2023 IJMB physics paper 3

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SECTION A

Answer All the Questions From this Section (40 Marks)

1. Assume an object of 7 cm height is placed at a distance of 12 cm from a convex lens of focal length 8 cm, find the height and nature of the image.

2. A ray of light travels from glass to water, such that the angle of incidence is 50°. Find the angle of refraction, take the refractive index of glass as 1.52 and that of water as 1.33.

3. An equilateral Pyrex glass prism has index of refraction 1.72. Find its angle of minimum deviation.

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4. When two capacitors are connected in parallel, the equivalent capacitance is 8μF and when connected in series, the equivalent capacitance is 15/8 μF. Find the values of the two capacitances.

5. An alternating voltage is given by V = 7.3sin(200πt − 1.2) Volts, find the maximum voltage and root mean square (rms) voltage.

6. Define electric flux density and state its unit. Find the velocity at which a conductor 75 mm long cut a magnetic field of flux density 0.6 T if an e.m.f. of 9 V is to be induced in it. Assume the conductor, the field and the direction of motion are mutually perpendicular.

7. In a d.c. potentiometer, a balance point is obtained at a length of 400 mm when using a standard cell of 1.0186 Volts. Determine the e.m.f. of a dry cell if the balance is obtained at 650 mm.

8. Give two examples each of the following terms: conductors, semiconductors or insulators. State the one effect of temperature on conductivity each on conductors and semiconductors.

9. Calculate the de Broglie wavelength each of electron and proton all having the same kinetic energy of 100 eV.

10. What is meant by the statement that ‘the work function of a metal is 0.48 eV’. Find the threshold wavelength for this metal.

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SECTION B — GEOMETRIC OPTICS

11. a(i) Consider the ray diagram of formation of an image by a certain mirror as shown in figure 1. From the ray diagram: State (with reasons) the type of the mirror used, identify/locate the image, the object and their nature (with reasons) also identify point B.

(Figure 1 — ray diagram showing a mirror with points A, C, F, B, D, P, E marked)

(ii) Assume that, an object of size 1 cm is placed at a distance of 15 cm from a concave mirror of focal length 10 cm. Find the size, nature and position of the image formed.

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11b. Consider an object placed at the following positions: 8.5 cm, 15 cm, 20 cm, 25 cm and 27.7 cm from a convex lens of radius of curvature 20 cm respectively. Determine the position of the object that produce (i) a real diminished image (ii) a magnified real image (iii) a magnified virtual image (iv) an image of the same size as the object.

12a(i). State the factor(s) upon which the refractive index of kerosene depends. Calculate the speed of light in flint glass of refractive index 1.63 when travelling from air.

12a (ii). What is meant by dispersive power? Calculate the dispersive power of crown glass. Given that, the refractive index for violent colour is 1.5230 and that of red colour is 1.5145.

12b (i). When a light ray travelling from a rectangular glass slab to water. The ray travels all the way through the slab to water and finally emerges into air. Sketch the ray diagram and compare clearly the angle of incidence with angle of refraction. Briefly discuss the possibility of total internal reflection to occur.

12b (ii). A lens has a power of +5 diopter in air. Find its power when completely immersed in water. (Take ₐnw = 4/3, and ₐns = 3/2.)

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SECTION C — ELECTRICITY AND MAGNETISM

13a (i) Define the terms: gold leaf electroscope and electric field strength. (ii) If fₑ is the electrostatic force and f_g the gravitational force, find the magnitude of the ratio fₑ/f_g between two electrons.

b(i). What is an electric current? Write down its mathematical form. (ii) Distinguish between direct and alternating current. Use sketch graphs where necessary.

(ii) What is a conductance? Four resistors of resistances R₁, R₂, R₃ and R₄ are connected in parallel across a battery V volts. Sketch the circuit and derive the expressions for the equivalent resistance and hence, write down the expressions for its equivalent conductance.

(iii) Two resistors of resistances 2Ω and 5Ω are connected in parallel and the combination is connected to 10Ω in series. Sketch the circuit diagram and determine the equivalent conductance of the circuit.

