Chapter 10: Mechanical Properties of Fluids
Physics XI : Complete NCERT Exercise Solutions
Master fluid mechanics with NCERT solutions for pressure, Bernoulli's principle, viscosity, surface tension, and practical applications in daily life.
Question 10.1
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 sea level
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area
(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 sea level
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area
Answer & Explanation:
(a) Blood pressure increases with depth according to P = P₀ + ρgh. Feet are lower than the brain, so the column of blood above them exerts additional pressure.
(b) Atmospheric density decreases exponentially with height. Most air mass is concentrated in the lower atmosphere, so pressure drops rapidly in the first few kilometers.
(c) Pressure is defined as the magnitude of the normal force per unit area. It has magnitude but no specific direction at a point in a fluid at rest, making it a scalar quantity.
(b) Atmospheric density decreases exponentially with height. Most air mass is concentrated in the lower atmosphere, so pressure drops rapidly in the first few kilometers.
(c) Pressure is defined as the magnitude of the normal force per unit area. It has magnitude but no specific direction at a point in a fluid at rest, making it a scalar quantity.
Question 10.2
Explain why:
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute
(b) Water spreads on clean glass while mercury forms drops
(c) Surface tension is independent of surface area
(d) Water with detergent has small contact angles
(e) A free liquid drop is spherical
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute
(b) Water spreads on clean glass while mercury forms drops
(c) Surface tension is independent of surface area
(d) Water with detergent has small contact angles
(e) A free liquid drop is spherical
Answer & Explanation:
(a) Mercury molecules have stronger cohesion than adhesion to glass, causing obtuse angle. Water has stronger adhesion, causing acute angle.
(b) Water wets glass due to adhesive forces > cohesive forces. Mercury doesn't wet glass due to cohesive > adhesive forces.
(c) Surface tension is force per unit length at the boundary, not dependent on total area.
(d) Detergents reduce surface tension and increase adhesion, decreasing contact angle.
(e) Sphere has minimum surface area for given volume, minimizing surface energy.
(b) Water wets glass due to adhesive forces > cohesive forces. Mercury doesn't wet glass due to cohesive > adhesive forces.
(c) Surface tension is force per unit length at the boundary, not dependent on total area.
(d) Detergents reduce surface tension and increase adhesion, decreasing contact angle.
(e) Sphere has minimum surface area for given volume, minimizing surface energy.
Question 10.3
Fill in the blanks:
(a) Surface tension of liquids generally ______ with temperature
(b) Viscosity of gases ______ with temperature, while liquids ______
(c) For solids, shearing force ∝ ______; for fluids, it ∝ ______
(d) Increased flow speed at constriction follows ______
(e) Turbulence in wind tunnel occurs at ______ speed than actual plane
(a) Surface tension of liquids generally ______ with temperature
(b) Viscosity of gases ______ with temperature, while liquids ______
(c) For solids, shearing force ∝ ______; for fluids, it ∝ ______
(d) Increased flow speed at constriction follows ______
(e) Turbulence in wind tunnel occurs at ______ speed than actual plane
Answer & Explanation:
(a) decreases
(b) increases; decreases
(c) shear strain; rate of shear strain
(d) conservation of mass (continuity equation)
(e) smaller
(b) increases; decreases
(c) shear strain; rate of shear strain
(d) conservation of mass (continuity equation)
(e) smaller
Question 10.4
Explain why:
(a) Blowing over paper keeps it horizontal
(b) Fast jets through fingers when closing tap
(c) Syringe needle controls flow better than thumb pressure
(d) Fluid outflow causes backward thrust
(e) Spinning cricket ball doesn't follow parabola
(a) Blowing over paper keeps it horizontal
(b) Fast jets through fingers when closing tap
(c) Syringe needle controls flow better than thumb pressure
(d) Fluid outflow causes backward thrust
(e) Spinning cricket ball doesn't follow parabola
Answer & Explanation:
(a) Bernoulli's principle: faster air above creates lower pressure, upward force balances weight.
(b) Continuity equation: same flow rate through smaller area = higher velocity.
(c) Poiseuille's law: flow rate ∝ r⁴, so needle radius controls flow more than pressure.
(d) Conservation of momentum: ejecting fluid forward pushes vessel backward.
(e) Magnus effect: spin creates pressure difference, causing curved trajectory.
(b) Continuity equation: same flow rate through smaller area = higher velocity.
(c) Poiseuille's law: flow rate ∝ r⁴, so needle radius controls flow more than pressure.
(d) Conservation of momentum: ejecting fluid forward pushes vessel backward.
