JEE-Advanced-2013 Paper 2 Code 0 - Amazon Web...

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VMC/Solutions 1 Paper 2 - Code 0 – JEE Advanced - 2013

JEE-Advanced-2013

Paper 2

Code 0

2 June, 2013 | 2:00 PM – 5:00 PM

Detailed solutions to the paper by Vidyamandir Classes

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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∆ = = ×∫ area under C T− graph

dQ dT

mCdt dt

=

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

dtis 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 22 1

1 1 1RZ

n nλ

= −

13

32

hL n

π= ⇒ =

2 21

0 0 03 9

4 5 24 5

na a . a a . Z

z z .= ⇒ = ⇒ = =

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2

2 22

1 1 1

3R( Z )

nλ

= −

29

25

n ,R

λ= =

29

132

n ,R

λ= =

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

θ in the range 0 to 90 . The wavelength λ is exactly known and the error in θ is constant for all values

of θ . As θ increases from 0 ,

(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

dsin

λλ

θ⇒ =

2

d cos ecλ

θ⇒ = .....(1)

2

d( cos ec cot )

∆ λθ θ

∆θ⇒ = −

2

d cos ec cot ( )λ

∆ θ θ ∆θ= − ...(2)

Equation 2/1 ( ) is constantd

cot ( )d

∆θ ∆θ ∆θ= −

both andd

dd

∆∆ are decreasing with θ .

4. Two non-conducting spheres of radii R1 and R2 carrying uniform volume charge densities and+ −ρ 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 03 3

O' P OPEρ

ρ ρ= −

∈ ∈

= 03

O' O( O' O OP O' P )

ρ+ =

∈

∵

So, E is constant in magnitude as well as direction.

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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 1 2 1( v w ) ( u ) v w u

f f f f( v w ) ( u ) v w u

+ − − + += = ⇒ >

+ − + + −

If the wind blows from observer to source

2 1 1 2 1( v w ) ( u ) v w u

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 4GM

L

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

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

221 GMm GM

mu u 22 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 =m

tk

π

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

3=

mt

k

π

(D) the time at which the particle passes through the equilibrium position for the second time is 5

3=

mt

k

π

Ans. (A,D)

x Asin t= ω

v (A )cos t= ω ω

00

uu cos t cos t 1/ 2 t / 3

2= ω ⇒ ω = ⇒ ω = π

K mt / 3 t

m 3 K

π⇒ = π ⇒ =

(time taken to reach wall from mean position)

m 2 mtime of passing equilibrium for first time 2

3 K 3 K

π π= × =

2 m 2 m 7 mtime of maximum compression

3 K 4 K 6 K

π π π= + =

2 m m 5 mtime of passing equilibrium for second time

3 K K 3 K

π 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, 210 )−=g m s .

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

(A) 15 −ms (B) 110 −ms (C) 110 3 −ms (D) 120 −ms

Ans. (B)

against fLoss of P.E.=gain of K.E + W

2

2

R 1mg mv 150

2 2

40 11 10 1 v 150 v 10m / 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)

20

2

At Q

mvN mg cos60

R

1 1 10N 1 10

2 40

5N 5 7.5N

2

− =

×⇒ − × × =

⇒ = + =


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 10 4 Ω km−. is 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 100Percentage 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

ss s

s

p p

s s

V N 1V 40000V

V N 10

At Stepdown V 200V

N V 40000200 :1

N V 200

= = ⇒ =

=

= = =


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

Qωπ

. 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) 2d

E. Rdt

φπ =

22d

E. R Rdt

Βπ π⇒ =

2

2E. R R1

Β − 0π π⇒ =

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2

RE

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) 2− BQRγ (B) 2

2−

BQRγ (C)

2

2

BQRγ (D) 2BQRγ

Ans.(B)

22

M L

.( t )

.(QER×1)

R.(Q. .R)

Q R.

