Physics भौतिकशास्त्र

Quick summary

This chapter studies circular and rotational motion. It covers uniform and non-uniform circular motion, banking of roads, motion in a vertical circle, moment of inertia, radius of gyration, the two theorems of moment of inertia, angular momentum and its conservation, and rolling motion.

Key points

• Circular motion is accelerated motion (direction of velocity keeps changing) and it is periodic.
• In uniform circular motion (UCM), speed is constant; only centripetal acceleration acts, directed towards the centre.
• Centrifugal force is a pseudo force. It is seen only in a rotating (non-inertial) frame.
• On a level road, friction alone gives the centripetal force: v(max) = √(μ r g).
• Roads are banked (outer edge raised) so that a component of the normal reaction gives the centripetal force. Then friction is less needed.
• Conical pendulum: the bob moves in a horizontal circle; the string traces a cone.
• In a vertical circle, the speed is not constant (non-uniform circular motion). Minimum speed at the top is √(r g) and at the bottom √(5 r g).
• Moment of inertia (MI) is the rotational analogue of mass. It depends on mass and how mass is distributed about the axis.
• Torque = rate of change of angular momentum. If net external torque is zero, angular momentum is conserved (e.g. ballet dancer, diver pulling arms in spin faster).
• A rolling body has both translational and rotational kinetic energy.

Important terms

• Centripetal force – force directed towards the centre that keeps a body in circular motion.
• Banking of road – raising the outer edge of a curved road above the inner edge.
• Moment of inertia (I) – sum of m r² for all particles about the axis; I = Σ m r².
• Radius of gyration (K) – distance from the axis at which the whole mass may be assumed to be concentrated: I = M K².
• Angular momentum (L) – moment of linear momentum: L = r × p = I ω.
• Rolling motion – rotation plus translation without slipping, v = r ω.

Formulas

Common MI values: ring about its axis M R²; disc about its axis ½ M R²; solid sphere about a diameter (2/5) M R²; thin rod about centre (1/12) M L².

5 important questions

Q1. Derive the expression for the most safe speed on a banked road (neglect friction).

Forces on the car: weight m g down, normal reaction N perpendicular to the road. Resolve N: N cos θ balances m g, and N sin θ gives the centripetal force m v²/r. Dividing, tan θ = v²/(r g). So v = √(r g tan θ). This speed does not depend on the mass of the vehicle.

Q2. State and prove the theorem of parallel axes.

Statement: MI about any axis = MI about a parallel axis through the centre of mass + M h². Outline: take a particle at distance x from the CM axis; its distance from the new axis is (x + h). Sum m(x + h)² = Σ m x² + 2h Σ m x + h² Σ m. Σ m x = 0 about the CM, so I(o) = I(c) + M h².

Q3. Why does a diver spin faster when he folds his body?

No external torque acts about his centre of mass, so L = I ω stays constant. Folding brings mass closer to the axis, so I decreases and ω increases.

Q4. A disc of mass 2 kg and radius 0.1 m rotates at 10 rad/s about its axis. Find its rotational KE.

I = ½ M R² = ½ × 2 × 0.01 = 0.01 kg m². E = ½ I ω² = ½ × 0.01 × 100 = 0.5 J.

Q5. Find the minimum speed at the highest point of a vertical circle of radius 1.6 m (g = 10 m/s²).

At the top, tension can be zero at the minimum speed: m g = m v²/r, so v = √(r g) = √16 = 4 m/s.

Exam tip

Learn the derivations of banking of road, conical pendulum, the two theorems and KE of rolling body; they are asked almost every year. Always draw the force diagram.

Quick summary

Fluids are substances that can flow, i.e. liquids and gases. This chapter covers pressure in fluids, Pascal's law, surface tension, capillarity, viscosity, Stokes' law and terminal velocity, the equation of continuity and Bernoulli's principle with its applications.

Key points

• Pressure at depth h in a liquid: P = P₀ + h ρ g. It does not depend on the shape of the vessel (hydrostatic paradox).
• Pascal's law: pressure applied to an enclosed fluid is transmitted equally in all directions. Used in hydraulic lift and hydraulic brakes.
• Surface tension (ST) arises because molecules at the surface feel a net inward pull from cohesive forces.
• Surface tension = surface energy per unit area (numerically). Its SI unit is N/m or J/m².
• Impurities like detergent lower ST; ST decreases as temperature rises.
• Angle of contact is acute for water–glass (water wets glass, rises in a capillary) and obtuse for mercury–glass (mercury falls).
• Excess pressure inside a drop is 2T/R and inside a soap bubble 4T/R (two surfaces).
• Viscosity is internal friction between layers of a moving fluid. Streamline flow happens at low speed; turbulent flow above a critical velocity (judged by Reynolds number).
• A falling sphere in a viscous fluid reaches a constant terminal velocity.
• Bernoulli: along a streamline, pressure energy + KE + PE per unit volume is constant. Faster flow means lower pressure.

Important terms

• Cohesive force – attraction between molecules of the same substance.
• Adhesive force – attraction between molecules of different substances.
• Angle of contact – angle between the tangent to the liquid surface and the solid surface, measured inside the liquid.
• Capillarity – rise or fall of a liquid in a narrow tube.
• Coefficient of viscosity (η) – tangential force per unit area per unit velocity gradient; unit N s/m² (Pa s).
• Terminal velocity – constant velocity when net force on a falling body becomes zero.
• Reynolds number – Rₑ = ρ v D/η; below about 2000 flow is usually streamline.

Formulas

(ρ = density of sphere, σ = density of fluid, r = radius, T = surface tension.)

5 important questions

Q1. Derive the expression for capillary rise.

Upward force due to ST along the contact line = 2π r T cos θ. Weight of liquid column = π r² h ρ g. At equilibrium they are equal: 2π r T cos θ = π r² h ρ g, so h = 2 T cos θ/(r ρ g). Narrower tube means higher rise.

Q2. Derive the expression for terminal velocity using Stokes' law.

Forces on the sphere: weight (4/3)π r³ ρ g down; buoyancy (4/3)π r³ σ g up; viscous drag 6π η r v up. At terminal velocity net force is zero: 6π η r v = (4/3)π r³ (ρ − σ) g, giving v = 2 r² (ρ − σ) g/(9 η).

Q3. Why does a soap bubble have excess pressure 4T/R, while a drop has 2T/R?

A soap bubble has two surfaces (inner and outer), so the surface tension effect doubles, giving 4T/R. A drop has only one free surface.

Q4. Water flows through a pipe of area 4 cm² at 2 m/s. Find its speed where the area is 1 cm².

A₁ v₁ = A₂ v₂, so v₂ = 4 × 2/1 = 8 m/s.

