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

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

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

The acceleration due to gravity on the surface of earth varies

The acceleration due to gravity on the surface of earth varies

What is the value of acceleration due to gravity on the surface of earth ?

The value of acceleration due to gravity at the surface of earth

The value of acceleration due to gravity at the surface of earth