JEE Advance 2015 Paper 1 Solution

JEE-Advance – 2015 (Hints & Solutions) Paper - 1

NARAYANA Group of Educational Institutions

-1-

NARAYANA IIT ACADEMY INDIA

PAPER - I CODE : 0 Time : 3 Hours Maximum Marks : 264

READ THE INSTRUCTIONS CAREFULLY

GENERAL : 1. This sealed booklet is your Question Paper. Do not break the seal till you are instructed to do so. 2. The question paper CODE is printed on the left hand top corner of this sheet and the right hand top

corner of the back cover of this booklet. 3. Use the Optical Response Sheet (ORS) provided separately for answering the questions. 4. The ORS CODE is printed on its left part as well as the right part. Ensure that both these codes are

identical and same as that on the question paper booklet. If not, contact the invigilator. 5. Blank spaces are provided within this booklet for rough work. 6. Write your name and roll number in the space provided on the back cover of this booklet. 7. After breaking the seal of the booklet, verify that the booklet contains 32 pages and that all the 60

questions along the with the options are legible. QUESTION PAPER FORMAT AND MARKING SCHEME : 8. The question paper has three parts : Physics, Chemistry and Mathematics. Each part has three

sections. 9. Carefully read the instructions given at the beginning of each section. 10. Section 1 contains 8 questions. The answer to each question is a single digit integer ranging from

0 to 9 (both inclusive) 11. Section 2 contains 10 multiple choice questions with one or more than one correct option. Marking Scheme : +4 for correct answer, 0 if not attempted and –2 in all other cases. 12. Section 3 contains 2 “match the following” type questions and you will have to match entries in

Column I with the entries in Column II. Marking Scheme : for each entry in Column I, +2 for correct answer, 0 if not attempted and –1 in

all other cases. OPTICAL RESPONSE SHEET : 13. The ORS consists of an original (top sheet) and its carbon-less copy (bottom sheet). 14. Darken the appropriate bubbles on the original by applying sufficient pressure. This will leave an

impression at the corresponding place on the carbon-less copy. 15. The original is machine-gradable and will be collected by the invigilator at the end of the

examination. 16. You will be allowed to take away the carbon-less copy at the end of the examination. 17. Do not tamper with or mutilate the ORS. 18. Write your name, roll number and the name of the examination center and sign with pen in the

space provided for this purpose on the original. Do not write any of these details anywhere else. Darken the appropriate bubble under each digit of your roll number.



-2-

JEE-ADVANCE SOLUTIONS PHYSICS

SECTION : 1 (Maximum Marks : 32)

This section contains EIGHT questions The answer to each question is a SINGLE DIGIT NUMBER 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 the bubble corresponding to the answer is darkened. 0 In all other cases. 1. Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm

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 of 15 cm from the mirror. Its erect image formed by this combination has magnification M1. When the set-up is kept is a medium of refractive index

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

1

M

M is

Ans. (7) Sol: For mirror u 15cm f 10cm

1 1 1

v u f

1 1 1

v 15 10

1 1 1

v 15 10

v = –30 cm For lens u 20cm f 10cm

1 1 1

v 20 10

1 1 1

v 10 20

v = 20 cm

1

30 20M 2

15 20

In second case, for mirror v 30 cm New focal length of lens



-3-

3 11 352 2f 10 10

3 6 9 21 12 7 7

for lens 1 1 2

v 20 35

1 2 1 1

v 35 20 140

v= 140 cm

2

30 140M 14

15 20

2

1

M 147

M 2 .

2. A Young’s double slit interference arraignment 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 = p2m22 – d2, where 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) Sol: For maxima path difference = n

2 2 4d x 1 n

3

2 2 4

d x 1 n3

2 2 2 2x 9n d p = 3 3. Two identical uniform discs roll without slipping on two different surface AB and CD (see

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



-4-

Sol: 2 21 1 2 2

3 3mv mgh mv mgh

4 4

2 22 1 1 2

3v v g h h

4

2 22 1v v 4g 40

22v 49

v2 = 7 m/s. 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 form the planet is vese = v N, then the value of N is (ignore energy loss due to atmosphere)

Ans. (2) Sol: At the maximum height, acceleration due to gravity is 1/4th hmax = R (radius of planet)

Now 21 GMm GMmCOE mv

2 R 2R

GMv

R

Now esc

2GMv v 2 N 2

R

5. 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

of their wavelengths A

and B at which the peaks occur in their respective radiation curves is Ans. (2)

Sol: 2

4

rp

(For spherical stars)

