MITIHANI POPOTE PRE NATIONAL EXAMINATION PHYSICS 2 - SERIES 4 (With Comprehensive Solutions)

PRE NATIONAL EXAMINATION PHYSICS 2 - SERIES 4

MITIHANI POPOTE EXAMINATION SERIES

FORM SIX SECONDARY SCHOOL

PRE-NATIONAL EXAMINATION SERIES

PHYSICS 2 - SERIES 4

131/02

TIME: 2:30 HRS

JANUARY - MAY, 2023

INSTRUCTIONS

This paper consists of Six (6) questions.
Answer any Five (5) questions.
Each question carries twenty (20) marks.
Mathematical tables and non-programmable calculators may be used.
Cellular phones and any unauthorized materials are not allowed in the examination room.
Write your Examination Number on every page of your answer booklet(s).
The following information may be useful:

(a) Acceleration due to gravity, \( (g) = 9.8 \, \text{m/s}^2 \)

(b) Pie, \( (\pi) = 3.14 \)

(c) Young's modulus of copper wire, \( (E) = 1.2 \times 10^{11} \, \text{N/m}^2 \)

(d) Young's modulus of rubber, \( (E) = 6 \times 10^8 \, \text{N/m}^2 \)

(e) Surface tension of soap solution, \( (y) = 2.5 \times 10^{-2} \, \text{N/m} \)

(f) \( \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 / \text{C}^2 \)

(g) Radius of the earth, \( (\tau_e) = 6.4 \times 10^6 \, \text{m} \)

(h) Permeability of free space, \( (\mu_0) = 4\pi \times 10^{-7} \, \text{H/m} \)

(i) Electronic charge, \( (e) = 1.6 \times 10^{-19} \, \text{C} \)

(j) Planck's constant, \( (h) = 6.63 \times 10^{-34} \, \text{Js} \)

(k) Avogadro's number, \( (N_A) = 6.02 \times 10^{23} \, \text{per mole} \)

(l) Speed of light in vacuum, \( (c) = 3 \times 10^8 \, \text{m/s} \)

(m) Permittivity of free space, \( (\varepsilon_0) = 8.85 \times 10^{-12} \, \text{F/m} \)

Answer any five (5) questions.

1. (20 marks)

(a) (i) What are the three applications of fluid dynamics in daily life (03 marks)

Solution 1(a)(i):

Three applications of fluid dynamics in daily life:

Aerodynamics of vehicles: Design of cars, airplanes, and trains to reduce drag and improve fuel efficiency.
Plumbing systems: Design of water supply and drainage systems in buildings.
Medical applications: Blood flow in arteries and veins, design of heart valves, and respiratory airflow.
Weather forecasting: Study of atmospheric flow patterns and prediction of weather systems.
Hydraulic systems: Operation of brakes, lifts, and excavators using pressurized fluids.

(ii) For the same applied force in honey and water, the rate of flow of honey will be less than that of water. Explain why (02 marks)

Solution 1(a)(ii):

Honey has higher viscosity than water.

Viscosity (η) is a measure of a fluid's resistance to flow.

According to Poiseuille's law for flow through a pipe: \( Q = \frac{\pi r^4 \Delta P}{8 \eta L} \)

For the same applied force (ΔP), length (L), and radius (r):

\( Q \propto \frac{1}{\eta} \)

Since η_honey > η_water, then Q_honey < Q_water

Honey has higher viscosity, creating more internal friction and resistance to flow.

(b) A cylindrical tank of 10 cm in radius rests on platform 5 m high. Initially, the tank was filled with water to a height of 5 m. If the plug of area \( 1 \times 10^{-4} \, \text{m}^2 \) is removed by an orifice on the side of the tank at the bottom. Calculate (06 marks)

(i) The initial speed with which water flows from the orifice. (03 marks)

Solution 1(b)(i):

Using Torricelli's theorem: \( v = \sqrt{2gh} \)

Where h = height of water above orifice = 5 m

g = 9.8 m/s²

\( v = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} = 9.899 \, \text{m/s} \)

Initial speed from orifice = 9.9 m/s

(ii) The speed with which water strikes the ground. (03 marks)

Solution 1(b)(ii):

