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

JEE-Advanced – 2015 (Hints & Solutions) Paper - 2
NARAYANA Group of Educational Institutions
-1-
NARAYANA IIT ACADEMY
INDIA
JEE – ADVANCED– PAPER – 2
CODE : 0
Time : 3 Hours. Maximum Marks : 240
READ THE INSTRUCTIONS CAREFULLY
GENERAL :
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identical and same as that on the question paper booklet. If not, contact the invigilator.
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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.
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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-
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 r
r 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 2
A
0
2 Kr 8 K R
I .4 r dr .r
3 R 18


 
  
 

R 5 5
2 2
B 5
0
2 Kr 8 KR
I 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 

-3-
3. For a radioactive material, its activity A and rate of change of its activity R are defined as
dN dA
A 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: t
0N N e 

 t
0
dN
A N e
dt

 
   &
2 t
0
dA
& R N e
dt

   

Now
 
 
P
Q
t2
P 0P P
t2
Q Q 0 Q
N eR
R N e






  Q P
2
tP
Q
1
e
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
 
o
sin 60 sin
n
sin r sin 60 r

 

2nsin r 3 … (1)  nsin 60 r sin  … (2)
3
sin r
2n


-4-
A = 600
(60 – r)
0
r
2
4n 3
cosr
2n


Putting values of sin r & cos r in equation (2)
2
3 4n 3 1 3
n . sin
2 2n 2 2n

  
    
  
2
3 4n 3 3 4sin   
differentiate both sides w.r.t. n
2
2n d
3 4cos .
dn2 4n 3

 

2
d 3n
dn cos 4n 3




Putting
o
n 3 & 60  
d 3 3 6
2
1dn 94 3 3
2
 
  
 
5. In the following circuit, the current through the resistor R (= 2 ) is I Amperes. The value of I
is
Ans. 1
Sol:

-5-
6.5
2
2 8
2
4
4
10
12
E D
F
B
C
A 1
ABCA is a balanced wheatstone’s bridge of equivalent resistance 2 . So
2
4
10
12
E D
A
F C
6
ACDEFA is also a balanced wheatstone’s bridge of equivalent resistance 4.5 . So
2
4.56.5
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 0pa (where 0a is the Bohr radius). The value of p is
Key. (2)
Sol. Given
3
2


h
mvr
3
2


h
mv
r
  debroglie =
2
2
3 3
 
 
h h r r
mv h
…(i)
Now radius of the electron in 3rd
orbit of 2+
Li ion
2 2
0 0
0
3
3
3
  
a n a
r a
Z
…(ii)
 Putting r from (ii) in (i)
 debroglie = 0
0
3
2 2
3
  
a
a
 p = 2.

-6-
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
 
  
 
M
m 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 2
163
  
Gm GmGMm GMm
l l ll

7
288
m 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 1
0.2 
  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
 t
E A
Taking Log on both sides and differentiating we get 2  
dE dA
dt
E A
Maximum % error
100 2 100 100
 
       
dE dA dt
t
E A t
=  2 1.25 0.2 5 1.5   
= 2.5 + 1.5
= 4 %.
Ans : 4.

-7-
 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
4
P 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
3
P V
(D) If 2 1 13 and 4 , 2V V T T then the heat supplied to the gas is 1 1
17
6
P V
Key. (B)
Sol. 1 1kx PA
 1 2k x x P A 
 2 1kx P P A 
 2
2 1
1 1
kx P P A.x
2 2
 
=  1 1
1 3
P P v
2 2
 
  
 
=  1
1
P1
v
2 2
 
 
 
1 1P v
4
3
2 2
    
nf
U R T nR T

-8-
2 2
1 1 1
( ) x ( A)
2 2 2

      
V
W kx P P V
A
2 2 2
2 1
1 1 1
  
P nRT T
P P
P nRT T
When 2 1 12 and 3 2V V T T
Energy stored in the spring = 2
1
2
P V
= 1
1 1 1 1
1
31 3
2 2
   
