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Vidyamandir Classes VMC/JEE Advanced-2014 1 Paper-2 JEE Advanced 2014 Paper 2 | Code 5 25 May, 2014 | 2:00 PM – 5:00 PM Answer key and solutions by Vidyamandir Classes

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JEE Advanced 2014 Paper 2 | Code 5

25 May, 2014 | 2:00 PM – 5:00 PM

Answer key and solutions by Vidyamandir Classes

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PART-A PHYSICS

1. A glass capillary tube is of the shape of a truncated cone with an apex angle α

so that its two ends have cross section of different radii. When dipped in water

vertically, water rises in it to a height h, where the radius of its cross section is b.

If the surface tension of water is S, its density is ρ , and its contact angle with

glass is θ , the value of h will be : (g is the acceleration due to gravity)

(A) ( )2S

cosb g

θ αρ

− (B) ( )2S

cosb g

θ αρ

+

(C) 2

2

Scos

b g

αθ

ρ

(D)

2

2

Scos

b g

αθ

ρ

+

Answer

(D)

0 0

2sP gh P

r− + =ρ

⇒ 2s

hr gρ

=

From triangle AOP

2

b r cosα

θ

= +

2

br

cosα

θ

=

+

⇒ 2

2

sh cos

b g

αθ

ρ

= +

2. A planet of radius R 1

10R = × (radius of Earth) has the same mass density as Earth. Scientists dig a

well of depth 5

R on it and lower a wire of the same length and of linear mass density 3 110 kgm− − into

it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in

place is : (Take the radius of Earth = 66 10 m× and the acceleration due to gravity of Earth is 210ms− )

(A) 96 N (B) 108 N (C) 120 N (D) 150 N

Answer

(B)

4

3

G xE kx

π ρ= =

dF Edxλ=∫ ∫

22

4 5

16

2 25

R

R

K RF .kxdx R

λλ

= = −

A B

O

b b

P

r r

θ

2

α

2

αθ +

O

x

dx

(centre)

5

R

4

5

R

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29

2 25

KRF

λ=

×

20 03

0

9 4 R = radius of earch10

M = Mass of earth2 25 3 10

GP Rπ− = × × ×

×

= 108 N

3. Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/2,

R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance

R from the centre of spheres 1, 2 and 3 are E1, E2 and E3 respectively, then :

(A) E1 > E2 > E3 (B) E3 > E1 > E2 (C) E2 > E1 > E3 (D) E3 > E2 > E1

Answer

(C)

1 204

QE

Rπ=

2 20

2

4

QE

Rπ=

( )3 2 230 0 0 0

4 2

43 3 8 42

3

R Q R Q QE

R RR

ρ

π ππ

= = = = ∈ ∈ ∈ ∈

Hence E2 > E1 > E3

4. If Cuλ is the wavelength of Kα X-ray line of copper (atomic number 29) and Moλ is the wavelength of

the Kα X-ray line of molybdenum (atomic number 42), then the ratio Cu Mo/λ λ is close to :

(A) 1.99 (B) 2.14 (C) 0.50 (D) 0.48

Answer

(B)

( )21

1zλ∝ −

2 2 21 42 1 41

2 141 29 1 28

cu mo

mo cu

Z.

z

λ

λ

− − = = = = − −

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5. A wire, which passes through the hold in a small bead is bent in the form of quarter of a circle.

The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of

the wire and its sides along the wire without friction. As the bead moves from A to B, the force it

applies on the wire :

(A) always radially outwards.

(B) always radially inwards.

(C) radially outwards initially and radially inwards later.

(D) radially inwards initially and radially outwards later.

Answer

(D)

After some time bead will tend to leave the wire.

6. A metal surface is illuminated by light of two different wavelength 248 mm and 310 mm. The

maximum speeds of the photoelectrons corresponding to these wavelength are 1 2andu u , respectively.

