**Useful Constants:**
- Charge on the electron (e) = −1.6 × 10⁻¹⁹ C
- Permittivity of free space (ε₀) = 8.85 × 10⁻¹² Fm⁻¹
- Velocity of light (c) = 3.0 × 10⁸ ms⁻¹
- Planck's constant (h) = 6.6 × 10⁻³⁴ Js
---
## **SECTION A**
**Attempt All Questions**
**1.** A monochromatic light propagating from medium 1 of refractive index n₁ to medium 2 of refractive index n₂. If the angle of incidence is î, and refracted angle is r̂. Under what condition or relationship between n₁ and n₂ will the refraction produce the following results:
- (i) î = r̂
- (ii) î < r̂
- (iii) î > r̂
**2.** Calculate the focal length of a combination of a concave lens of focal length 20 cm and a convex lens of focal length 30 cm in contact. Is the system a converging or diverging lens? (Ignore the thickness of the lenses).
**3.** Consider an object AB placed in front of a concave mirror (object is placed beyond the center of curvature C). Copy and complete the ray diagram showing the image formation of the object. What is the nature of the image formed?
**4.** How can you get an effective capacitance of 6μF from three capacitors each of capacitance 4μF? Show the calculations and draw the circuit diagram of the final arrangement.
**5.** State any four factors upon which resistance of a resistor depends. An electric kettle has a resistance of 30 Ω. What is the power rating of the kettle when it is connected to a 240V power supply?
**6.** The magnetic materials can be classified into three categories. Mention them and give one example for each. What is Curie temperature as applied to magnetism?
**7.** Given the instantaneous current in an a.c circuit, I = 1.2sin31.4t, find the maximum current, rms current and the frequency of the current.
**8.** Consider X-rays produced by 30 kV, find the maximum frequency and minimum wavelength of the most accelerated electrons.
**9.** If 1 g of radioactive substance takes 50 sec to lose 0.01 g. Find its half-life.
**10.** The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 5896 Å is 0.63 Volt. Calculate the maximum kinetic energy of photoelectrons.
---
## **SECTION B: GEOMETRIC OPTICS (20 marks)**
**Attempt Only One (1) Question from this Section**
**11a. (i)** Write down the image of the word **pqf** when viewed by a plane mirror. What is the name of this phenomena? Imagine the image of an object placed 12 cm in front of a plane mirror. Describe the nature and characteristics of the image.
**(ii)** Calculate the angle of minimum deviation and angle of incidence of a light ray on an equilateral triangular prism of refractive index √(7/3).
**(iii)** An object 2 cm high is placed 16 cm infront of a concave mirror which produced an image 3 cm high. Find the radius of curvature of the mirror.
**11b. (i)** Write down the necessary conditions for the total internal reflection to occur. Find the critical angle for a rectangular slab of glass, with refractive index 1.57 in air.
**(ii)** Distinguish between a Plane Mirror, a Concave Mirror and a Convex Mirror by viewing an object close to each mirror, without touching them (ray diagram is not needed). Also state one similarity for these mirrors.
**(iii)** A small object is placed 20 cm in front of (1) a concave mirror of radius of curvature 30 cm. (2) a convex mirror of focal length 12 cm. In each case find the position, nature and the magnification of the image. Also find the new position, nature and the magnification of the image when the object shifts towards the mirror by 7 cm.
**12a.** Consider a refraction of a light ray through a parallel sided rectangular glass slab of thickness 't' which incidents on the slab at an angle of incidence î with respect to the angle of refraction r̂. Finally the ray emerges out in the direction parallel to the initial ray. If refractive index of the glass used is μ.
**(i)** Sketch the ray diagram and indicate clearly the **lateral displacement**. Hence, prove that the expression for the lateral displacement (d) is given by:
**d = t sin(î − r̂) / cos r**
Define the underlined term, and deduce the expression for d near normal incidence in terms of î and μ (i.e at a very small angle of incidence). What would be the value of d when î = 90°?
**12b. (i)** Considering a concave reflector of a telescope of radius of curvature 250 cm. What is the focal length of the eyepiece to obtain a magnification 50 times?
**(ii)** Use ray diagram to briefly explain the nature of the focal length when parallel beam of light is incident on (A) convex lens (B) concave lens.
**(iii)** An object is placed 12 cm from (A) a converging lens (B) diverging lens: of focal length 18 cm. Find the nature of the image.
---
## **SECTION C: ELECTRICITY AND MAGNETISM (40 marks)**
**Attempt Only Two (2) Questions from this Section**
**13a. (i)** Determine the physical quantity whose SI unit is JC⁻¹, VA, and ΩA. In each case, state whether it is a scalar or a vector quantity. The Coulombic force on a small sphere of charge 0.8μC and 0.4μC in vacuum is 0.2 N. Calculate the separation between the spheres.
