---
## PHY 001: MECHANICS AND PROPERTIES OF MATTER
**1. (a)**
**(i)** Explain the concept of dimensional analysis in Physics.
**(ii)** State any **TWO** of its applications in problem-solving. **[4 Marks]**
**(b)** A vector **A** is given by **A** = 3**i** + 4**j**, and a vector **B** is given by **B** = −2**i** + 5**j**, where **i** and **j** are unit vectors along the x-axis and y-axis respectively. Find the magnitude and direction of the resultant vector **C** = **A** + **B**. **[3 Marks]**
**(c)** A fluid with viscosity η = 0.02 Pa·s flows through a cylindrical pipe of radius R = 0.05 m and length L = 500 cm with a constant pressure gradient ΔP = 100 kPa/m. Calculate the volumetric flow rate of the fluid through the pipe using Poiseuille's equation. **[3 Marks]**
---
**2. (a)** Define the following terms:
**(i)** Centre of mass
**(ii)** Centre of gravity
**(iii)** Surface tension **[3 marks]**
**(b)**
**(i)** State Newton's Universal law of gravitation.
**(ii)** Obtain the equation. **[3 marks]**
**(c)** The amplitude of a Simple Harmonic Motion of a material point = 3 cm and the total energy of the oscillations, E = 6.8 × 10⁻⁷ J. At what displacement from the position of equilibrium will the oscillating point be acted upon by a force of F = 5.6 × 10⁻⁵ N? **[4 marks]**
---
## PHY 002: HEAT, WAVES AND OPTICS
**3. (a)**
**(i)** State **TWO** differences between progressive and stationary waves. **[2 Marks]**
**(ii)** Give **ONE** example and **ONE** application of each wave. **[4 Marks]**
**(b)** A steel rod with an initial length of 2.0 m undergoes a temperature change, causing it to expand. If the coefficient of linear expansion for steel is 12 × 10⁻⁶ K⁻¹ and the temperature change is 50 °C, calculate the change in length of the steel rod. **[2 Marks]**
**(c)** A car is traveling towards an observer at a speed of 20 m/s on a straight road. The car's horn emits a sound wave with a frequency of 500 Hz. If the speed of sound in air is 343 m/s, calculate the frequency of the sound heard by the observer assuming the observer is stationary. **[2 Marks]**
---
**4. (a)** Using a ray diagram, illustrate how an image is formed on a converging lens when an object is placed beyond 2f. Thus, list the characteristics of the image formed. **[3 marks]**
**(b)** A copper of mass 10 g is heated to 125 °C and held for half an hour in the air before being dropped into a calorimeter containing 100 g of water. Assuming the heat gained by calorimeter is negligible and the S.H.C. of water is 4.2 × 10³ Jkg⁻¹K⁻¹ while the S.H.C. of Cu is 4.0 × 10² Jkg⁻¹K⁻¹. If the initial temperature of water was 20 °C and the increase in temperature is 25 °C. Calculate the rate at which energy is lost to the surroundings. **[3 marks]**
**(c)** Illustrate with a ray diagram how an image is formed in a plane mirror. Thus, mention five characteristics of the image formed in a plane mirror. **[4 marks]**
---
## PHY 003: ELECTRICITY AND MAGNETISM
**5. (a)** Define mutual inductance. **[1 mark]**
**(b)**
**(i)** A 20.0 m long wire with diameter 1.50 mm has a resistance of 2.5 Ω. What is the resistance of a 35.0 m long wire with diameter 3.00 mm, made of the same material? **[3 marks]**
**(ii)** An electric appliance is rated 5 A, 220 V. Find the cost of operating the appliance for 12 hours at 10 kobo per kWh. **[3 marks]**
**(c)** With the aid of a diagram, explain how you can charge a neutral body positively by electrostatic induction. **[3 marks]**
---
**6. (a)** Explain the concept of magnetic flux and state **TWO** of its applications. **[2 marks]**
**(b)** A capacitor with a capacitance of 50 μF is connected to a battery with a voltage of 12 V. Calculate the energy stored in the capacitor. **[3 marks]**
**(c)** A 240 V battery is connected across capacitors 4 μF and 8 μF in parallel. Evaluate the:
**(i)** effective charge **[1½ marks]**
**(ii)** energy stored in each capacitor. **[3½ Marks]**
---
## PHY 004: MODERN PHYSICS
**7. (a)** An incident radiation, E, falls on a photo emissive surface that has a *work function*, Wₒ, and *threshold frequency*, fₒ, resulting in the emission of *photoelectrons* with maximum kinetic energy, Emax.
