Class 11 : Physics (In English) – Chapter 9: Mechanical Properties of Fluids
EXPLANATION & SUMMARY
🔶 1. Introduction
This chapter studies how fluids (liquids and gases) behave when external forces act on them. Unlike solids, fluids flow and cannot resist shear stress. Their mechanical properties are different and are described using pressure, buoyancy, viscosity, surface tension, etc.
🔶 2. Pressure in a Fluid
Pressure (P) = Force / Area = F / A
Unit: Pascal (Pa) = N/m²
Pressure is a scalar quantity but acts equally in all directions at a point in a fluid.
🔶 3. Variation of Pressure with Depth
In a fluid of density ρ at depth h:
P = P₀ + ρgh
Where:
P = pressure at depth h
P₀ = atmospheric pressure at surface
ρ = density of fluid
g = acceleration due to gravity
h = depth from surface
This shows that pressure increases linearly with depth.
🔶 4. Pascal’s Law
Pascal’s Law: A change in pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and to the walls of the container.
📌 Applications:
Hydraulic lifts
Hydraulic brakes
Hydraulic press
🔶 5. Atmospheric Pressure
Standard atmospheric pressure: 1 atm = 1.013 × 10⁵ Pa
Measured using barometer
🔶 6. Gauge Pressure and Absolute Pressure
Absolute pressure = atmospheric pressure + gauge pressure
Gauge pressure = pressure measured above atmospheric pressure
(negative gauge pressure = vacuum)
🔶 7. Buoyancy and Archimedes’ Principle
A body immersed in a fluid experiences an upward force (buoyant force) equal to the weight of fluid displaced.
Buoyant force (Fb) = ρ × V × g
Where:
ρ = density of fluid
V = volume displaced
g = acceleration due to gravity
📌 Archimedes’ Principle: Upthrust = weight of displaced fluid
🔶 8. Conditions for Floating
Object floats if its density < fluid density
Object sinks if its density > fluid density
Weight of floating object = buoyant force
🔶 9. Streamline and Turbulent Flow
Streamline flow: Fluid particles move in smooth paths; velocity at a point remains constant.
Turbulent flow: Irregular motion with eddies and fluctuations.
Critical velocity (vₛ): Maximum velocity below which flow is streamline. Beyond this, it becomes turbulent.
🔶 10. Reynolds Number (Re)
Re = (ρ × v × D) / η
Where:
ρ = fluid density
v = velocity
D = diameter
η = viscosity
Flow is:
Streamline if Re < 2000
Turbulent if Re > 3000
Transition zone: 2000 < Re < 3000
🔶 11. Viscosity
Viscosity is the internal friction within a fluid resisting flow.
Viscous force (F) = η × A × (dv/dx)
η = coefficient of viscosity
SI unit of viscosity = N·s/m² = Pa·s
Stoke’s Law: For a sphere of radius r moving with velocity v in a fluid:
F = 6π η r v
🔶 12. Terminal Velocity
When a falling body in a fluid reaches constant velocity (net force = 0), it’s called terminal velocity.
Using Stoke’s law:
Vₜ = (2/9) × (r² × g × (ρ – σ)) / η
Where:
ρ = density of object
σ = density of fluid
η = viscosity
🔶 13. Bernoulli’s Theorem
Bernoulli’s principle: For an incompressible, non-viscous fluid in streamline flow:
P + (1/2)ρv² + ρgh = constant
Where:
P = pressure
v = speed of fluid
h = height from reference level
Interpretation: Increase in speed → decrease in pressure.
📌 Applications:
Flight of airplanes (lift)
Venturimeter
Atomizer and sprayer
🔶 14. Surface Tension
Surface tension (T) is the force per unit length acting along the surface of a liquid due to intermolecular forces.
T = F / L
Unit: N/m
Water forms droplets due to surface tension trying to minimize surface area.
🔶 15. Capillarity
Capillary rise (h) in a tube of radius r:
h = (2T cosθ) / (rρg)
Where:
T = surface tension
θ = angle of contact
ρ = fluid density
g = gravity
Water rises (concave meniscus), mercury falls (convex meniscus) due to contact angle.
