L-9: Properties of Fluids
Physics — Class 12 · NIOS Code 312 · Module 2 · Source: 312_Physics_Eng_Lesson9.pdf
Fluids All Around — Why This Lesson Matters
Fluids (liquids and gases) flow when conditions permit. Unlike solids, fluids cannot sustain shearing stress — they exert forces normal to surfaces. This lesson explains why dam walls thicken at the base, hydraulic jacks lift cars, mosquitoes walk on water, honey flows slowly, and cricket balls swing in air.
NIOS objectives: hydrostatic pressure; buoyancy and Archimedes' principle; Pascal's law and hydraulic devices; surface tension and surface energy; capillary rise; streamline vs turbulent flow; Reynolds number; viscosity; Bernoulli's principle and applications.
9.1 Hydrostatic Pressure
Pressure is thrust (normal force) per unit area — the same force has greater effect on a smaller area (sharp pin vs blunt rod).
F = Normal force / thrust (N)
A = Area (m²)
SI unit named after Blaise Pascal.
Fluids at rest exert hydrostatic pressure. Consider a vertical cylinder of area A and height h in a liquid of density ρ. Equilibrium of the liquid column gives:
ρ = Liquid density (kg·m⁻³)
g = 9.8 m·s⁻²
h = Depth below free surface (m)
Pressure increases linearly with depth — dam walls thicker at base.
At the open surface, add atmospheric pressure:
P_atm ≈ 1.01×10⁵ Pa (76 cm Hg barometer — Torricelli).
Pressure at given depth is independent of container shape (Fig. 9.5).
Example 9.1: Dam depth 100 m, ρ = 10³ kg/m³ → P = 9.8×10⁵ Pa. If 1 m wall withstands 10⁵ Pa, bottom wall needs ~9.8 m thickness.
9.1.2 Atmospheric Pressure
Earth's atmosphere extends ~200 km. O. von Guericke showed with evacuated copper hemispheres that 8 horses could barely pull them apart. Standard atmospheric pressure from mercury barometer:
ρ_Hg = 13600 kg·m⁻³
9.2 Buoyancy and Archimedes' Principle
Lifting an object underwater feels easier because the fluid exerts an upward buoyant force. Archimedes' principle: when a body is submerged wholly or partially, buoyant force equals the weight of displaced fluid.
- Floating: buoyant force = weight of object; object displaces fluid equal to its own weight.
- Sinking: object denser than fluid — net downward force.
- Hot-air balloon: less dense hot air → net upward buoyancy in cold air.
9.3 Pascal's Law
When pressure is applied at any point in an enclosed liquid, it is transmitted undiminished to every point and to the container walls. This is the law of transmission of liquid pressure.
PASCAL'S LAW — HYDRAULIC DEVICES
=================================
Small piston (A₁) Large piston (A₂)
F₁ applied ──liquid──► lifts load F₂
|
v
Same pressure P = F₁/A₁ = F₂/A₂
|
v
F₂ = F₁ × (A₂/A₁) → force multiplication
Driver's small force on brake pedal → large force on all four wheel cylinders simultaneously.
Intext check: Boy 25 kg on 0.05 m² vs elephant 5000 kg on 10 m² — ratio A₂/A₁ = 200, so boy can balance 25×200 = 5000 kg!
9.4 Surface Tension
Liquid drops are spherical; soap bubbles form easily but pure water bubbles do not — due to surface tension. Surface molecules experience net inward force (fewer neighbours above); work is needed to expand the surface → surface energy.
F = Force normal to line L on surface
W = Work to increase area by A
Surface acts like a stretched elastic membrane; decreases with temperature.
Cohesive forces — same substance. Adhesive forces — different substances (glue, water on glass). Water wets glass; mercury does not (strong cohesion).
Excess Pressure on Curved Surfaces
Example 9.3: r = 1 mm — soap bubble P = 100 Pa; water drop and air bubble in water P ≈ 144 Pa.
Applications: mosquitoes on water (2πrT cos θ balances mg); detergents lower T for cleaning; camphor-under-duck random motion.
9.5 Angle of Contact
The angle between the liquid surface tangent and container wall (measured inside the liquid) is the angle of contact θ.
- Acute θ → concave meniscus (water in glass).
- Obtuse θ → convex meniscus (mercury in glass, water in paraffin).
Determined by balance of cohesive (F_c) and adhesive (F_a) forces at the boundary.