Q5. Explain the working of an atomizer (sprayer) using Bernoulli's principle.

Air is blown fast across the top of a narrow tube dipped in liquid. Fast air has low pressure, so the higher atmospheric pressure on the liquid surface pushes liquid up the tube; the air stream breaks it into fine spray.

Exam tip

Units and derivations (capillary rise, terminal velocity, excess pressure, Bernoulli) carry marks. Remember: detergent and higher temperature both reduce surface tension.

Quick summary

The kinetic theory explains the behaviour of gases using the motion of molecules. The chapter derives the pressure of an ideal gas, links temperature to kinetic energy, introduces degrees of freedom and equipartition of energy, specific heats, and then studies thermal radiation: absorption, reflection, transmission, perfect blackbody, Kirchhoff's law, Wien's law and Stefan-Boltzmann law.

Key points

• Ideal gas assumptions: molecules are point-like, collisions are perfectly elastic, no intermolecular forces except during collisions, motion is random.
• Mean free path is the average distance a molecule travels between two collisions. It decreases with higher density and larger molecular size.
• Pressure of a gas P = (1/3) ρ v²(rms). Average KE per molecule = (3/2) k T, so temperature is a measure of average KE.
• Degrees of freedom (f): monatomic 3, diatomic 5 (rigid, at moderate temperature), add 2 if vibration is active.
• Law of equipartition: each degree of freedom has average energy ½ k T.
• γ = Cp/Cv: monatomic 5/3, diatomic 7/5. Mayer's relation: Cp − Cv = R.
• For radiation falling on a body: a + r + t = 1 (coefficients of absorption, reflection, transmission).
• A perfect blackbody absorbs all radiation falling on it (a = 1). Ferry's blackbody is a practical example.
• Kirchhoff's law: good absorbers are good emitters; ratio of emissive power to absorptivity is the same for all bodies at a given temperature.
• Wien's displacement law: λ(max) T = b (b = 2.897 × 10⁻³ m K). Hotter bodies peak at shorter wavelength.

Important terms

• rms speed – square root of the mean of squares of molecular speeds.
• Degree of freedom – an independent way in which a molecule can move or store energy.
• Emissive power – energy radiated per unit area per unit time.
• Emissivity (e) – ratio of emissive power of a body to that of a perfect blackbody at the same temperature.
• Absorptive power (a) – fraction of incident radiation absorbed.

Formulas

σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; k = 1.38 × 10⁻²³ J/K; R = 8.31 J/mol K.

5 important questions

Q1. Derive P = (1/3) ρ v²(rms) for an ideal gas (outline).

Take a cube of side L with N molecules. A molecule with x-velocity vₓ changes momentum by 2m vₓ per collision with a wall and returns after time 2L/vₓ, so force = m vₓ²/L. Sum over all molecules and use average vₓ² = v²/3 (random motion). Pressure = force/L² = N m v²(rms)/(3 V) = (1/3) ρ v²(rms).

Q2. Show that Cp/Cv = 5/3 for a monatomic gas.

f = 3, so U = (3/2) R T per mole and Cv = (3/2) R. Cp = Cv + R = (5/2) R. γ = (5/2)/(3/2) = 5/3.

Q3. State Kirchhoff's law of radiation and give one example.

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. Example: a black spot on a white plate looks brightest when both are heated in the dark, because a good absorber is a good emitter.

Q4. Find the rms speed of oxygen molecules at 300 K (M = 32 × 10⁻³ kg/mol).

v = √(3 R T/M) = √(3 × 8.31 × 300/0.032) = √(233719) ≈ 483 m/s.

Q5. The Sun's spectrum peaks at about 500 nm. Estimate its surface temperature.

T = b/λ = 2.897 × 10⁻³/(500 × 10⁻⁹) ≈ 5794 K, about 5800 K.

Exam tip

Learn the assumptions of kinetic theory as a list and the derivation of pressure. In radiation numericals, always convert temperature to kelvin.

Quick summary

Thermodynamics deals with heat, work and internal energy. This chapter covers the zeroth and first laws, thermodynamic processes (isothermal, adiabatic, isobaric, isochoric), work done from p-V diagrams, heat engines, refrigerators, the second law and the Carnot cycle.

Key points

• Zeroth law: if A and B are each in thermal equilibrium with C, then A and B are in equilibrium with each other. This defines temperature.
• Internal energy (U) is the total KE + PE of molecules. It is a state function.
• First law: Q = ΔU + W (heat given = rise in internal energy + work done by the system). It is the law of conservation of energy.
• Work done by a gas = area under the p-V curve. In a cyclic process, net work = area enclosed by the loop and ΔU = 0.
• Isothermal: T constant, ΔU = 0, Q = W. Adiabatic: Q = 0, W = −ΔU, P V^γ = constant.
• Isobaric: P constant, W = P ΔV. Isochoric: V constant, W = 0, Q = ΔU.
• Heat engine: takes Q_H from a hot reservoir, does work W, rejects Q_C to a cold reservoir. Efficiency η = W/Q_H = 1 − Q_C/Q_H.
• Refrigerator / heat pump: work is done to move heat from cold to hot. Coefficient of performance K = Q_C/W.
• Second law (Kelvin-Planck): no engine can convert all heat into work. (Clausius): heat cannot flow by itself from cold to hot.
• Carnot engine (two isothermal + two adiabatic, all reversible) has the maximum efficiency η = 1 − T_C/T_H.

Important terms

• Thermodynamic system – the part of the universe being studied; open, closed or isolated.
• State variables – P, V, T, U that describe the state.
• Quasi-static process – an extremely slow process passing through equilibrium states.
• Reversible process – one that can be retraced exactly with no net change in surroundings.
• Heat reservoir – a body so large that its temperature does not change when heat is added or removed.

Formulas

5 important questions

Q1. Derive the work done in an isothermal expansion of an ideal gas.

W = ∫ P dV from V₁ to V₂. For an ideal gas P = n R T/V and T is constant, so W = n R T ∫ dV/V = n R T ln(V₂/V₁).

Q2. State the first law of thermodynamics and apply it to an adiabatic process.

Q = ΔU + W. In an adiabatic process no heat is exchanged, Q = 0, so W = −ΔU. When the gas expands it does work and cools; when compressed, it heats up (e.g. bicycle pump gets hot).

Q3. Explain the Carnot cycle with its four steps.

1. Isothermal expansion at T_H, absorbs Q_H. 2. Adiabatic expansion, temperature falls to T_C. 3. Isothermal compression at T_C, rejects Q_C. 4. Adiabatic compression back to the start. Net work = Q_H − Q_C = area of the loop; η = 1 − T_C/T_H.

