2026 HSC Physics — Intuition Education Predicted Paper

Answers and marking notes

Section I — answer key

Section II — marking notes

Q21 (a). ΔΦ per turn = BA = 0.25 × (0.040)² = 4.0 × 10⁻⁴ Wb; ε = NΔΦ/Δt = 50 × 4.0 × 10⁻⁴ / 0.20 = 0.10 V. (b). Flux into the bench is decreasing, so the induced current acts to maintain it: clockwise viewed from above (right-hand grip: clockwise current produces flux into the bench through the coil). The mark is for the Lenz reasoning, not the bare direction.

Q22 (a). τ = nBIA = 20 × 0.30 × 1.5 × (0.050 × 0.080) = 20 × 0.30 × 1.5 × 4.0 × 10⁻³ = 3.6 × 10⁻² N m. (Trap: cm → m before computing A — the flagged 2025 Q23 error.) (b). |sin θ| shape: maximum 3.6 × 10⁻² N m at 0° (plane parallel to B), zero at 90° (plane perpendicular to B), maximum again at 180°. One mark for the correct shape, one for the labelled zero/maximum angles.

Q23 (a). u_y = 20 sin 30° = +10 m s⁻¹ (up), Δy = −25 m: −25 = 10t − 4.9t² ⇒ 4.9t² − 10t − 25 = 0 ⇒ t = [10 + √(100 + 490)]/9.8 = [10 + 24.3]/9.8 = 3.5 s. Full marks require the sign convention stated and the quadratic (or an explicit two-stage up-then-down calculation) — a symmetric-flight calculation (t = 2u_y/g ≈ 2.0 s) scores at most 1. (b). u_x = 20 cos 30° = 17.3 m s⁻¹; range = 17.3 × 3.5 = 61 m. (c). v_y = 10 − 9.8 × 3.5 = −24.3 m s⁻¹; v = √(17.3² + 24.3²) = √890 ≈ 30 m s⁻¹ at 55° below the horizontal (consistency check: v² = u² + 2gΔh = 400 + 490 = 890 ✓).

Q24 (a). GMm/r² = 4π²mr/T² (gravitational force supplies the centripetal force); cancel m, rearrange: r³/T² = GM/4π². Both force expressions and the cancellation must be shown. (b). T = 2.0 × 24 × 3600 = 1.728 × 10⁵ s (the days-to-seconds conversion is the first mark). M = 4π²r³/(GT²) = 39.48 × (3.8 × 10⁸)³ / (6.67 × 10⁻¹¹ × (1.728 × 10⁵)²) = 2.17 × 10²⁷ / 1.99 = 1.1 × 10²⁷ kg.

Q25 (a). E_photon = hf = 6.626 × 10⁻³⁴ × 6.5 × 10¹⁴ = 4.31 × 10⁻¹⁹ J = 2.69 eV. 2.69 eV > 2.1 eV (caesium emits); 2.69 eV < 4.3 eV (zinc does not). Both comparisons required. (b). Kmax = 2.69 − 2.1 = 0.59 eV = 0.59 × 1.602 × 10⁻¹⁹ = 9.4 × 10⁻²⁰ J. (c). Doubling intensity doubles the number of photons per second, each of unchanged energy hf; one photon releases one electron, so the number emitted per second doubles but Kmax = hf − φ is unchanged. "Brighter light gives faster electrons" is the target misconception.

Q26 (a). λ = d sin θ / m = 2.5 × 10⁻⁵ × sin 3.0° / 2 = 2.5 × 10⁻⁵ × 0.0523 / 2 = 6.5 × 10⁻⁷ m (654 nm). (b). sin θ = mλ/d: λ falls from 654 nm to 450 nm, so sin θ (and θ) decreases in proportion — every maximum moves closer to the central maximum. The proportionality must be used, not just "the pattern shrinks". (c). Decrease the slit separation d in the same proportion (sin θ = mλ/d restored), or increase the slit-to-screen distance L (fringe spacing on the screen ≈ λL/d restored). Justification via the relevant relationship for one of the two.

Q27 (a). t = 1.5 × 10⁴ / (0.995 × 3.0 × 10⁸) = 5.0 × 10⁻⁵ s (50 μs). (b). t = γt₀ = 10.0 × 2.2 μs = 22 μs. The proper lifetime is the muon's own 2.2 μs — swapping t and t₀ (giving 0.22 μs) is the flagged error. (c). In the muons' frame the atmosphere is length-contracted: L = 15 km / 10.0 = 1.5 km, so the journey takes 1.5 × 10³ / (2.985 × 10⁸) ≈ 5.0 μs ≈ 2.3 proper lifetimes — a substantial fraction survives, whereas the classical 50 μs ≈ 23 lifetimes would leave essentially none. Both frames predict the same survival; the mark is for a single-frame, internally consistent account with the calculation.

Q28 (a). qE = qvB ⇒ v = E/B = 4.0 × 10⁴ / 0.20 = 2.0 × 10⁵ m s⁻¹. (b). r = mv/qB = 1.673 × 10⁻²⁷ × 2.0 × 10⁵ / (1.602 × 10⁻¹⁹ × 0.20) = 1.0 × 10⁻² m (1.0 cm).

