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Question

A particle of mass m is projected with velocity v making an angle of 45 with the horizontal. Themagnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where g = acceleration due to gravity)


A

Zero

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B

mv3(42g)

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C

mv3(2g)

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D

mv22g

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Solution

The correct option is B

mv3(42g)


L=m u3 cosθ sin2 θ2g=mv3(42g) [As θ=45]


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