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

Answer

B

Explanation

The denisty of the sphere is `rho=(M)/(V)=(M)/((4pi)/(3)R^(3))`
`=(3M)/(4piR^(3))`
a = length of the side of the cube of maximum volume `=(2R)/(sqrt3)`
Mass of the cube is `m=rhoa^(3)=((3M)/(4piR^(3)))((8R^(2))/(3sqrt3))=(2M)/(sqrt3pi)`
The moment of inertia of the cube is
`I=(ma^(2))/(6)=(2M)/(sqrt3pi)xx(1)/(6)xx((2R)/(sqrt3))^(2)`
`=(8MR^(2))/(18sqrt3pi)=(4MR^(2))/(9sqrt3pi)`