... 1 ...
JEE Advanced Paper 1 – 2014Answers & Explanations
1 C,D 11 3 21 A 31 4 41 B,C 51 3
2 A,B,D 12 4 22 B,D 32 8 42 A,B,C 52 2
3 A,C 13 5 23 A,B,D 33 2 43 A,B,C 53 8
4 B,D 14 4 24 B,C 34 6 44 A,D 54 6
5 A,D 15 2 25 A,B,C 35 1 45 B,D 55 3
6 C,D 16 5 26 A 36 7 46 A,C 56 5
7 A,B,C 17 3 27 A,B,D 37 7 47 C,D 57 4
8 D 18 5 28 A,B,C 38 5 48 A,B 58 2
9 C 19 8 29 A,C,D 39 4 49 C 59 7
10 A,C,D 20 2 30 B,C,D 40 3 50 C,D 60 4
Physics Chemistry M athematics
6
... 2 ...
Part I – Physics1. C, D 2 cannot be zero.
if 1 = 0,taking torque about contact with ground
N1 l sin = mg l cos
2
N1 tan = mg2 (D)
if 1 20, 0
N1 = f2 N1 = 2 N2 mg
N1
N2
f2
f1
f1 + N2 = mgf1 = 1N1 1 2 N2 + N2 = mg
N2 = 1 2
mg1
2. A, B, DTaking potential across R2 to be zero.Current in both R1 and R3 must be the same
Hence, 1 2
1 3
V VR R
3. A, C For air to glass
1.4 1 0.4V R
…(1)
1
1.5 1.4 0.1f V R
…(2)
Adding (1) and (2)
1
1.5 0.5f R
f1 = 3R
For glass to air
1.4 1.5 0.1V R
…(3)
2
1 1.4 0.4f V R
…(4)
Adding (3) and (4)
2
1 1.5 0.5f R
f2 = 2R
4. B, D 1 2LRd
1
2 2RLR4(2d)
In series 2 2
S s11
V V tt t t 2sRR 22
In parallel 2 2
p p11
V V tt t t 0.5sRR 88
5. A, D 1
2
E 1E
1 1
2 2
Q C KQ C 2
0 0
01
K A 2 AC 2 K3d 3d
K AC K3d
6. C,D I = I0 cos t
q = 0II dt sin t
30max
Iq 2 10 C
at t = 76 I is negative hence anticlockwise
and 30Iq 1 10 C2
and when A is connected to D
–+
50 50
0100
upper plate of C is negativeVoltage across resistance = 100V
100I 10A10
Now the battery has to supply double the charge tore-charge the capacitor. Q = 2 × 10–3 C
... 3 ...
7. A, B, C
Dd
2 > 1 2 > 1
m = d
m1 > m2let mth minima of 1 overlaps with nth maxima of 2.
12
(2m 1) n2
2m 1 3n 1
hence 3rd maxima of A2 coincides with 5th minima ofA1.
8. D h 4h 1.4m4
V = fl = 1.4 × 244
V=RTM
VA = 7 640 448 m / s
10
VB = 6 590 354 m / s
10
VC = 9 590 331.87 m / s
16
VD = 17 640 340 m / s32
9. C 20 0 00 0
Q2 r 24 r
20 0Q 2 r & r
Let E1(r0) = E2(r0) = E3(r0) = E0
01 0
rE 4E2
02 0
rE 2E2
(C) is correct
02 0
rE 2E2
03. 0
rE E2
(D) is incorrect.
10. A, C, DValues of allowed,
4L 4L 4L 4L4L, , , , etc3 3 5 7
Comparing k = 2 ,
the values are satisfied by A and Cnow = 2f
Vf
allowed values of
50 250,3 3
11. 3x y z1 3 3 1M L T ML MT T
on comparing corresponding powers y = 3
12. 4 By Law of conservation of angular momentum,
2mvr = 12 MR2
m = 0.05 kgV = 9 m/sr = 0.25 mM = 0.45 kgR = 0.5 m = 4 rad/s
13. 5 Ig(G + S) = VIg = 0.006S = 4990V = 30 Von solvingG = 10 for ammeter
g
g
ISG I I
G = 10 ohmIg = 0.006I = 1.5 Aon solvingS = 5
14. 4
... 4 ...
