Physics depends on measurement. This chapter explains the SI system of units, how very large and very small lengths, masses and times are measured, dimensional analysis, and how to handle errors and significant figures in measured values.
महत्त्वाचे मुद्दे
A physical quantity = numerical value × unit (for example 5 m). A good unit is well defined, easy to reproduce and does not change with time or place.
The SI system has 7 fundamental (base) quantities: length (metre, m), mass (kilogram, kg), time (second, s), electric current (ampere, A), temperature (kelvin, K), amount of substance (mole, mol) and luminous intensity (candela, cd).
Derived quantities (area, velocity, force, energy) are expressed using base units, e.g. newton = kg m s⁻².
Large distances (planets, stars) are measured by the parallax method: D = b/θ, where b is the basis (distance between two observation points) and θ is the parallax angle in radian.
Units for large distances: astronomical unit (1 AU ≈ 1.496 × 10¹¹ m), light year (1 ly ≈ 9.46 × 10¹⁵ m), parsec (1 pc ≈ 3.08 × 10¹⁶ m).
Very small sizes are measured with tools like the electron microscope; atomic sizes are of the order of 10⁻¹⁰ m. Atomic masses are expressed in unified atomic mass unit, 1 u ≈ 1.66 × 10⁻²⁷ kg.
Time is measured very precisely with atomic (cesium) clocks.
Dimensions show how a quantity depends on M, L, T etc. Dimensional analysis is used to check equations, derive relations and convert units.
Limitations: it cannot find dimensionless constants (like ½ or 2π), cannot handle sums of trigonometric/exponential functions, and cannot tell if a quantity is scalar or vector.
Accuracy = closeness to the true value. Precision = closeness of repeated readings to each other (depends on least count).
Errors are systematic (instrumental, due to method, personal) or random. Mean of many readings reduces random error.
Significant figures show the reliability of a measured value.
महत्त्वाच्या संज्ञा
Fundamental quantity – a quantity that does not depend on other quantities, e.g. mass.
Derived quantity – a quantity expressed in terms of fundamental quantities, e.g. speed.
Dimensional formula – a quantity written in powers of M, L, T, e.g. force = [M¹L¹T⁻²].
Least count – the smallest value an instrument can measure.
Absolute error – difference between the true (mean) value and a measured value.
Relative error – mean absolute error divided by the mean value.
Parallax – apparent shift of an object when seen from two different points.
Product or quotient Z = AB or A/B: ΔZ/Z = ΔA/A + ΔB/B
Power Z = Aⁿ: ΔZ/Z = n (ΔA/A)
Common dimensions: velocity [L¹T⁻¹], acceleration [L¹T⁻²], force [M¹L¹T⁻²], work/energy [M¹L²T⁻²], pressure [M¹L⁻¹T⁻²], G [M⁻¹L³T⁻²]
५ महत्त्वाचे प्रश्न
Q1. State the seven SI base quantities with their units.
Length – metre (m), mass – kilogram (kg), time – second (s), electric current – ampere (A), temperature – kelvin (K), amount of substance – mole (mol), luminous intensity – candela (cd).
Q2. Check the correctness of v = u + at using dimensions.
[v] = [L¹T⁻¹]; [u] = [L¹T⁻¹]; [at] = [L¹T⁻²][T¹] = [L¹T⁻¹]. All terms have the same dimensions, so the equation is dimensionally correct.
Q3. Convert 1 newton into dyne using dimensional analysis.
Force = [M¹L¹T⁻²]. n₂ = n₁ (M₁/M₂)(L₁/L₂)(T₁/T₂)⁻² = 1 × (1000 g/1 g)(100 cm/1 cm)(1)⁻² = 10⁵. So 1 N = 10⁵ dyne.
Q4. Readings of a length are 2.63, 2.56, 2.42, 2.71 and 2.80 cm. Find the mean value, mean absolute error and percentage error.
Mean = 13.12/5 = 2.624 ≈ 2.62 cm. Absolute errors: 0.01, 0.06, 0.20, 0.09, 0.18 → mean absolute error = 0.54/5 ≈ 0.11 cm. Percentage error = (0.11/2.62) × 100 ≈ 4.2%. Result: (2.62 ± 0.11) cm.
Q5. Distinguish between accuracy and precision.
Accuracy tells how close a measurement is to the true value. Precision tells how close repeated measurements are to each other and how fine the instrument's least count is. A measurement can be precise but not accurate (for example with a zero error).
परीक्षेसाठी टीप
Learn the rules of combination of errors and practise one dimensional conversion — these come almost every year.
This chapter gives the maths tools used in physics: scalars and vectors, adding and resolving vectors, the dot and cross product, and a first introduction to differential and integral calculus.
महत्त्वाचे मुद्दे
A scalar has only magnitude (mass, distance, speed, work, energy). A vector has magnitude and direction (displacement, velocity, force).
A vector is shown by an arrow; its length shows magnitude. A unit vector has magnitude 1 and shows only direction: û = u⃗/|u⃗|. î, ĵ, k̂ are unit vectors along x, y, z.
Equal vectors have the same magnitude and direction. A negative vector has the same magnitude but opposite direction. A zero (null) vector has zero magnitude.
Vector addition follows the triangle law and the parallelogram law. Addition is commutative and associative.
Resolution: a vector can be split into components, A⃗ = Aₓî + A_yĵ, with Aₓ = A cos θ and A_y = A sin θ.
Scalar (dot) product gives a scalar: A⃗·B⃗ = AB cos θ. Example: work W = F⃗·s⃗.
Vector (cross) product gives a vector perpendicular to both: |A⃗ × B⃗| = AB sin θ; direction by the right-hand screw rule. Example: torque τ⃗ = r⃗ × F⃗.
A⃗ × B⃗ = −(B⃗ × A⃗): the cross product is not commutative.
Differential calculus gives the rate of change: velocity v = dx/dt, acceleration a = dv/dt.
Integral calculus is summation: displacement = ∫v dt; area under a graph.
महत्त्वाच्या संज्ञा
Resultant vector – single vector having the same effect as two or more vectors together.
Resolution of a vector – splitting a vector into two or more components.
Unit vector – vector of magnitude one used to show direction.
Derivative – instantaneous rate of change of one quantity with respect to another.
Integration – reverse process of differentiation; continuous summation.