4

24 B

A

10 400

A

B

2

6. A nuclear power plant 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 nT years, then the value of n is

Ans. (3) Sol: At t = 0 Prequired = Pavailable (12.5/100) = Pavailable/8 Pavailable = 8 Prequired After first half life, Pavailable = 4 Prequired After second half life, Pavailable = 2Prequired After third half life, Pavailable = Prequired Power required by the village can be met upto 3 half lives. n = 3



-5-

7. An infinitely long uniform line charge distribution of charge per unit length lies parallel to

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

z a2

(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 00

L(

n

permittivity of free space), then the value of n is

z

A B

C D a

L

O

x

y

3a

2

Ans. (6) From figure,

a / 2 1

tan3 3

a2

=

30o 2 = 60o Total flux through 360o =

in

o o

q L

Flux through 60o is equal to

o

oo o

L 60 L

360 6

n = 6

+ + + + + + + + + + + + + + + + + + + +

a a/2

D

a/2

y

x

A B

C

G H

E

(3/2)a

8. 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) Sol: Energy of the incident radiation = hc/ = 1242/90 = 13.8 eV

Total energy of electron in nth orbit = 13.6eV2n

2

13.613.8 10.4eV

n

2

13.613.8 10.4 3.4

n

2 13.6n 4

3.4 n = 2



-6-

SECTION : 2 (Maximum Marks : 40) This section contains TEN questions Each question has FOUR options (A), (B), (C) and (D). ONE OR ORE 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 one of the bubbles is darkened. 2 In all other cases. 9. The figures below depict two situations in which two infinitely long static line charges of

constant positive line charge density 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)

x +q

x q

(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) Sol: In situation 1, let the distance between 2 line charges be 2a. Now if +q is shifted to the left by a

small distance x , then

Fnet = 2K 2K

a x a x

(towards right)

= 2 2 2 2

2x 4K2K x

a x a x

Force is opposite to direction of displacement

Fnet = 2 2

4Kx

a x

Now as x << a, a2 – x2 = a2

Fnet = 2

4Kx

a

+q will execute SHM Whereas in situation 2 when q is displaced it will continue to move in the direction of

displacement. option (C) is correct.



-7-

10. 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

P S1 S2

50 cm d

(A) 60 cm (B) 70 cm (C) 80 cm (D) 90 cm Ans. (B) For refraction on curved surface of S1 1 = 1.5, 2 = 1, R = 10, u = 50

using 2 1 2 1

v u R

, we get v = 50 cm

P S1 S2

50 cm

20 cm

Now for refraction at curved surface of S2 v = (as rays inside S2 are parallel to the principal axis) 1 = 1, 2 = 1.5, R = +10, u = x

1.5 1 1.5 1

x 10

x = 20

d = 50 + 20 = 70, so option (B) is correct. 11. 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)

Sol: mF i dl B

In (A) part mF L R

If B is along z axis. In (B) part mF 0

If B is along x axis. In (C) part mF L R

If B is along y axis. All the result as according to above formula.



-8-

12. A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium 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 of 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 ½

(D) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/ 2 Ans. (A, B, D)

Sol: Average energy/ mole

3 5RT RT

2 2 2RT2

(B) Hemix

mix He

MSpeedof soundin mixture

Speedof soundin theHeliumgas M

RTv

M

i Pmix

i V

n C

n C

1 1Average molar massof mixture 4 2 2 1 3

2 2

3/ 2 4 6

3 5/ 3 5

(D) rms1

vM

2

rms He

rms H

v 2 1

v 4 2

13. In an aluminium (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 82.7 10 and

71.0 10 m , respectively. The electrical resistance between the two faces P and Q the composite bar is

(A) 2475

64 (B)

1875

64 (C)

1875

49 (D)

2475

132

Ans. (B) Sol: 3m50 10 l , 8

Al 2.7 10 , 7Fe 1.0 10

8 35Al

Al 6

2.7 10 50 10R 3.05 10

A 7 7 2 2 10

l



-9-

7 35Fe

Fe 6

1.0 10 50 10R 125 10

A 7 7 2 2 10

l

Now Fe Aleqn

Fe Al

R R 1875R

R R 64

14. For photo-electric effect with incident photon wavelength , the stopping potential is V0. Identify the correct variation(s) of V0 with and 1/

(A) (B)

(C) (D)

Ans. (A, C) Sol: By photoelectric equation 0 0hf W eV

0 0c

h W eV

01

V

15. Consider a Vernier calipers 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 calipers, 5 divisions of the Vernier scale coincide with 4 divisions 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 calipers, the least count of the screw gauge is 0.01 mm.

