Vidya Mandir classes study material

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JEE‐Advanced‐2013

Paper 2

Code 0

2 June, 2013 | 2:00 PM – 5:00 PM

Answer key & Detailed solutions to the paper by Vidyamandir Classes

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JEE Advanced 2013

Paper 2 Answer Key

Code 0

PHYSICS

CHEMISTRY

MATHS

1 A,B,C,D 21 C,D 41 B,C,D

2 A,C 22 A,B,D 42 A,B

3 D 23 A,C,D 43 C,D 4 C,D 24 A,B 44 A,B,C 5 A,D 25 B 45 A,D 6 A,B 26 C 46 B,D 7 B,D 27 B 47 B,D 8 A,D 28 B,D 48 A,C 9 B 29 A 49 A 10 A 30 D 50 C 11 B 31 B 51 B 12 A 32 A 52 D 13 B 33 C 53 B 14 B 34 B 54 C 15 C 35 A 55 A 16 A 36 A 56 D 17 A 37 A 57 C 18 C 38 D 58 A 19 D 39 D 59 B 20 C 40 A 60 A

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Part 1 : PHYSICS

Section – 1: (One or more options correct Type)

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct

1. The figure below shows the variation of specific heat capacity (C) of a solid as a function of temperature (T). The temperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume change, the following statement(s) is (are) correct to a reasonable approximation.

(A) the rate at which heat is absorbed in the range 0 – 100 K varies linearly with temperature T.

(B) heat absorbed in increasing the temperature from 0 – 100 K is less than the heat required for increasing the temperature from 400 – 500 K.

(C) there is no change in the rate of heat absorption in the range 400 – 500 K (D) the rate of heat absorption increases in the range 200 – 300 K.

Ans.(A,B,C,D)

Q mCdT m

Δ =

∫

= ×area under C T− graph

dQ dT

mC

dt = dt

if C increases rate of heat absorption increases. ( as dT

dt is constant)

2. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0, where a0 is the Bohr radius. Its orbital

angular momentum is 3 2

h

π . It is given that h is Planck constant and R is Rydberg constant. The possible

wavelength(s), when the atom de-excites, is /are

(A) 9 32R (B) 9 16R (C) 9 5R (D) 4 3R Ans.(A,C) 2 ₂ ₂ 2 1 1 1 1 RZ n n λ ⎡ ⎤ = ⎢ − ⎥ ⎢ ⎥ ⎣ ⎦ 1 3 3 2 h L n π = ⇒ = 2 2 1 0n 4 5 0 0 3 _{4 5}9 2 a a . a a . Z z z . = ⇒ = ⇒ = =

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2 2 2 2 1 1 1 3 R( Z ) n λ ⎡ ⎤ = ⎢ − ⎥ ⎢ ⎥ ⎣ ⎦ 2 2 ₅9 n , R λ = = 2 1 ₃₂9 n , R λ = =

3. Using the expression 2d sinθ =λ, one calculates the values of d by measuring the corresponding angles

θ in the range 0 to 90o. The wavelength λis exactly known and the error in θ is constant for all values of θ. As θ increases from 0o_,

(A) the absolute error in d remains constant (B) the absolute error in d increases (C) the fractional error in d remains constant (D) the fractional error in d decreases Ans.(D) 2d sinθ λ= ( is exactly known) 2 d sin λ _λ θ ⇒ = 2 d λcos ecθ ⇒ = ...(1) 2 d ( cos ec cot ) Δ λ _θ _θ Δθ ⇒ = − 2 d λcos ec cot ( ) Δ = − θ θ Δθ ...(2)

Equation 2/1 d

(

cot

)

( is constant)

d

Δ _{= −} _{θ Δθ} _Δθ

both dand d

d

Δ

Δ are decreasing with θ .

4. Two non-conducting spheres of radii R1 and R2 carrying uniform volume charge densities +ρ and−ρ,

respectively, are placed such that they partially overlap as shown in the figure. At all points in the overlapping region,

(A) the electrostatic field is zero (B) the electrostatic potential is constant

(C) the electrostatic field is constant in magnitude (D) the electrostatic field has same direction Ans. (C,D) 0 0 3 3 O' P OP E_ρ = ρ −ρ ∈ ∈ r r r = 0 3 O' O ( O' O OP O' P ) ρ + = ∈ r r r r Q

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5. A Steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed coaxilly inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is (are)

(A) In the region 0 < r < R, the magnetic field is non-zero

(B) In the region R < r < 2R, the magnetic field is along the common axis.

(C) In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centered on the axis. (D) In the region r > 2R, the magnetic field is non-zero.