Q4. A Carnot engine works between 527 °C and 127 °C. Find its efficiency.

T_H = 800 K, T_C = 400 K. η = 1 − 400/800 = 0.5 = 50%.

Q5. A gas absorbs 500 J of heat and does 200 J of work. Find the change in internal energy.

ΔU = Q − W = 500 − 200 = 300 J (increase).

Exam tip

Watch the sign convention: heat added and work done by the system are positive. Temperatures in Carnot problems must be in kelvin.

Quick summary

Oscillations are to-and-fro periodic motions about a mean position. The chapter studies simple harmonic motion (SHM), its equations, energy, the simple pendulum, angular SHM, composition of two SHMs, and damped, free and forced oscillations.

Key points

• Linear SHM: the restoring force is directly proportional to displacement and directed towards the mean position: F = −k x.
• Differential equation of SHM: d²x/dt² + ω² x = 0, where ω = √(k/m).
• Displacement x = A sin(ω t + α). Velocity is maximum (A ω) at the mean position and zero at the extremes.
• Acceleration is maximum (A ω²) at the extremes and zero at the mean position.
• Total energy in SHM is constant: E = ½ k A² = ½ m ω² A². KE and PE keep exchanging.
• Simple pendulum (small amplitude): T = 2π √(L/g). A second's pendulum has T = 2 s, L ≈ 1 m.
• Spring–mass system: T = 2π √(m/k). Springs in series: 1/k = 1/k₁ + 1/k₂; in parallel: k = k₁ + k₂.
• Two SHMs of the same period along the same line combine into an SHM; resultant amplitude depends on phase difference.
• Damped oscillations: amplitude decreases with time due to friction or air resistance.
• Free oscillations happen at the natural frequency; forced oscillations follow the driving frequency; resonance occurs when driving frequency equals natural frequency.

Important terms

• Amplitude (A) – maximum displacement from the mean position.
• Period (T) – time for one oscillation.
• Frequency (n) – number of oscillations per second (Hz).
• Phase – the state of oscillation (position and direction) at a given instant, given by (ω t + α).
• Epoch (α) – initial phase at t = 0.
• Force constant (k) – restoring force per unit displacement (N/m).

Formulas

5 important questions

Q1. Define linear SHM and obtain its differential equation.

SHM is motion in which the restoring force is proportional to displacement and acts towards the mean position. F = m a = −k x, so m d²x/dt² + k x = 0, i.e. d²x/dt² + ω² x = 0 with ω² = k/m.

Q2. Show that the total energy of a particle in SHM is constant.

KE = ½ m ω² (A² − x²) and PE = ½ m ω² x². Adding, E = ½ m ω² A². This has no x, so total energy is the same at every point.

Q3. Derive the period of a simple pendulum for small oscillations.

Restoring force = −m g sin θ ≈ −m g θ = −m g x/L. So acceleration a = −(g/L) x, which is SHM with ω² = g/L. Hence T = 2π √(L/g).

Q4. A body in SHM has amplitude 5 cm and period 2 s. Find its maximum velocity and maximum acceleration.

ω = 2π/T = π rad/s. v(max) = A ω = 0.05 × 3.14 = 0.157 m/s. a(max) = A ω² = 0.05 × 9.87 ≈ 0.49 m/s².

Q5. What is the length of a second's pendulum on Earth (g = 9.8 m/s²)?

T = 2 s. L = g T²/(4π²) = 9.8 × 4/39.48 ≈ 0.993 m, about 1 m.

Exam tip

Remember where velocity and acceleration are maximum and zero. Use ω = 2π/T in every numerical first.

Quick summary

When two or more waves meet, the resultant displacement is the sum of their displacements. This principle explains reflection effects, stationary (standing) waves on strings and in air columns, harmonics and overtones, the sonometer, end correction and beats.

Key points

• A progressive wave carries energy forward: y = A sin 2π(t/T − x/λ). Speed v = n λ.
• Reflection from a rigid (denser) boundary gives a phase change of π; from a free (rarer) end, no phase change.
• Principle of superposition: resultant displacement = vector sum of individual displacements.
• Stationary waves form when two identical waves travel in opposite directions. They have nodes (zero amplitude) and antinodes (maximum amplitude).
• Distance between two adjacent nodes (or antinodes) is λ/2; node to next antinode is λ/4.
• Energy is not transported in a stationary wave.
• String fixed at both ends: all harmonics are present, n_p = p v/(2L).
• Pipe closed at one end: only odd harmonics (n, 3n, 5n…). Pipe open at both ends: all harmonics (n, 2n, 3n…).
• End correction: the antinode forms slightly outside the open end, e = 0.3 d (d = inner diameter).
• Beats: periodic rise and fall of loudness when two sounds of slightly different frequencies combine. Beat frequency = n₁ − n₂.

Important terms

• Node – point of zero displacement in a stationary wave.
• Antinode – point of maximum displacement.
• Fundamental frequency – lowest frequency of vibration (first harmonic).
• Overtone – any frequency higher than the fundamental that is actually produced.
• Free vibration – vibration at natural frequency with no outside force.
• Forced vibration / resonance – vibration at the driving frequency; resonance when it equals the natural frequency.

Formulas

(T = tension, m = mass per unit length of the string, L = length.)

5 important questions

Q1. Show that only odd harmonics are present in a pipe closed at one end.

The closed end is a node and the open end an antinode. So L = λ/4, 3λ/4, 5λ/4 … Frequencies n = v/λ = v/(4L), 3v/(4L), 5v/(4L)… i.e. n, 3n, 5n: only odd harmonics.

Q2. State the laws of vibrating strings.

For the fundamental mode, n = (1/2L) √(T/m). Law of length: n ∝ 1/L (T, m constant). Law of tension: n ∝ √T. Law of linear density: n ∝ 1/√m.

Q3. What are beats? Give two applications.

Beats are the periodic waxing and waning of sound intensity due to superposition of two waves of nearly equal frequency. Uses: tuning musical instruments, finding an unknown frequency of a tuning fork, detecting harmful gases in mines.

Q4. A string 1 m long under tension vibrates with fundamental 100 Hz. What is the wave speed on it?

λ = 2L = 2 m. v = n λ = 100 × 2 = 200 m/s.

Q5. Two tuning forks of 256 Hz and 260 Hz are sounded together. How many beats are heard in 5 s?

Beat frequency = 260 − 256 = 4 per second. In 5 s: 20 beats.

Exam tip

Draw the node–antinode diagrams for the first three modes of strings, open pipes and closed pipes; the formulas follow directly from them.

