IIT JEE adv-2015 question paper-2, hints & solutions

JEE-Advanced – 2015 (Hints & Solutions) Paper - 2

NARAYANA Group of Educational Institutions

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NARAYANA IIT ACADEMY INDIA

JEE – ADVANCED– PAPER – 2

CODE : 0 Time : 3 Hours. Maximum Marks : 240

READ THE INSTRUCTIONS CAREFULLY GENERAL : 1. The sealed booklet is your Question Paper. Do not break the seal till you are told 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 with the options are legible. QUESTION PAPER FORMAT AND MARKING SCHEME : 8. The question paper has 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). Marking Scheme : +4 for correct answer and 0 in all other cases. 11. Section 2 contains 8 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 paragraph type questions. Each paragraph describes an experiment, a situation

or a problem. Two multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.

Marking scheme : +4 for correct answer, 0 if not attempted and –2 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.



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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. The densities of two solid spheres A and B of the same radii R vary with radial distance r as

5

A B

r rr k and r k ,

R R

respectively, where k is a constant. The moments of inertia

of the individual spheres about axes passing through their centres are IA and IB, respectively.

If B

A

I n,

I 10 the value of n is

Ans. 6

Sol: R 5

2 2A

0

2 Kr 8 K RI .4 r dr .r

3 R 18

R 5 5

2 2B 5

0

2 Kr 8 K RI 4 r dr .r

3 R 30

Now B

A

I 18 6

I 30 10

n 6

2. Four harmonic waves of equal frequencies and equal intensities Io have phase angles 0, /3, 2/3 and . When they are superposed, the intensity of the resulting wave is nIo. The value of n is

Ans. 3

Sol: From the given diagram

Resultant amplitude is 3a Resultant intensity 03I

/ 3

3a

/ 3

a / 3

a

a

a

n 3



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3. For a radioactive material, its activity A and rate of change of its activity R are defined as

dN dAA and R ,

dt dt where N(t) is the number of nuclei at time t. Two radioactive sources

P (mean life ) and Q (mean life 2) have the same activity at t = 0. Their rates of change of

activities at t = 2 are RP and RQ, respectively. If P

Q

R n,

R e then the value of n is

Ans. 2

Sol: t0N N e

t0

dNA N e

dt &

2 t0

dA& R N e

dt

Now

P

Q

t2P 0P P

t2Q Q 0 Q

N eR

R N e

Q P

2

tP

Q

1e

2

2

e

n 2 4. A monochromatic beam of light is incident at 60o on one face of an equilateral prism of

refractive index n and emerges from the opposite face making an angle (n) with the normal

(see the figure. For n = 3 the value of is 60o and d

m.dn

The value of m is

Ans. 2

Sol: Using Snell’s law at the

osin 60 sinn

sin r sin 60 r

2n sin r 3 … (1) nsin 60 r sin … (2)

3

sin r2n



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A = 600

(60 – r)0

r

24n 3

cos r2n

Putting values of sin r & cos r in equation (2)

23 4n 3 1 3

n . sin2 2n 2 2n

23 4n 3 3 4sin differentiate both sides w.r.t. n

2

2n d3 4cos .

dn2 4n 3

2

d 3n

dn cos 4n 3

Putting

on 3 & 60

d 3 3 6

21dn 94 3 32

5. In the following circuit, the current through the resistor R (= 2 ) is I Amperes. The value of I is

Ans. 1

Sol:



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6.5

2

2 82

4

4

1012

E D

F

B

C

A 1

ABCA is a balanced wheatstone’s bridge of equivalent resistance 2 . So

2

4

1012

E D

A

F C

6

ACDEFA is also a balanced wheatstone’s bridge of equivalent resistance 4.5 . So

2

4.56.5

Hence the current is 1 ampere 6. An electron in an excited state of 2+Li ion has angular momentum 3h / 2 . The de Broglie

wavelength of the electron in this state is 0pa (where 0a is the Bohr radius). The value of p is Key. (2)

Sol. Given 3

2

hmvr

3

2

hmv

r

debroglie = 2

23 3

h h r r

mv h …(i)

Now radius of the electron in 3rd orbit of 2+Li ion

2 2

0 00

33

3

a n ar a

Z …(ii)

Putting r from (ii) in (i)

debroglie = 00

32 2

3

aa

p = 2.



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7. A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length l and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point

mass nearer to M is at a distance r = 3l from M, the tension in the rod is zero for 288

Mm k .

The value of k is

M

r l

m m

x

Key. (7)

Sol. For Tension in rod to be zero forces on both m's should be equal

x M

3l l

1 2 m m

Force on mass - 1 = Force on mass - 2

2 2

2 2 2 2163

Gm GmGMm GMm

l l ll

7

288m M

K = 7. 8. The energy of a system as a function of time t is given as 2( ) exp( ) E t A t , where 10.2 s .

The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E(t) at t = 5 s is

Key. (4)

Sol. 2 e tE A

Taking Log on both sides and differentiating we get 2 dE dA

dtE A

Maximum % error

100 2 100 100

dE dA dtt

E A t

= 2 1.25 0.2 5 1.5

= 2.5 + 1.5 = 4 %. Ans : 4.



