are correct. NH 3 with NaOH.
• Effervescence of CO 2 gas with con.H 2SO 4
78. (b,c,d) Among the given species
• No reaction with water.
(a) H 2O and F are weak field ligands and do not
cause pairing of electrons in the central metal Thus, (NH 4)2CO 3 can be used to label all three
atom. As CN is a strong field ligand, it causes beakers.
pairing of electrons in the central metal atom. Thus, options (a, c) are correct.
Now,
80. (d) O
Oxidation state of :
Me Me
(b) Fe in [Fe(H 2O)6] Cl 3 = (+) 3and electronic Et
CO2COOH Et CH2—C— OH
∆ (100ºC) ∗
configuration of Fe3 + ions is 3d 5 4s 0 O
O
Fe3 + =
–CO2 ∗
H—O—C C—OH CH (Me) COOH
3d5 4s0 Me Chiral
(b)
i.e. have 5 unpaired electrons
Me Me
(c) Fe in K 3 [FeF6] Et CH2COOH Et CH2COOH
O ∆ (100ºC) ∗
Oxidation no. of Fe = (+) 3
–CO2
electronic configuration of Fe3 + is 3d 5 4s 0 HO—C CO2H CH2(COOH)
3+ 3d5 4s0 Chiral
Fe =
(c)
Me Me
i.e. have 5 unpaired electrons. Et CO2H Et H
∆ (100ºC) ∗
(d) Oxidation state of Mn in K 4 [MnF6] = (+) 2and
electronic configuration of Mn2 + is 3d 5 4s 0 Me
–CO2
Me
2+ O O
Mn =
Chiral
3d5 4s0
(d)
H H
Et CO2H Et H
i.e., have 5 unpaired electrons. ∆ (100ºC) ∗
Hence, options (b) (c) and (d) are the correct –CO2
options. O Me O Me
⇒ t2
dt
= a2 + t2 ⇒ 3a 2 − b 2 = 0
dx
⇒ 3a 2 = b 2
t2
⇒ dt = dx [variable separation] ∴ Eccentricity of the hyperbola is
a + t2
2
a2 + b2
t2 e=
⇒ ∫ (t 2 + a 2) dt = ∫ dx [integration] a
a + 3a 2
2
4a 2 2a
t2 + a2 − a2 = = = =2
⇒ ∫ (t 2 + a 2) dt = x + C′ a a a
[doing a 2 adding and subtracting] 12. (b) Let V be the volume of spherical ballon of
radius r.
a2
⇒ ∫ 1 − t 2 + a 2 dt = x + C′ Then, V =
4 3
πr
3
4π
a2 log V = log
1 dV 3
⇒ + 3log r ⇒ =0+
⇒ ∫ dt − ∫ t 2 + a 2 dt = x + C′ 3 V dr r
dt 1 dV 3 1 ∆V 3
⇒ t − a2∫ = x + C′ ⇒ = ⇒ =
t2 + a2 V dr r V ∆r r
∆V ∆r
1 t ⇒ =3
⇒ t − a2 tan− 1 = x + C ′ V r
a a
∆V ∆r
x + y ⇒ × 100 = 3 × 100
⇒ (x + y) − a tan− 1 V r
a ∆V
⇒ × 100 = 3 × 01 . = 0.3
= x + C ′[on putting the value of t] V
y − C′ x + y
⇒ = tan− 1
∴ Percentage increase in volume is 0.3%.
a a
13. (b) Let ∆ABC be a right angled triangle at B. Let
x+ y y − C′
⇒ = tan ∠A and ∠C be α and β
a a
A
x+ y y + C
⇒ = tan
a a α
[where, C = − C ′ is an arbitrary constant] a
ar2
y + C x + y
⇒ tan =
a a
β
B ar C
WB JEE (Engineering) Solved Paper 2019 35
Since, sides are in GP so sides are a, ar, ar 2 15. (c) (Im)
Now, AC 2 = AB2 + BC 2
(z)
⇒ (ar 2)2 = a 2 + (a ⋅ r)2
⇒ a 2r 4 = a 2 + a 2r 2 θ
(Re)
⇒ r − r −1 = 0
4 2
–π+θ
1+ 5
⇒ r2 = [r 2 > 0] (–z)
2
5+1
⇒ r= ∴ arg(z) − arg(− z) = θ − (− π + θ)
2
BC ar = π
∴ tanα = =
AB a 16. (b) We have,
5+1 (cosθ + i sinθ) (cos 2θ + i sin 2θ) ...
⇒ r=
2 (cos nθ + i sin nθ) = 1
and tanβ =
AB
=
a 1
= ⇒ e iθ ⋅ e i( 2θ) ⋅ e i( 3θ) ⋅ ... e i( nθ) = 1
BC ar r
⇒ e iθ(1 + 2 + 3 + K + n)
=1
1 5 −1 i n( n + 1) θ
= =
5+1 2 ⇒ e 2 =1
n(n + 1) n(n + 1)
⇒ cos θ + i sin
2
θ = 1 + 0i
2 2
14. (b) We have,
n(n + 1)
1 1 ⇒ cos θ = 1
log 62 + = log 2 2x + 8 2
2x
n(n + 1)
⇒ θ = 2kπ
1 1 2
⇒ 1 + log 32 + = log 2 2x + 8 4k
2x ⇒ θ= π
n(n + 1)
1
1
2x + 8 1+ 17. (d) Let f (x) = ax 2 + bx + c
⇒ =2 2x
3 ⇒ 1 =a+ b+ c<0
f ()
1 1 Again, f (x) has imaginary zeros. So, a < 0.
⇒ 2x + 8= 3⋅ 2⋅ 22x Also, f (0) = c. Since f (x) is downward parabola. So,
1 1 c < 0.
⇒ 2x + 8 = 6 ⋅ 22x Y
1
Let y= 22x
⇒ y2 + 8 = 6 y
X′ X
⇒ y2 − 6 y + 8 = 0
⇒ (y − 4) (y − 2) = 0 f (x)
⇒ y = 4, 2
1 1 Y′
⇒ 22x = 4 and 22x = 2
1 1 18. (a)
⇒ = 2 and =1 A B
2x 2x
1 1 4 2
⇒ x= ,
4 2 3 3
2 4
36 WB JEE (Engineering) Solved Paper 2019
∴ Total number of ways 1 1 1 3 0 0
= 6C4 × 6C2 + 6C3 × 6C3 + 6C2 × 6C4 = 1 1 1 − 0 3 0
6 × 5 6 × 5 6 × 5× 4 6 × 5× 4 1 1 1 0 0 3
= × + ×
2×1 2×1 3× 2×1 3× 2×1 − 2 1
1
+
6×5 6×5
× =1 −2 1
2×1 2×1 1 1 − 2
= 15 × 15 + 20 × 20 + 15 × 15 = [− 2(4 − 1) − 1(− 2 − 1) + 11
( + 2)]
= 225 + 400 + 225 = [− 2(3) − 1(− 3) + 1(3)]
= 850 = [− 6 + 3 + 3] = 0
19. (b) Required number of ways Now, det = 0
= 7C4 × D(3) So, matrix A − 3I 3 is non-invertible matrix.
1 1
= 7C3 × 3!1 − + 23. (c) Since, adj (M ′) = (adj M)′
1
−
1! 2! 3! = adj (M ′) − (adj M)′
Q n Cr = n Cn − r , and [Q adj (A′) = (adj A)′]
D(n) = n!1 − 1 + 1 − 1 + K = 0, a null matrix.
1! 2! 3!
24. (a) Given,
= 7C3 × 2 = 2(7 C3) 5 5x x
A = 0 x 5x
20. (d) We have,
0 0 5
72n + 16n − 1
∴ A2 = A ⋅ A
Put n = 1, we get
72 + 16 − 1 5 5x x 5 5x x
= 0 x 5x 0 x 5x
= 49 + 15
= 64 0 0 5 0 0 5
which is divisible by 64. 25 25x + 5x 2 5x + 25x 2 + 5x
1 1
84 =0 x2 5x 2 + 25
21. (b) Given, 38 + 54 0 0 25
84 − r r = 25(25x 2 − 0)
1 1
here, Tr + 1 = 84 Cr 38 54 = 25(25x 2)
Since, given that| A |2 = 25
84 − r r
Now, 25(25x 2) = 25
= 84
Cr 3 8 ⋅ 54
⇒ 25x 2 = 1
Since, Tr + 1 is rational for r = 4, 12, 20, 28, 36, 44, 1
⇒ x2 =
52, 60, 68, 76, 84 25
∴ no. of rational terms = 11 1
⇒ x=±
Here, total number of terms = 85 5
∴ Number of irrational terms = 85 − 11 1
⇒ x=
= 74 5
22. (c) Given, that A − 3I 3 25. (d) Since, product of two non-null matrix can be
a null matrix.
1 1 1 1 0 0 Therefore, may be
= 1 1 1 − 3 0 1 0 A ≠ O3, B ≠ O3.
1 1 1 0 0 1
WB JEE (Engineering) Solved Paper 2019 37
O (a, 0)
y − x 3 + 2+ 3 3 = 0
A
x1 y 35. (a) Let (α , β) be the given point, let Q(x , y) be the
∴ + 1 =1 foot of the perpendicular, and let O be the origin.
a b
The line can have any direction
Since, OAPB is a rectangle, therefore the coordinate
of P will be (a , b). ∠PQO = 90º
Hence, locus of P is Point Q lies on the circle having diameter OP.
x1 y The locus of point Q.
+ 1 =1
x y (x − 0) (x − α) + (y − 0) (y − β) = 0
⇒ x 2 − xα + y 2 − yβ = 0
34. (b) Given line,
⇒ x 2 + y 2 − αx − βy = 0
3x + y = 1
⇒ y=− 3x + c 36. (b) Let coordinate of the point be (α , − α)
We know that, m = − 3 [Q y = mx + c] Since, (α , − α) lie on 2ax + 4ay + c = 0
and 7bx + 3by − d = 0
m1 − m2 ∴ 2aα − 4aα + c = 0
tanθ =
1 + m1 m2 ⇒ − 2aα + c = 0
m1 − (− 3) ⇒ α=
c
…(i)
tan 60° = [Qθ = 60º]
1 + m1 (− 3) 2a
Also, 7bα − 3bα − d = 0
m1 + 3
3= m1 [tan 60º = 3] ⇒ 4bα − d = 0
1− 3 d
⇒ α= …(ii)
m1 + 3 4b
± 3=
1 − 3m1 From Eqs. (i) and (ii),
c d
=
taking + sign taking − sign 2a 4b
m1 + 3 m1 + 3 ⇒ 2ad = 4bc
+ 3= − 3= ad 4
1 − 3m1 1 − 3m1 ⇒ =
bc 2
3(1 − 3m1 ) = m1 + 3 ⇒− 3(1 − 3m1 ) ad 2
⇒ =
= m1 + 3 bc 1
⇒ 3 − 3m1 = m1 + 3 ⇒− 3 − 3m1 = m1 + 3 ⇒ ad : bc = 2 : 1
c b T S
60. (a) Since, Q is the point for which PQ is bisected 64. (a)
Y
perpendicularly by the initial line (X-axis).
Therefore, Q will be the image of P in X-axis x2 – 1 cos x
∴ Coordinates of Q is (2, π / 4)
Y X′ O X
Q (2, π/4)
Y′
X′ X It is clear from the graph that it intersects at exactly
two points.
n 1
π 1 x 1 (0, 1)
= − +
4(n + 1) n + 1 n 0 n + 1
[x n − 1 ⋅ tan− 1 x]1 − 1 tan− 1 x ⋅ (n − 1) x n − 2 dx (π/2, 0)
0 ∫0 (–1, 0)
π 1 π n −1
= − + −
4(n + 1) n(n + 1) 4(n + 1) n + 1
1 n−2 π /2
∫0 x tan− 1 x dx
0
∴ Require area = ∫− 1 (x + 1) dx + ∫0 cos x dx
π 1 n −1
⇒ In = − − In − 2 x2
0
2(n + 1) n(n + 1) n + 2 = + x + [sin x]π0 / 2
2 − 1
n −1 π 1
⇒ In + In − 2 = −
n+ 2 2(n + 1) n(n + 1) = (0) − − 1 + [1 − 0]
1
2
Put n = n + 2, we get
n+1 π 1 3
I n+ 2 + In = −
1 = + 1 = sq unit.
n+ 3 2(n + 3) (n + 2) (n + 3) 2 2
π 1 69. (b) We have,
⇒ (n + 3) I n + 2 + (n + 1) I n = −
2 n+ 2 x1 , x 2 be the roots of equation x 2 − 3x + a = 0
π 1 ∴ x1 + x 2 = 3 and x1 x 2 = a
∴ a n = n + 3, bn = n + 1, c n = −
2 n+ 2 Also, x 3, x 4 be the roots of equation
∴ a n and bn are in AP. x 2 − 12x + b = 0
∴ x 3 + x 4 = 12 and x 3 x 4 = b
67. (c) Let B travels x units, v = u + at
Again, x1 , x 2, x 3, x 4 are in GP
According to problem, ht = f (t + m)
∴ x1 = A, x 2 = AR, x 3 = AR2, x 4 = AR3
⇒ ht = ft + fm
Now, x1 + x 2 = 3
⇒ ht − ft = fm
⇒ A(1 + R) = 3 …(i)
⇒ t(h − f) = fm
h− f m and x 3 + x 4 = 12
⇒ = ⇒ AR2(1 + R) = 12 …(ii)
f t
f From Eqs. (i) and (ii), we have
⇒ t = m R2 = 4 ⇒ R = ± 2
h− f
When, R = 2, A = 1 and when, R = − 2, A = − 3
WB JEE (Engineering) Solved Paper 2019 45
∴ Numbers are either 1, 2, 4, 8 or − 3, 6, − 12, 24
But x1 < x 2 < x 3 < x 4 . 3
So, x1 = 1, x 2 = 2 , x 3 = 4, x 4 = 8 2
x –y=1
∴ ab = x1 x 2 x 3 x 4 1
1/2
= 1 × 2 × 4 × 8 = 64 1 2 3 4 5 6 7 8
6. If Young’s double slit experiment is done 12. In case of a simple harmonic motion, if the
with white light, which of the following velocity is plotted along the X -axis and the
statements will be true? displacement (from the equilibrium position)
(a) All the bright fringes will be coloured. is plotted along the Y -axis, the resultant
(b) All the bright fringes will be white. curve happens to be an ellipse with the ratio:
(c) The central fringe will be white. major axis (along X )
(d) No stable interference pattern will be visible. = 20 π
minor axis (along Y)
7. How the linear velocity v of an electron in the What is the frequency of the simple harmonic
Bohr orbit is related to its quantum number n? motion?
1 1 1
(a) v ∝ (b) v ∝ (a) 100 Hz (b) 20 Hz (c) 10 Hz (d) Hz
n n 2 10
1
(c) v ∝ (d) v ∝ n 13. A block of mass m2 is placed on a horizontal
n
table and another block of mass m1 is placed
8. If the half-life of a radioactive nucleus is on top of it. An increasing horizontal force
3 days, nearly what fraction of the initial F = αt is exerted on the upper block but the
number of nuclei will decay on the third day? lower block never moves as a result. If the
(Given, 3 0.25 ≈ 0 .63) coefficient of friction between the blocks is µ 1
and that between the lower block and the
(a) 0.63 (b) 0.5 (c) 0.37 (d) 0.13 table is µ 2 , then what is the maximum
9. An electron accelerated through a potential possible value of µ 1 / µ 2 ?
m2 m2 m1 m1
of 10000 V from rest has a de-Broglie wave (a) (b) 1 + (c) (d) 1 +
length λ. What should be the accelerating m1 m1 m2 m2
potential, so that the wavelength is doubled?
14. In a triangle ABC, the sides AB and AC are
(a) 20000 V (b) 40000 V (c) 5000 V (d) 2500 V
represented by the vectors 3 $i + $j + k$ and
10. In the circuit shown, inputs A and B are in $i + 2 $j + k$ , respectively. Calculate the angle
states 1 and 0 respectively. What is the only
∠ ABC.
possible stable state of the outputs X and Y ?
5 6
(a) cos −1 (b) cos −1
A 11 11
(1) X
5 5
(c) 90° − cos −1 (d) 180° − cos −1
11 11
(0)
Y
B 15. The velocity (v) of a particle (under a force F)
(a) X = 1, Y = 1 (b) X = 1, Y = 0 depends on its distance (x) from the origin
(c) X = 0, Y = 1 (d) X = 0, Y = 0 1
(with x > 0) v ∝ . Find how the magnitude
11. What will be the current flowing through the x
6 kΩ resistor in the circuit shown, where the of the force (F) on the particle depends on x?
breakdown voltage of the Zener is 6 V? 1 1 1
(a) F ∝ (b) F ∝ (c) F ∝ 2
(d) F ∝ x
x3 / 2 x x
6 kΩ
16. The ratio of accelerations due to gravity g1 : g2
10V on the surfaces of two planets is 5 : 2 and the
4 kΩ
ratio of their respective average densities
ρ1 : ρ2 is 2 : 1. What is the ratio of respective
escape velocities v1 : v2 from the surface of the
2 3
(a) mA (b) 1mA (c) 10 mA (d) mA planets?
3 2
(a) 5 : 2 (b) 5 : 2 (c) 5 : 2 2 (d) 25 : 4
WB JEE (Engineering) Solved Paper 2018 3
17. A spherical liquid drop is placed on a 22. For an ideal gas with initial pressure and
horizontal plane. A small disturbance causes volume p i and Vi respectively, a reversible
the volume of the drop to oscillate. The time isothermal expansion happens, when its
period of oscillation (T) of the liquid drop volume becomes V0 . Then, it is compressed to
depends on radius (r) of the drop, density (ρ) its original volume Vi by a reversible
and surface tension (S) of the liquid. Which adiabatic process. If the final pressure is p f ,
among the following will be a possible then which of the following statement(s)
expression for T (where, k is a dimensionless is/are true?
constant)? (a) pf = pi (b) pf > pi
ρr ρ2 r ρr 3 ρr 3 p p
(a) k (b) k (c) k (d) k (c) pf < pi (d) f = i
S S S S2 V0 Vi
18. The stress along the length of a rod (with 23. A point charge − q is carried from a point A to
rectangular cross-section) is 1% of the another point B on the axis of a charged ring
Young’s modulus of its material. What is the of radius r carrying a charge + q. If the point
approximate percentage of change of its 4
volume? (Poisson’s ratio of the material of A is at a distance r from the centre of the
3
the rod is 0.3.) 3
(a) 3% (b) 1% (c) 0.7% (d) 0.4% ring and the point B is r from the centre but
4
19. What will be the approximate terminal velocity on the opposite side, what is the net work
of a rain drop of diameter 1.8 × 10 −3 m, when that need to be done for this?
density of rain water ≈ 10 3 kgm −3 and the 7 q2 1 q2
(a) − ⋅ (b) − ⋅
5 4 πε0 r 5 4 πε0 r
coefficient of viscosity of air ≈ 1.8 × 10 −5 N-sm
−2 7 q2 1 q2
? (Neglect buoyancy of air) (c) ⋅ (d) ⋅
5 4 πε0 r 5 4 πε0 r
(a) 49 ms −1 (b) 98 ms −1 (c) 392 ms −1 (d) 980 ms −1
20. The water equivalent of a calorimeter is 10 g 24. Consider a region in free space bounded by
and it contains 50 g of water at 15°C. Some the surfaces of an imaginary cube having
amount of ice, initially at − 10 °C is dropped sides of length a as shown in the figure. A
in it and half of the ice melts till equilibrium charge + Q is placed at the centre O of the
is reached. What was the initial amount of cube. P is such a point outside the cube that
ice that was dropped (when specific heat of the line OP perpendicularly intersects the
surface ABCD at R and also OR = RP = a /2.
ice = 0 .5 cal gm −1 ° C−1 , specific heat of water
A charge + Q is placed at point P also. What is
. cal gm −1 ° C−1 and latent heat of melting
= 10 the total electric flux through the five faces
of ice = 80 cal gm −1 )? of the cube other than ABCD?
(a) 10 g (b) 18 g (c) 20 g (d) 30 g A
a
B
21. One mole of a monoatomic ideal gas
a/2 a/2
undergoes a quasistatic process, which is O P
depicted by a straight line joining points a +Q R +Q
D
(V0 , T0) and (2 V0 , 3T0) in a V-T diagram. What
is the value of the heat capacity of the gas at
a C
the point (V0 , T0)?
3 Q 5Q
(a) R (b) R (a) (b)
2 ε0 6ε0
(c) 2R (d) 0 10Q
(c) (d) zero
6 ε0
4 WB JEE (Engineering) Solved Paper 2018
25. Four equal charges of value + Q are placed at B at its centre. If instead, a circular loop of
any four vertices of a regular hexagon of side radius 2r , made of same material, having the
‘a’. By suitably choosing the vertices, what same cross-section is connected to the same
can be the maximum possible magnitude of voltage source, what will be the magnetic
electric field at the centre of the hexagon? field at its centre?
Q 2Q B B
(a) (b) (a) (b) (c) 2B (d) B
2 4
4 πε0 a2 4 πε0 a2
3Q 2Q 29. An alternating current is flowing through a
(c) (d)
4 πε0 a2 4 πε0 a2 series L-C-R circuit. It is found that the
current reaches a value of 1 mA at both
26. A proton of mass m moving with a speed v 200 Hz and 800 Hz frequency. What is the
(<< c, velocity of light in vacuum) completes resonance frequency of the circuit?
a circular orbit in time T in a uniform (a) 600 Hz (b) 300 Hz
magnetic field. If the speed of the proton is (c) 500 Hz (d) 400 Hz
increased to 2 v, what will be time needed to
30. An electric bulb, a capacitor, a battery and a
complete the circular orbit?
switch are all in series in a circuit. How does
(a) 2 T (b) T the intensity of light vary when the switch is
T T
(c) (d) turn on?
2 2
(a) Continues to increase gradually
27. A uniform current is flowing along the length (b) Gradually increases for sometime and then
of an infinite, straight, thin, hollow cylinder becomes steady
of radius R. The magnetic field B produced at (c) Sharply rises initially and then gradually
a perpendicular distance d from the axis of decreases
the cylinder is plotted in a graph. Which of (d) Gradually increases for sometime and then
gradually decreases
the following figures looks like the plot?
B Category-II (Q. Nos. 31 to 35)
(a) Only one answer is correct. Correct answer will
d
fetch full marks 2. Incorrect answer or any
B
R combination of more than one answer will fetch
1
− marks. No answer will fetch 0 marks.
(b) 2
d 31. A light charged particle is revolving in a
R
B circle of radius r in electrostatic attraction of
a static heavy particle with opposite charge.
(c) How does the magnetic field B at the centre
of the circle due to the moving charge
d depend on r ?
R
1 1
B (a) B ∝ (b) B ∝
r r2
1 1
(d) (c) B ∝ (d) B ∝
r3/ 2 r5/ 2
d 32. As shown in the figure, a rectangular loop of
R
a conducting wire is moving away with a
28. A circular loop of radius r of conducting wire constant velocity v in a perpendicular
connected with a voltage source of zero direction from a very long straight conductor
internal resistance produces a magnetic field carrying a steady current I. When the
WB JEE (Engineering) Solved Paper 2018 5
breadth of the rectangular loop is very small conductor with uniform surface charge
compared to its distance from the straight density σ is placed below it. What will be the
conductor, how does the emf. E induced in time period of the pendulum for small
the loop vary with time t ? amplitude oscillations?
v
L
I
m
1 1 1
(a) E ∝ (b) E ∝ (c) E ∝−ln(t ) (d) E ∝ σ
t2 t t3
L L
33. A solid spherical ball and a hollow spherical (a) 2 π (b)
mq mqσ
ball of two different materials of densities ρ1 g − g −
ε0 σ ε0
and ρ2 respectively have same outer radii and
same mass. What will be the ratio, the 1 L L
(c) (d) 2 π
moment of inertia (about an axis passing 2π qσ qσ
g − g −
through the centre) of the hollow sphere to ε0 m ε0 m
that of the solid sphere?
5 5
Category-III (Q. Nos. 36 to 40)
ρ2 ρ 3 ρ2 ρ2 3
(a) 1 − 2 (b) 1 − 1 −
ρ1 ρ1 ρ1 ρ1 One or more answer(s) is (are) correct. Correct
answer(s) will fetch full marks 2. Any combination
5 5
containing one or more incorrect answer will fetch
ρ2 ρ 3 ρ ρ1 3
(c) 1 − 1 (d) 2 1 − 1 − 0 marks. Also, no answer will fetch 0 marks. If all
ρ1 ρ2 ρ1 ρ2
correct answers are not marked and also no
incorrect answer is marked then score
34. The insulated plates of a charged parallel = 2 × number of correct answers marked ÷ actual
plate capacitor (with small separation number of correct answers.
between the plates) are approaching each
other due to electrostatic attraction. 36. A non-zero current passes through the
Assuming no other force to be operative and galvanometer G shown in the circuit when
no radiation taking place, which of the the key K is closed and its value does not
following graphs approximately shows the change when the key is opened. Then, which
variation with time ()
t of the potential of the following statement(s) is/are true?
difference (V) between the plates?
Ω
30
0
V V
0
20
(a) (b) K
t t
O O G
10
0
Ω
V V
(c) (d) 10 V
37. A ray of light is incident R 39. Which of the following statements(s) is/are
on a right angled true?
isosceles prism parallel Q “Internal energy of an ideal gas ……… .”
to its base as shown in S (a) decreases in an isothermal process.
the figure. Refractive (b) remains constant in an isothermal process.
index of the material of (c) increases in an isobaric process.
P
the prism is 2. Then, (d) decreases in an isobaric expansion.
which of the following statement(s) is/are true?
40. Two positive charges Q and 4Q are placed at
(a) The reflection at P is total internal.
points A and B respectively, where B is at a
(b) The reflection at Q is total internal.
distance d units to the right of A. The total
(c) The ray emerging at R is parallel to the ray
incident at S. electric potential due to these charges is
(d) Total deviation of the ray is 150°. minimum at P on the line through A and B.
What is (are) the distance (s) of P from A?
38. The intensity of a sound appears to an (a)
d
units to the right of A
observer to be periodic. Which of the 3
following can be the cause of it? d
(b) units to the left of A
(a) The intensity of the source is periodic 3
d
(b) The source is moving towards the observer (c) units to the right of A
(c) The observer is moving away from the source 5
(d) The source is producing a sound composed of (d) d units to the left of A
two nearby frequencies
Chemistry
Category-I (Q. Nos. 41 to 70) 44. The ease of hydrolysis in the compounds
Only one answer is correct. Correct will fetch full CH 3COCl(I),CH 3 CO O COCH 3 (II),
marks 1. Incorrect answer or any combination of CH 3COOC2H 5 (III) and CH 3CONH 2 (IV) is of
more than one answer will detch − 1/4 marks. No the order
answer will fetch 0 marks. (a) I > II > III > IV
(b) IV > III > II > I
41. Cl 2O 7 is the anhydride of (c) I > II > IV > III
(a) HOCl (b) HClO 2 (d) II > I > IV > III
(c) HClO 3 (d) HClO 4
45. CH 3 C ≡≡ C MgBr can be prepared by the
42. The main reason that SiCl 4 is easily reaction of
hydrolysed as compared to CCl 4 is that (a) CH3 C ≡≡ C Br with MgBr2
(a) Si Cl bond is weaker than C Cl bond (b) CH3 C ≡≡ CH with MgBr2
(b) SiCl 4 can form hydrogen bonds (c) CH3 C ≡≡ CH with KBr and Mg metal
(c) SiCl 4 is covalent
(d) CH3 C ≡≡ CH with CH3MgBr
(d) Si can extend its coordination number beyond
four 46. The number of alkene (s) which can produce
43. Silver chloride dissolves in excess of 2-butanol by the successive treatment of
ammonium hydroxide solution. The cation (i) B 2H 6 in tetrahydrofuran solvent and
present in the resulting solution is (ii) alkaline H 2O 2 solution is
(a) 1 (b) 2
(a) [Ag(NH3 )6 ]+ (b) [Ag(NH3 )4 ]+
(c) 3 (d) 4
(c) Ag + (d) [Ag(NH3 )2 ]+
WB JEE (Engineering) Solved Paper 2018 7
47. Identify ‘M’ in the following sequence of 51. If aniline is treated with conc. H 2SO 4 and
reactions heated at 200°C, the product is
CH3
(a) anilinium sulphate
Cl (b) benzenesulphonic acid
NH3 Br2
C8H6Cl2O C8H8ClNO (c) m-aminobenzenesulphonic acid
NaOH
M H2N (d) sulphanilic acid
O O 52. Which of the following electronic
configuration is not possible?
Cl C C
Cl Cl (a) n = 3, l = 0, m = 0
(a) (b)
(b) n = 3, l = 1, m = − 1
CH3 Cl (c) n = 2, l = 0, m = − 1
CH3 (d) n = 2, l = 1, m = 0
CHO Cl
53. The number of unpaired electrons in Ni
(c) (d) Cl
(atomic number = 28) are
Cl C
(a) 0 (b) 2 (c) 4 (d) 8
CH2Cl CH3
O
54. Which of the following has the strongest
48. Methoxybenzene on treatment with HI H-bond?
produces (a) O H ... S (b) S H ... O
(a) iodobenzene and methanol (c) F H ... F (d) F H ... O
(b) phenol and methyl iodide
55. The half-life of C14 is 5760 years. For a
(c) iodobenzene and methyl iodide
(d) phenol and methanol
200 mg sample of C14 , the time taken to
change to 25 mg is
K 2 Cr2 O7 I /NaOH
49. C 4 H10O → C 4 H 8O →
2
CHI3 (a) 11520 years (b) 23040 years
H 2 SO 4 Warm
N (c) 5760 years (d) 17280 years
Here, N is
56. Ferric ion forms a prussian blue precipitate
OH
(a) OH (b) due to the formation of
(a) K 4 [Fe(CN)6 ] (b) K 3 [Fe(CN)6 ]
(c) Fe(CNS)3 (d) Fe 4 [Fe(CN)6 ]3
(c) O (d)
OH
64
57. The nucleus 29 Cu accepts an orbital electron
50. The correct order of reactivity for the addition to yield,
reaction of the following carbonyl (a) 65
28 Ni (b) 64
30 Zn (c) 64
28 Ni (d) 65
30 Zn
compounds with ethylmagnesium iodide is
58. How many moles of electrons will weigh one
H H3C
kilogram?
C=
=O C=
=O 1
H H3C (a) 6.023 × 1023 (b) × 1031
(I) (II) 9108
.
6.023 1
(c) × 1054 (d) × 108
H (CH3)3C .
9108 9108
. × 6.023
C=
=O C=
=O
H3C (CH3)3C 59. Equal weights of ethane and hydrogen are
(III) (IV) mixed in an empty container at 25°C. The
(a) I > III > II > IV fraction of total pressure exerted by hydrogen
(b) IV > III > II > I is
(c) I > II > IV > III (a) 1 : 2 (b) 1 : 1
(d) III > II > I > IV (c) 1 : 16 (d) 15 : 16
8 WB JEE (Engineering) Solved Paper 2018
60. The heat of neutralisation of a strong base 66. Which of the following is present in
and a strong acid is 13.7 kcal. The heat maximum amount in ‘acid rain’?
released when 0.6 mole HCl solution is added (a) HNO 3 (b) H2SO 4
to 0.25 mole of NaOH is (c) HCl (d) H2CO 3
(a) 3.425 kcal (b) 8.22 kcal
67. Which of the set of oxides are arranged in the
(c) 11.645 kcal (d) 13.7 kcal
proper order of basic, amphoteric, acidic?
61. A compound formed by elements X and Y (a) SO 2 ,P2O 5 ,CO (b) BaO,Al 2O 3 ,SO 2
crystallises in the cubic structure, where X (c) CaO,SiO 2 ,Al 2O 3 (d) CO 2 ,Al 2O 3 ,CO
atoms are at the corners of a cube and Y
atoms are at the centre of the body. The
68. Out of the following outer electronic
configurations of atoms, the highest
formula of the compounds is
oxidation state is achieved by which one?
(a) XY (b) XY2
(c) X 2 Y3 (d) XY3 (a) (n − 1)d 8 ns 2 (b) (n − 1)d 5 ns 2
(c) (n − 1)d ns3 2
(d) (n − 1)d 5 ns1
62. What amount of electricity can deposit
1 mole of Al metal at cathode when passed 69. At room temperature, the reaction between
through molten AlCl 3 ? water and fluorine produces
(a) 0.3 F (b) 1 F (a) HF and H2O 2 (b) HF, O 2 and F2O 2
(c) 3 F (d) 1/3 F (c) F − , O 2 and H+ (d) HOF and HF
63. Given the standard half-cell potentials (E°) of 70. Which of the following is least thermally
the following as stable?
Zn → Zn 2+ + 2 e − ; E° = + 0 .76 V (a) MgCO 3 (b) CaCO 3
(c) SrCO 3 (d) BeCO 3
Fe → Fe2+ + 2 e − ; E° = 0 .41 V
Then the standard e.m.f. of the cell with the Category-II (Q. Nos. 71 to 75)
reaction Fe2+ + Zn → Zn 2+ + Fe is Only one answer is correct. Correct answer will
(a) − 0.35 V (b) + 0.35 V fetch full marks 2. Incorrect answer or any
(c) + 117
. V (d) − 117
. V combination of more than one answer will fetch −
64. The following equilibrium constants are 1/2 marks. No answer will fetch 0 marks.
given Br 2 NaNH 2
N 2 + 3H 2 c
2NH 3 ; K 1 71. [ P] → C2H 4Br2 →
NH
[ Q]
3
N2 + O 2
1
c
2NO;K 2 20% H 2 SO 4
[ Q] → [ R] → [ S]
Zn -Hg/HCl
H2 + O 2
2
c
H 2O; K 3 Hg 2+ , ∆
Mathematics
Category-I (Q. Nos. 1 to 50) 6. Let f :[ a , b] → R be such that f is
Only one answer is correct. Correct answer will differentiable in (a , b), f is continuous at x = a
fetch full marks 1. Incorrect answer or any and x = b and moreover f (a) = 0 = f (b). Then
combination of more than one answer will (a) there exists at least one point c in (a, b ) such that
fetch-1/4 marks. No answer will fetch 0 marks. f ′(c ) = f(c )
(b) f ′( x) = f( x) does not hold at any point in (a, b )
1. The approximate value of sin31° is (c) at every point of (a, b ), f ′( x) > f( x)
(a) > 0.5 (b) > 0.6 (d) at every point of (a, b ), f ′( x) < f( x)
(c) < 0.5 (d) < 0.4
7. Let f : R → R be a twice continuously
2. Let f1(x) = e x , f 2(x) = e f1 (x ), ……, differentiable function such that
f n + 1 (x) = e f n (x ) for all n ≥ 1. Then for any fixed f (0) = f ()
1 = f ′ (0) = 0 . Then
d (a) f ′ ′(0) = 0
n, f n(x) is (b) f ′ ′(c ) = 0 for some c ∈ R
dx
(c) if c ≠ 0, then f ′ ′(c ) ≠ 0
(a) fn ( x)
(d) f ′( x) > 0 for all x ≠ 0
(b) fn ( x)fn − 1( x)
x cos 3 x − sin x
(c) fn ( x)fn − 1( x)…f1( x) 8. If ∫ esin x ⋅ 2 dx = e
sin x
f (x) + c,
(d) fn ( x)… f1( x)e x cos x
1 − | x| where c is constant of integration, then f (x) is
3. The domain of definition of f (x) = is equal to
2 − | x|
(a) sec x − x (b) x − sec x
(a) (−∞, − 1) ∪ (2, ∞ ) (c) tan x − x (d) x − tan x
(b) [−1, 1] ∪ (2, ∞ ) ∪ (−∞, − 2 )
1
(c) (−∞, 1) ∪ (2, ∞ ) 9. If ∫ f (x) sin x cos x dx = log f (x) + c,
2(b − a 2) 2
(d) [−1, 1] ∪ (2, ∞ )
where c is the constant of integration, then
Here (a , b) ≡ { x : a < x < b} and
f (x) is equal to
[ a , b] ≡ { x : a ≤ x ≤ b} 2 2
(a) (b)
4. Let f :[ a , b] → R be differentiable on [ a , b] (b 2 − a2 )sin2 x ab sin2 x
2 2
and k ∈ R. Let f (a) = 0 = f (b). (c) (d)
(b 2 − a2 )cos 2 x ab cos 2 x
Also let J (x) = f ′ (x) + kf (x). Then
(a) J( x) > 0 for all x ∈[a, b ] π /2 π /4
cos x sin x cos x
(b) J( x) < 0 for all x ∈[a, b ]
10. If M = ∫ x +2
dx , N = ∫ (x + 1)2
dx , then
0 0
(c) J( x) = 0 has at least one root in (a, b )
the value of M − N is
(d) J( x) = 0 through (a, b ) π 2 2
(a) π (b) (c) (d)
5. Let f (x) = 3 x 10
− 7 x + 5 x − 21 x + 3 x − 7 .
8 6 3 2 4 π−4 π+ 4
f (1 − h) − f ()
1
tan −1 x
2014
Then lim
h→ 0 h + 3h
3 11. The value of the integral I = ∫ x
dx is
1 / 2014
50
(a) does not exist (b) is π π
3 (a) log 2014 (b) log 2014
53 22 4 2
(c) is (d) is 1
3 3 (c) πlog 2014 (d) log 2014
2
WB JEE (Engineering) Solved Paper 2018 11
π /3
sin x 19. Given that n numbers of arithmetic means
12. Let I = ∫ x
dx . Then
are inserted between two sets of numbers
π /4
a , 2 b and 2a , b where a , b ∈ R. Suppose further
1
(a) ≤ I≤1 (b) 4 ≤ I ≤ 2 30 that the mth means between these sets of
2
numbers are same, then the ratio a : b equals
3 2 2 3
(c) ≤ I≤ (d) 1 ≤ I ≤ (a) n − m + 1: m (b) n − m + 1: n
8 6 2
(c) n : n − m + 1 (d) m : n − m + 1
13. The value of 20. If x + log 10(1 + 2x ) = x log 10 5 + log 10 6 , then
5π / 2 tan −1 (sin x )
e
I= ∫ tan −1 (sin x ) tan −1 (cos x )
dx , is the value of x is
π /2 e +e (a)
1
(b)
1
π 2 3
(a) 1 (b) π (c) e (d) (c) 1 (d) 2
2
2 πr 2 πr 10
14. The value of 21. If Z r = sin − i cos , then ∑ Z r is equal
1 2 π 2π nπ 11 11 r=0
lim sec + sec 2 + ... + sec 2 is
n→ ∞ n 4n 4n 4 n to
π 4 (a) −1 (b) 0 (c) i (d) −i
(a) loge 2 (b) (c) (d) e
2 π
22. If z1 and z 2 be two non-zero complex
15. The differential equation representing the z1 z 2
numbers such that + = 1, then the
family of curves y 2 = 2 d(x + d), where d is a z 2 z1
parameter, is of origin and the points represented by z1 and z 2
(a) order 2 (b) degree 2 (a) lie on a straight line
(c) degree 3 (d) degree 4 (b) form a right angled triangle
(c) form an equilateral triangle
16. Let y(x) be a solution of (d) form an isosceles triangle
dy
(1 + x 2) + 2 xy − 4 x 2 = 0 and y(0) = − 1. Then 23. If b1b2 = 2(c1 + c2) and b1 , b2 , c1 , c2 are all real
dx
1 is equal to
y() numbers, then at least one of the equations
1 1 x 2 + b1 x + c1 = 0 and x 2 + b2 x + c2 = 0 has
(a) (b)
2 3 (a) real roots
1
(c) (d) −1 (b) purely imaginary roots
6 (c) roots of the form a + ib (a, b ∈ R, ab ≠ 0)
(d) rational roots
17. The law of motion of a body moving along a
1 24. The number of selection of n objects from 2n
straight line is x = vt. x being its distance
2 objects of which n are identical and the rest
from a fixed point on the line at time t and v are different, is
is its velocity there. Then (a) 2 n (b) 2 n − 1
(a) acceleration f varies directly with x (c) 2 n − 1 (d) 2 n − 1 + 1
(b) acceleration f varies inversely with x
(c) acceleration f is constant 25. If (2 ≤ r ≤ n), then nC r + 2 ⋅ nC r + 1 + nC r + 2 is
(d) acceleration f varies directly with t equal to
n+1
(a) 2 ⋅n C r + 2 (b) Cr +1
18. Number of common tangents of y = x 2 and n+ 2 n+1
(c) Cr + 2 (d) Cr
y = − x 2 + 4 x − 4 is
(a) 1 (b) 2 26. The number (101)100 − 1 is divisible by
(c) 3 (d) 4 (a) 104 (b) 106
(c) 108 (d) 1012
12 WB JEE (Engineering) Solved Paper 2018
27. If n is even positive integer, then the 33. On the set R of real numbers, the relation ρ is
condition that the greatest term in the defined by xρy , (x , y) ∈ R.
expansion of (1 + x)n may also have the
(a) If| x − y| < 2, then ρ is reflexive but neither
greatest coefficient, is symmetric nor transitive.
n n+2 n n+1 (b) If x − y < 2, then ρ is reflexive and symmetric but
(a) < x< (b) < x<
n+2 n n+1 n not transitive.
n+1 n+2 n+2 n+ 3 (c) If| x| ≥ y, then ρ is reflexive and transitive but not
(c) < x< (d) < x<
n+2 n+1 n+ 3 n+2 symmetric.
(d) If x >| y|, then ρ is transitive but neither reflexive
−1 7 0 13 −11 5 nor symmetric.
28. If 2 1 −3 = A, Then −7 −1 25 is
34. If f : R → R be defined by f (x) = e x and
3 4 1 −21 −3 −15
g : R → R be defined by g(x) = x 2 . The
(a) A 2 (b) A 2 − A + I3
mapping gof : R → R be defined by (gof ) (x)
(c) A 2 − 3 A + I3 (d) 3 A 2 + 5 A − 4I3
= g[ f (x)] ∀ x ∈ R. Then,
(I 3 denotes the det of the identity matrix of (a) gof is bijective but f is not injective
order 3) (b) gof is injective and g is injective
(c) gof is injective but g is not bijective
29. If a r = (cos 2rπ + i sin 2rπ)1/ 9, then the value of
(d) gof is surjective and g is surjective
a1 a2 a 3
a 4 a5 a 6 is equal to 35. In order to get a head at least once with
probability ≥ 0 .9 , the minimum number of
a 7 a8 a 9
times a unbiased coin needs to be tossed is
(a) 1 (b) −1 (a) 3 (b) 4 (c) 5 (d) 6
(c) 0 (d) 2
2r x n(n + 1) 36. A student appears for tests I, II and III. The
student is successful if he passes in tests I, II
30. If Sr = 6 r 2 − 1 y n (2 n + 3), then the
2
3 or I, III. The probabilities of the student
4 r − 2 nr z n3 (n + 1) passing in tests I, II and III are respectively
n
p , q and 1/2. If the probability of the student
value of ∑ Sr is independent of to be successful is 1/2. Then
r =1
(a) p(1 + q ) = 1 (b) q (1 + p) = 1
(a) only x (b) only y 1 1
(c) only n (d) x, y, z and n (c) pq = 1 (d) + =1
p q
31. If the following three linear equations have a 37. If sin 6θ + sin 4θ + sin 2θ = 0, then general
non-trivial solution, then
value of θ is
x + 4 ay + az = 0 nπ π nπ π
x + 3 by + bz = 0 (a) , nπ ± (b) , nπ ±
4 3 4 6
x + 2 cy + cz = 0 nπ π nπ π
(c) , 2 nπ ± (d) , 2 nπ ±
(a) a, b, c are in AP (b) a, b, c are in GP 4 3 4 6
(c) a, b, c are in HP (d) a + b + c = 0 (n is an integer)
39. Without changing the direction of the axes, 45. Let the eccentricity of the hyperbola
the origin is transferred to the point (2, 3). x2 y2
− = 1 be reciprocal to that of the ellipse
Then the equation x 2 + y 2 − 4 x − 6 y + 9 = 0 a2
b2
changes to x 2 + 9 y 2 = 9 , then the ratio a 2 : b2 equals
(a) x2 + y2 + 4 = 0
(a) 8 : 1 (b) 1 : 8
(b) x2 + y2 = 4 (c) 9 : 1 (d) 1 : 9
(c) x2 + y2 − 8 x − 12 y + 48 = 0
46. Let A , B be two distinct points on the
(d) x2 + y2 = 9
parabola y 2 = 4 x . If the axis of the parabola
40. The angle between a pair of tangents drawn touches a circle of radius r having AB as
from a point P to the circle diameter, the slope of the line AB is
x 2 + y 2 + 4 x − 6 y + 9 sin 2 α + (a) −
1
(b)
1
(c)
2
(d) −
2
r r r r
13 cos 2 α = 0 is 2α. The equation of the locus of
the point P is 47. Let P (at2 , 2at), Q, R(ar 2 , 2ar) be three points on
(a) x2 + y2 + 4 x + 6 y + 9 = 0 a parabola y 2 = 4 ax . If PQ is the focal chord
(b) x2 + y2 − 4 x + 6 y + 9 = 0 and PK, QR are parallel where the co-ordinates
(c) x2 + y2 − 4 x − 6 y + 9 = 0 of K is (2 a , 0), then the value of r is
(d) x2 + y2 + 4 x − 6 y + 9 = 0 t 1− t2 t2 + 1 t2 − 1
(a) (b) (c) (d)
1− t 2
t t t
41. The point Q is the image of the point P(1, 5)
about the line y = x and R is the image of the x 2 y2
48. Let P be a point on the ellipse + = 1 and
point Q about the line y = − x . The 9 4
circumcentre of the ∆PQR is the line through P parallel to the Y-axis
(a) (5, 1) (b) (−5, 1) meets the circle x 2 + y 2 = 9 at Q, where P , Q
(c) (1, − 5) (d) (0, 0)
are on the same side of the X -axis. If R is a
42. The angular points of a triangle are A(−1, − 7), PR 1
point on PQ such that = , then the locus
B(5 , 1) and C(1, 4). The equation of the bisector RQ 2
of the angle ∠ABC is of R is
(a) x = 7 y + 2 (b) 7 y = x + 2 x2 9 y2 x2 y2
(a) + =1 (b) + =1
(c) y = 7 x + 2 (d) 7 x = y + 2 9 49 49 9
43. If one of the diameter of the circle, given by x2 y2 9 x2 y2
(c) + =1 (d) + =1
the equation x 2 + y 2 + 4 x + 6 y − 12 = 0 , is a 9 49 49 9
chord of a circle S, whose centre is (2 , − 3), the 49. A point P lies on a line through Q(1, − 2, 3) and
radius of S is x y z
(a) 41 unit (b) 3 5 unit is parallel to the line = = . If P lies on
1 4 5
(c) 5 2 unit (d) 2 5 unit the plane 2 x + 3 y − 4 z + 22 = 0 , then segment
44. A chord AB is drawn from the point A(0 , 3) on PQ equals
(a) 42 units (b) 32 units
the circle x 2 + 4 x + (y − 3)2 = 0 , and is
(c) 4 units (d) 5 units
extended to M such that AM = 2 AB. The locus
of M is 50. The foot of the perpendicular drawn from the
(a) x +
2
y − 8x − 6y + 9 = 0
2 point (1, 8, 4) on the line joining the point
(b) x2 + y2 + 8 x + 6 y + 9 = 0 (0, −11, 4) and (2, −3, 1) is
(c) x2 + y2 + 8 x − 6 y + 9 = 0 (a) (4, 5, 2) (b) (−4, 5, 2)
(d) x2 + y2 − 8 x + 6 y + 9 = 0 (c) (4, −5, 2) (d) (4, 5, −2)
14 WB JEE (Engineering) Solved Paper 2018
52. For 0 ≤ p ≤ 1 and for any positive a , b; let 58. Let ρ be a relation defined on N, the set of
I (p) = (a + b)p , J (p) = a P + bP , then natural numbers, as
(a) I( p) > J( p) ρ = {(x , y) ∈ N × N : 2 x + y = 41}. Then
(b) I( p) ≤ J( p) (a) ρ is an equivalence relation
(b) ρ is only reflexive relation
(c) I( p) < J( p) in 0, and I( p) > J( p) in , ∞
p p
2 2 (c) ρ is only symmetric relation
p p (d) ρ is not transitive
(d) I( p) < J( p) in , ∞ and J( p) < I( p) in 0,
2 2
59. If the polynomial
→
$ →
53. Let α = i$ + j$ + k, β = i$ − j$ − k$ and (1 + x)a (2 + x)b 1
→ → f (x) = 1 (1 + x) (2 + x)b, then the
a
γ = − i$ + j$ − k$ be three vectors. A vector δ , in
(2 + x) (1 + x)a
b
→ → → 1
the plane of α and β , whose projection on γ
1 constant term of f (x) is
is , is given by (a) 2 − 3 ⋅ 2 b + 2 3 b (b) 2 + 3 ⋅ 2 b + 2 3 b
3
(c) 2 + 3 ⋅ 2 − 2
b 3b
(d) 2 − 3 ⋅ 2 b − 2 3 b
(a) − $i − 3 $j − 3k$ (b) $i − 3 $j − 3k$
[a and b are positive integers]
(c) − $i + 3 $j + 3k$ (d) $i + 3 $j − 3k$
60. A line cuts the X-axis at A(5, 0) and the Y-axis
→ → →
54. Let α , β , γ be the three unit vectors such that at B(0 , − 3). A variable line PQ is drawn
→ → → → →
perpendicular to AB cutting the X-axis at P
α ⋅ β = α ⋅ γ = 0 and the angle between β and and the Y-axis at Q. If AQ and BP meet at R,
→ → then the locus of R is
γ is 30°. Then α is (a) x2 + y2 − 5 x + 3 y = 0 (b) x2 + y2 + 5 x + 3 y = 0
→ → → →
(a) 2(β × γ ) (b) −2(β × γ ) (c) x2 + y2 + 5 x − 3 y = 0 (d) x2 + y2 − 5 x − 3 y = 0
→ → → →
(c) ±2(β × γ ) (d) (β × γ ) 61. Let A be the centre of the circle
x 2 + y 2 − 2 x − 4 y − 20 = 0 . Let B(1, 7) and
55. Let z1 and z 2 be complex numbers such that
D(4 , − 2) be two points on the circle such that
z1 ≠ z 2 and|z1| = |z 2|. If Re(z1) > 0 and tangents at B and D meet at C. The area of the
z + z2
Im(z 2) < 0 , then 1 is quadrilateral ABCD is
z1 − z 2 (a) 150 sq units (b) 50 sq units
(a) one (b) real and positive (c) 75 sq units (d) 70 sq units
(c) real and negative (d) purely imaginary
WB JEE (Engineering) Solved Paper 2018 15
Answers
Physics
1. (d) 2. (c) 3. (a) 4. (b) 5. (d) 6. (c) 7. (a) 8. (d) 9. (d) 10. (c)
11. (a) 12. (c) 13. (b) 14. (a) 15. (c) 16. (c) 17. (c) 18. (d) 19. (b) 20. (c)
21. (c) 22. (b) 23. (b) 24. (a) 25. (c) 26. (b) 27. (c) 28. (b) 29. (d) 30. (c)
31. (d) 32. (a) 33. (d) 34. (a) 35. (d) 36. (b,c,d) 37. (a,c) 38. (a,d) 39. (b) 40. (a)
Chemistry
41. (d) 42. (d) 43. (d) 44. (a) 45. (d) 46. (b) 47. (b) 48. (b) 49. (b) 50. (a)
51. (d) 52. (c) 53. (b) 54. (c) 55. (d) 56. (d) 57. (c) 58. (d) 59. (d) 60. (a)
61. (a) 62. (c) 63. (b) 64. (b) 65. (c) 66. (b) 67. (b) 68. (b) 69. (c) 70. (d)
71. (a) 72. (a) 73. (b) 74. (a) 75. (a) 76. (a,c) 77. (a, d) 78. (b,c) 79. (a,c,d) 80. (a,c,d)
Mathematics
1. (a) 2. (c) 3. (b) 4. (c) 5. (c) 6. (a) 7. (b) 8. (b) 9. (c) 10. (d)
11. (b) 12. (c) 13. (b) 14. (c) 15. (c) 16. (c) 17. (c) 18. (b) 19. (d) 20. (c)
21. (b) 22. (c) 23. (a) 24. (a) 25. (c) 26. (a) 27. (a) 28. (a) 29. (c) 30. (d)
31. (c) 32. (c) 33. (d) 34. (c) 35. (b) 36. (a) 37. (a) 38. (c) 39. (b) 40. (d)
41. (d) 42. (b) 43. (a) 44. (c) 45. (a) 46. (c,d) 47. (d) 48. (a) 49. (a) 50. (d)
51. (a) 52. (b) 53. (c) 54. (c) 55. (d) 56. (a) 57. (b) 58. (d) 59. (a) 60. (a)
61. (c) 62. (b) 63. (c) 64. (b) 65. (c) 66. (b) 67. (c,d) 68. (b) 69. (a,d) 70. (b,c)
71. (a,c) 72. (a,b) 73. (b) 74. (a,c) 75. (c)
Answer with Explanations
Physics
1. (d) For maximum equivalent resistance across the 3. (a) Ray diagram for the question,
diagonal of the square, the given resistors
Y
connected as
P(3,3)
Q
Ω
20
S(0,1)
0
0
30
P X′ X
O M Plane mirror
R
Ω
40
I(0,–1)
0
0
10
Ω
Y′
S
I is image of source S by the plane mirror placed
Resistance of PQR arm, R1 = 300 + 200 = 500 Ω perpendicularly along X-axis.
Resistance of PSR arm, R2 = 400 + 100 Ω = 500 Ω
Q SM = IM
The equivalent resistance between P and R.
1 1 1 ∴PM + MS = PM + MI = PI
= +
Req R1 R2
∴ PI = (3 − 0)2 + (3 + 1)2
1 1 1+1
= + = = 9 + 16 = 25 = 5
500 500 500
500 4. (b) The given combination of lenses
∴ Req = = 250 Ω
2 L2
2. (c) 100 Ω 1µF 400 Ω
f1
200 Ω 300 Ω 2 µF f1
L1 L3
6V K
f2
In steady state, arm having capacitors does not flow Here, f1 = focal length of equiconvex lenses of glass.
current. So, we can neglect them. f2 = focal length of lens formed by water (concave).
∴The given circuit reduces to The focal length of the combination.
1 1 1 1 1 1 1 2 1
200 Ω 400 Ω ∴ = + + = − + = −
F f1 f2 f3 f1 f2 f1 f1 f2
1 2 f2 − f1 f1 f2
= ⇒F= [Q f3 = f1 ]
6V F f1 f2 2 f2 − f1
f1
F= ...(i)
f
∴ Rnet = 200 + 400 = 600 Ω 2− 1
f2
∴Current in circuit,
1 2
I= =
V 6
= 0.01 A Here, = (µ g − 1) , for L1 and L3
R 600 f1 R
= (µ w − 1) − , for L2
= 10mA 1 2
and
f2 R
18 WB JEE (Engineering) Solved Paper 2018
Given, µ w < µ g Number of nuclei that will decay on the 3rd day,
f
Thus, 1 <1 N 3 = N 2 − N1 = 0.63 N 0 − 0.5N 0 = 013
. N0
f2 In the term, fraction is 0.13.
f1
So, F> ...(ii) 9. (d) QKinetic energy of a electron due to
2 accelerated by a potential V, KE = eV
From Eqs. (i) and (ii), we get 1
f1 f me v2 = eV
< F < f1 or < F< f (Q f1 = f) 2
2 2 1 × p2
⇒ = eV [Qp = mv]
5. (d) 2me
∴ p = 2eVme
Small air Solid glass
r
13. (b) According to the question, 15. (c) According to the question,
µ1 1
v∝
m1 F=α t
x
k
m2 or v= ...(i)
µ2 x
1
FBD of lower block of mass m2, dv d K − 1 − 2 − 1 dx
= ⋅ =k ⋅ ⋅x ⋅
N1 =m1g dt dt x 2 dt
3
k − 2 k dx
from Eq.(i)
k
fr1 a =− ⋅x ⋅ Q = v and v =
2 x dt x
fr 2 − k2 1
= ⋅ 2
2 x
N2
∴Force, F = Mass × |Acceleration |
m2 g
k2 1
∴ N 2 = m2 g + N1 = (m1 + m2) g =m ⋅ 2 (magnitude)
2 x
QLower block never moves.
k2 m
∴ fr2 ≥ fr1 F= ⋅
2 x2
µ 2 N 2 ≥ µ 1 N1 ⇒ µ 2(m1 + m2) g ≥ µ 1 m1 g 1
m1 + m2 µ 1 ∴ F∝ 2
⇒ ≥ x
m1 µ2
µ1 m 16. (c) QEscape velocity from a planet,
∴ ≤1 + 2
µ2 2GM 2GM
m1 ve = = R
R R2
µ1 m2
∴ =1 + = 2gR ...(i)
µ2 max
m1
According due to gravity
14. (a) C 4
G πR 3 ⋅ρ
GM 3 4
g= = = GπRρ
R2 R2 3
θ 3g
A B ∴ Radius R = ...(ii)
4πGρ
$ and AC = $i + 2$j + k
AB = 3$i + $j + k $
From Eqs. (i) and (ii), we get
∴ $)
BA = − (3$i + $j + k 3g 3 g2
ve = 2g ⋅ =
∠ABC is angle between BA and BC 4πGρ 2 π Gρ
g
Thus, ve ∝
ρ
θ
A B ve1 g1 ρ 5 1
∴ = × 2 = ×
BC = AC + BA ve 2 ρ1 g2 2 2
= AC − AB (Q BA = − AB) g1 5 ρ
$ $ ) = − 2$i + $j (given, = and 1 = 2 : 1)
= i + 2 j + k − (3i + j + k
$ $ $ $
g2 2 ρ2
∴ BA ⋅ BC =| BA|| BC|cosθ 5
$ ) ⋅ (− 2$i + $j) =
− (3$i + $j + k 2 2
20 WB JEE (Engineering) Solved Paper 2018
Electric flux through the five faces of the cube, Binside = 0 (for d < r)
φE =
Q µ i
Boutside = 0 ⋅ (for d ≥ r)
ε0 2π d
22 WB JEE (Engineering) Solved Paper 2018
So, graph between B and d. 1 q q
Fcentripetal = mrω2 = ⋅ 1 2
4 πε0 r 2
1 q q q1 q 2
1
B∝— ∴ ω2 = ⋅ 1 2 ⇒ ω=
B r 4 πε0 m r 3 4 πε0 mr 3
1
O R ω∝ ...(i)
r d r 3/ 2
Current produced due to moving q1 ,
28. (b) ω
i= q1 ...(ii)
2π
O r
Thus, magnetic field due to motion of q1 in circular
path at point O
µ i µ ωq
B= 0 ⋅ = 0 ⋅ 1 [using Eq. (ii)]
I 2 r 2 2πr
µ I ω
B= 0 ⋅ Hence, B ∝
2 r r
V
Q I1 = 1
Magnetic field B ∝ 5/ 2 [using Eq. (i)]
R1 r
Case (I) l1 = 2πr (length of the loop),
32. (a) E2=vB2l
l µ I B2
R1 = ρ 1 and B1 = 0 ⋅ 1 ...(i)
A 2 r b l v=constant
Case (II) l 2 = 2π ⋅ 2r = 2l1 B1
l 2l E1=vB1l
R2 = ρ ⋅ 2 = ρ ⋅ 1 = 2R1 y
A A
V V I
I2 = = = 1
R2 2R1 2 I
µ I µ I ∴Motional emf in the loop,
B2 = 0 ⋅ 2 = 0 ⋅ 1 ...(ii)
2 2r 2 2 ⋅ 2r E = E1 − E 2 = vl(B1 − B2)
From Eqs. (i) and (ii), µ I µ 0I
= vl 0 ⋅ −
µ0
⋅
I1
2π y 2 π( y + b)
B2 2 2⋅ 2r 1
= = µ y+ b − y µ 0 vlI
µ0 = 0 ⋅ I ⋅ vl = 2π ⋅ y(y + b) b
B1 I 4
⋅ 1
2 r 2π y( y + b)
Magnetic field B2 =
B1 Q b << y
4 µ vlI
∴ E= 0 ⋅ 2 b
29. (d) QResonance frequency, f0 = f1 f2 2π y
Qv = constant (∴ y = vt)
= 200 × 800 = 400 Hz
µ vlI
Thus, E= 0 ⋅ 2 2 b
30. (c) Initially, there will be no voltage drop across 2π v t
capacitor, so intensity of bulb will rise sharply and 1
gradually voltage drop across capacitor will Hence, E∝ 2
t
increase as a result voltage drop across bulb
decreases, so intensity of bulb will decreases. 33. (d) Solid sphere Hollow sphere
ρ1 ρ2
31. (d)
q1 OA=R O′B=R
r A O B O′ O′C=r
O
q2 C
WB JEE (Engineering) Solved Paper 2018 23
4 3 4
∴ M1 = ρ1 πR and M 2 = ρ2 π( R3 − r 3) 35. (d)
3 3
4 4 σ qE
∴ ρ1 πR3 = ρ2 π(R3 − r 3) E= —
ε0 m
3 3
mg
⇒ ρ1 R3 = ρ2(R3 − r 3) + + + + + + + +
ρ1 r3 σ
⇒ =1 − 3
ρ2 R ∴Apparent weight of the bob, w′ = mg − qE
1 mg′ = mg − qE
r ρ 3
∴ = 1 − 1 ...(i) ∴ g′ = g −
qE
R ρ2 m
Moment of inertia, QTime period of a pendulum,
2 R5 − r 5 L
M2 3 3
T = 2π
IH
=
5 R − r geff
IS 2
M1 R2 = 2π
L
= 2π
L
5 qE qσ
g− g−
4 R5 − r 5 m ε
ρ2 π (R3 − r 3) 3 3 0 m
=
3 (R − r )
4 3 2 36. (b, c, d)
ρ1 πR R
30
Ω
30
3
0
0
20
20
Ω
ρ2 R5 − r 5 ρ2 r
Ω
5
= I3
= 1 −
ρ1 R ρ1
5
R K IG
5
I H ρ2 ρ 3
10
G
1 − 1 − 1
10
∴ = G
0
0
ρ1 ρ2
Ω
Ω
IS
34. (a) A
10 V 10 V
Moving each other 10
∴ IG =
100 + G
B
10
Plates of a charged capacitor 300 I 3 = G
moving toward each other 100 + G
due to electrostatic attraction.
1
G
1 σ I3 = 100 + G
Force = ⋅ q 30
2 ε0
(300 + G)
10 + =
1 G
∴Relative acceleration of plates, ∴ I3 + IG =
2F (100 + G) 30 (100 + G) × 30
a rel =
M According to the problem,
Potential difference, V = Ed 200 300 G (300 + G)
...(i) 10 = +
1 3 300 + G (100 + G) 30
where, d = d0 − a rel t 2 (Qplates are moving)
2 [Q V = RI]
60000 + 1100 G 1 1
V = E d0 − a rel t 2
1 10 = × ×
∴
2 3 100 + G 30
This is a equation of a parabola with downward ⇒ 900(100 + G) = 60000 + 1100 G
concavity. ⇒ 30000 = 200 G
∴ G = 150 Ω
24 WB JEE (Engineering) Solved Paper 2018
IG =
10
=
10
= 40 mA ∴Partial reflection and refraction at Q.
100 + 150 250 As SP || QR , emergent ray at R is parallel to incident
200 300 ray at S.
As = ∴Net deviation =180°.
100 150
So, I 200 = I 300 38. (a, d) The intensity of a sound source appears to
be periodic due to
37. (a, c) µ=√2
(i) source intensity is periodic.
(ii) source is producing a sound composed of two
nearby frequencies.
Chemistry
41. (d) Cl 2O 7 is the anhydride of HClO 4 . It reacts with The hydrolysis of SiCl 4 occurs due to coordination
of OH with empty 3d-orbitals of Si-atom of SiCl 4
water slowly to give perchloric acid.
molecule.
Cl 2O 7 + H 2O → 2HClO 4 OH
Perchloric acid
Cl Cl Cl Cl Cl OH
Its structure is as follows: 2H2O
Si Si Si
O O O –2H+ –2Cl–
Cl Cl Cl Cl Cl OH
Cl 117.6° Cl OH 2H2O
O O O O –2H+
44. (a) The ease of hydrolysis of carbonyl compounds Step-I It involves the formation of an amide on
depend on the group attached to the carbonyl carbon. reaction of ammonia and acetyl chloride so, M must be
Among the given options, chlorine (— Cl) group is O
electron withdrawing group which makes the
carbonyl group electrophilic. Hence, hydrolysis can C
Cl
occur most easily. The order for electron withdrawing
tendency among given options is as follows: Cl
Cl > OCOCH 3 > OC2 H 5 > NH 2
CH3
So, the correct order for the ease of hydrolysis in
the given compounds is as follows: Step-II It involves the Hofmann-bromamide
reaction.
CH 3 COCl > CH 3 COOCOCH 3
I II
CH3
O–Na+ + CHI3
53. (b) The number of unpaired electrons in Ni
[C4H8O] (Yellow ppt.)
(atomic number=28) are 2. It can be easily
concluded from the electronic configuration of Ni.
50. (a) The correct order of reactivity for the addition Electronic configuration of
reaction of the given carbonyl compounds with Ni=1s 2 2 s 2 2p 6 3s 2 3p 6 4s 2 3d 8
ethylmagnesium iodide is I > III > II >IV.
3d8
H H H3C
=O >
C= C= =O > C==O Orbitals of Ni in 3d =
H H3C H3C Unpaired electrons
(I) (III) (II)
(CH3)3C 54. (c) The strongest H-bond is FH------ F because
> C==O in this case, a hydrogen is bonded to a most
(CH3)3C electronegative atom i.e. fluorine. In all other
(IV) options, hydrogen is bonded to oxygen and
sulphur that are less electronegative atom than
This order can be explained on the basis of
fluorine.
following two factors:
(i) Inductive effect Greater the number of alkyl The correct electronegativity order of elements are
(electron releasing) groups attached to carbonyl as follows: F > O > S
group, greater will be the electron density on
55. (d) Given,
carbonyl carbon. Thus, it lowers the attack of
nucleophile and hence, reactivity decreases. Half-life of C14 , t1 / 2 = 5760 years
(ii) Steric effect As the number of alkyl group Initial concentration of sample of C14 , N 0 = 200 mg
attached to carbonyl carbon increases, the attack
of nucleophile on carbonyl group becomes more Final concentration of sample of C14 , N t = 25mg
and more difficult due to steric hinderance. The given decay is radioactive and all radioactive
51. (d) When aniline is treated with conc. H 2SO 4 decay follows first order kinetics.
0.693
followed by heating at 200° C, the product obtained is t1 / 2 =
sulphanilic acid i.e. p-aminobenzene sulphonic acid λ
+ 0.693 0.693 −1
NH2 NH3HSO4
– ∴ λ= = yr
t1 / 2 5760
We know that, for first order reaction
+ Conc. H2SO4 2.303 N .
2303 [200]
t= log 0 = log
Aniline Anilinium hydrogen λ N t 0.693 [25]
sulphate 5760
+
NH3 NH2 .
2303 × 5760
= log 8
0.693
200°C
a =17, 286. 78 yrs ≈17, 280 yrs
Therefore, the time taken to change 200 mg to
– 25 mg is 17, 286.78 yr, which is very close to option
SO3 SO3H
Zwitter ion Sulphanilic acid
(d) i.e. 17280 yrs.
WB JEE (Engineering) Solved Paper 2018 27
C2H 6 H2
Alternative method Initial gram weight = w g wg
200 mg sample
w w
t1/2 = 5760 years Number of moles =
30 2
100 mg According to Henry’s law,
w
t1/2 = 5760 years pH 2 nH 2 2
17280 years = χH 2 = =
ptotal nH 2 + nC2 H 6 w
+
w
50 mg 2 30
w
t1/2 = 5760 years pH 2 2 w 30 15
= = × =
25 mg ptotal 15w + w 2 16 w 16
30
56. (d) Ferric ion forms a prussian blue precipitate So, the fraction of total pressure exerted by
due to the formation of Fe4 [Fe(CN)6]3. This hydrogen is 15 : 16.
complex is formed during the determine action of
presence of nitrogen in the given sample. In this 60. (a) Given, the heat of neutralisation of a strong
method, to portion of sodium fusion extract, base and a strong acid is 13.7 kcal.
freshly prepared ferrous sulphate, FeSO 4 solution The reaction of neutralisation is as follows
is added and warmed. Then about 2 to 3 drops of HCl + NaOH → NaCl + H 2O; ∆H = − 137 . kcal
FeCl 3 solution are added and acidified with 1 mol 1 mol
conc. HCl. The appearance of a prussian blue According to question,
colour indicate the presence of nitrogen.
HCl + NaOH → NaCl + H 2O … (i)
FeSO 4 + 2NaOH → Fe(OH)2 + Na 2SO 4
0.6 mol 0.25 mol
6NaCN + Fe(OH)2 → Na 4 [Fe(CN)6] + 2NaOH
Sodium ferrocyanide
In equation (i), NaOH acts as a limiting reagent. For
1 mole of NaOH and 1 mole of HCl, heat of
3 Na 4 [Fe (CN)6] + 4FeCl 3 → neutralisation = 137
. kcal.
Fe4 [Fe (CN)6]3 +12NaCl ∴For 0.25 mole of NaOH and 0.6 mole of HCl, heat
Ferric ferrocyanide of neutralisation = 137
. × 0.25 ⇒ 3.425 kcal
(prussian blue)
61. (a) Number of X atoms at the corners = 8
57. (c) The nucleus 64
29 Cu accepts an orbital electron 1
Number of X atoms per unit cell = 8 × = 1 atom
to yield 64
28 Ni. The atomic number of Cu is 29, 8
which is equal to the number of electrons and also Number of Y atoms at the centre of the body = 1 atom
equal to the number of protons. Hence, the formula of the compound is XY.
When 64 29 Cu accepts an orbital electron then
electrons subtract from the atomic number, i.e.
62. (c) The number of electrons involved in the
reaction are three as shown below
29 −1 = 28
64 Cu + e 0 → 64 Ni Al 3+ + 3e − → Al
29 −1 28 It means the conversion of every aluminium ion to
58. (d) Mass of an electron = 9.108 ×10−31 kg aluminium atom requires three electrons.
Mass of one mole of electron Therefore, the amount of electricity required for
= (9.108 × 10−31 × 6.023 × 1023) kg one mole of Al 3+ ions = 3F.
Mathematics
1. (a) We know that, sin 30° = 1 = 0.5 4. (c) We have, f : [a , b] → R be differentiable
2
on [a , b] and k ∈ R, also f (a) = 0 = f (b)
In 1st quadrant sin x is increasing function.
and J(x) = f '(x) + kf (x)
∴ sin 31° > sin 30°
Let g(x) = kxf (x) which is continuous in [a , b] and
⇒ sin 31° > 0.5 differentiable in (a , b) such that
2. (c) We have, f1 (x) = e x g(a) = 0 = g(b)
f2(x) = e f1 ( x ) Then, for every c ∈ (a , b), g'(c) = 0
(by Rolle’s theorem)
…… …… …… …… Now, g′(x) = kf (x) + kxf '(x)
…… …… …… …… ⇒ g'(c) = kf (c) + kcf '(c)
fn + 1 (x) = e f n( x ) ⇒ kf (c) + kcf '(c) = 0
Now, fn (x) = e f n−1 ( x ) ⇒ f (x) = 0,for every x = c ∈(a , b)
On taking log both sides, we get ∴ J(x) = 0 has atleast one root in (a , b).
5. (c) We have,
log { fn (x)} = fn − 1 (x) log e f (x) = 3x10 − 7 x 8 + 5x 6 − 21 x 3 + 3x 2 − 7
d d
⇒ log ( fn (x)) = fn − 1 (x) (Qlog e = 1) ∴ f (1 − h) = 31
( − h)10 − 71
( − h)8
dx dx
1 d + 51
( − h)6 − 211
( − h)3 + 31
( − h)2 − 7
⇒ fn (x) = fn′ − 1 (x)
fn (x) dx = 31
( − 10h + 45h2 − 120h3 + ...... + h10)
⇒
d
fn (x) = fn (x) fn′ − 1 (x) ... (i) − 71
( − 8h + 28h2 − 56h3 + ..... + h8)
dx + 51
( − 6h + 15h2 − 20h3 + ..... + h6)
Now, f1′ (x) = e x
− 211
( − 3h + 3h2 − h3)
and f2(x) = e f1 ( x )
+ 31
( − 2h + h2) − 7
⇒ log f2(x) = f1 (x) log e = f1 (x) ⇒ f (1 − h) = − 24 + 53h + h2(−46) + h3(− 47) + ....
1 and 1 = − 24
f()
⇒ ⋅ f2′ (x) = f1′(x)
f2(x) f (1 − h) − f ()
1
∴ lim
h→ 0 h3 + 3h
⇒ f2′(x) = f2(x) ⋅ f1′(x)
−24 + 53h + h2(− 46) + h3(− 47) + ... − ( − 24)
= f2(x) ⋅ e x (Q f1′(x) = e x ) = lim
h→ 0 h(h2 + 3)
= f2(x) ⋅ f1 (x) [Q e x = f1 (x)] ...(ii) 53h + h2(− 46) + h3(− 47) + K
= lim
From Eq. (i), h→ 0 h(h2 + 3)
d 53 + h (− 46) + h2 (− 47) + ... 53
fn (x) = fn (x) ⋅ fn −1 (x) .... f1 (x) [using Eq. (ii)] = lim =
dx h→ 0 h2 + 3 3
cos2 x π 2
sin x π 2 sin x
⇒ ∫e
sin x
(x cos x − sec x tan x) dx = e sin x
f (x) + c =
x + 2 0
− ∫0 −
(x + 2)2
dx
∫e
⇒ sin x
(x cos x − 1 + 1 − sec x tan x) dx sin x π 2
(x + 2)2
dx − ∫0
= esin x f (x) + c
sin π 2 1 2
= = =
⇒ ∫ [esin x cos x(x − sec x) + esin x (1 − sec x tan x)] dx π 2+ 2 π + 4 π + 4
= esin x f (x) + c 2
d sin x 11. (b) We have,
⇒ ∫ dx {e (x − sec x)}dx = e f (x) + c
sin x
2014 tan−1 x
⇒ esin x (x − sec x) = esin x f (x) + c
I= ∫1 2014 x
dx ... (i)
⇒ f (x) = x − sec x 1
Let x=
t
9. (c) We have,
−1
1 ⇒ dx = dt
∫ f (x) sin x cos x dx = 2(b 2 − a 2) log( f (x)) + c t2
1 2014 tan−1 (1 t) − 1
⇒ f (x)sin x cos x =
1
⋅
1
⋅ f ′(x)
Now, I= ∫2014 1t
2 dt
t
2(b 2 − a 2) f (x)
f ′(x) cot −1 t
2014
⇒ f (x) sin 2x = 2
1
⋅ = ∫1 2014
t
dt
b − a 2 f (x)
2014 cot −1 x
1 f ′(x) =∫ dx ... (ii)
⇒ sin 2x = 1 2014 x
b 2 − a 2 ( f (x))2
On adding Eqs. (i) and (ii), we get
1 f ′(x)
⇒ ∫ sin 2x dx = b 2 − a 2 ∫ ( f (x))2 dx 2I = ∫
2014 π 2 π
dx = ( log x )
2014
1 2014
1 2014 x 2
− cos 2x 1 −1 π
⇒ = 2 ⋅ = ( log 2014 − log1 2014)
2 b − a 2 f (x) 2
cos 2x (b 2 − a 2) 1 π 1
⇒ = ⇒ I = log 2014 − log
2 f (x) 4 2014
⇒ f (x) = 2
2 π
= (log 2014 + log 2014)
(b − a 2) cos 2x 4
π 2 π π
10. (d) Given, M = ∫ cos x dx = (2 log 2014) = log 2014
0 (x + 2) 4 2
π 4 sin x cos x
and N=∫ dx 12. (c) We have,
0 (x + 1)2 π 3 sin x
π / 41 sin 2x
I= ∫π 4 dx
=∫ ⋅ dx x
0 2 (x + 1)2
WB JEE (Engineering) Solved Paper 2018 33
sin x 16. (c) We have,
Since, is a decreasing function.
x dy
(1 + x 2) + 2xy − 4 x 2 = 0
π sin π 3 π sin π 4
∴ × ≤I≤ × dx
12 π3 12 π4 dy 2x 4x2
⇒ + 2
y=
3 2 dx 1+ x 1 + x2
⇒ ≤I≤
8 6 ∫
2x
1 + x2 2
Here, IF = e = elog(1 + x )
= 1 + x2
13. (b) Let
−1 2
4x
etan y (1 + x 2) = ∫ (1 + x 2) ×
(sin x )
5π 2 ∴ dx + C
I= ∫π 2 etan
−1
(sin x )
+ etan
−1
(cos x )
dx (1 + x 2)
⇒ y (1 + x 2) = ∫ 4 x 2dx + C
5π 2 tan −1 (sin x )
e
= ∫0 etan
−1
(sin x )
+ etan
−1
(cos x )
dx
⇒ y(1 + x 2) =
4x3
+ C
−1
3
π 2 etan (sin x )
4x3
−∫ −1 −1
dx ...(i) ⇒ y (1 + x 2) = −1 [y(0) = − 1]
0
etan (sin x )
+ etan (cos x )
3
−1
5π 2 etan (cos x ) 4x3 1
⇒ y= −
I= ∫0 e tan −1 (cos x )
+ etan
−1
(sin x )
dx
( + x 2) 1 + x 2
31
−1 4 1 1
π 2 etan (cos x ) ∴ 1 = − =
y()
−∫ −1 −1
dx ...(ii) 6 2 6
0
etan (cos x )
+ etan (sin x )
using, b f (x) dx = b f (a + b − x) dx
17. (c) We have, x = 1 vt
∫a ∫a
2
⇒ x=
1 dx
t Q v = dx
On adding Eqs. (i) and (ii), we get 2 dt dt
5π 2 π 2
2 dt dx
2I = ∫0 dx – ∫0 dx ⇒
t
=
x
⇒ 2I = (x)50π 2 − (x)π0 / 2 dt dx
⇒ 2⋅ ∫ = ∫
5π π t x
⇒ 2I = − = 2π ⇒ I= π
2 2 ⇒ 2log |t| + log |c| = log |x|
⇒ log (t 2 ⋅ c) = log x
14. (c) We have,
1 2 π 2π nπ ⇒ x = t 2c
lim sec + sec2 + .... + sec2
n→ ∞ n 4n 4n 4n ⇒
dx
= 2tc
n
rπ πx dt
sec2 = 2
1 1
= lim
n→ ∞
∑n 4n ∫0 sec 4
d2x
r =1 ⇒ = 2c
1 dt 2
πx
tan
4
= ⇒ acceleration f is constant.
π 4 0
4 4 18. (b) We have,
= ×1 =
π π equation of parabola y = x 2
Let P(α , α 2) is a point on the parabola,
15. (c) Given, y 2 = 2d (x + d) ... (i)
∴ y − α 2 = 2α (x − α)
⇒ 2y y1 = 2d ⇒ d = y y1
From Eq. (i),
dy dy
Q Q = 2x ⇒ = 2α
y 2 = 2y y1 (x + y y1 ) dx dx( α , α 2)
⇒ y 2 − 2y y1 x = y y1 ⋅ 2y y1
⇒ y = 2αx − α 2
⇒ (y − 2y y1 x) = 4(y y1)3
2 2
Also, given y = − x 2 + 4 x − 4
So, degree of above equation is 3.
34 WB JEE (Engineering) Solved Paper 2018
∴ − x 2 + 4 x − 4 = 2αx − α 2 or t − 2= 0
⇒ x + 2x (α − 2) + (4 − α ) = 0
2 2 ⇒ t=2 [Q neglect t = − 3]
⇒ 2x = 2 ⇒ x = 1
Discriminant = 0
4(α − 2)2 − 4 (4 − α 2) = 0 21. (b) We have, Zr = sin 2πr − i cos 2πr
11 11
⇒ (α − 2) − (4 − α ) = 0
2 2
2πr 2πr
⇒ α 2 − 4α + 4 − 4 + α 2 = 0 = − i cos + i sin
11 11
⇒ α 2 − 2α = 0 i 2πr
= − ie 11
⇒ α = 0, α = 2
10 10 i 2πr
19. (d) 2b = (n + 2)th form ∴ ∑ Zr = − i∑ e 11
r =0 r =0
= a + (n + 2 − 1) d
⇒ 2b = a + (n + 1) d =−i×0 =0
2b − a 22. (c) We know that, if z1 , z2 and z3 are the vertices
⇒ d=
n+1 of an equilateral triangle. Then,
2b − a z12 + z22 + z32 − z1 z2 − z2z3 − z3z1 = 0 …(i)
∴ m th mean = a + m
n+1 Now, but we have
and also, b = (n + 2)th form z1 z
+ 2 =1
= 2a + (n + 2 − 1) d z2 z1
= 2a + (n + 1) d ⇒ z12 + z22 = z1 z2
b − 2a
⇒ d= ⇒ z12 + z22 − z1 z2 = 0
n+1
Here, z3 = 0
b − 2a
∴ m th mean = 2a + m Hence, given points form an equilateral triangle.
n+1
23. (a) We have equations
According to the question,
x 2 + b1 x + c1 = 0
2b − a b − 2a
a + m = 2a + m D1 = b12 − 4c1
n+1 n+1
a m and x + b2 x + c 2 = 0
2
⇒ =
b n+1− m D2 = b22 − 4c 2
Now, D1 + D2 = b12 + b22 − 4(c1 + c 2)
20. (c) We have,
= b12 + b22 − 2b1 b2 [Qb1 b2 = 2(c1 + c 2)]
x + log10 (1 + 2x ) = x log10 5 + log10 6
= (b1 − b2) ≥ 02
⇒ log10 (1 + 2x ) = x log10 5 + log10 6 − x
⇒ At least one of D1 and D2 are non-negative
= log10 5x + log10 6 − x log10 10
real roots.
= log10 (5x ⋅ 6) − log10 10x
5x ⋅ 6
24. (a) Number of ways of selection of n objects from
⇒ log10 (1 + 2x ) = log10 x 2n objects, where as n objects are identical in out
10 of 2n objects.
5x ⋅ 6 6 n identical and no different object = 1 ways
⇒ 1 + 2x = = x
10x 2 = n C0
⇒ 2x (1 + 2x ) = 6 n −1 identical and 1 different object = 1 × n C1
⇒ t(1 + t) = 6 (let 2 = t)
x
n − 2 identical and 2 different object = 1 × n C2
⇒ t2 + t − 6 = 0 …………………………………………………
⇒ (t + 3) (t − 2) = 0 0 identical and n different objects = 1 × n Cn
⇒ t + 3= 0 = n C0 + n C1 + n C2 + ....... + n Cn = 2n
WB JEE (Engineering) Solved Paper 2018 35
25. (c) n Cr + 2 ⋅ n Cr + 1 + n
Cr + 13 −11 1
2
= 15 −7 −1 5 (R3 → R3 – R2)
= n Cr + n Cr + 1 + n Cr + 1 + n Cr + 2
0 0 −6
n+1 n+1
= Cr + 1 + Cr + 2 = 15 {0 − 0 − 6 (−13 − 77)}
n+1
(Qn Cr + n Cr + 1 = Cr + 1 ) = 15 {(− 6) (− 90)} = 90 × 90 ... (ii)
n+ 2 From Eqs. (i) and (ii),
= Cr + 2
B = A2
26. (a) (101)100 − 1 = (1 + 100)100 − 1
29. (c) We have,
= (1 + 100
C1 ⋅100 + 100
C21002 + ......) − 1 2rπi
= 100
C1100 + 100
C2(100)2 + a r = (cos 2rπ + i sin 2rπ)1 / 9 = e 9
2πi 4 πi 6πi
100
C3(100)3 + ........ + 100
C100 (100)100 e 9 e 9 e 9
a1 a2 a3 8 πi 10 πi 12πi
= 104 (1 + 100
C2 + 100
C3102 + ....
Now, a 4 a5 a6 = e 9 e 9 e 9
+ 100
C100 (100)98 14 πi 16 πi 18 πi
a7 a8 a9
e 9 e 9 e 9
= 10 (1 + an integer multiple of 10)
4
2πi 4 πi
27. (a) For greatest term of (1 + x)n , we have
1 e 9 e 9
n n+1 n 2πi 8 πi 2πi 4 πi
< < +1
2 1+ x 2 =e 9 ×e 9 1 e 9 e 9
14 πi 16πi 18 πi
n n+1 n+1 n
⇒ < and < +1 e 9 e 9 e 9
2 1+ x 1+ x 2
n+1 n+1 2πi 8 πi
⇒ 1+ x< and <1 + x
n2 n
+1 =e 9 ×e 9 ×0 (Q R1 and R2 are identical)
2 =0
n+1− n 2
⇒ x< 2r x n(n + 1)
n2
30. (d) We have, Sr = 6r − 1
2
y n2(2n + 3)
n + 1 − (n 2 − 1)
and < x 4r 3 − 2nr z n3(n + 1)
n+ 2
2 n
2∑ r x n(n + 1)
n+ 2 n
⇒ x< and < x r =1
n n+ 2 n n
⇒ ∑ Sr = ∑ (6r − 1) y n2(2n + 3)
2
n n+ 2
⇒ < x< r =1 r =1
n+ 2 n n
∑ (4r − 2nr) n3(n + 1)
3
z
28. (a) We have, r =1
−1 7 0 n(n + 1) x n(n + 1)
A= 2 1 −3 = n2(2n + 3) y n2(2n + 3)
3 4 1 n3(n + 1) z n3(n + 1)
tan A tan A
= tan−1 − tan−1
1 − tan A 1 − tan A
2 2
=0 Q(5,1)
R(–1,–5)
⇒ (x' + 2)2 + (y ' + 3)2 − 4(x' + 2) 42. (b) Here,AB = (5 + 1)2 + (1 + 7)2 = 36 + 64 = 10
− 6(y' + 3) + 9 = 0 A(–1,–7)
⇒ x' 2 + 4 + 4 x' + y ' 2 + 9
10 P
+ 6 y ' − 4 x' − 8 − 6 y ' − 18 + 9 = 0
⇒ x ′2 + y′2 − 4 = 0 B(5,1) C(1,4)
5
⇒ x 2 + y2 = 4
BC = (1 − 5)2 + (4 − 1)2 = 16 + 9 = 5
40. (d) We have equation of circle By angle bisector theorem,
x 2 + y 2 + 4 x − 6 y + 9sin2 α + 13cos2 α = 0 AP : CP = 10 : 5 = 2 : 1
Here, C ≡ (− 2, 3) 2 × 1 + 1 × (− 1) 2 × 4 + 1 × (− 7) 1 1
∴ P , = P ,
Radius = (− 2)2 + (3)2 − (9sin2 α + 13cos2 α) 2+ 1 2+ 1 3 3
⇒ h2 + k2 + 4h − 6k + 9 = 0
Hence, locus of a point is
x 2 + y 2 + 4 x + 6 y − 12 = 0
x 2 + y2 + 4 x − 6 y + 9 = 0
38 WB JEE (Engineering) Solved Paper 2018
2
46. (c, d) Centre of circle = t1 + t2 , (t1 + t2)
2
∴ C1 C2 = (2 + 2)2 + (−3 + 3)2
2
= (4)2 + (0)2 = 4
2
A(t1,2t1)
∴ Radius of circle, S Y
2
B(t2,2t2)
= (4) + (5) = 16 + 25 =
2 2
41 unit
r
44. (c) Given, AM = 2AB X′ X
y2=4x
Y′
(x1,y1) Since, circle touch the x-axis, so equation of tangent
A(0,3)
B M is y = 0
⇒ B is mid-point of AM. Q Radius = Perpendicular distance from centre to
0 + x1 3 + y1 the tangent
∴Coordinate of B is , ⇒ Radius = |t1 + t2| = r
2 2
2 2
x y + 3 Slope of AB = =
= 1 , 1 t1 + t2 ± r
2 2
Since, B lies on the circle x 2 + 4 x + (y − 3)2 = 0 Y
47. (d)
P(at ,2at)
2
2
y + 3
2 ar)
∴ 1 + 4 1 + 1 − 3 = 0 r2 ,2
x x
R(a
2 2 2
2
x12 y −
+ 2x1 + 1
3 F K(2a, 0)
⇒ =0 X′ X
4 2
x12 y 2 + 9 − 6 y1
⇒ + 2x1 + 1 =0
4 4 Q y 2 = 4ax
⇒ x12 + y12 + 8 x1 − 6 y1 + 9 = 0
Y′
Hence, locus of a point is a −2a
Here, coordinate of Q will be 2 , .
x 2 + y2 + 8 x − 6 y + 9 = 0 t t
2
45. (a) Given equation of ellipses is x 2 + 9 y2 = 9 Slope of QR =
1
x2 y2 r−
⇒ + =1 t
9 1 2at 2t
Slope of PK = 2 =
Here, a = 3, b = 1 at − 2a t 2 − 2
c = (3)2 − (1)2 = 8 Since, Slope of QR = Slope of PK
c 2 2t
∴Eccentricity of ellipse, e = ∴ =
1 t2 − 2
a r−
8 t
⇒ e=
3 t2 − 1
⇒ r=
3 t
∴Eccentricity of hyperbola =
8 2 2
2 48. (a) Since, point P on the ellipse x + y = 1
b 9 9 4
⇒ 1+ =
a2 8 ∴ P(3cosθ, 2sinθ)
b2 1 Now, equation of line parallel of Y-axis is
⇒ =
a2 8 x = 3cosθ
⇒ a2 : b2 = 8 :1 and above line meets circle at Q
∴ Q(3cosθ, 3sinθ)
WB JEE (Engineering) Solved Paper 2018 39
Given,
PR 1
= ⇒ 4λ + 2 + 64λ − 88 + 9λ + 9 = 0
RQ 2 ⇒ 77λ − 77 = 0
R(h, k) ⇒ λ =1
P Q ∴Required foot of perpendicular,
(3 cos q, 2 sin q) (3 cos q, 3 sin q)
P(4, 5, − 2)
3cosθ + 6 cosθ 3sinθ + 4sinθ
∴h= , k=
3 3 51. (a)
7 20 y
⇒ h = 3cosθ, k = sinθ
3
3k
⇒ cosθ = h 3, sinθ = x
7 x 2 + y 2 = (20)2 = 400
9k2
Now, cos θ + sin θ = h 9 +
2 2 2
=1 We have,
dy
= 2ft /sec
49 dt
x 2 9 y2 When x = 12
Hence, locus of a point is + =1
9 49 then (12)2 + y 2 = 400
49. (a) Equation of line through Q (1, − 2, 3) and ⇒ 144 + y 2 = 400
=0
Q R(x,
20
y) (1,2)
y–
A
–4
A(5
3x
,0)
X
X O D(4,–2)
P
Y ⇒ (x − a)2 + y 2 = a 2
2 ,2t)
Q(9,6) and equation of parabola is y 2 = ax , a > 0
R(t Y
A
X′ X x2 + y2 = 2ax
y2=4x X′ X
P(4,–4) (a,0)
Y′
y2 = ax
Perpendicular distance Y′
2t 2 − 2t − 12 2 Intersection points of circle and parabola
AR = = (t − 3) (t + 2)
5 5 ⇒ x 2 + ax = 2ax
1 ⇒ x 2 = ax
AR is Maximum, at t =
2
⇒ x 2 − ax = 0
∴R is , 1
1
4 ⇒ x(x − a) = 0
⇒ x = 0, a
1 3
66. (b) I = ∫ x cos 32x dx Intersecting points are (0, 0) and (a , a).
