Class 11 Physics — Important Board Questions with Answers

[v] = L T^-1, [u] = L T^-1, and [at] = (L T^-2)(T) = L T^-1. Every term has the same dimension L T^-1, so the equation is dimensionally consistent (homogeneous).

[a] = L T^-2, [t] = T

Three significant figures (4, 2 and the trailing 0). Leading zeros are not significant; a trailing zero after the decimal point is significant.

From F = G M m / r^2, we get G = F r^2 / (M m). Dimensions: [G] = ([F][r^2]) / ([M][m]) = (M L T^-2 * L^2) / (M * M) = M^-1 L^3 T^-2.

G = F r^2 / (M m)

Systematic errors have a definite cause and always shift readings in one direction (e.g. a zero error in an instrument, or a mis-calibrated scale); they can be minimised by correcting the instrument. Random errors occur due to unpredictable fluctuations and vary in size and sign; they are reduced by taking many readings and averaging.

For a product/power, percentage errors add with the powers as weights: (dP/P) = (dA/A) + 2 (dB/B) = 2% + 2(3%) = 2% + 6% = 8%.

dP/P = dA/A + 2 dB/B

Uses: (i) to check the dimensional consistency of an equation; (ii) to convert a quantity from one system of units to another; (iii) to derive the relation between physical quantities. Limitations: (i) it cannot find dimensionless constants (like 1/2 or 2 pi); (ii) it cannot be used if a quantity depends on more than three others, or on trigonometric/exponential/logarithmic functions.

At maximum height the velocity is zero. (a) h = u^2 / (2g) = (20)^2 / (2*10) = 400/20 = 20 m. (b) Time of flight T = 2u/g = (2*20)/10 = 4 s.

h = u^2/2g ; T = 2u/g

Distance is the total length of the actual path travelled; it is a scalar and is always positive. Displacement is the shortest straight-line vector from the initial to the final position; it is a vector and can be positive, negative or zero. Displacement magnitude is always less than or equal to the distance.

On a v-t graph for constant acceleration, the velocity rises linearly from u to v = u + a t. The displacement equals the area under the graph, which is a rectangle (height u, width t) plus a triangle (base t, height a t): s = (u * t) + (1/2)(t)(a t) = u t + (1/2) a t^2.

s = u t + (1/2) a t^2

Using v^2 = u^2 - 2 a s with v = 0: 0 = (20)^2 - 2(5) s, so 10 s = 400, giving s = 40 m.

v^2 = u^2 - 2 a s

Relative velocity of a body A with respect to B is the velocity of A as seen from B: v_AB = v_A - v_B. For the trains, v = 60 - 40 = 20 km/h, so the faster train appears to move ahead at 20 km/h relative to the slower one.

v_AB = v_A - v_B

Distance in the nth second: s_n = u + (a/2)(2n - 1). With u = 0, a = 2, n = 5: s_5 = 0 + (2/2)(2*5 - 1) = 1 * 9 = 9 m.

s_n = u + (a/2)(2n - 1)

Resolve u into ux = u cos(theta) and uy = u sin(theta). Vertical motion has acceleration -g. Time of flight: uy - g(T/2) = 0 at the top, so T = 2u sin(theta)/g. Maximum height: H = uy^2/2g = u^2 sin^2(theta)/2g. Horizontal range: R = ux * T = u cos(theta) * 2u sin(theta)/g = u^2 sin(2 theta)/g. R is maximum when theta = 45 degrees.

T = 2u sin(theta)/g ; H = u^2 sin^2(theta)/2g ; R = u^2 sin(2theta)/g

Parallelogram law: if two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, their resultant is represented by the diagonal from that point. Magnitude: R = sqrt(P^2 + Q^2 + 2 P Q cos(theta)), and the resultant makes an angle alpha with P where tan(alpha) = (Q sin(theta)) / (P + Q cos(theta)).

