Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?
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Since the pressure depends on the height of the water column in the two vessels of different shapes is the same. It is due to water on the base of each vessel is same. It is expected that there will be equal force acting on the two base areas. The water exerts force on walls of the vessel also. In case, the walls of the vessel are not perpendicular to the base, the force exerted by water on the walls has a net non zero vertical component which is more in first vessel than that second vessel. That is why, the two vessels filled with water to same Vertical height show different readings on a weighing machine.
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Solution: Since h is the same for both vessels, pressure =hρg is the same. Now, force = pressure * base area. Since the base area is the same, the force exerted by water on the base is the same for both vessels. Thus, Fᵃ=Fᵇ.
The liquid exerts pressure, and hence force, not only on the base but also on the sides of the vessel. This force is normal to the sides of the vessel. For vessel B, this force has no component in the downward direction. But for the vessel A, this force has a non-zero component vertically downwards. Hence vessel A gives a higher reading than vessel B when they are weighed on a weighing scale. Thus, Wᵃ>Wᵇ.
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