SECTION A

Answer all the questions from this section (40 marks)

1. (a). State the condition for a ray of light to be refracted in a triangular glass prism.
(b) A monochromatic ray of light is incident on one face of an equilateral glass prism of refractive index √2. It undergoes a minimum deviation and finally emerges from the opposite face. What is the angle of incidence?

2. Sketch the ray diagrams to illustrate the following terms: parallel beams, converging and diverging beams of light. Show the ray diagram for the formation of virtual image of a point object by a plane mirror.

3. An object 2 cm height is placed 10 cm in front of a concave lens of focal length 20 cm. What is the image size?

4. (i) What are the main sources of electric field and magnetic field? (ii) Consider 2 resistors, of resistance 3 and 6Ω connected in parallel across a 12 V battery. Determine the current flowing in the 3Ω resistor and the voltage across each resistor.

5. When a 2A current flows through a very long wire, Find the magnetic field at a point 2.3 cm from the wire.

6. Determine the capacitance that must be connected in series with a 30μF capacitor for the equivalent capacitance to be 12μF? Also find the charges through each capacitor when the arrangement is connected to 10V d.c supply.

7. An inductor is connected to a 150 V, 50 Hz supply, to a 300 VA power supply. Calculate the value of the inductance.

8. Show that, the Einstein equation that related the frequency (f) and wavelength (λ) of light wave by a photon has energy:
$E(eV) = \frac{12431}{\lambda(\text{Å})}$

9. What is electromagnetic wave (emw)? State the basic sources of emw. Write down any four examples of emw.

10. Cesium has a work function of 1.8eV. When light of wavelength 5000Å is incident on it, find the threshold frequency, cutoff wavelength and the maximum kinetic energy of the emitted electrons.

SECTION B — Geometric Optics

Answer only one (1) question from this section (20 marks)

11a (i) State any four applications of optical instruments that used lenses. Sketch ray diagrams of focal length by parallel beam of light striking the face each of converging (or convex) and diverging (concave) lenses. What is the nature of each focal length?
(ii). A small pin object is placed 15 cm from a concave lens with focal length 10 cm. Discuss the nature of the image formed.
(b) Consider an object 3.7 cm high placed 15 cm from the pole of a (i) concave mirror (ii) convex mirror of radius of curvature 40 cm. Find the nature of the image position, magnification and image height.

12a. (i) A ray of light travels from water to glass. Show the ray diagram and indicate clearly the angle of incidence (î), angle of refraction (r̂) and the angle of deviation (δ̂). Express δ̂ in terms of r̂ and î. (ii) If î = cos⁻¹(4/5), Calculate r̂ and δ̂, given that, refractive indices of water and glass are respectively 4/3 and 3/2.

12b(i) An object AB 10 cm from the mirror is viewed through a plane mirror at angle 30° as shown in the diagram. Copy and locate the image formed by the mirror.

12b(ii). Calculate the object distance from a convex lens of focal length 10 cm when the image is magnified two times (A) if the image is real (B) if the image is virtual.

SECTION C — Electricity and Magnetism

Answer only Two (2) questions from this section (40 marks)

13a (i) State the mathematical form of Coulomb’s law of electrostatics in terms of relative permittivity (εᵣ) and state any two properties each of electric charges and electric field line of force.
(ii) Write down (without proof) the expression in each of the following forms of electric field:

(A) The electric field due a point charge.
(B) The electric field uniformly distributed due to infinite line charge density (λ).
© The electric field uniformly distributed due to surface charge density (σ).
(D) The electric field uniformly distributed due to volume charge density (ρ).

13b. (i) Find the equivalent capacitance of the given network in figure 2.
(ii) If a 200 volt d.c supply is connected to the circuit, calculate the charge and energy stored in the circuit.

14. (i) Define the following terms as applied to electrical circuit: branch (give two examples), node and loop. Define the underlined term. (ii) State the Kirchhoff’s laws in words and copy the circuit in figure 3 and indicate the number of branches, nodes and loops. (iii) Find the current through each resistor in figure 3.

