Question

A seconds' pendulum is taken to a planet where the acceleration due to gravity is 4 times than on earth. What would be the time period of the pendulum on the planet?​

Answer 1

$T > {g}^{ - \frac{1}{2} } \\ \\ {T}^{0} > {(4g)}^{ - \frac{1}{2} } \\ \\ {T}^{0} > \frac{ {g}^{ - \frac{1}{2} } }{2} \\ \\ divide \: first \: and \: second \\ {T}^{0} = \frac{T}{2}$

hope it helps.