CLASS 11 PHYSICS CHAPTER 9 – MECHANICAL PROPERTIES OF FLUIDS
NCERT CLASS 11 PHYSICS
CHAPTER 9 – MECHANICAL PROPERTIES OF FLUIDS
IN-TEXT QUESTIONS AND ANSWERS
1. What are fluids?
Fluids are substances that can flow and do not have a definite shape of their own. Liquids and gases are called fluids.
Examples: Water, air, oil and petrol.
2. What is pressure?
Pressure is defined as the normal force acting per unit area of a surface.
Pressure = Force / Area
P = F/A
SI unit: Pascal (Pa) or N/m².
3. Why does a sharp knife cut better than a blunt knife?
A sharp knife has a smaller contact area than a blunt knife. For the same applied force, a smaller area produces greater pressure.
Therefore, a sharp knife cuts more easily.
4. What is density?
Density is defined as the mass per unit volume of a substance.
ρ = m/V
Where:
ρ = Density
m = Mass
V = Volume
SI unit: kg/m³.
5. What is relative density?
Relative density is the ratio of the density of a substance to the density of water at 4°C.
Relative density = Density of substance / Density of water
It has no unit because it is a ratio of two quantities having the same unit.
6. State Pascal's law.
Pascal's law states that when external pressure is applied to an enclosed fluid, the pressure is transmitted undiminished and equally in all directions throughout the fluid.
Applications:
Hydraulic lift
Hydraulic brakes
Hydraulic press
7. What is hydrostatic pressure?
Hydrostatic pressure is the pressure exerted by a fluid at rest due to its weight.
At a depth h below the surface of a liquid:
P = ρgh
Absolute pressure:
P = Pₐ + ρgh
Where Pₐ is atmospheric pressure.
8. What is the difference between absolute pressure and gauge pressure?
Absolute pressure is the total pressure measured with respect to a perfect vacuum.
Gauge pressure is the pressure measured above atmospheric pressure.
Gauge pressure = Absolute pressure − Atmospheric pressure
Pᵍ = P − Pₐ
9. What is atmospheric pressure?
Atmospheric pressure is the pressure exerted by the weight of the air surrounding the Earth.
At sea level, its average value is:
1 atm = 1.013 × 10⁵ Pa
10. What is a mercury barometer?
A mercury barometer is an instrument used to measure atmospheric pressure. It consists of a glass tube filled with mercury and inverted into a mercury trough.
At normal atmospheric pressure, the height of the mercury column is approximately 76 cm.
11. What is streamline flow?
Streamline flow is the type of fluid flow in which every fluid particle follows a smooth path and the velocity at any fixed point remains constant with time.
In streamline flow, two streamlines never intersect.
12. What is turbulent flow?
Turbulent flow is a type of fluid flow in which the motion of fluid particles is irregular and their velocities at a point change randomly with time.
It generally occurs when the flow speed is sufficiently high.
13. State the equation of continuity.
The equation of continuity is based on the law of conservation of mass.
For an incompressible fluid flowing through a pipe:
A₁v₁ = A₂v₂
Where:
A₁ and A₂ are the cross-sectional areas.
v₁ and v₂ are the flow speeds.
It means that the speed of a fluid increases when it passes through a narrower section of a pipe.
14. State Bernoulli's principle.
Bernoulli's principle states that for steady flow of an ideal, incompressible and non-viscous fluid, the sum of pressure energy, kinetic energy and potential energy per unit volume remains constant along a streamline.
P + ½ρv² + ρgh = Constant
15. What is viscosity?
Viscosity is the property of a fluid by which it opposes the relative motion between its adjacent layers.
It is sometimes described as the internal friction of a fluid.
Example: Honey has greater viscosity than water.
16. Define coefficient of viscosity.
The coefficient of viscosity is the ratio of shearing stress to the rate of shearing strain.
η = Shearing stress / Rate of shearing strain
SI unit: Pa s or N s/m².
17. State Stokes' law.
Stokes' law states that the viscous drag force acting on a small sphere moving slowly through a viscous fluid is directly proportional to the radius of the sphere, its speed and the coefficient of viscosity of the fluid.
F = 6πηrv
Where:
η = Coefficient of viscosity
r = Radius of sphere
v = Speed of sphere
18. What is terminal velocity?
Terminal velocity is the constant maximum speed acquired by a body falling through a viscous fluid when the net force acting on it becomes zero.
At terminal velocity:
Weight = Buoyant force + Viscous drag
19. What is surface tension?
Surface tension is the tangential force acting per unit length along an imaginary line drawn on the free surface of a liquid.
T = F/l
SI unit: N/m.
