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From a solid sphere of mass M and radius...

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

A

`(MR^(2))/(32 sqrt(2)pi)`

B

`(MR^(2))/(16 sqrt(2)pi)`

C

`(4MR^(2))/(9 sqrt(3)pi)`

D

`(4MR^(2))/(3sqrt(3)pi)`

Text Solution

Verified by Experts

The correct Answer is:
C
Let a be the side of the cube.
So, its diagonal = 2 R = `sqrt( a^(2) + a^(2) + a^(2)) = sqrt(3)a`
or, a = `(2)/(sqrt(3))R`
Hence, volume of the cube = `a^(3) = (8)/(3 sqrt(3)) R^(3)`
Now, density of the material of the sphere
= `(M)/((4)/(3)pi R^(3)) = (3M)/(4 pi R^(3))`
So the mass of the cube = `(3M)/(4 pi R^(3)) cdot (8)/(3 sqrt(3)) R^(3) = (2M)/(sqrt(3)pi)`
Now let `(2M)/(sqrt(3)pi) = M.`
Hence, required moment of inertia
`= (1)/(12) M. (a^(2) + a^(2)) = (1)/(6) cdot (2M)/(sqrt(3)pi) cdot ((2)/(sqrt(3))R)^(2) = (4MR^(2))/(9 sqrt(3)pi)`
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