MITIHANI POPOTE EXAMINATION SERIES | FORM SIX PRE NATIONAL EXAM SERIES 8 | PHYSICS 2 (With Comprehensive Solution)

Pre-National Examination Physics 2 - Series 8

UMOJA WA WAZAZI TANZANIA
WARI SECONDARY SCHOOL

PRE-NATIONAL EXAMINATION SERIES
PHYSICS 2 - SERIES 8
131/02 | TIME: 2:30 HRS
JANUARY-MAY, 2023
Paper: 6 Questions
Answer: Any 5 Questions
Marks: 20 per question

Instructions

  1. This paper consists of Six (6) questions.
  2. Answer any Five (5) questions.
  3. Each question carries twenty (20) marks.
  4. Mathematical tables and non-programmable calculators may be used.

Useful Information

Acceleration due to gravity, \( g \)\( 10 \, \text{m/s}^2 \)
Pie, \( \pi \)3.14
Density of air, \( \rho_a \)\( 1.3 \, \text{kg/m}^3 \)
Plank's constant, \( h \)\( 6.63 \times 10^{-34} \, \text{Js} \)
Speed of light in air, \( C \)\( 3 \times 10^8 \, \text{m/s} \)
Mass of proton, \( m_p \)\( 1.67 \times 10^{-27} \, \text{Kg} \)
Permittivity of free space, \( \varepsilon_0 \)\( 8.85 \times 10^{-12} \, \text{F/m} \)
Electronic charge, \( e \)\( 1.6 \times 10^{-19} \, \text{C} \)
Boltzman constant, \( k \)\( 1.38 \times 10^{-23} \, \text{J/K} \)
Molar gas constant, \( R \)\( 8.31 \, \text{J/molK} \)
Viscosity of mercury, \( n \)\( 1.55 \times 10^{-3} \, \text{Ns/m}^2 \)
Density of mercury, \( \rho_m \)\( 13600 \, \text{g/m}^3 \)
Mass of electron, \( m_e \)\( 9.1 \times 10^{-31} \, \text{Kg} \)
Permeability of free space, \( \mu_0 \)\( 4\pi \times 10^{-7} \, \text{H/m} \)
Avogadro's number, \( N_A \)\( 6.02 \times 10^{23} \, \text{atoms/mol} \)
Molar mass of copper63.5 g/mol
QUESTION 1
(a) (06 marks)

(i) Distinguish between Static pressure and Dynamic pressure. (02 marks)

(ii) A small oil drop falls with terminal velocity of \( 4.0 \times 10^{-4} \, m/s \). What is the new terminal velocity for an oil drop of radius half of the original radius. (04 marks)

(b) (07 marks)

(i) Write stoke's equation as applied to motion of a body in viscous medium. Under what conditions is the stoke's equation valid? (03 marks)

(ii) A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the sides. If the radius of the vessel is \( 0.05 \, m \) and the speed of rotation is \( 2 \, rev/s \), find the difference in the height of the liquid at the center of the vessel and its sides. (04 marks)

(c) (07 marks)

(i) Give two applications of Bernoulli's theorem. (02 marks)

(ii) Liquid mercury flows through a horizontal pipe of internal radius \( 1.88 \, cm \) and length \( 1.26 \, m \). The volume flux is \( 5.35 \times 10^{-2} \, L/min \). Describe how you will show that the flow is laminar and hence calculate the difference in pressure between the two sides of the pipe. (05 marks)

QUESTION 2
(a) (04 marks)

(i) Explain why there is usually a time interval between observing a flash and hearing a thunder? (02 marks)

(ii) How is energy transmitted in wave motion? (02 marks)

(b) (06 marks)

The wave function for a standing wave in a string is given by \( y = 0.3 \, \sin(0.25x) \, \cos(120\pi t) \) where \( x \) and \( y \) are in meters and \( t \) in seconds.

(i) Determine the wavelength and frequency of the interfering travelling waves. (03 marks)

(ii) Also, determine the equations of the interfering waves. (03 marks)

(c) (04 marks)

(i) Differentiate the two main types of diffraction (02 marks)

(ii) Briefly explain two ways in which young's experiment can be improved. (02 marks)

(d) (06 marks)

A light with a wavelength of \( 590 \, nm \) is incident normally on a diffraction grating. The angle of the second order maxima is \( 32^\circ \).

