Question

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 20 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 of 4.0 m. Compute the force necessary to keep the door closed.

Answer 1

Given, Base area of the tank = 1 m²
Area of the hinged door = 20 cm²
Height of both water and acid columns, h = 4 m
Density of the water, $\rho_w$ = 1000 kg/m³
Relative density of acid is 1.7
so, density of the acid, $\rho_a$= 1.7 × $\rho_w$ = 1700 kg/m³

Pressure, P = ρgh
Where, ρ = density
g = acceleration due to gravity
h = height of the column

∴ Pressure exerted by water, $P_w$ = 1000 × 9.81 × 4
= 3.92 × 10⁴ N/m²

Pressure exerted by the acid, $P_a$ = 1700 × 9.81 × 4
= 6.664 × 10⁴  N/m²

∴ Pressure difference, ΔP = $P_a-P_b$
= (6.664-3.92) × 10⁴ N/m²
= 2.744 × 10⁴ N/m²

Thus, Force exerted on the hinged plate, 
F = change in pressure × area of hinged door
= 2.744 × 10⁴ × 20 × 10^-4
= 54.88 N

Answer 2

Explanation:

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