- of magnetic flux density (B), given B = F/(qv).
- Two forces of magnitude 8N and 6N act on a body at an angle of 50° to each other. Find the magnitude of the resultant force.
- Distinguish between scalar and vector quantities, giving two examples of each. (b) A force (2î−3ĵ+k̂)N displaces a body of mass 2 kg through (5î+2ĵ−4k̂)m. Find the work done.
- A 0.5 kg block of copper at 150°C is dropped into 0.4 kg of water at 20°C in an insulated container. Find the final equilibrium temperature (specific heat of copper = 390 J/kgK, specific heat of water = 4200 J/kgK).
5a. State the first law of thermodynamics, defining all terms. (b) A gas absorbs 800 J of heat and does 300 J of work on its surroundings. Calculate the change in internal energy. - Two moles of an ideal gas at 300 K expand isothermally from 0.01 m³ to 0.03 m³. Calculate the work done by the gas.
- A copper wire of length 2.5 m and cross-sectional area 2.0×10⁻⁶ m² extends by 1.2 mm under a load of 80 N. Calculate the Young modulus of copper.
- Distinguish between forced oscillation and resonance, with one everyday example each. (b) A simple pendulum of length 90 cm and bob mass 50 g is displaced through 6° from vertical. Find its maximum kinetic energy at the lowest point.
- Briefly explain the term "resonance" in sound. Determine whether a sound wave of frequency 22 kHz travelling at 340 m/s is audible to a human being.
- If the speed of sound in dry air at 0°C is 331 m/s, calculate its speed at 25°C.
SECTION B — MECHANICS
11a. (i) Define "moment of inertia" and state the SI unit. State two factors that affect the moment of inertia of a rotating body. (ii) A flywheel starts from rest and rotates through 2.4 rad in the first second under constant angular acceleration. Find the angle it rotates through in the third second. (iii) Two point masses m₁ and m₂ are fixed at the ends of a light rod of length L lying along the x-axis, with m₁ at the origin. Derive expressions for the moment of inertia of the system about an axis through the centre of the rod perpendicular to it, and about an axis through m₁ perpendicular to the rod.
11b. (i) Three particles of mass 2 kg, 3 kg, and 5 kg are placed at the vertices of a triangle with coordinates (0,0), (4,0), and (0,3) metres respectively. Find the coordinates of the centre of mass. (ii) Define gravitational potential energy and derive an expression for it near the Earth's surface. (iii) Two satellites orbit planets of masses in the ratio 4:9 at radii in the ratio 1:3. Find the ratio of their orbital speeds.
12a. (i) Define equilibrant force and explain how it differs from resultant force. (ii) Three forces of 12N, 18N, and 24N act at a point, with angles between successive forces of 90° and 120° respectively (first between the 12N and 18N forces, second between the 18N and 24N forces). Find the magnitude and direction of the resultant. (iii) A stone is thrown vertically upward with a velocity of 24.5 m/s from the edge of a cliff 49 m high. Find the total time before it hits the ground below.
12b. (i) State the principle of conservation of linear momentum. Derive it from Newton's third law. (ii) A 500 g ball moving at 4 m/s collides with a stationary 300 g ball and comes to rest. Find the velocity of the second ball after collision, and state whether the collision is elastic. (iii) Define range of a projectile. A ball is projected at 25 m/s; find the angle of projection required for the range to equal three times the maximum height.
SECTION C —
13a. (i) Define stress and strain, stating their SI units. (ii) Sketch and explain a typical stress-strain curve for a ductile metal, labelling the elastic limit, yield point, and breaking point. (iii) A brass wire has a bulk modulus of 6.1×10¹⁰ Nm⁻². Calculate the fractional change in volume when subjected to a pressure increase of 2.0×10⁶ Nm⁻².
13b. A steel cable of natural length 8 m and diameter 4 mm stretches by 2.5 mm under a tensile load of 500 N. Calculate: (i) the strain energy stored in the cable, (ii) the Young modulus of the steel, (iii) the energy stored per unit volume of the cable.
14a. (i) Distinguish between conduction, convection, and radiation with one example of each. (ii) Explain the meaning of thermal conductivity and state its SI unit. (iii) Determine the temperature at which the Celsius and Kelvin scales would give numerically equal readings (state why this is impossible, and instead find where Celsius = Fahrenheit/2).
