Question

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

• A. 2.08 sec, slow
• B. 2.08 sec, fast
• C. 1.04 sec, slow
• D. 1.04 sec, fast

Answer 1

The correct option is C 1.04 sec, slow

$\Delta t=\frac{1}{2}\alpha \Delta \theta t=\frac{1}{2}\times \alpha \times \Delta \theta \times 86400$