Water emerges horizontally from orifice at platform height H = 5 m

Horizontal velocity v_x = 9.9 m/s (from part i)

Vertical motion: Initial vertical velocity v_y = 0, falling from height 5 m

Time to fall: \( t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2 \times 5}{9.8}} = \sqrt{1.0204} = 1.01 \, \text{s} \)

Vertical velocity at impact: \( v_y = gt = 9.8 \times 1.01 = 9.898 \, \text{m/s} \)

Resultant velocity: \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{9.9^2 + 9.898^2} = \sqrt{98.01 + 97.96} = \sqrt{195.97} = 14.0 \, \text{m/s} \)

Speed when striking ground = 14.0 m/s

(c) (i) Use the Poiseuille's theorem to show that the height of the liquid at any particular point of the volume at any time is \( h = h_0 e^{-ct} \), whereby \( h \) is the final height of the liquid, \( h_0 \) is the original height of the liquid, \( t \) is the time taken by the liquid to flow and \( c \) is constant. (04 marks)

Solution 1(c)(i):

Poiseuille's law: \( Q = \frac{\pi r^4 \Delta P}{8 \eta L} \)

For liquid flowing out of a tank: ΔP = ρgh

Flow rate Q = -A(dh/dt) (negative because h decreases)

Where A = cross-sectional area of tank

Equating: \( -A\frac{dh}{dt} = \frac{\pi r^4 \rho g h}{8 \eta L} \)

Let \( k = \frac{\pi r^4 \rho g}{8 \eta L A} \)

Then: \( \frac{dh}{dt} = -kh \)

Separating variables: \( \frac{dh}{h} = -k dt \)

Integrating: \( \ln h = -kt + C \)

At t=0, h = h₀: C = ln h₀

So: \( \ln h = -kt + \ln h₀ \)

\( \ln\left(\frac{h}{h₀}\right) = -kt \)

\( \frac{h}{h₀} = e^{-kt} \)

\( h = h₀ e^{-kt} \) where k is constant

Hence proved: \( h = h_0 e^{-ct} \) where c = k

(ii) Two tubes AB and BC of the same length and having radii of 2 mm and 1 mm respectively are joined end to end at B, a liquid flows through them and the difference between A and C is 340 mmHg. What is the difference in pressure between B and C? (05 marks)

Solution 1(c)(ii):

For series flow: Q = constant through both tubes

Poiseuille's law: \( Q = \frac{\pi r^4 \Delta P}{8 \eta L} \)

For tube AB (r₁ = 2 mm): \( Q = \frac{\pi (2 \times 10^{-3})^4 \Delta P_{AB}}{8 \eta L} \)

For tube BC (r₂ = 1 mm): \( Q = \frac{\pi (1 \times 10^{-3})^4 \Delta P_{BC}}{8 \eta L} \)

Since Q and L are same: \( \frac{\Delta P_{AB}}{r_1^4} = \frac{\Delta P_{BC}}{r_2^4} \)

\( \frac{\Delta P_{AB}}{\Delta P_{BC}} = \frac{r_1^4}{r_2^4} = \frac{(2)^4}{(1)^4} = 16 \)

So ΔP_AB = 16 ΔP_BC

Total ΔP_AC = ΔP_AB + ΔP_BC = 340 mmHg

16 ΔP_BC + ΔP_BC = 340

17 ΔP_BC = 340

ΔP_BC = 20 mmHg

Pressure difference between B and C = 20 mmHg

2. (20 marks)

(a) (i) Briefly explain why after being used for a long time, spring balances shows wrong readings. (02 marks)

Solution 2(a)(i):

Spring balances show wrong readings after long use due to:

Elastic fatigue: Spring loses its elasticity over time due to repeated stretching and compression.
Permanent deformation: Spring may not return to original length after load is removed.
Metal fatigue: Crystals in spring material rearrange, changing spring constant.
Corrosion: Rust or corrosion can affect spring properties.
Calibration drift: Gradual change in spring constant over time.

(ii) We often wet the end of the thread before putting it through the eye of the needle. Briefly explain why? (02 marks)

Solution 2(a)(ii):

Wetting the thread helps because:

Surface tension: Water forms a film around thread, holding fibers together.
Reduces fraying: Keeps loose fibers together, making thread stiffer.
Easier alignment: Stiffer thread is easier to guide through needle eye.
Lubrication: Water acts as lubricant, reducing friction with needle.

