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The acceleration due to gravity on the surface of the moon is `1.7ms^(-2)`. What is the time perioid of a simple pendulum on the surface of the moon, if its time period on the surface of earth is `3.5s ?` Take `g=9.8ms^(-2)` on the surface of the earth.A. 4.4 sB. 8.4 sC. 16.8 sD. |
Answer» Correct Answer - C For moon, `g_(m)=1.7ms^(-2)` for earth, `g_(e)=9.8ms^(-2),T_(e)3.5s` But, `T_(m)=2pisqrt((l)/(g_(m))) and T_(e)=2pisqrt((l)/(g_(e))) therefore(T_(m))/(T_(e))=sqrt((g_(e))/(g_(m)))` or `T_(m)=sqrt((g_(e))/(g_(m)))xxT_(e)=sqrt((9.8)/(1.7))xx3.5=8.4s` |
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