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(B) Let the height of the liquid in the cylindrical vessel be $h$ and the radius of the vessel be $r$. The density of the liquid is $\rho$.The force e

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Radius of the vessel

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(B) Let the height of the liquid in the cylindrical vessel be $h$ and the radius of the vessel be $r$. The density of the liquid is $ho$.The force exerted by the liquid on the bottom of the vessel is given by the product of pressure at the bottom and the area of the base:$F_{	ext{bottom}} = P_{	ext{bottom}} 	imes A_{	ext{base}} = (h ho g) 	imes (\pi r^2)$.The pressure on the vertical side wall varies linearly with depth from $0$ at the surface to $h ho g$ at the bottom. The average pressure on the side wall is $P_{	ext{avg}} = \frac{0 + h ho g}{2} = \frac{1}{2} h ho g$.The area of the side wall (lateral surface area) is $A_{	ext{side}} = 2 \pi r h$.The force exerted on the side wall is $F_{	ext{side}} = P_{	ext{avg}} 	imes A_{	ext{side}} = (\frac{1}{2} h ho g) 	imes (2 \pi r h) = \pi r h^2 ho g$.According to the problem,$F_{	ext{bottom}} = F_{	ext{side}}$:$h ho g \pi r^2 = \pi r h^2 ho g$.Dividing both sides by $\pi r h ho g$,we get:$r = h$.Thus,the height of the liquid must be equal to the radius of the vessel.

The height to which a cylindrical vessel must be filled with a homogeneous liquid,so that the average force with which the liquid presses the side of the vessel is equal to the force exerted by the liquid on the bottom of the vessel,is equal to:

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