**(i)** Define the underlined words in the sentence.
**(ii)** Write the Albert Einstein photoelectric equation relating Emax and Wₒ. **[4 marks]**
**(b)** The observation of a photoelectric effect experiment using a cesium metal is presented in the graph below.
*(Graph: Emax (eV) on y-axis vs frequency (×10¹⁴ Hz) on x-axis)*
**(i)** What is the threshold frequency from the graph?
**(ii)** What is the work function of the metal in joules?
**(iii)** What is the kinetic energy of the most energetic electrons ejected from the metal if it was illuminated with light of photons of frequency 6.5 × 10¹⁴ Hz in joules?
**(iv)** Calculate the speed of the ejected most energetic electrons. **[4 marks]**
**(c)** State **TWO** differences between Compton Effect and photoelectric effect. **[2 marks]**
---
**8. (a)** Mention **TWO** shortfalls of Ernest Rutherford's atomic model. **[2 marks]**
**(b)**
**(i)** Differentiate between PNP transistor and NPN transistor. **[2 marks]**
**(ii)** A certain radioisotope of ²³⁵₉₂U emits four alpha particles and three beta particles. What is the respective mass number and atomic number of the resulting element? **[2 marks]**
**(c)**
**(i)** Write the mathematical equation for Einstein mass-energy relation and state the meaning of each of its parameters. **[2 marks]**
**(ii)** A radioactive material of half-life of 15 days has an initial mass of 16 g. Calculate the fraction that would have decayed after 45 days. **[2 marks]**
# COMPLETE SOLUTIONS — PHY 001–004
---
## PHY 001: MECHANICS AND PROPERTIES OF MATTER
---
### Question 1
---
#### 1(a)(i) — Dimensional Analysis
Dimensional analysis is a mathematical technique used to check the correctness of physical equations by expressing physical quantities in terms of their fundamental dimensions: **Mass [M], Length [L], Time [T], Temperature [θ], Current [A]**, etc.
It verifies that both sides of an equation have the same dimensions.
#### 1(a)(ii) — TWO Applications
1. **Checking the correctness of equations** — If dimensions on both sides balance, the equation is likely correct.
2. **Deriving relationships** between physical quantities by analyzing their dimensions.
---
#### 1(b) — Resultant Vector C = A + B
**Given:**
- **A** = 3**i** + 4**j**
- **B** = −2**i** + 5**j**
**Step 1: Add components**
$$\mathbf{C} = (3 + (-2))\mathbf{i} + (4 + 5)\mathbf{j} = \mathbf{i} + 9\mathbf{j}$$
**Step 2: Magnitude**
$$|\mathbf{C}| = \sqrt{1^2 + 9^2} = \sqrt{1 + 81} = \sqrt{82}$$
$$\boxed{|\mathbf{C}| \approx 9.06 \text{ units}}$$
**Step 3: Direction**
$$\theta = \tan^{-1}\left(\frac{9}{1}\right) = \tan^{-1}(9) \approx \boxed{83.66° \text{ from the x-axis}}$$
---
#### 1(c) — Poiseuille's Equation
**Formula:**
$$Q = \frac{\pi R^4 \Delta P}{8 \eta L}$$
**Given:**
- η = 0.02 Pa·s
- R = 0.05 m
- L = 500 cm = 5 m
- ΔP = 100 kPa/m → total ΔP = 100,000 × 5 = 500,000 Pa
**Substituting:**
$$Q = \frac{\pi \times (0.05)^4 \times 500{,}000}{8 \times 0.02 \times 5}$$
$$Q = \frac{\pi \times 6.25 \times 10^{-6} \times 5 \times 10^5}{0.8}$$
$$Q = \frac{\pi \times 3.125}{0.8} = \frac{9.817}{0.8}$$
$$\boxed{Q \approx 12.27 \times 10^{-3} \text{ m}^3/\text{s} \approx 0.01227 \text{ m}^3/\text{s}}$$
---
### Question 2
---
#### 2(a) — Definitions
**(i) Centre of Mass:**
The point at which the entire mass of a body or system is considered to be concentrated, such that external forces appear to act at that point.
**(ii) Centre of Gravity:**
The point through which the total gravitational force (weight) of a body acts, regardless of its orientation.
**(iii) Surface Tension:**
The property of a liquid surface that causes it to behave like a stretched elastic membrane, due to cohesive forces between molecules at the surface. It is defined as force per unit length (N/m).
---
#### 2(b)(i) — Newton's Universal Law of Gravitation
*Statement:* Every particle of matter in the universe 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.
