: Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi … · 2019. 4. 23. · (1) Answers & Solutions for JEE (Advanced)-2013 Time : 3 hrs. Max. Marks: 180 DATE : 02/06/2013

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Answers & Solutionsforforforforfor

JEE (Advanced)-2013

Time : 3 hrs. Max. Marks: 180

DATE : 02/06/2013 CODE

8

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075Ph.: 011-47623456 Fax : 011-47623472

INSTRUCTIONS

Question Paper Format

The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part

consists of three sections.

Section 1 contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and

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

Section 2 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).

Section 3 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.

Marking Scheme

For each question in Section 1, you will be awarded 3 marks if you darken all the bubble(s)

corresponding to only the correct answer(s) and zero mark if no bubbles are darkened. In all other

cases, minus one (-1) mark will be awarded.

For each question in Section 2 and 3, you will be awarded 3 marks if you darken the bubble

corresponding to only the correct answer and zero mark if no bubbles are darkened. In all other

cases, minus one (-1) mark will be awarded.

PAPER - 2 (Code - 8)

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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 ofwhich ONE or MORE are correct.

1. A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placedcoaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steadycurrent 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

Answer (A, D)

Hint : Assuming BC for cylinder and BS for solenoid

I

R

I(A) R > r > 0

BS = μ0ni > 0 (Correct)

(B) 2R > r > R

= +2 2S CB B B not along the axis of cylinder, hence, wrong.

(C) Wrong, not in the plane of circle.

(D) r > 2R, B = BC

2. 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 observerin the other vehicle hears the frequency of the whistle to be f2. The speed of sound in still air is V. The correctstatement(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

Answer (A, B)

Hint : (A)− +

= >− −2 1 2 1( )V w uf f f f

V w u ∴ Correct.

u u

f1

(B) + += >

+ −2 1 2 1( )V w uf f f fV w u

∴ Correct.

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

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

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

(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 constantAnswer (B)

Hint :m

M 2L M

− × + =210 2 02

GMm mvL

⇒ =2 4GMvL

⇒ =4GMv

L

⇒ = 2 GMvL

4. A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a frictionlesshorizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from itsequilibrium position at time t = 0 with an initial velocity u0. When the speed of the particle is 0.5 u0, it collideselastically 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 = πmtk

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

43

mtk

(D) The time at which the particle passes through the equilibrium position for the second time is π=

53

mtk

Answer (A, D)

Hint :

u0 0.5u0

(A) is correct due to conservation of energy.

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(B) x = A sin ωt

= ω ω = = ω00cos cos

2udx A t u t

dt

⇒ ω =1cos2

t

⇒ π

ω =3

t

⇒ π=

ω3t

π πΔ = =

ω2 23 3

ktm , its wrong.

(C) Maximum compression will occur at time = π π+

23 2

k km m

= π76

km , hence, its wrong.

(D) Second time at equilibrium at = π+ π

23

k km m

= π53

mk , so correct.

5. 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.

C

100 200 300 400 500T k( )

(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.

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Answer (B, C, D)Hint : dQ = mCdT

=dQ mCdt

For 0 < T < 100 graph is not linear

Δ = ∫Q m CdT

= m area under C-T graphFor 400-500 K area is more than for 0-100 K

dQdt increases for 200 < T < 300 K as C increases in this region.

6. 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 π

32h

. 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

Answer (A, C)

Hint : × =π π

32 2h hn

⇒ n = 3

4.5a0 = 2

0naZ

⇒ Z = 2Possible transitions are 3 → 2, 3 → 1 and 2 → 1

For 3 → 2 : ⎡ ⎤= −⎢ ⎥λ ⎣ ⎦21 1 1(2)

4 9R ⎡ ⎤= =⎢ ⎥⎣ ⎦

9 – 4 5436 9

RR

⇒ λ =9

5R

3 → 1 : ⎡ ⎤= −⎢ ⎥λ ⎣ ⎦21 1(2) 1

9R ⎛ ⎞= =⎜ ⎟

⎝ ⎠

8 3249 9

RR

⇒ λ =9

32R

2 → 1 : ⎡ ⎤= −⎢ ⎥λ ⎣ ⎦21 1(2) 1

4R ⎛ ⎞= =⎜ ⎟

⎝ ⎠

34 34

R R

⇒ λ =1

3R

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

Answer (D)

Hint :λ

=θ2sin

d

λ= θ θ θ| ( )| |cosec cot |

2d d d

Absolute error in d decreases with increase in θ as cosec θ and cot θ decrease with increase in θ.

= θ θ( ) |cot |d d dd

Fractional error decreases with increase in θ.8. 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 overlappingregion,

ρ –ρ

R1 R2

(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

Answer (C, D)

Hint :c1 c2

r1 r2

E1

a

ρ ρ= =

ε ε1 1 2 20 0

–3 3

E r E r

ρ=

ε03E a

Electric field of constant magnitude and direction.

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SECTION - 2 : (Paragraph Type)This section contains 4 Paragraphs each describing theory, experiment, data etc. Eight questions relate to fourparagraphs with tow questions on each paragraph. Each question of a paragraph has only one correct answeramong the four choices (A), (B), (C) and (D).

Paragraph for Questions 9 and 10

A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 kmaway from the power plant for consumers' usage. It can be transported either directly with a cable of large currentcarrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawbackof the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is muchsmaller. In this method , a step-up transformer is used at the plant side so that the current is reduced to a smallervalue. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specifiedlower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are idealwith power factor unity. All the currents and voltages mentioned are rms values.

