If a uniform solid sphere of radius R and mass m rotates about a tangent and has moment of inertia `42 kg m^(2)`, then the moment of inertia of a solid sphere about an axis passing through its center and perpendicular to its plane will be

Options

A. `12 kg m^(2)`
B. `18kg m^(2)`
C. `300 kg m^(2)`
D. `24 kg m^(2)`

Answer

A

Explanation

`(I_(C ))/(I_(T))=((2)/(5)MR^(2))/((7)/(5)MR^(2))=(2)/(7)`
`I_(C )=(2)/(7)xx42`
`= 12 kg m^(2)`.