A sphere of outer radius `R` having some cavity inside is allowed to roll down on an incline. The incline is then made smooth by waxing and the sphere is allowed to slide without rolling and now the speed attained is `(5//4)v_0`. What is the radius of gyration of the sphere about an axis passing through its centre ?

Options

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

Answer

D

Explanation

(d) `(1)/(2) mv_0^2 + (1)/(2) I ((v_0)/( R))^2 = mgh`
`(1)/(2) m((5)/(4) v_0)^2 = mgh`
solve them to get : `I = (9)/(16) mR^2`
now `I = mK^2, K = sqrt((I)/(m)) = (3)/(4) R`.