A body is rolling down an inclined plane. If kinetic energy of rotation is `40 %` of kinetic energy in translatory state then the body is a

Options

A. ring
B. cylinder
C. hollow ball
D. solid ball

Answer

D

Explanation

(d) Rotational kinetic energy is
`K_R = (1)/(2) I omega^2 = (1)/(2) Mk^2 ((v)/(R))^2 (because I = Mk^2 and v = R omega)`
=`(1)/(2) Mv^2 ((k^2)/(R^2))`
Translational kinetic energy is
`K_T = (1)/(2) Mv^2`
As per question, `K_R = 40 % K_T`
:. `(1)/(2) Mv^2 ((k^2)/(R^2)) = 40 % (1)/(2) Mv^2` or `(k^2)/(R^2) = (40)/(100) =(2)/(5)`
For solid sphere, `(k^2)/(R^2) = (2)/(5)`
Hence, the body is solid ball,.