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)`.

Explanation

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`