Two immiscible liquids $A$ and $B$ are kept in an $U$-tube. If the density of liquid $A$ is smaller than the density of liquid $B$,then the equilibrium situation is

Explain the hydrostatic paradox.

Two liquids of densities $\rho_{1}$ and $\rho_{2}$ (where $\rho_{2} = 2\rho_{1}$) are filled behind a vertical wall of height $10 \; m$ and width $w$,as shown in the figure. Each liquid has a height of $h = 5 \; m$. The ratio of the force exerted by the upper liquid on the part $MN$ to the force exerted by the lower liquid on the part $NO$ is (Assume that the liquids do not mix):

An open-ended $U$-tube of uniform cross-sectional area contains water (density $10^3 \ kg \ m^{-3}$). Initially,the water level stands at $0.29 \ m$ from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density $800 \ kg \ m^{-3}$ is added to the left arm until its length is $0.1 \ m$,as shown in the schematic figure. The ratio $\left(\frac{h_1}{h_2}\right)$ of the heights of the liquid in the two arms is-

$A$ tank $5 \,m$ high is half filled with water and then is filled to the top with oil of density $0.85 \,g/cm^3$. The pressure at the bottom of the tank,due to these liquids is ........ $g/cm^2$.

What is $1$ Torr pressure? And what is $1$ bar?