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A rain drop of mass m, starts falling from rest and it collects water vapour and grows. If it gains lambda kg/s, find its velocity at any instant.

Answer» <html><body><p></p>Solution :`(d)/(dt) (<a href="https://interviewquestions.tuteehub.com/tag/mv-1082193" style="font-weight:bold;" target="_blank" title="Click to know more about MV">MV</a>) = mg , (dm)/(dt) = lambda or m= lambda t + <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>` <br/> where K is a constant when t=0 , `m=m_0 :. m=m_0 + lambda t ` <br/> so , ` (d)/(dt) {(m_0 + lambda t)v} = (m_0 + lambda t ) g` <br/> Integrating , `(m_0 + lambda t) v= int (m_0 + lambda t) g dt = (m_0 t + ( lambda t^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/(2) ) g + <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> ` where C is constant <br/> when t =0, v=0 ` :.` C =0 <br/> `:. (m_0 + lambda t) v = ( m_0 t + ( lambda t^2)/(2)) g , v=(g(m_0 t + (lambda t^2)/( 2)))/(m_0 + lambda t ) =(g (t+ (lambdat^(2))/(2m_0)))/( 1 + (lambda t)/( m_0))` <br/> This mean <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> changes w.r.t . time</body></html>


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