The acceleration due to gravity on the surface of the moon is `1.7ms^(-2)`. What is the time period 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.

Options

A. 4.4 s
B. 8.4 s
C. 16.8 s
D. `5 s`

Answer

B

Explanation

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`