If the ratio 1 2: 2:1u u = and 1240hc eV nm= the work function of the metal is nearly

(A) 3.7 eV (B) 3.2 eV (C) 2.8 eV (D) 2.5 eV

Answer

(A)

max

hcw K

λ= +

1 1

2 2

1

44

hcw

K hc

hcKw

λλ

λ

= = −−

⇒ 1 2

44

hc hcw w

λ λ− = −

⇒ 2 1

4 13w hc

λ λ

= −

⇒ 1240 4 1

3 310 248w

= −

⇒ 3 7w . eV≃

7. A tennis ball is dropped on a horizontal smooth surface it bounces back to its original position after

hitting the surface. The force on the ball during the collision is proportional to the length of

compression of the ball. Which of the following sketches described the variation of its kinetic energy K

with time t most appropriately ? The figures are only illustrative and not to the scale.

(A) (B)

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(C) (D)

Answer

(B)

8. During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is

pressed at 40.0 cm using a standard resistance of 90Ω, as shown in the figure. The least count of the

scale used in the metre bridge is 1 mm. The unknown resistance is :

(A) 60 0 15± Ω. (B) 135 0 56± Ω. (C) 60 0 25± Ω. (D) 135 0 23± Ω.

Answer

(C)

60 90 40R× = ×

⇒ 90 40

6060

= = Ω

Let at null point length of right part is x and left is y.

90Rx y=

90y

Rx

=

R y x

R y x

∆ ∆ ∆= +

1 1

0 2560 400 600

RR .

∆= + ⇒ ∆ =

⇒ 60 0 25R .= ±

9. Parallel rays of light of intensity 2912 −=I Wm are incident on a spherical black body kept in

surroundings of temperature 300 K. Take Stefan-Boltzmann constant 8 2 45 7 10− − −= ×. Wm Kσ and

assume that the energy exchange with the surroundings is only through radiation. The final steady state

temperature of the black body is close to :

(A) 300 K (B) 660 K (C) 990 K (D) 1550

Answer

(A)

Let steady state temp is T.

Power gain = Power lost

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( ) ( ) ( )2 2 4 406 4I R R T Tπ π= −ℓ

⇒ ( )8 4 4912 4 5 7 10 300. T

−= × × −

4 4 8300 40 10T − = ×

4 8121 10T = ×

331T K=

10. A point source S is placed a the bottom of a transparent block of height 10 mm and refractive index

2.72. It is immersed in a lower refractive index liquid as shown in the figure. I is found that the light

emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of

the block. The refractive index of the liquids is :

(A) 1.21 (B) 1.30 (C) 1.36 (D) 1.42

Answer

(C)

2 2

1 2 72= =sinC

.

µ µµ

r

tanCh

=

11 54 2 5 77

0 57710 10

. / ..= = =

( )22

0 5770 5

1 0 577

.sinC .

.

=+

220 5 1 36

2 72. .

.

µµ= ⇒ =

Paragraph for Questions 11 - 12

In the figure a container is shown to have a movable (without friction) piston on top.

The container and the piston are all made of perfectly insulating material allowing no

heat transfer between outside and inside the container. The container is divided into two

compartments by a rigid partition made of a thermally conducting material that allows

slow transfer of heat. The lower compartment of the container is filled with 2 moles of

an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an

ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are

3 5

2 2V PC R, C R= = , and those for an ideal diatomic gas are

5 7

2 2V PC R, C R= = .

10mm = h

s

r

C C

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11. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved,

the final temperature of the gases will be :

(A) 550 K (B) 525 K (C) 513 K (D) 490 K

Answer

(D)

Let final temp be T.

For lower compartment process is isochoric and for upper process is isobaric

lower upperQ Q−∆ = ∆

( ) ( )2 700 2 400V PC T C T× × − = × × −

( ) ( )3 72 700 2 400

2 2

RT R T⇒ × × − = × × −

490T K.⇒ =

12. Now consider the partition to be free to move without friction so that the pressure of gases in both

compartments is the same. The total work done by the gases till the time they achieve equilibrium will

be :

(A) 250 R (B) 200 R (C) 100 R (D) 100R−

Answer

(D)

Let T be final temperature

lower upper

p p

Q Q

C ( T ) C (T ) (Both processare isobaric )

R( T ) R(T )

T K

w R( ) R( )

R R R

−∆ = ∆

× × − = × −

× − = × −

⇒ =

= − + −

= − + = −

2 700 2 400

5 72 700 2 4002 2

525

2 525 700 2 525 400

350 250 100

Paragraph for Questions 13 - 14

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross

section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes

air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun

shown, the radii of the piston and the nozzle are 20 mm and 1mm respectively. The upper end of the container

is open to the atmosphere.