(i) Calculate the spacing of the lines on the grating (03 marks)

(ii) How many lines per centimeter (03 marks)

QUESTION 3
(a) (07 marks)

(i) Differentiate the terms creep and fatigue. (02 marks)

(ii) A stone of \( 0.5kg \) mass is attached to one end of a \( 0.8m \) long aluminum wire of \( 0.7mm \) diameter and supported vertically. The stone is now rotated in a horizontal plane at a rate such that the wire makes an angle of \( 85^\circ \) with the vertical. Find the increase in length of wire. (05 marks)

(b) (07 marks)

(i) What is meant by the term most probable velocity? (02 marks)

(ii) At what temperature will root mean square velocity of hydrogen molecules be doubled its value at N.T.P? (05 marks)

(c) (06 marks)

(i) Mention three evidences that proves that there is surface tension in water surfaces. (03 marks)

(ii) Estimate the radius of a single droplet when the rain drop of radius 0.5 mm strikes the surface and breaks to 125 droplets of equal size. (03 marks)

QUESTION 4
(a) (07 marks)

(i) State three factors affecting capacitance of a parallel capacitor. (03 marks)

(ii) A battery of 10V is connected to a capacitor of capacitance of 0.1F. The battery is now removed and this capacitor is connected to a second uncharged capacitor. If the charge distribute equally on these two capacitors, find the total energy stored in the two capacitors. Also compare this energy with the initial energy stored in the first capacitor. (04 marks)

(b) (04 marks)

Three particles are fixed in place in a horizontal plane, as shown in the figure below.

Diagram Placeholder

\( |Q_1| = 4 \times 10^{-6} C \) 37° 5 m \( Q_3 = -1 \times 10^{-6} C \) 53° \( Q_2 = 1.7 \times 10^{-6} C \)

Particle 3 at the top of the triangle has charge \( Q_3 \) of \(-1 \times 10^{-6} C\), and the electrostatic force F on it due to the charge on two other particles is measured to be entirely in the negative x-direction. The magnitude of the charge \( Q_1 \) on particle 1 is known to be \( 4 \times 10^{-6} C \), and the magnitude of the charge \( Q_2 \) on the particle 2 is known to be \( 1.7 \times 10^{-6} C \), but their signs are not known.

(i) Determine the signs of the charges \( Q_1 \) and \( Q_2 \).

(ii) Draw and label arrows to indicate the direction of the forces \( F_1 \) exerted by particle 1 on particle 3 and the force \( F_2 \) exerted by particle 2 on particle 3.

(iii) Calculate the magnitude of F, the electrostatic force on particle 3.

(c) (Question continues...)

An electron falls through a distance of 1.5cm in a uniform electric field of magnitude \( 2.0 \times 10^4 N/C \), figure (1). The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance figure (2). Calculate the time of fall in each case.

QUESTION 5
(a) (06 marks)

(i) Wearing a metal bracelet in a region of strong magnetic field could be hazardous. Discuss this statement. (02 marks)

(ii) A piece of aluminum is dropped vertically downward between the poles of an electromagnet. Does the magnetic field affect the velocity of the aluminum? (02 marks)

(iii) Does dropping a magnet down a copper tube produce the current in the tube? Explain. (02 marks)

(b) (06 marks)

(i) Briefly explain at which position of the rotating coil in the magnetic field, the induced e.m.f is maximum? (03 marks)

(ii) A coil of a wire of certain radius has 200 turns and inductance of 89mH. What will be the inductance of another similar coil with 120 turns? (03 marks)

(c) (08 marks)

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

(ii) An α - particle of mass 6.65 x \(10^{-27}\) kg is travelling at right angles to a magnetic field with a speed of 6 x \(10^5\) m/s. The strength of the magnetic field is 0.2T. Calculate the force on the α-particle and its acceleration. (05 marks)

QUESTION 6
(a) (05 marks)

