A solid sphere and a ring of same radius roll down an inclined plane without slipping . Both start from rest from the top of the inclined plane. If the sphere and the ring reach the bottom of the inclined plane with velocities `v_s` and `v_r` repectively. , `then (v_r^2)/(v_s^2)` is

Options

A. 0.2
B. 0.5
C. 0.7
D. 0.9

Answer

C

Explanation

When a body rolls down an inclined plane, energy conservation gives
`mgh =1/2 mv^2 + 1/2 I omega^2 = 1/2 mv^2 (1+ (k2)/(R^2))`
Where, `I= mk^2` and k= radius of gyration.
So, `v_("bottom")= sqrt(((2gh)/(1+k^2/R^2)))`
Now, for ting `(k^2)/(R^2)=1` and for sphere , `(k^2)/(R^2)= 2/5`
`therefore (v_("bottom ring")^(2))/(v_("bottom sphere")^(2))= (1+ (2)/(5))/(1+1) =0.7`