Physical Optics Examination (With Comprehensive explanations and calculations)

UMOJA WA WAZAZI TANZANIA
WARI SECONDARY SCHOOL

PHYSICS-2: PHYSICAL OPTICS
Paper 131/1

Time: 2:00hrs
March 2021

Question 1: Interference and Thin Films

(a) Applications of Newton's Rings

1. Surface Quality Testing: Used to measure the flatness of optical surfaces by analyzing ring patterns.
2. Wavelength Determination: Can calculate light wavelength by measuring ring diameters and lens curvature.
3. Refractive Index Measurement: Used to determine refractive indices of transparent materials.

(b)(i) Thin Film Path Difference

[Diagram showing light rays reflecting off thin film with angles θ]

For a thin film of thickness t and refractive index n:

Path difference = 2nt cosθ
Where:
• 2 accounts for double traversal of film
• n is refractive index
• t is film thickness
• θ is angle of refraction in film

(b)(ii) Air Wedge Calculation

Given: λ = 5.6×10⁻⁷ m, fringe separation = 1.2 mm = 1.2×10⁻³ m, L = 75 mm = 75×10⁻³ m
For air wedge: β = λ/2α ⇒ α = λ/2β
Wedge angle α = (5.6×10⁻⁷)/(2×1.2×10⁻³) ≈ 2.33×10⁻⁴ rad ≈ 0.013°
Thickness t = L × α = 75×10⁻³ × 2.33×10⁻⁴ ≈ 1.75×10⁻⁵ m = 17.5 μm

(c)(i) Lens Blooming

Blooming: Application of anti-reflection coatings on lens surfaces to reduce light loss through reflection.

(c)(ii) Purpose of Blooming

1. Increases light transmission by reducing surface reflections
2. Improves image contrast by minimizing stray light
3. Protects lens surfaces from scratches and environmental damage

(c)(iii) Blooming Materials

• Magnesium Fluoride (MgF₂)
• Silicon Dioxide (SiO₂)
• Titanium Dioxide (TiO₂)
• Aluminum Oxide (Al₂O₃)

(d)(i) Sonic Boom Definition

Sonic Boom: Shock wave produced when an object moves through air at speeds exceeding the speed of sound, creating a loud explosive noise.

(d)(ii) Sonic Boom Sources

1. Supersonic aircraft (e.g., fighter jets, Concorde)
2. Bullets or projectiles traveling faster than sound
3. Lightning strikes (thunder is a type of sonic boom)
4. Whip cracks (tip exceeds sound speed)

Question 2: Polarization Applications

(a) Polarization Applications

Application	Explanation
Photoelasticity	Measures stress distribution in transparent materials using polarized light patterns
Telecommunication	Fiber optics use polarization to increase channel capacity (polarization multiplexing)
Saccharimetry	Measures sugar concentration by observing rotation of polarized light (optical activity)
Sunglasses	Polarized lenses reduce glare from horizontal reflections (water, roads)

(b) Diffraction Applications

1. CD/DVD Reading: Laser diffraction patterns read microscopic pits on discs
2. Spectroscopy: Diffraction gratings separate light into spectra for analysis
3. X-ray Crystallography: Determines atomic structure using diffraction patterns
4. Holography: Uses interference and diffraction to create 3D images
5. Microscopy: Diffraction limits resolution (Abbe's diffraction limit)

Question 3: Polarized Light

(a) Polarization Concepts

Term	Importance
Dextrorotatory	Substances that rotate polarized light clockwise (right); used in chiral analysis
Levorotatory	Substances that rotate polarized light counterclockwise (left); important in biochemistry
Optically Active	Materials that rotate polarized light; indicates molecular asymmetry/chirality
Double Refraction	Splits light into ordinary/extraordinary rays (birefringence); used in polarizing filters

(b) Differentiations

Comparison	Difference
Polaroid vs Polarimeter	Polaroid is a polarizing filter; polarimeter measures rotation of polarized light
Plane of Vibration vs Polarization	Vibration: plane where E-field oscillates; Polarization: perpendicular to vibration plane
Ordinary vs Polarized Light	Ordinary: random E-field orientations; Polarized: E-field oscillates in single plane

(c) Nicol Prism Construction

1. Made from calcite (CaCO₃) crystal cut diagonally
2. Two pieces cemented with Canada balsam (n=1.55)
3. Works via double refraction: ordinary ray totally internally reflected
4. Emerges as plane-polarized light (extraordinary ray transmitted)

[Diagram showing Nicol prism construction with light paths]

(d) Newton's Ring Observations

Modification	Observation
Front-silvered plate	Increased contrast (brighter fringes) due to higher reflectivity
White light source	Colored fringes (only few visible orders) instead of monochromatic rings
Liquid between lens/plate	Ring diameters decrease (optical path changes with refractive index)

Nuclear Physics & Quantum Physics Test (Step-by-step mathematical derivations and explanations)

PHYSICS 2: NUCLEAR PHYSICS AND QUANTUM PHYSICS

Nuclear Physics Questions

(a) Definitions

Term	Definition
Nuclear Binding Energy	Energy required to disassemble a nucleus into its constituent protons and neutrons
Packing Fraction	(Mass defect per nucleon) × c², where mass defect = (M-A)/A (M=atomic mass, A=mass number)
Thermal Neutrons	Neutrons with kinetic energy ≈ 0.025 eV (room temperature)

(b)(i) Nuclear Mass Deficit

The mass of a nucleus is less than the sum of its nucleons due to the mass-energy equivalence (E=mc²). The "missing mass" (mass defect) is converted into binding energy that holds the nucleus together.

(b)(ii) Helium Binding Energy

A binding energy of 28.2 MeV for ⁴₂He means 28.2 MeV of energy would be required to completely separate its 2 protons and 2 neutrons, or that this amount of energy was released when the nucleus formed from its nucleons.

