This MCQ module is based on: NCERT Exercises and Solutions: Mechanical Properties of Fluids
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NCERT Exercises and Solutions: Mechanical Properties of Fluids
🎓 Class 11
Physics
CBSE
Theory
Ch 9 – Mechanical Properties of Fluids
⏱ ~8 min
🌐 Language: [gtranslate]
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NCERT Exercises and Solutions: Mechanical Properties of Fluids
Chapter 9 — Summary Table
Here is a concise reference for all the key physical quantities used in the chapter.
| Quantity | Symbol / Formula | Dimensions | SI Unit | Remarks |
|---|---|---|---|---|
| Pressure | \(P = F/A\) | \([ML^{-1}T^{-2}]\) | Pa (N/m²) | Scalar; 1 atm = 1.013×10⁵ Pa |
| Density | \(\rho = m/V\) | \([ML^{-3}]\) | kg/m³ | Water = 1000 kg/m³ |
| Specific gravity | \(\rho/\rho_w\) | dimensionless | — | Mercury = 13.6 |
| Gauge pressure | \(\rho gh\) | \([ML^{-1}T^{-2}]\) | Pa | P(absolute) − P₀ |
| Volume flow rate | \(Q = Av\) | \([L^{3}T^{-1}]\) | m³/s | Conserved along streamline |
| Coefficient of viscosity | \(\eta\) | \([ML^{-1}T^{-1}]\) | Pa·s | 1 Pa·s = 10 poise |
| Terminal velocity | \(v_t = \tfrac{2r^2(\rho-\sigma)g}{9\eta}\) | \([LT^{-1}]\) | m/s | Stokes regime |
| Surface tension | \(S = F/L\) | \([MT^{-2}]\) | N/m (= J/m²) | Water ≈ 0.073 N/m |
| Capillary rise | \(h = \tfrac{2S\cos\theta}{\rho g r}\) | \([L]\) | m | θ is angle of contact |
| Excess pressure (drop) | \(\Delta P = 2S/r\) | \([ML^{-1}T^{-2}]\) | Pa | Soap bubble: 4S/r |
| Reynolds number | \(R_e = \rho v d/\eta\) | dimensionless | — | <1000 laminar; >2000 turbulent |
Keywords
FluidAnything that flows: liquids + gases
PressureNormal force per unit area; scalar
Pascal (Pa)SI unit of pressure, 1 N/m²
Atmospheric Pressure1.013 × 10⁵ Pa at sea level
Gauge PressurePressure above atmospheric
Pascal's LawPressure transmitted equally in a fluid
Hydraulic LiftMultiplies force via Pascal's law
Streamline FlowSmooth, layered, steady flow
Turbulent FlowChaotic, high-Reynolds flow
ContinuityA₁v₁ = A₂v₂
Bernoulli's PrincipleP + ½ρv² + ρgh = const
Torricelli's Theoremv = √(2gh) speed of efflux
Venturi MeterMeasures flow via pressure drop
Dynamic LiftForce on aerofoil from pressure difference
Magnus EffectSideways force on spinning ball
Viscosity (η)Resistance to relative layer motion
PoiseCGS unit of viscosity; 10 P = 1 Pa·s
Stokes' LawF = 6πηrv
Terminal VelocityConstant-speed fall in fluid
Reynolds Numberρvd/η; laminar vs turbulent
Surface TensionForce per unit length at surface
Surface EnergyEnergy per unit area; = S
Angle of Contactθ between liquid and solid
CapillarityRise/fall in narrow tubes
MeniscusCurved liquid-air boundary
Wettingθ < 90°; liquid spreads on solid
NCERT Exercises — Full Solutions
All problems are solved step-by-step. Click "Show Solution" on each to reveal the working. Use \(g = 9.8\) m/s² unless stated otherwise; \(\rho_w = 1000\) kg/m³; atmospheric pressure \(P_0 = 1.01\times 10^5\) Pa; \(\rho_{Hg} = 13.6\times 10^3\) kg/m³.
Q 9.1 — Why is the blood pressure in humans higher in the feet than in the brain?
Blood is a liquid, and like any liquid at rest it obeys \(P = P_0 + \rho gh\). The feet are about 1.3 m below the heart, while the brain is roughly 0.5 m above it. The extra hydrostatic pressure in the feet is \(\Delta P = \rho g h \approx 1060\times 9.8\times 1.8 \approx 1.87\times 10^4\) Pa ≈ 140 mm Hg. The heart therefore pumps blood with enough pressure to reach the brain against gravity, while the same pump also creates a higher pressure in the lower body.
