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BIOLOGY, M1 EQ-Bank 2

  1. Describe the role of two cell organelles found in eukaryotic cells.   (2 marks)

  2. How do specialised organelles contribute to the efficient functioning of eukaryotic cells, compared to prokaryotic cells?   (2 marks)

Show Answers Only

a.    Answers could include any two of the following:

→ Nucleus – the nucleus contains the cell’s genetic material and regulates gene expression.

→ Mitochondria – generates ATP through cellular respiration, providing energy for the cell’s functions.

→ Endoplasmic reticulum (ER) – helps synthesise proteins (rough ER) and lipids (smooth ER), contributing to various cell processes.
 

b.    → Specialised organelles allow eukaryotic cells to compartmentalise tasks.

→ This increases metabolic efficiency and enables complex processes to occur within the cell.

→ In contrast, prokaryotic cells, lacking organelles, must perform all functions in the cytoplasm, limiting their ability to handle as many simultaneous or specialised activities as eukaryotic cells.

Show Worked Solution

a.    Answers could include any two of the following:

→ Nucleus – the nucleus contains the cell’s genetic material and regulates gene expression.

→ Mitochondria – generates ATP through cellular respiration, providing energy for the cell’s functions.

→ Endoplasmic reticulum (ER) – helps synthesise proteins (rough ER) and lipids (smooth ER), contributing to various cell processes.
 

b.    → Specialised organelles allow eukaryotic cells to compartmentalise tasks.

→ This increases metabolic efficiency and enables complex processes to occur within the cell.

→ In contrast, prokaryotic cells, lacking organelles, must perform all functions in the cytoplasm, limiting their ability to handle as many simultaneous or specialised activities as eukaryotic cells.

BIOLOGY, M1 EQ-Bank 1

  1. Compare the structural and functional differences between prokaryotic and eukaryotic cells.   (2 marks)

  2. Discuss how these differences influence the complexity and organisation of organisms that consist of these cell types.   (2 marks)

Show Answers Only

a.    Structural and functional differences:

→ Prokaryotic cells lack membrane-bound organelles and a true nucleus, with their DNA free-floating in the cytoplasm.

→ Eukaryotic cells, on the other hand, have a membrane-bound nucleus and various organelles such as mitochondria, which allow compartmentalisation of functions.

→ This structural difference enables eukaryotes to perform more complex processes, while prokaryotes are limited to simpler metabolic activities.
 

b.    → Eukarytic organisms can manage multiple, specialised functions simultaneously.

→ This supports greater complexity in multicellular organisms and leads to highly organised systems with differentiated tissues and organs.

→ Prokaryotic organisms remain unicellular or simple multicellular forms, with all functions occurring in the same space.

Show Worked Solution

a.    Structural and functional differences:

→ Prokaryotic cells lack membrane-bound organelles and a true nucleus, with their DNA free-floating in the cytoplasm.

→ Eukaryotic cells, on the other hand, have a membrane-bound nucleus and various organelles such as mitochondria, which allow compartmentalisation of functions.

→ This structural difference enables eukaryotes to perform more complex processes, while prokaryotes are limited to simpler metabolic activities.
 

b.    → Eukarytic organisms can manage multiple, specialised functions simultaneously.

→ This supports greater complexity in multicellular organisms and leads to highly organised systems with differentiated tissues and organs.

→ Prokaryotic organisms remain unicellular or simple multicellular forms, with all functions occurring in the same space.

BIOLOGY, M1 EQ-Bank 3 MC

Which of the following is a key structural difference between prokaryotic and eukaryotic cells?

  1. Prokaryotic cells have mitochondria, while eukaryotic cells do not.
  2. Eukaryotic cells have a nucleus, while prokaryotic cells lack a true nucleus.
  3. Eukaryotic cells have a cell wall, while prokaryotic cells do not.
  4. Prokaryotic cells contain membrane-bound organelles, while eukaryotic cells do not.
Show Answers Only

\(B\)

Show Worked Solution

→ Prokaryotic cells lack a membrane-bound nucleus, whereas eukaryotic cells have a well-defined nucleus that encloses their DNA.

