How many SCSA Physics questions cover Science Understanding?

AusGrader has 577 SCSA Physics questions on Science Understanding, all with instant AI grading and detailed marking feedback.

WACE Physics — Science Understanding Questions & Answers

Q12

2025

VCAA

7 marks

Q12

Denzil is using a demonstration hand-cranked generator.

A schematic diagram of the generator is shown in Figure 17. The generator contains a rectangular coil with side lengths of 5.0 cm and 2.5 cm, consisting of 20 turns of insulated copper wire. The coil is rotated between two bar magnets that provide a field strength of 0.60 T between the magnets.

Denzil rotates the coil at a frequency of 50 Hz.

Q12a

1 mark

State why the flux through the coil changes as the coil rotates.

Reveal Answer

The flux through the coil is determined by the angle between the plane of the coil and the magnetic field.

Marking Criteria

Descriptor	Marks
States that the angle or orientation between the plane of the coil and the magnetic field changes	1

Q12b

1 mark

Show that the change in flux as the coil rotates from a horizontal to a vertical position is $7.5 \times 10^{-4} \text{ Wb}$ .

Reveal Answer

$\Delta\phi = BA$

$\Delta\phi = 0.60 \times (0.05 \times 0.025)$

$\Delta\phi = 7.5 \times 10^{-4} \text{ Wb}$

Marking Criteria

Descriptor	Marks
Demonstrates correct substitution into the magnetic flux formula (e.g., $\Delta\phi = 0.60 \times (0.05 \times 0.025)$ )	1

Q12c

3 marks

Calculate the average EMF induced as the coil is rotated through a quarter turn from a horizontal to a vertical position.

Reveal Answer

$\varepsilon = n\frac{\Delta\phi}{\Delta t}$

$\varepsilon = 20 \times \frac{7.5 \times 10^{-4}}{0.0050}$

$\varepsilon = 3.0 \text{ V}$

Marking Criteria

Descriptor	Marks
Calculates the correct time for a quarter turn ( $\Delta t = 0.0050 \text{ s}$ )	1
Substitutes values correctly into Faraday's law formula ( $\varepsilon = 20 \times \frac{7.5 \times 10^{-4}}{0.0050}$ )	1
Calculates the correct average EMF of $3.0 \text{ V}$	1

Q12d

2 marks

State a change to the set-up in Figure 17 that could produce a DC output from the generator. Give a reason for your choice.

Reveal Answer

Replace the slip rings with a split-ring commutator. The split-ring commutator will reverse the connections to the loop every half turn to ensure a DC output.

Marking Criteria

Descriptor	Marks
Identifies replacing the slip rings with a split-ring commutator	1
Explains that the split-ring commutator reverses the connections to the loop every half turn to ensure a DC output	1

Q33

2023

NESA

9 marks

Q33

9 marks

Consider the following statement.

"The interaction of subatomic particles with fields, as well as with other types of particles and matter, has increased our understanding of processes that occur in the physical world and of the properties of the subatomic particles themselves."

Justify this statement with reference to observations that have been made and experiments that scientists have carried out.

Reveal Answer

Marking Criteria

Descriptor	Marks
Provides a reasoned detailed justification for the statement with explanation referencing at least TWO observations and at least TWO experiments	9
Provides a justification for the statement with explanation referencing at least TWO observations and at least TWO experiments	8
The student response meets all criteria of the 6-mark band, and additionally meets the majority of criteria in the 8-mark band.	7
Provides a justification for the statement with an explanation of an observation/experiment of particle–field interactions and an observation/experiment of particle–particle interactions	6
The student response meets all criteria of the 4-mark band, and additionally meets the majority of criteria in the 6-mark band.	5
Provides details of TWO experiments or observations and how they relate to the statement	4
The student response meets all criteria of the 2-mark band, and additionally meets the majority of criteria in the 4-mark band.	3
Relates the statement to an experiment or observation OR Describes relevant experiments or observations	2
Provides some relevant information	1
None of the above	0

Q10

2023

SCSA

4 marks

Q10

4 marks

Estimate the de Broglie wavelength for a standard men's basketball travelling at 10.0 m s $^{-1}$ .