📘 PART B: CRISP SUMMARY (~300 WORDS)
🔷 Summary – Mechanical Properties of Fluids
Fluids exert pressure: P = F / A
Pressure increases with depth: P = P₀ + ρgh
Pascal’s law: Pressure is transmitted equally in all directions.
Buoyant force = ρ × V × g
Object floats if its density is less than fluid’s.
Flow types:
Streamline: orderly
Turbulent: chaotic
Critical velocity marks the shift
Reynolds number predicts flow type:
Re < 2000: streamline
Re > 3000: turbulent
Viscosity is internal friction in fluids:
Stoke’s law: F = 6π η r v
Terminal velocity: velocity when net force = 0
Bernoulli’s equation:
P + (1/2)ρv² + ρgh = constant
Shows inverse relationship between pressure and speed.
Surface tension makes liquids form spherical drops.
Capillary action explains rise/fall of liquid in narrow tubes:
h = (2T cosθ) / (rρg)
Fluids behave differently from solids but obey physical laws related to force, energy, and motion. This chapter helps in understanding atmospheric, biological, and industrial fluid systems.
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QUESTIONS FROM TEXTBOOK
🔷 Question 9.1
Explain why:
(a) The blood pressure in humans 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 the sea level, though the height of the atmosphere is more than 100 km.
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.
🔶 Answer:
(a) ⬇️ Pressure increases with depth in fluids due to gravity. Feet are lower than the brain, so blood exerts more pressure at the feet:
📌 P = ρgh (greater h ⇒ greater pressure)
(b) 🌬️ Air is compressible. Most of the atmospheric mass is concentrated in the lower layers.
At 6 km, about 50% of atmospheric pressure is lost due to rapid decrease in density.
(c) 🧪 Though pressure = force/area, it acts equally in all directions in a fluid at rest.
No specific direction ⇒ scalar quantity ✅
🔷 Question 9.2
Explain why:
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops.
🔶 Answer:
(a)
🔹 Mercury has stronger cohesive forces than adhesive forces with glass ⇒ obtuse angle.
🔹 Water has stronger adhesive forces ⇒ acute angle with glass.
(b)
💧 Water wets glass → adhesive > cohesive
🌑 Mercury doesn’t wet glass → cohesive > adhesive
Hence, water spreads, mercury forms drops.
🔷 Question 9.3
Fill in the blanks using the word(s) from the list appended with each statement:
(a) Surface tension of liquids generally decreases with temperature.
(b) 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 shear strain, while for fluids it is proportional to rate of shear strain.
(d) For a fluid in steady flow, the increase in flow speed at a constriction follows Bernoulli’s principle.
(e) For the model of a plane in a wind tunnel, turbulence occurs at a greater speed for turbulence in an actual plane.
🔷 Question 9.4
Explain why:
(a) To keep a piece of paper horizontal, you should blow over, not under it.
(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
🔶 Answer:
(a) 💨 Air moves faster over the top → pressure above decreases → paper rises and stays horizontal (Bernoulli’s principle).
(b) 🚿 Smaller opening ⇒ faster speed of water ⇒ higher kinetic energy ⇒ strong jets escape between fingers.
(c) The size of the needle of a syringe controls flow rate better than the thumb.
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.
(e) A spinning cricket ball in air does not follow a parabolic trajectory.
🔶 Answer:
(c) ✒️ A needle has a much smaller area ⇒ velocity of flow increases significantly ⇒ pressure drops sharply ⇒ controlled flow.
(d) 🔁 Backward thrust is a reaction (Newton’s 3rd law) to forward momentum of fluid.
(e) 🏏 Spinning ball creates uneven airflow ⇒ pressure difference ⇒ Magnus effect ⇒ curved trajectory, not parabolic.
