## PHY 001: MECHANICS AND PROPERTIES OF MATTER
**1. (a)** Mention the **THREE** types of modulus.
**(b)** Briefly explain the following: **[5 Marks]**
- i. Elasticity
- ii. Plasticity
- iii. Ductility
- iv. Malleability
**(c)** A particle is moving along the x-axis such that its position is given by: **[4 Marks]**
**x = 4t² − 16t + 12**
- i. Find the instantaneous velocity when t = 5s
- ii. Find the instantaneous acceleration when t = 5s
- iii. At what time is the particle stationary?
**[4½ Marks]**
**[Total = 10 Marks]**
---
**2. (a)** i. State the Archimedes' Principle.
- ii. List **FOUR** characteristics of pressure in a fluid. **[3 Marks]**
**(b)** Distinguish between a compressible fluid and an incompressible fluid. **[2 Marks]**
**(c)** Two divers M and N are at a depth 60 m and 80 m respectively below the water surface of a sea. The pressure on M is P₁ and the pressure on N is P₂. If the atmospheric pressure is equivalent to 20 m of water, find the value of P₂/P₁. **[2 Marks]**
**(d)** Differentiate between conservative force and non-conservative force, giving one example each.
**[3 Marks]**
**[Total = 10 Marks]**
---
## PHY 002: HEAT, WAVES AND OPTICS
**3. (a)** State Huygens' principle. **[1 Mark]**
**(b)** i. Explain the term "electromagnetic spectrum". List any **THREE** of its components.
- ii. In a Young's slit experiment, the separation between the first and the fifth bright fringes is 2.5 mm when the wavelength used is 4.5 × 10⁻⁷ m. The distance from the slits to the screen is 0.9 m. Calculate the separation of the two slits. **[4 Marks]**
---
**(c)** i. Define internal energy.
- ii. State the second law of thermodynamics. **[2 Marks]**
**(d)** i. Using the first law of thermodynamics, write expressions for adiabatic and isochoric processes.
- ii. State **FOUR** factors which affect heat loss by convection. **[3 Marks]**
**[Total = 10 Marks]**
---
**4. (a)** State **TWO** similarities and **TWO** differences between image formed by a converging mirror and a converging lens. **[2 Marks]**
**(b)** i. Mention any **TWO** uses each of plane mirror, concave and convex mirror.
- ii. Copy and complete the table below for the image formed by a concave mirror for different positions of the object.
| Position of Object | Position of Image | Size of Image | Nature of Image |
|---|---|---|---|
| At infinity | | | |
| At C | | | |
| Between C and F | | | |
| At F | | | |
| Between F and P | | | |
**[8 Marks]**
**[Total = 10 Marks]**
---
## PHY 003: ELECTRICITY AND MAGNETISM
**5. (a)** i. Explain what is meant by relative permittivity.
- ii. State **TWO** physical desirable properties in a material considered for dielectric in a capacitor. **[3 Marks]**
**(b)** A long magnet is removed from the centre of a coil of 30 turns. The speed of the magnet is controlled to maintain an induced e.m.f. of 80 μV across the coil. Removing the magnet in this way takes 3 minutes. Calculate the change of flux through the coil. **[3 Marks]**
**(c)** An electron enters the region of a uniform electric field as shown in Figure 1 below, with v₀ = 3.00 × 10⁶ m/s and E = 200 N/C. The horizontal length of such plates is l = 0.100 m.
- i. Find the acceleration of the electron while it is in the electric field.
- ii. Assuming the electron enters the field at time t = 0, find the time at which it leaves the field. (mass of an electron = 1.67 × 10⁻³¹ kg, charge of an electron = 1.60 × 10⁻¹⁹ C)
**[4 Marks]**
**[Total = 10 Marks]**
---
**6. (a)** How can the motion of a moving charged particle be used to distinguish between a magnetic field and an electric field? **[2 Marks]**