सूत्रे
Magnitude of resultant (parallelogram law): R = √(P² + Q² + 2PQ cos θ)
Direction: tan α = Q sin θ / (P + Q cos θ), α measured from P⃗
Dot product: A⃗·B⃗ = AB cos θ = AₓBₓ + A_yB_y + A_zB_z ; î·î = 1, î·ĵ = 0
Cross product: |A⃗ × B⃗| = AB sin θ ; î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ, î × î = 0
Derivatives: d(xⁿ)/dx = n xⁿ⁻¹ ; d(sin x)/dx = cos x ; d(cos x)/dx = −sin x ; d(eˣ)/dx = eˣ
Integrals: ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C (n ≠ −1) ; ∫(1/x) dx = log x + C
५ महत्त्वाचे प्रश्न
Q1. Distinguish between scalar and vector quantities with two examples each.
Scalar: only magnitude, added by ordinary algebra – mass, work. Vector: magnitude and direction, added by vector laws – force, velocity.
Q2. Two forces of 3 N and 4 N act at 90°. Find the resultant.
R = √(3² + 4² + 2 × 3 × 4 × cos 90°) = √(9 + 16 + 0) = 5 N. tan α = 4/3, so α ≈ 53° from the 3 N force.
Q3. Find the angle between A⃗ = î + ĵ and B⃗ = î − ĵ.
A⃗·B⃗ = (1)(1) + (1)(−1) = 0. Since the dot product is zero, cos θ = 0 and θ = 90°. The vectors are perpendicular.
Q4. State any two properties of the vector product.
(1) It is not commutative: A⃗ × B⃗ = −B⃗ × A⃗. (2) The cross product of two parallel vectors is a null vector (sin 0° = 0). It is distributive over addition.
Q5. If x = 3t² + 2t, find velocity and acceleration at t = 2 s.
v = dx/dt = 6t + 2 = 14 m/s at t = 2 s. a = dv/dt = 6 m/s² (constant).
परीक्षेसाठी टीप
Remember: dot product → cos θ and scalar; cross product → sin θ and vector. Draw a neat parallelogram diagram in vector-addition answers.
This chapter studies motion in a straight line and in a plane (two dimensions): velocity, acceleration, equations of motion, relative velocity, projectile motion and uniform circular motion.
महत्त्वाचे मुद्दे
Rectilinear motion is motion along a straight line. Position, displacement, velocity and acceleration are vectors.
Average velocity = total displacement / total time. Instantaneous velocity v = dx/dt (slope of x–t graph).
Average acceleration = change in velocity / time. Instantaneous acceleration a = dv/dt (slope of v–t graph).
Area under a v–t graph gives displacement.
In two dimensions, r⃗ = xî + yĵ and motion along x and y can be treated separately.
Relative velocity of A with respect to B: v⃗_AB = v⃗_A − v⃗_B.
A projectile moves under gravity alone. Horizontal velocity stays constant; vertical motion has acceleration g downward. Its path is a parabola.
Maximum range is obtained at θ = 45°. Angles θ and (90° − θ) give the same range.
In uniform circular motion, speed is constant but velocity changes direction, so there is centripetal acceleration towards the centre.
Angular displacement is in radian; angular velocity ω = dθ/dt; v = rω.
महत्त्वाच्या संज्ञा
Projectile – a body thrown into the air that moves only under gravity.
Trajectory – path followed by a projectile (a parabola).
Time of flight – total time the projectile stays in air.
Range – horizontal distance covered by a projectile.
Centripetal acceleration – acceleration of a body in circular motion, directed towards the centre.
Period (T) – time for one revolution.
सूत्रे
Equations of motion (constant a): v = u + at ; s = ut + ½at² ; v² = u² + 2as
Distance in nth second: sₙ = u + a(2n − 1)/2
Time of flight T = 2u sin θ / g
Maximum height H = u² sin²θ / 2g
Horizontal range R = u² sin 2θ / g ; R_max = u²/g at 45°
Equation of path: y = (tan θ) x − g x² / (2u² cos²θ)
Projectile (speed u at angle θ to horizontal):
Circular motion: v = rω ; ω = 2π/T = 2πn ; centripetal acceleration a = v²/r = ω²r
Units: velocity m/s, acceleration m/s², ω rad/s
५ महत्त्वाचे प्रश्न
Q1. Show that the path of a projectile is a parabola.
x = (u cos θ) t gives t = x/(u cos θ). Vertically y = (u sin θ) t − ½gt². Substituting t: y = (tan θ)x − [g/(2u² cos²θ)]x². This is of the form y = ax − bx², which is a parabola.
Q2. Derive the expression for centripetal acceleration.
Let a particle move on a circle of radius r with speed v. In a small time Δt it turns through Δθ and the velocity vector also turns through Δθ, so |Δv⃗| = vΔθ. Acceleration a = vΔθ/Δt = vω. Using v = rω, a = v²/r = ω²r, directed towards the centre.
Q3. A ball is thrown with 20 m/s at 30° to the horizontal. Find time of flight, maximum height and range (g = 10 m/s²).
T = 2 × 20 × sin 30°/10 = 2 s. H = (20² × 0.25)/(2 × 10) = 5 m. R = 400 × sin 60°/10 = 40 × 0.866 ≈ 34.6 m.
Q4. Two trains move in the same direction at 60 km/h and 40 km/h. Find the velocity of the first relative to the second.
v_AB = 60 − 40 = 20 km/h in the direction of motion. If they move in opposite directions, v_AB = 60 − (−40) = 100 km/h.
Q5. A wheel makes 300 rotations per minute. Find its angular speed.
n = 300/60 = 5 rotations per second. ω = 2πn = 10π ≈ 31.4 rad/s.
परीक्षेसाठी टीप
For projectile sums, first write the horizontal and vertical components of velocity (u cos θ, u sin θ); every formula follows from them.
This chapter covers Newton's three laws, inertial and non-inertial frames, types of forces, work-energy theorem, conservation of momentum, collisions, impulse, torque, couple, equilibrium, centre of mass and centre of gravity.
महत्त्वाचे मुद्दे
Aristotle wrongly believed a force is needed to keep a body moving. Galileo and Newton showed a body keeps moving unless a net force acts.
First law (law of inertia): a body stays at rest or in uniform straight-line motion unless an external unbalanced force acts on it.
Second law: rate of change of momentum equals the applied force, F⃗ = dp⃗/dt = ma⃗ (for constant mass).
Third law: to every action there is an equal and opposite reaction; they act on different bodies.
Newton's laws hold in inertial frames (not accelerating). In a non-inertial (accelerating) frame we add a pseudo force = −ma₀.
Four fundamental forces: gravitational, electromagnetic, strong nuclear and weak nuclear.
Conservative force (gravity, spring force): work done does not depend on path and is zero in a closed path. Friction is non-conservative.
Work-energy theorem: work done by net force = change in kinetic energy.
Law of conservation of linear momentum: if no external force acts, total momentum of a system stays constant.