Quick summary

Wave optics treats light as a wave. The chapter covers Huygens' principle and the wave theory of reflection and refraction, polarization (including Brewster's law and Malus' law), interference in Young's double slit experiment, diffraction at a single slit, and the resolving power of optical instruments.

Key points

• Huygens' principle: every point on a wavefront acts as a source of secondary wavelets; the forward envelope of these wavelets gives the new wavefront.
• A wavefront is the locus of points vibrating in the same phase. Light rays are perpendicular to wavefronts.
• Using Huygens' construction we can prove the laws of reflection and Snell's law of refraction: n = sin i/sin r = v₁/v₂.
• When light enters a denser medium, its speed and wavelength decrease, but frequency stays the same.
• Polarization shows that light is a transverse wave. Polaroids allow vibrations in only one plane.
• Brewster's law: at the polarizing angle, reflected light is completely plane polarized and tan θ_B = n. Reflected and refracted rays are then perpendicular.
• Malus' law: intensity through an analyser I = I₀ cos² θ.
• Interference: bright fringes where path difference = n λ; dark fringes where it is (2n − 1) λ/2. Fringe width β = λ D/d.
• Diffraction is bending of light around small obstacles or apertures. Single slit: central maximum is twice as wide as other maxima.
• Resolving power: ability to show two close objects as separate; limited by diffraction (Rayleigh's criterion).

Important terms

• Coherent sources – sources with constant phase difference and same frequency.
• Fringe width (β) – distance between two consecutive bright or dark fringes.
• Plane polarized light – light with vibrations in only one plane.
• Fresnel diffraction – source or screen at finite distance; Fraunhofer – both effectively at infinity (using lenses).
• Limit of resolution – smallest angular or linear separation that can be resolved.

Formulas

(d = slit separation, D = slit-to-screen distance, a = slit width.)

5 important questions

Q1. Prove Snell's law of refraction using Huygens' principle (outline).

A plane wavefront AB strikes the boundary obliquely. While the wavelet from B travels BC = v₁t in medium 1, the wavelet from A travels AE = v₂t in medium 2. Tangent CE is the refracted wavefront. From triangles ABC and AEC: sin i = v₁t/AC and sin r = v₂t/AC. So sin i/sin r = v₁/v₂ = n (constant).

Q2. Derive the expression for fringe width in Young's double slit experiment.

For a point P at distance x from the centre, path difference = x d/D. Bright fringe: x d/D = n λ, so x(n) = n λ D/d. The distance between consecutive bright fringes β = x(n+1) − x(n) = λ D/d. Same for dark fringes.

Q3. State Brewster's law and show that reflected and refracted rays are perpendicular at the polarizing angle.

tan θ_B = n. Also n = sin θ_B/sin r. So sin θ_B/cos θ_B = sin θ_B/sin r, giving cos θ_B = sin r, so θ_B + r = 90°. Hence the angle between reflected and refracted rays is 90°.

Q4. In YDSE, d = 0.5 mm, D = 1 m and λ = 600 nm. Find the fringe width.

β = λ D/d = 600 × 10⁻⁹ × 1/(0.5 × 10⁻³) = 1.2 × 10⁻³ m = 1.2 mm.

Q5. Find the polarizing angle for glass of refractive index 1.732.

tan θ_B = 1.732 = √3, so θ_B = 60°.

Exam tip

Huygens' construction diagrams for reflection and refraction are frequently asked; label the incident and refracted wavefronts clearly.

Quick summary

This chapter deals with charges at rest. It covers Gauss' law and its applications, electric potential and potential energy, equipotential surfaces, dielectrics and polarisation, capacitors (parallel plate, series and parallel combinations), energy stored in a capacitor, and the Van de Graaff generator.

Key points

• Electric flux through a surface φ = E A cos θ. Gauss' law: total flux through a closed surface = q(enclosed)/ε₀.
• Field of an infinitely long charged wire: E = λ/(2π ε₀ r). Of an infinite charged plane sheet: E = σ/(2ε₀), independent of distance.
• Field outside a charged spherical shell acts as if all charge were at the centre; inside the shell E = 0.
• Electric potential V at a point = work done per unit positive charge in bringing it from infinity: V = q/(4π ε₀ r).
• Electric field is the negative potential gradient: E = −dV/dr.
• Equipotential surfaces are perpendicular to field lines; no work is done in moving a charge on them.
• A dielectric placed in a field gets polarised; it reduces the field inside by factor k (dielectric constant).
• Polar molecules have a permanent dipole moment (e.g. H₂O); non-polar molecules do not (e.g. O₂, CO₂).
• Capacitance of a parallel plate capacitor C = ε₀ A/d; with dielectric C = k ε₀ A/d.
• Van de Graaff generator builds up very high voltage using corona discharge and the fact that charge resides on the outer surface of a conductor.

Important terms

• Electric flux – number of field lines passing through a surface (N m²/C).
• Surface charge density (σ) – charge per unit area (C/m²).
• Electric potential energy – work done in assembling a system of charges.
• Capacitance – ratio of charge to potential difference: C = Q/V (farad, F).
• Dielectric strength – maximum field a dielectric can withstand without breaking down.

Formulas

ε₀ = 8.85 × 10⁻¹² C²/N m²; 1/(4π ε₀) = 9 × 10⁹ N m²/C².

5 important questions

Q1. Using Gauss' law, find the electric field due to an infinitely long charged wire.

Take a coaxial cylinder of radius r and length L as the Gaussian surface. Flux passes only through the curved surface: φ = E (2π r L). Charge enclosed = λ L. Gauss' law: E 2π r L = λ L/ε₀, so E = λ/(2π ε₀ r).

Q2. Derive the expression for the energy stored in a charged capacitor.

Work to move small charge dq at potential V = q/C is dW = (q/C) dq. Total work W = ∫ q/C dq from 0 to Q = Q²/(2C). This is stored as energy: U = Q²/(2C) = ½ C V².

Q3. Distinguish between polar and non-polar molecules.

Polar: centres of positive and negative charge do not coincide, permanent dipole moment (HCl, H₂O). Non-polar: centres coincide, no permanent dipole; dipole is induced only in an external field (H₂, N₂, CO₂).

Q4. Find the capacitance of a parallel plate capacitor with plate area 0.02 m² and separation 1 mm in air.

C = ε₀ A/d = 8.85 × 10⁻¹² × 0.02/10⁻³ = 1.77 × 10⁻¹⁰ F = 177 pF.

Q5. Three 6 μF capacitors are connected in series. Find the effective capacitance. What if they are in parallel?

Series: 1/C = 3/6, C = 2 μF. Parallel: C = 18 μF.

