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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 8 24 B,C 34 6 44 A,D 54 6

5 B,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 20 2 30 B,C,D 40 3 50 C,D 60 4

Physics Chemistry M athematics

6

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Part I – Physics

1. 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. B, D1

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

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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, CValues 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. 8

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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

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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

... 8 ...

at x = 0, 1, h(x) will not be differentiable

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