Roadmap: decode the 2024-25 sample paper, flag your weak concepts, and practise targeted problems to convert them into scoring strengths.
Sample paper snapshot – 33 questions, 5 sections (A–E), 70 marks, 3 h: our plan mirrors this pattern.
Source: CBSE Physics Sample Question Paper 2024-25 (Code 042)
More questions mean higher exam weightage. Target “low” chapters first, consolidate “med”, then polish “high”.
| Chapter | # Qs | Strength |
|---|---|---|
| Electric Charges & Fields | 3 | Low |
| Moving Charges & Magnetism | 3 | Low |
| Alternating Current | 1 | Low |
| Wave Optics | 3 | Low |
| Nuclei | 2 | Low |
| Electrostatic Potential & Capacitance | 2 | Med |
| Magnetism & Matter | 1 | Med |
| Electromagnetic Waves | 2 | Med |
| Dual Nature of Radiation & Matter | 3 | Med |
| Semiconductor Electronics | 2 | Med |
| Current Electricity | 4 | High |
| Electromagnetic Induction | 2 | High |
| Ray Optics & Optical Instruments | 3 | High |
| Atoms | 2 | High |
Gauss’s theorem: Net electric flux through a closed surface equals enclosed charge divided by \( \varepsilon_0 \). 1. Gaussian surface — coaxial cylinder of radius \( r \) and length \( L \). 2. Flux evaluation — \( E \) is radial & uniform on curved area; ends contribute zero, so \( \Phi_E = E(2\pi rL) \). 3. Apply law — \( E(2\pi rL) = \lambda L/\varepsilon_0 \). Result: \( E(r) = \dfrac{\lambda}{2\pi \varepsilon_0 r} \).
Pitfalls: choosing a spherical surface breaks symmetry; remember no flux through end-caps.
A 10 g ball carrying +10 mC slides freely inside a vertical, frictionless tube. The tube is moved horizontally from east to west through a uniform 2 T magnetic field. Find (i) the minimum tube speed that keeps the ball stationary relative to the tube, and (ii) the direction of the magnetic field.
Key idea: \(F = qvB\) for motion ⟂ \(B\).
Important formula: \(F = qvB = mg\)
Question: In a series LCR circuit \(V_R = V_C = V_L = 10\,\text{V}\) (resonance). The capacitor is short-circuited. Find the new voltage across the inductor.
Hints
1. Resonance ⇒ \(X_L = X_C\) and \(R = X_L\).
2. Without the capacitor, \(Z = \sqrt{R^{2}+X_L^{2}} = R\sqrt{2}\); current becomes \(I/\sqrt{2}\).
3. Hence \(V_L = IX_L/\sqrt{2} = 10/\sqrt{2} ≈ 7.1 V\).
Answer: \(10/\sqrt{2}\,\text{V}\).
Pitfalls: assuming current unchanged or forgetting \(V_C = 0\) after the short.
Key relation: \(Z = \sqrt{R^{2}+(\omega L)^{2}}\)
Q. In a Young’s double-slit set-up using λ1 = 400 nm and λ2 = 600 nm, find the smallest distance y from the central maximum where a common dark fringe appears.
Key relation \( \Delta = m\lambda_1 = \left(m+\tfrac{1}{2}\right)\lambda_2 \)
Pitfalls: mix-ups between bright & dark criteria, forgetting to convert nm to metres.
Question: Using the binding-energy per nucleon curve, decide which nucleus—W(190), X(90), Y(60) or Z(30)—is most likely to (i) undergo fission and (ii) undergo fusion. Give a brief reason.
Energy is released whenever a nucleus moves toward the curve’s peak at A≈60. Don’t confuse total binding energy with binding energy per nucleon.
Problem A parallel-plate capacitor (plate area \(A\), gap \(d\)) receives
(i) a dielectric slab of thickness \(t\) and permittivity \(\kappa\),
(ii) a conducting slab of thickness \(t\) ( \(t<d\) ).
Derive \(C\) in each case and state which insertion increases capacitance more.
Hint 1 Model dielectric case as two series sections:
air gaps \((d-t)\) and dielectric slab \(t/\kappa\).
Hint 2 Equivalent capacitance
\(C_{\text{dielectric}}=\dfrac{\varepsilon_0A}{d-t+t/\kappa}\).
Hint 3 In a metal, \(E=0\); effective separation becomes \(d-t\), so
\(C_{\text{metal}}=\dfrac{\varepsilon_0A}{d-t}\).
Because \(d-t < d-t+t/\kappa\), we get \(C_{\text{metal}} > C_{\text{dielectric}}\).
⚠︎ Common slips: adding capacitances in parallel, omitting \(\kappa\).
Capacitance grows as the effective plate separation shrinks.
Source: CBSE Physics Sample Question Paper (2024 – 25)
A solid wire of radius \(a\) carries uniform current \(I\).