Exam tip

For Gauss' law questions, always state the chosen Gaussian surface and why flux through some faces is zero.

Quick summary

This chapter studies electric circuits. It covers Kirchhoff's laws, the Wheatstone bridge and metre bridge, the potentiometer and its uses (comparing emfs, finding internal resistance), and the conversion of a galvanometer into an ammeter and a voltmeter.

Key points

• Kirchhoff's first law (current/junction law): algebraic sum of currents at a junction is zero. It follows from conservation of charge.
• Kirchhoff's second law (voltage/loop law): in a closed loop, sum of emfs = sum of IR drops. It follows from conservation of energy.
• Wheatstone bridge is balanced (no galvanometer current) when P/Q = R/S.
• A metre bridge is a practical Wheatstone bridge using a 1 m uniform wire: X = R (100 − l)/l (when the unknown is in the right gap).
• Potentiometer principle: for a uniform wire carrying steady current, potential difference across a length is proportional to that length (V ∝ l).
• A potentiometer measures emf without drawing current at balance, so it is more accurate than a voltmeter.
• Potential gradient K = V/L. A smaller gradient (longer wire) gives higher sensitivity.
• A galvanometer is converted into an ammeter by a low resistance shunt in parallel, and into a voltmeter by a high resistance in series.
• An ideal ammeter has zero resistance; an ideal voltmeter has infinite resistance.

Important terms

• Junction – point where three or more conductors meet.
• Potential gradient – fall of potential per unit length of potentiometer wire (V/m).
• Shunt – small resistance connected in parallel with a galvanometer.
• Internal resistance (r) – resistance offered by the cell itself.
• Null point / balance point – point on the wire where the galvanometer shows no deflection.

Formulas

(G = galvanometer resistance, I_g = full scale current, I = range of ammeter, V = range of voltmeter.)

5 important questions

Q1. Derive the balancing condition of a Wheatstone bridge using Kirchhoff's laws.

Resistances P, Q, R, S form the arms with galvanometer between B and D. At balance, I_g = 0, so the current in P equals that in Q (I₁) and in R equals that in S (I₂). Loop ABDA: I₁P − I₂R = 0. Loop BCDB: I₁Q − I₂S = 0. Dividing: P/Q = R/S.

Q2. How is the internal resistance of a cell found using a potentiometer?

First find the balancing length l₁ with the cell on open circuit (balances E). Then connect a resistance R across the cell and find l₂ (balances terminal voltage V). E/V = l₁/l₂ and E/V = (R + r)/R. So r = R (l₁ − l₂)/l₂.

Q3. Why is a potentiometer preferred over a voltmeter to measure emf?

At the null point no current is drawn from the cell, so the true emf is measured. A voltmeter always draws some current and reads the terminal voltage, which is less than the emf.

Q4. A galvanometer of resistance 50 Ω gives full deflection for 5 mA. Find the shunt needed to convert it into an ammeter of range 0–5 A.

S = I_g G/(I − I_g) = 0.005 × 50/(5 − 0.005) = 0.25/4.995 ≈ 0.05 Ω.

Q5. Two cells balance at 300 cm and 200 cm on a potentiometer wire. Find the ratio of their emfs.

E₁/E₂ = l₁/l₂ = 300/200 = 3 : 2.

Exam tip

Learn the circuit diagrams for the metre bridge and both potentiometer experiments; marks are given for correct diagrams.

Quick summary

Moving charges and electric currents produce magnetic fields and also experience forces in them. This chapter covers the magnetic (Lorentz) force, cyclotron motion, helical motion, force on a current-carrying conductor, torque on a coil, the moving coil galvanometer, the Biot-Savart law, Ampere's law and their applications.

Key points

• Force on a moving charge: F = q (v × B). It is perpendicular to both v and B, so it does no work.
• A charge moving perpendicular to B moves in a circle; radius r = m v/(q B). Its period does not depend on speed (cyclotron principle).
• If velocity has a component along B, the path is a helix.
• Force on a straight wire: F = I L × B, magnitude I L B sin θ.
• Torque on a current loop: τ = N I A B sin θ; magnetic dipole moment m = N I A.
• Moving coil galvanometer uses a radial magnetic field so that torque is maximum for every position; deflection θ ∝ I.
• Biot-Savart law gives the field due to a small current element: dB = (μ₀/4π) I dl sin θ/r².
• Field at centre of a circular loop: B = μ₀ N I/(2R). Field of a long straight wire: B = μ₀ I/(2π r).
• Ampere's law: line integral of B around a closed path = μ₀ × current enclosed.
• Two parallel wires carrying currents in the same direction attract; opposite directions repel.

Important terms

• Lorentz force – total force on a charge in electric and magnetic fields: F = q(E + v × B).
• Cyclotron frequency – f = q B/(2π m).
• Magnetic dipole moment – m = N I A (A m²).
• Current sensitivity – deflection per unit current, θ/I = N A B/c.
• Solenoid – long coil of wire; field inside is uniform.
• Toroid – solenoid bent into a ring.

Formulas

μ₀ = 4π × 10⁻⁷ T m/A.

5 important questions

Q1. Show that the period of a charged particle in a uniform magnetic field does not depend on its speed.

Magnetic force gives centripetal force: q v B = m v²/r, so r = m v/(q B). Period T = 2π r/v = 2π m/(q B). It has no v, so faster particles move in bigger circles but take the same time.

Q2. Use Ampere's law to find the magnetic field inside a long solenoid.

Choose a rectangular loop with one side of length L inside. B is along the axis inside and nearly zero outside; the perpendicular sides give zero. So ∮B·dl = B L = μ₀ (n L I), giving B = μ₀ n I.

Q3. Why is a radial magnetic field used in a moving coil galvanometer?

In a radial field the plane of the coil is always parallel to the field, so sin θ = 1 and torque = N I A B for every position. This makes the deflection proportional to current and gives a linear scale.

Q4. Find the field at the centre of a circular coil of 50 turns, radius 10 cm, carrying 2 A.

B = μ₀ N I/(2R) = 4π × 10⁻⁷ × 50 × 2/(0.2) = 6.28 × 10⁻⁴ T.

Q5. A proton moves at 3 × 10⁶ m/s perpendicular to a 0.5 T field. Find the radius of its path (m = 1.67 × 10⁻²⁷ kg).

r = m v/(q B) = 1.67 × 10⁻²⁷ × 3 × 10⁶/(1.6 × 10⁻¹⁹ × 0.5) ≈ 0.063 m = 6.3 cm.

Exam tip

Use the right-hand rules carefully for direction questions, and remember the magnetic force never changes a particle's speed.