14a (i). Define the terms Magnetic flux and Magnetic flux density. What is the flux density in a magnetic field of cross-sectional area 20 cm² having a flux of 3μWb?

(ii) Briefly explain the permeability of free space. Write down its mathematical expressions. Determine the magnetic field strength and the magnetomotive force (mmf) required to produce a flux density of 0.25 μT in an air gap of length 12 mm.

b(i). State the factors upon which the force on the current-carrying conductor in a magnetic field depends.

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(ii) Consider a conductor 50 mm carries a current of 20 A at right angle to a magnetic field of flux density 0.9 T. Calculate the force acting on the conductor. Also, Determine the value of the force if the conductor is inclined at an angle of 30° to the direction of the field.

15a (i) With the help of labelled diagram, explain the principle of electromagnetic induction.

(ii) State the Faraday’s and Lenz’s laws of electromagnetic induction.

b(i) An 0.58 H inductor has a current of 5 A flowing through it. Find the energy stored in the magnetic field of the inductor?

(ii) Use the circuit in Figure 2 to determine the equivalent inductance at terminals a-b.

(Figure 2 — circuit diagram showing inductors: 10 mH, 60 mH, 25 mH, 20 mH, 30 mH)

16a(i) The unit of capacitance is Farad, hence, define the term Farad. Consider three capacitors of Capacitances 3μF, 6μF and 12μF connected in series across a 250 V supply. Draw the circuit diagram and calculate the equivalent circuit capacitance, the charge on each capacitor and the p.d across each capacitor.

(ii) The p.d. at the terminals of a battery is 25 V when no load is connected and 24 V when a load taking 10 A is connected. Determine the internal resistance of the battery.

16b(i) 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 and inductance of the coil.

(ii). State the Kirchhoff’s laws in words. Use figure 3 to find i₁, i₂ and V and the power dissipated in the 6Ω resistor.

(Figure 3 — circuit showing 20V source, with i₁ through 5Ω, resistors 4Ω, 10Ω, 6Ω, and i₂)

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SECTION D — MODERN PHYSICS

17a(i). Balmer series is one of the five spectral series observed in the Hydrogen Spectrum corresponding to the five energy levels of the Hydrogen atom. Mention the other four.

(ii) Distinguish between the terms ‘ionization Potential and work function’. The photoelectric threshold of the photo electric effect of a certain metal is 2750 Å. Find the maximum velocity of the electrons ejected from the metal by light with a wavelength 1800 Å.

17b (i) Name three types of radiations emitted by radioactive sources in order of their relative most penetrating power. Determine, which of the radiations carries a negative charge, is similar to X-rays, is most easily absorbed,

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travels with the greatest speed, is not deflected by a magnetic field emitted when decays ²³⁸₉₂U decays to ²³⁴₉₀Th and is similar in nature to cathode rays.

(ii) Carbon-14 (¹⁴₆C) is radioactive isotope of carbon. It is a radioactive, decaying to nitrogen-14 (¹⁴₇C). Define the underlined term and write down the equations for the radioactive decay. How many protons and neutrons do ¹⁴₆C have?

18a. (i) with the help of 5abeled diagram, briefly discuss the production of x-rays in a Coolidge tube. Hence, explain the terms: soft and hard x-rays.

(ii) When X-ray tube produces a continuous spectrum of radiation with its short-wavelength end at 0.45 Å. What is the maximum energy (in eV) of a photon in the radiation? And accelerating voltage (for electrons) is required in such a tube?

18b.(i). Calculate the energy released in MeV when neutron breaks into a proton and electron.

(ii) Given that, the half-life of radon is 3.8 days. After how many days will one twentieth $\left(\frac{1}{20}th\right)$ of radon sample will remain?

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COMPLETE SOLUTIONS

SECTION A

Question 1

Object height h = 7 cm, object distance u = −12 cm, f = +8 cm (convex lens)

Lens formula:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{8} + \frac{1}{-12} = \frac{3-2}{24} = \frac{1}{24}$$

$$v = +24 \text{ cm}$$

Magnification:

$$m = \frac{v}{u} = \frac{24}{-12} = -2$$

Image height:

$$h' = m × h = -2 × 7 = \boxed{-14 \text{ cm}}$$

Nature: Real, inverted, magnified image formed 24 cm on the other side of the lens.

