(1)

SAFE HANDS & IIT-ian's PACE

MOCK TEST-02 (JEE) SOLUTIONS

1. (a) Sol. Let the retardation produced by instantaneous opposition

= v (where is a constant)

Net instantaneous acceleration = g - v

i.e., dv/dt = (g - v)

Integrating,

t

0

v

0

dtvg

dv

teg

vgeit

g

vg

.,.ln

i.e., v =

g (1 – e-t)

i.e., te1g

dt

dS

i.e., dS = te1g

dt

Integrating,

t

0

t

s

0

dte1g

dS

=

t

0

t

t

0

dteg

dtg

or S = 2

t

2

ge

gt

g

or S = 2

g

(e-t – 1 ) +

gt

2. (b)

Sol. Velocity before strike u = 2gh Component of acceleration along the inclined plane = g sin

and the perpendicular component = g cos Using S = ut + 1/2 at2 For vertical direction we get,

0 = v cos t – 1/2 g cost2 and For horizontal direction

x = u sint + 1/2 g sin t2

= u sin

g

u2Qt

g

u2sing

2

1

g

u22

= g

sinu4

g

sinu2

g

sinu2 222

= 4 × g

singh2 = 8h sin

3. (c) Sol. Due to buoyant force on the aluminium block the reading of spring balance A will be less than 2 kg but it increase the reading of balance B. 4. (c)

Sol. Thrust on the rocket

5. (c)

Sol. Moment of inertia of rod AB about its centre and

perpendicular to the length = = I

Now moment of inertia of the rod about the axis which is passing through O and perpendicular to the plane of

hexagon Irod= [From the theorem of parallel

axes]

Now the moment of inertia of system Isystem

=

Isystem = 5 (12 I) = 60 I [As ] 6. (a)

Sol. Young's modulus (As Y, L and F are

constant) From the graph it is clear that for same load elongation is minimum for graph OD. As elongation (l) is minimum therefore area of cross-section (A) is maximum. So thickest wire is represented by OD. 7. (c)

Sol. Specific gravity of alloy of water Density

alloyof Density

of water densityalloyof Volume

alloyof Mass

wp

mm

mm

2

2

1

1

21

2

2

1

1

21

2

2

1

1

21

// s

m

s

m

mm

mm

mm

ww

of water density

substanceof density substanceof gravityspecific As

8. (b)

Sol. At V = V0, P =2

P0

Ti =nR

PV=

R

)V(2

P0

0

R2

VP 00 (n = 1)

and at V = 2V0, P =5

P4 0 dt

udmF )40(105 4

N6102

12

2ml

Iml 122

22

12mx

ml

6

5

2

3

12

22

2 mllm

ml

6

566

2

rod

mlI

25ml

Iml 122

Al

FLY

Al

1

l l

l

x

O

B A

(2)

Tf =nR

PV=

R

5

P4)V2( 0

0

=R5

VP8 00

T = Tf – Ti =

2

1

5

8

R

VP 00 =R10

VP11 00

9. (a) Sol. Temp. of source T1 = 100 + 273 = 373 K. Temp of sink T2 = 0 + 273 = 273 K.

Efficiency of carnot engine = 1 –1

2

T

T= 1 –

373

273 =

373

100

To increase =373

100+

373

100×

5

1=

373

120

= 1 –373

T2 =373

120

or T2 =373

120373× 373 = 253 K.

.K253tempksinNew

10. (b) Sol.

6° 3°

A

B

C

Time taken by pendulum in going from A to B

= 4

T where T = 2

g

Time taken by pendulum in going from B to C

= 12

T

Time period of pendulum

=

12

T

4

T2

= 3

T2=

3

2.

5

=

15

2sec

Altier :

30°

T = 360

240. T

= T3

2

11. (d)

Sol. 2r4

P

= I for an isotropic point sound source.