**(ii)** State the necessary condition(s) for which Ohm's law is valid and define a resistance. A current of 6A passes through a copper wire of length 12 mm and 3 mm² cross section. Calculate the current density.
**(iii)** What is the charge and energy stored in the circuit shown in figure 2?
*(Figure 2: 12 μF and 6 μF capacitors in series, with 2 μF in parallel, connected to a 24 V supply)*
**13b. (i)** A 60 V battery is connected across an appliance having a conductance of 0.025 S. Determine the current flowing in the appliance, the power consumed, the energy dissipated and quantity of charges transferred in 10 minutes.
**(ii)** Consider a capacitor of capacitive reactance of 4Ω at 250 Hz. Find the capacitance of the circuit. Calculate the reactance at 200/π Hz and the rms current, when it is connected to a 220 V 50 Hz a.c supply.
**14a. (i)** Define the following terms: *Magnetic field intensity, Magnetic permeability* and *Magnetic susceptibility.*
**(ii)** Calculate the magnetic field strength and the magnetomotive force (mmf) required to produce a flux density of 0.25 mT in an air gap of length 12 mm.
**14b. (i)** A straight conductor carries a current of 5.3 A. What is the magnitude of magnetic field at point 20 cm from the conductor?
**(ii)** When the electron in a thermionic tube travels at 3×10⁷ ms⁻¹ inclined at cos⁻¹(0.168) to a field of flux density 18.7T. Calculate the force exerted on the electron in the field.
**(iii)** A flux of 0.025 Wb links with a 1500 turns coil and a current of 3 A passes through the coil. Calculate the inductance of the coil, the energy stored in the magnetic field, and the average e.m.f. induced if the current falls to zero in 0.150 μs.
**15a. (i)** Define the terms capacitor and capacitance of a capacitor. Four capacitors of capacitance C₁, C₂, C₃, C₄ are connected in series across a p.d V volt. Draw the circuit diagram and derive the expressions for the equivalent capacitance. Also find the effective capacitance when C₁ = C₄ = 8nF, C₂ = C₃ = 20nF.
**(ii)** The plates of a parallel plate capacitor have area 90 cm² each separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. How much electrostatic energy is stored by the capacitor?
**15b. (i)** In terms of resistance R, the temperature coefficient of resistance of a material when the temperature variations is not too large can be written as:
**α = (R₂ − R₁) / R₁ΔT**
Express α in terms of resistivity ρ.
**(ii)** A given wire has a resistance 2.1 Ω at 27.5 °C and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of the resistance of the wire.
**(iii)** Assume a straight copper wire 0.30 m long moves at a uniform speed of 4 ms⁻¹ perpendicular to a uniform magnetic field of flux density 1.25 T. What is the current flowing in the conductor when (A) its ends are open-circuited. (B) its ends are connected to a load of 2 Ω resistor?
**16a. (i)** State the necessary condition(s) for a resonance frequency (f₀) to occur in an L-R-C circuit. What happens to the current at resonance condition? Hence derive the expression for the resonance frequency (f₀).
**(ii)** A coil takes a current of 2 A from a 12 V d.c. supply. When connected to a 240 V, 50 Hz a.c supply the current is 20 A. Calculate the resistance, impedance, inductive reactance, Susceptance, admittance and inductance of the coil.
**16b.** Use Kirchhoff's laws to find the current in the 12Ω resistor in the circuit shown in figure 3.
*(Figure 3: Circuit with 8Ω and 6Ω resistors in one branch, 28V source, 12Ω and 4Ω resistors, and 10V source)*
---
## **SECTION D: MODERN PHYSICS (20 marks)**
**Attempt Only One (1) Question from this Section**
**17a. (i)** Briefly explain the concept of the de Broglie hypothesis. Hence, deduce the wavelength connections between particle and wave.
**(ii)** Assume the wavelength of light from the spectral emission of sodium is 589 nm. Calculate the kinetic energy (in eV) of an electron and a neutron at the same de Broglie wavelength.
**17b. (i)** Define the term natural radioactivity. Hence, state the radiations emitted by a radioactive element in the order of increasing ionization power.
**(ii)** The Radioactive decay law states that: *The activity of a radioactive element at any instant is directly proportional to the number of undecayed active atoms present at that instant.* Prove this law.
**(iii)** Considering a radioactive isotope G — decays sequence: ˣ_yG → G₁ →¹⁷⁶₅₁ G₂. When G decays to G₁, α-particle is released. When G₁ decays to G₂, β-particle is released. Determine the value of x and y. Also how many neutrons do the G₁-isotope have?
**18a. (i)** For hydrogen atom, the ground state energy is −13.6 eV. If an electron makes a transition from an energy level −1.51 eV to −3.4 eV. Calculate the wavelength of the spectral line emitted.
