1. **Problem Statement:** We have a tank with three layers: mercury (depth $m$), water (depth $2m$), and oil (depth $1.2$ m) with specific gravity $0.92$. We need to find the pressure intensity at: i. The interface of oil and water. ii. The interface of water and mercury. 2. **Relevant Formula:** Pressure intensity at a depth $h$ in a fluid is given by: $$P = \rho g h$$ where $\rho$ is the fluid density, $g$ is acceleration due to gravity, and $h$ is the depth. 3. **Important Notes:** - Specific gravity (SG) relates fluid density to water density: $\rho = \text{SG} \times \rho_{water}$. - Density of water $\rho_{water} = 1000 \text{ kg/m}^3$. - Density of mercury $\rho_{Hg} = 13600 \text{ kg/m}^3$. - Gravity $g = 9.81 \text{ m/s}^2$. - Pressure at the surface is atmospheric, taken as zero gauge pressure. 4. **Calculations:** **i. Pressure at oil-water interface:** - Depth of oil layer $= 1.2$ m. - Density of oil $= 0.92 \times 1000 = 920 \text{ kg/m}^3$. - Pressure due to oil: $$P_{oil} = 920 \times 9.81 \times 1.2 = 10822.6 \text{ Pa}$$ **ii. Pressure at water-mercury interface:** - Depth of oil + water layers $= 1.2 + 2m = 1.2 + 2m$ m. - Pressure at oil-water interface $= 10822.6$ Pa (from above). - Density of water $= 1000 \text{ kg/m}^3$. - Pressure due to water layer: $$P_{water} = 1000 \times 9.81 \times 2m = 19620m \text{ Pa}$$ - Total pressure at water-mercury interface: $$P = P_{oil} + P_{water} = 10822.6 + 19620m \text{ Pa}$$ 5. **Summary:** - Pressure at oil-water interface: $$10822.6 \text{ Pa}$$ - Pressure at water-mercury interface: $$10822.6 + 19620m \text{ Pa}$$