The acceleration due to gravity on the surface of the moon is 1.7ms^(-2). What is the time period of a simple pedulum on the surface of moon it its time period on the surface earth is 3.5s? (g on the surface earth 9.8ms^(-2))

The acceleration due to gravity on the surface of the moon is 1.7ms^(-

1.	The acceleration due to gravity on the surface of the moon is 1.7ms^(-2). What is the time period of a simple pedulum on the surface of moon it its time period on the surface earth is 3.5s? (g on the surface earth 9.8ms^(-2))
Answer» SOLUTION :Here `g_(m)=1.7ms^(-2),g_(e)=9.8ms^(_2),T_(e)=3.5s^(-1)` As `T_(e)=2pisqrt(L/(g_(e)))"and" T_(m)=2pisqrt(l/(g_(m)))` `:.(T_(m))/(T_(e))=SQRT((g_(e))/(g_(m)))"or" T_(m)=T_(e)sqrt((g_(e))/(g_(m)))=3.5sqrt(9.8/1.7)=8.4s`