**(ii)** What is mass defect? Determine the binding energy of an α-particle. Given that: Mass of the helium nucleus = ⁴₂He = 4.001265 u, mass of proton = 1.007277 u, mass of neutron = 1.008665 u.
**(iii)** Distinguish between: *Nuclear Fission* and *Nuclear Fusion.*
**18b. (i)** What do you understand by the terms: *thermionic Emission* and *field emission.*
**(ii)** In the process of photoelectric effects, briefly explain the effects of "*intensity and frequency of the radiation*" when light is shone on a metal surface.
**(iii)** In an experiment, a tungsten cathode which has a threshold wavelength of 2300 Å is irradiated by UV light of wavelength 1800 Å. Calculate the maximum energy of the emitted photoelectron and work function for tungsten in electron volt.
# 2024 IJMBE PHYSICS PAPER II — SOLUTIONS
## SECTION A
**QUESTION 1**
Using Snell's Law: n₁ sin î = n₂ sin r̂
**(i) î = r̂ when n₁ = n₂**
Both media have equal refractive indices (same medium).
**(ii) î < r̂ when n₁ > n₂**
Light travels from denser to less dense medium; it bends away from the normal.
**(iii) î > r̂ when n₁ < n₂**
Light travels from less dense to denser medium; it bends towards the normal.
---
**QUESTION 2**
Using lens combination formula:
1/f = 1/f₁ + 1/f₂
f₁ = −20 cm (concave), f₂ = +30 cm (convex)
1/f = 1/(−20) + 1/30 = −3/60 + 2/60 = −1/60
**f = −60 cm**
The negative sign indicates the system is a **diverging (concave) lens.**
---
**QUESTION 3**
For an object placed beyond C in front of a concave mirror:
**Ray diagram construction:**
- Ray 1: Parallel to principal axis → reflects through focal point F
- Ray 2: Through centre of curvature C → reflects back along same path
- Image forms between F and C, in front of mirror
**Nature of image:**
- Real
- Inverted
- Diminished (smaller than object)
---
**QUESTION 4**
We need 6μF from three 4μF capacitors.
**Arrangement:** Connect two capacitors in series, then connect the third in parallel with the series combination.
**Two in series:**
1/Cₛ = 1/4 + 1/4 = 2/4
Cₛ = 2μF
**Then parallel with third:**
C_total = 2 + 4 = **6μF ✓**
**Circuit diagram:**
```
┌──[4μF]──[4μF]──┐
────┤ ├────
└────[4μF]────────┘
```
---
**QUESTION 5**
**Four factors affecting resistance:**
1. Length (R ∝ L)
2. Cross-sectional area (R ∝ 1/A)
3. Nature/type of material (resistivity ρ)
4. Temperature
**Power of kettle:**
P = V²/R = (240)²/30 = 57600/30 = **1920 W = 1.92 kW**
---
**QUESTION 6**
**Three categories of magnetic materials:**
| Category | Example |
|---|---|
| Ferromagnetic | Iron (Fe) |
| Paramagnetic | Aluminium (Al) |
| Diamagnetic | Copper (Cu) |
**Curie Temperature:**
This is the temperature above which a ferromagnetic material loses its ferromagnetic properties and becomes paramagnetic. Beyond this temperature, thermal agitation destroys the magnetic domain alignment.