9. If the direct transmission method with a cable of resistance 0.4 Ω km–1 is used, the power dissipation (in %)during transmission is

(A) 20 (B) 30

(C) 40 (D) 50

Answer (B)

Hint : Total resistance = 0.4 × 20 = 8 Ω

P = Vi

600 × 103 = 4000 × i

= =600 150 A4

i

∴ P = i2R

= (150)2 × 8

= 22500 × 8

= 180000 watt

= 180 kW

= × =180% loss 100 30%600

10. In the method using the transformers, assume that the ratio of the number of turns in the primary to thatin the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformeris

(A) 200 : 1 (B) 150 : 1

(C) 100 : 1 (D) 50 : 1

Answer (A)

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Hint : =S S

P P

V NV N

⇒ = ⇒ =10 40,000 volt

4000 1S

SV V

=P P

S S

N VN V

=40,000

200= 200 : 1

Paragraphs for Questions 11 and 12A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius40 m. The block slides along the track without toppling and a frictional force acts on it in the direction oppositeto the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figurebelow, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)

y

x

Q

RP

R

O

30°

11. The speed of the block when it reaches the point Q is(A) 5 ms–1 (B) 10 ms–1

(C) –110 3 ms (D) 20 ms–1

Answer (B)Hint : Applying mechanical energy theorem

–150 = –1 × 10 × + × × 240 1 12 2

V

⇒ V2 = 100⇒ V = 10 m/s

12. 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

Answer (A)

Hint : 30°

N

60°5 N

f

×− =

1 100540

N

⇒ N = 7.5 N

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

11 H 1.007825 u 2

1 H 2.014102 u 31 H 3.016050 u 4

2 He 4.002603 u

63 Li 6.015123 u 7

3Li 7.016004 u 7030Zn 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)13. 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 34Zn and Se can undergo complete fusion

Answer (C)

Hint : ( ) ( )+ = +2 41 2H He 2.014102 4.002603m m

= 6.016705 u

( ) =63 Li 6.015123 um

m1 + m2 > M(A is wrong)

( ) ( )+ = +1 2091 83H Bi 1.007825 208.980388m m

= 209.988213 u

( ) =21084 Po 209.982876m u

m1 + m2 > M(B is wrong)

( ) ( )+ = +2 41 2H He 2.014102 4.002603m m

= 6.016705 u

=63 Li 6.015123 u

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(m3 + m4) > M ′

Correct (C), Deutron and alpha particle can go complete fusion.

( ) + = +70 8230 34Zn Se 69.925325 81.916709m

= 151.842034 u

=15264 Gd 151.919803u

m3 + m4 < M ′

(D is wrong)

14. 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

Answer (A)

Hint : = + + Δ210 206 484 82 2Po Pb He E

( ) =20682 Pb 205.974455 um

( ) =42 He 4.002603um

( ) ( )+ =206 482 2Pb He 209.977058um m

Δm = 209.977058 – 209.982876

= 0.005818 u

ΔE = 0.005818 × 931.5

= 5.419467 MeV = 5419.467 KeV

= 5419.5 keV

ΔE > KEα

Paragraph for Questions 15 and 16A 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 .2Qωπ

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 radiusof the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emfis defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. Itis known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum witha proportionality constant γ.15. 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) 4BR (B) 2

BR

(C) BR (D) 2BR

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Answer (B)

Hint : . d dBE dl Adt dtφ

= − = −∫ givendB Bdt

⎛ ⎞=⎜ ⎟⎝ ⎠∵

2.2R BE

Rπ

=π

⇒ 2BRE =

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

(A) 2BQR−γ (B)2

2BQR

−γ

(C)2

2BQR

γ (D) γBQR2

Answer (B)Hint : B along (+ Z axis)

M = γL = γ mωR2

(where m = mass of charged particle) R

QEind

B

= γ ω 2.mM R

M will change due to change in ω. Change in ω is given by 2Bdmθ

ω =

Change in M = −γ

γ =2

2.m .2 2BQ BQRRm

Negative sign shows change is opposite to direction of B.

SECTION - 3 : (Matching List Type)This section contains 4 multiple choice questions. Each question has matching lists. The codes for the listshave choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

17. Match List I of the nuclear processes with List II containing parent nucleus and one of the end products ofeach 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. 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

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Answer (C)Hint : In alpha decay, charge number decreases by 2 and mass number decreases by 4.

In B+ decay, charge number decreases by 1 and mass number remains same.In proton emission, charge no. decreases by 1 and mass no. decreases by 1.

18. 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.

FP

32P0

P0E H

V0

G

V

Match the paths in List I with the magnitudes of the work done in List II and select the correct answer usingthe codes given below the lists.