Ans. (A,D)

6. Two vehicles, each moving with speed u on the same horizontal straight road, are approaching each other. Wind blows the road with velocity w. One of these vehicles blows a whistle of frequency f1. An observer in the other

vehicle hearts the frequency of the whistle to be f2. The speed of sound in still air is V. The correct statement(s) is

(are)

(A) If the wind blows from the observer to the source, f2 > f1.

(B) If the wind blows from the source to the observer, f2 > f1

(C) If the wind blows from observer to the source, f2 < f1

(D) If the wind blows from the source to the observer, f2 < f1

Ans.(A,B)

If the wind blows from source to observer 2 ( v w ) ( u ) 1 v w u 2 1 f f f f ( v w ) ( u ) v w u + − − + + = = ⇒ > + − + + −

If the wind blows from observer to source

2 ( v w ) ( u ) 1 v w u 1 2 1 f f f f f ( v w ) ( u ) v w u − − − − + = = ⇒ > − − + − −

7. Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the midpoint of the line joining centres, perpendicular to the line. The gravitational constant is G. The correct statement(s) is (are)

(A) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 4 GM

L

(B) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 2 GM

L

(C) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 2GM

L

(D) The energy of the mass m remains constant.

Ans. (B,D) In order to escape to infinity T. E = 0 2 2 1_mu GMm _{u 2} GM 2 L L ⎛ ⎞ =_⎜ _⎟ ⇒ = ⎝ ⎠

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8. A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t = 0 with an initial velocity u0. when the speed of the particle is 0.5 u0, it collides elastically with

a rigid wall. After this collision,

(A) the speed of the particle when it returns to its equilibrium position is u0

(B) the time at which the particle passes through the equilibrium position for the first time is t= m k

π (C) the time at which the maximum compression of the spring occurs is 4

3

= m

t

k

π

(D) the time at which the particle passes through the equilibrium position for the second time is 5 3 = m t k π Ans. (A,D) x A sin t= ω v (A ) cos t= ω ω 0 0 u u cos t cos t 1/ 2 t / 3 2 = ω ⇒ ω = ⇒ ω = π K_t _{/ 3} _t m m 3 K π ⇒ = π ⇒ =

(time taken to reach wall from mean position)

m 2 m

time of passing equilibrium for first time 2

3 K 3 K

π π

= × =

2 m 2 m 7 m

time of maximum compression

3 K 4 K 6 K

π π π

= + =

2 m m 5 m

time of passing equilibrium for second time

3 K K 3 K

π π

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Section – 2: (Paragraph Type)

This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions related to four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer

among the four choices (A), (B), (C) and (D). Paragraph for Question 9-10

A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g=10m s−2).

9. The speed of the block when it reaches the point Q is.

(A) 5ms−1 (B) 10ms−1 (C) 10 3 ms−1 (D) 20ms−1

Ans. (B)

against f Loss of P.E.=gain of K.E + W

2 2 R 1 mg mv 150 2 2 40 1 1 10 1 v 150 v 10 m / s 2 2 = + × × = × × + ⇒ =

10. The magnitude of the normal reaction that acts on the block at the point Q is

(A) 7.5 N (B) 8.6 N (C) 11.5 N (D) 22.5 N Ans. (A) 2 0 2 At Q mv N mg cos 60 R 1 1 10 N 1 10 2 40 5 N 5 7.5 N 2 − = × ⇒ − × × = ⇒ = + =

Paragraph for Question 11-12

A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers’ usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers’ end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to

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assume that the power cable is purely resistive and the transformers are ideal with a power factor unity. All the currents and voltages mentioned are rms values.

11. If the direct transmission method with a cable of resistance 0 4 Ω km. −1is used, the power dissipation (in %)

during transmission is.

(A) 20 (B) 30 (C) 40 (D) 50 Ans.(B) P Vi (production)= 3 2 2 2 600 10 I 4000 I 150 amp Power dissipation I R 150 (0.4 20) 150 8 180 kW 180 100 Percentage 30% 600 ⇒ × = × ⇒ = = × = × × = × = × = =

12. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is.

(A) 200 : 1 (B) 150 : 1 (C) 100 : 1 (D) 50 : 1 Ans.(A) At step up p p s s s s p p s s V N ₁ V 40000 V V N 10 At Step down V 200 V N V ₄₀₀₀₀ 200 :1 N V 200 = = ⇒ = = = = =

Paragraph for Question 13-14

A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity ω. This can be considered as equivalent to a loop carrying a steady current

2

Qω

π . A uniform magnetic field along the positive

z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionally constant γ .

