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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

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$t' = 2 T'$

$t' > 2 T'$

$t' = \sqrt{2} T'$

$t' < 2 T'$

answer is A.

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Detailed Solution

For surface of earth time taken in falling h distance.

$t = \sqrt{\frac{2 h}{g}}$

$and T = 2 π \sqrt{\frac{I}{g}}$

Given t = 2T

$\frac{t}{T} = 2$

For surface of other planet

$g^{'} = \frac{g}{2}$

Time taken in falling h distance

$t^{'} = \sqrt{\frac{2 h}{g^{'}}} = \sqrt{2} t$

$and T^{'} = 2 π \sqrt{\frac{l}{g^{'}}} = \sqrt{2} T$

$Here \frac{t^{'}}{T^{'}} = \frac{\sqrt{2} t}{\sqrt{2} T} = 2$

$t^{'} = 2 T^{'}$

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