0 2+ x
πa 2 a
Here, −1 < cos 3x < 1 ∴ Required area = − ∫ axdx
4 0
⇒ − x 3 < x 3 cos 3x < x 3 a
πa 2 x3 2
−x −x
3
−x 3 3
x cos 3x 3 = − a
⇒ < < < 4 3 2 0
x2 x 2 + x2 2 + x2
πa 2 2a 2 π 2
x3 x3 x3 = − = a 2 −
< < < 2 4 3 4 3
2+ x 2
x x
WB JEE (Engineering) Solved Paper 2018 43
x+
y=
⇒ x 2cosec2 θ − y 2 sec2 θ = 1
y 2
44 WB JEE (Engineering) Solved Paper 2018
74. (a, c) f (x) = cos π 75. (c) Given, y = log a (x + x 2 + 1), a > 0, a ≠ 1
x
π −π π π ⇒ a y = (x + x 2 + 1)
⇒ f ′(x) = − sin 2 = 2 sin
x x x x ⇒ a− y =
1
For increasing function, f ′(x) > 0 x+ x2 + 1
π π
⇒sin > 0 ⇒ 2kπ < < (2k + 1) π = x2 + 1 − x
x x
a y − a− y
⇒
1
> x>
1 ⇒ a y − a − y = 2x ⇒ x =
2k 2k + 1 2
e y log a − e − y log a
For decreasing function, f ′(x) < 0 ⇒ f −1 (y) =
π 2
⇒ sin < 0 e x − e −x
x ⇒ f −1 (y) = sinh (y log a) Q = sinh (x)
π 1 1 2
⇒ ∈ [(2k + 1) π , (2k + 2) π] ⇒ x ∈ ,
x 2k + 2 2k + 1
WB JEE
Engineering Entrance Exam
SOLVED PAPER 2017
PHYSICS
Category-I (Q.1 to Q. 30) tension T. It can vibrate at frequencies ( v )
Direction Only one answer is correct. Correct given by the formula (where n = 1, 2, 3, ...)
answer will fetch full marks 1. Incorrect answer or n T n T
(a) v = (b) v =
any combination of more than one answer will 2 ML 2L M
1
fetch − marks. No answer will fetch 0 marks. (c) v =
1 T
(d) v =
n TL
4 2n ML 2 M
Unless otherwise specified in the question, the 3. A uniform capillary tube of length l and inner
following values should be used. radius r with its upper end sealed is
Mechanical equivalent of heat, J = 4.2 J cal − 1 submerged vertically into water. The outside
Acceleration due to gravity, g = 9.8 ms− 2 pressure is p 0 and surface tension of water is
Absolute zero temperature = − 273 °C γ. When a length x of the capillary is
The following symbols usually carry meaning submerged into water, it is found that water
as given below levels inside and outside the capillary
ε 0 : electric permittivity of free space coincide. The value of x is
µ 0 : magnetic permeability of free space l p r
(a) (b) l 1 − 0
R : universal gas constant p 0r 4γ
1 +
1. The velocity of a particle executing a simple 4γ
harmonic motion is 13 ms− 1 , when its distance p r l
(c) l 1 − 0 (d)
from the equilibrium position (Q ) is 3 m and 2γ p 0r
its velocity is 12 ms− 1 , when it is 5 m away 1 +
2γ
from Q. The frequency of the simple harmonic
motion is 4. A liquid of bulk modulus k is compressed by
5π 5 applying an external pressure such that its
(a) (b)
8 8π density increases by 0.01%. The pressure
8π 8 applied on the liquid is
(c) (d)
5 5π k k
(a) (b)
10000 1000
2. A uniform string of length L and mass M is
(c) 1000 k (d) 0.01 k
fixed at both ends while it is subject to a
2 WEST BENGAL (Engineering) Solved Paper 2017
16. Consider the circuit shown in the figure. (a) infinite distance above the lens
(b) 0.1 m above the center of the lens
R
(c) infinite distance below the lens
R X (d) 0.1 m below the center of the lens
E
21.
n
The value of the resistance X for which the
thermal power generated in it is practically
independent of small variation of its R
resistance is
R
(a) X = R (b) X =
3 x
R
(c) X = (d) X = 2R A parallel beam of light is incident on a glass
2 prism in the shape of a quarter cylinder of
17. A radius R = 0.05 m and refractive index
n = 1.5, placed on a horizontal table as shown
X in the figure. Beyond the cylinder, a patch of
light is found whose the nearest distance x
B
from the cylinder is
Consider the circuit shown in the figure where (a) ( 3 3 − 4) × 10− 2 m
all the resistances are of magnitude 1 kΩ. If (b) ( 2 3 − 2) × 10− 2 m
the current in the extreme right resistance X is
(c) ( 3 5 − 5) × 10− 2 m
1 mA, the potential difference between A and
B is (d) ( 3 2 − 3) × 10− 2 m
(a) 34 V (b) 21 V 22. The de-Broglie wavelength of an electron is
(c) 68 V (d) 55 V 0.4 × 10− 10 m when its kinetic energy is
18. The ratio of the diameter of the sun to the 1.0 keV. Its wavelength will be 1.0 × 10− 10 m,
distance between the earth and the sun is when its kinetic energy is
approximately 0.009. The approximate (a) 0.2 keV (b) 0.8 keV
diameter of the image of the sun formed by a (c) 0.63 keV (d) 0.16 keV
concave spherical mirror of radius of 23. When light of frequency v 1 is incident on a
curvature 0.4 m is metal with work function W (where hv 1 > W ),
(a) 4.5 × 10− 6 m (b) 4.0 × 10− 6 m then photocurrent falls to zero at a stopping
− 3
(c) 3. 6 × 10 m (d) 1.8 × 10− 3 m potential of V1 . If the frequency of light is
19. Two monochromatic coherent light beams A increased to v 2, the stopping potential
L changes to V 2. Therefore, the charge of an
and B have intensities L and , respectively. If
4 electron is given by
these beams are superposed, the maximum W( v 2 + v 1 ) W( v 2 + v 1 )
(a) (b)
and minimum intensities will be v 1 V 2 + v 2V1 v 1 V1 + v 2V 2
9L L
(a) ,
5L
(b) , 0 W( v 2 − v 1 ) W( v 2 − v 1 )
(c) (d)
4 4 4 v 1 V 2 − v 2V1 v 2V 2 − v 1 V1
5L L
(c) , 0 (d) 2L,
2 2 24. Radon-222 has a half-life of 3.8 days. If one
starts with 0.064 kg of radon-222, the
20. A point object is held above a thin equiconvex quantity of radon-222 left after 19 days will
lens at its focus. The focal length is 0.1 m and be
the lens rests on a horizontal thin plane (a) 0.002 kg (b) 0.062 kg
mirror. The final image will be formed at (c) 0.032 kg (d) 0.024 kg
4 WEST BENGAL (Engineering) Solved Paper 2017
26. When a semiconducting device is connected 31. A particle with charge Q coulomb, tied at the
in series with a battery and a resistance, a end of an inextensible string of length
current is found to flow in the circuit. If R metre, revolves in a vertical plane. At the
however, the polarity of the battery is centre of the circular trajectory, there is a
reversed, practically no current flows in the fixed charge of magnitude Q coulomb. The
circuit. The device may be mass of the moving charge M is such that
Q2
(a) a p-type semiconductor Mg = . If at the highest position of the
(b) a n-type semiconductor 4πε 0R 2
(c) an intrinsic semiconductor particle, the tension of the string just
(d) a p-n junction vanishes, the horizontal velocity at the lowest
27. The dimension of the universal constant of point has to be
gravitation, G is (a) 0 (b) 2 gR (c) 2gR (d) 5gR
(a) [ ML2T − 1 ] (b) [M− 1 L3T − 2]
32. A bullet of mass 4.2 × 10− 2 kg, moving at a
(c) [M− 1 L2T − 2] (d) [ML3T − 2]
speed of 300 ms − 1 , gets stuck into a block
28. Two particles A and B (both initially at rest) with a mass 9 times that of the bullet. If the
start moving towards each other under a block is free to move without any kind of
mutual force of attraction. At the instant, friction, the heat generated in the process
when the speed of A is v and the speed of B is will be
2v, the speed of the centre of mass is (a) 45 cal (b) 405 cal
(a) zero (b) v (c) 450 cal (d) 1701 cal
3v 3v
(c) (d) − 33. A particle with charge e and mass m, moving
2 2
along the X-axis with a uniform speed u,
29. Three vectors A = ai + j + k ; B = i + bj + k enters a region where a uniform electric field
and C = i + j + ck are mutually perpendicular E is acting along the Y-axis. The particle starts
(i, j and k are unit vectors along X, Y and Z- to move in a parabola. Its focal length
axes respectively). The respective values of a, (neglecting any effect of gravity) is
b and c are 2mu 2 eE mu mu 2
1 1 1 (a) (b) 2
(c) (d)
(a) 0, 0, 0 (b) − , − , − eE 2mu 2eE 2eE
2 2 2
1 1 1 34. A unit negative charge with mass M resides at
(c) 1, − 1, 1 (d) , , the mid-point of the straight line of length 2a
2 2 2
adjoining two fixed charges of magnitude + Q
30. A block of mass 1 kg starts from rest at x = 0 each. If it is given a very small displacement
and moves along the X-axis under the action x( x << a ) in a direction perpendicular to the
of a force F = kt, where t is time and straight line, it will
WEST BENGAL (Engineering) Solved Paper 2017 5
(a) come back to its original position and (b) 0.2 m from the 1st wire, towards the
stay there 2nd wire
(b) execute oscillations with frequency (c) 0.1 m from the 1st wire, away from the
1 Q 2nd wire
2π 4πε 0Ma 3 (d) 0.2 m from the 1st wire, away from the
2nd wire
(c) fly to infinity
(d) execute oscillations with frequency 38. If χ stands for the magnetic susceptibility of a
1 Q substance, µ for its magnetic permeability and
2π 4πε 0Ma 2 µ 0 for the permeability of free space, then
(a) for a paramagnetic substance: χ < 0,
35. Consider the circuit given here. The potential µ>0
difference V BC between the points B and C is (b) for a paramagnetic substance: χ > 0,
1 µF 2 µF µ > µ0
A C D (c) for a diamagnetic substance: χ > 0, µ < 0
B (d) for a ferromagnetic substance: χ > 1,
1kΩ 2kΩ µ > µ0
39. Let v n and E n be the respective speed and
3kΩ 3V
energy of an electron in the nth orbit of radius
rn , in a hydrogen atom, as predicted by Bohr’s
(a) 1 V (b) 0.5 V (c) 0 V (d) − 1 V model. Then.
E r
Category-III (Q. 36 to Q. 40) (a) plot of n n as a function of n is a
E1 r1
Direction One or more answer(s) is (are) straight line of slope 0
correct. Correct answer(s) will fetch full marks 2. r v
Any combination containing one or more (b) plot of n n as a function of n is a
r1 v 1
incorrect answer will fetch 0 marks. Also, no
straight line of slope 1
answer will fetch 0 marks. If all correct answers
r
are not marked and also no incorrect answer is (c) plot of ln n as a function of ln ( n ) is a
marked then score = 2 × number of correct r1
answers marked ÷ actual number of correct straight line of slope 2
answers. r E
(d) plot of ln n 1 as a function of ln ( n ) is
36. If the pressure, temperature and density of an E n r1
ideal gas are denoted by p, T and ρ, a straight line of slope 4
respectively, the velocity of sound in the gas is 40. A small steel ball bounces on a steel plate held
(a) proportional to p, when T is constant. horizontally. On each bounce the speed of the
(b) proportional to T. ball arriving at the plate is reduced by a factor
(c) proportional to p, when ρ is constant. e (coefficient of restitution) in the rebound, so
(d) proportional to T. that V upward = eV downward
37. Two long parallel wires separated by 0.1 m If the ball is initially dropped from a height of
carry currents of 1 A and 2 A, respectively in 0.4 m above the plate and if 10 seconds later
opposite directions. A third current-carrying the bouncing ceases, the value of e is
wire parallel to both of them is placed in the 2 3
(a) (b)
same plane such that it feels no net magnetic 7 4
force. It is placed at a distance of 13 17
(c) (d)
(a) 0.5 m from the 1st wire, towards the 18 18
2nd wire
CHEMISTRY
Category-I (Q.41 to 70) 1 1
(a) 4, 1, − 1, + (b) 4, 0 , 0 , +
Direction Only one answer is correct. Correct 2 2
will fetch full marks 1. Incorrect answer or any 1 1
(c) 3, 2, 0, − (d) 3, 2, − 2, +
combination of more than one answer will detch 2 2
− 1/4 marks. No answer will fetch 0 marks.
45. 0.126 g of an acid is needed to completely
41. neutralise 20 mL 0.1 (N) NaOH solution. The
equivalent weight of the acid is
(a) 53 (b) 40
(c) 45 (d) 63
log V T=const.
46. In a flask, the weight ratio of CH 4 (g ) and
SO 2(g ) at 298 K and 1 bar is 1 : 2. The ratio of
M1 M2
the number of molecules of SO 2(g ) and
log p CH 4 (g ) is
For same mass of two different ideal gases of (a) 1 : 4 (b) 4 : 1 (c) 1 : 2 (d) 2 : 1
molecular weights M1 and M2, Plots of log V vs 47. C 6H 5F18 is a F18 radio-isotope labelled organic
log p at a given constant temperature are compound. F18 decays by positron emission.
shown. Identify the correct option. The product resulting on decay is
(a) M1 > M2 (a) C 6H 5O18 (b) C 6H 5Ar19
(b) M1 = M2 (c) B12C 5H 5F (d) C 6H 5O16
(c) M1 < M2
(d) Can be predicted only if temperature is 48. Dissolving NaCN in de-ionised water will
known result in a solution having
(a) pH < 7 (b) pH = 7
42. Which of the following has the dimension if (c) pOH = 7 (d) pH > 7
[ ML0T − 2]?
49. Among Me3N, C 5H 5N and MeCN
(a) Coefficient of viscosity
(b) Surface tension (Me = methyl group) the electronegativity of
(c) Vapour pressure N is in the order
(d) Kinetic energy (a) MeCN > C 5H 5N > Me3N
(b) C 5H 5N > Me3N > MeCN
43. If the given four electronic configurations. (c) Me3N > MeCN > C 5H 5N
(i) n = 4, l = 1 (ii) n = 4, l = 0 (d) Electronegativity same in all
52. The number of unpaired electrons in 60. The correct order of acid strengths of
[ NiCl 4 ]2 − , Ni(CO)4 and [Cu(NH 3 )4 ]2 + benzoic acid (X), peroxybenzoic acid (Y) and
respectively are p-nitrobenzoic acid (Z) is
(a) 2, 2, 1 (b) 2, 0, 1 (a) Y > Z > X
(c) 0, 2, 1 (d) 2, 2, 0 (b) Z > Y > X
(c) Z > X > Y
53. Which of the following atoms should have the
(d) Y > X > Z
highest I st electron affinity?
(a) F (b) O (c) N (d) C 61. The yield of acetanilide in the reaction
(100% conversion) of 2 moles of aniline with
54. PbCl 2 is insoluble in cold water. Addition of
1 mole of acetic anhydride is
HCl increases its solubility due to
(a) 270 g (b) 135 g
(a) formation of soluble complex anions like
(c) 67.5 g (d) 177 g
[ PbCl 3]−
(b) oxidation of Pb(II) to PB(IV) 62. The structure of the product P of the following
(c) formation of [Pb(H 2O)6]2 + reaction is
(d) formation of polymeric lead complexes OH
55. Of the following compounds, which one is the (i) CO2 (high temperature
and high pressure)
strongest Bronsted acid in a aqueous + NaOH P
(ii) H3O+
solution?
(a) HClO 3 (b) HClO 2 (c) HOCl (d) HOBr
OMe
56. The correct basicity order of the following
lanthanide ions is OH OH
(a) La 3 + > Lu 3 + > Ce3 + > Eu 3 + CO2H CO2H
(b) Ce3 + > Lu 3 + > La 3 + > Eu 3 + (a) (b)
(c) Lu 3 + > Ce3 + > Eu 3 + > La 3 + CO2H HO2C
(d) La 3 + > Ce3 + > Eu 3 + > Lu 3 + OMe OMe
57. When BaCl 2 is added to an aqueous salt
solution, a white precipitate is obtained. The OH OH
anion among CO 23 − , SO 23 − and SO 24 − that was CO2H
present in the solution can be (c) (d)
(a) CO 23 − but not any of the other two CO2H
(b) SO 23 − but not any of the other two OMe OMe
(c) SO 24 − but not any of the other two
(d) Any of them 63. ADP and ATP differ in the number of
(a) phosphate units
58. In the IUPAC system, PhCH 2CH 2CO 2H is (b) ribose units
named as (c) adenine base
(a) 3-phenylpropanoic acid (d) nitrogen atom
(b) benzylacetic acid
(c) carboxyethylbenzene 64. The compound that would produce a
(d) 2-phenylpropanoic acid nauseating smell/odour with a hot mixture of
chloroform and ethanolic potassium
59. The isomerisation of 1-butyne to 2-butyne can hydroxide is
be achieved by treatment with (a) PhCONH 2
(a) hydrochloric acid (b) PhNHCH 3
(b) ammoniacal silver nitrate (c) PhNH 2
(c) ammoniacal cuprous chloride (d) PhOH
(d) ethanolic potassium hydroxide
8 WEST BENGAL (Engineering) Solved Paper 2017
65. For the reaction below The equilibrium constant for the reaction?
(i) PhMgBr, THF
H 2(g ) + CO 2(g ) - CO(g ) + H 2O(g )
(ii) H3O+
CN at 1200 K will be
the structure of the product Q is (a) 0.05 (b) 20
(c) 0.2 (d) 5.0
Ph Ph
(a) (b)
Category-II (Q.71 to Q.75)
OH O Direction Only one answer is correct. Correct
answer will fetch full marks 2. Incorrect answer or
any combination of more than one answer will
Ph Ph
(c) (d) fetch − 1/2 marks. No answer will fetch 0 marks.
CN
NH2 71. In a close-packed body-centred cubic lattice of
potassium, the correct relation between the
66. You are supplied with 500 mL each of 2N HCl atomic radius ( r ) of potassium and the
and 5N HCl. What is the maximum volume of edge-length (a) of the cube is
3M HCl that you can prepare using only these a a
two solutions? (a) r = (b) r =
2 3
(a) 250 mL (b) 500 nL 3 3
(c) 750 mL (d) 1000 mL (c) r = a (d) r = a
2 4
67. Which one of the following corresponds to a
72. Which of the following solutions will turn
photon of highest energy?
violet when a drop of lime juice is added to it?
(a) λ = 300 mm
(b) v = 3 × 108 s− 1 (a) A solution of NaI
(c) v = 30 cm − 1 (b) A solution mixture of KI and NaIO 3
(d) ε = 6. 626 × 10− 27 J (c) A solution mixture of NaI and KI
(d) A solution mixture of KIO 3 and NaIO 3
68. Assuming the compounds to be completely
73. The reaction sequence given below given
dissociated in aqueous solution, identify the
product R.
pair of the solutions that can be expected to be CO2Me (i) Ag2O
isotonic at the same temperature. HO2C R
(ii) Br2, CCl4
(a) 0.01 M urea and 0.01 M NaCl
(b) 0.02 M NaCl and 0.01 M Na 2SO 4 The structure of the product R is
(c) 0.03 M NaCl and 0.02 M MgCl 2
(a) Br
(d) 0.01 M sucrose and 0.02 M glucose CO2H
69. How many faradays are required to reduce
Br
1 mol of Cr2O72 − to Cr 3 + in acid medium?
(a) 2 (b) 3 CO2Me
(c) 5 (d) 6 (b) HO2C
13. The equation sin x(sin x + cos x ) = k has real 19. The common chord of the circles
solutions, where k is a real number. Then, x 2 + y 2 − 4x − 4y = 0 and 2x 2 + 2y 2 = 32
1+ 2 subtends at the origin an angle equal to
(a) 0 ≤ k ≤ π π
2 (a) (b)
(b) 2 − 3 ≤ k ≤ 2 + 3 3 4
π π
(c) 0 ≤ k ≤ 2 − 3 (c) (d)
6 2
1− 2 1+ 2
(d) ≤k≤ 20. The locus of the mid-points of the chords of
2 2
the circle x 2 + y 2 + 2x − 2y − 2 = 0, which
14. The possible values of x, which satisfy the make an angle of 90° at the centre is
trigonometric equation (a) x 2 + y 2 − 2x − 2y = 0
x − 1 − 1 x + 1 π (b) x 2 + y 2 − 2x + 2y = 0
tan − 1 + tan = are
x − 2 x + 2 4 (c) x 2 + y 2 + 2x − 2y = 0
1 (d) x 2 + y 2 + 2x − 2y − 1 = 0
(a) ± (b) ± 2
2 21. Let P be the foot of the perpendicular from
1
(c) ± (d) ± 2 x2 y 2
2 focus S of hyperbola 2 − 2 = 1 on the line
a b
15. Transforming to parallel axes through a point bx − ay = 0 and let C be the centre of the
( p, q), the equation hyperbola. Then, the area of the rectangle
2x 2 + 3xy + 4y 2 + x + 18y + 25 = 0 becomes
whose sides are equal to that of SP and CP is
2x 2 + 3xy + 4y 2 = 1. Then,
(a) 2ab (b) ab
(a) p = − 2, q = 3 (b) p = 2, q = − 3
( a 2 + b2 ) a
(c) p = 3, q = − 4 (d) p = − 4, q = 3 (c) (d)
2 b
16. Let A( 2 , − 3) and B( − 2 , 1) be two angular
points of ∆ABC. If the centroid of the triangle 22. B is an extremity of the minor axis of an ellipse
moves on the line 2x + 3y = 1, then the locus whose foci are S and S′. If ∠SBS ′ is a right
of the angular point C is given by angle, then the eccentricity of the ellipse is
1 1 2 1
(a) 2x + 3y = 9 (a) (b) (c) (d)
(b) 2x − 3y = 9 2 2 3 3
(c) 3x + 2y = 5 23. The axis of the parabola
(d) 3x − 2y = 3
x 2 + 2xy + y 2 − 5x + 5y − 5 = 0 is
17. The point P( 3, 6) is first reflected on the line (a) x + y = 0 (b) x + y − 1 = 0
y = x and then the image point Q is again (c) x − y + 1 = 0 (d) x − y =
1
reflected on the line y = − x to get the image 2
point Q′. Then, the circumcentre of the
24. The line segment joining the foci of the
∆PQQ′ is
hyperbola x 2 − y 2 + 1 = 0 is one of the
(a) (6, 3) (b) ( 6, − 3)
diameters of a circle. The equation of the
(c) ( 3, − 6) (d) (0, 0)
circle is
18. Let d1 and d 2 be the lengths of the (a) x 2 + y 2 = 4 (b) x 2 + y 2 = 2
perpendiculars drawn from any point of the (c) x 2 + y 2 = 2 (d) x 2 + y 2 = 2 2
line 7 x − 9y + 10 = 0 upon the lines
3x + 4y = 5 and 12x + 5y = 7, respectively. 25. The equation of the plane through (1, 2 , − 3)
Then, and ( 2 , − 2 , 1) and parallel to X-axis is
(a) d 1 > d 2 (b) d 1 = d 2 (a) y − z + 1 = 0 (b) y − z − 1 = 0
(c) d 1 < d 2 (d) d 1 = 2d 2 (c) y + z − 1 = 0 (d) y + z + 1 = 0
12 WEST BENGAL (Engineering) Solved Paper 2017
26. Three lines are drawn from the origin O with 33. Let f ( x ) = x13 + x11
direction cosines proportional to (1, − 1, 1), + x 9 + x7 + x 5+ x 3 + x + 19. Then, f ( x ) = 0
( 2 , − 3, 0) and (1, 0, 3). The three lines are has
(a) not coplanar (a) 13 real roots
(b) coplanar (b) only one positive and only two negative
(c) perpendicular to each other real roots
(d) coincident (c) not more than one real root
27. Consider the non-constant differentiable (d) has two positive and one negative real
function f of one variable which obeys the root
f (x) xp π
relation = f ( x − y ). If f ′ ( 0) = p and , if 0 < x ≤
f (y ) 34. Let f ( x ) = (sin x )q 2,
f ′ ( 5) = q, then f ′ ( − 5) is 0 , if x = 0
p2 q ( p, q ∈ R ). Then, Lagrange’s mean value
(a) (b)
q p theorem is applicable tof ( x ) in closed
p interval [ 0, x]
(c) (d) q
q (a) for all p, q
(b) only when p > q
28. If f ( x ) = log 5log 3 x, then f ′ ( e) is equal to (c) only when p < q
(a) e log e 5 (b) e log e 3 (d) for no value of p, q
1 1
(c) (d) 35. lim(sin x )2 tan x is equal to
e log e 5 e log e 3 x→ 0
− x (a) 2 (b) 1
29. Let F( x ) = e , G( x ) = e
x
and H( x ) = G( F( x )),
(c) 0 (d) does not exist
dH
where x is a real variable. Then, at x = 0
dx 36. ∫ cos(log x ) dx = F( x ) + C, where C is an
is arbitrary constant. Here, F( x ) is equal to
1
(a) 1 (b) − 1 (c) − (d) − e (a) x [cos (log x ) + sin (log x )]
e (b) x [cos (log x ) − sin (log x )]
x
30. If f ′ ′ ( 0) = k , k ≠ 0, then the value of (c) [cos (log x ) + sin (log x )]
2
2f ( x ) − 3f ( 2x ) + f ( 4x ) x
lim is (d) [cos (log x ) − sin (log x )]
x→ 0 x2 2
(a) k (b) 2k (c) 3k (d) 4k
x2 − 1
31. If y = e m sin − 1 x
then 37. ∫x + 3x 2 + 1
4
dx ( x > 0) is
d 2y dy 1
(1 − x ) 2 − x
2
− ky = 0, where k is (a) tan − 1 x + + C
dx dx x
equal to
1
(a) m 2 (b) 2 (c) − 1 (d) − m 2 (b) tan − 1 x − + C
x
32. The chord of the curve y = x 2 + 2ax + b, 1
joining the points where x = α and x = β, is x+ −1
(c) log e x +C
parallel to the tangent to the curve at abscissa 1
x is equal to x+ +1
x
a+b 2a + b 1
(a) (b) x− −1
2 3 x
2α + β α +β (d) log e +C
(c) (d) 1
3 2 x− +1
x
WEST BENGAL (Engineering) Solved Paper 2017 13
sin x 19
44. The integrating factor of the first order
38. Let I = ∫
1 + x8
dx. Then,
10 differential equation
dy
(a)| I| < 10− 9 (b)| I| < 10− 7 x 2( x 2 − 1) + x( x 2 + 1) y = x 2 − 1 is
− 5
(c)| I| < 10 (d)| I| > 10− 7 dx
1
n n (a) ex (b) x −
39. Let I1 = ∫0
[ x] dx and I2 = ∫
0
{ x} dx, where [x] x
1 1
and {x} are integral and fractional parts of x (c) x + (d) 2
and n ∈ N − {1}. Then, I1 / I2 is equal to x x
1 1
(a) (b) (c) n (d) n − 1 45. In a GP series consisting of positive terms,
n−1 n each term is equal to the sum of next two
terms. Then, the common ratio of this
40. The value of
GP series is
n n 1
lim + 2 + ... + is 5−1
n → ∞ n 2 + 12 n +2 2
2n (a) 5 (b)
2
nπ π 5 5+1
(a) (b) (c) (d)
4 4 2 2
π π
(c) (d) 46. If (log 5 x ) (log x 3x ) (log 3x y ) = log x x 3, then y
4n 2n
1
equals
41. The value of the integral ∫ ex dx
2
0
(a) 125 (b) 25
(c) 5/3 (d) 243
(a) is less than 1
(b) is greater than 1 (1 + i )n
47. The expression equals
(c) is less than or equal to 1 (1 − i )n − 2
(d) lies in the closed interval [1, e]
(a) − i n + 1 (b) i n +1
100
(c) − 2 i n + 1
∫
x − [ x] (d) 1
42. e dx is equal to
0
−1
e 100 48. Let z = x + iy , where x and y are real. The
(a) z+i
100 points ( x, y ) in the X-Y plane for which is
e100 − 1 z−i
(b)
e−1 purely imaginary, lie on
(c) 100( e − 1) (a) a straight line
e−1 (b) an ellipse
(d)
100 (c) a hyperbola
(d) a circle
dy
43. Solution of ( x + y )2 = a2 49. If p, q are odd integers, then the roots of the
dx
(‘a’ being a constant) is equation 2px 2 + ( 2p + q) x + q = 0 are
(x + y ) y +C (a) rational (b) irrational
(a) = tan , C is an arbitrary (c) non-real (d) equal
a a
constant 50. Out of 7 consonants and 4 vowels, words are
(b) xy = a tan Cx, C is an arbitrary constant formed each having 3 consonants and 2
x y vowels. The number of such words that can be
(c) = tan , C is an arbitrary constant
a C formed is
(d) xy = tan( x + C ), C is an arbitrary (a) 210 (b) 25200
constant (c) 2520 (d) 302400
14 WEST BENGAL (Engineering) Solved Paper 2017
0 0 n 2
56. If f ( x ) = ∫−1
| t| dt, then for any x ≥ 0, f ( x ) is
equal to
1
1 n 2n − 1 (a) (1 − x 2 ) (b) 1 − x 2
2
n+1 1
(d) 0 n2 (c) (1 + x 2 ) (d) 1 + x 2
2 2
0 n +1
0
2 57. Let for all x > 0, f ( x ) = lim n( x1 / n − 1), then
n→∞
a
Category-III (Q. 66 to Q.75) (a) a (b)
4
Direciton One or more answer(s) is/are a
correct. Correct answer(s) will fetch full marks 2. (c) (d) f ( a )
2
Any combination containing one or more
incorrect answer will fetch 0 marks. Also no 72. If the line ax + by + c = 0, ab ≠ 0, is a tangent
answer will fetch 0 marks. If all correct answers to the curve xy = 1 − 2x, then
are not marked and also no incorrect answer is (a) a > 0, b < 0
marked, then score = 2 × number of correct (b) a > 0, b > 0
answers marked ÷ actual number of correct (c) a < 0, b > 0
answers. (d) a < 0, b < 0
16 WEST BENGAL (Engineering) Solved Paper 2017
73. Two particles move in the same straight line 74. The complex number z satisfying the equation
starting at the same moment from the same | z − i| = | z + 1| = 1 is
point in the same direction. The first moves (a) 0
with constant velocity u and the second starts (b) 1 + i
from rest with constant acceleration f. Then, (c) − 1 + i
(a) they will be at the greatest distance at (d) 1 − i
u 75. On R, the set of real numbers, a relation ρ is
the end of time from the start
2f defined as ‘aρb if and only if 1 + ab > 0’. Then,
(b) they will be at the greatest distance at (a) ρ is an equivalence relation
u
the end of time from the start (b) ρ is reflexive and transitive but not
f symmetric
u2 (c) ρ is reflexive and symmetric but not
(c) their greatest distance is
2f transitive
u2 (d) ρ is only symmetric
(d) their greatest distance is
f
Physics
ANSWERS
1. (b) 2. (a) 3. (d) 4. (a) 5. (c) 6. (c) 7. (d) 8. (a) 9. (c) 10. (a)
11. (b) 12. (b) 13. (a) 14. (b) 15. (d) 16. (c) 17. (a) 18. (d) 19. (a) 20. (b)
21. (c) 22. (d) 23. (c) 24. (a) 25. (b) 26. (d) 27. (b) 28. (a) 29. (b) 30. (a)
31. (b) 32. (b) 33. (d) 34. (*) 35. (b) 36. (b,c) 37. (c) 38. (b,d) 39. (a,b,c,d) 40. (d)
Chemistry
41. (a) 42. (b) 43. (a) 44. (b) 45. (d) 46. (c) 47. (a) 48. (d) 49. (a) 50. (c)
51. (a) 52. (b) 53. (a) 54. (a) 55. (a) 56. (d) 57. (d) 58. (a) 59. (d) 60. (c)
61. (b) 62. (c) 63. (a) 64. (c) 65. (b) 66. (c) 67. (a) 68. (c) 69. (d) 70. (d)
71. (d) 72. (b) 73. (d) 74. (c) 75. (b) 76. (b,c,d) 77. (b, c) 78. (a,b) 79. (a) 80. (a,d)
Mathematics
1. (b) 2. (b) 3. (b) 4. (d) 5. (b) 6. (b) 7. (a) 8. (a) 9. (d) 10. (b)
11. (d) 12. (b) 13. (d) 14. (a) 15. (b) 16. (a) 17. (d) 18. (b) 19. (d) 20. (c)
21. (b) 22. (b) 23. (a) 24. (c) 25. (d) 26. (b) 27. (a) 28. (c) 29. (c) 30. (c)
31. (a) 32. (d) 33. (c) 34. (b) 35. (b) 36. (c) 37. (a) 38. (b) 39. (d) 40. (b)
41. (d) 42. (c) 43. (a) 44. (b) 45. (b) 46. (a) 47. (c) 48. (d) 49. (a) 50. (b)
51. (b) 52. (d) 53. (c) 54. (a) 55. (d) 56. (c) 57. (b) 58. (b) 59. (a) 60. (a)
61. (c) 62. (b) 63. (c) 64. (a) 65. (c) 66. (c) 67. (a, c) 68. (c) 69. (b) 70. (b)
71. (c) 72. (b, d) 73. (b,c) 74. (a,c) 75. (c)
HINTS & SOLUTIONS
Physics
1. (b) The speed of a particle executing simple 3. (d) When the sealed capillary tube is
harmonic motion is v = ω a 2 − x 2 submerged vertically into water, the pressure
inside the tube changes
where, a = Amplitude
ω = Angular frequency For the inside capillary,
x = Displacement p1 V1 = p 2V 2
or v 2 = ω 2( a 2 − x 2 ) ∴ p 0( lA ) = p ′ ( l − x ) A
According to the question, where p′ is pressure in capillary after being
v 12 = ω 2( a 2 − x12 ) submerged
v 22 = ω 2( a 2 − x 22 ) pl
∴ p′ = 0
v 1 − v 22 = ω 2( x 22 − x12 )
2 l−x
v 12 − v 22 According to question, since level of water
ω=
x 22 − x12 inside capillary coincides with outside.
2γ
Here v 1 = 13 m/s ∴ p′ − p0 =
r
v 2 = 12 m/s
p 0l 2γ l
x1 = 3 m ∴ − p0 = =
l−x r p 0r
x2 = 5 m 1 +
2γ
(13)2 − (12)2
ω= 4. (a) The Bulk modulus k
( 5)2 − ( 3)2
p
169 − 144 25 5 k=− … (i)
= = = ∆V / V
25 − 9 16 4 Where, p = Pressure
ω = 2πr ∆V = Change in volume
ω 5/ 4 V = Volume of liquid
v= =
2π 2π From (i)
5 p ∆V ∆p
v= =− =
8π k V p
2. (a) The formula of frequency of 3 vibrations k∆p
⇒ p=
produced in tensed wire is p
n T ∆p = 0 ⋅ 01% = 0 ⋅ 01 / 100
v=
2 m k
m p=
m = Mass per unit length of the wire = 10000
L
5. (c) Initial temperature of ideal gas,
M = Mass of wire
T1 = 273 + 27 = 300 K
L = Length of wire
when temperature of gas is raised by 6°C, the
n = Number of loops produced in wire final temperature of gas
n T n T T2 = 273 + 6 + 27
v= =
2L M / L 2 ML = 306 K
18 WEST BENGAL (Engineering) Solved Paper 2017
Let initial velocities are v rms1 , and v rms2 The coefficient of linear expansion along its
v rms ∝ T breadth = α 2
v rms1 Increase in length,
T1
= L t = l 0(1 + α1 ∆t )
v rms2 T2
Increase in breadth,
T2 Bt = b0(1 + α 2∆t 2 )
v rms2 = × v rms1
T1 Let coefficient of surface expansion is β
306 Area = length × breadth
= × v rms1
300 = l 0(1 + α1 ∆t ) × b0(1 + α 2∆t )
= 1 ⋅ 00 × v rms1 = l 0b0(1 + α1 ∆t )(1 + α 2∆t )
So, it will increase by 1% = S 0(1 + α1 ∆t + α 2∆t + . . . )
6. (c) The internal energy where, S 0 = l 0 ⋅ b0
∆U = nC v∆T = Initial area of surface
C v = Specific heat of gas at constant volume In state of expansion,
3R 4p 0V 0 p 0V 0 S t = Lt × Bt
⇒ ∆U = n ⋅ −
2 nR nR = l 0b0(1 + α1 ∆t ) (1 + α 2∆t )
3R 4p 0V 0 − p 0V 0 = S 0(1 + α1 ∆t + α 2∆t + . . . )
= n⋅ S t = S 0(1 + β∆t )
2 nR
3R 3p 0V 0 ∴ S 0(1 + β∆t ) = S 0 (1 + α1 ∆t + α 2∆t + . . . )
= n⋅ ⋅ β ⋅ ∆t = α1 ∆t + α 2∆t
2 nR
9 β = α1 + α 2
= p 0V 0 …(i)
2 8. (a) The positive charge Q is situated at the
Work done by the gas centre of a cube
V 3p 0V 0
W = ( 2p 0 + p 0 ) 0 = …(ii)
2 2
From first law of thermodynamics,
∆Q = dW + dU
3p 0V 0 9
= + p 0V 0 According to Gauss theorem, the flux from
2 2
any closed surface
[from Eqs. (i) and (ii)] Q
3p 0V 0 9 φE =
= + p 0V 0 ε0
2 2
12p 0V 0 Cube has 6 faces, so electric flux from any
= = 6p 0V 0 Q
2 face φ E =
6ε 0
7. (d) The coefficient of linear expansion along
its length = α1 9. (c) 1µF 2µF 5µF
C1 C2 C3
1 1 1 17 1
= + + = µF So, B∝
1 2 5 10 r2
10 So, the magnetic field due to current
C= µF
17 decreases as inverse square of distance of the
The charge will be same on all the capacitors point of observation.
in series 12. (b) In galvanometer I ∝ θ = Gθ
10 100 k
Q = CV = × 10 = Where, G=
17 17 NAB
The potential difference across 2.0µF When coil is in square of shape,
capacitor k k
Q 100 / 17 50 G= = … (i)
V′ = = = volt NAB Na 2B
C 2 17 When coil is in circular shape of radius
10. (a) Q1 Q2 a/ π
30 cm k k
G= =
Checking the given options one by one NAB Nπr 2 ⋅ B
(i) Q1 = Q 2 = 0 ⋅ 4 C k k
= 2
= … (ii)
The force on Q1 due to Q 2 a Na 2B
Nπ ⋅B
Q1 × Q 2 π
F = kp
30 × 10− 2 So, value of G is same for both type of shape
Q Q × 100 of coil. So, the torque e = NBIA sin θ will be
=k 1 2
30 same in both the case
0 ⋅ 4 × 0 ⋅ 4 × 100 e: e = 1:1
=k
30 13. (a) Velocity of proton = 106 m/s along
k × 0 ⋅ 16 × 100 y-direction
=
30 Y
(ii) When Q1 = 0 ⋅ 8C Q 2 ≈ 0
k × Q1Q 2
F= ≈0
30 × 10− 2 X
(iii) when Q1 ≈ 0, Q 2 = 0 ⋅ 8 C
Z
F=0
(iv) Q1 = 0 ⋅ 2 C Q 2 = 0 ⋅ 6 C Electric field = 2 × 104 V/m
k ⋅ 0⋅ 2 × 0⋅ 6 e
F= for proton = 10−8C / kg
30 × 10− 2 m
The magnetic field,
k × 0 ⋅ 12 × 102
= E
30 B= (Q qp vB = qp E )
v
Thus we find that, in option(a), the force on
2 × 104
Q1 will be maximum. = = 2 × 10− 2 T
106
∴ correct answer is Q1 = Q 2 = 0 ⋅ 4 C
The radius of circular path
11. (b) By Biot-Savart’s law, the magnetic field mv 10− 8 × 106
due to a current carrying wire is γ= =
qp ⋅ B 2 × 10− 2
µ Idl sin θ
B= 0 ⋅ 1
4π r2 = = 0⋅ 5 m
2
20 WEST BENGAL (Engineering) Solved Paper 2017
dSE
1 1 2kQ
= mv − ( M + m )V
2 2
=− ⋅x
2 2 ( x 2 + a 2 )3/ 2
1 1
= × 4.2 × 10−2 ( 300)2 − ( 4.2 × 10−2 2kQ
2 2 ∴ Fnet = − 3 ⋅ x
a
+ 9 × 4.2 × 1) ( 30)2
Frequency of oscillation
= 6.3 × 270
1 2kQ
= 1701 J =
2π Ma 3
1701
= Cal 1
4.2 2× ⋅Q
= 405 Cal 1 4πε 0
=
2π Ma 3
33. (d) From parabola
1 Q
Y =
2π 2πε 0Ma 3
E
⇒ (* None of the option matches)
35. (b)
1 µF 2µF
y
A C D
B
X 1kΩ 2kΩ
x=ut
3kΩ
1 Ee X 2 3V
y= × ×
2 m u 2 From circuit the current,
E⋅e E
= ⋅ X2 I=
2mu 2 R
As X = 4ay (for parabola) The equivalent resistance of above circuit
2 = 3+ 2+1
2mu
∴ X2 = ⋅y = 6k Ω
E⋅e
= 6 × 103 Ω
2mu 2 mu 2
a= = E = 3 volt
4E ⋅ e 2E ⋅ e
WEST BENGAL (Engineering) Solved Paper 2017 25
Chemistry
41. (a)For ideal gas, log
k
, log
k 42. (b) Q Surface tension ( T ) = Work done per
M2 M1 unit area
dW
∴ T=
dA
F ⋅ dx
= ( FL− 1 = MT − 2 )
dA
∴Dimension of surface tension = ML0T − 2
M2
M1
43. (a) According to ( n + l ) rule;
(i) More be the sum of ( n + l ) value, more be the
log p energy.
W
Q pV = RT …(i) (ii) For same value of sum, more be the value of n,
M more be the energy.
Let, WRT = K
For
Taking log of both the sides of Eq. (i)
K (i) n + l → 4 + 1 = 5
log p + log V = log
M (ii) n + l → 4 + 0 = 4
K (iii) n + l → 3 + 2 = 5
or log V = − log p + log …(ii)
M
(iv) n + l → 3 + 1 = 4
[On compairing Eq. (ii) with y = mx + C ]
K K Hence, order of energy will be
log > log or M1 > M2
M2 M1 (iv) < (ii) < (iii) < (i)
WEST BENGAL (Engineering) Solved Paper 2017 27
n1 WSO 2 MCH 4 F F F
= ×
n 2 MSO 2 WCH 4 The Xe show sp 3d 3-hybridisation in XeF5− ,
2 16 Thus, its geometry is
= ×
64 1 Pentagonal-bipyramidal.
n1 1
= , i.e. 1 : 2 8+ 5+1
But, number of electron pair = =7
n2 2 2
Also n ∝ N Number of bond pair = 52, number of lone
∴Ratio of number of molecules is 1 : 2. pair = 2
A lot of electrons are present at axil position
47. (a) 9 F18 → xE y + + 1 e0
and all bonds are in same plane. Hence, the
x = 8, y = 18, shape is planar.
∴ x E y
= 8O18
51. (a) According to molecular orbital theory,
QPosition has one unit of the charge and
outer configuration for
zero mass.
[C 2] = C12 → πp x = πp y
2 2
52. (b) Configuration for metal (M) in various 55. (a) Q Higher be the oxidation number of
complex. central atom in oxo-acids, more strongly it
3d 4s behave as Bronsted-acid.
Ni2+ : Oxidation number for Cl in HClO 3 → + 5
3d Oxidation number for Cl in HClO 2 → + 3
[NiCl4]2–= Oxidation number for Cl in HClO → + 1
4s
Oxidation number for Cl in HBrO → + 1
Hence, HClO 3 is the strongest Bronsted acid
in aqueous solution.
– – – –
Cl Cl Cl Cl
56. (d) As size of cation decreases, ionic character
Ni 2 + has 2 unpaired electrons decreases, thus basicity also decreases
3d
Hence, correct order is
[Ni(CO)4]=
La 3 + > Ce3 + > Eu 3 + > Lu 3 +
4s 4p
57. (d) Q If CO 23 − , SO 23 − and SO 24 − are present
alongwith BaCl 2, these can also show white
CO CO CO CO precipitate (as precipitate of all these are also
0 white).
Ni has zero unpaired electrons with
sp 3-hybridisation. 58. (a) IUPAC name of Ph ⋅ CH 2CH 2CO 2H is,
3d 3-phenylpropanoic acid
[Cu(NH3)4]2+=Cu2+= 3
CH2⋅ CH2⋅ COOH
4s 4p 2 1
Acid-base
+ H+ + NaOH reaction + H2O
+ MeCOOH OMe
(Product)
0 0
QO − is more reactive than OMe and
1 1 p-position is occupied by OMe, the
O O substitution occurs at ortho-position
w.r.t O − group.
C C is limiting reagent
As Me Me 63. (a) ADP is adenosine diphosphate
O
O (2-phosphate groups) and ATP is adenosine
triphosphate (3 phosphate groups)
NH C Me
Thus, they differ in number of phosphate units.
i.e. C8H9ON is 135 g/mol 64. (c) When primary amine react with chloroform
M. nt of
and ethanolic potassium hydroxide, they
O produce −isocyamides as main product.
The reaction is also known as carbyl amine
NH C Me reaction.
CHCl 3
Ph − NH 2 → Ph ⋅ N → C
As 1 mole of is formed KOH (ethanolic)
(Nauseating smell)
30 WEST BENGAL (Engineering) Solved Paper 2017
(ii) H2O/H+ C 2 = 0. 02
O
C NH
(On further
hydrolysis) C Thus, i1 × C1 = i 2 × C 2
Ph
Hence, are isotonic, i.e.
Ph
0. 06 = 0. 06
Ph In all other options, i1 × C1 ≠ i 2 × C 2
C Thus, are not isotonic.
O 69. (d) Cr2O72 − → Cr 3 + Change in + 3
66. (c) As per given relation,
Oxi no. = + 6
N1 V1 + N 2V 2 = N( V1 + V 2 )
Total change in number of electrons
N1 = 2 N 2 = 5
= 2 × 3 = 6 mole = 6 F
V1 = 500 V 2 = say-x
70. (d) For the given reactions,
Thus,
2 × 500 + 5x = 3( x + 500) or 2x = 500 2H 2O - 2H 2 + O 2, K1 = 6. 4 × 10− 8
i.e. x = 250 1
or, H 2O - H 2 + O 2, K ′1 = K1
Hence, maximum volume of 2
1
3M HCl = 500 + 250 = 750 mL or H 2 + O 2- H 2O, K1′′ = 1 / K1 … (i)
2
hc 1
67. (a)QE = hv = = hc ⋅ v Q = v 2CO 2 - 2CO + O 2, K 2 = 1.6 × 10− 6
λ λ
1
or CO 2 - CO + O 2, K ′ 2 = K 2 … (ii)
where E = energy of photon 2
c = velocity of photon (= light) Form (i) and (ii) [on adding]
λ = wavelength of photon K2 1.6 × 10− 6
h = plank’s constant. K= =
K1 6. 4 × 10− 8
∴ For (a)
102
E = 6 . 63 × 10− 34 × 3 × 108 / 300 × 10− 9 K= = 25 = 5
4
E = 1 . 98 × 10− 25 ⋅ J / 300 × 10− 9
1 . 98 × 10− 25 J 71. (d) Q For be packing,
∴ E= 3
300 × 10− 9 m 3a = 4r ∴ r = ⋅a
4
(a) → E = 6. 6 × 10− 19 J where a = edge length
For (b) E = hv = 6. 63 × 10−34 × 3 × 108 r = radius of lattice sphere
WEST BENGAL (Engineering) Solved Paper 2017 31
72. (b) When drop of line juice is added to (d) Fire can likely take place because
mixture of I− + IO −3 . The following reaction relations given given in (b) is highly
take place and violet colour appears due to exothermic.
formation of I2 ∴(b), (c) and (d) are correct.
I− + IO 3− + H + → I2 + H 2O
77. (b), (c) Among the given statements (b) and
73. (d) The reaction occurs as follows : (c) are correct these are the facts.
Mathematics
1. (b) Total number of 5-digit numbers having 5. (b) All the orthogonal matrix are non-singular
all the digits distinct = 10P5 − 9P4 matrix.
10 ! 9 ! 10 × 9 ! 9 ! ∴Q is proper subset of P.
= − = −
5! 5! 5! 5! 6. (b) We know that,
9! 9!
= (10 − 1) = ( 9) det ( AB) = det ( A ) det ( B)
5! ( 9 − 4)!
∴ det ( AB) = 0
= 9 × 9P4 ⇒ det ( A ) ⋅ det ( B) = 0
2. (b) ( p + 1) ( p + 2)( p + 3) . . . ( p + q) is the x + 2 3x x 0
product of q consecutive natural number ⇒ =0
3 x+2 5 x+2
( p, q ∈ N ). The product of q consecutive
natural number is always divisible by q !. ⇒ {( x + 2)2 − 9x} { x( x + 2) − 0} = 0
⇒ ( x 2 + 4x + 4 − 9x ) x( x + 2) = 0
3. (b) (1 + x + x 2 )9 = a 0 + a1 x + a 2x 2 +
⇒ x( x + 2) ( x 2 − 5x + 4) = 0
. . . + a18x18 ⇒ x( x + 2) ( x − 1) ( x − 4) = 0
Put x = − 1 , we get ⇒ x = 0, − 2 , 1, 4
(1 − 1 + 1)9 = a 0 − a1 + a 2 + . . . + a18
7. (a) We have,
⇒ 1 = a 0 − a1 + a 2 + . . . + a18 ...(i)
1 cos θ 0
Put x = 1, we get
| A | = − cos θ 1 cos θ
(1 + 1 + 1)9 = a 0 + a1 + a 2 + . . . + a18
–1 − cos θ 1
⇒ 39 = a 0 + a1 + a 2 + . . . + a18 ...(ii)
On adding Eqs. (i) and (ii), we get = 1 [1 − ( − cos θ ) (cos θ)]
39 + 1 = 2( a 0 + a 2 + . . . + a18 ) − cos θ [ − cos θ + cos θ] +0 (cos2 θ + 1)
39 + 1 = 1 + cos2 θ
⇒ a 0 + a 2 + a 4 + . . . + a18 =
2 Now, we know that
19683 + 1 − 1 ≤ cos θ ≤ 1
=
2 ⇒ 0 ≤ cos2 θ ≤ 1
=
19684 ⇒ 1 ≤ 1 + cos2 θ ≤ 2
2 ⇒ 1 ≤ | A |≤ 2
= 9842, which is even number. ∴ | A | ∈[1, 2]
4. (d) We have, 8x − 3y − 5z = 0, 8. (a) Since, f : R → R is injective and
5x − 8y + 3z = 0, f ( x ) f ( y ) = f ( x + y ), ∀ x, y ∈ R.
3x + 5y − 8z = 0 ∴ f (x) = ax
8 −3 −5 Again, f ( x ), f ( y ), f (z ) are in GP.
∴ D= 5 −8 3 ⇒ ( f ( y ))2 = f ( x ) ⋅ f ( y )
3 5 −8 ⇒ a 2y = a x ⋅ a z
⇒ a 2y = a x + z
= 8( 64 − 15) + 3( − 40 − 9) − 5( 25 + 24)
⇒ 2y = x + z
= 8 × 49 + 3 × ( − 49) − 5 × 49
∴ x, y, z are in AP.
=0
∴The system has infinitely many non-zero 9. (d) For every real number x, x 2 ≥ 0
solutions. ∴ ( x, x ) ∈ P
WEST BENGAL (Engineering) Solved Paper 2017 33
2x 2 − 4 17. (d)
⇒ =1 Y
−3
P(3, 6) y=x
⇒ 2x 2 − 4 = − 3
⇒ 2x 2 = 1
1 Q(6, 3)
⇒ x2 = X′ X
2
1
⇒ x=±
2
15. (b) Given equations are
Q′(–3,–6) y=–x
2x 2 + 3xy + 4y 2 + x + 18y + 25 = 0 ...(i) Y′
2x + 3xy + 4y + 1 = 0
2 2
...(ii) The coordinates of Q and Q′ will be ( 6, 3) and
Let the origin be transferred to ( p, q) axes ( − 3, − 6), respectively.
6− 3
being parallel to the previous axes; then the Now, Slope of PQ = = −1
equation (i) becomes. 3− 6
2( x ′ + p )2 + 3( x ′ + p ) ( y ′ + q) − 6− 3
and Slope of QQ′ = =1
+ 4( y ′ + q)2 + ( x ′ + p ) + 18( y ′ + q) + 25 = 0 − 3− 6
⇒ 2x ′ 2 + 2p 2 + 4x ′ p + 3x ′ y ′ + 3x ′ q + 3py ′ ∴ Slope of PQ × Slope of QQ′ = − 1 × 1 = − 1
+ 3pq + 4y ′ 2 + 4q2 + 8y ′ q + x ′ + p
∴∆PQQ′ is right angled triangle at Q.
+ 18y ′ + 18q + 25 = 0
∴Circumcentre will be the mid-point of
⇒ 2x ′ + 4y ′ + 3x ′ y ′ + ( 4p + 3q + 1)x ′
2 2
− 3 + 3 − 6 + 6
1 + ( 3p + 8q + 18)y ′ + 2p 2 + 3pq hypotenuse PQ′ = ,
2 2
+ 4q2 + p + 25 = 0
= ( 0, 0)
From Eq. (ii) coefficient of x′ and y ′ must be
zero. 18. (b) Let ( h, k ) be any point on the line
∴ 4p + 3q + 1 = 0 ...(iii) 7 x − 9y + 10 = 0, then 7 h − 9k + 10 = 0
3p + 8q + 18 = 0 ...(iv) ⇒ 7 h = 9k − 10
9k − 10
By solving Eqs. (iii) and (iv), we get ⇒ h= ...(i)
p = 2, q = − 3 7
Now, perpendicular distance from point
16. (a) Let the coordinates of C be (α, β). ( h, k ) to the line 3x + 4y = 5 is d1
∴Coordinates of centroid
3h + 4k − 5
2 − 2 + α − 3 + 1 + β d1 =
= , 32 + 42
3 3
3h + 4k − 5
α β − 2 ⇒ d1 = ...(ii)
= , 5
3 3
and perpendicular distance from ( h, k ) to
Since, centroid lie on 2x + 3y = 1. the line 12x + 5y = 7 is d 2
2α β − 2 12h + 5k − 7
∴ + 3 =1 ∴ d2 =
3 3
122 + 52
2α 3β − 6 12h + 5k − 7
⇒ + =1 ⇒ d2 = ...(iii)
3 3 13
⇒ 2α + 3β − 6 = 3 ⇒ 2α + 3β = 9 3h + 4k − 5 12h + 5k − 7
Now, d1 − d 2 = −
∴Locus of point C will be 2x + 3y = 9. 5 13
WEST BENGAL (Engineering) Solved Paper 2017 35
⇒ d1 − d 2 From figure,
13 ( 3h + 4k − 5) − 5 (12h + 5k − 7 ) OP = ( h + 1)2 + ( k − 1)2
=
65
In ∆OAP,
39h + 52k − 65 − 60h − 25k + 35
= OP
65 sin 45° =
OA
− 21 h + 27 k − 30
= 1 ( h + 1)2 + ( k − 1)2
65 ⇒ =
9k − 10 2 2
− 21 + 27 k − 30
7 On squaring both sides, we get
= [from Eq.(i)]
65 ( h + 1)2 + ( k − 1)2 = 2
− 27 k + 30 + 27 k − 30 ⇒ h 2 + k 2 + 2h − 2k = 0
= =0
65 ∴Locus of P will be
⇒ d1 − d 2 = 0 ⇒ d 1 = d 2 x 2 + y 2 + 2x − 2y = 0
19. (d) Given, equation of circles are 21. (b) Given, equation of hyperbola is
x 2 + y 2 − 4x − 4y = 0 and 2x 2 + 2y 2 = 32 x2 y 2
− =1
or x 2 + y 2 − 4x − 4y = 0 a 2 b2
and x 2 + y 2 = 16
b
∴Equation of common chord is y= a x
( x 2 + y 2 − 4x − 4y ) − ( x 2 + y 2 − 16) = 0 P
⇒ − 4x − 4y + 16 = 0
⇒ x+y =4 C S(ae, 0)
This common chord passes through (2, 2),
i.e. centre of first circle.
Also, (0, 0) is at the circumference of the first
circle.
π From figure,
∴Common chord will subtent angle at
2 abe
(0, 0). SP =
b2 + a 2
20. (c) Given, equation of circle is
abe
x 2 + y 2 + 2x − 2y − 2 = 0 = =b
ae
⇒ ( x + 1)2 + ( y − 1)2 = 4
and CS = ae
∴Centre ( − 1, 1) and radius = 2
Again, ∆SPC is right angled triangle at P.
Let ( h, k ) be the mid-point of chord.
∴ CP = CS 2 − SP 2
= a 2e2 − b2
b2
= a 2 1 + 2 − b2
O(–1, 1) a
= a 2 + b2 − b2 = a
⇒ a 5k P = q 2f ( x ) − 3f ( 2x ) + f ( 4x )
30. (c) lim
q x→ 0 x2
⇒ a 5k =
P 2f ′ ( x ) − 3f ′ ( 2x ) ⋅ 2 + f ′ ( 4x ) ⋅ 4
= lim
Now, f ′ ( − 5) = ka − 5k log a x→ 0 2x
=
k log a f ′ ( x ) − 3f ′ ( 2x ) + 2f ′ ( 4x )
= lim
a 5k x→ 0 x
p p2 f ′′( x ) − 3f ′′( 2x ) ⋅ 2 + 2f ′′( 4x ) ⋅ 4
= = = lim
q q x→ 0 1
p = lim f ′′( x ) − 6f ′′( 2x ) + 8f ′′( 4x )
x→ 0
log x
log dy −1 m
log 3 ⇒ = em sin x ⋅
= dx 1 − x2
log 5
dy −1
log log x − log log 3 ⇒ 1 − x2 = mem sin x
= dx
log 5
2 dy
1 1 1 ⇒ 1− x = my …(i)
∴ f ′ (x) = ⋅ ⋅ dx
log 5 log x x d 2y 1 dy dy
1 ⇒ 1 − x2 2
+ ( − 2x ) =m
∴ f ′ ( e) = dx 2 1− x 2 dx dx
e log 5 ⋅ log e
d 2y dy dy
1 ⇒ (1 − x 2 ) −x = m 1 − x2
= [Q log e = 1] dx 2 dx dx
e log 5 2
dy dy
1 ⇒ (1 − x 2 ) 2 − x = m ⋅ ( my )
= dx dx
e log e 5
[From Eq. (i)]
29. (c) We have, F( x ) = ex, G( x ) = e− x dy2
dy
⇒ (1 − x 2 ) 2 − x − m 2y = 0
∴ H( x ) = G( F( x )) dx dx
= G( ex ) ∴ k = m2
= e− e
x
y = α 2 + 2aα + b
dH and at x = β,
= − e0 ⋅ e− e
0
∴
dx x = 0 y = β 2 + 2aβ + b
= –1. e–1 ∴Slope of line joining (α, α 2 + 2aα + b) and
= − e− 1 (β, β 2 + 2aβ + b) is
−1 (α 2 + 2aα + b) − (β 2 + 2aβ + b)
= =
e α −β
38 WEST BENGAL (Engineering) Solved Paper 2017
=
α 2 + 2aα + b − β 2 – 2aβ – b 36. (c) Let I = ∫ cos (log x ) dx
α −β
Put log x = t
(α 2 – β 2 ) + 2a(α − β ) ⇒ x = et
=
α −β ∴ dx = et dt
= (α – β )(α + β ) + 2a(α – β )
∴ I= ∫e
t
cos t dt
α–β
t
e
= α + β + 2a = [cos t + sin t] + C
12 + 12
dy
Slope of given curve =
dx e ax
Q ∫ e cos bx dx = a2 + b 2 [ a cos bx + b sin bx] + C
ax
= 2x + 2a
Now, according to question, tangent is et
⇒ I= [cos t + sin t] + C
parallel to the chord. Therefore, 2
2x + 2a = α + β + 2a x
= [cos(log x ) + sin(log x] + C
α +β 2
⇒ 2x = α + β ⇒ x =
2 x
∴f ( x ) = [cos(log x ) + sin(log x )]
33. (c) We have, 2
f ( x ) = x13 + x11 + x 9 + x7 + x 5 + x 3 + x + 19 x2 − 1
⇒ f ′ ( x ) = 13x12 + 11x10 + 9x 8
37. (a) Let I = ∫ x 4 + 3x 2 + 1
dx
+ 7 x 6 + 5x 4 + 3x 2 + 1 1 − 1 / x2
∴f ′ ( x ) has no real root.
= ∫x 2
+ 3 + 1 / x2
dx
∴f ( x ) = 0 has not more than one real root. 1 − 1 / x2
34. (b) Since, Lagrange’s mean value theorem is
= ∫ 2 1
dx
x + 2 + 3
applicable on f ( x ). x
xP
∴ lim = f ( 0) 1 − 1 / x2
x → 0 (sin x )q = ∫ 1
2
dx
x + − 2 + 3
P
x
⇒ lim =0 x
x→ 0 (sin x )q
1 − 1 / x2
Above equation holds only when p > q. = ∫ 1
2
dx
35. (b) Let y = lim(sin x ) 2 tan x
x + + 1
x→ 0 x
⇒ log y = lim log(sin x )2 tan x 1
x→ 0
Let x + =t
= 2 lim tan x log sin x x
x→ 0
log sin x 1
= 2 lim ⇒ 1 − 2 dx = dt
x→ 0 cot x x
1 dt
⋅ cos x ∴ I= ∫t 2
+1
= 2 lim sin x
x → 0 − cosec 2x
= tan − 1 t + C
= 2 lim( − sin x cos x ) 1 1
x→ 0 = tan − 1 x + + C Q t = x + x
= 2 × 0= 0 x
∴ log y = 0 ⇒ y = e0 = 1
WEST BENGAL (Engineering) Solved Paper 2017 39
( n − 1) ( n − ( n − 1)) ⇒ ex ≤ e
= 1 + 2 + 3 + K + ( n − 1)
1 1
∫ e ∫
x2
⇒ dx ≤ e dx
0 0
( n − 1) ( n − 1 + 1) n( n + 1)
= Q Σn =
1
∫ e
x2
2 2 ⇒ dx ≤ e
0
n( n − 1) 1
∫ e
2
= ∴ x
∈[1, e]
0
2
100
∫ ex − [ x]dx
n n
42. (c) Let I =
Now, I2 = ∫ 0
{ x} dx = ∫0
x − [ x] dx 0
1
= 100 ∫ ex − [ x] dx
n n
=∫ x dx − ∫ [ x] dx 0
0 0
0 and ∫ f ( x ) dx = m ∫ f ( x ) dx , where T is
0 0
n 2 n( n − 1) period of f ( x )]
= − 1
2 2 = 100 ∫ ex dx [Qx − [ x] = x for 0 < x < 1]
0
n2 − n2 + n
= = 100 [ ex]10
2
n = 100 [ e1 − e0]
=
2 = 100 ( e − 1)
n( n − 1) 43. (a) We have,
I1 2
∴ = dy
I2 n ( x + y )2 = a2
dx
2
Let x+y =v
= ( n − 1)
40 WEST BENGAL (Engineering) Solved Paper 2017
⇒ 1+
dy dv
= Put x = 0,
dx dx ∴ 1= − A
dy dv ⇒ A = −1
⇒ = −1
dx dx Put x = 1,
dv ∴ 2 = 2B
∴ v 2 − 1 = a 2
dx
⇒ B=1
dv
⇒ v2 = v 2 + a2 Put x = − 1,
dx
∴ 2 = 2C
dv v 2 + a 2
⇒ = ⇒ C =1
dx v2
v 2
x +1
2
−1 1 1
⇒ dv = dx ∴ = + +
v 2 + a2 x( x − 1) ( x + 1) x x−1 x+1
−1 1 1
On integrating both sides, we get ∫ + + dx
x x−1 x + 1
v2 ∴ IF = e
∫ v 2 + a 2 dv = ∫ dx = e[ − log x + log( x − 1 ) + log( x + 1)]
x2 − 1
a2 log
x2 − 1
∫ 1 − v 2 + a 2 dv = x + C ′
⇒ =e x
=
x
a2 v =x−
1
⇒ v− tan − 1 = x + C ′
a a x
−1 x + y 45. (b) Let a n be the general term of a GP whose
⇒ x + y − a tan = x + C′
a first term is a and common ratio is r.
x+y Now according to the question,
⇒ y = a tan − 1 + C′
a an = an + 1 + an + 2
y – C′ x+y ⇒ ar n − 1 = ar n + ar n + 1
= = tan –1
a a ⇒ rn − 1 = rn + rn + 1
x + y y + C rn rn + 1
⇒ = tan , ⇒ 1 = n −1 + n −1
a a r r
where − C ′ = C. ⇒ 1= r + r 2
log a x2 + y 2 − 1
[Q log b a = and log a m = m log a] ⇒ =0
log b x 2 + ( y − 1)2
⇒
log y
=3 [Q log a a = 1] ⇒ x2 + y 2 = 1
log 5 ∴( x, y ) lies on a circle.
⇒ log y = 3 log 5 49. (a) Given equation,
⇒ log y = log 53 2px 2 + ( 2p + q) x + q = 0
⇒ y = 53 = 125 ∴D = ( 2p + q)2 − 4( 2p ) ( q)
(1 + i ) n
(1 + i )n = 4p 2 + q2 + 4pq − 8pq
47. (c) =
(1 − i )n − 2 (1 − i )n (1 − i )− 2 = 4p 2 + q2 − 4pq
n = ( 2p − q)2
1 + i
= (1 − i )
2
= a perfect square
1 − i
∴Given equation has rational roots
n
1 + i 1 + i 50. (b) Out of 7 consonants, the number of ways
= × (1 + i − 2i )
2
1 − i 1 + i of selecting 3 consonants = 7C 3
n
1 + 2i + i 2 Similarly, number of ways of selecting
= (1 + i − 2i )
2
2 vowels out of 4 vowels = 4C 2
1 − i2
n
∴Total number of words formed
1 + 2i − 1 = 7C 3 × 4C 2 × 5P5
= (1 − 1 − 2i )
1 − ( − 1) 7 × 6× 5 4× 3
= × × 5!
[Q i 2 = − 1] 3× 2×1 2×1
n
2i = 7 × 5 × 2 × 3 × 120 = 25200
= ( − 2i )
2 1 1 1
= i ( − 2i ) = − 2i
n n +1
51. (b) We have, A = 0 1 1
z + i ( x + iy ) + i 0 0 1
48. (d) =
z − i ( x + iy ) − i ∴ A2 = A ⋅ A
x + (1 + y ) i 1 1 1 1 1 1
=
x + ( y − 1) i = 0 1 1 0 1 1
x + ( y + 1) i x − ( y − 1) i 0 0 1 0 0 1
= ×
x + ( y − 1) i x − ( y − 1) i
1 + 0 + 0 1 + 1 + 0 1 + 1 + 1
=
x 2 − x( y − 1) i + x( y + 1) i − ( y + 1) ( y − 1) i 2 = 0 + 0 + 0 0 + 1 + 0 0 + 1 + 1
x 2 − ( y − 1)2 i 2
0 + 0 + 0 0 + 0 + 0 0 + 0 + 1
x + i[ − xy + x + xy + x] + ( y 2 − 1)
2
= 2( 2 + 1)
x 2 + ( y − 1)2 1 2 3 1 2
2
[Q i 2 = − 1 ] = 0 1 2 = 0 1 2
x +y2 2
−1 2x 0 0 1 0 0 1
= + i
x2 + (y − 1)2 x 2 + ( y − 1)2
z+ i Again, A 3 = A 2 ⋅ A
Now, is purely imaginary.
z− i 1 2 3 1 1 1
z + i = 0 1 2 0 1 1
∴
Re =0 0 0 1 0 0 1
z − i
42 WEST BENGAL (Engineering) Solved Paper 2017
a a+1 a−1
A (1, 1)
⇒ − b b+1 b−1
c c−1 c+1 √5
n + 2
a + 1 a – 1 ( − 1) a C
2
B(2, 3)
+ b + 1 b − 1 ( − 1)n + 1 b = 0
c − 1 c + 1 ( − 1)n c
∴ AB = ( 2 − 1)2 + ( 3 − 1)2 = 1 + 4 = 5
a a+1 a−1
∴ AC = AB2 + BC 2 = ( 5 )2 + ( 2)2
⇒ − b b+1 b−1
= 5+ 4= 9 = 3
c c−1 c+1
+ 2 ∴Radius of required circle = 3
( − 1)n a a+1 a−1
+ ( − 1)n +1
b b+1 b−1 = 0 55. (d) (h, k)
M
( − 1) c
n
c−1 c+1 Y
a(1 + ( − 1)n + 2 ) a + 1 a − 1
⇒ b( − 1 + ( − 1)n + 1 ) b + 1 b − 1 = 0
c(1 + ( − 1)n ) c−1 c+1 A(–1, 0) θ 2θ
X
(0, 0) B(2, 0)
∴n is any odd integer. Y′
WEST BENGAL (Engineering) Solved Paper 2017 43
2 / (1 + h ) 1
58. (b) I = ∫ 0
1 − cos 2x dx
⇒ =
(1 + h )2 − k 2 2 − h 100π
(1 + h )2
= ∫
0
2 sin 2 x dx
2 (1 + h )2 1 100π
⇒ × = = 2∫ sin x | dx
1 + h (1 + h )2 − k 2 2 − h 0
π
2(1 + h ) 1 = 2 × 100 ∫ sin x | dx|
⇒ = 0
(1 + h ) − k
2 2
2− h [| sin x | has period of π]
⇒ 1 + h 2 + 2h − k 2 = 2(1 + h ) ( 2 − h ) π
⇒ 1 + h 2 + 2h − k 2
= 100 2 ∫
0
sin x dx
B Now, ( x × $i )2 = ( x × $i ) ⋅ ( x × $i )
5
P 2, 3 = ( − βk$ + γ $j ) ⋅ ( − βk$ + γ $j )
= β2 + γ 2
Similarly, ( x × $j )2 = α 2 + γ 2
C F2(–2, 0) O F1(2, 0) A
and ( x × k$ )2 = α 2 + β 2
D ∴( x × $i )2 + ( x × $j )2 + ( x × k$ )2
Equation of tangent at the end of = β2 + γ 2 + α2 + γ 2 + α2 + β2
latusrectum P is = 2(α 2 + β 2 + γ 2 ) = 2| x |2
WEST BENGAL (Engineering) Solved Paper 2017 45
Physics
Category I (Q.1 to Q.30) Only one answer is correct. Correct answer will fetch full mark 1. Incorrect
1
answer or any combination of more than one answer will fetch - mark.
4
1. Equivalent capacitance between A and B in 4. The potential difference V required for
the figure is accelerating an electron to have the
4mF 4mF de-Broglie wavelength of 1 Å is
A B (a) 100 V (b) 125 V
4mF (c) 150 V (d) 200 V
5. The work function of Cesium is 2.27 eV. The
4mF 4mF cut-off voltage which stops the emission of
C electrons from a cesium cathode irradiated
with light of 600 nm wavelength is
(a) 20 mF (b) 8 mF
(a) 0.5 V (b) -0. 2 V
(c) 12 mF (d) 16 mF
(c) -0.5 V (d) 0.2 V
2. Two wires of same radius having lengths l 1
6. The number of de-Broglie wavelengths
and l2 and resistivities r 1 and r2 are contained in the second Bohr orbit of
connected in series. The equivalent hydrogen atom is
resistivity will be (a) 1 (b) 2
r1l2 + r 2 l1 r1l1 + r 2 l2
(a) (b) (c) 3 (d) 4
r1 + r 2 l1 + l2
r1l1 - r 2 l2 r1l2 + r 2 l1 7. The wavelength of second Balmer line in
(c) (d)
l1 - l2 l1 + l2 hydrogen spectrum is 600 nm. The
wavelength for its third line in Lyman series
3. A hollow metal sphere of radius R is charged is
with a charge Q. The electric potential and (a) 800 nm (b) 600 nm
intensity inside the sphere are respectively (c) 400 nm (d) 200 nm
Q Q
(a) and 8. A ray of light strikes a glass plate at an angle
4p e0 R 2 4p e0 R
Q
of 60°. If the reflected and refracted rays are
(b) and zero perpendicular to each other, the refractive
4p e0 R
index of glass is
(c) zero and zero
4p e0Q Q 3 3 1
(d) and (a) (b) (c) (d) 3
R 4p e0 R 2 2 2 2
2 WB JEE (Engg.) Solved Paper 2016
9. Light travels through a glass plate of 17. A particle vibrating simple harmonically has
thickness t and having refractive index m. If c an acceleration of 16 cms -2 when it is at a
be the velocity of light in vacuum, time taken distance of 4 cm from the mean position. Its
by the light to travel through this thickness time period is
of glass is (a) 1 s (b) 2.572 s
t tc mt (c) 3.142 s (d) 6.028 s
(a) (b) (c) (d) m tc
mc m c
18. Work done for a certain spring when
2
10. If x = at + bt , where x is in metre ( m ) and t is stretched through 1 mm is 10 joule. The
in hour ( h ), then unit of b will be amount of work that must be done on the
m2 spring to stretch it further by 1 mm is
(a) (b) m
h (a) 30 J (b) 40 J
m m (c) 10 J (d) 20 J
(c) (d)
h h2 19. If the rms velocity of hydrogen gas at a
11. The vectors A and B are such that certain temperature is c, then the rms
| A + B| = | A - B|. The angle between the two velocity of oxygen gas at the same
vectors will be temperature is
(a) 0° (b) 60° (c) 90° (d) 45° c c
(a) (b)
8 10
12. At a particular height, the velocity of an c c
ascending body is u. The velocity at the same (c) (d)
4 2
height while the body falls freely is
(a) 2 u (b) -u (c) u (d) -2 u 20. For air at room temperature, the atmospheric
pressure is 1.0 ´ 105 Nm -2 and density of air
13. Two bodies of masses m 1 and m 2 are
separated by a distance R. The distance of the is 1.2 kgm -3 . For a tube of length 1.0 m,
centre of mass of the bodies from the mass closed at one end, the lowest frequency
m 1 is generated is 84 Hz. The value of g
m2 R m1R (ratio of two specific heats) for air is
(a) (b)
m1 + m2 m1 + m2 (a) 2.1 (b) 1.5 (c) 1.8 (d) 1.4
m1m2 m + m2
(c) R (d) 1 R 21. A gas bubble of 2 cm diameter rises through a
m1 + m2 m1
liquid of density 1.75 g cm -3 with a fixed
14. The velocity of sound in air at 20°C and 1 atm speed of 0.35 cms -1. Neglect the density of the
pressure is 344.2 m/s. At 40°C and 2 atm gas. The coefficient of viscosity of the liquid is
pressure, the velocity of sound in air is (a) 870 poise (b) 1120 poise
approximately (c) 982 poise (d) 1089 poise
(a) 350 m/s (b) 356 m/s 22. The temperature of the water of a pond
(c) 363 m/s (d) 370 m/s
is 0 °C while that of the surrounding
15. The perfect gas equation for 4 g of hydrogen atmosphere is -20 °C. If the density of ice
gas is is r, coefficient of thermal conductivity is k
(a) pV = RT (b) pV = 2 RT and latent heat of melting is L, then the
1
(c) pV = RT (d) pV = 4RT thickness Z of ice layer formed increases as a
2 function of time t is
16. If the temperature of the Sun gets doubled, (a) Z 2 =
60k
t (b) Z =
40k
t
the rate of energy received on the Earth will rL rL
increase by a factor of 40k 40k
(c) Z 2 = t (d) Z 2 = t
(a) 2 (b) 4 (c) 8 (d) 16 rL rL
WB JEE (Engg.) Solved Paper 2016 3
23. 1000 droplets of water having 2 mm 27. Two coils of self-inductances 6 mH and
diameter each coalesce to form a single drop. 8 mH are connected in series and are
Given the surface tension of water is adjusted for highest coefficient of coupling.
0.072 Nm -1. The energy loss in the process is Equivalent self-inductance L for the assembly
(a) 8146
. ´ 10-4 J (b) 4.4 ´ 10-4 J is approximately
(c) 2.108 ´ 10-5
J . ´ 10-1 J
(d) 47 (a) 50 mH
(b) 36 mH
24. A Zener diode having break-down voltage (c) 28 mH
5.6 V is connected in reverse bias with a (d) 18 mH
battery of emf 10 V and a resistance of 100 W 28. A 1 mF capacitor C is connected to a battery of
in series. The current flowing through the
10 V through a resistance 1 MW. The voltage
Zener diode is
across C after 1 s is approximately
(a) 88 mA (b) 0.88 mA
(c) 4.4 mA (d) 44 mA (a) 5.6 V (b) 7.8 V
(c) 6.3 V (d) 10 V
25. In case of a bipolar transistor b = 45. The
29. Two equal resistances, 400 W each, are
potential drop across the collector resistance
connected in series with a 8 V battery. If the
of 1 kW is 5V. The base current is
resistance of first one increases by 0.5%, the
approximately
change required in the resistance of the
(a) 222 m A (b) 55 m A (c) 111m A (d) 45 m A
second one in order to keep the potential
26. An electron enters an electric field having difference across it unaltered is to
intensity E = 3 i$ + 6 $j + 2 k$ Vm -1 and (a) increase it by 1W
magnetic field having induction (b) increase it by 2 W
(c) increase it by 4 W
B = 2 i$ + 3 $j T with a velocity v = 2 i$ + 3 $j ms -1.
(d) decrease it by 4 W
The magnitude of the force acting on the
electron is 30. Angle between an equipotential surface and
(Given, e = - 1.6 ´ 10 -19 C) electric lines of force is
(a) 0° (b) 90°
(a) 2.02 ´ 10-18 N . ´ 10-16 N
(b) 516 (c) 180° (d) 270°
. ´ 10-17 N
(c) 372 (d) 4.41 ´ 10-18 N
Category II (Q. 31 to Q. 35) Only one answer is correct. Correct answer will fetch full marks 2. If correct
1
answer are any combination of more than one answer will fetch - mark.
2
31. A current I = I0e - lt is flowing in a circuit (d) Both source and screen should be at finite
consisting of a parallel combination of distance
resistance R and capacitance C. The total 33. The temperature of a blackbody radiation
charge over the entire pulse period is enclosed in a container of volume V is
I
(a) 0
2I
(b) 0 increased from 100 °C to 1000 °C. The heat
l l required in the process is
(c) I0 l (d) e I 0 l . ´ 10-4 cal
(a) 479 . ´ 10-5 cal
(b) 921
-4
(c) 2.17 ´ 10 cal (d) 7.54 ´ 10-4 cal
32. For Fraunhoffer diffraction to occur
(a) Light source should be at infinity 34. A mass of 1 kg is suspended by means of a
(b) Both source and screen should be at infinity thread. The system is (i) lifted up with an
(c) Only the source should be at finite distance acceleration of 4.9 ms -2 . (ii) lowered with an
4 WB JEE (Engg.) Solved Paper 2016
acceleration of 4.9 ms -2 . The ratio of tension between the same two points, when the link
in the first and second case is AB is removed, is
(a) 3 : 1 (b) 1 : 2 B
(c) 1 : 3 (d) 2 : 1
A
35. The effective resistance between A and B in
7
the figure is W if each side of the cube has
12 7 5 7 5
(a) W (b) W (c) W (d) W
1W resistance. The effective resistance 12 12 5 7
Category III (Q. 36 to Q. 40) One or more answer (s) is (are) correct. Correct answers will fetch marks 2.
Any combination containing one or more incorrect answer (s) will fetch 0 mark. If all correct answers are
not marked and also no incorrect answer is marked then score = 2 ´ number of correct answers
marked/actual number of correct answers.
36. A charged particle of mass m 1 and charge q 1 38. A train moves from rest with acceleration a
is revolving in a circle of radius r. Another and in time t 1 covers a distance x. It then
charged particle of charge q 2 and mass m 2 is decelerates to rest at constant retardation b
situated at the centre of the circle. If the for distance y in time t2 . Then,
x b b t1
velocity and time period of the revolving (a) = (b) =
particle be v and T respectively, then y a a t2
x bt 1
q1q 2 r (c) x = y (d) =
(a) v = y at 2
4p e0 m1
(b) v =
1 q1q 2 39. A drop of water detaches itself from the exit
m1 4 p e0 r of a tap when (s = surface tension of water,
3 r = density of water, R = radius of the tap
16 p e0 m12 r 3
(c) T = exit,r = radius of the drop)
q1q 2 1/ 3
æ 2 Rs ö
16 p e0 m2 r 3
3 (a) r > ç ÷
(d) T = è 3 rg ø
q1q 2
2 s
(b) r =
37. The distance between a light source and 3 rg
photoelectric cell is d. If the distance is 2s
(c) > atmospheric pressure
d r
decreased to , then 2/ 3
2 æ 2 Rs ö
(d) r > ç ÷
(a) the emission of electron per second will be four è 3 rg ø
times
(b) maximum kinetic energy of photoelectrons will be 40. A rectangular coil carrying-current is placed
four times in a non-uniform magnetic field. On that
(c) stopping potential will remain same coil, the total
(d) the emission of electrons per second will be (a) force is non-zero (b) force is zero
doubled (c) torque is zero (d) torque is non-zero
Chemistry
Category I (Q. 1 to Q. 30). Only one answer is correct. Correct answer will fetch full mark 1. Incorrect
1
answer or any combination of more than one answer will fetch – mark.
4
1. Amongst the following compounds, the one 6. Which of the following reactions will not
that will not respond to Cannizzaro reaction result in the formation of carbon-carbon
upon treatment with alkali is bonds?
(a) Cl 3CCHO (b) Me 3CCHO (a) Cannizzaro reaction
(c) C 6H5CHO (d) HCHO (b) Wurtz reaction
(c) Reimer-Tiemann reaction
2. Which of the following compounds would
(d) Friedel-Crafts acylation
not react with Lucas reagent at room
temperature? 7. Point out the false statement.
(a) H2C== CHCH2OH (a) Colloidal sols are homogeneous
(b) C 6H5CH2OH (b) Colloids carry + ve or - ve charges
(c) CH3CH2CH2OH (c) Colloids show Tyndall effect
(d) (CH3 )3 COH (d) The size range of colloidal particles is 10-1000 Å
3. Amongst the following compounds, the one 8. The correct structure of the drug paracetamol is
which would not respond to iodoform test is OH OH
(a) CH3CH(OH)CH2CH3
(a) (b)
(b) ICH2COCH2CH3
(c) CH3COOH
CONH2 NHCOCH3
(d) CH3CHO
Cl Cl
4. Which of the following will be dehydrated
most readily in alkaline medium? (c) (d)
O O OH
CONH2 COCH3
(a) (b)
OH 9. Which of the following statements regarding
Lanthanides is false?
