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IIT - JEE 2013 (Advanced)

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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 : 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 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.

1. (A), (B), (C), (D) dQ = mc dT

dQdT

= m (aT + b) in the range (0 − 100 k)

∴ option (A) is correct. Considering (0 − 100 k), the graph to be linear using the reasonable 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 is 3h2π

. 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) 932R

(B) 916R

(C) 95R

(D) 43R

2. (A), (C)

rn = 2

0n az

4.5a0 = 2

0n az

∴ 2n

z = 4.5

∴ 9z

= 4.5 and z = 2

L = nh2π

L = 3h2π

∴ n = 3

100 200 300 400 500

C

T (K)


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Wavelength is given by

1λ

= 22 2f i

1 1Rzn n

⎛ ⎞−⎜ ⎟⎝ ⎠

1λ

= 2 2f i

1 14Rn n

⎛ ⎞−⎜ ⎟⎝ ⎠

For transition, n = 3 → 1

λ = 114R 19

⎛ ⎞−⎜ ⎟⎝ ⎠

= 932R

…(A)

For transition n = 3 → 2

λ = 11 14R4 9

⎛ ⎞−⎜ ⎟⎝ ⎠

= 1 364R 5

××

= 95R

… (C)

3. Using the expression 2d sin θ = λ, one calculates the values of d by measuring the corresponding angles θ in the range 0 to 90º. The wavelength λ is 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.

3. (D) 2d sin θ = λ ⇒ δ (2d sin θ) = δλ But δλ = 0 ⇒ (2 sin θ) δd + 2d cos θ δθ = 0

⇒ ddδ + cot θ δθ = 0

⇒ ddδ = − cot θ (dθ)

δθ = constant absolute error in d is |δd| = d cot θ (δθ)

⇒ |δd| = 2cos | d |

2 sinλ θ θ

θ

at θ increase from 0 to 90°, |δd| 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 +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.


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4. (C), (D)

1 10

E O P3

ρ=∈

2 20

E PO3

ρ=∈

Net field at P 1 2E E E= +

1 20

[O P PO ]3

ρ= +∈

1 20

E O O3

ρ=∈

∴ 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 = NL

Bd∫ = μ0I

B1 = 0I2 Rμπ

along axis of cylinder

B2 = μ0nI 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 = 0I2 rμπ

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

Bsolenoid = μ0nI 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 > 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.

O1

E2

O2

E1P


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6. (A), (B) uSW = u − w uOW = u + w

f2 = v u wv (u w)

+ +− −

f1 ⇒ f2 = v u wv u w

+ +⎛ ⎞⎜ ⎟− +⎝ ⎠

f1

Clearly f2 > f1 irrespective 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 is GM4L

.

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

bodies is GM2L

.

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

bodies is 2GML

.

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

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

2e

2GMm 1 mvL 2

− + = 0

⇒ ve = GM2L

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 = mk

π .

(C) the time at which the maximum compression of the spring occurs is t = 4 m3 kπ .

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

time is t = 5 m3 kπ .

8. (A), (D) The impulse of the wall merely reverses the direction of motion is thout changing particle

speed. Hence (A)

S Ou u

W


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Displacement from mean position be x

x = 0uw

⎛ ⎞⎜ ⎟⎝ ⎠

sin (ωt) ⇒ x = u0 cos (ωt)

when particle speed is 0u2

⎛ ⎞⎜ ⎟⎝ ⎠

cos (ωt) = 12

⇒ ωt = 3π [for first time]

⇒ t = 3πω

Time after which it passes the mean position = 2t = 33uπ = 2 m

3 Kπ

Hence (B) is incorrect.

Time after which particle crosses mean position for second time = 23

π π+ω ω

= 53

πω

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 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 = 10 ms−2).

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

(A) 5 ms−1 (B) 10 ms−1 (C) 10 3 ms−1 (D) 20 ms−1 9. (B)

Kinetic energy of block at Q = mgR sin 30° − 150 J

⇒ 21 mv2

= 10 × 40 × 12

− 150 = 50 J

⇒ r = 2 501

× = 10 ms−1


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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 1002 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 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 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 0.4Ω km−1 is used, the power dissipation (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)2 × 0.4 × 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 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

12. (A) Step up transformer 1 : 10 4000 V → 40,000 V Step 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 ω. This can be considered as equivalent to a loop carrying a steady current Q2

ωπ

. 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 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) BR4

(B) BR2

(C) BR (D) 2BR

13. (B)

Magnitude of induced emf is given by = A dB2 R dtπ

= 2R B

2 Rππ

= BR2

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) −γBQR2 (B) 2BQR

2−γ (C)

