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JEE (ADVANCED) 2015 PAPER – 1

PART – I : PHYSICS

SECTION – 1 : (Maximum Marks : 32)

This section contains EIGHT questions

The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both

inclusive

For each question, darken the bubble corresponding to the correct integer in the ORS

Marking scheme :

+4 If only the bubble corresponding to the answer is darkened

0 In all other cases

1. An infinitely long uniform line charge distribution of charge per unit length λ lies parallel

to the y-axis in the y-z plane at 3

2a (see figure). If the magnitude of the flux of the electric

field through the rectangular surface ABCD lying in the x-y plane with its centre at the

origin is 0

λL

nε( 0ε = permittivity of free space), then the value of n is:

Ans. 6

Solution:

Flux from total cylindrical surface (angle = 2π )

= in

0

Q

ε

Flux from cylindrical surface AB = flux from the given surface

http://www.pioneermathematics.com/



2.

= in

0 0

Q λ

6ε 6ε n = 6

3. Consider a hydrogen atom with its electron in the nth orbital. An electromagnetic

radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the

ejected electron is 10.4 eV, then the value of n is (hc = 1242 eV nm)

Ans. 2

Solution:

2

hC 113.6eV. 10.4

λ n

2

1242eV 13.610.4

90 n

2 2 2

41.4 13.6 13.6 13.610.4 13.8 10.4 3.4

3 n n n

2n 4 n 2

4. A Bullet is fired vertically upwards with velocity v from the surface of a spherical planet.

When it reaches its maximum height, its acceleration due to the planet's gravity is 1/4th of

its value at the surface of the planet. If the escape velocity from the planet is escv v n ,

then the value of N is (ignore energy loss due to atmosphere)

Ans. 2

Solution:

When it reaches its maximum height, its acceleration due to the planet's gravity is 1/4th of

its value at the surface of the planet.

2 2

GM 1 GM

r 4 R

r = 2R

By conservation of mechanical energy




2GMm 1 GMmmv 0

R 2 r

21 GMmmv

2 2R

sec

2GMv v N

R

N = 2

5. Two identical uniform discs roll without slipping on two different surfaces AB and CD (see

figure) starting at A and C with linear speeds v1 and v2, respectively, and always remain in

contact with this surfaces. If they reach B and D with the same linear speed and v1 = 3 m/s,

then v2 in m/s is (g = 10 m/s2)

Ans. 7

Solution:




Final kinetic energy of both discs is same

2 2

2

3 1 3 1m 3 mg 30 mv mg 27

2 2 2 2

22

3 3. 9 300 v 270

4 4

22

27 330 v

4 4

22 2v 9 40 v 7

6. Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of

B and A emits 104 times the power emitted from B. The ratio A

B

λ

λ

for their wavelengths

A Bλ and λ at which the peaks occur in their respective radiation curves is :

Ans. 2

Solution:

According to Wien’s displacement law

A Bm A m Bλ T λ T

Ratio of energy radiated per unit time 4

A A A4

B B B

E σT A

E σT A

2 44A

2 4B

σ 4π 400r T10 E

E σ 4π r T

4

4B A

A B

λ λ2 2

λ λ

7. A nuclear power planet supplying electrical power to a village uses a radioactive material

of half life T years as the fuel. The amount of fuel at the beginning is such that the total

power requirement of the village is 12.5% of the electrical power available from the plant

at that time. If the plant is able to meet the total power needs of the village for a maximum

period of n T years, then the value of n is.

Ans. 3

Solution:

12.5E E

100

E'E

8 (E = Power requirement to the village, E = Power of plant)




3

E'E

2

Number of half life = 3

So total time required = 3 T years

8. A young's double slit interference arrangement with slits S1 and S2 is immersed in water

(refractive index = 4/3) as shown in the figure. The positions of maxima on the surface of

water are given by x2 =p2m2x2λ 2, where λ is the wavelength of light in air (refractive

index = 1), 2d is the separation between the slits and m is an integer.

The value of p is

Ans. 3

Solution:

For constructive interference x mλ

2 2 2 24d x d x mλ

3

2 21d x mλ

3

2 2 2 2x 9m λ d

p = 3

9. Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10cm

each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure.




An object is placed at a distance of15 cm from the mirror. Its erect image formed by this

combination has magnification M1. When the set-up is kept in a medium of refractive

index 7/6, the magnification becomes M2. The magnitude 2

1

M

M

Ans. 7

Solution:

For mirror

M= f V

f u u

10 VM

10 15 15

M = - 2

v = – 30 cm

For lens

M’ = f 10

1f u 10 20

M1 = 2

In liquid

0

f ' μ 1 7

f 4μ1

μ

(f’ is the focal length of lens in medium of refractive index 0

7μ

6 }

70f ' cm

4

70f ' 4M' 7

70f u 204

M2 = 14

1

2

M7

M




Section – 2 : (Maximum Marks : 40)

This section contains TEN questions

Each quetstion has Four options (A), (B), (C) and (D). ONE OR MORE THAN ONE of thse

option (s) is (are) correct

For each question, darken the bubble(s) corresponding to all the correct option(s) in the

ORS

Marking scheme :

+4 If only the bubble (s) corresponding to all the correct option (s) is (are) darkened 0 If

none of the bubbles is darkened

– 2 In all other cases

10. Consider a vernier callipers in which each 1 cm on the main scale is divided into 8 equal

divisions and a screw gauge with 100 divisions on its circular scale. In the vernier

callipers, 5 divisions of the vernier scale coincide with 4 division on the main scale and in

the screw gauge, one complete rotation of the circular scale moves it by two divisions on

the linear scale. Then,

(a) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least

count of the screw gauge is 0.01mm.

