Band 4

PHYSICS, M6 2025 HSC 6 MC

In the diagram, an electron enters the shaded region where it is subjected to an external magnetic field that causes it to move in a circular arc, as indicated by the dotted line.

Which magnetic field could produce the motion of the electron shown?

Show Answers Only

$A$

Show Worked Solution

The electron curves in a circular path, so a magnetic force must act at right angles to its velocity.
When it first enters the shaded region, the force on the electron is upward.
For a positive charge, the right-hand palm rule $(F = qvB)$ would give a magnetic field into or out of the page depending on the direction of the force.
Since an electron is negatively charged, the actual direction of the magnetic field is the opposite of what the right-hand rule predicts.
Reversing the direction gives a magnetic field out of the page as it is the only field direction that produces the observed upward force on a negative charge.

$\Rightarrow A$

PHYSICS, M5 2025 HSC 5 MC

A planet orbits a star in an elliptical orbit as shown.

At which point in its orbit is the planet's kinetic energy increasing?

$W$
$X$
$Y$
$Z$

Show Answers Only

$D$

Show Worked Solution

As the planet gets closer to the sun in its orbit, its gravitational potential energy decreases.
The Law of Conservation of Energy means that as the planet’s potential energy decreases, its kinetic energy increases.
This occurs at point $Z$, noting that the planet is moving perpendicular to the sun at $W$ and $Y$.

$\Rightarrow D$

PHYSICS, M8 2025 HSC 4 MC

Which row in the table identifies the particle with the shortest de Broglie wavelength?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \textit{Particle}\quad\rule[-1ex]{0pt}{0pt}& \quad\textit{Velocity}\quad \\
\hline
\rule{0pt}{2.5ex}\text{Electron}\rule[-1ex]{0pt}{0pt}&0.1 c\\
\hline
\rule{0pt}{2.5ex}\text{Electron}\rule[-1ex]{0pt}{0pt}& 0.9 c\\
\hline
\rule{0pt}{2.5ex}\text{Proton}\rule[-1ex]{0pt}{0pt}& 0.1 c \\
\hline
\rule{0pt}{2.5ex}\text{Proton}\rule[-1ex]{0pt}{0pt}& 0.9 c \\
\hline
\end{array}
\end{align*}

Show Answers Only

$D$

Show Worked Solution

Using $\lambda = \dfrac{h}{p}$, the large momentum $(p = mv)$ of a fast, heavy particle makes $\lambda$ smallest.
The proton at 0.9$c$ is the heaviest and fastest option. Since this has the greatest momentum, it has the shortest de Broglie wavelength.

$\Rightarrow D$

PHYSICS, M5 2025 HSC 29

A mass moves around a vertical circular path of radius $r$, in Earth's gravitational field, without loss of mechanical energy. A string of length $r$ maintains the circular motion of the mass.

When the mass is at its highest point $B$, the tension in the string is zero.

Show that the speed of the mass at the highest point, $B$, is given by $v=\sqrt{r g}$. (2 marks)

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Compare the speed of the mass at point $A$ to that at point $B$. Support your answer using appropriate mathematical relationships. (3 marks)

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a. $\text {At point} \ B \ \Rightarrow \ F_c=F_{\text {net}}$

$\dfrac{mv_B^2}{r}$	$=T+mg$
$v_B^2$	$=rg \ \ (T=0)$
$v_B$	$=\sqrt{r g}$

b. $\text {Total} \ ME=E_k+GPE$

$\text{At point} \ B :$

$\text {Total} \ ME$	$=\dfrac{1}{2} m v_B^2+mg(2r)$
	$=\dfrac{1}{2} m \times r g+2 mrg$
	$=\dfrac{5}{2} m r g$

$\text{At point} \ A :$

$\text {Total} \ ME=\dfrac{1}{2} mv_A^2+mrg$

$\text{Since \(ME$ is conserved:}\)

$\dfrac{5}{2} mrg$	$=\dfrac{1}{2} m v_A^2+m r g$
$\dfrac{1}{2} mv_A^2$	$=\dfrac{3}{2} m r g$
$v_A^2$	$=3 rg$
$v_A$	$=\sqrt{3rg}=\sqrt{3} \times v_B$

Show Worked Solution

a. $\text {At point} \ B \ \Rightarrow \ F_c=F_{\text {net}}$

$\dfrac{mv_B^2}{r}$	$=T+mg$
$v_B^2$	$=rg \ \ (T=0)$
$v_B$	$=\sqrt{r g}$

b. $\text {Total} \ ME=E_k+GPE$

$\text{At point} \ B :$

$\text {Total} \ ME$	$=\dfrac{1}{2} m v_B^2+mg(2r)$
	$=\dfrac{1}{2} m \times r g+2 mrg$
	$=\dfrac{5}{2} m r g$

$\text{At point} \ A :$

$\text {Total} \ ME=\dfrac{1}{2} mv_A^2+mrg$

$\text{Since \(ME$ is conserved:}\)

$\dfrac{5}{2} mrg$	$=\dfrac{1}{2} m v_A^2+m r g$
$\dfrac{1}{2} mv_A^2$	$=\dfrac{3}{2} m r g$
$v_A^2$	$=3 rg$
$v_A$	$=\sqrt{3rg}=\sqrt{3} \times v_B$

PHYSICS, M6 2025 HSC 26

The starting position of a simple AC generator is shown. It consists of a single rectangular loop of wire in a uniform magnetic field of 0.5 T. This loop is connected to two slip rings and the slip rings are connected via brushes to a voltmeter.

The loop is rotated at a constant rate through an angle of 90 degrees from the starting position in the direction indicated, in 0.1 seconds.
Calculate the magnitude of the average emf generated during this rotation. (2 marks)

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The same coil was then rotated at 10 revolutions per second from the starting position. The voltage varies with time, as shown in the graph.

On the same axes, sketch a graph that shows the variation of voltage with time if the rotational speed is 20 revolutions per second in the opposite direction, beginning at the original starting position. (3 marks)

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a. $\text{Using}\ \ \phi=BA \ \ \text{and} \ \ \varepsilon=\dfrac{\Delta \phi}{\Delta t}$

$\varepsilon=\dfrac{\Delta(B A)}{\Delta t}=\dfrac{B \Delta A}{\Delta t}=\dfrac{0.5 \times 0.4 \times 0.3}{0.1}=0.6\ \text{V}$

Show Worked Solution

a. $\text{Using}\ \ \phi=BA \ \ \text{and} \ \ \varepsilon=\dfrac{\Delta \phi}{\Delta t}$

$\varepsilon=\dfrac{\Delta(B A)}{\Delta t}=\dfrac{B \Delta A}{\Delta t}=\dfrac{0.5 \times 0.4 \times 0.3}{0.1}=0.6\ \text{V}$

PHYSICS, M7 2025 HSC 25

A student conducts an experiment to determine the work function of potassium.

The following diagram depicts the experimental setup used, where light of varying frequency is incident on a potassium electrode inside an evacuated tube.

For each frequency of light tested, the voltage in the circuit is varied, and the minimum voltage (called the stopping voltage) required to bring the current in the circuit to zero is recorded.

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \textit{Frequency of incident light} \ \ & \quad \quad \quad \textit{Stopping voltage} \quad \quad \quad \\
\left(\times 10^{15} \ \text{Hz}\right) \rule[-1ex]{0pt}{0pt}& \text{(V)}\\
\hline
\rule{0pt}{2.5ex}0.9 \rule[-1ex]{0pt}{0pt}& 1.5 \\
\hline
\rule{0pt}{2.5ex}1.1 \rule[-1ex]{0pt}{0pt}& 2.0 \\
\hline
\rule{0pt}{2.5ex}1.2 \rule[-1ex]{0pt}{0pt}& 2.5 \\
\hline
\rule{0pt}{2.5ex}1.3 \rule[-1ex]{0pt}{0pt}& 3.0 \\
\hline
\rule{0pt}{2.5ex}1.4 \rule[-1ex]{0pt}{0pt}& 3.5 \\
\hline
\rule{0pt}{2.5ex}1.5 \rule[-1ex]{0pt}{0pt}& 4.0 \\
\hline
\end{array}

Construct an appropriate graph using the data provided, and from this, determine the threshold frequency of potassium. (3 marks)

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Using the particle model of light, explain the features shown in the experimental results. (3 marks)

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b. Features shown in experiment results:

Light is made of discrete energy packets called photons.
A photon’s energy is directly proportional to its frequency: $E = hf$.
When photons hit a metal surface, they transfer their energy to electrons.
If the photon energy is greater than the metal’s work function $(\phi)$, the threshold frequency is exceeded and electrons are ejected with kinetic energy equal to the excess energy according to the equation $KE = hf-\phi$.
Increasing the frequency increases the photon energy which results in ejected electrons with greater kinetic energy.
A larger stopping voltage is therefore needed to halt these more energetic electrons.

Show Worked Solution

b. Features shown in experiment results:

Light is made of discrete energy packets called photons.
A photon’s energy is directly proportional to its frequency: $E = hf$.
When photons hit a metal surface, they transfer their energy to electrons.
If the photon energy is greater than the metal’s work function $(\phi)$, the threshold frequency is exceeded and electrons are ejected with kinetic energy equal to the excess energy according to the equation $KE = hf-\phi$.
Increasing the frequency increases the photon energy which results in ejected electrons with greater kinetic energy.
A larger stopping voltage is therefore needed to halt these more energetic electrons.

PHYSICS, M8 2025 HSC 27

Outline TWO ways in which Schrödinger’s model of electron behaviour is different from electron behaviour in the atomic models of Rutherford and Bohr. (3 marks)

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Answers could include two of the following:

Electron location

Bohr/Rutherford: Electrons move in fixed paths (or orbits) around the nucleus.
Schrödinger: Electrons exist in orbitals, which are regions where they are likely to be found.

Nature of the electron

Bohr/Rutherford: Electrons are treated mainly as particles.
Schrödinger: Electrons behave as waves, following de Broglie’s wave ideas.

Certainty vs probability (extra option)

Bohr/Rutherford: The position of an electron can be predicted exactly in its orbit.
Schrödinger: Only the probability of an electron’s position can be known, not its exact location.

Show Worked Solution

Answers could include two of the following:

Electron location

Bohr/Rutherford: Electrons move in fixed paths (or orbits) around the nucleus.
Schrödinger: Electrons exist in orbitals, which are regions where they are likely to be found.

Nature of the electron

Bohr/Rutherford: Electrons are treated mainly as particles.
Schrödinger: Electrons behave as waves, following de Broglie’s wave ideas.

Certainty vs probability (extra option)

Bohr/Rutherford: The position of an electron can be predicted exactly in its orbit.
Schrödinger: Only the probability of an electron’s position can be known, not its exact location.

PHYSICS, M5 2025 HSC 24

Two satellites, $A$ and $B$, are in stable circular orbits around the Earth. The radius of satellite $A$'s orbit is three times that of satellite $B$'s orbit. Both satellites have the same kinetic energy.

