CODE : 8

IIT - JEE 2015 (Advanced)

(2) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

PART I : PHYSICS

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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 the bubble corresponding to the answer is darkened
0 In all other cases

1. The energy of a system as a function of time t is given as E(t) = A 2 exp(t), where
 = 0.2 s1. 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
1. [4]
E(t) = A2 exp (t)
(E) = 2A (A) exp (t) + A2  exp ( t)
 (E) = 2A exp ( t)  A + A2  exp (t) (t)
 (E) = 2A exp (t)  A + A2 exp (t) tt





(E)
 A 
 2
  (t)20.01250.20.0155
E
 A 
(E)

= 0.0250 + 0.015 = 0.040
 % error = 4%
(E)
2. The densities of two solid spheres A and B of the same radii R vary with radial distance r



5

r 
r
as A(r) = k   and B(r) = k   , respectively, where k is a constant. The moments
R
R of inertia of the individual spheres about axes passing through their centres are IA and IB,
I
n respectively. If B =
, the value of n is
IA
10
2. [6]
x
x

R
V = 4 x2  x
 (m) = 4(x) x2 . x
I =

2
2
(m)x 2  4(x)  x 4 x
3
3

R

 I

 x (x)dx
4

0

R

1 x 9  dx
5 
R 0

1
 R5
IB
6

 10


R
1 5
IA
10
1
R x 5  dx
6
R
0

2

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (3)
3. Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0,
/3, 2/3 and . When they are superposed, the intensity of the resulting wave is nI0. The value of n is
3. [3]
Some Intensity  some amplitude


3
 Resultant amplitude = 2asin
 I


=
3

3a

= 3I0

4. For a radioactive material, its activity A and rate of change of its activity R are defined as dN dA
A=  and R = 
, where N(t) is the number of nuclei at time t. Two radioactive dt dt sources P (mean life ) and Q (mean life 2) have the same activity at t = 0. Their rates of
R
n change of activities at t = 2 are RP and RQ, respectively. If P = , then the value of n
RQ
e is 4. [2]
P and Q have same activity at t = 0.
If NPO and NQO are the number of nuclei at t= 0
Then A =P  NPO = Q  NQO
N PO  Q
=
N QO  P

1
1
and Q =

2

N
1
1
 Q=
 PO =
N QO 2
P 2
dN
A=
= N dt dA d 2 N dN R=
= 2 =
= 2 N dt dt dt 2
R
P N
 P= 2  P
R Q Q NQ
P =

At t = 2
N
N
NP = PO and NQ = QO e e2
N P N PO e
1

= 2 
=
e
N QO 2e
NQ
RP
1 2

= 4 =
 n=2
RQ
2e e

3

(4) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

5. A monochromatic beam of light is incident at 60° on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle (n) with the d normal (see the figure). For n = 3 the value of  is 60° and
= m. The value of m is dn 5. [1] sin 60° = 3sinr1 r1 = 30°
 r2 = 30° n sin r2 = sin n sin =
2
d 1
=
cos  dn 2 d 1
=
=1 dn 2cos 60

60


r1

r2



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

6. [1]
The given Ckt reduces to simple Ckt like this
2

2

1

2
6

2

2

6

4



6.5
12

10

10
6.5

4

12
4

4

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (5)
I

2

2

2



6

4.5



6.5
12

6.5 V
4

 I = 1A
I=1
7. An electron in an excited state of Li2+ ion has angular momentum 3h/2. The de Broglie wavelength of the electron in this state is pa0 (where a0 is the Bohr radius). The value of p is
7. [2]
3h
=
 electron in quantum state n = 3
2
In the light of de-Broglie's hypothesis n = 2 rn n2 n = 2 a0 z n  = 2 a 0 z 3
 Pa0 = 2 a 0
3
P=2
8. 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
 M  the rod is zero for m = k 
 . The value of k is
 288 

8. [7]
For zero tension
Tensile force = Compressive force
F1 = 2F3 + F2
Gmm
m 2 GMm
 2G 2 
9 2
16 2
M
M
 2m 
9
16
7M
M
 2 K
144
288
K=7

F1

M
3

5

F3

F3 F2 m


(6) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

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Section 2 (Maximum Marks : 32)
This section contains EIGHT questions.
Each questions 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 bubbles is darkened
2 In all other cases

9. 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
C
ratio 2 is
C1

(A) 6/5
9. (D)