(b) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least

count of the screw gauge is 0.005mm.

(c) If the least count of the linear scale of the screw gauge is twice the least count of the

Vernier callipers, the least count of the screw gauge is 0.01 mm.

(d) If the least count of the linear scale of the screw gauge is twice the least count of the

vernier callipers, the least count of the screw gauge is 0.005 mm.

Ans. (B, C)

Solution:

For Vernier calipers

1MSD= 1

8cm

5 VSD = 4 MSD

1VSD=4

5MSD =

4 1 1cm

5 8 10




LC of vernier calliper = 1 1

cm cm 0.025 cm8 10

(A) & (B)

pitch of screw gauge = 2 (0.025) = 0.05 cm

Leastcount of screw gauge = 0.05

100= 0.005 mm

(C) & (D) Least count of linear scale of screw gauge = 0.05 cm

pitch = 0.05 2cm = 0.1 cm

Leastcount of screw gauge =0.1

100 =0.01 mm

11. Planck's constant h, speed of light c and gravitational constant G are used to from a unit of

length L and a unit of mass M. Then the correct options(s) is (are)

(a) M c (b) M G (c) L h (d) L G

Ans. (A, C, D)

Solution: x y zM h c G

x y z

2 1 1 1 3 2M ML T LT M L T

x – z = 1

2x + y + 3z = 0

– x – y – 2z = 0

1 1 1x , y , z

2 2 2

1Mα h c

G

For L

x – z = 0

2x + y + 3z = 1

–x – y – 2z = 0

1 3 1x y z

2 2 2

3/2

1L α h G

C

12. Two independent harmonic oscillators of equal mass are oscillating about the origin with

angular frequencies 1 2ω and ω and have total energies E1 and E2, respectively.

The variations of their momenta p with positions x are shown in figures. If




2a an and n,

b R and then the correct equation(s) is (are):

(a) 1 1 2 2E ω E ω (b) 22

1

ωn

ω (c) 2

1 2ω ω n (d) 1 2

1 2

E E

ω ω

Ans. (B, D)

Solution:

For first oscillator For Second oscillator

1b maω

2

1

a 1n

b mω

1

11

mω

22

1

ωn

ω Ans. B

2 21 1

1E m ω a

2

2 22 2

1E mω R

2

2 221 1 1 2

2 22 2 2 1

E ω ω ωn

E ω ω ω

1 1 1 2

2 2 1 2

E ω E E

E ω ω ω Ans. D

13. A ring of mass M and radius R is rotating with angular speed ω about a fixed vertical axis

passing through its centre O with two point masses each of mass M

8 at rest at 0. These

masses can move radially outwards along two massless rods fixed on the ring as shown in

the figure. At some instant the angular speed of the system is 8ω

9 and one of the masses is

ata distance of 3

5R from O. At this instant the distance of the other mass from 0 is :




(a) 2

R3

(b) 1

R3

(c) 3

R5

(d) 4

R5

Ans. (D)

Solution:

By conservation of angular momentum

2 2

2 2 M 9R Md 8ωMR ω MR

8 25 8 9

2 2 22 200R 9R 25d 8

R8 25 9

225 R2 – 209 R2 = 25 d2

d = 216R

25

4Rd

5

14. The figures below depict two situations in which two infinitely long static line charges of

constant positive line charge density λ are kept parallel to each other. In their resulting

electric field, point charges q and – q are kept in equilibrium between them. The point

charges are confined to move in the x-direction only. If they are given a small

displacement about their equilibrium positions, then the correct statement(s) is (are):




(a) Both charges execute simple harmonic motion.

(b) Both charges will continue moving in the direction of their displacement.

(c) Charge +q executes simple harmonic motion while charge -q continues moving in the

direction of its displacement.

(d) Charge -q executes simple harmonic motion while charge +q continues moving in the

direction of its displacement.

Ans. (C)

Solution:

As +q is displaced towards right, the repulsion of right side wire will dominate and the net

force on +q will be towards left, and vice versa

restoring

2kλ 2kλF q

d x d x

restoring 2 2 2

2kλ 2x q 4kλqF x

d x d

Hence SHM

For –q, as it is displaced towards right the attraction of right side wire will dominate,

which forces the -q charge to move in the same direction of displacement similarity for

other side

Hence it is not SHM.

15. Two identical glass rods S1 and S2 (refractive index = 1.5) have one convex end of radius of

curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in the

figure, with their axes (shown by the dashed line) aligned. When a point source of light P

is placed inside rod S1 on its axis at a distance of 50 cm from the curved face, the light rays

emanating from it are found to be parallel to the axis inside S2. The distance d is :

(a) 60 cm (b) 70 cm (c) 80 cm (d) 90 cm

Ans. (B)

Solution:




at glass rod S2

1 n refraction

2

n 1 n 1

u 10

2u 20

at glass rod S1

for n 1 refraction

V1 = d – 20

1 n 1 n

d 20 50 10

1 n 1

d 20 50 20

d = 70 cm

16. A conductor (shown in the figure) carrying constant current I is kept in the x-y plane in a

uniform magnetic field B. IF F is the magnitude of the total magnetic force acting on the

conductor, then the correct statement(s) is (are):

(a) If B is along z, F L R

(b) If B is along x, F 0

(c) If B is along y, F L R

(d) If B is along z, F 0

Ans. (A, B, C)

Solution:




F i B

= i{2 L R i B}

If B is along z

F i 2 L R B] j

If B is along x

F 0

If B is along y

F i 2 L R B)k

Alternate Solution

dF i d B

In uniform magnetic field

dF i d B i d B

F i PQ B

(a) F [i2 L R B] 2iB L R

(b) F = 0

(c) F [i 2 L R B] 2iB L R

(d) F [i 2 L R B] 2iB L R

17. A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium

in equilibrium at temperature T. Assuming the gases are ideal, the correct statement(s) is

(are)

(a) The average energy per mole of the gas mixture is 2RT.




