**1a.** Define fundamental quantities. Give three examples of fundamental quantities with their units.
**b.** Determine the dimensions of the constants 'a' and 'b' in Vander Waal's equation using the theory of dimension: $\left(p + \dfrac{a}{V^2}\right)(V - b) = RT$
**2.** An object of mass 6 kg is suspended from a spring balance which is attached to a ceiling of a lift. If the spring balance is calibrated in SI unit of force, what will be the reading on the balance when (a) the lift is stationary (b) the lift is moving with an acceleration of 0.3 ms⁻² upwards. (c) if incidentally the lift cable breaks and the lift falls freely under gravity.
---
## 2023 LIMBE PHYSICS I
**3.** An aeroplane flying horizontally with speed of 200 ms⁻¹ releases a bomb at a height of 500 m from the ground — when and where will the bomb strike the ground?
**4.** A human heart pumps 8 × 10⁻⁵ m³/s of blood through the arteries under a pressure 12.0 cm Hg. Calculate the power of the heart in watts. (Density of mercury = 1.36 × 10³ kgm⁻³).
**5.** A fan turns at a rate of 15 Hz. Calculate the (i) angular speed of any one of the fan blades (ii) the tangential speed of the tip of the blade if the distance from the center of the tip is 0.2 m.
**6.** An ideal gas at STP undergoes a reversible thermal expansion. If the work done by the gas in the process is 2000 J, what is the change in entropy?
**7.** An engine of 20 kW is used to pump water from a river which is 100 m deep. Calculate the quantity of water which it can pump in 20 minutes.
**8.** Two turning forks A and B give 8 beats per second. From turning fork "A" resonant with a closed column of air 16 cm long and "B" resounds with an open column of 33 cm long. Evaluate their frequencies.
**9.** One end of a 1 m copper rod and 0.02 m in radius is immersed in boiling water at 100 °C, while the other end is immersed in ice at 0 °C. Evaluate the rate at which the ice melts. Given that, thermal conductivity of copper is 380 Wm⁻¹K⁻¹.
**10.** A man standing between two parallel hills fires a gun. He hears the first echo after 2.4 s and the second after 4 s. Find the distance between the two hills.
---
## SECTION B — MECHANICS (20 Marks)
### Answer one (1) question only from this section
**11a.** (i) What are vector and scalar quantities? To which class do: momentum, current, electrical energy, acceleration due to gravity and mass belong? (ii) Explain the term 'relative velocity'. (iii) At the instant the traffic lights turn green, an auto mobile that has been waiting at the intersection of roads accelerates with 3 ms⁻². At the same instant, a truck travelling with a constant velocity of 12 ms⁻¹ overtakes and passes the automobile. How far beyond its starting point will the automobile overtake the truck and how fast will it be travelling at that time?
**b.** (i) Define centripetal force and derive its expression. (ii) A bend in a level road has a radius of 100 m. Find the maximum speed which a car turning this bend may have without skidding if the coefficient of friction between the road and the car is 0.25.
**12a.** (i) Define amplitude and angular frequency of a simple harmonic motion and write down an expression relating them, explaining all the symbols used. (ii) A mass of 0.12 kg is suspended from a spring of force constant 25 Nm⁻¹. Calculate the extension of the spring assuming that Hooke's law is obeyed. If the mass is now pulled a further distance of 10 mm and released. Show that the motion is simple harmonic and calculate the period. Also evaluate the maximum kinetic energy of the mass during its motion.
---
## Page 3 — 2023 LIMBE PHYSICS I contd.
*(Section B continued / Section C begins)*
**b.** (i) Define moment of inertia of a body and state its units. Derive the expression for the rotational kinetic energy of a body in terms of moment of inertia. (ii) An electric motor exerts a constant torque of 15 Nm on a grindstone mounted on its shaft. The moment of inertia of grindstone is 2.0 kgm². If the system starts from rest, find the work done by the motor in 10 seconds and the kinetic energy at the end of this time. Evaluate the average power delivered by the motor.
