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A pendulum clock (fitted with a small heavy bob that is connected with a metal rod) is 5 seconds fast each day at a temperature of $15^{o} C$ and 10 seconds slow at a temperature of $30^{o} C$ . The temperature at which it is designed to give correct time, is

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$18^{o} C$

$20^{o} C$

$24^{o} C$

$25^{o} C$

answer is B.

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Detailed Solution

$Fractional loss of time per second = \frac{1}{2} α ∆ T$

$\begin{array}{l} \frac{1}{2} α (T_{0} - 15) \times 24 hr = 5 ........... (i) \\ \frac{1}{2} α (30 - T_{0}) \times 24 hr = 10 .......... (ii) \end{array}$

Solving equations (i) and (ii), we get $T_{0} = 20^{0} C$

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