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SECTION : 2 (Maximum Marks : 32) This section contains EIGHT 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. An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as

shown in the figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed state. The gas is then heated very slowly to temperature T2, pressure P2 and volume V2. During this process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct statement(s) is(are)

(A) If 2 1 12 and 3 , 2V V T T then the energy stored in the spring is 1 1

1

4P V

(B) If 2 1 12 and 3 , 2V V T T then the change in internal energy is 1 13P V

(C) If 2 1 13 and 4 , 2V V T T then the work done by the gas is 1 1

7

3P V

(D) If 2 1 13 and 4 , 2V V T T then the heat supplied to the gas is 1 1

17

6P V

Key. (B) Sol. 1 1kx P A

1 2k x x P A

2 1kx P P A

22 1

1 1kx P P A.x

2 2

= 1 1

1 3P P v

2 2

= 11

P1v

2 2

1 1P v

4

3

2 2

nfU R T nR T



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2 2

1 1 1( ) x ( A)

2 2 2

VW kx P P V

A

2 2 22 1

1 1 1

P nRT T

P PP nRT T

When 2 1 12 and 3 2V V T T

Energy stored in the spring = 2

1

2P V

= 11 1 1 1

1

31 3

2 2

TP V P V

T

1 1 1 1

32 3 3

2 U nR T nRT P V

When 2 1 2 13 , 4 V V T T

Work done = 22 1 1

1

1 12

2 2

TP V P V

T

= 11 1 1 1

1

412 4

2

TP V P V

T

1 1

3 3 93

2 2 2 U nR T nR T nRT = 1 1

9

2P V

1 1 1 1 1 1

9 174

2 2 Q U W P V P V P V

Correct option is (B). 10. A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in

air. When two dielectrics of different relative permittivities 1( 2 and 2 4) are introduced

between the two plates as shown in the figure, the capacitance becomes C2. The ratio 2

1

C

Cis

+ -

d/2

d

2

1 S/2

S/2

(A) 6/5 (B) 5/3 (C) 7/5 (D) 7/3 Key. (D)

Sol. Initially 01

C AC

d



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1 2 2 4

1 2

Ca Cb

Cc After the insertion of di-electrics

Where 0

02 22

2

a

AA

Cd d

0

04 42

2

b

AA

Cd d

0

02

2

c

AA

Cd d

So, 2 a b cC C SC C

=

a bc

a b

C CC

C C

= 07

3

A

d

2

1

7

3

C

C

(D). 11. A spherical body of radius R consists of a fluid of constant density and is in equilibrium under

its own gravity. If P(r) is the pressure at r(r< R), then the correct option (s) is (are)

(A) P r 0 0 (B)

P r 3R / 4 63

P r 2R / 3 80

(C)

P r 3R / 5 16

P r 2R / 5 21

(D)

P r R / 2 20

P r R / 3 27

Ans. (B, C)

Sol: 2 2

.

.4

R

r

G m dmdp

r r

3 2

2 2

44

3.4

p r r drP G

r r

5

4

4.

3

R

r

rP GP dr

r

2 24

3

R

r rP G R r (Pressure as a function

rdr

dp



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of r)

22

2

222

3 4 93 255 16 16

21 212 4 43 255

R RP r G R

R

RR RG RP r

22

2 22

3 97 9 634 16

2 4 16 5 803 9

r

r

R RP R RR R R

P R

12. In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement (s) is (are)

(A) P has more tensile strength than Q (B) P is more ductile than Q (C) P is more brittle than Q (D) The Young’s modulus of P is more than that of Q Ans. (A, B) Sol: (A) Correct from graph plastic range of P is more than Q (B) Correct from graph plastic range of P is more than Q (C) Q is more brittle because plastic range of P is more than Q

(D) Young’s modulus ‘Y’stress

strain

So, Y (Young’s modulus ) of Q is more than P 13. Consider a uniform spherical charge distribution of radius R1 centred at the origin O. In this

distribution, a spherical cavity of radius R2, centred at P with distance OP = a = R1 – R2 (see

figure) is made. If the electric field inside the cavity at position r is E r , then the correct

statement (s) is (are)



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(A) E is uniform, its magnitude is independent of R2 but its direction depends on r (B) E is uniform, its magnitude depends on of R2 and its direction depends on r

(C) E is uniform, its magnitude is independent of a but its direction depends on a

(D) E is uniform and both its magnitude and direction depends on a Ans. (D)

Sol: Field at Q due to whole sphere 10

OQE

3

Field at Q due to removed sphere 20

PQE

3

Hence Net field at O 1 2E E E

0

OQ PQ3

0

OP3

0

E a3

O

R1

PQ

14. In terms of potential difference V, electric current I, permittivity 0 , permeability 0 and speed of light c, the dimensionally correct equation(s) is (are)

(A) 2 20 0I V (B) 2

0 0I V

(C) 0I cV (D) 0 0cI V

Ans. (A,C) Sol: Use these formula to find 0, 0 & V

0 1 2

4

I IF

l r

for 2 2

0 0 MLA T

1 22

04

q qF

r

for 1 3 2 4

0 0 M L A T

W

Vq

2 1 3V ML A T

0 0

1c

Substituting above data with all the choices we get (C) is correct. Rest all are incorrect 15. Two spheres P and Q of equal radii have densities 1 and 2

respectively. The spheres are connected by a massless string and placed in liquid L1 and L2 of densities 1 and 2 and viscosities 1

and 2 , respectively. They float in equilibrium with the sphere P and L1 and sphere Q in L2 and the string being taut (see figure). If sphere P alone in L2 has terminal velocity PV and Q alone in L1 has terminal velocity QV , then

(A) P

1

Q 2

V

V

(B) P

2

Q 1

V

V



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(C) P QV V 0 (D) P QV V 0 Ans. (A, D) Sol.

