Band 4 – Page 3

Financial Maths, STD1 EQ-Bank 11

The table shows JobSeeker Payment rates per fortnight.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Your situation} \rule[-1ex]{0pt}{0pt} & \textbf{Maximum fortnightly payment} \\
\hline
\rule{0pt}{2.5ex} \text{Single, no children} \rule[-1ex]{0pt}{0pt} & $793.60 \\
\hline
\rule{0pt}{2.5ex} \text{Single, with a dependent child or children} \rule[-1ex]{0pt}{0pt} & $849.90 \\
\hline
\rule{0pt}{2.5ex} \text{Partnered} \rule[-1ex]{0pt}{0pt} & $726.50 \\
\hline
\end{array}

James is single with a dependent child and receives the maximum JobSeeker Payment. How much does he receive per fortnight? (1 mark)
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James earns $280 per fortnight from casual work. His JobSeeker Payment is reduced by 50 cents for each dollar earned over $150 per fortnight. Calculate James's reduced JobSeeker Payment. (1 mark)
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a. $$849.90$

b. $$784.90\ \text{per fortnight}$

Show Worked Solution

a. $\text{From table: Single, with a dependent child} = $849.90$

b. $\text{Income over free area} = 280-150=$130$

$\text{Reduction in payment} = 0.50\times 130=$65$

$\text{Reduced JobSeeker Payment} = 849.90-65=$784.90$

$\therefore\ \text{James receives }$784.90\ \text{per fortnight}$

Financial Maths, STD2 EQ-Bank 15 MC

Margaret is single and receives the maximum Age Pension payment of $1,178.70 per fortnight.

She earns $280 per fortnight from part-time work. Her pension is reduced by 50 cents for each dollar earned over $218 per fortnight.

What is Margaret's total fortnightly income (including her Age Pension and earnings)?

$1,427.70
$1,447.70
$1,458.70
$1,489.70

Show Answers Only

$A$

Show Worked Solution

$\text{Maximum Age Pension} = $663.30$

$\text{Income over free area} = 280-218 = $62$

$\text{Reduction in payment} = 0.50\times 62 = $31$

$\text{Age Pension received} = 1178.70-31 = $1,147.70$

$\therefore\ \text{Total fortnightly income} = 1147.70+280 = $1,427.70$

$\Rightarrow A$

Financial Maths, STD2 EQ-Bank 14 MC

Emma is 20 years old, single with no children, and lives away from her parents' home to study.

She receives the maximum Youth Allowance payment of $663.30 per fortnight.

Emma earns $420 per fortnight from part-time work. Her Youth Allowance is reduced by 50 cents for each dollar earned over $236 per fortnight.

What is Emma's total fortnightly income (including her Youth Allowance payment and earnings)?

$899.30
$991.30
$1011.30
$1083.30

Show Answers Only

$B$

Show Worked Solution

$\text{Maximum Youth Allowance} = $663.30$

$\text{Income over free area} = 420-236 = $184$

$\text{Reduction in payment} = 0.50\times 184 = $92$

$\text{JobSeeker Payment received} = 663.30-92 = $571.30$

$\therefore\ \text{Total fortnightly income} = 571.30+420 = $991.30$

$\Rightarrow B$

Financial Maths, STD2 EQ-Bank 13 MC

Sarah is single with no children and receives the maximum JobSeeker Payment of $793.60 per fortnight.

She earns $350 per fortnight from casual work. The JobSeeker Payment is reduced by 50 cents for each dollar earned over $150 per fortnight.

What is Sarah's total fortnightly income (including her JobSeeker Payment and earnings)?

$893.60
$943.60
$993.60
$1043.60

Show Answers Only

$D$

Show Worked Solution

$\text{Maximum JobSeeker Payment (single, no children)} = $793.60$

$\text{Income over free area} = 350-150 = $200$

$\text{Reduction in payment} = 0.50\times 200 = $100$

$\text{JobSeeker Payment received} = 793.60-100 = $693.60$

$\therefore\ \text{Total fortnightly income} = 693.60+350 = $1043.60$

$\Rightarrow D$

Functions, EXT1′ F1 2007 HSC 3a*

The diagram shows the graph of $y = f(x)$. The line $y = x$ is an asymptote.

Draw separate one-third page sketches of the graphs of the following:

$f(\abs{x})$. (2 marks)

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$f(x)-x$. (2 marks)

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ii.

Show Worked Solution

MARKER’S COMMENT: In part (i), a significant number of students graphed $y=\abs{f(x)}$.

ii.

Functions, EXT1′ F1 2019 HSC 12d

Consider the function $f(x) = x^3-1$.

Sketch the graph $y = \abs{f(x)}$. (1 mark)

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Sketch the graph $y = \dfrac{1}{f(x)}$. (2 marks)

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i. $y = \abs{x^3-1}$

ii. $y = \dfrac{1}{x^3-1}$

Show Worked Solution

i. $y = \abs{x^3-1}$

ii. $y = \dfrac{1}{x^3-1}$

Functions, EXT1′ F1 2016 HSC 11dii

The diagram shows the graph of `y = f(x).`

Draw a graph of `y = 1/(f(x))`, showing all asymptotes and intercepts. (3 marks)

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Show Worked Solution

Functions, EXT1′ F1 2009 HSC 3ai

The diagram shows the graph `y = f(x).`

Draw the graphs of `y = 1/(f(x)) .` (2 marks)

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Show Worked Solution

`text(Vertical asymptotes at)\ x=0\ \text{and}\ x=4.`

`text(Horizontal asymptote at)\ y=-1/3.`

Functions, EXT1 EQ-Bank 07

A cubic function is given by $f(x)=\left(2 x^2+3 x-5\right)(x+2)$

Find the zeros of $f(x)$. (1 mark)

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Hence, or otherwise, solve $f(x) \geqslant 0$, giving your answer in set notation. (2 marks)

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a. $\text{Zeros at} \ \ x=-\dfrac{5}{2}, x=1 \ \ \text{and}\ \ x=-2$

b. $x \in\left[-\frac{5}{2},-2\right] \cup\ x \in[1, \infty)$

Show Worked Solution

a. $f(x)=\left(2 x^2+3 x-5\right)(x+2)=(2 x+5)(x-1)(x+2)$

$\text{Zeros at} \ \ x=-\dfrac{5}{2}, x=1 \ \ \text{and}\ \ x=-2$

b. $\text{Find} \ x \ \text{such that} \ \ f(x) \geqslant 0.$

$\text{At} \ \ x=0: \ (5)(-1)(2)<0$

$\therefore \text{Graph}\ (f(x)) \geqslant 0 \ \ \text{for} \ \ x \in\left[-\frac{5}{2},-2\right] \cup\ x \in[1, \infty)$

Functions, 2ADV EQ-Bank 2

Graph $y=4$ and $y=|2 x+2|$ on the same number plane. (2 marks)

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Hence, solve $|2 x+2|=4$. (1 mark)

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b. $\text{Solutions occur at graph intersections:}$

$\Rightarrow x=1 \ \ \text{or} \ \ x=-3.$

Show Worked Solution

b. $\text{Solutions occur at graph intersections:}$

$\Rightarrow x=1 \ \ \text{or} \ \ x=-3.$

Functions, 2ADV EQ-Bank 10

The function $y=f(x)$ is defined by:

\begin{align*}
f(x)= \begin{cases}|x+1|-2, & \text { for }\ x \leqslant 1 \\ x^2-4, & \text { for }\ x>1\end{cases}
\end{align*}

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

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For what values of $x$ is $f(x)=0$? (1 mark)

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a.

b. $f(x)=0\ \ \text{when}\ \ x=-3,2$.

Show Worked Solution

a.

b. $f(x)=0\ \ \text{when}\ \ x=-3,2$.