Q 9.2 — Atmospheric density at sea level is 1.29 kg/m³ and the atmosphere reaches up to roughly 8 km. Yet atmospheric pressure is ~1.013×10⁵ Pa. Discuss in terms of \(P = \rho g h\).
If the atmosphere had uniform density 1.29 kg/m³, a column 8 km tall would give \(P = \rho g h = 1.29\times 9.8\times 8000 \approx 1.01\times 10^5\) Pa — consistent with observed atmospheric pressure. In reality, air density falls rapidly with altitude (the atmosphere is compressible, unlike liquids), so the simple formula is only approximate. The real atmosphere extends to several hundred kilometres, but most of the mass is within the first 8 km — which is why the simple model works.
Q 9.3 — In a U-tube of uniform cross-section, a liquid fills both arms. If a second, immiscible, less dense liquid is poured into one arm, what happens to the levels?
By Pascal's principle, the pressure at the same horizontal level in a connected fluid must be equal. Let the denser liquid stand at height \(h_1\) in arm 1, and the lighter liquid at height \(h_2\) above a common lower level in arm 2. Then \(\rho_1 g h_1 = \rho_2 g h_2\) at the common level. Since \(\rho_2 < \rho_1\), the lighter liquid's column \(h_2\) must be taller — the level in arm 2 rises higher than the level in arm 1. The heights are inversely proportional to densities.
Q 9.4 — A vertical off-shore structure needs to withstand the water pressure at a depth of 3 km below sea level. Is the structure suitable if its walls are designed for pressures of up to 10⁹ Pa? Sea-water density = 1030 kg/m³.
\[P = P_0 + \rho g h = 1.01\times 10^5 + 1030\times 9.8\times 3000\]
\[= 1.01\times 10^5 + 3.028\times 10^7 = 3.04\times 10^7\text{ Pa}\]
Design limit = \(10^9\) Pa ≈ 33× the expected pressure. Yes — the structure is vastly over-engineered for the 3 km depth (safety factor ≈ 33).
Q 9.5 — A hydraulic car lift has an input piston of radius 5 cm. What load (in kg) can be lifted on the output piston (radius 15 cm) if a force of 1350 N is applied to the input? Take \(g = 10\) m/s².
\(A_1 = \pi(0.05)^2 = 0.00785\) m², \(A_2 = \pi(0.15)^2 = 0.0707\) m². Ratio \(A_2/A_1 = 9\).
\(F_2 = F_1(A_2/A_1) = 1350 \times 9 = 12150\) N.
Maximum mass \(M = F_2/g = 12150/10 = \boxed{1215\text{ kg}}\).
\(F_2 = F_1(A_2/A_1) = 1350 \times 9 = 12150\) N.
Maximum mass \(M = F_2/g = 12150/10 = \boxed{1215\text{ kg}}\).
Q 9.6 — A mercury manometer reads 760 mm in open air and 800 mm when connected to a gas cylinder. What is the gauge pressure and the absolute pressure of the gas?
Actually the question asks for the difference in the mercury levels in the two arms.
Gauge pressure \(= \rho_{Hg}\,g\,\Delta h = 13600\times 9.8\times (0.800 - 0.760) = 13600\times 9.8\times 0.040\)
\(= 5.33\times 10^3\) Pa ≈ \(\boxed{5.33\text{ kPa}}\).
Absolute pressure \(= P_0 + 5.33\) kPa = 101.3 + 5.33 = \(\boxed{106.6\text{ kPa}}\).
Gauge pressure \(= \rho_{Hg}\,g\,\Delta h = 13600\times 9.8\times (0.800 - 0.760) = 13600\times 9.8\times 0.040\)
\(= 5.33\times 10^3\) Pa ≈ \(\boxed{5.33\text{ kPa}}\).
Absolute pressure \(= P_0 + 5.33\) kPa = 101.3 + 5.33 = \(\boxed{106.6\text{ kPa}}\).
Q 9.7 — A plane is moving at 200 m/s. Air over the upper wing surface flows at 240 m/s, and under the lower surface at 160 m/s. If the wing area is 2 m² and density of air is 1.2 kg/m³, find the lift. Will it support an aircraft of mass 500 kg? \(g = 10\) m/s².
\[\Delta P = \tfrac{1}{2}\rho(v_\text{top}^2 - v_\text{bot}^2) = \tfrac{1}{2}\times 1.2\times(240^2 - 160^2)\]
\[= 0.6\times(57600 - 25600) = 0.6\times 32000 = 19200\text{ Pa}\]
Lift \(= \Delta P \times A = 19200\times 2 = 38400\) N.