(B) 5/3

(C) 7/5

d/2
C

C
C3

S/2

s s C  2 0 ,C  4 0 d d s s
4 0 2 0 d d  4  s
These are in series, their equivalent  s s 3 0d
4 0 2 0 d d
2 0 s 0 s
C3 

2d
d
4 s s 7 s s
C2  0 0  0

and
C1 0
3
d d 3 d d
C2 7


…(D)
C1 3

6

(D) 7/3

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (7)
10. 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)

1
P1V1
4
(B) If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3P1V1
7
(C) If V2 = 3V1 and T2 = 4T1, then the work done by the gas is P1V1
3
17
(D) If V2 = 3V1 and T2 = 4T1, then the heat supplied to the gas is
P1V1
6
10. (B)
P1V1
PV
= 2 2
T1
T2
P1V1
2V
3
(A)
= P2  1
 P2 = P1
 S = Cross section Area
3T1
T1
2
P2S = Kx  P2Sx = kx2
1
1
1
 kx 2 = P2Sx = P2 (V2 V2 )
2
2
2
1 3
3
 Energy is spring =  P1 (2V1V2 ) = P1V1
[A is incorrect]
2 2
4
3
3
(B) U = nR(T2 T1 ) = (P2V2 P1V1 )
2
2
P1V1
P2 2V1
3
=
 P2 = P1
T1
3T1
2
3 3

U =  P1.2V1P1V1  = 3P1V1
[B is correct]
2 2

PV
3V
4
(C) 1 1 = P2 1
 P2 = P1
T1
4T1
3
Sx
1
P2S = Kx
 P2
= kx 2 = w
2
2
1 4
 w =  P1 (V2 V1 )
2 3
2
4
[C is incorrect]
P1 (3V1  V1 ) = P1V1
3
3
4
3 4
4
3
4
9

(D) P1V1 (P2 V2 P1V1 ) = P1V1  P13V1P1V1  = P1V1 P1V1
3
2
3
2
3
2 3

8P V 27P1V1
35
= 1 1
= P1V1
[D is incorrect]
6
6
(A) If V2 = 2V1 and T2 = 3T1, then the energy stored in the spring is

7

(8) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

11. A fission reaction is given by

236
92

94
U 140 Xe 38 Sr  x  y, where x and y are two particles.
54

236
Considering 92 U to be at rest, the kinetic energies of the products are denoted by KXe,
KSr, Kx (2 MeV) and Ky (2 MeV), respectively. Let the binding energies per nucleon of
140
236
94
92 U, 54 Xe and 38 Sr be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s) is (are)
(A) x = n, y = n, KSr = 129 MeV, KXe = 86 MeV
(B) x = p, y = e, KSr = 129 MeV, KXe = 86 MeV
(C) x = p, y = n, KSr = 129 MeV, KXe = 86 MeV
(D) x = n, y = n, KSr = 86 MeV, KXe = 129 MeV
11. (A)
Q Value of the reaction
Q = 94  8.5 + 140  8.5  236  7.5
Q = 219 MeV

X and Y share 4 MeV together
 remaining 215 MeV will be shared between Xe and Sr nucleus. Heavier particle will have less kinetic energy.
 x = n and y = n and KSr = 129 MeV and KXe = 86 MeV
…(A)
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
12. (A), (B)
P
Strain

Strain

Q
P

Q

Stress

Stress

8

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (9)
For same strain stress in Q is more than P.
 Young's module of Q > Young's modulus of P.
 (D) is not correct.
For same stress strain in P is more.  P is more ductile
 (A), (B)
13. 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)
P(r3R / 4) 63
(A) P(r = 0) = 0
(B)

P(r2R / 3) 80
P(r3R / 5) 16
P(rR / 2) 20
(C)
(D)


P(r2R / 5) 21
P(rR / 3) 27
13. (B), (C) dp  g(r) dr g(r) 

Gm(r) r 2

r

4
;m(r)  r 3
3

4
3

 g(r)  Gr


dp 4 2
  Gr dr 3

 p(r)  p(R) =

4 2  R 2  r2 
 G 
 2 

3



 p(r)  p(R) =

2 2
 G R 2  r 2
3





put p(R) = 0 p(r) =



2 2
 G R 2  r 2
3



9 p(3R / 4)
9 7 63
 16   
4 16 5 80 p(2R / 3) 1 
9
1

9
1
p(3R / 5)
25  16

p(2R / 5) 1  4 21
25

1
1
p(R / 2)
4  9  3  27

1 4 8 32 p(R / 3) 1 
9

9

(10) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

14. 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 liquids L1 and L2 of densities 1 and 2 and viscosities 1 and 2, respectively. They float in equilibrium with the sphere P in L1 and sphere Q in L2 and the string being taut (see figure). If sphere P alone in L2 has terminal velocity VP and Q alone in L1 has terminal velocity VQ , then

