A circular disc of radius `R` and thickness `R//6` has moment of inertia `I` about an axis passing through its centre and perpendicular to its plane. It is melted and recast into a solid sphere. The `M.I` of the sphere about its diameter as axis of rotation is

Options

A. `(I)/(100)`
B. `(2I)/(8)`
C. `I`
D. `(I)/(5)`

Answer

D

Explanation

Since mass and density are not changed,
Volume of the disc = Volume of the sphere
`piR^(2)h=piR^(2)xx(R)/(6)=(4)/(3)pir^(3)`
where r is the radius of the sphere.
`therefore r^(3)=(R^(3))/(6)xx(3)/(4)=(R^(3))/(8)" "therefore r=(R)/(2)`
`"M.I. of the disc I"=(MR^(2))/(2)`
M.I. of the sphere I'
`=(2)/(5)Mr^(2)=(2)/(5)M((R)/(2))^(2)=(2)/(5)(MR^(2))/(4)`
`therefore I=(1)/(5)((1)/(2)MR^(2))=(I)/(5)`