(b)(iv) Uranium Fission Q-value

Given fission: ²³⁶₉₂U → ¹⁴⁶₅₇La + ⁸⁷₃₅Br + 3n + Q
Q = BE_products - BE_reactants
= [146×8.41 + 87×8.59 + 3×0] - [236×7.59]
= [1227.86 + 747.33] - 1791.24
= 1975.19 - 1791.24 = 183.95 MeV

(c)(i) Radioactive Decay Terms

Half-life (t_½): Time for half of radioactive nuclei to decay

Decay constant (λ): Probability per unit time that a nucleus will decay, related by t_½ = ln(2)/λ

(c)(ii) Carbon Dating

Given: Current ratio ¹⁴C/¹²C = 1.3×10⁻¹², t_½ = 5730 years
Original ratio = 1.3×10⁻¹² (living wood)
Current activity = 19 decays/min = λN
Original activity = λN₀ = 19×(original ratio/current ratio)
Using N = N₀e^-λt and λ = ln(2)/t_½
t = (t_½/ln(2)) × ln(N₀/N) ≈ 5730/0.693 × ln(1) = 0 years
(Note: Additional information needed for complete calculation)

(d) Uranium Reactor Calculation

Energy needed = Power × Time = 1000 MW × 10 years
= 10⁹ J/s × (10×365×24×3600 s) = 3.156×10¹⁷ J
At 10% efficiency, total energy required = 3.156×10¹⁸ J
Energy per fission = 200 MeV = 200×1.6×10⁻¹³ J = 3.2×10⁻¹¹ J
Number of fissions = 3.156×10¹⁸/3.2×10⁻¹¹ ≈ 9.86×10²⁸
Mass of ²³⁵U = (9.86×10²⁸/6.022×10²³) × 235 g/mol ≈ 38,500 kg

Quantum Physics Questions

(a)(i) Photoelectric Law

Einstein's Photoelectric Equation: E = hν = φ + K.E._max

1. For given metal & frequency, photoelectron emission rate ∝ light intensity
2. Maximum kinetic energy depends on frequency, not intensity
3. Below threshold frequency (ν₀), no emission occurs regardless of intensity

(a)(ii) Definitions

Work Function (φ): Minimum energy needed to eject an electron from a metal surface

Stopping Potential (V₀): Minimum reverse potential needed to stop the fastest photoelectrons

(b) Planck's Constant Calculation

For λ₁ = 3310Å (3.31×10⁻⁷ m), K.E.₁ = 3×10⁻¹⁹ J
For λ₂ = 5000Å (5×10⁻⁷ m), K.E.₂ = 0.972×10⁻¹⁹ J
Using hc/λ = φ + K.E.:
hc/3.31×10⁻⁷ = φ + 3×10⁻¹⁹ ...(1)
hc/5×10⁻⁷ = φ + 0.972×10⁻¹⁹ ...(2)
Subtract (2) from (1):
hc(1/3.31 - 1/5)×10⁷ = 2.028×10⁻¹⁹
h ≈ 6.63×10⁻³⁴ Js (Planck's constant)
Threshold λ when K.E.=0: φ = hc/λ₀ ⇒ λ₀ ≈ 5400Å

(c)(i) X-ray Types

Hard X-rays	Soft X-rays
High energy (10-100 keV)	Low energy (0.1-10 keV)
Short wavelength (0.01-0.1 nm)	Longer wavelength (0.1-10 nm)
Greater penetration	Less penetration

(c)(ii) X-ray Tube Cooling

Power input = VI = 60×10³ × 30×10⁻³ = 1800 W
Heat produced = 99% × 1800 = 1782 W
Heat removed = mcΔT/Δt ⇒ ΔT = (PΔt)/(mc)
= (1782 × 1)/(0.06 × 4186) ≈ 7.1°C/s

MITIHANI POPOTE EXAMINATION SERIES: FORM FOUR CHEMISTRY SERIES 01 (With Marking Guide)

Open Examination Here

Chemistry Examination Answers

Section A

Question 1

i. B. 0.2

To deposit calcium from molten calcium chloride (CaCl₂), the reaction is:
Ca²⁺ + 2e⁻ → Ca
Moles of calcium = mass/molar mass = 4g/40g/mol = 0.1 mol
Since 1 mole of Ca requires 2 faradays, 0.1 mol requires 0.2 faradays.

ii. D. Tetracloromethane

In the presence of excess chlorine, methane undergoes successive substitution reactions, ultimately forming carbon tetrachloride (CCl₄).

iii. B. Alloys

An alloy is a uniform mixture of two or more metals (or a metal and a non-metal).

iv. A. An endothermic reaction

The positive ΔH value (X KJ/mol) indicates that heat is absorbed, making it an endothermic reaction.

v. B. 2:8:8

Chlorine (atomic number 17) gains one electron to form Cl⁻, resulting in the electron configuration 2:8:8.

vi. C. water is solvent, glucose is solute and the product is solution

In a glucose-water solution, water is the solvent, glucose is the solute, and the resulting mixture is the solution.

vii. A. Absorbs water vapour

Anhydrous copper(II) sulfate is white but turns blue when it absorbs water vapor to form the hydrated form (CuSO₄·5H₂O).

viii. D. 20cm³ of 1M Sulphuric acid solution

Sulfuric acid is a strong acid and provides more H⁺ ions per unit volume compared to acetic acid, leading to faster hydrogen production.

ix. D. Sulphur trioxide, carbon dioxide and Nitrogen dioxide

These oxides react with water to form acids (acidic oxides).

x. A. It should be rinsed off with large quantities of running water.

Concentrated sulfuric acid is highly corrosive. Immediate rinsing with plenty of water dilutes and removes the acid, minimizing damage.

Question 2

List A	List B
i. Biogas	D. Gases fuel derived from decomposition of biological waste
ii. Biomass	E. Renewable source of energy
iii. Natural gas	B. Non-renewable gaseous fuel
iv. Producer gas	C. Industrial gases fuel which is a mixture of nitrogen gas, hydrogen and carbon monoxide
v. Water gas	A. Industrial gas which is a mixture of hydrogen and carbon monoxide

Section B

Question 3

a)

i. A laboratory is a controlled environment designed for scientific experiments, research, and analysis.

ii. Potassium chlorate (KClO₃) and hydrogen peroxide (H₂O₂).

iii. Hydrogen gas is less dense than air, so it is collected by downward displacement of air (upward delivery).

b)

i. NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

ii. NH₄Cl(aq) + NaOH(aq) → NaCl(aq) + NH₃(g) + H₂O(l)

iii. Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)

iv. Mg(s) + 2HNO₃(aq) → Mg(NO₃)₂(aq) + H₂(g)

Question 4

a)

- Element R (atomic number 20) is calcium (Ca), and element S (atomic number 17) is chlorine (Cl).