Reveal Answer

Using de Broglie's equation,

\begin{align*} \lambda &= \frac{h}{mv} \end{align*}

Taking the mass of a standard men's basketball as approximately

\begin{align*} m &= 0.62\ \text{kg} \end{align*}

then

\begin{align*} \lambda &= \frac{6.63\times10^{-34}}{(0.62)(10.0)} \\ &= 1.07\times10^{-34}\ \text{m} \end{align*}

So the estimated de Broglie wavelength is approximately

\begin{align*} \lambda &\approx 1.1\times10^{-34}\ \text{m} \end{align*}

Marking Criteria

Descriptor	Marks
Estimates mass of basketball	1
Substitutes $mv$ for $p$ in equation (using 0.60 kg)	1
Calculates answer	1
2 significant figures	1

Q4

2024

SCSA

3 marks

Q4

3 marks

According to the Big Bang theory, the strong nuclear force separated from the electromagnetic and weak forces around $10^{-36}$ s after the expansion of the universe began.

Explain how this separation enabled the formation of hadrons.

Reveal Answer

The strong nuclear force is mediated by the exchange of gluons. Quarks could now exchange gluons and interact, and hadrons are formed by quarks exchanging gluons and binding together.

Marking Criteria

Descriptor	Marks
States that the strong nuclear force is mediated by the exchange of gluons	1
Describes that quarks could now exchange gluons and interact	1
Explains that hadrons are formed by quarks exchanging gluons and binding together	1

Q13

2022

QCAA

Paper 1

1 mark

Q13

1 mark

A rectangular coil of 3000 turns and dimensions $0.1 \text{ m} \times 0.2 \text{ m}$ is rotated in a uniform magnetic field of 2 mT.
Calculate the minimum number of revolutions per second required to produce an average EMF of 6 V.

A

1

B

3

C

13

D

50

Reveal Answer

A

1

This rotation speed is too slow. Using the formula $\varepsilon_{avg} = 4NBAf$ , a frequency of $1 \text{ rev/s}$ would only produce an average EMF of $0.48 \text{ V}$ .

B

3

This value is insufficient. Substituting $f=3$ into the average EMF equation yields approximately $1.44 \text{ V}$ , which is less than the required $6 \text{ V}$ .

C

13

Correct Answer

The average EMF for a rotating coil is $\varepsilon_{avg} = 4NBAf$ . Solving for frequency: $f = \frac{6}{4(3000)(2 \times 10^{-3})(0.02)} = 12.5 \text{ rev/s}$ . The closest integer option is 13.

D

50

This frequency is too high. At $50 \text{ rev/s}$ , the generated average EMF would be $24 \text{ V}$ , far exceeding the required $6 \text{ V}$ .

Q5

2021

QCAA

Paper 2

4 marks

Q5

4 marks

An alpha particle with a charge of $+3.2 \times 10^{-19}$ C moves through an electric field, accelerating from rest through a potential difference of 240 V.

Determine the velocity of the particle at the end of its acceleration, expressing your answer in scientific notation. (m/s to 2 significant figures)

Reveal Answer

The change in potential energy of an electric charge moving through an electric field is equivalent to the work done on the charge.
$V = \frac{\Delta U}{q}$
$\Delta U = Vq$
$= 240 \times 3.2 \times 10^{-19}$
$= 7.68 \times 10^{-17} J = W$

The work done on an object is equal to the change in kinetic energy.
$E_k = \frac{1}{2}mv^2$
$7.68 \times 10^{-17} = \frac{1}{2} \times 6.64 \times 10^{-27} \times v^2$
$v^2 = \frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}$
$v = \sqrt{\frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}}$
Velocity = $1.5 \times 10^5 \text{ m s}^{-1}$ (to 2 significant figures)

Marking Criteria

Descriptor	Marks
Recognises the scenario relates to work done on a moving charge in an electric field	1
Identifies that work done on the charge equates to its kinetic energy	1
Provides appropriate mathematical reasoning	1
Determines the velocity	1

Q5

2023

QCAA

Paper 2

4 marks

Q5

4 marks

Describe what happens when light is shone onto a metallic surface in the context of the photoelectric effect.