2BQR2

γ (D) γBQR2

14. (B)

ωf − ωi = α t = ( )2q ER 1MR

− = ( )qE 1MR

−

E = BR2

Δω = q BRMR 2

⎛ ⎞− ⎜ ⎟⎝ ⎠

= qB2M

−

μ = γL = γIω

Δμ = ( )2MRγ Δω = 2 qBMR2M−⎛ ⎞γ ⎜ ⎟

⎝ ⎠ =

2qBR2

γ−


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 :

11H 1.007825 u 2

1 Η 2.014102 u 31 H 3.016050 u 4

2 He 4.002603 u 63 Li 6.015123 u 7

3 Li 7.016004 u 7030 Zn 69.925325 u 82

34 Se 81.916709 u 15264 Gd 151.919803 u 206

82 Pb 205.974455 u 20983 Bi 208.980388 u 210

84 Po 209.982876 u

(1 u = 932 MeV/c2)


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15. The correct statement is (A)The nucleus 6

3 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 8234Se can undergo complete fusion.

15. (C) 3 2 4

6 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 1

84 83 1Po Bi P⎯⎯→ + M = 209.982876 m1 = 208.980388 m2 = 001.007825

m1 + m2 = 209.988213 M < m1 + m2 (not possible) (C) 2 4 6

1 2 3H H Li+ ⎯⎯→ (m1 + m2) > M′ Possible (D) 70 82 152

30 34 64Zn Se Gd+ ⎯⎯→ m1 = 69.925325 m2 = 81.916709

m1 + m2 = 141.832034 M′ = 151.919803 m1 + 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 4

84 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

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


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2 2

1 2

P P E2m 2m

+ =

2P E

2=

μ

2

1 2

1 2

Em mP2 m m

=+

2

21

1 1 2

Emp E(206) 103k E2m m m 210 105

= = = =+

1103k (5422) 5319105

= =

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 → E

and E → F → H → E as shown in the PV diagram. The processes involved are purely isochoric, 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. G→E 1. 160 P0V0 1n2 Q. G→H 2. 36 P0V0

R. F→H 3. 24 P0 V0 S. F→G 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 Vγ γ=

0 0 0 H32P V P Vγ γ=

VH = 1

032 Vγ

= ( )3 50 032 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

FHP 8V 32P VP V P VW

51 13

−−= =− γ −

= 0 00 0

24P V 36P V 223

( )−

= →−

(A) G 0FG F F 0 0

F 0

V 32VW P V n 32P V nV V

= =

= 0 032P V n 32

= ( )50 032 P V n 2 = 0 0160P 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 15

8 7O N ..... → +

Q. β+ 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 μ1 is placed in a rectangular block of refractive index μ2, which is surrounded by a medium of refractive index μ3, as shown in the figure. A ray of light 'e' enters the rectangular block at normal incidence. Depending upon the relationships between μ1, μ2 and μ3, 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. μ1 > 2 μ2 Q. e → g 2. μ2 > μ1 and μ2 > μ3 R. e → h 3. μ1 = μ2 S. e → i 4. μ2 < μ1 < 2 μ2 and μ2 > μ3

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 ⇒ μ2 > μ1 For 2 → 3 always from normal ⇒ μ3 < μ2 i.e., (2)

(Q) e → g → no deviation any where μ1 = μ2 = μ3, i.e., (3)

(R) e → h → 1 → 2 away from normal. μ2 < μ1 2 → 3 away from normal. ⇒ μ3 < μ2 And also at 1 − 2 no I.R.

⇒ sin 45 < 2

1

μμ

⇒ 1 22μ < μ

⇒ 2 1 22μ < μ < μ → i.e., (4)

(S) e → i → TIR at 1 → 2


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⇒ sin 45 > 2

1

μμ

⇒ 1 22μ > μ

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. [ML2T−1] Q. Coefficient of viscosity 2. [ML−1T−1] R. Planck constant 3. [MLT−3T−1] S. Thermal conductivity 4. [ML2T−2K−1]

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

BML T K

K[ ]

[ ][ ]

−=

[KB] = 2 2 1ML T K 4[ ]− − →

(Q) F = 6π rη V

[η] = 2

1 11

F MLT ML T 2rv L LT

[ ]( )

−− −

−

⎡ ⎤⎡ ⎤ = = →⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

(R) E = hν

[h] = 2 2

2 11

E ML T ML T 1T

[ ] ( )−

−−

⎡ ⎤⎡ ⎤ = = →⎢ ⎥⎢ ⎥ν⎣ ⎦ ⎣ ⎦

(S) Qt

= 2 1KA T T( )−

[K] = 2 3

2ML T LL 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), (C) and (D) out of which ONE or MORE are correct. 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) ΔH is dependent on T (B) K is independent of the initial amount of CaCO3