(e) Magnus effect: spin creates pressure difference, causing curved trajectory.
Question 10.5
A 50 kg girl wearing high heels balances on single heel of diameter 1.0 cm. What pressure does heel exert on floor?
Answer & Explanation:
Given: m = 50 kg, d = 1.0 cm = 0.01 m, g = 9.8 m/s²
Area: A = π(d/2)² = π(0.005)² = 7.854 × 10⁻⁵ m²
Force: F = mg = 50 × 9.8 = 490 N
Pressure: P = F/A = 490 / 7.854 × 10⁻⁵ = 6.24 × 10⁶ Pa ≈ 6.24 MPa
Answer: Pressure ≈ 6.24 × 10⁶ Pa (62.4 times atmospheric pressure!)
Area: A = π(d/2)² = π(0.005)² = 7.854 × 10⁻⁵ m²
Force: F = mg = 50 × 9.8 = 490 N
Pressure: P = F/A = 490 / 7.854 × 10⁻⁵ = 6.24 × 10⁶ Pa ≈ 6.24 MPa
Answer: Pressure ≈ 6.24 × 10⁶ Pa (62.4 times atmospheric pressure!)
Question 10.6
Torricelli used mercury (ρ = 13.6×10³ kg/m³). Pascal used wine (ρ = 984 kg/m³). What height of wine column gives normal atmospheric pressure?
Answer & Explanation:
Given: Pₐ = 1.013 × 10⁵ Pa, ρwine = 984 kg/m³, g = 9.8 m/s²
Formula: P = ρgh ⇒ h = P/(ρg)
Calculation:
h = 1.013 × 10⁵ / (984 × 9.8)
= 1.013 × 10⁵ / 9643.2 ≈ 10.51 m
Answer: Wine column height ≈ 10.5 m (vs mercury's 0.76 m)
Formula: P = ρgh ⇒ h = P/(ρg)
Calculation:
h = 1.013 × 10⁵ / (984 × 9.8)
= 1.013 × 10⁵ / 9643.2 ≈ 10.51 m
Answer: Wine column height ≈ 10.5 m (vs mercury's 0.76 m)
Question 10.7
Can a structure withstand 10⁹ Pa be put on ocean floor (depth ≈ 3 km)? Ignore currents.
Answer & Explanation:
Given: h = 3 km = 3000 m, ρsea ≈ 1030 kg/m³, g = 9.8 m/s², Pₐ = 1.01 × 10⁵ Pa
Pressure at depth: P = Pₐ + ρgh
= 1.01 × 10⁵ + (1030 × 9.8 × 3000)
= 1.01 × 10⁵ + 3.03 × 10⁷ = 3.04 × 10⁷ Pa
Comparison: Structure can withstand 10⁹ Pa > 3.04 × 10⁷ Pa
Answer: Yes, structure is suitable (withstands ~30 times required strength).
Pressure at depth: P = Pₐ + ρgh
= 1.01 × 10⁵ + (1030 × 9.8 × 3000)
= 1.01 × 10⁵ + 3.03 × 10⁷ = 3.04 × 10⁷ Pa
Comparison: Structure can withstand 10⁹ Pa > 3.04 × 10⁷ Pa
Answer: Yes, structure is suitable (withstands ~30 times required strength).
Question 10.8
Hydraulic lift lifts max 3000 kg. Load piston area = 425 cm². What max pressure must smaller piston bear?
Answer & Explanation:
Given: m = 3000 kg, A₂ = 425 cm² = 0.0425 m², g = 9.8 m/s²
Force on large piston: F₂ = mg = 3000 × 9.8 = 29400 N
Pressure in fluid (Pascal's law): P = F₂/A₂
= 29400 / 0.0425 ≈ 6.92 × 10⁵ Pa
Answer: Maximum pressure ≈ 6.92 × 10⁵ Pa (≈ 6.8 atm)
Force on large piston: F₂ = mg = 3000 × 9.8 = 29400 N
Pressure in fluid (Pascal's law): P = F₂/A₂
= 29400 / 0.0425 ≈ 6.92 × 10⁵ Pa
Answer: Maximum pressure ≈ 6.92 × 10⁵ Pa (≈ 6.8 atm)
Question 10.9
U-tube: mercury columns level with 10.0 cm water and 12.5 cm spirit. Find spirit's specific gravity.
