A solid sphere of mass M and radius R, rolling down a smooth inclined plane, without slipping, reaches the bottom with a velocity v. What is the height of the inclined plane in terms of the velocity v ?

Options

A. `(2v^(2))/(5g)`
B. `(3v)/(5g)`
C. `(7v)/(10g)`
D. `(7v^(2))/(10g)`

Answer

D

Explanation

The M.I. of the solid sphere about its diameter is given by `I=(2)/(5)MR^(2)`
and the velocity of the sphere when it reaches the bottom is given by
`v=sqrt([(2gh)/((1+(1)/(MR^(2))))])=sqrt((2gh)/(1+(2)/(5)(MR^(2))/(MR^(2))))`
`=sqrt((2gh)/((7)/(5)))=sqrt((10gh)/(7))`
`therefore v^(2)=(10gh)/(7) therefore h=(7v^(2))/(10g)`