Question

A clock pendulum having coefficient of linear expansion, $\alpha =9\times 10^{-7}{/}^{°}{C}^{-1}$ has a period of $0.5s$ 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, $T_0=05s$, 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,
$\Rightarrow \frac{\Delta T}{0.5}=\frac{1}{2}\left[\frac{l\times 9\times 10^{-7}}{l}(30-20)\right]$
$\Rightarrow \Delta T=\frac{0.5\times 2\times 10^{-7}\times 10}{2}$
$\Rightarrow \Delta T=2.86\times 10^{-6}s$