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 centre and perpendicular to one of its faces is :

Options

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

Answer

B

Explanation

`d=2R=asqrt(3)`
`implies a=(2)/(sqrt(3))R`
`(M)/(M.)=((4)/(3)piR^(3))/(((2)/(sqrt(3))R)^(3))=sqrt(3)/(2)pi`
`implies M.=(2M)/(sqrt(3pi))`
`I=(M.a^(2))/(6)=(2M)/(sqrt(3pi))xx(4)/(3)R^(2)xx(1)/(6)`
`I=(4MR^(2))/(9sqrt(3pi))`
`v=(3000)/(60)=50Hz therefore omega=2piv=100pi" rad/s"`
Angular acceleration `a=(omega-omega_(0))/(t)=(100pi-60pi)/(20)=2omega" rad/s"`