Grade 12 — Term 1
14 study notes · 188 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 14 lessons. 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
Scientific Method & Scope of Physics
What you must know
- Hypothesis vs Theory vs Law: A Hypothesis is a tentative testable proposition; a Theory is a well-substantiated overarching explanation; a Law is a concise mathematical description of universal behavioral regularity.
- Repeatability vs Reproducibility: Repeatability refers to consistent results by the same observer with identical apparatus; Reproducibility refers to agreement across different observers, laboratories, and methods.
- Principle of Falsifiability (Popper): Any scientific proposition must make empirically testable predictions capable of being proven false.
Try one question
In the scientific method applied to empirical physics, which statement correctly distinguishes a scientific hypothesis from an established physical law?
- A. A hypothesis mathematically unifies all four fundamental forces, whereas a physical law is purely qualitative without mathematical equations
- B. A physical law can never be refuted or modified by any future observational evidence, whereas a hypothesis cannot be tested experimentally
- C. A hypothesis is universally true under all cosmic conditions, whereas a physical law is only valid in microscopic quantum systems
- D. A hypothesis is a tentative proposed explanation awaiting experimental verification, whereas a physical law summarizes consistent, universally validated empirical relationships
Show the answer
Answer: D. A hypothesis is a tentative proposed explanation awaiting experimental verification, whereas a physical law summarizes consistent, universally validated empirical relationships. In physics, a hypothesis is a falsifiable, tentative proposition based on initial observations. When repeatedly validated across diverse conditions without exception, concise mathematical descriptions of regularities become established physical laws (e.g., Newton's laws, Ohm's law).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Physical Quantities & SI Units
What you must know
- Unit formatting rules: Never use plural forms for unit symbols (write 10 kg, not 10 kgs); use a small space between numeric value and unit (5 m, not 5m).
- Celsius to Kelvin conversion: T(K) = θ(°C) + 273.15; note that temperature interval 1 K is identically equal to 1 °C (ΔT = Δθ).
- Proper derived unit expansion: Always reduce derived units to SI base units when checking equivalence (e.g., 1 Pa = 1 N/m² = 1 kg·m⁻¹·s⁻²).
Try one question
Which of the following expresses the SI base units of Planck's constant (h)?
- A. kg m² s⁻¹
- B. kg m⁻¹ s⁻¹
- C. kg m² s⁻²
- D. kg m s⁻²
Show the answer
Answer: A. kg m² s⁻¹. From the photon energy equation E = h·f, Planck's constant is h = E / f. The unit of energy is Joule (J = kg·m²·s⁻²), and the unit of frequency is s⁻¹. Therefore, [h] = (kg·m²·s⁻²) / s⁻¹ = kg·m²·s⁻¹ (equivalent to J·s).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Dimensions & Dimensional Analysis
What you must know
- Core Dimensions to memorize: Force = [M L T⁻²], Work/Energy = [M L² T⁻²], Power = [M L² T⁻³], Pressure = [M L⁻¹ T⁻²], Viscosity η = [M L⁻¹ T⁻¹], Surface Tension γ = [M T⁻²].
- Dimensionless quantities: Strain (ΔL/L), Refractive index, Relative density, Reynolds number, Angle (rad = arc/radius) all have dimension [M⁰ L⁰ T⁰] = 1.
- Dimensional correctness is a necessary but INSUFFICIENT condition for physical truth (e.g., s = ut + at² is dimensionally homogeneous but physically wrong by a factor of 1/2).
Try one question
According to Poiseuille's formula and Stokes' law, what are the dimensions of dynamic viscosity (η)?
- A. [M L⁻¹ T⁻¹]
- B. [M L T⁻²]
- C. [M L² T⁻¹]
- D. [M L⁻¹ T⁻²]
Show the answer
Answer: A. [M L⁻¹ T⁻¹]. From Stokes' Law, viscous force is F = 6πηrv, which gives η = F / (6πrv). Dimensional formula: [η] = [F] / ([r][v]) = [M L T⁻²] / ([L][L T⁻¹]) = [M L⁻¹ T⁻¹].
A model question written by Idasara on the exam pattern. It is not from a past paper.
Measuring Instruments & Errors
What you must know
- Minimizing percentage error: To minimize percentage error in experiments, choose measuring instruments such that the measured dimension is large compared to the least count (e.g., measuring thickness of 100 sheets instead of 1).
- Micrometer ratchet mechanism: Prevents overtightening and deformation of specimen by slipping when uniform pressure is reached.
- Backlash Error: Caused by wear in screw threads; eliminated by always rotating the thimble in one continuous direction during measurement.
