An incompressible liquid flows through a horizontal tube LMN as shown in the figure, then the velocity V of the liquid through the tube N is

Since the liquid is incompressible, The rate of volume entering into pipes will be the rate of volume leaving the pipes

${\mathrm{A}}_1{\mathrm{V}}_1+{\mathrm{A}}_2{\mathrm{V}}_2={\mathrm{A}}_3{\mathrm{V}}_3$

$\frac{\mathrm{A}}{2}\times 4+\mathrm{A}\times \mathrm{V}=2\mathrm{A}\times 4\mathrm{V}\Rightarrow \mathrm{V}=6\mathrm{m}/\mathrm{s}$