\(\Rightarrow B\)

BIOLOGY, M6 2020 VCE 29 MC

A small sample of DNA was obtained from a fossil. Polymerase chain reaction (PCR) was used to amplify the amount of DNA obtained from the sample.

Which one of the following is a correct statement regarding the PCR process?

  1. DNA polymerase catalyses the pairing of primers with complementary nucleotides.
  2. RNA polymerase catalyses the additions of nucleotides to a DNA strand.
  3. Annealing and extension of the DNA occur at different temperatures.
  4. The number of copies of the DNA is quadrupled in each cycle.
Show Answers Only

\(C\)

Show Worked Solution

Features of the PCR process:

→ the annealing step, where the primers bind to the template DNA, occurs at a low temperature

→ the extension step, where DNA polymerase adds nucleotides to the growing DNA strand, occurs at a higher temperature.

\(\Rightarrow C\)

BIOLOGY, M6 2020 VCE 27*

Tasmanian devils (Sarcophilus harrisii) were originally broadly distributed across Australia. When sea levels rose 12 000 years ago, an island, now referred to as Tasmania, was formed. The small number of Tasmanian devils on Tasmania was cut off from the Australian mainland populations. The population in Tasmania showed less genetic variation than the mainland populations. Mainland populations became extinct approximately 3000 years ago.

Over the last 20 years, the total Tasmanian devil population on Tasmania has halved. Many of the deaths have been the result of Tasmanian devil facial tumour disease (DFTD). Scientists have taken some Tasmanian devils that do not have DFTD to mainland Australia to set up a conservation program. The scientists have shown that greater genetic diversity among offspring in this program is observed when the Tasmanian devils are kept in isolated male-female pairs rather than in larger groups.

Giving reasons, describe if the conservation program for Tasmanian devils is an example of

  1. allopatric speciation   (1 mark)

  2. selective breeding   (1 mark)

  3. natural selection   (1 mark)

Show Answers Only

i.    Allopatric speciation – No

→  Involves the formation of new species due to geographic isolation whereas the program focuses on preserving the existing Tasmanian devil species.
 

ii.   Selective breeding – Yes

→ Scientists are intentionally choosing which Tasmanian devils to breed together based on their genetic diversity.
 

iii.  Natural selection – No

→ Natural selection is the differential survival and reproduction based on inherited traits.

→ The program involves intentional breeding decisions by scientists.

Show Worked Solution

i.    Allopatric speciation – No

→  Involves the formation of new species due to geographic isolation whereas the program focuses on preserving the existing Tasmanian devil species.
 

ii.   Selective breeding – Yes

→ Scientists are intentionally choosing which Tasmanian devils to breed together based on their genetic diversity.
 

iii.  Natural selection – No

→ Natural selection is the differential survival and reproduction based on inherited traits.

→ The program involves intentional breeding decisions by scientists.

BIOLOGY, M6 2021 VCE 20 MC

The diagram below shows the chromosomes from two different but related plant species.
 

Compare the chromosomes of the two plant species.

The differences seen in the chromosomes of Plant species 2 compared to the chromosomes of Plant species 1 can be described as

  1. aneuploidy.
  2. polyploidy.
  3. triploidy.
  4. trisomy.
Show Answers Only

\(B\)

Show Worked Solution

→ Polyploidy – the cells of an organism have more than one pair of (homologous) chromosomes.

\(\Rightarrow B\)

CHEMISTRY, M8 EQ-Bank 5

A chemical reaction occurs at a constant temperature.

Describe the effect on the yield if

  1. the value of the equilibrium expression is higher.   (2 marks)
  2. the activation energy of the reaction is decreased.   (2 marks)
Show Answers Only

a.   Equilibrium expression is higher:

→ The equilibrium expression shows the ratio of reactants to products in a chemical equation.

→ A higher value for the equilibrium expression suggests a higher proportion of products / increased product concentration compared to reactants, and hence a higher yield.
 

b.   The activation energy is decreased:

→ A decrease in activation energy increases the reaction rate and allows the system to get to equilibrium faster.

