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(C) Let the velocities of the liquid exiting from the upper hole $A$ and the lower hole $B$ be $v_A$ and $v_B$ respectively.According to the principle

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$2\rho agh$

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(C) Let the velocities of the liquid exiting from the upper hole $A$ and the lower hole $B$ be $v_A$ and $v_B$ respectively.According to the principle of conservation of momentum,the force exerted by the liquid jet is equal to the rate of change of momentum,$F = \frac{dp}{dt} = \rho a v^2$.Since the holes are on opposite sides,the forces act in opposite directions. The net force $F$ required to keep the tank in equilibrium is:$F = F_B - F_A = \rho a v_B^2 - \rho a v_A^2 = \rho a (v_B^2 - v_A^2)$Applying Bernoulli's theorem between the two holes,where $h$ is the vertical distance between them:$P_A + \frac{1}{2}\rho v_A^2 + \rho g h_A = P_B + \frac{1}{2}\rho v_B^2 + \rho g h_B$Taking the pressure at both holes to be atmospheric pressure $(P_A = P_B = P_{atm})$,we get:$\frac{1}{2}\rho v_A^2 + \rho g h = \frac{1}{2}\rho v_B^2 + 0$ (taking the lower hole as reference level $0$)$v_B^2 - v_A^2 = 2gh$Substituting this into the force equation:$F = \rho a (2gh) = 2\rho agh$.

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