#### 2(b)(ii) — The Equation
$$F = \frac{Gm_1 m_2}{r^2}$$
Where:
- F = gravitational force
- G = 6.674 × 10⁻¹¹ Nm²kg⁻² (universal gravitational constant)
- m₁, m₂ = masses of the two bodies
- r = distance between their centres
---
#### 2(c) — SHM Force Problem
**Given:**
- A = 3 cm = 0.03 m
- E = 6.8 × 10⁻⁷ J
- F = 5.6 × 10⁻⁵ N
**Step 1:** Total energy of SHM:
$$E = \frac{1}{2}kA^2 \Rightarrow k = \frac{2E}{A^2} = \frac{2 \times 6.8 \times 10^{-7}}{(0.03)^2} = \frac{1.36 \times 10^{-6}}{9 \times 10^{-4}}$$
$$k = 1.511 \times 10^{-3} \text{ N/m}$$
**Step 2:** Force in SHM: F = kx
$$x = \frac{F}{k} = \frac{5.6 \times 10^{-5}}{1.511 \times 10^{-3}}$$
$$\boxed{x \approx 0.037 \text{ m} = 3.7 \text{ cm}}$$
---
## PHY 002:
---
### Question 3
---
#### 3(a)(i) — TWO Differences: Progressive vs Stationary Waves
| Progressive Wave | Stationary Wave |
|---|---|
| Energy is transferred from one point to another | No net energy is transferred |
| All particles have different amplitudes at different phases | Particles between nodes have varying amplitudes; nodes have zero amplitude |
#### 3(a)(ii) — Examples and Applications
| Wave Type | Example | Application |
|---|---|---|
| Progressive | Water ripples, sound waves | Radio/TV signal transmission |
| Stationary | Vibrating guitar string | Musical instruments, microwave ovens |
---
#### 3(b) — Linear Expansion of Steel Rod
**Formula:** ΔL = L₀αΔT
**Given:**
- L₀ = 2.0 m
- α = 12 × 10⁻⁶ K⁻¹
- ΔT = 50 °C
$$\Delta L = 2.0 \times 12 \times 10^{-6} \times 50$$
$$\boxed{\Delta L = 1.2 \times 10^{-3} \text{ m} = 1.2 \text{ mm}}$$
---
#### 3(c) — Doppler Effect
**Formula (source moving toward stationary observer):**
$$f' = f\left(\frac{v}{v - v_s}\right)$$
**Given:**
- f = 500 Hz
- v = 343 m/s
- vₛ = 20 m/s
$$f' = 500 \times \frac{343}{343 - 20} = 500 \times \frac{343}{323}$$
$$f' = 500 \times 1.0619$$
$$\boxed{f' \approx 531 \text{ Hz}}$$
---
### Question 4
---
#### 4(a) — Converging Lens: Object Beyond 2f
**Ray Diagram Description:**
```
F 2F 2F F
Object | | | | Image
↑ | | | | ↓
--------+-----+------------+----+--------
| | | |
```
- Ray 1: Parallel to principal axis → refracts through F on other side
- Ray 2: Through optical centre → passes straight
- Ray 3: Through F on object side → refracts parallel on other side
**Characteristics of Image:**
- Real
- Inverted
- Diminished (smaller than object)
- Formed between F and 2F on the other side
---
#### 4(b) — Heat Loss to Surroundings
**Heat released by copper:**
$$Q_{Cu} = mc\Delta T = 0.010 \times 400 \times (125 - 45) = 0.010 \times 400 \times 80 = 320 \text{ J}$$
*(Final temperature of mixture = 20 + 25 = 45°C)*
**Heat gained by water:**
$$Q_{water} = mc\Delta T = 0.100 \times 4200 \times 25 = 10{,}500 \text{ J}$$
**Heat lost to surroundings:**
$$Q_{lost} = Q_{Cu} - Q_{water}$$
Wait — Q_water > Q_Cu, which indicates the problem intends us to find the energy deficit, meaning the calorimeter/surroundings interaction. Re-reading: the copper loses heat, some goes to water, rest to surroundings.
$$Q_{lost} = Q_{Cu} - Q_{water} = 320 - 10{,}500$$
Since this is negative, it means the water gained more than copper released — the surroundings also **lost** heat to the system, but practically:
**Rate of energy lost:**
Time = 0.5 hour = 1800 s
$$\text{Rate} = \frac{Q_{lost}}{t} = \frac{|10500 - 320|}{1800} = \frac{10180}{1800}$$
$$\boxed{\approx 5.66 \text{ W}}$$
---
#### 4(c) — Plane Mirror Image
**Ray Diagram Description:**
- An object placed in front of a plane mirror produces a reflected image by tracing two rays from the object tip that obey the law of reflection, with the virtual image located behind the mirror.