9.6 Capillary Action
Rise or fall of liquid in a narrow tube due to surface tension is capillarity. Water rises in blotting paper, plant stems, and fine glass tubes; mercury is depressed.
T = Surface tension
θ = Angle of contact
r = Radius of capillary tube
ρ = Liquid density
Fine bore → greater rise. θ = 0° (water in clean glass) maximises rise.
Pressure deficit under concave meniscus (2T/R) drives liquid up until hρg = 2T/R with R = r/cos θ.
9.7 Viscosity
Adjacent fluid layers exert tangential drag opposing relative motion — viscosity. Water stirs differently near walls vs centre; glycerin flows slower than water.
η = Coefficient of viscosity (N·s·m⁻²)
A = Layer area
dv/dx = Velocity gradient
Negative sign: opposes motion. 1 poise = 0.1 N·s·m⁻².
Viscosity of liquids decreases with temperature; gases show opposite trend. Wall layer of pipe assumed stationary; velocity maximum at centre (parabolic profile for laminar flow).
9.8 Types of Liquid Flow
9.8.1 Streamline (Laminar) vs Turbulent
Line of flow: path of a fluid particle. Streamline: curve whose tangent gives velocity direction. Streamlines never cross (unique velocity at a point).
- Laminar: velocity < critical velocity v_c — layers slide over each other smoothly.
- Turbulent: v > v_c — mixing, zig-zag paths (flood river vs city water pipe).
9.8.2 Equation of Continuity
Mass entering per second = mass leaving.
Narrow section → higher speed.
9.8.3 Critical Velocity and Reynolds Number
R < 1000 → laminar
1000–2000 → unsteady
R > 2000 → turbulent
Also R = ρ v d / η
9.9 Stokes' Law and Terminal Velocity
For a smooth sphere radius r moving at speed v in viscous fluid:
Derived via dimensional analysis: F ∝ η r v.
A sphere falling in viscous liquid reaches terminal velocity when weight = buoyancy + viscous drag:
σ = Density of fluid
r = Radius
Applications: parachutes, raindrop size limit.
9.10 Bernoulli's Principle
Where fluid velocity is high, pressure is low; where velocity is low, pressure is high. Governs chimneys, burrows, convertible tops bulging, umbrella lifting in storms.
Flowing fluid has three energy forms: kinetic (½ρv²), gravitational potential (ρgh), and pressure energy (P).
1. Incompressible (constant ρ)
2. Non-viscous (no friction losses)
3. Streamline (steady) flow
Along a tube of flow between two points.
BERNOULLI APPLICATIONS MAP
==========================
High v → Low P → Lift / suction
|
+-------+-------+-------+-------+
| | | | |
Venturi Aerofoil Atomizer Cricket Efflux
meter (plane) /spray swing √(2gH)
| | | | |
Flow Wing Paint Spin Tank
rate lift gun curve hole
Key Applications
- Torricelli efflux: v = √(2gH) — speed of water leaving hole at depth H (from Bernoulli, P_atm at surface and jet).
- Venturimeter: narrow section → higher v → lower P; measures flow rate V ∝ √(ΔP).
- Aerofoil: crowded streamlines above → lower pressure → lift.
- Atomizer / spray gun / Bunsen burner / carburetor: fast air → low P → sucks liquid/fuel.
- Cricket swing: spin makes air speed asymmetric above/below ball → pressure difference → curved path.
- Pressed hose: constriction increases v, decreases P — jet travels farther.
Example: Tank height 2.5 m → v = √(2×9.8×2.5) = 7 m/s.
Quick Revision
- Pressure: P = F/A; hydrostatic P = ρgh; absolute = P_atm + ρgh.
- Archimedes: buoyancy = weight of displaced fluid.
- Pascal: F₂ = F₁(A₂/A₁) — hydraulic machines.
- Surface tension: T = F/L; drop P = 2T/r; soap bubble P = 4T/r.
- Capillary: h = 2T cos θ/(rρg).
- Viscosity: F = −ηA(dv/dx); Stokes: F = 6πηrv.
- Continuity: A₁v₁ = A₂v₂; Reynolds number classifies flow.
- Bernoulli: P + ½ρv² + ρgh = constant; high v → low P.
