The moment of inertial of a thin circular disc about an axis passing through its centre and perpendicular to its plane is I. Then, the moment of inertia of the disc about an axis parallel to its diameter and touching the edge of the rim is

Options

A. I
B. 2I
C. `(3)/(2)I`
D. `(5)/(2)I`

Answer

(d)

Explanation

The expression for moment of inertia of a circular disc is given as,
`I=(1)/(2)mR^(2)`
`implies MR^(2)=2I`
The expression for moment of inertia of the disc about tangent, perpendicular to the diameter
`I_(5)/(4) MR^(2)`
`=(5)/(4)(2I)`   (because From equation (1))
`(10)/(4)I`
`(5)/(2)I`.