Answer & Explanation:
Given: hw = 10.0 cm, hs = 12.5 cm, ρw = 1000 kg/m³
Pressure balance: ρwghw = ρsghs
⇒ ρs = ρw(hw/hs) = 1000 × (10.0/12.5)
= 800 kg/m³
Specific gravity = ρs/ρw = 800/1000 = 0.8
Answer: Specific gravity of spirit = 0.8
Pressure balance: ρwghw = ρsghs
⇒ ρs = ρw(hw/hs) = 1000 × (10.0/12.5)
= 800 kg/m³
Specific gravity = ρs/ρw = 800/1000 = 0.8
Answer: Specific gravity of spirit = 0.8
Question 10.10
In above, add 15.0 cm more water and spirit. Find mercury level difference. (ρHg = 13.6 g/cm³)
Answer & Explanation:
Given: Extra h = 15.0 cm, ρw = 1 g/cm³, ρs = 0.8 g/cm³, ρHg = 13.6 g/cm³
Pressure difference: ΔP = ρwgh - ρsgh = gh(ρw - ρs)
= g × 15.0 × (1 - 0.8) = 3g cm of water
Mercury difference: ρHgghHg = 3g ⇒ hHg = 3/13.6 ≈ 0.221 cm
Answer: Mercury level difference ≈ 0.22 cm
Pressure difference: ΔP = ρwgh - ρsgh = gh(ρw - ρs)
= g × 15.0 × (1 - 0.8) = 3g cm of water
Mercury difference: ρHgghHg = 3g ⇒ hHg = 3/13.6 ≈ 0.221 cm
Answer: Mercury level difference ≈ 0.22 cm
Question 10.11
Can Bernoulli's equation describe water flow through rapids? Explain.
Answer & Explanation:
Answer: No
Reason: Bernoulli's equation applies to steady, incompressible, non-viscous, irrotational flow along a streamline. Rapids involve:
• Turbulent flow (not steady)
• Significant viscous effects (energy dissipation)
• Air entrainment (not single fluid)
• Rotational motion and eddies
These violate Bernoulli's assumptions. For rapids, Navier-Stokes equations with turbulence modeling are needed.
Reason: Bernoulli's equation applies to steady, incompressible, non-viscous, irrotational flow along a streamline. Rapids involve:
• Turbulent flow (not steady)
• Significant viscous effects (energy dissipation)
• Air entrainment (not single fluid)
• Rotational motion and eddies
These violate Bernoulli's assumptions. For rapids, Navier-Stokes equations with turbulence modeling are needed.
Question 10.12
Does using gauge vs absolute pressure matter in Bernoulli's equation? Explain.
Answer & Explanation:
Answer: No, it doesn't matter
Reason: Bernoulli's equation: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
If we use gauge pressures: Pg = Pabs - Patm
Both sides subtract same Patm, so difference remains same.
Example: P1,abs - P2,abs = (P1,g + Patm) - (P2,g + Patm) = P1,g - P2,g
So pressure differences (what matters in Bernoulli) are identical.
Reason: Bernoulli's equation: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
If we use gauge pressures: Pg = Pabs - Patm
Both sides subtract same Patm, so difference remains same.
Example: P1,abs - P2,abs = (P1,g + Patm) - (P2,g + Patm) = P1,g - P2,g
So pressure differences (what matters in Bernoulli) are identical.
Question 10.13
Glycerine flows through tube: L = 1.5 m, r = 1.0 cm, flow rate = 4.0×10⁻³ kg/s. Find pressure difference. (ρ = 1.3×10³ kg/m³, η = 0.83 Pa·s)
Answer & Explanation:
Given: L = 1.5 m, r = 0.01 m, ṁ = 4.0×10⁻³ kg/s, ρ = 1300 kg/m³, η = 0.83 Pa·s
Volume flow rate: Q = ṁ/ρ = 4×10⁻³/1300 ≈ 3.077×10⁻⁶ m³/s
Poiseuille's law: Q = (πΔPr⁴)/(8ηL)
⇒ ΔP = (8ηLQ)/(πr⁴)
= (8 × 0.83 × 1.5 × 3.077×10⁻⁶) / (π × (0.01)⁴)
= (3.065×10⁻⁵) / (3.142×10⁻⁸) ≈ 975.6 Pa
Check Reynolds number: v = Q/(πr²) ≈ 0.0098 m/s, Re = ρvd/η ≈ 0.31 (laminar ✓)
Answer: ΔP ≈ 976 Pa
Volume flow rate: Q = ṁ/ρ = 4×10⁻³/1300 ≈ 3.077×10⁻⁶ m³/s
Poiseuille's law: Q = (πΔPr⁴)/(8ηL)
⇒ ΔP = (8ηLQ)/(πr⁴)
= (8 × 0.83 × 1.5 × 3.077×10⁻⁶) / (π × (0.01)⁴)
= (3.065×10⁻⁵) / (3.142×10⁻⁸) ≈ 975.6 Pa
Check Reynolds number: v = Q/(πr²) ≈ 0.0098 m/s, Re = ρvd/η ≈ 0.31 (laminar ✓)
Answer: ΔP ≈ 976 Pa
Question 10.14
Wind tunnel: wing upper speed = 70 m/s, lower = 63 m/s, area = 2.5 m², ρair = 1.3 kg/m³. Find lift.
