There are two identical small holes of 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 $h$. Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is
Diffcult
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(c)Net force (reaction) = $F = {F_B} - {F_A}$$ = \frac{{d{p_B}}}{{dt}} - \frac{{d{p_A}}}{{dt}}$
$ = a{v_B}\rho \times {v_B} - a{v_A}\rho \times {v_A}$
$F = a\rho \left( {v_B^2 - v_A^2} \right)$…(i)
According to Bernoulli's theorem
${p_A} + \frac{1}{2}\rho v_A^2 + \rho gh = {p_B} + \frac{1}{2}\rho v_B^2 + 0$
==> $\frac{1}{2}\rho \left( {v_B^2 - v_A^2} \right) = \rho gh$==> $v_B^2 - v_A^2 = 2gh$
From equation (i), $F = 2a\rho gh.$
$ = a{v_B}\rho \times {v_B} - a{v_A}\rho \times {v_A}$
$F = a\rho \left( {v_B^2 - v_A^2} \right)$…(i)
According to Bernoulli's theorem
${p_A} + \frac{1}{2}\rho v_A^2 + \rho gh = {p_B} + \frac{1}{2}\rho v_B^2 + 0$
==> $\frac{1}{2}\rho \left( {v_B^2 - v_A^2} \right) = \rho gh$==> $v_B^2 - v_A^2 = 2gh$
From equation (i), $F = 2a\rho gh.$
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