## PHY 001: MECHANICS AND PROPERTIES OF MATTER
### Question 1
**(a)** State the units and dimensions of the following quantities:
- **(i)** Surface tension [1 mark]
- **(ii)** Frequency [1 mark]
- **(iii)** Show that the expression V² = V₀² + 2aS is dimensionally correct, where V and V₀ represent final and initial velocities, a is acceleration, and S is displacement [2 marks]
**(b)** A race car moves such that its position is given as X = 0.75t² + 5.0t + 1. Find:
- **(i)** The position at t = 4.00 s [2 marks]
- **(ii)** The instantaneous velocity of the car at t = 4.00 s [2 marks]
- **(iii)** The average velocity for the time interval t = 2.00 s to t = 7.00 s [2 marks]
---
### Question 2
**(a)** State Pascal's principle. [1 mark]
**(b)** State two applications each of:
- **(i)** Surface tension [1 mark]
- **(ii)** Viscosity [1 mark]
- **(iii)** The electromagnetic Poynting vector **S** is defined as **S** = **E** × **H**, where **E** and **H** are the electric and magnetic fields respectively. Given **E** = 10.10**i** + 0.20**j** − 0.60**k** and **H** = 0.40**i** + 9.80**j** + 0.10**k**, calculate **S**. [3 marks]
**(c)** The mass of an object in air is 50 g and it appears to have a mass of 35 g when immersed in water. Find the:
- **(i)** Relative density of the substance [2 marks]
- **(ii)** Density of the substance [2 marks]
---
## PHY 002: HEAT, WAVES AND OPTICS
### Question 3
**(a)** State any three assumptions of the kinetic theory of gases. [3 marks]
**(b)** What is the pressure of 3 moles of an ideal gas at a temperature of 27°C, having a volume of 5 litres? [2½ marks]
**(c)** A piece of copper of mass 0.04 kg at 160°C is transferred into a copper calorimeter of mass 0.06 kg containing 0.05 kg of water at 20°C. What will be the final temperature of the mixture? Specific heat capacities of copper and water are 400 J/kg/K and 4200 J/kg/K respectively. (Neglect heat losses to surroundings.) [4½ marks]
---
### Question 4
**(a)**
- **(i)** State the principle of superposition of waves. [1 mark]
- **(ii)** Briefly describe Huygens' principle with the aid of an appropriate diagram. [2 marks]
**(b)** The manufacturer's manual of a violin shows that the heaviest and lightest strings have linear densities of 6.0 and 0.58 kg/m respectively. Assuming the strings are of the same material, determine the ratio of their radii. [3 marks]
**(c)** The voltage from an electromagnetic wave travelling on a transmission line is given by:
**V(x, t) = 10e^(−αx) sin(4π × 10⁹t − 30πx) V**
where x is the distance in metres from the transmitter.
- **(i)** Find the frequency, wavelength, and phase velocity of the wave. [2 marks]
- **(ii)** Find the voltage at x = 2.1 × 10⁻² cm and t = 0.32 s. [1 mark]
- **(iii)** If the amplitude of the wave is measured to be 2 V, find α. [1 mark]
---
## PHY 003: ELECTRICITY AND MAGNETISM
### Question 5
**(a)** Define electromotive force. [1 mark]
**(b)** A cell of e.m.f. E and internal resistance r is connected in series with two external resistors A (8 Ω) and B (2 Ω). A high-resistance voltmeter across A reads 8 V. When a third resistor C (8 Ω) is connected in parallel with A, the voltmeter across A and C reads 6 V.
