Using Gauss’s law, derive the electric field near a uniformly charged infinite plane sheet of surface charge density \( \sigma \).

Place a symmetrical “pillbox” cylinder of face area \( A \) cutting the sheet. Field is normal and equal on both sides, so flux passes only through the two flat faces.

Gauss’s law: \( 2EA = \dfrac{\sigma A}{\varepsilon_0} \) ⇒ \( E = \dfrac{\sigma}{2\varepsilon_0} \). Direction: outward, opposite on the two faces.

A 10 g sphere carrying +10 mC moves freely inside a vertical pipe. The pipe is dragged westward through a 2 T uniform field. Find the minimum speed \(v\) so that magnetic force balances the weight, and state the direction of \(\mathbf B\).

Balance forces: \(qvB = mg\) (90° between \( \mathbf v \) and \( \mathbf B \)).
Insert values: \(v = \dfrac{0.01 \times 9.8}{0.01 \times 2} = 4.9\ \text{m s}^{-1}\).
Use Fleming’s left-hand rule: force needed upward, velocity West → \( \mathbf B \) must point North.

• Forgetting the \( \sin\theta =1 \) term when \( \mathbf v \perp \mathbf B \).
• Using grams or millicoulombs directly—always convert to kg and C.
• Choosing wrong field direction by applying the hand rule for a negative charge.

A series LCR circuit has \(V_R = V_C = V_L = 10\text{ V}\). If the capacitor is short-circuited, find the new voltage across the inductor.

At resonance, \(V_s = V_R = 10\text{ V}\) and \(X_L = R\).
After shorting C, impedance \(Z = \sqrt{R^2 + X_L^2}=R\sqrt{2}\).
Current \(I' = V_s/Z = 10/(R\sqrt{2})\).
Hence \(V_L' = I'X_L = 10/\sqrt{2}\text{ V} \approx 7.1\text{ V}\).

• Adding 10 V + 10 V algebraically and getting 20 V.
• Assuming current stays 10 V / R and predicting \(10\sqrt{2}\) V.
Always use phasors and recalculate current after altering the circuit.

In Young’s double-slit set-up, wavelengths \(400\,\text{nm}\) and \(600\,\text{nm}\) interfere. Find the nearest distance from the central maximum where a dark fringe appears.

Common dark fringe occurs when \(m\lambda_{600}= \bigl(m+\tfrac12\bigr)\lambda_{400}\). Smallest integer \(m=1\) gives path difference \(=600\,\text{nm}\). Hence distance from centre \(y=\dfrac{600D}{d}= \dfrac{3}{2}\beta\) where \(\beta=\dfrac{400D}{d}\). Predict fringe position quickly by comparing path differences, not fringe orders.

Students often equate fringe numbers instead of path differences, mix up bright-dark conditions, or forget that \(D\) and \(d\) cancel, leaving the answer in terms of \(\beta\).

From the binding-energy-per-nucleon graph, decide which of W, X, Y, Z is likely to undergo (i) fission and (ii) fusion, with reasoning.

Low-BE/A heavy nuclei (left of iron-56) gain energy by splitting; low-BE/A light nuclei (far right) gain energy by merging. Therefore, W (A≈190) favours fission, while Z (A≈30) favours fusion.

Picking the nucleus with the highest BE/A (X or Y) – these are already near the peak, so neither fission nor fusion releases extra energy. Also, confusing the heavy and light sides of the curve.

For a parallel-plate capacitor \(A, d\): derive \(C\) when (a) a dielectric slab of thickness \(t\) and constant \(k\), (b) a metal slab of thickness \(t<d\) is inserted. Which yields larger \(C\)?

Model two regions in series. Dielectric: \(C=\varepsilon_0 A /(d-t+t/k)\). Metal: gap becomes \(d-t\), so \(C=\varepsilon_0 A /(d-t)\). Because \(d-t < d-t+t/k\), the metal slab gives the higher capacitance.

Treating the two regions as parallel, skipping the \(t/k\) term, or assuming a dielectric always boosts \(C\) more than a conductor.

Parallel-plate capacitor: \(A = 1.0\times10^{-3}\,\text{m}^2\), \(d = 1.0\times10^{-4}\,\text{m}\). Voltage rises at \(dV/dt = 1.0\times10^{8}\,\text{V\,s}^{-1}\). Find the displacement current \(I_d\).

