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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 change in velocity as it reaches a position where the string is horizontal ( $g$ being acceleration due to gravity)is:
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Verified Answer
The correct answer is:
$\sqrt{2\left(u^2-g l\right)}$
According to the question
Total energy at $\mathrm{A}=$ Total energy at $\mathrm{B}$
$\begin{array}{l}
\Rightarrow \frac{1}{2} m u^2 =\frac{1}{2} m v^2+m g l \\
\Rightarrow v^2 =\sqrt{u^2-2 g l}
\end{array}$
$\therefore$ Change in magnitude of velocity
$=\sqrt{u^2+v^2}=\sqrt{2\left(u^2-g l\right)}$
Total energy at $\mathrm{A}=$ Total energy at $\mathrm{B}$
$\begin{array}{l}
\Rightarrow \frac{1}{2} m u^2 =\frac{1}{2} m v^2+m g l \\
\Rightarrow v^2 =\sqrt{u^2-2 g l}
\end{array}$
$\therefore$ Change in magnitude of velocity
$=\sqrt{u^2+v^2}=\sqrt{2\left(u^2-g l\right)}$
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