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JEE ADVANCED: 2014 (PAPER – I)

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PAPER -1 : PHYSIC, CHEMISTY & MATHEMATICS

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C. Question Paper Format

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

Section – 1 contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE

OR MORE THAN ONE are correct.

Section – 2 contains 10 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9 (both

inclusive).

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answer and zero mark if no bubble is darkened. No negative marks will be awarded for incorrect answer in this section.

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PART-I: PHYSICS Section – 1 : (One Or More Than One Options Correct Type)

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

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

1. Let E1(r), E2(r) and E3(r) be the respective electric fields at a distance r from a point charge Q, an

infinitely long wire with constant linear charge density , and an infinite plane with uniform

surface charge density . If E1(r0) = E2(r0) = E3(r0) at a given distance r0, then

(A) 2

0Q 4 r (B) 0r

2

(C) E1(r0/2)=2E2(r0/2) (D) E2(r0/2)=4E3(r0/2)

Sol. 1 2

0

1 QE (r)

4 r

2

0

1 2E

4 r

,

3

0

E2

2

0 0 0 0 0

Q 2 2

4 r 4 r 4

2

0Q 2 r (i)

0r

(ii)

01 2 2

0 0 o0

r 1 Q QE ( )

2 4 rr

2

Also, 02

0 0 0 0

r 2 2 22E

2 2 r r

C is correct

02

0 0

rE

2 r

03

0 0 0 0 0

r 4 2 24E

2 2 r r

only C is correct

2. Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to

raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two

wires of the same material, each of length L and diameter 2d. The way these wires are connected

is given in the options. How much time in minutes will it take to raise the temperature of the

same amount of water by 40 K ?

(A) 4 if wires are in parallel (B) 2 if wires are in series

(C) 1 if wires are in series (D) 0.5 if wires are in parallel

Sol. Resistance of single wire

L

d

L

d

3

1 2

4 LR

d

Resistance of New Single wire

L

2d

1

2

R4 LR '

4d 4

series

12

RR

2

parallel

12

RR

8

Initially 2

1

V4 mS

R … (i)

Finally 2

2

2

Vt mS Q

R

2

1 2

t4

R R

Putting s

1 22

1 1

R 2t4R

2 R R

t2 = 2 (B)

Putting 12,P

RR ,

8

t2 = 0.5 (D)

3. A transparent thin film of uniform thickness and refractive index n1=1.4 is coated on the convex

spherical surface of radius R at one end of a long solid glass cylinder of refractive index n2=1.5,

as shown in figure. Rays of light parallel to the axis of the cylinder traversing through the film

from air to glass get focused at distance f1 from the film, while rays of light traversing from glass

to air get focused at distance f2 from the film. Then

Air

(A) |f1| = 3R

(B) |f1| = 2.8 R

(C) |f2| = 2R

(D) |f2| = 1.4 R

Sol.

4

n1

n2

n0

For IInd focus

1 02 2 1

1

n nn n n

f R R

1

1.5 0.4 0.1

f R R

1f 3R

For f2

2

1 0.4 0.1 0.5

f R R R

f2 = -2 R

|f2| = 2A

So, A & C are correct.

4. A student is performing an experiment using a resonance column and a tuning fork of frequency

244 s-1

. He is told that the air in the tube has been replaced by another gas (assume that the

column remains filled with the gas). If the minimum height at which resonance occurs is (0.350

0.005)m, the gas in the tube is

(Useful information : 1/ 2 1/ 2 1/ 2 1/ 2167RT 640J mole ; 140RT 590J mole . The molar masses M in

grams are given in the options. Take the values of 10

M for each gas as given there.)

(A) Neon (M = 20, 10 7

20 10 ) (B) Nitrogen (M = 28,

10 3

28 5 )

(C) Oxygen (M = 32, 10 9

32 16 ) (D) Argon (M = 36,

10 17

36 32 )

Sol. (0.35 0.005)4

(0.35 0.005)

RT

v 244 4(0.35 0.005)M

= (346.48, 336.72)

m D1.67, 1.4

m

167RT 640 64 10 64 10v

100M M10 M M 10 10

D

140RT 590 59 10 59 10v

100M M10 M M 10 10

5

10 17

32M

64 7

1010

64 3

510

kg

1 101000

M10

10

64 10M

(i) 7

64 10 44810

(ii) 17

64 10 34032

10 3

59 10 59 10 354M 5

Only Argon gas is satisfying the equation. Hence D is correct option.

5. In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium

making an angle with the horizontal floor. The coefficient of friction between the wall and the

ladder is 1 and that between the floor and the ladder is 2 . The normal reaction of the wall on

the ladder is N1 and that of the floor is N2. If the ladder is about to slip, then

(A) 1 2 2

mg0 0 and N tan

2

(B) 1 2 1

mg0 0 and N tan

2

(C) 1 2 2

1 2

mg0 0 and N

1

(D) 1 2 1

mg0 0 and N tan

2

Sol. If 1 20, 0

2 1 1N N mg

2 2 1N N

2 1 2 2N N mg

2

1 2

mgN

(1 )

If 1 20, 0; Taking torque about A

1

mgN sin cos

2

mg

N2

N1B

A

6

1

mgN tan

2

So, C & D

6. Two ideal batteries of emf V1 and V2 and three resistances R1, R2 and R3 are connected as shown

in the figure. The current in resistance R2 would be zero if

(A) V1 = V2 and R1 = R2 = R3 (B) V1 = V2 and R1 = 2R2 = R3

(C) V1 = 2V2 and 2R1 = 2R2 = R3 (D) 2V1 = V2 and 2R1 = R2 = R3

Sol. If current in R2 is zero, we can remake the figure as

V 1

A B

R 3

R 1

V2

R2

V 1

A B

R 3

R 1

V2

And VA = VB

1 2

1 2

V VI

(R R )

