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A stone tied to a string $L$ 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 as its lowest position and has a speed $u$. the magnitude of the change in its velocity as it reaches a position where the string is horizontal is
$\sqrt {{u^2} - 2gL} $
$\sqrt {2gL} $
$\sqrt {{u^2} - gL} $
$\sqrt {2\left( {{u^2} - gL} \right)} $
Solution
As velocity is vector quantity
$\Delta \mathrm{v}=\sqrt{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}-2 \mathrm{v}_{1} \mathrm{v}_{2} \cos \theta}$ as $\left[\theta=90^{\circ}\right]$
$\Delta v=\sqrt{v_{1}^{2}+v_{2}^{2}}$
By applying $W.E.T.$
$\frac{1}{2} m v_{2}^{2}-\frac{1}{2} m u^{2}=-m g l$
$\mathrm{u}_{2}^{2}=\mathrm{u}^{2}-2 \mathrm{g} \ell$
$\left[\mathrm{v}_{1}=\mathrm{u}\right] \quad \Delta \mathrm{v}=\sqrt{\mathrm{u}^{2}+\mathrm{u}^{2}-2 \mathrm{g} \ell}$
$\Delta v=\sqrt{2\left(u^{2}-g l\right)}$