(b) A string under tension is observed to vibrate with a frequency of 40 Hz in its fundamental mode, when the supports are 0.6 m apart. The amplitude of the note is 3 cm. If the string has a mass of 30g, calculate (06 marks)

(i) The velocity of propagation of the transverse wave in string (03 marks)

Solution 2(b)(i):

For fundamental mode: \( f = \frac{v}{2L} \)

Where f = 40 Hz, L = 0.6 m

\( v = 2fL = 2 \times 40 \times 0.6 = 48 \, \text{m/s} \)

Wave velocity = 48 m/s

(ii) The tension in the string. (03 marks)

Solution 2(b)(ii):

Wave velocity on string: \( v = \sqrt{\frac{T}{\mu}} \)

Where T = tension, μ = mass per unit length

Mass m = 30 g = 0.03 kg, Length L = 0.6 m

\( \mu = \frac{m}{L} = \frac{0.03}{0.6} = 0.05 \, \text{kg/m} \)

From (i): v = 48 m/s

\( T = \mu v^2 = 0.05 \times (48)^2 = 0.05 \times 2304 = 115.2 \, \text{N} \)

Tension in string = 115.2 N

(c) (i) By means of equation or expression show how the temperature and density affect the velocity of sound waves in air. (06 marks)

Solution 2(c)(i):

Velocity of sound in gas: \( v = \sqrt{\frac{\gamma P}{\rho}} \)

Where γ = adiabatic index, P = pressure, ρ = density

From ideal gas law: \( P = \frac{\rho RT}{M} \) where M = molar mass

Substituting: \( v = \sqrt{\frac{\gamma \rho RT}{\rho M}} = \sqrt{\frac{\gamma RT}{M}} \)

Thus: \( v \propto \sqrt{T} \) (velocity increases with temperature)

At constant temperature: \( v = \sqrt{\frac{\gamma P}{\rho}} \propto \frac{1}{\sqrt{\rho}} \) (velocity decreases with increasing density)

For air: \( v = 331.4 + 0.6T_c \, \text{m/s} \) where T_c is temperature in °C

Velocity increases with √T and decreases with √ρ

(ii) Ultra sound of frequency \( f \) is incident to a blood vessel of a patient at an angle \( \theta \). The sound is from a transducer that sends Ultrasound and detects the sound reflected from a blood vessel of a diameter d. Derive an expression for the volume flow rate \( Q \) of the blood in patient's blood vessel. (04 marks)

Solution 2(c)(ii):

Using Doppler effect for ultrasound flow measurement:

Frequency shift: \( \Delta f = \frac{2fv \cos\theta}{c} \)

Where v = blood velocity, c = speed of sound in blood

Thus: \( v = \frac{c \Delta f}{2f \cos\theta} \)

For laminar flow in cylindrical vessel: Average velocity \( \bar{v} = \frac{v_{max}}{2} \)

Volume flow rate: \( Q = A \bar{v} = \frac{\pi d^2}{4} \times \frac{v}{2} = \frac{\pi d^2 v}{8} \)

Substituting v: \( Q = \frac{\pi d^2}{8} \times \frac{c \Delta f}{2f \cos\theta} = \frac{\pi d^2 c \Delta f}{16f \cos\theta} \)

\( Q = \frac{\pi d^2 c \Delta f}{16f \cos\theta} \)

3. (20 marks)

(a) (i) Why work is said to be done in stretching wire? (02 marks)

Solution 3(a)(i):

Work is done in stretching a wire because:

A force is applied to stretch the wire.
The point of application of force moves in the direction of force.
Energy is transferred to the wire, stored as elastic potential energy.
Work done = Force × Displacement in direction of force.
For small extension dx: dW = F dx
Total work: \( W = \int F \, dx \)

(ii) A uniform copper wire of length 0.6 m and cross sectional area of \(1.8 \times 10^6 \, \text{m}^2\) is extended 0.8 mm. What is the strain energy in the wire if the elastic limit is not exceeded? What is the strain energy per unit volume of the wire? (04 marks)

Solution 3(a)(ii):

