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(C) Let the heights of the two holes from the free surface of the liquid be $h_1$ and $h_2$. Given that the difference in height is $h = h_2 - h_1$.Ac

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$2g h \rho a$

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(C) Let the heights of the two holes from the free surface of the liquid be $h_1$ and $h_2$. Given that the difference in height is $h = h_2 - h_1$.According to Torricelli's law,the velocity of efflux from the upper hole is $v_1 = \sqrt{2gh_1}$ and from the lower hole is $v_2 = \sqrt{2gh_2}$.The thrust force exerted by the liquid jet exiting from each hole is given by $F = \frac{dm}{dt} v = (a \rho v) v = a \rho v^2$.Thus,the force from the upper hole is $F_1 = a \rho v_1^2 = a \rho (2gh_1) = 2a \rho gh_1$.The force from the lower hole is $F_2 = a \rho v_2^2 = a \rho (2gh_2) = 2a \rho gh_2$.Since the holes are on opposite sides,the forces act in opposite directions. The net force on the tank is $F_{net} = F_2 - F_1 = 2a \rho g (h_2 - h_1)$.Substituting $h = h_2 - h_1$,we get $F_{net} = 2a \rho gh$.To keep the tank in equilibrium,an external force equal to $F_{net}$ must be applied.

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