JEE-2015 : Advanced Paper – 2 Answers and Explanations

CODE #
JEE-2015 : Advanced
Paper – 2
Answers and Explanations
Physics
Chemistry
Mathematics
1
6
11
A
21
2
31
C
41
8
51
D
2
3
12
B,D
22
6
32
A
42
4
52
A,D
3
2
13
A,C
23
6
33
B
43
2
53
B,C,D
4
2
14
D
24
3
34
A
44
9
54
A,B
5
1
15
A,B
25
9
35
B,C
45
7
55
A,B,D
6
2
16
B,C
26
8
36
C,D
46
9
56
A,C
7
7
17
A,D
27
4
37
C
47
4
57
A,B
8
4
18
A,C
28
4
38
D
48
9
58
C,D
9
D
19
A,C
29
B
39
A
49
B,C
59
A,B,C
10
A,B,C
20
D
30
B,C,D
40
B
50
B
60
C,D
PART - I : PHYSICS
1. 6
dI =
2. 3
The phase diagram for amplitudes will be
2
dm r 2
3
A
dm = 4πr 2 ⋅ ρ dr
for sphere A
60°
R
A
6
5
8 πk 5
r dr = 8πk ⋅ R = 4πk R
IA = ∫
3R
3R 6
9
0
R
IB =
8πk
∫ 3R5 r
0
IB 3 6
= =
IA 5 10
n=6
9
A
120°
A
10
5
dr = 8πk ⋅ R = = 4π k R
15
3R5 10
Ares = J3 A
∴ Ires = 3 I0
3. 2
A = A0 e–t/τ
A
dA
= R = 0 e− t / τ
τ
dt
Both sources have same A0
Rρ
at (t = τ) R
Q
... 1 ...
=
τQ e−2τ / τ Rρ 2
⋅
=
τρ e−2τ / 2τ ; RQ e
4. 2
sini
=n,
sinr1
∴ sinr1 =
6. 2
sini
n
λ=
h
p
nh
2π
mv rn =
sini
 dr 
cosr1  1  = − 2
dn
n


At second surface
nh
P = mv = 2πr
n
sinr2 1
=
sin θ n
h 2πrn 2πn2r0
=
=
p
n
nz
∴ sin θ = nsinr2
λ = 2nπ
dθ
 dr 
= sinr2 + ncosr2  2 
∴ cos θ
dn
 dθ 
here
dθ
1 
 dr2  
∴ dn = cos θ sinr2 + ncosr2  dn  



r0
z
3.h 

mvr = 2π 


n=3
z=3
r2 = A - r1
λ=
dθ
1 
 −dr1  
∴ dn = cos θ sinr2 + ncosr2  dn  



7. 7
6πr0
,P=2
z
For tension to be zero,
=
ncosr2 sini 
1 
sinr2 +

cos θ 
cosr1n2 
GMm  1 1  2Gm2
 − =
l2  9 T6 
l2
=
cosr2 sini 
1 
sinr2 +

cos θ 
cosr1n 
m=
8. 4
3 
 1
= 2 +
=2
 2 2 × 3 
2
E(t) = A2e–αt
log E(t) = 2 log A – αt
100 ×
1
A
5. 1
7M
,k=7
288
∆E
 ∆A 
= 2
 100 + α∆t × 100
E
 ∆ 
∆t
× 100 = 1.5%
T
8
2
6
10
12
∴ ∆t =
B
4
1.5 × 5
100
Solving we get
C
9. D
CA =
There is a wheatstone bridge between A & B whose
equivalent is 2Ω.
There is again a wheat stone bridge between A & C
whose equivalent is 4.5
D=
∆E
× 100 = 4%
E
2 εA /1
d/2 C = 4 εA /2
A
d/2
2
4
=
6.5
=1A
2 + 4.5
2
CA =
... 2 ...
2 εA /2
=C
d
CA = 2C, CB = 4C, CC = C
∴ equivalent capacitance
Option C:
4C
 4×2 
7C
=C+ 
C = C + 3 =
4
2
+
3


