Physics Guide

Class 11 & 12 — mechanics, electricity, waves, optics, and modern physics, explained with simulations and solved problems.

1. Kinematics

Core Idea

Describing Motion

Kinematics is the study of motion without asking why it happens. Three key quantities describe any motion:

Displacement ($s$): Change in position (vector — direction matters)
Velocity ($v$): Rate of change of displacement
Acceleration ($a$): Rate of change of velocity

Equations of Uniformly Accelerated Motion

When acceleration is constant, these four equations fully describe the motion:

$$v = u + at$$ $$s = ut + \tfrac{1}{2}at^2$$ $$v^2 = u^2 + 2as$$ $$s = \tfrac{1}{2}(u + v)\,t$$

Where $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $s$ = displacement, $t$ = time.

Memory trick: "SUVAT" — name the five variables and pick the equation that contains the four you know/want, leaving out the one you don't need.

Example 1 — Car braking to rest

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A car moving at 72 km/h applies brakes and stops in 50 m. Find the deceleration.

Convert speed: $u = 72 \times \frac{1000}{3600} = 20\ \text{m/s}$, $v = 0$

Use $v^2 = u^2 + 2as$: $\;0 = 400 + 2a(50)$

Solve: $a = \dfrac{-400}{100} = -4\ \text{m/s}^2$

Answer: Deceleration = $4\ \text{m/s}^2$. The negative sign confirms the car is slowing down.

Projectile Motion

Key Insight

Horizontal and Vertical are Independent

A projectile has constant horizontal velocity ($a_x = 0$) and constant downward acceleration ($a_y = -g$) simultaneously. Treat them separately.

$$x = u\cos\theta \cdot t \qquad y = u\sin\theta \cdot t - \tfrac{1}{2}g\,t^2$$ $$\text{Range}\ R = \frac{u^2\sin 2\theta}{g} \qquad \text{Max Height}\ H = \frac{u^2\sin^2\theta}{2g}$$ $$\text{Time of flight}\ T = \frac{2u\sin\theta}{g}$$

Projectile Motion Simulator

Speed: 30 m/s Angle: 45°

Set speed & angle, then click Launch.

Example 2 — Ball thrown at 30°

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A ball is projected at $30\ \text{m/s}$ at $30°$ to the horizontal. Find the range and max height. ($g = 10\ \text{m/s}^2$)

Components: $u_x = 30\cos 30° = 15\sqrt{3}\ \text{m/s}$, $u_y = 30\sin 30° = 15\ \text{m/s}$

Max height: $H = \dfrac{u_y^2}{2g} = \dfrac{225}{20} = 11.25\ \text{m}$

Range: $R = \dfrac{u^2\sin 60°}{g} = \dfrac{900\times\frac{\sqrt{3}}{2}}{10} = 45\sqrt{3} \approx 77.9\ \text{m}$

2. Laws of Motion

Newton's Laws

Three Fundamental Laws

First Law (Inertia): A body remains at rest or in uniform motion unless acted upon by an external force.
Second Law: $\vec{F} = m\vec{a}$. The net force equals mass times acceleration.
Third Law: For every action, there is an equal and opposite reaction. Forces always come in pairs.

$$\vec{F}_{net} = m\vec{a} \qquad \text{Impulse} = \vec{F}\,\Delta t = \Delta\vec{p}$$ $$\text{Friction:}\quad f_s \leq \mu_s N \qquad f_k = \mu_k N$$

Free Body Diagram (FBD) Strategy: Isolate the object, draw every force acting on it (not forces it exerts), choose a convenient coordinate system, then apply $\Sigma F = ma$ along each axis.

Example 3 — Block on inclined plane

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A 5 kg block rests on a 37° incline. Coefficient of static friction $\mu_s = 0.75$. Is the block stationary? ($g = 10\ \text{m/s}^2$)

Weight components: Along incline: $mg\sin 37° = 5\times 10\times 0.6 = 30\ \text{N}$

Normal force: $N = mg\cos 37° = 5\times 10\times 0.8 = 40\ \text{N}$

Max static friction: $f_{s,max} = \mu_s N = 0.75\times 40 = 30\ \text{N}$

Since $f_{s,max} = 30\ \text{N} \geq 30\ \text{N}$ (gravity component), the block is just on the verge of slipping — it remains stationary.

