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 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. $h^{\frac{3}{2}}$
• D. $h^2$

Answer 1

The correct option is B $h$

There are two identical small holes on the opposite sides of a tank, hence velocities of liquid flowing out of both the holes are
$v_1=\sqrt{2g(h+x)}$ 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 A$\rho (v_2{)}^2$ and
The force exerted at the lower hole is A$\rho (v_1{)}^2$
The net force on the tank is
$F=A\rho \left[\right(v_1{)}^2-\left(v_2{)}^2\right]$
$F=A\rho \left[2g\right(h+x)-2gx]$
$F=2A\rho gh$
$\Rightarrow F\propto h$