Quick summary

This chapter explains why materials behave magnetically. It covers torque on a magnetic dipole, the origin of magnetism from electron motion (orbital magnetic moment and Bohr magneton), magnetization and magnetic intensity, the three types of magnetic materials, hysteresis, and permanent magnets and electromagnets.

Key points

• A magnetic dipole (bar magnet) in a uniform field experiences a torque τ = m × B, but no net force.
• Potential energy of a dipole in a field U = −m B cos θ; it is minimum when m is parallel to B (stable equilibrium).
• Magnetism of atoms comes from orbital motion and spin of electrons.
• The orbital magnetic moment of an electron is related to its angular momentum: M₀ = (e/2m) L. Gyromagnetic ratio = e/2m.
• Bohr magneton μ_B = e h/(4π m) ≈ 9.27 × 10⁻²⁴ A m² is the smallest unit of electron magnetic moment.
• Diamagnetic materials (bismuth, copper, water) are weakly repelled; susceptibility is small and negative.
• Paramagnetic materials (aluminium, platinum, oxygen) are weakly attracted; susceptibility small and positive; obeys Curie's law χ = C/T.
• Ferromagnetic materials (iron, cobalt, nickel) are strongly attracted due to domains; above the Curie temperature they become paramagnetic.
• Hysteresis: magnetization lags behind magnetic intensity. The loop area gives energy loss per cycle.
• Soft iron (low retentivity, low coercivity, narrow loop) suits electromagnets and transformer cores; steel (high coercivity) suits permanent magnets.

Important terms

• Magnetization (M) – net magnetic moment per unit volume (A/m).
• Magnetic intensity (H) – magnetizing field, H = B₀/μ₀ (A/m).
• Magnetic susceptibility (χ) – χ = M/H, no unit.
• Permeability (μ) – μ = μ₀ (1 + χ); relative permeability μᵣ = 1 + χ.
• Retentivity – magnetization left when H is reduced to zero.
• Coercivity – reverse H needed to bring magnetization to zero.
• Magnetic shielding – using a soft iron enclosure to keep a region nearly free of magnetic field.

Formulas

5 important questions

Q1. Obtain the expression for the orbital magnetic moment of an electron revolving round the nucleus.

The electron forms a current loop: I = e/T = e v/(2π r). Area A = π r². Magnetic moment M₀ = I A = e v r/2. Since L = m v r, M₀ = (e/2m) L.

Q2. Distinguish between diamagnetic, paramagnetic and ferromagnetic substances (any three points).

Dia: weakly repelled, χ small negative, unaffected by temperature. Para: weakly attracted, χ small positive, χ ∝ 1/T. Ferro: strongly attracted, χ very large positive, become paramagnetic above Curie temperature.

Q3. Why is soft iron used for the core of an electromagnet?

Soft iron has high permeability, low retentivity and low coercivity. It gets strongly magnetized when current flows and loses magnetism quickly when current is switched off; its narrow hysteresis loop means less energy loss.

Q4. A bar magnet of moment 2 A m² is placed at 30° to a field of 0.4 T. Find the torque on it.

τ = m B sin θ = 2 × 0.4 × 0.5 = 0.4 N m.

Q5. The magnetic intensity in a material is 2000 A/m and the magnetization is 600 A/m. Find susceptibility and relative permeability.

χ = M/H = 600/2000 = 0.3. μᵣ = 1 + χ = 1.3.

Exam tip

Make a three-column comparison table for dia, para and ferro materials; it is an easy 3-mark question.

Quick summary

A changing magnetic flux induces an emf in a circuit. This chapter covers Faraday's and Lenz's laws, motional emf, induced emf in a rotating coil (AC generator), eddy currents, self-induction and mutual induction, energy stored in an inductor and the transformer.

Key points

• Magnetic flux φ = B A cos θ (unit weber, Wb).
• Faraday's laws: an emf is induced whenever flux linked with a circuit changes, and its magnitude equals the rate of change of flux: e = −N dφ/dt.
• Lenz's law: the induced current opposes the change that causes it. It follows from conservation of energy (the minus sign).
• Motional emf in a rod of length l moving with speed v perpendicular to B: e = B l v. Fleming's right-hand rule gives the direction of current.
• A coil rotating in a field produces sinusoidal emf: e = N A B ω sin ω t. This is the principle of the AC generator.
• Eddy currents are induced loops of current in bulk metal. They cause heating and loss; reduced by laminating cores. Used in induction furnaces, electromagnetic brakes, speedometers.
• Self-induction: a changing current in a coil induces emf in the same coil, e = −L dI/dt.
• Mutual induction: a changing current in one coil induces emf in a nearby coil, e₂ = −M dI₁/dt.
• Energy stored in an inductor = ½ L I².
• A transformer works on mutual induction and changes AC voltage. Step-up: Nₛ > Nₚ; step-down: Nₛ < Nₚ. Losses: copper, eddy current, hysteresis, flux leakage.

Important terms

• Induced emf – emf produced due to change in magnetic flux.
• Self-inductance (L) – flux linked per unit current, unit henry (H).
• Mutual inductance (M) – flux linked with secondary per unit current in primary.
• Back emf – induced emf in a motor coil that opposes the applied voltage.
• Laminated core – core made of thin insulated sheets to reduce eddy currents.

Formulas

5 important questions

Q1. Derive the expression for motional emf in a rod moving in a uniform magnetic field.

A rod of length l slides with speed v on rails in field B. In time dt the area swept is l v dt, so change in flux dφ = B l v dt. Induced emf e = dφ/dt = B l v. (Alternative: force on charges q v B balances q E, so E = v B and e = E l = B l v.)

Q2. State Lenz's law and show that it follows from the law of conservation of energy.

Induced current opposes the cause producing it. When a magnet's N-pole moves towards a coil, the near face of the coil becomes N and repels it; we must do work to push the magnet, and this work appears as electrical energy. If the induced current helped the motion, energy would be created from nothing, which is impossible.

Q3. Derive the expression for energy stored in an inductor.

To increase current by dI against induced emf L dI/dt, work dW = e I dt = L I dI. Total W = ∫ L I dI from 0 to I = ½ L I².

Q4. A coil of 200 turns has its flux changed from 0.02 Wb to 0.005 Wb in 0.1 s. Find the induced emf.

e = N Δφ/Δt = 200 × 0.015/0.1 = 30 V.

Q5. A transformer steps down 220 V to 11 V. The primary has 1000 turns. Find the secondary turns.

Nₛ = Nₚ Vₛ/Vₚ = 1000 × 11/220 = 50 turns.

Exam tip

Always mention the energy conservation argument with Lenz's law, and list the transformer losses with ways to reduce them.

