· for e to be independent of r 2 2 2 2 oo q aa q 0a 4 r 2 a 11. (2) sol. resistance of the lamp =

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JEE-MAINS-2016 SOLUTIONS

PHYSICS 1. (2) Sol: Applying conservation of energy

2 2

2 2 2 2 2

1

1 2A 1 A 1m 9 m A m A

2 3 2 3 2

1

7AA

3

2. (4)

Sol: 1

,so1 1

is incorrect

Also, as >> 1. Therefore 2

21

1

3. (3)

Sol: Mean value of time 90 91 95 92

92s4

Also absolute errors are 92 90 2 92 91 92 95 92 92 0

mean absolute error = 2 1 3 0

1.54

s but since the least count of the clock is 1s so it

can’t count 1.5s. hence it will count 2 s as error. (92 2) s

4. (1) Sol: In OR gate, if all the inputs are low (zero), only. Then the output is low (zero). Hence the correct option is (1). 5. (2, 4)

Sol: L r p

Hence option 2&4 are incorrect. 6. (3)

Sol: Carrier wave

Audio signal


Modulated wave In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in

proportion to the amplitude of audio signal. 7. (3)

Sol: 2

o

hc 1w mv

2

…(1)

2

o 1

4 hc 1w mv

3 2

…(ii)

From (i) & (ii)

2 2

o o 1

4 1 1w mv w mv

3 2 2

2 2o1

w1 4 1mv mv

2 3 3 2

1

4v v

3 .

8. (2)

Sol: o oA

I IB

2.2

o

B

I cos45 cos45B 4

4

8

= oI 2.8

l

2

A

B

B

B 8 2

.

9. (2)

Sol: 1

v vf f

24

2

2 10. (3) Sol: Using gauss theorem for dotted Gaussian surface

r

2 2

ao

1 AE.4 r Q .4 r dr

r

rQ

ab

2 2

2 2

o o

Q 1E .2 A r a

4 r 4 r

= 2

2 2

o o o

Q A Aa

4 r 2r


For E to be independent of r

2

2 2 2

o o

Q Aa Q0 A

4 r 2 a

11. (2)

Sol. Resistance of the lamp = 80

810

L

~

(220 V, 50 Hz)

Lamp

Now VR = 80 V

Here 2 2

L RV V V 2 2

L220 V 80

VL = 10 420 V

Now VR = IR = 80 ... (1)

VL = IXL = 10 420 ... (2)

On dividing (1) and (2),

L

R 80

X 10 420

8 8

2 fL 420

On solving, L = 0.065 H.

12. (3) Sol. The equation of the process is y = mx + c

oo

o

PP V 3P

V

Now, PV = nRT

P = nRT

V

oo

o

PnRTV 3P

V V

2o

o

o

PnRT V 3P V

V

2o o

o

P 3PT V V

nRV nR

… (1)

Now for maximum T,

dT

0dV


o o

o

2P 3PdTV 0

dV nRV nR

o

3V V

2

Putting in (1), we get

2o oo o

o

P 3P9 3T V V

nRV 4 nR 2

= o o o oP V P V9 9

2 nR 4 nR

o omax

P V9T

4 nR

13. (2)

Sol. Total potential energy

U = mgh × 1000 = 10 × 9.8 × 1 × 1000

Since 20% efficiency rate, hence

3.8 × 107 × 20

m100

= 9.8 × 104

m = 4

6

9.8 10

3.8 2 10

= 12.89 × 10–3 kg

14. (1)

Sol. o

hPQ 2h 4m

sin 30

Given that work done against friction along track PQ and QR are equal. mg cos 30o 4 = mg x

4 3

x2

x = 2 3 2 1.732 = 3.46 3.5 m

Loss in PE = work done against friction along PQ and QR

mg 2 = mg 3

2 4 + mg (3.5)

On solving = 0.29

15. (1) Sol. For metals like Cu, resistance increases linearly with temperature.

T

R


For intrinsic semiconductors like Si, resistance decreases exponentially with temperature

increases.

T

R

16. (2)

Sol. yellow blue Radiowave X ray

As E = hc/

yellow blue radiowave X rayE E E E

17. (3) Sol. G = 100 Ig = 10-3 A The shunt resistance (S) required to convert the galvanometer to an ammeter of full scale

deflection current of 10 A is

S = 3

g

3

g

I G 0 100

I I 10 10

110 1

10 100

0.01

18. (2)

Sol. No. of atoms left after n half lives is 0

n

N

2 (where n is the number of half lives)

Number of atoms of A left = o

4

N

2 = oN

16

(4 half lives t = 80 min, t1/2 = 20 min, n = 80/20 = 4)

Number of atoms of B left = o 0

2

N

2 4

(Number of half lives = 2)

Hence ratio of decayed number of A and B will be

o oo

o oo

N 15NN

516 16N 3N 4

N4 4

19. (3) Sol. Theoretical 20. (1) Sol.

