CBSE Physics Study Guide

Overview

This guide covers the CBSE Class 11 and 12 Physics syllabus (NCERT). It is organised by topic with key concepts, equations, and exam-focused advice.

The CBSE Class 12 Physics board exam carries 70 marks (theory) + 30 marks (practical). The theory paper includes MCQs, very-short-answer (1-mark), short-answer (2-3 marks), and long-answer (5-mark) questions. Numerical problems carry approximately 15—20 marks.

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1. Mechanics

1.1 Laws of Motion

Newton’s Laws:

1. An object remains at rest or in uniform motion unless acted upon by a net external force.
2. $\vec{F}_{\text{net}} = m\vec{a}$
3. Every action has an equal and opposite reaction.

Friction. Static friction: $f_s \leq \mu_s N$. Kinetic friction: $f_k = \mu_k N$.

Connected body problems. Draw free-body diagrams for each mass separately. Apply Newton’s second law to each, then use the constraint (common acceleration for connected strings).

1.2 Work, Energy, and Power

Work: $W = \vec{F} \cdot \vec{d} = Fd\cos\theta$

Work-energy theorem: $W_{\text{net}} = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

Kinetic energy: $K = \frac{1}{2}mv^2$

Potential energy (gravity): $U = mgh$

Conservation of mechanical energy: $K + U = \text{constant}$ (when only conservative forces act)

Power: $P = \frac{dW}{dt} = \vec{F} \cdot \vec{v}$

1.3 Rotational Motion

Moment of inertia (key results):

Body	$I$ (about given axis)
Thin rod (centre)	$\frac{1}{12}ML^2$
Thin rod (end)	$\frac{1}{3}ML^2$
Solid cylinder/disc	$\frac{1}{2}MR^2$
Hollow cylinder	$MR^2$
Solid sphere	$\frac{2}{5}MR^2$
Hollow sphere	$\frac{2}{3}MR^2$

Parallel axis theorem: $I = I_{\text{cm}} + Md^2$

Perpendicular axis theorem (2D lamina): $I_z = I_x + I_y$

Torque: $\vec{\tau} = \vec{r} \times \vec{F}$

Angular momentum: $\vec{L} = I\vec{\omega}$; conservation: $\vec{\tau}_{\text{net}} = 0 \Rightarrow \vec{L} = \text{const}$

Rolling without slipping: $v = R\omega$

1.4 Gravitation

Newton’s law: $F = \frac{Gm_1 m_2}{r^2}$

Gravitational field: $\vec{g} = -\frac{GM}{r^2}\hat{r}$

Escape velocity: $v_e = \sqrt{\frac{2GM}{R}}$

Orbital velocity: $v_o = \sqrt{\frac{GM}{r}}$

Kepler’s third law: $T^2 \propto r^3$

1.5 Fluid Mechanics

Bernoulli’s equation: $P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$

Continuity equation: $A_1 v_1 = A_2 v_2$

Surface tension: $F = \gamma L$; capillary rise: $h = \frac{2\gamma\cos\theta}{\rho g r}$

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2. Thermodynamics

2.1 Thermal Properties

Linear expansion: $\Delta L = \alpha L_0 \Delta T$

Heat transfer mechanisms:

• Conduction: $\frac{dQ}{dt} = -kA\frac{dT}{dx}$
• Convection: transfer by fluid motion
• Radiation: Stefan-Boltzmann: $P = \sigma A T^4$

Specific heat capacity: $Q = mc\Delta T$

Latent heat: $Q = mL$ (phase change at constant temperature)

2.2 Laws of Thermodynamics

Zeroth law: If $A$ is in thermal equilibrium with $B$ and $C$, then $B$ and $C$ are in thermal equilibrium.

First law: $\Delta Q = \Delta U + \Delta W$

• $\Delta Q$: heat supplied to the system
• $\Delta U$: change in internal energy
• $\Delta W$: work done by the system; for isobaric process: $W = P\Delta V$

Second law:

• Kelvin-Planck: No process converts heat entirely into work.
• Clausius: Heat cannot flow from cold to hot without external work.

