A vessel with square cross-section and height of 6 m is vertically partitioned. A small window of 100 cm$$^2$$ with hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density $$1.5 \times 10^3$$ kg/m$$^3$$. What force one needs to apply on the hinged door so that it does not get opened?

(Acceleration due to gravity = 10 m/s$$^2$$)

The window has area $$A = 100\ \text{cm}^2$$.
Since $$1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$$, we get
$$A = 100 \times 10^{-4}\ \text{m}^2 = 0.01\ \text{m}^2$$.

Depth of the window below the liquid surfaces is $$h = 3\ \text{m}$$.

Pressure at depth $$h$$ inside a liquid of density $$\rho$$ is given by
$$P = \rho g h\quad$$(hydrostatic pressure formula).

Side-1 is filled with water: $$\rho_1 = 1000\ \text{kg m}^{-3}$$.
Pressure on this side:
$$P_1 = \rho_1 g h = 1000 \times 10 \times 3 = 3.0 \times 10^{4}\ \text{Pa}$$.

Side-2 is filled with the heavier liquid: $$\rho_2 = 1.5 \times 10^{3}\ \text{kg m}^{-3}$$.
Pressure on this side:
$$P_2 = \rho_2 g h = 1.5 \times 10^{3} \times 10 \times 3 = 4.5 \times 10^{4}\ \text{Pa}$$.

The window experiences a pressure difference
$$\Delta P = P_2 - P_1 = 4.5 \times 10^{4} - 3.0 \times 10^{4} = 1.5 \times 10^{4}\ \text{Pa}$$.

Resultant force on the window due to this pressure difference is
$$F = \Delta P \times A = 1.5 \times 10^{4} \times 0.01 = 150\ \text{N}$$.

Therefore, one must apply a force of $$150\ \text{N}$$ on the hinged door to keep it from opening.