OH O
OH
(a) All lanthanides are solid at room temperature
(c) (d) (b) Their usual oxidation state is +3
(c) They can be separated from one another by
5. The correct order of basicity of the following ion-exchange method
compounds is (d) Ionic radii of trivalent lanthanides steadily
increases with increase in atomic number
NH2 NH
1 2 10. Nitrogen dioxide is not produced on heating
NH2 NH2 (a) KNO 3 (b) Pb(NO 3 )2
(c) Cu(NO 3 )2 (d) AgNO 3
NH H2N NH 11. The boiling points of HF, HCl, HBr and HI
3 4
follow the order
(a) 1 < 2 < 3 < 4 (b) 1 < 2 < 4 < 3 (a) HF >HCl >HBr >HI (b) HF >HI >HBr >HCl
(c) 2 < 1 < 3 < 4 (d) 4 < 3 < 2 < 1 (c) HI >HBr >HCl >H (d) HCl >HF >HBr >HI
6 WB JEE (Engg.) Solved Paper 2016
12. In the solid state, PCl5 exists as 20. The number of s and p bonds between two
(a) [PCl 4 ]- and [PCl 6 ]+ ions carbon atoms in calcium carbide are
(b) covalent PCl 5 molecules only (a) one s, one p (b) one s, two p
(c) [PCl 4 ]+ and [PCl 6 ]- ions 1
(c) two s, one p (d) one s, 1 p
(d) covalent P2Cl10 molecules only 2
13. Which statement is not correct for ortho and 21. An element E loses one a and two b particles
para hydrogen? in three successive stages. The resulting
(a) They have different boiling points element will be
(b) Ortho-form is more stable than para-form (a) an isobar of E (b) an isotone of E
(c) They differ in their nuclear spin (c) an isotope of E (d) E itself
(d) The ratio of ortho to para hydrogen changes with 22. An element X belongs to fourth period and
change in temperature
fifteenth group of the periodic table. Which
14. The acid in which O ¾ O bonding is present of the following statements is true?
is (a) It has completely filled s-orbital and a partially
(a) H2S2O 3 (b) H2S2O 6 filled d-orbital
(c) H2S2O 8 (d) H2S4O 6 (b) It has completely filled s-and p -orbitals and a
partially filled d-orbital
15. The metal which can be used to obtain (c) It has completely filled s- and p-orbitals and a half
metallic Cu from aqueous CuSO4 solution is filled d-orbital
(a) Na (b) Ag (d) It has a half filled p -orbital and completely filled
(c) Hg (d) Fe s-and d-orbitals
16. If radium and chlorine combine to form 23. Which of the following plots represent an
radium chloride, the compound would be exothermic reaction?
(a) half as radioactive as radium
(b) twice as radioactive (a) In Kp (b) In Kp
(c) as radioactive as radium
(d) not radioactive 1/T 1/T
17. Which of the following arrangements is
correct in respect of solubility in water? (c) In Kp (d) In Kp
(a) CaSO 4 >BaSO 4 >BeSO 4 >MgSO 4 >SrSO 4
(b) BeSO 4 >MgSO 4 >CaSO 4 >SrSO 4 >BaSO 4 1/T 1/T
(c) BaSO 4 >SrSO 4 >CaSO 4 >MgSO 4 >BeSO 4 24. If p° and p are the vapour pressure of the pure
(d) BeSO 4 >CaSO 4 >MgSO 4 >SrSO 4 >BaSO 4 solvent and solution and n 1 and n2 are the
18. The energy required to break one mole of moles of solute and solvent respectively in
hydrogen-hydrogen bonds in H2 is 436 kJ. the solution then the correct relation
What is the longest wavelength of light between p and p° is
required to break a single hydrogen- é n1 ù é n2 ù
(a) p° = p ê ú (b) p° = p ê ú
hydrogen bond? ë n1 + n2 û ë n1 + n2 û
(a) 68.5 nm (b) 137 nm é n2 ù é n1 ù
(c) 274 nm (d) 548 nm (c) p = p° ê ú (d) p = p° ê ú
ë n1 + n2 û ë n1 + n2 û
19. The correct order of O ¾ O bond length in
25. Ionic solids with Schottky defect may
O2 , H2O2 and O3 is
contain in their structure
(a) O 2 >O 3 >H2O 2
(a) cation vacancies only
(b) H2O 2 >O 3 >O 2
(b) cation vacancies and interstitial cations
(c) O 3 >O 2 >H2O 2
(c) equal number of cation and anion vacancies
(d) O 3 >H2O 2 > O 2
(d) anion vacancies and interstitial anions
WB JEE (Engg.) Solved Paper 2016 7
26. The condition for a reaction to occur 29. Ozonolysis of an alkene produces only one
spontaneously is dicarbonyl compound. The structure of the
(a) DH must be negative alkene is
(b) DS must be negative
(c) (DH - TDS ) must be negative (a) H3C CH CH CH3
(d) (DH + TDS ) must be negative
(b)
27. The order of equivalent conductances at
infinite dilution for LiCl, NaCl and KCl is (c)
(a) LiCl >NaCl >KCl (d) CH3 CH CH CH CH2
(b) KCl >NaCl >LiCl
(c) NaCl >KCl >LiCl
(d) LiCl >KCl >NaCl 30. From the following compounds, choose the
one which is not aromatic.
28. The molar solubility (in mol L-1) of a
sparingly soluble salt MX4 is ‘S’. The (a) (b)
corresponding solubility product is Ksp . S in +
terms of ‘Ksp ’ is given by the relation
1/ 4 1/ 5
æK ö æK ö
(a) S = ç sp ÷ (b) S = ç sp ÷ (c) (d)
è 128 ø è 256 ø
–
(c) S =(256 Ksp )1/ 5 (d) S =(128 Ksp )1/ 4 Me2N
••
CH3
Category II (Q. 31 to Q. 35) Only one answer is correct. Correct answer will fetch full marks 2. Incorrect
1
answer or any combination of more than one answer will fetch - mark.
2
31. Identify X in the following sequence of 32. Compound X is tested and the results are
reactions. shown in the table.
CH3 ¾ C H ¾ C H ¾ CH2 ¾ CH2 ¾ CH3
Test Result
½ ½
Br Br (i) NaNH2 Aqueous sodium Gas given off which turns
¾¾¾¾¾¾® X hydroxide is added, damp red litmus paper
(ii) Na in liquid NH3
then heated gently. blue.
(a) CH3 ¾ CH ¾ CH ¾ CH2CH2CH3
Dilute hydrochloric Effervescence, gas given
½ ½
Br NH2 acid is added. off which turns lime water
milky and acidified
H 3C H K 2Cr2O 7 paper green.
34. Amongst the following , which should have 35. The major products obtained during
the highest rms speed at the same ozonolysis of 2, 3-dimethyl-1-butene and
temperature? subsequent reductions with Zn and H2O are
(a) SO 2 (a) methanoic acid and 2-methyl-2-butanone
(b) CO 2 (b) methanal and 3-methyl-2-butanone
(c) O 2 (c) methanol and 2, 2 -dimethyl-3-butanone
(d) H2 (d) methanoic acid and 2-methyl-3-butanone
Category III (Q. 36 to Q. 40) One or more answer (s) is (are) correct. Correct answers will fetch marks 2.
Any combination containing one or more incorrect answer (s) will fetch 0 mark. If all correct answers are
not marked and also no incorrect answer is marked then score = 2 ´ number of correct answers
marked/actual number of correct answers.
36. Choose the correct statement(s) among the 38. Of the following molecules, which have
following. shape similar to CO2 ?
H 3C H H CH3 (a) HgCl 2 (b) SnCl 2 (c) C 2H2 (d) NO 2
(a) C C and C C are enantiomers 39. In which of the following mixed aqueous
CH3
solutions, pH = pKa at equilibrium?
H H 3C H
(b) CH3CHO on reaction with HCN gives
(1) 100 mL of 0.1 M CH3COOH + 100 mL of
racemic mixture 0.1 M CH3COONa
C2H5 C2H5 (2) 100 mL of 0.1 M CH3COOH + 50 mL of 0.1
M NaOH
(c) CH3 C H and H C OH are enantiomers
(3) 100 mL of 0.1 M CH3COOH + 100 mL of
OH CH3 0.1 M NaOH
(d) CH3 CH NOH shows geometrical isomerism (4) 100 mL of 0.1 M CH3COOH + 100 mL of
0.1 M NH3
(a) (1) is correct
37. Which of the following statement(s) is (are) (b) (2) is correct
(c) (3) is correct
correct when a mixture of NaCl and K2Cr2O7
(d) Both (1) and (2) are correct
is gently warmed with conc. H2SO4 ?
(a) A deep red vapour is evolved 40. Amongst the following compounds, the
(b) The vapour when passed through NaOH solution, one(s) which readily react with
gives a yellow solution ethanolic KCN.
(c) Chlorine gas is also evolved (a) Ethyl chloride (b) Chlorobenzene
(d) Chromyl chloride is formed (c) Benzaldehyde (d) Salicylic acid
Mathematics
Category I (Q. 1 to Q. 50) Only one answer is correct. Correct answer will fetch full mark 1. Incorrect
answer or any combination of more than one answer will fetch -1 / 4 mark.
1. If the solution of the differential equation 7. The cosine of the angle between any two
dy diagonals of a cube is
x + y = xe x be xy = e x f( x ) + C , then f ( x )
dx 1 1
(a) (b)
is equal to 3 2
(a) x + 1 (b) x - 1 2 1
(c) (d)
(c) 1 - x (d) x 3 3
2. The order of the differential equation of all 8. If x is a positive real number different from 1
parabolas whose axis of symmetry along such that log a x, log b x, log c x are in AP,
X-axis is then
(a) 2 (b) 3 a+c
(a) b = (b) b = ac
(c) 1 (d) None of these 2
(c) c 2 = (ac )log a b (d) None of these
3. The line y = x + l is tangent to the ellipse
2 x2 + 3 y2 = 1. Then, l is 9. If a, x are real numbers and |a| < 1,| x| < 1,
then 1 + (1 + a ) x + (1 + a + a2 )x2 + K ¥ is
(a) -2 (b) 1
equal to
5 2
(c) (d) 1 1
6 3 (a) (b)
(1 - a) (1 - ax) (1 - a) (1 - x)
4. The area enclosed by y = 5 - x2 and (c)
1
(d)
1
(1 - x) (1 - ax) (1 + ax) (1 - a)
y = | x - 1| is
5p 5p - 2 ö 10. If log 0.3 ( x - 1 ) < log 0.09 ( x - 1 ), then x lies in
(a) æç - 2 ö÷ sq units (b) æç ÷ sq units
è 4 ø è 2 ø the interval
æ 5p 1 ö p
(c) ç - ÷ sq units (d) æç - 5ö÷ sq units (a) (2, ¥) (b) (1, 2 )
è 4 2ø è2 ø (c) (-2, - 1) (d) None of these
13
5. Let S be the set of points, whose abscissae
and ordinates are natural numbers. Let P Î S,
11. The value of å ( i n + i n + 1 ), i = -1 is
n =1
such that the sum of the distance of P from
(a) i (b) i - 1
(8, 0 ) and (0, 12 ) is minimum among all (c) 1 (d) 0
elements in S. Then, the number of such
points P in S is 12. If z 1, z2 , z3 are imaginary numbers such
(a) 1 (b) 3 (c) 5 (d) 11 1 1 1
that| z 1| = | z2| = | z3| = + + = 1,
6. Time period T of a simple pendulum of z 1 z2 z3
l then| z 1 + z2 + z3| is
length l is given by T = 2p . If the length is
g (a) equal to 1 (b) less than 1
increased by 2%, then an approximate (c) greater than 1 (d) equal to 3
change in the time period is 13. If p, q are the roots of the equation
(a) 2% (b) 1% x2 + px + q = 0, then
1
(c) % (d) None of these (a) p = 1, q = - 2 (b) p = 0, q = 1
2
(c) p = - 2, q = 0 (d) p = - 2, q = 1
10 WB JEE (Engg.) Solved Paper 2016
20. If x, y and z are greater than 1, then the value 27. Let A and B be two events such that
1 31 7
1 log x y log x z P( A Ç B) = , P( A È B) = and P( B ) = ,
6 45 10
of log y x 1 log y z is
then
log z x log z y 1 (a) A and B are independent
(a) log x × log y × log z (b) A and B are mutually exclusive
A 1
(b) log x + log y + log z (c) P æç ö÷ <
è Bø 6
(c) 0
B 1
(d) 1 - {(log x) × (log y) × (log z)} (d) P æç ö÷ <
è Aø 6
21. Let A be a 3 ´ 3 matrix and B be its adjoint 1° 1°
28. The value of cos 15° cos 7 sin 7 is
matrix. If| B| = 64, then| A| is equal to 2 2
(a) ± 2 (b) ± 4 1 1 1 1
(a) (b) (c) (d)
(c) ± 8 (d) ± 12 2 8 4 16
WB JEE (Engg.) Solved Paper 2016 11
29. The smallest positive root of the equation 36. The equation of a line parallel to the line
tan x - x = 0 lies in 3 x + 4 y = 0 and touching the circle
p p x2 + y2 = 9 in the first quadrant, is
(a) æç 0, ö÷ (b) æç , p ö÷
è 2ø è2 ø (a) 3 x + 4 y = 15 (b) 3 x + 4 y = 45
3p ö 3p
(c) æç p, ÷ (d) æç , 2 p ö÷ (c) 3 x + 4 y = 9 (d) 3 x + 4 y = 27
è 2 ø è2 ø
37. A line passing through the point of
30. If in a DABC, AD, BE and CF are the altitudes intersection of x + y = 4 and x - y = 2 makes
and R is the circumradius, then the radius of æ3 ö
an angle tan -1 ç ÷ with the X-axis. It
the circumcircle of DDEF is è4 ø
R 2R
(a) (b) intersects the parabola y2 = 4( x - 3 ) at points
2 3
R ( x 1, y 1 ) and ( x2 , y2 ), respectively. Then,
(c) (d) None of these | x 1 - x2| is equal to
3
16 32
(a) (b)
31. The points ( -a, - b ), (a, b ), (0, 0 ) and 9 9
(a2 , ab ), a ¹ 0, b ¹ 0 are always (c)
40
(d)
80
(a) collinear 9 9
(b) vertices of a parallelogram 38. The equation of auxiliary circle of the ellipse
(c) vertices of a rectangle 16 x2 + 25 y2 + 32 x - 100 y = 284 is
(d) lie on a circle
(a) x2 + y2 + 2 x - 4 y - 20 = 0
32. The line AB cuts off equal intercepts 2a from
(b) x2 + y2 + 2 x - 4 y = 0
the axes. From any point P on the line AB
perpendiculars PR and PS are drawn on the (c) ( x + 1)2 + ( y - 2 )2 = 400
axes. Locus of mid-point of RS is (d) ( x + 1)2 + ( y - 2 )2 = 225
a
(a) x - y = (b) x + y = a 39. If PQ is a double ordinate of the hyperbola
2
(c) x2 + y2 = 4a2 2 2
(d) x - y = 2 a 2 x2 y2
- = 1 such that DOPQ is equilateral. O
a2 b2
33. x + 8 y - 22 = 0, 5 x + 2 y - 34 = 0,
being the centre. Then, the eccentricity e
2 x - 3 y + 13 = 0 are the three sides of a
satisfies
triangle. The area of the triangle is 2 2
(a) 1 < e < (b) e =
(a) 36 sq units 3 2
(b) 19 sq units
3 2
(c) 42 sq units (c) e = (d) e >
(d) 72 sq units 2 3
2
34. The line through the points andIf the vertex of the conic y - 4 y = 4 x - 4a
(a, b ) 40.
( -a, - b ), passes through the point always lies between the straight lines
x + y = 3 and 2 x + 2 y - 1 = 0, then
(a) (1, 1) (b) (3a, - 2 b )
1
(c) (a2 , ab ) (d) (a, b ) (a) 2 < a < 4 (b) - < a<2
2
1 3
35. The locus of the point of intersection of the (c) 0 < a < 2 (d) - < a <
x y x y 1 2 2
straight lines + = K and - = ,
a b a b K 41. A straight line joining the points (1, 1, 1 ) and
where K is a non-zero real variable, is given (0, 0, 0) intersects the plane 2 x + 2 y + z = 10
by at
(a) a straight line (b) an ellipse (a) (1, 2, 5) (b) (2, 2, 2)
(c) a parabola (d) a hyperbola (c) (2, 1, 5) (d) (1, 1, 6)
12 WB JEE (Engg.) Solved Paper 2016
Category II (Q. 51 to Q. 65) Only one answer is correct. Correct answer will fetch full marks 2. Incorrect
1
answer or any combination of more than one answer will fetch - mark.
2
51. The sum of n terms of the following series (d) a2 x2 + (b 2 + 2 ac ) x + c 2 = 0
13 + 33 + 53 + 73 + K is 53. If w is an imaginary cube root of unity, then
(a) n2 (2 n2 - 1) (b) n3 (n - 1) the value of
(c) n3 + 8n + 4 (d) 2 n4 + 3n2 (2 - w) (2 - w2 ) + 2(3 - w) (3 - w2 )
52. If a and b are roots of ax2 + bx + c = 0, then + . . . + ( n - 1 ) ( n - w) ( n - w2 ) is
the equation whose roots are a 2 and b2 , is n2 n2
(a) (n + 1)2 - n (b) (n + 1)2 + n
(a) a2 x2 - (b 2 - 2 ac ) x + c 2 = 0 4 4
n2 n2
(b) a2 x2 + (b 2 - ac ) x + c 2 = 0 (c) (n + 1)2 (d) (n + 1) - n
4 4
(c) a2 x2 + (b 2 + ac ) x + c 2 = 0
WB JEE (Engg.) Solved Paper 2016 13
n
54. If C r - 1 = 36, n C r = 84 and n
C r + 1 = 126, (a) 0 (b) 1 (c) 2 (d) 3
Chemistry
1. (a) 2. (c) 3. (c) 4. (b) 5. (c) 6. (a) 7. (a) 8. (b) 9. (d) 10. (a)
11. (b) 12. (c) 13. (b) 14. (c) 15. (d) 16. (c) 17. (b) 18. (c) 19. (b) 20. (b)
21. (c) 22. (d) 23. (a) 24. (c) 25. (c) 26. (c) 27. (b) 28. (b) 29. (b) 30. (b)
31. (b) 32. (d) 33. (c) 34. (d) 35. (b) 36. (b,d) 37. (a,b,d ) 38. (a,c) 39. (a,b,d) 40. (a,c)
Mathematics
1. (b) 2. (a) 3. (c) 4. (c) 5. (b) 6. (b) 7. (a) 8. (c) 9. (c) 10. (a)
11. (b) 12. (a) 13. (a) 14. (d) 15. (c) 16. (b) 17. (a) 18. (b) 19. (a) 20. (c)
21. (c) 22. (c) 23. (c) 24. (c) 25. (c) 26. (c) 27. (a) 28. (b) 29. (c) 30. (a)
31. (a) 32. (b) 33. (b) 34. (c) 35. (d) 36. (a) 37. (b) 38. (a) 39. (d) 40. (b)
41. (b) 42. (c) 43. (c) 44. (c) 45. (c) 46. (d) 47. (a) 48. (c) 49. (b) 50. (a)
51. (a) 52. (a) 53. (a) 54. (c) 55. (b) 56. (b) 57. (b) 58. (c) 59. (b) 60. (c)
61. (d) 62. (b) 63. (a) 64. (a) 65. (d) 66. (b,d) 67. (a,b) 68. (a,c) 69. (a,c) 70. (a,b)
71. (a,c) 72. (b,d) 73. (b,d) 74. (a,b) 75. (b,c)
(*) No option is correct.
Solutions
Physics
1 1 1 1 1 2
1. = + = + =
C C1 C 2 4 4 4
V=constant
C = 2µF
2 µF will be parallel to 4µF, so the resultant capacitance R
4mF 4mF
A B E=0
–u
i = 60° i
air 90°
glass Let there is no air resistance when body falls freely in
downward direction, the direction of falling body will
90°–i opposite to the ascending body.
∴ The velocity at the same height while the body falls
sin i freely = − u
µ =
sin r
13. The bodies are separated by a distance R, so the
sin i = µ sin r coordinates of m1 and m2 will be ( 0, 0) and ( R, 0)
here, r = 90 − i Xcm
sin i = µ sin( 90° − i ) = µ cos i
sin i m2
µ = = tan i = tan 60° m1
cos i
(0, 0)
µ = 3 (R, 0)
R
18 WB JEE (Engg.) Solved Paper 2016
From the formula of centre of mass, 17. In SHM, the acceleration of vibrating particle is
m x + m2 x2 proportional to displacement
Xcm = 1 1
m1 + m2 |a| = ω 2 x.
m1 × 0 + m2 × R
= Here a = 16 cm / s 2
m1 + m2
m2 ⋅ R = 16 × 10 −2 m / s 2
Xcm =
m1 + m2 Displacement, x = 4cm = 4 × 10 −2 m
∴ 16 × 10 − 2 = ω 2 ( 4 × 10 −2 )
14. T1 = 273 + 20 = 293 K
ω2 = 4
T2 = 273 + 40 = 313 K
The velocity of sound wave in gases or air is given by ω = ± 2 rad/s
γ R T1 2π
ω=
v = T
M
2π 2π
where, γ = Ratio of C p / C V T = = = π = 3142
. s
ω 2
R = gas constant
1 2
γ R T1 18. The work done in stretching the spring, W = Kx
∴ v1 = 2
M K = force constant of spring
γ R T2 x = extension in length of the spring
v2 =
M 1
W1 = Kx12
v1 T1 293 2
= =
v2 T2 313 1
W 2 = Kx22
Given, v1 = 344.2 m/s 2
313 x1 = 1
∴ v 2 = v1 x2 = 1 + 1 = 2
293
1
= 344. 2 × 1068
. W1 K (1)2 1
= 344. 2 × 103
. m/s = 2 =
W2 1 2 4
K (2 )
= 35575
. m/s 2
−− 356 m/s
~ W 2 = 4W1
15. Mass of gas = 4g (given) W1 = 10 J
W 2 = 40 J
From perfect gas equation
∴ More work required = 40 J − 10 J = 30 J
pV = nRT
where n = number of moles 3RT
19. vrms =
M
here, n = = 2
4
2 R = gas constant
∴ pV = 2 RT T = absolute temperature
It will be the perfect gas equation for 4g of hydrogen. M = mass of gas
3RT
16. Suppose temperature of Sun = T vrms of H 2 =
MH 2
When it is doubled, T ′ = 2T 3RT
From law of radiation (Stefan’s law) vrms of O 2 =
MO 2
E ∝ T4 = σ T4
vrms .H 2 MO 2
σ = Stefan’s constant. =
vrms .O 2 MH1
E1 ∝ T 4
Given vrms H 2 = c
E 2 ∝ (2T )4 MO 2
c 32
⇒ = = = 16 = 4
E1 T4 vrms O 2 MH1 2
=
E 2 2 4 .T 4
c
∴ vrms = .
E 2 = 16 E1. 4
WB JEE (Engg.) Solved Paper 2016 19
4 3
20. The tube is closed at one end. So, it will act like a 6 πηrvt = πr ρe ⋅ g
closed organ pipe. 3
24 π ⋅ r 3ρe ⋅ g 2 r 2 ρe ⋅ g
The lowest frequency produced in the tube η= = ⋅
3 × 6 πrvT 9 vT
here, D = 2 cm, r = 1 cm = 1 × 10 − 2 m
λ
l 4 ρe = 1.75 g / cm3 = 1.75 × 10 +3 kg / m3
g = 10 m / s 2
vT = terminal velocity
λ = 0.35 × 10 − 2 m / s
l= ,λ = 4
4
−2 2 +3
l = 1m 2 (10 ) × 1.75 × 10 × 10
η= × −2
f = 84 Hz
9 0.35 × 10
−4
v = nλ = 84 × 4 2 10 × 1.75 × 10 3 × 10
= ×
When wave propagate in air, the velocity 9 0.35 × 10 − 2
γp 2 × 175 × 10 × 10
v = = × 10 poise
ρ 9 × 35
Here, p = 1.0 × 10 5 N / m2 − 1089 poise
~
ρ = 1.2 kg/m 3
22. Temperature of water = 0° C
γp Temperature of surrounding atmosphere = − 20° C
v2 =
ρ
Density of ice = ρ
v 2. ρ
γ = air –20°C
p
( 84 × 4)2 × 1.2 ice x
=
1.0 × 10 5
135472.2 0°C dx
= = 1.354
10 5 water
≈ 1.4 Coefficient of thermal conductivity = k
21. The gas bubble is rising through the liquid of density latent heat of melting = L
. g / cm3 with constant speed. The acting forces on
175
dQ kA(T1 − T2 )
bubble are The rate of heat flow H = =
dt x
kA[( 0 − ( − 20)]
=
x
Fb kA × 20
=
x
dQ dm
= ⋅L (Q mL = Q )
dt dt
dm = ρA ⋅ dx
Fv
dQ dx kA ⋅ 20
= ρA ⋅ L=
dt dt x
Buoyancy force of liquid = Fb (in upward direction) Z 20k t 40k
∫0 ⋅ =
ρ L ∫0
⇒ =
2
x dx dt Z t
Viscous force due to liquid = FV (in downward direction) ρL
∴ FV = Fb So, thickness Z increases as a function of time t
For free falling in a viscous medium according to the above equation.
FV = 6 πηrvT
23. Let the radius of single drop = r
η = viscosity coefficient of the liquid
r = radius of bubble 2mm
Radius of small drop =
Fb = mg = v × ρ. g 2
4 = 1 mm
= π r 3ρe . g
3 = 1 × 10 − 3 m
20 WB JEE (Engg.) Solved Paper 2016
Energy loss in the formation of large drop 26. Given, E = 3$i + 6$j + 2 k$ V /m
∆E = U f − U i B = 2 $i + 3$j T
= s ⋅ 4 π100r 2 − 1000s ⋅ 4 πr 2 v = 2 $i + 3$j m/s
= s ⋅ πr 2( 400 − 4000) e = − 1.6 × 10 − 19C
= − 3600s ⋅ πr 2
The magnitude of electric field
= − 3600 × 0.72 × 3.14 × (1 × 10 − 3 )2 | E | = ( 3)2 + ( 6)2 + (2 )2
From catenation
= 9 + 36 + 4
= 813.888 × 10 − 6
= 49 = 7
= 81388
. × 10 − 4 J
The force acting on electron due to electric field
= 8146
. × 10 − 4 J F = qE
| F | = | q || E |
24. 100 W 5.6 V
= 1.6 × 10 − 19 × 7 N
10 V = 11.2 × 10 − 19 N
= 1.12 × 10 − 18 N
The time constant of the circuit 33. Information in the statement of the questions is
τ = C. R insufficient to calculate heat required to raise the
temperature from 100°C to 1000°C.
Here, C = 1 µF = 1 × 10 −6
R = 1 MΩ = 1 × 10 6 Ω 34. (i) When the body of mass m lifted up the forces acting
−6 on the body
τ = 1 × 10 × 1 × 10 = 1 s
6
T1
We know that in one (1) time constant 63% charging is
done.
63
∴ × q max (q = CV )
100 a=4.9 m/s2
63
= × 1 × 10 −6 × 10
100
63
= × 1 µF × 10 = 6.3 µC
100 mg
q 6.3 × 10 −6C
⇒ V = = = 6.3 V mg
C 1 × 10 −6 F T1 − mg =
2
mg 3mg
29. In series combination, T1 = mg + = ...(i)
2 2
R = R1 + R 2
(ii) When the body lowered the acting forces
Same current will pass in R1 and R 2. According to
question, R1 is increased by 0.5%. T2
∴ Increment in the resistance
0.5
= × 400
100 a=4.9 m/s2
= 2.0 Ω
So to keep the potential unchanged in second
resistance the change required will be2.0 Ω increment. mg
30. The electric lines of force are always perpendicular to
mg
the equpotential surface ( mg − T2 ) =
2
θ = 90°
mg mg
T2 = mg − = ...(i)
2 2
On dividing Eq. (i) from Eq. (ii) we get
T1 3 mg / 2 3
Electric lines ∴ = =
T2 mg / 2 1
of force
= 3:1
Equipotential
surface
22 WB JEE (Engg.) Solved Paper 2016
Chemistry
1. For a carbonyl group to give Cannizaro reaction, it
should not have an a-hydrogen atom. Among the given –
O OH O OH – O
compound Ne 3 CCHO, C 6H 5C HO and HCHO respond –
4. OH –OH
to Cannizaro reaction but Cl 3CCHO do not due to the –H+
following reason.
– H
O O O Cl
OH
Cl Cl +– Cl
H H OH H OH
– Cl Cl Cl In alkaline medium the acidic proton present in the
HO Cl Cl molecule is abstracted to give enolate ion. In the next
step hydroxide ion if present at b-position leaves giving
O Cl O Cl a -b-unsaturated carbonyl compound. In (a), (c) and
H
–
Cl –
+ Cl (d), dehydration cannot take place due to improper
H O H O placement of groups.
Cl Cl
5. The correct order is 2<1<3<4. Since, more the
When hydroxyl group attacks the carbonyl group a number of 2° — NH — group, more is the basic nature.
tetrahedral intermediate forms. This tetrahedral (i) Thus, 3 and 4 are more basic than 1 and 2.
intermediate will revert to a carbonyl compound by
! (ii) Between 3 and 4, 4 has one two ¾ NH 2 group
expelling the best leaving group. The C Cl 3 carbonium while 3 has only one ¾ NH 2 group.
is resonance stabilised therefore it is better leaving \ 4 is more basic than 3.
group than OH. These initial products exchange a
(iii) Between 1 and 2, 1 has one ¾ NH 2 group with no
proton to give ion of carboxylic acid and
trichloromethane (CHCl 3 ). resonance, thus is more basic than 2.
2. Primary (1°)-alcohols do not give Lucas reagent test at 6. C ¾ C bond is not formed in Cannizaro reaction while
room temperature. CH 3CH 2CH 2OH does not undergo other reactions result in the formation of C ¾ C bond.
SN 1 or SN 2 at room temperature. Cannizaro reaction
3. Iodoform test is used to detect the presence of
CH 3 ù O O
é ù é
ê ú ê ½ ú
ê ¾ C ¾ CH 3 ú or ê ¾ C ¾ H ú in a molecule.
–
2 H OH OH + OH
ê ½½ ú ê ½ ú
êë O úû ê OH ú
ë û
In the given compounds only CH 3 COOH does not fulfill Wurtz reaction
the structural requirement of iodoform test hence, it Å !
does not respond to iodoform test. 2 R - X + 2Na ¾® R - R + 2 Na X
24 WB JEE (Engg.) Solved Paper 2016
Reimer-Tiemann reaction 12. PCl 5 exists as [PCl 4 ]+ and [PCl 6 ]- in solid state.
CHCl3 H
13. (i) Ortho and para hydrogens have different
3KOH nuclear spin.
In H 2 molecule, two protons in two H atoms with parallel
spin are called ortho-hydrogen
Friedel-Crafts acylation
O p p
9. The option (d) is false as lonic radii of trivalent 14. H 2S 2O 8 shows ¾O ¾O ¾ (peroxy) bonding as follows:
lanthanides decreases with increase in atomic number O
O
due to lanthanoid contraction.
10. Only KNO 3 on heating do not giveNO 2 but all other give HO S O O S OH
(Peroxide
D
(a) 2KNO 3 ¾¾ ® 2KNO 2 + O 2 O bond) O
D Peroxodisulphuric acid
(b) 2Pb(NO 3 )2 ¾¾ ® 2PbO + 4NO 2 + O 2
D
or
(c) 2Cu(NO 3 )2 ¾¾ ® 2CuO + 4NO 2 + O 2 Marshall’s acid
D
(d) 2 AgNO 3 ¾ ¾¾ ® 2 Ag + O 2 + 2NO 2
Red hot
Heavy metal nitrates liberate nitrogen dioxide on heating. 15. Iron (Fe) is used to obtain Cu-metal from CuSO 4 ( aq )
11. Due to hydrogen bonding, HF shows highest boiling because Fe is more reactive than Cu (and also easily
point. available and is cheaper) thus, replace the copper from
CuSO 4 to give copper (Cu).
Thereafter van der Waals’ force decide the boiling
point. CuSO 4 ( aq ) + Fe( s ) ¾® FeSO 4 ( aq ) + Cu( s )
Larger the size or molecules mass, greater is the van der 16. Radium in isolated form and in its chloride (i.e. Ra + )
Waals’ forces. Hence, higher is the boiling point. only differ in the number of orbital electron.
Q van der Waals’ forces is more for HI than HBr, which is But, radioactivity is independent of chemical environment
in turn is more than HCl. of an ion or atom, i.e. radioactivity is the phenomenon
Hence, correct order is HF > HI > HBr > HCl. which does not depend on the orbital electrons but
depends only on the composition of nucleus.
WB JEE (Engg.) Solved Paper 2016 25
17. The correct order of solubility of sulphates in water is 22. The element belonging to 4th period and 15th group is
BeSO 4 > MgSO 4 > CaSO 4 > SrSO 4 > BaSO 4 arsenic (As). The outer electronic configuration of 15th
group element is ns 2np3.
Solubility of 2nd group sulphates decreases as we
Since arsenic belongs to 4th period, therefore it has
move down the group due to less release of hydration
filled ‘d’ orbital too. Thus, the electronic configuration
energy. Be 2+ < Mg 2+ < Ca 2+ < Sr 2+ < Ba 2+ (Ionic
of As is [Ar ]4 s 2 3 d 10 4 p3.
Size) As hydration energy decreases more rapidly than
latice energy, the solubility decreases down the group. So, it has filled ‘s’ and ‘d’ orbital and half-filled p-orbital.
Hydration Energy µ
Charge 23. For exothermic reaction,
Size 1
Q Kp µ
While lattice energy almost remains constant. Hence, T
solubility decreases. 1
Thus, as increases (i.e. T decreases), K p increases.
T
18. Given, energy of one mole of H 2 bonds = 436 kJ
n × hc Q DH for exothermic reactions is negative.
Q E=
l 24. Given,
(energy for one H ¾H bond) p° = vapour pressure of pure solvent
(n =number of H ¾H bonds) p = vapour pressure of solution
n × hc
\ l= n1 = moles of solute
E
n2 = moles of solvent
6.634 ´ 10 - 34 Js ´ 3 ´ 10 8 ms -1 ´ 6.023 ´ 10 23
l= m Thus, according to Raoult’s law.
436 ´ 10 3 J p° - p n1
= x1 =
= 0.2736 ´ 10 - 34 + 8 + 23 - 3
p° n1 + n2
-6
= 0.2736 ´ 10 m (x1 = mole fraction of solute)
= 273.6 nm » 274 nm p n1
or, 1- =
p° n1 + n2
19. Correct order of O ¾ O bond length is p n1
= 1-
H 2O 2 > O 3 > O 2 p° n1 + n2
(i) Between O 3 and O 2 n1 + n2 - n1
Þ
QBond order of O 2 > O 3 n1 + n2
Thus, bond length of O 3 > O 2. p n2
=
æ 1 ö p° n1 + n2
çQ Bond order µ ÷
è Bond length ø æ n2 ö
Þ p = p° çç ÷
÷
(ii) Between O 3 and H 2O 2 è n1 + n 2 ø
Q O 3 has p-bonds in resonance so, has shorter
bond length than O ¾O in H 2O 2.
25. In Schottky defect, equal number of cations and anions
are missing creating equal number of cation and anion
\ Hence, (b) is the answer. vacancies.
20. Calcium carbide is calcium acetylide, having structure 26. For a spontaneous reaction, DG = negative and
È È D S = positive
Ca 2 + × C ºº C
Q DG = D H - TDS = - ve
Thus, it has one sigma ( s ) and two pi (p) bonds.
Hence, DH - TDS = - ve only when D S = + ve and
21. Let Z = atomic number DH = -ve.
M = atomic mass 27. QKCl is more ionic than NaCl while NaCl is more ionic
-a - 2b M- 4
than LiCl.
M
E ¾¾¾® EM - 4 ¾¾¾® E Also, more the ionic nature, more is the conductivity
Z Z -2 Z
thus, order of equivalent conductance at infinite dilution
M
Hence, ZE and 2 E M - 4 has same atomic number with is KCl > NaCl > LiCl.
different atomic mass thus, they are isotopes.
26 WB JEE (Engg.) Solved Paper 2016
28. QFor a compound A x By 32. (i) It is a test for ammonia.
D
x y x+ y
Ksp = x × y × S . (where,S = solubility) NH 4Cl + NaOH ¾¾® NH 3 + NaCl + H 2O
For MX 4 (Basic)
Ksp = 1 ´ ( 4)4 × S 1 + 4 ½
½
Ksp = 256 ´ S 5 ¯
1/ 5 Turns red litmus blue
æK ö
\ S = ç sp ÷
è 256 ø (ii) It is a test for SO 2-
4 ions (Orange)
CH3 CH3
+ –
NaNH2
4 3 2 H
H 3C H CH3 CH C O+ O C
H
C C 1
CH3 CH3 (HCHO)
H CH2CH2CH3 methanal
3-methyl 2-butanone
trans-alkene
Na in liquid NH 3 is a Birch reagent. It is used to alkyne 36. (a) These are cis and trans isomers not enantiomers
into trans alkene only. (Thus, false).
WB JEE (Engg.) Solved Paper 2016 27
(b) CH 3CHO on reaction with HCN gives racemic 39. pH of buffer solutions is given by
mixture of cyanohydrin (Thus, true). [salt ]
pH = pK a + log
OH H H OH [acid]
H
C O + HCN C + C (1) CH 3 COOH and CH 3 COONa
Cl2
H 3C CN H3C CN 100 ´ 0 .1 100 ´ 0 .1
37. (A) When NaCl and K 2Cr2O 7 warmed with H 2SO 4 Final =0 0 10 10
(i.e. in acidic medium), they produce deep red
[salt] = 10 and [acid] = 0,
vapours of chromyl chloride (CrO 2Cl 2).
10
(B) When NaOH is passed, the product CrO 3 formed pH = pK a + log
0
in step (a) will react with NaOH and gives yellow
colour solution. \ pH ¹ pK a
(C) Chlorine gas is not evolved, thus false. (4) CH 3COOH + NH 3 ¾® CH 3COONH 4
(D) In the given reaction, chromyl chloride (CrO 2Cl 2) is Initial 10 10 ¾® No buffer action
formed. All reactions are as follows: Final Due to weak CH 3COONH 4 salt
(i) K 2Cr2O 7 + 2H 2SO 4 ¾® 2KHSO 4 + 2CrO 3 + H 2O \ pH ¹ pK a
(ii) NaCl + H 2SO 4 ¾® NaHSO 4 + HCl (a) Ethyl chloride gives ethyl cyanide, when it reacts
40.
(iii) CrO 3 + 2HCl ¾® CrO 2Cl 2 with KCN (ethanolic),
from (i) from (ii) Red vapours
Ethanolic
C 2H 5Cl + KCN ¾¾¾¾® C 2H 5 ¾ CN
(iv) CrO 2Cl 2 + 2NaOH ¾® Na 2CrO 4 + 2HCl
Yellow solution
(b) It does not react with KCN.
38. HgCl 2 and C 2H 2 both show linear shape as of CO 2.
(c) Benzaldehyde : KCN gives HCN in alcoholic
(Qboth have zero lone pair of electrons) medium and HCN gives cyanohydrin with
(a) HgCl 2 ¾® Cl ¾Hg ¾Cl (Linear) benzaldehyde.
OH
(b) SnCl2 Sn (Bent) C 6H 5CHO + HCN ¾® C 6H 5 ¾CH
CN
Cl Cl
(c) C2H2 OH
H C C H (Linear)
(d) Salicylic acid does not react.
COOH
(d) NO2 N (Bent)
O O
Mathematics
dy
1. Given, x + y = xe x Y
dx
dy y y=–x+1
Þ + = ex …(i) y=(x–1)
dx x
dy (–1, 2)
On comparing Eq. (i) by + Py = Q, we get
dx
1 (2, 1)
P = and Q = e x
x
1
X¢ X
ò x
dx O (1, 0) y = √5–x2
\ IF = e
= elog x = x
Hence, solution of differential equation,
y × x = ò xe x dx + C
Þ xy = xe x - ò e x dx + C
Y¢
Þ xy = xe x - e x + C
Let A be the area bounded by the given curves. Then,
Þ xy = e x ( x - 1) + C … (ii) 1 2
A= ò- 1 ( 5 - x 2 + x - 1) dx + ò1 ( 5 - x 2 - x + 1) dx
Q xy = e x f ( x ) + C [given]…(iii)
1 2
On comparing Eqs. (ii) and (iii), we get = ò- 1 5 - x 2 dx + ò1 5 - x 2 dx
f( x ) = ( x - 1) 1 2
+ ò-1 ( x - 1) dx + ò1 (- x + 1) dx
2. The general equation of the parabola is x = ay 2 + b, 1 2
2 é x2 ù é x2 ù
where a and b are arbitrary constants. = ò- 1 5 - x 2 dx + ê - xú + ê - + xú
ë2 û- 1 ë 2 û1
So, the differential equation is of order 2. 2
é1 5 x ù 5
3. The equation of the line is y = x + l …(i) = x 5 - x 2 + sin- 1 -
ëê 2 2 5 ûú - 1 2
On comparing it with y = mx + c, we get 5 2 5 1 5
= 1+ sin- 1 + 1 + sin- 1 -
m = 1 and c = l. 2 5 2 5 2
The equation of the ellipse is 1 5 æ 2 1 1 4ö
2 x 2 + 3 y2 = 1 =- + sin- 1 ç ´ 1- + 1- ÷
2 2 è 5 5 5 5ø
x2 y2 1 5 5p 1
or 2
+ 2
=1 =- + sin- 1(1) = - sq units
æ 1 ö æ 1 ö 2 2 4 2
ç ÷ ç ÷
è 2ø è 3ø
5. Sum of distances of point P from point (8, 0) and (0, 12)
x2 y2 will be minimum, if points are collinear.
On comparing it with + = 1, we get
a2 b2 \ Equation of line at point (8, 0) and (0, 12).
1 1
a 2 = and b 2 = . x
+
y
=1
2 3 8 12
If the line touches the ellipse, then y x
1 1 5 Þ = 1-
c 2 = a 2m2 + b 2 Þ l2 = ×1 + = 12 8
2 3 6 x
5 5 Þ y = 12 - ´ 12
Þ l2 = Þl= 8
6 6 3
Þ y = 12 - x
4. The graphs of y = | x - 1| and y = 5 - x 2 are shown in 2
\ Points ( x, y) º (2, 9), ( 4, 6) and (6, 3).
the figure and the shaded region is the required region
bounded by the two curves. Hence, the number of such points P in S is 3.
WB JEE (Engg.) Solved Paper 2016 29
l 1 1 a
6. Given, T = 2 p = × -
g (1 - a ) (1 - x ) (1 - a ) (1 - ax )
On differentiating w.r.t. l, we get 1
=
dT 2p 1 (1 - ax ) (1 - x )
= ×
dl g 2 l
10. Given, log 0. 3( x - 1) < log 0. 09( x - 1)
dT p æ 2l ö
\ DT = Dl = ×ç ÷ Þ log 0. 3( x - 1) < log( 0. 3)2 ( x - 1)
dl gl è 100 ø
Þ log 0. 3( x - 1)2 < log 0. 3( x - 1)
l 1 T
= 2p × = Þ ( x - 1)2 > x - 1 [Q 0.3 < 1]
g 100 100
2
DT 1 Þ x + 1- 2x - x + 1> 0
Þ = = 1%
T 100 Þ x2 - 3 x + 2 > 0
Hence, approximate change in the time period is 1%. Þ ( x - 1) ( x - 2 ) > 0
Þ x < 1, x > 2 Þ x > 2 [Q x <| 1]
7. Let the direction ratio’s of two diagonals of a cube are
\ (2, ¥)
(1, 1, 1) and ( - 1, 1, 1).
a1a2 + b1b 2 + c1c 2 Hence, x lies in the interval (2, ¥).
\ cos q =
a12 + b12 + c12 a22 + b 22 + c 22 11. We have,
13
(1) ´ ( - 1) + (1) ´ (1) + (1) ´ (1) S (i n + i n + 1)
= n=1
2 2 2 2 2 2
(1) + (1) + (1) ( - 1) + (1) + (1)
13 13
- 1+ 1+ 1 1 = S in + S in+ 1
= = n=1 n=1
3 3 3
æ 1 - i 13 ö 13
2 æ1 - i ö
= iç ÷+ i ç ÷
8. Given, log a x, log b x and logc x are in AP. è 1- i ø è 1- i ø
\ 2 log b x = log a x + logc x æ1 - i ö æ1 - i ö
1 1 = iç ÷-ç ÷
Þ 2 log b x = + è1 - i ø è1 - i ø
log x a log x c
= i -1
2 log x ac
Þ =
log x b log x a log x c 12. We have,
log x b | z1 | = | z2 | = | z3 | = 1
Þ 2 log x c = (log x ac )
log x a Þ | z1 |2 = | z2 |2 = | z3 |2 = 1
Þ 2 log x c = log a b × log x ac Þ z1 z1 = z2 z2 = z3 z3 = 1
Þ log x c 2 = log x ( ac )log a b 1 1 1
Þ = z1, = z2, = z3
z1 z2 z3
\ c 2 = ( ac )log a b
1 1 1
9. We have, Now, + + =1
z1 z2 z3
1 + (1 + a ) x + (1 + a + a 2 ) x 2 + K ¥
Þ | z1 + z2 + z3 | = 1
¥
= S (1 + a + a2 + K + an - 1 ) xn - 1 Þ | z1 + z2 + z3 | = 1
n=1
¥ nö
\ | z1 + z2 + z3 | = 1
æ1 - a n-1
= S ç ÷ x 13. Given, p, q are roots of equation x 2 + px + q = 0.
n=1 è 1- a ø
¥
xn - 1 an xn - 1
¥ \ p + q = - p Þ2p + q = 0 … (i)
= S - S and pq = q
n = 11 - a n=1 1- a
¥ ¥ Þ p=1
1 a
= S xn - 1 - S ( ax)n - 1 Put p = 1 in Eq. (i), we get
1- a n=1 1- a n=1
2 ´ 1+ q = 0
1
= [(1 + x + x 2 + K ¥)] Þ q +2 =0
1- a
Þ q = -2
a
- [{1 + ax + ( ax )2 + K ¥}] Hence, p = 1, q = - 2
1- a
30 WB JEE (Engg.) Solved Paper 2016
n n
14. Given equation is q q q
= æç2 cos ö÷ Re æçcos + i sin ö÷
x2 - 3 x + k = 0 è 2ø è 2 2ø
n
b 3 q nq q
As, - = Ï ( 0, 1) = æç2 cos ö÷ Re æçcos + i sin ö÷
2a 2 è 2ø è 2 2ø
Since, both roots cannot lie between 0 and 1. [by De moivre’s theorem]
So, no value of k is possible. n
q nq
= æç2 cos ö÷ × cos
15. Given word is ‘ARRANGE’. è 2ø 2
If R’s occur together, number of letters in word = 6 1 log x y log x z
(h , k ) (2a, 0)
X¢ X
p R A
–p x= 3p (2h, 0)
x= Y¢ x=
2 2 2
30. Let circumradius of DDEF be R ¢. We know,
ÐFDE = 180 ° - 2 A and FE = R sin2 A.
A
Y¢
Since, from the given points ( a 2, ab ) and ( a, b ) are 37. Given equations are
satisfy the Eq. (i), but ( a 2, ab ) is the required answer x+ y=4 …(i)
because (a, b) is already given in question. and x- y=2 …(ii)
From Eqs. (i) and (ii), we get
35. Given equations of straight lines
x = 3 and y = 1
x y
+ =K …(i) 3
a b The line through this point making an angle tan- 1 with
x y 1 4
and - = …(ii) the X-axis is
a b K
3 éQ m = 3 ù
Let the point of intersection be ( a, b ). ( y - 1) = ( x - 3)
4 êë 4 úû
So, from Eqs. (i) and (ii), we get
a b 3x 5 3x - 5
+ =K Þ y= - = …(iii)
a b 4 4 4
a b 1 Since, this line intersects the parabola
and - =
a b K y 2 = 4( x - 3)at points( x1, y1 ) and( x2, y2 ), respectively.
2 2 2 2
Þ æ aö - æb ö = 1 Þ a - b = 1 3x - 5
ç ÷ ç ÷ \ Putting y = in equation of parabola, we get
èaø èbø a2 b2 4
2
x2 y2 æ 3x - 5ö
\ Locus - = 1, which is the equation of a ç ÷ = 4( x - 3)
a 2
b 2 è 4 ø
hyperbola. Þ 9 x 2 - 94 x + 217 = 0
36. Equation of circle is x 2 + y 2 = 9 …(i) 94 217
Þ x1 + x2 = and x1 x2 =
9 9
And equation of line is 3 x + 4 y = 0 …(ii)
\ | x1 - x2| = ( x1 + x2 )2 - 4 x1 x2
Equation of line parallel to the line (ii),
2
3x + 4y = k = æ 94 ö - 4 ´ 217
ç ÷
- 3x k è 9ø 9
Þ y= + …(iii)
4 4 32
=
9
34 WB JEE (Engg.) Solved Paper 2016
38. Given equation of ellipse is \ Vertex = ( a - 1, 2 )
16 x 2 + 25 y 2 + 32 x - 100 y = 284
x2 y2 From figure,
39. Given equation of hyperbola is
=1 -
a b2 2 3
clearly, - < a-1< 1
and DOPQ is an equilateral triangle. PQ is double 2
ordinate of the hyperbola. 1
\ - < a<2
2
Y P (a secq, b tanq)
41. Equation of line joining the points (1, 1, 1) and
(0, 0, 0) is
x-0 y-0 z-0
= = =l (say)
1- 0 1- 0 1- 0
X′ X
(0, 0) O Þ x= y= z=l
So, the point is ( l, l, l ).
The point intersects the plane 2 x + 2 y + z = 10.
Q
\ 2( l ) + 2( l ) + l = 10
Y′ (a secq, – b tanq) Þ 5l = 10
Þ l =2
Let the coordinates of P and Q be ( a sec q, b tan q) and Hence, the point is (2, 2, 2).
( a sec q, - b tan q), respectively.
In DOPQ ,
42. Equations of planes are
x + y + 2z - 6 = 0 …(i)
OP = PQ
and 2x - y + z - 9 = 0 …(ii)
Þ a 2sec 2q + b 2 tan2 q = (2 b tan q)2
a1a2 + b1b 2 + c1c 2
Þ a 2sec 2 q = 3b 2 tan2 q \ cos q =
a1 + b12 + c12 a22 + b 22 + c 22
2
a2
Þ sin2 q = 1 ´ 2 + 1 ´ ( - 1) + 2 ´ 1
3b 2 =
1+ 1+ 4 4 + 1+ 1
Now, sin2 q < 1
2 - 1+ 2 3 1
a2 = = =
Þ <1 6´ 6 6 2
2
3b
p
b2 1 Þ cos q = cos æç ö÷
Þ > è 3ø
a2 3
p
b2 4 \ q=
Þ 1+ > 3
a2 3
2 4 2 43. Given,
\ e > Þe >
3 3 y = (1 + x )(1 + x 2 ) (1 + x 4 ) + K + (1 + x 2n )
Þ log y = log(1 + x ) + log(1 + x 2 )
40. Equation of the vertex of conic
y 2 - 4 y = 4 x - 4a + log(1 + x 4 ) + K + log(1 + x 2n )
Þ y 2 - 4 y + 4 = 4 x - 4a + 4 1 dy 1 2x 3 x2 2 nx 2n - 1
Þ = + + + ... +
y dx 1 + x 1 + x 2 1 + x 3 1 + x 2n
Þ ( y - 2 )2 = 4[ x - ( a - 1)]
WB JEE (Engg.) Solved Paper 2016 35
dy é 1 2x 3 x2 2 nx 2n - 1 ù 48. Let I = ò 2 x [f ¢ ( x ) + f( x ) log 2 ] dx
Þ = yê + + +K+ ú
dx ë1 + x 1 + x
2
1+ x 3
1 + x 2n û
Consider g ( x ) = 2 x f( x )
æ dy ö æ 1 ö Þ g ¢ ( x ) = 2 x f ¢ ( x ) + 2 x f( x ) log 2
\ ç ÷ = yç ÷=1
è dx ø x = 0 è1 + 0 ø Þ g ¢ ( x ) = 2 x [f ¢ ( x ) + f( x ) log 2 ]
x
44. Given that f( x ) is an odd differentiable function. \ I= ò g ¢( x) dx = g ( x) + C = 2 f( x ) + C
Then, f( - x ) = - f( x ) 1 æ1 ö
Þ - f ¢( - x) = - f ¢( x)
49. Let I= ò0 log çè x - 1÷ø dx
Þ f ¢( - x) = f ¢( x) …(i) 1 æ1 - xö
Put x = 3 in Eq. (i), we get = ò0 log çè ÷ dx
x ø
f ¢ ( - 3) = f ¢ ( 3) 1 æ x ö
\ f ¢ ( - 3) = 2 [Q f ¢ ( 3) = 2] Þ I= ò0 log çè 1 - ÷ dx = - I
xø
1- x b b
æ1 + x ö 1- x [Q òa f( x) dx = òa f(a + b - x ) dx]
45. We have, lim ç ÷
x ®1 è 2 + x ø
\ 2I = 0 Þ I = 0
1- x
é ù
æ 1 + x ö (1 + x )(1 - x) n+ 1+ n+2 +K+ 2 n - 1ú
= lim ç ÷ 50. Q lim ê 3
x ®1 è 2 + x ø n®¥ ê ú
ë n2 û
1
æ 1 + x ö (1 + x) é 1 2 n - 1ù 1
= lim ç ÷ = lim ê 1 + + 1+ +K+ 1+ ú
x ®1 è 2 + x ø n®¥
ë n n n û n
1 1 n-1
1 r
æ 1 + 1 ö (1 + 1) æ 2 ö 2
=ç ÷ =ç ÷ =
2 = lim
n®¥
å n
1+
n
r =1
è 2 + 1ø è 3ø 3
1
1 2 2
é æ e öù = ò0 1 + x dx = é (1 + x )3/ 2 ù = (2 2 - 1)
êë 3 úû
ê log çè x 2 ÷ø ú é 3 + 2 log x ù 0 3
46. Given, f( x ) = tan- 1 ê 2 ú
+ tan-1 ê ú
ê log(ex ) ú ë 1 - 6 log x û 51. We have,
êë úû 13 + 3 3 + 5 3 + 7 3 + K
é log e - 2 log x ù - 1 é 3 + 2 log x ù Now, Tn = (2 n - 1)3
= tan- 1 ê ú + tan ê 1 - 6 log x ú
ë log e + 2 log x û ë û = 8n3 - 3 (2 n)2 (1) + 3 (2 n) (1)2 - (1)3
Let 2 log x = tan q, we get = 8n3 - 12 n2 + 6n - 1
é 1 - tan q ù é 3 + tan q ù \ S n = S Tn
= tan-1 ê ú + tan
-1
ê 1 - 3 tan q ú 8n2 ( n + 1)2 12 n( n + 1)(2 n + 1)
ë 1 + tan q û ë û = -
4 6
é p ù
= tan- 1 ê tan æç - qö÷ ú + tan- 1 [tan tan-1( 3 + q)] +
6n ( n + 1)
-n
ë è 4 ø û 2
p p = 2 n2 ( n + 1)2 - 2 n ( n + 1) (2 n + 1) + 3n ( n + 1) - n
= - q + tan- 1 3 + q = + tan- 1 3
4 4 = n [2 n ( n + 1)2 - 2 ( n + 1) (2 n + 1) + 3 ( n + 1) - 1]
Q f( x ) = constant = n [2 n ( n2 + 1 + 2 n) - 2 (2 n2 + 3n + 1)
\ f ¢ ¢( x) = 0 + 3n + 3 - 1]
47. Let I = ò
log x
dx = n [2 n3 + 2 n + 4n2 - 4n2 - 6n - 2 + 3n + 2 ]
3x
1 = n [2 n3 - n] = n2 (2 n2 - 1)
Again, let log x = z Þ dx = dz
2x 52. Given, a and b are roots of ax 2 + bx + c = 0.
2z 2
\ I=ò dz = ò z dz -b
3 3 \ a+b=
a
2 z2 1
= × + C = (log x )2 + C c
3 2 3 and ab =
a
36 WB JEE (Engg.) Solved Paper 2016
Now, if the roots are a2 and b 2, then 56. Given equation is
a2 + b 2 = ( a + b )2 - 2 ab x 3 - yx 2 + x - y = 0
2
- bö 2c b 2 2c Þ x 2 ( x - y) + ( x - y) = 0
= æç ÷ - = 2 - …(i)
è a ø a a a Þ ( x 2 + 1) ( x - y) = 0
2 2
And
c c
a2b 2 = ( ab )2 = æç ö÷ = 2 Now, x2 + 1 ¹ 0
è aø a and x- y=0
The equation whose roots are a2 and b 2, is \ x= y
æ b 2 2c ö c2 So, the equation represents a straight line.
x2 - ç 2 - ÷ x+ 2 =0
èa aø a
57. Let ( h, k ) be the coordinates of the mid-point of a chord,
\ a 2 x 2 - ( b 2 - 2 ac ) x + c 2 = 0. which subtends a right angle at the origin.
Then, equation of the chord is
53. Given,
hx + ky - 1 = h2 + k 2 - 1 [using T = S ¢]
(2 - w ) (2 - w 2 ) + 2 ( 3 - w ) ( 3 - w 2 ) + K +
Þ hx + ky = h2 + k 2
( n - 1)( n - w ) ( n - w 2 )
The combined equation of the pair of lines joining the
Now, Tn = ( n - 1) ( n - w ) ( n - w 2 )
origin to the points of intersection of x 2 + y 2 = 1 and
= ( n - 1) ( n2 - nw - nw 2 + w 3 ) hx + ky = h2 + k 2 is
= ( n - 1) ( n2 + n + 1) = n3 - 1 æ hx + ky ö
2
x 2 + y2 - 1ç 2 ÷ =0
\ S n = STn èh + k ø
2
n2 ( n + 1)2
= Sn3 - S1 = -n Lines given by the above equation are at right angle.
4
Therefore, coefficient of x 2 + coefficient of y 2 = 0
54. Given,
1
n
C r - 1 = 36 …(i) i.e. h2 + k 2 =
2
n
C r = 84 …(ii) 1
\ x 2 + y2 =
n 2
C r + 1 = 126 …(iii)
On dividing Eq. (i) by Eq. (ii), we get 58. Let P ( at 2, 2 at ) be a moving point on the parabola
r
=
36 y 2 = 4 ax and S( a, 0) be its focus. Let Q ( h, k ) be the
n - r + 1 84 mid-point of SP.
Þ 84r = 36n - 36r + 36 at 2 + a
Then, h=
Þ 120r = 36n + 36 …(iv) 2
Also, on dividing Eq. (ii) by Eq. (iii), we get 2 at + 0
and k=
r+1 84 2
=
n - r 126 2h
Þ = t2 + 1
Þ 126r + 126 = 84n - 84r a
Þ 210r = 84n - 126 …(v) k
and t =
a
On solving Eqs. (iv) and (v), we get
n = 9and r = 3 2h k2
Þ = 2 +1
\ n
C 8 = 9C 8 = 9 a a
Þ k 2 = 2 ah - a 2 [on eliminatingt ]
55. Number of males = 14
Thus, the locus of ( h, k ) is y = 2 ax - a 2
2
and females = 6
Total = 14 + 6 = 20 Now, y 2 = 2 ax - a 2
a
Males above 40 yr = 8 Þ ( y - 0)2 = 2 a æç x - ö÷
è 2ø
and females above 40 yr = 3
Now, P (selected person is female and above 40 yr) The equation of the directrix of this parabola is
3 1 a a
= = x - = - , i.e. x = 0.
6 2 2 2
WB JEE (Engg.) Solved Paper 2016 37
59. Let, 63. Given that, the ordinate decreases at the same rate at
2 2 which the abscissa increases, therefore
I= ò0 x [ x ] dx
dy dx
1 2 2
=- …(i)
x 2 ´ 1 dx dt dt
= ò0 x × 0 dx + ò1 Also, equation of ellipse
2
é x3 ù 8 1 7 16 x 2 + 9 y 2 = 400 …(ii)
=ê ú =é - ù=
ê ú On differentiating w.r.t. t , we get
ë 3 û1 ë 3 3 û 3
dx dy
16 ´ 2 x + 9 ´ 2y =0
60. Graph of f( x ), dt dt
Y dx dy
Þ 16 x + 9y =0
b dt dt
dy dx
Þ 9y = - 16 x
dt dt
dy dy
–a a X Þ 9y = - 16 x æç - ö÷ [using Eq. (i)]
dt è dt ø
16
Þ 9 y = 16 x Þ y= x ..(iii)
9
Using Eq. (iii) in Eq. (ii), we get
There are two sharp turns. Hence, f( x ) cannot be 2
16
16 x 2 + 9 æç x ö÷ = 400
differentiable at two points. è9 ø
61. Given, |a + b| < |a - b| (16)2 2
Þ 16 x 2 + x = 400
9
Now,
16 ù
|a + b|2 < |a - b|2 Þ 16 x é x + x = 400
êë 9 úû
Þ (a + b ) × (a + b ) < (a - b ) (a - b ) 25 400 ´ 9
Þ 16 x é x ù = 400 Þ x 2 =
Þ |a |2 + |b|2 + 2 a × b < |a |2 + |b|2 - 2 a × b êë 9 úû 25 ´ 16
Þ 4a ×b < 0 Þ x2 = 9 Þ x=± 3
Þ a ×b < 0 16 ö 16 ö
\ Required points are ç 3, ÷ and æç - 3, -
æ
÷.
Þ |a ||b| cos q < 0 è 3ø è 3ø
Þ cos q < 0
Þ q Îobtuse angle
64. In a dictionary, the words at each stage are arranged in
alphabetical order.
\ a and b are inclined at an obtuse angle.
In the given problem, we must consider the words
62. Given, beginning with C, C, H, I, N, O in order. So,
dy Number of words starting with CC = 4!
y + by 2 = a cos x, 0 < x < 1 …(i)
dx Number of words starting with CH = 4!
Let y2 = z Number of words starting with CI = 4!
dy dz Number of words starting with CN = 4!
Þ 2y =
dx dx Next word starting with CO i.e. COCHIN = 1
dy 1 dz \ Required number of words = 4 ´ 4! = 96
Þ y = …(ii)
dx 2 dx 65. We have,
1 dz
\ + by 2 = a cos x [using Eq. (ii)] é2 0 0ù
2 dx A = ê 0 2 0ú
dz ê ú
Þ + 2 by 2 = 2 a cos x êë2 0 2 úû
dx
Now, IF = e 2b ò dx = e 2bx é2 0 0ù é2 0 0ù é 4 0 0ù
\ A 2 = ê 0 2 0ú ê 0 2 0ú = ê 0 4 0ú
\ z × e 2bx = ò 2 a cos x × e 2bx dx ê ú ê ú ê ú
êë2 0 2 úû êë2 0 2 úû êë 8 0 4úû
2a
Þ y 2 e 2bx = (sin x + 2 b cos x ) e 2bx + C é 22 0 0ù
4b 2 + 1 ê ú
=ê 0 22 0 ú
Þ ( 4b 2 + 1) y 2 = 2 a (sin x + 2 b cos x ) + Ce - 2bx ê2 × 2 2 2
0 2 ú
ë û
38 WB JEE (Engg.) Solved Paper 2016
é 2n 0 0ù 68. Given,
n ê ú
Now, A =ê 0 2n 0ú x 2 + y 2 - 10 x + 21 = 0
ê n ×2 n 0 n
2 ú Þ x 2 - 10 x + y 2 + 21 = 0
ë û
éa 0 0ù Now, for real roots of x, D ³ 0
But An = ê 0 a 0ú [given] 100 - 4 ( y 2 + 21) ³ 0
ê ú
êë b 0 a úû Þ y2 £ 4 Þ - 2 £ y £ 2
n
é a 0 0ù é 2 0 0ù Also,
\ ê 0 a 0ú = ê 0 2 n
0ú
ú
y 2 = - x 2 + 10 x - 21
ê ú ê n n
êë b 0 a úû êë n × 2 0 2 ú
û Now, for real roots of y,
Hence, a = 2n -4( x 2 - 10 x + 21) ³ 0
and b = n 2 n. - x 2 + 10 x - 21 ³ 0
Þ ( x - 7 ) ( x - 3) £ 0 \ 3 £ x £ 7
66. Given equation of ellipse is
4 x 2 + 9 y2 = 1 ...(i) 69. Given, z = sin q - i cos q
æ pö
On differentiating, we get p p i çq - ÷
Þ 8 x + 18 yy¢ = 0 = cos æç q - ö÷ + i sin æç q - ö÷ = e è 2ø
è 2 ø è 2 ø
-8 x
Þ y¢ = = m, …(ii) Now,
18 y æ np ö
i ç nq - ÷
np ö np ö
Also, 8 x = 9 y [equation of line] zn = e è 2ø
= cos æç nq - æ
÷ + i sin ç nq - ÷
è 2 ø è 2 ø
On differentiating, we get
æ np ö
Þ 8 = 9 y¢ 1 i ç - nq÷
np ö np ö
8 Þ =e = cos æç nq -
è 2 ø æ
÷ - i sin ç nq - ÷
Þ y¢ = = m, …(iii) zn è 2 ø è 2 ø
9 np ö
1
zn + n = 2 cos æç nq - æ np - nqö
[Qtangents are parallel to this line] \ ÷ = 2 cos ç ÷
z è 2 ø è2 ø
So, by Eqs. (ii) and (iii), we get 1 np
- 8x 8 and zn - n = 2 i sin æç nq - ö.
÷
= z è 2 ø
18 y 9
Þ - x = 2y 70. Given, f[f( x )] = x
Þ x = - 2y Now, f -1( x ) = f( x )
On substituting x = - 2 y in Eq. (i), we get i.e. f( x ) is bijective.
4 ( - 2 y)2 + 9 y 2 = 1 Hence, f( x ) has to be one-one and onto.
Þ 16 y 2 + 9 y 2 = 1 Þ 25 y 2 = 1
71. We know,
1 2
Þ \ x=m y=± P ( A È B) = P( A ) + P( B) - P ( A Ç B)
5 5
Þ P ( A ) + P ( B) = P ( A È B) + P ( A Ç B) …(i)
æ 2 1ö æ 2 1ö 3
So, required points are ç , - ÷ and ç - , ÷. Given, £ P( A È B) £ 1
è5 5ø è 5 5ø 4
3000 æ 2016 ö 1 3
67. ò ç å f(t - r ¢ )f(t - 2016) ÷ dt and £ P( A Ç B) £
-3000 è ø 8 8
r ¢= 2014
7 11
3000 So, £ P( A È B) + P( A Ç B) £
= ò-3000 [f(t - 2014)f(t - 2016) + f(t - 2015)f(t - 2016) 8 8
7 11
+ f(t - 2016)f(t - 2016)]dt \ P ( A ) + P( B) ³ and P( A ) + P( B) £
3000 8 8
= ò-3000 f(t - 2016)[f(t - 2014) + f(t - 2015)
72. Q a, b and c are respectively the AM, GM and HM of first
+ f(t - 2016)]dt and (2 n - 1)th terms.
and also, we know
2016 2017 3000
= ò-3000 0 dt + ò2016 1× (0 + 0 + 1)dt + ò2017 0 dt AM ³ GM ³ HM and AM × HM = GM2
Þ ± ( k - 1) = 1 Þ 40 = 4( a 2 + a ) + 16 ( a 2 + a )
Þ k = 0 and 2 Þ 2 = a 2 + a Þ a = 1, - 2
\ h = - 1, -. 3
75. Given,
Hence, the required points are ( -1, 0) and (-3, 2 ).
f ¢ ( x ) = ( x - 1)2 ( 4 - x )
74. Given, equation of the parabola The sign scheme of f ¢( x )
x 2 = ay
+ + –
x2 1 4
i.e. y= …(i)
a Clearly, f( x ) is increasing, for x Î ( -¥, 4) and
and the equation of line decreasing, for n Î ( 4, ¥).
y - 2x = 1 Since, f ¢ ( x ) = 0 at x = 4.
i.e. y = 2x + 1 …(ii) So, x = 4 is a critical point.
Solved Paper 2015
W B JEE
(Engineering Entrance Exam)
Physics
1. Two particles of mass m 1 and m 2 ; approach 5. A simple pendulum of length L swings in a
each other due to their mutual gravitational vertical plane. The tension of the string
attraction only. Then, when it makes an angle q with the vertical
(a) accelerations of both the particles are equal and the bob of mass m moves with a speed v
(b) acceleration of the particle of mass m1 is is (g is the gravitational acceleration)
proportional to m1 (a) mv 2 / L (b) mg cos q + mv 2 / L
(c) acceleration of the particle of mass m1 is (c) mg cos q - mv 2 / L (d) mg cos q
proportional to m2
(d) acceleration of the particle of mass m1 is inversely 6. The length of a metal wire is L 1 when the
proportional to m1 tension is T1 and L2 when the tension is T2 .
2. Three bodies of the same material and The unstretched length of wire is
having masses m , m and 3m are at L1 + L2
(a) (b) L1L2
temperatures 40° C, 50° C and 60° C, 2
respectively. If the bodies are brought in T L - TL T2 L1 + T1L2
(c) 2 1 1 2 (d)
thermal contact, the final temperature T2 - T1 T2 + T1
will be
7. The line AA¢ is on charged infinite A
(a) 45°C (b) 54°C conducting plane which is
(c) 52°C (d) 48°C
perpendicular to the plane of the θ B
3. A satellite has kinetic energy K, potential paper. The plane has a surface
energy V and total energy E. Which of the density of charge s and B is ball of A′
following statements is true? mass m with a like charge of
(a) K = - V /2 (b) K = V /2 magnitude q. B is connected by string from a
(c) E = K /2 (d) E = - K /2 point on the line AA¢. The tangent of angle (q )
formed between the line AA¢ and the string is
4. An object is located 4 m from the first of two qs qs qs qs
thin converging lenses of focal lengths 2 m (a) (b) (c) (d)
2 Î0 mg 4pe0 mg 2 pe0 mg e0 mg
and 1 m, respectively. The lenses are
separated by 3 m. The final image formed by 8. The current I is in the circuit shown is
the second lens is located from the source at –
a distance of 2V +
Source 2Ω 2Ω 2Ω
+ +
4m 3m 2V – I 2V –
(a) 8.0 m (b) 5.5 m (a) 1.33 A (b) zero (c) 2.00 A (d) 1.00 A
(c) 6.0 m (d) 6.5 m
2 WB JEE (Engineering) Solved Paper 2015
9. A hollow sphere of external radius R and 15. A straight conductor 0.1 m long moves in a
thickness t ( << R) is made of a metal of uniform magnetic field 0.1T. The velocity of
density r. The sphere will float in water, if the conductor is 15 m/s and is directed
(a) t £
R
(b) t £
R
(c) t £
R
(d) t ³
R perpendicular to the field. The emf induced
r 3r 2r 3r between the two ends of the conductor is
(a) 0.10 V (b) 0.15 V (c) 1.50 V (d) 15.00 V
10. A metal wire of circular cross-section has a
resistance R1. The wire is now stretched 16. A ray of light is incident at an angle i on a
without breaking, so that its length is doubled glass slab of refractive index m. The angle
and the density is assumed to remain the between reflected and refracted light is 90°.
same. If the resistance of the wire now Then, the relationship between i and m is
becomes R2 , then R2 : R1 is æ 1ö
(a) i = tan- 1 ç ÷ (b) tan i = m
(a) 1 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4 èm ø
(c) sin i = m (d) cos i = m
11. Assume that each diode as shown in the
figure has a forward bias resistance of 50 W 17. Two particles A and B are moving as shown
and an infinite reverse bias resistance. The in the figure.
current through the resistance 150 W is 6.5 kg 2.2 m/s
50 Ω A 3.6 m/s
1.5 m
B
100 Ω O 2.8 m 3.1 kg
14. A large number of particles are placed 20. A photon of wavelength 300 nm interacts
around the origin, each at a distance R from with a stationary hydrogen atom in ground
the origin. The distance of the center of mass state. During the interaction, whole energy of
of the system from the origin is the photon is transferred to the electron of
the atom. State which possibility is correct.
(a) equal to R
(b) less than equal to R
(Consider, Plank constant = 4 ´ 10 -15 eVs ,
(c) greater then R velocity of light = 3 ´ 108 m /s, ionisation
(d) greater than equal to R energy of hydrogen = 13.6 eV)
WB JEE (Engineering) Solved Paper 2015 3
B
20
Y
B
C
(a) A + B + C
60° (b) ( A + B) C
X
(c) A + B + C
A 10 m/s
(d) A + B + C
(a) 10 m/s along X-axis
(b) 10 3 m/s along Y-axis (perpendicular to X-axis) 26. Two particles A and B having different
(c) 10 5 m/s along the bisection of the velocities of masses are projected from a tower with same
A and B speed. A is projected vertically upward and
(d) 30 m/s along negative X-axis B vertically downward. On reaching the
ground
22. When light is refracted from a surface, which (a) velocity of A is greater than that of B
of its following physical parameters does not (b) velocity of B is greater than that of A
change? (c) both A and B attain the same velocity
(a) Velocity (d) the particle with the larger mass attains higher
(b) Amplitude velocity
(c) Frequency
(d) Wavelength 27. The work function of metals is in the range of
2 eV to 5 eV. Find which of the following
23. A solid maintained at t°1 C is kept in an wavelength of light cannot be used for
evacuated chamber at temperature t°2 C photoelectric effect? (Consider, Plank
( t2 >> t 1 ). The rate of heat absorbed by the constant = 4 ´ 10 - 15 eVs, velocity of light
body is proportional to = 3 ´ 108 m /s)
(a) t 24 - t 14 (a) 510 nm (b) 650 nm
(b) (t 24 + 273) - (t 14 + 273) (c) 400 nm (d) 570 nm
(c) t 2 - t 1 28. A thin plastic sheet of refractive index 1.6 is
(d) t 22 - t 12 used to cover one of the slits of a double slit
arrangement. The central point on the screen
24. Block B lying on a table weighs W. The
is now occupied by what would have been the
coefficient of static friction between the
7th bright fringe before the plastic was used. If
block and the table is m. Assume that the cord
the wavelength of light is 600 nm, what is the
between B and the knot is horizontal. The
thickness (in mm) of the plastic sheet?
maximum weight of the block A for which
(a) 7 (b) 4
the system will be stationary is
(c) 8 (d) 6
4 WB JEE (Engineering) Solved Paper 2015
29. The length of an open organ pipe is twice the 33. In the circuit shown below, the switch is
length of another closed organ pipe. The kept in position a for a long time and is then
fundamental frequency of the open pipe is thrown to position b. The amplitude of the
100 Hz. The frequency of the third harmonic resulting oscillating current is given by
of the closed pipe is R E
(a) 100 Hz (b) 200 Hz
(c) 300 Hz (d) 150 Hz a
C Switch
30. A 5 mF capacitor is connected in series with a b
10 mF capacitor. When a 300 V potential L
difference is applied across this
combination, the total energy stored in the (a) E L /C (b) E / R (c) infinity (d) E C / L
capacitors is
(a) 15 J (b) 1.5 J 34. A charge q is placed at one corner of a cube.
(c) 0.15 J (d) 0.10 J The electric flux through any of the three
faces adjacent to the charge is zero. The flux
31. A cylinder of height h is filled with water through any one of the other three faces is
and is kept on a block of height h /2. The (a) q /3 Î0 (b) q /6 Î0 (c) q /12 Î0 (d) q /24 Î0
level of water in the cylinder is kept
constant. Four holes numbered 1, 2, 3 and 4 35. Two cells A and B of emf 2V and 1.5 V
are at the side of the cylinder and at heights respectively, are connected as shown in
0, h / 4, h /2 and 3 h / 4, respectively. When all figure through an external resistance 10 W.
four holes are opened together, the hole from The internal resistance of each cell is 5W.
which water will reach farthest distance on The potential difference EA and EB across
the plane PQ is the hole number. the terminals of the cells A and B
respectively are
4
A 2V, 5Ω
h 3
2
1
10 Ω
h/2
P Q
B 1.5V
(a) 1 (b) 2 5Ω
(c) 3 (d) 4 (a) EA = 2.0 V, EB = 1.5 V
(b) EA = 2.125 V, EB = 1.375 V
32. The pressure p, volume V and temperature T
(c) EA = 1.875 V, EB = 1.625V
AT - BT2 (d) EA = 1.875 V, EB = 1.375 V
for a certain gas are related by p = ,
V 36. Two charges + q and - q are placed at a
where A and B are constants. The work done distance a in a uniform electric field. The
by the gas when the temperature changes dipole moment of the combination is
from T1 to T2 while the pressure remains 2qa (cos q i$ + sin q $j ), where, q is the angle
constant, is given by between the direction of the field and the
(a) A (T2 - T1 ) + B (T22 - T12 ) line joining the two charges.
37. Find the right condition(s) for Fraunhoffer 39. A circular disc rolls on a horizontal floor
diffraction due to a single slit. without slipping and the centre of the disc
(a) Source is at infinite distance and the incident moves with a uniform velocity v. Which of
beam has converged at the slit the following values of the velocity at a point
(b) Source is near to the slit and the incident beam is
on the rim of the disc can have?
parallel
(c) Source is at infinity and the incident beam is parallel (a) v
(d) Source is near to the slit and the incident beam (b) -v
has converged at the slit (c) 3v
(d) Zero
38. The conducting loop in the form of a circle is
placed in a uniform magnetic field with its 40. Consider two particles of different masses. In
plane perpendicular to the direction of the which of the following situations the heavier
field. An emf will be induced in the loop, if of the two particles will have smaller
(a) it is translated parallel to itself de-Broglie wavelength?
(b) it is rotated about one of its diameters (a) Both have a free fall through the same height
(c) it is rotated about its own axis which is parallel to (b) Both move with the same kinetic energy
the field (c) Both move with the same linear momentum
(d) the loop is deformed from the original shape (d) Both move with the same speed
Chemistry
41. Match the flame colours of the alkaline earth 44. CH3
HBr (1 equiv.)
metal salts in the Bunsen burner. CH2
H2C
A. Calcium p. Brick red
B. Strontium q. Apple green The major product of the above reaction is
C. Barium r. Crimson CH3
CH3
A B C (a)
(a) p r q
(b) r p q Br
(c) q r p CH3
(d) p q r
(b) H C
42. Extraction of gold (Au) involves the 3 Br
formation of complex ions X and Y. H 3C
Roasting Zn
Gold ore ¾¾¾¾® HO - + X ¾® Y + Au CH2
-
CN , H2O, O2 (c)
H 3C
Br
X and Y respectively are
H3 C
(a) Au(CN)-2 and Zn(CN)2-
4
(b) Au(CN)3-
4 and Zn(CN)4
2-
(d)
(c) Au(CN)-3 and Zn(CN)4- H 3C Br
6
(d) Au(CN)-4 and Zn(CN)-3 45. Cl
43. The atomic number of cerium (Ce) is 58. The
NH3
correct electronic configuration of Ce 3 + ion is →
1 1 EtOH
(a) [Xe] 4f (b) [Kr] 4f Br
(c) [Xe] 4f13 (d) [Kr] 4d 1
The product of the above reaction is
6 WB JEE (Engineering) Solved Paper 2015
65. The units of surface tension and viscosity of 72. At temperature of 298 K, the emf of the
liquids respectively are following electrochemical cell
(a) kg m-1s -1, Nm-1 (b) kg s -2 , kg m-1s -1 Ag (s )|Ag + (0.1 M)||Zn2 + (0.1 M)|Zn (s )
(c) Nm-1, kg m-1s -2 (d) kg s -1,kg m-2 s -1
will be (given E °cell = - 1.562 V)
66. The ratio of volumes of CH3COOH 0.1 (N) to (a) -1.532 V (b) -1.503 V
CH3COONa 0.1 (N) required to prepare a (c) 1.532 V (d) – 3.06 V
buffer solution of pH 5.74 is
73. For the reaction, X2 Y4 ( l ) ¾® 2 XY2 (g ) at
(given pKa of CH3COOH is 4.74)
300 K, the values of DU and DS are 2 kcal and
(a) 10 : 1 (b) 5 : 1 (c) 1 : 5 (d) 1 : 10
20 cal K -1 respectively. The value of DG for
67. The reaction of methyltrichloroacetate the reaction is
(Cl3CCO2 Me) with sodium methoxide (a) - 3400 cal
(NaOMe) generates (b) 3400 cal
(a) carbocation (b) carbene (c) -2800 cal
(c) carbanion (d) carbon radical (d) 2000 cal
8 WB JEE (Engineering) Solved Paper 2015
74. The total number of aromatic species 78. The increase in rate constant of a chemical
generated in the following reaction is reaction with increasing temperature is (are)
due to the fact(s) that
(i) Cl + SbCl5 (a) the number of collisions among the reactant
molecules increases with increasing temperature
(b) the activation energy of the reaction decreases
(ii) THF
+ Sodium metal with increasing temperature
(c) the concentration of the reactant molecules
increases with increasing temperature
Br H
(d) the number of reactant molecules acquiring the
(iii) + H2O activation energy increases with increasing
temperature
79. Within the list shown below, the correct pair
H2N
of structures of alanine in pH ranges 2-4 and
9-11 is
(iv) HNO2
I. H3 N+ ¾ CH(CH3 )CO2 H
II. H2 N ¾ CH(CH3 )CO2-
(a) zero (b) 2
(c) 3 (d) 4
III. H3 N+ ¾ CH(CH3 )CO2-
IV. H2 N ¾ CH(CH3 )CO2 H
75. Roasted copper pyrite on smelting with sand (a) I and II
produces (b) I and III
(a) FeSiO 3 as fusible slag and Cu2S as matte (c) II and III
(b) CaSiO 3 as infusible slag and Cu2O as matte (c) III and IV
(c) Ca 3 (PO 4 )2 as fusible slag and Cu2S as matte
(d) Fe 3 (PO 4 )2 as infusible slag and Cu2S as matte 80. Identify the correct method for the synthesis
of the compound shown below from the
76. Ionisation potential values of noble gases following alternatives.
decrease down the group with increase in
atomic size. Xenon forms binary fluorides by
CH3
the direct reaction of elements. Identify the
correct statement(s) from below. O2N
(a) Only the heavier noble gases form such
compounds
CH3CH2CH2CH2Cl HNO3
(b) It happens because the noble gases have higher (a)
ionisation energies AlCl3 H2SO4
98. Given that, x is a real number satisfying 105. Which of the following is not always true?
5 x2 - 26 x + 5 (a)|a + b|2 = |a|2 + |b|2 , if a and b are
< 0, then
3 x2 - 10 x + 3 perpendicular to each other
1 1 (b)|a + lb| ³ |a| for all l ÎR, if a and b are
(a) x < (b) < x<3 perpendicular to each other
5 5
1 1 (c)|a + b|2 + |a - b|2 = 2(|a|2 + |b|2 )
(c) x > 5 (d) < x < or 3 < x < 5
5 3 (d)|a + lb| ³ |a| for all l ÎR, if a is parallel to b
99. The value of l such that the following system 106. If the four points with position vectors
of equations has no solution, is -2 i + j + k , i + j + k , j - k and lj + k are
2 x - y - 2 z = 2; x - 2 y + z = - 4; coplanar, then l is equal to
x + y + lz = 4 (a) 1 (b) 2 (c) -1 (d) 0
(a) 3 (b) 1 (c) 0 (d) -3 107. The least positive value of t, so that the lines
1 x x = t + a, y + 16 = 0 and y = ax are
concurrent, is
100. If f(x) = 2x x( x - 1 )
(a) 2 (b) 4 (c) 16 (d) 8
3 x( x - 1 ) x ( x - 1 )( x - 2 )
x +1 108. If in a DABC , a2 cos 2 A - b2 - c2 = 0, then
p p p
( x + 1 )x . (a) < A< (b) < A<p
4 2 2
( x + 1 )x( x - 1 ) p p
(c) A = (d) A <
2 4
Then, f(100 ) is equal to
é 3p ù
(a) 0 (b) 1 (c) 100 (d) 10 109. {x Î R : cos x ³ sin x} I ê0, is equal to
2 2 2 ë 2 úû
æ 1ö æ 1ö æ 1 ö
101. Let x n = ç1 - ÷ ç1 - ÷ ç1 - ÷ ... p 3p 3p ù
(a) éê 0, ùú È éê ,
p p 3p
(b) éê 0, ùú È éê , ùú
è 3ø è 6ø è 10 ø ë 4 û ë 4 2 úû ë 4 û ë2 2 û
2 p 5p 3p ù 3p ù
æ ö (c) ê 0, ùú È éê
é , (d) éê 0,
ç 1 ÷ ë 4 û ë 4 2 úû ë 2 úû
ç1 - ÷ , n ³ 2. Then, the value of
ç n( n + 1 ) ÷ æ x2 x3 x4 ö p
è 2 ø 110. If sin -1 ç x - + - + . . . ÷ = , where
è 2 4 8 ø 6
lim x n is
n ®¥ | x| < 2, then the value of x is
(a) 1/3 (b) 1/9 (c) 1/81 (d) 0 2 3 2 3
(a) (b) (c) - (d) -
102. The variance of first 20 natural numbers is 3 2 3 2
(a) 133/4 (b) 279/12 111. The area of the region bounded by the curve
(c) 133/2 (d) 399/4 y = x3 , its tangent at (1,1) and X-axis is
103. A fair coin is tossed a fixed number of times. (a)
1
sq unit (b)
1
sq unit
If the probability of getting exactly 3 heads 12 6
equals the probability of getting exactly 2 2
(c) sq unit (d) sq unit
5 heads, then the probability of getting 17 15
exactly one head is 112. If log 0.2 ( x - 1 ) > log 0.04 ( x + 5 ), then
(a) 1/64 (b) 1/32
(c) 1/16 (d) 1/8 (a) - 1 < x < 4 (b) 2 < x < 3
(c) 1 < x < 4 (d) 1 < x < 3
104. If the letters of the word ‘PROBABILITY’ are
written down at random in a row, then 113. The number of real roots of equation
probability that two B’s are together, is log e x + ex = 0 is
2 10 3 6 (a) 0 (b) 1
(a) (b) (c) (d) (c) 2 (d) 3
11 11 11 11
WB JEE (Engineering) Solved Paper 2015 11
128. If (2 + i ) and ( 5 - 2 i ) are the roots of the 136. If the point (2 cos q, 2 sin q ) for 0 Î(0, 2 p ) lies
equation ( x2 + ax + b ) ( x2 + cx + d ) = 0, in the region between the lines x + y = 2 and
where a, b, c and d are real constants, then x - y = 2 containing the origin, then q lies in
product of all the roots of the equation is p 3p
(a) æç q, ö÷ È æç , 2 p ö÷ (b) [0, p ]
è 2ø è 2 ø
(a) 40 (b) 9 5 (c) 45 (d) 35
æ p 3p ö p p
(c) ç , ÷ (d) é , ù
129. If f : [0, p / 2 ) ® R is defined as è2 2 ø êë 4 2 úû
143. A person goes to office by car or scooter or 149. Let a and b be two distinct roots of
bus or train, probability of which are 1/7, 3/7, a cos q + b sin q = c, where a, b, c are three
2/7 and 1/7, respectively. Probability that he real constants and q Î[0, 2 p ]. Then, a + b is
reaches office late, if he takes car, scooter, also a root of the same equation, if
bus or train is 2/9, 1/9, 4/9, and 1/9, (a) a + b = c (b) b + c = a
respectively. Given that he reached (c) c + a = b (d) c = a
office in time, the probability that he æ1 0 0 ö
travelled by a car, is ç ÷
150. For a matrix A = ç2 1 0 ÷, if U 1, U2 and U3
(a) 1/7 (b) 2/7 ç3 2 1 ÷
(c) 3/7 (d) 4/7 è ø
( x - 2 ) dx are 3 ´ 1 column matrices satisfying
144. The value of ò is
{( x - 2 )2 ( x + 3 )7 }1 / 3 æ1 ö æ2 ö æ2 ö
ç ÷ ç ÷ ç ÷
3 æ x -2ö
4/ 3
3 æ x -2ö
3/ 4 AU 1 = ç0 ÷, AU2 = ç3 ÷, AU3 = ç3 ÷ and U is
(a) ç ÷ +C (b) ç ÷ +C ç0 ÷ ç0 ÷ ç1 ÷
20 è x + 3 ø 20 è x + 3 ø è ø è ø è ø
5 æ x -2ö
4/ 3
3 æ x -2ö
5/ 3
3 ´ 3 matrix whose columns are U 1, U2 and
(c) ç ÷ +C (d) ç ÷ +C U3 . Then, sum of the elements of U -1 is
12 è x + 3 ø 20 è x + 3 ø
(a) 6 (b) 0 (c) 1 (d) 2/3
145. Let f : N ® R be such that f(1 ) = 1 151. Let f be any continuously differentiable
and f (1 ) + 2 f (2 ) + 3 f (3 ) +. . . + nf ( n ) = function on [a, b ] and twice differentiable on
n( n + 1 ) f ( n ), for all n Î N, n ³ 2, where N is (a, b ) such that f (a ) = f ¢ (a ) = 0 and f ( b ) = 0.
the set of natural numbers and R is the set of Then,
real numbers. Then, the value of f(500 ) is (a) f ¢ ¢ (a) = 0
(a) 1000 (b) 500 (b) f ¢( x) = 0 for some x Î(a, b )
(c) 1/500 (d) 1/1000 (c) f ¢ ¢( x) = 0 for some x Î(a, b )
(d) f ¢ ¢ ¢( x) = 0 for some x Î(a, b )
146. If 5 distinct balls are placed at random into 5
cells, then the probability that exactly one 152. A relation r on the set of real number R is
cell remains empty, is defined as follows: x ry if and only if xy > 0.
