Mechanical Properties of Fluids NCERT Solutions Class 11 Physics - Solved Exercise Question 10.21

Question 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.
Solution:
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
Pressure due to water is given as:
P1 = h1ρ1g
= 4 X 103 X 9.8  =  3.92 X 104 Pa
Pressure due to acid is given as:
P2 = h2ρ2g
= 4 X 1.7 X 103 X 9.8  =  6.664 X 104 Pa
Pressure difference between the water and acid columns:
ΔP = P2 - P1
= 6.664 X 104 - 3.92 X 104
= 2.744 X 104 Pa
Hence, the force exerted on the door = ΔP × a
= 2.744 × 104 × 20 × 10–4
= 54.88 N
Therefore, the force necessary to keep the door closed is 54.88 N.

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