15a. (i) Distinguish between the terms: cell and battery (as applied to electricity), electromotive force and terminal potential difference.
(ii) A copper thick coil of 300 turns is wound uniformly on a non-magnetic ring. If the mean circumference of ring is 40 cm and a uniform cross sectional area of 4 cm². When the current in the coil is 5 A, calculate the magnetic field strength, the flux density, the total magnetic flux in the ring and reluctance. Also define the underlined term.

15b. (i) With the help of suitable diagram, briefly explain the phenomena that: “magnetic monopole does not exist” (i.e a magnet with a single N-pole or S-pole can not be formed).
(ii) The magnetic field at a point on the axis of a circular current-carrying coil is given by:
$B = \frac{1}{2} \frac{\mu_0 Nir^2}{(r^2 + x^2)^{\frac{3}{2}}}$
where N is the number of turns, r is the radius of the coil and x is the distance of the observation point from the centre of the coil. Determine the B: At the centre of the loop, and at place when the observation point is far far away from the coil (x >> r).

15c. Calculate the value of the energy stored when a current of 30 mA is flowing in a coil of inductance 400 mH.

16a(i). What are Ohmic and non Ohmic conductors? Support your answers with suitable sketches graphs. State the factors upon which the resistance and resistivity of on electrical conductors depend and the effect of each factor. (ii) The resistance of a 5 m length of wire is 600 Ω. Determine the resistance of an 8 m length of the same wire, and the length of the same wire when the resistance is 420 Ω.

16b(i) State two advantages of a.c over d.c and one advantage of d.c over a.c. (ii) A series circuit of resistance 60 ohm and 75 mH inductance are connected to a 110 V, 60 Hz a.c supply. Calculate the power dissipated and the power factor.

(iii) By substituting the values μ₀ and ε₀, Show that physical quantity $\frac{1}{\sqrt{\mu_0\varepsilon_0}}$ represents the speed of light in vacuum. And find the value of $\sqrt{\frac{\mu_0}{\varepsilon_0}}$ (has a unit of Ω).

SECTION D — Modern Physics

Answer only one (1) question from this Section (20 marks)

17a. (i) Briefly explain the concept of photoelectric emission. Write down the Einstein photoelectric relations in terms of frequency of the emitted radiation, wavelength, stopping potential.
(ii) What are the factors affecting the photoelectric emissions? State the effects of each factor.
(iii) Assume the stopping potential for a certain metal is 3V when is illuminated by radiation of wavelength 0.5μm. What would be the stopping potential of the metal when the wavelength is 600nm?

17b. State the natural radiations emitted by a radioactive element. Assume a radioactive element X eject each one of the radiation, explain with help of the disintegration reaction, what would be the resulting nucleus?
(i) A radioactive source at any instant has its disintegration rate of 5000 disintegrations per minute. After 5 minutes, the rate is 1250 disintegrations per minute. Calculate the decay constant (per minute).

18a. (i) Briefly explain the terms “mass defect and nuclear binding energy”. How are they related?
(ii). Consider the Energy released in fission fragments when Barium, Krypton and neutrons are released with high velocities according to the reaction:
$^1_0n + ^{235}_{92}U \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + 3^1_0n + \gamma$
Find the energy released due to γ emission. Given that:
$^{235}_{92}U = 235.0439u, \quad ^1_0n = 1.0087u, \quad ^{141}_{56}Br = 140.9129u, \quad ^{92}_{36}Kr = 91.8973u$

18b(i) State the Heisenberg’s Uncertainty Principle in words and write down (without proof) its mathematical form, state clearly the meaning of each symbol for a particle having: a momentum, for a particle having an energy, and for a particle moving round a circular orbit.

(ii). Find the de Broglie wavelength of an electron (A) moving with velocity 1000 ms⁻¹ (B) having the kinetic energy of 100 eV.

