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 = 2 pi sqrt((l )/(g ))`
Take the time of seconds pendulum as `.T_E ` and *`T_M`. on the surface of the Earth and the moon, respectively. Take the values of acceleration due to gravity on the surface of the Earth and the moon as .ge. and 8M , respectively.
Given that `g_E = 1/6 g_M`
Now ` T_E = 2 pi sqrt((l )/(g_E))` and ` T_M`
` T_M = 2 pi sqrt((l )/( g_M))`
Take ` T_E = 2s `
Substitute the value of (1) in (2) and find the value of `T_M`