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Question 2

Angle of incidence i = 50°, n_glass = 1.52, n_water = 1.33

Snell’s Law:

$$n_g \sin i = n_w \sin r$$

$$\sin r = \frac{n_g \sin i}{n_w} = \frac{1.52 × \sin 50°}{1.33} = \frac{1.52 × 0.766}{1.33} = \frac{1.164}{1.33} = 0.875$$

$$r = \sin^{-1}(0.875) = \boxed{61.05°}$$

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Question 3

Equilateral prism: A = 60°, n = 1.72

Using: $n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

$$1.72 = \frac{\sin\left(\frac{60° + D_m}{2}\right)}{\sin 30°} = \frac{\sin\left(\frac{60° + D_m}{2}\right)}{0.5}$$

$$\sin\left(\frac{60° + D_m}{2}\right) = 1.72 × 0.5 = 0.86$$

$$\frac{60° + D_m}{2} = \sin^{-1}(0.86) = 59.32°$$

$$60° + D_m = 118.64°$$

$$\boxed{D_m = 58.64°}$$

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Question 4

Parallel: C₁ + C₂ = 8 μF …(1)

Series: $\frac{C_1 C_2}{C_1 + C_2} = \frac{15}{8}$ μF

$$C_1 C_2 = \frac{15}{8} × 8 = 15 \text{ μF}^2 \quad ...(2)$$

From (1): C₂ = 8 − C₁

Substituting into (2):

$$C_1(8 - C_1) = 15$$

$$C_1^2 - 8C_1 + 15 = 0$$

$$(C_1 - 3)(C_1 - 5) = 0$$

$$\boxed{C_1 = 3 \text{ μF}, \quad C_2 = 5 \text{ μF}}$$

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Question 5

V = 7.3 sin(200πt − 1.2) Volts

Maximum voltage:

$$V_{max} = \boxed{7.3 \text{ V}}$$

RMS voltage:

$$V_{rms} = \frac{V_{max}}{\sqrt{2}} = \frac{7.3}{\sqrt{2}} = \frac{7.3}{1.414} = \boxed{5.16 \text{ V}}$$

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Question 6

Electric flux density (D): Electric flux per unit area perpendicular to the field.

$$D = \frac{Q}{A} = \varepsilon_0 E \quad \text{Unit: Cm}^{-2}$$

Velocity of conductor:

L = 75 mm = 0.075 m, B = 0.6 T, EMF = 9 V

$$\text{EMF} = BLv$$

$$v = \frac{\text{EMF}}{BL} = \frac{9}{0.6 × 0.075} = \frac{9}{0.045} = \boxed{200 \text{ ms}^{-1}}$$

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Question 7

Standard cell: E₁ = 1.0186 V at L₁ = 400 mm

Dry cell: E₂ = ? at L₂ = 650 mm

Potentiometer principle:

$$\frac{E_1}{E_2} = \frac{L_1}{L_2}$$

$$E_2 = \frac{E_1 × L_2}{L_1} = \frac{1.0186 × 650}{400} = \boxed{1.655 \text{ V}}$$

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Question 8

Conductors: Copper, Aluminium

Semiconductors: Silicon, Germanium

Insulators: Rubber, Glass

Effect of temperature:

• Conductors: Conductivity decreases with increasing temperature (resistance increases)

• Semiconductors: Conductivity increases with increasing temperature (more charge carriers released)

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Question 9

KE = 100 eV = 100 × 1.6 × 10⁻¹⁹ = 1.6 × 10⁻¹⁷ J

de Broglie wavelength: $\lambda = \frac{h}{\sqrt{2mKE}}$

For electron (mₑ = 9.0 × 10⁻³¹ kg):

$$\lambda_e = \frac{6.6×10^{-34}}{\sqrt{2 × 9.0×10^{-31} × 1.6×10^{-17}}}$$

$$= \frac{6.6×10^{-34}}{\sqrt{2.88×10^{-47}}} = \frac{6.6×10^{-34}}{1.697×10^{-23.5}}$$

$$= \frac{6.6×10^{-34}}{5.366×10^{-24}} = \boxed{1.23 × 10^{-10} \text{ m} = 1.23 \text{ Å}}$$