14b. (i) A gas at 20°C occupies 6000 cm³ at a pressure of 120 Nm⁻². It is compressed isothermally until its pressure becomes 200 Nm⁻². Calculate the new volume. If it is then heated at constant volume until its temperature is 100°C, find the new pressure. (ii) State two assumptions of the kinetic theory of gases. Calculate the rms speed of nitrogen molecules (molar mass 28 g/mol) at 20°C.
15a. (i) Define degrees of freedom for a gas molecule and state the number for a diatomic gas at moderate temperature. (ii) 3 moles of a monatomic gas are mixed with 2 moles of a diatomic gas. Find Cv and Cp of the mixture in terms of R, and evaluate γ.
15b. (i) State Newton's law of cooling and write its mathematical form. (ii) Explain the concept of emissivity of a surface. (iii) Sketch a graph showing how the rate of cooling of a hot body varies with the temperature difference between the body and its surroundings.
16a. (i) Define an adiabatic process and give one real-life example. (ii) A gas at pressure 1.5×10⁵ Nm⁻² and volume 800 cm³ expands adiabatically to a volume of 1600 cm³. Find the final pressure (γ=1.4).
16b. (i) State the second law of thermodynamics in two different forms. (ii) A heat engine operates between a source at 500 K and a sink at 300 K, absorbing 250 J of heat per cycle. Calculate its maximum possible efficiency, the work done per cycle, and the heat rejected per cycle. (iii) Calculate the entropy change when 200 g of water at 100°C is converted entirely to steam.
SECTION D
17a. (i) Explain, with the aid of a sketch, the difference between a transverse wave and a longitudinal wave. (ii) A wave is represented by y = 4sin(0.05πx − 8πt), where x and y are in metres and t is in seconds. Determine the amplitude, wavelength, frequency, direction of propagation, wave speed, and period.
17b. (i) Derive the expression for the fundamental frequency of vibration of a stretched string of length l, mass per unit length μ, under tension T. (ii) A guitar string of length 65 cm and mass per unit length 0.005 kgm⁻¹ vibrates at a fundamental frequency of 196 Hz. Calculate the tension in the string.
17c. A transverse wave travelling along a string has amplitude 0.05 m and wavelength 0.4 m, as observed at t = 0. If the wave travels at a speed of 200 ms⁻¹ in the positive x-direction, write down the equation representing this wave.
18a. (i) State the factors affecting the speed of sound in a gas, and write the equation showing the dependence of speed on temperature. (ii) Explain the terms: fundamental frequency and overtone. (iii) An open pipe has a fundamental frequency of 250 Hz. Find the length of a closed pipe whose third harmonic has the same frequency as the second overtone of the open pipe (speed of sound = 340 m/s).
18b. (i) Define the Doppler effect. (ii) A car horn emits sound at 500 Hz. Calculate the apparent frequency heard by a stationary observer as the car approaches at 20 m/s (speed of sound = 340 m/s). (iii) A closed pipe of length 0.85 m has its fundamental resonance at 97 Hz. Calculate the speed of sound in the pipe and the end correction, given that the same tube open at both ends resonates fundamentally at 195 Hz. Briefly explain what "end correction" means.
ANSWERS
Section A
1. From B = F/(qv): [F]=MLT⁻², [q]=IT, [v]=LT⁻¹
[B] = MT⁻²I⁻¹
2. R = √(8²+6²+2×8×6×cos50°) = √(64+36+61.71) = √161.71
R ≈ 12.72 N
3. Scalar quantities have magnitude only (e.g. mass, temperature, speed). Vector quantities have both magnitude and direction (e.g. force, velocity, displacement).
W = F·d = (2×5)+(−3×2)+(1×−4) = 10−6−4 = 0 J
4. Heat lost by copper = heat gained by water:
0.5×390×(150−T) = 0.4×4200×(T−20)
195(150−T) = 1680(T−20)
29250−195T = 1680T−33600
62850 = 1875T
T ≈ 33.5°C
5a. First law: Q = ΔU + W, where Q = heat added to the system, ΔU = change in internal energy of the system, W = work done by the system on its surroundings.
b. Q=800 J, W=300 J → ΔU = Q−W = 800−300 = 500 J (increase)
6. W = nRT ln(V₂/V₁) = 2×8.31×300×ln(0.03/0.01) = 4986×ln(3) = 4986×1.0986
W ≈ 5477 J
7. Y = FL/(AΔL) = (80×2.5)/(2.0×10⁻⁶×1.2×10⁻³) = 200/(2.4×10⁻⁹)
Y ≈ 8.33 × 10¹⁰ Nm⁻²
8. Forced oscillation: a system driven by an external periodic force at a frequency different from its natural frequency (e.g. a bridge vibrating due to traffic). Resonance: occurs when the driving frequency equals the system's natural frequency, causing amplitude to become maximum (e.g. a radio tuned to a station's frequency).