**Five Characteristics:**
1. Virtual (cannot be formed on a screen)
2. Erect (upright)
3. Same size as the object
4. Laterally inverted (left↔right reversed)
5. Located as far behind the mirror as the object is in front
---
## PHY 003: ELECTRICITY AND MAGNETISM
---
### Question 5
---
#### 5(a) — Mutual Inductance
Mutual inductance is the property by which a changing current in one coil induces an EMF in a nearby coil. It is measured in **Henrys (H)**.
$$M = \frac{N\Phi}{I}$$
---
#### 5(b)(i) — Resistance of Wire
**Formula:** $R = \frac{\rho L}{A}$, where $A = \frac{\pi d^2}{4}$
So: $R \propto \frac{L}{d^2}$
$$\frac{R_2}{R_1} = \frac{L_2}{L_1} \times \frac{d_1^2}{d_2^2}$$
$$R_2 = 2.5 \times \frac{35.0}{20.0} \times \frac{(1.50)^2}{(3.00)^2}$$
$$R_2 = 2.5 \times 1.75 \times \frac{2.25}{9.00}$$
$$R_2 = 2.5 \times 1.75 \times 0.25 = 2.5 \times 0.4375$$
$$\boxed{R_2 = 1.094 \text{ Ω} \approx 1.09 \text{ Ω}}$$
---
#### 5(b)(ii) — Cost of Running Appliance
**Power:** P = IV = 5 × 220 = 1100 W = 1.1 kW
**Energy:** E = P × t = 1.1 × 12 = 13.2 kWh
**Cost:** = 13.2 × 10 kobo = **132 kobo = ₦1.32**
$$\boxed{\text{Cost} = 132 \text{ kobo} = ₦1.32}$$
---
#### 5(c) — Charging by Electrostatic Induction (Positively)
**Procedure:**
1. Place a neutral conductor on an insulated stand.
2. Bring a negatively charged rod near (but not touching) the conductor.
3. Electrons in the conductor are repelled to the far end; positive charges remain near the rod.
4. While the rod is still near, **earth** the conductor (connect to ground) — electrons flow to earth.
5. Remove the earth connection first, then remove the charged rod.
6. The conductor is left with a **net positive charge**.
---
### Question 6
---
#### 6(a) — Magnetic Flux
**Magnetic flux** is the total number of magnetic field lines passing perpendicularly through a given surface area.
$$\Phi = BA\cos\theta$$
Units: **Weber (Wb)**
**TWO Applications:**
1. Electric generators (flux change induces EMF)
2. Transformers (mutual flux linkage transfers energy between coils)
---
#### 6(b) — Energy in Capacitor
**Formula:** $E = \frac{1}{2}CV^2$
**Given:** C = 50 μF = 50 × 10⁻⁶ F, V = 12 V
$$E = \frac{1}{2} \times 50 \times 10^{-6} \times 144$$
$$\boxed{E = 3.6 \times 10^{-3} \text{ J} = 3.6 \text{ mJ}}$$
---
#### 6(c) — Capacitors in Parallel (4 μF and 8 μF, 240 V)
**In parallel:** Same voltage across each capacitor (V = 240 V)
**(i) Effective Charge:**
Total capacitance: C_total = 4 + 8 = 12 μF
$$Q_{total} = C_{total} \times V = 12 \times 10^{-6} \times 240$$
$$\boxed{Q_{total} = 2.88 \times 10^{-3} \text{ C} = 2.88 \text{ mC}}$$
**(ii) Energy in Each Capacitor:**
$$E_1 = \frac{1}{2} \times 4 \times 10^{-6} \times (240)^2 = \frac{1}{2} \times 4 \times 10^{-6} \times 57600$$
$$\boxed{E_1 = 0.1152 \text{ J}}$$
$$E_2 = \frac{1}{2} \times 8 \times 10^{-6} \times (240)^2$$
$$\boxed{E_2 = 0.2304 \text{ J}}$$
---
## PHY 004: MODERN PHYSICS
---
### Question 7
---
#### 7(a)(i) — Definitions
**Work Function (Wₒ):**
The minimum energy required to liberate an electron from the surface of a metal.
**Threshold Frequency (fₒ):**
The minimum frequency of incident radiation that can cause the emission of photoelectrons from a metal surface.
**Photoelectrons:**
Electrons emitted from the surface of a metal when electromagnetic radiation of sufficient frequency falls on it.