(i) What are the main two differences between Rutherford's and Bohr's model of an atom. (02 marks)

(ii) By using Bohr's model, find the orbital speed of an electron in the atom of hydrogen when n=1, 2 and 3 levels (03 marks)

(b) (08 marks)

(i) Why is a neutron most effective as a bullet in nuclear reaction? (02 marks)

(ii) What is the power output of a \( U^{235} \) reactor if it takes 30days to fuse up 2kg of fuel and if each fission gives 185MeV of usable energy? \( N_A = 6.02 \times 10^{26} \) per kilomole. (06 marks)

(c) (07 marks)

The half life of strontium – 90 is 32 years. Find

(i) The initial activity of 5g of strontium (03 marks)

(ii) How much activity will be remaining after 6 half-lives? (04 marks)

Comprehensive Solutions - Physics 2 Exam Series 8

COMPREHENSIVE SOLUTIONS

PRE-NATIONAL EXAMINATION - PHYSICS 2 SERIES 8
WARI SECONDARY SCHOOL | JANUARY-MAY 2023
QUESTION 1: FLUIDS AND FLOW
20 MARKS
(a) Pressure Concepts (06 marks)
(i) Static vs Dynamic Pressure:
Static Pressure (Pstatic): Pressure exerted by a fluid at rest
Dynamic Pressure (Pdynamic): Pressure due to fluid motion, Pdynamic = ½ρv²
Key Difference: Static pressure acts in all directions, while dynamic pressure is directional and depends on fluid velocity.
(ii) Terminal Velocity Calculation:
Terminal velocity vt ∝ r² (from Stokes' Law: vt = 2r²(ρ-σ)g/9η)
Given: Original radius = r, New radius = r/2
Original vt = 4.0 × 10⁻⁴ m/s
Since vt ∝ r²:
New vt = (r/2)²/r² × Original vt
= (1/4) × 4.0 × 10⁻⁴
= 1.0 × 10⁻⁴ m/s
Answer:
New terminal velocity = 1.0 × 10⁻⁴ m/s
(b) Viscous Flow (07 marks)
(i) Stokes' Equation:
F = 6πηrv
Where: η = viscosity, r = radius, v = velocity
Conditions for validity:
  1. Spherical body
  2. Laminar flow (Re < 1)
  3. Infinite fluid medium
  4. No wall effects
(ii) Rotating Cylinder:
Δh = ω²r²/(4g)
Where: ω = angular velocity, r = radius, g = 10 m/s²
Given: r = 0.05 m, f = 2 rev/s
ω = 2πf = 2 × 3.14 × 2 = 12.56 rad/s
Δh = (12.56)² × (0.05)²/(4 × 10)
= (157.75 × 0.0025)/40
= 0.3944/40 = 0.00986 m ≈ 9.86 mm
Answer:
Height difference = 9.86 mm
(c) Bernoulli's Theorem (07 marks)
(i) Applications:
  1. Venturi meter for fluid flow measurement
  2. Aerofoil design for aircraft wings
  3. Atomizers and sprayers
(ii) Mercury Flow Analysis:
Given: r = 1.88 cm = 0.0188 m, L = 1.26 m
Volume flux = 5.35 × 10⁻² L/min = 8.92 × 10⁻⁷ m³/s

Step 1: Check for laminar flow
Calculate Reynolds number: Re = ρvd/η
v = Q/A = 8.92×10⁻⁷/(π×(0.0188)²) = 0.000803 m/s
Re = (13600×0.000803×0.0376)/(1.55×10⁻³)
= 0.410/0.00155 ≈ 264.5
Since Re < 2000, flow is laminar.

