Answer:
9v
Explanation:
Given that the water flows at speed v in a pipe of radius r.
Let [tex]v_2[/tex] is the speed of water flow where the radius of the pipe is r/3.
By using the continuity equation, the mass flow rate of water is the same at all the cross-sections of the pipe, wh have
[tex]\rho_1A_1_v_1 =\rho_2A_2_v_2[/tex]
Where [tex]\rho_1, A_1[/tex] and [tex]v_1[/tex] are the density, area of the cross-section and speed at cross-section 1 and [tex]\rho_2, A_2[/tex] and [tex]v_2[/tex] are the density, area of the cross-section and speed at cross-section 2.
As the density of the water remains constant. so [tex]\rho_1=\rho_2[/tex].
[tex]\Rightarrow A_1_v_1 =A_2_v_2 \\\\\Rightarrow (\pi r^2)v=(\pi (r/3)^2v_2 \\\\\Rightarrow r^2v=\frac {r^2}{9}\times v_2 \\\\\Rightarrow v= \frac {v_2}{9} \\\\\Rightarrow v_2=9v.[/tex]
Hence, the speed of the water flow where the radius is r/3 is [tex]9v[/tex].