13. If the piston is pushed at a speed of 15 mms− , the air comes out of the nozzle with a speed of :

(A) 10 1. ms− (B) 11 ms− (C) 12 ms− (D) 18 ms−

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Answer

(C)

AV A V

v

v m / s

=

⇒ π( ) × × = π( )

⇒ =

1 1 2 2

2 3 220 5 10 1

2

14. If the density of air is aρ and that of the liquid ρℓ , then for a given piston speed the rate (volume per

unit time) at which the liquid is sprayed will be proportional to :

(A) aρ

ρℓ (B) aρ ρℓ (C)

a

ρ

ρℓ (D) ρℓ

Answer

(A) 2 3→

For air

22 a a 0

1P V P 0

2+ ρ = +

For liquid

4 2→

20 2

10

2+ = + ℓ ℓP P Vρ

⇒ =ℓℓ

aaV V .

ρ

ρ

Paragraph for Questions 15 - 16

The figure shows a circular loop of radius a with two long parallel wires

(numbered 1 and 2) all in the plane of the paper. The distance of each

wire from the centre of the loop is d. The loop and the wires are carrying

the same current I. The current in the loop is in the counterclockwise

direction if seen from above.

15. When d a≈ but wires are not touching the loop, it is found that the net magnetic field on the axis of

the loop is zero at a height h above the loop. In that case :

(A) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h a≈

(B) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h a≈

(C) current in wire 1 and wire2 is the direction PQ and SR, respectively and 1 2h . a≈

(D) current in wire 1 and wire 2 is the direction PQ and RS, respectively and 1 2h . a=

Answer (C)

( ) ( )2

0 0

3 22 2 2 2

22

42/

Ia Ia

a h a h

µ µ πππ

× = ×+ +

1 2h . a⇒ ≃

1

2

3

4

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16. Consider d a>> , and the loop is rotated its diameter parallel to the wires by 30° from the position

shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at

its new position will be (assume that the net field due to the wires is constant over the loop)

(A) 2 2

0I a

d

µ (B)

2 20

2

I a

d

µ (C)

2 203 I a

d

µ (D)

2 203

2

I a

d

µ

Answer (B)

0 22

IB

d

µπ

= ×

( )180 30MB sinτ = −

( )2 0 12

2 2

Ia Id

µπ

π

= × ×

2 20

2

a I

d

µ=

17. Four changes Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x axis at 2= − − +x a, a, a and

2+ a, respectively. A positive charge q is placed on the positive y axis at a distance 0>b . Four options

of the signs of these charges are given in List I. The direction of the forces on the charge q 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) Q1, Q2, Q3, Q4 all positive 1. +x

(Q) Q1, Q2 positive; Q3, Q4 negative 2. −x

(R) Q1, Q4 positive; Q2, Q3 negative 3. + y

(S) Q1, Q3 positive; Q2, Q4 negative 4. −y

P Q R S P Q R S

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

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

Answer

(A)

Option – P

18. Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces

is r and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal

length in List II and select the correct answer using the code given below the lists.