(C) K is dependent on the pressure of CO2 at a given T (C) ΔH is independent of the catalyst, if any 22. (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 O3 is(are) (A) O−O bond lengths are equal (B) Thermal decomposition of O3 is endothermic (C) O3 is diamagnetic in nature (D) O3 has a bent structure

23. (A), (C), (D) 3 22O 3O 68 k calΔ ⎯⎯→ + i.e. Exothermic 24. In the nuclear transmutation

9 84 4Be X Be Y+ ⎯⎯→ +

(X, Y) is (are)

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

24. (A), (B) (A) 9 8 1

4 4 0Be Be n + γ ⎯⎯→ +

(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 + γ ⎯⎯→ + (False)


16

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 : 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 26. (C)

NaOH

SO3H

OH

2aq. Br ⎯⎯⎯⎯→

Br

OHBrBr

(Q)

23

Br (1mol)CH COOH (p)⎯⎯⎯⎯⎯→


17

27. The Ksp of Ag2CrO4 is 1.1 × 10−12 at 298 J. The solubility (in mol / L) of Ag2CrO4 in a 0.1 M AgNO3 solution is (A) 1.1 × 10−11 (B) 1.1 × 10−10 (C) 1.1 × 10−12 (D) 1.1 × 10−9 27. (B) 2

2 4(s) 4Ag CrO 2Ag (aq) CrO (aq)+ −+

2s + 0.1 s = 0.1 2 12

spK (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

OHCHO

major

2H O←⎯⎯⎯

CH3

OCHO

⎯⎯→

O O

HC3 CHCl2(minor)


18



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 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 H2O2 in an aqueous NaOH medium.

29. The precipitate P contains (A) Pb2+ (B) 2

2Hg + (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) Na2CrO4 2

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+ + ⎯⎯→ +

Using the following Paragraph , Solve Q. No. 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.


19

31. (B)

C

C

H

H

COOH

COOH

2H OΔ

−⎯⎯⎯→ C

C

C

C

O

H

H

O

O

(P)(cis)

cyclic anhydride

C

C

H

H

COOH

HOOC(Q)

H OHH OH

COOH

COOH

(S)(meso)

4dil. KMnOP ⎯⎯⎯⎯⎯→

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

(A) (B)

(C) (D)

4dil. KMnOQ ⎯⎯⎯⎯⎯→H OHOH H

COOH

COOH

(T)

OH HH OH

COOH

COOH

+

(U)


20

32. (A)

2H /Ni(Q) Δ ⎯⎯⎯→

3anhy. AlCl ⎯⎯⎯⎯⎯→

CH2

CH2

CO

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

Using the following Paragraph , Solve Q. No. 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 (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 Using the following Paragraph , Solve Q. No. 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.


21

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 , PCl5 and H3PO4 (B) SO2Cl2 , PCl3 and H3PO3 (C) SOCl2 , PCl3 and H3PO2 (D) SOCl2 , PCl5 and 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+ ⎯⎯→ +


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 YΝ + 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. 3CH C KOHYX

ΟΟΗ + 3. Conductivity increases and then does not change much

S. NaOH HIYX

+ 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 37. (A) Factual

38. The standard reduction potential data at 25° C is given below . E° (Fe3+ , Fe2+) = + 0.77 V; E° (Fe2+ , Fe) = − 0.44 V E° (Cu2+ , Cu) = + 0.34 V E° (Cu+ , Cu) = + 0.52 V E° [O2(g) + 4H+ + 4e− → 2H2O] = + 1.23 V;


22

E° [O2(g) + 2H2O + 4e− → 4OH−] = + 0.40 V E° (Cr3+ , Cr) = −0.74 V ; E° (Cr2+ , Cr) = −0.91 V Match E° of the redox pair in List I with the values given in List II and select the correct

answer using the code given below the lists :

List I List II P. E° (Fe3+ , Fe) 1. −0.18 V Q. E° 2(4H O 4H 4OH )+ − + 2. −0.4 V

R. E° (Cu2+ + Cu → 2Cu+) 3. −0.04 V S. 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 0

3 3 1 1 2 2n F E n FE n FE= +

⇒ 0 0

0 1 1 23

3

n E n EEn+=

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

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→ ⎯⎯→ +

NO2S XeF Xe 2NOF→ ⎯⎯⎯→ +


23

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), (C) and (D) out of which ONE or MORE are correct.

