Physics — Complete Formula Sheet

$v$: Linear speed (m s⁻¹) · $\omega$: Angular speed (rad s⁻¹) · $r$: Radius of the circular path (m) · $T$: Period of revolution (s) · $n$: Frequency of revolution (Hz = rev s⁻¹)

Valid for any point at distance r from the axis. In vector form v⃗ = ω⃗ × r⃗. ω is in rad/s; n is the frequency in rev/s (Hz).

$a_r$: Centripetal acceleration (m s⁻²) · $v$: Linear speed (m s⁻¹) · $\omega$: Angular speed (rad s⁻¹) · $r$: Radius (m)

Directed towards the centre. Present even in uniform circular motion because the direction of velocity keeps changing. Unit: m s⁻².

$a_t$: Tangential acceleration (m s⁻²) · $\alpha$: Angular acceleration (rad s⁻²) · $a_r$: Radial (centripetal) acceleration (m s⁻²)

When the speed changes, a tangential component a_t appears in addition to a_r. The two are perpendicular, so they add as a right-angled triangle.

$F$: Centripetal force (N) · $m$: Mass of the body (kg) · $v$: Linear speed (m s⁻¹) · $r$: Radius (m)

The net force towards the centre required to keep a body on a circular path. It is provided by tension, friction, gravity or a component of the normal reaction. Unit: N.

${\mu }_s$: Coefficient of static friction (no unit) · $r$: Radius of the curve (m) · $g$: Acceleration due to gravity (m s⁻²)

On an unbanked road the only horizontal force is static friction, so μ_s·mg ≥ mv²/r. The safe speed does not depend on the mass of the vehicle.

$\theta$: Angle of banking (with the horizontal) · $v$: Most safe speed (m s⁻¹) · $r$: Radius of the curve (m)

At this speed no friction is needed. The horizontal component of the normal reaction alone provides the centripetal force. Independent of the vehicle's mass.

${\mu }_s$: Coefficient of static friction · $\theta$: Angle of banking

At v_max the friction acts down the slope. The minimum speed (friction up the slope) is v_min = √[rg(tan θ − μ_s)/(1 + μ_s tan θ)]. Putting μ_s = 0 gives back tan θ = v²/rg.

$T$: Period of revolution (s) · $L$: Length of the string (m) · $\theta$: Angle made by the string with the vertical · $r$: Radius of the horizontal circle = L sin θ (m)

The bob moves in a horizontal circle of radius r = L sin θ. L cos θ is the vertical height of the point of suspension above the circle. Period does not depend on the mass of the bob.

$r$: Radius of the vertical circle (m) · $g$: Acceleration due to gravity (m s⁻²)

These are the minimum speeds for a body on a string (or a vehicle in a loop) to just complete the vertical circle. v_mid is at the horizontal position, level with the centre.

$T_{bottom}$: Tension at the lowest point (N) · $T_{top}$: Tension at the highest point (N) · $m$: Mass of the body (kg)

Holds for any speed, as long as the string stays taut over the full circle. Here T means tension, not period.

$I$: Moment of inertia about the given axis (kg m²) · $m_i$: Mass of the i-th particle (kg) · $r_i$: Perpendicular distance of the i-th particle from the axis (m) · $M$: Total mass of the body (kg) · $K$: Radius of gyration (m)

Moment of inertia is the rotational analogue of mass. The radius of gyration K is the distance from the axis at which the whole mass could be placed to give the same I. SI unit of I: kg m².

$M$: Mass of the body (kg) · $R$: Radius of ring, disc or sphere (m) · $L$: Length of the rod (m)

Axes: ring and disc about their own axis (through the centre, perpendicular to the plane); thin rod about a perpendicular axis through its centre; solid sphere about a diameter. Thin hollow sphere about a diameter: ⅔MR².

$I_O$: MI about the given axis through O (kg m²) · $I_C$: MI about the parallel axis through the centre of mass (kg m²) · $h$: Distance between the two parallel axes (m)

One axis MUST pass through the centre of mass C; the other is parallel to it at a distance h. I_C is the minimum MI among all parallel axes. Example: rod about one end = ML²/12 + M(L/2)² = ML²/3.

$I_Z$: MI about the axis perpendicular to the plane (kg m²) · $I_X$: MI about the X axis in the plane (kg m²) · $I_Y$: MI about the Y axis in the plane (kg m²)

Valid only for a plane lamina (ring, disc, plate). X and Y lie in the plane of the lamina, Z is perpendicular to it, and all three meet at one point. Example: disc about a diameter = ½ × ½MR² = ¼MR².

$E_{rot}$: Rotational kinetic energy (J) · $I$: Moment of inertia (kg m²) · $\omega$: Angular speed (rad s⁻¹)

The rotational analogue of ½mv². Doubling ω makes the rotational energy four times larger. Unit: J.

$\vec{L}$: Angular momentum (kg m² s⁻¹) · $\vec{r}$: Position vector from the axis (m) · $\vec{p}$: Linear momentum (kg m s⁻¹)

L = Iω applies to a rigid body rotating about a fixed axis. SI unit: kg m² s⁻¹ (equivalently J s).

$\tau$: Torque (N m) · $I$: Moment of inertia (kg m²) · $\alpha$: Angular acceleration (rad s⁻²)

Rotational analogue of F = ma. τ = Iα holds when I is constant. Unit of torque: N m.