- **(i)** Draw the circuit diagrams of the two arrangements. [2 marks]
- **(ii)** Calculate the internal resistance of the cell. [5 marks]
- **(iii)** Calculate the e.m.f. of the cell. [1 mark]
**(c)** Explain electrostatic induction and mention ONE method of producing electrostatic charges. [1 mark]
---
### Question 6
**(a)** What is electrostatics? [1 mark]
**(b)** Explain, with the aid of a diagram, how you can charge a gold leaf electroscope positively using the method of charging by induction. [4 marks]
**(c)** Two charges are located on the positive x-axis. Charge q₁ = 2 × 10⁻⁹ C is 2 cm from the origin, and charge q₂ = 3 × 10⁻⁹ C is 4 cm from the origin. What is the magnitude of the total force exerted by these two charges on a charge q₃ = 5 × 10⁻⁹ C located at the origin? [5 marks]
---
## PHY 004: MODERN PHYSICS
### Question 7
**(a)** Calculate the total binding energy per nucleon of an alpha particle. The masses of the neutron, proton, and alpha particle are 1.008665 u, 1.007825 u, and 4.004603 u respectively. [3 marks]
**(b)**
- **(i)** Radium with an atomic mass of 226 has a half-life of 800 years. For 0.5 g of radium, calculate the number of decays per second. [4 marks]
- **(ii)** Define the half-life of a radioactive sample. [1 mark]
**(c)** Which of the following radiations — alpha-rays, beta-rays, and gamma-rays:
- **(i)** Are similar to X-rays? [½ mark]
- **(ii)** Are most easily absorbed by matter? [½ mark]
- **(iii)** Travel with the greatest speed? [½ mark]
- **(iv)** Are similar in nature to cathode rays? [½ mark]
---
### Question 8
**(a)**
- **(i)** State four properties of X-rays. [2 marks]
- **(ii)** State four uses of X-rays. [2 marks]
**(b)**
- **(i)** Calculate the minimum wavelength of X-rays that can be produced by an electron accelerated through a potential difference of 20 kV. [2 marks]
- **(ii)** Write down the mathematical form of Bragg's law and explain each term. [2 marks]
- **(iii)** Determine the wavelength of the X-ray that was Bragg-diffracted by a cobalt crystal of interatomic spacing 4.07 × 10⁻¹⁰ m, if the first-order scattering angle is 24°. [2 marks]
---
---
# ANSWERS
## PHY 001
### Q1(a) — Units and Dimensions
**(i) Surface tension:**
Unit: **N/m** | Dimensions: **[M T⁻²]**
**(ii) Frequency:**
Unit: **Hz (s⁻¹)** | Dimensions: **[T⁻¹]**
**(iii) Dimensional check of V² = V₀² + 2aS:**
- LHS: V² → (m/s)² = **[L² T⁻²]**
- RHS: V₀² → [L² T⁻²]; 2aS → [L T⁻²][L] = **[L² T⁻²]**
All three terms have dimensions **[L² T⁻²]** ✓ — the equation is dimensionally correct.
---
### Q1(b) — Race Car Motion: X = 0.75t² + 5.0t + 1
**(i) Position at t = 4.00 s:**
$$X = 0.75(4)^2 + 5.0(4) + 1 = 12 + 20 + 1 = \boxed{33.0 \text{ m}}$$
**(ii) Instantaneous velocity at t = 4.00 s:**
Differentiate: v = dX/dt = 1.5t + 5.0
$$v = 1.5(4) + 5.0 = 6.0 + 5.0 = \boxed{11.0 \text{ m/s}}$$
**(iii) Average velocity from t = 2 s to t = 7 s:**
- X at t = 2: X = 0.75(4) + 10 + 1 = **14.0 m**
- X at t = 7: X = 0.75(49) + 35 + 1 = 36.75 + 36 = **72.75 m**
$$v_{avg} = \frac{72.75 - 14.0}{7 - 2} = \frac{58.75}{5} = \boxed{11.75 \text{ m/s}}$$
---
### Q2(b)(iii) — Poynting Vector S = E × H
**E** = 10.10**i** + 0.20**j** − 0.60**k**
**H** = 0.40**i** + 9.80**j** + 0.10**k**
Using the determinant method:
$$\mathbf{S} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 10.10 & 0.20 & -0.60 \\ 0.40 & 9.80 & 0.10 \end{vmatrix}$$
- **i** component: (0.20×0.10) − (−0.60×9.80) = 0.02 + 5.88 = **5.90**