P = I.4pr2

= (0.008w/m2)(4..10

2)

= 10 watt. 12. (b)

Potential at the centre of first ring

220

2

0

1

44 RR

Q

R

QVA

Potential at the centre of second ring

220

1

0

2

44 RR

Q

R

QVB

Potential difference between the

two centres 24

))(12(

0

21

R

QQVV BA

Workdone24

))(12(

0

21

R

QQqW

13. (c) Initially the potential at centre of sphere is

VC =

After the sphere grounded, potential at centre becomes zero. Let the net charge on sphere finally be q.

+ = 0 or q =

The charge flowing out of sphere is

14. (c)

Here angle between and is 0º

U = – PE cos 0 = – P = –

15. (d) Sol. Magnetic field at P

Due to wire 1, d

B)8(2

.4

01

and due to wire 2, d

B)16(2

.4

02

2

0

2

022

21

12.

4

16.

4

ddBBBnet

dd

00 510

2

4

16. (b)

x

Q3

4

1

x

Q2

4

1

x

Q

4

1

000

04

1

r

q

x

Q3

4

1

0 x

Qr3

x

Qr3

P

E

+ –

–q

P

E

q

0 0

P

8 A

6 A

(0, 0, d)

X

Y

B2

B1

P

(3)

Sol. By using

2/3

2

2

1

r

x

B

B

axis

centre ,

given r = R and centreaxis BB8

1

2/3

2

2

18

R

x

32/1

2

22 1)2(

R

x

2/1

2

2

12

R

x

2

2

14R

x Rx 3

17. (b) Sol. When the slider moves towards B, the resistance of the circuit (bigger loop) decreases. Therefore, the current in the bigger loop increases. The increasing current results in

increasing flux ( i) linked by the smaller coil. Consequently, induced emf will be generated in the smaller loop causing an induced current so as to oppose the increase in flux. Therefore the current flows anticlockwise in the inner loop. Hence (B) is correct. 18. (b) Sol.

1 sin 600 = sin 300

= 19. (c)

Sol. Y = BA

111

001

010

000

YBA

= AND

20. (b)

Sol. eVs =

= 3.1 – 1.9 eVs = 1.2 eV Vs = 1.2 V

21. Ans. 5 Sol. m ur = m (v0 + v) a

u = as v0 =

also mu2 – = m (v0 +v)

2 –

22. Ans. 6 Sol.

We can assume that 9V battery with 1Ω internal resistance is

in parallel with 0V battery with internal resistance 2Ω.

Eeq =

2

1

1

12

0

1

9

= 9 2

3= 6 volt

4 1 1 3 6V B A

3V

P.d. across 2 = 6 V

23. Ans. 2 Sol.

IL

IC INet = IL – IC

= 0.8 – 0.6 = 0.2 Amp.

24. Ans. 2 Sol. Shortest wavelength of Brackett corresponds to n = 4 and

n = and shortest wavelength of Balmer series corresponds to n

= 2 and n =

(Z2)

16

6.13=

4

6.13 Z = 2

25. Ans. 5

Sol. d = f1 ~ f2

26. (c)

Sol. It is 2, 4, 6-trinitrophenol

27. (d)

Sol.

+ CHCl3 + 3KOH + 3KCl + 3H2O

28. (c)

Sol. In aryl halides the C–X bond has partial double bond

character due to resonance so it will not give SN reaction

29. (b)

Sol.

intra-molecular H-bonding

30. (d)

Sol.

31. (a)

Sol.

30°

30° 60°

=1

3

9.14000

12400

2

e

r4

aGM5

a

GM e

2

1

r

mGMe

2

1

a

mGMe

NH2

CH3

N C

CH3

=

(4)

+

32. (c)

Sol.

+ CH2 = CHCH2Cl

33. (d)

Sol. Follow conditions of geometrical isomerism.

34. (b)

Sol.

(I) +

(II) + H2NOH

(III)

(IV)

+

35. (a)

Sol. (i) 4p, (ii) 4s, (iii) 3d, (iv) 3p

The order of increasing energy,

(iv)< (ii)< (iii)< (i)

36. (a)

Sol. The vapour pressure increase with increase in

intermolecular forces. When the forces are weak, the liquid

has high volatility and maximum vapour pressure diethyl

ether has higher vapoour pressure while water has lowest

vapour pressure.

