Grade 12 — Term 2
14 study notes · 206 practice questions · Aligned with NIE syllabus
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This term in the Idasara Academy app has 14 lessons (NIE Units 3, 4). 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.
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Lessons in This Term
Simple Harmonic Motion
What you must know
- Small Angle Approximation for Pendulum: Period T = 2π √(L/g) is valid strictly for small oscillations where sin θ ≈ θ in radians (amplitude < 10°); for large angles, period increases.
- Spring Combinations: Two springs of stiffness k1 and k2: In Parallel: k_eq = k1 + k2; In Series: 1/k_eq = 1/k1 + 1/k2.
- Phase Difference in SHM: Velocity leads displacement by π/2 radians (90°); Acceleration leads displacement by π radians (180°).
Try one question
A particle executes simple harmonic motion (SHM) with amplitude A and angular frequency ω. At what displacement from the central equilibrium position is the kinetic energy of the particle exactly equal to its potential energy?
- A. x = A / 4
- B. x = A / √2
- C. x = A / 2
- D. x = A / √3
Show the answer
Answer: B. x = A / √2. Total mechanical energy in SHM is E = ½ m ω² A². Kinetic energy is KE = ½ m ω² (A² - x²) and Potential energy is PE = ½ m ω² x². Setting KE = PE gives A² - x² = x² => 2x² = A² => x = A / √2.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Wave Motion
What you must know
- Speed of waves in strings: Speed of transverse wave on stretched string v = √(T / μ), where T is string tension (N) and μ is mass per unit length (kg/m).
- Speed of sound in gases (Laplace correction): v = √(γ P / ρ) = √(γ R T / M); speed is independent of gas pressure P (because P/ρ remains constant at constant temperature) but increases proportionally to √T.
- Intensity of a Wave: Average power transmitted per unit area perpendicular to wave direction: I = P / A = (1/2) ρ v ω² A²; Intensity is proportional to square of amplitude (I ∝ A²) and square of frequency (I ∝ f²).
Try one question
A progressive transverse wave is represented by the wave function y = 0.05 sin(20π t - 2π x), where x and y are in meters and t is in seconds. What are the frequency and wavelength of this wave?
- A. Frequency = 10 Hz, Wavelength = 2.0 m
- B. Frequency = 10 Hz, Wavelength = 1.0 m
- C. Frequency = 20 Hz, Wavelength = 2.0 m
- D. Frequency = 5 Hz, Wavelength = 0.5 m
Show the answer
Answer: B. Frequency = 10 Hz, Wavelength = 1.0 m. Comparing with the standard equation y = A sin(ω t - k x): ω = 20π rad s⁻¹ and k = 2π rad m⁻¹. Frequency f = ω / (2π) = 20π / 2π = 10 Hz. Wavelength λ = 2π / k = 2π / 2π = 1.0 m.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Wave Phenomena (Reflection, Refraction, Diffraction, Interference)
What you must know
- Young's Double Slit Formula: Fringe width β = (λ D) / d, where D is screen distance and d is slit separation; fringe width increases with longer wavelength (red > blue) and larger screen distance.
- Diffraction Grating Equation: d sin θ = n λ, where d is grating element (d = 1 / N lines per meter) and n is spectral order (n = 0, 1, 2, ...); maximum observable order n_max ≤ d / λ.
- Conditions for Sustained Interference: Sources must be coherent, monochromatic, have equal/comparable amplitudes, and be positioned close together with small path differences.
Try one question
What fundamental condition must be satisfied by two wave sources to produce an observable, steady and sustained interference pattern?
- A. Sources must be mutually coherent with a constant phase difference and identical frequency
- B. Sources must emit random bursts of white light at unequal velocities
- C. Sources must have widely differing wavelengths and orthogonal polarization
- D. Sources must be separated by a distance much larger than 10,000 wavelengths
Show the answer
Answer: A. Sources must be mutually coherent with a constant phase difference and identical frequency. Sustained interference requires coherent wave sources—meaning they emit waves of the same frequency and wavelength with a constant (time-invariant) phase difference. Without coherence, rapid random phase shifts wash out the constructive and destructive nodes into uniform illumination.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Stationary Waves & Resonance
What you must know
- End Correction (c) in Resonance Tubes: The displacement antinode forms slightly outside the open end of a tube of internal diameter D: End correction c ≈ 0.3 * D; for open pipe effective length L_eff = L + 2c; for closed pipe L_eff = L + c.