---
## SECTION C: HEAT AND PROPERTIES OF MATTER (40 MARKS)
### Answer any two (2) questions only from this section
**13a.** (i) Derive an expression for the terminal speed of a sphere falling in a viscous fluid in terms of the sphere radius 'r', density 'ρ' and fluid viscosity 'η'. Assuming that, the flow is laminar and Stoke's law holds. (ii) The $k = 3.5 \times 10^{-4}$ Nm⁻² s. In a Millikan's oil drop experiment, the terminal speed of an oil drop of density 920 kgm⁻³ was measured to be $1.57 \times 10^{-5}$ ms⁻¹, when there was no electric field. The same oil drop was held stationary by an electric field of intensity $3.2 \times 10^{5}$ Fm⁻¹. Calculate (A) the charge on the drop (B) the radius of the drop (C) the weight of the drop.
**b.** (i) Explain the term 'Bernoulli's principle' with its significance and give two simple applications of it.
(ii) Water flows steadily along a horizontal pipe at a volume rate of $10^{-2}$ m³s⁻¹, if the area of cross-section of the pipe is 50 cm². (A) Calculate the flow velocity of the water. (B) Find the total pressure in the horizontal pipe if the static pressure in the horizontal pipe is $2.0 \times 10^{5}$ Nm⁻². State any assumption you have made. (C) What is the new flow velocity if the total pressure is $3.6 \times 10^{5}$ Nm⁻²?
**14a.** (i) What do you understand by Newton's law of cooling? (ii) 500 g of water and equal volume of alcohol of mass 250 g are placed successively in the same calorimeter and cooled from 70 °C to 35 °C in 390 seconds and 201 seconds respectively. Find the specific heat capacity of the alcohol. Water equivalent of the calorimeter is 10.
**b.** (i) Explain the term 'specific heat capacity' and 'specific latent heat' of a substance. (ii) The densities of two substances 'A' and 'B' are in the ratio 5:6 and their specific heat capacities are in the ratio 4:5. Compare their thermal capacities per unit volume.
**15a.** (i) State briefly five assumptions of kinetic theory of gas and write down the expression for the pressure of the gas. (ii) Explain why the equation of state for an ideal gas differs from that of real gas.
(iii) Helium gas occupies a volume of 0.04 m³ at a pressure of $2.0 \times 10^{5}$ Nm⁻² and at 300 K. Calculate the mass of Helium, r.m.s. speed of its molecules, r.m.s. speed at 432 K when gas is heated at constant pressure to this temperature, r.m.s speed of hydrogen molecules at 432 K if the molecular mass of helium and hydrogen are 4 g and 2 g respectively.
---
## Page 4 — 2023 LIMBE PHYSICS I contd.
**15b.** (i) Show that, the molar heat capacities of a gas are related by $C_p - C_v = R$. When R is the gas constant. (ii) A refrigerator operates with its cold reservoir at 5 °C and rejects heat to its surroundings at 27 °C. If 200 g of water are turned to ice at 0 °C, evaluate the total work done by the motor, assuming there is maximum efficiency.
**16a.** (i) State the second law of thermodynamics and explain its significance. (ii) A Carnot engine takes 5000 J of heat from a reservoir at 600 K, does some work and discards some heat to a reservoir at 400 K. (A) How much does it do? (B) How much heat is discarded? (C) What is the efficiency of the engine?
**b.** (i) Explain the term 'entropy' and write down an expression for it. For a reversible isothermal expansion, evaluate an expression for entropy. (ii) 2 kg of ice is melted and converted to water at 0 °C. Evaluate the change in entropy.
---
## SECTION D — VIBRATION AND WAVES (20 marks)
### Answer one (1) question only from this section
**17a.** (i) State how the velocity of sound in air depends on the temperature, pressure and humidity. Prove and defend your statements. (ii) The velocity of sound in air at 14 °C is 340 ms⁻¹. What will be the velocity when the pressure of the gas is tripled and its temperature raised to 200 °C?
**b.** (i) Explain the terms 'fundamental note, overtone and harmonic frequencies'. (ii) A wire of length 100 cm when stretched by a load of 10 kg, vibrates with frequency of 300 vibrations per second. Evaluate the mass of the wire. (iii) State the conditions for two sources 'A' and 'B' to produce stationary wave.
**18a.** (i) Briefly explain the concept of Doppler's effect. (ii) A whistle is giving out 600 Hz moves away from a stationary observer in a direction towards and perpendicular to a flat wall with a velocity of 2 ms⁻¹. Evaluate the apparent shift in frequency.