(A)

VP
VQ



1
2

(B)

VP
VQ



2
1

(C) VP  VQ 0

14. (A), (D)

(UP  WP)

(  1 )
(  2 )
P  1
 Q  a

2
1

p

     2
 1  1
a  2  2  1

…(i)

(D) VP  VQ 0

T

P

For Eq.
UP  WP = WQ  UQ
T

1  1
= 2  2

1  2
1  2
...(ii)
=
From (i) and (ii) option (A) follows
Also it is obvious from (i) & (ii)  p &1 have opposite signs 

(WQ  UQ)

D is true.

15. 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) 0I2 = 0V2
(B) 0I = 0V
(C) I = 0cV
(D) 0cI = 0V
15. (A), (C)
Correct equations are (dimensionally)
(A) and (C)
16. 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)
(A) E is uniform, its magnitude is independent of R2 but its direction depends on r
(B) E is uniform, its magnitude depends on R2 and its direction depends on r
(C) E is uniform, its magnitude is independent of  but its direction depends on 
(D) E is uniform and both its magnitude and direction depend on 

10

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (11)
16. (D)
E = E1  E 2


=
OA 
AP
3 0
3 0


=
OP
OAAP 
3 0
3 0



E

A

r



P
R2
O

R1

 a 3 0

 = Charge density.
 Option (D) only is correct.
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 value im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im.

17. For two structures namely S1 with n1 = 45 / 4 and n2 =3/2, and S2 with n1 = 8/5 and n2 = 7/5 and taking the refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is (are)
(A) NA of S1 immersed in water is the same as that of S2 immersed in a liquid of
16
refractive index
.
3 15
6
(B) NA of S1 immersed in liquid of refractive index is the same as that of S2
15
immersed in water.
(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.
17. (A), (C)
 > c n2  sin  = > sin c
n  n3  sin  >  2 
…(i)

 n1 

i n1 
2

11

(12) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution



Also n3 sin i = n1sin   = n1 cos 
2

2
2
2
2
 n3 sin i = n1  1  sin 
…(ii)


2

n 
2
2
2
1sin 1 2   n1 1sin 2  n1 n 2
 n1 
From (ii) & (iii) it follow
2
2
2
n3 sin 2 in1 n 2
2

2 n1 n 2
2
n3

 Numerical aperture =

45 3

16 4 =
Option A :
NA1 =
4
3
64 49

NA2 = 25 25 =
16
3 15
NA1 = NA2  Option (A) is correct.

9
16 = 9
4
16
3
15
5 = 153 = 9
16
165 16
3 15

Option (B) :

NA1

NA2

=

3
4 = 3  15 =
6
4 6
15

=

15
5 = 3 15
4
20
3

 Option (B) is incorrect.
Option (C) :
NA1

NA2

3
3
= 4 =
1
4

=

15
5 = 3
4
4
15

 Option (C) is correct.
Option (D) : is incorrect at (C) is correct.

12

15
8

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (13)
18. If two structures of same crosssectional area, but different numerical apertures NA1 and
NA2 (NA2 < NA1) are joined longitudinally, the numerical aperture of the combined structure is
NA1NA 2
(A)
(B) NA1  NA 2
(C) NA1
(D) NA2
NA1  NA 2
18. (D) i1 i1 i2 The numerical aperture is limited by second slab. [smaller NA]
 sin i2 < NA2 sin i1 > sin i2
 NA of combination = smaller NA

Section 3 (Maximum Marks : 16)






This section contains TWO paragraph.
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 none of the bubbles is darkened
2 In all other cases
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 , and d, respectively. A uniform magnetic field B is applied 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 electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.