Weight of aircraft \(= 500\times 10 = 5000\) N. The lift (38.4 kN) far exceeds the weight (5 kN), so yes — the plane easily stays aloft.
Answer: Lift \(\boxed{= 3.84\times 10^4\text{ N}}\); sufficient.
Weight of aircraft \(= 500\times 10 = 5000\) N. The lift (38.4 kN) far exceeds the weight (5 kN), so yes — the plane easily stays aloft.
Answer: Lift \(\boxed{= 3.84\times 10^4\text{ N}}\); sufficient.
Q 9.8 — Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. The volume flow rate is 4.0 × 10⁻³ m³/s. Find the pressure difference between the ends of the tube. Use \(\eta = 0.83\) Pa·s.
Using Poiseuille's formula: \(Q = \pi \Delta P\,r^4/(8\eta L)\), so:
\[\Delta P = \frac{8\eta L Q}{\pi r^4} = \frac{8\times 0.83\times 1.5\times 4.0\times 10^{-3}}{\pi\times(0.01)^4}\]
\[= \frac{0.0399}{3.14\times 10^{-8}} = 1.27\times 10^6\text{ Pa}\]
Answer: \(\Delta P \approx \boxed{1.27\times 10^6\text{ Pa} \approx 12.5\text{ atm}}\). This is a huge pressure — glycerine's high viscosity makes pipe-flow very lossy.
Q 9.9 — Water flows in a horizontal tube of varying cross-section. At one section the area is 10 cm² and the velocity is 1 m/s. At a narrower section the velocity is 2 m/s. Find the area at the narrow section and the pressure difference between the sections.
Continuity: \(A_2 = A_1 v_1/v_2 = 10\times 1/2 = 5\) cm² \(= 5\times 10^{-4}\) m².
Bernoulli (horizontal): \(\Delta P = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}\times 1000\times (4 - 1) = 1500\) Pa.
Answer: \(A_2 = \boxed{5\text{ cm}^2}\); \(P_1 - P_2 = \boxed{1500\text{ Pa}}\).
Bernoulli (horizontal): \(\Delta P = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}\times 1000\times (4 - 1) = 1500\) Pa.
Answer: \(A_2 = \boxed{5\text{ cm}^2}\); \(P_1 - P_2 = \boxed{1500\text{ Pa}}\).
Q 9.10 — A cylindrical tank stands on a horizontal floor and is 5.0 m high. Water fills the tank to a depth of 3.0 m. A hole is drilled at the side, 1.0 m above the base. With what speed does water emerge, and at what horizontal distance from the tank does it strike the floor?
Depth of water above the hole: \(h = 3.0 - 1.0 = 2.0\) m.
Torricelli: \(v = \sqrt{2gh} = \sqrt{2\times 9.8\times 2.0} = \sqrt{39.2} = 6.26\) m/s.
Time to fall 1.0 m: \(t = \sqrt{2\times 1.0/9.8} = 0.452\) s.
Horizontal range \(R = v t = 6.26\times 0.452 = 2.83\) m.
Answer: \(v = \boxed{6.3\text{ m/s}}\), \(R \approx \boxed{2.8\text{ m}}\).
Torricelli: \(v = \sqrt{2gh} = \sqrt{2\times 9.8\times 2.0} = \sqrt{39.2} = 6.26\) m/s.
Time to fall 1.0 m: \(t = \sqrt{2\times 1.0/9.8} = 0.452\) s.
Horizontal range \(R = v t = 6.26\times 0.452 = 2.83\) m.
Answer: \(v = \boxed{6.3\text{ m/s}}\), \(R \approx \boxed{2.8\text{ m}}\).
Q 9.11 — A sphere of radius 1.0 mm falls through a tank of oil. Density of the sphere = 7800 kg/m³, density of oil = 900 kg/m³, viscosity of oil = 2.0 Pa·s. Find the terminal velocity.
\(r = 1.0\times 10^{-3}\) m, \(\rho - \sigma = 7800 - 900 = 6900\) kg/m³.
\[v_t = \frac{2 r^2(\rho - \sigma)g}{9\eta} = \frac{2\times 10^{-6}\times 6900\times 9.8}{9\times 2.0}\]
\[= \frac{0.1352}{18} = 7.51\times 10^{-3}\text{ m/s} \approx 7.5\text{ mm/s}\]
Answer: \(v_t \approx \boxed{7.5\times 10^{-3}\text{ m/s}}\).
