1. Continuity of air flow

Let the piston speed be $u$.

• Radius of piston: $R=20\,\text{mm}$
• Radius of nozzle: $r=1\,\text{mm}$

By continuity for air, $A_p u = A_n v_a$ where $v_a$ is the air speed in the nozzle.

Thus, $v_a = \frac{A_p}{A_n}u = \frac{\pi R^2}{\pi r^2}u = \left(\frac{R}{r}\right)^2 u$

Since $R/r=20$, $v_a = 400u$ So for a given piston speed, $v_a$ is fixed and does not depend on densities.

2. Pressure drop at the nozzle due to fast air

The air moving rapidly over the top of the vertical tube creates a lower pressure there.

Using Bernoulli principle for air, $\Delta P \sim \frac{1}{2}\rho_a v_a^2$

Since $v_a$ is fixed for a given piston speed, $\Delta P \propto \rho_a$

3. Liquid rises and is drawn into the nozzle

This pressure drop accelerates the liquid from the container into the nozzle.

If $v_\ell$ is the speed of the liquid in the thin tube/nozzle, then by Bernoulli for the liquid, $\Delta P \sim \frac{1}{2}\rho_\ell v_\ell^2$

Hence, $v_\ell \propto \sqrt{\frac{\Delta P}{\rho_\ell}}$

Using $\Delta P \propto \rho_a$, $v_\ell \propto \sqrt{\frac{\rho_a}{\rho_\ell}}$

4. Liquid flow rate

The liquid volume flow rate is $Q_\ell = A_t v_\ell$ where $A_t$ is the cross-sectional area of the thin tube.

Since the tube area is fixed, $Q_\ell \propto v_\ell \propto \sqrt{\frac{\rho_a}{\rho_\ell}}$

5. Check options

• A: $\sqrt{\dfrac{\rho_a}{\rho_\ell}}$ ✅
• B: $\sqrt{\rho_a\rho_\ell}$ ❌
• C: $\sqrt{\dfrac{\rho_\ell}{\rho_a}}$ ❌
• D: $\rho_\ell$ ❌

Therefore, the correct option is $\boxed{A}$