Five equal resistances each of resistance `R` are connected as shown in the figure. A battery of `V` volts is connected between `A` and `B`. The current flowing in `AFCEB` will be

Answer

`10^(5) m`

Explanation

`COLM rArr m u = (m + rho A x)v rArr v = (m u)/(m + rhoAx)`
`rArr (dx)/(dt) = (m u)/(m + rhoAx) rArr underset(0)overset(150)int (m + rhoAx)dx = underset(0)overset(150)int m u dt`
`rArr (mx + rho(Ax^(2))/(2)) = m ut`
`rArr 10^(-2) x + 10^(-3) xx (10^(-4))/(2)x^(2) = 10^(-2) xx 10^(3) xx 150 x = 10^(5) m`