... 1 ...

JEE Main 2015Answers & Explanations

1 4 16 1 31 3 46 4 61 4 76 2

2 4 17 4 32 2 47 1 62 3 77 4

3 3 18 1 33 3 48 2 63 4 78 4

4 4 19 1 34 2 49 4 64 1 79 3

5 4 20 1 35 1 50 2 65 3 80 1

6 2 21 3 36 3 51 2 66 3 81 2

7 4 22 1 37 1 52 2 67 1 82 3

8 1 23 3 38 1 53 3 68 1 83 4

9 4 24 1 39 2 54 3 69 1 84 2

10 2 25 1 40 3 55 2 70 2 85 4

11 3 26 4 41 4 56 1 71 2 86 3

12 4 27 1 42 2 57 1 72 3 87 2

13 1 28 3 43 1 58 1 73 3 88 4

14 4 29 3 44 1 59 2 74 4 89 2

15 4 30 1 45 1 60 75 4 90 3

Physics Mathematics Chemistry

... 2 ...

PART A – PHYSICS

1. 4 Centre of mass of a solid cone is 3h4

from

the vertex. (factual)

2. 2 2

PowerI

4 r=

π

2o

(rms)

I 1E C

2 2= ∈

rmsE 2E=

3. 3Mg

' 1AY

= +

T 2g

= π … (1)

m'

T 2g

= π … (2)

2m2

T '

T=

2m2

T Mg1

AYT= +

2mT A

Y 1T Mg

= −

4. 4 P.E = 21kx

2

( )2 21K.E k A x

2= −

5. 4 % change in frequency

s s

s

V VV V V V

100V

V V

−− +

= ×

−

s

s

2V100

V V= ×

+

2 20100 12%

320 20×= × =

+

6. 2 dV

neAvR

=

RA

= ρ

d

VneV

ρ =

51.6 10 m−= × Ω

7. 4

Fm g

θ

T sin Fθ =Tcos mgθ =

20I

tan2 dmg

µθ =

π

d 2Lsin , m= θ = λ

0

gLI 2sin

cosπλ= θ

µ θ

... 3 ...

8. 1

2

loop 1

9

loop 2

I – i 1

i 1

3

1

For loop 1

15i i 9+ = … (1)

For loop 2

13i 4i 6− = … (2)

On solving (1) and (2)

3i 0.13 from Q to P

23−= =

9. 41.22 x

D 25 cmλθ = =

1.22 500nm 25 cmx

0.25 2 cm× ×∴ =

×

30 m= µ

10. Here i0 = 0.1 AUpon closing K2

t /0i i e− τ=

here L

0.2 msR

τ = =

at t = 1 ms1/ 0.2

0i i e−=

50i i e−=

0.1i mA 0.67 mA

150= =

11. 3 This system is equivalent to dampedoscillation of spring & mass.

mK

m - corresponds to L- of electric circuitK - corresponds to C- of electric circuitIf m is very massive, the spring will keep onmoving for long time & the system will oscillatefor longer time.Similarly, If L is greater, system will dampslowly.∴ Option (3).

12. 4 2EC

V3 C

=+

∴ The appropriate graph is (4).

13. 1 For cube of max. volume,

3 2R=

Moment of inertia of cube

2cM

6=

3c

3

M 2MM

4 3R3

= × =ππ

o

24MRM I

9 3=

π

14.g 2 T

gg T

∆ ∆ ∆= +

31 10 m−∆ = ×

220 10 m−= ×

1T 0.01 s

100∆ = =

90T 0.9

100= =

3

2

g 0.01 10100 2 100 3%

g 0.9 20 10

−

−

∆× = × + × = ×

... 4 ...

15. 4 The light is moving horizontally the refractiveindex will not change and it will continue tomoving straight line.For example in Mirage formation,

Ray 1

Ray 2

Light rays which moves horizontally keep onmoving horizontally whereas one that goesdownward, keep on deviating.

