A stone tied to a string of length L is whirled in a vertical circle with the other end of string at the centre.
At a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude of its velocity , as it reach a position where the string is horizontal, is

Options

A. `sqrt(u^2-2gL)`
B. `sqrt(2gL)`
C. `sqrt(u^2-gL)`
D. `sqrt(2(u^2-gL))`

Answer

A

Explanation

Let mass of the stone is m ,
At lowest point of the circular path,
potential energy of the stone=0 and kinetic energy `=1/2 mu^2`
When the string is horizontal , the stone is at height L from its lowest point.
Hence, potential energy =mgL
and kinetic energy =`1/2 mv^2`
[v=velocity of the stone at L]
From the law of conservation of energy, we get,
`0+1/2 mu^2=mgL+1/2 mv^2`
or, `v^2=u^2-2gL or, v=sqrt(u^2-2gL)`