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25 objectives • 250 MCQs • 100 short answers

Student Revision Notes

Term 3 Physics Quick Review

250 MCQs • 100 Short Answers
Adapted Question Credit
50 MCQs are adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank. Adapted questions show a small citation inside their answer box.

Grade 9 Advanced Physics — Term 3 Revision Notes

1. Waves

A wave transfers energy from one place to another without transferring matter permanently.

Types of waves

  • Transverse wave: particles vibrate perpendicular to the direction of travel. Examples: rope waves, light waves.
  • Longitudinal wave: particles vibrate parallel to the direction of travel. Example: sound in air.
  • Surface wave: occurs at the boundary between two media, such as water ripples.

Important wave quantities

  • Amplitude: maximum displacement from equilibrium. Larger amplitude usually means greater energy.
  • Wavelength, \(\lambda \): distance between two consecutive crests or compressions.
  • Period, T: time for one complete oscillation.
  • Frequency, f: number of oscillations per second, measured in hertz.
  • Wave speed: speed of wave travel.

Key equations:

  • \(f = 1/T\)
  • \(v = \lambda f\)

Reflection of pulses

  • At a fixed end, a reflected pulse is inverted.
  • At a free end, a reflected pulse remains upright.

Superposition

When two waves overlap, the resultant displacement is the algebraic sum of the individual displacements.

  • Same direction: constructive interference.
  • Opposite direction: destructive interference.

Standing waves

Standing waves form when two waves of the same frequency travel in opposite directions and interfere.

  • Node: point of zero/minimum displacement.
  • Antinode: point of maximum displacement.
  • For a string fixed at both ends: \(\lambda_n = 2L/n\).

2. Sound

Sound is a mechanical longitudinal wave. It travels through matter as pressure oscillations.

Sound properties

  • Sound travels as compressions and rarefactions.
  • Sound needs a medium; it cannot travel through a perfect vacuum.
  • Sound generally travels fastest in solids, slower in liquids, and slowest in gases.
  • In air, sound speed increases as temperature increases.

Pitch and loudness

  • Pitch depends on frequency. Higher frequency means higher pitch.
  • Loudness depends mainly on amplitude/intensity. Larger amplitude means louder sound.
  • Sound level is measured in decibels, dB.

Doppler effect

The Doppler effect is the change in detected frequency due to relative motion between source and observer.

  • Approaching source/observer: detected frequency increases.
  • Moving apart: detected frequency decreases.

Resonance

Resonance occurs when a system is driven at its natural frequency, causing a large increase in amplitude. Musical instruments use resonance in strings and air columns.

3. Light and Illumination

Light sources and media

  • Luminous source: produces its own light, such as the Sun or a lamp.
  • Non-luminous object: seen by reflected light, such as the Moon or a book.
  • Transparent: allows clear transmission of light.
  • Translucent: allows light through but scatters it.
  • Opaque: blocks most light.

Light quantities

  • Luminous flux, P: total visible light output, unit lumen (lm).
  • Illuminance, E: light falling on unit area, unit lux (lx).
  • Luminous intensity: light emitted in a direction, unit candela (cd).

Useful equations:

  • \(E = P/A\)
  • For a point source: \(E = P/(4\pi r^2)\)

Illuminance follows an inverse-square relationship:

  • Double distance → illuminance becomes one-quarter.
  • Triple distance → illuminance becomes one-ninth.

4. Diffraction and Color

Diffraction

Diffraction is the bending or spreading of waves as they pass an edge or opening. It is most noticeable when the gap size is similar to the wavelength.

Color of light

Color depends on wavelength and frequency.

  • Red light: longer wavelength, lower frequency.
  • Violet light: shorter wavelength, higher frequency.

Additive color mixing

Primary colors of light:

  • Red
  • Green
  • Blue

Combinations:

  • Red + green = yellow
  • Green + blue = cyan
  • Red + blue = magenta
  • Red + green + blue = white

Object color depends on which wavelengths are present in incident light and which wavelengths are reflected or absorbed.

Pigments and dyes

Pigments use subtractive color mixing.

  • Cyan absorbs red.
  • Magenta absorbs green.
  • Yellow absorbs blue.

Mixing pigments usually absorbs more wavelengths and gives darker colors.

5. Polarization and Malus's Law

Polarization is the restriction of light vibrations to one direction. Only transverse waves can be polarized.

Polarization by filtering

A polarizing filter allows one vibration direction to pass and blocks the perpendicular component.

Polarization by reflection

Reflected glare can be partially polarized. Polarizing sunglasses reduce glare by blocking much of this reflected polarized light.

Malus's law

For polarized light passing through an analyzer:

\(I_2 = I_1 \cos^2\theta \)

Where:

  • \(I_1\) is intensity before the analyzer.
  • \(I_2\) is intensity after the analyzer.
  • \(\theta \) is the angle between the polarizer and analyzer axes.

Special cases:

  • \(\theta \) = 0° → maximum transmitted intensity.
  • \(\theta \) = 90° → zero/minimum transmitted intensity.
  • For unpolarized light passing through the first ideal polarizer: \(I_1 = I_0/2\).

Exam Tips

  • Always check units: Hz, m/s, m, s, dB, lm, lx, cd.
  • For formula questions, write the formula first, substitute values, then calculate.
  • For wave diagrams, mark the direction of travel and direction of particle vibration.
  • For superposition, add displacements with signs.
  • For Malus's law, decide whether the light entering the first polarizer is unpolarized or already polarized.

Waves • pages 115-116

OBJ01: Differentiate between transverse, longitudinal, and surface waves and give examples.

10 MCQs • 4 Short Answers
G9T3_OBJ01_MCQ01 MCQ Easy Identify • 4 marks

Which statement correctly describes a transverse wave?

Transverse vs Longitudinal Waves Transverse: vibration ⟂ travel travel Longitudinal: vibration ∥ travel
Wave Types
AIt never needs a medium
BIt travels only in gases
CParticles vibrate perpendicular to the direction of wave travel
DParticles vibrate parallel to the direction of wave travel
Answer
C. Particles vibrate perpendicular to the direction of wave travel
In a transverse wave, vibration is perpendicular to energy transfer, such as a wave on a rope.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ01_MCQ02 MCQ Easy Identify • 4 marks

When particles vibrate parallel to the wave direction, the wave is:

Transverse vs Longitudinal Waves Transverse: vibration ⟂ travel travel Longitudinal: vibration ∥ travel
Wave Types
AOnly electromagnetic
BTransverse
COnly surface
DLongitudinal
Answer
D. Longitudinal
Longitudinal waves form compressions and rarefactions; the vibration is parallel to wave travel.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ01_MCQ03 MCQ Easy Identify • 4 marks

A student shakes one end of a rope up and down while the wave travels horizontally. What type of wave is produced?

Transverse vs Longitudinal Waves Transverse: vibration ⟂ travel travel Longitudinal: vibration ∥ travel
Wave Types
ALongitudinal wave
BSound wave
CSurface wave
DTransverse wave
Answer
D. Transverse wave
The disturbance is perpendicular to the direction of travel.
Source: Original generated question
G9T3_OBJ01_MCQ04 MCQ Easy Classify • 4 marks

Which example is best classified as a longitudinal mechanical wave?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
AA ripple on the surface of water
BLight travelling through vacuum
CSound travelling through air
DA wave on a rope shaken vertically
Answer
C. Sound travelling through air
Sound in air consists of compressions and rarefactions parallel to the direction of travel.
Source: Original generated question
G9T3_OBJ01_MCQ05 MCQ Easy Complete • 4 marks

In a longitudinal wave, the particles of the medium vibrate ____ the direction of energy transfer.

Transverse vs Longitudinal Waves Transverse: vibration ⟂ travel travel Longitudinal: vibration ∥ travel
Wave Types
Aparallel to
Bat 90° to
Copposite only to
Dperpendicular to
Answer
A. parallel to
Longitudinal motion is parallel to the wave direction.
Source: Original generated question
G9T3_OBJ01_MCQ06 MCQ Medium Explain • 4 marks

A surface water wave is different from a pure transverse wave because the water particles near the surface usually move in:

Transverse vs Longitudinal Waves Transverse: vibration ⟂ travel travel Longitudinal: vibration ∥ travel
Wave Types
Aonly straight-line vertical motion
Bno motion at all
Ccombined up-down and back-forth motion
Donly straight-line horizontal motion
Answer
C. combined up-down and back-forth motion
Surface waves have both transverse and longitudinal components.
Source: Original generated question
G9T3_OBJ01_MCQ07 MCQ Medium Compare • 4 marks

Which statement correctly compares transverse and longitudinal waves?

Transverse vs Longitudinal Waves Transverse: vibration ⟂ travel travel Longitudinal: vibration ∥ travel
Wave Types
ABoth have vibrations parallel to travel.
BBoth require vacuum to travel.
CTransverse waves have compressions only.
DTransverse vibrations are perpendicular; longitudinal vibrations are parallel.
Answer
D. Transverse vibrations are perpendicular; longitudinal vibrations are parallel.
The comparison is based on the direction of particle vibration relative to propagation.
Source: Original generated question
G9T3_OBJ01_MCQ08 MCQ Medium Match • 4 marks

Which pair correctly matches wave type and example?

Transverse vs Longitudinal Waves Transverse: vibration ⟂ travel travel Longitudinal: vibration ∥ travel
Wave Types
ATransverse: sound in air; longitudinal: light wave
BTransverse: compression wave; longitudinal: water surface only
CTransverse: echo; longitudinal: shadow
DTransverse: light wave; longitudinal: sound wave in air
Answer
D. Transverse: light wave; longitudinal: sound wave in air
Light is a transverse electromagnetic wave; sound in air is longitudinal.
Source: Original generated question
G9T3_OBJ01_MCQ09 MCQ Medium Reason • 4 marks

A wave has crests and troughs clearly visible on a stretched string. Which property identifies it as transverse?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
AThe wave needs no medium.
BThe wave has no wavelength.
CThe string displacement is perpendicular to the direction of travel.
DThe particles compress together in the same direction.
Answer
C. The string displacement is perpendicular to the direction of travel.
Crests and troughs occur when the vibration is perpendicular to the travel direction.
Source: Original generated question
G9T3_OBJ01_MCQ10 MCQ Easy Identify • 4 marks

A slinky is pushed and pulled along its length. Regions of compression move down the spring. The wave is:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Atransverse
Blongitudinal
Csurface
Dpolarized light
Answer
B. longitudinal
Compressions moving along a slinky show particle motion parallel to the wave direction.
Source: Original generated question
G9T3_OBJ01_SA01 Short Answer 4 marks

Compare transverse and longitudinal waves using direction of particle vibration and give one example of each.

Expected Answer

States transverse vibration is perpendicular to direction of travel. States longitudinal vibration is parallel to direction of travel. Gives a valid transverse example such as rope wave or light. Gives a valid longitudinal example such as sound in air.

Mark Scheme
  • 1 mark: States transverse vibration is perpendicular to direction of travel.
  • 1 mark: States longitudinal vibration is parallel to direction of travel.
  • 1 mark: Gives a valid transverse example such as rope wave or light.
  • 1 mark: Gives a valid longitudinal example such as sound in air.
Source: Original generated question
G9T3_OBJ01_SA02 Short Answer 4 marks

Explain why a water ripple is usually classified as a surface wave.

Expected Answer

States it occurs at a boundary/interface. Mentions air-water or water surface. Explains motion has both up-down and back-forth components. States energy travels across the surface.

Mark Scheme
  • 1 mark: States it occurs at a boundary/interface.
  • 1 mark: Mentions air-water or water surface.
  • 1 mark: Explains motion has both up-down and back-forth components.
  • 1 mark: States energy travels across the surface.
Source: Original generated question
G9T3_OBJ01_SA03 Short Answer 4 marks

A slinky is pushed and pulled along its length. Identify the wave type and justify.

Expected Answer

Identifies the wave as longitudinal. Mentions compressions and rarefactions. States particle vibration is parallel to wave direction. Gives a clear link to the slinky motion.

Mark Scheme
  • 1 mark: Identifies the wave as longitudinal.
  • 1 mark: Mentions compressions and rarefactions.
  • 1 mark: States particle vibration is parallel to wave direction.
  • 1 mark: Gives a clear link to the slinky motion.
Source: Original generated question
G9T3_OBJ01_SA04 Short Answer 4 marks

List three wave types and one example for each.

Expected Answer

Names transverse wave with valid example. Names longitudinal wave with valid example. Names surface wave with valid example. Examples are correctly matched to wave types.

Mark Scheme
  • 1 mark: Names transverse wave with valid example.
  • 1 mark: Names longitudinal wave with valid example.
  • 1 mark: Names surface wave with valid example.
  • 1 mark: Examples are correctly matched to wave types.
Source: Original generated question

Waves • pages 118-120

OBJ02: Calculate frequency from period and vice versa; apply v = λf to calculate speed, wavelength, or frequency.

10 MCQs • 4 Short Answers
G9T3_OBJ02_MCQ01 MCQ Medium Identify • 4 marks

A wave has wavelength \(\lambda \) = 0.5 m and frequency \(f = 8 Hz. Calculate its speed.\)

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
A2 m/s
B40 m/s
C8 m/s
D4 m/s
Answer
D. 4 m/s
Use \(v = \lambda f\), so \(v = 0.5\)× 8 = 4 m/s.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ02_MCQ02 MCQ Medium Identify • 4 marks

A wave has wavelength \(\lambda \) = 2 m and frequency \(f = 3 Hz. Calculate its speed.\)

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
A12 m/s
B6 m/s
C3 m/s
D60 m/s
Answer
B. 6 m/s
Use \(v = \lambda f\), so \(v = 2\)× 3 = 6 m/s.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ02_MCQ03 MCQ Medium Calculate • 4 marks

A wave has a period of 0.2 s. What is its frequency?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
A10 Hz
B0.2 Hz
C5 Hz
D2.5 Hz
Answer
C. 5 Hz
Use \(f = 1/T\) = 1/0.2 = 5 Hz.
Source: Original generated question
G9T3_OBJ02_MCQ04 MCQ Medium Calculate • 4 marks

A wave has a period of 0.5 s. What is its frequency?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
A0.5 Hz
B1 Hz
C2 Hz
D4 Hz
Answer
C. 2 Hz
Use \(f = 1/T\) = 1/0.5 = 2 Hz.
Source: Original generated question
G9T3_OBJ02_MCQ05 MCQ Medium Calculate • 4 marks

A wave has a period of 0.025 s. What is its frequency?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
A40 Hz
B80 Hz
C0.025 Hz
D20 Hz
Answer
A. 40 Hz
Use \(f = 1/T\) = 1/0.025 = 40 Hz.
Source: Original generated question
G9T3_OBJ02_MCQ06 MCQ Medium Calculate • 4 marks

A wave has a period of 4 s. What is its frequency?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
A4 Hz
B0.25 Hz
C0.125 Hz
D0.5 Hz
Answer
B. 0.25 Hz
Use \(f = 1/T\) = 1/4 = 0.25 Hz.
Source: Original generated question
G9T3_OBJ02_MCQ07 MCQ Medium Calculate • 4 marks

A wave has wavelength 2 m and frequency 8 Hz. What is its speed?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
A8 m/s
B32 m/s
C16 m/s
D4 m/s
Answer
C. 16 m/s
Use \(v = \lambda f\) = 2 × 8 = 16 m/s.
Source: Original generated question
G9T3_OBJ02_MCQ08 MCQ Medium Calculate • 4 marks

A wave has wavelength 0.75 m and frequency 20 Hz. What is its speed?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
A7.5 m/s
B26.6667 m/s
C15 m/s
D30 m/s
Answer
C. 15 m/s
Use \(v = \lambda f\) = 0.75 × 20 = 15 m/s.
Source: Original generated question
G9T3_OBJ02_MCQ09 MCQ Medium Calculate • 4 marks

A wave has wavelength 1.5 m and frequency 4 Hz. What is its speed?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
A12 m/s
B3 m/s
C2.66667 m/s
D6 m/s
Answer
D. 6 m/s
Use \(v = \lambda f\) = 1.5 × 4 = 6 m/s.
Source: Original generated question
G9T3_OBJ02_MCQ10 MCQ Easy Recall • 4 marks

Which equation is used to find wave speed from wavelength and frequency?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
A\(v = f/\lambda \)
B\(\lambda \) = vf
C\(v = \lambda f\)
D\(f = \lambda /v\)
Answer
C. \(v = \lambda f\)
Wave speed equals wavelength times frequency.
Source: Original generated question
G9T3_OBJ02_SA01 Short Answer 4 marks

A wave has period 0.20 s and wavelength 1.5 m. Calculate frequency and wave speed.

Expected Answer

Uses \(f = 1/T\). Calculates \(f = 5 Hz. Uses\)\(v = \lambda f\). Calculates \(v = 7.5 m/s with units.\)

Mark Scheme
  • 1 mark: Uses \(f = 1/T\).
  • 1 mark: Calculates \(f = 5 Hz.\)
  • 1 mark: Uses \(v = \lambda f\).
  • 1 mark: Calculates \(v = 7.5 m/s with units.\)
Source: Original generated question
G9T3_OBJ02_SA02 Short Answer 4 marks

Explain the difference between period and frequency and state their relationship.

Expected Answer

Defines period as time for one cycle. Defines frequency as cycles per second. States \(f = 1/T\) or \(T = 1/f. Uses correct units\): s and Hz.

Mark Scheme
  • 1 mark: Defines period as time for one cycle.
  • 1 mark: Defines frequency as cycles per second.
  • 1 mark: States \(f = 1/T\) or \(T = 1/f.\)
  • 1 mark: Uses correct units: s and Hz.
Source: Original generated question
G9T3_OBJ02_SA03 Short Answer 4 marks

A wave travels at 24 m/s with frequency 6 Hz. Find wavelength.

Expected Answer

Writes \(v = \lambda f\). Rearranges \(\lambda \) = v/f. Substitutes 24/6. Gives \(\lambda \) = 4 m.

Mark Scheme
  • 1 mark: Writes \(v = \lambda f\).
  • 1 mark: Rearranges \(\lambda \) = v/f.
  • 1 mark: Substitutes 24/6.
  • 1 mark: Gives \(\lambda \) = 4 m.
Source: Original generated question
G9T3_OBJ02_SA04 Short Answer 4 marks

Describe how wavelength changes when frequency increases while wave speed remains constant.

Expected Answer

States \(v = \lambda f\). States v is constant in the same medium. Explains wavelength and frequency are inversely related. Concludes wavelength decreases.

Mark Scheme
  • 1 mark: States \(v = \lambda f\).
  • 1 mark: States v is constant in the same medium.
  • 1 mark: Explains wavelength and frequency are inversely related.
  • 1 mark: Concludes wavelength decreases.
Source: Original generated question

Waves • pages 121-122

OBJ03: Describe that a mechanical wave is inverted if reflected from a fixed end and remains upright if reflected from a free end.