→ This has no effect on the equilibrium yield which remains the same.

Show Worked Solution

a.   Equilibrium expression is higher:

→ The equilibrium expression shows the ratio of reactants to products in a chemical equation.

→ A higher value for the equilibrium expression suggests a higher proportion of products / increased product concentration compared to reactants, and hence a higher yield.
 

b.   The activation energy is decreased:

→ A decrease in activation energy increases the reaction rate and allows the system to get to equilibrium faster.

→ This has no effect on the equilibrium yield which remains the same.

v1 Algebra, STD2 A4 2020 HSC 24

There are two tanks at an industrial plant, Tank A and Tank B. Initially, Tank A holds 2520 litres of liquid fertiliser and Tank B is empty.

  1. Tank A begins to empty liquid fertiliser into a transport vehicle at a constant rate of 40 litres per minute.

     

    The volume of liquid fertiliser in Tank A is modelled by  \(V=1400-40t\)  where \(V\) is the volume in litres and  \(t\) is the time in minutes from when the tank begins to drain the fertiliser.

     

    On the grid below, draw the graph of this model and label it as Tank A.   (1 mark)
     

     

  2. Tank B remains empty until  \(t=10\)  when liquid fertiliser is added to it at a constant rate of 60 litres per minute.
     By drawing a line on the grid (above), or otherwise, find the value of  \(t\)  when the two tanks contain the same volume of liquid fertiliser.  (2 marks)

  3. Using the graphs drawn, or otherwise, find the value of  \(t\)  (where  \(t > 0\)) when the total volume of liquid fertiliser in the two tanks is 1400 litres.  (1 mark)

Show Answers Only
  1.  \(\text{Tank} \ A \ \text{will pass through (0, 1400) and (35, 0)}\)
      
  2. \(20 \ \text{minutes}\)
  3. \(30 \ \text{minutes}\)
Show Worked Solution

a.     \(\text{Tank} \ A \ \text{will pass through (0, 1400) and (35, 0)}\)
 

 

b.   \(\text{Tank} \ B \ \text{will pass through (10, 0) and (30, 1200)}\)  
 

 

\(\text{By inspection, the two graphs intersect at} \ \ t = 20 \ \text{minutes}\)

c.   \(\text{Strategy 1}\)

\(\text{By inspection of the graph, consider} \ \ t = 30\)

\(\text{Tank A} = 200 \ \text{L} , \ \text{Tank B} =1200 \ \text{L}\)

\(\therefore\ \text{Total volume = 1400 L when  t = 30}\)
  

\(\text{Strategy 2}\)

\(\text{Total Volume}\) \(=\text{Tank A} + \text{Tank B}\)
\(1400\) \(=1400-40t+(t-10)\times 60\)
\(1400\) \(=1400-40t+60t-600\)
\(20t\) \(= 600\)
\(t\) \(= 30 \ \text{minutes}\)

♦♦ Mean mark part (c) 22%.

v1 Algebra, STD2 A4 2019 HSC 36

A small business makes and dog kennels.

Technology was used to draw straight-line graphs to represent the cost of making the dog kennels \((C)\) and the revenue from selling dog kennels \((R)\). The \(x\)-axis displays the number of dog kennels and the \(y\)-axis displays the cost/revenue in dollars.
 


 

  1. How many dog kennels need to sold to break even?  (1 mark)

  2. By first forming equations for cost `(C)` and revenue `(R)`, determine how many dog kennels need to be sold to earn a profit of $2500.  (3 marks)

Show Answers Only
  1. \(20\)
  2. \(145\)
Show Worked Solution

a.  \(20\ \ (x\text{-value at intersection})\)

  

b.   \(\text{Find equations of both lines}:\)

\((0, 400)\ \text{and}\ (20, 600)\ \text{lie on}\ \ C\)

\(\text{gradient}_C = \dfrac{600-400}{20-0}=10\)

\(\rightarrow\ C=400+10x\)
   

\((0,0)\ \text{and}\ (20, 600)\ \text{lie on}\ \ R\)

