Two liquids having densities $d_1$ and $d_2$ are mixed in such a way that both have the same mass. The density of the mixture is ............

Two points $A$ and $B$ are at the same depth in a liquid at rest. Then the pressure on them $P_A$ ........... $P_B$.

Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so,why do the vessels filled with water to that same height give different readings on a weighing scale?

$A$ hollow sphere of radius $R$ is filled completely with an ideal liquid of density $\rho$. The sphere is moving horizontally with an acceleration $2g,$ where $g$ is the acceleration due to gravity. If the minimum pressure of the liquid is $P_0$,then what is the pressure at the centre of the sphere?

Consider a vessel filled with a liquid up to height $H$. The bottom of the vessel lies in the $X-Y$ plane passing through the origin. The density of the liquid varies with the $Z$-axis as $\rho(z) = \rho_0 \left[ 2 - \left( \frac{z}{H} \right)^2 \right]$. If $P_1$ and $P_2$ are the pressures at the bottom surface and top surface of the liquid respectively, the magnitude of $(P_1 - P_2)$ is:

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