IIT - JEE 2013 (Advanced) - Vidyalankar Classes

IIT - JEE 2013 (Advanced)

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

2

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

3

PART I : PHYSICS

SECTION I : (Only One option correct Type) This section contains 10 multiple choice questions. Each question has four choices (A), (B),

(C) and (D) out of which ONLY ONE is correct.

1. The work done on a particle of mass m by a force,

( )3/2 2 2 3/22 2

x yˆ ˆK i j(x y )x y

⎡ ⎤+⎢ ⎥++⎢ ⎥⎣ ⎦

(K being a constant of appropriate dimension), when the particle is taken from the point

(a, 0) to the point (0, a) along a circular path of radius a about the origin in the x-y plane is

(A) 2K

a

π (B)

K

a

π (C)

K

2a

π (D) 0

1. (D)

3

kF r F

r= ⇒ is conservative

112 2 2 2 2 2

11ˆ ˆi iF K Ux

y(x y )(x y )

⎡ ∂ ⎤⎛ ⎞ ∂⎢ ⎥⎜ ⎟ += − = −∇∂⎜ ⎟ ∂⎢ ⎥+⎝ ⎠ +⎢ ⎥⎣ ⎦

1

2 2 2

KU

(x y )

=

+

[potential energy]

U(0, a) = U(a, 0) ⇒ W = 0

2. Two rectangular block, having identical dimensions, can be arranged either in

configuration I or in configuration II as shown in the figure. One of the block has thermal

conductivity k and the other 2k. The temperature difference between the ends along the x-

axis is the same in both the configurations. It takes 9 s to transport a certain amount of

heat from the hot end to the cold end in the configuration I. The time to transport the same

amount of heat in the configuration II is

(A) 2.0 s (B) 3.0 s (C) 4.5 s (D) 6.0 s

2. (A)

Thermal resistance of individual block are

2R and R.

For config 1; 1R 3R ;= For config 2; 2

2R R

3=

Ratio of heat currents 1

2

Q 1 2 2R

Q 3R 3 9= ⋅ =

Q1 t1 = Q2 t2 ⇒ 12 1 s sec

2

Q 2t t 9 2

Q 9= ⋅ = × =


4

3. two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The

ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is

4 : 3. The ratio of their densities is

(A) 1 : 4 (B) 1 : 2 (C) 6 : 9 (D) 8 : 9

3. (D)

1 1 2 2P V P V=

1 2 1

2 1 2

P M

P M

ρ=ρ

1

2

4 28 / 9

3 3

ρ = × =ρ

4. A particle of mass m is projected from the ground with an initial speed u0 at an angle α

with the horizontal. At the highest point of its trajectory, it makes a completely inelastic

collision with another identical particle, which was thrown vertically upward from the

ground with the same initial speed u0. The angle that the composite system makes with the

horizontal immediately after the collision is

(A) 4

π (B)

4

π + α (C) 2

π − α (D) 2

π

4. (A)

2

2uH sin

2g= α

Speed the particle that was thrown vertically at the

highest point be “v”.

2 2

2 2 00

u sinv u 2g

2g

α− = − × ⇒ v = u0 cos α

∴ y

x

ptan 1

pθ = =

5. A pulse of light of duration 100 ns is absorbed completely by a small object initially at

rest. Power of the pulse is 30 mW and the speed of light is 3 × 108 ms

−1. The final

momentum of the object is

(A) 0.3 × 10−17

kg ms−1

(B) 1.0 × 10−17

kg ms−1

(C) 3.0 × 10−17

kg ms−1

(D) 9.0 × 10−17

kg ms−1

5. (B)

P h / d= or E = PC

2 7

17

8

3 10 10P 1 10

3

− −−× ×= = ×

×10

6. In the Young’s double slit experiment using a monochromatic light of wavelength λ, the

path difference (in terms of an integer n) corresponding to any point having half the peak

intensity is

(A) ( )2n 12

λ+ (B) (2n 1)

4

λ+ (C) (2n 1)8

λ+ (D) (2n 1)16

λ+

6. (B)

20I 4I cos

2

δ⎛ ⎞= ⎜ ⎟⎝ ⎠

If I = 2I0

mu0 cos α

θ

mu0 cos α


5

∴ 2 1 1cos cos

22 2 2

δ δ⎛ ⎞ ⎛ ⎞= ⇒ = ±⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ n2 4

δ π= π ±

2 x

2n2 4

π Δ π⋅ = π ±λ

⇒ 1

x 2n4

⎛ ⎞Δ = λ±⎜ ⎟⎝ ⎠

7. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the

lens, is real and is one-third the size of the object. The wavelength of light inside the lens

is 2

3 times the wavelength in free space. The radius of the curved surface of the lens is

(A) 1 m (B) 2 m (C) 3 m (D) 6 m

7. (C)

v = 8 m, μ = 3

2

m = − 1

3

1 1 1

v u f− = ⇒

v v1 m

f u− = =

⇒ f v 1

mf 3

− = = −

⇒ f 8 1

f 3

− = − ⇒ f = 6

f = ( )R

1μ − ⇒ 2R = 6 ⇒ R = 3

8. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end

of another horizontal thin copper wire of length L and radius R. When the arrangement is

stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that

in the thick wire is

(A) 0.25 (B) 0.50 (C) 2.00 (D) 4.00

8. (C)

2 1

F F

4A A

2L L

=Δ Δ

⇒ 2 1

2 1

4=

Δ Δ

⇒ 2

1

1

2

Δ =Δ

⇒ 1

2

2Δ =Δ

9. A ray of light travelling in the direction ( )1 ˆ ˆi 3j2

+ is incident on a plane mirror. After

reflection, it travels along the direction ( )1 ˆ ˆi 3j2

− . The angle of incidence is

(A) 30° (B) 45° (C) 60° (D) 75°

2L

2R F F

R

L


6

9. (A)

