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 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

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

The length of seconds pendulum at a place wl}ere g = 4.9 m/ s^2 is

The length of seconds pendulum at a place wl}ere g = 4.9 m/ s^2 is

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 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 ?

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

A second's pendulum is taken to a height where the value of g is 4.36ms^(-2) . What will be its new time period?

Find the length of seconds pendulum at a place where g = 10 m/s^2 .

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