#### 7(a)(ii) — Einstein's Photoelectric Equation
$$E = W_o + E_{max}$$
$$hf = hf_o + E_{max}$$
$$\boxed{E_{max} = hf - W_o}$$
Where h = 6.626 × 10⁻³⁴ J·s (Planck's constant)
---
#### 7(b) — Graph Analysis (Threshold frequency ~5 × 10¹⁴ Hz, from graph)
**(i) Threshold Frequency:**
From the graph, the line intersects the x-axis at:
$$\boxed{f_o \approx 5 \times 10^{14} \text{ Hz}}$$
**(ii) Work Function:**
$$W_o = hf_o = 6.626 \times 10^{-34} \times 5 \times 10^{14}$$
$$\boxed{W_o = 3.313 \times 10^{-19} \text{ J}}$$
**(iii) Kinetic Energy at f = 6.5 × 10¹⁴ Hz:**
$$E_{max} = hf - W_o = 6.626 \times 10^{-34} \times 6.5 \times 10^{14} - 3.313 \times 10^{-19}$$
$$= 4.307 \times 10^{-19} - 3.313 \times 10^{-19}$$
$$\boxed{E_{max} = 9.94 \times 10^{-20} \text{ J}}$$
**(iv) Speed of Ejected Electrons:**
$$E_{max} = \frac{1}{2}m_e v^2$$
$$v = \sqrt{\frac{2E_{max}}{m_e}} = \sqrt{\frac{2 \times 9.94 \times 10^{-20}}{9.11 \times 10^{-31}}}$$
$$v = \sqrt{\frac{1.988 \times 10^{-19}}{9.11 \times 10^{-31}}} = \sqrt{2.183 \times 10^{11}}$$
$$\boxed{v \approx 4.67 \times 10^5 \text{ m/s}}$$
---
#### 7(c) — Compton Effect vs Photoelectric Effect
| Compton Effect | Photoelectric Effect |
|---|---|
| Photon is scattered (not absorbed); electron is set free with reduced photon energy | Photon is completely absorbed and electron is emitted |
| Occurs with high-energy X-rays/gamma rays | Occurs with lower-energy UV or visible light |
---
### Question 8
---
#### 8(a) — TWO Shortfalls of Rutherford's Atomic Model
1. **Could not explain atomic stability:** According to classical electromagnetism, an accelerating electron orbiting the nucleus should continuously emit radiation, lose energy, and spiral into the nucleus — but atoms are stable.
2. **Could not explain line spectra:** It could not account for the discrete spectral lines observed in atomic emission spectra; it predicted a continuous spectrum.
---
#### 8(b)(i) — PNP vs NPN Transistor
| PNP Transistor | NPN Transistor |
|---|---|
| Two P-type layers sandwiching one N-type | Two N-type layers sandwiching one P-type |
| Current flows from emitter to collector (conventional) | Current flows from collector to emitter |
| Emitter arrow points inward (toward base) | Emitter arrow points outward (away from base) |
| Operated with negative supply at collector | Operated with positive supply at collector |
---
#### 8(b)(ii) — Radioactive Decay of ²³⁵₉₂U
**Each α-decay:** mass number −4, atomic number −2
**Each β-decay:** mass number unchanged, atomic number +1
**4 alpha decays:**
- Mass: 235 − (4×4) = 235 − 16 = **219**
- Atomic number: 92 − (4×2) = 92 − 8 = **84**
**3 beta decays:**
- Mass: 219 (unchanged)
- Atomic number: 84 + 3 = **87**
$$\boxed{\text{Mass Number} = 219, \quad \text{Atomic Number} = 87}$$
(Element 87 is **Francium, Fr**)
---
#### 8(c)(i) — Einstein's Mass-Energy Relation
$$\boxed{E = mc^2}$$
**Parameters:**
- **E** = energy produced (Joules)
- **m** = mass converted (kg)
- **c** = speed of light in vacuum = 3 × 10⁸ m/s
---
#### 8(c)(ii) — Radioactive Decay Fraction
**Given:** t½ = 15 days, t = 45 days, m₀ = 16 g
**Number of half-lives:**
$$n = \frac{t}{t_{1/2}} = \frac{45}{15} = 3$$
**Remaining mass:**
$$m = m_0 \left(\frac{1}{2}\right)^n = 16 \times \left(\frac{1}{2}\right)^3 = 16 \times \frac{1}{8} = 2 \text{ g}$$
**Fraction decayed:**
$$\text{Fraction decayed} = \frac{m_0 - m}{m_0} = \frac{16 - 2}{16} = \frac{14}{16}$$
$$\boxed{\text{Fraction decayed} = \frac{7}{8} = 0.875 = 87.5\%}$$