Q 9.12 — A capillary tube of internal diameter 0.6 mm is dipped vertically in water. Find the height to which water rises. S = 7.3 × 10⁻² N/m, θ = 0°.
\(r = 0.30\) mm \(= 3.0\times 10^{-4}\) m.
\[h = \frac{2S\cos\theta}{\rho g r} = \frac{2\times 7.3\times 10^{-2}\times 1}{1000\times 9.8\times 3.0\times 10^{-4}}\]
\[= \frac{0.146}{2.94} = 0.0497\text{ m} \approx 4.97\text{ cm}\]
Answer: \(h \approx \boxed{5.0\text{ cm}}\).
Q 9.13 — A soap bubble of radius 2 cm is blown at the end of a narrow tube. If the surface tension of the soap solution is 0.030 N/m, calculate (a) the excess pressure inside the bubble and (b) the work done to blow it.
(a) \(\Delta P = 4S/r = 4\times 0.030/0.02 = 6.0\) Pa.
(b) Work done = S × (change in surface area). A soap bubble has two surfaces, so: \[W = S\times 2\times 4\pi r^2 = 0.030\times 8\pi\times (0.02)^2\] \[= 0.030\times 8\pi\times 4\times 10^{-4} = 3.02\times 10^{-4}\text{ J}\] Answer: \(\Delta P = \boxed{6\text{ Pa}}\); \(W \approx \boxed{3.0\times 10^{-4}\text{ J}}\).
(b) Work done = S × (change in surface area). A soap bubble has two surfaces, so: \[W = S\times 2\times 4\pi r^2 = 0.030\times 8\pi\times (0.02)^2\] \[= 0.030\times 8\pi\times 4\times 10^{-4} = 3.02\times 10^{-4}\text{ J}\] Answer: \(\Delta P = \boxed{6\text{ Pa}}\); \(W \approx \boxed{3.0\times 10^{-4}\text{ J}}\).
Q 9.14 — Water flows through a horizontal pipe of diameter 2.0 cm at 2.0 m/s. Find the Reynolds number and state whether the flow is laminar or turbulent. Take \(\rho = 1000\) kg/m³ and \(\eta = 1.0\times 10^{-3}\) Pa·s.
Reynolds number:
\[R_e = \frac{\rho v d}{\eta} = \frac{1000\times 2.0\times 0.020}{1.0\times 10^{-3}} = \frac{40}{10^{-3}} = 4.0\times 10^4\]
Since \(R_e \gg 2000\), the flow is turbulent.
Rule of thumb: \(R_e < 1000\) — laminar; \(1000 < R_e < 2000\) — transitional; \(R_e > 2000\) — turbulent.
Answer: \(R_e = \boxed{4\times 10^4}\); flow is turbulent.
Rule of thumb: \(R_e < 1000\) — laminar; \(1000 < R_e < 2000\) — transitional; \(R_e > 2000\) — turbulent.
Answer: \(R_e = \boxed{4\times 10^4}\); flow is turbulent.
Q 9.15 — Eight identical spherical drops of mercury, each of radius 1 mm, combine to form a single large drop. If surface tension of mercury is 0.465 N/m, calculate the energy released in the process.
Volume conservation: \(8\times \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi R^3 \Rightarrow R = 2r = 2\) mm.
Initial surface area: \(A_i = 8\times 4\pi r^2 = 32\pi r^2\).
Final surface area: \(A_f = 4\pi R^2 = 16\pi r^2\).
Decrease: \(\Delta A = 16\pi r^2 = 16\pi\times (10^{-3})^2 = 5.03\times 10^{-5}\) m².
Energy released: \(E = S\,\Delta A = 0.465\times 5.03\times 10^{-5} = 2.34\times 10^{-5}\) J.
Answer: \(E \approx \boxed{2.3\times 10^{-5}\text{ J}}\). This appears as a slight temperature rise of the merged drop.
Initial surface area: \(A_i = 8\times 4\pi r^2 = 32\pi r^2\).
Final surface area: \(A_f = 4\pi R^2 = 16\pi r^2\).
Decrease: \(\Delta A = 16\pi r^2 = 16\pi\times (10^{-3})^2 = 5.03\times 10^{-5}\) m².