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List I List II

(P)

1. 2r

(Q)

2. r/2

(R)

3. −r

(S)

4. r

P Q R S P Q R S

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

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

Answer

⇒ 1

1 1=

f R

⇒ 2

1 0 5=

.

f R

⇒ 3

1 0 5= −

.

f R

⇒ 1 1

1 1 1 2= + =

f f f R ⇒

2=

Rf

⇒ 2 2

1 1 1 1= + =

f f f R ⇒ f = R

⇒ 3 3

1 1 1 1−= + =

f f f R ⇒ = −f R

⇒ 1 3

1 1 1 0 5= + =

.

f f f R ⇒ f = 2R

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19. A block of mass 1 1m kg= another mass 2 2m kg= , are placed together (see figure) on an inclined plane

with angle of inclinationθ . Various value of θ are given in List I. The coefficient or friction between

the block 1m and the plane is always zero. The coefficient of static dynamic friction between the block

2m and the plane are equal to 0 3. .µ = In List II expressions for the friction on block 2m are given.

Match the correct expression of the friction in List II with the angles in list I, and choose the correct

option. The acceleration due to gravity is denoted by g.

List I List II

P. 5θ = ° 1. 2m g sinθ

Q. 10θ = ° 2. ( )1 2m m g sinθ+

R. 15θ = ° 3. 2m g cosµ θ

S. 20θ = ° 4. ( )1 2m m g cosµ θ+

[Useful information : ( ) ( ) ( )5 5 0 1 ; 11 5 0 2 ; 16 5 0 3tan . . tan . . tan . .° ≈ ° ≈ ° ≈ ]

Code :

(A) P – 1, Q – 1, R – 1, S – 3 (B) P – 2, Q – 2, R – 2, S – 3

(C) P – 2, Q – 2, R – 2, S – 4 (D) P – 2, Q – 2, R – 3, S – 3

Answer

(D)

For No slipping

2 2 1> +m g cos m g sin m g sinµ θ θ θ

2

3tan

µθ <

⇒ ( ) 0 2tan .θ <

⇒ 11 5.θ <

For 5P θ⇒ = ° ⇒ No slipping

For Q ⇒ ( )1 2f m m g sin θ= + Hence Option → 2

For R ⇒ 2mf g cosµ θ= ⇒ slipping.

⇒ 2f m g cosµ θ= Hence option → 3

For 20S θ ° ⇒ slipping

⇒ ( )2f m g cosµ θ= Hence Option → 3

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20. A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the

lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from

the person. In the following, state of the lift’s motion is given in List-I and the distance where the water

jet hits the floor of the lift is given in List-II. Match the statements from List-I with those in List-II and

select the correct answer using the code given below the lists.

List-I List-II

(P) Lift is accelerating vertically up 1. d = 1.2 m

(Q) Lift is acceleration vertically down with an acceleration less than

the gravitational acceleration.

2. d > 1.2 m

(R) Lift is moving vertically up with constant speed. 3. d < 1.2 m

(S) Lift is falling freely. 4. No water leaks

out of the jar

Code:

(A) P -2, Q -3, R – 2, S - 4 (B) P – 2, Q – 3, R – 1, S - 4

(C) P – 1, Q – 1, R – 1, S – 4 (D) P – 2, Q – 3, R – 1, S – 1

Answer

22h

tg a

=+

⇒ d vt=

1 24h h=

Which is independent of a

Hence 1 1P I , Q , R→ → →

For S, V = 0 (as ( )a g= −

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PART-B CHEMISTRY

21. Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the

figure.

The correct order of their boiling point is :

(A) I > II > III (B) III > II > I (C) II > III > I (D) III > I > II

Answer

(B)

III > II > I , Branching leads to decrease in B.P.

22. The major product in the following reaction is :

(A) (B) (C) (D)

Answer

(D)

23. The acidic hydrolysis of ether (X) shown below is fastest when :

(A) One phenyl group is replaced by a methyl group

(B) One phenyl group is replaced by a para-methoxyphenyl group

(C) Two phenyl groups are replaced by two para-methoxyphenyl group

(D) No structural change is made to X

Answer

(C)

Methoxy group stabilizes the carbocation.