∴
C2 7
∴ C =3
1
P2 V2 P1V1
=
T2
T1
P2 × 3V1 P1V1
=
4T1
T1
4
P1 .
3
i.e. work done by gas is stored in potential energy
of spring = ½kx2. Plus the work done v against
atmospheric pressure = P1x
∴ work done by gas = P1xA + ½kx2
V2 = V1 + Ax = 3V1
∴ {Ax = 2V1}
P2A = P1A + kx
∴ P2 =
10. A, B, C
T 1 ,P 1 ,V 1
P1
1–n
T 2 ,P 2 ,V 2
Ideal monoatomic
P2A = P1A + kx
P1V1 = µRT1
P2V2 = µRT2
∴
4
AP1 = P1A + kx
3
⇒
1
P1A = kx
3
work done by gas =
=
1
(kx)x + P1xA
2
P1V1 P2 V2
=
T1
T2
=
1 1

P1A  x + P1xA
2  3

3 

∴ P2 = P1 
2 

=
an equilibrium P2A = P1A + kx →
7
7
7
P1Ax = P1 2V1 = P1V1
6
6
3
Change in intend energy = ncr(T2 – T1)
= ncr3T1
P1V1 = nRT1
3
P,
2
A = P1A + kx
P1V1 9R 9
×
= P1V1
R
2
2
∆v = ∆Q – ∆W
1
P1A = kx
2
further V1 + Ax = V2 = 2V1
∴ Ax = V1
∴ nT1 =
∴
Energy stored =
∴ ∆Q =
1
1 1
 V 
(kx) × x = ×  P1A  ×  1 
2
2 2
 A
P1V1
4
Change in intend energy = ncr(T2 – T1)
= ncr2T1
But, P1V1 = nRT1
1 2
kx + P1xA
2
=
9
7
21 + 14
P1V1 + P1V1 =
(P1V1 )
2
3
6
35
(P1V1 )
6
=
11. A
P1V1
3
× 2 × R = 3P1V1
R
2
∴ option (B) is correct.
∴ (nT1)×2cr =
... 3 ...
94
U → 140
54 Xe + 38 Sr + x + y
Q = kxe + ksr + kx + ky = ksr + kxe + 4
= 8 : 5 (234) – 236(7.5) = 1989 – 1770 = 219
∴ kxe + ksr = 219 – 4 = 215 mev
∴ Mass number = 236 – (140 + 94) = 2(X ∪ Y)
Atomic number = 92 – (54 + 38) = 0 (x ∪ Y)
Only option is (A).
236
92
12. B, D
dp 4πr2 = r (4π r2dr) ρ
ρ at depth r will be directly proportional to r.
∴ dp = – krdr.
Integration
n1
σ1
P1
L1
0
∫ dp = –k rdr ]
R
r
P
L2
P = k (R2 – r2)
Where k is some cosnt.
solving we get,
option (B) & (C) are correct
σ2
P2
n2
Since the rope is in tension force on P will be
upward, force on Q will be downward.
∴ they will more in opposite direction if revosed.
17. A, D
ev dB =
∴ VP .VQ < 0 {D}
Since the 2 spheres are in equilibrium,
∴ T = Net force due to wt & liquid
= 6πnrV
vd =
ev
w
I
IB
ne wd ⇒ v = ne d
if B, n & I remain same
1 v 2 = d1
;
d v1 d2
for d1 = 2d2
v2 = 2v1
for d1 = d2 v1 = v2
n1 | Vp |
∴ n | V | =1
2
q
v∝
| Vp | n2
∴ | V | = n {B}
q
1
18. A, C
if I & w remain same
13. A, C
By dimensional analysis.
Only option A & C are correct.
v∝
Pa
14. D Electric field inside cority = 3ε
0
B
n
v 2 B2 n1
=
×
v1 B1 n2
Only option is (D).
if
15. A, B
(A) (B) (from stress strain graph)
Tenside strength is the stress at which metal breaks.
B1 = 32 & n1 = 2n2
v2 = 2v1
B = 2B2 & n1 = n2
v2 = 0.5 v1
if
(A)
(C)
19. A, C
16. B, C
The situation is similar to that of earth where
pressure keeps on increasing as we go down.
c
i
p
for S1
p+d p
sin c =
Let us there a shell at r distance, the pressure on 2
sides is (p + dp) & P. the difference in pressure dp
can be found by dpA = net attractive force on shell
dpA = drx (ρ)
... 4 ...
2
5
for S1 in water & S2 in
cos c =
16
3 15
,
1
5
K HX 10 −4
=
= 10 −3
K HY 10 −1
45 2
×
4
5
4
sin im =
3
KHX = 10–3KHY
in KHX – log KHY = – 3
(– log KHX) – (– log KHY) = 3
pKHX – pKHY = 3
9
sin im =
16
&
16
sin im =
3 15
sin im =
8
15
×
5
8
9
16
92
U238 
→ 82Pb 206 + x ( 2 He 4 ) + y ( −1 eo )
92 = 82 + 2x – y
2x – y = 10
238 = 206 + 4x + 0
4x = 32
x=8
y=6
(A)
for S1 in air and S2 in
1 × sin im =
25. 9
4
15
3
(for S1)
4
Pf × V 9RT
=
Pi × V 1RT
&
Pf
=9
Pi
4
8
15
sin im = ×
15
5
8
→
26. 8 5 [Fe(H2 O)2 (C2 O 4 )2 ] + 3MnO 4− + 24H+ 
2−
3
sin im =
4
(C)
5Fe3 + + 10H2 O + 10 CO2 + 3Mn2 +
n
Rate H+
24
= H =
=8
−
Rate MnO 4 nMnO
3
20. D if NA1 < NA2
we will take smaller numerical question.
PART - II : CHEMISTRY
+
−
4
+
H

