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Thermodynamics

# A pendulum clock is made of iron rod $\;(\alpha_{iron}={10^{-5}}^{0}C^{-1})\;.$ At room temperature $\;(15^{0}C)\;,$ it works correctly . The error when the temperature rises to $\;24^{0}C\;$

$(a)\;7.77\;sec\;per\;day\qquad(b)\;77.7\;sec\;per\;day\qquad(c)\;3.88\;sec\;per\;day\qquad(d)\;38.8\;sec\;per\;day$

Answer : (c) $\;3.88\;sec\;per\;day$
Explanation :
$T=2 \pi \sqrt{\large\frac{l_{0}}{g}}=2 \pi \sqrt{\large\frac{l_{0}(1+\alpha \bigtriangleup \theta)}{g}}=2 \pi \sqrt{\large\frac{l_{0}}{g}}\;(1+\large\frac{\alpha \bigtriangleup \theta}{2})$
$T=T_{0}\;(1+\large\frac{\alpha}{2} \bigtriangleup \theta)$
$\large\frac{\bigtriangleup T}{T_{0}}=\large\frac{\alpha}{2} \bigtriangleup \theta$
$=\large\frac{10^{-5}}{2}\times9=4.5 \times 10^{-5}$