Question

A clock with a metallic pendulum gains $6\text{s}$ each day when the temperature is $20{}^{\circ }\text{C}$ and loses $6\text{s}$ when the temperature is $40{}^{\circ }\text{C}$. Find the coefficient of linear expansion of the metal.

• A. $1.4\times {10}^{-5}{}^{\circ }{\text{C}}^{-1}$
• B. $2.8\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$
• C. $1.4\times {10}^{-4}{}^{\circ }{\text{C}}^{-1}$
• D. $2.8\times {10}^{-5}{}^{\circ }{\text{C}}^{-1}$

Answer 1

The correct option is A $1.4\times {10}^{-5}{}^{\circ }{\text{C}}^{-1}$

Time period of pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

$\Rightarrow T^2=4{\pi }^2\times \frac{L}{g}$

Taking $\log$ both side and then differentiating, we get

$2\frac{\Delta T}{T}=\frac{\Delta L}{L}=\alpha \Delta \theta$

$\Delta T\to \text{change in time}$
$T\to \text{original time period}$
$\Delta \theta \to \text{change in temperature}$
$\alpha \to \text{coefficient of linear expansion}$

Let $\theta$ be the temperature at which the clock runs correctly

For gain in time,

$\frac{\Delta T_1}{T}=\frac{1}{2}\times \alpha \times (\theta -{20}^{\circ }\text{C})......\left(1\right)$

For loss in time,

$\frac{\Delta T_2}{T}=\frac{1}{2}\times \alpha \times ({40}^{\circ }\text{C}-\theta )........\left(2\right)$

Dividing equation $\left(1\right)$ and $\left(2\right)$,

$\Rightarrow \frac{6}{6}=\frac{\theta -20}{40-\theta }$

$\Rightarrow \theta -20=40-\theta$

$\Rightarrow \theta =30{}^{\circ }\text{C}$

Substituting these values in eq $\left(1\right)$,

$\Rightarrow \frac{6}{24\times 3600}=\frac{1}{2}\times \alpha \times (30-20)$

$\therefore \alpha =1.4\times {10}^{-5}{}^{\circ }{\text{C}}^{-1}$

Hence, option $\left(\text{A}\right)$ is correct.