13. The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is.

(A) 4 BR (B) 2 BR (C) BR (D) 2BR Ans.(B) E.2 R d dt φ π = 2 2 d E. R R dt Β π π ⇒ = 2 2 E. R R 1 Β − 0 π π ⇒ =

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2

R

E Β

⇒ =

14. The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change is.

(A) −γ BQR2 (B) 2 2 −γ BQR (C) 2 2 BQR γ (D) γBQR2 Ans.(B) 2 2 M L .( t ) .(QER×1) R .(Q. .R) Q R . Δ γΔ γ τ Δ = γ Β = γ Β = γ 2 = = ×

Since angular momentum increases in downward direction (-ve z) Hence, ΔM = γ.Q RΒ 2

2 − Paragraph for Question 15-16 The mass of a nucleus A

ZX is less than the sum of the masses of (A – Z) number of neutrons and Z number of

protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m1 and m2 only if

(m1 + m2) < M. Also two light nuclei of masses m3 and m4 can undergo complete fusion and form a heavy

nucleus of mass M’ only if (m3 + m4) > M’. The masses of some neutral atoms are given in the table below:

15. The correct statement is.

(A) The nucleus 63Li can emit an alpha particle (B) The nucleus 84210Po can emit a proton

(C) Deuteron and alpha particle can undergo complete fusion (D) The nuclei 7030Znand 8234Secan undergo complete fusion Ans.(C)

6 4 2

3Li→2 He+1H is not feasible as

4.002603+2.014102=6.016705u i.e. total mass of product is more than 6.015123 u ie the mass of reactant. Similarly reaction mention in option (B) and (D) are not feasible.

On this basis, the only feasible reaction is that deuteron and alpha particle can undergo complete fusion. 16. the kinetic energy (in keV) of the alpha particle, when the nucleus ₈₄210Po at rest undergoes alpha decay, is

(A) 5319 (B) 5422 (C) 5707 (D) 5818 Ans.(A) 210 206 4 84 Po→82 Pb+2He m 209.982876 (205.974455 4.002603) 0.005818 Δ = − + =

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Po Po

Po Po Po Po

k k E (i)

p p (from conservation of momentum) 2m k 2m k m k m k (ii) α α α α α α + = = ⇒ = ⇒ =

From (i) and (ii)

po po m ₂₀₆ K E xE 5319 KeV m m 210 α α = = = +

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Section – 3: (Matching list Type)

This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 17. One mole of a monatomic ideal gas is taken along two cyclic processes

and

→ → → → → →

E F G E E F H Eas shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal of adiabatic.

List -I List -II

(P) _G_→_E (1) 160 P0V0 ln2 (Q) G→H (2) 36 P0V0 (R) F→H (3) 24 P0V0 (S) F→G ₍₄₎ 31 P₀V₀ Code: P Q R S (A) 4 3 2 1 (B) 4 3 1 2 (C) 3 1 2 4 (D) 1 3 2 4 Ans.(A) WGE =P (V0 0−V )G ForFG, P VF F =P VG G⇒VG =32V0 0 0 32 0 31 0 0 GE W =P (V − V )= − P V 0 8 0 32 0 24 0 0 GH W =P ( V − V )= − P V for F to H r PV =const 0 0 32 0 _H ( P )(V )γ P (V )γ ⇒ = 5 3 5 3 0 0 0 32 2 2 8 8 / H H H V V V V V V ⎛ ⎞ ⇒_⎜ _⎟ = = ⇒ = = ⇒ = ⎝ ⎠ 2 2 FH F H F F H H f f W = nR( T −T )= ( P V −P V ) 0 0 0 0 3 32 8 2( P V P V ) = − =36P V_{0 0} 2 1 FH V W PV ln V ⎛ ⎞ = _⎜ _⎟ ⎝ ⎠

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0 0 0 0 32 32P V ln V V ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ =160P V ln( )0 0 2 4 3 2 1 P⎯⎯→ Q⎯⎯→ R⎯⎯→ S⎯⎯→

18. Match List-I of the nuclear processes with List-II containing parent nucleus and one of the end products of each process and then the correct answer using the codes given below the lists:

List-I List -II

(P) Alpha decay (1) 15 15 9 O→7 N+... (Q) _β+_decay (2) 238 234 92 U→90 Th+... (R) Fission (3) 185 184 83 Bi→82 Pb ...+ (S) Proton emission (4) 239 140 94 Pu→57 La ...+ Code: P Q R S (A) 4 2 1 3 (B) 1 3 2 4 (C) 2 1 4 3 (D) 4 3 2 1 Ans.(C) 238 234 4 92 90 2 P : Alpha Decay: u→ Th+ α 15 15 0 8 7 1 239 140 99 94 94 37 185 184 1 83 82 1 Q : Decay : O N e R : Fission : Pu La X S :Pr oton Emission : Bi Pb p P 2, Q 1, R 4, S 3 + + β → + → + → + → → → →

19. A right angled prism of refractive index μ is placed in a rectangular block of refractive index ₁ μ , which is ₂ surrounded by a medium of refractive index μ , as shown in the figure. A ray of light 'e' enters the rectangular 3 block at normal incidence. Depending upon the relationships between μ μ and 1, 2 μ , it takes one of the four 3 possible paths 'ef', 'eg', 'eh' or 'ei'.

Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists :

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List I List II [P] e→ f 1. μ1> 2 μ2 [Q] e→g 2. μ2>μ1 and μ2 >μ3 [R] e→h 3. μ1=μ2 [S] e→i 4. μ₂<μ₁< 2μ and ₂ μ₂ >μ₃ Codes : P Q R S P Q R S (A) 2 3 1 4 (B) 1 2 4 3 (C) 4 1 2 3 (D) 2 3 4 1 Ans.(D) e→ ⇒ μ > μf ₂ ₁ andμ > μ₂ ₃

20. Match List I and List II and select the correct answer using the codes given below the lists :

List I List II [P] Boltzmann constant 1. ⎡_⎣ML T2 −1⎤_⎦ [Q] Coefficient of viscosity 2. ⎡_⎣ML T−1 −1⎤_⎦ [R] Planck constant 3. ⎡_⎣MLT−3K−1⎤_⎦ [S] Thermal conductivity 4. ⎡_⎣ML T2 −3K−1⎤_⎦ Codes : P Q R S P Q R S (A) 3 1 2 4 (B) 3 2 1 4 (C) 4 2 1 3 (D) 4 1 2 3 Ans.(C) 1 2 2 1 P : Boltzmann cons tan t_{= ⎣}⎡M L T K− − ⎤_⎦

1 1 Q : coefficient of vis cos ity_{= ⎣}⎡ML T− − ⎤_⎦

2 1 R : Planck cons tan t_{= ⎣}⎡ML T− ⎤_⎦

3 1 S : Thermal conductivity_{= ⎣}⎡MLT K− − ⎤_⎦ P→4, Q→2, R→1, S→3 1 2 2 1 2 2 2 3 0 1 2 0 1 1 2 e g e h and e i C 45 sin 45 2 P 2, Q 3, R 4, S 1 − → ⇒ μ = μ → ⇒ μ < μ < μ μ > μ → ⇒ < ⎛μ ⎞ ⇒ ⎜_⎜ ⎟_⎟< μ ⎝ ⎠ ⇒ μ > μ → → → →

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Part 2 : CHEMISTRY

Section – 1: (One or more options correct Type)

21. The carbon-based reduction method is NOT used for the extraction of

(A) tin from SnO₂ (B) iron from Fe O_{2 3}

(C) aluminium from Al O_{2 3} (D) magnesium from MgCO .CaCO₃ ₃

Ans.(C,D)Strong metals like Na, Mg, Al are extracted by electrolytic reduction method but not by carbon reduction.

22. The thermal dissociation equilibrium of CaCO s3( )is studied under different conditions.

3( ) ( ) 2( )

CaCO s CaO s +CO g

For this equilibrium, the correct statement (s) is (are) (A) Δ is dependent on T H

(B) K is independent of the initial amount of CaCO3

(C) K is dependent on the pressure of CO2 at a given T

(D) Δ is independent of the catalyst, if any H Ans.(A,B,D) (A) ΔH is dependent on T

(B) K is independent of the initial amount of CaCO3

(D) A catalyst does not changes ΔH

23. The correct statement(s) about O₃ is (are) (A) O O− bond lengths are equal

(B) Thermal decomposition of O3 is endothermic.

(C) O3is diamagnetic in nature.

(D) O3 has a bent structure.

Ans.(A,C,D) (A) In ozone, bond lengths are equal because of resonance. (B) Thermal decomposition of O3 is exothermic.

(C) Ozone is diamagnetic in nature. (D) Structure of ozone is bent.

24. In the nuclear transmutation

9 8 4Be+X ⎯⎯→4Be Y+ (X ,Y) is(are) (A) (γ,n) (B) (p,D) (C) (n,D) (D) (γ, p) Ans.(A,B) 9 0 8 1 4Be 0 4Be 0n ( (n) + γ ⎯⎯→ + γ)

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9 1 8 2 4Be 1H 4Be 1H

(p (D)

+ ⎯⎯→ +

)

25. The major product(s) of the following reaction is(are)

(A) P (B) Q (C) R (D) S

Ans.(B)

26. After completion of the reactions (I and II), the organic compounds(s) in the reaction mixtures is(are)

(A) Reaction I: P and Reaction II: P

(B) Reaction I: U, acetone and Reaction II: Q, acetone (C) Reaction I: T, U, acetone and Reaction II: P