$I_1,I_2$: Initial and final moment of inertia (kg m²) · ${\omega }_1,{\omega }_2$: Initial and final angular speed (rad s⁻¹ or rpm, same unit on both sides)

If no external torque acts, L stays constant. A spinning dancer or diver who pulls the arms in reduces I and so spins faster. Rotational KE is not conserved in this process.

$m$: Mass (kg) · $v$: Speed of the centre of mass (m s⁻¹) · $K$: Radius of gyration about the axis through the centre of mass (m) · $R$: Radius of the rolling body (m)

For rolling without slipping, v = Rω. K²/R² = 1 (ring), ½ (disc or solid cylinder), ⅖ (solid sphere), ⅔ (hollow sphere).

$a$: Linear acceleration along the incline (m s⁻²) · $\theta$: Angle of inclination · $h$: Vertical height descended (m)

Smaller K²/R² means larger acceleration. So a solid sphere reaches the bottom first, then a disc, and a ring last, whatever their masses and radii.

$P$: Pressure (Pa) · $F$: Normal force (N) · $A$: Area (m²)

Normal force per unit area. Pressure is a scalar. SI unit: pascal (1 Pa = 1 N m⁻²). 1 atm ≈ 1.013 × 10⁵ Pa.

$h$: Depth below the free surface (m) · $\rho$: Density of the liquid (kg m⁻³) · $g$: Acceleration due to gravity (m s⁻²)

Depends only on the depth h, not on the shape or cross-section of the vessel (hydrostatic paradox).

$P$: Absolute pressure at depth h (Pa) · $P_0$: Atmospheric pressure (Pa) · $P_{gauge}$: Gauge pressure (Pa)

Absolute pressure includes atmospheric pressure P₀. Gauge pressure is the excess over atmospheric, which is what a tyre gauge reads.

$F_1,A_1$: Force (N) and area (m²) of the small piston · $F_2,A_2$: Force (N) and area (m²) of the large piston

Pressure applied to an enclosed fluid is transmitted undiminished. A small force on a small piston gives a large force on a large piston. Energy is not multiplied, because the large piston moves through a smaller distance.

$T$: Surface tension (N m⁻¹) · $F$: Force due to the surface film (N) · $l$: Length of the line on which the force acts (m)

Force per unit length acting along the surface, perpendicular to a line drawn on it. For a film on a wire frame, l = 2 × wire length (two surfaces). SI unit: N m⁻¹ (= J m⁻²). Decreases as temperature rises.

$W$: Work done = increase in surface energy (J) · $\Delta A$: Total increase in surface area (m²)

Work done in increasing the surface area isothermally is stored as surface energy. Surface energy per unit area is numerically equal to surface tension.

$R$: Radius of the bubble (m) · $T$: Surface tension of the soap solution (N m⁻¹)

A bubble of radius R has two surfaces of area 4πR² each, so ΔA = 8πR² when it is blown from nothing. For enlarging from R₁ to R₂: W = 8πT(R₂² − R₁²).

$R$: Radius of the big drop (m) · $n$: Number of identical droplets formed

Volume is conserved: nr³ = R³, so r = R/n^{1/3}. The total surface area increases, so work must be done. When droplets merge into one drop, the same amount of energy is released.

$P_i-P_o$: Excess pressure, inside minus outside (Pa) · $T$: Surface tension (N m⁻¹) · $R$: Radius of the drop (m)

Pressure inside a curved liquid surface is greater on the concave side. Smaller drops have larger excess pressure. For an air bubble inside a liquid there is also only one surface, so the result is 2T/R as well.

$R$: Radius of the soap bubble (m)

Twice the value for a drop, because a soap bubble in air has two surfaces (inner and outer).

$h$: Height of capillary rise (m) · $\theta$: Angle of contact · $r$: Radius of the capillary bore (m) · $\rho$: Density of the liquid (kg m⁻³)

θ is the angle of contact. For θ < 90° (water in glass) the liquid rises; for θ > 90° (mercury in glass) cos θ < 0 and the liquid is depressed. h ∝ 1/r (Jurin's law).

$\eta$: Coefficient of viscosity (Pa s) · $A$: Area of the layer (m²) · $dv/dx$: Velocity gradient (s⁻¹)

The viscous force between layers is proportional to the area and to the velocity gradient. It opposes relative motion. SI unit of η: N s m⁻² = Pa s. CGS unit: poise (1 Pa s = 10 poise).

$F$: Viscous force (N) · $r$: Radius of the sphere (m) · $v$: Speed of the sphere relative to the fluid (m s⁻¹)

Viscous drag on a small sphere moving slowly through a fluid (streamline flow). The force grows with speed, which is why a terminal velocity exists.

$v_t$: Terminal velocity (m s⁻¹) · $\rho$: Density of the material of the sphere (kg m⁻³) · $\sigma$: Density of the fluid (kg m⁻³) · $\eta$: Coefficient of viscosity of the fluid (Pa s)

ρ is the density of the sphere and σ the density of the fluid. v_t ∝ r². If σ > ρ, v_t is negative, meaning the body rises (as an air bubble in water does).