13

(14) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

19. Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are w1 and w2 and thicknesses 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 = w2 and d1 = 2d2, then V2 = 2V1
(B) If w1 = w2 and d1 = 2d2, then V2 = V1
(C) If w1 = 2w2 and d1 = d2, then V2 = 2V1
(D) If w1 = 2w2 and d1 = d2, then V2 = V1
19. (A), (D)
I
I
J=
= nevj
 evd = mwd wd
IB
 FB = evdB = mwd V eV E=
 FG = w w
Under equation condition
IB
eV
IB
=
 V= nwd w nde  Vd is same in two strips.
V1d1 = V2d2
20. Consider two different metallic strips (1 and 2) of same dimensions (length , width w and thickness d) with carrier densities n1 and n2 respectively. Strip 1 is placed in magnetic field B1 and strip 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
(D) If B1 = 2B2 and n1 = n2, then V2 = V1
20. (A), (C)
IB1
IB2
V1 =
V2 = n1de n 2de n1V1 n V
= 2 2
B1
B2

PART II : CHEMISTRY





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 the bubble corresponding to the answer is darkened
0 In all other cases

14

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (15)
21. Among the following, the number of reaction(s) that produce(s) benzaldehyde is

I.

II.

III.

IV.

21. [4]
CHO

(I)

CO, HCl


Anhydrous AlCl /CuCl
3

CHO
Cl

(II)

OH

H2O

CH Cl 
100C

CH OH

(III)

H2

Cl 
Pd BaSO 4

C H
O

O

(IV)

C

2

O

O
C



H O

O

DIBAL  H

Me 
Toluene,  78C H2O

C

H

+ MeOH

22. In the complex acetylbromidodicarbonylbis(triethylphosphine)iron(II), the number of
FeC bond(s) is
22. [3]
23. Among the complex ions, [Co(NH2CH2CH2NH2)2Cl2]+, [CrCl2(C2O4)2]3,
[Fe(H2O)4(OH)2]+, [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

15

(16) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

23. [6]
NH2

Cl
NH2
NH2

NH2

NH2

NH2

+3
Co

NH2

Cl trans +3
Co

NH2
Cl

Cl cis 24. Three moles of B2H6 are completely reacted with methanol. The number of moles of boron containing product formed is
24. [6]

B2H6  6MeOH  2B(OMe)3  6H2

Diborane reacts with methanol to give hydrogen and trimethoxyborate ester.
25. 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    0  , the
X

Y

difference in their pKa values, pKa(HX)  pKa(HY), is (consider degree of ionization of both acids to be << 1)
25. [3]
2 = 10 1 pKa1 = pH1  log 1 pKa2 = pH2  log 2 pKa1  pKa2 = pH1  pH2 = 3
26. A closed vessel with rigid walls contains 1 mol of
238
Considering complete decay of 92 U to pressure of the system at 298 K is

206
82 Pb,

238
92 U and

1 mol of air at 298 K.

the ratio of the final pressure to the initial

26. [9]
238
4
0
 206
92 U 82 Pb8 2He6 1

Initial no. of moles of gas = 1.
Final no. of moles of gases = 1 + 8 = 9
27. In dilute aqueous H2SO4, the complex diaquodioxalatoferrate(II) is oxidized by MnO .
4
For this reaction, the ratio of the rate of change of [H+] to the rate of change of [MnO ] is
4
27. [8]
Rate of change of [H+] is 8 times the rate of change of  MnO  

4

16

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (17)
28. The number of hydroxyl group(s) in Q is

28. [4]
H

H



 H


HO
CH3

CH3

 H 2O

H2O
 CH3


CH3

CH3

CH3
1,2 methyl shift

OH
CH3
HO

Aq.dilute
KMnO4 (excess)O C



OH
CH3

OH




CH3

CH3

H

CH3


CH3

(P)

Section 2 (Maximum Marks : 32)





This section contains EIGHT questions.
Each questions 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 bubbles is darkened
2 In all other cases

29. 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
29. (B)
30. 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
30. (A), (B), (C), (D)

17

(18) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

31. One mole of a monoatomic 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)

(B)

(C)

(D)

31. (C)
32. In the following reactions, the product S is

(A)

(B)

(C)

(D)

32. (A)
CH3

CH3

CHO

(i) O3


(ii) Zn,H2O

CH3
NH

3



CH2 CHO

18

N

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (19)
33. The major product U in the following reactions is

(A)

(B)

(C)

(D)

33. (B)
CH3

CH3

CH CH3
CH CHCH ,H

2
3


High pressure,

CH3 C

O O H

Radical nitiator, O

2



Cumene

Cumene hydroperoxide

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

(A)

(B)

(C)

(D)

19

(20) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

34. (A)

N2Cl
NH2

OH
HO

NaNO ,HCl

2


O C

N

N

35. 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)
35. (B), (C)
(i) H  O  Cl

..
..
..