In elastic collisions both momentum and kinetic energy are conserved; in inelastic collisions only momentum is conserved. In a perfectly inelastic collision the bodies stick together.
Coefficient of restitution e: e = 1 (perfectly elastic), 0 < e < 1 (partially elastic), e = 0 (perfectly inelastic).
Torque is the turning effect of a force. A couple has two equal, opposite, non-collinear forces; its moment is the same about every point.
Equilibrium: net force = 0 (translational) and net torque = 0 (rotational). Equilibrium can be stable, unstable or neutral.
Centre of mass is the point where the whole mass may be taken to be concentrated. Centre of gravity is where the resultant gravitational force acts; they coincide in a uniform gravitational field.
महत्त्वाच्या संज्ञा
Momentum – product of mass and velocity, p = mv (kg m/s).
Impulse – product of force and time, equal to change in momentum (N s).
Pseudo force – imaginary force used in an accelerating frame.
Torque – moment of force about an axis, τ⃗ = r⃗ × F⃗ (N m).
Coefficient of restitution – ratio of relative speed of separation to relative speed of approach.
सूत्रे
F = dp/dt = ma ; p = mv
Impulse J = F × Δt = mv − mu
Work-energy theorem: W = ½mv² − ½mu²
Momentum conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
e = (v₂ − v₁)/(u₁ − u₂)
Perfectly inelastic: common velocity v = (m₁u₁ + m₂u₂)/(m₁ + m₂)
Torque τ = rF sin θ ; moment of couple = F × perpendicular distance
Centre of mass (two particles): x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂)
५ महत्त्वाचे प्रश्न
Q1. State Newton's laws of motion.
(1) A body continues in its state of rest or uniform motion in a straight line unless acted upon by an external unbalanced force. (2) The rate of change of momentum is proportional to the applied force and takes place in the direction of force. (3) Every action has an equal and opposite reaction.
Q2. Distinguish between elastic and inelastic collisions.
Elastic: momentum and kinetic energy both conserved; e = 1; forces are conservative (e.g. collisions of gas molecules). Inelastic: momentum conserved but kinetic energy not conserved; e < 1; part of KE becomes heat/sound.
Q3. A 50 g bullet moving at 400 m/s hits and stops in a wooden block in 0.01 s. Find the average force.
Change in momentum = 0.05 × (0 − 400) = −20 kg m/s. F = Δp/Δt = 20/0.01 = 2000 N (opposing the motion).
Q4. A 2 kg ball at 3 m/s hits a 1 kg ball at rest and they stick together. Find the common velocity and loss in KE.
v = (2 × 3 + 0)/(2 + 1) = 2 m/s. Initial KE = ½ × 2 × 9 = 9 J. Final KE = ½ × 3 × 4 = 6 J. Loss = 3 J.
Q5. Show that the moment of a couple does not depend on the point about which it is taken.
Let forces F and −F act at points A and B, with perpendicular distances x and (x + d) from any point O. Total moment = F(x + d) − Fx = Fd. It depends only on F and the arm d, not on the position of O.
परीक्षेसाठी टीप
In collision problems, always write the momentum equation first, then use e or the KE condition. Show direction with + and − signs.
This chapter explains Kepler's laws, Newton's universal law of gravitation, acceleration due to gravity and how it changes, gravitational potential energy, escape velocity, and the motion of earth satellites.
महत्त्वाचे मुद्दे
Kepler's 1st law (orbits): planets move in elliptical orbits with the Sun at one focus.
Kepler's 2nd law (areas): the line joining a planet to the Sun sweeps equal areas in equal times, so a planet moves faster when nearer the Sun.
Kepler's 3rd law (periods): T² ∝ a³, where a is the semi-major axis.
Newton's law: every two masses attract each other with F = Gm₁m₂/r². G = 6.67 × 10⁻¹¹ N m² kg⁻²; it is a universal constant, first measured by Cavendish.
Acceleration due to gravity on the earth's surface: g = GM/R² ≈ 9.8 m/s².
g decreases with height and with depth, and is zero at the centre of the earth.
g is least at the equator and greatest at the poles, because of the earth's rotation and its flattened shape.
Gravitational potential energy of mass m at distance r from earth's centre: U = −GMm/r (taken zero at infinity).
Escape velocity is the minimum speed needed to escape the earth's gravity; for the earth it is about 11.2 km/s. It does not depend on the mass of the body.
A satellite needs a horizontal speed (critical velocity) to stay in orbit. Gravity provides the centripetal force.
A geostationary satellite has period 24 h and orbits about 36,000 km above the equator.
Astronauts in an orbiting satellite feel weightless because they and the satellite are both in free fall.
महत्त्वाच्या संज्ञा
Gravitational constant (G) – force between two 1 kg masses kept 1 m apart.
Escape velocity – minimum velocity to escape from a planet's gravitational field.
Critical (orbital) velocity – horizontal velocity needed to keep a satellite in a circular orbit.
Binding energy – minimum energy needed to free a satellite from earth's gravitational pull.
Weightlessness – state in which apparent weight is zero, as in free fall.
सूत्रे
F = Gm₁m₂/r²
g = GM/R²
With height h: g_h = g R²/(R + h)² ≈ g(1 − 2h/R) for h ≪ R
With depth d: g_d = g(1 − d/R)
Latitude: g_φ = g − Rω² cos²φ
Potential energy U = −GMm/r ; potential V = −GM/r
Escape velocity vₑ = √(2GM/R) = √(2gR)
Critical velocity vc = √(GM/(R + h)) ; near the surface vc = √(gR)
Period of satellite T = 2π √((R + h)³/GM)
Binding energy of orbiting satellite = GMm/2(R + h)
५ महत्त्वाचे प्रश्न
Q1. State Kepler's three laws of planetary motion.
(1) Each planet moves in an elliptical orbit with the Sun at one focus. (2) The radius vector sweeps equal areas in equal intervals of time. (3) The square of the period is proportional to the cube of the semi-major axis (T² ∝ a³).
Q2. Derive the expression for escape velocity.
Total energy at the surface = ½mvₑ² − GMm/R. To just escape, total energy must be zero (at infinity KE = 0, PE = 0). So ½mvₑ² = GMm/R, giving vₑ = √(2GM/R) = √(2gR), since GM = gR².
Q3. Find the escape velocity from the earth (g = 9.8 m/s², R = 6400 km).
Q4. At what height above the surface does g become one-fourth of its value on the surface?
g_h/g = R²/(R + h)² = 1/4, so (R + h)/R = 2 and h = R ≈ 6400 km.
Q5. Why does an astronaut feel weightless in an orbiting satellite?