24. Hydrogen peroxide in its reaction with KIO4 and NH2OH respectively, is acting as a :

(A) reducing agent, oxidising agent (B) reducing agent, reducing agent

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(C) oxidising agent, oxidising agent (D) oxidising agent, reducing agent

Answer

(A)

2 2 4 2 2(Re ducing Agent)

H O IO O I+ → +Θ

1

2 2 2 2 2(Oxidisin g Agent)

H O N H OH H O N−

+ → +

25. The product formed in the reaction of SOCl2 with white phosphorus is :

(A) PCl3 (B) SO2Cl2 (C) SCl2 (D) POCl3

Answer

(A)

4 2 3 2 2 2P 8SOCl 4PCl 4SO 4S Cl+ + → + +

26. For the identification of β -naphthol using dye test, it is necessary to use :

(A) Dichloromethane solution of β -naphthol

(B) Acidic solution of β -naphthol

(C) Neutral solution of β -naphthol

(D) Alkaline solution of β -naphthol

Answer

(D)

27. For process 2 2H O( ) H O(g)→ℓ at T 100 C= ° and 1 atmosphere pressure, the correct choice is :

(A) system surroundingsS 0 and S 0∆ > ∆ > (B) system surroundingsS 0 and S 0∆ > ∆ <

(C) system surroundingsS 0 and S 0∆ < ∆ > (D) system surroundingsS 0 and S 0∆ < ∆ <

Answer

(B)

systemS ↑ because 2H O change its state from liquid to gas and surroundingS ↓ because heat is supplied to

the system from the surroundings.

⇒ Heat is lost by the surrounding.

28. For the elementary reaction M N→ , the rate of disappearance of M increases by a factor of 8 upon

doubling the concentration of M. The order of the reaction with respect to M is :

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

Answer

(B)

d[M] d[N]r

dt dt= = + ,

3

3

as [M] 2[M] r [M]

r k[M]them r 8r

→ ⇒ ∝⇒ =→

⇒ Order of reaction w.r.t M = 3

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29. Under ambient conditions, the total number of gases released as products in the final step of the

reaction scheme shown below is :

(A) 0 (B) 1 (C) 2 (D) 3

Answer

(C) _

2 23H O OH / H O disproportionation 46 3 4 6 2

gas gas

XeF 6HF XeO HXeO XeO Xe O− −→ + → → + +

30. Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is :

(A) Be2 (B) B2 (C) C2 (D) N2

Answer

(C)

If 2s-2p mixing is NOT operative them even the specises with 14 or less 14 electrons will have the normal

molecular orbital energy levels as shown below. Clearly, Only C2 among the given options will be

paramagnetic if the following diagram is followed for fill electrons.

Paragraph for Questions 31 - 32

Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major products

formed in each step for both the schemes.

31. The product X is :

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(A) (B) (C) (D)

Answer

32. The correct statement with respect to product Y is :

(A) It gives a positive Tollens test and is a functional isomer of X

(B) It gives a positive Tollens test and is a geometrical isomer of X

(C) It gives a positive iodoform test and is a functional isomer of X

(D) It gives a positive iodoform test and is a geometrical isomer of X

Answer

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Paragraph for Question 33 - 34

An aqueous solution of metal ion M1 reacts separately with regents Q and R in excess to give tetrahedral and

square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral

complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate

which dissolves in express of S. The reactions are summarized in the scheme given below :

SCHEME :

33. M1, Q and R respectively are :

(A) 2Zn , KCN and HCl+ (B) 2Ni , HCl and KCN+

(C) 2Cd , KCN and HCl+ (D) 2Co , HCl and KCN+

34. Reagent S is :

(A) ( )4 6K Fe CN (B) 2 4Na HPO (C) 2 4K CrO (D) KOH

Answer

Paragraph for Question 35 - 36

X and Y are two volatile liquids with molar weights of 10 1g mol− and 140 g mol− respectively. Two cotton

plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length

L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a

temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm

from the plug soaked in X. Take X and Y to have equal molecular diameter and assume ideal behaviour for the

inert gas and the two vapours.