COMPLETE SOLUTIONS

SECTION A (40 marks)

Q1(a) Condition for Refraction in a Triangular Glass Prism

The ray of light must:

Travel from one medium to another of different optical density
Strike the surface at an angle less than the critical angle
The two refracting surfaces must be non-parallel

Q1(b) Angle of Incidence — Equilateral Prism, n = √2

For minimum deviation in an equilateral prism:

Prism angle A = 60°
At minimum deviation: r = A/2 = 30°

Using Snell’s law:
$n = \frac{\sin i}{\sin r}$

$\sqrt{2} = \frac{\sin i}{\sin 30°} = \frac{\sin i}{0.5}$

$\sin i = 0.5\sqrt{2} = 0.7071$

$\boxed{i = 45°}$

Q2 Ray Diagrams (Descriptions)

Parallel beams: Rays travel parallel to each other and to the principal axis.

Converging beams: Rays meet at a point (focus) after refraction/reflection.

Diverging beams: Rays spread out from a point after refraction/reflection.

Virtual image by plane mirror:

Object in front of mirror → image formed behind the mirror
Image is: virtual, erect, same size, same distance behind mirror as object is in front

Q3 Concave Lens — Image Size

Given: h = 2 cm, u = -10 cm, f = -20 cm (concave lens)

Using lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-20} + \frac{1}{-10}$

$\frac{1}{v} = -\frac{1}{20} - \frac{2}{20} = -\frac{3}{20}$

$v = -\frac{20}{3} = -6.67 \text{ cm}$

Magnification:
$m = \frac{v}{u} = \frac{-6.67}{-10} = 0.667$

$\text{Image size} = m \times h = 0.667 \times 2 = \boxed{1.33 \text{ cm}}$

Image is virtual, erect, diminished.

Q4(i) Sources of Electric and Magnetic Fields

Electric field sources:

Static electric charges
Time-varying magnetic fields

Magnetic field sources:

Moving electric charges (current)
Time-varying electric fields
Permanent magnets

Q4(ii) Parallel Resistors — 3Ω and 6Ω across 12V

Voltage across each resistor = 12V (parallel connection)

Current through 3Ω:
$I_3 = \frac{V}{R} = \frac{12}{3} = \boxed{4 \text{ A}}$

Current through 6Ω:
$I_6 = \frac{12}{6} = 2 \text{ A}$

Voltage across each resistor = 12V

Q5 Magnetic Field near Long Wire

Given: I = 2A, r = 2.3 cm = 0.023 m

$B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 2}{2\pi \times 0.023}$

$B = \frac{8\pi \times 10^{-7}}{0.046\pi} = \frac{8 \times 10^{-7}}{0.046}$

$\boxed{B = 1.74 \times 10^{-5} \text{ T}}$

Q6 Capacitors in Series

Given: C₁ = 30μF, C_eq = 12μF, V = 10V

$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$

$\frac{1}{12} = \frac{1}{30} + \frac{1}{C_2}$

$\frac{1}{C_2} = \frac{1}{12} - \frac{1}{30} = \frac{5-2}{60} = \frac{3}{60} = \frac{1}{20}$

$\boxed{C_2 = 20\mu F}$

Charge on each capacitor (series — same charge):
$Q = C_{eq} \times V = 12 \times 10^{-6} \times 10$

$\boxed{Q = 1.2 \times 10^{-4} \text{ C} = 120\mu\text{C}}$

Q7 Inductance Calculation

Given: V = 150V, f = 50Hz, Power = 300VA

$I = \frac{P}{V} = \frac{300}{150} = 2\text{ A}$

Inductive reactance:
$X_L = \frac{V}{I} = \frac{150}{2} = 75\text{ Ω}$

$X_L = 2\pi f L$

$L = \frac{X_L}{2\pi f} = \frac{75}{2\pi \times 50} = \frac{75}{314.16}$

$\boxed{L = 0.239 \text{ H} \approx 239 \text{ mH}}$

Q8 Einstein Equation Proof

Energy of photon:
$E = hf = \frac{hc}{\lambda}$

Converting to eV (divide by e = 1.6×10⁻¹⁹):
$E(eV) = \frac{hc}{e\lambda}$

Substituting: h = 6.6×10⁻³⁴, c = 3×10⁸, e = 1.6×10⁻¹⁹:

$E(eV) = \frac{6.6\times10^{-34} \times 3\times10^8}{1.6\times10^{-19} \times \lambda}$

$= \frac{19.8\times10^{-26}}{1.6\times10^{-19} \times \lambda} = \frac{1.2375\times10^{-6}}{\lambda}$

Converting λ from metres to Ångströms (1Å = 10⁻¹⁰m):

$E(eV) = \frac{1.2375\times10^{-6} \times 10^{10}}{\lambda(\text{Å})} \approx \boxed{\frac{12431}{\lambda(\text{Å})}} \checkmark$

Q9 Electromagnetic Waves

Definition: EMW are waves produced by oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation. They travel at speed c = 3×10⁸ m/s in vacuum and require no medium.

Basic Sources:

Oscillating electric charges
Accelerating charged particles
Nuclear transitions
Thermal radiation

Four Examples:

Radio waves
X-rays
Visible light
Gamma rays

Q10 Cesium Photoelectric Effect

Given: Work function φ = 1.8eV, λ = 5000Å

Threshold frequency:
$f_0 = \frac{\phi}{h} = \frac{1.8 \times 1.6\times10^{-19}}{6.6\times10^{-34}}$

$\boxed{f_0 = 4.36 \times 10^{14} \text{ Hz}}$

Cutoff wavelength:
$\lambda_0 = \frac{c}{f_0} = \frac{3\times10^8}{4.36\times10^{14}}$

$\boxed{\lambda_0 = 6.88 \times 10^{-7} \text{ m} = 6880\text{ Å}}$

Energy of incident photon:
$E = \frac{12431}{5000} = 2.486 \text{ eV}$

Maximum KE of emitted electrons:
$KE_{max} = E - \phi = 2.486 - 1.8$

$\boxed{KE_{max} = 0.686 \text{ eV} = 1.098\times10^{-19} \text{ J}}$

SECTION B — GEOMETRIC OPTICS

Q11a(i) Applications of Optical Instruments Using Lenses

Microscope — magnifies tiny/microscopic objects
Telescope — views distant astronomical objects
Camera — captures images on film/sensor
Spectacles/Eyeglasses — corrects vision defects

Converging (Convex) lens: Parallel rays converge to a real focal point in front. Focal length is positive (+f)

Diverging (Concave) lens: Parallel rays diverge and appear to come from a virtual focal point behind the lens. Focal length is negative (-f)

Q11a(ii) Object 15 cm from Concave Lens, f = -10 cm

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-10} + \frac{1}{-15} = -\frac{3+2}{30} = -\frac{5}{30}$

$v = -6 \text{ cm}$

Nature: Image is virtual, erect, diminished, formed on same side as object at 6 cm.

Q11b Concave and Convex Mirror — Object 3.7 cm high, u = -15 cm, R = 40 cm → f = 20 cm

(i) Concave Mirror (f = -20 cm using sign convention)

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-20} - \frac{1}{-15} = -\frac{1}{20} + \frac{1}{15} = \frac{-3+4}{60} = \frac{1}{60}$

$v = 60 \text{ cm}$

Magnification:
$m = -\frac{v}{u} = -\frac{60}{-15} = +4$

Image height = 4 × 3.7 = 14.8 cm

Nature: Image is virtual, erect, magnified, formed at 60 cm behind mirror (object placed inside focal length).

(ii) Convex Mirror (f = +20 cm)

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{20} - \frac{1}{-15} = \frac{1}{20} + \frac{1}{15} = \frac{3+4}{60} = \frac{7}{60}$

$v = \frac{60}{7} = 8.57 \text{ cm}$

$m = -\frac{v}{u} = -\frac{8.57}{-15} = +0.571$

Image height = 0.571 × 3.7 = 2.11 cm

Nature: Virtual, erect, diminished, 8.57 cm behind mirror.

