Physics (Theory)

Divide the ring into small elements. For every element there is an identical element diametrically opposite to it.
Both are at the same distance R from the centre, so their fields at the centre are equal in magnitude but opposite in direction – they cancel.
Adding over the whole ring, the net field at the centre is zero.
Common mistake: the potential at the centre is NOT zero (V = λ·2πR/(4πε₀R) = λ/(2ε₀)); only the field vanishes.

In parallel, capacitances add: C = 10 × 1 µF = 10 µF = 10⁻⁵ F.
Energy U = ½CV² = ½ × 10⁻⁵ × (100)² = ½ × 10⁻⁵ × 10⁴ = 0.05 J = 5.0 × 10⁻² J.

Two cells (emf ε₁, ε₂, internal/series resistance r₁, r₂) in parallel with the same orientation are equivalent to one cell of emf
\( \varepsilon_{eq} = \dfrac{\varepsilon_1/r_1 + \varepsilon_2/r_2}{1/r_1 + 1/r_2} \).
Here ε₁/r₁ = 12/1 = 12, ε₂/r₂ = 6/0.5 = 12; 1/r₁ + 1/r₂ = 1 + 2 = 3.
ε_eq = 24/3 = 8 V. No current is drawn from A–B externally, so V_AB = 8 V.
Check: circulating current = (12 − 6)/(1 + 0.5) = 4 A; V = 12 − 4×1 = 8 V ✓.

Net force on a closed loop in a uniform field is zero, so the loop does not translate (A, B wrong).
Force on each element: dF = I dl × B. Take B along +z (viewed from +z). At the top of a clockwise loop current flows along +x: x̂ × ẑ = −ŷ, i.e. towards the centre. The same happens at every element – all forces point inward.
Hence the loop tends to shrink. (Its magnetic moment is antiparallel to B, an unstable orientation.)

\( \vec F = I\vec L \times \vec B \), with \( \vec L = 0.1\,\hat j \) m.
The ĵ part of B is parallel to the wire – no force.
\( \vec F = 1.0 \times 0.1\,\hat j \times (-8\times10^{-3}\,\hat k) = -0.8\times10^{-3}(\hat j\times\hat k) = -0.8\times10^{-3}\,\hat i \) N.
F = −(0.8 mN) î.

I_g = 0.001 I, so shunt current = 0.999 I.
Shunt S: I_g G = (I − I_g)S ⇒ S = 0.001G/0.999 = G/999.
Ammeter resistance = G ∥ S = GS/(G + S) = G·(G/999)/(G + G/999) = G/(999 + 1) = G/1000.
Shortcut: the whole I flows through R_A while I_g flows through G at the same voltage: R_A = I_g G/I = 0.001G.
Common mistake: G/999 is the shunt, not the ammeter resistance.

\( X_C = \dfrac{1}{2\pi\nu C} \).
New: \( X' = \dfrac{1}{2\pi(3\nu)(2C)} = \dfrac{X}{6} \).

Displacement current \( i_d = \varepsilon_0 \dfrac{d\Phi_E}{dt} \) exists only where the electric field changes with time.
Only in Region I does E depend on t (through ωt). Regions II, III and IV have fields that vary with position but are constant in time, so i_d = 0 there.

Balmer series: all transitions end on n_f = 2.
First line (Hα): 3 → 2; second line (Hβ): 4 → 2; the series limit is ∞ → 2.
So the second line is n_i = 4 → n_f = 2.

As is pentavalent (group 15). Four of its valence electrons form covalent bonds with neighbouring Ge atoms; the fifth is loosely bound and easily becomes a free (conduction) electron.
So As is a donor, giving an n-type semiconductor with more conduction electrons (holes actually decrease slightly due to recombination). The lattice is not distorted because the dopant is in tiny amount and of similar size.

K_max = hν − φ₀.
Beam A: 3.3 − 2.3 = 1 eV; Beam B: 11.3 − 2.3 = 9 eV.
K = ½mv² ⇒ v ∝ √K.
v_A/v_B = √(1/9) = 1/3.

λ = h/p. Equal λ ⇒ equal momentum p.
Since masses differ, their speeds (v = p/m) and kinetic energies (K = p²/2m) are different.

A is false: K_max = hν − φ₀ depends on frequency, not on intensity. Higher intensity only means more photons per second.
R is false: for light above the threshold frequency, the photocurrent depends on the intensity (number of photons), not on the wavelength.
So both are false.