41. Let ω be a complex cube root of unity with ω ≠ 1 and P = [pij] be a n × n matrix with pij = ωi + j. Then P2 ≠ 0, 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

⎡ ⎤ ⎡ ⎤ω ω ω ω⎢ ⎥ ⎢ ⎥

ω ω ⋅ ω ω⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ω ω ω ω⎣ ⎦ ⎢ ⎥⎣ ⎦

= 0 0 00 0 00 0 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

∴ P2 ≠ 0, when n = 55, 56, 58.

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) 23

− (C) 2 (D) 23


24

42. (A), (B) Given, f(x) = 2|x| + |x + 2| − ||x + 2| − 2|x|| Sign scheme of |x + 2| − 2|x| is

For 2 x 23

− ≤ ≤

f (x) = 4|x|

For x < 23

− ∪ x > 2

f(x) = 2 |x + 2|

Now, f(x) = 4 |x|, 23

− ≤ x ≤ 2

= 2|x + 2|, x < 23

− ∪ x > 2

The point of maxima or minima are −2, 23

− , 0.

43. Let 3 iw2+ = and P = {wn : n = 1, 2, 3, … }. Further H1 = 1z : Re z

2⎧ ⎫ ∈ > ⎨ ⎬⎩ ⎭

and

21H z : Re z ,

2−⎧ ⎫ = ∈ < ⎨ ⎬

⎩ ⎭where is the set of all complex numbers. If z1 ∈ P ∩ H1,

z2 ∈ P ∩ H2 and O represents the origin, then ∠ z1 O z2 =

(A) 2π (B)

6π (C) 2

3π (D) 5

6π

43. (C), (D)

ω = 3 i2 2

+ = cos i sin6 6π π⎛ ⎞ ⎛ ⎞+ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

ωn = n ncos i sin6 6π π⎛ ⎞ ⎛ ⎞+ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(by De Moivre’s Theorem)

If z1 ∈ P ∩ H1, then ncos6π⎛ ⎞ ⎜ ⎟

⎝ ⎠ > 1

2

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

−2/3 2

8/38

y

x

−2/3 2

−ve −ve +ve

x

−2 2

8/38

y

−2/3 x

−2 2

8/38

y

−2/3 x


25

If z2 ∈ P ∩ H2, then ncos6π⎛ ⎞

⎜ ⎟⎝ ⎠

< 12− , which is true for n = 5, 6, 7.

∴ ∠ z1 oz2 can be 23π or 5 .

6π

44. If 3x = 4x−1, then x =

(A) 3

3

2 log 22log 2 1−

(B) 2

22 log 3−

(C) 4

11 log 3−

(D) 2

2

2 log 32log 3 1−

44. (A), (B), (C) 3x = 4x−1 2x log 3 = (x−1) 2log 4 x 2log 3= 2x − 2 x (2 − 2log 3 ) = 2

∴ x = 2

22 log 3−

(B)

3x = 4x−1

x 4log 3 = x − 1

⇒ x = 4

11 log 3−

(C)

3x = 4x−1 x 3log 3 = (x − 1) 3log 4 x = (x − 1) log3 4 = (x − 9) 2. log3 2

∴ x = 3

3

2.log 22.log 2 1−

(A)

45. Two lines 1 2y z y zL : x 5, and L : x ,

3 2 1 2= = = α =

− α − − − α are coplanar. Then α

can take values (s) (A)1 (B) 2 (C) 3 (D) 4 45. (A), (D)

L1 : x 50− = y

3 − α = Z

2− Line L1 passes (x1, y1, z1) ≡ (5, 0, 0)

L2 : x

0 − α = y

1− = z

2 − α Line L2 passes (x2, y2, z3) ≡ (α, 0, 0)

As Line co-planar ⇒ Shortest distance = 0

3 1,2 2

⎛ ⎞− ⎜ ⎟⎝ ⎠

3 1,2 2

⎛ ⎞ ⎜ ⎟

⎝ ⎠

3 1,2 2

⎛ ⎞− − ⎜ ⎟⎝ ⎠

3 1,2 2

⎛ ⎞− ⎜ ⎟

⎝ ⎠

(−1, 0) (1, 0)


26

⇒ 2 1 2 1 2 1

1 1 1

2 2 2

x x y y z zm nm n

− − − = 0

⇒ 5 0 0

0 3 20 1 2

α − − α −− − α

= 0 ⇒ (α − 5) (α − 4) (α − 1) = 0

α = 1, 4, 5

46. In a triangle PQR, P is the largest angle and 1cos P3

= . 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

46. (B), (D)

cos p = 2 2 2(4x 4) (4x 2) (4x 6)

2(4x 4) (4x 2) + + + − +

+ +

13

= 216x 16

16(x 1) (2x 1) −

+ +

13

= 2

2x 1

2x 3x 1 −

+ +

⇒ 3x2 − 3 = 2x2 + 3x + 1 ⇒ 3x2 − 3x − 4 = 0 x = 4 or −1. Hence x = 4 So, possible lengths of sides are 18, 22.

