Question

A second's [pendulum clock having steel wire is calibrated at ${20}^{\circ }C$. When temperature is increased to ${30}^{\circ }C$ , then how much time does the clock lose or gain in one week ?

Given: ${\alpha }_{steel}=1.2\times {10}^{-5}{(}^{0}C{)}^{-1}$

• A. 0.3628 s
• B. 3.626 s
• C. 362.8 s
• D. 36.28 s

Answer 1

The correct option is D 36.28 s

Let $T_0$ be the initial time period=$2\pi \sqrt{\frac{l_0}{g}}$

Hence, $T_1=2\pi \sqrt{\frac{l_0(1+\alpha t)}{g}}$

$⟹T_1-T_0=T_0\left(\frac{\alpha t}{2}\right)$

This much time is lost in one second, since period $T_0$ is 1s.

Hence total time lost=$\frac{\alpha t}{2}\times 7\times 24\times 60\times 60=36.28s$