20. Why are small liquid drops spherical?
A liquid surface tends to have minimum surface area because of surface tension. For a given volume, a sphere has the minimum surface area.
Therefore, small liquid drops are spherical in shape when external forces are negligible.
21. What is angle of contact?
The angle of contact is the angle between the tangent to the liquid surface and the solid surface at the point of contact, measured inside the liquid.
Water on clean glass generally has an acute angle of contact, whereas mercury on glass has an obtuse angle of contact.
22. What is capillary action?
Capillary action is the rise or fall of a liquid in a narrow tube due to surface tension and the forces of adhesion and cohesion.
Examples:
Rise of water in a narrow glass tube.
Absorption of water by a towel.
Movement of water through the roots and stems of plants.
EXERCISE QUESTIONS AND ANSWERS
Q9.1 Explain why:
(a) The blood pressure in humans is greater at the feet than at the brain.
Answer:
The pressure in a fluid increases with depth.
P = P₀ + ρgh
The feet are at a lower level than the brain. Therefore, the height of the blood column above the feet is greater than that above the brain.
Hence, blood pressure is greater at the feet than at the brain.
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at sea level, though the height of the atmosphere is more than 100 km.
Answer:
Air is compressible and its density is not uniform throughout the atmosphere.
The density of air is maximum near the Earth's surface and decreases rapidly with increasing height.
A large portion of the atmosphere's mass is concentrated near the Earth's surface. Therefore, atmospheric pressure decreases considerably with height and becomes nearly half at a height of about 6 km.
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.
Answer:
Pressure is defined as the normal component of force per unit area.
P = F/A
The force acting on a surface has a direction, but pressure itself has no fixed direction. At a point in a fluid at rest, pressure acts equally in all directions.
Therefore, hydrostatic pressure is a scalar quantity.
Q9.2 Explain why:
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
Answer:
The angle of contact depends on the relative strengths of adhesive and cohesive forces.
In mercury, the cohesive force between mercury molecules is greater than the adhesive force between mercury and glass. Therefore, mercury does not wet glass and forms an obtuse angle of contact.
In water, the adhesive force between water and glass is greater than the cohesive force between water molecules. Therefore, water wets glass and forms an acute angle of contact.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops.
Answer:
Water has a greater attraction towards clean glass than towards itself. Therefore, it spreads over the glass surface.
Mercury has stronger cohesive forces between its molecules than adhesive forces with glass. Therefore, it does not spread and tends to form drops.
(c) Surface tension of a liquid is independent of the area of the surface.
Answer:
Surface tension is a property of the liquid and depends mainly on the nature of the liquid and its temperature.
At a fixed temperature, the force per unit length along the surface remains constant, irrespective of the total surface area.
Therefore, surface tension is independent of the area of the surface.
(d) Water with detergent dissolved in it should have small angles of contact.
Answer:
Detergent reduces the surface tension of water and improves its wetting ability.
As a result, water spreads more easily over surfaces and forms a smaller angle of contact.
(e) A drop of liquid under no external forces is always spherical in shape.
Answer:
Surface tension tries to reduce the surface area of a liquid drop.
For a given volume, a sphere has the minimum surface area among all possible shapes.
Therefore, in the absence of external forces, a liquid drop takes a spherical shape.
Q9.3 Fill in the blanks using the correct words:
(a) Surface tension of liquids generally increases/decreases with temperature.
Answer: Decreases.
Surface tension generally decreases as temperature increases.
(b) Viscosity of gases increases/decreases with temperature, whereas viscosity of liquids increases/decreases with temperature.
Answer: Viscosity of gases increases with temperature, whereas viscosity of liquids decreases with temperature.
(c) For solids with elastic modulus of rigidity, the shearing force is proportional to ______, while for fluids it is proportional to ______.
Answer:
For solids, shearing force is proportional to shear strain.
For fluids, shearing force is proportional to rate of shear strain.
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows conservation of mass/Bernoulli's principle.
Answer: Conservation of mass.
The equation of continuity explains why the speed of an incompressible fluid increases at a narrow section.
(e) For the model of a plane in a wind tunnel, turbulence occurs at a greater/smaller speed than the speed for turbulence for an actual plane.
Answer: Greater.
A smaller model generally requires a greater speed to reach the same Reynolds number as the actual aeroplane.
Q9.4 Explain why:
(a) To keep a piece of paper horizontal, you should blow over, not under, it.
Answer:
When air is blown over the upper surface of the paper, the speed of air above it increases.
According to Bernoulli's principle, an increase in fluid speed is associated with a decrease in pressure.