(ii) HOClO
(iii) HOClO
O

O
(iv) HOClO

O
(A) False (B) True

(C) True

(D) False

36. 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+
36. (C) , (D)
Group2 ions are Hg+2, Pb+2, Bi+3, Cu+2, Cd+2, As+3, Sb+3, Sn+2, and Sn+4

Section 3 (Maximum Marks : 16)






This section contains TWO paragraph.
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 none of the bubbles is darkened
2 In all other cases

20

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (21)

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 (Expt. 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 to 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 all solutions as
1.0 g mL1)
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
37. (A)
38. The pH of the solution after Expt. 2 is
(A) 2.8

(B) 4.7

(C) 5.0

(D) 7.0

38. (B)
200 m.eq. + 100 m.eq Base  100 m.eq. salt
Salt
pH = pKa + log
Acid
100
= 2 105  log
= 5  log2 = 5  0.3010 = 4.7
100
PARAGRAPH 2
In the following reactions

39. Compound X is

(A)

(B)

(C)

39. (C)

21

(D)

(22) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

40. The major compound Y is
(A)

(B)

(C)

(D)

40. (D)
Solution : Q. 39 & 40
C
CH

CH

Pd.BaSO
H2

4


CH2

CH2

OH

B H

2 6


H O ,NaOH,H O
2 2

2

(X)

H2O , HgSO4 , H2SO4
O
C

CH2

OH
CH3

Et

EtMgBr,H O

2 


C

CH3

CH3 CH

CH

CH3

Heat




PART III - MATHEMATICS





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 the bubble corresponding to the answer is darkened
0 In all other cases

41. 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
41. [9]
Let the first term is a and common difference is d.
S7
6
=
S11 11
7
2a  6d
6
2
=
11 
2a  10d 11
2
2a  6d
6
=
2a  10d
7

22

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (23)
14a + 42 d = 12a + 60d
2a = 18 d a = 9d
130 < a7 < 140
130 < a + 6d < 140
130 < 15d < 140 d=9 42. The coefficient of x9 in the expansion of (1 + x) (1 + x2) (1 + x3) …. (1 + x100) is
42. [8]
(1 + x) (1 + x2) (1 + x3)…. (1 + x100) co-eff of x9 is number of ways sum of power of x is 9.
Of the form 1  1 way
Form 2 {(1, 8) (2, 7) (3, 6) (4, 5)} Total 4 ways
Form 3 {(1, 2, 6) (1, 3, 5) (2, 3, 4)} Total 3 ways. x 2 y2
43. Suppose that the foci of the ellipse

 1 are (f1, 0) and (f2, 0) where f1 > 0 and
9
5 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
 1

the slope of T2, then the value of  2  m2  is
2
 m1

43. [4] x 2 y2

1
9
5

5 2
9 3
2
3. , 0 )
3

e = 1 

Focus ( ae, 0) = (

= ( 2, 0)

f1 = (2, 0), f2 = (2, 0)
Parabola P1 = y2 = 4.2 x y2 = 8x

… (1)

Parabola P2 = y2 =  4(4)x y2 = 16x

… (2)

Tangent y2 = 8x is y = m1 x +

2 m1 Given it passes (4, 0)
0 = 4 m1 +
 m12 

(4, 0)

2 m1 1
2

Tangent on y2 =  16x is y = m2x +

4 m2 P2

Given it passess (2, 0)
 0 =  2 m2 +

4 m2 m2 = 2
2

So,

f1 (2,
0)

T1

1
 m2
2
2 m1 =2+2=4

23

f1 (2, 0)

P1

(24) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

 cos n 

e
e
e
    then the
44. Let m and n be two positive integers greater than 1. If lim m 
0 

2



 m value of is n
44. [2]
 
 cos  n   e  e ecos  1  1 e e lim = lim 
=
 =  = 0 m m
0 

2
2


n

e ecos 
= lim
0
m

 n 1



e ecos  1  1   n   e e
=  = lim m
 cos   1 = 
2 0   cos  n   1
2
 n



n 
n 
e 2 sin 2
e 2 sin 2
  n 2

n 2 e 

2  
2     =  e lim = lim
  =  = 0
 
2
2
0
 2 
 2 
2
2
 n 
 n 
m .  
m   
 2 
 2 

e 2n  m e ,
=
0 2
2

= lim

= 2n  m = 0
=

m
=2
n





 12  9x 2 
1

 dx , where tan x takes only principal values, then the
1 x2 

0
3 

value of  log e 1     is
4 

45. [9]
1

45. If    e9x 3tan

Let

1

x

t = 9x + 3 tan1x
3
1 x2

dt = 9 + dt =

12  9x 2
1 x2
9  3

=



4

e t dt

0

9  3

 = e t  0
  loge |1 + | 

3
4

4

= e93/4  1


= loge 1  e

93


4

1 

3 3
3
= 9 
=9
4
4
4

24

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (25)



