Question

In the figure, the cross-sectional area of the smaller tube is $a$ and that of the larger tube is $2a$. A block of mass $m$ is kept in the smaller tube having the same base area $a$, as that of the tube. The difference between water levels of the two tubes is

A. $\dfrac{{{P_0}}}{{\rho g}} + \dfrac{m}{{a\rho }}$
B. $\dfrac{{{P_0}}}{{\rho g}} + \dfrac{m}{{2a\rho }}$
C. $\dfrac{m}{{a\rho }}$
D. $\dfrac{m}{{2a\rho }}$

Answer 1

Use Pascal's law. Determine the pressure of the liquid in the larger tube at depth h from the surface of the larger tube and the pressure in the liquid at the same horizontal level at depth h in the smaller tube. Equate the values of both pressures and determine the difference between the water levels of the tube which is the required answer.

Formulae used:
The pressure-depth relation is given by
$P = {P_0} + \rho gh$ …… (1)
Here, ${P_0}$ is the atmospheric pressure, $\rho$ is the density of liquid, $g$ is acceleration due to gravity and $P$ is the pressure of the liquid at depth $h$.
The pressure $P$ is given by
$P = \dfrac{F}{A}$ …… (2)
Here, $F$ is the force and $A$ is the area on which force is applied.

Complete step by step answer:
We can redraw the given diagram as follows:

In the above diagram, ${h_1}$ is the height of the liquid for the horizontal level at which the pressure in both the tubes are equal. We know that according to Pascal’s law of pressure, the pressure of the liquid at the same horizontal level is the same. The pressure ${P_L}$ of the liquid at depth $h$ from the surface of the tube is given by equation (1).
${P_L} = {P_0} + \rho gh$

The pressure ${P_S}$ of the liquid at depth ${h_1}$ from the surface of the tube is the sum of the pressure ${P_{{h_1}}}$ given by equation (1) and the pressure ${P_b}$ due to weight of the block.
${P_S} = {P_{{h_1}}} + {P_b}$ …… (3)
Rewrite equation (1) for the pressure ${P_{{h_1}}}$ of liquid at depth ${h_1}$ from the surface of the smaller tube.
${P_{{h_1}}} = {P_0} + \rho g{h_1}$
Rewrite equation (2) for pressure ${P_b}$ due to weight of the block.
${P_b} = \dfrac{{mg}}{a}$
Substitute ${P_0} + \rho g{h_1}$ for ${P_{{h_1}}}$ and $\dfrac{{mg}}{a}$ for ${P_b}$ in equation (3).
${P_S} = {P_0} + \rho g{h_1} + \dfrac{{mg}}{a}$

According to Pascal’s law, the pressures ${P_L}$ and ${P_S}$ are equal.
${P_L} = {P_S}$
Substitute ${P_0} + \rho gh$ for ${P_L}$ and ${P_0} + \rho g{h_1} + \dfrac{{mg}}{a}$ for ${P_S}$ in the above equation.
${P_0} + \rho gh = {P_0} + \rho g{h_1} + \dfrac{{mg}}{a}$
$\Rightarrow \rho gh = \rho g{h_1} + \dfrac{{mg}}{a}$
$\Rightarrow \rho gh - \rho g{h_1} = \dfrac{{mg}}{a}$
$\Rightarrow \rho g\left( {h - {h_1}} \right) = \dfrac{{mg}}{a}$
$\therefore h - {h_1} = \dfrac{m}{{a\rho }}$
Therefore, the pressure difference between the water levels of the two tubes is $\dfrac{m}{{a\rho }}$.

Hence, the correct option is C.

Note: The students should not forget to include the pressure due to the weight of the block in the pressure of liquid in the smaller tube at depth h1 from the upper edge of the smaller tube. If this value is not included in the pressure of liquid in the smaller tube then the obtained pressure at depth h1 will not be equal to the pressure of liquid in the larger tube at depth h. Hence, the final answer will also be incorrect.