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Where required, take the speed of light = 3.0 × 108 m/s and the speed of sound in air = 340 m/s unless otherwise stated.
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Question 1 -- Sound Waves
Total: 12 marks
(a)[2]
State what is meant by a longitudinal wave. Give one example of a longitudinal wave.
Model Answer -- 1(a)
A longitudinal wave is a wave in which the oscillations / vibrations of the particles are parallel to the direction of energy transfer / wave travel [1]
Example: sound waves (accept: ultrasound, P-waves / seismic waves, compression waves in a spring) [1]
⚠ If you missed marks here: "The particles move backwards and forwards" leaves out what they move relative to, so the first mark goes — the vibrations must be stated as parallel to the direction of energy transfer. Offering light or water waves as the example scores nothing, since both are transverse.
Mark 1 -- correct definition (vibrations parallel to direction of energy transfer)
Mark 2 -- correct example of a longitudinal wave
(b)[3]
Explain the terms compression and rarefaction as they apply to sound waves. The diagram below shows a representation of a sound wave. Identify the compressions and rarefactions.
Model Answer -- 1(b)
A compression is a region in a longitudinal wave where the particles are pushed close together / the pressure is higher than normal [1]
A rarefaction is a region in a longitudinal wave where the particles are spread further apart / the pressure is lower than normal [1]
Clear diagram reference: compressions and rarefactions alternate along the wave, with compressions shown as closely packed particles and rarefactions as widely spaced particles [1]
⚠ If you missed marks here: "A compression is where the wave is squashed up" describes the drawing rather than the air — say the particles are closer together and the pressure is above normal. Calling a rarefaction "a gap with no particles" is wrong: the particles are still there, spread further apart at lower pressure.
Mark 1 -- correct explanation of compression (particles close together / high pressure)
Mark 2 -- correct explanation of rarefaction (particles spread apart / low pressure)
Mark 3 -- correctly identifies compression and rarefaction on diagram or describes alternating pattern
(c)[3]
A musician at a concert in Bangalore plays a note on a sitar. The sound wave has a frequency of 440 Hz and the speed of sound in air is 340 m/s.
Calculate the wavelength of the sound wave.
Model Answer -- 1(c)
v = f × λ, so λ = v / f
Correct formula stated or rearranged [1]
λ = 340 / 440
Correct substitution [1]
λ = 0.77 m (or 0.773 m)
Correct answer with unit: 0.77 m (accept 0.773 m) [1]
⚠ If you missed marks here: Dividing the wrong way, 440 / 340, gives 1.29 m instead of 0.77 m, and multiplying gives 149 600 m, which is absurd for a note you can hear across a room. Rearrange to λ = v / f before any numbers go in.
Mark 1 -- correct formula (λ = v / f)
Mark 2 -- correct substitution
Mark 3 -- correct answer with unit (0.77 m)
(d)[4]
Explain what happens to the pitch and loudness of a sound when:
(i) the frequency is increased
(ii) the amplitude is increased
A student says: "A louder sound travels faster than a quiet sound." State whether this is correct and explain your answer.
Model Answer -- 1(d)
(i) When the frequency is increased, the pitch increases (the sound becomes higher-pitched). The loudness does not change [1]
(ii) When the amplitude is increased, the loudness increases (the sound becomes louder). The pitch does not change [1]
The student's statement is incorrect [1]
The speed of sound in air depends on the medium (e.g. temperature and density of the air), not on the amplitude / loudness of the sound. A louder sound has greater amplitude but travels at the same speed as a quiet sound through the same medium [1]
⚠ If you missed marks here: Raising the frequency changes pitch only; writing "it gets louder" mixes the two and loses a mark, as does saying a bigger amplitude raises the pitch. The last two marks are separate: state that the student is incorrect, then give the reason — speed depends on the medium and its temperature, never on how loud the sound is.
Mark 1 -- increasing frequency increases pitch
Mark 2 -- increasing amplitude increases loudness
Mark 3 -- student's statement is incorrect
Mark 4 -- speed depends on medium, not amplitude/loudness
Question 2 -- Refraction and Dispersion of Light
Total: 11 marks
(a)[2]
State what happens to a ray of light when it passes from air into glass at an angle to the normal.