46. Let f :

be a continuous odd function, which vanishes exactly at one point and x x

1 f 1  . Suppose that F  x    f  t  dt for all x  [1, 2] and G  x    t f  f  t   dt for
2
1
1
Fx 1
 , then the value of x 1 G  x 
14

all x  [1, 2]. If lim

1 f   is
2

46. [7] x  f (t) dt

1

Lim

=

x 1 x

 t | f (t) | dt

f (x) x f (f (x))

=

1
14

1

1
2 = 1

 1  14 f 
2

1
 f  =7
2

47. Suppose that p,q and r are three non-coplanar vectors in 3 . 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
47. [9]
 (i) s = 4p  3q  5r

s =  p  q  r  x  p  q  r  y  p  q  r z

s = (x + y  z) p + (x  y  z) q + (x + y + z) r
x + y  z = 4 xyz=3 x+y+z=5
9
7 x = 4, y = , z = 
2
2
2x + y + z = 9

 (ii)
 (iii)
 (iv)
 (v)

 k 
 k 
48. For any integer k, let a k  cos    i sin   , where i  1 . The value of the
 7 
 7 
12

 k 1  k expression k 1

3

 4k 1  4k 2

is

k 1

48. [4]
12



3

i(K 1) 
K
i
7
e
e 7

K 1 i(4K 1) 
(4K  2) 
 i
7
e e 7



K 1

12

e

=

i

K
7

 e

i


7 1

K 1
3



i(4K 1)  e 7

K 1

i
 e 7 1

12



=

K 1
3



K 1

25

i e 7 1 i e 7 1

=

i
12 e 7 1

3

i e 7 1

=4

(26) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

Section 2 (Maximum Marks : 32)





This section contains EIGHT questions.
Each questions 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 bubbles is darkened
2 In all other cases

49. Let f, g : [1, 2]  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=2 x = 1 x = 0 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)
49. (B), (C)
Let h(x) = f(x)  3g(x) h(1) = 3 h (0) = 3 h (2) = 3 h(x) = f(x)  3g(x)
 f(x)  3 g(x) = 0 has exactly one solution is (1, 0)
& f(x)  3g(x) = 0 has exactly one solution is (0, 2)

1

0

2

  
50. Let f(x) = 7tan8x + 7tan6x  3tan4x  3tan2x for all x    ,  . Then the correct
 2 2 expression(s) is (are)
4

(A)



0
4

(C)

1 x f  x  dx 
12
1

 x f  x  dx  6

4

(B)

 f  x  dx  0

0
4

(D)

0

 f  x  dx  1
0

50. (A), (B) f(x) = 7tan6x . sec2x  3 tan2 (x) . sec2(x)

26

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (27)
4



4

f (x)dx =

0

 (7 tan

6

(x)3 tan 2 (x))sec 2(x)dx

0
4

=  tan 7 xtan 3 x 0
4



=0

4

xf (x)dx

=

0

 xsec

2

(x)(7 tan 6 x 3tan 2 x)dx
.

0
4

=  x(tan xtan x) 0   (tan 7 (x)tan 3(x)).1dx

7

4

3

0
4

=

 tan

4
3

 tan

(x)(1tan x)dx =
4

0

3

(x)(1tan 2 x)sec 2 (x)dx
.

0
4

 tan 4 (x) tan 6 (x) 
1 1
1
=

 =  =
6 0
4 6 12
 4

192x 3
51. Let f(x) = for all x 
2sin 4 x possible values of m and M are

1 with f   = 0. If m 
2

1

 f (x)dxM , then the t/2 1
1
,M=
4
2
(D) m = 1, M = 12

(A) m = 13, M = 24

(B) m =

(C) m = 11, M = 0
51. (D)

192x 3

2sin 4 (x)
1
Also, as x increases from to1 ,

2 f(x) increases from 8 to 96 f (1)f (1 2)
 8
96
1
1
2
 4 < f(1) < 48
1
1
1
1
1
  4  f (x)dx  48
2
2
2
2
12
f(x) =

1

 1  f (x)dx12

 (D) are correct.

