CQ1. A 10 kg crate rests on a horizontal floor. The coefficient of static friction is 0.50 and g = 10 m·s⁻². What is the maximum horizontal force that can be applied before the crate starts sliding?
CQ2. A 2.0 kg ball moves at 5.0 m·s⁻¹ on a smooth track. Using K = ½mv², what is its kinetic energy?
CQ3. A student slowly lifts a 5.0 kg bag vertically by 2.0 m (g = 10 m·s⁻²). How much work is done against gravity?
CQ4. Calculate gauge pressure at depth h = 5.0 m in water (ρ = 1000 kg·m⁻³, g = 10 m·s⁻²) using P = ρgh.
CQ5. A hydraulic jack has input piston area A₁ = 2 cm² and output area A₂ = 50 cm². If F₁ = 100 N is applied on the small piston, what output force F₂ does Pascal's law give (F₂ = F₁ × A₂/A₁)?
CQ6. A Carnot engine operates between source temperature T₁ = 600 K and sink T₂ = 300 K. Using η = 1 − T₂/T₁, what is its maximum efficiency?
CQ7. A sound wave has frequency f = 500 Hz and wavelength λ = 0.68 m. Using v = fλ, what is the wave speed?
CQ8. Two point charges q₁ = q₂ = 1.0 μC are separated by r = 0.10 m in vacuum (k = 9 × 10⁹ N·m²·C⁻²). What is the magnitude of force between them (F = kq₁q₂/r²)?
CQ9. A parallel-plate capacitor has C = 10 μF when plate separation is d. If d is doubled (area unchanged), how does capacitance change (C ∝ 1/d)?
CQ10. A copper wire of resistance 4.0 Ω is stretched so its length doubles while volume stays constant (area halves). What is the new resistance (R ∝ l/A)?
CQ11. An electric heater draws I = 5.0 A from a 220 V mains supply. Using P = VI, what is the power consumed?
CQ12. A long straight wire carries I = 10 A. At perpendicular distance r = 0.10 m, what is B (B = μ₀I/(2πr), μ₀ = 4π × 10⁻⁷ T·m·A⁻¹)?
CQ13. A step-up transformer has N_p = 100 turns and N_s = 500 turns. If primary voltage V_p = 220 V, what is secondary voltage (V_s/V_p = N_s/N_p)?
CQ14. A capacitor C = 10 μF is connected to an AC source with angular frequency ω = 100 rad·s⁻¹. Find X_C = 1/(ωC).
CQ15. In Young's experiment, λ = 600 nm, slit separation d = 0.60 mm, screen distance D = 1.0 m. Position of 3rd bright fringe: x₃ = 3λD/d.
CQ16. For hydrogen (Z = 1), what is the energy of the n = 3 level using E_n = −13.6/n² eV?
CQ17. An electron is accelerated through V = 100 V. Using λ = 12.3/√V Å (from lesson), what is its de Broglie wavelength?
CQ18. A radioactive sample has decay constant λ = 0.10 year⁻¹. What is its half-life T₁/₂ = 0.693/λ?
CQ19. Water escapes through a hole at depth H = 5.0 m below the free surface (g = 10 m·s⁻²). Using Torricelli's law v = √(2gH), what is efflux speed?
CQ20. A spring of constant k = 200 N·m⁻¹ is compressed by x = 0.10 m. Stored elastic energy U_s = ½kx² equals:
2–3 marks · Apply formulas to new situations · Try first, then show model answer.
AQ1. A cricket ball (m = 0.15 kg) arrives at 30 m·s⁻¹ and is caught in 0.03 s, coming to rest. (a) Find impulse on the ball. (b) Find average force on the hands. (c) Why does a fielder withdraw hands while catching?
Model Answer
(a) Impulse = Δp = 0 − 0.15×30 = −4.5 N·s (magnitude 4.5 N·s). (b) F_avg = Δp/Δt = 4.5/0.03 = 150 N. (c) Withdrawing hands increases Δt, so for same Δp the average force on palms decreases — less injury (L3 impulse-momentum).
AQ2. A 20 kg block is pulled at constant speed on a rough horizontal floor by a horizontal force of 60 N. Find μ_k and explain why acceleration is zero.
Model Answer
At constant speed, net force = 0, so applied force equals kinetic friction: f_k = 60 N. Normal reaction F_N = mg = 20×10 = 200 N. μ_k = f_k/F_N = 60/200 = 0.30. Zero acceleration means ΣF = 0 (Newton's first law, L3).
AQ3. A 1 kg object slides from rest down a frictionless ramp, dropping 4 m vertically, then compresses a spring (k = 400 N·m⁻¹). Find maximum spring compression.
