Physics is built on measurement. Every measured value has a number, a unit and some uncertainty. This topic covers SI units, dimensions, dimensional analysis, and the types of errors in measurement and how they combine.
Key points
A physical quantity = numerical value × unit (e.g. 5 m).
The SI system has 7 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 (force, energy, pressure) are made from base quantities.
Dimensions show how a quantity depends on M, L, T, etc. Force = [M L T⁻²].
Principle of homogeneity: every term in a correct physical equation has the same dimensions.
Uses of dimensional analysis: check equations, convert units, derive relations between quantities.
Limits of dimensional analysis: cannot find dimensionless constants (like 2π or ½), cannot handle sums of several terms, fails for trigonometric, log and exponential relations.
Systematic errors come from faulty instruments (zero error), wrong method or personal bias. They act in one direction and can be corrected.
Random errors vary by chance from reading to reading. Taking the mean of many readings reduces them.
Least count is the smallest value an instrument can measure (vernier calliper 0.01 cm, screw gauge 0.001 cm usually).
Significant figures are the reliable digits plus the first doubtful digit.
Important terms
Accuracy – how close a measured value is to the true value.
Precision – how close repeated measurements are to each other (अचूकता vs. सुस्पष्टता).
Absolute error – difference between the measured value and the true (mean) value.
Relative error – absolute error divided by the true value.
Percentage error – relative error × 100.
Dimensionless quantity – has no dimensions, e.g. strain, refractive index, angle.
Formulas
Mean value: a_mean = (a₁ + a₂ + … + aₙ)/n
Absolute error of i-th reading: Δaᵢ = |a_mean − aᵢ|
Sum or difference Z = A ± B: ΔZ = ΔA + ΔB (absolute errors add)
Product or quotient Z = AB or A/B: ΔZ/Z = ΔA/A + ΔB/B (relative errors add)
Power Z = Aⁿ: ΔZ/Z = n (ΔA/A)
Unit conversion: n₂ = n₁ [M₁/M₂]ᵃ [L₁/L₂]ᵇ [T₁/T₂]ᶜ
5 important questions
Q1. State the principle of homogeneity of dimensions.
In a correct physical equation, the dimensions of every term on both sides must be the same. For example, in v = u + at, each term has dimensions [L T⁻¹].
Q2. Distinguish between systematic and random errors.
Systematic errors are in one direction (always more or always less), come from instrument or method, and can be removed by correction. Random errors change irregularly between readings and are reduced by taking the mean of many readings.
Q3. Using dimensions, derive the time period of a simple pendulum.
Assume T depends on mass m, length l and g: T = k mᵃ lᵇ gᶜ.
In error questions, remember: absolute errors add for sums and differences; relative (percentage) errors add for products, quotients and powers. Always write units and the final answer as value ± error.
Many physical quantities (force, velocity, field) have both size and direction, so physics needs vectors. This topic covers unit vectors, components, the scalar (dot) product, the vector (cross) product, the scalar and vector triple products, and simple uses such as work, torque and area. These tools are used in almost every later topic of B.Sc. physics.
Key points
A scalar has only magnitude (mass, time, work). A vector has magnitude and direction and adds by the parallelogram (or triangle) law (velocity, force).
In Cartesian form A = Ax î + Ay ĵ + Az k̂, where î, ĵ, k̂ are unit vectors along x, y, z. Magnitude |A| = √(Ax² + Ay² + Az²).
Unit vector  = A ÷ |A|; it gives only direction.
Dot product A·B = AB cos θ = AxBx + AyBy + AzBz. It is a scalar and is commutative. If A·B = 0 (both non-zero), A ⟂ B.
Cross product A×B = AB sin θ n̂. It is a vector perpendicular to both A and B (right-hand rule). It is not commutative: A×B = −(B×A). If A×B = 0, A ∥ B.
B×C is ⟂ to the plane of B and C, so A×(B×C) lies in that plane: A×(B×C) = mB + nC.
Dotting with A gives 0 = m(A·B) + n(A·C), so m = λ(A·C), n = −λ(A·B).
Check with unit vectors, e.g. A = B = î, C = ĵ: left side = î×k̂ = −ĵ; right side = î(0) − ĵ(1) = −ĵ, so λ = 1.
Hence A×(B×C) = B(A·C) − C(A·B).
Q5. Show that A = î − 2ĵ + 3k̂, B = 2î + ĵ − k̂ and C = 3î − ĵ + 2k̂ are coplanar.
A·(B×C) = 1[(1)(2) − (−1)(−1)] − (−2)[(2)(2) − (−1)(3)] + 3[(2)(−1) − (1)(3)] = 1(2 − 1) + 2(4 + 3) + 3(−2 − 3) = 1 + 14 − 15 = 0. The scalar triple product is zero, so the volume is zero and the vectors are coplanar.
Exam tip
Always write the cross product as a 3×3 determinant and remember the minus sign on the ĵ term. Check your answer: A×B must give zero when dotted with A or B.
Kinematics describes motion without asking what causes it. This topic covers displacement, velocity and acceleration in one dimension, the equations of uniformly accelerated motion (derived by calculus), motion under gravity, and projectile motion in two dimensions with its time of flight, maximum height, range and path.
Key points
Displacement is the change in position (a vector); distance is the total path length (a scalar).
Instantaneous velocity v = dx/dt; instantaneous acceleration a = dv/dt = d²x/dt² = v dv/dx.
Area under a v–t graph = displacement; slope of a v–t graph = acceleration; slope of an x–t graph = velocity.
For constant acceleration, integrating a = dv/dt and v = dx/dt gives the three equations of motion.
Freely falling bodies have a = g ≈ 9.8 m s⁻² downward (air resistance neglected).
A projectile has constant horizontal velocity (no horizontal force) and uniformly accelerated vertical motion (a = −g). The two motions are independent.
The path of a projectile is a parabola.
Range is maximum at θ = 45°; angles θ and (90° − θ) give the same range.
At the highest point the vertical velocity is zero but the horizontal velocity u cos θ remains.
Important terms
Uniform motion – equal displacements in equal intervals of time.
Relative velocity – v_AB = v_A − v_B, velocity of A as seen by B.
Time of flight – total time a projectile stays in air.
Horizontal range – horizontal distance covered during the time of flight.
Angle of projection – angle between the initial velocity and the horizontal.
Formulas
v = u + at; s = ut + ½at²; v² = u² + 2as (u = initial velocity, a = constant acceleration, s = displacement, m)
Distance in the nth second: s_n = u + (a/2)(2n − 1)
Time of flight T = 2u sin θ / g
Maximum height H = u² sin²θ / 2g
Range R = u² sin 2θ / g; R_max = u²/g at θ = 45°
Path: y = x tan θ − g x² / (2u² cos²θ)
Projectile (speed u, angle θ, g in m s⁻²):
Horizontal projection from height h: T = √(2h/g), R = u√(2h/g)
5 important questions
Q1. Derive v = u + at and s = ut + ½at² using calculus.
a = dv/dt, so dv = a dt. Integrate from (t = 0, v = u) to (t, v): v − u = at, so v = u + at.
v = dx/dt = u + at, so dx = (u + at) dt. Integrate from 0 to t: s = ut + ½at².
Using a = v dv/dx: ∫v dv = ∫a dx gives ½(v² − u²) = as, so v² = u² + 2as.
Q2. A body starts from rest with acceleration 2 m s⁻². Find the distance covered in the 5th second.
Q3. Show that the path of a projectile is a parabola and derive its range.
Horizontal: x = (u cos θ) t, so t = x / (u cos θ).
Vertical: y = (u sin θ) t − ½ g t².
Substituting t: y = x tan θ − g x² / (2u² cos²θ). This has the form y = bx − cx², which is a parabola.
y = 0 again when t = T = 2u sin θ / g. Then R = u cos θ × T = 2u² sin θ cos θ / g = u² sin 2θ / g.
R is largest when sin 2θ = 1, i.e. θ = 45°, giving R_max = u²/g.
Q4. A ball is thrown at 20 m s⁻¹ at 30° to the horizontal. Find T, H and R (g = 10 m s⁻²).
u sin θ = 20 × 0.5 = 10 m s⁻¹; u cos θ = 20 × 0.866 = 17.32 m s⁻¹
T = 2 × 10 / 10 = 2 s
H = (10)² / (2 × 10) = 5 m
R = 17.32 × 2 = 34.6 m (check: u² sin 60° / g = 400 × 0.866 / 10 = 34.6 m)
Q5. Prove that for the same speed, angles θ and (90° − θ) give equal range. Which one gives more height?
R₁ = u² sin 2θ / g. For (90° − θ): R₂ = u² sin(180° − 2θ) / g = u² sin 2θ / g = R₁. Heights: H₁ = u² sin²θ / 2g and H₂ = u² cos²θ / 2g. The larger angle gives the greater height and a longer time of flight. Also H₁ × H₂ = R²/16 (a useful result).
Exam tip
Always split the projectile's velocity into u cos θ (horizontal, constant) and u sin θ (vertical, changes with g). Write the units of every answer.
Newton's three laws explain how forces change motion. From them come the great conservation laws: conservation of linear momentum, of energy and of angular momentum. This topic also introduces inertial and non-inertial frames and the work–energy theorem.
Key points
First law (law of inertia): a body stays at rest or in uniform motion in a straight line unless an external force acts on it.
Second law: the rate of change of momentum equals the applied force, F = dp/dt. For constant mass, F = ma.
Third law: to every action there is an equal and opposite reaction; the two forces act on different bodies.
An inertial frame is one in which Newton's first law holds (not accelerating). A frame that accelerates is non-inertial; in it we must add pseudo forces (e.g. centrifugal force).
Conservation of linear momentum: if the net external force on a system is zero, total momentum stays constant. Examples: recoil of a gun, rocket motion.
Work–energy theorem: work done by the net force equals the change in kinetic energy.
A force is conservative if work done by it does not depend on the path (gravity, spring force, electrostatic force). Friction is non-conservative.
For a conservative force, F = −dU/dx (force is the negative gradient of potential energy).
Conservation of mechanical energy: with only conservative forces, K + U = constant.
Elastic collision: momentum and kinetic energy both conserved. Inelastic: only momentum conserved.
Conservation of angular momentum: if net external torque is zero, L stays constant (a skater pulling in arms spins faster).
