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.