A simple pendulum of length one meter has a bob of mass 100g. It is displaced through an angle 60° from the vertical and then released. Kinetic energy of the bob when it passes through the mean position (g = 10 m/s²)

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0.05J

10J

0.5J

answer is D.

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Detailed Solution

$\begin{array}{l} V = \sqrt{2 gl (1 - \cos 0)} \\ ∴ K . E = \frac{1}{2} {mV}^{2} = mgl (1 - \cos θ) \\ \Rightarrow K \cdot E = 0 .1 \times 10 \times 1 \times (1 - \frac{1}{2}) = 0 .5 J \end{array}$

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