The satellite and the astronaut both fall freely towards the earth with the same acceleration (equal to g at that height). The floor gives no reaction force on the astronaut, so the apparent weight is zero.
परीक्षेसाठी टीप
Remember GM = gR² — it converts almost every gravitation formula into one with g and R that you can calculate easily.
This chapter explains how solids behave when forces deform them: elasticity, stress and strain, Hooke's law, elastic moduli, the stress–strain curve, strain energy, hardness and friction in solids.
महत्त्वाचे मुद्दे
Elasticity is the property of a body to regain its original shape and size when the deforming force is removed. Plasticity is the opposite (e.g. putty, clay).
Inside a deformed body, a restoring force per unit area develops; this is stress.
Three kinds of strain: longitudinal (change in length), volume (change in volume) and shearing (change in shape).
Hooke's law: within the elastic limit, stress is directly proportional to strain.
Young's modulus (Y) is for change in length, bulk modulus (K) for change in volume and modulus of rigidity (η) for change in shape.
Compressibility is the reciprocal of bulk modulus. Liquids and gases have only bulk modulus.
Poisson's ratio = lateral strain / longitudinal strain; it has no unit.
Ductile materials (copper, aluminium) have a large plastic region and can be drawn into wires. Brittle materials (glass, cast iron) break soon after the elastic limit.
Work done in stretching a wire is stored as strain energy.
Hardness is resistance to permanent deformation such as scratching or denting.
Friction opposes relative motion between surfaces in contact. Its origin is the interlocking and adhesion of tiny irregularities on surfaces.
Static friction > kinetic (sliding) friction > rolling friction.
महत्त्वाच्या संज्ञा
Stress – restoring force per unit area (N/m²).
Strain – change in dimension divided by original dimension (no unit).
Elastic limit – maximum stress up to which a body remains elastic.
Breaking stress – stress at which the wire breaks.
Coefficient of friction (μ) – ratio of friction force to normal reaction.
Strain energy per unit volume = ½ × stress × strain
Friction: f_s ≤ μ_s N ; f_k = μ_k N
Units: stress and all moduli in N/m² (Pa)
५ महत्त्वाचे प्रश्न
Q1. State Hooke's law and define Young's modulus.
Hooke's law: within the elastic limit, stress is directly proportional to strain. Young's modulus is the ratio of longitudinal stress to longitudinal strain within the elastic limit.
Q2. A wire 2 m long and 1 mm² in cross-section stretches by 1 mm under a 100 N load. Find Young's modulus.
Q3. Show that strain energy per unit volume = ½ × stress × strain.
Work done in stretching a wire by l with final force F is ½Fl (average force × extension). Dividing by volume AL: U/V = ½ (F/A)(l/L) = ½ × stress × strain.
Q4. Distinguish between ductile and brittle materials.
Ductile materials have a large plastic region between the elastic limit and the breaking point, so they can be drawn into wires (copper, gold). Brittle materials break soon after the elastic limit with little plastic flow (glass, ceramics).
Q5. What is the origin of friction? Name the types.
When two surfaces touch, their microscopic bumps interlock and atoms of the two surfaces form temporary bonds (adhesion). Breaking these needs force, which appears as friction. Types: static, kinetic (sliding) and rolling friction.
परीक्षेसाठी टीप
Draw and label the stress–strain curve neatly; it is a favourite 3-mark question.
This chapter deals with temperature and heat, thermometers and temperature scales, the ideal gas equation, thermal expansion, specific heat, calorimetry, change of state, latent heat, modes of heat transfer and Newton's law of cooling.
महत्त्वाचे मुद्दे
Heat is energy that flows from a hotter body to a colder one because of temperature difference. Temperature tells how hot or cold a body is.
Celsius, Fahrenheit and Kelvin scales are related by C/100 = (F − 32)/180 = (T − 273.15)/100.
Absolute zero (0 K = −273.15 °C) is the lowest possible temperature.
Ideal gas equation PV = nRT combines Boyle's law and Charles' law. R = 8.31 J mol⁻¹ K⁻¹.
Solids expand on heating: linear, areal and volume expansion, with β = 2α and γ = 3α.
Water shows anomalous expansion: it contracts when heated from 0 °C to 4 °C, and has maximum density at 4 °C.
Specific heat capacity is heat needed to raise the temperature of 1 kg by 1 K. Water has a high specific heat.
For gases, Cp > Cv and Cp − Cv = R (Mayer's relation).
Principle of calorimetry: heat lost by the hot body = heat gained by the cold body (no losses).
During change of state (melting, boiling) temperature stays constant; the heat used is latent heat.
A phase diagram shows solid, liquid and vapour regions; the triple point is where all three phases coexist (for water, 273.16 K and 0.006 atm).
Heat transfer: conduction (solids, no movement of particles as a whole), convection (fluids, by movement of the medium) and radiation (no medium needed).
Newton's law of cooling: rate of loss of heat is proportional to the temperature difference between the body and surroundings (for small differences).
महत्त्वाच्या संज्ञा
Coefficient of linear expansion (α) – increase in length per unit original length per degree rise in temperature.
Heat capacity – heat needed to raise the temperature of a whole body by 1 K (J/K).
Latent heat – heat absorbed or released per unit mass during change of state at constant temperature (J/kg).
Sublimation – direct change from solid to vapour (e.g. camphor, dry ice).
Thermal conductivity (k) – rate of heat flow per unit area per unit temperature gradient.
C/100 = (98.6 − 32)/180 → C = 66.6/1.8 = 37 °C. T = 37 + 273.15 = 310.15 K.
Q2. Explain the anomalous behaviour of water.
When water is heated from 0 °C to 4 °C it contracts instead of expanding; above 4 °C it expands normally. So water is densest at 4 °C. That is why lakes freeze from the top, and fish survive in the water at 4 °C below the ice.
Q3. How much heat is needed to convert 1 kg of ice at 0 °C into water at 20 °C? (L = 3.34 × 10⁵ J/kg, c = 4186 J/kg K)
Melting: 1 × 3.34 × 10⁵ = 334000 J. Heating: 1 × 4186 × 20 = 83720 J. Total ≈ 4.18 × 10⁵ J.
Q4. State the three modes of heat transfer with one example each.
Conduction – a metal spoon in hot tea becomes hot. Convection – sea breeze, boiling water. Radiation – heat from the Sun reaching the earth through empty space.
Q5. State Newton's law of cooling and its limitation.
The rate of loss of heat of a body is directly proportional to the difference between its temperature and that of the surroundings. It is valid only for small temperature differences, and when heat is lost mainly by convection.
परीक्षेसाठी टीप
Learn the relation between α, β and γ and practise one calorimetry problem — write the "heat lost = heat gained" line clearly.