Answer & Explanation:
Given: v₁ = 63 m/s (below), v₂ = 70 m/s (above), A = 2.5 m², ρ = 1.3 kg/m³
Bernoulli's principle (simplified): ΔP = ½ρ(v₂² - v₁²)
ΔP = 0.5 × 1.3 × (70² - 63²)
= 0.65 × (4900 - 3969) = 0.65 × 931 = 605.15 Pa
Lift force: F = ΔP × A = 605.15 × 2.5 ≈ 1512.9 N
Answer: Lift ≈ 1513 N (can support ~154 kg mass)
Bernoulli's principle (simplified): ΔP = ½ρ(v₂² - v₁²)
ΔP = 0.5 × 1.3 × (70² - 63²)
= 0.65 × (4900 - 3969) = 0.65 × 931 = 605.15 Pa
Lift force: F = ΔP × A = 605.15 × 2.5 ≈ 1512.9 N
Answer: Lift ≈ 1513 N (can support ~154 kg mass)
Question 10.15
Figures 10.23(a) and (b) show steady non-viscous liquid flow. Which is incorrect? Why?
Answer & Explanation:
Answer: Figure (b) is incorrect
Reason: According to equation of continuity (Av = constant):
• Where streamlines are closer, area is smaller, velocity is higher
• In Fig (b), at the constriction, streamlines should be closer together (higher velocity)
• But Fig (b) shows streamlines farther apart at constriction, violating continuity
Fig (a) correctly shows streamlines converging at constriction (higher velocity).
Reason: According to equation of continuity (Av = constant):
• Where streamlines are closer, area is smaller, velocity is higher
• In Fig (b), at the constriction, streamlines should be closer together (higher velocity)
• But Fig (b) shows streamlines farther apart at constriction, violating continuity
Fig (a) correctly shows streamlines converging at constriction (higher velocity).
Question 10.16
Spray pump: tube area = 8.0 cm², 40 holes each d = 1.0 mm, flow inside = 1.5 m/min. Find ejection speed.
Answer & Explanation:
Given: Atube = 8.0 cm² = 8×10⁻⁴ m², n = 40 holes, d = 1 mm = 0.001 m, vtube = 1.5 m/min = 0.025 m/s
Flow in tube: Q = Atubevtube = 8×10⁻⁴ × 0.025 = 2×10⁻⁵ m³/s
Area per hole: Ahole = π(d/2)² = π×(0.0005)² ≈ 7.854×10⁻⁷ m²
Total hole area: Atotal = 40 × 7.854×10⁻⁷ ≈ 3.142×10⁻⁵ m²
Ejection speed: v = Q/Atotal = 2×10⁻⁵ / 3.142×10⁻⁵ ≈ 0.637 m/s
Answer: Ejection speed ≈ 0.64 m/s
Flow in tube: Q = Atubevtube = 8×10⁻⁴ × 0.025 = 2×10⁻⁵ m³/s
Area per hole: Ahole = π(d/2)² = π×(0.0005)² ≈ 7.854×10⁻⁷ m²
Total hole area: Atotal = 40 × 7.854×10⁻⁷ ≈ 3.142×10⁻⁵ m²
Ejection speed: v = Q/Atotal = 2×10⁻⁵ / 3.142×10⁻⁵ ≈ 0.637 m/s
Answer: Ejection speed ≈ 0.64 m/s
Question 10.17
Soap film on U-wire supports weight 1.5×10⁻² N including slider (L = 30 cm). Find surface tension.
Answer & Explanation:
Given: W = 1.5×10⁻² N, L = 0.30 m (Note: soap film has 2 surfaces)
Surface tension force: F = 2 × S × L (2 sides of film)
This balances weight: 2SL = W
Surface tension: S = W/(2L) = 1.5×10⁻² / (2 × 0.30)
= 1.5×10⁻² / 0.60 = 2.5×10⁻² N/m
Answer: Surface tension S = 0.025 N/m
Surface tension force: F = 2 × S × L (2 sides of film)
This balances weight: 2SL = W
Surface tension: S = W/(2L) = 1.5×10⁻² / (2 × 0.30)
= 1.5×10⁻² / 0.60 = 2.5×10⁻² N/m
Answer: Surface tension S = 0.025 N/m
Question 10.18
Fig (a) shows film supporting 4.5×10⁻² N. What weight supported in (b) and (c)? Explain physically.
