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. $\sqrt{h}$
• B. $h$
• C. $h^{3/2}$
• D. $h^2$

Answer 1

Answer: (B)

There are two identical small holes on the opposite sides of a tank, hence velocities of liquid flowing out of both the holes are
$v_1=\sqrt{2g(h+x)}\text{ and }{v}_2=\sqrt{2gx}$
Now,
Volume of the liquid discharged per second at a hole is
$Av,(v=v_1$ or $v{}_2)$
Mass of liquid discharged per second is $Av\rho$,
Momentum of liquid discharged per second is $A{v}^2\rho$
The force exerted at the upper hole is $Ap{\left(v_2\right)}^2$ and
The force exerted at the lower hole is $Ap{\left(v_1\right)}^2$
The net force on the tank is
$F=A\rho \left[{\left(v_1\right)}^2-{\left(v_2\right)}^2\right]$
$F=A\rho [2g(h+x)-2gx]$
$F=2Apgh$
$\Rightarrow F\propto h$