∆ γ∆γ τ ∆

= γΒ

= γ

Β= γ

2

=

= ×

Since angular momentum increases in downward direction (-ve z)

Hence, ∆M = 2Q R

.Β

γ2

−


The mass of a nucleus AZ X 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 63 Li can emit an alpha particle

(B) The nucleus 21084 Po can emit a proton

(C) Deuteron and alpha particle can undergo complete fusion

(D) The nuclei 70 8230 34andZn Se can undergo complete fusion

Ans.(C)

6 4 23 2 1Li He H→ + 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 21084 Po at rest undergoes alpha decay, is

(A) 5319 (B) 5422 (C) 5707 (D) 5818

Ans.(A)

210 206 484 82 2Po Pb He→ +

m 209.982876 (205.974455 4.002603) 0.005818∆ = − + =

Total energy released = 0.005818 x 932 MeV=E

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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 206K E xE 5319 KeV

m m 210α

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 E as 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) 0 0GE GW P (V V )= −

For 032F F G G GFG, P V P V V V= ⇒ =

0 0 0 0 032 31GEW P (V V ) P V= − = −

0 0 0 0 08 32 24GHW P ( V V ) P V= − = −

for F to H

rPV const=

0 032 0 H( P )(V ) P (V )γ γ⇒ =

5 35 3

00 0

32 2 2 8 8

/

H HH

V VV V

V V

⇒ = = ⇒ = = ⇒ =

2 2

FH F H F F H Hf f

W nR(T T ) ( P V P V )= − = −

0 0 0 03

32 82( P V P V )= − 0 036P V=

2

1FH

VW PV ln

V

=

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00 0

0

3232

VP V ln

V

=

0 0160 2P V ln( )=

4 3 2 1P 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) 15 159 7→ +O N .....

(Q) decay+β (2) 238 23492 90→ +U Th .....

(R) Fission (3) 185 18483 82→ +Bi Pb .....

(S) Proton emission (4) 239 14094 57→ +Pu 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)

238 234 492 90 2P : Alpha Decay: u Th→ + α

15 15 08 7 1

239 140 9994 94 37

185 184 183 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 1µ is placed in a rectangular block of refractive index 2µ , which is

surrounded by a medium of refractive index 3µ , as shown in the figure. A ray of light 'e' enters the rectangular

block at normal incidence. Depending upon the relationships between 1 2,µ µ and 3µ , it takes one of the four

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. 1 22µ µ>

[Q] e g→ 2. 2 1µ µ> and 2 3µ µ>

[R] e h→ 3. 1 2µ µ=

[S] e i→ 4. 2 1 22< <µ µ µ and 2 3µ µ>

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) 2 1 2 3e f and→ ⇒µ > µ µ > µ

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. 2 1

ML T−

[Q] Coefficient of viscosity 2. 1 1ML T

− −

[R] Planck constant 3. 3 1MLT K− −

[S] Thermal conductivity 4. 2 3 1

ML T K− −

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 1P : Boltzmann cons tan t M L T K− − =

1 1

Q : coefficient of vis cos ity ML T− − =

2 1R : 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 02

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




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

(A) tin from 2SnO (B) iron from 2 3Fe O

(C) aluminium from 2 3Al O (D) magnesium from 3 3MgCO .CaCO

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 3( )CaCO s is studied under different conditions.

3 2( ) ( ) ( )CaCO s CaO s CO g+

For this equilibrium, the correct statement (s) is (are)

(A) H∆ is dependent on T

(B) K is independent of the initial amount of 3CaCO

(C) K is dependent on the pressure of 2CO at a given T

(D) H∆ is independent of the catalyst, if any

Ans.(A,B,D) (A) H∆ is dependent on T

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

(D) A catalyst does not changes H∆

23. The correct statement(s) about 3O is (are)

(A) O O− bond lengths are equal

(B) Thermal decomposition of 3O is endothermic.

(C) 3O is diamagnetic in nature.

(D) 3O 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 84 4Be X Be Y+ → +

( )X ,Y is(are)

(A) ( ),nγ (B) ( )p,D (C) ( )n,D (D) ( ), pγ

Ans.(A,B) 9 0 8 14 0 4 0Be Be n

( (n)

+ γ → +

γ)

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9 1 8 24 1 4 1Be H Be H

(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 : 2Br (1mol)3 3

NaOHCH C CH− − →

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

3 3

O||

CH C O Na , CHBr− +− − and remaining acetone

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Reaction 2 : 2

3

Br3 3 3 2

CH COOH

O O|| ||

CH C CH CH C CH Br− − → − − −

27. The spK of 2 4Ag CrO is 1.1 × 1210

− at 298 K. The solubility (in mol/L) of 2 4Ag CrO in a 0.1 M 3AgNO solution

is

(A) 111 1 10.

−× (B) 101 1 10.

−×

(C) 12

1 1 10.−× (D)

91 1 10.

−×

Ans.(B) 22 4 (aq) 4 (aq)Ag CrO (s) 2Ag CrO+ −+

2 2sp 4K [Ag ] [CrO ]+ −=

12 21.1 10 (0.1) (s)−× =

⇒ 10s 1.1 10−= ×

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

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

Ans.(BD)

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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 2 2H O in an aqueous NaOH medium.