16. 1 Carrier wave frequency = 2000 KHzSignal frequency = 5 KHz.Frequency of resultant wave

= 1 2 1 2( ), , ( )ν + ν ν ν − ν∴ 2005, 2000, 1995 KHz.

17. 4 Entropy like internal energy is a state function.It only depends upon initial and final state &not on the process or path.∴ The change in entropy in the two conditionswill be same.Only option is (4).But actually none of the options are truebecause temp is given in celsius & Not Kelvin.

∴ Ans should be n473

373

.

18. 11 U U

P or PV3 V 3

= = But for ideal gas, PV = nRT = VT4

V = Volume 34

V r3

∴ = π

3 44PV r T nRT

3= π =

1rT const., thus, T

r ∴ = ∝

19. 1 Initially the move with const. velocity relativeto each other, & later one of the mass comesto rest while the other is still moving.

21

1y 10t gt

2= −

22

1y 40t gt

2= −

2 1y y 30t− = when both were moving.

When only 2nd body is moving

22

1y 40t gt

2= −

1y 240 m= −

22 1

1y y 40t gt 240

2∴ − = − +

Which is a parabolic.

20. 1 Variation of potential in a solid sphere is

V = Vsphere

2 2

3

3R r

2R

−

Inside the surface

surface1

V Vr

= × Outside the surface

Where Vsurface = 0

Q4πε

at centre, r = 0, V = 03V

2RSimilarly we can find R2, R3 & R4.Solving we get R2 < (R4 – R3)

21.

... 5 ...

For light to jjust come out r2 must be less

cθ .

1c

1sin−

θ = µ

But 1 2r r A+ =

1 2r A r= −

12

1r sin−

< µ

11

1r A sin−

> − µ

But 1

sinsinr

θ = µ

( )11sin sinr−θ = µ

1 1 1sin sin A sin− −

∴θ > µ − µ

22. 1 Bι = µ × µ is along the direction of right

hand thumb.

In (b) || B→ µ∴ Stable.

In (d) → µ Antiparallel to B .∴ Unstable.

23. 3I 1

I 2

The magnetic field of Inner solenoid is onlyinside it.∴ Exerts no force on solenoid outside.From Newton’s 3rd low, outer solenoid fieldwill also exert No force on solenoid inside.

24. 1 f 1 2P P P= +

2 2f 1 2 fP P P 2 2 mV 3mV= + = =

f2 2

V V3

∴ =

( )

22 2 2

2 2i

1 1 1 2 2m(2V) (2m)V (3m) V

2 2 2 3K1 1K m 2V (2m)V2 2

+ − ∆ ∴ =

+

= 56%

25. 1λτ =ν λ = mean free path.

ν = Average velocity

2

1

2 ndλ =

πn = no. of molecules per unit volume

22 Nd

νλ =π

Tν ∝

V

T∴ τ ∝

But in adiabatic expansion1TV Kγ− =

12V V

γ− ×τ ∝

r 12V+

τ ∝

26. 4

... 6 ...

total smallV V V= +

total smallV V V∴ = −

22

3

GM R 3GMV 3R

R22R 2 82

= − − + × ×

Solving, we get V = GMR

−.

27. 1

20

20

20

20 100

y

f 120N∴ =

28. 3 Electric line of lence originate from +ve charge& end at –ve charge.They will be continious & will not form kinks.∴ Only option is (3).

29. 3 21K mV

2=

2keU

r−=

T = –KAs particle Jumps to ground state

V KE↑ ∴ ↑

r U , T↓ ∴ ↓ ↓

30. 1 Option (1).