Model Answer -- 2(a)
The ray of light bends / changes direction (refracts) towards the normal [1]
The ray slows down as it enters the glass (glass is optically denser than air) [1]
⚠ If you missed marks here: "Bends away from the normal" describes light leaving glass, not entering it, and scores zero here — on the way into the denser medium the ray bends towards the normal. The second mark is for the speed decreasing; "light speeds up in glass" is the same misconception the other way round.
Mark 1 -- bends towards the normal
Mark 2 -- slows down / glass is optically denser
(b)[3]
A ray of light enters a glass prism with an angle of incidence of 50°. The refractive index of the glass is 1.52.
Calculate the angle of refraction.
Model Answer -- 2(b)
n = sin i / sin r, so sin r = sin i / n
Correct formula stated or rearranged [1]
sin r = sin 50° / 1.52 = 0.7660 / 1.52 = 0.5039
Correct substitution [1]
r = sin−1(0.5039) = 30.3°
Correct answer: r = 30.3° (accept 30°) [1]
⚠ If you missed marks here: Multiplying instead of dividing gives sin r = sin 50° × 1.52 = 1.16, and since a sine can never exceed 1 that result is itself telling you the rearrangement is wrong. Writing 0.5039 as the final answer forgets the inverse sine that turns it into 30.3°.
Mark 1 -- correct formula (n = sin i / sin r)
Mark 2 -- correct substitution
Mark 3 -- correct answer (30.3° or 30°)
(c)[3]
Explain what is meant by the dispersion of white light. State the colours seen and the order in which they appear.
Model Answer -- 2(c)
Dispersion is the splitting of white light into its component colours / into a spectrum when it passes through a prism [1]
This occurs because different colours (wavelengths) of light are refracted by different amounts. Violet light is refracted the most and red light the least [1]
The order of colours is: Red, Orange, Yellow, Green, Blue, Indigo, Violet (ROYGBIV) -- red is deviated least, violet is deviated most [1]
⚠ If you missed marks here: Saying red is refracted most reverses the spectrum — red has the longest wavelength and is deviated least, violet most. The prism does not add colour: the white light already contains every colour, and the mark is for separating them, not for creating them.
Mark 1 -- dispersion is splitting of white light into spectrum
Mark 2 -- different wavelengths refracted by different amounts
Mark 3 -- correct order of colours (ROYGBIV)
(d)[3]
Draw a labelled diagram showing a ray of white light entering a triangular glass prism and being dispersed into a spectrum. Label the colours at each end of the spectrum.
Model Answer -- 2(d)
Correct triangular prism shape drawn with incident ray of white light entering one face [1]
Spectrum of colours shown emerging from the other face, spread into a fan / band [1]
Red labelled at one end (least deviated) and violet at the other end (most deviated) [1]
⚠ If you missed marks here: A diagram showing the colours splitting only at the first face misses the mark — the fan widens at both faces and the emerging rays must diverge, not run parallel. Both ends need labelling, red where the deviation is least and violet where it is greatest; an unlabelled rainbow scores no more than one.
Mark 1 -- correct prism shape with incident white light ray
Mark 2 -- spectrum of colours emerging and spreading out
Mark 3 -- red and violet correctly labelled at each end
Question 3 -- Waves in a Ripple Tank
Total: 11 marks
(a)[2]
Define the amplitude of a wave.
Model Answer -- 3(a)
The amplitude is the maximum displacement of a point on the wave from its rest position / equilibrium position [1]
It is measured from the undisturbed position to the crest (or trough), not from crest to trough [1]
⚠ If you missed marks here: Measuring from crest to trough gives double the amplitude and loses the mark. Amplitude runs from the undisturbed (rest) position to the crest, and it is a maximum displacement, not a distance measured along the wave — that is wavelength.
