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If the K.E. of a body is increased by 300%, its momentum will increase by

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$100 %$

$150 %$

$\sqrt{300} %$

$175 %$

answer is A.

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Detailed Solution

$P = \sqrt{2 m K}, and P' = \sqrt{2 m (K + 3 K)} = 2 P$

$\frac{P' - P}{P} x 100 = \frac{2 P - P}{P} x 100 = 100 %$

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