🔷 Question 9.5
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
🔶 Answer:
Given:
m = 50 kg, d = 1.0 cm = 0.01 m
Radius r = 0.005 m ⇒ A = πr² = 3.14 × (0.005)² = 7.85 × 10⁻⁵ m²
P = F/A = (mg)/A = (50 × 9.8) / 7.85×10⁻⁵ = 6.24 × 10⁵ Pa
🔷 Question 9.6
Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg/m³. Determine the height of the wine column for normal atmospheric pressure.
🔶 Answer:
Given:
P = 1.01×10⁵ Pa, ρ = 984 kg/m³, g = 9.8 m/s²
P = ρgh ⇒ h = P / (ρg) = 1.01×10⁵ / (984 × 9.8) = 10.44 m
🔷 Question 9.7
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 depth = 3 km, ignore ocean currents.
🔶 Answer:
Given:
h = 3×10³ m, ρ (water) ≈ 1000 kg/m³
P = ρgh = 1000 × 9.8 × 3000 = 2.94 × 10⁷ Pa
Stress limit = 10⁹ Pa > pressure ⇒ ✅ Structure is safe.
🔷 Question 9.8
A hydraulic automobile lift is designed to lift cars with a max mass of 3000 kg. The area of the piston is 425 cm². What is the max pressure the piston must bear?
🔶 Answer:
Given:
m = 3000 kg, A = 425 cm² = 4.25×10⁻² m²
F = mg = 3000 × 9.8 = 29400 N
P = F/A = 29400 / 4.25×10⁻² = 6.91 × 10⁵ Pa
🔷 Question 9.9
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:
📌 Principle: Mercury levels are equal ⇒ pressure by water = pressure by spirit
Let ρₛ = density of spirit
ρₛ × g × 0.125 = 1000 × g × 0.10
⇒ ρₛ = (1000 × 0.10)/0.125 = 800 kg/m³
✅ Specific gravity = ρₛ / 1000 = 0.8
🔷 Question 9.10
If 15.0 cm of water and spirit (of SG = 0.8) are poured into the respective arms of a U-tube, what is the difference in mercury levels?
🔶 Answer:
📌 Pressure balance:
P_water = ρ × g × h = 1000 × g × 0.15 = 150g
P_spirit = 0.8 × 1000 × g × h = 120g
🧮 Pressure difference = 30g ⇒ balanced by mercury column
Let h = difference in mercury levels:
ρHg × g × h = 30g ⇒ h = 30 / 13.6 = 2.21 cm
🔷 Question 9.11
Can Bernoulli’s equation be used for flow of water through a rapid in a river? Explain.
🔶 Answer:
❌ No, because Bernoulli’s equation is valid only for steady, incompressible, non-viscous, and streamlined flow.
🌊 Rapids cause turbulent flow → invalid conditions ⇒ Bernoulli’s equation not applicable.
🔷 Question 9.12
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
🔶 Answer:
🟢 No, it doesn’t matter.
✅ Bernoulli’s equation involves pressure differences, and both absolute and gauge pressures differ by a constant (atmospheric pressure), which cancels out in differences.
🔄 So, result remains same.
🔷 Question 9.13
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, what is the pressure difference between the two ends?
(Density = 1.3 × 10³ kg/m³, viscosity η = 0.83 Pa·s)
🔶 Answer:
📌 Use Poiseuille’s equation:
Q = (πR⁴ΔP)/(8ηlρ) ⇒ Rearranged:
ΔP = (8ηlQρ) / (πR⁴)
Given:
R = 0.01 m, l = 1.5 m, η = 0.83 Pa·s, Q = 4.0×10⁻³ kg/s
⇒ Volume flow rate = Q/ρ = 4.0×10⁻³ / 1300 = 3.08×10⁻⁶ m³/s
ΔP = (8 × 0.83 × 1.5 × 3.08×10⁻⁶) / (π × (0.01)⁴)
= 1227 Pa (approx.)
🔷 Question 9.14
In a wind tunnel, flow speeds on upper and lower surfaces of wing are 70 m/s and 63 m/s. If wing area is 2.5 m², density of air is 1.3 kg/m³, what is the lift?