Mark 1 -- maximum displacement from rest/equilibrium position
Mark 2 -- measured from rest position to crest (not crest to trough)
(b)[3]
Describe an experiment using a ripple tank to demonstrate diffraction. Include details of the apparatus and what is observed.
Model Answer -- 3(b)
Set up a ripple tank with a straight vibrating bar to produce plane (straight) wavefronts. Place two barriers in the water with a gap between them. The gap width should be similar to (or smaller than) the wavelength of the waves [1]
The plane waves approach the gap and pass through it. On the other side, the waves spread out (diffract) and become curved / circular wavefronts. This spreading is called diffraction [1]
A lamp above the tank projects the wave pattern onto a white screen below. Diffraction is most noticeable when the gap width is approximately equal to the wavelength. If the gap is much wider than the wavelength, there is very little spreading [1]
⚠ If you missed marks here: A glass plate creating a shallow region belongs to the refraction experiment and earns nothing here; diffraction needs two barriers with a gap between them. The gap has to be described as about the same width as the wavelength, otherwise the waves pass almost straight through and there is nothing to observe.
Mark 1 -- correct apparatus: ripple tank, straight bar, barriers with gap
Mark 2 -- waves spread out / become circular after passing through gap
Mark 3 -- most diffraction when gap width similar to wavelength / observation method
(c)[3]
In a ripple tank experiment, water waves of wavelength 2.0 cm travel at 0.30 m/s in deep water. When they enter shallow water, their speed decreases to 0.20 m/s.
Calculate the new wavelength in the shallow region.
Model Answer -- 3(c)
First, find the frequency (which stays constant):
f = v / λ = 0.30 / 0.020 = 15 Hz
Correct calculation of frequency: f = 15 Hz [1]
Then, find the new wavelength in shallow water:
λ = v / f = 0.20 / 15
Correct substitution using new speed and same frequency [1]
λ = 0.013 m (or 1.3 cm)
Correct answer with unit: λ = 0.013 m (accept 1.3 cm) [1]
⚠ If you missed marks here: The wavelength is 2.0 cm, so it enters the formula as 0.020 m — using 2.0 gives f = 0.15 Hz and a final answer of 1.3 m instead of 0.013 m. The whole method rests on the frequency being the same in both regions; recalculating a new frequency from the slower speed leaves you with nothing to work with.
Mark 1 -- correct frequency calculation (f = 15 Hz)
Mark 2 -- correct substitution using new speed and same frequency
Mark 3 -- correct answer with unit (0.013 m or 1.3 cm)
(d)[3]
Explain, using the concept of wavefronts, why waves change direction when they travel from deep water to shallow water at an angle.
Model Answer -- 3(d)
When plane wavefronts approach the boundary between deep and shallow water at an angle, one end of each wavefront reaches the shallow water first [1]
The part of the wavefront that enters the shallow water first slows down, while the part still in deep water continues at the original (faster) speed. This causes the wavefront to pivot / rotate [1]
As a result, the direction of travel of the wave changes -- the wave bends towards the normal when entering the shallower (slower) region. The wavelength also decreases in shallow water but the frequency remains the same [1]
⚠ If you missed marks here: "The wave slows down so it bends" earns only the middle idea — the marks want wavefronts: one end reaches the shallow water first, that end slows while the other end carries on, and the wavefront swings round. Adding that the frequency falls contradicts the physics; the source sets the frequency, so only speed and wavelength change.
Mark 1 -- one end of wavefront enters shallow water first
Mark 2 -- that end slows down while other end continues faster, causing wavefront to pivot
Mark 3 -- wave bends towards normal / wavelength decreases / frequency constant
Question 4 -- Critical Angle and Total Internal Reflection
Total: 12 marks
(a)[2]
State what is meant by the critical angle.
Model Answer -- 4(a)
The critical angle is the angle of incidence (in the optically denser medium) for which the angle of refraction is exactly 90° [1]
At angles of incidence greater than the critical angle, total internal reflection occurs (the light is completely reflected back into the denser medium) [1]
⚠ If you missed marks here: Defining it as "the angle at which light is totally internally reflected" is circular and picks up one mark at best — the definition needs an angle of refraction of exactly 90°. Say too that the angle of incidence is measured in the denser medium, or the definition is incomplete.
