Q. 213.6( 8 Votes )

# A tank with a square base of area 1.0 m^{2} is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm^{2}. 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 close.

Answer :

Given,

Base area of the tank = 1 m^{2}

Area of the hinged door = 20 cm^{2}

Height of both water and acid columns, h = 4 m

Density of the water, ρ_{w} = 1000 kg/m^{3}

Density of the acid, ρ_{a} = 1.7 × ρ_{w} = 1700 kg/m^{3}

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 × 10^{4} pa

Pressure exerted by the acid, P_{a} = 1700 × 9.81 × 4

= 6.664 × 10^{4} pa

∴ Pressure difference, ΔP = P_{a} – P_{w}

= (6.664-3.92) × 10^{4} pa

= 2.744 × 10^{4} pa

Thus, Force exerted on the hinged plate,

⇒ = 54.88 N

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