A thick walled hollow sphere has outer radius `R`. It down an inclined plane without slipping and its -speed the bottom is `v`. If the inclined plane is frictionless and the sphere slides down without rolling, its speed at the bottom `5v//4`. What is the radius of gyration of the sphere?

Options

A. `R/(sqrt(2))`
B. `R//2`
C. `3R//4`
D. `(sqrt3R)/4`

Answer

C

Explanation

Case 1 `mgh=1/2mv^(2)+1/2mk^(2)omega^(2)`
Case 2 `mgh=1/2mxx((5v)/4)^(2)`
`1/2mk^(2) (v^(2))/(R^(2))+1/2mv^(2)=1/2mxx(25v^(2))/16`
` (k^(2))/(R^(2))+1=25/16impliesk=(3R)/4`