This chapter treats sound as a mechanical wave. It covers types of waves, the wave equation, speed of sound (Newton's formula and Laplace's correction), superposition, echo, reverberation, acoustics, qualities of sound and the Doppler effect.
महत्त्वाचे मुद्दे
A wave carries energy from one place to another without the medium moving along with it. Sound is a mechanical wave and needs a medium.
Transverse waves: particles vibrate perpendicular to the direction of the wave (waves on a string); they have crests and troughs.
Longitudinal waves: particles vibrate along the direction of the wave (sound in air); they have compressions and rarefactions.
Wave speed v = nλ. Amplitude, wavelength, frequency, period and phase describe a wave.
Speed of a transverse wave on a string depends on tension and mass per unit length.
Newton assumed sound travels isothermally, giving v = √(P/ρ) ≈ 280 m/s in air, which is too low.
Laplace said compressions and rarefactions happen fast, so the process is adiabatic: v = √(γP/ρ) ≈ 331 m/s at 0 °C, which matches experiment.
Speed of sound does not change with pressure (at constant temperature), increases with temperature (v ∝ √T) and with humidity (moist air is less dense).
Superposition: when waves overlap, the resultant displacement is the vector sum of individual displacements.
An echo is heard when the reflected sound reaches us at least 0.1 s later, so the obstacle must be at least about 17 m away.
Reverberation is the persistence of sound in a hall due to repeated reflections. Acoustics is the design of halls for good hearing.
Qualities of sound: pitch (depends on frequency), loudness (depends on intensity/amplitude) and quality or timbre (depends on overtones/harmonics).
Doppler effect: the observed frequency changes when the source or listener moves. It is higher when they approach and lower when they move apart.
महत्त्वाच्या संज्ञा
Wavelength (λ) – distance between two successive particles in the same phase.
Frequency (n) – number of vibrations per second (Hz).
Intensity – energy passing per unit area per unit time.
Reverberation time – time for sound to die down to inaudible level after the source stops.
Doppler effect – apparent change in frequency due to relative motion of source and listener.
सूत्रे
v = nλ ; T = 1/n
Wave equation: y = A sin(ωt − kx), with ω = 2πn and k = 2π/λ
Transverse wave on string: v = √(T/m), T = tension, m = mass per unit length
Newton's formula: v = √(P/ρ) ; Laplace's correction: v = √(γP/ρ) (γ = 1.41 for air)
Temperature: v ∝ √T (T in kelvin) ; v_t ≈ v₀ + 0.61 t m/s
Source approaching stationary listener: n = n₀ v/(v − v_s)
Listener approaching stationary source: n = n₀ (v + v_L)/v
Both moving towards each other: n = n₀ (v + v_L)/(v − v_s)
Doppler effect (v = speed of sound, v_s source, v_L listener):
५ महत्त्वाचे प्रश्न
Q1. Explain Laplace's correction to Newton's formula.
Newton assumed sound travels isothermally (PV = constant), giving v = √(P/ρ) ≈ 280 m/s, about 16% less than the measured value. Laplace argued that compressions and rarefactions are so fast that heat cannot be exchanged, so the change is adiabatic (PV^γ = constant). Then v = √(γP/ρ) ≈ 331 m/s, which agrees with experiment.
Q2. Distinguish between transverse and longitudinal waves.
Transverse: particles vibrate perpendicular to wave direction; crests and troughs; e.g. string waves, light. Longitudinal: particles vibrate parallel to wave direction; compressions and rarefactions; e.g. sound in air.
Q3. A tuning fork of 512 Hz produces waves of wavelength 0.66 m in air. Find the speed of sound.
v = nλ = 512 × 0.66 ≈ 338 m/s.
Q4. A train approaches a station at 20 m/s blowing a 600 Hz whistle. What frequency does a person on the platform hear? (v = 340 m/s)
n = n₀ v/(v − v_s) = 600 × 340/320 = 637.5 Hz.
Q5. Why is sound heard better on a rainy day? Why does pressure not affect speed of sound?
Humid air is less dense than dry air, so speed of sound is greater. At constant temperature, P/ρ stays constant (Boyle's law), so changing pressure does not change v = √(γP/ρ).
परीक्षेसाठी टीप
For Doppler sums, remember: approaching → frequency increases. Use this to check if your sign is right.
This chapter covers the nature of light and ray optics: reflection at plane and curved mirrors, refraction, total internal reflection, refraction at spherical surfaces and lenses, dispersion by a prism, natural phenomena like the rainbow, defects of lenses and optical instruments.
महत्त्वाचे मुद्दे
Light behaves both as a wave and as a particle (photon). Ray optics treats light as rays travelling in straight lines.
Cartesian sign convention: distances are measured from the pole/optical centre; along the incident light they are positive, against it negative; heights above the axis positive.
Laws of reflection: angle of incidence = angle of reflection, and the incident ray, reflected ray and normal lie in one plane.
For a spherical mirror, focal length f = R/2.
Refraction: Snell's law n₁ sin i = n₂ sin r. Light bends towards the normal when entering a denser medium.
Total internal reflection happens when light goes from denser to rarer medium and the angle of incidence exceeds the critical angle. Uses: optical fibres, sparkling of diamond, mirage, totally reflecting prisms.
Lens maker's equation links focal length to refractive index and radii of curvature.
Power of a lens P = 1/f (f in metre), unit dioptre (D). Convex lens power is positive, concave negative.
A prism deviates light; at minimum deviation, the ray passes symmetrically through the prism.
Dispersion is splitting of white light into colours because refractive index is different for each colour (violet deviates most, red least).
Rainbow (dispersion + total internal reflection in raindrops), blue sky (scattering, which is more for shorter wavelengths) and red sunrise/sunset are due to sunlight.
Lens defects: spherical aberration and chromatic aberration (corrected by an achromatic doublet).
Optical instruments: simple microscope, compound microscope and telescope.
महत्त्वाच्या संज्ञा
Critical angle – angle of incidence in the denser medium for which the refracted ray grazes the surface (angle of refraction 90°).
Refractive index – ratio of speed of light in vacuum to speed in the medium, n = c/v.
Angular dispersion – difference in deviation of violet and red rays.
Dispersive power – ratio of angular dispersion to mean deviation.
Magnifying power – ratio of angle subtended by the image at the eye to that subtended by the object.
Prism: A + δ = i + e ; n = sin((A + δm)/2) / sin(A/2)
Thin prism: δ = (n − 1)A
Dispersive power ω = (nᵥ − n_r)/(n − 1)
Simple microscope: M = 1 + D/f (D = 25 cm)
Telescope (normal adjustment): M = fₒ/fₑ
५ महत्त्वाचे प्रश्न
Q1. What is total internal reflection? State its conditions.
