Question

To what height $h$ should a cylindrical vessel of diameter $d$ be filled with a liquid so that the total force on the vertical surface of the vessel be equal to the force on the bottom-
A. h= $d$
B. h= $2d$
C. h= $3d$
D. h= $\dfrac{d}{2}$

Answer 1

To solve these types of problems we got to understand the relationship between force and pressure. Now we see that we are supposed to find two forces, one at the vertical surface and the other at the bottom, so we will find both and equate them and we get the required height.

Complete step by step answer:
First let us consider a cylindrical vessel whose diameter is $D$. Thus, the radius of the vessel will be $D/2$. Let the density of the liquid be $\rho$. We know that the magnitude of pressure is obtained by the formula:
$P = h\rho g$
Where P is pressure, h is depth to where the pressure is to be determined and g is the acceleration due to gravity.
Let us consider the depth as y here (say). Now we know that Force and Pressure are related according to the formula:
$F = P \times A$
where $A$ is the area of the applied pressure.

Now let us consider a small area of cross section of the beaker wall: $dA$
The small force on the small area of cross section be $dF$,
Thus $dF$= P. $dA$
Integrating the above equation to get the total force:
$F = \int {dF}$ = $\int {P.dA}$
$\Rightarrow F = P \int {dA}$
Now Area(A) of the beaker is $\pi \times {(radius)^2}$ = $\pi \times {\left( {\dfrac{D}{2}} \right)^2}$
Thus $dA$= $2\pi \dfrac{d}{2}.dy$
Thus F = P $\int\limits_0^h {2\pi \dfrac{d}{2}.dy}$
$F = \int\limits_0^h {\rho gy.2\pi \dfrac{d}{2}.dy}$

On integrating:
$F = 2\rho g\pi \dfrac{d}{2}\left( {\dfrac{{{y^2}}}{2}} \right)_0^h$
$\Rightarrow F = \dfrac{{\rho g\pi d{h^2}}}{2}$
So, the force on vertical surface is found, now let us find the pressure at the bottom:
${P_2} = \rho gh$
Thus Force at bottom :
${F_2}$ = $\rho gh$ $\times \dfrac{{\pi {D^2}}}{4}$
Now as both the forces are same thus:
$\dfrac{{\rho g\pi d{h^2}}}{2}$ = $\rho gh \times \dfrac{{\pi {D^2}}}{4}$
which upon solving we get
$\therefore h = \dfrac{d}{2}$

Therefore, the correct answer is option D.

Note: We need to just equate the forces to get the answer, note that the integration part is the most important one as there we have to consider a general depth at first. Remember the formulas and relations between force area and pressure.