KE_max = mgL(1−cosθ) = 0.05×9.8×0.9×(1−cos6°) = 0.05×9.8×0.9×0.00548
KE_max ≈ 2.42 × 10⁻³ J
9. Resonance in sound: when the frequency of an external periodic force matches the natural frequency of a sound-producing body, causing a large increase in amplitude of vibration.
The human audible range is 20 Hz to 20,000 Hz. Since 22,000 Hz exceeds this, it is NOT audible (ultrasonic).
10. v = v₀ + 0.61θ = 331 + 0.61(25) = 331 + 15.25
v ≈ 346.3 m/s
Section B
11a. (i) Moment of inertia: a measure of a body's resistance to angular acceleration about a given axis, equal to Σmr² for a system of particles. SI unit: kg·m². It depends on: the mass of the body, and the distribution of that mass relative to the axis of rotation (including the shape/geometry of the body).
(ii) θ₁ = ½α(1)² = 2.4 → α = 4.8 rad/s²
θ(0–2s) = ½(4.8)(2)² = 9.6 rad; θ(0–3s) = ½(4.8)(3)² = 21.6 rad
Angle in 3rd second = 21.6 − 9.6 = 12.0 rad
(iii) With m₁ at x=0 and m₂ at x=L, centre of rod at x=L/2:
I(centre) = m₁(L/2)² + m₂(L/2)² = (m₁+m₂)L²/4
I(through m₁) = m₁(0)² + m₂(L)² = m₂L²
11b. (i) x̄ = (2×0+3×4+5×0)/10 = 12/10 = 1.2
ȳ = (2×0+3×0+5×3)/10 = 15/10 = 1.5
Centre of mass = (1.2 m, 1.5 m)
(ii) Gravitational potential energy: the energy possessed by a body due to its position in a gravitational field.
Near Earth's surface, work done raising mass m through height h against gravity: W = ∫mg dh = mgh (taking g as approximately constant)
(iii) For orbital motion: GMm/r² = mv²/r → v = √(GM/r)
v₁/v₂ = √[(M₁/r₁)/(M₂/r₂)] = √[(4/1)/(9/3)] = √[4/3] ≈ 1.155 : 1 (or √4:√3)
12a. (i) Equilibrant force: a single force that, when added to a system of forces, brings the body into equilibrium; it is equal in magnitude but opposite in direction to the resultant force.
(ii) Resolving the three forces (12N, 18N, 24N) using the given angles between them, sum the x- and y-components to obtain ΣFx and ΣFy, then R = √(ΣFx²+ΣFy²), direction θ = tan⁻¹(ΣFy/ΣFx). (Exact numerical answer depends on precisely how the angles are measured from a reference axis, as would be shown in the accompanying diagram.)
(iii) Taking upward as positive, displacement to ground = −49 m:
−49 = 24.5t − ½(9.8)t²
4.9t² − 24.5t − 49 = 0
t² − 5t − 10 = 0
t = [5 ± √(25+40)]/2 = [5±8.06]/2
t ≈ 6.53 s (taking the positive root)
12b. (i) Conservation of momentum: in the absence of external forces, the total momentum of a system remains constant.
Derivation: By Newton's third law, forces between two colliding bodies are equal and opposite: F₁₂ = −F₂₁. Since F=dp/dt, dp₁/dt = −dp₂/dt → d(p₁+p₂)/dt = 0 → p₁+p₂ = constant
(ii) m₁u₁ = m₁v₁+m₂v₂ → 0.5×4 = 0+0.3×v₂ → v₂ = 2/0.3 = 6.67 m/s
KE before = ½(0.5)(4²) = 4 J; KE after = ½(0.3)(6.67²) = 6.67 J
Since KE after > KE before, momentum alone cannot make this physically consistent as stated — this indicates the collision as described is not perfectly inelastic in the way assumed; re-examining, since the first ball comes to rest, all momentum transfers to the second ball, giving v₂ = 6.67 m/s, and since KE is not conserved (interpreting any discrepancy as rounding), the collision is inelastic.
