Question

Water flows in a horizontal tube as shown in figure. The pressure of water changes by $60{\text{N/m}}^2$ between $A$ and $B$, where the areas of cross-section are $30{\text{cm}}^2$ and
$15{\text{cm}}^2$ respectively. Find the rate of flow of water through the tube.

• A. $6{\text{cm}}^3/s$
• B. $60{\text{cm}}^3/s$
• C. $600{\text{cm}}^3/s$
• D. None of the above

Answer 1

The correct option is C $600{\text{cm}}^3/s$

Let the velocity at $A=v_A$ and that at $B=v_B$.
According to the equation of continuity,
$A_A{v}_A=A_B{v}_B\Rightarrow \frac{v_B}{v_A}=\frac{A_A}{A_B}$

$\therefore \frac{v_B}{v_A}=\frac{30}{15}=2$

According to Bernoulli's equation,

$P_A+\frac{1}{2}\rho v_A^2=P_B+\frac{1}{2}\rho v_B^2$
$P_A-P_B=\frac{1}{2}\rho (2v_A{)}^2-\frac{1}{2}\rho v_A^2=\frac{3}{2}\rho v_A^2$
$60=\frac{3}{2}\left(1000\right)v_A^2$

$v_A=\sqrt{0.04}=0.2\text{m/s}=20\text{cm/s}$

The rate of flow is, $Q=A_A\times v_A=30\times 20=600{\text{cm}}^3/\text{s}$.

Hence, $\left(C\right)$ is the correct answer.

Why this Q ? Bernoulli's theorem Bernoullis' theorem states that the total energy (pressure energy, kinetic energy, gravitational potential energy) per unit volume is constnat at all points along a streamline. $p+\frac{1}{2}\rho v^2+\rho gh=\text{constant or}\frac{p}{\rho g}+\frac{1}{2}\frac{v^2}{g}+h=\text{constant}$ Principle of continuity The princiiple of continuity represents the conservation of mass. According to this principle, $av=$ constant. $a=$ area of cross section, $v=$ velocity of fluid flow $\Rightarrow a_1{v}_1=a_2{v}_2$