R = sqrt(P^2 + Q^2 + 2 P Q cos(theta))

In uniform circular motion, centripetal acceleration is the acceleration directed towards the centre of the circle that continuously changes the direction of velocity. Its magnitude is a_c = v^2 / r = omega^2 r, and it always points radially inward (towards the centre).

a_c = v^2 / r = omega^2 r

R = u^2 sin(2 theta) / g = (30)^2 * sin(60 deg) / 10 = 900 * 0.866 / 10 = 77.9 m (about 78 m).

R = u^2 sin(2 theta) / g

Resolution of a vector is the process of splitting it into two or more components, usually along mutually perpendicular directions. For a vector A at angle theta to the x-axis, the rectangular components are Ax = A cos(theta) along the x-axis and Ay = A sin(theta) along the y-axis, with A = sqrt(Ax^2 + Ay^2).

Ax = A cos(theta) ; Ay = A sin(theta)

Impulse is the product of a force and the time for which it acts: J = F * (delta t). By the impulse-momentum theorem it equals the change in momentum: J = delta p = m v - m u. SI unit: N s (= kg m/s).

J = F * delta t = delta p

Force of friction f = mu * m g = 0.2 * 2 * 10 = 4 N. Net force = applied - friction = 10 - 4 = 6 N. Acceleration a = net force / mass = 6 / 2 = 3 m/s^2.

f = mu m g ; a = (F - f)/m

Newton's second law: the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force. F = dp/dt = d(mv)/dt. For constant mass, F = m (dv/dt) = m a.

F = dp/dt = m a

Newton's third law: to every action there is an equal and opposite reaction, and the two forces act on different bodies. Example: when we walk, the foot pushes the ground backward (action) and the ground pushes the foot forward (reaction), which moves us ahead.

F(AB) = - F(BA)

By conservation of linear momentum (initial total momentum = 0): m_bullet * v_bullet = m_gun * v_gun. So 0.020 * 400 = 4 * v_gun, giving v_gun = 8 / 4 = 2 m/s (opposite to the bullet).

m1 v1 = m2 v2

Static friction acts on a body at rest and is self-adjusting up to a maximum value (limiting friction) f_s(max) = mu_s N. Kinetic (sliding) friction acts on a moving body and has a nearly constant value f_k = mu_k N. For the same surfaces, mu_k < mu_s, so it is harder to start motion than to keep it going.

f_s(max) = mu_s N ; f_k = mu_k N

On a smooth incline the acceleration along the plane is a = g sin(theta) = 10 * sin(30 deg) = 10 * 0.5 = 5 m/s^2, directed down the incline.

a = g sin(theta)

Roads are banked at curves so that the horizontal component of the normal reaction provides the centripetal force needed to turn, reducing reliance on friction and the risk of skidding. On a frictionless banked road, the ideal (safe) speed is v = sqrt(r g tan(theta)).

v = sqrt(r g tan(theta))

When the lift accelerates upward, the apparent weight (normal reaction) is N = m (g + a) = 50 (10 + 2) = 50 * 12 = 600 N. (At rest it would be m g = 500 N, so the person feels heavier.)

N = m (g + a)

The work-energy theorem states that the work done by the net force on a body equals the change in its kinetic energy: W_net = (1/2) m v^2 - (1/2) m u^2. If net work is positive the body speeds up; if negative, it slows down.

W_net = (1/2)m v^2 - (1/2)m u^2

By conservation of energy, KE at the ground = loss in potential energy = m g h = 5 * 10 * 20 = 1000 J. (Check: v = sqrt(2gh) = 20 m/s, KE = (1/2)(5)(20^2) = 1000 J.)

KE = m g h

Work done by a constant force is the dot product of force and displacement: W = F s cos(theta), where theta is the angle between them. (a) W = 0 when theta = 90 degrees (force perpendicular to displacement, e.g. centripetal force). (b) W is negative when theta is obtuse (90 to 180 degrees), e.g. friction opposing motion.

W = F s cos(theta)

A conservative force does work that depends only on the initial and final positions, not on the path, and the work done in a closed loop is zero (example: gravity, spring force). A non-conservative force does path-dependent work and dissipates energy (example: friction, viscous drag).

Work done W = m g h = 100 * 10 * 10 = 10000 J. Power P = W / t = 10000 / 5 = 2000 W = 2 kW.