---
**QUESTION 7**
Given: I = 1.2 sin 31.4t
Comparing with I = I₀ sin ωt:
**Maximum current (I₀):**
I₀ = **1.2 A**
**RMS current:**
I_rms = I₀/√2 = 1.2/√2 = 1.2/1.414 = **0.849 A**
**Frequency:**
ω = 31.4 rad/s
f = ω/2π = 31.4/(2 × 3.14) = 31.4/6.28 = **5 Hz**
---
**QUESTION 8**
Given: V = 30 kV = 30,000 V
**Maximum frequency:**
eV = hf_max
f_max = eV/h = (1.6×10⁻¹⁹ × 30,000)/(6.6×10⁻³⁴)
f_max = (4.8×10⁻¹⁵)/(6.6×10⁻³⁴)
**f_max = 7.27 × 10¹⁸ Hz**
**Minimum wavelength:**
λ_min = c/f_max = (3×10⁸)/(7.27×10¹⁸)
**λ_min = 4.13 × 10⁻¹¹ m = 0.0413 nm**
---
**QUESTION 9**
Given: Initial mass = 1 g, mass lost = 0.01 g in 50 sec
Remaining mass = 1 − 0.01 = 0.99 g
Using: N = N₀(1/2)^(t/t½)
0.99 = 1 × (1/2)^(50/t½)
ln(0.99) = (50/t½) × ln(0.5)
−0.01005 = (50/t½)(−0.6931)
t½ = (50 × 0.6931)/0.01005
**t½ = 3448 seconds ≈ 3.45 × 10³ s**
---
**QUESTION 10**
Given: λ = 5896 Å = 5896×10⁻¹⁰ m, V₀ = 0.63 V
Maximum kinetic energy of photoelectrons:
KE_max = eV₀
KE_max = 1.6×10⁻¹⁹ × 0.63
**KE_max = 1.008 × 10⁻¹⁹ J**
In electron volts: **KE_max = 0.63 eV**
---
## SECTION B: GEOMETRIC OPTICS
---
**QUESTION 11a**
**(i) Image of "pqf" in plane mirror:**
The image is **"bqd"** (lateral inversion)
The phenomenon is called **Lateral Inversion.**
**Characteristics of image in plane mirror (object 12 cm in front):**
- Virtual
- Erect (upright)
- Same size as object
- Located 12 cm behind the mirror
- Laterally inverted
**(ii) Equilateral prism, μ = √(7/3)**
For equilateral prism: A = 60°
Using: μ = sin[(A + D_min)/2] / sin(A/2)
√(7/3) = sin[(60 + D_min)/2] / sin 30°
√(7/3) = sin[(60 + D_min)/2] / 0.5
sin[(60 + D_min)/2] = 0.5 × √(7/3) = 0.5 × 1.528 = 0.7638
(60 + D_min)/2 = sin⁻¹(0.7638) = 49.8°
60 + D_min = 99.6°
**D_min = 39.6° ≈ 40°**
**Angle of incidence:**
i = (A + D_min)/2 = (60 + 40)/2 = **50°**
**(iii) Concave mirror, object height h₁ = 2 cm, image height h₂ = 3 cm, u = 16 cm**
Magnification: m = h₂/h₁ = 3/2 = 1.5
Also m = −v/u → v = −1.5 × 16 = −24 cm (real image, same side)
Using mirror formula:
1/f = 1/v + 1/u = 1/(−24) + 1/(−16)
1/f = −1/24 − 1/16 = −2/48 − 3/48 = −5/48
f = −48/5 = −9.6 cm
**Radius of curvature R = 2f = 2 × 9.6 = 19.2 cm**
---
**QUESTION 11b**
**(i) Conditions for Total Internal Reflection:**
1. Light must travel from a denser to a less dense medium
2. The angle of incidence must be greater than or equal to the critical angle
**Critical angle for glass (n = 1.57) in air:**
sin C = 1/n = 1/1.57 = 0.6369
**C = sin⁻¹(0.6369) = 39.6°**
**(ii) Distinctions:**
| Feature | Plane Mirror | Concave Mirror | Convex Mirror |
|---|---|---|---|
| Object close to mirror | Virtual, erect, same size | Virtual, erect, magnified | Virtual, erect, diminished |
| Surface | Flat | Curved inward | Curved outward |
| Focus | No real focus | Real focus | Virtual focus |
**Similarity:** All three mirrors obey the law of reflection.
**(iii)**
**Case 1: Concave mirror, R = 30 cm → f = 15 cm, u = −20 cm**
1/v = 1/f − 1/u = 1/(−15) − 1/(−20) = −1/15 + 1/20 = −4/60 + 3/60 = −1/60
v = −60 cm
m = −v/u = −(−60)/(−20) = **−3**
Image: Real, Inverted, Magnified (×3), 60 cm in front of mirror
**When object shifts 7 cm closer → u = −13 cm:**
1/v = 1/(−15) − 1/(−13) = −1/15 + 1/13 = (−13+15)/195 = 2/195
v = +97.5 cm (behind mirror)
m = −97.5/(−13) = +7.5
Image: Virtual, Erect, Magnified (×7.5)
---
**Case 2: Convex mirror, f = +12 cm, u = −20 cm**
1/v = 1/12 − 1/(−20) = 1/12 + 1/20 = 5/60 + 3/60 = 8/60
v = +7.5 cm
m = −v/u = −7.5/(−20) = **+0.375**
Image: Virtual, Erect, Diminished, 7.5 cm behind mirror
**When object shifts 7 cm closer → u = −13 cm:**
1/v = 1/12 + 1/13 = 13/156 + 12/156 = 25/156
v = +6.24 cm
m = −6.24/(−13) = +0.48
Image: Virtual, Erect, Diminished (×0.48)
---
**QUESTION 12a**
**(i) Lateral displacement:**
From geometry of the refracted ray inside the glass slab:
The path length inside the slab = t/cos r
The lateral displacement d = (t/cos r) × sin(î − r̂)
**∴ d = t sin(î − r̂) / cos r ✓**
**Definition:** Lateral displacement is the perpendicular distance between the incident ray and the emergent ray.
**Near normal incidence (small angle approximation):**
sin(î − r̂) ≈ î − r̂, cos r ≈ 1
From Snell's law: sin î ≈ î = μ sin r ≈ μr → r = î/μ
d ≈ t(î − î/μ) = tî(1 − 1/μ)
**d = tî(μ − 1)/μ**
**When î = 90°:**
d = t sin(90° − r)/cos r = t cos r/cos r = **d = t**
The lateral displacement equals the thickness of the slab.