Model Answer
Initial PE = mgh = 1×10×4 = 40 J; initial K = 0. At maximum compression, K = 0 and all energy is spring PE: ½kx² = 40 → x² = 80/400 = 0.2 → x = 0.45 m (≈ 45 cm). Uses conservation K + U = constant (L6).
AQ4. A wooden block of volume 0.002 m³ and density 600 kg·m⁻³ floats in water (ρ = 1000 kg·m⁻³). What fraction of the block is submerged?
Model Answer
Weight = ρ_wood V g = 600×0.002×10 = 12 N. Floating: buoyant force = weight. Submerged volume V_sub = weight/(ρ_water g) = 12/(1000×10) = 0.0012 m³. Fraction submerged = V_sub/V = 0.0012/0.002 = 0.60 (60%). Archimedes: weight of displaced fluid equals weight of body (L9).
AQ5. Air flows faster over the curved upper surface of an aircraft wing than below. Applying Bernoulli's equation, explain lift generation.
Model Answer
Bernoulli: P + ½ρv² + ρgh = constant along a streamline. Upper surface: higher speed v → lower pressure P. Lower surface: slower air → higher pressure. Pressure difference (high below, low above) gives upward net force — lift. Assumes streamline, incompressible flow (L9).
AQ6. A gas receives 500 J of heat and does 300 J of work on the surroundings in an isobaric expansion. Find ΔU and state the sign convention used in the lesson.
Model Answer
First law: ΔQ = ΔU + ΔW. ΔQ = +500 J (heat to system), ΔW = +300 J (work done by system). ΔU = ΔQ − ΔW = 500 − 300 = 200 J. Internal energy increases by 200 J (L11 sign convention).
AQ7. Why must the work input to a refrigerator be greater than the heat removed from the cold reservoir, even for an ideal Carnot refrigerator?
Model Answer
Second law: heat cannot flow spontaneously from cold to hot. Work must be supplied. For Carnot refrigerator, coefficient of performance COP = T₂/(T₁−T₂); finite work is needed to transfer heat from sink T₂ to source T₁. Real refrigerators also have irreversibilities (L11).
AQ8. A stationary listener hears a siren of frequency 600 Hz from an ambulance approaching at 20 m·s⁻¹ (v_sound = 340 m·s⁻¹). Find observed frequency using n′ = n(v − v₀)/(v − vₛ) with v₀ = 0.
Model Answer
Source approaches: n′ = 600 × 340/(340 − 20) = 600 × 340/320 = 637.5 Hz. Pitch rises as source approaches — Doppler effect (L14).
AQ9. A +2 μC charge is at the origin. Find (a) electric field at (0.2 m, 0) and (b) potential at the same point. Which is a scalar?
Model Answer
(a) E = kq/r² = 9×10⁹ × 2×10⁻⁶ / (0.2)² = 4.5×10⁵ N·C⁻¹, directed radially outward. (b) V = kq/r = 9×10⁹ × 2×10⁻⁶ / 0.2 = 9×10⁴ V. Potential V is scalar; E is vector (L15, L16).
AQ10. Two capacitors C₁ = 6 μF and C₂ = 3 μF are connected in series to 12 V. Find equivalent capacitance, charge on each, and voltage across each.
Model Answer
1/C_s = 1/6 + 1/3 = 1/2 → C_s = 2 μF. Charge Q = C_s V = 2×12 = 24 μC (same on each in series). V₁ = Q/C₁ = 24/6 = 4 V; V₂ = Q/C₂ = 24/3 = 8 V; V₁ + V₂ = 12 V (L16 series rule).
AQ11. In a balanced Wheatstone bridge, arms are P = 10 Ω, Q = 20 Ω, R = 4 Ω. Find unknown S using P/Q = R/S.
Model Answer
P/Q = R/S → 10/20 = 4/S → S = 8 Ω. At balance, galvanometer current is zero — no current through central branch (L17).
AQ12. A potentiometer wire of length 4 m balances a 2 V cell when contact is at 2.5 m. Find emf per unit length and length needed to balance a 1.5 V cell.
Model Answer
k = E/l₁ = 2/2.5 = 0.8 V·m⁻¹. For 1.5 V: l₂ = E₂/k = 1.5/0.8 = 1.875 m. Potentiometer compares emf without drawing current at balance (L17).
AQ13. A 0.50 m wire carrying 8 A is placed perpendicular to a uniform B = 0.25 T field. Find magnetic force (F = BIL sin θ).