- Sonometer Laws: 1) Law of Length: f ∝ 1/L (constant T, μ); 2) Law of Tension: f ∝ √T (constant L, μ); 3) Law of Mass: f ∝ 1/√μ (constant L, T).
- Acoustic Resonance Condition: Occurs when the frequency of an external periodic driving force matches a natural resonant frequency of the oscillating system, causing maximum transfer of energy and maximum amplitude.
Try one question
A sonometer wire of length 0.50 m and mass per unit length 2.0 × 10⁻³ kg m⁻¹ vibrates in its fundamental mode under a tension of 80 N. What is the fundamental frequency of vibration?
- A. 100 Hz
- B. 200 Hz
- C. 400 Hz
- D. 50 Hz
Show the answer
Answer: B. 200 Hz. The wave speed on a stretched string is v = √(T / m) = √(80 / (2.0 × 10⁻³)) = √40000 = 200 m s⁻¹. For the fundamental mode of a wire fixed at both ends, length L = λ / 2 => λ = 2L = 2 × 0.50 m = 1.0 m. Fundamental frequency f = v / λ = 200 m s⁻¹ / 1.0 m = 200 Hz.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Sound & Doppler Effect
What you must know
- Sign conventions in Doppler formula: Numerator: Use (+) when observer approaches source, (-) when moving away; Denominator: Use (-) when source approaches observer, (+) when moving away.
- Doppler Shift in Wavelength: When source moves at speed v_s toward observer, the emitted wavelength is physically compressed: λ' = (v - v_s) / f; when observer moves, emitted wavelength λ is unchanged.
- Reflection from moving boundary: If sound reflects off a moving reflector (e.g., bat echolocation or radar gun), Doppler shift occurs twice: once upon reception at moving reflector, and once upon emission back to observer.
Try one question
A police siren emits sound of frequency 600 Hz. The police car moves towards a stationary pedestrian at a speed of 30 m s⁻¹. Taking the speed of sound in air as 330 m s⁻¹, what apparent frequency is heard by the pedestrian?
- A. 660 Hz
- B. 720 Hz
- C. 545 Hz
- D. 600 Hz
Show the answer
Answer: A. 660 Hz. Using the Doppler effect formula for a moving source approaching a stationary observer: f' = f [v / (v - v_s)]. Here v = 330 m s⁻¹, v_s = 30 m s⁻¹, f = 600 Hz. Therefore, f' = 600 × [330 / (330 - 30)] = 600 × (330 / 300) = 600 × 1.1 = 660 Hz.
A model question written by Idasara on the exam pattern. It is not from a past paper.
EM Waves & LASER
What you must know
- Spontaneous vs Stimulated Emission: Spontaneous emission emits photons randomly in arbitrary direction and phase (incoherent, e.g., incandescent bulbs); Stimulated emission produces photons with identical frequency, phase, direction, and polarization as trigger photon.
- Energy of Photon: E = h f = (h c) / λ where Planck's constant h = 6.626 × 10⁻³⁴ J·s; 1 eV = 1.602 × 10⁻¹⁹ J.
- Optical Resonator Cavity: Consists of one fully reflecting mirror and one partially transmitting mirror; provides positive optical feedback to sustain laser oscillations.
Try one question
Which combination of optical properties distinguishes LASER radiation from conventional thermal light sources?
- A. Longitudinal polarization with isotropic omnidirectional divergence
- B. High temporal and spatial coherence, monochromaticity, and extreme directional collimation
- C. Zero photon momentum and zero energy density
- D. Broad continuous spectral emission, complete incoherence, and high divergence
Show the answer
Answer: B. High temporal and spatial coherence, monochromaticity, and extreme directional collimation. LASER (Light Amplification by Stimulated Emission of Radiation) light is produced by stimulated emission in a population-inverted gain medium. It exhibits extraordinary spatial and temporal coherence, high monochromaticity (narrow linewidth), and low beam divergence (collimation).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Temperature & Thermometers
What you must know
- Thermometer Discrepancy: Different types of thermometers calibrated at 0 °C and 100 °C generally disagree slightly at intermediate temperatures because thermometric properties do not vary strictly linearly with thermodynamic temperature.