M = N₂Φ₂/I₁ – it is the flux linked with one coil per unit current in the other.
When the coils are wound over each other, almost all the flux of one coil passes through the other (minimum flux leakage), so flux linkage is maximum and therefore M is maximum.
R is exactly the reason for A.

When two parallel wires are joined in series with a battery, the current goes down one wire and returns up the other, so the currents are antiparallel.
Antiparallel currents repel (F/l = μ₀I₁I₂/2πd, repulsive), so the wires move apart.
R is true and correctly explains A.

A is false: if a converging beam (virtual object behind the mirror) falls on a plane mirror, the reflected rays meet in front of it – a real image. A convex mirror can also give a real image of a virtual object lying within its focal length.
R is false: a virtual image can serve as an object for another optical element, e.g. the virtual image formed by one lens becomes the object for the eye lens, which forms a real image on the retina.
So both are false.

R_T = R₀[1 + α(T − T₀)], T₀ = 20 °C.
Given R_T = 2R₀ ⇒ 2 = 1 + α(T − 20).
T − 20 = 1/α = 1/(4.0 × 10⁻³) = 250.
T = 270 °C ≈ 543 K.

Frequency does not change on reflection or refraction.
(i) Reflected light stays in air: λ = c/ν = (3 × 10⁸)/(5.0 × 10¹⁴) = 6 × 10⁻⁷ m = 600 nm.
(ii) In the medium speed v = c/n, so λ' = λ/n = 600/1.5 = 400 nm (4 × 10⁻⁷ m).
Common mistake: changing the frequency instead of the wavelength.

Lens maker's formula: \( \dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right) \).
For a plano-convex lens R₁ = R, R₂ = ∞:
1/f = (n − 1)/R ⇒ R = (n − 1)f = 0.4 × 16 = 6.4 cm.

f = −R/2 = −20 cm, u = −30 cm (sign convention).
(i) Mirror formula: 1/v + 1/u = 1/f ⇒ 1/v = 1/f − 1/u = −1/20 + 1/30 = −1/60 ⇒ v = −60 cm.
Image is 60 cm in front of the mirror – real.
(ii) m = −v/u = −(−60)/(−30) = −2: inverted and twice the size.

Photon: E = hc/λ_p ⇒ λ_p = hc/E.
Neutron: p = √(2mE) ⇒ λ_n = h/√(2mE).
\( \dfrac{\lambda_n}{\lambda_p} = \dfrac{h}{\sqrt{2mE}} \times \dfrac{E}{hc} = \sqrt{\dfrac{E}{2mc^2}} \).

Graph (V on x-axis, I on y-axis): starting at the origin the current first rises almost linearly with V, then bends, reaches a maximum, then decreases while V keeps increasing, and finally rises again.
(a) Negative-resistance region: the falling part of the curve after the peak, where dI/dV < 0 (current decreases as voltage increases).
(b) Ohm's law region: the initial straight-line portion near the origin, where I ∝ V.
Marks: correct shape of the graph + both regions labelled.

E is along x only, so only the two faces perpendicular to x contribute; flux through the other four faces is zero.
Area of each face A = (0.1)² = 0.01 m².
Left face (x = 0.1 m): E = 50 N/C, outward normal is −x: φ_L = −50 × 0.01 = −0.5 N m²/C.
Right face (x = 0.2 m): E = 100 N/C, outward normal +x: φ_R = +100 × 0.01 = +1.0 N m²/C.
(a) Net flux φ = −0.5 + 1.0 = 0.5 N m² C⁻¹.
(b) Gauss's law: q = ε₀φ = 8.854 × 10⁻¹² × 0.5 ≈ 4.4 × 10⁻¹² C.

Current density: current flowing per unit area held normal to the direction of flow, j = I/A. It is a vector (direction of flow of positive charge).
Acceleration of an electron: a = −eE/m. Average drift velocity: v_d = −(eE/m)τ.
In time Δt, charge crossing area A: IΔt = neA|v_d|Δt ⇒ I = neA·(eEτ/m) = (ne²τ/m)AE.
j = I/A = (ne²τ/m)E. Since electrons drift opposite to E but carry negative charge, j is along E:
\( \vec j = \dfrac{ne^2\tau}{m}\vec E = \alpha\vec E \) (this is Ohm's law, α = conductivity σ).