P

QL2

L1

V = Volume of both spheres

P

1 1B V g 2 2B V g

1m gT

T

Q

2m g

1 1V g V g T …(i)

– 2 2V g V g T …(ii)

1 2 1 2V g V g

1 2 1 2 …(iii)

2

1 2P

2

r2V g

9

…(iv)

2

2 1Q

1

r2V g

9

…(v)

P 1 2 1

Q 2 1 2

V

V

P 1

Q 2

V1

V

P QV & V are opposite

P

1

Q 2

V

V

P QV V 0

16. A fission reaction is given by 236 140 9492 54 38Xe Sr x y; U where x and y are two particles.

Considering 23692 U to be at rest, the kinetic energies of the products are denoted by

XeK KSr, Kx (2 MeV) and Ky (2 MeV), respectively. Let the binding energies per nucleon

of 236 14092 54U, Xe and 94

38Sr be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering

different conservation laws, the correct option(s) is (are)



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(A) Sr Xex n, y n,K 129 MeV,K 86MeV

(B) Sr Xex p, y e ,K 129 MeV,K 86MeV

(C) Sr Xex p, y n,K 129 MeV,K 86MeV

(D) Sr Xex n, y n,K 86 MeV,K 129MeV

KEY: (A) SOL:

SECTION : 3 (Maximum Marks : 16) This section contains TWO paragraphs Based on each paragraph, there will be TWO 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 one of the bubbles is darkened. 2 In all other cases. Paragraph – 1

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular values im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im.

i

Cladding n2

n2n1 >

Air

n1

Core

17. For two structures namely S1 with 1n 45 / 4 and 2n 3 / 2, and S2 with 1n 8 / 5 and

2n 7 / 5 and taking the refractive index of water to be 4/3 and that of air to be 1, the

correct options(s) is (are)

(A) NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive

index 16

3 15

(B) NA of S1 immersed in liquid of refractive index 6

15 is the same as that of S2 immersed

in water



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(C) NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index 4

15(D) NA of S1 placed in air is the same as that of S2 placed in water.

KEY: (A,C) SOL: m 1 mnsin i n sin r

mr 90

So, 2

1 2m 2

1

n nsin i 1

n n

Put the value and check the answer

example

for 1S

m

45 94 4sin i 14 453 16

9

16

For 2S

m

8 495 25sin i 1

16 64253 15

= 9

16

18. If two structures of same cross-sectional area, but different numerical apertures NA1 and NA2 (NA2 < NA2) are joined longitudinally, the numerical aperture of the combined structure is

(A) 1 2

1 2

NA NA

NA NA (B) 1 2NA NA (C) NA1 (D)NA2

KEY: (D) SOL: conceptual

Paragraph -2

In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are l, w and d, respectively.

A uniform magnetic field B is supplied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the



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electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.

d

l

K

M

RwIS

I

P

y

z

x

Q 19. Consider two different metallic strips (1 and 2) of the same material. Their lengths are

the same, widths are w1 and w2 and thickness are d1 and d2 respectively. Two points K and M are symmetrically located on the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences between K and M in strips 1 and 2 respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the correct statement(s) is (are)

(A) If w1 = w 2 and d1 = 2d2, then V2 = 2V1

(B) If w 1 = w 2 and d1 = 2d2, then V2 = V1

(C) If w 1 =2 w 2 and d1 = d2, then V2 = 2V1

(D) If w 1 = 2 w 2 and d1 = d2, then V2 = V1

KEY: (A,D)

SOL:

d

I VV B E B

ne Wd W

1

Vd

independent of ‘w’

Also V d = constant

2 1

1 2

V d

V d

correct options are (A) and (D)

20. Consider two difference metallic strips (1 and 2) of same dimensions (length l, width w and thickness d) with carrier densities n1 and n2 respectively. Strips 1 is placed in magnetic field B1 and strips 2 is placed in magnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences developed between K and M in strips 1 and 2 respectively. Assuming that the current I is the same for both the strips, the correct option(s) is (are)

(A) If B1 = B2 and n1 = 2n2, then V2 = 2V1

(B) If B1 = B2 and n1 = 2n2, then V2 = V1

(C) If B1 = 2B2 and n1 = n2, then V2 = 0.5V1



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(D) If B1 = 2B2 and n1 = n2, then V2 = V1

KEY: (A,C) SOL: For same dimension and current

B

Vn

and nV

constantB

B1 = B2

1 2

2 1

n V

n V

option (A) correct

For same ‘n’

