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 − = 0.04 128
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 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