Q1. The SI unit of pressure is:
Q2. Hydrostatic pressure at depth h in a liquid of density ρ is:
Q3. According to Archimedes' principle, buoyant force equals:
Q4. Pascal's law states that pressure in an enclosed fluid:
Q5. Excess pressure inside a soap bubble of radius r is:
Q6. Capillary rise h is inversely proportional to:
Q7. Coefficient of viscosity has SI unit:
Q8. For streamline (laminar) flow, Reynolds number R should be:
Q9. According to Bernoulli's principle, where fluid velocity is high:
Q10. Torricelli's law for efflux speed from a tank at depth H is:
PYQ — Previous Year Questions
Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L9 — Properties of Fluids only. Use Model Answer for marking points; Explanation for concept clarity.
Chapter 9 — Properties of Fluids (L9)
20 questions · Section A (MCQ & objective) + Section B (short/long) · Sources: 312/TUS/104A, 312/MAY/204A–C, 68/ESS/1-312-A, Marking Scheme
Section A — Multiple Choice (1 mark)
PYQ1. Pressure due to a liquid column does not depend on — (A) its density (B) its viscosity (C) its height (D) acceleration due to gravity
Model Answer
Answer: (B) its viscosity
Hydrostatic pressure P = hρg depends on height, density and g — not on viscosity.
Explanation
Viscosity matters for flowing fluids (drag, Poiseuille flow). In a static column at rest, pressure at depth h is P = hρg (L9 §9.1).
PYQ2. Two solid spheres of the same metal (masses M and 8M) fall together in a viscous liquid. If terminal velocities are v and nv, then n is — (A) 2 (B) 4 (C) 8 (D) 16
Model Answer
Answer: (B) 4
Stokes: v_t ∝ r². Same metal ⇒ m ∝ r³ ⇒ r ∝ m^(1/3). Hence v_t ∝ m^(2/3). For 8M: n = 8^(2/3) = 4.
Explanation
v₀ = 2r²g(ρ−σ)/(9η). Radius doubles when mass is 8× (volume ×8). Speed scales as r² → factor 4.
PYQ3. Two spherical drops of the same liquid have volumes in ratio 1 : 8. The ratio of excess pressures inside them is — (A) 8 : 1 (B) 2 : 1 (C) 1 : 1 (D) 1 : 2
Model Answer
Answer: (B) 2 : 1
V ∝ r³ ⇒ volume ratio 1:8 gives r ratio 1:2. ΔP = 2T/r ⇒ smaller drop (r₁) has higher pressure: P₁/P₂ = r₂/r₁ = 2 : 1.
Explanation
Liquid drop (one surface): ΔP = 2T/r. Smaller radius ⇒ larger excess pressure. r₂/r₁ = 2 ⇒ P₁/P₂ = 2.
PYQ4. Capillarity is due to — (A) Cohesion only (B) Adhesion only (C) Cohesion and adhesion both (D) Neither
Model Answer
Answer: (C) Cohesion and adhesion both
Explanation
Cohesion (liquid–liquid) and adhesion (liquid–wall) together fix the contact angle and capillary rise/fall: h = 2T cos θ/(rρg).
PYQ5. For non-viscous incompressible steady flow, if pipe area is halved, velocity becomes — (A) 4× (B) 3× (C) 2× (D) unchanged
Model Answer
Answer: (C) Doubled
Equation of continuity: A₁v₁ = A₂v₂. If A₂ = A₁/2, then v₂ = 2v₁.
Explanation
Mass conservation for incompressible flow. Narrower tube ⇒ faster flow — basis of Bernoulli applications (L9 §9.10).
PYQ6. Excess pressure inside two soap bubbles of diameters in ratio 4 : 1 is — (A) 1 : 4 (B) 2 : 1 (C) 1 : 2 (D) 4 : 1
Model Answer
Answer: (A) 1 : 4
Soap bubble: ΔP = 4T/r. Pressure ∝ 1/r. Diameter ratio 4:1 ⇒ radius ratio 4:1 ⇒ pressure ratio 1:4.
Explanation
Bubble has two surfaces → P = 4T/r (L9). Larger bubble has smaller excess pressure.
PYQ7. Match device with principle (any two): (a) Barometer — ? (b) Hydraulic brake — ? Options: (i) Upthrust = weight of displaced fluid (ii) P = hdg (iii) F₁/A₁ = F₂/A₂ (iv) F = 6πηrv
Model Answer
(a) Barometer ↔ (ii) P = hdg
(b) Hydraulic brake ↔ (iii) F₁/A₁ = F₂/A₂ (Pascal's law)
(i) ↔ Archimedes · (iv) ↔ Stokes' law
Explanation
Barometer measures atmospheric pressure from mercury column height. Hydraulic systems transmit pressure equally — small piston force amplified on large piston.
