1. No category

CODE #
JEE-2015 : Advanced
Paper – 1
Answers and Explanations
Physics
Chemistry
1
2
11
A,C
21
8
31
A
41
8
51
A,D,C
2
7
12
B,C
22
4
32
B,D
42
5
52
C,D
3
2
13
A,C
23
4
33
A
43
2
53
B,C
4
3
14
B,D
24
3
34
D
44
0
54
B,D
5
3
15
D
25
4
35
A
45
4
55
A,B
6
7
16
C
26
1
36
C
46
3
56
A,D
7
6
17
B
27
2
37
A,B,C
47
8
57
A,C
8
2
18
A,B,C
28
9
38
B,C,D
48
4
58
B,C
9
A,B,D
19
See Sol.
29
A,C,D
39
See Sol.
49
A,D
59
See Sol.
10
B
20
See Sol.
30
C
40
See Sol.
50
A,B,C
60
See Sol.
PART - I : PHYSICS
1. 2
⇒
g
4
⇒h=R
If g =
v=
GM
R
(
)
3
m v 22 − v12 = mg(3)
4
⇒ v 22 − v12 = 40
v 22 = 40 + 9 = 49
v2 = 7 m/s.
1
GMm GMm
− mv 2 −
=
2
R
2R
2. 7
Mathematics
3. 2
P ∝ R2 T4
PA = 104 PB
∴ vesc = v 2.
R A .TA = 10 4 RB TB4
For a disc rolling without slipping
R 
TA
= 10 ×  B 
TB
 Ra 
2
4
2
1/ 2
2
1
1  MR2   v 
mv 2 + 
 ⋅  = K ⋅E
2
2  2   R 
1/ 2
TA
 1 
= 10 × 

TB
 400 
3
mv 2 = K ⋅ E
4
Now,
⇒
=
1
2
Now according to wien’s law
λ A TB
=
=2
λB TA
3
3
mv12 + mg(30) = mv 22 + mg(27)
4
4
... 1 ...
4. 3
5. 3
Fuel available ∝ power
initial power = 8 × Required power
∴ after 3 half lives, the power output will become ≤
power required
∴n=3
M2 14
=
= 7.
M1
2
7. 6
30° 30°
Ö3a/2
For wave traveling in water the equivalent distance
is µ times the geometrical
∴ For maxima
a/2
The flux passing through the plane is 1/6th of the
total flux emitted by the wire.
λL
∴ φ = 6∈
0
n = 6.
µ d2 + x 2 − d2 + x 2 = mλ
d2 + x2 (µ − 1) = mλ
∴ d2 + x2 = 9 m2 λ2
x2 = 32 m2 λ2 – d2
Hence, P = 3
6. 7
8. 2
For air as medium Mirror
hc
−φ =E
λ
φ=
1 1 1
+ =
v u f
f = 10, u = 15
v = 30 cm
|m1| = 2
For lens
1242
− 10.4 = 3.4
90
13.6
= 3.4
n2
n = 2.
9. A, B, D
One mole of H2 + one mole of helium.
at temp T
1 1 1
− =
v u f
f = +10 u = –20
m2 = 1
Total magnification
m1m2 = M1 = 2
For medium as µ = 7/6
No change for mirror
but for the lens
U=
1× C v1 × T + 1× C v 2 × T 3 / 2RT + 5 / 2RT
=
2
2
8 RT
= 2RT
4
(A) is correct
=
1 3×6 
=
− 1 (x )
f ′  2 × 7  g
… (i)
1 3 
=
− 1 (x )
f  2  g
… (ii)
(B) → In gas mixture v =
C
P1 + CP2
γ RT
M
5 + 7 12 4 × 3 3
=
=
=
= .
v= C
C
v1 + v 2 3 + 5 8 4 × 2 2
From (i) and (ii)
35
cm
2
for lens
u = –20 cm
f′ =
M=
+35
cm
2
∴ v = 140 cm
f=
140
=7
20
M2 = 2 × 7 = 14
| m |=
... 2 ...
1× 2 + 1× 4
=3
1+ 1
vmix =
3 RT
2×3
vHe =
5RT
3×4
1 cm
vmix
3 RT × 3 × 4
6
=
=
vHe
2 × 3 × 5RT
5
(B) is correct
(C) vrms =
8RT
8RT
8RT
, vHe =
, vH2 =
µM
µ4
µ24
vHe
8RT × π × 2
=
=
vH
µ × 4 × 8RT
2
each division of main scale =
5 division of vc = 4div of main scale
1
2
1
 1
5 × (distance of vc) = 4 ×   cm = cm
2
8
∴ (D) is correct
10. B R1 =
=
=
 1

