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`

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 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 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

A second pendulum is moved to moon where acceleration dur to gravity is 1/6 times that of the earth, the length of the second pendulum on moon would be

A pendulum clock Which ticks 60 times is one minute on earth is placed on the moon where the acceleration due to gravity is only one - sixth of that one the earth . How many times the clock will tick over there in one minute ?

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 time period of a simple pendulum is given by T = 2pi sqrt((l)/(g)) where 'l' is the length of the pendulum and 'g' is the acceleration due to gravity at that place. (a) Express 'g' in terms of T (b) What is the length of the seconds pendulum ? (Take g = 10 m s^(-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.

A second's pendulum on the earth is taken to the moon, where acceleration due to gravity is nearly 1/6 th that on the earth. If it is to act as second's pendulum there too on the moon, its length should be