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A rain drop of mass m_(0) starts falling from rest and it collects water vapour and grows . If it gains lambda kg//s,find its velocity at any instant. |
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Answer» Solution :`d/(dt) (MV) =MG,(DM)/(dt)=lambda "or" m=lambdat+K` where K is a constant when `t=0, m=m_(0) :. M=m_(0)+lambdat` so,`d/(dt){(m_(0)+lambdat)v}=(m_(0)+lambdat)G` Integrating, `(m_(0)+lambdat)v=int(m_(0)+lambdat)gdt =(m_(0)t+(lambdat^(2))/2)g+C ` where c is a constant when `t=0, v=0``:. C=0` `:. (m_(0)+lambdat) v=(m_(0)t+(lambdat^(2))/2)g , v =(g(m_(0)t+(lambdat^(2))/2)/(m_(0)+lambdat)=(g(t+(lambdat^(2))/(2m_(0))/(1+(lambdat)/m_(0)))` Thismean velocitychanges w.r.t.time |
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