4 F

3 F

9 F

4 F 12 F

V1 V2

8 V Now V1 + V2 = 8V and 4V1 = 12V2 On solving, V1 = 6 V, V2 = 2 V


Hence Q1 = 24 C, Q2 = 18 c Total Q = 42 C

Hence, E = 2

o

1 Q

4 r

=

9 6

2

9 10 42 10

30

= 420 N/C

21. (2)

Sol: o

GMV

R

2

e

GMV

R

Increase e oV V

2 1GM

R

2

2 1gR

R h

2 1 gR

22. (4)

Sol: L.C. 0.5

0.0150

mm

error is negative error = 5 × 0.01 = 0.05 Reading = 0.5 + 25 × 0.01 + 0.05 = 0.5 + 0.25 + 0.05 = 0.80 mm 23. (4) Sol:

Initially C.M. will move straight, hence its distances from left rail will decrease.

Normal reaction at left contact N1 will increase. Horizontal component of left N1x will also increase and continue to increase. Hence the roller will tend to move towards left direction.

24. (2) Sol: Hysteresis loss for electromagnets and transformers are less because it can magnetize or

demagnetize easily

N 1 N y 1N y 2 N 2

N x 1 N x2

A

B

C

D


25. (2) Sol:

bqa

sinb nq

tan sinb a

Lq q

a b a

L

2ab a L

2 0a ab L

2 4b L

min 4b L

24a L a L

2

aL

26. (1)

Sol: xg

v xg

dx

xgdt

0

l dxg t

x

1/2

0

lx dx g t

1/22x g t

2

20 tg

2 2 t 27. (4)

Sol: nPV constant

Then processC for the above polytropic process is known

1

process V

RC C C

n


1

V

RC C

n

1V

Rn

C C

Finally after solving

V

V

C C Rn

C C

[as P VC C R ]

P

V

C Cn

C C

28. (2) Sol: In case of telescopes, magnifying power = 20 means that the object appears 20 times nearer. 29. (4) Sol:

35°r 1

r2

e

A

35i

40

79e i = 35o

e = 79o

o40

o4 i e A 74

with above set of data, as refractive index increases emerging ray will have tendency to suffer TIR.

For i = 35o

o o

o1 1

sin 35 sin 90

sin r sin 74 r

Using reversibility principle For i = 79o

o o

o2 2

sin 79 sin 90

sin r sin 74 r

From 2nd case o

2r 37

1.63

Hence Answer is (4)

30. (3)

Sol: 1 1

2 2

T lT

T l


1 1

1 124 3600

2 2

TT T

T

112 12 3600T .........(i)

24 12 3600T ........(ii)

1

2

03

20

T T

T T

3 60 40T T

25T C

51.85 10 / C

PART : B : MATHEMATICS

31. (4)

Sol. 2 2 2 2y 2x, x y 4x x 2x x 0, 2

Required area = 2

2

0

4x x 2x dx

= 2

2

0

4 x 2 2 x dx

=

2

2 1 3/2

0

x 2 x 2 2 24x x 2sin x

2 2 3

= 8

0 0 0 23 2

= 8

3

32. (1)

Sol. 1

f x 2f 3xx

… (1)

Replace 1

xx

1 3

f 2f xx x

1 6

2f 4f xx x

… (2)

6

3f x 3xx

From (1) and (2)

2

f x xx


2

f x xx

2 2

f x f x x xx x

242x x 2 x 2

x

33. (4) Sol.

12 9

35 3

2x 5xdx

x x 1

= 3 6

3

2 5

2 5dx

x x

1 11

x x

Let 2 5

1 11 t

x x

3 6

2 5dx dt

x x

3 2

dt 1I C

t 2t

= 2

2 5

1C

1 12 1

x x

=

10

25 3

xC

2 x x 1

34. (4)

Sol. f x log 2 sin x

g x f f x

g x log 2 sin log 2 sin x

g x log 2 sin log 2 sin x in neighbourhoodof x 0

g ' 0 cos log 2 sin 0 . cos0

= cos log 2

35. (2) Sol. Circles touch externally and required circle touches x-axis. Centre (h, k) and radius = | k |

1 2 1 2c c r r

2 2 2

h 4 k 4 6 | k |

2 2 2h 2.4h 16 k 8k 16 36 k 12| k |

2h 8h 8k 12 | k | 4

2x 8x 8y 12 y 4

It represents two-semi parabola 36. (3)