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

( )

( ) ( ) ( ) ( )

a a a

a 1x

1 2 .... n 1lim60n 1 na 1 na 2 ... na n−→∞

+ + +=

+ + + + + +⎡ ⎤⎣ ⎦

Then a =

(A) 5 (B) 7 (C) 152

− (D) 172

−

47. (B), (D)

Let, L = ( ) ( )( )

a a a

a 1n

1 2 ...... nlimn 1 na 1 (na 2) ... (na n)−→∞

+ + + + + + + + + +

= ( )

a a aa

n a 1 2

1 2 nn ...n n nlim

n(n 1)n 1 n a2

→∞ −

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ + + +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

+⎛ ⎞+ +⎜ ⎟⎝ ⎠

= ( )

ana

r 1

n a 1

1 rn nn n

limn 1n 1 n na

2

=

→∞ −

⎛ ⎞⎛ ⎞⋅ ⋅ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠+⎛ ⎞+ +⎜ ⎟

⎝ ⎠

∑ =

( )

1a

a0

a 1n

x dxnlim n 1n 1 na

2−→∞

++ +

∫

P

Q R

2x 2x

2x + 4

2x + 42x + 2

2x + 2


27

Using limit of a sum we get an

r 1

1 rn n =

⎛ ⎞ ⎜ ⎟⎝ ⎠

∑ = 1

a

0

x dx∫

∴ L = ( )

a

a 1n

1n a 1lim n 1n 1 na

2−→∞

+ ++ + = a 1n

11 a 1lim 1 11 a1 2 2nn

−→∞+ ⋅

⎛ ⎞ + ++⎜ ⎟⎝ ⎠

= ( )2

(a 1) 2a 1+ + Given L = 1

60

∴ (a + 1) (2a + 1) = 120 ⇒ 2a2 + 3a − 119 = 0

⇒ a = 7 or a = 172

−

∴ Answer is (B), (D)

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

48. (A), (C) Let centre, C = (3, α) and BC = 2 2 cα − = 2 7 ⇒ α2 − c = 7 …(1) Agai α2 = α2 + 9 − c ⇒ c = 9 …(2) From (1), α2 − 9 = 7 ⇒ α = ± 4 Eq is x2 + y2 − 6x ± 8y + 9 = 0



Paragraph for Question 49 and 50 Let f : [0, 1] → R (the set of all real numbers) be a function. Suppose the function f is twice differentiable, f(0) = f(1) = 0 and satisfies ( ) ( ) ( ) [ ]xf '' x 2f ' x f x e , x 0,1− + ≥ ∈ . 49. Which of the following is true for 0 < x < 1?

(A) ( )0 f x< < ∞ (B) ( )1 1f x2 2

− < < (C) ( )1 f x 14

− < < (D) ( )f x 0−∞ < <

49. (D)

C

B

e

α

A 0


28

50. If the function e−x f(x) assumes its minimum in the interval [0, 1] at 1x4

= , which of the

following is true?

(A) ( ) ( ) 1 3f ' x f x , x4 4

< < < (B) ( ) ( ) 1f ' x f x , 0 x4

> < <

(C) ( ) ( ) 1f ' x f x , 0 x4

< < < (D) ( ) ( ) 3f ' x f x , x 14

< < <

50. (C) Let g(x) = e−x f(x) ⇒ g′(x) = e−x (f′(x) − f(x))

If x = 14

is point of minima, then

e−x (f′(x) − f(x)) < 0 ∀ x ∈ 10,4

⎛ ⎞ ⎜ ⎟⎝ ⎠

⇒ f′(x) < f(x) ∀ x ∈ 10,4

⎛ ⎞ ⎜ ⎟⎝ ⎠

Paragraph for Question 51 and 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 y = 2x + a, a > 0. 51. Length of chord PQ is (A) 7a (B) 5a (C) 2a (D) 3a 51. (B)

P + T Given point T lies on y = 2x + a ⇒ a(t1 + t2) = 2at1t2 + a

t1 + t2 = −2 + 1 (as focal chord t1t2 = −1) t1 + t2 = −1

∴ ( )1 2t t− = ( )21 2 1 2t t 4t t+ − = 5

So, PQ = ( ) ( )2 22 2 2

1 2 1 2a t t 2at 2at− + −

= ( ) ( )2 22 2

1 2 1 2a t t 4 t t− + − = ( ) ( )2 2a ( 1) 5 4 5− ⋅ + = a 5 20 + = 5a

52. If chord PQ subtends an angle θ at the vertex of y2 = 4ax, then tan θ =

(A) 2 73

(B) 2 73

− (C) 2 53

(D) 2 53

−

52. (D)