Q29 (a). qV = ½mv² ⇒ v = √(2qV/m) = √(2 × 1.602 × 10⁻¹⁹ × 500 / 9.109 × 10⁻³¹) = √(1.76 × 10¹⁴) = 1.33 × 10⁷ m s⁻¹. (Trap: uppercase V is voltage, lowercase v speed.) (b). r = mv/qB = 9.109 × 10⁻³¹ × 1.33 × 10⁷ / (1.602 × 10⁻¹⁹ × 1.5 × 10⁻³) = 5.0 × 10⁻² m (5.0 cm). (c). r ∝ m/q at fixed v and B: the proton's radius is ≈ 1836 times larger (≈ 92 m), and its path curves in the opposite sense because its charge is opposite.

Q30 (a). Vs = 240/20 = 12 V; Pin = Pout = 24 W ⇒ Ip = 24/240 = 0.10 A. (b). Secondary power doubles to 48 W; for an ideal transformer Pin must equal Pout, so with Vp fixed at 240 V the primary current doubles to 0.20 A. The mark is for invoking conservation of energy to carry the change from secondary to primary — describing only the secondary caps at 1.

Q31 (a). T = b/λmax = 2.898 × 10⁻³ / 5.80 × 10⁻⁷ = 5.0 × 10³ K. (λmax in metres; answer in kelvin.) (b). The line is shifted to a longer wavelength (redshift), so star Q is receding from Earth. (c). As the star rotates, one limb approaches (its absorption blueshifted) while the other recedes (redshifted); light from the whole disc is received together, so each line is smeared over a range of wavelengths — broadened symmetrically. Faster rotation gives broader lines. (Rotation about the star's own axis, not orbital revolution — the flagged 2023 error.)

Q32 (a). Y: red giant (post-main-sequence); Z: white dwarf (final stage for a Sun-like star, fusion ceased). (b). X is on the main sequence fusing hydrogen to helium in its core via the proton–proton chain (Sun-like core temperature). Y has exhausted core hydrogen: it fuses helium (to carbon/oxygen) in its core and/or hydrogen in a surrounding shell, at a higher core temperature. Both stars named, both processes distinguished. (c). Luminosity depends on both temperature and surface area (L ∝ R²T⁴): Z's radius is tiny, so despite the higher surface temperature its radiating area — and hence luminosity — is far smaller.

Q33 (a). Mass halves every 8.0 days (100 → 50 g at 8.0 d; 50 → 25 g at 16 d). The value must be read from the graph. (b). λ = ln 2 / t½ = 0.693 / (8.0 × 24 × 3600 s) = 0.693 / 6.91 × 10⁵ = 1.0 × 10⁻⁶ s⁻¹. (Days-to-seconds conversion is the second mark.) (c). 5.0 = 100 e^(−λt) ⇒ t = ln(20)/λ = 3.00 / 0.0866 day⁻¹ = 35 days (equivalently 3.00 / 1.0 × 10⁻⁶ s⁻¹ ≈ 3.0 × 10⁶ s). Requires solving with natural logarithms, not stepping half-lives (4.32 half-lives is not an integer count).

Q34 (a). Δm = 209.98286 − (205.97446 + 4.00260) = 209.98286 − 209.97706 = 0.00580 u. E = 0.00580 × 931.5 = 5.40 MeV = 5.40 × 10⁶ × 1.602 × 10⁻¹⁹ = 8.7 × 10⁻¹³ J. (Traps: early rounding of the masses; MeV–joule conversion.) (b)(i). The nucleus is initially at rest, so total momentum is zero: the alpha particle and lead nucleus recoil with equal and opposite momenta. For equal momentum magnitudes, KE = p²/2m — the lighter particle carries the larger share of the kinetic energy, so the alpha (4 u vs 206 u) takes almost all of it. (b)(ii). KE_α = E × m_Pb/(m_Pb + m_α) = 5.40 × 205.97/209.98 = 5.30 MeV (≈ 98% of the released energy; KE_Pb ≈ 0.10 MeV).

Q35 (a). E = −GMm/2r = −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 800)/(2 × 7.0 × 10⁶) = −3.20 × 10¹⁷ / 1.4 × 10⁷ = −2.3 × 10¹⁰ J. The negative sign is required — it signifies a bound orbit. (b). E₂ = −3.20 × 10¹⁷ / 2.8 × 10⁷ = −1.14 × 10¹⁰ J. Energy supplied = E₂ − E₁ = −1.14 × 10¹⁰ − (−2.29 × 10¹⁰) = +1.1 × 10¹⁰ J. (Trap: a less negative total energy is an increase.) (c). Moving outwards, kinetic energy decreases but potential energy (U = −GMm/r) increases by twice as much; the total mechanical energy therefore rises, and that difference must be supplied by the thrusters. "It slows down so it loses energy" is the target misconception.

Q36 (a). v = 2πrf = 2π × 0.10 × 12 = 7.5 m s⁻¹. (b). Full chain (one mark per link band): as the disc rotates, the magnetic flux through each region of the disc near the magnet changes → an emf is induced (Faraday) → eddy currents circulate in the conducting disc → by Lenz's law these currents flow so that the magnetic forces on them oppose the relative motion, producing a retarding torque on the disc → the disc's rotational kinetic energy is transformed into heat in the disc's resistance. Answers that say only "Lenz's law opposes the motion" without the intermediate emf/current steps cap at 2. (c). The gradient of the solid-disc curve decreases in magnitude as the disc slows: the rate of flux change is proportional to the disc's speed, so the induced emf, eddy currents and retarding torque all shrink as it slows — the braking effect weakens, and the disc approaches zero asymptotically rather than stopping abruptly. (d). The radial slots interrupt the circulating paths of the eddy currents, so only much smaller currents can flow; the opposing forces are correspondingly smaller and the disc keeps most of its kinetic energy over the same interval.