Also, On

VA – VB = -V1 + IR1

VA – VB =0 V1 = IR1 and VA – VB = V2 –IR3

VA – VB = 0 V2 = IR3

1 1

2 3

V R

V R

If R1 = R3, V1 = V2 and 2RI 0 for all values of R2

(A) and B are Correct

If 1 1

2 3

V R2 2

V R

R1 = 2R3

If 1 1

2 3

V R1 1

V 2 R 2

V 1

V A B

V 2

R 3

R 1

I

B

I

7

2R1=R3

(D) is also correct

A, B and D are correct

7. A light source, which emits two wavelengths 1 400nm and

2 600nm , is used in a Young’s

double slit experiment. If recorded fringe widths for 1 and

2 are 1 and

2 and the number of

fringes for them within a distance y on one side of the central maximum are m1 and m2,

respectively, then

(A) 2 1 (B) m1 > m2

(C) From the central maximum, 3rd

maximum of 2 overlaps with 5

th minimum of

1

(D) The angular separation of fringes for 1 is greater than

2

Sol.

2 1

Since the fringe width of 2 is more than

1 so number of fringes of 1 will be more than

2 in a

given separation.

So, M1 > M2

Position of 3rd

max. of 2 is given by

3

D Dy 3 600 1800

d d .

Position of 5th

minima of 1

5

1 D Dy 5 400 1800

2 d d

y3 = y5, So (C) is also correct,

A, B, C are correct.

8. At time t =0, terminal A in the circuit shown in the figure is connected to B by a key an a

alternating current I(t) = Iocos (t), with Io = 1A an = 500 ra s-1

starts flowing in it with the

initial direction shown in the figure. At 7

t ,6

the key is switched from B to D. Now onwards

only A and are connected. A total charge Q flows from the battery to charge the capacitor fully.

If C = 20F, R = 10 and the battery is ideal with emf of 50 V, identify the correct statement

(s).

(A) Magnitude of the maximum charge on the capacitor before 37t is1 10 C

6

.

(B) The current in the left part of the circuit just before 7

t6

is clockwise.

(C) Immediately after A is connected to D, the current in R is 10A.

(D) 3Q 2 10 C.

8

Ans. C, D

Sol: oQ i dt i cos t dt

oiQ sin t C

at t 0, Q 0

C = 0

oiQ sin t

at t ,2

Q will be maximum

3omax

i 1Q 2 10 C

500

(hence (A) is incorrect)

Since i = io cos t

after t2

, cos t becomes “–ve”

current changes its direction i.e. anticlockwise [hence (B) is incorrect]

7

at t6

3oi 7Q sin 1000 C 1 10 C

2

P.D. across capacitor at this instant V = -50 V

50 50

i 10A10

[hence (C) is correct]

Find charge on the capacitor Q1 = CV

Q1 = 50 × 20 × 10

–6 = 1 × 10

–3 C

initial charged was Q = –1 × 10–3

C

So charge given by battery Q = 2 × 10–3

C [hence (D) is correct]

9. One end of a taut string of length 3m along the x axis is fixed at x = 0. The speed of the waves in

the string is 100 ms-1

. The other end of the string is vibrating in the y direction so that stationary

waves are set up in the strring. The possible waveforem (s) of these stationary waves is (are)

(A) x 50 t

y t Asin cos6 3

(B)

x ty t Asin cos

3 3

(C) x t

y t Asin cos6 3

(D)

xy t Asin cos t

6

Ans. A, C, D

Sol: Since at x = 3, antinode should be there, only B does not satisfy the third condition

[hence A, C, D are correct options]

10. A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that

covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor

is C while that of the portion with dielectric in between in C1. When the capacitor is charged, the

plate area covered by the dielectric gets charge Q1 and the rest of the area gets charge Q2. The

9

electric field in the dielectric is E1 and that in the other portion is E2. Choose the correct

option/options, ignoring edge effect.

(A) 1

2

E1

E (B) 1

2

E 1

E K (C) 1

2

Q 3

Q K (D)

1

C 2 K

C K

Ans. A, D

Sol: K

A / 3

2A / 3

o1

A KC

3d

o2

2AC

3d

o1 2

AC C C 2 k

3d

1

C 2 K

C 2

[Hence (D) is correct]

In Parallel ‘V’ is same hence E1 = E2 [(A), is correct]

1 1

2 2

Q C K

Q C 2 [(C) is incorrect]

Section -2 : (One Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one int3eger

from 0 to 9 (both inclusive).

