Question

There are two identical holes of the area of cross section $a$ on the opposite sides of a tank containing a liquid of density $\rho$. The difference in height between the holes is $h$. The tank is resting on a smooth horizontal surface. The horizontal force which will have to be applied on them to keep it in equilibrium is

• A. $gh\rho a$
• B. $\frac{2gh}{\rho a}$
• C. $2\rho agh$
• D. $\frac{\rho gh}{a}$

Answer 1

The correct option is A $gh\rho a$

Force due to efflux $=\frac{1}{2}{\rho V}^2\left(A\right)$

Now,

force due to upper orifice $=\frac{1}{2}\rho {\left(\sqrt{2gx}\right)}^{2}a$

$Fu=\rho gax$

Where,

$x-$ distance of upper orifice from top layer of liquid

Similarly,

$F_{lower}=\frac{1}{2}\rho {\left(\sqrt{2g(h+x)}\right)}^{2}a$

$F_{lower}=\rho gha+\rho ghx$

Now the net force on the tank

$F_{net}=F_{lower}-F_{upper}$

$F_{net}=\rho gha$