Question

A seconds pendulum clock made up of a wire shows correct time at $25{}^{\circ }\text{C}$. How much time does the clock lose or gain, in one week, when the temperature is increased to $35{}^{\circ }\text{C}$?
$[\alpha =1.2\times {10}^{-5}/{}^{\circ }\text{C}]$

• A. $26.29\text{s}$
• B. $66.29\text{s}$
• C. $56.29\text{s}$
• D. $36.29\text{s}$

Answer 1

The correct option is D $36.29\text{s}$

Given:
Time period of clock pendulum, $T=1\text{s}$
Change in temperature, $\Delta \theta =35{}^{\circ }\text{C}-25{}^{\circ }\text{C}=10{}^{\circ }\text{C}$
Coefficient of linear expansion of wire, $\alpha =1.2\times {10}^{-5}/{}^{\circ }\text{C}$
Time interval, $t=1$ week $=1\times 7\times 24\times 60\times 60=604800\text{s}$
To find:
Time lose or gain by clock, $\Delta t=?$

We know that,
Fractional change in time period is given by
$\frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta \theta$
$=\frac{1}{2}\times 1.2\times {10}^{-5}\times 10=6\times {10}^{-5}$
$\Rightarrow \Delta T=6\times {10}^{-5}T=6\times {10}^{-5}\times 1=6\times {10}^{-5}$ [$t=1\text{s}$ for seconds pendulum]
Hence, time lost in 1 week
$\Delta t=t\times \Delta T$
$=604800\times 6\times {10}^{-5}=36.29\text{s}$

We know time period of a pendulum is given by
$T=2\pi \sqrt{\frac{l}{g}}$
$\Rightarrow T\propto l^{1/2}$
Length increases with increase in temperature, therefore time period increases hence time is lost.