Functions, 2ADV EQ-Bank 9

Consider the function

\begin{align*}
f(x)=\begin{cases}2^x, & \text {for }\ x<0 \\ m x+c, & \text {for }\ 0 \leq x \leq 2 \\ \dfrac{8}{x}, & \text {for }\ x>2\end{cases}
\end{align*}

Given that $f(x)$ is continuous at both $x =0$ and $x =2$:

Find the values of $m$ and $c$. (2 marks)

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Identify any asymptotes of $y=f(x)$. (1 mark)

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a. $f(x)=\left\{\begin{array}{cl}2^x & \text {for } x<0 \\ m x+c & \text {for } 0 \leq x \leq 2 \\ \dfrac{8}{x} & \text {for } x>2\end{array}\right.$

$\text {Continuous at}\ \ x=0:$

$2^\circ=m(0)+c \ \Rightarrow \ c=1$

$\text {Continuous at}\ \ x=2:$

$2 m+1=\dfrac{8}{2} \ \Rightarrow \ m=\dfrac{3}{2}$

b. $\text{Asymptotes:}$

$\text{As} \ x \rightarrow-\infty, 2^x \rightarrow 0^{+}$

$\text{As} \ x \rightarrow \infty, \dfrac{8}{x} \rightarrow 0^{+}$

$\text{Asymptote at} \ \ y=0.$

$\text{There are no vertical asymptotes.}$

Show Worked Solution

a. $f(x)=\left\{\begin{array}{cl}2^x & \text {for } x<0 \\ m x+c & \text {for } 0 \leq x \leq 2 \\ \frac{8}{x} & \text {for } x>2\end{array}\right.$

$\text {Continuous at}\ \ x=0:$

$2^\circ=m(0)+c \ \Rightarrow \ c=1$

$\text {Continuous at}\ \ x=2:$

$2 m+1=d\dfrac{8}{2} \ \Rightarrow \ m=\dfrac{3}{2}$

b. $\text{Asymptotes:}$

$\text{As} \ x \rightarrow-\infty, 2^x \rightarrow 0^{+}$

$\text{As} \ x \rightarrow \infty, \dfrac{8}{x} \rightarrow 0^{+}$

$\text{Asymptote at} \ \ y=0.$

$\text{There are no vertical asymptotes.}$

Functions, 2ADV EQ-Bank 8

The temperature $T$ (in °C) in a greenhouse follows the pattern:

\begin{align*}
T(h)= \begin{cases}10+2 h, & \text {for }\ 0 \leqslant h<6 \\ 22, & \text {for }\ 6 \leqslant h \leqslant 18 \\ 58-2 h, & \text {for }\ 18<h \leqslant 24\end{cases}
\end{align*}

where $h$ is the number of hours after midnight.

Sketch the graph of $T(h)$ for $0 \leqslant h \leqslant 24$ (2 marks)

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At what time(s) during the day is the temperature exactly 18 °C? (2 marks)

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b. $\text{Temperature is 18° at 4 am and 8 pm.}$

Show Worked Solution

b. $\text{Temperature}=18^{\circ} \ \text{twice (see graph)}$

$10+2 h=18 \ \Rightarrow \ h=4$

$58-2 h=18 \ \Rightarrow \ h=20$

$\therefore \ \text{Temperature is 18° at 4 am and 8 pm.}$

Functions, 2ADV EQ-Bank 7

Consider the function $h(x)=\begin{cases}\dfrac{x^2-9}{x-3}, & \text {for } x \neq 3 \\ k, & \text {for } x=3\end{cases}$

For what value of $k$ is $h(x)$ continuous at $x =3$? (2 marks)

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Sketch $y=h (x)$ for this value of $k$. (2 marks)

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a. $\text{Since} \ \ \dfrac{x^2-9}{x-3}=\dfrac{(x+3)(x-3)}{x-3}=x+3$

$\text{As} \ \ x \rightarrow 3, x+3 \rightarrow 6$

$h(x) \ \text {is continuous when}\ \ k=6$

Show Worked Solution

a. $\text{Since} \ \ \dfrac{x^2-9}{x-3}=\dfrac{(x+3)(x-3)}{x-3}=x+3$

$\text{As} \ \ x \rightarrow 3, x+3 \rightarrow 6$

$h(x) \ \text {is continuous when}\ \ k=6$

Financial Maths, STD2 EQ-Bank 18

Michael is a children's book author. His publisher printed 25000 copies of his new book and after 6 months there are 9200 copies left in stock. Michael receives 12% of the retail price as royalties.

How many copies of Michael's book were sold? (1 mark)
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Calculate Michael's royalty payment if the retail price is $19.95. (2 marks)
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The publisher decides to sell the remaining stock at a reduced price of $15.00 per book. Calculate Michael's total royalty payment if all the remaining books are sold at this price. (2 marks)
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a. $15\,800\ \text{copies}$

b. $$37\,825.20$

c. $$54\,385.20$

Show Worked Solution

a. $\text{Copies sold}\ =25\,000-9200\ =15\,800\ \text{copies}$

b. $\text{Total sales}\ =15\,800\times 19.95=$315\,210$

$\text{Royalty payment}$	$=12\% \times 315\,210$
	$= 0.12\times 315\,210$
	$=$37\,825.20$

c. $\text{Sales from remaining books}\ =9200\times 15.00\ =$138\,000$

$\text{Total royalty on remaining books}\ =0.12\times 138\,000=$16\,560$

$\text{Total royalty when all books sold}\ = 37\,825.20 + 16\,560\ =$54\,385.20$

$\therefore\ \text{Michael’s total royalty payment is }$54\,385.20$

Financial Maths, STD2 EQ-Bank 24

Sophie is a songwriter and receives a royalty of 9.5% on the sales of her music album.

Calculate Sophie's royalty payment if 15600 albums were sold at $22.95 each. (2 marks)

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$$34\,011.90$

Show Worked Solution

$\text{Total sales}\ =15\,600 \times 22.95\ =$358\,020$

$\text{Royalty payment}$	$=9.5\% \times 358\,020$
	$= 0.095\times 358\,020$
	$=$34\,011.90$

$\therefore\ \text{Sophie’s royalty payment is }$34\,011.90$

Financial Maths, STD2 EQ-Bank 10 MC

An author receives a 15% royalty on book sales. In the first quarter, 12000 copies were printed and 8400 copies were sold at $32.50 each.

What is the author's royalty payment for the first quarter?

$40 950
$41 400
$58 500
$59 800

Show Answers Only

$A$

Show Worked Solution

$\text{Copies sold} = 8400$

$\text{Total sales}\ =8400 \times 32.50\ =$273\,000$

$\text{Royalty payment}=15\% \times 273\,000= 0.15\times 273\,000=$40\,950$

$\Rightarrow A$

Financial Maths, STD2 EQ-Bank 17

Roberto works as a furniture assembler and is paid $18.50 for each bookshelf he completes. In one week he worked 40 hours and earned $1110.

How many bookshelves did Roberto complete that week? (1 mark)
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Calculate his hourly rate of pay. (2 marks)
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a. $60\ \text{bookshelves}$

b. $$27.75\text{/hour}$

Show Worked Solution

a.	$\text{Number of Bookshelves}$	$=\dfrac{1110}{18.50}$
		$=60\ \text{bookshelves}$

b.	$\text{Hourly rate}$	$=\dfrac{\text{Total earnings}}{\text{Hours worked}}$
		$=\dfrac{1110}{40}=$27.75$

$\therefore\ \text{Roberto’s hourly rate is }$27.75$

Financial Maths, STD2 EQ-Bank 23

Jessica works in a shoe factory and is paid $5.40 for each pair of shoes she completes. Last week she worked 9 hours per day for 4 days and completed 180 pairs of shoes.