(b) The ratio speed of sound in the gas mixture to that in helium gas is 6 /5 .

(c) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2.

(d) The ratio of the rms speed of helium atoms to that of hydrogen molecules is1/ 2 .

Ans. (A, B, D)

Solution:

Total Energy = 3 5

RT RT 4RT2 2

Average energy per moles of mixture

= 4RT

2RT2

sound

γRTv

M

mix mix

5 71 1 3 1 2 1 42 2γ , M 3

3 5 2 21 12 2

sound,mix mix He

sound,He He mix

v γ M 6

v γ M 5

rms

3RTv

M

2

rms, He

rms H

v 2 1

v , 4 2

18. In an aluminum (Al) bar of square cross section, a square hole is drilled and is filled with

iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are 8 72.7 10 m and 1.0 10 m, respectively. The electrical resistance between the two

faces P and Q of the composite bar is




(a) 2475

μ64

(b) 1875

μ64

(c) 1875

μ49

(d) 2475

μ132

Ans. (B)

Solution:

Af FeR and R are Parallel to each other

ef fe

1 1 1

R R R

8

2 2

2.7 10 .051;

R .007 .002

7

fe 2

1.0 10 .05R

.002

8 45

t 6

2.7 10 5 10R 3 10

10 45 100

72 3

Fe 6

10 5R 5 10 10

4 10 4

5 3

5 3ef

1 1 4 10 4 10

R 3 10 5 10 3 5

5 3 5ef

1 15 15

R 5 10 12 10 5.12 10

ef

1 150 1875μ μ

R 5.12 64

19. For photo-electric effect with incident photon wavelength λ , the stopping potentials is 0V .

Identity the correct variation (s) of V0 with λ and 1/λ.




Ans. (A, C)

Solution:

0

hceV

λ

0

hc 1V

e λ


This section contains TWO questions

Each question contains two columns, Column I and Column II

Column I has four entries (A), (B), (C) and (D)

Column II has five entries (P), (Q), (R), (S) and (T)

Match the entries in Column I with the entries in Column II

One or more entries in Column I may watch with one or more entries in Column II

The ORS contains a 4 5 matrix whose layout will be similar to the one shown below :

For each entry in Column I, darken the bubbles of all the matching entries. For example, if

entry (A) in Column I matches with entries (Q), (R) and (T), then darken these three

bubbles in the ORS. Similarly, for entries (B), (C) and (D).




Marking scheme :

For each entry in Column I

+2 In only the bubbles (s) corresponding to all the correct match (es) is (are) darkened

0 If none of the bubbles is darkened

– 1 In all other cases

20. Match the nuclear processes given in Column I with the appropriate option (s) in Column

II.

Column-I Column-II

(A) Nuclear fusion (P) Absorption of thermal neutrons by 23592 U

(B) Fission in a nuclear reactor (Q) 6027Co nucleus

(C) β decay (R) Energy production in stars via hydrogen

conversion to helium

(D) γ ray emission (S) Heavy water

(T) Neutrino emission

Ans. (A) R B P, S; C Q, T; D Q, R, T

21. A particle of unit mass is moving along the x-axis under the influence of a force and its

total energy is conserved. Four possible forms of the potential energy of the particle are

given in column I (a and U0 are constants). Match the potential energies in column I to the

corresponding statements (s) in column II.

Column-I Column-II

(A)

22

01

U xU x 1

2 a

(P) the force acting on the particle is zero at

x = a.

(B)

2

02

U xU x

2 a

(Q) the force acting on the particle is zero at

x = 0.

(C)

2 2

03

U x xU x exp

2 a a

(R) the force acting on the particle is zero at x =

– a.

(D)

3

04

U x 1 xU x

2 a 3 a

(S) The particle experiences an attractive force

towards x = 0 in the region x a

(T) The particle with total energy 0U

4can

oscillate about the point x = – a.

Ans. (A) P, Q, R, T (B) Q, S; (C)P, Q, R, S ; (D) P, R, T

Solution:




(A) 0x 3

2UdUF [x a] [x][x a]

dx a

A P Q R T

(B) x 0

dU xF U

dx a

(C) x

dUF

dx

=

2 2x /a

0 3

eU [x][x a][x a]

a




(D) x

dUF

dX

= 03

U[ x a x a ]

2a

P, R, T

PART – II : CHEMISTRY

Atomic masses : [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al =

27, Si = 28, P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40, Cr = 52, Mn= 55, Fe = 56, Cu = 63.5, Zn




= 65, As = 75, Br = 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]

Section – 1 : (Maximum Marks : 32)



inclusive


Marking scheme :

+4 If the bubble corresponding to the answer is darkened


22. The total number of stereoisomers that can exist for M is

Ans. 2

Solution:

Total number of stereoisomers = 2

This molecule can not show geometrical isomerism so only mirror image will be other

stereoisomer.