Angle between I and R is

cos δ = ( )( )

( ) ( )i 3 j i 3 jI R 1 3 1

2 2 4 2I R

ˆ ˆ.

| || |

+ − −= = = −

δ = 120°

180 − 2 i = 120°

2i = 60°

i = 30°

10. The diameter of a cylinder is measured using a Vernier caliper with no zero error. It is

found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main

scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th

division of the

Vernier scale exactly coincides with one of the main scale divisions. The diameter of the

cylinder is

(A) 5.112 cm (B) 5.124 cm (C) 5.136 cm (D) 5.148 cm

10. (B)

M.S.R. = 5.1 cm

1 V. C. = 2 45

0 049cm50

..=

L.C. = 0.050 − 0.049

= 0.001 cm

V.S.R. = 24

⇒ d = (5.1) cm + 24 × 0.001 = 5.124 cm

SECTION II : (One or more options correct Type) This section contains 5 multiple choice questions. Each question has four choices (A), (B),

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

11. In the circuit shown in the figure, there are two parallel plate capacitors each of

capacitance C. The switch S1 is pressed first to fully charge the capacitor C1 and then

released. The switch S2 is then pressed to charge the capacitor C2. After some time, S2 is

released and then S3 is pressed. After some time,

(A) the charge on the upper plate of C1 is 2CV0.

(B) the charge on the upper plate of C1 is CV0.

(C) the charge on the upper plate of C2 is 0.

(D) the charge on the upper plate of C2 is −CV0.

11. (B), (D)

After S1 is released :

δ

RI

C2

+2CV0

−2CV0

C1


7

After S2 is released

After S3 is released

12. A particle of mass M and positive charge Q, moving with a constant velocity

1u = 1ˆ4i ms− , enters a region of uniform static magnetic field normal to the x-y plane.

The region of the magnetic field extends from x = 0 to x = L for all values of y. After

passing through this region, the particle emerges on the other side after 10 milliseconds

with a velocity 2u = ( )ˆ ˆ2 3i j + ms−1

. The correct statement(s) is (are)

(A) The direction of the magnetic field is −z direction.

(B) The direction of the magnetic field is +z direction.

(C) The magnitude of the magnetic field 50 M

3Q

π units.

(D) The magnitude of the magnetic field is 100 M

3Q

π units.

12. (A), (C)

Magnetic field is clearly in −z axis.

y

x

v 1tan

v 3θ = = ⇒

6

πθ =

mvr

qB=

r sin Lθ =

⇒ mvsin L

qBθ = ⇒ mv

B sinqL

= θ

Time of travel r vt

t rv

θ= = ⇒ =θ

∴ vt

sin Lθ =θ

∴ 3

mv m 50 mB .

q v t 6 3qq −π π= θ = =

×10×10

−CV0

+CV0

C2

+CV0

−CV0

C1

+CV0

−CV0

C1

+CV0

−CV0

C2


8

13. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according

to the equation, y(x, t) = (0.01 m) sin [(62.8 m−1

)x]. cos[(628 s−1

)t]. Assuming π = 3.14,

the correct statement(s) is (are)

(A) The number of nodes is 5.

(B) The length of the string is 0.25 m.

(C) The maximum displacement of the midpoint of the string, from its equilibrium

position is 0.01 m.

(D) The fundamental frequency is 100 Hz.

13. (B), (C)

y = 0.01 sin [(62.8 m−1

) x] cos [(628 s−1

) t]

= 0.01 sin (20π x) cos (200 πt)

k = 20 π ⇒ λ = 2 1

m 10cmK 10

π = =

ω = 200 π ⇒ ν = 1100s2

−ω =π

v = 10m/sk

ω =

in 5th

harmonic.

ν = 5v

2 ⇒

525cm

2= λ = … (B)

⇒ 6 nodes ⇒ A is wrong.

Mid-point is antinode.

⇒ sin (20 πx) = 1

⇒ y = 0.01 cos (200 π t)

⇒ ymax = 0.01 m … (C)

0

v 10

122

4

ν = =⎛ ⎞⎜ ⎟⎝ ⎠

= 20 hz. … (D) is wrong.

14. A solid sphere of radius R and density ρ is attached to one end of a mass-less spring of

force constant k. The other end of the spring is connected to another solid sphere of radius

R and density 3ρ. The complete arrangement is placed in a liquid of density 2ρ and is

allowed to reach equilibrium. The correct statement(s) is (are)

(A) the net elongation of the spring is 34 R g

3k

π ρ

(B) the net elongation of the spring is 38 R g

3k

π ρ.

(C) the light sphere is partially submerged.

(D) the light sphere is completely submerged.

14. (A), (D)

B = 2ρ vg

3 mg = 3ρ vg

⇒ Kx + B = 3 mg

⇒ Kx = ρ vg1

3m

Kx B

3 mg


9

⇒ x =

34R g

vg 3

K K

ρ πρ = = 34 R g

3K

π ρ

B′ = mg + Kx= ρ vg + ρ vg

B′ = 2ρ vg

2ρ Δvg = 2ρ vg

⇒ v = Δv

⇒ Sphere is completely submerged

15. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge

densities ρ1 and ρ2 respectively, touch each other. The net electric field at a distance 2R

from the centre of the smaller sphere, along the line joining the centres of the spheres, is

zero. The ratio 1

2

ρρ

can be

(A) −4 (B) 32

25− (C)

32

25 (D) 4

15. (B), (D)

|Q1 | = 1 Vρ

|Q2 | = 2 V8ρ

Case I : If P is between A and B then 1e

2

Vρ + ρ

Case II : If P is left of A then 1

2

veρ −ρ

Case II Case I

⇒ 1 2

2 20 0

Q Q

4 (2R) 4 (5R)=

πε πε ⇒ 1 2

2 30 0

Q Q (R)

4 (2R) 4 (2R)=

πε πε

⇒ 1

2

Q 4

Q 25= ⇒ 1 2Q Q

4 8=

⇒ 1

2

V 4

8 V 25

ρ =

ρ ⇒ 1

2

Q 1

Q 2=

⇒ 1

2

32

25

ρ =

ρ ⇒ 1

2

V 1

8 V 2

ρ =ρ

⇒ 1

2

32

25

ρ −=ρ

⇒ 1

2

4ρ =ρ

Kx

B′

mg

A P B 2R R

ρ1V 8 ρ2V

2R

−ve +ve

R 2R

P


10

SECTION III : (Integer value correct Type) This section contains 5 questions. The answer to each question is a single digit integer,

ranging from 0 to 9 (both inclusive).

