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

Given that $g=9.8m/s^2$ and $△g=-0.012m/s^2$

Time period of pendulum is given by :

$T=2\pi \sqrt{\frac{L}{g}}$

We get,

$\frac{△T}{T}=\frac{1}{2}(\frac{△L}{L}-\frac{△g}{g})\to \left(L\right)$

using linear expansion of steel rod,

$\frac{△L}{L}=\alpha △T$ ($△T:$change in temperature)

So, maintain its time $(i.e.△T=0)$ we get-

$\frac{△L}{L}=\frac{△g}{g}$

$\Rightarrow \alpha △T=\frac{△g}{g}$

$\Rightarrow (T-20)=\frac{-0.012}{9.8\times 12\times {10}^{-6}}$

$\Rightarrow T-20\approx 102.04\Rightarrow -{82.04}^{0}c$