P = W / t = m g h / t

In an elastic collision both linear momentum and kinetic energy are conserved (e.g. collisions between hard steel balls, or gas molecules). In an inelastic collision momentum is conserved but kinetic energy is not - some is lost as heat, sound or deformation; in a perfectly inelastic collision the bodies stick together.

The restoring force at extension x' is F = k x'. The work done in stretching the spring from 0 to x is W = integral of (k x') dx' from 0 to x = (1/2) k x^2. This work is stored as elastic potential energy: U = (1/2) k x^2.

U = (1/2) k x^2

The coefficient of restitution (e) is the ratio of the relative velocity of separation after collision to the relative velocity of approach before collision: e = (velocity of separation) / (velocity of approach). For a perfectly elastic collision e = 1; for a perfectly inelastic collision e = 0.

e = (v2 - v1) / (u1 - u2)

At the highest point, gravity alone must provide the centripetal force, so m g = m v_top^2 / r, giving the minimum speed at the top v_top = sqrt(g r). Using energy conservation between the lowest and highest points, the minimum speed at the bottom is v_bottom = sqrt(5 g r).

v_top = sqrt(g r) ; v_bottom = sqrt(5 g r)

Moment of inertia is the rotational analogue of mass: I = sum of (m_i r_i^2). It measures a body's opposition to a change in its rotational motion. SI unit: kg m^2. It depends on (i) the mass and its distribution and (ii) the position/orientation of the axis of rotation.

I = sum(m_i r_i^2)

Parallel axes theorem: I = I_cm + M d^2, where I_cm is the moment of inertia about a parallel axis through the centre of mass and d is the distance between the axes. For a rod, I_cm = M L^2 / 12 and d = L/2, so I_end = M L^2/12 + M (L/2)^2 = M L^2/12 + M L^2/4 = M L^2/3.

I = I_cm + M d^2

Torque is the turning effect of a force about an axis, equal to the product of the force and the perpendicular distance of its line of action from the axis: tau = r F sin(theta) = r x F (cross product). SI unit: newton metre (N m). It is a vector along the axis of rotation.

tau = r F sin(theta)

When the net external torque on a system is zero, its total angular momentum L = I omega remains constant. A spinning skater experiences almost no external torque, so I omega is constant. On pulling the arms in, the moment of inertia I decreases, so the angular speed omega increases to keep L constant - the skater spins faster.

I1 omega1 = I2 omega2

Perpendicular axes theorem (for a planar body): the moment of inertia about an axis perpendicular to the plane equals the sum of the moments of inertia about two mutually perpendicular axes in the plane meeting at the same point: Iz = Ix + Iy. For a ring, Iz (about central axis) = M R^2, and by symmetry Ix = Iy = I(diameter). So M R^2 = 2 I(diameter), giving I(diameter) = M R^2 / 2.

Iz = Ix + Iy

The radius of gyration K is the distance from the axis at which the whole mass of the body can be assumed to be concentrated so as to give the same moment of inertia: I = M K^2, hence K = sqrt(I / M). Its SI unit is the metre.

K = sqrt(I / M)

For a ring about its central axis, I = M R^2 = 2 * (0.5)^2 = 2 * 0.25 = 0.5 kg m^2.

I = M R^2

A rolling body has both translation and rotation. Its total kinetic energy is KE = (1/2) m v^2 + (1/2) I omega^2, where the first term is the translational KE of the centre of mass and the second is the rotational KE about the centre. Using v = R omega and I = m K^2, this becomes KE = (1/2) m v^2 (1 + K^2/R^2).

