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(C) Let $H$ be the height of the liquid in the tank and $A_h$ be the area of the hole. The area of cross-section of the tank is $A_t = N A_h$.The velo

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$\frac{g}{N}$

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(C) Let $H$ be the height of the liquid in the tank and $A_h$ be the area of the hole. The area of cross-section of the tank is $A_t = N A_h$.The velocity of efflux $v$ of the liquid from the hole at the center (at depth $h = H/2$) is given by Torricelli's law: $v = \sqrt{2gh} = \sqrt{2g(H/2)} = \sqrt{gH}$.The force $F$ exerted on the tank due to the efflux of the liquid is given by $F = \rho A_h v^2$,where $\rho$ is the density of the liquid.Substituting $v^2 = gH$,we get $F = \rho A_h (gH)$.The mass of the liquid in the tank is $M = \rho V = \rho (A_t H) = \rho (N A_h H)$.The initial acceleration $a$ of the tank is given by Newton's second law: $a = \frac{F}{M}$.$a = \frac{\rho A_h g H}{\rho N A_h H} = \frac{g}{N}$.

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