What would be the time period of a seconds pendulum constructed on the earth if it is taken to the surface of the moon. The acceleration due to gravity on the surface of the moon is `1//6g_("earth")`

Explanation

(i) Find the length and time period of a seconds pendulum on the surface of the earth.
Then, the formula for time period (T) is given by, `T = 2pi sqrt((l)/(g))`
Take the time of seconds pendulum as `'T_(E)' and 'T_(M)'` on the surface of the earth and the earth and the moon, respectively.
Take the values of acceleration due to gravity on the surface of the earth and the moon as `g_(E)' and 'g_(M)'` respectively.
Given that `g_(E) = 1//6 g_(M)`
Now, `T_(E) = 2pi sqrt((l)/(g_(E))) and T_(M)`
`= 2pi sqrt((l)/(g_(M)))`
Take, `T_(E) = 2s`
Substitute the value of (1) in (2) and find the value of `T_(M)`.
(ii) `4.9s`