A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 10 cm2. 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?

Question

Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 2)

Answer 1

Concept

Pressure at the depth of h m in a liquid,

P = ρgh

The difference between the pressure of two sides of the partition multiplied by the area of the hinged door is equal to the force.

Calculation:

Given: ρ₁ = 1000 kg/m³, ρ₂ = 1.7 × 1000 kg/m³, h = 3 m and g = 9.8 m/s²

The pressure at the bottom of the tank on the side of the water.

⇒ P₁ = 1000 × 9.8 × 3 = 29400 Pa

The pressure at the bottom of the tank is at the side of the acid,

⇒ P₂ = 1.7 × 1000× 9.8 × 3 = 49980 Pa

The force necessary to keep the door closed

F = ΔP × A

⇒ F = (P₂ -P₁) × A

⇒ F = (49980 - 29400) × 10 × 10^-4 N

⇒ F = 20580 × 10^-3 N = 20.58N

⇒ F = 20.5N

Hence the force required to keep the door closed will be 20.5N