11. A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990

resistance, it can be converted into a voltmeter of range 0 – 30 V. If connected to a 2n

249

resistance, it becomes an ammeter of range 0 – 1.5 A. The value of n is

Ans. 5

10

Sol:

G

V

RgI

G

RI gI

gI I S

V= Ig × (G + R) g

g

IS G

I I

30 = 0.006 × [G + 4990] ….(1) 2n 0.006

G249 1.5 0.006

…(2)

By solving (1) & (2)

n = 5

12. To find the distance d over which a single can be seem clearly in foggy conditions, a railways

engineer uses dimensional analysis and assumes that the distance depends on the mass density

of the fog, intensity (power/area) S of the light from the signal and its frequency f. The engineer

finds that d is proportional to S1/n

. The value n is

Ans. 3

Sol: a b cd s f

a b cd k s f

a b C

3 3 1L ML MT T

O 1 o a b 3a 3b cM LT M L T

a b 0 …(1)

3a 1 …(2)

By (1) & (2) 1

a3

1

b3

Hence n = 3

13. During Searle’s experiment, zero of the Vernier scale lies between 3.20 ×10–2

m and 3.25 × 102

m of the main scale. The 20th division of the Vernier scale exactly coincides with one of the

main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the

Vernier scale still lies between 3.20 × 10-2 m and 3.25 × 10_2 m of the main scale but now the

45th division of Vernier scale coincides with one of the main scale divisions. The length of the

thin metallic wire is 2 m and its cross-sectional area is 8 × 10–7

m2. The least count of the Vernier

scale is 1.0 × 10–5

m. The maximum percentage error in the Young’s modulus of the wire is

Ans. 4

Sol: 2 5 2

1L 3.20 10 20 10 3.220 10 m

2 5 2

2L 3.20 10 45 10 3.245 10 m

Elongation l = 0.025 × 10–2

m

F.L

YlA

11

Y l

100 100Y l

= 5

2

1 10100

0.025 10

= 4%

14. A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis.

Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are attached to the

platform at a distance 0.25 m from the centre on its either sides along its diameter (see figure).

Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in

opposite directions. After leaving the platform, the balls have horizontal speed of 9 ms–1

with

respect to the ground. The rotational speed of the platform in rad s–1

after the balls leave the

platform is

Ans. 4

Sol: mvr mvr I 0

2mvr I

20.45 0.5

2 0.05 9 0.252

4rad / s 14. A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to

rotate about its axis. Two massless spring toy-guns, each carrying a steel ball

of mass 0.05 kg are attached to the platform at a distance 0.25 m from the

centre on its either sides along its diameter (see figure). Each gun

simultaneously fires the balls horizontally and perpendicular to the diameter

in opposite directions. After leaving the platform, the balls have horizontal

speed of 9 m/s with respect to the ground. The rotational speed of the

platform in rad/s after the balls leave the platform is

Key: (4)

Sol. From conservation of angular momentum about the central axis

mvr + mvr + I 0

2mvr = I

20.45 (0.5)

2 0.05 9 0.252

4 .

12

15. A uniform circular disc of mass 1.5 kg and radius 0 . 5 m is

initially at rest on a horizontal frictionless surface. Three forces of

equal magnitude F = 0 . 5 N are applied simultaneously along

the three sides of an equilateral triangle XYZ with its vertices on

the perimeter of the disc (see figure). One second after applying

the forces, the angular speed of the disc in rad/s is

F

F

F

X

Y Z

O

Key. (2)

Sol. Given OZ = R

In OZA, oOAsin 30

R

OA = R

2

Torque on the disc about O due to all forces is

= OA F 3 = (R/2)3F

F

F

F

X

Y Z

O

A 30°

= I

=

2

R / 2 3F 3F 3 0.52

MRI MR 1.5 .5

2

= t = 2 1 = 2 rad/s

16. Consider an elliptically shaped rail PQ in the vertical plane with

OP = 3 m and OQ = 4m, A block of mass 1 kg is pulled along

the rail from P to Q with a force of 18 N, which is always

parallel to line PQ (see the figure given). Assuming no frictional

losses, the kinetic energy of the block when it reaches Q is

(n 10) Joules. The value of n is (take acceleration due to

gravity = 10 m/s2)

3 m O P

Q

4 m

90o

ds F

Key: (5)

Sol: FW F.ds

Let ds ds ds where

ds is component of ds along F and

ds is component of ds to F

W = F. ds ds F.ds F.ds F ds

= F PQ 18 5 90 J

Now E = Work done by external agent

FK U W 90

K 90 U 90 mgh 90 1 10 4 50 J

Kinetic energy at point Q = 50 J = n 10

n = 5.

13

17. A thermodynamic system is taken from an initial slate i with internal

energy Ui = 100 J to the final state f along two different paths iaf and ibf,

as schematically shown in the figure. The work done by the system along

the paths af, ib and bf are Waf= 200J, Wib =50 J and Wbf = 100 J

respectively. The heat supplied to the system along the path iaf, ib and bf

are Qiaf, Qib and Qbf respectively. If the internal energy of the system in

the state b is Ub = 200 J and Qiaf, = 500J. the ratio Qbf/Qib is

P

i

V

a f

b

Key: (2)

Sol: For process ib For process iaf

IB IBQ U W ib

Q U W iaf iaf iaf

Q 100 50 ib 500 (U 100) 200 f

Q 150 ib U 100 500 f

U 400f

For process bf

Q W U bf bf bf

= 100 + 200

= 300

Ratio Q 300

2Q 150

bf

ib

18. Two parallel wires in the plane of the paper are distance Xo apart. A point charge is moving with speed u

between the wires in the same plane at a distance X1 from one of the wires When the wires carry current of

magnitude I in the same direction, the radius of curvature of the path of the point charge is R1. In contrast, if

the currents I in the two wires have directions opposite to each other, the radius of curvature of the path is

R2. if o

1

X3

X , the value of 1

2

R

R is

Key: (3)

Sol:

