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The time peri...

Question

The time period of a simple pendulum on a freely moving artificial satellite is

A

Zero

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B

2 sec

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C

3 sec

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D

Infinite

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Solution

The correct option is D Infinite
$T = 2 π \sqrt{\frac{1}{g^{'}}} = 2 π \sqrt{\frac{1}{0}} = \infty [A s g^{'} = 0 i n t h e s a t e l l i t e]$

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