- Standard Thermometer: The Constant-Volume Hydrogen Gas Thermometer is the internationally accepted standard against which other practical thermometers are calibrated.
- Thermocouple Advantages: Fast response time (very low thermal capacity), wide range (-200 °C to 1500 °C), can measure temperature at a pinpoint location.
Try one question
In a constant-volume gas thermometer, the pressure of a fixed mass of gas is 80.0 kPa at the ice point (0 °C) and 108.0 kPa at the steam point (100 °C). What is the temperature in °C when the gas pressure reads 96.8 kPa?
- A. 80.0 °C
- B. 60.0 °C
- C. 70.0 °C
- D. 50.0 °C
Show the answer
Answer: B. 60.0 °C. For a constant-volume gas thermometer on the Celsius scale: θ = [(P_θ - P₀) / (P₁₀₀ - P₀)] × 100 °C. Substituting gives: θ = [(96.8 - 80.0) / (108.0 - 80.0)] × 100 = (16.8 / 28.0) × 100 = 0.60 × 100 = 60.0 °C.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Thermal Expansion
What you must know
- Anomalous Expansion of Water: Water contracts as heated from 0 °C to 4 °C (density increases), reaching maximum density of 1000 kg/m³ at 4 °C; expands when heated above 4 °C; enables aquatic life survival under frozen lake surfaces.
- Bimetallic Strip: Composed of two metals with different expansivities (e.g., brass α_b > iron α_i) riveted together; bends toward the metal with lower expansivity upon heating; used in thermostats.
- Density variation with temperature: Density at temperature T: ρ = ρ0 / (1 + γ ΔT) ≈ ρ0 (1 - γ ΔT).
Try one question
A glass flask of volume 500 cm³ is completely filled with mercury at 20 °C. The linear expansivity of the glass is α = 8.0 × 10⁻⁶ K⁻¹ and the real volume expansivity of mercury is γ_r = 1.80 × 10⁻⁴ K⁻¹. If heated to 100 °C, what volume of mercury overflows from the flask?
- A. 8.12 cm³
- B. 7.20 cm³
- C. 6.24 cm³
- D. 5.40 cm³
Show the answer
Answer: C. 6.24 cm³. Volume expansivity of glass container is γ_g = 3α = 3 × 8.0 × 10⁻⁶ = 2.4 × 10⁻⁵ K⁻¹. Apparent volume expansivity of mercury is γ_a = γ_r - γ_g = 1.80 × 10⁻⁴ - 0.24 × 10⁻⁴ = 1.56 × 10⁻⁴ K⁻¹. Overflow volume ΔV = V₀ γ_a ΔT = 500 cm³ × (1.56 × 10⁻⁴ K⁻¹) × (100 - 20) K = 500 × 1.56 × 10⁻⁴ × 80 = 6.24 cm³.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Gas Laws & Kinetic Theory
What you must know
- Translational Kinetic Energy: Average translational KE per molecule = (3/2) k_B T; depends SOLELY on absolute temperature T, independent of gas type or molecular mass.
- Dalton's Law of Partial Pressures: Total pressure exerted by a non-reacting mixture of gases equals the sum of partial pressures of individual components: P_total = P1 + P2 + ... + Pn.
- Real Gas Deviation: Real gases behave ideally only under LOW pressure and HIGH temperature (molecules far apart and moving fast, minimizing intermolecular attraction and finite volume effects).
Try one question
According to the kinetic theory of an ideal gas, if the absolute temperature of an ideal gas is increased fourfold from 300 K to 1200 K, by what factor does the root-mean-square (rms) speed of its molecules increase?
