## PHY 001: Mechanics and Properties of Matter
### QUESTION 1
A steel cable of original length 4.0 m and cross-sectional area 2.5 × 10⁻⁴ m² is used to support a load of 5000 N. If the Young's modulus of steel is 2.0 × 10¹¹ Pa:
**(a)** Calculate the tensile stress in the cable. [3 marks]
**Solution:**
Stress = Force / Area
Stress = 5000 / (2.5 × 10⁻⁴)
**Stress = 2.0 × 10⁷ Pa (20 MPa)**
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**(b)** Determine the extension produced in the cable. [4 marks]
**Solution:**
E = (F × L) / (A × e)
Rearranging: e = (F × L) / (A × E)
e = (5000 × 4.0) / (2.5 × 10⁻⁴ × 2.0 × 10¹¹)
e = 20000 / (5.0 × 10⁷)
**e = 4.0 × 10⁻⁴ m (0.4 mm)**
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**(c)** If the cable is replaced with another of the same material but twice the cross-sectional area, what would be the new extension under the same load? Explain. [3 marks]
**Solution:**
From e = (F × L) / (A × E), extension is inversely proportional to area (e ∝ 1/A).
If area doubles, extension halves:
New extension = 4.0 × 10⁻⁴ / 2 = **2.0 × 10⁻⁴ m (0.2 mm)**
---
### QUESTION 2
A spherical metal ball of radius 2.0 mm and density 7800 kg/m³ is dropped into a tall cylinder containing glycerine of viscosity 1.5 Pa·s and density 1260 kg/m³.
**(a)** Explain what is meant by terminal velocity. [2 marks]
**Solution:**
Terminal velocity is the constant maximum velocity attained by a falling object when the viscous drag force plus upthrust exactly equals the weight of the object, resulting in zero net force and zero acceleration.
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**(b)** Calculate the terminal velocity of the ball in the glycerine. (g = 10 m/s²) [5 marks]
**Solution:**
Using Stokes' Law at terminal velocity:
6πηrv = (4/3)πr³(ρ_ball − ρ_fluid)g
Simplifying:
v = 2r²(ρ_ball − ρ_fluid)g / 9η
Given:
r = 2.0 × 10⁻³ m, ρ_ball = 7800 kg/m³, ρ_fluid = 1260 kg/m³, η = 1.5 Pa·s, g = 10 m/s²
v = [2 × (2.0 × 10⁻³)² × (7800 − 1260) × 10] / (9 × 1.5)
v = [2 × 4.0 × 10⁻⁶ × 6540 × 10] / 13.5
v = 523200 × 10⁻⁶ / 13.5
**v ≈ 0.0387 m/s (3.87 cm/s)**
---
**(c)** State two factors that would increase the terminal velocity of the ball. [3 marks]
1. Increasing the radius (size) of the ball
2. Increasing the density of the ball
3. Decreasing the viscosity of the fluid
4. Decreasing the density of the fluid
*(Any two of the above)*
---
## PHY 002: Heat, Waves, and Optics
### QUESTION 3
A progressive wave travelling along a string is represented by:
y = 0.05 sin(40πt − 0.25πx)
where y and x are in metres and t is in seconds.
**(a)(i)** Amplitude [2 marks]
Comparing with y = A sin(ωt − kx):
**A = 0.05 m (5 cm)**
---
**(a)(ii)** Frequency [2 marks]
Angular frequency ω = 40π rad/s
Since ω = 2πf:
f = 40π / 2π
**f = 20 Hz**
---
**(a)(iii)** Wavelength [3 marks]
Wave number k = 0.25π m⁻¹
Since k = 2π/λ:
λ = 2π / (0.25π) = 2 / 0.25
**λ = 8 m**
---
**(b)** Calculate the speed of the wave. [3 marks]
Method 1: v = fλ = 20 × 8 = **160 m/s**
Method 2: v = ω/k = 40π / (0.25π) = **160 m/s**
---
### QUESTION 4
Monochromatic light of wavelength 589 nm is incident normally on a diffraction grating.