Therefore, pressure above the paper becomes lower than the pressure below it. The greater pressure below the paper supports it and helps keep it horizontal.
(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
Answer:
When the openings are reduced, the cross-sectional area available for water flow decreases.
According to the equation of continuity:
Av = Constant
The speed of water increases as it passes through the narrow openings.
Therefore, fast jets of water come out between the fingers.
(c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.
Answer:
For laminar flow through a narrow cylindrical tube, Poiseuille's law gives:
Q = πr⁴ΔP/(8ηl)
Where:
Q = Volume flow rate
r = Radius of the needle
ΔP = Pressure difference
η = Coefficient of viscosity
l = Length of the needle
The flow rate depends on the fourth power of the radius of the needle.
Therefore, even a small change in the radius can produce a large change in the flow rate. Hence, the size of the needle controls the flow rate very effectively.
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.
Answer:
When fluid flows out of a hole, it carries momentum in the direction of the outgoing jet.
According to the law of conservation of momentum, the vessel experiences an equal and opposite reaction.
Therefore, the vessel experiences a backward thrust.
(e) A spinning cricket ball in air does not follow a parabolic trajectory.
Answer:
A spinning cricket ball produces different air speeds and pressures on its two sides.
According to Bernoulli's principle, the pressure is lower on the side where the air speed is higher.
This pressure difference produces a sideways force on the ball. This effect is known as the Magnus effect.
Therefore, a spinning cricket ball does not follow a simple parabolic trajectory.
Q9.5 A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter of 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Answer:
Given:
Mass of girl, m = 50 kg
Diameter of heel = 1.0 cm
Radius:
r = 0.5 cm = 0.005 m
g = 9.8 m/s²
Step 1: Calculate the force exerted by the girl.
F = mg
F = 50 × 9.8
F = 490 N
Step 2: Calculate the area of the heel.
A = πr²
A = 3.14 × (0.005)²
A = 7.85 × 10⁻⁵ m²
Step 3: Calculate pressure.
P = F/A
P = 490/(7.85 × 10⁻⁵)
P ≈ 6.24 × 10⁶ Pa
Final Answer:
The pressure exerted by the heel on the floor is approximately:
6.24 × 10⁶ Pa
Q9.6 Torricelli's barometer used mercury. Pascal duplicated it using French wine of density 984 kg m⁻³. Determine the height of the wine column for normal atmospheric pressure.
Answer:
Given:
Density of wine, ρ = 984 kg/m³
Atmospheric pressure, P = 1.013 × 10⁵ Pa
g = 9.8 m/s²
Using the barometer relation:
P = ρgh
Therefore:
h = P/(ρg)
Substituting:
h = (1.013 × 10⁵)/(984 × 9.8)
h ≈ 10.51 m
Final Answer:
The height of the wine column is approximately:
10.5 m
Q9.7 A vertical off-shore structure is built to withstand a maximum stress of 10⁹ Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.
Answer:
Given:
Maximum stress = 10⁹ Pa
Depth of ocean, h = 3 km = 3000 m
Density of seawater, ρ ≈ 1.03 × 10³ kg/m³
g = 9.8 m/s²
Pressure due to seawater:
P = ρgh
P = 1.03 × 10³ × 9.8 × 3000
P ≈ 3.03 × 10⁷ Pa
Comparing:
Maximum stress the structure can withstand = 10⁹ Pa
Pressure at 3 km depth ≈ 3.03 × 10⁷ Pa
Since:
3.03 × 10⁷ Pa is less than 10⁹ Pa,
the structure can withstand the hydrostatic pressure at this depth.
Final Answer:
Yes, the structure is suitable based on the given hydrostatic-pressure calculation. The pressure at 3 km depth is approximately 3.03 × 10⁷ Pa, which is much smaller than the maximum stress of 10⁹ Pa.
Q9.8 A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm². What maximum pressure would the smaller piston have to bear?
Answer:
Given:
Mass of car, m = 3000 kg
Area of larger piston:
A = 425 cm²
A = 425 × 10⁻⁴ m²
A = 0.0425 m²
g = 9.8 m/s²
Step 1: Calculate the weight of the car.
F = mg
F = 3000 × 9.8
F = 29,400 N
Step 2: Calculate pressure.
According to Pascal's law, the pressure is transmitted equally throughout the enclosed fluid.
P = F/A
P = 29,400/0.0425
P ≈ 6.92 × 10⁵ Pa
Final Answer:
The maximum pressure that the smaller piston must produce is:
6.92 × 10⁵ Pa
Q9.9 A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?
Answer:
Given:
Height of water column, h₁ = 10 cm
Height of spirit column, h₂ = 12.5 cm
The mercury levels are at the same height.