List I List IIP. G → E 1. 160 P0V0 ln2Q. G → H 2. 36 P0V0

R. F → H 3. 24 P0V0

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

Answer (A)Hint : FG is Isothermal

FP

32P0

P0E H

V0

G

32T0

8T0

8V0

32T0

32V0

V

P0V RT0 0=FH is Adiabatic

5 53 3

0 0 032GP V P V=⎛ ⎞γ = + =⎜ ⎟⎝ ⎠

mono2 51

3f

⇒ ( )35

0 032 8GV V V= =

( )0 0 0 0 032 31GEW P V V P VΔ = − = −

4P →

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

3Q →

(13)

( )0 0 0 08 32513

FH

P V P VW

−Δ =

−

0 00 0

24 3623

P V P V−= =

−

2R →

032 ln32FGW RTΔ =

5032 ln2RT=

0160 ln 2RT=

1S →

19. 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–3K–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

Answer (C)

Hint : [k] = [ML2T–2K–1]

(P) → (4)

[n] = [ML–1T–2] [T]

[n] = [ML–1T–1]

(Q) → (2)

[h] = [ML2T–2] [T]

= [ML2T–1]

(R) → (1)

[k] = −2 3[ML T ]

[L][K]

= [ML1T–3K–1]

(S) → (3)

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20. A right angled prism of refractive index μ1 is placed in a rectangular block of refractive index μ2, which issurrounded by a medium of refractive index μ3, as shown in the figure. A ray of light e enters the rectangularblock at normal incidence. Depending upon the relationships between μ1, μ2 and μ3, it takes one of the fourpossible paths 'ef ', 'eg', 'eh' or 'ei'.

e

ih

g

f

45°

μ1

μ2 μ3

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

List I List II

P. e → f 1. μ > μ1 22

Q. e → g 2. μ > μ2 1 and μ > μ2 3

R. e → h 3. μ = μ1 2

S. e → i 4. μ < μ < μ2 1 22 and μ > μ2 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

Answer (D)

Hint : For path e-f, μ2 > μ1; μ3 < μ2.

(P) → (2)

For path e-g μ1 = μ2 (No bending)

(Q) → (3)

For e-h, μ2 < μ1, μ3 < μ2. Also μ1 < 2 μ2 (No Total Internal Reflection)

(R) → (4)

For e-i, Total Internal Reflection occurs. i.e., μ° >

μ2

1sin45 ⇒ μ > μ1 22

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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 ofwhich ONE or MORE are correct

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

OH

SO H3

aqueous Br (3.0 equivalents)2 ?

OH OH OH OH

SO H3 Br Br SO H3

Br Br

Br BrBr

Br Br

Br Br

P Q R S

Br

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

Answer (B)

Hint :

OH

SO H3

aqueousBr (3.0 equivalents)2

OH

Br

BrBr

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

Reaction I : H C3

O

CH3(1.0 mol)

Br (1.0 mol)2

aqueous NaOH

Reaction II : H C3

O

CH3(1.0 mol)

Br (1.0 mol)2

CH COOH3

H C3

O

CH Br2 H C3

O

CBr3 Br C3

O

CBr3 BrH C2

O

CH Br2 H C3

O

ONa CHBr3

P Q R S T U(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

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Answer (C)

Hint : Reaction I :CH3

O

CH3(1.0 mol)

Br (1.0 mol)2

aqueous NaOH CHBr + CH COONa3 3C

Reaction II :Br (1.0 mol)2

CH COOH3CH3

O

CH3(1.0 mol)

CCH3

O

CH2BrC

23. The Ksp of Ag2CrO4 is 1.1 × 10–12 at 298 K. The solubility (in mol/L) of Ag2CrO4 in a 0.1 M AgNO3 solutionis(A) 1.1 × 10–11 (B) 1.1 × 10–10

(C) 1.1 × 10–12 (D) 1.1 × 10–9

Answer (B)

Hint : The solubility of Ag2CrO4 = ⎡ ⎤⎣ ⎦

–24CrO

∴ −

+

×⎡ ⎤ = = = ×⎣ ⎦ ⎡ ⎤⎣ ⎦

–12sp–2 –10 1

4 2 2K 1.1 10CrO 1.1 10 mol L

(0.1)Ag

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

OH

CH3

CHCl3

OH– ?

OH O OH OH

CH3 H C3 CH3

CHOOHC

P Q R S

CHO

CHCl2 H C3 CHCl2

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

Answer (B, D)

Hint :

OH

CH3

CHCl3

OH–

OH

CH3

CHO+

O

H3C CHCl2

(S)(Major product)

(Q)(Minor product)

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25. 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

Answer (C, D)

Hint : Al2O3 and MgCO3.CaCO3 are separately reduced by electrolytic reduction.

26. 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

(D) ΔH is independent of catalyst, if any

Answer (A, B, D)

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

=2COK P

K is only dependent on temperature and it is independent of the amount of reactant or product.

ΔH is dependent on temperature according to Kirchoff's equation but independent of addition of catalyst.

27. 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

Answer (A, C, D)

Hint :O

O O

⊕O

OO

⊕

Ozone is diamagnetic in nature and both the O – O bond length are equal. It has a bent structure.

28. 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)

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Answer (A, B)

Hint : + γ → +9 8 14 4 0Be Be n

+ → +9 1 8 24 1 4 1Be H Be H



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

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

COOH

OHH

OHH

COOHS

COOH

OHH

HHO

COOHT

COOH

OHHO H

H

COOHU

29. 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

Answer (B)

Hint :

COOH

OHH

OHH

COOH(S)

H COOH

C

C

H COOH

(P)

dil. alk. KMnO4

H COOH

C

C

HHOOC(Q)

COOH

OHH

HHO

COOH(T)

dil. alk. KMnO4

+

COOH

OHHO H

H

COOH(U)

(19)

30. In the following reaction sequences V and W are respectively

Δ⎯⎯⎯→2H /NiQ V

+ VAlCl (anhydrous)3 1. Zn-Hg/HCl

2. H PO3 4

W

(A) O and

O

OVOW

(B)CH OH2

CH OH2

V

and

W

(C) O and

O

OVW

(D) and

W

HOH C2

CH OH2V CH OH2

Answer (A)

Hint :

H /2 Ni

O

O

O(V)

H COOH

C

C

HHOOC(Q)

Δ

O

O

O(V)

+ AlCl3

anhydrous

HO

O

O

Zn(Hg)

HCl

HO

O

H PO3 3

O

(W)


An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) anda filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, whentreated with H2S in a dilute mineral acid medium. However, it gave a precipitate (R) with H2S in an ammoniacalmedium. The precipitate R gave a coloured solution (S), when treated with H2O2 in an aqueous NaOH medium.