Show that the mass of $A$ is three times the mass of $B$. (3 marks)

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$\text{Show}\ \ m_A=3 \times m_B$

$\text{Since \(A$ and $B$ are in stable orbits:}\)

$F_C$	$=F_G$
$\dfrac{m v^2}{r}$	$=\dfrac{G Mm}{r}$
$\dfrac{1}{2} m v^2$	$=\dfrac{G M m}{2 r}$

$\text{Since}\ \ K=\dfrac{1}{2} m v^2 \ \ \text{and} \ \ K_A=K_B \ \text{(given)}:$

$\dfrac{GMm_{A}}{2r_A}$	$=\dfrac{GMm_{B}}{2r_B}$
$\dfrac{m_A}{r_A}$	$=\dfrac{m_B}{r_B}$
$\dfrac{m_A}{3r_B}$	$=\dfrac{m_B}{r_B}\left(r_A=3 \times r_B\right)$
$m_A$	$=3 \times m_B \ \text{… as required}$

Show Worked Solution

$\text{Show}\ \ m_A=3 \times m_B$

$\text{Since \(A$ and $B$ are in stable orbits:}\)

$F_C$	$=F_G$
$\dfrac{m v^2}{r}$	$=\dfrac{G Mm}{r}$
$\dfrac{1}{2} m v^2$	$=\dfrac{G M m}{2 r}$

$\text{Since}\ \ K=\dfrac{1}{2} m v^2 \ \ \text{and} \ \ K_A=K_B \ \text{(given)}:$

$\dfrac{GMm_{A}}{2r_A}$	$=\dfrac{GMm_{B}}{2r_B}$
$\dfrac{m_A}{r_A}$	$=\dfrac{m_B}{r_B}$
$\dfrac{m_A}{3r_B}$	$=\dfrac{m_B}{r_B}\left(r_A=3 \times r_B\right)$
$m_A$	$=3 \times m_B \ \text{… as required}$

PHYSICS, M6 2025 HSC 23

A wire loop is carrying a current of 2 A from $A$ to $B$ as shown. The length of wire within the magnetic field is 5 cm. The loop is free to pivot around the axis. The magnetic field is of magnitude $3 \times 10^{-2}\ \text{T}$ and at right angles to the wire.

Determine the torque produced on the wire loop due to the motor effect. (3 marks)

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Both the current and the magnetic field were changed, and the torque was observed to be in the same direction but twice the magnitude.
What changes to the magnitude of BOTH the current and the magnetic field are required to produce this result? (2 marks)

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Show Answers Only

a. $\tau=6 \times 10^{-4} \ \text{Nm clockwise from side} \ B$

b. $\text{Torque is doubled in same direction (given)}$

$\text{Using}\ \ \tau=BILd:$

$\text{If current is halved}\ \ \left(I \rightarrow \frac{1}{2} I\right) \ \ \text{and the magnetic field is}$

$\text {increased by a factor of} \ 4 \ (B \rightarrow 4B),\ \text{torque is doubled.}$

$\tau_{\text{new}}=4B \times \frac{1}{2} I \times Ld=2 \times BILd=2 \tau_{\text {orig }}$

Show Worked Solution

a. $\text{Convert units:}\ \ 5 \ \text{cm} \ \Rightarrow \ \dfrac{5}{100}=0.05 \ \text{m}$

$F=BIL=3 \times 10^{-2} \times 2 \times 0.05=0.003 \ \text{N}$

$\text{Convert units:}\ \ 20\ \text{cm} \ \Rightarrow \ \ 0.2 \ \text{m}$

$\tau$	$=F d$
	$=0.003 \times 0.2$
	$=6 \times 10^{-4} \ \text{Nm clockwise from side} \ B$

b. $\text{Torque is doubled in same direction (given)}$

$\text{Using}\ \ \tau=BILd:$

$\text{If current is halved}\ \ \left(I \rightarrow \frac{1}{2} I\right) \ \ \text{and the magnetic field is}$

$\text {increased by a factor of} \ 4 \ (B \rightarrow 4B),\ \text{torque is doubled.}$

$\tau_{\text{new}}=4B \times \frac{1}{2} I \times Ld=2 \times BILd=2 \tau_{\text {orig }}$

Financial, STD2 EQ-Bank 02

Kylie is a financial broker who earns a salary of $\$93\,600$ per annum.
She has 35% of her salary deducted for tax.
Show that her net weekly pay is $1170. (1 marks)
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Ben, Kylie's partner, works as a carpenter.
He works 36 hours per week at a normal rate of $40 per hour and averages 6 hours overtime at time-and-a-half.
Show that his average weekly pay before tax is $1800. (2 marks)
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Kylie and Ben drew up the weekly budget below.
They need to save $400 per week for an overseas holiday and also want to continue to save for a home.
Ben has 25% of his gross wage deducted for taxation.
Comment on the budget as it appears in the table, indicating where they may have made an error, and suggest some practical ways to make the budget work for them. (3 marks)
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a.	$\text{Kylie’s net salary}$	$=\dfrac{$93\,600\times 0.65}{52}$
		$=$1170$

b.	$\text{Ben’s gross pay}$	$=$40\times (36+1.5\times 6)$
		$=$1800$

c.	$\text{Ben’s net pay}$	$=1800\times 0.75$
		$=$1350$
	$\text{Total net income}$	$=1170+1350=$2520$
	$\text{Total expenses}$	$=1000+360+210+250+250+400+500$
		$=$2720$

$\text{They have mistakenly used Ben’s gross salary in their}$

$\text{calculations which will leave them short of their savings}$

$\text{target of }$500\ \text{per week by } $200. $

$\text{Option 1. Reduce savings by }$200\ \text{per week.}$

$\text{Option 2. Reduce entertainment by by }$100 $

$\text{per week and reduce savings by }$100\ \text{per week.}$

Show Worked Solution

a.	$\text{Kylie’s net salary}$	$=\dfrac{93\,600\times 0.65}{52}$
		$=$1170$

b.	$\text{Ben’s gross pay}$	$=40\times (36+1.5\times 6)$
		$=$1800$

c.	$\text{Ben’s net pay}$	$=1800\times 0.75$
		$=$1350$
	$\text{Total net income}$	$=1170+1350=$2520$
	$\text{Total expenses}$	$=1000+360+210+250+250+400+500$
		$=$2720$

$\text{They have mistakenly used Ben’s gross salary in their}$

$\text{calculations which will leave them short of their savings}$

$\text{target of }$500\ \text{per week by } $200. $

$\text{Option 1. Reduce savings by }$200\ \text{per week.}$

$\text{Option 2. Reduce entertainment by by }$100 $

$\text{per week and reduce savings by }$100\ \text{per week.}$

BIOLOGY, M7 2025 HSC 21

The diagram shows components of the innate immune system in humans.

State the role of TWO components that protect against infection. (2 marks)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \textit{Component} \quad \quad \rule[-1ex]{0pt}{0pt} & \quad \quad \textit{How it protects against infection}\quad \quad \rule[-1ex]{0pt}{0pt}\\
\hline
\ & \\
\ & \\
\ & \\
\ & \\
\ & \\
\hline
\ & \\
\ & \\
\ & \\
\ & \\
\ & \\
\hline
\end{array}

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Any TWO of the following components

\begin{array} {|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Component} \quad \quad \rule[-1ex]{0pt}{0pt} & \textit{How it protects against infection}\quad\rule[-1ex]{0pt}{0pt}\\
\hline
\text{Skin} & \text{Acts as a physical barrier}\\
\ & \text{preventing pathogen entry into} \\
\ & \text{tissue}\\
\hline
\text{Stomach} &\text{Destroys ingested pathogens} \\
\text{acid} & \text{through low pH chemical}\\
\ & \text{environment.}\\
\hline
\text{Mucus} &\text{Traps pathogens and prevents} \\
\text{lining} & \text{their entry into underlying}\\
\ & \text{tissues.}\\
\hline
\text{Nasal} &\text{Filters and traps airborne} \\
\text{hair} & \text{pathogens, preventing}\\
\ & \text{respiratory entry.}\\
\hline
\text{Tear glands} &\text{Produce lysozyme enzyme that} \\
\ & \text{destroys bacterial cell walls.}\\
\hline
\text{Urinary} &\text{Flushes pathogens from the} \\
\text{tract} & \text{urethra preventing infection.}\\
\hline
\end{array}

Show Worked Solution

Any TWO of the following components

BIOLOGY, M8 2025 HSC 24

The following flow chart represents the control of body temperature in humans.

Complete the flow chart to give an example of mechanism A and an example of mechanism B. (2 marks)
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Outline how mechanism B maintains homeostasis. (2 marks)
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a. Mechanism A (Decreases temperature):

Sweating/perspiration → Vasodilation

Mechanism B (Increases temperature):

Shivering → Vasoconstriction

b. Maintaining homeostasis

When body temperature drops below normal range, thermoreceptors detect the change.
The hypothalamus (control centre) activates mechanism B responses like shivering and vasoconstriction.
Shivering generates heat through muscle contractions whilst vasoconstriction reduces heat loss.
Body temperature increases back to normal range, restoring homeostasis.

Show Worked Solution

a. Mechanism A (Decreases temperature):

Sweating/perspiration → Vasodilation

Mechanism B (Increases temperature):

Shivering → Vasoconstriction

b. Maintaining homeostasis

When body temperature drops below normal range, thermoreceptors detect the change.
The hypothalamus (control centre) activates mechanism B responses like shivering and vasoconstriction.
Shivering generates heat through muscle contractions whilst vasoconstriction reduces heat loss.
Body temperature increases back to normal range, restoring homeostasis.

BIOLOGY, M5 2025 HSC 31

Congenital amegakaryocytic thrombocytopenia (CAMT) is a rare, inherited disorder where bone marrow no longer makes platelets that are important for clotting and preventing bleeding. The pedigree below shows the inheritance of CAMT in a family.

What type of inheritance is shown in the pedigree above? Justify your answer? (3 marks)

Type of Inheritance:

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A CAMT mutation was found to produce the following amino acid sequence:
1. Glutamine – Tyrosine – Isoleucine – Aspartic acid.
The same DNA fragment has been sequenced from an unaffected individual.
Template strand GTC ATA CAG CTG.
The following codon chart displays all the codons and corresponding amino acids. The chart translates mRNA sequences into amino acids.
Use the codon chart shown to explain the type of mutation which causes CAMT. (3 marks)

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a. Type of inheritance: Autosomal recessive

Both males and females are affected equally, ruling out sex-linked inheritance.
Two affected parents (10 and 11) produce only affected offspring (17, 18, 19). This is consistent with autosomal recessive inheritance (aa × aa = all aa).
The disorder skips generations. Unaffected carriers can pass on the recessive allele without expressing the phenotype.
Affected individual 5 and unaffected individual 6 produce affected child 13, confirming individual 6 is a heterozygous carrier.

b. Type of mutation causing CAMT

The template strand transcribes to mRNA CAG UAU GUC GAC, which translates to Glutamine-Tyrosine-Valine-Aspartic acid in unaffected individuals.
In CAMT, Isoleucine replaces Valine at position 3. This results from a single nucleotide substitution changing the codon from GUC to an Isoleucine codon.
This is a point mutation (missense mutation). This causes one amino acid replacement, which affects protein function and leads to impaired platelet production.