PYQ8. True or False (any two): (a) Kerosene rises in a lantern wick due to surface tension (b) Raindrop hits ground at terminal velocity (c) Time to reach terminal velocity in air depends on air density (d) Excess pressure in soap bubble of radius r is 2T/r
Model Answer
- (a) True — capillary action driven by surface tension
- (b) True — viscous drag balances weight at terminal speed
- (c) True — drag depends on fluid density
- (d) False — soap bubble has two surfaces: ΔP = 4T/r
Explanation
2T/r is for a liquid drop with one free surface. A bubble in air has inner and outer surfaces → factor of 2 extra.
PYQ9. Passage — Bernoulli spray gun: In P + ½ρv² + ρgh = const, for unit volume A, B, C are — ? If piston speed 5 mm/s (r=20 mm), nozzle air speed (r=1 mm) is — ?
Model Answer
(i) (c) pressure energy, kinetic energy, potential energy (per unit volume: P, ½ρv², ρgh)
(ii) (c) 2 m·s⁻¹
Continuity: π(20)²×5 = π(1)²×v₂ ⇒ v₂ = 2000 mm/s = 2 m/s
Explanation
Bernoulli + continuity explain spray guns: fast air at nozzle lowers pressure, sucking liquid up the tube (L9 application).
PYQ10. Match: (i) SI unit of coefficient of viscosity — ? (ii) CGS unit — ? Options: P. N·s·m⁻² Q. poise R. N·m⁻²
Model Answer
(i) ↔ P. N·s·m⁻² (pascal-second)
(ii) ↔ Q. poise (1 poise = 0.1 N·s·m⁻²)
Explanation
From F = −ηA(dv/dx). SI unit of η is N·s·m⁻². CGS uses poise (dyne·s·cm⁻²).
PYQ11. What is Reynolds number? How does it help decide the nature of flow?
Model Answer
Reynolds number: R = ρvd/η — dimensionless ratio of inertial to viscous forces in fluid flow.
Significance: R < 1000 → laminar (streamline); 1000–2000 → transition; R > 2000 → turbulent. Predicts whether flow stays orderly or becomes chaotic.
Explanation
Critical velocity v_c = Rη/(ρd). Same as L9 §9.8.3 — used in pipe design, aircraft, blood flow.
PYQ12. What is terminal velocity? Write the expression for a sphere of radius r and density ρ falling in a fluid of viscosity η and density σ.
Model Answer
Terminal velocity: Constant maximum speed when weight equals viscous drag (net force zero).
Expression: v₀ = 2r²g(ρ − σ) / (9η)
Set weight = Stokes drag: (4/3)πr³ρg = 6πηrv₀.
Explanation
Valid for small spheres in viscous fluid at low R. Raindrops, Millikan oil-drop experiment use same idea.
PYQ13. What is Reynolds number? What is its significance? (Raynold's number)
Model Answer
Same as PYQ11: R = ρvd/η. Low R → viscous forces dominate (laminar); high R → inertia dominates (turbulent). Engineers use it to avoid turbulent losses or ensure mixing.
Explanation
Alternate spelling "Raynold" in paper — same dimensionless number named after Osborne Reynolds.
PYQ14. Give in brief the principle of working of a parachute.
Model Answer
Large canopy increases cross-sectional area → air resistance (viscous + pressure drag) increases sharply. When drag = weight, acceleration becomes zero and descent continues at safe terminal velocity.
Explanation
Fluid drag on bluff bodies depends on area and speed. Parachute trades fast fall for survivable terminal speed — viscosity/air resistance application.
PYQ15. Water flows in a horizontal pipe of non-uniform cross-section. At v = 0.2 m·s⁻¹, pressure = 20 mm of Hg. Find pressure where v = 1.5 m·s⁻¹. (ρ_water = 10³ kg·m⁻³)
Model Answer
Bernoulli (horizontal): P₁ + ½ρv₁² = P₂ + ½ρv₂²
P₁ − P₂ = ½ρ(v₂² − v₁²) = 500 × (2.25 − 0.04) = 1105 Pa
P₁ = 20 mm Hg ≈ 2666 Pa ⇒ P₂ ≈ 1561 Pa ≈ 11.7 mm Hg
Explanation
Faster flow → lower pressure (Bernoulli). Use 1 mm Hg ≈ 133 Pa. Assumes ideal fluid, no height change.
