A tank with a square base of area 2 ${\mathrm{m}}^2$ is divided into two compartments by a vertical partition in the middle. There is a small hinged door of face area 20 ${\mathrm{cm}}^2$ at the bottom of the partition. Water is filled in one compartment and an acid of relative density 1.5 in the other, both to a height of 4 m. The force necessary to keep the door closed is (Take g = 10 ${\mathrm{ms}}^{-2}$)

The situation is as shown in the figure.

For compartment containing water,
H = 4m, ${\mathrm{\rho }}_{\mathrm{w}} = {10}^3 \mathrm{kg} {\mathrm{m}}^{-3}$
Pressure exerted by the water at the door at the bottom is
${\mathrm{P}}_{\mathrm{w}} = \mathrm{\rho hg} = {10}^3 \mathrm{kg} {\mathrm{m}}^{-3} \times 4 \mathrm{m} \times 10 {\mathrm{ms}}^{-2}$

       = $4\times {10}^4 \mathrm{N} {\mathrm{m}}^{-2}$

For compartment containing acid,
$\mathrm{\rho }$_a = 1.5 x ${10}^3 \mathrm{kg} {\mathrm{m}}^{-3}$, h = 4m
Pressure exerted by the acid at the door at the bottom is
${\mathrm{P}}_{\mathrm{a}} = {\mathrm{\rho }}_{\mathrm{a}}\mathrm{hg} = 1.5\times {10}^3 \mathrm{kg} {\mathrm{m}}^{-3} \times 4 \mathrm{m} \times 10 {\mathrm{ms}}^{-3}$

      = $(6\times {10}^4){\mathrm{Nm}}^{-2}$

$\therefore \mathrm{Net} \mathrm{pressure} \mathrm{on} \mathrm{the} \mathrm{door} = {\mathrm{P}}_{\mathrm{a}}-{\mathrm{P}}_{\mathrm{w}}$

$=(6\times {10}^4-4\times {10}^4){\mathrm{Nm}}^{-2}$

$= 2\times {10}^4 \mathrm{N} {\mathrm{m}}^{-2}$
Area of the door = 20 ${\mathrm{cm}}^2$ = 20 x ${10}^{-4}$ ${\mathrm{m}}^2$
Force = Pressure x Area = 40 N
Thus, to keep the door closed the force of 40 N must be applied horizontally from the water side.