Small sphere of mass m and radius r is kept on fixed inclined rough surface. Surface makes an angle with the horizontal. Initially centre of sphere is at a height h above the horizontal floor. Coefficient of friction between the sphere and surface is u. Sphere is released from the state of rest.

Assume that friction is sufficiet enough to provide pure rolling. Speed of the sphere as it reaches the bottom is

Options

A. `sqrt(2gh)`
B. `sqrt((2gh)/(5))`
C. `sqrt((10gh)/(7))`
D. `sqrt((5gh)/(7))`

Answer

C

Explanation

If it is a case of pure rolling thon point of contact remains at rest and thus work done by friction remains zero and total mechanical energy of the system remains conserved. Let speed guined by sphere is u when it reaches the bottom then its kinetic energy can be written as follows:
`K.E. =(1)/(2) I_(cm) omega^(2)+(1)/(2) mv^(2)`
`=(1)/(2)((2)/(5) mr^(2))((v)/(r))^(2)+(1)/(2) mv^(2)=(7)/(10) mv^(2)`
According to conservation of mechanical energy we can write the following:
Gain in KE,Loss of gravitational potential energy
`rArr (7)/(10) mv^(2)= mgh rArr v= sqrt((10 gh)/(7))`
Hence option (c) is correct