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Two bodies of different masses are dropped simultaneously from same height. If air friction acting on them is directly proportional to the square of their mass, then ,

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lighter body reaches the ground earlier

heavier body reaches the ground earlier

both reach the ground at the same time

any of the above

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detailed solution

Correct option is A

f = k.m2 where k = constant..: a = mg-f/m = (g-km)Again , t = √(2h/a)Therefore , If 'm' increases , 'a' decreases and 't' increases.

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