Answer & Explanation:
Answer: All support same weight: 4.5×10⁻² N
Reason: Surface tension force depends on:
• Surface tension coefficient (S)
• Length of contact line (L)
• Number of surfaces (2 for film)
In all three figures:
- Same liquid ⇒ same S
- Same total contact length ⇒ same L
- Two surfaces in all cases
Weight supported = 2SL = constant regardless of shape.
Reason: Surface tension force depends on:
• Surface tension coefficient (S)
• Length of contact line (L)
• Number of surfaces (2 for film)
In all three figures:
- Same liquid ⇒ same S
- Same total contact length ⇒ same L
- Two surfaces in all cases
Weight supported = 2SL = constant regardless of shape.
Question 10.19
Mercury drop: r = 3.00 mm at 20°C, S = 4.65×10⁻¹ N/m, Patm = 1.01×10⁵ Pa. Find inside pressure and excess pressure.
Answer & Explanation:
Given: r = 3.00 mm = 0.003 m, S = 0.465 N/m, Patm = 1.01×10⁵ Pa
Excess pressure (liquid drop): ΔP = 2S/r
= 2 × 0.465 / 0.003 = 310 Pa
Inside pressure: Pin = Patm + ΔP
= 1.01×10⁵ + 310 = 1.0131×10⁵ Pa
Answer: Excess pressure = 310 Pa, Inside pressure ≈ 1.013×10⁵ Pa
Excess pressure (liquid drop): ΔP = 2S/r
= 2 × 0.465 / 0.003 = 310 Pa
Inside pressure: Pin = Patm + ΔP
= 1.01×10⁵ + 310 = 1.0131×10⁵ Pa
Answer: Excess pressure = 310 Pa, Inside pressure ≈ 1.013×10⁵ Pa
Question 10.20
Soap bubble: r = 5.00 mm, S = 2.50×10⁻² N/m. Find excess pressure. If air bubble at 40.0 cm depth in soap solution (ρ = 1.20 g/cm³), find pressure inside.
Answer & Explanation:
Part 1 - Soap bubble:
r = 0.005 m, S = 0.025 N/m
Excess pressure (bubble has 2 surfaces): ΔP = 4S/r
= 4 × 0.025 / 0.005 = 20 Pa
Part 2 - Air bubble in liquid:
h = 0.40 m, ρ = 1200 kg/m³, g = 9.8 m/s², Patm = 1.01×10⁵ Pa
Liquid pressure at depth: Pliquid = Patm + ρgh
= 1.01×10⁵ + (1200×9.8×0.40) ≈ 1.01×10⁵ + 4704 ≈ 1.057×10⁵ Pa
Excess pressure for air bubble (1 surface): ΔP = 2S/r = 10 Pa
Inside bubble: Pin = Pliquid + ΔP ≈ 1.057×10⁵ + 10 ≈ 1.0571×10⁵ Pa
Answer: Soap bubble ΔP = 20 Pa; Air bubble Pin ≈ 1.057×10⁵ Pa
r = 0.005 m, S = 0.025 N/m
Excess pressure (bubble has 2 surfaces): ΔP = 4S/r
= 4 × 0.025 / 0.005 = 20 Pa
Part 2 - Air bubble in liquid:
h = 0.40 m, ρ = 1200 kg/m³, g = 9.8 m/s², Patm = 1.01×10⁵ Pa
Liquid pressure at depth: Pliquid = Patm + ρgh
= 1.01×10⁵ + (1200×9.8×0.40) ≈ 1.01×10⁵ + 4704 ≈ 1.057×10⁵ Pa
Excess pressure for air bubble (1 surface): ΔP = 2S/r = 10 Pa
Inside bubble: Pin = Pliquid + ΔP ≈ 1.057×10⁵ + 10 ≈ 1.0571×10⁵ Pa
Answer: Soap bubble ΔP = 20 Pa; Air bubble Pin ≈ 1.057×10⁵ Pa
Question 10.21
Tank: base 1.0 m², divided vertically. Bottom door area = 20 cm². One side: water (4.0 m), other: acid (ρ = 1.7 g/cm³, 4.0 m). Find force to keep door closed.