For proton (mₚ = 1.67 × 10⁻²⁷ kg):

$$\lambda_p = \frac{6.6×10^{-34}}{\sqrt{2 × 1.67×10^{-27} × 1.6×10^{-17}}}$$

$$= \frac{6.6×10^{-34}}{\sqrt{5.344×10^{-44}}} = \frac{6.6×10^{-34}}{2.312×10^{-22}}$$

$$= \boxed{2.85 × 10^{-12} \text{ m} = 0.0285 \text{ Å}}$$

The electron has a much larger de Broglie wavelength than the proton for the same KE.

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Question 10

Work function φ = 0.48 eV = 0.48 × 1.6 × 10⁻¹⁹ = 7.68 × 10⁻²⁰ J

Threshold wavelength:

$$\phi = \frac{hc}{\lambda_0}$$

$$\lambda_0 = \frac{hc}{\phi} = \frac{6.6×10^{-34} × 3×10^8}{7.68×10^{-20}}$$

$$= \frac{1.98×10^{-25}}{7.68×10^{-20}} = \boxed{2.578 × 10^{-6} \text{ m} = 2578 \text{ nm}}$$

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SECTION B — GEOMETRIC OPTICS

Question 11a(i)

From Figure 1, the rays converge after reflection:

• Type of mirror: Concave mirror — because it converges reflected rays to a real focus

• Point B: Centre of curvature © — where the two reflected rays cross

• Object: Located beyond C (real object)

• Image: Real, inverted, diminished — formed between F and C on the same side as the object

Question 11a(ii)

Object size = 1 cm, u = −15 cm, f = −10 cm (concave mirror)

Mirror formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

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

$$v = -30 \text{ cm}$$

Magnification:

$$m = -\frac{v}{u} = -\frac{-30}{-15} = -2$$

Image size: |m| × 1 = 2 cm

Results:

• Position: 30 cm in front of mirror (real)

• Nature: Real, inverted, magnified

• Size: 2 cm

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Question 11b

Convex lens: R = 20 cm → f = R/2 = 10 cm

Using lens formula for each position:

| Object distance (u) | Image position (v) | Nature |

|—|---|—|

| 8.5 cm (< f) | Virtual, behind object | Magnified virtual image |

| 15 cm | v = +30 cm | Real, magnified |

| 20 cm (= 2f) | v = +20 cm | Real, same size |

| 25 cm (> 2f) | v = +16.7 cm | Real, diminished |

| 27.7 cm | v = +15.8 cm | Real, diminished |

(i) Real diminished image: Object at 25 cm or 27.7 cm (beyond 2f = 20 cm)

(ii) Magnified real image: Object at 15 cm (between f and 2f)

(iii) Magnified virtual image: Object at 8.5 cm (inside focal length)

(iv) Same size image: Object at 20 cm (at 2f)

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Question 12a(i)

Factors affecting refractive index of kerosene:

• Nature/colour (wavelength) of incident light

• Temperature of the medium

Speed of light in flint glass (n = 1.63):

$$n = \frac{c}{v} \Rightarrow v = \frac{c}{n} = \frac{3.0×10^8}{1.63} = \boxed{1.84 × 10^8 \text{ ms}^{-1}}$$

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Question 12a(ii)

Dispersive power (ω): Ratio of angular dispersion to mean deviation. It measures the ability of a prism to spread light into its spectrum.

$$\omega = \frac{n_v - n_r}{n_{mean} - 1}$$

n_v = 1.5230, n_r = 1.5145

$$n_{mean} = \frac{n_v + n_r}{2} = \frac{1.5230 + 1.5145}{2} = 1.5188$$

$$\omega = \frac{1.5230 - 1.5145}{1.5188 - 1} = \frac{0.0085}{0.5188} = \boxed{0.01638}$$

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Question 12b(i)

When light travels from glass slab → water → air:

• At glass-water interface: angle of refraction > angle of incidence (going less dense)

• At water-air interface: further bending away from normal

• Total internal reflection is possible at the water-air interface if the angle of incidence exceeds the critical angle for water-air boundary (≈ 48.75°)

• It is NOT possible at glass-water boundary since their refractive indices are close, so critical angle is large

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Question 12b(ii)

Power in air P_air = +5 D, ₐnw = 4/3, ₐns = 3/2

Focal length in air:

$$f_{air} = \frac{1}{P_{air}} = \frac{1}{5} = 0.2 \text{ m}$$

Using lensmaker’s equation ratio:

$$\frac{P_{water}}{P_{air}} = \frac{\left(\frac{n_s}{n_w} - 1\right)}{\left(\frac{n_s}{n_a} - 1\right)} = \frac{\left(\frac{3/2}{4/3} - 1\right)}{\left(\frac{3/2}{1} - 1\right)} = \frac{\left(\frac{9}{8} - 1\right)}{\left(\frac{1}{2}\right)} = \frac{\frac{1}{8}}{\frac{1}{2}} = \frac{1}{4}$$

$$P_{water} = \frac{1}{4} × 5 = \boxed{+1.25 \text{ D}}$$

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SECTION C — ELECTRICITY AND MAGNETISM

Question 13a

(i)

• Gold leaf electroscope: An instrument used to detect and measure electric charge, consisting of a metal rod with two thin gold leaves that diverge when charged.

• Electric field strength (E): The force per unit positive charge at a point in a field: $E = \frac{F}{q}$, units: NC⁻¹ or Vm⁻¹

(ii) Ratio fₑ/f_g between two electrons:

r = separation, mₑ = 9.0 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C

$$f_e = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} = \frac{9×10^9 × (1.6×10^{-19})^2}{r^2}$$

$$f_g = \frac{Gm_e^2}{r^2} = \frac{6.67×10^{-11} × (9.0×10^{-31})^2}{r^2}$$

$$\frac{f_e}{f_g} = \frac{9×10^9 × (1.6×10^{-19})^2}{6.67×10^{-11} × (9.0×10^{-31})^2}$$

$$= \frac{9×10^9 × 2.56×10^{-38}}{6.67×10^{-11} × 8.1×10^{-61}}$$

$$= \frac{2.304×10^{-28}}{5.403×10^{-71}} = \boxed{4.26 × 10^{42}}$$

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Question 13b(i)

Electric current: The rate of flow of electric charge through a conductor.

$$I = \frac{Q}{t} \text{ (Amperes)}$$

DC vs AC:

• Direct Current (DC): Flows in one direction only; constant magnitude. Graph: straight horizontal line

• Alternating Current (AC): Periodically reverses direction; sinusoidal. Graph: sine wave

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Question 13b(ii)

Conductance (G): Reciprocal of resistance: $G = \frac{1}{R}$, units: Siemens (S)

Four resistors in parallel:

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$$

Equivalent conductance:

$$G_{eq} = G_1 + G_2 + G_3 + G_4$$

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Question 13b(iii)

2Ω and 5Ω in parallel:

$$R_{parallel} = \frac{2×5}{2+5} = \frac{10}{7} = 1.429 \text{ Ω}$$

Total with 10Ω in series:

$$R_{total} = 1.429 + 10 = 11.429 \text{ Ω}$$

Equivalent conductance:

$$G = \frac{1}{R_{total}} = \frac{1}{11.429} = \boxed{0.0875 \text{ S}}$$

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Question 14a(i)

• Magnetic flux (Φ): Total magnetic field lines passing through a surface: $\Phi = BA\cos\theta$, units: Weber (Wb)

• Magnetic flux density (B): Magnetic flux per unit area: $B = \frac{\Phi}{A}$, units: Tesla (T)

Flux density:

A = 20 cm² = 20 × 10⁻⁴ m², Φ = 3μWb = 3 × 10⁻⁶ Wb

$$B = \frac{\Phi}{A} = \frac{3×10^{-6}}{20×10^{-4}} = \frac{3×10^{-6}}{2×10^{-3}} = \boxed{1.5 × 10^{-3} \text{ T} = 1.5 \text{ mT}}$$