37. (a)

Sol. Electrical work obtained FnEGW cell0

max

JJ 212300)1.1(965002 kJ3.212 38. (a)

Sol. Adding the first two reactions, we get the third equation

and using the free energy concept, we have 03

02

01 GGG ; FEnFEnFEn 0

33022

011

(n = number of electron involved)

Vn

EnEnE 741.0

3

)900.0(2)424.0(1

3

022

0110

3

39. (b)

Sol. s = )AgCl(

10

m

3

=

)6763(

106.210 6–3

= 2 × 10

–5 (M)

Ksp = ( 2 × 10–5

)2 = 4 × 10

–10.

40. (b)

S o l . 1 M H2SO4 = 2g eq. of H2SO4

hence, y = 2x or x = 2

1 y.

41. (a)

S o l . Wnet = area enclosed

V = P

nRT

VA = 1

R600

1

300R2

VB = 1

R300

2

300R2

; Vc =

1

R400

2

400R2

VD = 1

R800

WAB = –nRT TAln

A

B

V

V= –2R (300)

2

1ln= 600 Rln2

WBC = –2(400 –300) R= – 200 R

WCD = – 2 R(400)ln

C

D

V

V= – 800 Rln2

WAD = –1(600 R – 800 R) = 200 R

W Total dqy = WAB + WBC + WCD + WAD =– 200 Rln2

= – 100 Rln4

42. (c)

S o l . STHG 156.30 molkJH ; 11066.0 KkJmolS ; 0G at

equilibrium; ?T STH or 066.056.30 T

KT 463

43. (a)

Sol. We know that Antipyretics are those compounds which

are applied to reduce the body temperature in fever.

Ex. Aspinic (acyl salicylic acid) paracetamol, phenacetin,

novalgin, & analgin

44. (c)

Sol. Both Be and Al are rendered passive due to formation of

inert, insoluble and impervious oxide layer on their surface.

45. (a)

Fe/Br

2

3AlCl

(5)

Sol. The process is truly adsorption as gangue particles are

wetted with water and sulphide ore particles are wetted with

pine oil.

INEGER TYPE QUESTIONS

46. 6 47. 7 48. 8 49. 6 50. 2

51. (c)

Sol. }4,,|),{( 22 yxZyxyxR

)2,0(),2,0(),1,0)(1,0(),1,1(),0,1(),0,2{( R )}0,2(),1,1(),0,1(

Hence, Domain of }2,1,0,1,2{ R .

52. (d)

Sol. Because, inequality is not applicable for a complex

number.

53. (c)

Sol. tr = r

1 (1 + 2 + 3 ….. + r) =

2

1.

r

1r (r + 1)

S = tr =

n

1r

)1r(2

1 = 115

54. (a)

Sol. Let f(x) = ax2 – bx + 1.

Now f(0) = 1 and roots are imaginary, f(x) > 0 x R

f(–1) = a + b + 1 > 0

55. (c)

Sol. Each group will have m

mn= n students.

Also, there is no distinction between groups.

Required no. of ways = !m),timesm......!n!n(

!mn

= )!m()!n(

!mnm

56. (c)

Sol. Let P the image of P in the line x + 3y = 7.

Then x + 3y = 7 will be perpendicular bisector of PP.

Equation of PP is y – 8 = 3(x-3)

3x –y – 1 = 0.

On sol v ing 3x-y-1 = 0 and x + 3y = 7,we get the

coordinates of M as (1, 2). Since M is mid point of PP’,

coordinates of P are (-1, -4)

57. (a)

Sol. centre(3, –4), radius = 5

Shortest distance = OM – radius

= – 5 = – 5 =

58. (a)

Sol. Centre of conic is

22

,hab

afgh

hab

bghf

Here, ,14a 2h , 11b , 22g , 29f , 71c Centre

22 )2()11)(14(

)29)(14()2)(22(,

)2()11)(14(

)22)(11()29)(2(Centre )3,2(

59. (d)

S o l .