**b.** (i) State the four characteristics of a progressive wave. (ii) A man standing in front of a mountain at a certain distance, beats a drum at regular intervals. The drumming rate is gradually increased and he finds that the echo is not heard distinctly when the rate becomes 40 per minute. He then moves nearer to the mountain by 90 m and finds that, the echo is again not heard when the drumming rate becomes 40 per minute. Evaluate the distance between the mountain and the initial position of man and the velocity of sound.
# 2023 LIMBE PHYSICS I — Complete Solutions
---
## SECTION A
---
### Question 1a
**Fundamental quantities** are physical quantities that cannot be derived from other quantities.
| Quantity | Unit |
|----------|------|
| Length | metre (m) |
| Mass | kilogram (kg) |
| Time | second (s) |
---
### Question 1b
$$\left(p + \frac{a}{V^2}\right)(V - b) = RT$$
**For 'a':** Since $\frac{a}{V^2}$ must have dimensions of pressure:
$$[a] = [p][V^2] = ML^{-1}T^{-2} \times L^6 = ML^5T^{-2}$$
**For 'b':** Since b is subtracted from V, b has dimensions of volume:
$$[b] = [V] = L^3$$
---
### Question 2
m = 6 kg, g = 10 ms⁻²
**(a) Stationary:**
$$W = mg = 6 \times 10 = 60 \text{ N}$$
**(b) Accelerating upward, a = 0.3 ms⁻²:**
$$W = m(g + a) = 6(10 + 0.3) = 61.8 \text{ N}$$
**(c) Free fall (a = g):**
$$W = m(g - g) = 0 \text{ N}$$
---
### Question 3
u = 200 ms⁻¹, h = 500 m, g = 10 ms⁻²
**Time to hit ground:**
$$h = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 500}{10}} = 10 \text{ s}$$
**Horizontal distance:**
$$x = ut = 200 \times 10 = 2000 \text{ m from release point}$$
---
### Question 4
Q = 8×10⁻⁵ m³/s, P = 12 cmHg, ρ_Hg = 1.36×10³ kgm⁻³
$$P = \rho g h = 1.36 \times 10^3 \times 10 \times 0.12 = 1632 \text{ Pa}$$
$$\text{Power} = PQ = 1632 \times 8 \times 10^{-5} = 0.131 \text{ W}$$
---
### Question 5
f = 15 Hz, r = 0.2 m
**(i) Angular speed:**
$$\omega = 2\pi f = 2\pi \times 15 = 94.25 \text{ rads}^{-1}$$
**(ii) Tangential speed:**
$$v = \omega r = 94.25 \times 0.2 = 18.85 \text{ ms}^{-1}$$
---
### Question 6
W = 2000 J, T = 273 K (STP)
$$\Delta S = \frac{Q}{T} = \frac{W}{T} = \frac{2000}{273} = 7.33 \text{ JK}^{-1}$$
---
### Question 7
P = 20 kW = 20000 W, h = 100 m, t = 20 min = 1200 s, g = 10 ms⁻²
$$W = Pt = 20000 \times 1200 = 2.4 \times 10^7 \text{ J}$$
$$W = mgh \Rightarrow m = \frac{W}{gh} = \frac{2.4 \times 10^7}{10 \times 100} = 24000 \text{ kg}$$
$$\text{Volume} = \frac{m}{\rho} = \frac{24000}{1000} = 24 \text{ m}^3$$
---
### Question 8
Beats = 8/s, closed column for A: L_A = 16 cm = 0.16 m, open column for B: L_B = 33 cm = 0.33 m
For closed pipe: $f_A = \frac{v}{4L_A}$
For open pipe: $f_B = \frac{v}{2L_B}$
$$\frac{f_A}{f_B} = \frac{v/4L_A}{v/2L_B} = \frac{2L_B}{4L_A} = \frac{2 \times 0.33}{4 \times 0.16} = \frac{0.66}{0.64} = \frac{33}{32}$$
