The loss of pressure head in travelling a distance l is seen from the middle section to be `h_2-h_1=10cm`. Since `h_2-h_1=h_1` in our problem and `h_3-h_2=15cm=5+h_2-h_1`, we see that a pressure head of `5cm` remains incompensated and must be converted into kinetic eneryg, the liquid flowing out. Thus
`(rhov^2)/(2)=rhogDeltah` where `Delta h=h_3-h_2`
Thus `v=sqrt(2gDelta h)~~1m//s`