KE = (1/2) m v^2 + (1/2) I omega^2

The centre of mass is the point at which the entire mass of the system can be considered to be concentrated, and where an applied external force produces the same acceleration as on the whole system. For two particles of masses m1 and m2 at positions x1 and x2, x_cm = (m1 x1 + m2 x2) / (m1 + m2).

x_cm = (m1 x1 + m2 x2) / (m1 + m2)

Since g = G M / (R + h)^2, increasing the height h increases the distance from the centre of the Earth, so g decreases. For small heights, g' = g (1 - 2h/R) approximately.

g' = g (1 - 2h/R)

Escape velocity v_e = sqrt(2 g R) = sqrt(2 * 9.8 * 6.4 x 10^6) = sqrt(1.2544 x 10^8) = 1.12 x 10^4 m/s = 11.2 km/s.

v_e = sqrt(2 g R)

1) Law of orbits: every planet moves in an ellipse with the Sun at one focus. 2) Law of areas: the line joining a planet to the Sun sweeps out equal areas in equal intervals of time (so a planet moves faster when nearer the Sun); this follows from conservation of angular momentum. 3) Law of periods: the square of the orbital period is proportional to the cube of the semi-major axis, T^2 is proportional to a^3.

T^2 = k a^3

For an orbit close to the surface the required centripetal force is provided by gravity, so v_o = sqrt(g R) = sqrt(9.8 * 6.4 x 10^6) = sqrt(6.272 x 10^7) = 7.92 x 10^3 m/s = 7.92 km/s (approximately 8 km/s).

v_o = sqrt(g R)

A geostationary satellite appears stationary relative to the Earth because it revolves in step with the Earth's rotation. Conditions: (i) its orbital period must equal 24 hours (equal to Earth's rotation period); (ii) it must orbit in the equatorial plane, in the same (west-to-east) sense as the Earth, at a height of about 36000 km.

Orbital velocity near the surface: v_o = sqrt(g R). Escape velocity: v_e = sqrt(2 g R). Dividing, v_e / v_o = sqrt(2 g R) / sqrt(g R) = sqrt(2). Hence v_e = sqrt(2) * v_o (about 1.41 times).

v_e = sqrt(2) * v_o

U = - G M m / r. It is negative because the gravitational force is attractive: zero potential energy is taken at infinity, and as the mass is brought closer, work is done by gravity, lowering the energy below zero. The negative sign shows the mass is in a bound state.

U = - G M m / r

With depth d, g decreases as g' = g (1 - d/R), assuming uniform density, because only the mass within the smaller inner sphere attracts the body. At the centre (d = R) the value becomes zero, since the mass is symmetrically distributed all around.

g' = g (1 - d/R)

Every particle of matter attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, directed along the line joining them: F = G m1 m2 / r^2, where G is the universal gravitational constant (6.67 x 10^-11 N m^2 / kg^2).

F = G m1 m2 / r^2

Young's modulus is the ratio of longitudinal (tensile) stress to longitudinal strain, within the elastic limit: Y = (F/A) / (delta L / L). SI unit: N/m^2 (pascal, Pa).

Y = (F L) / (A * delta L)

Hooke's law: within the elastic limit, the stress developed in a body is directly proportional to the strain produced, i.e. stress / strain = a constant (the modulus of elasticity). The elastic limit is the maximum stress up to which a body returns to its original shape and size on removing the deforming force; beyond it, permanent (plastic) deformation occurs.

stress = E * strain

Y = (F L) / (A * delta L). Here F = 10 N, L = 2 m, A = 1 mm^2 = 1 x 10^-6 m^2, delta L = 1 mm = 1 x 10^-3 m. Y = (10 * 2) / (1 x 10^-6 * 1 x 10^-3) = 20 / (1 x 10^-9) = 2 x 10^10 N/m^2.

Y = (F L) / (A * delta L)

Young's modulus (Y) measures resistance to change in length (longitudinal stress/strain). Bulk modulus (B) measures resistance to change in volume (volume stress/strain) under uniform pressure. Shear (rigidity) modulus (G) measures resistance to change in shape (shearing stress/strain).

For the streamline (steady, non-viscous, incompressible) flow of a fluid, the sum of the pressure energy, kinetic energy and potential energy per unit volume is constant: P + (1/2) rho v^2 + rho g h = constant. Thus where the speed is high, the pressure is low.

P + (1/2) rho v^2 + rho g h = constant

Pascal's law: a pressure applied to an enclosed incompressible fluid is transmitted equally and undiminished to every part of the fluid and to the walls of the container. Application: the hydraulic lift/brake, where a small force on a small piston produces a large force on a large piston because pressure is the same throughout.

