Question

If the density of air is ${\rho }_a$ and that of the liquid ${\rho }_l$, then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to

• A. $\sqrt{\frac{{\rho }_a}{{\rho }_l}}$
• B. $\sqrt{{\rho }_a{\rho }_l}$
• C. $\sqrt{\frac{{\rho }_l}{{\rho }_a}}$
• D. ${\rho }_l$

Answer 1

The correct option is A $\sqrt{\frac{{\rho }_a}{{\rho }_l}}$

Pressure at point in nozzle $\Rightarrow P$

$P=P_0-\frac{1}{2}{\rho }_a{V}_1^2$ ...(1)

$P=P_0-\frac{1}{2}{\rho }_l{V}_2^2-{\rho }_{l}gh$ ...(2)

Equation (1) and (2)

$\frac{1}{2}{\rho }_a{V}_1^2=\frac{1}{2}{\rho }_l{V}_2^2+{\rho }_{l}gh$

Neglecting the terms $r_{e}gh$

$V_2=\sqrt{\frac{{\rho }_a}{{\rho }_l}{V}_1}$

Rate of liquid flow $=A{V}_2$

$=A{V}_1\sqrt{\frac{{\rho }_a}{{\rho }_l}}$

Rate of liquid flow $\propto \sqrt{\frac{{\rho }_a}{{\rho }_l}}$