Use \(I_d = \varepsilon_0 A \frac{1}{d}\frac{dV}{dt}\). Substitute \(\varepsilon_0 = 8.85\times10^{-12}\,\text{F m}^{-1}\). Hence \(I_d = 8.85\times10^{-4}\,\text{A}\) (0.885 mA).

• Omitting the \(1/d\) factor. • Mixing SI units and prefixes. • Using \(I_d = C\,dV/dt\) without first finding \(C\), causing power-of-ten slips.

Three photoelectric I–V curves A, B and C are shown. Identify which curves share equal light intensity and which share equal photon frequency.

Check saturation current: identical heights ⇒ equal intensity. Check stopping potential \(V_s\): identical intercepts ⇒ equal frequency. Result: A & B same intensity; B & C same frequency.

Do not use slope or initial current to judge intensity, and never link frequency to curve gradient; only saturation current and \(V_s\) matter.

Identify blocks X & Y in a centre-tapped full-wave rectifier, sketch load waveforms, and predict change when the centre tap shifts toward D1.

X = two diodes (D1 & D2). Y = load R_L (plus filter C). Ideal output: equal positive half-cycles, frequency 2f. Moving tap toward D1 makes secondary voltages unequal—pulses via D1 peak higher, via D2 lower—raising ripple and lowering average DC.

• Drawing negative half-cycles—output is always positive.
• Thinking frequency reverts to f; it stays at 2f.
• Ignoring drop in average DC with asymmetry.

A \(100\text{ V}\) battery with \(r = 1\,\Omega\) sends \(10\text{ A}\) through a heater at \(20^{\circ}\text{C}\). At \(320^{\circ}\text{C}\) find power lost inside the battery. Given \(\alpha = 3.7\times10^{-4}\,^{\circ}\text{C}^{-1}\).

Initial \(R_0 = V/I = 10\,\Omega\). Temperature rise \(\Delta T = 300^{\circ}\text{C}\). New \(R = R_0\bigl(1+\alpha\Delta T\bigr)=10\!\left(1+3.7\times10^{-4}\times300\right)=11.11\,\Omega\).

Total resistance \(R_{\text{tot}} = R + r = 11.11 + 1 = 12.11\,\Omega\). Current \(I = \dfrac{100}{12.11} \approx 8.26\text{ A}\).

Internal dissipation \(P = I^{2}r = (8.26)^2 \times 1 \approx 68\text{ W}\). Learning outcome achieved: temperature change & internal resistance combined.

AC generator: N-turn coil (area A) spins with angular speed ω in field B. Derive \( \varepsilon(t) \) and cite the energy source.

Flux: \( \Phi = N B A \cos \omega t \).

Differentiate to get \( \varepsilon = -\dfrac{d\Phi}{dt} = N B A \omega \sin \omega t \).

Peak emf \( \varepsilon_0 = N B A \omega \).

Mechanical work of the prime mover becomes electrical energy.

Omitting the minus sign or using cos instead of sin in \( \varepsilon(t) \) breaks the correct phase relation.

Telescope with \(f_o = 15\,\text{m}\) and \(f_e = 1\,\text{cm}\). (a) Find angular magnification in normal adjustment. (b) Determine the Moon’s image diameter formed by the objective.

Use \(M = f_o / f_e\). So \(M = 15\,\text{m} / 0.01\,\text{m} = 1500\). Moon’s angular size \( \theta = D/R \approx 9.16\times10^{-3}\,\text{rad}\). Image diameter on focal plane: \(y = f_o \theta = 15 \times 9.16\times10^{-3} \approx 0.137\,\text{m}\) (≈ 13.7 cm). Large \(f_o\) links big angular magnification to a sizeable real image.

Avoid using linear magnification. Always link image size to \(y = f_o \theta\); the eyepiece only alters viewing angle, not the physical image on the focal plane.

Hydrogen has only one electron, yet its emission spectrum shows many lines. Explain this observation.

A discharge lamp holds billions of hydrogen atoms. Each atom’s electron may occupy a different excited level. When it falls to a lower level, one photon of that gap’s wavelength is emitted. All gaps are sampled across the ensemble, so the combined light displays every allowed wavelength—hence many spectral lines.

Assuming one atom emits all lines simultaneously. In reality, a single hydrogen atom can release only one photon per transition; spectral richness arises from many atoms making different transitions.