F1 / A1 = F2 / A2

For the steady flow of an incompressible fluid, the product of area of cross-section and speed is constant along the tube: A1 v1 = A2 v2 (A v = constant). It is a statement of conservation of mass - where a pipe is narrow the fluid flows faster, and where it is wide it flows slower.

A1 v1 = A2 v2

Surface tension is the property by which the free surface of a liquid behaves like a stretched elastic membrane; it is the force per unit length acting along the surface (SI unit: N/m). Example: small water droplets and mercury beads become spherical (minimum surface area), and some insects can walk on water.

T = F / L

Viscosity is the internal friction between adjacent layers of a fluid moving with different velocities. Stokes' law: the viscous drag on a small sphere of radius r moving with speed v through a fluid of viscosity eta is F = 6 pi eta r v. Terminal velocity is the constant maximum velocity attained by a body falling through a fluid when the net force (weight minus buoyancy and viscous drag) becomes zero.

F = 6 pi eta r v

Total pressure P = P_atm + rho g h = 1.0 x 10^5 + (1000)(10)(10) = 1.0 x 10^5 + 1.0 x 10^5 = 2.0 x 10^5 Pa.

P = P_atm + rho g h

v_t = (2 r^2 (rho - sigma) g) / (9 eta), where r is the radius of the sphere, rho its density, sigma the density of the fluid, g the acceleration due to gravity and eta the coefficient of viscosity. It is obtained by balancing the weight against the buoyant force and the viscous (Stokes) drag.

v_t = 2 r^2 (rho - sigma) g / (9 eta)

The first law is the law of conservation of energy for a thermodynamic system: delta Q = delta U + delta W. Here delta Q is the heat supplied to the system (positive if absorbed), delta U is the increase in internal energy, and delta W is the work done by the system (positive if the gas expands).

delta Q = delta U + delta W

Isothermal: temperature stays constant, so delta U = 0; obeys P V = constant; occurs slowly with good thermal contact; heat can flow in or out. Adiabatic: no heat is exchanged (Q = 0); obeys P V^gamma = constant; occurs quickly or with perfect insulation; any work done changes the internal energy (delta U = -delta W).

Isothermal: PV = const ; Adiabatic: P V^gamma = const

A Carnot cycle has four reversible steps: (1) isothermal expansion at the source temperature T_h (heat Q_h absorbed); (2) adiabatic expansion (temperature falls from T_h to T_c); (3) isothermal compression at the sink temperature T_c (heat Q_c rejected); (4) adiabatic compression (temperature rises from T_c back to T_h). The efficiency is eta = 1 - Q_c/Q_h = 1 - T_c/T_h, where temperatures are in kelvin. Efficiency depends only on the two temperatures and is always less than 1.

eta = 1 - T_c/T_h

The zeroth law states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. It leads to the concept of temperature - a property that is the same for all bodies in thermal equilibrium.

At constant volume all the heat supplied goes to increase the internal energy (no work is done). At constant pressure the gas also expands and does external work, so extra heat is needed for the same temperature rise; hence Cp > Cv. The relation (Mayer's relation) is Cp - Cv = R, where R is the universal gas constant.

Cp - Cv = R

Kelvin-Planck statement: it is impossible to construct an engine that, working in a cycle, converts all the heat absorbed from a source completely into work with no other effect. (Equivalently, Clausius statement: heat cannot flow of its own accord from a colder body to a hotter body.)

For an isothermal process, W = n R T ln(V2/V1) = 1 * 8.31 * 300 * ln(2) = 8.31 * 300 * 0.693 = 1727 J (approximately 1.73 kJ).