In case 1 Let Magnetic field be B1

B1 = 0 10 0 0 0

1 0 1 1 0 1 1 0 1

2I I I I1 1

2 2 ( ) 2 2

x x

x x x x x x x x x

In case 2 Let Magnetic field be B2

B2 =

0 0 0 0

1 0 1 1 0 1

I I I x

2 2 ( ) 2

x x x x x x

Since, R mv

qB

14

1 0 10 01 2

12 1 0 1 0 11 0 1

0

R B 1

R B 2 21 2

x x xx x

xx x x xx x x

x

1 3

31 1

1 23

19. Airplanes A and B are flying with constant velocity in the same

vertical plane at angles 30o and 60

o with respect to the horizontal

respectively as shown in figure The speed of A is 100 3 m/s. At

time t = 0 s, an observer in A finds B at a distance of 500 m. This

observer sees B moving with a constant velocity perpendicular to

the line of motion of A. If at t = to, A just escapes being hit by B,

to in seconds is

B A

30o

60o

Key. (5)

Sol. o o

Aˆ ˆv 100 3 cos30 i 100 3sin30 j

= 3 1ˆ ˆ ˆ ˆ100 3 i 100 3 j 150i 50 3 j

2 2

o o BB B B B

v 3ˆ ˆ ˆ ˆv v cos60 i v sin 60 j i v j2 2

B BBA B A

v vˆ ˆv v v 150 i 3 50 J2 2

Given BA A BA Av v v .v 0

B Bv v150 150 50 3. 3 50 0

2 2

75vB – (150)2 + 75vB – 150 50 = 0 150vB = (150)[(150) + 50] = 150 200

vB = 200 m/s

Putting vB = 200 m/s BA

00 200ˆ ˆ ˆ ˆv 150 i 3 50 j 50i 50 3 j2 2

22 2

BA| v | 50 50 3 50 1 3 100m / s

As BAv = constant

airplane B is moving along a straight line w.r.t. airplane A.

Motion of B w.r.t. A

x

y

B

100 m/s 500m

A

time after which B just grazes A is to = 500/100 = 5 s.

15

20. A rocket is moving in a gravity free space with a constant

acceleration of 2 m/s along +x direction (see figure). The length of

a chamber inside the rocket is 4 m. A ball is thrown from the left

end of the chamber in + x direct ton with a speed of 0.3 m/s

relative to the rocket. At the same time, another ball is thrown in

x direction with a speed of 0.2 m/s from its right end relative to

the rocket. The time in seconds when the two balls hit each other is

0.3 m/s

0.2 m/s

a = 2 m/s2

x

4m

Key: (8)

Sol: From the reference frame of the rocket the magnitude of the relative velocity of one of the ball

w.r.t. other is (0.2 + 0.3) = 05 m/s.

So time taken = Relative separation

Relative veloc

4

ty 5i 0. 8 m/s.

PART - II: CHEMISTRY

Section – 1 : (One Or More Than One Options Correct Type)

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

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

21. Upon heating with Cu2S, the reagent(s) that give copper metal is/are

(A) CuFeS2 (B) CuO

(C) Cu2O (D) CuSO4

Key. (C)

Sol. Cu2S + 2Cu2O 6Cu(s) +SO2(g)

22. In a galvanic cell, the salt bridge

(A) does not participate chemically in the cell reaction

(B) stops the diffusion of ions from one electrode to another.

(C) is necessary for the occurrence of the cell reaction

(D) ensure mixing of the two electrolytic solutions.

Key. (A)

Sol. Conceptual

23. Hydrogen bonding plays a central role in the following phenomena:

(A) Ice floats in water.

(B) Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions

(C) Formic acid is more acidic than acetic acid

(D) Dimerisation of acetic acid in benzene.

Key. (ABD)

Sol. Ice has lower density due to the H-bonding, giving it open cage like structures.

3R N H

will have greater hydration than 3R N H

in general due to larger extent of H-bonding in

former.

O H O

C

OHO

CH3CCH3

+

Dimer of acetic acid in benzene

16

24. The correct combination of name for isomeric alcohols with molecular formula C4H10O is/are

(A) tert-butanol and 2-methylpropan-2-ol

(B) tert-butanol and 1, 1-dimethylethan-1-ol

(C) n-butanol and butan-1-ol

(D) isobutyl alcohol and 2-methylpropan-1-ol

Key. (ACD)

Sol. Conceptual

25. The reactivity of compound Z with different halogens under appropriate conditions is given

below:

OH

C(CH 3)3

mono halo substituted derivative when X2 = I

2

di halo substituted derivative when X2 = Br

2

X2

Z

tri halo substituted derivative when X2 = Cl

2

The observed pattern of electrophilic substitution can be explained by

(A) the steric effect of the halogen

(B) the steric effect of the tert-butyl group

(C) the electronic effect of the phenolic group

(D) the electronic effect of the tert-butyl group

Key. (ABC)

Sol.

OH

C(CH3)3

OH

C(CH3)3

I

EAS

When X2= I2

X2= Br - Br

EAS

OH

C(CH3)3

Br

Br

X2= Cl2

OH

C(CH3)3

Cl Cl

Cl

17

26. The pair (s) of reagents that yield paramagnetic species is/are

(A) Na and excess of NH3 (B) K and excess of O2

(C) Cu and dilute HNO3 (D) O2 and 2-ethylanthraquinol

Key. (BC)

Sol. The colour of the dilute solution is blue and it is paramagnetic due to the presence of unpaired

electrons but in excess of NH3, blue colour changes to bronze colour and becomes diamagnetic.

K in excess of O2 gives superoxide i.e. KO2 which is paramagnetic

Cu in dil. HNO3 gives NO and NO is paramagnetic

27. For the reaction:

3 2 4 4 2I ClO H SO Cl HSO I

The correct statement(s) in the balanced equation is/are

(A) Stoichiometric coefficient of 4HSO is 6. (B) Iodide is oxidized.