When light travelling from a denser to a rarer medium falls at an angle greater than the critical angle, it is completely reflected back into the denser medium. Conditions: (1) light must go from denser to rarer medium; (2) angle of incidence must be greater than the critical angle.
Q2. Find the critical angle for glass of refractive index 1.5.
sin C = 1/1.5 = 0.667, so C ≈ 41.8°.
Q3. A convex lens of focal length 20 cm forms a real image of an object kept 30 cm away. Find the image distance.
1/v − 1/u = 1/f with u = −30 cm, f = +20 cm: 1/v = 1/20 − 1/30 = 1/60, so v = +60 cm (real, inverted, magnification −2).
Q4. Explain why the sky appears blue.
Sunlight is scattered by air molecules. Scattering is inversely proportional to the fourth power of wavelength, so blue (short wavelength) light is scattered much more than red. This scattered blue light reaches our eyes from all directions, so the sky looks blue.
Q5. State the lens maker's equation and find the power of a lens of focal length 50 cm.
1/f = (n − 1)(1/R₁ − 1/R₂). Power P = 1/f = 1/0.5 m = +2 D (convex lens).
परीक्षेसाठी टीप
Always write the sign of u, v and f before substituting — most marks are lost on signs, not on the formula.
Electrostatics is the study of charges at rest. This chapter covers properties of charge, Coulomb's law, superposition, electric field and lines of force, electric flux, Gauss' law, electric dipole and continuous charge distribution.
महत्त्वाचे मुद्दे
There are two kinds of charge, positive and negative. Like charges repel and unlike charges attract.
Charge is additive (total charge = algebraic sum), conserved (cannot be created or destroyed) and quantized: q = ne, where e = 1.6 × 10⁻¹⁹ C.
Coulomb's law: force between two point charges is proportional to the product of charges and inversely proportional to the square of the distance.
In a medium, the force reduces by the factor K (dielectric constant / relative permittivity).
Principle of superposition: net force on a charge = vector sum of forces due to all other charges.
Electric field E at a point = force per unit positive test charge. It is a vector, unit N/C.
Electric lines of force start on positive charges and end on negative charges; they never cross; they are closer where the field is stronger.
Electric flux is the number of field lines passing through a surface: φ = E⃗·A⃗ = EA cos θ.
Gauss' law: total flux through a closed surface = (charge enclosed)/ε₀.
An electric dipole is two equal and opposite charges separated by a small distance. Dipole moment p = q × 2l, directed from −q to +q.
A dipole in a uniform field feels no net force but a torque τ = pE sin θ.
Field of a dipole on the axis is twice the field at the same distance on the equator (for a short dipole), and opposite in direction.
Continuous charge distributions are described by linear (λ), surface (σ) and volume (ρ) charge densities.
महत्त्वाच्या संज्ञा
Coulomb – SI unit of charge.
Permittivity (ε) – property of a medium that decides the force between charges; ε₀ = 8.85 × 10⁻¹² C²/N m².
Electric field intensity – force on a unit positive charge at a point.
Electric flux – total number of lines of force through a surface (N m²/C).
Electric dipole moment – product of one charge and distance between the charges (C m).
सूत्रे
Coulomb's law: F = (1/4πε₀) q₁q₂/r² ; 1/4πε₀ = 9 × 10⁹ N m²/C²
In a medium: F = (1/4πε₀K) q₁q₂/r² ; K = ε/ε₀
Electric field: E = F/q₀ ; due to point charge E = (1/4πε₀) q/r²
Flux: φ = EA cos θ ; Gauss' law: φ = q/ε₀
Dipole moment p = q(2l)
Torque on dipole: τ = pE sin θ
Short dipole: axial field E = (1/4πε₀) 2p/r³ ; equatorial field E = (1/4πε₀) p/r³
Charge densities: λ = q/L, σ = q/A, ρ = q/V
५ महत्त्वाचे प्रश्न
Q1. State the basic properties of electric charge.
(1) Charges are additive. (2) Charge is conserved. (3) Charge is quantized, q = ±ne. (4) Like charges repel, unlike attract.
Q2. Two charges of +2 μC and +3 μC are 0.3 m apart in air. Find the force between them.
The total electric flux through any closed surface is equal to 1/ε₀ times the net charge enclosed by the surface: φ = q/ε₀. Charges outside the surface contribute zero net flux.
Q4. Derive the expression for torque on an electric dipole in a uniform field.
Forces +qE and −qE act on the two charges. They are equal, opposite and parallel, so they form a couple. Perpendicular distance between them = 2l sin θ. Torque τ = qE × 2l sin θ = pE sin θ. In vector form τ⃗ = p⃗ × E⃗.
Q5. State any four properties of electric lines of force.
(1) They start from positive and end on negative charge. (2) Two lines never intersect. (3) The tangent at a point gives the field direction. (4) They are crowded where the field is strong and spread out where it is weak.
परीक्षेसाठी टीप
Convert μC to C (×10⁻⁶) before substituting in Coulomb's law — this is the most common mistake.
This chapter explains electric current, how current flows through a conductor (drift velocity), Ohm's law and its limits, resistors and their combinations, resistivity, emf and combinations of cells, and types of cells.
महत्त्वाचे मुद्दे
Electric current is the rate of flow of charge, I = q/t. Its SI unit is ampere (A). Conventional current flows opposite to electron flow.
In a metal, free electrons move randomly. When a field is applied they slowly drift opposite to the field; this average speed is the drift velocity (very small, about 10⁻⁴ m/s).
Current is related to drift velocity by I = nAv_d e.
Ohm's law: at constant temperature, current through a conductor is directly proportional to the potential difference, V = IR.
Ohm's law does not hold for all devices. Diodes, thermistors and electrolytes are non-ohmic (V–I graph not a straight line).
Resistors are made of carbon or wire; carbon resistor values are read using a colour code (Black 0, Brown 1, Red 2, Orange 3, Yellow 4, Green 5, Blue 6, Violet 7, Grey 8, White 9).
A rheostat is a variable resistor used to change current in a circuit.
In series, the same current flows and resistances add. In parallel, the same voltage acts and reciprocals add.
Resistance depends on length, area and material: R = ρL/A. Resistivity (ρ) depends only on material and temperature.
Resistance of metals increases with temperature; that of semiconductors decreases.
The emf of a cell is the work done per unit charge in taking it once round the circuit. Terminal voltage is less than emf when current flows, because of internal resistance.
Cells in series give more emf; cells in parallel give more current capacity at the same emf.