29. The precipitate P contains

(A) 2Pb

+ (B) 22Hg +

(C) Ag+ (D) 2Hg +

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) 4 3Fe( SO ) (B) 4CuSO

(C) 4ZnSO (D) 2 4Na CrO

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

2 2H O / NaOH3 2 4Cr(OH) Na CrO→ (Yellow solution)

(Try to reject other options)


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 (∵ it forms anhydride)

Q must be fumaric acid.

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(cis) 4P dil. KMnO Meso (S)+ →

(Trans) 4Q dil. KMnO Racemic (T U)+ → +

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

(A)

(B)

(C)

(D)

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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

Ans.(B) Isochoric processes are those in which volume remain constant. So isochoric processes are L to M and N to K.

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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 3NaOH 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) 2 2 2 2SO Cl SO Cl+ →

2 2 5 25SO Cl 2P 2PCl 5O+ → +

5 2 3 4PCl H O H PO 5HCl+ → +

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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. 2 5 3 3(C H ) N CH COOH

X Y

+ 1. Conductivity decreases and then increases

Q. 3KI(0.1M) AgNO (0.01M)

X Y

+ 2. Conductivity decreases and then does not change much

R. 3

(CH COOH) KOH

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.

3CH COOH(aq)(aq) (aq) 2 (aq) 3 (aq)K OH H O( ) 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( ) 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

E (Fe ,Fe ) 0.77V;+ + = +

Vidyamandir Classes


0 2

0 2

0

02 2

02 2

0 3

0 2

E (Fe ,Fe) 0.44V

E (Cu ,Cu) 0.34V;

E (Cu ,Cu) 0.52V

E O (g) 4H 4e 2H O 1.23V;

E O (g) 2H O 4e 4OH 0.40V

E (Cr ,Cr) 0.74V;

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

2Q. 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) 3 2Fe e Fe

+ − ++ → E 0.77° = +

2

3

Fe 2e Fe E 0.44___________________

Fe 3e Fe E ?

+ −

+ −

+ → ° = −

+ → ° =

( )0.77 1 2 0.44 0.11

E 0.04V3 3

× + × − −° = = = −

So, P 3→ and hence Answer is (D).

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. ?

2 2 4 4 2PbO H SO PbSO O other product+ → + + 1. NO

Q ?

2 2 3 2 4Na S O H O NaHSO other product+ → + 2. I2

R ?

2 4 2N H N Other product→ + 3. Warm

S ?

2XeF Xe Other product→ + 4. Cl2

Vidyamandir Classes


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 2 2

2 3 4 6S O S O− −→ So Cl2 is required for

22 3 4S O HSO

− −→

[R] 2 4 2 2N H 2I N 4HI+ → + [S] 2XeF 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)

Vidyamandir Classes


Part 3 : MATHS




41. Let ω be a complex cube root of unity with 1ω ≠ and ijP p = be a n n× matrix with i jijp ω += .

Then 20P ≠ , when n =

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

Ans.

42. The function ( ) 2 2 2 2f x x x x 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

iw

+= and 1 2 3nP w : n , , , . . .= = . Further 1

1

2H z C : Re z

= ∈ >

and 2

1

2H z C : Re z

− = ∈ <

,

where C is the set of all complex numbers. If 1 1 2 2z P H , z P H∈ ∩ ∈ ∩ and O represents the origin, then

1 2z O z∠ =

(A) 2

π (B)

6

π (C)

2

3

π (D)

5

6

π

Ans.

Vidyamandir Classes


44. If 1

3 4x x−= , then x =

(A) 3

3

2 2

2 2 1

log

log − (B)

2

2

2 3log− (C)

4

1

1 3log− (D)

2

2

2 3

2 3 1

log

log −

Ans.

45. Two lines L1 : 53 2

y zx ,

α= =

− − and L2 :

1 2

y zx ,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

3cos 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) :

(A) 16 (B) 18 (C) 24 (D) 22

Ans.

Vidyamandir Classes


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

( )

( ) ( ) ( ) ( )1

1 2 1

601 1 2

a a a

an

. . . nlim

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) 2 2 6 8 9 0x y x y+ − + + = (B) 2 2 6 7 9 0x y x y+ − + + =

(C) 2 2 6 8 9 0x y x y+ − − + = (D) 2 2 6 7 9 0x y x y+ − − + =

Ans.

