Question

A pendulum clock gives correct time at ${20}^o$ (at a place where $g=10m/s^2$) The pendulum consists of a light steel rod connected to a heavy ball. If it is taken to a different place $(whereg=10)01m/s^2$ at what temperature the pendulum gives correct time $\left(\alpha ofsteelis{10}^{-5{/}^{o}C}\right)$ [2007 E ]

• A. ${30}^{o}C$
• B. ${60}^{o}C$
• C. ${1000}^{o}C$
• D. ${120}^{o}C$

Answer 1

The correct option is D ${120}^{o}C$

We know that,

For a pendulum clock

$T=2n\sqrt{\frac{l}{g}}\Rightarrow T=2\pi \left(l{)}^{1/2}\right(g{)}^{-1/2}$

So, $\frac{\Delta T}{T}=\frac{\Delta l}{2l}-\frac{\Delta g}{2g}$

Here, $\Delta T=0$ and $\Delta l=l\propto \Delta T$(Thermal expansion)

So,

$\frac{O}{T}=\frac{\Delta l}{2l}-\frac{\Delta g}{2g}$

$\Rightarrow \frac{\Delta g}{g}=\frac{\Delta l}{l}$

$\Rightarrow \frac{\Delta g}{g}=\frac{\alpha e\Delta T}{e}$

$\Rightarrow \frac{10.01-10}{10}=\left({10}^{-5}\right)(T-20)$

$\Rightarrow \frac{{10}^{-3}}{{10}^{-5}}=T-20$

$\Rightarrow T={10}^2+20={120}^{o}C$

Option D is correct.