Types of cells: primary (not rechargeable, e.g. dry cell) and secondary (rechargeable, e.g. lead-acid, Li-ion). Fuel cells convert chemical energy of fuel directly into electricity.
महत्त्वाच्या संज्ञा
Drift velocity – average velocity of free electrons in a conductor under an electric field.
Current density – current per unit area, J = I/A.
Resistivity – resistance of a conductor of unit length and unit cross-sectional area (Ω m).
Conductance – reciprocal of resistance (siemens, S).
Internal resistance (r) – resistance offered by the electrolyte of a cell.
emf (ε) – potential difference across the terminals of a cell in an open circuit.
Cell with internal resistance: I = ε/(R + r) ; terminal voltage V = ε − Ir
n identical cells in series: I = nε/(R + nr) ; in parallel: I = ε/(R + r/n)
५ महत्त्वाचे प्रश्न
Q1. Define drift velocity and derive I = nAv_d e.
Drift velocity is the average velocity of free electrons under an applied field. In time t, electrons move a distance v_d t, so all electrons in a volume A v_d t pass a section. Charge passing = nA v_d t e. Current I = charge/time = nAv_d e.
Q2. Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel. Find the equivalent resistance.
1/R = 1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = 1, so R = 1 Ω. In series they would give 11 Ω.
Q3. A cell of emf 2 V and internal resistance 0.5 Ω is connected to a 3.5 Ω resistor. Find the current and terminal voltage.
I = ε/(R + r) = 2/4 = 0.5 A. V = ε − Ir = 2 − 0.25 = 1.75 V.
Q4. Find the resistance of a carbon resistor with bands brown, black, red and gold.
Brown = 1, black = 0, red = multiplier 10². R = 10 × 100 = 1000 Ω = 1 kΩ, gold tolerance ±5%.
Q5. State the limitations of Ohm's law.
(1) It holds only when temperature and other physical conditions stay constant. (2) It does not apply to non-ohmic devices such as diodes, transistors, thermistors and electrolytic cells, where V–I graph is not linear or depends on direction of V.
परीक्षेसाठी टीप
Learn the colour code with a memory line such as "B B ROY of Great Britain had a Very Good Wife" (Black, Brown, Red, Orange, Yellow, Green, Blue, Violet, Grey, White).
This chapter introduces magnetic lines of force and magnetic flux, the bar magnet as a magnetic dipole, magnetic field on its axis and equator, Gauss' law of magnetism, and the earth's magnetism (declination, dip and magnetic maps).
महत्त्वाचे मुद्दे
Magnetic lines of force show the direction of a magnetic field; a compass needle sets along them. Their crowding shows the field strength.
Number of lines of force through an area is magnetic flux (φ), unit weber (Wb). Magnetic field B = φ/A, unit Wb/m² or tesla (T); 1 T = 10⁴ gauss.
A bar magnet has poles of strength +q_m (N) and −q_m (S), separated by magnetic length 2l. It is a magnetic dipole with moment m = q_m × 2l, directed from S to N.
Magnetic length ≈ (5/6) × geometric length.
Pole strength unit: A m. Magnetic moment unit: A m².
Magnetism can be studied by analogy with electrostatics: q → q_m, E → B, 1/4πε₀ → μ₀/4π, p → m.
For a short magnet at the same distance r, the field on the axis is twice the field on the equator; the equatorial field is opposite to m.
Gauss' law of magnetism: net magnetic flux through any closed surface is zero. This means isolated magnetic poles (monopoles) do not exist.
The earth behaves like a huge magnet. Its magnetic north pole lies near the geographic south (below Antarctica) and its magnetic south pole is below north Canada.
The magnetic equator passes through India near Thiruvananthapuram.
Magnetic elements of the earth: declination (α), angle of dip (φ) and horizontal component (B_H). They change with place and time.
At the magnetic equator dip is 0°; at the magnetic poles dip is 90°.
Magnetic maps join places of equal values: isodynamic lines (equal B_H), isogonic lines (equal declination), isoclinic lines (equal dip).
महत्त्वाच्या संज्ञा
Magnetic dipole moment – product of pole strength and magnetic length.
Magnetic meridian – vertical plane passing through the magnetic axis at a place.
Declination – angle between the geographic meridian and the magnetic meridian.
Angle of dip (inclination) – angle made by the earth's total magnetic field with the horizontal.
Agonic line – line joining places of zero declination.
Field at any point: B = (μ₀/4π)(m/r³) √(3cos²θ + 1)
Earth's field: B_H = B cos φ ; B_V = B sin φ ; tan φ = B_V/B_H ; B = √(B_H² + B_V²)
μ₀ = 4π × 10⁻⁷ T m/A (μ₀/4π = 10⁻⁷)
५ महत्त्वाचे प्रश्न
Q1. State Gauss' law of magnetism. What does it show?
The net magnetic flux through any closed surface is always zero. It shows that magnetic poles always exist in pairs; an isolated magnetic monopole does not exist.
Q2. A short magnet has m = 0.5 A m². Find the magnetic field at 20 cm on its axis and equator.
Declination is the angle between the geographic meridian and the magnetic meridian at a place. Angle of dip is the angle made by the direction of the earth's resultant magnetic field with the horizontal in the magnetic meridian.
Q4. At a place B_H = 0.3 G and B_V = 0.3 G. Find the dip and total field.
tan φ = B_V/B_H = 1, so φ = 45°. B = √(0.09 + 0.09) = 0.3√2 ≈ 0.42 G.
Q5. Write any four differences/analogies between electric and magnetic dipoles.
Electric dipole: charges ±q, moment p = q(2l) from −q to +q, constant 1/4πε₀, Gauss' flux = q/ε₀. Magnetic dipole: poles ±q_m, moment m = q_m(2l) from S to N, constant μ₀/4π, Gauss' flux always zero (no monopole).
परीक्षेसाठी टीप
Use μ₀/4π = 10⁻⁷ directly in numericals — it saves time and avoids π mistakes.
This chapter introduces electromagnetic (EM) waves, their sources and properties, the EM spectrum from radio waves to gamma rays, how EM waves travel through the atmosphere, and the basic parts of a communication system including modulation.
महत्त्वाचे मुद्दे
EM waves are produced by accelerated (oscillating) charges. They contain electric and magnetic fields vibrating perpendicular to each other and to the direction of travel, so they are transverse.
EM waves need no medium. In vacuum all travel at c = 3 × 10⁸ m/s; c = 1/√(μ₀ε₀).
They carry energy and momentum, and show reflection, refraction, interference, diffraction and polarisation.
EM spectrum in increasing frequency (decreasing wavelength): radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays.