(a) 48/125 (b) 12/125 Then, which of the following is/are true?
(c) 8/125 (d) 1/125 (a) r is reflexive and symmetric
(b) r is symmetric but not reflexive
147. A survey of people in a given region showed (c) r is symmetric and transitive
that 20% were smokers. The probability of (d) r is an equivalence relation
death due to lung cancer, given that a person 153. If cos x and sin x are solutions of the
smoked, was 10 times the probability of differential equation
death due to lung cancer, given that a person
d2 y dy
did not smoke. If the probability of death due a0 2
+ a1 + a2 y = 0 ,
to lung cancer in the region is 0.006. What is dx dx
the probability of death due to lung cancer where a0 , a 1 and a2 are real constants, then
given that a person is a smoker? which of the following is/are always true?
(a) A cos x + B sin x is a solution, where A and B are
(a) 1/140 (b) 1/70
real constants
(c) 3/140 (d) 1/10
p
(b) A cos æç x + ö÷ is a solution, where A is a real
148. In a DABC, if ÐC = 90°, r and R are the è 4ø
inradius and circumradius of the DABC constant
respectively, then 2 ( r + R) is equal to (c) A cos x sin x is a solution, where A is a real
constant
(a) b + c
p p
(b) c + a (d) A cos æç x + ö÷ + B sin æç x - ö÷ is a solution, where
è 4 ø è 4ø
(c) a + b
(d) a + b+c A and B are real constants
14 WB JEE (Engineering) Solved Paper 2015
154. Which of the following statements is /are (c) A quadratic equation with irrational coefficients
p has zero or two rational roots
correct for 0 < q < ? (d) A quadratic equation with integer coefficients has
2
zero or two irrational roots
q 3q
(a) (cos q)1/ 2 £ cos (b) (cos q)3 / 4 ³ cos
2 4 158. If the straight line (a - 1 )x - by + 4 = 0 is
5q 7q
(c) cos ³ (cos q)5 / 6 (d) cos £ (cos q)7 / 8 normal to the hyperbola xy = 1, then which
6 8 of the following does not hold?
155. Let 16 x2 - 3 y2 - 32 x - 12 y = 44 represents (a) a > 1, b > 0
a hyperbola. Then, (b) a > 1, b < 0
(c) a < 1, b < 0
(a) length of the transverse axis is 2 3
(d) a < 1, b > 0
(b) length of each latusrectum is 32 / 3
(c) eccentricity is 19 / 3 159. Suppose a machine produces metal parts
(d) equation of a directrix is x =
19 that contain some defective parts with
3 probability 0.05. How many parts should be
é 1 ù produced in order that the probability of
156. For the function f ( x ) = ê ú, where [ x ] atleast one part being defective is 1/2 or
ë [x]û more? (Given that, log10 95 = 1.977 and
denotes the greatest integer less than or log10 2 = 0.3)
equal to x, which of the following statements (a) 11
are true? (b) 12
(a) The domain is (-¥, ¥) (c) 15
(b) The range is {0} È {-1} È {1} (d) 14
(c) The domain is (-¥, 0) È [1, ¥)
160. Let f : R ® R be such that f (2 x - 1 ) = f ( x ) for
(d) The range is {0} È {1}
all x Î R. If f is continuous at x = 1 and
157. Which of the following is / are always false? f(1 ) = 1, then
(a) A quadratic equation with rational coefficients has (a) f(2 ) = 1
zero or two irrational roots (b) f(2 ) = 2
(b) A quadratic equation with real coefficients has (c) f is continuous only at x = 1
zero or two non-real roots (d) f is continuous at all points
Answers
Physics
1. (c) 2. (b) 3. (a) 4. (b) 5. (b) 6. (c) 7. (a) 8. (a) 9. (b) 10. (c)
11. (b) 12. (c) 13. (b) 14. (b) 15. (b) 16. (b) 17. (a) 18. (d) 19. (a) 20. (d)
21. (b) 22. (c) 23. (c) 24. (b) 25. (c) 26. (c) 27. (b) 28. (a) 29. (c) 30. (c)
31. (a) 32. (c) 33. (c) 34. (d) 35. (d) 36. (a) 37. (c) 38. (b) 39. (d) 40. (d)
Chemistry
41. (a) 42. (a) 43. (a) 44. (b) 45. (c) 46. (a) 47. (a) 48. (a) 49. (c) 50. (c)
51. (b) 52. (b) 53. (b) 54. (c) 55. (a) 56. (a) 57. (a) 58. (a) 59. (c) 60. (b)
61. (c) 62. (a) 63. (a) 64. (a) 65. (b) 66. (d) 67. (b) 68. (d) 69. (c) 70. (a)
71. (a) 72. (a) 73. (c) 74. (c) 75. (a) 76. (a,c) 77. (a,b,d) 78. (a,d) 79. (a) 80. (b)
Mathematics
81. (d) 82. (a) 83. (*) 84. (c) 85. (a) 86. (c) 87. (a) 88. (a) 89. (d) 90. (d)
91. (a) 92. (b) 93. (b) 94. (b) 95. (c) 96. (b) 97. (b) 98. (d) 99. (d) 100. (a)
101. (b) 102. (a) 103. (b) 104. (a) 105. (d) 106. (a) 107. (d) 108. (b) 109. (a) 110. (a)
111. (a) 112. (c) 113. (b) 114. (c) 115. (b) 116. (c) 117. (a) 118. (d) 119. (b) 120. (c)
121. (a) 122. (b) 123. (d) 124. (d) 125. (c) 126. (c) 127. (d) 128. (c) 129. (c) 130. (c)
131. (b) 132. (c) 133. (d) 134. (a) 135. (d) 136. (c) 137. (c) 138. (a) 139. (b) 140. (a)
141. (c) 142. (a) 143. (a) 144. (a) 145. (d) 146. (a) 147. (c) 148. (c) 149. (d) 150. (b)
151. (b,c) 152. (b,c) 153. (a,b,d) 154. (a,c) 155. (a,b,c) 156. (b, c) 157. (c) 158. (a, c) 159. (c,d) 160. (c)
Solutions
Physics
1. The gravitational force acting between the two Þ 3 ms (60 - q) = ms ( q - 50 ) + ms ( q - 40 )
masses m1 and m2 is given by Þ 3 (60 - q) = q - 50 + q - 40
Gm1m2 Þ 180 - 3 q = 2 q - 90
FG =
r2 270
q= = 54° C
Force on mass m1, 5
Gm1m2 3. As we know that for the earth and satellite
F1 = = m1a1
r2 system,
where, a1 = acceleration GMm
Kinetic energy, K =
Gm 2a
Þ a1 = 2 2
r GMm
Potential energy, V = -
Þ a1 µ m2 a
GMm
and similarly, a2 µ m1 and total energy, E = -
2a
2. Let the final temperature after the masses in
thermal contact is q, then from the principle of where, a = radius of the orbit of the satellite and
calorimetry. m = mass of the satellite
Heat lost = Heat gained On the basis of above three expressions for the
energies, we have
16 WB JEE (Engineering) Solved Paper 2015
æ GMm ö T1 T2
GMm ç ÷ V A = A
K= = - ç- a ÷ = -
2a 2 2 DL1 DL2
ç ÷
è ø L L
4. From the lens formula (for first lens) where, A is an cross-section of the wire. assume
1 1 1 1 1 1 to be same at all the situations.
= - Þ = - T L T L
f1 v1 u1 2 v1 ( - 4 ) Þ 1´ = 2 ´
A DL1 A DL2
1 1 1 3
Þ + = = …(i) T1 T2
2 4 v1 4 Þ =
( L1 - L ) ( L2 - L )
4 4
Þ v1 = , u2 = 3 - = 5 / 3 T L -T L
3 3 T1( L2 - L ) = T2 ( L1 - L ); L = 2 1 1 2
T2 - T1
This image will be treated as the source for
second lens, then again from lens formula, we 7. The diagram is as follows
have A
1 1 1
= -
f2 v2 u2
θ B qσ
1 1 F= —
Þ = +5 /3 [By equation (i)] 2∈0
1 v2
1
Þ 1 -5 / 3 = Þ -3 / 2 A′ mg
v2
This is the final image distance from 2nd lens. So, The electric field due to charged infinite
the overall distance of image from the primary s
conducting sheet is E = .
source (or object) 2 Î0
Let d = 4 + 3 - 1.5 = 5 .5 Now, force (electric force) on the charged ball is
qs
5. The situation is given below F = qE =
2Î0
The resultant of electric force and mg balance the
θ L v
L T tension produced in the string.
m F qs qs
So, tan q = e = =
mg 2 Î0 2 Î0 mg
θ
m mg cos θ mg
mg
8. The circuit diagram can be redrawn as The
For motion along a vertical circular track, the potential between A and B is
required centripetal force is along the radius and A
D E
towards the centre of the circle is given by
– 2V
mv2 2Ω + 2Ω
T - mg cos q =
L 2V +
+ 2Ω –2V
mv2 –
Þ T= + mg cos q I
L C F
I′ B I–I′
6. Let the initial length of the metal wire is L. For the loop ABCDA,
The strain at tension T1 is DL1 = L1 - L +2 - 2 I ¢+2 - 2 I = 0 …(i)
The strain at tension T2 is DL2 = L2 - L For the loop ABFEA,
Suppose, the young’s modulus of the wire is Y, 2 - 2 I + 2 - 2( I - I ¢ ) = 0
then 4 - 2 I - 2 I + 2 I ¢= 0
WB JEE (Engineering) Solved Paper 2015 17
Now, radii of the circular paths followed by two 17. Total angular momentum about O is given as,
changes is given by L = L1 + L2 = m1v1r1 + m2 v2 r2
mV = -6.5 ´ 2.2 ´ 15
. + 3.1 ´ 3.6 ´ 2.8
R1 = 1
qB . + 31248
= -2145 . = 9.8 kgm 2 /s
m2V R1 m1
and R2 = Þ = 18. The height raised by liquid in capillary tube
qB R2 m2
2 l cos q
h=
14. As large number of particles is situated at a rgh
distance R from the origin. If particles are
uniformly distributed and make a circular As in freely falling platform a body experience
boundary around the origin, then centre of mass weight lessness.
will be at the origin. So, the liquid will rise upto to length of the
While if the particles are not uniformly capillary.
distributed, then centre of mass will lie between i.e. height raised by the liquid will be 20 cm.
particle and origin. This implies the distance
between centre of mass and origin is always less 19. We have,
than equal to R. Apparent frequency,
15. Given, length of conductor, l = 0.1 m æ v + u0 ö æ 333 + 33 ö
n=ç ÷ n0 = ç ÷ ´ 1000
Mangetic field, B = 0.1 T è v - us ø è 333 - 33 ø
366
Velocity of conductor, V = 15 m/s = ´ 1000 = 1220 Hz
300
The angle between V and B is 90°
20. The energy of the photon,
When V and B are mutually perpendicular, then hc
emf (induced) is given by E=
l
e = VBl
15 4 ´ 10 - 15 eVs ´ 3 ´ 108 m /s
= 15 ´ 0.1 ´ 0.1 = = 0.15 V =
100 300 ´ 10 - 9 m
2
4 ´ 10
16. As situation can be diagrammatically as below = eV
300
4
= eV = 1.33 eV
i θ 3
The ionisation energy is 13.6 eV which is greater
90° than energy of photon, so atom can not come into
r excited state and will remain in ground state.
21. The component of velocity of B along x-direction
VBx = 20 cos 60 °
From law of reflection, 1
i=q = 20 ´ = 10 m/s
2
Now, q + r + 90° = 180° VA = 10 $i
Þ i + r + 90 ° = 180 °
V = 10 $i + 10 3 $j
B
r = 90° - i
VBA = VB - VA = 10 $i + 10 3 $j - 10 $i = 10 3 $j
Also, from Snell’s law
sin i 22. When light is refracted its frequency will remain
=m
sin r unchanged.
sin i sin i 23. As we know that, the rate of cooling is
Þ = =m
sin (90° - i ) cos i dq
= bA ( q - q0 )
Þ tan i = m dt
WB JEE (Engineering) Solved Paper 2015 19
26. The situation is diagrammatically as below 30. According to question the figure can be drawn as
A below
5 µF 10 µF
X
B
Tower
– +
When A is projected with a vertical speed the after 300 V
sometime it comes on the same level with same
speed as it was projected. Now, the downward The equivalent capacitance,
speeds of A and B at the level-X is same. So, on 1 1 1 1 1
reaching the ground, velocity of A and B are same. = + = +
C eq C1 C2 5 10
27. Given that, 2 eV £ f £ 5 eV 2+1 3
Wavelength corresponding to minimum and = =
10 10
maximum values of work function are 10
hc 4 ´ 10 - 15 ´ 3 ´ 108 Þ C eq = mF
l max = = 3
E 2
20 WB JEE (Engineering) Solved Paper 2015
B = 15
. - 0.025 ´ 5
= A (T2 - T1 ) - [(T2 )2 - (T1 )2 ] = 15
. - 0.0125 = 1.375 V
2
33. When switch is in position a, then capacitor will 36. The situation can be diagrammatically as,
be changed. E
+q F = qE
When switch is in position b, then circuit
becomes an L C oscillator with frequency, a θ
E
1 1
n= . F = –qE –q
2 p LC
E
In two situation the net reactance or impedance
of the circuit is zero. The dipole moment,
This implies the current, i =
E
i.e. infinite P = 2qa (cos q $i + sin q $j)
zero
current through the circuit even without any The net force on the dipole is always zero while
applied emf. net torque on the dipole is not zero.
34. Consider the diagram, 37. The source is at infinity and incident beam is
parallel.
38. Whenever, there is a change in the magnetic flux
q crossing through the loop, an emf induced in it
and hence there is an amount of induced current
through the loop. If loop is rotated about one of its
diameter, the flux through it varies and causing
the emf induced in it.
In above the flux coverage of three face is shown 39. In this case (pure rolling) the velocity of point of
in figure. Which is like a quadrant in any plane. contact is zero.
So, flux will be
WB JEE (Engineering) Solved Paper 2015 21
Chemistry
41. Flame colours are produced from the movement 45. It is a substitution reaction, so the product
of the electrons in the metal ions present in the formed is
compounds. Cl NH2
Calcium gives brick red colour. Strontium gives
crimson colour. →
NH3
CH3 F
1
CH2 + H Br– →
3 +
4
H2C 2
Xe
H H
—
—
+ +
CH2 ==C—CH—CH2 ! CH2—C == CH—CH2 F
—
—
49. CH2 == C == CH ¾C H ¾C ºº CH
CH3 CH3
½
1,2-addition 1,4-addition C ºº N
(Minor product) (Major product) sp
50. Rate of reaction is doubled when concentration of 57. Cold FeSO4 solution on absorption of NO
A is doubled. Again, rate of reaction becomes four develops brown colour due to the formation of
times when concentration of A is increased by [Fe(H2O)5 (NO)] SO4 .
four times. It is clearly shown that there is no FeSO4 + NO + 5H2O ¾®
effect on rate of reaction on increasing the
[Fe(H2O)5 (NO)]SO4
concentration of B. Brown ring complex
Thus, order with respect to A is 1 and order with
This complex has 3.89 BM magnetic moment
respect to B is 0. Total order of reaction = 1
shows that it has 3 unpaired electrons.
51.
58. Group II A Group III A
Osmotic pressure
Be B
Mg Al
RT The electronic configuration of boron is 2 s 2 2 p1.
And electronic configuration of B+ is 2 s 2 . Hence,
O Concentration it is difficult to remove second electron from
2 s 2 shell because half-filled and full-filled orbitals
Slope, RT = 291 R or T = 291 K are more stable than others.
\Temperature = 291 - 273 = 18 ° C 59. 15% will form racemic mixture with another 15%.
3 RT Hence, the enantiomeric excess is
52. As from the formula, vrms = = (85 - 15 ) = 70%
M
Given that (v rms )CO = 1000 m/ s 60. CH3 CH3
(Temp.)CO = 27° C = 300 K
∆
(Temp.) N2 = 600 K + AlCl3 + HCl →
Now putting the values, we get CH3
( v rms )CO 3 R TCO M N2
= ´ ´ CH3
( v rms ) N2 3 R MCO TN2 1, 4-dimethyl benzene 1, 3-dimethyl benzene
-
or = 2.303 log K = 2.303 67. NaOMe r Na +
+ OMe
or K = 101 = 10
O–
Unit of K = (atm)Dn = (atm)2 - 3 = (atm)-1 O
– →
\ K = 10 atm-1 Cl3C — C + OMe Cl3C — C — OMe
OMe
64. As we know that, OMe
→
Equivalent conductance, (l)
specific conductance (K ) ´ 1000 O
=
concentration –
CCl2 + Cl–
→
l 1000 CCl3 + C
or, = Carbene
K conc. MeO OMe
l 1000 W -1 cm2eq -1
or, =
K 0.01 W -1 cm-1 68. Best reagent for nuclear iodination of aromatic
compounds is I2 / HNO3 .
(Given, conc. = 0.01 N)
l →
or, = 105 cm3 eq -1 HNO3 Nitronium ion
K I2
–
I I+ +I
65. From the formula, Aromatic compounds
→
Surface tension, Electrophilic substitution
reaction
DW J kg m2s -2
(g ) = = 2= = kg s -2
DA m m2
69. To a part of sodium extract, FeSO4 solution is
dV
Form the formula, F = h × A × added and the contents are warmed. A few drops
dx of FeCl3 solution are then added and resulting
or, h (coefficient of viscosity) solution is acidified with conc. HCl. The
F N appearance of bluish green or Prussian blue
= = -1
= Nm-2 × s
dV ms colouration confirms the presence of nitrogen.
A× m2 × The reactions that occur during this test are as
dx m
or kg ms -2 × m-2 × s = kg m-1 s -1 2NaCN + FeSO4 ¾® Na 2SO4 + Fe(CN)2
Fe(CN)2 + 4NaCN ¾® Na 4 [Fe(CN)6 ]
66. Given, that pH = 5.74, pK a =4.74 Sodium ferrocyanide
Suppose that volume of acid solution = x L 3 Na 4 [Fe(CN)6 ] + 4 FeCl3 ¾®
Volume of salt solution = y L Fe4 [Fe(CN)6 ]3 + 12NaCl
From Henderson equation. Ferric ferrocyanide
(Prussian blue)
[Salt]
pH = pK a + log 70.
[Acid] OC2H5 H
[CH3COONa]
or, pH - pK a = log RMgBr+H — C — OC2H5 R—C— OC2H5
[CH3COOH]
[CH3COONa]
or, 5.74 - 4.74 = 1 = log OC2H5 OC2H5
[CH3COOH] (Acetal)
CH3COONa]
→
or = 10 H3O+
[CH3COOH]
0.1 x
O
[CH3COOH] 1 x+ y
or = =
[CH3COONa] 10 0.1 y R—C—H
Aldehyde
x+ y (P)
x 1
Thus, =
y 10
24 WB JEE (Engineering) Solved Paper 2015
71. The sequence of reactions are 74. Criteria for aromaticity (Huckel rule)
2AgNO3 + Na2S2O3
→ Ag S O +2NaNO The cyclic p molecular orbital (electron cloud)
2 2 3 3
Silver Sodium (X) formed by overlap of p-orbitals must contain
nitrate thiosulphate (White ppt.) (4 n + 2 ) p-electrons, where n = integer 0, 1, 2, 3,
es 3
etc. This is known as Huckel rule.
xc 2 O
(e 2 S
s)
Na
Na3[Ag(S2O3)2] (Y) Not soluble
Soluble (i) Cl + SbCl5
in water
H2O/∆ r –
SbCl6
Ag2S (Z) 2πe– system
Silver sulphide (aromatic)
Hence, X = Ag 2S2O3
(ii) Na / THF
Y = Na 3 [Ag(S2O3 )2 ] Z = Ag 2S s
72. From the given cell, the cell reaction is 6πe– system
2+ (aromatic)
2Ag (s ) + Zn (0.1 M ) ¾®
2Ag + (0.1 M) + Zn(s ) H Br
H2O
The Nernst equation is (iii) r
0.0591 [Ag + ]2
Ecell = E °cell - log Tropylium cation
n [Zn2 + ]
6πe– system
2
(0.0591) (0.1) (aromatic)
or, Ecell = ( - 1.562) - log
2 (0.1) H2N
(where, E°cell = -1.562 V)
0.0591 (iv)
or, Ecell = ( - 1.562) - log 10 -1 HNO2
2
0.0591 4πe– system
= -1562
. +
2 (non-aromatic)
= - 1562
. + 0.02955
75. Roasted copper pyrite on smelting with sand
= -1532
. V produces FeSiO3 as fusible slag and Cu2S as
73. For the reaction, X 2Y4 (l ) ¾® 2 X Y2 (g ) matte.
Dng = number of gaseous products For removing the gangue, FeS, silica present in
- number of gaseous reactants the lining of the Bessemer converter, acts as a flux
and forms slag (iron silicate) on reaction with
Dng = 2 - 0 = 2 FeO.
Given that, DU = 2 kcal
FeS + 3O2 ¾® 2FeO + 2SO2
DS = 20 cal K -1 Gangue
Mathematics
81. From option (d), we have Also, g is the root of x2 + px + 1 = 0.
a-b =c -d …(i) \ g 2 + pg + 1 = 0 Þ g 2 = - pg - 1
and a - b2 = c2 - d2
2
…(ii)
Now, ( a + g )(b + g ) = ab + ag + bg + g2
Consider, a2 - b2 = c2 - d2 = 1 + g(a + b)- pg - 1 = g ( a + b - p)
Þ ( a + b )( a - b ) = ( c - d)( c + d) = g ´0 = 0 [Q a + b = p]
Þ a+ b =c + d …(iii) a
[using Eq. (i)] Since, g = - a or -b
On adding Eqs. (i) and (iii), we get \ ( a + g )(b + g ) = 0
2a = 2c Þ a = c
83. General term of (31/5 + 71/3 )100 is given by
Þ b=d [using Eq. (iii)]
Tr + 1 = 100C r (31/5 )100 - r (71/3 )r
Thus, an + b n = c n + dn for all n ÎN.
100 - r r
82. Since, a and b are the roots of x2 - px + 1 = 0. = 100C r × 3 5
× 73
\ a + b = p and ab =1 100 - r r
For a rational term, and must be integer.
5 3
26 WB JEE (Engineering) Solved Paper 2015
Þ 0
ke = 0 éQ af ( x ) dx = 0 ù lim 3 x2
ëê òa ûú = x ®2
[using L’ Hospital’s rule]
1
Þ k =0 [Q e° = 1]
\ f ( x ) = 0 Þ f (log e 5 ) = 0 = 3 ´ (2 )2 = 12
2x x kx
x2 - x + 4 93. Given, cot + tan = cos ec
88. Let y = 3 3 3
x2 + x + 4
x
Þ x2 y + xy + 4 y = x2 - x + 4 Let q = , then
3
WB JEE (Engineering) Solved Paper 2015 27
2
Taking common x, ( x -1) from R2 and R3 1æ 2ö
respectively, we get Þ ç1 + ÷xn =
9 è nø
1 x x +1 1 1
f ( x ) = x( x - 1) 2 x -1 x +1 Þ lim xn = (1 + 0 )2 =
n®¥ 9 9
3x x ( x - 2 ) ( x + 1) x
102. Variance of n natural numbers
Taking common ( x + 1) from C3 , we get n2 - 1 (20 )2 - 1
1 x 1 = = [Q n = 20]
12 12
f ( x ) = x( x - 1)( x + 1) 2 x -1 1
400 - 1 399 133
3x x( x - 2 ) x = = =
12 12 4
Taking common x from R3 , we get
103. Let the coin be tossed n times.
1 x 1
Let getting head is consider to be success.
f ( x ) = x2 ( x - 1) ( x + 1) 2 x - 1 1 1 1 1
3 x -2 1 \ p = , q = 1- p = 1- =
2 2 2
Applying R1 ® R1 - R2 , R2 ® R2 - R3 , It is given that,
we get P ( X =3 ) = P( X = 5 )
-1 1 0 3 n -3 5 n -5
n æ 1ö æ 1ö æ 1ö æ 1ö
2
f ( x ) = x ( x - 1)( x + 1) -1 1 0 =0 Þ C3 ç ÷ ç ÷ = nC5 ç ÷ ç ÷
è2 ø è2 ø è2 ø è2 ø
3 x -2 1 Þ n
C3 = nC5
[Q R1 is identical with R2 ] Þ n = 3 +5 [Q nC x = nC y Þ x + y = n]
Þ f (100 ) = 0
Þ n =8
é 1 1
101. We have, xn = ê æç1 - ö÷ æç1 - ö÷
1 8 -1
æ 1ö æ 1ö
ëè 3 øè 6ø Now, P ( X = 1) = 8C1 ç ÷ ç ÷
è2 ø è2 ø
2
æ 1ö æ 2 öù æ 1ö 1
8
ç1 - ÷ ... ç1 - ÷ú = 8C1 ´ ç ÷ =
è 10 ø è n( n + 1) øû è2 ø 32
2
é n æ n2 + n - 2 ö ù 104. Total number of ways in which the letters of the
Þ xn = ê Õ ç ÷ú word ‘PROBABILITY’ can be arranged in a row
êë n=2 è n ( n + 1) ø úû 11!
2 =
é n æ ( n + 2 )( n - 1) ö ù 2! 2!
= êÕ ç ÷ú
ë n=2 è n ( n + 1) ø û Number of ways in which two B’s are together
n n 2 10 !
é æn + 2ö æ n - 1ö ù =
= êÕ ç ÷ × Õç ÷ ú 2!
ë n=2 è n + 1 ø n =2 è n ø û \ Required probability
2 2
é n æ n + 2 ö ù é n æ n - 1ö ù 10 !
= êÕ ç ÷ú êÕ ç ÷ú 10 ! ´ 2 ! 2
ë n=2 è n + 1 ø û ë n=2 è n ø û = 2! = =
11! 11! 11
2
æ4 5 6 n + 2ö 2! 2!
Þ xn = ç × × × ... × ÷
è3 4 5 n + 1ø
2
105. Option (a) If a and b are perpendicular to each
æ1 2 3 n - 1ö other, then a × b = 0
ç × × × ... × ÷
è2 3 4 n ø Now consider,
2 2 |a + b|2 = (a + b ) × (a + b )
æn + 2ö æ1ö
Þ xn = ç ÷ ç ÷
è 3 ø è nø = |a|2 + |b|2
2 So, option (a) is always true.
1 æn + 2ö
Þ xn = ç ÷
9 è n ø
WB JEE (Engineering) Solved Paper 2015 29
Þ 3 (0 + 2 l - 2 ) = 0 æ ö
ç ÷ é
-1 ç x ÷=p a ù
Þ l =1 Þ sin êQ S¥ = ú
ç æ -xö ÷ 6 ë 1 - rû
107. Consider the given equation of lines, ç 1 - çè ÷÷
è 2 øø
x - (t + a ) = 0 …(i)
æ 2x ö p
y + 16 = 0 …(ii) Þ sin-1 ç ÷=
è2 + x ø 6
and - ax + y = 0 …(iii)
2x p 2x 1
Since, these lines are concurrent, therefore the Þ = sin Þ =
2+ x 6 2+ x 2
system of equation is consistent.
2
Þ 4x =2+ x Þ 3x =2 Þ x =
3
30 WB JEE (Engineering) Solved Paper 2015
1 4 dy
135. 2 cot -1 - cot -1 139. Given, = - (3 x2 tan-1 y - x3 )(1 + y2 )
2 3 dx
3 dy
= 2 tan-1 2 - tan-1 Þ = + x3 (1 + y2 ) - 3 x2
4 dx
= p + tan-1
4
- tan-1
3 (tan-1 y )(1 + y2 )
-3 4 1 dy
Þ × = x3 - 3 x2 tan-1 y
-1 4 -1 3 (1 + y2 ) dx
= p - tan - tan
3 4
1 dy
ì 4 3 ü Þ × + 3 x2 tan-1 y = x3
ï + ï 1 + y2 dx
= p - í tan-1 3 4 ý 1 dy dt
4 3 Put tan-1 y = t Þ × =
ï 1- × ï 2
î 3 4þ 1 + y dx dx
p p dt
-1
= p - tan ¥ = p - = \ + 3 tx2 = x3
2 2 dx
136. which is linear differential equation in t.
(0,2) 3 x 2 dx 3
x–y=2 Now, IF = eò = ex
(2,0)
O 140. Given, y = e- x cos2 x …(i)
x+y=2 dy
(0,–2) \ = e- x ( - sin 2 x ) 2
dx
2 2
(2 cos q, 2 sin q) will lie on the circle x + y = 4 + cos 2 x × e- x ( -1)
(from the above figure). Since, point lies on the dy
region containing origin. Þ = - 2 sin 2 x × e- x - y
dx
So, point will be on the shaded region. dy
Þ + y = - 2 sin 2 x × e- x
æ p 3p ö dx
\ q Îç , ÷
è2 2 ø d2 y dy
Þ + = - 2 [sin 2 x × e- x ( -1)
137. Let the point be ( h, k ). dx2 dx
+ e- x 2 cos 2 x]
½h - 2k + 1 ½
Then, ½ ½= 5 = 2 sin 2 x × e- x - 4 y [from Eq. (i)]
2 2
½ 1 + ( - 2 ) ½ d2 y dy dy
Þ + =- - y -4y
Þ h - 2k + 1 = ± 5 …(i) dx2 dx dx
½2 h + 3 k - 1½ d2 y dy
Also, ½ ½ = 13 Þ 2
+2 + 5y =0
2 2 dx dx
½ 2 +3 ½
Þ 2 h + 3 k - 1 = ± 13 …(ii) 141. Given, f (0 ) = 0 and f (0 ) = 2
On solving Eqs. (i) and (ii), we get four points. So, 1
\ lim [ f ( x ) + f (2 x ) + f (3 x ) + K + f (2015 x )]
there are such 4 points. x ®0x
f ¢( x ) + 2 f ¢(2 x ) + 3 f ¢(3 x )
138. Q P( x ) = ( x - 3 )( x - 5 )
+ K + 2015 f ¢(2015 x )
= Q( x ) + ( ax + b ) = lim
x ®0 1
Given, P(3 ) = 10 and P(5 ) = 6 [applying L’ Hospital’s rule]
Þ 3 a + b = 10 …(i)
and 5a + b = 6 …(ii) 2 + 2 ´ 2 + 3 ´ 2 + K + 2015 ´ 2
=
On solving Eqs. (i) and (ii), we get 1
a = - 2and b = 16 = 2[1 + 2 + 3 + K + 2015]
\Remainder = - 2 x + 16
34 WB JEE (Engineering) Solved Paper 2015
T
æ a1 ö æ 1ö é æ1 ö ù é -1 -7 9 ù é -1 -2 0 ù
ç ÷ ç ÷ êQ AU = ç0 ÷ ú adj U = ê -2 -5 6 ú = ê -7 -5 -3 ú
Þ ç 2 a1 + b1 ÷ = ç0 ÷ ê 1 ç ÷ú ê ú ê ú
ç ÷ ç ÷ ç ÷ êë 0 -3 3 úû êë 9 6 3 úû
è3 a1 + 2 b1 + c1 ø è0 ø êë è0 ø úû
\ a1 = 1, 2 a1 + b1 = 0 é 1 2 ù
ê- 3 - 3 0 ú
Þ 2 + b1 = 0 ê ú
7 5
Þ b1 = - 2 Þ U -1 = ê - - -1ú
and 3 a1 + 2 b1 + c1 = 0 ê 3 3 ú
ê 3 2 1ú
Þ 3 + 2( -2 ) + c1 = 0 ê ú
Þ 3 - 4 + c1 = 0 ë û
Þ c1 = 1 Hence, the sum of all elements of U -1 is 0.
æ 1 0 0 ö æ a2 ö 151. We have, f is continuous and differentiable
ç ÷ç ÷ function on [a, b].
Similarly, AU2 = ç2 1 0 ÷ ç b2 ÷
ç ÷ç ÷ Also, f ( a) = f ( b ) = 0.
è3 2 1 ø è c2 ø
By Rolle’s theorem $ c Î(a, b)
æ a2 ö æ2 ö é æ2 ö ù such that f ¢( c ) = 0
ç ÷ ç ÷ êQ AU = ç3 ÷ ú
Þ ç 2 a2 + b2 ÷ = ç3 ÷ ê 2 ç ÷ú Thus, $ x Î(a, b) such that f ¢( x ) = 0
ç ÷ ç ÷ ç ÷
è3 a2 + 2 b2 + c2 ø è0 ø êë è0 ø úû Let at x = c Î(a,b), f ¢( c ) = 0
\ a2 = 2 and 2 a2 + b2 = 3 Now, f is continuously differentiable on [a, b].
Þ 2 ´ 2 + b2 = 3 Þ f ¢ is continuous on [a, b]
Þ 4 + b2 = 3 Also, f is twice differentiable on (a, b)
Þ b2 = - 1 \ f ¢ is differentiable on (a, b)
and 3 a2 + 2 b2 + c2 = 0 and f ¢( a) = 0 = f ¢( c )
Þ 3 ´ 2 + 2( -1) + c2 = 0 By Rolle’s theorem $ k Î(a, c) such that f ¢¢( k ) = 0
Þ 6 - 2 + c2 = 0 Thus, $ x Î(a, c) such that f ¢¢( x ) = 0
Þ c2 = - 4 So, $ x Î(a, b) such that f ¢¢( x ) = 0
æ 1 0 0 ö æ a3 ö Let us consider, f ( x ) = ( x - a)2 ( x - b ),
ç ÷ç ÷ where f ( a) = f ( b ) = f ¢( a) = 0 but
Now, AU3 = ç2 1 0 ÷ ç b3 ÷
ç ÷ç ÷ f ¢¢( a) ¹ 0 and f ¢¢¢( x ) ¹ 0 for any x Î(a, b)
è3 2 1 ø è c3 ø
æ a3 ö æ2 ö é æ2 ö ù 152. We have, xry : xy > 0
ç ÷ ç ÷ êQ AU = ç3 ÷ ú (i) Reflexive suppose xrx Î R (Relation)
Þ ç 2 a3 + b3 ÷ = ç3 ÷ ê 3 ç ÷ú
ç ÷ ç ÷ ç ÷
è3 a3 + 2 b3 + c3 ø è 1 ø êë è1 ø úû Þ x2 > 0
\ a3 = 2 and 2 a3 + b3 = 3 which is not true when x = 0.
Þ 2(2 ) + b3 = 3 Hence, relation is not reflexive.
Þ 4 + b3 = 3 (ii) Symmetric xry Î R.
Þ b3 = - 1 Þ xy > 0
and 3 a3 + 2 b3 + c3 = 1 Þ yx > 0
Þ 3 ´ 2 + 2( -1) + c3 = 1 Þ yrx Î R
Þ 6 - 2 + c3 = 1 If ( x, y ) Î R, then hence, relation is symmetric.
Þ c3 = - 3 (iii) Transitive xry Î R
æ1 2 2ö Þ xy > 0
ç ÷ and yrz Î R
\ U = ç -2 -1 -1 ÷
ç ÷ Þ yz > 0
è 1 -4 -3 ø
Þ xy2 z > 0
|U| = 1 (3 - 4 ) - 2 (6 + 1) + 2 (8 + 1)
= - 1 - 14 + 18 = 3 Þ xz > 0
WB JEE (Engineering) Solved Paper 2015 37
Þ ( x, z ) Î R p
154. 0 < q < ; cos q is strictly decreasing function.
If ( x, y ) Î R, then ( y, z ) Î R 2
Þ ( x, z ) Î R q
Option (a) When q > , then
Hence, relation is transitive. 2
q
153. Option (a) Let f ( x ) = cos x and g( x ) = sin x cos q £ cos
2
Consider the Wronskian of f ( x ) and g( x ), q
\ cos q £ cos [correct]
f ( x ) g( x ) 2
W=
f ¢( x ) g ¢( x ) 3q
Option (b) When q > , then
cos x sin x 4
= 3q
- sin x cos x cos q £ cos
4
= cos2 x + sin2 x = 1 =/ 0 3 /4 3q
\ (cos q) £ cos [incorrect]
Thus, the functions are linearly independent. So, 4
the general solution of given differential equation 5q
Option (c) When q > , then
is given by y = A cos x + B sin x, where A and B 6
are real constants. 5q
cos q £ cos
[Q if y1 and y2 are linearly independent solutions 6
of the differential equation ay ¢¢ + by ¢ + c = 0, 5 /6 5q
then the general solution is y = c1 y1 + c2 y2 , \ (cos q) £ cos [correct]
6
where c1 and c2 are constants]. 7q
Hence, option (a) is true. Option (d) When < q, then
8
æ pö 7q
Option (b) Let y = A cos ç x + ÷ cos q £ cos
è 4ø 8
æ p pö 7 /8 7q
= A çcos x × cos - sin x × sin ÷ But (cos q) £ cos [incorrect]
è 4 4ø 8
A 155. Given equation of hyperbola is
= (cos x - sin x )
2 16 x2 - 3 y2 - 32 x - 12 y = 44
A æ Aö
= cos x + ç - ÷ sin x Þ 16 x2 + 16 - 32 x - 3 y2 - 12 - 12 y
2 è 2ø
= 44 + 4
which is in the form of general solution.
Þ 16( x - 1)2 - 3( y + 2 )2 = 48
Hence, option (b) is true.
Option (c) Let y = A cos x sin x, which can not be ( x - 1)2 ( y + 2 )2
Þ - =1
expressed in the form of general solution. 3 16
æ pö On comparing the equation with standard
Option (d) Let y = A cos ç x + ÷ equation of hyperbola,
è 4ø
pö we get a = 3, b = 4
æ
+ B sin ç x - ÷
è 4ø Now, length of transverse axis
æ 1 1 ö = 2a = 2 3
= A çcos x × - sin x × ÷
è 2 2ø and length of latusrectum
æ 1 1 ö 2 b2 2 ´ 16 32
+ B çsin x × - cos x × ÷ = = =
è 2 2ø a 3 3
æ A B ö æ B Aö b2
= cos x ç - ÷ + sin x ç - ÷ which is in \ Eccentricity ( e) = 1 +
è 2 2ø è 2 2ø a2
the form of general solution. 16 19
= 1+ =
Hence, option (d) is true. 3 3
38 WB JEE (Engineering) Solved Paper 2015
WB JEE
Engineering Entrance Exam
Physics
Category I
Directions (Q. Nos. 1-45) Carry one mark each, for which only one option is correct. Any wrong answer will
lead to deduction of 1/3 mark.
1. Consider three vectors A = i$ + $j - 2k $, 5. The output Y of the logic circuit given below is
$ $ $ $ $ $
B = i - j + k and C = 2i - 3 j + 4 k. A vector X of
A Y
the form aA + bB (a and b are numbers) is
perpendicular to C. The ratio of a and b is B
(a) 1 : 1 (b) 2 : 1 (a) A + B (b) A
(c) -1 : 1 (d) 3 : 1 (c) ( A + B)×A (d) ( A + B)×A
2. Three capacitors 3 mF, 6 mF and 6 mF are 6. One mole of an ideal monoatomic gas is heated
connected in series to a source of 120 V. The at a constant pressure from 0°C to 100°C. Then
potential difference, in volt, across the 3 mF the change in the internal energy of the gas is
capacitor will be (Given, R = 8.32 J mol-1K -1 )
(a) 24 (b) 30 (c) 40 (d) 60 (a) 0.83 ´ 10 3 J (b) 4.6 ´ 10 3 J
3
3. A galvanometer having internal resistance (c) 2.08 ´ 10 J (d) 1.25 ´ 10 3 J
10 W requires 0.01 A for a full scale deflection.
To convert this galvanometer to a voltmeter of 7. The ionization energy of hydrogen is 13.6 eV.
full-scale deflection at 120 V, we need to The energy of the photon released when an
connect a resistance of electron jumps from the first excited state
( n = 2) to the ground state of a hydrogen atom is
(a) 11990 W in series
(b) 11990 W in parallel (a) 3.4 eV (b) 4.53 eV
(c) 12010 W in series (c) 10.2 eV (d) 13.6 eV
(d) 12010 W in parallel 8. A parallel plate capacitor is charged and then
4. A whistle whose air column is open at both disconnected from the charging battery. If the
ends has a fundamental frequency of 5100 Hz. plates are now moved farther apart by pulling
If the speed of sound in air is 340 ms-1, the at them by means of insulating handles, then
length of the whistle, in cm, is (a) the energy stored in the capacitor decreases
(a) 5/3 (b) 10/3 (b) the capacitance of the capacitor increases
(c) the charge on the capacitor decreases
(c) 5 (d) 20/3
(d) the voltage across the capacitor increases
2 WB JEE (Engineering) · Solved Paper 2014
18. A metal rod is fixed rigidly at two ends so as to (a) 2.16 ´ 10 -23 (b) 3.21 ´ 10 -22
-24
prevent its thermal expansion. If L, a and Y (c) 3.21 ´ 10 (d) 1.26 ´ 10 -23
respectively denote the length of the rod,
coefficient of linear thermal expansion and 23. A very small circular loop of radius a is initially
Young’s modulus of its material, then for an (at t = 0) coplanar and concentric with a much
increase in temperature of the rod by DT, the larger fixed circular loop of radius b. A
longitudinal stress developed in the rod is constant current I flows in the larger loop. The
(a) inversely proportional to a smaller loop is rotated with a constant angular
(b) inversely proportional to Y speed w about the common diameter. The emf
(c) directly proportional to DT / Y induced in the smaller loop as a function of
(d) independent of L time t is
pa 2m 0 I pa 2m 0 I
19. A particle is moving uniformly in a circular (a) w cos ( wt ) (b) w sin ( w2 t 2 )
2b 2b
path of radius r. When it moves through an 2
pa m 0 I pa 2m 0 I
angular displacement q, then the magnitude of (c) w sin ( wt ) (d) w sin2 ( wt )
the corresponding linear displacement will be 2b 2b
æq ö æq ö 24. A luminous object is separated from a screen
(a) 2 r cos ç ÷ (b) 2 r cot ç ÷
è2 ø è2 ø by distance d. A convex lens is placed between
æq ö æq ö the object and the screen such that it forms a
(c) 2 r tan ç ÷ (d) 2 r sin ç ÷
è2 ø è2 ø distinct image on the screen. The maximum
possible focal length of this convex lens is
20. A uniform rod is suspended horizontally from (a) 4d (b) 2d
its mid-point. A piece of metal whose weight is d d
(c) (d)
w is suspended at a distance l from the 2 4
mid-point. Another weight W1 is suspended on
the other side at a distance l1 from the 25. An infinite sheet carrying a uniform surface
mid-point to bring the rod to a horizontal charge density s lies on the xy-plane. The work
position. When w is completely immersed in done to carry a charge q from the point
$ ) to the point B = a( i$ - 2$j + 6k
A = a( $i + 2$j + 3k $)
water, w1 needs to be kept at a distance l2 from
the mid-point to get the rod back into (where a is a constant with the dimension of
horizontal position. The specific gravity of the length and e0 is the permittivity of free space)
metal piece is is
w wl1 3saq 2saq
(a) (b) (a) (b)
w1 wl - w 1l2 2 e0 e0
l1 l 5saq 3saq
(c) (d) 1 (c) (d)
l1 - l2 l2 2 e0 e0
21. A drop of some liquid of volume 0.04 cm3 is 26. The intensity of magnetization of a bar magnet
placed on the surface of a glass slide. Then is 5.0 ´ 104 Am-1. The magnetic length and the
another glass slide is placed on it in such a way area of cross-section of the magnet are 12 cm
that the liquid forms a thin layer of area 20 cm2 and 1 cm2 respectively. The magnitude of
between the surfaces of the two slides. To magnetic moment of this bar magnet is
separate the slides a force of 16 ´ 105 dyne has (in SI unit)
to be applied normal to the surfaces. The (a) 0.6 (b) 1.3 (c) 1.24 (d) 2.4
surface tension of the liquid is (in dyne-cm-1 )
27. A particle moves with constant acceleration
(a) 60 (b) 70 (c) 80 (d) 90
along a straight line starting from rest. The
22. An electron in a circular orbit of radius 0.05 nm percentage increase in its displacement during
performs 1016 revolutions per second. The the 4th second compared to that in the
magnetic moment due to this rotation of 3rd second is
electron is (in Am2 ) (a) 33% (b) 40% (c) 66% (d) 77%
4 WB JEE (Engineering) · Solved Paper 2014
28. A proton of mass m and charge q is moving in a 35. If n denotes a positive integer, h the Planck’s
plane with kinetic energy E. If there exists a constant, q the charge and B the magnetic
uniform magnetic field B, perpendicular to the é nh ù
field, then the quantity ê ú has the
plane of the motion, the proton will move in a ë 2pqB û
circular path of radius dimension of
2 Em 2 Em Em 2 Eq (a) area (b) length
(a) (b) (c) (d)
qB qB 2qB mB (c) speed (d) acceleration
29. An artificial satellite moves in a circular orbit 36. In a transistor output characteristics
around the earth. Total energy of the satellite commonly used in common emitter
is given by E. The potential energy of the configuration, the base current I B ,the collector
satellite is current IC and the collector-emitter voltage
2E 2E VCE have values of the following orders of
(a) -2 E (b) 2 E (c) (d) -
3 3 magnitude in the active region
(a) IB and IC both are in mA and VCE in volt
30. In which of the following phenomena, the heat (b) IB is in mA and IC is in mA and VCE in volt
waves travel along straight lines with the (c) IB is in mA and IC is in mA and VCE in mV
speed of light? (d) IB is in mA and IC is in mA and VCE in mV
(a) Thermal conduction (b) Forced convection
(c) Natural convection (d) Thermal radiation 37. The displacement of a particle in a periodic
æ tö
motion is given by y = 4 cos 2 ç ÷ sin (1000t ).
31. In the bandgap between valence band and è 2ø
conduction band in a material is 5.0 eV, then This displacement may be considered as the
the material is result of superposition of n independent
(a) semiconductor (b) good conductor harmonic oscillations. Here n is
(c) superconductor (d) insulator
(a) 1 (b) 2 (c) 3 (d) 4
32. A uniform solid spherical ball is rolling down a 38. Consider a black body radiation in a cubical
smooth inclined plane from a height h. The
box at absolute temperature T. If the length of
velocity attained by the ball when it reaches
each side of the box is doubled and the
the bottom of the inclined plane is v. If the ball
temperature of the walls of the box and that of
is now thrown vertically upwards with the
the radiation is halved, then the total energy
same velocity v, the maximum height to which
(a) halves (b) doubles
the ball will rise is
(c) quadruples (d) remains the same
5h 3h 5h 7h
(a) (b) (c) (d)
8 5 7 9 39. Consider two concentric spherical metal shells
of radii r1 and r2 (r2 > r1 ). If the outer shell has a
33. Two coherent monochromatic beams of charge q and the inner one is grounded, the
intensities I and 4I respectively are charge on the inner shell is
superposed. The maximum and minimum -r2
intensities in the resulting pattern are (a) q (b) zero
r1
(a) 5 I and 3 I (b) 9 I and 3 I
-r
(c) 4 I and I (d) 9 I and I (c) 1 q (d) -q
r2
34. In the circuit shown assume the diode to be
ideal. When Vi increases from 2V to 6 V, the 40. Four cells, each of emf E and internal
change in the current is (in mA) resistance r, are connected in series across an
external resistance R. By mistake one of the
Vi 150 W +3V cells is connected in reverse. Then the current
in the external circuit is
2E 3E 3E 2E
(a) (b) (c) (d)
(a) zero (b) 20 4r + R 4r + R 3r + R 3r + R
(c) 80/3 (d) 40
WB JEE (Engineering) · Solved Paper 2014 5
41. Same quantity of ice is filled in each of the two 43. A car is moving with a speed of 72 km-h -1
metal containers P and Q having the same towards a roadside source that emits sound at
size, shape and wall thickness but made of a frequency of 850 Hz. The car driver listens to
different materials. The containers are kept in the sound while approaching the source and
identical surroundings. The ice in P melts again while moving away from the source after
completely in time t1 whereas in Q takes a time crossing it. If the velocity of sound is 340 ms -1,
t2. The ratio of thermal conductivities of the the difference of the two frequencies, the driver
materials of P and Q is hears is
(a) t 2 : t 1 (a) 50 Hz (b) 85 Hz (c) 100 Hz (d) 150 Hz
(b) t 1 : t 2
44. The energy of gamma (g ) ray photon is E g and
(c) t 12 : t 22
that of an X-ray photon is E X . If the visible
(d) t 22 : t 12 light photon has an energy of Ev , then we can
say that
42. For the radioactive nuclei that undergo either a or (a) EX > Eg > Ev (b) Eg > Ev > E x
b decay, which one of the following cannot occur? (c) Eg > EX > Ev (d) EX > Ev > Eg
(a) Isobar of original nucleus is produced
45. The intermediate image formed by the
(b) Isotope of the original nucleus is produced
objective of a compound microscope is
(c) Nuclei with higher atomic number than that of the
(a) real, inverted and magnified
original nucleus is produced
(b) real, erect and magnified
(d) Nuclei with lower atomic number than that of the (c) virtual, erect and magnified
original nucleus is produced (d) virtual, inverted and magnified
Category II
Directions (Q. Nos. 46-55) Carry two marks each, for which only one option is correct. Any wrong answer will
lead to deduction of 2/3 mark.
46. A solid uniform sphere resting on a rough consecutive layers, respectively. The integer
horizontal plane is given a horizontal impulse m = 0, 1, 2, 3, K denotes the numbers of the
directed through its centre so that it starts successive layers.
sliding with an initial velocity v0. When it A ray of light from the 0th layer enters the 1st
finally starts rolling without slipping the speed layer at an angle of incidence of 30°. After
of its centre is undergoing the mth refraction, the ray
2 3 5 6
(a) v0 (b) v0 (c) v 0 (d) v 0 emerges parallel to the interface. If m = 1.5 and
7 7 7 7 Dm = 0.015, the value of m is
47. The de-Broglie wavelength of an electron is the m –mD m
same as that of a 50 keV X-ray photon. The
ratio of the energy of the photon to the kinetic
energy of the electron is (the energy equivalent
of electron mass is 0.5 MeV)
(a) 1 : 50 (b) 1 : 20 m –2D m
(c) 20 : 1 (d) 50 : 1 m –D m
48. A glass slab consists of thin uniform layers of m
progressively decreasing refractive indices RI
°
30
Category III
Directions (Q. Nos. 56-60) Carry two marks each, for which one or more than one options may be correct.
Marking of correct options will lead to a maximum mark of two on pro rata basis. There will be no negative
marking for these questions. However, any marking of wrong option will lead to award of zero mark against the
respective questions - irrespective of the number of correct options marked.
56. A heating element of resistance r is fitted (b) Einstein analysis gives a threshold frequency
inside an adiabatic cylinder which carries a above which no electron can be emitted
frictionless piston of mass m and cross-section (c) The maximum kinetic energy of the emitted
A as shown in diagram. The cylinder contains photoelectrons is proportional to the frequency of
one mole of an ideal incident radiation
diatomic gas. The piston (d) The maximum kinetic energy of electrons does
current flows through not depend on the intensity of radiation
the element such that the 59. Half of the space between the plates of a
temperatures rises with parallel-plate capacitor is filled with a
1 r
time t as DT = at + bt 2 dielectric material of dielectric constant K. The
2 remaining half contains air as shown in the
(a and b are constants), while pressure remains figure. The capacitor is now given a charge Q.
constant. The atmospheric pressure above the Then
piston is P0. Then
(a) electric field in the dielectric-filled region is higher
(a) the rate of increase in internal energy is than that in the air-filled region
5
R(a + bt ) (b) on the two halves of the bottom plate the charge
2 densities are unequal
5
(b) the current flowing in the element is R(a + bt ) (c) charge on the half of the top plate above the
2r Q
air-filled part is
(c) the piston moves upwards with constant K+1
acceleration (d) capacitance of the capacitor shown above is
(d) the piston moves upwards with constant speed C
(1 + K ) 0 , where C 0 is the capacitance of the
57. A thin rod AB is held horizontally so that it can 2
same capacitor with the dielectric removed.
freely rotate in a vertical plane about the end A
as shown in the figure. The potential energy of 60. A stream of electrons and protons are directed
the rod when it hangs vertically is taken to be towards a narrow slit in a screen (see figure).
zero. The end B of the rod is released from rest The intervening
from a horizontal position. At the instant the region has a B G
rod makes an angle q with the horizontal uniform electric E
(a) the speed of end B is proportional to sin q field E
(b) the potential energy is proportional to (1 - cos q) (vertically downwards) and a uniform
(c) the angular acceleration is proportional to cos q magnetic field B (out of the plane of the figure)
(d) the torque about A remains the same as its initial as shown. Then
value | E|
(a) electrons and protons with speed will pass
| B|
58. Find the correct A
q
B
through the slit
statement(s) about
photoelectric effect. (b) protons with speed| E|/| B | will pass through the
slit, electrons of the same speed will not
(a) There is no significant
(c) neither electrons nor protons will go through the
time delay between the
slit irrespective of their speed
absorption of a suitable
(d) electrons will always be deflected upwards
radiation and the emission of electrons
irrespective of their speed
Chemistry
Category I
Directions (Q. Nos. 1-45) Carry one mark each, for which only one option is correct. Any wrong answer will
lead to deduction of 1/3 mark.
1. The emission spectrum of hydrogen discovered 7. The reagents to carry out the following
first and the region of the electromagnetic conversion are
spectrum in which it belongs, respectively are Me Me
(a) Lyman, ultraviolet (b) Lyman, visible Me
(c) Balmer, ultraviolet (d) Balmer, visible O
2. The electronic configuration of Cu is (a) HgSO 4 /dil ×H2SO 4
(a) [Ne] 3s 2 , 3 p6 , 3d 9 , 4s 2 (b) BH3 ; H2O 2 /NaOH
(b) [Ne] 3s 2 , 3 p6 , 3d 10 , 4s1 (c) OsO 4 ; HIO 4
(d) NaNH2 / CH3I; HgSO 4 / dil ×H2SO 4
(c) [Ne] 3s 2 , 3 p6 , 3d 3 , 4s 2 , 4 p6
(d) [Ne] 3s 2 , 3 p6 , 3d 5 , 4s 2 , 4 p4 8. The correct order of decreasing H¾C¾H
angle in the following molecule is
3. As per de-Broglie’s formula a macroscopic H H H H
particle of mass 100 g and moving at a velocity
of 100 cm s -1 will have a wavelength of
H H H H
(a) 6.6 ´ 10-29 cm (b) 6.6 ´ 10-30 cm
I II III
(c) 6.6 ´ 10-31 cm (d) 6.6 ´ 10-32 cm
(a) I > II > III
4. For one mole of an ideal gas, the slope of V vs. T (b) II > I > III
curve at constant pressure of 2 atm is X L (c) III > II > I
mol-1 K -1. The value of the ideal universal gas (d) I > III > II
constant ‘R ’ in terms of X is 9. During the emission of a positron from a
-1 -1 X nucleus, the mass number of the daughter
(a) X L atm mol K (b) L atm mol -1 K -1
2 element remains the same but the atomic
(c) 2 X L atm mol -1 K -1 (d) 2 X atm L-1mol -1 K -1 number
5. At a certain temperature the time required for (a) is decreased by 1 unit
(b) is decreased by 2 units
the complete diffusion of 200 mL of H2 gas is
(c) is increased by 1 unit
30 min. The time required for the complete (d) remains unchanged
diffusion of 50 mL of O2 gas at the same
temperature will be 10. b-emission is always accompanied by
(a) 60 min (b) 30 min (a) formation of antineutrino and a-particle
(c) 45 min (d) 15 min (b) emission of a-particle and g-ray
(c) formation of antineutrino and g-ray
6. The IUPAC name of the following molecule is (d) formation of antineutrino and positron
Me Me Me 11. Four gases P, Q, R and S have almost same
values of ‘b’ but their ‘a’ values (a, b are van der
Me Waals’ constants) are in the order
Q < R < S < P. At a particular temperature,
(a) 5,6-dimethylhept-2-ene among the four gases, the most easily
(b) 2,3-dimethylhept-5-ene liquefiable one is
(c) 5,6-dimethylhept-3-ene
(a) P (b) Q (c) R (d) S
(d) 5-iso-propylhex-2-ene
WB JEE (Engineering) · Solved Paper 2014 9
246
12. Among the following structures the one which 17. 98 Cf was formed along with a neutron when
is not a resonating structure of others is an unknown radioactive substance was
O O
–
O O
bombarded with 6C12. The unknown substance
was
Me – Me
O O 234 234 235 238
Me Me (a) 91Pa (b) 90 Th (c) 92 U (d) 92 U
I II
–
18. The rate of a certain reaction is given by,
O O O O rate = k [H+ ]n.
Me Me – The rate increases 100 times when the pH
O Me O
changes from 3 to 1. The order (n ) of the
III IV reaction is
(a) I (b) II (a) 2 (b) 0 (c) 1 (d) 1.5
(c) III (d) IV
19. The values of DH and DS of a certain reaction
13. The compound that will have a permanent are -400 kJ mol-1 and -20 kJ mol-1K -1
dipole moment among the following is respectively. The temperature below which the
H Cl Cl H reaction is spontaneous, is
(a) 100 K (b) 20°C (c) 20 K (d) 120°C
H Cl H Cl
20. The correct statement regarding the following
I II compounds is
Cl Cl OH OH OH
Br Br
Cl Cl
III IV
OH OH OH
(a) I (b) II I II III
(c) III (d) IV (a) all three compounds are chiral
14. The correct order of decreasing length of the (b) only I and II are chiral
(c) I and III are diastereomers
bond as indicated by the arrow in the following
(d) only I and III are chiral
structures is
21. The correct statement regarding the following
energy diagrams is
+ + +
+
I II III
(a) I > II > III (b) II > I > III
(c) III > II > I (d) I > III > II
Reactant
E E
15. An atomic nucleus having low n/p ratio tries to Reactant
Product Product
find stability by
(a) the emission of an a-particle Reaction M Reaction N
(b) the emission of a positron
(c) capturing an orbital electron (K-electron capture) (a) Reaction M is faster and less exothermic than
(d) emission of a b-particle reaction N
(b) Reaction M is slower and less exothermic than
16. ( 32Ge76 , 34Se76 ) and (14Si30 , 16S32) are examples of reaction N
(a) isotopes and isobars (c) Reaction M is faster and more exothermic than
(b) isobars and isotones reaction N
(c) isotones and isotopes (d) Reaction M is slower and more exothermic than
(d) isobars and isotopes reaction N
10 WB JEE (Engineering) · Solved Paper 2014
22. In the following reaction, the product E is 28. Among the following compounds, the one(s)
CHO that gives (give) effervescence with aqueous
(i) NaOH
E NaHCO3 solution is (are)
(ii) H+
CHO (CH3CO)2O CH3COOH PhOH CH3COCHO
CH2OH CHO I II III IV
(a) (b) (a) I and II (b) I and III
CHO CO2H (c) Only II (d) I and IV
CH2OH CO2H 29. The 4th higher homologue of ethane is
(c) (d) (a) butane (b) pentane (c) hexane (d) heptane
CO2H CO2H
30. In case of heteronuclear diatomics of the type
23. If Cl2 is passed through hot aqueous NaOH, AB, where A is more electronegative than B,
the products formed have Cl in different bonding molecular orbital resembles the
oxidation states. These are indicated as character of A more than that of B. The
(a) -1and +1 (b) -1and +5 statement
(c) +1 and +5 (d) -1and +3 (a) is false
(b) is true
24. Commercial sample of H2O2 is labeled as 10 V. (c) cannot be evaluated since data is not sufficient
Its % strength is nearly (d) is true only for certain systems
(a) 3 (b) 6 (c) 9 (d) 12
31. The hydrides of the first elements in groups
25. The enthalpy of vaporisation of a certain liquid 15-17, namely NH3 , H2O and HF respectively
at its boiling point of 35°C is 24.64 kJ mol-1. show abnormally high values for melting and
The value of change in entropy for the process boiling points. This is due to
is (a) small size of N, O and F
(a) 704 J K -1mol -1 (b) 80 J K -1mol -1 (b) the ability to form extensive intermolecular
(c) 24.64 J K -1mol -1 (d) 7.04 J K -1mol -1 H-bonding
(c) the ability to form extensive intramolecular
26. Given that H-bonding
(d) effective van der Waals’ interaction
C + O2 ¾® CO2; DH ° = - x kJ
32. The quantity of electricity needed to separately
2CO + O2 ¾® 2CO2; DH ° = - y kJ
electrolyze 1 M solution of ZnSO4 , AlCl3 and
The heat of formation of carbon monoxide will AgNO3 completely is in the ratio of
be (a) 2 : 3 : 1 (b) 2 : 1 : 1
y - 2x 2x - y (c) 2 : 1 : 3 (d) 2 : 2 : 1
(a) (b) y + 2 x (c) 2 x - y (d)
2 2
33. The amount of electrolytes required to
27. The intermediate J in the following Wittig coagulate a given amount of AgI colloidal
reaction is solution (-ve charge) will be in the order
+
(a) NaNO 3 > Al 2 (NO 3 )3 >Ba(NO 3 )2
(i) n-BuLi
PPh3Br– [J ] (b) Al 2 (NO 3 )3 >Ba(NO 3 )2 >NaNO 3
(ii) CH2==O (c) Al 2 (NO 3 )3 >NaNO 3 >Ba(NO 3 )2
(d) NaNO 3 >Ba(NO 3 )2 > Al 2 (NO 3 )3
PPh3 PPh3
(a) O (b) 34. The value of DH for cooling 2 mole of an ideal
O monoatomic gas from 225°C to 125°C at
5
O O constant pressure will be [given C p = R]
+ 2
(c) PPh3 (b) PPh3
(a) 250 R (b) -500 R
(c) 500 R (d) -250 R
WB JEE (Engineering) · Solved Paper 2014 11
35. An amine C3 H9N reacts with benzene 41. Metal ion responsible for the Minimata disease
sulphonyl chloride to form a white precipitate is
which is insoluble in aq. NaOH. The amine is (a) Co 2+ (b) Hg 2+
Me Me Me (c) Cu2+ (d) Zn2+
(a) N (b) N Me 42. The reagent with which the following reaction
H
is best accomplished is
Me
Me Me
Me
(c) Me NH2 (b)
Me NH2
39. The system that contains the maximum 44. The reaction of aniline with chloroform under
number of atoms is alkaline conditions leads to the formation of
(a) 4.25 g of NH3 (b) 8 g of O 2 (a) phenylcyanide
(c) 2 g of H2 (d) 4 g of He (b) phenylisonitrile
(c) phenylcyanate
40. Among the following observations, the correct (d) phenylisocyanate
one that differentiates between SO32- and SO2-
4
is
45. The two half-cell reactions of an
electrochemical cell is given as
(a) both form precipitate with BaCl 2 ,SO 2-
3 dissolves in
HCl but SO 2- Ag + + e- ¾® Ag ; Eo = - 0.3995 V
4 does not Ag + /Ag
2+ 3+ -
(b) SO 2-
3 forms precipitate with BaCl 2 ,SO 2-
4 does not
Fe ¾® Fe + e ; E°Fe 3+/ Fe2+ = - 0.7120 V
(c) SO 2-
4
2-
forms precipitate with BaCl 2 ,SO 3 does not The value of cell EMF will be
(d) both form precipitate with BaCl 2 ,SO 2-
4 dissolves in (a) -0.3125 V (b) 0.3125 V
HCl but SO 2-
3 does not (c) 1.114 V (d) -1.114 V
Category II
Directions (Q. Nos. 46-55) Carry two marks each, for which only one option is correct. Any wrong answer will
lead to deduction of 2/3 mark.
46. The compressibility factor (Z ) of one mole of a van der Waals’ gas of negligible ‘a’ value is
bp bp bp
(a) 1 (b) (c) 1 + (d) 1 -
RT RT RT
12 WB JEE (Engineering) · Solved Paper 2014
47. When phenol is treated with D2SO4 / D2O, some 50. At 25°C, the molar conductance of 0.007 M
of the hydrogens get exchanged. The final hydrofluoric acid is 150 mho cm2mol-1 and its
product in this exchange reaction is L°m = 500 mho cm2mol-1. The value of the
dissociation constant of the acid at the given
OD OD concentration at 25°C is
D D D H (a) 7 ´ 10-4 M (b) 7 ´ 10-5 M
(a) (b) (c) 9 ´ 10-3 M (d) 9 ´ 10-4 M
H H H H
51. To observe an elevation of boiling point of
D D
0.05°C, the amount of a solute (mol. wt. = 100)
OD OD to be added to 100 g of water (K b = 0.5) is
(a) 2 g (b) 0.5 g (c) 1 g (d) 0.75 g
D H H H
(c) (d) 52. The volume of ethyl alcohol (density 1.15 g/cc)
H D D D that has to be added to prepare 100 cc of 0.5 M
H H ethyl alcohol solution in water is
(a) 1.15 cc (b) 2 cc (c) 2.15 cc (d) 2.30 cc
48. The most likely protonation site in the
following molecule is 53. The bond angle in NF3 (102.3°) is smaller than
3 4 NH3 (107.2°). This is because of
Me 1 5 (a) large size of F compared to H
2 (b) large size of N compared to F
Me 6
8 7 (c) opposite polarity of N in the two molecules
(d) small size of H compared to N
(a) C-1 (b) C-2
(c) C-3 (d) C-6 54. A piece of wood from an archaeological sample
has 5.0 counts min -1 per gram of C-14, while a
49. The order of decreasing ease of abstraction of
fresh sample of wood has a count of
hydrogen atoms in the following molecule
15.0 min -1 g -1. If half-life of C-14 is 5770 yr,
Ha the age of the archaeological sample is
Me
(a) 8,500 yr (b) 9,200 yr
Hb
(c) 10,000 yr (d) 11,000 yr
55. The structure of XeF6 is experimentally
Hc determined to be distorted octahedron. Its
structure according to VSEPR theory is
(a) Ha > Hb > Hc
(a) octahedron
(b) Ha > Hc > Hb
(b) trigonal bipyramid
(c) Hb > Ha > Hc
(c) pentagonal bipyramid
(d) Hc > Hb > Ha
(d) tetragonal bipyramid
WB JEE (Engineering) · Solved Paper 2014 13
Category III
Directions (Q. Nos. 56-60) Carry two marks each, for which one or more than one options may be correct.
Marking of correct options will lead to a maximum mark of two on pro rate basis. There will be no negative
marking for these questions. However, any marking of wrong option will lead to award of zero mark against the
respective question - irrespective of the number of correct options marked.
56. Two gases X (molecular weight M X ) and Y 59. Cupric compounds are more stable than their
(molecular weight MY ; MY > M X ) are at the cuprous counterparts in solid state. This is
same temperature T in two different because
containers. Their root mean square velocities (a) the endothermic character of the 2nd IP of Cu is
are C X and CY respectively. If the average not so high
kinetic energies per molecule of two gases X (b) size of Cu2+ is less than Cu+
and Y are E X and EY respectively then which of (c) Cu2+ has stabler electronic configuration as
the following relation(s) is(are) true? compared to Cu+
(a) EX > EY (d) the lattice energy released for cupric compounds
(b) C X > CY is much higher than Cu+
(c) EX = EY = (3 / 2 ) RT
(d) EX = EY = (3/2 ) kBT
60. Among the following statements about the
molecules X and Y, the one(s) which correct is
57. For a spontaneous process, the correct (are)
statement(s) is (are)
CHO CHO
(a) (DGsystem ) T , p > 0
(b) (DSsystem ) + (DSsurroundings ) > 0 H OH HO H
(c) (DGsystem )T , p < 0 HO H H OH
(d) (DUsystem ) T , V > 0
H OH HO H
58. The formal potential of Fe3 + / Fe2+ in a
H OH HO H
sulphuric acid and phosphoric acid mixture
(E° = + 0.61 V ) is much lower than the
CH2OH CH2OH
standard potential (E° = + 0.77 V ). This is due
to X Y
+
(a) formation of the species [FeHPO 4 ] (a) X and Y are diastereomers
(b) lowering of potential upon complexation (b) X and Y are enantiomers
(c) formation of the species [FeSO 4 ]+ (c) X and Y are both aldohexoses
(d) high acidity of the medium (d) X is a D-sugar and Y is an L-sugar
Mathematics
Category I
Directions (Q. Nos. 1-60) Carry one mark each, for which only one option is correct. Any wrong answer will
lead to deduction of 1/3 mark.
12. Let f (x) be a differentiable function in [2, 7]. If 20. Let f (x) = x + 1 / 2. Then, the number of real
f (2) = 3 and f ¢(x) £ 5 for all x in (2, 7), then the values of x for which the three unequal terms
maximum possible value of f (x) at x = 7 is f (x), f (2x), f (4x) are in HP is
(a) 7 (b) 15 (a) 1 (b) 0
(c) 28 (d) 14 (c) 3 (d) 2
WB JEE (Engineering) · Solved Paper 2014 15
24. The equation of the common tangent with 31. The solution of the differential equation
positive slope to the parabola y2 = 8 3 x and é æ y2 ö ù
the hyperbola 4x2 - y2 = 4 is ê 2 f ç 2÷ ú
dy y èx øú
(a) y = 6 x + 2 (b) y = 6 x- 2 y =xê 2 + is (where, c is a
dx êx æ y2 ö ú
(c) y = 3 x + 2 (d) y = 3 x- 2 ê f¢ ç 2 ÷ ú
êë è x ø úû
25. The point on the parabola y2 = 64x which is
constant)
nearest to the line 4x + 3 y + 35 = 0 has æ y2 ö æ y2 ö
coordinates (a) f çç 2 ÷÷ = cx (b) xf çç 2 ÷÷ = c
èx ø èx ø
(a) (9, - 24) (b) (1, 81)
(c) (4, - 16) (d) (-9, - 24) æ y2 ö æ y2 ö
(c) f çç 2 ÷÷ = cx2 (d) x2 f çç 2 ÷÷ = c
èx ø èx ø
26. Let z1 , z2 be two fixed complex numbers in the
argand plane and z be an arbitrary point
32. Suppose that the equation f (x) = x2 + bx + c = 0
satisfying| z - z1 | + | z - z2| = 2| z1 - z2|. Then, has two distinct real roots a and b. The angle
the locus of z will be between the tangent to the curve y = f (x) at the
(a) an ellipse æa + b æ a + böö
point ç ,f ç ÷ and the positive
(b) a straight line joining z1 and z2 è 2 è 2 ø ÷ø
(c) a parabola direction of the x-axis is
(d) a bisector of the line segment joining z1 and z2 (a) 0° (b) 30° (c) 60° (d) 90°
16 WB JEE (Engineering) · Solved Paper 2014
33. The function f (x) = x2 + bx + c, where b and c 40. Let R be the set of all real numbers and
real constants, describes f : [-1, 1] ® R be defined by
ìï 1
(a) one-to-one mapping x sin , x ¹ 0
f (x) = í x . Then,
(b) onto mapping
(c) not one-to-one but onto mapping îï0, x=0
(d) neither one-to-one nor onto mapping (a) f satisfies the conditions of Rolle’s theorem on
[-1, 1]
34. Let n ³ 2 be an integer, (b) f satisfies the conditions of Lagrange’s mean
é cos (2p / n ) sin (2p / n ) 0ù value theorem on [-1, 1]
A = ê - sin (2p / n ) cos (2p / n ) 0ú and I is the (c) f satisfies the conditions of Rolle’s theorem on [0, 1]
ê 0 0 1úû (d) f satisfies the conditions of Lagrange’s mean
ë
identity matrix of order 3. Then, value theorem on [0, 1]
(a) A n = I and A n - 1 ¹ I 41. If a , b and c are positive numbers in a GP, then
(b) A m ¹ I for any positive integer m the roots of the quadratic equation
(c) A is not invertible (log e a ) x2 - (2 log e b) x + (log e c) = 0 are
(d) A m = O for a positive integer m loge c
(a) -1and
35. Ram is visiting a friend. Ram knows that his loge a
friend has 2 children and 1 of them is a boy. loge c
(b) 1 and -
Assuming that a child is equally likely to be a loge a
boy or a girl, then the probability that the other (c) 1 and log a c
child is a girl, is (d) -1and logc a
(a) 1/2 (b) 1/3 (c) 2/3 (d) 7/10
42. There is a group of 265 persons who like either
36. The value of the sum singing or dancing or painting. In this group
(nC1 )2 + (nC 2)2 + (nC3 )2 + K + (nC n )2 is 200 like singing, 110 like dancing and 55 like
(a) ( 2 n C n )2 (b) 2n
Cn painting. If 60 persons like both singing and
dancing, 30 like both singing and painting and
(c) 2 n C n + 1 (d) 2n
Cn - 1
10 like all three activities, then the number of
37. The remainder obtained when persons who like only dancing and painting is
1 ! + 2 ! + 3 ! + K + 11 ! is divided by 12 is (a) 10 (b) 20 (c) 30 (d) 40
(a) 9 (b) 8 (c) 7 (d) 6 æ p2 ö
43. The range of the function y = 3 sin çç - x2 ÷ is
38. Out of 7 consonants and 4 vowels, the number ÷
è 16 ø
of words (not necessarily meaningful) that can
be made, each consisting of 3 consonants and (a) [0, 3 / 2 ] (b) [0, 1]
2 vowels, is (c) [0, 3 / 2 ] (d) [0, ¥)
(a) 24800 (b) 25100 x2
ò0 cos (t 2) dt
(c) 25200 (d) 25400 44. The value of lim is
x®0 x sin x
2 22 n 23 n
39. Let S = nC 0 + C1 + C2 (a) 1 (b) -1
1 2 3
(c) 2 (d) loge 2
n+1
2 n
+K+ C n. Then, S equals 45. Let f (x) be a differentiable function and
n+1
f (4) - f (x2)
2n + 1 - 1 3n + 1 - 1 f ¢(4) = 5. Then, lim equals
(a) (b) x®2 x-2
n+1 n+1
n n (a) 0 (b) 5
3 -1 2 -1
(c) (d) (c) 20 (d) -20
n n
WB JEE (Engineering) · Solved Paper 2014 17
¥
æ n! p ö 52. Suppose that f (x) is a differentiable function
46. The sum of the series å sin ç
è 720 ø
÷ is
such that f ¢(x) is continuous, f ¢(0) = 1 and f ¢¢(0)
n =1
p ö does not exist. Let g (x) = xf ¢(x). Then,
(a) sin æç æ p ö + sin æ p ö
÷ + sin ç ÷ ç ÷ (a) g ¢(0) does not exist (b) g ¢(0) = 0
è 180 ø è 360 ø è 540 ø
p p p ö æ p ö (c) g ¢(0) = 1 (d) g ¢(0) = 2
(b) sin æç ö÷ + sin æç ö÷ + sin æç ÷ + sin ç ÷
è 6ø è 30 ø è 120 ø è 360 ø 53. Let z1 be a fixed point on the circle of radius 1
p p p ö
(c) sin æç ö÷ + sin æç ö÷ + sin æç ÷ centred at the origin in the argand plane and
è 6ø è 30 ø è 120 ø
z1 ¹ ± 1. Consider an equilateral triangle
p ö æ p ö
+ sin æç ÷ + sin ç ÷ inscribed in the circle with z1 , z2, z3 as the
è 360 ø è 720 ø vertices taken in the counter clockwise
p ö æ p ö
(d) sin æç ÷ + sin ç ÷ direction. Then, z1z2z3 is equal to
è 180 ø è 360 ø
(a) z12 (b) z13
47. Let I denote the 3 ´ 3 identity matrix and P be a (c) z14 (d) z1
matrix obtained by rearranging the columns of
I. Then, 54. Suppose that z1 , z2, z3 are three vertices of an
(a) there are six distinct choices for P and det (P) = 1 equilateral triangle in the argand plane. Let
1
(b) there are six distinct choices for P and det(P) = ± 1 a = ( 3 + i ) and b be a non-zero complex
(c) there are more than one choices for P and some 2
of them are not invertible number. The points az1 + b , az2 + b , az3 + b will
(d) there are more than one choices for P and P -1 = I be
in each choice (a) the vertices of an equilateral triangle
(b) the vertices of an isosceles triangle
48. The coefficient of x3 in the infinite series (c) collinear
2 (d) the vertices of a scalene triangle
expansion of , for| x| < 1, is
(1 - x) (2 - x)
55. The curve y = (cos x + y)1/ 2 satisfies the
1 15 1 15
(a) - (b) (c) - (d) differential equation
16 8 8 16 2
d2y dy
49. For every real number x, (a) (2 y - 1) 2
+ 2 æç ö÷ + cos x = 0
dx è dx ø
2
x 3 2 7 3 15 4 d2y dy
let f (x) = + x + x + x + K . Then, (b) 2 - 2 y æç ö÷ + cos x = 0
1! 2! 3! 4! dx è dx ø
the equation f (x) = 0 has 2 2
d y dy
(c) (2 y - 1) 2 - 2 æç ö÷ + cos x = 0
(a) no real solution dx è dx ø
(b) exactly one real solution 2 2
d y dy
(c) exactly two real solutions (d) (2 y - 1) 2 - æç ö÷ + cos x = 0
dx è dx ø
(d) infinite number of real solutions
50. Let S denote the sum of the infinite series 56. In the argand plane, the distinct roots of
8 21 40 65 1 + z + z3 + z 4 = 0 (z is a complex number)
1+ + + + + K . Then, represent vertices of
2! 3! 4! 5!
(a) S < 8 (b) S > 12 (a) a square (b) an equilateral triangle
(c) 8 < S < 12 (d) S = 8 (c) a rhombus (d) a rectangle
51. Let [x] denote the greatest integer less than or 57. In a DABC , a , b, c are the sides of the triangle
equal to x for any real number x. Then, opposite to the angles A, B, C, respectively.
[n 2 ] Then, the value of a3 sin (B - C )
lim is equal to 3 3
+ b sin (C - A ) + c sin ( A - B) is equal to
n® ¥ n
(a) 0 (b) 1
(a) 0 (b) 2
(c) 3 (d) 2
(c) 2 (d) 1
18 WB JEE (Engineering) · Solved Paper 2014
Category II
Directions (Q. Nos. 61-75) Carry two marks each, for which only one option is correct. Any wrong answer will
lead to deduction of 2/3 mark.
61. The solution of the differential equation 65. The equation of hyperbola whose coordinates
dy y 1 of the foci are (±8, 0) and the length of
+ =
dx x log e x x latusrectum is 24 units, is
(a) 3 x2 - y2 = 48 (b) 4 x2 - y2 = 48
under the condition y = 1 when x = e is 2 2
1 (c) x - 3 y = 48 (d) x2 - 4 y2 = 48
(a) 2 y = loge x +
loge x 66. A student answers a multiple choice question
2 with 5 alternatives, of which exactly one is
(b) y = loge x +
loge x correct. The probability that he knows the
(c) y loge x = loge x + 1
correct answer is p, 0 < p < 1. If he does not
(d) y = loge x + e
know the correct answer, he randomly ticks
one answer. Given that he has answered the
62. Let f (x) = max { x + | x|, x - [x]}, where [x] question correctly, the probability that he did
denotes the greatest integer £ x. Then, the not tick the answer randomly, is
3 3p 5p
value of ò f (x) dx is (a) (b)
-3 4p + 3 3p + 2
(a) 0 (b) 51/2 5p 4p
(c) (d)
(c) 21/2 (d) 1 4p + 1 3p + 1
1 2p 4p 6p
63. Let X n = ìí z = x + iy :| z |2 £ üý for all integers 67. cos + cos + cos
î nþ 7 7 7
¥ (a) is equal to zero (b) lies between 0 and 3
n ³ 1. Then, Ç X n is (c) is a negative number (d) lies between 3 and 6
n =1
69. For any two real numbers q and f, we define 73. Let tn denotes the nth term of the infinite
qRf, if and only if sec2 q - tan 2 f = 1. The series
1 10 21 34 49
+ + + + +K . Then,
relation R is 1! 2! 3! 4! 5!
(a) reflexive but not transitive lim tn is
(b) symmetric but not reflexive n® ¥
(c) both reflexive and symmetric but not transitive (a) e (b) 0 (c) e 2 (d) 1
(d) an equivalence relation 74. A particle starting from a point A and moving
70. The minimum value of 2sin x + 2cos x is with a positive constant acceleration along a
straight line reaches another point B in time T.
1 - 1/ 2 1 + 1/ 2
(a) 2 (b) 2 Suppose that the initial velocity of the particle
2
(c) 2 (d) 2 is u > 0 and P is the mid-point of the line AB. If
the velocity of the particle at point P is v1 and if
71. We define a binary relation ~ on the set of all T
3 ´ 3 real matrices as A ~ B, if and only if there the velocity at time is v2, then
2
exist invertible matrices P and Q such that 1
B = PAQ -1. The binary relation ~ is (a) v1 = v 2 (b) v1 > v 2 (c) v1 < v 2 (d) v1 = v2
2
(a) neither reflexive nor symmetric
(b) reflexive and symmetric but not transitive 75. A poker hand consists of 5 cards drawn at
(c) symmetric and transitive but not reflexive random from a well-shuffled pack of 52 cards.
(d) an equivalence relation Then, the probability that a poker hand
consists of a pair and a triple of equal face
72. Let a, b denote the cube roots of unity other
n
values (for example, 2 sevens and 3 kings or
302
æ aö 2 aces and 3 queens, etc.) is
than 1 and a ¹ b. Let S = å (-1)n ç ÷ . Then,
èbø 6 23
n=0 (a) (b)
4165 4165
the value of S is
1797 1
(a) either -2w or -2 w2 (b) either -2w or 2 w2 (c) (d)
4165 4165
(c) either 2w or -2 w2 (d) either 2w or 2 w2
Category III
Directions (Q. Nos. 76-80) Carry two marks each, for which one or more than one options may be correct.
Marking of correct options will lead to a maximum mark of two on pro rata basis. There will be no negative
marking for these questions. However, any marking of wrong option will lead to award of zero mark against the
respective question-irrespective of the number of correct options marked.
76. If u (x) and v(x) are two independent solutions of 77. The angle of intersection between the curves
the differential equation y = [|sin x| + |cos x|] and x2 + y2 = 10, where
d 2y dy [x] denotes the greatest integer £ x, is
+b + cy = 0, (a) tan-1 3 (b) tan-1(-3)
dx2 dx
(c) tan-1 3 (d) tan-1(1 / 3 )
then additional solution(s) of the given
differential equation is(are) ì x |1 - t | dt , x > 0
(a) y = 5u( x) + 8v( x) ïò
78. Let f (x) = í 0 1 . Then,
(b) y = c1{u( x) - v( x)} + c 2 v( x),c1 and c 2 are arbitrary ïx - , x£1
constants î 2
(c) y = c1u( x) v( x) + c 2u( x) / v( x), c1 and c 2 are (a) f( x) is continuous at x = 1
arbitrary constants (b) f( x) is not continuous at x = 1
(d) y = u( x) v( x) (c) f( x) is differentiable at x = 1
(d) f( x) is not differentiable at x = 1
20 WB JEE (Engineering) · Solved Paper 2014
79. If the circle x2 + y2 + 2 gx + 2 fy + c = 0 cuts the 80. For two events A and B, let P ( A ) = 0.7 and
2 2
three circles x + y - 5 = 0, P (B) = 0.6. The necessarily false statement(s)
x2 + y2 - 8x - 6 y + 10 = 0 and is/are
x2 + y2 - 4x + 2 y - 2 = 0 at the extremities of (a) P( A Ç B) = 0.35
their diameters, then (b) P( A Ç B) = 0.45
(a) c = - 5 (b) fg = 147 / 25 (c) P( A Ç B) = 0.65
(c) g + 2 f = c + 2 (d) 4f = 3g (d) P( A Ç B) = 0.28
Answers
Physics
1. (a) 2. (d) 3. (a) 4. (b) 5. (b) 6. (d) 7. (c) 8. (d) 9. (b) 10. (c)
11. (a) 12. (c) 13. (d) 14. (a) 15. (c) 16. (b) 17. (a) 18. (c) 19. (d) 20. (c)
21. (c) 22. (d) 23. (c) 24. (d) 25. (a) 26. (a) 27. (b) 28. (b) 29. (a) 30. (d)
31. (d) 32. (c) 33. (d) 34. (b) 35. (a) 36. (b) 37. (c) 38. (d) 39. (c) 40. (a)
41. (a) 42. (b) 43. (c) 44. (c) 45. (a) 46. (c) 47. (c) 48. (d) 49. (d) 50. (b)
51. (a) 52. (a) 53. (b) 54. (d) 55. (b) 56. (a) 57. (a,c) 58. (a,d) 59. (b,c,d) 60. (c,d)
Chemistry
1. (d) 2. (b) 3. (c) 4. (c) 5. (b) 6. (a) 7. (d) 8. (b) 9. (a) 10. (c)
11. (a) 12. (d) 13. (a) 14. (c) 15. (b) 16. (b) 17. (c) 18. (c) 19. (c) 20. (d)
21. (c) 22. (c) 23. (b) 24. (a) 25. (b) 26. (a) 27. (a) 28. (a) 29. (c) 30. (b)
31. (b) 32. (a) 33. (d) 34. (b) 35. (b) 36. (d) 37. (d) 38. (d) 39. (c) 40. (a)
41. (b) 42. (a) 43. (a) 44. (b) 45. (b) 46. (c) 47. (a) 48. (a) 49. (b) 50. (d)
51. (c) 52. (b) 53. (c) 54. (b) 55. (c) 56. (b,d) 57. (b,c) 58. (a,b,d) 59. (a,b,d) 60. (b,c,d)
Mathematics
1. (b) 2. (c) 3. (a) 4. (b) 5. (c) 6. (c) 7. (d) 8. (c) 9. (a) 10. (b)
11. (a) 12. (c) 13. (b) 14. (a) 15. (b) 16. (c) 17. (a) 18. (a) 19. (b) 20. (a)
21. (c) 22. (a) 23. (d) 24. (a) 25. (a) 26. (a) 27. (a) 28. (c) 29. (a) 30. (a)
31. (c) 32. (a) 33. (d) 34. (a) 35. (c) 36. (d) 37. (a) 38. (c) 39. (b) 40. (d)
41. (c) 42. (a) 43. (c) 44. (a) 45. (d) 46. (c) 47. (b) 48. (b) 49. (b) 50. (c)
51. (c) 52. (c) 53. (b) 54. (a) 55. (a) 56. (b) 57. (a) 58. (a) 59. (c) 60. (d)
61. (a) 62. (c) 63. (a) 64. (c) 65. (a) 66. (c) 67. (c) 68. (d) 69. (d) 70. (a)
71. (d) 72. (a) 73. (b) 74. (b) 75. (a) 76. (a,b) 77. (a) 78. (a,d) 79. (d) 80. (c,d)
Hints & Solutions
Physics
1. Vector X of the form aA + bB 5. A A+A.B
A NOT OR Y
X = aA + bB
= a ( $i + $j - 2 k$ ) + b ( $i - $j + k$ ) A
AND
X = i$ ( a + b ) + $j ( a - b ) + k$ ( -2a + b ) B A .B
By absorption laws
A vector X is perpendicular to C, i.e., X ×C = 0
Y = A + A ×B
[ $i ( a + b ) + $j ( a - b )
or Y = ( A ×B ) + A = A ×( B + 1) = A ×1 = A
+ k$ ( -2a + b )]×[2$i - 3$j + 4k$ ) = 0
or 2 ( a + b ) - 3( a - b ) + 4( -2a + b ) = 0 6. We know that DU = nCv DT
or 2a + 2b - 3a + 3b - 8a + 4b = 0 where DT = (T2 - T1)
or -9a + 9b = 0 or a = b T1 = 0° C = 273 K
a 1 T2 = 100° C = 373 K
or = or a : b = 1 : 1
b 1 n = 1 (monoatomic gas)
3
2. The combination of three charges in series R = 8.32 J mol -1K -1 = 1 ´ R ´ ( 373 - 273)
2
1 1 1 1
= + + 3
C C1 C2 C3 = 1 ´ ´ 8.32 ´ 100 = 3 ´ 8.32 ´ 50
2
1 1 1 1 6
= + + Þ C = = 1.5 mF Þ 1248 kJ
C 3 6 6 4
or = 1.25 ´ 103 J
The charge of this circuit
q = CV = 1.5´120 7. Energy required to removed electron in the
q = 180 mC ( n = 2) stage
The potential difference across the 3 mF The energy of the photon
q = CV æ1 1ö æ 1 1ö
= R ç 2 - 2 ÷ Þ 13.6 eV ç - ÷
q 180 è1 n ø è1 4ø
V = = = 60 V
C 3 æ 4 - 1ö 3 40.8
= 13.6 eV ç ÷ = 13.6 eV ´ = eV
V 120 è 4 ø 4 4
3. Resistance R = -G = - 10
ig 0.01 = 10.2 eV
The different of the two frequency, From Eqs. (ii) and (i), we get
N Approach - N Separation E photon hc / l hc × l2 ´ 2m
= 2 =
æ 360 ö æ 320 ö K electron h / 2m × l2 h2 × l
= 850 ç ÷ - 850 ç ÷
è 340 ø è 340 ø
where m = 0.5 MeV = 5 ´ 105 eV
850 850
= ´ 40 = = 100 Hz h 2mlc
340 8.5 = 50 ´ 103 eV =
lc h
44. E g ³ 100 keV 2m 2 ´ 5 ´ 105 æ hö
= = çQ m = ÷
E X = 100 eV to 100 keV h / lc 50 ´ 103 è cl ø
E v = 2.48 eV = 20 : 1
So, we can say that 48. By Snell’s law,
E g > E X > Ev m sin i = constant
45. The intermediate image formed by the objective m sin i = ( m - m Dm ) sin r
of a compound microscope is real, inverted and where m = 1.5, i = 30°, r = 90°, Dm = 0.015
magnified.
