Question

A pendulum clock gives correct time at ${20}^{\circ }C$ at a place where g = 9.800 $m{s}^{-2}$. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where g 9.788 $m{s}^{-2}$. At what temperature will it give correct time ? Coefficient of linear expansion of steel $=12\times {10}^{-6}{}^{\circ }{C}^{-1}$

Answer 1

$g_1=9.8m/s^2,$

$g_2=9.788m/s^2$

$T_1=\frac{2\pi \sqrt{l_1}}{\sqrt{g_1}}$

$T_2=\frac{2\pi \sqrt{l_2}}{\sqrt{g_2}}$

$=\frac{2\pi \sqrt{l_2(1+\alpha T)}}{\sqrt{g_2}}$

${\alpha }_{steel}=12\times {10}^{-6}{/}^{\circ }C$

$Q_1={20}^{\circ }$

$Q_2=?$

$T_1=T_2$

$\Rightarrow \frac{2\pi \sqrt{l_1}}{\sqrt{g_1}}=\frac{2\pi \sqrt{l_1(1+\alpha \Delta T)}}{\sqrt{g_2}}$

$\Rightarrow \sqrt{\left(\frac{l_1}{g_1}\right)}=\frac{\sqrt{l_1(1+\alpha \Delta T)}}{\sqrt{g_2}}$

$\Rightarrow \frac{1}{9.8}=\frac{1+12\times {10}^{-6}\times \Delta T}{9.788}$

$\Rightarrow \frac{9.788}{9.8}=1+12\times {10}^{-6}\times \Delta T$

$\Rightarrow \frac{9.788}{9.8}-1=12\times {10}^{-6}\times \Delta T$

$\Rightarrow \Delta T=\frac{-0.00122}{12\times {10}^{-6}}$

$\Rightarrow T_2-20=-102.4$

$\Rightarrow T_2=-102.4+20$

$=-82.4\approx -{82}^{\circ }C$