Explanation

(c) For a simple pendulum time period
$T = 2\pi\sqrt{\frac{l}{g}}$
or $\frac{2\pi}{T} = \sqrt{\frac{g}{l}}$
$\therefore$ $\omega = \sqrt{\frac{g}{l}}$
$\omega^{2} = \frac{g}{l}$ …(i)
Amplitude, when angular displacement is $60{^\circ}$
$= \frac{2\pi l}{360} \times 60 = \frac{2\pi l}{6}$
Therefore, displacement when angular displacement is $30{^\circ}$
$= \frac{1}{2}\ \left( \frac{2\pi l}{6} \right)$
$y = \frac{\pi l}{6}$ …(ii)
Acceleration $(\alpha) = - \omega^{2}y$
Using Eqs. (i) and (ii), we get
$\alpha = - \frac{g}{l} \times \frac{\pi l}{6} = - \frac{10 \times 3.14}{6}$
$= - 5.2\ ms^{- 2} = - 5\ ms^{- 2}$