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  4. There are two identical small holes of area of cross-section textitA on the opposite sides of a tank containing a liquid of density ρ . The difference in height between the holes is textith . Tank is resting on a smooth horizontal surface, horizontal force which has to be applied on the tank to keep it in equilibrium is <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-vvfsufisk9ma.png />

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

Question

NTA AbhyasNTA Abhyas 2020

Solution:

Rate of change of momentum $=\frac{d p}{d t}$
$F=F_{Q}-F_{P}=\frac{d p_{Q}}{d t}-\frac{d p_{P}}{d t}=\left(\frac{d m}{d t}\right)_{Q}\times V_{Q}-\left(\frac{d m}{d t}\right)_{P}\times V_{P}$ $$
$=A\rho v_{Q}\times v_{Q}-A\rho v_{P}\times v_{P}$
$=A\rho \left(v_{Q} ^{2} - v_{P} ^{2}\right)$ ...(i)

Solution
$P_{P}+\frac{1}{2}\rho v_{P}^{2}+\rho gh=P_{Q}+\frac{1}{2}\rho v_{Q}^{2}+0 \, \left(\because P_{P} = P_{Q}\right)$
$\Rightarrow \frac{1}{2}\rho \left(v_{Q}^{2} - v_{P}^{2}\right)=\rho gh$
$\therefore F=2A\rho gh$ (from (i))