10 MCQs • 4 Short Answers
G9T3_OBJ03_MCQ01 MCQ Easy Identify • 4 marks

When a wave pulse reflects from a fixed end, what happens to it?

Reflection of Mechanical Pulses fixed end → inverted free end → upright
Reflection Fixed Free
AIt is inverted
BIt disappears without reflection
CIt always remains upright
DIt stops carrying energy
Answer
A. It is inverted
A fixed end causes an inverted reflection because the end cannot move freely.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ03_MCQ02 MCQ Easy Identify • 4 marks

When a wave pulse reflects from a free end, what happens to it?

Reflection of Mechanical Pulses fixed end → inverted free end → upright
Reflection Fixed Free
AIt is always inverted
BIt remains upright
CIts direction becomes perpendicular
DIt becomes light
Answer
B. It remains upright
At a free end, the end can move, so the pulse reflects without inversion.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ03_MCQ03 MCQ Easy Identify • 4 marks

A pulse reaches a wall where the end of the rope is fixed. The reflected pulse is:

Reflection of Mechanical Pulses fixed end → inverted free end → upright
Reflection Fixed Free
Aupright
Bchanged into light
Cabsorbed completely
Dinverted
Answer
D. inverted
Reflection from a fixed end produces an inverted pulse.
Source: Original generated question
G9T3_OBJ03_MCQ04 MCQ Easy Identify • 4 marks

A pulse reaches a loose ring that can move freely on a vertical rod. The reflected pulse is:

Reflection of Mechanical Pulses fixed end → inverted free end → upright
Reflection Fixed Free
Aupright
Binverted
Ccancelled forever
Dconverted into a compression
Answer
A. upright
Reflection from a free end is not inverted.
Source: Original generated question
G9T3_OBJ03_MCQ05 MCQ Medium Explain • 4 marks

Why does a fixed end produce an inverted reflected pulse?

Reflection of Mechanical Pulses fixed end → inverted free end → upright
Reflection Fixed Free
AThe boundary exerts an opposite force on the rope.
BThe rope becomes transparent.
CThe wave speed becomes zero everywhere.
DThe wave loses all energy.
Answer
A. The boundary exerts an opposite force on the rope.
A fixed boundary cannot move, so the reaction reverses the displacement.
Source: Original generated question
G9T3_OBJ03_MCQ06 MCQ Medium Predict • 4 marks

A crest pulse returns from a fixed end. What shape returns?

Reflection of Mechanical Pulses fixed end → inverted free end → upright
Reflection Fixed Free
Aa circular pulse
Ba crest pulse
Ca trough pulse
Dno pulse
Answer
C. a trough pulse
The reflection is inverted: crest becomes trough.
Source: Original generated question
G9T3_OBJ03_MCQ07 MCQ Medium Predict • 4 marks

A trough pulse returns from a free end. What shape returns?

Reflection of Mechanical Pulses fixed end → inverted free end → upright
Reflection Fixed Free
Aa transverse light wave
Ba trough pulse
Ctwo crests
Da crest pulse
Answer
B. a trough pulse
Free-end reflection remains upright, so the trough remains a trough.
Source: Original generated question
G9T3_OBJ03_MCQ08 MCQ Medium Infer • 4 marks

A mechanical pulse reflects from a boundary and returns upside down. The boundary was most likely:

Reflection of Mechanical Pulses fixed end → inverted free end → upright
Reflection Fixed Free
Afree
Bfixed
Ctransparent
Dvacuum only
Answer
B. fixed
Inversion is the sign of reflection from a fixed end.
Source: Original generated question
G9T3_OBJ03_MCQ09 MCQ Easy Compare • 4 marks

Which statement is true for both fixed-end and free-end reflection?

Reflection of Mechanical Pulses fixed end → inverted free end → upright
Reflection Fixed Free
AThe wave is always inverted.
BThe wave always disappears.
CThe wave changes direction after reflection.
DThe wave becomes electromagnetic.
Answer
C. The wave changes direction after reflection.
Both types reflect the pulse back along the medium.
Source: Original generated question
G9T3_OBJ03_MCQ10 MCQ Easy Identify • 4 marks

A pulse on a spring reflects upright. Which diagram label should be assigned to the boundary?

Reflection of Mechanical Pulses fixed end → inverted free end → upright
Reflection Fixed Free
Aopaque surface
Bnode only
Cfixed end
Dfree end
Answer
D. free end
Upright reflection is associated with a free boundary.
Source: Original generated question
G9T3_OBJ03_SA01 Short Answer 4 marks

Describe how a pulse reflects from a fixed end and from a free end.

Expected Answer

Fixed-end reflection is inverted. Free-end reflection remains upright. Both reflected pulses travel back in the opposite direction. Uses correct terms fixed and free boundary.

Mark Scheme
  • 1 mark: Fixed-end reflection is inverted.
  • 1 mark: Free-end reflection remains upright.
  • 1 mark: Both reflected pulses travel back in the opposite direction.
  • 1 mark: Uses correct terms fixed and free boundary.
Source: Original generated question
G9T3_OBJ03_SA02 Short Answer 4 marks

A crest pulse reflects from a fixed wall. Draw/describe the reflected pulse.

Expected Answer

Identifies reflected pulse as a trough. Mentions inversion. States direction is reversed. Sketch/description shows correct displacement sign.

Mark Scheme
  • 1 mark: Identifies reflected pulse as a trough.
  • 1 mark: Mentions inversion.
  • 1 mark: States direction is reversed.
  • 1 mark: Sketch/description shows correct displacement sign.
Source: Original generated question
G9T3_OBJ03_SA03 Short Answer 4 marks

Explain the force reason for inversion at a fixed end.

Expected Answer

Fixed end cannot move. Boundary exerts opposite reaction force. Displacement reverses sign. Reflected pulse is inverted.

Mark Scheme
  • 1 mark: Fixed end cannot move.
  • 1 mark: Boundary exerts opposite reaction force.
  • 1 mark: Displacement reverses sign.
  • 1 mark: Reflected pulse is inverted.
Source: Original generated question
G9T3_OBJ03_SA04 Short Answer 4 marks

A pulse reflects upright. What can you infer about the boundary?

Expected Answer

Boundary is free/movable. Pulse is not inverted. Reflected pulse travels back along the medium. Gives a correct physical reason or example.

Mark Scheme
  • 1 mark: Boundary is free/movable.
  • 1 mark: Pulse is not inverted.
  • 1 mark: Reflected pulse travels back along the medium.
  • 1 mark: Gives a correct physical reason or example.
Source: Original generated question

Waves • pages 116-118

OBJ04: Describe wave properties: amplitude, energy, wavelength, speed, phase, period, and frequency.

10 MCQs • 4 Short Answers
G9T3_OBJ04_MCQ01 MCQ Easy Identify • 4 marks

What is the amplitude of a wave?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
AThe maximum displacement from equilibrium
BThe distance between two consecutive crests
CThe time for one vibration
DThe number of vibrations per second
Answer
A. The maximum displacement from equilibrium
Amplitude measures how far a particle is displaced from equilibrium and is related to wave energy.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ04_MCQ02 MCQ Medium Identify • 4 marks

If the amplitude of a mechanical wave increases while other factors remain constant, what happens to the wave energy?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
AIt becomes zero
BIt decreases
CIt increases
DThere is no relationship
Answer
C. It increases
Wave energy increases with amplitude; a larger-amplitude wave carries more energy.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ04_MCQ03 MCQ Easy Define • 4 marks

Which wave property is the distance between two consecutive crests?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
APeriod
BWavelength
CFrequency
DAmplitude
Answer
B. Wavelength
Wavelength is the distance for one complete cycle.
Source: Original generated question
G9T3_OBJ04_MCQ04 MCQ Easy Define • 4 marks

Which property tells how many complete oscillations occur each second?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
AWavelength
BPeriod
CFrequency
DAmplitude
Answer
C. Frequency
Frequency is cycles per second, measured in hertz.
Source: Original generated question
G9T3_OBJ04_MCQ05 MCQ Medium Interpret • 4 marks

Two waves have the same wavelength, but wave A has twice the amplitude of wave B. Which statement is correct?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
AWave A has no period.
BThey must have different frequencies.
CWave B carries more energy.
DWave A carries more energy.
Answer
D. Wave A carries more energy.
For mechanical waves, greater amplitude usually means greater energy.
Source: Original generated question
G9T3_OBJ04_MCQ06 MCQ Medium Define • 4 marks

Points on two identical waves are in phase when they:

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
Atravel in different media only
Bhave opposite displacements only
Care at the same part of the cycle at the same time
Dhave no wavelength
Answer
C. are at the same part of the cycle at the same time
In-phase points move together through the same stage of oscillation.
Source: Original generated question
G9T3_OBJ04_MCQ07 MCQ Medium Identify • 4 marks

Which pair of wave properties are reciprocals?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
AAmplitude and wavelength
BCrest and trough
CSpeed and phase
DFrequency and period
Answer
D. Frequency and period
Frequency and period satisfy \(f = 1/T\).
Source: Original generated question
G9T3_OBJ04_MCQ08 MCQ Easy Recognize • 4 marks

A wave with larger amplitude appears:

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
Alower speed only
Bcloser crests
Ctaller from the equilibrium line
Dshorter period only
Answer
C. taller from the equilibrium line
Amplitude is maximum displacement from equilibrium.
Source: Original generated question
G9T3_OBJ04_MCQ09 MCQ Easy Define • 4 marks

What is wave speed?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
AThe distance a wave pattern travels per unit time
BThe mass of the medium
CThe number of crests only
DThe height of a crest
Answer
A. The distance a wave pattern travels per unit time
Speed describes how fast the disturbance travels.
Source: Original generated question
G9T3_OBJ04_MCQ10 MCQ Hard Infer • 4 marks

If the period of a wave increases while its speed is unchanged, what happens to frequency?

Wave Properties amplitude wavelength λ period T = time for one cycle frequency f = cycles per second
Wave Properties
AIt increases.
BIt remains double.
CIt decreases.
DIt becomes equal to wavelength.
Answer
C. It decreases.
Frequency is inversely proportional to period.
Source: Original generated question
G9T3_OBJ04_SA01 Short Answer 4 marks

Define amplitude, wavelength, period, and frequency.

Expected Answer

Amplitude is maximum displacement. Wavelength is distance for one complete cycle. Period is time for one cycle. Frequency is cycles per second.

Mark Scheme
  • 1 mark: Amplitude is maximum displacement.
  • 1 mark: Wavelength is distance for one complete cycle.
  • 1 mark: Period is time for one cycle.
  • 1 mark: Frequency is cycles per second.
Source: Original generated question
G9T3_OBJ04_SA02 Short Answer 4 marks

Explain how amplitude is related to wave energy.

Expected Answer

Larger amplitude means larger displacement. Larger amplitude waves carry more energy. Comparison assumes same medium/other factors similar. Gives a valid example such as louder sound or bigger rope wave.

Mark Scheme
  • 1 mark: Larger amplitude means larger displacement.
  • 1 mark: Larger amplitude waves carry more energy.
  • 1 mark: Comparison assumes same medium/other factors similar.
  • 1 mark: Gives a valid example such as louder sound or bigger rope wave.
Source: Original generated question
G9T3_OBJ04_SA03 Short Answer 4 marks

Describe what it means for two points on a wave to be in phase.

Expected Answer

Points are at the same stage of oscillation. They move in the same way at the same time. Examples: crest with crest or trough with trough. Uses phase terminology correctly.

Mark Scheme
  • 1 mark: Points are at the same stage of oscillation.
  • 1 mark: They move in the same way at the same time.
  • 1 mark: Examples: crest with crest or trough with trough.
  • 1 mark: Uses phase terminology correctly.
Source: Original generated question
G9T3_OBJ04_SA04 Short Answer 4 marks

A wave has frequency 10 Hz. What does this mean physically?

Expected Answer

Frequency is cycles per second. 10 Hz means 10 oscillations each second. Unit Hz equals s^-1. Relates to the vibrating source/medium.

Mark Scheme
  • 1 mark: Frequency is cycles per second.
  • 1 mark: 10 Hz means 10 oscillations each second.
  • 1 mark: Unit Hz equals s^-1.
  • 1 mark: Relates to the vibrating source/medium.
Source: Original generated question

Waves • pages 122-123

OBJ05: Sketch snapshots for superposition of two overlapping pulses and find resultant amplitude.

10 MCQs • 4 Short Answers
G9T3_OBJ05_MCQ01 MCQ Medium Identify • 4 marks

A +5 cm pulse meets a −5 cm pulse. What is the resultant amplitude at complete overlap?

Superposition of Pulses Before overlap Resultant amplitude = algebraic sum
Superposition
A10 cm
B-10 cm
C5 cm
D0 cm
Answer
D. 0 cm
Amplitudes add algebraically: +5 + (−5) = 0, a complete destructive interference.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ05_MCQ02 MCQ Medium Identify • 4 marks

Constructive interference occurs when:

Superposition of Pulses Before overlap Resultant amplitude = algebraic sum
Superposition
AA crest meets an equal trough
BEnergy always becomes zero
CThe wave only refracts
DTwo crests or disturbances in the same direction overlap
Answer
D. Two crests or disturbances in the same direction overlap
In constructive interference amplitudes add and the resultant disturbance increases.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ05_MCQ03 MCQ Medium Calculate • 4 marks

Two pulses of +2 cm and +3 cm overlap at the same point. What is the resultant displacement?

Superposition of Pulses Before overlap Resultant amplitude = algebraic sum
Superposition
A+1 cm
B+5 cm
C−5 cm
D0 cm
Answer
B. +5 cm
Superposition adds displacements algebraically.
Source: Original generated question
G9T3_OBJ05_MCQ04 MCQ Medium Calculate • 4 marks

A +4 cm pulse overlaps a −4 cm pulse. What is the resultant amplitude?

Superposition of Pulses Before overlap Resultant amplitude = algebraic sum
Superposition
A0 cm
B4 cm
C−8 cm
D8 cm
Answer
A. 0 cm
Equal and opposite displacements cancel.
Source: Original generated question
G9T3_OBJ05_MCQ05 MCQ Easy Identify • 4 marks

Which condition produces constructive interference?

Superposition of Pulses Before overlap Resultant amplitude = algebraic sum
Superposition
AAn upward and equal downward pulse overlap.
BTwo upward pulses overlap.
CA pulse reflects from a wall only.
DNo waves meet.
Answer
B. Two upward pulses overlap.
Same-direction displacements add to make a larger displacement.
Source: Original generated question
G9T3_OBJ05_MCQ06 MCQ Easy Identify • 4 marks

Which condition produces destructive interference?

Superposition of Pulses Before overlap Resultant amplitude = algebraic sum
Superposition
ATwo upward pulses overlap.
BTwo crests overlap.
CA single pulse travels alone.
DA crest and trough overlap.
Answer
D. A crest and trough overlap.
Opposite displacements subtract.
Source: Original generated question
G9T3_OBJ05_MCQ07 MCQ Medium Explain • 4 marks

During superposition, after two pulses pass through each other, they generally:

Superposition of Pulses Before overlap Resultant amplitude = algebraic sum
Superposition
Alose all wavelength
Bcontinue moving with their original shapes
Cstop permanently
Dcombine into one fixed pulse
Answer
B. continue moving with their original shapes
In a linear medium, pulses pass through and continue.
Source: Original generated question
G9T3_OBJ05_MCQ08 MCQ Medium Calculate • 4 marks

A +6 cm pulse and a −2 cm pulse overlap completely. What is the resultant displacement?

Superposition of Pulses Before overlap Resultant amplitude = algebraic sum
Superposition
A+4 cm
B0 cm
C−4 cm
D+8 cm
Answer
A. +4 cm
Add algebraically: +6 + (−2) = +4 cm.
Source: Original generated question
G9T3_OBJ05_MCQ09 MCQ Easy Recall • 4 marks

Which principle allows the resultant displacement to be found by adding individual displacements?

Superposition of Pulses Before overlap Resultant amplitude = algebraic sum
Superposition
APrinciple of superposition
BInverse-square law
CDoppler effect
DMalus's law
Answer
A. Principle of superposition
The superposition principle states that overlapping wave displacements add.
Source: Original generated question
G9T3_OBJ05_MCQ10 MCQ Medium Calculate • 4 marks

Two identical downward pulses each have amplitude −3 cm. At complete overlap, the resultant is:

Superposition of Pulses Before overlap Resultant amplitude = algebraic sum
Superposition
A−3 cm
B−6 cm
C0 cm
D+6 cm
Answer
B. −6 cm
Same-direction downward displacements add.
Source: Original generated question
G9T3_OBJ05_SA01 Short Answer 4 marks

Two pulses +5 cm and −3 cm overlap. Find resultant displacement and name the interference type.

Expected Answer

Adds displacements algebraically. Calculates +2 cm. Identifies partial destructive interference. Includes correct sign/direction.

Mark Scheme
  • 1 mark: Adds displacements algebraically.
  • 1 mark: Calculates +2 cm.
  • 1 mark: Identifies partial destructive interference.
  • 1 mark: Includes correct sign/direction.
Source: Original generated question
G9T3_OBJ05_SA02 Short Answer 4 marks

Explain the principle of superposition for overlapping pulses.

Expected Answer

Resultant displacement is found point by point. Individual displacements are added algebraically. Applies only while pulses overlap. Pulses continue after passing through in a linear medium.

Mark Scheme
  • 1 mark: Resultant displacement is found point by point.
  • 1 mark: Individual displacements are added algebraically.
  • 1 mark: Applies only while pulses overlap.
  • 1 mark: Pulses continue after passing through in a linear medium.
Source: Original generated question
G9T3_OBJ05_SA03 Short Answer 4 marks

Sketch/describe complete constructive interference of two equal upward pulses.

Expected Answer

Shows two upward pulses overlap. Resultant amplitude is sum of amplitudes. Resultant is larger than either pulse. Pulses reappear after overlap.

Mark Scheme
  • 1 mark: Shows two upward pulses overlap.
  • 1 mark: Resultant amplitude is sum of amplitudes.
  • 1 mark: Resultant is larger than either pulse.
  • 1 mark: Pulses reappear after overlap.
Source: Original generated question
G9T3_OBJ05_SA04 Short Answer 4 marks

Sketch/describe complete destructive interference of equal opposite pulses.

Expected Answer

Shows crest and equal trough overlap. Resultant displacement is zero at complete overlap. Identifies destructive interference. Pulses continue after crossing.

Mark Scheme
  • 1 mark: Shows crest and equal trough overlap.
  • 1 mark: Resultant displacement is zero at complete overlap.
  • 1 mark: Identifies destructive interference.
  • 1 mark: Pulses continue after crossing.
Source: Original generated question

Waves • pages 123

OBJ06: Define nodes and antinodes and describe how they are formed.