16. A bob of mass m, suspended by a string of length l1, is given a minimum velocity required

to complete a full circle in the vertical plane. At the highest point, it collides elastically

with another bob of mass m suspended by a string of length l2, which is initially at rest.

Both the strings are mass-less and inextensible. If the second bob, after collision acquires

the minimum speed required to complete a full circle in the vertical plane, the ratio 1

2

is

16. [5]

1 2m g m 5g=

1 25=

1

2

5=

17. A particle of mass 0.2 kg is moving in one dimension under a force that delivers a

constant power 0.5 W to the particle. If the initial speed (in ms−1

) of the particle is zero,

the speed (in ms−1

) after 5 s is

17. [5]

P = 2d 1mv

dt 2

⎛ ⎞⎜ ⎟⎝ ⎠

2d

dtν =

2P

m

d

2dt

νν = 2P

m;

0

dv

νν ∫ =

t

0

Pdt

m ∫

2

2

ν =

Pt

m

ν2 =

2Pt

m =

2 0.55

0.2

×× = 25

ν = 5 m/s ν = 5

18. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the

slope of the stopping potential versus frequency plot for Silver to that of Sodium is

18. [1]

Vs = h

e e

φν −

Slope = h

e for both silver and Na

∴ Ratio of slope = 1

2

1

1

m

25g

m


11

19. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103

disintegrations per second. Given that n 2 = 0.693, the fraction of the initial number of

nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after

preparation of the sample is

19. [4]

N = t0N e−λ

t

0

Ne

N

−λ=

∴ t

0

N1 1 e

N

−λ− = −

( )t

0

N1 e

N

−λ⎛ ⎞−Δ = −⎜ ⎟⎝ ⎠

λ t << 1 ⇒ t1 e t−λ− ≈ λ

0

Nt

N

⎛ ⎞−Δ = λ⎜ ⎟⎝ ⎠

= 1 2

n 2t

t /

= 0 693

80 0 041386

..

⎛ ⎞ =⎜ ⎟⎝ ⎠

⇒ 0

N100 4

N

−Δ × = ⇒ |ΔN| % = 4 %

20. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular

velocity of 10 rad s−1

about its own axis, which is vertical. Two uniform circular rings,

each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in

such a manner that they are touching each other along the axis of the disc and are

horizontal. Assume that the friction is large enough such that the rings are at rest relative

to the disc and the system rotates about the original axis . The new angular velocity

(in rad s−1

) of the system is

20. [8]

R = 0.4

M = 50 km

ω = 10

mR = 6.25 kg

r = 0.2

21mR

2ω = ( )2 2

R

12 2m R mR

2

⎡ ⎤ ′+ ω⎢ ⎥⎣ ⎦

1

50 0 16 102

.× × × = 1

4 6 25 0 04 50 0 162

. . .⎛ ⎞ ′× × + × × ω⎜ ⎟⎝ ⎠

25 0 16 10.× × = ( )25 0 04 25 0 16. . ′× + × ω

4 × 10 = 5 ′ω

′ω = 8

R


12

PART II : CHEMISTRY SECTION I : (Only One option correct Type)

This section contains 10 multiple choice questions. Each question has four choices (A), (B),


21. Consider the following complex ions, P, Q and R.

P = [FeF6]−3

, Q = [V(H2O)6]2+

and R = [Fe(H2O)6]2+

The correct order of the complex ions, according to their spin−only magnetic moment

values (in B.M.) is

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

21. (B)

[ ]3

6P Fe F−= i.e. [ ]3 5Fe Ar 3d+ =

Magnetic moment = ( )n n 2 BM+

= 35 BM

( ) 2

2 6Q V H O

+⎡ ⎤= ⎣ ⎦ [ ]2 3V Ar 3d+ =

Magnetic moment = 15 BM

( ) 2

2 6R Fe H O

+= i.e. [ ]2 6Fe Ar 3d+ =

Magnetic moment = ( )4 4 2+

= 24 BM

P > R > Q

22. The arrangement of X− ions around A

+ ion in

solid AX is given in the figure (not drawn to

scale). If the radius of X− is 250 pm, the radius

of A+ is

(A) 104 pm (B) 125 pm

(C) 183 pm (D) 57 pm

22. (A)

A A

x x

r r0.414

r r

+ +

− −= =

Ar 0.414 250 pm+ = × = 104 pm

23. Sulfide ores are common for the metals

(A) Ag, Cu and Pb (B) Ag, Cu and Sn (C) Ag, Mg and Pb (D) Al, Cu and Pb

23. (A)

Factual

24. The standard enthalpies of formation of CO2(g), H2O ( ) and glucose (s) at 25 °C are

−400 kJ/mol, −300 kJ/mol and −1300 kJ/mol, respectively. The standard enthalpy of

combustion per gram of glucose at 25 °C is

(A) + 2900 kJ (B) − 2900 kJ (C) − 16.11 kJ (D) + 16.11 kJ

24. (C)

6 12 6 2 2 2C H O O 6CO 6H O + ⎯⎯→ +

2CO ⎯⎯→ − 400 kJ/ mol


13

2 2 2Zn(OH) 2NaOH NaZnO 2H O + ⎯⎯→ +

H2O = −300 kJ / mol

Glucose = −1300 kJ / mol

ΔH = ΔHP − ΔHR

= 2 2 6 12 6f (CO f (H O) f C H )6 H 6H H) ( Ο + − Δ Δ

= (6 × −400) + 6(−300) − (−1300)

= −2400 − 1800 + 1300 = −2900 kJ / mol

ΔH = 2900

180

− = −16.11 kJ / gm

25. Upon treatment with ammoniacal H2S, the metal ion that precipitates as a sulfide is

(A) Fe(III) (B) Al(III) (C) Mg(II) (D) Zn(II)

25. (D)

White ppt of zinc sulfide by adding ammoniacal H2S to Zn(II).

