Moment of inertial of a uniform horizontal solid cylinder of mass M about an axis passing through its edge and perpendicular to the axis of the cylinder when its length is 6 times its radius R is

Options

A. `(39MR^(2))/(4)`
B. `(30MR^(2))/(4)`
C. `(49MR)/(4)`
D. `(49MR^(2))/(4)`

Answer

(d)

Explanation

From parallel asix theorem,
Moment of inertia about AB,
`I=M((R^(2))/(4)+(l^(2))/(3))=M[(R^(2))/(4)+((6R^(2)))/(3)]`
`=M[(R^(2))/(4)+(36R^(2))/(3)]=(49MR^(2))/(4)`