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Properties of Bulk Matter - Pressure in a Fluid

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Fluid Pressure: Pressure PP is defined as the normal force (thrust) acting per unit area AA. It is a scalar quantity. P=FAP = \frac{F}{A}.

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Pressure at a Point in a Fluid: The pressure exerted by a liquid column of height hh and density ρ\rho is given by P=hρgP = h \rho g.

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Pascal's Law: Pressure applied to any part of an enclosed incompressible fluid is transmitted undiminished in all directions to every portion of the fluid and to the walls of the vessel.

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Atmospheric Pressure: The pressure exerted by the atmosphere at sea level is approximately 1.013×105 Pa1.013 \times 10^5 \text{ Pa} or 1 atm1 \text{ atm}. It is measured using a mercury barometer.

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Gauge Pressure: The difference between the absolute pressure at a point and the atmospheric pressure is called gauge pressure: Pg=P−Pa=hρgP_g = P - P_a = h \rho g.

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Hydrostatic Paradox: The pressure at a certain depth in a fluid depends only on the vertical depth hh and the density ρ\rho, and is independent of the shape or cross-sectional area of the container.

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Archimedes' Principle: When a body is partially or wholly immersed in a fluid, it experiences an upward force (upthrust) equal to the weight of the fluid displaced by it: FB=VρlgF_B = V \rho_l g.

📐Formulae

P=FAP = \frac{F}{A}

P=Pa+hρgP = P_a + h \rho g

Pgauge=hρgP_{gauge} = h \rho g

F1A1=F2A2 (Hydraulic Lift Principle)\frac{F_1}{A_1} = \frac{F_2}{A_2} \text{ (Hydraulic Lift Principle)}

U=VinρfgU = V_{in} \rho_{f} g

Relative Density=Density of substanceDensity of water at 4∘C\text{Relative Density} = \frac{\text{Density of substance}}{\text{Density of water at } 4^\circ C}

💡Examples

Problem 1:

Calculate the pressure exerted by a column of water of height 10 m10 \text{ m} at its bottom. Take the density of water ρ=1000 kg m−3\rho = 1000 \text{ kg m}^{-3} and g=9.8 m s−2g = 9.8 \text{ m s}^{-2}. Also find the total pressure if atmospheric pressure is 1.013×105 Pa1.013 \times 10^5 \text{ Pa}.

Solution:

  1. Gauge Pressure: Pg=hρg=10×1000×9.8=9.8×104 PaP_g = h \rho g = 10 \times 1000 \times 9.8 = 9.8 \times 10^4 \text{ Pa}.
  2. Total Pressure: Ptotal=Pa+Pg=1.013×105+0.98×105=1.993×105 PaP_{total} = P_a + P_g = 1.013 \times 10^5 + 0.98 \times 10^5 = 1.993 \times 10^5 \text{ Pa}.

Explanation:

The pressure due to the liquid column is calculated using hρgh \rho g, and the total (absolute) pressure is the sum of the atmospheric pressure and the liquid pressure.

Problem 2:

In a hydraulic lift, the area of the smaller piston is 5 cm25 \text{ cm}^2 and that of the larger piston is 150 cm2150 \text{ cm}^2. If a force of 25 N25 \text{ N} is applied to the smaller piston, what is the force exerted on the larger piston?

Solution:

According to Pascal's Law: F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}. Given: F1=25 NF_1 = 25 \text{ N}, A1=5 cm2A_1 = 5 \text{ cm}^2, A2=150 cm2A_2 = 150 \text{ cm}^2. F2=F1×(A2A1)=25×(1505)=25×30=750 NF_2 = F_1 \times \left(\frac{A_2}{A_1}\right) = 25 \times \left(\frac{150}{5}\right) = 25 \times 30 = 750 \text{ N}.

Explanation:

Pascal's Law allows a small force applied over a small area to be transmitted as a larger force over a larger area, acting as a force multiplier.