- **j** component: −[(10.10×0.10) − (−0.60×0.40)] = −[1.01 + 0.24] = **−1.25**
- **k** component: (10.10×9.80) − (0.20×0.40) = 98.98 − 0.08 = **98.90**
$$\boxed{\mathbf{S} = 5.90\mathbf{i} - 1.25\mathbf{j} + 98.90\mathbf{k}}$$
---
### Q2(c) — Relative Density and Density
Mass in air = 50 g; apparent mass in water = 35 g
Upthrust = 50 − 35 = **15 g**
**(i) Relative density:**
$$RD = \frac{\text{Mass in air}}{\text{Upthrust}} = \frac{50}{15} = \boxed{3.33}$$
**(ii) Density:**
$$\rho = RD \times \rho_{water} = 3.33 \times 1000 = \boxed{3333 \text{ kg/m}^3}$$
---
## PHY 002
### Q3(b) — Ideal Gas Pressure
n = 3 mol, R = 8.314 J/mol/K, T = 27 + 273 = 300 K, V = 5 L = 0.005 m³
$$P = \frac{nRT}{V} = \frac{3 \times 8.314 \times 300}{0.005} = \boxed{1.496 \times 10^6 \text{ Pa} \approx 14.8 \text{ atm}}$$
---
### Q3(c) — Calorimetry
Heat lost by hot copper = heat gained by calorimeter + water:
$$m_{Cu}c_{Cu}(T_i - T_f) = (m_{cal}c_{Cu} + m_w c_w)(T_f - T_i^{cold})$$
$$0.04 \times 400 \times (160 - T_f) = (0.06 \times 400 + 0.05 \times 4200)(T_f - 20)$$
$$16(160 - T_f) = (24 + 210)(T_f - 20)$$
$$2560 - 16T_f = 234T_f - 4680$$
$$250T_f = 7240$$
$$\boxed{T_f \approx 28.96°C}$$
---
### Q4(b) — Violin String Radii
Linear density: μ = ρ × A = ρ × πr²
Since both strings are same material (same ρ):
$$\frac{\mu_1}{\mu_2} = \frac{r_1^2}{r_2^2} \Rightarrow \frac{r_1}{r_2} = \sqrt{\frac{\mu_1}{\mu_2}} = \sqrt{\frac{6.0}{0.58}} = \sqrt{10.34}$$
$$\boxed{\frac{r_{heavy}}{r_{light}} \approx 3.22}$$
---
### Q4(c) — Transmission Line Wave
From V(x,t) = 10e^(−αx) sin(4π × 10⁹t − 30πx):
Comparing to standard form V = V₀e^(−αx) sin(ωt − βx):
- ω = 4π × 10⁹ rad/s
- β = 30π rad/m
**(i) Wave parameters:**
$$f = \frac{\omega}{2\pi} = \frac{4\pi \times 10^9}{2\pi} = \boxed{2 \times 10^9 \text{ Hz} = 2 \text{ GHz}}$$
$$\lambda = \frac{2\pi}{\beta} = \frac{2\pi}{30\pi} = \boxed{\frac{1}{15} \approx 0.0667 \text{ m}}$$
$$v_p = \frac{\omega}{\beta} = \frac{4\pi \times 10^9}{30\pi} = \boxed{1.33 \times 10^8 \text{ m/s}}$$
**(ii) Voltage at x = 2.1 × 10⁻² cm = 2.1 × 10⁻⁴ m, t = 0.32 s:**
$$V = 10e^{-\alpha(2.1\times10^{-4})} \sin(4\pi \times 10^9 \times 0.32 - 30\pi \times 2.1\times10^{-4})$$
Assuming α = 0 (not yet given): the exponential term ≈ 1 for small x.
Phase argument = (4π × 10⁹ × 0.32) − (30π × 2.1 × 10⁻⁴)
= 4.021 × 10⁹ × π − 6.3π × 10⁻³ ≈ large multiple of 2π, so sin value depends on remainder. For a complete answer, α is needed (see part iii).
**(iii) Finding α when amplitude = 2 V at x = 2.1 × 10⁻² cm:**
$$2 = 10e^{-\alpha \times 2.1\times10^{-4}}$$
$$e^{-\alpha \times 2.1\times10^{-4}} = 0.2$$
$$\alpha = \frac{-\ln(0.2)}{2.1\times10^{-4}} = \frac{1.609}{2.1\times10^{-4}} = \boxed{7.66 \times 10^3 \text{ m}^{-1}}$$
---
## PHY 003
### Q5(b) — Internal Resistance and EMF
**Case 1:** R_ext = 8 + 2 = 10 Ω; V across A (8 Ω) = 8 V
$$I_1 = \frac{V_A}{R_A} = \frac{8}{8} = 1 \text{ A}$$
$$E = I_1(R_{ext} + r) = 1(10 + r) \quad \cdots (1)$$
**Case 2:** A ∥ C = (8×8)/(8+8) = 4 Ω; total R_ext = 4 + 2 = 6 Ω; V across parallel combination = 6 V
$$I_2 = \frac{V_{AC}}{R_{AC}} = \frac{6}{4} = 1.5 \text{ A}$$
$$E = I_2(R_{ext} + r) = 1.5(6 + r) \quad \cdots (2)$$
**Solving (1) and (2):**
$$10 + r = 1.5(6 + r) = 9 + 1.5r$$
$$1 = 0.5r$$
$$\boxed{r = 2 \text{ Ω}}$$
**(iii) EMF:**
$$E = 10 + 2 = \boxed{12 \text{ V}}$$
---
### Q6(c) — Coulomb Forces on q₃
k = 9 × 10⁹ N m²/C², q₁ = 2 × 10⁻⁹ C at 2 cm, q₂ = 3 × 10⁻⁹ C at 4 cm, q₃ = 5 × 10⁻⁹ C at origin.