\(\text{gradient}_R =\dfrac{600-0}{20-0}=30\)

\(\rightarrow\ R=30x\)
 

\(\text{Profit} = R-C\)

\(\text{Find}\ \ x\ \text{when Profit }= $2500:\)

\(2500\) \(=30x-(400+10x)\)
\(20x\) \(=2900\)
\(x\) \(=145\)

  
\(\therefore\ 145\ \text{dog kennels need to be sold to earn }$2500\ \text{profit}\)


♦♦ Mean mark 28%.

v1 Algebra, STD2 A4 2023 HSC 21

Electricity provider \(A\) charges 30 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $90.

  1. Complete the table showing Provider \(A\)'s monthly charges for different levels of electricity usage.   (1 mark)

    \begin{array} {|l|c|}
    \hline
    \rule{0pt}{2.5ex} \textit{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ 0 \ \ & \ \ 400 \ \ & \ \ 1000 \ \ \\
    \hline
    \rule{0pt}{2.5ex} \textit{Monthly Charge (\$)} \rule[-1ex]{0pt}{0pt} & \ \ 90 \ \ & \ \ 210 \ \ & \ \ 390 \ \ \\
    \hline
    \end{array}

Provider \(B\) charges 52.5 cents per kWh, with no fixed monthly charge. The graph shows how Provider \(B\)'s charges vary with the amount of electricity used in a month.
 

 
  1. On the grid above, graph Provider \(A\)'s charges from the table in part (a).   (1 mark)
  2. Use the two graphs to determine the number of kilowatt hours per month for which Provider \(A\) and Provider \(B\) charge the same amount.   (1 mark)

  3. A customer uses an average of 600 kWh per month.
  4. Which provider, \(A\) or \(B\), would be the cheaper option and by how much?   (2 marks)

Show Answers Only

a.    \(\text{When kWh} =400\)

\(\text{Monthly charge}\ =$90+0.30\times 400=$210\)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ 0 \ \ & \ \ 400 \ \ & \ 1000 \ \\
\hline
\rule{0pt}{2.5ex} \textit{Monthly Charge (\$)} \rule[-1ex]{0pt}{0pt} & \ \ 90 \ \ & \ \ 210 \ \ & \ \ 390 \ \ \\
\hline
\end{array}

b.    
         

c.    \(\text{400 kWh}\)

d.    \(\text{Provider}\ A\ \text{is cheaper by \$45.}\)

Show Worked Solution

a.   \(\text{When kWh} =400\)

\(\text{Monthly charge}\ =$90+0.30\times 400=$210\)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ 0 \ \ & \ \ 400 \ \ & \ 1000 \ \\
\hline
\rule{0pt}{2.5ex} \textit{Monthly Charge (\$)} \rule[-1ex]{0pt}{0pt} & \ \ 90 \ \ & \ \ 210 \ \ & \ \ 390 \ \ \\
\hline
\end{array}

b. 
          
 

c.    \(A_{\text{charge}} = B_{\text{charge}}\ \text{at intersection.}\)

\(\therefore\ \text{Same charge at 400 kWh}\)
 

d.    \(\text{Cost at 600 kWh:}\)

\(\text{Method 1: Using graph}\rightarrow\ $315-270=$45\)

\(\text{Method 2: Algebraically}:\)

\(\text{Provider}\ A: \ 90 + 0.30 \times 600 = $270\)

\(\text{Provider}\ B: \ 0.525 \times 600 = $315\)

\(\therefore \text{Provider}\ A\ \text{is cheaper by \$45.}\)

v1 Algebra, STD2 A4 2018 HSC 27d

The graph displays the cost (\($c\)) charged by two companies for the hire of a jetski for \(x\) hours.
 


  

Both companies charge $450 for the hire of a jetski for 5 hours.

