Grade 13 — Term 3
6 study notes · 78 practice questions · Aligned with NIE syllabus
Counted in the Idasara Academy app, which you open with a free account. The free notes, revision, practice and library pages for each lesson are linked below.
This term in the Idasara Academy app has 6 lessons (NIE Unit 10). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
Is this what my class is doing now?
Maybe not. The Idasara app groups lessons into its own terms, and your school may teach them in a different order. Where we know a lesson's NIE unit, it is shown on the lesson. Find the unit your class is on, and start there.
Lessons in This Term
Stress, Strain & Elasticity
What you must know
- Stress-Strain Curve Features: 1) Proportional Limit (Hooke's law strictly obeyed); 2) Elastic Limit / Yield Point (beyond which permanent plastic deformation sets in); 3) Ultimate Tensile Strength (UTS, maximum stress material can bear); 4) Breaking / Fracture Point.
- Ductile vs Brittle Materials: Ductile materials (copper, mild steel) exhibit extensive plastic deformation range between yield point and fracture; Brittle materials (glass, cast iron) fracture abruptly with negligible plastic strain just past elastic limit.
- Elastic Hysteresis: Energy dissipated as heat per unit volume during a cyclic stress-strain loading and unloading cycle, represented by the enclosed loop area on stress-strain plot (prominent in rubber, used in shock absorbers).
Try one question
A steel wire of length 2.0 m and uniform cross-sectional area 1.0 × 10⁻⁶ m² is stretched within its elastic limit by a tensile force of 200 N. If the Young's modulus of steel is Y = 2.0 × 10¹¹ N m⁻², what is the extension (ΔL) produced in the wire?
- A. 2.0 mm (2.0 × 10⁻³ m)
- B. 0.5 mm (0.5 × 10⁻³ m)
- C. 4.0 mm (4.0 × 10⁻³ m)
- D. 1.0 mm (1.0 × 10⁻³ m)
Show the answer
Answer: A. 2.0 mm (2.0 × 10⁻³ m). From Hooke's Law and the definition of Young's Modulus: Y = (F / A) / (ΔL / L) = F L / (A ΔL). Rearranging for extension: ΔL = F L / (A Y) = (200 N × 2.0 m) / (1.0 × 10⁻⁶ m² × 2.0 × 10¹¹ N m⁻²) = 400 / (2.0 × 10⁵) = 2.0 × 10⁻³ m = 2.0 mm.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Young’s Modulus & Elastic Potential Energy
What you must know
- Searle's Apparatus for Young's Modulus: Employs two identical parallel wires (Experimental wire and Reference/Dummy wire) suspended from the same rigid support; eliminates errors due to support yielding and thermal expansion.
- Work done in breaking vs storing: Area under entire F-ΔL curve up to fracture represents total work done to break the wire; area under unloading curve represents recoverable elastic potential energy; difference is dissipated permanently as heat.
- Dependence on dimensions: Young's Modulus Y depends ONLY on the material and temperature; doubling wire length L or area A does NOT change Y, but changes elongation ΔL and stiffness k.
Try one question
What is the elastic strain energy stored per unit volume (u) in a stretched elastic wire in terms of its Young's modulus Y and tensile strain ε?
- A. u = ½ Y² ε
- B. u = Y ε²
- C. u = 2 Y ε
- D. u = ½ Y ε²
Show the answer
Answer: D. u = ½ Y ε². Energy stored per unit volume is u = ½ × stress × strain. Since stress σ = Y × strain ε, substituting gives u = ½ × (Y ε) × ε = ½ Y ε².
A model question written by Idasara on the exam pattern. It is not from a past paper.
Viscosity & Fluid Resistance
What you must know
- Terminal velocity dependence on radius: Terminal velocity is strictly proportional to the square of radius (v_t ∝ r²); large raindrops fall significantly faster than tiny mist droplets.
- Boundary wall effect in Stokes' experiment: Stokes' law assumes an infinitely wide fluid container; in narrow laboratory cylinders, container walls increase drag, requiring Faxen correction.
- Temperature sensitivity: Viscosity of liquids decreases rapidly with rising temperature (loosened cohesive intermolecular forces), whereas gas viscosity increases with temperature.
Try one question
A small spherical lead pellet of radius r and density ρ falls vertically under gravity through a viscous liquid of density σ and dynamic viscosity η. When the pellet achieves its constant terminal velocity v_t, which expression gives v_t according to Stokes' Law?
- A. v_t = 9 r² (ρ - σ) g / (2 η)
- B. v_t = 2 r (ρ - σ) g / (9 η)
- C. v_t = 4 r² (ρ - σ) g / (3 η)
- D. v_t = 2 r² (ρ - σ) g / (9 η)
Show the answer
Answer: D. v_t = 2 r² (ρ - σ) g / (9 η). At terminal velocity, the net downward force is zero: Weight = Upthrust + Viscous drag => (4/3) π r³ ρ g = (4/3) π r³ σ g + 6πη r v_t => 6πη r v_t = (4/3) π r³ (ρ - σ) g => v_t = 2 r² (ρ - σ) g / (9 η).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Surface Tension & Capillarity
What you must know
- Capillary Rise (Jurin's Law): In a capillary tube of internal radius r: Height of liquid column h = (2 γ cos θ) / (r ρ g); liquid rises when θ < 90° (concave meniscus) and depresses when θ > 90° (mercury, convex meniscus); h ∝ 1/r (narrower tube gives higher rise).
- Insufficient Length Tube: If a capillary tube is shorter than calculated equilibrium height h (tube length L < h), liquid DOES NOT overflow; instead, meniscus flattens, increasing its radius of curvature R such that h * R = Constant.
