Explanation

(c) The bob possess kinetic energy at its mean position which gets converted to potential energy at height$\ \ h$. But the total energy remains converted.
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Hence, we have
KE=PE
Let velocity of the bob at mean position be $v\ and\ m$ be its mass, then we have
$\frac{1}{2}mv^{2} = mgh$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \sqrt{2gh}$
Putting g$\ = 9.8\ ms^{- 2},\ h = 0.1\ m$
$\therefore$ $v = \sqrt{2 \times 9.8 \times 0.1} = 1.4\ ms^{- 1}$