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If momentum is increased by 20% then K.E increases by(A) 44%(B) 55%(C) 66%(D) 77% |
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Answer» Answer is (A) 44% K = p2/2m K’ = [(1200/100)P]2/2m = 1.44 p2/2m = 1.44 K ∆K = K’ – K = 0.44 K = 44 % K |
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