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
$v _{1}=\sqrt{2 g ( h + x )} \text { and } v _{2}=\sqrt{2 g x }$
Now,
Volume of the liquid discharged per second at a hole is
$Av ,\left( v = v _{1}\right.$ or $v\left._{2}\right)$
Mass of liquid discharged per second is $Av\rho$,
Momentum of liquid discharged per second is $Av ^{2} \rho$
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
$F=A \rho\left[\left( v _{1}\right)^{2}-\left( v _{2}\right)^{2}\right]$
$F=A \rho[2 g ( h + x )-2 gx ]$
$F=2 Apgh$
$\Rightarrow F \propto h$