**(b)** Consider the circuit diagram in Figure 2. Calculate the current in each resistor.
*(Circuit diagram: Two batteries — 11.0 V with 20.0 Ω resistor, and 12.0 V with 17.0 Ω resistor — connected in a network with a 10.0 Ω resistor)*
**[5 Marks]**
**(c)** A proton is moving in a circular orbit of radius 14 cm in a uniform 0.35 T magnetic field perpendicular to the velocity of the proton. Determine the speed of the proton. **[3 Marks]**
---
## PHY 004: MODERN PHYSICS
**7. (a)** Calculate the de Broglie wavelength for a particle moving with a speed of 2.30 × 10⁸ m/s if the particle is: **[3 Marks]**
- i. an electron
- ii. a proton
- iii. a 200 g bullet
**(b)** i. Give the properties of α, β, and γ radiation in terms of charge, mass, ionizing effect, and field effect.
- ii. Explain the effect of temperature and pressure on the rate of disintegration of a radioactive nucleus.
**[7 Marks]**
**[Total = 10 Marks]**
**(c)** i. State Heisenberg's Uncertainty Principle.
- ii. List **FOUR** practical applications of X-rays. **[3 Marks]**
**(d)** A photon has a wavelength of 1Å. Calculate the:
- i. energy of the photon in electron volts
- ii. momentum of the photon
*(1Å = 10⁻¹⁰ m)* **[3 Marks]**
**(e)** i. Define binding energy.
- ii. Calculate the average binding energy per nucleon of ¹⁵⁶N which has a mass number of 56.0930 u. (mass of proton = 1.007825 u; mass of neutron = 1.008665 u; 1u = 931.5 MeV)
# SECTION B: PHYSICS ESSAY — COMPLETE SOLUTIONS
---
## PHY 001: MECHANICS AND PROPERTIES OF MATTER
### Question 1
---
### (a) THREE Types of Modulus
1. **Young's Modulus (E)** — ratio of tensile stress to tensile strain
2. **Bulk Modulus (K)** — ratio of volumetric stress to volumetric strain
3. **Shear (Rigidity) Modulus (G)** — ratio of shear stress to shear strain
---
### (b) Brief Explanations
**i. Elasticity:**
The property of a material that enables it to regain its original shape and size after the removal of an applied force (deforming force).
**ii. Plasticity:**
The property of a material that causes it to permanently deform and not return to its original shape after the deforming force is removed.
**iii. Ductility:**
The property of a material that allows it to be drawn into thin wires without breaking when a tensile force is applied (e.g., copper, gold).
**iv. Malleability:**
The property of a material that allows it to be rolled or hammered into thin sheets without breaking (e.g., aluminium, lead).
---
### (c) x = 4t² − 16t + 12
**i. Instantaneous velocity when t = 5s**
$$v = \frac{dx}{dt} = 8t - 16$$
At t = 5:
$$v = 8(5) - 16 = 40 - 16 = \boxed{24 \text{ m/s}}$$
---
**ii. Instantaneous acceleration when t = 5s**
$$a = \frac{dv}{dt} = 8 \text{ m/s}^2$$
Acceleration is constant = **8 m/s²** (independent of t)
$$\boxed{a = 8 \text{ m/s}^2}$$
---
**iii. Time when particle is stationary (v = 0)**
$$8t - 16 = 0$$
$$8t = 16$$
$$\boxed{t = 2 \text{ s}}$$
---
### Question 2
### (a) i. Archimedes' Principle
*"When a body is wholly or partially immersed in a fluid, it experiences an upthrust (buoyant force) equal to the weight of the fluid displaced."*
