Question

Consider a rectangular tank of size $(l\times b)$ filled with a liquid of density $\rho$, to a height $H$, as shown in the figure. Find the ratio of the force acting on the base to that on the walls of the tank.

• A. $\frac{l}{H}$
• B. $\frac{2l}{H}$
• C. $\frac{l}{2H}$
• D. $\frac{2l}{3H}$

Answer 1

The correct option is B $\frac{2l}{H}$

Whenever a liquid comes in contact with solid boundaries, it exerts a force on it. The force on the boundary may be obtained by integrating the pressure over the entire area of the boundary. The variation of liquid pressure, acting on the base (diagram $\left(a\right)$ and on the wall (diagram $\left(b\right)$, are shown below.


(i) Force at the base

Since the pressure is uniform at the base, force acting at the base is given by,

$F=P\times$ ( area of the base )

Since $P=\rho gH$

$F=\rho gH(l\times b)=\rho g\left(lbH\right)$

Since $lbH=V$ ( Volume of the liquid ), so

$F=\rho gV$ = weight of the liquid inside the tank


(ii) Force acting on the vertical wall

Pressure acting on the vertical wall is not uniform but increases linearly with depth. Pressure at a depth $h$, from the free surface, is $P=\rho gh$.


Force $dF$ acting on a differential element of height $dh$ is,
$dF=P\left(bdh\right)=\left(\rho gh\right)\left(bdh\right)=\rho gbhdh$

The total force is $F=\rho gb{\int }_0^{H}hdh=\frac{1}{2}\rho gb{H}^2$

The total force acting per unit width of the vertical wall is,

$\frac{F}{b}=\frac{1}{2}\rho g{H}^2$

the ratio of forces is, $=\frac{\rho g\left(lbH\right)}{\frac{1}{2}\rho gb{H}^2}=\frac{2l}{H}$

Hence, $\left(B\right)$ is the correct answer.

$\begin{array}{c}\text{Why this Question ?}\\ \text{Thrust at the vertical surface :}\\ \text{On a vertical surface in contact with the liquid, the pressure is not the same at all points. Points at greater depth experience the larger pressures and hence greater thrust.}\\ \frac{F}{b}=\frac{1}{2}\left(base\right)\left(height\right)=\frac{1}{2}\left(\rho gH\right)\left(H\right)\\ \text{F [Pressure at CG of vertical surface}\times \text{[Area of surface]}\end{array}$