Discussion Forum

A spray gun is shown in the figure where a piston pushes air out of the nozzle. A thin tube of the uniform cross-section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out.

 For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1mm respectively. The upper end of the container is open to the atmosphere.

  If the piston is pushed  at a speed of 5 mms^-1, the air comes out of the nozzle with a speed of 

Answer:

Explanation:

From continuity equation

 $A_{1}v_{1}=A_{2}v_{2}$

 Here, A₁  =400 A₂ because

r₁=20r₂  and   $A=\pi r^{2}$

$\therefore $   $  v_{2}=\frac{A_{1}}{A_{2}}(v_{1})=400v_{1}$

  = 400(5)  mm/s

    =2000 mm/s= 2 m/s