Reveal Answer

Light with energy equivalent to $hf$ has the ability to produce photoelectrons from a metallic surface.

When the frequency of light is below the threshold frequency for the metallic surface, the light will be reflected with no transfer of energy.

When the frequency of light is above the threshold frequency for the metallic surface, the energy of the photons will be absorbed and photoelectrons with kinetic energy proportional to the excess energy will be released.

The intensity of incident light is proportional to the number of photoelectrons for frequencies greater than the threshold frequency.

Marking Criteria

Descriptor	Marks
Identifies incident light has energy equivalent to $hf$	1
Describes transfer of energy when frequency of light is below the threshold frequency	1
Describes transfer of energy when frequency of light is above the threshold frequency	1
Identifies relationship between intensity of incident light and resultant photoelectrons	1

Q7

2021

VCAA

7 marks

Q7

The generator of an electrical power plant delivers $500 \text{ MW}$ to external transmission lines when operating at $25 \text{ kV}$ . The generator's voltage is stepped up to $500 \text{ kV}$ for transmission and stepped down to $240 \text{ V}$ $100 \text{ km}$ away (for domestic use). The overhead transmission lines have a total resistance of $30.0 \text{ } \Omega$ . Assume that all transformers are ideal.

Q7a

2 marks

Explain why the voltage is stepped up for transmission along the overhead transmission lines.

Reveal Answer

Students were required to identify that stepping up the voltage allowed the current to be reduced while maintaining constant power. The reason for reducing the current is that the power lost is related to the transmission current by: $P = I^2R$ .

Marking Criteria

Descriptor	Marks
Identifies that stepping up the voltage allows the current to be reduced while maintaining constant power	1
Relates the reduced current to a reduction in power lost during transmission ( $P = I^2R$ )	1

Q7b

2 marks

Calculate the current in the overhead transmission lines. Show your working.

Reveal Answer

$P = VI$
$I = \frac{500 \times 10^6}{500 \times 10^3}$
$I = 1000 \text{ A or } 1.0 \text{ kA}$

Marking Criteria

Descriptor	Marks
Correct substitution into $P = VI$	1
Correct final answer of $1000 \text{ A}$ or $1.0 \text{ kA}$	1

Q7c

3 marks

Determine the maximum power available for domestic use at $240 \text{ V}$ . Show all your working.

Reveal Answer

This solution has two steps. The first is to calculate the power lost:
$P = I^2R$
$P = 1000^2 \times 30$
$P = 30 \times 10^6 \text{ W (30 MW)}$
This was then subtracted from the power delivered by the generator:
$P_{avail} = 500 \times 10^6 - 30 \times 10^6$
$P_{avail} = 470 \text{ MW}$

Marking Criteria

Descriptor	Marks
Calculates the power lost in the lines ( $30 \text{ MW}$ )	1
Subtracts the power lost from the total power delivered by the generator	1
Calculates the correct available power ( $470 \text{ MW}$ )	1

Q13

2023

NESA

1 mark

Q13

1 mark

Nucleus X has a greater binding energy than nucleus Y.

What can be deduced about X and Y?

A

X is more stable than Y.

B

Y is more stable than X.

C

X has a greater mass defect than Y.

D

Y has a greater mass defect than X.

Reveal Answer

A

X is more stable than Y.

Stability is determined by binding energy per nucleon, not total binding energy, so we cannot determine which nucleus is more stable without knowing their nucleon numbers.

