IITJEE 2014 Advanced Paper II & Solution

8/12/2019 IITJEE 2014 Advanced Paper II & Solution

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IIT - JEE 2013 (Advanced)

CODE : O


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(2) Vidyalankar : IIT JEE 2013 Advanced : Question Paper & Solution

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IIT JEE 2013 Advanced : Question Paper & Solution (Paper II) (3)

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PART I : PHYSICSSECTION 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 MOREare correct.

1. The figure below shows the variation of specific heat capacity (C) of solid as a function of

temperature (T). The temperature is increased continuously from 0 to 500 K at a constantrate. 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 withtemperature T.

(B) heat absorbed in increasing the temperature from 0 100 K is less than the heatrequired 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.

1. (A), (B), (C), (D)

dQ = mc dT

dQ

dT

= m (aT + b) in the range (0 100 k)

option (A) is correct. Considering (0 100 k), the graph to be linear using thereasonable approximation given in the question.

During (400k 500k) C is the maximum. option (B) is correct. Option (C) is correct as C is constant in the range (400k 500k)and option (D) is also correct.

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 is3h

2. 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

2. (A), (C)

rn =2

0

na

z

4.5a0 =2

0

na

z

2n

z= 4.5

9

z = 4.5 and z = 2

L =nh

2

L =3h

2

n = 3

100 200 300 400 500

C

T (K)


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Wavelength is given by

1

= 2

2 2

f i

1 1Rz

n n

1

=

2 2

f i

1 14R

n n

For transition, n = 3 1

=1

14R 1

9

=9

32R (A)

For transition n = 3 2

=1

1 14R

4 9

=1 36

4R 5

=

9

5R (C)

3. Using the expression 2d sin = , one calculates the values of d by measuring thecorresponding angles in the range 0 to 90. The wavelength is exactly known and theerror in is constant for all values of . As 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.

3. (D)

2d sin = (2d sin ) = But = 0

(2 sin ) d + 2d cos = 0

d

d

+ cot = 0

d

d

= cot (d)

= constantabsolute error in d is |d| = d cot ()

|d| =2

cos| d |

2 sin

at increase from 0 to 90, |d| decreases.Hence (B) is not true.By the same fractional error decreases.

4. Two non-conducting spheres of radii R1 and R2 and carrying uniform volume charge

densities +and , respectively, are placed such that they partially overlap, as shown inthe 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.


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4. (C), (D)

1 10

E O P3

=

2 20

E PO3

=

Net field at P

1 2E E E= +

1 20

[O P PO ]3

= +

1 20

E O O3

=

E is constant (C)And option (D) is also correct.

5. A steady current I flows along an infinitely long hollow cylindrical conductor of radius R.

This cylinder is placed coaxially 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.

5. (A), (D)

n =N

L

Bd = 0I

B1 =0I

2 R

along axis of cylinder

B2 = 0nI along axis of solenoid

(i) 0 < r < R, magnetic field

At this point, there will be field only due to solenoid.

(ii) R < r < 2R

B = 0I

2 r

by the rod. (Along the tangent to the circular part around the cylinder)

Bsolenoid= 0nI along axis

6. Two vehicles, each moving with speed u on the same horizontal straight road, are

approaching each other. Wind blows along the road with velocity w. One of these vehicles

blows a whistle of frequency f1. An observer in the other vehicle hears the frequency of

the whistle to be f2. The speed of sound in still air is V. The correct statements(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> f

1.

(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.

O1

E2

O2

E1P


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6. (A), (B)

uSW= u wuOW= u + w

f2 = v u wv (u w)+ +

f1 f2= v u wv u w+ + + f1

Clearly f2> f1irrespective of sign of w.

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 their 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 isGM

4L

.

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

bodies isGM

2L

.

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

bodies is2GM

L.

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

7. (B), (D)

(D) is clearly correct, since gravitational field is conservative.

2

e

2GMm 1

mvL 2 + = 0

ve=GM

2L

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 .

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

3 k

.

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

time is t =5 m

3 k

.

8. (A), (D)

The impulse of the wall merely reverses the direction of motion is thout changing particlespeed. Hence (A)

S Ou u

W


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Displacement from mean position be x

x = 0u

w

sin (t) x = u0cos (t)

when particle speed is 0u

2

cos (t) =1

2 t =

3 [for first time]

t =3

Time after which it passes the mean position = 2t =3

3u

=

2 m

3 K

Hence (B) is incorrect.