(B) If the pitch of the screw gauge is twice the least count o the Vernier calipers, the least count of the screw gauge is 0.005 mm

(C) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier calipers, 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 calipers, the least count of the screw gauge is 0.005 mm

Ans. (B, C) Sol: L.C. of vernier caliper = 1 MSD – 1 VSD



-10-

1 1

cm cm8 2 5

10 8 2 1

cm 0.025cm 0.25mm80 80 40

L.C. of screw gauge pitch

totalno.of division on circular scale

Pitch = linear distance moved on main scale in one rotation of circular scale. 16. Planck’s constant h, speed of light c and gravitational constant G are used to form a unit of

length L and a unit of mass M. Then the correct option(s) is (are) (A) M c (B) M G (C) L h (D) L G Ans. (A,C, D) Sol. [h] = ML2T-1 ...(i) [G] = M-1L3T-2 .....(ii) [C]= LT-1 ...(iii) From (i) and (ii) [hG] = L5T-3 = L2 =L2(L3 T-3) L2 [LT-1]3

2

3

hGL

C

L G

L h

3 2hc ML T

3 2

1 3 2

hc ML T

G M L T

2hc

mG

hc

MG

m c

Option (A), (C) and (D) are correct. 17. Two independent harmonic oscillators of equal mass are oscillation about the origin with

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

variations of their momenta p with positions x are shown in the figure. If 2an

b and

an,

R

then the correct equations (s) is (are) p

Energy = E1

xa

b

p

Energy = E2

xR



-11-

(A) 1 1 2 2E E (B) 22

1

n

(C) 21 2 n (D) 1 2

1 2

E E

Ans. (B,D) Sol.

Given 2a an , n

n R

2

2p 1E kx

2m 2

p

Energy = E1

xa

b

p

Energy = E2

xR

From figure

2

211

mE a

2

2 22 2

mE R

2

2 22

21 1 1

2 2 2

E a.n

E R

.....(i)

Also,

21 2 2

1 1222 2

pE p b2m

pE p R2m

Or,

2

12

22

E a 1aE nnn

.....(ii)

From (i) and (ii)

2

1 14 2

2 2

1 1

n n

22

1

n

And 1 1

2 2

E

E

1 2

1 2

E E

Hence, (B) and (D) are correct options



-12-

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

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

8 at rest at O. 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 8

9 and one of the masses is at a

distance of 3

R5

from O. At this instant the distance of the other mass from O is

O

(A) 2

R3

(B) 1

R3

(C) 3

R5

(D)4

R5

Ans. (D) Sol. Using conservation of angular momentum We have

2 2 2 2M 9 M 8MR MR R x

8 25 8 9

2 2

2 29 9R xR R

8 200 8

2

2 21 9 xR R

8 200 8

2 216x R

25

4

x R5

O



-13-

SECTION : 3 (Maximum Marks : 16) 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 match with one or more entries in column II. The ORS contains a 4 5 matrix. 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 If only the bubble(s) corresponding to all the correct match(es) is(are) darkened. 0 If one of the bubbles is darkened. 1 In all other cases. 2 In all other cases. 19. Match the nuclear process 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) 6027 Co nucleus

(C) -decay (R) Energy production in stars via hydrogen conversion to helium

(D) ray emission (S) Heavy water (T) Neutrino emission

Ans. AR,T; BP,S,T; CQ,T; DR,T Sol. Based on theory



-14-

20. 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 statement (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. AP,Q,R,T; BQ,S; CP,Q,R,S; DP,R,T.

Sol. 2

011 2 2

2U xdu xF x 1

dx a a

02 2

U 2xF x

2 a

2

2 22

x 2x /a0 a

3 2 2 2

U 2x x 2xF x e e

2 a a a

2

2

x 20 a

3 2 2

2U xF xe 1

a a

2

04 3

U 1 1 3xF

2 a 3 a



-15-

PART II: CHEMISTRY

SECTION : 1 (Maximum Marks : 32) This section contains EIGHT questions The answer to each question is a SINGLE DIGIT NUMBER 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 the bubble corresponding to the answer is darkened. 0 In all other cases. 21. 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 –0.0558oC, the number of chloride(s) in the coordination sphere of the complex is

Key: (1) Sol. f fT i K m

0.0558 i 1.86 0.01

0.0558

i 30.0186

3 3CoCl .x NH

If i = 3, then total ions = 3 No. of Cl– ions in the coordination sphere = 1

The complex could be 3 2xCo NH Cl Cl

22. All the energy released from the reaction o 1X Y, G 193kJ mol

is used for oxidizing M+ as 3 oM M 2e ,E 0.25V .

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]

Key: (4) Sol. Energy released when one mole of X converts into Y = 193 kJ.