$$f_A - f_B = 8$$
$$f_A = \frac{33}{32}f_B$$
$$\frac{33}{32}f_B - f_B = 8 \Rightarrow \frac{f_B}{32} = 8 \Rightarrow f_B = 256 \text{ Hz}$$
$$f_A = 256 + 8 = 264 \text{ Hz}$$
---
### Question 9
L = 1 m, r = 0.02 m, T₁ = 100°C, T₂ = 0°C, k = 380 Wm⁻¹K⁻¹, L_f = 3.34×10⁵ J/kg
$$A = \pi r^2 = \pi (0.02)^2 = 1.257 \times 10^{-3} \text{ m}^2$$
$$\frac{dQ}{dt} = \frac{kA\Delta T}{L} = \frac{380 \times 1.257 \times 10^{-3} \times 100}{1} = 47.77 \text{ W}$$
$$\text{Rate of melting} = \frac{dQ/dt}{L_f} = \frac{47.77}{3.34 \times 10^5} = 1.43 \times 10^{-4} \text{ kgs}^{-1}$$
---
### Question 10
v = 340 ms⁻¹, t₁ = 2.4 s, t₂ = 4 s
$$d_1 = \frac{v \times t_1}{2} = \frac{340 \times 2.4}{2} = 408 \text{ m}$$
$$d_2 = \frac{v \times t_2}{2} = \frac{340 \times 4}{2} = 680 \text{ m}$$
$$\text{Distance between hills} = d_1 + d_2 - d_1 = d_2$$
Total distance = d₁ + d₂ but man is between hills:
$$D = d_1 + d_2 - D \Rightarrow \text{Actually: } d_1 = x,\ d_2 = D - x$$
$$x + (D-x) = D,\quad t_1 = \frac{2x}{v},\quad t_2 = \frac{2(D-x)}{v}$$
$$D = \frac{v(t_1+t_2)}{2} = \frac{340(2.4+4)}{2} = \frac{340 \times 6.4}{2} = \boxed{1088 \text{ m}}$$
---
## SECTION B — MECHANICS
---
### Question 11a
**(i)**
- **Vector quantities** have both magnitude and direction; **scalar quantities** have magnitude only.
- Momentum → **Vector**
- Current → **Scalar**
- Electrical energy → **Scalar**
- Acceleration due to gravity → **Vector**
- Mass → **Scalar**
**(ii)** Relative velocity is the velocity of one object as observed from another moving object. If A moves at v_A and B at v_B, relative velocity of A with respect to B = v_A − v_B.
**(iii)**
Automobile: u = 0, a = 3 ms⁻²; Truck: v = 12 ms⁻¹ (constant)
Automobile overtakes truck when displacements are equal:
$$\frac{1}{2}at^2 = vt \Rightarrow \frac{1}{2}(3)t^2 = 12t \Rightarrow t = 8 \text{ s}$$
$$s = vt = 12 \times 8 = 96 \text{ m}$$
Speed of automobile at that instant:
$$v_a = at = 3 \times 8 = 24 \text{ ms}^{-1}$$
---
### Question 11b
**(i) Centripetal force** is the force directed toward the center of a circular path that keeps an object in circular motion.
Derivation:
For uniform circular motion, centripetal acceleration:
$$a = \frac{v^2}{r}$$
$$F_c = ma = \frac{mv^2}{r}$$
**(ii)**
r = 100 m, μ = 0.25, g = 10 ms⁻²
$$F_c = \mu mg \Rightarrow \frac{mv^2}{r} = \mu mg$$
$$v = \sqrt{\mu g r} = \sqrt{0.25 \times 10 \times 100} = \sqrt{250} = 15.81 \text{ ms}^{-1}$$
---
### Question 12a
**(i)**
- **Amplitude (A):** Maximum displacement from equilibrium position.
- **Angular frequency (ω):** Rate of change of phase angle; $\omega = 2\pi f$
$$x = A\sin(\omega t + \phi)$$
Where x = displacement, A = amplitude, ω = angular frequency, t = time, φ = phase constant.