$R_n$: Reynolds number (no unit) · $\rho$: Density of the fluid (kg m⁻³) · $v_c$: Critical velocity of the fluid (m s⁻¹) · $d$: Diameter of the pipe (m)

A pure number (no unit) that indicates the type of flow. Textbook values: streamline for R_n < 1000, turbulent for R_n > 2000, unsteady in between. Rearranged, the critical velocity is v_c = R_n·η/(ρd).

$A_1,A_2$: Cross-sectional areas at two sections (m²) · $v_1,v_2$: Speeds of flow at those sections (m s⁻¹)

For an incompressible fluid in steady flow, the volume flow rate Av is constant. A narrower section means faster flow. For circular pipes, v ∝ 1/r², not 1/r.

$P$: Pressure (Pa) · $\rho$: Density of the fluid (kg m⁻³) · $v$: Speed of flow (m s⁻¹) · $h$: Height above a reference level (m)

Applies along a streamline for steady, incompressible, non-viscous flow. It is energy conservation per unit volume: pressure energy + kinetic energy + potential energy. Where speed is high, pressure is low.

$v$: Speed of efflux (m s⁻¹) · $h$: Depth of the hole below the free surface (m)

Speed of liquid coming out of a small hole at depth h below the free surface of an open tank. It equals the speed of a body falling freely through height h.

$P$: Pressure (Pa) · $V$: Volume (m³) · $n$: Number of moles · $R$: Universal gas constant = 8.314 J mol⁻¹ K⁻¹ · $N$: Number of molecules · $k_B$: Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹ · $T$: Absolute temperature (K)

n = number of moles, N = number of molecules. k_B = R/N_A = 1.38 × 10⁻²³ J K⁻¹. A real gas behaves like an ideal gas at low pressure and high temperature.

$\lambda$: Mean free path (m) · $d$: Diameter of a molecule (m) · $N/V$: Number of molecules per unit volume (m⁻³)

Average distance travelled by a molecule between two successive collisions. It is smaller for a denser gas and for bigger molecules.

$m$: Mass of one molecule (kg) · $v_{rms}$: Root mean square speed (m s⁻¹) · $\rho$: Density of the gas (kg m⁻³)

Pressure comes from the change in momentum of molecules colliding with the walls. Here ρ = Nm/V is the density of the gas.

$M_0$: Molar mass of the gas (kg mol⁻¹) · $m$: Mass of one molecule (kg) · $T$: Absolute temperature (K)

v_rms ∝ √T and ∝ 1/√M₀. Lighter gases move faster at the same temperature. M₀ must be in kg mol⁻¹.

$k_B$: Boltzmann constant (J K⁻¹) · $T$: Absolute temperature (K)

The average translational KE of a molecule depends only on the absolute temperature, not on the nature of the gas. For one mole: E = (3/2)RT. At T = 0 K the translational KE would be zero.

$k_B$: Boltzmann constant (J K⁻¹) · $T$: Absolute temperature (K)

In thermal equilibrium, each degree of freedom of a molecule has average energy ½k_BT (per mole: ½RT). A vibrational mode counts twice (kinetic + potential), contributing k_BT.

$C_P$: Molar specific heat at constant pressure (J mol⁻¹ K⁻¹) · $C_V$: Molar specific heat at constant volume (J mol⁻¹ K⁻¹) · $R$: Universal gas constant (J mol⁻¹ K⁻¹)

C_P and C_V are molar specific heats. C_P > C_V because, at constant pressure, part of the heat supplied goes into work done in expansion. If heat is measured in calories, C_P − C_V = R/J.

$\gamma$: Adiabatic ratio (no unit)

Always greater than 1. Its value depends on the atomicity of the gas through the number of degrees of freedom.

Monatomic gases (He, Ne, Ar) have 3 translational degrees of freedom, so U = (3/2)RT per mole.

Rigid diatomic molecule (O₂, N₂ at ordinary temperatures): 3 translational + 2 rotational degrees of freedom. If vibration is also excited: C_V = (7/2)R, C_P = (9/2)R, γ = 9/7 ≈ 1.29.

$a$: Coefficient of absorption (absorptive power) · $r$: Coefficient of reflection (reflectance) · $t_r$: Coefficient of transmission (transmittance)

a, r, t_r are the fractions of the incident radiant energy that are absorbed, reflected and transmitted. Perfect blackbody: a = 1, r = t_r = 0. Athermanous (opaque) body: t_r = 0, so a + r = 1.

$R$: Emissive power of the body (W m⁻²) · $Q$: Radiant energy emitted (J) · $A$: Surface area (m²) · $t$: Time (s) · $R_B$: Emissive power of a perfect blackbody at the same temperature (W m⁻²) · $e$: Coefficient of emission or emissivity (no unit)

R is the radiant energy emitted per unit time per unit area (W m⁻²). Do not confuse it with the gas constant. e is the emissivity: e = 1 for a perfect blackbody and 0 < e < 1 for ordinary bodies. Here t is time (the transmission coefficient is written t_r).

$a$: Coefficient of absorption of the body · $e$: Coefficient of emission of the body

At a given temperature, the ratio of emissive power to absorptive power is the same for all bodies and equals the emissive power of a perfect blackbody. Good absorbers are good emitters.

${\lambda }_{max}$: Wavelength at which the blackbody emits the most energy (m) · $T$: Absolute temperature of the blackbody (K) · $b$: Wien's constant = 2.897 × 10⁻³ m K

As temperature increases, the wavelength of maximum emission shifts to shorter wavelengths. b = 2.897 × 10⁻³ m K. Used to estimate the surface temperature of stars.

