The length of a second's pendulum on the earth is L .What would be its length at a planet where the acceleration due g to gravity is `(g)/(4)` ?

The length of a second's pendulum on the surface of Earth in 1m. What will be the length of a second's pendulum on the moon?

The length of a second's pendulum at the equator is L. The length of second's pendulum at the North pole will be

The period of a conical pendulum in terms of its length (l) , semivertical angle ( theta ) and acceleration due to gravity (g) is

What would be the time period of a seconds pendulum constructed on the earth if it is taken to the surface of the moon. The acceleration due to gravity on the surface of the moon is 1//6g_("earth")

Time period of simple pendulum of length l at a place, where acceleration due to gravity is g and period T. Then period of simple pendulum of the same length at a place where the acceleration due to gravity 1.02 g will be

A man can jump 2 m high on the surface of the earth. How high he can jump on a planet where the acceleration due to gravity is (g)/(12) , where g is the acceleration due to gravity on the surface of the earth ?

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's surface =9.8 ?

Calculate the length of the second's pendulum on the surface of moon where acceleration due to gravity is one sixth as that on earth.

Let g be the acceleration due to gravity on the earth's surface.