Therefore, the pressure exerted by the two liquid columns is equal.
ρ₁gh₁ = ρ₂gh₂
Cancelling g:
ρ₁h₁ = ρ₂h₂
Specific gravity of spirit:
S = ρ₂/ρ₁
S = h₁/h₂
S = 10/12.5
S = 0.8
Final Answer:
The specific gravity of spirit is:
0.8
Q9.10 In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)
Answer:
Given:
Additional height of water = 15 cm
Additional height of spirit = 15 cm
Specific gravity of spirit = 0.8
Specific gravity of mercury = 13.6
From the previous question, the specific gravity of spirit is 0.8.
Step 1: Calculate the pressure difference produced by the additional liquid columns.
Pressure difference in terms of water column:
ΔP = ρw g × 15 − ρs g × 15
Since:
ρs = 0.8ρw
ΔP = ρw g × 15 − 0.8ρw g × 15
ΔP = 0.2ρw g × 15
ΔP = ρw g × 3 cm
Thus, the pressure difference is equivalent to a 3 cm column of water.
Step 2: Convert this pressure difference into a mercury column.
Let h be the difference in mercury levels.
ρHg gh = ρw g × 3 cm
h = 3 × (ρw/ρHg)
h = 3/13.6
h = 0.2206 cm
Converting into millimetres:
h = 2.206 mm
Final Answer:
The difference in mercury levels is approximately:
2.2 mm
Q9.11 Can Bernoulli's equation be used to describe the flow of water through a rapid in a river? Explain.
Answer:
Bernoulli's equation is applicable to steady, non-viscous and incompressible fluid flow along a streamline.
Water flowing through a rapid in a river generally has irregular and turbulent motion. Viscous effects and energy losses are also important.
Therefore, Bernoulli's equation cannot be applied accurately to describe the entire flow of water through a rapid.
Q9.12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli's equation? Explain.
Answer:
No, it does not matter, provided the same atmospheric pressure is used at all points.
Bernoulli's equation is:
P + ½ρv² + ρgh = Constant
Gauge pressure is:
Pᵍ = P − Pₐ
Substituting gauge pressure:
Pᵍ + Pₐ + ½ρv² + ρgh = Constant
Since atmospheric pressure Pₐ is the same at all points, it is a common constant and can be removed from both sides.
Therefore, Bernoulli's equation can be applied using either absolute pressure or gauge pressure.
Q9.13 Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10⁻³ kg s⁻¹, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 10³ kg m⁻³ and viscosity of glycerine = 0.83 Pa s.) Also check whether the assumption of laminar flow is correct.
Answer:
Given:
Length of tube, l = 1.5 m
Radius, r = 1.0 cm = 0.01 m
Mass flow rate = 4.0 × 10⁻³ kg/s
Density, ρ = 1.3 × 10³ kg/m³
Viscosity, η = 0.83 Pa s
Step 1: Calculate the volume flow rate.
Mass flow rate = ρQ
Q = Mass flow rate/ρ
Q = (4.0 × 10⁻³)/(1.3 × 10³)
Q ≈ 3.08 × 10⁻⁶ m³/s
Step 2: Apply Poiseuille's law.
Q = πr⁴ΔP/(8ηl)
Rearranging:
ΔP = 8ηlQ/(πr⁴)
Substituting:
ΔP = [8 × 0.83 × 1.5 × 3.08 × 10⁻⁶]/[3.14 × (0.01)⁴]
ΔP ≈ 9.75 × 10⁴ Pa
Step 3: Check whether the flow is laminar.
Average speed:
v = Q/(πr²)
v = (3.08 × 10⁻⁶)/(3.14 × 10⁻⁴)
v ≈ 9.8 × 10⁻³ m/s
Reynolds number:
Re = ρv(2r)/η
Re = (1.3 × 10³ × 9.8 × 10⁻³ × 0.02)/0.83
Re ≈ 0.31
Since the Reynolds number is much smaller than the usual critical value for pipe flow, the flow is laminar.
Final Answer:
Pressure difference ≈ 9.75 × 10⁴ Pa
Reynolds number ≈ 0.31
Therefore, the assumption of laminar flow is correct.
Q9.14 In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s⁻¹ and 63 m s⁻¹ respectively. What is the lift on the wing if its area is 2.5 m²? Take the density of air to be 1.3 kg m⁻³.