10 MCQs • 4 Short Answers
G9T3_OBJ06_MCQ01 MCQ Easy Identify • 4 marks

What is the name of a point that has almost no motion in a standing wave?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
ACrest
BRarefaction
CAntinode
DNode
Answer
D. Node
A node has nearly zero displacement, while an antinode has maximum displacement.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ06_MCQ02 MCQ Medium Identify • 4 marks

For a string fixed at both ends, what is the relation between string length L and wavelength for harmonic n?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Aλₙ = nL/2
Bλₙ = L/n²
Cλₙ = 2L/n
Dλₙ = 2nL
Answer
C. λₙ = 2L/n
Allowed wavelengths on a string fixed at both ends are λₙ = 2L/n.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ06_MCQ03 MCQ Easy Define • 4 marks

In a standing wave, a node is a point where:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Adisplacement is always zero or minimum
Bdisplacement is maximum
Cspeed of sound is zero
Dfrequency changes continuously
Answer
A. displacement is always zero or minimum
Nodes remain nearly still in a standing wave.
Source: Original generated question
G9T3_OBJ06_MCQ04 MCQ Easy Define • 4 marks

An antinode is a point where:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Afrequency is zero
Bdisplacement is zero
Cthe wave is absent
Ddisplacement is maximum
Answer
D. displacement is maximum
Antinodes are points of maximum vibration.
Source: Original generated question
G9T3_OBJ06_MCQ05 MCQ Medium Explain • 4 marks

How are standing waves formed?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
ABy light changing color
BBy a single pulse stopping forever
CBy reducing all amplitude to zero
DBy interference of two waves of the same frequency travelling in opposite directions
Answer
D. By interference of two waves of the same frequency travelling in opposite directions
Standing waves result from superposition of incident and reflected waves.
Source: Original generated question
G9T3_OBJ06_MCQ06 MCQ Easy Identify • 4 marks

In a string fixed at both ends, the fixed ends are:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Aantinodes
Brarefactions
Cnodes
Dcompressions
Answer
C. nodes
Fixed ends cannot move, so they are nodes.
Source: Original generated question
G9T3_OBJ06_MCQ07 MCQ Hard Recall • 4 marks

The distance between two adjacent nodes in a standing wave is:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
A
B\(\lambda \)/2
C\(\lambda \)/4
D\(\lambda \)
Answer
B. \(\lambda \)/2
Adjacent nodes are half a wavelength apart.
Source: Original generated question
G9T3_OBJ06_MCQ08 MCQ Easy Identify • 4 marks

A point on a standing wave moves with maximum amplitude. It is called:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Aa node
Ban antinode
Ca wavelength
Da fixed end
Answer
B. an antinode
Maximum displacement occurs at antinodes.
Source: Original generated question
G9T3_OBJ06_MCQ09 MCQ Medium Identify • 4 marks

Which statement is true for a standing wave pattern?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
AFrequency is measured in lux.
BThe pattern cannot have nodes.
CAll points travel forward with the wave.
DNodes and antinodes remain in fixed positions.
Answer
D. Nodes and antinodes remain in fixed positions.
The standing pattern does not travel through the medium.
Source: Original generated question
G9T3_OBJ06_MCQ10 MCQ Medium Interpret • 4 marks

Between two adjacent antinodes, there is usually:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Aone node
Bonly a compression
Cthree complete wavelengths
Dno node
Answer
A. one node
A node lies midway between adjacent antinodes.
Source: Original generated question
G9T3_OBJ06_SA01 Short Answer 4 marks

Define node and antinode in a standing wave.

Expected Answer

Node has zero/minimum displacement. Antinode has maximum displacement. Both belong to a standing wave pattern. Uses correct terminology.

Mark Scheme
  • 1 mark: Node has zero/minimum displacement.
  • 1 mark: Antinode has maximum displacement.
  • 1 mark: Both belong to a standing wave pattern.
  • 1 mark: Uses correct terminology.
Source: Original generated question
G9T3_OBJ06_SA02 Short Answer 4 marks

Explain how a standing wave is formed on a string.

Expected Answer

Incident and reflected waves overlap. Waves have same frequency. They travel in opposite directions. Superposition forms fixed nodes and antinodes.

Mark Scheme
  • 1 mark: Incident and reflected waves overlap.
  • 1 mark: Waves have same frequency.
  • 1 mark: They travel in opposite directions.
  • 1 mark: Superposition forms fixed nodes and antinodes.
Source: Original generated question
G9T3_OBJ06_SA03 Short Answer 4 marks

Describe the standing wave pattern for the first harmonic on a string fixed at both ends.

Expected Answer

Nodes at both ends. One antinode in the middle. Length equals half a wavelength. States \(\lambda \) = 2L for first harmonic.

Mark Scheme
  • 1 mark: Nodes at both ends.
  • 1 mark: One antinode in the middle.
  • 1 mark: Length equals half a wavelength.
  • 1 mark: States \(\lambda \) = 2L for first harmonic.
Source: Original generated question
G9T3_OBJ06_SA04 Short Answer 4 marks

What is the distance between adjacent nodes in terms of wavelength?

Expected Answer

States adjacent node spacing is \(\lambda \)/2. Mentions adjacent antinode spacing is also \(\lambda \)/2 if included. Explains using standing wave geometry. Uses correct symbol \(\lambda \).

Mark Scheme
  • 1 mark: States adjacent node spacing is \(\lambda \)/2.
  • 1 mark: Mentions adjacent antinode spacing is also \(\lambda \)/2 if included.
  • 1 mark: Explains using standing wave geometry.
  • 1 mark: Uses correct symbol \(\lambda \).
Source: Original generated question

Waves • pages 124

OBJ07: Determine wavelengths for the first several harmonics on a string under tension: λn = 2L/n.

10 MCQs • 4 Short Answers
G9T3_OBJ07_MCQ01 MCQ Medium Identify • 4 marks

A string of length L = 1.2 m is fixed at both ends. What is the wavelength for harmonic n = 1?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
A2.4 m
B24 m
C4.8 m
D1.2 m
Answer
A. 2.4 m
Use λₙ = 2L/n = 2(1.2)/1 = 2.4 m.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ07_MCQ02 MCQ Medium Identify • 4 marks

A string of length L = 1.2 m is fixed at both ends. What is the wavelength for harmonic n = 2?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
A1.2 m
B2.4 m
C0.6 m
D12 m
Answer
A. 1.2 m
Use λₙ = 2L/n = 2(1.2)/2 = 1.2 m.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ07_MCQ03 MCQ Medium Calculate • 4 marks

A string fixed at both ends has length L = 2 m. What wavelength is required for harmonic n = 1?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
A4 m
B8 m
C2 m
D2 m
Answer
A. 4 m
Use λₙ = 2L/n = 2(2)/1 = 4 m.
Source: Original generated question
G9T3_OBJ07_MCQ04 MCQ Medium Calculate • 4 marks

A string fixed at both ends has length L = 2 m. What wavelength is required for harmonic n = 2?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
A4 m
B2 m
C1 m
D4 m
Answer
B. 2 m
Use λₙ = 2L/n = 2(2)/2 = 2 m.
Source: Original generated question
G9T3_OBJ07_MCQ05 MCQ Medium Calculate • 4 marks

A string fixed at both ends has length L = 2 m. What wavelength is required for harmonic n = 4?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
A0.5 m
B2 m
C8 m
D1 m
Answer
D. 1 m
Use λₙ = 2L/n = 2(2)/4 = 1 m.
Source: Original generated question
G9T3_OBJ07_MCQ06 MCQ Medium Calculate • 4 marks

A string fixed at both ends has length L = 1.5 m. What wavelength is required for harmonic n = 3?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
A4.5 m
B0.5 m
C1 m
D2 m
Answer
C. 1 m
Use λₙ = 2L/n = 2(1.5)/3 = 1 m.
Source: Original generated question
G9T3_OBJ07_MCQ07 MCQ Medium Calculate • 4 marks

A string fixed at both ends has length L = 0.8 m. What wavelength is required for harmonic n = 2?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
A1.6 m
B1.6 m
C0.8 m
D0.4 m
Answer
C. 0.8 m
Use λₙ = 2L/n = 2(0.8)/2 = 0.8 m.
Source: Original generated question
G9T3_OBJ07_MCQ08 MCQ Easy Recall • 4 marks

For a string fixed at both ends, which harmonic has wavelength 2L?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
ASecond harmonic
BFirst harmonic
CFourth harmonic
DThird harmonic
Answer
B. First harmonic
For n = 1, λ₁ = 2L.
Source: Original generated question
G9T3_OBJ07_MCQ09 MCQ Medium Apply • 4 marks

For a fixed string, the third harmonic wavelength is:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
A6L
B3L/2
C2L/3
DL/3
Answer
C. 2L/3
Substitute n = 3 into λₙ = 2L/n.
Source: Original generated question
G9T3_OBJ07_MCQ10 MCQ Medium Infer • 4 marks

As harmonic number n increases on the same string, the allowed wavelength:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Abecomes zero for all n
Bdecreases
Cincreases
Dstays always 2L
Answer
B. decreases
λₙ = 2L/n, so wavelength is inversely proportional to n.
Source: Original generated question
G9T3_OBJ07_SA01 Short Answer 4 marks

For a fixed string of length 1.2 m, calculate wavelengths of the first three harmonics.

Expected Answer

Uses \(\lambda_n = 2L/n\). λ1 = 2.4 m. λ2 = 1.2 m. λ3 = 0.8 m.

Mark Scheme
  • 1 mark: Uses \(\lambda_n = 2L/n\).
  • 1 mark: λ1 = 2.4 m.
  • 1 mark: λ2 = 1.2 m.
  • 1 mark: λ3 = 0.8 m.
Source: Original generated question
G9T3_OBJ07_SA02 Short Answer 4 marks

Derive/justify the relationship \(\lambda_n = 2L/n\) for a string fixed at both ends.

Expected Answer

States both ends are nodes. A whole number of half-wavelengths fits on the string. Writes L = nλ/2. Rearranges to \(\lambda_n = 2L/n\).

Mark Scheme
  • 1 mark: States both ends are nodes.
  • 1 mark: A whole number of half-wavelengths fits on the string.
  • 1 mark: Writes L = nλ/2.
  • 1 mark: Rearranges to \(\lambda_n = 2L/n\).
Source: Original generated question
G9T3_OBJ07_SA03 Short Answer 4 marks

Explain what happens to wavelength as harmonic number increases.

Expected Answer

States \(\lambda_n = 2L/n\). Harmonic number is in denominator. Therefore wavelength decreases. Gives example n=1 vs n=2.

Mark Scheme
  • 1 mark: States \(\lambda_n = 2L/n\).
  • 1 mark: Harmonic number is in denominator.
  • 1 mark: Therefore wavelength decreases.
  • 1 mark: Gives example n=1 vs n=2.
Source: Original generated question
G9T3_OBJ07_SA04 Short Answer 4 marks

A string length is 0.90 m. Find the wavelength of the third harmonic.

Expected Answer

Identifies n=3. Uses λ3 = 2L/3. Substitutes 2(0.90)/3. Gives 0.60 m.

Mark Scheme
  • 1 mark: Identifies n=3.
  • 1 mark: Uses λ3 = 2L/3.
  • 1 mark: Substitutes 2(0.90)/3.
  • 1 mark: Gives 0.60 m.
Source: Original generated question

Sound • pages 132

OBJ08: Define sound as pressure oscillation transmitted through matter; explain that sound is a longitudinal wave.

10 MCQs • 4 Short Answers
G9T3_OBJ08_MCQ01 MCQ Easy Identify • 4 marks

Which example is a longitudinal wave?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
ARed light in vacuum
BA sound wave in air
CPolarized light
DA rope moving up and down
Answer
B. A sound wave in air
Sound in air travels as compressions and rarefactions, so it is longitudinal.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ08_MCQ02 MCQ Easy Identify • 4 marks

Which description of sound is most accurate?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
ALow-frequency light
BParticles traveling from the source to the ear only
CAn electromagnetic wave that needs no medium
DA mechanical longitudinal wave that travels through matter
Answer
D. A mechanical longitudinal wave that travels through matter
Sound needs a material medium and travels as compressions and rarefactions.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ08_MCQ03 MCQ Easy Define • 4 marks

Sound is best described as:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Aparticles permanently moving from source to ear
Bpressure oscillations transmitted through matter
Clight oscillations in vacuum
Dcolor changes in air
Answer
B. pressure oscillations transmitted through matter
Sound is a mechanical pressure wave.
Source: Original generated question
G9T3_OBJ08_MCQ04 MCQ Medium Explain • 4 marks

Why can sound not travel through a perfect vacuum?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
AIts frequency becomes too high.
BLight blocks it.
CThere are no particles to transmit pressure oscillations.
DVacuum is always opaque.
Answer
C. There are no particles to transmit pressure oscillations.
Sound needs a material medium.
Source: Original generated question
G9T3_OBJ08_MCQ05 MCQ Easy Identify • 4 marks

In air, sound travels mainly as:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Amagnetic pulses only
Bpolarized light bands
Ccrests and troughs only
Dcompressions and rarefactions
Answer
D. compressions and rarefactions
Sound in air is a longitudinal pressure wave.
Source: Original generated question
G9T3_OBJ08_MCQ06 MCQ Easy Define • 4 marks

The particle motion in a sound wave in air is:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Aparallel to the direction of wave travel
Bin a circular orbit only
Cperpendicular to the direction of travel
Dzero everywhere
Answer
A. parallel to the direction of wave travel
Sound is longitudinal in air.
Source: Original generated question
G9T3_OBJ08_MCQ07 MCQ Medium Identify • 4 marks

A compression in a sound wave is a region of:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Awhite light
Bhigher pressure and higher particle density
Clower pressure only
Dzero matter
Answer
B. higher pressure and higher particle density
Compressions are regions where particles are closer together.
Source: Original generated question
G9T3_OBJ08_MCQ08 MCQ Medium Identify • 4 marks

A rarefaction in a sound wave is a region of:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Ano frequency
Bmaximum density
Clower pressure and lower particle density
Dhigher light intensity
Answer
C. lower pressure and lower particle density
Rarefactions are spread-out particle regions.
Source: Original generated question
G9T3_OBJ08_MCQ09 MCQ Easy Recall • 4 marks

Which property does sound share with other waves?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
AIt has no amplitude.
BIt always travels in vacuum.
CIt has frequency, wavelength, and speed.
DIt cannot reflect.
Answer
C. It has frequency, wavelength, and speed.
Sound is a wave and has standard wave properties.
Source: Original generated question
G9T3_OBJ08_MCQ10 MCQ Medium Explain • 4 marks

When a tuning fork vibrates, it produces sound because it:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Aemits visible light only
Bcreates pressure variations in the surrounding air
Cremoves air particles permanently
Dstops molecules moving
Answer
B. creates pressure variations in the surrounding air
Vibration pushes and pulls on air to make pressure oscillations.
Source: Original generated question
G9T3_OBJ08_SA01 Short Answer 4 marks

Explain why sound is called a longitudinal pressure wave.

Expected Answer

Sound consists of pressure oscillations. It has compressions and rarefactions. Particle motion is parallel to wave travel. It requires matter/a medium.

Mark Scheme
  • 1 mark: Sound consists of pressure oscillations.
  • 1 mark: It has compressions and rarefactions.
  • 1 mark: Particle motion is parallel to wave travel.
  • 1 mark: It requires matter/a medium.
Source: Original generated question
G9T3_OBJ08_SA02 Short Answer 4 marks

Define compression and rarefaction in a sound wave.

Expected Answer

Compression is high pressure/high density. Rarefaction is low pressure/low density. They alternate as sound travels. They transmit sound energy through matter.

Mark Scheme
  • 1 mark: Compression is high pressure/high density.
  • 1 mark: Rarefaction is low pressure/low density.
  • 1 mark: They alternate as sound travels.
  • 1 mark: They transmit sound energy through matter.
Source: Original generated question
G9T3_OBJ08_SA03 Short Answer 4 marks

Why can sound not travel in a perfect vacuum?

Expected Answer

Sound is mechanical. It needs particles/matter. Vacuum has no medium to vibrate. Therefore no pressure oscillations can be transmitted.

Mark Scheme
  • 1 mark: Sound is mechanical.
  • 1 mark: It needs particles/matter.
  • 1 mark: Vacuum has no medium to vibrate.
  • 1 mark: Therefore no pressure oscillations can be transmitted.
Source: Original generated question
G9T3_OBJ08_SA04 Short Answer 4 marks

List three wave quantities that sound has and define one.

Expected Answer

Names frequency. Names wavelength. Names speed or amplitude. Defines one correctly with unit if relevant.

Mark Scheme
  • 1 mark: Names frequency.
  • 1 mark: Names wavelength.
  • 1 mark: Names speed or amplitude.
  • 1 mark: Defines one correctly with unit if relevant.
Source: Original generated question

Sound • pages 133-134

OBJ09: Explain that the speed of sound varies with different mediums and temperatures.

10 MCQs • 4 Short Answers
G9T3_OBJ09_MCQ01 MCQ Medium Identify • 4 marks

What usually happens to the speed of sound in air when temperature increases?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
AIt increases
BIt decreases
CIt becomes zero
DIt is never affected
Answer
A. It increases
As temperature rises, air molecules have greater kinetic energy, so the disturbance travels faster.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ09_MCQ02 MCQ Easy Identify • 4 marks

In which medium does sound usually travel fastest?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
AIt does not vary with medium
BSolids
CGases
DVacuum
Answer
B. Solids
Particles in solids are closer and more strongly connected, so sound usually travels faster in solids.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ09_MCQ03 MCQ Easy Recall • 4 marks

In general, sound travels fastest in:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Agases
Bvacuum
Cempty space
Dsolids
Answer
D. solids
Sound usually travels fastest in solids because particles are closely linked.
Source: Original generated question
G9T3_OBJ09_MCQ04 MCQ Medium Interpret • 4 marks

The speed of sound in air increases when:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Afrequency becomes visible
Btemperature increases
Cair is removed completely
Dtemperature decreases to zero
Answer
B. temperature increases
Warmer air transmits sound faster.
Source: Original generated question
G9T3_OBJ09_MCQ05 MCQ Medium Order • 4 marks

Which ordering is generally correct for speed of sound?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Agas > liquid > solid
Bsolid > liquid > gas
Cliquid > vacuum > solid
Dvacuum > solid > gas
Answer
B. solid > liquid > gas
Sound is generally fastest in solids, slower in liquids, slowest in gases.
Source: Original generated question
G9T3_OBJ09_MCQ06 MCQ Medium Explain • 4 marks

Why does sound travel faster in water than in air?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
AWater has no particles.
BSound becomes light in water.
CParticles in water are closer together than in air.
DAir is always a vacuum.
Answer
C. Particles in water are closer together than in air.
Closer particles can transfer vibrations more quickly.
Source: Original generated question
G9T3_OBJ09_MCQ07 MCQ Easy Identify • 4 marks

What is required for the speed of sound to be defined in a medium?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
AA perfect vacuum
BA luminous source
CA material medium that can transmit vibrations
DA polarizing filter
Answer
C. A material medium that can transmit vibrations
Sound speed depends on the medium's properties.
Source: Original generated question
G9T3_OBJ09_MCQ08 MCQ Easy Predict • 4 marks

If sound enters a warmer region of air, its speed usually:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Ais converted into color
Bbecomes zero
Cdecreases
Dincreases
Answer
D. increases
Higher temperature generally increases sound speed in air.
Source: Original generated question
G9T3_OBJ09_MCQ09 MCQ Medium Identify • 4 marks

Which statement is true?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
ASound speed is always \(3.0 \times 10^{8}\) m/s.
BSound speed depends on the medium.
CSound cannot travel through steel.
DSound speed is the same in every material.
Answer
B. Sound speed depends on the medium.
Unlike light in vacuum, sound speed varies widely with material.
Source: Original generated question
G9T3_OBJ09_MCQ10 MCQ Medium Apply • 4 marks

Two students clap underwater and in air. The sound reaches a nearby sensor faster through:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Awater
Bair
Cvacuum
Dneither medium
Answer
A. water
Sound generally travels faster in liquids than in gases.
Source: Original generated question
G9T3_OBJ09_SA01 Short Answer 4 marks

Explain how the speed of sound depends on medium.