🔶 Answer:
📌 Use Bernoulli’s principle:
ΔP = ½ ρ(v²_upper – v²_lower)
= 0.5 × 1.3 × (70² – 63²) = 0.65 × (4900 – 3969)
= 0.65 × 931 = 605.15 Pa
Lift = ΔP × A = 605.15 × 2.5 = 1513 N
🔷 Question 9.15
Figures 9.20 (a) and (b) refer to steady flow of a (non-viscous) liquid. Which figure is incorrect? Why?
🔶 Answer:
✅ Figure (b) is incorrect.
➡ In a narrow section, speed increases (continuity), so pressure decreases (Bernoulli).
➡ But in (b), height of fluid is higher in narrow part ⇒ higher pressure, which contradicts Bernoulli’s principle.
❌ So, (b) violates physics of flow.
🔷 Question 9.16
Can Bernoulli’s equation be applied to the flow of water through a hole in a large tank? Explain briefly.
🔶 Answer:
✅ Yes, it can be applied.
📌 Conditions satisfied:
• Flow is steady
• Water behaves like an incompressible, non-viscous fluid
• Streamline flow exists at the hole
🧪 From Bernoulli’s equation:
v = √(2gh), where h is height of water above the hole
Hence, applicable and gives accurate speed of outflow.
🔷 Question 9.17
A fluid of viscosity 1.5 × 10⁻³ Pa·s and density ρ = 1.5 × 10³ kg/m³ flows through a pipe of radius 1.0 cm. The flow is streamlined. What is the max velocity with which the fluid can flow so that the flow remains laminar? (Reynolds number = 2000)
🔶 Answer:
📌 Use: Re = (ρvd)/η ⇒ v = Re × η / (ρ × d)
Given: Re = 2000, η = 1.5×10⁻³, ρ = 1.5×10³, d = 2 × 1.0 cm = 0.02 m
v = (2000 × 1.5×10⁻³) / (1.5×10³ × 0.02)
= 3.0 / 30 = 0.1 m/s
🔷 Question 9.18
A tank is 2.0 m deep. Water is filled up to 1.5 m. An oil layer of 0.5 m (SG = 0.8) is above the water. Find pressure at bottom of tank due to liquids.
(g = 9.8 m/s²)
🔶 Answer:
Pressure due to water:
P₁ = ρg h = 1000 × 9.8 × 1.5 = 14700 Pa
Pressure due to oil:
P₂ = 800 × 9.8 × 0.5 = 3920 Pa
Total Pressure = P₁ + P₂ = 18620 Pa
🔷 Question 9.19
What is the speed of efflux of water (through a sharp-edged hole) at a depth of 2.5 m below the free surface?
🔶 Answer:
📌 By Torricelli’s Law:
v = √(2gh) = √(2 × 9.8 × 2.5) = √49 = 7.0 m/s
🔷 Question 9.20
A vessel has two identical small holes at depths of 10 cm and 40 cm. Find ratio of speeds of efflux.
🔶 Answer:
📌 From Torricelli’s Law:
v = √(2gh)
v₁/v₂ = √h₁ / √h₂ = √10 / √40 = 1 / 2 ⇒ Ratio = 1:2
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OTHER IMPORTANT QUESTIONS FOR EXAMS
(CBSE MODEL QUESTIONS PAPER)
ESPECIALLY MADE FROM THIS LESSON ONLY
⚙️ SECTION A — Multiple Choice Questions (Q1–Q18)
Question 1:
Pressure in a liquid at rest depends on
🔵 (A) Depth only
🟢 (B) Density and depth
🟠 (C) Volume
🔴 (D) Surface area
Answer: (B) Density and depth
Question 2:
The SI unit of pressure is
🔵 (A) dyne/cm²
🟢 (B) N/m²
🟠 (C) erg/cm³
🔴 (D) N/cm²
Answer: (B) N/m²
Question 3:
Which of the following is a scalar quantity?
