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

Options

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

Answer

C

Explanation

`d=2R = asqrt(3) rArr a=2/sqrt(3) R`

`M/M^(') = (4/3 pi R^(2))/(2/sqrt(3)R)^(2) = sqrt(3)/2 pi rArr M^(') = (2M)/(sqrt(3)pi)`
`I =(M^(')a^(2))/6 = (2M)/(sqrt(3)pi) xx 4/3 R^(2) xx 1/6, I =(4MR^(2))/(9sqrt(3)pi)`.