(C) Sulphur is reduced (D) H2O is one of the products.

Key. (ABD)

Sol. 3 2 4 2 2 2 46I ClO 3H SO I 3I Cl 3H O 6HSO

28. In the reaction shown below, the major product(s) formed is/are

(A) (B)

(C) (D)

Ans. A

Sol:

2NH

2C NH

O

3 2

2 2

CH CO O

CH Cl

3N C CH

H O

2C NH

O

3CH COOH

29. An ideal gas in a thermally insulated vessel at internal pressure = P1, volume = V1 and absolute

temperature = T1 expands irreversibly against zero external pressure, as shown in the diagram.

18

The final internal pressure, volume and absolute temperature of the gas are P2, V2 and T2,

respectively. For this expansion,

Ans. A, B, C

Sol: No work done during given process so no change in internal energy i.e. U = 0 So T1 = T2.

As the system is insulated q = 0.

During change pressure decreases and proportionally volume increases in such a way that overall

product of pressure and volume remains same as n and T are same

It is not an adiabatic process.

30. The correct statement(s) for orthoboric acid is/are

(A) It behaves as a weak acid in water due to self ionization.

(B) Acidity of its aqueous solution increases upon addition of ethylene glycol.

(C) It has a three dimensional structure due to hydrogen bonding.

(D) It is a weak electrolyte in water.

Ans. (B), (D)

Sol: H3BO3 does not give H+ ions of its own when dissolved in water rather it furnishes H

+ ion due to

hydrolysis as 3 3 2 4H BO H O H B OH

By adding glycol in orthoboric acid its acidic strength increases.

H O B

O

O

H H O

H H O

chelation H O B

O

O

+

ethylene glycol More acidic than just H3BO3.

Hydrogen bonding is responsible for two dimensional structure of orthoboric acid.

Section -2 : (One Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one int3eger

from 0 to 9 (both inclusive).

31. The total number of distinct naturally occurring amino acids obtained by complete acidic

hydrolysis of the peptide shown below is

19

Ans. (1)

Sol: Distinct naturally occurring amino acid (glycine)

N

N

O

CH2

N

O

N

OHO

N

H

O

N

H H2C

N

N

O

CH3

H

O

O

HH (1)

(2)

(3) (4) (5) (6) (7)

(8)

(9)

Acidic hydrolysis

Hydrolysis 1, 2, 3, 8, 9 gives

NH2

O

OH2

glycine

HO

O

NH2

CH2

CHO

O

NH2

3

32. MX2 dissociates into M

+2 and X

– ions in an aqueous solution, with a degree of dissociation () of

0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of

the depression of freezing point in the absence of ionic dissociation is

Ans. 2

20

Sol: 2

2MX M 2X

0.5

After dissociation 1f fT k .m i

1f f

i 1T k .m

n 1

i 1

0.53 1

i 2 Before dissociation or in absence of ionic dissociation

2f fT k .m

Ratio is

1

2

f f

f f

T k .m 2

T k .m 1

= 2.0

33. If the value of Avogadro number is 6.023 × 1023

mol-1

and the value of Boltzmann constant is

1.380 × 10–23

JK–1

, then the number of significant digits in the calculated value of the universal

gas constant is

Ans. 4

Sol: R = K. No

= 23 231.380 10 6.023 10

8.31174

No of significant figures = 4

34. A compound H2X with molar weight of 80 g is dissolved in a solvent having density of 0.4 g/ml.

Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is

Key. (8)

Sol. 3.2 M solution = 3.2 mole of solute in 1000 ml solution

As there is no change in volume, hence volume of solution is same as volume of solvent.

Density of solvent = 0.4 gm/ml

Weight of solvent = 0.4 1000 = 400 gm

Molality = 3.2

1000 8400

35. In an atom, the total number of electrons having quantum numbers n = 4, |ml| = 1 and ms = 1

2

is

Key. (6)

Sol. n = 4 (4th

shell) includes 4s, 4p, 4d and 4f subshells.

For |ml| = 1 i.e +1 and 1 values of m, there are two orbitals each in 4p, 4d and 4f subshells.

Total 6 orbitals having 12 electrons but only 6 electrons will have spin quantum number of 1/2.

36. Consider the following list of reactions:

Acidified K2Cr2O7, alkaline KMnO4, CuSO4, H2O2, Cl2, O3, FeCl3, HNO3 and Na2S2O3.

The total number of reagents that can oxidize aqueous iodide to iodine is

21

Key. (7)

Sol. 2 2 7K Cr O 6KI + 7H2SO4 Cr2(SO4)3 + 4K2SO4 + 3I2

+ 7H2O

H2O + 2KMnO4 + KI KIO3+ 2MnO2 + 2KOH

2CuSO4 + 4KI Cu2I2 + I2 + 2K2SO4

H2O2 + 2KI 2KOH + I2

Cl2 + 2KI 2KCl + I2

2KI + O3 + H2O 2KOH + I2 + O2

2FeCl3 + 2KI I2 + 2FeCl2 + 2KCl

(dil.)8HNO3 + 6 KI 3I2 + 4H2O + 2NO + 6KNO3

Na2S2O3 + 2KI No reaction

37. A list of species having the formula XZ4 is given below:

XeF4, SF4, SiF4, 4BF , 4BrF , [Cu(NH3)4]

2+, [FeCl4]

2, [CoCl4]

2 and [PtCl4]

2.