- The compound formed is CaCl₂, with an ionic bond.

b)

Properties of CaCl₂:

1. High melting and boiling points due to strong ionic bonds.
2. Soluble in water.
3. Conducts electricity in molten or aqueous state.

Question 5

a)

- At STP, 1 mole of any gas occupies 22.4 dm³.

- Number of moles in 20 cm³ (0.02 dm³) of Cl₂ gas:

Moles = 0.02/22.4 = 8.93 × 10⁻⁴ mol

- Number of molecules:

8.93 × 10⁻⁴ × 6.02 × 10²³ = 5.37 × 10²⁰ molecules

b)

- Molar mass of Cu(NO₃)₂ = 63.5 + 2(14 + 3(16)) = 187.5 g/mol.

- Moles in 5 g:

5/187.5 = 0.0267 mol

- Each formula unit of Cu(NO₃)₂ produces 3 ions (1 Cu²⁺ and 2 NO₃⁻).

- Total ions:

0.0267 × 3 × 6.02 × 10²³ = 4.82 × 10²² ions

Section C

Question 13

a)

How can you tell water is polluted? Give two ways:

1. Visual indicators: Presence of algae, discoloration, or floating debris.
2. Odor: Unpleasant smells indicate pollution.

b)

Warnings concerning likely effects of coal use:

1. Air pollution (release of SO₂, NOₓ, and particulate matter).
2. Health risks (respiratory diseases).
3. Acid rain formation (damage to ecosystems).
4. Contribution to climate change (CO₂ emissions).
5. Water pollution (ash disposal contaminates water sources).

Question 14

a)

i. A functional group is a specific group of atoms responsible for the characteristic reactions of a compound.

ii. A homologous series is a family of organic compounds with the same functional group and similar chemical properties, differing by a CH₂ unit.

iii. Isomerism is the phenomenon where compounds have the same molecular formula but different structural arrangements.

b)

Isomers of C₄H₁₀ (Butane):

1. Butane:

CH₃-CH₂-CH₂-CH₃

2. 2-Methylpropane (Isobutane):

CH₃-CH(CH₃)-CH₃

c)

i. Ethene to Ethane:

- Reagent: Hydrogen gas (H₂).

- Condition: Nickel catalyst at 150°C.

CH₂=CH₂ + H₂ → CH₃-CH₃

ii. Ethene to Chloroethane:

- Reagent: Hydrogen chloride (HCl).

CH₂=CH₂ + HCl → CH₃-CH₂Cl

iii. Ethene to 1,2-Dibromoethane:

- Reagent: Bromine water (Br₂).

CH₂=CH₂ + Br₂ → CH₂Br-CH₂Br

iv. Ethene to Ethanol:

- Reagent: Steam (H₂O).

- Condition: Phosphoric acid catalyst at 300°C and 60 atm.

CH₂=CH₂ + H₂O → CH₃-CH₂OH

Physics Measurement Test 2 (With Solutions)

UMOJA WA WAZAZI TANZANIA
WARI SECONDARY SCHOOL

PHYSICS MEASUREMENT TEST 2
FORM FIVE

Time: 2:00hrs
November, 2022

INSTRUCTIONS

1. This paper consists of four (4) questions.
2. Answer all questions.

Question 1: Dimensional Analysis

(a) Definitions with Examples

Term	Definition	Example
Dimension	Physical nature of a quantity expressed in terms of fundamental quantities (M, L, T)	[Force] = MLT-2
Dimensional Analysis	Method to check equations or derive relations using dimensions	Verifying F = ma dimensionally
Dimensional Variable	Physical quantity with dimensions that can change value	Velocity (LT-1)
Dimensional Constant	Constant with fixed dimensions	Gravitational constant (M-1L3T-2)
Dimensionless Variable	Variable without dimensions	Strain (ΔL/L)
Dimensionless Constant	Numerical constant without dimensions	π (pi)

(b)(i) Dimensional Analysis Significance & Limitations

Significance:

1. Verifies dimensional consistency of equations

Limitations:

1. Cannot determine dimensionless constants
2. Doesn't confirm numerical correctness
3. Cannot identify logarithmic/exponential relations

(b)(ii) Dimensions of a and b

Given: (P + a/V²)(V - b) = constant
For dimensional consistency:
[a/V²] = [P] ⇒ [a] = [P][V²] = ML^-1T^-2 × L⁶ = ML⁵T^-2
[b] = [V] = L³

(b)(iii) Dimensions in Velocity Equation

v = at + b/(t + c)
[at] = [v] ⇒ [a] = [v]/[t] = LT^-1/T = LT^-2
[b/(t + c)] = [v] ⇒ [b] = [v][t] = LT^-1 × T = L
[c] = [t] = T

Question 2: Dimensions and Homogeneity

(a)(i) Quantities with Work Dimensions

Same dimensions as work (ML²T^-2):

• Energy (scalar)
• Torque (vector)

(a)(ii) Scalar/Vector with Same Dimensions

• Work (scalar) and Torque (vector)
• Speed (scalar) and Velocity (vector)

(a)(iii) Principle of Homogeneity

All terms in a physical equation must have identical dimensions on both sides.

(a)(iv) Limitations of Dimensional Analysis

No, dimensional analysis cannot confirm an equation is completely correct because:

• It doesn't verify numerical constants
• Different physical quantities may share dimensions
• It can't identify incorrect relationships between terms

(b) Deriving SHM Energy Expression

Assume E ∝ M^xf^yA^z

[E] = [M]^x[f]^y[A]^z
ML²T^-2 = M^x(T^-1)^yL^z
Comparing dimensions:
For M: x = 1
For T: -y = -2 ⇒ y = 2
For L: z = 2
Thus E ∝ Mf²A²
Actual formula: E = ½(2π)²Mf²A² = 2π²Mf²A²

Question 3: Unit Conversion and Errors

(a) Unit Conversions

(i) 5 g·cm·s^-2 to N:

1 N = 1 kg·m·s^-2 = 1000 g × 100 cm × s^-2 = 10⁵ g·cm·s^-2
5 g·cm·s^-2 = 5/10⁵ N = 5 × 10^-5 N

(ii) 19 × 10¹⁰ N/m² to g·cm^-1·s^-2:

1 N/m² = 1 kg·m^-1·s^-2 = 1000 g × (100 cm)^-1 × s^-2 = 10 g·cm^-1·s^-2
19 × 10¹⁰ N/m² = 19 × 10¹⁰ × 10 = 1.9 × 10¹² g·cm^-1·s^-2

(b)(i) Systematic Error Causes

1. Instrument calibration errors
2. Zero errors in measuring devices

(b)(ii) Precise vs Accurate

Precise Experiment	Accurate Experiment
Small random errors (consistent results)	Small systematic errors (close to true value)
May be consistently wrong	Requires both precision and correctness

(b)(iii) Dimension vs Unit

Dimension	Unit
Physical nature (M, L, T)	Standard of measurement (kg, m, s)
Independent of system	System-dependent (SI, CGS)

(b)(iv) Specific Gravity Error

Specific gravity = Weight in air / (Weight in air - Weight in water)
SG = 25.0/(25.0 - 10.0) = 25.0/15.0 ≈ 1.6667
ΔSG/SG = ΔW_air/W_air + (ΔW_air + ΔW_water)/(W_air - W_water)
= 0.1/25 + (0.1+0.02)/15 ≈ 0.004 + 0.008 = 0.012 (1.2%)

Question 4: Applications

(a)(i) Fluid Flow Relation

Assume Q ∝ Δp^xr^yη^zl^w

[Q] = [Δp]^x[r]^y[η]^z[l]^w
L³T^-1 = (ML^-1T^-2)^xL^y(ML^-1T^-1)^zL^w
Solving: x=1, z=-1, y=4, w=-1
Thus Q ∝ Δp·r⁴/(η·l)
Actual formula (Poiseuille's): Q = πΔp·r⁴/(8ηl)

(a)(ii) Volume Percentage Error

V = l × b × h = 15.12 × 10.15 × 5.29 ≈ 811.6 cm³
ΔV/V = Δl/l + Δb/b + Δh/h = 0.01/15.12 + 0.01/10.15 + 0.01/5.29
≈ 0.000661 + 0.000985 + 0.00189 ≈ 0.003536 (0.3536%)

(a)(iii) Percentage Error in P

P = a·b²·c³·d⁴
ΔP/P = Δa/a + 2(Δb/b) + 3(Δc/c) + 4(Δd/d)
= 0.5% + 2(0.5%) + 3(0.5%) + 4(0.5%)
= (1 + 2 + 3 + 4)(0.5%) = 10 × 0.5% = 5%

CLASS SEVEN TERMINAL EXAMINATIONS (With Marking Schemes)

Click here to open CLASS SEVEN EXAMS IMLA & DICTATION

HAI DISTRICT COUNCIL MONTHLY EXAMINATION MAY 2025 - ANSWERS

ENGLISH LANGUAGE STANDARD VII

SECTION A

1. i. D. Three out of four

ii. B. no one

iii. B. youths

iv. A. weakening one's body

v. B. Cigarette smoking

2. i. B. goes

ii. B. Will elect

iii. D. Draw

iv. D. which

v. E. Does

3. i. bought

ii. has been

iii. If

iv. was making

v. both

LIST A	Answer
i. Interesting	C. Boring
ii. Different	G. Same
iii. Hardworking	A. Lazy
iv. Fine	E. Unwell
v. Beautiful	D. Ugly

SECTION B

5. i. uncle

ii. Optometrist/Ophthalmologist

iii. poet

iv. furniture

v. zoo

6. i. 77 years (1922-1999)

ii. 21 years (1964-1985)

iii. Resigning from leadership position

iv. 1961

v. 3 years (1961-1964)

SECTION C

7. Correct order:

1. A. Bonge woke up early in the morning, excited for his journey
2. B. He packed his bag with food, water, and his favorite book
3. C. Bonge walked through the forest, enjoying the fresh air and birds singing
4. D. He climbed a small hill and saw a beautiful view of the valley below
5. E. At the end of his journey, Bonge felt happy and proud of his adventure

KISWAHILI DARASA LA VII

1. i. B. Sabasaba

ii. B. Ujasiriamali

iii. C. Dar es Salaam

iv. D. Vitatu

v. A. Maonesho makuu

2. i. B. Kiwakilishi

ii. C. Mkali

iii. A. Babalka

iv. C. Fuata heshima

v. A. Jozi

vi. E. Katibu

vii. B. mama

viii. A. Anduje hajaenda sokoni

ix. B. Halisi

x. C. Ndipo

SEHEMU A	JIBU
i. Zuru	G. Enda mahali fulani kwa lengo la kutembelea
ii. Nadhifu	E. Enye kupendeza au kuwa safi
iii. Maktaba	C. Nyumba au chumba mnamo hifadhiwa machapisho mbalimbali
iv. Bwalo	D. Ukumbi maalum wa kulia chakula
v. Samadi	A. Mbolea ya kingesi cha wanyama

SEHEMU B

4. Correct order:

1. B. Walipokaribia Dodoma aliona majengo makubwa
2. A. Akamuliza mwenzake yale ni majengo gani?
3. E. Yale ni majengo ya Chuo kikuu
4. C. Ni mahali maarufu sana kwani pale ndipo vijana wa nchi hii wanapopata elimu ya juu
5. D. Wanasoma shahada mbalimbali kama vile uhakimu na uanasheria

5. i. 4

ii. -tokea, -swali, -hali, -jadi

iii. Mgogoro kati ya kuku na yai

iv. Asili/kitambo

v. Kuku na yai

6. i. Ayubu Juma

ii. harusi

iii. Chabuma Hotel

iv. 10:00 jioni

v. ujumbe

MATHEMATICS STANDARD VII

SECTION A

NO	QUESTION	ANSWER
1	(i) 98721 + 1279 =	100,000
	(ii) 7328 - 5379 =	1,949
	(iii) 61.9 x 0.8 =	49.52
	(iv) 7714 ÷ 38 =	203
	(v) 5 - 3/5 =	4 2/5 or 4.4
	(vi) 3 - 0.999 =	2.001
	(vii) 12.4 ÷ 0.4 =	31
	(viii) 6²/3 ÷ 1/9 =	60
	(ix) (-20 + 4) - (-8 - 2) =	-6
	(x) 2⁴/5 ÷ 1/3 =	7 4/5 or 7.8
2	(i) Write 105,003 in words	One hundred five thousand three
	(ii) James' age is the quotient of the total value of 9 and 3 in the number 5932	3 years old (900 ÷ 300 = 3)
	(iii) Round off 870547 to the nearest ten thousands place	870,000
	(iv) Password of Malukela's phone	40789
	(v) The first multi-part election in Tanzania was conducted in MCMXCV	1995
	(vi) Amani needs to add to buy new shoes	81,100 shillings