**(ii)**
m = 0.12 kg, k = 25 Nm⁻¹, extra distance = 10 mm = 0.01 m
**Extension:**
$$e = \frac{mg}{k} = \frac{0.12 \times 10}{25} = 0.048 \text{ m}$$
**SHM proof:** Restoring force F = −kx ∝ −x → SHM confirmed.
**Period:**
$$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.12}{25}} = 2\pi \times 0.0693 = 0.435 \text{ s}$$
**Maximum KE** (at equilibrium, amplitude = 0.01 m):
$$KE_{max} = \frac{1}{2}kA^2 = \frac{1}{2} \times 25 \times (0.01)^2 = 1.25 \times 10^{-3} \text{ J}$$
---
### Question 12b (Rotational)
**(i)**
**Moment of inertia (I)** is the rotational analogue of mass; it is the sum of products of each particle's mass and the square of its distance from the axis:
$$I = \sum mr^2 \quad \text{Units: kgm}^2$$
**Rotational KE derivation:**
$$KE = \sum \frac{1}{2}m_i v_i^2 = \sum \frac{1}{2}m_i(\omega r_i)^2 = \frac{1}{2}\omega^2\sum m_i r_i^2 = \frac{1}{2}I\omega^2$$
**(ii)**
τ = 15 Nm, I = 2.0 kgm², t = 10 s, ω₀ = 0
**Angular acceleration:**
$$\alpha = \frac{\tau}{I} = \frac{15}{2} = 7.5 \text{ rads}^{-2}$$
**Angular velocity at t = 10 s:**
$$\omega = \alpha t = 7.5 \times 10 = 75 \text{ rads}^{-1}$$
**Angular displacement:**
$$\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2} \times 7.5 \times 100 = 375 \text{ rad}$$
**Work done:**
$$W = \tau\theta = 15 \times 375 = 5625 \text{ J}$$
**Kinetic energy:**
$$KE = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 2 \times 75^2 = 5625 \text{ J}$$
**Average power:**
$$P = \frac{W}{t} = \frac{5625}{10} = 562.5 \text{ W}$$
---
## SECTION C — HEAT AND PROPERTIES OF MATTER
---
### Question 13a(i) — Terminal Velocity Derivation
At terminal velocity, net force = 0:
$$\text{Weight} = \text{Upthrust} + \text{Viscous drag}$$
$$\frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \rho_f g + 6\pi\eta r v_t$$
$$v_t = \frac{2r^2(\rho - \rho_f)g}{9\eta}$$
---
### Question 13a(ii)
η = 3.5×10⁻⁴ Nm⁻²s, v_t = 1.57×10⁻⁵ ms⁻¹, ρ = 920 kgm⁻³, E = 3.2×10⁵ Fm⁻¹, ρ_air ≈ 1.2 kgm⁻³
**Radius (from terminal velocity):**
$$v_t = \frac{2r^2(\rho-\rho_{air})g}{9\eta}$$
$$r^2 = \frac{9\eta v_t}{2(\rho-\rho_{air})g} = \frac{9 \times 3.5\times10^{-4} \times 1.57\times10^{-5}}{2 \times (920-1.2) \times 10}$$
$$r^2 = \frac{4.945\times10^{-8}}{18396} = 2.69\times10^{-12}$$
$$r = 1.64\times10^{-6} \text{ m}$$
**Weight:**
$$W = \frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi(1.64\times10^{-6})^3 \times 920 \times 10$$
$$W = \frac{4}{3}\pi \times 4.41\times10^{-18} \times 9200 = 1.70\times10^{-13} \text{ N}$$
**Charge (at equilibrium: qE = W):**
$$q = \frac{W}{E} = \frac{1.70\times10^{-13}}{3.2\times10^5} = 5.31\times10^{-19} \text{ C}$$
---
### Question 13b(i) — Bernoulli's Principle
**Bernoulli's Principle:** For a steady, non-viscous, incompressible flow, the total mechanical energy along a streamline is constant:
$$P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$$
**Applications:**
1. Aerofoil/aircraft lift — faster flow over top creates lower pressure
2. Venturi meter — measures fluid flow rate using pressure difference
---
### Question 13b(ii)
Q = 10⁻² m³s⁻¹, A₁ = 50 cm² = 50×10⁻⁴ m², P₁ = 2.0×10⁵ Nm⁻², ρ_water = 1000 kgm⁻³
**(A) Flow velocity:**
$$v_1 = \frac{Q}{A_1} = \frac{10^{-2}}{50\times10^{-4}} = 2 \text{ ms}^{-1}$$
**(B) Total pressure (Bernoulli):**
$$P_{total} = P_1 + \frac{1}{2}\rho v_1^2 = 2.0\times10^5 + \frac{1}{2}\times1000\times4 = 2.0\times10^5 + 2000 = 202000 \text{ Nm}^{-2}$$
*Assumption: horizontal pipe, no height difference, incompressible fluid.*
**(C) New velocity when P_total = 3.6×10⁵ Nm⁻²:**
$$\frac{1}{2}\rho v_2^2 = P_{total} - P_2$$
Using Bernoulli between two points; since total pressure is now higher:
$$v_2 = \sqrt{\frac{2(P_{total}-P_1)}{\rho}} = \sqrt{\frac{2(3.6\times10^5 - 2.0\times10^5)}{1000}} = \sqrt{\frac{3.2\times10^5}{1000}} = \sqrt{320} = 17.9 \text{ ms}^{-1}$$
---
### Question 14a
**(i) Newton's Law of Cooling:** The rate of heat loss of a body is directly proportional to the temperature difference between the body and its surroundings, provided the temperature difference is small:
$$\frac{dQ}{dt} \propto (\theta - \theta_0)$$
**(ii)**
Let c_alcohol = specific heat of alcohol.