W = n R T ln(V2/V1)

A refrigerator is a heat engine working in reverse: it uses external work W to extract heat Q_c from a cold body and reject a larger heat Q_h to the hot surroundings, so Q_h = Q_c + W. Its coefficient of performance is beta = Q_c / W = Q_c / (Q_h - Q_c). A good refrigerator has a high beta.

beta = Q_c / (Q_h - Q_c)

A quasi-static process is one carried out infinitely slowly so that the system stays in thermal and mechanical equilibrium with its surroundings at every stage. It is important because only for such (reversible) processes can the state variables (P, V, T) be defined throughout, and the work done can be represented as an area on a P-V diagram.

v_rms = sqrt(3 R T / M) = sqrt(3 * 8.31 * 300 / 0.032) = sqrt(7479 / 0.032) = sqrt(233719) = 483 m/s (approximately).

v_rms = sqrt(3 R T / M)

From kinetic theory, P = (1/3) rho <v^2>, where rho is the density and <v^2> is the mean-square speed of the molecules. Since the kinetic energy per unit volume is (1/2) rho <v^2>, we get P = (2/3) * (kinetic energy per unit volume). Thus pressure is two-thirds of the translational KE per unit volume.

P = (1/3) rho <v^2>

Law of equipartition: in thermal equilibrium, the total energy is shared equally among all degrees of freedom, each contributing (1/2) k T of energy per molecule. A monatomic molecule has 3 degrees of freedom (translational only); a diatomic molecule at ordinary temperatures has 5 (3 translational + 2 rotational).

Energy per degree of freedom = (1/2) k T

Average translational KE per molecule = (3/2) k T = (3/2)(1.38 x 10^-23)(300) = 1.5 * 1.38 x 10^-23 * 300 = 6.21 x 10^-21 J.

KE = (3/2) k T

The mean free path is the average distance a gas molecule travels between two successive collisions. It increases when the number density of molecules is low and when the molecular diameter is small; that is, it is inversely proportional to the number density and to the square of the molecular diameter.

lambda = 1 / (sqrt(2) pi d^2 n)

For a bob displaced by a small angle, the restoring force is F = -m g sin(theta). For small theta, sin(theta) ~ theta = x/L, so F = -(m g / L) x. This is of the form F = -k x with k = m g / L, hence the motion is SHM. Then omega^2 = k/m = g/L, so the time period T = 2 pi / omega = 2 pi sqrt(L/g).

T = 2 pi sqrt(L / g)

Simple harmonic motion is an oscillation in which the restoring force (or acceleration) is directly proportional to the displacement from the mean position and is always directed towards it. Defining equation: a = - omega^2 x, where omega is the angular frequency and x the displacement.

a = - omega^2 x

For SHM of amplitude A, kinetic energy KE = (1/2) m omega^2 (A^2 - x^2) and potential energy PE = (1/2) m omega^2 x^2. Adding, total energy E = KE + PE = (1/2) m omega^2 A^2. This is constant - it does not depend on x - and equals (1/2) m omega^2 A^2. Energy shuttles between KE (maximum at the mean position) and PE (maximum at the extremes).

E = (1/2) m omega^2 A^2

T = 2 pi sqrt(m / k). Since T is proportional to sqrt(m), quadrupling the mass (m -> 4m) makes sqrt(4m) = 2 sqrt(m), so the time period doubles.

T = 2 pi sqrt(m / k)

Angular frequency omega = 2 pi / T = 2 pi / 2 = pi rad/s. Maximum velocity v_max = A omega = 0.05 * pi = 0.05 * 3.14 = 0.157 m/s (about 15.7 cm/s).

v_max = A omega

Forced oscillations occur when a body is made to oscillate under an external periodic force with the frequency of that force. Resonance is the special case when the driving frequency equals the body's natural frequency; the amplitude of oscillation then becomes maximum. Example: a child's swing pushed at its natural frequency.

Velocity v = dx/dt = A omega cos(omega t), with magnitude v = omega sqrt(A^2 - x^2); it is maximum (A omega) at the mean position (x = 0) and zero at the extremes. Acceleration a = dv/dt = - A omega^2 sin(omega t) = - omega^2 x; its magnitude is maximum (A omega^2) at the extreme positions (x = +/- A) and zero at the mean position.

v = omega sqrt(A^2 - x^2) ; a = - omega^2 x

In a transverse wave the particles of the medium vibrate perpendicular to the direction of wave propagation (example: a wave on a stretched string, light waves). In a longitudinal wave the particles vibrate parallel to the direction of propagation, forming compressions and rarefactions (example: sound waves in air).

v = f * lambda

In one time period T, a wave advances by one wavelength lambda. So wave speed v = distance / time = lambda / T. Since frequency f = 1/T, we get v = f lambda. This holds for all progressive waves.

v = f lambda

A progressive wave transfers energy continuously in one direction, and every particle has the same amplitude but oscillates with a phase lag. A stationary wave is formed by two identical waves travelling in opposite directions; it does not transfer net energy, has fixed nodes (zero amplitude) and antinodes (maximum amplitude), and the amplitude varies from point to point.