$\sigma$: Stefan–Boltzmann constant = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ · $T$: Absolute temperature of the body (K) · $e$: Emissivity of the body

The energy radiated per unit time per unit area is proportional to T⁴. Doubling the absolute temperature multiplies the emissive power by 16. Total power from area A: P = eσAT⁴.

$Q/t$: Net rate of loss of heat (W) · $T$: Temperature of the body (K) · $T_0$: Temperature of the surroundings (K) · $A$: Surface area of the body (m²)

A body at temperature T in surroundings at T₀ emits eσAT⁴ and absorbs eσAT₀⁴. If T < T₀ the result is negative, meaning a net gain of heat.

$W$: Work done by the gas (J) · $P$: Pressure of the gas (Pa) · $V_i,V_f$: Initial and final volume (m³)

Area under the P–V curve. Positive for expansion, negative for compression. Work depends on the path, not only on the initial and final states. Unit: J.

$Q$: Heat supplied to the system (J); negative if heat is given out · $\Delta U$: Change in internal energy (J) · $W$: Work done by the system (J); negative if work is done on it

Energy conservation for a thermodynamic system. Heat supplied (Q) partly raises the internal energy (ΔU) and partly does external work (W). Equivalent form: ΔU = Q − W.

$n$: Number of moles · $C_V$: Molar specific heat at constant volume (J mol⁻¹ K⁻¹) · $\Delta T$: Change in temperature (K)

Internal energy of an ideal gas depends only on temperature. This formula holds for ANY process, not only at constant volume.

$T$: Constant absolute temperature (K) · $V_i,V_f$: Initial and final volume (m³ or L, same unit for both)

Temperature is constant, so ΔU = 0 and Q = W. Because P_iV_i = P_fV_f, also W = 2.303 nRT log₁₀(P_i/P_f). Isothermal changes are slow and take place in a container with conducting walls.

$\Delta V$: Change in volume (m³) · $C_P$: Molar specific heat at constant pressure (J mol⁻¹ K⁻¹)

Pressure stays constant. All three terms of the first law are non-zero: Q = nC_PΔT, ΔU = nC_VΔT and W = nRΔT.

Volume is constant, so the gas does no work. All the heat supplied goes into internal energy, which raises the temperature and the pressure.

$\gamma$: Ratio of specific heats C_P/C_V

No heat exchange (Q = 0), so ΔU = −W. Expansion cools the gas; compression heats it. Happens in insulated containers or in very rapid processes. γ = C_P/C_V.

$P_i,V_i,T_i$: Initial pressure, volume and temperature · $P_f,V_f,T_f$: Final pressure, volume and temperature

Positive when the gas expands and cools (T_f < T_i). Negative for adiabatic compression, where work is done on the gas.

The system returns to its initial state, so the internal energy is unchanged. The net heat absorbed equals the net work done in one cycle.

A gas expands suddenly into a vacuum in an insulated container. No work is done because there is no opposing pressure. For an ideal gas the temperature does not change. It is not a quasi-static process.

$\eta$: Efficiency (fraction or %) · $Q_H$: Heat absorbed from the hot reservoir (J) · $|Q_C|$: Magnitude of the heat rejected to the cold reservoir (J) · $W$: Net work done per cycle (J)

The engine absorbs Q_H from the hot reservoir, does work W, and rejects heat |Q_C| to the cold reservoir, with W = Q_H − |Q_C|. In the textbook sign convention Q_H is positive and Q_C (heat rejected) is negative, so use its magnitude. The second law (Kelvin–Planck) says Q_C can never be zero, so η < 1.

$K$: Coefficient of performance (no unit) · $Q_C$: Heat extracted from the cold reservoir (J) · $|W|$: Magnitude of the work done on the refrigerant (J)

Heat extracted from the cold region per unit work done on the refrigerant. For a refrigerator Q_C > 0 while Q_H and W are negative, so their magnitudes are used. K can be greater than 1. For a heat pump the useful output is the heat delivered to the hot side, so its coefficient of performance is |Q_H|/|W|.

$T_H$: Temperature of the hot reservoir (K) · $T_C$: Temperature of the cold reservoir (K)

Holds only for a reversible (Carnot) cycle: isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression.

$T_H$: Source temperature (K) · $T_C$: Sink temperature (K)

The maximum possible efficiency of any engine working between T_H and T_C. It depends only on the reservoir temperatures (in kelvin), not on the working substance. It is 100% only if T_C = 0 K, which cannot be reached.

The smaller the temperature difference between inside and outside, the higher the coefficient of performance. Temperatures in kelvin.

$F$: Restoring force (N) · $k$: Force constant (N m⁻¹) · $x$: Displacement from the mean position (m)

The restoring force is proportional to the displacement from the mean position and acts towards it (hence the minus sign). k is the force constant, unit N m⁻¹.

$\omega$: Angular frequency (rad s⁻¹) · $m$: Mass of the oscillating particle (kg)

Any system whose equation of motion reduces to this form performs SHM with angular frequency ω. Unit of ω: rad s⁻¹.

$A$: Amplitude (m) · $\varphi$: Initial phase or epoch (rad) · $t$: Time (s)

A is the amplitude (maximum displacement). (ωt + φ) is the phase and φ is the initial phase (epoch). If the particle starts at the mean position moving in the positive direction, φ = 0; if it starts at the positive extreme, φ = π/2.