Q12a Ray — Water to Glass

(i) δ̂ = r̂ - î (deviation = refraction angle minus incidence angle, since glass is denser)

(ii) Given: î = cos⁻¹(4/5), n_water = 4/3, n_glass = 3/2

$\cos\hat{i} = \frac{4}{5} \Rightarrow \sin\hat{i} = \frac{3}{5} = 0.6$

Snell’s law:
$n_w \sin\hat{i} = n_g \sin\hat{r}$

$\frac{4}{3} \times \frac{3}{5} = \frac{3}{2} \times \sin\hat{r}$

$\frac{4}{5} = \frac{3}{2}\sin\hat{r}$

$\sin\hat{r} = \frac{4}{5} \times \frac{2}{3} = \frac{8}{15} = 0.5333$

$\boxed{\hat{r} = 32.23°}$

$\hat{i} = \sin^{-1}(0.6) = 36.87°$

$\boxed{\hat{\delta} = \hat{r} - \hat{i} = 32.23° - 36.87° = -4.64°}$

(Negative sign indicates ray bends toward normal)

Q12b(ii) Convex Lens — Magnification = 2, f = 10 cm

(A) Real Image (m = -2)

$m = \frac{v}{u} \Rightarrow v = -2u$

Lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{-2u} - \frac{1}{u} = \frac{1}{10}$

$\frac{-1-2}{2u} = \frac{1}{10}$

$\frac{-3}{2u} = \frac{1}{10} \Rightarrow u = -15\text{ cm}$

$\boxed{u = 15 \text{ cm (object distance)}}$

(B) Virtual Image (m = +2)

$v = 2u$

$\frac{1}{2u} - \frac{1}{u} = \frac{1}{10}$

$\frac{1-2}{2u} = \frac{1}{10}$

$\frac{-1}{2u} = \frac{1}{10} \Rightarrow u = -5\text{ cm}$

$\boxed{u = 5 \text{ cm}}$

SECTION C — ELECTRICITY AND MAGNETISM

Q13a(i) Coulomb’s Law

$F = \frac{1}{4\pi\varepsilon_0\varepsilon_r} \cdot \frac{Q_1 Q_2}{r^2}$

Where εᵣ = relative permittivity of the medium.

Properties of Electric Charges:

Like charges repel; unlike charges attract
Charge is conserved (cannot be created or destroyed)

Properties of Electric Field Lines:

They originate from positive charges and terminate on negative charges
They never cross each other

Q13a(ii) Electric Field Expressions

(A) Point charge:
$E = \frac{Q}{4\pi\varepsilon_0 r^2}$

(B) Infinite line charge density (λ):
$E = \frac{\lambda}{2\pi\varepsilon_0 r}$

© Surface charge density (σ):
$E = \frac{\sigma}{2\varepsilon_0}$

(D) Volume charge density (ρ):
$E = \frac{\rho r}{3\varepsilon_0}$

Q13b Equivalent Capacitance — Figure 2

From the network (3μF, 2μF, 3μF arrangement):

Top row: 3μF, 3μF, 3μF in series:
$\frac{1}{C_{top}} = \frac{1}{3}+\frac{1}{3}+\frac{1}{3} = 1 \Rightarrow C_{top} = 1\mu F$

Middle row: 2μF, 2μF in series:
$\frac{1}{C_{mid}} = \frac{1}{2}+\frac{1}{2} = 1 \Rightarrow C_{mid} = 1\mu F$

Bottom: 3μF + 3μF in series = 1.5μF, with 3μF parallel:

All rows in parallel:
$C_{eq} = 1 + 1 + 1.5 = \boxed{3.5\mu F}$

(ii) Charge and Energy at 200V:

$Q = C_{eq} \times V = 3.5\times10^{-6} \times 200 = \boxed{7\times10^{-4}\text{ C} = 0.7\text{ mC}}$

$E = \frac{1}{2}C_{eq}V^2 = \frac{1}{2}\times3.5\times10^{-6}\times200^2$

$\boxed{E = 0.07 \text{ J} = 70\text{ mJ}}$

Q14 Kirchhoff’s Laws — Figure 3

(i) Definitions:

Branch: A single element or series of elements connected between two nodes. Examples: a resistor branch, a battery branch
Node: A point where two or more circuit elements meet
Loop: Any closed path in a circuit

(ii) Figure 3 has:

Branches: 3
Nodes: 2 (points a,b / d,e)
Loops: 2

Kirchhoff’s Laws:

KCL: The algebraic sum of currents at any node equals zero (currents in = currents out)
KVL: The algebraic sum of voltages around any closed loop equals zero

(iii) Finding currents — Figure 3:

Let I₁ through 1Ω (2V branch), I₂ through 2Ω (3V branch), I₃ through 5Ω

By KCL: I₁ + I₂ = I₃

Loop 1 (top):
$2 - I_1(1) - I_3(5) = 0$
$2 = I_1 + 5I_3 \quad ...(1)$

Loop 2 (bottom):
$3 - I_2(2) - I_3(5) = 0$
$3 = 2I_2 + 5I_3 \quad ...(2)$

Since I₃ = I₁ + I₂, substituting:

From (1): I₁ = 2 - 5(I₁+I₂) → 6I₁ + 5I₂ = 2 …(3)
From (2): 5I₁ + 7I₂ = 3 …(4)

Solving (3) and (4):
Multiply (3)×7: 42I₁ + 35I₂ = 14
Multiply (4)×5: 25I₁ + 35I₂ = 15
Subtract: 17I₁ = -1 → I₁ = -0.059 A

From (3): I₂ = (2 - 6(-0.059))/5 = 0.47 A

I₃ = I₁ + I₂ = -0.059 + 0.47 = 0.41 A

Q15a(i) Definitions

Term	Definition
Cell	A single unit that converts chemical energy to electrical energy
Battery	A combination of two or more cells connected together
EMF	Total energy supplied per unit charge by the source (including internal resistance)
Terminal PD	Actual voltage across the terminals when current flows (EMF minus voltage drop across internal resistance)

Q15a(ii) Toroidal Coil — N=300, circumference=40cm, A=4cm², I=5A

Magnetic field strength H:
$H = \frac{NI}{l} = \frac{300 \times 5}{0.40} = \boxed{3750 \text{ A/m}}$

Flux density B (non-magnetic, μᵣ = 1):
$B = \mu_0 H = 4\pi\times10^{-7} \times 3750$

$\boxed{B = 4.71\times10^{-3} \text{ T}}$

Total magnetic flux:
$\Phi = B \times A = 4.71\times10^{-3} \times 4\times10^{-4}$

$\boxed{\Phi = 1.885\times10^{-6} \text{ Wb}}$

Reluctance:
$\mathcal{R} = \frac{l}{\mu_0 A} = \frac{0.40}{4\pi\times10^{-7}\times4\times10^{-4}}$

$\boxed{\mathcal{R} = 7.96\times10^8 \text{ A/Wb}}$

Reluctance is the opposition offered by a magnetic circuit to the establishment of magnetic flux.

Q15b(ii) Magnetic Field on Axis of Coil

$B = \frac{1}{2}\frac{\mu_0 Nir^2}{(r^2+x^2)^{3/2}}$

At centre of loop (x = 0):
$(r^2+0)^{3/2} = r^3$

$\boxed{B = \frac{\mu_0 Ni}{2r}}$

When x >> r (far from coil):
$(r^2+x^2)^{3/2} \approx x^3$

$\boxed{B = \frac{\mu_0 Nir^2}{2x^3}}$

Q15c Energy Stored in Inductor

Given: I = 30 mA = 0.03 A, L = 400 mH = 0.4 H

$E = \frac{1}{2}LI^2 = \frac{1}{2}\times0.4\times(0.03)^2$

$\boxed{E = 1.8\times10^{-4} \text{ J} = 0.18 \text{ mJ}}$

Q16a(i) Ohmic vs Non-Ohmic Conductors

Ohmic conductors: Obey Ohm’s law; V∝I (linear relationship). Examples: resistors, metallic wires at constant temperature.

