The vessel shown in the figure has two sections of areas of cross-section A₁ and A₂. A liquid of density p fills both the sections, up to a height h in each. Neglect atmospheric pressure.

$\mathrm{a}) \mathrm{The} \mathrm{pressure} \mathrm{at} \mathrm{the} \mathrm{base} \mathrm{of} \mathrm{the} \mathrm{vessel} \mathrm{is} 2\mathrm{h} \mathrm{p} \mathrm{g}.$

$$\begin{gathered} \mathrm{b}) \mathrm{The} \mathrm{force} \mathrm{exerted} \mathrm{by} \mathrm{the} \mathrm{liquid} \mathrm{on} \mathrm{the} \mathrm{base} \mathrm{of} \mathrm{the} \\ \mathrm{vessel} \mathrm{is} 2\mathrm{h} \mathrm{p} {\mathrm{gA}}_2 \end{gathered}$$

$\mathrm{c}) \mathrm{The} \mathrm{weight} \mathrm{of} \mathrm{the} \mathrm{liquid} \mathrm{is} <2\mathrm{h} \mathrm{p} {\mathrm{gA}}_2.$

$$\begin{gathered} \mathrm{d}) \mathrm{The} \mathrm{walls} \mathrm{of} \mathrm{the} \mathrm{vessel} \mathrm{at} \mathrm{the} \mathrm{level} \mathrm{X} \mathrm{exert} \mathrm{a} \\ \mathrm{downward} \mathrm{force} \mathrm{hp} \mathrm{g}({\mathrm{A}}_2-{\mathrm{A}}_1) \mathrm{on} \mathrm{the} \mathrm{liquid} \end{gathered}$$

The entire height of the water column is calculated as h + h = 2h because we know that the same liquid is filled. The pressure at the bottom is calculated to be $\mathrm{\rho g}\left(2\mathrm{h}\right)=2\mathrm{\rho gh}$.

Force becomes $\mathrm{F}=\mathrm{PA}$ because, according to Pascal's rule of pressure, pressure is provided by $\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}$. The column's base now has an area of ${\mathrm{A}}_2$. Thus, the force at the bottom is ${\mathrm{PA}}_2=2{\mathrm{\rho ghA}}_2$.

Water in both the top and bottom of the column will exert weight, and it is obvious that the liquid's net weight will be${\mathrm{\rho ghA}}_1+{\mathrm{\rho ghA}}_2$, which is $<2{\mathrm{hpgA}}_2$

${\mathrm{F}}_{\mathrm{at} \mathrm{level} \mathrm{X}}=\mathrm{\rho gh}({\mathrm{A}}_2-{\mathrm{A}}_1)$, since Force is equal to pressure multiplied by area. 

Hence the correct answer is all are correct.