$v$: Velocity at displacement x (m s⁻¹) · $v_{max}$: Maximum speed (m s⁻¹)

Maximum at the mean position (x = 0) and zero at the extreme positions (x = ±A).

$a$: Acceleration at displacement x (m s⁻²) · $a_{max}$: Maximum acceleration (m s⁻²)

Always directed towards the mean position. Zero at the mean position and maximum (Aω²) at the extremes. Dividing a_max by v_max gives ω.

$T$: Period (s) · $n$: Frequency (Hz) · $m$: Mass (kg) · $k$: Force constant (N m⁻¹)

Useful general form: T = 2π√(displacement/acceleration), using magnitudes. The period does not depend on the amplitude.

$R$: Resultant amplitude (m) · $A_1,A_2$: Amplitudes of the two SHMs (m) · ${\varphi }_1,{\varphi }_2$: Initial phases of the two SHMs (rad)

For two SHMs of the same period along the same path: x₁ = A₁ sin(ωt + φ₁) and x₂ = A₂ sin(ωt + φ₂). The resultant is also an SHM of the same period. In phase: R = A₁ + A₂. Opposite phase: R = |A₁ − A₂|.

$\delta$: Initial phase of the resultant SHM (rad)

Comes from dividing R sin δ by R cos δ. Check the signs of the numerator and denominator to place δ in the correct quadrant.

$E_K$: Kinetic energy (J)

Maximum (½kA²) at the mean position, zero at the extremes.

$E_P$: Potential energy (J)

Equal to the work done against the restoring force in displacing the particle from 0 to x. Zero at the mean position, maximum at the extremes.

$E$: Total mechanical energy (J) · $n$: Frequency (Hz)

Constant throughout the motion, independent of x. Proportional to the square of the amplitude and the square of the frequency.

$L$: Length of the pendulum (m) · $g$: Acceleration due to gravity (m s⁻²)

Valid for small amplitudes. Independent of the mass of the bob and of the amplitude. L is measured from the point of suspension to the centre of the bob.

A pendulum with a period of 2 s, so each swing from one extreme to the other takes 1 s. Its length depends on the local value of g (here g = 9.8 m s⁻²).

$\tau$: Restoring torque (N m) · $c$: Torque per unit angular displacement (N m rad⁻¹) · $\theta$: Angular displacement (rad) · $I$: Moment of inertia (kg m²)

The restoring torque is proportional to the angular displacement. c is the torque per unit angular displacement (N m rad⁻¹). Compare with linear SHM: m ↔ I, k ↔ c.

$I$: Moment of inertia of the magnet about the axis of rotation (kg m²) · $\mu$: Magnetic dipole moment of the magnet (A m²) · $B$: Magnetic field (T)

For small angular displacements, the restoring torque is μB sin θ ≈ μBθ, so c = μB. A stronger field or a stronger magnet gives faster oscillations.

$A_0$: Initial amplitude (m) · $b$: Damping constant (kg s⁻¹) · ${\omega }^{\prime }$: Angular frequency of damped oscillation (rad s⁻¹)

A damping force −bv makes the amplitude decay exponentially, and the oscillation is slightly slower than the undamped one (ω' < ω). With b = 0 you get back ideal SHM.

$y$: Displacement of the particle at position x and time t (m) · $A$: Amplitude (m) · $n$: Frequency (Hz) · $\lambda$: Wavelength (m) · $\omega$: Angular frequency = 2πn (rad s⁻¹) · $k$: Propagation constant (wave number) = 2π/λ (rad m⁻¹)

Wave travelling along the +x direction. For a wave travelling along −x, use (nt + x/λ). The Maharashtra textbook uses n for frequency (NCERT writes ν).

$v$: Wave speed (m s⁻¹) · $T$: Period = 1/n (s)

The speed depends on the medium; the frequency is fixed by the source. When a wave enters another medium, n stays the same and λ changes. Unit: m s⁻¹.

$\Delta \varphi$: Phase difference (rad) · $\Delta x$: Path difference (m)

A path difference of one wavelength corresponds to a phase difference of 2π rad. Points λ/2 apart on a progressive wave are in opposite phase.

$A$: Amplitude of each component wave (m) · $x$: Position along the medium (m)

Formed by two identical waves travelling in opposite directions. Every particle performs SHM of frequency n, but the amplitude 2A cos(2πx/λ) changes from point to point. There is no (nt ± x/λ) term, so the disturbance does not travel.

For y = 2A cos(2πx/λ) sin(2πnt): nodes (zero amplitude) at x = λ/4, 3λ/4, 5λ/4 …, antinodes (amplitude 2A) at x = 0, λ/2, λ … Nodes and antinodes alternate.

$L$: Vibrating length of the string (m) · $T$: Tension in the string (N) · $m$: Linear density — mass per unit length (kg m⁻¹)

String fixed at both ends, vibrating in one loop. Both ends are nodes. √(T/m) is the speed of transverse waves on the string.

$p$: Number of loops (harmonic number)

String vibrating in p loops: L = pλ/2. All harmonics (n, 2n, 3n …) are present. The pth harmonic is the (p − 1)th overtone.

Law of length (T, m constant), law of tension (L, m constant) and law of linear density (L, T constant). On a sonometer with a fixed tension, nL = constant.

