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(A) Let the height of the liquid in the vessel be $h$. The velocity of the emerging liquid is given by Torricelli's law as $v = \sqrt{2gh}$.The force

Vedclass · Accepted Answer

$\frac{2a}{A}$

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(A) Let the height of the liquid in the vessel be $h$. The velocity of the emerging liquid is given by Torricelli's law as $v = \sqrt{2gh}$.The force exerted by the emerging liquid on the vessel (thrust force) is $F_{thrust} = \rho a v^2 = \rho a (2gh) = 2 \rho agh$.For the vessel to not slide,the frictional force $f$ must balance this thrust force. Thus,$f \geq F_{thrust}$.The maximum frictional force is $f_{max} = \mu N$,where $N$ is the normal reaction from the surface. Since the vessel is light (mass is negligible),the normal reaction $N$ is equal to the weight of the liquid,$N = mg = (\rho A h) g$.Therefore,$\mu (\rho A h g) \geq 2 \rho a g h$.Simplifying this,we get $\mu \geq \frac{2a}{A}$.Thus,the minimum coefficient of friction is $\frac{2a}{A}$.

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