A tank with a square base of area 1.0 $$\text{m}^2$$​ is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 10 $$\text{cm}^2$$​. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height 3.0 m. Calculate the force necessary to keep the door close?

Correct option is C

​​$$\textbf{Given:} \\\rho_1 = 1000 \, \text{kg/m}^3, \quad\rho_2 = 1.7 \times 1000 \, \text{kg/m}^3 = 1700 \, \text{kg/m}^3, \\h = 3 \, \text{m}, \quad g = 9.8 \, \text{m/s}^2 \\[10pt]\text{The pressure at the bottom of the tank on the side of the water:} \\P_1 = \rho_1 \cdot g \cdot h = 1000 \times 9.8 \times 3 = 29400 \, \text{Pa} \\[10pt]\text{The pressure at the bottom of the tank on the side of the acid:} \\P_2 = \rho_2 \cdot g \cdot h = 1700 \times 9.8 \times 3 = 49980 \, \text{Pa} \\[10pt]\text{The force necessary to keep the door closed:} \\F = \Delta P \cdot A = (P_2 - P_1) \cdot A \\[5pt]Given Area (A) = 10 \times 10^{-4} \\F = (49980 - 29400) \cdot 10 \times 10^{-4} \\F = 20580 \cdot 10^{-3} = 20.58 \, \text{N} \\[10pt]\boxed{F = 20.5 \, \text{N}} \\[10pt]\text{Hence, the force required to keep the door closed is } \boldsymbol{20.5 \, \text{N}}.$$​​