$v$: Speed of sound in air (m s⁻¹) · $L$: Corrected length of the air column, L = l + e (m) · $l$: Length of the pipe (m) · $e$: End correction (m)

Closed end is a node, open end is an antinode, so the simplest mode has L = λ/4. L = l + e is the corrected length of the air column; use it whenever the diameter of the pipe is given (put e = 0 only if told to ignore end correction).

$p$: Overtone number (p = 0 for the fundamental)

As in the textbook, p is the overtone number (p = 0 is the fundamental). Only odd harmonics are present: n, 3n, 5n … The first overtone (p = 1) is the 3rd harmonic, the second overtone (p = 2) is the 5th harmonic.

Both ends are antinodes, so the simplest mode has L = λ/2. End correction is applied at both open ends, so the corrected length is L = l + 2e. For the same length, an open pipe has twice the fundamental frequency of a closed pipe.

$p$: Overtone number (p = 0 for the fundamental)

p is the overtone number (p = 0 is the fundamental). All harmonics are present: n, 2n, 3n … The first overtone (p = 1) is the 2nd harmonic. Richer in harmonics than a closed pipe, so its sound quality is different.

$e$: End correction (m) · $d$: Inner diameter of the pipe (m)

d is the inner diameter of the pipe (equivalently e = 0.6r). Corrected length of the air column: L = l + e for a closed pipe, L = l + 2e for an open pipe (l = length of the pipe).

$n_1,n_2$: Fundamental frequencies of the two pipes (Hz) · $l_1,l_2$: Lengths of the two pipes (m)

Two closed pipes of the same diameter, lengths l₁ and l₂, with fundamental frequencies n₁ and n₂. Since v is the same, 4n₁(l₁ + e) = 4n₂(l₂ + e). For two pipes open at both ends the same method gives e = (n₂l₂ − n₁l₁)/[2(n₁ − n₂)].

$N$: Beats per second (Hz) · $n_1,n_2$: Frequencies of the two sources (Hz)

Number of beats per second equals the difference of the two frequencies (n₁ > n₂). Beats can be heard only when this difference is small (practically less than about 6–7 Hz for the normal human ear).

${\lambda }_1,{\lambda }_2$: Wavelengths of the two waves, λ₁ < λ₂ (m)

Two waves of wavelengths λ₁ < λ₂ in the same medium produce N beats per second. Comes from v/λ₁ − v/λ₂ = N.

Basis for laws of reflection and refraction. Wavelets propagate at wave speed.

y = distance of fringe from central maximum. D ≫ d assumed.

n = 0, ±1, ±2, ... Central maximum at n = 0.

Uniform separation between consecutive bright (or dark) fringes.

φ = phase difference = (2π/λ)Δx. Maximum I_max = (√I₁+√I₂)².

Useful when amplitudes differ. Reduces to (a₁+a₂)²/(a₁−a₂)².

a = slit width. Central maximum has angular half-width sin⁻¹(λ/a).

Twice the width of any secondary maximum. On a screen at distance D.

Intensity of light through two polarisers with axes at angle θ. Maharashtra specific 2-mark derivation.

At Brewster's angle, reflected light is completely polarised perpendicular to plane of incidence.

Higher μ (immersion oil) and larger θ → better resolution.

Larger aperture D → better angular resolution.

${\vec{F}}_{12}$: Force on q₁ due to q₂ (N) · ${\epsilon }_0$: Permittivity of free space = 8.854×10⁻¹² C²·N⁻¹·m⁻² · $q_1,q_2$: Interacting point charges (C) · $r$: Separation between charges (m)

Force on charge 1 due to charge 2. r̂₁₂ points from 2 to 1. Same sign ⇒ repulsion, opposite sign ⇒ attraction.

Force per unit positive test charge in the limit q₀ → 0. Vector, radially outward from +q.

$2l$: Length of the dipole (m) · $q$: Magnitude of either charge (C)

Vector directed from −q to +q. Magnitude p = q(2l). SI unit: C·m.

Direction: parallel to p. HSC Board derivation question — start by writing E₊ and E₋ separately.

Direction: antiparallel to p. Axial field is twice the equatorial field.

Zero at θ = 0 (stable) and θ = π (unstable). Maximum at θ = 90°.

Total electric flux through any closed (Gaussian) surface = (net charge enclosed)/ε₀.

Cylindrical Gaussian surface, λ = linear charge density (C/m).

Independent of distance from the sheet. σ = surface charge density.

$A$: Common plate area (m²) · $d$: Plate separation (m)

Depends only on plate area, separation, and medium. Not on Q or V.

t = thickness of slab, K = dielectric constant. Reduces to Kε₀A/d when t = d.

All three forms are equivalent — pick the form that avoids the unknown quantity.

Energy per unit volume between capacitor plates. HOTS 2-marker.

Applies to ohmic conductors at constant temperature. R depends on material and geometry.

$\rho$: Resistivity (Ω·m) · $l$: Length of conductor (m) · $A$: Area of cross-section (m²)

ρ is material-specific and depends on temperature. l = length, A = area of cross-section.

n = number density of free electrons, v_d = drift velocity. Explains microscopic origin of current.

Drift velocity per unit electric field. Unit: m²·V⁻¹·s⁻¹.

Reciprocal of resistivity. Higher mobility ⇒ higher conductivity.

