

InterviewSolution
Saved Bookmarks
1. |
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. 8.4 sB. 8.2 sC. 7.4 sD. 6.4 s |
Answer» Correct Answer - A 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` |
|