A tank with a square base of area 2 m2 is divided into two compartments by a vertical partition in the middle. There is a small hinged door of face area 20 cm2 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 ms-2)

The situation is as shown in the figure.For compartment containing water,H = 4m, ρw = 103 kg m-3Pressure exerted by the water at the door at the bottom isPw = ρhg = 103 kg m-3 × 4 m × 10 ms-2       = 4×104 N m-2For compartment containing acid,ρa = 1.5 x 103 kg m-3, h = 4mPressure exerted by the acid at the door at the bottom isPa = ρahg = 1.5×103 kg m-3 ×4 m × 10 ms-3      = (6×104)Nm-2∴ Net pressure on the door = Pa-Pw=(6×104-4×104)Nm-2= 2×104 N m-2Area of the door = 20 cm2 = 20 x 10-4 m2Force = Pressure x Area = 40 NThus, to keep the door closed the force of 40 N must be applied horizontally from the water side.