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")

What would be the time period of a seconds pendulum constructed on the

1.	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")
Answer» Solution :(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`