(D) Reaction I: R, acetone and Reaction II: S, acetone

Ans.(C) Reaction 1 : ₃ ₃ Br (1mol)2 NaOH CH − −C CH ⎯⎯⎯⎯⎯→

1/3rd of acetone reacts with 1 mole of Br2, then products are [Bromoform reaction]

3 3

O ||

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Reaction 2 : 2 3 Br 3 3 _{CH COOH} 3 2 O O || || CH − −C CH ⎯⎯⎯⎯⎯→ CH − −C CH −Br

27. The K_spof Ag CrO₂ ₄ is 1.1 × 10−12at 298 K. The solubility (in mol/L) of Ag CrO₂ ₄in a 0.1 M AgNO₃solution is

(A) 1 1 10. × −11 (B) 1 1 10. × −10

(C) 1 1 10. × −12 (D) 1 1 10. × −9

Ans.(B) Ag CrO (s)₂ ₄ 2Ag_(aq)+ +CrO2_{4 (aq)}− 2 2 sp 4 K =[Ag ] [CrO ]+ − 12 2 1.1 10× − =(0.1) (s) ⇒ s 1.1 10= × −10

28. In the following reaction, the product(s) formed is (are)

(A) P (major) (B) Q (minor) (C) R (minor) (D) S (major)

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Section – 2: (Paragraph Type)

among the four choices (A), (B), (C) and (D). Paragraph for Questions 29 and 30

An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and a filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with H2S in a

dilute mineral acid medium. However, it gave a precipitate (R) with H2S in an ammoniacal medium. The precipitate R gave a coloured solution (S), when treated with H O2 2 in an aqueous NaOH medium.

29. The precipitate P contains

(A) Pb2+ (B) Hg22+

(C) Ag+ (D) Hg2+

Ans.(A) Its clear that P contains Pb+2 as PbCl2 does not dissolve in cold water but in hot water it dissolves.

30. The coloured solution S contains

(A) Fe( SO )4 3 (B) CuSO4

(C) ZnSO4 (D) Na CrO2 4

Ans.(D) R is Cr(OH)3

2 2

H O / NaOH

3 2 4

Cr(OH) ⎯⎯⎯⎯⎯⎯→ Na CrO (Yellow solution) (Try to reject other options)

Paragraph for Questions 31 and 32

P and Q are isomers of dicarboxylic acid C4H4O4. Both decolorize Br2/H2O. On heating, P forms the cyclic anhydride.

Upon treatment with dilute alkaline KMnO4, P as well as Q could produce one or more than one from S, T and U.

31. Compounds formed from P and Q are, respectively (A) Optically active S and optically active pair (T, U) (B) Optically inactive S and optically inactive pair (T, U) (C) Optically active pair (T, U) and optically active S (D) Optically inactive pair (T, U) and optically inactive S Ans.(B) P must be maleic acid (_Q it forms anhydride)

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(cis) 4

P +dil. KMnO ⎯⎯→ Meso (S)

(Trans) 4

Q +dil. KMnO ⎯⎯→ Racemic (T U)+

32. In the following reaction sequences V and W are, respectively

(A)

(B)

(C)

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Ans.(A)

Paragraph for questions 33 and 34

A fixed mass ‘m’ of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure

33. The succeeding operations that enable this transformation of states are (A) Heating, cooling, heating, cooling

(B) Cooling, heating, cooling, heating (C) Heating, cooling, cooling, heating (D) Cooling, heating, heating, cooling Ans.(C) Using ideal gas equation

PV = nRT

From K to L, heating occurs From L to M, cooling occurs From M to N, cooling occurs From N to K, heating occurs

34. The pair of isochoric processes among the transformation of states is (A) K to L and L to M

(B) L to M and N to K (C) L to M and M to N (D) M to N and N to k

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Paragraph for Questions 35 and 36

The reactions of Cl2 gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids

of chlorine, P and Q, respectively. The Cl2 gas reacts with SO2 gas, in presence of charcoal, to give a product R. R reacts

with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus, T. 35. P and Q, respectively, are the sodium slats of

(A) hypochlorus and chloric acids (B) hypochlorus and chlorus acids (C) chloric and perchloric acids (D) chloric and hypochlorus acids

Ans.(A) 2 3

NaOH+Cl ⎯⎯→ NaOCl + NaClO Salt of Salt of Chloric acid hypochlorous acid

36. R, S and T, respectively, are (A) SO2Cl2, PCl5 and H3PO3 (B) SO2Cl2, PCl3 and H3PO3 (C) SOCl2, PCl3 and H3PO2 (D) SOCl2, PCl5 and H3PO4 Ans.(A) SO2 +Cl2 ⎯⎯→SO Cl2 2 2 2 5 2 5SO Cl +2P ⎯⎯→ 2 PCl +5O 5 2 3 4 PCl +H O ⎯⎯→H PO +5HCl

