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. `(l)/(10)`
B. `(2l)/(6)`
C. `l`
D. `(l)/(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)`
`"Ml of the disc l"=(MR^(2))/(2)`
`"Ml of the disc l'"=(2)/(5)Mr^(2)=(2)/(5)M((R)/(2))^(2)`
`=(2)/(5)(MR^(2))/(4)`
`therefore" "l'=(1)/(5)((1)/(2)MR^(2))=(l)/(5)`