Three identical vessels are filled to the same height with three different liquids $A, B$ and $C$ $(\rho_A > \rho_B > \rho_C)$. The pressure at the base will be

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_1$ and $d_2$ are filled in the tube. Each liquid subtends a $90^o$ angle at the centre. The radius joining their interface makes an angle $\alpha$ with the vertical. The ratio $\frac{d_1}{d_2}$ is

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

Three liquids of densities $d, 2d$,and $3d$ are mixed in equal volumes. Then the density of the mixture is

The pressure at half the depth of a lake is equal to two-third pressure at the bottom of the lake. So the depth '$h$' of the lake is ($\rho$ = density of water in the lake, $g$ = acceleration due to gravity, $P_0$ = atmospheric pressure).

$A$ $U$-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are initially in level with $10.0 \; cm$ of water in one arm and $12.5 \; cm$ of spirit in the other. If $15.0 \; cm$ of water and spirit each are further poured into the respective arms of the tube,what is the difference in the levels (in $cm$) of mercury in the two arms? (The specific gravity of spirit is $0.8$.)