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. 8.4 s
B. 8.2 s
C. 7.4 s
D. 6.4 s

Answer

A

Explanation

Given, `g_(m)=1.7 ms^(-2),g_(e)=9.8 ms^(-2)`
`T_(m) = ? and T = 3.5 s`
As `T_(e)=2pi sqrt((l)/(g_(e)))`
and `T_(m)=2pi sqrt((l)/(g_(m)))`
`therefore" "(T_(m))/(T_(e))=sqrt(g_(e)/(g_(m))) or T_(m)=T_(e) sqrt((g_(e))/(g_(m)))`
`rArr" "T_(m) = 3.5 sqrt((9.8)/(1.7))=8.4 s`