Show Worked Solution

a. Type of inheritance: Autosomal recessive

Both males and females are affected equally, ruling out sex-linked inheritance.
Two affected parents (10 and 11) produce only affected offspring (17, 18, 19). This is consistent with autosomal recessive inheritance (aa × aa = all aa).
The disorder skips generations. Unaffected carriers can pass on the recessive allele without expressing the phenotype.
Affected individual 5 and unaffected individual 6 produce affected child 13, confirming individual 6 is a heterozygous carrier.

b. Type of mutation causing CAMT

The template strand transcribes to mRNA CAG UAU GUC GAC, which translates to Glutamine-Tyrosine-Valine-Aspartic acid in unaffected individuals.
In CAMT, Isoleucine replaces Valine at position 3. This results from a single nucleotide substitution changing the codon from GUC to an Isoleucine codon.
This is a point mutation (missense mutation). This causes one amino acid replacement, which affects protein function and leads to impaired platelet production.

BIOLOGY, M6 2025 HSC 34

The following graph shows the changes in allele frequencies in two separate populations of the same species. Each line represents an introduced allele.

Explain why the fluctuations of the allele frequencies are more pronounced in the small population, compared to the larger population. (2 marks)

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Evaluate the effects of gene flow on the gene pools of the two populations. (4 marks)
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a. Genetic Drift and Population Size

Small populations are more susceptible to genetic drift because each individual represents a larger proportion of the gene pool.
Random events cause greater fluctuations, while larger populations buffer these changes, resulting in stable allele frequencies.

b. Evaluation Statement

Gene flow is highly effective for maintaining genetic diversity in the larger population but shows limited effectiveness in the smaller population.

Population Size Differences

The large population (2000) demonstrates strong effectiveness in maintaining stable allele frequencies around 0.5.
This occurs because gene flow introduces consistent genetic material that prevents random loss of alleles.
The population size allows introduced alleles to establish without being lost through drift.

Vulnerability in Small Populations

The small population (20) shows limited benefit from gene flow.
Despite introduction of new alleles, genetic drift overwhelms the stabilising effect.
Multiple alleles are lost completely, demonstrating that population size critically determines whether gene flow can maintain genetic diversity.

Final Evaluation

Overall, gene flow proves highly effective in large populations for maintaining diversity,.
However, it demonstrates insufficient effectiveness in small populations where stochastic processes dominate.

Show Worked Solution

a. Genetic Drift and Population Size

Small populations are more susceptible to genetic drift because each individual represents a larger proportion of the gene pool.
Random events cause greater fluctuations, while larger populations buffer these changes, resulting in stable allele frequencies.

b. Evaluation Statement

Gene flow is highly effective for maintaining genetic diversity in the larger population but shows limited effectiveness in the smaller population.

Population Size Differences

The large population (2000) demonstrates strong effectiveness in maintaining stable allele frequencies around 0.5.
This occurs because gene flow introduces consistent genetic material that prevents random loss of alleles.
The population size allows introduced alleles to establish without being lost through drift.

Vulnerability in Small Populations

The small population (20) shows limited benefit from gene flow.
Despite introduction of new alleles, genetic drift overwhelms the stabilising effect.
Multiple alleles are lost completely, demonstrating that population size critically determines whether gene flow can maintain genetic diversity.

Final Evaluation

Overall, gene flow proves highly effective in large populations for maintaining diversity,.
However, it demonstrates insufficient effectiveness in small populations where stochastic processes dominate.

BIOLOGY, M8 2025 HSC 26

The diagram shows the steps in LASIK (Laser-Assisted In Situ Keratomileusis) surgery.

Compare the LASIK technology shown with ONE other technology that can be used to treat a named visual disorder. (4 marks)

Visual disorder:

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Visual disorder: Myopia (short-sightedness)

Similarities:

Both LASIK surgery and corrective spectacles correct refractive errors by changing light focus.
Both technologies enable clear vision at distance for myopia patients.

Differences:

LASIK permanently reshapes the cornea using laser ablation to correct vision.
Spectacles use external convex or concave lenses to refract light without altering eye structure.
LASIK requires surgical procedure with recovery time whilst spectacles require no invasive procedure.
LASIK provides permanent correction whereas spectacles require continuous wear for vision correction.

Show Worked Solution

Visual disorder: Myopia (short-sightedness)

Similarities:

Both LASIK surgery and corrective spectacles correct refractive errors by changing light focus.
Both technologies enable clear vision at distance for myopia patients.

Differences:

LASIK permanently reshapes the cornea using laser ablation to correct vision.
Spectacles use external convex or concave lenses to refract light without altering eye structure.
LASIK requires surgical procedure with recovery time whilst spectacles require no invasive procedure.
LASIK provides permanent correction whereas spectacles require continuous wear for vision correction.

BIOLOGY, M7 2025 HSC 28

Alpha-gal syndrome (AGS) is a tick-borne allergy to red meat caused by tick bites. Alpha-gal is a sugar molecule found in most mammals but not humans, and can also be found in the saliva of ticks. The diagram shows how a tick bite might cause a person to develop an allergic reaction to red meat.

The flow chart shows the process of antibody production following exposure to alpha-gal.
Describe the role of X, Y and Z in the process of antibody production. (4 marks)

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An allergic reaction to alpha-gal sugar is similar to a secondary immune response.
Describe the features of antibody production shown in the graph. (2 marks)

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Explain the role of memory cells in the immune response. (3 marks)

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a. Antibody Production Process

X is a Helper T-cell that recognises the alpha-gal antigen presented by macrophages on MHC-II molecules.
Helper T-cells activate and coordinate the adaptive immune response through cytokine release.
Y is a B-cell that has receptors specific to the alpha-gal antigen.
B-cells are activated by Helper T-cells and undergo clonal expansion.
Some B-cells differentiate into memory cells for long-term immunity.
Z is a Plasma cell, which is a differentiated B-cell specialised for antibody production.
Plasma cells produce large quantities of antibodies specific to alpha-gal that circulate in the bloodstream.

b. Features of Antibody Production

Initial tick bite produces low antibody concentration with slow, gradual increase over time, representing primary immune response.
Subsequent meat consumption triggers rapid elevation to higher antibody concentration, demonstrating secondary immune response with accelerated, amplified production.

c. Role of Memory Cells

Memory cells are produced during primary exposure and remain in circulation for years, maintaining immunological memory.
Upon re-exposure, memory cells rapidly recognise the specific antigen, which triggers immediate clonal expansion.
This results in faster and stronger antibody production because memory cells bypass the initial activation phase. Hence, providing enhanced immune protection against subsequent infections.

Show Worked Solution

a. Antibody Production Process

X is a Helper T-cell that recognises the alpha-gal antigen presented by macrophages on MHC-II molecules.
Helper T-cells activate and coordinate the adaptive immune response through cytokine release.
Y is a B-cell that has receptors specific to the alpha-gal antigen.
B-cells are activated by Helper T-cells and undergo clonal expansion.
Some B-cells differentiate into memory cells for long-term immunity.
Z is a Plasma cell, which is a differentiated B-cell specialised for antibody production.
Plasma cells produce large quantities of antibodies specific to alpha-gal that circulate in the bloodstream.

b. Features of Antibody Production

Initial tick bite produces low antibody concentration with slow, gradual increase over time, representing primary immune response.
Subsequent meat consumption triggers rapid elevation to higher antibody concentration, demonstrating secondary immune response with accelerated, amplified production.

c. Role of Memory Cells

Memory cells are produced during primary exposure and remain in circulation for years, maintaining immunological memory.
Upon re-exposure, memory cells rapidly recognise the specific antigen, which triggers immediate clonal expansion.
This results in faster and stronger antibody production because memory cells bypass the initial activation phase. Hence, providing enhanced immune protection against subsequent infections.

PHYSICS, M6 2025 HSC 22

The diagram represents the parts of the AC system used to transfer energy from a power station to people's houses.

Describe the energy transformations that take place in the transformers, and in the transmission line. (4 marks)

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Transformers $A$ and $B$: Some electrical energy is always lost, so each transformer outputs less electrical energy than it receives.
Core losses in transformers: Part of the input electrical energy is transformed into heat in the iron core due to eddy currents and magnetic effects.
Resistive losses in transformers: Some electrical energy becomes heat in the coils because of their resistance. Small amounts may also become sound/vibration energy.
Transmission lines: Electrical energy is also transformed into heat in the wires due to their resistance.

Show Worked Solution

Transformers $A$ and $B$: Some electrical energy is always lost, so each transformer outputs less electrical energy than it receives.
Core losses in transformers: Part of the input electrical energy is transformed into heat in the iron core due to eddy currents and magnetic effects.
Resistive losses in transformers: Some electrical energy becomes heat in the coils because of their resistance. Small amounts may also become sound/vibration energy.
Transmission lines: Electrical energy is also transformed into heat in the wires due to their resistance.

Functions, 2ADV EQ-Bank 10

Express $y=x^2-4 x+6$ in the form $y=(x-a)^2+c$. (1 mark)

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Graph the parabola, labelling its vertex and $y$-intercept. (2 marks)

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a. $y=(x-2)^2+2$

b. $\text {Vertex}=(2,2)$

$y \text{-intercept}=(0,6)$

Show Worked Solution

a.	$y$	$=x^2-4 x+6$
		$=x^2-4 x+4+2$
		$=(x-2)^2+2$

b. $\text {Vertex}=(2,2)$

$y \text{-intercept}=(0,6)$

Functions, 2ADV EQ-Bank 09

Using the discriminant, or otherwise, justify why the graph of $f(x)=-x^2+2 x-2$ lies entirely below the $x$-axis. (2 marks)

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$\Delta=b^2-4 a c=2^2-4(-1)(-2)=-4$

$\text{Since \(\ \Delta<0, \ y=-x^2+2 x-2\ $ does not intersect the $x$-axis.}\)

$\text{Since \(\ a=-1<0, f(x)$ is an upside down parabola.}\)

$\Rightarrow f(x)\ \text{must lie entirely below} \ x\text{-axis.}$

Show Worked Solution

$\Delta=b^2-4 a c=2^2-4(-1)(-2)=-4$

$\text{Since \(\ \Delta<0, \ y=-x^2+2 x-2\ $ does not intersect the $x$-axis.}\)

$\text{Since \(\ a=-1<0, f(x)$ is an upside down parabola.}\)

$\Rightarrow f(x)\ \text{must lie entirely below} \ x\text{-axis.}$

Functions, 2ADV EQ-Bank 02

Find the equation of the line that passes through $(2,1)$ and $(-3,4)$. (2 marks)

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Determine whether $(7,-2)$ lies on the line. (1 mark)

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a. $y=\dfrac{4-1}{-3-2}=-\dfrac{3}{5}$

b. $\text {Substitute}\ (7,-2) \ \text{into equation:}$

$-2$	$=-\dfrac{3}{5} \times 7+\dfrac{11}{5}$
$-2$	$=-\dfrac{21}{5}+\dfrac{11}{5}$
$-2$	$=-2 \ \text{(correct)}$

$\therefore (7,-2) \text{ lies on line.}$

Show Worked Solution

a. $(2,1),(-3,4)$

$\text{Gradient}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{4-1}{-3-2}=-\dfrac{3}{5}$

$\text{Find equation with} \ \ m=-\dfrac{3}{5} \ \ \text{through}\ \ (2,1):$

$y-1$	$=-\dfrac{3}{5}(x-2)$
$y$	$=-\dfrac{3}{5} x+\dfrac{11}{5}$

b. $\text {Substitute}\ (7,-2)\ \text{into equation:}$

$-2$	$=-\dfrac{3}{5} \times 7+\dfrac{11}{5}$
$-2$	$=-\dfrac{21}{5}+\dfrac{11}{5}$
$-2$	$=-2 \ \text{(correct)}$

$\therefore (7,-2) \text{ lies on line.}$

Functions, 2ADV EQ-Bank 05

Simplify $\dfrac{p+1}{q-q^3} \ ÷ \ \dfrac{p^3+p^2}{q^2-q}$. (2 marks)

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$\dfrac{1}{p^2(1+q)}$

Show Worked Solution

$\dfrac{p+1}{q-q^3} \ ÷ \ \dfrac{p^3+p^2}{q^2-q}$	$=\dfrac{p+1}{q\left(1-q^2\right)} \times \dfrac{q\left(1-q\right)}{p^2(p+1)}$
	$=\dfrac{(1-q)}{(1+q)(1-q) p^2}$
	$=\dfrac{1}{p^2(1+q)}$

Functions, 2ADV EQ-Bank 9

The quantity $y$ varies inversely with the square of $x$.