Mark 1 -- angle of incidence where angle of refraction is 90°
Mark 2 -- above this angle, total internal reflection occurs
(b)[3]
A ray of light travels from water (refractive index n = 1.33) into air.
Calculate the critical angle for the water-air boundary.
Model Answer -- 4(b)
sin c = 1 / n
Correct formula stated [1]
sin c = 1 / 1.33 = 0.7519
Correct substitution [1]
c = sin−1(0.7519) = 48.8°
Correct answer: c = 48.8° (accept 48.7° to 48.8°) [1]
⚠ If you missed marks here: Typing sin−1(1.33) produces a calculator error, which is the sign that you inverted n too late — sin c = 1 / 1.33 = 0.7519 comes first, then the inverse sine gives 48.8°. An answer near 42° means you used glass (n = 1.5) instead of water.
Mark 1 -- correct formula (sin c = 1/n)
Mark 2 -- correct substitution
Mark 3 -- correct answer (48.8°)
(c)[4]
Draw a ray diagram showing three rays of light hitting a water-air boundary from inside the water:
(i) one at an angle of incidence less than the critical angle
(ii) one at the critical angle
(iii) one at an angle of incidence greater than the critical angle
Label each case clearly.
Model Answer -- 4(c)
Case (i): Ray at angle below critical angle -- both a refracted ray (bending away from normal into air) and a weak partial reflection shown [1]
Case (ii): Ray at the critical angle -- refracted ray travels along the boundary surface (angle of refraction = 90°) [1]
Case (iii): Ray above critical angle -- total internal reflection occurs, ray reflects back into water obeying law of reflection (angle of incidence = angle of reflection). No refracted ray [1]
All three cases correctly labelled with normal lines drawn at the point of incidence [1]
⚠ If you missed marks here: In case (i) the refracted ray bends away from the normal, because the light is leaving the denser medium; drawing it towards the normal loses that mark. Case (ii) is drawn wrong more often than any other — the refracted ray runs along the water surface, 90° from the normal, not straight up out of it. Case (iii) must show no refracted ray at all.
Mark 1 -- Case (i) correct: refracted ray bending away from normal
Mark 2 -- Case (ii) correct: refracted ray along surface at 90°
Mark 3 -- Case (iii) correct: total internal reflection, no refracted ray
Mark 4 -- all cases labelled with normals drawn
(d)[3]
Describe how total internal reflection is used in a periscope that uses prisms instead of mirrors. State one advantage of using prisms over mirrors.
Model Answer -- 4(d)
A prism periscope uses two right-angled (45°-45°-90°) glass prisms. Light enters the top prism through one face and hits the hypotenuse (longest face) at 45°, which is greater than the critical angle for glass (approximately 42°) [1]
Total internal reflection occurs at the hypotenuse, turning the light through 90° downwards. The light then enters the bottom prism and is again totally internally reflected at the hypotenuse, turning 90° towards the observer's eye [1]
Advantage of prisms over mirrors: Prisms produce a brighter / clearer image because 100% of the light is reflected (no energy is absorbed). Mirrors absorb some light and can also tarnish / degrade over time, reducing image quality [1]
⚠ If you missed marks here: "The prisms act like mirrors" is exactly the answer the question is testing against — the light is turned by total internal reflection, and the mark needs 45° set against the critical angle of about 42° for glass. "Prisms are cheaper" or "stronger" is not the advantage; it is that no light is lost, so the image is brighter, and prisms do not tarnish.
Mark 1 -- two right-angled prisms, light hits hypotenuse at 45° which exceeds critical angle
Mark 2 -- TIR at each prism turns light 90°, directing it to observer
Distinguish between a real image and a virtual image.
Model Answer -- 5(a)
A real image is formed where light rays actually converge / meet. It can be projected onto a screen [1]
A virtual image is formed where light rays appear to come from (but do not actually converge). It cannot be projected onto a screen and can only be seen by looking through the lens / mirror [1]
⚠ If you missed marks here: "A real image is upside down and a virtual image is the right way up" gives properties instead of definitions and scores nothing — the distinction is whether the rays actually meet. Real: the rays converge and the image can be caught on a screen. Virtual: they only appear to come from that point, so no screen will show it.
