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. `(1)/(5)`
B. `(1)/(6)`
C. `(1)/(32)`
D. `(1)/(64)`

Answer

A

Explanation

According to question,
Moment of inertia of disc is given by `l=(MR^(2))/(2)`
[Symbols have their usual meanings.]
When the disc is remoulded into solid sphere, then volume remains same.
i.e., Volume of disc = Volume of solid sphere
`"i.e. "piR^(2)xx(R)/(6)=(4)/(2)pir^(3)`
`rArr" "r^(3)=(R^(3))/(8) rArr r=(R)/(2)`
Now, moment of inertia of solid sphere is given by `(2)/(5)mr^(2).`
`=(2)/(5)xx m xx(R^(2))/(4)=(mR^(2))/(10)`
`"i.e, "=(l)/(5)`
[ l = moment of inertia of disc]