Both q₁ and q₂ are positive and located on the **positive** x-axis, so they both repel q₃ (also positive) **toward the negative x-direction** (same direction):
**Force from q₁:**
$$F_1 = \frac{kq_1 q_3}{r_1^2} = \frac{9\times10^9 \times 2\times10^{-9} \times 5\times10^{-9}}{(0.02)^2} = \frac{9\times10^{-8}}{4\times10^{-4}} = 2.25 \times 10^{-4} \text{ N}$$
**Force from q₂:**
$$F_2 = \frac{kq_2 q_3}{r_2^2} = \frac{9\times10^9 \times 3\times10^{-9} \times 5\times10^{-9}}{(0.04)^2} = \frac{1.35\times10^{-7}}{1.6\times10^{-3}} = 8.44 \times 10^{-5} \text{ N}$$
**Total force (same direction):**
$$F_{total} = F_1 + F_2 = 2.25\times10^{-4} + 8.44\times10^{-5} = \boxed{3.09 \times 10^{-4} \text{ N}}$$
---
## PHY 004
### Q7(a) — Binding Energy of Alpha Particle
Alpha particle: 2 protons + 2 neutrons
**Mass defect:**
$$\Delta m = [2(1.007825) + 2(1.008665)] - 4.004603$$
$$= [2.015650 + 2.017330] - 4.004603$$
$$= 4.032980 - 4.004603 = 0.028377 \text{ u}$$
**Total binding energy** (1 u = 931.5 MeV):
$$E_B = 0.028377 \times 931.5 = 26.43 \text{ MeV}$$
**Binding energy per nucleon:**
$$\frac{E_B}{A} = \frac{26.43}{4} = \boxed{6.61 \text{ MeV/nucleon}}$$
---
### Q7(b)(i) — Radioactive Decay Rate of Radium
**Decay constant:**
$$\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{800 \times 365 \times 24 \times 3600} = \frac{0.693}{2.524 \times 10^{10}} = 2.746 \times 10^{-11} \text{ s}^{-1}$$
**Number of atoms:**
$$N = \frac{0.5}{226} \times 6.022 \times 10^{23} = 2.212 \times 10^{-3} \times 6.022 \times 10^{23} = 1.332 \times 10^{21} \text{ atoms}$$
**Activity:**
$$A = \lambda N = 2.746\times10^{-11} \times 1.332\times10^{21} = \boxed{3.66 \times 10^{10} \text{ decays/s}}$$
---
### Q7(c) — Radiation Properties
| | Answer |
|---|---|
| **(i) Similar to X-rays** | **Gamma rays** (both are high-energy electromagnetic radiation) |
| **(ii) Most easily absorbed by matter** | **Alpha rays** (stopped by a sheet of paper) |
| **(iii) Greatest speed** | **Gamma rays** (travel at speed of light, c) |
| **(iv) Similar to cathode rays** | **Beta rays** (both are streams of electrons) |
---
### Q8(b)(i) — Minimum X-ray Wavelength
At 20 kV, all kinetic energy converts to photon energy:
$$\lambda_{min} = \frac{hc}{eV} = \frac{6.626\times10^{-34} \times 3\times10^8}{1.6\times10^{-19} \times 20000} = \frac{1.988\times10^{-25}}{3.2\times10^{-15}} = \boxed{6.21 \times 10^{-11} \text{ m} = 0.062 \text{ nm}}$$
---
### Q8(b)(ii) — Bragg's Law
$$\boxed{n\lambda = 2d\sin\theta}$$
Where:
- **n** = order of diffraction (integer: 1, 2, 3…)
- **λ** = wavelength of the X-ray
- **d** = interplanar spacing of the crystal
- **θ** = glancing angle between the X-ray beam and the crystal plane
---
### Q8(b)(iii) — Bragg Diffraction Wavelength
d = 4.07 × 10⁻¹⁰ m, θ = 24°, n = 1
$$\lambda = \frac{2d\sin\theta}{n} = 2 \times 4.07\times10^{-10} \times \sin(24°)$$
$$= 8.14\times10^{-10} \times 0.4067 = \boxed{3.31 \times 10^{-10} \text{ m} = 0.331 \text{ nm}}$$