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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 = 2T. The entire set up is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as t’ and T’. Then we can say

1. t’ = 2T’
2. t’ = T’
3. t’ < 2T’
4. t’ > 2T’

Correct answer: t’ = 2T’

Solution

The time of fall, t, depends on the acceleration due to gravity, g, which is proportional to the mass of the planet. On the new planet, g is halved, so the time of fall t' will increase by a factor of √2. Similarly, the time period T' of the pendulum also increases by √2. Thus, the ratio t'/T' remains the same, and t' = 2T'.

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