Question

Water flows through the tube as shown in the figure. The areas of cross-section of the wide and the narrow portions of the tube are $5\,c{m^2}$ and $2\,c{m^2}$ respectively. The rate of flow of water through the tube is $500\,\dfrac{{c{m^3}}}{s}$. Find the difference of mercury levels in the U-tube.

Answer 1

Hint: In this question, we need to determine the difference of mercury levels in the U-tube. We will determine the area and velocity by using the continuous equation. Then by using Bernoulli’s equation we will determine the equation for $h$. Then, substitute the values and evaluate to determine the difference of mercury levels.

Complete step by step answer:
It is given that, the rate of flow of water through the tube=$500\,\dfrac{{c{m^3}}}{s}$
We know that, the rate of flow of water=$area(A) \times velocity(V)$
Now, by using continuity equation,
${A_1}{V_1} = {A_2}{V_2}$
It is given that the areas of cross-section of the wide and the narrow portions of the tube are $5\,c{m^2}$ and $2\,c{m^2}$ respectively.
Therefore, $5{V_1} = 2{V_2} = 500\,\dfrac{{c{m^3}}}{s}$
Then,${V_1} = \dfrac{{500}}{5}$
Therefore, ${V_1} = 100\,\dfrac{{cm}}{s}$ $ = 0.1\,\dfrac{m}{s}$
Also, ${V_2} = \dfrac{{500}}{2}$
${V_2} = 250\,\dfrac{{cm}}{s}$ $ = 2.5\,\dfrac{m}{s}$
Let the pressure at the wide section=${p_A}$
And, the pressure at narrow section=${p_B}$
Now, by using Bernoulli’s equation, we have,
${p_A} + \dfrac{1}{2}{\rho _W}{V_A}^2 = {p_B} + \dfrac{1}{2}{\rho _W}{V_B}^2$
$\Rightarrow{p_A} - {p_B} = \dfrac{1}{2}{\rho _W}\left( {{V_B}^2 - {V_A}^2} \right)$
Since, the difference in pressure at two points i.e.,${p_A} - {p_B} = {\rho _{{H_g}}}gh$
Let the difference in height levels of mercury=$h$
And,${\rho _{{H_g}}}$ is the density of mercury
${\rho _{{H_g}}}gh = \dfrac{1}{2}{\rho _W}\left( {{V_2}^2 - {V_1}^2} \right)$
Therefore, $h = \dfrac{{{\rho _W}\left( {{V_2}^2 - {V_1}^2} \right)}}{{2{\rho _{{H_g}}}g}}$
Now, by substituting the values, we have,
$h= \dfrac{{{{10}^3}\left( {6.25 - 1} \right)}}{{2 \times 13.6 \times {{10}^3} \times 9.8}}$
$\Rightarrow h = 0.0196\,m$
$\therefore h = 1.96\,cm$

Hence, the difference of mercury levels in the U-tube is $1.96\,cm$.

Note: It is important here to note that, whenever these types of problems are given first, determine the area and velocity then, apply Bernoulli’s equation and solve it. The continuous equation is nothing but it describes the transport of some quantity i.e., the product of cross-sectional area and the velocity which is a constant.