(distance of vc) =  cm  .
 10

∴ L.C = 0.125 – 0.10 cm = 0.025 cm
Option A: pitch = 2 × 0.025 = 0.050 cm
2.7 × 10−8 × 50 × 10−3
(72 − 4) × 10−6
2.7 × 50 × 10−8 × 10−3
45 × 10−6
=
2.7 × 50 × 10−5
45
0.050cm
= 0.0005cm = 0.005 mm
100
∴ Option B is correct
least count of linear scale = 0.050 cm
distance of pitch = 0.100 cm
∴ LC =
27 × 5 × 10−5 27 × 5 × 10−5
=
= 3 × 10−5
9
9
R2 =
1
cm
8
1× 107 × 50 × 10−3
0.100
= 0.001cm = 0.01mm
100
∴ Option C is correct.
∴ L.C =
4 × 10−6
 250 
× 10−5
12.5 × 10–4 = 

 2 
∴ equvialant resistance
R1R2
(3 × 250 / 2) × 10−5 × 10−5
=
= R +R
(3 + 250 / 2) × 10−5
1
2
13. A, C
h = ML2T–1
c = LT–1
FL2
= [M−1L3 T −2 ]
M2
solving we get
G=
 750 
−5  7500 
−6 1875µΩ
=
 × 10 =  256  × 10 =
256
64




a=
11. A, C
hv = ev0 + φ
−1
1
1
,c=
,b=
2
2
2
1
1
−1
∴ M = kh 2 .c 2 .G 2
option (A) is correct.
Similarly if
L = kha.cb.Gc = [ML2T–1]a[LT–1]b[M–1L3T–2]c
a–c=0
2a + b + 3c = 1
a + b + 2c = 0
hc
= ev 0 + φ
λ
hc
−φ
λ
1
graph of vs v0 will be a st line (option C)
λ
v0 vs λ will be a decreasing curve option (A)
ev0 =
solving we get a =
12. B, C
1 cm main scale is divided in 8 equal divisions.
screw gauge 100 divisions on circular scale
5 division of vc = 4 division on main scale.
1
1
−3
−3
1
1
,c = ,b =
2
2
2
∴ L = kh 2 .c 2 .G 2
∴ Option C is correct.
... 3 ...
14. B, D
17. B
1
E1 = m(aω1 )2
2
b = maω1
1
E2 = m(Rω2 )2
2
a  1 
2
=
=n
b  mω1 
E1 a2  ω1 
=
 
E2 R2  ω2 
s2
d
for light rays to become parallel on the 2nd refracting
surface.
1.5 1 1.5 − 1
− =
∞ u
10
R = m(Rω2)
−10
= −20 cm
+0.5
∴ Image of P is 20 cm away from refrating surface
s 2.
1 1.5 1 − 1.5
=
Image P will be formed at +
−10
v 50
∴ u=
1
= 1 , mω = 1
2
mω2
a (mω2 )
=
= n2
b (mω1 )
  ω1 
 =ω 
  2
 E1 E2 
=


 ω1 ω2 
∴
1
1 1.5
=
−
v 20 50
∴
u 50 − 30
=
v
1000
1000
= 50
20
∴ d = 50 + 20 = 70 cm
∴ v=
15. D, C
Angular momentum will remain conserved.
Initial angular momentum = MR2w
18. A, B, C
Magnitude of force is proportional to the length of
the wire between the 2 ends in a uniform magnetic
field.
∴ (A) is correct, (B) is correct, (C) is correct
2