Energy needed for oxidation of 1 mole of M+ under standard condition = onFE

2 96500 0.25 J

48.25kJ

No. of moles of M+ oxidized 193

448.25

23. For the octahedral complexes of Fe3+ in SCN– (thicyanato-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]

Key: 4 Sol. Fe3+ with SCN– (weak filed ligand) High spin octahedral complex

3

6Fe SCN



-16-

3 3 2Fe tg eg

I 5 5 2 35 BM

3Fe with CN (strong field ligand) Low spin octahedral complex

3

6Fe CN

3 52Fe t g

II 1 1 2 3 BM

I II 35 3

5.91 1.73 4.18 4 BM

24. The total number of lone pairs of electrons in 2 3N O is

Key: (8)

Sol.

N NO

O

O

(Asymmetric)

N

O ON

O

(symmetric)

Total number of lone pairs are 8 in 2 3N O molecule.

25. Among the triatomic molecules/ions, BeCl2, 3N , 2N O , 2NO , 3O , 2SCl , 2ICl , 3I and XeF2,

the total number of linear molecule(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]

Key: (4) Sol. 2BeCl : sp

3N : sp

2N O : sp

2NO : sp

Other linear molecules/ions involve d orbitals. 26. 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 Key. (3) Sol. The possible excited states are 1s12 1

xp

1s11y2p

1s2 1z2p



-17-

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

H3C CH3

CH3O

M Key. (2) Sol.

H3C CH3

CH3O

M

12

3

45

6

78

9

The compound M has two stereogenic carbons, hence 1R4R, 1S4S, 1R4S and 1S4R

configurations are possible. Due to cage like structure it is not possible to have R configuration at C1 and S-configuration at C4. So total number of stereoisomers are 2.

28. The number of resonance structures for N is

OHNaOH N

Key. (9) Sol.

OO O

O

O

O

O O

O



-18-

SECTION : 2 (Maximum Marks : 40) This section contains TEN questions Each question has FOUR options (A), (B), (C) and (D). ONE OR ORE 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 one of the bubbles is darkened. 2 In all other cases. 29. The major product of the reaction is

H3C CO2H

CH3 NH2

2o

NaNO , aqueous HCl

0 C

(A)

H3C NH2

CH3 OH (B)

H3C CO2H

CH3 OH

(C)

H3C CO2H

CH3 OH (D)

H3C NH2

CH3 OH Key. (B) Sol.

H3C CO2H

CH3 NH2

2o

NaNO , aqueous HCl

0 C

H3C CO2H

CH3

1N N

2N

H3C COOH

CH3

2H OH

H3C CO2H

CH3 OH

H3C CO2H

CH3 OH 30. 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 d4 electronic configuration (D) When Cr2+ is used as a reducing agent, the chromium ion attains d5 electronic configuration Key. (A), (B), (C) Sol.

3 1 3

2 3

2g eg 2g

Cr Cr

t t (more stable)



-19-

3 2

4 5

Mn e Mn

3d 3d (stable electronicconfiguration)

31. 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 Key. (B, C, D) Sol. Anodes are of impure Cu and Cu-strip is cathode. The electrolyte is acidified CuSO4 and the net

process is,

Anode: 2Cu Cu 2e

Cathode: 2Cu 2e Cu s

Impurities from the blister Cu deposited as anode mud. 32. Fe3+ is reduced to Fe2+ by using (A) H2O2 in presence of NaOH (B) Na2O2 in water (C) H2O2 in presence of H2O4 (D) Na2O2 in presence of H2SO4 Key. (A, B) Sol. In acidic medium, 2 2 4 2 4 2 4 23

H O 2FeSO H SO Fe SO 2H O

In Basic medium, 2 2 3 4 2 26 6H O 2K Fe CN 2KOH 2K Fe CN 2H O O

22 2 2 2H O 2Fe 2OH 2Fe 2H O O

Since, 2 2 2 2 2Na O 2H O 2NaOH H O

Therefore, Na2O2 in water also represent same reaction as that of basic H2O2. 33. The % yield of ammonia as a function of time in the reaction

2 2 3N g 3H g 2NH g , H 0

At (P, T1) 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



-20-

(A) (B)

(C) (D) Key. (B)

Sol. 2 2 3N g 3H g 2NH g , H 0

With referenced to the graph given at )P, T1) Initially at high temperature yield will be more but after some interval it will be constant with type

34. 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

Key. (A) Sol. In cubic close packing unit cell = fcc O2– form ccp lattice (Z=4) Al3+ occupy m fraction of OV = 4m Mg2+ occupy n fraction of TV = 8n Charge should be neutral Hence, 4m ×(3) + 8n × (2) = 4 × (2) 12 m + 16 n = 8 3 m + 4n =

That is satisfied by when 1 1

m , n2 8



-21-

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

(A)

(B)

(C)

(D)

Key. (B, D) Sol. Both (B) (D) will form optically inactive product.