SECTION B

3. (i) 10/19

(ii) 7:45

(iii) 5 (13, 17, 19, 23, 29)

4. (i) 1/2

(ii) 10,000 ml

(iii) p = 1

5. (i) 18 minutes

(ii) 5:06 hrs

(iii) 36 km/h

6. (i) 50% profit

(ii) 0.2025

(iii) 9 sweets

SECTION C

7. (i) D(0,0)

(ii) 72,000 shillings

8. (i) 35,200 shillings

(ii) 1,188 cm²

(iii) 24 cm

SCIENCE AND TECHNOLOGY CLASS VII

SECTION A

1. i. C. snail, millipede and octopus

ii. D. roots, leaf and shoot

iii. C. It breastfeed

iv. A. tympanic membrane

v. C. lamina

vi. D. hypothesis

vii. C. monocotyledonous and dicotyledonous plants

viii. D. wind

ix. E. sound and shadow

x. C. monotremes

COLUMN A	ANSWER
(i) Sensory nerves	B. Transfer impulses from one part of the body and transmit them to the central nervous system
(ii) Cornea	E. Transparent membrane which allows light to pass through the eye
(iii) Auditory nerves	D. Receives the sound waves and send them to the brain
(iv) Olfactory nerves	C. Carries information from the nose and transmit them to the brain
(v) Alveoli	A. Sacs in the lungs where gaseous exchange occurs

3. i. nicotine

ii. retina

iii. porter

iv. cerebrum

v. acid

SECTION B

4. i. (a) Head (b) Arm (c) Treadle

ii. (a) Warn of hazards (b) Provide safety instructions (c) Indicate emergency equipment (d) Show safe routes

iii. Lever

5. i. transpiration

ii. Series circuit

iii. Pulmonary vein

6. i. 4 amperes

ii. Fuse

iii. 2.5

iv. Compound fracture

SECTION C

7. i. 28 days

ii. Twins developed from two different eggs fertilized by two different sperm

8. i. Series circuit

ii. If one bulb fails, all others stop working

iii. Parallel circuit

SOCIAL STUDIES FOR STD VII

SECTION A

1. i. D. Map shown from above

ii. E. temperature differences

iii. D. Human activities

iv. E. loving foreigners and supporting them

v. B. Ismila

vi. D. price

vii. E. oral testimonies

viii. A. Orphan family

ix. E. weaving

x. D. private parts

xi. A. NBS

xii. E. hygrometer

xiii. B. Seven

xiv. D. it need perseverance

xv. E. areas with less population

LIST A	ANSWER
i. Customs	F. Are beliefs about what is important in life
ii. Traditions	B. Inherited ways of thinking
iii. Norms	C. Acceptable ways of doing things
iv. Language	D. Ways of communication
v. Attire	A. Dressing style

SECTION B

3. i. Road transport

ii. Carl Peters

iii. a) Buildings b) Machines

4. i. Micro-enterprise

ii. 16°C

iii. Improved farming tools and weapons

5. a)

Archive	Museum
Collection of documents and records	Collection of artifacts and historical objects

b) i) Geita ii) Mara

6. a) Nuclear family

b) Oral tradition

SECTION C

7. i. Oil

ii. Serengeti National Park

iii. Lake Victoria

iv. Laundry

v. Hygiene

CIVIC AND MORAL EDUCATION FOR STD VII

SECTION A

1. i. e. discipline teacher

ii. a. Embezzlement

iii. e. village assembly

iv. c. Must be Tanzanian from Tanzania mainland

v. a. our flag flies a half must

vi. c. Electing school leaders

vii. d. EAC

viii. a. Child praising

ix. c. prostitution

x. a. Truancy

LIST A	ANSWER
i. New year	C. 1st of January each year
ii. The Union day	E. 26th of April each year
iii. Revolution day	D. 12th of January each year
iv. Karume day	B. 7th April each year
v. Labour day	A. 1st of May each year

3. i. Counselor

ii. Court of Appeal of Tanzania

iii. Chief Justice

iv. Secretary

v. By-laws

SECTION B

4. i. 500 shillings

ii. Giraffe

iii. 1994

iv. Julius Nyerere

v. Bank of Tanzania

5. i. a) HIV/AIDS b) Gender equality

ii. a) Health complications b) Psychological trauma

iii. Because it's easier to gather all citizens in one place at village level

iv. Early marriage is before legal age but with some consent, forced marriage is against full consent

v. Attorney General

SECTION C

6. i. Freedom of citizens to elect leaders of their choices

ii. 2 (direct and indirect)

iii. Prefects

iv. It helps us to promote human rights and understand more about responsibilities of citizens

v. a) Written constitution b) Unwritten constitution

Statistics Questions and Answers

Statistics Questions and Answers (1-10)

Question 1

In a certain school, the scores of 24 students in one of the Chemistry tests were as follows:

49, 64, 38, 46, 60, 68, 46, 42, 62, 38, 68, 57, 63, 76, 72, 73, 51, 55, 66, 63, 58, 47, 59 and 54

(a) Summarize the above data in a frequency distribution table using class size 5 and lowest limit of 35.

(b) Determine median score (Give your answer to two significant figures).

(c) Determine the score obtained by many students (Write your answer in two significant figures).

Answer 1

(a) Frequency distribution table:

Class Interval	Frequency
35-39	3
40-44	2
45-49	4
50-54	3
55-59	4
60-64	4
65-69	2
70-74	2
75-79	1

(b) Median score: 58

(c) Mode (score obtained by many students): 46 and 63 (both appear twice, bimodal)

Question 2

Display the results below in a histogram using 10 class interval size whereby the first class mark x be 5.5 and the last class mark x be 75.5

15 25 30 40 60 40 25 15 15 20 30 75 20 30 20 15 10 20 35 45 35 30 20 10 55 25 10 5 15 30 30 40 15 5 10 45 30 20 20 15 65 40 15 5 45 20 30 20 30 10

Answer 2

Histogram data (class intervals and frequencies):

Class Interval	Class Mark	Frequency
0.5-10.5	5.5	7
10.5-20.5	15.5	15
20.5-30.5	25.5	16
30.5-40.5	35.5	7
40.5-50.5	45.5	5
50.5-60.5	55.5	1
60.5-70.5	65.5	2
70.5-80.5	75.5	1

Note: A histogram would be drawn with these class intervals on the x-axis and frequencies on the y-axis.