2 2ZnS 2HCl ZnCl H S + ⎯⎯→ +

2 2ZnCl 2NaOH Zn(OH) 2NaCl + ⎯⎯→ +

2 2 2white ppt.

Na ZnO H S ZnS 2NaOH

+ ⎯⎯→ +

26. Methylene blue, from its aqueous solution is adsorbed on activated charcoal at 25° C. for

this process, the correct statement is

(A) The adsorption requires activation at 25° C.

(B) The adsorption is accompanied by a decrease in enthalpy.

(C) The adsorption increases with increase of temperature.

(D) The adsorption is irreversible.

26. (B)

Factual

27. KI in acetone, undergoes SN2 reaction with each of P, Q, R and S. The rates of the

reaction vary as

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

27. (B)

SN2 reactions are through back attack of attacking Nu− and not on the basis of stability of

carbocation.

28. In the reaction,

P Q R S + ⎯⎯→ +

the time taken for 75% reaction of P is twice the time taken

for 50% reaction of P. The concentration of Q varies with

reaction time as shown in the figure. The overall order of the

reaction is

(A) 2 (B) 3

(C) 0 (D) 1


14

28. (D)

For P, t75% = 2t50% Ist order wrt ‘P’

For Q, [Q] = [Q0] − kt. i.e. zero order.

i.e. overall order of a reaction = 1

29. Concentrated nitric acid, upon long standing, turns yellow − brown due to the formation of

(A) NO (B) NO2 (C) N2O (D) N2O4

29. (B)

Sunlight

3 2 2 24HNO 4NO 2H O O ⎯⎯⎯⎯→ + +

30. The compound that does NOT liberate CO2, on treatment with aqueous sodium

bicarbonate solution, is

(A) Benzoic acid (B) Benzenesulphonic acid

(C) Salicylic acid (D) Carbolic acid (Phenol)

30. (D)

Compound with COOH gr or SO3H group can liberate CO2 with NaHCO3 solution.



31. The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100th

of that of a strong acid (HX, 1M), at 25°C. the Ka of HA is

(A) 1 × 10−4

(B) 1 × 10−5

(C) 1 × 10−6

(D) 1 × 10−3

31. (A)

Ka = Cα2 = 1 × (0.01)

2 = 1 × 10

−4

32. The hyperconjugative stabilities of tert − butyl cation and 2 − butene, respectively, are due to

(A) σ → p (empty) and σ → π* electron delocalisations.

(B) σ → σ* and σ →π electron delocalisations.

(C) σ → p (filled) and σ → π electron delocalisations.

(D) p(filled) → σ* and σ → π*

electron delocalisations.

32. (A)

CH3 − CH = CH2 (σ − π* electron) delocalization.

33. The pair(s) of coordination complexes / ions exhibiting the same kind of isomerism is (are)

(A) [Cr(NH3)5Cl]Cl2 and [Cr(NH3)4Cl2]Cl

(B) [Co(NH3)4Cl2]+ and [Pt(NH3)2(H2O)Cl]

+

(C) [CoBr2Cl2]2−

and [PtBr2Cl2]2−

(D) [Pt(NH3)3(NO3)]Cl and [Pt(NH3)3Cl]Br

33 . (B) , (D)

Factual.

2 3CH C CH − −

CH3

H+

σ − p(empty) delocalisation


15

34. Among P, Q, R and S, the aromatic compound(s) is /are

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

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

N

H

⊕OH

⊕

All are Aromatic except (R)

(P) (Q) (R) (S)

35. Benzene and naphthalene from an ideal solution at room temperature. For this process, the

true statement(s) is(are)

(A) ΔG is positive (B) ΔSsystem is positive

(C) ΔSsurroundings = 0 (D) ΔH = 0

35. (B) , (C) , (D)

Conditions for Ideal Solution.



36. The atomic masses of He and Ne are 4 and 20 a.m.u, respectively. The value of the de

Broglie wavelength of He gas at −73°C is “M” times that of the de Broglie wavelength of

Ne at 727°C. M is

36. [5]

He Ne Ne

Ne He He

M .T 20 10005

M .T 4 200

λ ×= = =λ ×

37. EDTA4−

is ethylenediaminetetraacetate ion. The total number of N − Co − O bond angles

in [Co(EDTA)]1−

complex ion is

37. [8]

Total no. of N − Co − O bond angle in [Co (EDTA)]−1

= 8


16

OO

C

O

C

OO

3H O

Heat

+⎯⎯⎯→

O

C

C

O

3

2 2

O

H O⎯⎯⎯→. .

. .

O

O

OOH

OH

O

C

C

CH2

NH2

COOH

Glycine =

CH

NH2

COOH

CH3

Alanine =

CH

NH2

COOH

Valine (CH3)2CH

CH

NH2

COOH

CH2

Phphenyl alanine

N

N

N

NH2

NH2

NH2. . . .

. .. .

. .