Quick summary

This chapter studies alternating current (AC) circuits. It covers rms values, phasors, AC through a resistor, inductor and capacitor, the series LCR circuit and its impedance, resonance, sharpness (Q factor), power in AC circuits, wattless current and LC oscillations.

Key points

• AC changes magnitude and direction periodically: e = e₀ sin ω t, i = i₀ sin(ω t + φ).
• Average value of AC over a full cycle is zero; over a half cycle it is 2i₀/π = 0.637 i₀.
• rms (effective) value i(rms) = i₀/√2 = 0.707 i₀. AC meters read rms values.
• A phasor is a rotating vector whose projection gives the instantaneous value.
• Pure resistor: current and voltage are in phase.
• Pure inductor: current lags voltage by π/2; inductive reactance X_L = ω L.
• Pure capacitor: current leads voltage by π/2; capacitive reactance X_C = 1/(ω C).
• Series LCR: impedance Z = √(R² + (X_L − X_C)²); tan φ = (X_L − X_C)/R.
• At resonance X_L = X_C, Z = R (minimum), current is maximum, frequency f₀ = 1/(2π √(LC)).
• Average power P = V(rms) I(rms) cos φ; cos φ is the power factor. For pure L or C, cos φ = 0, so the current is wattless.
• In an LC circuit, energy oscillates between the capacitor's electric field and the inductor's magnetic field at f = 1/(2π √(LC)).

Important terms

• Reactance – opposition to AC by an inductor or capacitor (ohm).
• Impedance (Z) – total opposition to AC in a circuit (ohm).
• Power factor – cos φ, ratio of true power to apparent power.
• Wattless current – current component that consumes no power (I sin φ).
• Quality factor (Q) – sharpness of resonance, Q = ω₀ L/R = 1/(ω₀ C R).
• Choke coil – inductor used to reduce AC current with almost no power loss.

Formulas

5 important questions

Q1. Show that in a pure inductor the current lags behind the emf by π/2.

Applied emf e = e₀ sin ω t must equal L di/dt. So di = (e₀/L) sin ω t dt. Integrating, i = −(e₀/ω L) cos ω t = i₀ sin(ω t − π/2), with i₀ = e₀/(ω L). Thus current lags by π/2 and X_L = ω L.

Q2. Obtain the expression for impedance of a series LCR circuit using a phasor diagram.

Take current along the reference axis. V_R is in phase with I, V_L leads by 90°, V_C lags by 90°. Net voltage e = √(V_R² + (V_L − V_C)²) = I √(R² + (X_L − X_C)²). So Z = √(R² + (X_L − X_C)²) and tan φ = (X_L − X_C)/R.

Q3. Why is a choke coil preferred over a resistor to control AC current?

A choke has high inductance and very low resistance. Its power factor is nearly zero, so it reduces current with almost no energy loss, while a resistor wastes energy as heat (I²R).

Q4. Find the resonant frequency of an LCR circuit with L = 0.1 H and C = 10 μF.

f₀ = 1/(2π √(0.1 × 10⁻⁵)) = 1/(2π × 10⁻³) ≈ 159 Hz.

Q5. The peak value of an AC is 10 A. Find its rms value and its average value over a half cycle.

i(rms) = 10/1.414 ≈ 7.07 A. Average over half cycle = 0.637 × 10 = 6.37 A.

Exam tip

Draw phasor diagrams with the current as reference. Remember "ELI the ICE man": in L, E leads I; in C, I leads E.

Quick summary

Light sometimes behaves like a wave and sometimes like a stream of particles (photons). Matter particles like electrons can also behave like waves. This chapter covers the photoelectric effect, its experimental results, Einstein's photoelectric equation, photocells, de Broglie's matter waves and the Davisson-Germer experiment.

Key points

• Photoelectric effect: emission of electrons from a metal surface when light of suitable frequency falls on it.
• For each metal there is a threshold frequency ν₀. Below it, no electrons are emitted, however intense the light.
• Photoelectric emission is instantaneous (no time lag).
• Photoelectric current is proportional to intensity of light (for ν above ν₀).
• Maximum KE of photoelectrons depends on frequency, not intensity. Stopping potential V₀ measures it: ½ m v²(max) = e V₀.
• The wave theory cannot explain these results. Einstein explained them using photons of energy h ν.
• Einstein's equation: h ν = φ₀ + KE(max), where φ₀ = h ν₀ is the work function.
• A graph of stopping potential vs frequency is a straight line with slope h/e and intercept ν₀.
• de Broglie: every moving particle has a wave of wavelength λ = h/p = h/(m v).
• Davisson and Germer's electron diffraction from a nickel crystal confirmed matter waves. Electron microscopes use this wave nature.

Important terms

• Photon – a packet (quantum) of light energy E = h ν.
• Work function (φ₀) – minimum energy needed to free an electron from a metal surface.
• Threshold frequency (ν₀) – minimum frequency for photoemission; threshold wavelength λ₀ = c/ν₀.
• Stopping potential (V₀) – negative potential at which photocurrent becomes zero.
• Photocell – device that converts light into electric current (used in automatic doors, light meters, burglar alarms).

Formulas

h = 6.63 × 10⁻³⁴ J s; 1 eV = 1.6 × 10⁻¹⁹ J.

5 important questions

Q1. State the experimental observations of the photoelectric effect.

(1) Emission occurs only above a threshold frequency. (2) Photocurrent is proportional to light intensity. (3) Maximum KE increases linearly with frequency and is independent of intensity. (4) Emission is instantaneous.

Q2. Explain why the wave theory of light fails to explain the photoelectric effect.

Wave theory says energy depends on intensity, so bright light of any frequency should eject electrons, and dim light should need time to build up energy. Experiments show a threshold frequency, KE independent of intensity, and no time lag, which only the photon picture explains.

Q3. Derive the de Broglie wavelength of an electron accelerated through a potential difference V.

KE gained = e V = p²/(2m), so p = √(2 m e V). λ = h/p = h/√(2 m e V). Putting values, λ ≈ 1.227/√V nm.

Q4. The work function of a metal is 2 eV. Light of wavelength 400 nm falls on it. Find the maximum KE of photoelectrons.

Photon energy = h c/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J ≈ 3.1 eV. KE(max) = 3.1 − 2 = 1.1 eV.

Q5. Find the de Broglie wavelength of an electron accelerated through 100 V.

λ = 1.227/√100 = 0.1227 nm.

Exam tip

Practise the conversion between joules and eV and use h c = 1240 eV nm for quick photon energy calculations.

Quick summary

This chapter traces atomic models from Thomson and Rutherford to Bohr's model of the hydrogen atom, explains hydrogen spectral series, and then studies the nucleus: its size, composition, mass defect, binding energy, radioactive decay, half-life, and nuclear fission and fusion.