Mark 1 -- real image: rays actually converge / can be projected on screen
Mark 2 -- virtual image: rays appear to come from / cannot be projected on screen
(b)[4]
An object 3.0 cm tall is placed 30 cm from a converging lens of focal length 20 cm.
Draw a ray diagram to locate the image. State the nature, position, orientation, and size of the image.
Model Answer -- 5(b)
Correct ray diagram with at least two correctly drawn rays: (1) ray parallel to principal axis refracts through focal point on other side; (2) ray through optical centre passes straight through [1]
The object at 30 cm lies between F (20 cm) and 2F (40 cm), so the two construction rays meet beyond 2F on the opposite side of the lens. Reading the position off a diagram drawn to scale gives about 60 cm from the lens [1]
Measuring the image on the scale diagram, magnification = image height / object height = 2, so the image is about 6.0 cm tall [1]
Image is: real, inverted, magnified, on the opposite side of the lens to the object, beyond 2F [1]
⚠ If you missed marks here: There is no formula to fall back on in 0625 — the image position is read off a diagram drawn to scale, so a bare number with no construction rays earns nothing. The key step is placing the object at 30 cm, between F (20 cm) and 2F (40 cm); reading 30 cm merely as "further than the focal length" leads to "diminished", when the image is magnified, about 6.0 cm tall and beyond 2F.
Mark 1 -- correct ray diagram with at least two construction rays
Mark 2 -- image located beyond 2F, about 60 cm from the lens
Mark 3 -- image height = 6.0 cm (magnification = 2)
Mark 4 -- image described as real, inverted, magnified
(c)[3]
A diverging lens always produces a certain type of image regardless of where the object is placed.
State the type of image and three properties of this image.
Model Answer -- 5(c)
A diverging lens always produces a virtual image [1]
The image is always upright (erect / the same way up as the object) [1]
The image is always diminished (smaller than the object) and located on the same side of the lens as the object, between the lens and the focal point [1]
⚠ If you missed marks here: "It depends where the object is placed" contradicts the question and scores nothing — a diverging lens gives the same kind of image every time. The commonest wrong answer is "real and inverted"; it is virtual, upright and diminished, and the third mark is safe only if you also place it on the same side as the object.
Mark 1 -- virtual image
Mark 2 -- upright / erect
Mark 3 -- diminished / smaller than object
(d)[3]
Explain how a magnifying glass works. State where the object must be placed relative to the focal point and describe the image produced.
Model Answer -- 5(d)
A magnifying glass is a converging (convex) lens. The object must be placed between the focal point (F) and the lens (i.e. closer to the lens than the focal length) [1]
When the object is in this position, the lens produces an image that is virtual, upright, and magnified (larger than the object) [1]
The image appears on the same side of the lens as the object. The eye looks through the lens and sees the magnified virtual image. The closer the object is to the focal point, the larger the magnification [1]
⚠ If you missed marks here: Placing the object between F and 2F is the standard error — that arrangement makes a real, inverted image on a screen, which is a projector, not a magnifying glass. The object goes inside the focal length, and the image is then virtual, upright and magnified on the same side of the lens.
Mark 1 -- object placed between F and the lens
Mark 2 -- image is virtual, upright, and magnified
Mark 3 -- image on same side as object / closer to F gives greater magnification
Question 6 -- Electromagnetic Spectrum and Communication
Total: 12 marks
(a)[3]
Name three types of electromagnetic radiation that have a shorter wavelength than visible light. For each, state one practical use.