Step 2: Pressure difference (Poiseuille's Law)
ΔP = 8ηLQ/(πr⁴)
= (8 × 1.55×10⁻³ × 1.26 × 8.92×10⁻⁷)/(π×(0.0188)⁴)
= (1.392×10⁻⁸)/(π×1.247×10⁻⁸)
= 1.392/(π×1.247) ≈ 0.355 Pa
Answer:
Reynolds number = 264.5 (laminar)
Pressure difference = 0.355 Pa
QUESTION 2: WAVES AND OPTICS
20 MARKS
(a) Wave Properties (04 marks)
(i) Flash vs Thunder: Light travels much faster (3×10⁸ m/s) than sound (~340 m/s). The time interval equals distance divided by speed difference.
(ii) Energy Transmission: Energy is transmitted through wave motion via oscillations of particles, transferring energy without net mass transport.
(b) Standing Waves (06 marks)
Given: y = 0.3 sin(0.25x) cos(120πt)
(i) Wavelength and Frequency:
From sin(0.25x): k = 0.25 rad/m
λ = 2π/k = 2π/0.25 = 8π ≈ 25.12 m

From cos(120πt): ω = 120π rad/s
f = ω/(2π) = 120π/(2π) = 60 Hz
(ii) Interfering Waves:
Standing wave formed by two waves traveling in opposite directions:
y₁ = 0.15 sin(0.25x - 120πt)
y₂ = 0.15 sin(0.25x + 120πt)
Answer:
Wavelength = 25.12 m, Frequency = 60 Hz
(c) Diffraction (04 marks)
(i) Types of Diffraction:
  1. Fresnel Diffraction: Source and screen at finite distances
  2. Fraunhofer Diffraction: Source and screen at infinite distances (parallel rays)
(ii) Improving Young's Experiment:
  1. Use monochromatic light source to reduce chromatic effects
  2. Increase slit-to-screen distance for clearer pattern
(d) Diffraction Grating (06 marks)
Grating equation: d sinθ = nλ
(i) Line Spacing:
n = 2, θ = 32°, λ = 590 nm = 5.9×10⁻⁷ m
d = nλ/sinθ = (2 × 5.9×10⁻⁷)/sin32°
= (1.18×10⁻⁶)/0.5299
= 2.227×10⁻⁶ m
(ii) Lines per cm:
Lines/m = 1/d = 1/(2.227×10⁻⁶) = 449,000 lines/m
Lines/cm = 449,000/100 = 4490 lines/cm
Answer:
d = 2.227×10⁻⁶ m, Lines/cm = 4490
QUESTION 3: MATERIALS & KINETIC THEORY
20 MARKS
(a) Material Properties (07 marks)
(i) Creep vs Fatigue:
Creep: Time-dependent deformation under constant stress at high temperature
Fatigue: Progressive structural damage under cyclic loading
(ii) Wire Extension Calculation:
Given: m = 0.5 kg, L = 0.8 m, d = 0.7 mm = 7×10⁻⁴ m
θ = 85°, Young's modulus for Al = 7×10¹⁰ Pa

Step 1: Find tension T
T sinθ = mv²/r (centripetal force)
T cosθ = mg (vertical equilibrium)
From vertical: T = mg/cosθ = 0.5×10/cos85°
= 5/0.08716 = 57.37 N

Step 2: Calculate extension
Area A = π(d/2)² = π(3.5×10⁻⁴)² = 3.848×10⁻⁷ m²
ΔL = (T × L)/(A × Y) = (57.37 × 0.8)/(3.848×10⁻⁷ × 7×10¹⁰)
= 45.896/(2.6936×10⁴) = 1.704×10⁻³ m
Answer:
Increase in length = 1.70 mm
(b) Kinetic Theory (07 marks)
(i) Most Probable Velocity: The speed most likely to be possessed by molecules in a gas at given temperature.
(ii) Temperature Calculation:
vrms = √(3RT/M)
vrms ∝ √T
v2 = 2v1
√T2 = 2√T1
T2 = 4T1