Thermodynamic processes:

Process	$\Delta U$	$\Delta Q$	$\Delta W$
Isothermal	$0$	$nRT\ln(V_f/V_i)$	$nRT\ln(V_f/V_i)$
Adiabatic	$nC_V\Delta T$	$0$	$-\Delta U$
Isobaric	$nC_V\Delta T$	$nC_P\Delta T$	$P\Delta V$
Isochoric	$nC_V\Delta T$	$nC_V\Delta T$	$0$

2.3 Kinetic Theory of Gases

Ideal gas equation: $PV = nRT$

Root mean square speed: $v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3RT}{M}}$

Mean free path: $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$

Degrees of freedom and specific heats:

• Monatomic: $f = 3$, $C_V = \frac{3}{2}R$, $C_P = \frac{5}{2}R$, $\gamma = \frac{5}{3}$
• Diatomic: $f = 5$, $C_V = \frac{5}{2}R$, $C_P = \frac{7}{2}R$, $\gamma = \frac{7}{5}$

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3. Electrostatics

3.1 Coulomb’s Law

$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$

where $\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9\;\text{N m}^2\text{ C}^{-2}$.

3.2 Electric Field and Potential

Electric field: $\vec{E} = \frac{\vec{F}}{q}$

Field due to a point charge: $\vec{E} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{r}$

Electric potential: $V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$

Relation: $\vec{E} = -\nabla V$; in one dimension: $E = -\frac{dV}{dx}$

Gauss’s law: $\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}$

3.3 Capacitance

Parallel plate capacitor: $C = \frac{\varepsilon_0 A}{d}$

With dielectric: $C = \frac{\varepsilon_0 \kappa A}{d}$ where $\kappa$ is the dielectric constant.

Energy stored: $U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$

Series: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$

Parallel: $C_{\text{eq}} = C_1 + C_2$

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4. Current Electricity

4.1 Ohm’s Law and Resistance

$V = IR$

Resistivity: $R = \rho\frac{L}{A}$

Temperature dependence: $R_T = R_0[1 + \alpha(T - T_0)]$

4.2 Kirchhoff’s Laws

1. Junction rule: $\sum I_{\text{in}} = \sum I_{\text{out}}$ (conservation of charge)
2. Loop rule: $\sum V = 0$ around any closed loop (conservation of energy)

4.3 Wheatstone Bridge

At balance: $\frac{P}{Q} = \frac{R}{S}$ (galvanometer reads zero current).

4.4 Measuring Instruments

Ammeter: very low resistance, connected in series.

Voltmeter: very high resistance, connected in parallel.

Conversion:

• Ammeter from galvanometer: $R_s = \frac{I_g R_g}{I - I_g}$ (shunt resistance)
• Voltmeter from galvanometer: $R = \frac{V}{I_g} - R_g$ (series resistance)

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5. Magnetism and Electromagnetic Induction

5.1 Biot-Savart Law

$d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{l} \times \hat{r}}{r^2}$

Field due to a long straight wire: $B = \frac{\mu_0 I}{2\pi r}$

Field at centre of circular loop: $B = \frac{\mu_0 I}{2R}$

5.2 Ampere’s Law

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$

Inside a solenoid: $B = \mu_0 n I$ where $n = N/L$ is the number of turns per unit length.

5.3 Electromagnetic Induction

Faraday’s law: $\mathcal{E} = -\frac{d\Phi_B}{dt}$

Magnetic flux: $\Phi_B = \vec{B} \cdot \vec{A} = BA\cos\theta$

Lenz’s law: The induced emf opposes the change that produces it.

Self-inductance: $\mathcal{E} = -L\frac{dI}{dt}$

Energy stored in inductor: $U = \frac{1}{2}LI^2$

5.4 AC Circuits

RMS voltage and current: $V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$; $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$

Impedance in series RLC: $Z = \sqrt{R^2 + (X_L - X_C)^2}$

where $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$.

Resonance: $X_L = X_C \Rightarrow \omega_0 = \frac{1}{\sqrt{LC}}$

Power: $P = V_{\text{rms}}I_{\text{rms}}\cos\phi$ where $\cos\phi = R/Z$ is the power factor.

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6. Optics

6.1 Reflection and Refraction

Law of reflection: $\theta_i = \theta_r$

Snell’s law: $n_1\sin\theta_1 = n_2\sin\theta_2$

Total internal reflection: occurs when $\theta_1 > \theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right)$ for $n_1 > n_2$.

