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The vessel shown in the figure has two sections. The lower part is a rectangular vessel with area of cross-section $A$ and height $h$. The upper part is a conical vessel of height $h$ with base area $‘A’$ and top area $‘a’$ and the walls of the vessel are inclined at an angle $30^o$ with the vertical.A liquid of density $\rho$ fills both the sections upto a height $2h$. Neglecting atmospheric pressure.
The force $F $ exerted by the liquid on the base of the vessel is $2h\rho g$$\frac{{(A + a)}}{2}$
the pressure $P $ at the base of the vessel is $2h\rho g $$\frac{A}{a}$
the weight of the liquid $W $ is greater than the force exerted by the liquid on the base
the walls of the vessel exert a downward force $(F-W)$ on the liquid.
Solution
$F=\rho g 2 h A$
Pressure at bottom is $P=\rho g 2 h$
Weight of liquid is $W=\rho g(2 A h-(A-a) h / 2)=\rho g(3 A h / 2+a h / 2)$
We can clearly see $F>W$
Downward forces due to wall is excess forces apart from weight $=F-W$
