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's surface =9.8 ?

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 moon is one-sixth of that on the earth, what will be the length of a second pendulum there ? Take g = 9.8 ms^-2

Assertion: The acceleration due to gravity on the moon is one-sixth that on the earth. Reason : The law of gravitation is the same on both the moon and the earth. .

Assertion: The acceleration due to gravity on the moon is one-sixth that on the earth. Reason : The law of gravitation is the same on both the moon and the earth.

The acceleration due to gravity on the surface of moon is 1.7 ms ^-2 . What is the time period of a simple pendulum on the surface of moon if its time period of the surface of earth is 3.5 s ? (g on the surface of the earth is 9.8 ms^-2