31. The precipitate P contains

(A) Pb2+ (B) Hg22+

(C) Ag+ (D) Hg2+

(20)

Answer (A)

Hint : Lead salts give white precipitate of PbCl2 with dil. HCl which is soluble in hot water.

32. The coloured solution S contains

(A) Fe2(SO4)3 (B) CuSO4

(C) ZnSO4 (D) Na2CrO4

Answer (D)

Hint : The filtrate on treatment with ammonical H2S gives a precipitate which dissolves in aqueous NaOHcontaining H2O2 giving a coloured solution. It contains Cr3+ ion.

+ ⊕+ → ↓ +34 3 4

(Green)Cr 3NH OH Cr(OH) 3NH

2Cr(OH)3 + 3H2O2 + 4NaOH → 2Na2CrO4 + 8H2O(yellow colour)


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 productR. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus, T.

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

(A) Hypochlorus and chloric acids (B) Hypochlorus and chlorus acids

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

Answer (A)

Hint : + ⎯⎯⎯→ + +cold2 2(P)

Cl 2NaOH(dil.) NaOCl NaCl H O

+ ⎯⎯⎯→ + +hot2 3 2

(Q)3Cl 6NaOH(conc.) NaClO 5NaCl 3H O

(P) and (Q) are salts of HOCl and HClO3 respectively.

34. 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

Answer (A)

Hint : + ⎯⎯⎯⎯→Charcoal2 2 2 2Catalyst

(R)SO Cl SO Cl

+ → +2 2 4 5 2(R) (S)

10SO Cl P 4PCl 10SO

+ → +5 2 3 4(S) (T)

PCl 4H O H PO 5HCl

(21)

Paragraph for Questions 35 and 36A 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 shownin the figure

N M

K L

Volume

Pressure

35. 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

Answer (C)

36. 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

Answer (B)

Hint :K L

N M

Volume

Pre

ssur

e

As PV product increases temperature increases as PV product decreases temperature decreases.PV product increases (K → L, N → K)PV product decreases (M → N, L → M)Isochoric process are (L → M, N → K)


37. The standard reduction potential data at 25ºC is given below.Eº(Fe3+, Fe2+) = + 0.77 V;E°(Fe2+, Fe) = – 0.44 VEº(Cu2+, Cu) = + 0.34 V;Eº(Cu+, Cu) = + 0.52 VEº[O2(g) + 4H+ + 4e– → 2H2O] = + 1.23 V;Eº[O2(g) + 2 H2O + 4e– → 4OH–] = + 0.40 V

(22)

Eº(Cr3+, Cr) = – 0.74 V;

Eº(Cr2+, Cr) = – 0.91 V

Match Eº of the redox pair in List I with the values given in List II and select the correct answer using thecode given below the lists :

List I List II

P. Eº(Fe3+, Fe) 1. –0.18 V

Q. Eº(4H2O 4H+ + 4OH–) 2. –0.4 V

R. Eº(Cu2+ + Cu → 2Cu+) 3. –0.04 V

S. Eº(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

Answer (D)

Hint : (P) Fe + e+3

Fe+2 ΔºG = – 1F × 0.771

Fe + 2e+2

Fe ΔºG = + 2F × 0.772

Fe + 3e+3

Fe Δº ºG = – 3F × E3 Fe /Fe+3

Δ Δ Δº º ºG = G + G3 1 2

– 3F × EFe /Fe +3º = – 0.77 F + 0.88 F

– 3EFe /Fe +3º = 0.11 (V)

EFe /Fe +3º = –

0.11 (V)

3= 0.036 (V)–

(Q) 2H O2 O + 4H + 4e2

+E = – 1.23 Vº

4e + O + 2H O2 2 OH–

E = + 0.40 Vº

4 H O2 4H+ –

+ 4OH E = – 0.83 Vº

(R) Cu + 2e2+

Cu E = + 0.34 Vº

2Cu Cu2+

+ 2eu E = – 0.52 Vº

Cu + Cu2+

2Cu+

Eº = – 0.18 V

(23)

(S) Cr + 3e+3

Cr ΔºG = + 3 × F × 0.741

Cr Cr+2

+ 2e ΔºG = + 2 × F ×0.912

Cr + e+3

Cr+2

E = – 0.4 VºCr /Cr+3 +2

38. An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List I. The variation inconductivity of these reactions is given 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. (C2H5)3N + CH3COOH 1. Conductivity decreases and then increases

X Y

Q. KI (0.1M) + AgNO3 (0.01M) 2. Conductivity decreases and then does not

X Y change much

R. CH3COOH + KOH 3. Conductivity increases and then does not

X Y change much

S. NaOH + HI 4. Conductivity does not change much and

X Y 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

Answer (A)

Hint : (P)

Con

du

ctiv

ity

(C H ) N2 5 3

Firstly it decreases due to nutrilization of CH3COOH and replacement of H+ by (C H ) NH2 5 3+ but thereafter buffer

formation takes place and [H+] becomes constant and (C H ) NH2 5 3+ increases hence conductivity increases but

after equivalence point (C2H5)3N is not ionized due to much higher concentration of (C H ) N2 5 3+ in solution.