1.5 sin 30° = ( 1.5 - m ´ 0.15) sin 90°
46. Let, the final velocity be v 3 1
´ = ( 1.5 - m ´ 0.015) ´ 1
2 2
v0 3 3 15m
= -
4 2 1000
So, angular momentum will remain conserved æ3 3ö
15 m = ç - ÷ ´ 1000
along point of contact è2 4ø
By conservation of angular momentum 6 -3 3 1000
15m = ´ 1000 Þ m = ´
Angular momentum will remain conserved along 4 4 15
point of contact 3000
m= Þ m = 50
Iw = constant 60
2 æ vö
mv 0 r = mvr + mr 2 ´ w çQ w = ÷ 49. We know that
5 è rø
2 æv ö
mv 0 r = mvr + mr 2 ç ÷
5 èrø P
2 90°
v0 = v + v
5 30°
d3
60° 2
7 5
v0 = v Þ v = v0 60°
5 7
h ém i ù
47. de-Broglie wavelength l = Bnet = 2 ê 0 (sin q1 + sin q2 ) ú
2mK ë 4 pr û
The kinetic energy of the electron é ù
êm ú
1 h2 =2 ê 0 ´
i
´ (sin 90° + sin 30° ) ú
K electron = × ...(i)
2m l2 ê 4p d 3 ú
where h = Planck constant êë 2 úû
l = wavelength ém 2i æ 1ö ù
=2 ê 0 ´ ´ ç1 + ÷ ú
The photon energy ë 4 p d 3 è 2 øû
hc ém 2i 3ù 3 m 0i
E photon = ...(ii) =2 ê 0 ´ ´ ú=
l ë 4p d 3 2 û 2 pd
28 WB JEE (Engineering) · Solved Paper 2014
where f1 = 10 cm, u = - 30 cm dQ dT 7
\ = nC p ´ Þ i 2 r = R ´ [ a + bt ]
v =? dt dt 2
Screen Screen 7R
Object i = ( a + bt )
image image 2r
pV = nRT
nRT nR æ 1 2ö
or V = = ç at + bt ÷
30 cm 15 cm 45 cm p p è 2 ø
nR æ 1 2ö
X = ç at + bt ÷
pA è 2 ø
For the second condition when concave lens is
placed nR
v = ( a + bt )
v ¢ = ( 15 + 45) cm = 60 cm pA
1 1 1 nR
= - and acceleration = ´b
F v¢ u pA
(where F = focal length of combination) So, the rate of increase in internal energy is
1 1 1 5
\ = + R ( a + bt ) and the piston moves upwards with
F 60 30 2
60 constant acceleration.
F = cm = 20 cm
3 57. Loss in potential energy = gain in kinetic energy
The magnitude of focal length of concave lens
q
1 1 1 1 1 1
= + Þ = + Þ f2 = - 20 cm
F f1 f2 20 10 f2
L sin q
(Negative sign for concave lens) 2
55. We know that
1 2 q
K = Iw
2 L cos q
where K = kinetic energy 2
CM
I = moment of inertia mg
w = angular speed
1 L 1
So, w∝ ⇒ mg sin q = Iw2
I 2 2
1 1 1 So, w ∝ sin q
w1 : w2 : w3 = : :
I1 I2 I3 æ 1 2ö
and v ∝ sin q çQ K = Iw ÷
1 1 1 è 2 ø
= : : = 2 : 2 :1
1 1 2 The speed of end B is proportional to sin q
56. We know that and
nfRT 5R æ 1 2ö L
Internal energy, U = = ç at + bt ÷ U = mgh = mg ( 1 - sin q) (Q t = Ia )
2 2 è 2 ø 2
Differentiate with respect to t L mI 2
⇒ mg cos q = ´a
dU 5R 2 3
= [ a + bt ]
dt 2 a ∝ cos q
But, dQ = nC pdT The angular acceleration is proportional to cos q.
30 WB JEE (Engineering) · Solved Paper 2014
Chemistry
1. In 1885, Balmer for the first time showed that and h = 6.6 ´ 10 -34 J - s
the wave numbers of spectral lines present in = 6.6 ´ 10 -27 erg-s
the visible region in hydrogen spectrum are
given by On substituting values, we get
6.6 ´ 10 -27
æ 1 1ö l= = 6.6 ´ 10 -31 cm
n ( cm-1) = 109677 ç 2 - 2 ÷ 100 ´ 100
è ( 2) n ø
Here, n = 3, 4, 5, K 4. Ideal gas equation is
Thus, Balmer spectrum of hydrogen was nRT
pV = nRT or V = +C
discovered first and it lies in the visible region. p
RT
2. Atomic number of Cu is 29. or V = +0 (for 1 mol)
p
Thus, its electronic configuration can be written as
Cu29 = 1s2, 2s2, 2p 6, 3s2, 3p 6, 3d 10, 4s1 When V vs T is plotted, a straight line is
obtained, slope of which is given by R / p . Thus,
(Since, fully filled orbitals are more stable, so one
R R
electron from 4s orbital transfers into 3d orbital.) Slope = Þ X =
p 2
The electronic configuration of neon is
or R = 2 X L atm mol -1 K -1
Ne10 = 1s2, 2s2, 2p 6
Thus, using inert gas, the configuration of Cu 5. According to Graham’s law of diffusion,
can also be represented as 1 V
Rate of diffusion, r ∝ =
Cu29 = [ Ne] 3s2, 3p 6, 3d 10, 4s1 M t
29. The molecular formula of ethane is C2H 6. To get Thus, the order of amount of electrolytes
its fourth homologue, we have to add required is in the order
4 ´ CH2 = C 4H 8 to it. NaNO3 > Ba(NO3 )2 > Al2(NO3 )3
C2H 6 + C 4H 8 = C 6H14 Note In paper, the formula of aluminium nitrate
C 6 H14 is the molecular formula of hexane. is incorrect. Correct formula of aluminium nitrate
is Al2(NO3 )3 .
Thus, hexane is the fourth homologue of ethane.
5
30. The given statement is true as electrons are 34. DH = - nC p DT = - 2 ´ R ( 225 - 125)
2
shifted more towards the more electronegative
= - 5R ( 100) = - 500R
atom.
31. Hydrides of N, O and F because of the small size 35. Since, the amine gives white precipitate with
and high electronegativity of elements, have benzene sulphonyl chloride which is insoluble in
ability to form extensive intermolecular (i.e., aq. NaOH, so it must be a secondary amine (i.e.,
between two molecules) hydrogen bonding. amine having RNHR group).
Thus, a large amount of energy is required to
Among the given compound, only
break these bonds, i.e., the melting and boiling Me
points of hydrides of these elements are N Me is a secondary amine, so it
abnormally high. ½
H
32. From the Faraday’s IInd law, we know that represents the structure of given amine.
E
W ∝ Zit, W∝ it 36. Combination of four amino acid units gives a
96500 linear tetrapeptide, in which three peptide
1 1 bonds are present.
or it ∝ or electricity ∝
E E O H CH3
Thus, the amount of electricity required for ½½ ½ ½
ZnSO 4, AlCl3 and AgNO3 is in the ratio of H2 N ¾ CH2 ¾ C ¾ OH + H ¾ N ¾ CH ¾COOH
2 : 3 : 1. (Here, 2, 3 and 1 are the valencies of O O
metal atom in the given compounds). ½½ ½½
Alternatively +H2 N ¾ CH2 ¾ C ¾ OH + H2 N ¾ CH2 ¾C ¾ OH
In ZnSO 4, O H CH3 O H
Zn2 + + 2e - ¾® Zn ½½ ½ ½ ½½ ½
1 mol 2 mol H2 N ¾ CH2 1
¾4C2 N3
¾4 ¾ CH ¾ C4¾2N4
¾ CH2
or 2 Faraday 1 3
Peptide bond Peptide bond
In AlCl 3 ,
Al3+
+ 3e -
¾® Al O O
1 mol 3F ½½ ½½
¾ C ¾ N ¾CH2 ¾ C ¾ OH
In AgNO 3 , ½
Ag + + e - ¾® Ag 1424 H
3
1 mol 1F
Peptide bond
Thus, the ratio of amounts of electricity required (Tetrapeptide)
is 2 : 3 : 1 37. Among the given gases, N2O, CO2 and CH 4 have
33. Since, AgI is a negatively charged sol, so the ability to trap thermal radiation but O2 does
not. That’s why it is not a green house gas.
according to Hardy Schulze law, higher the
positive charge on the metal atoms, more is the Note O2 is the gas, presence of which is
tendency of metal atom to cause coagulation of essential for life and in absence of which life is
negatively charged colloid. not possible.
The order of positive charge is 38. 10-4 M KOH = 10-4 M [OH -]
Na + < Ba2 + < Al3 + We know that
[H + ] [OH - ] = 1 ´ 10 -14
WB JEE (Engineering) · Solved Paper 2014 35
[H + ] ´ 10 -4 = 1 ´ 10 -14
1 ´ 10 -14 O
[H + ] = = 1 ´ 10 -10 M
1 ´ 10 -4 5 CH2
O Base-I
+ -10 4
\ pH = - log [H ] = - log ( 1 ´ 10 ) H H 1
= 10 H H
3 2
mass O OH
39. Number of atoms = ´ NA
molar mass HO—P==O
´ Number of atoms in 1 mole
\ Number of atoms in 4.25 g NH3 O
4.25 5CH
2 O Base-II
= ´ NA ´ 4
17 4 1
H H
= NA H 3 2 H
Number of atoms in 8 g
O OH
8 N
O2 = ´ NA ´ 2 = A
32 2 HO—P==O
Number of atoms in 2 g O
2
H2 = ´ N A ´ 2 = 2 N A
2 44. Aniline is a primary amine (contains ¾NH2
Number of atoms in 4 g group). When it is treated with chloroform under
4 alkaline conditions, it results in the formation of
He = ´ N A ´ 1 = N A a bad smelling compound, known by the name
4
isonitrile or carbylamine. This reaction is known
Thus, 2 g H2 contains the maximum number of
as carbylamine reaction.
atoms among the given.
NH2
40. SO2- 2-
3 and SO 4 when treated with BaCl2 , give
white ppt of BaSO3 and BaSO 4 respectively. + CHCl3+Alc. KOH
BaCl2 + SO23 - ¾® BaSO3 + 2Cl - NC
Barium Aniline
sulphide
BaCl2 + SO24- ¾® BaSO 4 + 2Cl - +KCl+H2O
Barium sulphate
Phenyl
Out of these two, SO2- 2-
3 is soluble in HCl but SO 4 isonitrile
does not.
41. Eating animal poisoned with mercury ( Hg 2+ ) 45. Species with more negative E ° (standard
causes deformity, known as minamata reduction potential) generally acts as reducing
(minimata) disease which is characterised by agent while with less negative E ° acts as
diarrhoea, hemolysis, impairment of various oxidising agent. Thus, the overall reaction is
senses, meningitis and death. Ag + + Fe2+ ¾® Fe3+ + Ag
42. H3PO2 is the reagent that converts ¾ N+
2 Cl
-
DE ° = E ° OA - E °RA
group into H. During this process N2 and HCl
gases are released. = - 0.3995 - ( -0.7120) V
43. In DNA, the consecutive deoxynucleotides are = - 0.3995 + 0.7120 V
linked through phosphodiester linkage. = + 0.3125 V
36 WB JEE (Engineering) · Solved Paper 2014
46. van der Waals’ equation is Addition of proton at any other carbon atom,
æ a ö interupt in the delocalisation of p electrons by
ç p + 2 ÷ (V - b ) = RT (for 1 mole) disturbing planarity of molecules and hence,
è V ø
a makes it less stable. Thus, these all are less
If a is negligible, p + 2 » p reactive towards protonation.
V 3 4
p (V - b ) = RT Þ pV - pb = RT Me 1 Me
2 5 H+ +
pV pb pV pb Me 6
Me
or - =1 Þ =1+ 8 H
7
RT RT RT RT Aromatic
47. In acidic medium, phenol exists in following 49. Abstraction of ( H a ) hydrogen atom results in the
resonating structures. formation of 3° allylic carbocation whereas
OH + + abstraction of Hc gives 2° allylic carbocation
O—H O—H (which is less stable as compared to former one).
– However, abstraction of H b gives only a
2° carbocation which is less stable as compared
to former two. More the stability of generated
–
carbocation, higher is the tendency of
+
O—H OH abstraction of hydrogen.
Thus, the order of decreasing ease of abstraction
–
of hydrogen atoms from the given molecule is
H a > Hc > H b
l°c 150
Since, o/p positions are electron rich sites, so 50. Degree of dissociation, a = = = 0.3
electrophile will attack on these site, i.e., l° m 500
hydrogen of these sites get exchanged by Given, C = 0.007 M
D (deuterium). Hence, the final product of the Hydrofluoric acid dissociates in the following
OD manner
D D HF ¾® H + + F -
reaction is which is formed as C 0 0 Initially
H H C -Ca Ca Ca At time t
= C(1 - a)
D
Dissociation constant,
OH O O
[H + ] [F - ] C a × Ca Ca2
H H D Ka = = =
D2SO4 [HF ] C (1 - a) (1 - a)
6 H D
D2O On substituting values, we get
0.007 ´ ( 0.3)2 63 ´ 10 -3 ´ 10 -2
OD OD Ka = =
( 1 - 0.3) 0.7
D D D
Repeat at = 9 ´ 10 -4 M
6 the o and
p-position 51. Elevation of boiling point,
D w ´ K b ´ 1000
DTb =
M ´W
48. Addition of a proton at C1 results in the
(Here, w and W = weights of solute and solvent
formation of tropylium carbocation (an aromatic
respectively,
species with more stability due to delocalisation
of p electrons), thus, it is the most reactive site M = molecular weight of solute and
towards protonation. K b = constant)
WB JEE (Engineering) · Solved Paper 2014 37
59. Electronic configurations of cuprous ( Cu+ ) and cupric ( Cu2+ ) ions are as follows
Cu+ = [ Ar] 3d 104s 0
Cu2 + = [ Ar] 3d 94s 0
Thus, electronic configuration of Cu+ is more stable but it is less stable because in Cu2+ due to its small size
the nuclear charge is sufficient to hold 27 electrons but in Cu+ such a condition is not true.
Further, the IInd IP of Cu is not very high as compared to its Ist IP. Consequently a large amount of lattice
energy is released for cupric compounds as compared to Cu+ compounds.
60. Both the given molecules contain 6 carbon atoms with an aldehyde group, so these are called aldohexoses.
In X, the —OH attached to second last carbon in on right hand side while in Y it is on left hand side, so X is
D-sugar and Y is L-sugar. X and Y are the mirror images of each other, i.e., these are enantiomers.
Mathematics
1. Given equation is = x 2 + y 2 + ( x - 3)2 + y 2 + x 2 + ( y - 1)2
x + 1 - x - 1 = 4x - 1 ...(i) = x 2 + y 2 + x 2 - 6x + 9 + y 2 + x 2 + y 2
On squaring both sides, we get + 1 - 2y
( x + 1) + ( x - 1) - 2 x 2 - 1 = 4x - 1 2 2
= 3x + 3y - 6x - 2y + 10
2
Þ 2x - 2 x - 1 = 4x - 1 æ 2 1ö
= 3( x 2 - 2x + 1) + 3 çy 2 - y + ÷
⇒ 2
-2 x - 1 = 2 x - 1 è 3 9ø
1
Again, squaring both sides, we get + 10 - 3 -
3
4 ( x 2 - 1) = 4x 2 + 1 - 4x
2
æ 1ö 20
⇒ -4 = + 1 - 4x = 3( x - 1)2 + 3 çy - ÷ +
è 3ø 3
⇒ 4x = 5
1
5 It is minimum, when x - 1 = 0 and y - =0
⇒ x = 3
4 1
\ x = 1 and y =
5 3
But, when we put x = in Eq. (i), we get
4 1
\ z =1+ i
5 5 5 3
+1- -1 = 4 ´ -1
4 4 4 ì2x 2 + 1, x £ 1
3. Given, f ( x ) = í 3
9 1 3 1 î4x - 1, x > 1
⇒ - = 5 -1 ⇒ - =2
4 4 2 2 2 1 2
\ ò0 f ( x ) dx = ò f ( x ) dx + ò1 f ( x ) dx
⇒ 1 = 2, which is not true. 0
1 2
Hence, no value of x satisfy the given equation. = ò ( 2x 2 + 1) dx + ò1 (4x
3
- 1) dx
0
2. Let z = x + iy 1 2
é 2x 3 ù é 4x 4 ù
=ê + xú + ê - xú
\ | z |2 + | z - 3 |2 + | z - i |2 ë 3 û0 ë 4 û1
= | x + iy |2 + |( x - 3) + iy |2 2
= ( 1)3 + 1 - ( 0 + 0)
3
+ | x + i ( y - 1)|2 + [( 2) 4 - 2 - {( 1) 4 - 1}]
WB JEE (Engineering) · Solved Paper 2014 39
47 where, P = 2
and Q =
= sq units 1+ x 1 + x2
3
\ Integrating factor, IF = e ò
P dx
2a sin x - sin 2x
4. lim 1
x ®0 tan3 x ò dx
1 + x2 -1
æ 3 5
ö æ 3 ö 5 =e = e tan x
x x (2x ) (2x )
2a ç x - + -K÷ - ç2x - + -K÷
= lim
è 3! 5! ø è 3! 5! ø 7. Given, y = cos -1 x ⇒ y = (cos -1 x )2
x ®0 3
æ x3 2 5 ö On differentiating both sides w.r.t. x, we get
çx + + x + K÷
è 3 15 ø dy -1
= 2 (cos -1 x ) ´
æ 2a 23 ö æ 2a 25 ö dx 1 - x2
(2a - 2) x + ç - + ÷ x 3 + ç - ÷ x 5 + K
è 3 ! 3 ! ø è 5! 5! ø
= lim 3
Again, differentiating both sides w.r.t. x, we get
x ®0
3æ x2 2 4 ö
é -1 ù
x ç1 + + x + K÷ 2
è 3 15 ø ê 1- x ´ 2 ú
ê 1- x ú
Since, it is given that given limit is exist, then ê -1 æ 1 ö ( -2 x ) ú
- cos x ´ ç ÷
2a - 2 = 0 ⇒ a = 1 d 2y ê è 2 ø ( 1 - x 2 )1/ 2 ú
= - 2 ê ú
5. Given, log 101 log 7 ( x + 7 + x ) = 0 dx 2 ê ( 1 - x 2 )2 ú
ê ú
\ log 7 ( x + 7 + x ) = ( 101) 0 ê ú
ê ú
⇒ log 7 ( x + 7 + x) =1 ê
ë
ú
û
⇒ ( x +7+ x ) = 71 é x cos -1 x ù
ê -1 + ú
⇒ x +7+ x =7 ( 1 - x 2 )1/ 2 ú
= -2 ê 2
ê (1 - x ) ú
On squaring both sides, we get ê ú
ë û
( x + 7) + x + 2 x 2 + 7x = 49
é 2x cos -1 x ù
⇒ 2x - 42 = - 2 x 2 + 7x 2 ê2 - ú
dy ê ( 1 - x 2 )1/ 2 ú
=
⇒ x - 21 = - x 2 + 7x dx 2 ê (1 - x 2 ) ú
ê ú
Again, squaring both sides, we get ë û
x 2 + 441 - 42x = x 2 + 7x d 2y dy
⇒ (1 - x 2 ) =2 + x
⇒ 49x = 441 dx 2 dx
2
d y dy
441 ⇒ (1 - x 2 ) 2 - x =2
⇒ x = ⇒ x =9 dx dx
49
But, it is given
6. Given equation is d 2y dy
(1 - x 2 ) 2 - x =c
dy -1
dx dx
(1 + x 2 )
+ y = e tan x
dx \ c =2
or it can be rewritten as 8. Let y = 20301
tan -1 x
dy 1 e
+ y = Number of digits = Integral part of
dx ( 1 + x 2 ) 1 + x2 ( 301 log 10 20) + 1
40 WB JEE (Engineering) · Solved Paper 2014
13. Given, number of elements of A and B are p and 16. Given equation of ellipse is
q, respectively. x2 y2
+ =1
\ The number of relations from the set A to the 144 25
set B is 2 pq . Here, a2 = 144 and b 2 = 25
14. Given equation can be rewritten as b2 25
r2 Now, e = 1- = 1-
x2 - x + 1=0 a2 144
pq
119 119
r2 = =
\ tan A + tan B = 144 12
pq \ Foci of an ellipse = ( ± ae, 0)
and tan A tan B = 1 æ 119 ö
We know, A + B + C = 180° = ç ±12 ´ , 0÷
è 12 ø
⇒ ( A + B ) = 180° - C
= ( ± 119, 0)
⇒ tan ( A + B ) = tan ( 180° - C )
Since, the circle with centre ( 0, 2 ) and passing
tan A + tan B
⇒ = - tan C through foci ( ± 119, 0) of the ellipse.
1 - tan A tan B
\ Radius of circle = ( 119 - 0)2 + ( 0 - 2 )2
r 2 / pq
⇒ = - tan C
1-1 = 119 + 2 = 121 = 11
⇒ ¥ = - tan C ⇒ C = 90° 17. Given lines are
Hence, DABC is a right angled triangle. x + y =0 ...(i)
15. Given equation of parabola is 5x + y = 4 ...(ii)
and x + 5y = 4 ...(iii)
y 2 = 12x ...(i)
On solving above equations pairwise, we get
On differentiating both sides w.r.t. x, we get
æ2 2ö
dy dy 6 A( 1, - 1), B ç , ÷, C ( -1, 1)
2y = 12 ⇒ = è3 3ø
dx dx y
2 2
Since, the normal to the curve is parallel to the æ2 ö æ2 ö
Now, AB = ç - 1÷ + ç + 1÷
line y = 4x + 3. è3 ø è3 ø
13 1 2x + 1
Let cosec -1 =y 20. Given, f (x ) = x + =
12 2 2
13 12 1
Then, cosec y = ⇒ sin y = \ f ( 2x ) = 2x +
12 13 2
cos y = 1 - sin2 y 4x + 1
\ ⇒ f ( 2x ) =
2
2
æ 12 ö 144 1 8x + 1
= 1- ç ÷ = 1- and f ( 4x ) = 4x + ⇒ f ( 4x ) =
è 13 ø 169 2 2
25 5 5 Since, f ( x ), f ( 2x ) and f ( 4x ) are in HP.
= = ⇒ y = cos -1 1 1 1
169 13 13 \ , and are in AP.
Eq. (i) becomes, f ( x ) f ( 2x ) f ( 4x )
æx ö æ5ö p 1 1
sin-1 ç ÷ + cos -1 ç ÷ = ...(i) +
è 13 ø è 13 ø 2 1 f ( x ) f ( 4x )
⇒ =
f ( 2x ) 2
We know that
p 2 2
sin-1 q + cos -1 q = +
2 2x + 1 8x + 1
2 ⇒ =
\ Both angles of Eq. (i) should be same. 4x + 1 2
x =5 2 10x + 2
\ ⇒ =
4x + 1 ( 2x + 1) ( 8x + 1)
19. Given curve is
( 7x + 5)2 + ( 7y + 3)2 = l2( 4x + 3y - 24)2 ⇒ ( 2x + 1) ( 8x + 1) = ( 5x + 1) ( 4x + 1)
22. Given, a, b are the roots of ax 2 + bx + c = 0 and 24. Equation of tangent in slope form of parabola
a + h, b + h are the roots of px + qx + r = 02 y 2 = 8 3 x is
b c y = mx + c ...(i)
\ a+b=- , ab =
a a a
where, c=
q r m
and a + h + b + h = - , ( a + h ) ( b + h ) =
p p 2 3
\ c= ...(ii)
Now, ( a + h ) - (b + h ) = a - b m
⇒ [( a + h ) - ( b + h )]2 = ( a - b )2 Also, tangent to the hyperbola
4x 2 - y 2 = 4
⇒ [( a + h ) + ( b + h )]2 - 4( a + h ) ( b + h )
x2 y 2
= ( a + b )2 - 4ab or - = 1 is
1 4
q 2 4r b 2 4c
⇒ - = 2 - c 2 = a2 m2 - b 2
p2 p a a
c 2 = 1m2 - 4
q 2 - 4pr b 2 - 4ac
⇒ = 2
p2 a2 æ2 3 ö
⇒ ç ÷ = m2 - 4 [from Eq. (ii)]
b 2 - 4ac a2 è m ø
\ = 12
q 2 - 4pr p 2 ⇒ = m2 - 4
m2
Hence, the ratio of the square of their ⇒ m 4 - 4m2 - 12 = 0
discriminants is a2 : p 2 .
⇒ m 4 - 6m2 + 2m2 - 12 = 0
23. Since, a is a root of ⇒ m2( m2 - 6) + 2( m2 - 6) = 0
2 2 2
x + 3p x + 5q = 0 ⇒ ( m2 + 2) ( m2 - 6) = 0
2 2 2
\ a + 3p a + 5q = 0 ...(i) ⇒ m - 6 = 0 and m2 + 2 ¹ 0
2
and b is a root of ⇒ m2 = 6
2 2 2
x + 9p x + 15q = 0 ⇒ m=± 6
\ b2 + 9p 2b + 15q 2 = 0 ...(ii) i.e., m = 6 as m is positive slope.
Let f ( x ) = x 2 + 6p 2 x + 10q 2 \ From Eq. (i),
Then, f ( a ) = a2 + 6p 2 a + 10q 2 2 3
y = 6x +
2 2 2
= ( a + 3p a + 5q ) + 3p a + 5q 2 2 6
4 æ 1 ö
13
Also, slope of the given line is - . Now, the general term in çax - 2 ÷ is
3 è bx ø
4 32
\ - = ⇒ y = - 24 13 13 - r æ 1 ö
r
3 y T ¢r + 1 = C r ( ax ) ç- 2 ÷
è bx ø
From Eq. (i), ( -24)2 = 64x ⇒ x = 9
= 13C r a13 - r ´ b - r ´ ( x )13 - 3r ( -1) r
\ Hence, the required point is ( 9, - 24).
For coefficient of x -8, put 13 - 3r = - 8
26. We know that
⇒ 3r = 21 ⇒ r = 7
| z - z1 | + | z - z2 | = k will represent an ellipse, if
| z1 - z2 | < k. \ T ¢8 = ( -1)7 13C7a13 - 7 b -7 x -8
Hence, the equation | z - z1 | + | z - z2 | = ( -1)7 13C7a 6b -7 x -8
= 2 | z1 - z2 | represent an ellipse. According to the given condition,
13
ì é p ùü æ 1 ö
tan í p ê x - úý Coefficient of x 8 in çax 2 + ÷
î ë 2 ûþ è bx ø
27. Given, f ( x ) =
2 + [ x ]2 æ 1 ö
13
= Coefficient of x -8 in çax - 2 ÷
é pù è bx ø
Since, ê x - ú is an integer for all x, therefore
ë 2û \ 13C 6 a7 b -6 = -13 C7a 6b -7
é pù
p ê x - ú is an integral multiple of p for all x. a7 a6
ë 2û ⇒ 13
C7 = -13
C 7
b6 b7
ì é p ùü
Hence, tan í p êë x - 2 úûý = 0 for all x a7 b6 1
î þ ⇒ 6
=- 7 ⇒ a=- ⇒ ab + 1 = 0
a b b
Also, 2 + [ x ]2 ¹ 0 for all x 2 4
30. Given, I = ò e x ( x - a) dx = 0
0
Hence, f ( x ) = 0 for all x.
2 x4 2 x4
Hence, f ( x ) is continuous and derivable for all x. ⇒ ò0 e x dx = ò0 e a dx
28. Given, f ( x ) = a sin| x | + be| x | 2 4
Here, we see that ò e x x dx gives us an area
0
We know that sin| x | and e| x | is not 4
between two curves e x and x from x = 0 to x = 2.
differentiable at x = 0. 2 x4
Similarly, ò0 e a dx gives us a same area, so a
Therefore, for f ( x ) to differentiable at x = 0, we
should lies between 0 to 2, i.e., a Î( 0, 2).
must have a = b = 0.
31. Given differential equation can be rewritten as
\ a + b =0
æ y2 ö
13 x f ç 2÷
1 ö
29. The general term in æçax 2 + ÷ is dy y
= +
èx ø
è bx ø dx x æ y2 ö
r y f¢ ç 2 ÷
13 2 13 - r æ 1 ö èx ø
Tr + 1 = C r ( ax ) ç ÷
è bx ø dy dv
Put y = vx ⇒ =v + x
= 13C r a13 - r ´ b - r ( x )26 - 3r dx dx
For coefficient of x 8, put 26 - 3r = 8 \ Given equation becomes,
æ v 2x 2 ö
⇒ 3r = 18 ⇒ r = 6 x fç 2 ÷
dv vx è x ø
\ T7 = 13C 6 a 13 - 6b -6x 8 v +x = +
dx x æ v 2x 2 ö
= 13C 6 a7 b -6x 8 vx f¢ ç 2 ÷
è x ø
WB JEE (Engineering) · Solved Paper 2014 45
dv f (v 2 ) 34. Given,
⇒ x =
dx v f¢ ( v 2 ) é æ 2p ö æ 2p ö ù
ê cos çè n ÷ø sin çè n ÷ø 0ú
v f¢ ( v 2 ) dx ê ú
⇒ dv = æ 2p ö æ 2p ö
f (v 2 ) x A = ê - sin ç ÷ cos ç ÷ 0ú
ê è n ø è n ø ú
On integrating both sides, we get ê 0 0 1ú
1 ê ú
log f ( v 2 ) = log x + log c1 ë û
2 é æ 2p ö æ 2p ö ù
⇒ log f ( v 2 ) = 2 log xc1 ê cos çè n ÷ø sin çè n ÷ø 0ú
ê ú
æ y2 ö æ 2p ö æ 2p ö
⇒ f ( v 2 ) = ( xc1)2 ⇒ f ç 2 ÷ = x 2c12 Now, A ´ A = ê - sin ç ÷ cos ç ÷ 0ú ´
ê è n ø è n ø ú
èx ø ê 0 0 1ú
æ y2 ö ê ú
⇒ f ç 2 ÷ = x 2c [put c12 = c ] ë û
èx ø é æ 2p ö æ 2p ö ù
ê cos çè n ÷ø sin çè n ÷ø 0ú
32. Since, a and b are the roots of ê ú
f ( x ) = x 2 + bx + c ê - sin æç 2p ö÷ cos æç 2p ö÷ 0ú
ê è n ø è n ø ú
\ a + b = - b and ab = c ê 0 0 1ú
2 ê ú
æ a + bö æ a + bö æ a + bö ë û
\ f ç ÷=ç ÷ +bç ÷+c
è 2 ø è 2 ø è 2 ø
é 2 æ 2p ö 2 æ 2p ö
2 ê cos ç ÷ - sin ç ÷
æ bö æ bö è n ø è n ø
= ç- ÷ + b ç- ÷ + c ê
è 2ø è 2ø æ 2 p ö æ 2
ê - sin ç ÷ cos ç ÷ - sin ç 2p ö÷ cos æç 2p ö÷
p ö æ
b2 b2 b2 ê è n ø è n ø è n ø è n ø
= - +c=- +c ê 0
4 2 4 ê
dy ë
Now, = f ¢ ( x ) = 2x + b
dx æ 2p ö æ 2p ö æ 2p ö æ 2p ö ù
cos ç ÷ sin ç ÷ + sin ç ÷ cos ç ÷ 0ú
æa + b æ a + böö è n ø è n ø è n ø è n ø
At point ç ,f ç ÷ , ú
è 2 è 2 ø ÷ø æ 2 p ö æ 2 p ö
- sin2 ç ÷ + cos2 ç ÷ 0ú
è n ø è n ø ú
æ b b2 ö
ú
i.e., ç - , - + c ÷, 0
è 2 4 ø 1ú
û
dy æ bö é æ 2p ö æ 2p ö æ 2p ö ù
= 2 ç- ÷ + b = 0 cos ç2 ´ ÷ 2 sin ç ÷ cos ç ÷ 0ú
dx è 2ø ê è ø è ø è ø
n n n
ê ú
Hence, the slope of the tangent to the curve and ê æ 2p ö æ 2p ö æ 2p ö
= -2 sin ç ÷ cos ç ÷ cos ç2 ´ ÷ 0ú
the positive direction of x-axis is 0°. ê è n ø è n ø è n ø ú
ê 0 0 1ú
33. Given function is ê ú
f ( x ) = x 2 + bx + c ë û
é æ 2p ö æ 2p ö ù
It is a quadratic equation in x. ê cos çè2 ´ n ÷ø sin çè2 ´ n ÷ø 0ú
So, we will get a parabola either downward or ê ú
æ 2p ö æ 2p ö
upward. = ê - sin ç2 ´ ÷ cos ç2 ´ ÷ 0ú
ê è n ø è n ø ú
Hence, it is a many one mapping and not onto ê 0 0 1ú
mapping. ê ú
Hence, it is neither one-to-one nor onto mapping. ë û
46 WB JEE (Engineering) · Solved Paper 2014
and n n
( x + 1) = C 0 x + C1x n n n -1 2n + 1 n
+K+ Cn - 0
n+1
+ nC2 x n - 2 + K + nC n ...(ii)
2 n 22 n 23 n 2n + 1 n
On multiplying Eqs. (i) and (ii), we get ⇒ C0 + C1 + C2 + K + Cn
1 2 3 n+1
( 1 + x )2n = ( nC 0 + nC1x + nC2 x 2 + K + nC nx n ) 3n + 1 - 1
=
´ ( nC 0 x n + nC1x n - 1 + nC2 x n - 2 + K + nC n ) n+1
Coefficient of x n in RHS ì 1
ï x sin , x ¹ 0
40. Given, f (x ) = í x
= ( nC 0 )2 + ( nC1)2 + K + ( nC n )2
îï 0, x =0
and coefficient of x n in LHS = 2nC n Continuity at x = 0,
n 2 n 2 2n !
n 2 1
\ ( C 0 ) + ( C1 ) + K + ( C n ) = LHL = lim x sin =0
n!n! x ® 0- x
( 2n ) ! 1
⇒ ( nC1)2 + K + ( nC n )2 = -1 RHL = lim x sin =0
n!n! x ®0 + x
= 2nC n - 1 and f ( 0) = 0
WB JEE (Engineering) · Solved Paper 2014 47
45. Given, f ¢ ( 4) = 5 é0 1 0ù é0 0 1ù
2
ê0 0 1ú , ê0 1 0ú
f ( 4) - f ( x ) é0 ù ê 1 0 0ú ê 1 0 0ú
Now, lim êë 0 form úû ë û ë û
x ®2 x -2
who will give a determinant either -1 or 1.
0 - f ¢ ( x 2 ) × 2x Hence, there are six distinct choices for P and
= lim
x ®2 1 det ( P ) = ± 1.
-f ¢ ( 4) × 2 ´ 2
= 48. Given expression can be rewritten as
1
-1
= - ( 5) ´ 4 = - 20 æ xö
2 ( 1 - x ) -1 ´ 2 -1 ç 1 - ÷
è 2ø
46. We know that
= [1 + x + x 2 + x 3 + ...]
sin np = 0, ∀ n Î N .
é 2 3 ù
¥
æn!pö æ p ö æ2! p ö æxö æxö æxö
ê1 + ç ÷ + ç ÷ + ç ÷ + Kú
\E = å sin ç ÷ = sin ç
è 720 ø
÷ + sin ç
è 720 ø
÷
è 720 ø êë è 2 ø è 2 ø è 2 ø úû
n =1
æpö æpö ( 2 4 - 1) 4
+ sinç ÷ + sinç ÷ + x +K
è 30 ø è6ø 4!
2x ( 2x )2 ( 2x )3 ( 2x ) 4
é 1 0 0ù = + + + +K
47. Given, I = ê0 1 0ú 1! 2! 3! 4!
ê0 0 1ú æx x2 x3 x 4 ö
ë û -ç + + + + K÷
Then, det ( I ) = 1 è 1! 2 ! 3 ! 4 ! ø
If we take I as 2x ( 2x )2 ( 2x )3 ( 2x ) 4
=1+ + + + +K
é0 1 0ù 1! 2! 3! 4!
A1 = ê 1 0 0ú
ê0 0 1ú æ x x2 x3 x 4 ö
ë û - ç1 + + + + + K÷
è 1! 2 ! 3 ! 4 ! ø
Then, det ( I 1) = - 1
Similarly, there are four other possibilities, ⇒ f ( x ) = e 2x - e x
é 1 0 0ù é0 0 1ù When we put x = 0, we get
ê0 0 1ú , ê 1 0 0ú , f ( 0) = e 0 - e 0 = 1 - 1 = 0
ê0 1 0ú ê0 1 0ú
ë û ë û Hence, exactly one real solution exists.
WB JEE (Engineering) · Solved Paper 2014 49
8 21 40 65 \ z2 = z1w
50. Let S = 1 + + + + +K
2! 3! 4! 5! and z3 = z1w2
Again, let S1 = 1 + 8 + 21 + 40 + 65 + K + Tn \ z1z2z3 = z1 ´ z1w ´ z1w2
and S1 = + 1 + 8 + 21 + 40 + K + Tn
= z13 w3
0 = 1 + 7 + 13 + 19 + 25 + K - Tn
Tn = 1 + 7 + 13 + 19 + 25 + K + n terms = z13 [Q w3 = 1]
n 54. Since, z1, z2 and z3 are the vertices of an
= [2( 1) + ( n - 1) 6]
2 equilateral triangle, therefore
= n [1 + 3( n - 1)] = n( 3n - 2) | z1 - z2 | = | z2 - z3 |
n( 3 n - 2 )
\ S=å = | z3 - z1 | = k (say)
n! 1
3n - 2 Also, a = ( 3 + i)
=å 2
( n - 1) ! 1 1
Þ | a| = 3 + 1 = ´2 =1
3n - 3 + 1 2 2
=å
( n - 1) ! Let A = az1 + b, B = az2 + b
3 1 and C = az3 + b
S=å + å
( n - 2) ! ( n - 1) ! Now, | AB | = | az2 + b - ( az1 + b )|
é 1 1 ù = | a ( z2 - z1)|
= 3e + e êëQ e = 1 + 1 ! + 2 ! + Kúû = | a || z2 - z1 |
= 4e = | 1|| z2 - z1 |
We know 2 <e <3 = 1| z2 - z1 |
\ 8 < 4e < 12 = | z2 - z1 | = k
⇒ 8 < S < 12 Similarly, BC = CA = k
Hence, the points az1 + b, az2 + b and az3 + b
51. We have, are the vertices of an equilateral triangle.
n 2 - 1 < [ n 2] £ n 2 [Q x - 1 £ [ x ] £ x ] 55. Given curve is
1 y = (cos x + y )1/ 2
⇒ 2 - < [ n 2] £ 1
n
On differentiating both sides w.r.t. x, we get
\ By Sandwich theorem,
dy 1 æ dy ö
æ 1ö = ´ ç - sin x + ÷
lim ç 2 - ÷ = 2 dx 2(cos x + y )1/ 2 è dx ø
n®¥ è nø
æ dy ö
52. Given, f ¢ ( 0) = 1 and f ¢ ¢( 0) does not exist. ç - sin x + ÷
è dx ø
=
Also, given g ( x ) = xf ¢ ( x ) 2y
\ g ¢ ( x ) = xf ¢ ¢ ( x ) + f ¢ ( x ) dy dy
⇒ 2y = - sin x +
Put x = 0, we get dx dx
g ¢ ( 0) = 0 f ¢ ¢ ( 0) + f ¢ ( 0) dy
⇒ ( 2y - 1) = - sin x
= 0 + 1= 1 dx
Again, differentiating both sides w.r.t. x, we get
53. Given, z1 ¹ ± 1
d 2y dy æ dy ö
Since, z2 and z3 can be obtained by rotating ( 2y - 1) + ç2 ÷ = - cos x
dx 2 dx è dx ø
2p 4p
vector representing through and , d 2y dy æ dy ö
3 3 ⇒ ( 2y - 1) + ç2 ÷ + cos x = 0
respectively dx 2 dx è dx ø
50 WB JEE (Engineering) · Solved Paper 2014
éì 1 ü
= k3 êísin2 A ´ (cos 2C - cos 2B )ý 60. Q Sum of roots, a + b = - p and ab = q
ëî 2 þ
1 \ ( a3 + b3 ) = ( a + b )3 - 3ab ( a + b )
+ sin2 B ´ (cos 2A - cos 2C )
2 = ( - p )3 - 3q ( - p )
2 1 ù = - p 3 + 3pq
+ sin C ´ (cos 2B - cos 2A ) ú
2 û
and a 4 + a2b2 + b 4 = ( a 4 + b 4) + ( ab )2
k3
= [sin2 A ( 1 - 2 sin2 C - 1 + 2 sin2 B ) = ( a2 + b2 )2 - ( ab )2
2
+ sin2 B ( 1 - 2 sin2 A - 1 + 2 sin2 C ) = [( a + b )2 - 2ab]2 - ( ab )2
k3 = [ p 2 - 2q ]2 - 3q 2
= [ -2 sin2 A sin2 C + 2 sin2 A sin2 B
2 = p 4 - 4p 2q + 4q 2 - q 2
- 2 sin2 B sin2 A + 2 sin2 B sin2 C = p 4 - 4p 2q + 3q 2
2 2 2 2
- 2 sin C sin B + 2 sin C sin A]
= p 4 - 3p 2q - p 2q + 3q 2
3
k
= [0] = 0 = p 2( p 2 - 3q ) - q ( p 2 - 3q )
2
2
= ( p 2 - q ) ( p 2 - 3q )
58. Given, a and b are the roots of x - x - 1 = 0
\ a + b =1
61. Given, differential equation is
dy y 1
and ab = - 1 + =
dx x log e x x
Now, consider option (a),
It is a linear equation of the form
Sn + Sn - 1 = Sn + 1
dy
Put n = 2, we get, S2 + S1 = S3 + Py = Q
dx
\ ( a2 + b2 ) + ( a + b ) = a3 + b3 1 1
where P= and Q =
[QS n = a n + b n, given] x log e x x
WB JEE (Engineering) · Solved Paper 2014 51
h æE ö æE ö 1
sin-1 P ç ÷ = 1 and P ç ÷ =
1 2 è E2 ø è E1 ø 5
\ lim q = lim
h®q + 2 h ® 0+ h
2 \ The probability that student did not tick the
answer randomly
1 1
= ´1= = The probability that student tick the answer
2 2
correctly
65. Let equation of hyperbola be
æE ö
x2 y 2 P( E2 ) P ç ÷
- =1 ...(i) è E2 ø
a2 b 2 =
æE ö æE ö
Given, foci, ( ±8, 0) = ( ± ae, 0) P ( E 1) P ç ÷ + P ( E 2 ) P ç ÷
è E1 ø è E2 ø
⇒ ae = 8 ...(ii)
p ( 1)
2b 2 =
and length of latusrectum = 1
( 1 - p ) + p ( 1)
a 5
2b 2 p 5p
\ 24 = = =
a 1 - p + 5p 1 + 4p
Þ b 2 = 12a ...(iii) 5
f ( 1 - h ) - f ( 1) ⇒ -2( 4g + 3f ) = 5 + 10
LHD = lim
h®0 -h 15
1 æ 1ö ⇒ 4g + 3f = - ...(i)
(1 - h ) - - ç1 - ÷ 2
2 è 2ø -h and 2 [ - g ( 2) + ( - f ) ( -1)] = c - 2
= lim = lim =1
h®0 -h h ® 0 -h
⇒ 2 [ -2 g + f ] = 5 - 2
f ( 1 + h ) - f ( 1)
RHD = lim 3
h®0 h ⇒ -2 g + f = ...(ii)
2
( 1 + h )2 æ 1ö
- (1 + h ) + 1 - ç1 - ÷ On solving Eqs. (i) and (ii), we get
2 è 2ø
= lim 9 12
h®0 h f =- and g = -
10 10
1 + h2 + 2h 1 -9 12 27
-1- h + 1- \ fg = ´- =
= lim 2 2 10 10 25
h®0 h -9
and 4f = 4 ´
1 h2 1 10
+ +h-h-
= lim 2 2 2 -36
h®0 h ⇒ 4f =
10
h2 ⇒ 4f = 3g
= lim =0
h ® 0 2h
\ LHD ¹ RHD
80. Given, P ( A ) = 0.7 and P ( B ) = 0.6
Hence, f ( x ) is not differentiable at x = 1. We know that
P( A Ç B) £ P( B)
79. Centres and constant terms in the circles
2 2 2 2 \ P ( A Ç B ) £ 0.6 ...(i)
x + y - 5 = 0, x + y - 8x - 6y + 10 = 0 and
From Booley’s inequality
x 2 + y 2 - 4x + 2y - 2 = 0 are C1¢( 0, 0), c1 = - 5, n
æ n ö
C2¢ ( 4, 3), c2 = 10 and C3¢ ( 2, - 1), c3 = - 2 P ç Ç A i ÷ ³ å P ( A i ) - ( n - 1)
èi =1 ø i =1
Also, centre and constant of circle
x 2 + y 2 + 2gx + 2fy + c = 0 \ P ( A Ç B ) ³ [( P ( A ) + P ( B ) - ( 2 - 1)]
⇒ P ( A Ç B ) ³ [0.7 + 0.6 - 1]
is C 4¢ ( - g , - f ) and c 4 = c
Since, the first circle intersect all the three at the ⇒ P ( A Ç B ) ³ 0.3 ...(ii)
extremities of diameter, therefore they are From Eqs. (i) and (ii), we get
orthogonal to each other. 0.3 £ P ( A Ç B ) £ 0.6
\ 2( g1g2 + f1f2 ) = c1 + c Hence, options (c) and are (d) always incorrect.
\ 2 [ - g ( 0) + ( - f ) ( 0)] = c - 5
⇒ c =5
2 [ - g ( 4) + ( - f ) ( 3)] = c + 10
Engineering Entrance Exam
WB JEE
Solved Paper 2013
Physics
Category I
Directions (Q. Nos. 1 to 45) [Carry one mark each, for which only one option is correct. Any
wrong answer will lead to deduction of 1/3 mark.]
1. The r.m.s speed of the molecules of a gas at case, the inputs to the gates are A = 0, B = 0,
100°C is v. The temperature at which the C = 0. In the second case, the inputs are
r.m.s. speed will be 3v is A = 1, B = 0, C = 1. The output D in the first
(a) 546°C (b) 646°C case and second case respectively are
(c) 746°C (d) 846°C
A
B
2. The equation of state of a gas is given by C D
æ p + a ö ( V - b2 ) = cT, where p, V , T are
ç ÷
è v3 ø (a) 0 and 0 (b) 0 and 1
pressure, volume and temperature respec- (c) 1 and 0 (d) 1 and 1
tively and a, b, c are constants. The dimens- 5. Two soap bubbles of radii r and 2r are
ions of a and b are respectively connected by a capillary tube-valve
(a) [ML 8 T -2 and L 3 / 2 ] arrangement shown in the diagram. The
(b) [ ML 5 T -2 and L 3 ] valve is now opened. Then which one of the
(c) [ ML 5 T -2 and L 6 ] following will result.
(d) [ ML 6 T -2 and L3 / 2 ]
6. The velocity of a car travelling on a straight 11. Four small objects each of mass m are fixed
road is 3.6 kmh -1 at an instant of time. Now at the corners of a rectangular wire-frame of
travelling with uniform acceleration for 10 s, negligible mass and of sides a and b (a > b).
the velocity becomes exactly double. If the If the wire frame is now roated about an axis
wheel radius of the car is 25 cm, then which passing along the side of length b, then the
of the following is the closest to the number moment of inertia of the system for this axis
of revolutions that the wheel makes during of rotation is
this 10 s? (a) 2 ma2 (b) 4 ma2
(a) 84 (b) 95 (c) 2 m (a2 + b 2 ) (d) 2 m (a2 - b 2 )
(c) 126 (d) 135
12. The de-Broglie wavelength of an electron
7. The ionization energy of the hydrogen atom (mass = 1 ´ 10- 30 kg, charge = 16
. ´ 10-19 C)
is 13.6 eV. The potential energy of the with a kinetic energy of 200 eV is (Planck’s
electron in n = 2 state of hydrogen atom is constant = 6.6 ´ 10-34 Js
(a) + 3.4 eV (b) - 3.4 eV
(a) 9.60 ´ 10-11 m (b) 8.25 ´ 10-11 m
(c) + 6.8 eV (d) - 6.8 eV
(c) 6.25 ´ 10-11 m (d) 5.00 ´ 10-11 m
8. In the electrical circuit shown in the figure,
13. The number of atoms of a radioactive
the current through the 4W resistor is
substance of half-life T is N 0 at t = 0. The
3Ω 2Ω time necessary to decay from N 0 / 2 atoms to
9V N 0 / 10 atoms will be
5
8Ω 4Ω (a) T (b) Tlog 5
2
5 T
(c) Tlog é ù (d) log 5
3Ω 2Ω
êë 2 úû 2
(a) 1A (b) 0.5 A 14. A mass M at rest is broken into two pieces
(c) 0.25 A (d) 0.1 A having masses m and ( M - m). The two
masses are then separated by a distance r.
9. A current of 1 A is flowing along positive The gravitational force between them will be
x-axis through a straight wire of length the maximum when the ratio of the masses
0.5 m placed in a region of a magnetic field [ m : ( M - m)] of the two parts is
given by B = (2 i + 4 j )T. The magnitude and (a) 1:1 (b) 1:2 (c) 1:3 (d) 1:4
the direction of the force experienced by the
wire respectively are 15. A bullet of mass m travelling with a speed v
(a) 18 N, along positive z-axis hits a block of mass M initially at rest and
(b) 20 N, along positive x-axis gets embedded in it. The combined system is
(c) 2 N, along positive z-axis free to move and there is no other force
(d) 4 N, along positive y-axis acting on the system. The heat generated in
the process will be
10. S1 and S2 are the coherent point sources of
light located in the xy-plane at points (0,0) (a) Zero
mv 2
and (0, 3l) respectively. Here l is the (b)
wavelength of light. At which one of the 2
Mmv 2
following points (given as coordinates), the (c)
intensity of interference will be maximum? 2 (M - m)
16. A planet moves around the sun in an 21. At two different places, the angles of dip are
elliptical orbit with the sun at one of its foci. respectively 30° and 45°. At these two places
The physical quantity associated with the the ratio of horizontal component of earth’s
motion of the planet that remains constant magnetic field is
with time is (a) 3 : 2 (b) 1 : 2 (c) 1: 2 (d) 1 : 3
(a) velocity (b) centripetal force
(c) linear momentum (d) angular momentum 22. An equilateral triangle is made by uniform
wires AB, BC, CA. A current I enters at A
17. A particle of mass M and charge q is and leaves from the mid point of BC. If the
released from rest in a region of uniform lengths of each side of the triangle is L, the
electric field magnitude E. After a time t, the magnetic field B at the centroid O of the
distance travelled by the charge is S and the triangle is
kinetic energy attained by the particle is T.
Then, the ratio T / S A
(a) remains constant with time t
(b) varies linearly with the mass M of the particle O
(c) is independent of the charge q
(d) is independent of the magnitude of the electric B C
field E m0 æ 4 L ö m0 æ 4 L ö
(a) ç ÷ (b) ç ÷
4p è L ø 2 pè L ø
18. The specific heat C of a solid at low m 2 Lö
temperature shows temperature dependence (c) 0 æç ÷ (d) zero
4pè L ø
according to the relation C = DT 3, where D is
a constant and T is the temperature is 23. A particle is moving with a uniform speed v
kelvin. A piece of this solid of mass m kg is in a circular path of radius r with the centre
taken and its temperature is raised from at O. When the particle moves from a point P
20 K to 30 K. The amount of heat required in to Q on the circle such that Ð POQ = q, then
the process in energy units is the magnitude of the change in velocity is
(a) 5x10 4 Dm (b) (33/4)x10 4 Dm
(a) 2 v sin (2 q) (b) zero
(c) (65/4)x10 4 Dm (d) (5/4)x10 4 Dm
q q
(c) 2 v sin æç ö÷ (d) 2 v cos æç ö÷
19. The least distance of vision of a long sighted è2 ø è2 ø
person is 60 cm. By using a spectacle lens,
this distance is reduced to 12 cm. The power 24. A capacitor of capacitance C0 is charged to a
of the lens is potential V0 and is connected with another
capacitor of capacitance C as shown. After
(a) + 5.0 D (b) + (20/3) D
closing the switch S, the common potential
(c) -(10/3) D (d) +2.0 D
across the two capacitors becomes V. The
20. A particle of mass M and charge q, initially capacitance C is given by
at rest, is accelerated by a uniform electric S
field E through a distance D and is then
allowed to approach a fixed static charge Q of
the same sign. The distance of the closest
approach of the charge q will then be V0 C0 C
qQ Q C (V - V ) C (V - V0 )
(a) (b) (a) 0 0 (b) 0
4 p e0 D 4 p e0 ED V0 V0
qQ Q C 0 (V + V0 ) C 0 (V0 - V )
(c) (d) (c) (d)
2 p e0 D 2 4 p e0 E V V
4 | WB JEE (Engineering) l Solved Paper 2013
25. As shown in figure below, a charge +2 C is 30. Water is flowing in streamline motion
situated at the origin O and another charge through a horizontal tube. The pressure at a
+5 C is on the x-axis at the point A. The later point in the tube is p where the velocity of
charge from the point A is then brought to a flow is v. At another point, where the
point B on the y-axis. The work done is pressure is p / 2, the velocity of flow is
1 [density of water = r]
(given = 9 ´ 109 m/F) p p
4 p e0 (a) v 2 + (b) v 2 -
r r
2p 2p
B(0,2)m (c) v 2 + (d) v 2 -
r r
27. A bar magnet has a magnetic moment of 33. An alpha particle (4 He) has a mass of
200 Am2 . The magnet is suspended in a 4.00300 amu. A proton has a mass of
magnetic field of 0.30 NA -1m -1. The torque 1.00783 amu and a neutron has a mass of
required to rotate the magnet from its 1.00867 amu respectively. The binding
equilibrium position through an angle of energy of alpha estimated from these data is
30°, will be the closest to
(a) 30 N-m (b) 30 3 N-m (a) 27.9 MeV (b) 22.3 MeV
(c) 60 N-m (d) 60 3 N-m (c) 35.0 MeV (d) 20.4 MeV
28. An ideal mono-atomic gas of given mass is 34. The equivalent resistance between the
heated at constant pressure. In this process, points a and b of the electrical network
the fraction of supplied heat energy used for shown in the figure is
the increase of the internal energy of the gas is
(a) 3/8 (b) 3/5 (c) 3/4 (d) 2/5 r
29. The glass prisms P1 and P2 are to be r r
combined together to produce dispersion a b b
without deviation. The angles of the prisms
r r
P1 and P2 are selected as 4° and 3° r
respectively. If the refractive index of prism
P1 is 1.54, then that of P2 will be (a) 6 r (b) 4 r
(a) 1.48 (b) 1.58 (c) 1.62 (d) 1.72 (c) 2 r (d) r
WB JEE (Engineering) l Solved Paper 2013 | 5
35. An object placed at a distance of 16 cm from a 41. Four identical plates each of area a are
convex lens produces an image of separated by a distance d. The connection is
magnification m ( m > 1). If the object is shown below. what is the capacitance
moved towards the lens by 8 cm, then again between P and Q?
an image of magnification m is obtained. The
numerical value of the focal length of the P Q
lens is
(a) 12 cm (b) 14 cm (c) 18 cm (d) 20 cm (a) 2 ae0 / d (b) ae0 / (2 d )
(c) ae0 / d (d) 4ae0 / d
36. A travelling acoustic wave frequency 500 Hz
is moving along the positive x-direction with 42. A particle is acted upon by a constant
a velocity of 300 ms -1. The phase difference power. Then, which of the following
between two points x1 and x2 is 60°. Then physical quantity remains constant ?
the minimum separation between the two (a) Speed
points is (b) Rate of change of acceleration
(a) 1 mm (b) 1 cm (c) 10 cm (d) 10 mm (c) Kinetic energy
(d) Rate of change of kinetic energy
37. A shell of mass 5M, acted upon by no
external force and initially at rest, bursts 43. In an n- p -n transistor
into three fragments of masses M, 2M and (a) the emitter has higher degree of doping
2M respectively. The first two fragments compared to that of the collector
move in opposite directions with velocities of (b) the collector has higher degree of doping
magnitudes 2v and v respectively. The third compared to that of the emitter
fragment will (c) both the emitter and collector have same degree
of doping
(a) move with a velocity v in a direction perpendicular
(d) the base region is most heavily doped
to the other two
(b) move with a velocity2v in the direction of velocity of 44. The vectors are given by A$ = $i + 2 $j + 2 k$ and
the first fragment $ . Another vector C has the
(c) be at rest B = 3 $i + 6 $j + 2 k
(d) move with velocity v in the direction of velocity of same magnitude as B but has the same
the second fragment direction as A. Then which of the following
vectors represents C?
38. A particle moves along x-axis and its
7 $
displacement at any time is given by (a) (i + 2 $j + 2 k)
$
3
x( t) = 2t 3 - 3 t2 + 4 t in SI units. The velocity of 3
the particle when its acceleration is zero, is (b) ($i - 2 $j + 2 k)
$
7
(a) 2.5 ms -1 (b) 3.5 ms -1 (c) 4.54 ms -1 (d) 8.5 ms -1 7
(c) ($i - 2 $j + 2 k)
$
9
39. The fundamental frequency of a closed pipe 9
(d) ($i + 2 $j + 2 k)
$
is equal to the frequency of the second 7
harmonic of an open pipe. The ratio of their
lengths is 45. A car moving at a velocity of 17 ms -1
(a) 1 : 2 (b) 1 : 4 (c) 1 : 8 (d) 1 : 16
towards an approaching bus that blows a
horn at a frequency of 640 Hz on a straight
40. An alternating current in a circuit is given track. The frequency of this horn appears to
by I = 20 sin(100 p t + 005
. p) A. The r.m.s. be 680 Hz to the car driver. If the velocity of
value and the frequency of current sound in air is 340 ms -1, then the velocity of
respectively are the approaching bus is
(a) 10 A and 100 Hz (b) 10 A and 50 Hz (a) 2 ms -1 (b) 4 ms -1
(c) 10 2 A and 50 Hz (d) 10 2 A and 100 Hz (c) 8 ms -1 (d) 10 ms -1
6 | WB JEE (Engineering) l Solved Paper 2013
Category II
Directions (Q. Nos. 46 to 55) [Carry two marks each, for which only one option is correct. Any
wrong answer will lead to deduction of 2/3 mark.]
46. A small mass m, attached to one end of a 50. A magnetic field B = 2t + 4 t2 (where, t =
spring with a negligible mass and an time) is applied perpendicular to the plane
unstretched length L, executes virtual of a circular wire of radius r and resistance
oscillation with angular frequency w0 . When R. If all the units are in SI the electric charge
the mass is rotated with an angular speed w that flows through the circular wire during
by holding the other end of the spring at a t = 0 s to t = 2 s is
fixed point, the mass moves uniformly in a 6pr 2 20p r 2
circular path in a horizontal plane. Then the (a) (b)
R R
increase in length of the spring during the 32 pr 2 48pr 2
rotation is (c) (d)
R R
w2 L w20 L
(a) (b)
w20 -w 2
w - w20
2
51. Two simple harmonic motions are given by
2
w L w20 L x1 = a sin wt + a cos wt and
(c) (d) a
w20 w 2 x2 = a sin wt + cos wt
3
47. A sphere of radius R has a volume density of The ratio of the amplitudes of first and
charge r = kr, where r is the distance from second motion and the phase difference
the centre of the sphere and k is constant. between them are respectively
The magnitude of the electric field which 3 p 3 p
(a) and (b) and
exists at the surface of the sphere is given by 2 12 2 12
(e 0 = permittivity of free space) 2 p 3 p
(c) and (d) and
4 pk R 4 kR 4 pk R k R2 3 12 2 6
(a) (b) (c) (d)
3 e0 3 e0 e0 4 e0
52. A cylindrical block floats vertically in a liquid
of density r1 kept in a container such that the
48. A body is projected from the ground with a
velocity v = (3 i$ + 10 $j ) ms -1. The maximum fraction of volume of the cylinder inside the
height attained and the range of the body liquid is x1. Then some amount of another
respectively are (given g = 10 ms -2 ) immiscible liquid of density r2 (r2 < r1) is
(a) 5 m and 6 m (b) 3 m and 10 m
added to the liquid in the container so that
(c) 6 m and 5 m (d) 3 m and 5 m the cylinder now floats just fully immersed in
the liquids with x2 fraction of volume of the
49. A cell of emf E is connected to a resistance R1 cylinder inside the liquid of density r1. The
for time t and the amount of heat generated ratio r1 / r2 will be
in it is H. If the resistance R1 is replaced by
1 - x2 1 - x1 x1 - x2 x2
another resistance R2 and is connected to (a) (b) (c) (d) -1
x1 - x2 x1 + x2 x1 + x2 x1
the cell at the same time t, the amount of
heat generated in R2 is 4H. Then internal
53. A particle of mass M and charge q is at rest
resistance of the cell is
at the mid point between two other fixed
2 R1 + R 2 2 R2 - R1
(a) (b) R1R 2 similar charges each of magnitude Q placed
2 R 2 - 2 R1
a distance 2d apart. The system is collinear
R 2 - 2 R1 R2 - R1 as shown in the figure. The particle is now
(c) R1R 2 (d) R1R 2
2 R2 - R1 R2 + R1 displaced by a small amount x ( x < < d)
along the joining the two charges and is left
WB JEE (Engineering) l Solved Paper 2013 | 7
to itself. It will now oscillate about the mean constant and e be the charge of an electron,
position with a time period (e 0 = permittivity then the frequency of light in the second
of free space) case is
e e
d d (a) n1 - (V2 + V1 ) (b) n1 + n (V2 + V1 )
h h
q e e
Q Q (c) n1 - (V2 - V1 ) (d) n1 + (V2 - V1 )
h h
p 3 Me0d p 2 Me0d 3
(a) 2 (b) 2 55. 3 moles of mono-atomic gas ( g = 5 / 3) is mixed
Qq Qq
with 1 mole of a diatomic gas ( g = 7 / 3). The
p 3 Me0d 3 p 3 Me0 value of g for the mixture will be
(c) 2 (d) 2
Qq Qqd 3 9
(a)
11
54. The stopping potential for photoelectrons 11
from a metal surface is V1 when (b)
7
monochromatic light of frequency nn is 12
incident on it. The stopping potential (c)
7
becomes V2 when monochromatic light of 15
another frequency is incident on the (d)
7
same metal surface. If h be the Planck’s
Category III
Directions (Q. Nos. 56 to 60) [Carry two marks each, for which only one or more than one option
may be correct. Marking of correct option will lead to a maximum mark of two on pro data basis.
There will be no negative marking for these questions. However, any marking of wrong option
will lead to award of zero mark against the respective question - irrespective of the number of
correct options marked.]
56. An electron of charge e and mass m is
moving in a circular of radius r with a
uniform angular speed w. Then which of the
following statements are correct?
(a) The equivalent current flowing in the circular path
is proportional to r 2
(b) The magnetic moment due to circular current loop
is independent of m F
m
(c) The magnetic moment due to circular current loop
equal to2e / m times the angular momentum of the (a) The tension in the string is F
electron (b) The tension in the string is 3 N
(d) The angular momentum of the particle is (c) The work done by the tension on the block is 20 J
proportional to the areal velocity of electron. during this 1s
(d) The work done against the force of gravity is 10 J
57. A block of mass m ( = 01
. kg) is hanging over a
frictionless light fixed pulley by an 58. If E and B are the magnitudes of electric and
inextensible string of negligible mass. The magnetic fields respectively in some region
other end of the string is pulled by a constant of space, then the possibilities for which a
force F in the vertically downward direction. charged particle may move in that space
The linear momentum of the block increases with a uniform velocity of magnitude v are
by 2 kgms -1 in 1s after the block starts from (a) E = vB (b) E ¹ 0, B = 0
rest. Then, (given g = 10 ms -2 ) (c) E = 0, B ¹ 0 (d) E ¹ 0, B ¹ 0
8 | WB JEE (Engineering) l Solved Paper 2013
59. A bar of length marrying a small mass m at (c) The total distance travelled by the mass in air is
one of its ends rotates with a uniform proportional to w2
(d) The total distance travelled by the mass in air and
angular speed win a vertical plane about the
its total time of flight are both independent on its
mid point of the bar. During the rotation, at
mass.
some instant of time when the bar is
horizontal, the mass is detached from the 60. A biconvex lens of focal length f and radii of
bar but the bar continues to rotate with curvature of both the surfaces R is made of a
some w. The mass moves vertically up, comes material of refractive index n1. This lens is
back and reaches the bar at the same point. placed in a liquid of refractive index n2 . Now
At that place, the acceleration due to gravity this lens will behave like
is g. (a) either as a convex or as a concave lens
w2 l depending solely on R
(a) This possible if the quantity is an integer
2pg (b) a convex lens depending on n1 and n2
(c) a concave lens depending on n1 and n2
(b) The total time of flight of the mass is proportional
(d) a convex lens of same focal length irrespective of
to w2
R, n1 and n2
Chemistry
Category I
Directions (Q. Nos. 1 to 45) [Carry one mark each, for which only one option is correct. Any
wrong answer will lead to deduction of 1/3 mark.]
1. At 25°C, the solubility product of salt of MX2 4. (+)-2-chloro-2-phenylethane in toluene
type is 3.2 ´ 10-8 in water. The solubility racemises slowly in the presence of small
(in mol/L) of MX2 in water at the same amount of SbCl2 , due to the formation of
temperature will be (a) carbanion (b) carbene
(a) 1.2 ´ 10-3 (c) free-radical (d) carbocation
(b) 2 ´ 10-3
5. Acid catalysed hydrolysis of ethyl acetate
(c) 3.2 ´ 10-3
follows a pseudo-first order kinetics with
(d) 1.75 ´ 10-3 respect to ester. If the reaction is carried out
with large excess of ester, the order with
2. The IUPAC name of the compound X is
respect to ester will be
O CN (a) 1.5 (b) 0
(X = C C ) (c) 0.5 (d) 1
CH3
CH3 CH2 CH3 6. The different colours of litmus in acidic,
(a) 4-cyano-4-methyl-2-oxopentane neutral and basic solutions are, respectively.
(b) 2-cyano-2-methyl-4-oxopentane (a) red, orange and blue
(c) 2, 2-dimethyl-4-oxopentanenitrile (b) blue, violet and red
(d) 4-cyano-4-methyl-2-pentanone (c) red, colourless and blue
(d) red, violet and blue
3. In SOCl2 , the Cl ¾ S ¾ Cl and Cl ¾ S ¾ O
bond angles are 7. Baeyer’s reagent is
(a) 130° and 115° (a) alkaline potassium permanganate
(b) 106° and 96° (b) acidified potassium permanganate
(c) 107° and 108° (c) neutral potassium permanganate
(d) 96° and 106° (d) alkaline potassium manganate
WB JEE (Engineering) l Solved Paper 2013 | 9
8. The correct order of equivalent conductances 14. Chlorine gas reacts with red hot calcium
at infinite dilution in water at room oxide to give
temperature for H + , K + , CH 3COO- and HO- (a) bleaching powder and dichlorine monoxide
ions is (b) bleaching powder and water
(a) HO - > H+ > K + > CH3COO - (c) calcium chloride and chlorine dioxide
(b) H+ > HO - > K + > CH3COO - (d) calcium chloride and oxygen
(c) H+ > K + > HO - > CH3COO -
(d) H+ > K + > CH3COO - > HO - 15. For a chemical reaction at 27°C, the
activation energy is 600 R. The ratio of the
9. Nitric acid can be obtained from ammonia rate constants at 327°C to that of at 27°C
via the formations of the intermediate will be
compounds (a) 2 (b) 40 (c) e (d) e 2
(a) nitric oxide and nitrogen dioxide
(b) nitrogen and nitric oxide 16. 2-methylpropane on monochlorination
(c) nitric oxide and dinitrogen pentoxide under photochemical condition give
(d) nitrogen and nitrous oxide (a) 2-chloro-2-methylpropane as major product
(b) (1 : 1) mixture of 1-chloro-2-methylpropane and
10. In the following species, the one which is
2-chloro-2-methylpropane
likely to be the intermediate during benzoin (c) 1-chloro-2-methylpropane as a major product
condensation of benzaldehyde is (d) (1 : 9) mixture of 1-chloro-2-methylpropane and
Å Å OH 2-chloro-2-methylpropane
(a) Ph ¾ C ºº O (b) Ph ¾ C
CN 17. The half-life for decay of 14
C by b-emission is
s OH s 5730 yr. The fraction of 14 C decays, in a
(c) Ph ¾ C (d) Ph ¾ C == O
sample that is 22920 yr old, would be
CN
1 1 7 15
(a) (b) (c) (d)
11. In O2 and H2O2 , the O ¾ O bond lengths are 8 16 8 16
o
1.21 and 1.48 A respectively. In ozone, the 18. A van der Waals’ gas may behave ideally when
average O ¾ O bond length is (a) the volume is very low
(a) 1.28 Å (b) 1.18 Å (c) 1.44 Å (d) 1.52 Å (b) the temperature is very high
(c) the pressure is very low
12. The change of entropy (dS) is defined as (d) the temperature, pressure and volume all are very
dq dH high
(a) dS = (b) dS =
T T
dqrev dH - dG 19. The optically active molecule is
(c) dS = (d) dS =
T T
COOMe COOMe
13. Correct pair of compounds which gives blue
colouration/precipitate and white precipitate, HO H D OH
respectively, when their Lassaigne’s test is (a) (b)
HO H D OH
separately done is
(a) NH2NH2 , HCl and ClCH2COOH COOMe COOMe
(b) NH2CSNH2 and PhCH2Cl
COOMe COOH
(c) NH2NH2 , COOH and NH2CONH2
H Me H OH H OH
N COOH (c) (d)
H OH H OH
Cl
(d) and COOH COOH
10 | WB JEE (Engineering) l Solved Paper 2013
20. In diborane, the number of electrons that 26. When aniline is nitrated with nitrating
accounts for bonding in the bridges is mixture in ice cold condition, the major
(a) six (b) two product obtained is
(c) eight (d) four (a) p-nitroaniline (b) 2, 4-dinitroaniline
(c) o-nitroaniline (d) m-nitroaniline
21. The reaction of nitroprusside anion with
sulphide ion gives purple colouration due to 27. The measured freezing point depression for
the formation of a 0.1 m aqueous CH 3COOH solution is
(a) the tetranionic complex of iron (II) coordinating to 0.19°C. The acid dissociation constant
one NOS - ion K a at this concentration will be (Given, K f
(b) the dianoinic complex of iron (II) coordinating to the molal cryoscopic constant = 1.86 K
one NCS - ion kg mol -1)
(c) the trianoionic complex of iron (III) coordinating to
one NOS - ion (a) 4.76 ´ 10-5 (b) 4 ´ 10-5
-5
(d) the tetranionic complex of iron (III) coordinating to (c) 8 ´ 10 (d) 2 ´ 10-5
one NCS - ion
28. The ore chromite is
22. At 25°C, pH of a 10-8 M aqueous KOH (a) FeCr2O 4 (b) CoCr2O 3
solution will be (c) CrFe 2O 4 (d) FeCr2O 3
(a) 6.0 (b) 7.02
(c) 8.02 (d) 9.02 29. ‘Sulphan’ is
(a) a mixture of SO 3 and H2SO 5
23. An optically active compound having (b) 100% conc. H2SO 4
molecular formula C2H16 on ozonolysis gives (c) a mixture of gypsum and conc. H2SO 4
acetone as one of the products. The structure (d) 100% oleum (a mixture of 100% SO 3 in 100%
of the compound is H2SO 4 )
H3 C CH3 H3 C H
30. Pressure-volume (pV) work done by an
(a) C==C H (b) C==C CH3 ideal gaseous system at constant volume is
H 3C C H 3C C (where E is internal energy of the system)
H5C2 H H5C2 H
(a) -Dp / p (b) zero (c) -VDp (d) -DE
H3CH2HC CH3 H3 C CH2CH3 31. Amongst [Ni(H2O) 6 ] 2 +, [Ni(PPh 3)2 Cl2 ],
(c) C==C CH3 (d) C==C CH3 [Ni(CO)4 ] and [Ni(CN4 ) ] 2 - the paramagnetic
H C H 3C C species are
H3 C H H H
(a) [NiCl 4 ]2- , [Ni(H2O)6 ]2+ , [Ni(PPh3 )2 Cl 2 ]
(b) [Ni(CO)4 ], [Ni(PPh3 )2 Cl 2 ], [NiCl 4 ]2-
24. Mixing of two different ideal gases under (c) [Ni(CN)4 ]2- , [Ni(H2O)6 ]2+ , [NiCl 4 ]2-
isothermal reversible condition will lead to (d) [NI(PPh3 )2 Cl 2 ], [Ni(CO)4 ], [Ni(CN)4 ]2-
(a) increase of Gibbs free energy of the system
(b) no change of entropy of the system 32. Ribose and 2-deoxyribose can be
(c) increase of entropy of the system differentiated by
(d) increase of enthalpy of the system
(a) Fehling’s reagent (b) Tollen’s reagent
(c) Barfoed’s reagent (d) Osazone formation
25. The ground state electronic configuration of
CO molecule is 33. Number of hydrogen ions present in
(a) 1s2 2 s2 1p 4 3s2 10 milionth part of 1.33 cm 3 of pure water
(b) 1s2 2 s2 3s2 1p 2 2 p 2 at 25°C is
(c) 1s2 2 s2 1p 2 3s2 2 p 2 (a) 6.023 million (b) 60 million
(d) 1s2 1p 2 2 s2 2 s2 (c) 8.01 million (d) 80.23 million
WB JEE (Engineering) l Solved Paper 2013 | 11
34. The correct order of acid strength of the 41. The condition of spontaneity of a process is
following substituted phenols in water at (a) lowering of enthropy at constant temperature and
28°C is pressure
(a) p-nitrophenol < p-fluorophenol < p-chlorophenol (b) lowering of Gibbs free energy of system at
(b) p-chlorophenol < p-fluorophenol < p-nitrophenol constant temperature and pressure
(c) p-fluorophenol < p-chlorophenol < p-nitrophenol (c) increase of entropy of system at constant
(d) p-flurophenol < p-nitrophenol < p-chlorophenol temperature and pressure
(d) increase of Gibbs free energy of the universe at
35. For isothermal expansion of an ideal gas, constant temperature and pressure
the correct combination of the
thermodynamic parameters will be 42. The increasing order of O ¾ N ¾ O bond
(a) DU = 0, Q = 0, W ¹ 0 and DH ¹ 0 angle in the species NO2 , NO2 + and NO2 - is
(b) DU ¹ 0, Q ¹ 0, W ¹ 0 and DH = 0 (a) NO 2 + < NO 2 < NO 2 - (b) NO 2 < NO -2 < NO+2
(c) DU = 0, Q ¹ 0, W = 0 and DH ¹ 0 (c) NO+2 < NO -2 < NO 2 (d) NO 2 < NO+2 < NO -2
(d) DU = 0, Q ¹ 0, W ¹ 0 and DH = 0
43. The correct structure of the dipeptide gly-ala is
36. Addition of excess potassium iodide solution
to a solution of mercuric chloride gives the CH3 O H
O
halide complex (a) H2N CHC NH CC
(a) tetrahedral K 2 [HgI4 ]
OH
(b) trigonal K[HgI3 ] H
(c) linear Hg 2I2 CH2SH O
(d) square planar K 2 [HgCl 2I2 ]
(b) NH2 CH C NH CH2 COH
37. Amongst the following, the one which can O
exist in free state as a stable compound is
H O O
(a) C 7H9O (b) C 8H12O (c) C 6H12O (d) C10H17O
–
(c) H2NCCNH CH C—OH
38. A conductivity cell has been calibrated with
a 0.01 M 1 : 1 electrolyte solution (specific H CH3
conductance, k = 1.25 ´ 10-3 S cm -1) in the H O O
cell and the measured resistance was 800 W
at 25°C. The cell constant will be (d) H2N CCNHCHC—OH
(a) 1.02 cm -1 (b) 0.102 cm -1
H CH2SH
(c) 1.00 cm -1 (d) 0.5 cm -1
39. The orange solid on heating gives a 44. Equivalent conductivity at infinite dilution
colourless gas and a green solid which can be for sodium-potassium oxalate
reduced to metal by aluminium powder. The [(COO- )2 Na +K + ] will be [given molar
orange and the green solids are, respectively conductivities of oxalate, K + and Na + ions at
infinite dilution are 148.2, 50.1,
(a) NH4Cr2O 7 and Cr2O 3 (b) Na 2Cr2O 7 and Cr2O 3
73.5 S cm2 mol -1 respectively]
(c) K 2Cr2O 7 and CrO 3 (d) (NH4 )2 CrO 4 and CrO 3
(a) 271.8 S cm 2 eq -1 (b) 67.95 S cm 2 eq -1
40. The best method for the preparation of (c) 543.6 S cm 2 eq -1 (d) 135.9 S cm 2 eq -1
2, 2-dimethylbutane is via the reaction of
45. For BCl 3, AlCl 3 and GaCl 3 the increasing
(a) Me 3CBr and MeCH2Br in Na/ether
order of ionic character is
(b) (Me 3C)2 CuLi and MeCH2Br
(a) BCl 3 < AlCl 3 <GaCl 3 (b) GaCl 3 < AlCl 3 <BCl 3
(c) (MeCH2 )2 CuLi and Me 3CBr
(d) Me 3CMgI and MeCH2I (c) BCl 3 <GaCl 3 < AlCl 3 (d) AlCl 3 <BCl 3 <GaCl 3
12 | WB JEE (Engineering) l Solved Paper 2013
Category II
Directions (Q. Nos. 46 to 55) [Carry two marks each, for which only one option is correct. Any
wrong answer will lead to deduction of 2/3 mark.]
46. In borax the number of B ¾ O ¾ B links and O CBr3 O CH2Br
B ¾ OH bonds present are, respectively C C
(a) five and four (b) four and five
(c) three and four (d) five and five (c) (d)
Category III
Directions [Q. Nos. 56 to 60] [Carry two marks each, for which only one or more than one
option may be correct. Marking of correct option will lead to a maximum mark of two on pro
data basis. There will be no negative marking for these questions. However any marking of
wrong option will lead to award of zero mark against the respective question - irrespective of
the number of correct options marked.]
56. In basic medium the amount of Ni2 + in a 58. Tautomerism is exhibited by
solution can be estimated with the (a) (Me3CCO)3CH (b) O
dimethylglyoxime reagent. The correct
statement(s) about the reaction and the O
product is (are) (c) O (d)
O O
(a) in a ammoniacal solution Ni 2+ salts give
cherry-red precipitate of nickel(II) O
dimethylglyoximate 59. The important advantage(s) of Lintz and
(b) two dimethylglyoximate units are bound to one Donawitz (L.D.) process for the manufacture
Ni 2+ of steel is (are)
(c) in the complex two dimethylglyoximate units are (a) the process is very quick
hydrogen bonded to each other (b) operating costs are low
(d) each dimethylglyoximate unit forms a six (c) better quality steel is obtained
membered chelate ring with Ni 2+ (d) scrap iron can be used
60. Consider the following reaction for
57. Correct statement(s) in cases of n-butanol 2NO2 ( g) + F2 ( g) ¾¾® 2NO2F( g). The
and t-butanol is (are) expression for the rate of reaction in terms of
(a) both are having equal solubility in water the rate of change of partial pressure of
(b) t-butanol is more soluble in water than n-butanol reactant and product is/are
(c) boiling point of t-butanol is lower than n-butanol 1 é dp (NO 2 )ù 1 é dp (NO 2 )ù
(a) rate = - (b) rate =
(d) boiling point of n-butanol is lower than t-butanol 2 êë dt úû 2 êë dt úû
1 é dp (NO 2F )ù 1 é dp (NO 2F )ù
(c) rate = - úû (d) rate = 2 êë
2 êë dt dt úû
Mathematics
Category I
Directions (Q. Nos. 1 to 60) [Carry one mark each, for which only one option is correct. Any
wrong answer will lead to deduction of 1/3 mark.]
1. Each of a and b can take values 1 or 2 with 2. Cards are drawn one-by-one without
equal probability. The probability that the replacement from a well shuffled pack of 52
equation ax2 + bx + 1 = 0 has real roots, is cards. Then, the probability that a face card
equal to (jack, queen or king) will appear for the first
1 1 1 1 time on the third turn is equal to
(a) (b) (c) (d)
2 4 8 16 300 36 12 4
(a) (b) (c) (c)
2197 85 85 51
14 | WB JEE (Engineering) l Solved Paper 2013
3. There are two coins, one unbiased with 7. A point P lies on the circle x2 + y2 = 169. If
1 Q = (5, 12) and R = ( -12, 5), then the ÐQPR
probaility or getting heads and the other
2 is
3
one is biased with probability of getting (a)
p
(b)
p
(c)
p
(d)
p
4 6 4 3 2
heads. A coin is selected at random and
tossed. It shows heads up. Then, the 8. A point moves, so that the sum of squares of
probability that the unbiased coin was its distance from the points (1, 2) and ( -2, 1)
selected is is always 6. Then, its locus is
2 3 3 1
(a) (b) (a) the straight line y - = -3 æç x + ö÷
3 5 2 è 2ø
1 2
(c) (d) 1 3 1
2 5 (b) a circle with centre æç - , ö÷ and radius
è 2 2ø 2
4. Lines x + y = 1 and 3 y = x + 3 intersect the (c) a parabola with focus (1, 2) and directrix passing
ellipse x2 + 9 y2 = 9 at the points P, Q and R. through (-2, 1)
The area of the DPQR is (d) an ellipse with foci (1, 2) and (-2, 1)
36 18
(a) (b) 9. A circle passing through (0, 0), (2, 6),
5 5
9 1 (6, 2) cut the x-axis at the point P ¹ (0, 0).
(c) (d)
5 5 Then, the lenght of OP, where O is the
origin, is
5. For the variable , the locus of the point of 5 5
(a) (b) (c) 5 (d) 10
intersection of the lines 3 tx - 2 y + 6t = 0 and 2 2
3 x + 2ty - 6 = 0 is
x 2 y2 10. For the variable t, the locus of the points of
(a) the ellipse + =1 1
4 9 intersection of lines x - 2 y = t and x + 2 y = is
2 2 t
x y
(b) the ellipse + =1 (a) the straight line x = y
9 4
(b) the circle with centre at the origin and radius 1
x 2 y2
(c) the hyperbola - =1 (c) the ellipse with centre at the origin and one focus
4 9
æ 2 ö
x 2 y2 ç , 0÷
(d) the hyperbola - =1 è 5 ø
9 4
(d) the hyperbola with centre at the origin and one
6. The locus of the mid-points of the chords of æ 5 ö
focus ç , 0÷
an ellipse x2 + 4 y2 = 4 that are drawn from è 2 ø
the positive end of the minor axis, is
1 11. The number of onto functions from the set
(a) a circle with centre æç ,0ö÷ and radius 1
è2 ø {1, 2, ..., 11 } to the set {1, 2, ..., 10} is
1 11 !
(b) a parabola with focus æç ,0ö÷ and directrix x = -1 (a) 5 ´ 11 ! (b) 10 ! (c) (d) 10 ´ 11 !