When $x = 4, \ y = 10$.

Find the value of $y$ when $x = 8$. (2 marks)

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$y=\dfrac{5}{2}$

Show Worked Solution

$y \propto \dfrac{1}{x^2}\ \ \Rightarrow\ \ y= \dfrac{k}{x^2}$

$\text{When}\ \ x=4, \ y=10:$

$10= \dfrac{k}{4^2}\ \ \Rightarrow\ \ k=160$

$\text{Find}\ y\ \text{when}\ x=8:$

$y=\dfrac{160}{8^2}=\dfrac{5}{2}$

L&E, 2ADV EQ-Bank 5

The mass `M` kg of a baby pig at age `x` days is given by `M = A(1.1)^x` where `A` is a constant. The graph of this equation is shown.

What is the value of `A`? (1 mark)

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What is the daily growth rate of the pig’s mass? Write your answer as a percentage. (1 mark)

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i. `1.5\ text(kg)`

ii. `10text(%)`

Show Worked Solution

i. `text(When)\ x = 0:`

♦ Mean mark (i) 48%.
♦♦♦ Mean mark part (ii) 6%!

`1.5`	`= A(1.1)^0`
`:. A`	`= 1.5\ text(kg)`

ii. `text(Daily growth rate)\ = 0.1 = 10text(%)`

L&E, 2ADV EQ-Bank 4

The number of bacteria in a culture grows from 100 to 114 in one hour.

What is the percentage increase in the number of bacteria? (1 mark)

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The bacteria continue to grow according to the formula `n = 100(1.14)^t`, where `n` is the number of bacteria after `t` hours.

What is the number of bacteria after 15 hours? (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Time in hours $(t)$} \rule[-1ex]{0pt}{0pt} & \;\; 0 \;\; & \;\; 5 \;\; & \;\; 10 \;\; & \;\; 15 \;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of bacteria ( $n$ )} \rule[-1ex]{0pt}{0pt} & \;\; 100 \;\; & \;\; 193 \;\; & \;\; 371 \;\; & \;\; ? \;\; \\
\hline
\end{array}

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Use the values of `n` from `t = 0` to `t = 15` to draw a graph of `n = 100(1.14)^t`.

Use about half a page for your graph and mark a scale on each axis. (2 marks)

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Using your graph or otherwise, estimate the time in hours for the number of bacteria to reach 300. (1 mark)

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i. `text(14%)`

ii. `714`

iii. `text(Proof)\ \ text{(See Worked Solutions)}`

iv. `text(8.4 hours)`

Show Worked Solution

i. `text(Percentage increase)`

COMMENT: Common ADV/STD2 content in new syllabus.

`= (114 -100)/100 xx 100`

`= 14text(%)`

ii. `n = 100(1.14)^t`

`text(When)\ \ t = 15,`

`n= 100(1.14)^15= 713.793\ …\ = 714\ \ \ text{(nearest whole)}`

iii.

iv. `text(Using the graph)`

`text(The number of bacteria reaches 300 after ~ 8.4 hours.)`

Functions, 2ADV EQ-Bank 6

Identify where the graph $f(x)=\dfrac{\abs{x}}{x}$ is not continuous. (1 mark)

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Sketch the graph of $f(x)$. (2 marks)

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a. $\text {Denominator} \neq 0$

$f(x)\ \text{is not continuous when} \ \ x=0$

b. $\text{If} \ \ x>0 \ \Rightarrow \ f(x)=\dfrac{x}{x}=1$

$\text{If} \ \ x<0 \ \Rightarrow \ f(x)=-\dfrac{x}{x}=-1$

Show Worked Solution

a. $\text {Denominator} \neq 0$

$f(x)\ \text{is not continuous when} \ \ x=0$

b. $\text{If} \ \ x>0 \ \Rightarrow \ f(x)=\dfrac{x}{x}=1$

$\text{If} \ \ x<0 \ \Rightarrow \ f(x)=-\dfrac{x}{x}=-1$

Functions, 2ADV EQ-Bank 5

Identify where the graph $f(x)=\dfrac{x^2-1}{x-1}$ is not continuous. (1 mark)

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Sketch the graph of $f(x)$. (2 marks)

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a. $f(x)=\dfrac{x^2-1}{x-1}=\dfrac{(x+1)(x-1)}{(x-1)}=x+1$

$\text{Since denominator} \neq 0$

$f(x) \ \ \text{is not continuous when} \ \ x=1.$

Show Worked Solution

a. $f(x)=\dfrac{x^2-1}{x-1}=\dfrac{(x+1)(x-1)}{(x-1)}=x+1$

$\text{Since denominator} \neq 0$

$f(x) \ \ \text{is not continuous when} \ \ x=1.$

Functions, 2ADV EQ-Bank 4

Consider the function $y=f(x)$ where

$f(x)= \begin{cases}x^2+6, & \text { for } x \leqslant 0 \\ 6, & \text { for } 0<x \leqslant 3 \\ 2^x, & \text { for } x>3\end{cases}$

Sketch $y=f(x)$ (3 marks)

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For what value of $x$ is $y=f(x)$ NOT continuous? (1 mark)

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b. $f(x)\ \ \text {is NOT continuous at}\ \ x=3.$

Show Worked Solution

b. $f(x)\ \ \text {is NOT continuous at}\ \ x=3.$

Measurement, STD2 EQ-Bank 06

The distance between planets is measured in Astronomical Units (AU).

$1\ \text{AU}\ \approx 1.496\times 10^8$ kilometres.

Given Venus is approximately 0.7 AU from Earth, calculate this distance in kilometres giving your answer in scientific notation, correct to 3 significant figures. (2 marks)

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$1.03\times 10^8\ \text{(3 sig. fig.)}$

Show Worked Solution

$\text{Distance}$	$=0.7\times \left(1.496\times 10^8\right)$
	$=102\,830\,000$
	$=1.0283\times 10^8$
	$=1.03\times 10^8\ \text{(3 sig. fig.)}$

Statistics, STD2 EQ-Bank 03 MC

For the data below, which is the correct five figure summary?

$12,\ \ 16, \ \ 4,\ \ 6, \ \ 4, \ \ 5, \ \ 22, \ \ 20, \ \ 12, \ 8\ $

$4,\ \ 5, \ \ 8,\ \ 16, \ \ 21$
$4,\ \ 5, \ \ 8,\ \ 16, \ \ 22$
$4,\ \ 5, \ \ 10,\ \ 16, \ \ 21$
$4,\ \ 5, \ \ 10,\ \ 16, \ \ 22$

Show Answers Only

$D$

Show Worked Solution

$\text{Ordered data set}\ \rightarrow\ \ 4,\ \ 4, \ \ 5,\ \ 6, \ \ 8, \ \ 12, \ \ 12, \ \ 16, \ \ 20, \ 22\ $

$\text{Five number Summary}$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 4\\
\hline
\rule{0pt}{2.5ex} \ Q_1 \rule[-1ex]{0pt}{0pt} & 5 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & \dfrac{8+12}{2}=10\\
\hline
\rule{0pt}{2.5ex} \ Q_3 \rule[-1ex]{0pt}{0pt} & 16 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 22\\
\hline
\end{array}

$\Rightarrow D$

Statistics, STD2 EQ-Bank 02 MC

The results of a test are displayed in the box-and-whisker plot below.

Which of the following statements is false?

The median is 155
The range is 60
The interquartile range is 50
25% of the scores are below 150

Show Answers Only

$C$

Show Worked Solution

$\text{Five number Summary}$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 140\\
\hline
\rule{0pt}{2.5ex} \ Q_1 \rule[-1ex]{0pt}{0pt} & 150 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & 155\\
\hline
\rule{0pt}{2.5ex} \ Q_3 \rule[-1ex]{0pt}{0pt} & 190 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 200\\
\hline
\end{array}

$\text{Median}\ =\ 155\ \checkmark$

$\text{Range}\ =\ 200-140=60\ \checkmark$

$\text{IQR}\ =\ 190-150=40\ \text{not}\ 50\ $X

$\text{Q1}\ =\ 150\ \therefore\ 25\%\ \text{of scores lie below 150}\ \checkmark$

$\Rightarrow C$

Measurement, STD2 EQ-Bank 04 MC

Kathmandu is 30$^{\circ}$ west of Perth. Using longitudinal distance, what is the time in Kathmandu when it is noon in Perth?

10:00 am
11:30 am
12:30 pm
2:00 pm

Show Answers Only

$A$

Show Worked Solution

$15^{\circ}\ =\text{1 hour time difference}$

$\text{Longitudinal distance}$	$=30^{\circ}$
$\therefore\ \text{Time Difference}$	$=\dfrac{30}{15}$
	$=2\ \text{hours}$

$\text{Time in Perth}\ =\ 12\ \text{pm}$

$\therefore\ \text{Time in Kathmandu}$	$ =\ 12\ \text{pm}\ -\ 2\ \text{hours}$
	$=\ 10:00\ \text{am}$

$\Rightarrow A$

Measurement, STD2 EQ-Bank 03

The table shows the approximate coordinates of two cities.

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{City} \rule[-1ex]{0pt}{0pt} & \textit{Latitude} \rule[-1ex]{0pt}{0pt} & \textit{Longitude}\\
\hline
\rule{0pt}{2.5ex} \text{Buenos Aires} \rule[-1ex]{0pt}{0pt} & 35^{\circ}\ \text{S} \rule[-1ex]{0pt}{0pt} & 60^{\circ}\ \text{W} \\
\hline
\rule{0pt}{2.5ex} \text{Adelaide} \rule[-1ex]{0pt}{0pt} & 35^{\circ}\ \text{S} \rule[-1ex]{0pt}{0pt} & 140^{\circ}\ \text{E} \\
\hline
\end{array}

What is the time difference between Adelaide and Buenos Aires? (2 marks)
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Roy lives in Adelaide and his cousin Juan lives in Buenos Aires. Roy wants to telephone Juan at 7 pm on Friday night, Buenos Aires time.
At what time, and on what day, should Roy make the call? (2 marks)
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a. $13\ \text{hours}\ 20\ \text{minutes}$

b. $8:20\ \text{am on Saturday}$

Show Worked Solution

a. $15^{\circ}\ =\text{1 hour time difference}$

$\text{Angular distance}$	$=60+140=200^{\circ}$
$\therefore\ \text{Time Difference}$	$=\dfrac{200}{15}$
	$=13.\dot{3}$
	$=13\ \text{hours}\ 20\ \text{minutes}$

b. $\text{Time in Buenos Aires}\ =\ 7\ \text{pm Friday night}$

$\therefore\ \text{Time in Adelaide}$	$ =\ 7\ \text{pm}\ +\ 13\ \text{hours}\ 20\ \text{minutes}$
	$=\ 8:20\ \text{am on Saturday}$

Statistics, STD2 EQ-Bank 01

A Physics class of 12 students is going on a 4 day excursion by bus.