Water: m_w = 500 g, Alcohol: m_a = 250 g, W_cal = 10 g (water equivalent)
ΔT = 70 − 35 = 35°C, c_water = 4200 Jkg⁻¹K⁻¹ = 1 cal/g°C
Using equal cooling rates (Newton's law applied):
$$\frac{(m_w c_w + W)}{t_w} = \frac{(m_a c_a + W)}{t_a}$$
$$\frac{(500\times1 + 10)}{390} = \frac{(250\times c_a + 10)}{201}$$
$$\frac{510}{390} = \frac{250c_a + 10}{201}$$
$$1.3077 \times 201 = 250c_a + 10$$
$$262.85 = 250c_a + 10$$
$$c_a = \frac{252.85}{250} = 1.011 \text{ cal g}^{-1}{}^{\circ}\text{C}^{-1} \approx 4246 \text{ Jkg}^{-1}\text{K}^{-1}$$
---
### Question 14b
**(i)**
- **Specific heat capacity:** Heat required to raise the temperature of 1 kg of a substance by 1 K.
- **Specific latent heat:** Heat required to change the state of 1 kg of a substance without temperature change.
**(ii)**
ρ_A/ρ_B = 5/6, c_A/c_B = 4/5
Thermal capacity per unit volume = ρc
$$\frac{\rho_A c_A}{\rho_B c_B} = \frac{5}{6} \times \frac{4}{5} = \frac{4}{6} = \frac{2}{3}$$
Thermal capacity per unit volume of A : B = **2 : 3**
---
### Question 15a(i) — Kinetic Theory Assumptions
1. Gas consists of a large number of identical molecules in continuous random motion.
2. The volume of molecules is negligible compared to the volume of the gas.
3. Intermolecular forces are negligible except during collisions.
4. Collisions between molecules and walls are perfectly elastic.
5. The duration of a collision is negligible compared to time between collisions.
**Pressure of an ideal gas:**
$$P = \frac{1}{3}\rho\langle c^2\rangle = \frac{1}{3}\frac{Nm}{V}\langle c^2\rangle$$
---
### Question 15a(ii)
For ideal gas, molecules have no volume and no intermolecular forces → PV = nRT holds exactly.