12

52. Let S be the set of all nonzero real numbers  such that the quadratic equation
x2  x +  = 0 has two distinct real roots x1 and x2 satisfying the inequality |x1  x2| < 1.
Which of the following intervals is (are) a subset(s) of S?
1 
 1
 1

(A)   ,
(B)  
,0 

5
5 
 2


 1 
(C)  0, 
5


 1 1
(D) 
, 
 5 2

27

(28) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

52. (A), (D)
x2  x +  = 0
D>0
 1  4.. > 0



1
  2 0
4
1
2  0
4
1
1
 
2
2

x1 + x2 =

1


, x1 x2 = 1

 x1  x 2 

x1  x 2 =

2

 4x1x 2

=

1
4
2

as x1  x 2  1
1
4 < 1
2
1
4 1
2
1
5
2
1
  2
5



2

 1 
 0
 5
1

 2  

<

5

or  >

1
5

Take intersection
 1
 2

   ,

1   1 1
, 
 
5   5 2

6
4
53. If  = 3 sin1   and  = 3 cos1   , where the inverse trigonometric functions take
 11 
9
only the principal values, then the correct option(s) is (are)
(A) cos  > 0
(B) sin  > 0
(C) cos ( + ) > 0
(D) cos  < 0
53. (B), (C), (D)
1 6
1
6
 = 3sin 1   as
 
 11 
2 11
2

6 
 sin 1   
6
 11  4

 6  3
 3sin 1   
2
 11  4

3


2
4
4 1
4
 = 3cos 1   as

9 2
9

28

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (29)

4
1
cos1    cos 1  
2
9
4  cos 1   
9 3
4
 3cos1    
9
 cos  < 0, sin  < 0, cos  < 0, cos  < 0, sin  > 0 cos ( + ) = cos  cos   sin  sin  > 0
54. 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 x2 + (y  1)2 = 2. The straight line x + y = 3 touches the curves S, E1 and E2 at P, Q and R, respectively.
2 2
Suppose that PQ = PR =
. If e1 and e2 are the eccentricities of E1 and E2,
3
respectively, then the correct expression(s) is (are)
7
43
2
(A) e1 e2 
(B) e1 e2 
2
40
2 10
3
5
2
(C) e1 e2 
(D) e1 e 2 
2
4
8
54. (A), (B) x 2 y2 x 2 y2
Let ellipse 2  2  1  E1 (a > b) and E2  2  2  1 (a < b) a b a b
2
2
Circle s = x + (y  1) = 2
Tangent, x + y = 3
 (1)
Normal of circle x  y = K
Given it passes (0, 1)
 K = 1
Normal to circle x  y = k
 (2) eq. (1) and (2) solving P  (1, 2)
For point Q and R,
2 2 x 1 y2 =
=
1
1
3

2
2
5 4
1 8
 Q   ,  and R   , 
3 3
3 3 x 2 y2
As normal y =  x + 3 to 2  2  1 is a b c2 = a2 m2 + b2
 9 = a2 + b2
 (3)
2
2 x y
5 4
As point Q  ,  lies E1 : 2  2 = 1 a b
3 3
25 16

=1

9a 2 9b2
 a2 = 5

29

(30) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

From (3), b2 = 4 e1 = 1 

4 b2 = 1
2
5 a =

1
5

Similarly, e2 = 1 

1
8

=

7
2 2

7
2 10
1 7
8  35
43
2
=
= e1  e2 = 
2
5 8
40
40
1 7
8  35
27
2
2
= 
| e1  e2 | =  =
5 8
40
40

 e1 e2 =

55. Consider the hyperbola H: x2  y2 = 1 and a circle S with center 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 ( , m) is the centroid of the triangle
 PMN, then the correct expression(s) is(are) x1 dm d 1

forx1  1
(A)
(B)
1 2 forx1  1 dx1 3 x 2  1 dx1 3x1



(C)

d
1
1 2 forx1  1 dx1 3x1

(D)



dm 1
 fory1  0 dy1 3

55. (A), (B), (D) x2  y2 = 1
2x  2y y = 0 x y =
 (1) y x1 y  1
= 1 y x1  x 2
 x2 = 2x1
 (2) y  y1 x Equation of PM is
= 1 x  x1 y1 2 x 2  y1
1
Put y = 0, x =
=
x1 x 1 
 M   , 0
N  (2x1, 0)
 x1 
1
  = x1 +
3x1
d
1
=1 2 dx1 3x1 dm y
1
m = 1