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35. The value of d in cm (shown in the figure), as estimated from Graham’s law is :

(A) 8 (B) 12 (C) 16 (D) 20

36. The experimental value of d is found to be smaller than the estimate obtained using Graham’s law.

This is due to :

(A) Larger mean free path for X as compared to that of Y

(B) Larger mean free path for Y as compared to that of X

(C) Increased collision frequency of Y with the inert gas as compared to that of X with the inert gas

(D) Increased collision frequency of X with the inert gas as compared to that of Y with the inert gas

Answer

37. Different possible thermal decomposition pathways for peroxyesters are shown below. Match each

pathway from List 1 with an appropriate structure from List 2 and select the correct answer using the

code given below the lists.

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List I List II

(P) Pathway P 1.

(Q) Pathway Q 2.

(R) Pathway R 3.

(S) Pathway S 4.

P Q R S P Q R S

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

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

Answer

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38. Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction schemes

(I, II, III, IV) provided in List II and select the correct answer using the code given below the lists :

List I List II

(P) 1 Scheme I

4 2

2 3

(i) KMnO ,HO ,heat (ii) H ,H O(iii) SOCl (iv) NH

7 6 2 3? C H N O

− +

(Q) 2 Scheme II

3 2 4

3 2 4

(i) Sn / HCl (ii) CH COCl (iii) conc. H SO

(iv) HNO (v) dil. H SO ,heat (vi) HO

6 6 2 2? C H N O−

(R) 3 Scheme III

3 2 4

2 3 2 2 4

(i) red hot iron, 873 K (ii) fu min g HNO , H SO ,heat(iii) H S.NH (iv) NaNO , H SO (v) hydrolysis

6 5 3? C H NO→

(S) 4 Scheme IV

2 4

3 2 4 2 4

(i) conc. H SO , 60 C(ii) conc. HNO ,conc.H SO (iii) dil. H SO ,heat

6 5 4? C H NO

°

P Q R S P Q R S

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

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

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Answer

39. Match each coordination compound in List-I with an appropriate pair of characteristics from List-II

and select the correct answer using the code given below the lists.

2 2 2 2en H NCH CH NH ;= atomic numbers : Ti 22;Cr 24;Co 27;Pt 78= = = =

List I List II

(P) 3 4 2[Cr(NH ) Cl ]Cl 1 Paramagnetic and exhibits ionization isomerism

(Q) 2 5 3 2[Ti(H O) Cl](NO ) 2 Diamagnetic and exhibits cis-trans isomerism

(R) 3 3[Pt(en)(NH )Cl]NO 3 Paramagnetic and exhibits cis-trans isomerism

(S) 3 4 3 2 3[Co(NH ) (NO ) ]NO 4 Diamagnetic and exhibits ionization isomerism

P Q R S P Q R S

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

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

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Answer

40. Match the orbital overlap figures shown in List-I with the description given in List-II and select the

correct answer using the code given below the lists.

List I List II

(P) 1. p – d π antibonding

(Q) 2. d – d σ bonding

(R) 3. p – d π bonding

(S) 4. d – d σ antibonding

P Q R S P Q R S

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

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

Answer

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PART-C MATHEMATICS

41. Coefficient of 11x in the expansion of ( ) ( ) ( )4 7 122 3 41 1 1+ + +x x x is :

(A) 1051 (B) 1106 (C) 1113 (D) 1120

Answer

42. Let f : [ ]0 2, → ℝ be a function which is continuous on [0, 2] and is differentiable on (0, 2) with

( )0 1f = . Let ( )2

0

x

F x f t dt= ∫ for [ ]0 2x ,∈ . If ( ) ( )F x f x′ ′= for all ( )0 2x ,∈ , then F(2) equals :

(A) 2 1e − (B) 4 1e − (C) 1e − (D) 4e

Answer

43. The function ( )y f x= is the solution of the differential equation

4

2 2

2

1 1

dy xy x x

dx x x

++ =

− −

in ( )1 1,− satisfying ( )0 0f = . Then : ( )

3

2

3

2

f x dx

∫ is :

(A) 3

3 2

π− (B)

3

3 4

π− (C)

3

6 4

π− (D)

3

6 2

π−

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Answer

44. The following integral ( )2

17

4

2cosec x dx

π

π∫ is equal to :