Real gases have finite molecular volume and intermolecular attractive forces, so they deviate — hence Van der Waals equation:
$$\left(P + \frac{a}{V^2}\right)(V-b) = RT$$
---
### Question 15a(iii)
P = 2.0×10⁵ Nm⁻², V = 0.04 m³, T₁ = 300 K, M_He = 4 g/mol = 0.004 kg/mol, R = 8.314 Jmol⁻¹K⁻¹
**Mass of Helium:**
$$PV = nRT \Rightarrow n = \frac{PV}{RT} = \frac{2\times10^5 \times 0.04}{8.314 \times 300} = \frac{8000}{2494.2} = 3.208 \text{ mol}$$
$$m = nM = 3.208 \times 0.004 = 0.01283 \text{ kg} \approx 12.83 \text{ g}$$
**r.m.s speed at 300 K:**
$$v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3\times8.314\times300}{0.004}} = \sqrt{\frac{7482.6}{0.004}} = \sqrt{1870650} = 1368 \text{ ms}^{-1}$$
**r.m.s speed of He at 432 K:**
$$v_{rms} = \sqrt{\frac{3\times8.314\times432}{0.004}} = \sqrt{2693376} = 1641 \text{ ms}^{-1}$$
**r.m.s speed of H₂ at 432 K** (M = 0.002 kg/mol):
$$v_{rms} = \sqrt{\frac{3\times8.314\times432}{0.002}} = \sqrt{5386752} = 2321 \text{ ms}^{-1}$$
---
### Question 15b(i) — Show Cp − Cv = R
At constant volume: $Q_v = nC_v\Delta T$, so $\Delta U = nC_v\Delta T$
At constant pressure: $Q_p = nC_p\Delta T$
By first law: $Q_p = \Delta U + W = nC_v\Delta T + P\Delta V$
From ideal gas: $P\Delta V = nR\Delta T$
$$nC_p\Delta T = nC_v\Delta T + nR\Delta T$$
$$\boxed{C_p - C_v = R}$$
---
### Question 15b(ii)
T_cold = 5°C = 278 K, T_hot = 27°C = 300 K, m = 200 g = 0.2 kg, L_f = 3.34×10⁵ J/kg
**Heat removed from cold reservoir:**
$$Q_c = mL_f = 0.2 \times 3.34\times10^5 = 66800 \text{ J}$$
**COP (max = Carnot):**
$$COP = \frac{T_c}{T_h - T_c} = \frac{278}{300-278} = \frac{278}{22} = 12.636$$
$$COP = \frac{Q_c}{W} \Rightarrow W = \frac{Q_c}{COP} = \frac{66800}{12.636} = 5290 \text{ J}$$
---
### Question 16a
**(i) Second Law of Thermodynamics:** Heat cannot spontaneously flow from a colder body to a hotter body. Equivalently, no heat engine can be 100% efficient.
**Significance:** Sets limits on efficiency of heat engines; defines direction of natural processes.
**(ii)**
Q_h = 5000 J, T_h = 600 K, T_c = 400 K
**Efficiency:**
$$\eta = 1 - \frac{T_c}{T_h} = 1 - \frac{400}{600} = \frac{1}{3} = 33.3\%$$
**(A) Work done:**
$$W = \eta Q_h = \frac{1}{3} \times 5000 = 1666.7 \text{ J}$$
**(B) Heat discarded:**
$$Q_c = Q_h - W = 5000 - 1666.7 = 3333.3 \text{ J}$$
**(C) Efficiency = 33.3%**
---
### Question 16b
**(i) Entropy (S):** A measure of disorder or randomness of a system.
$$\Delta S = \frac{\Delta Q}{T}$$
For reversible isothermal expansion:
$$\Delta S = \frac{Q}{T} = \frac{nRT\ln(V_2/V_1)}{T} = nR\ln\frac{V_2}{V_1}$$
**(ii)**
m = 2 kg, L_f = 3.34×10⁵ J/kg, T = 273 K
$$\Delta S = \frac{mL_f}{T} = \frac{2 \times 3.34\times10^5}{273} = \frac{668000}{273} = 2447.3 \text{ JK}^{-1}$$
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## SECTION D — VIBRATION AND WAVES
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### Question 17a(i)
**Velocity of sound in air depends on:**
- **Temperature:** $v \propto \sqrt{T}$ — increases with temperature
- **Pressure:** At constant temperature, pressure has no effect (since $\rho \propto P$, keeping $v = \sqrt{\gamma P/\rho}$ constant)
- **Humidity:** Humid air is less dense → speed increases with humidity
**Proof:**
$$v = \sqrt{\frac{\gamma P}{\rho}}$$
From ideal gas: $P = \frac{\rho RT}{M}$, so:
$$v = \sqrt{\frac{\gamma RT}{M}} \propto \sqrt{T}$$
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### Question 17a(ii)
v₁ = 340 ms⁻¹ at T₁ = 14°C = 287 K
P tripled, T₂ = 200°C = 473 K
Since pressure has no effect:
$$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{473}{287}} = \sqrt{1.648} = 1.284$$
$$v_2 = 340 \times 1.284 = 436.4 \text{ ms}^{-1}$$
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### Question 17b(i)
- **Fundamental note:** Lowest frequency (first harmonic) produced by a vibrating body.