**(a)** If the second-order maximum is observed at 30° to the normal, calculate the number of lines per millimetre. [5 marks]
**Solution:**
d sin θ = nλ
n = 2, θ = 30°, λ = 589 × 10⁻⁹ m
d × sin 30° = 2 × 589 × 10⁻⁹
d × 0.5 = 1178 × 10⁻⁹
d = 2356 × 10⁻⁹ m = 2.356 × 10⁻⁶ m
Lines per metre = 1/d = 1/(2.356 × 10⁻⁶) = 424,406 lines/m
**Lines per mm ≈ 424 lines/mm**
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**(b)** Determine the total number of maxima observable. [3 marks]
**Solution:**
Maximum order when sin θ = 1:
n_max = d/λ = (2.356 × 10⁻⁶) / (589 × 10⁻⁹) = 4.0
Total maxima = 2n_max + 1 (both sides plus central maximum)
= 2(4) + 1 = **9 maxima**
---
**(c)** Explain why there is a limit to the number of observable maxima. [2 marks]
The limit exists because the angle of diffraction cannot exceed 90° — that is, sin θ cannot exceed 1. Beyond this, the diffracted rays would need to travel backwards, which is physically impossible.
---
## PHY 003: Electricity and Magnetism
### QUESTION 5
Three point charges: Q₁ = +4.0 μC at x = 0 cm, Q₂ = −6.0 μC at x = 3.0 cm, Q₃ = +2.0 μC at x = 7.0 cm.
**(a)** Define electric potential at a point in an electric field. [2 marks]
Electric potential at a point in an electric field is the work done per unit positive charge in bringing a small test charge from infinity to that point. Equivalently, it is the electric potential energy per unit charge at that point.
---
**(b)** Calculate the electric potential at point P (x = 5.0 cm) due to all three charges. (k = 9.0 × 10⁹ N·m²/C²) [6 marks]
**Solution:**
V = kQ/r for each charge; total V = V₁ + V₂ + V₃
Distances from P at x = 5.0 cm:
r₁ = |5.0 − 0| = 5.0 cm = 0.05 m
r₂ = |5.0 − 3.0| = 2.0 cm = 0.02 m
r₃ = |7.0 − 5.0| = 2.0 cm = 0.02 m
V₁ = (9.0 × 10⁹ × 4.0 × 10⁻⁶) / 0.05 = **+720,000 V**
V₂ = (9.0 × 10⁹ × (−6.0 × 10⁻⁶)) / 0.02 = **−2,700,000 V**
V₃ = (9.0 × 10⁹ × 2.0 × 10⁻⁶) / 0.02 = **+900,000 V**
Total V = 720,000 − 2,700,000 + 900,000
**V = −1,080,000 V = −1.08 MV**
---
**(c)** Determine the work done in bringing a charge of +1.0 μC from infinity to point P. [2 marks]
**Solution:**
W = qV = 1.0 × 10⁻⁶ × (−1.08 × 10⁶)
**W = −1.08 J**
(The negative sign indicates work is done by the electric field, not against it.)
---
### QUESTION 6
A parallel plate capacitor has plate area 0.04 m², plate separation 2.0 mm, and dielectric of relative permittivity 4.5.
**(a)** Calculate the capacitance. (ε₀ = 8.85 × 10⁻¹² F/m) [4 marks]
**Solution:**
C = ε₀εᵣA / d
C = (8.85 × 10⁻¹² × 4.5 × 0.04) / (2.0 × 10⁻³)
C = (1.593 × 10⁻¹²) / (2.0 × 10⁻³)
**C ≈ 7.97 × 10⁻¹⁰ F (796.5 pF)**
---
**(b)(i)** Charge stored on each plate [2 marks]
Q = CV = 7.965 × 10⁻¹⁰ × 250
**Q ≈ 1.99 × 10⁻⁷ C (199 nC)**
---
**(b)(ii)** Energy stored [2 marks]
E = ½CV²
E = 0.5 × 7.965 × 10⁻¹⁰ × (250)²
E = 0.5 × 7.965 × 10⁻¹⁰ × 62500
**E ≈ 2.49 × 10⁻⁵ J (24.9 μJ)**
---
**(c)** What happens to the stored energy if the dielectric is removed while the capacitor remains connected to the 250 V supply? [2 marks]
With the capacitor still connected to the supply, the voltage remains constant at 250 V. Removing the dielectric reduces the capacitance by a factor of εᵣ = 4.5. Since E = ½CV² and V is fixed, the stored energy decreases by the same factor:
New energy = 24.9 μJ / 4.5 ≈ **5.53 μJ**
The stored energy **decreases** because capacitance decreases while voltage remains constant.
