| Chapter | Questions / Marks | Your Score (%) | Priority Action |
|---|---|---|---|
| Electrostatics | 4 / 10 | 85 | Revise formulas |
| Magnetism | 5 / 12 | 78 | More MCQs |
| EMI & AC | 6 / 14 | 60 | Solve numericals |
| EM Waves | 2 / 4 | 90 | Quick revision |
| Optics | 8 / 16 | 55 | Concept videos |
| Dual Nature | 4 / 8 | 70 | Past papers |
| Atoms & Nuclei | 2 / 5 | 88 | Maintain pace |
| Semiconductors | 2 / 5 | 65 | Clarify basics |
Source: CBSE Physics Sample Paper 2024-25 & Your Mock Test Analytics
Class 12 Physics • CBSE
Using Gauss’s law, derive the electric field \(E\) at a distance \(r\) from an infinitely long straight wire carrying uniform line charge density \(\lambda\).
Try on your own first. Reveal hints one by one if stuck.
Choose a cylindrical Gaussian surface of radius \(r\) and length \(L\) coaxial with the wire.
By cylindrical symmetry, \(E\) is radial and constant on the curved surface; flux through flat end-caps is zero.
\(E(2\pi rL)=\lambda L/\varepsilon_0\) ⇒ \(E=\dfrac{\lambda}{2\pi\varepsilon_0 r}\).
Net electric flux through a closed surface equals enclosed charge divided by \(\varepsilon_0\). Powerful for high-symmetry cases.
For an infinite line charge, field magnitude depends only on radial distance; direction is radial outward (or inward for negative \(\lambda\)).
CBSE Class 12 Physics
An electron describes a circle of radius \(r_0\) with speed \(v_0\) and period \(T_0\) in a uniform magnetic field \(B\). Predict qualitatively how the radius \(r\) and period \(T\) change if its speed is doubled to \(2v_0\).
Try first, then reveal hints one at a time.
A charge moving with velocity \( \mathbf{v}\) in a field \( \mathbf{B}\) experiences \( \mathbf{F}=q\,\mathbf{v}\times\mathbf{B}\) perpendicular to motion.
When \( \mathbf{F}\) is always perpendicular to \( \mathbf{v}\), speed stays constant and the particle follows a circle with constant angular frequency.
Series LCR • Impedance & Reactance
In a series LCR circuit, the RMS voltages across R, C and L are each 10 V. The capacitor is suddenly short-circuited. Estimate the new RMS voltage across the inductor.
Try to solve first. Reveal hints as needed.
At resonance \(X_L = X_C\); impedance is purely resistive and minimum, equal to \(R\).
Removing a reactive element adds its reactance vectorially: \(Z = \sqrt{R^{2}+X^{2}}\). Current and element voltages adjust accordingly.
CBSE Class 12 • Physics
In a Young’s double-slit experiment, light of wavelengths 400 nm and 600 nm is used together. Find the smallest distance from the central bright fringe where both colours give a dark (minimum-intensity) fringe simultaneously.
Reveal hints one at a time only if you need them.
In a YDS experiment, dark fringes occur when \( \Delta = m\lambda \) with m = 1, 2, 3… .
Two wavelengths give a common dark fringe when integer orders satisfy \( m\lambda_{1}=n\lambda_{2} \).
Binding Energy per Nucleon & Nuclear Stability
From the binding-energy curve, nuclei W (190), X (90), Y (60) and Z (30) are marked. Which one favours (i) fission and (ii) fusion? State why.
Binding-energy curve (schematic)
Solve first, then reveal hints in order.
The BE/A curve peaks near A ≈ 56 with about 8.8 MeV per nucleon.
Nuclei with A > 140 gain stability by splitting into fragments nearer the peak.
Very light nuclei (A < 10) release energy when they fuse toward A ≈ 56.
Higher BE/A means stronger binding and greater nuclear stability. The maximum lies near iron (A≈56).
A nucleus will fission or fusion if products sit higher on the BE/A curve, releasing the difference as energy.
Class 12 Physics • CBSE
A parallel-plate capacitor has plate area \(A\) and separation \(d\). Derive its capacitance when (i) a dielectric slab of relative permittivity \(\kappa\) and thickness \(t\), and (ii) a metallic slab of thickness \(t\,(<d)\) are fully inserted between the plates. Which insertion yields the larger capacitance? Give reason.
Reveal one hint at a time.
When different media fill the space between plates along the field, treat each region as a separate capacitor connected in series.
The dielectric constant \(\kappa\) multiplies permittivity, reducing electric field and increasing capacitance by a factor \(\kappa\) for the same geometry.
CBSE Class 12 Physics
A parallel-plate capacitor of area 0.001 m² and plate gap 0.0001 m is being charged so that the voltage across it rises at \(10^{8}\,\text{V s}^{-1}\). Estimate the displacement current flowing through the gap.
Reveal hints step-by-step only if needed.
