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Mechanical Properties of Fluids - Pressure (Density, Atmospheric and Gauge Pressure, Hydraulic Machines)

Grade 11CBSEPhysics

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

🔑Concepts

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Density (ρ\rho) is defined as the mass per unit volume of a substance: ρ=mV\rho = \frac{m}{V}. The SI unit is kg/m3kg/m^3.

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Relative Density (Specific Gravity) is the ratio of the density of a substance to the density of water at 4∘C4^\circ C. It is a dimensionless quantity.

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Pressure (PP) is the normal force (thrust) exerted by a fluid per unit area: P=FAP = \frac{F}{A}. The SI unit is Pascal (PaPa), where 1Pa=1N/m21 Pa = 1 N/m^2.

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Pascal's Law states that pressure applied to any point of an enclosed fluid at rest is transmitted equally and undiminished to every other point of the fluid and to the walls of the container.

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Atmospheric Pressure (PaP_a) is the pressure exerted by the weight of the Earth's atmosphere. At sea level, Pa≈1.013×105PaP_a \approx 1.013 \times 10^5 Pa.

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Gauge Pressure is the difference between the absolute pressure (PP) at a point and the atmospheric pressure (PaP_a): Pg=P−PaP_g = P - P_a.

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Hydrostatic Pressure at a depth hh in a fluid of density ρ\rho is given by P=Pa+ρghP = P_a + \rho gh, where gg is the acceleration due to gravity.

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Hydraulic Machines, such as hydraulic lifts and brakes, work on Pascal's Law. A small force F1F_1 applied to a small area A1A_1 creates a pressure that results in a large force F2F_2 on a larger area A2A_2.

📐Formulae

ρ=mV\rho = \frac{m}{V}

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

P=Pa+ρghP = P_a + \rho g h

Pgauge=ρghP_{gauge} = \rho g h

F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}

Relative Density=ρsubstanceρwater at 4∘CRelative\ Density = \frac{\rho_{substance}}{\rho_{water\ at\ 4^\circ C}}

💡Examples

Problem 1:

Calculate the pressure at a depth of 1000 m1000\ m in the ocean. Take the density of sea water to be 1030 kg/m31030\ kg/m^3, g=9.8 m/s2g = 9.8\ m/s^2, and atmospheric pressure Pa=1.01×105 PaP_a = 1.01 \times 10^5\ Pa.

Solution:

Using the formula for absolute pressure: P=Pa+ρghP = P_a + \rho gh P=1.01×105+(1030×9.8×1000)P = 1.01 \times 10^5 + (1030 \times 9.8 \times 1000) P=1.01×105+100.94×105P = 1.01 \times 10^5 + 100.94 \times 10^5 P=101.95×105 PaP = 101.95 \times 10^5\ Pa

Explanation:

The total pressure at a depth consists of the pressure due to the water column (gauge pressure) plus the pressure of the atmosphere acting on the surface.

Problem 2:

In a hydraulic lift, the smaller piston has a radius of 5 cm5\ cm and the larger piston has a radius of 15 cm15\ cm. If a force of 50 N50\ N is applied to the smaller piston, what is the force exerted on the larger piston?

Solution:

Given r1=0.05 mr_1 = 0.05\ m and r2=0.15 mr_2 = 0.15\ m. Areas are A1=πr12A_1 = \pi r_1^2 and A2=πr22A_2 = \pi r_2^2. According to Pascal's Law: F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2} F2=F1×A2A1=F1×πr22πr12F_2 = F_1 \times \frac{A_2}{A_1} = F_1 \times \frac{\pi r_2^2}{\pi r_1^2} F2=50×(0.150.05)2F_2 = 50 \times \left(\frac{0.15}{0.05}\right)^2 F2=50×32=50×9=450 NF_2 = 50 \times 3^2 = 50 \times 9 = 450\ N

Explanation:

The pressure remains constant throughout the fluid. Since the area of the output piston is 9 times larger than the input piston, the output force is also 9 times larger.

Problem 3:

A tank contains water up to a height of 5 m5\ m. If the atmospheric pressure is 101325 Pa101325\ Pa, find the absolute pressure and gauge pressure at the bottom. (Take ρwater=1000 kg/m3\rho_{water} = 1000\ kg/m^3, g=10 m/s2g = 10\ m/s^2)

Solution:

Gauge Pressure: Pg=ρgh=1000×10×5=50000 PaP_g = \rho gh = 1000 \times 10 \times 5 = 50000\ Pa Absolute Pressure: P=Pa+PgP = P_a + P_g 101325+50000151325\begin{array}{r} 101325 \\ + 50000 \\ \hline 151325 \end{array} P=151325 PaP = 151325\ Pa

Explanation:

Gauge pressure is simply the pressure due to the liquid column, whereas absolute pressure is the sum of gauge pressure and atmospheric pressure.