Calculate her hourly rate of pay. (2 marks)

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$$27.00\text{/hour}$

Show Worked Solution

$\text{Total earnings}=5.40\times 180= $972$

$\text{Total hours worked}=9 \times 4= 36\ \text{hours}$

$\text{Hourly rate}=\dfrac{972}{36}=$27.00\text{/hour}$

$\therefore\ \text{Jessica’s hourly rate is }$27.00$

Financial Maths, STD2 EQ-Bank 15 MC

A garment worker is paid $4.20 for each shirt she completes. In a particular week she worked 8 hours per day for 5 days and earned $1176.

How many shirts per hour did she complete on average?

Show Answers Only

$C$

Show Worked Solution

$\text{Number of shirts}=\dfrac{1176}{4.20}= 280\ \text{shirts}$

$\text{Total hours worked}=8 \times 5= 40\ \text{hours}$

$\text{Shirts per hour}=\dfrac{280}{40}=7$

$\Rightarrow C$

Financial Maths, STD2 EQ-Bank 4 MC

Sam is paid $2.50 for each toy he assembles in a factory. In one week he worked 35 hours and assembled 168 toys.

What is Sam's hourly rate of pay?

$10.00
$12.00
$14.00
$16.00

Show Answers Only

$B$

Show Worked Solution

$\text{Total earnings}=2.50 \times 168= $420 $

$\text{Hourly rate}=\dfrac{\text{Total earnings}}{\text{Hours worked}}= \dfrac{420}{35}=$12.00$

$\Rightarrow B$

Functions, EXT1 EQ-Bank 5

Solve the inequality $\left(2 x^2+3 x\right)(1-x) \geqslant 0$, expressing your answer in set notation. (3 marks)

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$x \in\left(-\infty,-\dfrac{3}{2}\right]\ \cup\ x \in [0,1] $

Show Worked Solution

$\left(2 x^2+3 x\right)(1-x)=x(2 x+3)(1-x)$

$x(2 x+3)(1-x) \geqslant 0$

$\text{Zeros at} \ \ x=0, x=-\dfrac{3}{2} \ \ \text{and} \ \ x=1$

$\text{At} \ \ x=-1: \ (-1)(1)(2)<0$

$\therefore \text{Graph} \geqslant 0 \ \ \text{for} \ \ x \in\left(-\infty,-\dfrac{3}{2}\right]\ \cup\ x \in [0,1] $

Functions, EXT1 EQ-Bank 3 MC

For which values of $x$ is $(2x^2-3x-5)(x+2)<0$ ?

$x \in(-1,-2) \ \cup\ x \in\left(\dfrac{5}{2}, \infty\right]$
$x \in[-\infty,-2)\ \cup\ x \in\left(-1, \dfrac{5}{2}\right)$
$x \in(-1,-2)\ \cup\ x \in\left(\dfrac{5}{2}, \infty\right)$
$x \in(-\infty,-2)\ \cup\ x \in\left(-1, \dfrac{5}{2}\right)$

Show Answers Only

$D$

Show Worked Solution

$(2x^2-3x-5)(x+2)=(2x-5)(x+1)(x+2) $

$\text {Zeros at} \ \ x=\dfrac{5}{2}, \ x=-1,\ \ \text {and} \ \ x=-2$

$\text{At} \ \ x=0: \ (-5)(1)(2)<0$

$\therefore \ \text{Graph }<0\ \text{ for }\ x \in(-\infty,-2)\ \cup\ x \in\left(-1, \frac{5}{2}\right) $

$\Rightarrow D$

Functions, 2ADV EQ-Bank 6

The function $g(x)=\left\{\begin{array}{ll}a x+b, & \text {for } x \leq 2 \\ x^2-1, & \text {for } x>2\end{array}\right.$

$g(x)$ is continuous at $x =2$ and passes through the point $(0,-5)$.

Find the values of $a$ and $b$ (2 marks)

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Evaluate $g(3)-g(-1)$ (1 mark)

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a. $a=4, b=-5$

b. $17$

Show Worked Solution

a. $\text{Since} \ f(x) \ \text{is continuous at} \ \ x=2:$

$a(2)+b$	$=2^2-1$
$2 a+b$	$=3\ \ldots\ (1)$

$\text{Since} \ g(x) \ \text{passes through}\ (0,-5):$

$a(0)+b=-5 \ \ \Rightarrow\ \ b=-5$

$\text{Substitute}\ \ b=-5 \ \ \text{into (1):}$

$2 a-5=3 \ \ \Rightarrow\ \ a=4$

b.	$g(3)-g(-1)$	$=\left(3^2-1\right)-[4(-1)-5]$
		$=8+9$
		$=17$

Functions, 2ADV EQ-Bank 5

Consider the function

\begin{align*}
f(x)=\begin{cases}-x^2+4, & \text {for }\ x<1 \\ 2 x+1, & \text {for} \ 1 \leq x<3 \\ 7, & \text {for }\ x \geq 3\end{cases}
\end{align*}

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

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State the range of $f(x)$ (1 mark)

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b. $\text {Range:} \ \ y \in(-\infty, 7]$

Show Worked Solution

b. $\text {Range:} \ \ y \in(-\infty, 7]$

CHEMISTRY, M8 2025 HSC 37

Compound A has the molecular formula $\ce{C5H10}$. The information below shows some chemical reactions beginning with this compound.

Compound A reacts with $\ce{H+/H2O}$ to produce compounds B and C .
Compound B does not react with $\ce{H+/Cr2O7^{2-}}$.
Compound C reacts with $\ce{H+/Cr2O7^{2-}}$ to produce compound D .
Compound D does not react with $\ce{Na2CO3(aq)}$.

Determine the structure of compound A. Justify your answer using all the data given. (4 marks)

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The molecular formula $\ce{C5H10}$ corresponds to a 5-carbon alkene.
Compound A reacts with $\ce{H+/H2O}$ (hydration) to produce compounds B and C, which must be alcohols.
Compound B does not react with acidified dichromate, so it cannot be oxidised. Therefore, B is a tertiary alcohol.
Compound C reacts with acidified dichromate to form compound D. This means C can be oxidised and is either a primary or secondary alcohol.
Compound D does not react with $\ce{Na2CO3(aq)}$, so D is not a carboxylic acid. Therefore, D must be a ketone.
Since primary alcohols oxidise to carboxylic acids, compound C must be a secondary alcohol (which oxidises to a ketone).
For compound A to produce both a tertiary and secondary alcohol, the $\ce{C=C}$ double bond must be positioned so that one carbon is bonded to three other carbons and the other to two carbons.

$\Rightarrow$ Compound A is 2-methylbut-2-ene.

Show Worked Solution

The molecular formula $\ce{C5H10}$ corresponds to a 5-carbon alkene.
Compound A reacts with $\ce{H+/H2O}$ (hydration) to produce compounds B and C, which must be alcohols.
Compound B does not react with acidified dichromate, so it cannot be oxidised. Therefore, B is a tertiary alcohol.
Compound C reacts with acidified dichromate to form compound D. This means C can be oxidised and is either a primary or secondary alcohol.
Compound D does not react with $\ce{Na2CO3(aq)}$, so D is not a carboxylic acid. Therefore, D must be a ketone.
Since primary alcohols oxidise to carboxylic acids, compound C must be a secondary alcohol (which oxidises to a ketone).
For compound A to produce both a tertiary and secondary alcohol, the $\ce{C=C}$ double bond must be positioned so that one carbon is bonded to three other carbons and the other to two carbons.

$\Rightarrow$ Compound A is 2-methylbut-2-ene.

CHEMISTRY, M8 2025 HSC 36

Use the data sheet provided and the information in the table to answer this question.

Consider the molecule shown.

For each of the following instrumental techniques, predict the expected features of the spectra produced.

Refer to the structural features of the molecule in your answer.