A Wheatstone bridge is an arrangement of four resistances R₁, R₂, R₃, R₄ in a quadrilateral ABCD, with a cell across one diagonal (A–C) and a galvanometer G across the other (B–D). It is used to find an unknown resistance.
Let R₁ = AB, R₂ = BC, R₃ = AD, R₄ = DC, currents I₁ in AB, I₃ in AD, I_g in BD.
At balance I_g = 0, so by the junction rule the current in BC is I₁ and in DC is I₃.
Loop ABDA: I₁R₁ + I_gG − I₃R₃ = 0 ⇒ I₁R₁ = I₃R₃ … (1)
Loop BCDB: I₁R₂ − I₃R₄ − I_gG = 0 ⇒ I₁R₂ = I₃R₄ … (2)
Dividing (1) by (2): R₁/R₂ = R₃/R₄ – the balance condition (B and D are at the same potential).

(a) K = ½mv² ⇒ v = √(2K/m). With m = 1.673 × 10⁻²⁷ kg:
v = √(2 × 1.3384 × 10⁻¹⁴ / 1.673 × 10⁻²⁷) = √(1.6 × 10¹³) = 4.0 × 10⁶ m/s.
(b) v ⊥ B, so F = qvB and a = (q/m)vB = 1.0 × 10⁸ × 4 × 10⁶ × 2 × 10⁻³ = 8 × 10¹¹ m/s².
(c) r = mv/(qB) = v/[(q/m)B] = 4 × 10⁶ /(1.0 × 10⁸ × 2 × 10⁻³) = 20 m.

Let v = v_m sin ωt and the current i = i_m sin(ωt + φ), φ = phase difference.
Instantaneous power p = vi = v_m i_m sin ωt sin(ωt + φ) = (v_m i_m/2)[cos φ − cos(2ωt + φ)].
The average of cos(2ωt + φ) over a cycle is zero.
P_avg = (v_m i_m/2) cos φ = (v_m/√2)(i_m/√2) cos φ = V I cos φ = I²Z cos φ = I²R (since cos φ = R/Z).
Resonance: current is maximum when Z is minimum, i.e. X_L = X_C:
ω₀L = 1/(ω₀C) ⇒ ω₀ = 1/√(LC) ⇒ ν₀ = 1/(2π√(LC)).

(a) Any two:
• Gamma rays (λ ~ 10⁻¹⁴–10⁻¹⁵ m) come from atomic nuclei, whose size is ~10⁻¹⁴ m.
• X-rays (λ ~ 10⁻¹⁰ m) are emitted by inner-shell transitions in heavy atoms (atomic size ~10⁻¹⁰ m).
• Radio waves are radiated by accelerated charges in antennas; an antenna radiates best at a wavelength comparable to its own length.
(b)(i) Short-wave radio is reflected back by the ionosphere, so signals can travel far beyond the horizon by sky-wave propagation.
(ii) Visible light and radio waves pass through the atmosphere (atmospheric 'windows'), but X-rays are absorbed by the atmosphere, so X-ray telescopes must be above it, on satellites.

Drawbacks of Rutherford's model:
1. An orbiting electron is accelerated and, by classical theory, should radiate energy continuously, spiral inwards and fall into the nucleus – the atom would be unstable.
2. Its frequency of revolution would change continuously, so it would emit a continuous spectrum, whereas atoms show line spectra.
Bohr's remedy: electrons revolve only in certain stationary orbits (angular momentum mvr = nh/2π) in which they do not radiate; radiation is emitted only when an electron jumps from a higher to a lower orbit, with hν = E_i − E_f – this explains both stability and line spectra.
Spacing: r_n = n²h²ε₀/(πme²) = n²a₀, so r ∝ n².
r_{n+1} − r_n = a₀[(n+1)² − n²] = a₀(2n + 1), which increases with n. E.g. r₁ = a₀, r₂ = 4a₀, r₃ = 9a₀: gaps 3a₀, 5a₀… – not equal.