Important terms
Momentum (संवेग) – p = mv, a vector, unit kg m s⁻¹.
Impulse – force × time = change in momentum, J = FΔt = Δp.
Inertia – the tendency of a body to resist change in its state.
Pseudo force – an imaginary force −ma added in an accelerated frame.
Coefficient of restitution (e) – relative speed of separation ÷ relative speed of approach (e = 1 elastic, e = 0 perfectly inelastic).
Formulas
F = dp/dt = ma (N)
p = mv (kg m s⁻¹); J = FΔt = Δp (N s)
Work W = F·s = Fs cos θ (J)
K = ½mv²; gravitational U = mgh (near Earth); spring U = ½kx²
Work–energy theorem: W_net = K_f − K_i
Rocket thrust: F = u (dm/dt); velocity v = v₀ + u ln(m₀/m) (no gravity)
A frame of reference in which Newton's first law holds, i.e. a body with no net force stays at rest or moves with constant velocity. Any frame moving with constant velocity relative to an inertial frame is also inertial.
Q2. Show that impulse equals change in momentum.
From F = dp/dt, F dt = dp. Integrating from time t₁ to t₂: ∫F dt = p₂ − p₁. The left side is impulse J. So J = Δp.
Q3. State and prove the law of conservation of linear momentum for two colliding bodies.
Bodies A and B collide; A exerts F_AB on B, B exerts F_BA on A.
By third law, F_BA = −F_AB.
Change in momentum of A: Δp_A = F_BA Δt; of B: Δp_B = F_AB Δt.
Adding: Δp_A + Δp_B = (F_BA + F_AB)Δt = 0.
So total momentum before = total momentum after: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
Q4. A 4 kg gun fires a 20 g bullet with speed 400 m/s. Find the recoil velocity of the gun.
Initial momentum = 0 (both at rest).
Final: m_b v_b + m_g v_g = 0.
0.02 × 400 + 4 × v_g = 0 → 8 + 4v_g = 0.
v_g = −2 m/s. The gun recoils backward at 2 m/s.
Q5. A ball of mass 2 kg moving at 6 m/s hits a ball of 1 kg at rest in a 1-D elastic collision. Find their velocities after collision.
Check momentum: before 2 × 6 = 12; after 2 × 2 + 1 × 8 = 12 ✓.
Check KE: before ½ × 2 × 36 = 36 J; after ½ × 2 × 4 + ½ × 1 × 64 = 4 + 32 = 36 J ✓.
Exam tip
Always state the condition with each conservation law (zero external force, zero external torque, only conservative forces). Checking momentum and KE at the end of a collision numerical earns marks and catches mistakes.
Friction is the force that opposes relative motion between two surfaces in contact. Circular motion needs a force directed towards the centre (centripetal force). This topic covers the laws of friction, the angle of friction and angle of repose, motion on an inclined plane, uniform circular motion, banking of roads and motion in a vertical circle.
Key points
Static friction adjusts itself up to a maximum value fs(max) = μs N. Kinetic friction fk = μk N, and usually μk is less than μs.
Laws of limiting friction: friction is independent of the area of contact, depends on the nature of the surfaces, and is proportional to the normal reaction N.
Rolling friction is much smaller than sliding friction; that is why wheels and ball bearings are used.
Angle of friction λ: tan λ = μ. Angle of repose α: the largest incline angle at which a body stays at rest; tan α = μs. So α = λ.
In uniform circular motion speed is constant but velocity changes direction, so there is an acceleration v²/r towards the centre.
Centripetal force is not a new kind of force: tension, friction, gravity or a normal reaction supplies it.
Centrifugal force is a pseudo force seen only in the rotating (non-inertial) frame.
On a banked road, the horizontal component of the normal reaction supplies the centripetal force, so vehicles need less friction.
In a vertical circle, speed is not constant; the minimum speed at the top is √(gr) and at the bottom √(5gr).
Important terms
Coefficient of friction (μ) – ratio of limiting friction to normal reaction.
Angular velocity (ω) – rate of change of angular position, ω = dθ/dt = v/r (rad s⁻¹).
Banking angle (θ) – angle by which the outer edge of a curved road is raised.
Conical pendulum – a bob moving in a horizontal circle while the string traces a cone.
Formulas
f = μN; tan λ = μ; tan α = μs
Body sliding down a rough incline: a = g(sin θ − μk cos θ)
Centripetal acceleration a = v²/r = ω²r; force F = mv²/r
Banking without friction: tan θ = v²/(rg)
Banking with friction μ: v_max = √[rg(μ + tan θ)/(1 − μ tan θ)]
Conical pendulum: tan θ = v²/(rg), period T = 2π√(L cos θ/g)
5 important questions
Q1. Define angle of repose and show that it equals the angle of friction.
At the angle of repose α the body is just about to slide. Along the incline: mg sin α = fs(max) = μs N. Perpendicular: N = mg cos α. Dividing, tan α = μs. Since tan λ = μs too, α = λ.
Q2. Derive the expression for centripetal acceleration.
Let a particle move with speed v on a circle of radius r. In a short time Δt it turns through Δθ, and the velocity vector turns by the same angle, so |Δv| = vΔθ. Then a = |Δv|/Δt = v(Δθ/Δt) = vω = v²/r, directed towards the centre.
Q3. Derive the angle of banking for a road (no friction) and solve a problem.
Forces: weight mg down, normal reaction N perpendicular to the road.
Vertical: N cos θ = mg
Horizontal: N sin θ = mv²/r
Dividing: tan θ = v²/(rg)
Example: r = 100 m, v = 72 km h⁻¹ = 20 m s⁻¹, g = 10 m s⁻². tan θ = 400/1000 = 0.4, so θ ≈ 21.8°.
Q4. A block slides down a 30° incline with μk = 0.2. Find its acceleration (g = 9.8 m s⁻²).
a = g(sin θ − μk cos θ) = 9.8(0.5 − 0.2 × 0.866) = 9.8(0.5 − 0.1732) = 9.8 × 0.3268 ≈ 3.2 m s⁻².
Q5. Discuss motion in a vertical circle and find the minimum speed at the top and the bottom.
At the top: T + mg = mv_top²/r. The string just stays tight when T = 0, so v_top(min) = √(gr).
Energy conservation between bottom and top (height 2r): ½mv_b² = ½mv_top² + mg(2r).
With v_top² = gr: v_b² = gr + 4gr = 5gr, so v_bottom(min) = √(5gr).
Tension difference: T_b − T_t = 6mg.
Example: r = 0.9 m, g = 10 m s⁻²: v_top = 3 m s⁻¹, v_bottom = √45 ≈ 6.7 m s⁻¹.
Exam tip
Always draw a free-body diagram first. In circular motion, find which real force points to the centre; do not add a "centripetal force" as an extra arrow.
Rotational motion is the turning of a rigid body about an axis. Moment of inertia plays the role that mass plays in straight-line motion. This topic covers angular quantities, torque, the theorems of parallel and perpendicular axes, and the moment of inertia of standard bodies.
Key points
A rigid body keeps the distance between its particles fixed.
Linear and angular quantities are linked: v = rω, a_t = rα.
Moment of inertia (जडत्व आघूर्ण) I = Σ mr² depends on the mass and how far it is spread from the axis.
Radius of gyration K: I = MK², the distance at which the whole mass could be placed to give the same I.
Torque τ = r × F causes angular acceleration: τ = Iα.
Parallel axis theorem: I = I_cm + Md², where d is the distance between the axis and a parallel axis through the centre of mass.
Perpendicular axis theorem (only for flat laminas): I_z = I_x + I_y, where x and y lie in the plane and z is perpendicular to it.
Rotational KE = ½Iω². Angular momentum L = Iω.
A body rolling without slipping has both translational and rotational KE: KE = ½Mv²(1 + K²/R²).
A flywheel is a heavy wheel with large I used to keep speed steady in engines.
Important terms
Axis of rotation – the fixed line about which the body turns.
Torque (moment of force) – turning effect of a force, unit N m.
Couple – two equal, opposite, parallel forces not on the same line.
Radius of gyration – K = √(I/M).
Formulas
Body
Axis
I
Thin ring (M, R)
through centre, ⟂ plane
MR²
Disc (M, R)
through centre, ⟂ plane
½MR²
Thin rod (M, L)
through centre, ⟂ rod
ML²/12
Thin rod (M, L)
through one end, ⟂ rod
ML²/3
Solid sphere (M, R)
diameter
(2/5)MR²
Hollow sphere (M, R)
diameter
(2/3)MR²
Solid cylinder (M, R)
own axis
½MR²
τ = Iα (N m); L = Iω (kg m² s⁻¹); KE_rot = ½Iω² (J)
Rolling down an incline of height h: v = √(2gh/(1 + K²/R²))
Flywheel: mgh = ½mv² + ½Iω² + n₁W_f (friction work for n₁ turns)
5 important questions
Q1. State the parallel axis theorem.
The moment of inertia of a body about any axis equals its moment of inertia about a parallel axis through its centre of mass plus Md², where M is its mass and d the distance between the two axes: I = I_cm + Md².
Q2. Define radius of gyration. What is it for a disc about its central perpendicular axis?
Radius of gyration K is the distance from the axis at which the whole mass, if concentrated, gives the same moment of inertia: I = MK². For a disc, I = ½MR², so K = R/√2.
Q3. Derive the moment of inertia of a thin uniform rod about an axis through its centre perpendicular to its length.
Rod: mass M, length L; mass per unit length λ = M/L.
Take element dx at distance x from centre; mass dm = λ dx.
dI = x² dm = λx² dx.
I = ∫ from −L/2 to L/2 of λx² dx = λ [x³/3] = λ (L³/24 + L³/24) = λL³/12.
Substitute λ = M/L: I = ML²/12.
About one end (parallel axis, d = L/2): I = ML²/12 + ML²/4 = ML²/3.
Q4. Find the moment of inertia of a disc about its diameter using the perpendicular axis theorem.
About the central perpendicular axis, I_z = ½MR².
By symmetry, I about any two perpendicular diameters are equal: I_x = I_y = I_d.
Q5. A solid sphere of mass 2 kg and radius 0.1 m rotates about its diameter at 10 rad/s. Find its moment of inertia, rotational KE and angular momentum.
I = (2/5)MR² = 0.4 × 2 × 0.01 = 0.008 kg m².