V

constantB

1 1

2 2

V B

V B

option (C) is also correct

correct options are (A) & (C)



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PART II: CHEMISTRY


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. In the complex acetylbromidodicarbonylbis (triethylphosphine) iron (II), the number of Fe–C

bond(s) is Key: (3)

Sol: 2

3 2 52 3 2Fe CH CO Br CO C H P

Out of the given donors, only CO donates through carbon. Hence, number of Fe – Carbon bonds = 2 22. Among the complex ions, {Co(NH2.CH2.CH2.NH2)2Cl2]+, {CrCl2(C2O4)2]3–, [Fe(H2O)4(OH2)]+,

[Fe(NH3)2(CN)4]–, [Co(NH2.CH2.CH2.NH2)2(NH3)Cl]2+ and [Co(NH3)4(H2O)Cl]2+, the number of complex ion(s) that show(s) cis-trans isomerism is

Key: (6) Sol: All six complexes will exhibit Cis-trans isomerism. 23. Three moles of B2H6 are completely reacted with methanol. The number of moles of boron

containing product formed is Key: (6) Sol: 2 6 3 3 23

B H 6CH OH 2B OCH 6H

One mole B2H6 when completely reacted with methanol produces 2 moles of Boron containing product.

3 moles of B2H6 produce 6 moles of such product. 24. The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the

molar conductivity of a solution of a weak acid HY (0.10 M). If 0 0X Y , the difference in their

pKa values, pKa(HX) – pKa(HY), is (consider degree of ionization of both acids to be <<1) Key: (3)

Sol: Assume H X I

Assume HY II

II

o

IIII

o

I I

II II

1

10

22

3I I I2

II II II

Ka C 0.01 110

Ka C 0.1 10

pKa I– pKa II 3log10 0 3



-18-

25. A closed vessel with rigid walls contains 1 mol of 23892 U and 1 mol of air at 298 K. Considering

complete decay of 23892 U to 206

82 Pb , the ratio of the final pressure to the initial pressure of the system at 298 K is

Key: (9) Sol: 238 206 4

92 82 2U Pb 8 He 6

1 mol Uranium –238 gives 8 mols He. Number of moles of gases = 8 + 1 = 9 Ratio of final pressure to initial pressure

f f

i i

P n 99

P n 1

26. In dilute aqueous H2SO4, the complex diaquodioxalatoferrate (II) is oxidized by 4MnO . For

this reaction, the ratio of the rate of change of [H+] to the rate of change of 4MnO is

Key: (8) Sol. 3 2

2 2 4 4 2 22 2Fe H O C O MnO 8H Fe 6H O 4CO Mn

4

Rate of change of H 88

1Rate of change MnO

27. The number of hydroxyl group(s) in Q is

OH

H

CH3 CH3

HHeat

4

o

aqueous dilute KMnO excess

0 CP Q

Key: (4) Sol.

OHCH3 CH3

H

CH3 CH3 CH3 CH3 CH3

CH3H

CH3

CH3

4o

aq. dilute KMnO excess

0 C

P

OH

OH

CH3 OHCH3

Q

2H O

H

2H O

1,2 methylshift

HO

28. Among the following, the number of reaction(s) that produce(s) benzaldehyde is

I. 3

CO, HClAnhydrous AlCl /CuCl

II. CHCl2

2o

H O

100 C



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III. COCl

2

4

HPd BaSO

IV. CO2Me

o

2

DIBAL H

Toluene, 78 CH O

Key: (4)

Sol. I. 3

CO, HClAnhydrous AlCl /CuCl

CHO

II.

CHCl 2

2o

H O

100 C

C OH

H

OH

2H O C H

O

III. COCl

2

4

HPd BaSO

CHO

HCl

IV. CO2Me

o

2

DIBAL H

Toluene, 78 CH O

C H

O

3CH OH

SECTION : 2 (Maximum Marks : 32) This section contains EIGHT 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. When O2 is adsorbed on a metallic surface, electron transfer occurs from the metal to O2. The

TRUE statement(s) regarding this adsorption is(are) (A) O2 is physisorbed (B) Heat is released (C) occupancy of *

2p of O2 is increased (D) bond length of O2 is increased

Key: (B), (C), (D) Sol. When electron is transferred from metal to O2 then the bond is ionic, so, we consider this as

chemical adsorption. In adsorption, H = ve, electron goes to *2p of O2, so, bond length increases.

30. Under hydrolytic conditions, the compounds used for preparation of linear polymer and for

chain termination, respectively, are (A) CH3SiCl3 and Si(CH3)4 (B) (CH3)2SiCl2 and (CH3)3SiCl (C) (CH3)2SiCl2 and CH3SiCl3 (D) SiCl4 and (CH3)3SiCl Key: (B) Sol.