23. The number of resonance structures for N is

Ans. (9)




Solution:

24. The total number of lone pairs of electrons in N2O3 is

Ans. 8

Solution:

N2O3

25. For the octahedral complexes of Fe3+ in SCN– (thiocyanato-S) and in CN– ligand

environments, the difference between the spin-only magnetic moments in Bohr

magnetons (when approximated to the nearest integer) is [Atomic number of Fe = 26]

Ans. 4

Solution:

SCN– is weak field effect (WFE) ligand whereas CN– is strong field effect (SFE) ligand.

Spin only magnetic moment = 5 5 2 35 BM




Spin only magnetic moment = 1 1 2 3 BM

Difference = 35 3 4

26. Among the triatomic molecules/ions, 2 3 2 2 3 2 2 3 2BeCl , N ,N O,NO , O ,SCl , ICl , I and XeF , and

XeF2, the total number of linear molecules(s)/ion(s) where the hybridization of the

central atom does not have contribution from the d-orbital(s) is

[Atomic number: S = 16, Cl = 17, I = 53 and Xe = 54]

Ans. 4

Solution :

27. Not considering the electronic spin, the degeneracy of the second excited state (n = 3) of

H atom is 9, while the degeneracy of the second excited state of H– is

Ans. 3

Solution:

Energy order of orbitals of H is decided by only principle quantum number (n)

while energy order of H– is decided by (n + ) rule :

Electronic configuration of 'H–' is – 1s2 its Energy order is decided by n+ rule. 2 0 0H 1s 2s 2p

Its 2nd excited state is 2p

and degenery 2p is ‘3’

28. All the energy released from the reaction X Y, 0 1rG 193kJ mol is used for

oxidizing M+ as M+ M3+ + 2e– , E° = – 0.25 V.

Under standard conditions, the number of moles of M+ oxidized when one mole of X is

converted to Y is [F = 96500 C mol–1]

Ans. 4

Solution: 3m m 2e

o 0G nFE for 1 mole of m+

oG 2 96500 0.25 J

= + 48.25 KJ/mole

Energy released by conversion of 1 mole of




x y G 193 KJ

Hence mole of m+ convert

1934

48.25

29. If the freezing point of a 0.01 molal aqueous solution of a cobalt(III) chloride-ammonia

complex (which behaves as a strong electrolyte) is –0.0558°C, the number of chloride(s)

in the coordination sphere of the complex is

[Kf of water = 1.86 K kg mol–1]

Ans. (1)

Solution:

f fT K i m

0.0558 = 1.86 I 0.01

i = 3

Given complex behaves as a strong electrolyte

a =100%

n = 3 (no. of particles given by complex)

complex is [Co(NH3)5CI]CI2

no. of Cl– ions in the co-ordination sphere of the complex = 1


This section contains TEN questions

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these

four option (s) is (are) correct.

For each question, darken the bubble (s) corresponding to all the correct option (s) in the

ORS

Marking scheme :

+ 4 If only the bubble (s) corresponding to all the correct option (s) is (are) darkened

0 If none of the bubble (s) corresponding to all the correct option (s) is (are)

darkened

–2 In all other cases

30. Compound (s) that on hydrogenation produce (s) optically inactive compound(s) is (are)

Ans. (B, D)




Solution :

31. The major product of the following reaction is

Ans. (A)




Solution:

32. In the following reaction, the major product is

Ans. (D)

Solution:




33. The structure of D–(+)-glucose is

The structure of L–(–)–glucose is

Ans. (A)

Solution:

The structure of L–(–)–glucose is

34. The major product of the reaction is

Ans. (c)

Solution:

Treating with nitrous acid (diazotisation) gives a hydroxyl group in the place of the amino




group. The reaction occurs stereospecifically with retention in configuration.

35. The correct statement(s) about Cr2+ and Mn3+ is (are)

[Atomic numbers of Cr = 24 and Mn = 25]

(a) Cr2+ is a reducing agent

(b) Mn3+ is an oxidizing agent

(c) Both Cr2+ and Mn3+ exhibit da electronic configuration.

(d)When Cr2+ is used as a reducing agent, the chromium ion attains d5 electronic

configuration.

Ans. (A, B, C)

Solution: 2 3 3 2Cr Mn Cr Mn

Cr2+ is a reducing agent Mn3+ is an oxidising agent both Cr2+ & Mn3+ exhibit d4 electronic

configuration.

36. Copper is purified by electrolytic refining of blister copper. The correct statement(s)

about this process is (are):

(a) Impure Cu strip is used as cathode

(b) Acidified aqueous CuSO4 is used as electrolyte

(c) Pure Cu deposits at cathode

(d) Impurities settle as anode-mud

Ans. (B, C, D)

Solution:

Impure Cu is used as anode pure Cu deposited at cathode. Electrolyte is acidified solution

of CuSO4. impurities settle as anode mud.




37. Fe3+ is reduced to Fe2+ by using

(A) H2O2 in presence of NaOH (B) Na2O2 in water

(C) H2O2 in presence of H2SO4 (D) Na2O2 in presence of H2SO4

Ans. (C) & (D)

Solution:

In Basic medium, Fe3+ will be precipitated as Fe(OH)3.

38. The % yield of ammonia as a function of time in the reaction :

N2(g) 4 3H2(g) 2NH3(g), H < 0

at (P1T1) is given below.

