Question

A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at ${20}^{\circ }C$, how fast or slow will it go in $24hours$ at ${40}^{\circ }C$? Coefficient of linear expansion of iron $=1.2\times {10}^{-6}{/}^{\circ }C$.

• A. $10.4sec$.
• B. $1.04sec$.
• C. $0.52sec$.
• D. $5.02sec$.

Answer 1

Answer: (B)

$1.04sec$.

Time period in a pendulum is given as $T=2\pi \sqrt{\frac{l}{g}}$,

When temperature is increased by $\theta$ new length is $l=l_o(1+\alpha \theta )$. As $T\propto \sqrt{l}$ So new Time-period, $T=T_o\sqrt{(1+\alpha \theta )}=T_o(1+\alpha \theta /2)$, as $\alpha \theta <<1$ is small.

So change in Time for $T_o=24hours=86400seconds$ is $T-T_o=T_o\alpha \theta /2=86400\times (40-20)\times 1.2\times {10}^{-6}/2=1.04sec$