The students are asked to each pack one bag for the trip. The bags are weighed, and the weights (in kg) are listed in order as follows:

$8,\ \ 9, \ \ 10,\ \ 10, \ \ 15, \ \ 18, \ \ 22, \ \ 25, \ \ 29, \ \ 35, \ \ 38, \ \ 41 $

Use the above data to produce a five number summary for the weights of the bags. (2 marks)
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Using your five number summary from part (a), calculate the interquartile range of the weights. (2 marks)
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a. $\text{Five number Summary}$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 8\\
\hline
\rule{0pt}{2.5ex} \ Q_1 \rule[-1ex]{0pt}{0pt} & 10 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & 20\\
\hline
\rule{0pt}{2.5ex} \ Q_3 \rule[-1ex]{0pt}{0pt} & 32 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 41\\
\hline
\end{array}

b. $22$

Show Worked Solution

a. $\text{Five number Summary}$

b.	$\text{IQR}$	$=Q_3-Q_1$
		$=32-10=22$

Algebra, STD2 EQ-Bank 05

Jordan visits Italy on his holidays. He pays €180 (180 euros) for a pair of Italian leather boots.

How much is €180 in Australian dollars if AUD1 is worth €0.58? (2 marks)

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Show Answers Only

$$310.34$

Show Worked Solution

$0.58\ $€ $\ =\ \text{AUD 1}$

$\rightarrow\ 1\ $€$\ =$$\ \text{AUD }$$\dfrac{1}{0.58}$

$\rightarrow\ 180\ $€$=180\times\dfrac{1}{0.58}=310.344…$

$\therefore\ \text{Jordan’s boots cost }$310.34\ \text{in Australian dollars.}$

Algebra, STD2 EQ-Bank 04 MC

The graph shows the tax payable against taxable incomes up to $60 000 in a proposed tax system.

How much of each dollar earned over $$30\,000$ is payable in tax?

10 cents
12 cents
20 cents
23 cents

Show Answers Only

$C$

Show Worked Solution

$\text{The gradient of line represents the tax payable per dollar.}$

$\text{Tax payable per dollar}:$

$= \dfrac{\text{rise}}{\text{run}}$

$= \dfrac{7000-1000}{60\ 000-30\ 000}$

$=\dfrac{1}{5}=0.20$

$\therefore\ \text{20 cents per dollar is payable in tax after }$30\, 000.$

$\Rightarrow C$

Algebra, STD2 EQ-Bank 02 MC

If $w=2y^3-1$, what is the value of $y$ when $w=13$?

$\dfrac{\sqrt[3]{14}}{2}$
$\sqrt[3]{6}$
$\sqrt[3]{7}$
$\sqrt[3]{14}$

Show Answers Only

$C$

Show Worked Solution

$\text{When}\ w=13:$

$w$	$=2y^3-1$
$13$	$=2y^3-1$
$2y^3$	$=14$
$y^3$	$=\dfrac{14}{2}=7$
$y$	$=\sqrt[3]{7}$

$\Rightarrow C$

Algebra, STD2 EQ-Bank 01

Jerico is the manager of a weekend market in which there are 220 stalls for rent. From past experience, Jerico knows that if he charges $d$ dollars to rent a stall. then the number of stalls, $s$, that will be rented is given by:

$s=220-4d$

How many stalls will be rented if Jerico charges $7.50 per stall . (1 mark)

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Complete the following table for the function $s=220-4d$. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad d\quad \rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex} \quad 10\quad\rule[-1ex]{0pt}{0pt} & \rule{0pt}{2.5ex} \quad 30\quad & \rule{0pt}{2.5ex} \quad 50\quad \\
\hline
\rule{0pt}{2.5ex} \quad s\quad \rule[-1ex]{0pt}{0pt} & \ & \ & \\
\hline
\end{array}

Using an appropriate vertical scale and labelled axes, graph the function $s=220-4d$ on the grid below. (2 marks)

--- 0 WORK AREA LINES (style=lined) ---

Does it make sense to use the formula $s=220-4d$ to calculate the number of stalls rented if Jerico charges $60 per stall? Explain your answer. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

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a. $190\ \text{stalls will be rented}$

d. $\text{When}\ d=60, s=220-4\times 60=-20$

$\therefore\ \text{It does not make sense to charge }$60\ \text{ per stall}$

$\text{as you cannot have a negative number of stalls.}$

Show Worked Solution

a.	$s$	$=220-4d$
		$=220-4\times 7.50$
		$=190$

$190\ \text{stalls will be rented}$

d. $\text{When}\ d=60, s=220-4\times 60=-20$

$\therefore\ \text{It does not make sense to charge }$60\ \text{ per stall}$

$\text{as you cannot have a negative number of stalls.}$

Functions, 2ADV EQ-Bank 03

Find $a$ and $b$ such that $a$ and $b$ are real and $\dfrac{2\sqrt{3}+2}{\sqrt{6}-\sqrt{2}} = a\,\sqrt{2} + b\,\sqrt{6}$. (2 marks)

Show Answers Only

$a=2\ \ \text{and}\ \ b=1 $

Show Worked Solution

$\dfrac{2\sqrt{3}+2}{\sqrt{6}-\sqrt{2}} \times \dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}} $

$=\dfrac{(2\sqrt{3}+2)(\sqrt{6}+\sqrt{2})}{6-2}$

$=\dfrac{2\sqrt{18}+2\sqrt{6}+2\sqrt{6}+2\sqrt{2}}{4}$

$=\dfrac{6\sqrt{2}+4\sqrt{6}+2\sqrt{2}}{4}$

$=\dfrac{8\sqrt{2}+4\sqrt{6}}{4}$

$=2\sqrt{2}+\sqrt{6}$

$\therefore a=2\ \ \text{and}\ \ b=1 $

Functions, 2ADV EQ-Bank 01

Find $x$ and $y$ such that $x$ and $y$ are real and $\dfrac{\sqrt{2}+1}{\sqrt{6}-\sqrt{3}} = x\,\sqrt{3} + y\,\sqrt{6}$. (2 marks)

Show Answers Only

$x=1\ \ \text{and}\ \ y=\dfrac{2}{3} $

Show Worked Solution

$\dfrac{\sqrt{2}+1}{\sqrt{6}-\sqrt{3}} \times \dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}} $

$=\dfrac{(\sqrt{2}+1)(\sqrt{6}+\sqrt{3})}{6-3}$

$=\dfrac{\sqrt{12}+2\sqrt{6}+\sqrt{3}}{3}$

$=\dfrac{3\sqrt{3}+2\sqrt{6}}{3}$

$= \sqrt{3}+\dfrac{2}{3} \sqrt{6}$

$\therefore x=1\ \ \text{and}\ \ y=\dfrac{2}{3} $

CHEMISTRY, M6 2025 HSC 31

Hydrazine is a compound of hydrogen and nitrogen. The complete combustion of 1.0 L of gaseous hydrazine requires 3.0 L of oxygen, producing 2.0 L of nitrogen dioxide gas and 2.0 L of water vapour. All volumes are measured at 400°C.

Use the chemical equation for the combustion of hydrazine to show that the molecular formula for hydrazine is $\ce{N2H4}$. (2 marks)

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The relationship between the acid equilibrium constant $\left(K_a\right)$ and the corresponding conjugate base equilibrium constant $\left(K_b\right)$ is shown.
1. 1. $K_a \times K_b=K_w$
Use a relevant chemical equation to calculate the pH of a 0.20 mol L$^{-1}$ solution of $\ce{N2H5+}$ using the following data:

- the $K_b$ of hydrazine is $1.7 \times 10^{-6}$ at 25°C
- $\ce{N2H5+}$ is the conjugate acid of $\ce{N2H4}$. (4 marks)

Show Answers Only

a. Using Avogadro’s law:

At the same temperature and pressure, gas volumes are proportional to moles (i.e. the volume ratio will be equal to the mole ratio in a balanced equation).
$\ce{N2H4(g) + 3O2(g) -> 2NO2(g) + 2H2O(g)}$
Mole ratio $\text{Hydrazine} : \ce{O2} : \ce{NO2} : \ce{H2O} = 1:3:2:2\ \ \Rightarrow\ $ matches the volume ratios given.
Therefore $\ce{N2H4}$ is the correct molecular formula for hydrazine.

b. $\text{pH} = 4.46$

Show Worked Solution

a. Using Avogadro’s law:

At the same temperature and pressure, gas volumes are proportional to moles (i.e. the volume ratio will be equal to the mole ratio in a balanced equation).
$\ce{N2H4(g) + 3O2(g) -> 2NO2(g) + 2H2O(g)}$
Mole ratio $\text{Hydrazine} : \ce{O2} : \ce{NO2} : \ce{H2O} = 1:3:2:2\ \ \Rightarrow\ $ matches the volume ratios given.
Therefore $\ce{N2H4}$ is the correct molecular formula for hydrazine.

b. $K_a(\ce{N2H5+}) = \dfrac{K_w}{K_b(\ce{N2H4})} = \dfrac{1 \times 10^{-14}}{1.7 \times 10^{-6}} = 5.88235 \times 10^{-9}$

The ionisation of $\ce{N2H5+}$ is given the chemical equation below:
$\ce{N2H5+(aq) + H2O(l) \leftrightharpoons N2H4(aq) + H3O+(aq)}$
$K_a = \dfrac{\ce{[H3O+][N2H4]}}{\ce{[N2H5+]}}$
Using an Ice Table where all numbers are in mol L$^{-1}$.

\begin{array} {|c|c|c|c|}
\hline & \ce{[N2H5+]} & \ce{[N2H4]} & \ce{[H3O+]} \\
\hline \text{Initial} & 0.20 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 0.20 -x & x & x \\
\hline \end{array}

Substituting into the $K_a$ expression:

$\dfrac{x^2}{0.20-x}$	$=5.88235 \times 10^{-9}$
$\dfrac{x^2}{0.20}$	$=5.88235 \times 10^{-9}$, as $x$ is really small
$x$	$=3.42997 \times 10^{-5}$

$\text{pH}$	$=-\log_{10}(\ce{[H3O+]})$
	$=-\log_{10}(3.42997 \times 10^{-5})$
	$=4.46$

CHEMISTRY, M8 2025 HSC 30

Phosgene is used in industry as a starting material to synthesise useful polymers. Phosgene $\ce{(Cl2CO)}$ is a gas at room temperature and is highly toxic.