Slope of OP = 121

2 atat =

1

2t

= m1

Slope of OQ = 2

2t

= m2

tanθ = 1 2

1 2

m m1 m m

−+

= 1 2

1 2

2 2t t

2 21t t

−

+ ⋅ = 2 1

1 2

t t2t t 4

−+

= 521 4− +

= 2 53

−

as ‘θ’ is 2nd quadrant.

( )( )1 2 1 2T at t , a t t +

( )21 1P at , 2at

( )22 2Q at , 2at


29

Paragraph for Question 53 and 54 Let 1 2 3S S S S= ∩ ∩ , where

{ }1S z C : z 4= ∈ < , 2z 1 3 iS z C : Im 0

1 3 i

⎧ ⎫⎡ ⎤− +⎪ ⎪= ∈ >⎨ ⎬⎢ ⎥−⎪ ⎪⎣ ⎦⎩ ⎭

and

{ }3S z C : Re z 0= ∈ > 53. Area of S =

(A) 103π (B) 20

3π (C) 16

3π (D) 32

3π

53. (B) S1 = {z ∈ |z| < 4}

S2 = z 1 3 iz c : Im 01 3 i

⎧ ⎫⎡ ⎤ − + ⎪ ⎪ ∈ > ⎨ ⎬⎢ ⎥ − ⎪ ⎪⎣ ⎦⎩ ⎭

Let z = x + iy

z 1 3 i1 3 i − +

− = (x iy) 1 3 i

1 3 i + − +

− =

( )(x 1) i y 3 (1 3 i)1 3 i (1 3 i)

⎡ ⎤ − + + + ⎣ ⎦ × − +

= { } { }(x 1) 3 (y 3) i 3(x 1) (y 3

4

− − + + − + +

z 1 3 iIm1 3 i

⎛ ⎞ − + ⎜ ⎟⎜ ⎟ − ⎝ ⎠

= 3 x y4

+

z 1 3 iIm 01 3 i

⎛ ⎞ − + > ⎜ ⎟⎜ ⎟ − ⎝ ⎠

3 x y 04

+ >

3 x + y > 0

y

y′

xx′ A

B

C

D (0, −4)

(4, 0)

(0, 4)

(−4, 0)

x O


30

S3 = {z ∈ c, Re(z) > 0} x > 0

S1 ∩ S2 ∩ S3

Area of sector is = π × (4)2 × 5 / 62π

π = 16 5

6 2π × ×

× = 20

3π

54. min 1 3i zz S − − =∈

(A) 2 32

− (B) 2 32

+ (C) 3 32

− (D) 3 32

+

54. (C) min |1 − 3i − z| z ∈ s

|1 − 3i − z| = |z − (3i − 1)| = |z − (−1 + 3i)|

PM = 3 32

− (Perpendicular distance is minimum distance)

y

xO(4, 0)

(0, 4) x2 + y2 = 42

y + 3 x = 0

(2, −2 3 )

π/3

y

xO (4, 0)

(0, 4)

(2, −2 3 )

y′

x′ M

P(1, −3)


31

Paragraph for Question 55 and 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) 82648

(B) 90648

(C) 558648

(D) 566648

55. (A) 1W 3R 2B

2W 3R 4B

3W 4R 5B

E1 = one white from each box E2 = one red from each box E3 = one black from each box

B1 B2 B3 E1, E2, E3 are mutually exclusive events P(E) = P(E1) + P(E2) + P(E3)

= 1 2 3 3 3 4 2 4 56 9 12 6 9 12 6 9 12

⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × + × × + × ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 82648

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) 116181

(B) 126181

(C) 65181

(D) 55181

56. (D)

2BPE

⎛ ⎞⎜ ⎟⎝ ⎠

=

1 13 6

1 1 1 1 1 23 5 3 6 3 11

×

⎛ ⎞ ⎛ ⎞ ⎛ ⎞× + × + ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= 55181

P(B1) = 13

P(B2) = 13

P(B3) = 13

2

E 1PB 6

⎛ ⎞ =⎜ ⎟⎝ ⎠

1

E 1PB 5

⎛ ⎞ =⎜ ⎟⎝ ⎠

3

E 2PB 11

⎛ ⎞ =⎜ ⎟⎝ ⎠


32


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, b and c is 2. Then the volume of the parallelepiped determined by vectors

