Question

Water flows in a horizontal tube (see figure). The pressure of water changes by $700{\text{Nm}}^{-2}$ between $A$ and $B$, where the areas of cross-section are $40{\text{cm}}^2\text{and}20{\text{cm}}^2$, respectively. Find the rate of flow of water through the tube. (density of water $=1000{\text{kgm}}^{-3}$)

• A. $3020{\text{cm}}^3/\text{s}$
• B. $2720{\text{cm}}^3/\text{s}$
• C. $2420{\text{cm}}^3/\text{s}$
• D. $1810{\text{cm}}^3/\text{s}$

Answer 1

The correct option is B $2720{\text{cm}}^3/\text{s}$

According to the question,

Area of cross-section at $A$ is,

$A_a=40{\text{cm}}^2$ and

at $B$, $A_b=20{\text{cm}}^2$

Let the velocity of liquid at $A$, $V_a\text{and at B,}=V_b$

Using equation of continuity $A_a{V}_a=A_b{V}_b$

$40V_a=20V_b$

$\Rightarrow 2V_a=V_b$

Now, using Bernoulli's equation

$P_a+\frac{1}{2}\rho V_a^2=P_b+\frac{1}{2}\rho V_b^2$

$\Rightarrow P_a-P_b=\frac{1}{2}\rho (V_b^2-V_a^2)$

$\Rightarrow \Delta P=\frac{1}{2}\times 1000\times (V_b^2-\frac{V_b^2}{4})$

$\Rightarrow \Delta P=500\times \frac{3V_b^2}{4}$

$\Rightarrow V_b=\sqrt{\frac{\left(\Delta P\right)\times 4}{1500}}$

$=\sqrt{\frac{\left(700\right)\times 4}{1500}}=1.37\text{m/s}=1.37\times {10}^2\text{cm/s}$

Volume flow rate, $Q=A_b\times V_b$

$=20\times 1.37\times {10}^2$

$=2732{\text{cm}}^3/\text{s}\approx 2720{\text{cm}}^3/\text{s}$

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