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Question 14a(ii)

Permeability of free space (μ₀): The measure of resistance offered by free space (vacuum) against the formation of a magnetic field. It relates B to H in free space:

$$B = \mu_0 H \Rightarrow \mu_0 = 4\pi × 10^{-7} \text{ Hm}^{-1}$$

Magnetic field strength:

B = 0.25 μT = 0.25 × 10⁻⁶ T, l = 12 mm = 0.012 m

$$H = \frac{B}{\mu_0} = \frac{0.25×10^{-6}}{4\pi×10^{-7}} = \frac{0.25×10^{-6}}{1.257×10^{-6}} = \boxed{0.199 \text{ Am}^{-1}}$$

MMF:

$$MMF = H × l = 0.199 × 0.012 = \boxed{2.387 × 10^{-3} \text{ A} = 2.387 \text{ mA-turns}}$$

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Question 14b(i)

Factors affecting force on current-carrying conductor:

1. Magnitude of current (I)

2. Length of conductor (L)

3. Magnetic flux density (B)

4. Angle between conductor and field (θ)

$$F = BIL\sin\theta$$

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Question 14b(ii)

L = 50 mm = 0.05 m, I = 20 A, B = 0.9 T

At right angles (θ = 90°):

$$F = BIL\sin90° = 0.9 × 20 × 0.05 × 1 = \boxed{0.9 \text{ N}}$$

At θ = 30°:

$$F = BIL\sin30° = 0.9 × 20 × 0.05 × 0.5 = \boxed{0.45 \text{ N}}$$

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Question 15a(i)

Electromagnetic Induction: When a conductor moves through a magnetic field (or when the magnetic flux through a circuit changes), an EMF is induced in the conductor.

Key elements in diagram:

• Magnet (N-S poles)

• Conducting coil/loop

• Galvanometer showing induced current when flux changes

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Question 15a(ii)

Faraday’s Law: The induced EMF in a circuit is directly proportional to the rate of change of magnetic flux linkage through the circuit:

$$\varepsilon = -N\frac{d\Phi}{dt}$$

Lenz’s Law: The direction of the induced current is such that it opposes the change in magnetic flux that produced it (negative sign in Faraday’s law).

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Question 15b(i)

L = 0.58 H, I = 5 A

$$E = \frac{1}{2}LI^2 = \frac{1}{2} × 0.58 × 5^2 = \frac{1}{2} × 0.58 × 25 = \boxed{7.25 \text{ J}}$$

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Question 15b(ii)

From Figure 2 inductors:

• 10 mH and 60 mH appear in series: 10 + 60 = 70 mH

• 70 mH in parallel with 20 mH:

$$L_{parallel} = \frac{70 × 20}{70 + 20} = \frac{1400}{90} = 15.56 \text{ mH}$$

• 15.56 mH in series with 25 mH and 30 mH:

$$L_{total} = 25 + 15.56 + 30 = \boxed{70.56 \text{ mH}}$$

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Question 16a(i)

Farad: A capacitor has capacitance of one Farad when a charge of one Coulomb raises its potential by one Volt: $C = \frac{Q}{V}$

Three capacitors in series: C₁ = 3μF, C₂ = 6μF, C₃ = 12μF, V = 250 V

Equivalent capacitance:

$$\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} + \frac{1}{12} = \frac{4+2+1}{12} = \frac{7}{12}$$

$$C_{eq} = \frac{12}{7} = \boxed{1.714 \text{ μF}}$$

Total charge (same for series):

$$Q = C_{eq} × V = 1.714 × 10^{-6} × 250 = \boxed{4.286 × 10^{-4} \text{ C} = 428.6 \text{ μC}}$$

Voltage across each:

$$V_1 = \frac{Q}{C_1} = \frac{428.6}{3} = \boxed{142.9 \text{ V}}$$

$$V_2 = \frac{Q}{C_2} = \frac{428.6}{6} = \boxed{71.4 \text{ V}}$$

$$V_3 = \frac{Q}{C_3} = \frac{428.6}{12} = \boxed{35.7 \text{ V}}$$

Check: 142.9 + 71.4 + 35.7 = 250 V ✓

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Question 16a(ii)