15. 2 Ui = 100 J, Ub = 200 JWaf = 200 J W ib = 50 J Wbf = 100 JQiaf = 500 JQib + Qbf = W ib + Wbf + Uf – UiQib = W ib + Ub – Ui= 50 + Ub – 100Qib + Qbf = 150 + Uf – 100 = 500Uf = 400Qib + Qbf = 450Qib = 50 + 200 – 100 = 150Qbf = 300
bf
ib
Q 300 2Q 150
16. 5 AG ˆ ˆV 150i 50 3 j
BGV V 3ˆ ˆV i j2 2
ABV V 3ˆ ˆV 150 i 50 3 j2 2
AB AGV .V 0
V = 200 m/s
AB ˆ ˆV 50i 50 3j
ABV = 100 m/s
t = 500 5s100
17. 3 X0 = 3X1
X1 = 0X3
X2 = 02X3
01
1
iB2 X
,
02
2
iB2 X
B1 = B, 2BB2
r1 = mu
Bq2
r2 = mu3Bq2
1
2
r 3r 1
18. 5 WF + Wmg = KF – Ki18 × 5 + Ui – Uf = Kf – KiKf = 50Jn = 5
19. 8 Urel = 0.5 m/sarel = 0
t = rel
S 4 8sU 0.5
20. 2 3 × 0.5 ×0.52 =
1.5 0.5 0.52
= 2 rad/s2
= t = 2 rad/s
Part II – Chemistry21. A
(A)
NH2
NH2
O
C
CCH3
OC
OCH3
O
+
NH
NH2
O
+ CH COOH3
C CH3
O22. B, D
(D) 3 3 2 4H BO H O B(OH) H
23. A, B, DBalanced reaction is:
3 2 4 4 2 26I ClO 6H SO Cl 6HSO 3I 3H O
24. B, C(B) KO2 is paramagnetic(C) In this reaction, NO is formed which is
paramagnetic.
25. A, B, C
... 5 ...
26. A Salt bridge is an easy path for flow of ions however itdo not involves in a chemical reaction.
27. A, B, DDue to H-bonding it attains open cage like structurewith lower density.
28. A, B, CBoth steric and electronic factors affects the reactivityof compound.
29. A, C, D1st one is trivial name of alcohols and second one istheir IUPAC name.
30. B, C, D
31. 4P K
Na or R = K × Na
R = 6.023×1023 × 1.380×10–23
R = 6.023×1.380 = 8.31174
32. 8 M = 3.2 = 3.2 moles of solute
1000 ml.solution
m = 3.2 81000 0.4
1000
33. 2 i = 1 – + ni = 1 – 0.5 + 3×0.5 = 2
34. 6 There can be six electrons having givenconfiguration.
35. 1
36. 7 PbS, CuS, HgS, NiS, CoS, Ag2S, SnS2Bi2S3 – yellow colourMnS – Pink (buff) colour
37. 7
38. 5
39. 4 XeF4, 4BrF , 2 23 4Cu(NH ) , PtCl
40. 3
Part III – Mathematics41. B,C x2 + y2 + 2gx + 2fy + c = 0
1 + 2f + c = 0 (i)
(0,1)
(1,0)
Circle is orthogonal tox2 + y2 – 1 = 0 and x2 + y2 – 2x – 15 = 0 c – 1 = 0, c = 12[g(–1)] = c – 15– 2g = 1 – 15g = 71 + 2f + 1 = 0, f = – 1Centre (–g, –f) (–7, 1)
22 2 2r g f c 7 1 1 7
42. A,B,C
x . y y . z z . x 2 2 cos 13
a / / x y z
1a k y – z
2b k z – x
2 22y z y z 2y . z 2 2 – 2 2
y z 2
a
a y z2
b
b z x2
... 6 ...