🔵 (A) Pressure
🟢 (B) Force
🟠 (C) Torque
🔴 (D) Momentum
Answer: (A) Pressure
Question 4:
If pressure at a point inside a liquid increases linearly with depth, the liquid must be
🔵 (A) Non-viscous
🟢 (B) Ideal and incompressible
🟠 (C) Compressible
🔴 (D) Inviscid
Answer: (B) Ideal and incompressible
Question 5:
Pascal’s law states that
🔵 (A) Pressure at a point is same in all directions
🟢 (B) Force per unit area is constant
🟠 (C) Liquids flow from higher to lower pressure
🔴 (D) Density depends on pressure
Answer: (A) Pressure at a point is same in all directions
Question 6:
The pressure due to a liquid column of height h is given by
🔵 (A) ρgh
🟢 (B) gh/ρ
🟠 (C) ρ/hg
🔴 (D) h/ρg
Answer: (A) ρgh
Question 7:
The atmospheric pressure at sea level is approximately
🔵 (A) 10⁴ N/m²
🟢 (B) 10⁵ N/m²
🟠 (C) 10⁶ N/m²
🔴 (D) 10⁷ N/m²
Answer: (B) 10⁵ N/m²
Question 8:
Hydraulic brakes in automobiles work on
🔵 (A) Archimedes’ principle
🟢 (B) Pascal’s law
🟠 (C) Bernoulli’s theorem
🔴 (D) Conservation of energy
Answer: (B) Pascal’s law
Question 9:
When an object is immersed in a liquid, it experiences an upward force equal to
🔵 (A) Weight of liquid displaced
🟢 (B) Its own weight
🟠 (C) Density of liquid
🔴 (D) Volume of liquid
Answer: (A) Weight of liquid displaced
Question 10:
The buoyant force depends on
🔵 (A) Volume of body
🟢 (B) Volume of liquid displaced
🟠 (C) Mass of object
🔴 (D) Shape of container
Answer: (B) Volume of liquid displaced
Question 11:
A floating body displaces liquid
🔵 (A) Equal to its volume
🟢 (B) Equal to its weight
🟠 (C) Less than its weight
🔴 (D) Equal to the density of liquid
Answer: (B) Equal to its weight
Question 12:
Streamline flow means
🔵 (A) Random motion of molecules
🟢 (B) Flow of fluid where velocity at a point is constant
🟠 (C) Flow with irregular motion
🔴 (D) None of these
Answer: (B) Flow of fluid where velocity at a point is constant
Question 13:
Equation of continuity is based on
🔵 (A) Law of conservation of momentum
🟢 (B) Law of conservation of mass
🟠 (C) Law of conservation of energy
🔴 (D) Archimedes’ principle
Answer: (B) Law of conservation of mass
Question 14:
Bernoulli’s theorem is based on
🔵 (A) Conservation of energy
🟢 (B) Conservation of mass
🟠 (C) Conservation of momentum
🔴 (D) Conservation of volume
Answer: (A) Conservation of energy
Question 15:
For an ideal fluid, viscosity is
🔵 (A) Very high
🟢 (B) Zero
🟠 (C) One
🔴 (D) Infinite
Answer: (B) Zero
Question 16:
The unit of coefficient of viscosity is
🔵 (A) N·m⁻²·s
🟢 (B) N·s·m⁻²
🟠 (C) N·m/s
🔴 (D) m²/s
Answer: (B) N·s·m⁻²
Question 17:
Poiseuille’s formula gives the rate of flow of
🔵 (A) Ideal fluid
🟢 (B) Real viscous fluid
🟠 (C) Compressible gas
🔴 (D) None of these
Answer: (B) Real viscous fluid
Question 18:
The terminal velocity of a spherical body in a viscous medium depends on
🔵 (A) Radius of body
🟢 (B) Density of body and fluid
🟠 (C) Viscosity of medium
🔴 (D) All of these
Answer: (D) All of these
⚙️ SECTION B — Very Short / Short Answer Questions (Q19–Q23)
Question 19:
State Pascal’s Law.
Answer:
💡 Pascal’s Law: Pressure applied to a confined fluid is transmitted equally and undiminished in all directions throughout the fluid and on the walls of the container.