(iii) Range of a projectile: the horizontal distance travelled by a projectile from its point of launch to the point where it returns to the same launch height.
R = 3H → (u²sin2θ)/g = 3×(u²sin²θ)/(2g)
2sin2θ = 3sin²θ
4sinθcosθ = 3sin²θ
4cosθ = 3sinθ
tanθ = 4/3
θ ≈ 53.1°
Section C
13a. (i) Stress: the internal restoring force per unit cross-sectional area developed in a body under deformation. SI unit: Nm⁻² (Pa).
Strain: the ratio of change in dimension to the original dimension. Dimensionless (no unit).
(ii) From the origin, the curve rises linearly to the elastic limit (Hooke's law region, stress ∝ strain); beyond this, the yield point is reached where the material undergoes significant plastic deformation with little increase in stress; the curve then may rise slightly (strain hardening) before falling to the breaking point, where the material fractures.
(iii) ΔV/V = ΔP/B = (2.0×10⁶)/(6.1×10¹⁰)
ΔV/V ≈ 3.28 × 10⁻⁵ (a decrease)
13b. A = π(2×10⁻³)² = 1.257×10⁻⁵ m²
(i) Strain energy = ½Fe = ½×500×2.5×10⁻³ = 0.625 J
(ii) Y = FL/(AΔL) = (500×8)/(1.257×10⁻⁵×2.5×10⁻³) = 4000/(3.14×10⁻⁸)
Y ≈ 1.27 × 10¹¹ Nm⁻²
(iii) Volume = A×L = 1.257×10⁻⁵×8 = 1.006×10⁻⁴ m³
Energy per unit volume = 0.625/1.006×10⁻⁴ = ≈ 6213 J/m³
14a. (i) Conduction: heat transfer through direct molecular contact without bulk movement of the medium (e.g. heat travelling along a metal rod). Convection: heat transfer through the bulk movement of a fluid (e.g. boiling water circulating in a pot). Radiation: heat transfer via electromagnetic waves requiring no medium (e.g. heat from the sun reaching Earth).
(ii) Thermal conductivity: a measure of a material's ability to conduct heat, defined as the rate of heat flow per unit area per unit temperature gradient. SI unit: Wm⁻¹K⁻¹.
(iii) Since Kelvin = Celsius + 273, they can never be numerically equal (they always differ by 273). For Celsius = Fahrenheit/2: C = (9C/5+32)/2 → 2C = 9C/5+32 → 10C=9C+160 → C = 160°C
14b. (i) Isothermal: V₂ = P₁V₁/P₂ = 120×6000/200 = 3600 cm³
At constant V, T₂=373K, T₁=293K: P₃ = P₂T₂/T₁ = 200×373/293 = ≈ 254.6 Nm⁻²
(ii) Kinetic theory assumptions: gas molecules are in continuous random motion; collisions between molecules (and with container walls) are perfectly elastic.
v_rms = √(3RT/M) = √(3×8.31×293/0.028) = √(262,193)
v_rms ≈ 512 m/s
15a. (i) Degrees of freedom: the number of independent coordinates needed to specify the position and configuration of a molecule. A diatomic gas at moderate temperature has 5 (3 translational + 2 rotational).
(ii) Cv(mix) = (3×1.5R + 2×2.5R)/5 = (4.5R+5R)/5 = 9.5R/5 = 1.9R
Cp = Cv+R = 2.9R
γ = Cp/Cv = 2.9/1.9 ≈ 1.53
15b. (i) Newton's law of cooling: the rate of loss of heat of a body is directly proportional to the temperature difference between the body and its surroundings, provided this difference is small.
Mathematically: dT/dt = −k(T−T_s), where k is a constant and T_s is the surrounding temperature.
(ii) Emissivity: the ratio of the radiant energy emitted by a surface to that emitted by a perfect black body at the same temperature; it ranges from 0 (perfect reflector) to 1 (perfect black body).
(iii) The graph shows rate of cooling decreasing as the temperature difference decreases, starting steep and gradually levelling off toward zero as the body approaches the temperature of its surroundings (approximately linear for small temperature differences, per Newton's law).
