A rectangular container has a square base of side ‘a’ and water is filled into it up to height ‘h’ such that hydrostatic force on the base and side wall are equal in magnitude, then ‘h’ is equal to (neglect atmospheric pressure) ?
Answer
599.7k+ views
Hint: Since, the atmospheric pressure is to be neglected, therefore the pressure at the bottom of the tank will only be due to the weight of water, that is, no term for atmospheric pressure will be added to it. So, we will use this condition and the data given in the question to get the required solution of our question.
Complete answer:
Let us first define some useful terms, that we are going to use in our solution.
Let the force at the bottom of the container due to the weight of liquid be given by ${{F}_{B}}$ .
Now, the pressure at the bottom surface due to this force is ${{P}_{B}}$ .
Also, let the normal force due to water weight on the sides of the wall be given by ${{F}_{SW}}$ .
Then, the pressure corresponding to this force will be denoted by ${{P}_{SW}}$
It has been given to us in the problem that the hydrostatic force is equal on base and the side walls. Mathematically, this could be written as:
$\Rightarrow {{F}_{B}}={{F}_{SW}}$
In terms of pressure and area, this equation could be written as:
$\Rightarrow {{P}_{B}}{{(a)}^{2}}={{P}_{SW}}(h.a)$ [Let this expression be equation number (1)]
Where,
${{a}^{2}}$ is the area of base. And,
$h.a$ is the area of the side wall.
Since, we are to ignore the atmospheric pressure, therefore the pressure due to the water column on all the sides should be equal.
Thus, we can say that:
$\Rightarrow {{P}_{B}}={{P}_{SW}}$
Using this relation in equation number (1), we get:
$\begin{align}
& \Rightarrow {{a}^{2}}=h.a \\
& \Rightarrow a.a=h.a \\
& \therefore h=a \\
\end{align}$
Hence, the height of the water column comes out to be the same as, the base of the container that is (a).
Note:
These questions are just mere explanations of basic definitions. The fact that, how would a system behave under no atmospheric effect is the base of the problem. As we can see, there isn’t too much calculation in the solution, so one should always keep in mind all the basic definitions and concepts to solve these problems quickly and correctly.
Complete answer:
Let us first define some useful terms, that we are going to use in our solution.
Let the force at the bottom of the container due to the weight of liquid be given by ${{F}_{B}}$ .
Now, the pressure at the bottom surface due to this force is ${{P}_{B}}$ .
Also, let the normal force due to water weight on the sides of the wall be given by ${{F}_{SW}}$ .
Then, the pressure corresponding to this force will be denoted by ${{P}_{SW}}$
It has been given to us in the problem that the hydrostatic force is equal on base and the side walls. Mathematically, this could be written as:
$\Rightarrow {{F}_{B}}={{F}_{SW}}$
In terms of pressure and area, this equation could be written as:
$\Rightarrow {{P}_{B}}{{(a)}^{2}}={{P}_{SW}}(h.a)$ [Let this expression be equation number (1)]
Where,
${{a}^{2}}$ is the area of base. And,
$h.a$ is the area of the side wall.
Since, we are to ignore the atmospheric pressure, therefore the pressure due to the water column on all the sides should be equal.
Thus, we can say that:
$\Rightarrow {{P}_{B}}={{P}_{SW}}$
Using this relation in equation number (1), we get:
$\begin{align}
& \Rightarrow {{a}^{2}}=h.a \\
& \Rightarrow a.a=h.a \\
& \therefore h=a \\
\end{align}$
Hence, the height of the water column comes out to be the same as, the base of the container that is (a).
Note:
These questions are just mere explanations of basic definitions. The fact that, how would a system behave under no atmospheric effect is the base of the problem. As we can see, there isn’t too much calculation in the solution, so one should always keep in mind all the basic definitions and concepts to solve these problems quickly and correctly.
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