Answer & Explanation:
Given: h = 4.0 m, Adoor = 20 cm² = 0.002 m², ρw = 1000 kg/m³, ρacid = 1700 kg/m³, g = 9.8 m/s²
Pressure difference at door: ΔP = (ρacid - ρw)gh
= (1700 - 1000) × 9.8 × 4.0 = 700 × 39.2 = 27440 Pa
Force on door: F = ΔP × Adoor = 27440 × 0.002 = 54.88 N
Answer: Force needed ≈ 54.9 N (toward water side)
Pressure difference at door: ΔP = (ρacid - ρw)gh
= (1700 - 1000) × 9.8 × 4.0 = 700 × 39.2 = 27440 Pa
Force on door: F = ΔP × Adoor = 27440 × 0.002 = 54.88 N
Answer: Force needed ≈ 54.9 N (toward water side)
Question 10.22
Manometer readings: (a) gas pressure = 76 + 20 = 96 cm Hg, (b) after removing gas = 76 - 18 = 58 cm Hg. (a) Give absolute/gauge pressures. (b) If 13.6 cm water added to right limb, new levels?
Answer & Explanation:
(a) Pressures:
Case (a): Absolute = 96 cm Hg, Gauge = 20 cm Hg
Case (b): Absolute = 58 cm Hg, Gauge = -18 cm Hg (vacuum)
(b) With 13.6 cm water:
ρHg = 13.6 g/cm³, ρw = 1 g/cm³
13.6 cm water = 13.6/13.6 = 1 cm Hg equivalent
Right limb rises 1 cm relative to left.
New difference: Left higher by (18 + 1) = 19 cm
Gas pressure: Pgas = 76 - 19 = 57 cm Hg absolute
Case (a): Absolute = 96 cm Hg, Gauge = 20 cm Hg
Case (b): Absolute = 58 cm Hg, Gauge = -18 cm Hg (vacuum)
(b) With 13.6 cm water:
ρHg = 13.6 g/cm³, ρw = 1 g/cm³
13.6 cm water = 13.6/13.6 = 1 cm Hg equivalent
Right limb rises 1 cm relative to left.
New difference: Left higher by (18 + 1) = 19 cm
Gas pressure: Pgas = 76 - 19 = 57 cm Hg absolute
Question 10.23
Two vessels: same base area, different shapes. First holds twice water of second to same height. Is force on base same? Why different scale readings?
Answer & Explanation:
Force on base: Yes, same force
Reason: Pressure at base depends only on height: P = ρgh
Force = P × A (same A, same h) ⇒ same force
Scale reading: Different
Reason: Scale measures total weight of vessel + water
Different shapes ⇒ different volumes ⇒ different water weights
Even with same base force, total weight differs due to extra water in sides.
Key: Hydrostatic paradox - pressure depends on height, not shape.
Reason: Pressure at base depends only on height: P = ρgh
Force = P × A (same A, same h) ⇒ same force
Scale reading: Different
Reason: Scale measures total weight of vessel + water
Different shapes ⇒ different volumes ⇒ different water weights
Even with same base force, total weight differs due to extra water in sides.
Key: Hydrostatic paradox - pressure depends on height, not shape.
Question 10.24
Blood transfusion: vein gauge pressure = 2000 Pa. How high must container be placed? (ρblood = 1.06×10³ kg/m³)
Answer & Explanation:
Given: ΔP = 2000 Pa, ρ = 1060 kg/m³, g = 9.8 m/s²
Hydrostatic pressure: ΔP = ρgh ⇒ h = ΔP/(ρg)
h = 2000 / (1060 × 9.8) = 2000 / 10388 ≈ 0.1925 m
Answer: Container must be ≥ 0.193 m (≈ 19.3 cm) above vein entry point.
Hydrostatic pressure: ΔP = ρgh ⇒ h = ΔP/(ρg)
h = 2000 / (1060 × 9.8) = 2000 / 10388 ≈ 0.1925 m
Answer: Container must be ≥ 0.193 m (≈ 19.3 cm) above vein entry point.
Question 10.25
(a) Max average velocity for laminar flow in artery (d = 2×10⁻³ m)? (b) Do dissipative forces become more important with velocity?
