Question

A pendulum clock loses $12s$ a day if the temperature is $40^{\circ }C$ and gains $4s$ 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
$\Delta t=\left(\frac{\Delta T}{T}\right)t=\frac{1}{2}\alpha .\Delta \theta .t$
$\therefore 12=\frac{1}{2}\alpha \left(40-{\theta }_0\right)\times 1d$ .....(1)
$4=\frac{1}{2}\alpha \left({\theta }_0-20\right)\times 1d$ .....(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)
$\alpha =\frac{8}{5\times 29\times 60\times 60}=$