(24)

(Q)C

ond

uct

ivit

y

VKI

+ ⎯⎯→ +3 3AgNO KI AgI(S) KNO

Initially only Ag+ is replaced by K+ hence conductivity remain the same thereafter equivalence point [K+]increases hence conducitity increases.

(R)

Con

du

ctiv

ity

VCH COOH3

Initially conducitity decreases due to replacement of OH– by CH3COO– and then almost constant due tobuffer formation

(S)

Con

du

ctiv

ity

VNaOH

Decreases due to removal of H+ by Na+ then increases due to OH–.39. 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. Cl 1. (i) Hg(OAc)2; (ii) NaBH4

Q. ONa OEt 2. NaOEt

R.OH

3. Et-Br

S.

OH

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

(25)

Answer (A)

Hint : (P) ClNaOEt

(Q) ONa + Et-Br OEt

(R) (1) Hg(OAc)2

(2) NaBH4

OH

(S)(1) BH3

(2) H O /NaOH2 2

OH

40. The unbalanced chemical reactions given in List I show missing reagent or condition (?) which are providedin 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 2?PbO + H SO PbSO + O + other product⎯⎯→ 1. NO

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

R. 2 4 2?N H N other product⎯⎯→ + 3. Warm

S. 2?XeF 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

Answer (D)

Hint : (P) + ⎯⎯⎯⎯→ + +Warm2 2 4 4 2 22PbO 2H SO 2PbSO 2H O O

(Q) + + ⎯⎯→ +2 2 3 2 2 4Na S O 5H O 4Cl 2NaHSO 8HCl

(R) + ⎯⎯→ +2 4 2 2N H 2I N 4HI

(S) + ⎯⎯→ +2XeF 2NO Xe 2NOF

(26)

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 ofwhich ONE or MORE are correct

41. Two lines L1 : x = 5, 3 2y z

=− α −

and L2 : x = α, 1 2y z

=− − α

are coplanar. Then α can take value(s)

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

Answer (A, D)Hint :

−= =

− α −15:

0 3 2x y zL

− α= =− − α2 :

0 1 2x y zL

As the two lines are coplanar so− α

− α −− − α

5 0 00 3 20 1 2

= 0

( ) ( ) ( )⎡ ⎤− α α − α − − =⎣ ⎦5 2 3 2 0

( ) ⎡ ⎤− α α − α + − =⎣ ⎦25 5 6 2 0

( ) 25 5 4 0⎡ ⎤− α α − α + =⎣ ⎦

5, 4,1α =

42. In a triangle PQR, P is the largest angle and cos P = 13 . 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 consecutiveeven integers. Then possible length(s) of the side(s) of the triangle is (are)(A) 16 (B) 18(C) 24 (D) 22

Answer (B, D)Hint : QR is largest side

PM = PN = 2kRM = RL = 2k + 4QL = QN = 2k + 2

L 2 + 2k

2 + 2k

Q

P

R

M

N2 + 4k

2 + 4k

2k

2k

∴ QR = 4k + 6RP = 4k + 4QP = 4k + 2

cosP = ( ) ( ) ( )

( ) ( )

2 2 2

2PQ PR QR

PQ PR+ −

(27)

13 =

( ) ( ) ( )( ) ( )

2 2 24 2 4 4 4 62 4 2 4 4

k k kk k

+ + + − +

+ +

( ) ( ) ( )( ) ( )

2 2 22 1 4 1 2 313 4 2 1 1

k k kk k

+ + + − +=

+ +

13 =

( ) ( ) ( )( ) ( )

24 1 4 4 24 2 1 1

k kk k

+ − +

+ +

13 =

( ) ( )( ) ( )

21 – 2 11 2 1

k kk k+ +

+ +

(k + 1) (2k + 1) = 3(k – 1) (k + 1)k = – 1 or 2k + 1 = 3k – 34 = kSo,PQ = 18QR = 22RP = 20

43. For a ∈ (the set of all real numbers), a ≠ –1, 1(1 2 ... ) 1lim

60( 1) [( 1) ( 2) ( )]

a a a

an

nn na na na n−→∞

+ + +=

+ + + + +…+ +.

Then a =(A) 5 (B) 7

(C) 152

− (D) 172

−

Answer (B, D)

Hint :

( )

( ) ( )

1

1

1

1Lt60

1

n a

rnn a

r

r

n na r

=

→∞ −

=

=⎡ ⎤

+ +⎢ ⎥⎢ ⎥⎣ ⎦

∑

∑

( ) 1

1Lt601

aa

n a

rnn

rn n an

→∞ −

⎛ ⎞⎜ ⎟⎝ ⎠ =

⎛ ⎞+ +⎜ ⎟⎝ ⎠

∑

∑

= 1

1

1

1 1 1Lt6011

an

ra nn

r

rn

n rann

=−→∞

=

⎛ ⎞⎜ ⎟⎝ ⎠

=⎛ ⎞⎛ ⎞ ++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

∑

∑

=

1

01

0

160

ax

a x=

+

∫

∫

(28)

110

12

0

160

( 1)2

ax

xa ax

+

=⎡ ⎤⎛ ⎞⎢ ⎥+ +⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

1 01 111 602

a a

⎡ ⎤⎢ ⎥−

=⎢ ⎥+ ⎢ ⎥+

⎢ ⎥⎣ ⎦

( ) ( )2 1

2 1 1 60a a=

+ +

2a2 + 3a + 1 = 1202a2 + 3a – 119 = 0

( )± +=

–3 9 8 1194

a

= ±–3 9614

= –3 314±

177, –2

a =

44. Circle(s) touching x – axis at a distance 3 from the origin and having an intercept of length 2 7 on y – axisis (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