Infrared (IR) (Ignore any absorptions due to $\ce{C - C}$ or $\ce{C - H}$ )
Carbon-13 NMR
Proton NMR
Mass spectrometry (7 marks)

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Infrared (IR)

Strong broad $\ce{OH}$ signal at approximately 2750 cm$^{-1}$
Strong sharp $\ce{CO}$ signal at approximately 1700 cm$^{-1}$

$^{13}\text{C NMR}$

3 distinct carbon environments
1 signal at 5−40 ppm $\ce{(CH3)}$
1 signal at 20−50 ppm $\ce{(CH2)}$
1 signal at 160−185 ppm $\ce{(acid CO)}$

$^{1}\text{H NMR (Proton NMR)}$

3 distinct hydrogen environments
1 signal at 0.7−2.1 ppm $\ce{(CH3)}$, triplet splitting pattern, integration of 3
1 signal at 2.1−4.5 ppm $\ce{(CH2)}$, quartet splitting pattern, integration of 2
1 signal at 9.0−13.0 ppm $\ce{(COOH)}$, singlet, integration of 1

Mass Spectrometry

Molecular ion at 74 m/z (molar mass of propanoic acid ~74 g/mol)

Show Worked Solution

Infrared (IR)

Strong broad $\ce{OH}$ signal at approximately 2750 cm$^{-1}$
Strong sharp $\ce{CO}$ signal at approximately 1700 cm$^{-1}$

♦ Mean mark 62%.

$^{13}\text{C NMR}$

3 distinct carbon environments
1 signal at 5−40 ppm $\ce{(CH3)}$
1 signal at 20−50 ppm $\ce{(CH2)}$
1 signal at 160−185 ppm $\ce{(acid CO)}$

$^{1}\text{H NMR (Proton NMR)}$

3 distinct hydrogen environments
1 signal at 0.7−2.1 ppm $\ce{(CH3)}$, triplet splitting pattern, integration of 3
1 signal at 2.1−4.5 ppm $\ce{(CH2)}$, quartet splitting pattern, integration of 2
1 signal at 9.0−13.0 ppm $\ce{(COOH)}$, singlet, integration of 1

Mass Spectrometry

Molecular ion at 74 m/z (molar mass of propanoic acid ~74 g/mol)

CHEMISTRY, M5 2025 HSC 35

A purple solution at 25°C contains a mixture of two different cobalt$\text{(II)}$ complexes which are at equilibrium.

$\underset{\text{(blue)}}{\ce{CoCl4^{2-}(aq)}} \ce{+ 6H2O(l)} \rightleftharpoons
\underset{\text{(pink)}}{\ce{Co(H2O)6^{2+}(aq)}} \ce{+ 4Cl^{-}(aq)}$

The results of heating and cooling a sample of this solution are given in the table.

\begin{array}{|l|c|c|}
\hline\rule{0pt}{2.5ex} \textit{Temperature} \ \text{(°C)} \rule[-1ex]{0pt}{0pt}& 80 & 0 \\
\hline \rule{0pt}{2.5ex}\textit{Colour of solution} \rule[-1ex]{0pt}{0pt} & \quad \text{blue} \quad & \quad \text{pink} \quad \\
\hline
\end{array}

The energy profile diagram for this reaction is shown.

How do collision theory and Le Chatelier's principle account for the colour change to pink when the solution is cooled? Refer to the energy profile diagram in your answer. (5 marks)

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Show Answers Only

Energy Profile Diagram

The energy profile diagram shows that the forward reaction is exothermic, as the products have lower enthalpy than the reactants. The forward activation energy $(\text{E}_{a1})$ is smaller than the reverse activation energy $(\text{E}_{a2})$.

Le Chatelier’s Principle

Le Chatelier’s Principle says that if something disrupts a system in dynamic equilibrium, the system will shift in a way that works against that change.
Cooling the solution from 80 °C to 0 °C causes the system to favour the forward exothermic reaction, producing heat in response to the lowered temperature.
This shifts the equilibrium to the right, increasing $\left[ \ce{Co(H2O)6^{2+}}\right]$ and turning the solution pink.

Collision Theory

Collision theory states that for a reaction to occur, particles must collide with sufficient energy (above the activation energy) and correct orientation.
When the solution is cooled, the average kinetic energy of all particles decreases, reducing both the collision frequency and the proportion of successful collisions.
In this way, both forward and reverse reaction rates decrease.
However, the reverse reaction has a higher activation energy $(\text{E}_{a2})$ than the forward reaction $(\text{E}_{a1})$, as shown in the energy profile diagram.
Cooling causes a greater proportion of particles to fall below $(\text{E}_{a2})$ than $(\text{E}_{a1})$, so the reverse reaction rate decreases more significantly than the forward rate.
This causes a net shift toward products, increasing $\left[ \ce{Co(H2O)6^{2+}}\right]$ and changing the solution colour to pink.

Show Worked Solution

Energy Profile Diagram

The energy profile diagram shows that the forward reaction is exothermic, as the products have lower enthalpy than the reactants. The forward activation energy $(\text{E}_{a1})$ is smaller than the reverse activation energy $(\text{E}_{a2})$.

Le Chatelier’s Principle

Le Chatelier’s Principle says that if something disrupts a system in dynamic equilibrium, the system will shift in a way that works against that change.
Cooling the solution from 80 °C to 0 °C causes the system to favour the forward exothermic reaction, producing heat in response to the lowered temperature.
This shifts the equilibrium to the right, increasing $\left[ \ce{Co(H2O)6^{2+}}\right]$ and turning the solution pink.

Collision Theory

Collision theory states that for a reaction to occur, particles must collide with sufficient energy (above the activation energy) and correct orientation.
When the solution is cooled, the average kinetic energy of all particles decreases, reducing both the collision frequency and the proportion of successful collisions.
In this way, both forward and reverse reaction rates decrease.
However, the reverse reaction has a higher activation energy $(\text{E}_{a2})$ than the forward reaction $(\text{E}_{a1})$, as shown in the energy profile diagram.
Cooling causes a greater proportion of particles to fall below $(\text{E}_{a2})$ than $(\text{E}_{a1})$, so the reverse reaction rate decreases more significantly than the forward rate.
This causes a net shift toward products, increasing $\left[ \ce{Co(H2O)6^{2+}}\right]$ and changing the solution colour to pink.

CHEMISTRY, M6 2025 HSC 33

Chalk is predominantly calcium carbonate. Different brands of chalk vary in their calcium carbonate composition.

The table shows the composition of three different brands of chalk.

\begin{array}{|l|c|c|c|}
\hline \rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \ \ \textit{Brand X} \ \ & \ \ \textit{Brand Y} \ \ & \ \ \textit{Brand Z} \ \ \\
\hline \rule{0pt}{2.5ex}\ce{CaCO3(\%)} \rule[-1ex]{0pt}{0pt}& 85.5 & 83.9 & 82.4 \\
\hline
\end{array}

The following procedure was used to determine the calcium carbonate composition of a chalk sample.

A sample of chalk was crushed in a mortar and pestle.
A 3.00 g sample of the crushed chalk was placed in a conical flask.
100.0 mL of 0.550 mol L$^{-1} \ \ce{HCl(aq)}$ was added to the sample and left to react completely, resulting in a clear solution.
Four 20 mL aliquots of this mixture were then titrated with 0.10 mol L$^{-1} \ \ce{KOH}$ .