Hydrogenation

Optically inactive

Hydrogenation

Optically inactive

36) The major product of the following reaction is

O

O

3CH2. ,

. ,

i KOH H O

ii H heat

(A)

O

3CH

(B)

O

3CH

(C)

3CH

O

(D)

3CHO

KEY: (A)

SOL:

O

OCH3

OH O

OCH3

OH+O

CH3

OHCH3

OCH3

O



-22-

37) In the following reaction, the major product is

1 equivalent HBr

2H C

3CH

2CH

(A)

3CH

2H C

3CH

Br

(B)

3H C

3CH

Br

(C)

3CH

2H C Br (D)

3CH

2H CBr

KEY: (D)

SOL:

2CH

3CH

2CH

C+

3CH

3CH

3CH

C

Br

2CH

Br3CH

C

CH3

CH3

CH2

+

Br–

C

CH3

CH3

CH Br2

CH

CH2

H+

C

Thermodynamically more stable product

C

Since temperature is not given, therefore, option (B) may also be correct.



-23-

38) The structure of D-(+) glucose is

CHO

CH OH2

HHO

HH

OHHOHOH

The structure of L- (-) glucose is

(A)

CHO

CH OH2

HOH

HOHO

HOHHH

(B)

CHO

CH OH2

HHO

HHO

OHHOHH

(C)

CHO

CH OH2

HOHO

HHO

HHOHH

(D)

CHO

CH OH2

HOHOHO

H

HHHOh

KEY: (A)

SOL: L (–) glucose should be mirror image of D(+) glucose hence

The correct answer should be

CHO

CH OH2

HOH

HOHO

HOHHH

Hence correct answer is (A)



-24-

SECTION : 3 (Maximum Marks : 16) 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 match with one or more entries in column II. The ORS contains a 4 5 matrix. 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 If only the bubble(s) corresponding to all the correct match(es) is(are) darkened. 0 If one of the bubbles is darkened. 1 In all other cases. 2 In all other cases.

39) 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

KEY: AP,Q,S; BT; CR,Q, DR

SOL:

(P) Siderite - FeCO3

(Q) Malachite - CuCO3.Cu(OH)2

(R) Bauxite - AlOx(OH)3-2x

(S) Calamine - ZnCO3

(T) Argentite - Ag2S



-25-

40) Match the thermodynamics process 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 vacuum under isolated conditions (Q) w = 0

(C) Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container

(R) sysS 0

(D) Reversible heating of H2(g) at 1 atm from 300 K to 600 K, followed by reversible cooling to 300 K at 1 atm

(S) U 0

(T) G 0

KEY: AR,T; BP,Q,S; CP,Q,S; DP,Q,S,T

SOL:

(A) For freezing of water at 273K and 1 atm

S 0, G 0 (it is at equilibrium)

w is not zero as there will be expansion in freezing

(B) Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions

Q = 0, w = 0, U 0

(C) Mixing of equal volume of two ideal gases at constant temperature and pressure in an isolated container

Q = 0, w = 0, U 0

(D) Reversible heating of H2(g) at 1 atm from 300 K to 600 K, followed by reversible cooling to 300 K at 1 atm

Q = 0, w = 0, U 0 , G 0



-26-

MATHEMATICS

41. Let

2x6

2

x

F x 2cos tdt

for all x R and 1

f : 0, 0,2

be a continuous function. For

1a 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 Key. (3)

Sol.

2x6

2

x

F x 2cos t dt

2a6

2

a

F a 2cos t dt

2 2 2F' a 2cos a .2a 2cos a.16

2 2 2F' a 2 4a cos a 2cos a 26

= 2 2 24a cos a 2sin a6

a

2 2 2

0

f x dx 4a cos a 2sin a6

2 2 2 2f a 4 a 2cos a sin a .2a cos a .1 4sin a cosa6 6 6

f 0 3

42. A cylindrical container is to be made from certain solid material with the following constraints: It has a 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 a 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

250 is

Ans. (4) Sol. Let inner radius = r Height of cylinder = h Volume of inner cylinder = v

2v r h

2

vh

r

Now, Material used

2 22M r 2 h r h r 2 .2



-27-

= 2 2

2

vr 2 . v 2 r 2

r

M = 2

2

v 4 41 v 2 r 2

r r

M is the function of r

2 3

v 4 8f ' r 0 4 r 2 0

r r

when r = 10,

v 40 8

48 01000 1000

48v

481000

v

4250

43. 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) Sol. Clearly n 6!5! Now, to calculate value of m We make three cases

Case 1 : first four places filled by girls fifth by boy. This can be done in 25

4P .5 5! 5. 5!