Answer (A, C)Hint : Clearly centre is (3, α) and radius is |α|

(3, )α

(3, 0)

So, 2 2 6 2 0x y x y c+ − − α + =

Radius 2 29 c+ α − = α

c = 9

Intercept on y axis, 22 2 7cα − =

2 9 7α − =

4α = ±

So equation becomes 2 2 6 8 9 0x y x y+ − ± + =

(29)

45. Let ω be a complex cube root of unity with ω ≠ 1 and P = [pij] be a n × n matrix with pij = ωi+j. ThenP2 ≠ 0, when n =(A) 57 (B) 55(C) 58 (D) 56

Answer (B, C, D)

Hint : ij n nP P

×⎡ ⎤= ⎣ ⎦

2 0P ≠

ˆ ˆi jijP += ω

2 2

2

2 2

1 1 ...1 1 ...

1 ...... ... ... ... ... ... n n

P

×

⎡ ⎤ω ω ω⎢ ⎥

ω ω ω⎢ ⎥= ⎢ ⎥ω ω ω ω⎢ ⎥

⎢ ⎥⎣ ⎦

2 2 2 2

2 2 2

2 2 2 2

1 1 ... 1 1 ...1 1 ... 1 1 ...

1 ... 1 ...... ... ... ... ... ... ... ...... ... ... ...

P

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ω ω ω ω ω ω⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= ω ω ω ω ω ω⎢ ⎥ ⎢ ⎥ω ω ω ωω ω ω ω⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

4 2 4 2( 1 ) ( 1 ) ....... 0ω + +ω + ω + +ω + =Only when n is a multiple of 3.

∴ n can be 55, 58, 56 ( 2 0P ≠ )46. The function f(x) = 2|x| + |x + 2| – | |x + 2| – 2 |x| | has a local minimum or a local maximum at x =

(A) – 2 (B)2

3−

(C) 2 (D)23

Answer (A, B)Hint :

x < –2, f(x) = –2x – 4223

x− ≤ < − , f(x) = 2x + 4

2 03

x− ≤ ≤ , f(x) = –4x

0 2x≤ < , f(x) = 4x

2x ≥ , f(x) = 2x + 4

8/3

–2 23–

2

Clearly, x = –2 and x = 0 are point of minima.23

x = − is point of maxima.

(30)

47. Let w = 32

i+ and P = {wn : n = 1, 2, 3, …}. Further H1 = 1: Re2

z z⎧ ⎫∈ >⎨ ⎬⎩ ⎭

C and H2 = 1: Re

2z z −⎧ ⎫∈ <⎨ ⎬⎩ ⎭

C ,

where C is the set of all complex numbers. If z1 ∈ P ∩ H1, z2 ∈ P ∩ H2 and O represents the origin, then∠z1Oz2 =

(A) 2π (B) 6

π

(C)23π

(D)56π

Answer (C, D)Hint : Note that |ω| = 1

β1

β3

β2

α1

α2

α3

30°30°30°

30°

αi are possible value of z1

βi are possible value of z2

(i = 1,2,3)

32 2

iω = +

6i

eπ

ω =

2 3i

eπ

ω =

3 2i

eπ

ω =

24 3i

eπ

ω =

55 6

ie

π

ω =

So, 1 2z oz∠ can be 2 5,3 6π π

⇒

(31)

48. If 3x = 4x–1, then x =

(A) 3

3

2log 22log 2 1− (B)

2

22 log 3−

(C)4

11 log 3− (D) 2

2

2log 32log 3 1−

Answer (A, B, C)Hint :

13 4x x−=

2 2log 3 ( 1)log 4x x= −

2log 3 ( 1) 2x x= − ⋅

22 log 3 2x x− =

22 log 3 2x − =⎡ ⎤⎣ ⎦

2

22 log 3

x =− ⇒ (B)

⇒ 3

33

2 log 221 2log 2 12

log 2

=−−

⇒ (A)

⇒4

2

1 11 1 log 31 log 32

=−−

⇒ (C)

So, (A), (B), (C)


Paragraph for Questions 49 and 50Let PQ be a focal chord of the parabola y2 = 4ax. The tangents to the parabola at P and Q meet at a point lyingon the line y = 2x + a, a > 0.49. Length of chord PQ is

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

Answer (B)

Hint : 1 2( )a t t a+ =

P at at( , 2 )1 12

Q at at( , 2 )2 22

x = –a

(– , – )a a O1 2 1t t+ =

21 2( )PQ a t t= −

21 2 1 2( ) 4a t t t t⎡ ⎤= + −⎣ ⎦

[1 4] 5a a= + = use (1)

(32)

50. 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

Answer (D)Hint : t1 + t2 = 1

( , 2 )at at1 12

( , 2 )at at2 22

θ

2/t1

2/t2

t1 t2 = –1

11 5

2t ±= , t > 0

∴ 11 5

2t +=

1 2

1 2

2 2

tan 41

t t

t t

−θ =

+

= 2 1

1 2

2( ) 2(1 (1 5)) 2 54 3 3

t tt t

− − += = −

+


Let →: [0,1]f (the set of all real numbers) be a function. Suppose the function f is twice differentiable,f(0) = f(1) = 0 and satisfies f″(x) – 2f′(x) + f(x) ≥ ex, x ∈ [0, 1].51. Which of the following is true for 0 < x < 1?