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2H OCl Si Cl HO Si OH

CH3

CH3

CH3

CH3

xO Si O

x

CH3

CH3

3 3H C Si O Si O Si CH

3 3CH Si OH

CH3

CH3

CH3

CH3

CH3

CH3

Polymerization

Chain

termination

x

31. The pair(s) of ions where BOTH the ions are precipitated upon passing H2S gas in presence of dilute HCl, is (are)

(A) Ba2+, Zn2+ (B) Bi3+, Fe3+ (C) Cu2+, Pb2+

(D) Hg2+, Bi3+

Ans. (C, D) Sol: dilHCl2

2Black

Cu H S CuS 2H

dilHCl22

BlackPb H S PbS 2H

dilHCl22

BlackHg H S HgS 2H

dilHCl32 2 3

Blackor Brown

Bi H S Bi S 2H

dilHCl22 2 2Zn H S ZnS ZnCl soluble H S

3 22

Soluble MilkywhiteFe H S Fe 2H S

32. The correct statement(s) regarding, (i) HClO, (ii) HClO2, (iii) HClO3 and (iv) HClO4, is (are) (A) The number of Cl = O bonds in (ii) and (iii) together is two (B) The number of lone pairs of electrons on Cl in (ii) and (iii) together is three (C) The hybridization of Cl in (iv) is sp3

(D) Amongst (i) to (iv), the strongest acid is (i) Ans. (B, C) Sol: Sum of lone pair of electron in (2) and (3) = 3

OH Cl

OH Cl

O

(i) (ii)

OH Cl

OOH Cl

O

(iii) (iv)

OO

sp3



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33. In the following reactions, the major product W is

(A) (B)

(C) (D)

Ans. (A)

Sol:

N Cl2

V =

N N=

W =

OH

-naphthol favours E attack at ortho position due to greater stability of -complex. 34. The major product U in the following reactions is

(A) (B)

(C) (D)

Ans. (B)



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Sol:

C – H

CH3

T = CH3,

C – O – O – H

CH3

U = CH3

35. In the following reactions, the product S is

(A) (B)

(C) (D)

Ans. (A)

Sol: CHO C – HH C3 H C3

CH – C = O2 CH2

HH

O

C = N – H..

H C3 N–H O2NH3

36. One mole of monatomic real gas satisfies the equation p(V b) RT where b is a constant.

The relationship of interatomic potential V(r) and interatomic distance r for the gas is given by

(A)

V(r)0

r

(B)

V(r)0

r

(C)

V(r)0

r

(D)

V(r)0

r

KEY: (C) SOL: Since the equation suggest that a = 0, hence intermolecular force of attraction is zero. In such a

situation when molecules come close to each other there will be no decrease in potential energy. When the distance between the molecules is equal to the sum of their radii, the force of repulsion becomes too strong and potential energy shows a steep rise.

Hence the answer is (C)



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SECTION : 3 (Maximum Marks : 16)

This section contains TWO paragraphs Based on each paragraph, there will be TWO 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 one of the bubbles is darkened. 2 In all other cases. PARAGRAPH – 1

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment could be used to measure the calorimeter constant. In a second experiment (Experiment 2) 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions of Expt. 1) where a temperature rise of 5.6°C was measured.

(Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of solutions as 1.0 g mL–3)

37. Enthalpy of dissociation ( in kJ mol–1) of acetic acid obtained from the Expt. 2 is

(A) 1.0 (B) 10.0 (C) 24.5 (D) 51.4

KEY: (A) SOL: Standard enthalpy of neutralization = -57 kJ/mol

Amount of acid used = 0.1 mole

Amount of base used = 0.1 mole

Amount of heat released on neutralization = 5.7 kJ

Change in temperature = 5.7oC

Let the calorimeter constant be x

Now, 5700 J = 200 × 4.2 × 5.7 + x × 5.7

x = 160 J

For acetic acid and NaOH.

Mole of acetic acid neutralized = 0.1

Energy released in neutralization = 200 × 5.6 × 4.2 + 160 × 5.6 = 5600 J

Energy released per mole = 5600

56000J 56kJ0.1

Enthalpy of ionization of acetic acid = 1 kJ.

Hence answer is (A)



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38. The pH of the solution after Expt. 2 is

(A) 2.8 (B) 4.7 (C) 5.0 (D) 7.0

KEY: (B)

SOL: No. of moles of acetic acid = 100 2

0.21000

No. of moles of NaOH = 100 1

0.11000

Conc. of remaining acetic acid = 0.1

1000 0.5200

Conc. of 3CH COONa = 0.1

1000 0.5200

The solution is a buffer, hence

pH= 5salt 0.5

pka log log 2 10 logacid 0.5

pH=4.7

PARAGRAPH – 2

In the following reactions

C H8 6

Pd-BaSO 4

H2C H8 8

i B H2 6

H2ii O2NaOH, H O2

X

H2O

HgSO4 2 4, H SO

C H8 8

i EtMgBr, H O2

H , heat+ii Y

39. Compound X is

(A) H3C

O

(B) H3C

OH

(C)

OH

(D) HOC

KEY: (C) SOL:

Pd-BaSO4

H2

i B H2 6

H2ii O2 NaOH H O2

C CH CH CH2 CH2OHCH2

(X) Hence Answer is (C)



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40. The major compound Y is

(A) H3C

(B) H3C

(C)

H3CH2C

(D)

H3C

H3C

KEY: (D) SOL:

H2O HgSO4 2 4, H SO EtMgBr, H O2

C CHC CH3

O

C CH3

OH

CH3CH2

H , heat+ C CH3

CH3CH

(Y) Hence Answer is (D)



