A pendulam which keeps correct time at a place where g= 9.81`ms^(-2)` it taken to a place where g=9.8 `ms^(-2)`. Calculate by how much it will lose or gain in a day.

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.

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.

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

A second's pendulum is taken from a place where g=9.8 ms^(-2) to a place where g=9.7 ms^(-2) .How would its length be changed in order that its time period remains unaffected ?

A simplependulam beats seconds on the earth (g=9.8 ms^(-2) ) is taken to the moon where g=3.96 ms^(-2) . Find the number of seconds it will lose per day.

A second's pendulum is oscillating at a place where g = 9.8 m//s^(2) . If the pendulum is shifted to a place where g = 9.6 m//s^(2) , then to keep this time period unchanged, how much change of length should be introduced ?

A seconds pendulum is shifted from a place where g = 9.8 m//s^(2) to another place where g = 9.78 m//s^(2) . To keep period of oscillation constant its length should be

Calculate the length of a seconds' pendulum at a place where g= 9.8 m s^(-2)

The weight of a body is 9.8 N at the place where g=9.8 ms^(-2) . Its mass is

A second pendulum is shifted from a plane where g = 9.8 m//s^(2) to another place where g = 9.82 m//s^(2) . To keep period of oscillation constant its length should be