2018-10-01_Physics_For_You

10. (b) : For the given axis,
    IMdisc 5 RIring MR
    4

3

2

(^22) , ,
kRdisc R

5

4

5

2

(^2) , kRR
ring==

3

2

3

2

2

Thus,

k

k

R

R

disc

ring







52

32

5

6

56

/

/

:

11. (d) : F GMm
    R

x

1 ==() 2 2 4 , where

GMm

R

2 x

F

GMm

R

GM m

R

x



















3

2

8

(^229418)

(/)

(/)

(as Mc is the mass of the cavity)

Thus, FF 21 F xx x

418

7

36

 

or F

F

x

x

2

1

7

36

47

9



12. (c) :

V

V

Pr l

Pr l

′

=

πη′

πη

4

4

8

8

/

/



 

















r

r

r

r

(^4411)
14641

.

.

Thus, VV

V

V

V



















(^1001)  100
= (1.4641–1)100 = 46%

13. (d) : As dU = CV dT,

dU =

5

2

R(Tf – T 0 ) +^3

2

7

3 0

RT()f− T as CV for nitrogen

is (5/2)R and for helium its value is (3/2)R.

Put dU = 0 and obtain TTf 3

2 0

14. (b) : Z 1 =500p, v 1 = 250; Z 2 = 506p, v 2 = 253
    Number of beats/s = 3
    Further, A 1 = 4 and A 2 = 2
    Amplitude ratio, r A
    A

1

2

2

?

I

I

r

r

max

min









()

()

1

1

9

2

2

15. (a) : Let T be the temperature of the mixture. As
    U = U 1 + U 2 ,
    f
    nnRT

f

nRT

f

nRT

222

() 12  10  20 () 2

or (2 + 4)T = 2T 0 + 8T 0 (as n 1 = 2, n 2 = 4)

or T = (5/3)T 0

(^) 

5. (d) : As yg==ttt=

1

2

5

1

2

(^22) ,( 10 )ors 1
Further, as vball t = 20 m or vball==−

20

1

20 1

m

s

ms

and as vbullet t = 100 m, vbullet==−

100

1

100 1

m

s

ms

Applying the law of conservation of momentum,

(0.01 kg) v = (0.01 kg) (100 m s–1) + (0.2 kg) (20 m s–1),

hence v = 500 m s–1

6. (c) : When there is no friction, a = g sinq.
   When there is friction, net downwards force acting
   on the body, i.e.,
   F = mg sinq – f = mg sinq – μR = mg sin q – μ mg cosq
   Acceleration acting on the body,
   a

F

m

mg

m

 

(sin cos) or ac = g (sinq – μcosq)

If s is the distance through which the block slides,

satat

1

2

1

2

(^22) or a
a
t
 t







2

2 4 (as tc = 2t)

or g

g

sin

(sin cos)

θ

θμθ−

= 4 or 4sinq – 4μcosq = sinq

or μθ==^3 =

4

3

4

tan( 10 ). 75 [' q = 45°]

7. (c) : pf = pi + (1/2)pi = (3/2)pi or

p

p

f

i

=

3

2

Thus,

K

K

p

p

f

i

f

i













 

2

9

4

or KKfi

9

4

or KKfiKi

5

4

Thus, percentage increase in kinetic energy

=

KK

K

fi

i

−

×=100 125%

8. (c) : As 0.5 × 2 + 1 × 0 = (0.5 + 1)v
   (assuming the body of mass 1.00 kg to be rest)

(^) v==^2 −−
3
ms^1106. 7 ms
Energy loss = Ki – Kf =

1

2

052

1

2

(.)()(^22 − 15 06.)(. 7 )J

= 1 J – 0.33 J = 0.67 J

9. (d) : When the centre of mass of the rod rises
   through h, increase in PE = mgh.

Kinetic energy of the rod =

1

2

1

23

I^2  ml^22









.

Thus, mgh ml













2

2

6

 or h

l

g



1

6

(^22)