PART B – MATHEMATICS

31. 31

(a b) c | b || c | a3

× × =

1(a.c)b – (b.c)a | b || c | a

3=

a.c 0=

and 1

| b || c | –b.c3

=

– | b || c | cos= θ

1cos

3θ = −

2 2sin

3∴ θ =

32. 2Q

O

(x1x )2P 3

Let P(h1k) and (x1, y1)

1 1x yh and k

4 4= =

21x = 8y1

(4h)2 = 8(4k)

∴ h2 = 2kx2 = 2y

33. 3 E

h

x yB C z30° 45° 60°

A D

... 7 ...

htan60

z° =

hz

3=

htan45

y z° =

+

h hy h ( 3 1)

3 3= − = −

htan30

x y z° =

+ +

x h h 3+ =

x h( 3 1)= −

AB x 3BC y 1

= =

34. 2

x + y = 41

(1, 39)

(1, 1)(39,1)

(39,0) (40,0) (41,0)(1,0)(0,0)

(0,41)

Total points = 1 + 2 + ...39

39 40780

2×= =

35. 1 2x – 5y + z – 3 + λ (x + y + 4z – 5) = 0

(2 )x ( – 5)y (4 1)z (3 5 ) 0+ λ + λ + λ + − + λ =

2 5 4 1k

1 3 6+ λ λ − λ += = =

3( 2) 5λ + = λ −

2 11λ = −

112

λ = −

5 4 1 3 5x y z 0

2 2 2λ − λ + + λ∴ + + − =

+ λ + λ + λ

x + 3y + 6z

113 5( )

2 011

22

+ −− =

−

x + 3y + 6z – 7 = 0

36. 3 Total elements in A × B = 2 × 4 = 8

8C C C0 1 8

8 8 ...... 8 2+ + =

∴ 8C C C C C C3 4 8 0 1 2

8 8 ...... 8 2 (8 8 8 )+ + = − + +

= 28 – (1 + 8 + 28)= 256 – 37= 219

37. 1 2x – 3y + 4 = 0Point of intersection (1, 2)x – 2y + 3 = 0

∴ radius = 2 2(2 1) (3 2) 2− + − =

38. 1x 0

(1 cos2x)(3 cos x)lim

x tan4x→

− +

= 2

x 0 2

2sin x(3 cos x)lim

tan4x4x

4x→

+

= 1

4 22

× =

39. 2x 2 y 1 z 2

3 4 12− + −= = = λ

(3λ + 2, 4λ – 1, 12λ + 2)3λ + 2 – (4λ – 1) + 12λ + 2 = 1611λ = 11, λ = 1∴ (5, 3 , 14)

∴ Distance = 2 2 2(5 1) 3 (14 2) 13− + + − =

... 8 ...

40. 3

50 2C C C0 1 2

(1 2 x) 50 2.50 x 2 .50 x............)− = − +

350 = 2C C C0 1 2

50 2.50 2 .50 ...............+ + +

1 = 2C C C0 1 2

50 2.50 2 .50 ...............− + +

50

2 50C C C0 2 50

3 150 2 .50 .............. 2 .50

2+ = + + +

41. 43 3 3

n1 2 ...... n

t1 3 5 .....+ + +=+ + +

2

2

n(n 1)2

n

+ =

21(n 1)

4= +

tn 21

(n 1)4

= +

S9 2 2 2 21

(2 3 .....10 )4

= + +

1 10 11 211

4 6× × = −

= 96

42. 2y = 2x2

12

,1)(

y = 4x – 1

18

12

( , )

(4x – 1)2 = 2x16x2 – 8x + 1 – 2x = 016x2 – 10x + 1 = 016x2 – 8x – 2x + 1 = 0(8x – 1)(2x – 1) = 0

1 1x ,

2 8=

at 1 1

x , y 4 1 12 2

= = × − =

at 1 1 1

x , y 4 18 8 2

= = × − = −

1 2

1/ 2

y 1 ydy

4 2−

+ − ∫

12 3

1/ 2

y y y 98 4 6 32

−

= + − =

43. 1

44. 11 2

1 2

z 2z1

2 z z

−=

−

|z1 – 2z2|2 = |2 – z1

z |2

(z1 – 2z2) 1 2(z 2z )− = 2z1 2z ) 1 2(2 z z )−

2 21 2 1 2 1 2z 4 z 2z z 2z z+ − −

2 21 2 1 2 1 24 z z 2z z 2z z= + − −

2 2 2 21 2 1 2z 4 z 4 z z+ = +

2 21 2 1z 4 z (| z 4) 0− − − =

2 21 2( z 4)(1 | z | ) 0− − =

|z1|2 – 4 = 0

|z1| = 2

45. 1

x2 + y2 – 4x – 6y – 12 = 0x2 + y2 + 6x + 18y + 26 = 0A(2, 3) r1 = 8

... 9 ...