- A. 16 times
- B. 2 times
- C. 4 times
- D. √2 times
Show the answer
Answer: B. 2 times. The root-mean-square speed of ideal gas molecules is given by c_rms = √(3RT / M). Since c_rms is proportional to √T: c_new / c_old = √(T_new / T_old) = √(1200 / 300) = √4 = 2.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Calorimetry & Specific Heat
What you must know
- High Specific Heat of Water: Water has an exceptionally high specific heat capacity (c_water ≈ 4186 J·kg⁻¹·K⁻¹); acts as an excellent coolant and thermal buffer, regulating maritime climate and biological body temperature.
- Experimental Precautions in Calorimetry: Highly polished calorimeter surface (minimizes radiation loss), lid with stirrer (prevents convection and evaporation, ensures uniform temperature), lagging/insulation (minimizes conduction).
- Electrical Method for Specific Heat: Energy supplied by electrical heater E = V * I * t = (m c + C_cal)(T_final - T_initial).
Try one question
A 0.10 kg block of copper at 100 °C is dropped into 0.20 kg of water at 20 °C in a copper calorimeter of mass 0.05 kg. Specific heat capacity of water is 4200 J kg⁻¹ K⁻¹ and copper is 400 J kg⁻¹ K⁻¹. Assuming no heat loss to the surroundings, what is the final equilibrium temperature?
- A. 21.8 °C
- B. 23.6 °C
- C. 28.4 °C
- D. 25.0 °C
Show the answer
Answer: B. 23.6 °C. Heat lost by hot copper = Heat gained by water and calorimeter. Heat capacity of copper block = 0.10 × 400 = 40 J K⁻¹. Heat capacity of water + calorimeter = (0.20 × 4200) + (0.05 × 400) = 840 + 20 = 860 J K⁻¹. Equation: 40 × (100 - θ) = 860 × (θ - 20) => 4000 - 40θ = 860θ - 17200 => 900θ = 21200 => θ = 21200 / 900 ≈ 23.56 °C (approx 23.6 °C).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Latent Heat & Phase Change
What you must know
- Regelation of Ice: Melting point of ice decreases under increased pressure; when pressure is relieved, water refreezes (demonstrated by wire weighted across an ice block).
- Effect of Impurities: Dissolved impurities lower the freezing point (depression of freezing point, e.g., salting roads in winter) and elevate the boiling point.
- Supercooling: Phenomenon where a liquid cools below its normal freezing point without solidifying, due to lack of nucleation sites; instantly crystallizes upon disturbance.
Try one question
How much thermal energy is required to completely melt 0.050 kg of ice at 0 °C and raise the resulting water temperature to 40 °C? (Specific latent heat of fusion of ice L_f = 3.34 × 10⁵ J kg⁻¹, specific heat capacity of water c = 4200 J kg⁻¹ K⁻¹)
- A. 8.4 kJ
- B. 25.1 kJ
- C. 16.7 kJ
- D. 33.5 kJ
Show the answer
Answer: B. 25.1 kJ. Total heat Q = Q_melting + Q_heating = m L_f + m c ΔT = (0.050 kg × 3.34 × 10⁵ J kg⁻¹) + (0.050 kg × 4200 J kg⁻¹ K⁻¹ × 40 K) = 16,700 J + 8,400 J = 25,100 J = 25.1 kJ.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Heat Transfer
What you must know
- Wien's Displacement Law: The wavelength λ_max corresponding to peak radiant intensity of a black body is inversely proportional to absolute temperature: λ_max * T = b (constant b ≈ 2.898 × 10⁻³ m·K).
- Prevost's Theory of Heat Exchange: All bodies at any temperature above absolute zero (0 K) continuously emit and absorb thermal radiation simultaneously.
- Thermos (Vacuum) Flask Design: Double glass walls with vacuum between them eliminates conduction and convection; silvered interior walls reflect thermal radiation back, minimizing radiation loss.
Try one question
Two solid metal rods of identical length and cross-sectional area are joined end-to-end in series between two thermal reservoirs maintained at 100 °C and 0 °C. Rod A has thermal conductivity k_A = 300 W m⁻¹ K⁻¹ and Rod B has k_B = 100 W m⁻¹ K⁻¹. What is the steady-state temperature at the joint junction between the two rods? (Rod B is in contact with the 100 °C reservoir and Rod A with the 0 °C reservoir.)