Justify a suitable precaution when using phosgene. (2 marks)

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Phosgene is synthesised by the reaction of carbon monoxide $\ce{(CO)}$ and chlorine $\ce{(Cl2)}$ in the gas phase.
1. 1. $\ce{Cl2(g) + CO(g) \rightleftharpoons Cl2CO(g)}$
Explain why an excess of carbon monoxide and a catalyst are used in the industrial synthesis of phosgene. (3 marks)

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a. Precaution when using phosgene inside a fume hood:

As phosgene is a highly toxic gas at room temperature. working in a fume hood prevents inhalation by safely removing the gas from the breathing zone and venting it outside the laboratory.
This measure significantly reduces the risk of poisoning and exposure.

b. Role of excess carbon monoxide:

An excess of carbon monoxide is used to increase the rate of production of phosgene.
According to Le Chatelier’s principle, increasing the concentration of $\ce{CO}$ shifts the equilibrium to the right, favouring the formation of $\ce{Cl2CO}$ and increasing the overall yield of phosgene.

Role of catalyst:

A catalyst is used to increase the rate of reaction by providing an alternative reaction pathway with a lower activation energy.
This allows phosgene to be produced more rapidly and efficiently without affecting the equilibrium position.
In industrial settings, this increases production speed and reduces energy costs.

Show Worked Solution

a. Precaution when using phosgene inside a fume hood:

As phosgene is a highly toxic gas at room temperature. working in a fume hood prevents inhalation by safely removing the gas from the breathing zone and venting it outside the laboratory.
This measure significantly reduces the risk of poisoning and exposure.

b. Role of excess carbon monoxide:

An excess of carbon monoxide is used to increase the rate of production of phosgene.
According to Le Chatelier’s principle, increasing the concentration of $\ce{CO}$ shifts the equilibrium to the right, favouring the formation of $\ce{Cl2CO}$ and increasing the overall yield of phosgene.

Role of catalyst:

A catalyst is used to increase the rate of reaction by providing an alternative reaction pathway with a lower activation energy.
This allows phosgene to be produced more rapidly and efficiently without affecting the equilibrium position.
In industrial settings, this increases production speed and reduces energy costs.

CHEMISTRY, M7 2025 HSC 28

Kevlar and polystyrene are two common polymers.

A section of their structures is shown.

Kevlar is produced through a reaction of two different monomers, one of which is shown. Draw the missing monomer in the box provided. (1 mark)

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Kevlar chains are hard to pull apart, whereas polystyrene chains are not.
With reference to intermolecular forces, explain the difference in the physical properties of the two polymers. (3 marks)

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b. The physical differences between the two polymers are:

Kevlar chains are very hard to pull apart because the polymer contains many amide groups that can form strong hydrogen bonds between neighbouring chains. These strong forces hold the chains tightly together, making Kevlar rigid and very strong.
The close packing of the chains also gives Kevlar a high melting point, because a large amount of energy is required to break the hydrogen bonds.
Polystyrene, on the other hand, does not contain groups that can form hydrogen bonds. Its polymer chains are mostly non-polar, so the only forces between the chains are weak dispersion forces.
These weaker attractions mean the chains can slide past each other, making polystyrene much softer, brittle, and it also has a lower melting point than Kevlar. Because the forces between chains are weak, polystyrene is much easier to pull apart compared to Kevlar.

Show Worked Solution

b. The physical differences between the two polymers are:

Kevlar chains are very hard to pull apart because the polymer contains many amide groups that can form strong hydrogen bonds between neighbouring chains. These strong forces hold the chains tightly together, making Kevlar rigid and very strong.
The close packing of the chains also gives Kevlar a high melting point, because a large amount of energy is required to break the hydrogen bonds.
Polystyrene, on the other hand, does not contain groups that can form hydrogen bonds. Its polymer chains are mostly non-polar, so the only forces between the chains are weak dispersion forces.
These weaker attractions mean the chains can slide past each other, making polystyrene much softer, brittle, and it also has a lower melting point than Kevlar. Because the forces between chains are weak, polystyrene is much easier to pull apart compared to Kevlar.

CHEMISTRY, M7 2025 HSC 27

Mixtures of hydrocarbons can be obtained from crude oil by the process of fractional distillation. Examples include petrol, diesel and natural gas.

Outline an environmental implication for a use of a named hydrocarbon mixture that is obtained from crude oil. (2 marks)

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Ethene is a simple hydrocarbon obtained from crude oil.
Using structural formulae, write the chemical equation for the conversion of ethene to ethanol, including any other necessary reagents. (3 marks)

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When ethanol is reacted with ethanoic acid, ethyl ethanoate is formed, as shown by the equation.
1. 1. $\text{ethanol} \ + \ \text{ethanoic acid } \ \rightleftharpoons \ \text{ethyl ethanoate} \ +\ \text{water}$
The graph below shows the concentration of ethanol from the start of the reaction, $t_0$, up to a time $t_1$.
At time $t_1$, an additional amount of ethanol is added to the system.
Sketch on the graph the changes that occur in the concentration of ethanol between time $t_1$, and when the system reaches a new equilibrium before time $t_2$. (3 marks)
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a. Environmental implication:

The combustion of petrol, a hydrocarbon mixture obtained from crude oil, leads to the release of large amounts of carbon dioxide.
Increased $\ce{CO2}$ levels intensify the enhanced greenhouse effect, contributing to global warming, climate change, and associated environmental impacts such as rising sea levels and extreme weather patterns.
A typical component of petrol, octane, burns according to:
$\ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(g)}$

Show Worked Solution

a. Environmental implication:

The combustion of petrol, a hydrocarbon mixture obtained from crude oil, leads to the release of large amounts of carbon dioxide.
Increased $\ce{CO2}$ levels intensify the enhanced greenhouse effect, contributing to global warming, climate change, and associated environmental impacts such as rising sea levels and extreme weather patterns.
A typical component of petrol, octane, burns according to:
$\ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(g)}$

There will be a sudden increase at $t_1$ and then the concentration of ethanol will decrease smoothly until a new equlibrium concentration (greater than the original equilibrium concentration) is reached.

CHEMISTRY, M6 2025 HSC 26

A hydrogen atom on the methyl group of ethanoic acid can be replaced with a single halogen atom. A general formula for these haloethanoic acids is shown.

$\ce{X-CH2COOH \quad (X = F,Cl,Br or I)}$

The $p K_a$ values of the four haloethanoic acids are given in the table.

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Acid} \rule[-1ex]{0pt}{0pt}& \quad \quad \ce{X} \quad \quad & \quad \quad p K_a \quad \quad\\
\hline
\rule{0pt}{2.5ex}\text{Fluoroethanoic acid } \quad \rule[-1ex]{0pt}{0pt}& \ce{F} & 2.6 \\
\hline
\rule{0pt}{2.5ex}\text{Chloroethanoic acid } \rule[-1ex]{0pt}{0pt}& \ce{Cl} & 2.9 \\
\hline
\rule{0pt}{2.5ex}\text{Bromoethanoic acid } \rule[-1ex]{0pt}{0pt}& \ce{Br} & 2.9 \\
\hline
\rule{0pt}{2.5ex}\text{Iodoethanoic acid } \rule[-1ex]{0pt}{0pt}& \ce{I}& 3.2 \\
\hline
\end{array}

Construct an appropriate graph for the four haloethanoic acids, showing their $p K_a$ values and the identity of the halogen $\ce{X}$ in each molecule, in the order provided in the table. (3 marks)

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Describe the trend in the relative strengths of the haloethanoic acids. (2 marks)

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b. The strength of the haloethanoic acids can be explained by the relationship between $pK_a$, $K_a$ and the degree of ionisation.

Since $pK_a$ is inversely related to $K_a$ ($pK_a = -\log_{10}(K_a)$), a lower $pK_a$ corresponds to a higher $K_a$ and therefore greater ionisation in water.
Fluoroethanoic acid, with the lowest $pK_a$ value, has the highest $K_a$ and ionises the most, making it the strongest acid.
As you move down the halogen group from $\ce{F}$ to $\ce{I}$, $pK_a$ increases and $K_a$ decreases, so the acids ionise less and become weaker, with iodoethanoic acid being the weakest.

Show Worked Solution

b. The strength of the haloethanoic acids can be explained by the relationship between $pK_a$, $K_a$ and the degree of ionisation.

Since $pK_a$ is inversely related to $K_a$ ($pK_a = -\log_{10}(K_a)$), a lower $pK_a$ corresponds to a higher $K_a$ and therefore greater ionisation in water.
Fluoroethanoic acid, with the lowest $pK_a$ value, has the highest $K_a$ and ionises the most, making it the strongest acid.
As you move down the halogen group from $\ce{F}$ to $\ce{I}$, $pK_a$ increases and $K_a$ decreases, so the acids ionise less and become weaker, with iodoethanoic acid being the weakest.

CHEMISTRY, M7 2025 HSC 25

A student produced the ester propyl butanoate in the school laboratory, by refluxing 0.267 mol of propan-1-ol and 0.298 mol of butanoic acid with a catalyst.

Use the data in the table to calculate the percentage yield of the ester. (3 marks)

\begin{array}{|l|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Product} & \textit{Volume produced} &\quad \textit{Density} \quad & \quad \textit{Molar mass } \quad \\
\text{} \rule[-1ex]{0pt}{0pt}& \text{(mL)} &(\text{g mL}^{-1}) &(\text{g moL}^{-1}) \\
\hline
\rule{0pt}{2.5ex}\text{propyl butanoate} \rule[-1ex]{0pt}{0pt}& 12.2 & 0.873 & 130.2 \\
\hline
\end{array}

--- 10 WORK AREA LINES (style=lined) ---

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$30.6\%$

Show Worked Solution

The reflux reaction that takes place between propan-1-ol and butanoic acid is
$\ce{C3H8O + C4H8O2 \leftrightharpoons C7H14O2 + H2O}$
As the reaction occurs in $1:1:1$ mole ratio, propan-1-ol is the limiting reagent of the reaction. Hence the theortical number of moles of propyl butanoate produced is 0.267 mol.
The actual moles of propyl butanoate that are produced are:
$\text{mass = density \(\times$ volume}\ = 12.2 \times 0.873 = 10.6506\ \text{g}\)
$n_{\text{actual}} = \dfrac{m}{MM}= \dfrac{10.6506}{130.2} = 0.0818\ \text{mol}$

Hence the percentage yield of the ester $=\dfrac{0.0818}{0.267} \times 100 = 30.6$%.

CHEMISTRY, M7 2025 HSC 24

65.0 g of ethyne gas reacts with an excess of gaseous hydrogen chloride to produce chloroethene.