KE = ½Iω² = ½ × 0.008 × 100 = 0.4 J.
L = Iω = 0.008 × 10 = 0.08 kg m² s⁻¹.
Exam tip
Learn the I-table with axes; most numericals simply pick one formula and add Md² or use the perpendicular axis theorem. Draw a neat figure showing the axis — it earns a mark.
Gravitation is the force of attraction between any two masses. This topic covers Newton's law of gravitation, gravitational field and potential, variation of g with height and depth, Kepler's three laws of planetary motion, orbital velocity, escape velocity and geostationary satellites.
Key points
Newton's law: F = G m₁m₂/r², along the line joining the masses; always attractive. G = 6.67 × 10⁻¹¹ N m² kg⁻².
Gravitational field E = F/m = GM/r² (N kg⁻¹); it points towards the mass.
Gravitational potential V = −GM/r (J kg⁻¹); it is zero at infinity and negative everywhere else. E = −dV/dr.
Potential energy of two masses U = −Gm₁m₂/r.
At the Earth's surface g = GM/R² ≈ 9.8 m s⁻².
g decreases with height: g_h = g R²/(R + h)² ≈ g(1 − 2h/R) for small h.
g decreases with depth: g_d = g(1 − d/R); g = 0 at the centre.
Kepler's laws: (1) orbits are ellipses with the Sun at one focus; (2) the line from Sun to planet sweeps equal areas in equal times; (3) T² ∝ a³.
The second law is a consequence of conservation of angular momentum, because gravity is a central force (torque = 0).
Escape velocity from Earth ≈ 11.2 km s⁻¹; it does not depend on the mass of the body or the direction of throw.
A geostationary satellite has period 24 h, moves west to east above the equator, at about 36,000 km height.
Important terms
Central force – a force always directed along the line joining the body to a fixed point.
Areal velocity – area swept per unit time by the radius vector, dA/dt = L/2m.
Orbital velocity – speed needed to move in a circular orbit.
Binding energy – energy needed to free a satellite from the Earth's pull, = GMm/2r for a circular orbit.
Formulas
F = G m₁m₂/r²; g = GM/R²
V = −GM/r; U = −GMm/r
Orbital velocity v₀ = √(GM/r) = √(gR²/r); near the surface v₀ = √(gR) ≈ 7.9 km s⁻¹
Total energy of a satellite E = −GMm/2r (KE = GMm/2r, PE = −GMm/r)
Areal velocity dA/dt = L/(2m) = constant
5 important questions
Q1. State Kepler's three laws of planetary motion.
Law of orbits: every planet moves in an elliptical orbit with the Sun at one focus.
Law of areas: the radius vector sweeps equal areas in equal times (areal velocity is constant).
Law of periods: the square of the period is proportional to the cube of the semi-major axis, T² ∝ a³.
Q2. Show that Kepler's second law follows from conservation of angular momentum.
In time dt the radius vector r sweeps a small triangle of area dA = ½|r × v dt|. So dA/dt = ½|r × v| = L/(2m). Gravity is a central force, so torque τ = r × F = 0, and L stays constant. Hence dA/dt is constant: equal areas in equal times.
Q3. Derive the expression for escape velocity and calculate it for Earth.
A body of mass m escapes if its kinetic energy can carry it to infinity, where its PE becomes zero: ½mv_e² − GMm/R = 0, so v_e = √(2GM/R) = √(2gR). With g = 9.8 m s⁻² and R = 6.4 × 10⁶ m: v_e = √(2 × 9.8 × 6.4 × 10⁶) = √(1.254 × 10⁸) ≈ 1.12 × 10⁴ m s⁻¹ = 11.2 km s⁻¹.
Q4. Derive Kepler's third law for a circular orbit and find the height of a geostationary satellite.
Gravity gives the centripetal force: GMm/r² = mv²/r, so v² = GM/r. With v = 2πr/T: 4π²r²/T² = GM/r, so T² = (4π²/GM) r³, i.e. T² ∝ r³. For T = 86,400 s and GM = gR² = 9.8 × (6.4 × 10⁶)² ≈ 4.01 × 10¹⁴ m³ s⁻²: r³ = GM T²/4π² ≈ 4.01 × 10¹⁴ × 7.46 × 10⁹ / 39.48 ≈ 7.59 × 10²² m³, so r ≈ 4.23 × 10⁷ m. Height h = r − R ≈ 4.23 × 10⁷ − 0.64 × 10⁷ ≈ 3.6 × 10⁷ m (about 36,000 km).
Q5. Derive the variation of g with depth.
At depth d, only the inner sphere of radius (R − d) attracts the body (the outer shell gives zero net force). Taking uniform density ρ, mass inside = (4/3)π(R − d)³ρ. g_d = G(4/3)π(R − d)ρ, while g = G(4/3)πRρ. Dividing: g_d = g(1 − d/R). At the centre (d = R), g = 0.
Exam tip
Remember v_e = √2 × v₀ (near the surface): 11.2 ≈ 1.414 × 7.9. Potential and potential energy are always negative in gravitation; do not drop the minus sign.
Elasticity is the property of a body to regain its original shape and size when the deforming force is removed. This topic covers stress, strain, Hooke's law, the three elastic moduli, Poisson's ratio, their relations, and the bending of beams and torsion of a wire.
Key points
Stress = restoring force per unit area (N m⁻², Pa). Strain = change in dimension ÷ original dimension (no unit).
Hooke's law: within the elastic limit, stress ∝ strain. Stress/strain = modulus of elasticity.
Three moduli: Young's modulus Y (length change), bulk modulus K (volume change), modulus of rigidity η (shape change, shear).
Poisson's ratio σ = lateral strain ÷ longitudinal strain. Theoretical range −1 < σ < 0.5; for real materials it lies between 0 and 0.5.
On a stress–strain curve: proportional limit, elastic limit, yield point, ultimate stress, breaking point.
Work done in stretching a wire is stored as elastic potential energy.
A beam fixed at one end and loaded at the other is a cantilever. Bending moment depends on Y and the geometrical moment of inertia I_g.
For a rectangular cross-section, I_g = bd³/12; for a circular one, I_g = πr⁴/4.
Torsion: twisting a wire fixed at one end produces a restoring couple proportional to the twist angle.
Ductile materials (copper) show large plastic region; brittle materials (glass) break soon after the elastic limit.
Important terms
Elastic limit (स्थितिस्थापकता मर्यादा) – maximum stress up to which the body fully recovers.
Plasticity – the property of not regaining the original shape.
Elastic fatigue – loss of elastic strength after repeated stress cycles.
Neutral axis – the line in a bent beam whose length does not change.
Torsional rigidity (C) – couple per unit twist of a wire.
Energy per unit volume = ½ × stress × strain; total work W = ½ F Δl
Bending moment = YI_g/R (R = radius of curvature of neutral axis)
Cantilever depression at free end: δ = WL³/(3YI_g)
Twisting couple per unit twist: C = πηr⁴/(2l); couple = Cθ
Torsional pendulum: T = 2π√(I/C)
5 important questions
Q1. State Hooke's law and define Young's modulus.
Within the elastic limit, stress is directly proportional to strain. Young's modulus is the ratio of longitudinal (tensile) stress to longitudinal strain: Y = Fl/(AΔl). Its SI unit is N m⁻² (Pa).
Q2. What is Poisson's ratio? Why can it not exceed 0.5?
Poisson's ratio is lateral strain divided by longitudinal strain. From Y = 3K(1 − 2σ), if σ > 0.5, K would be negative, which is not physical (volume would increase under compression). So σ < 0.5.
Q3. Derive the expression for the work done in stretching a wire.
Wire: length l, area A, modulus Y. At extension x, restoring force F = YAx/l.
Work for small extra extension dx: dW = F dx = (YA/l) x dx.
Total work for extension Δl: W = ∫₀^Δl (YA/l) x dx = YA(Δl)²/(2l).
Since F = YAΔl/l, W = ½ F Δl.
Per unit volume (divide by Al): W/V = ½ stress × strain.
Q4. A steel wire of length 2 m and cross-section 1 mm² is stretched by a load of 10 kg. If Y = 2 × 10¹¹ N m⁻², find the extension (g = 9.8 m s⁻²).
F = 10 × 9.8 = 98 N; A = 1 mm² = 1 × 10⁻⁶ m²; l = 2 m.
Δl = Fl/(AY) = (98 × 2)/(10⁻⁶ × 2 × 10¹¹).
Denominator = 2 × 10⁵.
Δl = 196/(2 × 10⁵) = 9.8 × 10⁻⁴ m ≈ 0.98 mm.
Q5. For a material, Y = 2 × 10¹¹ Pa and η = 0.8 × 10¹¹ Pa. Find Poisson's ratio and bulk modulus.
Y = 3K(1 − 2σ) → K = Y/(3(1 − 0.5)) = 2 × 10¹¹/1.5.
K ≈ 1.33 × 10¹¹ Pa.
Exam tip
Learn the four relations among Y, K, η and σ together — numericals often give two of them and ask the other two. Remember mm² = 10⁻⁶ m² when converting.
Fluids (liquids and gases) can flow and take the shape of their container. This topic covers pressure in a fluid, Pascal's law and the hydraulic press, Archimedes' principle, types of flow, the equation of continuity, Bernoulli's theorem and its applications such as the venturimeter, Torricelli's theorem and the lift on an aircraft wing.
Key points
Pressure P = F/A (N m⁻² = Pa). In a liquid at rest it acts equally in all directions at a point.
Pressure at depth h: P = P₀ + ρgh. It depends only on depth, not on the shape of the vessel.
Pascal's law: pressure applied to an enclosed fluid is transmitted equally to every part of it. Used in hydraulic lifts and brakes.
Archimedes' principle: a body in a fluid feels an upward buoyant force equal to the weight of fluid displaced.
Streamline flow: each particle follows the path of the one before it; velocity at a point does not change with time. Turbulent flow: irregular, with eddies.
Equation of continuity (from conservation of mass): A₁v₁ = A₂v₂ for an incompressible fluid. Narrow pipe → faster flow.
Bernoulli's theorem (from conservation of energy): for ideal streamline flow, P + ½ρv² + ρgh = constant along a streamline.
Where speed is high, pressure is low. This explains the venturimeter, atomiser, lift of wings and roofs blown off in storms.
Bernoulli's theorem assumes the fluid is non-viscous, incompressible and in steady streamline flow.
