There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density ρ . The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is

The correct answer is 2ρagh.

Let us analyze the situation step-by-step to find the horizontal force required to keep the tank in equilibrium.

When a liquid of density $\rho$ flows out of a hole of cross-sectional area $a$, it exerts a thrust force (recoil force) on the tank in the opposite direction of the flow. This thrust force is due to the rate of change of momentum of the ejected liquid.

Let the velocity of efflux of the liquid from a hole at depth $y$ from the free surface be $v$. According to Torricelli's Law:
$v=\sqrt{2gy}$

The mass flow rate of the liquid emerging from the hole is given by:
$\frac{dm}{dt}$

The thrust force $F$ exerted on the tank by the outgoing liquid is:
$F=v\frac{dm}{dt}$

Substituting the expression for $v$:
$F=\rho a\left(2gy\right)=2\rho agy$

Now, let the first hole be at a depth $h_1$ from the liquid surface. The thrust force due to this hole is:
$F_1=2\rho ag{h}_1$

Let the second hole on the opposite side be at a depth $h_2$ from the liquid surface. The thrust force due to this hole is:
$F_2=2\rho ag{h}_2$

Since the two holes are on opposite sides, the two forces $F_1$ and $F_2$ act in opposite directions. The net horizontal force acting on the tank due to the flowing liquid is the difference between these two thrust forces:
$F_{\text{net}}=F_2-F_1$
$F_{\text{net}}=2\rho ag{h}_2-2\rho ag{h}_1$
$F_{\text{net}}=2\rho ag(h_2-h_1)$

We are given that the difference in height between the two holes is $h$. Therefore, we can write:
$h_2-h_1=h$

Substituting this back into the net force equation yields:
$F_{\text{net}}=2\rho agh$

To keep the tank in equilibrium, an equal and opposite external force must be applied to balance this net thrust. Thus, the required horizontal force is:
$F=2\rho agh$