A solid cylinder of mass 0.1 kg having radius 0.2m rolls down an inclined plane of height 0.6m without slipping. The linear velocity the cylinder at the bottom of the inclined plane is

Options

A. `28 ms^(-1)`
B. `2.8 ms^(-1)`
C. `280 ms^(-1)`
D. `0.28 ms^(-1)`

Answer

B

Explanation

According to the law of conservation of energy.
`(3)/(4)mv^(2)= mgh rArr v = sqrt((4gh)/(3)) = sqrt((4(9.8)(0.6))/(3))`
`= (28)/(10)= 2.8 ms^(-1)`.