Time after which particle crosses mean position for second time =2

3

+

=

5

3

Hence (D).

SECTION 2 : (Paragraph Type)

This section contains 4 paragraphs each describing theory, experiment, data etc. Eight

questions relate 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 Questions 9 and 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 inovercoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the

acceleration due to gravity, g = 10 ms2).

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

(A) 5 ms1 (B) 10 ms1 (C) 10 3 ms1 (D) 20 ms1

9. (B)

Kinetic energy of block at Q = mgR sin 30150 J

21

mv2

= 10 40 1

2150 = 50 J

r =2 50

1

= 10 ms1


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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

10. (A)

N mg cos 60 =2mv

r

N =mg cos 60+2mv

r

N =10 1 100

2 40

+ = 7.5 N


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 the two ends. The drawback of thedirect 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

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 0.4km1is used, the powerdissipation (in%) during transmission is

(A) 20 (B) 30 (C) 40 (D) 50

11. (B)P = 600 kW

V = 4000 V

I =3600 10

4000

= 150A

Power Loss = I2R

= (150)20.4 20 = 180 kW

% Power Loss =180

100600

= 30%

12. In the method using the transformers, assume that the ratio of the number of turns in theprimary 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

12. (A)

Step up transformer

1 : 10

4000 V 40,000 VStep down transform

40,000 V 200

turns ratio = 40000200

= 200 : 1

30

60

mg


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A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular

velocity . This can be considered as equivalent to a loop carrying a steady currentQ

2

. 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 proportionality constant.

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)BR

4 (B)

BR

2 (C) BR (D) 2BR

13. (B)

Magnitude of induced emf is given by =A dB

2 R dt=

2RB

2 R

=

BR

2

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) BQR2 (B)2BQR

2 (C)

2BQR

2 (D)BQR2

14. (B)

fi = t = ( )2q ER

1

MR

= ( )qE

1

MR

E =BR

2

=q BR

MR 2

=

qB

2M

= L = I

= ( )2MR = 2qB

MR2M

=

2qBR

2


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 m1and m2only if (m1+ m2) < M. Also two light nuclei

of masses m3and m4can undergo complete fusion and form a heavy nucleus of mass Monlyif (m3+ m4) > M. The masses of some neutral atoms are given in the table below :

11H

1.007825 u 21 2.014102 u

31 H

3.016050 u 42 He

4.002603 u

63 Li

6.015123 u 73 Li

7.016004 u 7030 Zn

69.925325 u 8234 Se

81.916709 u

15264 Gd 151.919803 u 20682 Pb 205.974455 u 20983 Bi 208.980388 u 21084 Po 209.982876 u

(1 u = 932 MeV/c2)


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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

30 Zn and

82

34Se can undergo complete fusion.15.(C)

3 2 46 1 2Li H He +

M 6.015123= m1= 2.014102

m2= 4.002603

m1+ m2= 6.016705

m1+ m2> M (not possible)

(B) 210 209 184 83 1Po Bi P +

M = 209.982876m1= 208.980388

m2= 001.007825

m1+ m2= 209.988213

M < m1+ m2 (not possible)

(C) 2 4 61 2 3H H Li+

(m1+ m2) > MPossible

(D) 70 82 15230 34 64Zn Se Gd+

m1= 69.925325

m2= 81.916709

m1+ m2= 141.832034

M= 151.919803m1+ m2< M (not possible)

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

16.(A)210 206 484 82 2Po Pb He +

M 209.982876=

1m 205.974455=

2m 4.002603=

m1+ m2= 209.977058

1 2M (m m ) 209.982876 + =

209.977058

0.0058182

E mc (0.005818)932= = 5422= 1 2K K E+ =


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11

2 2

1 2

P PE

2m 2m+ =

2PE

2=

2 1 2

1 2

Em mP2 m m

=+

22

11 1 2

Emp E(206) 103k E

2m m m 210 105= = = =

+

1

103k (5422) 5319

105= =

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 E F G Eand E F H E as shown in the PV diagram. The processes involved are purelyisochoric, isobaric, isothermal or adiabatic.

