Question

A tank with a square base of area $2\,{{\text{m}}^2}$ is divided into two components by a vertical partition in the middle. There is a small hinged door of the face area $20\,{\text{c}}{{\text{m}}^2}$ at the bottom of the partition. Water is filled in one compartment and an acid of relative density $1.53 \times {10^3}\,{\text{kg}}{{\text{m}}^{ - 3}}$ in the other, both to a height of $4\,{\text{m}}$. The force necessary to keep the door closed is (Take $g = 10\,{\text{m}}{{\text{s}}^{ - 2}}$ )
A. $10\,{\text{N}}$
B. $20\,{\text{N}}$
C. $40\,{\text{N}}$
D. $80\,{\text{N}}$

Answer 1

First of all, we will find out the individual pressures of both the water and acid column. Then we will find the net pressure. Then by using the correct formula, we will put the values as we have pressure and area. We will manipulate accordingly and obtain the result.

Complete step by step answer:
In the given question, we are supplied with the following data:
The base of the tank has an area of $2\,{{\text{m}}^2}$ .
There is a small hinged door in at the bottom of the tank, in the partition, whose area is $20\,{\text{c}}{{\text{m}}^2}$ .
The height of both the water and the acid column is given as $4\,{\text{m}}$ .
The relative density of acid is given as $1.53 \times {10^3}\,{\text{kg}}{{\text{m}}^{ - 3}}$ .
We know, the relative density of water is ${10^3}\,{\text{kg}}{{\text{m}}^{ - 3}}$ .

For the case of water,
Pressure exerted by the water column at the door at the bottom of the tank is given by the formula:
${P_w} = {\rho _w}hg$ …… (1)
Where,
${P_w}$ indicates the pressure exerted by the water column.
${\rho _w}$ indicates density of water.
$h$ indicates the height of the water column.
$g$ indicates acceleration due to gravity.
Substituting the required values in equation (1), we get:

$${P_w} = {\rho _w}hg \\\ \Rightarrow {P_w} = {10^3} \times 4 \times 10 \\\ \Rightarrow {P_w} = 4 \times {10^4}\,{\text{N}}{{\text{m}}^{ - 2}} \\\$$

Pressure exerted by the acid column at the door at the bottom of the tank is given by formula:
${P_a} = {\rho _a}hg$ …… (2)
Where,
${P_a}$ indicates the pressure exerted by the acid column.
${\rho _a}$ indicates density of acid.
$h$ indicates height of the acid column.
$g$ indicates acceleration due to gravity.
Substituting

$${P_a} = {\rho _a}hg \\\ \Rightarrow {P_a} = 1.5 \times {10^3} \times 4 \times 10 \\\ \Rightarrow {P_a} = 6 \times {10^4}\,{\text{N}}{{\text{m}}^{ - 2}} \\\$$

Net pressure on the door can be found out by substituting the pressure exerted by acid and water column. Mathematically we can write:
$\Delta P = {P_a} - {P_w}$ …… (3)
Substituting the required values in equation (3), we get:

$$\Delta P = \left( {6 \times {{10}^4} - 4 \times {{10}^4}} \right)\,{\text{N}}{{\text{m}}^{ - 2}} \\\ \Rightarrow \Delta P = 2 \times {10^4}\,{\text{N}}{{\text{m}}^{ - 2}} \\\$$

Area of the door is given as $20\,{\text{c}}{{\text{m}}^2}$
Converting ${\text{c}}{{\text{m}}^2}$ to ${{\text{m}}^2}$ ,

$$\Rightarrow 20\,{\text{c}}{{\text{m}}^2} \\\ \Rightarrow 20 \times 1\,{\text{cm}} \times 1\,{\text{cm}} \\\ \Rightarrow 20 \times {10^{ - 2}}{\text{m}} \times {10^{ - 2}}{\text{m}} \\\ \Rightarrow 20 \times {10^{ - 4}}{{\text{m}}^2} \\\$$

We know, the relationship between force and pressure, which is given as follows:
$P = \dfrac{F}{A}$
$F = P \times A$ …… (4)
Where,
$F$ indicates applied force.
$P$ indicates pressure.
$A$ indicates area.
Substituting the required values in the equation (4), we get:

$$\Rightarrow F = 40\,{\text{N}} \\\$$

Hence, the force necessary to keep the door closed is $40\,{\text{N}}$ .
The correct option is C.

Note: In this given problem, the relative density of acid is more than that of water, so the acid column will exert pressure on the door to diffuse in the water column. Higher the density, higher is the pressure exerted by it on the other column.