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A particle of mass 0.5 kg is projected horizontally with velocity 10 m/s from the top of a tower of height 40 m. The angular momentum of the particle about the base of the tower just after projection is

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$50 k g - m^{2} / s$

$400 k g - m^{2} / s$

$100 k g - m^{2} / s$

$200 k g - m^{2} / s$

answer is D.

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Detailed Solution

(4)

$L = m v h = 0.5 \times 10 \times 40 k g - m^{2} / s = 200 k g - m^{2} / s$

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