- Temperature and Impurities: Surface tension decreases with increasing temperature (reaching zero at critical temperature); highly soluble inorganic salts (e.g., NaCl) increase surface tension, while surfactants and detergents significantly decrease surface tension.
Try one question
A clean narrow glass capillary tube of internal radius r = 0.50 mm is dipped vertically into water of surface tension T = 0.070 N m⁻¹ and density ρ = 1000 kg m⁻³. If the contact angle is θ = 0° and g = 10 m s⁻², what height h does the water column rise in the tube?
- A. 5.6 cm (0.056 m)
- B. 1.4 cm (0.014 m)
- C. 7.0 cm (0.070 m)
- D. 2.8 cm (0.028 m)
Show the answer
Answer: D. 2.8 cm (0.028 m). From the capillary rise equation: h = (2 T cos θ) / (r ρ g). With r = 0.50 × 10⁻³ m, cos 0° = 1: h = (2 × 0.070 N m⁻¹ × 1) / (0.50 × 10⁻³ m × 1000 kg m⁻³ × 10 m s⁻²) = 0.140 / 5.0 = 0.028 m = 2.8 cm.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Pressure in Fluids, Upthrust & Archimedes’ Principle
What you must know
- Center of Buoyancy (B): The point through which the resultant upward buoyant force acts, coinciding with the center of gravity of the displaced fluid.
- Apparent Loss of Weight: When submerged in fluid: Apparent Weight W_app = True Weight W - Upthrust U = V (ρ_solid - ρ_fluid) g.
- Hydrometer Operating Principle: Constant-weight variable-immersion hydrometer sinks until displaced liquid weight equals hydrometer weight; stem calibration scale is non-linear (divisions spaced farther apart near top / lower density).
Try one question
A solid metal alloy block weighs 50.0 N in air and 38.0 N when completely immersed in water (density 1000 kg m⁻³). Taking g = 10 m s⁻², what is the volume and density of the metal alloy block?
- A. Volume = 2.4 × 10⁻³ m³, Density = 2083 kg m⁻³
- B. Volume = 0.6 × 10⁻³ m³, Density = 8333 kg m⁻³
- C. Volume = 1.2 × 10⁻³ m³, Density = 4167 kg m⁻³
- D. Volume = 1.0 × 10⁻³ m³, Density = 5000 kg m⁻³
Show the answer
Answer: C. Volume = 1.2 × 10⁻³ m³, Density = 4167 kg m⁻³. Upthrust U = Weight in air - Apparent weight in water = 50.0 N - 38.0 N = 12.0 N. By Archimedes' Principle: U = V ρ_w g => V = U / (ρ_w g) = 12.0 N / (1000 kg m⁻³ × 10 m s⁻²) = 1.2 × 10⁻³ m³. Mass of block m = W / g = 50.0 N / 10 m s⁻² = 5.0 kg. Density ρ = m / V = 5.0 kg / (1.2 × 10⁻³ m³) ≈ 4166.7 kg m⁻³.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Fluid Dynamics (Bernoulli, Continuity, Flow Rate)
What you must know
- Venturi Effect: When fluid flows into a constricted throat (smaller A), velocity increases (v2 > v1) and static pressure drops (P2 < P1); Pressure differential ΔP = P1 - P2 = (1/2) ρ (v2² - v1²); utilized in carburetors, aspirators, and flow meters.
- Pitot-Static Tube: Measures flow airspeed by comparing Stagnation Pressure P_stag = P_static + (1/2) ρ v² at forward-facing tube tip (where fluid is brought to rest v = 0) with Static Pressure P_static from side flush holes: Airspeed v = √[ (2 (P_stag - P_static)) / ρ ].
- Magnus Effect (Spinning Ball): A spinning cylinder or ball drags boundary air layer around it, creating higher velocity and lower pressure on one side, resulting in sideways aerodynamic deflection force (curveballs in cricket and football).
Try one question
A horizontal Venturi meter is installed in a pipeline carrying liquid of density ρ. The main pipe cross-section is A₁ and the throat is constricted to area A₂ (where A₁ > A₂). If the manometer indicates a static pressure difference ΔP between the inlet and the constriction, what is the volumetric flow rate Q through the pipe?
- A. Q = A₁ A₂ √[ ΔP / (2 ρ (A₁² - A₂²)) ]
- B. Q = √[ 2 ρ ΔP / (A₁² - A₂²) ]
- C. Q = (A₁ + A₂) √[ 2 ΔP / ρ ]
- D. Q = A₁ A₂ √[ 2 ΔP / (ρ (A₁² - A₂²)) ]
Show the answer
Answer: D. Q = A₁ A₂ √[ 2 ΔP / (ρ (A₁² - A₂²)) ]. Combining the continuity equation Q = A₁ v₁ = A₂ v₂ with Bernoulli's equation for horizontal flow P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂² gives: ΔP = P₁ - P₂ = ½ ρ (v₂² - v₁²) = ½ ρ Q² (1/A₂² - 1/A₁²) = ½ ρ Q² [(A₁² - A₂²) / (A₁² A₂²)]. Solving for Q: Q = A₁ A₂ √[ 2 ΔP / (ρ (A₁² - A₂²)) ].
A model question written by Idasara on the exam pattern. It is not from a past paper.
Before you sign up
What does it cost?
I already go to a Physics tuition class
Then you are paying for Physics by the subject. Whether an Idasara plan would cost you less depends on how many classes you go to, so put in your own numbers.
Work it out on the pricing page →