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Law of Conservation of Mechanical Energy

The bob of si...

Question

The bob of simple pendulum is released when the string is horizontally straight. If $E$ is kinetic energy of bob when string makes an angle $60 °$ with vertical, then its kinetic energy at mean position is:

A

\sqrt{2} E

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B

\sqrt{3} E

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C

2 E

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D

2 \sqrt{2} E

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Solution

The correct option is C $2 E$
Work Energy Theorem: K.E is equal to work done by gravity
at $60^{0}$ with vertical,
$(W . D)_{G} = \frac{m g R}{2} = E$
Mean position of the pendulum is it's initial position.
At, $90^{0}, (W . D)_{G} = m g R = 2 E$

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