If the acceleration due to gravity on the moon is one-sixth of that on the earth. What will be the change in length of a second pendulum there so that it may beat a second there ? Take acceleration due to gravity on earth surface =9.8 ms^(-2).

If the acceleration due to gravity on the moon is one-sixth of that on

1.	If the acceleration due to gravity on the moon is one-sixth of that on the earth. What will be the change in length of a second pendulum there so that it may beat a second there ? Take acceleration due to gravity on earth surface =9.8 ms^(-2).
Answer» Solution :On earth `g=9.87 m//s^(2)`. Let L be the LENGTH of second pendulum . Then `T=2s` `T=2pi sqrt((l)/(g))` (or) `l=(T^(2)g)/(4pi^(2))=(2^(2)xx9.8)/(4XX(22//7)^(2))` `=0.9921 m =99.21 CM` On moon, `g_(m)=(1)/(6)g=(9.8)/(6) ms^(-2)` `l'=(T^(2)g_(m))/(4pi^(2))=(2^(2)xx(9.8//6))/(4xx(22//7)^(2))` `=.1654 m=16.54 m` change4 in length `=l-l'=99.21-16.54` `=82.67 cm`