The results of the titrations are recorded.

\begin{array}{|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Burette volume}\text{(mL)} \rule[-1ex]{0pt}{0pt}& \textit{Trial 1} & \textit{Trial 2} & \textit{Trial 3} & \textit{Trial 4} \\
\hline
\rule{0pt}{2.5ex}\text{Final} \rule[-1ex]{0pt}{0pt}& 7.80 & 14.90 & 22.10 & 29.25 \\
\hline
\rule{0pt}{2.5ex}\text{Initial} \rule[-1ex]{0pt}{0pt}& 0.00 & 7.80 & 14.90 & 22.10 \\
\hline
\rule{0pt}{2.5ex}\text{Total used} \rule[-1ex]{0pt}{0pt}& 7.80 & 7.10 & 7.20 & 7.15 \\
\hline
\end{array}

Determine the brand of the chalk sample. Include a relevant chemical equation in your answer. (7 marks)

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$\text{Exclude the outlier (Trial 1):}$

$\text{Average volume} \ \ce{(KOH)} =\dfrac{7.10+7.20+7.15}{3}=0.00715 \ \text{L}$

$\ce{HCl(aq) + KOH(aq) \rightarrow KCl(aq) + H2O(l)}$

$\text{moles} \ \ce{KOH=0.10 \times 0.00715=0.000715 mol }$

$\text{Ratio}\ \ \ce{HCl:KOH=1: 1}$

$\ce{0.000715 mol HCl} \ \text{for each sample}$

$\ce{0.000715 \times 5=0.003575 mol}\ \text{total in sampled solution}$

$\text{Initial} \ \ \ce{n(HCl)=0.550 \times 0.1000=0.0550 mol}$

$\ce{n(HCl)}\ \text{that reacted with} \ \ce{CaCO3=0.0550-0.003575=0.051425 mol}$

$\ce{2HCl(aq) + CaCO3(s) \rightarrow CaCl2(aq) + H2O(l) + CO2(g)}$

$\text{Ratio}\ \ \ce{HCl:CaCO3=2:1}$

$\ce{n(CaCO3)}=\dfrac{0.051425}{2}=0.0257125\ \text{mol}$

$\ce{MM(CaCO3)}=40.08+12.01+3 \times 16=100.09$

$\text{Mass} \ \ce{CaCO3} =0.0257125 \times 100.09=2.5735641\ \text{g}$

$\% \ce{CaCO3}=\dfrac{2.5735641}{3.00} \times 100=85.7854 \% \approx 85.8 \%$

$\text{Chalk sample has to be Brand X.}$

Show Worked Solution

$\text{Exclude the outlier (Trial 1):}$

$\text{Average volume} \ \ce{(KOH)} =\dfrac{7.10+7.20+7.15}{3}=0.00715 \ \text{L}$

$\ce{HCl(aq) + KOH(aq) \rightarrow KCl(aq) + H2O(l)}$

$\text{moles} \ \ce{KOH=0.10 \times 0.00715=0.000715 mol }$

$\text{Ratio}\ \ \ce{HCl:KOH=1: 1}$

$\ce{0.000715 mol HCl} \ \text{for each sample}$

$\ce{0.000715 \times 5=0.003575 mol}\ \text{total in sampled solution}$

♦ Mean mark 55%.

$\text{Calculate}\ \ce{HCl}\ \text{that reacted with}\ \ce{CaCO3}:$

$\text{Initial} \ \ \ce{n(HCl)=0.550 \times 0.1000=0.0550 mol}$

$\ce{n(HCl)}\ \text{that reacted with} \ \ce{CaCO3=0.0550-0.003575=0.051425 mol}$

$\ce{2HCl(aq) + CaCO3(s) \rightarrow CaCl2(aq) + H2O(l) + CO2(g)}$

$\text{Ratio}\ \ \ce{HCl:CaCO3=2:1}$

$\ce{n(CaCO3)}=\dfrac{0.051425}{2}=0.0257125\ \text{mol}$

$\ce{MM(CaCO3)}=40.08+12.01+3 \times 16=100.09$

$\text{Mass} \ \ce{CaCO3} =0.0257125 \times 100.09=2.5735641\ \text{g}$

$\% \ce{CaCO3}=\dfrac{2.5735641}{3.00} \times 100=85.7854 \% \approx 85.8 \%$

$\text{Chalk sample has to be Brand X.}$

PHYSICS, M7 2025 HSC 19 MC

A system consists of a sealed glass jar containing some oxygen and a small strip of magnesium.

The magnesium reacts with the oxygen to produce magnesium oxide as a product. Energy is released from the system in this reaction.

The mass of the system will

increase because oxygen is added to the magnesium.
decrease because energy is removed from the system.
increase because energy is added to the system by the reaction.
decrease because magnesium and oxygen are lost in the reaction.

Show Answers Only

$B$

Show Worked Solution

Option $B$ is correct.

The system is sealed, so no matter enters or leaves – the magnesium and oxygen simply rearrange into magnesium oxide inside the jar.
Energy is released from the system and according to Einstein’s mass-energy equivalence $(E = mc^2)$, this lost energy corresponds to a decrease in mass.

Other options:

$A$ is incorrect. No oxygen is added – it was already in the sealed jar.
$C$ is incorrect. Energy is released (removed), not added to the system.
$D$ is incorrect. No magnesium or oxygen is lost – they’re converted to magnesium oxide within the sealed system.

$\Rightarrow B$

PHYSICS, M6 2025 HSC 35

A hollow copper pipe is placed upright on an electronic balance, which shows a reading of 300 g. A 50 g magnet is suspended inside the pipe and subsequently released.

It was observed that the readings on the balance began to increase after the magnet began to fall, and that the reading reached a constant maximum of 350 g before the magnet reached the bottom of the tube.

Explain these observations. (4 marks)

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The balance reading changes in two stages:

Stage 1 – The Magnet Accelerates

As the magnet begins to fall, it speeds up, causing a growing change in magnetic flux through the copper pipe.
This induces eddy currents that produce an upward electromagnetic force on the magnet, opposing the constant gravitational force.
By Newton’s third law, the pipe experiences an equal downward force, so the balance reading increases as the magnet accelerates.

Stage 2 – The Magnet Reaches a Constant Speed

Eventually the magnet reaches a constant speed, where the upward electromagnetic force balances its weight.
At this point, the magnet pushes down on the pipe with a force equal to its own weight, so the balance reads the combined weight of the pipe (300 g) plus the magnet (50 g), giving a constant maximum reading of 350 g.

Show Worked Solution

The balance reading changes in two stages:

Mean mark 51%.

Stage 1 – The Magnet Accelerates

As the magnet begins to fall, it speeds up, causing a growing change in magnetic flux through the copper pipe.
This induces eddy currents that produce an upward electromagnetic force on the magnet, opposing the constant gravitational force.
By Newton’s third law, the pipe experiences an equal downward force, so the balance reading increases as the magnet accelerates.

Stage 2 – The Magnet Reaches a Constant Speed

Eventually the magnet reaches a constant speed, where the upward electromagnetic force balances its weight.
At this point, the magnet pushes down on the pipe with a force equal to its own weight, so the balance reads the combined weight of the pipe (300 g) plus the magnet (50 g), giving a constant maximum reading of 350 g.

PHYSICS, M7 2025 HSC 34

The diagram shows a model of the orbits of Earth, Jupiter and Io, including their orbital direction and periods of orbit. In this model, it is assumed that the orbits of Earth, Jupiter and Io are circular.

A method to determine the speed of light using this model is described below.

When Earth was at position $P$, the orbital period of Io was measured, and the time that Io was at position $R$ was recorded.

Six months later, Io had orbited Jupiter 103 times, and Earth had reached position $Q$. The orbital period of Io was used to predict when it would be at position $R$. Assume that Jupiter has not moved significantly in its orbit around the Sun.

The time for Io to reach position $R$ was measured to be $1.000 \times 10^3$ seconds later than predicted, due to the time it takes light to cross the diameter of Earth's orbit from $P$ to $Q$.

Use the measurements provided in the model to calculate the speed of light. (2 marks)

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Consider a modification to this model in which the Earth's orbit is elliptical.
Explain how this modification will affect the determination of the speed of light. (3 marks)

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

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a. $d=2 \times 1.471 \times 10^{11} = 2.942 \times 10^{11}\ \text{metres} $

$v=\dfrac{d}{t}=\dfrac{2.942 \times 10^{11}}{1 \times 10^3} = 2.942 \times 10^8\ \text{m s}^{-1}$

b. Model modifications:

The model now assumes an elliptical orbit and (importantly) the time for Io to reach position $R$ is assumed to again be $1.000 \times 10^3$ seconds later.

