From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

Options

A. `(13 MR^(2))/(32)`
B. `(11 MR^(2))/(32)`
C. `(9 MR^(2))/(32)`
D. `(15MR^(2))/(32)`

Answer

A

Explanation

The mass of the removed part,
M. = `(M)/(pi R^(2)) xx pi ((R )/(2))^(2) = (M)/(4)`
The moment of inertia of the disc about the perpendicular axis passing through the centre,
` I = (MR^(2))/(2)`
The moment of inertia of the removed part about XX.
`I. = I_("cm") + M. d^(2)`
= `((M)/(4) ((R )/(2))^(2) )/(2) + (M)/(4) ((R )/(2))^(2)`
= `(MR^(2))/(32) + (MR^(2))/(16) = (3MR^(2))/(32)`
`therefore` The moment of inertia of the remaining part of the disc about XX. .
` I.. = I - I. = (MR^(2))/(2) -(3MR^(2))/(32) = (13 MR^(2))/(32)`