V_open = 25 V (EMF), V_load = 24 V, I = 10 A

Terminal voltage: V = EMF − Ir

$$24 = 25 − 10r$$

$$r = \frac{25 - 24}{10} = \boxed{0.1 \text{ Ω}}$$

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Question 16b(i)

DC: V = 12 V, I = 2 A → Resistance:

$$R = \frac{V}{I} = \frac{12}{2} = \boxed{6 \text{ Ω}}$$

AC: V = 240 V, I = 20 A, f = 50 Hz → Impedance:

$$Z = \frac{V}{I} = \frac{240}{20} = \boxed{12 \text{ Ω}}$$

Inductive reactance:

$$X_L = \sqrt{Z^2 - R^2} = \sqrt{12^2 - 6^2} = \sqrt{144-36} = \sqrt{108} = \boxed{10.39 \text{ Ω}}$$

Inductance:

$$X_L = 2\pi fL \Rightarrow L = \frac{X_L}{2\pi f} = \frac{10.39}{2\pi × 50} = \frac{10.39}{314.16} = \boxed{0.033 \text{ H} = 33 \text{ mH}}$$

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Question 16b(ii)

Kirchhoff’s Laws:

• KVL: The algebraic sum of all voltages around any closed loop equals zero

• KCL: The algebraic sum of currents at any junction equals zero

From Figure 3: (20V source, 5Ω, 4Ω, 10Ω, 6Ω)

By KCL at junction: $i_1 = i_2 + i_r$ where $i_r$ flows through 10Ω

Loop 1 (outer left):

$$20 = 5i_1 + 10(i_1 - i_2)$$

$$20 = 15i_1 - 10i_2 \quad ...(1)$$

Loop 2 (right):

$$0 = -4i_2 - 6i_2 + 10(i_1 - i_2)$$

$$0 = 10i_1 - 20i_2$$

$$i_1 = 2i_2 \quad ...(2)$$

Substituting (2) into (1):

$$20 = 15(2i_2) - 10i_2 = 30i_2 - 10i_2 = 20i_2$$

$$\boxed{i_2 = 1 \text{ A}}$$

$$\boxed{i_1 = 2 \text{ A}}$$

Voltage V across 10Ω:

$$V = 10(i_1 - i_2) = 10(2-1) = \boxed{10 \text{ V}}$$

Power in 6Ω:

$$P = i_2^2 × 6 = 1^2 × 6 = \boxed{6 \text{ W}}$$

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SECTION D — MODERN PHYSICS

Question 17a(i)

The five hydrogen spectral series:

1. Lyman series (UV region) — transitions to n=1

2. Balmer series (visible) — transitions to n=2

3. Paschen series (IR) — transitions to n=3

4. Brackett series (IR) — transitions to n=4

5. Pfund series (far IR) — transitions to n=5

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Question 17a(ii)

Ionization potential: Minimum energy (in eV) required to completely remove an electron from an atom.

Work function: Minimum energy required to liberate an electron from the surface of a metal.

Maximum velocity of ejected electrons:

λ₀ = 2750 Å = 2750 × 10⁻¹⁰ m (threshold), λ = 1800 Å = 1800 × 10⁻¹⁰ m

Work function:

$$\phi = \frac{hc}{\lambda_0} = \frac{6.6×10^{-34} × 3×10^8}{2750×10^{-10}} = \frac{1.98×10^{-25}}{2.75×10^{-7}} = 7.2×10^{-19} \text{ J}$$

Einstein’s photoelectric equation:

$$\frac{1}{2}m_e v^2 = \frac{hc}{\lambda} - \phi$$

$$= \frac{6.6×10^{-34} × 3×10^8}{1800×10^{-10}} - 7.2×10^{-19}$$

$$= \frac{1.98×10^{-25}}{1.8×10^{-7}} - 7.2×10^{-19}$$

$$= 1.1×10^{-18} - 7.2×10^{-19} = 3.8×10^{-19} \text{ J}$$

$$v = \sqrt{\frac{2 × 3.8×10^{-19}}{9.0×10^{-31}}} = \sqrt{\frac{7.6×10^{-19}}{9.0×10^{-31}}} = \sqrt{8.44×10^{11}}$$

$$\boxed{v = 9.19 × 10^5 \text{ ms}^{-1}}$$

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Question 17b(i)