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PART III: MATHEMATICS


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. 41. The coefficient of x9 in the expansion of (1 + x)(1 + x2) (1 + x3) ….. (1 + x100) is Ans. 8

Sol: 2 3 1001 x 1 x 1 x ........... 1 x

116 3 5 6 7 91 x 1 x 1 x 1 x 1 x 1 x 1 x

116 3 5 6 7 91 x 1 x 1 x 1 x 1 x 1 x 1 x

Now coefficient of x9 in

16 2 3 4 3 5 6 7 91 x 1 x x x x .......... 1 x 1 x 1 x 1 x 1 x

Coefficient of x9 in

16 2 3 5 6 7 8 91 x 1 x x ....... 1 x x x x x 2x

= 8

42. Suppose that the foci of the ellipse 2 2x y

19 5 are (f1, 0) and (f2, 0) where f1 > 0 and

f2 < 0. Let P1 and P2 be two parabolas with a common vertex at (0, 0) and with foci at (f1, 0) and (2f2, 0), respectively. Let T1 be a tangent to P1 which passes through (2f2, 0) and T2 be a tangent to P2 which passes through (f1, 0). If m1 is the slope of T1 and m2 is the slope of T2, then

the value of 222

1

1m

m

is

Ans. 4

Sol: 2 2x y

19 5

2

22

be 1

a

5

19

= 4

9

2

e3



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1

2

f ,0 ae,0 2,0

f ,0 ae,0 2,0

P1 parabola is y2 = 8x P2 parabola is y2 = –16x Equation of tangent to P1

T1 is 1 11 1

a 2y m x y m x

m m

T1 passes trough (–4, 0)

11

20 4m

m

21 1

1

1 12m m

m 2

T2 is 2 22 2

a 4y m x y m

m m

Passes trough (2, 0) 0 = 22

42m

m 2

2m 2

Now, 222

1

1m 2 2 4

m

43. Let m and n be two positive integers greater than 1. If

ncos

m0

e e elim

2

Then the value of m

n is

Ans. (2)

Sol ncos

m0

e elim

=

ncos 1

n

n m0

e e 1lim cos 1

cos 1 .

=

n

n2cos 1

n m0

2sine e 12lim

cos 1

=

n

2n

cos 12n

n mn0

sine e 1 e2lim4. 2cos 1

2

When 2n = m m

2n



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44. If 1

1 29x 3tan x

20

12 9xe dx

1 x

Where tan–1 x takes only principal values, then the value of e

3log |1 | is

4

Ans. 9

Sol: 1

19x 3tan x

20

3e 9 dx

1 x

Let 19x 3tan x t

2

39 dx dt

1 x

39

34 9t 4

0

e dt e 1

3

941 e

e

3log 1 9

4

e

3log 1 9

4

45. Let f: R R be a continuous odd function, which vanishes exactly at one point and 1

f 1 .2

Suppose that F(x) = x

1

f t dt

for all x 1,2 and x

1

G x t | f f t | dt

for all x 1,2 . If

x 1

F x 1lim ,

G x 14 then the value of

1f

2

is

Ans. (7)

Sol.

x 1

F x 1Lim

G x 14

x 1

F' x 1Lim F 1 G 1 0

G ' x 14

x 1

f x 1Lim

14x f f x

x 1

f x 1Lim

14x f f x

f 1 1

14f f 1

1

f 72

1f 1

2



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*46. Suppose that p,q and r are three non-coplanar vectors in R3. Let the components of a vector s along p,q and r be 4, 3 and 5, respectively. If the components of this vector s along

p q r , p q r and p q r are x, y and z, respectively, then the value of 2x + y + z

is Ans. (9/DEL) Sol. s=4p+3q+5r

Also, s=x p q r y p q r z p q r

Hence, x + y – z = 4 x – y – z = 3 x + y + z = 5 Solving above equations, x = 4 and y + z = 1 Hence 2x + y + z = 9

47. For any integer k, let k = k k

cos isin7 7

, where i 1 . The value of the expression

12

k 1 kk 1

3

4k 1 4k 2k 1

| |

| |

is

Ans. (4)

Sol. k = 2k 2k

cos isin7 7

12

k 1 kk 1

3

4k 1 4k 2k 1

| | 12 2sin14 4

3 2sin| |14

[ length of each side of a regular polygon is 2 sin (/n). 48. Suppose that all the terms of an arithmetic progression (A. P.) are natural numbers. If the

ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A. P. is

Key: (9)

Sol. According to question,

72a 6d 62

11 112a 10d2

7a 21d

6a 5d

a = 9d

130 < T7 < 140 130 < a + 6d < 140 130 < 15d < 140

130 140

d15 15

d = 9 (since terms of A. P. are natural numbers)



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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.

