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Question
A pendulum clock that keeps correct time on the earth is taken to the moon. It will run
Options
at correct rate
6 times faster
\[\sqrt{6}\] times faster
\[\sqrt{6}\] times slower
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Solution
(d)\[\sqrt{6}\] times slower
The acceleration due to gravity at moon is g/6.
Time period of pendulum is given by,
\[T = 2\pi\sqrt{\frac{l}{g}}\]
Therefore, on moon, time period will be :
Tmoon = \[2\pi\sqrt{\frac{l}{g_{moon}}} = 2\pi\sqrt{\frac{l}{( \frac{g}{6})}} = \sqrt{6}(2\pi\sqrt{\frac{l}{g}}) = \sqrt{6}T\]
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