At NTP: T1 = 273 K
T2 = 4 × 273 = 1092 K
Answer:
Required temperature = 1092 K
(c) Surface Tension (06 marks)
(i) Evidence of Surface Tension:
  1. Water droplets form spherical shapes
  2. Insects can walk on water surface
  3. Capillary rise in thin tubes
(ii) Droplet Radius Calculation:
Volume conservation: Vtotal = n × Vdroplet
(4/3)πR³ = 125 × (4/3)πr³
R³ = 125r³
r = R/∛125 = R/5
R = 0.5 mm, so r = 0.5/5 = 0.1 mm
Answer:
Radius of each droplet = 0.1 mm
QUESTION 4: ELECTROSTATICS & CAPACITANCE
20 MARKS
(a) Capacitance (07 marks)
(i) Factors affecting capacitance:
  1. Area of plates (C ∝ A)
  2. Distance between plates (C ∝ 1/d)
  3. Dielectric constant (C ∝ εr)
(ii) Energy Calculation:
Step 1: Initial energy
C₁ = 0.1 F, V = 10 V
Q = C₁V = 0.1 × 10 = 1 C
E₁ = ½C₁V² = ½ × 0.1 × 100 = 5 J

Step 2: After connection
Equal charge distribution: Q/2 on each capacitor
For each: V' = (Q/2)/C₁ = 0.5/0.1 = 5 V
E₂ (each) = ½C₁(5)² = ½ × 0.1 × 25 = 1.25 J
Total energy = 2 × 1.25 = 2.5 J

Step 3: Comparison
Energy ratio = 2.5/5 = 0.5
Answer:
Initial energy = 5 J, Final total energy = 2.5 J
Energy is halved (50% remains)
(b) Electrostatic Forces (04 marks)
Force Analysis:
Q₃ is negative. For net force to be in -x direction:
1. Q₁ must be positive (attracts Q₃ with force having +x and +y components)
2. Q₂ must be negative (repels Q₃ with force having -x and +y components)
Force diagram would show:
F₁ (from Q₁ on Q₃): directed toward Q₁ (attraction)
F₂ (from Q₂ on Q₃): directed away from Q₂ (repulsion)
Force magnitude calculation:
F₁ = k|Q₁Q₃|/r₁₃² = (9×10⁹×4×10⁻⁶×1×10⁻⁶)/(5²)
= 36×10⁻³/25 = 1.44×10⁻³ N

F₂ = k|Q₂Q₃|/r₂₃² = (9×10⁹×1.7×10⁻⁶×1×10⁻⁶)/(5²)
= 15.3×10⁻³/25 = 0.612×10⁻³ N

Net force components:
Fnet,x = -F₁cos53° - F₂cos37°
= -(1.44×0.6 + 0.612×0.8)×10⁻³
= -(0.864 + 0.490)×10⁻³ = -1.354×10⁻³ N
Answer:
Q₁ = positive, Q₂ = negative
Fnet = 1.354 × 10⁻³ N in -x direction
(c) Charged Particle Motion (09 marks)
Time of Fall Calculation:
a = qE/m, s = ½at²
For electron:
ae = eE/me = (1.6×10⁻¹⁹×2×10⁴)/(9.1×10⁻³¹)
= 3.2×10⁻¹⁵/9.1×10⁻³¹ = 3.516×10¹⁵ m/s²
te = √(2s/a) = √(2×0.015/3.516×10¹⁵)
= √(3×10⁻²/3.516×10¹⁵) = √(8.53×10⁻¹⁸)
= 2.92×10⁻⁹ s

For proton:
ap = eE/mp = (1.6×10⁻¹⁹×2×10⁴)/(1.67×10⁻²⁷)
= 3.2×10⁻¹⁵/1.67×10⁻²⁷ = 1.916×10¹² m/s²
tp = √(2×0.015/1.916×10¹²)
= √(3×10⁻²/1.916×10¹²) = √(1.566×10⁻¹⁴)
= 1.25×10⁻⁷ s
Answer:
Electron: 2.92 ns, Proton: 0.125 μs
QUESTION 5: ELECTROMAGNETISM
20 MARKS
(a) Magnetic Effects (06 marks)
(i) Metal Bracelet Hazard: Changing magnetic field induces eddy currents, causing heating and potential burns. Strong fields can also cause violent attraction/repulsion.
(ii) Aluminum in Magnetic Field: Yes, eddy currents induced in aluminum create opposing magnetic field, reducing velocity (magnetic damping).
(iii) Magnet in Copper Tube: Yes, changing flux induces eddy currents in copper (Lenz's Law), creating opposing magnetic field.
(b) Electromagnetic Induction (06 marks)
(i) Maximum EMF Position: When coil plane is parallel to magnetic field (θ = 90° or 270°), rate of flux change is maximum.
(ii) Inductance Calculation:
L ∝ N² (for same geometry)
L₁/L₂ = (N₁/N₂)²
89×10⁻³/L₂ = (200/120)² = (5/3)² = 25/9
L₂ = (89×10⁻³)×(9/25) = 0.089×0.36 = 0.03204 H = 32.04 mH
Answer:
New inductance = 32.0 mH
(c) Magnetic Materials & Forces (08 marks)
(i) Magnetic Materials:
  1. Diamagnetic: μr < 1 (slightly repelled)
  2. Paramagnetic: μr > 1 (slightly attracted)
  3. Ferromagnetic: μr ≫ 1 (strongly attracted)
(ii) Alpha Particle in Magnetic Field:
F = qvB sinθ, a = F/m
q = 2e = 3.2×10⁻¹⁹ C, v = 6×10⁵ m/s, B = 0.2 T
F = (3.2×10⁻¹⁹)×(6×10⁵)×0.2 = 3.84×10⁻¹⁴ N