Expected Answer

Sound generally travels fastest in solids. Slower in liquids. Slowest in gases. Explains using particle closeness/elastic properties.

Mark Scheme
  • 1 mark: Sound generally travels fastest in solids.
  • 1 mark: Slower in liquids.
  • 1 mark: Slowest in gases.
  • 1 mark: Explains using particle closeness/elastic properties.
Source: Original generated question
G9T3_OBJ09_SA02 Short Answer 4 marks

Describe how increasing air temperature affects speed of sound.

Expected Answer

States speed of sound in air increases. Warmer particles move faster/have more kinetic energy. Vibrations transfer more quickly. Conclusion linked to temperature.

Mark Scheme
  • 1 mark: States speed of sound in air increases.
  • 1 mark: Warmer particles move faster/have more kinetic energy.
  • 1 mark: Vibrations transfer more quickly.
  • 1 mark: Conclusion linked to temperature.
Source: Original generated question
G9T3_OBJ09_SA03 Short Answer 4 marks

Order air, water, and steel from lowest to highest speed of sound and justify.

Expected Answer

Air is lowest. Water is intermediate. Steel/solid is highest. Justification refers to medium structure.

Mark Scheme
  • 1 mark: Air is lowest.
  • 1 mark: Water is intermediate.
  • 1 mark: Steel/solid is highest.
  • 1 mark: Justification refers to medium structure.
Source: Original generated question
G9T3_OBJ09_SA04 Short Answer 4 marks

Why is there no sound transmission through space between planets?

Expected Answer

Space is nearly vacuum. Sound requires matter. No pressure oscillations can propagate. Contrasts with light/electromagnetic waves if relevant.

Mark Scheme
  • 1 mark: Space is nearly vacuum.
  • 1 mark: Sound requires matter.
  • 1 mark: No pressure oscillations can propagate.
  • 1 mark: Contrasts with light/electromagnetic waves if relevant.
Source: Original generated question

Sound • pages 135

OBJ10: Define sound pitch and relate it to frequency.

10 MCQs • 4 Short Answers
G9T3_OBJ10_MCQ01 MCQ Easy Identify • 4 marks

Sound pitch is directly related to:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
AAmplitude only
BSource color
CDistance only
DFrequency
Answer
D. Frequency
A higher-pitched sound has a higher frequency.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ10_MCQ02 MCQ Easy Identify • 4 marks

A sound of 800 Hz compared with 400 Hz is:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
ALower in pitch
BHigher in pitch
CAlways slower
DNecessarily louder
Answer
B. Higher in pitch
Pitch depends on frequency; higher frequency gives higher pitch.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ10_MCQ03 MCQ Easy Define • 4 marks

Pitch of a sound is mainly related to:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Adistance only
Bspeed only
Camplitude
Dfrequency
Answer
D. frequency
Higher frequency is perceived as higher pitch.
Source: Original generated question
G9T3_OBJ10_MCQ04 MCQ Easy Identify • 4 marks

A high-pitched sound has:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Ahigh frequency
Bno wavelength
Clarge amplitude only
Dlow frequency
Answer
A. high frequency
Pitch increases with frequency.
Source: Original generated question
G9T3_OBJ10_MCQ05 MCQ Easy Compare • 4 marks

A 900 Hz tone and a 300 Hz tone are played at the same loudness. Which has higher pitch?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
A300 Hz tone
B900 Hz tone
Cboth have zero pitch
Dpitch depends only on amplitude
Answer
B. 900 Hz tone
Higher frequency gives higher pitch.
Source: Original generated question
G9T3_OBJ10_MCQ06 MCQ Medium Infer • 4 marks

If a string instrument note becomes higher in pitch, its frequency has:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Adecreased
Bbecome zero
Cchanged to lux
Dincreased
Answer
D. increased
Pitch and frequency are directly related.
Source: Original generated question
G9T3_OBJ10_MCQ07 MCQ Easy Recall • 4 marks

Which unit is used to measure the physical quantity that controls pitch?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Alumen
Bhertz
Ccandela
Ddecibel
Answer
B. hertz
Frequency is measured in hertz.
Source: Original generated question
G9T3_OBJ10_MCQ08 MCQ Medium Apply • 4 marks

Two sounds have equal amplitude but different frequencies. What perceptual difference is most likely?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
AThey have no sound level.
BThey have different pitch.
CThey have different brightness.
DThey have different color only.
Answer
B. They have different pitch.
Frequency affects pitch, not brightness.
Source: Original generated question
G9T3_OBJ10_MCQ09 MCQ Easy Identify • 4 marks

A low-pitched sound corresponds to:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Ahigher frequency
Blarger luminous flux
Clower frequency
Dshorter period always with high frequency
Answer
C. lower frequency
Low pitch means fewer oscillations per second.
Source: Original generated question
G9T3_OBJ10_MCQ10 MCQ Medium Calculate • 4 marks

A vibrating object makes 440 oscillations each second. The sound frequency is:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
A440 Hz
B440 dB
C1/440 Hz
D440 lx
Answer
A. 440 Hz
Frequency is the number of oscillations per second.
Source: Original generated question
G9T3_OBJ10_SA01 Short Answer 4 marks

Define pitch and state its relationship with frequency.

Expected Answer

Pitch is how high or low a sound seems. Pitch depends on frequency. Higher frequency gives higher pitch. Lower frequency gives lower pitch.

Mark Scheme
  • 1 mark: Pitch is how high or low a sound seems.
  • 1 mark: Pitch depends on frequency.
  • 1 mark: Higher frequency gives higher pitch.
  • 1 mark: Lower frequency gives lower pitch.
Source: Original generated question
G9T3_OBJ10_SA02 Short Answer 4 marks

Compare 250 Hz and 1000 Hz sounds in terms of pitch.

Expected Answer

1000 Hz has higher frequency. Therefore 1000 Hz sound has higher pitch. 250 Hz has lower pitch. Frequency unit Hz identified.

Mark Scheme
  • 1 mark: 1000 Hz has higher frequency.
  • 1 mark: Therefore 1000 Hz sound has higher pitch.
  • 1 mark: 250 Hz has lower pitch.
  • 1 mark: Frequency unit Hz identified.
Source: Original generated question
G9T3_OBJ10_SA03 Short Answer 4 marks

A guitar string vibrates faster after tightening. Predict the pitch change.

Expected Answer

Faster vibration means higher frequency. Pitch increases. Mentions relationship pitch-frequency. Answer is clearly linked to the scenario.

Mark Scheme
  • 1 mark: Faster vibration means higher frequency.
  • 1 mark: Pitch increases.
  • 1 mark: Mentions relationship pitch-frequency.
  • 1 mark: Answer is clearly linked to the scenario.
Source: Original generated question
G9T3_OBJ10_SA04 Short Answer 4 marks

Explain why amplitude alone does not determine pitch.

Expected Answer

Amplitude relates mainly to loudness. Frequency relates to pitch. Two sounds can have same amplitude but different pitch. Uses correct distinction.

Mark Scheme
  • 1 mark: Amplitude relates mainly to loudness.
  • 1 mark: Frequency relates to pitch.
  • 1 mark: Two sounds can have same amplitude but different pitch.
  • 1 mark: Uses correct distinction.
Source: Original generated question

Sound • pages 135

OBJ11: Define loudness and relate it to amplitude of a sound wave.

10 MCQs • 4 Short Answers
G9T3_OBJ11_MCQ01 MCQ Easy Identify • 4 marks

Sound loudness is mainly related to:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
AWavelength only
BEar shape only
CAmplitude of the sound wave
DColor of the vibrating object
Answer
C. Amplitude of the sound wave
Greater amplitude means more energy reaches the ear, so the sound is perceived as louder.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ11_MCQ02 MCQ Medium Identify • 4 marks

When does resonance occur?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
AWhen frequency is only halved
BWhen amplitude is always zero
CWhen a periodic force acts at nearly the natural frequency of a system
DWhen sound travels in vacuum
Answer
C. When a periodic force acts at nearly the natural frequency of a system
Resonance causes a large amplitude increase when the driving frequency matches the natural frequency.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ11_MCQ03 MCQ Easy Define • 4 marks

Loudness of a sound is mainly related to:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Afrequency
Bamplitude
Ccolor
Dwavelength only
Answer
B. amplitude
Greater amplitude usually produces louder sound.
Source: Original generated question
G9T3_OBJ11_MCQ04 MCQ Easy Identify • 4 marks

Which sound wave is likely to be louder?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
AThe wave with zero pressure variation
BThe wave with no medium
CThe wave with lower amplitude
DThe wave with greater amplitude
Answer
D. The wave with greater amplitude
Loudness depends on the size of pressure oscillations.
Source: Original generated question
G9T3_OBJ11_MCQ05 MCQ Medium Infer • 4 marks

If a speaker vibrates air with larger pressure variations, the sound is perceived as:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Alouder
Bhigher pitch only
Clower frequency only
Dtransparent
Answer
A. louder
Larger pressure amplitude gives greater loudness.
Source: Original generated question
G9T3_OBJ11_MCQ06 MCQ Medium Compare • 4 marks

Two tones have the same frequency but different amplitudes. They differ mainly in:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Apolarization
Bloudness
Ccolor of light
Dpitch
Answer
B. loudness
Equal frequency means same pitch; amplitude changes loudness.
Source: Original generated question
G9T3_OBJ11_MCQ07 MCQ Easy Recognize • 4 marks

A whisper and a shout may have similar pitch but different:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Acolor
Bamplitude
Cluminous flux
Dspeed of light
Answer
B. amplitude
A shout has larger sound amplitude.
Source: Original generated question
G9T3_OBJ11_MCQ08 MCQ Easy Predict • 4 marks

What happens to the perceived loudness when sound amplitude decreases significantly?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
AIt becomes faster than light.
BIt becomes quieter.
CIt becomes light.
DIt becomes higher pitch.
Answer
B. It becomes quieter.
Lower amplitude means lower sound energy reaching the ear.
Source: Original generated question
G9T3_OBJ11_MCQ09 MCQ Medium Interpret • 4 marks

Which graph would represent the louder of two sounds at the same frequency?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
AThe graph with no oscillations
BThe graph with lower wavelength only
CThe graph with taller pressure peaks
DThe graph with smaller amplitude
Answer
C. The graph with taller pressure peaks
Taller peaks indicate greater amplitude.
Source: Original generated question
G9T3_OBJ11_MCQ10 MCQ Medium Explain • 4 marks

Loudness is a perception related to sound intensity, but in simple wave models it is strongly linked to:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Amedium color
Bphase only
Camplitude
Dperiod only
Answer
C. amplitude
Amplitude gives a simple indicator of perceived loudness.
Source: Original generated question
G9T3_OBJ11_SA01 Short Answer 4 marks

Define loudness and relate it to wave amplitude.

Expected Answer

Loudness is perception of how strong a sound is. It is related to amplitude/intensity. Greater amplitude means louder sound. Smaller amplitude means quieter sound.

Mark Scheme
  • 1 mark: Loudness is perception of how strong a sound is.
  • 1 mark: It is related to amplitude/intensity.
  • 1 mark: Greater amplitude means louder sound.
  • 1 mark: Smaller amplitude means quieter sound.
Source: Original generated question
G9T3_OBJ11_SA02 Short Answer 4 marks

Compare two sounds with the same frequency but different amplitudes.

Expected Answer

Same frequency means same pitch. Larger amplitude sound is louder. Smaller amplitude sound is quieter. Uses amplitude/loudness terms correctly.

Mark Scheme
  • 1 mark: Same frequency means same pitch.
  • 1 mark: Larger amplitude sound is louder.
  • 1 mark: Smaller amplitude sound is quieter.
  • 1 mark: Uses amplitude/loudness terms correctly.
Source: Original generated question
G9T3_OBJ11_SA03 Short Answer 4 marks

Explain how a pressure-time graph shows a louder sound.

Expected Answer

Louder sound has larger pressure variation. Graph has taller peaks/deeper troughs. Frequency can remain unchanged. Links larger amplitude to loudness.

Mark Scheme
  • 1 mark: Louder sound has larger pressure variation.
  • 1 mark: Graph has taller peaks/deeper troughs.
  • 1 mark: Frequency can remain unchanged.
  • 1 mark: Links larger amplitude to loudness.
Source: Original generated question
G9T3_OBJ11_SA04 Short Answer 4 marks

Why does moving away from a sound source usually reduce loudness?

Expected Answer

Sound energy spreads out. Less energy reaches each unit area. Amplitude/intensity at ear decreases. Perceived loudness decreases.

Mark Scheme
  • 1 mark: Sound energy spreads out.
  • 1 mark: Less energy reaches each unit area.
  • 1 mark: Amplitude/intensity at ear decreases.
  • 1 mark: Perceived loudness decreases.
Source: Original generated question

Sound • pages 135

OBJ12: Describe sound level and define decibel (dB) as a unit of measuring sound level.

10 MCQs • 4 Short Answers
G9T3_OBJ12_MCQ01 MCQ Medium Identify • 4 marks

What is the unit used for sound level?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Alumen lm
Bdecibel dB
Cnewton N
Dhertz Hz
Answer
B. decibel dB
Sound level is measured in decibels dB, while frequency is measured in hertz Hz.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ12_MCQ02 MCQ Medium Identify • 4 marks

What does it usually mean that the decibel scale is logarithmic?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
AdB is a unit of time
BA small numerical increase can represent a large physical intensity change
CdB measures wavelength only
DEvery 1 dB increase means sound stops
Answer
B. A small numerical increase can represent a large physical intensity change
Sound level in decibels is not linear with intensity; it uses a logarithmic scale.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ12_MCQ03 MCQ Easy Recall • 4 marks

Sound level is commonly measured in:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Ahertz
Bdecibels
Clux
Dnewtons
Answer
B. decibels
The unit of sound level is the decibel, dB.
Source: Original generated question
G9T3_OBJ12_MCQ04 MCQ Easy Define • 4 marks

The symbol dB represents:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Adensity balance
Bdecibel
Cdouble beat
Ddiffraction band
Answer
B. decibel
dB is the abbreviation for decibel.
Source: Original generated question
G9T3_OBJ12_MCQ05 MCQ Medium Interpret • 4 marks

A sound level meter gives a reading of 85 dB. What does this value describe?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Asound level
Bwavelength
Cluminous intensity
Dfrequency
Answer
A. sound level
Decibels describe sound level.
Source: Original generated question
G9T3_OBJ12_MCQ06 MCQ Easy Compare • 4 marks

Which quantity is measured in hertz, not decibels?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Aloudness level
Bsound level
Cfrequency
Dnoise exposure level
Answer
C. frequency
Frequency uses hertz; sound level uses decibel.
Source: Original generated question
G9T3_OBJ12_MCQ07 MCQ Easy Apply • 4 marks

Which is the most appropriate unit for comparing noise near a construction site?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Acd
BdB
Clx
Dlm
Answer
B. dB
Noise level is measured using decibels.
Source: Original generated question
G9T3_OBJ12_MCQ08 MCQ Easy Interpret • 4 marks

A larger dB reading usually indicates:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Aa lower frequency only
Ba higher light intensity
Ca greater sound level
Da lower sound level always
Answer
C. a greater sound level
Higher decibel value corresponds to greater sound level.
Source: Original generated question
G9T3_OBJ12_MCQ09 MCQ Hard Explain • 4 marks

Why is dB useful for sound levels?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
AIt gives the speed of light.
BIt gives exact object color.
CIt measures wavelength directly.
DIt represents the large range of sound intensities in a manageable scale.
Answer
D. It represents the large range of sound intensities in a manageable scale.
The decibel scale compresses a wide range of intensities.
Source: Original generated question
G9T3_OBJ12_MCQ10 MCQ Easy Match • 4 marks

Which pair is correctly matched?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Ailluminance — dB
Bsound level — dB
Cpitch — lux
Dluminous flux — Hz
Answer
B. sound level — dB
Sound level is measured in decibels.
Source: Original generated question
G9T3_OBJ12_SA01 Short Answer 4 marks

Define sound level and state its unit.

Expected Answer

Sound level compares sound intensity/loudness level. Unit is decibel. Symbol is dB. Uses correct context such as noise measurement.

Mark Scheme
  • 1 mark: Sound level compares sound intensity/loudness level.
  • 1 mark: Unit is decibel.
  • 1 mark: Symbol is dB.
  • 1 mark: Uses correct context such as noise measurement.
Source: Original generated question
G9T3_OBJ12_SA02 Short Answer 4 marks

Distinguish between hertz and decibel.

Expected Answer

Hertz measures frequency. Frequency relates to pitch. Decibel measures sound level. Sound level relates to loudness/noise.

Mark Scheme
  • 1 mark: Hertz measures frequency.
  • 1 mark: Frequency relates to pitch.
  • 1 mark: Decibel measures sound level.
  • 1 mark: Sound level relates to loudness/noise.
Source: Original generated question
G9T3_OBJ12_SA03 Short Answer 4 marks

Explain why the decibel scale is useful.

Expected Answer

Sound intensities cover a very wide range. dB gives manageable numbers. Used for comparing noise/sound levels. Higher dB generally means greater sound level.

Mark Scheme
  • 1 mark: Sound intensities cover a very wide range.
  • 1 mark: dB gives manageable numbers.
  • 1 mark: Used for comparing noise/sound levels.
  • 1 mark: Higher dB generally means greater sound level.
Source: Original generated question
G9T3_OBJ12_SA04 Short Answer 4 marks

Give two examples where measuring sound level in dB is important.

