(b) the prongs of the tuning fork were kept in a

horizontal plane above the resonance tube

(c) the length of the air-column at the first

resonance was somewhat shorter than 1/4th of

the wavelength of the sound in air

(d) Both (a) and (c)

SOLUTIONS

1. (d) : vv 21 LT LT

2

22

1

11

=⇒α −−=^12

β

α

β

[][] ...(i)

a 2 = a 1 DE Ÿ [L 2 T 2 –2] = [L 1 T 1 –2]DE ...(ii)

vv 21 LT LT

2

22

1

11

=⇒α −−=^12

β

α

β

[][]

and[F ][ ]

F

2 =⇒^1 ML 22 TM 2 −−^2 =× 11 LT 12 1

αβ αβ

...(iii)

Dividing eqn (iii) by eqn (ii), we get

M

MM

2

11

==()αβαβ αβ 22

Squaring eqn (i) and dividing by eqn (ii), we get

LL 21

3

= 3

α

β

Dividing eqn (i) by eqn (ii), we getTT 21 = α 2

β

2. (c) : u 1 cos D = u 2 cos E

30 3

2

30 3

3

2

2

2

2

2

=⇒=

u

g

u

g

sin β

Ÿ u^22 = 600 or u 2 = 10 6

t

u

g

2 2 t 2

260 210632

10

= 18

°

=⇒=

sin ()/

u 1

10 660

30

= 10 2

°

°

=

cos

cos

−=h 10 2301 °− 8 1

2

sin( 10 18 )^2

Ÿ –h = 30 – 90 Ÿ h = 60 m

3. (c) : Block slips on plank

30 N m 1 10 m (^2) 10 N
a 1 = 10 m s–2, a 2 = 2 m s–2
arel = 8 m s–2
Time taken by m 1 to fall over plank
sarelr= elt

1

2

(^2) Ÿ 64 1
2
=× 84 tt^2 or = s
Displacement of plank =

1

2

×× 24 ×= 416 m

4. (b) : Kinetic energy of bullet is converted into heat.
   25% of heat is absorbed by obstacle.
   75% of heat is absorbed by lead. is heat raises
   temperature of lead to 327° C and then melts it.

75 1

2

%×^2







Mv =+ms∆TmL

or^75

100 2

2

×=mv ms ()∆TL+

s

kg

=

×°

=

×

×°

=

××

− °×

(^00300342)
10

00342 1000

3

. ..

()

cal ..

gC

J

kg C

J

C

L=

×°

=

×

(^6) − =× × −

642

10

(^36421031)
cal
2 C kg

.J

.Jkg

?

3

42

2

×

v

= 126 × (327 – 27) + 25200 = 37800 + 25200

or v^2

8

3

= × 63000 v =168000 or v = 409.8 m s–1

5. (a) : Density of carpet =

M

πRh^2

or D = M/(pR^2 h)

R R/ 2

? Mass of unrolled carpet = M c

? ′=







M  ×=×=

R

h

Rh M

Rh

M

π

π

(^24) π 4
(^22)
density 2
Energy is conserved.
? Loss in potential energy = MgR – ′







Mg 

R

2

= MgR –

MgRMgR

42

7

× 8

= ...(i)

Gain in KE = Translational KE + Rotational KE

=   +=

1

24

1

22

M vI 22 v

c R

ωωwhere c

/

=+  













M

v

MR v

R

c

c

8

1

2

1

(^2422)
2
(^22)
(/)^2

M

v

Mv Mv

c

cc

816

3

16

2

22

+ ...(ii)

?

3

16

7

8

72

3

14

3

2

Mvc MgR v 2 gR gR

=⇒c=

××

=

or v

gR

c

2 14

3

= or v

Rg

c=

14

3

6. (b) : BC is the crater of moon. B denotes bottom
   of crater.
   Depth of crater = R/100