**12b. (i)** Concave reflector telescope, R = 250 cm
Focal length of objective mirror:
f₀ = R/2 = 250/2 = 125 cm
Magnification = f₀/fₑ
50 = 125/fₑ
**fₑ = 125/50 = 2.5 cm**
**(ii)**
- **(A) Convex lens:** Parallel rays converge to a **real focal point** on the other side of the lens. Focal length is **positive.**
- **(B) Concave lens:** Parallel rays diverge; they appear to come from a **virtual focal point** on the same side as the incident light. Focal length is **negative.**
**(iii) Object at u = −12 cm, f = 18 cm**
**A) Converging lens (f = +18 cm):**
1/v = 1/f + 1/u = 1/18 + 1/(−12) = 2/36 − 3/36 = −1/36
v = −36 cm
Image is **virtual, erect, magnified** (on same side as object)
**B) Diverging lens (f = −18 cm):**
1/v = 1/(−18) + 1/(−12) = −1/18 − 1/12 = −2/36 − 3/36 = −5/36
v = −7.2 cm
Image is **virtual, erect, diminished** (on same side as object)
---
## SECTION C: ELECTRICITY AND MAGNETISM
---
**QUESTION 13a**
**(i) Physical quantities:**
| SI Unit | Physical Quantity | Scalar/Vector |
|---|---|---|
| JC⁻¹ | Electric potential (Voltage) | Scalar |
| VA | Power (Volt-Ampere) | Scalar |
| ΩA | Voltage/EMF | Scalar |
**Separation between spheres:**
F = kq₁q₂/r²
0.2 = (9×10⁹ × 0.8×10⁻⁶ × 0.4×10⁻⁶)/r²
0.2 = (9×10⁹ × 3.2×10⁻¹³)/r²
0.2 = 2.88×10⁻³/r²
r² = 2.88×10⁻³/0.2 = 0.0144
**r = 0.12 m = 12 cm**
**(ii) Conditions for Ohm's law:**
1. Temperature must remain constant
2. Physical conditions of the conductor must not change
**Definition:** Resistance is the opposition offered by a conductor to the flow of electric current. R = V/I (unit: Ohm, Ω)
**Current density:**
J = I/A = 6/(3×10⁻⁶) = **2×10⁶ A/m²**
**(iii) Charge and energy in Figure 2:**
12 μF and 6 μF in series:
1/Cₛ = 1/12 + 1/6 = 1/12 + 2/12 = 3/12
Cₛ = 4 μF
Total capacitance (4 μF parallel with 2 μF):
C_total = 4 + 2 = **6 μF**
**Charge:** Q = CV = 6×10⁻⁶ × 24 = **1.44×10⁻⁴ C = 144 μC**
**Energy:** E = ½CV² = ½ × 6×10⁻⁶ × 24² = ½ × 6×10⁻⁶ × 576 = **1.728×10⁻³ J = 1.728 mJ**
---
**QUESTION 13b**
**(i) G = 0.025 S, V = 60 V, t = 10 min = 600 s**
R = 1/G = 1/0.025 = 40 Ω
**Current:** I = V/R = 60/40 = **1.5 A**
**Power:** P = VI = 60 × 1.5 = **90 W**
**Energy dissipated:** E = Pt = 90 × 600 = **54,000 J = 54 kJ**
**Charge transferred:** Q = It = 1.5 × 600 = **900 C**
**(ii) Xc = 4Ω at f = 250 Hz**
Capacitance:
Xc = 1/(2πfC)
C = 1/(2π × 250 × 4) = 1/6283.2 = **1.59×10⁻⁴ F = 159 μF**
**Reactance at f = 200/π Hz:**
Xc = 1/(2π × (200/π) × 159×10⁻⁶)
= 1/(2 × 200 × 159×10⁻⁶)
= 1/(0.0636) = **15.72 Ω**
**RMS current at 220V, 50 Hz:**
Xc = 1/(2π × 50 × 159×10⁻⁶) = 1/0.04995 = 20.02 Ω
I_rms = V/Xc = 220/20.02 = **10.99 A ≈ 11 A**
---
**QUESTION 14a**
**(i) Definitions:**
- **Magnetic Field Intensity (H):** The magnetizing force per unit length that produces a magnetic field in a medium. H = B/μ (unit: A/m)