Expected Answer

Gives a valid example such as construction noise. Gives a second valid example such as headphones/traffic/aircraft. Explains health/safety or comfort relevance. Uses dB correctly.

Mark Scheme
  • 1 mark: Gives a valid example such as construction noise.
  • 1 mark: Gives a second valid example such as headphones/traffic/aircraft.
  • 1 mark: Explains health/safety or comfort relevance.
  • 1 mark: Uses dB correctly.
Source: Original generated question

Sound • pages 138

OBJ13: Apply the Doppler effect equation to calculate detected frequencies and velocities.

10 MCQs • 4 Short Answers
G9T3_OBJ13_MCQ01 MCQ Medium Identify • 4 marks

When an ambulance approaches a stationary observer, how does the observed frequency change?

Doppler Effect source higher flower f
Doppler
AIt decreases
BIt increases
CIt always remains equal to the source frequency
DIt becomes zero
Answer
B. It increases
When approaching, wavefronts are compressed in front of the source, so the observer hears a higher frequency.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ13_MCQ02 MCQ Hard Identify • 4 marks

A sound source of frequency 500 Hz is moving toward the observer at 20 m/s. Assuming sound speed 343 m/s, what is the approximate observed frequency?

Doppler Effect source higher flower f
Doppler
A531 Hz
B472 Hz
C500 Hz
D425 Hz
Answer
A. 531 Hz
Use the appropriate Doppler relation for relative motion. The approximate result is 531 Hz.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ13_MCQ03 MCQ Easy Predict • 4 marks

A stationary observer hears a siren from a car moving toward them. Compared with the source frequency, the detected frequency is:

Doppler Effect source higher flower f
Doppler
Alower
Bzero
Chigher
Dunchanged in all cases
Answer
C. higher
Approaching source compresses wavefronts and increases detected frequency.
Source: Original generated question
G9T3_OBJ13_MCQ04 MCQ Easy Predict • 4 marks

A siren source moves away from a stationary observer. The detected frequency is:

Doppler Effect source higher flower f
Doppler
Ahigher than the source frequency
Balways equal to zero
Cmeasured in lux
Dlower than the source frequency
Answer
D. lower than the source frequency
Moving away stretches wavefronts, decreasing frequency.
Source: Original generated question
G9T3_OBJ13_MCQ05 MCQ Hard Set up • 4 marks

Using \(v = 340 m/s\), a 500 Hz source moves toward a stationary detector at 20 m/s. Which expression is correct?

Doppler Effect source higher flower f
Doppler
A500 × 340/(340 − 20)
B500 × (340 − 20)/340
C500 × 340/(340 + 20)
D500 × 20/340
Answer
A. 500 × 340/(340 − 20)
For a source moving toward a stationary detector, denominator is v − v_s.
Source: Original generated question
G9T3_OBJ13_MCQ06 MCQ Hard Calculate • 4 marks

A 600 Hz source moves away from a stationary observer at 40 m/s. Take \(v = 340 m/s. What is the approximate observed frequency\)?

Doppler Effect source higher flower f
Doppler
A600 Hz
B679 Hz
C537 Hz
D75 Hz
Answer
C. 537 Hz
f_d = f_s v/(v+v_s)=600×340/380≈537 Hz.
Source: Original generated question
G9T3_OBJ13_MCQ07 MCQ Hard Calculate • 4 marks

A detector moves toward a stationary 400 Hz source at 34 m/s. Take \(v = 340 m/s. What is the detected frequency\)?

Doppler Effect source higher flower f
Doppler
A400 Hz
B360 Hz
C440 Hz
D34 Hz
Answer
C. 440 Hz
f_d = f_s(v+v_d)/\(v = 400\)×374/340 = 440 Hz.
Source: Original generated question
G9T3_OBJ13_MCQ08 MCQ Hard Calculate • 4 marks

A detector moves away from a stationary 800 Hz source at 17 m/s. Take \(v = 340 m/s. What is the detected frequency\)?

Doppler Effect source higher flower f
Doppler
A17 Hz
B800 Hz
C840 Hz
D760 Hz
Answer
D. 760 Hz
f_d = 800×(340−17)/340 = 760 Hz.
Source: Original generated question
G9T3_OBJ13_MCQ09 MCQ Medium Identify • 4 marks

In the Doppler effect for sound, v usually represents:

Doppler Effect source higher flower f
Doppler
Aluminous intensity
Bspeed of light
Cspeed of sound in the medium
Damplitude of sound
Answer
C. speed of sound in the medium
The equation uses sound speed in the medium.
Source: Original generated question
G9T3_OBJ13_MCQ10 MCQ Medium Reason • 4 marks

Which situation gives the highest detected frequency?

Doppler Effect source higher flower f
Doppler
Asource and observer moving apart
Bobserver moving away from source
Csource and observer moving toward each other
Dboth stationary
Answer
C. source and observer moving toward each other
Relative approach increases frequency the most.
Source: Original generated question
G9T3_OBJ13_SA01 Short Answer 4 marks

Explain the Doppler effect using an approaching ambulance.

Expected Answer

Relative motion between source and observer occurs. Wavefronts are compressed in front. Observed frequency/pitch increases while approaching. Observed frequency/pitch decreases after moving away.

Mark Scheme
  • 1 mark: Relative motion between source and observer occurs.
  • 1 mark: Wavefronts are compressed in front.
  • 1 mark: Observed frequency/pitch increases while approaching.
  • 1 mark: Observed frequency/pitch decreases after moving away.
Source: Original generated question
G9T3_OBJ13_SA02 Short Answer 4 marks

A 500 Hz source moves away from a stationary observer at 20 m/s. Write the correct Doppler expression using \(v = 340 m/s.\)

Expected Answer

Identifies source moving away. Uses denominator v + vs. Writes fd = 500 × 340/(340 + 20). States observed frequency is lower than 500 Hz.

Mark Scheme
  • 1 mark: Identifies source moving away.
  • 1 mark: Uses denominator v + vs.
  • 1 mark: Writes fd = 500 × 340/(340 + 20).
  • 1 mark: States observed frequency is lower than 500 Hz.
Source: Original generated question
G9T3_OBJ13_SA03 Short Answer 4 marks

Calculate detected frequency for a stationary source of 600 Hz when observer moves toward it at 34 m/s. Use \(v = 340 m/s.\)

Expected Answer

Uses fd = fs(v + vd)/v. Substitutes 600(340 + 34)/340. Calculates 660 Hz. Includes higher frequency due to approach.

Mark Scheme
  • 1 mark: Uses fd = fs(v + vd)/v.
  • 1 mark: Substitutes 600(340 + 34)/340.
  • 1 mark: Calculates 660 Hz.
  • 1 mark: Includes higher frequency due to approach.
Source: Original generated question
G9T3_OBJ13_SA04 Short Answer 4 marks

State two factors that determine whether detected frequency increases or decreases.

Expected Answer

Direction of relative motion: approaching or moving apart. Speed of source/observer affects size of shift. Speed of sound in medium is relevant. Correctly links approaching to higher and receding to lower frequency.

Mark Scheme
  • 1 mark: Direction of relative motion: approaching or moving apart.
  • 1 mark: Speed of source/observer affects size of shift.
  • 1 mark: Speed of sound in medium is relevant.
  • 1 mark: Correctly links approaching to higher and receding to lower frequency.
Source: Original generated question

Sound • pages 141

OBJ14: Recall resonance as amplitude increase at natural frequency; explain resonance in air columns and instruments.

10 MCQs • 4 Short Answers
G9T3_OBJ14_MCQ01 MCQ Easy Identify • 4 marks

Which example is related to resonance in air columns?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
AThe sound of a flute or pipe
BPolarization by sunglasses
CColor of an object under red light
DRefraction of light in a prism
Answer
A. The sound of a flute or pipe
Wind instruments use resonance of air columns to produce specific notes.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ14_MCQ02 MCQ Medium Identify • 4 marks

In a string instrument, changing the effective string length changes:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
ASpeed of light
BThe unit decibel
COnly string color
DThe natural frequency of the note
Answer
D. The natural frequency of the note
String length affects allowed wavelengths and therefore produced frequencies.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ14_MCQ03 MCQ Easy Define • 4 marks

Resonance occurs when a system is driven at:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Athe speed of light
Bany random frequency
Czero frequency only
Dits natural frequency
Answer
D. its natural frequency
Resonance happens when driving frequency matches natural frequency.
Source: Original generated question
G9T3_OBJ14_MCQ04 MCQ Easy Identify • 4 marks

The main effect of resonance is:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Alarge increase in amplitude
Bcomplete loss of vibration
Cdecrease of all energy to zero
Dchange of sound into light
Answer
A. large increase in amplitude
Resonance transfers energy efficiently and increases amplitude.
Source: Original generated question
G9T3_OBJ14_MCQ05 MCQ Easy Apply • 4 marks

Which instrument uses resonance in an air column?

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Amirror
Bpolarizing filter
Clamp
Dflute
Answer
D. flute
Flutes and pipes produce notes by air-column resonance.
Source: Original generated question
G9T3_OBJ14_MCQ06 MCQ Medium Explain • 4 marks

A tuning fork makes a nearby air column sound loudly when the length is adjusted. This is due to:

Sound as a Longitudinal Wave compressionrarefactioncompression
Sound Wave
Aopaque reflection
Bpolarization only
Cresonance
Ddiffraction only
Answer
C. resonance
The air column resonates at the fork frequency.
Source: Original generated question
G9T3_OBJ14_MCQ07 MCQ Medium Explain • 4 marks

Why do musical instruments have specific shapes and lengths?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Ato stop all vibrations
Bto make light luminous
Cto support particular resonance frequencies
Dto remove pitch
Answer
C. to support particular resonance frequencies
Geometry controls resonant frequencies.
Source: Original generated question
G9T3_OBJ14_MCQ08 MCQ Easy Identify • 4 marks

A child pumps a swing at the right time and the amplitude grows. This is an example of:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Aresonance
BMalus's law
Cillumination
Ddiffraction
Answer
A. resonance
Periodic driving at natural frequency increases amplitude.
Source: Original generated question
G9T3_OBJ14_MCQ09 MCQ Medium Recall • 4 marks

In a resonating air column, the standing wave pattern contains:

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
Aonly luminous flux
Bno vibration
Conly pigments
Dnodes and antinodes
Answer
D. nodes and antinodes
Standing waves in air columns have nodes and antinodes.
Source: Original generated question
G9T3_OBJ14_MCQ10 MCQ Medium Identify • 4 marks

Which statement about resonance is correct?

Standing Wave on a String nodenodeantinode For fixed ends: λₙ = 2L/n
Standing Wave
AIt only happens to light.
BIt requires vacuum.
CIt always decreases amplitude.
DIt occurs when energy is added at the natural frequency.
Answer
D. It occurs when energy is added at the natural frequency.
At resonance, periodic driving efficiently transfers energy.
Source: Original generated question
G9T3_OBJ14_SA01 Short Answer 4 marks

Define resonance and give one everyday example.

Expected Answer

Resonance is large amplitude vibration. Occurs when driving frequency matches natural frequency. Gives valid example such as swing, tuning fork, air column, instrument. Explains amplitude increase.

Mark Scheme
  • 1 mark: Resonance is large amplitude vibration.
  • 1 mark: Occurs when driving frequency matches natural frequency.
  • 1 mark: Gives valid example such as swing, tuning fork, air column, instrument.
  • 1 mark: Explains amplitude increase.
Source: Original generated question
G9T3_OBJ14_SA02 Short Answer 4 marks

Explain resonance in an air column of a wind instrument.

Expected Answer

Air column has natural frequencies. Sound waves reflect and form standing waves. Resonance occurs at allowed frequencies. This produces strong notes/pitches.

Mark Scheme
  • 1 mark: Air column has natural frequencies.
  • 1 mark: Sound waves reflect and form standing waves.
  • 1 mark: Resonance occurs at allowed frequencies.
  • 1 mark: This produces strong notes/pitches.
Source: Original generated question
G9T3_OBJ14_SA03 Short Answer 4 marks

Why does changing the length of a flute or pipe change the note?

Expected Answer

Length changes allowed wavelengths. Allowed wavelengths change resonant frequencies. Frequency determines pitch. Therefore note changes.

Mark Scheme
  • 1 mark: Length changes allowed wavelengths.
  • 1 mark: Allowed wavelengths change resonant frequencies.
  • 1 mark: Frequency determines pitch.
  • 1 mark: Therefore note changes.
Source: Original generated question
G9T3_OBJ14_SA04 Short Answer 4 marks

Describe a laboratory demonstration of resonance using a tuning fork and tube.

Expected Answer

Tuning fork provides periodic driving sound. Air column length is adjusted. Loud sound occurs at resonance. Conclusion: air column natural frequency matches tuning fork.

Mark Scheme
  • 1 mark: Tuning fork provides periodic driving sound.
  • 1 mark: Air column length is adjusted.
  • 1 mark: Loud sound occurs at resonance.
  • 1 mark: Conclusion: air column natural frequency matches tuning fork.
Source: Original generated question

Light • pages 157-158

OBJ15: Differentiate luminous and non-luminous sources; differentiate opaque, translucent, and transparent media.

10 MCQs • 4 Short Answers
G9T3_OBJ15_MCQ01 MCQ Easy Identify • 4 marks

Which of the following is a luminous source?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
AAn unlit mirror
BA book on a table
CThe Sun
DThe Moon
Answer
C. The Sun
A luminous source produces its own light; the Moon reflects sunlight.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ15_MCQ02 MCQ Easy Identify • 4 marks

Which description matches a transparent object?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
AIt transmits most light and allows clear viewing
BIt transmits no light
CIt produces light by itself
DIt transmits some light but blurs images
Answer
A. It transmits most light and allows clear viewing
A transparent material transmits light so objects behind it can be seen clearly.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ15_MCQ03 MCQ Easy Identify • 4 marks

Which object is luminous?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Aa white wall
Ba mirror
Ca glowing lamp
Dthe Moon
Answer
C. a glowing lamp
A luminous object produces its own light.
Source: Original generated question
G9T3_OBJ15_MCQ04 MCQ Easy Explain • 4 marks

A non-luminous object is seen because it:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Aabsorbs no light
Breflects light from another source
Chas no interaction with light
Dproduces all light itself
Answer
B. reflects light from another source
Non-luminous objects are visible by reflected light.
Source: Original generated question
G9T3_OBJ15_MCQ05 MCQ Easy Classify • 4 marks

A clear glass window is best described as:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Aluminous
Bopaque
Ctransparent
Dnon-transparent
Answer
C. transparent
Transparent materials allow light through clearly.
Source: Original generated question
G9T3_OBJ15_MCQ06 MCQ Easy Classify • 4 marks

Frosted glass allows light through but does not form a clear image. It is:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Atransparent
Bopaque
Ctranslucent
Dluminous
Answer
C. translucent
Translucent materials transmit light but scatter it.
Source: Original generated question
G9T3_OBJ15_MCQ07 MCQ Easy Classify • 4 marks

A wooden door is:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Atransparent
Btranslucent
Copaque
Dluminous
Answer
C. opaque
Opaque materials do not transmit light clearly.
Source: Original generated question
G9T3_OBJ15_MCQ08 MCQ Medium Match • 4 marks

Which pair is correct?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
ABoth produce their own light
BBoth are opaque only
CSun — luminous; Moon — non-luminous
DSun — non-luminous; Moon — luminous
Answer
C. Sun — luminous; Moon — non-luminous
The Sun emits light; the Moon reflects sunlight.
Source: Original generated question
G9T3_OBJ15_MCQ09 MCQ Easy Identify • 4 marks

Which material allows the clearest view of an object behind it?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Atranslucent material
Bopaque material
Ctransparent material
Dblack pigment
Answer
C. transparent material
Transparent media transmit light with little scattering.
Source: Original generated question
G9T3_OBJ15_MCQ10 MCQ Easy Define • 4 marks

Which statement describes an opaque medium?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
AIt blocks most light from passing through.
BIt transmits all light.
CIt produces light.
DIt always allows a sharp image.
Answer
A. It blocks most light from passing through.
Opaque media do not transmit significant light.
Source: Original generated question
G9T3_OBJ15_SA01 Short Answer 4 marks

Differentiate luminous and non-luminous sources with examples.

Expected Answer

Luminous source emits its own light. Non-luminous source does not emit its own light. Gives valid luminous example such as Sun/lamp. Gives valid non-luminous example such as Moon/book.

Mark Scheme
  • 1 mark: Luminous source emits its own light.
  • 1 mark: Non-luminous source does not emit its own light.
  • 1 mark: Gives valid luminous example such as Sun/lamp.
  • 1 mark: Gives valid non-luminous example such as Moon/book.
Source: Original generated question
G9T3_OBJ15_SA02 Short Answer 4 marks

Compare transparent, translucent, and opaque materials.

Expected Answer

Transparent transmits light clearly. Translucent transmits light but scatters/blurred image. Opaque blocks most light. Gives at least one correct example.

Mark Scheme
  • 1 mark: Transparent transmits light clearly.
  • 1 mark: Translucent transmits light but scatters/blurred image.
  • 1 mark: Opaque blocks most light.
  • 1 mark: Gives at least one correct example.
Source: Original generated question
G9T3_OBJ15_SA03 Short Answer 4 marks

Why can we see a non-luminous object?

Expected Answer

It is illuminated by a luminous source. It reflects/scatters light into our eyes. It does not produce its own light. Example provided.

Mark Scheme
  • 1 mark: It is illuminated by a luminous source.
  • 1 mark: It reflects/scatters light into our eyes.
  • 1 mark: It does not produce its own light.
  • 1 mark: Example provided.
Source: Original generated question
G9T3_OBJ15_SA04 Short Answer 4 marks

Classify clear glass, frosted glass, and wood using light transmission.

Expected Answer

Clear glass is transparent. Frosted glass is translucent. Wood is opaque. Classification linked to transmission/scattering/blocking.

Mark Scheme
  • 1 mark: Clear glass is transparent.
  • 1 mark: Frosted glass is translucent.
  • 1 mark: Wood is opaque.
  • 1 mark: Classification linked to transmission/scattering/blocking.
Source: Original generated question

Light • pages 158-161

OBJ16: Define luminous flux, illuminance, and luminous intensity with SI units.

10 MCQs • 4 Short Answers
G9T3_OBJ16_MCQ01 MCQ Medium Identify • 4 marks

What is the SI unit of illuminance?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Adecibel dB
Blumen lm
Ccandela cd
Dlux lx
Answer
D. lux lx
Illuminance is in lux; luminous flux is in lumens; luminous intensity is in candela.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ16_MCQ02 MCQ Hard Identify • 4 marks

A point source has luminous flux \(P = 1000 lm. Calculate the illuminance on a surface at r\)= 1 m approximately.