T
P V P V
T
1 1 1 1
3
2 3 3
2
    U nR T nRT P V
When 2 1 2 13 , 4 V V T T
Work done = 2
2 1 1
1
1 1
2
2 2
   
T
P V P V
T
= 1
1 1 1 1
1
41
2 4
2
   
T
P V P V
T
1 1
3 3 9
3
2 2 2
     U nR T nR T nRT = 1 1
9
2
P V
1 1 1 1 1 1
9 17
4
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
C
is
+ -
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 0
1 
C A
C
d

-9-
1 2  2 4 
1 2 
Ca Cb
Cc
After the insertion of di-electrics
Where
0
0
2
22
2


 a
A
A
C
d d
0
0
4
42
2


 b
A
A
C
d d
0
0
2
2
 
 c
A
A
C
d d
So,  2  a b cC C SC C
= 

a b
c
a b
C C
C
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 dm
dp
r r
 
3 2
2 2
4
4
3
.4
p r r dr
P G
r r
 

 
 
5
4
4
.
3
R
r
r
P GP dr
r
 
 2 24
3
R
r r
P G R r    (Pressure as a function
r
dr
dp

-10-
of r)
2
2
2
22
2
3 4 9
3 255 16 16
21 212 4 4
3 255
R R
P r G R
R
RR R
G RP r
 
 
  
     
     
  
    
   
2
2
2 2
2
3 9
7 9 634 16
2 4 16 5 80
3 9
r
r
R RP R
R
R 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)

-11-
(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 1
0
OQ
E
3



Field at Q due to removed sphere 2
0
PQ
E
3



Hence Net field at O
1 2E E E 
 0
OQ PQ
3

 

0
OP
3



0
E a
3



O
R1
P
Q
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 2
0 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 2
2
04
q q
F
r



for 1 3 2 4
0 0 M L A T 
      
W
V
q
  2 1 3
V ML A T 
   
0 0
1
c



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




-12-
(C) P QV V 0  (D) P QV V 0 
Ans. (A, D)
Sol.
P
Q
L2
L1
V = Volume of both spheres
P
1 1B V g 
2 2B V g 
1m g
T
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 2
P
2
r2
V g
9
  


…(iv)
 2
2 1
Q
1
r2
V g
9
  


…(v)
 
 
P 1 2 1
Q 2 1 2
V
V
  

  
 P 1
Q 2
V
1
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 94
92 54 38Xe Sr x y;   U where x and y are two particles.
Considering 236
92 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 236 140
92 54U, Xe and 94
38 Sr be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering
different conservation laws, the correct option(s) is (are)

-13-
(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:
 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
 Marking scheme:
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

-14-
(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 mnsini n sin r
mr 90 
So,
2
1 2
m 2
1
n n
sini 1
n n
  
Put the value and check the answer
example
for 1S
m
45 9
4 4sini 1
4 45
3 16
 
9
16

For 2S
m
8 49
5 25sini 1
16 64
253 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

-15-
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
RwI
S
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 V
V B E B
ne Wd W
  

1
V
d

 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

-16-
(D) If B1 = 2B2 and n1 = n2, then V2 = V1
KEY: (A,C)
SOL: For same dimension and current
B
V
n
 and
nV
constant
B
 
B1 = B2
1 2
2 1
n V
n V
 
 option (A) correct
For same ‘n’
V
constant
B

1 1
2 2
V B
V B
 
 option (C) is also correct
 correct options are (A) & (C)

-17-
PART II: CHEMISTRY
inclusive.
 Marking scheme:
21. In the complex acetylbromidodicarbonylbis (triethylphosphine) iron (II), the number of Fe–C
bond(s) is
Key: (3)
Sol:       
2
3 2 52 3 2
Fe 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 0
X 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:
AssumeH X I
AssumeHY II


I
I
o




II
II
o




I I
II II
1
10



 