Draw the full structural formula of ethyne and identify the shape of the molecule. (2 marks)

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \text{Structural formula } \quad \quad \rule[-1ex]{0pt}{0pt}& \quad \quad\text{ Shape of molecule } \quad \quad\\
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\hline
\end{array}

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The molar masses of the compounds in the reaction are provided.

\begin{array}{|l|c|}
\hline \rule{0pt}{2.5ex}\text{Compound} \rule[-1ex]{0pt}{0pt}& \text{ Molar mass } \\
\hline \rule{0pt}{2.5ex}\text{Ethyne} \rule[-1ex]{0pt}{0pt}& 26.04 \\
\hline \rule{0pt}{2.5ex}\text{Hydrogen chloride} \rule[-1ex]{0pt}{0pt}& 36.46 \\
\hline \rule{0pt}{2.5ex}\text{Chloroethene} \rule[-1ex]{0pt}{0pt}& 62.50 \\
\hline
\end{array}

Calculate the mass of chloroethene produced, using the molar masses provided. Include a relevant chemical equation in your answer. (3 marks)

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a. Structural formula:

The shape of ethyne is linear.

b. 156 grams

Show Worked Solution

a. Structural formula:

The shape of ethyne is linear.

b. The chemical equation for the reacton occuring is shown below:

$\ce{C2H2(g) + HCl(g) -> C2H3Cl(g)}$

$n\ce{(C2H2)} = \dfrac{m}{MM} = \dfrac{65.0}{26.04} = 2.496\ \text{mol}$
As the reactants and products are in a $1:1:1$ ratio, $n\ce{(C2H2)}_{\text{reacted}} = n\ce{(C2H3Cl)}_{\text{produced}}$
$m\ce{(C2H3Cl)} = n \times MM = 2.496 \times 62.50 = 156\ \text{grams (3 sig.fig.)}$

CHEMISTRY, M8 2025 HSC 23

A student attempted to determine the % w/w of sulfate in a sample of solid fertiliser. They used the procedure described below.

Weigh a clean, dry beaker.
Add fertiliser to the beaker and weigh again.
Add 250 mL of distilled water and stir thoroughly.
Add 20 mL of 0.1 mol L$^{-1} \ \ce{BaCl2}$ solution.
Filter out the $\ce{BaSO4}$ precipitate, using distilled water to ensure all of the solid is transferred from the beaker to the filter paper.
Put the filter paper and precipitate onto a weighed watch glass and leave them to dry for 20 minutes in the sun.
Weigh the watch glass, the filter paper and the precipitate.
Calculate the % w/w.

Justify TWO changes that can be made to the procedure to ensure more accurate results. (3 marks)

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Change 1: Dry the precipitate in an oven or desiccator instead of in the sun.

Drying in the sun is inconsistent and may leave moisture in the precipitate, causing an overestimation of mass.
Using an oven at a controlled temperature ensures all water is removed, giving a more accurate mass of precipitate.
This would also prevent impurities from the air such as dust or particles mixing with the sample.

Change 2: Measure the mass of the filter paper initially without the precipitate.

Weighing the filter paper, precipitate and watch glass together without knowing the mass of the precipitate (the watch glass has been weighed) inflates the mass of the $\ce{BaSO4}$ causing inaccuracy in the results.

CHEMISTRY, M7 2025 HSC 21

Consider the following organic reaction.

In the space provided, identify reaction condition $X$ and name the organic product. (2 marks)

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \quad \textit{Reaction condition X} \quad \quad \rule[-1ex]{0pt}{0pt}& \quad \textit{IUPAC name of organic product } \quad\\
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt}& \\
\hline
\end{array}

--- 0 WORK AREA LINES (style=lined) ---

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Reaction condition $X$: UV light
IUPAC name: 2-bromobutane

Show Worked Solution

Reaction condition $X$: UV light
IUPAC name: 2-bromobutane

CHEMISTRY, M5 2025 HSC 13 MC

Which row of the table correctly shows the expected signs of the enthalpy $(\Delta H)$ and entropy $(\Delta S)$ changes for the complete combustion of octane above 100°C?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \Delta H \quad \quad\rule[-1ex]{0pt}{0pt}& \quad \quad\Delta S \quad \quad\\
\hline
\rule{0pt}{2.5ex}>0\rule[-1ex]{0pt}{0pt}&<0\\
\hline
\rule{0pt}{2.5ex}<0\rule[-1ex]{0pt}{0pt}& <0\\
\hline
\rule{0pt}{2.5ex}<0\rule[-1ex]{0pt}{0pt}& >0 \\
\hline
\rule{0pt}{2.5ex}>0\rule[-1ex]{0pt}{0pt}& >0 \\
\hline
\end{array}
\end{align*}

Show Answers Only

$C$

Show Worked Solution

When the complete combustion of octane occurs at above 100°C, all $\ce{H2O}$ will be gas. Therefore the chemical equation for the reaction is:
- $\ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(g)}$
A combustion reaction will always be an exothermic reaction $(\Delta H < 0)$.
As there are 34 gas molecules produced from 25 gas molecules and 2 liquid molecules, the entropy (disorder of the system) increases, hence $\Delta S > 0$.

$\Rightarrow C$

CHEMISTRY, M5 2025 HSC 12 MC

Consider the following reaction.

$\ce{3AgNO3(aq) + FeCl3(aq) \rightleftharpoons 3AgCl(s) + Fe(NO3)3(aq)}$

What is the correct equilibrium expression for this reaction?

$\dfrac{\left[\ce{Fe(NO3)3}\right]}{\left[\ce{AgNO3}\right]\left[\ce{FeCl3}\right]}$
$\dfrac{\left[\ce{Fe(NO3)3}\right]}{\left[\ce{AgNO3}\right]^3\left[\ce{FeCl3}\right]}$
$\dfrac{\left[\ce{AgNO3}\right]^3 \left[\ce{FeCl3}\right]}{\ce{[AgCl]^3}\left[ \ce{Fe}\ce{(NO3)_3}\right]}$
$\dfrac{\ce{[AgCl]^3}\left[ \ce{Fe}\ce{(NO3)_3}\right]}{\left[\ce{AgNO3}\right]^3\left[\ce{FeCl3}\right]}$

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$B$

Show Worked Solution

The equilibrium expression does not include reactants or products that are in the solid or liquid state, hence $\ce{3AgCl(s)}$ is not included in the equilibrium expression.
$\therefore K_{eq} = \dfrac{\ce{[Fe(NO3)3]}}{\ce{[AgNO3]^3[FeCl3]}}$

$\Rightarrow B$

CHEMISTRY, M7 2025 HSC 10 MC

Which of the following options lists ALL the forces that are present between molecules of butanoic acid?

Covalent bonding
Dispersion and dipole-dipole
Covalent bonding and hydrogen bonding
Dispersion, dipole-dipole and hydrogen bonding

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$D$

Show Worked Solution

Hydrogen bonding occurs due to the $\ce{OH}$ group in the carboxylic acid, which can form strong hydrogen bonds between molecules.
Dipole-dipole forces arise from the polar $\ce{C=O}$ bond in the carboxyl group (carbonyl oxygen is highly electronegative).
Dispersion forces are always present between molecules due to temporary dipoles, especially from the nonpolar hydrocarbon chain.

$\Rightarrow D$

CHEMISTRY, M7 2025 HSC 9 MC

Some Torres Strait Islander Peoples pound the leaves of the vine Derris uliginosa to extract a chemical called saponin. Saponin is a relatively large molecule that contains both a water-soluble carbohydrate chain and a fat-soluble side chain.

Which is the most likely use of saponin?

As a cleaning agent
As a neutraliser of insect bites
As a dye for pigmentation and painting
As an additive to food for flavouring and tenderising

Show Answers Only

$A$

Show Worked Solution

Saponin has both water-soluble and fat-soluble parts, allowing it to dissolve fats in water and act like a soap or detergent.
The fat-soluble side chain bonds with dirts and fats while the water soluble component bonds with the water molecules to form micelle.
The micelles can then be washed away leaving surfaces clean.

$\Rightarrow A$

CHEMISTRY, M8 2025 HSC 8 MC

An alkene $\text{X}$ with only one $\text{C} \ = \ \text{C}$ bond undergoes an addition reaction with an unknown substance to produce $\text{Y}$ .

$\text{X}\ +\ \text{unknown substance} \rightarrow \text{Y}$

The following table shows the molecular ion peaks for $\text{X}$ and $\text{Y}$.

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\text{X} \rule[-1ex]{0pt}{0pt}& \text{Y} \\
\hline
\rule{0pt}{2.5ex} \quad \quad 70 \ \text{m/z} \quad \quad\rule[-1ex]{0pt}{0pt}& \quad \quad 90 \ \text{m/z} \quad \quad \\
\hline
\end{array}

Which of the following can be the unknown substance?

Water
Fluorine
Hydrogen
Hydrogen fluoride

Show Answers Only

$\Rightarrow D$

Show Worked Solution

The molecule ion peaks for $\text{X}$ and $\text{Y}$ correspond to their molecular masses.
By the law of conservation of mass, the mass of the unknown substance must be the difference in mass between the substances $\text{X}$ and $\text{Y}$.
Hence the molecular mass of the unknown substance $=90-70 = 20\ \text{g mol}^{-1}$.
This corresponds to the molecular mass of $\ce{HF} = 1.008 + 19.00 = 20.008\ \text{g mol}^{-1}$.

$\Rightarrow D$

CHEMISTRY, M8 2025 HSC 7 MC

Copper ions can form coloured complexes with water molecules and with chloride ions in dilute aqueous solutions.

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Complex ion} \rule[-1ex]{0pt}{0pt}& \quad \quad \text{Colour} \quad \quad\\
\hline
\rule{0pt}{2.5ex} \left[\ce{Cu}\left(\ce{H2O}\right)_6\right]^{2+} \rule[-1ex]{0pt}{0pt}& \text{Blue} \\
\hline
\rule{0pt}{2.5ex} \left[ \ce{CuCl4}\right]^{2-} \rule[-1ex]{0pt}{0pt}& \text{Green} \\
\hline
\end{array}

Which of the following analytical techniques would be most suitable to distinguish between these two complexes?

Infrared spectrophotometry
Carbon-13 NMR spectroscopy
UV-visible spectrophotometry
Atomic absorption spectroscopy

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$C$

Show Worked Solution

The complexes $\left[\ce{Cu}\left(\ce{H2O}\right)_6\right]^{2+}$ (blue) and $\left[ \ce{CuCl4}\right]^{2-}$ (green) differ in colour because each absorbs light of different wavelengths in the visible region of the electromagnetic spectrum.
UV-visible spectrophotometry measures how much light a compound absorbs at each wavelength in the UV–visible range, so it can clearly distinguish between the two complexes by their distinct absorption spectra.

$\Rightarrow C$

CHEMISTRY, M6 2025 HSC 6 MC

What is the pH of a 0.25 mol L$^{-1}$ solution of hydrochloric acid?

$-0.60$
$-0.25$
$0.25$
$0.60$

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$D$

Show Worked Solution

Hydrochloric acid is a strong acid and will fully ionise in solution.
Therefore $\ce{[HCl] = [H+] = 0.25\ \text{mol L}^{-1}}$.

$\text{pH}$	$=-\log_{10}(\ce{[H+]})$
	$=-\log_{10}(0.25)$
	$=0.60$

$\Rightarrow D$

CHEMISTRY, M7 2025 HSC 2 MC

Consider this reaction.

Which reaction type is shown?

Addition
Oxidation
Reduction
Substitution

Show Answers Only

$B$

Show Worked Solution

The reaction shows butan-2-ol being converted to butanone.
This occurs via oxidation where a secondary alcohol is converted to a ketone.

$\Rightarrow B$

Mechanics, EXT2 M1 2025 HSC 8 MC

The graph shows the velocity of a particle as a function of its displacement.

Which of the following graphs best shows the acceleration of the particle as a function of its displacement?