  1. What is the hourly rate charged by Company A?  (1 mark)

  2. Company B charges an initial booking fee of $80.

     

    Write a formula, in the of  \(c=b+mx\), for the cost of hiring a jetski from Company B for \(x\) hours.  (2 marks)

  3. A jetski is hired for 7 hours from Company B.

     

    Calculate how much cheaper this is than hiring from Company A.  (2 marks)

Show Answers Only
  1. \($90\)
  2. \(c=80+74x\)
  3. \($32\)
Show Worked Solution
i.    \(\text{Hourly rate}\ (A)\) \(=\dfrac{450}{5}\)
    \(=$90\)

 

ii.   \(m=\text{hourly rate}\)

\(\text{Find}\ m,\ \text{given}\ c = 450,\ \text{when}\ \ x = 5\ \text{and}\ \ b = 80\)

\(450\) \(=80+m\times 5\)
\(5m\) \(=370\)
\(m\) \(=\dfrac{370}{5}=74\)

\(\therefore\ c=80+74x\)
 

iii.    \(\text{Cost}\ (A)\) \(=90\times 7=$630\)
  \(\text{Cost}\ (B)\) \(=80+74\times 7=$598\)

 
\(\therefore\ \text{Company}\ B’\text{s hiring cost is }$32\ \text{cheaper.}\)

v1 Algebra, STD2 A4 SM-Bank 6 MC

A computer application was used to draw the graphs of the equations

\(x-y=-5\)  and  \(x+y=5\)

Part of the screen is shown.

What is the solution when the equations are solved simultaneously?

  1. \(x=5,\ y=5\)
  2. \(x=5,\ y=0\)
  3. \(x=-5,\ y=0\)
  4. \(x=0,\ y=5\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Solution occurs at the intersection of the two lines.}\)

\(\Rightarrow D\)

v1 Algebra, STD2 A2 2010 HSC 27c

The graph shows tax payable against taxable income, in thousands of dollars.
  

  1. Use the graph to find the tax payable on a taxable income of \($18\ 000\).  (1 mark)

  2. Use suitable points from the graph to show that the gradient of the section of the graph marked  \(A\)  is  \(\dfrac{7}{15}\).    (1 mark)

  3. How much of each dollar earned between  \($18\ 000\)  and  \($33\ 000\) is payable in tax? Give your answer correct to the nearest whole number.   (1 mark)

  4. Write an equation that could be used to calculate the tax payable, \(T\), in terms of the taxable income, \(I\), for taxable incomes between  \($18\ 000\)  and  \($33\ 000\).   (2 marks)

Show Answers Only
  1. \($3000\ \ \text{(from graph)}\)
  2. \(\text{See worked solution}\)
  3. \(46\frac{2}{3}\approx  47\ \text{cents per dollar earned}\)
  4. \(\text{Tax payable →}\ T=\dfrac{7}{15}I-5400\)
Show Worked Solution
i.   

\(\text{Income on}\ $18\ 000=$3000\ \ \text{(from graph)}\)

  

ii.  \(\text{Using the points}\ (18, 3)\ \text{and}\ (33, 10)\)

\(\text{Gradient at}\ A\) \(=\dfrac{y_2-y_1}{x_2-x_1}\)
  \(=\dfrac{10\ 000-3000}{33\ 000-18\ 000}\)
  \(=\dfrac{7000}{15\ 000}\)
  \(=\dfrac{7}{15}\ \ \ \ \text{… as required}\)

♦♦ Mean mark (ii) 25%.

iii.  \(\text{The gradient represents the tax applicable on each dollar}\)

\(\text{Tax}\) \(=\dfrac{7}{15}\ \text{of each dollar earned}\)
  \(=46\frac{2}{3}\approx 47\ \text{cents per dollar earned (nearest whole number)}\)

♦♦♦ Mean mark (iii) 12%!
MARKER’S COMMENT: Interpreting gradients is an examiner favourite, so make sure you are confident in this area.

iv.  \(\text{Tax payable up to }$18\ 000 = $3000\)

\(\text{Tax payable on income between }$18\ 000\ \text{and }$33\ 000\)

\(=\dfrac{7}{15}(I-18\ 000)\)

\(\therefore\ \text{Tax payable →}\ \ T\) \(=3000+\dfrac{7}{15}(I-18\ 000)\)
  \(=3000+\dfrac{7}{15} I-8400\)
  \(=\dfrac{7}{15}I-5400\)

♦♦♦ Mean mark (iv) 15%.
STRATEGY: The earlier parts of this question direct students to the most efficient way to solve this question. Make sure earlier parts of a question are front and centre of your mind when devising strategy.

v1 Algebra, STD2 A2 2014 HSC 27b

Clara is comparing the costs of two different ways of travelling to work.