49. Let f(x) = 7 tan8x + 7 tan6x – 3 tan4x – 3 tan2x for all x ,2 2

. Then the correct

expression(s) is(are)

(A) /4

0

1xf x dx

12

(B) /4

0

f x dx 0

(C) /4

0

1xf x dx

6

(D) /4

0

f x dx 1

Key: (A), (B) Sol. f(x) = 7 tan8x + 7 tan6x – 3 tan4x – 3 tan2x

6 2 2 2 6 2 2f x 7tan x sec x 3tan x sec x 7 tan x 3tan x sec x

/4 1

6 2

0 0

f x dx 7z 3z dz 0

/4 /4

/46 2 2 7 3 7 3

00 0

x 7 tan x 3tan x sec xdx x tan x tan x tan x tan x dx

= /4

3 2 2

0

0 tan x tan x 1 sec xdx

= /46 4

0

tan x tan x 1 1 3 2 1

6 4 4 6 12 12

50. Let f, g : 1,2 R be continuous functions which are twice differentiable on the interval

1,2 . Let the values of f and g at the points 1, 0 and 2 be as given in the following table.

x = 1 x = 0 x = 2 f(x) 3 6 0 g(x) 0 1 1

In each of the intervals (1, 0) and (0, 2) the function (f 3g) never vanishes. Then the correct statement(s) is(are)

(A) f ' x 3g ' x 0 has exactly three solutions in 1,0 0,2

(B) f ' x 3g ' x 0 has exactly one solution in (1, 0)

(C) f ' x 3g ' x 0 has exactly one solution in (0, 2)

(D) f ' x 3g ' x 0 has exactly two solutions in 1,0 and exactly two solutions in (0, 2)

Key: (B), (C)



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Sol. Let, h(x) = f(x) – 3g(x) Since, h(x) is differentiable and h(1) = h(0) = h(2)

h ' x f ' x 3g ' x 0 must have at least one root in (1, 0) and (0, 2) each.

By using Rolle’s theorem, h '' x 0 must have at least one root in 1,2 .

so, therefore, h '' x 0 at x = 0

Hence, h ' x = 0 must have exactly one root in 1,0 and other root in (0, 2)

51. The option(s) with the values of a and L that satisfy the following equation is (are)

4t 6 4

0

t 6 4

0

e sin at cos at dt

L?

e sin at cos at dt

(A) 4e 1

a 2,Le 1

(B)

4e 1a 2,L

e 1

(C) 4e 1

a 4,Le 1

(D)

4e 1a 4,L

e 1

Ans. (A, C)

Sol: 4

t 41

0

I e sin at cos at dt

4

t 4 2 4

0

e sin at 1 cos at cos at dt

4

t 2 2 4 2

0

e 1 2sin at cos at sin at cos at dt

4

t 2 2 2

0

e 1 sin at cos at 2 sin at dt

24t

0

sin 2at 1 cos 2ate 1 2 dt

4 2

4 4 4

2t t 2 t 4

0 0 0

1 1 11e dt e sin 2at dt e sin 2at 1 cos 2at dt e 1

2 8 16

t 42

0

I e sin at cos at dt

t

0

11e dt

16

(as same way as above)

11

e 116

4

1

2

I e 1

I e 1

Whether a = 2 or 4



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52. Consider the hyperbola H : x2 – y2 = 1 and a circle S with centre N(x2, 0). Suppose that H and S touch each other at a point P(x1, y1) with x1 > 1 and y1 > 0. The common tangent to H and S at P intersects the x-axis at point M. If (l, m) is the centroid of the triangle PMN then the correct expression (s) is (are)

(A) 121 1

d 11 for x 1

dx 3x

l

(B) 1

11

21

xdmfor x 1

dx 3 x 1

(C) 121 1

d 11 for x 1

dx 3x

l

(D) 11

dm 1for y 0

dy 3

Ans. (A, B, D) Sol: Tangent at P (x1, y1) is xx1 – yy1 = 1

P(x y )1 1

M N(x , 0)2

1

M ,0x

Now normal 11 1

1

yy y x x

x

Passes through 2N x ,0

Hence 11 2 1

1

y0 y x x

x

2 1x 2x

For centroid 21 111 1

1x 2x

x 1x y, m

3 3 3

l

12 2

1 1 1 1 1

xd 1 dm 1 dm1 , ,

dx 3x dy 3 dx 3 x 1

l



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53. Let E1 and E2 be two ellipses whose centers are at the origin. The major axes of E1 and E2 lie

along the x-axis and the y-axis, respectively. Let S be the circle 22x y 1 2 . The straight

line x y 3 touches the cures S, E1 and E2 at P, Q and R, respectively. Suppose that

2 2PQ = PR =

3. If e1 and e2 are the eccentricities of E1 and E2, respectively, then the correct

expression (s) is (are)

(A) 2 21 2

43e e

40

(B) 1 2

7e e

2 10

(C) 2 21 2

5| e e |

8

(D) 1 2

3e e

4

Key. (A, B)

Sol. 1, 2P ; 2 2

,3 3

a bQ

; 2 2

,3 3

A BR

Also 2 2 2 29a b A B

2 2

3PQ

2 8

9PQ

2 22 29 8

1 23 3 9

a a

or 2 22 23 3 8 a a

22 3 4 a

2 3 2a 2 2 25,1 5, 1a a A

2

21 2

9 5 11 1

5 5

be

a

2

22 2

1 71 1

8 8

Ae

B

2 21 2

1 7 8 35 43

5 8 40 40

e e

2 21 2

7

40e e

1 2

7

2 10e e

2 21 2

1 7 27

5 8 40 e e



-34-

54. If 1 63sin

11

and 1 43cos

9

, where the inverse trigonometric functions take only the

principal values, then the correct option(s) is (are) (A) cos 0 (B) sin 0

(C) cos 0 (D) cos 0

Key. (B, C, D)

Sol. 1 063sin 90

11

1 043cos 180

9

0270

cos 0 cos 0 sin 0

cos 0

55. Let S be the set of all non zero real numbers such that the quadratic equation 2ax x 0 has two distinct real roots x1 and x2 satisfying the inequality 1 2| x x | 1 . Which

of the following intervals is (are) a subset(s) of S?