Model Answer
θ = 90°, sin θ = 1. F = 0.25 × 8 × 0.50 = 1.0 N. Direction given by Fleming's left-hand rule — perpendicular to both I and B (L18).
AQ14. A north pole of a bar magnet is pushed toward a closed coil. Predict induced current direction and explain using Lenz's law.
Model Answer
Flux through coil increases (more field lines enter). Induced current opposes this increase — coil's near face acts as north pole, repelling approaching north pole. By right-hand rule, current direction as viewed from magnet side is anticlockwise (L19 Lenz's law).
AQ15. An LCR series circuit has L = 0.20 H, C = 50 μF. Find resonant frequency ν_r = 1/(2π√LC). What happens to impedance at resonance?
Model Answer
LC = 0.20 × 50×10⁻⁶ = 10⁻⁵. √LC = 0.00316. ν_r = 1/(2π × 0.00316) ≈ 50.3 Hz. At resonance X_L = X_C, Z = R (minimum) and current is maximum (L19).
AQ16. White light passes through a prism and forms a spectrum. Why is violet deviated more than red? Relate to μ and wavelength.
Model Answer
For small-angle prism δ = (μ − 1)A. Refractive index μ increases as wavelength decreases — violet has shorter λ, higher μ_v than μ_r. Hence δ_v > δ_r and violet bends more (L21 dispersion). Interference needs coherent sources; dispersion separates colours by refraction (L22).
AQ17. Light is incident on glass (μ = 1.5). Find Brewster angle i_p where reflected and refracted rays are perpendicular (tan i_p = μ).
Model Answer
tan i_p = 1.5 → i_p = tan⁻¹(1.5) ≈ 56.3°. At this angle reflected light is completely plane-polarised perpendicular to plane of incidence (L22).
AQ18. Hydrogen atom jumps from n = 3 to n = 2. Find photon energy (E₃ − E₂) and wavelength region (Balmer series).
Model Answer
E₃ = −13.6/9 = −1.51 eV; E₂ = −13.6/4 = −3.40 eV. ΔE = 1.89 eV ≈ 3.0×10⁻¹⁹ J. λ = hc/ΔE ≈ 656 nm — red visible line (Balmer, L24). Photon energy quantization links to photoelectric threshold concept (L25).
AQ19. Light of frequency 8 × 10¹⁴ Hz falls on a metal with work function φ₀ = 2.0 eV. Will photoelectrons be emitted? (h = 6.6×10⁻³⁴ J·s, 1 eV = 1.6×10⁻¹⁹ J)
Model Answer
Photon energy hν = 6.6×10⁻³⁴ × 8×10¹⁴ = 5.28×10⁻¹⁹ J = 3.3 eV. Since hν > φ₀, emission occurs. K_max = hν − φ₀ = 1.3 eV. Below threshold frequency, no emission regardless of intensity (L25).
AQ20. Explain why ²³⁵U fission releases ~200 MeV per event while ²H + ²H fusion releases ~24 MeV, yet fusion is proposed for long-term energy.
Model Answer
Fission: heavy nucleus splits — large energy per fission (~200 MeV) but products radioactive. Fusion: light nuclei combine — less energy per reaction (~24 MeV) but fuel abundant (deuterium in water), no long-lived radioactive waste, and binding energy per nucleon peaks near iron — fusion of light nuclei also moves toward more stable configuration (L26, L27).
5 marks · Compare, contrast and evaluate · Deep understanding of physics concepts.
ZQ1. Compare work done by friction and gravity when a block slides down a rough incline and is lifted back vertically. Which forces are conservative? How does mechanical energy change?
Model Answer
Gravity is conservative: work lifting = −work lowering; PE = mgh is path-independent. Kinetic friction is non-conservative: work depends on path length; mechanical energy lost as heat. On rough incline, mgh = K_final + W_friction. Round trip on same path, gravity does net zero work; friction always dissipates energy — total mechanical energy decreases (L3 friction, L6 conservative forces).
ZQ2. Two identical carts collide on a track. Contrast a perfectly elastic collision with a perfectly inelastic one. Which quantities are conserved in each?
Model Answer
Both conserve total momentum (isolated system, L3). Elastic: kinetic energy also conserved — ½m₁v₁i² + ½m₂v₂i² = ½m₁v₁f² + ½m₂v₂f² (L6). Inelastic: KE not conserved; some converts to heat/deformation. Perfectly inelastic: bodies stick, maximum KE loss consistent with momentum conservation. Real collisions lie between extremes.
ZQ3. Bernoulli's equation assumes ideal fluid flow; Carnot cycle assumes reversible quasi-static processes. Analyse similarities and limitations of both idealisations.
