Question

A pendulum clock lose $12sa$ day if the temperature is ${40}^{\circ }C$ and gains $4sa$ day if the temperature is ${20}^{\circ }C$ The temperature at which the clock will show correct time and the co-efficient of linear expansion $\left(\alpha \right)$ of the metal of the pendulum shaft are respectively

• A. ${60}^{\circ }C;\alpha =1.85\times {10}^{-4}{/}^{0}C$
• B. ${30}^{\circ }C;\alpha =1.85\times {10}^{-3}{/}^{0}C$
• C. ${55}^{\circ }C;\alpha =1.85\times {10}^{-2}{/}^{0}C$
• D. ${25}^{\circ }C;\alpha =1.85\times {10}^{-5}{/}^{0}C$

Answer 1

The correct option is D ${25}^{\circ }C;\alpha =1.85\times {10}^{-5}{/}^{0}C$

Time gain or loss is given by,

$\Delta t=\left(\frac{\Delta T}{T}\right)t=\frac{1}{2}\alpha \theta t12=\frac{1}{2}\alpha (40-{\theta }_o)\times id-----\left(1\right)4=\frac{1}{2}\alpha ({\theta }_o-20)\times id-----\left(2\right)$

$\frac{\left(1\right)}{\left(2\right)}$ gives $\left(3\right)=\frac{(40-{\theta }_o)}{({\theta }_o-20)}$

Solving, ${\theta }_o={25}^{\circ }$ and putting in (2), we get,

$\alpha =\frac{8}{5\times 29\times 60\times 60}=1.85\times {10}^{-5}{℃}^{-1}$