Answer:
Given:
Speed above the wing, v₁ = 70 m/s
Speed below the wing, v₂ = 63 m/s
Area of wing, A = 2.5 m²
Density of air, ρ = 1.3 kg/m³
According to Bernoulli's principle:
P + ½ρv² = Constant
At the upper and lower surfaces:
P₁ + ½ρv₁² = P₂ + ½ρv₂²
Rearranging:
P₂ − P₁ = ½ρ(v₁² − v₂²)
The lift force is:
F = (P₂ − P₁)A
Therefore:
F = ½ρA(v₁² − v₂²)
Substituting:
F = ½ × 1.3 × 2.5 × (70² − 63²)
F = 1.625 × (4900 − 3969)
F = 1.625 × 931
F ≈ 1513 N
Final Answer:
The lift on the wing is approximately:
1.51 × 10³ N
Q9.15 Figures 9.20(a) and (b) refer to the steady flow of a non-viscous liquid. Which of the two figures is incorrect? Why?
Answer:
Figure 9.20(b) is incorrect.
According to the equation of continuity:
Av = Constant
When a liquid passes through a narrow section of a pipe, its speed increases.
According to Bernoulli's principle:
P + ½ρv² = Constant
An increase in speed produces a decrease in pressure.
Therefore, the pressure at the narrow section must be lower than the pressure at the wider section. The liquid column in the vertical tube connected to the narrow section should consequently be lower.
Figure 9.20(a) shows this correctly, whereas Figure 9.20(b) shows the opposite pressure difference.
Final Answer:
Figure 9.20(b) is incorrect because the pressure at the narrow section should be lower, not higher.
Q9.16 The cylindrical tube of a spray pump has a cross-section of 8.0 cm², one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min⁻¹, what is the speed of ejection of the liquid through the holes?
Answer:
Given:
Area of tube, A₁ = 8.0 cm²
A₁ = 8.0 × 10⁻⁴ m²
Number of holes = 40
Diameter of each hole = 1.0 mm
Radius of each hole:
r = 0.5 mm = 5 × 10⁻⁴ m
Speed inside tube:
v₁ = 1.5 m/min
v₁ = 1.5/60
v₁ = 0.025 m/s
Step 1: Calculate the total area of the holes.
Area of one hole:
A = πr²
A = 3.14 × (5 × 10⁻⁴)²
A = 7.85 × 10⁻⁷ m²
Total area of 40 holes:
A₂ = 40 × 7.85 × 10⁻⁷
A₂ = 3.14 × 10⁻⁵ m²
Step 2: Apply the equation of continuity.
A₁v₁ = A₂v₂
v₂ = A₁v₁/A₂
v₂ = (8.0 × 10⁻⁴ × 0.025)/(3.14 × 10⁻⁵)
v₂ ≈ 0.637 m/s
Final Answer:
The speed of ejection of liquid through the holes is approximately:
0.64 m/s
Q9.17 A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10⁻² N, which includes the small weight of the slider. The length of the slider is 30 cm. What is the surface tension of the film?
Answer:
Given:
Weight supported, F = 1.5 × 10⁻² N
Length of slider, l = 30 cm = 0.30 m
A soap film has two surfaces. Therefore, the total force due to surface tension is:
F = 2Tl
Where T is the surface tension.
Rearranging:
T = F/(2l)
Substituting:
T = (1.5 × 10⁻²)/(2 × 0.30)
T = (1.5 × 10⁻²)/0.60
T = 0.025 N/m
Final Answer:
The surface tension of the soap film is:
2.5 × 10⁻² N/m
Q9.18 Figure 9.21(a) shows a thin liquid film supporting a small weight of 4.5 × 10⁻² N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.
Answer:
Given:
Weight supported in Fig. (a) = 4.5 × 10⁻² N
Length of the movable wire = 40 cm in all three cases.
Surface tension depends on the nature of the liquid and its temperature. Since the liquid and temperature are the same, the surface tension remains the same.
The force exerted by a liquid film is:
F = 2Tl
The factor 2 appears because the film has two surfaces.
In all three figures, the length of the movable boundary is 40 cm. Therefore, the force due to surface tension is the same.
Final Answers:
Weight supported in Fig. (b) = 4.5 × 10⁻² N
Weight supported in Fig. (c) = 4.5 × 10⁻² N
Explanation:
The shape of the film does not change the surface tension. Since the length of the movable boundary and the liquid temperature are unchanged, the force remains the same.
Q9.19 What is the pressure inside a drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10⁻¹ N m⁻¹. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.
Answer:
Given:
Radius of mercury drop:
r = 3.00 mm = 3 × 10⁻³ m
Surface tension:
T = 4.65 × 10⁻¹ N/m
Atmospheric pressure:
Pₐ = 1.01 × 10⁵ Pa
For a liquid drop, excess pressure is:
ΔP = 2T/r
Substituting:
ΔP = (2 × 4.65 × 10⁻¹)/(3 × 10⁻³)
ΔP = 0.93/0.003
ΔP = 310 Pa
Step 2: Calculate the total pressure inside the drop.