. .Melamine

38. The total number of carboxylic acid groups in the product P is

38. [2]

Total no. of carboxylic group = 2

39. A tetrapeptide has − COOH group on alanine. This produces glycine (Gly), valine (Val),

phenyl alanine (Phe) and alanine(Ala), on complete hydrolysis. For this tetrapeptide, the

number of possible sequences (primary structures) with −NH2 group attached to a chiral

center is

39. [4]

No. of possible sequences (primary structures) with −NH2 group attached to a chiral

center is 4

40. The total number of lone − pairs of electrons in melamine is

40. [6]

Lone Pairs → 6.


17

PART III : MATHEMATICS SECTION I : (Only One option correct Type)

This section contains 10 multiple choice questions. Each question has four choices (A), (B),


41. For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines

ax + by + c = 0 and bx + ay + c = 0 is less than 2 2. Then

(A) a + b − c > 0 (B) a − b + c < 0 (C) a − b + c > 0 (D) a + b − c < 0

41. (A) Given equations are

ax + by + c = 0

bx + ay + c = 0

Now,

x

bc ac − =

y

bc ac − =

2 2

1

a b−

∴ x = ( )

2 2

b a c

a b

−−

= c

a b

−+

y = 2 2

(b a)c

a b

− −

= c

a b

−+

Now

2 2c c

1 1a b a b

⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ < 2 2

2

2

2(a b c)

(a b)

+ ++

< 2 2

2(a b c)

a b

+ ++

< 2 2

(a + b + c) < 2a + 2b

a + b − c > 0

42. The area enclosed by the curves y = sin x + cos x and y = | cos x − sin x | over the interval

0,2

π⎡ ⎤ ⎢ ⎥⎣ ⎦ is

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

42. (B)

∵ y = 2 sin x4

π⎛ ⎞ + ⎜ ⎟⎝ ⎠

y = 2 sin x4

π⎛ ⎞ − ⎜ ⎟⎝ ⎠

Area of shaded part =

0

2 sin x 2 sin x dx4 4

π/4 ⎡ ⎤π π⎛ ⎞ ⎛ ⎞ + − − ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

∫

/2

/4

2 sin x 2 sin x dx4 4

π

π

⎡ ⎤π π⎛ ⎞ ⎛ ⎞+ + − − ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

∫

4

π


18

=

/4 /2

0 /4

2 sin x sin x dx 2 sin x sin x4 4 4 4

π π

π

⎡ ⎤⎛ π π ⎞ π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + − + + − −⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦∫ ∫

= 2 − 2 + 2 − 2

= 4 − 2 2

43. The number of points in (−∞, ∞), for which x2 − x sin x − cos x = 0, is

(A) 6 (B) 4 (C) 2 (D) 0

43. (C)

f(x) = x2 − x sin x − cos x

f′(x) = 2x − sin x − x cos x + sin x

= x (2 − cos x)

So, f′(x) =

0 x 0

0 x 0

0 x 0

> ∀ > ⎧⎪ = ⎨⎪< ∀ < ⎩

For x = 0, f(x) = −1

Hence two solutions

44. The value of 23 n

1

n 1 k 1

cot cot 1 2k−

= =

⎛ ⎞⎛ ⎞ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∑ ∑ is

(A) 23

25 (B)

25

23 (C)

23

24 (D)

24

23

44. (B)

cot 23 n

1

n 1 k 1

cot 1 2k−

= =

⎛ ⎞⎛ ⎞+⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

∑ ∑

= cot ( )23

1

n 1

n 1cot 1 2n

2

−

=

⎛ ⎞⎛ + ⎞+⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

∑ = cot ( )23

1 2

n 1

cot n n 1−

=

⎛ ⎞+ +⎜ ⎟

⎝ ⎠∑

= cot 23

1

n 1

1tan

1 n(n 1)

−

=

⎛ ⎞ ⎜ ⎟+ +⎝ ⎠

∑ = cot23

1

n 1

(n 1) ntan

1 (n 1).n

−

=

⎛ ⎞+ −⎜ ⎟+ +⎝ ⎠∑

= cot ( )23

1 1

n 1

tan (n 1) tan n− −

=

⎡ ⎤+ −⎢ ⎥

⎣ ⎦∑ = cot 1 1tan 24 tan 1− −⎡ ⎤ − ⎣ ⎦

= cot 1 24 1tan

1 24.1

−⎡ ⎤−⎛ ⎞⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦

= cot 1 25cot

23

−⎡ ⎤⎢ ⎥⎣ ⎦

= 25

23

45. A curve passes through the point 1, .6

π⎛ ⎞ ⎜ ⎟⎝ ⎠

Let the slope of the curve at each point (x, y)

be y y

sec , x 0.x x

⎛ ⎞+ > ⎜ ⎟⎝ ⎠

Then the equation of the curve is

(A) y 1

sin log xx 2

⎛ ⎞ = +⎜ ⎟⎝ ⎠

(B) y

cosec log x 2x

⎛ ⎞ = + ⎜ ⎟⎝ ⎠

(C) 2y

sec log x 2x

⎛ ⎞ = + ⎜ ⎟⎝ ⎠

(D) 2y 1

cos log xx 2

⎛ ⎞ = +⎜ ⎟⎝ ⎠

(0, −1)


19

45. (A)

dy

dx =

y ysec

x x

⎛ ⎞+ ⎜ ⎟⎝ ⎠

let y

x= v

y = vx

dy

dx = v + x

dv

dx

v + x dv

dx= v + sec v

cos v dv ∫ = dx

x∫

sin v = lnx + lnC

Given it passes (1, 6

π)