---
### (a) ii. FOUR Characteristics of Pressure in a Fluid
1. Pressure at a point in a fluid acts equally in **all directions**.
2. Pressure increases with **depth** (P = ρgh).
3. Pressure depends on the **density** of the fluid.
4. Pressure acts **perpendicular** to any surface in contact with the fluid.
---
### (b) Compressible vs Incompressible Fluid
| Compressible Fluid | Incompressible Fluid |
|---|---|
| Density changes with applied pressure | Density remains constant regardless of pressure |
| Volume decreases under pressure | Volume does not change under pressure |
| Example: gases (air, steam) | Example: liquids (water, oil) |
---
### (c) Find P₂/P₁
Atmospheric pressure = 20 m of water
Pressure at depth h: **P = P_atm + ρgh** → in metres of water: **P = (20 + h)**
- P₁ (diver M at 60 m) = 20 + 60 = **80 m** of water
- P₂ (diver N at 80 m) = 20 + 80 = **100 m** of water
$$\frac{P_2}{P_1} = \frac{100}{80} = \boxed{1.25}$$
---
### (d) Conservative vs Non-Conservative Force
| Conservative Force | Non-Conservative Force |
|---|---|
| Work done is **independent of path** | Work done **depends on path** |
| Total mechanical energy is conserved | Energy is lost (usually as heat) |
| Work done in a closed loop = 0 | Work done in a closed loop ≠ 0 |
| **Example:** Gravitational force, elastic spring force | **Example:** Friction force, air resistance |
---
## PHY 002: HEAT, WAVES AND OPTICS
### Question 3
### (a) Huygens' Principle
*"Every point on a wavefront acts as a source of secondary wavelets that spread out in the forward direction with the same speed as the wave. The new wavefront is the tangent (envelope) to all these secondary wavelets."*
---
### (b) i. Electromagnetic Spectrum
**Definition:** The electromagnetic spectrum is the complete range of all electromagnetic waves arranged in order of increasing frequency (or decreasing wavelength). These waves all travel at the speed of light (3 × 10⁸ m/s) in a vacuum.
**THREE components:**
1. Radio waves
2. X-rays
3. Visible light (gamma rays, microwaves, infrared, ultraviolet also acceptable)
---
### (b) ii. Young's Double Slit — Separation of Slits
**Given:**
- Fringe separation between 1st and 5th bright fringe: distance spans **4 fringe widths**
- Total distance = 2.5 mm = 2.5 × 10⁻³ m
- λ = 4.5 × 10⁻⁷ m
- D = 0.9 m
**Fringe width:**
$$\beta = \frac{2.5 \times 10^{-3}}{4} = 6.25 \times 10^{-4} \text{ m}$$
**Formula:** $\beta = \frac{\lambda D}{d}$
$$d = \frac{\lambda D}{\beta} = \frac{4.5 \times 10^{-7} \times 0.9}{6.25 \times 10^{-4}}$$
$$d = \frac{4.05 \times 10^{-7}}{6.25 \times 10^{-4}}$$
$$\boxed{d = 6.48 \times 10^{-4} \text{ m} \approx 0.648 \text{ mm}}$$
---
### (c) i. Internal Energy
**Internal energy** is the total sum of the kinetic and potential energies of all the molecules (particles) making up a substance. It includes translational, rotational, and vibrational energies of molecules.
---
### (c) ii. Second Law of Thermodynamics
*"Heat cannot spontaneously flow from a colder body to a hotter body without the expenditure of external work."*
OR (Kelvin-Planck statement): *"It is impossible to construct a heat engine that operates in a cycle and converts all its heat input into work with no other effect."*
---
### (d) i. First Law Expressions
First Law: **ΔU = Q − W**
**Adiabatic process** (no heat exchange, Q = 0):
$$\Delta U = -W \quad \Rightarrow \quad W = -\Delta U$$
**Isochoric process** (constant volume, W = 0 since W = PΔV and ΔV = 0):
$$\Delta U = Q$$
---
### (d) ii. FOUR Factors Affecting Heat Loss by Convection
1. **Temperature difference** between the surface and the surrounding fluid — greater difference increases convection.
2. **Surface area** — larger area allows more heat loss.
3. **Density and viscosity of the fluid** — less dense, less viscous fluids convect more easily.
4. **Velocity/flow of the fluid** — faster fluid movement (forced convection) increases heat loss.
---
### Question 4
### (a) Similarities and Differences: Converging Mirror vs Converging Lens
**TWO Similarities:**
1. Both can produce **real and inverted images** when the object is beyond the focal point.
2. Both produce **virtual, erect, and magnified images** when the object is between the focal point and the optical centre/pole.