= dy1 3
3

1

(x1, y1)
N
M

P  (x1, y1)

30

(x2, 0)

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (31)
Also, m =

1
1 2 y1 = x1  1
3
3 x1 dm
=
2 dx1 3 x1  1

56. The option(s) with the values of a and L that satisfy the following equation is(are)
4

 e (sin t 6

atcos 4 at)dt

 e (sin

6

atcos at)dt

0


= L? t 4

0

e4   1 e  1 e4   1
(C) a = 4, L =  e 1
56. (A), (C)

e4   1 e  1 e4   1
(D) a = 4, L =  e 1

(A) a = 2, L =

(B) a = 2, L =

4

 e (sin t 6

(at)  cos 4 (at)) dt

6

(at)  cos 4 (at))dt

0


 e (sin t 0

Let f(t) = et (sin6 (at) + cos6 (at))
F(k + t) = ek + t (sin6 (a (k + t)) + cos6 (a (k + t))
= ek f(t)
4





 f (t) dt
0


… (for even values of a)

(1  e   e 2   33 )  f (t) dt

=



 f (t) dt

 f (t) dt

0

0

e4   1
=  e 1

0

Section 3 (Maximum Marks : 16)






This section contains TWO paragraph.
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 none of the bubbles is darkened
2 In all other cases

31

(32) Vidyalankar : IIT JEE 2015  Advanced : Question Paper & Solution

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
(C) f(x)  0 for any x  (1, 3)

(B) f(2) < 0
(D) f(x) = 0 for some x  (1, 3)

57. (A), (B), (C) f (x) = x F(x) f(x) = F(x) + x F(x)
 f(1) = F(1) + F(1) = F(1) < 0
( F(x) < 0  x (1/2, 3) f (2) = 2 F (2) < 0
( F (2) < 0) f(x) = F(x) + x F(x) < 0  x  (1, 3)
( F (x) < 0 and x F (x) < 0 for all x  (1, 3))
 f(x)  0 for any x  (1, 3)
Hence (A), (B), (C) are correct.
3

58. If

3

 x F'(x) dx = 12 and  x F"(x) dx = 40, then the correct expression(s) is(are)
2

3

1

1
3

(B)  f (x)dx = 12

(A) 9f(3) + f(1)  32 = 0

1
3

(D)`  f (x)dx = 12

(C) 9f(3)  f(1) + 32 = 0

1

58. (C), (D)
3

2

 3 2
 x   x    x  F  x  dx
 f  x  dx =  x F  x  dx =  F 2 1 2

1
1
1

3

3

=
3

9
1
(4)  0   12 
2
2

 x F  x  = 40
3

= 12
3

3
2
  x F  x  1   3x F  x 
3

 27 F(3)  F(1)
Also,

= 40

1

1

= 4 (1)

f(3) = F(3) + 3F(3) f(1) = F(1)

(C)

 9 f(3)  f(1) + 32
= 9 F(3) + 27 F(3)  F(1) + 32
= 27 F(3)  F(1)  4 = 0

( F(3) = 4)

32

IIT JEE 2015 Advanced : Question Paper & Solution (Paper – II) (33)
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, box 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
1
was drawn from box II is , then the correct option(s) with the possible values of n1, n2,
3
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
59. (A), (B)
P(B1) = 1/2

P(B2) = 1/2

P(R/B1) =

n1 n1  n 2

P(R/B2) =

B  1
P 2  =
R 3
1  n3 
2  n3  n4 
1


=
1  n 3  1  n1  3

2  n 3  n 4  2  n1  n 2 





n3 n3  n4

60. A ball is drawn at random from box I and transferred to box II. If the probability of
1
drawing a red ball from box I, after this transfer, is , then the correct option(s) with the
3
possible values of n1 and n2 is(are)
(A) n1 = 4 and n2 = 6
(B) n1 = 2 and n2 = 3
(C) n1 = 10 and n2 = 20
(D) n1 = 3 and n2 = 6
60. (C), (D)
P (E) =

1
3

n1
 n1   n1  1   n 2  
 1
 n  n   n  n 1   n  n   n  n 1 = 3
2  1
2
2  1
2
 1
  1

Among the given options, (C) and (D) satisfies the equation.


33