B(–3, –9) r2 = 8

AB = r1 + r2

46. 4 Case I: Four digit number3 × 4 × 3 × 2 = 72

3 4 3 2

6,7,8Case II: Five digit number

5 120=∴ Total = 72 + 120 = 192

47. 1dy

(x logx) y 2x logxdx

+ =

dy y2

dx xlogx+ =

e1

elog (logx)x logxdx

eI.F e log x=∫

= =

∴ y.logex e2 log xdx= ∫

ylogex = 2xlogex – 2x + cy(e) = 2e – 2e + 2 = 2at x = 1c = 2

48. 2

49. 4 2m = l + nl, G1, G2, G3, n G.P.

14n

rl

=

4 4 41 2 3G 2G G+ + = (lr)4 + 2(lr2)4 + (lr3)4

= nl3 + 2n2l2 + ln3

= nl[l+n]2

= nl(2m)2

=4lm2n

50. 2

51. 23

5 44

dx

1x (1 )

x+

∫

Let 4

4

11 t

x+ =

35

4dx 4t dt

x− =

3

3

t dt

t= −∫= –t + c

14

4

1(1 ) C

x= − + +

144

4

x 1–( ) C

x

+= +

52. 2

(1, 1)

x – y = 0

x + 3y = 0

(x + 3y)(x – y) = 0

53. 3 tan–1 = tan–1x + tan–12

2x

1 x−= tan–1x + 2tan–1x= 3tan–1x

31

2

3x xtan

1 3x− −=

−

54. 3 g(3–) = g(3+)2k = 3m + 2g’(3–1) = g’(3+)

km

4=

... 10 ...

k = 4m8m = 3m + 2

2m

5=

k + m = 4m + m = 5m = 2

55. 2

56. 1

4 2

2 22

logxI dx

logx log(6 x)=

+ −∫ =

4 2

2 22

log(6 x)dx

log(6 x) logx

−− +∫

4

2

2I dx 2= =∫

I 1=

57. 110 10 8 8

10 89 9

9

a 2a 2( )2a 2( )

− α β − α β=α β

8 2 8 8

9 9

( 2) ( 2)

2( )

α α − − β β −=α β

8 8

9 9

(6 ) (6 )

2( )

α α − β β=α − β

= 3

58. 1 2x 0

f(x)lim 2

x→=

f(x) = ax4 + bx3 + cx2

c 2∴ =f '(x) = 4ax3 + 3bx2 + 4x

f '(1) f '(2) 0= =∴ 4a + 3b + 4 = 032a + 12b + 8 = 08a + 3b + 2 = 0

1a

2= +

4 + 3b + 2 = 0b = –2

4 3 21f(2) (2 ) – 2(2 ) 2.2 0

2= + =

59. 2 Q

OR P

(ae, b2

a)

S

Equation of PQ

2

2 2

byxae a 1

a b+ =

x y1

a ae

+ =

Area of quadrilateral = 4 × Area of OPQ∆

14 OP OQ

2= × × ×

a2 a

e= ×

2a2

e=

92 27

23

= × =

5 2e 1

9 3= − =

60.

... 11 ...

PART C – CHEMISTRY

61. 4

CH 3

CH 3

(i) O 3

(ii) Zn /H 2OCH – C – CH – CH – CH – CHO3 2 2

O

O

62. 3 Vitamin C is water soluble.