Given: L = 0.6 m, A = 1.8 × 10⁶ m² (likely 1.8 × 10⁻⁶ m²), ΔL = 0.8 mm = 8 × 10⁻⁴ m

Young's modulus for copper E = 1.2 × 10¹¹ N/m²

Strain energy: \( U = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume} \)

Stress = E × strain = \( E \times \frac{\Delta L}{L} \)

Strain = \( \frac{\Delta L}{L} = \frac{8 \times 10^{-4}}{0.6} = 1.333 \times 10^{-3} \)

Stress = 1.2 × 10¹¹ × 1.333 × 10⁻³ = 1.6 × 10⁸ N/m²

Volume V = A × L = (1.8 × 10⁻⁶) × 0.6 = 1.08 × 10⁻⁶ m³

Strain energy U = ½ × stress × strain × volume

U = ½ × 1.6 × 10⁸ × 1.333 × 10⁻³ × 1.08 × 10⁻⁶ = 0.1152 J

Strain energy per unit volume = ½ × stress × strain = ½ × 1.6 × 10⁸ × 1.333 × 10⁻³ = 1.066 × 10⁵ J/m³

Strain energy = 0.115 J, Strain energy per unit volume = 1.07 × 10⁵ J/m³

(b) A rubber cord of a catapult has a cross sectional area of 2 mm² and an initial length of 0.2 m is stretched to 0.24 m to fire a small object of mass 10 g. Assuming that the elastic limit is not exceeded, calculate the initial velocity of the object when it is released. (04 marks)

Solution 3(b):

Given: A = 2 mm² = 2 × 10⁻⁶ m², L₀ = 0.2 m, L = 0.24 m, m = 10 g = 0.01 kg

Young's modulus for rubber E = 6 × 10⁸ N/m²

Extension ΔL = 0.24 - 0.2 = 0.04 m

Strain = ΔL/L₀ = 0.04/0.2 = 0.2

Stress = E × strain = 6 × 10⁸ × 0.2 = 1.2 × 10⁸ N/m²

Force F = stress × A = 1.2 × 10⁸ × 2 × 10⁻⁶ = 240 N

Strain energy U = ½ F ΔL = ½ × 240 × 0.04 = 4.8 J

This energy converts to kinetic energy: ½ mv² = U

½ × 0.01 × v² = 4.8

v² = (4.8 × 2)/0.01 = 960

v = √960 = 30.98 m/s

Initial velocity of object = 31 m/s

Solution 3(c)(i):

Three evidences of surface tension:

Water droplets: Water forms spherical droplets due to surface tension minimizing surface area.
Floating objects: Small insects (water striders) and needles can float on water surface.
Capillary action: Water rises in thin tubes against gravity.
Soap bubbles: Formation of thin soap films that maintain spherical shape.
Meniscus formation: Curved surface at edges of water in containers.

(ii) Calculate the work done against surface tension forces in blowing a soap bubble of 1 cm in diameter. (03 marks)

Solution 3(c)(ii):

For soap bubble: Two surfaces (inside and outside)

Surface tension γ = 2.5 × 10⁻² N/m (soap solution)

Diameter d = 1 cm = 0.01 m, Radius r = 0.005 m

Increase in surface area = 2 × 4πr² = 8πr² (for two surfaces)

Work done W = γ × increase in area = γ × 8πr²

W = 2.5 × 10⁻² × 8 × 3.14 × (0.005)²

W = 2.5 × 10⁻² × 8 × 3.14 × 2.5 × 10⁻⁵

W = 2.5 × 10⁻² × 6.28 × 10⁻⁴ = 1.57 × 10⁻⁵ J

Work done = 1.57 × 10⁻⁵ J

(iii) Estimate the radius of a single droplet when the rain drop of radius 0.5mm strikes the surface and breaks to 125 droplets of equal size. (04 marks)

Solution 3(c)(iii):

Volume conserved: Volume of big drop = Total volume of small droplets

Let R = radius of big drop = 0.5 mm = 5 × 10⁻⁴ m

Let r = radius of each small droplet

Number of droplets n = 125

\( \frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \)

\( R^3 = n r^3 \)

\( r^3 = \frac{R^3}{n} = \frac{(5 \times 10^{-4})^3}{125} = \frac{1.25 \times 10^{-10}}{125} = 1 \times 10^{-12} \)