P = Pₐ + ΔP
P = 1.01 × 10⁵ + 310
P = 101,310 Pa
P ≈ 1.0131 × 10⁵ Pa
Final Answers:
Excess pressure inside the drop = 310 Pa
Total pressure inside the drop ≈ 1.0131 × 10⁵ Pa
Q9.20 What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 × 10⁻² N m⁻¹? If an air bubble of the same dimension were formed at a depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 10⁵ Pa.)
Answer:
Given:
Radius of bubble:
r = 5.00 mm = 5 × 10⁻³ m
Surface tension:
T = 2.50 × 10⁻² N/m
Depth:
h = 40 cm = 0.40 m
Relative density of soap solution = 1.20
Atmospheric pressure:
Pₐ = 1.01 × 10⁵ Pa
g = 9.8 m/s²
Part 1: Excess pressure inside a soap bubble
A soap bubble has two surfaces.
Therefore:
ΔP = 4T/r
Substituting:
ΔP = (4 × 2.50 × 10⁻²)/(5 × 10⁻³)
ΔP = 0.10/0.005
ΔP = 20 Pa
Answer: Excess pressure inside the soap bubble = 20 Pa.
Part 2: Pressure inside an air bubble at a depth of 40 cm
First, calculate the density of the soap solution.
ρ = Relative density × Density of water
ρ = 1.20 × 1000
ρ = 1200 kg/m³
Pressure outside the air bubble at depth h:
Pₒ = Pₐ + ρgh
Pₒ = 1.01 × 10⁵ + 1200 × 9.8 × 0.40
Pₒ = 1.01 × 10⁵ + 4704
Pₒ = 105,704 Pa
An air bubble inside a liquid has only one liquid-air surface.
Therefore, excess pressure inside it is:
ΔP = 2T/r
ΔP = (2 × 2.50 × 10⁻²)/(5 × 10⁻³)
ΔP = 0.05/0.005
ΔP = 10 Pa
Pressure inside the air bubble:
P = Pₒ + ΔP
P = 105,704 + 10
P = 105,714 Pa
Final Answers:
Excess pressure inside the soap bubble = 20 Pa.
Pressure inside the air bubble at 40 cm depth ≈ 1.057 × 10⁵ Pa.
IMPORTANT QUESTIONS AND ANSWERS
1. Define pressure and write its SI unit.
Answer:
Pressure is the normal force acting per unit area of a surface.
P = F/A
SI unit: Pascal (Pa) or N/m².
2. State Pascal's law and mention two applications.
Answer:
Pascal's law states that pressure applied to an enclosed fluid is transmitted undiminished and equally in all directions.
Applications:
Hydraulic lift.
Hydraulic brakes.
3. Derive the expression for pressure at a depth h inside a liquid.
Answer:
Consider a liquid column of height h and cross-sectional area A.
Let:
ρ = Density of liquid
g = Acceleration due to gravity
P₀ = Pressure at the surface
P = Pressure at depth h
The volume of the liquid column is:
V = Ah
Its mass is:
m = ρAh
Its weight is:
W = mg = ρAhg
For equilibrium of the liquid column:
PA = P₀A + ρAhg
Dividing by A:
P = P₀ + ρgh
Therefore:
P = P₀ + ρgh
This is the expression for pressure at depth h in a liquid.
4. What is the difference between streamline and turbulent flow?
Answer:
Streamline flow is smooth and regular. The velocity at a fixed point does not change with time.
Turbulent flow is irregular and contains random changes in velocity.
Streamline flow generally occurs at lower speeds, while turbulent flow is more likely at higher speeds.
5. State and explain the equation of continuity.
Answer:
The equation of continuity is based on the conservation of mass.
For an incompressible fluid:
ρA₁v₁ = ρA₂v₂
Since density is constant:
A₁v₁ = A₂v₂
This means that the product of cross-sectional area and flow speed remains constant.
A fluid moves faster through a narrow section and slower through a wider section.
6. State Bernoulli's principle and write its mathematical expression.
Answer:
Bernoulli's principle states that for steady flow of an ideal, incompressible and non-viscous fluid, the sum of pressure energy, kinetic energy and potential energy per unit volume remains constant along a streamline.
P + ½ρv² + ρgh = Constant
7. Derive Bernoulli's equation.
Answer:
Consider an ideal fluid flowing steadily through a pipe of varying cross-section.
