Revise notes Class 11 mechanical properties of fluids

Revise notes
Class 11 Physics
Mechanical Properties of Fluids

Topics

1. Introduction:Fluids
2. Pressure
3. Pressure in Fluids
4. Pascals Law
5. Variation of pressure with depth
6. Hydrostatic Paradox
7. Atmospheric Pressure
8. Gauge Pressure
9. Absolute Pressure
10. Pascals law for transmission of fluid pressure
11. Applications : Hydraulic Lift
12. Application : Hydraulic Brakes
13. Types of Fluid flow: Steady Flow
14. Equation of Continuity
15. Turbulent Flow
16. Bernoullis Principle
17. Torricellis law
18. Venturimeter
19. Dynamic Lift
20. Magnus Effect
21. Viscosity
22. Stokes Law
23. Terminal Velocity
24. Reynolds Number
25. Liquid Surfaces
26. Surface Energy
27. Surface Tension
28. Surface tension, energy: practical applications
29. Water stick to glass but not mercury
30. How detergents work
31. Angle of Contact
32. Drops and Bubbles
33. Capillary Rise

Absolute Pressure

Absolute pressure is defined as the pressure above the zero value of pressure.
It is the actual pressure which a substance has.
It is measured against the vacuum.
Absolute pressure is measured relative to absolute zero pressure.
It is sum of atmospheric pressure and gauge pressure.
P =P_a+hρg where P = pressure at any point, P_a = atmospheric pressure and hρg= gauge pressure.
Therefore P =P_a + Gauge Pressure. Where P = absolute pressure.
It is measured with the help of barometer.

Problem: The density of theatmosphere at sea level is 1.29 kg/m³.Assume that it does not change withaltitude. Then how high would theatmosphere extend?

Answer:

From equation: - P=P_a+ρgh

ρgh = 1.29 kg m^–3 × 9.8 m s² × h m = 1.01 × 10⁵ Pa

∴ h = 7989 m ≈ 8 km

In reality the density of air decreases withheight. So does the value of g. The atmospheric

cover extends with decreasing pressure over100 km. We should also note that the sea level

atmospheric pressure is not always 760 mm ofHg. A drop in the Hg level by 10 mm or more isa sign of an approaching storm.

Problem:- At a depth of 1000 m in anocean (a) what is the absolute pressure?(b) What is the gauge pressure? (c) Findthe force acting on the window of 20 cm × 20 cm of a submarine at this

depth, the interior of which is maintainedat sea-level atmospheric pressure. (Thedensity of sea water is 1.03 × 10³ kg m^-3,g = 10m s^–2.)

Answer:

Here h = 1000 m and ρ = 1.03 × 10³ kg m^-3.

(a) From Eq. P₂ − P₁= ρgh, absolute pressure

P = P_a + ρgh

= 1.01 × 10⁵ P_a+ 1.03 × 10³ kg m^–3 × 10 m s^–2 × 1000 m

= 104.01 × 10⁵ Pa

≈ 104 atm

(b) Gauge pressure is P − P_a = ρgh = P_g

P_g = 1.03 × 10³ kg m^–3 × 10 ms² × 1000 m

= 103 × 10⁵ Pa

≈ 103 atm

(c) The pressure outside the submarine isP = P_a + ρgh and the pressure inside it isP_a. Hence, the net pressure acting on thewindow is gauge pressure, P_g = ρgh. Sincethe area of the window is A = 0.04 m², theforce acting on it is

F = P_g A = (10³ × 10⁵ Pa) × 0.04 m² = 4.12 × 10⁵ N

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