Question

A clock with a metallic pendulum at ${15}^{\circ }C$ runs faster by $5s$ each day and at ${30}^{\circ }C$, runs slow by $10s$. Find the coefficient of linear expansion of the metal. (nearly in ${10}^{-6}{/}^{\circ }C$)

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is A 2

Time period $T=2\pi \sqrt{\frac{l}{g}}$
or $T^2=4{\pi }^2\left(l/g\right)$
or $2TdT=\frac{4{\pi }^2}{g}dl=\left(T^2/l\right)dl$
or $\frac{dT}{T}=\frac{1}{2}\frac{dl}{l}=\frac{1}{2}\frac{l\alpha \Delta t}{l}$
Loss or gain per day $=\frac{1}{2}\alpha \Delta t\times T=\frac{1}{2}\alpha \Delta t\times 86400$
At ${15}^{o}C$ the gain : $5=\frac{\alpha }{2}(t-15)\times 86400...\left(1\right)$
At ${30}^{o}C$ the loss : $10=\frac{\alpha }{2}(30-t)\times 86400...\left(2\right)$ where t is the temperature at which clock gives correct time.
$\left(2\right)/\left(1\right)\Rightarrow 2=\frac{30-t}{t-15}$ or $t={20}^{o}C$
putting the value of t in (1) we get, $\alpha =2.31\times {10}^{-5}{/}^{o}C$