M 3  M
8
MR2 w =  MR2 + ×  R  + × x 2  w ×


8
5
8
9





9
x2  8
R2 =  R2 +
R2 +  ×
8 × 25
8  9

∴
– 20
S oc
2
 n2 × 1   1
E1 a2m2 ω2
= 2 2 2 =  4 = 2
E2 R m ω2  n   n
1 – Su
19.
9R2
9
x2
R2 =
− R2 −
8
8 × 25
8
1
9  2 x2
1
−

R =
8  25 
8
20.
(A)
A
B
C
D
→
→
→
→
R, T
P, T, S
Q, T, R
P, Q, R
P, Q, R, T
U  x2 
U1 (x) = 0 1 − 2 
2  a 
16 2
R
∴ x =
25
2
4 

∴ x = 5 R 


2
 x 2  2x
dU1(x) −U0
=
× 2 × 1 − 2  × 2
dx
2
 a  a
16. C +q charge will more in SHM,
–q charge will more along direction of its
displacement because it will experience a net
attractive force towards one of the wire.
... 4 ...
U0 4x  x 2  2U0 x  x 2 
F = 2 × a2 1 − a2  = a2 1 − a2 




at x = 0,
x = a,
x = – a,
F=0
F=0
F=0
for x > 0, F > 0
∴ force will be away from equibrium
If x = – a, F = 0, U = 0
at x = 0, U =
U0 
U0
1
−1 +  = −
2 
3
3
x > – a,
F4 (x) = – ve,
x < – a,
F4 (x) = + ve
∴ Particle oscillate about x = – a,
where its potential energy.
=
U0
2
U0
4
it will never reach origin, oscillate about x = – a,
because at x = – a, it is at equilibrium & force is
restoring.
Since the energy of particle is
PART - II : CHEMISTRY
21. 8
=N –N
2
(B)
(C)
U0  x 
U 2x
,F2 (x) = – 0 × 2


2 a
2 a
∴ x = 0, F2 (x) = 0,
particle will oscillate about origin
Q (Q), & (S)
U2 (x) −
22. 4 Fe3 + + 6SCN− →[Fe(SCN)6 ]3 −
Fe3 + + 6CN− 
→[Fe(CN)6 ]3 −
Spin only magnetic moment =
PQRS
For [Fe(SCN)6 ]3 − , No. of unpaired electrons
2
2
 U 2x
U x2
2x 2 2 
F3  − 0 × 2 × e − x / a + 0 × 2 × + 2 e − x / a 
2
2
a
a
a


F3 = –
=
U0 − x2 / a2  x 2 
e
x 1 − 2 
2a2
 a 
F4 (x) =
U0
2
1(1 + 2) = 3 = 1.732
Difference = 6 – 1.73 ; 4.27 ; 4
23. 4 BeCl2, N2O, NO2+ , N3−
24. 3
25. 4 M+ 
→ M3 + + 2e−
∆G° = −nFE°
∆G° = −2 × 96500 × − 0.25 = 48250 J
= 48.25 kJ
193
=4
So, no. of moles of M+ oxidized =
48.25
 1 3x 2 
 − 3
 a 3a 
U0  x 2 
1 − 
2a  a2 
x = 0,
x = a,
x = – a,
35 ; 6
=
P, R, T
F4 (x) = −
5(5 + 2) =
For [Fe(CN)6 ]3 − , No. of unpaired electrons
x = 0,
F3 = 0
x = a,
F3 = 0
x = – a,
F3 = 0
Since Fα – x, ∴ particle experience as an attractive
force towards x = 0.
at x = – a, U ≠ 0,
∴ particle will not oscillate about x = – a.
(D)
n(n + 2) B.M.
26. 1 As
So,
F4 = 0
F4 = 0
F4 ≠ 0
Now,
∆Tf = T° – T
∆Tf = 0 – (– 0.0558) = 0.0558
m = 0.01
Kf = 1.86
∆Tf = iKfm
0.0558
=3
1.86 × 0.01
So formula is [Co(NH3)5Cl]Cl2
for x > 0, F(x) = –ve 

for x < 0, F(x) = –ve 
∴ force is not always towards x = 0.
i=
3
U0 
1  −a  
x
a,U
1
=
−
=
−
−
×


