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The acceleration due to gravity on the s...

The acceleration due to gravity on the surface of the moon is 1.7 `ms^(-2)`. The time period of a simple pendulum on the moon if its time period on the earth is 3.5 s is (Given, `g=9.8ms^(-2)`)2.2 s

Text Solution

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`g_(m)= 1.7ms^(-2), g_(e)= 9.8 ms^(-1) , T_(e)= 3.5 s^(-1)`
As `T_(e)= 2pisqrt((l)/(g_(e)))` and `T_(m)= 2pisqrt((l)/(g_(m)))`
`:. (T_(m))/(T_(e))= sqrt((g_(e))/(g_(m)))` or `T_(m)= T_(e) sqrt((g_(e))/(g_(m)))= 3.5 sqrt((9.8)/(1.7))= 8.4 s`
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