The moment of inertia of a sphere of mass M and radius R about an axis passing through its centre is `(2)/(5) MR^(2)` . The radius of gyration of the sphere about a parallel axis to the above and tangent to the sphere is

Options

A. `(7)/(5)R`
B. `(3)/(5)R`
C. `(sqrt((7)/(4)))`
D. `(sqrt((3)/(5)))`

Answer

C

Explanation

Given, `l=(2)/(5)MR^(2)`
Using the theorem of paraell axes, moment of inetria of the sphere about a parallel axis tangential to the sphere is
`l'=l+MR^(2)=(2)/(5)MR^(2)+MR^(2)=(7)/(5)MR^(2)`
`therefore " "l'=MK^(2)=(7)/(5)MR^(2)`
`K=(sqrt((7)/(5)))R` (where, K is radius of gyrations)