m = 6.65×10⁻²⁷ kg
a = F/m = 3.84×10⁻¹⁴/6.65×10⁻²⁷ = 5.774×10¹² m/s²
Answer:
Force = 3.84 × 10⁻¹⁴ N
Acceleration = 5.77 × 10¹² m/s²
QUESTION 6: ATOMIC & NUCLEAR PHYSICS
20 MARKS
(a) Atomic Models (05 marks)
(i) Rutherford vs Bohr:
  1. Bohr introduced quantized orbits; Rutherford had continuous orbits
  2. Bohr explained atomic spectra; Rutherford couldn't
(ii) Electron Orbital Speeds:
vn = (e²)/(2ε₀hn)
For n=1: v₁ = 2.188×10⁶ m/s
For n=2: v₂ = v₁/2 = 1.094×10⁶ m/s
For n=3: v₃ = v₁/3 = 7.293×10⁵ m/s
Answer:
v₁ = 2.19×10⁶ m/s, v₂ = 1.09×10⁶ m/s, v₃ = 7.29×10⁵ m/s
(b) Nuclear Reactions (08 marks)
(i) Neutron Effectiveness: Neutrons are neutral, so not repelled by nucleus, allowing closer approach for reaction.
(ii) Reactor Power Output:
Step 1: Number of U-235 atoms
Molar mass = 235 g/mol = 0.235 kg/kmol
Moles in 2 kg = 2/0.235 = 8.51 kmol
Atoms = 8.51 × 6.02×10²⁶ = 5.124×10²⁷ atoms

Step 2: Total energy released
Energy per fission = 185 MeV = 185×10⁶ eV
= 185×10⁶ × 1.6×10⁻¹⁹ = 2.96×10⁻¹¹ J

Total energy = 5.124×10²⁷ × 2.96×10⁻¹¹
= 1.517×10¹⁷ J

Step 3: Power output
Time = 30 days = 30×24×3600 = 2.592×10⁶ s
Power = Energy/Time = 1.517×10¹⁷/2.592×10⁶
= 5.85×10¹⁰ W = 58.5 GW
Answer:
Power output = 58.5 GW
(c) Radioactivity (07 marks)
Strontium-90 Activity:
A = λN, λ = ln2/T½
(i) Initial activity:
T½ = 32 years = 1.009×10⁹ s
λ = ln2/1.009×10⁹ = 6.87×10⁻¹⁰ s⁻¹

Molar mass Sr-90 = 90 g/mol
Moles in 5 g = 5/90 = 0.0556 mol
Atoms N = 0.0556 × 6.02×10²³ = 3.346×10²²

A₀ = λN = 6.87×10⁻¹⁰ × 3.346×10²²
= 2.299×10¹³ Bq

(ii) After 6 half-lives:
A = A₀(½)⁶ = 2.299×10¹³ × (1/64)
= 3.592×10¹¹ Bq
Answer:
Initial activity = 2.30×10¹³ Bq
Activity after 6 half-lives = 3.59×10¹¹ Bq

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