Key points

• Thomson's model: positive charge spread like a sphere with electrons embedded (plum pudding). It could not explain alpha-scattering.
• Rutherford's alpha-scattering: most alphas passed straight, a few deflected at large angles. So the atom has a tiny, dense, positive nucleus.
• Rutherford's model could not explain atomic stability or line spectra.
• Bohr's postulates: (1) electrons move in certain stable orbits without radiating; (2) in these orbits angular momentum = n h/2π; (3) radiation is emitted or absorbed only when an electron jumps between orbits: h ν = E₂ − E₁.
• For hydrogen, radius rₙ ∝ n² (r₁ = 0.053 nm) and energy Eₙ = −13.6/n² eV.
• Spectral series of hydrogen: Lyman (to n = 1, UV), Balmer (to n = 2, visible), Paschen (n = 3), Brackett (n = 4), Pfund (n = 5) – the last three in infrared.
• de Broglie explained Bohr's quantum condition: an orbit is allowed if its circumference fits a whole number of electron wavelengths, 2π r = n λ.
• Nuclear radius R = R₀ A^(1/3); nuclear density is nearly the same for all nuclei.
• Mass of a nucleus is less than the sum of masses of its nucleons; this mass defect appears as binding energy (E = Δm c²; 1 u = 931.5 MeV).
• Radioactive decay law: N = N₀ e^(−λ t). Half-life T½ = 0.693/λ; mean life τ = 1/λ.
• Fission: a heavy nucleus splits (U-235 with a neutron) releasing about 200 MeV. Fusion: light nuclei join (as in the Sun) releasing energy.

Important terms

• Atomic number (Z) – number of protons; mass number (A) – number of nucleons.
• Isotopes – same Z, different A.
• Mass defect – difference between total mass of nucleons and actual nuclear mass.
• Binding energy per nucleon – measure of nuclear stability; maximum near iron (about 8.8 MeV).
• Half-life – time in which half the radioactive nuclei decay.
• Activity – rate of decay, A = λ N (unit becquerel, Bq).

Formulas

5 important questions

Q1. State Bohr's postulates of the hydrogen atom.

1. An electron revolves in certain circular orbits (stationary orbits) without radiating energy. 2. In these orbits its angular momentum is an integral multiple of h/2π. 3. When it jumps from a higher orbit E₂ to a lower orbit E₁, a photon of energy h ν = E₂ − E₁ is emitted.

Q2. Derive the expression for the radius of the nth Bohr orbit.

Electrostatic force gives centripetal force: m v²/r = Z e²/(4π ε₀ r²). Bohr's condition: m v r = n h/2π, so v = n h/(2π m r). Substituting and solving: rₙ = ε₀ n² h²/(π m Z e²). So r ∝ n².

Q3. Explain Bohr's second postulate using de Broglie's hypothesis.

An electron in an orbit is a standing wave. For a stable orbit, the circumference must hold a whole number of wavelengths: 2π r = n λ = n h/(m v). This gives m v r = n h/2π, which is Bohr's quantum condition.

Q4. Find the wavelength of the first line of the Balmer series (R = 1.097 × 10⁷ m⁻¹).

1/λ = R (1/4 − 1/9) = R × 5/36 = 1.524 × 10⁶ m⁻¹. λ ≈ 656 nm (red line).

Q5. The half-life of a sample is 10 days. What fraction remains after 30 days?

30 days = 3 half-lives. N/N₀ = (1/2)³ = 1/8.

Exam tip

Remember the five series with their lower levels and regions, and practise one binding energy numerical using 1 u = 931.5 MeV.

Quick summary

This chapter deals with applications of p-n junctions and transistors. It covers half-wave and full-wave rectifiers, filter circuits, special diodes (Zener diode, photodiode, solar cell, LED), the bipolar junction transistor and its use as an amplifier and switch, and logic gates.

Key points

• A p-n junction diode conducts in forward bias and blocks in reverse bias, so it can rectify AC.
• Half-wave rectifier: one diode, output only during one half cycle. Ripple frequency = supply frequency.
• Full-wave rectifier (centre-tap or bridge): output in both half cycles. Ripple frequency = twice the supply frequency. Efficiency is higher.
• Filters (capacitor, inductor, π filter) smooth the pulsating DC output.
• Zener diode is heavily doped and works in reverse breakdown. Its voltage stays nearly constant, so it is used as a voltage regulator.
• Photodiode works in reverse bias; light increases reverse current. Used in light detectors.
• Solar cell works without bias; light creates electron-hole pairs and an emf (photovoltaic effect).
• LED emits light in forward bias when electrons and holes recombine. Colour depends on band gap of the material.
• A transistor (n-p-n or p-n-p) has emitter, base and collector. Emitter is heavily doped, base is thin and lightly doped, collector is large.
• In a common emitter (CE) amplifier, a small base current controls a large collector current; output is 180° out of phase with input.
• Logic gates: OR, AND, NOT are basic gates; NAND and NOR are universal gates (any gate can be made from them).

Important terms

• Rectifier – circuit converting AC into DC.
• Ripple – AC component left in rectified output.
• Current gain α (CB) – α = I_C/I_E (slightly less than 1).
• Current gain β (CE) – β = I_C/I_B (large, e.g. 50–300).
• Truth table – table showing outputs for all input combinations.
• XOR gate – output is 1 only when inputs are different.

Formulas

5 important questions

Q1. Explain the working of a full-wave (centre-tap) rectifier with a circuit diagram.

The secondary of a centre-tap transformer feeds two diodes D₁ and D₂. In the first half cycle, D₁ is forward biased and conducts through the load. In the next half cycle, D₂ conducts. Current through the load is always in the same direction, so output is obtained in both halves.

Q2. How is a Zener diode used as a voltage regulator?

The Zener is connected in reverse bias across the load with a series resistor. If input voltage rises, Zener current rises and the extra drop appears across the series resistor; the voltage across the Zener (and load) stays at V_Z. Similarly for a fall in input.

Q3. Why are NAND and NOR gates called universal gates?

Any basic gate (OR, AND, NOT) can be made using only NAND gates or only NOR gates. Example: joining both inputs of a NAND gate gives a NOT gate.

Q4. In a transistor, I_E = 10 mA and I_B = 0.2 mA. Find I_C, α and β.

I_C = 10 − 0.2 = 9.8 mA. α = 9.8/10 = 0.98. β = 9.8/0.2 = 49.

Q5. Write the truth table of a NOR gate.

Exam tip

Practise drawing rectifier circuits with input and output waveforms, and memorise truth tables of all six gates.