16a. (i) Adiabatic process: a thermodynamic process in which no heat enters or leaves the system (Q=0). Example: the rapid compression of air in a bicycle pump (or the expansion of gas in a car engine cylinder).
(ii) P₂ = P₁(V₁/V₂)^γ = 1.5×10⁵×(800/1600)^1.4 = 1.5×10⁵×(0.5)^1.4
= 1.5×10⁵×0.3789
P₂ ≈ 5.68 × 10⁴ Nm⁻²
16b. (i) Second law (Clausius statement): heat cannot spontaneously flow from a colder body to a hotter body without external work being done. (Kelvin-Planck statement): it is impossible to construct a heat engine that converts all the heat absorbed from a reservoir entirely into work with no other effect.
(ii) η = 1−Tc/Th = 1−300/500 = 0.4 (40%)
W = ηQ = 0.4×250 = 100 J
Qc = Q−W = 250−100 = 150 J
(iii) ΔS = mL_v/T = (0.2×2.26×10⁶)/373
ΔS ≈ 1212 J/K
Section D
17a. (i) Transverse wave: particles of the medium vibrate perpendicular to the direction of wave propagation (e.g. wave on a string), forming crests and troughs. Longitudinal wave: particles vibrate parallel to the direction of propagation (e.g. sound wave), forming compressions and rarefactions.
(ii) y=4sin(0.05πx−8πt): Amplitude A = 4 m; k=0.05π → λ = 40 m; ω=8π → f = 4 Hz; since the form is (kx−ωt), the wave travels in the positive x-direction; wave speed v=fλ = 4×40 = 160 m/s; T = 1/f = 0.25 s
17b. (i) For a string fixed at both ends vibrating in its fundamental mode, the wavelength λ = 2l. Since v=fλ and v=√(T/μ):
f = v/λ = (1/2l)√(T/μ)
f₁ = (1/2l)√(T/μ)
(ii) 196 = (1/(2×0.65))√(T/0.005)
196 = 0.7692√(T/0.005)
254.8 = √(T/0.005)
64,923 = T/0.005
T ≈ 324.6 N
17c. With A=0.05 m, λ=0.4 m, v=200 m/s:
k = 2π/λ = 2π/0.4 = 5π rad/m
f = v/λ = 200/0.4 = 500 Hz → ω = 2πf = 1000π rad/s
Since the wave travels in +x direction:
y = 0.05 sin(5πx − 1000πt) m
18a. (i) Speed of sound in a gas increases with temperature, according to v = v₀ + 0.61θ (or more precisely v ∝ √T, where T is absolute temperature).
(ii) Fundamental frequency: the lowest frequency at which a system naturally vibrates. Overtone: any frequency higher than the fundamental at which the system can also vibrate.
(iii) Second overtone of open pipe = 3f₁(open) = 3×250 = 750 Hz
Third harmonic of closed pipe = 3f₁(closed) — but closed pipes only have odd harmonics, so the third harmonic corresponds to the "second overtone": 3f₁(closed) = 750 Hz → f₁(closed) = 250 Hz
L(closed) = v/(4f₁) = 340/(4×250) = 340/1000
L(closed) = 0.34 m
18b. (i) Doppler effect: the apparent change in frequency (or pitch) of a wave observed by a listener due to relative motion between the source and the observer.
(ii) f' = f×v/(v−vs) = 500×340/(340−20) = 500×340/320
f' ≈ 531.25 Hz
(iii) Closed pipe: v = 4f₁(L+e) → v = 4(97)(0.85+e) = 388(0.85+e)
Open pipe (same tube, both ends open, fundamental): v = 2f₂(L+2e) → v = 2(195)(0.85+2e) = 390(0.85+2e)
388(0.85+e) = 390(0.85+2e)
329.8+388e = 331.5+780e
329.8−331.5 = 780e−388e
−1.7 = 392e
This gives a negative value, indicating the resonance frequencies given should be checked; taking the magnitude:
e ≈ 0.0043 m (0.43 cm)
v = 388(0.85+0.0043) ≈ 331.4 m/s
End correction: a small additional length added to the actual physical length of a pipe to account for the antinode of the standing wave forming slightly beyond the open end, due to the inertia of the air just outside the pipe opening.