\( r = (1 \times 10^{-12})^{1/3} = 1 \times 10^{-4} \, \text{m} = 0.1 \, \text{mm} \)

Radius of each small droplet = 0.1 mm

4. (20 marks)

(a) (i) Briefly explain the importance of dielectric material that are inserted in between the plates of a capacitor. (02 marks)

Solution 4(a)(i):

Dielectric materials are important because:

Increase capacitance: C = κε₀A/d, where κ > 1 for dielectrics.
Increase breakdown voltage: Can withstand higher voltages before breakdown.
Provide mechanical support: Keep plates separated and prevent contact.
Reduce size: For same capacitance, capacitor with dielectric can be smaller.

(ii) A parallel air plate capacitor is charged to a p.d of 300 V and then connected in parallel with another capacitor of equal dimension but with ebonite as a dielectric, the p.d is found to be 75 V. Calculate the dielectric constant of the ebonite. (05 marks)

Solution 4(a)(ii):

Let C₁ = capacitance of air capacitor

Initial charge Q = C₁ × 300

When connected in parallel with ebonite capacitor C₂ = κC₁ (κ = dielectric constant)

Total capacitance C_total = C₁ + κC₁ = C₁(1 + κ)

Charge conserved: Q = C₁ × 300 = C₁(1 + κ) × 75

Canceling C₁: 300 = (1 + κ) × 75

1 + κ = 300/75 = 4

κ = 4 - 1 = 3

Dielectric constant of ebonite = 3

(b) (i) Assuming that the earth to be an isolated conducting sphere. What is its capacitance? (03 marks)

Solution 4(b)(i):

For isolated conducting sphere: C = 4πε₀R

Earth radius R = 6.4 × 10⁶ m

ε₀ = 8.85 × 10⁻¹² F/m

C = 4 × 3.14 × 8.85 × 10⁻¹² × 6.4 × 10⁶

C = 4 × 3.14 × 5.664 × 10⁻⁵ = 7.12 × 10⁻⁴ F

Earth's capacitance = 712 μF

(ii) In a Van de Graff generator, the shell electrode is at \(25 \times 10^5 \, \text{V}\). If the dielectric strength of the gas surrounding the electrode is \(5 \times 10^7 \, \text{V/m}\), calculate the minimum radius of the spherical shell. (03 marks)

Solution 4(b)(ii):

For sphere: Electric field at surface E = \( \frac{Q}{4\pi\varepsilon_0 R^2} \) = \( \frac{V}{R} \)

Given V = 25 × 10⁵ V, E_max = 5 × 10⁷ V/m

\( E_{max} = \frac{V}{R_{min}} \)

\( R_{min} = \frac{V}{E_{max}} = \frac{25 \times 10^5}{5 \times 10^7} = 5 \times 10^{-2} \, \text{m} = 5 \, \text{cm} \)

Minimum radius = 5 cm

(i) Find the initial discharge current. (02 marks)

Solution 4(c)(i):

Initial voltage V₀ = 50 V, Resistance R = 500 Ω

Initial current I₀ = V₀/R = 50/500 = 0.1 A

Initial discharge current = 0.1 A

(ii) How long it will take for the p.d across the capacitor to fall to 2% of its original value. (03 marks)

Solution 4(c)(ii):

Discharge equation: V = V₀e^{-t/RC}

C = 300 pF = 300 × 10⁻¹² F, R = 500 Ω

RC = 500 × 300 × 10⁻¹² = 1.5 × 10⁻⁷ s

V/V₀ = 0.02 = e^{-t/RC}

ln(0.02) = -t/RC

t = -RC × ln(0.02) = -1.5 × 10⁻⁷ × (-3.912) = 5.868 × 10⁻⁷ s

Time = 0.587 μs

(iii) What is the value of time constant? (02 marks)

Solution 4(c)(iii):

Time constant τ = RC = 500 × 300 × 10⁻¹² = 1.5 × 10⁻⁷ s

Time constant = 150 ns

5. (20 marks)

(a) (i) Distinguish between diamagnetic, paramagnetic and ferromagnetic materials on the basis of relative permeability μᵣ. (03 marks)

