Electric Charges & Fields – Toughest Question (Board-style Application)
Locate the Zero-Force Point
Two +3 μC charges are fixed 0.5 m apart. Where should a −2 μC charge experience zero net electrostatic force?
Apply superposition and the inverse-square law to locate the point where forces cancel.
Key Points:
- Sketch charges on a straight line; choose origin at left +3 μC.
- Let −2 μC be \(x\) metres from the left charge, outside the 0.5 m pair.
- Set magnitudes equal: \( \frac{1}{x^{2}} = \frac{1}{(0.5 + x)^{2}} \).
- Solve: \( x = 0.25\ \text{m} \) left of the first + charge.
- Between the + charges, forces add; zero force cannot occur there.
Moving Charges & Magnetism – Toughest Question
Helical motion analysis
A proton enters a 0.3 T field at 60° with speed \(2\times10^{6}\,\text{m s}^{-1}\). Find the pitch of its helical path.
Lorentz magnetic force bends the charged particle into a circle for \(v_{\perp}\) while \(v_{\parallel}\) stays uniform along the field.
Key Points:
- Resolve velocity: \(v_{\parallel}=v\cos60^{\circ}\), \(v_{\perp}=v\sin60^{\circ}\).
- Magnetic force on \(v_{\perp}\) gives circular motion; none acts on \(v_{\parallel}\).
- Time period \(T=\frac{2\pi m}{qB}\) for a proton.
- Pitch \(p=v_{\parallel}T\approx0.22\,\text{m}\).
- Use \(v_{\parallel}\), not total \(v\), when computing pitch.
Alternating Current – Toughest Question
Transformer & losses
Ideal transformer: \(N_p = 500\), \(V_p = 220\,\text{V}\). Refrigerator needs \(V_s = 110\,\text{V}\), \(I_s = 4\,\text{A}\). Find \(N_s\), \(I_p\) and efficiency with 5 % copper loss.
AC-circuit relations: \( \frac{V_p}{V_s} = \frac{N_p}{N_s} \). Power balance: \( V_p I_p = V_s I_s \). Efficiency: \( \eta = \frac{P_{\text{out}}}{P_{\text{in}}} \).
Key Points:
- Turns ratio gives \(N_s = 250\) turns.
- Power conservation ⇒ \(I_p = 2.0\,\text{A}\).
- 5 % copper loss ⇒ \(\eta = 95\%\). Avoid mixing primary & secondary.
Wave Optics – Toughest Question
YDSE with medium change
Two-slit interference: \(d = 0.25\,\text{mm}\), \(\lambda = 600\,\text{nm}\), measured fringe width \(\beta = 2.4\,\text{mm}\).
Use \( \beta = \frac{\lambda D}{d} \) to find screen distance \(D\). Replace \(\lambda\) by \( \lambda' = \lambda/\mu \) to predict new fringe width in water.
Key Points:
- Screen distance: \(D = \frac{\beta d}{\lambda} = 1.0\,\text{m}\).
- In water, new wavelength \( \lambda' \approx 451\,\text{nm}\).
- New fringe width: \(\beta' = \beta/\mu \approx 1.8\,\text{mm}\) (reduced by factor 1/1.33).
- Predict fringe shift by comparing original and altered interference patterns.
Nuclei – Toughest Question
Secular equilibrium
Activity ratio in secular equilibrium
Radioactivity chains reach secular equilibrium when the parent’s half-life is far longer than the daughter’s.
Using the given half-lives, show that the activity of \(^{234}\text{Th}\) tracks \(^{238}\text{U}\) for \(4.5\times10^{9}\,\text{y}\).
Key Points:
- Equilibrium condition: \(A_{\text{U}} = A_{\text{Th}}\).
- Decay constant: \(\lambda = \ln 2 / T_{1/2}\).
- After \(4.5\times10^{9}\,\text{y}\): \(e^{-\lambda_{\text{Th}}t}\!\approx\!0\), \(e^{-\lambda_{\text{U}}t}\!\approx\!0.5\).
- Thus \(A_{\text{Th}}/A_{\text{U}} = 1\); equilibrium persists.