3. Work, Energy & Power

Definition

Work Done by a Force

Work is done when a force causes displacement. Only the component of force along the displacement does work.

$$W = \vec{F}\cdot\vec{s} = Fs\cos\theta$$ $$\text{Kinetic Energy:}\quad KE = \tfrac{1}{2}mv^2$$ $$\text{Work-Energy Theorem:}\quad W_{net} = \Delta KE$$ $$\text{Power:}\quad P = \frac{W}{t} = \vec{F}\cdot\vec{v}$$

Conservative Forces & Potential Energy

Energy Conservation

Mechanical Energy is Conserved

For conservative forces (gravity, spring), the total mechanical energy $E = KE + PE$ remains constant.

$$PE_{gravity} = mgh \qquad PE_{spring} = \tfrac{1}{2}kx^2$$ $$KE_1 + PE_1 = KE_2 + PE_2$$

Example 4 — Ball dropped from height

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A 2 kg ball is dropped from 20 m height. Find its velocity just before hitting the ground. ($g = 10\ \text{m/s}^2$)

Initial state: $KE_1 = 0$ (at rest), $PE_1 = mgh = 2\times10\times20 = 400\ \text{J}$

Final state: $PE_2 = 0$ (ground level), $KE_2 = \frac{1}{2}\times2\times v^2 = v^2$

Conservation: $400 = v^2 \Rightarrow v = 20\ \text{m/s}$

4. Gravitation

Newton's Law of Gravitation

Universal Gravitational Force

Every mass attracts every other mass. The force is proportional to the product of masses and inversely proportional to the square of the distance between them.

$$F = G\frac{m_1 m_2}{r^2} \qquad G = 6.674\times10^{-11}\ \text{N\,m}^2\text{kg}^{-2}$$ $$g = \frac{GM_E}{R_E^2} \approx 9.8\ \text{m/s}^2$$

Orbital Motion & Escape Velocity

$$v_{orbit} = \sqrt{\frac{GM}{r}} \qquad T^2 = \frac{4\pi^2}{GM}\,r^3\ \text{(Kepler's 3rd Law)}$$ $$v_{escape} = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} \approx 11.2\ \text{km/s (Earth)}$$

Escape velocity is the minimum speed needed to escape a planet's gravity without further propulsion. It does not depend on the mass or direction of the object.

Example 5 — Orbital period of a satellite

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A satellite orbits Earth at 400 km above the surface. Find its orbital period. ($R_E = 6400\ \text{km}$, $g = 9.8\ \text{m/s}^2$)

Orbital radius: $r = 6400 + 400 = 6800\ \text{km} = 6.8\times10^6\ \text{m}$

Use $T^2 = \dfrac{4\pi^2 r^3}{GM_E}$ and $GM_E = gR_E^2 = 9.8\times(6.4\times10^6)^2 = 4.014\times10^{14}\ \text{m}^3\text{s}^{-2}$

$T = 2\pi\sqrt{\dfrac{r^3}{GM_E}} = 2\pi\sqrt{\dfrac{(6.8\times10^6)^3}{4.014\times10^{14}}} \approx 5559\ \text{s} \approx 92.7\ \text{min}$

5. Oscillations & Waves

Simple Harmonic Motion

SHM — Restoring Force Proportional to Displacement

Any system where the restoring force obeys $F = -kx$ exhibits Simple Harmonic Motion.

$$x(t) = A\cos(\omega t + \phi) \qquad \omega = \sqrt{\frac{k}{m}} = \frac{2\pi}{T}$$ $$T_{spring} = 2\pi\sqrt{\frac{m}{k}} \qquad T_{pendulum} = 2\pi\sqrt{\frac{L}{g}}$$ $$v_{max} = A\omega \qquad a_{max} = A\omega^2$$

Simple Pendulum Simulator

Length: 1.5 m Amplitude: 20°

Period: — s

Wave Properties

Key Relationships

Wave Speed, Frequency, Wavelength

$$v = f\lambda \qquad f = \frac{1}{T} \qquad k = \frac{2\pi}{\lambda}$$ $$y(x,t) = A\sin(kx - \omega t) \quad \text{(travelling wave)}$$ $$v_{sound} = \sqrt{\frac{\gamma P}{\rho}} \approx 340\ \text{m/s in air at 20°C}$$