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A\(3.2 \times 10^{2}\) lx
B1000 lx
C\(1.6 \times 10^{2}\) lx
D80 lx
Answer
D. 80 lx
For a point source: \(E = P/(4\pi r^2)\) = 1000/(4π×1²) ≈ 80 lx.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ16_MCQ03 MCQ Easy Recall • 4 marks

Luminous flux is measured in:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Alux (lx)
Blumen (lm)
Cdecibel (dB)
Dcandela (cd)
Answer
B. lumen (lm)
Luminous flux has SI unit lumen.
Source: Original generated question
G9T3_OBJ16_MCQ04 MCQ Easy Recall • 4 marks

Illuminance is measured in:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Ahertz (Hz)
Bcandela (cd)
Clux (lx)
Dlumen (lm)
Answer
C. lux (lx)
Illuminance is luminous flux per unit area, measured in lux.
Source: Original generated question
G9T3_OBJ16_MCQ05 MCQ Easy Recall • 4 marks

Luminous intensity is measured in:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Anewton (N)
Blux (lx)
Ccandela (cd)
Dlumen (lm)
Answer
C. candela (cd)
The SI base unit for luminous intensity is candela.
Source: Original generated question
G9T3_OBJ16_MCQ06 MCQ Medium Define • 4 marks

Which quantity describes luminous flux per unit area on a surface?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Afrequency
Bpolarization
Cilluminance
Dluminous intensity
Answer
C. illuminance
Illuminance describes how much light falls on each square metre.
Source: Original generated question
G9T3_OBJ16_MCQ07 MCQ Easy Match • 4 marks

Which light-quantity pair is correctly matched?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Afrequency — lux
Billuminance — lux
Csound level — lumen
Dluminous flux — candela
Answer
B. illuminance — lux
Illuminance is measured in lux.
Source: Original generated question
G9T3_OBJ16_MCQ08 MCQ Medium Identify • 4 marks

A lamp emits a total visible light output. This total output is best described as:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Aluminous flux
BMalus angle
Csound level
Dilluminance
Answer
A. luminous flux
Luminous flux is the total perceived light output.
Source: Original generated question
G9T3_OBJ16_MCQ09 MCQ Medium Compare • 4 marks

Which unit belongs to luminous intensity rather than illuminance?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Acandela
Blux
Cdecibel
Dhertz
Answer
A. candela
Candela is the SI unit of luminous intensity.
Source: Original generated question
G9T3_OBJ16_MCQ10 MCQ Medium Infer • 4 marks

If the same luminous flux spreads over a larger area, illuminance:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Aincreases
Bdecreases
Cbecomes frequency
Dstays always constant
Answer
B. decreases
\(E = P/A\), so increasing area lowers illuminance.
Source: Original generated question
G9T3_OBJ16_SA01 Short Answer 4 marks

Define luminous flux, illuminance, and luminous intensity with units.

Expected Answer

Luminous flux is total visible light output in lumen (lm). Illuminance is light per unit area in lux (lx). Luminous intensity is light in a direction in candela (cd). All units correctly stated.

Mark Scheme
  • 1 mark: Luminous flux is total visible light output in lumen (lm).
  • 1 mark: Illuminance is light per unit area in lux (lx).
  • 1 mark: Luminous intensity is light in a direction in candela (cd).
  • 1 mark: All units correctly stated.
Source: Original generated question
G9T3_OBJ16_SA02 Short Answer 4 marks

Explain why illuminance changes when the same light spreads over a larger area.

Expected Answer

Illuminance is luminous flux per unit area. Same flux over larger area gives smaller E. Uses \(E = P/A\). Conclusion: illuminance decreases.

Mark Scheme
  • 1 mark: Illuminance is luminous flux per unit area.
  • 1 mark: Same flux over larger area gives smaller E.
  • 1 mark: Uses \(E = P/A\).
  • 1 mark: Conclusion: illuminance decreases.
Source: Original generated question
G9T3_OBJ16_SA03 Short Answer 4 marks

A source emits 600 lm uniformly onto 3 m². Calculate average illuminance.

Expected Answer

Uses \(E = P/A\). Substitutes 600/3. Calculates 200. Uses unit lux.

Mark Scheme
  • 1 mark: Uses \(E = P/A\).
  • 1 mark: Substitutes 600/3.
  • 1 mark: Calculates 200.
  • 1 mark: Uses unit lux.
Source: Original generated question
G9T3_OBJ16_SA04 Short Answer 4 marks

Match units lm, lx, and cd to the correct light quantities.

Expected Answer

lm matched to luminous flux. lx matched to illuminance. cd matched to luminous intensity. No incorrect sound units included.

Mark Scheme
  • 1 mark: lm matched to luminous flux.
  • 1 mark: lx matched to illuminance.
  • 1 mark: cd matched to luminous intensity.
  • 1 mark: No incorrect sound units included.
Source: Original generated question

Light • pages 162

OBJ17: Apply the illuminance equation for a point source to numerical problems.

10 MCQs • 4 Short Answers
G9T3_OBJ17_MCQ01 MCQ Hard Identify • 4 marks

A point source has luminous flux \(P = 1000 lm. Calculate the illuminance on a surface at r\)= 2 m approximately.

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A40 lx
B250 lx
C80 lx
D20 lx
Answer
D. 20 lx
For a point source: \(E = P/(4\pi r^2)\) = 1000/(4π×2²) ≈ 20 lx.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ17_MCQ02 MCQ Hard Identify • 4 marks

A point source has luminous flux \(P = 500 lm. Calculate the illuminance on a surface at r\)= 1 m approximately.

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A500 lx
B40 lx
C\(1.6 \times 10^{2}\) lx
D80 lx
Answer
B. 40 lx
For a point source: \(E = P/(4\pi r^2)\) = 500/(4π×1²) ≈ 40 lx.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ17_MCQ03 MCQ Hard Calculate • 4 marks

A point source emits luminous flux \(P = 800 lm. What is the illuminance at distance r\)= 1 m? Use \(E = P/(4\pi r^2)\) and \(\pi \) ≈ 3.14.

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A64 lx
B32 lx
C128 lx
D800 lx
Answer
A. 64 lx
\(E = P/(4\pi r^2)\) = 800/(4π×1²) ≈ 64 lx.
Source: Original generated question
G9T3_OBJ17_MCQ04 MCQ Hard Calculate • 4 marks

A point source emits luminous flux \(P = 800 lm. What is the illuminance at distance r\)= 2 m? Use \(E = P/(4\pi r^2)\) and \(\pi \) ≈ 3.14.

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A8 lx
B32 lx
C1600 lx
D16 lx
Answer
D. 16 lx
\(E = P/(4\pi r^2)\) = 800/(4π×2²) ≈ 16 lx.
Source: Original generated question
G9T3_OBJ17_MCQ05 MCQ Hard Calculate • 4 marks

A point source emits luminous flux \(P = 1200 lm. What is the illuminance at distance r\)= 2 m? Use \(E = P/(4\pi r^2)\) and \(\pi \) ≈ 3.14.

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A24 lx
B48 lx
C12 lx
D2400 lx
Answer
A. 24 lx
\(E = P/(4\pi r^2)\) = 1200/(4π×2²) ≈ 24 lx.
Source: Original generated question
G9T3_OBJ17_MCQ06 MCQ Hard Calculate • 4 marks

A point source emits luminous flux \(P = 600 lm. What is the illuminance at distance r\)= 3 m? Use \(E = P/(4\pi r^2)\) and \(\pi \) ≈ 3.14.

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A1800 lx
B5.3 lx
C10.6 lx
D2.65 lx
Answer
B. 5.3 lx
\(E = P/(4\pi r^2)\) = 600/(4π×3²) ≈ 5.3 lx.
Source: Original generated question
G9T3_OBJ17_MCQ07 MCQ Hard Calculate • 4 marks

A point source emits luminous flux \(P = 1600 lm. What is the illuminance at distance r\)= 4 m? Use \(E = P/(4\pi r^2)\) and \(\pi \) ≈ 3.14.

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A4 lx
B8 lx
C6400 lx
D16 lx
Answer
B. 8 lx
\(E = P/(4\pi r^2)\) = 1600/(4π×4²) ≈ 8 lx.
Source: Original generated question
G9T3_OBJ17_MCQ08 MCQ Medium Recall • 4 marks

Which formula is suitable for illuminance from a point source spreading uniformly in all directions?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A\(E = 4\pi Pr^2\)
B\(E = r^2/P\)
C\(E = Pr^2\)
D\(E = P/(4\pi r^2)\)
Answer
D. \(E = P/(4\pi r^2)\)
The light spreads over a sphere of area 4πr².
Source: Original generated question
G9T3_OBJ17_MCQ09 MCQ Medium Calculate • 4 marks

A lamp is moved from 1 m to 2 m away. If it was 80 lx at 1 m, what is it at 2 m?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A20 lx
B320 lx
C160 lx
D40 lx
Answer
A. 20 lx
Doubling distance reduces illuminance to one quarter.
Source: Original generated question
G9T3_OBJ17_MCQ10 MCQ Medium Calculate • 4 marks

A surface receives 1000 lm over 5 m². What is the average illuminance?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A5 lx
B1005 lx
C5000 lx
D200 lx
Answer
D. 200 lx
\(E = P/A\) = 1000/5 = 200 lx.
Source: Original generated question
G9T3_OBJ17_SA01 Short Answer 4 marks

Calculate illuminance from a point source of 1000 lm at 2.0 m using \(E = P/(4\pi r^2)\).

Expected Answer

Substitutes \(P = 1000 lm and r\)= 2.0 m. Calculates denominator 4πr² = 16π. Calculates E ≈ 19.9 lx. Gives correct unit lux.

Mark Scheme
  • 1 mark: Substitutes \(P = 1000 lm and r\)= 2.0 m.
  • 1 mark: Calculates denominator 4πr² = 16π.
  • 1 mark: Calculates E ≈ 19.9 lx.
  • 1 mark: Gives correct unit lux.
Source: Original generated question
G9T3_OBJ17_SA02 Short Answer 4 marks

A lamp gives 90 lx at 1.0 m. Predict illuminance at 3.0 m.

Expected Answer

Uses inverse-square relationship. Distance triples so illuminance divided by 9. Calculates 90/9 = 10 lx. Includes unit.

Mark Scheme
  • 1 mark: Uses inverse-square relationship.
  • 1 mark: Distance triples so illuminance divided by 9.
  • 1 mark: Calculates 90/9 = 10 lx.
  • 1 mark: Includes unit.
Source: Original generated question
G9T3_OBJ17_SA03 Short Answer 4 marks

A surface receives 1500 lm over 6 m². Find average illuminance.

Expected Answer

Uses \(E = P/A\). Substitutes 1500/6. Calculates 250. Unit lx.

Mark Scheme
  • 1 mark: Uses \(E = P/A\).
  • 1 mark: Substitutes 1500/6.
  • 1 mark: Calculates 250.
  • 1 mark: Unit lx.
Source: Original generated question
G9T3_OBJ17_SA04 Short Answer 4 marks

Explain why the point-source formula contains 4πr².

Expected Answer

Light spreads in all directions. It forms a spherical wavefront/surface. Area of sphere is 4πr². Illuminance is flux divided by area.

Mark Scheme
  • 1 mark: Light spreads in all directions.
  • 1 mark: It forms a spherical wavefront/surface.
  • 1 mark: Area of sphere is 4πr².
  • 1 mark: Illuminance is flux divided by area.
Source: Original generated question

Light • pages 158-161

OBJ18: Define quantities of light such as luminous flux and illuminance with SI units.

10 MCQs • 4 Short Answers
G9T3_OBJ18_MCQ01 MCQ Hard Identify • 4 marks

A point source has luminous flux \(P = 2000 lm. Calculate the illuminance on a surface at r\)= 2 m approximately.

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A\(1.6 \times 10^{2}\) lx
B80 lx
C40 lx
D500 lx
Answer
C. 40 lx
For a point source: \(E = P/(4\pi r^2)\) = 2000/(4π×2²) ≈ 40 lx.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ18_MCQ02 MCQ Hard Identify • 4 marks

A point source has luminous flux \(P = 1500 lm. Calculate the illuminance on a surface at r\)= 3 m approximately.

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A40 lx
B27 lx
C13 lx
D\(1.7 \times 10^{2}\) lx
Answer
C. 13 lx
For a point source: \(E = P/(4\pi r^2)\) = 1500/(4π×3²) ≈ 13 lx.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ18_MCQ03 MCQ Easy Recall • 4 marks

Which quantity has unit lumen?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Asound level
Bfrequency
Cilluminance
Dluminous flux
Answer
D. luminous flux
Luminous flux is measured in lumens.
Source: Original generated question
G9T3_OBJ18_MCQ04 MCQ Easy Recall • 4 marks

Which quantity has unit lux?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Aluminous flux
Bpitch
Cluminous intensity
Dilluminance
Answer
D. illuminance
Illuminance is measured in lux.
Source: Original generated question
G9T3_OBJ18_MCQ05 MCQ Easy Recall • 4 marks

Which quantity has unit candela?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Awavelength
Bluminous intensity
Cilluminance
Dluminous flux
Answer
B. luminous intensity
Candela is the SI unit for luminous intensity.
Source: Original generated question
G9T3_OBJ18_MCQ06 MCQ Medium Define • 4 marks

Which definition best matches illuminance?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Afrequency per wavelength
Btotal sound energy
Cluminous flux incident per unit area
Dlight vibration direction
Answer
C. luminous flux incident per unit area
Illuminance is light received per square metre.
Source: Original generated question
G9T3_OBJ18_MCQ07 MCQ Medium Match • 4 marks

Which symbol-unit pair is correct?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
AI — Hz
BP — dB
CE — lx
Df — lx
Answer
C. E — lx
Illuminance E is measured in lux.
Source: Original generated question
G9T3_OBJ18_MCQ08 MCQ Easy Interpret • 4 marks

A lamp has a high luminous flux. This means it:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Ahas a high sound level
Bmust be opaque
Cemits a large amount of visible light
Dhas a high frequency sound
Answer
C. emits a large amount of visible light
Luminous flux is total visible light output.
Source: Original generated question
G9T3_OBJ18_MCQ09 MCQ Easy Select • 4 marks

Which unit is not a light quantity unit?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Alumen
Blux
Ccandela
Ddecibel
Answer
D. decibel
Decibel is for sound level.
Source: Original generated question
G9T3_OBJ18_MCQ10 MCQ Medium Calculate • 4 marks

If 400 lm falls uniformly on 2 m², the illuminance is:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A2 lx
B200 lx
C398 lx
D800 lx
Answer
B. 200 lx
\(E = P/A\) = 400/2 = 200 lx.
Source: Original generated question
G9T3_OBJ18_SA01 Short Answer 4 marks

State the SI units for luminous flux, illuminance, and luminous intensity.

Expected Answer

Luminous flux: lumen (lm). Illuminance: lux (lx). Luminous intensity: candela (cd). Correct spelling/symbols.

Mark Scheme
  • 1 mark: Luminous flux: lumen (lm).
  • 1 mark: Illuminance: lux (lx).
  • 1 mark: Luminous intensity: candela (cd).
  • 1 mark: Correct spelling/symbols.
Source: Original generated question
G9T3_OBJ18_SA02 Short Answer 4 marks

A student writes that illuminance is measured in lumens. Correct the mistake.

Expected Answer

States illuminance is measured in lux. Lumen is unit of luminous flux. Explains illuminance is flux per unit area. Gives \(E = P/A\) if appropriate.

Mark Scheme
  • 1 mark: States illuminance is measured in lux.
  • 1 mark: Lumen is unit of luminous flux.
  • 1 mark: Explains illuminance is flux per unit area.
  • 1 mark: Gives \(E = P/A\) if appropriate.
Source: Original generated question
G9T3_OBJ18_SA03 Short Answer 4 marks

Describe a practical situation where illuminance is more useful than luminous flux.

Expected Answer

Gives valid situation such as desk/classroom lighting. Explains illuminance tells light falling on surface. Mentions lux measurement. Contrasts with source's total flux.

Mark Scheme
  • 1 mark: Gives valid situation such as desk/classroom lighting.
  • 1 mark: Explains illuminance tells light falling on surface.
  • 1 mark: Mentions lux measurement.
  • 1 mark: Contrasts with source's total flux.
Source: Original generated question
G9T3_OBJ18_SA04 Short Answer 4 marks

Calculate average illuminance when 900 lm falls uniformly on 9 m².

Expected Answer

Uses \(E = P/A\). Substitutes 900/9. Calculates 100. Uses unit lx.

Mark Scheme
  • 1 mark: Uses \(E = P/A\).
  • 1 mark: Substitutes 900/9.
  • 1 mark: Calculates 100.
  • 1 mark: Uses unit lx.
Source: Original generated question

Light • pages 159

OBJ19: Show inverse-square relation of illuminance with distance and direct relation with luminous flux.

10 MCQs • 4 Short Answers
G9T3_OBJ19_MCQ01 MCQ Medium Identify • 4 marks

If the distance between a lamp and a surface doubles, what happens to illuminance?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
AIt doubles
BIt becomes one-quarter
CIt stays constant
DIt becomes half
Answer
B. It becomes one-quarter
Illuminance from a point source follows an inverse-square relation: doubling r gives \(E = 1/4.\)
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ19_MCQ02 MCQ Medium Identify • 4 marks

The distance becomes 2 times the original distance. What is the new illuminance relative to the original?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A0.5
B4
C0.25
D2
Answer
C. 0.25
Since E ∝ 1/r², the ratio = 1/(2)² = 0.25.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ19_MCQ03 MCQ Easy Recall • 4 marks

Illuminance from a point source varies with distance according to:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Alinear increase with distance
Binverse-square relationship
Cdirect-square relationship
Dno relationship
Answer
B. inverse-square relationship
E ∝ 1/r² for a point source.
Source: Original generated question
G9T3_OBJ19_MCQ04 MCQ Medium Calculate • 4 marks

If distance from a point source triples, illuminance becomes:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Athree times
Bone-third
Cone-ninth
Dnine times
Answer
C. one-ninth
E ∝ 1/r², so 3² = 9.
Source: Original generated question
G9T3_OBJ19_MCQ05 MCQ Medium Calculate • 4 marks

If distance is halved, illuminance becomes:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Afour times larger
Bunchanged
Cone-quarter as large
Dhalf as large
Answer
A. four times larger
Halving r gives E multiplied by 4.
Source: Original generated question
G9T3_OBJ19_MCQ06 MCQ Medium Infer • 4 marks

If luminous flux doubles and distance is unchanged, illuminance:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Abecomes one-quarter
Bhalves
Cdoubles
Ddoes not change
Answer
C. doubles
E is directly proportional to P.
Source: Original generated question
G9T3_OBJ19_MCQ07 MCQ Medium Calculate • 4 marks

A lamp gives 120 lx at 1 m. What is the illuminance at 2 m?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
A60 lx
B240 lx
C480 lx
D30 lx
Answer
D. 30 lx
Doubling distance reduces illuminance to 120/4 = 30 lx.
Source: Original generated question
G9T3_OBJ19_MCQ08 MCQ Easy Select • 4 marks

Which change gives the greatest illuminance?