**TWO Differences:**
| Converging Mirror (Concave) | Converging Lens |
|---|---|
| Works by **reflection** of light | Works by **refraction** of light |
| Image forms on the **same side** as the object | Image forms on the **opposite side** from the object (for real image) |
---
### (b) i. Uses of Mirrors
**Plane Mirror:**
1. Used in dressing/bathrooms to see one's reflection
2. Used in periscopes and kaleidoscopes
**Concave Mirror:**
1. Used in shaving/makeup mirrors (magnified upright image)
2. Used as reflectors in torches, car headlamps, and satellite dishes
**Convex Mirror:**
1. Used as driving/rear-view mirrors in vehicles (wide field of view)
2. Used in supermarkets and road junctions as security/surveillance mirrors
---
### (b) ii. Completed Table — Concave Mirror
| Position of Object | Position of Image | Size of Image | Nature of Image |
|---|---|---|---|
| At infinity | At F (focal point) | Highly diminished (point) | Real, Inverted |
| At C | At C | Same size as object | Real, Inverted |
| Between C and F | Beyond C | Magnified (enlarged) | Real, Inverted |
| At F | At infinity | Infinitely large | Real, Inverted |
| Between F and P | Behind the mirror (same side as object) | Magnified (enlarged) | Virtual, Erect |
---
## PHY 003: ELECTRICITY AND MAGNETISM
### Question 5
### (a) i. Relative Permittivity
**Relative permittivity (εᵣ)**, also called the **dielectric constant**, is the ratio of the permittivity of a material (ε) to the permittivity of free space (ε₀):
$$\varepsilon_r = \frac{\varepsilon}{\varepsilon_0}$$
It indicates how much more charge a capacitor can store when a dielectric material is inserted compared to vacuum.
---
### (a) ii. TWO Physical Desirable Properties of Dielectric in a Capacitor
1. **High dielectric constant (εᵣ)** — to increase the capacitance.
2. **High dielectric strength** — ability to withstand large electric fields without breaking down (conducting).
---
### (b) Change of Flux Through the Coil
**Given:**
- N = 30 turns
- EMF (ε) = 80 μV = 80 × 10⁻⁶ V
- Time (t) = 3 minutes = 180 s
**Faraday's Law:**
$$\varepsilon = N \frac{\Delta\Phi}{\Delta t}$$
$$\Delta\Phi = \frac{\varepsilon \cdot \Delta t}{N} = \frac{80 \times 10^{-6} \times 180}{30}$$
$$\Delta\Phi = \frac{14.4 \times 10^{-3}}{30}$$
$$\boxed{\Delta\Phi = 4.8 \times 10^{-4} \text{ Wb}}$$
---
### (c) Electron in Uniform Electric Field
**Given:**
- v₀ = 3.00 × 10⁶ m/s
- E = 200 N/C
- l = 0.100 m
- mₑ = 9.11 × 10⁻³¹ kg *(standard value; note: question gives 1.67 × 10⁻³¹ which is closer to proton — using standard electron mass)*
- q = 1.60 × 10⁻¹⁹ C
**i. Acceleration of electron:**
$$F = qE = 1.60 \times 10^{-19} \times 200 = 3.20 \times 10^{-17} \text{ N}$$
$$a = \frac{F}{m} = \frac{3.20 \times 10^{-17}}{9.11 \times 10^{-31}}$$
$$\boxed{a = 3.51 \times 10^{13} \text{ m/s}^2}$$
---
**ii. Time to leave the field:**
The electron travels horizontal distance l = 0.100 m at constant horizontal velocity v₀:
$$t = \frac{l}{v_0} = \frac{0.100}{3.00 \times 10^6}$$
$$\boxed{t = 3.33 \times 10^{-8} \text{ s}}$$
---
### Question 6
### (a) Distinguishing Magnetic and Electric Fields Using a Moving Charged Particle
- In an **electric field**, a charged particle experiences a force **parallel (or anti-parallel) to the field direction**, regardless of whether the particle is moving or stationary.
- In a **magnetic field**, a charged particle experiences a force **only when it is moving**, and the force is always **perpendicular to both the velocity and the magnetic field** (F = qv × B).