Three types of radiation in order of penetrating power (least to most):

1. Alpha (α) particles — least penetrating

2. Beta (β) particles — moderately penetrating

3. Gamma (γ) rays — most penetrating

| Property | Radiation |

|—|---|

| Carries negative charge | Beta (β) |

| Similar to X-rays | Gamma (γ) |

| Most easily absorbed | Alpha (α) |

| Travels with greatest speed | Gamma (γ) — speed of light |

| Not deflected by magnetic field | Gamma (γ) |

| Emitted when ²³⁸₉₂U → ²³⁴₉₀Th | Alpha (α) (mass drops by 4, Z drops by 2) |

| Similar to cathode rays | Beta (β) |

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Question 17b(ii)

Radioactive isotope: An atom with an unstable nucleus that spontaneously emits radiation to become more stable.

Decay equation for Carbon-14:

$$^{14}_{6}C \rightarrow ^{14}_{7}N + ^{0}_{-1}\beta + \bar{\nu}_e$$

¹⁴₆C has:

• Protons: 6 (atomic number)

• Neutrons: 14 − 6 = 8 neutrons

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Question 18a(i)

Production of X-rays in a Coolidge tube:

Components: heated cathode (electron source), anode (tungsten target), evacuated tube, high voltage supply.

Process: Electrons accelerated from cathode to anode; upon striking target they decelerate rapidly → X-rays emitted (Bremsstrahlung) or knock out inner electrons → characteristic X-rays.

• Soft X-rays: Low frequency, long wavelength, low penetrating power (low accelerating voltage)

• Hard X-rays: High frequency, short wavelength, high penetrating power (high accelerating voltage)

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Question 18a(ii)

λ_min = 0.45 Å = 0.45 × 10⁻¹⁰ m

Maximum photon energy:

$$E_{max} = \frac{hc}{\lambda_{min}} = \frac{6.6×10^{-34} × 3×10^8}{0.45×10^{-10}}$$

$$= \frac{1.98×10^{-25}}{0.45×10^{-10}} = 4.4×10^{-15} \text{ J}$$

In eV:

$$E = \frac{4.4×10^{-15}}{1.6×10^{-19}} = \boxed{27,500 \text{ eV} = 27.5 \text{ keV}}$$

Accelerating voltage:

$$eV = E_{max}$$

$$V = \frac{E_{max}}{e} = \frac{4.4×10^{-15}}{1.6×10^{-19}} = \boxed{27,500 \text{ V} = 27.5 \text{ kV}}$$

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Question 18b(i)

Neutron → Proton + Electron

Masses:

• Neutron: 1.008665 u

• Proton: 1.007276 u

• Electron: 0.000549 u

Mass defect:

$$\Delta m = 1.008665 - (1.007276 + 0.000549) = 1.008665 - 1.007825 = 0.000840 \text{ u}$$

Energy released:

$$E = \Delta m × 931 \text{ MeV/u} = 0.000840 × 931 = \boxed{0.782 \text{ MeV}}$$

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Question 18b(ii)

T½ = 3.8 days, fraction remaining = 1/20

$$\frac{N}{N_0} = \left(\frac{1}{2}\right)^n = \frac{1}{20}$$

where n = number of half-lives

$$\left(\frac{1}{2}\right)^n = \frac{1}{20}$$

Taking logarithm:

$$n\ln\left(\frac{1}{2}\right) = \ln\left(\frac{1}{20}\right)$$

$$-n\ln 2 = -\ln 20$$

$$n = \frac{\ln 20}{\ln 2} = \frac{2.996}{0.693} = 4.322$$

Total time:

$$t = n × T_{1/2} = 4.322 × 3.8 = \boxed{16.42 \text{ days}}$$