(a) Any two: the nucleus is positively charged; it is made of protons and neutrons (nucleons); its radius R = R₀A^{1/3} (R₀ ≈ 1.2 fm); nuclear density is independent of mass number; nuclear forces are short-range and very strong.
(b) Almost the entire mass of the atom is concentrated in the nucleus, whose radius (~10⁻¹⁵ m) is about 10⁵ times smaller than the atom (~10⁻¹⁰ m). The atom is mostly empty space, so the nucleus is far denser.
(c) Mass of nucleus ≈ mA (m = mass of one nucleon). Volume = (4/3)πR³ = (4/3)πR₀³A.
ρ = mA / [(4/3)πR₀³A] = 3m/(4πR₀³) – A cancels, so the density is the same (~2.3 × 10¹⁷ kg/m³) for all nuclei.

(i) 1/f = (n − 1)(1/R − 1/(−R)) = 2(n − 1)/R ⇒ P = 2(n − 1)/R → (A).
(ii) Each half is plano-convex: P' = (n − 1)/R = half of 2(n − 1)/R → P/2 → (D).
(iii) Two lenses in contact: P/2 + P/2 = P → (B).
(iv) Cutting along the axis does not change the curvature of the faces, so each half still has power P (only the aperture is halved). Two such halves in contact coaxially: P + P = 2P → (C).
Common mistake: in (iv) thinking each half has power P/2 – that happens only for a cut perpendicular to the axis.

(i) The full-wave output is |V₀ sin ωt|; squaring removes the sign, so its rms value equals that of the input: V₀/√2 → (A).
(ii) Each diode conducts only when it is forward biased, i.e. during one half of each cycle → (B).
(iii) The two ends of the centre-tapped secondary are always in opposite phase, so at any instant one diode is forward biased and the other reverse biased → (C).
(iv) A half-wave rectifier gives one output pulse per input cycle, so the ripple frequency equals the input frequency, 50 Hz → (B). (Full-wave would give 100 Hz.)

(i) Torque on the dipole in uniform E: τ = pE sin θ (tends to align p with E).
Work done by an external torque in rotating it slowly from θ₀ to θ₁:
W = ∫ pE sin θ dθ (θ₀ → θ₁) = pE(cos θ₀ − cos θ₁).
This work is stored as potential energy. Choosing U = 0 at θ₀ = 90°:
U(θ) = −pE cos θ = −\( \vec p\cdot\vec E \).
(1) Maximum U = +pE when θ = 180° (p antiparallel to E) – unstable equilibrium.
(2) Minimum U = −pE when θ = 0° (p parallel to E) – stable equilibrium.
(ii) Dipole moment points from −q to +q: \( \vec p = q\vec d = 1.0\times10^{-12}(3\hat i + 4\hat j)\times10^{-3} = (3\hat i + 4\hat j)\times10^{-15} \) C m (|p| = 5 × 10⁻¹⁵ C m).
\( \vec\tau = \vec p \times \vec E = (3\hat i + 4\hat j)\times10^{-15} \times 1000\,\hat i = 4\times10^{-12}(\hat j\times\hat i) = -4\times10^{-12}\,\hat k \) N m.
Magnitude 4 × 10⁻¹² N m (check: pE sin θ = 5 × 10⁻¹⁵ × 1000 × 0.8), directed along −z.

(i) −q is at x = −a, +q at x = +a. At point P (x, 0):
V = (1/4πε₀)[q/(x − a) − q/(x + a)] = (1/4πε₀)·q·2a/(x² − a²) = (1/4πε₀)·p/(x² − a²), with p = 2qa.
For x ≫ a, a² is negligible: V = p/(4πε₀x²). Since \( \vec p = p\hat i \), p = \( \vec p\cdot\hat i \), so V = (1/4πε₀)(\( \vec p\cdot\hat i \))/x².
(ii) Initial charges Q = σ·4πr²:
Q₁ = (2/π × 10⁻⁹)(4π × 10⁻⁴) = 8 × 10⁻¹³ C; Q₂ = (2/π × 10⁻⁹)(4π × 9 × 10⁻⁴) = 72 × 10⁻¹³ C.
Total Q = 80 × 10⁻¹³ C (conserved).
When joined, they reach a common potential: kQ₁'/r₁ = kQ₂'/r₂ ⇒ Q₁'/Q₂' = r₁/r₂ = 1/3.
Q₁' = Q/4 = 20 × 10⁻¹³ = 2 × 10⁻¹² C (and Q₂' = 6 × 10⁻¹² C).
Note: charge flows from S₂ to S₁ even though σ was equal, because S₂ was at the higher potential (V = σr/ε₀).