22
3I I I
2
II II II
Ka C 0.01 1
10
Ka C 0.1 10


 
   
 
 pKa I– pKa II 3
log10 0 3
   

-18-
25. A closed vessel with rigid walls contains 1 mol of 238
92 U and 1 mol of air at 298 K. Considering
complete decay of 238
92 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 9
9
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 2
Fe H O C O MnO 8H Fe 6H O 4CO Mn   
        

 
 4
Rate of change of H 8
8
1Rate of change MnO


 
27. The number of hydroxyl group(s) in Q is
OH
H
CH3
CH3
H
Heat

  4
o
aqueous dilute KMnO excess
0 C
P Q
Key: (4)
Sol.
OH
CH3
CH3
H

CH3
CH3 CH3
CH3 CH3
CH3
H
CH3
CH3
 4
o
aq. dilute KMnO excess
0 C

 P
OH
OH
CH3 OH
CH3
 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, HCl
Anhydrous AlCl /CuCl

II.
CHCl2
2
o
H O
100 C


-19-
III.
COCl
2
4
H
Pd BaSO

IV.
CO2Me
o
2
DIBAL H
Toluene, 78 C
H O



Key: (4)
Sol. I.
3
CO, HCl
Anhydrous AlCl /CuCl

CHO
II.
CHCl 2
2
o
H O
100 C

C OH
H
OH
2H O C H
O
III.
COCl
2
4
H
Pd BaSO

CHO
HCl
IV.
CO2Me
o
2
DIBAL H
Toluene, 78 C
H O



C H
O
3CH OH
 Marking scheme:
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.

-20-
2H O
Cl Si Cl HO Si OH    
CH3
CH3
CH3
CH
3
x
O Si O   
x
CH3
CH3
3 3H C Si O Si O Si CH     
 3 3
CH 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
2
Black
Cu H S CuS 2H 
  
dilHCl2
2
Black
Pb H S PbS 2H 
  
dilHCl2
2
Black
Hg H S HgS 2H 
  
dilHCl3
2 2 3
BlackorBrown
Bi H S Bi S 2H 
   
 dilHCl2
2 2 2Zn H S ZnS ZnCl soluble H S
    
3 2
2
Soluble Milkywhite
Fe 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
O
H Cl
O
H Cl
O
(i) (ii)
O
H Cl
O
O
H Cl
O
(iii) (iv)
O
O
sp3

-21-
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)

-22-
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
H
H
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)

-23-
 Marking scheme:
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 = 5.7o
C
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 56kJ
0.1
 
Enthalpy of ionization of acetic acid = 1 kJ.
Hence answer is (A)

-24-
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
0.2
1000


No. of moles of NaOH =
100 1
0.1
1000


Conc. of remaining acetic acid =
0.1
1000 0.5
200
 
Conc. of 3CH COONa =
0.1
1000 0.5
200
 
The solution is a buffer, hence
pH=
  5salt 0.5
pka log log 2 10 log
acid 0.5
  
         
pH=4.7
PARAGRAPH – 2
In the following reactions
C H8 6
Pd-BaSO4
H2
C H8 8
i B H2 6
H2
ii 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)

-25-
40. The major compound Y is
(A)
H3C
(B)
H3C
(C)
H3C
H2C
(D)
H3C
H3C
KEY: (D)
SOL:
H2O HgSO4 2 4, H SO
EtMgBr, H O2
C CH
C CH3
O
C CH3
OH
CH3
CH2
H , heat
+ C CH3
CH3
CH
(Y) H
ence Answer is (D)

-26-
PART III: MATHEMATICS
inclusive.
 Marking scheme:
41. The coefficient of x9 in the expansion of (1 + x)(1 + x2) (1 + x3) ….. (1 + x100) is
Ans. 8
Sol:      2 3 100
1 x 1 x 1 x ........... 1 x   
       
