The acceleration due to gravity on a planet is `3//2` times that on the earth. If length of a seconds pendulum on earth is `1m`, length of second pengulum on surface of planet is

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

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

If acceleration due to gravity on the moon is one-sixth that on the earth, then the length and time period of seconds pendulum on the surface of moon are

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 surface =9.8 ms^(-2) .

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 ?