Answer & Explanation:
(a) Critical velocity:
For flow in tubes, turbulence begins at Re ≈ 2000
Re = ρvd/η ⇒ vcrit = (Re·η)/(ρd)
For blood: ρ ≈ 1060 kg/m³, η ≈ 4×10⁻³ Pa·s
vcrit = (2000 × 4×10⁻³)/(1060 × 2×10⁻³)
= 8 / 2.12 ≈ 3.77 m/s
(b) Dissipative forces: Yes, more important
• Viscous drag ∝ velocity (Stokes) or v² (high Re)
• Turbulence increases energy dissipation dramatically
• Bernoulli's equation becomes less accurate
• More pressure drop needed to maintain flow
For flow in tubes, turbulence begins at Re ≈ 2000
Re = ρvd/η ⇒ vcrit = (Re·η)/(ρd)
For blood: ρ ≈ 1060 kg/m³, η ≈ 4×10⁻³ Pa·s
vcrit = (2000 × 4×10⁻³)/(1060 × 2×10⁻³)
= 8 / 2.12 ≈ 3.77 m/s
(b) Dissipative forces: Yes, more important
• Viscous drag ∝ velocity (Stokes) or v² (high Re)
• Turbulence increases energy dissipation dramatically
• Bernoulli's equation becomes less accurate
• More pressure drop needed to maintain flow
Question 10.26
(a) Max average velocity for laminar flow in artery (r = 2×10⁻³ m)? (b) Corresponding flow rate? (ηblood = 2.084×10⁻³ Pa·s)
Answer & Explanation:
Given: r = 2×10⁻³ m ⇒ d = 4×10⁻³ m, η = 2.084×10⁻³ Pa·s, ρ = 1060 kg/m³, Recrit ≈ 2000
(a) Critical velocity:
vcrit = (Re·η)/(ρd) = (2000 × 2.084×10⁻³)/(1060 × 4×10⁻³)
= 4.168 / 4.24 ≈ 0.983 m/s
(b) Flow rate: Q = Av = πr²v
= π × (2×10⁻³)² × 0.983 = π × 4×10⁻⁶ × 0.983
≈ 1.235×10⁻⁵ m³/s = 12.35 mL/s
Answer: (a) vmax ≈ 0.98 m/s, (b) Q ≈ 1.24×10⁻⁵ m³/s (12.4 mL/s)
(a) Critical velocity:
vcrit = (Re·η)/(ρd) = (2000 × 2.084×10⁻³)/(1060 × 4×10⁻³)
= 4.168 / 4.24 ≈ 0.983 m/s
(b) Flow rate: Q = Av = πr²v
= π × (2×10⁻³)² × 0.983 = π × 4×10⁻⁶ × 0.983
≈ 1.235×10⁻⁵ m³/s = 12.35 mL/s
Answer: (a) vmax ≈ 0.98 m/s, (b) Q ≈ 1.24×10⁻⁵ m³/s (12.4 mL/s)
Question 10.27
Plane level flight: each wing area = 25 m², vlower = 180 km/h, vupper = 234 km/h, ρair = 1 kg/m³. Find plane mass.
Answer & Explanation:
Given: A = 25 m² per wing (total 50 m²), v₁ = 180 km/h = 50 m/s, v₂ = 234 km/h = 65 m/s, ρ = 1 kg/m³
Pressure difference: ΔP = ½ρ(v₂² - v₁²)
= 0.5 × 1 × (65² - 50²) = 0.5 × (4225 - 2500) = 0.5 × 1725 = 862.5 Pa
Lift force: F = ΔP × Atotal = 862.5 × 50 = 43125 N
Plane mass: m = F/g = 43125 / 9.8 ≈ 4400.5 kg
Answer: Plane mass ≈ 4400 kg (4.4 metric tons)
Pressure difference: ΔP = ½ρ(v₂² - v₁²)
= 0.5 × 1 × (65² - 50²) = 0.5 × (4225 - 2500) = 0.5 × 1725 = 862.5 Pa
Lift force: F = ΔP × Atotal = 862.5 × 50 = 43125 N
Plane mass: m = F/g = 43125 / 9.8 ≈ 4400.5 kg
Answer: Plane mass ≈ 4400 kg (4.4 metric tons)
Question 10.28
Millikan's experiment: oil drop r = 2.0×10⁻⁵ m, ρ = 1.2×10³ kg/m³, ηair = 1.8×10⁻⁵ Pa·s. Find terminal speed and viscous force. Neglect buoyancy.
