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

Options

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

Answer

D

Explanation

From energy conservation.
`v^2 = u^2 - 2gL`
Now, since the two velocity vectors shown in the figure are mutually perpendicular, the magnitude of change of velocity will be given by
`|Delta vecv| = sqrt(u^2 + v^2)`
Substituting the value of `v^2` from Eq. (i) , `|Delta vecv| = sqrt(u^2 + u^2 - 2gL) = sqrt(2(u^2 - gL))`.