- Pitfalls: treating 24.1 d as long, or confusing activity with mass.
Electrostatic Potential & Capacitance – Toughest Question
Concentric Shells: Potential Profile
Isolated shells: \(r_1=10\,\text{cm}, Q_1=+8\,\text{nC}\); \(r_2=20\,\text{cm}, Q_2=-4\,\text{nC}\).
Use Gauss law to find \(E\). Then apply \(V(r)=kQ/r\) for each shell, superpose, sketch \(V(r)\) and obtain \(\Delta V\).
Key Points:
- Inside any conductor \(E=0\) ⇒ flat potential regions.
- Evaluate three zones: \(r<10\text{ cm}\), \(10<r<20\text{ cm}\), \(r>20\text{ cm}\).
- For \(r>20\text{ cm}\), \(V(r)=k\left(\frac{Q_1}{r}+\frac{Q_2}{r}\right)\).
- Potential difference: \(\Delta V=V_{10}-V_{20}=360\,\text{V}\) (inner shell higher).
Magnetism & Matter – Toughest Question
Atomic moment estimation
A toroid with \( \mu_r = 2000 \), mean radius \( r = 0.15\,\text{m} \) and 600 turns carries a current of 3 A.
Find (a) the magnetic field \( B \) inside the core and (b) the magnetic moment per iron atom when the atomic density is \( 8.5\times10^{28}\,\text{m}^{-3} \).
Key Points:
- \( B = \mu_0 \mu_r \dfrac{NI}{2\pi r} \) — include relative permeability.
- \( H = \frac{NI}{2\pi r} \), \( M = \frac{B}{\mu_0}-H \) connects field and magnetisation.
- Atomic moment \( m = \frac{M}{n} \approx 2.1\,\mu_B \).
- Core relation \( B = \mu_0(H+M) \) links macroscopic \( B \) to microscopic \( m \).
- Pitfalls: ignoring permeability or the \( H \) term overestimates magnetisation.
Electromagnetic Waves – Toughest Question
Displacement Current via Maxwell’s Correction
A parallel-plate capacitor (area 50 cm², gap 2 mm) is driven by \(V = 200\sin(1.5\times10^{5}t)\,\text{V}\). Find the displacement current between its plates.
Solution showcases Maxwell’s corrective term in the Ampère–Maxwell law.
Key Points:
- \(A = 50\,\text{cm}^{2} = 5.0\times10^{-3}\,\text{m}^{2}\).
- Capacitance \(C = \varepsilon_{0}A/d\).
- Electric flux \( \Phi_E = C V / \varepsilon_{0}\).
- Displacement current \(I_d = \varepsilon_{0}\,d\Phi_E/dt = C\,dV/dt\) (Ampère–Maxwell).
- \(dV/dt = 200(1.5\times10^{5})\cos(1.5\times10^{5}t)\).
- Use SI units; do not confuse with conduction current.
Dual Nature – Toughest Question
Work function deduction
Zn photocell stops current at \(V_s = 1.3\,\text{V}\) for \(\lambda = 250\,\text{nm}\). Determine the work function \(\phi\) and predict \(V_s\) for \(\lambda = 300\,\text{nm}\).
Key Points:
- Use \(eV_s = h\nu - \phi\) – Einstein photoelectric equation.
- Convert wavelength to frequency: \(\nu = \frac{c}{\lambda}\).
- First data ➜ \(\phi = h\nu - eV_s\).
- Insert \(\phi\) into second wavelength to obtain new \(V_s\).
- Keep energies in eV; avoid mixing Joules and eV.
Semiconductor Electronics – Toughest Question
Plot load line & set Q-point
Question: For a Si BJT in CE mode with \(V_{CC}=12\,\text{V}\), \(R_C=2\,\text{k}\Omega\), \(R_B=200\,\text{k}\Omega\) and \(\beta =120\), find the Q-point and comment on thermal runaway.
Solution: \(I_B=\frac{V_{CC}-V_{BE}}{R_B}= \frac{12-0.7}{200\,\text{k}\Omega}=56\,\mu\text{A}\). Hence \(I_C=\beta I_B \approx 6.8\,\text{mA}\), but the load-line limit is \(I_{C\,sat}= \frac{V_{CC}}{R_C}=6\,\text{mA}\). The transistor therefore saturates; Q-point is near \((I_C \approx 6\,\text{mA},\, V_{CE} \approx 0\,\text{V})\).
