Discussion Forum

A pendulum clock loses  12 s a day if the temperature is 40° C and gains 4 s a day if the temperature is 20° C. The temperature at which the clock will show the correct time and the coefficient of linear expansion α of the metal of the pendulum shaft are, respectively

Answer:

Explanation:

$T_{0}=2\pi\sqrt{\frac{L}{g}}$

$T^{'}=T_{0}+\triangle T=2\pi\sqrt{\frac{L+\triangle L}{g}}$

$T^{'}=T_{0}+\triangle T=2\pi\sqrt{\frac{L(1+\propto\triangle\theta) }{g}}$

$= \left\{2\pi \sqrt{\frac{L}{g}}\right\} (1+\alpha \triangle \theta)^{\frac{1}{2}}=T_{0}(1+\frac{\alpha\triangle\theta}{2})$

$\therefore$       $\triangle T=T'-T_{0}=\frac{\alpha\triangle \theta T_{0}}{2}$

$        \frac{\triangle T_{1}}{\triangle T_{2}}=\frac{\alpha \triangle \theta_{1}T_{0}}{\alpha \triangle \theta_{2}T_{0}}$

$\Rightarrow$   $\frac{12}{4}=\frac{40^{0}-\theta}{\theta-20^{0}}$

$\Rightarrow$   $3(\theta -20^{0})=40^{0}-\theta$

$\Rightarrow$     $4\theta=100^{0}$

$\Rightarrow$     $\theta=25^{0}C$

 Time gained or lost given by

    $\triangle T= (\frac{\triangle T}{T_{0}+\triangle T})t=\frac{\triangle t}{T_{0}}t$

From Eq.(i0.

   $\frac{\triangle T}{T_{0}}=\frac{\alpha \triangle \theta}{2}$

$\triangle t= \frac{\alpha (\triangle \theta)t}{2}$

$12= \frac{\alpha(40^{0}-25^{0})(24\times 3600)}{2}$

$\alpha =1.85\times 10^{-5}/^{0}C$