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

Answer

C

Explanation

Let `rho` be the density of sphere diagonal of cube
`d = 2R = a sqrt(3) :. a= (2)/(sqrt(3))R`

Mass of sphere `(M) = (4)/(3)piR^(3) rho`
Mass of cube `(M') = ((2)/(sqrt(3))R)^(3) rho`
`(M)/(M') = ((4)/(3)pi R^(3) rho)/(((2)/(sqrt(3))R)^(3)rho) = (sqrt(3))/(2) pi or M' = (2)/(sqrt(3)) (M)/(pi)`,
`:.` Moment of Inertia `I = (M'a^(2))/(6)`
`= (2M)/(sqrt(3)pi) xx (4)/(3)R^(2) xx (1)/(6) = (4MR^(2))/(9sqrt(3)pi)`