⇒ Sin 6

π = ln1 + lnC

⇒ lnC = 1

2

So, curve is sin v = lnx+ 1

2

sin y

x

⎛ ⎞⎜ ⎟⎝ ⎠

= lnx + 1

2

46. Let f : 1

,12

⎡ ⎤ → ⎢ ⎥⎣ ⎦(the set of all real numbers) be a positive, non-constant and

differentiable function such that f (x) 2f (x)′ < and 1

f 1.2

⎛ ⎞ = ⎜ ⎟⎝ ⎠

Then the value of

1

1/2

f (x) dx ∫ lies in the interval

(A) (2 e − 1, 2 e) (B) (e − 1, 2e − 1) (C) e 1

, e 12

−⎛ ⎞ −⎜ ⎟⎝ ⎠

(D) e 1

0,2

−⎛ ⎞ ⎜ ⎟⎝ ⎠

46. (D)

f′(x) < 2f(x)

f (x)2

f (x)

′ <

f (x)dx 2dx

f (x)

′ < ∫ ∫

{ }n f (x) 2x c < +

f(x) < e2x + c

f(1/2) < e1 + c

1 < e1 + c

….(f(1/2)) = 1)

e0 < e

1 + c

0 < 1 + c

c > −1 ….(i)


20

1 12x c

1/2 1/2

f (x)dx e dx + < ∫ ∫

11

2x c

1/21/2

1f (x)dx e

2

+ ⎡ ⎤ < ⎣ ⎦∫

12 c 1 c

1/2

1f (x)dx e e

2

+ + ⎡ ⎤ < − ⎣ ⎦∫

[ ]1

1 c

1/2

1f (x)dx e e 1

2

+ < − ∫

Maximum value of

1

1/2

1f (x)dx is [e 1]

2 − ∫

And minimum value of

1

1/2

f (x) dx∫ is greater than zero because it is positive, non-

constant function.

47. Let ˆ ˆ ˆPR 3i j 2k = + − and SQ = ˆ ˆ î 3j 4k− − determine diagonals of a parallelogram PQRS

and ˆ ˆ ˆPT i 2j 3k= + + be another vector. Then the volume of the parallelepiped determined

by the vectors PT, PQ and PS is

(A) 5 (B) 20 (C) 10 (D) 30

47. (C)

PR = ˆ ˆ ˆ3i j 2k + − and SQ = ˆ ˆ î j 4k − 3 − are diagonals of a parallelogram.

PT = ˆ ˆ î j 3k + 2 + is another vector.

Volume of the parallelepiped by the vector PT, PQ & PS

ˆ ˆ â b 3i j 2k

ˆ ˆ â b i j 4k

ˆ ˆ â 2i j 3k

ˆ ˆ ˆb i j k

+ = + −

− = − 3 −

= − −

= + 2 +

Volume of parallelepiped formed by PT, PQ & RS is given by scalar triple product of

ˆ ˆ î j 3k + 2 + , ˆ ˆ ˆ2i j 3k − − & ˆ ˆ î j k + 2 +

Volume =

1 2 3

2 1 3

1 2 1

− − = 1(5) + 2(−5) + 3(5) = 5 − 10 + 15 = 10

48. Perpendiculars are drawn from points on the line x 2 y 1 z

2 1 3

+ += =−

to the plane

x + y + z = 3. The feet of perpendiculars lie on the line

(A) x y 1 z 2

5 8 13

− −= =−

(B) x y 1 z 2

2 3 5

− −= =−

(C) x y 1 z 2

4 3 7

− −= =−

(D) x y 1 z 2

2 7 5

− −= =−

P Q

R S

a

b

a

b


21

48. (D)

Any point on x 2

2

+ =

y 1

1

+−

= z

3is (2r − 2, −r−1, 3r).

Let P(x, y) be feet of perpendicular then,

1x x

a

− = 1y y

b

−= 1z z

c

− = − 1 1 1

2 2 2

ax by cz d

a b c

+ + +⎛ ⎞⎜ ⎟+ +⎝ ⎠

x (2r 2)

1

− −=

( )y r 1)

1

− − −=

z 3r

1

−= −

4r 6

3

−

So, x = 2r − 2 − 4r 6

3

− =

2r

3;

3 7ry

3

− = ; 5r 6

z3

+=

Obviously 2r 3 7r 5r 6

, ,3 3 3

− +⎛ ⎞ ⎜ ⎟⎝ ⎠

satisfy x y 1 z 2

2 7 5

− −= =−

49. Four persons independently solve a certain problem correctly with probabilities

1 3 1 1, , , .

2 4 4 8 Then the probability that the problem is solved correctly by at least one of

them is

(A) 235

256 (B)

21

256 (C)

3

256 (D)

253

256

49. (A)

Required Probability = 1 − P(No one solves the problem correctly)

= 1 − 1 1 3 7

2 4 4 8× × × =

235

256

50. Let complex numbers α and 1

α lie on circles (x − x0)

2 + (y − y0)

2 = r

2 and (x − x0)

2 +

(y − y0)2 = 4r

2, respectively. If z0 = x0 + iy0 satisfies the equation 2 |z0|

2 = r

2 + 2, then |α| =

(A) 1

2 (B)

1

2 (C)

1

7 (D)

1

3

50. (C)

| α − Z0| = r

(1)

and 0

1Z−

α= 2r (2)

(1) ⇒ (α− Z0) ( )0Zα − = r2

⇒ |α|2 − α 0 0Z Z− α +

2r

2+ 1 = r

2 (∵, |Z0|

2 =

2r

2+1)

⇒ α 0Z + Z0 α = | α |2+ 1 −

2r

2 (3)

(2) ⇒ 01 Z− α = 2r α

⇒ ( ) ( )001 Z . 1 Z− α − α = 4r2 α α

⇒ 1 − α Z0 − α 0Z + |α|2

2r1

2

⎛ ⎞+⎜ ⎟⎜ ⎟

⎝ ⎠ = 4 r

2 |α|

2

⇒ α Z0 + α 0Z = 1 + |α|2 −

27r

2|α|

2 (4)

