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.

Options

A. 6.4 s
B. 7.4 s
C. 8.4 s
D. 9.4 s

Answer

D

Explanation

Here, `g_(m)=1.7m s^(-2),g_(e)=9.8m s^(-2),T_(e)=3.5 s`
As `T_(e)=2pisqrt((l)/(g_(e)))andT_(m)=2pisqrt((l)/(g_(m)))`
`therefore (T_(m))/(T_(e))=sqrt((g_(e))/(g_(m)))orT_(m)=T_(e)sqrt((g_(e))/(g_(m)))=3.5sqrt((9.8)/(1.7))=8.4s`