Match the paths in List I with the magnitudes of the work done in List II and select the

correct answer using the codes given below the lists.

List I List II

P. GE 1. 160 P0V01n2

Q. GH 2. 36 P0V0R. FH 3. 24 P0V0S. FG 4. 31 P0V0

Codes :

P Q R S

(A) 4 3 2 1

(B) 4 3 1 2

(C) 3 1 2 4

(D) 1 3 2 4

17. (A)

F F G GP V P V= 0 032P V = 0 GP V


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G 0V 32V=

F F H HP V P V =

0 0 0 H32P V P V =

VH=

1

032 V

= ( )3 5

0 032 V 8V/

=

(P) ( ) ( )GE 0 E G 0 0 0W P V V P V 32V= =

= 0 031P V 4.

(Q) ( )GH 0 0 0 0 0W P 8V 32V 24P V= = (3)

(R)( )0 0 0 0H H F F

FH

P 8V 32P VP V P VW

511

3

= =

= 0 0 0 024P V 36P V 22

3

( ) =

(A) G 0FG F F 0 0F 0

V 32VW P V n 32P V n

V V= =

= 0 032P V n 32

= ( )5

0 032 P V n 2 = 0 0160 P V n 2 1( )

(P) 4; (Q) (3); (R) (2); (S) (1)

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 select the correct answer using the codes given

below the lists:

List I List II

P. Alpha decay 1. 15 158 7O N ..... +

Q. +decay 2. 238 23492 90U Th ..... +

R. Fission 3. 185 18483 82Bi Pb ..... +

S. Proton emission 4. 239 14094 57Pu La ..... +

Codes :

P Q R S

(A) 4 2 1 3

(B) 1 3 2 4

(C) 2 1 4 3

(D) 4 3 2 1

18. (C)

P Q R S

2 1 4 3


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19. A right angled prism of refractive index 1 is placed in a rectangular block of refractiveindex 2, which is surrounded by a medium of refractive index 3, as shown in the figure.A ray of light 'e' enters the rectangular block at normal incidence. Depending upon the

relationships between 1, 2and 3, 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:

List I List II

P. e f 1. 1> 2 2Q. e g 2. 2> 1and 2> 3R. e h 3. 1= 2S. e i 4. 2< 1< 2 2and 2> 3

Codes :

P Q R S

(A) 2 3 1 4

(B) 1 2 4 3

(C) 4 1 2 3

(D) 2 3 4 1

19. (D)

(P) e f for 1 2 towards normal 2> 1For 2 3 always from normal 3< 2i.e., (2)

(Q) e g no deviation any where1= 2= 3, i.e., (3)

(R) e h 1 2 away from normal.2< 12 3 away from normal. 3< 2And also at 1 2 no I.R.

sin 45 < 2

1

1 22 <

2 1 22 < < i.e., (4)

(S) e i TIR at 1 2


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sin 45 > 2

1

1 22 >

i.e., 1

(P) 2; (Q) 3; (R) (4); (S) (1).

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

lists:

List I List II

P. Boltzmann constant 1. [ML2T1]

Q. Coefficient of viscosity 2. [ML1T1]

R. Planck constant 3. [MLT3T

1]

S. Thermal conductivity 4. [ML2T2K1]

Codes :

P Q R S(A) 3 1 2 4

(B) 3 2 1 4

(C) 4 2 1 3

(D) 4 1 2 3

20.(C)

(P) u = B3

K T2

[u] = BK T[ ]

2 2

B

ML TK

K

[ ][ ]

[ ]

=

[KB] =2 2 1ML T K 4[ ]

(Q) F = 6rV

[] =2

1 1

1

F MLTML T 2

rv L LT[ ]

( )

= =

(R) E = h[h] =

2 22 1

1

E ML TML T 1

T[ ] ( )

= =

(S)Q

t = 2 1

KA T T( )

[K] =2 3

2

ML TL

L K

[ ][ ]

[ ][ ]

= 3 1MLT K 3[ ] ( )

(P) 4; (Q) (2); (R) (1); (S) (3)


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PART II : CHEMISTRY

SECTION 1 : (One or more options correct Type)This section contains 8 multiple choice questions. Each question has four choices (A), (B),


21. The carbon-based reduction method is NOT used for the extraction of(A) tin from SnO2

(B) iron from Fe2O3

(C) aluminium from Al2O3

(D) magnesium from MgCO3. CaCO3

21.(C), (D)

The oxides of the less electropositive methods are reduced by strongly heating them with

coke.