(A) b
b . z z x z x . z z x2
b b
2 1 z x z x b2 2
(B) a
a . y y z y z . y y z2
a
y z a2
(C) a b a b
a . y b . z .2 2 2
a b
a . b y z . z x2
a b a b
1 1 2 12 2
43. A, B, Cf(x) = [log (sec x + tan x)]3f(x) = [log(secx – tanx)]3
=
3sec x tanx sec x tanx
logsec x tax
3
31log log sec x tan xsec tan x
= – [log(secx + tanx)]3 = –f(x)odd functionf’(x) = 3 [log(sec x + tan x)]2
2sec x tan x sec xsec x tan x
= 3 [log(secx + tanx)]2 sec x > 0function is one-one & onto [Range will be R]
44. A, D
45. B, Df’(x) = 5x4 – 5= 5 (x + 1) (x – 1) (x2 + 1)
f’(x)+ –
1–1
++
For three roots, f(–1) f(1) < 0(– 1 + 5 + a) (1 – 5 + a) < 0–4 < a < 4For one root f(–1) f(1) > 0a < – 4 or a > 4
46. A, C
0, x a
g x f x , a x b0 x b
New f(x) 1
g a 0
g a f a 1
and g b f b 1
g b 0
47. C, D
1x tt
1x
dtf x et
11x tt
x
1 dtf ex t
Let 1x uu
21x
1 dut , e u.4 u
1x uu
1x
dueu
1x tt
1x
dtet
= –f(x)
1f x f 0x
x 12 t
x t
x2
dtf 2 et
x xf 2 f 2
1f x f 0x
Odd function
... 7 ...
1 1x x
' x x2
1 1f x e . e .xx x
1xx2 e 0
x
48. A, B
49. C
50. C, D
51. 3
0
2 3 4
10 xy10
y = f(x) = cos–1(cos x)
10 xy10
total point = 3
52. 2
1 x1 x
x 1
ax sin x 1 a 1limx sin x 1 1 4
1 x
x 1
sin x 1 a x 1 1limsin x 1 x 1 4
Let x – 1 =
1 1
0
sin a 1limsin 4
1 1
0
sin a 1lim sin 41
21 a 12 4
1 a 12 2
a = 0, 2
53. 8 (y – x5)2 = x (1 + x2)2
25 4 2 2dy2 y x 5x 1 x 2x 1 x 2x
dx
at (1, 3)
2dy2 3 1 5 1 1 2 2 2dx
dy 8dx
54. 6
x + y = 0
2 2
2 22
2x – y = 0
(h,k)
h k h k2 42 2
2 2 2h 4 2
2 h 2 2
Area = 2 22 2 2 6
55. 3
–1
0
g(x)=x + 12
f(x) = |x|+1
at x = 0, 1, h(x) will not be differentiable
... 8 ...
56. 5 n2n C n
n n 12n
2
4 = n – 1, n = 5
57. 41a . b b . c c . a2
a b b c pa qb rc
b. a b b c b pa qb rc
p rq 02 2
p + r = – 2q (1)
a. a b b c a. pa qb rc
q rp a b c2 2
q r 3p2 2 4
(2)
Similarly p q 3r2 2 4 (3)
4p + 2q + 2r = 32p + 2q + 4r = 3
3p r2
3q2
2 2 2
2p 2q r 4
q
58. 2 1 43 2
0
d4x 5 1 x 2x dxdx
1 43 2
0
d40 x x 1 x dxdx
1 14 43 2 2 2
0 0
40 x x 1 x 40 3x .x x x dx
1 43 2
0
120 x 1 x dx
2 43 2
0
120 sin cos cos d
Let x = sin
23 9
0
120 sin cos d
2.8.6.4.212012.10.8.6.4.2
= 2
59.7 n1 + n2 + n3 + n4 + n5 = 20
1 2 3 4 5n n n n n1 2 3 4 101 2 3 5 91 2 3 6 81 2 4 5 81 2 4 6 71 3 4 5 72 3 4 5 6
60. 4 a, ar, ar2
2a ar ar ar 23
ar2 – 2ar + a = 6
2 6r 2r 1 0a
62 4 4 1ar
2
61a
r is integer a = 6r = 2
2 2a a 14 6 6 14 4a 1 6 1