A pendulum clock giving correct time at a place where `g=9.800 ms^-2 is taken to another place where it loses 2 seconds during 24 hours. Find the value of g at this new place.

Answer

`((3600)/(3601))^(2) g = 9.794 m//s^(2)`

Explanation

`T = 2pisqrt((l)/(g)), T' = 2pisqrt((l)/(g'))`
`(T)/(T') = sqrt((g')/(g)) rArr (24 xx 3600)/(24 xx 3600 + 24) = sqrt((g')/(g)) rArr g' = ((3600)/(3601))^(2) g = 9.794 m//s^(2)`