From a solid sphere of M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendiular to one of its faces is:

Options

A. `(4MR^(2))/(9sqrt(3)pi)`
B. `(4MR^(2))/(3sqrt(3)pi)`
C. `(MR^(2))/(32sqrt(2)pi)`
D. `(MR^(2))/(16sqrt(2)pi)`

Answer

A

Explanation

Here `a=(2)/sqrt(3)R`
Now, `(M)/(M.)=((4)/(3)piR^(3))/(a^(3))=((4)/(3)piR^(3))/(((2)/(sqrt(3))R)^(3))=(sqrt(3))/(2)pi`
`M.=(2M)/(sqrt(3)pi)`
Moment of inertia of the cube about the given axis,
`I=(M.a^(2))/(6)=((2M)/(sqrt(3)pi)xx((2)/(sqrt(3))R)^(2))/(6)=(4MR^(2))/(9sqrt(3)pi)`