Wave Type	Description	Example
Transverse	Particles vibrate perpendicular to wave travel	Light, water waves
Longitudinal	Particles vibrate parallel to wave travel	Sound waves
Standing Wave	Superposition of two opposite-direction waves	Guitar string vibration

6. Thermodynamics

Laws of Thermodynamics

Four Fundamental Laws

Zeroth Law: If A is in thermal equilibrium with B, and B with C, then A is in equilibrium with C. (Defines temperature.)
First Law: $\Delta U = Q - W$. Energy is conserved; heat added = increase in internal energy + work done by system.
Second Law: Heat cannot spontaneously flow from a colder body to a hotter body. Entropy of an isolated system never decreases.
Third Law: Entropy approaches a constant minimum as temperature approaches absolute zero.

Thermodynamic Processes

Process	Condition	Work Done	$\Delta U$
Isothermal	$T = \text{const}$	$nRT\ln\dfrac{V_2}{V_1}$	$0$
Adiabatic	$Q = 0$	$\dfrac{P_1V_1 - P_2V_2}{\gamma-1}$	$-W$
Isobaric	$P = \text{const}$	$P\Delta V$	$Q - P\Delta V$
Isochoric	$V = \text{const}$	$0$	$Q$

$$\text{Carnot Efficiency:}\quad \eta = 1 - \frac{T_L}{T_H} \qquad \text{(temperatures in Kelvin)}$$ $$PV = nRT \qquad \text{(Ideal Gas Law)}$$

Example 6 — Carnot Engine efficiency

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A Carnot engine operates between $727°C$ and $27°C$. Find its efficiency and the work done if it absorbs 1000 J per cycle.

Convert: $T_H = 727 + 273 = 1000\ \text{K}$, $T_L = 27 + 273 = 300\ \text{K}$

Efficiency: $\eta = 1 - \dfrac{300}{1000} = 1 - 0.3 = 0.7 = 70\%$

Work per cycle: $W = \eta \times Q_H = 0.7\times1000 = 700\ \text{J}$

1. Electrostatics

Coulomb's Law

Force Between Two Point Charges

The electrostatic force between two charges is proportional to the product of the charges and inversely proportional to the square of the separation.

$$F = k_e\frac{q_1 q_2}{r^2} \qquad k_e = \frac{1}{4\pi\epsilon_0} = 9\times10^9\ \text{N\,m}^2\text{C}^{-2}$$ $$\epsilon_0 = 8.854\times10^{-12}\ \text{F/m}$$

Electric Field & Potential

$$\vec{E} = \frac{\vec{F}}{q_0} = k_e\frac{q}{r^2}\hat{r}$$ $$V = k_e\frac{q}{r} \qquad E = -\frac{dV}{dr}$$ $$U = k_e\frac{q_1 q_2}{r} \qquad W = q(V_A - V_B)$$

Gauss's Law

Electric Flux & Enclosed Charge

The total electric flux through any closed surface equals the enclosed charge divided by $\epsilon_0$.

$$\oint \vec{E}\cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$$

Use Gauss's Law to find $E$ for symmetric charge distributions:

Point charge: $E = \dfrac{kq}{r^2}$
Infinite line charge (linear density $\lambda$): $E = \dfrac{\lambda}{2\pi\epsilon_0 r}$
Infinite plane (surface density $\sigma$): $E = \dfrac{\sigma}{2\epsilon_0}$

Capacitors

$$C = \frac{Q}{V} \qquad C_{parallel} = \epsilon_0\frac{A}{d}$$ $$C_{series}: \frac{1}{C_{eq}} = \sum\frac{1}{C_i} \qquad C_{parallel}: C_{eq} = \sum C_i$$ $$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{QV}{2}$$

Electric Field Lines Visualizer

Click a configuration to visualize electric field lines.

Example 1 — Force between two charges

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Two charges $q_1 = +3\ \mu\text{C}$ and $q_2 = -4\ \mu\text{C}$ are placed 0.2 m apart. Find the force.