If this reaction is conducted at (P, T2), with T2 > T1, the % yield of ammonia as a function

of time is represented by

Ans. (B)

Solution:

Initially on increasing temperature rate of reaction will increase, so % yield will also

increase with time. But at equilibrium % yield at high temperature (T2) would be less




than at T1 as reaction is exothermic so the graph is

39. If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m

fraction of octahedral holes occupied by aluminium ions and n fraction of tetrahedral

holes occupied by magnesium ions, m and n, respectively, are

(a) 1 1

,2 8

(b) 1

1,4

(c) 1 1

,2 2

(d) 1 1

,4 8

Ans. (A)

Solution:

In ccp, O2– ions are 4.

Hence total negative charge = –8

Let Al3+ ions be x, and Mg2+ ions be y.

Total positive charge = 3x + 2y

3x + 2y = 8

This relation is satisfied only by x = 2 and y = 1.

Hence number of Al3+ = 2.

and number of Mg2+ = 1.

n = fraction of octahedral holes occupied byAl3+

= 2 1

4 2

and m = fraction of tetrahedral holes occupied by Mg2+

= 1

8

Hence, answer is (A)





Column II has five entries (P), (Q), (R), (S) and (T)


One or more entries in Column I may match with one or more entries in Column II




The ORS contains a 4 5 matrix whose layout will be similar to the one shown below:




Marking scheme :


+2 If only the bubble (s) corresponding to all the correct match(es) is(are) darkened



40. Match the anionic species given in Column I that are present in the ore (s) given in

Column II.

Column I Column II

(A) Carbonate (P) Siderite

(B) Sulphide (Q) Malachite

(C) Hydroxide (R) Bauxite

(D) Oxide (S) Calamine

(T) Argentite

Ans. AP, Q, S; BT; CQ, R; DR

Solution:

(P) Siderite – FeCO3

(Q) Malachite – CuCO3. Cu(OH)2

(R) Bauxite – AlOx(OH)3–2x

(S) Calamine – ZnCO3

(T) Argentite – Ag2S

41. Match the thermodynamic processes given under Column I with the expressions given

under Column II.

Column I Column II

(A) Freezing of water at 273 K and 1 atm (P) q = 0

(B) Expansion of 1 mol of an ideal gas into a (Q) w = 0

(C) Mixing of equal volume of two ideal gases at (R) sysS 0




(D) Reversible heating of H2(g) at 1 atm from (S) U 0

(T) G 0

Ans. (A-R, T) (B-P, Q, S) (C-P, Q, S) (D-P, Q, S, T)

Solution:

(A) 2 2H O H O s at 273 K, & 1 atm

H ve q

sysS 0, G 0.

w 0 (as water expands on freezing), U 0

(B) Free expansion of ideal gas. q = 0

w = 0

U 0

sysS 0

G 0

(C) Mixing of equal volume of ideal gases at constant pressure & temp in an isolated

container

q = 0, w = 0, sysU 0, S 0, G 0

(D) Reversible Reversible2 Heating, 1atm Cooling, 1 atm

H g 300K 600K 300 K.

q = 0, w = 0, sysU 0, G 0, S 0

PART-III-MATHEMATICS

SECTION -1 : (Maximum Marks : 32)



inclusive


Marking scheme :

+4 If the bubble corresponding to the answer is darkened


42. The number of distinct solutions of the equation

2 4 4 6 65cos 2x cos x sin x cos x sin x 2

4 in the interval [0, 2π ] is

Ans. 8

Solution:

2 4 4 6 65cos 2x cos x sin x cos x sin x 2

4




2 2 25 1 3cos 2x 1 sin 2x 1 sin 2x 2

4 2 4

2 2cos 2x sin 2x

2tan 2x 1

Now 2x [0, 4π] π 3π 5π 7π 9π 11π 13π 15π

x , , , , , , ,8 8 8 8 8 8 8 8

so number of solution = 8

43. Let the curve C be the mirror image of the parabola y2 = 4x with respect to the line

x + y + 4 = 0. If A and B are the points of intersection of C with the line y = – 5, then the

distance between A and B is

Ans. 4

Solution:

let P(t2, 2t) be a point on the curve y2 = 4x and Q(h, k) be it's image in x + y + 4 = 0

22 2 t 2t 4h t k 2t

1 1 2

h 2t 4

2k t 4

Now k = – 5

so t = 1

hence h = – 2, – 6

so A, B are (–2, – 5) & (–6, - 5)

Hence AB = 4

44. The minimum number of times a fair coin needs to be tossed, so that the probability of

getting at least two heads is at least 0.96, is

Ans. 8

Solution:

Let coin is tossed n times

P(least value heads) = n n

n2

1 11 C . 0.96

2 2

n

4 n 1

100 2

n

n 1 1

2 25

n225

n 1

least value of n is 8.

45. Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a

way that all the girls stand consecutively in the queue. Let m be the number of ways in

which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand

consecutively in the queue. Then the value of m

n is




Ans. 5

Solution:

n = 5! 6!

m = 5! 6C2 5C4. 2!. 4!

m 5! 15 2 5!5

n 6!