---
## PHY 004: Modern Physics
### QUESTION 7
Light of wavelength 450 nm is incident on a photosensitive surface with work function 2.1 eV.
**(a)** State Einstein's photoelectric equation. [2 marks]
**hf = φ + KE_max**
where hf is the energy of the incident photon, φ is the work function of the surface, and KE_max is the maximum kinetic energy of the emitted photoelectron.
---
**(b)(i)** Energy of incident photons in joules [3 marks]
E = hc/λ
E = (6.63 × 10⁻³⁴ × 3.0 × 10⁸) / (450 × 10⁻⁹)
E = (19.89 × 10⁻²⁶) / (4.50 × 10⁻⁷)
**E = 4.42 × 10⁻¹⁹ J**
---
**(b)(ii)** Maximum kinetic energy of emitted photoelectrons [3 marks]
Convert work function:
φ = 2.1 × 1.6 × 10⁻¹⁹ = 3.36 × 10⁻¹⁹ J
KE_max = E − φ = 4.42 × 10⁻¹⁹ − 3.36 × 10⁻¹⁹
**KE_max = 1.06 × 10⁻¹⁹ J (0.66 eV)**
---
**(c)** Determine the threshold wavelength. [2 marks]
At threshold: hc/λ₀ = φ
λ₀ = hc/φ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸) / (3.36 × 10⁻¹⁹)
**λ₀ = 5.92 × 10⁻⁷ m = 592 nm**
---
### QUESTION 8
**(a)** Distinguish between intrinsic and extrinsic semiconductors, giving one example of each. [4 marks]
**Intrinsic semiconductor:**
A pure semiconductor material containing no significant impurities. Its electrical conductivity depends entirely on thermal excitation of electrons across the band gap. Equal numbers of electrons and holes are present.
*Example: Pure silicon (Si) or pure germanium (Ge)*
**Extrinsic semiconductor:**
A semiconductor that has been doped with controlled amounts of impurity atoms to significantly enhance its electrical conductivity. Electrons and holes are present in unequal numbers, with one type dominating as the majority carrier.
*Example: Silicon doped with phosphorus (n-type) or silicon doped with boron (p-type)*
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**(b)(i)** Doping with pentavalent impurity atoms [3 marks]
When a pentavalent impurity (5 valence electrons) such as phosphorus (P), arsenic (As), or antimony (Sb) is added to silicon:
- Four of the five valence electrons form covalent bonds with neighbouring silicon atoms.
- The fifth electron is loosely bound and is easily liberated as a free electron.
- This produces an **n-type semiconductor**, with electrons as the majority carriers and holes as minority carriers.
- The impurity atoms are called **donor atoms** because they donate free electrons to the lattice.
---
**(b)(ii)** Doping with trivalent impurity atoms [3 marks]
When a trivalent impurity (3 valence electrons) such as boron (B), aluminium (Al), or gallium (Ga) is added to silicon:
- The three valence electrons form covalent bonds with three neighbouring silicon atoms.
- The fourth covalent bond is incomplete, creating a **hole** — an absence of an electron that acts as a positive charge carrier.
- This produces a **p-type semiconductor**, with holes as the majority carriers and electrons as minority carriers.
- The impurity atoms are called **acceptor atoms** because they accept electrons from the lattice.
---
**(c)** Balmer series wavelength for transition n = 4 → n = 2. [5 marks]
**Solution:**
Rydberg formula: 1/λ = R(1/n₁² − 1/n₂²)
With n₁ = 2, n₂ = 4, R = 1.097 × 10⁷ m⁻¹:
1/λ = 1.097 × 10⁷ × (1/4 − 1/16)
1/λ = 1.097 × 10⁷ × (4/16 − 1/16)
1/λ = 1.097 × 10⁷ × (3/16)
1/λ = 1.097 × 10⁷ × 0.1875
1/λ = 2.057 × 10⁶ m⁻¹
λ = 1 / (2.057 × 10⁶)
**λ = 4.86 × 10⁻⁷ m = 486 nm**