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
AMove the same lamp closer.
BMove the lamp farther away.
CReduce luminous flux.
DIncrease distance and reduce flux.
Answer
A. Move the same lamp closer.
Illuminance increases when distance decreases.
Source: Original generated question
G9T3_OBJ19_MCQ09 MCQ Hard Interpret • 4 marks

The graph of illuminance E against 1/r² should be:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Aa circle
Ba horizontal line always
Ca straight line through the origin
Da random curve
Answer
C. a straight line through the origin
E is directly proportional to 1/r².
Source: Original generated question
G9T3_OBJ19_MCQ10 MCQ Hard Calculate • 4 marks

A lamp emits twice the luminous flux but is placed twice as far away. Compared with the original illuminance, the new illuminance is:

Point-Source Illuminance P distance r E = P / (4πr²)
Illuminance
Afour times
Bone-half
Cunchanged
Dtwice
Answer
B. one-half
Flux doubles × distance effect 1/4, so net = 1/2.
Source: Original generated question
G9T3_OBJ19_SA01 Short Answer 4 marks

Show mathematically why illuminance from a point source follows an inverse-square law.

Expected Answer

Starts with \(E = P/(4\pi r^2)\). For constant P, E ∝ 1/r². Explains distance is squared. Concludes doubling distance gives one-quarter illuminance.

Mark Scheme
  • 1 mark: Starts with \(E = P/(4\pi r^2)\).
  • 1 mark: For constant P, E ∝ 1/r².
  • 1 mark: Explains distance is squared.
  • 1 mark: Concludes doubling distance gives one-quarter illuminance.
Source: Original generated question
G9T3_OBJ19_SA02 Short Answer 4 marks

A lamp gives 240 lx at 1 m. Find illuminance at 2 m and 4 m.

Expected Answer

At 2 m, divide by 4 = 60 lx. At 4 m, divide by 16 = 15 lx. Uses inverse-square relationship. Correct units.

Mark Scheme
  • 1 mark: At 2 m, divide by 4 = 60 lx.
  • 1 mark: At 4 m, divide by 16 = 15 lx.
  • 1 mark: Uses inverse-square relationship.
  • 1 mark: Correct units.
Source: Original generated question
G9T3_OBJ19_SA03 Short Answer 4 marks

Explain the direct relationship between luminous flux and illuminance.

Expected Answer

Uses \(E = P/(4\pi r^2)\). At constant distance, E ∝ P. Doubling P doubles E. Halving P halves E.

Mark Scheme
  • 1 mark: Uses \(E = P/(4\pi r^2)\).
  • 1 mark: At constant distance, E ∝ P.
  • 1 mark: Doubling P doubles E.
  • 1 mark: Halving P halves E.
Source: Original generated question
G9T3_OBJ19_SA04 Short Answer 4 marks

A lamp has twice the luminous flux but is moved three times farther away. Find the factor change in illuminance.

Expected Answer

Flux factor is ×2. Distance factor is ÷9. Net factor is 2/9. States illuminance becomes 2/9 of original.

Mark Scheme
  • 1 mark: Flux factor is ×2.
  • 1 mark: Distance factor is ÷9.
  • 1 mark: Net factor is 2/9.
  • 1 mark: States illuminance becomes 2/9 of original.
Source: Original generated question

Light • pages 165

OBJ20: Define diffraction as bending of a wave as it passes the edge of a barrier.

10 MCQs • 4 Short Answers
G9T3_OBJ20_MCQ01 MCQ Easy Identify • 4 marks

What is diffraction?

Diffraction at an Edge/Opening waves spread after a narrow gap
Diffraction
AOnly increasing light intensity
BConversion of light into sound
CReflection of light from a smooth surface
DBending of a wave as it passes an edge or through an opening
Answer
D. Bending of a wave as it passes an edge or through an opening
Diffraction is evidence of wave nature; it occurs when a wave spreads around an edge or opening.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ20_MCQ02 MCQ Medium Identify • 4 marks

Diffraction is more noticeable when the opening width is:

Diffraction at an Edge/Opening waves spread after a narrow gap
Diffraction
AUnrelated to wavelength
BComparable to the wavelength
CExactly zero
DAlways much larger than wavelength
Answer
B. Comparable to the wavelength
Diffraction is noticeable when the opening or obstacle size is comparable to wavelength.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ20_MCQ03 MCQ Easy Define • 4 marks

Diffraction is:

Diffraction at an Edge/Opening waves spread after a narrow gap
Diffraction
Afiltering vibration direction
Bsound level in dB
Cbending/spreading of a wave as it passes an edge or opening
Dabsorption of all light
Answer
C. bending/spreading of a wave as it passes an edge or opening
Diffraction is a wave bending/spreading effect.
Source: Original generated question
G9T3_OBJ20_MCQ04 MCQ Medium Explain • 4 marks

Diffraction is strongest when the gap size is:

Diffraction at an Edge/Opening waves spread after a narrow gap
Diffraction
Aunrelated to wavelength
Bsimilar to the wavelength
Cmillions of times larger than wavelength
Dzero for all waves
Answer
B. similar to the wavelength
Waves spread noticeably when openings are comparable to wavelength.
Source: Original generated question
G9T3_OBJ20_MCQ05 MCQ Easy Identify • 4 marks

Which situation shows diffraction?

Diffraction at an Edge/Opening waves spread after a narrow gap
Diffraction
Aa siren changing pitch due to motion
Ba mirror forming an image
Cwater waves spreading after passing through a narrow gap
Da lamp producing light
Answer
C. water waves spreading after passing through a narrow gap
Spreading after a gap is diffraction.
Source: Original generated question
G9T3_OBJ20_MCQ06 MCQ Medium Recall • 4 marks

Diffraction can occur with:

Diffraction at an Edge/Opening waves spread after a narrow gap
Diffraction
Aall types of waves
Bonly water waves and not light
Conly sound
Donly light
Answer
A. all types of waves
Diffraction is a general wave behavior.
Source: Original generated question
G9T3_OBJ20_MCQ07 MCQ Easy Identify • 4 marks

At the edge of a barrier, a wave may bend into the shadow region. This is called:

Diffraction at an Edge/Opening waves spread after a narrow gap
Diffraction
Aresonance
Bloudness
Cluminous flux
Ddiffraction
Answer
D. diffraction
Bending around an obstacle is diffraction.
Source: Original generated question
G9T3_OBJ20_MCQ08 MCQ Medium Reason • 4 marks

Which observation supports the wave model of light?

Diffraction at an Edge/Opening waves spread after a narrow gap
Diffraction
Aa battery heating
Ba book falling
Clight spreading slightly after passing through a narrow slit
Da mass increasing
Answer
C. light spreading slightly after passing through a narrow slit
Diffraction is evidence that light behaves as a wave.
Source: Original generated question
G9T3_OBJ20_MCQ09 MCQ Medium Infer • 4 marks

If a slit becomes much wider compared with wavelength, diffraction generally:

Diffraction at an Edge/Opening waves spread after a narrow gap
Diffraction
Aturns into Doppler effect
Bdecreases
Cbecomes total internal reflection
Dincreases strongly
Answer
B. decreases
Very wide openings produce less noticeable spreading.
Source: Original generated question
G9T3_OBJ20_MCQ10 MCQ Easy Apply • 4 marks

A sound can sometimes be heard around a corner because sound waves:

Diffraction at an Edge/Opening waves spread after a narrow gap
Diffraction
Ado not need a medium
Bdiffract around edges
Care always visible
Dhave no wavelength
Answer
B. diffract around edges
Sound waves can bend around obstacles.
Source: Original generated question
G9T3_OBJ20_SA01 Short Answer 4 marks

Define diffraction and give one example.

Expected Answer

Diffraction is bending/spreading of a wave. Occurs at an edge/opening/barrier. Gives valid example such as water waves through gap or sound around corner. Identifies it as wave behavior.

Mark Scheme
  • 1 mark: Diffraction is bending/spreading of a wave.
  • 1 mark: Occurs at an edge/opening/barrier.
  • 1 mark: Gives valid example such as water waves through gap or sound around corner.
  • 1 mark: Identifies it as wave behavior.
Source: Original generated question
G9T3_OBJ20_SA02 Short Answer 4 marks

Explain why diffraction is more noticeable through a narrow gap.

Expected Answer

Diffraction depends on gap size relative to wavelength. More noticeable when gap is similar to wavelength. Wave spreads after passing the gap. Correctly uses wavelength concept.

Mark Scheme
  • 1 mark: Diffraction depends on gap size relative to wavelength.
  • 1 mark: More noticeable when gap is similar to wavelength.
  • 1 mark: Wave spreads after passing the gap.
  • 1 mark: Correctly uses wavelength concept.
Source: Original generated question
G9T3_OBJ20_SA03 Short Answer 4 marks

Describe how diffraction supports the wave model of light.

Expected Answer

Light can spread/bend through a small opening. Diffraction is a property of waves. Therefore light shows wave behavior. Gives a slit/edge example.

Mark Scheme
  • 1 mark: Light can spread/bend through a small opening.
  • 1 mark: Diffraction is a property of waves.
  • 1 mark: Therefore light shows wave behavior.
  • 1 mark: Gives a slit/edge example.
Source: Original generated question
G9T3_OBJ20_SA04 Short Answer 4 marks

Compare diffraction at a wide opening and a narrow opening.

Expected Answer

Wide opening gives little spreading. Narrow opening gives greater spreading. Comparison uses wavelength size. Mentions wavefront bending/spreading.

Mark Scheme
  • 1 mark: Wide opening gives little spreading.
  • 1 mark: Narrow opening gives greater spreading.
  • 1 mark: Comparison uses wavelength size.
  • 1 mark: Mentions wavefront bending/spreading.
Source: Original generated question

Light • pages 166

OBJ21: Describe that color of light is related to wavelength and frequency.

10 MCQs • 4 Short Answers
G9T3_OBJ21_MCQ01 MCQ Easy Identify • 4 marks

Which visible light color generally has the longer wavelength?

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
AViolet
BRed
CBlue
DGreen
Answer
B. Red
Red light has a longer wavelength and lower frequency than violet light.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ21_MCQ02 MCQ Medium Identify • 4 marks

Why does an object appear red under white light?

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
ABecause it reflects all colors equally
BBecause it always emits red light
CBecause color does not depend on incident light
DBecause it mainly reflects red light and absorbs most other colors
Answer
D. Because it mainly reflects red light and absorbs most other colors
Object color depends on wavelengths in incident light and which wavelengths the object absorbs or reflects.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ21_MCQ03 MCQ Easy Compare • 4 marks

In visible light, red light has:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Ashorter wavelength than violet light
Blonger wavelength than violet light
Cno frequency
Dthe highest frequency
Answer
B. longer wavelength than violet light
Red has long wavelength and lower frequency.
Source: Original generated question
G9T3_OBJ21_MCQ04 MCQ Easy Compare • 4 marks

Violet light has:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Athe same wavelength as red always
Bno wavelength
Clower frequency than red light
Dhigher frequency than red light
Answer
D. higher frequency than red light
Violet has shorter wavelength and higher frequency.
Source: Original generated question
G9T3_OBJ21_MCQ05 MCQ Medium Recall • 4 marks

For light waves in vacuum, frequency and wavelength are related by:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Ac = \(\lambda \)/f
B\(f = c\lambda \)
C\(\lambda \) = cf
Dc = λf
Answer
D. c = λf
Wave speed equals wavelength times frequency.
Source: Original generated question
G9T3_OBJ21_MCQ06 MCQ Medium Infer • 4 marks

If wavelength decreases for light travelling in the same medium, frequency:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Abecomes zero
Bincreases
Cdecreases
Dis unrelated
Answer
B. increases
For fixed wave speed, \(\lambda \) and f are inversely related.
Source: Original generated question
G9T3_OBJ21_MCQ07 MCQ Easy Recall • 4 marks

Which visible color generally has the shortest wavelength?

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Aviolet
Byellow
Corange
Dred
Answer
A. violet
Violet light has the shortest wavelength in the visible spectrum.
Source: Original generated question
G9T3_OBJ21_MCQ08 MCQ Easy Recall • 4 marks

Which visible color generally has the lowest frequency?

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Agreen
Bred
Cblue
Dviolet
Answer
B. red
Red has the longest wavelength and lowest frequency.
Source: Original generated question
G9T3_OBJ21_MCQ09 MCQ Easy Define • 4 marks

Color of monochromatic light is determined mainly by its:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Amass
Bloudness
Cwavelength/frequency
Ddecibel level
Answer
C. wavelength/frequency
Visible color corresponds to wavelength and frequency.
Source: Original generated question
G9T3_OBJ21_MCQ10 MCQ Medium Explain • 4 marks

Green light has a different color from blue light because it has a different:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Asound level
Bwavelength and frequency
Cluminous source type only
Dmass of photons in this grade model
Answer
B. wavelength and frequency
Different colors correspond to different wave properties.
Source: Original generated question
G9T3_OBJ21_SA01 Short Answer 4 marks

Explain how color is related to wavelength and frequency.

Expected Answer

Different colors have different wavelengths. Different colors have different frequencies. In the same medium, c = λf. Longer wavelength corresponds to lower frequency.

Mark Scheme
  • 1 mark: Different colors have different wavelengths.
  • 1 mark: Different colors have different frequencies.
  • 1 mark: In the same medium, c = λf.
  • 1 mark: Longer wavelength corresponds to lower frequency.
Source: Original generated question
G9T3_OBJ21_SA02 Short Answer 4 marks

Compare red and violet light in terms of wavelength and frequency.

Expected Answer

Red has longer wavelength. Red has lower frequency. Violet has shorter wavelength. Violet has higher frequency.

Mark Scheme
  • 1 mark: Red has longer wavelength.
  • 1 mark: Red has lower frequency.
  • 1 mark: Violet has shorter wavelength.
  • 1 mark: Violet has higher frequency.
Source: Original generated question
G9T3_OBJ21_SA03 Short Answer 4 marks

If a visible light wave has shorter wavelength than another in the same medium, what happens to frequency?

Expected Answer

Uses c = λf or constant speed. Frequency increases when wavelength decreases. States inverse relationship. Applies to color difference.

Mark Scheme
  • 1 mark: Uses c = λf or constant speed.
  • 1 mark: Frequency increases when wavelength decreases.
  • 1 mark: States inverse relationship.
  • 1 mark: Applies to color difference.
Source: Original generated question
G9T3_OBJ21_SA04 Short Answer 4 marks

Arrange red, green, and violet from highest to lowest frequency.

Expected Answer

Violet highest. Green middle. Red lowest. Order linked to wavelength/frequency relationship.

Mark Scheme
  • 1 mark: Violet highest.
  • 1 mark: Green middle.
  • 1 mark: Red lowest.
  • 1 mark: Order linked to wavelength/frequency relationship.
Source: Original generated question

Light • pages 167

OBJ22: Describe primary, secondary, and complementary colors of light and explain object color by absorption/reflection.

10 MCQs • 4 Short Answers
G9T3_OBJ22_MCQ01 MCQ Easy Identify • 4 marks

Primary colors of light in additive mixing are:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
ACyan, magenta, and yellow
BWhite, black, and gray
CRed, yellow, and blue
DRed, green, and blue
Answer
D. Red, green, and blue
Light mixing uses RGB: red, green, and blue.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ22_MCQ02 MCQ Medium Identify • 4 marks

Glare reflected from a road can be reduced using:

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
AA sound filter
BA thermometer
COnly a stronger light source
DA polarizing filter
Answer
D. A polarizing filter
Reflected glare is partially polarized, so polarizing filters reduce it.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ22_MCQ03 MCQ Easy Recall • 4 marks

The primary colors of light are:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Ared, yellow, and blue pigments
Bred, green, and blue
Cblack, white, and gray
Dcyan, magenta, and yellow
Answer
B. red, green, and blue
Additive primary light colors are RGB.
Source: Original generated question
G9T3_OBJ22_MCQ04 MCQ Medium Apply • 4 marks

Red light plus green light produces:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Acyan light
Bmagenta light
Cyellow light
Dblack
Answer
C. yellow light
In additive mixing, red + green = yellow.
Source: Original generated question
G9T3_OBJ22_MCQ05 MCQ Medium Apply • 4 marks

Green light plus blue light produces:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Ared light
Bcyan light
Cyellow light
Dblack
Answer
B. cyan light
In additive color mixing, green + blue = cyan.
Source: Original generated question
G9T3_OBJ22_MCQ06 MCQ Medium Apply • 4 marks

Red light plus blue light produces:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Amagenta light
Byellow light
Cgreen light
Dcyan light
Answer
A. magenta light
In additive mixing, red + blue = magenta.
Source: Original generated question
G9T3_OBJ22_MCQ07 MCQ Easy Recall • 4 marks

Combining red, green, and blue light in suitable intensities produces:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Awhite light
Bblack
Cbrown
Dcyan only
Answer
A. white light
RGB add to white in additive color mixing.
Source: Original generated question
G9T3_OBJ22_MCQ08 MCQ Medium Explain • 4 marks

An object appears blue under white light because it mainly:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Areflects blue and absorbs many other colors
Babsorbs blue only
Cemits sound
Dreflects no light
Answer
A. reflects blue and absorbs many other colors
Object color depends on reflected wavelengths.
Source: Original generated question
G9T3_OBJ22_MCQ09 MCQ Hard Reason • 4 marks

A red apple under pure blue light may appear dark because:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Ablue light has no wavelength
Bthe apple becomes transparent
Cthere is little red light available for it to reflect
Dall blue light becomes sound
Answer
C. there is little red light available for it to reflect
A red object reflects red; under blue light it absorbs much of the incident light.
Source: Original generated question
G9T3_OBJ22_MCQ10 MCQ Easy Define • 4 marks

Complementary colors of light combine to produce:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Asound
Bwhite light
Cblack pigment only
Dzero frequency
Answer
B. white light
Complementary light colors add to white.
Source: Original generated question
G9T3_OBJ22_SA01 Short Answer 4 marks

State the primary colors of light and describe what happens when all three combine.

Expected Answer

Primary light colors are red, green, blue. They combine additively. Suitable intensities produce white light. Uses additive color terminology.

Mark Scheme
  • 1 mark: Primary light colors are red, green, blue.
  • 1 mark: They combine additively.
  • 1 mark: Suitable intensities produce white light.
  • 1 mark: Uses additive color terminology.
Source: Original generated question
G9T3_OBJ22_SA02 Short Answer 4 marks

Explain why a blue object appears blue under white light.

Expected Answer

White light contains blue and other wavelengths. Object reflects blue wavelengths. Object absorbs many other wavelengths. Reflected blue reaches eye.

