**Paragraph:**
A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of 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 1 mm respectively. The upper end of the container is open to the atmosphere. (IIT JEE 2014)

**Question 1:**
If the piston is pushed at a speed of 5 mm/s, the air comes out of the nozzle with a speed of

- 0.1 m/s
- 1 m/s
- 2 m/s
- 8 m/s

**Solution:**
The speed of air inside the piston (spray gun) is equal to the speed of the piston i.e., $v_1=5$ mm/s. The area of the piston is $A_1=\pi r_1^2$ and that of the nozzle is $A_2=\pi r_2^2$, where $r_1=20$ mm and $r_2=1$ mm. The continuity equation, $A_1v_1=A_2 v_2$, gives speed of the air at the nozzle as
\begin{align}
v_2&=\frac{A_1 v_1}{A_2}=\frac{\pi(20)^2 (5)}{\pi(1)^2}\\
&=2000\, \mathrm{mm/s}=2\, \mathrm{m/s}\nonumber
\end{align}
Thus, option (C) is correct.

**Question 2:**
If the density of air is $\rho_a$ and that of the liquid is $\rho_l$, then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to

- $\sqrt{{\rho_a}/{\rho_l}}$
- $\sqrt{{\rho_a}{\rho_l}}$
- $\sqrt{{\rho_l}/{\rho_a}}$
- $\rho_l$

**Solution:**
Let the speed of the piston be $v_{a,1}$. The continuity equation gives speed of air at the nozzle, $v_{a,2}=\frac{A_1}{A_2}v_{a,1}$, a constant for fixed $v_{a,1}$. The air expands when it comes out to the atmosphere (pressure $P_0$) and finally its speed reduces to zero.

Let pressure of the air at the nozzle be $P_2$. Bernoulli's equation for the air gives \begin{align} \label{wxb:eqn:2:1} P_2+\tfrac{1}{2}\rho_a v_{a,2}^2=P_0. \end{align} The liquid rises due to pressure difference $(P_0-P_2)$. Apply Bernoulli's equation for the liquid in the vertical tube (assume height to be negligibly small) to get \begin{align} \label{wxb:eqn:2:2} P_0=\tfrac{1}{2}\rho_l v_l^2+P_2. \end{align} Eliminate $(P_0-P_2)$ from above equations to get \begin{align} \tfrac{1}{2}\rho_l v_l^2=\tfrac{1}{2}\rho_a v_{a,2}^2, \nonumber \end{align} which gives, \begin{align} v_l=v_{a,2} \sqrt{{\rho_a}/{\rho_l}}. \end{align} If the cross-section area of the tube is $A$ then the volume flow rate of the liquid is \begin{align} A v_l=A v_{a,2} \sqrt{{\rho_a}/{\rho_l}}. \end{align}

**Problem:**
Consider a horizontally oriented syringe containing water located at a height of 1.25 m above the ground. The diameter of the plunger is 8 mm and the diameter of the nozzle is 2 mm. The plunger is pushed with a constant speed of 0.25 m/s. Find the horizontal range of water stream on the ground.

**Solution:**
Apply continuity equation, $A_1v_1=A_2v_2$, on the two ends of the syringe with $A_1=\pi (d_1/2)^2=16\pi\,\mathrm{mm^2} $, $v_1=0.25$ m/s and $A_2=\pi (d_2/2)^2=\pi\,\mathrm{mm^2}$, to get $v_2=4$ m/s.

After coming out of the syringe, water falls like a projectile thrown horizontally with a speed $v_2=4$ m/s from a height $h=1.25$ m. The time taken by the projectile is \begin{align} t=\sqrt{2h/g}=0.5\,\mathrm{s}. \end{align} The horizontal distance travelled by the projectile in this time is \begin{align} x=v_2 t=2\mathrm{m}. \end{align}