Q173 / 200 · Manipal (MET) 2012 · Physics
The length of the second's pendulum is decreased by 0.3 cm when it is shifted to Chennai from London. If the acceleration due to gravity at London is $981 \mathrm{~cm} / \mathrm{s}^2$, the acceleration due to gravity at Chennai is
- A$981 \mathrm{~cm} / \mathrm{s}^2$
- Correct answerB$978 \mathrm{~cm} / \mathrm{s}^2$
- C$984 \mathrm{~cm} / \mathrm{s}^2$
- D$975 \mathrm{~cm} / \mathrm{s}^2$
Solution
$L_1=\frac{g_1 T^2}{4 \pi^2}=\frac{g_1}{\pi^2} ; L_2=\frac{g_2 T^2}{4 \pi^2}=\frac{g_2}{\pi^2}$
Since, length is decreased, $g_2$ is less than $g_1$. $\therefore \quad L_1-L_2=\frac{g_1-g_2}{\pi^2}$ or $\quad\left(L_1-L_2\right) \pi^2=g_1-g_2$ or $\quad 0.3 \times 10=g_1-g_2$ $\therefore \quad g_2=981-3=978 \mathrm{~cm} / \mathrm{s}^2$
Since, length is decreased, $g_2$ is less than $g_1$. $\therefore \quad L_1-L_2=\frac{g_1-g_2}{\pi^2}$ or $\quad\left(L_1-L_2\right) \pi^2=g_1-g_2$ or $\quad 0.3 \times 10=g_1-g_2$ $\therefore \quad g_2=981-3=978 \mathrm{~cm} / \mathrm{s}^2$
Hence, the correct answer is option: 'B'.