- **Magnetic Permeability (μ):** The ability of a material to support the formation of a magnetic field within itself. μ = B/H (unit: Hm⁻¹)
- **Magnetic Susceptibility (χ):** The ratio of magnetization of a material to the applied magnetic field intensity. χ = M/H (dimensionless)
**(ii) B = 0.25 mT = 0.25×10⁻³ T, l = 12 mm = 12×10⁻³ m**
**Magnetic field strength:**
H = B/μ₀ = (0.25×10⁻³)/(4π×10⁻⁷)
H = (0.25×10⁻³)/(1.257×10⁻⁶)
**H = 198.9 A/m ≈ 199 A/m**
**MMF:**
MMF = H × l = 199 × 12×10⁻³
**MMF = 2.39 A-turns**
---
**QUESTION 14b**
**(i) Magnetic field from straight conductor:**
B = μ₀I/(2πr)
B = (4π×10⁻⁷ × 5.3)/(2π × 0.20)
B = (4π×10⁻⁷ × 5.3)/(2π × 0.20)
B = (2×10⁻⁷ × 5.3)/0.20
**B = 5.3×10⁻⁶ T = 5.3 μT**
**(ii) Electron in thermionic tube:**
v = 3×10⁷ ms⁻¹, B = 18.7 T, θ = cos⁻¹(0.168)
cos θ = 0.168 → sin θ = √(1 − 0.168²) = √(1 − 0.02822) = √0.9718 = 0.9858
F = qvB sin θ
F = 1.6×10⁻¹⁹ × 3×10⁷ × 18.7 × 0.9858
F = 1.6×10⁻¹⁹ × 3×10⁷ × 18.433
F = 1.6×10⁻¹⁹ × 5.53×10⁸
**F = 8.85×10⁻¹¹ N**
**(iii) Coil: Φ = 0.025 Wb, N = 1500, I = 3A, t = 0.150 μs**
**Inductance:**
L = NΦ/I = (1500 × 0.025)/3 = 37.5/3 = **12.5 H**
**Energy stored:**
E = ½LI² = ½ × 12.5 × 3² = ½ × 12.5 × 9 = **56.25 J**
**Average EMF:**
EMF = L × ΔI/Δt = 12.5 × 3/(0.150×10⁻⁶)
**EMF = 2.5×10⁸ V = 250 MV**
---
**QUESTION 15a**
**(i) Definitions:**
**Capacitor:** A device that stores electric charge (and energy) consisting of two conducting plates separated by an insulating material (dielectric).
**Capacitance:** The ability of a capacitor to store charge per unit potential difference. C = Q/V (unit: Farad, F)
**Series combination derivation:**
For capacitors C₁, C₂, C₃, C₄ in series:
- Same charge Q on each
- Total voltage: V = V₁ + V₂ + V₃ + V₄
- V = Q/C₁ + Q/C₂ + Q/C₃ + Q/C₄
∴ **1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ + 1/C₄**
**When C₁ = C₄ = 8nF, C₂ = C₃ = 20nF:**
1/C_eq = 1/8 + 1/20 + 1/20 + 1/8 (all in nF)
1/C_eq = 2/8 + 2/20 = 1/4 + 1/10 = 5/20 + 2/20 = 7/20
**C_eq = 20/7 = 2.86 nF**
**(ii) Parallel plate capacitor:**
A = 90 cm² = 90×10⁻⁴ m², d = 2.5 mm = 2.5×10⁻³ m, V = 400 V
C = ε₀A/d = (8.85×10⁻¹² × 90×10⁻⁴)/(2.5×10⁻³)
C = (79.65×10⁻¹⁶)/(2.5×10⁻³) = 31.86×10⁻¹³ = **3.186×10⁻¹² F = 3.186 pF**
**Energy stored:**
E = ½CV² = ½ × 3.186×10⁻¹² × 400²
E = ½ × 3.186×10⁻¹² × 160,000
**E = 2.549×10⁻⁷ J = 254.9 nJ**
---
**QUESTION 15b**
**(i) Express α in terms of ρ:**
Since R = ρL/A, then R₂ = ρ₂L/A and R₁ = ρ₁L/A
α = (R₂ − R₁)/(R₁ΔT) = (ρ₂ − ρ₁)/(ρ₁ΔT)
**∴ α = Δρ/(ρ₁ΔT)**
**(ii) R₁ = 2.1Ω at T₁ = 27.5°C, R₂ = 2.7Ω at T₂ = 100°C**
ΔT = 100 − 27.5 = 72.5°C
α = (R₂ − R₁)/(R₁ΔT) = (2.7 − 2.1)/(2.1 × 72.5)
α = 0.6/152.25
**α = 3.94×10⁻³ °C⁻¹**
**(iii) Copper wire: L = 0.30 m, v = 4 ms⁻¹, B = 1.25 T**
Induced EMF:
ε = BLv = 1.25 × 0.30 × 4 = **1.5 V**
**(A) Open circuit:**
No complete circuit → **I = 0 A**