B

Y is more stable than X.

Stability depends on binding energy per nucleon, which cannot be determined from total binding energy alone.

C

X has a greater mass defect than Y.

Correct Answer

Binding energy is directly proportional to mass defect according to the mass-energy equivalence equation $E = \Delta m c^2$ . Therefore, a greater binding energy means a greater mass defect.

D

Y has a greater mass defect than X.

Since nucleus X has a greater binding energy, it must have a greater mass defect than Y, not the other way around.

Q6

2022

QCAA

Paper 1

1 mark

Q6

1 mark

After coherent light has been passed through a double slit, the observation of an interference pattern on a screen is explained by the

A

wave nature of light.

B

equal width of the slits.

C

discrete packets of photons.

D

distance from the slits to the screen.

Reveal Answer

A

wave nature of light.

Correct Answer

Interference is a fundamental property of waves where overlapping wavefronts add constructively or destructively; this experiment is the classic evidence for the wave nature of light.

B

equal width of the slits.

While slit width affects the diffraction envelope and contrast, the interference pattern itself is caused by wave superposition, which can occur even if the slits are not perfectly equal in width.

C

discrete packets of photons.

Discrete packets (photons) refer to the particle nature of light; if light behaved strictly as classical particles, it would form two distinct bands rather than an interference pattern.

D

distance from the slits to the screen.

The distance to the screen affects the spacing of the fringes (scale), but it is not the fundamental cause of the interference phenomenon itself.

Q10

2024

QCAA

Paper 1

1 mark

Q10

1 mark

An experiment was conducted to determine the force experienced by an 85 cm wire with a 2.4 A current flowing through it in an external magnetic field. It was rotated through varying angles within the magnetic field such that data analysis identified the relationship $F = 0.0306 \sin \theta$ .

What is the order of magnitude of the strength of the external magnetic field?

A

$10^{-4}$ T

B

$10^{-2}$ T

C

$10^{2}$ T

D

$10^{4}$ T

Reveal Answer

A

$10^{-4}$ T

This value is too small. The calculated magnetic field strength is approximately $0.015 \text{ T}$ , which is two orders of magnitude larger than $10^{-4} \text{ T}$ .

B

$10^{-2}$ T

Correct Answer

Comparing the experimental relationship $F = 0.0306 \sin \theta$ to the theoretical formula $F = ILB \sin \theta$ , we see that $ILB = 0.0306$ . Solving for $B$ gives $B = \frac{0.0306}{2.4 \times 0.85} \approx 0.015 \text{ T}$ , which is on the order of $10^{-2} \text{ T}$ .

C

$10^{2}$ T

This value is significantly larger than the actual magnetic field. The calculation yields $B \approx 0.015 \text{ T}$ , whereas $10^2 \text{ T}$ represents an extremely strong magnetic field not supported by the data.

D

$10^{4}$ T

This value is far too large. The calculated field strength is approximately $1.5 \times 10^{-2} \text{ T}$ , which is six orders of magnitude smaller than $10^4 \text{ T}$ .

Q7

2023

QCAA

Paper 1

1 mark

Q7

1 mark

An electron and positron can annihilate into a photon, producing another electron and positron pair in the process. An outcome of this interaction is that

A

total mass decreases.

B

fewer baryons will be produced.

C

the lepton number does not change.

D

the number of particles will decrease.

Reveal Answer

A

total mass decreases.

Total energy is conserved in the interaction, and since the final state contains the same types of particles as the initial state, the sum of the rest masses remains constant rather than decreasing.

B

fewer baryons will be produced.

Electrons and positrons are leptons, not baryons. Since there are no baryons in the initial or final state, the baryon number remains constant at zero.

C

the lepton number does not change.

Correct Answer

Lepton number is conserved in this interaction. The electron has a lepton number of $+1$ and the positron $-1$ , resulting in a net lepton number of $0$ both before and after the event.

D

the number of particles will decrease.