( ) ( ) ( )2 a b ,3 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 a and b is 20. Then the area of the triangle with adjacent sides

determined by vectors ( ) ( )2a 3b and a b+ − is 3. 24

S

Area of a parallelogram with adjacent sides determined by vectors a and 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 57. (C) (P) Given volume of parallelepiped formed by a,b,c is equal to 2 ⇒ [ a b c ] = 2 (1) ∴ volume of parallelepiped formed by 2 ( )a b× , 3 ( )b c× and c a× is given by

v = ( ) ( )2 a b , 3 b c , c a⎡ ⎤× × ×⎣ ⎦

or v = 6 [ a b c ]2 ⇒ v = 24 ∴ p ⇒ 3

(Q) given [ a b c ] = 5 ∴ volume of parallelepiped formed by 3 ( )a b+ , b c+ , and 2 ( )c a+ is given by

v = ( ) ( )3 a b , b c, 2 c a⎡ ⎤+ + +⎣ ⎦

v = 6 a b, b c, c a⎡ ⎤+ + +⎣ ⎦

v = 12 [ a b c ] ⇒ v = 60 Q ⇒ 4


33

(R) given ( )1 a b2

× = 20

⇒ a b× = 40 (1) Area of triangle with adjacent sides ( )2a 3b+ and ( )a b− is given as

Δ = ( ) ( )1 2a 3b a b2

+ × −

Δ = ( )1 5 a b2

− × = 100

∴ R → 1

(S) given a b× = 30

∴ area of parallelogram with sides a b+ and a is given as

area = ( )a b a+ × = b a× = 30

∴ S → 2 ∴ P Q R S 3 4 1 2 Answer is (C)

58. Consider the lines 1 2x 1 y z 3 x 4 y 3 z 3L : , L :

2 1 1 1 1 2− + − + += = = =

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

58. (A)

Let L1 : x 1

2− = y

1− = z 3

1+ = t

and L2 : x 4

1− = y 3

1+ = z 3

2+ = s

∴ any point on L1 is M1 (2t + 1, −t, t−3) and any point on L2 is M2 (s + 4, s − 3, 2s − 3)


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at the point on integration 2t + 1 = s + 4 …(1)

−t = s − 3 …(2) t − 3 = 2s − 3 …(3)

Solving (1), (2) and (3) we get t = 2 and s = 1

∴ M1 ≡ M2 ≡ (5, −2, −1) Also if the plane is perpendicular to P1and P2 then

7a + b + 2c = 0 …(4) and 3a + 5b − 6c = 0 …(5) ∴ b = −3a; c = −2a hence the plane becomes,

ax − 3ay − 2az = d …(6) using M1 or M2 in (6) we get

5a + 6a + 2a = d ⇒ d = 13a ∴ the plane becomes

ax − 3ay − 2az = 13a or x − 3y − 2z = 13 (a ≠ 0) ∴ (P) a = 1 P → 3

(Q) b = −3 Q → 2 (R) c = −2 R → 4 (S) d = 13 S → 1

∴ P Q R S

3 2 4 1

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

List I List II

P ( ) ( )( ) ( )

1 221 14

2 1 1

cos tan y ysin tan y1 yy cot sin y tan sin y

− −

− −

⎛ ⎞⎛ ⎞+⎜ ⎟⎜ ⎟ +⎜ ⎟⎜ ⎟+⎜ ⎟⎝ ⎠⎝ ⎠

takes value 1. 1 52 3

Q If cos x + cos y + cos z = 0 = sin x + sin y + sin z then possible

value of x ycos2− is 2. 2

R If cos x cos 2x sin x sin 2x sec x cos x sin 2x sec x

4

cos x cos 2x4

π⎛ ⎞− + =⎜ ⎟⎝ ⎠π⎛ ⎞+ +⎜ ⎟

⎝ ⎠

then possible value of sec x is

3. 12

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

then possible value of x is 4. 1


35

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

59. (B) (P) Now,

( ) ( )( ) ( )

1/21 1

42 1 1

cos tan y y sin tan y1 yy cot sin y tan sin y

− −

− −

⎛ ⎞⎛ ⎞+ ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎝ ⎠⎝ ⎠

=

1/221 1

2 24

2 2 112

1 ycos cos y sin sin1 y 1 y1 y

y y1 y tancot cot tany 1 y

− −

−−

⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟+ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟+ + ⎝ ⎠ ⎝ ⎠⎜ ⎟+⎜ ⎟⎜ ⎟⎛ ⎞ ⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟+ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟−⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠

=

1/222

2 24

2 2

2

1 y

1 y 1 y1 yy 1 y y

y 1 y

⎛ ⎞⎛ ⎞⎜ ⎟+⎜ ⎟⎜ ⎟+ +⎜ ⎟ +⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟+⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠

= 1

∴ P → 4

(Q) From the given equations we have cos z = − (cos x + cos y) …(1) sin z = −(sin x + sin y) …(2)

Squaring and adding (1) and (2) we get 1 = cos2x + cos2y + 2 cos x cos y + sin2x + sin2y + 2 sin x sin y

⇒ 2 + 2 cos (x − y) = 1

⇒ cos (x − y) = 12

−

⇒ 2 x y2 cos 12−⎛ ⎞ −⎜ ⎟

⎝ ⎠ = 1

2−

⇒ x ycos2−⎛ ⎞

⎜ ⎟⎝ ⎠

= 12

Q → 3

(R) The given equation can be simplified as

2sin sin x cos 2x4π = sin 2x sec x (cos x sin x) −

⇒ 2 sin x cos 2x = sin 2x sec x (cos x sin x) −

⇒ 1 sin 2x cos 2x2

= ( )sin 2x cos x sin x −


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⇒ ( )cos 2xsin 2x cos x sin x2 ⎛ ⎞ − − ⎜ ⎟

⎝ ⎠ = 0

⇒ ( )2 2cos x sin xsin x cos x sin x

2⎛ ⎞ − 2 − − ⎜ ⎟⎝ ⎠

= 0

⇒ cos x sin xsin 2x (cos x sin x) 12

+ ⎛ ⎞ − − ⎜ ⎟⎝ ⎠

= 0

⇒ ( )sin 2x cos x sin x sin x 14

⎛ π ⎞⎛ ⎞ − + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ = 0

⇒ x = 4π ∴ sec x = 2

R → 2 (S) Now,

( )1 2cot sin 1 x− − = 12

xcot cot1 x

−⎛ ⎞ ⎜ ⎟−⎝ ⎠

∴ ( )1 2cot sin 1 x− − = 2

x

1 x − …(1)

and ( )( )1sin tan x 6− = 12

x 6sin sin1 6x

−⎛ ⎞⎜ ⎟⎜ ⎟+⎝ ⎠

∴ ( )( )1sin tan x 6− = 2

x 6

1 6x+ …(2)

From (1) and (2) we get

2

x

1 x− =

2

x 6

1 6x + (∵ x ≠ 0)

Solving this we x2 = 512

or x = 1 52 3

S → 1

∴ P Q R S

4 3 2 1 60. A line L : y = mx + 3 meets y − axis at E (0, 3) and the arc of the parabola

y2 = 16x, 0 ≤ y ≤ 6 at the point F(x0, y0). The tangent to the parabola at F(x0, y0) 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 and List II and select the correct answer using the code given below the lists :


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List I List II

P m = 1. 12

Q Maximum area of ΔEFG is 2. 4 R y0 = 3. 2 S y1 = 4. 1

Codes : P Q R S (A) 4 1 2 3

(B) 3 4 1 2

(C) 1 3 2 4

(D) 1 3 4 2 60. (A) Tangent at P(x0, y0) to y2 = 16x is yy0 = 2 × 4(x + x0)

S0, G = 0

0

8x0,y

⎛ ⎞⎜ ⎟⎝ ⎠

As F (x0, y0) lies on y2 = 16x

⇒ 20y = 16x0 … (1)

Δ = 12

× EG × x0

= 2

0 0

0

8x y1 32 y 16

⎛ ⎞× − ×⎜ ⎟⎝ ⎠

= 00 0

y1 [3y 8x ]2 16

− ×

= 2 30 0

1 [6y y ]64

− … (2)

F(x0, y0) = (1, 4)(0, y1) = 0

0

8x0,y

⎛ ⎞⎜ ⎟⎝ ⎠

E

G

y2 = 16x


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Δ′ = 20 0

1 [12y 3y ]64

−

Δ′ = 0 ⇒ 3y0 (4 − y0) = 0

y0 = 0 or y0 = 4

0y 4=

′′Δ = 01 [12 6y ]64

− < 0

⇒ local maximum when y0 = 4

So, put in (2) y0 = 4,

Δmax = 2 31 [6 4 4 ]64

× − = 1 [6 4]4

− = 12

From (1), 20y = 16x0

⇒ 16 = 16x0

⇒ x0 = 1

y1 = 0

0

8xy

= 8.14

= 2

Line GF, y 3x− = 4 3

1 0−−

⇒ y = x + 3

So, slope m = 1

Hence, (A) option correct.


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