Non-Ohmic conductors: Do NOT obey Ohm’s law; V-I graph is non-linear. Examples: diodes, transistors, thermistors, filament bulbs.

Factors affecting Resistance:

Length (L): R ∝ L (longer wire = more resistance)
Cross-sectional area (A): R ∝ 1/A (thicker wire = less resistance)
Temperature: Resistance increases with temperature for metals
Material/Resistivity (ρ): R = ρL/A

Q16a(ii) Wire Resistance Problems

Given: R₁ = 600Ω for L₁ = 5m

For L₂ = 8m:
$R_2 = R_1 \times \frac{L_2}{L_1} = 600 \times \frac{8}{5} = \boxed{960 \text{ Ω}}$

For R₃ = 420Ω:
$L_3 = L_1 \times \frac{R_3}{R_1} = 5 \times \frac{420}{600} = \boxed{3.5 \text{ m}}$

Q16b(i) AC vs DC

Advantages of AC over DC:

Easily stepped up/down using transformers for efficient transmission
Can be generated more easily and cheaply

Advantage of DC over AC:

Required for charging batteries and electroplating

Q16b(ii) Series RL Circuit

Given: R = 60Ω, L = 75mH = 0.075H, V = 110V, f = 60Hz

Inductive reactance:
$X_L = 2\pi fL = 2\pi \times 60 \times 0.075 = 28.27\text{ Ω}$

Impedance:
$Z = \sqrt{R^2 + X_L^2} = \sqrt{60^2 + 28.27^2} = \sqrt{3600+799.2} = \sqrt{4399.2}$

$Z = 66.33\text{ Ω}$

Current:
$I = \frac{V}{Z} = \frac{110}{66.33} = 1.659\text{ A}$

Power factor:
$\cos\phi = \frac{R}{Z} = \frac{60}{66.33} = \boxed{0.905}$

Power dissipated:
$P = I^2R = (1.659)^2 \times 60 = \boxed{165.1 \text{ W}}$

Q16b(iii) Speed of Light Proof

$\frac{1}{\sqrt{\mu_0\varepsilon_0}} = \frac{1}{\sqrt{4\pi\times10^{-7}\times8.85\times10^{-12}}}$

$= \frac{1}{\sqrt{111.5\times10^{-19}}} = \frac{1}{\sqrt{1.115\times10^{-17}}}$

$= \frac{1}{1.056\times10^{-9}} \approx \boxed{3\times10^8 \text{ m/s} = c} \checkmark$

$\sqrt{\frac{\mu_0}{\varepsilon_0}} = \sqrt{\frac{4\pi\times10^{-7}}{8.85\times10^{-12}}} = \sqrt{1.131\times10^5} = \boxed{377 \text{ Ω}}$

SECTION D — MODERN PHYSICS

Q17a(i) Photoelectric Emission

Photoelectric emission is the ejection of electrons from a metal surface when electromagnetic radiation of sufficient frequency is incident on it.

Einstein photoelectric relations:

$hf = \phi + KE_{max}$

In terms of wavelength:
$\frac{hc}{\lambda} = \phi + eV_s$

Where:

h = Planck’s constant
f = frequency of radiation
φ = work function
V_s = stopping potential
KE_max = eV_s = maximum kinetic energy

Q17a(ii) Factors Affecting Photoelectric Emission

Factor	Effect
Frequency of radiation	Higher frequency → more KE of emitted electrons
Intensity of radiation	Higher intensity → more electrons emitted (not more KE)
Work function of metal	Higher work function → harder to eject electrons
Nature of metal surface	Cleaner surface → easier emission

Q17a(iii) Stopping Potential at λ = 600nm

Given: V_s1 = 3V at λ₁ = 0.5μm = 500nm = 5000Å

Using the difference method with Einstein’s equation:

$e(V_{s1} - V_{s2}) = hc\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)$

$V_{s2} = V_{s1} - \frac{hc}{e}\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)$

