Ncert Exercises & Solutions

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.

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

Pressure due to water is:

$P_{1} = h_{1} ρ_{1} g$
$= 4 \times 10^{3} \times 9.8$
$= 3.92 \times 10^{4} Pa$

Pressure due to acid is:

$P_{2} = h_{2} ρ_{2} g$
$= 4 \times 1.7 \times 10^{3} \times 9.8$
$= 6.664 \times 10^{4} Pa$

The pressure difference between the water and acid columns:

$∆ P = P_{2} - P_{1}$
$= 6.664 \times 10^{4} - 3.92 \times 10^{4}$
$= 2.744 \times 10^{4} Pa$

The force required to keep the door closed=the force exerted on the door = ΔP × A

= 2.744 × 10⁴ × 20 × 10^–4

= 54.88 N

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