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)`)

Explanation

Mass of tank 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)`