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10.21 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 close.

0 2 Views | Posted 5 months ago
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    Answered by

    Vishal Baghel | Contributor-Level 10

    5 months ago

    Base area of the given tank, A = 1.0 m2

    Area of the hinged door, a = 20 cm2 = 20 ×10-4 m2

    Density of water, ρ1 = 103 kg/ m3

    Density of acid, ρ2 = 1.7 ×103 kg/ m3

    Height of the water column, h1 = 4 m

    Height of the acid column, h2 = 4 m

    Acceleration due to gravity, g = 9.8 m/ s2

    The pressure due to Water column, P1 = ρ1gh1 = 103×9.8×4 = 3.92 ×104 Pa

    The pressure due to Acid column, P2 = ρ2gh2 = 1.7 ×103×9.8×4 = 6.64 ×104 Pa

    Pressure difference between two columns ΔP = P2 - P1 = 2.774 ×104 Pa

    Force exerted on the sma

    ...more

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