Solution 5(a)(i):

Based on relative permeability μᵣ = μ/μ₀:

Diamagnetic: μᵣ < 1 (slightly less than 1), typically 0.999 to 0.9999
Paramagnetic: μᵣ > 1 (slightly greater than 1), typically 1.0001 to 1.001
Ferromagnetic: μᵣ >> 1 (much greater than 1), typically 10³ to 10⁵ or more

(ii) What are the three independent quantities which are conventionally used to describe the magnetic field on the earth's surface. (03 marks)

Solution 5(a)(ii):

Three quantities describing Earth's magnetic field:

Magnetic declination (D): Angle between geographic north and magnetic north.
Magnetic inclination/dip (I): Angle between horizontal plane and magnetic field.
Horizontal component (H): Horizontal component of Earth's magnetic field.

(b) (i) Why the material used for making the core of a transformer should have narrow hysteresis loop? (02 marks)

Solution 5(b)(i):

Transformer core should have narrow hysteresis loop because:

Reduces energy loss: Area of hysteresis loop represents energy loss per cycle.
Reduces heating: Less energy loss means less heat generation.
Improves efficiency: More energy transferred, less wasted.
Easier magnetization: Requires less energy to magnetize and demagnetize.

(ii) A specimen of iron is uniformly magnetized by the magnetizing field of 300 Am⁻¹. If the magnetic flux density in the specimen is 0.4 Wbm⁻², find the relative permeability, susceptibility and the permeability of the specimen. (06 marks)

Solution 5(b)(ii):

Given: H = 300 A/m, B = 0.4 Wb/m²

μ₀ = 4π × 10⁻⁷ H/m

B = μH = μ₀μᵣH

μᵣ = B/(μ₀H) = 0.4/(4π × 10⁻⁷ × 300)

μᵣ = 0.4/(3.7699 × 10⁻⁴) = 1061.0

Permeability μ = μ₀μᵣ = 4π × 10⁻⁷ × 1061 = 1.333 × 10⁻³ H/m

Susceptibility χ = μᵣ - 1 = 1061 - 1 = 1060

Relative permeability = 1061, Susceptibility = 1060, Permeability = 1.33 × 10⁻³ H/m

(c) Two identical wires R and S lie parallel in a horizontal plane, the axis being 10 cm apart. A current of 10 A flows in R and in the opposite direction to a current of 30 A flows in S. Neglecting the effect of the earth's magnetic flux density, calculate the magnitude of B at a point P in the plane of the wire, if P is; (06 marks)

(i) Midway between R and S. (03 marks)

Solution 5(c)(i):

Distance between wires = 10 cm = 0.1 m

At midpoint: distance from each wire = 0.05 m

Field due to wire R: \( B_R = \frac{\mu_0 I_R}{2\pi r} = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 0.05} = 4 \times 10^{-5} \, \text{T} \)

Direction: Using right-hand rule, both fields are in same direction (into page or out of page depending on current directions)

Field due to wire S: \( B_S = \frac{\mu_0 I_S}{2\pi r} = \frac{4\pi \times 10^{-7} \times 30}{2\pi \times 0.05} = 12 \times 10^{-5} \, \text{T} \)

Total B = B_R + B_S = (4 + 12) × 10⁻⁵ = 16 × 10⁻⁵ T = 1.6 × 10⁻⁴ T

B at midpoint = 1.6 × 10⁻⁴ T

(ii) 5 cm from R and 15 cm from S. (03 marks)

Solution 5(c)(ii):

Distance from R = 0.05 m, from S = 0.15 m

\( B_R = \frac{\mu_0 I_R}{2\pi r_R} = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 0.05} = 4 \times 10^{-5} \, \text{T} \)

\( B_S = \frac{\mu_0 I_S}{2\pi r_S} = \frac{4\pi \times 10^{-7} \times 30}{2\pi \times 0.15} = 4 \times 10^{-5} \, \text{T} \)

Since currents are opposite, fields are in opposite directions

Net B = |B_R - B_S| = |4 - 4| × 10⁻⁵ = 0 T

B at this point = 0 T (fields cancel exactly)

6. (20 marks)

(a) (i) Explain how nuclear reactor produces electricity. (03 marks)

Solution 6(a)(i):

Nuclear reactor produces electricity through:

Nuclear fission: Uranium-235 nuclei split when hit by neutrons, releasing energy.
Heat generation: Fission releases heat in reactor core.
Coolant system: Heat is transferred to coolant (water, gas, or liquid metal).
Heat exchanger: Hot coolant produces steam in secondary loop.
Turbine: Steam drives turbine connected to generator.
Generator: Turbine rotates electromagnetic to produce electricity.
Condenser: Steam is cooled back to water and recycled.

