00:01
Question number 68 talks about a simple pendulum of length l, of length l and the time period of that pendulum is given as t on earth, that is taken onto the surface of the moon.
00:13
So how should the length of the pendulum be changed and the most of the time period remains constant and the gravity on the earth is six times the gravity on the moon.
00:22
All right, so first we should know that the formula for the time period is 2 pi square root of l over g.
00:29
Now if we talk about earth, so let's put a subscript of that's put an e over here to represent that this is for the earth.
00:42
So time period will be 2 pi square root of l over g .e.
00:46
And if we take this on the moon, then the time period of the moon is going to be 2 pi square root of let's call this l1 and let's call this l2 because we have to change the length.
00:59
Now, these two should be same because the time period is constant.
01:05
So t is equal to t m.
01:07
It means that 2 pi square root of l1 over g e is going to be same as 2 pi square root of l2 over gm...