question_answer

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

Answer:

                \[T=2\pi \sqrt{\frac{l}{g}}\] In both the cases, T is same so that \[l\propto g\] On the moon, the value of acceleration due to gravity is one-sixth of that on the surface of earth. So the length of second?s pendulum is \[\frac{1}{6}\]m.