(A) ( )( ) 161 2

02

logu ue e du

+ −+∫ (B) ( )( ) 171 2

0

logu ue e du

+ −+∫

(C) ( )( ) 171 2

0

logu ue e du

+ −−∫ (D) ( )( ) 161 2

02

logu ue e du

+ −−∫

Answer

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45. The quadratic equation ( ) 0=p x with real coefficient has purely imaginary roots. Then the equation

( )( ) 0=p p x has :

(A) only purely imaginary roots

(B) all real roots

(C) two real and two purely imaginary roots

(D) neither real nor purely imaginary roots

Answer

46. For ( )0x ,π∈ , the equation 2 2 3 3sin x sin x sin x+ − = has :

(A) infinity many solutions (B) three solutions

(C) one solution (D) no solution

Answer

47. In a triangle the sum of two sides is x and the product of the same two sides is y. If 2 2x c y− = , where

c is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is :

(A) ( )3

2

y

x x c+ (B)

( )3

2

y

c x c+ (C)

( )3

4

y

x x c+ (D)

( )3

4

y

c x c+

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Answer

48. Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl

is at least one more than the number of girls ahead of her, is :

(A) 1

2 (B)

1

3 (C)

2

3 (D)

3

4

Answer

49. The common tangents to the circle 2 2 2x y+ = and the parabola 2 8y x= touch the circle at the points

P,Q and the parabola at the points R, S . Then the area of the quadrilateral PQRS is :

(A) 3 (B) 6 (C) 9 (D) 15

Answer

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50. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that

each envelope contains exactly one card and no card is placed in the envelope bearing the same number

and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways

it can be done is :

(A) 264 (B) 265 (C) 53 (D) 67

Answer

Paragraph for Questions 51 - 52

Let a, r, s, t be nonzero real numbers. Let P (at2, 2at), Q, R (ar

2, 2ar) and S (as

2, 2as) be distinct points on the

parabola y2 = 4ax. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point

(2a, 0).

51. The value of r is :

(A) 1

t− (B)

2 1t

t

+ (C)

1

t (D)

2 1t

t

Answer

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52. If st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is.

(A) ( )22

3

1

2

t

t

+ (B)

( )22

3

1

2

a t

t

+ (C)

( )22

3

1a t

t

+ (D)

( )22

3

2a t

t

+

Answer

Paragraph for Questions 53 - 54

Given that for each ( ) ( )1

1

0

0 1 1

haa

h h

a , , lim t t dt+

−−−

→∈ −∫ exists. Let this limit be g (a). In addition, it is given that

the function g (a) is differentiable on (0, 1).

53. The value of 1

2g

is :

(A) π (B) 2π (C) 2

π (D)

4

π

Answer

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54. The value of 1

2g'

is :

(A) 2

π (B) π (C)

2

π− (D) 0

Answer

Paragraph for Questions 55 - 56

Box 1 contains three cards bearing numbers 1, 2, 3 ; box 2 contains five cards bearing number 1, 2, 3, 4, 5 ; and

box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let ix

be the number on the card drawn from the thi box, 1 2 3i , ,= .

55. The probability that 1 2 3x x x+ + is odd, is :

(A) 29

105 (B)

53

105 (C)

57

105 (D)

1

2

Answer

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56. The probability that 1 2 3x , x , x are in an arithmetic progression, is :

(A) 9

105 (B)

10

105 (C)

11

105 (D)

7

105

Answer

57. Let 2 2

1 2 910 10

k

k kZ cos i sin ; k , ,.....,

π π = + =

.