PYQ16. Calculate capillary rise of liquid (ρ = 1000 kg·m⁻³) in tube l = 0.05 m, r = 0.2×10⁻³ m. (T = 7.27×10⁻² N·m⁻¹, g = 10 m·s⁻², θ ≈ 0°)
Model Answer
h = 2T cos θ / (rρg) = 2 × 7.27×10⁻² × 1 / (0.2×10⁻³ × 1000 × 10)
= 0.1454 / 2 = 0.0727 m ≈ 7.3 cm (< 5 cm tube length — rise is physically limited by tube length in real setup).
Explanation
Jurin's law. Narrower tube (smaller r) gives greater rise — explains water creeping up wick/paper.
PYQ17. A 50 kg body stands on the small piston (A = 0.1 m²) of a hydraulic lift. Large piston A = 10 m². Find the weight of the car that can be lifted.
Model Answer
Pascal: F₁/A₁ = F₂/A₂
F₁ = mg = 500 N
F₂ = F₁ × A₂/A₁ = 500 × 10/0.1 = 50 000 N
Mass lifted = 50 000/10 = 5000 kg (5 tonne)
Explanation
Pressure transmitted undiminished through fluid. Area ratio 100 ⇒ force multiplied by 100 — hydraulic jack/lift principle (L9 §9.3).
PYQ18. Excess pressure inside a soap bubble of radius 4 cm. (T = 25×10⁻³ N·m⁻¹)
Model Answer
ΔP = 4T/r = 4 × 25×10⁻³ / 0.04 = 2.5 Pa
Explanation
Soap film has two surfaces (inner + outer) → twice the drop formula 2T/r. Excess pressure makes bubble slightly higher pressure than atmosphere.
PYQ19. Using Bernoulli's theorem, explain how a spray gun sucks liquid from a container when air is pushed through a narrow nozzle (piston r = 20 mm, nozzle r = 1 mm).
Model Answer
Fast air at narrow nozzle (v ≈ 2 m·s⁻¹ by continuity) ⇒ low pressure (Bernoulli). Pressure at liquid surface = atmospheric; pressure in tube at nozzle < atmospheric ⇒ liquid is pushed up the tube and atomised in the fast air stream.
Explanation
P + ½ρv² + ρgh = constant. High v at throat ⇒ reduced P. Same principle as perfume sprayers and carburettors.
PYQ20. State Pascal's law. A hydraulic lift has pistons 0.1 m² and 10 m². If 500 N is applied on the small piston, what force is available on the large piston?
Model Answer
Pascal's law: Pressure applied to enclosed fluid is transmitted undiminished in all directions.
F₂ = F₁ × A₂/A₁ = 500 × 10/0.1 = 50 000 N
Explanation
Same calculation as PYQ17 (50 kg person). Force multiplication equals area ratio — hydraulic press, brake, and lift designs.
Problem Solving — L9 Properties of Fluids
Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.
Find pressure due to a 5.0 m column of water (ρ = 10³ kg·m⁻³, g = 10) at the bottom. What is gauge pressure and absolute pressure if atmospheric pressure is 10⁵ Pa?
Solution — step by step with formulas
- P_gauge = hρg = 5×1000×10 = 5.0×10⁴ Pa.
- P_abs = P_atm + P_gauge = 1.5×10⁵ Pa.
Final answer: Gauge 5×10⁴ Pa; absolute 1.5×10⁵ Pa
Formulas used in this problem
Textbook formal language
Hydrostatic pressure at depth h in an incompressible fluid of density ρ is hρg (gauge). Absolute pressure adds atmospheric pressure.
Working formula set for this problem: P = F/A; P = hρg. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Deeper water presses harder: 5 m of water adds 50 kPa. Total push including air is 150 kPa.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Fluid pressure
Pressure acts equally in all directions at a point (Pascal). Same depth ⇒ same pressure regardless of container shape (hydrostatic paradox).
Link to chapter notes (L9 — Fluid pressure): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: P = F/A; P = hρg. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write P = F/A; P = hρg before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
In a hydraulic lift, small piston area 0.01 m², large 0.20 m². What force on small piston lifts a 2000 N load on the large piston?
Solution — step by step with formulas
- F₁ = F₂ (A₁/A₂) = 2000×(0.01/0.20) = 100 N.
Final answer: F₁ = 100 N
Formulas used in this problem
Textbook formal language
Pascal’s law: pressure applied to an enclosed fluid is transmitted undiminished. Hence F/A is equal on both pistons.