For a string of length L, linear density mu and tension T, the fundamental frequency (first harmonic) is f1 = (1/2L) sqrt(T/mu), because the fundamental mode fits half a wavelength in the length L. Harmonics (overtones) are the higher allowed frequencies, which are integer multiples of the fundamental: fn = n f1 (n = 1, 2, 3, ...).

fn = (n / 2L) sqrt(T / mu)

Beats are the periodic rise and fall in the loudness of sound produced when two waves of slightly different frequencies superpose. The beat frequency equals the difference of the two frequencies: 260 - 256 = 4 beats per second.

f_beat = |f1 - f2|

For a source approaching a stationary observer, f' = f * v / (v - v_s) = 340 * 340 / (340 - 34) = 340 * 340 / 306 = 377.8 Hz (about 378 Hz). The pitch appears higher.

f' = f v / (v - v_s)

Newton assumed sound propagation in a gas is isothermal, giving v = sqrt(P / rho), where P is pressure and rho density; this gave a value about 15% lower than the measured value. Laplace corrected this by treating the process as adiabatic (compressions and rarefactions are too rapid for heat exchange), giving v = sqrt(gamma P / rho), where gamma is the ratio of specific heats; this matches experiment.

v = sqrt(gamma P / rho)

In an open pipe (open at both ends) both even and odd harmonics are present; the fundamental is f1 = v/(2L) and the harmonics are f1, 2 f1, 3 f1, ... In a closed pipe (closed at one end) only odd harmonics are present; the fundamental is f1 = v/(4L) and the harmonics are f1, 3 f1, 5 f1, ... So an open pipe of the same length gives a higher fundamental and a richer set of harmonics.

Open: fn = n v/2L ; Closed: fn = (2n-1) v/4L

The coefficient of linear expansion (alpha) is the fractional increase in length per degree rise in temperature: alpha = (delta L) / (L * delta T). The increase in length is delta L = L alpha (delta T), and the new length is L' = L (1 + alpha delta T). SI unit of alpha: per kelvin (K^-1).

delta L = L alpha delta T

delta L = L alpha (delta T) = 1 * (1.2 x 10^-5) * 100 = 1.2 x 10^-3 m = 1.2 mm.

delta L = L alpha delta T

Specific heat capacity is the heat required to raise the temperature of 1 kg of a substance by 1 K (Q = m c delta T; unit J/kg/K) - the temperature changes. Latent heat is the heat required to change the state of 1 kg of a substance at constant temperature (Q = m L; unit J/kg) - the temperature stays constant during the phase change.

Q = m c delta T ; Q = m L

Q = m L = 0.1 kg * 3.36 x 10^5 J/kg = 3.36 x 10^4 J = 33600 J. The temperature stays at 0 degrees C throughout the melting.

Q = m L

Conduction: heat flows through a material from the hotter to the cooler region by molecular vibrations, without bulk movement of the material (e.g. a metal spoon heating up). Convection: heat is carried by the actual movement of the heated fluid (e.g. boiling water, sea breeze). Radiation: heat travels as electromagnetic waves and needs no medium (e.g. heat from the Sun).

Newton's law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings, provided this difference is small: (- dQ/dt) is proportional to (T - T_surroundings).

dQ/dt = -k (T - T0)

Stefan's (Stefan-Boltzmann) law: the total energy radiated per unit area per unit time by a black body is proportional to the fourth power of its absolute temperature, E = sigma T^4. Wien's displacement law: the wavelength at which the emission is maximum is inversely proportional to the absolute temperature, lambda_max * T = constant (= 2.9 x 10^-3 m K).

E = sigma T^4 ; lambda_max T = b