The variation of the square of the time period ( T^2 ) with the length ( l ) of a simple pendulum is plotted for two different planets, A and B. The resulting graphs are straight lines passing through the origin, making angles of 30^ and 60^ respectively with the l -axis. If a block is dropped from the same height h on both planets, what is the ratio of the time of fall on planet A to that on planet B?

Step-by-step solution

For a simple pendulum, the time period T is given by T = 2 l g . Squaring both sides, we get T^2 = ( 4 ^2 g )l . The graph of T^2 versus l is a straight line with slope m = = 4 ^2 g . Thus, the acceleration due to gravity g 1 . For planet A: g_A 1 30^ = 3 . For planet B: g_B 1 60^ = 1 3 . The ratio of accelerations due to gravity is g_A g_B = 3 1/ 3 = 3 . The time of fall from a height h is t = 2h g , which means t 1 g . Therefore, the ratio of the times of fall is t_A t_B = g_B g_A = 1 3 = 1 3 . The ratio is 1 : 3