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Section – 3: (Matching list Type)

This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 37. An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List I. The variation in

conductivity of these reactions is given in List II. Match List I with List II and select the correct answer using the code given below the lists:

List I List II

P. (C H ) N CH COOH2 5 3 3

X Y

+

1. Conductivity decreases and then increases Q. KI(0.1M) AgNO (0.01M)3

X Y

+

2. Conductivity decreases and then does not change much R. (CH COOH) KOH3

X Y

+

3. Conductivity increases and then does not changes much

S. NaOH HI

X Y

+

4. Conductivity does not change much and then increases Codes P Q R S (a) 3 4 2 1 (b) 4 3 2 1 (c) 2 3 4 1 (d) 1 4 3 2

Ans.(A) When Et3N solution is added to CH3COOH, ' 'α of CH3COOH will increase, so though H+ ions ions will be

consumed by Et3N but overall no. of ions in the solution will increase and hence conductivity increases initially.

KI (aq)

(aq) 3(aq) (aq) 3(aq)

Ag+ +NO− ⎯⎯⎯⎯→ AgI(s)↓ + K+ +NO+

Number of ions remain the same so, conductivity does not change much. After all Ag+_{ions have been precipitated}

out, then the conductivity will increase.

3

CH COOH (aq)

(aq) (aq) 2 (aq) 3 (aq)

K+ +OH− ⎯⎯⎯⎯⎯⎯⎯→ H O( )_l + K+ +CH COO−

Number of ions remains the same, but number of OH− ions decreases and hence conductivity decreases.

NaOH

2 (aq) (aq)

H+ +I− ⎯⎯⎯⎯→ H O( )_l + Na+ +I−

Number of ions remains the same, but number of H+ decreases and hence conductivity decreases. 38. The standard reduction potential data at 25oC is given below.

0 3 2

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0 2 0 2 0 0 2 2 0 2 2 0 3 0 2 E (Fe , Fe) 0.44 V E (Cu ,Cu) 0.34 V; E (Cu ,Cu) 0.52 V E O (g) 4H 4e 2H O 1.23V; E O (g) 2H O 4e 4OH 0.40 V E (Cr ,Cr) 0.74 V; E (Cr ,Cr) 0.91V + + + + − − − + + = − = + = + ⎡ ₊ ₊ _→ ⎤_{= +} ⎢ ⎥ ⎣ ⎦ ⎡ ₊ ₊ _→ ⎤_{= +} ⎢ ⎥ ⎣ ⎦ = − = −

Match E0_{of the redox pair in List I with the values given in List II ad select the correct answer using the code}

Given below the lists:

List I List II 0 3 P. E (Fe , Fe)+ 1. -0.18 V 0 2 Q. E (4H O 4H++4OH )− 2. -0.4 V 0 2 R. E (Cu ++Cu→2Cu )+ 3. -0.04 V 0 3 2 S. E (Cr ,Cr )+ + 4. -0.83 V Codes P Q R S (a) 4 1 2 3 (b) 2 3 4 1 (c) 1 2 3 4 (d) 3 4 1 2 Ans.(D) Fe+3+e− ⎯⎯→ Fe+2 E° = +0.77 2 3 Fe 2e Fe E 0.44 ___________________ Fe 3e Fe E ? + − + − + ⎯⎯→ ° = − + ⎯⎯→ ° =

(

)

0.77 1 2 0.44 0.11 E 0.04V 3 3 × + × − − ° = = = −

So, P→ and hence Answer is (D). 3

39. The unbalanced chemical reactions given in List I show missing reagent or condition (?) which are

provided in List II. Match List I with List II and select the correct answer using the code given below the lists:

List I List II

P. PbO₂+H SO₂ ₄⎯⎯?→PbSO₄+O₂+other product 1. NO Q Na S O_{2 2 3}+H O₂ ⎯⎯?→NaHSO₄+other product 2. I2

R N H_{2 4}⎯⎯?→N₂+Other product 3. Warm

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Codes P Q R S (a) 4 2 3 1 (b) 3 2 1 4 (c) 1 4 2 3 (d) 3 4 2 1

Ans.(D) [P] Its disproportionation reaction, so just heating will do.