Question 3

By grouping the data below using 10 class interval size with class mark of 5.5, 15.5, 25.5, ... find the mean, median and mode using assumed mean method where applicable taking A = 45.5.

15 25 30 40 60 40 25 15 15 20 30 75 20 30 20 15 10 20 35 45 35 30 20 10 55 25 10 5 15 30 30 40 15 5 10 45 30 20 20 15 65 40 15 5 45 20 30 20 30 10

Answer 3

Using class marks: 5.5, 15.5, 25.5, 35.5, 45.5, 55.5, 65.5, 75.5

Mean (using assumed mean method, A = 45.5): 28.9

Median: 25.5 (middle value of the ordered data set)

Mode: 15.5 (most frequent class mark)

Question 4

The marks below were obtained by 32 students in a civics test.

52 48 37 45 38 29 32 46 50 42 35 32 28 34 37 64 42 37 54 36 48 52 58 71 67 48 62 54 57 34 64 58

a) Prepare a frequency distribution table by grouping the marks using the class marks: 26.5, 32.5, 38.5, ...

b) Determine the mode from the histogram.

c) Calculate: i. the mean mark ii. The median mark

d) Prepare the pie chart from the frequency distribution table prepared in (a) above

Answer 4

a) Frequency distribution table:

Class Mark	Class Interval	Frequency
26.5	23.5-29.5	2
32.5	29.5-35.5	5
38.5	35.5-41.5	6
44.5	41.5-47.5	5
50.5	47.5-53.5	5
56.5	53.5-59.5	5
62.5	59.5-65.5	3
68.5	65.5-71.5	1

b) Mode from histogram: Approximately 38.5 (highest bar in the histogram)

c) i. Mean mark: 47.2

ii. Median mark: 46.5

d) Pie chart would show the percentage distribution of marks across the class intervals.

Question 5

Mr. Ndukeki timed all the phone calls his lovely wife made during august 2020. This list gives the times in minutes and seconds (e.g. 1.23 stands for 1 minute and 23 seconds).

6.24 5.12 3.04 6.26 5.01 3.52 4.14 7.19 6.34 2.05 5.40 4.48 5.17 2.58 3.48 2.00 1.46 0.48 6.06 0.50 2.45 6.00 3.03 4.24 3.34 1.20 7.48 2.44 2.03 3.45 3.53 1.34 0.45 3.54 5.40 4.47 7.41 6.50 6.43 5.23 4.49 2.34 3.49 2.00 4.09 3.48 5.23 2.47 5.23 1.40 6.05 5.10 7.09 5.45 5.07 6.45 7.53 6.09 4.06 5.45 5.32 0.56 4.57 3.48 4.25 4.40 7.10 0.54 6.10 0.36

a) Prepare a frequency distribution table by grouping the marks using the class intervals: 0 - 0.59, 1 - 1.59, 2 - 2.59 etc.

b) Draw the histogram for this data

Answer 5

a) Frequency distribution table:

Class Interval (minutes)	Frequency
0 - 0.59	4
1 - 1.59	5
2 - 2.59	11
3 - 3.59	13
4 - 4.59	10
5 - 5.59	12
6 - 6.59	9
7 - 7.59	6

b) Histogram would show call durations on the x-axis and frequencies on the y-axis, with bars for each class interval.

Question 6

In a basic mathematics test, the following marks were awarded:

48 47 67 73 50 76 47 44 44 42 56 66 67 45 43 71 48 64 52 49 34 35 46 89 37 47 54 45 60 57 58 54 45 42 54 62 32 64 44 58

Prepare a frequency distribution table starting with 32 having class size of 10, and then find:

(i) Mean using assumed mean of 56.5

(ii) Mode of the distribution

Answer 6

Frequency distribution table:

Class Interval	Frequency
32-41	4
42-51	14
52-61	8
62-71	9
72-81	4
82-91	1

(i) Mean using assumed mean (56.5): 53.8

(ii) Mode: 45 (most frequent value)

Question 7

The data below represents the Mass in kilograms(kg) of 20 bags of maize.

67 59 52 74 69 63 62 58 70 62 73 56 68 65 70 61 64 54 62 59

(a) Construct a frequency distribution table whose class marks are 53, 58, 63, ...

(b) Use the information in the table constructed in (a) above to calculate the mean mass using the class mark of the modal class as the assumed mean.

(c) Draw a histogram and a frequency polygon on same pair of axes

Answer 7

(a) Frequency distribution table:

Class Mark	Class Interval	Frequency
53	50.5-55.5	2
58	55.5-60.5	3
63	60.5-65.5	6
68	65.5-70.5	6
73	70.5-75.5	3

(b) Mean using modal class (63) as assumed mean: 63.8 kg

(c) Histogram and frequency polygon would show mass intervals on x-axis and frequencies on y-axis.

Question 8

The number of patients who attended maternity clinic daily in June 2017 in a certain village was recorded as follows:

52 61 42 27 38 44 56 36 73 20 41 48 77 30 46 43 72 63 43 76 47 53 38 55 60 51 47 58 33 37

(a) Make the frequency distribution table by grouping the number of patient in the class size of 10.

(b) By using the frequency distribution table obtained in (a), calculate the mean number of patient per day

(c) Construct a pie chart for the frequency distribution obtained in part (a)

Answer 8

(a) Frequency distribution table:

Class Interval	Frequency
20-29	2
30-39	7
40-49	9
50-59	6
60-69	3
70-79	3

(b) Mean number of patients per day: 48.6

(c) Pie chart would show the percentage distribution of patients across the class intervals.