The interaction begins with two particles (an electron and a positron) and ends with two particles (an electron and a positron), so the total number of particles remains unchanged.

Q17

2021

VCAA

1 mark

Q17

1 mark

Which one of the following is closest to the de Broglie wavelength of a $663 \text{ kg}$ motor car moving at $10 \text{ m s}^{-1}$ ?

A

$10^{-37} \text{ m}$

B

$10^{-36} \text{ m}$

C

$10^{-35} \text{ m}$

D

$10^{-34} \text{ m}$

Reveal Answer

A

$10^{-37} \text{ m}$

Correct Answer

Correct. Using the de Broglie wavelength formula $\lambda = \frac{h}{mv}$ , we calculate $\lambda = \frac{6.63 \times 10^{-34} \text{ J s}}{663 \text{ kg} \times 10 \text{ m s}^{-1}} = 10^{-37} \text{ m}$ .

B

$10^{-36} \text{ m}$

Incorrect. This result is off by a factor of 10, which would occur if the velocity was $1 \text{ m s}^{-1}$ instead of $10 \text{ m s}^{-1}$ .

C

$10^{-35} \text{ m}$

Incorrect. This answer is off by a factor of 100, likely due to a miscalculation of the momentum denominator $mv = 6630 \text{ kg m s}^{-1}$ .

D

$10^{-34} \text{ m}$

Incorrect. This is approximately the value of Planck's constant ( $6.63 \times 10^{-34} \text{ J s}$ ), which means the momentum $mv$ was incorrectly treated as $1 \text{ kg m s}^{-1}$ .

Q11

2024

VCAA

1 mark

Q11

1 mark

In Victoria, the electrical energy generated at the Loy Yang A power station is transmitted to Melbourne, approximately $170 \text{ km}$ away, using $500 \text{ kV}$ transmission lines.

Which one of the following best describes the reason for the use of high-voltage transmission of electrical energy over long distances?

A

Transformers can be used to increase the voltage.

B

High voltages reduce energy losses in the transmission lines.

C

High voltages can easily carry the large power required by cities.

D

High voltages reduce the overall total resistance in the transmission lines.

Reveal Answer

A

Transformers can be used to increase the voltage.

While transformers are indeed used to step up the voltage, this explains how high voltages are achieved, not why they are beneficial for long-distance transmission.

B

High voltages reduce energy losses in the transmission lines.

Correct Answer

For a given amount of power, transmitting at a higher voltage reduces the current ( $P=VI$ ). A lower current significantly reduces the power lost as heat in the transmission lines ( $P_{\text{loss}} = I^2R$ ).

C

High voltages can easily carry the large power required by cities.

While high voltages are used to transmit large amounts of power, the fundamental reason for stepping up the voltage is to minimize power loss during transmission, not just to increase capacity.

D

High voltages reduce the overall total resistance in the transmission lines.

The resistance of a transmission line is determined by its physical properties (material, length, and cross-sectional area), not by the voltage applied to it.

Q22

2023

QCAA

Paper 1

3 marks

Q22

3 marks

Particles move at a rate of $1.3\times10^6$ times per second around a circular particle accelerator with a radius of 35 m.

Calculate the average speed of the particles. Show your working.

Average speed = ______ $ms^{-1}$ (two significant figures)

Reveal Answer

$C = 2\pi r = 2 \times \pi \times 35 \approx 219.91 \text{ m}$

$f = \frac{1}{T}$

$\therefore v = \frac{2\pi r}{T} = 219.91 \times 1.3 \times 10^6 = 2.86 \times 10^8 \text{ m s}^{-1}$

Average speed $= 2.9 \times 10^8 \text{ m s}^{-1}$ (to two significant figures)

Marking Criteria

Descriptor	Marks
Recognises the scenario relates to average speed of objects in uniform circular motion	1
Provides appropriate mathematical reasoning	1
Calculates the average speed of the particles	1

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