22.The thermal dissociation equilibrium of CaCO3(s) is studied under different conditions.

3 2CaCO (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 CaCO3(C) K is dependent on the pressure of CO2at a given T

(C) H is independent of the catalyst, if any22.(A), (B), (D)

K Equilibrium constant is independent of the partial pressure of CO2.

It is independent of the catalyst.

23. The correct statement(s) about O3is(are)

(A) OO bond lengths are equal(B) Thermal decomposition of O3is endothermic(C) O3is diamagnetic in nature

(D) O3has a bent structure

23.(A), (C), (D)

3 22O 3O 68 k cal + i.e. Exothermic

24. In the nuclear transmutation

9 84 4Be X Be Y+ +

(X, Y) is (are)

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

24.(A), (B)

(A) 9 8 14 4 0Be Be n+ +

(B) 9 1 8 24 1 4 1Be H Be D+ +

(C) 9 1 8 24 0 4 1Be n Be D+ + (False)

(D) 9 8 14 4 1Be Be H+ + (False)


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2||

B r (1m o l )3 3 N a O H( 1 m o l )

O

C H C C H (T ) ( U ) +

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

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

25.(B)

26. After completion of the reactions (I and II), the organic compound(s) in the reaction

mixtures is(are)

(A) Reaction I: Pand 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

26.(C)

NaOH

SO3H

OH

2aq. Br

Br

OHBrBr

(Q)

2

3

Br (1mol)

CH COOH(p)


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27. The Kspof Ag2CrO4 is 1.1 1012at 298 J. The solubility (in mol / L) of Ag2CrO4 in a

0.1 M AgNO3solution is

(A) 1.1 1011 (B) 1.1 1010 (C) 1.1 1012 (D) 1.1 10927.(B)

22 4(s) 4Ag CrO 2Ag (aq) CrO (aq)

+ +

2s + 0.1 s= 0.1

2 12spK (0.1) s 1.1 10

= =

10s 1.1 10 M=

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

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

28. (B), (D)

CCl2

..

CH3

O O

CH3

H

CCl2

O

CH3

CHCl2

OH

CH3

OH

CHO

major

2H O

CH3

OCHO

O O

HC3 CHCl2(minor)


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18

SECTION 2 : (Paragraph Type)

This section contains 4 paragraphs each describing theory, experiment, data etc. Eight

questions relate 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)..

Using the following Paragraph , Solve Q. No. 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 Pwas 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 H2O2in an aqueous NaOH medium.

29.The precipitate P contains

(A) Pb2+ (B) 22Hg+ (C) Ag+ (D) Hg2+

29.(A)2

2(p)

Pb HCl PbCl+ + soluble in hot water

30.The coloured solution S contains

(A) Fe2(SO4)3 (B) CuSO4 (C) ZnSO4 (D) Na2CrO42

2(p)

Pb HCl PbCl+ + soluble in hot water

30.(D)

3 2 4 2(S)(R )

2Cr(OH) 4NaOH 3[O] 2Na CrO 5H O+ + +


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.


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20

32.(A)

2H /Ni(Q)

3anhy.AlCl

CH2

CH2

C

O

C

O

O

2 2C CH CH COOH

O

O

3 4H PO

2 2 2CH CH CH COOH

+ (V)

(V)

(W)

Zn-Hg/conc.HCl


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 (C) Cooling , heating cooling, heating

(C) Heating, cooling, cooling, heating (D) Cooling, heating, heating, cooling

33.(C)

Factual

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

34.(B)

Factual


The reactions of Cl2 gas with cold dilute and hot concentrated NaOH in water givesodium salts of two (different) oxoacids of chlorine, P and Q respectively. The Cl2gas reacts

with SO2gas , 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.


