A stone is tied to a string of length l and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is

Options

A. (a) `sqrt(u^2-2gL)`
B. (b) `sqrt(2gL)`
C. (c) `sqrt(u^2-gL)`
D. (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 figure are mutually perpendicular, the magnitude of change of velocity will be given by
`|Deltavecv|=sqrt(u^2+v^2)`
Substituting the value of `v^2` from Eq. (i),
`|Deltavecv|=sqrt(u^2+u^2-2gL)=sqrt(2(u^2-gL))`