Sol. 2x 4x 60

2x 5x 5 1


ba 1 , when a is 1 b R

a is not zero b = 0 a is -1 b even integer

2x 5x 5 1

x 1, 4

2x 4x 60 0 ; x 10,6 ; a 0

2x 5x 5 1 ,

2x 4x 60 even integer

x 2, 3 at x = 2, 2x 4x 60 48 ( even integer )

at x = 3, 9 12 60 odd

solutions are x 1, 4, 10, 6, 2

sum = 3 37. (4) Sol. Given terms of A.P. are

2T a d ; 5T a 4d ; 9T a 8d

If 2 5 9T , T & T are in G.P.

2

a 4d a d a 8d

2 2 2a 16d 8ad a ad 8ad 8d

28d ad d 8d a 0 8d a d 0

Now common ratio is a 4d 12d 4

a d 9d 3

38. (1)

Sol. Given 22b

8 & 2b aea

2 2 24b a e ; 2 2 2 2 2

4a e 1 a e e3

39. (??) Sol. Remark : Here total number of terms will be 2n + 1 which cannot be equal to 28. 40. (1) Sol.

p q p q p q p q p q q p q q p q p q

T T F F F F T T T

T F F T T F T T T

F T T F F T T T T

F F T T F F F F F

So p q q p q is equivalent to p q


41. (4)

Sol. y = 1

x xcos sin

2 2tanx x

cos sin2 2

y = 1 x

tan tan4 2

y = x

4 2

dy

dx =

1

2

Hence, slope of normal –2 Equation of normal becomes

y4 12

= –2 x6

The point 2

0,3

satisfying the equation.

42. (4)

Sol. Let, y =

1/n

2n

(n 1)(n 2)....3n

n

log y = 1 1 2 2n

log 1 1 .... 1n n n n

log y = 2n

r 1

1 rlog 1

n n

nlim log y

= 2n

nr 1

1 rlim log 1

n n

nlim log y

= 2

0log(1 x) dx = 3log 3 2

y = 3log3 2e

= 2

27

e

43. (4)


AC = 50 25 = 75

44. (2)

P(E1) = 6

36 =

1

6

P(E2) = 6

36 =

1

6

P(E3) = 18

36 =

1

2

1 2P(E E ) = 1

36

2 3P(E E ) = 1

12

1 3P(E E ) = 1

12 and 1 2 3P(E E E ) = 0

Hence, E1, E2 and E3 are not independent.

45. (2)

Sol. Z = 2 3i sin

1 2i sin

q

q

For Z to be purely imaginary, Z = –Z

2 3i sin

1 2i sin

q

q =

2 3i sin

1 2i sin

q

q

sin q = 1

3

q = 1 1

sin3

46. (4)

Sol. Here, Tn =

28 (n 1)4

5

Tn = 216(n 1)

25

Sn = 16 (n 1)(n 2)(2n 3)

125 6

S10 = 16 11 12 23

125 6


= 16

[506 1]25

= 16

50525

= 16

5 × 101 m = 101

47. (2) Sol. D = 0

1 1

1 1

1 1

= 0

21( 1) ( 1) 1( 1) = 0

31 1 = 0

3 = 0

2( 1) = 0

= 0, 1, –1

(2) exactly three value of .

48. (2) As per given condition, 2 m = 3 … (i)

and 3 2m = 5 … (ii) = 1 and m = –1

Hence, 2 + m2 = 2

49. (2) Sol. A, L, L, M, S

No. of words starting with A = 4!

2! = 12

No. of words starting with L = 4! = 24

No. of words starting with M = 4!

2! = 12

No. of words starting with S A = 3!

2! = 3

No. of words starting with S L = 3! = 6 and next word formed will be SMALL.

Hence, rank of the word SMALL = 12 + 24 + 12 + 3 + 6 + 1 = 58

50. (4)

Sol. x = 16 a

4

Variance = (3.5)2 = 12.25

Variance =

2

i 2x

(x)n

2

ix = 134 + a2


12.25 =

22134 a 16 a

4 4

49

4 =

2 2536 4a 256 a 32a

16

196 = 280 + 3a2 – 32a 3a2 – 32a + 84 = 0 51. (1)

Sol. 4x 2 r 2

2x r 1 ...(i)

2 2A x r

22 1 2x

A x

22 1

x 1 2x

For maximum / minimum

dA

0dx

1

2x 4 1 2x 0

2

x4

2

2

d A 82 0

dx

For 2

x4

Area is minimum

21 2

14r

( 4)

x

2r

x = 2r 52. (1)

Sol. 1

2 2x

x 0

P lim 1 tan x

2

x 0

tan xlim

2xe

2

x 0

1 tan xlim

2 xe


1

2e

1

log p2

53. (3) Sol.