α = temperature coefficient of resistance. Positive for metals, negative for semiconductors.

Algebraic sum of currents at a junction is zero. Consequence of conservation of charge.

Algebraic sum of potential differences around a closed loop is zero. Consequence of conservation of energy.

No current through galvanometer at balance. P, Q, R, S are the four arm resistances.

For a uniform wire carrying constant current, potential drop is proportional to length. Compare two EMFs by comparing balancing lengths.

l₁ = balancing length with open circuit, l₂ = balancing length when cell shunted by R.

Series adds EMFs; parallel reduces internal resistance.

Three equivalent forms. Choose the one that avoids the unknown.

${\mu }_0$: Permeability of free space = 4π×10⁻⁷ T·m/A · $Id\vec{l}$: Current element (A·m) · $r$: Distance from element to field point (m)

Field due to a small current element I dl at position r from the element. Direction by right-hand rule.

For a circular loop of radius R carrying current I. Axis of loop is normal to plane.

x is distance from centre along axis. Reduces to μ₀I/(2R) at x = 0.

Line integral of B around a closed Amperian loop = μ₀ × current enclosed.

By Ampere's law with a circular Amperian loop of radius r around the wire.

n = number of turns per unit length. Field is uniform inside a long solenoid.

Perpendicular to both v and B. No work is done on the charge — speed does not change.

θ is angle between L and B. Direction by Fleming's left-hand rule.

Independent of speed and radius — enables resonance in the cyclotron.

R = maximum radius (dee radius). Sets a limit on ion energy.

N = number of turns. Direction from right-hand rule.

Ratio of magnetic moment to angular momentum of the orbital electron. Maharashtra-specific.

$m$: Magnetic dipole moment (A m²) · $B$: Uniform magnetic field (T) · $\theta$: Angle between m and B

Torque tends to align the dipole with the field. Maximum (mB) at θ = 90°, zero at θ = 0° or 180°. A uniform field gives no net force, only a torque. Unit: N m.

Minimum (−mB) when aligned with B (stable equilibrium). Maximum (+mB) when antiparallel (unstable equilibrium). Work done in rotating from θ₁ to θ₂: W = mB(cosθ₁ − cosθ₂). Unit: J.

$e$: Electronic charge = 1.6×10⁻¹⁹ C · $v$: Orbital speed of the electron (m s⁻¹) · $r$: Radius of the orbit (m) · $m_e$: Mass of electron = 9.1×10⁻³¹ kg · $L_{orb}$: Orbital angular momentum (kg m² s⁻¹ = J s)

Because the electron is negative, m_orb points opposite to its orbital angular momentum: m_orb = −(e/2mₑ)L_orb as vectors. The ratio e/2mₑ is the gyromagnetic ratio.

$h$: Planck's constant = 6.63×10⁻³⁴ J s

The smallest orbital magnetic moment of an electron (n = 1). Used as the natural unit of atomic magnetic moments. Equivalent unit: J T⁻¹.

$M$: Magnetisation (A m⁻¹) · $m_{net}$: Net magnetic dipole moment of the sample (A m²) · $V$: Volume of the sample (m³)

Net magnetic dipole moment per unit volume of the material. Unit: A m⁻¹ (same as H). It is a vector along the net dipole moment.

$H$: Magnetic intensity (A m⁻¹) · $B_0$: Applied magnetic field in vacuum (T) · ${\mu }_0$: Permeability of free space = 4π×10⁻⁷ T m A⁻¹ · $n$: Number of turns per unit length (m⁻¹)

Depends only on the free current, not on the material. B₀ is the field the current would produce in vacuum. Unit: A m⁻¹.

The field inside the material is the applied part μ₀H plus the contribution μ₀M of the material's own dipoles. Unit of B: tesla.

$\chi$: Magnetic susceptibility (no unit)

Dimensionless. Diamagnetic: small, negative. Paramagnetic: small, positive. Ferromagnetic: large, positive (and not constant — depends on H).

$\mu$: Permeability of the material (T m A⁻¹)

μ measures how easily a material lets magnetic field lines pass through it. Unit: T m A⁻¹ (= H m⁻¹).

${\mu }_r$: Relative permeability (no unit)

Dimensionless. μᵣ slightly less than 1 (diamagnetic), slightly more than 1 (paramagnetic), much greater than 1 (ferromagnetic).

Filling a solenoid or toroid with a material changes the field by the fraction χ. Multiply by 100 for percentage increase (paramagnetic) or decrease (diamagnetic).

$\Phi$: Magnetic flux (Wb) · $A$: Area of cross-section (m²)

Magnetic flux through the cross-section divided by area. Convert cm² to m² (1 cm² = 10⁻⁴ m²).

$C$: Curie constant (depends on the material) · $B$: Applied magnetic field, B = μ₀H (T) · $T$: Absolute temperature (K)

Magnetisation of a paramagnetic material is directly proportional to the applied field B and inversely proportional to the absolute temperature T. Putting B = μ₀H and χ = M/H gives χ = Cμ₀/T. T must be in kelvin. Valid for paramagnetic materials when the field is not too strong (far from saturation).

$N$: Number of atomic dipoles per unit volume (m⁻³) · $m_{atom}$: Dipole moment of each atom (A m²)

Maximum possible magnetisation, reached when every atomic dipole is aligned with the field. N is the number of atoms (dipoles) per unit volume.

