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Thermodynamics

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

$(a)\;{2.6 \times 10^{-5}} ^{0}{C^{-1}}\qquad(b)\;{1.6 \times 10^{-5}} ^{0}{C^{-1}}\qquad(c)\;{2.7 \times 10^{-5}} ^{0}{C^{-1}}\qquad(d)\;{1.7 \times 10^{-5}} ^{0}{C^{-1}}$

Answer : (d) $\;{1.7 \times 10^{-5}} ^{0}{C^{-1}}$
Explanation :
$T_{0}=2 \pi \sqrt{\large\frac{L_{0}}{g}}$
$T=2 \pi \sqrt{\large\frac{L}{g}}=2 \pi \sqrt{\large\frac{l_{0}(1+\alpha \bigtriangleup \theta)}{g}} \approx 2 \pi \sqrt{\large\frac{l_{0}}{g}}(1+\large\frac{\alpha}{2}\bigtriangleup \theta)$
For $\; \alpha \bigtriangleup \theta \to 0$
$T=T_{0}(1+\large\frac{\alpha \bigtriangleup theta}{2})-----(1)$
It gains 15 sec/24 hrs =>$\; 15 sec / 360\times 24$
From $\;(1) \to \large\frac{\bigtriangleup T}{T_{0}}=\large\frac{15}{3600 \times 24 } \times \large\frac{\alpha \bigtriangleup \theta}{2}\;,\bigtriangleup \theta=20$
$\alpha=\;{1.7 \times 10^{-5}} ^{0}{C^{-1}}$