Equation of normal at P y + tx = 4t + 2t3 As it passes through (0, -6) -6 = 4t + 2t3 or 2t3 + 4t + 6 = 0 or t3 + 2t + 3 = 0 or (t + 1) (t2 – t + 3) = 0 t = - 1 P (2, 4)

Equation of required circle is (x – 2)2 + (y + 4)2 = 4 + 4 or x2 + y2 – 4x + 8y + 12 = 0 54. (2) Sol. y (1 + xy ) dx = x dy xy2 dx = - (ydx – x dy)

2

ydx xdy xxdx d

yy

Integrating both sides

2x xC

2 y

As curve passes through the point (1, -1)

1

1 C2

1

C2


2x x 1

y 2 2

2

xy

1 x

2 2

2

xf (x)

1 x

2 2

11 2f

1 12

2 8

142

5 5

8

55. (2)

Sol. 3

(a.c)b (a.b)c b c2

3

a.b2

3

a . b cos2

q

3

1.1.cos2

q

5

6

q

56. (4)

Sol. 5a b

A3 2

A (Adj A) = A AT

T2| A | .I A A

or 1 0 5a b 5a 3

(10a 3b)0 1 3 2 b 2

or 2 210a 3b 0 25a b 15a 2b

0 10a 3b 15a 2b 13

15a – 2b = 0 ... (i) 10 a + 3b = 13 ...(ii) and 10a + 3b = 25a2 + b2 ...(iii) from (i) and (ii)


2

a b 35

which satisfies (iii) equation also

2

5a b 5 3 55

57. (2) Sol.

Let velocity is V, AB

V10

o

BO

BO BO PO cot 60t 10 10

V AB AB

o

POAO

tan 30

AB + BO = PO Cot 30° AB = PO Cot 30° - PO Cot 60°

o

BO o o

POcot 60t 10

POcot 30 POcot 60

1

310

13

3

= 5 minutes

58. (4) Sol.

A(1,-5, 9)

B

x 1 y 5 z 9

1 1 1

x-y+z=5

Co-ordinates of point B can be taken as (r + 1, r – 5, r + 9) As this lie on plane x – y + z – 5 = 0 (r + 1) – (r – 5) + (r + 9) – 5 = 0 or r + 10 = 0 r = - 10 B (-9, -15, -1)

2 2 2AB (1 9) ( 5 15) (9 1)

3 100 10 3


59. (1) Sol.

Equation of AC

1

y 2 (x 1)2

x + 2y + 5 = 0

Solving, this equation with 7x – y – 5 = 0 we get 1 8

C ,3 3

60. (1) Sol. cos x + cos 2x + cos 3 x + cos 4x = 0 or 2 cos 2x cos x + 2 cos 3x cos x = 0 or 2 cos x (cos 2x + cos 3x) = 0

or 5x x

2cos x 2cos cos 02 2

or 5x x

4cos x cos cos 02 2

3 3 7 9

x , , , , , ,5 2 5 2 5 5

CHEMISTRY

61. (1) Sol. Initially there were equal moles of ideal gas in both bulbs. Now when temperature of second bulb is raised to T2 and both are still connected it means

momentarily (until both the bulbs comes to thermal equilibrium) some moles of gas must have been got shifted to first bulb to equalize pressure of both bulbs. Suppose n1 and n2 are the moles of gases present in first and second bulbs then

1 2

f 1 1 f 2 2 i 1

n np V n RT p V n RT p V RT

2

i f f

1 1 2

2p V p V p V

RT RT RT

or 2f i

1 2

Tp 2p

T T


62. (1) Sol. Water molecules have extensive intermolecular hydrogen bonding and no intramolecular

hydrogen bonding. 63. (2)

Sol. 2 2 2 3 2R CONH Br 4NaOH RNH 2NaBr NaCO 2H O

64. (2) Sol. The first ionization energy of d-block elements are greater than the first ionization energy of

alkali metals. Therefore, The order is

1IE : Na : = 495.8 kJ/mol

IE1 : Sc : = 631.0 kJ/mol So, Scandium has highest ionization energy.