è2 ø 2
1
(c) an ellipse with centre æç 0, ö÷ , major axis 1 and 12. Let p ( x) be a quadratic polynomial with
è 2ø
constant term 1. Suppose p ( x), when divided
1
minor axis by x - 1 leaves remainder 2 and when
2
1 divided by x + 1 leaves remainder 4. Then,
(d) a hyperbola with centre æç 0, ö÷ , transverse axis 1 the sum of the roots of p ( x) = 0 is
è 2ø
1 1 1
and conjugate axis (a) -1 (b) 1 (c) - (d)
2 2 2
WB JEE (Engineering) l Solved Paper 2013 | 15
16. Five numbers are in HP. The middle term is 22. If f ( x) = 2100 x + 1, g ( x) = 3100 x + 1, then the
1 and the ratio of the second and the fourth set of real numbers x such that f { g( x)} = x is
terms is 2 : 1. Then, the sum of the first three (a) empty
terms is (b) a singleton
11 (c) a finite set with more than one element
(a) (b) 5 (d) infinite
2
14
(c) 2 (d) ì1 1ü
3 23. The limit of í 1 + x - 1 + 2 ý as x ® 0
îx x þ
æcos p p æ 1 ö
ç - sin ö÷ ç ÷ (a) does not exist (b) is equal to
1
17. If p = ç 4 4 and X = ç 2 ÷. Then, 2
p p ÷ 1
çsin cos ÷ ç ÷ (c) is equal to 0 (d) is equal to 1
è 4 4 ø è 2ø
P 3 X is equal to 24. The value of cos2 75° + cos2 45° + cos2 15°
æ- 1 ö - cos2 30° - cos2 60° is
0 ç ÷ (a) 0 (b) 1
(a) æç ö÷ (b) ç 2 ÷ 1 1
è1 ø ç 1 ÷ (c) (d)
è 2 ø 2 4
æ- 1 ö
-1 ç ÷ 25. The maximum and minimum values of
(c) æç ö÷ (d) ç 2 ÷ cos 6 q + sin 6 q are respectively
è 0ø ç- 1 ÷ 1
è 2ø (a) 1and
4
(b) 1and 0
18. If a and b are roots of x2 - x + 1 = 0, then the
(c) 2 and 0
value of a 2013 + b2013 is 1
(d) 1 and
(a) 2 (b) -2 (c) -1 (d) 1 2
16 | WB JEE (Engineering) l Solved Paper 2013
27. If a, b and c are in AP, then the straight line 34. The value of determinant
ax + 2by + c = 0 will always pass through a
fixed point whose coordinates are 1 + a2 - b2 2ab -2b
(a) (1, - 1) (b) (-1, 1) 2ab 1 - a2 + b2 2a is
(c) (1, - 2 ) (d) (-2, 1) 2 2
2b -2a 1- a - b
28. The equation 2x2 + 5xy - 12 y2 = 0 represents (a) 0 (b) (1 + a2 + b 2 )
a (c) (1 + a2 + b 2 )2 (d) (1 + a2 + b 2 )3
(a) circle
(b) pair of non-perpendicular intersecting straight 35. If a , b are the roots of the quadratic equation
lines x2 + ax + b = 0, ( b ¹ 0), then the quadratic
(c) pair of perpendicular straight lines 1 1
equation whose roots are a - , b - , is
(d) hyperbola b a
29. If one end of a diameter of the circle (a) ax2 + a(b - 1)x + (a - 1)2 = 0
3 x2 + 3 y2 - 9 x + 6 y + 5 = 0 is (1, 2), then the (b) bx2 + a(b - 1) x + (b - 1)2 = 0
other end is (c) x2 + ax + bv = 0
(a) (2, 1) (b) (2, 4) (c) (2, - 4) (d) (-4, 2 ) (d) abx2 + bx + a = 0
30. The line y = x intersects the hyperbola 36. If the distance between the foci of an ellipse
x2 y2 is equal to the length of the latusrectum,
- = 1 at the points P and Q. The
9 25 then its eccentricity is
eccentricity of ellipse with PQ as major axis (a)
1
( 5 - 1) (b)
1
( 5 + 1)
5 4 2
and minor axis of length is
2 1
(c) ( 5 - 1)
1
(d) ( 5 + 1)
5 5 5 2 2 2 4
(a) (b) (c) (d)
3 3 9 9
37. The equation of the circle passing through
æ 1ö the point (1, 1) and the points of intersection
31. The limit of x sin çç e x ÷÷ as x ® 0 of x2 + y2 - 6x - 8 = 0 and x2 + y2 - 6 = 0 is
è ø
(a) x2 + y2 + 3 x - 5 = 0 (b) x2 + y2 - 4 x + 2 = 0
(a) is equal to 0 (b) is equal to 1 2 2
(c) x + y + 6 x - 4 = 0 (d) x2 + y2 - 4 y - 2 = 0
e
(c) is equal to (d) does not exist
2 38. The number of lines which pass through the
point (2, - 3) and are at a distance 8 from the
32. The value of
point ( -1, 2) is
é 1 1 1 1 ù (a) infinite (b) 4 (c) 2 (d) 0
1000 ê + + +...+ ú
ë1 ´ 2 2 ´ 3 3 ´ 4 999 ´ 1000 û
39. Six positive numbers are in GP, such that
is
their product is 1000. If the fourth term is 1,
(a) 1000 (b) 999 then the last term is
1
(c) 1001 (d) 1 1
999 (a) 1000 (b) 100 (c) (d)
100 1000
WB JEE (Engineering) l Solved Paper 2013 | 17
1000
40. If a and b are the roots of the quadratic 46. The limit of å ( -1) n x n as x ® ¥
equation ax2 + bx + c = 0 and 3 b2 = 16ac, n=1
then
(a) does not exist
(a) a = 4 b or b = 4 a (b) a = - 4 b or b = - 4 a (b) exists and equals to 0
(c) a = 3 b or b = 3 a (d) a = - 3 b or b = - 3 a (c) exists and approaches to + µ
(d) exists and approaches -¥
41. In the set of all 3 ´ 3 real matrices a relation
is defined as follows. A matrix A is related to 47. Let f (q ) = (1 + sin2 q ) (2 - sin2 q). Then, for
a matrix B, if and only if there is a all values of q
non-singular 3 ´ 3 matrix P, such that 9
(a) f(q) > (b) f(q) < 2
B = P -1 AP. This relation is 4
(a) reflexive, symmetric but not transitive 11 9
(c) f(q) > (d) 2 £ f(q) £
(b) reflexive, transitive but not symmetric 4 4
(c) symmetric, transitive but not reflexive
(d) an equivalence relation 48. If f ( x) = ex ( x - 2)2 , then
(a) f is increasing in (- µ, 0) and (2, µ) and decreasing
42. For any two real numbers a and b, we define in (0, 2 ).
a R b if and only if sin2 a + cos2 b = 1. The (b) f is increasing in (- µ, 0) and decreasing in (0, µ).
relation R is (c) f is increasing in (2, µ) and decreasing in (- µ, 0).
(a) reflexive but not symmetric (d) f is increasing in (0, 2 ) and decreasing in (- µ, 0)
(b) symmetric but not transitive and (2, µ).
(c) transitive but not reflexive
(d) an equivalence relation 49. Let n be a positive even integer. If the ratio
of the largest coefficient and the 2nd largest
43. For the curve x2 + 4 xy + 8 y2 = 64 the coefficient in the expansion of (1 + x) n is
tangents are parallel to the x-axis only at the 11 : 10. Then, the number of terms in the
points expansion of (1 + x) n is
(a) (0, 2 2 ) and (0, - 2 2 ) (a) 20 (b) 21
(b) (8, - 4) and (-8, 4) (c) 10 (d) 11
(c) (8 2 , - 2 2 ) and (-8 2 , 2 2 )
(d) (9, 0) and (-8, 0) 50. Five numbers are in AP with common
difference ¹ 0. If the 1st, 3rd and 4th terms
ì x3 - 3 x + 2 , x < 2,
44. If f ( x) = í 3 2 are in GP, then
î x - 6x + 9 x + 2, x³2
(a) the 5th term is always 0.
then (b) the 1st term is always 0.
(a) lim f( x) does not exist (c) the middle term is always 0.
x® 2 (d) the middle term is always -2.
(b) f is not continuous at x = 2
(c) f is continuous but not differentiable at x = 2 51. Let exp (x) denote the exponential function
(d) f is continuous and differentiable at x = 2 æ 1ö
p ex . If f ( x) = exp çç x x ÷÷, x > 0, then the
45. The (tan n + 1 x) dx è ø
value of I= ò
4
0
1
p minimum value of f in the interval [2, 5] is
+ ò 2 tan n + 1 æ x ö dx is
ç ÷ æ 1ö æ 1ö
2 0 è2ø
(a) exp çe e ÷ (b) exp ç2 2 ÷
1 n+2 ç ÷ ç ÷
(a) (b) è ø è ø
n 2n + 1 æ 1ö æ 1ö
2n - 1 2n - 3 (c) exp ç 5 5 ÷ (d) exp ç 3 3 ÷
(c) (d) ç ÷ ç ÷
n 3n - 2 è ø è ø
18 | WB JEE (Engineering) l Solved Paper 2013
52. The minimum value of the function 56. The value of the integral
f ( x) = 2|x - 1|+ |x - 2|is x + 1ö
2 xæ
(a) 0 (b) 1 (c) 2 (d) 3 ò1 e çlog e x +
è x ø
÷ dx is
Category II
Directions [(Q. Nos. 61 to 75) Carry two marks each, for which only one option is correct. Any
wrong answer will lead to deduction of 2/3 mark.]
61. An objective type test paper has 5 questions. 5 3 3 3
(a) (b) (c) (d)
Out of these 5 questions, 3 questions have 32 128 256 64
four options each (a, b, c, d) with one option 62. The solution of the differential equation
being the correct answer. The other dy
2 questions have two options each, namely ( y2 + 2x) = y satisfies x = 1, y = 1. Then,
dx
true and false. A candidate randomly ticks the solution is
the options. Then, the probability that (a) x = y2 (1 + loge y) (b) y = x2 (1 + loge x)
he/she will tick the correct option in atleast 2
(c) x = y (1 - loge y) (d) y = x2 (1 - loge x)
four questions, is
WB JEE (Engineering) l Solved Paper 2013 | 19
Chemistry
1. (b) 2. (c) 3. (d) 4. (d) 5. (b) 6. (d) 7. (a) 8. (b) 9. (a) 10. (c)
11. (a) 12. (c) 13. (d) 14. (d) 15. (c) 16. (c) 17. (d) 18. (c) 19. (c) 20. (d)
21. (a) 22. (b) 23. (b) 24. (c) 25. (a) 26. (a) 27. (b) 28. (a) 29. (d) 30. (b)
31. (a) 32. (d) 33. (c) 34. (c) 35. (d) 36. (a) 37. (b) 38. (c) 39. (b) 40. (b)
41. (b) 42. (*) 43. (c) 44. (d) 45. (c) 46. (a) 47. (b) 48. (b) 49. (a) 50. (c)
51. (d) 52. (b) 53. (a) 54. (d) 55. (d) 56. (a, b, c,d) 57. (b,c) 58.(a, b, d) 59.(a, c, d) 60. (a, d)
According to Kirchhoff’s second law in closed Moment of inertia of the system about side of
circuit BCDEB length b say CD is
2 I1 + 4I1 + 2 I1 - 8 ( I - I1 ) = 0 = M.I. of mass at A about CD + M.I. of mass at B
Þ 16 I1 - 8 I = 0 about CD + M . I. of mass at C about CD + M.I. of
8I 1 mass ot D about CD
Þ I1 = Þ I1 = I …(i) = m(a)2 + m(a)2 + m(0 )2 + m(0 )2
16 2
In closed circuit ABEFA = 2 ma2
- 9 + 3 I + 8 (I - I1 ) + 2 I = 0 h
12. The de-Broglie wavelength, l =
æ1 ö 2 mk
Þ 13 I - 8 I1 = 9 Þ 13 I - 8 ç I ÷ = 9
è2 ø
Given, h = 6.6 ´ 10 -34 J -s
9
Þ I = = 1A …(ii) m = 1 ´ 10 -30 kg
9
1 K = 200 eV = 200 ´ 1.6 ´ 10 -19 J
So current through 4 W resistor I1 = ´ 1 = 0 .5 A
2 Substituting all these values
9. The force experienced by the wire placed in 6.6 ´ 10 -34
l=
magnetic field, F = Bil 2 ´ 1 ´ 10 -30 ´ 200 ´ 1.6 ´ 10 -19
^ ^ æ ^ 1ö
= (2 i + 4 j ) ç1 i ´ ÷ = 0.825 ´ 10 -10 = 8.25 ´ 10 -11 m
è 2ø
1^ ^ N(t )
^ ^ ^
= i ´ i + 4 ´ (i ´ j ) = 2 k N 13. We have, = e - lt or N(t ) = N0 e - lt
2 N0
Direction of this force can be find out by \For the given condition,
Fleming’s left hand rule which is along pasitive N0 N
z-axis. = N0 e - lt 1 and 0 = N0 e - lt 2
2 10
10. The given situation can be show as 1 1
Þ = e - lt 1 and = e - lt 2
2 10
S2 or e lt 1 = 2 and e lt 2 = 10
(0, 3λ)
Taking log on both sides,
5λ
log 2
lt1 = log 2 Þ t1 =
l
S1 log 2 ´ T
(0, 0) (4λ, 0) or t1 =
log2
The intensity will be maximum at those given
points where the path difference between the and lt 2 = log 10
two interfering waves is an integral multiple of log 10
Þ t2 =
wavelength i . e ., Dx = nl l
For the the given points, the intensity will be log 10 ´ T
or t2 =
maximum for (4l, 0 ). log 2
11. The given situation can be shown as é log 10 ù é log 10 - log 2 ù
\ (t 2 - t1 ) = T ê - 1ú = T ê ú
A a Dm ë log 2 û ë log 2 û
m
é log 5 ù
b = Tê ú
ë log 2 û
m C é 5ù
B m Þ ( t 2 - t1 ) = T log ê ú
ë2û
24 | WB JEE (Engineering) l Solved Paper 2013
14. The gravitation force between two masses, 16. A planet revolves around the sun, is an elliptical
Gm1m2 orbit under the effect of gravitational pull on the
F= planet.
r
Planet
here, m1 = m
and m2 = (M - m)
Gm (M - m) Fg
\ F=
r2
dF Sun
For maximum gravitational force, =0
dm
d
\ [m(M - m)] = 0
dm
M So, torque, C = r ´ F = r F sin 180° n = 0
By solving, we get m = dL
2 As C = ; so L = a constant
m 1 dt
So, =
( M - m) 1 Þ Angular momentum is constant.
17. Given, mass of the particle = M,
15. Mass of bullet = m1 = m
Charge on the particle = q
Initial speed of bullet = u1 = v
Electric field = E
Mass of block = m2 = M
Initial velocity, u = 0
Initial speed of block = u 2 = 0
F qE
Let the common velocity of the bodies after \ Acceleration, a = =
M M
collision = V
\Distance travelled in electric field,
According to conservation of linear momentum 1
S = ut + at 2
m ´ v + M ´ 0 = (m + M ) V 2
mv 1 æ qE ö
V= S = ç ÷ t2
(m + M ) 2 èMø
2
\Heat generated = loss in KE 1 æ qEt ö
Also, kinetic energy T = Mç ÷
= Initial KE - Final KE 2 è M ø
2
1 1 1 æ qEt ö
= mv 2 - (m + M ) v 2 Mç ÷
2 2 T 2 è M ø
So, = = qE
2 S 1 æ qE ö 2
1 1 æ mv ö ç ÷t
= mv 2 - (m + M ) ç ÷ 2 èMø
2 2 è m + Mø
T
Þ Ratio of remains constant with time t.
1 1 m2 v 2 S
= mv 2 -
2 2 (m + M ) 18. Amount of heat required, Q = ò dQ
1 æ m ö T2 = 30
= mv 2 ç1 - ÷ =ò mcdT …(i)
2 è m + Mø T1 = 20
1 æ m + M - mö Given, C = DT 3
= mv 2 ç ÷ 30 30
2 è m+ M ø Q = ò m DT 3 dT = mD 3
\
20 ò20 T dT
1 mM 1 65
= v2 = mD [(30 )2 - (20 )2 ] = ´ 10 4 mD
2 (m + M ) 4 4
WB JEE (Engineering) l Solved Paper 2013 | 25
21. The horizontal component of earth’s magnetic 26. The drift velocity is given as
field is given by, H = R cos d i E
vd = =
where, d is the angle of dip so H1 = R cos 30° neA R ´ neA
and H2 = R cos 45° E´A E
= =
H1 R cos 30° 3/2 3 rl ´ neA Ine
\ = = =
H2 R cos 45° 1 / 2 2 When length of wire changed to 2l, the new drift
velocity,
22. The given situation can be shown as
E
vd ¢ =
I
r ´ 2 l ´ ne
A vd ¢ E / r2 lne 1
\ = =
vd E / r lne 2
vd
Þ vd ¢ =
O 2
B C
27. Given, M = 200 A -m 2 B = 0.30 NA -1M-1
and q = 30°
I
We know that the Torque,
The magnetic field at centroid O = Magnetic field t =M´B
due to left part + Magnetic field due to right part
1
\ B = B1 + B2 = 0 Þ | t | = MB sin q = 200 ´ 0.3 ´
2
(Qdirection ofB due to both parts is different of O) = 100 ´ 0.3 = 30 N-m
26 | WB JEE (Engineering) l Solved Paper 2013
37. Let the velocity of third fragment is v¢. Then by the From figure, it is clear that we have two
conservation of linear momentum capacitors C1 and C2 . Positive plates of C1 are
5 (0 ) = M ´ 2 v - 2 M ´ v + 2 M ´ v ¢ connected to positive plate of C2 and negative
plate to negative. Therefore C1 and C2 are in
Þ v¢ = 0 parallel.
i . e ., the third fragment will be at rest. 2e a
So, Cp = C1 + C2 = 2C = 0
38. Given, x (t ) = 2 t 3 - 3 t 2 + 4 t d
dx 42. We have
So, velocity, v = = (6 t 2 - 6 t + 4)
dt dW
Power = Rate of doing work =
dv dt
and acceleration a = = (12 t - 6)
dt dW
or, P= = rate of change in KE
when acceleration is zero, i . e .,(12 t - 6) = 0 dt
6 1 dW d (KE )
or t = = s i . e ., P= = = constant
12 2 dt dt
\Velocity of the particle at zero acceleration is 43. In an n-p-n or p-n-p transistor, the left hand side
2
æ 1ö æ 1ö thick layer of the transistor is heavily doped
v = 6 ç ÷ - 6 ç ÷ 4 = 2.5 m/s known as emitter and right hand side thick layer
è2ø è2ø
of the transistor is moderately doped known as
39. Let the length of closed pipe is L1 and that of pipe collector.
is L2 . 44. (a) Given, A = $i + 2 $j + 2 k$ and B = 3 $i + 6$j + 2k$
v
Fundamental frequency of closed pipe, n1 = $i + 2 $j + 2k$
4L1 So, C= ´ 32 + 62 + 2 2
1+ 4 + 4
and frequency of second harmonic of open pipe,
v $i + 2 $j + 2k$ 7
n2 = 2 ´ = ´ 49 = ( $i + 2 $j + 2k$ )
2 L2 3 3
Given, n1 = n 2 45. Given, velocity of sound, v = 340 m/s
v 2v L 1
i . e ., = Þ 1 = Velocity of listner, vL = 17 m/s
4 L1 2 L2 L2 4
Velocity of source = vS
40. Given, I = 20 sin (100pt + 0.05 p ) Frequency of horn emitted
The root mean square value of alternating v = 640 Hz
I 20
current = Irms = 0 = = 10 2 A
2 2 Car Bus
Þ 2 pf = 100 p or f = 50 Hz
41. Suppose the pair of plates is connected to The apparent frequency
positive terminal of the battery and the pair of ( v + vL )
plates Q is connected to the negative terminal of n¢ = n
v - vs
the battery.
æ 340 + 17 ö
+ + + + + + + + d 680 = 640 ç ÷
è 340 - vS ø
P Q
+ + + + + + + + d On solving we get vS = 4 m/s
28 | WB JEE (Engineering) l Solved Paper 2013
Kx sin θ R2 4 R1
\ =
(R2 + r )2 (R1 + r )2
mg Þ R2 (R1 + r ) = 2 R1 (R2 + r )
2
From figure, Kx sin q = mw (L + x ) sin q R1R2 [ R1 - 2 R2 ]
On solving, r=
2 2 R1 - R2
Þ Kx = mw (L + x ) …(i)
Also,
K
= w0
50. Given, B = 2 t + 4 t 2
m
at t = 0, B1 = 0
Þ K = mw20 and at t = 2, B2 = 2 ´ 2 + 4(2 )2
Substituting the value in Eq. (i) = 4 + 16 = 20Wb/m 2
mw20 x 2
= mw (L + x )
Df pr 2 (B2 - B1 )
wL 2 We have, DQ = =
Þ x= R R
w20 - w2 pr 2 [20 - 0 ] 20 pr 2
= =
47. Given, r = K × r R R
By Gauss’s theorem 51. The given situation can be shown as
2 a
E (4pr 2 ) =
ò r ´ 4pr dr a/ 3
e0
2 2a and 2a 3
=
ò Kr ´ 4pr dr
e0 π/4 π/6
a a
2
Kr
Þ E= Fon second SHM
4e 0
a1 3
Here r =R Ratio of amplitude = =
a2 2
KR 2 p p p
So, E= and phase difference, - =
4eo 4 6 12
48. (a) Given, v = (3 $i + 10 $j ) m/s 52. The given situation is shown as
Þ v x = 3 and vy = 10
(1- x1 ) (1- x2 ) l
vy2 2
\Maximum height attained, H = l x2
2g x1 1 l
1
10 ´ 10
= = 5m As the Bouyant force in both the cases are same
2 ´ 10
\ r1 x1g = r1 x2 g + r 2 (1 - x2 ) g
2 vy 3 ´ 2 ´ 10
Range = v x ´ T = v x ´ = = 6m r1 (1 - x2 )
g 10 On solving =
r 2 ( x1 - x2 )
WB JEE (Engineering) l Solved Paper 2013 | 29
Chemistry
1. For MX2 type salt, 7. Baeyer’s reagent is 1% cold dilute alkaline
3 -8 3 potassium permanganate. It is used to identify
Solubility product, Ksp = 4s ; 3 . 2 ´ 10 = 4s
unsaturation. All unsaturated compounds lose
3.2 ´ 10 -8 its purple colour.
or s=3 = 2 ´ 10 -3 mol/L
4 8. Equivalent conductance (Leq ) is defined as the
1 conducting power of all the ions produced by
2. O CN
—
—
10. When benzaldehyde is heated with aqueous ethanolic NaCN or KCN, it dimerises to form an a-hydroxy
ketone called benzoin, and this reaction is formed as benzoin condensation.
It involves self condensation of an aromatic aldehyde in the presence of CN - as catalyst.
O
O H O H HO CN H
C C C N C C
Proton
exchange Slow
C N + +
Intermediate
CN O H O
C C C C
+ CN–
OH H OH
benzoin
WB JEE (Engineering) l Solved Paper 2013 | 31
20. H H H H H H 25. 2p s4 –
2p
B B or B B π*
H H H H H H (3s2) s3
Structure of B2H6
Boron electron (1π4)
Hydrogen electron
π
In diborone, odd-electron bonds are found in the (2s2) s2
bridges. From figure, it is clear that 4 electrons 2s
are present for bonding in the bridges. C-atom 2s
O-atom
21. Na 2S + Na 2 [Fe(CN)5 NO] ¾® (1s2) s1
sodium nitropruside'
CO
Na 4 [Fe(CN )5 [NOS ] (Carbon Monoxide)
+ purple colour
Na 4 [Fe(CN)5 NOS] r 4Na 26. Direct nitration of aniline with nitric acid gives a
tetra anionic complex of iron (II) co-ordinating
complex mixture of mono, di- and tri-nitro
+ [Fe(CN)5 NOS]4- compounds and oxidation products. If —NH 2
to one NOS - ions. group is protected by acetylation and then
22. NaOH r Na + OH + -
[OH - ]=10 -8 M nitrated with nitrating mixture, p-isomer is the
main product.
+
H O r H + OH
2
-
[OH - ]=10 -7M NH2 NHCOCH3 NHCOCH3 NH2
\[OH - ]total =(10 -8 + 10 -7 )M CH3COCl HNO3 NaOH/
-7
= 10 (1.1)
NO2 NO2
= 1.1 ´ 10 -7
p-nitroaniline
\ pOH = log 1.1 ´ 10 -7 ~
- 6.98 27. Q DTf = i ´ K f ´ m
\ pH = 14 - 6.98 DTf 0.19
\ i= = = 1.02
= 7.02 K f ´ m 1.86 ´ 0.1
23. H3C i - 1 1. 02 - 1
H O3 Again from, a = =
n -1 2 -1
C C « CH3 Zn/H2O
H 3C C = 0.02 = 2.0 ´ 10 -2
- +
H5C2 H for CH 3COOH q CH COO + H 3
optically active Ka = Ca 2
compound
H3C H = 0.1 ´ (2 ´ 10 -3 )2
C O+ O C CH3 = 4 ´ 10 -5
H3C C
28. Ore chomite is FeCr2O 4 .
acetone H5C2 H
29. H 2SO 4 saturated with SO 3 is called oleum or
24. On mixing of two different ideal gases under
isothermal reversible conditions, DSmix is always sulphan.
positive i . e ., increasing. H 2SO 4 + SO 3 ¾® H 2S 2O 7
WB JEE (Engineering) l Solved Paper 2013 | 33
30. From first law of thermodynamics 34. The order of electron withdrawing tendency from
DE = q + W benzene ring i.e.,
where, work do net (W) = p DV — F < — Cl < — NO 2
= p´V (Q DV = 0) \ Correct order of acidic strength of substituted
phenols will be
=0
OH OH OH
31. Species having unpaired electrons are
paramagnetic < <
Ni 2 + = [Ar ] 3 d 8
F Cl NO2
(i) In [NiCl 4 ]2- and [Ni (H 2O )6 ]2+ ligands Cl - and 35. For isothermal expansion of an ideal gas
H 2O are weak ligands, therefore no pairing will be DT = 0
possible. Thus, there are two unpained electrons. \From DU = nC v DT ,
(ii) In [Ni (Ph 3 )2 Cl 2 ], although PPh 3 has DU = 0 and, from
d-acceptance nature but presence of Cl, makes DH = nC p Dt = 0
electrons unpaired.
From first law of thermodynamics,
32. Ribose forms osazone with Ph—NH—NH 2
DU = Q + W
whereas in deoxyribose, one — OH group is
missing, due to which, it cannot form osazone. as DU = 0 Þ Q ¹ 0
CHO CH N—NH—Ph and W¹0
\ Parameters are
H—C—OH C N—NH—Ph
DU = 0, Q ¹ 0 ; W ¹ 0
H—C—OH Ph—NH—NH2 H—C—OH
and DH = 0
H—C—OH H—C—OH 36. HgCl 2 + 4 KI ¾® K 2 [Hg I4 ] + 2KCl
CH2OH CH2OH Hg = [Xe ] 4f 14 , 5 d 10 6s2
ribose osazone
Hg 2 + = [Xe] 4f 14 , 5 d 10 , 6s°
-7
33. 10 million = 10
5d 6s 6p
\10 millionth part of 1.33 cm 3 = 1.33 ´ 10 -7 cm 3
= 1.33 ´ 10 -7 mL
HgI4 = × × × ×
For pure water, [H+ ] = 10 -7 mol/L
sp3-hybridisation tetrahedral
Þ 1 L water contains [H + ] = 10 -7 mol
37. Molecules having whole number for degree of
10 -7
or 1 mL water contains [H + ] = = 10 -10 mol unsaturation can exist in free state as stable
1000 compounds.
or 10 millionth part of 1× 33 cm 3 water contains Sn ( v - 2 )
Degree of unsaturation = +1
[H + ] = 1× 33 ´ 10 -7 ´ 10 -10 mol 2
where, n = number of atoms of a particular type
= 1× 33 ´ 10 -17 mol
v = valency of the atom.
\Number of H + ions = 1.33 ´ 10 -17 ´ NA
(a) For C 7H 9O,
= 1.33 ´ 10 -17 ´ 6.022 ´ 10 23 7 (4 - 2 ) + 9 (1 - 2 ) + 1 (2 - 2 )
DU = + 1 = 3× 5
= 8.009 ´ 10 6 = 8.01million 2
34 | WB JEE (Engineering) l Solved Paper 2013
1 1 NO -2 < NO 2 < NO +2
r= = S -1 cm 115° 132° 180°
k 1.25 ´ 10 -3
l 43. The structure of dipeptide gly-ala is
From R =r O
O
A
H
1 l H2N—CH2—C —N—CH—C—OH
800 = ´
1.25 ´ 10 -3 A (Gly)
CH3
l (Ala)
(where, = cell constant)
A
l 44. l¥m = l¥m (oxalate) + l¥m (Na + ) + l¥m (K + )
\ = 800 ´ 1.25 ´ 10 -3
A = (148 × 2 + 73 × 5 + 50 × 1)
= 1 cm -1 = 271.85 cm 2 mol -1
39. Orange solid is ammonium dichromate. l¥m 271.8
Q l¥eq = =
D
(NH 4 )2 Cr2O 7 ¾¾ ® N2 + Cr2O 3 + 4 H 2O n × factor 2
orange solid colourless
gas
green solid = 135.9 Scm 2 eq -1
48. CH 3COO - + H + ¾® CH 3COOH subshell and the shape of the orbital occupied
Initially 0 × 01 0 × 001 0 × 01 by the electron.
at equi. 0 × 01 - 0 × 001 0 × 01 + 0 × 001 m describes the preferred orientation of orbitals
= 0 × 009 = 0 × 011 in space.
(salt) 0.009 s describes the spining of an electron on its axis.
pH = pK a + log = 4.75 + log
(acid) 0.011
56. Estimation of Ni 2+ is carried out as.
= 4.66 D
Filtrate of group III + NH 4OH + NH 4Cl ¾¾ ®
D
49. 3HClO 3 ¾¾® HClO 4 + Cl 2 + 2O 2 + H 2O passing H 2S gas ® black ppt of NiS.
Me 3C This black ppt of NiS is soluble in conc. HCl in
50. Mg + Cl — CN ¾® presence of oxidising agent like KClO 3
Br Cl NiS + 2HCl + O ¾® NiCl 2 + H 2 . O+S
Me 3C—CN + Mg Conc
Br Now this NiCl 2 , in basic medium, treated
51. O CH3 O CH2Br with dimethyl glyoxime, cherry red ppt of nickel (II)
C C dimethyl glyoximate is obtained, in which two
Bromination dimethyl glyoximate units are hydrogen bonded
CH3COOH to each other and each unit forms a
six-membered chelate ring with Ni 2+ .
52. DG° = - 2.303 RT log K NiCl2 H3C—C NOH
= - 2.303 ´ 8.314 ´ 298 ´ log 0.15 in +2
NH4OH H3C—C NOH
= - 2.303 ´ 8.314 ´ 298 ´ (- 0.82 ) dimethyl glyoxime
= 4678.7 J = 4.67 kJ
O H—O
53. Silicon oil in a polymer of trimethylchloro
silane and dimethyldichloro silane. These H3C—C N N C—CH3
Ni
are useful as broad spectrum antifoaming H3C—C N N C—CH3
agents.
Polymerisation O—H O
Si (Mel 3 ) Cl + Si (Me )2 Cl 2 + H 2O ¾¾¾¾¾®
nickel (II) dimethyl glyoximate
Me Me Me Me cherry red ppt
CH 3
—Si—O—Si—O—Si—O—Si— |
57. CH 3CH 2CH 2CH 2OH CH 3 —C —CH 3
Me Me Me Me |
n-butanol
54. F OH
D D
t - butanol
LiN. NH3
+ NaNH2
More branching results high solubility and low
NH2 H
D H D boiling point.
NH2
+ 58. Availability of acidic a - H- atoms at (*) marked
positions, enable the compounds to show keto-
This reaction proceeds via benzyne mechanism
enol tautomerism.
D
with intermediate as . O
MeCCO O
55. All the quantum numbers (n, l , m, s) are required to MeCCO— CH*
describe an electron of an atom completely. MeCCO O O
e.g., n describes the position and energy of the (a) (b) (d)
electron in an orbit or shell. l is used to describe
36 | WB JEE (Engineering) l Solved Paper 2013
59. Lintz and Donawitz (L.D.) process is an iron quickly remaining pure iron is mixed with
important process for manufacturing of steel. In spiegeleisen (alloy Fe, Mn, C) to produce better
India, Rourkela steel plant is based upon this quality steel.
process. 60. 2NO 2 (g ) + F2 (g ) ¾® 2 NO 2F(g )
This process is carried out in a converter similar
to bessemer converter having lining (usually 1 é dp (NO 2 )ù
Rate of reaction = -
refractory made of MgO and (CaO ). 2 êë dt úû
Mathematics
1. The given equation 3. Let E ® Event of head showing up
2
ax + bx + 1 = 0 …(i) E1 ® Event of biased coin chosen
has real roots. E2 ® Event of unbiased coin chosen
1 1
\Discriminant (D ) ³ 0 Now, P(E2 ) = and P(E1 ) =
2 2
Þ b2 - 4a ³ 0 …(ii)
æ Eö 1 æEö 3
Also, P ç ÷ = and P ç ÷ =
From Eq. (ii), we observe that è E2 ø 2 è E1 ø 4
a has to be 1 and b has to be 2. (by conditional probability)
1 1 1
So, the required probability = ´ = By Baye’s theorem
2 2 4
æ Eö
2. Ist turn Total number of face card = 12 P(E2 )× P ç ÷
æE ö è E2 ø
Pç 2 ÷ =
Total number of elements in sample space, è Eø æ Eö æEö
P(E2 )× P ç ÷ + P(E1 )× P ç ÷
n(s) = 52 è E2 ø è E1 ø
52 - 12 40 1 1
\P1 (no face card in first turn) = = ´
52 52 2 2 2
= =
IInd turn P2 (no face card in second turn) 1 1 1 3 5
´ + ´
(52 - 13) 39 2 2 2 4
= =
(52 - 1) 51 4. Given equation of ellipse
IIIrd turn P3 (face card in third turn) x 2 + 9y 2 = 9
=
(52 - 40 ) 12
= x2 y2
Þ + =1 …(i)
(51 - 1) 50 9 1
\Required P (face card on third turn) and equation of lines
= P1 ´ P2 ´ P3 x + y = 1and 3 y = x + 3
40 39 12 12 x y
= ´ ´ = or + =1
52 51 50 85 (-3) 1
WB JEE (Engineering) l Solved Paper 2013 | 37
y
Þ 9 (4 - x 2 ) = 4 y 2
3y = x + 3
Þ 36 - 9 x 2 = 4 y 2
(0, 1)P
(–3, 0)
Q Þ 9 x 2 + 4 y 2 = 36
x′ x
O A x2 y2
R Þ + = 1 which represents an ellipse.
x+y = 1 4 9
9, – 4
y
5 5 6.
y′
0 1 1 B (0, 1)
1
\Area of DPQR = -3 0 1 (h, k)
2 9 / 5 -4 / 5 1 x′ x
M O A (2, 0)
1 é 24 12 ù P (x, y)
= +
2 êë 5 5 úû
y′
1 36 18
= ´ = Given equation of an ellipse is
2 5 5
x 2 + 4y 2 = 4
5. Given equation of lines are
x2 y2
3t x - 2 y + 6t = 0 …(i) Þ + =1 …(i)
4 1
and 3 x + 2 ty - 6 = 0 …(ii)
\ Coordinate of positive end of minor axis is
On multiplying Eq. (i) by t and then adding in
B(0, 1.) Let mid-point of the chord BP is M(h, k )
Eq. (ii), we get
æ 0 + x 1+ yö
(3 t 2 + 3)x + 6 t 2 - 6 = 0 Then, (h, k ) = ç
è 2
, ÷
2 ø
2(1 - t 2 ) x
Þ x= Þ h= Þ x = 2h
(1 + t 2 ) 2
Þ x + xt 2 = 2 - 2t 2 and k=
1+ y
Þ y = 2k - 1
2
Þ ( x + 2 ) t 2 = (2 - x )
\ P( x, y ) º {2 h, (2 k - 1}
)
2-x
Þ t2 = …(iii) Since, the point ‘P’ lies cllipes so form Eq. (i), we
2+ x
get
On multiplying Eq. (ii) by t and then subtract (2 h )12 + 4(2 k - 1)2 = 4
from Eq. (i), we get
Þ 4h 2 + 4(2 k - 1)2 = 4
(-2 - 2 t 2 )y + 6 t + 6 t = 0
Þ h 2 + 4k 2 + 1 - 4k = 1
12 t = 2(1 + t 2 )y
Þ h 2 + 4k 2 - 4k = 0
On squaring both sides, we get
144 t 2 = 4 y 2 (1 + t 2 )2 Thus, required locus is an ellipse whose
2
equation is
æ2 - x ö 2 æ 2- xö
Þ 144 ç ÷ = 4 y ç1 + ÷ [form Eq. (iii)] x 2 + 4y 2 - 4y = 0
è2 + x ø è 2 + xø
2
2 æ 1ö
æ2 - x ö 2 æ 4 ö 2 çy - ÷
Þ 36ç ÷=y ç ÷ ( x - 0) è 2ø
è2 + x ø è2 + x ø Þ + =1
1 æ 1ö
ç ÷
(2 - x ) 16 y 2 è 4ø
Þ 36 =
(2 + x ) (2 + x )2 æ 1ö 1
whose centre ç 0, ÷ and major and minor axis
Þ 2
36 (4 - x ) = 16 y 2 è 2ø 2
38 | WB JEE (Engineering) l Solved Paper 2013
and focus 14. When eleven apples are distributed among a girl
5 5 and a boy, then the girl receives atleast 14
= (± ae , 0 ) = ± 1 × , 0 = ± , 0 apples or the boys receives atleast 8 apples.
2 2
(by hypothesis)
and centre = (0, 0 ) 15. Given, z1 = 2 + 3 i and z2 = 3 + 4i
11. Let A = {1, 2, ..., 11}
Now, we have
∴ n( A) = 11 and B = {1, 2, ..., 10} | z − z1|2 + | z − z2|2 = | z1 − z2|2 (let z = x + iy )
∴ n(B) = 10 ⇒ |( x + iy ) − (2 + 3i )|2 + |( x + iy ) − (3 + 4 i )|2
∴Hence number of onto function = |(2 + 3 i ) − (3 + 4 i )|2
= n( A )
Cn( B ) × n(B)! × n(B) ⇒ |( x − 2 ) + i ( y − 3)|2 + |( x − 3) + i ( y − 4)|2
= C10 × 10 ! × 10
11
= |−1 − i|2
= (11 × 10 !) × 10 = 11! × 10 ⇒ ( x − 2 ) + ( y − 3)2 + ( x − 3)2 + ( y − 4)2 = 1 + 1
2
1+ x 1 −3 3 3
23. lim − 1+ 2 ⇒ ≤ cos 4 θ ≤
x→ 0 8 8 8
x x
5 3 5 3 5 3
1 + x − 1 + x 2 0 ⇒ − ≤ + cos 4θ ≤ +
8 8 8 8 8 8
= lim form
x→ 0 x 0 1
⇒ ≤ f(θ ) ≤ 1 [from Eq. (i)]
4
1 x
− So, the maximum value is 1and minimum value is
2 1+ x 1 + x2 1
= lim (Use L-Hospital rule) .
x→ 0 1 4
1 1 26. Given, z = x + iy
= −0=
2 1+ 0 2
z − 1 ( x + iy ) − 1
Now, =
24. Given z − i ( x + iy ) − i
cos 2 75° + cos 2 45° + cos 2 15° − cos 2 30 ° ( x − 1) + iy x − i ( y − 1)
= ×
− cos 60 °
2 x + i ( y − 1) x − i ( y − 1)
x( x − 1) + ixy − i ( x − 1) ( y − 1) + y( y − 1)
3 1 1
2
3 1
2 2
=
= − + + + x 2 + ( y − 1)2
2 2 2 2 2 2 2 2 2
( x 2 + y 2 − x − y ) + i ( xy − xy + y + x − 1)
3
2
1
2 =
− − ( x 2 + y 2 + 1−2 y )
2 2
x2 + y2 − x − y x + y −1
3 + 1 − 2 3 1 3 + 1 + 2 3 3 1 = 2 +i 2 ...(i )
= + + − − x + y − 2 y + 1
2
x + y − 2 y + 1
2
8 2 8 4 4
z −1
Given, is real.
1 3 1 1 3 3 1 3 1 z−i
= − + + + − − = − 1=
2 4 2 2 4 4 4 2 2 So, its imaginary part should be zero.
25. Let f(θ ) = sin 6 θ + cos 6 θ x + y −1
i . e ., =0
⇒ f(θ ) = (sin 2 θ )3 + (cos 2 θ )3 x2 + y2 − 2 y + 1
= (sin 2 θ + cos 2 θ ) ⇒ x + y =1
(Q x 2 + y 2 − 2 y + 1 ≠ 0 )
(sin 4 θ + cos 4 θ − sin 2 θ ⋅ cos 2 θ )
which represent a straight line.
[Q a3 + b3 = (a + b) (a2 + b2 − ab)]
27. Given that, a, b and c are in AP
= 1⋅ {(sin 2 θ + cos 2 θ )2 − 3 sin 2 θ ⋅ cos 2 θ}
∴ 2b = a + c
3
= 1⋅ 1 − ⋅ 4 sin 2 θ ⋅ cos 2 θ ⇒ c = 2b − a
4 …(i)
3 and equation of straight line is
= 1− (sin 2θ)2 (Qsin 2 A = 2 sin A cos A)
4 ax + 2 by + c = 0
3 ⇒ ax + 2 by + (2 b − a) = 0
= 1 − (1 − cos 4 θ )
8 ⇒ a( x − 1) + b(2 y + 2 ) = a ⋅ 0 + b ⋅ 0
3 3
= 1 − + cos 4 θ On comparing, we get
8 8 x − 1= 0 ⇒ x = 1
5 3
f(θ ) = + ⋅ cos 4 θ …(i) and 2y + 2 = 0
8 8
⇒ y = −1
Q −1 ≤ cos 4 θ ≤ 1
Hence, the required fixed point is (1, − 1).
42 | WB JEE (Engineering) l Solved Paper 2013
⇒ sin a + cos b = 1
2 2
…(i) = (2 )3 − 6 + 2 = 8 − 6 + 2 = 4
and sin 2 b + cos 2 c = 1 …(ii) RHL = f (2 + 0 ) = lim (2 + h )3 − 6 (2 + h )2
h→ 0
On adding Eqs. (i) and (ii), we get + 9 (2 + h ) + 2
sin 2 a + (sin 2 b + cos 2 b) + cos 2 c = 2 = (2 )3 − 6 (2 )2 + 9 (2 ) + 2
⇒ sin 2 a + cos 2 c + 1 = 2 = 8 − 24 + 18 + 2 = 4
⇒ sin a + cos c = 1
2 2 Q LHL = RHL
∴ lim f ( x ) exist
Hence, R is transitive also. x→ 2
1 + 25
C2 x 2 + K + 25
C25 x 25 ]dx
⇒ log g( x ) = log x
x On integrating w.r.t. x , taking limits 0 to x, we get
On differentiating w.r.t. x, we get x
1 (1 + x )26
x ⋅ − log x 26
1
⋅ g′ ( x ) = x 0
g( x ) x2 x
x2 x3 x 26
1 − log x = 25 C0 x + 25
C1 ⋅ + 25
C2 +K+ C25 ⋅
25
=
x2
2 3 26 0
48 | WB JEE (Engineering) l Solved Paper 2013
(Q cos θ + i sin θ = e iθ ) x + 1
⇒ log 3 ⋅ ( x + 5) = log 3 3
x + 5
= e − iP + e i(Q − R ) + e − i ( P + Q + R )
π π (Q log m + log n = log mn and log n n = 1)
= e −iπ / 2 + e i( 0 ) + e −iπ (Q = R = P= )
2 2 ⇒ ( x + 1) = 3 ⇒ x = 2
= − i + 1 + (−1) = − i So, only one solution is possible.
WB JEE (Engineering) l Solved Paper 2013 | 49
58. Given, ⇒ x( x − 4) − 1 ( x − 4) = 0
P = 1+
1
+
1
+K ⇒ ( x − 1)( x − 4) = 0 or x = 1, 4
2 × 2 3 × 22 then from is (ii) y = 2, 5
1 1 1 4
and Q= + + +K ∴ Required area = ∫ {( x + 1) − ( x 2 − 4 x + 5)} dx
1× 2 3 × 4 5 × 6 1
4
P (1 / 2 )1 (1 / 2 )2 (1 / 2 )3 4 − x 3 5x 2
Now, = + + +K = ∫ (− x 2 + 5 x − 4)dx = + − 4x
2 1 2 3 1
3 2 1
1 2 3
1 / 2 (1 / 2 ) (1 / 2 ) 64 1 5
⇒ − p / 2 = − − − ... = − + 40 − 16 + − + 4
1 3 3 3 2
1 5
= (−21 − + 28) =
9
⇒ − P / 2 = loge 1 −
2 2 2
P 1 60. Given,
⇒ − = loge
2 2 π 3π
f ( x ) = sin x + 2 cos 2 x, x ∈ ,
⇒ P = 2 loge 2 …(i) 4 4
1 1 1 1 1 ∴ f ′ ( x ) = cos x − 4 cos x ⋅ sin x
Now, Q = 1 − + − + − + K
2 3 4 5 6 and f ′ ′ ( x ) = − sin x − 4 cos 2 x
⇒ Q = loge 2 …(ii) For maximum or minimum of f ( x )
Now, from Eqs. (i) and (ii), we get Put f′ ( x) = 0
P = 2Q ⇒ cos x − 4 cos x ⋅ sin x = 0
59. Given equation of parabola is ⇒ cos x(1 − 4 sin x ) = 0
π 1
y = x 2 − 4x + 5 ⇒ cos x = 0 = cos and sin x ≠
2 4
⇒ y = ( x − 2 )2 + 1 π 3π π
Q x∈ , ⇒ x=
⇒ ( x − 2 ) = ( y − 1)
2
…(i) 4 4 2
and equation of line is π π
Now, f′′ = − sin − 4 cos π
2 2
y=x+1
⇒ x − y = −1 …(ii) = − 1 + 4 = 3 > 0 (min)
π
y
R So, f ( x ) is minimum at x =
(4, 5) 2
and its minimum value is
π π π
(0, 5) f = sin + 2 cos 2 = 1 − 2 × 0 = 1
P 2 2 2
(1, 2)
(0, 1) (2, 1) 61. Total sample space, n(S ) = 43 ⋅ 2 2 and total
x′ x
(1, 0) (2, 0) number of favourable cases
n(E ) = (3 C1 ⋅ 3 + 2C1 ⋅ 1) + 1
y′ n(E )
∴ Required probability =
On putting the value of ( y − 1) from Eqs. (ii) in (i), n(S )
we get
C1 ⋅ 3 + 2 C1 ⋅ 1 + 1 3 ⋅ 3 + 2 + 1
3
( x − 2 )2 = x ⇒ x 2 + 4 − 4x = x = =
43 ⋅ 2 2 43 ⋅ 4
⇒ x 2 − 5x + 4 = 0 12 3
= =
⇒ x 2 − 4x − x + 4 = 0 64 ⋅ 4 64
50 | WB JEE (Engineering) l Solved Paper 2013
3 3 3− 5 + 3+ 5
θ = tan −1 is ( y − 1) = ( x − 3) =
4 4 2
⇒ 4y − 4 = 3x − 9 {(cos θ + secθ )2 − 3 cos θ ⋅ secθ}
⇒ 3x − 4y = 5 …(i)
= 3 ⋅ {(3)2 − 3} = 3 (9 − 3) = 3 × 6 = 18
Now for the intersection point of the line (i) with
3 x − 5 π/2 π /2 1
parabola y 2 = 4( x − 3) . Put y = , 67. Let I = ∫ [sin x ⋅ cos x ]dx = ∫ sin 2 x dx
4 −π / 2 −π / 2 2
(3 x − 5)2 y
then we get = 4( x − 3)
16
⇒ 9 x 2 + 25 − 30 x = 64 x − 192
x′ x
⇒ 9 x 2 − 94 x + 217 = 0 –π O +π
1 2 2 2 3 3 3
= + + + K + + + + K
6 1! 1! 2 ! 1! 2 ! 3 !
1 1 1 1 1 y=x2/3
= 2 1 + + + K + 3 + + K
6 1! 2 ! 1! 2 ! x′
O
x
1
= {2e + 3(e − 1)}
6
x=1 x=8
1
= {2e + 3 e − 3} y′
6
1 5e 1 8
= (5 e − 3) = − ∴ Required area A = ∫ ( x − x 2 / 3 ) dx
6 6 2 x =1
WB JEE (Engineering) l Solved Paper 2013 | 53
8
x 2 3 5/ 3 3 1 3 75. Given function is
= − x = 32 − × 32 − −
2 5 1 5 2 5 F( x ) = ∫
x cos t
dt , 0 ≤ x ≤ 2 π
0 (1 + t2)
2 (5 − 6) 64 1 On differentiation w.r.t. x., (apply Leibnitz rule)
= 32 × − = +
5 10 5 10 cos x cos x
F′ ( x ) = × 1= , where (1 + x 2 ) > 0
128 + 1 129 1 + x2 1 + x2
= =
10 10
π 3π
Here, cos x > 0 ⇒ x ∈ 0, ∪ , 2 π
73. Given function is 2 2
1 1 1 π 3π
f( x) = x + + , x > 1 and cos x < 0 ⇒ x ∈ ,
x − 1 x x + 1 2 2
2x 1 2 x2 π 3π
= x 2 + = 2 +1 So, F is increasing in 0, and , 2 π and
x − 1 x x − 1 2 2
π 3 π
decreasing in ,
2 2 2
= + 1 > 3 (Q x > 1)
1 −
1 76. Radius (r ) = perpendicular distance on line
x2 4 x + 3 y = 12 from centre
∴ f( x) > 3 y
Q
x′ x
= O (r, 0) 4x+3y=12
x′ x
O F(a, 0)
x=–a y′
77. Equation of parabola is y 2 = x and line y = mx and 2 sin α ⋅ cos α = x2c [from Eq. (iii)]
For intersection point of both curves put x = y , 2 ⇒ sin 2α = 2c (Q − 1 ≤ sin 2α ≤ 1)
1
we get ⇒ 2c ≤ 1 ⇒ c ≤
y = my 2 ⇒ y(my − 1) = 0 2
1 79. Given system of equations is
⇒ y = 0 or y =
m x+ y+ z=0
1 αx + βy + γ z = 0
Then, x = 0 or x = 2
m α 2 x + β2 y + γ 2 z = 0
1 1
∴Intersection points are (0, 0 ) and P 2 , 1 1 1
m m
The coefficient matrix, A = α β γ
y y = mx 2
α β γ2
2
P
y2 = x 1 1 1
Now,| A| = α β γ = (α − β ) (β − γ ) (γ − α )
O
) 1 , 1
m2 m ) α 2 β2 γ2
x′ x
(0, 0) (i) The system of equations has a unique
solution, if α , β and γ are distinct
i . e ., | A| ≠ 0
y′ (ii) The system of equations has infinite number
of solutions, if any two of α , β and γ are
∴Required area
equal.
1/ m
1/ m y 2 y2 y3 | A| = 0
=∫ − y dy = − i . e .,
0 m 2 m 3 0
80. we know that,
1 1 1 1 if f (− x ) = f ( x ), then function is even and if
= − = = (given) f (− x ) = − f ( x ), then function is odd.
2 m3 3 m3 6 m3 48
1 1 (a) f ( x ) = x 3 sin x
⇒ =± ⇒ m3 = ± 8
6 m3 48 f (− x ) = (− x )3 sin(− x )
Now, if m3 = 8 = − x 3 (− sin x )
⇒ m3 = (2 )3 ⇒ m = 2 = x 3 sin x = f ( x )
If m = −8
3
So, f ( x ) is even.
⇒ m3 = (−2 )3 ⇒ m = − 2 (b) f ( x ) = x 2 cos x
78. Given equation is f (− x ) = (− x )2 cos(− x ) = x 2 cos x = f ( x )
x 2 − bx + c = 0 …(i) So, f ( x ) is even.
and roots are sin α and cos α (c) f ( x ) = e x x 3 sin x
Also sin α + cos α = b …(ii)
f (− x ) = e − x (− x )3 sin(− x )
and sin α ⋅ cos α = c …(iii)
Q sin 2 α + cos 2 α = (sin α + cos α )2 = e − x x 3 sin x ≠ f ( x )
− 2 sin α ⋅ cos α ∴ f ( x ) is not even.
⇒ 1 = b2 − 2c, [from Eqs. (ii) and (iii)] (d) f ( x ) = x − [ x ]
Q − 1 + 1 ≤ (sin α + cos α ) ≤ 1 + 1 f (− x ) = (− x ) − [− x ] ≠ f ( x )
∴ f ( x ) is not even.
⇒ − 2 ≤b≤ 2 [ from Eq. (ii)]
WB JEE (Engineering) l Solved Paper 2012 | 1
10. What is the phase difference between two This maximum acceleration of the car, for the
simple harmonic motions represented by box to remain stationary, is.
p (a) 8 ms -2 (b) 6 ms -2 (c) 4 ms -2 (d) 2 ms -2
x1 = A sinæç wt + ö÷ and x2 = A cos(wt ) ?
è 6ø
p p p 2p 16. The dimension of angular momentum is
(a) (b) (c) (d)
6 3 2 3 (a) [M0L1T-1 ] (b) [M1L2 T-2 ] (c) [M1L2 T-1 ] (d) [M2L1T-2 ]
Y N
V 2C
(a) T1 > T2 (b) T1 < T2 (c) T1 = T2 (d) T1 = 2 T2
(a) 10 V (b) 15 V (c) 20 V (d) 30 V
27. In a slide caliper, ( m + 1) number of vernier 33. A body, when fully immersed in a liquid of
divisions is equal to m number of smallest specific gravity 1.2 weighs 44 gwt. The same
main scale divisions. If d unit is the magnitude body when fully immersed in water weighs 50
of the smallest main scale division, then the gwt. The mass of the body is
magnitude of the vernier constant is
(a) 36 g (b) 48 g (c) 64 g (d) 80 g
d d
(a) unit (b) unit
(m + 1) m 34. When a certain metal surface is illuminated
md (m + 1)d with light of frequency n, the stopping potential
(c) unit (d) unit
(m + 1) m for photoelectric current is V0. When the same
n
surface is illuminated by light of frequency ,
28. From the top of a tower, 80 m high from the 2
ground, a stone is thrown in the horizontal the stopping potential
direction with a velocity of 8 ms-1. The stone V
is 0 . The threshold frequency for
reaches the ground after a time t and falls at a 4
distance of d from the foot of the tower. photoelectric, emission is
Assuming, g = 10 m/s2, the time t and distance n n 2n 4n
(a) (b) (c) (d)
d are given respectively by 6 3 3 3
4 | WB JEE (Engineering) l Solved Paper 2012
35. Three blocks of masses 4 kg, 2 kg, 1 kg 38. 22320 cal of heat is supplied to 100 g of ice at
respectively, are in contact on a frictionless 0° C. If the latent heat of fusion of ice is
table as shown in the figure. If a force of 14 N is 80 cal g -1 and latent heat of vaporisation of
applied on the 4 kg block, the contact force water is 540 cal g -1, the final amount of water
between the 4 kg and the 2 kg block will be thus obtained and its temperature
respectively are
14 N (a) 8 g, 100° C (b) 100 g, 90°C
4 kg 2 kg 1 kg
(c) 92 g, 100° C (d) 82 g, 100° C
(a) 2 N (b) 6 N
39. A progressive wave moving along x-axis is
(c) 8 N (d) 14 N
(c) æç ö÷ pA (d) æç ö÷ pA
wire will be the maximum? A 2A 3 2
(a) (b)
(a) L = 200 cm, d = 0.5 mm 3 3p è 4ø è 3ø
(b) L = 300 cm, d = 10
. mm
40. Two radioactive substances A and B have
(c) L = 50 cm, d = 0.05 mm
decay constants 5l and l, respectively. At
(d) L = 100 cm, d = 02
. mm
t = 0, they have the same number of nuclei.
37. An object placed in front of a concave mirror at a The ratio of number of nuclei of A to that of B
2
distance of x cm from the pole gives a 3 times æ1ö
will be ç ÷ after a time interval of
magnified real image. If it is moved to a distance è eø
of ( x + 5) cm, the magnification of the image 1 1 1 1
becomes 2. The focal length of the mirror is (a) (b) (c) (d)
l 2l 3l 4l
(a) 15 cm (b) 20 cm (c) 25 cm (d) 30 cm
Chemistry
1. Li occupies higher position in the (b) two unpaired electrons is p MO
*
6. The well known compounds, ( + )-lactic acid and 12. Which of the following will show a negative
( - )-lactic acid, have the same molecular deviation from Raoult's law?
formula, C 3H6O 3. The correct relationship (a) Acetone-benzene
between them is (b) Acetone-ethanol
(a) constitutional isomerism (c) Benzene-methanol
(b) geometrical isomerism (d) Acetone-chloroform
(c) identicalness
(d) optical isomerism
13. In a reversible chemical reaction at
equilibrium, if the concentration of any one of
7. The stability of Me2C==CH2 is more than that the reactants is doubled, then the equilibrium
of MeCH2CH == CH2 due to constant will
(a) inductive effect of the Me groups (a) also be doubled
(b) resonance effect of the Me groups (b) be halved
(c) hyperconjugative effect of the Me groups (c) remains the same
(d) resonance as well as inductive effect of the Me (d) becomes one-fourth
groups
14. Identify the correct statement from the
8. Which one of the following characteristics following in a chemical reaction.
belongs to an electrophile?
(a) The entropy always increases.
(a) It is any species having electron deficiency which (b) The change in entropy alongwith suitable change
reacts at an electron rich C-centre. in enthalpy decides the rate of reaction.
(b) It is any species having electron enrichment, that (c) The enthalpy always decreases.
reacts at an electron deficient C-centre. (d) Both the enthalpy and the entropy remain constant.
(c) It is cationic in nature.
(d) It is anionic in nature. 15. Which one of the following is wrong about
molecularity of a reaction?
9. Which one of the following methods is used to
prepare Me3COEt with a good yield? (a) It may be whole number or fractional.
(b) It is calculated from reaction mechanism
(a) Mixing EtONa with Me 3 CCl (c) It is the number of molecules of the reactants
(b) Mixing Me 3 CONa with EtCl taking part in a single step chemical reaction.
(c) Heating a mixture of (1 : 1) EtOH and Me 3 COH in (d) It is always equal to the order of elementary
the presence of conc. H 2 SO4 . reaction.
(d) Treatment of Me 3 COH with EtMgI
16. Which of the following does not represent the
10. 58.4g of NaCl and 180g of glucose were mathematical expression for the Heisenberg
separately dissolved in 1000 mL of water. uncertainty principle?
Identify the correct statement regarding the
(a) Dx × DP ³ h / (4p ) (b) Dx × Dv ³ h / (4pm)
elevation of boiling point (bp) of the resulting
(c) DE × Dt ³ h / (4p ) (d) DE × Dx ³ h / (4p )
solutions.
(a) NaCl solution will show higher elevation of boiling 17. The stable bivalency of Pb and trivalency of
point. Bi is
(b) Glucose solution will show higher elevation of (a) due to d contraction in Pb and Bi
boiling point. (b) due to relativistic contraction of the 6s-orbitals of
(c) Both the solutions will show equal elevation of Pb and Bi, leading to inert pair effect
boiling point. (c) due to screening effect
(d) The boiling point elevation will be shown by neither (d) due to attainment of noble liquid configuration
of the solutions.
18. The equivalent weight of K 2Cr2O7 in acidic
11. Equal weights of CH4 and H2 are mixed in an medium is expressed in terms of its molecular
empty container at 25°C. The fraction of the weight (M) as
total pressure exerted by H2 is
(a) M/3 (b) M/4
(a) 1/9 (b) 1/2 (c) 8/9 (d) 16/17 (c) M/6 (d) M/7
6 | WB JEE (Engineering) l Solved Paper 2012
19. Which of the following is correct? 27. Which one of the following properties is
(a) Radius of Ca 2+
< Cl < S - 2- exhibited by phenol?
(b) Radius of Cl - < S2 - < Ca2 + (a) It is soluble in aq. NaOH and evolves CO2 with
(c) Radius of S2 - = Cl - = Ca2 + aq. NaHCO3 .
(d) Radius of S2 - < Cl - < Ca2 + (b) It is soluble in aq. NaOH and does not evolve CO2
with aq. NaHCO3 .
20. CO is practically non-polar since (c) It is not soluble in aq. NaOH but evolves CO2 with
(a) the s-electron drift from C to O is almost nullified aq. NaHCO3 .
by the p-electron drift from O to C (d) It is insoluble in aq. NaOH and does not evolve
(b) the s-electron drift from O to C is almost nullified CO2 with aq. NaHCO3 .
by the p-electron drift from C to O
(c) the bond moment is low
28. The basicity of aniline is weaker in comparison
(d) there is a triple bond between C and 0 to that of methyl amine due to
(a) hyperconjugative effect of Me-group in MeNH 2
21. The number of acidic protons in H3PO 3 are (b) resonance effect of phenyl group in aniline
(a) 0 (b) 1 (c) 2 (d) 3 (c) lower molecular weight of methyl amine as
compared to that of aniline
22. When H2O 2 is shaken with an acidified (d) resonance effect of ¾ NH 2 group in MeNH 2
solution of K 2Cr2O7 in the presence of ether, the
ethereal layer turns blue due to the formation 29. Under identical conditions, the SN 1 reaction
of will occur most efficiently with
34. By passing excess Cl 2( g ) in boiling toluene, 37. Which of the following is correct?
which one of the following compounds is (a) Evaporation of water causes an increase in
exclusively formed? disorder of the system.
Me Me (b) Melting of ice causes a decrease in randomness
Cl Cl of the system.
Cl (c) Condensation of steam causes an increase in
(a) (b)
disorder of the system.
Cl (d) There is practically no change in the randomness
Cl Cl of the system, when water is evaporated.
CCl3 CCl3
Cl 38. On passing C ampere of current for time t sec
Cl through 1 L of 2 (M) CuSO 4 solution
(c) (d) (Atomic weight of Cu = 63.5), the amount m of
Cu (in gram) deposited on cathode will be
Cl (a) m = Ct / (63.5 ´ 96500)
(b) m = Ct / (3125
. ´ 96500)
35. An equimolar mixture of toluene and
(c) m = (C ´ 96500) / (3125
. ´t )
chlorobenzene is treated with a mixture of
(d) m = (3125
. ´ C ´ t ) / 96500
conc. H2SO 4 and conc. HNO 3. Indicate the
correct statement from the following.
(a) p-nitrotoluene is formed in excess. 39. If the first ionisation energy of H atom is 13.6
(b) Equimolar amounts of p-nitrotoluene and eV, then the second ionisation energy of He
p-nitrochlorobenzene are formed. atom is
(c) p-nitrochlorobenzene is formed in excess. (a) 27.2 eV (b) 40.8 eV
(d) m-nitrochlorobenzene is formed in excess. (c) 54.4 eV (d) 108.8 eV
36. Among the following carbocations 40. The weight of oxalic acid that will be required
Ph 2C +CH2Me (I), PhCH 2CH 2CH + Ph (II), to prepare a 1000 mL (N/20) solution is
Ph 2 CHCH + Me (III) and Ph 2C(Me)CH2 (IV),
the order of stability is (a) 126/100 g (b) 63/40 g
(c) 63/20 g (d) 126/20 g
(a) IV > II > I > III (b) I > II > III > IV
(c) II > I > IV > III (d) I > IV > III > II
Mathematics
1. If ( a + b ) and ( a - b ) are the roots of the (a) 0 (b) 1
equation x 2 + px + q = 0, where a, b, p and q (c) 2 (d) 3
are real, then the roots of the equation
( p2 - 4q)( p2x 2 + 4 px ) - 16q = 0 are
3. The sum of the series
1n 1 1
æ1 1 ö æ1 1 ö 1+ C1 + nC2 + nCn is equal to
(a) çç + ÷÷ and çç - ÷ n+1
b ÷ø
a a
2 3
è b ø è
æ 1 1ö æ 1 1ö 2n + 1 - 1 3(2 n - 1)
(b) ç + ÷ and ç - ÷ (a) (b)
è a bø è a bø n+1 2n
æ 1 1 ö æ 1 1 ö 2n + 1 2n + 1
(c) çç + ÷ and çç - ÷÷ (c) (d)
è a b ÷ø è a b ø n+1 2n
(d) ( a + b ) and ( a - b ) ¥ 1 + 2 + K + ( r - 1)
4. The value of S
2. The number of solutions of the equation r = `2 r!
log 2( x 2 + 2x - 1) = 1 is e 3e
(a) e (b) 2e (c) (d)
2 2
8 | WB JEE (Engineering) l Solved Paper 2012
5. If P = éê
1 2 1ù 13. Two coins are available, one fair and the other
, Q = PP T , then the value of the
ë1 3 1úû two-headed. Choose a coin and toss it once;
determinant of Q is assume that the unbiased coin is chosen with
3
(a) 2 (b) -2 (c) 1 (d) 0 probability . Given that the outcome is head,
4
6. The remainder obtained when 1 ! + 2 ! K + 95 ! the probability that the two-headed coin was
is divided by 15 is chosen, is
(a) 14 (b) 3 (c) 1 (d) 0 3 2
(a) (b)
5 5
7. If P, Q and R are angles of DPQR, then the (c)
1
(d)
2
-1 cos R cos Q 5 7
value of cos R -1 cos P is equal to 14. Let R be the set of real numbers and the
cos Q cos P -1 functions f : R ® R and g : R ® R be defined
1 by f ( x ) = x 2 + 2x - 3 and g( x ) = x + 1. Then, the
(a) -1 (b) 0 (c) (d) 1 value of x for which f ( g( x )) = g( f ( x )) is
2
(a) -1 (b) 0
8. The number of real values of a for which the (c) 1 (d) 2
system of equations
x + 3 y + 5 z = ax 15. If a, b and c are in arithmetic progression, then
the roots of the equation ax 2 - 2bx + c = 0 are
5x + y + 3 z = ay
c 1
3 x + 5 y + z = az (a) 1 and (b) - and -c
a a
has infinite number of solutions is c c
(a) 1 (b) 2 (c) 4 (d) 6 (c) -1and - (d) -2 and -
a 2a
9. The total number of injections (one one into 16. The equation y2 + 4x + 4 y + k = 0 represents a
mappings) from { a1, a 2, a 3, a 4 } to parabola whose latusrectum is
{ b1, b2, b3, b4, b5, b6, b7 } is
(a) 1 (b) 2
(a) 400 (b) 420 (c) 800 (d) 840 (c) 3 (d) 4
10
10. Let (1 + x )10 = S cr xr and 17. If the circles x 2 + y2 + 2x + 2ky + 6 = 0 and
r =0
7 5 x 2 + y2 + 2ky + k = 0 intersect orthogonally,
(1 + x )7 = S dr xr . If P = S c2r and then k is equal to
r =0 r =0
3 P 3 3
Q = S d2r + 1, then is equal to (a) 2 or - (b) -2 or -
r =0 Q 2 2
3 3
(a) 4 (b) 8 (c) 16 (d) 32 (c) 2 or (d) -2 or
2 2
11. Two decks of playing cards are well shuffled 18. If four distinct points (2k, 3 k), (2, 0), (0, 3),
and 26 cards are randomly distributed to a (0, 0) lie on a circle, then
player. Then, the probability that the player
(a) k < 0 (b) 0 < k < 1
gets all distinct cards is
(c) k = 1 (d) k >1
(a) 52
C 26 / 104
C 26 (b) 2 ´ 52C 26 / 104
C 26
(c) 2 13
´ C 26 /
52 104
C 26 (d) 2 26
´ C 26 /
52 104
C 26 19. The line joining a( b cos a , b sin a ) and
B ( a cos b, a sin b ), where a ¹ b, is produced to
12. An urn contains 8 red and 5 white balls. Three the point M ( x, y) so that AM : MB = b : a.
balls are drawn at random. Then, the a +b a +b
then, x cos + y sin is equal to
probability that balls of both colours are drawn 2 2
is
(a) 0 (b) 1
40 70 3 10
(a) (b) (c) (d) (c) -1 (d) a2 + b 2
143 143 13 13
WB JEE (Engineering) l Solved Paper 2012 | 9
x2 28. Maximum value of the function f ( x ) =
x 2
+ on
20. Let the foci of the ellipse + y2 = 1 subtend a
9 8 x
right angle at a point P. Then, the locus of P is the interval [1, 6] is
(a) x + y = 1
2 2
(b) x + y = 2
2 2 9 13 17
(a) 1 (b) (c) (d)
8 12 8
(c) x + y = 4
2 2
(d) x + y = 8
2 2
p 3p
29. For - < x < , the value of
21. The general solution of the differential 2 2
dy x + y+1 d ì -1 cos x ü
equation = is ítan ý is equal to
dx 2x + 2 y + 1 dx î 1 + sin x þ
where the first digit is not zero. Then, the total (a) for no value of a
number of vehicles with distinct registration (b) for exactly one value of a
numbers is (c) for exactly two values of a
(a) 262 ´ 104 (b) 26
P2 ´ 10 P4 (d) for exactly three values of a
(c) 26
P2 ´ 9 ´ 10
P3 (d) 262 ´ 9 ´ 103 45. If 64, 27, 36 are the Pth Qth and Rth terms of a
GP, then P + 2Q is equal to
36. The number of words that can be written using
all the letters of the word ‘IRRATIONAL’ is (a) R (b) 2R
10! 10! 10! (c) 3R (d) 4R
(a) (b) (c) (d) 10! 3p
(2 !)3 (2 !)2 2! 46. If sin -1 x + sin -1 y + sin -1 z = , then the
2
37. Four speakers will address a meeting where 1
value of x 9 + y9 + z 9 - is equal to
speaker Q will always speak P. Then, the x 9 y9 z 9
number of ways in which the order of speakers
(a) 0 (b) 1 (c) 2 (d) 3
can be prepared is
(a) 256 (b) 128 (c) 24 (d) 12 47. Let p, q and r be the sides opposite to the
angles P, Q and R, respectively in a DPQR. If
38. The number of diagonals in a regular polygon r 2 sin P sin Q = pq, then the triangle is
of 100 sides is
(a) equilateral
(a) 4950 (b) 4850 (c) 4750 (d) 4650 (b) acute angled but not equilateral
(c) obtuse angled
39. Let the coefficients of powers of x in the 2nd,
(d) right angled
3rd and 4th terms in the expansion of (1 + x )n ,
where n is a positive integer, be in arithmetic 48. Letp, q andr be the sides opposite to the
progression. Then, the sum of the coefficients
angles P, Q and R respectively in a DPQR
of odd powers of x in the expansion is
æ P - Q + Rö
(a) 32 (b) 64 (c) 128 (d) 256 Then,2 pr sin ç ÷ equals
è 2 ø
40. Let f ( x ) = ax + bx + c, g( x ) = px 2 + qx + r
2
(a) p2 + q 2 + r 2 (b) p2 + r 2 - q 2
such that f (1) = g(1), f (2) = g(2) and
f (3 ) - g(3 ) = 2. Then f (4) - g(4) is (c) q 2 + r 2 - p2 (d) p2 + q 2 - r 2
(a) 4 (b) 5 (c) 6 (d) 7 49. Let P(2, - 3 ), Q ( -2, 1) be the vertices of the
DPQR. If the centroid of DPQR lies on the line
41. The sum 1 ´ 1 ! + 2 ´ 2 ! + K + 50 ´ 50 ! equals
2x + 3 y = 1, then the locus of R is
(a) 51! (b) 51! - 1
(a) 2 x + 3 y = 9 (b) 2 x - 3 y = 7
(c) 51! + 1 (d) 2 ´ 51!
(c) 3 x + 2 y = 5 (d) 3 x - 2 y = 5
42. Six numbers are in AP such that their sum is 3. px - 1
The first term is 4 times the third term. Then, 50. lim
the fifth term is
x ®0 1 + x -1
(a) -15 (b) -3 (c) 9 (d) -4 (a) does not exist (b) equals loge (p 2 )
(c) equals 1 (d) lies between 10 and 11
43. The sum of the infinite series
1 1 ×3 1 ×3 ×5 1 ×3 ×5 × 7 51. If f is a real-valued differentiable function
1+ + + + +K such that f ( x )f ¢ ( x ) < 0 for all real x, then
3 3 × 6 3 × 6 × 9 3 × 6 × 9 × 12
is equal to (a) f( x) must be an increasing function
(b) f( x) must be a decreasing function
3 1
(a) 2 (b) 3 (c) (d) (c)|f( x)| must be an increasing function
2 3 (d)|f( x)| must be a decreasing function
WB JEE (Engineering) l Solved Paper 2012 | 11
52. Rolle’s theorem is applicable in the interval 59. The number of integer values of m , for which
[ -2, 2] for the function the x-coordinate of the point of intersection of
(a) f( x) = x3 (b) f( x) = 4 x4 the lines 3 x + 4 y = 9 and y = mx + 1 is also an
integer, is
(c) f( x) = 2 x3 + 3 (d) f( x) = p| x|
(a) 0 (b) 2
2
d y dy (c) 4 (d) 1
53. The solution of 25 - 10 + y = 0, y(0) = 1,
dx 2 dx
60. If a straight line passes through the point ( a, b)
y(1) = 2e1/ 5 is
and the portion of the line intercepted between
(a) y = e 5 x + e -5 x (b) y = (1 + x)e 5 x the axes is divided equally at that point, then
(c) y = (1 + x)e x / 5 (d) y = (1 + x)e - x / 5 x y
+ is
a b
54. Let P be the mid point of a chord joining the (a) 0 (b) 1 (c) 2 (d) 4
vertex of the parabola y2 = 8x to another point
10
on it. Then, the locus of P is 61. The coefficient of x in the expansion of
(a) y = 2 x
2
(b) y = 4 x
2 1 + (1 + x ) + K + (1 + x )20 is
x2 y2 (a) 19 C 9 (b) 20
C10
(c) + y2 = 1 (d) x2 + =1 21 22
4 4 (c) C11 (d) C12
55. The line x = 2 y intersects the ellipse 62. The system of linear equations :
x2 lx + y + z = 3
+ y = 1 at the points P and Q. The equation
2
4
x - y - 2z = 6
of the circle with PQ as diameter is
1 - x + y + z = m has
(a) x2 + y2 = (b) x2 + y2 = 1
2 (a) infinite number of solutions for l ¹ -1and all m
5
(c) x2 + y2 = 2 (d) x2 + y2 = (b) infinite number of solutions for l = - 1 and m = 3
2 (c) no solution for l ¹ -1
56. The eccentric angle in the first quadrant of a (d) unique solution for l = - 1and m = 3
x 2 y2 63. Let A and B be two events with P( A C ) = 0.3,
point on the ellipse + = 1 at a distance 3
10 8 P( B) = 0.4 and P( A Ç BC ) = 0.5. Then,
units from the centre of the ellipse is P( B| A È BC ) is equal to
p p p p 1 1 1 2
(a) (b) (c) (d) (a) (b) (c) (d)
6 4 3 2 4 3 2 3
57. The transverse axis of a hyperbola is along the 64. Let p, q and r be the altitudes of a triangle with
X-axis and its length is 2a. The vertex of the area S and perimeter 2 t. Then, the
hyperbola bisects the line segment joining the
1 1 1
centre and the focus. The equation of the value of + + is
hyperbola is p q r
(a) 6 x2 - y2 = 3a2 (b) x2 - 3 y2 = 3a2 S t S 2S
(a) (b) (c) (d)
(c) x - 6 y = 3a
2 2 2
(d) 3 x - y = 3a
2 2 2 t S 2t t
58. A point moves in such a way that the difference 65. Let C1 and C2 denote the centres of the circles
of its distance from two points x 2 + y2 = 4 and ( x - 2)2 + y2 = 1 respectively
(8, 0) and (-8, 0) always remains 4. Then, the and let P and Q be their points of
locus of the point is intersection.Then, the areas of DC1PQ and
D C2PQ are in the ratio
(a) a circle (b) a parabola
(c) an ellipse (d) a hyperbola (a) 3 : 1 (b) 5 : 1
(c) 7 : 1 (d) 9 : 1
12 | WB JEE (Engineering) l Solved Paper 2012
66. A straight line through the point of 73. The value of the integral
intersection of the lines x + 2 y = 4 and 5
2x + y = 4 meets the coordinate axes at A and B. ò1 [|x - 3| + |1 - x|] dx is equal to
The locus of the mid-point of AB is
(a) 4 (b) 8
(a) 3( x + y) = 2 xy (b) 2( x + y) = 3 xy (c) 12 (d) 16
(c) 2( x + y) = xy (d) x + y = 3 xy
74. If f ( x ) and g( x ) are twice differentiable
67. Let P and Q be the points on the parabola functions on (0, 3) satisfying
y2 = 4x, so that the line segment PQ subtends f ¢ ¢ ( x ) = g ¢ ¢ ( x ), f ¢ (1) = 4, g¢ (1) = 6 , f (2) = 3,
right angle at the vertex. If PQ intersects the g(2) = 9, then f (1) - g(1) is
axis of the parabola at R, then the distance of
(a) 4 (b) -4
the vertex from R is
(c) 0 (d) -2
(a) 1 (b) 2 (c) 4 (d) 6
75. Let [ x ] denote the greatest integer less than or
68. The incentre of an equilateral triangle is (1, 1) equal to x, then the value of the integral
and the equation of one side is 3 x + 4 y + 3 = 0. 1
Then, the equation of the circumcircle of the ò-1(| x| - 2[ x])dx is equal to
triangle is (a) 3 (b) 2
(a) x2 + y2 - 2 x - 2 y - 2 = 0 (c) -2 (d) -3
(b) x2 + y2 - 2 x - 2 y - 14 = 0 76. The points representing the complex number z
(c) x2 + y2 - 2 x - 2 y + 2 = 0 æ z -2ö p
for which arg ç ÷
ç z + 2÷ = 3
lie on
(d) x + y - 2 x - 2 y + 14 = 0
2 2
è ø
1 (a) a circle (b) a straight line
( n !) n (c) an ellipse (d) a parabola
69. The value of lim is
n ®¥ n
77. Let a, b, c, p, q and r be positive real numbers
1 1 1
(a) 1 (b) (c) (d) such that a, b and c are in GP andq p = bq = cr .
e2 2e e
Then,
70. The area of the region bounded by the curves (a) p, q , r are in GP (b) p, q , r are in AP
1 (d) p2 , q 2 , r 2 are in AP
y = x 3, y = , x = 2 is (c) p, q , r are in HP
x
1
(a) 4 - loge 2 (b) + loge 2 78. Let Sk be the sum of an infinite GP series
4
15 whose first term is k and common ratio is
(c) 3 - loge 2 (d) - loge 2 k ¥ ( - 1 )k
4 ( k > 0). Then, the value of S is
k+1 k =1 S
k
71. Let y be the solution of the differential
dy y2 equal to
equation x = satisfying y(1) = 1.
dx 1 - y log x (a) loge 4 (b) loge 2 - 1
(c) 1 - loge 2 (d) 1 - loge 4
Then, y satisfies
-1 79. The quadratic equation
(a) y = xy (b) y = xy
(c) y = x y +1
(d) y = x y + 2 2x 2 - ( a 3 + 8a - 1)x + a 2 - 4a = 0 possesses
roots of opposite sign. Then,
72. The area of the region, bounded by the curves (a) a £ 0 (b) 0 < a < 4
y = sin -1 x + x(1 - x ) and (c) 4 £ a < 8 (d) a ³ 8
y = sin-1 x - x(1 - x) in the first quadrant, is 80. If log e( x 2 - 16) £ log e(4x -`11), then
1 1 1
(a) 1 (b) (c) (d) (a) 4 < x £ 5 (b) x < -4 or x > 4
2 3 4
(c) -1 £ x £ 5 (d) x < -1or x > 5
WB JEE (Engineering) l Solved Paper 2012 | 13
Answers
Physics
1. (d) 2. (a) 3. (c) 4. (a) 5. (a) 6. (a) 7. (d) 8. (c) 9. (d) 10. (b)
11. (a) 12. (d) 13. (b) 14. (b) 15. (d) 16. (c) 17. (a) 18. (b) 19. (c) 20. (c)
21. (b) 22. (b) 23. (a) 24. (c) 25. (d) 26. (b) 27. (a) 28. (c) 29. (d) 30. (c)
31. (b) 32. (d) 33. (d) 34. (b) 35. (b) 36. (c) 37. (d) 38. (a) 39. (d) 40. (b)
Chemistry
1. (a) 2. (d) 3. (a) 4. (c) 5. (c) 6. (d) 7. (c) 8. (a) 9. (b) 10. (a)
11. (c) 12. (d) 13. (c) 14. (b) 15. (a) 16. (d) 17. (b) 18. (c) 19. (a) 20. (a)
21. (c) 22. (d) 23. (c) 24. (d) 25. (a) 26. (d) 27. (b) 28. (b) 29. (a) 30. (d)
31. (b) 32. (a) 33. (b) 34. (d) 35. (a) 36. (b) 37. (a) 38. (d) 39. (c) 40. (c)
Mathematics
1. (a) 2. (c) 3. (a) 4. (c) 5. (a) 6. (b) 7. (b) 8. (a) 9. (d) 10. (b)
11. (d) 12. (d) 13. (b) 14. (a) 15. (a) 16. (d) 17. (a) 18. (c) 19. (a) 20. (d)
21. (d) 22. (d) 23. (d) 24. (d) 25. (c) 26. (d) 27. (a) 28. (d) 29. (b) 30. (c)
31. (c) 32. (d) 33. (b) 34. (c) 35. (d) 36. (a) 37. (d) 38. (b) 39. (b) 40. (c)
41. (b) 42. (d) 43. (b) 44. (b) 45. (c) 46. (c) 47. (d) 48. (b) 49. (a) 50. (b)
51. (d) 52. (b) 53. (c) 54. (b) 55. (d) 56. (b) 57. (d) 58. (d) 59. (b) 60. (c)
61. (c) 62. (b) 63. (a) 64. (b) 65. (c) 66. (b) 67. (c) 68. (b) 69. (d) 70. (b)
71. (b) 72. (c) 73. (c) 74. (b) 75. (a) 76. (a) 77. (c) 78. (d) 79. (b) 80. (c)
e = 22 V X
O A (a, 0)
3. The velocity of water below, which the flow
remains streamline flow is known as critical Q
W = qDV and V =
velocity. 4pe0 r
4. Given, v = 3 ´ 10 8 ms -1, d = 10 m and m = 15
. æ Q Q ö
Hence, W = q çç - ÷÷
d 10 ´ 10 -2 è 4pe0 b 4pe0 a ø
Here, t = =
c 2 ´ 10 8 Qq æ a - b ö
m = ç ÷
4pe0 è ab ø
14 | WB JEE (Engineering) l Solved Paper 2012
2
6. From KCL, we have æV ö
(i) ç ÷ 2
6 kW 4 kW è 3ø = V
So, H¢ =
2R 18R
H
72 V 3 kW 2 kW H¢ =
18
12. Given, ZA : ZB = 1 : 2
72
i= = 9 ´ 10 -3 A We know that,
8 ´ 10 3
n µ Z2
i ´ 6 = (9 - i ) ´ 3
Hence,
i = 3 mA
n A (1)2
7. Given, E = (2 $i + 3$j + k$ )NC -1 =
n B (2 )2
and S = 10 $i m2 nA : nB = 1: 4
We know that,
13. de-Broglie wavelength,
f = E×S h
l=
f = (2 $i + 3$j + k$ )× (10 $i ) mv
f = 20Nm 2C -1 Here,
-1 -1 h h
8. Given, v = 340ms , u s = 20ms and le = and l p =
c mpc
v0 = 640 Hz me
2
From Doppler’s law, Given, le = l p
æ 340 ö h h
n = çç ÷÷ 640 So, =
è 340 - 20 ø me
c mpc
n = 680 Hz 2
me
9. Given, i = 10 A, B = 0.15 T, q = 45° and l = 2 m =2
mp
Here,
Ratio of KE,
F = ilB sin q = 10 ´ 2 ´ 0.15 sin 45°
1
3 me ve2
= N Ke 2 1
2 = =
Kp 1 2 2
mp vp
10. Given, 2
æ pö s s
x1 = A sin ç wt + ÷ 14. E = +
è 6ø 2 e0 2 e0
æ pö s
x2 = A cos(wt ) = A cos t ç wt + ÷ E = , towards the negatively charged plane.
è 2ø e0
Phase difference, 15. Given, m = 2 kg, m = 0.2 and g = 10ms -2
Df = f2 - f1
Here, ma = m mg
p p 3p - p p
Df = - = = Þ a=m g
2 6 6 3
Þ a = 0. 2 ´ 10
V2
11. We know that, H = a = 2 m /s2
R
V 16. Dimension of angular momentum,
Here, in second condition V¢ becomes and R¢
3 L = r ´p
becomes 2R. = [L][MLT -1 ] = [ML2 T -1 ]
WB JEE (Engineering) l Solved Paper 2012 | 15
v2
17. Here, triangle is 23.
30º
q 30º
4 kg 1 kg
A v1
C
Along, Y-axis momentum remains zero.
Here,
B
4v1 sin 30 ° = v2 sin 60 °
æ| C| ö
÷÷ = cos -1 æç ö÷
3
q = cos -1 çç v1
=
3
è| A| ø è 5ø v2 4
18. Given, x = 37 + 27t - t 3 24. Given, binary number = 1011001
dx
v= = 27 - 3 t 2 Its equivalent decimal number
dt
= 2° + 0 + 0 + 23 + 24 + 26
According to problem,
v = 0 Þ 27 - 3 t 2 = 0 = 1 + 8 + 16 + 64 = 89
Here, t = 3s (2 n - 1)
25. Here, v1 = v
x = 37 + 27 ´ 3 - (3)2 = 91 m 4l1
n
1
19. At ground, E = mu 2 v2 = v
3 2l2
1 Hence, v1 = v2
At highest point, E ¢ = m (u cos 60 ° )2
2 l1 3
E =
E¢ = l2 4
4
26. R increases with temperature and slope of V-I
20. Given, s = 30 cm = 30 ´ 10 -2 m and loss in graph gives resistance, so T1 < T2 .
velocity is 50%
27. (m + 1) vernier division = m main scale division
Equation of motion
v 2 = u 2 + 2 as æ m ö
One division on vernier scale = çç ÷÷ division
u2 è m + 1ø
Here, = u 2 + 2 a ´ 30 ´ 10 -2 …(i)
4 on main scale
u2 æ m ö d
0= + 3a ´ x …(ii) Vernier constant = çç1 - ÷÷d = unit
4 è m + 1 ø m +1
On solving Eqs. (i) and (ii), we get
28. Given, h = 80 m, v = 8ms -1 and g = 10 ms -2
x = 10 cm
1 2 1
21. For 1st situation, Here, h= gt Þ 80 = ´ 10 ´t 2
2 2
1
E= k(10 ´ 10 -2 )2 …(i) t = 4s
2
\ x = vt = 8 ´ 4 = 32 m
For 2nd situation
1 29. In balanced Wheatstone bridge,
E ¢ = K (20 ´ 10 -2 )2 …(ii) 1 1 1
2 = +
On comparing Eqs. i and ii, we get 10 x 30
E ¢ = 4E x = 15W
22. By Kepler’s law, 30. We know that ,
T 2 µ R3 V2
R=
3/ 2 P
æL ö
Hence, T = D çç 2 ÷÷ Given, V = 200 V, P = 50 W
è L1 ø
and V ¢ = 100V
16 | WB JEE (Engineering) l Solved Paper 2012
V 2 200 ´ 200 14 - N = 8
Hence, R= =
P 50 N = 6N
V ¢2 100 ´ 100 ´ 50
P¢ = = = 12.5 W 36. We know that,
R 200 ´ 200 F ´L 4L
DL = =
31. Given, work done = W and q = 60 ° A ´ Y pd 2
We know that, 4
Here, is a constant.
W = MB(1 - cos q) p
MB L
W = MB(1 - cos 60 ° ) = Hence, DL µ 2
2 d
Hence,| t| = MB sin 60 ° = 3W 37. Given, v1 = 3 x and v2 = 2( x + 5)
32. Capacitors in series, In Ist case,
2C1 ´ V1 = C ´ V2 1 1 1
+ = …(i)
2V1 = V2 …(i) - x -3 x f
1 1 1
and V1 + V2 + V1 = 60 …(ii) + = …(ii)
-( x + 5) -2( x + 5) f
Solving Eqs. (i) and (ii), we get
V2 = 30 V On solving Eqs. (i) and (ii), we get f = 30 cm
33. Balancing equation, 38. Heat required to convert ice to water at 100°C,
W = mg - Vrg Q = m ´ L + msDT = 18000 cal
Hence, 44 = m - 12.V Amount of heat left = 4320 cal
50 = m - V m ´ L = 4320
On solving Eqs. (i) and (ii), we get m = 8 g steam
m = 80 g é 2p ù
39. y = A sin ê (vt - x )ú
34. In Ist case, ë l û
hn = hn 0 + eV0 …(i) 2p
v= VA
hn eV l
= hn 0 + 0 …(ii)
2 4 (vp )max = 3v
Solving Eqs. (i) and (ii), we get 2p
l= A
n 3
V0 =
3 40. We know that,
35. We know that, F = ma N = N0e -lt
F 14
a= = = 2ms -2 For A,
m 7 NA = N0e -5 lt
For B,
NB = N0e - lt
N 14 N N 1
Given, A = 2
NB e
N 1
=
NB e 4 lt
4 kg
1
Hence, from the figure t =
2l
14 - N = 4a
WB JEE (Engineering) l Solved Paper 2012 | 17
Chemistry
1. In the electrochemical series, metals are 8. Electrophiles are electron deficient species
arranged in increasing order of their standard (neutral or cationic). They attack on electron rich
reduction potential. The standard reduction C-atom.
potentials of Li + /Li and Cu 2+ /Cu are -3.05 V and
EtCl
+ 0.34 V respectively. Therefore, Li occupies 9. Me3CONa ¾¾¾® Me3COEt
higher position in series as compared to Cu. (1 °alkyl halide)
- NaCl
2. 11 Na
24
¾® 12 Mg
24
+ -1b
0
(Stable)
This reaction is known as Williamson’s synthesis.
