Two vessels A and B of different shapes have the same base area and are filled with water up to the same height h (see figure). The force exerted by water on the base is $F_{A}$ for vessel A and $F_{B}$ for vessel B. The respective weights of the water filled in vessels are $W_{A}$ and $W_{B}$ Then:

F_{A} > F_{B}; W_{A} > W_{B}

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F_{A} = F_{B}; W_{A} > W_{B}

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F_{A} = F_{B}; W_{A} < W_{B}

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F_{A} > F_{B}; W_{A} = W_{B}

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Solution

The correct option is B $F_{A} = F_{B}; W_{A} > W_{B}$
since h is same for both vessels,pressure $= ρ g h$ is the same.Now $f o r c e = p r e s s u r e \times$ base area.since the base area is the same,the force exerted by water on the base is the same for both vessels.thus $F_{A} = F_{B}$
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 side of the vessel.for vessel B,this force has no component in the downward direction.but for 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 the weighing scale.thus $W_{A} > W_{B}$

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Two vessels A and B of different shapes have the same base area and are filled with water up to the same height h (see figure). The force exerted by water on the base is FA for vessel A and FB for vessel B. The respective weights of the water filled in vessels are WA and WB Then: