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(C) Let $h_1$ be the depth of the upper hole from the free surface and $h_2$ be the depth of the lower hole from the free surface. Given $h_2 - h_1 =

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$2g h \rho a$

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(C) Let $h_1$ be the depth of the upper hole from the free surface and $h_2$ be the depth of the lower hole from the free surface. Given $h_2 - h_1 = h$.According to Torricelli's law,the velocity of efflux for the upper hole is $v_1 = \sqrt{2gh_1}$ and for the lower hole is $v_2 = \sqrt{2gh_2}$.The force exerted by the liquid jet from the upper hole is $F_1 = \rho a v_1^2 = \rho a (2gh_1) = 2\rho agh_1$.The force exerted by the liquid jet from the lower hole is $F_2 = \rho a v_2^2 = \rho a (2gh_2) = 2\rho agh_2$.Since the holes are on opposite sides,the forces act in opposite directions. The net force on the tank is $F_{net} = F_2 - F_1 = 2\rho ag(h_2 - h_1)$.Substituting $h_2 - h_1 = h$,we get $F_{net} = 2\rho agh$.To keep the tank in equilibrium,an external force equal to $2\rho agh$ must be applied.

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