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There are two identical small holes of area of cross-section on the opposite sides of a tank containing a liquid of density, The difference in height between the holes is $h$.
Tank is resting on a smooth horizontal surface. Horizontal force which will has to be
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Tank is resting on a smooth horizontal surface. Horizontal force which will has to be
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The correct answer is:
2 pagh
Net force $\left(\right.$ reaction) $=F=F_B-F_A=\frac{d p_B}{d t}-\frac{d p_A}{d t}$
$=a v_B \rho \times v_B-a v_A \rho \times v_A$
$F=a \rho\left(v_B^2-v_A^2\right)$
$\therefore$ According to Bernoulli's theorem
$p_A+\frac{1}{2} \rho v_A^2+\rho g h=p_B+\frac{1}{2} \rho v_B^2+0$
$\Rightarrow \frac{1}{2} \rho\left(v_B^2-v_A^2\right)=\rho g h \Rightarrow v_B^2-v_A^2=2 g h$
From equation (i), $F=2 a p g h$.
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