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A suspended simple pendulum of length l is making an angle $θ$ with the vertical. On releasing, its velocity at lowest point will be :-

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$\sqrt{2 gl}$

$\sqrt{2 glsinθ}$

$\sqrt{2 gl (1 + cosθ)}$

$\sqrt{2 gl (1 - cosθ)}$

answer is C.

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Detailed Solution

$\begin{array}{l} mgl (1 - \cos θ) = \frac{1}{2} {mv}^{2} \\ \Rightarrow v = \sqrt{2 gl (1 - \cos θ)} \end{array}$

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