**(B) Connected to 2Ω load:**
(Assuming wire resistance is negligible)
I = ε/R = 1.5/2 = **0.75 A**
---
**QUESTION 16a**
**(i) Resonance in LRC circuit:**
**Conditions:**
- Inductive reactance equals capacitive reactance: X_L = X_C
- i.e., ωL = 1/ωC
**At resonance:** Impedance Z is minimum (Z = R only), therefore current is **maximum.**
**Derivation of resonance frequency:**
X_L = X_C
2πf₀L = 1/(2πf₀C)
(2πf₀)² = 1/(LC)
4π²f₀² = 1/(LC)
**f₀ = 1/(2π√LC)**
**(ii) Coil: I_dc = 2A, V_dc = 12V; I_ac = 20A, V_ac = 240V, f = 50Hz**
**Resistance:**
R = V_dc/I_dc = 12/2 = **6 Ω**
**Impedance:**
Z = V_ac/I_ac = 240/20 = **12 Ω**
**Inductive reactance:**
X_L = √(Z² − R²) = √(144 − 36) = √108 = **10.39 Ω**
**Inductance:**
L = X_L/(2πf) = 10.39/(2π × 50) = 10.39/314.16 = **0.033 H = 33 mH**
**Susceptance:**
B = X_L/Z² = 10.39/144 = **0.0722 S**
**Admittance:**
Y = 1/Z = 1/12 = **0.0833 S**
---
**QUESTION 16b — Kirchhoff's Laws (Figure 3)**
Circuit: 28V source with 8Ω and 6Ω; 10V source with 12Ω and 4Ω
Let I₁ = current through 28V branch, I₂ = current through 10V branch, I₃ = current through 12Ω
**KCL:** I₁ + I₂ = I₃
**KVL Loop 1 (28V branch and 12Ω):**
28 = 8I₁ + 6I₁ + 12I₃
28 = 14I₁ + 12I₃ ... (1)
**KVL Loop 2 (10V branch and 12Ω):**
10 = 4I₂ + 12I₃ ... (2)
Substituting I₂ = I₃ − I₁ into (2):
10 = 4(I₃ − I₁) + 12I₃
10 = −4I₁ + 16I₃ ... (3)
From (1): 14I₁ + 12I₃ = 28 → I₁ = (28 − 12I₃)/14 ... (4)
Substituting (4) into (3):
10 = −4(28 − 12I₃)/14 + 16I₃
10 = (−112 + 48I₃)/14 + 16I₃
140 = −112 + 48I₃ + 224I₃
252 = 272I₃
**I₃ = 252/272 = 0.926 A**
**Current through 12Ω resistor = 0.926 A**
---
## SECTION D: MODERN PHYSICS
---
**QUESTION 17a**
**(i) de Broglie Hypothesis:**
de Broglie proposed that just as radiation (light) has both wave and particle properties, all moving material particles also have an associated wave (matter wave). The wavelength of this matter wave is:
**λ = h/mv = h/p**
where h = Planck's constant, m = mass, v = velocity, p = momentum.
**Derivation:**
For a photon: E = hf = hc/λ → λ = hc/E
Also E = pc (for photon) → λ = h/p
By analogy for a particle: **λ = h/p = h/mv**
**(ii) λ = 589 nm = 589×10⁻⁹ m**
**For electron (m_e = 9.11×10⁻³¹ kg):**
p = h/λ = 6.6×10⁻³⁴/(589×10⁻⁹) = 1.121×10⁻²⁷ kg m/s
KE = p²/2m = (1.121×10⁻²⁷)²/(2 × 9.11×10⁻³¹)
KE = (1.256×10⁻⁵⁴)/(1.822×10⁻³⁰)
KE = 6.894×10⁻²⁵ J
**KE = 6.894×10⁻²⁵/1.6×10⁻¹⁹ = 4.31×10⁻⁶ eV = 4.31 μeV**
**For neutron (m_n = 1.675×10⁻²⁷ kg):**
KE = p²/2m_n = (1.256×10⁻⁵⁴)/(2 × 1.675×10⁻²⁷)
KE = (1.256×10⁻⁵⁴)/(3.35×10⁻²⁷)
KE = 3.749×10⁻²⁸ J
**KE = 3.749×10⁻²⁸/1.6×10⁻¹⁹ = 2.34×10⁻⁹ eV = 2.34 neV**
---
**QUESTION 17b**
**(i) Natural Radioactivity:**
Natural radioactivity is the spontaneous disintegration of the nucleus of an atom of a heavy element with the emission of radiation, without any external influence.
