A hollow sphere of outer radius R is rolling down an inclined plane without slipping and attains a speed `v_0` at the bottom. Now the inclined plane is made smooth and the sphere is allowed to slide without rolling. Now it attains a speed `(5v_0)/4`. What is the radius of gyration of sphere?

Options

A. `sqrt(2/5)R`
B. `3/4R`
C. `4/5R`
D. `sqrt(2/3)R`

Answer

B

Explanation

Let the sliding acceleration be a. So rolling acceleration `=a/(1+K^2//R^2)`,
where K is the radius of gyration.
Using `v^2-u^2=2as`
`rArr (5/4v_0)^2=2aS`….(i)
`v_0^2=(2aS)/(1+K^2//R^2)` …(ii)
From (i) and (ii) we get
`1+K^2/R^2=25/16` or `K^2/R^2=25/16-1`
`rArr K^2=(9R^2)/16` or `K=(3R)/4`