Therefore, if a stationary charge is undeflected but a moving charge is deflected in a direction perpendicular to its motion, the field is **magnetic**. If the charge is deflected regardless of motion, the field is **electric**.
---
### (b) Circuit Analysis — Current in Each Resistor
**Given circuit (from Figure 2):**
- Battery 1: 11.0 V with 20.0 Ω (top branch)
- Battery 2: 12.0 V with 17.0 Ω (middle branch)
- 10.0 Ω resistor (bottom branch, no battery)
Using **Kirchhoff's Voltage Law (KVL)** — assign mesh currents I₁ (top loop) and I₂ (bottom loop):
**Mesh 1 (top loop):**
$$11.0 = 20I_1 + 10(I_1 - I_2)$$
$$11.0 = 30I_1 - 10I_2 \quad \text{...(1)}$$
**Mesh 2 (bottom loop):**
$$12.0 = 17I_2 + 10(I_2 - I_1)$$
$$12.0 = -10I_1 + 27I_2 \quad \text{...(2)}$$
**Solving simultaneously:**
From (1): $11 = 30I_1 - 10I_2$ → multiply by 2.7:
$$29.7 = 81I_1 - 27I_2 \quad \text{...(3)}$$
Add (2) + (3):
$$12.0 + 29.7 = -10I_1 + 81I_1$$
$$41.7 = 71I_1$$
$$I_1 = \frac{41.7}{71} = 0.587 \text{ A}$$
Substitute into (1):
$$11 = 30(0.587) - 10I_2$$
$$11 = 17.61 - 10I_2$$
$$10I_2 = 6.61$$
$$I_2 = 0.661 \text{ A}$$
**Currents:**
- **20 Ω resistor:** I₁ = **0.587 A**
- **17 Ω resistor:** I₂ = **0.661 A**
- **10 Ω resistor:** I₁ − I₂ = 0.587 − 0.661 = **−0.074 A** (flows in direction of I₂)
$$\boxed{I_{20\Omega} = 0.587 \text{ A}, \quad I_{17\Omega} = 0.661 \text{ A}, \quad I_{10\Omega} = 0.074 \text{ A}}$$
---
### (c) Speed of Proton in Circular Orbit
**Given:**
- r = 14 cm = 0.14 m
- B = 0.35 T
- mₚ = 1.67 × 10⁻²⁷ kg
- q = 1.60 × 10⁻¹⁹ C
**Using:** $r = \frac{mv}{qB}$
$$v = \frac{qBr}{m} = \frac{1.60 \times 10^{-19} \times 0.35 \times 0.14}{1.67 \times 10^{-27}}$$
$$v = \frac{7.84 \times 10^{-21}}{1.67 \times 10^{-27}}$$
$$\boxed{v = 4.69 \times 10^6 \text{ m/s}}$$
---
## PHY 004: MODERN PHYSICS
### Question 7
### (a) de Broglie Wavelength: λ = h/mv
**Given:** v = 2.30 × 10⁸ m/s, h = 6.626 × 10⁻³⁴ J·s
**i. Electron** (mₑ = 9.11 × 10⁻³¹ kg):
$$\lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 2.30 \times 10^8}$$
$$\lambda = \frac{6.626 \times 10^{-34}}{2.095 \times 10^{-22}}$$
$$\boxed{\lambda_e = 3.16 \times 10^{-12} \text{ m}}$$
**ii. Proton** (mₚ = 1.67 × 10⁻²⁷ kg):
$$\lambda = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 2.30 \times 10^8}$$
$$\lambda = \frac{6.626 \times 10^{-34}}{3.841 \times 10^{-19}}$$
$$\boxed{\lambda_p = 1.72 \times 10^{-15} \text{ m}}$$
**iii. 200 g bullet** (m = 0.200 kg):
$$\lambda = \frac{6.626 \times 10^{-34}}{0.200 \times 2.30 \times 10^8}$$
$$\lambda = \frac{6.626 \times 10^{-34}}{4.60 \times 10^7}$$
$$\boxed{\lambda_{bullet} = 1.44 \times 10^{-41} \text{ m}}$$
*(Negligibly small — confirming macroscopic objects show no wave behaviour)*
---
### (b) i. Properties of α, β, γ Radiation
| Property | Alpha (α) | Beta (β) | Gamma (γ) |
|---|---|---|---|
| **Charge** | +2 (positive) | −1 (negative) | 0 (neutral) |
| **Mass** | 4 u (heavy) | ~1/1836 u (light) | 0 (massless photon) |
| **Ionizing Effect** | Strongly ionizing | Moderately ionizing | Weakly ionizing |
| **Field Effect** | Deflected by E and B fields (toward −ve plate) | Deflected (toward +ve plate) | Not deflected |
| **Penetrating Power** | Least (stopped by paper) | Moderate (stopped by Al) | Most (reduced by thick Pb) |
---
### (b) ii. Effect of Temperature and Pressure on Radioactive Disintegration
**Temperature:** Radioactive decay is a **nuclear process** (occurs in the nucleus). Temperature changes affect only the electron energy levels (chemical changes) and have **no effect** on the rate of radioactive disintegration.