Key Points:
- Subtract \(V_{BE}\) when computing base bias current.
- Draw the \(I_C\!-\!V_{CE}\) load line to visualise active, cut-off and saturation regions.
- Select a Q-point in the active region for bias stability; extremes cause distortion.
- Near saturation, \(P = V_{CE} I_C\) is low, so thermal runaway is unlikely.
- Bias stabilisation safeguards BJT amplifiers against temperature rise.
Current Electricity – Toughest Question
Potentiometer calibration
A 1 m potentiometer wire (10 Ω) is fed by a 2.5 V DC source with internal resistance 0.5 Ω.
Null deflection occurs at 65 cm for the unknown DC cell.
Using the null-deflection method we obtain \(E_x \approx 1.55\,\text{V}\).
Key Points:
- \(k=\frac{E_D}{R+r}\times\frac{R}{L}\approx2.38\,\text{V m}^{-1}\); therefore \(E_x=k\times0.65=1.55\,\text{V}\).
- Sensitivity rises when the wire is longer or the current is reduced with a series resistance.
- Use driver cell emf \(E_D\), not terminal voltage; always include internal \(r\).
- Null deflection keeps the test cell in an open circuit, avoiding loading error in DC circuits.
Electromagnetic Induction – Toughest Question
Moving Loop Problem
A square loop of side 10 cm (R = 1 Ω) is pulled out of a 0.4 T region at 3 m s⁻¹, leaving the field in 0.2 s.
Determine \(I(t)\) and the mechanical work done. Relate the changing magnetic flux to induced emf and work.
Key Points:
- Faraday law: \(\varepsilon=-\frac{d\Phi}{dt}=-B\,\frac{dA}{dt}\).
- Uniform pull ⇒ \(dA/dt=-l v\); motional emf constant: \(\varepsilon=Bvl=0.12\;\text{V}\).
- Induced current: \(I(t)=\varepsilon/R=0.12\;\text{A}\;(0\le t\le0.2\text{ s})\), zero afterwards.
- Work–energy: \(W=\int_0^{0.2}I^{2}R\,dt\approx2.9\times10^{-3}\,\text{J}\).
- Avoid using constant flux; keep Lenz’s-law sign for direction.
Ray Optics – Toughest Question
Compound microscope
Objective \(f_o = 0.8\;\text{cm}\), eyepiece \(f_e = 2.5\;\text{cm}\), tube length 15 cm. Object is 1 cm from objective. Calculate total magnifying power for near-point viewing.
Goal: design the required magnification by chaining lens formulae and the microscope relation.
Key Points:
- Use \( \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}\) with \(u_o = -1\;\text{cm}\).
- Match instrument tube: \(v_o + v_e = 15\;\text{cm}\).
- \(\frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e}\) gives eyepiece object distance.
- Total magnification \(M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)\) with \(L = 15\;\text{cm},\; D = 25\;\text{cm}\).
- Computed value: \(M \approx 150\).
- Avoid sign errors; never drop the \((1 + D/f_e)\) factor.
Atoms – Toughest Question
Bohr model limits
Use Bohr theory to find the electron speed in hydrogen for \(n = 2\).
Compute the photon wavelength when the electron drops from \(n = 2\) to \(n = 1\).
Discuss how these results fit the observed spectra and where the model falls short.
Key Points:
- Speed: \(v_2=\frac{e^{2}}{2\varepsilon_0 h}\frac{1}{2}\).
- Energy gap: \(\Delta E = 13.6\,\text{eV}\left(1-\frac{1}{4}\right)=10.2\,\text{eV}\).
- Wavelength: \(\lambda = \frac{hc}{\Delta E}\approx 1.22\times10^{-7}\,\text{m}\) (Lyman-α line).
- Limitations: no fine-structure, fails for multi-electron atoms.
- Common errors: ignore J–eV conversion, apply non-relativistic \(v\) at high \(n\).