The net electric flux through any closed surface equals the enclosed charge divided by \( \varepsilon_0 \). Use this law to spot symmetries and predict electric-field patterns.
Wrap a coaxial cylinder (radius \( r \), length \( \ell \)) around the wire. Flux \( = E(2\pi r\ell) \). Enclosed charge \( = \lambda \ell \). Hence \( E = \dfrac{\lambda}{2\pi \varepsilon_0 r} \), radially outward.
Goal: derive \(E(r)\) for an infinite line charge using Gauss’s law — essential derivation practice for Q28.
Infinite wire has cylindrical symmetry; choose a coaxial cylinder of radius \(r\) and length \(l\).
Linear density \( \lambda \) gives \( q_{\text{enc}} = \lambda l \) inside the surface.
Field is radial and uniform on curved surface, zero on ends → \( \Phi = E(2\pi r l) \).
Set \( \Phi = q_{\text{enc}}/\varepsilon_0 \) so \( E(2\pi r l) = \lambda l / \varepsilon_0 \).
\[ E(r) = \frac{\lambda}{2\pi \varepsilon_0 r} \] directed radially outward (inward if \( \lambda < 0 \)).
Remember \(E \propto 1/r\); the same five-step logic solves any cylindrical symmetry problem swiftly.
Within the Drude model, free electrons undergo random collisions yet experience a net force \(eE\). The average time between collisions, called relaxation time \( \tau\), gives a steady drift speed \(v_d = \dfrac{eE \tau}{m}\). Because \(J = n e v_d\), a longer \( \tau\) directly increases both drift velocity and current.
In a wire of area \(A\): \(I = n e A v_d\). The \(v_d\!-\!\tau\) plot is a straight line through the origin.
Source: Drude free-electron model
Goal: predict average electron drift velocity for AC current (Q21 II).
In a conductor \(v_d \propto I\); precisely \(v_d = I\!/\!(n e A)\).
Given \(I = I_0 \sin 2\pi\nu t\), so \(v_d(t) = v_0 \sin 2\pi\nu t\) where \(v_0 = I_0\!/\!(n e A)\).
\(\langle v_d \rangle = \frac{1}{T}\!\int_{0}^{T}\! v_0 \sin 2\pi\nu t\,dt = 0\). Electrons merely oscillate; net drift is zero.
The integral of any sine or cosine over its full period is always zero.
A charge q moving with velocity v in a field B experiences \( \mathbf F = q\,\mathbf v \times \mathbf B \), always perpendicular to v; speed stays unchanged while direction bends.
An electron entering a 0.1 T field at 3 × 106 m s⁻¹ (perpendicular) travels in a circle of radius ≈ 0.17 mm and period ≈ 3.6 ns.
Goal – find how radius & time change when electron speed doubles so you can answer parts (i) & (ii) of Case Q29.
For circular motion in \(B\): \(r=\frac{mv}{qB}\Rightarrow r\propto v\). Period \(T=\frac{2\pi m}{qB}\); it is speed-independent.
Speed doubles: \(v' = 2v_0\). Hence \(r' = 2r_0\). Radius doubles.
Since \(T\) is independent of \(v\), doubling speed leaves \(T\) unchanged: \(T' = T_0\).
For Q29 pick (i) 2 r₀ and (ii) T₀ — radius doubles, time stays the same.
In Young’s double-slit set-up, fringe spacing is \( \beta = \lambda D / d \). A dark fringe appears when the path difference is \((m + ½)\lambda\). Using two wavelengths overlaps their patterns, forming slow ‘beat’ fringes useful in mixed-colour analysis.
Quiz: What happens to \( \beta \) if slit separation \( d \) is increased?
Source: CBSE Sample Question Paper 2024-25
Goal → locate the nearest dark fringe when 400 nm and 600 nm beams interfere (Q18 I).
Simultaneous dark bands need \(m_1\lambda_1=(m_2+\tfrac12)\lambda_2\).
Insert \( \lambda_1=400\text{ nm}, \lambda_2=600\text{ nm}\). Smallest integers giving half-wavelength shift: \(m_1=3, m_2=2\).
Distance from central maximum: \(y=m_1\beta_1=3\beta_1\) where \(\beta_1=D\lambda_1/d\). Thus the least dark fringe is three 400 nm spacings away.
Use the LCM mindset—search lowest integer pair satisfying odd-even rule before plugging into \(y=m\beta\).
A photon of energy \(h\nu\) liberates an electron by spending the metal’s work function \(\Phi\); the rest appears as maximum kinetic energy \(K_{\max}\), giving \(K_{\max}=h\nu-\Phi\).
Goal: compute sodium’s work function so you can solve Q30 (III).
Given threshold wavelength λ₀ = 500 nm; work function formula \( \Phi = \dfrac{hc}{\lambda_0} \).
Use \( h = 6.63\times10^{-34}\,\text{J·s} \), \( c = 3.0\times10^{8}\,\text{m s}^{-1} \), \( \lambda_0 = 5\times10^{-7}\,\text{m} \).
\( \Phi = 3.98\times10^{-19}\,\text{J} \approx 4\times10^{-19}\,\text{J} \).
Shorter threshold wavelength means higher work function—keep units consistent.
It converts AC to DC using two PN-junction diodes and a centre-tapped transformer. In alternate half-cycles, D₁ or D₂ conducts, giving consecutive positive pulses across load RL. A shunt capacitor smooths ripple.
Now, sketch the full-wave output and label D₁, D₂ and RL on the circuit.
Source: CBSE Physics SQP 2024-25
Goal: solve Q22—predict waveform when centre tap moves toward D₁.
With equal halves each diode conducts alternately, producing equal-height positive pulses—full-wave DC.
Moving the tap toward D₁ shortens its secondary segment. Voltage feeding D₁ drops while that for D₂ rises.
Draw alternate pulses: lower peaks for D₁, higher peaks for D₂. The train stays unipolar but now asymmetric.
Average output remains positive; ripple increases because peaks are unequal.
On the BE/A–A curve, binding energy per nucleon climbs rapidly, peaks near A≈60 and then declines slowly. Mid-mass nuclei (Fe, Ni) sit at this peak and are most stable. Fusion of light nuclei and fission of heavy nuclei push products toward the peak, increasing BE/A and liberating energy.
Q20 asks: From nuclei W(190), X(90), Y(60), Z(30), decide which one fissions and which one fuses.
Using curve reasoning: BE / A rises to a peak near A≈60, then declines. Energy is gained by splitting heavy nuclei or fusing very light ones.
W has A = 190, far right of the peak, so splitting it raises BE / A and releases energy. W is most likely to undergo nuclear fission.
Z has A = 30, well left of the peak. Fusing two such light nuclei moves toward A≈60, increasing BE / A. Hence Z is most likely to undergo nuclear fusion. X and Y lie near the peak, so little energy gain either way.
To justify choices in Q20, always compare each nucleus’s position to the 60-amu peak of the binding-energy curve.
Recall key formulas and avoid common pitfalls before the test.
Use symmetry: \(E = \lambda / 2\pi\varepsilon_0 r\).
Across one pure AC cycle, average electron drift equals zero.
Radius \(r \propto v\); period \(T = 2\pi m / qB\) stays constant.
First common dark fringe appears at path difference equal to LCM of wavelengths.
\(K_{\text{max}} = h\nu - \Phi;\; V_s = K_{\text{max}}/e\).
Centre tap must sit exactly midway to produce equal positive pulses.
Heavy nuclei favor fission, light nuclei favor fusion—energy gain follows the curve.