Question

A pendulum clock loses $12\,s$ a day if the temperature is $40^{\circ}C$ and gains $4\,s$ a day if the temperature is $20^{\circ}C$. The temperature at which the clock will show correct time, and the co-efficient of linear expansion $(\alpha)$ of the metal of the pendulum shaft are respectively :

• A. $25^{\circ} C ; \alpha = 1.85 \times 10^{-5} / {^{\circ}C}$
• B. $60^{\circ} C ; \alpha = 1.85 \times 10^{-4} / {^{\circ}C}$
• C. $30^{\circ} C ; \alpha = 1.85 \times 10^{-3} / {^{\circ}C}$
• D. $55^{\circ} C ; \alpha = 1.85 \times 10^{-2} / {^{\circ}C}$

Answer 1

Answer: (A)

Time loss or gain is given by
$\Delta t = \left(\frac{\Delta T}{T}\right)t = \frac{1}{2} \alpha. \Delta\theta.t $
$\therefore \, 12 = \frac{1}{2} \alpha\left(40 - \theta_{0}\right) \times1d$ .....(1)
$ 4 = \frac{1}{2} \alpha\left( \theta_{0} - 20\right)\times1d $ .....(2)
$\frac{(1)}{(2)}$ gives
$3 = \frac{\left(40 - \theta_{0}\right)}{\left(\theta_{0} - 20\right)} $
Solving $\theta_{0} = 25^{\circ} C$ and putting in (2)
$ \alpha = \frac{8}{5 \times 29\times 60\times 60} = $