(A) 0 < f(x) < ∞ (B) − < <1 1( )2 2

f x

(C) − < <1 ( ) 14

f x (D) –∞ < f(x) < 0

Answer (D)

Hint : − + ≥2

2 2 xd y dy y edxdx

⇒ − − −− + ≥2

2 2 1x x xd y dye e e ydxdx

⇒ − ≥2

2 ( ) 1xd yedx

⇒ ye–x is concave up

0 1

ye–x

Hence, –∞ < f(x) < 0

(33)

52. If the function e–x f(x) assumes its minimum in the interval [0, 1] at =14

x , which of the following is true?

(A) ′ < < <1 3( ) ( ),4 4

f x f x x (B) ′ > < <1( ) ( ), 04

f x f x x

(C) ′ < < <1( ) ( ), 04

f x f x x (D) ′ < < <3( ) ( ), 14

f x f x x

Answer (C)

Hint :

O 11/4

−( )xd yedx

is an increasing function.

< <104

x >14

x

− <( ) 0xd yedx

− >( ) 0xd yedx

− −− < 0x xdye e ydx

− −− > 0x xdye e ydx

<dy ydx

>dy ydx

f′(x) < f(x) f′(x) > f(x)

Paragraph for Questions 53 and 54A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red ballsand 4 black balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls.53. 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)90648

(C)558648 (D)

566648

Answer (A)Hint :

B1 B2 B3 1W 3R 2B

2W 3R 4B

3W 4R 5B

P(WWW + RRR + BBB)

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

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

= 6 36 40 82

648 648+ +

=

(34)

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

(A)116181 (B)

126181

(C)65

181 (D)55

181

Answer (D)

Hint :

( )

( ) ( ) ( )

222

1 2 31 2 3

. . .

WRP P BBB

PWR WR WR WRP P B P P B P P B

B B B

⎛ ⎞×⎜ ⎟

⎛ ⎞ ⎝ ⎠=⎜ ⎟⎛ ⎞⎛ ⎞ ⎛ ⎞⎝ ⎠ + + ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=

2 31 19

21 3 2 3 3 4

1 1 1 1 1 19 9 12

2 2 2

13

1 1 13 3 3

C CC

C C C C C CC C C

××

⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×× + × + ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=

2 39 4

3 2 2 3 3 4 26 5 9 4 12 11

××

× × × ×+ +× × ×

= 6 5 6 11

36 181× × ×

×

= 55

181


Let S = S1 ∩ S2 ∩ S3, where

= ∈ <1 { :| | 4}S z z , ⎧ ⎫⎡ ⎤− +⎪ ⎪= ∈ >⎢ ⎥⎨ ⎬

−⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭2

1 3: Im 01 3

z iS zi

and = ∈ >3 { : Re 0}S z z .

55. Area of S =

(A)π10

3 (B)π20

3

(C)π16

3 (D)π32

3

(35)

Answer (B)Hint : S1 represent circle with centre (0, 0) and radius 4

A

C

B

S

60°Oy x + 3 = 0

S1 : |z| < 4 ⇒ x2 + y2 < 16

⎡ ⎤− +>⎢ ⎥

−⎢ ⎥⎣ ⎦2

1 3: Im 01 3

z iSi

⎛ ⎞− + + +>⎜ ⎟⎜ ⎟

⎝ ⎠

[( 1) ( 3 )][1 3 ]Im 02

x y i i

≡ + >2 3 0S y x

S3 Re(z) > 0, i.e., x > 0

S = S1 ∩ S2 ∩ S3

Area of shaded region is OAB + OBC = π + × π2

2(4) 60 (4)4 360

= π

π +164

6

= π

π +843

= π20

3

56.∈

− − =min|1 3 |z S

i z

(A) −2 32 (B) +2 3

2

(C) −3 32

(D) +3 32

Answer (C)

Hint : min|z – (1 – 3i)|

Minimum distance of z from (1, –3)

From question, minimum distance of (1, –3) from + =3 0y x is − + −

=3 3 3 3

2 2

(36)


57. Consider the lines 11 3:

2 1 1x y zL − +

= =−

, 24 3 3:

1 1 2x y zL − + +

= = and the planes 1 : 7 2 3,P x y z+ + =

2 3 5 6 4P x y z= + − = . Let ax by cz d+ + = be the equation of the plane passing through the point ofintersection 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-IIP. a = 1. 13Q. b = 2. –3R. c = 3. 1S. d = 4. –2Codes :

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

Answer (A)

Hint :− +

≡ = = =−1 1

1 32 1 1

x y zL t

and − + +≡ = = =2 2

4 3 31 1 2

x y zL t

For finding point of intersection, we have1 + 2t1 = 4 + t2 ...(i)

and –t1 = –3 + t2 ...(ii)Solving, we get

t1 = 2, t2 = 1i.e., point of intersection is (5, –2, –1).Now, equation of plane P be

− + +=

−

5 2 17 1 2 03 5 6

x y z

⇒ (–16)(x – 5) + 48(y + 2) + 32(z + 1) = 0⇒ (x – 5) – 3(y + 2) – 2(z + 1) = 0⇒ x – 3y – 2z = 13i.e., a = 1, b = –3, c = –2, d = 13

(37)