Try one question
A micrometer screw gauge has a pitch of 0.50 mm and 50 circular scale divisions. When closed without any object, the circular scale zero is 2 divisions above the reference datum line (-0.02 mm negative zero error). In measuring the diameter of a wire, the main scale reads 2.50 mm and the circular scale reads 36 divisions. What is the corrected diameter of the wire?
- A. 2.84 mm
- B. 2.88 mm
- C. 2.90 mm
- D. 2.86 mm
Show the answer
Answer: B. 2.88 mm. Least count = Pitch / divisions = 0.50 mm / 50 = 0.01 mm. Observed reading = Main scale + (circular division × least count) = 2.50 mm + (36 × 0.01 mm) = 2.86 mm. Corrected reading = Observed reading - Zero error = 2.86 mm - (-0.02 mm) = 2.88 mm.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Vectors & Resolution
What you must know
- Orthogonal Unit Vectors: î · î = ĵ · ĵ = k̂ · k̂ = 1, while î · ĵ = 0; For cross products: î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ (cyclic), and î × î = 0.
- Relative Velocity: Velocity of A relative to B: v_AB = v_A - v_B; solve vectorially using vector triangles or resolved orthogonal components.
- Equilibrium of three coplanar concurrent forces: If three forces acting at a point keep it in equilibrium, they can be represented in magnitude and direction by the sides of a closed triangle taken in order (Lami's Theorem: P/sin α = Q/sin β = R/sin γ).
Try one question
Two coplanar forces of equal magnitude F act on a particle at an angle of 120° to each other. What is the magnitude of their resultant force?
- A. 2F
- B. F / 2
- C. √3 F
- D. F
Show the answer
Answer: D. F. Using the parallelogram law of vectors: R = √(F₁² + F₂² + 2F₁F₂ cos θ). With F₁ = F₂ = F and θ = 120°: R = √(F² + F² + 2F² cos 120°) = √(2F² + 2F²(-1/2)) = √(2F² - F²) = √F² = F.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Graphs & Data Interpretation
What you must know
- Graph scaling rules: Graph must occupy more than 50% of the available grid space along both axes; use simple, convenient scale intervals (1, 2, 5, 10 units), never clumsy intervals like 3 or 7.
- Gradient calculation points: Points chosen to calculate gradient MUST lie directly on the drawn line of best fit and be far apart; NEVER use experimental data points unless they happen to lie exactly on the line.
- Units of gradient: Always state units for gradients: Units of m = (Units of y-axis) / (Units of x-axis).
Try one question
In a rectilinear motion experiment with uniform constant acceleration a and initial velocity u, a graph of v² against displacement s is plotted. What do the gradient and intercept on the vertical (v²) axis represent?
- A. Gradient = a, Intercept = u
- B. Gradient = a/2, Intercept = u²
- C. Gradient = 2a, Intercept = u²
- D. Gradient = 2a, Intercept = 2u
Show the answer
Answer: C. Gradient = 2a, Intercept = u². From the kinematic equation v² = u² + 2as, comparing with the linear equation y = mx + c (where y = v² and x = s) shows that the gradient m = 2a and the y-intercept c = u².
A model question written by Idasara on the exam pattern. It is not from a past paper.
Linear Motion & Projectile Motion
What you must know
- Maximum Range Condition: Range R is maximized at launch angle θ = 45° (R_max = u²/g).
- Complementary Angles of Projection: Launch angles θ and (90° - θ) yield identical horizontal ranges R for the same initial speed u.
- Air Resistance Reality: In air, maximum range occurs at θ < 45° (~38°-42°), maximum height is reduced, and the trajectory is asymmetric (steeper descent).
Try one question
A projectile is launched from ground level with speed u at an elevation angle θ to the horizontal. If air resistance is negligible, what is the maximum height H achieved by the projectile?
- A. H = (u² cos² θ) / (2g)
- B. H = (u² sin² θ) / (2g)
- C. H = (u² sin 2θ) / g
- D. H = (u sin θ) / g
Show the answer
Answer: B. H = (u² sin² θ) / (2g). At the apex of projectile motion, the vertical velocity component becomes zero: v_y = 0. Using v_y² = u_y² - 2gH with u_y = u sin θ: 0 = (u sin θ)² - 2gH, giving H = (u² sin² θ) / (2g).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Forces & Newton’s Laws
What you must know
- Action-Reaction Criteria: 1) Equal magnitude; 2) Opposite direction; 3) Same nature of force (both gravitational, both contact, etc.); 4) Act on TWO DIFFERENT BODIES simultaneously.