P ··
H OH Acidic ; CH3NH2
Acidic H proton
Methyl amine
proton Aniline
The lone pair of electrons, present in N-atom in 35. CH3 CH3 CH3
aniline involve in ring resonance. So, it is less HNO3 NO2
/H2SO4
basic than methyl amine. +
29. SN 1reaction is most effective in 3° carbon. Toluene o-nitro toluene
NO2
p-nitrotoluene
(CH3 )3 ¾ C ¾ Cl CH3CH2CH2CH2Cl (excess)
tert -butyl chloride(3 °) 1-chlorobutane(1 °)
Cl Cl Cl
Cl ¾ CH2 ¾ CH(CH3 )¾ CH3 HNO3 NO2
2-methyl -1-chloropropane(1 °) /H2SO4
+
CH3CH2CH(Cl)CH3
o-nitro chloro
2-chloro butane (2 °) NO2
benzene
p-nitro chloro
30. O C O + (CH3 )3 ––C––MgI benzene
Dry ice (Very low yield)
C(CH3 )3
As toluene and chlorobenzene, both can react
HOH
O C––OMgI with mixture of conc. H2SO 4 and conc. HNO 3 ,
OH but being more reactive, toluene will yield
(CH3 )3 C––COOH + Mg
I p-nitrotoluene in excess.
36. The order of stability of carbocation is 3° benzylic
31. CH3COOH + NaOH
> 2° benzylic > 2° > 1°.
Initial 20 ´ 0.1 = 2 mmol 10 ´ 0.1 = 1 mmol Thus, the correct order of stability will be
At time t( 2 - 1) = 1 mmol ( 1 - 1) = 0 mmol Ph H
¾® CH3COONa + H2O ½ ½
From Henderson’s equation 1m mol Ph ¾ C ¾ CH2CH3 > PhCH2CH2 ¾ C ¾ Ph
½ ½
[CH3COONa] + +
pH = pK a + log (I) (II)
[CH3COOH]
1 H Me
= 474
. + log = 474 . ½ ½
1
> Ph 2CH ¾ C + > Ph ¾ C ¾ CH2
32. In Fe complex, NO behaves as NO + ½ ½
O H O Me Ph
33. (III) (IV)
CH3 C C C OC2H5
37. Water evaporates into steam. In this physical
H state its disorder or randomness increases.
O H O
63.5
CH3 C C C OC2H5 ´C ´t
ECt
38. m = = 2
H F 96500
34. CH3 CH2Cl . ´C ´t
3175
=
96500
Cl2 Cl2
(excess) (excess) 22
39. Second ionisation energy = 13.6 ´ = 54.4 eV
12
CHCl2 CCl3
NE V 1 63 ´ 1000 63
40. w = = ´ = g
Cl2 1000 20 1000 20
(excess)
20 | WB JEE (Engineering) l Solved Paper 2012
Mathematics
1. Since, (a + b ) and (a - b ) are the roots of (n + 1) n(n - 1) ù
+ + ....+ 1ú
the equation x + px + q = 0
2 3! û
1 n+1
\ Sum of roots = - p = [ C1 + n + 1C2 + K n + 1Cn + 1 ]
n+1
Þ (a + b ) + (a - b ) = - p 1
p = [ n + 1C0 + n + 1C1 + n + 1C2 + ...+ n + 1Cn + 1 - 1]
Þ 2a = - p Þ a = - n+1
2 [n + 1C0 = 1]
and product of roots = q 1
(a + b )(a - b ) = q = (2 n + 1 - 1)
n+1
a2 - b = q ¥ (r - 1)r ¥ 1
Þ 4. S = S
r = 2 2r ! r = 2 2(r - 2 )!
2
æ pö p2
b = a2 - q = ç- ÷ - q = -q 1é 1 1 1 ù 1
è 2ø 4 = ê + + + K ¥ú = e
2 ë 0 ! 1! 2 ! û 2
Þ p2 - 4q = 4b
5. Given, P = éê ù
\The given equation 1 2 1
( p2 - 4 q ) ( p2 x 2 + 4 px ) - 16q = 0 ë1 3 1úû
é 1 1ù
4 b( p2 x 2 + 4 px ) - 16(a 2 - b ) = 0 é1 2 1ù ê
\ Q = PPT = ê ú 2 3ú
ë1 3 1û ê 1 1ú
[a 2 - b = q ] ë û
b(4 a 2 x 2 - 8 ax ) - 4(a 2 - b ) = 0 é1 ´ 1 + 2 ´ 2 + 1 ´ 1 1 ´ 1 + 2 ´ 3 + 1 ´ 1ù
=ê
a 2bx 2 - 2 abx + b = a 2 ë1 ´ 1 + 3 ´ 2 + 1 ´ 1 1 ´ 1 + 3 ´ 3 + 1 ´ 1úû
é1 + 4 + 1 1 + 6 + 1ù é 6 8 ù
Þ (ax b - b )2 = a 2 =ê =
ë1 + 6 + 1 1 + 9 + 1ûú ëê 8 11ûú
Þ ax b - b = ± a é6 8 ù
\The determinant of Q = ê = 66 - 64 = 2
1 1 ë 8 11úû
\ x= ±
a b 6. Since, 1! = 1, 2 ! = 2, 3 ! = 6, 4 ! = 24, 5 ! = 120 Since, all
æ1 1 ö æ1 1 ö terms from 5! onwards are divisible by 15 and
Þ x = çç + ÷ and ç -
÷ ç
÷
èa bø èa b ÷ø 1! + 2 ! + 3 ! + 4 ! = 33
\The required remainder after dividing by 15 will be 3.
2. Given, log 2 ( x 2 + 2 x - 1) = 1
-1 cos R cos Q
Þ log 2 ( x 2 + 2 x - 1) = log 2 2 7. Let D = cos R -1 cos P
Þ x2 + 2 x - 1 = 2 cos Q cos P -1
Þ x2 + 2 x - 3 = 0 Multiplying C1 by p and then doing
Þ x + 3x - x - 3 = 0
2 C1 ® C1 + q C2 + rC3 , we get
-p cos R cos Q
Þ x( x + 3) - 1( x + 3) = 0 D=
1
p cos R -1 cos P
Þ ( x + 3)( x - 1) = 0 p p cos Q cos P -1
Þ x = 1, - 3 - p + q cos R + r cos Q cos R cos Q
Since, x = 1 and x = - 3 are satisfy the given 1
= p cos R - q + r cos P -1 cos P
equation therefore the number of solutions of p p cos Q + q cos P - r cos P -1
the equation are two.
0 cos R cos Q
1 1 1 n 1
3. 1 + nC1 + nC2 + K Cn = 0 -1 cos P = 0
2 3 n+1 p 0 cos P -1
1 é (n + 1) n Using the projection formula
= ê (n + 1) +
n + 1ë 2! p = q cos R + r cos Q
WB JEE (Engineering) l Solved Paper 2012 | 21
8. The system of equations are 11. Since, these are 52 distinct cards in decks and
x + 3 y + 5 z = ax each distinct card is 2 in number. Therefore, 2
decks will also contain only 52 distinct cards two
5 x + y + 3 z = ay each.
3 x + 5 y + z = az \Probability that the player gets all distinct cards
Þ (1 - a) x + 3 y + 5 z = 0 …(i) 52
C26 ´ 2 26
= 104
5 x + (1 - a )y + 3 z = 0 …(ii) C26
3 x + 5 y + (1 - a )z = 0 …(iii)
12. Total number of selections of three balls in which
For infinite number of solutions, we must have balls of both colours are drawn
1- a 3 5
= 2 red balls and 1 white ball
5 1- a 3 =0
3 5 1- a +1red ball and 2 white balls
= 8C2 ´5 C1 + 8C1 ´5 C2 = 140 + 80 = 220
9- a 3 5
Þ 9 - a 1- a 3 = 0, \Required probability
9- a 5 1- a 220 220 ´ 6 10
= 13 = =
C1 ® C1 + C2 + C3 C3 13 ´ 12 ´ 11 13
Taking (9 - a) common from the first column, 13. Let F denotes fair coin, T denotes two headed, H
we get denotes head occurs
1 3 5
3 3 1
Þ (9 - a ) 1 1 - a 3 =0 \ P(F ) = , P(T ) = 1 - =
1 5 1- a 4 4 4
æ Hö
1 3 5 P ç ÷. P(T )
Þ (9 - a ) 0 - a - 2 -2 =0 æT ö èT ø
\ Pç ÷ =
0 2 - a -4 è Hø æ Hö æ Hö
P ç ÷. P(T ) + P ç ÷. P(F )
a+2 2 èT ø èFø
Þ (9 - a ) =0
-2 a+4 (By Baye’s theorem)
Þ (9 - a)(a 2 + 6 a + 8 + 4) = 0 1
1×
Þ (a - 9)(a 2 + 6 a + 12 ) = 0 4 2
= =
1 1 3 5
Þ a = 9 is the only real value of a. 1× + ×
4 2 4
9. Let A = { a1, a2 , a3 , a4 } 14. According to question,
B = { b1, b2 , b3 , b4 , b5 , b6 , b7} f (g ( x )) = g(f ( x ))
\n( A) = 4, n(B) = 7 Þ f ( x + 1) = g( x 2 + 2 x - 3)
\The total number of infections = 7P4
Þ ( x + 1)2 + 2( x + 1) - 3 = x 2 + 2 x - 3 + 1
7!
= = 7 × 6 × 5 × 4 = 840 Þ x + 1 + 2 x + 2 x + 2 - 3 = x2 + 2 x - 2
2
3!
5
Þ x 2 + 4x = x 2 + 2 x - 2
10. P = S C2 r Þ x 2 + 4x - x 2 - 2 x + 2 = 0
r =0
Þ 2x + 2 = 0Þ 2x = -2
210 Þ x = -1
= 10C0 + 10
C2 + ...... + C10 =
10
= 29
2
3 15. Since, a, b and c are in AP.
Q = S d 2 r + 1 = d1 + d 3 + d 5 + d 7 \ 2b = a + c
r =0
æ 3 x - 1ö d æ 3 x - 1ö 2 2 2
ò-2 (1 + 2 sin x )e dx = ò e xdx + 2 ò sin xe| x|dx
d | x|
= çç x ÷ sin x + sin x ç ÷ 30.
÷ ç 3 x + 1÷ -2 -2
è 3 + 1ø dx dx è ø
(Since, first function is even and second
1 d
+ (1 + x ) function is odd)
1 + x dx 2
æ 3x - 1ö = 2 ò e xdx + 2, 0 = 2[e x ]20 = 2(e 2 - 1)
= çç x ÷ cos x + sin x 0
÷
è 3 + 1ø 31. We have the identity,
d d
(3 x + 1) (3 x - 1) - (3 x - 1) (3 x + 1) | z| = ½z + - ½ £ ½z + ½ + ½ ½
2 2 2 2
dx dx 1
+ ½ z z½ ½ z½ ½z½
(3 x + 1)2 1+ x
2
æ 3 x - 1ö Þ | z| £ 2 + Þ| z|2 £ 2| z| + 2
= çç x ÷ cos x + sin x(3 x + 1)(3 x loge 3)
÷ | z|
è 3 + 1ø
Þ | z|2 - 2| z| + 1 £ 3 Þ (| z| - 1)2 £ 3
-(3 x - 1)(3 x loge 3) 1
+ Þ - 3 £| z| - 1 £ 3
(3 x + 1)2 1+ x
æ dy ö 1 Þ - 3 + 1 £| z| £ 3 + 1
\ ç ÷ =0+0+ =1
è dx ø at x =0 1+ 0 Hence, the maximum value = 3 + 1
x 2 3 3 9 3 12
28. f ( x ) = + 32. Let z = +i and r = + = = 3
8 x 2 2 4 4 4
1 2 x 2 - 16 æ 3ö
\ f ¢( x) = - 2 = ç ÷
8 x 8x 2 æ 1 ö p
q = tan -1 ç 2 ÷ = tan -1 ç ÷=
For maximum or minimum f ¢( x )must be vanished ç 3 ÷ è 3ø 6
ç 2 ÷
x 2 - 16 è ø
\ f ¢( x) = 0 Þ =0
8x 2 3 i 3 ip
\ + = 3e
Þ x = 4, - 4 and x Î[1, 6] ¹ - 4 2 2 6
Also, in [1, 4], f ¢ ( x ) < 0 50
æ3 ö æ ip ö
50
Þ f ( x ) is decreasing \ ç + i 3÷ = ç 3e ÷
ç2 2 ÷ø è 6ø
In [4, 6], f ¢ ( x ) > 0 Þ f ( x ) is increasing è
1 2 17 æ ip ö 50 i 50 p
f(1) = + = = ( 3) 50 ç
e6÷ = 325 e 6
8 1 8 ç ÷
6 2 3 1 13 è ø
f(8) = + = + = i 25 p
50
8 6 4 3 12 æ 3 i 3ö
Þ ç + ÷ = 325 e 3
17 ç2 2 ÷ø
Hence, maximum value of f ( x ) in [1, 6] is . è
8
æ 25p 25p ö
d ì -1 cos x ü 1 d cos x = 325 çcos + i sin ÷
29. ítan ý= è 3 3 ø
dx î 1 + sin x þ æ cos x ö dx 1 + sin x
2
1 + çç ÷÷ = 325 (cos 1500 + i sin 1500 )
è 1 + sin x ø = 325 [cos(360 ´ 4 + 60 ) + i sin(360 ´ 4 + 60 )]
(1 +`sin x )2
= = 325 (cos 60 + i sin 60 )
1 + sin x + 2 sin x + cos 2 x
2
50
(1 + sin x )(- sin x ) - cos(cos x ) æ3 3 ö÷ æ1 3 ö÷
Þ çç + i = 325 çç + i
(1 + sin x )2 è2 2 ÷ø è2 2 ÷ø
1 According to question,
= ´ [-(sin x + sin 2 x + cos 2 x )]
2 + 2 sin x æ3 ö
50
ç + i 3÷ = 325 ( x + iy )
=
1
´ -(1 + sin x ) = -
1 ç2 ÷
è 2 ø
2(1 + sin x ) 2
WB JEE (Engineering) l Solved Paper 2012 | 25
æ1 3 ö÷ = 9 ´ 10 ´ 10 ´ 10
325 çç + i ÷ = 2 ( x + iy )
25
-1/ 2
Þ 3(3 a + b) + c - 3(3 p + q ) - r = 2 æ2ö
3
æ 2ö
ç ÷ .... ¥ = ç1 - ÷ = 31/ 2 = 3
Þ c - r =2 …(iv) è ø
3 è 3ø
(3a + b = 3 p + q)
From Eq. (i) we get 44. Let a be the common roots
(a - p) + (b - q ) + (c - r ) = 0 \ a2 + a + a = 0 …(i)
Þ ( a - p) + ( b - q ) + 2 = 0 …(v) and a + aa + 1 = 0
2
…(ii)
From Eq. (ii), we get a2 a 1
= = [a ¹ 1]
4(a - p) + 2 )(b - q ) + c - r = 0 1 - a2 a - 1 a - 1
Þ 2(a - p) + (b - q ) + 1 = 0 …(vi) Eliminating a, we get
Subtracting Eq. (v) from Eq. (vi), we get (a - 1)2 = (1 - a2 )(a - 1)
( a - p) - 1 = 0
Þ (a - 1) = 1 - a2
a- p=1
\From Eq. (v), b - q = - 3 Þ a2 + a - 2 = 0
Now, Þ a2 + 2 a - a - 2 = 0
f (4) - g(4) = (16 a + 4b + c ) - (16 p + 4q + r ) Þ a(a + 2 ) - 1(a + 2 ) = 0
= 16 (a - p) + 4(b - q ) + (c - r ) …(vii)
Þ (a + 2 )(a - 1) = 0
Substituting the values of (a - p), (b - q ), (c - r ) Þ a = -2 [Qa ¹ 1]
from above in Eq. (vii), we get
f (4) - g(4) = 16 ´ 1 + 4(-3) + 2 = 16 - 12 + 2 = 6 45. Let a be the first term and r be the common ratio
of a GP.
41. 1 ´ 1! + 2 ´ 2 ! + K + 50 ´ 50 ! \Pth, Qth and Rth terms of a GP are
= (2 - 11) ! + (3 - 1)2 ! + (4 - 1)3 ! + K + (51 - 1)50 ! respectively ar P - 1, ar Q - 1 and ar R - 1
= (2 ! - 1!) + (3 ! - 2 !) + (4 ! - 3 !) + .. + (51! - 50 !) According to question,
= 51! - 1! = 51! - 1
ar P - 1 = 64 …(i)
42. Let the numbers be ar Q -1
= 27 …(ii)
a - 5d , a - 3d , a - d , a + d , a + 3d , a + 5d R -1
ar = 36 …(iii)
\ a - 5d + a - 3d + a - d + a + d
+ a + 3d + a + 5d = 3 (\Sum = 3) Dividing Eq, (i) by Eq (ii) we get
3
1 æ 4ö
Þ 6 a = 3Þ a = r P -Q = ç ÷ …(iv)
2 è 3ø
Also, given T1 = 4T3 , where T1, T3 are respectively, Dividing Eq. (ii) by Eq. (iii), we get
first and third terms of AP. 3
3 rQ - R =
Þ a -`5d = 4(a -`d ) Þ d = - 3 a = - 4
2 3
æ 3ö
\The fifth term Þ r 3Q - 3 R = ç ÷ …(v)
1 æ 3ö 1 9 è 4ø
a + 3d = + 3ç - ÷ = - = - 4
2 è 2ø 2 2 Multiplying Eq. (iv) and Eq. (v), we get
r P - Q ´ r 3Q - 3 R = 1
43. Given series,
1 1× 3 1× 3 × 5 1× 3 × 5 × 7 Þ r P - Q + 3Q - 3 R = 1
1+ + + + + K¥
3 3 × 6 3 × 6 × 9 3 × 6 × 9 × 12 Þ r P + 2Q - 3 R = r 0
1 3 1 3 5 Þ P + 2Q - 3R = 0 Þ P + 2Q = 3R
. 2 . . 3
1 2 2 æ2ö æ2ö
= 1+ + ç ÷ + 2 2 2 ç ÷ .... ¥ p
3 1× 2 è 3 ø 1× 2 × 3 è 3 ø 46. We know that|sin -1 x| £
2
1æ1 ö 1æ1 öæ 1 ö
ç + 1÷ 2 ç + 1÷ ç + 2 ÷ Hence, from the given relation we observe that
1 2 2 è2 ø æ2ö 2 è2 ø è2 ø p
= 1+ . + ç ÷ + each of sin -1 x, sin -1 y and sin -1 z will be
2 3 2! è 3ø 3! 2
WB JEE (Engineering) l Solved Paper 2012 | 27
p æ h -2 + k ö
So that x = y = z = sin =1 2 x + 3 y = 1, therefore the point ç , ÷
2 è3 3 ø
1
\ x 9 + y9 + z9 - 9 9 9
Must be satisfy the equation of the given line
x y z æ hö æ -2 + k ö 2 h - 6 + 3k
1 i.e, 2 ç ÷ + 3 ç ÷ =1Þ =1
= (1)9 + (1)9 + (1)9 - è 3ø è 3 ø 3
(1)9 (1)9 (1)9 Þ 2 h - 6 + 3k = 3 Þ 2 h + 3k = 9
1 Replace h = x and k = y, we get the required
= 1+ 1+ 1- = 3 -1= 2
1 ´1 ´1 locus of R .
47. We know that in D ABC i.e, 2 x + 3y = 9
a
=
b
=
c
= 2R px - 1 é0 ù
50. lim êë 0 from úû
sin A sin B sin C x ®0 1+ x - 1
\ r 2 sin P sin Q = pq p x loge p
p q = lim = lim 2 1 + x (p x loge p )
Þ r 2. . = pq, where R1 is circumradius of x ®0 1 x ®0
2 R1 2 R1 2 1+ x
DPQR.
= 2 1(p 0 loge p ) = 2 loge p =`loge p 2
Þ r 2 = 4R12 Þ r = 2 R1
Þ 2 R1 sin R = 2 R1
51. Given, f ( x )f ¢ ( x ) < 0
Þ R1 = 90 ° Þ f ( x ) and f ¢( x ) must be of opposite sign.
\DPQR is right angled. (i) Let f ( x ) = e - x
48. In DPQR, P + Q + R = 180 ° \ f ¢( x) = - e - x
æ P - Q + Rö 180 ° - Q - Q Þ f ( x ) > 0 and f ¢ ( x ) < 0, " x Î R
\2 pr sin ç ÷ = 2 pr sin
è 2 ø 2 (ii) Let f ( x ) = - e - x
= 2 pr sin(90 ° - Q ) = 2 pr cos Q \ f ¢( x) = e - x
æ 2ö
ç In DPQR cos Q = P + r -`q ÷
2 2
Þ f ( x ) < 0 and f ¢ ( x ) > 0 " x Î R
ç 2 pr ÷ But| f ( x )| = |± e - x| = e - x in both cases
è ø
æ p2 + r 2 - q 2 ö d
\ | f ( x )| = |-e - x| < 0 in both case "x Î R
= 2 pr çç ÷ = p2 + r 2 - q 2
÷ dx
è 2 pr ø
Þ | f ( x )| must be a decreasing function.
49. Let the vertices of R be (h, k )
52. If we take f ( x ) = 4 x 4 , then
P (2, –3)
(i) f ( x ) is continuous in (-2, 2 )
(ii) f ( x ) is differentiable in (-2, 2 )
(iii) f (-2 ) = f (2 )
So, f ( x ) = 4 x 4 satisfies all the conditions of
Rolle’s theorem therefore $ a point c such that
(–2, 1) Q R (h, k) f ¢ (c ) = 0
\Centroid of triangle is given by Þ 16c 3 = 0 Þ c = 0 Î (-2, 2 )
æ x1 + x2 + x3 y1 + y2 + y3 ö 53. Let y = e mx be the solution of given differential
ç , ÷
è 3 3 ø equation,
æ 2 + h - 2 -3 + 1 + k ö æ h -2 + k ö d 2y
i.e ç , ÷ Þç , ÷ Þ
dy
= me mx Þ 2 = m2e mx
è 3 3 ø è3 3 ø dx dx
æ h -2 + k ö
Since, the point ç , ÷ lies on the line
è3 3 ø
28 | WB JEE (Engineering) l Solved Paper 2012
Þ m= ,
1 1 56. Let P( 10 cos q, 8 sin q) be the required point
5 5 x2 y2
on + =1
Since, roots are real and equal, . 10 8
\General solution is y = (c1 + c 2 x )e x / 5 …(i) Whose distance from centre (0, 0) is 3 units.
y(0 ) = 1Þ c1 = 1 \10 cos 2 q + 8 sin 2 q = 9 = 9(sin 2 q + cos 2 q)
y(1) = 2e 1/ 5 Þ 2e 1/ 5 = (c1 + c 2 )e 1/ 5 p
Þ tan 2 q = 1 Þ q =
Þ c1 + c 2 = 2 4
Þ c1 = 1 57. Let e be the eccentricity of hyperbola and length
Putting the value of c1 and c 2 in Eq. (i), we get of conjugate axis be 2b. Since, vertex (a, 0 )
particular solution bisects the join of centre (0, 0 ) and focus (ae , 0 )
y = (1 + x )e x / 5 ae + 0
a= Þ e =2
2
54. Let the chord be MN. b2 = a2 (e 2 - 1) = 3a2
Let other end of chord joining from vertex to it
x2 y2
lying on y 2 = 8 x be y2=4ax \Required hyperbola is 2 - 2 = 1
2 N a b
N(2 t , 4t ) 2 (2 t , 4t) x2 y2
\Mid-point of Þ - = 1 Þ 3 x 2 - y 2 = 3 a2
(0, 0) P(t2, 2 t) a2 3 a2
MN = (t 2 , 2 t )
M
\ x =`t 2 , y = 2 t
58. We know by the definition of hyperbola, ‘‘A
hyperbola is the locus of a point which moves in
2
æ yö such a way that the difference between two fixed
Þ x = ç ÷ Þ y 2 = 4x
è2ø points remains constant,
is the required locus of P. i.e| PS - PS ¢| = 2 a
y = mx + 1
=
and …(ii)
px
æ B ö P(B Ç ( A È B ))
c 66. Given lines are, x + 2 y = 4 …(i)
63. P ç c ÷
=
è AÈB ø P( A È B )
C and 2x + y = 4 …(ii)
P( A Ç B) Solving Eqs. (i) and (ii), we get
= 4 4
P( A) + P(Bc ) - P( A Ç Bc ) x =` , y =
3 3
P( A) - P( A Ç BC )
= x y
P( A) + P(Bc ) - P( A Ç Bc ) Let AB : + = 1 …(iii)
a b
0.7 - 0.5 0.2 1
= = = Meets X-axis at A and Y - axis at B.
0.7 + 0.6 - 0.5 0.8 4 If the straight line (iii) pases through the
30 | WB JEE (Engineering) l Solved Paper 2012
æ 4 4ö 4 4 3(1) + 4(1) + 3
Point ç , ÷ , then + =1 = =2
è 3 3ø 3a 3b 32 + 42
1 1 3
Þ + = …(iv) AG = 2 GD = 4
a b 4
\Equation of circumcircle with centre at (1, 1) and
Let the mid-point ofAB be (h, k ) radius = 4 units
a+0 0+b
\ h= ,k = Þ a = 2 h, b = 2 k ( x - 1)2 + ( y - 1)2 = 42
2 2
1 1 3 Þ x 2 + y 2 - 2 x - 2 y - 14 = 0
\From Eq. (iv), + = Þ 2 h + 2 k = 3hk 1/ n
2h 2k 4 (n !)1/ n æ n !ö
69. lim = lim ç n ÷
Replace x, y from h, k respectively, we get x ®¥ n n ®¥ è n ø
required locus n ! 1× 2 × 3 .... n
i.e 2( x + y ) = 3 xy We have, n =
n n × n × n .... n
67. According to question, it is given that slope of PX ì n !ü
1/ n
ì1 2 3 r nü
1/ n
and QX are perpendicular to each other, i.e, \í ný = í . . .... K ý
în þ în n n n nþ
Slope of PX ´ slope of XQ = - 1 1/ n 1/ n
2 t 2m ì n !ü æ1 2 3 r nö
Þ ´ = - 1 Þ tm = - 4 Þ lim í n ý = lim ç . . K K ÷
n ®¥î n þ n ®¥ è n n n n nø
t 2 m2
1/ n
x -t 2 y - 2t ì n !ü
PQ : 2 = Let A = lim í n ý
t - m2 2 t - 2 m n ®¥î n þ
1/ n
ì1 2 3 r nü
y2=4x Then, A = lim í . . . K K ý
n ®¥î n n n n nþ
P(t2, 2t)
1 ærö 1
(0, 0) Þ log A = lim S log ç ÷ = ò log x dx
n ®¥ n è nø 0
X 90º R
1
é 1 ù
= ê x log x -`ò . x dx ú
ë x û0
Q (m 2, 2m)
Integrating by parts
Let PQ meets axis of parabola i.e. X-axis at = [ x log x - x ]10 = - 1
R(a , 0 ), then 1
Þ A = e -1 =
a -t2 0 -2t e
= Þ a - t 2 = t 2 - tm
(t + m)(t - m) 2 (t - m) 70. First of all we draw the graph,
= - t 2 + 4 Þ a = 4Þ AR = 4 1
y = x 3, y = , x =2
x
68. Since, triangle is equilateral therefore incentre (1,
1) lies on the centroid of the DABC. y=1/x y=x3
\GD = Length of perpendicular from the point
G(1, 1) to the line 3 x + 4 y + 3 = 0
(1, 1)
A M
O
(0, 0) N P
x=2
G (1, 1)
D
B C
76. Let z = x + iy 1 1 1
78. S1 = 1 + + + K¥ = =2
z -2 x + iy - 2 2 22 1 -`
1
\ = 2
z + 2 x + iy + 2
2
( x - 2 ) + iy ( x + 2 ) - iy 2 æ2ö 2
= ´ S 2 = 2 + 2. + 2 ç ÷ + .... ¥ = =6
( x + 2 ) + iy ( x + 2 ) - iy 3 è 3ø 1-
2
3
( x - 2 )( x + 2 ) + iy( x + 2 ) - iy( x - 2 ) - i 2 y 2
= æ 3ö æ 3ö
2
3
( x + 2 )2 - (iy )2 S 3 = 3 + 3ç ÷ + 3ç ÷ + K ¥ = = 12
è 4ø è 4ø 1-
3
x 2 - 4 + ixy + 2 iy - ixy + 2 iy + y 2 4
=
( x + 2 )2 + y 2 æ 4ö æ 4ö
2
4
S4 = 4 + 4 ç ÷ + 4 ç ÷ + K ¥ = = 20
( x + y - 4) + 4iy
2 2
è ø
5 è ø
5 4
= 1-
( x + 2 )2 + y 2 5
¥ (-1)k 1 1 1 1
x2 + y2 - 4 4iy \ S =- + - + - .... ¥
= + 2 k =`1 S k S1 S 2 S 3 S 4
( x + 4 + 4x + y ) x + 4 + 4x + y 2
2 2
1 1 1 1
æ z -2 ö 4y =- + - + -K ¥
\ arg çç ÷÷ = tan -1 2 2 6 12 20
è z + 2ø x + 4 + 4x + y 2 1 1 1 1
=- + - + - .... ¥
x 2 + 4 + 4x + y 2 p 1× 2 2 × 3 3 × 4 4 × 5
´ = (given)
x2 + y2 - 4 3 æ 1ö æ 1 1ö æ 1 1ö æ 1 1ö
= - ç1 - ÷ + ç - ÷ - ç - ÷ + ç - ÷ - K ¥
p 4y è 2 ø è 2 3ø è 3 4ø è 4 5ø
\ tan = 2
3 x + y2 - 4 æ1 1 1 1 ö
= - 1 + 2ç - + - + K ¥ ÷
4y è 2 3 4 5 ø
3= 2
x + y2 - 4 æ 1 1 1 1 ö
= - 1 - 2 ç1 - + - + - .... ¥ ÷ + 2 ¥
4 è 2 3 4 5 ø
x2 + y2 - 4 - y = 0, which represents
3 = - 1 - 2 loge 2 + 2 = 1 - loge 4
a circle. 79. Since, roots are opposite sign.
77. Let ap = bq = c r = k \Product of the roots < 0
\ a=k 1/ p
,b=k 1/ q
,c = k 1/ r Þ a2 - 4 a < 0
Practice Set 1
Physics
Category I (Q. Nos. 1 to 30) 3. When the momentum of a proton is changed
Carry 1 marks each and only one option is by an amount p 0 , the corresponding change
correct. In case of incorrect answer or any in the de-Broglie wavelength is found to be
combination of more than one answer, 1/4 mark 0.25%. Then, the original momentum of the
will be deducted. proton was
(a) p0 (b) 100 p0
1. A ray of light is incident on a transparent (c) 400 p0 (d) 4 p0
glass slab of refractive index 1.62. If the 4. Difference between nth and (n + 1)th Bohr’s
reflected and refracted rays are mutually radius H-atom is equal to its (n − 1)th Bohr’s
perpendicular, then what is the angle of radius. The value of n is
incidence?
(a) 1 (b) 2 (c) 3 (d) 4
5. The circuit has two oppositely connected
Medium 1 ideal diodes in parallel. What is the current
r
flowing in the circuit?
4Ω
90°
D1 D2
r′
Medium 2 12V
2Ω 2Ω
(µ1 > µ2)
(a) 58.3° (b) 85.3°
(c) 60° (d) 65° (a) 1.71 A (b) 2 A
(c) 2.31 A (d) 1.33 A
2. For a substance, the average life for
α-emission is 1620 yr and for β-emission is 6. A truth table is given below. Which of the
1 following has this type of truth table?
405 yr. After how much time, the th of the
4 A B Y
material remains after α and β-emission?
0 0 1
(a) 1500 yr
1 0 0
(b) 300 yr
(c) 449 yr 0 1 0
(d) 810 yr 1 1 0
4 WB JEE (Engineering) Practice Set 1
(a) NOR gate (b) OR gate 12. The planet neptune travels around the sun
(c) AND gate (d) NAND gate with a period of 165 yr. What is the radius of
7. The correct dimensional formula for pressure orbit approximately, if the orbit is considered
is given by as circular?
(a) [ML−1T−1 ] (b) [ML−1T−2 ] (a) 20 R1 (b) 30 R1 (c) 25 R1 (d) 35 R1
(c) [ML2 T−2 ] (d) [MLT−2 ] 13. A steel wire of length 4m and diameter 5 mm
is stretched by 5 kg-wt. The increase in its
8. The velocity of transverse wave in a string is length, if the Young’s modulus of steel wire
v=
T
, where T is the tension in the string is 2.4 × 10 12 dyne cm −2 is
M (a) 0.003 m (b) 0.0041 cm
and M is mass per unit length. If T = 3.0 kgf, (c) 0.00041 cm (d) 0.005 cm
mass of string is 2.5 g and length of string is
14. The terminal velocity of a copper ball of radius
1.00 m, then the percentage error in the 2 mm falling through a tank of oil at 20°C is
measurement of velocity is
6.5 cms −1 . The viscosity of the oil at 20°C is
(a) 0.5 (b) 0.7 (c) 2.3 (d) 3.6
[Take, density of oil = 15. × 10 3 kgm −3 ,
9. Three equal weight A, B and C of mass 2 kg density of copper = 8 .9 × 10 3 kgm −3 ]
each are hanging on a string passing over a (a) 3.3 × 10−1 kg m −1s −1 (b) 6.3 × 10−2 kg m −1s −1
fixed frictionless pulley as shown in the −3 −1 −1
. × 10
(c) 92 kg m s (d) 9.9 × 10−1 kg m −1s −1
figure. The tension in the string connecting
weight B and C is
15. A black body at 227°C radiates heat at a rate
of 7 cal/cm 2 s. At a temperature of 727°C, the
rate of heat radiated in the same units will be
(a) 112 (b) 105 (c) 101 (d) 89
19. A copper ball 1 cm in diameter is immersed 23. A proton, a deuteron and an α-particle
in oil with a density 800 kg m −3 . What is the moving with same kinetic energy enter a
charge of the ball, if in a homogeneous region of uniform magnetic field moving at
electric field, it is suspended in oil? The right angles to the direction of field. Ratio of
electric field is directed vertically upwards radii of circular paths travelled by these
and its intensity E = 3600 V cm −1 . The particle is
density of copper is 8600 kg m −3 (a) 2 : 2 : 1 (b) 1 : 2 : 1
. × 10−7 C
(a) 110 . × 10−8 C
(b) 110 (c) 2 : 1 : 1 (d) 1 : 2 : 2
. × 10−7 C
(c) 111 . × 10−8 C
(d) 111
24. A vertical wire carries a current in upward
20. Two charges of magnitude ‘Q’ are located at direction. An electron beam sent horizontally
distance ‘r’ as shown in figure. A third charge towards the wire will be deflected
q is placed on the line joining the above two (a) upwards (b) downwards
charges such that all the three charges are in (c) towards left (d) towards right
equilibrium.
Q Q 25. The supply voltage to room is 120 V. The
A r B resistance of the lead wires is 6 Ω. A 60 W
bulb is already switched ON. What is the
What is the magnitude, sign and position of
decrease of voltage across the bulb, when a
the charge?
240 W heater is switched ON in parallel to
(a) Q / 4 and it is located exactly midway A and B the bulb?
(b) − Q / 4 and it is located exactly midway A and B
(a) 2.9 V (b) 13.3 V (c) zero (d) 10.04 V
(c) Q / 2 and it is located exactly midway A and B
(d) None of the above 26. In the Wheatstone’s network figure P = 10 Ω,
Q = 20 Ω, R = 15Ω, S = 30 Ω. The current
21. The magnetic moment of an electron orbiting passing through battery of negligible
in a circular orbits of radius r with a speed v resistance is
is equal to B
(a) evr / 2 (b) evr R
(c) er / 2 v (d) None of these P
A C
22. The magnetic field at the centre of the G
circular loop as shown in figure, when a
single wire is bent to form a circular loop and Q
S
also extends to form straight section is
D
(a) T1 T2 ∝ R (b) T1 T2 ∝ R 2 R
V
R + –
Chemistry
Category I (Q. Nos. 41 to 70) 42. Among the following, the least stable
Carry 1 marks each and only one option is resonance structure is
correct. In case of incorrect answer or any r
r O s
r O
combination of more than one answer, 1/4 mark (a) s N (b) r N
will be deducted.
O O
s s
41. In the following structure, which is the better s
site of protonation? s
r O r O
(c) r N (d) N
O N–H
O O
s s
(a) Nitrogen (b) Oxygen
(c) Double bond (d) All of the above
8 WB JEE (Engineering) Practice Set 1
43. Correct order of nucleophilicity is 49. The equilibrium constant for the reaction,
(a) CH−3 < NH−2 < OH− < F − H 2 (g) + S(s)
3 H 2S(g) is 18.5 at 925 K and
(b) F − < OH− < CH−3 < NH−2 9.25 at 1000 K respectively. The enthalpy of
(c) OH− < NH−2 < F − < CH−3 the reaction is
(d) F − < OH− < NH−2 < CH−3 (a) 2 kJ mol −1 (b) + 71kJ mol −1
−1
h (c) − 71kJ mol (d) 75 kJ mol −1
44. If is written as h direc-h, the orbital
2π
angular momentum of a d-subshell will be 50. NaOH is a strong base. What will be pH of
5.0 × 10 −2 M NaOH solution? (log 2 = 0 .3)
h h
(a) (b) (a) 13.70 (b) 12.70 (c) 14.00 (d) 13.00
6 2
(c) 6 h (d) 2 6 h 51.
O
45. In which case racemic mixture is obtained on
mixing its mirror image is 1 : 1 molar ratio? H2/Ni
X=?
(a) [Ni(dmg)2 ] (b) [Cr(en)3 ]3+
(c) cis − [Cu(gly)2 ] (d) All of these OH O
NaNO2 CuCN OH
A B
HCl, 278K ∆
(c) (d) None of these
Compound A and B respectively are
(a) fluorobenzene and phenol
(b) benzene diazonium chloride and benzonitrile 52. For the process,
(c) phenol and bromobenzene l → H 2O ( g)
H 2O()
(d) nitrobenzene and chlorobenzene (1 bar, 273 K) (1 bar, 373 K),
The correct set of thermodynamic
47. The term anomers of glucose refers to parameters is
(a) isomers of glucose that differ in configurations (a) ∆G = 0, ∆S = + ve
at carbons one and four (C − 1 and C − 4) (b) ∆G = 0, ∆S = − ve
(b) a mixture of (D) - glucose and (L) - glucose (c) ∆G = + ve, ∆S = 0
(c) enantiomers of glucose (d) ∆G = − ve , ∆S = + ve
(d) isomers of glucose that differ in configuration at
carbon one (C − 1) 53. Which has the smallest bond angle
(X —S—X ) in the given molecules?
48. A compound liberates CO 2 with NaHCO 3 and (a) OSBr2 (b) OSCl 2
also gives colour with neutral FeCl 3 solution. (c) OSF2 (d) OSl 2
The compound can be
54. For an isomerisation reaction A 3B, the
OH OH
temperature dependence of equilibrium
COOCH3
constant is given by
(a) (b)
2000
log e K = 4 .0 −
OH T
CH—COOH CH2—COOH The value of ∆S° at 300 K is therefore,
(c) OH (d) (a) 4R (b) 5 R
(c) 400 R (d) 2000 R
WB JEE (Engineering) Practice Set 1 9
55. In van der Waals’ equation of state for a 63. Which of the following will produce
non-ideal gas, the term that accounts for isopropyl amine?
NH 2OH LiAlH 4
intermolecular force is I. (CH 3)2 CO → X → ?
(a) (V − b ) (b) (RT )−1 NH 3 LiAlH 4
II. CH 3 CH 2 CHO → X → ?
(c) p + 2
a Heat
(d) RT
V NH 3
III. (CH 3)2 CH OH + PCl 5 → X → ?
56. Which one of the following electronic IV. CH 3 CH 2 CH 2 NH 2 → ?
Heat
arrangements is absurd?
(a) I, II (b) I, III (c) II, III (d) III only
(a) n = 3, l = 2, m = − 3 (b) n = 4, l = 3, m = 3
(c) n = 4, l = 3, m = 2 (d) n = 2, l = 1, m = 0 64. How many coulombs of charge is required to
convert 12.3 gm of nitrobenzene to aniline?
57. Maximum freezing point will be for 1 molal (a) 57900 C (b) 59700 C
solution of assuming equal ionisation in each (c) 75900 C (d) 95700 C
care
(a) [Fe(H2O)6 ]Cl 3 65. If excess of AgNO 3 solution is added to 100
(b) [Fe(H2O)5 Cl]Cl 2 ⋅ H2O ml of a 0.024 M solution of dichloro bis
(c) [Fe(H2O)4 Cl 2 ]Cl ⋅ 2H2O (ethylenediamine) cobalt (III) chloride, how
(d) [Fe(H2O)3 Cl 3 ]⋅ 3H2O many moles of AgCl be precipitated?
(a) 0.024 mol (b) 0.012 mol
58. The correct order of increasing ionic (c) 0.016 mol (d) 0.0048 mol
character is
(a) BaCl 2 < CaCl 2 < MgCl 2 < BeCl 2 66. FeCr2O 4
I→ Na 2CrO 4
II
→ Cr2O 3
III
→ Cr
(b) BeCl 2 < MgCl 2 < CaCl 2 < BaCl 2 I, II and III are
(c) BeCl 2 < BaCl 2 < MgCl 2 < CaCl 2
I II III
(d) BaCl 2 < CaCl 2 < MgCl 2 < BeCl 2
(a) Na 2CO 3 / air H+ / NH4Cl Al
59. The correct order of electron affinity of the (b) NaOH / air C C
elements of oxygen family in the periodic (c) Na 2CO 3 / air C C
table is
(d) NaOH / air Al C
(a) O > S > Se (b) S > Se > O
(c) S > O > Se (d) Se > O > S 67. An inorganic compound on strong heating
gave a blackish brown powder and two
60. The second ionisation energy of the following
oxides of sulphur. The powder was dissolved
elements follows the order.
in HCl and a yellow solution was obtained
(a) C > N > O > F (b) O > N > F > C which gave a blood red coloured solution
(c) O > F > N > C (d) F > O > N > C with thiocyanate ions. The inorganic
61. Under S.T.P. 1 mol of N 2 and 3 mol of H 2 will compound may be
form on complete reaction (a) CuSO 4 (b) FeSO 4 (c) ZnSO 4 (d) NiSO 4
(a) 22.4 L of NH3 (b) 4 moles of NH3 68. Which is/are correct order of ionic mobility?
(c) 44.8 L of NH3 (d) 8 moles of NH3 (a) Li + < Na + < K + (b) Al 3+ < Mg 2 + < Na +
∆
62. 4AgNO 3 + 2H 2O + H 3PO 2 → 4Ag + X + Y (c) Both (a) and (b) (d) None of these
77. The product of the reaction D-glyceraldehyde (a) (NH2 )2 C == 0 (b) (NH2 )2 C == S
Me CO H C ==CHMgCl O (c) p NH2C 6H4SO 3H (d) C 6H5SO 3H
→
2
→
2
→
3
is/are?
HCl 79. Which of the following options are correct for
CHO CHO [Fe(CN)6 ]3− complex ?
H OH H H (a) d 2 sp3 hybridisation (b) sp3d 2 hybridisation
(a) H OH (b) H OH (c) Diamagnetic (d) Paramagnetic
CH2OH CH2OH
80. The reaction of CH 3 C == CHCH 3
CH2OH CHO
CH 3
H OH HO H
(c) HO H (d) H OH with NaIO 4 or boiling KMnO 4 produces
CH2OH CH2OH (a) CH3 C CH3 (b) CH3 CHO
78. Which of the following compounds may give O
blood red colouration while performing
(c) CH3 C OH (d) CH3 CH2 OH
Lassaigne’s test for nitrogen?
O
Mathematics
Category I (Q. Nos. 1 to 50) 4. Mean of 9 observations is 100 and mean of
Only one answer is correct. Correct answer will 6 observations is 80, then the mean of 15
fetch full marks 1. Incorrect answer or any observation is
combination of more than one answer will (a) 29 (b) 92 (c) 184 (d) 90
fetch−1/4 marks. 5. If [ x ] denotes the greatest integer less than or
equal to x, then [log 10 87213
. ] is equal to
1. If A = { x : 2cos 2 x + sin x ≤ 2} and (a) 3 (b) 4 (c) 5 (d) 6
π 3π
B = x : ≤ x ≤ , then A ∩ B is equal to 6. If an error of 1° is made in measuring the
2 2
angle of a sector of radius 60 cm, then the
π 5π 3π
(a) x : ≤ x ≤ (b) x : π ≤ x ≤
approximate error in its area is
2 6 2 (a) 2.5 π cm 2 (b) 10 π cm 2 (c) 50 π cm 2 (d) 25 π cm 2
π 5π 3π
(c) x : ≤ x ≤ or π ≤ x ≤ [ x] + [ x 2] + [ x 3] + … + [ x 2n + 1] + n + 1
2 6 2 7. lim ,
x → 0− 1 + [ x 2 ] + | x| + 2 x
(d) None of the above
n∈N
2. For any integer n, the argument of (a) 0 (b) 1 (c) 2 n + 1 (d) n
( 3 + i)4 n + 1
z= is 8. ∫ sec x − 1 dx is equal to
(1 − i 3)4 n x x 1
(a) − 2 log cos + cos 2 − + C
π π 2π 2 2 2
(a) (b) (c) (d) All of these
6 3 3 x 2 x 1
(b) 2 log cos + cos − +C
3. Ten coins are tossed. The probability of 2 2 2
getting atleast 8 tails is x x 1
(c) log cos + cos 2 − + C
(a)
3
(b)
7
(c)
5
(d)
1 2 2 2
256 128 256 64 (d) None of the above
12 WB JEE (Engineering) Practice Set 1
39. The normal at an end of a latus rectum of the 46. Radius of the circle passing through the foci
x2 y2 x2 4 2
ellipse 2
+ 2
= 1 passes through an end of of the ellipse + y = 1 having centre at
a b 4 7
the minor axis if 1
, 2 is
(a) e 2 + e = 1 (b) e 4 + e 2 = 1 2
(c) e 3 + e 2 = 1 (d) e 3 + e = 1 7
(a) 5 (b) 2 2 (c) 4 (d)
2
40. Consider the family of lines
(x + y − 1) + λ (2 x + 3 y − 5) = 0 and xe x
47. If ∫ dx = f (x) 1 + e x − 2 log| g(x)| + C ,
(3 x + 2 y − 4) + µ (x + 2 y − 6) = 0 , equation of a 1 + ex
straight line that belongs to both the families
is then
(a) x − 2 y + 8 = 0 (b) 2 x − y − 8 = 0 (a) f( x) = 3( x − 2 ) (b) f( x) = x − 1
(c) x − y + 8 (d) 2 x + y + 8 = 0 1+ e − 1
x
1+ ex + 1
(c) g ( x) = (d) g ( x) =
41. If u, v , w , z are positive real numbers such 1+ e + 1
x
1+ ex − 1
that u + v + w + z = 2, then m = (u + v)(w + z) 1 − cos(ax 2 + bx + c)
satisfies the relation 48. The value of lim where
x→α (x − α)2
(a) 3 < m ≤ 4 (b) 1 ≤ m ≤ 2
(c) 2 ≤ m ≤ 3 (d) 0 < m ≤ 1 α and β are the roots of ax 2 + bx + c = 0 is
cosec 2 x − 2005 (a − b )2
(b) (α − β )2
∫
(a)
42. 2005
dx 2
cos x 1
− cot x tan x (c) a2 (α − β )2 (d) None of these
(a) +C (b) +C 2
(cos x)2005 (cos x)2005
− tan x cot x 49. If the graph of y = ax 3 + bx 2 + cx + d is
(c) +C (d) +C
(cos x)2005 (cos x)2005 symmetric about the line x = k, then the
value of a + k is
43. Domain of f (x) = sin −1[2 − 4 x 2] is; where [⋅] −c −c
(a) (b) (c) c − 2 b (d) c + 2d
denotes the greatest integer function is a 2b
(a) (−2, 2 )
50. The equation of circle having centre as its
(b) [−1, 1]
origin and passing through the vertices of an
3 3
(c) − , 0 ∪ 0, equilateral triangle whose median is of 6 a is
2 2
(a) x2 + y2 = 4a2 (b) x2 + y2 = 9a2
(d) None of the above
(c) x + y = 16 a
2 2 2
(d) None of these
44. If x1 , x 2 , x 3 are roots of
x−a x−b b a Category II (Q. No. 51 to 65)
+ = + , given (a , b > 0),
b a x−a x−b Carry 2 marks each and only one option is
x1 > x 2 > x 3 and x1 − x 2 − x 3 = c, then a , b, c correct. In case of incorrect answer or any
are in combination of more than one answer, 1/2 mark
(a) GP (b) HP will be deducted.
(c) AP (d) None of these
51. If a , b, c and d are unit vectors, then
45. On the parabola y = x 2, the point at least |a − b|2 + |b − c |2 + |c − d|2 + |d − a|2
distance from the straight line y = 2 x − 4 is
+ |c − a |2 + |b − d|2
(a) (0, 0) (b) (0, 1)
(c) (1, 1) (d) (−1, 0) does not exceed
(a) 2 (b) 8 (c) 12 (d) 6
WB JEE (Engineering) Practice Set 1 15
52. If z is a point on the argand plane such that 58. The lengths of the intercepts made by any
z−2 circle on the coordinate axes are equal if the
|z − 1| = 1, then equals
z centre lies on line represented by
(a) cot(arg z) (b) i tan(arg z) (a) x + y = 1 (b) x − y = 1
(c) tan(arg z) (d) i cot(arg z) (c) x + y + 1 = 0 (d) x2 − y2 = 0
53. The largest term common to the sequences 1, 59. The number of solutions of| x| = cos x is
11, 21, 31 …… to 100 terms and 31, 36, 41, (a) 3 (b) 2 (c) 1 (d) 0
46 …… to 100 terms is
(a) 471 (b) 521 60. The largest term in the sequence
(c) 421 (d) 371 n2
an = is given by
3 1 n + 200
3
54. If P = 2 2 , A = 1 1 and Q = PAP T , (a)
8
(b)
49
0 1
−1 3 49 529
8
2 2 (c) (d) None of these
89
then P T (Q2005)P equals
3
61. If a , b, c are three non coplanar, non zero
1 2005
(a) 1 2 (b) vectors then
0 0 1
1 (a ⋅ a)b × c + (a ⋅ b)c × a + (a ⋅ c)a × b is
1 2005 1 2005 equal to
(d) 3
1
(c)
1 0 2
(a) [c a b] b (b) [a b c ] c
(c) [b c a ] a (d) [a b c ] b
Chemistry
41. (b) 42. (a) 43. (d) 44. (c) 45. (b) 46. (b) 47. (d) 48. (d) 49. (c) 50. (b)
51. (a) 52. (d) 53. (c) 54. (a) 55. (c) 56. (a) 57. (d) 58. (b) 59. (b) 60. (c)
61. (c) 62. (d) 63. (b) 64. (a) 65. (a) 66. (a) 67. (b) 68. (c) 69. (d) 70. (d)
71. (c) 72. (b) 73. (d) 74. (b) 75. (c) 76. (b, d) 77. (a, d) 78. (b, c) 79. (a, d) 80. (a, c)
Mathematics
1.(c) 2.(a) 3.(b) 4.(b) 5.(a) 6.(b) 7.(a) 8.(a) 9.(c) 10.(c)
11.(b) 12.(d) 13.(d) 14.(b) 15.(c) 16.(a) 17.(d) 18.(a) 19.(d) 20.(c)
21.(b) 22.(a) 23.(a) 24.(c) 25.(d) 26.(b) 27.(d) 28.(a) 29.(a) 30.(c)
31.(d) 32.(b) 33.(b) 34.(d) 35.(d) 36.(d) 37.(a) 38.(a) 39.(b) 40.(a)
41.(d) 42.(a) 43.(c) 44.(b) 45.(c) 46.(a) 47.(c) 48.(c) 49.(b) 50.(c)
51.(c) 52.(b) 53.(b) 54.(b) 55.(c) 56.(a) 57.(a) 58.(d) 59.(b) 60.(d)
61.(c) 62.(a) 63.(b) 64.(c) 65.(a) 66.(b, c, d) 67.(b,c) 68.(a,b,d) 69.(c,d) 70.(a,c,d)
71.(b,d) 72.(a,b,c,d) 73.(a,b,c,d) 74.(c) 75.(b,c)
Practice Set 2
Physics
Category I (Q. Nos. 1 to 30) (Take, atomic mass of nitrogen = 14 .0076 u)
Carry 1 marks each and only one option is (a) 0.01 nm (b) 0.09 nm
(c) 0.03 nm (d) 0.2 nm
correct. In case of incorrect answer or any
combination of more than one answer, 1/4 mark 4. Taking the Bohr radius as a 0 = 53 pm, the
will be deducted. radius of Li + + ion in its ground state, on the
1. A ray of light from a denser medium strikes a basis of Bohr’s model, will be about
rarer medium at angle of incidence i. The (a) 53 pm (b) 27 pm
reflected and refracted rays make an angle of (c)18 pm (d) 13 pm
reflection and refraction are r and r′, 5. In a Zener regulated power supply, a Zener
respectively. The critical angle is diode with Vz = 6 .0 V is used for regulation.
The load current is to be 4.0 mA and the
Denser unregulated input 10.0 V. The value of series
r
resistor RS must be
I
r′
RS
Rarer
Unregulated IL
(a) sin−1(tan r ) (b) sin−1(cot i ) voltage IZ
(c) tan−1(sin r ) (d) tan−1(sin i ) (Regulated
(Load) VZ
RL voltage)
2. The half-life of At is 100 µs. The time
215
7. The correct dimensional formula for linear 13. A uniform rod of length (L) and area of
momentum is given by cross-section (A) is subjected to tensile load
(a) ML2 T−2 (b) MLT−1 (F). If σ be the poisson’s ratio and Y be the
(c) ML2 T−1 (d) MLT−2 Young’s modulus of the material of the rod,
then find the volumetric strain produced in
8. A wire has a mass (0.3 ± 0.003) g, radius the rod.
(0 .5 + 0 .005) mm and length (6 ± 0 .06) cm. The 5 F
(a) (1 + 2 σ ) (b) (1 − 2 σ )
maximum percentage error in the AY AY
measurement of its density is (c) Zero (d) None of these
(a) 1 (b) 2 (c) 3 (d) 4
14. Suppose a fluid like oil enclosed between two
9. If the surface is smooth, the acceleration of glass plates as shown in figure. The bottom
the block m2 will be plate is fixed, while the top plate is moved
with a constant velocity v relative to fixed
m1 plate.
∆x=v∆t
F
B E C F v
m2 l
m2 g 2 m2 g
(a) (b)
4m1 + m2 4m1 + m2
2 m1g 2 m1g A D
(c) (d)
m1 + 4m2 m1 + m2
If oil is replaced by honey, then which one
has greater viscous force?
10. The work done in time t on a body of mass m
which is accelerated from rest to a speed v in (a) Honey > Oil (b) Honey < Oil
(c) Honey = Oil (d) Honey ≥ Oil
time t1 as a function of time t is given by
(a)
1 mv 2
t (b)
mv 2
t 15. A black body radiate energy at rate of X W/m 2
2 t1 t1 at a high temperature of T kelvin. When
1 mv 2 1 mv 2 2 T
(c) t (d) t temperature is reduced to kelvin, the
2 t1 2 t 12 2
radiant energy is
11. The speed of a projectile u reduces by 50% on X X X
reaching maximum height. What is the range (a) (b) (c) (d) 2 X
16 4 2
on the horizontal plane?
u2 3 u2 u2 4 u2 16. Certain amount of heat is given to 100 g of
(a) × (b) (c) ×3 (d) copper to increase its temperature by 21°C. If
g 2 g 2g g
the same amount of heat is given to 50 g
12. The time period of a satellite of the earth is water, then the rise in its temperature is
5h. If the separation between the earth and (Take, specific heat capacity of
the satellite is increased to 4 times the copper = 400 Jkg −1 K −1 and that of
previous value, then the new time period will water = 4200 Jkg −1 K −1 )
become
(a) 4°C (b) 5.25°C
(a) 10 h (b) 80 h (c) 8°C (d) 6°C
(c) 40 h (d) 20 h
20 WB JEE (Engineering) Practice Set 2
17. The top surface of an incompressible liquid is 22. A long straight solid metal wire of radius R
open to the atmosphere. The pressure at a carries a current i uniformly distributed over
depth h1 below the surface is p1 . How does its circular cross-section. Find the magnetic
the pressure p 2 at depth h2 = 2 h1 compare field inside the wire at a distance r from the
with p1 ? axis of wire.
µ 0µ r ir
(a) p2 > 2 p1 (b) p2 = 2 p1 (a) B =
2 πR 2
(c) p2 < 2 p1 (d) p2 = p1 µ 0µ r ir
(b)B =
18. A network of four capacitors each of 12 µF 4 πR 2
capacitance, if connected to a 500 V supply as µ 0µ r ir
(c) B =
shown in figure. What is the total amount of 8 πR 2
µ 0µ r ir
charge stored? (d) B =
C2 πR 2
23. A current i ampere flows along an infinitely
long straight thin walled tube, then the
magnetic induction at any point inside tube is
C1 C3
(a) infinite
(b) zero
µ 2i
(c) 0 ⋅ tesla
C4 4π r
2i
(d) tesla
r
500V 24. A wire has a non-uniform cross-sectional
(a) 6000 µC (b) 5000 µC (c) 3000 µC (d) 8000 µC area as shown in figure. A steady current i
flows through it. Which one of the following
19. Four particles, each having a charge q are
statement is correct?
placed on the four corners A, B, C and D of a
regular pentagon ABCDE. The distance of A B
each corner from the centre is a. Find the
electric field at the centre of the pentagon.
q q (a) The drift speed of electron is constant
(a) along OE (b) along OC
4 π ε0 a 2
4 π ε0 a 2 (b) The drift speed increases on moving from A to B
q q (c) The drift speed decreases on moving from A to B
(c) along OD (d) along OA
4 π ε0 a 2 4 π ε0 a 2 (d) The drift speed varies randomly
20. Two point charges placed at certain distance r 25. A battery of internal resistance 4 Ω is
in air exert a force F on each other. Then, the connected to the network of resistance as
distance r at which, these charges will exert shown in the figure. In order to give the
the same force in a medium of dielectric maximum power to the network, the value of
constant K is given by R (in Ω) should be
R R
(a) r (b) r / K (c) r / K (d) r K
21. If magnetic field B is present at a place E R 6R R
alongwith an electric field of intensity E,
then force experienced by a moving charge q
in that region is given by R 4R
(a) F = (qE + v × B), where v is velocity of charge
(b) F = q (E + B) × v, where v is velocity of charge
(c) F = E + q (v × B), where v is velocity of charge
(d) F = q (E + v × B), where v is velocity of charge
(a) 4 / 9 (b) 8 / 9 (c) 2 (d) 18
WB JEE (Engineering) Practice Set 2 21
26. In the circuit, if no current flows through the 29. In a Young’s double slit experiment, the
galvanometer when the key K is closed, the fringe width obtained is 0.4 cm, when light
bridge is balanced. The balancing condition of wavelength 5400 Å is used. If the distance
for bridge is between the screen and the slit is reduced to
B half, then what should be the wavelength of
R1 R2 light used to obtain fringes 0.0048 m wide?
K (a) 12960 Å (b) 1300 Å
A C (c) 1400 Å (d) 1500 Å
G
28. A car is fitted with a convex side-view mirror 33. A block of copper having mass 2 kg is heated
of focal length 20 cm. A second car 2.8 m to a temperature of 500°C and then placed
behind the first car is overtaking the first car in a large block of ice at 0°C. What is the
at relative speed 15 m/s. The speed of the maximum amount of ice that can melt?
image of the second car as seen in the mirror The specific heat of copper is 400 J kg −1 C −1
of the first one is and latent heat of fusion of water is
(a)
1
m/s (b) 10 m/s 3.5 × 10 5 J kg −1 .
15 4 6
1 (a) kg (b) kg
(c) 15 m/s (d) m/s 3 5
10 8 10
(c) kg (d) kg
7 9
22 WB JEE (Engineering) Practice Set 2
p
34. An iron rod of length L and magnetic moment
m is bent in the form of a semi-circle. Now, its
magnetic moment will be I
2m
(a) m (b) IV
π
m A II
(c) (d) mπ
π
B
35. In the given circuit, calculate the Q-factor of III
the circuit.
200mH C V
10W
(a) Change in internal energy is same in IV and III
cases, but not in I and II
(b) Change in internal energy is same in all the four
cases
(c) Work done is maximum in case I
(d) Work done is minimum in case II
50Hz
(a) 6.32
38. A particle is hurled into air from a point on
(b) 5.32
the horizontal ground at an angle with the
(c) 4.00 vertical. If the air exerts a constant resistive
(d) 3.32 force,
(a) the path of projectile will be parabolic path
Category III (Q. Nos. 36 to 40) (b) the time of ascent will be equal to time of
descent
Carry 2 marks each and one or more option(s) (c) the total energy of the projectile is not conserved
is/are correct. If all correct answers are not marked (d) at the highest point, the velocity of projectile is
and also no incorrect answer is marked then score horizontal
= 2 × number of correct answers marked ÷ actual
number of correct answers. If any wrong option is 39. Temperature dependence of resistivity ρ(T) of
marked or if any combination including a wrong semiconductors, insulators and metals is
option is marked, the answer will be considered significantly based on the following factors:
wrong, but there is no negative marking for the (a) Number of charge carriers can change with
same and zero marks will be awarded. temperature T
(b) Time interval between two successive collisions
36. In a photoelectric experiment, the can depend on T
wavelength of the incident light is decreased (c) Length of material can be function of T
from 6000 Å to 4000 Å, while the intensity of (d) Mass of carriers is a function of T
radiation remains the same, 40. Magnetic field due to a current carrying wire
(a) the cut-off potential will increase
loop along its axis can be calculated by
(b) the cut-off potential will decrease
(a) finding electric field of coil and then by using
(c) the kinetic energy of the emitted photoelectron
E/B=c
will increase
(b) finding electric field of a small element of
(d) the photoelectric current will increase
current carrying coil and then by using E / B = c
37. Figure shows the p-V diagram of an ideal gas (c) suming up the magnetic field of small elements
of the coil
undergoing a change of state from A to B.
(d) finding magnetic field of a wire of length l and
Four different paths I, II, III and IV as shown then by substituting l = 2 πr, r being radius of
in the figure may lead to the same changes of loop
state.
Chemistry
Category I (Q. Nos. 41-70) 48. The uncertainty in position of a particle of
Carry 1 mark each and only one option is correct. 25 g in space is 10 −5m. Hence, its uncertainty
In case of incorrect answer or any combination of in velocity (in ms −1 ) is
more than one answer, 1/ 4 mark will be deducted. (Given : Planck’s constant h = 6 .6 × 10 −34 J-s)
41. The angular momentum for an electron (a) 2.1 × 10−28 (b) 2.1 × 10−20
−34
revolving in s-subshell will be (c) 2.1 × 10 (d) 2.1 × 1012
1 h h h N
(a) . (b) zero (c) (d) 2 . 49. 5 mL of N.HCl, 20 mL of H 2SO 4 and 30 mL
2 2π 2π π
2
42. Atomic number of vanadium (V), chromium N
of HNO 3 are mixed and volume made to 1 L.
(Cr), maganese (Mn) and Iron (Fe) are 3
respectively 23, 24, 25 and 26. Which of these The normality of the resulting solution will be
way be expected to have highest II nd (a)
N
(b)
N
(c)
N
(d)
N
ionisation enthalpy? 5 10 20 40
(a) V (b) Cr (c) Mn (d) Fe 50. Which of the following formula does not
43. The first ionisation potential of Na is 5.1 eV, correctly represent the bonding capacity of
the atoms involved?
the value of electron gain enthalpy of Na +
+
will be H
(a) − 2.55 eV (b) − 5.1 eV (a) H—P—H (b) F F
(c) − 10.2 eV (d) + 2.55 eV O
H
44. Normality of ‘30 volume’ H 2O 2 solution is O O
(c) O N O (d) H—C—C
(a) 1.6 (b) 91.07 (c) 10.72 (d) 5.36 O
—H
45. Among the following complex ions, the
species whose central atom does not have ‘d’ 51. Hydrogen can not be produced by the action
electrons is of dil.H 2SO 4 on
(a) [MnO 4 ]− (b) [Co(NH3 )6 ]3+ (a) Cu (b) Zn (c) Fe (d) Al
(c) [Fe(CN)6 ] 3−
(d) [Cr(H2O)6 ] 3+
52. The oxidation state of Cr in [Cr(NH 3)4 Cl 2 ]+ is
(a) 0 (b) + 1 (c) + 2 (d) + 3
46. The molecular shape of SF4 , CF4 and XeF4 are:
(a) different with 1, 0 and 2 lone-pair of electrons
53. The pK a of a weak acid [HA] is 4.5 the pOH
on the central atom, respectively of an aqueous buffered solution of HA in
(b) different with 0, 1 and 2 lone-pair of electrons
which 50% of it is ionised is?
on the central atom, respectively (a) 4.5 (b) 2.5 (c) 9.5 (d) 7.0
(c) same with 1, 1 and 1 lone-pair of electrons on 54. For a reaction
the central atom, respectively
CO(g) + Cl 2 (g) ] COCl 2(g), the K p / K c is
(d) same with 2, 0 and 1 lone-pair of electrons on
equal to
the central atom, respectively
1
(a) (b) RT (c) RT (d) (RT )2
47. A gas is initially at 1 atm pressure, to RT
1 55. A spontaneous change is one, in which the
compress it to th of its initial volume.
4 system suffers
Pressure to be applied is (a) an increase in internal energy
1 (b) lowering in entropy (c) lowering in free energy
(a) 1 atm (b) 2 atm (c) 4 atm (d) atm
4 (d) no change in energy
24 WB JEE (Engineering) Practice Set 2
56. When a mole of sodium chloride is dissolved in 63. Which of the following nomenclature is not
water at 298 K, the free energy change will be according to IUPAC system?
(Given, lattice energy of NaCl (a) Br CH2 CH == CH2
(1-bromo prop 2-ene)
= − 777.8 kJmol −1 , Hydration energy of
CH3
NaCl = − 774.1 kJ mol −1 and ∆S at 298K
= 0 .043 kJmol −1 ) (b) CH3 CH2 C CH2 CH CH3
(a) − 37
. kJ (b) − 12.814 kJ Br3 CH3
(c) − 9114
. kJ (d) − 16.54 kJ (4-bromo 2,4, dimethyl hexane)
57. Graphite is a good conductor of electricity. (c) CH3 CH CH CH2 CH3
because it contain
CH3 Ph
(a) bonded electrons (b) mobile electrons
(2-methyl 3-phenyl pentane)
(c) strong C—C bonds (d) strong C==C bonds
(d) CH3 C CH2 CH2 CH2 COOH
58. Which of the following does not form M 3+
ion? O
(a) B (b) Al (c) Ga (d) In (5-oxohexanoic acid)
59. Sodium is made by the electrolysis of molten 64. Which of the following energy level diagram
mixture of about 40% NaCl and 60% CaCl 2 , for [FeF6 ]3− is correct on the basis of crystal
because field theory?
(a) Ca 2+ can reduce NaCl to Na eg
(b) CaCl 2 helps in conduction of electricity
(c) the mixture has lower melting point than NaCl (a)
(d) Ca 2+ ion can displace Na from NaCl 5
3d
t2g
60. The compound which does not exhibit optical
isomerism is
(a) CH3CHBrCOOH (b) CH2ClCH2COOH t2g
(c) CH3CH.OH.COOC 2H5 (d) CH3CHOHCOOH
H CH3 H H
(a) (b) t2g
H H H H
(d)
H CH3 5
3d
CH3 CH3 eg
CH3 H
65. In XeF2 , XeF4 and XeF6 , the number of
(c) (d)
lone-pairs on the Xe-atom are respectively.
H H H H (a) 2, 3, 1 (b) 1, 2, 3
H H H CH3 (c) 3, 1, 2 (d) 3, 2, 1
WB JEE (Engineering) Practice Set 2 25
66. The salts of Cu in + 1 oxidation are unstable (b) X = path of reaction with catalyst,
because Y = path of reaction without catalyst
(a) Cu+ and 3d 10 configuration (c) X = energy of activation with catalyst,
Y = energy of activation without catalyst
(b) Cu+ disproportionates easily to Cu(0) and Cu2+
(d) X = energy of endothermic reaction,
(c) Cu+ disproportionates easily to Cu2+ and Cu3+
Y = energy of exothermic reaction
(d) Cu+ is easily reduced to Cu2+
67. Number of P—O—P bonds present in cyclic 72. The amount of metal deposited, when a
metaphosphoric acid are current of 12 ampere with 75% efficiency is
passed through the cell for 3 H .
(a) four (b) three (c) two (d) one
(Given, Z = 4 × 10 −4 ).
68. Which of the following changes take place (a) 32.4 g (b) 38.8 g
during roasting? (c) 36.0 g (d) 22.4 g
(i) Impurities are removed as their volatile
73. Molar conductivity of NH 4OH can be
oxides.
calculated by the equation
(ii) Ore is converted in to oxide.
(a) Λ°NH 4 OH = Λ°Ba(OH) 2 + Λ°NH 4 Cl − Λ°BaCl 2
(iii) Changes like oxidation, chlorination, etc.
can take place. (b) Λ°NH 4 OH = Λ°BaCl 2 + Λ°NH 4 Cl − Λ°Ba(OH) 2
(a) (i) and (ii) only (b) (ii) and (iii) only Λ°Ba(OH) 2 + 2 Λ°NH 4 Cl − Λ°BaCl 2
(c) (i) and (iii) only (d) (i), (ii) and (iii) (c) Λ°NH 4 OH =
2
69. The movement of dispersion medium under Λ°NH 4 Cl + Λ°Ba(OH) 2
(d) Λ°NH 4 Cl =
the influence of electric field is known as 2
(a) electrodialysis (b) electrophoresis
(c) electroosmosis (d) cataphoresis
74. Halogen acids react with alcohols to form
alkyl halides. The reaction follows a
70. The half-life period of a first order reaction is nucleophilic substitution mechanism. What
10 min, what percentage of the reaction will will be the major product of the following
be completed in 100 min? reaction
(a) 25% (b) 50% (c) 75% (d) 99.9% CH 3 CH CH CH 3 + HCl →
Category II (Q. Nos. 71-75) CH 3 OH
Carry 2 marks each and only one option is correct. (a) CH3 CH CH CH3
In case of incorrect answer or any combination of
more than one answer, 1/2 mark will be deducted. CH3 Cl
(b) CH3 CH CH CH3
71. The graph of the effect of catalyst on
activation energy is given below. Choose the Cl CH3
blanks X and Y with appropriate statements. CH3
(c) CH3 C CH2 CH3
Cl
Potential energy
Category III (Q. Nos. 76-80) 78. Which of the following has same
hybridisation of the central atom?
Carry 2 marks each and one or more option(s)
is/are correct. If all correct answers are not (a) H2SO 4 (b) NH3 (c) H2O (d) PCl 5
marked and also no incorrect answer is marked 79. Which of the following solution show same
then score = 2 × number of correct answers number of moles of solute in the given
marked ÷ actual number of correct answers. If any solution?
wrong option is marked or if any combination (a) 100 mL solution of 0.1 M HCl
including a wrong option is marked, the answer (b) 50 mL solution of 0.2 M NaOH
will considered wrong, but there is no negative (c) 200 mL solution of 0.01 M HCl
marking for the same and zero mark will be (d) 75 mL solution of 0.025 M NaOH
awarded. NH2
76. Which of the following has same oxidation 80. on reaction with HNO 2 forms
number for the central atom?
(a) KMnO 4 (b) KClO 3 NH2 OH
(c) K 2Cr2O 7 (d) CrO 5 (a) (b)
77. Which of the following are used to convert NO2
R CHO into RCH 2OH ? OH
(a) H2 /Pd
(b) LiAlH4
(c) NaBH4 (c) (d)
(d) Reaction with RMgX followed by hydrolysis HO
Mathematics
Category I (Q. Nos. 1 to 50) π
3. If sin −1 x − cos −1 x = , then x is
Only one answer is correct. Correct answer will 6
fetch full marks 1. Incorrect answer or any 1 3 1
(a) (b) (c) (d) 2
combination of more than one answer will fetch 2 2 2
−1/4 marks. π /4
∫ (tan x + tan n − 2 x)
n
4. The value of
| x|
3 3
x
1. lim − , a > 0 where [ x ] denotes 0
−
x→a a
a
[ x ] [ x ]2 [ x ]3
the greatest integer less than or equal to x, is d x − + − + …
1! 2! 3!
(a) a2 − 3 (b) a2 − 1
(c) a 2
(d) None of these where [ x ] is greatest integer function, is
1 1 1 1
x2 (a) (b) (c) (d)
2. Let f (x) = ∫ (1 + x 2){1 + 1 + x2}
dx and n n+2 n−1 n−2
x
1 − x
f (0) = 0 . Then, f ()
1 is equal to 5. The function f (x) = ∫ log dx is
π 1 + x
(a) loge (1 + 2) (b) loge (1 + 2) − 0
4 (a) an even function (b) an odd function
π
(c) loge (1 + 2) + (d) None of these (c) a periodic function (d) None of these
4
WB JEE (Engineering) Practice Set 2 27
π /6
larger angle is twice the smallest, then the
common ratio r satisfies the inequality
(a) I1 = 2 I2 (b) I1 = 3I2 (a) 0 < r < 2 (b) 1 < r < 2
(c) 2 I1 = I2 (d) None of these (c) 1 < r < 2 (d) None of these
2π
7. The value of ∫|cos x − sin x|dx is 14. If 9 log 3(log 2 x ) = log 2 x − (log 2 x)2 + 1, then x is
0 equal to
1
1 (a) 1 (b)
(a) (b) 2 2 2
2
2 (c) 3 (d) None of these
(c) (d) 4 2
2 15. Let Z1 and Z 2 be nth roots of unity which
subtend a right angle at the origin. Then, n
8. The value must be of the form
2 −2
1 + x 1 − x (a) 4K + 1 (b) 4K + 2
∫ p ln1 − x + q ln1 + x + r dx depends
(c) 4K + 3 (d) 4K
−2
50. If f (x) and g(x) are differentiable function for 57. Consider the system of linear equations
0 ≤ x ≤ 1 such that f (0) = 10 , g(0) = 2, f ()
1 = 2, x1 + 2 x 2 + x 3 = 3; 2 x1 + 3 x 2 + x 3 = 3
1 = 4 , then in the interval (0, 1)
g() 3 x1 + 5 x 2 + 2 x 3 = 1
(a) f ′( x) = 0 for all x
The system has
(b) f ′( x) + 4g ′( x) = 0 for atleast one x
(c) f ′( x) = 2 g ′( x) for atmost one x (a) a unique solution
(d) None of the above (b) no solution
(c) infinite number of solutions
(d) exactly three solutions
Category II (Q. Nos. 51 to 65)
Carry 2 marks each and only one option is correct. 58. The value of parameter α, for which the
In case of incorrect answer or any combination of function f : R → R given by f (x) = 1 + αx ,
more than one answer, 1/2 mark will be deducted. α ≠ 0 is the inverse of itself is
(a) −2 (b) −1 (c) 1 (d) 2
n r
51. The value of ∑ ∑ n
C r ⋅ C p is equal to
r
59. The function f :[0 , 3] → [1, 29], defined by
r=0 p=0
f (x) = 2 x 3 − 15 x 2 + 36 x + 1, is
(a) 3n − 2 n (b) 3n − 2 n − 2
(a) one-one and onto
(c) 3n − 2 n + 2 (d) None of these
(b) onto but not one-one
52. The coefficient of x 20 in the expansion of (c) one-one but not onto
−5 (d) neither one-one nor onto
1
(1 + x 2)40 x 2 + 2 + 2 is
x 60. If the line passing through the points (5, 1, a)
(a) 30
C10 (b) 30
C 25 and (3 , b, 1) crosses the yz-plane at the point
17 13
(c) 1 (d) None of these 0 , , − , then
2 2
53. If a = i$ + $j + k$ , a ⋅ b = 1 and a × b = $j − k$ ,
(a) a = 6, b = 4 (b) a = 8,b = 2
then b (c) a = 2, b = 8 (d) a = 4,b = 6
(a) $i − $j + k$ (b) 2 $j − k$
(c) $i (d) 2 $i 61. The coordinates of the point of intersection
of tangents drawn to the hyperbola
54. If θ is the angle between the line x 2 y2
− = 1 at the points where it is
r = (i$ + 2 $j − k$) + λ ($i − $j + k$) a 2 b2
and the plane r ⋅ (2 $i − $j + k$) = 4, then cosθ = intersected by the line lx + my + n = 0 are
a2 l b2 m a2 l b 2 m
1 2 2 (a) ,− (b) − ,
(a) (b) n n n n
2 2 3
1 a2 l b 2 m a2 l b2 m
(c) (d) None of these (c) , (d) − ,−
3 n n n n
WB JEE (Engineering) Practice Set 2 31
66. For a ∈ R (the set of all real numbers),a ≠ −1 , 72. The locus of the point of intersection of the
(1 + 2 + … + n )
a a a
1 lines x cos α + y sin α = a and
lim a −1
= x sin α − y cos α = b, where α is a variable is
n→ ∞ (n + 1) [(na + 1) + (na + 2) + 60
(a) x2 + y2 = a2 − b 2
… (na + n)] (b) x2 − y2 = a2 − b 2
Then, a is equal to (c) x2 + y2 = a2 + b 2
15 17
(a) 5 (b) 7 (c) − (d) − (d) None of the above
2 2
32 WB JEE (Engineering) Practice Set 2
2
73. The locus of a point p(α , β) having under the (a) 2 (b)
3
condition that the line y = α x + β is a tangent 1
(c) (d) None of these
x 2 y2 2
to the hyperbola 2 − 2 = 1 is
a b
75. The equation sin x + x cos x = 0 has at least
(a) a hyperbola (b) a parabola
(c) a circle (d) an ellipse one root in the interval
π
(a) − , 0 (b) (0, π )
74. Let f and g be differentiable functions 2
satisfying g ′ (a) = 2, g(a) = b and fog = I π π
(c) − , (d) None of these
(Identity function). The f ′(b) is equal to 2 2
Answers
Physics
1. (a) 2. (a) 3. (c) 4. (c) 5. (a) 6. (a) 7. (b) 8. (d) 9. (a) 10. (d)
11. (a) 12. (c) 13. (b) 14. (a) 15. (a) 16. (a) 17. (c) 18. (d) 19. (a) 20. (c)
21. (d) 22. (a) 23. (b) 24. (c) 25. (c) 26. (b) 27. (a) 28. (a) 29. (a) 30. (b)
31. (b) 32. (a) 33. (c) 34. (b) 35. (a) 36. (a, c) 37. (b, c) 38. (a, c, d) 39. (a, b) 40. (c)
Chemistry
41. (b) 42. (b) 43. (b) 44. (d) 45. (a) 46. (a) 47. (c) 48. (a) 49. (d) 50. (d)
51. (a) 52. (d) 53. (c) 54. (a) 55. (c) 56. (c) 57. (b) 58. (a) 59. (c) 60. (b)
61. (d) 62. (b) 63. (a) 64. (c) 65. (d) 66. (b) 67. (b) 68. (d) 69. (c) 70. (d)
71. (c) 72. (b) 73. (c) 74. (c) 75. (c) 76. (c, d) 77. (a, b, c) 78. (a, b, c) 79. (a, b) 80. (b, c, d)
Mathematics
1. (c) 2. (b) 3. (b) 4. (c) 5. (a) 6. (d) 7. (d) 8. (c) 9. (d) 10. (a)
11. (c) 12. (c) 13. (b) 14. (d) 15. (d) 16. (d) 17. (c) 18. (c) 19. (c) 20. (b)
21. (b) 22. (d) 23. (b) 24. (d) 25. (a) 26. (d) 27. (a) 28. (a) 29. (d) 30. (b)
31. (b) 32. (b) 33. (a) 34. (d) 35. (d) 36. (b) 37. (b) 38. (c) 39. (a) 40. (b)
41. (a) 42. (a) 43. (d) 44. (b) 45. (d) 46. (c) 47. (c) 48. (b) 49. (c) 50. (b)
51. (d) 52. (b) 53. (c) 54. (c) 55. (b) 56. (c) 57. (b) 58. (b) 59. (b) 60. (a)
61. (b) 62. (d) 63. (c) 64. (d) 65. (c) 66. (b, d) 67. (b, d) 68. (a, d) 69. (b,c) 70. (a, d)
71. (b) 72. (c) 73. (a) 74. (c) 75. (b)
Practice Set 3
Physics
Category I (Q. Nos. 1 to 30) 4. An electron of a stationary hydrogen atom
Carry 1 marks each and only one option is passes from the fifth energy-level of the
correct. In case of incorrect answer or any ground level. The velocity, that the atom
combination of more than one answer, 1/4 mark acquired as a result of photon emission will
will be deducted. be (m is the mass of the electron, R = Rydberg
constant and h = Planck’s constant)
1. The angle of minimum deviation for a glass 25 hR 24 m 24 hR 25 hR
(a) ⋅ (b) ⋅ (c) ⋅ (d) ⋅
prism with µ = 3 equals the refracting angle 24 m 25 hR 25 m 24 m
of the prism. What is the angle of the prism? 5. A Zener diode, having breakdown voltage equal
(a) 60° (b) 30° to 15 V is used in a voltage regulator circuit
(c) 45° (d) 90° shown in figure. The current through the
diode is
2. In the chemical analysis of a rock, the mass
ratio of two radioactive isotopes is found to +
250 Ω
be 100 : 1. The mean times of the two
isotopes are 4 × 10 9 yr and 2 × 10 9 yr
20 V Z =15 V 1k Ω
respectively. If it is assumed that at the time
of formation, the atoms of both the isotopes
were is equal proportion. Calculate the age of –
the rock. Ratio of the atomic weight of the
two isotopes is 1.02 : 1. (a) 15 mA (b) 10 mA
(c) 5 mA (d) 20 mA
(a) 1834
. × 10 yr
10
. × 1010 yr
(b) 184 6. The circuit
. × 1010 yr
(c) 181 NOR NAND NOT
. × 1010 yr
(d) 18
7. The dimensional formula of torque is 13. A 5 cm length of the cube has its upper face
(a) [ML−2 T−2 ] (b) [ML2 T−2 ] displaced by 0.2 cm by a tangential force of
(c) [MLT−2 ] (d) [ML−1T−2 ] 8 N. The modulus of rigidity of the material
of cube is
8. The length of a simple pendulum is about (a) 5 × 104 Nm−2 (b) 6 × 104 Nm−2
100 cm known to an accuracy of 1 mm. Its
(c) 7 × 104 Nm−2 (d) 8 × 104 Nm−2
period of oscillation is 2 s determined by
measuring the time for 100 oscillations using 14. A square plate of 10 cm side moves parallel to
a clock of 0.1 s resolution. What is the
accuracy in the determined of g? another plate with a velocity of 10 cm s −1 ;
both the plates immersed in water. If the
(a) 2% (b) 0.5% (c) 0.1% (d) 0.2%
viscous force is 200 dyne and viscosity of
9. The pulleys and strings shown in figure are water is 0.01 poise, what is their separation
smooth and of negligible mass. For the distance?
system to the under equilibrium, the angle (a) 0.05 cm (b)1cm (c) 0.07 cm (d) 7 cm
θ should be
15. In the above problem, if the sun radiates as
an ideal black body, what is the temperature
of its surface?
(a) 6803 K (b) 5603 K
(c) 5803 K (d) 5503 K
19. Equal charges q each are placed at the magnitude of magnetic field induction at the
vertices A and B of an equilateral triangle centre of the loop is
ABC of side a. The magnitude of the electric i
field intensity at the point C is A C B
(a)
1
⋅
q
(b)
1 q 2
⋅ r θ
4 πε0 a2 4 πε0 a2
O
1 q 3 1 2q
(c) ⋅ (d) ⋅
4 πε0 a2 4 πε0 a2 D
31. In the circuit shown here, the point C is kept 35. In a series L-C circuit, L = 4 H and C = 25 µF.
connected to a point A till the current
If the frequency is twice of resonant
flowing through the circuit becomes
frequency, the net reactance of circuit is
constant. Afterward, suddenly point C is
disconnected from point A and connected to (a) zero (b) 100 Ω (c) 200 Ω (d) 600 Ω
point B at time t = 0. Ratio of the voltage
across resistance R and the inductor L at Category III (Q. Nos. 36 to 40)
t = L / R will be equal to Carry 2 marks each and one or more option(s)
A C R is/are correct. If all correct answers are not marked
and also no incorrect answer is marked then score
= 2 × number of correct answers marked ÷ actual
L
number of correct answers. If any wrong option is
B marked or if any combination including a wrong
option is marked, the answer will be considered
wrong, but there is no negative marking for the
e 1− e same and zero marks will be awarded.
(a) (b) 1 (c) − 1 (d)
1− e e
WB JEE (Engineering) Practice Set 3 37
50. The compressibility factor for a real gas at 57. 1 L of 1.0 M CuSO 4 solution is electrolysed by
high pressure is passing 1.5 F charge. Final molarity of the
(a) 1 +
RT
(b) 1 +
pb
(c) 1 −
pb
(d) 1 solution will be [atomic weight of Cu = 63.5 ]
pb RT RT (a) 0.10 M (b) 0.20 M
(c) 0.25 M (d) 0.30 M
51. 29.5 mg of an organic compound containing
nitrogen was digested according to Kjeldahl’s 58. Potassium permaganate acts as an oxidant in
method and the evolved ammonia was neutral, alkaline as well as acidic medium.
absorbed in 20 mL of 0.1 M HCl solution. The The final products obtained from it in the
excess of the acid required 15 mL of 0.1 M three conditions are, respectively.
NaOH solution for complete neutralisation. (a) MnO 22 − , Mn3+,Mn2+ (b) MnO, MnO 2− 3+
4 , Mn
The percentage of nitrogen in the compound
(c) MnO 2 , MnO 2 , Mn2+ (d) MnO 2 , MnO 2−
4 , Mn
3+
is
(a) 59.0 (b) 47.4 (c) 23.7 (d) 29.5 59. The absolute configuration of
52. The correct order of increasing basicity of the H O 2C C O 2H
given conjugate bases (R == CH 3) is
– – – – HO H H OH
(a) RCOO < HC ≡≡ C < R < NH2
– – – –
(b) R < HC ≡≡ C < RCOO <NH2 (a) S, S (b) R, R
– – – – (c) R, S (d) S, R
(c) RCOO <N H2 < HC ≡≡ C < R
– – – – 60. A gaseous hydrocarbon gives 0.72 g of water
(d) RCOO <HC ≡≡ C <NH2 < R and 3.08 g of CO 2 upon combustion. The
empirical formula of the hydrocarbon is
53. The two form of d-glucopyranose obtained
(a) C 3 H4 (b) C 6H5
from the solution of d-glucose are called
(c) C 7H8 (d) C 2H4
(a) epimer (b) anomer
(c) enantiomer (d) isomer 61. A blue colouration is not obtained when
(a) ferric chloride reacts with potassium
54. The correct stability order for the following ferrocyanide
species is (b) anhydrous CuSO 4 is dissolved in water
+ + (c) ammonium hydroxide dissolves in copper
O sulphate
(I) (II) (d) copper sulphate solution reacts withK 4 [Fe(CN)6 ]
65. Rusting follows the following reactions ketone with molecular formula C8H 8O.
1 Which shows positive iodoform test. The
(i) 2H + + O 2 + 2 e − → H 2O; E° = 123
. V structure of (A) is
2
(a) C 2H5COOC 6H5 (b) C 6H5COOC 2H5
(ii) Fe2+ + 2 e − → Fe ; E° = − 0 .44 V (c) H3COCH2COC 6H5 (d) CH3 COOC 2H5
The net work done is 72. The energy required to break one mole of
(a) 152 kJ (b) 322 kJ Cl—Cl bonds in Cl 2 is 242 kJ mol −1 . The
(c) 132 kJ (d) 233 kJ
longest wavelength of light capable of
66. At pH of 0.1 M solution of following increases in breaking a single Cl—Cl bond is
the order (a) 594 nm (b) 694 nm
(a) NaCN < NH4Cl < NaCl < HCl (c) 600 nm (d) 494 nm
(b) HCl < NaCl < NaCN < NH4Cl
(c) NaCl < NH4Cl < NaCN < HCl 73. Arrange the following compounds in the
(d) HCl < NH4Cl < NaCl < NaCN order of decreasing acidity.