- A. 25.0 °C
- B. 50.0 °C
- C. 33.3 °C
- D. 75.0 °C
Show the answer
Answer: A. 25.0 °C. In steady state, the rate of heat conduction through rod A equals that through rod B: (dQ/dt)_A = (dQ/dt)_B. Since lengths L and areas A are equal: k_A (100 - θ) = k_B (θ - 0) => 300 (100 - θ) = 100 θ => 30000 - 300θ = 100θ => 400θ = 30000 => θ = 30000 / 400 = 75 °C if heat flows from A to B. But here rod A is connected to 100 °C reservoir and rod B to 0 °C reservoir: 3(100 - θ) = 1(θ) => 300 - 3θ = θ => 4θ = 300 => θ = 75 °C. If rod B is connected to 100 °C and rod A to 0 °C: 1(100 - θ) = 3(θ) => 4θ = 100 => θ = 25 °C. For the standard arrangement with rod B (low k) at 100 °C: θ = 25.0 °C.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Humidity & Vapour Pressure
What you must know
- Dew Point: The temperature to which unsaturated air must be cooled at constant pressure to become saturated with water vapor (at dew point, actual vapor pressure becomes the SVP).
- Wet and Dry Bulb Hygrometer: Evaporation from wet muslin cools the wet bulb; larger temperature difference (depression) indicates drier air and lower relative humidity.
- Pressure Cooker Principle: Increasing pressure above liquid raises its boiling point (water boils at ~120 °C at 2 atm), cooking food much faster.
Try one question
On a particular day, the dew point of the atmosphere is 15.0 °C when the ambient room temperature is 25.0 °C. The saturated vapour pressure (SVP) of water is 1.70 kPa at 15.0 °C and 3.40 kPa at 25.0 °C. What is the relative humidity (RH) of the air?
- A. 40.0%
- B. 50.0%
- C. 60.0%
- D. 75.0%
Show the answer
Answer: B. 50.0%. Relative humidity is defined as RH = (Actual vapour pressure at room temperature / SVP at room temperature) × 100%. Since actual vapour pressure equals the SVP at the dew point: RH = (SVP at dew point / SVP at room temperature) × 100% = (1.70 kPa / 3.40 kPa) × 100% = 0.50 × 100% = 50.0%.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Thermodynamic Processes
What you must know
- P-V Indicator Diagrams: Work done during any process equals the area under the curve on a P-V diagram; for a cyclic process, net work done equals enclosed area (positive for clockwise/engine cycle, negative for anticlockwise/refrigerator).
- Adiabatic vs Isothermal Slopes: The slope of an adiabatic curve on a P-V diagram is steeper than an isothermal curve by a factor of γ: (dP/dV)_adiab = γ (dP/dV)_isotherm.
- Second Law of Thermodynamics (Kelvin-Planck & Clausius): Heat cannot spontaneously flow from cold to hot body without external work; No heat engine can be 100% efficient (Carnot efficiency η = 1 - T_cold / T_hot).
Try one question
During an adiabatic expansion of an ideal gas, which equation correctly describes the relationship between the heat exchanged with the surroundings (ΔQ), change in internal energy (ΔU), and work done by the gas (ΔW)?
- A. ΔQ = 0 and ΔW = -ΔU (Work is done entirely at the expense of internal energy)
- B. ΔU = 0 and ΔQ = ΔW (Internal energy remains strictly constant)
- C. ΔW = 0 and ΔQ = ΔU (No mechanical work is performed)
- D. ΔQ = ΔU + 2ΔW
Show the answer
Answer: A. ΔQ = 0 and ΔW = -ΔU (Work is done entirely at the expense of internal energy). By definition of an adiabatic process, there is no heat exchange with the environment (ΔQ = 0). Applying the First Law of Thermodynamics (ΔQ = ΔU + ΔW) yields 0 = ΔU + ΔW => ΔW = -ΔU. During adiabatic expansion (ΔW > 0), the internal energy decreases (ΔU < 0), causing the gas to cool.
A model question written by Idasara on the exam pattern. It is not from a past paper.
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