- Elastic vs Inelastic Collisions: In ALL collisions, total momentum is conserved; Kinetic Energy is conserved ONLY in Elastic collisions (e = 1); In perfectly inelastic collisions, bodies stick together (e = 0, maximum KE loss).
- Apparent weight in elevators: Accelerating upward with 'a': R = m(g + a); Accelerating downward with 'a': R = m(g - a); Free fall (a = g): R = 0 (weightlessness).
Try one question
A 0.20 kg rubber ball traveling horizontally at 15 m s⁻¹ strikes a vertical wall normally and rebounds horizontally with a speed of 10 m s⁻¹. If the duration of contact is 0.05 s, what is the average force exerted by the wall on the ball?
- A. 80 N
- B. 20 N
- C. 60 N
- D. 100 N
Show the answer
Answer: D. 100 N. Taking the rebound direction as positive: initial velocity u = -15 m s⁻¹, final velocity v = +10 m s⁻¹. Impulse J = m(v - u) = 0.20 kg × (10 - (-15)) m s⁻¹ = 0.20 × 25 = 5.0 N s. Average force F = J / Δt = 5.0 N s / 0.05 s = 100 N.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Moments & Equilibrium
What you must know
- Three Non-Parallel Forces in Equilibrium: The lines of action of three non-parallel forces in equilibrium MUST be concurrent (intersect at a single common point) and coplanar.
- Center of Mass vs Center of Gravity: Center of Mass (COM) depends purely on mass distribution; Center of Gravity (CG) coincides with COM only in a uniform gravitational field.
- Stability of Equilibrium: Stable (CG rises on small displacement), Unstable (CG falls), Neutral (CG height remains constant).
Try one question
A uniform horizontal beam AB of length 4.0 m and weight 200 N is pivoted at end A and supported horizontally by a vertical cable attached at end B. A 600 N weight is placed on the beam at a distance of 1.0 m from A. What is the tension in the vertical supporting cable?
- A. 250 N
- B. 300 N
- C. 200 N
- D. 400 N
Show the answer
Answer: A. 250 N. Taking moments about pivot A in static equilibrium: Sum of clockwise moments = Sum of anticlockwise moments. Clockwise moments: Weight of beam at center of gravity (2.0 m): 200 N × 2.0 m = 400 N m. Added weight at 1.0 m: 600 N × 1.0 m = 600 N m. Total clockwise moment = 1000 N m. Anticlockwise moment = T × 4.0 m. Hence, 4.0 T = 1000 => T = 250 N.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Work, Energy & Power
What you must know
- Conservative vs Non-Conservative Forces: For conservative forces (gravity, electrostatic, spring), work done around a closed path is zero (∮ F · dr = 0) and path-independent; Non-conservative forces (friction, drag) dissipate energy as heat.
- Law of Conservation of Mechanical Energy: In an isolated system governed strictly by conservative forces: E_total = KE + PE = Constant.
- Power in vehicle motion: At maximum terminal velocity on level road: Engine driving force F equals total resistive resistance R (F = R), hence P = R * v_max.
Try one question
An electric locomotive pulls a train of total mass 2.0 × 10⁵ kg up an incline of 1 in 100 (sin θ = 0.01) at a constant steady speed of 20 m s⁻¹. Total resistive friction forces amount to 1.5 × 10⁴ N. Taking g = 10 m s⁻², what output power must the locomotive develop?
- A. 300 kW
- B. 400 kW
- C. 1000 kW
- D. 700 kW
Show the answer
Answer: D. 700 kW. Tractive force F must balance the downhill component of gravity plus frictional resistance: F = mg sin θ + F_friction = (2.0 × 10⁵ kg × 10 m s⁻² × 0.01) + 1.5 × 10⁴ N = 2.0 × 10⁴ N + 1.5 × 10⁴ N = 3.5 × 10⁴ N. Output Power P = F × v = 3.5 × 10⁴ N × 20 m s⁻¹ = 7.0 × 10⁵ W = 700 kW.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Circular Motion
What you must know
- Vertical Circular Motion (mass on light string of length L): Critical speed at highest point to prevent slack (T ≥ 0): v_top = √(gL); Minimum speed at lowest point to complete vertical circle: v_bottom = √(5gL); Tension difference: T_bottom - T_top = 6mg.
- Centrifugal Force in Non-Inertial Frames: A fictitious inertial force directed radially outward: F_inertial = m ω² r, introduced ONLY when solving problems in a rotating frame of reference.
- Conical Pendulum: Tension T = mg / cos θ; Period of revolution T_period = 2π √(h/g) = 2π √((L cos θ)/g).