Consider the long axis of the ellipse in line with $PQ:$

Along this “longer” eliptical axis, light travels further than $2.942 \times 10^{11}\ \text{m} $.
The longer distance travelled in the same time would result in a faster calculation of the speed of light.

Consider the short axis of the ellipse in line with $PQ:$

Along this axis, light travels less than $2.942 \times 10^{11}\ \text{m} $.
The shorter distance travelled in the same time would result in a slower calculation of the speed of light.

Show Worked Solution

a. $d=2 \times 1.471 \times 10^{11} = 2.942 \times 10^{11}\ \text{metres} $

$v=\dfrac{d}{t}=\dfrac{2.942 \times 10^{11}}{1 \times 10^3} = 2.942 \times 10^8\ \text{m s}^{-1}$

Mean mark (a) 51%.

b. Model modifications:

The model now assumes an elliptical orbit and (importantly) the time for Io to reach position $R$ is assumed to again be $1.000 \times 10^3$ seconds later.

Consider the long axis of the ellipse in line with $PQ:$

Along this “longer” eliptical axis, light travels further than $2.942 \times 10^{11}\ \text{m} $.
The longer distance travelled in the same time would result in a faster calculation of the speed of light.

Consider the short axis of the ellipse in line with $PQ:$

Along this axis, light travels less than $2.942 \times 10^{11}\ \text{m} $.
The shorter distance travelled in the same time would result in a slower calculation of the speed of light.

♦ Mean mark (b) 46%.

Algebra, STD2 EQ-Bank 26

The amount of water ($W$) in litres used by a garden irrigation system varies directly with the time ($t$) in minutes that the system operates.

This relationship is modelled by the formula $W=kt$, where $k$ is a constant.

The irrigation system uses 96 litres of water when it operates for 24 minutes.

Show that the value of $k$ is 4. (1 mark)
--- 3 WORK AREA LINES (style=lined) ---
The water tank for the irrigation system contains 650 litres of water. Calculate how many minutes the irrigation system can operate before the tank is empty. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---

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a. $W=kt$

$\text{When } W = 96 \text{ and } t = 24:$

$96$	$=k \times 24$
$k$	$=\dfrac{96}{24}=4\ \text{(as required)}$

b. $162.5\ \text{minutes}$

Show Worked Solution

a. $W=kt$

$\text{When } W = 96 \text{ and } t = 24:$

$96$	$=k \times 24$
$k$	$=\dfrac{96}{24}=4\ \text{(as required)}$

b. $W = 4t$

$\text{When } W = 650:$

$650$	$=4\times t$
$t$	$ =\dfrac{650}{4}=162.5\ \text{minutes}$

$\therefore\ \text{The irrigation system can operate for 162.5 minutes.}$

Algebra, STD2 EQ-Bank 23

The cost ($C$) of copper wire varies directly with the length ($L$) in metres of the wire.

This relationship is modelled by the formula $C = kL$, where $k$ is a constant.

A 250 metre roll of copper wire costs $87.50.

Show that the value of $k$ is 0.35. (1 mark)
--- 3 WORK AREA LINES (style=lined) ---
A builder has a budget of $140 for copper wire. Calculate the maximum length of wire that can be purchased. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---

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a. $C=kL$

$\text{When } C = 87.50 \text{ and } L = 250:$

$87.50$	$=k \times 250$
$k$	$=\dfrac{87.50}{250}$
$k$	$=0.35\ \text{(as required)}$

b. $400\ \text{m}$

Show Worked Solution

a. $C=kL$

$\text{When } C = 87.50 \text{ and } L = 250:$

$87.50$	$=k \times 250$
$k$	$=\dfrac{87.50}{250}$
$k$	$=0.35\ \text{(as required)}$

b. $C = 0.35L$

$\text{When } C = 140:$

$140$	$=0.35\times L$
$L$	$ =\dfrac{140}{0.35}=400\ \text{m}$

$\therefore\ \text{The builder can purchase 400 metres of wire.}$

Algebra, STD2 EQ-Bank 8 MC

The energy ($E$) required to heat water varies directly with the mass ($m$) of the water.

It takes 2100 joules of energy to heat 500 grams of water by 1°C.

Which equation represents the relationship between $E$ and $m$?

$E = 4.2m$
$E = 0.24m$
$m = 4.2E$
$m = 0.24E$

Show Answers Only

$A$

Show Worked Solution

$E \propto m$

$E=km$

$\text{When } E = 2100 \text{ and } m = 500:$

$2100$	$=k \times 500$
$k$	$=\dfrac{2100}{500}=4.2$

$\therefore\ E=4.2m$

$\Rightarrow A$

Algebra, STD2 EQ-Bank 7 MC

The distance ($d$) a spring stretches varies directly with the force ($F$) applied to it.

When a force of 18 newtons is applied, the spring stretches 27 mm.

What is the value of the constant of variation ($k$)?

$\dfrac{2}{5}$
$\dfrac{5}{2}$
$\dfrac{2}{3}$
$\dfrac{3}{2}$

Show Answers Only

$D$

Show Worked Solution

$d \propto F$

$d=kF$

$\text{When } d = 27 \text{ and } F = 18:$

$27$	$=k \times 18$
$k$	$=\dfrac{27}{18}=\dfrac{3}{2}$

$\Rightarrow D$

PHYSICS, M8 2025 HSC 15 MC

Two stars, $X$ and $Y$, are identified on the Hertzsprung-Russell diagram.

In what way are these two stars different?

$X$ has a higher luminosity than $Y$.
$X$ is a red star, and $Y$ is a blue star.
$X$ has a lower core temperature than $Y$.
$X$ has a higher surface temperature than $Y$.

Show Answers Only

$C$

Show Worked Solution

$C$ is correct: The $x$-axis shows surface temperature, not core temperature. $Y$ is a giant and $X$ is a main-sequence star; giants have higher core temperatures due to heavier-element fusion, so this statement is correct.

Other options:

$A$ is incorrect. Luminosity increases up the $y$-axis; since $X$ is below $Y$, it is less luminous.
$B$ is incorrect. Colour depends on surface temperature ($x$-axis). Since $X$ and $Y$ are located above the same $x$-value, they have the same temperature and colour.
$D$ is incorrect. $X$ and $Y$ share the same $x$-position, so they have the same surface temperature.

PHYSICS, M5 2025 HSC 13 MC

A mass is attached by a light, inextensible string of length $l$ to a vertical rigid support.

The mass rotates with uniform speed, $v$, in a horizontal circle as shown.

What is the acceleration of the mass?

$g$
$\dfrac{v^2}{l}$
$\dfrac{v^2}{l\, \sin \theta}$
$\dfrac{v^2}{l\, \cos \theta}$

Show Answers Only

$C$

Show Worked Solution

Mass moves in a horizontal circle $\Rightarrow\ \ F_c=\dfrac{mv^2}{r} $
Radius of circle $(r) = l\,\sin\,\theta$
$ma=\dfrac{mv^2}{l\,\sin\,\theta}\ \ \Rightarrow \ \ a=\dfrac{v^2}{l\,\sin\,\theta} $

$\Rightarrow C$

PHYSICS, M7 2025 HSC 9 MC

The diagram shows the absorption spectra of two different stars in the Milky Way galaxy.

Based on the information in the diagram, which of the following statements about the two stars is true?

Star $A$ has a lower density than Star $B$.
Star $A$ has a greater rotational velocity than Star $B$.
Star $A$ has a greater translational velocity than Star $B$.
Star $A$ and Star $B$ have different chemical compositions.

Show Answers Only

$A$

Show Worked Solution

Broader spectral lines occur if a star has a higher density or a higher rotational velocity.

$\Rightarrow A$

PHYSICS, M5 2025 HSC 8 MC

A projectile is launched from point $P$, with speed $u$, at angle $\theta$ to the horizontal. It lands at point $Q$.

The time of flight of the projectile is $t$.

