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(A) The force exerted by the water jet exiting the hole is given by $F = \rho A v^2$,where $A$ is the area of the hole and $v$ is the velocity of the

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$\sqrt{\frac{2\mu M}{\pi \rho H}}$

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(A) The force exerted by the water jet exiting the hole is given by $F = \rho A v^2$,where $A$ is the area of the hole and $v$ is the velocity of the water.Using Torricelli's Law,the velocity of the water at the bottom is $v = \sqrt{2gH}$.Thus,the force is $F = \rho A (2gH)$.For the vessel to begin moving,this force must overcome the limiting friction force,$f_s = \mu Mg$.Equating the two: $\rho A (2gH) = \mu Mg$.Substituting $A = \frac{\pi D^2}{4}$,we get $\rho \left( \frac{\pi D^2}{4} \right) (2gH) = \mu Mg$.Simplifying the equation: $\frac{\rho \pi D^2 g H}{2} = \mu Mg$.Solving for $D$: $D^2 = \frac{2\mu M}{\pi \rho H}$.Therefore,$D = \sqrt{\frac{2\mu M}{\pi \rho H}}$.

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