Important terms
Pressure head P/ρg, velocity head v²/2g, gravitational head h – the three terms of Bernoulli's equation in units of length.
Venturimeter – a tube with a narrow throat used to measure flow rate.
Pitot tube – device to measure the speed of fluid flow.
Efflux – liquid flowing out through an opening.
Formulas
P = P₀ + ρgh (ρ = density in kg m⁻³, h = depth in m)
Q1. State Pascal's law and explain the hydraulic press.
Pressure applied to an enclosed liquid is transmitted undiminished to every point. In a hydraulic press a small force F₁ on a small piston A₁ creates pressure F₁/A₁, which acts on the large piston A₂, giving F₂ = F₁(A₂/A₁). A small force lifts a heavy load.
Q2. Water flows through a pipe of diameter 4 cm at 2 m s⁻¹. Find its speed where the diameter is 2 cm.
A₁v₁ = A₂v₂ and A ∝ d², so v₂ = v₁(d₁/d₂)² = 2 × (4/2)² = 8 m s⁻¹.
Q3. State and prove Bernoulli's theorem.
Take a tube of flow. At section 1: area A₁, pressure P₁, speed v₁, height h₁; at section 2: A₂, P₂, v₂, h₂.
In time Δt, volume ΔV = A₁v₁Δt = A₂v₂Δt (continuity) flows through each section; mass Δm = ρΔV.
Net work done by pressure = P₁ΔV − P₂ΔV.
This work equals the gain in KE and PE: (P₁ − P₂)ΔV = ½ρΔV(v₂² − v₁²) + ρΔVg(h₂ − h₁).
Rearranging: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂, i.e. P + ½ρv² + ρgh = constant.
Q4. Derive Torricelli's theorem and solve: water stands 5 m above a small hole. Find the speed of efflux (g = 9.8 m s⁻²).
Apply Bernoulli between the top surface (P₀, v ≈ 0, height h) and the hole (P₀, v, height 0): P₀ + ρgh = P₀ + ½ρv². So v = √(2gh), the same as free fall from height h. v = √(2 × 9.8 × 5) = √98 ≈ 9.9 m s⁻¹.
Q5. Explain the working of a venturimeter.
A venturimeter has a wide inlet (area A₁) and a narrow throat (A₂). By continuity, speed increases at the throat; by Bernoulli (same height), pressure falls there. A manometer measures P₁ − P₂. Combining A₁v₁ = A₂v₂ with P₁ − P₂ = ½ρ(v₂² − v₁²) gives v₁ = A₂√[2(P₁ − P₂)/ρ(A₁² − A₂²)], and the flow rate Q = A₁v₁. It is used to measure flow in water mains and fuel lines.
Exam tip
Continuity comes from conservation of mass and Bernoulli from conservation of energy — examiners like this one-line link. Always convert diameters to areas (A ∝ d²) before using A₁v₁ = A₂v₂.
Surface tension explains why a liquid surface behaves like a stretched membrane — drops are round and water rises in thin tubes. Viscosity is the internal friction of a fluid that opposes relative motion between its layers. This topic covers molecular theory, capillary rise, excess pressure, Newton's law of viscosity, Stokes' law and Poiseuille's equation.
Key points
Cohesive forces act between like molecules; adhesive forces between unlike molecules.
A molecule at the surface feels a net inward pull, so the surface tends to contract to minimum area.
Surface tension T = force per unit length acting along the surface, perpendicular to a line drawn on it. Unit N m⁻¹.
Surface energy = work done per unit area to increase the surface; numerically equal to T.
Angle of contact θ: less than 90° for water–glass (liquid wets glass, rises in capillary); more than 90° for mercury–glass (falls).
Surface tension decreases as temperature rises; detergents lower it.
Viscosity (श्यानता): the property by which a fluid opposes relative motion of its layers.
Streamline flow happens at low speed; turbulent flow at high speed. The change is decided by Reynolds number.
Stokes' law gives the drag on a small sphere moving slowly in a viscous fluid.
Terminal velocity: the constant speed reached when weight = buoyancy + viscous drag.
Viscosity of liquids decreases with temperature; of gases it increases.
Important terms
Capillarity – rise or fall of liquid in a narrow tube.
Coefficient of viscosity (η) – tangential force per unit area per unit velocity gradient. SI unit N s m⁻² (Pa s); CGS unit poise (1 Pa s = 10 poise).
Velocity gradient – dv/dx, change of velocity per unit distance between layers.
Critical velocity – the speed above which flow becomes turbulent.
Formulas
T = F/l (N m⁻¹); surface energy W = T × ΔA
Excess pressure in a liquid drop: P = 2T/r; in a soap bubble (two surfaces): P = 4T/r
Capillary rise (Jurin's law): h = 2T cos θ/(rρg)
Newton's law of viscosity: F = −ηA (dv/dx)
Stokes' law: F = 6πηrv
Terminal velocity: v = 2r²(ρ − σ)g/(9η) (ρ = density of sphere, σ = density of fluid)
Poiseuille's equation: V/t = πPr⁴/(8ηl)
Reynolds number: Re = ρvD/η (low Re → streamline)
5 important questions
Q1. Why does mercury fall in a glass capillary while water rises?
For water–glass, adhesion is stronger than cohesion; θ < 90°, cos θ is positive, so h is positive (rise). For mercury, cohesion is stronger; θ > 90°, cos θ is negative, so h is negative (fall).
Q2. State Stokes' law and give the meaning of the terms.
A small sphere of radius r moving with small velocity v through a fluid of viscosity η feels a viscous drag F = 6πηrv. It holds for streamline flow only.
Q3. Derive the excess pressure inside a liquid drop.
Drop of radius r, surface tension T, excess pressure P inside.
Let radius increase by dr due to P.
Work done by pressure = P × 4πr² dr.
Increase in surface area = 8πr dr; increase in surface energy = T × 8πr dr.
Equate: P × 4πr² dr = 8πrT dr → P = 2T/r.
For a soap bubble with two surfaces, P = 4T/r.
Q4. Water rises 3 cm in a capillary of radius 0.5 mm. Find the surface tension of water (θ = 0°, ρ = 1000 kg m⁻³, g = 9.8 m s⁻²).
T = hrρg/(2 cos θ).
h = 0.03 m, r = 0.5 × 10⁻³ m.
T = 0.03 × 0.5 × 10⁻³ × 1000 × 9.8/2.
Numerator = 0.03 × 0.5 × 9.8 = 0.147; T = 0.147/2 = 0.0735 N m⁻¹.
Q5. Derive the terminal velocity of a sphere falling in a viscous liquid.
Forces: weight W = (4/3)πr³ρg down; buoyancy B = (4/3)πr³σg up; drag F = 6πηrv up.
At terminal velocity, net force = 0: W = B + F.
(4/3)πr³(ρ − σ)g = 6πηrv.
v = 2r²(ρ − σ)g/(9η).
So v ∝ r²; bigger drops fall faster.
Exam tip
Remember: drop = 2T/r, bubble = 4T/r (two surfaces). In viscosity numericals convert poise to Pa s (divide by 10) and mm to m before substituting.
The kinetic theory explains the behaviour of gases using the random motion of their molecules. From simple assumptions it gives the pressure of a gas, links temperature to molecular kinetic energy, and explains gas laws, degrees of freedom, mean free path and the Maxwell distribution of speeds.
Key points
Assumptions for an ideal gas: molecules are tiny hard spheres; their size is negligible compared with distances between them; no forces except during collisions; collisions are perfectly elastic; motion is random; time of collision is negligible.
Pressure is due to momentum transferred by molecules hitting the walls.
Temperature is a measure of the average translational KE of molecules: average KE per molecule = (3/2)kT.
The ideal gas equation PV = nRT follows from kinetic theory.
Degrees of freedom (f): number of independent ways a molecule can store energy. Monatomic f = 3; rigid diatomic f = 5 (3 translational + 2 rotational).
Law of equipartition of energy: each degree of freedom gets ½kT of energy on average.
Mean free path λ: average distance a molecule travels between two collisions.
Maxwell's distribution: molecular speeds are spread out; the curve rises, reaches a peak and falls; at higher temperature the peak moves to the right and flattens.
Real gases deviate from ideal behaviour at high pressure and low temperature. Van der Waals' equation corrects for molecular size (b) and attraction (a).
Energy per mole = (f/2)RT; Cv = (f/2)R; Cp = Cv + R; γ = Cp/Cv = 1 + 2/f
γ: monatomic 5/3 ≈ 1.67; diatomic 7/5 = 1.4
Mean free path: λ = 1/(√2 π d² n) (d = molecular diameter, n = molecules per unit volume)
Van der Waals: (P + a/V²)(V − b) = RT (for one mole)
5 important questions
Q1. State the law of equipartition of energy.
In thermal equilibrium, the total energy of a molecule is shared equally among all its degrees of freedom, each getting an average of ½kT.
Q2. Show that γ = 1.4 for a rigid diatomic gas.
For a diatomic gas f = 5. Cv = (5/2)R, Cp = Cv + R = (7/2)R. γ = Cp/Cv = 7/5 = 1.4.
Q3. Derive the expression for pressure exerted by a gas.
Cube of side l contains N molecules of mass m. A molecule has velocity components (cx, cy, cz).
On hitting wall perpendicular to x, momentum change = 2mcx.
Time between successive hits on same wall = 2l/cx, so force from one molecule = 2mcx ÷ (2l/cx) = mcx²/l.
Total force = (m/l) Σcx²; pressure = force/l² = (m/l³)Σcx².
By randomness Σcx² = (1/3)Σc² = (1/3)N c_rms².
P = (1/3)(mN/V)c_rms² = (1/3)ρc_rms².
Q4. Find the rms speed of oxygen molecules at 27 °C (M = 32 × 10⁻³ kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹).
T = 27 + 273 = 300 K.
c_rms = √(3RT/M) = √(3 × 8.31 × 300/0.032).
3 × 8.31 × 300 = 7479; 7479/0.032 = 233719.
c_rms = √233719 ≈ 483 m/s.
Q5. At what temperature will the rms speed of a gas be double its value at 27 °C?
c_rms ∝ √T, so c₂/c₁ = √(T₂/T₁).
2 = √(T₂/300) → T₂/300 = 4 → T₂ = 1200 K.
