**(a)** The M. I of a thin uniform disc about an axis passing through its centre and perpendicular to its plane is given by,

IC =

()MR2 i. According to theorem of parallel axis,

IT =

Id +

Mh2 =

Id+MR2 [∵ h = R]

But

Id =

MR2/4

∴

IT =

+MR2 ∴

IT =

MR2 Now , radius of gyration is given by,

K =

√ ∴ K =

√ ∴ K =

R

ii. Applying theorem of parallel axis,

IT =

I0+Mh2 =

I0+MR2 [∵ h = R]

But

I0 =

MR2/2

IT =

+MR2 =

MR2 Now, radius of gyration is given by,

K =

√ ∴ K =

√ ∴ K =

√ R

**(b)** Given: N = 21, x = 4,

nF = 2

nL To find:

i. Frequency of first fork (

nF)

ii. Frequency of tenth fork (

n10)

Formula:

nL =

nF - (N – 1)x

Calculation:-

i. When tuning forks are arranged in the decreasing order of frequencies, the frequency of the pth tuning fork is,

nL =

nF - (N - 1)x =

n1 - (21 - 1) 4

∴

nL =

nF - 80 ............(1)

As frequency of first fork is an octave of last,

∴

nF = 2nL

∴

nL =

nF/2

From equation (1),

nF/2 =

nF - 80

∴

nF - (

nF/2) = 80

∴

nF/2 = 80

∴

nF = 160 Hz

The frequency of the first fork is 160 Hz.

ii. For 10th fork,

n10 =

n1 - (10 - 1)x

= 160 - 9 * 4 = 160 - 36

n10 = 124Hz

The frequency of the tenth fork is 124 Hz.