→
heat
27. 4
21. 2
HO
CH3
+
CH3
CH3
CH3
22. 6
1, 2 methyl

→
shift
23. 6 3B 2H6 + 6CH3 OH → 6(CH3 O)3 B + 6H2
CH3
24. 3 ∧HX = x
C = 0.01 M
α1 =
+
∧HY = 10x
C = 0.1 M
∧HX
0
∧HX
α2 =
aq.dil.

→
kMnO (excess)
4
HO
CH3
OH
OH CH3
28. 4
α1 = 0.1 α2
KHX = Cα12 = 0.01 α12
KHX = 0.01(0.1 α2)2
KHY = 0.1 α 22
KHX = 10–4 α 2
KHY = 10–1 α 22
2
CH3
OH
∧HY
0
∧HY
α1
X
1
=
=
α 2 10x 10
CH3
CH3
29. B
30. B, C, D
As oxygen chemically participates with metal and
metal donates electrons to antibonding π-orbitrals
of oxygen.
... 5 ...
31. C
35. B, C
OH
CH3
→
32. A
O
CH3
C –H
Cl
Cl
H C lO 2
(ii)
H C lO 3
(iii)
O
Cl
O
H C lO 4
(iv)
For questions no. 37 and 38.
O
C
C –H
NH

→
CH
OH
OH
36. C, D

→
CH2
CH3
O
H
C = O + H 2N H
CH3
O
HO
(i) O3
(ii) Zn, H2 O
O
O
CH
(i)
CH
CH2
Pd − BaSO
→
H
4
2
C H 2O H
H
CH3
N –H
CH
+
N

→
2
C
CH3
−
(i) B H , (ii) H O ,OH

→
6
2
2
CH
COCH3
CH3
CH
(ii)
2+
+
Hg ,H

→
+
CH2 = CH − CH3 , H

→
High Pr essure, ∆
33. B
CH3 HC
CH3
CH3
C
C
CH3
+
(i) C2H5 MgBr,H2 O (ii) H , heat

→
OOH
(M ajor)
O radical initiator

→
2
37. C
+
NH2
34. A
N = NCl
NaNO / HCl

→
0° C
38. D
–
+
2
39. A
40. B
O–
N=N
OH

→
CH3 COOH + NaOH 
→ CH3 COONa + H2 O
Initial
0.2
0.1
0
Final
0.1
0
0.1
0.1
−5
pH = − log(2 × 10 ) + log
0.1
= – [0.3010 – 5]
= 4.6990
... 6 ...
PART - III : MATHEMATICS
lim
F (x )
x →1 G
41. 8
42. 4
Coeff of ×9 in (1 + x) (1 + x2) L (1 + x100)
= Coeff of x9 in (1 + x) (1 + x2) L (1 + x9)
f (1)
f ( f ( x ))
1
x2 y2
+
= 1 , e = F1 (2, 0), F2 (–2, 0)
2
9
5
F1 (2, 0) 2F2 (–4, 0)
P1 : y2 = 8x
P2 : y2 = –16x
Tangent to P1 (2t2, 4t) : ty = x +2t2 , it passes
through (–4, 0)
0 = –4 + 2t2 ⇒ t2 = 2
m1 =
46. 9
47. 4
Tangent to P2 at (–4t2, 8t) ty = – x + 4 + 2,
it passes through (2, 0)
f ( f ( x ))
1
14
S = 4i$ + 3j$ + 5k
αk +1 − αk = const (r )
k =1
3
∑ α4k −1 − α4k −2
=
12r
=4
3r
k =1
43. 2
( ) −e0
cos an
48. 9
0
 