θ is angle between B and the area vector A. SI unit: weber (Wb) = T·m².

Induced EMF equals negative rate of change of flux. Negative sign expresses Lenz's law.

Rod of length l moving with velocity v perpendicular to a magnetic field B.

$L$: Self-inductance (H)

L depends on geometry and material. SI unit: henry (H).

n = turns per unit length, A = cross-sectional area, l = length. Independent of current.

M is symmetric: M₁₂ = M₂₁. Basis of transformer action.

Inner solenoid area A, common length l, turn densities n₁ and n₂.

Analogous to (1/2)CV² for capacitor. Energy density = B²/(2μ₀).

N = number of turns, B = field, A = area, ω = angular frequency of rotation.

Natural angular frequency of undamped LC circuit. Analogous to √(k/m) in SHM.

Undamped harmonic; energy oscillates between C (electric) and L (magnetic).

0 ≤ k ≤ 1. k = 1 for perfect coupling (ideal transformer).

$e_0$: Peak emf (V) · $i_0$: Peak current (A) · $\omega$: Angular frequency (rad s⁻¹) · $f$: Frequency (Hz) · $\varphi$: Phase angle by which the current lags the emf (rad)

e₀ and i₀ are peak values. Here φ is the angle by which the current lags the emf; a negative φ means the current leads (capacitive circuit). Indian mains: f = 50 Hz, ω = 100π ≈ 314 rad s⁻¹.

Taken over a half cycle. Over a full cycle the mean value of a sinusoidal current is zero, because the positive and negative halves cancel.

The steady (DC) current that would produce the same heat in a resistor in the same time. AC ammeters and voltmeters read rms values.

Current and emf are in phase (φ = 0). Ohm's law holds for instantaneous, peak and rms values.

$X_L$: Inductive reactance (Ω) · $L$: Self-inductance (H)

In a pure inductor the current lags the emf by π/2. X_L grows with frequency: an inductor blocks high-frequency AC but passes DC freely. Unit: Ω.

$X_C$: Capacitive reactance (Ω) · $C$: Capacitance (F)

In a pure capacitor the current leads the emf by π/2. X_C falls as frequency rises: a capacitor blocks DC (f = 0, X_C infinite) but passes high-frequency AC. Unit: Ω.

$Z$: Impedance (Ω) · $R$: Resistance (Ω)

Impedance is the total opposition to AC. Resistance and the net reactance (X_L − X_C) add like perpendicular vectors, not like numbers. Unit: Ω.

φ is the angle by which the emf leads the current. X_L > X_C: inductive, current lags. X_L < X_C: capacitive, current leads. X_L = X_C: resonance, φ = 0.

Only the component of current in phase with the emf delivers power. The product e_rms·i_rms (without cos φ) is the apparent power. Unit: W.

Pure resistor: cos φ = 1 (maximum power). Pure inductor or capacitor: cos φ = 0 (no power). Series LCR at resonance: cos φ = 1.

The textbook defines wattless (idle) current as the current through a pure inductor or ideal capacitor, which consumes no power (φ = 90°). In a general circuit, the component i_rms sin φ, 90° out of phase with the emf, is the wattless part; the in-phase component i_rms cos φ is the one that does work.

At resonance X_L = X_C, so Z = R (minimum), current is maximum and in phase with the emf: an acceptor circuit. In a parallel LC circuit (negligible resistance) the same frequency gives maximum impedance and minimum line current: a rejector circuit.

$Q$: Quality factor (no unit) · ${\omega }_r$: Resonant angular frequency (rad s⁻¹) · ${\omega }_1,{\omega }_2$: Half-power angular frequencies; ω₁ − ω₂ = 2Δω is the bandwidth (rad s⁻¹)

Textbook definition: resonant frequency divided by the bandwidth ω₁ − ω₂, where ω₁ and ω₂ are the frequencies on either side of resonance at which the current falls to 1/√2 of its maximum (half-power points). For a series LCR circuit the bandwidth equals R/L, which gives Q = ω_rL/R. Q also equals the ratio of the voltage across L (or C) at resonance to the applied voltage. Small R gives large Q and a sharp resonance curve: good selectivity, as in a radio tuning circuit. Q has no unit.

$q_0$: Initial (maximum) charge on the capacitor (C) · $U$: Total energy of the circuit (J)

A charged capacitor discharging through an ideal inductor. Energy passes back and forth between the electric field of C and the magnetic field of L; the total stays constant if R = 0. Peak current i₀ = ω₀q₀.

$h$: Planck's constant = 6.626×10⁻³⁴ J·s · $\nu$: Frequency of radiation (Hz) · $\lambda$: Wavelength (m)

Photon carries a quantum of energy equal to Planck's constant times frequency.

Photon has zero rest mass but non-zero momentum. Basis for de Broglie hypothesis.

Minimum energy needed to eject an electron from a metal surface.

K_max = maximum KE of emitted electron. V₀ = stopping potential.

Linear graph of V₀ vs ν has slope h/e and x-intercept ν₀.

Matter wave: every particle of momentum p has an associated wavelength.

V in volts, λ in ångströms. Handy shortcut for electron microscopy.

Below ν₀ (or above λ₀), no photoelectrons no matter how intense the light.

P = radiant power (W). Useful when told the source power.

Characteristic length of the electron. Appears in Compton scattering.