Explanation

(b) The time period of a pendulum of length $l$, is
$T = 2\pi\sqrt{\frac{l}{g}}$
$\Rightarrow$ $l = g\ \frac{T^{2}}{4\pi^{2}}$
Since, $T = 2\ s$ (for second’s pendulum)
$\therefore$ $l_{1} = \frac{g_{1}(2)^{2}}{4\pi^{2}} = \frac{g_{1}}{\pi^{2}};l_{2} = \frac{g_{2}(2)^{2}}{4\pi^{2}} = \frac{g_{2}}{\pi^{2}}$
Since, length is decreased, $g_{2}\ is\ less\ than\ g_{1}$
$\therefore$ $l_{1} - l_{2} = \frac{g_{1} - g_{2}}{\pi^{2}}$
$\Rightarrow$ $\left( l_{1} - l_{2} \right)\pi^{2} = g_{1} - g_{2}$
$0.3 \times 10 = g_{1} - g_{2}$
$\therefore$ $g_{2} = 981 - 3 = 978\ cms^{- 2}$