Defining shape on the basis of the location of X and Z atoms, the total number of species having

a square planar shape is

Key. (4)

Sol. Species Hybridization Shape

1. XeF4 sp3d

2 Square planar

2. SF4 sp3d See-saw(distorted T. B. P.)

3. 4BF sp

3 Tetrahedral

4. 4BrF sp

3d

2 Square planar

5. [Cu(NH3)4]2+

dsp2 Square planar

Cu+2

– 3d9

3d9

3d

4s

4p

4so

4po

dsp2

forward push

of electron

6. [FeCl4]

2 sp

3 Tetrahedral

Fe+2

3d6

3d6

4s

4p

sp3

Cl

being a weak field ligand cannot show back pairing.

7. [CoCl4]2

sp3 Tetrahedral

Same reason as in case of [FeCl4]2

8. [PtCl4]2

dsp2

Square planar

Pt+2

– 5d8

In case of elements of 4d and 5d series with C.N = 4 complex formed will always be square

planar (low spin complex) irrespective of nature of ligand.

22

38. Consider all possible isomeric ketones, including Stereoisomers of M.W. = 100. All these

isomers are independently reacted with NaBH4 (Note: Stereoisomers are also reacted separately).

The total number of ketones that give a racemic product(s) is/are

Key. (5)

With mol. wt. 100, ketone is C6H12O and its possible isomeric ketones are following:

(1)

O

(2)

O

(3)

O

*

(4)

O

(5) O

(6) O

When each of them react independently with NaBH4 and form alcohols, isomers 1, 2, 4, 5 and 6

form racemic products. Isomer 3 has one stereogenic centre whose reduction product has 2

stereogenic centres hence form diastereomers which are not racemic mixtures.

Compound (3) has two stereoisomers i.e.

H3C H

O

(i)

H CH3

O

(ii) on reduction isomer (i) gives two products they are

H3C H

H OH

(a)

H3C H

OH H

(b) on reduction isomer (2) gives two products they are

H CH3

H OH

(c)

H CH3

OH H

(d) Here (a) and (b) are diastereomers and (c) and (d) are also diastereomers to each other.

39. The total number(s) of stable conformers with non-zero dipole moment for the following

compound is(are)

23

Cl

CH3

Br CH3

Br Cl

Key. (3)

Sol. Newman projections (conformers) of given Fisher projection are

Cl

CH3

Br CH3

Br Cl

BrBr

CH3

CH3

Cl

Cl

Most Eclipsed

(Unstable)

o60

rotation

Br

Br

Cl

Cl

CH3

CH3

Cl

Cl

Br

Br

CH3CH3

not stable (eclipsed)

o60

rotation

Br CH3

CH3

BrCl

Clo60

rotationCH3

CH3

ClCl

Br

Br

o60

rotation

CH3CH3

Br Br

Cl Cl

stable conformer 0 ustable (eclipsed)

stable conformer 0

o60

rotation CH3

CH3Cl

Cl

BrBr

ustable (Eclipsed)

o60 rotation

stable 0

40. Among PbS, CuS, HgS, MnS, Ag2S, NiS, CoS, Bi2S3 and SnS2, the total number of BLACK

coloured sulphides is

Key. (7)

Sol. PbS, CuS, HgS, Ag2S, NiS, CoS and Bi2S3 all are Black.

MnS is buff coloured and SnS2 is yellow.

24

PART III: MATHEMATICS

SECTION – 1 (One or More Than One option correct Type) This section contains 10 multiple choice questions. Each question has four choices (A) (B), (C) and (D)

out of which ONE or MORE THAN ONE are correct.

41. For every pair of continuous functions, , : 0,1f g such that

max : 0,1 max : 0,1 ,f x x g x x

the correct statement(s) is(are):

(A) 2 2

3 3f c f c g c g c for some 0,1c

(B) 2 2

3f c f c g c g c for some 0,1c

(C) 2 2

3f c f c g c g c for some 0,1c

(D) 2 2

f c g c for some 0,1c

Key. (A, D)

Sol. Let maxm value of f x occurs at x1

10 1x

maxm value of g x occurs at x2

20 1x

Consider

h x f x g x

1 0h x

2 0h x

0h c for some 0,1c

Option (A) and (D) is correct.

(B) and (C) are not always correct for each pair of continuous function f and g for e.g consider

f x k and g x k

42. A circle S passes through the point (0, 1) and is orthogonal to the circles 2 21 16x y

and 2 2 1.x y Then

(A) radius of S is 8 (B) radius of S is 7

(C) centre of S is (7, 1) (D) centre of S is (8, 1)

Key. (B, C)

Sol. Let equation of circle is

2 2 2 2 0x y gx fy c

We have

1 2 0f c ………..(i)

2 1 2 0 15g f c

2 15g c ……….(ii)

1c ……….(iii)

25

7g

1

12

cf

equation of circle is

2 2 14 2 1 0x y x y

Centre=(7, 1) 49 1 1 7r

43. From a point , , ,P perpendiculars PQ and PR are drawn respectively on the lines

, 1y x z and , 1.y x z If P is such that QPR is a right angle then the possible value(s) of

is(are)

(A) 2 (B) 1

(C) – 1 (D) 2

Key. (C)

Sol. 1 : , 1L y x z

Symmetric form 0 0 1

1 1 0

x y z

2 : , 1L y x z

Symmetric form 0 0 1

1 1 0

x y z

Foot of perpendicular form P to L1 (Q) will be , ,1 and foot of perpendicular from P to L1(R)

will be (0, 0, 1). Now PQ and PR are perpendicular so (1 ) ( + 1) = 0

2 1 0 1

For = 1 points P and Q will coincide = 1.