$F = k_e\dfrac{|q_1||q_2|}{r^2} = 9\times10^9\times\dfrac{3\times10^{-6}\times4\times10^{-6}}{(0.2)^2}$

$F = 9\times10^9\times\dfrac{12\times10^{-12}}{0.04} = 9\times10^9\times3\times10^{-10} = 2.7\ \text{N}$

Since charges are opposite, the force is attractive.

2. Current Electricity

Ohm's Law & Resistance

V = IR

Current through a conductor is proportional to the potential difference across it (for ohmic materials).

$$I = \frac{Q}{t} \qquad V = IR \qquad R = \rho\frac{L}{A}$$ $$P = VI = I^2R = \frac{V^2}{R}$$

Kirchhoff's Laws

Circuit Analysis

KCL and KVL

KCL (Junction Rule): $\sum I_{in} = \sum I_{out}$ — Charge is conserved at every junction.
KVL (Loop Rule): $\sum V = 0$ — The algebraic sum of potential differences around any closed loop is zero.

$$R_{series} = R_1 + R_2 + \cdots \qquad \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$$ $$\text{EMF:}\quad \varepsilon = I(R + r) \qquad V_{terminal} = \varepsilon - Ir$$

Example 2 — Resistors in a mixed circuit

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Two resistors $R_1 = 6\ \Omega$ and $R_2 = 3\ \Omega$ are in parallel, and this combination is in series with $R_3 = 4\ \Omega$. Battery EMF = 10 V, internal resistance = 1 Ω. Find total current.

Parallel combination: $R_{12} = \dfrac{6\times3}{6+3} = 2\ \Omega$

Total external resistance: $R_{ext} = R_{12} + R_3 = 2 + 4 = 6\ \Omega$

Total current: $I = \dfrac{\varepsilon}{R_{ext}+r} = \dfrac{10}{6+1} = \dfrac{10}{7} \approx 1.43\ \text{A}$

3. Magnetic Effects of Current

Biot-Savart & Ampere's Law

Magnetic Field from Currents

$$d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2} \qquad \mu_0 = 4\pi\times10^{-7}\ \text{T\,m/A}$$ $$\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc} \quad \text{(Ampere's Law)}$$

Common Magnetic Field Formulas

Configuration	Magnetic Field
Long straight wire (distance $r$)	$B = \dfrac{\mu_0 I}{2\pi r}$
Circular loop (centre, radius $R$)	$B = \dfrac{\mu_0 I}{2R}$
Solenoid (n turns/m, inside)	$B = \mu_0 nI$
Toroid (N turns, radius $r$)	$B = \dfrac{\mu_0 NI}{2\pi r}$

Force on a Charge & Current

Lorentz Force Law

$$\vec{F} = q(\vec{E} + \vec{v}\times\vec{B}) \qquad F_{wire} = I\vec{L}\times\vec{B}$$ $$\text{Cyclotron radius:}\quad r = \frac{mv}{qB}$$

4. Electromagnetic Induction

Faraday's & Lenz's Laws

Induced EMF from Changing Flux

$$\Phi_B = \vec{B}\cdot\vec{A} = BA\cos\theta$$ $$\varepsilon = -\frac{d\Phi_B}{dt} = -N\frac{d\Phi}{dt} \quad \text{(Faraday's Law)}$$ $$\varepsilon_{motional} = Bvl \qquad \varepsilon_{self} = -L\frac{dI}{dt}$$ $$U_L = \frac{1}{2}LI^2 \qquad L_{solenoid} = \mu_0 n^2 V$$

Lenz's Law: The induced current always flows in a direction that opposes the change in flux that caused it — a direct consequence of energy conservation.

Example 3 — EMF in a moving rod

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A rod of length 0.5 m moves at 4 m/s perpendicular to a magnetic field of 0.2 T. Find the induced EMF.

$\varepsilon = Bvl = 0.2\times4\times0.5 = 0.4\ \text{V}$

If the rod has resistance $R = 2\ \Omega$, the induced current $I = \varepsilon/R = 0.2\ \text{A}$.