at
2 
3  a  
... 5 ...
38. B, C, D
In electrolysis process to obtain copper, impure
copper is made anode while pure copper is made
cathode using CuSO4 solution.
27. 2 Two chiral center
So stereoisomers = 21 = 2
28. 9
29. A, C, D
39.
Fe3 + + H2 O2 + OH− 
→ Fe 2 + + H+ + O 2 + H2 O
30. C As the reaction is exothermic. So increase of
temperature yields in less production of NH3.
31. A In CCP arrangement O2– = 4
So no. of Al3+ = 2
So no. of Mg2+ = 1
40.
→ T
B 
Malachite → Cu2CO3(OH)2
→ Q, R
C 
Bauxite → Al(OH)3 or Al2O3
→ R
D 
Calamine → ZnCO3
Argentite → Ag2S
→ R, T
A 
→ P, Q, S
C 
→ P, Q, S, T
D 
1
8
PART - III : MATHEMATICS
32. B, D
41. 8 Let x be Bernoulli RV, Probability to achieve J
heads in ntrials = P [x = j] – nCJ pj (1 – p)Aj
P [x ≥ 2] = 1 – P [x < 2]
O
33. A
n
CH3
O
⇒
CH3
CH3
O
O
n
 1
 1
= 1 – P [0] – P [1] = 1 −   − n   ≥ 0.96
2
2
(–)
HO
Siderite → FeCO3
→ P, Q, S
B 
2 1
So octahederal holes ocupied by Al = =
4 2
3+
So tetrahederal holes ocupied by Mg2+ =
→ P, Q, S
A 
1+ n
2n
f (n ) =
HT , ∆
→
≤ 0.04
1+ n
n
2
,f (1 + n ) =
2+n
2n+1
−n
2 + n 1+ n
−
=
<0
n+1
n
2
2
2n+1
f(n) in dereving function f ‘n’
⇒ f (1 + n ) − f (n ) =
CH3
→
34. D C H 2 = C – C H = C H 2 
Now f(7) =
CH3
C H 3 – C = C H – C H 2B r
Br– is a weak base, so it follows 1, 4 addition.
f (8 ) =
8
> 0.04
128
9
< 0.04
256
nmin = 8
35. A
36. C
37. A, B, C
Cr2+ is reducing agent and itself oxidises to Cr3+.
Mn3+ is an oxidizing agent and reduces itself to
Mn2+.
42. 5 n = 6 C5 × 5 5
... 6 ...
m = 6 C5 × 5 C4 4 × 5C1 × 5
m
=5
n
43.2
Equation of normal at (x1,y1) to parabola y2 = 4ax
−1
= 2
Normal at (1, – 2) to y2 = 4x
2
x–y–3=0
r=
1
−1
0
1
=
1 1 1
− =
2 4 4
4 I – 1 = 4×
1
−1= 0
4
∫ 0 dx + ∫ 0 +
 x2 
= 
 4 1
1+ 1
( −2)
0
=
3−2−3
y+2= −
∫ 2 + f ( x + 1) − dx
I=
2
y – 2 = − ( x − 1)
2
x+y–3=0
r=
( )
xf x 2
2
−y
y − y1 = 1 ( x – x1 )
2a
∴ at (1, 2) to y2 = 4x
( x − 1)
3+2−3
1+ 1
2
2
∫
x
dx +
2
0
∫0
2
45. 4
2 2 π
2
46. 3 F′ ( x ) = 2x × 2cos  x +  − 2cos x
6

π

= 4x × cos2  x 2 +  − 2cos2 x
6

= 2
r2 = 2
a
[x ], x ≤ 2
44. 0 f ( x ) = 
0 x > 2
2 2
x
2 2
0
∫ f ( x ) dx = 4a cos
 x2  , x2 ≤ 2

