First slide

Thermal expansion

Question

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 l5°C and l0 seconds slow at a temperature of 30°C. The temperature at which it is designed to given correct time, is

Moderate

A

$18 ° C$
B

$20 ° C$
C

$24^{0} C$
D

$25^{0} C$

Solution

Fractional loss of time per second = $\frac{1}{2} α ∆ T$
Therefore $\frac{1}{2} α (T_{0} - 15) \times (24 hrs) = 5$
and $\frac{1}{2} α (30 - T_{0}) \times (24 hrs) = 10$
on solving $T_{0} = 20^{0} C$

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