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$A$

Show Worked Solution

$\text{By elimination:}$

$v(x)\ \Rightarrow\ \text{degree 3 (see graph)},\ \ \dfrac{dv}{dx}\ \Rightarrow\ \text{degree 2}$

$a=v \cdot \dfrac{dv}{dx}\ \Rightarrow\ \text{degree 5 (eliminate C and D)}$

$\text{At}\ \ x=0,\ \ v(x)>0\ \ \text{and}\ \ \dfrac{dv}{dx}<0\ \ \Rightarrow\ \ v \cdot \dfrac{dv}{dx} \neq 0\ \text{(eliminate B)}$

$\Rightarrow A$

Complex Numbers, EXT2 N2 2025 HSC 15c

Show that
$\dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \dfrac{\theta}{2}.$ (3 marks)

--- 12 WORK AREA LINES (style=lined) ---
Use De Moivre's theorem to show that the sixth roots of $-1$ are given by
$\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right)$ for $k=0,1,2,3,4,5$. (2 marks)

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Hence, or otherwise, show the solutions to $\left(\dfrac{z-1}{z+1}\right)^6=-1$ are
$z=i \cot \left(\dfrac{\pi}{12}\right), i \cot \left(\dfrac{3 \pi}{12}\right), i \cot \left(\dfrac{5 \pi}{12}\right), i \cot \left(\dfrac{7 \pi}{12}\right), i \cot \left(\dfrac{9 \pi}{12}\right)$, and $i \cot \left(\dfrac{11 \pi}{12}\right)$. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

i. $\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\dfrac{\theta}{2}\right)$

$\text{LHS}$	$=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}$
	$=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}$
	$=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}$
	$=i \cot \left(\frac{\theta}{2}\right)$

ii. $z=\cos \theta+i \sin \theta$

$\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}$

$z^6=\cos (6 \theta)+i \sin (6 \theta)=-1$

$\cos (6 \theta)$	$=-1 \ \text{and} \ \ \sin (6 \theta)=0$
$6 \theta$	$=\pi, 3 \pi, 5 \pi, \ldots$
$\theta$	$=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5$

$\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for} \ \ k=0,1, \ldots, 5$

iii. $\left(\dfrac{z-1}{z+1}\right)^6=-1$

$\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1$

$\text {Using part (ii):}$

$\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)$

$\alpha=\dfrac{z-1}{z+1}\ \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}$

$\text{Consider} \ \ \alpha=\operatorname{cis}\left(\dfrac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}$

$z=\dfrac{1+\operatorname{cis}\left(\dfrac{\pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{\pi}{12}\right) \quad \text{(using part (i))}$

$\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{3 \pi}{12}\right)$

$\therefore z=i \cot \left(\dfrac{\pi}{12}\right), \, i \cot \left(\dfrac{3 \pi}{12}\right), \, i \cot \left(\dfrac{5 \pi}{12}\right), \ldots, i \cot \left(\dfrac{11 \pi}{12}\right)$

Show Worked Solution

i. $\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\dfrac{\theta}{2}\right)$

$\text{LHS}$	$=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}$
	$=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}$
	$=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}$
	$=i \cot \left(\frac{\theta}{2}\right)$

ii. $z=\cos \theta+i \sin \theta$

$\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}$

$z^6=\cos (6 \theta)+i \sin (6 \theta)=-1$

$\cos (6 \theta)$	$=-1 \ \text{and} \ \ \sin (6 \theta)=0$
$6 \theta$	$=\pi, 3 \pi, 5 \pi, \ldots$
$\theta$	$=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5$

$\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for} \ \ k=0,1, \ldots, 5$

iii. $\left(\dfrac{z-1}{z+1}\right)^6=-1$

$\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1$

$\text {Using part (ii):}$

$\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)$

$\alpha=\dfrac{z-1}{z+1}\ \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}$

$\text{Consider} \ \ \alpha=\operatorname{cis}\left(\dfrac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}$

$z=\dfrac{1+\operatorname{cis}\left(\dfrac{\pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{\pi}{12}\right) \quad \text{(using part (i))}$

$\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{3 \pi}{12}\right)$

$\therefore z=i \cot \left(\dfrac{\pi}{12}\right), \, i \cot \left(\dfrac{3 \pi}{12}\right), \, i \cot \left(\dfrac{5 \pi}{12}\right), \ldots, i \cot \left(\dfrac{11 \pi}{12}\right)$

Mechanics, EXT2 M1 2025 HSC 15b

A particle moves in simple harmonic motion about the origin with amplitude $A$, and it completes two cycles per second. When it is $\dfrac{1}{4}$ metres from the origin, its speed is half its maximum speed.

Find the maximum positive acceleration of the particle during its motion. (4 marks)

--- 12 WORK AREA LINES (style=lined) ---

Show Answers Only

$a_{\text{max}}=\dfrac{8 \pi^2}{\sqrt{3}}$

Show Worked Solution

$v^2=-n^2\left(x^2-A^2\right)$

$\operatorname{Period}\ (T)=\dfrac{1}{2} \ \Rightarrow \ \dfrac{2 \pi}{n}=\dfrac{1}{2} \ \Rightarrow \ n=4 \pi$

$\text{Max velocity occurs at}\ \ x=0:$

$v_{\text{max}}^2=-(4 \pi)^2\left(0-A^2\right)=16 \pi^2 A^2$

$v_{\text{max}}=\sqrt{16 \pi^2 A^2}=4 \pi A$

$\text{At} \ \ x=\dfrac{1}{4}, \ v=\dfrac{1}{2} \times 4 \pi A=2 \pi A$

$(2 \pi A)^2$	$=-(4 \pi)^2\left(\dfrac{1}{16}-A^2\right)$
$\dfrac{A^2}{4}$	$=A^2-\dfrac{1}{16}$
$\dfrac{3 A^2}{4}$	$=\dfrac{1}{16}$
$A^2$	$=\dfrac{1}{12}$
$A$	$=\dfrac{1}{\sqrt{12}}$

$\text{Max positive acceleration occurs at} \ \ x=-\dfrac{1}{\sqrt{12}}:$

$a_{\text{max}}=-n^2 x=-(4 \pi)^2 \times-\dfrac{1}{\sqrt{12}}=\dfrac{8 \pi^2}{\sqrt{3}}$

Proof, EXT2 P1 2025 HSC 14d

Positive real numbers $a, b, c$ and $d$ are chosen such that $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ and $\dfrac{1}{d}$ are consecutive terms in an arithmetic sequence with common difference $k$, where $k \in \mathbb{R} , k>0$.

Show that $b+c<a+d$. (3 marks)

--- 7 WORK AREA LINES (style=lined) ---

Show Answers Only

$\text{AP:} \ \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}, \dfrac{1}{d} \ \ \text{where common difference}=k\ \ (k>0)$

$\text{Show} \ \ b+c<a+d$

$\dfrac{1}{a}<\dfrac{1}{a}+k(k>0)$

$\dfrac{1}{a}<\dfrac{1}{b}$

$\text{Similarly for}\ \ \dfrac{1}{b}<\dfrac{1}{c}\ \ldots\ \text{such that}$

$\dfrac{1}{a}<\dfrac{1}{b}<\dfrac{1}{c}<\dfrac{1}{d} \ \ \Rightarrow\ \ a>b>c>d\ \ldots\ (1)$

$\dfrac{1}{b}=\dfrac{1}{a}+k \ \ \Rightarrow \ \ k=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{a b}\ \ \Rightarrow \ \ a-b=kab$

$\text {Similarly:}$

$\dfrac{1}{d}=\dfrac{1}{c}+k \ \ \Rightarrow \ \ k=\dfrac{c-d}{cd}\ \ \Rightarrow\ \ c-d=kcd$

$\text{Using \(\ a>b>c>d\ $ (see (1) above):}\)

$a-b>c-d$

$b+c<a+d$

Show Worked Solution

$\text{AP:} \ \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}, \dfrac{1}{d} \ \ \text{where common difference}=k\ \ (k>0)$

$\text{Show} \ \ b+c<a+d$

$\dfrac{1}{a}<\dfrac{1}{a}+k\ \ (k>0)$

$\dfrac{1}{a}<\dfrac{1}{b}$

$\text{Similarly for}\ \ \dfrac{1}{b}<\dfrac{1}{c}\ \ldots\ \text{such that}$

$\dfrac{1}{a}<\dfrac{1}{b}<\dfrac{1}{c}<\dfrac{1}{d} \ \ \Rightarrow\ \ a>b>c>d\ \ldots\ (1)$

$\dfrac{1}{b}=\dfrac{1}{a}+k \ \ \Rightarrow \ \ k=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{a b}\ \ \Rightarrow \ \ a-b=kab$

$\text {Similarly:}$

$\dfrac{1}{d}=\dfrac{1}{c}+k \ \ \Rightarrow \ \ k=\dfrac{c-d}{cd}\ \ \Rightarrow\ \ c-d=kcd$

$\text{Using \(\ a>b>c>d\ $ (see (1) above):}\)

$a-b>c-d$

$b+c<a+d$

\(\dfrac{mv_B^2}{r}\)	\(=T+mg\)
\(v_B^2\)	\(=rg \ \ (T=0)\)
\(v_B\)	\(=\sqrt{r g}\)

\(\text {Total} \ ME\)	\(=\dfrac{1}{2} m v_B^2+mg(2r)\)
	\(=\dfrac{1}{2} m \times r g+2 mrg\)
	\(=\dfrac{5}{2} m r g\)

\(\dfrac{5}{2} mrg\)	\(=\dfrac{1}{2} m v_A^2+m r g\)
\(\dfrac{1}{2} mv_A^2\)	\(=\dfrac{3}{2} m r g\)
\(v_A^2\)	\(=3 rg\)
\(v_A\)	\(=\sqrt{3rg}=\sqrt{3} \times v_B\)

\(\dfrac{mv_B^2}{r}\)	\(=T+mg\)
\(v_B^2\)	\(=rg \ \ (T=0)\)
\(v_B\)	\(=\sqrt{r g}\)

\(\text {Total} \ ME\)	\(=\dfrac{1}{2} m v_B^2+mg(2r)\)
	\(=\dfrac{1}{2} m \times r g+2 mrg\)
	\(=\dfrac{5}{2} m r g\)

\(F_C\)	\(=F_G\)
\(\dfrac{m v^2}{r}\)	\(=\dfrac{G Mm}{r}\)
\(\dfrac{1}{2} m v^2\)	\(=\dfrac{G M m}{2 r}\)

\(\dfrac{GMm_{A}}{2r_A}\)	\(=\dfrac{GMm_{B}}{2r_B}\)
\(\dfrac{m_A}{r_A}\)	\(=\dfrac{m_B}{r_B}\)
\(\dfrac{m_A}{3r_B}\)	\(=\dfrac{m_B}{r_B}\left(r_A=3 \times r_B\right)\)
\(m_A\)	\(=3 \times m_B \ \text{… as required}\)

\(F_C\)	\(=F_G\)
\(\dfrac{m v^2}{r}\)	\(=\dfrac{G Mm}{r}\)
\(\dfrac{1}{2} m v^2\)	\(=\dfrac{G M m}{2 r}\)

\(\dfrac{GMm_{A}}{2r_A}\)	\(=\dfrac{GMm_{B}}{2r_B}\)
\(\dfrac{m_A}{r_A}\)	\(=\dfrac{m_B}{r_B}\)
\(\dfrac{m_A}{3r_B}\)	\(=\dfrac{m_B}{r_B}\left(r_A=3 \times r_B\right)\)
\(m_A\)	\(=3 \times m_B \ \text{… as required}\)