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35.P and Q respectively, are the sodium salts of

(A) hypochlorus and chloric acids (B) hypoclorus and chlorus acids

(C) chloric and perchloric acids (D) chloric and hypochlorus acids

35.(A)Cold

2 22NaOH Cl NaCl NaOCl(P) H O+ + +

hot2 3 26NaOH 3Cl 5NaCl NaClO (Q) 3H O+ + +

36.R, S and T , respectively are

(A) SO2Cl2, PCl5and H3PO4 (B) SO2Cl2, PCl3and H3PO3

(C) SOCl2, PCl3and H3PO2 (D) SOCl2, PCl5and H3PO4

36.(A)

2 2 2 2SO Cl SO Cl (R)+

4 2 2 5 2P 10SO Cl 4PCl 10 SO+ +

5 2 3 4PCl 4H O H PO 5HCl+ +

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.2 5 3 3(C H ) CH COOH

X Y

+ 1. Conductivity decreases and then increases

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

X Y

+ 2. Conductivity decreases and then does notchange much

R.3CH C KOH

YX

+ 3. Conductivity increases and then does notchange much

S. NaOH HIYX

+ 4. Conductivity does not change much and thenincreases

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

37.(A)

Factual

38. The standard reduction potential data at 25C is given below .E(Fe3+, Fe2+) = + 0.77 V;E(Fe2+, Fe) = 0.44 V

E(Cu2+

, Cu) = + 0.34 VE(Cu+, Cu) = + 0.52 VE[O2(g) + 4H

++ 4e2H2O] = + 1.23 V;


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E[O2(g) + 2H2O + 4e 4OH] = + 0.40 V

E(Cr3+, Cr) = 0.74 V ;E(Cr2+, Cr) = 0.91 VMatch Eof the redox pair in List I with the values given in List II and select the correctanswer using the code given below the lists :

List I List II

P. E(Fe3+, Fe) 1. 0.18 VQ.

E 2(4H O 4H 4OH )+ +

2. 0.4 V

R. E(Cu2++ Cu 2Cu+) 3. 0.04 VS. E(Cr3+, Cr2+) 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

38.(D)

Use formula 0 0 03 3 1 1 2 2n F E n FE n FE= +

0 0

0 1 1 23

3

n E n EE

n

+=

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 usingthe 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

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

39.(D)warm

2 2 4 4 2 2P PbO H SO PbSO O H O + + +

2Cl2 2 3 2 4Q Na S O H O NaHSo NaCl HCl + + +

2I2 4 2R N H N 4HI + NO

2S XeF Xe 2NOF +


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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 :

List I List II

P. 1. (i) Hg(OAc)2,; (ii) NaBH4

Q. 2. NaOEt

R. 3. Et Br

S. 4. (i) BH3; (ii) H2O2/ NaOH

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

40.(A)

P = 2, Q = 3, R = 1, S = 4

PART III : MATHEMATICS

SECTION 1 : (One or more options correct Type)

This section contains 8 multiple choice questions. Each question has four choices (A), (B),


41.Let be a complex cube root of unity with 1 and P = [pij] be a n n matrix withpij=

i + j. Then P20, when n =(A) 57 (B) 55 (C) 58 (D) 56

41. (B), (C), (D)P2 = O only when n is multiple of 3.

Ex :

2 2

2 2

2 2

1 1

1 1

1 1

=

0 0 0

0 0 0

0 0 0

P20, when n = 55, 56, 58.

42.The function f(x) = 2 | x | + | x + 2 | | | x + 2 | 2 | x | | has a local minimum or a localmaximum at x =

(A)2 (B)2

3

(C) 2 (D)

2

3


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42. (A), (B)

Given, f(x) = 2|x| + |x + 2| ||x + 2| 2|x||

Sign scheme of |x + 2| 2|x| is

For2

x 23

f (x) = 4|x|

For x 2

f(x) = 2 |x + 2|

Now, f(x) = 4 |x|,2

3 x 2

= 2|x + 2|, x 2

The point of maxima or minima are 2,2

3 , 0.

43. Let3 i

w2

+= and P = {wn : n = 1, 2, 3, }. Further H1 =

1z : Re z

2

>

and

2

1H z : Re z ,

2

=

1

2

which is true for n = 0, 1, 11.

2/3 2

8/38

y

x

2/3 2

veve +ve

x

2 2

8/38

y

2/3 x

2 2

8/38

y

2/3 x


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If z2P H2, thenn

cos6

IITJEE 2014 Advanced Paper II & Solution

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