Question

A rubber pipe with a diameter of $10cm$ is connected to a nozzle $2cm$ in diameter. Water flowing through the pipe at a speed of $0.6{ms}^{-1}$ comes out like a jet through the nozzle. The backward force of the nozzle is about:

• A. $7.7N$
• B. $67.9N$
• C. $Zero$
• D. $2.8N$

Answer 1

The correct option is B $67.9N$

Given : Diameter of pipe , $D=10cm=0.1m$ ,

diameter of nozzle , $d=2cm=0.02m$ ,

velocity through the pipe ,$v_1=0.6m/s$ ,

Let ,velocity through the nozzle is $v_2$ .

We have , equation of continuity ,

$A_1{v}_1=A_2{v}_2$ ,

$\pi (D/2{)}^2\times 0.6=\pi (d/2{)}^2{v}_2$ ,

or $v_2=\frac{(0.05{)}^2\times 0.6}{(0.01{)}^2}=15m/s$ ,

hence , backward force exerted by nozzle ,

$F=m^{\prime }\Delta v$ ,

where , $m^{\prime }=$ mass flow rate ,

$\Delta v=$ change in velocity ,

now, $F=\rho A_1{v}_1(15-0.6)$ ,($\rho =1000kg/m^3$, density of water) ,

or $F=1000\times \pi \times (D/2{)}^2\times 0.6\times 14.4$ ,

or $F=1000\times 3.14\times (0.1/2{)}^2\times 0.6\times 14.4=67.9N$