Case 2 : Four girls are in a row having both the side a boy this can be done in

25 5

4 25 P P 4! 20. 5!

Case 3 : Last four places are filled by girls and 6th place by a boy this can be done in

25

4P .5.5! 5. 5!

2

m 30. 5!

So, m

5n

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) Sol. Let coin is tossed ‘m’ times, then probability of getting at least 2 heads is

m m

m1 m

m 11 11 C 1

2 2 2

Given that

m

m 11 0.96

2

m

m 1 10.04

2 25



-28-

Let m

m 1f m

2

3 1 5 6

f 1 1, f 2 , f 3 , f 4 , f 54 2 16 32

,

7 8 9 10

f 6 , f 7 , f 8 , f 9 ,....64 128 256 512

So minimum value of m is 8 45. If the normals of the parabola y2 = 4x drawn at the end points of its latus rectum are

tangents to the circle 2 2 2x 3 y 2 r , then the value of r2 is

Ans. (2) Sol. 2y 4x , end points of latus rectum are 1, 2

Equation of normal at (1, 2) is y x 3 , it touches the circle

So 2

3 2 3r r 2 r 2

1 1

46. Let f : RR be a function defined by x , x 2

f (x) ,0, x 2

Where [x] is the greatest integer less than or equal to x. If 22

-1

xf(x )I = dx,

2+ f(x+1)then the value

of (4I – 1) is Key. (0)

Sol. 22

-1

xf(x )I = dx,

2+ f(x+1)

1 2

0 10 0

2 0

0

-1

xI = dx+ dx dx +

3 4

320 0 dx+ dx

2 2

1( )1 1 1. 2 1

2 2 4 4

xI =

4 -1= 0.I 47. The number of distinct solutions of the equation

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

4 x x x x x

In the interval 0,2 is

Key. (8)

Sol. 2 2 2 2 25cos 2 1 2sin cos 1 3sin cos 2

4 x x x x x

2 2 25 1 3cos 2 sin 2 sin 2 0

4 2 4 x x x

2 2cos 2 sin 2 0 x x

cos4 0x Since, 0,2 x

Hence no. of distinct solution = 8.



-29-

48. Let the curve C be the mirror image of the parabola 2 4y x with respect to the line

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

Key. (4) Sol. Let a Point 2P(t , 2t) on 2 4y x

Image of 'P' w.r.t. the line 4 0 x y can be given as: 2

2y 2tt 2t 4

1 1

x - t

2 4; x t 2 4. y t

Hence, Curve C becomes

2

44

x+4

- y.

Since, it intersect with the line y = -5 2( 4) 16 20 x

2( 4) 4 x

4 2 x 4 2 x OR 4 2 x = (-2, -5)A ; = (-6, -5)B AB = 4. 49. Let P and Q be distinct points on the parabola 2 2y x such that a circle with PQ as

diameter passes through the vertex O of the parabola. If P lies in the first quadrant 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

Key. (A, D) Sol.

O

21 1P at , 2at

22 2Q at , 2at

1 2t t 4 (Since, PQ is the diameter of circle)

area of 21 122 2

0 0 11

at 2at 12

at 2at 1

OPQ =

11

43 2 t

t

1t 2,2 2 (Since P lies in first Quadrant)

P 1, 2 , 4,2 2



-30-

50. Let y(x) be a solution of the differential equation 1 e y ye 1 'x x . If y(0) = 2, then which of

the following statements is (are) true? (A) ( 4) 0 y (B) ( 2) 0 y (C) ( )y x has a critical point in the interval ( 1,0) (D) ( )y x has no critical point in the interval ( 1,0) Key. (A,C)

Sol. 1 e y.e 1 x xdy

dx

d

y e 1 1 x

dx

y e 1 c x x

(0) 2y 2(2) c c = 4

Hence, (e 1) 4 xy x .

y( 4) 0

4

ye 1

x

x

x

2x

e 1 4 e

e 1

xxdy

dx

For critical point xe x 3 1

Let: x 1g x e

x 3

Now, 1 1

g 1 0e 2

And 1

g 0 1 03

Hence, g 0x must has one real root in (-1,0).

