Questions
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
detailed solution
Correct option is B
Fractional loss of time per second = 12α∆TTherefore 12α(T0-15)×(24 hrs) = 5 and 12α(30-T0)×(24 hrs) = 10on solving T0 = 200CTalk to our academic expert!
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A rotating solid metal sphere is heated by . If then percentage change in its moment of inertia is
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