Question

There are two identical small holes of area of cross-section $\text{A}$ on the opposite sides of a tank containing a liquid of density $\rho$ . The difference in height between the holes is $\text{h}$ . Tank is resting on a smooth horizontal surface, horizontal force which has to be applied on the tank to keep it in equilibrium is

• A. $\rho ghA$
• B. $2gh/\rho A$
• C. $2\rho ghA$
• D. $\rho gh/A$

Answer 1

Answer: (C)

Rate of change of momentum $=\frac{dp}{dt}$
$F=F_Q-F_P=\frac{d{p}_Q}{dt}-\frac{d{p}_P}{dt}={\left(\frac{dm}{dt}\right)}_Q\times V_Q-{\left(\frac{dm}{dt}\right)}_P\times V_P$
$=A\rho v_Q\times v_Q-A\rho v_P\times v_P$
$=A\rho \left(v_Q^2-v_P^2\right)$ ...(i)


$P_P+\frac{1}{2}\rho v_P^2+\rho gh=P_Q+\frac{1}{2}\rho v_Q^2+0\left(\because P_P=P_Q\right)$
$\Rightarrow \frac{1}{2}\rho \left(v_Q^2-v_P^2\right)=\rho gh$
$\therefore F=2A\rho gh$ (from (i))