The length of a second's pendulum on the surface of the moon, where g is `1//6^(th)` if the value of g on the surface of the

The length of a seconds pendulum on the surface of the earth is

The value of g on the surface of the moon

The length of a second's pendulum on the surface of the earth, where g=9.8 m//s^(2) , is approximately equal to

The acceleration due to gravity on the moon is (1)/(6) th the acceleration due to gravity on the surface of the earth. If the length of a second's pendulum is 1 m on the surface of the earth, then its length on the surface of the moon will be

Where is the maximum value of g on the surface of Earth. Why?

The length of a second’s pendulum at the surface of earth is 1 m. The length of second’s pendulum at the surface of moon where g is 1/6th that at earth’s surface is