Question

There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is $h$. As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to -

• A. $h^{1/2}$
• B. $h$
• C. $h^{3/2}$
• D. $h^2$

Answer 1

The correct option is B $h$

Let the velocity of efflux from lower and upper hole is $v_1$ and $v_2$ respectively and depth of upper hole from surface of water is $x$


Velocity of efflux from lower hole,
$v_1=\sqrt{2g(h+x)}$
$\because v=\sqrt{2gH},$ where $H$ is the depth of hole from surface of water

Velocity of efflux from upper hole,
$v_2=\sqrt{2gx}$
Let $\rho =$ density of the liquid.
Let $A=$ cross-sectional area of each hole (Identical holes).
Rate of discharge of volume of liquid through hole is,
$\Rightarrow \frac{dV}{dt}=Q=Av$
Rate of discharge of mass of liquid is given by,
$\frac{dm}{dt}=\rho \frac{dV}{dt}=\rho Av$
Rate of change of momentum of liquid is,
$\frac{dP}{dt}=\frac{d\left(mv\right)}{dt}=v\frac{dm}{dt}$
$\Rightarrow \frac{dP}{dt}=v\left(\rho Av\right)=\rho A{v}^2$
According to Newton's second law,
$F=\frac{dP}{dt}$
From Newton's $3^{\text{rd}}$ law, the force on tank will be in a direction opposite to that of velocity of efflux.
$\Rightarrow$The force exerted on tank at the upper hole towards right is,
$F_2=\rho A{v}_2^2$
& the force exerted on tank at the lower hole towards left is,
$F_1=\rho A{v}_1^2$
Net force on the tank (in horizontal direction) will be,
$F_{\text{net}}=F_1-F_2$
$\because v_1>v_2$ hence $F_1>F_2$
$\Rightarrow F_{\text{net}}=\rho A[v_1^2-v_2^2]$
$\Rightarrow F_{\text{net}}=\rho A\left[2g\right(h+x)-2gx]$
$\therefore F_{\text{net}}=2\rho Agh$
Hence, $F_{\text{net}}\propto h$, since $\rho ,A,g$ are the constant terms here.