(3) = (4) ⇒ |α|2 + 1 −

2r

2= |α|

2+ 1 −

7

2r

2|α|

2 ⇒ |α| =

1

7


22



51. A line passing through the origin is perpendicular to the lines

1ˆ ˆ ˆ: (3 t)i ( 1 2t) j (4 2t)k, t + + − + + + − ∞ < < ∞

2ˆ ˆ ˆ: (3 2s)i (3 2s) j (2 s)k, s + + + + + − ∞ < < ∞

Then, the coordinate (s) of the point (s) on 2 at a distance of 17 from the point of

intersection of and 1 is (are)

(A) 7 7 5

, ,3 3 3

⎛ ⎞⎜ ⎟⎝ ⎠

(B) ( 1, 1,0)− − (C) (1,1,1) (D) 7 7 8

, ,9 9 9

⎛ ⎞ ⎜ ⎟⎝ ⎠

51. (B), (D)

1ˆ ˆ ˆ ˆ ˆ ˆ: r (3i j 4k) t(i 2 j 2k) = − + + + +

2ˆ ˆ ˆ ˆ ˆ ˆ: r (3i 3j 2k) s(2i 2 j k) = + + + + +

Vector perpendicular to ˆ ˆ ˆi 2 j 2k+ + and ˆ ˆ ˆ2i 2 j k+ + is

ˆ ˆ ˆi j k

ˆ ˆ ˆ1 2 2 2i 3j 2k

2 2 1

= − + −

So, line is ˆ ˆ ˆr k ( 2i 3j 2k) = − + −

For point of intersection of and 1

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆk ( 2i 3j 2k) 3i j 4k t (i 2 j 2k) − + − = − + + + +

ˆ ˆ ˆ ˆ ˆ ˆ( 2k 3)i (3k 1) j ( 2k 4)k t (i 2 j 2k)− − + + + − − = + +

−2k − 3 = t … (i)

and 3k + 1 = 2t … (ii)

−2k − 4 = 2t … (iii)

(ii) ÷ (iii) 3k 1

12k 4

+ =− −

⇒ 3k + 1 = −2k − 4 ⇒ k = −1

From (i), −2 (−1) − 3 = t

⇒ t = −1. So, : ˆ ˆ ˆr 2i 3j 2k = − +

So, point of intersection of and 1 is A ≡ (2, −3, 2).

The line 2 can be written as x 3 y 3 z 2

s2 2 1

− − −= = =

Any point on 2 (2s + 3, 2s + 3, s + 2) = B

Given, AB = 17

⇒ (2s + 1)2 + (2s + 6)

2 + s

2 = 17

⇒ 9s2 + 28s + 20 = 0

⇒ (s + 2) (9s + 10) = 0

s = −2 or s = 10

9−

So, point B ( 1, 1, 0) and ≡ − − 7 7 8, ,

9 9 9

⎛ ⎞ ⎜ ⎟⎝ ⎠


23

52. Let f(x) = x sin πx, x > 0. Then for all natural numbers n, ( )f ' x vanishes at

(A) a unique point in the interval 1

n, n2

⎛ ⎞ +⎜ ⎟⎝ ⎠

(B) a unique point in the interval 1

n ,n 12

⎛ ⎞+ +⎜ ⎟⎝ ⎠

(C) a unique point in the interval (n, n + 1)

(D) two points in the interval (n, n + 1)

52. (B), (C)

f(x) = x sinπx

f′(x) = sinπx + (x) π(cosπx)

sinπx + xπ cosπx = 0

sinπx = −πx cos πx

tanπx = −πx

53. Let Sn =

k(k 1)4n2

k 1

( 1)

+

=−∑ k

2. Then Sn can take value(S)

(A) 1056 (B) 1088 (C) 1120 (D) 1332

53. (A), (D)

Sn =

k(k 1)4n 2 2

k 1

( 1) .k

+

=−∑

= (−1)1.1

2 + (−1)

3.2

2 + (−1)

6.3

2+ (−1)

10.4

2 + ….. upto 4n terms

= − 12 − 2

2 + 3

2 + 4

2 − 52 − 6

2 + …. + (4n)

2

= − 1 + 5 − 9 + 13 − 17 + ….. + 16n2 (2n − 1 + 2 terms)

= 2

n times

4 4 4 16n

+ + +

= 16n2 + 4n

54. For 3 × 3 matrices M and N, which of the following statement(s) is (are) NOT correct?

(A) NT M N is symmetric or skew symmetric, according as M is symmetric or skew

symmetric

(B) M N − N M is skew symmetric for all symmetric matrices M and N

(C) M N is symmetric for all symmetric matrices M and N

(D) (adj M) (adj N) = adj (M N) for all invertible matrices M and N

x

y

−1 1 3

2−

3

2 01

2−

1

2


24

54. (C), (D)

(C) is wrong as (MN)T = N

T M

T = NM ≠ MN

(D) is wrong as adj (MN) = adj (N) adj (M) for all invertible M and N.

55. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is

converted into an open rectangular box by folding after removing squares of equal area

from all four corners. If the total area of removed squares is 100, the resulting box has

maximum volume. Then the lengths of the sides of the rectangular sheet are

(A) 24 (B) 32 (C) 45 (D) 60

55. (A), (C)

Let the sides be 8C and 15C where C is a parameter given 2 (8C + 15 C) = K (constant)

⇒ C = K

46

Now,

∴ Volume is given by

V = (8C − 2x) (15C − 2x) x

or V = 4x3 − 46x

2C + 120C

2x

⇒ dv

dx= 12x

2 − 92 xC + 120 C

2 … (1)

area of all squares = 4x2 = 100

⇒ x = 5

∴ dv

0dx

= for maximum volume at x = 5

∴ using x = 5 in (1) we get

120 C2 − 460 C + 300 = 0

⇒ 6C2 − 23C + 15 = 0

⇒ C = 3 or 5/6

but C = 5/6 and x = 5 makes sides negative. Thus C = 3 is the only value of the

parameter.