116 3 5 6 7 9
1 x 1 x 1 x 1 x 1 x 1 x 1 x

      
        116 3 5 6 7 9
1 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 9
1 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 9
1 x 1 x x ....... 1 x x x x x 2x          
= 8
42. Suppose that the foci of the ellipse
2 2
x y
1
9 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 2
22
1
1
m
m
 
 
 
is
Ans. 4
Sol:
2 2
x y
1
9 5
 
2
2
2
b
e 1
a
 
5
1
9
 
=
4
9
2
e
3


-27-
     
     
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 1
1 1
a 2
y m x y m x
m m
    
T1 passes trough (–4, 0)
1
1
2
0 4m
m
  
2
1 1
1
1 1
2m m
m 2
  
T2 is 2 2
2 2
a 4
y m x y m
m m

    
Passes trough (2, 0)  0 = 2
2
4
2m
m
  2
2m 2
Now, 2
22
1
1
m 2 2 4
m
 
    
 
43. Let m and n be two positive integers greater than 1. If
 n
cos
m0
e e e
lim
2


       
   
 
Then the value of
m
n
is
Ans. (2)
Sol
 n
cos
m0
e e
lim




=
 
   
 
n
cos 1
n
n m0
e e 1
lim cos 1
cos 1 .
 


 
  
=
 n
n
2cos 1
n m0
2sine e 1
2lim
cos 1
 




  
=
 
 
n
2n
cos 1
2n
n mn0
sine e 1 e2lim
4. 2cos 1
2
 

 
    
  
   
 
 
When 2n = m 
m
2
n


-28-
44. If  1
1 2
9x 3tan x
2
0
12 9x
e dx
1 x

  
   
 

Where tan–1 x takes only principal values, then the value of e
3
log |1 | is
4
 
   
 
Ans. 9
Sol:
1
1
9x 3tan x
2
0
3
e 9 dx
1 x

  
   
 

Let 1
9x 3tan x t
 
2
3
9 dx dt
1 x
 
   
 
3
9
34
9
t 4
0
e dt e 1




   
3
9
4
1 e


  
 e
3
log 1 9
4

   
 e
3
log 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 1
lim ,
G x 14
 then the value of
1
f
2
 
 
 
is
Ans. (7)
Sol.
 
 x 1
F x 1
Lim
G x 14


 
 
    x 1
F' x 1
Lim F 1 G 1 0
G ' x 14
  

 
  x 1
f x 1
Lim
14x f f x


 
   x 1
f x 1
Lim
14x f f x


 
  
f 1 1
14f f 1


1
f 7
2
 
 
 
 
1
f 1
2
 
 
 

-29-
*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 isin
7 7
    
   
   
, where i 1  . The value of the expression
12
k 1 k
k 1
3
4k 1 4k 2
k 1
| |
| |


 

  
  


is
Ans. (4)
Sol. k =
2k 2k
cos isin
7 7
    
   
   

12
k 1 k
k 1
3
4k 1 4k 2
k 1
| | 12 2sin
14 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,
 
 
7
2a 6d
62
11 112a 10d
2



7a 21d
6
a 5d



 a = 9d
130 < T7 < 140
130 < a + 6d < 140
130 < 15d < 140
130 140
d
15 15
   d = 9 (since terms of A. P. are natural numbers)

-30-
 This section contains TEN questions
 Marking scheme:
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
1
xf x dx
12

 (B)  
/4
0
f x dx 0


(C)  
/4
0
1
xf x dx
6

 (D)  
/4
0
f x dx 1


Key: (A), (B)
Sol. f(x) = 7 tan8
x + 7 tan6
x – 3 tan4
x – 3 tan2
x
       6 2 2 2 6 2 2
f 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
/4
6 2 2 7 3 7 3
0
0 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)

-31-
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)
 
 
4
t 6 4
0
t 6 4
0
e sin at cos at dt
L?
e sin at cos at dt







(A)
4
e 1
a 2,L
e 1



 

(B)
4
e 1
a 2,L
e 1



 

(C)
4
e 1
a 4,L
e 1



 

(D)
4
e 1
a 4,L
e 1



 

