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A clock with a metal pendulum beating seconds keep correct time at $0^{\circ}C$ . If it loses 12.5 s a day at $25^{\circ}C$ the coefficient of linear expansion of metal pendulum is
Option: 1
$\frac{1}{86400}/^{\circ}C$

Option: 2
$\frac{1}{43200}/^{\circ}C$

Option: 3
$\frac{1}{14400}/^{\circ}C$

Option: 4
$\frac{1}{28800}/^{\circ}C$

Answers (1)

best_answer

As we learned

Effect of temperature on Simple Pendulum -

$\frac{\Delta T}{T}=\frac{1}{2}\;\alpha\; \Delta \theta$

- wherein

$\Delta \theta=\theta-\theta_{0}$

$\Delta \theta=$ Change in temperature

$\frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta \Theta$ or $\Delta T=\frac{1}{2}\alpha \Delta \Theta \cdot T$

$\Rightarrow 12.5 = \frac{1}{2}\times \alpha \times 25\times 86400$

or $\alpha =\frac{1}{86400}/^{\circ}C$

Posted by

Gautam harsolia

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