Vidyamandir Classes






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 0f f= = and satisfies ( ) ( ) ( ) [ ]2 0 1xf x f x f x e , x ,′′ ′− + ≥ ∈ .

49. Which of the following is true for 0 < x < 1 ?

(A) ( )0 f x< < ∞ (B) ( )1 1

2 2f x− < < (C) ( )

11

4f x− < < (D) ( ) 0f x−∞ < <

Ans.

Given : ( ) ( ) ( )2 xf x f x f x e′′ ′− + ≥

⇒ ( ) ( ) ( ) ( ) xf x f x f x f x e′′ ′ ′ − − − ≥

Multiply by x

e−

( ) ( ) ( ) ( ) 1x xe f x f x e f x f x− −′′ ′ ′ − − − ≥

( ) ( ) 1x xd de f x e f x

dx dx

− − ′ − ≥

( ) ( )( ) 1xde f x f x

dx

− ′ − ≥

( )2

21xd

e f xdx

− ≥

⇒ Function ( ) ( )xe x f x− is concave upwards

( ) ( )0 0 1f f= = ⇒ ( ) 0xe f x− = at 0 1x ,=

⇒ ( ) [ ]0 0 1xe f x x ,− ≤ ∀

As 0xe x R− ≥ ∀ ∈

⇒ ( ) [ ]0 0 1f x x ,≤ ∀ ∈

Hence option (D) is correct.

50. If the function ( )xe f x− assumes its minimum in the interval [0, 1] at 1

4x = , which of the following is true ?

(A) ( ) ( )1 3

4 4f x f x , x′ < < < (B) ( ) ( )

10

4f x f x , x′ > < <

(C) ( ) ( )1

04

f x f x , x′ < < < (D) ( ) ( )3

14

f x f x , x′ < < <

Ans.

Vidyamandir Classes



Let PQ be a focal chord of the parabola 2 4y ax= . The tangents to the parabola at P and Q meet at a point lying on the line

2 0y 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 2 4y ax= , then tanθ =

(A) 2

73

(B) 2

73

− (C)

25

3 (D)

25

3

−

Ans.

Vidyamandir Classes



Let 1 2 3S S S S= ∩ ∩ , where

1 2

1 34 0

1 3

z iS z C : z , S z C : Im

i

− + = ∈ < = ∈ >

− and 3 0S z C : Re z= ∈ > .

53. Area of S =

(A) 10

3

π (B)

20

3

π (C)

16

3

π (D)

32

3

π

Ans.

54. 1 3z Smin i z∈

− − =

(A) 2 3

2

− (B)

2 3

2

+ (C)

3 3

2

− (D)

3 3

2

+

Ans.

Vidyamandir Classes



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.

Vidyamandir Classes





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 , b

and c is 2. Then the

volume of the parallelepiped determined by vectors ( ) ( )2 3a b , b c× ×

and

( )c a×

is

1. 100

(Q)

Volume of parallelepiped determined by vectors a , b

and c is 5. Then the

volume of the parallelepiped determined by vectors ( ) ( )3 a b , b c+ +

and

( )2 c a+

is

2. 30

(R)

Area of a triangle with adjacent sides determined by vectors aand b is 20. Then

the area of the triangle with adjacent sides determined by vectors ( )2 3a b+

and

( )a b−

is

3. 24

(S)

Area of a parallelogram with adjacent sides determined by vectors aand b is 30.

Then the area of the parallelogram with adjacent sides determined by vectors

( )a b+

and a is

4. 60

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

Vidyamandir Classes


Ans.

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 : 7 2 3x y z+ + = ,

P2 : 3 5 6 4x y z+ − = . 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

Vidyamandir Classes


Ans.

Vidyamandir Classes


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 22

1 1

4

2 1 1

1 cos tan y y sin tan yy

y cot sin y tan sin y

− −

− −

+ + +

takes value 1. 1 5

2 3

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

x ycos

− is 2. 2

(R)

If

2 2 2 24 4

cos x cos x sin x sin x sec x cos x sin x sec x cos x cos xπ π − + = + +

then

possible value of sec x is

3. 1

2

(S) If ( )( )1 2 11 6 0cot sin x sin tan x , x− − − = ≠

, 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

Ans.

Vidyamandir Classes


60. A line L : 3y mx= + meets Y-axis at E(0, 3) and the arc of the parabola 2 16 0 6y x, y= ≤ ≤ at the point

( )0 0F x , y . The tangent to the parabola at ( )0 0F x , y intersects the Y-axis at ( )10G , y . 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 EFG∆ 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.

Vidyamandir Classes


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,C,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 D

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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