(ii) Why is a neutron most effective as a bullet in nuclear reactions. (03 marks)

Solution 6(a)(ii):

Neutrons are most effective because:

No charge: Neutrons are electrically neutral, not repelled by positively charged nuclei.
Easy penetration: Can approach and enter nuclei without Coulomb barrier.
Slow neutrons: Thermal neutrons have high probability of causing fission.
Chain reaction: Fission releases more neutrons, sustaining chain reaction.
Moderation: Can be slowed by moderators to optimal energy for fission.

(b) (i) Briefly explain why alpha particle and gold foil were used in Rutherford scattering experiment (03 marks)

Solution 6(b)(i):

Alpha particles were used because:

Have positive charge (2+) - interact with nucleus via Coulomb force.
Relatively massive (4 amu) - not easily deflected by electrons.
Available from radioactive sources (e.g., radium).

Gold foil was used because:

Can be made very thin (few atoms thick) - reduces multiple scattering.
High atomic number (Z=79) - strong Coulomb interaction with alpha.
Malleable - can be made into extremely thin foils.

(ii) Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 9750 nm. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them, assuming the ionization energy of hydrogen atom is 13.6 eV. (05 marks)

Solution 6(b)(ii):

Energy of photon: E = hc/λ

λ = 9750 nm = 9.75 × 10⁻⁶ m

h = 6.63 × 10⁻³⁴ Js, c = 3 × 10⁸ m/s

E = (6.63×10⁻³⁴ × 3×10⁸)/(9.75×10⁻⁶) = 2.04 × 10⁻²⁰ J

Convert to eV: 1 eV = 1.6×10⁻¹⁹ J, so E = 0.1275 eV

Ground state energy E₁ = -13.6 eV

Excited state energy E_n = E₁ + 0.1275 = -13.4725 eV

For hydrogen: E_n = -13.6/n² eV

-13.6/n² = -13.4725

n² = 13.6/13.4725 = 1.0095, n ≈ 1.0047 ≈ 1

This suggests the photon energy is too small to excite from n=1.

If we assume the atom is excited to some level n, then:

Number of spectral lines = n(n-1)/2

Longest wavelength corresponds to smallest energy transition (n to n-1)

Assuming excitation to n=2: 1 line possible, λ = 121.6 nm (Lyman alpha)

Solution 6(c)(i):

Nuclear fission:

Heavy nucleus splits into lighter nuclei.
Neutron-induced reaction.
Releases ~200 MeV per fission.
Used in nuclear reactors and atomic bombs.
Example: \( ^{235}_{92}U + n → ^{141}_{56}Ba + ^{92}_{36}Kr + 3n \)

Nuclear fusion:

Light nuclei combine to form heavier nucleus.
Requires extremely high temperature and pressure.
Releases more energy per nucleon than fission.
Occurs in stars, hydrogen bombs.
Example: \( ^{2}_{1}H + ^{3}_{1}H → ^{4}_{2}He + n + 17.6 MeV \)

(ii) Assuming that the energy released by the fission of a single \(_{92}U^{235}\) nucleus is 200MeV. Calculate the number of fission per second required to produce 1 watt of power. (04 marks)

Solution 6(c)(ii):

Energy per fission = 200 MeV = 200 × 10⁶ eV

1 eV = 1.6 × 10⁻¹⁹ J

Energy per fission in joules = 200 × 10⁶ × 1.6 × 10⁻¹⁹ = 3.2 × 10⁻¹¹ J

Power = 1 W = 1 J/s

Number of fissions per second = Power / Energy per fission

= 1 / (3.2 × 10⁻¹¹) = 3.125 × 10¹⁰ fissions/s

3.125 × 10¹⁰ fissions per second

MITIHANI POPOTE

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