In Celsius: 1200 − 273 = 927 °C.
Exam tip
Always convert °C to kelvin and molar mass from g to kg before using √(3RT/M). Remember the order c_mp < c_avg < c_rms — it is a favourite short question.
Real gases do not obey the ideal gas equation PV = RT at high pressure and low temperature. Van der Waals corrected the ideal gas equation for the finite size of molecules and the attraction between them. This topic covers the deviation of real gases, Andrews' experiment on CO₂, Van der Waals' equation, critical constants and the reduced equation of state.
Key points
Ideal gas assumptions that fail for real gases: (1) molecules have no volume; (2) there are no intermolecular forces.
Andrews' experiment (CO₂ isotherms): above 31.1 °C CO₂ cannot be liquefied by pressure alone; below it, the isotherm has a flat part where liquid and vapour coexist.
Critical temperature Tc: the temperature above which a gas cannot be liquefied however large the pressure.
Volume correction: molecules occupy space, so the free volume is (V − b). b ≈ 4 × the actual volume of the molecules.
Pressure correction: attraction by inner molecules reduces the force with which molecules hit the wall, so the real pressure is P + a/V².
Van der Waals equation (one mole): (P + a/V²)(V − b) = RT.
At the critical point, the isotherm has a point of inflection: dP/dV = 0 and d²P/dV² = 0.
Critical constants: Vc = 3b, Pc = a/27b², Tc = 8a/27Rb, and RTc/(PcVc) = 8/3 ≈ 2.67 (critical coefficient).
Real gases give critical coefficients of about 3.4 to 3.7, so Van der Waals' equation is only approximately correct.
Boyle temperature T_B = a/Rb: near it, a real gas behaves almost ideally over a wide range of pressure.
Important terms
Critical pressure (Pc) – pressure needed to liquefy the gas at Tc.
Critical volume (Vc) – volume of one mole at Tc and Pc.
Gas vs vapour – a vapour is the gaseous state below its critical temperature (can be liquefied by pressure); a gas is above Tc.
Law of corresponding states – gases with the same reduced pressure and temperature have the same reduced volume.
Formulas
Van der Waals: (P + a/V²)(V − b) = RT (a in Pa m⁶ mol⁻², b in m³ mol⁻¹)
For n moles: (P + an²/V²)(V − nb) = nRT
Vc = 3b; Pc = a/(27b²); Tc = 8a/(27Rb)
a = 3PcVc², b = Vc/3; also a = 27R²Tc²/(64Pc), b = RTc/(8Pc)
Critical coefficient RTc/(PcVc) = 8/3
Reduced equation: (Pr + 3/Vr²)(3Vr − 1) = 8Tr
Boyle temperature T_B = a/(Rb) = (27/8) Tc
5 important questions
Q1. Define critical temperature, critical pressure and critical volume.
Tc is the highest temperature at which a gas can be liquefied by pressure alone. Pc is the pressure needed to liquefy it at Tc. Vc is the volume of one mole at Tc and Pc. For CO₂, Tc ≈ 31.1 °C (304 K) and Pc ≈ 73 atm.
Q2. Explain the two corrections made by Van der Waals.
Volume: molecules have finite size, so the space available to move is V − b, not V.
Pressure: a molecule about to hit the wall is pulled back by the other molecules. The reduction is proportional to (number striking) × (number pulling) ∝ ρ² ∝ 1/V², so the ideal pressure is P + a/V².
Q3. Derive the critical constants in terms of a and b.
Write P = RT/(V − b) − a/V².
At the critical point dP/dV = 0: RTc/(Vc − b)² = 2a/Vc³ … (i)
d²P/dV² = 0: 2RTc/(Vc − b)³ = 6a/Vc⁴ … (ii)
Dividing (i) by (ii): (Vc − b)/2 = Vc/3, so Vc = 3b.
Put in (i): RTc = 2a(2b)²/(27b³) = 8a/(27b), so Tc = 8a/(27Rb).
Put in P: Pc = R(8a/27Rb)/(2b) − a/(9b²) = 4a/(27b²) − 3a/(27b²) = a/(27b²).
Then RTc/(PcVc) = (8a/27b) ÷ (3b × a/27b²) = 8/3.
Q4. For CO₂, a = 0.364 Pa m⁶ mol⁻² and b = 4.27 × 10⁻⁵ m³ mol⁻¹. Find Tc (R = 8.314 J mol⁻¹ K⁻¹).
Tc = 8a/(27Rb) = (8 × 0.364)/(27 × 8.314 × 4.27 × 10⁻⁵) = 2.912/(9.585 × 10⁻³) ≈ 304 K, which matches the measured 31 °C.
Q5. Describe Andrews' experiment and its result.
CO₂ was compressed in a capillary tube at fixed temperatures, with air in a similar tube as a pressure gauge. Isotherms (P vs V) were plotted. At 13.1 °C and 21.5 °C the isotherm shows a gas region, a horizontal part (liquefaction at constant pressure) and a steep liquid part. The horizontal part shrinks as temperature rises and becomes a point at 31.1 °C — the critical point. At 48.1 °C the curve is close to a rectangular hyperbola (nearly ideal). Result: every gas has a critical temperature above which it cannot be liquefied by pressure alone.
Exam tip
Learn the three critical constants and the 8/3 ratio by heart — they come every year. Always state that the experimental ratio (about 3.5) differs, showing the limits of the equation.
Thermodynamics studies heat, work and internal energy and the rules that connect them. The zeroth law defines temperature, the first law is energy conservation, and the second law sets the direction of natural processes and limits engine efficiency. Entropy and the Carnot cycle are central ideas.
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.
First law: heat given to a system = increase in internal energy + work done by the system: dQ = dU + dW.
Internal energy U of an ideal gas depends only on temperature. U is a state function; Q and W depend on the path.
Isothermal process: T constant, ΔU = 0, so Q = W.
Adiabatic process: no heat exchange, Q = 0, so W = −ΔU; PVᵞ = constant.
Isobaric (P constant) and isochoric (V constant, W = 0) processes.
Second law (Kelvin–Planck): no engine working in a cycle can turn all the heat taken from a source into work.
Second law (Clausius): heat cannot flow by itself from a colder body to a hotter body.
Carnot engine: ideal reversible engine with two isothermal and two adiabatic steps. No engine working between the same two temperatures can be more efficient (Carnot's theorem).
Entropy (एन्ट्रॉपी) measures disorder. In a reversible cycle total entropy change is zero; in an irreversible process the entropy of the universe increases.
Third law: entropy of a perfect crystal approaches zero as T → 0 K.
Important terms
System and surroundings – the part studied and everything outside it.
Reversible process – can be retraced exactly, passing through equilibrium states.
Heat engine – converts heat into work in a cycle; a refrigerator does the reverse using work.
Coefficient of performance (COP) – heat removed from cold body ÷ work done (refrigerator).
Formulas
dQ = dU + dW; dW = P dV
Isothermal work (ideal gas): W = nRT ln(V₂/V₁) = 2.303 nRT log₁₀(V₂/V₁)
Q1. State the Kelvin–Planck and Clausius statements of the second law.
Kelvin–Planck: it is impossible to build an engine that works in a cycle and converts all heat taken from a source into work with no other effect. Clausius: it is impossible for heat to flow from a colder body to a hotter body without external work.
Q2. Why is ΔU = 0 in an isothermal process of an ideal gas?
The internal energy of an ideal gas depends only on temperature. In an isothermal process T does not change, so U does not change and ΔU = 0. Hence all heat supplied is used as work.
Q3. Derive the work done in an isothermal expansion of an ideal gas.
For n moles at temperature T: P = nRT/V.
Work dW = P dV = (nRT/V) dV.
W = ∫ from V₁ to V₂ of nRT dV/V = nRT ln(V₂/V₁).
Since P₁V₁ = P₂V₂, W = nRT ln(P₁/P₂).
In log base 10: W = 2.303 nRT log₁₀(V₂/V₁).
Q4. A Carnot engine works between 527 °C and 127 °C. Find its efficiency. If it takes 1000 J from the source per cycle, how much work does it do?
T₁ = 527 + 273 = 800 K; T₂ = 127 + 273 = 400 K.
η = 1 − T₂/T₁ = 1 − 400/800 = 0.5 = 50%.
W = ηQ₁ = 0.5 × 1000 = 500 J.
Heat rejected Q₂ = 1000 − 500 = 500 J.
Q5. Find the entropy change when 1 kg of ice melts at 0 °C (latent heat L = 3.36 × 10⁵ J kg⁻¹).
Melting is at constant temperature T = 273 K.
Q = mL = 1 × 3.36 × 10⁵ J.
ΔS = Q/T = 3.36 × 10⁵/273 ≈ 1231 J K⁻¹.
Entropy increases because liquid water is more disordered than ice.
Exam tip
Always use kelvin in efficiency, COP and entropy formulas. Learn the four Carnot steps in order (isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression) with a labelled P–V diagram.
Maxwell's Thermodynamic Relations and Joule–Thomson Effect
Quick summary
Maxwell's thermodynamic relations connect quantities that are easy to measure (P, V, T) with ones that are hard to measure (entropy S). They come from the four thermodynamic potentials. This topic covers the potentials, the four Maxwell relations, the TdS equations, the Clausius–Clapeyron equation and the Joule–Thomson (throttling) effect used to cool and liquefy gases.
Key points
The four thermodynamic potentials: internal energy U, enthalpy H = U + PV, Helmholtz free energy F = U − TS, Gibbs free energy G = H − TS.
Their differentials (using dU = TdS − PdV): dU = TdS − PdV, dH = TdS + VdP, dF = −SdT − PdV, dG = −SdT + VdP.
Each is an exact differential, so the cross partial derivatives are equal. This gives the four Maxwell relations.
Clausius–Clapeyron equation: dP/dT = L/[T(V₂ − V₁)]. Since ice contracts on melting, pressure lowers its melting point (regelation).
Joule–Thomson effect: when a gas expands slowly through a porous plug from high to low pressure, with no heat exchange, its temperature changes. Enthalpy stays constant (isenthalpic process).
For an ideal gas there is no JT effect. Real gases cool because work is done against intermolecular attraction.
Every gas has an inversion temperature Ti: below it the gas cools on throttling, above it the gas warms. For hydrogen and helium Ti is very low, so they warm at room temperature and must be pre-cooled.