Uses: radio waves – radio, TV; microwaves – radar, satellite communication, microwave ovens; infrared – remote controls, night vision, heating; visible – seeing; UV – sterilising, but harmful to skin (ozone layer absorbs it); X-rays – medical imaging; gamma rays – cancer treatment, from nuclear reactions.
Ground (surface) waves travel along the earth's surface; used for low frequencies (up to a few MHz), e.g. AM radio.
Space waves travel in straight lines (line of sight) from transmitter to receiver or via satellites; used for TV, FM, mobile, satellite (above 40 MHz).
Sky waves (about 3–30 MHz) are reflected back by the ionosphere; used for long-distance short-wave radio.
A communication system has three basic elements: transmitter, communication channel and receiver.
Low-frequency audio signals cannot be sent far directly, so they are carried by a high-frequency carrier wave. This is modulation.
Types: amplitude modulation (AM), frequency modulation (FM) and phase modulation (PM).
AM: simple, cheap, long range, but noisy and power-inefficient. FM: more complex, but less noise and higher quality.
महत्त्वाच्या संज्ञा
Transducer – device that converts one form of energy into another (microphone, loudspeaker).
Signal – information converted into electrical form.
Noise – unwanted signals that disturb the transmission.
Bandwidth – range of frequencies needed to send a signal.
Attenuation – loss of signal strength while travelling.
Modulation – changing a property of the carrier wave according to the signal.
सूत्रे
c = νλ = 3 × 10⁸ m/s
c = 1/√(μ₀ε₀) ; speed in a medium v = 1/√(με)
E₀/B₀ = c
Range of a TV tower (height h): d = √(2Rh)
५ महत्त्वाचे प्रश्न
Q1. State any four characteristics of EM waves.
(1) They are transverse; E and B are perpendicular to each other and to the direction of travel. (2) They need no material medium. (3) In vacuum they travel at 3 × 10⁸ m/s. (4) They carry energy and momentum and are not deflected by electric or magnetic fields.
Q2. Arrange the EM spectrum in order of increasing wavelength and give one use of each.
Waves of about 3 to 30 MHz go up and are reflected back to the earth by the ionosphere, the charged layer of the upper atmosphere. By repeated reflection between the ionosphere and the ground, they cover very long distances. Higher frequencies pass through the ionosphere.
Q4. Why is modulation needed?
(1) Audio signals have low frequency and cannot travel far or need very long antennas. (2) Different transmitters can be given different carrier frequencies, so their signals do not mix. (3) It allows efficient transmission with practical antenna size.
Q5. Compare AM and FM.
AM: amplitude of carrier changes; simple and cheap; large range; more affected by noise. FM: frequency of carrier changes; more complex; less noise; better sound quality, so used for high-quality broadcasting.
परीक्षेसाठी टीप
Learn the order of the EM spectrum with one use each — it appears as MCQs and short answers every year.
This chapter explains how solids conduct electricity using band theory, the difference between conductors, insulators and semiconductors, intrinsic and extrinsic (n-type, p-type) semiconductors, the p-n junction diode and its biasing, and uses of semiconductor devices such as the thermistor.
महत्त्वाचे मुद्दे
In a solid, energy levels of atoms combine into bands: the valence band (filled) and conduction band, separated by the forbidden energy gap Eg.
Conductors: bands overlap (no gap). Insulators: large gap (more than about 3 eV, diamond about 5–6 eV). Semiconductors: small gap (Si about 1.1 eV, Ge about 0.7 eV).
In a pure (intrinsic) semiconductor, at room temperature some electrons jump to the conduction band, leaving holes. Number of electrons = number of holes.
Resistance of a semiconductor decreases as temperature rises.
Doping means adding a small amount of impurity to increase conductivity; the result is an extrinsic semiconductor.
An n-type or p-type semiconductor is still electrically neutral as a whole.
In a p-n junction, electrons and holes diffuse across and recombine, forming a depletion region with a potential barrier (about 0.7 V for Si, 0.3 V for Ge).
Forward bias (p to +, n to −): barrier decreases, depletion layer narrows, large current flows. Reverse bias: barrier increases, only a tiny leakage current.
So a diode conducts mainly in one direction; it is used as a rectifier.
Advantages of semiconductor devices: small, light, cheap, low power, no heating filament, long life. Disadvantages: sensitive to heat and to over-voltage/current.
A thermistor is a temperature-sensitive resistor. NTC: resistance falls with temperature (temperature sensors). PTC: resistance rises with temperature (resettable fuse).
महत्त्वाच्या संज्ञा
Energy band – range of closely spaced energy levels in a solid.
Forbidden gap – energy gap between valence and conduction bands where no electron can exist.
Hole – vacancy left by an electron in the valence band; acts like a positive charge.
Depletion region – region near the junction with no free charge carriers.
Potential barrier – potential difference across the depletion region that opposes further diffusion.
सूत्रे
Intrinsic semiconductor: nₑ = n_h = nᵢ
Current in semiconductor: I = Iₑ + I_h
Mass action law: nₑ n_h = nᵢ²
Energy gap in eV: 1 eV = 1.6 × 10⁻¹⁹ J
५ महत्त्वाचे प्रश्न
Q1. Distinguish between conductors, insulators and semiconductors on the basis of band theory.
Conductors: valence and conduction bands overlap, many free electrons. Insulators: large forbidden gap (> 3 eV), conduction band empty. Semiconductors: small gap (~1 eV), a few electrons cross the gap at room temperature.
Q2. Distinguish between n-type and p-type semiconductors.
n-type: doped with pentavalent (donor) impurity; majority carriers are electrons. p-type: doped with trivalent (acceptor) impurity; majority carriers are holes. Both are neutral overall.
Q3. Explain the formation of the depletion region in a p-n junction.
When p and n regions are joined, electrons diffuse from n to p and holes from p to n, where they recombine. Near the junction, fixed positive ions remain on the n side and negative ions on the p side. This thin region without free carriers is the depletion region, and the field of the ions creates a potential barrier.
Q4. What happens in forward and reverse bias?
Forward bias: external voltage opposes the barrier, depletion region becomes thin and a large current flows (mA). Reverse bias: external voltage adds to the barrier, depletion region widens and only a very small current (μA) due to minority carriers flows.
Q5. In a pure semiconductor nᵢ = 1.5 × 10¹⁶ m⁻³. After doping, n_h = 4.5 × 10²² m⁻³. Find nₑ.
nₑ = nᵢ²/n_h = (2.25 × 10³²)/(4.5 × 10²²) = 5 × 10⁹ m⁻³. It is a p-type semiconductor.
परीक्षेसाठी टीप
Draw the p-n junction with the depletion region and the battery connections for both biases — diagrams carry marks.