(i) The same current i = i_m sin(ωt + φ) flows through R and C. Voltage across R (v_R = i_mR) is in phase with i; voltage across C (v_C = i_mX_C) lags i by 90°.
Phasor addition: v_m² = v_R² + v_C² = i_m²(R² + X_C²).
Z = v_m/i_m = √(R² + X_C²) = √(R² + 1/(ω²C²)), and tan φ = X_C/R (current leads voltage).
(ii) An inductor acts as a pure conductor for direct current: X_L = ωL = 2πνL and for dc ν = 0, so X_L = 0. (Equivalently, for steady current dI/dt = 0, so the back emf L dI/dt = 0.) Only its small wire resistance remains.
(iii) Lamp resistance R = 110/11 = 10 Ω. Required current 11 A from 220 V rms: Z = 220/11 = 20 Ω.
X_L = √(Z² − R²) = √(400 − 100) = √300 = 10√3 ≈ 17.3 Ω.
L = X_L/(2πν) = 17.3/(2π × 50) = 17.3/314 ≈ 0.055 H (55 mH).

(i) Diagram: a laminated soft-iron core with a primary coil of few turns (N_p, connected to ac input) and a secondary coil of many turns (N_s > N_p, connected to the load).
Principle: mutual induction. The alternating primary current produces an alternating flux in the core which links the secondary and induces an emf in it. With the same flux per turn, V_s/V_p = N_s/N_p.
Energy losses (any three): (a) flux leakage – not all primary flux links the secondary; (b) copper (I²R) loss in the resistance of the windings; (c) eddy currents in the core (reduced by laminating); (d) hysteresis loss in repeated magnetisation of the core; (e) humming/vibration.
(ii) No. For an ideal transformer V_pI_p = V_sI_s; when the voltage is stepped up, the current is stepped down in the same ratio, so output power never exceeds input power.
(iii) (1) V_s = V_p(N_s/N_p) = 90 × 3000/200 = 90 × 15 = 1350 V.
(2) I_p = I_s(N_s/N_p) = 2.0 × 15 = 30 A (ideal transformer).

(i) Graph of δ (y-axis) against i (x-axis): a U-shaped curve – δ first decreases as i increases, reaches a minimum δ_m, and then increases again.
Angle of minimum deviation: the smallest deviation produced by the prism; at this position the angle of incidence equals the angle of emergence (i = e) and the ray inside the prism is parallel to the base.
(ii) At the first face incidence is normal, so the ray enters undeviated (r₁ = 0). Since r₁ + r₂ = A, it meets the second face at r₂ = A.
It emerges at angle e. All the deviation occurs at the second face: δ = e − r₂ = e − A ⇒ e = A + δ.
Snell's law at the second face (glass to air): n sin A = sin e = sin(A + δ) ⇒ n = sin(A + δ)/sin A.
(iii) (1) \( n = \dfrac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}} \) ⇒ sin((60° + δ_m)/2) = √2 × sin 30° = 1/√2 ⇒ (60° + δ_m)/2 = 45° ⇒ δ_m = 30°.
(2) At minimum deviation i = (A + δ_m)/2 = 45°.

(i) Huygens' principle: every point on a wavefront acts as a source of secondary wavelets that spread out in all directions with the speed of the wave; the common forward tangent (envelope) to these wavelets gives the new wavefront at a later time.
Construction: plane wavefront AB meets the mirror MN at A, at angle i. While the disturbance from B travels BC = vt to reach the mirror at C, the wavelet from A spreads to a sphere of radius AE = vt. The tangent CE from C to this sphere is the reflected wavefront.
In right triangles ABC and AEC: AC is common, BC = AE = vt, ∠ABC = ∠AEC = 90°, so the triangles are congruent ⇒ ∠BAC = ∠ECA ⇒ i = r.
(ii) Coherent sources: sources whose phase difference stays constant with time (and having the same frequency).
Two independent sodium lamps cannot be coherent: light from each comes from random, independent emissions by millions of atoms lasting ~10⁻⁸ s, so the phase difference between the two lamps changes randomly and rapidly; no steady interference pattern is seen.
(iii) Position of nth bright fringe y = nλD/d. Coincidence: 4λD/d = 5 × 520 × D/d ⇒ λ = 5 × 520/4 = 650 nm.