Ans. (A, C)
Sol:  
4
t 4
1
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

    
 
24
t
0
sin 2at 1 cos2at
e 1 2 dt
4 2
   
    
   

     
4 4 4
2t t 2 t 4
0 0 0
1 1 11
e dt e sin 2at dt e sin 2at 1 cos2at dt e 1
2 8 16
  

       
 t 4
2
0
I e sin at cos at dt


 
t
0
11
e dt
16

 
  
 
 (as same way as above)
 11
e 1
16

 

4
1
2
I e 1
I e 1





Whether a = 2 or 4

-32-
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) 12
1 1
d 1
1 for x 1
dx 3x
  
l
(B)
 1
1
1
2
1
xdm
for x 1
dx 3 x 1
 

(C) 12
1 1
d 1
1 for x 1
dx 3x
  
l
(D) 1
1
dm 1
for 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 ,0
x
 
 
 
Now normal  1
1 1
1
y
y y x x
x
   
Passes through  2N x ,0
Hence  1
1 2 1
1
y
0 y x x
x
   
 2 1x 2x
For centroid
21 1
11 1
1
x 2x
x 1x y
, m
3 3 3
 

  l
1
2 2
1 1 1 1 1
xd 1 dm 1 dm
1 , ,
dx 3x dy 3 dx 3 x 1
   

l

-33-
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  
22
x 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 2
PQ = PR =
3
. If e1 and e2 are the eccentricities of E1 and E2, respectively, then the correct
expression (s) is (are)
(A) 2 2
1 2
43
e e
40
 
(B) 1 2
7
e e
2 10

(C) 2 2
1 2
5
| e e |
8
 
(D) 1 2
3
e e
4

Key. (A, B)
Sol.  1, 2P  ;
2 2
,
3 3
a b
Q
 
  
 
;
2 2
,
3 3
A B
R
 
  
 
Also 2 2 2 2
9a b A B   
2 2
3
PQ 
2 8
9
PQ 

2 22 2
9 8
1 2
3 3 9
a a   
      
   
or    
2 22 2
3 3 8   a a
 
22
3 4 a
2
3 2a   
2 2 2
5,1 5, 1a a A   
2
2
1 2
9 5 1
1 1
5 5
 
     
 
b
e
a
2
2
2 2
1 7
1 1
8 8
 
     
 
A
e
B
2 2
1 2
1 7 8 35 43
5 8 40 40

    e e
2 2
1 2
7
40
e e
1 2
7
2 10
e e
2 2
1 2
1 7 27
5 8 40
   e e

-34-
54. If 1 6
3sin
11
  
   
 
and 1 4
3cos
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 06
3sin 90
11
 
 
1 04
3cos 180
9
 
 
0
270  
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
2
ax 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
, 0
5
 
 
 
(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
1
0 4 1

  
21 1
5 4
 
1 1 1 1
, ,
2 25 5

   
      
   
56. Let  
3
4
192x
f x
2 sin x
 
 
for all x  with
1
f 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 ,M
4 2
 
(C) m = – 11, M = 0 (D) m = 1, M = 12
Key (D)
Sol  
 
3
4
192x
f x
2 sin x
 
 

-35-
1
x 1
2
 
  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 48
2
 
     
 
 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 1/2
1/2
4x 4x f x dx 48x 48x         
 
1
1/ 2
1 f x dx 12 
m 1,M 12 
 Marking scheme: +4 If only the bubble(s) corresponding to all the correct option(s) is(are)
darkened.
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 

-36-
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 1
2 12x F x xF x dx      
     
3
1
9 3 1 2 12F F f x dx   
   
3 3
1 1
36 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     
    
33
2 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
1
2 1
1 1 3
2 2
n
n n
P
nn
n n n n

 

 
 
 
3
1
3 4 3
1 2
1
3
n
n
n n n
n n

 

   3 1 2 1 3 42   n n n n n n
 1 2 3 1 42n n n n n 

-37-
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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