51. 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) 2

P Q 1 x y y y (D) 2

P Q x y y y

Key. (B, C) Sol. Equation of family of circles is 2 2 2 2 0x y x y



-31-

dy

y xdx

……..(i)

22

21

d y dyy

dx dx

…….(ii)

2" ' 1

"

yy y

y

Putting the value of in (i)

2 2" ' 1 " ' 1

'" "

y y y y y yy y x

y y

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

P y x

2

' ' 1Q y y

2

1 ' 'P Q x y y y

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

xg x , x 0

| x |f x

0, x 0

And |x|h x e for all x . Let f oh 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) f o h is differentiable at x = 0 (D) h o f is differentiable at x = 0 Key. (A, D)

Sol.

0

0 0' 0 lim

h

f h ff

h

0

0

limh

hg h

h

h

0lim 0h

g h

h

(As ' 0g exists and g 0 0 )

0

0 0lim 0h

g h g

h

0lim 0h

g h

h

0

0 0' 0 lim

h

f h ff

h

0

0

lim

h

hg h

h

h

0limh

g h

h



-32-

1 0 0

function f is differentiable at x = 0

0

0

x

x

e xh x

e x

0

'.0

x

x

e xh x

e x

' 0 1h ; ' 0 1h

h is not differentiable at x = 0

xf h x f e

| | | | x

x x

x

eg e g e

e

f h x is not differentiable at x =0 as 1 0 g

0

0 0

xh g x x

xh f x

h x

0

1 0

xg x

xe x

x

0

1 0

g xe x

x

h f x is continuous at x = 0

Also h f x is differentiable at x = 0.

53. Let f x sin sin sin x6 2

for all x and g x sin x

2

for all x . Let f og 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 f o g is 1 1

,2 2

(C)

x 0

f xlim

g x 6

(D) There is an x such that gof x 1

Key. (A, B, C)

Sol. sin sin6 2

f x Sin x

x R sin2 2 2

x

1 sin sin 12

x

sin sin6 6 2 6

x



-33-

1 1

sin sin sin x2 6 2 2

Range of f is 1 1

,2 2

sin2

f g x f x

sin sin sin sin6 2 2

x

Range of f g x is 1 1

,2 2

0

sin sin sin6 2

lim6 6

sin sin6 2

x

x

x

sin sin6 2

g f x g x

sin sin sin2 6 2

x

1 1

sin sin sin2 6 2 2

x

sin sin sin4 2 6 2 4

x

for no x R ; 1g f x

54. 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)

2c

a 122

(B)

2c

a 302

(C) a b c a 48 3 (D) a .b 72

Key. (A, C, D) Sol. a QR , b RP , c PQ

12a , 4 3b , . 24b c

As 0a b c

b c a

2 2 2

2 .b c b c a

or 248 48 144c

248c

2

4812 12

2 2

ca

2

4812 36

2 2

ca



-34-

a b c

22 2

2 .a b a b c

2 . 48 144 48a b

. 72a b

2a b c a a b

2 a b 22 2

2 . a b a b

2 144 48 72 72

48 3 55. Let X and Y be two arbitrary, 3×3, non zero, skew-symmetric matrices and Z be an

arbitrary 3×3, 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 Key. (C, D) Sol. TX X

TY Y TZ Z

3 4 4 3 3 4 4 3T T TY Z Z Y Y Z Z Y

4 3 3 4T T T TZ Y Y Z

3 44 3 3 4 4 3 Z Y Y Z Y Z Z Y

44 4444 44 T T TX Y X Y

44 44

T TX Y

44 44

X Y

44 44X Y

4 3 3 4 4 3 3 4T T TX Z Z X X Z Z X

3 4 4 3

T T T TZ X X Z

3 4 4 3

Z x X Z

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

23 2323 23 T T TX Y X Y

23 23

X Y 23 23X Y

56. Which of the following values of satisfy the equation

2 2 2

2 2 2

2 2 2

1 1 2 1 3

2 2 2 2 3 648

3 3 2 3 2

(A) –4 (B) 9 (C) –9 (D) 4 Key (B, C)



-35-

Sol.

2 2 2

2 2 2

2 2 2

1 2 1 4 4 1 9

4 4 4 4 8 4 9 72

9 6 9 4 12 9 9 18

c

Then by simplifying

2 2

2

1 2 3 2 1

2 3 2 6

5 2 0

2 2 32 6 10 4 2 8 648

2 81

9

57. In R2 consider the planes P1 : y = 0 and P2 : x+z =1. Let P3 be a plane different from P1 and P2 passes through the intersection of P1 and P2. If the distance of the point (0,1,0) from P2 is 1

and the distance of a point , , from P3 is 2, then which of the following relations is

