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A rectangular tank of mass $\;m_{0}\;$ and charge Q over it is placed over a smooth horizontal floor. A horizontal electric field E exist in the region . Rain drops are falling vertically in the tank at the constant rate of n drops per second . Mass of each drop is m . find velocity of tank as function of time .

$(a)\;\large\frac{2QEt}{m_{0}+m n t}\qquad(b)\;\large\frac{QEt}{m_{0}+m n t}\qquad(c)\;\large\frac{QEt}{2(m_{0}+m n t)}\qquad(d)\;None$

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Answer : (b) $\;\large\frac{QEt}{m_{0}+m n t}$
Explanation :
Two forces will act on the tank .
(a) Electrostatic force (b) Thrust force
Let v be the velocity at any instant . Then
$F_{net}=QE-mvn$
$(m_{0}+mnt)\;\large\frac{dv}{dt}=QE-mnv$
$\int_{0}^{v}\;\large\frac{dV}{QE-mnv}=\int_{0}^{t}\;\large\frac{dt}{m_{0}+mnt}$
$ln(\large\frac{QE}{QE-mnv})=ln(\large\frac{m_{0}+mnt}{m_{0}})$
$m_{0}QE=m_{0}QE-m_{0}mnv+QEmnt-m^2n^2vt$
$v=\large\frac{QEt}{m_{0}+mnt}\;.$
answered Feb 18, 2014 by yamini.v
 

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