Question

The length of second's pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity (g) on moon is 1/6 th of that on the earth’s surface]

(A) 1/6 m     

(B) 6 m     

(C) 1/36 m     

(D) \(\frac{1}{\sqrt{6}}m\)  

Answer 1

A) 1/6 m 

Its length on the surface of moon should be 1/6 m