**Radiations in order of increasing ionization power:**
1. Gamma (γ) rays — least ionizing
2. Beta (β) particles
3. Alpha (α) particles — most ionizing
**(ii) Proof of Radioactive Decay Law:**
Let N = number of undecayed atoms at time t.
By the decay law: dN/dt ∝ −N
∴ dN/dt = −λN (where λ = decay constant)
Separating variables:
dN/N = −λ dt
Integrating both sides:
ln N = −λt + C
At t = 0, N = N₀:
C = ln N₀
∴ ln N − ln N₀ = −λt
ln(N/N₀) = −λt
**N = N₀e^(−λt) ✓**
**(iii) Decay sequence: ˣ_yG → G₁ →¹⁷⁶₅₁G₂**
**G → G₁ (α emission):**
α particle = ⁴₂He
Mass number: x − 4 = A₁
Atomic number: y − 2 = Z₁
**G₁ → G₂ (β emission):**
β particle = ⁰₋₁e
Mass: A₁ = 176 (unchanged)
Atomic number: Z₁ + 1 = 51 → Z₁ = 50
From G → G₁:
A₁ = 176 → x − 4 = 176 → **x = 180**
Z₁ = 50 → y − 2 = 50 → **y = 52**
**G₁ isotope:** Mass = 176, Atomic number = 50
**Neutrons in G₁ = 176 − 50 = 126 neutrons**
---
**QUESTION 18a**
**(i) Hydrogen atom transition: −1.51 eV → −3.4 eV**
Energy emitted:
ΔE = −1.51 − (−3.4) = 1.89 eV = 1.89 × 1.6×10⁻¹⁹ = 3.024×10⁻¹⁹ J
Using E = hc/λ:
λ = hc/E = (6.6×10⁻³⁴ × 3×10⁸)/(3.024×10⁻¹⁹)
λ = (19.8×10⁻²⁶)/(3.024×10⁻¹⁹)
**λ = 6.55×10⁻⁷ m = 655 nm (Red — Hα line)**
**(ii) Mass defect and binding energy of α-particle:**
⁴₂He: 2 protons + 2 neutrons
Mass of constituents:
= 2(1.007277) + 2(1.008665)
= 2.014554 + 2.017330
= 4.031884 u
Mass of helium nucleus = 4.001265 u
**Mass defect:**
Δm = 4.031884 − 4.001265 = **0.030619 u**
**Binding energy:**
E = Δm × 931.5 MeV/u
E = 0.030619 × 931.5
**E = 28.52 MeV**
**(iii) Nuclear Fission vs Nuclear Fusion:**
| Feature | Nuclear Fission | Nuclear Fusion |
|---|---|---|
| Definition | Splitting of heavy nucleus into lighter nuclei | Combining of light nuclei to form heavier nucleus |
| Example | U-235 split by neutron | H + H → He |
| Energy released | Large | Much larger |
| Temperature needed | Low (room temp with neutrons) | Extremely high (millions of °C) |
| Radioactive waste | High | Minimal |
---
**QUESTION 18b**
**(i) Definitions:**
**Thermionic Emission:** The emission of electrons from the surface of a metal when it is heated to a sufficiently high temperature. The thermal energy gives electrons enough energy to escape the metal surface.
**Field Emission:** The emission of electrons from the surface of a metal due to the application of a strong external electric field, without heating. The electric field lowers the potential barrier allowing electrons to tunnel out.
**(ii) Effects of intensity and frequency in photoelectric effect:**
**Intensity:**
- Increasing intensity increases the number of photons incident per second
- This increases the number of photoelectrons emitted (larger photocurrent)
- But does NOT increase the maximum kinetic energy of photoelectrons
**Frequency:**
- Increasing frequency increases the energy of each photon (E = hf)
- This increases the maximum kinetic energy of emitted photoelectrons
- Below the threshold frequency, no electrons are emitted regardless of intensity
**(iii) Tungsten: λ₀ = 2300 Å, λ = 1800 Å**
**Work function:**
φ = hc/λ₀ = (6.6×10⁻³⁴ × 3×10⁸)/(2300×10⁻¹⁰)
φ = (19.8×10⁻²⁶)/(2300×10⁻¹⁰) = 8.61×10⁻¹⁹ J
**φ = 8.61×10⁻¹⁹/1.6×10⁻¹⁹ = 5.38 eV**
**Energy of incident photon:**
E = hc/λ = (6.6×10⁻³⁴ × 3×10⁸)/(1800×10⁻¹⁰)
E = (19.8×10⁻²⁶)/(1800×10⁻¹⁰) = 1.1×10⁻¹⁸ J = **6.875 eV**
**Maximum KE of photoelectron:**
KE_max = E − φ = 6.875 − 5.38 = **1.495 eV ≈ 1.5 eV**