[Q] I2 does reaction as S O2 32− ⎯⎯→S O4 6−2 So Cl2 is required for S O2 3−2 ⎯⎯→ HSO4−

[R] N H2 4 +2I2 ⎯⎯→ N2+4HI [S] XeF2 + NO ⎯⎯→Xe

40. Match the chemical conversions in List I with the appropriate reagents in List II and select the correct answer using the code given below the lists:

Codes P Q R S (a) 2 3 1 4 (b) 3 2 1 4 (c) 2 3 4 1 (d) 3 2 4 1 Ans.(A)

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Part 3 : MATHS

Section – 1: (One or more options correct Type)

41. Let ω be a complex cube root of unity with ω ≠ and 1 P= ⎣ ⎦ be a ⎡p_ij⎤ n× matrix with n p_ij=ωi+ j. Then P2≠ , when n = 0

(A) 57 (B) 55 (C) 58 (D) 56

Ans.

42. The function f x

( )

=2 x + x+2 − x+2 −2 x has a local minimum or a local maximum at x =

(A) −2 (B) 2 3 − (C) 2 (D) 2 3 Ans. 43. Let 3 2 i w= + and P=

{

w : nn =1 2 3, , , . . .

}

. Further ₁ 1 2 H =⎧⎨z∈C : Re z> ⎫⎬ ⎩ ⎭ and 2 1 2 H =⎧_⎨z∈C : Re z< − ⎫_⎬

⎩ ⎭, where C is the set of all complex numbers. If z1∈ ∩P H , z1 2∈ ∩P H2 and O represents the origin, then ∠z Oz1 2=

(A) 2 π (B) 6 π (C) 2 3 π (D) 5 6 π

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Ans. 44. If 3x =4x−1, then x = (A) 3 3 2 2 2 2 1 log log − (B) ₂ 2 2−log 3 (C) ₄ 1 1−log 3 (D) 2 2 2 3 2 3 1 log log − Ans. 45. Two lines L1 : 5 3 2 y z x , α = = − − and L2 : 1 2 y z x α, α = =

− − are coplanar. Then α can take value(s) :

(A) 1 (B) 2 (C) 3 (D) 4

Ans.

46. In a triangle PQR, P is the largest angle and 1 3

cos P= . Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is(are) :

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Ans.

47. For a∈R (the set of all real numbers), a≠ −1,

(

)

(

)

1

(

) (

)

(

)

1 2 ₁ 60 1 1 2 a a a a n . . . n lim n − na na . . . na n → ∞ + + + = ⎡ ⎤ + _⎣ + + + + + + _⎦ Then a = (A) 5 (B) 7 (C) 15 2 − (D) 17 2 − Ans.

48. Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length 2 7 on y-axis is(are) : (A) x2+y2−6x+8y+ =9 0 (B) x2+y2−6x+7y+ =9 0

(C) x2+y2−6x−8y+ =9 0 (D) x2+y2−6x−7y+ =9 0 Ans.

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Section – 2: (Paragraph Type)

among the four choices (A), (B), (C) and (D). Paragraph for Questions 49 - 50

Let f :

[ ]

0 1, → R (the set of all real numbers) be a function. Suppose the function f is twice differentiable,

( )

0

( )

1 0

f = f = and satisfies f′′

( )

x −2f′

( )

x + f x

( )

≥e , xx ∈

[ ]

0 1, .

49. Which of the following is true for 0 < x < 1 ? (A) 0< f x

( )

< ∞ (B) 1

( )

1 2 f x 2 − < < (C) 1

( )

1 4 f x − < < (D) −∞ < f x

( )

<0 Ans.

50. If the function e−xf x

( )

assumes its minimum in the interval [0, 1] at 1

4

x= , which of the following is true ?

(A)

( )

1 3 4 4 f′ x < f x , < <x (B)

( )

0 1 4 f′ x > f x , < <x (C)

( )

0 1 4 f′ x < f x , < <x (D)

( )

3 1 4 f′ x < f x , < <x Ans.

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Paragraph for Questions 51 - 52

Let PQ be a focal chord of the parabola y2=4ax. The tangents to the parabola at P and Q meet at a point lying on the line

2 0

y= x+a, a> .

51. Length of chord PQ is :

(A) 7a (B) 5a (C) 2a (D) 3a

Ans.

52. If chord PQ subtends an angle θ at the vertex of y2 =4ax, then tanθ= (A) 2 7 3 (B) 2 7 3 − (C) 2 5 5 (D) 2 5 3 − Ans.

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Paragraph for Questions 53 - 54 Let S=S1∩S2∩S3, where

{

}

1 2 1 3 4 0 1 3 z i S z C : z , S z C : Im i ⎧ ⎡ − + ⎤ ⎫ ⎪ ⎪ = ∈ < =⎨ ∈ ⎢ ⎥> ⎬ − ⎢ ⎥ ⎪ _⎣ _⎦ ⎪ ⎩ ⎭ and S₃ =

{

z∈C : Re z>0

}

. 53. Area of S = (A) 10 3 π (B) 20 3 π (C) 16 3 π (D) 32 3 π Ans.