5. Optics

Ray Optics

Mirror & Lens Formulas

Sign Convention: distances measured from the optical centre / pole

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \qquad m = -\frac{v}{u} = \frac{h_i}{h_o}$$ $$\text{Mirror: } f = \frac{R}{2} \qquad \text{Lens Maker's: } \frac{1}{f} = (n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

Refraction & Total Internal Reflection

Snell's Law

$$n_1\sin\theta_1 = n_2\sin\theta_2$$ $$n = \frac{c}{v} \qquad \sin\theta_c = \frac{n_2}{n_1} \quad (n_1 > n_2)$$

Refraction Simulator

Incident angle: 40° n₂: 1.5

Wave Optics

Interference & Diffraction

Young's Double-Slit Experiment (YDSE)

$$\text{Bright fringes: } d\sin\theta = m\lambda \quad (m = 0,\pm1,\pm2,\ldots)$$ $$\text{Dark fringes: } d\sin\theta = (m+\tfrac{1}{2})\lambda$$ $$\text{Fringe width: } \beta = \frac{\lambda D}{d}$$

Example 4 — Fringe width in YDSE

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In a YDSE, slit separation $d = 0.5\ \text{mm}$, screen distance $D = 1\ \text{m}$, wavelength $\lambda = 600\ \text{nm}$. Find fringe width.

$\beta = \dfrac{\lambda D}{d} = \dfrac{600\times10^{-9}\times1}{0.5\times10^{-3}}$

$\beta = \dfrac{6\times10^{-7}}{5\times10^{-4}} = 1.2\times10^{-3}\ \text{m} = 1.2\ \text{mm}$

6. Modern Physics

Dual Nature of Matter & Radiation

Wave-Particle Duality

Photon energy: $E = hf = \dfrac{hc}{\lambda}$, where $h = 6.626\times10^{-34}\ \text{J\,s}$
Photoelectric effect: $KE_{max} = hf - \phi$ (work function $\phi$)
de Broglie wavelength: $\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$
Heisenberg Uncertainty: $\Delta x\,\Delta p \geq \dfrac{h}{4\pi}$

Atomic Structure — Bohr Model

$$r_n = \frac{n^2 a_0}{Z} \qquad a_0 = 0.529\ \text{Å} \quad \text{(Bohr radius)}$$ $$E_n = -\frac{13.6\,Z^2}{n^2}\ \text{eV} \qquad \frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

$R = 1.097\times10^7\ \text{m}^{-1}$ (Rydberg constant). Spectral series: Lyman ($n_1=1$), Balmer ($n_1=2$), Paschen ($n_1=3$).

Nuclear Physics

Radioactivity & Nuclear Energy

Decay Law & Mass-Energy Equivalence

$$N(t) = N_0 e^{-\lambda t} \qquad t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}$$ $$E = mc^2 \qquad \text{Binding Energy} = \Delta m\cdot c^2$$ $$\text{Q-value} = (m_{reactants} - m_{products})\cdot c^2$$

Radiation	Symbol	Charge	Penetration
Alpha ($\alpha$)	$^4_2\text{He}$	+2	Stopped by paper
Beta ($\beta^-$)	$e^-$	-1	Few mm of aluminium
Gamma ($\gamma$)	$h\nu$	0	Thick lead / concrete

Example 5 — Radioactive decay

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A radioactive sample has a half-life of 5 years. How much of an 80 g sample remains after 20 years?

Number of half-lives: $n = \dfrac{20}{5} = 4$

$N = N_0\left(\dfrac{1}{2}\right)^n = 80\times\left(\dfrac{1}{2}\right)^4 = 80\times\dfrac{1}{16} = 5\ \text{g}$

Example 6 — Photoelectric effect

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Light of wavelength 300 nm falls on a metal with work function 2 eV. Find the maximum kinetic energy of ejected electrons. ($h = 6.63\times10^{-34}\ \text{J\,s}$, $c = 3\times10^8\ \text{m/s}$)

Photon energy: $E = \dfrac{hc}{\lambda} = \dfrac{6.63\times10^{-34}\times3\times10^8}{300\times10^{-9}} = 6.63\times10^{-19}\ \text{J} = 4.14\ \text{eV}$

$KE_{max} = E - \phi = 4.14 - 2 = 2.14\ \text{eV}$