The JT effect is used in the Linde process to liquefy air.
Important terms
Throttling – slow expansion through a narrow opening or porous plug, at constant enthalpy.
JT coefficient μ = (∂T/∂P) at constant H; positive μ means cooling on expansion.
Inversion temperature – the temperature at which μ = 0 and the effect changes sign.
Exact differential – a differential whose integral depends only on the end states.
Formulas
(∂T/∂V)_S = −(∂P/∂S)_V
(∂T/∂P)_S = (∂V/∂S)_P
(∂S/∂V)_T = (∂P/∂T)_V
(∂S/∂P)_T = −(∂V/∂T)_P
First TdS equation: TdS = Cv dT + T(∂P/∂T)_V dV
Cp − Cv = −T (∂V/∂T)_P² (∂P/∂V)_T; for an ideal gas this gives Cp − Cv = R
Clausius–Clapeyron: dP/dT = L/[T(V₂ − V₁)]
Van der Waals gas: inversion temperature Ti = 2a/(Rb); JT cooling ΔT ≈ (2a/RT − b) ΔP/Cp
5 important questions
Q1. Derive the Maxwell relation (∂S/∂V)_T = (∂P/∂T)_V.
Helmholtz function F = U − TS, so dF = dU − TdS − SdT = −SdT − PdV. Hence (∂F/∂T)_V = −S and (∂F/∂V)_T = −P. Since dF is exact, ∂²F/∂V∂T = ∂²F/∂T∂V, giving −(∂S/∂V)_T = −(∂P/∂T)_V, i.e. (∂S/∂V)_T = (∂P/∂T)_V.
Q2. Why does an ideal gas show no Joule–Thomson effect?
In throttling, H is constant. For an ideal gas H = U + PV = U + RT depends only on T (U depends only on T). If H is constant, T must be constant. So there is no cooling or heating. A real gas cools because its U also depends on the distance between molecules.
Q3. Derive the Clausius–Clapeyron equation and use it for ice.
Use (∂S/∂V)_T = (∂P/∂T)_V during a phase change, where T and P stay fixed. The entropy change is L/T for volume change (V₂ − V₁), so dP/dT = L/[T(V₂ − V₁)]. Example: ice at 273 K, L = 3.36 × 10⁵ J kg⁻¹, V(water) − V(ice) = 1.000 × 10⁻³ − 1.091 × 10⁻³ = −9.1 × 10⁻⁵ m³ kg⁻¹. dT/dP = T(V₂ − V₁)/L = 273 × (−9.1 × 10⁻⁵)/(3.36 × 10⁵) ≈ −7.4 × 10⁻⁸ K Pa⁻¹. For 1 atm (1.013 × 10⁵ Pa) the melting point falls by about 0.0075 K.
Q4. Describe the porous plug experiment and give its results.
A gas is forced at constant high pressure P₁ through a porous plug (cotton or silk) in an insulated tube to a region of lower pressure P₂. Thermometers on both sides record the temperatures. Results: (1) at room temperature most gases (air, CO₂) cool; (2) hydrogen and helium warm up; (3) the change is proportional to the pressure drop; (4) every gas cools if its initial temperature is below its inversion temperature. Work done on the gas P₁V₁ − work done by it P₂V₂ = change in U, so U₁ + P₁V₁ = U₂ + P₂V₂: enthalpy is constant.
Q5. Using the Van der Waals constants for nitrogen (a = 0.137 Pa m⁶ mol⁻², b = 3.87 × 10⁻⁵ m³ mol⁻¹), estimate its inversion temperature.
Ti = 2a/(Rb) = (2 × 0.137)/(8.314 × 3.87 × 10⁻⁵) = 0.274/(3.217 × 10⁻⁴) ≈ 852 K. (The measured value is about 620 K; the Van der Waals model gives only an estimate.) Room temperature is far below Ti, so nitrogen cools on throttling.
Exam tip
To remember Maxwell's relations, start from the potential and its natural variables (U: S, V; H: S, P; F: T, V; G: T, P) and take cross derivatives. The relations from U and G carry a minus sign; those from H and F do not — check signs by writing dU or dG first.
An oscillation is a to-and-fro motion repeated about a mean position. This topic covers simple harmonic motion (SHM), its differential equation and energy, and then real oscillations: damped oscillations, forced oscillations and resonance.
Key points
SHM: restoring force is proportional to displacement and directed towards the mean position: F = −kx.
Differential equation of SHM: d²x/dt² + ω²x = 0, with ω² = k/m.
Solution: x = A sin(ωt + φ), where A is amplitude and φ the initial phase (epoch).
Velocity is maximum at the mean position and zero at the extreme positions; acceleration is the opposite.
Total energy in SHM = ½kA², constant; KE and PE keep exchanging.
Examples: simple pendulum (small angles), mass on a spring, compound pendulum, LC circuit.
Damped oscillation: amplitude decreases with time because of friction or resistance. Three cases: underdamped (oscillates with falling amplitude), critically damped (returns fastest without oscillating), overdamped (returns slowly without oscillating).
Forced oscillation: a body is driven by an external periodic force; it finally oscillates at the driver's frequency.
Resonance: when the driving frequency equals the natural frequency, amplitude becomes maximum. Sharpness depends on damping (less damping → sharper resonance).
Quality factor Q measures how slowly the energy is lost; high Q means sharp resonance.
Important terms
Time period (आवर्तकाल) T – time for one complete oscillation.
Frequency n – number of oscillations per second, n = 1/T, unit hertz.
Angular frequency ω – ω = 2πn = 2π/T.
Phase – the state of motion at an instant, given by the angle (ωt + φ).
Relaxation time τ – time in which the energy of a damped oscillator falls to 1/e of its value.
Formulas
x = A sin(ωt + φ); v = Aω cos(ωt + φ) = ω√(A² − x²); a = −ω²x
v_max = Aω; a_max = Aω²
Spring: T = 2π√(m/k); simple pendulum: T = 2π√(l/g)
Compound pendulum: T = 2π√((K² + l²)/(lg))
KE = ½k(A² − x²); PE = ½kx²; E = ½kA² = ½mω²A²
Damped: m d²x/dt² + b dx/dt + kx = 0; underdamped amplitude falls as A₀e^(−bt/2m)
Damped angular frequency ω' = √(ω₀² − (b/2m)²)
Q = ω₀/(2β) where β = b/2m (for light damping)
Series springs: 1/k = 1/k₁ + 1/k₂; parallel: k = k₁ + k₂
5 important questions
Q1. Define SHM and write its differential equation.
SHM is periodic motion in which the restoring force is directly proportional to displacement from the mean position and acts towards it. Equation: d²x/dt² + ω²x = 0, where ω = √(k/m).
Q2. At what displacement are the KE and PE of a particle in SHM equal?
KE = PE means ½k(A² − x²) = ½kx². So A² = 2x², x = A/√2 ≈ 0.707A.
Q3. Show that the total energy of a particle in SHM is constant.
x = A sin ωt, v = Aω cos ωt.
KE = ½mv² = ½mA²ω² cos² ωt.
PE = ½kx² = ½mω²A² sin² ωt (since k = mω²).
E = KE + PE = ½mω²A²(cos² ωt + sin² ωt) = ½mω²A².
This does not depend on t, so E is constant.
Q4. A body of mass 0.5 kg on a spring of k = 200 N/m oscillates with amplitude 0.1 m. Find period, maximum velocity and total energy.
ω = √(k/m) = √(200/0.5) = √400 = 20 rad/s.
T = 2π/ω = 2 × 3.1416/20 ≈ 0.314 s.
v_max = Aω = 0.1 × 20 = 2 m/s.
E = ½kA² = ½ × 200 × 0.01 = 1 J.
Check: ½mv_max² = ½ × 0.5 × 4 = 1 J ✓.
Q5. Explain damped and forced oscillations and the condition of resonance.
Damped: a resistive force −b(dx/dt) acts; amplitude decays exponentially; frequency is slightly less than natural.
Cases: underdamped, critically damped, overdamped (with sketches of x vs t).
Forced: external force F₀ sin pt acts; after transients die out the body oscillates at frequency p.
Amplitude depends on (ω₀ − p) and damping; maximum near p = ω₀ — resonance.
Examples: tuning a radio, soldiers breaking step on a bridge, sonometer resonance.
Exam tip
Draw the three damping curves (x against t) and the resonance curves for low and high damping — diagrams fetch easy marks. In numericals, find ω first; everything else follows from it.
When a particle is acted on by two simple harmonic motions (SHMs) at the same time, its resulting motion is a combination of the two. This topic covers the superposition of two SHMs along the same line (with equal and slightly different frequencies, giving beats) and two perpendicular SHMs, whose paths are called Lissajous figures. It also explains how a CRO uses these figures to compare frequencies and phases.
Key points
Principle of superposition: the resultant displacement is the vector sum of the individual displacements.
Two collinear SHMs of the same frequency combine into one SHM of the same frequency with amplitude R = √(a₁² + a₂² + 2a₁a₂ cos δ).
R is maximum (a₁ + a₂) when δ = 0 and minimum |a₁ − a₂| when δ = π.
Two collinear SHMs of slightly different frequencies give beats: the amplitude rises and falls with beat frequency (n₁ − n₂).
Two perpendicular SHMs of the same frequency give an ellipse in general. Special cases: δ = 0 or π → straight line; δ = π/2 with equal amplitudes → circle; δ = π/2 with unequal amplitudes → ellipse with axes along x and y.
For a frequency ratio such as 1:2, 2:3 etc. the figure is more complex (figure-of-eight for 1:2) but stays a closed curve if the ratio is a whole-number ratio.
Frequency ratio from the figure: draw a box around the figure; f_y/f_x = (points touching the top edge) ÷ (points touching a side edge).
Lissajous figures are seen on a CRO by applying one signal to the X plates and another to the Y plates.
Important terms
Phase difference (δ) – difference in the phase angles of two SHMs.
Beats – periodic rise and fall of loudness when two close frequencies combine.
Lissajous figure – path traced by a point under two perpendicular SHMs.
CRO – cathode ray oscilloscope, an instrument that displays voltage signals.
