Question

A liquid flows through a horizontal tube. The velocities of the liquid in the two sections, which have have areas of cross section $A_1$ and $A_2$, are $v_1$ and $v_2$ respectively. The difference in the levels of the liquid in the two vertical tubes is $h$. Then :

• A. The volume of the liquid flowing through the tube in unit time is $A_1{v}_1$.
• B. $v_2-v_1=\sqrt{2gh}$
• C. $v_2^2-v_1^2=2gh$
• D. The energy per unit volume of the liquid is the same in both sections of the tube.

Answer 1

Answer: (A), (C), (D)

The energy per unit volume of the liquid is the same in both sections of the tube.

Since there are no sources and sinks in between the two sections of the tube, by using the equation of continuity, we can say that $A_1{v}_1=A_2{v}_2$ = constant. Thus, option A is correct.
Using Bernoulli's equation for the two sections of the tube we have
$\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+h_1=\frac{P_2}{\rho g}+\frac{v_2^2}{2g}+h_2.....\left(1\right)$
As the tube is horizontal, we take $h_1=h_2$.
$\therefore P_1-P_2=\frac{\rho }{2}(v_2^2-v_1^2).....\left(2\right)$
where $P_1$ and $P_2$ are the pressures at section 1 and section 2.
But, $P_1-P_2=\rho gh.....\left(3\right)$
(from the difference in heights of the liquid columns)
Thus, from $\left(2\right)$ and $\left(3\right)$, we have
$\rho gh=\frac{\rho }{2}(v_2^2-v_1^2)$
or, $v_2^2-v_1^2=2gh$.
Thus, option (c) is correct.
$\because$ Bernouli's theorem plays the role of law of conservation of energy.
From $\left(1\right)$, we can deduce that there is no change in the energy per unit volume between the two sections. Thus, option (d) is also correct.
Hence, options (a), (c) and (d) are the correct answers.