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(D) Let point $1$ be the top surface of the water and point $2$ be the hole.Applying Bernoulli's equation between points $1$ and $2$:$P_1 + \frac{1}{2

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$4$

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(D) Let point $1$ be the top surface of the water and point $2$ be the hole.Applying Bernoulli's equation between points $1$ and $2$:$P_1 + \frac{1}{2} ho v_1^2 + ho g h_1 = P_2 + \frac{1}{2} ho v_2^2 + ho g h_2$Here,$P_1 = P_0 + \frac{F}{A} = P_0 + \frac{mg}{A}$,where $m = 50 \ kg$,$g = 10 \ m/s^2$,and $A = 0.5 \ m^2$.$P_1 = P_0 + \frac{50 	imes 10}{0.5} = P_0 + 1000 \ Pa$.$P_2 = P_0$ (atmospheric pressure).Height difference $h = h_1 - h_2 = 1.6 \ m - 0.9 \ m = 0.7 \ m$.Since the tank area is large,$v_1 \approx 0$.Substituting values:$(P_0 + 1000) + 0 + ho g (0.7) = P_0 + \frac{1}{2} ho v_2^2$$1000 + 1000 	imes 10 	imes 0.7 = \frac{1}{2} 	imes 1000 	imes v_2^2$$1000 + 7000 = 500 	imes v_2^2$$8000 = 500 	imes v_2^2$$v_2^2 = 16$$v_2 = 4 \ m/s$.

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