(are) true ? (A) 2 2 2 0 (B) 2 2 4 0

(C) 2 2 10 0 (D) 2 2 8 0

Key (B, D) Sol Given P1 = y = 0 and x + z – 1 = 0 P2

= x + z– 1 = 0

P3 = P1 + 2P = 0

1 0x z y

The perpendicular distance from (0, 1, 0) = 1

2

11

2

1

2

3 2 2 2 0P x y z

The distance from , , from P3 = 2

2 2 2

23

2 2 4 0 and 2 2 8 0

58. In R3, let L be a straight line passing through the origin. Suppose that all the points an L are at a constant distance from the two planes P1 : x + 2y – z + 1 = 0: 2x – y + z –2 = 0. Let M be the locus of the feet of the perpendicular drawn from the points on L to the plane P1. Which of the following points lie(s) of 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



-36-

Key (A, B) Sol Equation of angle bisectors

2 1 2 1

6 6

x y z x y z

3 2 2 0 3 0 x y z and x y

Hence plane 3x+y = 0 taken, as passes through origin. Line through origin on 3x + y = 0 and parallel to

ˆˆ ˆ1 2 1 3 5

2 1 1

i j k

i j k is 1 3 5

x y z any point on this line is , 3 , 5

Foot of perpendicular from , 3 , 5 to x + 2y –z + 1 = 0

13 5

1 2 1 6

x y z

Point is 1 1 1

, 3 , 56 3 6

, which is satisfied by A, B

59. Column – I Column – II

(A) In R2, if the magnitude of the projection vector of the vector ˆ ˆi j on ˆ ˆ3 i j is 3 and if 2 3 , then

possible value(s) of | | is (are)

(P) 1

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

2

2

3 2, 1

, 1

ax xf x

bx a x

is differential for all .x R Then

possible value (s) of a is (are)

(Q) 2

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

4 323 3 2

n

4 322 3 3

n

4 323 2 3 0,

n

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 value(s) of | |q a is

(are)

(S) 4

(T) 5 Key (A-P,Q, B-PQ, C-P,Q,S,T, D-Q,T)

Sol (A) ˆ ˆ. 3

32

i j i j

3 2 3



-37-

23

, compare with 2 3

0, 3

Then 1,2

(B) 6

'ax

f xb

1x for differentiable function

6 .... 1a b

For continuous function, 23 2 6a a a ….(2) From equation (1) & (2) 2,1a

(C) 4 3 4 3 4 32 2 23 3 2 2 3 3 3 2 3

n n n

2 4 3 8 63 2 3 1 0 n n

23 2 3 0

21 0 n n Hence n is not multiple of 3 n = 1, 2, 4, 5

(D) Given 2

4,

, 5, , . .

ab

a ba q b are in A P

…(1)

10q a , 5

2

bq

10

2 5

a q

b q

…(2)

From (1) and (2)

22 23 60 0q q

15

4,2

q for q = 4, a = 6

Then q = 15/2, a = 5/2 | q | 2,5a



-38-

60. 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. 2 2 22 a b c

and sin

,sin

X Y

Z

then possible values of n for which

cos 0n 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 2 2cos 2 2sin sin ,X Y X Y then possible value(s) of a

b

is (are)

(Q) 2

(C) In R2, let ˆ ˆ ˆ ˆ3 , 3i j i j and ˆ ˆ1i 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 | | is

(are)

(R) 3

(D) Suppose that F(x) denotes the area of the region bounded by x = 0, x = 2, y2 = 4x and | 1| | x 2 | x,y x where

{0,1}. Then the value(s) of 8

2,3

F when 0 and

1, is (are)

(S) 5

(T) 6 Key (A-P,R,S; B – P; C- P; D – S,T)

Sol (A) Given 2 2 22 a b c and sin

,sin

X Y

Z

2 2 22 sin sin sin X Y Z

2 sin sin X Y Zz

2 1 sin 0 Z

1

2

cos 02

n

n = odd number (B) Given 1 cos 2 2cos 2 2sin sin X Y X Y

2 21 1 2sin 2 4sin 2sin sin X Y X Y



-39-

2 22b a ab

2

2 0a a

b b

1a

b

(C) Internal angular bisector is ˆ ˆ ˆ ˆ3 3

2

i j i j

Angular bisector is 0y x

The perpendicular distance ,1 is 3

2

|1 2 | 3

,2 2

| | 1 2or

(D) for 0 , | 1| | x 2 | xy x

y = 3

x=0 (2,2 2)

x=L

(2,–2 2)

y=3

2 2 2

0

3 2 2 24

yF dy

8 2

63

8 2

63

F

For 1 , 3 0 1

| 1| | x 2 | x1 1 2

x xy x

x x

y=2y=2 x

x=1 x=2

8 2 8 2

6 1 53 3

F

8

2 53

F

Documents

JEE Advance 2015 Paper 1 Solution