Find \(\displaystyle \frac{B_{\,a/2\;{\text{below}}}}{B_{\,a/2\;{\text{above}}}}\).
Hints
1. Ampère’s law → \(B_{\text{in}}=\mu_0 I r /(2\pi a^{2})\), \(B_{\text{out}}=\mu_0 I /(2\pi r)\).
2. Radii: inside point \(r=a/2\); outside point \(r=1.5a\).
3. Ratio \(= (a/2)/(1.5a)=1/3\).
Expected answer : \(1:3\).
Watch for mixing inside/outside formulas or using surface distances instead of center distances. Remember: inside \(B\propto r\); outside \(B\propto 1/r\).
Source: CBSE Physics Sample Paper 2024 – Q3
Match each electromagnetic wave with its typical production mechanism: Infra-red, Radio, Light and Microwave.
Hints
Task → From the three photoelectric I–V plots (A, B, C), rank the incident beams by intensity and by frequency.
Hint 1: Higher saturation current ⇒ higher intensity.
Hint 2: More negative stopping potential ⇒ higher photon frequency (energy).
Hint 3: Compare both features to classify A, B, C.
Pitfall – Stopping potential is independent of intensity.
Einstein’s Photoelectric Equation: \(eV_0 = h\nu - \phi\). Thus \(K_{\text{max}} = eV_0\).
Question : Draw energy-band diagrams for a P-type semiconductor at 0 K and room temperature. Decide if the bulb glows when the switch is (a) open, (b) closed in the ideal-diode circuit.
Hints
• 0 K: Fermi level lies just above valence band, near acceptor level.
• Room T: holes populate valence band; \(E_F\) rises slightly.
• Switch open — diode reverse-biased; circuit broken; bulb off.
• Switch closed — diode forward-biased; current flows; bulb on.
Watch out : Misplacing \(E_F\) in conduction band or assuming reverse leakage in an ideal diode.
P-N Junction: Forward bias lowers the depletion barrier, allowing majority carriers to conduct.
A heating element on a 100 V battery (internal resistance 1 Ω) draws 10 A at 20 °C. After warming to 320 °C, find the power dissipated inside the battery. Temperature coefficient α = 3.7 × 10−4 °C−1.
Remember: apply α only to the element, not to the constant internal resistance. Power lost inside a source is always \(I^{2}r\).
Key formula: \(R_T = R_0(1 + \alpha\,\Delta T)\)
Question – Derive the instantaneous emf induced in an N-turn coil of area A rotating with angular speed \( \omega \) in a uniform field \( B \).
1. Magnetic flux: \( \Phi = N B A \cos \theta \), where \( \theta = \omega t \).
2. Faraday’s law: \( e = -\dfrac{d\Phi}{dt} \).
3. Differentiate ⇒ \( e = N B A \omega \sin \omega t \).
Peak value: \( e_{\text{max}} = N B A \omega \).
Negative sign shows Lenz’s law; we usually quote magnitude only.
Energy comes from the mechanical work done by the prime mover rotating the coil.
Task: Draw a ray diagram of a Cassegrain telescope and state two advantages over a refractor.
• Primary concave and secondary convex mirrors fold the optical path.
• Secondary reflects rays through the central hole in the primary to the final focus.
• Advantages: no chromatic aberration; long focal length with a short, light tube.
Avoid: forgetting the hole in the primary mirror or claiming lower spherical aberration without proof.
Source: CBSE Physics SQP 2024-25 Q24
A discharge lamp contains an enormous number of hydrogen atoms. Each one-electron atom can be excited to different quantum levels. When electrons in different atoms return to lower levels, they emit photons with energies equal to the level gaps, \(E = 13.6\,\text{eV}\,(1/n_f^{2} - 1/n_i^{2})\). The many possible \(n_i \rightarrow n_f\) transitions across the ensemble create the full set of discrete spectral lines, even though every atom supplies only one electron.
Common mistake: imagining one electron makes all lines at once. In reality, many atoms, each in a single state, together produce the spectrum.
• Exploit symmetry before using Gauss’s or Ampere’s law.
• Levitation needs \(qvB = mg\).
• At resonance \(V_L = V_C\); shorting \(C\) changes impedance.
• Interference minima: path difference = \((m+\tfrac12)\lambda\).
• Peak energy release near \(A \approx 60\).
• Capacitance rises as effective gap shrinks.
• Wire field: \(B \propto r\) inside, \(B \propto 1/r\) outside.
• Photon energy shifts stopping potential, not saturation current.
• Forward bias narrows depletion layer, enabling current.
• I²R losses heat conductors—monitor current.
• Generator emf \(e_{\text{max}} = NBA\omega\); mechanical work → electricity.
• Reflecting telescopes avoid chromatic aberration and stay compact.
• Many atoms yield multiple spectral lines, not one electron.
Next steps: revisit weaker chapters first, then drill derivations and numericals daily.