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Q.

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

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a

18°C

b

20°C

c

240C

d

250C

answer is B.

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Detailed Solution

Fractional loss of time per second = 12α∆TTherefore 12α(T0-15)×(24 hrs) = 5  and 12α(30-T0)×(24 hrs) = 10on solving T0 = 200C
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