➡️ Mathematically:
P₁ = P₂ = P₃ = constant
✔️ Applications:
Hydraulic brakes and lifts
Hydraulic presses
Question 20:
What is the difference between gauge pressure and absolute pressure?
Answer:
✔️ Gauge pressure: It is the pressure measured above atmospheric pressure.
➡️ P_g = P_abs − P_atm
✔️ Absolute pressure: It is the total pressure including atmospheric pressure.
➡️ P_abs = P_g + P_atm
💡 Example: A tire gauge reading of 2×10⁵ Pa means its absolute pressure is 3×10⁵ Pa (since 1×10⁵ Pa = atmospheric).
Question 21:
State Archimedes’ Principle.
Answer:
💧 When a body is fully or partially immersed in a fluid, it experiences an upward force equal to the weight of the fluid displaced by the body.
➡️ Buoyant force, F_b = ρ_f × V × g
where ρ_f = density of fluid, V = volume displaced, g = acceleration due to gravity.
✔️ Used to determine density and purity of solids or liquids.
Question 22:
Define viscosity.
Answer:
💡 Viscosity: It is the property of a fluid that opposes the relative motion between its layers.
➡️ For a velocity gradient (dv/dx), viscous force F = ηA(dv/dx)
where η = coefficient of viscosity.
✔️ Unit: N·s·m⁻²
✔️ Fluids with higher η (like honey) flow slowly.
Question 23:
What is meant by streamline flow?
Answer:
💡 Streamline flow: It is a steady flow of fluid in which velocity at every point remains constant in time.
✔️ Streamlines never cross each other.
✔️ Kinetic energy per unit mass is constant along a streamline (Bernoulli’s principle).
⚡ SECTION C — Mid-Length Numericals / Theory (Q24–Q27)
Question 24:
Derive the expression for pressure at a depth h in a liquid.
Answer:
✏️ Step 1:
Consider a liquid column of height h, density ρ, and base area A.
✏️ Step 2:
Weight of column = mg = ρVg = ρ(Ah)g
✏️ Step 3:
Pressure = Force / Area = (ρAhg)/A = ρgh
✔️ Therefore:
➡️ P = ρgh
💡 Pressure increases linearly with depth and acts equally in all directions.
Question 25:
Explain Bernoulli’s Theorem and write its expression.
Answer:
💡 Statement:
For a steady, incompressible, and non-viscous fluid, the total mechanical energy per unit volume (pressure energy + kinetic energy + potential energy) remains constant along a streamline.
➡️ P + ½ρv² + ρgh = constant
✔️ Terms:
P → Pressure energy
½ρv² → Kinetic energy
ρgh → Potential energy
✔️ Applications:
Airplane lift
Flow of blood in arteries
Venturimeter
Question 26:
State and derive the Equation of Continuity.
Answer:
💡 Statement:
For an incompressible fluid flowing in a tube, the mass flow rate remains constant.
✏️ Step 1:
Mass flow rate = ρAV
✏️ Step 2:
For steady flow: ρ₁A₁v₁ = ρ₂A₂v₂
✏️ Step 3:
For incompressible liquid (ρ constant):
➡️ A₁v₁ = A₂v₂ = constant
✔️ Meaning: Narrow tube ⇒ higher velocity (inverse relation).
Question 27:
What is terminal velocity? Derive its expression using Stokes’ Law.
Answer:
💡 Definition:
Terminal velocity is the constant velocity attained by a body falling through a viscous medium when net force = 0.
✏️ Forces acting:
Downward weight = mg = (4/3)πr³ρg
Upward buoyant force = (4/3)πr³ρ_f g
Upward viscous drag = 6πηrv
✏️ At terminal velocity:
mg = 6πηrv_t + buoyant force
✏️ Simplify:
(4/3)πr³(ρ − ρ_f)g = 6πηr v_t
➡️ v_t = [2r²(ρ − ρ_f)g] / [9η]
✔️ Conclusion:
Heavier, larger spheres fall faster in less viscous fluids.
⚙️ SECTION D — Long Answer Questions (Q28–Q31)
Question 28:
Derive Bernoulli’s Equation from the principle of conservation of energy.