Model Answer -- 6(a)
Ultraviolet (UV) -- used for sterilisation / detecting forged banknotes / causing fluorescence / treating skin conditions [1]
X-rays -- used for medical imaging (seeing bones / detecting fractures) / airport security scanning [1]
Gamma rays -- used for sterilising medical equipment / treating cancer (radiotherapy) / detecting cracks in metal [1]
⚠ If you missed marks here: Naming infrared or microwaves loses the mark outright: both have wavelengths longer than visible light, so only ultraviolet, X-rays and gamma rays qualify. Every mark also carries a use, and "used in hospitals" is too vague — name the job, such as imaging bone or sterilising equipment.
Mark 1 -- UV named with a correct use
Mark 2 -- X-rays named with a correct use
Mark 3 -- Gamma rays named with a correct use
(b)[3]
Explain how satellite communication works. Include the role of microwaves and geostationary satellites in your answer. You may refer to India's satellite communication network.
Model Answer -- 6(b)
A ground station (transmitter) sends a signal using microwaves up to a communication satellite in orbit. Microwaves are used because they can pass through the atmosphere with minimal absorption [1]
The satellite receives the signal, amplifies it, and retransmits it back down to a receiving station on a different part of the Earth's surface. India's ISRO operates the GSAT / INSAT series of satellites for broadcasting and telecommunications across the subcontinent [1]
Geostationary satellites orbit above the equator at a height of approximately 36,000 km. They have an orbital period of 24 hours, so they remain above the same point on Earth's surface. This means ground-based dish antennas do not need to track the satellite -- they stay pointed in a fixed direction [1]
⚠ If you missed marks here: "The satellite reflects the signal back down" is not what happens and costs the second mark: it receives the signal, amplifies it and retransmits it, usually on a different frequency. "Geostationary means the satellite stays still" is wrong too — it orbits above the equator with a period of 24 hours, which is why the dish can stay pointed in one fixed direction.
Mark 1 -- microwaves sent from ground station to satellite / pass through atmosphere
Mark 2 -- satellite amplifies and retransmits to receiving station
Mark 3 -- geostationary: 24-hour orbit, stays above same point, fixed dish
(c)[3]
A radio station broadcasts at a frequency of 100 MHz.
Calculate the wavelength of the radio waves. (Speed of all electromagnetic waves = 3.0 × 108 m/s)
Model Answer -- 6(c)
λ = v / f
Correct formula stated [1]
λ = 3.0 × 108 / 100 × 106
Correct substitution (converting 100 MHz to 1.0 × 108 Hz or equivalent) [1]
λ = 3.0 m
Correct answer with unit: λ = 3.0 m [1]
⚠ If you missed marks here: Dividing by 100 rather than by 1.0 × 108 Hz gives 3.0 × 106 m, a wavelength longer than India — convert the megahertz first. Note also that the speed to use is 3.0 × 108 m/s for every part of the spectrum; slipping in the speed of sound wrecks the whole answer.
Mark 1 -- correct formula (λ = v / f)
Mark 2 -- correct substitution
Mark 3 -- correct answer (3.0 m)
(d)[3]
State one danger of each of the following types of electromagnetic radiation, and for each state one precaution that can be taken to reduce the risk:
(i) X-rays
(ii) Gamma rays
(iii) Microwaves
Model Answer -- 6(d)
(i) X-rays: Danger -- can cause cell damage / mutations / cancer with prolonged or excessive exposure. Precaution -- radiographers stand behind lead screens / wear lead aprons / limit exposure time / monitor exposure with film badges [1]
(ii) Gamma rays: Danger -- can cause cell death / cancer / radiation sickness (most penetrating form of EM radiation). Precaution -- use thick lead or concrete shielding / handle sources with tongs / keep exposure time short / keep maximum distance from source [1]
(iii) Microwaves: Danger -- can cause internal heating of body tissue (especially water-containing tissue). Precaution -- microwave ovens have metal mesh screens in the door to prevent leakage / maintain safe distance from transmitters [1]
⚠ If you missed marks here: Every mark needs a danger and a matching precaution, so listing three hazards and stopping scores one at best. "Microwaves cause cancer" is wrong physics: they are non-ionising, and the hazard is internal heating of tissue. For gamma rays a lead apron is too weak — thick lead or concrete shielding, distance and short exposure are the credited answers.