Let:
P₁ and P₂ = Pressures at two sections
v₁ and v₂ = Speeds at the two sections
h₁ and h₂ = Heights of the two sections
ρ = Density of fluid
ΔV = Volume of fluid displaced
The work done by pressure forces is:
W = P₁ΔV − P₂ΔV
W = (P₁ − P₂)ΔV
According to the work-energy theorem, this work is equal to the change in kinetic energy plus the change in potential energy.
(P₁ − P₂)ΔV = ½ρΔV(v₂² − v₁²) + ρgΔV(h₂ − h₁)
Dividing by ΔV:
P₁ − P₂ = ½ρ(v₂² − v₁²) + ρg(h₂ − h₁)
Rearranging:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Therefore:
P + ½ρv² + ρgh = Constant
This is Bernoulli's equation.
8. What are the applications of Bernoulli's principle?
Answer:
Important applications include:
Working of an aeroplane wing.
Atomiser and spray pump.
Venturimeter.
Motion of a spinning cricket ball.
Efflux of liquid from a small hole.
9. What is viscosity? Define coefficient of viscosity.
Answer:
Viscosity is the property of a fluid by which it opposes the relative motion between its adjacent layers.
Coefficient of viscosity is the ratio of shearing stress to the rate of shearing strain.
η = Shearing stress / Rate of shearing strain
SI unit: Pa s.
10. State Stokes' law and write its mathematical expression.
Answer:
Stokes' law states that the viscous drag force on a small sphere moving slowly through a fluid is directly proportional to the radius of the sphere, its speed and the coefficient of viscosity.
F = 6πηrv
11. What is terminal velocity? Derive its expression for a small spherical body falling through a viscous fluid.
Answer:
Terminal velocity is the constant speed attained by a body falling through a viscous fluid when the net force acting on it becomes zero.
Consider a sphere of radius r and density ρₛ falling through a fluid of density ρ and viscosity η.
At terminal velocity, the downward weight is balanced by the upward buoyant force and viscous drag.
Weight:
W = (4/3)πr³ρₛg
Buoyant force:
Fᵦ = (4/3)πr³ρg
Viscous force according to Stokes' law:
Fᵥ = 6πηrvₜ
At terminal velocity:
W = Fᵦ + Fᵥ
(4/3)πr³ρₛg = (4/3)πr³ρg + 6πηrvₜ
Rearranging:
(4/3)πr³g(ρₛ − ρ) = 6πηrvₜ
Therefore:
vₜ = 2r²g(ρₛ − ρ)/(9η)
This is the expression for terminal velocity.
12. Define surface tension and write its SI unit.
Answer:
Surface tension is the tangential force acting per unit length along an imaginary line drawn on the free surface of a liquid.
T = F/l
SI unit: N/m.
13. Why does surface tension decrease with an increase in temperature?
Answer:
When temperature increases, the kinetic energy of liquid molecules increases.
The effective cohesive forces at the surface become less significant, so the surface tension decreases.
At the critical temperature, the distinction between liquid and gas disappears and surface tension becomes zero.
14. Derive the expression for excess pressure inside a liquid drop.
Answer:
Consider a spherical liquid drop of radius r and surface tension T.
Let the excess pressure inside the drop be ΔP.
If the radius increases by a small amount dr, the increase in volume is:
dV = 4πr²dr
The work done by the excess pressure is:
Work = ΔP × 4πr²dr
The increase in surface area is:
dA = 8πrdr
Increase in surface energy:
dU = T × dA
dU = 8πTrdr
Equating work done to the increase in surface energy:
ΔP × 4πr²dr = 8πTrdr
Cancelling common terms:
ΔP = 2T/r
This is the excess pressure inside a liquid drop.
15. Derive the expression for excess pressure inside a soap bubble.
Answer:
A soap bubble has two surfaces: an inner surface and an outer surface.
For a spherical bubble of radius r, the increase in total surface area is:
dA = 16πrdr
Increase in surface energy:
dU = 16πTrdr
Work done by excess pressure:
Work = ΔP × 4πr²dr
Equating work done and increase in surface energy:
ΔP × 4πr²dr = 16πTrdr
Therefore:
ΔP = 4T/r
This is the excess pressure inside a soap bubble.
16. What is capillary rise? Write its expression.
Answer:
Capillary rise is the rise of a liquid in a narrow tube due to surface tension.
For a capillary tube of radius r:
h = 2T cosθ/(ρgr)
Where:
T = Surface tension
θ = Angle of contact
ρ = Density of liquid
g = Acceleration due to gravity
r = Radius of capillary tube
For water in clean glass, the angle of contact is approximately zero, so cosθ ≈ 1.