Answer:
💡 Concept:
For a non-viscous, incompressible, steady-flowing fluid, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline.
✏️ Step 1: Consider two points A and B
At A: P₁, v₁, h₁ At B: P₂, v₂, h₂
✏️ Step 2: Work done by pressure forces
Work = (P₁ − P₂)V
✏️ Step 3: Change in energy
ΔKE = ½ ρV (v₂² − v₁²)
ΔPE = ρVg (h₂ − h₁)
✏️ Step 4: Apply Work–Energy Theorem
(P₁ − P₂)V = ½ ρV (v₂² − v₁²) + ρVg (h₂ − h₁)
➡️ P₁ + ½ ρv₁² + ρgh₁ = P₂ + ½ ρv₂² + ρgh₂ = constant
✔️ Bernoulli’s Equation: P + ½ ρv² + ρgh = constant
Question 29:
Explain Torricelli’s Theorem and derive its expression.
Answer:
At the free surface: P₁ = atmospheric, v₁ ≈ 0, h₁ = h
At the orifice: P₂ = atmospheric, v₂ =?, h₂ = 0
Apply Bernoulli’s equation:
P₁ + ½ ρv₁² + ρgh₁ = P₂ + ½ ρv₂² + ρgh₂
Cancel P₁ = P₂, neglect v₁ ⇒ ρgh = ½ ρv₂²
➡️ v₂ = √(2gh)
✔️ Torricelli’s Theorem: Speed of efflux through depth h equals speed of free fall through height h.
Question 30:
Derive expression for viscous force and state Stokes’ Law.
Answer:
💡 Concept:
When a sphere moves through a viscous medium, a drag force acts opposite to its motion.
✏️ Step 1: According to Stokes,
➡️ F = 6 π η r v
✏️ Step 2: At terminal velocity (vₜ):
(4⁄3) π r³ ρ g = (4⁄3) π r³ ρ_f g + 6 π η r vₜ
Simplify ⇒ vₜ = [ 2 r² ( ρ − ρ_f ) g ] ⁄ 9 η
✔️ Stokes’ Law: F ∝ r v η
💡 Applications: Motion of raindrops, measurement of viscosity by falling-sphere method.
Question 31:
What is the Equation of Continuity and how does it show mass conservation?
Answer:
💡 Statement: For steady incompressible flow, mass flow rate is constant.
ρ₁ A₁ v₁ = ρ₂ A₂ v₂
If ρ₁ = ρ₂ = ρ (constant), then ➡️ A₁ v₁ = A₂ v₂ = constant
Hence, when A ↓ ⇒ v ↑ to conserve mass.
💡 Example: River narrows ⇒ flow speed increases.
🌊 SECTION E — Case / Application Based Questions (Q32–Q33)
Question 32:
A horizontal pipe has area A₁ = 4 cm², v₁ = 2 m/s. Find v₂ when A₂ = 2 cm².
Answer:
✏️ Step 1: Use continuity equation: A₁ v₁ = A₂ v₂
(4×10⁻⁴)(2) = (2×10⁻⁴) v₂
v₂ = 4 m/s
✔️ Result: v₂ = 4 m/s
💡 As area halves, velocity doubles.
Question 33:
A steel ball of radius r = 1 mm falls through glycerine (η = 1.2 N·s·m⁻²). ρ = 7800 kg/m³, ρ_f = 1260 kg/m³. Find terminal velocity vₜ.
Answer:
✏️ Step 1: Formula from Stokes’ Law:
vₜ = [ 2 r² ( ρ − ρ_f ) g ] ⁄ 9 η
✏️ Step 2: Substitute values:
vₜ = [ 2 (1×10⁻³)² (7800 − 1260) (9.8) ] ⁄ [ 9 × 1.2 ]
= (2×10⁻⁶ × 6540 × 9.8) ⁄ 10.8
= 0.0119 m/s
✔️ vₜ = 1.19 × 10⁻² m/s
💡 Dense small spheres quickly reach constant speed in viscous fluids.
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