Which row in the table best describes the time to reach maximum height and the speed of the projectile at $Q$?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
\ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{3.2ex}\textbf{A.}\rule[-1.5ex]{0pt}{0pt}\\
\rule{0pt}{3.2ex}\textbf{B.}\rule[-1.5ex]{0pt}{0pt}\\
\rule{0pt}{3.2ex}\textbf{C.}\rule[-1.5ex]{0pt}{0pt}\\
\rule{0pt}{3.2ex}\textbf{D.}\rule[-1.5ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Time to reach}& \textit{Speed of projectile} \\
\ \textit{maximum height}\ \rule[-1ex]{0pt}{0pt}& \textit{at Q} \\
\hline
\rule{0pt}{2.5ex}>\dfrac{t}{2}\rule[-1ex]{0pt}{0pt}&<u\\
\hline
\rule{0pt}{2.5ex}>\dfrac{t}{2}\rule[-1ex]{0pt}{0pt}& =u\\
\hline
\rule{0pt}{2.5ex}=\dfrac{t}{2}\rule[-1ex]{0pt}{0pt}& =u\\
\hline
\rule{0pt}{2.5ex}=\dfrac{t}{2}\rule[-1ex]{0pt}{0pt}& <u \\
\hline
\end{array}
\end{align*}

Show Answers Only

$A$

Show Worked Solution

Max height occurs at $\frac{t}{2}$ only if the projectile lands at the same height as it is launched. Here, since point $Q$ is higher than point $P$, time to reach max > $\frac{t}{2}$ .
Since the projectile gains GPE at $Q$ and by the Law of Conservation of Energy, it must have a lower speed at $Q$.

$\Rightarrow A$

PHYSICS, M5 2025 HSC 7 MC

A satellite in a circular orbit around Earth at an altitude of 500 km is moved to a new circular orbit at a higher altitude of 35 800 km .

Which statement correctly compares properties of the satellite in the higher orbit with its properties in the lower orbit?

Its period is greater, and its acceleration is the same.
Its kinetic energy is less, and its acceleration is less.
Its orbital velocity is less, and its potential energy is positive.
Its escape velocity is greater, and the centripetal force is less.

Show Answers Only

$B$

Show Worked Solution

Option $B$ is correct:

Orbital velocity decreases when altitude increases $\left( v=\sqrt{\frac{GM}{r}}\right) $
Kinetic energy decreases $\left(K=\frac{1}{2}mv^{2}\right)$
Acceleration decreases $\left(a_c=\frac{v^{2}}{r}\right) $

Other options:

$A$ is incorrect. Acceleration does not stay the same.
$C$ is incorrect. Gravitational potential energy is always negative.
$D$ is incorrect. Escape velocity is lower $\left( v_{esc}=\sqrt{\dfrac{2GM}{r}} \right)$

$\Rightarrow B$

PHYSICS, M8 2025 HSC 30

A beam of electrons travelling at $4 \times 10^3 \ \text{m s}^{-1}$ exits an electron gun and is directed toward two narrow slits with a separation, $d$, of 1 $\mu\text{m}$. The resulting interference pattern is detected on a screen 50 cm from the slits.

Show that the wavelength of the electrons in this experiment is 182 nm. (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
An interference fringe occurs on the screen where constructive interference takes place.

Determine the distance between the central interference fringe $A$ and the centre of the next bright fringe $B$. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Determine the potential difference acting in the electron gun to accelerate the electrons in the beam from rest to $4 \times 10^3 \ \text{m s}^{-1}$. (2 marks)

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

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a. $\text {Using} \ \ \lambda=\dfrac{h}{m v}:$

$\lambda=\dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 4 \times 10^3}=1.82 \times 10^{-7} \ \text{m}=182 \ \text{nm}$

b. $x=9.1 \ \text{cm}$

c. $V=4.5 \times 10^{-5} \ \text{V}$

Show Worked Solution

a. $\text {Using} \ \ \lambda=\dfrac{h}{m v}:$

$\lambda=\dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 4 \times 10^3}=1.82 \times 10^{-7} \ \text{m}=182 \ \text{nm}$

b. $\text {Using}\ \ x=\dfrac{\lambda m L}{d}$

$\text{where} \ \ x=\text{distance between middle of adjacent bright spots}$

$x=\dfrac{182 \times 10^{-9} \times 1 \times 0.5}{1 \times 10^{-6}}=0.091 \ \text{m}=9.1 \ \text{cm}$

c. $\text{Work done by field}=\Delta K=K_f-K_i$

$qV$	$=\dfrac{1}{2} m v^2-0$
$V$	$=\dfrac{m v^2}{2 q}=\dfrac{9.109 \times 10^{-31} \times\left(4 \times 10^3\right)^2}{2 \times 1.602 \times 10^{-19}}=4.5 \times 10^{-5} \ \text{V}$

♦ Mean mark (c) 45%.

PHYSICS, M7 2025 HSC 32

Analyse the consequences of the theory of special relativity in relation to length, time and motion. Support your answer with reference to experimental evidence. (8 marks)

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

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Overview Statement

Special relativity predicts that time dilation, length contraction and relativistic momentum arise from the principle that the speed of light is constant for all observers.
These effects change how time, distance and motion are measured and each consequence is supported by experimental evidence.

Time–Length Relationship in Muon Observations

Muon-decay experiments show how time dilation and length contraction depend on relative motion.
Muons created high in the atmosphere have such short lifetimes that, in classical physics, they should decay long before reaching Earth’s surface. Yet far more muons are detected at ground level than predicted.
In Earth’s frame of reference, the muons experience time dilation, so they “live longer” and travel further.
In the muon’s frame of reference, the atmosphere is length-contracted according to the equation $l=l_0 \sqrt{\left(1-\dfrac{v^{2}}{c^{2}}\right)}$, so the distance they travel is much shorter.
Both viewpoints are valid within their own reference frames, showing that time and length are not absolute but depend on relative motion.

Momentum–Energy Relationship in Particle Accelerators

Relativistic momentum explains how objects behave as they approach light speed.
Particle accelerators show that enormous increases in energy produce only small increases in speed at high velocities.
Particles act as though their mass increases, so each additional acceleration requires disproportionately more energy.
This makes it impossible for any object with mass to reach the speed of light, since doing so requires infinite energy.
Thus, relativistic momentum preserves light speed as a universal limit.

Implications and Synthesis

The consequences of these observations reveal that space and time form a single, interconnected framework rather than separate absolute quantities.
At high velocities, motion fundamentally alters measurements of time, distance and momentum.
Together, these consequences confirm that classical physics fails at relativistic speeds and that special relativity accurately describes the behaviour of fast-moving objects.

Show Worked Solution

Overview Statement

Special relativity predicts that time dilation, length contraction and relativistic momentum arise from the principle that the speed of light is constant for all observers.
These effects change how time, distance and motion are measured and each consequence is supported by experimental evidence.

♦ Mean mark 53%.

Time–Length Relationship in Muon Observations

Muon-decay experiments show how time dilation and length contraction depend on relative motion.
Muons created high in the atmosphere have such short lifetimes that, in classical physics, they should decay long before reaching Earth’s surface. Yet far more muons are detected at ground level than predicted.
In Earth’s frame of reference, the muons experience time dilation, so they “live longer” and travel further.
In the muon’s frame of reference, the atmosphere is length-contracted according to the equation $l=l_0 \sqrt{\left(1-\dfrac{v^{2}}{c^{2}}\right)}$, so the distance they travel is much shorter.
Both viewpoints are valid within their own reference frames, showing that time and length are not absolute but depend on relative motion.

Momentum–Energy Relationship in Particle Accelerators

Relativistic momentum explains how objects behave as they approach light speed.
Particle accelerators show that enormous increases in energy produce only small increases in speed at high velocities.
Particles act as though their mass increases, so each additional acceleration requires disproportionately more energy.
This makes it impossible for any object with mass to reach the speed of light, since doing so requires infinite energy.
Thus, relativistic momentum preserves light speed as a universal limit.

