Three bodies, a ring, a solid disc and a solid sphere roll down the same inclined plane without slipping. The radii of the bodies are identical and they start from rest. If `V_(S), V_(R)` and `V_(D)` are the speeds of the sphere, ring and disc, respectively when they reach the bottom, then the correct option is

Options

A. `V_(S) gt V_(R) gt V_(D)`
B. `V_(D) gt V_(S) gt V_(R)`
C. `V_(R) gt V_(D) gt V_(S)`
D. `V_(S) gt V_(D) gt V_(R)`

Answer

D

Explanation

Given, radii of ring, soild disc and solid sphere are identical, say R.
When a body of mass m starting from rest rolls down an inclined plane, then the velocity of body when it reaches at the ground is given by
`v=sqrt((2gh)/(1+k^(2)/R^(2)))`
where, h = height of inclined plane,
R = radius of body,
and k = radius of gtration
For a ring, `k^(2)=R^(2)`
`:.` Speed of the ring, `v_(R)=sqrt((2gh)/(1+R^(2)/R^(2)))=sqrt(g h)` ...(i)
For a solid sphere, `k^(2)=(2 R^(2))/5`
`:.` Speed of the solid sphere, `v_(S)=sqrt((2gh)/(1+2/5))`
`implies sqrt((10 g h)/7)=1.19 sqrt(g h)` ...(ii)
For a solid disc, `k^(2)=R^(2)/2`
`:.` Speed of the solid disc, `v_(D)=sqrt((2gh)/(1+1/2))`
`implies sqrt((4gh)/3)=1.15 sqrt(g h)` ...(iii)
From Eqs. (i), (ii) and (iii), it is clear that `V_(S) gt V_(D) gt V_(R)`.