46. If the normals of the parabola y2 = 4x drawn at the end points of its latus rectum are

tangents to the circle (x – 3)2 + (y + 2)2 = r2, then the value of r2 is

Ans. 2

Solution:

Equation of normals at points (1, ±2) are

y = – x + 3 & y = x – 3

x + y – 3 = 0 & x – y – 3 = 0

Now 23 2 3r r 2

1 1

47. Let f: R R be a function defined by f(x) = [x], x 2

0, x 2

where [x] is the greatest integer

less than or equal to x. If I =

22

1

xf xdx

2 f x 1

, then the value of (4I –1) is

Ans. 0

Solution: 2 2 0 1 22 2

1 1 1 0 1

x[x ] x[x ] 0 0 x.1I dx dx dx dx dx

2 [x 1] 3 [x 1] 3 1 3 0 3 1

=

22

1

1 x 2 1 1

4 2 8 4

4I – 1 = 0

48. A cylindrical container is to be made from certain solid material with the following

constraints: It has fixed inner volume of V mm3, has a 2 mm thick solid wall and is open at

the top. The bottom of the container is solid circular disc of thickness 2 mm and is of

radius equal to the outer radius of the container.

If the volume of the material used to make the container is minimum when the inner

radius of the container is 10 mm, then the value of V

250πis

Ans. 4

Solution:

Volume of material 2V πr h




2 2 2

1V π r 2 2 π r 2 h πr h

2

1V 2π r 2 πh 4 4r

2

1V 2π r 2 4πh r 1

2

1 2

2 r 1 VV 2π r 2

πr

12 3

dV 2v 1 22π 2 r 2 0

dr π r r

3

2V 2 1024 0

π 10

3

24v24

10 π

2v 10 π

v

4250π

49. Let F(x) =

2 πx

62

x

2cos t

dt for all 1

x R and f : 0, 0,2

be a continuous function.

For a 1

0,2

, if F’(a) + 2 is the area of the region bounded by x = 0, y = 0, y = f(x) and x =

a, then f(0) is

Ans. (3)

Solution:

2 πx

62

x

F x 2cos dt

2

2 2πF' x 2 cos x 2x 2cos x

6

a

0

F' a 2 f x dx

2 a

2 2

0

π2 cos a 2a 2cos a 2 f x dx

6

2 2 2 2π π π4cos a 4a 2cos a . sin a 2a 4 cosa sin a f a

6 6 6




2

3f 0 4 3

2


This section contain TEN questions

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these

four option (s) is (are) correct

For each question, darken the bubble (s) corresponding to all the correct option (s) in the

ORS

Marking scheme :

+ 4 If only the bubble (s) corresponding to all the correct option (s) is(are) darkened



50. Let X and Y be two arbitrary, 33, non-zero, skew-symmetric matrices and Z be an

arbitrary 33, non–zero, symmetric matrix. Then which of the following matrices is (are)

skew symmetric ?

(a) 3 4 4 3Y Z Z Y (b) 44 44X Y (c) 4 3 3 4X Z Z X (d) 23 23X Y

Ans. (C, D)

Solution: (C)

T T T

4 3 3 4 4 3 3 4x Z Z X X Z Z X

= 3 4 3

T T T TZ X X Z

= 3 4 4 3Z X X Z

(D) T

23 23 23 23X Y X Y 23 23X Y is skew-symmetric

51. Which of the following values of a satisfy the equation

2 2 2

2 2 2

2 2 2

1 α 1 2α 1 3α

2 α 2 2α 2 3α 648α?

3 α 3 2α 3 3α

Ans. (B, C)

Solution :

3 3 2 2 2 1R R R ,R R R

2 2 2

1 α 1 2α 1 3α

3 2α 3 4α 3 6α 648α

5 2α 5 4α 5 6α




3 3 2R R R

2 2 2

1 α 1 2α 1 3α

3 2α 3 4α 3 6α 648α

2 2 2

3 3 2 2 2 1C C C , C C C

2

1 α α 2 3α α 2 5α

3 2α 2α 2α 648α

2 0 0

2 22α 2 3α 2α 2 5α 324α

34α 324α α 0, 9

52. In R3, consider the planes P1 : y = 0 and P2: x + z = 1. Let P3 be a plane, different from

P1 and P2, which passes through the intersection of P1 and P2. If the distance of the point

(0, 1, 0) from P3 is 1 and the distance of a point (α, β, γ) from P3 is 2, then which of the

following relation is (are) true ?

(a) 2α β 2γ 2 0 (b) 2α β 2γ 4 0

(c) 2α β 2γ 10 0 (d) 2α β 2γ 8 0

Ans. (B, D)

Solution:

Let 3P be 2 1P λP 0 x λy z 1 0

Distance from (0, 1, 0) is 1

2

0 λ 0 11

1 λ 1

1λ

2

Equation of 3P is 2x – y + 2z – 2 = 0

Dist. From α, β, γ is 3

2α β 2γ 2

23

2α β 2γ 2 6

(B, D)

53. In R3, let L be a straight line passing through the origin. Suppose that all the points on L

are at a constant distance from the two planes P1 : x + 2y – z + 1 = 0 and P2 : 2x – y + z – 1

= 0. Let M be the locus of the feet of the perpendiculars drawn from the points on L to the

plane P1. Which of the following points lie(s)on M?