List I List II

(P) For each zk there exists a zj such that 1⋅ =k jz z 1. True

(Q) There exists 1 2 9∈a k , ,....., such that 1 ⋅ = kz z z has no solution z in

the set of complex numbers

2. False

(R) 1 2 91 1 1

10

− − −z z .... z equals

3. 1

(S) 9

1

21

10=

∑k

kcos equals

4. 2

P Q R S P Q R S

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

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

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Answer

58. MATCH THE FOLLOWING LISTS :

List I List II

(P) The number of polynomials f (x) with non-negative integer

coefficients of degree 2≤ , satisfying ( ) ( )1

00 0 and 1f f x dx= =∫ , is

1. 8

(Q) The number of points in the interval 13 13, − at which

f (x) = sin (x2) + cos (x

2) attains its maximum value, is

2. 2

(R)

( )2

2

2

3

1 x

xdx

e− +∫ equals

3. 4

(S) 1

21

2

1

20

12

1

12

1

xcos x log dx

x

xcos x log dx

x

+ − + −

∫equals

4. 0

P Q R S P Q R S

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

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

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Answer

59. MATCH THE FOLLOWING LISTS :

List I List II

(P) Let ( ) ( ) [ ]1 3

3 1 12

y x cos cos x , x , , x−= ∈ − ≠± . Then

( ) ( ) ( ) ( )22

2

11

d y x dy xx x

y x dxdx

− +

equals

1. 1

(Q) Let A1, A2, …, An (n > 2) be the vertices of a angular polygon on

n sides with its centre at the origin. Let ka

be the position vector

of the point Ak, k = 1, 2,…,n. If

( ) ( )1 1

1 1

1 1k k k k

n na a a · a

k k+ +

− −× =

= =∑ ∑

, then the minimum

value of n is

2. 2

(R) If the normal from the point P (h, 1) on the ellipse

2 2

16 3

x y+ =

is perpendicular to the line x + y = 8, then the value of h is

3. 8

(S) Number of positive solutions satisfying the equation

1 1 1

2

1 1 2

2 1 4 1tan tan tan

x x x

− − − + = + +

4. 9

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P Q R S P Q R S

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

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

Answer

60. Let [ )1 2 3: : 0 :f , f , , f→ ∞ → →ℝ ℝ ℝ ℝ ℝ and [ )4 : 0f ,→ ∞ℝ be defined by :

( )1

if 0

if 0x

x xf x

e x

<=

≥ ; ( ) 2

2f x x= ;

( )3

if 0

if 0

sin x xf x

x x

<=

≥ and ( )

( )( )( )( )

2 1

4

2 1

if 0

1 if 0

f f x xf x

f f x x

<=

− ≥

List I List II

(P) 4f is 1. Onto but not one-one

(Q) 3f is 2. Neither continuous nor one-one

(R) 2 1f of is 3. Differentiable but not one-one

(S) 2f is 4. Continuous and one-one

Codes :

P Q R S P Q R S

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

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

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Answer

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Q.No. 1 D Q.No.21 B Q.No.41 C

Q.No. 2 B Q.No.22 D Q.No.42 B

Q.No. 3 C Q.No.23 C Q.No.43 B

Q.No. 4 B Q.No.24 A Q.No.44 A

Q.No. 5 D Q.No.25 A Q.No.45 D

Q.No. 6 A Q.No.26 D Q.No.46 D

Q.No. 7 B Q.No.27 B Q.No.47 B

Q.No. 8 C Q.No.28 B Q.No.48 A

Q.No. 9 A Q.No.29 C Q.No.49 D

Q.No. 10 C Q.No.30 C Q.No.50 C

Q.No. 11 D Q.No.31 A Q.No.51 D

Q.No. 12 D Q.No.32 C Q.No.52 B

Q.No. 13 C Q.No.33 B Q.No.53 A

Q.No. 14 A Q.No.34 D Q.No.54 D

Q.No. 15 C Q.No.35 C Q.No.55 B

Q.No. 16 B Q.No.36 D Q.No.56 C

Q.No. 17 A Q.No.37 A Q.No.57 C

Q.No. 18 B Q.No.38 C Q.No.58 D

Q.No. 19 D Q.No.39 B Q.No.59 A

Q.No. 20 C Q.No.40 C Q.No.60 D

JEE Advanced 2014

Paper 225 May 2014 | 2:00PM - 5:00 PM

Code - 5

Answer keyPHYSICS CHEMISTRY MATHS