f x2 =  
 0 x 2 > 2
( )
π
2
 a + 6  − 2cos a + 2


∫ f ( x ) dx = 4a cos
0
π
2
 x + 6  − 2cos x + 2



π 
π  


f ( x ) = x 8cos  x 2 +  .  − sin  x 2 +   2x 
6 
6  



 0
x<− 2
 2
=  x  , − 2 ≤ x ≤ 2
 

x> 2
 0
π

−4cos x ( − sin ) + 4cos2  x 2 + 
6

2
f (0 ) = 4cos2
0, −1 ≤ x ≤

= 0 , 1 < x ≤ 2
0,
x> 2

[x + 1],x + 1 ≤ 2
f ( x + 1) = 
x +1> 2
 0,
0 − 1 ≤ x < 0

= 1 0 < x ≤ 1
0 x > 1

47.8
⇒
5
cos2 2x + sin4 x + cos4 x + sin6 x + cos6 x = 2
4
(
)
2
5
cos2 2x + sin2 x + cos2 x − 2sin2 x cos2 x
4
(
+ sin2 x + cos2 x
⇒
... 7 ...
 3
π
= 4×
=3
 2 
6


(
)
3
(
)
− 3 sin2 x cos2 x sin2 x + cos2 x = 2
)
5
1
3
1 − sin2 2x + 1 − sin2 2x + 1 − sin2 2x = 2
4
2
4
5
5 1 3
− sin2 2x  + +  = 0
4
4 2 4
g ( f ( x )) =
1
x ∈ [0, 2π]
2
total solutions are 8.
 π
π
1 π
1
π
    −π
sin  sin  .sin  sin x    ∈  .sin , .sin 
2
6
2
2
2
2
2





 
sin2 2x =
51. A, D, C
P
48. 4 Image of y = – 5 about x + y + 4 = 0 is, x = 1
y2 = 4x
c
at x = 1, y = ±2
distance between A & B = Distance between
(1, 2) & (1, –2) = 4
49.A,D
Q
R
a
2
2
 g ( x ) x > 0 then f ' ( x ) =  g' ( x ) x > 0


f (x ) =  0 x = 0

 −g ( x ) x < 0
 −g' ( x ) x < 0


a+b+c = 0 ⇒ a+b
 e x x > 0 then h' ( x ) =  e x x > 0
h (x ) = 

 e− x x < 0
 e− x x < 0
b . c = 24 ⇒ b c cos θ = 24 ⇒ θ = 30°
( )
( )
2
b + c + 2b.c = a
(
)
( )
( )
⇒ c =4 3
θ
120 °
 ef(x) f ( x ) x > 0 then h f x 1  f ' ( x ).e( fx )f ( x ) > 0
( ( )) = 
h ( f ( x )) = 
 −f ' x .e− f ( x )f x < 0
 e− f ( x ) f ( x ) x < 0
( )

 ( )
(
2
)
30°
30°
(
a×b + c ×a = a× b − c
50. A, B, C
(
π
π
π

f ( x ) = sin  sin  sin x   & g ( x ) = sin x
6
2
2



= 2 a = b sin (30° )
= 48 3
π
π
π
    −1 1 
f (g ( x )) = sin  sin  .sin  sin x    ∈  , 
2
   2 2 
2
6
(x )
= lim
52. C, D
53. B, C
C2 → C2 → C1
C3 → C3 → C1
f ' (x )
x →0 g'
)
= a × 2b + a  = 2 a × b
 −1 1 
⇒ f (x ) ∈  , 
 2 2
x →0 g
)
= a × b − −a − b 


π
π
 −π π 
π
  −π π 
sin x ∈  ,  ⇒ .sin  sin x  ∈  , 
2
2
2
6
2



  6 6
f (x )
= a
a . b = a b cos 150° = – 72
x
 g ex
1 
x > 0 then f (h ( x )) =  e x g' e x > 0