Formulas
Collinear, same ω: x = R sin(ωt + φ), R² = a₁² + a₂² + 2a₁a₂ cos δ, tan φ = a₂ sin δ/(a₁ + a₂ cos δ)
Beat frequency = n₁ − n₂
Perpendicular, same ω (x = a sin ωt, y = b sin(ωt + δ)): x²/a² + y²/b² − (2xy/ab) cos δ = sin²δ
Frequency ratio from a Lissajous figure: f_y / f_x = (number of tangent points on a horizontal edge) ÷ (number of tangent points on a vertical edge)
5 important questions
Q1. What are Lissajous figures? Give two uses.
They are closed curves traced by a particle acted on by two perpendicular SHMs whose frequencies are in a simple whole-number ratio. Uses: (1) to compare or measure an unknown frequency against a known one on a CRO; (2) to measure the phase difference between two signals of the same frequency.
Q2. Two collinear SHMs of amplitudes 3 cm and 4 cm and the same frequency differ in phase by 90°. Find the resultant amplitude.
R = √(3² + 4² + 2 × 3 × 4 × cos 90°) = √(9 + 16 + 0) = 5 cm.
Q3. Derive the equation of the path for two perpendicular SHMs of the same frequency.
x = a sin ωt → sin ωt = x/a, cos ωt = √(1 − x²/a²).
y = b sin(ωt + δ) = b(sin ωt cos δ + cos ωt sin δ), so y/b − (x/a) cos δ = √(1 − x²/a²) sin δ.
Square both sides: y²/b² − (2xy/ab) cos δ + (x²/a²) cos²δ = sin²δ − (x²/a²) sin²δ.
Rearranging: x²/a² + y²/b² − (2xy/ab) cos δ = sin²δ, an ellipse.
Cases: δ = 0 → y = (b/a)x, a straight line; δ = π → y = −(b/a)x; δ = π/2 → x²/a² + y²/b² = 1; if also a = b, a circle.
Q4. Find the resultant of two collinear SHMs with the same frequency and derive the amplitude.
Let x₁ = a₁ sin ωt and x₂ = a₂ sin(ωt + δ). Then x = (a₁ + a₂ cos δ) sin ωt + (a₂ sin δ) cos ωt. Put R cos φ = a₁ + a₂ cos δ and R sin φ = a₂ sin δ. Then x = R sin(ωt + φ): an SHM of the same frequency. Squaring and adding: R² = a₁² + a₂² + 2a₁a₂ cos δ; dividing: tan φ = a₂ sin δ/(a₁ + a₂ cos δ).
Q5. A Lissajous figure on a CRO touches the horizontal edge at 2 points and the vertical edge at 1 point. The signal on the X plates is 100 Hz. Find the Y frequency.
f_y / f_x = (touches on the horizontal edge) ÷ (touches on the vertical edge) = 2/1, so f_y = 2 × 100 = 200 Hz. This is the horizontal figure-of-eight (∞) traced by x = a sin ωt, y = b sin 2ωt.
Exam tip
Draw the five standard figures for equal frequency (δ = 0, π/4, π/2, 3π/4, π). Examiners usually ask for the derivation of the ellipse equation and these special cases.
Real oscillators lose energy to friction and air resistance, so their amplitude dies away (damped oscillations). If a periodic external force keeps driving them, they settle into forced oscillations at the driver's frequency, and the amplitude becomes very large near the natural frequency (resonance). This topic covers the differential equations, the three cases of damping, logarithmic decrement, relaxation time, quality factor, forced oscillations and sharpness of resonance.
Key points
Damping force is taken proportional to velocity: F_d = −b v (b = damping constant, kg s⁻¹).
Equation of a damped oscillator: m d²x/dt² + b dx/dt + kx = 0, or d²x/dt² + 2γ dx/dt + ω₀²x = 0, where 2γ = b/m and ω₀² = k/m.
Underdamped (γ less than ω₀): oscillates with decaying amplitude, x = A e^(−γt) cos(ω′t + φ), ω′ = √(ω₀² − γ²). Example: a pendulum in air.
Critically damped (γ = ω₀): returns to rest fastest without oscillating. Example: moving-coil galvanometer needle, car shock absorbers.
Overdamped (γ greater than ω₀): returns slowly without oscillating. Example: pendulum in thick oil.
Energy of an underdamped oscillator decays as e^(−2γt).
Forced oscillations: m d²x/dt² + b dx/dt + kx = F₀ sin ωt. After the transient dies, the body oscillates at the driving frequency ω.
Amplitude resonance occurs at ω_r = √(ω₀² − 2γ²) ≈ ω₀ for small damping. Less damping → higher and sharper peak.
At resonance in velocity (ω = ω₀), the driving force is in phase with velocity and power absorbed is maximum.
Examples: tuning a radio (electrical resonance), soldiers breaking step on bridges, Tacoma Narrows bridge collapse (1940).
Important terms
Logarithmic decrement (λ) – natural log of the ratio of two successive amplitudes on the same side.
Relaxation time (τ) – time in which energy falls to 1/e of its initial value; τ = 1/(2γ) = m/b.
Quality factor (Q) – measure of sharpness of resonance; high Q means low damping.
Sharpness of resonance – how fast the amplitude falls when ω moves away from resonance.
Q1. Distinguish between free, damped and forced oscillations.
Free: no damping, constant amplitude, natural frequency. Damped: amplitude falls exponentially due to resistive forces, frequency slightly less than natural. Forced: an external periodic force keeps the body oscillating at the force's frequency; amplitude depends on how close the driver is to the natural frequency.
Q2. The amplitude of a damped oscillator falls from 10 cm to 8 cm in one oscillation. Find the logarithmic decrement and the amplitude after 3 more oscillations.
λ = ln(10/8) = ln 1.25 = 0.223. Each oscillation multiplies the amplitude by 0.8, so after 3 more: 8 × 0.8³ = 8 × 0.512 = 4.1 cm.
Q3. Solve the equation of a damped oscillator and discuss the three cases.
Try x = e^(αt): α² + 2γα + ω₀² = 0, so α = −γ ± √(γ² − ω₀²).
γ greater than ω₀ (overdamped): both roots real and negative → x = e^(−γt)(C₁e^(βt) + C₂e^(−βt)), β = √(γ² − ω₀²); no oscillation, slow return.
Steady-state trial solution: x = A sin(ωt − θ). Substitute and equate coefficients of sin and cos.
This gives A = (F₀/m)/√[(ω₀² − ω²)² + 4γ²ω²] and tan θ = 2γω/(ω₀² − ω²).
A is maximum when the denominator is minimum: d/dω[(ω₀² − ω²)² + 4γ²ω²] = 0 → ω² = ω₀² − 2γ². So ω_r = √(ω₀² − 2γ²).
At ω = ω₀ the phase lag θ = 90°: displacement lags the force by a quarter cycle.
Q5. A tuning fork of frequency 256 Hz has its energy fall to 1/e in 4 s. Find its quality factor.
τ = 4 s, ω₀ = 2π × 256 = 1608 rad s⁻¹. Q = ω₀τ = 1608 × 4 ≈ 6.4 × 10³. A high Q means very low damping, so the fork rings for a long time.
Exam tip
Always state the three damping cases with one real example each. In numericals, check whether the question gives amplitude ratio (use λ) or energy (use τ); energy decays twice as fast as amplitude.
A wave carries energy from one place to another without carrying matter. This topic covers the differential equation of a wave, transverse and longitudinal waves, the speed of waves on a string and in a gas (Newton's formula and Laplace's correction), phase and group velocity, standing waves, and the Doppler effect for sound and light.
Key points
Transverse wave: particles vibrate perpendicular to the direction of travel (waves on a string, light). Longitudinal wave: particles vibrate along the direction of travel (sound in air).
A progressive wave moving along +x: y = a sin(ωt − kx), where k = 2π/λ and v = ω/k = nλ.
Every travelling wave satisfies the wave equation ∂²y/∂t² = v² ∂²y/∂x².
Speed on a stretched string v = √(T/μ), where T is tension and μ is mass per unit length.
Newton assumed sound travels isothermally: v = √(P/ρ) ≈ 280 m s⁻¹ in air, which is too low.
Laplace's correction: compressions and rarefactions are fast, so the process is adiabatic: v = √(γP/ρ) ≈ 332 m s⁻¹ at 0 °C.
Speed of sound in a gas is proportional to √T and does not depend on pressure (at constant temperature).
Phase velocity v_p = ω/k; group velocity v_g = dω/dk is the speed of the energy (wave packet). In a non-dispersive medium v_g = v_p.
Standing waves form when two equal waves travel in opposite directions; nodes are λ/2 apart.
Doppler effect: the apparent frequency changes when the source or the observer moves. Approach → higher pitch; recession → lower pitch.
Uses: police radar speed guns, Doppler ultrasound of blood flow, red shift of galaxies (expanding universe).
Important terms
Wavelength (λ) – distance between two nearest points in the same phase.
Wave number (k) – 2π/λ, in rad m⁻¹.
Dispersive medium – one where wave speed depends on frequency.
Node / antinode – points of zero / maximum amplitude in a standing wave.
Red shift – increase in wavelength of light from a receding source.
Formulas
v = nλ = ω/k
Wave equation: ∂²y/∂t² = v² ∂²y/∂x²
String: v = √(T/μ); fundamental of a string fixed at both ends n = (1/2L)√(T/μ)
Gas (Laplace): v = √(γP/ρ) = √(γRT/M); v_t = v₀√(T/273)
Group velocity v_g = dω/dk = v_p − λ dv_p/dλ
Doppler (sound, speed v in medium): n′ = n (v + v_o)/(v − v_s), with v_o positive when the observer moves towards the source and v_s positive when the source moves towards the observer
Doppler (light, u much less than c): Δλ/λ = u/c
5 important questions
Q1. What is Laplace's correction to Newton's formula?
Newton took sound propagation as isothermal, giving v = √(P/ρ) ≈ 280 m s⁻¹ in air at STP, about 15% less than the measured 332 m s⁻¹. Laplace argued that compressions are too quick for heat exchange, so the change is adiabatic and the elasticity is γP. With γ = 1.4: v = √(1.4 × 1.013 × 10⁵/1.293) ≈ 331 m s⁻¹, which agrees with experiment.
Q2. Derive the relation between group velocity and phase velocity.
v_g = dω/dk. Since ω = v_p k: v_g = v_p + k dv_p/dk. With k = 2π/λ, dk = −(2π/λ²)dλ, so k dv_p/dk = −λ dv_p/dλ. Hence v_g = v_p − λ dv_p/dλ. In a non-dispersive medium dv_p/dλ = 0 and v_g = v_p.