OH OH OH OH
67. In which of the following molecules, the
van der Waals’ forces is likely to be the most
important in determining the melting and
boiling points?
(a) H2S (b) Br2 Cl CH3 NO2 OCH3
(c) HCl (d) CO (I) (II) (III) (IV)
68. The quantum numbers for the outer electrons of (a) III > I > II > IV (b) IV > III > I > II
an atom are given by (c) II > IV > I > III (d) I > II > III > IV
1
n = 2, l = 0 , m = 0 , s = + 74. Which one of the following has an optical
2 isomer?
(a) hydrogen(b) lithium (c) beryllium (d) boron (a) [Co(en)3 ]3+ (b) [Zn(en)2 ]2+
3+
69. X litres of carbon monoxide is present at S.T.P. It (c) [Co(H2O)4 (en)] (d) [Zn(en)(NH3 )2 ]2+
is completely oxidised to CO 2 . The volume of CO 2
formed is 11.207 L at S.T.P. What is the value of 75. The oxidation numbers of phosphorus is
X in litres? Ba(H 2PO 2)2 and xenon in Na 4 XeO 6 are
respectively
(a) 12.414 L (b) 44.828 L
(c) 11.207 L (d) 8.6035 L (a) + 2 and + 6
(b) + 3 and + 4
70. Which of the following salts undergoes anionic (c) + 1 and + 8
hydrolysis? (d) − 1 and − 6
(a) CuSO 4 (b) FeCl 3
(c) NH4Cl (d) Na 2CO 3 Category III (Q. Nos. 76 to 80)
Carry 2 marks each and one or more option(s)
Category II (Q. Nos. 71 to 75 ) is/are correct. If all correct answers are not
marked and also no incorrect answer is
Carry 2 marks each and only one option is correct. In
case of incorrect answer or any combination of more marked then score = 2 × number of correct
than one answer, 1/2 mark will be deducted. answers marked ÷ actual number of correct
answers. If any wrong option is marked or if
71. An ester (A) with molecular formula, C9H10O 2 any combination including a wrong option is
was treated with excess of CH 3 MgBr and the marked, the answer will considered wrong,
complex so formed was treated with H 2SO 4 to but there is no negative marking for the same
give an olefin (B). Ozonolysis of (B) gave a and zero mark will be awarded.
WB JEE (Engineering) Practice Set 3 41
76. Which of the following statements is/are 78. Which statement among the following is/are
correct about half-life period for a incorrect?
reaction? (a) Value of ∆G ° f cannot be determined
1 (b) Absolute value of heat content of the system
(a) For a zero order reaction t 1/ 2 ∝
a can be easily determined by calorimetery
1 (c) Absolute value of entropy cannot be known
(b) For a second order reaction t 1/ 2 ∝ (d) Absolute value of internal energy cannot be known
a
1 79. Which of the following, the hybrid orbitals of
(c) For third order reaction t 1/ 2 ∝
a the central atom have the same s-character?
(d) Time taken for 75% completion of a first order (a) XeO 3 (b) CH4 (c) [Ni(CN)4 ]2− (d) Ni(CO)4
reaction is twice to t 1/ 2
80. NaNO 3 is heated in a closed vessel, O 2 is liberated
77. Which among the following have regular and NaNO 2 is left behind. At equilibrium,
geometry? (a) increasing pressure favours reverse reaction
(a) BF4− (b) NF3 (b) addition of NaNO 2 favours reverse reaction
(c) BF3 (d) PF3 (c) increasing temperature favours forward reaction
(d) addition of NaNO 2 favours forward reaction
Mathematics
Category I (Q. Nos. 1 to 50) sin2 x
(b) x cos x + sin x − +C
4
Only one answer is correct. Correct answer will sin2 x
fetch full marks 1. Incorrect answer or any (c) sin x + x cos x − +C
2
combination of more than one answer will fetch sin2 x
−1/4 marks. (d) cos x + x sin x − +C
2
x2 π
a − a2 − x 2 −
1. Let L = lim 4 , a > 0. If L is 2
x 2 cos x
x→0 x 4 4. The value of ∫ 1 + ex
dx is equal to
π
finite, then −
2
1 1
(a) a = 2 (b) a = 1 (c) L = (d) L = π2 π2
64 32 (a) −2 (b) +2
4 4
2. If ∫ f (x) dx = ψ(x), then ∫ x 5 f (x 3) dx is equal to (c) π 2 − e − π / 2 (d) π 2 + e π / 2
1 x + y
(a) [ x3 ψ( x3 ) − ∫ x2 ψ( x3 ) dx] + C 2
3 ( ∫ e t dt)2
1 0
(b) x3 ψ( x3 ) − 3∫ x3 ψ ( x3 )dx + C 5. The value of lim is equal to
x→∞ x + y
3 2t2
1
(c) x3 ψ( x3 ) − ∫ x2 ψ( x3 ) dx + C
∫ e dt
0
3
1 (a) 0 (b) 1
(d) [ x3 ψ( x3 ) − ∫ x3 ψ ( x3 ) dx] + C xcos t 2
3 ∫
(c) lim 0 −1 (d) − 1
3. The value of ∫ (e log x + sin x) cos x dx x→ 0
x
cos 2 x
(a) x sin x + cos x − +C
4
42 WB JEE (Engineering) Practice Set 3
21. The number of rational terms in the 28. Let W denote the words in the English
expansion of (1 + 2 + 3 3)6 is dictionary, define the relation R by
R = {(x , y) ∈ N × W : the words x and y have
(a) 6 (b) 7 (c) 5 (d) 8 at least one letter is common}. Then R is
(a) not reflexive, symmetric and transitive
cos 2 θcos θ sin θ
22. If E(θ) = and θ and φ (b) reflexive, symmetric and not transitive
cos θ sin θ sin 2 θ (c) reflexive, not symmetric and transitive
π (d) reflexive, symmetric and transitive
differ by an odd multiple of , then E(θ) ⋅ E(φ)
2 29. Two numbers are selected randomly from the
is a set S = {1, 2, 3, 4, 5, 6} without replacement
(a) null matrix (b) unit matrix one by one. The probability that the
(c) diagonal matrix (d) None of these minimum of two numbers is less than 4 is
1 14
(a) (b)
23. Let M be a 3 × 3 non-singular matrix with 15 15
det (M) = α. If M −1 adj (adj M) = kI , then the (c)
1
(d)
4
value of k is 5 5
(a) 1 (b) α (c) α 2 (d) α 3 30. A random variable X takes the values 0, 1, 2, 3
and its mean is 1.3. If P(X = 3) = 2 P(X = 1)
24. If α is a non-real cube of −2, then the value of and P(X = 2) = 0 .3, then P(X = 0) is
1 2α 1
(a) 0.1 (b) 0.2
α 2
1 3α 2 , is (c) 0.3 (d) 0.4
2 2α 1 31. If a and b are distinct positive real numbers
(a) − 11 (b) −12 (c) −13 (d) 0 such that a , a1 , a 2 , a 3 , a 4 , a 5 , b are in AP,
a , b1 , b2 , b3 , b4 , b5 , b are in GP and
1 0 1 0 a , c1 , c2 , c3 , c4 , c5 , b are in HP, then the roots of
25. If A = and I = 0 1, then which one
1 1 a 3 x 2 + b3 x + c3 = 0 are
of the following holds for all n ≥ 1, by the (a) real and distinct (b) real and equal
principle of mathematical induction (c) imaginary (d) None of these
(a) A n = 2 n − 1 A + (n − 1)I (b) A n = nA + (n − 1)I 32. If the angles A , B and C of a triangle are in
(c) A n = 2 n − 1 A − (n − 1)I (d) A n = nA − (n − 1)I arithmetic progression and if a , b and c
denote the lengths of the sides opposite to
4
26. If A = {(x , y) : y = , x ≠ 0 } and A , B and C respectively, then the value of the
x a c
expression sin 2C + sin 2 A is
B = {(x , y) : x 2 + y 2 = 8 , x , y ∈ R}, then c a
(a) A ∩ B = φ (a)
1
(b)
3
(b) A ∩ B contains one point only 2 2
(c) A ∩ B contains two points only (c) 1 (d) 3
(d) A ∩ B contains 4 points only
33. Let A be the fixed point (0, 4) and B(2t, 0) be a
27. Let the function f : R − { − b} → R − {1} be moving point. Let M be the mid-point of AB
x+a and let the perpendicular bisector of AB meet
defined by f (x) = , a ≠ b, then the Y -axis at R. The locus of the mid-point P
x+b
of MR is
(a) f is one-one but not onto 1
(b) f is onto but not one-one (a) x2 + y2 = (b) ( y − 2 )2 − x2 = 4
4
(c) f is both one-one and onto (c) y + x2 = 2 (d) 3 x2 + y2 = 8
(d) None of the above
44 WB JEE (Engineering) Practice Set 3
34. Let a , b, c and d be non-zero numbers. If the 41. The equation of a tangent to the hyperbola
point of intersection of the lines 16 x 2 − 25 y 2 − 96 x + 100 y − 356 = 0 , which
4 ax + 2 ay + c = 0 and 5 bx + 2 by + d = 0 lies in π
the fourth quadrant and is equidistant from makes an angle with the transverse axis, is
the two axes, then 4
(a) 3bc − 2 ad = 0 (b) 3bc + 2 ad = 0 (a) y = x + 2 (b) y = x − 5 (c) y = x + 3 (d) x = y + 2
(c) 2 bc − 3ad = 0 (d) 2 bc + 3ad = 0
42. The slope of a common tangent to the ellipse
35. If a variable line passes through the point of x2 y2
intersection of the lines x + 2 y − 1 = 0 and 2
+ = 1 and a concentric circle of radius r is
2 x − y − 1 = 0 and meets the coordinate axes a b2
in A and B, then the locus of the mid-point of r 2 − b2 r 2 − b2
(a) tan−1 (b)
AB is a −r
2 2
a2 − r 2
(a) x + 3 y = 0 (b) x + 3 y = 10 r 2 − b2 a2 − r 2
(c) x + 3 y = 10 xy (d) None of these (c) (d)
a −r
2 2
r 2 − b2
36. Let the perpendiculars from any point on the
line 2 x + 11 y = 5 upon the lines 43. The equation of the plane containing the line
24 x + 7 y − 20 = 0 and 4 x − 3 y − 2 = 0 have x +1 y − 3 z + 2
= = and the point (0, 7, −7), is
the lengths P1 and P2 , respectively. Then, −3 2 1
(a) 2 P1 = P2 (b) P1 = P2 (a) x + y + z = 1 (b) x + y + z = 2
(c) P1 = 2 P2 (d) None of these (c) x + y + z = 0 (d) None of these
37. The locus of the centre of the circles which 44. Let a , b and c be three real numbers satisfying
touch both the circles x 2 + y 2 = a 2 and 1 9 7
x 2 + y 2 = 4 ax externally has the equation [ a b c] 8 2 7 = [0 0 0 ].
(a) 12( x − a)2 − 4 y2 = 3a2 (b) 9( x − a)2 − 5 y2 = 2 a2 7 3 7
(c) 8 x2 − 3( y − a)2 = 9a2 (d) None of these
If the point P(a , b, c) lies on the plane
38. The range of values of a for which the point 2 x + y + z = 1, then the value of 7 a + b + c is
(a , 4) is outside the circles x 2 + y 2 + 10 x = 0 (a) 0 (b) 12 (c) 7 (d) 6
and x 2 + y 2 − 12 x + 20 = 0 , is 45. If [⋅] denotes the greatest integer function,
(a) (−∞, − 8) ∪ (−2, 6) ∪ (6, ∞ ) tan([ −2 π 2 ] x 2) − x 2 tan[ −2 π 2 ]
(b) (−∞, − 2 ) then lim is
(c) (−∞, − 8) ∪ (−2, ∞ )
x→0 sin 2 x
(d) None of the above equal to
(a) − 20 + tan20° (b) 20 + tan20°
39. A hyperbola, having the transverse axis of (c) 20 (d) None of these
length 2sinθ, is confocal with the ellipse
1
3 x 2 + 4 y 2 = 12. Then, its equation is 46. If f (x) = , then the derivation of the
1− x
(a) x2cosec 2θ − y2sec 2θ = 1
composite function f [ f { f (x)}] is equal to
(b) x2sec 2θ − y2cosec 2θ = 1 1
(a) 0 (b) (c) 1 (d) 2
(c) x2 sin2 θ − y2 cos 2 θ = 1 2
(d) x2 cos 2 θ − y2 sin2 θ = 1
47. If the period of the function
sin(sin(nx))
40. The number of values of c such that the f (x) = , n ∈ N is 6 π, then n is
straight line y = 4 x + c touches the curve x
tan
x2 n
+ y 2 = 1, is
4 equal to
(a) 0 (b) 1 (c) 2 (d) infinite (a) 3 (b) 2
(c) 1 (d) None of these
WB JEE (Engineering) Practice Set 3 45
48. Let the sequence < bn > of real numbers 54. Let a = 2$i + $j − 2k$ and b = $i + $j. If c is a vector
satisfy the recurrence relation such that a ⋅ c =|c|,|c − a| = 2 2 and the angle
1 125 between a × b and c is 30°. Then,|(a × b) × c|
bn + 1 = 2 bn + 2 , bn ≠ 0, then lim bn is
3 bn n→ ∞ is equal to
2 3
equal to (a) (b) (c) 2 (d) 3
3 2
(a) 0 (b) ∞
2 55. Let S1 , S2 ,… be squares such that for each
(c) 5 (d)
3
n ≥ 1, the length of a side of Sn equals the
49. The set of all points where the function length of a diagonal of Sn + 1 . If the length of
f (x) = 3
x 2 | x| is differentiable is a side of S1 is 10 cm, then for which of the
following values of n is the area of Sn less
(a) [0, ∞ ) (b) (0, ∞ )
than 1 sq cm?
(c) (−∞, ∞ ) (d) (−∞, 0) ∪ (0, ∞ )
(a) 7 (b) 8 (c) 5 (d) 6
50. Let f be twice differentiable function
56. The value of k for which the inequality
1 = 1, f (2) = 4 , f (3) = 9 , then
satisfying f ()
|Re(z)| + |1m (z)| ≤ λ |z| is true for all z ∈ C , is
(a) f ′( x) = 2 for all x ∈ R
(a) 2 (b) 2
(b) f ′( x) = 5 = f ′′( x), for some x ∈[1, 3]
(c) 1 (d) None of these
(c) there exists at least one x ∈(1, 3) such that
f ′′( x) = 2 57. Given, 2 x − y + 2z = 2; x − 2 y + z = − 4;
(d) None of the above
x + y + λz = 4, then the value of λ such that
the given system of equations has no
Category II (Q. No. 51 to 65)
solution, is
Carry 2 marks each and only one option is (a) 3 (b) 1 (c) 0 (d) −3
correct. In case of incorrect answer or any
combination of more than one answer, 1/2 mark 58. Let X and Y be two non-empty sets and
will be deducted. f : X → Y be a function such that
f (C) = { f (x) : x ∈ C } for C ⊆ X
51. If C 0 , C1 , C 2 ,…, C n denote the binomial and f −1 (D) = { x : f (x) ∈ D} for D ⊆ Y
coefficient in the expansion of (1 + x)n, then
If A ⊆ X and B ⊆ Y , then
C1 C 3 C 5
+ + + … is equal to (a) f −1(f( A)) = A
2 4 6
(b) f −1(f( A)) = A only if f( X ) = Y
2n − 1 2n
(a) (b) (c) f(f −1(B)) = B only if B ⊆ f( X )
n+1 n+2
(d) f(f −1(B)) = B
2n − 1 2n
(c) (d)
n n+1 59. The function f : (−∞ , − 1] → (0 , e 5] defined by
3
− 3x + 2
52. If I n = ∫ (log x)n dx , then I n + n ⋅ I n − 1 is equal to f (x) = e x is
(a) ( x log x)n (b) x (log x)n (a) one-one and onto (b) one-one and into
(c) n(log x)n (d) (log x)n − 1 (c) many one and onto (d) many one and onto
53. Let a = $i + 2$j + k$ , b = $i − $j + k$ , c = $i + $j − k$ . A 60. The incentre of the triangle with vertices
(1, 3), (0 , 0) and (2, 0) is
vector caplanar to a and b has a projection
3
(b) ,
1 2 1
along c of magnitude , then the vector is (a) 1,
3 2 3 3
2 3
(d) 1,
(a) 4$i − $j + 4k$ (b) 4$i + $j − 4k$ 1
(c) ,
3 2 3
(c) 2 $i + $j + k$ (d) None of these
46 WB JEE (Engineering) Practice Set 3
Answers
Physics
1.(a) 2.(a) 3.(a) 4.(c) 5.(c) 6.(c) 7.(b) 8.(d) 9.(c) 10.(c)
11.(a) 12.(a) 13.(d) 14.(a) 15.(c) 16.(b) 17.(c) 18.(a) 19.(c) 20.(d)
21.(d) 22.(b) 23.(a) 24.(b) 25.(a) 26.(d) 27.(c) 28.(c) 29.(b) 30.(a)
31.(c) 32.(a) 33.(a) 34.(b) 35.(d) 36.(c, d) 37.(a, c, d) 38.(b, c) 39.(a, d) 40.(a, b, d)
Chemistry
41. (c) 42. (a) 43. (c) 44. (d) 45. (d) 46. (c) 47. (c) 48. (c) 49. (d) 50. (b)
51. (c) 52. (d) 53. (b) 54. (d) 55. (a) 56. (a) 57. (c) 58. (c) 59. (b) 60. (c)
61. (d) 62. (c) 63. (a) 64. (c) 65. (b) 66. (d) 67. (b) 68. (b) 69. (c) 70. (d)
71. (b) 73. (d) 73. (a) 74. (a) 75. (c) 76. (b, d) 77. (a, c) 78. (a, b) 79. (b, c, d) 80. (a, c)
Mathematics
1. (c) 2. (c) 3. (a) 4. (a) 5. (a) 6. (c) 7. (a) 8. (b) 9. (c) 10. (b)
11. (a) 12. (a) 13. (c) 14. (b) 15. (a) 16. (c) 17. (b) 18. (c) 19. (c) 20. (d)
21. (b) 22. (a) 23. (b) 24. (c) 25. (d) 26. (c) 27. (c) 28. (b) 29. (d) 30. (d)
31. (c) 32. (d) 33. (c) 34. (a) 35. (c) 36. (b) 37. (a) 38. (a) 39. (a) 40. (c)
41. (a) 42. (b) 43. (c) 44. (d) 45. (a) 46. (c) 47. (a) 48. (c) 49. (d) 50. (c)
51. (a) 52. (b) 53. (a) 54. (b) 55. (b) 56. (b) 57. (b) 58. (c) 59. (b) 60. (d)
61. (c) 62. (a) 63. (a,d) 64. (c) 65. (c) 66. (a,b) 67. (b,c,d) 68. (a,c) 69. (a,d) 70. (a,c,d)
71. (c) 72. (a) 73. (b) 74. (a) 75. (a)
Practice Set 4
Physics
Category I (Q. Nos. 1 to 30) 4. Ionisation potential of hydrogen atom is
Carry 1 mark each and onlyone option is correct. 13.6 eV. The least energy of photon of Balmer
In case of incorrect answer or any combination of series is
more than one answer, 1/ 4 mark will be deducted. (a) 3.4 eV (b) 1.89 eV (c) 10.2 eV (d) 8.5 eV
9. A particle of mass m collides with another 16. If C s is the velocity of sound in air and C is
stationary particle of mass m. If the particle the rms velocity, then
m stops just after collision. The coefficient of (a) C s < C (b) C s = C
restitution is equal to γ
1/ 2
m m M−m (c) C s = C (d) None of these
(a) (b) (c) 1 (d) 3
M+ m M M+ m
17. At constant temperature, the volume of gas is
10. A machine gun fire 360 bullets per minute to be decreased by 4%. The pressure must be
with a velocity of 600 ms −1 . If the power of increased by
gun is 5.4 kW. Mass of each bullet is (a) 4% (b) 4.16% (c) 8% (d) 3.86%
(a) 5 kg (b) 0.5 kg (c) 5 g (d) 0.5 g
18. Find the equivalent capacitance of the system
11. A range of projectile fixed at an angle of 15° across the terminals A and B. All the
is 50 m. If it is fired with the same speed at capacitors have equal capacitances.
an angle of 45°. Its range will be
C
(a) 25 m (b) 50 m (c) 100 m (d) 77.6 m A
23. The magnetic susceptibility of a 29. A beam of light of wavelength 600 nm from a
paramagnetic material at − 73 °C is 0.0075. Its distance source falls on a single slit 1.0 mm
value at − 173 °C will be wide and the resulting diffraction pattern is
(a) 0.0045 (b) 0.0030 (c) 0.015 (d) 0.0075 observed on a screen 2 m away. What is the
distance between first dark fringe on either
24. The magnetic moment µ of a revolving side of the central bright fringe.
electron around the nucleus varies with (a) 1.5 mm (b) 1 mm (c) 1.2 mm (d) 2.4 mm
principal quantum numbers n as
1 1 30. In young’s double slit experiment, the length
(a) µ ∝ n (b) µ ∝ (c) µ ∝ n2 (d) µ ∝ 2 of band is 1 mm. The ring width is 1.021 mm.
n n
The number of fringe is
25. The value of i1 is the circuit diagram will be (a) 45 (b) 46 (c) 47 (d) 48
1 3 3
(a) 1 A (b) A (c) A (d) A
2 4 2 E L
R1
26. Under what condition will the strength of
S R
current in a wire of resistance R be the same
for connection is n-series or in parallel of n
identical cells each of the internal resistance
12 −3t
r1 when (a) e V (b) 6(1 − e −t / 0. 2 ) V
t
(a) R = nr (b) R = r / n (c) 12e −5t V (d) 6e −5t V
(c) R = r (d) R → ∞, r → 0
32. Water flows through a horizontal pipe of
27. In an L-R circuit, the value of L is 0.4 / π and variable cross-section at rate of 20 L per min.
the value of R is 30 Ω. If in the circuit, an What will be velocity of water at a point
alternating emf of 200 V at 50 cycle per where diameter is 4 cm?
second is connected, the impedance of the (a) 0.2651 ms −1 (b) 0.5639 ms −1
circuit and current will be (c) 0.4639 ms −1 (d) 0.3639 ms −1
(a) 11.4 Ω, 17.5 A (b) 30.7 Ω, 6.5 Ω
33. Water falls from a height of 500 m. What is
(c) 40.4 Ω, 5A (d) 50 Ω, 4A
the rise in temperature of water at the bottom,
28. Two identical glass (µ g = 3 / 2) equiconvex if whole energy is used up in heating water?
lenses of focal length f are kept in contact. (a) 0.96°C (b) 1.02°C (c) 1.16°C (d) 0.23°C
The space between the two lenses is filled 34. Instantaneous displacement current of 1.0 A
with water (µ w = 4 / 3). The focal length of in the space between the parallel plates of
the combination is 1 µF capacitor can be established by changing
f 4f 3f potential difference of
(a) f (b) (c) (d)
2 3 4 (a) 10−6 Vs −1 (b) 106 Vs −1 (c) 10−8 Vs −1 (d) 108 Vs −1
WB JEE (Engineering) Practice Set 4 51
35. The self inductance of the motor of an (a) Heat is absorbed by the gas in all the three
electric fan is 10 H. In order to impart paths
(b) Heat absorbed/released by the gas is maximum
maximum power at 50 Hz. It should be
in path 2
connected to a capacitor of
(c) Temperature of the gas increases first and then
(a) 3 × 10−6 F (b) 2 × 10−6 F decreased continuously in path 2
−6
(c) 10 F (d) 10−6 F (d) Change in internal energy of the gas is same
along all the three paths
Category III (Q. Nos. 36 to 40)
38. A particle is projected from a point A with a
Carry 2 mark each and onlyone option is/are velocity v at an angle of elevation θ. At a
correct. If all correct answers are not marked and certain point B, the particle moves at right
also no incorrect answer is marked then score angle to its initial direction. Then,
= 2 × number of correct answers marked ÷ actual
(a) velocity of particle at B is v sin θ
number of correct answers. If any wrong option is (b) velocity of particle at B is v cot θ
marked or if any combination including a wrong (c) velocity of particle at B is v tan θ
option is marked, the answer will be considered v
(d) velocity of flight from A to B is
wrong, but there is not negative marking for the g sin θ
same and zero mark will be awarded.
39. The potential difference between the points A
36. When photons of energy 4.25 eV strike the and B in the circuit shown in figure is 16 V.
surface of a metal, the ejected photoelectrons Which is/are the correct statements out of
have a maximum kinetic energy E A eV and the following?
de-Broglie wavelength λ A . The maximum
kinetic energy of a photoelectrons diberated 9V 1Ω 3V 1Ω
4Ω + – 3Ω
from another metal B by photons of energy A + – B
4.70 eV is E B = (E A − 150
. ) eV. If the 1Ω
de-Broglie wavelength of the these 2Ω
photoelectron is λ B = 2 λ A , then
(a) The current through the 2Ω resistor is 3.5 A
(a) the work function of A is 2.25 eV
(b) The current through the 4Ω resistor is 2.5 A
(b) the work function of B is 4.20 eV
(c) The current through the 3Ω resistor is 1.5 A
(c) EA = 2.5 eV (d) EB = 2.75 eV
(d) The potential difference between the terminals
37. A gas undergoes a change in its state from of the 9 V battery is 7 V
position A to position B via three different
40. A charged particle P leaves the origin with
paths as shown in the figure. Select the
speed v = v0 , at some inclination with the
correct statement.
X -axis. There is uniform magnetic field B
Y
along the X -axis. P strikes a fixed target T on
the X -axis for a minimum value of B = B0 . P
1 will also strikes T, if
p
2 (a) B = 2 B0 , v = 2 v 0
A
3
B (b) B = 2 B0 , v = v 0
(c) B = B0 , v = 2 v 0
(d) B = B0 / 2, v = v 0 / 2
X
O V
Chemistry
Category I (Q. Nos. 41 to 70) 47. KF combines with HF to form KHF2 . The
compound contains the species.
Carry 1 mark each and only one option is correct.
(a) K + , F − and H+ (b) K + , F − and HF
In case of incorrect answer or any combination of
(c) K + and [HF2 ]− (d) [KHF]+ and F −
more than one answer, 1/4 mark will be deducted.
48. Which of the following factors is most
41. How many moles of magnesium phosphate important in making the fluoride as strongest
[Mg 3 (PO 4)2 ] will contain 0.25 moles of oxidising agent?
oxygen atoms? (a) Electron affinity
(a) 0.02 (b) 3125
. × 10−2 (b) Ionisation enthalpy
(c) 1.25 × 10−2 (d) 2.5 × 10−2 (c) Hydration enthalpy
(d) Bond dissociation energy
42. 250 mL of sodium carbonate solution 49. The electrons can be identified by quantum
contains 2.65 g of Na 2CO 3 . If 10 ml of this numbers n and l, the correct order of
solution is diluted to 500 mL, the increasing energy for the following orbitals is
concentration of the diluted solution will be 1. n = 4, l = 1 2. n = 4 , l = 0
(a) 0.01 M (b) 0.001 M 2. n = 3 , l = 2 4. n = 3 , l = 1
(c) 0.05 M (d) 0.002 M
(a) (3) < (4) < (2) < (1) (b) (4) < (2) < (3) < (1)
43. As temperature is raised from 20°C to 40°C, (c) (2) < (4) < (1) < (3) (d) (1) < (3) < (2) < (4)
the average kinetic energy of neon atoms 50. Hydrolysis of trichloromethane with aqueous
changes by a factor of which of the following KOH gives
1 313 (a) methanol (b) acetic acid
(a) (b) (c) ethanol (d) formic acid
2 293
313
(c) (d) 2 51. For the reaction,
293
∆/Cu
—N2Cl —Cl + N2
44. A blue liquid formed by equimolar mixture of
two gases at − 30 °C is (A)
(a) N2O (b) N2O 3
(c) N2O 4 (d) N2O 5
the half life is independent of concentration
of (A). After 10 min. volume of N 2 gas is 10 L
45. Which of the following oxide is neutral in and after complete reaction is 50 L. Hence,
nature? rate constant is
2.303 2.303
(a) CO (b) SnO 2 (a) log 5 min−1 (b) . min−1
log 125
(c) ZnO (d) SO 2 10 10
2.303 2.303
(c) log 2 min−1 (d) log 4 min−1
46. Which of the following is the correct 10 10
statement about cryolite?
(a) Cryolite is NaAl 3F6 and is used in electrolysis of
52. Which of the following pair of compounds
alumina for decreasing the electrical conductivity. cannot exist together in a solution?
(b) Cryolite is Na 3 AlF6 and is used in electrolysis of (a) NaHCO 3 and NaOH (b) NaHCO 3 and H2O
alumina for lowering the melting point of alumina. (c) NaHCO 3 and NaCl (d) Na 2CO 3 and NaOH
(c) Cryolite is Na 3 AlF6 and is used in electrolysis of 53. An acid, that cannot suitable for preparation
alumina for freezing the alumina.
of H 2 by the action of metals is
(d) Cryolite is NaAl 3F6 and is used in electrolysis of
alumina to produce Na, Al and fluorine. (a) HCl (b) CH3COOH
(c) HNO 3 (d) H2SO 4
WB JEE (Engineering) Practice Set 4 53
54. During electrolysis of fused calcium hydride, (a) III > IV > II > I (b) III > II > IV > I
hydrogen is produced at (c) III > II > I > IV (d) II > III > IV > I
(a) cathode 62. The decreasing order of rate of bromination
(b) anode of the following compounds is
(c) hydrogen will not liberate ⊕ ⊕
(d) H2 produced will react with oxygen and give H2O (I) ph NMe3 (II) ph CH 2 N Me3
55. Assuming that each salt is 90% dissociated. (III) ph Me (IV) ph N.Me2
Which of the following will have highest (a) (I) >(II) >(III) > (IV)
osmotic pressure? (b) (IV) >(III) >(II) > (I)
(a) Decinormal Al 2 (SO 4 )3 (b) Decinormal BaCl 2 (c) (III) >(IV) >(I) > (II)
(c) Decinormal Na 2SO 4 (d) Decinormal Urea. (d) (III) >(IV) >(II) > (I)
61. The decreasing order of reactivity of the 67. Which of the following undergoes
following compounds with HBr is nucleophilic substitution exclusively by
SN1 mechanism?
OH H3C—
I. II. (a) Benzyl chloride (b) Ethylchloride
OH (c) Chloro propane (d) Isopropyl chloride
OH
III. MeO— IV. 68. Hydrolysis of sucrose is called
(a) inversion (b) esterification
OH OMe (c) hydration (d) saponification
54 WB JEE (Engineering) Practice Set 4
(b) — )n
( CH2—CH— 74. Which of the following is a polar molecule
(a) SF4 (b) SiF4 (c) BaCl 2 (d) BF3
(c) — )n
( CH2—CH—
70. Which one of the following is employed as a Category III (Q Nos. 76 to 80)
tranquilizer drug?
Carry 2 marks each and one or more options(s)
(a) Promethazine (b) Valium
is/are correct. If all correct answers are not
(c) Naproxen (d) Mifepristone
marked and also no incorrect answer is marked
then score = 2 × number of correct answers
Category II (Q Nos. 71 to 75)
marked ÷ actual number of correct answers. If any
Carry 2 marks each and only one option is wrong option is marked or if any combination
correct. In case of incorrect answer or any including a wrong option is marked, the answer
combination of more than one answer, 1/2 mark will considered wrong, but there is no negative
will be deducted. marking for the same and zero mark will be
71. In the following reaction, identify product (C) awarded.
ph NH 2 HNO
2→ (A) HF
→ B
D
→ (C) 76. Which of the following graphs is correct for a
0°C BF3
+
zero order reaction?
(a) pH N BF4
Concentration
Reaction rate
(b) F
of reactant
(a) (b)
(c) F F
Concentration
of reactant
O slope = – k
(i) CH3NH2 (c) (d)
Product is
(ii)LiAlH4 (iii)H2O
O—NHCH3 Time Time
(a) (b)
77. Which of the following are cyclic compounds ?
NHCH3
(a) Borazole
NHCH3 NHCH3 (b) Pyrrole
(c) (d) (c) Anthracene
(d) Isobutylene
OH O
78. Which of the following species are
73. For the reaction, isoelectronic as well as isostructural?
2NO 2(g) - 2NO(g) + O 2 (g). (a) CH4 (b) NH+4 (c) HF (d) NO −3
WB JEE (Engineering) Practice Set 4 55
79. Which of the following options are not in 80. Which of the following reactions would give
accordance with the property mentioned propanal?
against them? (a) Ethyl propanoate + DBAH (diisobutyl aluminium
(a) F2 >Cl 2 >Br2 >I2 Oxidising power hydride) [(i - C 4H9 )2 AlH] at –70°C followed by
(b) MI > MBr > MCl > MF Ionic character of metal halide hydrolysis
(c) F2 >Cl 2 >Br2 >I2 Bond dissociation enthalpy (b) Propyl propanoate + DIBAL-H/H2O
(d) HI <HBr <HCl <HF Hydrogen-halogen bond (c) Ethyl cyanide + DBAH
strength (d) Propanoyl chloride + LBAH
Mathematics
Category I (Q. Nos. 1 to 50) 6. If ∫ f (x)dx = 2{ f (x)} 3 + c, then f (x) is
Only one answer is correct. Correct answer will fetch x 1 x
full marks 1. Incorrect answer or any combination of (a) (b) x3 (c) (d)
2 x 3
more than one answer will fetch−1/4 marks.
400 π
1. lim
log[ x ]
, where [ x ] denotes the greatest
7. The value of the integral ∫ 1 − cos 2 x dx is
x→∞ x 0
integer less than or equal to x, is (a) 200 2 (b) 400 2
(a) 0 (b) 1 (c) 800 2 (d) None of these
(c) −1 (d) Non-existent 1 1 2 4 1
8. lim sec 2 + sec 2 + … sec 2 1°
2. If u = ∫ e sin bx dx and v = ∫ e cos bx dx , then n→ ∞ n2 2 2 2
ax ax n n n n
equals
(u + v )(a + b ) is equal to
2 2 2 2
1
(a) tan1° (b) tan1°
(a) 2e ax (b) e 2 ax (c) 2e 2 ax (d) bxe ax 2
1 1
3. The equation of a curve passing through the (c) cosec1° (d) sec 1°
3
2 2
point (0, 1) be given by y = ∫x ⋅ e x dx . If the
2
9. The general solution of the differential
equation of the curve be written in the form equation (x + y)(dx − dy) = dx + dy is
x = f (y), then f (y) is equal to (a) ( x − y) = ke x − y (b) x + y = ke x + y
(a) loge (3 y − 2 ) (b) 3 loge (3 y − 2 ) (c) ( x + y) = k( x − y) (d) x + y = ke x − y
(c) 3 loge (2 − 3 y) (d) None of these
10. Solution of the differential equation
1
π π
4. If ∫ t f ()
t dt = 1 − sin x for all x ∈ 0 , , then
2
cos x dy = y(sin x − y)dx , 0 < x < is
sin x 2 2
1 (a) y tan x = sec x + c (b) tan x = (sec x + c )y
f is equal to (c) sec x = (tan x + c )y (d) ysec x = tan x + c
3
(a) 3 (b) 3 11. If the chords of tangents from two points
(c)
1
(d) None of these
(x1 , y1) and (x 2 , y 2) to the hyperbola
3 x 2 y2 x x
2
− 2 = 1 are at right angles, then 1 2
2 a b y1 y 2
5. The value of ∫ xd([ x ] − x) is equal to
0
a2 b2 b4 a4
1 (a) − (b) − (c) − (d) −
(a) (b) 1 (c) −1 (d) 0 b 2
a 2
a 4
b4
2
56 WB JEE (Engineering) Practice Set 4
12. If errors of 1% each are made in the base 20. For n ∈ N , 32n + 2 − 23 n − 9 is divisible by
radius and height of a cylinder, then the (a) 3 (b) 9 (c) 64 (d) 81
percentage error in its volume is
(a) 1% (b) 2% 21. The sum of the rational terms in the
(c) 3% (d) None of these expansion of ( 2 + 5 3)10 is
(a) 32 (b) 9
13. A GP consists of 2n terms. It the sum of the
(c) 41 (d) None of these
terms occupying the add places is S1 , and that
S 22. If A satisfies the equation
of the terms in the even places is S2 , then 2 is
S1 x 3 − 5 x 2 + 4 x + λ = 0 , then A −1 exists if
(a) independent of a (a) λ ≠ 1 (b) λ ≠ 2 (c) λ ≠ − 1 (d) λ ≠ 0
(b) independent of r
(c) independent of a and r 23. If B is a non-singular matrix and A is a
(d) dependent on r square matrix, then det (B−1 AB) is equal to
14. If log (sin x + 2 2 cos x) ≥ 2, −2 π ≤ x ≤ 2 π, (a) det ( A −1 ) (b) det (B−1 )
3
(c) det ( A) (d) det (B)
then the number of values of x is
(a) 0 (b) 3 x + λ x x
(c) infinite (d) None of these 24. If f (x) = x x+λ x ,
15. If arg (z) < 0, then arg (− z) − arg(z) is equal to x x x + λ
(a) π (b) − π then f (3 x) − f (x) is equal to
π π (a) 3 xλ2 (b) 6 xλ2
(c) − (d)
2 2 (c) xλ2 (d) None of these
28. Let R be the relation over the set of all 35. Let A(2, − 3) and B(−2, 1) be vertices of a
straight lines in a plane such that triangle ABC. If the centroid of this triangle
l1 Rl2 ⇔ l1 ⊥ l2 . Then, R is moves on the line 2 x + 3 y = 1, then the locus
(a) symmetric (b) reflexive of the vertex C is the line
(c) transitive (d) an equivalence relation (a) 3 x − 2 y = 3 (b) 2 x − 3 y = 7
(c) 3 x + 2 y = 5 (d) 2 x + 3 y = 9
29. A biased coin with probability P, 0 < P < 1, 36. The value of λ, for which the lines joining the
of heads is tossed with a head appears for point intersection of curves C1 and C 2 to the
the first time. If the probability that the origin are equally inclined to the axis of X .
2 C1 : λx 2 + 3 y 2 − 2 λxy + 9 x = 0
number of tosses required is even is , then P
5 C 2 :3 x 2 − 4 y 2 + 8 xy − 3 x = 0
is equal to
4
2 1 1 1 (a) λ = (b) λ = 12
(a) (b) (c) (d) 3
3 2 3 4
(c) λ = 1 (d) None of these
30. A random variable X takes values − 1, 0 , 1, 2 37. The locus of the centre of a circle which cuts
1 + 3P 1 − P 1 + 2P 1 − 4 P off an intercept of constant length on the
with probabilities , , ,
4 4 4 4 X -axis and which passes through a fixed
respectively, where P varies over R. Then the point on the Y -axis is
minimum and maximum values of the mean (a) a circle (b) a parabola
of X are respectively (c) an ellipse (d) a hyperbola
7 1 1 5
(a) − and (b) − and 38. If one of the diameter of the circle
4 2 16 16
7 5 1 5 x 2 + y 2 − 2 x − 6 y + 6 = 0 is a chord to the circle
(c) − and (d) − and
4 16 16 4 with centre (2 , 1,) then the radius of circle is
(a) 3 (b) 2 (c) 3 (d) 2
31. The solution of the equation
5log a x
+ 5 ×log a 5 = 3, a > 0 is 39. The product of the lengths of perpendiculars
log a 5 −log 5 a drawn from any point on the hyperbola
(a) 2 (b) 2
(c) 2 −log a 5 (d) 2log 5 a
x 2 − 2 y 2 − 2 = 0 to its asymptotes is
1 2 3
(a) (b) (c) (d) 2
32. The sides of a ∆ABC are in AP such that a 2 3 2
minimum { b, c}. Then cos A may be equal to
40. The radius of the circle passing through the
3c − 4 b 3c − 4 b foci of the ellipse 9 x 2 + 16 y 2 = 144 and
(a) (b)
2b 2c having its centre at (0, 3) is
4c − 3 b 4c − 3 b 7
(c) (d) (a) 4 (b) 3 (c) 12 (d)
2b 2c 2
6 6 41. The equation of the tangent to the curve
33. A variable line through the point , cuts
5 5 4 x 2 − 9 y 2 = 1 which is parallel to 4 y = 5 x + 7 is
the coordinate axes in the points A and B. If (a) 24 y − 30 x = 17 (b) 30 y − 24 x = ± 161
the point P divides AB internally in the ratio (c) 24 y − 30 x = ± 161 (d) None of these
2 : 1, then the locus of P is 2
πx + sin 4 πx + x − [ x ]
(a) xy = 2 x + y (b) 5 xy = 2 x + y 42. The function f (x) = 3sin ,
(c) 5 xy = 2(2 x + y) (d) 5 xy = 2( x + 2 y) where [ x ] denotes the greatest integer less
than or equal to x, is
34. The bisector of the acute angle formed
(a) a periodic function with period 1
between the lines 4 x − 3 y + 7 = 0 and
(b) a periodic function with period 2
3 x − 4 y + 14 = 0 has the equation 1
(c) a periodic function with period
(a) x + y + 3 = 0 (b) x − y − 3 = 0 2
(c) x − y + 3 = 0 (d) 3 x + y − 7 = 0 (d) not a periodic function
58 WB JEE (Engineering) Practice Set 4
44. The equation of the plane through the line Category II (Q. Nos. 51 to 65)
x + y + z − 3 = 0 = 2 x − y + 3 z + 1 and parallel
x y z Carry 2 marks each and only one option is
to the line = = , is correct. In case of incorrect answer or any
1 2 3
combination of more than one answer, 1/2 mark
(a) x − 5 y + 3 z = 7 (b) x − 5 y + 3 z = − 7
(c) x + 5 y + 3 z = 7 (d) x + 5 y + 3 z = − 7
will be deducted.
∞ ∞
1 π2 1
45. If α and β are the roots of the equation 51. If ∑ (2r − 1)2 = 8
, then ∑ 2 is equal to
r
ax 2 + bx + c = 0 , then r =1 r =1
1 π2 π2
(a) (b)
x −α
lim (1 + ax 2 + bx + c) is 24 3
x→α π2
(c) (d) None of these
(a) e a( α − β ) (b) e a( β − α ) 6
(c) 1 (d) None of these
52. If I n = ∫ (log x)n dx , then I n + n ⋅ I n − 1 is
46. Let f and g be differentiable functions
equal to
satisfying g ′ (a) = 2, g(a) = b and fog = 1
(identity function). Then, f ′(b) is equal to (a) ( x log x)n (b) x (log x)n (c) n(log x)n (d) (log x)n − 1
2
(a) 2 (b) 53. Given,|a| = |b| = 1 and|a + b| = 3. If c be
3
1 vector such that c − a − 2 b = 3(a × b), then
(c) (d) None of these
2 c ⋅ b is equal to
1 1 3 5
(a) − (b) (c) (d)
47. A line makes the same angle θ, with each of 2 2 2 2
the X and Z-axes. If the angle β, which it $ β = b$i + c$j + ak$ and
makes with Y -axis, is such that 54. Let α = a$i + b$j + ck,
sin 2 β = 3 sin 2 θ, then cos 2 θ equals γ = c$i + a$j + bk$ be three coplanar vectors with
(a)
2
(b)
1
(c)
3
(d)
2 a ≠ b, and v = $i + $j + k$ . Then v is
5 5 5 3 perpendicular to
x (a) α (b) β
2 (c) γ (d) All of these
48. lim where [⋅] denotes the greatest
π log (sin x)
x→ e 55. A particle P starts from the point Z 0 = 1 + 2i,
2
integer function) where i = − 1 . It moves first horizontally
(a) does not exist (b) equals 1 away from origin by 5 units and then
(c) equals 0 (d) equals −1 vertically away from origin by 3 units to
reach a pont Z1 . From Z1 , the particle moves
49. Let f be a function satisfying f ′ (x) = f (x). 2 units in the direction of the vector $i + $j
Then, ( f −1)′′ (x) is equal to π
1 1
and then it moves through an angle in
(a) − (b) − (c) f( x) (d) f −1( x) 2
x3 x2 anticlockwise direction on a circle with
WB JEE (Engineering) Practice Set 4 59
56. The set of points Z in the complex plane 64. If the graph of the function y = f (x) is
satisfying|Z − i|Z|| = |Z + i|Z|| is contained or symmetrical about the line x = 2, then
equal to the set of points Z satisfying (a) f( x) = − f(− x) (b) f(2 + x) = f(2 − x)
(a) Im (Z ) = 0 (b) Im (Z ) ≤ 1 (c) f( x) = f(− x) (d) f( x + 2 ) = f( x − 2 )
(c)|Re(Z )| ≤ 2 (d)|Z| ≤ 3
65. The altitude of a cone is 20 cm and its semi-
57. If the system of lines equations vertical angle is 30°. If the semi-vertical angle is
x + 4 ay + az = 0; x + 3b + bz = 0; x + 2cy + cz = 0 increasing at the rate of 2° per second, then the
have a non-trivial solution, then a , b, c are in radius of the base is increasing at the rate of
160
(a) HP (b) GP (a) 30 cm/sec (b) cm/sec
(c) AP (d) None of these 3
(c) 10 cm/sec (d) 160 cm/sec
58. Let the function f : R − { − b} → R − {1} be
x+a Category III (Q. Nos. 66 to 75)
defined by f (x) = , a ≠ b, then
x+b Carry 2 marks each and one or more option(s)
(a) f is one-one but not onto is/are correct. If all correct answers are not marked
(b) f is onto but not one-one and also no incorrect answer is marked then score
(c) f is both one-one and × number of correct answers marked ÷ actual
(d) None of the above number of correct answer. If any wrong option is
marked or if, any combination including a wrong
59. Let f : R → R be given by option is marked, the answer will be considered
f (x) = [ x 2 ] + [ x + 1] − 3 wrong, but there is no negative marking for the
Where [ x ] denotes the greatest integer less same and zero marks will be awarded.
than or equal to x. Then, f (x) is 98 k +1
k +1
(a) many-one and onto (b) many-one and into 66. If ∑ ∫ x (x + 1)
dx , then
(c) one-one and onto (d) one-one and onto k=1 k
(a) I > loge 99 (b) I < loge 99
60. The area of the triangle formed by the origin, 49 49
the point P(x , y) and its reflection in X -axis is (c) I < (d) I >
50 50
1
(a) xy (b) 2| x ⋅ y| (c) | xy| (d)| xy|
2 67. Let f : R → R, g : R → R and h : R → R be
differentiable functions such that
61. The foci of a hyperbola are (−5, 18) and f (x) = x 3 + 3 x + 2, g( f (x)) = x and
(10, 20) and it touches the Y -axis. The length h(g(g(x)) = x for all x ∈ R. Then,
of its transverse axis is 1
89 (a) g ′(2 ) = (b) h′(1) = 666
(a) 100 (b) (c) 89 (d) 50 15
2 (c) h(0) = 16 (d) h(g (3)) = 36
πx
tan
a 2a 68. Area of the region bounded by the curve
62. The value of lim 2 − is y = e x and lines x = 0 and y = e is
x→a x
e
−
1
2/ π −2/ π 1/ π
(a) e − 1 (b) ∫ ln(e + 1 − y) dy
(a) e π (b) e (c) e (d) e 1
1 e
63. Let P(x) be a real polynomial of least degree (c) e − ∫ e dyx
(d) ∫ ln y dy
which has a local maximum at x = 1 and a 0 1
60 WB JEE (Engineering) Practice Set 4
Answers
Physics
1. (a) 2. (b) 3. (d) 4. (b) 5. (b) 6. (a) 7. (d) 8. (b) 9. (b) 10. (c)
11. (c) 12. (d) 13. (a) 14. (b) 15. (b) 16. (c) 17. (b) 18. (a) 19. (c) 20. (d)
21. (d) 22. (d) 23. (c) 24. (a) 25. (a) 26. (c) 27. (d) 28. (d) 29. (c) 30. (c)
31. (c) 32. (a) 33. (c) 34. (b) 35. (c) 36. (a, b) 37. (a, d) 38. (b, d) 39. (a, d) 40. (a, d)
Chemistry
41. (b) 42. (d) 43. (c) 44. (b) 45. (a) 46. (b) 47. (c) 48. (c) 49. (b) 50. (d)
51. (b) 52. (a) 53. (c) 54. (b) 55. (a) 56. (a) 57. (a) 58. (b) 59. (c) 60. (c)
61. (c) 62. (b) 63. (b) 64. (b) 65. (b) 66. (b) 67. (a) 68. (a) 69. (a) 70. (b)
71. (d) 72. (b) 73. (d) 74. (a) 75. (c) 76. (a, d) 77. (a, b, c) 78. (a, b) 79. (b, c) 80. (a,b,c,d)
Mathematics
1. (a) 2. (b) 3. (b) 4. (a) 5. (b) 6. (d) 7. (c) 8. (a) 9. (d) 10. (c)
11. (d) 12. (c) 13. (d) 14. (d) 15. (a) 16. (d) 17. (b) 18. (d) 19. (c) 20. (c)
21. (c) 22. (d) 23. (c) 24. (b) 25. (d) 26. (c) 27. (b) 28. (a) 29. (c) 30. (d)
31. (b) 32. (d) 33. (c) 34. (c) 35. (d) 36. (b) 37. (b) 38. (c) 39. (b) 40. (a)
41. (c) 42. (a) 43. (b) 44. (a) 45. (a) 46. (c) 47. (c) 48. (c) 49. (b) 50. (b)
51. (c) 52. (b) 53. (d) 54. (d) 55. (d) 56. (a) 57. (a) 58. (c) 59. (b) 60. (d)
61. (c) 62. (c) 63. (b) 64. (b) 65. (b) 66. (b,d) 67. (b,c) 68. (b,c,d) 69. (b,c) 70. (a,b,c)
71. (d) 72. (b) 73. (c) 74. (b) 75. (c)
Practice Set 5
Physics
Category I (Q. Nos. 1 to 30) 4. An element with atomic number Z = 11 emits
Carry 1 marks each and only one option is kα X-ray of wavelength λ. The atomic number of
correct. In case of incorrect answer or any element which emits kα X-ray of wavelength
combination of more than one answer, 1/4 mark 4λ, then
(a) 6 (b) 4 (c) 11 (d) 44
will be deducted.
5. A Zener diode of power rating 1 W is to be used
1. Consider a ray of light incident from air into a as a voltage regulator. If Zener has a breakdown
slap of glass (refractive index) of width d, at an of 5 V and it has to regulate voltage which
angle θ. The phase difference between the ray fluctuated between 3 V and 7 V, what should be
reflected by the top surface of the glass and the the value of RS for safe operation?
bottom surface is
− 1/ 2
2 πnd 1 2
(a) 1 − 2 sin θ + π
λ n
Unregulated Regulated
1/ 2
2 πd 1 2 voltage voltage
(b) 1 − 2 sin θ
λ n
1/ 2
2 πd 1 − 1 sin2 θ + π / 2
(c) (a) 5 Ω (b) 6 Ω (c) 8 Ω (d) 10 Ω
λ n2
1/ 2
2 πd 1 − 1 sin2 θ + 2 π 6. The combination of NAND gates shown here
(d)
λ n2 under figure, are equivalent to
2. A source contains two phosphorous radio A
32
nuclides 15 P (T1 / 2 = 14.3 days) and 15
33
P
(T1 / 2 = 25.3 days). C
33
Initially, 10% of the decay come from 15 P. B
How long one must wait until 90% do so?
(a) 250 days (b) 295 days
(c) 305 days (d) 208 days A
C
3. The wavelength of characteristic X-rays λ kα B
7. The correct dimensional formula for power is same speed as on Earth and the density of
given by planet is same as that of Earth. (Take, escape
(a) ML2T −2 (b) MLT −1 (c) ML2T -3 (d) MLT −2 speed on the surface of the Earth to be 11.2
km/s and radius of Earth to be 6400 km.)
8. The relative density of material of a body is the (a) 2 km (b) 2.2 km
ratio of its weight in air and the loss of its (c) 2.5 km (d) 3 km
weight in water. By using a spring balance, the
weight of the body in air is measured to be 13. The energy stored per unit volume in copper
5.00 ± 0.05 N. The weight of the body in water is wire, which produces longitudinal strain of
measured to be 4.00 ± 0.05 N. Then, the 0.1% is (Young’s modulus = 1.1 × 1011 Nm−2)
maximum possible percentage error in relative (a) 11 × 10 4 Jm−3 (b) 11 × 10 3 Jm−3
density is (c) 5.5 × 10 4 Jm−3 (d) 5.5 × 10 3 Jm−3
(a) 11% (b) 10% (c) 9% (d) 7%
14. A 50 kg girl wearing high heel shoes balance on
9. The two bodies of mass m1 and m2(m1 > m2) a single heel. The heel is circular with the
respectively are tied to the ends of a massless diameter 1.0 cm. What is the pressure exerted
string, which passes over a light and frictionless on the horizontal floor?
pulley. The masses are initially at rest and then (a) 6.24 × 10 6 Pa (b) 9 × 10 3 Pa
released. Then, acceleration of the centre of (c) 3 × 10 6 Pa (d) 2 × 10 4 Pa
mass of the system is
15. An ideal gas heat engine operates in Carnot
cycle between 227°C and 127°C. It absorbs
6 × 104 cal of heat of higher temperature.
Amount of heat converted to work is
(a) 1.2 × 10 4 cal (b) 6 × 10 4 cal
T
(c) 4.8 × 10 4 cal (d) 2.4 × 10 4 cal
a2
m2
16. In an adiabatic process, when pressure is
T Cp 3
2
increased by %. If = , then the volume
a1 m 1 3 CV 2
2 2 decreases by about
m − m2 m − m2
(a) 1 g (b) 1 9 2 4
m1 + m2 m1 + m2 (a) % (b) % (c) % (d) 4%
4 3 9
(c) g (d) zero
10. If r1 and r2 are the position vectors of two 17. An ideal refrigerator has a freezer at a
positive charges q1 and q2 with respect to same temperature of − 13°C. The coefficient of
origin, then force on q2 by q1 is directed in the performance of the engine is 5. The
direction given by unit vector temperature of the air (to which heat is
r2 − r1 r1 − r2 rejected) will be
(a) $r21 = (b) $r21 = (a) 320°C (b) 39°C (c) 325°K (d) 325°C
| r2 − r1| | r2 − r1|
(c) $r21 = r2 = r1 (d) r$21 = r1 − r2 18. In the circuits shown in the figure, the potential
difference across the 4.5 µF capacitor is
11. A boat crosses a river of width 1 km by shortest
path in 15 min. If the speed of boat in still 3µF
water is 5 kmh −1 , then what is the speed of the 4.5µF
river?
(a) 4 kmh−1 (b) 12 kmh−1
(c) 5 kmh−1 (d) 3 kmh−1
6µF
12. A man can jump upto a height of h0 = 1 m on
the surface of the Earth. What should be the 12V
radius of a spherical planet, so that the man
makes a jump on its surface and escape out of 8
its gravity? Assume that the man jumps with (a) 8V (b) 4V (c) 6V (d) V
3
WB JEE (Engineering) Practice Set 5 63
19. In a region, the potential is represented by 24. Three infinite straight wires A, B and C carry
V (x , y , z) = 6 x − 8 xy − 8 y + 6 yz, where V is in currents as shown in the figure. The net force
volts and x , y , z are in metres. The electric force on the wire B is directed
experienced by a charge of 2 C situated at point
(1, 1, 1) is
(a) 4 35 N (b) 30 N
(c) 24 N (d) 6 5 N 1A 2A 3A
21. A current I = 5.0 A flows along a thin wire 25. The current i drawn from the 5 V source will be
shaped as shown in figure. The radius of 10Ω
the curved part of the wire is equal to
R = 120 mm. The angle 2φ = 90°, find the
magnetic induction of the field at the point O. 5Ω 10Ω 20Ω
10Ω
5V
+ –
O
I
90° R (a) 0.17 A (b) 0.33 A (c) 0.5 A (d) 0.67 A
26. An ideal battery of emf 2V and a series
A B resistance R, are connected in the primary
−5 circuit of a potentiometer of length 1 m and
(a) 2 × 10 T (b) 2.8 × 10 T
5
resistance 5Ω. The value of R, to give a potential
(c) 2.8 × 10 −5 T (d) 2 × 10 5 T
difference of 5 mV across 10 cm of
22. An electron after being accelerated through a potentiometer, where is
potential difference of 104 V enters a uniform (a) 180 Ω (b) 190 Ω (c) 195 Ω (d) 200 Ω
magnetic field of 0.04 T perpendicular to its
27. A coil of inductance 300 mH and resistance 2Ω
direction of motion. Calculate the radius of
curvature of its trajectory. is connected to source of voltage 2V. The current
(a) 8.4 mm (b) 8.43 mm reaches half of its steady state value in
(c) 8 mm (d) 8.2 mm (a) 0.05 s (b) 0.1 s (c) 0.15 s (d) 0.3 s
23. A current path shaped as shown in figure 28. A concave lens forms the image of an object
produces a magnetic field at P, the centre of the such that distance between the object and
arc. If the arc subtends an angle of 30° and the image is 10 cm; and the magnification
radius of the arc is 0.6 m. What is the magnitude 1
produced is . The focal length of the lens will
of the field at P, if the current is 3.0 A? 4
A be
(a) 8.6 cm (b) 6.2 cm (c) 10 cm (d) 4.4 cm
C
29. A parallel beam of monochromatic light of
wavelength 450 nm passes through a slit of
P 30° width 0.2 mm. Find the angular divergence in
which most of light is diffracted.
D (a) 4.5 × 10 −3rad (b) 5.5 × 10 −3 rad
(c) 3.2 × 10 −2 rad (d) 3.5 × 10 −3 rad
E
(a) 2.62 × 10 −6 T (b) 2.62 × 10 −7 T 30. White light is used to illuminate the two slits in
(c)3.62 × 10 −9 T (d) 2.62 × 10 −8 T
a Young’s double slit experiment. The
separation between the slits is b and the screen
64 WB JEE (Engineering) Practice Set 5
is at a distance D > > > b from the slit. At a point Category III (Q. Nos. 36 to 40)
on the screen directly in front of one of the
slits. Find the missing wavelength. Carry 2 marks each and one or more option(s)
is/are correct. If all correct answers are not marked
b2 b2 b2 b2 b2 b2
(a) , , ...... (b) , , ......
and also no incorrect answer is marked then score
D 3D 5D 2 D 4D 5D
b2 b2 b2
= 2 × number of correct answers marked ÷ actual
(c) , , ...... (d) None of the above number of correct answers. If any wrong option is
3D 5D 6D
marked or if any combination including a wrong
Category II (Q. Nos. 31 to 35) option is marked, the answer will be considered
Carry 2 marks each and only one option is correct. wrong, but there is no negative marking for the
In case of incorrect answer or any combination of same and zero marks will be awarded.
more than one answer, 1/2 mark will be deducted. 36. In which of the following situations are heavier
31. An inductor (L = 100 mH), a resistor of the two particles has smaller de-Broglie
(R = 100 Ω) and a battery (E = 100 V) are initially wavelength?
connected in series as shown in figure. After The two particles
sometime, the battery is disconnected after (a) move with same speed
short circuiting the points A and B. The current (b) move with same KE
(c) move with same linear momentum
in the circuit 1 ms after the short circuit is (d) have fallen through the same height
L 37. An ideal gas is taken from the state A(p , V) to
R the state B (p / 2, 2V) along a straight line path
as shown in figure. Select the correct statement
A B
from the following.
E p
1
(a) eA (b) 0.1 A (c) 1 A (d) A
e p
A
32. The diameter of a pipe at two points, where a
venturimeter is connected is 8 cm and 5 cm; and p/2
B
the difference of levels in it is 4 cm. The volume
V V 2V V0
of water flowing through the pipe per second is
−1 −1
(a) 1889 ccs (b) 1520 ccs (a) Work done by the gas in going from A to B
(c) 1321 ccs −1 (d) 1125 ccs −1
exceeds the work done in going from A to B under
33. A rectangular block is heated from 0°C to 100°C. isothermal conditions
The percentage increase in its length is 0.2%. (b) In the T-V diagram, path AB would become a
What is the percentage increase in its volume? parabola
(a) 0.6% (b) 0.10% (c) 0.2% (d) 0.4% (c) In the p-T diagram, path AB would be part of
hyperbola
34. The incident intensity on a horizontal surface at (d) In going from A to B, the temperature T of gas first
a sea level from Sun is about 1 kW m −2. increases to a maximum value 1 and then
Assuming that 50 per cent of this intensity is decreases
reflected and 50 per cent is absorbed, determine
the radiation pressure on this horizontal surface. 38. Two particles are projected in air with speed v 0
− 11
(a) 5 × 10 Pa −6
(b) 5 × 10 Pa at angles θ1 and θ 2 (both acute) to the
(c) 1 × 10 −6 Pa (d) 1 × 10 −11 Pa horizontal, respectively. If the height reached
by the first particle is greater than that of the
35. A 20 W, 50 V filament is connected in series to second, then tick the right choices.
an AC mains of 250 V-Hz. Calculate the value of (a) Angle of projection : θ 1 > θ 2
the capacitor required to run the lamp. (b) Time of flight : T1 > T2
(a) 6.6 × 10 −5 F . × 10 −6 F
(b) 51 (c) Horizontal range : R1 > R 2
−6
(c) 4.3 × 10 F (d) 302 × 10 −5 F (d) Total energy : U1 > U 2
WB JEE (Engineering) Practice Set 5 65
(b) VB − VD = 0
39. Figure shows a network of resistances and
(c) VB − VD = 2 V
batteries. Select the correct statements out of
(d) No current flows in the branch BD
the following.
40. Consider a wire carrying a steady current
B
I placed in a uniform magnetic field B
4Ω 4Ω perpendicular to its length. Consider the
+ C
charges inside the wire. It is known that
A
– 2V magnetic forces do no work. This implies that,
(a) motion of charges inside the conductor is
2Ω 2Ω unaffected by B, since they do not absorb energy
D (b) some charges inside the wire move to the surface
as a result of B
+ –
(c) if the wire moves under the influence of B, no work
6V is done by the force
(d) if the wire moves under the influence of B, no work
(a) The circuit satisfies the condition of a balanced is done by the magnetic force on the ions,
Wheatstone bridge assumed fixed within the wire
Chemistry
Category I (Q. Nos. 41 to 70) 45. What is the value of ∆H for a reversible
Carry 1 marks each and only one option is isothermal evaporation of 180 g of water
correct. In case of incorrect answer or any at 100°C ? Latent heat of evaporation of water
combination of more than one answer, 1/4 mark = 539.7 cal K −1 g −1 .
will be deducted. (a) 97.146 kcal (b) 5.397 kcal
10 × 373 × 5397
. . × 373
5397
41. A compound with haemoglobin like structure (c) kcal (d) kcal
1000 1000
contain one atom of iron per molecule. If it
contain 4.6% iron, its approximate molecular 46. The right order of the solubility of sulphates of
mass is alkaline earth metals in water is
−1 −1 (a) Be > Ca > Mg > Ba > Sr
(a) 100 g mol (b) 1200 g mol
(c) 1600 g mol −1 (d) 1400 g mol −1 (b) Mg > Be > Ba > Ca > Sr
(c) Be > Mg > Ca > Sr > Ba
42. Sulphur oxides which are responsible for major (d) Mg > Ca > Ba > Be > Sr
air pollution are caused by
(a) burning of coal and refining of petroleum 47. The correct order of stability of conformations
(b) using indoor combustion devices like cooking gas of NH2 CH2 CH2 OH is
(a) gauche > eclipsed > anti
(c) combustion of fuels containing carbon and
(b) gauche > anti > eclipsed
hydrogen (c) anti > eclipsed > gauche
(d) burning of fuels in automobiles (d) eclipsed > gauche > anti
43. 1, 2-benzpyrene is 48. The rise in boiling point of a solution containing
(a) a polynuclear hydrocarbon 1.8 g glucose in 100 g of solvent is 0.1°C. The
(b) carcinogenic in nature
molal elevation constant of the liquid is
(c) an aromatic hydrocarbon
(d) Both (a) and (b) (a) 1.0 (b) 2.0 (c) 3.0 (d) 2.5
44. The bond angles of NH3, NH+4 and NH−2 are in 49. 1 mole heptane (v.p. = 92 mm of Hg) is mixed
with 4 mol. Octane (v.p. = 31 mm of Hg), form
the order
an ideal solution. What will be the vapour
(a) NH −2 > NH 3 > NH+4 (b) NH+4 > NH 3 > NH −2 pressure of the solution?
(c) NH 3 > NH 2− > NH+4 (d) NH > NH+4 > NH −2 (a) 41.2 mm of Hg (b) 43.2 mm of Hg
(c) 40.2 mm of Hg (d) 23.2 mm of Hg
66 WB JEE (Engineering) Practice Set 5
50. If the standard half cell reduction potentials are 56. Calcium crystallises in a face centred cubic unit
0.522 V for Cu + /Cu and 0.3402 V for Cu 2+ /Cu. cell with a = 0.556 nm. Calculate the density if it
The standard half cell reduction potential for contained 0.1% Schottky defects.
Cu + /Cu 2+ is (a) 1.5463 g/cm3 (b) 14962
. g/cm3
(a) 0.158 V (b) 0.20 V (c) 0.80 V (d) 0.40 V (c) 15448
. g/cm3 (d) 15943
. g/cm3
51. Identify Z in the following reaction series, 57. Limiting molar conductivity for some ions is
NaOH( ) Al 2 O3 given below (in S cm2 mol −1 ) Na+ = 50.1,
CH3CH2CH2Br → X → Y
aq
∆ Cl − = 76.3, H+ = 349.6, CH3COO − = 40.9,
Ca2+ = 119.0.
HOCl
→ Z
Cl OH What will be the limiting molar conductivities
(a) Mixture of H3C and H3C (λ°m) of CaCl 2, CH3COONa and NaCl respectively?
(a) 2716 . and 126.4 S cm2 mol −1
. , 910
Cl Cl . and 242.8 S cm2 mol −1
(b) 97.65, 1110
(c) 119.0, 1024.5 and 9.2 S cm2 mol −1
OH (d) 111.0, 97.65 and 119.0 S cm2 mol −1
(b) H3C 58. The decreasing order of boiling points of the
following hydrides is
Cl (a) H 2O > PH 3 > AsH 3 > SbH 3 > NH 3
(b) H 2O > SbH 3 > NH 3 > AsH 3 > PH 3
Cl (c) H 2O > NH 3 > SbH 3 > AsH 3 > PH 3
(c) H3C (d) H 2O > SbH 3 > AsH 3 > PH 3 > NH 3
59. Consider the following bromides :
HO Me Me
Me Br Me
Cl
(A) Br Br
(d) H3C (B) (C)
Cl
The correct order of SN1 reactivity is
(a) C > B > A (b) A > B > C
52. In the reaction, (c) B > C > A (d) B > A > C
4NH3( g )+ 5O 2( g ) → 4NO ( g )+ 6H2O ()
l 60. What is the reagent X in the following
reaction?
When 1 mol of ammonia and 1 mol of O 2 are
made to react to completion then C 6H5 N(CH3)2 + X → (CH3)2N C 6H4
(a) all the oxygen will consumed CO C 6H4 N(CH3)2
(b) all the ammonia will be consumed (a) CO (b) CO 2
(c) 1.0 mol of NO will be produced (c) COCl 2 (d) OC(OC 2H 5 )2
(d) 1.0 mol of H 2O is produced
Br Br
53. Which of the following finds use as a superior
thermometer liquid for high temperature 61. CH2 CH == CH CH2 Zn
→ A
(dust)
63. 2, 2, 6, 6-tetramethyl cyclohexanol is treated 70. The rate constant of a reaction will be equal to
with an acid. An alkene is formed after the pre-exponential factor when
rearrangement. The structure of the alkene is (a) the absolute temperature is infinity
(b) the absolute temperature is zero
(c) temperature in centigrade is zero
(d) None of the above
(a) (b)
Category III (Q. Nos. 76 to 80) (a) (NH 2 )2C == O (b) (NH 2 )2 C == S
(c) P − NH 2C 6H 4SO 3H (d) C 6H 5SO 3H
Carry 2 marks each and one or more option(s)
78. Pottassium cyanide is used for separating
is/are correct. If all correct answers are not marked
(a) Ba 2+ and Ca 2+ (b) Co 2+ and Ni 2+
and also no incorrect answer is marked then score (c) Cu2+ and Cd 2+
= 2 × number of correct answers marked ÷ actual (d) Mn2+ and Zn2+
number of correct answers. If any wrong option is
79. Proteins can be classified into two types on the
marked or if any combination including a wrong
basis of their molecular shape, i.e., fibrous
option is marked, the answer will considered proteins and globular proteins. Examples of
wrong, but there is no negative marking for the globular proteins are
same and zero marks will be awarded. (a) insulin (b) keratin
(c) myosin (d) albumin
76. For particles having same K.E., the de-Broglie
wavelength is 80. Which of the following reactions should be
(a) unpredictable balanced in basic medium?
(b) inversely proportional to its velocity 4 → MnO 2 + NO 2
(a) NH 3 + MnOs
(c) independent of its mass and velocity
(b) Cr(OH) + I → Cr(OH) + 2Is
2 2 2
(d) directly proportional to its velocity
(c) HNO 3 + Fe 2 + → Fe 3 + + NO 2
77. Which of the following compounds may give
blood red colouration while performing (d) H 2O 2 + Fe 3+ → O 2 + Fe 2+
Lassaigne’s test for nitrogen
Mathematics
Category I (Q. Nos. 1 to 50) (a) 7 y = 5 x (b) x = 5 y
(c) 7 x = 5 y (d) y = 5 x
Only one answer is correct. Correct answer will
5. There are 7 seats in a row. three persons take
fetch full marks 1. Incorrect answer or any
seats at random. The probability that the
combination of more than one answer will fetch middle seat is always occupied and no two
−1/4 marks. persons are consecutive is
6 9 4 8
a −1 + n
b (a) (b) (c) (d)
1. The value of lim a, b > 0 is 35 22 35 21
n→∞ a π cos 2 x
(a) 2 a (b) a b (c) b a (d) 4 b 6. The value of ∫ dx, a > 0 is
−π 1 + ax
x
2. The point of intersection of f1 (x) = ∫ (2t − 5) dt (a)
2
(b)
π
(c)
π
(d)
a
2
x π a 2 π
and f 2(x) = ∫ 2t dt is
0 ax + b
7. If S.D of variable x is ∀x, then the SD of ,
(a) ,
6 36
(b) , (c) , (d) ,
2 4 1 1 1 1
p
5 25 3 9 3 9 5 25
∀ a, b, p ∈ R is
3. If the normal at point P(θ) to the ellipse
p a p a
5 x 2 + 14 y 2 = 70 intersects it again at the point (a) σx (b) σx (c) σx (d) σx
Q (2θ), then cosθ is equal to a p a p
2 2 1 1
(a) (b) − (c) (d) − 8. The complex numbers z1 , z2 and z3 satisfying
3 3 3 3
z1 − z3 1 − 3i
4. In ∆ ABC, equation of the right bisectors of the = are the vertices of a triangle
z2 − z3 2
sides AB and AC are x + y = 0 and x − y = 0,
respectively. If A ≡ (5, 7), then equation of side which is
BC is (a) equilateral (b) right angled
(c) right angled isosceles (d) obtuse angled isosceles
WB JEE (Engineering) Practice Set 5 69
10. Let f (x + y) + f (x − y) = 2 f (x) ⋅ f (y), ∀ x , y ∈ R 18. In a ∆ABC, ∠B = 90°, AC = h and the length of
the perpendicular from B to AC is p such that
f (0) ≠ 0, then f (x) is
h = 4 p. If AB < BC, then ∠C has the measure
(a) periodic (b) odd π 7π π π
(c) even (d) None of these (a) (b) (c) (d)
4 12 12 6
11. Solution of the differential equation 19. If (x 2 + y 2)dy = xydx and y(x 0) = e , y()
1 = 1, then
xf (y / x)
xdy = y + dx is x 0 is
f ′ (y / x) (a) e 3 (b) 2e 2 − 1 / 2
(a) f( y / x ) = c x , c > 0
e2 + 1
(b) f( y / x ) = x + c, c < 0 (c) e 2 − 1 / 2 (d)
2
(c) f( x / y) = x + c, c > 0
(d) f( x / y) = c x , c < 0 20. If A and B are different matrices satisfying
A 3 = B 3 and A 2B = B 2 A , then
12. Median of 2n C 0 , 2nC1 , 2nC 2 ,....,2n C n (a) at least one det ( A 2 + B2 ) or det ( A − B) must be
(When n is odd) is zero.
1 2n (b) det ( A 2 + B2 ) as well as det ( A − B) must be zero
(a) C n −1 + 2n
C n +1 (b) 2n
C n/ 2
2 2 2
(c) det ( A 2 + B2 ) must be zero.
(c) 2n
Cn (d) None of these (d) det ( A 2 − B2 ) must be zero.
13. If the sum of the coefficients in the expansion 21. If a line is tangent to one point and normal at
of (x − 2 y + 3z) is 128, then the greatest
n another point on the curve x = 4 t 2 + 3,
coefficient in the expansion of (1 + x)n is y = 8t 3 − 1, then slope of each line is
(a) 35 (b) 20 (c) 15 (d) 10 (a) ± 2 (b) ± 3 (c) ±1 (d) ± 6
π π2 π2 π2
sin θ
2
1 + cos θ2
4 sin 4θ (a) (b) (c) (d)
ab 2 ab ab a + b2
2
sin 2 θ cos 2 θ 1 + 4 sin 4θ
π 5π 9π 7π 42. The locus of the point which divides the
(a) (b) (c) (d)
24 24 24 24 x2 y 2
∞ double ordinates of the ellipse + = 1 in
2n a2 b2
34. The sum of series Σ is
n =1 (2n + 1)! the ratio 1 : 2 internally, is
(a) e −1 (b) e (c) 2e (d) e 3 x2 y2 x2 9 y2
(a) 2
+ 2
=1 (b) 2
+ =1
a b a b2
35. The function f satisfies the functional equation
x2 y2 9 x2 y2
x + 59 (c) + =9 (d) + =1
3 f (x) + 2 f = 10 x + 30 for all real x ≠ 1. a2 b2 a2 b2
x −1
WB JEE (Engineering) Practice Set 5 71
(a) 5 x 2 − 5 y 2 − 72 x + 54 y + 225 = 0
43. The point on the axis of X, whose perpendicular
x y (b) 5 x 2 + 5 y 2 − 72 x + 54 y + 225 = 0
distance from the straight line + = 1 is a, are (c) 5 x 2 + 5 y 2 + 72 x − 54 y + 225 = 0
y b
(d) 5 x 2 + 5 y 2 − 72 x − 54 y − 225 = 0
(b) ( b ± a 2 + b 2 ), 0
b a
(a) ( a ± a 2 + b 2 , 0)
a b
b a Category II (Q. No. 51 to 65)
(c) ( a + b, 0) (d) ( a ± a 2 + b 2 , 0)
a b
Carry 2 marks each and only one option is correct.
44. The Length of major axis of ellipse In case of incorrect answer or any combination of
(3 x − 4 y + 7)2 more than one answer, 1/2 mark will be deducted.
(5 x − 10)2 + (5 y + 15)2 = is
4
20 10 5 51. The three vectors $i + $j, $j + k,
$ k$ + $i taken two at
(a) (b) (c) (d) 2
3 3 3 a time form three planes. The three unit vectors
drawn perpendicular to three planes form a
45. If f (x) = (x + 1)2 − 1, (x ≥ −1) Then, the set parallelopiped of volume.
S = { x : f (x) = f −1 (x)} is (a)
4
cubic unit (b) 4 cubic unit
−3 + i 3 −3 − i 3 3 3
(a) 0, − 1, , 1 3 3
2 2 (c) cubic unit (d) cubic unit
2 4
(b) { 0, 1, − 1}
(c) { 0, − 1} 52. The normal at a variable point A on the ellipse
(d) Empty set
x2 y 2
+ = 1 of eccentricity e meets the axes of
46. The equation of tangent to the curve a2 b2
n n
x y the ellipse at S and T . Then locus of midpoint of
+ = 2 at (a, b) is
a b ST is a conic with eccentricity e′ such that
x y x y 1 (a) e ′ = e 2 (b) e ′ = e
(a) + =2 (b) + =
a b a b 2 1
(c) e ′ = (d) None of these
x y
(c) − = 2 (d) ax + by = 2 e
b a
53. Let x , y , z be such that y(x + 3) ≠ 0. If
47. If A (z1), B(z2) and C(z3) are the vertices of ∆ABC x x + 1 x −1 x +1 y +1 z −1
π AB −y y + 1 y −1 + x −1 y −1 z +1
in which ∠ABC = and = 2 then z2
4 BC
z z − 1 z + 1 (−1)n + 2 x (−1)n + 1 y (−1)n z
equals
(a) z3 + i ( z1 + z3 ) (b) z3 − i ( z1 − z3 ) = 0, then n equals
(c) z3 + i ( z1 − z3 ) (d) None of these (a) zero (b) Any integer
r −1 r −1 r −1
(c) Any odd integer (d) Any even integer
2 2.3 4.5 n
48. If D r = α β γ , then ∑ D r
54. If z1 and z2 are two complex numbers such that
r =1 z1 = z2 + z1 − z2 , then
2n − 1 3n − 1 5n − 1 z z z
(a) Re 1 = 0 (b) Re 1 = Im 1
equals z2 z2 z2
(a) 0 (b) αβγ z1
(c) α + β + γ (d) α.2 n + β.3 n + r.4 n (c) Im = 0 (d) None of these
z2
49. The numbers of terms in the expansion of
55. Suppose a, b and c are in AP and a2 , b2 and c 2
(a + b + c)n will be
3
(a) n + 1 (b) n + 3 are in GP. If a < b < c and a + b + c = , then a
( n + 1) ( n + 2 ) z
(c) (d) None of these
2 equals
1 1
(a) (b)
50. The locus of a point P which moves such that 2 3 3 2
2PA = 3PB where coordinates of points A and B 1 1 1 1
(c) − (d) −
are (0, 0) and (4, −3) is 2 2 3 2
72 WB JEE (Engineering) Practice Set 5
56. The vertices of a variable triangle are (3, 4), 63. If α and β are the roots of ax 2 + bx + c = 0, then
(5 cos θ , 5sin θ) and (5sin θ , − 5 cos θ), where θ ∈ R. the equation ax 2 − bx(x − 1) + c(x − 1)2 = 0 has
The Locus of its othocenter is
roots
(a) ( x + y − 7 )2 + ( x − y + 1)2 = 100
α β 1− α 1− β
(b) ( x − y + 7 )2 + ( x − y − 1)2 = 100 (a) , (b) ,
1− α 1− β α β
(c) ( x + y − 7 )2 + ( x + y − 1)2 = 25
α β α +1 β+1
(d) ( x + y − 1)2 + ( x − y − 7 )2 = 25 (c) , (d) ,
α +1 β+1 α β
rα rα
57. If zr = cos + i sin 2 , where r = 1, 2, 3... n , then
n2 n 64. The value of
lim z1 z2... zn equals
n→∞
2(x)1 / 2 + 3(x)1 / 3 + 4(x)1 / 4 + ....+ n(x)1 / n
lim is
α α x →∞ (2 x − 3)1 / 2 + (2 x − 3)1 / 3 + ....+ (2 x − 3)1 / n
(b) cos − i sin
3
(a) e iα
z z (a) 2 (b) 2 (c) −1 (d) 0
(c) cos α − i sinα (d) e iα / 2
65. tanα and tanβ are the roots of the equation
π π
58. If f (x) = sin 2 x + sin 2 x + + cos x cos x + x 2 + ax + b = 0, then the value of
3 3
sin 2(α + β) + a sin(α + β)cos(α + β)
and g(5 / 4) = 1, then gof (x) is equal to
(a) 1 (b) −1 (c) 2 (d) −2 + b cos 2(α + β) is equal to
(a) ba (b) a (c) b 2a (d) b
59. The maximum value of the function.
(1 + x)0. 6
f (x) = in the interval [0, 1] is Category III (Q. Nos. 66 to 75)
1 + x 0. 6
Carry 2 marks each and one or more option(s)
(a) 2 0. 6 (b) 2 −0. 4 (c) 2 0 (d) 0
is/are correct. If all correct answers are not marked
60. A man is moving away from a tower 41.8m and also no incorrect answer is marked then score
high at a rate of 2m / s. If the eye level of the = 2 × number of correct answers marked ÷ actual
man is 1.8 m above the ground, then the rate at number of correct answer. If any wrong option is
which the angle of elevation of the top of the marked or if, any combination including a wrong
tower changes, when he is 40 m from the foot option is marked, the answer will be considered
of the tower is wrong, but there is no negative marking for the
−1 −2
(a) rad/s (b) rad/s same and zero marks will be awarded.
40 625
−4
(c) rad/s (d) None of these 66. A(z1), B(z2), C(z3) and D(z4) are four complex
125
numbers representing the vertices of a rhombus
61. Let A (1, 1) and B(3, 2) be two points. If C is a on the complex plane, then
point on x-axis such that AC + BC is minimum, z1 − z4 z − z4
(a) amp = amp 2
then the coordinates of C are z2 − z4 z3 − z4
(a) , 0 (b) , 0 z1 −
5 1 z3
3 3 (b) is purely imaginary
z2 − z4
(c) ( 3, 0) (d) None of these z − z4
(c) 1 is purely real
62. If a, b, c are the pth, qth, rth terms of an HP and z2 − z3
(d) Not necessary that z1 − z3 ≠ z2 − z4
u = (q − r)$i + (r − p)$j + (p − q )k$ and
$i $j k$ n n +1 n+2
v = + + , then 67. Let f (x) = n Pn n+1
Pn + 1 n+ 2
Pn + 2
a b c
n+1 n+ 2
(a) uxv = $i + $j + k$ n
Cn Cn + 1 Cn + 2
(b) u,v are parallel vectors
(c) u,v are orthogonal vectors Then f (x) is divisible by
(d) u.v = 1 (a) n2 + n + 1 (b) ( n + 1)!
(c) n! (d) None of these
WB JEE (Engineering) Practice Set 5 73
(b) there is no director circle of the hyperbola 73. Let f (x) + f (y) = f (x 1 − y 2 + y 1 − x 2). Then,
(c) centre of the director circle is (0, 0) (a) f(2 x 1 − x 2 ) = 2 f( x )
(d) Length of latusrectum of the hyperbola = 12 (b) f( 4 x 3 − 3 x ) + 3f( x ) = 0
69. The parabolas y 2 = 4 x and x 2 = 4 y divide (c) f( 4 x 3 − 3 x ) = 3f( x )
the square region bounded by the lines x = 4, (d) f(2 x 1 − x 2 ) + 2 f( x ) = 0
y = 4 and the coordinate axes. If A1 , A2 and A3
74. If the tangent at any point A (4 m2 , 8m3) of
are respectively, the areas of these parts
x 3 − y 2 = 0 is also a normal to the curve
numbered from top to bottom, then
A2 A1 1 x 3 − y 2 = 0, then m equals
(a) =1 (b) = −3
A3 A2 2 (a)
A 1 A 2
(c) 3 = (d) 1 = 1 3
A2 2 A2 (b)
2
− 2
70. The roots of the equation (c)
3
x 5 − 40 x 4 + Px 3 + Qx 2 + Rx − S = 0 are in GP If 2
(d)
the sum of their reciprocals is 10, then the 3
value of S can be equal to
1 75. If a1 , a2 , ... an are in AP with common difference
(a) −32 (b)
32 d, then
−1
(c) (d) 32 1 + a1 a2 1 + a2a3
32 cot −1 + cot −1
d d
71. f (x) = x 2 − 3 x + 2 , then which of the 1 + a a 1 + an an − 1
+ cot −1 3 4
+ ... + cot −1
following is/are true ? d d
(a) f ′ ( x ) = 2 x − 3 for x ∈ ( 0, 1) ∪ (2, ∞ ) is equal to
(b) f ′ ( x ) = 2 x + 3 for x ∈ ( −∞, − 2 ) ∪ ( −1, 0) (a) tan−1 an − tan−1 a1
(c) f ′ ( x ) = − 2 x − 3 for x ∈ ( −2, − 1)
(b) cot −1 a1 + cot −1 an
(d) None of the above
(c) cot −1 a1 − cot −1 an
(d) None of these
Answers
Physics
1. (a) 2. (d) 3. (b) 4. (a) 5. (d) 6. (a) 7. (c) 8. (a) 9. (a) 10. (a)
11. (d) 12. (c) 13. (c) 14. (a) 15. (a) 16. (c) 17. (b) 18. (a) 19. (a) 20. (b)
21. (c) 22. (b) 23. (b) 24. (a) 25. (c) 26. (c) 27. (b) 28. (d) 29. (a) 30. (a)
31. (d) 32. (a) 33. (a) 34. (b) 35. (b) 36. (a,b, d) 37. (a, b) 38. (a, b) 39. (a, c) 40. (b, d)
Chemistry
41. (b) 42. (a) 43. (d) 44. (b) 45. (a) 46. (c) 47. (b) 48. (a) 49. (b) 50. (a)
51. (b) 52. (a) 53. (b) 54. (a) 55. (d) 56. (c) 57. (a) 58. (b) 59. (c) 60. (c)
61. (d) 62. (d) 63. (a) 64. (a) 65. (a) 66. (a) 67. (a) 68. (d) 69. (b) 70. (a)
71. (d) 72. (b) 73. (d) 74. (d) 75. (a) 76. (d) 77. (b, c) 78. (b, c) 79. (a, d) 80. (a, b)
Mathematics
1. (b) 2. (a) 3. (b) 4. (a) 5. (c) 6. (c) 7. (b) 8. (a) 9. (b) 10. (c)
11. (a) 12. (a) 13. (a) 14. (a) 15. (a) 16. (c) 17. (d) 18. (c) 19. (a) 20. (a)
21. (a) 22. (b) 23. (a) 24. (a) 25. (c) 26. (c) 27. (a) 28. (a) 29. (b) 30. (c)
31. (a) 32. (a) 33. (d) 34. (a) 35. (a) 36. (b) 37. (b) 38. (a) 39. (d) 40. (a)
41. (c) 42. (b) 43. (b) 44. (b) 45. (c) 46. (a) 47. (c) 48. (a) 49. (c) 50. (b)
51. (a) 52. (b) 53. (c) 54. (c) 55. (c) 56. (a) 57. (d) 58. (a) 59. (c) 60. (a)
61. (a) 62. (c) 63. (c) 64. (b) 65. (d) 66. (a, b, c, d) 67. (a, c) 68. (a, b, d) 69. (a, d) 70. (a, d)
71. (a, b, c) 72. (a, b, c, d) 73. (a, b) 74. (c, d) 75. (a, c)