Mark Scheme
  • 1 mark: White light contains blue and other wavelengths.
  • 1 mark: Object reflects blue wavelengths.
  • 1 mark: Object absorbs many other wavelengths.
  • 1 mark: Reflected blue reaches eye.
Source: Original generated question
G9T3_OBJ22_SA03 Short Answer 4 marks

Predict the appearance of a red object under pure green light and explain.

Expected Answer

Red object mainly reflects red. Green light contains little/no red. Green is mostly absorbed. Object appears dark/blackish.

Mark Scheme
  • 1 mark: Red object mainly reflects red.
  • 1 mark: Green light contains little/no red.
  • 1 mark: Green is mostly absorbed.
  • 1 mark: Object appears dark/blackish.
Source: Original generated question
G9T3_OBJ22_SA04 Short Answer 4 marks

Give examples of secondary colors produced by adding primary light colors.

Expected Answer

Red + green = yellow. Green + blue = cyan. Red + blue = magenta. Any two correct combinations accepted for relevant marks.

Mark Scheme
  • 1 mark: Red + green = yellow.
  • 1 mark: Green + blue = cyan.
  • 1 mark: Red + blue = magenta.
  • 1 mark: Any two correct combinations accepted for relevant marks.
Source: Original generated question

Light • pages 168

OBJ23: Describe primary and secondary pigments and effects of mixing pigments or dyes.

10 MCQs • 4 Short Answers
G9T3_OBJ23_MCQ01 MCQ Medium Identify • 4 marks

Mixing pigments differs from mixing light because pigments usually:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
AIncrease the speed of light
BTurn light into sound
CAlways add all wavelengths
DAbsorb some wavelengths and subtract them from reflected light
Answer
D. Absorb some wavelengths and subtract them from reflected light
Pigments mix subtractively; they absorb certain colors and reflect others.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ23_MCQ02 MCQ Medium Identify • 4 marks

Primary pigment colors are usually:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
ARed, green, and blue
BWhite, red, and black
CCyan, magenta, and yellow
DViolet, orange, and green
Answer
C. Cyan, magenta, and yellow
Pigments mix subtractively, commonly using CMY.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ23_MCQ03 MCQ Easy Define • 4 marks

Pigments mix by:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
AMalus's law
BDoppler shift
Csubtractive color mixing
Dadditive color mixing only
Answer
C. subtractive color mixing
Pigments absorb selected wavelengths and reflect the remainder.
Source: Original generated question
G9T3_OBJ23_MCQ04 MCQ Easy Recall • 4 marks

The primary pigments are commonly:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Ablack, white, and gray
Bred, orange, and violet
Cred, green, and blue
Dcyan, magenta, and yellow
Answer
D. cyan, magenta, and yellow
Primary pigments for subtractive mixing are CMY.
Source: Original generated question
G9T3_OBJ23_MCQ05 MCQ Medium Explain • 4 marks

A cyan pigment appears cyan because it absorbs mainly:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Ared light
Bblue light only
Call light equally
Dgreen light only
Answer
A. red light
Cyan reflects green and blue and absorbs red.
Source: Original generated question
G9T3_OBJ23_MCQ06 MCQ Medium Explain • 4 marks

A magenta pigment absorbs mainly:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Ared light only
Bgreen light
Cwhite light only
Dblue light only
Answer
B. green light
Magenta reflects red and blue, absorbing green.
Source: Original generated question
G9T3_OBJ23_MCQ07 MCQ Medium Explain • 4 marks

A yellow pigment absorbs mainly:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Agreen light only
Bred light only
Call red and green
Dblue light
Answer
D. blue light
Yellow reflects red and green, absorbing blue.
Source: Original generated question
G9T3_OBJ23_MCQ08 MCQ Medium Explain • 4 marks

Mixing many pigments often produces a darker color because:

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Asound frequency increases
Blight speed becomes zero
Cmore wavelengths are added
Dmore wavelengths are absorbed
Answer
D. more wavelengths are absorbed
Subtractive mixing removes more light from reflection.
Source: Original generated question
G9T3_OBJ23_MCQ09 MCQ Medium Define • 4 marks

Which statement best describes a dye?

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
AIt always emits its own light.
BIt is always a sound wave.
CIt absorbs some wavelengths and transmits or reflects others.
DIt makes light longitudinal.
Answer
C. It absorbs some wavelengths and transmits or reflects others.
Dyes and pigments affect color by selective absorption.
Source: Original generated question
G9T3_OBJ23_MCQ10 MCQ Hard Apply • 4 marks

What happens when cyan and yellow pigments are mixed ideally?

Color of Light short λ, high flong λ, low f Color depends on wavelength and frequency.
Colors
Awhite light is emitted
Bgreen is mainly reflected
Cred is mainly reflected
Dblue and red are reflected only
Answer
B. green is mainly reflected
Cyan removes red; yellow removes blue; green remains.
Source: Original generated question
G9T3_OBJ23_SA01 Short Answer 4 marks

Explain subtractive color mixing using pigments.

Expected Answer

Pigments absorb selected wavelengths. They reflect/transmit remaining wavelengths. Mixing pigments removes more wavelengths. Result often becomes darker.

Mark Scheme
  • 1 mark: Pigments absorb selected wavelengths.
  • 1 mark: They reflect/transmit remaining wavelengths.
  • 1 mark: Mixing pigments removes more wavelengths.
  • 1 mark: Result often becomes darker.
Source: Original generated question
G9T3_OBJ23_SA02 Short Answer 4 marks

State the primary pigments and one color they absorb.

Expected Answer

Primary pigments are cyan, magenta, yellow. Cyan absorbs red. Magenta absorbs green. Yellow absorbs blue.

Mark Scheme
  • 1 mark: Primary pigments are cyan, magenta, yellow.
  • 1 mark: Cyan absorbs red.
  • 1 mark: Magenta absorbs green.
  • 1 mark: Yellow absorbs blue.
Source: Original generated question
G9T3_OBJ23_SA03 Short Answer 4 marks

Explain why mixing cyan and yellow pigments ideally gives green.

Expected Answer

Cyan absorbs red. Yellow absorbs blue. Green is not absorbed by either ideally. Green is reflected/transmitted.

Mark Scheme
  • 1 mark: Cyan absorbs red.
  • 1 mark: Yellow absorbs blue.
  • 1 mark: Green is not absorbed by either ideally.
  • 1 mark: Green is reflected/transmitted.
Source: Original generated question
G9T3_OBJ23_SA04 Short Answer 4 marks

Compare mixing colored light with mixing pigments.

Expected Answer

Light mixing is additive. Pigment mixing is subtractive. RGB primary lights differ from CMY primary pigments. Explains absorption/reflection in pigments.

Mark Scheme
  • 1 mark: Light mixing is additive.
  • 1 mark: Pigment mixing is subtractive.
  • 1 mark: RGB primary lights differ from CMY primary pigments.
  • 1 mark: Explains absorption/reflection in pigments.
Source: Original generated question

Light • pages 169

OBJ24: Explain polarization of light by filtering and by reflection.

10 MCQs • 4 Short Answers
G9T3_OBJ24_MCQ01 MCQ Easy Identify • 4 marks

What does a polarizing filter do?

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
AIt converts light to sound
BIt increases light speed in vacuum
CIt transmits light vibrations mainly in one direction
DIt removes all colors except red
Answer
C. It transmits light vibrations mainly in one direction
Polarization restricts the vibration direction of the light’s electric field.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ24_MCQ02 MCQ Hard Identify • 4 marks

If two polarizing filters are crossed at 90°, why is transmitted light nearly zero?

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
ABecause the vibration direction passed by the first is blocked by the second
BBecause light always fully turns into heat
CBecause filters increase frequency
DBecause wavelength becomes infinite
Answer
A. Because the vibration direction passed by the first is blocked by the second
By Malus’s law, at 90° cos²θ is approximately 0.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ24_MCQ03 MCQ Easy Define • 4 marks

A polarizing filter mainly allows light vibrations in:

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
Aonly the sound direction
Ball directions equally
Cone direction
Dno direction
Answer
C. one direction
A polarizer transmits one component of the electric field.
Source: Original generated question
G9T3_OBJ24_MCQ04 MCQ Medium Recall • 4 marks

Only which type of wave can be polarized?

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
Atransverse waves
Blongitudinal waves only
Cpressure waves only
Dsound waves in air only
Answer
A. transverse waves
Polarization requires vibration directions perpendicular to travel.
Source: Original generated question
G9T3_OBJ24_MCQ05 MCQ Medium Define • 4 marks

Unpolarized light contains vibrations:

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
Aparallel to travel only
Bin many random directions perpendicular to travel
Cwith no electric field
Donly in one fixed direction
Answer
B. in many random directions perpendicular to travel
Unpolarized light has random polarization directions.
Source: Original generated question
G9T3_OBJ24_MCQ06 MCQ Easy Predict • 4 marks

After unpolarized light passes through an ideal polarizer, it becomes:

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
Aa compression wave
Bsound
Cpolarized light
Dopaque material
Answer
C. polarized light
The filter selects one vibration direction.
Source: Original generated question
G9T3_OBJ24_MCQ07 MCQ Medium Explain • 4 marks

Polarizing sunglasses reduce glare mainly because reflected glare is often:

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
Alongitudinal sound
Bopaque
Cpartly polarized
Dwithout frequency
Answer
C. partly polarized
Light reflected from surfaces can become polarized.
Source: Original generated question
G9T3_OBJ24_MCQ08 MCQ Medium Predict • 4 marks

If two polarizing filters have perpendicular axes, transmitted light is:

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
Amaximum
Bconverted to sound
Calways doubled
Dminimum or nearly zero
Answer
D. minimum or nearly zero
Crossed polarizers block the transmitted polarization direction.
Source: Original generated question
G9T3_OBJ24_MCQ09 MCQ Easy Identify • 4 marks

Which example demonstrates polarization by filtering?

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
Alamp dimming with distance
Blight passing through polarizing sunglasses
Csound heard around a corner
Dsiren pitch changing
Answer
B. light passing through polarizing sunglasses
Sunglasses use polarizing filters.
Source: Original generated question
G9T3_OBJ24_MCQ10 MCQ Hard Explain • 4 marks

Why is sound in air not polarized in the usual way?

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
AIt has no frequency.
BIt is always visible.
CIt travels faster than light.
DIt is longitudinal, so vibration is along the direction of travel.
Answer
D. It is longitudinal, so vibration is along the direction of travel.
Longitudinal waves do not have transverse vibration directions to select.
Source: Original generated question
G9T3_OBJ24_SA01 Short Answer 4 marks

Explain polarization by filtering.

Expected Answer

Unpolarized light has many vibration directions. Polarizing filter transmits one vibration direction. Emerging light is polarized. Blocked components reduce intensity.

Mark Scheme
  • 1 mark: Unpolarized light has many vibration directions.
  • 1 mark: Polarizing filter transmits one vibration direction.
  • 1 mark: Emerging light is polarized.
  • 1 mark: Blocked components reduce intensity.
Source: Original generated question
G9T3_OBJ24_SA02 Short Answer 4 marks

Describe polarization by reflection and give an application.

Expected Answer

Reflected light can become partially polarized. Glare from horizontal surfaces is often polarized. Polarizing sunglasses reduce glare. Application correctly linked to reflection.

Mark Scheme
  • 1 mark: Reflected light can become partially polarized.
  • 1 mark: Glare from horizontal surfaces is often polarized.
  • 1 mark: Polarizing sunglasses reduce glare.
  • 1 mark: Application correctly linked to reflection.
Source: Original generated question
G9T3_OBJ24_SA03 Short Answer 4 marks

Why can light be polarized but sound in air cannot be polarized in the same way?

Expected Answer

Light is transverse. Polarization selects transverse vibration direction. Sound in air is longitudinal. Longitudinal vibration is along travel direction.

Mark Scheme
  • 1 mark: Light is transverse.
  • 1 mark: Polarization selects transverse vibration direction.
  • 1 mark: Sound in air is longitudinal.
  • 1 mark: Longitudinal vibration is along travel direction.
Source: Original generated question
G9T3_OBJ24_SA04 Short Answer 4 marks

Two polarizing filters are crossed at 90°. Describe the transmitted light.

Expected Answer

First filter polarizes the light. Second filter axis is perpendicular. Very little/no light is transmitted. Explanation refers to vibration direction.

Mark Scheme
  • 1 mark: First filter polarizes the light.
  • 1 mark: Second filter axis is perpendicular.
  • 1 mark: Very little/no light is transmitted.
  • 1 mark: Explanation refers to vibration direction.
Source: Original generated question

Light • pages 170

OBJ25: Apply Malus’s law to light filtered by polarizer and analyzer filters.

10 MCQs • 4 Short Answers
G9T3_OBJ25_MCQ01 MCQ Medium Identify • 4 marks

According to Malus’s law, when the angle between two polarizers is 90°, the transmitted intensity is approximately:

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
AHalf the original intensity
BAll original intensity
CZero
DFour times the intensity
Answer
C. Zero
Malus’s law: \(I = I_0 \cos^2\theta . At 90\)°, cos 90° = 0, so \(I = 0.\)
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ25_MCQ02 MCQ Hard Identify • 4 marks

Polarized light with initial intensity I₀ = 100 units passes through an analyzer at 0°. Calculate transmitted intensity.

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
A0 units
B100 units
C50 units
DNot applicable
Answer
B. 100 units
Malus’s law: \(I = I_0 \cos^2\theta \)= 100 × cos²(0°) = 100.
Source: Adapted from Abdul Elah Sheik Omar’s G 9 A Q-Bank
G9T3_OBJ25_MCQ03 MCQ Easy Define • 4 marks

Malus's law relates transmitted intensity through an analyzer to:

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
Athe object mass
Bthe angle between polarizer and analyzer axes
Cluminous flux only
Dthe speed of sound
Answer
B. the angle between polarizer and analyzer axes
Malus's law uses \(I_2 = I_1 \cos^2\theta \).
Source: Original generated question
G9T3_OBJ25_MCQ04 MCQ Medium Calculate • 4 marks

If polarized light of intensity I₁ passes through an analyzer at \(\theta \) = 0°, I₂ is:

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
AI₁
BI₁/2
C2I₁
D0
Answer
A. I₁
cos²0° = 1, so I₂ = I₁.
Source: Original generated question
G9T3_OBJ25_MCQ05 MCQ Medium Calculate • 4 marks

If polarized light passes through an analyzer at \(\theta \) = 90°, the transmitted intensity is:

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
A2I₁
BI₁/2
C0
DI₁
Answer
C. 0
cos²90° = 0.
Source: Original generated question
G9T3_OBJ25_MCQ06 MCQ Hard Calculate • 4 marks

If I₁ = 80 W/m² and \(\theta \) = 60°, what is I₂?

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
A20 W/m²
B40 W/m²
C80 W/m²
D60 W/m²
Answer
A. 20 W/m²
cos60° = 0.5, so cos²60° = 0.25; I₂ = 80×0.25 = 20.
Source: Original generated question
G9T3_OBJ25_MCQ07 MCQ Hard Calculate • 4 marks

If I₁ = 100 W/m² and \(\theta \) = 45°, what is I₂?

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
A100 W/m²
B50 W/m²
C25 W/m²
D0 W/m²
Answer
B. 50 W/m²
cos²45° = 0.5, so I₂ = 50 W/m².
Source: Original generated question
G9T3_OBJ25_MCQ08 MCQ Medium Recall • 4 marks

Unpolarized light of intensity I₀ passes through the first ideal polarizer. The intensity after the first polarizer is:

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
A2I₀
BI₀
CI₀/2
D0
Answer
C. I₀/2
An ideal polarizer transmits half of unpolarized light intensity.
Source: Original generated question
G9T3_OBJ25_MCQ09 MCQ Hard Calculate • 4 marks

Unpolarized light has I₀ = 120 W/m². After the first polarizer, then an analyzer at 60°, what is the final intensity?

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
A120 W/m²
B60 W/m²
C15 W/m²
D30 W/m²
Answer
C. 15 W/m²
After first polarizer I₁ = 60. Then I₂ = 60×cos²60° = 60×0.25 = 15.
Source: Original generated question
G9T3_OBJ25_MCQ10 MCQ Medium Define • 4 marks

In Malus's law, \(\theta \) is measured between:

Polarization and Malus's Law I₂ = I₁ cos²θ
Polarization
Athe sound wave and air molecules
Bthe transmission axes of the two filters
Cred and violet wavelengths
Dthe ray and the wall only
Answer
B. the transmission axes of the two filters
\(\theta \) is the angle between polarizing axes.
Source: Original generated question
G9T3_OBJ25_SA01 Short Answer 4 marks

State Malus's law and define the angle used in it.

Expected Answer

Writes I2 = I1 cos²θ. I1 is intensity after first polarizer/before analyzer. I2 is intensity after analyzer. \(\theta \) is angle between polarizing axes.

Mark Scheme
  • 1 mark: Writes I2 = I1 cos²θ.
  • 1 mark: I1 is intensity after first polarizer/before analyzer.
  • 1 mark: I2 is intensity after analyzer.
  • 1 mark: \(\theta \) is angle between polarizing axes.
Source: Original generated question
G9T3_OBJ25_SA02 Short Answer 4 marks

Polarized light of intensity 80 W/m² passes through an analyzer at 60°. Calculate final intensity.

Expected Answer

Uses I2 = I1 cos²θ. cos60° = 0.5. cos²60° = 0.25. I2 = 20 W/m².

Mark Scheme
  • 1 mark: Uses I2 = I1 cos²θ.
  • 1 mark: cos60° = 0.5.
  • 1 mark: cos²60° = 0.25.
  • 1 mark: I2 = 20 W/m².
Source: Original generated question
G9T3_OBJ25_SA03 Short Answer 4 marks

Unpolarized light of intensity 100 W/m² passes through a polarizer and then an analyzer at 45°. Find final intensity.

Expected Answer

After first polarizer \(I_1 = I_0/2\) = 50 W/m². Uses I2 = I1 cos²45°. cos²45° = 0.5. I2 = 25 W/m².

Mark Scheme
  • 1 mark: After first polarizer \(I_1 = I_0/2\) = 50 W/m².
  • 1 mark: Uses I2 = I1 cos²45°.
  • 1 mark: cos²45° = 0.5.
  • 1 mark: I2 = 25 W/m².
Source: Original generated question
G9T3_OBJ25_SA04 Short Answer 4 marks

Describe how transmitted intensity changes as analyzer angle changes from 0° to 90°.

Expected Answer

At 0°, intensity is maximum I1. As \(\theta \) increases, intensity follows cos²θ. At 90°, intensity is zero/minimum. Explanation linked to alignment of polarizing axes.

Mark Scheme
  • 1 mark: At 0°, intensity is maximum I1.
  • 1 mark: As \(\theta \) increases, intensity follows cos²θ.
  • 1 mark: At 90°, intensity is zero/minimum.
  • 1 mark: Explanation linked to alignment of polarizing axes.
Source: Original generated question