A solid sphere of mass m and radius r is released from rest at the top of smooth inclined plane of inclination`theta`.
The ratio of acceleration when it rolls down the plane without slipping and when it slipping down the plane is

Options

A. `7/5`
B. `5/7`
C. `3/7`
D. `8/7`

Answer

B

Explanation

When the spehre rolls down the plane without slipping, the acceleration down the plane is
`a_(1)=(g sin theta)/(1+K^(2)//R^(2))=(g sin theta)/(1+(2//5r^(2))/(r^(2))`
`=(g sin theta)/(((5+2)/4))`
`=5/7 g sin theta [:' K^(2)=2/5r^(2)]`
When the sphere slides down the plane, the acceleration is `a_(2)=g sin theta`
`:. (a_(1))/(a_(2)0=5/7 xx(g sin theta)/(g sin theta)=5/7`