A mass falls from a height ' h ' and its time of fall ' t ' is recorded in terms of time period T of a simple pendulum. On the surface of earth it is found that t=2 ~T . The entire set u is taken on the surface of another planet whose mass is half of earth and radius the same. Same experiment is repeated and corresponding times noted as t ^ and T ^ .

Step-by-step solution

The distance covered by the mass falling from height ' h ' during its time of fall ' t ' is given by array r s = h = ut + 1 2 gt ^2 As, u =0 h = 1 2 gt ^2 t = 2 ~h ~g array The time period of simple pendulum is T =2 l g where, lis the length of the pendulum. From Eq. (i) and (ii), since ' h ' and ' l ' are constant so, we can conclude that, t l g and T l g t T = l Thus, the ratio of time of fall and time period of pendulum is independent of value of gravity ( g ) or any other parameter like mass and radius of the p