Hence sides are 24, 45.



56. Consider the set of eight vectors { }V ai b j ck : a, b,c { 1,1}= + + ∈ − . Three non−coplanar

vectors can be chosen from V in 2p ways. Then p is,

56. [5]

Total number of vectors are, ˆ ˆ î j k + + ˆ ˆ î j k + −

− ˆ ˆ î j k + + − ˆ ˆ î j k+ −

ˆ ˆ î j k − + ˆ ˆ î j k − −

− ˆ ˆ î j k − + − ˆ ˆ î j k − −

Total no. of ways choosing non−coplanar vectors

= ( ) ( ) ( )4 2 4 3 42 1 1 2 3C C C C C × + × + 2 = 12 + 12 + 8 = 32

2P = 32

P = 5

8C − 2x

x x x x

15C − 2x


25

57. Of the three independent events E1, E2 and E3, the probability that only E1 occurs is α,

only E2 occurs is β and only E3 occurs is γ. Let the probability p that none of events E1, E2

or E3 occurs satisfy the equations (α − 2β) p = αβ and (β − 3γ) p = 2βγ. All the given

probabilities are assumed to lie in the interval (0, 1).

Then 1

3

Pr obability of occurrence of E

Pr obability of occurrence of E=

57. [6]

P (E1) = x

P (E2) = y

P (E3) = z

Then x (1 − y) (1 − z) = α

y (1 − z) (1 − x) = β

z (1 − x) (1 − y) = γ

(1 − x) (1 − y) (1 − z) = P

Now,

(α − 2β) P = αβ

⇒ [x (1 − y) (1 − z) − 2y (1 − z) (1 − x)] (1 − x) (1 − y) (1 − z)

= xy (1 − y) (1 − z)2 (1 − x)

⇒ x (1 − y) − 2y (1 − x) = xy

⇒ x − xy − 2y + 2xy = xy

⇒ x = 2y

⇒ x

2(1)y

=

Also,

[y (1 − z) (1 − x) − 3z (1 − x) (1 −y)] (1 − x) (1 − y) (1 − z) = 2yz (1 − z) (1 − y) (1 − x)2

⇒ y (1 − z) − 3z (1 − y) = 2yz

⇒ y − 3z = 0 ⇒ y

3z

=

∴ x

6z

=

58. The coefficients of three consecutive terms of (1 + x)n + 5

are in the ratio 5 : 10 : 14.

Then n =

58. [6]

Let the coefficients of three consecutive terms be n 5rC+ , n 5

r 1C++ and n 5

r 2C++

Given n 5rC+ : n 5

r 1C++ : n 5

r 2C++ : : 5 : 10 : 14

Solving n 5rC+ : n 5

r 1C++ : : 5 : 10

We get 3r = n + 3 …(1)

likewise solving n 5

r 1n 5

r 2

C

C

++

++

= 10

14 we get

12r = 5n + 6 …(2)

solving (1) & (2) we get n = 6.

59. A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are

removed from the pack and the sum of the numbers on the remaining cards is 1224. If the

smaller of the numbers on the removed cards is k, then k − 20 =


26

59. [5]

The sum of first n cards is, S = n(n 1)

2

+.

if two consecutive cards numbered k and k + 1 are removed then

s′ = n(n 1)

(k k 1)2

+ − + + = 1224 (given)

⇒ n(n + 1) = 2450 + 4k, which is satisfied for n = 50

and k = 25 in the given limits

∴ k = 25 and k − 20 = 5.

60. A vertical line passing through the point (h, 0) intersects the ellipse 2 2x y

14 3

+ = at the

points P and Q. Let the tangents to the ellipse at P and Q meet at the point R.

If Δ(h) = area of the triangle PQR, ( )1

maxh

1/ 2 h 1Δ = Δ

≤ ≤ and , ( )2

minh

1/ 2 h 1Δ = Δ

≤ ≤then

1 2

88

5Δ − Δ =

60. [9]

Let, line x = h

For P and Q points 2 2h y

14 3

+ =

⇒ 2

2 hy 3 1

4

⎛ ⎞= −⎜ ⎟⎜ ⎟

⎝ ⎠

2 23y (4 h )

4= −

23y 4 h

2 = ± −

Tangent at P,

23y 4 h

xh 2 14 3

⋅ −+ = . So,

4R , 0

h

⎛ ⎞ ⎜ ⎟⎝ ⎠

Δ P Q R = ( )21 4

h3 4 h2 h

⎛ ⎞⋅ ⋅ −⎜ ⎟− ⎝ ⎠ =

223 4 h

4 h2 h

− ⋅ − ⋅ = 2 3/23 (4 h )

2 h

−⋅

Δ′ = 2

2

4 h6 3 0 h

h

−− ⋅ < ∀ ⇒ Δ is decreasing

So, Δ1 = 1

h 12

1max (h) at h

2

⎛ ⎞ Δ = ⎜ ⎟⎝ ⎠≤ ≤

=

3/2

3/21

43 154

312 4

2

⎛ ⎞−⎜ ⎟ ⎛ ⎞⎝ ⎠⋅ = ⋅⎜ ⎟⎝ ⎠

Δ2 = 3/23min (h) (at h 1) 3

2 Δ = = ⋅

So, 1 2

88

5 Δ − Δ =

3/23/21 15 3

8 3 34 25

⎡ ⎤⎛ ⎞⋅ ⋅ −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

= 5 9 9

88 2

×⎡ ⎤−⎢ ⎥⎣ ⎦ = 9

R (h, 0)

23P h, 4 h

2

⎛ ⎞ −⎜ ⎟⎝ ⎠

Q23

h, 4 h2

⎛ ⎞− −⎜ ⎟⎜ ⎟

⎝ ⎠


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IIT - JEE 2013 (Advanced) - Vidyalankar Classes

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