Question

A pendulum clock having copper rod keeps correct time at ${20}^{\circ }$C. It gains 15 seconds per day if cooled to $0^{\circ }$C. Calculate the coefficient of linear expansion of copper.

• A. $1.7\times {10}^{-5}{}^{\circ }{C}^{-1}$
• B. $2.7\times {10}^{-5}{}^{\circ }{C}^{-1}$
• C. $3.7\times {10}^{-5}{}^{\circ }{C}^{-1}$
• D. $4.7\times {10}^{-5}{}^{\circ }{C}^{-1}$

Answer 1

The correct option is A $1.7\times {10}^{-5}{}^{\circ }{C}^{-1}$

Time period at any temperature T is
$t_T=2\pi \sqrt{\frac{l_T}{g}};t_0=2\pi \sqrt{\frac{l_0}{g}}\frac{t_T}{t_0}=\sqrt{\frac{l_T}{l_0}}=\sqrt{\frac{l_0[1+\alpha (\Delta T\left)\right]}{l_0}}\Rightarrow \frac{t_T}{t_0}\sqrt{1+\alpha T}\approx 1+\frac{\alpha }{\alpha }T\frac{t_T}{t_0}-1=\frac{\alpha }{2}T\Rightarrow \frac{t_T-t_0}{t_0}=\frac{\alpha T}{\alpha }\Rightarrow \frac{t_{20}-t_0}{t_0}=\frac{\alpha T}{\alpha }=\alpha \times {10}^{\circ }C$
Time gained in 24 hours, $\Delta T=24hrs\times (\alpha .10)$
$\Rightarrow 15s=24\times 60\times 60\times \alpha \times 10\Rightarrow \alpha =1.7\times {10}^{-5}{}^{\circ }{C}^{-1}$