17. Why does water rise in a narrow glass tube while mercury falls in it?
Answer:
Water wets clean glass because the adhesive force between water and glass is greater than the cohesive force between water molecules. Therefore, water rises in a capillary tube.
Mercury does not wet glass because its cohesive force is greater than its adhesive force with glass. Therefore, mercury is depressed in a capillary tube.
18. What is the Magnus effect?
Answer:
The Magnus effect is the sideways force experienced by a spinning object moving through a fluid.
The rotation produces different fluid speeds and pressures on opposite sides of the object. This pressure difference causes the object to follow a curved path.
Example: A spinning cricket ball curves while moving through air.
19. Differentiate between a liquid drop and a soap bubble in terms of excess pressure.
Answer:
For a liquid drop, there is only one liquid-air surface.
Excess pressure:
ΔP = 2T/r
For a soap bubble, there are two surfaces.
Excess pressure:
ΔP = 4T/r
Therefore, for the same radius and surface tension, the excess pressure inside a soap bubble is twice that inside a liquid drop.
20. What is Reynolds number? Explain its significance.
Answer:
Reynolds number is a dimensionless quantity used to predict whether fluid flow is likely to be laminar or turbulent.
Re = ρvD/η
Where:
ρ = Density of fluid
v = Characteristic speed
D = Characteristic diameter
η = Coefficient of viscosity
A low Reynolds number generally indicates laminar flow, while a high Reynolds number indicates a greater possibility of turbulent flow.
IMPORTANT FORMULAE – CHAPTER 9
Pressure:
P = F/A
Density:
ρ = m/V
Relative density:
Relative density = Density of substance / Density of water
Hydrostatic pressure:
P = ρgh
Absolute pressure:
P = Pₐ + ρgh
Gauge pressure:
Pᵍ = P − Pₐ
Pressure in a mercury barometer:
Pₐ = ρgh
Pascal's law:
F₁/A₁ = F₂/A₂
Equation of continuity:
A₁v₁ = A₂v₂
Bernoulli's equation:
P + ½ρv² + ρgh = Constant
Volume flow rate:
Q = Av
Poiseuille's law:
Q = πr⁴ΔP/(8ηl)
Coefficient of viscosity:
η = Shearing stress / Rate of shearing strain
Stokes' law:
F = 6πηrv
Terminal velocity:
vₜ = 2r²g(ρₛ − ρ)/(9η)
Surface tension:
T = F/l
Force due to a soap film:
F = 2Tl
Excess pressure inside a liquid drop:
ΔP = 2T/r
Excess pressure inside a soap bubble:
ΔP = 4T/r
Excess pressure inside an air bubble in a liquid:
ΔP = 2T/r
Capillary rise:
h = 2T cosθ/(ρgr)
Reynolds number:
Re = ρvD/η
Surface energy:
Surface energy = T × Increase in surface area
IMPORTANT SI UNITS
Pressure – Pascal (Pa)
Density – kg/m³
Relative density – No unit
Force – Newton (N)
Volume – m³
Flow speed – m/s
Volume flow rate – m³/s
Coefficient of viscosity – Pa s
Surface tension – N/m
Terminal velocity – m/s
Reynolds number – No unit
Surface energy – Joule (J)
IMPORTANT POINTS TO REMEMBER
Liquids and gases are called fluids because they can flow.
Pressure is force per unit area.
Pressure is a scalar quantity.
The SI unit of pressure is pascal.
Pressure in a liquid increases with depth.
Pressure at the same horizontal level in a stationary liquid is the same.
Pascal's law is used in hydraulic lifts and hydraulic brakes.
A mercury barometer measures atmospheric pressure.
Streamline flow is smooth and regular.
Turbulent flow is irregular and generally occurs at higher speeds.
The equation of continuity follows the law of conservation of mass.
Bernoulli's principle follows from conservation of mechanical energy for ideal fluid flow.
Where fluid speed increases, pressure generally decreases under Bernoulli's conditions.
Viscosity is the internal friction of a fluid.
Stokes' law describes the viscous drag on a small sphere moving slowly through a fluid.
Terminal velocity is reached when the net force on a falling body becomes zero.
Surface tension tends to minimise the surface area of a liquid.
Surface tension generally decreases with increasing temperature.
A soap bubble has two surfaces, while an ordinary liquid drop has one.
Excess pressure inside a soap bubble is 4T/r.
Excess pressure inside a liquid drop is 2T/r.
Capillary rise depends inversely on the radius of the tube.
Reynolds number helps distinguish laminar and turbulent flow.
The Magnus effect explains the curved path of a spinning ball.
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