(a) 5 2

0, ,6 3

(b)

1 1 1, ,

6 3 6

(c) 5 1

,0,6 6

(d) 1 2

,0,3 3

Ans. (A, B)

Solution:

Let v be the vector along L

then

i j k

v 1 2 1 i 3j 5k

2 1 1

So any point on line L is A λ, 3λ, 5λ

Foot of perpendicular from A to P, is

λ 6λ 5λ 1h λ k 3λ 5λ 1

1 2 1 1 4 1 6

1 1 1h λ , k 3λ , 5λ

6 3 6

so foot is 1 1 1

λ , 3λ , 5λ6 3 6

So, (A, B)

54. Let P and Q be distinct points on the parabola y2 = 2x such that a circle with PQ as

diameter passes through the vertex O of the parabola. If P lies in the first quad rant and

the area of the triangle OPQ is 3 2 'then which of the following is (are) the coordinates

of P ?

(a) ( 4,2 2 ) (b) 9, 3 2 (c) 1 1

,4 2

(d) 1, 2

Ans. (A, D)

Solution:

OP OQ 1 2t t 4

Now 1

OP,OQ 3 22




4 4

2 21 21 2

1 t tt t 3 2

2 4 4

2 2

1 2t 4 t 41.4. 3 2

2 4 4

2 2

1 216 4 t t 164. 9 2

16

2 21 28 t t 18

2 21 2t t 10 0

4 21 1t 10t 16 0

21t 2,8

55. Let y(x) be a solution of the differential equation (1 + ex)y' + yex = 1. If y(0) = 2, then

which of the following statements is (are) true ?

(a) y(–4) = 0

(b) y(–2) = 0

(c) y(x) has a critical point in the interval (–1, 0)

(d) y(x) has no critical point in the interval (–1, 0)

Ans. (A, C)

Solution:

x xdy1 e ye 1

dx

x

x x

dy e 1y

dx 1 e 1 e

I.F =

x

x xx

edx n 1 e 1 e

1 ee e




complete solution

xy. 1 e 1dx

x1 e y x c

x = 0, y = 2 c = 4

x1 e y x 4

x

x 4y

e 1

x = – 4, y = 0

x = – 2, y = 2

2

x 1

x x

2x

e 1 . 1 x 4 edy

dx e 1

x

2x

e x 3 1

e 1

dy

0dx

xx 3 e

x 1e

x 3

56. Consider the family of all circles whose centers lie on the straight line y = x. If this family

of circles is represented by the differential equation Py" + Qy' +1=0, where P, Q are

functions of x, y and y' (here 2

2

dy d yy' , y"

dx dx ), then which of the following statements is

(are) true?

(a) P = y + x

(b) P = y – x

(c) P + Q = 1 – x + y + y’ + (y’)2

(d) P – Q = x + y – y’ – (y’)2




Ans. (B, C)

Solution:

2 2 2x α y α r

2 2 2 2x y 2αx 2αy 2α r 0

2x 2yy' 2α 2αy' 0 ..(i)

x yy'

α1 y'

..(ii)

again diff. w. r. t.

2 + 2(y’) 2 + 2yy” – 2α y” = 0

2 x yy'

1 y' yy' y" 01 y'

2 3

1 y' y' y' yy" yy'y' xy" yy'y" 0

2y x y" 1 y y' y' 1 0

2

P y x, Q 1 y' y'

Ans. (B, C)

Note : P & Q will not be unique function as

Py” + Qy’ + Ry’ – Ry’ + 1 = 0

Py' Qy'

1 01 Ry" 1 Ry'

57. Let g : RR be a differentiable function with g(0) = 0, g’(0) = 0 and g’(1) 0.

Let f(x) =

xg x , x 0

x

0, x 0

and h(x) = x

e for all x R. Let (foh) (x)denote f(h(x)) and

(hof)(x) denote h(f(x)). Then which of the following is (are) true?

(a) f is differentiable at x = 0 (b) h is differentiable at x = 0

(c) foh is differentiable at x = 0 (d) hof is differentiable at x = 0

Ans. (A, D)

Solution:

g(0) = 0, g’(0) = 0 g’(1) 0

x

g x ; x 0

f x g x ; x 0 h x e

0 ; 0

g xxf h x g e , h f x e




x 10

g x g 0R f ' 0 lim g' 0 0

x 0

x 10

g x g 0I f ' 0 lim g' 0 0

x 0

R h' 0 1 & L h' 0 1 So h(x) is non derivable, at x = 0

Now x

x 0 x 0

g e g 1f h x f h 0lim lim

x x

R f ' h x g' 1 Hence f(h(x)) is non derivable at x = 0

Since, x = 0 is repeated root of g(x) So g x

e is differentiable at x= 0

hence (A), (D)

58. Let f(x) = sin π π π

sin sin x for all x R and g x6 2 2

sin x for all x R. Let (fog) (x)

denote f(g(x) and (gof) (x)denote g(f(x)). Then which of the following is (are) true?

(a) Range of f is 1 1

,2 2

(b) Range of fog is 1 1

,2 2

(c) x 0

f x πlim

g x 6 (d) There is an x R such that (gof) (x) = 1

Ans. (A, B, C)

Solution:

π π

f x sin sin sin x6 2

Let π

sin x θ2

π π

θ ,2 2

π

f x sin sin θ6

Let π

sin θ6

π π

,6 6

1 1

f x sin ,2 2

A

Now fog(x) = sin π π π

sin sin sin x6 2 2

Clearly, range of fog is also 1 1

,2 2

B




Now, x 0

π πsin sin sin x

6 2lim

πsin x

2

= x 0

π π π πsin sin sin x sin sin x

6 22 6 2lim

sin xπ ππ xsin sin xx6 2

= x 0

π πsin sin x sin x2 π 2 2limππ 6 xsin x2

1 π π

3 2 6 C

Now, gof(x) = 1

π π π

sin sin sin x 12 2 2

π π 2 2 1

sin sin sin x6 2 λ 3.14 2

(D)

59. Let PQR be a triangle. Let a QR, b RP and c PQ . If a 12, b 4 3 and b. c 24,

then which of the following is (are) true?

