If A is the area of the surface, $$\overline {\text{x}} $$ is the depth of its C.G. from the surface of the water and $$\omega $$ is the density of water, then A. Total pressure on the surface is equal to $$\omega \overline {\text{x}} $$ B. Depth of the point at which total pressure acts is equal to its moment of inertia divided by $${\text{A}}\overline {\text{x}} $$ C. Depth of the centre of pressure is $$\frac{2}{3}{\text{H}}\,$$ vertically below the surface D. All the above

The correct answer is D. All the above.

The total pressure on a surface submerged in a fluid is equal to the weight of the fluid above the surface. The depth of the point at which total pressure acts is equal to the depth of the centroid of the fluid above the surface. The depth of the centroid of a surface submerged in a fluid is equal to $\frac{2}{3}H$, where $H$ is the depth of the surface.

To prove that the total pressure on a surface submerged in a fluid is equal to the weight of the fluid above the surface, consider a rectangular surface of area $A$ and depth $h$ submerged in a fluid of density $\rho$. The weight of the fluid above the surface is equal to $\rho Ah$. The pressure at any point on the surface is equal to the weight of the fluid above that point divided by the area of the surface, or $\rho g h$. The total pressure on the surface is therefore equal to $\rho g Ah$, which is the weight of the fluid above the surface.

To prove that the depth of the point at which total pressure acts is equal to the depth of the centroid of the fluid above the surface, consider a rectangular surface of area $A$ and depth $h$ submerged in a fluid of density $\rho$. The centroid of the fluid above the surface is located at a depth of $\frac{2}{3}h$. The total pressure on the surface acts at a depth of $\frac{2}{3}h$, because the weight of the fluid above the surface is equal to the weight of the fluid above the centroid of the fluid above the surface.

To prove that the depth of the centroid of a surface submerged in a fluid is equal to $\frac{2}{3}H$, consider a rectangular surface of area $A$ and depth $h$ submerged in a fluid of density $\rho$. The centroid of the surface is located at a depth of $\frac{2}{3}h$. The centroid of the fluid above the surface is also located at a depth of $\frac{2}{3}h$. Therefore, the depth of the centroid of the surface is equal to the depth of the centroid of the fluid above the surface, which is equal to $\frac{2}{3}H$.