Mark 1 -- X-rays: correct danger and precaution
Mark 2 -- Gamma rays: correct danger and precaution
Mark 3 -- Microwaves: correct danger and precaution
Question 7 -- Mixed Waves Application
Total: 10 marks
(a)[2]
State two uses of ultrasound.
Model Answer -- 7(a)
Medical imaging / pre-natal scanning -- used to produce images of a baby in the womb (safe because no ionising radiation) [1]
Industrial flaw detection -- used to detect cracks or defects inside metal structures / castings without cutting them open. (Also accept: sonar / echo sounding to measure depth of water, cleaning delicate equipment such as jewellery, physiotherapy, breaking kidney stones) [1]
⚠ If you missed marks here: Giving "sonar" and "measuring the depth of the sea" is one use written twice and earns one mark — choose two different fields, such as pre-natal scanning and finding cracks inside metal. If you mention the scan, say why ultrasound is chosen: it is not ionising, unlike X-rays.
Mark 1 -- first correct use of ultrasound
Mark 2 -- second correct use of ultrasound (different from first)
(b)[3]
An ultrasound scanner is used in a hospital in Bangalore to examine a patient. A pulse is sent into the body and reflects off an organ 0.08 m below the skin surface. The speed of ultrasound in body tissue is 1540 m/s.
Calculate the time between sending the pulse and receiving the echo.
Model Answer -- 7(b)
The pulse travels to the organ and back, so the total distance is:
total distance = 2 × 0.08 = 0.16 m
Correct total distance (there and back): 0.16 m [1]
time = distance / speed
Correct formula [1]
time = 0.16 / 1540 = 1.04 × 10−4 s
Correct answer: t = 1.04 × 10−4 s (accept 0.000104 s or 0.104 ms) [1]
⚠ If you missed marks here: Leaving out the doubling gives 5.2 × 10−5 s instead of 1.04 × 10−4 s — the pulse has to return before it can be detected. Dividing the wrong way, 1540 / 0.16, gives 9625 and no sensible unit; time = distance / speed.
Mark 1 -- total distance = 2 × 0.08 = 0.16 m
Mark 2 -- correct formula (time = distance / speed)
Mark 3 -- correct answer (1.04 × 10−4 s)
(c)[2]
State two differences between ultrasound and infrasound.
Model Answer -- 7(c)
Ultrasound has a frequency above 20,000 Hz (above the upper limit of human hearing), while infrasound has a frequency below 20 Hz (below the lower limit of human hearing) [1]
Ultrasound has a very short wavelength (good for imaging / detecting small objects), while infrasound has a very long wavelength (can travel long distances / through the Earth). Neither can be heard by humans [1]
⚠ If you missed marks here: "Ultrasound is high and infrasound is low" has the idea but no numbers, and the marks expect the boundaries: above 20 000 Hz and below 20 Hz. Do not claim infrasound cannot travel far — its very long wavelength is the reason it carries over great distances.
Mark 1 -- ultrasound > 20 kHz, infrasound < 20 Hz
Mark 2 -- second valid difference (wavelength / uses / detection)
(d)[3]
A wave has a speed of 12 m/s and a period of 0.50 s.
Calculate:
(i) the frequency of the wave
(ii) the wavelength of the wave
Model Answer -- 7(d)
(i) Frequency:
f = 1 / T = 1 / 0.50
f = 2.0 Hz
Correct answer: f = 2.0 Hz [1]
(ii) Wavelength:
λ = v / f = 12 / 2.0
Correct substitution [1]
λ = 6.0 m
Correct answer with unit: λ = 6.0 m [1]
⚠ If you missed marks here: Feeding 0.50 straight in as a frequency is the usual slip: 0.50 s is the period, so f = 1 / 0.50 = 2.0 Hz, and using 0.50 Hz would give λ = 24 m instead of 6.0 m. Both parts must end with a unit — "6.0" on its own does not take the final mark.
Mark 1 -- correct frequency (f = 2.0 Hz)
Mark 2 -- correct substitution for wavelength
Mark 3 -- correct wavelength (6.0 m)
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