(a) 2|c|

|a| 122

(b) 2|c|

|a| 302

(c) |a b c a| 48 3 (d) a. b 72

Ans. (A, C, D)

Solution:

a b c 0

b c a




2

48 c 48 144

2

c 48

2

c 4 3

2

ca 24 12 12

2 Ans. (A)





Column II has five entries (P), (Q), (R), (S), and (T)


One or more entries in Column I may match with one or more entries in Column II

The ORS contains a 4 5 matrix whose layout will be similar to the one shown below:




Marking scheme :


+2 If only the bubble (s) corresponding to all the correct match (es) is (are) darkened



60.

Column I Column – II

(A) In R2, if the magnitude of the projection vector of the vector

αi βj on 3 i j is 3 and if α 2 3 β, then possible value

(s) of α is (are)

(P) 1

(B) Let a and b be real numbers such that the function (Q) 2




2

2

3ax , x 1f x

bx a , x 1

is differentiable for all x R.

(C) Let ω 1 be a complex cube root of unity.

If 4n 3 4n 3 4n 3

2 2 23 3ω 2ω 2 3ω 3ω 3 2ω 3ω 0,

then possible value (s) of n is (are)

(R) 3

(D) Let the harmonic mean of two positive real numbers a and b be

4. If q is a positive real number such that a, 5, q, b is an

arithmetic progression, then the values (s) of q a is (are)

(S) 4

(T) 5

Ans. A P,Q; B P, Q; C P, Q, S, T; D Q,T

Solution:

(A) 3i jαi βj . 3

2

3α β 2 3

α 23α 2 3

3

3α α 2 6 4α 8, 4 α 2, 1

A P, Q

(B) Continuous 23a 2 b a

differentiable 26a b 6a a 3a 2

2a 3a 2 0 a 1,2

B P,Q

(D) 2ab

4a b

ab = 2a + 2b ..(i)

q = 10 – a and 2q = 5 + b

20 – 2a = 5 + b 15 = 2a+ b ..(ii)

From (i) and (ii) a (15 – 2a) = 2a + 2(15 – 2a)

15a – 2a2 = – 2a + 30 2a2 – 17a + 30 = 0 5

a 6,2

15

q 4,2

q a 2,5

D Q,T

(C) Let 2a 3 3ω 2ω




2aω 3ω 3ω 2 2 2aω 3ω 3 2ω

Now 4n 34n 3 4x 3 2a 1 ω ω 0

n should not be a multiple of 3

Hence P, Q, S, T

61.

Column I Column – II

(A) In a triangle XYZ, let a, b and c be the lengths of the sides

opposite to the angles X, Y and Z, respectively. If 2(a2 – b2) = c2

and sin X Y

λsinZ

, then possible values of n for which

cos nπλ 0 is (are)

(P) 1

(B) In a triangle XYZ, let a, b and c be the lengths of the sides

opposite to the angles X, Y and Z, respectively. If

1 + cos 2X–2cos 2Y = 2sin X sin Y, then possible value (s) of a

b

is (are)

(Q) 2

(C) In R2, let 3 i j , i 3 j and βi 1 β j be the position vectors

of X, Y and Z with respect to the origin O, respectively, if the

distance of Z from the bisector of the acute angle of OX and OY

is 3

2, then possible value (s) of β is (are)

(R) 3

(D) Suppose the F α denotes the area of the region bounded by

x = 0, x = 2, y2 = 4x and y = αx 2 αx , where α {0,1}. Then

the value (s) of F 8

α 2,3

when α 0 and α 1, is are

(S) 5

(T) 6

Ans. A P,R,S; B P; C P,Q; D S,T

Solution: (A)




Given 2 2 22 a b c

2(sin2x – sin2y) = sin2z

2sin(x + y) sin (x – y) = sin2z

2sin (π – z) sin (x – y) = sin2z

z sin z

sin x y2

..(i)

also given,

sin x y 1λ

sin z 2

Now, cos nπλ 0

nπ

cos 02

n = 1, 3, 5 A P, R, S

(B)

1 + cos 2x – 2 cos 2y = 2 sin x sin y

2cos2x – 2cos2y = 2sin x sin y

1 – sin2x – 1 + 2sin2y = sin x sin y

sin2x + sin x sin y = 2 sin2y

sin (sin x + sin y) = 2sin2y sin x = ak, sin y = bk

a(a + b) = 2b2

a2 + ab – 2b2 = 0 2

a a2 0

b b




a2,1

b

a1

b B P

(C)

Hence equation of acute angle bisector of OX and OY is y = x

Hence x – y = 0

Now, distance of βi 1 β j Z β, 1 β from x – y is β 1 β 3

2 2

2β 1 3

2β 1 3

2β 4, 2

β 2, 1

β 2,1 Ans. (P, Q)

(D)

For α 1

3 x ; x 1

y x 1 x 1 x 1 x ; 1 x 2

3x 3 ; x 2




2

0

1 1A 2 3 1 2 3 1 2 xdx

2 2

8A 5 2

3

8

F 1 2 53

For α 0, y 1 2 3

2

0

A 6 2 xdx 8

A 6 23

8

F 0 2 63

D s, t


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