$= 3 - \left(\frac{12431}{5000} - \frac{12431}{6000}\right)$

$= 3 - 12431\left(\frac{6000-5000}{30000000}\right)$

$= 3 - 12431 \times \frac{1}{30000} = 3 - 0.414$

$\boxed{V_{s2} = 2.586 \text{ V} \approx 2.59\text{ V}}$

Q17b Radioactive Disintegration

Natural radiations: Alpha (α), Beta (β), Gamma (γ)

For element X (mass A, atomic number Z):

Alpha decay: $^A_Z X \rightarrow ^{A-4}_{Z-2}Y + ^4_2He$
Beta decay: $^A_Z X \rightarrow ^A_{Z+1}Y + ^0_{-1}e + \bar{\nu}$
Gamma decay: $^A_Z X^* \rightarrow ^A_Z X + \gamma$

Q17b(i) Decay Constant

Given: N₀ = 5000/min, N = 1250/min, t = 5 min

Using: N = N₀e^(-λt)

$\frac{N}{N_0} = \frac{1250}{5000} = 0.25 = e^{-5\lambda}$

$\ln(0.25) = -5\lambda$

$-1.386 = -5\lambda$

$\boxed{\lambda = 0.277 \text{ min}^{-1}}$

Q18a(i) Mass Defect and Binding Energy

Mass defect (Δm): The difference between the sum of masses of individual nucleons and the actual mass of the nucleus.
$\Delta m = (Zm_p + Nm_n) - M_{nucleus}$

Nuclear binding energy: The energy required to completely separate a nucleus into its constituent nucleons.
$E_b = \Delta m \times c^2$

Relationship: Greater mass defect → Greater binding energy → More stable nucleus.

Q18a(ii) Energy Released in Fission

$^1_0n + ^{235}_{92}U \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + 3^1_0n + \gamma$

Mass of reactants:
$= 1.0087 + 235.0439 = 236.0526\text{ u}$

Mass of products:
$= 140.9129 + 91.8973 + 3(1.0087)$
$= 140.9129 + 91.8973 + 3.0261 = 235.8363\text{ u}$

Mass defect:
$\Delta m = 236.0526 - 235.8363 = 0.2163\text{ u}$

Energy released:
$E = \Delta m \times 931 \text{ MeV/u} = 0.2163 \times 931$

$\boxed{E = 201.4 \text{ MeV}}$

Q18b(i) Heisenberg’s Uncertainty Principle

Statement: It is impossible to simultaneously determine both the position and momentum (or energy and time) of a particle with absolute precision.

Mathematical forms:

For a particle with momentum:
$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$

For a particle with energy:
$\Delta E \cdot \Delta t \geq \frac{h}{4\pi}$

For a particle in circular orbit:
$\Delta L \cdot \Delta\theta \geq \frac{h}{4\pi}$

Where: Δx = uncertainty in position, Δp = uncertainty in momentum, ΔE = uncertainty in energy, Δt = uncertainty in time.

Q18b(ii) de Broglie Wavelength

$\lambda = \frac{h}{mv} \quad \text{or} \quad \lambda = \frac{h}{\sqrt{2mKE}}$

(A) v = 1000 ms⁻¹

$\lambda = \frac{h}{m_e v} = \frac{6.6\times10^{-34}}{9.0\times10^{-31}\times1000}$

$\boxed{\lambda = 7.33\times10^{-7} \text{ m} = 733\text{ nm}}$

(B) KE = 100 eV = 100 × 1.6×10⁻¹⁹ = 1.6×10⁻¹⁷ J

$\lambda = \frac{h}{\sqrt{2m_e KE}} = \frac{6.6\times10^{-34}}{\sqrt{2\times9.0\times10^{-31}\times1.6\times10^{-17}}}$

$= \frac{6.6\times10^{-34}}{\sqrt{2.88\times10^{-47}}} = \frac{6.6\times10^{-34}}{1.697\times10^{-23}}$

$\boxed{\lambda = 3.89\times10^{-11} \text{ m} = 0.389\text{ Å}}$

2025 IJMB physics paper 2