44. Let ,x y and z be three vectors each of magnitude 2 and the angle between each pair of them

is .3

If a is a nonzero vector perpendicular to x and y z and b is a nonzero vector

perpendicular to y and ,z x then

(A) .b b z z x (B) .a a y y z

(C) . . .a b a y b z (D) .a a y z y

Key. (A, B, C)

Sol. Since a is perpendicular to x and y z

So a x y z

. .x z y x y z

y z since . . 1x z x y

Now take dot product with y

. . .a y y y y z

2 1

.a y

26

So, .a a y y z

Similarly b is perpendicular to y and z x

So, b y z x

. .b y x z y z x

b z x since . . 1y x y z

Now take dot product with z

. . .b z z z z x

2 1

.b z

.b b z z x

Now take dot product of ,a b gives

. .a y b z

Option (A), (B) and (C) are correct

45. Let : ,2 2

f

be given by

3

log sec tanf x x x

Then

(A) f x is an odd function (B) f x is a one-one function

(C) f x is an onto function (D) f x is an even function

Key. (A, B, C)

Sol. 3

log sec tanf x x x

3

log sec tanf x x x

3

1log sec tanx x

3

log sec tanx x

3

log sec tanx x

f x f x is an odd function

3

log sec tanf x x x

22 sec tan sec

' 3 log sec tansec tan

x x xf x x x

x x

2

3 log sec tan secx x x

For ,2 2

x

' 0f x f x is S.I function

27

one-one function

Also when 2

x

3

log sec tanx x

When 3

log sec tan2

x x x

Range = co domain = R

function is onto

46. Let : , 1,f a b be a continuous function and let :g be defined as

0 ,

,

,

x

a

b

a

if x a

g x f t dt if a x b

if x bf t dt

Then

(A) g x is continuous but not differentiable at a

(B) g x is differentiable on

(C) g x is continuous but not differentiable at b

(D) g x is continuous and differentiable at either a or b but not both

Key. (A, C)

Sol.

0

'

0

x a

g x f x a x b

x b

As 1,f x for ,a b

'g x does not exist at x a and x b

g a g a g a

And g b g b g b

g x is continuous at x = a and b

47. Let : 0,f be given by

1

1 .tx

t

x

dtf x e

t

Then

(A) f x is monotonically increasing on [1, ) (B) f x is monotonically decreasing on (0, 1)

(C) 1

0,f x fx

for all 0,x (D) 2xf is an odd function of x on

Key. (A, C, D)

28

Sol. 1

2'

xxf x e

x

For 0x f x is strictly Increasing

1

11 x t

t

x

dtf e f x

x t

12

2

2

x

x

tx t dt

g x f et

12

2

x

x

tt dt

g x et

g x

48. Let M be a 2 2 symmetric matrix with integer entries. Then M is invertible if

(A) The first column of M is the transpose of the second row of M

(B) The second row of M is the transpose of the first column of M

(C) M is a diagonal matrix with nonzero entries in the main diagonal

(D) The product of entries in the main diagonal of M is the square of an integer

Key. (C,D)

Sol. a b

Mb c

For option (A) transpose of [b c] = b

c

By the given statement b a

a b cc b

So, |M| = ac – b2 = 0 M is none invertible

So option (A) is wrong in the similar way option (B) is also wrong By the given statement

For option (C)

If a 0

M0 c

| M | ac 0 a and c are nonzero

So, M is invertible

For option ‘D’

If a b

Mb c

If ac 2 for some I, then det(M) 0

Hence M is invertible

49. Let a R and let f : R R be given by 5f x x 5x a . Then

(A) f (x) has three real roots if a 4 (B) f (x) has only one real root if a 4

(C) f (x) has three real roots if a 4 (D) f (x) has three real roots if 4 a 4

Key. (B, D)

29

Sol. Consider the graph of y = x5 – 5x

local maxima.at x 1 & Minima at x = 1

1

45 , 0

1, 0

0, 0

0, 4

1, 01

45 , 0

0, 4

X

Y

From this graph we can get graph of y = f(x) = x

5 – 5x + a

It is clear from the graph that f(x) has only one real root if a > 4 & f(x) has three real roots if

4 a 4

50. Let M and N be two 3 3 matrices such that MN NM further. If 2M N and

2 4M N , then

(A) Determinant of 2 2M MN is 0

(B) There is a 3 3 non -zero matrix U such that 2 2M MN U is the zero matrix

(C) Determinant of 2 2M MN 1

(D) For a 3 3 and matrix U. If 2 2M MN U equals the zero matrix then U is the zero matrix

Key. (A, B)

Sol. 2 2MN NM N M NMN MN

and 2 4M N 0

2 2M N M N 0 MN NM … (1)

2det M N 0

[if det(M + N2) 0 then by multiplying with (M + N

2)–1

in (1),it gives M – N2 = 0]

2 2det M MN 0

from (1) option B is correct & D is wrong.

30

SECTION – 2 : (One Integer Value correct Type) This section contains 10 questions. Each question, when worked out will result in one integer from 0 to

9 (both inclusive).

