Question

The figure shows a model of perfume atomizer. When the bulb $\text{A}$ is compressed, air having density ${\rho }_a$ flows through the narrow tube. Consequently, pressure at the position of the vertical tube reduces. The liquid (perfume) of density $\rho$ rises through the vertical tube and emerges through the end. If the excess pressure applied to the bulb in this process be $\Delta p$, the minimum speed of air in the tube to lift the perfume is

• A. $\sqrt{\frac{2(\Delta p+\rho gh)}{{\rho }_a}}$
• B. $\sqrt{\frac{2(\Delta p-\rho gh)}{{\rho }_a}}$
• C. $\sqrt{\frac{\Delta p+\rho gh}{{\rho }_a}}$
• D. None of these

Answer 1

The correct option is A $\sqrt{\frac{2(\Delta p+\rho gh)}{{\rho }_a}}$


Applying Bernoulli’s principle at $1$ and $2$, we obtain

$P_1+\frac{1}{2}{\rho }_a{v}_1^2=P_2+\frac{1}{2}\rho v_2^2$

putting, $v_1=0$ since the air is at rest inside the bulb,

Also, $P_1=P_0+\Delta p$

And, $P_2=P_0-\rho gh$

where $\rho$ = density of liquid (perfume)

${\rho }_a$ = density of air

$v_2$ = minimum velocity of air to lift the liquid through the height $h$

$p_0$ = atmospheric pressure

$\Rightarrow P_0+\Delta p+0=P_0-\rho gh+\frac{1}{2}\rho v_2^2$

$\Rightarrow v_2=\sqrt{\frac{2(\Delta p+\rho gh)}{{\rho }_a}}$

Hence, option $\left(\text{A}\right)$ is correct.