A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity `v m//s`. If it is to climb the inclined surface then `v` should be :

Options

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

Answer

A

Explanation

From conservation of energy Potential energy = translational kinetic energy + rotational kinetic energy
`mgh=(1)/(2)mv^(2)+(1)/((2)/(5))mR^(2) cdot (v^(2))/(R^(2))`
`rArr" "(7)/(10) mv^(2)=mgh rArr v ge sqrt((10)/(7))gh.`