##### An object is released from rest. The time it takes to fall through a distance h and the speed of the object as it falls through this distance are measured with a pendulum clock. The entire apparatus is taken on the moon and the experiment is repeated

The time period of a simple Pendulum is given by the Relation,

On the moon, g changes thus Time period of the Pendulum changes.

Therefore, actual time, times in the fall will not be equal.

Therefore, Option (c). is incorrect.

Also,

V2 - u2 = 2gh

Now, Velocity will also be different on the moon. Hence, Actual speeds will not be equal.

Therefore, Option (d) is incorrect.

Now, the measured time and speed will be same. This is because, if the particle covers h distance in time t, then it will cover the h distance on moon in time t.

Let's find the time taken by the body to fall freely, this is what measured by the Clock.

We have the equation,

The time a pendulum measure,

As previously the time taken by body to fall freely is equal to the time measured by the simple pendulum

And from here we can see that the time is changing by the same factor in both the cases so, the time that the clock measures will always be equal to the time taken by the particle to fall freely perfect irrespective of acceleration due to gravity.

Therefore, Option (a). and (b). both are correct.

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