In free space \( \nabla \times \mathbf{B} = \mu_0\left(\mathbf{J} + \varepsilon_0 \dfrac{\partial \mathbf{E}}{\partial t}\right) \). The extra term represents displacement current, ensuring continuity in time-varying fields.
Between plates, no real charges move. A changing electric field substitutes with \(I_d\), making Ampère’s law valid for the loop linking the capacitor.
Photoelectric Effect · Einstein Equation
Photoelectric I–V curves A, B and C show different saturation currents and stopping potentials. Relate these differences to changes in (a) light intensity and (b) light frequency for the same metal surface.
Reveal hints one by one only when needed.
Energy of one photon \(h\nu\) is partitioned into work function \( \phi \) and kinetic energy \( eV_0 \). Thus \(V_0\) tracks frequency.
Current rises quickly, saturates when every emitted electron is collected, then reverses sign past stopping potential.
CBSE Class 12 Physics
Sketch the energy-band diagram of a P-type semiconductor at (a) 0 K and (b) room temperature. For the circuit shown, will the bulb glow when the switch is (i) open and (ii) closed? Justify.
Circuit with PN diode, switch and bulb
Try on your own first. Reveal hints one by one if needed.
Acceptor level lies just above valence band. Ionisation creates holes, shifting Fermi level slightly above the valence band as temperature rises.
Under forward bias, majority carriers cross the junction and current flows; reverse bias widens depletion region and blocks current.
Class 12 • CBSE Physics
A heater draws 10 A from a 100 V battery whose internal resistance is 1 Ω at 20 °C. When the heater reaches 320 °C, calculate the power dissipated inside the battery. Take the temperature coefficient of resistance \( \alpha = 3.7 \times 10^{-4}\,\text{°C}^{-1} \).
Solve first, then reveal hints one by one.
For small ranges, \(R = R_0(1 + \alpha \Delta T)\); resistance rises with temperature for metals.
A real battery behaves as \(E\) in series with \(r\); heat inside it equals \(I^{2} r\), reducing useful output.
CBSE Class 12 Physics
In an AC generator, a coil with N turns and area A spins at angular speed ω in a uniform magnetic field B. Derive the instantaneous emf and name the energy source that keeps the generator running.
[No diagram provided]
Reveal hints one at a time as you need them.
Write flux through the rotating coil: \( \Phi(t)=N B A \cos \omega t \).
Use \( \varepsilon(t)= -\dfrac{d\Phi}{dt} \) to differentiate the flux with respect to time.
After differentiating, \( \varepsilon(t)=N B A \omega \sin \omega t\). Peak emf \( \varepsilon_0=N B A \omega\). The coil is driven by mechanical energy from a turbine or engine.
Induced emf is proportional to the negative time rate of change of magnetic flux through a circuit.
A coil rotating in a uniform field produces a sinusoidal emf because the projected area facing the field varies as \(\cos \omega t\).
Ray Optics – Telescope Calculations
A refracting telescope has an objective of focal length 15 m and an eyepiece of 1 cm.
(a) Find its angular magnification in normal adjustment.
(b) Calculate the diameter of the Moon’s image formed by the objective lens.
No diagram provided
Try to solve the problem on your own first. If you need help, reveal the hints one by one.
Convert \(f_e = 1\,\text{cm}\) to metres and remember the final image is at infinity in normal adjustment.
Use \(M = f_o / f_e\). With \(f_o = 15\,\text{m}\) and \(f_e = 0.01\,\text{m}\), find \(M\).
The Moon’s angular diameter is \(\theta \approx D/R = 3.48\times10^{6} / 3.8\times10^{8}\,\text{rad}\). Image diameter: \(y = f_o \theta\).
For an astronomical telescope in normal adjustment, \(M = -\dfrac{f_o}{f_e}\). The sign indicates an inverted image; magnitude is used.
A distant object subtends angle \( \theta \). The objective forms a real image of height \(y = f_o \theta\) at its focal plane.
Bohr Model & Atomic Spectra
Why does hydrogen, with only one electron, exhibit many spectral lines in its emission spectrum?
Try to solve the problem on your own first. If you need help, reveal the hints one by one.
A discharge tube holds countless hydrogen atoms, each with its own electron.
Different atoms get excited to different Bohr levels \(n = 2,3,4,\dots\).
Each electron can drop between two levels, releasing photons of energy \(E = 13.6\,(1/n_1^2-1/n_2^2)\:\text{eV}\), creating many lines.
Allowed orbits have fixed energies; electrons jump between them by absorbing or emitting photons.
Downward jumps produce emission lines; energy difference sets photon frequency \( \nu = \Delta E/h \).
Write c, h, e, μ0, ε0 and NA on the answer sheet margin for instant recall.
Always state the equation before substitution to secure partial marks.
Verify unit consistency to spot errors quickly, especially in MCQs.
45 min (A+B), 60 min (C), 35 min (D), 40 min (E); keep last 10 min for review.
Start with sure-shot one-markers to build momentum and confidence.
Practice quick powers-of-ten estimates; calculators are not allowed.