A body of mass 2kg slides down a curved track which is quadrant of a circle of radius 1 metre . All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is

Options

A. `4.43ms^(-1)`
B. `2ms^(-1)`
C. `0.5ms^(-1)`
D. `19.6ms^(-1)`

Explanation

Equating gain of kinetic energy with loss of gravitational potential energy, we get
`(1)/(2)mv^(2)=mgh`
or `v=sqrt2gh`
or `v=sqrt(2xx.9.8xx1)" "ms^(-1)
`v=sqrt(19.6)" "ms^(-1)=4.43ms^(-1)`