Implications and Synthesis

The consequences of these observations reveal that space and time form a single, interconnected framework rather than separate absolute quantities.
At high velocities, motion fundamentally alters measurements of time, distance and momentum.
Together, these consequences confirm that classical physics fails at relativistic speeds and that special relativity accurately describes the behaviour of fast-moving objects.

PHYSICS, M8 2025 HSC 31

Experiments have been carried out by scientists to investigate cathode rays.

Assess the contribution of the results of these experiments in developing an understanding of the existence and properties of electrons. (5 marks)

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Judgment Statement

The cathode ray experiments were highly valuable in establishing both the existence and properties of electrons through definitive experimental evidence and quantitative measurements.

Demonstrating Particle Nature

Electric field deflection experiments produced significant results by proving cathode rays were negatively charged particles rather than electromagnetic radiation.
This was highly effective because it eliminated the competing theory that cathode rays were electromagnetic waves, since waves are not deflected by electric fields.
The consistent deflection pattern across many experiments provided strong evidence for the particle nature of electrons.

Quantifying Electron Properties

By adjusting electric and magnetic field strengths within experiments, the charge-to-mass ratio of the electron was determined.
This measurement proved highly effective as it provided the first quantitative property of electrons.
The e/m ratio demonstrated considerable value by revealing electrons were much lighter than atoms, indicating subatomic particles existed.

Overall Assessment

Assessment reveals these experiments achieved major significance in atomic theory development.
The combined results produced measurable, reproducible data that definitively established electrons as fundamental charged particles with specific properties.
Overall, these contributions proved essential for understanding atomic structure.

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*}

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$D$

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, M8 2025 HSC 3 MC

The diagram shows lines in the emission spectrum of hydrogen.

The production of this spectrum can be explained by applying the atomic model developed by which scientist?

Balmer
Bohr
Planck
Rutherford

Show Answers Only

$B$

Show Worked Solution

Bohr’s atomic model correctly explains hydrogen’s emission spectrum by proposing that electrons occupy quantised energy levels and emit photons when transitioning between them.

$\Rightarrow B$

PHYSICS, M5 2025 HSC 28

A bicycle rider jumps from one ramp to a second ramp separated by 16 m as shown. The ramps are inclined at 20° and are 2 m high.

What minimum speed is required for the rider to land on the second ramp? (4 marks)

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$u=15.6 \ \text{ms}^{-1}$

Show Worked Solution

$\text {Find time of flight}\ \ \left(S_y=0\right)$

$S_y$	$=u \sin \theta t+\dfrac{1}{2} a t^2$
$0$	$=u \sin 20^{\circ} t-4.9 t^2$
$0$	$=t\left(u \sin 20^{\circ}-49 t\right)$
$t$	$=\dfrac{u \sin 20^{\circ}}{4.9}$

♦ Mean mark 47%.

$\text{Since flight distance = 16 m:}$

$S_x$	$u \, \cos \theta \ t$
$16$	$=\dfrac{u \,\cos 20^{\circ} \times u \, \sin 20^{\circ}}{4.9}$
$u^2$	$=\dfrac{16 \times 4.9}{\cos 20^{\circ} \times \sin 20^{\circ}}$
$u$	$=\sqrt{\dfrac{16 \times 4.9}{\cos 20^{\circ} \times \sin 20^{\circ}}}=15.6 \ \text{ms}^{-1}$

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.

♦ Mean mark 47%.

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

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

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

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

Mean mark (a) 51%.

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

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

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

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

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

♦ Mean mark (b) 45%.

Financial Maths, STD2 EQ-Bank 2

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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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, 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)

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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, M5 2025 HSC 33

The following diagram shows the cell division processes occurring in two related individuals.

Compare the cell division processes carried out by cells $R$ and $S$ in Individual 1. (3 marks)

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Explain the relationship between Individuals 1 and 2. (2 marks)
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$A$ and $B$ are two separate mutations. Analyse how mutations $A$ and $B$ affect the genetic information present in cells $U$, $V$, $W$ and $X$. (4 marks)
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a. Similarities:

Both cell R and cell S undergo cell division to produce daughter cells.

Differences:

Cell R undergoes mitosis producing two genetically identical diploid somatic cells (T and U).
Cell S undergoes meiosis producing four genetically different haploid gametes.
Mitosis in R maintains chromosome number for growth and repair.
Meiosis in S reduces chromosome number by half for sexual reproduction and genetic variation.

b. Relationship between Individuals 1 and 2

Individual 2 is the offspring of Individual 1.
This is because Individual 1’s germ-line cell S produces a gamete (sperm) which fertilises an egg to form the zygote that develops into Individual 2.

c. Mutation’s affect on genetic information

Mutation A occurs in the germ-line pathway after zygote Q. This means that mutation A is present in cell S and is passed to cell X.
Mutation A does not affect cells U, V or W because it occurred after the R/S split, so the R lineage and Individual 2’s somatic cells lack it.
Mutation B occurs in Individual 2’s somatic pathway. This results in mutation B being present in cells V and W only.
The significance is that only mutation A can be inherited by offspring, while mutation B cannot.

Show Worked Solution

a. Similarities:

Both cell R and cell S undergo cell division to produce daughter cells.

Differences:

Cell R undergoes mitosis producing two genetically identical diploid somatic cells (T and U).
Cell S undergoes meiosis producing four genetically different haploid gametes.
Mitosis in R maintains chromosome number for growth and repair.
Meiosis in S reduces chromosome number by half for sexual reproduction and genetic variation.

b. Relationship between Individuals 1 and 2

Individual 2 is the offspring of Individual 1.
This is because Individual 1’s germ-line cell S produces a gamete (sperm) which fertilises an egg to form the zygote that develops into Individual 2.

♦ Mean mark (b) 52%.

c. Mutation’s affect on genetic information

Mutation A occurs in the germ-line pathway after zygote Q. This means that mutation A is present in cell S and is passed to cell X.
Mutation A does not affect cells U, V or W because it occurred after the R/S split, so the R lineage and Individual 2’s somatic cells lack it.
Mutation B occurs in Individual 2’s somatic pathway. This results in mutation B being present in cells V and W only.
The significance is that only mutation A can be inherited by offspring, while mutation B cannot.

♦ Mean mark (c) 52%.

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 $\text{X}$, $\text{Y}$ and $\text{Z}$ 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. In this way the process provides 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.

♦ Mean mark (a) 45%.

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. In this way, the process provides enhanced immune protection against subsequent infections.

\(\text{Royalty payment}\)	\(=12\% \times 315\,210\)
	\(= 0.12\times 315\,210\)
	\(=$37\,825.20\)

\(\text{Royalty payment}\)	\(=9.5\% \times 358\,020\)
	\(= 0.095\times 358\,020\)
	\(=$34\,011.90\)

a.	\(\text{Number of Bookshelves}\)	\(=\dfrac{1110}{18.50}\)
		\(=60\ \text{bookshelves}\)

b.	\(\text{Hourly rate}\)	\(=\dfrac{\text{Total earnings}}{\text{Hours worked}}\)
		\(=\dfrac{1110}{40}=$27.75\)

\(a(2)+b\)	\(=2^2-1\)
\(2 a+b\)	\(=3\ \ldots\ (1)\)

b.	\(g(3)-g(-1)\)	\(=\left(3^2-1\right)-[4(-1)-5]\)
		\(=8+9\)
		\(=17\)

\(96\)	\(=k \times 24\)
\(k\)	\(=\dfrac{96}{24}=4\ \text{(as required)}\)

\(96\)	\(=k \times 24\)
\(k\)	\(=\dfrac{96}{24}=4\ \text{(as required)}\)

\(650\)	\(=4\times t\)
\(t\)	\( =\dfrac{650}{4}=162.5\ \text{minutes}\)