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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 drops is m. Find the time in second taken by tank to reach to half the maximum speed. (given m_(0)=30kg, m=1mg and n=10^(4)) |
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Answer» Solution :Mass oftank at time `t` `M=m_(0)+nmt` Let VELOCITY of TANK be `v` `M(dv)/(DT)=QE-vnm` For `v_(max)`, `(dv)/(dt)=0` `RARR v_(max)=(QE)/(nm)` from `(i)` `(m_(0)+nmt)(dv)/(dt)=QE-vnm` `rArrint_(0)^(v)(dv)/(QE-vnm)=int_(0)^(t)(dt)/(m_(0)+nmt)` ` rArr-(1)/(nm)[LN(QE-nmv)]_(0)^(v)=(1)/(nm)[ln(m_(0)+nmt)]_(0)^(t)` `rArr ln((QE-nmv)/(QE))=-ln((m_(0)+nmt)/(m_(0)))` ` rArr(QE-nmv)/(QE)=(m_(0))/(m_(0)+nmv)` `rArr1-(nmv)/(QE)=(m_(0))/(m_(0)+nmt)` `rArr(nmv)/(QE)=1-(m_(0))/(m_(0)+nmt)=(nmt)/(m_(0)+nmt)` `rArrt=(m_(0))/((QE)/(v)-nm)` required time when `v=v_(max)//2=(m_(0))/((2QE)/(v_(max))-nm)=(m_(0))/(nm)` 
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