What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of $10\,cm$ ($g = 9.8\, m/s^2$) ..... $m/s$
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(c) According to the principle of conservation of energy, $\frac{1}{2}m{v^2} = mgh$ or $v = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 0.1} $ $ = 1.4\,m/s.$
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