Working formula set for this problem: F₁/A₁ = F₂/A₂. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Same pressure, bigger pad multiplies force. Area ratio 20 ⇒ force ratio 20; 2000 N needs only 100 N on the small side.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Pascal’s law / hydraulic lift
Ideal hydraulic machines ignore friction and fluid compressibility.
Link to chapter notes (L9 — Pascal’s law / hydraulic lift): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: F₁/A₁ = F₂/A₂. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write F₁/A₁ = F₂/A₂ before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
A 0.5 kg object fully immersed displaces 200 cm³ of water. Find buoyant force (g=10) and apparent weight.
Solution — step by step with formulas
- V = 200×10⁻⁶ m³ = 2×10⁻⁴ m³.
- F_B = 2×10⁻⁴×1000×10 = 2.0 N.
- True weight = 5.0 N; apparent weight = 5.0 − 2.0 = 3.0 N.
Final answer: F_B = 2 N; apparent weight 3 N
Formulas used in this problem
Textbook formal language
Archimedes’ principle: upthrust equals weight of fluid displaced. Apparent weight is true weight minus upthrust when fully immersed.
Working formula set for this problem: F_B = V_displaced ρ_fluid g. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Water pushes up with the weight of the water kicked aside—here 2 N. Scale reading drops by 2 N.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Buoyancy
Floatation: weight = upthrust ⇒ average density ≤ fluid density.
Link to chapter notes (L9 — Buoyancy): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: F_B = V_displaced ρ_fluid g. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write F_B = V_displaced ρ_fluid g before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
State Bernoulli’s equation and explain why airspeed over an aeroplane wing being higher than below can produce lift (qualitative).
Solution — step by step with formulas
- Along a streamline for steady, incompressible, non-viscous flow: P + ½ρv² + ρgh = const.
- Higher v above wing ⇒ lower P above than below ⇒ net upward force (lift).
Final answer: Higher speed → lower pressure above wing → lift
Formulas used in this problem
Textbook formal language
Bernoulli’s theorem equates mechanical energy per unit volume along a streamline under ideal-flow assumptions. Pressure and kinetic terms trade off when height is fixed.
Working formula set for this problem: P + ½ρv² + ρgh = constant. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Fast air means lower push. Wing shape makes air rush over the top, so bottom pressure is higher and the plane is pushed up.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Bernoulli’s principle
Assumptions fail for turbulent or highly viscous flows; real wings also use angle of attack and circulation.
Link to chapter notes (L9 — Bernoulli’s principle): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: P + ½ρv² + ρgh = constant. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write P + ½ρv² + ρgh = constant before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Water flows in a pipe of area 4.0 cm² at 0.50 m·s⁻¹. Find speed where area narrows to 1.0 cm².
Solution — step by step with formulas
- v₂ = v₁(A₁/A₂) = 0.50×4 = 2.0 m·s⁻¹.
Final answer: v₂ = 2.0 m·s⁻¹
Formulas used in this problem
Textbook formal language
For incompressible steady flow, volume flux is conserved: Av = constant.
Working formula set for this problem: A₁v₁ = A₂v₂. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Same amount of water each second must go through. Narrower pipe ⇒ faster flow.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Continuity equation
Combine with Bernoulli for Venturi effect: narrow region has higher speed, lower pressure.
Link to chapter notes (L9 — Continuity equation): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: A₁v₁ = A₂v₂. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write A₁v₁ = A₂v₂ before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Explain terminal velocity of a raindrop falling in air qualitatively using force balance.
Solution — step by step with formulas
- Initially gravity dominates ⇒ acceleration.
- Viscous drag (and buoyancy) grow with speed until net force ≈ 0 ⇒ constant terminal speed.
Final answer: v_t when weight = drag + buoyancy
Formulas used in this problem
Textbook formal language
At terminal velocity, downward gravitational force is balanced by upward buoyant force and viscous drag; net force and acceleration vanish.
Working formula set for this problem: F_d = 6πηrv (Stokes); mg = 6πηrv_t + buoyancy (terminal). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Drop falls faster until air resistance catches up with weight; then it stops speeding up and falls at steady speed.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Viscous drag and terminal speed
Stokes’ law applies to small spheres at low Reynolds number; larger drops deform and use different drag laws.
Link to chapter notes (L9 — Viscous drag and terminal speed): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: F_d = 6πηrv (Stokes); mg = 6πηrv_t + buoyancy (terminal). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write F_d = 6πηrv (Stokes); mg = 6πηrv_t + buoyancy (terminal) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).