Try one question
A car negotiates an unbanked flat circular road curve of radius 50 m. The coefficient of static friction between the tires and the dry road is 0.50. Taking g = 10 m s⁻², what is the maximum speed at which the car can round the bend without skidding outward?
- A. 15.8 m s⁻¹ (approx 16 m s⁻¹)
- B. 5.0 m s⁻¹
- C. 25.0 m s⁻¹
- D. 10.0 m s⁻¹
Show the answer
Answer: A. 15.8 m s⁻¹ (approx 16 m s⁻¹). On an unbanked circular path, the centripetal force is provided solely by static friction: f_s = m v² / r. Maximum friction without slipping is f_max = μ_s mg. Equating gives m v_max² / r = μ_s mg => v_max = √(μ_s g r) = √(0.50 × 10 × 50) = √250 ≈ 15.81 m s⁻¹.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Fluid Statics
What you must know
- Atmospheric Pressure measurements: Standard atmospheric pressure = 1 atm = 760 mmHg = 1.013 × 10⁵ Pa; Fortin's Barometer uses mercury because of its high density (compact column height) and very low vapor pressure.
- Apparent weight of immersed body: W_apparent = W_true - Upthrust = V ρ_body g - V ρ_fluid g = V g (ρ_body - ρ_fluid).
- Manometer differential equation: P_gas = P_atm ± h ρ g depending on liquid level difference.
Try one question
A vertical U-tube containing mercury (density 13,600 kg m⁻³) has water (density 1,000 kg m⁻³) poured into one arm to a height of 27.2 cm. What is the difference in mercury levels between the two arms?
- A. 2.0 cm
- B. 27.2 cm
- C. 1.36 cm
- D. 4.0 cm
Show the answer
Answer: A. 2.0 cm. At the horizontal datum level passing through the water-mercury interface: Pressure in water arm = Pressure in mercury arm. Hence, h_w ρ_w g = h_m ρ_m g => h_m = h_w (ρ_w / ρ_m) = 27.2 cm × (1,000 / 13,600) = 27.2 / 13.6 = 2.0 cm.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Fluid Dynamics
What you must know
- Torricelli's Law of Efflux: Speed of efflux from an orifice at depth h below open liquid surface: v = √(2gh).
- Venturi Meter: Measures volume flow rate in pipe: Q = A1 A2 √[ (2 ΔP) / (ρ (A1² - A2²)) ] where ΔP = h ρ_m g.
- Aerodynamic Lift: Higher fluid velocity over curved upper airfoil surface creates reduced static pressure, producing upward net lift force.
Try one question
Water flows steadily through a horizontal pipe of non-uniform cross-section. At point A where the cross-sectional area is 20 cm², the speed is 2.0 m s⁻¹. What is the flow speed at a constriction B where the cross-sectional area narrows to 5.0 cm²?
- A. 8.0 m s⁻¹
- B. 10.0 m s⁻¹
- C. 4.0 m s⁻¹
- D. 0.5 m s⁻¹
Show the answer
Answer: A. 8.0 m s⁻¹. For an incompressible fluid, the continuity equation applies: A₁ v₁ = A₂ v₂. Therefore, v₂ = (A₁ / A₂) v₁ = (20 cm² / 5.0 cm²) × 2.0 m s⁻¹ = 4 × 2.0 m s⁻¹ = 8.0 m s⁻¹.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Fluids in Motion (Hydrodynamics)
What you must know
- Dependence on radius in Poiseuille flow: Q is extremely sensitive to tube radius: doubling radius increases flow rate 16-fold (r⁴ dependence).
- Temperature dependence of Viscosity: For liquids, viscosity decreases with increasing temperature (weakened intermolecular cohesive bonds); for gases, viscosity INCREASES with temperature (increased molecular collisions).
- Reynolds Number (Re): Dimensionless flow regime predictor: Re = (ρ v d) / η; Re < 2000 indicates laminar flow, Re > 3000 indicates turbulent flow.
Try one question
According to Bernoulli's principle for the streamline flow of an incompressible, non-viscous fluid, which physical quantity remains constant along a streamline?
- A. P + ½ ρ v² + ρ g h (Total mechanical energy per unit volume)
- B. P / v + ρ g h
- C. ½ ρ v² - P
- D. P · v · A
Show the answer
Answer: A. P + ½ ρ v² + ρ g h (Total mechanical energy per unit volume). Bernoulli's equation states that for an ideal fluid in steady streamline flow, the sum of pressure energy per unit volume (P), kinetic energy per unit volume (½ ρ v²), and gravitational potential energy per unit volume (ρ g h) is constant along any streamline.
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 →