51. For a point P in the plane, let 1d P and 2d (P) be the distances of the point P from the lines

x y 0 and x y 0 respectively. The area of the region R consisting of points P lying in the

first quadrant of the plane and satisfying 1 22 d P d P 4 , is

Key. 6

Sol. 1 22 d P d P 4

2 2 | x y | | x y | 4 2

x 2

y 2

y 2 2y xy x

x 2 2

x

y

Required Area = 2 21

4 2 2 24

=6

52. Let a,b and c be three non-coplanar unit vectors such that the angle between every pair of them

is 3

. If a b b c pa qb rc , where p, q and r are scalars, then the value of

2 2 2

2

p 2q r

q

is.

Key. 4

Sol. Given that | a | | b | | c | 1

And 1

a b b.c c.a 1.1.cos3 2

Now 2

1 11

2 2a.a a.b a.c1 1 1 1

[a b c] b.a b.b b.c 1 a b c2 2 2 2

c.a c.b c.c 1 11

2 2

Now

a b b c pa qb rc ….. (1)

31

Apply dot product with a,b and c

20 [abc] p | a | qa.b r a.c

20 0 pa.b q | b | r b.c

2[a b c] 0 pa.c q b.c r | c |

q r 1

p2 2 2

p r

q 02 2

& p q 1

r2 2 2

1 1

p ,q ,2 2

1

r2

So, the value of 2 2 2

2

p 2q r4

q

53. The largest value of the non-negative integer a for which

1 x

1 x

x 1

ax sin x 1 a 1lim

x sin x 1 1 4

is

Key. (2)

Sol.

1 x

1 x

x 1

a 1 x sin x 1 1lim

x 1 sin x 1 4

1 x 1 x

1 x

x 1

sin x 11 xa

x 1 x 1 1lim

sin x 1 4x 1

x 1 x 1

21 a 1

2 4

2

1 a 1

a 0,2

largest Non-negative value of a = 2

54. Let f: [0,4 ] [0, ] be defined by f (x) = cos–1

(cos x). The number of points

x 0,4 satisfying the equation 10 x

f (x)10

is:

Key. (3)

Sol. Number of points is given by number of solution of equation

32

cos–1

(cos x) 10 x

10

( x 0,4 )

Hence, number of points are 3.

55. Let f: R R and g: R R be respectively given by f (x) = x 1 and g (x) = x2 + 1. Define h:

R R by

max if( )

min if 0,

, 0,

,

f x g x x

f x xg xh x

The number of points at which h (x) is not differentiable is:

Key: (3)

Sol.

2

2

1: 1

1: 1 0( )

1: 0 1

1: 1

x x

x+ xh x

x x

x+ x

Hence, f (x) is not differentiable at three points

x = –1, 0, 1

56. Let n1 < n2 < n3 < n4 < n5 be positive integers such that

n1 + n2 + n3 + n4 + n5 = 20. Then the number of such distinct arrangements

(n1 < n2 < n3 < n4 < n5) is:

Key. (7)

Sol. Required no of solution = Total no of positive integral solution of

5n1 + 4x + 3y + 2z + t = 20

is equal to 7.

57. Let n 2 be an integer. Take n distinct points on a circle and join each pair of points by a line

segment. Colour the line segment joining every pair of adjacent points by blue and the rest by

red. If the number of red and blue segments are equal then the value of n is :

33

Key. (5)

Sol. Blue line segments = n

Red line segments = C

2n n

C

2n n= n

n= 5 58. The slope of the tangent to the curve (y – x

5)2 = x(1 + x

2)2 at the point (1, 3) is:

Key. (8)

Sol. 2 2

5 21 y x x x

2

5 4 2 22 5 2 1 2 1

dyy x x x x x+ x

dx

For the point (1, 3)

5 2 2 2 1 42 3 –

dy

dx 5 2 1

dy

dx

Hence, slope of tangent = 8.

59. Let a, b, c be positive integers such that b

ais an integer. If a, b, c are in geometric progression

and the arithmetic mean of a, b, c is b + 2, then the value of 2 –14

1

a a

a+is:

Key. (4)

Sol. a, ar, ar2

2

3

a+ar +arar +2

2 3a+ar+ar ar+6 ;

2 2 6 0ar – ar+a –

22 4 – 4 ( 6)

2

a a a a –r =

a ;

6a ar =

a

r is integer Hence, a must be equal to 6

a = 6 26 6 14

46 1

60. The value of 1 2

53 2

2

0

4 1

d

x x dxdx

is:

Key. (2)

Sol. I = 1 1

5 53 2 2 2

0 0

d d4x 1 x 12x . 1 x dx

dx dx

I = 11

4 43 2 2 2

00

4x 5 1 x 2x 12x 5 1 x 2x dx

= 0 + 60 1 1

41 4

0 0

y 1 y dy 60 1 y y dy 2

34

ANSWER KEY

PHYSICS CHEMISTRY MATHEMATICS

Q. No. Answer Key Q. No. Answer Key Q .No. Answer Key

1 C 21 C 41 A,D

2 B,D 22 A 42 B,C

3 A,C 23 A,B,D 43 C

4 D 24 A,C,D 44 A,B,C

5 C,D 25 A,B,C 45 A,B,C

6 A,B,D 26 B,C 46 A,C

7 A,B,C 27 A,B,D 47 A,C,D

8 C,D 28 A 48 C,D

9 A,C,D 29 A,B,C 49 B,D

10 A,D 30 B,D 50 A,B

11 5 31 1 51 6

12 3 32 2 52 4

13 4 33 4 53 2

14 4 34 8 54 3

15 2 35 6 55 3

16 5 36 7 56 7

17 2 37 4 57 5

18 3 38 5 58 8

19 5 39 3 59 4

20 8 40 7 60 2

35