6.2 Lenses and Optical Instruments

Lens maker’s equation: $\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

Thin lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Magnification: $m = \frac{v}{u}$

Power of a lens: $P = \frac{1}{f}$ (in dioptres when $f$ is in metres)

Magnifying glass: $M = 1 + \frac{D}{f}$ (at near point)

Compound microscope: $M = \frac{v_o}{u_o} \times \frac{D}{u_e}$

Astronomical telescope: $M = \frac{f_o}{f_e}$

6.3 Wave Optics

Interference (Young’s double slit):

$y = \frac{n\lambda D}{d} \quad \text{(bright fringes)} \qquad y = \left(n + \frac{1}{2}\right)\frac{\lambda D}{d} \quad \text{(dark fringes)}$

Fringe width: $\beta = \frac{\lambda D}{d}$

Diffraction: Single slit minima at $a\sin\theta = n\lambda$; central maximum width $= 2\beta$.

Polarisation: Malus’s law: $I = I_0\cos^2\theta$

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7. Modern Physics

7.1 Dual Nature of Radiation and Matter

Photoelectric effect (Einstein’s equation):

$K_{\max} = h\nu - \phi$

where $\phi = h\nu_0$ is the work function and $\nu_0$ is the threshold frequency.

De Broglie wavelength: $\lambda = \frac{h}{p} = \frac{h}{mv}$

7.2 Atoms

Bohr model:

• $mvr = \frac{nh}{2\pi}$ (quantised angular momentum)
• $r_n = \frac{n^2 h^2}{4\pi^2 m k Z e^2} = 0.529\frac{n^2}{Z}\;\text{\AA}$ (Bohr radius)
• $E_n = -13.6\frac{Z^2}{n^2}\;\text{eV}$

7.3 Nuclei

Mass-energy equivalence: $E = \Delta m\, c^2$

Nuclear binding energy: the energy equivalent of the mass defect of a nucleus.

Radioactive decay: $N = N_0 e^{-\lambda t}$; half-life: $t_{1/2} = \frac{\ln 2}{\lambda}$

Fission: heavy nucleus splits, releasing energy (used in nuclear reactors).

Fusion: light nuclei combine, releasing energy (powers stars).

7.4 Semiconductor Devices

• Intrinsic semiconductor: equal electrons and holes; conductivity increases with temperature
• Extrinsic (n-type): donor impurities add free electrons
• Extrinsic (p-type): acceptor impurities add holes
• p-n junction: diode (rectification), LED, photodiode
• Transistor: n-p-n or p-n-p; used for amplification and switching
• Logic gates: AND, OR, NOT, NAND, NOR

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8. Key Equations

Topic	Equation
Newton’s second law	$\vec{F} = m\vec{a}$
Kinetic energy	$K = \frac{1}{2}mv^2$
Gravitational force	$F = Gm_1 m_2/r^2$
First law of thermodynamics	$\Delta Q = \Delta U + \Delta W$
Ideal gas law	$PV = nRT$
Coulomb’s law	$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$
Ohm’s law	$V = IR$
Faraday’s law	$\mathcal{E} = -d\Phi_B/dt$
Lens formula	$1/v - 1/u = 1/f$
Photoelectric effect	$K_{\max} = h\nu - \phi$
De Broglie wavelength	$\lambda = h/mv$
Bohr energy levels	$E_n = -13.6\,Z^2/n^2\;\text{eV}$
Radioactive decay	$N = N_0 e^{-\lambda t}$

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9. Exam Tips

1. Draw diagrams for every problem. Free-body diagrams, circuit diagrams, and ray diagrams earn marks and clarify your thinking.
2. Show dimensional analysis checks. If you are unsure of a derived formula, substitute SI units to verify dimensional consistency.
3. Use SI units consistently. Convert everything to metres, kilograms, seconds, amperes, kelvin before calculating. Very common source of lost marks.
4. State assumptions. In mechanics problems, explicitly note “frictionless surface”, “no air resistance”, etc. This demonstrates understanding.
5. Memorise standard values. $g = 9.8\;\text{m/s}^2$, $e = 1.6 \times 10^{-19}\;\text{C}$, $c = 3 \times 10^8\;\text{m/s}$, $h = 6.63 \times 10^{-34}\;\text{J s}$.
6. Practise numericals from NCERT exemplar. CBSE numericals are in most cases NCERT-level with minor modifications.
7. Label circuit diagrams evidently. Show current direction, polarity of cells, and junction points.