63. 4 BeSO4 has low lattice enthalpy than hydrationenthalpy hence it is fairly soluble in water.

64. 1N a N O /H C l2

0 -5 °C

N H 2

C H 3

N C l2

C H 3

+ –

C u C N /K C N

∆

C N

C H 3

+ N 2

65. 3 In b.c.c. lattice,

4r 3a=

3 3 1.732r a 4.29A 4.29A

4 4 4= = × ° = × °

1.857A 1.86A= ° ≈ °

66. 3 Colour of Zn2[Fe(CN)6] is white while all othersare of yellow colour.

67. 1 Energy of 2nd excited state of hydrogen is

2

13.6

2

−

3.4 eV/atom= −

68. 1

69. 1 Ionic radii increases with increase in negativecharge.So correct answer is:N3– > 02– > F–

70. 2 Na3AlF6 acts as solvent for Aluminium Oxide.

71. 2 K M N O 4

C H 3 C O O H C O C l

S O C l2

C H O

H /P d2

B aSO 4

72. 3 Higher order (>3) reactions are rare becauseof low probability of simultaneous collisoin ofall the reacting species.

73. 3 H

C H6 5

C = C

H

C H 3

shows geometrical isomerism as cis- andtrans-forms

74. 4

75. 4

76. 2 2(C8H7SO3Na) + Ca2+ → (C8H7SO3)2Ca +2Na+

2 × 206 g of resin is reacted with = 1 mole ofCa+2

1 ..................................... = 1

2 206×

= 1

gm/mole412

77. 4 2e– + Cu2+ Cu→By passing 2F charge, 1 mole of copper isdeposited.

78. 4

C l Py

NH 3 NH OH2

Pt

+

,

C l Py

NH OH 2 NH 3

P t

+

,

C l NH OH2

NH 3 Py

Pt

+

... 12 ...

79. 3 Percentage of Bromine

80 Mass of AgBrformed100

188 Massof subs tance= × ×

3

3

80 141 10 4512100 24%

188 188250 10

−

−×= × × = =×

80. 1 It is probably due to charge transfer fromhigher orbital of Mn–O to lower Manganeseorbital. It is a type of Ligand to Metal chargetransfer.

81. 2 Swarts reaction is used for the formation ofalkyl flourides.

82. 3 Milliequivalent of CH3COOH = 50 × 0.06 = 3Left out m.eq. of CH3COOH = 50 × 0.042= 2.1Adsorbed m.eq. of CH3COOH = 3 – 2.1= 0.9Mass of CH3COOH (Adsorbed)= (0.9 × 60) mg. = 54 mg.

Adsorbed Per gram of charcoal = 543

= 18 mg.

83. 4s

s

p p np N

° −=

1.2185 183 m

10018358

− =

2 1.2 58183 m 100

= ×

m = 63.684

84. 2 Interhalogen compounds are more reactivethan halogens itself.

85. 4 G –2.303RTlogK∆ ° =2494.2 = – 2.303 × 8.314 × 300 logK

2494.2logK – –0.434

2.303 8.314 300= =

× ×K = Antilog(–0.434) = 0.368

Now 2

1 12 2Q 112

×= =

As Q K,> Reverse reaction takes place.

86. 3 Nitrogen and oxygen react together at hightemperature to form oxides of nitrogen but inatmosphere, oxides are not formed.

87. 2 Xe has highest boiling point due to strongforces as compared to other noble gases.

88. 4 Glyptal is used in paints and lacquers.

89. 2 ∆G° = – 2.303 RT log Kp= – RT ln Kp

– RT ln Kp = [2 × 0NO2

G∆ ] – [1 × 0 + 2 ×

86.6 × 1000]

x = pR 298 ln K

866002

×− +

x = 86600 12R 298 ln 1.6 10

2× × ×−

x = 0.5[2 × 86600 – R × 298 ln 1.6 × 1012]

90. 3 H2O2 can act as both oxidizing as well asreducing agent.