Three bodies, a ring , a solid disc and a solid 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 a disc respectively when they reach the bottom , they 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

A ring, a solid disc and a solid shere of identical radii roll down the same inclined plane without slipping as shown in figure.

Let radius be R,
Then, according to' law of conservation of mechanical energy,
`mgh=1/2mv^2(1+k^2/R^2)`
`v^2=(2gh)/(1+k^2/R^2)`
`V=sqrt((2gh)/(1+k^2/R^2))`
Where, ,br> k - Radius of gyration
For a sphere , `k^2=(2R^2)/5`
`V_S=sqrt((2gh)/(1+((2R^2)/5)/R^2))`
`=sqrt((2gh)/(7/5))`
`=sqrt((10gh)/7)=sqrt(10/7)sqrt(gh)approx 1.19 sqrt(gh)`
For a ring , `k^2=R^2`
`rArr V_R=sqrt((2gh)/(1+k^2/k^2))`
`=sqrt((2gh)/2)`
`=sqrt(gh)`
For a disc, `k^2=R^2/2`
`V_0=sqrt((2gh)/(1+(R^2/2)/R^2))`
`=sqrt((2gh)/(3/2))`
`=sqrt((4gh)/3)=sqrt(4/3)sqrt(gh)=1.15 sqrt(gh)`
`therefore V_S=V_D > V_R`