Energy released: \(E = S\,\Delta A = 0.465\times 5.03\times 10^{-5} = 2.34\times 10^{-5}\) J.
Answer: \(E \approx \boxed{2.3\times 10^{-5}\text{ J}}\). This appears as a slight temperature rise of the merged drop.
Q 9.16 — Advanced: A mercury drop of radius R splits into n equal-sized smaller drops. Show that the work done equals \(4\pi R^2 S (n^{1/3} - 1)\).
Let \(r\) = radius of each small drop. Volume conservation: \(\tfrac{4}{3}\pi R^3 = n\cdot\tfrac{4}{3}\pi r^3 \Rightarrow r = R/n^{1/3}\).
Total initial surface area: \(A_i = 4\pi R^2\).
Total final surface area: \(A_f = n\times 4\pi r^2 = 4\pi n(R/n^{1/3})^2 = 4\pi R^2 n^{1/3}\).
Increase in area: \(\Delta A = A_f - A_i = 4\pi R^2(n^{1/3} - 1)\).
Work done \(= S\,\Delta A = \boxed{4\pi R^2 S\,(n^{1/3} - 1)}\). □
Total initial surface area: \(A_i = 4\pi R^2\).
Total final surface area: \(A_f = n\times 4\pi r^2 = 4\pi n(R/n^{1/3})^2 = 4\pi R^2 n^{1/3}\).
Increase in area: \(\Delta A = A_f - A_i = 4\pi R^2(n^{1/3} - 1)\).
Work done \(= S\,\Delta A = \boxed{4\pi R^2 S\,(n^{1/3} - 1)}\). □
Q 9.17 — Advanced: Two soap bubbles of radii 3 cm and 5 cm are connected by a narrow tube. Air flows from one to the other. Find the radius of the resultant curved surface at the tube junction. S = 0.030 N/m.
Excess pressure inside smaller bubble (r₁ = 3 cm): \(P_1 - P_0 = 4S/r_1\).
Excess pressure inside larger bubble (r₂ = 5 cm): \(P_2 - P_0 = 4S/r_2\).
The common curved film at the junction experiences a net pressure \(P_1 - P_2 = 4S(1/r_1 - 1/r_2)\), which equals \(4S/r\) where \(r\) is the radius of curvature of that film: \[\frac{1}{r} = \frac{1}{r_1} - \frac{1}{r_2} = \frac{r_2 - r_1}{r_1 r_2} = \frac{5-3}{3\times 5} = \frac{2}{15}\text{ cm}^{-1}\] \[r = \frac{15}{2} = 7.5\text{ cm}\] Answer: \(r = \boxed{7.5\text{ cm}}\). The curved surface bulges into the larger bubble.
Excess pressure inside larger bubble (r₂ = 5 cm): \(P_2 - P_0 = 4S/r_2\).
The common curved film at the junction experiences a net pressure \(P_1 - P_2 = 4S(1/r_1 - 1/r_2)\), which equals \(4S/r\) where \(r\) is the radius of curvature of that film: \[\frac{1}{r} = \frac{1}{r_1} - \frac{1}{r_2} = \frac{r_2 - r_1}{r_1 r_2} = \frac{5-3}{3\times 5} = \frac{2}{15}\text{ cm}^{-1}\] \[r = \frac{15}{2} = 7.5\text{ cm}\] Answer: \(r = \boxed{7.5\text{ cm}}\). The curved surface bulges into the larger bubble.
Frequently Asked Questions - NCERT Exercises and Solutions: Mechanical Properties of Fluids
What are the key NCERT exercise types in Chapter 9 Mechanical Properties of Fluids?
NCERT Class 11 Physics Chapter 9 Mechanical Properties of Fluids exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question in the chapter, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Mechanical Properties of Fluids?
For numerical problems in NCERT Class 11 Physics Chapter 9 Mechanical Properties of Fluids: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 9?
From NCERT Class 11 Physics Chapter 9 (Mechanical Properties of Fluids), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 11 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula (M^a L^b T^c). For Chapter 9 Mechanical Properties of Fluids problems, write the dimensions of each quantity, substitute, and simplify. If dimensions match the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 9 exercises?
Common mistakes in NCERT Class 11 Physics Chapter 9 Mechanical Properties of Fluids exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for vectors, (3) forgetting to consider all forces in free-body diagrams, (4) algebra errors during derivations, and (5) misreading the problem statement. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 11 Physics Chapter 9 Mechanical Properties of Fluids solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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