(A) 1 1

,2 5

(B) 1

, 05

(C) 1

0,5

(D) 1 1

,25

Key. (A, D) Sol. 1 2 1x x

2

1 20 1x x

2

1 2 1 20 4 1x x x x

2

10 4 1

21 1

5 4

1 1 1 1

, ,2 25 5

56. Let 3

4

192xf x

2 sin x

for all x with

1f 0

2

. If 1

1/ 2

m f x dx M , then the possible

values of m and M are

(A) m = 13, M = 24 (B) 1 1

m ,M4 2

(C) m = – 11, M = 0 (D) m = 1, M = 12 Key (D)

Sol

3

4

192xf x

2 sin x



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1

x 12

8 f x 96

x x x

1/ 2 1/ 2 1/ 2

8dx f x dx 96dx

1

8x 4 f x f 96x 482

8x 4 f x 96x 48

1 1 1

1/ 2 1/ 2 1/ 2

8x 4 dx f x dx 96x 48 dx

1

1 12 2

1/ 2 1/ 21/ 2

4x 4x f x dx 48x 48x

1

1/ 2

1 f x dx 12

m 1,M 12

SECTION : 3 (Maximum Marks : 16) This section contains TWO paragraphs Based on each paragraph, there will be TWO 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 one of the bubbles is darkened. 2 In all other cases. PARAGRAPH 1

Let F: be a thrice differentiable function. Suppose that F(1) = 0, F(3) = –4 and F x 0 for

all x 1/ 2, 3 . Let f x xF x for all x .

57. The correct statement(s) is (are) (A) f 1 0 (B) f(2) < 0

(C) f x 0 for any x 1,3 (D) f x 0 for some x 1, 3

Key. (A, B, C) Sol. ' 'f x F x xF x

' 1 1 ' 1f F F

0 ive ive

2 2 2 0f F as F x decreasing for x > 1

' ' 0 1, 3f x F x xF x x

as 0F x for 1x and ' 0F x



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58. If 3

2

1

x F x dx 12 and 3

3

1

x F x dx 40 , then the correct expression(s) is (are)

(A) 9f 3 f 1 32 0 (B) 3

1

f x dx 12

(C) 9f 3 f 1 32 0 (D) 3

1

f x dx 12

Key. (C, D)

Sol. 3 2

1' 12x F x dx

332

1 12 12x F x xF x dx

3

19 3 1 2 12F F f x dx

3 3

1 136 2 12 12f x dx f x dx

3 3

1" 40x F x dx

333 2

1 1' 3 ' 40x F x x F x dx

332 2

1 1' 3 ' 40x xF x x F x dx

9 ' 3 3 1 ' 1 1 3 12 40f F f F

9 ' 3 ' 1 32 0f f

PARAGRAPH 2 Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of red and black balls, respectively, in box II.

59. One of the two boxes I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from

box II is 1

3, then the correct option(s) with the possible values of n1, n2, n3 and n4 is(are)

(A) n1 = 3, n2 = 3, n3 = 5, n4 = 15 (B) n1 = 3, n2 = 6, n3 = 10, n4 = 50 (C) n1 = 8, n2 = 6, n3 = 5, n4 = 20 (D) n1 = 6, n2 = 12, n3 = 5, n4 = 20 Key. (A, B)

Sol. 1

2

n red

n black 3

4

n red

n black

Box I Box II

3

3 4

31

1 2 3 4

12 1

1 1 32 2

n

n nP

nn

n n n n

3

13 4 3

1 2

1

3

nn

n n nn n

3 1 2 1 3 42 n n n n n n

1 2 3 1 42n n n n n



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is satisfied by (A), (B) 60. A ball is drawn at random from box I and transferred to box II. If the probability of drawing

a red ball from box I, after this transfer, is 1

3, then the correct option (s) with the possible

values of n1 and n2 is (are) (A) n1 = 4 and n2 = 6 (B) n1 = 2 and n2 = 3 (A) n1 = 10 and n2 = 20 (D) n1 = 3 and n2 = 6 Key (C, D)

Sol. 1 1 2 1

1 2 1 2 1 2 1 2

1 1. .

1 1 3

n n n n

n n n n n n n n

1 1 2 1 2 1 23 1 1n n n n n n n

1 1 23 n n n

1 22n n

IIT JEE adv-2015 question paper-2, hints & solutions

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Transcript of IIT JEE adv-2015 question paper-2, hints & solutions