Q3. Derive the expression for the Doppler effect when the source moves towards a stationary observer.
The source emits n waves per second; in 1 s the first wave travels v and the source moves v_s forward.
These n waves are squeezed into length (v − v_s), so the apparent wavelength λ′ = (v − v_s)/n.
The observer receives n′ = v/λ′ = n v/(v − v_s), which is greater than n.
If the observer also moves towards the source with v_o, he meets (v + v_o)/λ′ waves per second: n′ = n(v + v_o)/(v − v_s).
Q4. An ambulance siren of 500 Hz approaches a standing person at 20 m s⁻¹ and then moves away. Find the frequencies heard (v = 340 m s⁻¹).
Approaching: n′ = 500 × 340/(340 − 20) = 500 × 340/320 = 531 Hz. Receding: n″ = 500 × 340/(340 + 20) = 500 × 340/360 = 472 Hz. The drop in pitch as it passes is about 59 Hz.
Q5. Show that y = a sin(ωt − kx) satisfies the wave equation, and find the speed of a wave on a string with T = 100 N and μ = 0.01 kg m⁻¹.
∂²y/∂t² = −aω² sin(ωt − kx) and ∂²y/∂x² = −ak² sin(ωt − kx). So ∂²y/∂t² = (ω²/k²) ∂²y/∂x² = v² ∂²y/∂x² with v = ω/k. For the string: v = √(T/μ) = √(100/0.01) = √10⁴ = 100 m s⁻¹.
Exam tip
In Doppler problems, write the sign convention first and check the answer: approach must raise the frequency, recession must lower it. Remember the Doppler effect for sound is not symmetric — a moving source and a moving observer at the same speed give different results.
Acoustics is the science of sound, especially how sound behaves inside halls and rooms. Ultrasonics deals with sound waves above 20 kHz that humans cannot hear. This topic covers intensity and decibels, reverberation and Sabine's formula, good acoustic design, and the production, detection and uses of ultrasonic waves.
Key points
Sound is a longitudinal mechanical wave; it needs a medium.
Audible range for humans: about 20 Hz to 20 kHz. Below 20 Hz is infrasonic; above 20 kHz is ultrasonic.
Intensity of sound = energy flowing per second per unit area (W m⁻²). Loudness is our sensation of it and is roughly logarithmic (Weber–Fechner law).
Reverberation: sound persisting in a hall after the source stops, due to repeated reflections.
Reverberation time: time for sound intensity to fall to one-millionth (10⁻⁶) of its initial value, i.e. a drop of 60 dB.
Sabine's formula relates reverberation time to hall volume and total absorption.
A good hall has suitable reverberation time, uniform loudness, no echoes, no focusing, and low outside noise. Too little reverberation makes sound dull; too much makes it unclear.
Absorbing materials: curtains, carpets, perforated boards, people in the audience.
Production of ultrasonics: magnetostriction method (a ferromagnetic rod changes length in an alternating magnetic field; good for lower ultrasonic frequencies) and piezoelectric method (a quartz crystal vibrates under alternating voltage; for higher frequencies).
Uses: SONAR, echo sounding of sea depth, non-destructive testing of metals for flaws, medical scanning (sonography), cleaning, drilling, homogenising milk.
Important terms
Decibel (dB) – unit of intensity level on a log scale.
Absorption coefficient (a) – fraction of sound energy absorbed by a surface; open window = 1.
Sabine – unit of absorption: 1 m² of open window (perfect absorber).
Piezoelectric effect (दाब-विद्युत परिणाम) – voltage across a crystal when pressed; the inverse effect makes it vibrate under voltage.
Magnetostriction – change in length of a ferromagnetic rod when magnetised.
Formulas
Intensity level: L = 10 log₁₀(I/I₀) dB, I₀ = 10⁻¹² W m⁻²
Sabine's formula: T = 0.161 V/A (SI; V in m³, A in m² sabine), with A = Σ aᵢSᵢ
Average absorption coefficient: a_avg = Σ aᵢSᵢ / ΣSᵢ
Piezoelectric oscillator frequency: n = (1/2t)√(Y/ρ) (t = crystal thickness)
Magnetostriction oscillator: n = (1/2l)√(Y/ρ) (l = rod length)
SONAR depth: d = vt/2 (t = time for echo)
5 important questions
Q1. Define reverberation time.
It is the time taken for the intensity of sound in a hall to fall to one-millionth of its original value after the source is stopped (a fall of 60 dB).
Q2. The intensity of a sound is 10⁻⁶ W m⁻². Find its intensity level.
Q3. Explain the factors affecting acoustics of buildings and their remedies.
Reverberation time: keep it optimum by adding absorbing materials or audience seats.
Loudness: should be uniform; use reflecting boards and proper seating.
Focusing: avoid curved (concave) walls and domes, or cover them with absorbers.
Echoes: cover distant reflecting walls with absorbers.
Echelon effect (regular steps/railings): cover with carpet.
Noise: use double doors, sound-proof windows, and keep machinery isolated.
Q4. A hall has volume 2000 m³ and total absorption 200 m² sabine. Find the reverberation time. If 100 people (absorption 0.5 each) enter, what is the new time?
T = 0.161V/A = 0.161 × 2000/200 = 1.61 s.
Extra absorption = 100 × 0.5 = 50; new A = 250.
New T = 0.161 × 2000/250 = 322/250 ≈ 1.29 s.
Q5. Describe the piezoelectric method of producing ultrasonic waves.
A quartz crystal plate is placed between two metal plates connected to an oscillator circuit (tank circuit with L and C).
Alternating voltage applied to the crystal makes it expand and contract (inverse piezoelectric effect).
Adjust C so the circuit frequency equals the crystal's natural frequency n = (1/2t)√(Y/ρ).
At resonance the crystal vibrates strongly and emits ultrasonic waves.
Advantages: very high frequencies (up to hundreds of MHz), stable output.
Exam tip
Learn Sabine's formula with its constant 0.161 (SI). In acoustics long answers, list the defects and their remedies as a neat two-column table.
A blackbody absorbs all radiation falling on it and, when heated, emits radiation of all wavelengths. Classical physics could not explain the observed blackbody spectrum. Planck's quantum hypothesis (1900) solved this problem and began quantum physics. This topic covers Kirchhoff's law, Stefan–Boltzmann law, Wien's law, Rayleigh–Jeans law and Planck's law.
Key points
A perfect blackbody absorbs all incident radiation (absorptivity = 1). A small hole in a hollow cavity with blackened inner walls is a good practical blackbody (Ferry's blackbody).
Kirchhoff's law: at a given temperature and wavelength, the ratio of emissive power to absorptive power is the same for all bodies and equals the emissive power of a blackbody. Good absorbers are good emitters.
Stefan–Boltzmann law: total energy radiated per unit area per second by a blackbody is proportional to T⁴.
Energy distribution curves: for each temperature, emitted energy rises with wavelength, reaches a peak at λ_m and then falls. As T rises, the area under the curve increases and λ_m shifts to shorter wavelengths.
Wien's displacement law: λ_m T = constant.
Rayleigh–Jeans law (classical) fits only at long wavelengths. At short wavelengths it predicts infinite energy — the ultraviolet catastrophe.
Wien's distribution law fits only at short wavelengths.
Planck's hypothesis: oscillators in the cavity walls can have only discrete energies nhν; energy is emitted or absorbed in packets (quanta) of hν.
Planck's law agrees with experiment at all wavelengths, and gives Wien's and Rayleigh–Jeans laws as limits.
Important terms
Emissive power (eλ) – energy emitted per unit area per second per unit wavelength range.
Absorptive power (aλ) – fraction of incident radiation absorbed.
Quantum (पुंज) – smallest packet of energy, E = hν.
Ultraviolet catastrophe – failure of classical theory at short wavelengths.
Formulas
Stefan–Boltzmann: E = σT⁴, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; net loss E = σ(T⁴ − T₀⁴); for non-black body E = eσT⁴
Wien's displacement: λ_m T = b, b ≈ 2.898 × 10⁻³ m K
Energy of a quantum: E = hν = hc/λ, h = 6.63 × 10⁻³⁴ J s
Average energy of a Planck oscillator: ε = hν/(e^(hν/kT) − 1)
5 important questions
Q1. What is the ultraviolet catastrophe?
The Rayleigh–Jeans law says energy density grows as 1/λ⁴. So at very short (ultraviolet) wavelengths it predicts infinite energy, which is not seen. This failure of classical physics is called the ultraviolet catastrophe.
Q2. State Wien's displacement law and give one use.
The wavelength of maximum emission of a blackbody is inversely proportional to its absolute temperature: λ_m T = b. It is used to estimate the surface temperature of the Sun and stars from their spectrum.
Q3. State the assumptions of Planck's quantum theory and write Planck's radiation law.
A blackbody cavity has atomic oscillators of all frequencies.
An oscillator of frequency ν can have only energies 0, hν, 2hν, … nhν.
Energy is emitted or absorbed only in whole quanta of hν, not continuously.
Average energy of an oscillator: ε = hν/(e^(hν/kT) − 1), not kT.
Multiply by the number of modes 8π/λ⁴ per unit volume per unit wavelength: u_λ = (8πhc/λ⁵)/(e^(hc/λkT) − 1).
Q4. Show that Planck's law reduces to the Rayleigh–Jeans law at long wavelengths.
For long λ, x = hc/λkT is very small.
Then e^x ≈ 1 + x, so e^x − 1 ≈ hc/λkT.
u_λ ≈ (8πhc/λ⁵) × (λkT/hc) = 8πkT/λ⁴.
This is the Rayleigh–Jeans law. (For short λ, e^x ≫ 1 and we get Wien's law.)
Q5. The Sun's spectrum peaks near 500 nm. Estimate its surface temperature (b = 2.898 × 10⁻³ m K).
T = b/λ_m = 2.898 × 10⁻³/(500 × 10⁻⁹).
= 2.898 × 10⁻³/(5 × 10⁻⁷) = 0.5796 × 10⁴.
T ≈ 5796 K, i.e. about 5800 K.
Exam tip
Draw the energy-distribution curves for three temperatures with λ_m shifting left — it is asked almost every year. Know how Planck's law gives both Wien's and Rayleigh–Jeans laws as limits.