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A pendulum clock is 5 sec. Slow at a temperature `30^(@)C` and `10` sec. fast at a temperature of `15^(@)C`, At what temperature does it give the correct time-A. `18^(@)C`B. `20^(@)C`C. `25^(@)C`D. None of these |
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Answer» Correct Answer - C Let correct time at `theta^(@)C` & `15^(@) lt theta lt 30^(@)` apply `Deltat = 1/2 alpha Deltatheta.t` for slow: `5 = 1/2 alpha xx (30^(@) - theta).t "….."(1)` for fast : `10 = 1/2 alpha xx (theta - 15)t "……"(2)` By `(1)(2) : theta = 25^(@)C` |
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