Question 9

(NECTA CSEE 2021), The following are the marks obtained by 40 students in one of the Basic mathematics examination:

48 47 57 56 71 62 46 45 50 76 58 66 48 32 89 60 42 47 54 67 64 49 37 64 67 44 45 45 42 34 47 44 73 44 58 43 54 35 54 52

(a) Prepare a frequency distribution using the information: Number of classes = 8, Size of each class = 8 and The lower limit of the first class interval = 32.

(b) Use the frequency distribution obtained in part (a) to find the actual mean, when the assumed mean is 83.5

(c) Calculate the difference between the actual mean and the median of this distributions. Hence comment on the difference obtained.

Answer 9

(a) Frequency distribution table:

Class Interval	Frequency
32-39	4
40-47	11
48-55	9
56-63	5
64-71	6
72-79	3
80-87	1
88-95	1

(b) Actual mean (using assumed mean 83.5): 53.4

(c) Median: 52.5

Difference between mean and median: 0.9

Comment: The small difference suggests the data is approximately symmetric.

Question 10

The following were scores of 35 students in a Mathematics mock examination.

07 19 78 53 43 67 12 54 27 22 33 80 25 58 50 36 65 33 16 19 34 20 55 27 37 41 04 32 48 28 70 31 61 08 35

(a) Prepare a frequency distribution table using class marks 4.5, 14.5, 24.5, ...

(b) Which class interval has more students?

(c) Calculate the (i) Median mark (ii) Mode mark

Answer 10

(a) Frequency distribution table:

Class Mark	Class Interval	Frequency
4.5	0-9	3
14.5	10-19	5
24.5	20-29	6
34.5	30-39	8
44.5	40-49	3
54.5	50-59	5
64.5	60-69	2
74.5	70-79	2
84.5	80-89	1

(b) Class interval with most students: 30-39 (8 students)

(c) (i) Median mark: 34

(ii) Mode mark: 27 and 33 (both appear twice, bimodal)

MITIHANI POPOTE EXAMINATIONS SERIES FORM FOUR BIOLOGY EXAMINATION SERIES 1 (With Marking Scheme)

MITIHANI POPOTE EXAMINATIONS SERIES

FORM FOUR BIOLOGY EXAMINATION SERIES 1

Time: 3 Hours

Instructions:

• Answer all questions in Sections A and B.
• Answer two (2) questions from Section C.
• Use neat diagrams where applicable.

SECTION A (16 Marks)

1. Multiple Choice Questions (10 Marks)

Choose the correct answer:

1. Which of the following is a characteristic feature of Phylum Annelida?
   • a) Jointed appendages
   • b) Segmented body
   • c) Radial symmetry
   • d) Exoskeleton
2. In genetics, the term "allele" refers to:
   • a) A type of chromosome
   • b) A variant form of a gene
   • c) A cell organelle
   • d) A protein molecule
3. The theory of natural selection was proposed by:
   • a) Jean-Baptiste Lamarck
   • b) Charles Darwin
   • c) Gregor Mendel
   • d) Louis Pasteur
4. HIV primarily attacks which type of immune cells?
   • a) B cells
   • b) T-helper cells
   • c) Red blood cells
   • d) Platelets
5. Which of the following is an opportunistic infection associated with AIDS?
   • a) Malaria
   • b) Tuberculosis
   • c) Influenza
   • d) Common cold
6. PLWHA stands for:
   • a) People Living With Health Awareness
   • b) Patients Living With Human Allergy
   • c) People Living With HIV and AIDS
   • d) Protection of Life With Health Assistance
7. Which hormone is primarily responsible for growth in plants?
   • a) Insulin
   • b) Auxin
   • c) Adrenaline
   • d) Testosterone
8. The presence of homologous structures in different species is evidence of:
   • a) Convergent evolution
   • b) Divergent evolution
   • c) Artificial selection
   • d) Genetic drift
9. In humans, albinism is caused by:
   • a) A dominant allele
   • b) A recessive allele
   • c) A viral infection
   • d) A chromosomal abnormality
10. The class Insecta belongs to which phylum?
    • a) Mollusca
    • b) Arthropoda
    • c) Annelida
    • d) Chordata

2. Matching Items (6 Marks)

Match the items in List A with the correct responses in List B:

List A	List B
(i) Rhizoids	A. Female reproductive organ in moss
(ii) Sporophyte	B. Thread-like structure that anchors the moss to the ground
(iii) Capsule	C. Spore-producing structure found in the sporophyte
(iv) Protonema	D. Initial filamentous stage of moss growth from spores
(v) Archegonium	E. Male reproductive organ in moss
(vi) Antheridium	F. The diploid generation in moss life cycle responsible for spore production

SECTION B (54 Marks)

3. Phylum Chordata (10 Marks)

1. Describe three general characteristics of Phylum Chordata. (3 marks)
2. Classify the following organisms to their respective classes under Phylum Chordata:
   1. Frog
   2. Shark
   3. Pigeon
   4. Human
   (4 marks)
3. State the economic importance of any two classes under Phylum Chordata. (3 marks)

4. Genetics and Codominance (10 Marks)

1. Define the following genetic terms:
   1. Genotype
   2. Phenotype
   3. Codominance
   (3 marks)
2. In a certain plant species, red flower color (R) is codominant with white flower color (W). A cross between a red-flowered plant and a white-flowered plant produces plants with pink flowers.
   1. What is the genotype of the pink-flowered plants?
   2. If two pink-flowered plants are crossed, what would be the phenotypic ratio of their offspring? Show your working.
   (7 marks)

5. Organic Evolution (10 Marks)

1. Explain the concept of organic evolution. (2 marks)
2. Differentiate between homologous and analogous structures, giving one example of each. (4 marks)
3. State two evidences that support the theory of organic evolution. (4 marks)

6. HIV Life Cycle and AIDS (9 Marks)

1. Describe the life cycle of HIV in the human body. (5 marks)
2. Explain how the destruction of T-helper cells by HIV leads to the development of AIDS. (4 marks)

7. Opportunistic Infections and PLWHA Challenges (9 Marks)

1. Define the term "opportunistic infection" and give two examples common in PLWHA. (3 marks)
2. Discuss three challenges faced by PLWHA in society. (6 marks)

8. Human Growth and Development (9 Marks)

1. Outline the stages of human growth and development from infancy to adulthood. (5 marks)
2. Explain two internal and two external factors that affect growth in humans. (4 marks)

SECTION C (30 Marks)

Answer any two (2) questions.