**Pressure:** Similarly, pressure affects the electron cloud and intermolecular distances but **cannot affect the nucleus**. Therefore, pressure also has **no effect** on the rate of radioactive decay.
*This distinguishes radioactive decay from ordinary chemical reactions, which are affected by both temperature and pressure.*
---
### (c) i. Heisenberg's Uncertainty Principle
*"It is impossible to simultaneously determine with perfect accuracy both the position and the momentum (or velocity) of a particle."*
Mathematically:
$$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$
Where Δx = uncertainty in position, Δp = uncertainty in momentum.
---
### (c) ii. FOUR Practical Applications of X-rays
1. **Medical diagnosis** — used in radiography to detect bone fractures and internal injuries.
2. **Cancer treatment (Radiotherapy)** — high-energy X-rays destroy cancerous tumours.
3. **Airport security** — X-ray scanners inspect luggage for contraband items.
4. **Crystallography** — X-ray diffraction is used to determine the crystal structure of materials.
---
### (d) Photon with wavelength λ = 1 Å = 10⁻¹⁰ m
**i. Energy of photon in electron volts:**
$$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{10^{-10}}$$
$$E = \frac{1.988 \times 10^{-25}}{10^{-10}} = 1.988 \times 10^{-15} \text{ J}$$
Converting to eV (1 eV = 1.6 × 10⁻¹⁹ J):
$$E = \frac{1.988 \times 10^{-15}}{1.6 \times 10^{-19}}$$
$$\boxed{E = 12,425 \text{ eV} \approx 12.4 \text{ keV}}$$
---
**ii. Momentum of the photon:**
$$p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{10^{-10}}$$
$$\boxed{p = 6.626 \times 10^{-24} \text{ kg·m/s}}$$
---
### (e) Binding Energy of ⁵⁶Fe (Iron-56)
*(Note: The question states mass number 56 with 56.0930 u — this corresponds to Iron-56, ²⁶Fe₅₆ with 26 protons and 30 neutrons)*
**Given:**
- Mass of nucleus = 56.0930 u *(as given)*
- mₚ = 1.007825 u
- mₙ = 1.008665 u
- 1 u = 931.5 MeV
- Z = 26 (protons), N = 30 (neutrons)
**i. Binding Energy Definition:**
Binding energy is the minimum energy required to completely separate all the nucleons (protons and neutrons) in a nucleus from each other, or equivalently, the energy released when nucleons combine to form the nucleus.
**ii. Calculation:**
**Mass of constituents:**
$$M_{constituents} = 26(1.007825) + 30(1.008665)$$
$$= 26.20345 + 30.25995 = 56.46340 \text{ u}$$
**Mass defect:**
$$\Delta m = 56.46340 - 56.0930 = 0.37040 \text{ u}$$
**Total binding energy:**
$$BE = \Delta m \times 931.5 = 0.37040 \times 931.5$$
$$BE = 345.13 \text{ MeV}$$
**Average binding energy per nucleon:**
$$BE/A = \frac{345.13}{56}$$
$$\boxed{BE/nucleon \approx 6.16 \text{ MeV/nucleon}}$$