f (h ( x1)) = 
 −x
−x x < 0
x
g x
e g' e x < 0
h
b
(x )
π   x
  π
π
 π
cos   sin  sin x    . .cos  sin x  . cos x
π
  6
2
 2
6   2
=
=
π
6
cos x
2
... 8 ...
(1 + α )2 α (2 + 3d) 2α (2 + 4α )
(2 + α )2 α ( 4 + 3d) 2α ( 4 + 4α )
(3 + α )2 α (6 + 3d) 2α (6 + 4α )
56. A,D
(1 + α )
2
2
2 + 3α 2 + 4α
⇒ 2α 2 (2 + α )
2
P
4 + 3α 4 + 4α
(3 + α )2 6 + 3α
6 + 4α
O
C3 → C3 → C1
(1 + α )2
2 + 3α α
= 2α 2 (2 + α )
4 + 3α α
2
(3 + α )2 6 + 3α
2
t2 ,t
2
2
(Slope of OP) × (Slope of OQ) = – 1
t1t2 = – 4
Q
α
C2 → C2 –3 C1
Area of ∆OPQ = 3 2
(1 + α )
2
2
2 α
= 2α2 (2 + α )
4 α
2
(3 + α )
2
= 4α
(1 + α )2
11
(2 + α )
2 1
(3 + α )
2
 t4
  t4

 1 + t12   2 + t 22  = 72
4
 4




 t2
  t2

+ 1  2 + 1 = 72
 4

4


( t1t 2 )2 
3 1
2
  t2

+ 1  2 + 1 = 72
 4

4


2 ( 4 + 2α ) 2 0
= 4α

16 
 + 16 = 72
16 + 4  t12 +

t12 

[6 + 4α − 8 − 4α ]
t12 +
= −8 α 3
−8α3 = −648α
(
2
1
(t22 + 4)(t22 + 4) = 72
( t1t2 )2 + 4 ( t12 + t22 ) + 16 = 72
11
= 4α3 3 + 2α 1 0
3
1
 t2
( −4 )2 
R2 → R2 –R1
R3 → R3 –R1
(1 + α )
 t2 
 t2 
 1  + t12  2  + t 22 = 3 2
2
2
 
 
1
2
6 α
2
3
t1 ,t
1
2
16
t12
= 10
t14 − 10t12 + 16 = 0
)
8α α − 81 = 0
(t12 − 2)(t12 − 8) = 0
α = 0, 9, −9
t1 = ± 2, ± 2 2
54. B, D
(
)(
P 1, 2 , 4,2 2
55. A, B
... 9 ...
)
57. A, C
(1 + e ) y '+ ye
x
x
 −3ax 2 − 2 x < 1
(B) f ( x ) = 
 bx + a2 x ≥ 1
for continuity – 3a – 2 = b + a2
= 1, If
x dx
=e
(1+ ex )
∫ e 1+ ex
= eloge
= 1e x
 ex
y '+ 
 1 + ex

(
 −6ax x < 1
f ' (x ) = 
 b x ≥1
for differentiability b = – 6a
a2 – 6a = – 3a – 2
⇒ a2 – 3a + 2 = 0
⇒ a = 1, 2

1
y =

1 + ex

)
⇒ y. 1 + ex = x + k at x = 0, y = 2
⇒k=4
y=
x+4
1+ e
dy
=
dx
(
2ab
=4
a+b
⇒ ab = 2(a + b)
a + q = 10
a+b=5+q
ab = 2 (5 + q)
(D)
x
)
1 + ex − ( x + 4 ) ex
(1 + ex )
2
Let h(x) = 1 + ex – xex – 4ex
h’ (x) = ex – 4ex – ex – xex
= – (x + 4) ex < 0 for (–1, 0)
h (–1) = 1 + e–1 + e–1 – 4e–1
⇒ b=
⇒a=
A→Q
B → P, Q
C → P, Q, R, S, T
D → Q, T
(A) Projection of αi + β i on
3i + j
2
)=
5
15
,6 q =
,4
2
2
(a − q | = 2,5
58. B, C
(
10 + 2q
=5+q
a
a
⇒ a + 10 + 2q = 5a + aq
a2 + 10 + 2 (10 – a) = 5a + a(10 – a)
2a2 – 17a + 30 = 0
2
>0
e
h(0) = 1 + 1 – 4 < 0
then h (x) has a root in (–1, 0)
3i + j = (αi + βi ).
a+
2
= 1−
59.
2 (5 + q )
(1)
(2)
(3)
3α + β
= 3
2
60.
α = 2 + 3β
⇒ α = 2 + 3  2 3 − 3α 


= 2 + 6 – 3α ⇒ α = 2
... 10 ...
A → P, R, S
B→P
C → P, Q
D → S, T