9. Evidences for Organic Evolution (15 Marks)

Explain seven evidences for organic evolution.

10. Mendelian Inheritance (15 Marks)

Explain the principles of Mendelian inheritance and how they apply to monohybrid crosses. Use appropriate examples to illustrate your answer.

11. Impact of HIV/AIDS and ART (15 Marks)

Analyze the impact of HIV/AIDS on the immune system and the importance of antiretroviral therapy (ART) in managing the disease. Include the role of community support for PLWHA.

MITIHANI POPOTE EXAMINATIONS SERIES

FORM FOUR BIOLOGY EXAMINATION SERIES 1 – ANSWERS

Time: 3 Hours

SECTION A (16 Marks)

1. Multiple Choice Questions (10 Marks)

1. Which of the following is a characteristic feature of Phylum Annelida?
   • b) Segmented body
2. In genetics, the term "allele" refers to:
   • b) A variant form of a gene
3. The theory of natural selection was proposed by:
   • b) Charles Darwin
4. HIV primarily attacks which type of immune cells?
   • b) T-helper cells
5. Which of the following is an opportunistic infection associated with AIDS?
   • b) Tuberculosis
6. PLWHA stands for:
   • c) People Living With HIV and AIDS
7. Which hormone is primarily responsible for growth in plants?
   • b) Auxin
8. The presence of homologous structures in different species is evidence of:
   • b) Divergent evolution
9. In humans, albinism is caused by:
   • b) A recessive allele
10. The class Insecta belongs to which phylum?
    • b) Arthropoda

2. Matching Items (6 Marks)

List A	List B
(i) Rhizoids	B. Thread-like structure that anchors the moss to the ground
(ii) Sporophyte	F. The diploid generation in moss life cycle responsible for spore production
(iii) Capsule	C. Spore-producing structure found in the sporophyte
(iv) Protonema	D. Initial filamentous stage of moss growth from spores
(v) Archegonium	A. Female reproductive organ in moss
(vi) Antheridium	E. Male reproductive organ in moss

SECTION B (54 Marks)

3. Phylum Chordata (10 Marks)

1. Characteristics of Phylum Chordata:
   • Notochord: Flexible rod supporting the body.
   • Dorsal hollow nerve cord: Develops into the spinal cord.
   • Pharyngeal gill slits: Present at some developmental stage.
2. Classification:
   1. Frog: Class Amphibia
   2. Shark: Class Chondrichthyes
   3. Pigeon: Class Aves
   4. Human: Class Mammalia
3. Economic Importance:
   • Class Mammalia: Provide food (e.g., milk, meat) and labor (e.g., cattle).
   • Class Aves: Source of food (e.g., eggs, poultry) and pest control (e.g., birds eating insects).

4. Genetics and Codominance (10 Marks)

1. Definitions:
   1. Genotype: Genetic makeup of an organism (e.g., RR).
   2. Phenotype: Observable traits of an organism (e.g., red flowers).
   3. Codominance: Both alleles in a heterozygote are fully expressed (e.g., pink flowers in RW).
2. Codominance Cross:
   1. Genotype of pink-flowered plants: RW
   2. Phenotypic ratio of offspring (RW x RW):
      • Punnett Square: R (RW) x W (RW) yields 1 RR (red), 2 RW (pink), 1 WW (white).
      • Phenotypic ratio: 1 red : 2 pink : 1 white.

5. Organic Evolution (10 Marks)

1. Organic Evolution: Gradual change in species over time through genetic variation and natural selection.
2. Homologous vs. Analogous Structures:
   • Homologous: Same origin, different function (e.g., human arm and bat wing).
   • Analogous: Different origin, similar function (e.g., bird and insect wings).
3. Evidences for Evolution:
   • Fossil records: Show transitional forms (e.g., Archaeopteryx).
   • Comparative embryology: Similar embryonic stages (e.g., gill slits in vertebrates).

6. HIV Life Cycle and AIDS (9 Marks)

1. HIV Life Cycle:
   • HIV attaches to T-helper cells via CD4 receptors.
   • Viral RNA integrates into host DNA using reverse transcriptase.
   • Replicates, producing new virions.
   • Infected cells burst, releasing viruses.
   • Cycle repeats, spreading infection.
2. Destruction of T-helper Cells:
   • HIV destroys T-helper cells, reducing immune coordination.
   • Low T-helper count weakens immunity.
   • Opportunistic infections proliferate.
   • Progresses to AIDS, causing severe immune failure.

7. Opportunistic Infections and PLWHA Challenges (9 Marks)

1. Opportunistic Infection: Diseases exploiting weakened immunity.
   • Examples: Tuberculosis, Pneumocystis pneumonia.
2. Challenges for PLWHA:
   • Stigma: Social rejection and discrimination.
   • Access to healthcare: Limited ARV availability or cost.
   • Emotional distress: Anxiety and depression from diagnosis.

8. Human Growth and Development (9 Marks)

1. Stages:
   • Infancy: Rapid growth, motor development.
   • Childhood: Steady growth, cognitive development.
   • Adolescence: Puberty, rapid physical changes.
   • Adulthood: Physical maturity, slower growth.
   • Old age: Decline in physical functions.
2. Factors Affecting Growth:
   • Internal: Hormones (e.g., growth hormone), genetics.
   • External: Nutrition (e.g., protein intake), environment (e.g., stress).

SECTION C (30 Marks)

Answers for Questions 9 and 11.

9. Evidences for Organic Evolution (15 Marks)

• Fossil records: Show transitional forms (e.g., Tiktaalik).
• Homologous structures: Indicate common ancestry (e.g., vertebrate forelimbs).
• Analogous structures: Show convergent evolution (e.g., bird and insect wings).
• Vestigial structures: Remnants of ancestral traits (e.g., human appendix).
• Comparative embryology: Similar embryonic stages (e.g., gill slits).
• Biogeography: Species distribution reflects evolution (e.g., Galápagos finches).
• Molecular biology: DNA similarities indicate shared ancestry (e.g., cytochrome c).

11. Impact of HIV/AIDS and ART (15 Marks)

• Impact on Immune System: HIV destroys T-helper cells, leading to immune failure and AIDS.
• ART Importance: Inhibits HIV replication, restores immunity, reduces transmission.
• Community Support: Reduces stigma, provides emotional support, ensures treatment access.