m
a →0
a
( ).(− sinan ).(nan−1 )
−e
=
cos an
e
mam−1
a→0
44. 9
(
)
−1
1
3  ax + tan x

α = ∫9 +
.e
dx
2

+

1
x
0
9x + tan−1 x
=e
1
0
=e
loge (1 + α ) = 9 +
F (x ) =
x
∫
−1
9+
3π
4
d=9
−1
49. B, C
h(x) = f(x) – 3g (x)
h(–1) = f(–1) – 3g (–1) = 3
h (0) = 6 – 3 = 3
h(2) = 3
h(–1) = h(0) = h(2)
∴ h’(x) = 0 has one root in (–1, 0) & one root in
(0, 2)
3π
4
f ( t ) dt ⇒ F (1) =
1
∫ f ( x ) dt = 0
−1
x
1
−1
−1
7
[2a + 6d] 6
2
=
11
11
2a + 10d]
[
2
⇒ 7 (2a + 6d) = 6 (2a + 10d)
14a + 41d = 12a + 60d
2a = 18d
29 = 18d
⇒ a = 9d
T7 = a + 6d = 15d
130 < 15d < 140
2
m = 2n
G (x ) =
x →1 x.
=
12
m22 = 2
45. 7
(x )
f (x )
1
 1
⇒ f  = 7
14
2
∑ αk +1 − αk
1
0=–2+4+2 ⇒ t =
2
= lim
x →1 G'
= lim
→
2
lim
=
F' ( x )
x=–4+3+5=4
y=4–3+5=6
z=–4–3+5=–2
1
1
⇒
= t2 = 2
2
t
m1
e
(x )
= lim
∫ t f (f (t )) df ⇒ G (1) = ∫ x. f (f (t )) dt = 0
... 7 ...
{h'' ( x ) ≠ 0}
50. B
53. B, C, D
f(x) = (7tan6x – 3 tan2x) sec2x
π
4
π
4
0
0
54. A, B
∫ f ( x ) dx = ∫ (7 tan
6
)
x − 3 tan2 x sec 2 xdx
55. A, B, D
56. A, C
1
Let tan x = t =
∫ (7t
6
0
)
1
− 3t 2 dt =  t7 − t3  = 0

0
π
4
π
1
(
57. A, B
)
7
3
∫ xf ( x ) dx =  x f ( x ) dx  02 − ∫ t − t dt
0
0
1
x1 − x2 < 1
( x1 − x2 )2 − 4x1x2
α2
<1
n3 
1  n1
+


2  n1 + n2 n3 + n4 
58. C, D
n1
n −1
n2
n1
1
+
=
. 1
.
n1 + n2 n1 + n2 n1 + n2 n1 + n2 − 1 3
−4 <1
n1
1
=
n1 + n2 3
<5
1
< α2
5
α2 <
II
1
 II 
=
P  =
 R  1  n1  1  n3  3


+ 
2  n1 + m  2  n3 + n4 
52. A, D
1
I
1  n3 


2  n3 + n4 
51. D
α2
n3 R
n1 B
P (R ) =
 t8 t 4 
1 1 1
=0− −  = − + =
8 4 8
 8 4 0
1
n1 R
n2 B
−1
5
or α >
1
59. A, B, C
f(x) = x.F(x)
f(2) = 2. F(2)
5
1 
F’(x) < 0 for x ∈  ,3 
2 
therefore F(2) < F(1)
⇒ f(2) < 0
f’(x) = F(x) + xF’(x) < 0 for x ∈ (1, 3)
f’(1) = F(1) + F’(1) < 0
1
1
1
1
or
∴− < α < −
<α<
2
2
5
5
1
1

2
2
1 − 4α > 0 4α − 1 < 0 − 2 < α < 2 


60. C, D
... 8 ...