3

2 5dx

xx

xI

Put tx 32 dtdx

dxxx

xdt

tt

tI

2

3

3

2 5

5)(

5

5 and

3

2

3

2

15

52 dxdx

xx

xxI

1][2 22 xI 2/1I

60. (a)

Sol. = x

= dx

= dx

put t = sin x – cos x

dt = (cos x + sin x) dx

also t2 = 1 –sin 2x

I =

= sin–1

t

= sin–1

(sin x –cos x) + c

61. (b)

Sol. We have , I. F. = sin x = sin x = log

sin x P = (log sin x) = cot x.

Hence (B) is the correct answer

62. (a)

Sol.

.

63. (b)

Sol. P(B) = 1 – P ( ) = 1 -

and P(AUB) = P(A) + P(B) – P(AB)

or,

P(A) =

P(A) P(B) =

Hence, A and B are independent.

64. (d)

Sol. The given expression is equal to

= 1+ 4 +1 9 = 15

x+3y = 7

P

M

P(3, 8)

5

25169

5

32

5

7

O(3, –4)

M

xcos

xsin

xsin

xcos

xcosxsin

xsinxcos

2

x2sin

xsinxcos

2 2t1

dt

2

2

dxPe dxP

dx

d

2

4

npq

np

8,2

1,

2

1 npq

256

28

2

1.28

2

1

2

1)2(

8

62

28

CXp

B2

1

2

1

3

1

2

1)A(P

6

5

3

2

)BA(P3

1

2

1x

3

2

2121 3cotcot12tantan1

(6)

65. (c)

Sol.

– sin–1

x

sin–1

x + sin–1

y + sin– 1

z =

sin–1

x = sin–1

y = sin– 1

z =

x = y = z = 1

Also f(p + q) = f(p). f(q) p, q R …(1)

Given f(1) = 1

from (1),

F(1 + 1) = f(1). F(1) f(2) =12 = 1

from (2), f(2 + 1) = f(2) . f(1)

f(3) = 12. 1 = 1

3 = 1

Now given expression = 3 – = 2

66. (d)

Sol. We know that . We have i.e.

and

67. (c)

Sol. The given expression is equal to

68. (b)

Sol. Required distance =

metre

69. (b)

Sol. )34).(34(|34| bababa

baba .24||9||16 22

.122

14324144144

70. (b)

Sol. Angle between two diagonals of a cube = cos–1

3

1

71. Ans. 1

Sol. Tr = tan–1

22

22

)1r(r1

r)1r(

= tan

–1 (r + 1)

2 – tan

–1 r

2

S = Tr = tan–1

(n +1)2 –

4

tan

4)1n(tanlim 21

n= tan

42 = (1)

72. Ans. 1

Sol. 1

73. Ans. 2

Sol.

– +

2

1

f (x) = n2x + 1 – 1 = n2x

f2

1

2

1

, f

2

e

2

e

ne

2

e = 0

Range

0,

2

1 a + b = 2

74. Ans. 9

Sol. Equation of plane is a

z

a

y

a

x = 1

It passes through the point (1,1,1).

So a

1

a

1

a

1 = 1 a = 3

hence A(3, 0, 0), B(0, 3, 0), C(0, 0, 3)

volume of tetrahedron = 6

1

2

9

300

030

003

= 2

9 2 = 9

75. Ans. 2

Sol. z = 8

16

2

122

161

81

62

62

81

)z(|z|

|z||z|

z

z

z

z

arg(z) = – 8 arg )z( 1 – 6arg )z( 2 + 2k, kI

= 8 arg (z1) + 6 arg (z2) + 2k

= 8.6

+ 6.

4

+ 2k =

2

3

3

4

– 2 =

6

5

4 sin = 4 sin 6

5= 2

2

2

2

3

2

3

3

||

)(1

A

AadjA

)(11 AadjK

A

|| AK

110

121

423

K )1(4)1(2)3(3 11429

222

222

22

22

acbbc2

bc2acb

cba

acb

2

Acot

Acos1

Acos1

Acosbc2bc2

bc2Acosb2 2

o15cot60

13

1360

15°

15°

60

x