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, calculate the total force experienced by the tank.

• A. $2\rho Agh$
• B. $\frac{\rho Agh}{2}$
• C. $\frac{3}{2\rho Agh}$
• D. $\rho Agh$

Answer 1

The correct option is A $2\rho Agh$


If $x$ is the distance of upper hole from the top, then velocities of discharge from respective holes will be

$v_1=\sqrt{2g(h+x)}$
$v_2=\sqrt{2gx}$

Now,
The force exerted at the upper hole is $A\rho (v_2{)}^2$ and the force exerted at the lower hole is $A\rho (v_1{)}^2$, then net force on the tank will be
$=\rho A(v_1^2-v_2^2)$
$=\rho A\left[2g\right(h+x)-2gx]=2\rho Agh$
$\therefore \left(a\right)$ is the correct answer.