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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
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The product of time period in seconds and frequency of oscillation in rad/sec of a simple pendulum of length $l$ suspended inside a rocket moving vertically with an acceleration 'a' just above the earth is:

$\begin{array}{1 1}(a)\;2 * \pi \\(b)\;\frac{\sqrt{(a+g)}}{I} \\( c)\;\frac{\sqrt{(g-a)}}{I} \\(d)\;\frac{\sqrt{(a-g)}}{I} \end{array}$
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Time period of frequency $=2 \pi$
Hence a is the correct answer.
answered Dec 11, 2013 by meena.p

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