Model Answer
Both are ideal models ignoring dissipation — Bernoulli neglects viscosity and turbulence; Carnot neglects friction and rapid irreversibility. Bernoulli: P + ½ρv² + ρgh = constant gives speed-pressure trade-off (venturi, lift). Carnot sets maximum η = 1 − T₂/T₁. Real systems always fall short — Reynolds number marks flow regime change (L9); real engines have η below Carnot (L11).
ZQ4. Distinguish interference and diffraction of waves using conditions and superposition principle from both lessons.
Model Answer
Interference: superposition of waves from coherent sources with constant phase relation — Young's slits give equally spaced bright/dark fringes (Δ = nλ or (n+½)λ, L22). Diffraction: bending around obstacles/apertures — single slit produces broader central maximum. Both follow superposition (L14): resultant amplitude from phase difference. Interference needs distinct sources; diffraction is interference of wavelets from same wavefront (Huygens, L22).
ZQ5. Analyse how electric field E, potential V and current I relate when charge moves through a conductor. Why is E zero inside a conductor in electrostatic equilibrium but non-zero during current flow?
Model Answer
Electrostatic equilibrium (L15): excess charge on surface, E = 0 inside, V constant throughout conductor. Potential V = W/q₀ — work per unit charge (L16). Current flow (L17): maintained potential difference drives drift velocity v_d; E = V/l inside wire balances collisions; I = nAev_d. Ohm's law V = IR links field-driven drift to terminal voltage. Capacitor stores energy ½CV² without steady current; resistor dissipates I²R in steady state.
ZQ6. Compare force on a current-carrying conductor in a magnetic field with motional emf when the same conductor moves in the field. Link to energy conversion.
Model Answer
Magnetic force F = BIL sin θ (L18) — external supply must do work to move conductor against force; can drive motors (electrical → mechanical). Motional emf ε = Blv (flux change, L19) — mechanical motion induces current; generators convert mechanical → electrical. Lenz's law: induced effects oppose cause — back emf in motors, opposition to flux change. Same B and I link both phenomena via Lorentz force on charges.
ZQ7. Explain why the sky appears blue at noon but reddish at sunset, connecting Rayleigh scattering (L21) with wave nature of light (L22).
Model Answer
Rayleigh scattering intensity ∝ 1/λ⁴ — shorter wavelengths (blue/violet) scatter more strongly by air molecules. At noon, scattered blue reaches observer from all directions. At sunset, longer path through atmosphere scatters away shorter wavelengths; transmitted light enriched in red/orange. Scattering selects wavelengths without prism dispersion. Wave model: light as transverse EM wave interacting with particles smaller than λ (L21, L22).
ZQ8. Bohr quantised angular momentum in orbits; de Broglie proposed matter waves. Analyse how de Broglie wavelength explains Bohr's stationary orbits.
Model Answer
Bohr postulate: mvr = nh/(2π) — quantised orbits (L24). de Broglie: λ = h/p = h/(mv) (L25). For orbit to be stationary, circumference must fit integer wavelengths: 2πr = nλ → 2πr = nh/(mv) → mvr = nh/(2π). Matter-wave constructive interference in closed orbit gives Bohr condition without ad hoc postulate. Both link discreteness to h.
ZQ9. Using binding energy per nucleon (B/A) curve, analyse why both fission of heavy nuclei and fusion of light nuclei release energy, yet iron-56 is most stable.
Model Answer
B/A rises from light nuclei to ~8.8 MeV/nucleon near ⁵⁶Fe, then falls for heavy nuclei (L26). Fission: ²³⁵U splits — fragments closer to iron have higher B/A; mass defect converts to ~200 MeV (L27). Fusion: ²H + ²H → ⁴He — product more tightly bound; ~24 MeV released. Iron peak means neither fission nor fusion of iron releases net energy. Nuclear power exploits moving toward maximum B/A.
ZQ10. Trace energy transformations from thermal power plant (steam turbine) to household AC appliance. Identify where second law of thermodynamics and transformer principle apply.
Model Answer
Boiler: chemical/nuclear → thermal (Q to steam, L11). Turbine: thermal → mechanical (W); Carnot limit sets maximum fraction convertible — rest rejected to condenser (η = 1 − T₂/T₁). Generator: mechanical → electrical via EMI (ε = −dφ/dt, L19). Transformer steps voltage (V_s/V_p = N_s/N_p) for transmission. Home: P = V_rms I_rms cos φ (L19); device uses current at rated power (L17). Entropy increases at each irreversible step — second law governs engine ceiling.