58. Match List-I with List-II and select the correct answer using the code given below the listsList-I List-II

P. Volume of parallelepiped determined by vectors , and a b c is 2. 1. 100Then the volume of the parallelepiped determined by vectors2( ),3( ) and ( )a b b c c a× × × is

Q. Volume of parallelepiped determined by vectors , and a b c is 5. 2. 30Then the volume of the parallelepiped determined by vectors3( ),( ) and 2( )a b b c c a+ + + is

R. Area of a triangle with adjacent sides determined by vectors a 3. 24and b is 20. Then the area of the triangle with adjacent sides

determined by vectors (2 3 )a b+ and ( )a b− isS. Area of a parallelogram with adjacent sides determined by vectors 4. 60

a and b is 30. Then the area of the parallelogram with adjacent

sides determined by vectors ( )a b+ and a isCodes

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

Answer (C)

Hint : (P) Given, =[ ] 2a b c

Now, V = × × ×[2( ) 3( ) ]a b b c c a

= 26[ ]a b c

= 24

(Q) Given, =[ ] 5a b c

Now, V = + + +[3( ) 2( )]a b b c c a

= + + +6[ ]a b b c c a

= 12[ ]a b c= 60

(R) Given, × =| | 40a b

Now, A = + × −1|(2 3 ) ( )|2

a b a b

= ⋅ +1 5| |2

a b

= ×5 402

= 100

(38)

(S) Given, × =| | 30a b

Now, A = + ×|( ) |a b a

= +| |a b= 30

59. 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 pointF(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 Lis 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 = 1.12

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

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

Answer (A)Hint : Parabola is y2 = 16x

and line is y = mx + 3Solving,

(mx + 3)2 = 16x⇒ m2x2 + (6m – 16)x + 9 = 0 ...(i)Also, tangent at F is

y y x2 = 16

F x y( , )0 0

EG

x

yy0 = 8(x +x0)

So, = 01

0

8xyy

...(ii)

Now, area of ΔEFG,

Δ = − ⋅1 01 (3 )2

y x

= ⎛ ⎞

−⎜ ⎟⎝ ⎠

00

0

81 32

x xy

= ⎛ ⎞

−⎜ ⎟⎜ ⎟⎝ ⎠

20

00

81 32

xxy

= ⎛ ⎞⎜ ⎟−⎜ ⎟⎝ ⎠

20

00

81 32 4

xxx

(39)

For Δ is to be maximum,

Δ=

00d

dx

⇒ − × =033 2 02

x

⇒ x0 = 1So, y0 = 4

= =01

0

8 2xyy

From y0 = mx0 + 3 ⇒ m = 160. 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 sin tan1

cot sin tan sin

y y yy

y y y

− −

− −

⎛ ⎞⎛ ⎞+⎜ ⎟⎜ ⎟ +⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ takes value 1.

1 52 3

Q. If cosx + cosy + cosz = 0 = sinx + siny + sinz then 2. 2

possible value of cos2

x y−is

R. If π⎛ ⎞−⎜ ⎟

⎝ ⎠cos

4x cos 2x + sinx sin 2x secx = cosx sin2x secx + 3. 1

2

π⎛ ⎞+⎜ ⎟⎝ ⎠

cos4

x cos 2x then possible value of secx is

S. If ( ) ( )( )=–1 2 –1cot sin 1 – sin tan 6x x , x ≠ 0, 4. 1

then possible value of x isCodes :

P Q R S

(A) 4 3 1 2

(B) 4 3 2 1

(C) 3 4 2 1

(D) 3 4 1 2Answer (B)Hint :

P.

1/221 14

2 1 11 cos(tan ) sin(tan )

cot(sin ) tan(sin )y y y y

y y y

− −

− −

⎧ ⎫⎛ ⎞+⎪ ⎪+⎜ ⎟⎨ ⎬⎜ ⎟+⎪ ⎪⎝ ⎠⎩ ⎭

(40)

=

1/222

2 24

2 2

2

11 111

1

yy y y

y y yy y

⎧ ⎫⎛ ⎞⎪ ⎪⎜ ⎟+⎪ ⎪⎜ ⎟+ +⎪ ⎪⎜ ⎟ +⎨ ⎬⎜ ⎟⎪ ⎪−

+⎜ ⎟⎪ ⎪⎜ ⎟−⎪ ⎪⎝ ⎠⎩ ⎭

=

1/222 2

42

1 111

y y y yy

⎧ ⎫⎛ ⎞+ −⎪ ⎪⎜ ⎟ +⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

= { }1/24 41 1y y− + =

Q. If cos cos cos sin sin sin 0x y z x y z+ + = + + =

then possible value of x – y is 23π

± i.e. 1cos

2 2x y−⎛ ⎞ =⎜ ⎟

⎝ ⎠.

R. Given equation can be written is

( )cos2 cos cos sin2 1 tan4 4

x x x x x⎧ ⎫π π⎛ ⎞ ⎛ ⎞

− − + = −⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎩ ⎭

⇒ ( )cos2 2sin sin sin2 1 tan4

x x x xπ⋅ ⋅ = −

⇒1 cos2 cos sin2

x x x= −

⇒1 (cos sin ) 12

x x+ =

⇒ 4x π=

So, sec 2x =

S. Given equation is 2 2

61 1 6

x xx x

=− +

either x = 0

or 21 6 6 6x x2+ = −

⇒1 52 3

x = ±

Documents

: Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi … · 2019. 4. 23. · (1) Answers & Solutions for JEE (Advanced)-2013 Time : 3 hrs. Max. Marks: 180 DATE : 02/06/2013