Question

A clock pendulum having coefficient of linear expansion, $\alpha = 9 \times 10^{-7}/^°C^{-1}$ has a period of $0.5\, s$ at $20^°C$. If the clock is used in a climate, where the temperature is $30^°C$, how much time does the clock lose in each oscillation? ($g =$ constant)

• A. $25 \times 10^{-7}\, s$
• B. $5 \times 10^{-7}\, s$
• C. $1.125 \times 10^{-6}\, s$
• D. $2.25 \times 10^{-6}\, s$

Answer 1

Answer: (D)

Given, the coefficient of linear expansion,
$\alpha=9 \times 10^{-7}{ }^{\circ} C ^{-1}$
Initial time period, $T_{0}=05 s$, initial temperature $T_{i}=20^{\circ} C$ and final temperature, $T_{f}=30^{\circ} C$
Expansion in length, $\Delta l=l \alpha\left(T_{f}-T_{i}\right)$
$\Rightarrow \Delta l=l \times 9 \times 10^{-7}(30-20)$
Now, the time period of pendulum,
$T=2 \pi \sqrt{\frac{l}{g}}$
Error in time period,
$\frac{\Delta t}{T}=\frac{1}{2} \frac{\Delta l}{l}+\frac{1}{2} \frac{\Delta g}{g}$
Since, $\Delta g=0$
$\Rightarrow \frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta l}{l}$
Now, substituting values in the above equation, we get,
$\Rightarrow \frac{\Delta T}{0.5}=\frac{1}{2}\left[\frac{l \times 9 \times 10^{-7}}{l}(30-20)\right]$
$\Rightarrow \Delta T=\frac{0.5 \times 2 \times 10^{-7} \times 10}{2}$
$\Rightarrow \Delta T=2.86 \times 10^{-6} s$