65. (1) Sol. The permissible limit for Nitrate to be present in drinking water is 50 parts per million. 66. (2)

Sol. Equation I s 2 g 2 g

C O CO H 393.5

Equation II g 2 g 2 g

1CO O CO H 283.5

2

Subtracting equation II from equation I we get

s 2 g g

1C O CO

2

H 393.5 283.5 110.0 kJ / mol

67. (1)

Sol. A B C D

Initial 1 1 1 1 Equilibrium 1-x 1-x 1+x 1+x

c

C DK 100

A B

1 x 1 x100

1 x 1 x

1 x

101 x

1 x 10 10x 11x 9

9

x 0.81811

(D) at equilibrium = 1 + x = 1 + 0.818 = 1.818 68. (4)


Sol.

H

H Cl

OH

COOH

CH3

1

2

3

4

‘S’

‘R’

69. (1) Sol. Freundlich adsorption 1/n isotherm equation is

1/nx

kpm

where k & n are constants.

So, x 1

log log k log pm n

(y = mx+c form)

So slope = 1/n and intercept = log k 70. (2) Sol. Glycerol boils at pretty high temperature of 290oC. To prevent it from decomposition it is made

to distill at lower temperature by reducing external pressure during its isolation from spent-lye. 71. (4)

Sol: Anionic detergents mainly contains 4SO as polar group.

72. (3)

Sol: In 2NO, Number of valence electrons = 16

No lone pairs on central atom So, it is sp hybridized. 73. (1) Sol:

SH OH

NH2

O

Cysteine 74. (1) Sol: Galena (PbS) is a sulphide ore hence it is concentrated by froth flotation process. 75. (2) Sol: High density polythene is used for manufacturing buckets and dustbins. 76. (4) Sol: CrO2 is attracted very strongly by magnetic field. 77. (4) Sol:

2 2 3

1. NBS/h

2. H O/K CO

Br2 3 2K CO /H O

Nucleophilic substitution byOH

OH


78. (4) Sol: Hottest part of the flame is at the tip part of the inner curve (in blue flame). 79. (None of them is correct) Remark : The given data does not correspond to any one of the options given. 80. (3) Sol: Orthophosphorous acid H3PO3 and pyrophosphorous acids H4P2O5

81. (4)

Sol. Cl2 Cl is added first, hence possible intermediate among the given option will be

3 2H C C H CH|Cl

82. (3)

Sol. Under given condition both SN and elimination is possible.

3 3

3

CH O / CH OH

3 2 2 2 5 2 3

3 3

Cl H CH| | |

H C CH CH C CH C H CH C OCH| | |H CH CH

2 5 2 2 2 5 3

3 3

C H CH C CH C H CH C CH| |

CH CH

83. (4)

Sol.

Co

Cl

Cl

en

en

Co

Cl

Cl

en

en

Enantiomer

84. (2)

Sol. 2 24Li O 2Li O oxide

2 2 22Na O Na O peroxide

2 2K O KO superoxide

85. (1)

Sol. We assume that the vapour pressure has to be found out at 100oC.

Vapour pressure of pure water at 100oC is 760 Torr.


w = 18 gm; W = 178.2 gm

O S

S

18

P P n w 180 0.0118 178.2P N n

180 18

O

S

P1 0.01

P

O

S

P1.01

P

S

760P 752.47

1.01

86. (3)

Sol. 3 3 2 224Zn s 10HNO dil 4Zn NO 5H O N O

3 3 2 22Zn s 4HNO conc. Zn NO 2H O 2NO

87. (4)

Sol. 2 2 2 2

1H O H O O

2

t1/2 = 25 min

2rate K H O

3 1

1/ 2

0.693 0.6930.05 0.05 1.386 10 mole lit min

t 25

Expression for rate will be,

2 2 2d H O d O

rateof reaction 2dt dt

2 2d O d H O 1

dt dt 2

3

11.386 10mol lit min

2

4 16.93 10 mollit min

88. (4)

Sol. [Cr(H2O)6]2+ and [Fe(H2O)6]2+ has same no. of unpaired electron (i.e. four) in Cr2+ and Fe2+ in

respectively. Thus they have same magnetic moment i.e. 24 B.M.

89. (2)

Sol. Galvanization is the process of applying a protective ‘Zn’ coating to steel or iron to prevent rusting.


90. (2)

Sol. de Broglie equation

h

mv ...(1)

21

K.E. mv m2

...(2)

2

mv 2K.E.m

Mv 2K.E.m ...(3)

For a charged particle moving under acceleration by applying potential ‘V’ the EK (kinetic energy) of charged particle will be

KE eV ...(4)

mv 2meV ...(5)

h

2meV ...(6)

h

2meV

· for e to be independent of r 2 2 2 2 oo q aa q 0a 4 r 2 a 11. (2) sol. resistance of the lamp =

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