a
(a) Tension in the string when bob passes through lowest point
$T = mg + \frac{{m{v^2}}}{r} = mg + mv\omega $ ( $v = r\omega$)
putting $v = \sqrt {2gh} $ and $\omega= \frac{{2\pi }}{T} = \frac{{2\pi }}{2} = \pi $
we get $T = m\;(g + \pi \sqrt {2gh} )$