- **Overtone:** Any frequency above the fundamental produced simultaneously.
- **Harmonic frequencies:** Integer multiples of the fundamental frequency (f, 2f, 3f...).
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### Question 17b(ii)
L = 100 cm = 1 m, load = 10 kg, f = 300 Hz
Tension: T = mg = 10×10 = 100 N
For fundamental mode of string:
$$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$
$$300 = \frac{1}{2\times1}\sqrt{\frac{100}{\mu}}$$
$$600 = \sqrt{\frac{100}{\mu}}$$
$$360000 = \frac{100}{\mu}$$
$$\mu = \frac{100}{360000} = 2.78\times10^{-4} \text{ kgm}^{-1}$$
$$\text{Mass} = \mu L = 2.78\times10^{-4} \times 1 = 2.78\times10^{-4} \text{ kg} = 0.278 \text{ g}$$
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### Question 17b(iii) — Conditions for Stationary Waves
1. Two sources must have the same frequency.
2. Two sources must have the same amplitude.
3. The waves must travel in opposite directions.
4. The waves must be coherent (constant phase difference).
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### Question 18a(i) — Doppler Effect
The **Doppler effect** is the apparent change in frequency of a wave due to relative motion between the source and observer. If source approaches observer, frequency appears higher; if receding, frequency appears lower:
$$f' = f\left(\frac{v \pm v_o}{v \mp v_s}\right)$$
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### Question 18a(ii)
f = 600 Hz, v_s = 2 ms⁻¹, v = 340 ms⁻¹
The whistle moves toward the wall. Wall acts as a reflector.
**Frequency received at wall (wall as observer, source moving toward):**
$$f_1 = f\frac{v}{v - v_s} = 600 \times \frac{340}{340-2} = 600 \times \frac{340}{338} = 603.55 \text{ Hz}$$
**Frequency of echo heard by observer (source = wall, observer moving away from wall):**
$$f_2 = f_1 \times \frac{v - v_s}{v} = 603.55 \times \frac{338}{340} = 599.99 \approx 600 \text{ Hz}$$
Wait — observer is stationary; source (whistle) moves away from observer:
$$f_{direct} = f\frac{v}{v+v_s} = 600\times\frac{340}{342} = 596.49 \text{ Hz}$$
**Apparent frequency shift:**
$$\Delta f = f_1 - f_{direct} = 603.55 - 596.49 = \boxed{7.06 \text{ Hz}}$$
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### Question 18b(i) — Characteristics of Progressive Wave
1. Energy is transferred in the direction of wave propagation.
2. All particles vibrate with the same amplitude and frequency.
3. There is a phase difference between adjacent particles.
4. Waveform moves through the medium.
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### Question 18b(ii)
Drumming rate at which echo is indistinct = 40/min → time between beats = 60/40 = 1.5 s → echo returns in 1.5 s
Let d₁ = initial distance from mountain, v = speed of sound
$$\frac{2d_1}{v} = 1.5 \quad \Rightarrow \quad d_1 = \frac{1.5v}{2} \quad \cdots (1)$$
After moving 90 m closer: d₂ = d₁ − 90, echo still indistinct at 40/min (same time 1.5 s):
Since the echo is again not heard at same rate after moving closer, the time must now equal 1.5 s again, implying a different harmonic is involved. The first echo is at t, second position at t − Δ:
$$\frac{2d_2}{v} = \frac{1.5}{2} = 0.75 \text{ s (next harmonic position)}$$
$$d_2 = \frac{0.75v}{2} \quad \cdots (2)$$
$$d_1 - d_2 = 90$$
$$\frac{1.5v}{2} - \frac{0.75v}{2} = 90$$
$$\frac{0.75v}{2} = 90 \Rightarrow v = \frac{180}{0.75} = 240 \text{ ms}^{-1}$$
$$d_1 = \frac{1.5 \times 240}{2} = 180 \text{ m}$$
**Distance = 180 m; Speed of sound = 240 ms⁻¹**