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Common Pitfalls

1. Forgetting to convert units. Using cm instead of m, or grams instead of kg, throws off the entire calculation.
2. Incorrect sign conventions in lenses and mirrors. Always state your convention (e.g., distances measured from the lens/mirror along the incident ray direction are positive).
3. Mixing up emf and terminal voltage. Terminal voltage $V = \mathcal{E} - Ir$ for a discharging cell; internal resistance $r$ reduces the terminal voltage.
4. Applying Gauss’s law without symmetry. The law is only directly solvable when the field has sufficient symmetry (spherical, cylindrical, or planar).
5. Ignoring Lenz’s law in EMI problems. The induced current must oppose the change in flux; getting the direction wrong loses marks.
6. Confusing $B$ and $H$ fields. $B$ is magnetic flux density (tesla), $H = B/\mu_0$ is magnetic field strength (A/m). In most CBSE problems, $B$ is what is used.
7. Writing “heat is lost” instead of specifying the mechanism. In thermodynamics, distinguish between heat transfer by conduction, convection, and radiation.

Worked Examples

Example 1: Kirchhoff’s Laws Problem

Problem: Two cells of emf 2V (internal resistance 0.5 ohm) and 4V (internal resistance 1 ohm) are connected in parallel across a 5 ohm external resistor. Find the current through each cell. Solution: Let I1 flow from the 2V cell and I2 from the 4V cell. By Kirchhoff’s junction rule: I1 + I2 = I (current through 5 ohm resistor). Loop equation for 2V cell: 2 - 0.5I1 = 5(I1 + I2). Loop equation for 4V cell: 4 - I2 = 5(I1 + I2). From second equation: 4 - I2 = 5I1 + 5I2 => 4 = 5I1 + 6I2. From first: 2 - 0.5I1 = 5I1 + 5I2 => 2 = 5.5I1 + 5I2. Subtracting: 2 = -0.5I1 + I2 => I2 = 2 + 0.5I1. Substituting into 4 = 5I1 + 6(2 + 0.5I1) = 5I1 + 12 + 3I1 = 8I1 + 12. So 8I1 = -8, I1 = -1A. The 2V cell is being charged. I2 = 2 + 0.5(-1) = 1.5A. Current through 5 ohm: 0.5A.

Example 2: Young’s Double Slit Calculation

Problem: Light of wavelength 600 nm passes through double slits separated by 0.1 mm. The screen is 1.5 m away. Find the fringe width. Solution: Fringe width beta = lambda * D / d = (600 x 10^-9)(1.5) / (0.1 x 10^-3) = 9 x 10^-3 m = 9 mm. The bright fringes are 9 mm apart.

Example 3: Photoelectric Effect

Problem: Light of wavelength 400 nm falls on a metal with work function 2.0 eV. Find the maximum kinetic energy of emitted photoelectrons. (h = 6.63 x 10^-34 J s, c = 3 x 10^8 m/s, 1 eV = 1.6 x 10^-19 J) Solution: E_photon = hc/lambda = (6.63 x 10^-34)(3 x 10^8) / (400 x 10^-9) = 4.97 x 10^-19 J = 3.11 eV. K_max = E_photon - phi = 3.11 - 2.0 = 1.11 eV. Since K_max > 0, photoelectrons are emitted.

Summary

CBSE Physics covers mechanics (Newton’s laws, work-energy, rotational motion, gravitation, fluids), thermodynamics (laws, kinetic theory), electrostatics (Coulomb’s law, Gauss’s law, capacitance), current electricity (Ohm’s law, Kirchhoff’s laws, Wheatstone bridge), magnetism and electromagnetic induction (Biot-Savart, Faraday’s law, AC circuits), optics (reflection, refraction, wave optics), and modern physics (photoelectric effect, atomic structure, nuclear physics, semiconductors). The theory paper is 70 marks and practicals are 30 marks. Key exam strategies include drawing diagrams, using SI units consistently, showing dimensional analysis checks, and practising NCERT exemplar numericals.