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54. 1 3 z S min i z ∈ − − = (A) 2 3 2 − (B) 2 3 2 + (C) 3 3 2 − (D) 3 3 2 + Ans.

Paragraph for Questions 55 - 56

A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red balls and 4 black

balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls.

55. If 1 ball is drawn from each of the boxes B1, B2 and B3, the probability that all 3 drawn balls are of the same

colour is : (A) 82 648 (B) 90 648 (C) 558 648 (D) 566 648 Ans.

56. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2 is :

(A) 116 181 (B) 126 181 (C) 65 181 (D) 55 181 Ans.

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Section – 3: (Matching list Type)

This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 57. Match List I with List II and select the correct answer using the code given below the lists :

List I List II

(P) Volume of parallelepiped determined by vectors a , bur ur and cur is 2. Then the volume of the parallelepiped determined by vectors 2

(

aur ur×b ,

) (

3 bur ur×c

)

and

(

cur ur×a

)

is

1. 100

(Q) Volume of parallelepiped determined by vectors a , bur ur and cur is 5. Then the volume of the parallelepiped determined by vectors 3

(

aur ur+b ,

) (

3 bur+ cur

)

and

(

)

2 cur ur+a is

2. 30

(R) Area of a triangle with adjacent sides determined by vectors aurand bur is 20. Then the area of the triangle with adjacent sides determined by vectors

(

2aur+3bur

)

and

(

aur ur−b

)

is

3. 24

(S) Area of a parallelogram with adjacent sides determined by vectors aurand bur is 30. Then the area of the parallelogram with adjacent sides determined by vectors

(

aur ur+b

)

and aur is

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Codes : P Q R S (A) 4 2 3 1 (B) 2 3 1 4 (C) 3 4 1 2 (D) 1 4 3 2 Ans.

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58. Consider the line, L1 : 2 1 3 4 3 3 2 1 1 1 1 2 x y z x y z , L − + − + + = = = = =

− and the planes P1 : 7x+ +y 2z=3, P2 : 3x+5y−6z=4. Let ax+by+cz=d be the equation of the plane passing through the point of intersection of

lines L1 and L2, and perpendicular to planes P1 and P2. Match List I with List II and select the correct answer

using the code given below the lists :

List I List II [P] a = 1. 13 [Q] b = 2. −3 [R] c = 3. 1 [S] d = 4. −2 Codes : P Q R S P Q R S (A) 3 2 4 1 (B) 1 3 4 2 (C) 3 2 1 4 (D) 2 4 1 3 Ans.

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59. Match List I with List II and select the correct answer using the code given below the lists : List I List II (P)

(

)

(

)

(

) (

)

1 2 2 1 1 4 2 1 1

1 cos tan y y sin tan y y

y cot sin y tan sin y

− − − − ⎛ _⎛ _⎞ ⎞ + ⎜ _⎜ ⎟ ⎟ + ⎜ _⎜ _⎟ ⎟ ⎜ ⎜ + ⎟ ⎟ ⎜ _⎝ _⎠ ⎟ ⎝ ⎠ takes value 1. 1 5 2 3

(Q If cos x+cos y+cos z= =0 sin x+sin y+sin z then possible value of 2

x y

cos − is 2. 2

(R) If cos 4 x cos x2 sin x sin x sec x2 cos sin x sec x2 cos 4 x cos x2

π π

⎛ ⎞ ⎛ ⎞

− + = + +

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ then

possible value of sec x is

3. 1 2

(S) If cot sin⎜⎛ −1 1−x2⎞⎟=sin tan

(

−1

( )

x 6

)

, x≠0

⎝ ⎠ , then possible value of x is 4. 1

Codes :

P Q R S P Q R S

(A) 4 3 1 2 (B) 4 3 2 1

(C) 3 4 2 1 (D) 3 4 1 2

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60. A line L : y=mx+ meets y-axis at E(0, 3) and the arc of the parabola 3 y2 =16 0x, ≤ ≤ at the point y 6

(

0 0

)

F x , y . The tangent to the parabola at F x , y

(

₀ ₀

)

intersects the y-axis at G

(

0, y₁

)

. The slope m of the line L is chosen such that the area of the triangle EFG has a local maximum. Match List I with List II and select the correct answer using the code given below the lists :

List I List II

[P] m = 1. 1/2

[Q] Maximum area of EFGΔ is 2. 4

[R] y0 = 3. 2 [S] y1 = 4. 1 Codes : P Q R S P Q R S (A) 4 1 2 3 (B) 3 4 1 2 (C) 1 3 2 4 (D) 1 3 4 2 Ans.

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