Answer & Explanation:
Given: r = 2.0×10⁻⁵ m, ρ = 1200 kg/m³, η = 1.8×10⁻⁵ Pa·s, g = 9.8 m/s², neglect buoyancy (σ = 0)
Terminal velocity (Stokes): vt = 2r²ρg/(9η)
= 2 × (2×10⁻⁵)² × 1200 × 9.8 / (9 × 1.8×10⁻⁵)
= (2 × 4×10⁻¹⁰ × 11760) / (1.62×10⁻⁴)
= (9.408×10⁻⁶) / (1.62×10⁻⁴) ≈ 0.0581 m/s
Viscous force at vt: F = 6πηrvt
= 6π × 1.8×10⁻⁵ × 2×10⁻⁵ × 0.0581
≈ 6π × 2.0916×10⁻¹⁰ ≈ 3.94×10⁻⁹ N
Check: Also equals weight mg = (4/3)πr³ρg ≈ 3.94×10⁻⁹ N ✓
Answer: vt ≈ 5.81 cm/s, Fviscous ≈ 3.94×10⁻⁹ N
Terminal velocity (Stokes): vt = 2r²ρg/(9η)
= 2 × (2×10⁻⁵)² × 1200 × 9.8 / (9 × 1.8×10⁻⁵)
= (2 × 4×10⁻¹⁰ × 11760) / (1.62×10⁻⁴)
= (9.408×10⁻⁶) / (1.62×10⁻⁴) ≈ 0.0581 m/s
Viscous force at vt: F = 6πηrvt
= 6π × 1.8×10⁻⁵ × 2×10⁻⁵ × 0.0581
≈ 6π × 2.0916×10⁻¹⁰ ≈ 3.94×10⁻⁹ N
Check: Also equals weight mg = (4/3)πr³ρg ≈ 3.94×10⁻⁹ N ✓
Answer: vt ≈ 5.81 cm/s, Fviscous ≈ 3.94×10⁻⁹ N
Question 10.29
Mercury-glass: θ = 140°, tube r = 1.00 mm, S = 0.465 N/m, ρHg = 13.6×10³ kg/m³. Find capillary depression.
Answer & Explanation:
Given: θ = 140°, r = 0.001 m, S = 0.465 N/m, ρ = 13600 kg/m³, g = 9.8 m/s²
Note: cos140° = cos(180°-40°) = -cos40° ≈ -0.7660
Capillary depression: h = (2S cosθ)/(ρgr)
= (2 × 0.465 × -0.7660) / (13600 × 9.8 × 0.001)
= -0.7124 / 133.28 ≈ -0.005345 m
Answer: Depression ≈ 5.35 mm below outside level
Note: cos140° = cos(180°-40°) = -cos40° ≈ -0.7660
Capillary depression: h = (2S cosθ)/(ρgr)
= (2 × 0.465 × -0.7660) / (13600 × 9.8 × 0.001)
= -0.7124 / 133.28 ≈ -0.005345 m
Answer: Depression ≈ 5.35 mm below outside level
Question 10.30
U-tube: bores d₁ = 3.0 mm, d₂ = 6.0 mm, water S = 7.3×10⁻² N/m, θ = 0°, ρ = 1000 kg/m³. Find level difference.
Answer & Explanation:
Given: r₁ = 1.5 mm = 0.0015 m, r₂ = 3.0 mm = 0.003 m, S = 0.073 N/m, θ = 0° ⇒ cosθ = 1, ρ = 1000 kg/m³, g = 9.8 m/s²
Capillary rise in each: h = (2S cosθ)/(ρgr)
h₁ = (2 × 0.073 × 1) / (1000 × 9.8 × 0.0015) = 0.146 / 14.7 ≈ 0.00993 m
h₂ = 0.146 / (1000 × 9.8 × 0.003) = 0.146 / 29.4 ≈ 0.00497 m
Level difference: Δh = h₁ - h₂ = 0.00993 - 0.00497 ≈ 0.00496 m
Answer: Level difference ≈ 4.96 mm (water higher in narrower tube)
Capillary rise in each: h = (2S cosθ)/(ρgr)
h₁ = (2 × 0.073 × 1) / (1000 × 9.8 × 0.0015) = 0.146 / 14.7 ≈ 0.00993 m
h₂ = 0.146 / (1000 × 9.8 × 0.003) = 0.146 / 29.4 ≈ 0.00497 m
Level difference: Δh = h₁ - h₂ = 0.00993 - 0.00497 ≈ 0.00496 m
Answer: Level difference ≈ 4.96 mm (water higher in narrower tube)
📘 Exam Preparation Tip:
These exercise questions will help you understand fluid behavior and properties. You'll learn to calculate pressure variations with depth, apply Pascal's law and Bernoulli's principle to practical situations, analyze viscous flow using Poiseuille's and Stokes' laws, solve surface tension problems involving capillary action and bubbles, and apply fluid mechanics to real-world scenarios like aircraft lift, blood flow, and hydraulic machines. Perfect for building strong fundamentals for board exams and engineering entrance tests.