Clara’s motor scooter uses one litre of fuel for every 22 km travelled. The cost of fuel is $2.24/L and the distance from her home to the work car park is 33 km. The cost of travelling by bus and light rail is $35.80 for 10 single trips.

Which way of travelling is cheaper and by how much? Support your answer with calculations.  (2 marks)

Show Answers Only

\(\text{Motor scooter is }$0.22 \text{ cheaper per one-way trip.}\)

Show Worked Solution

\(\text{Compare cost of a one-way trip}\)

\(\text{Motor scooter}\)

\(\text{Fuel used}=\dfrac{33}{22}=1.5\ \text{L}\)

\(\text{Cost}=1.5\times 2.24=$3.36\)
  

\(\text{Bus and light rail}\)

\(\text{Cost}=\dfrac{35.80}{10}=$3.58\)
 

\(\text{Difference}=$3.58-3.36=$0.22\)
  

\(\therefore\ \text{Motor scooter is }$0.22 \text{ cheaper per one-way trip.}\)

v1 Algebra, STD2 A2 2016 HSC 26c

Bonn’s car uses fuel at the rate of 6.1 L /100 km for country driving and 8.3 L /100 km for city driving. On a trip, he drives 350 km in the country and 40 km in the city.

Calculate the amount of fuel he used on this trip.  (2 marks)

Show Answers Only

\(24.67\ \text{L}\)

Show Worked Solution

\(\text{Fuel used in country}\)

\(=350\times \dfrac{6.1}{100}\)

\(=21.35\ \text{L}\)

\(\text{Fuel used in city}\)

\(=40\times \dfrac{8.3}{100}\)

\(=3.32\ \text{L}\)

  

\(\therefore\ \text{Total fuel used}\)

\(=21.35+3.32\)

\(=24.67\ \text{L}\)

v1 Algebra, STD2 A2 2016 HSC 29e

The graph shows the life expectancy of people born between 1900 and 2010.
 


  1. According to the graph, what is the life expectancy of a person born in 1968?  (1 mark)

  2. With reference to the value of the gradient, explain the meaning of the gradient in this context.  (2 marks)

Show Answers Only
  1. \(\text{76 years}\)
  2. \(\text{After 1900, life expectancy increases by 0.38 years for}\)
    \(\text{each year later that someone is born.}\)
Show Worked Solution

i.    \(\text{76 years}\)

ii.    \(\text{Using (2000, 88) and (1900, 50):}\)

\(\text{Gradient}\) \(= \dfrac{y_2-y_1}{x_2-x_1}\)
  \(= \dfrac{88-50}{2000-1900}\)
  \(= 0.38\)

 
\(\text{After 1900, life expectancy increases by 0.38 years for}\)

\(\text{each year later that someone is born.}\)

♦♦ Mean mark (ii) 33%.

v1 Algebra, STD2 A2 2018 HSC 5 MC

The driving distance from Burt's home to his work is 15 km. He drives to and from work five times each week. His car uses fuel at the rate of 12 L/100 km.

How much fuel does he use driving to and from work each week?

  1. 15 L
  2. 18 L
  3. 27 L
  4. 36 L
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Total distance travelled each week}\)

\(=5\times 2\times 15\)

\(=150\ \text{km}\)
 

\(\therefore\ \text{Total fuel used}\)

\(=\dfrac{150}{100}\times 12\ \text{L}\)

\(=18\ \text{L}\)

\(\Rightarrow B\)

v1 Algebra, STD2 A2 2017 HSC 3 MC

The graph shows the relationship between infant mortality rate (deaths per 1000 live births) and life expectancy at birth (in years) for different countries.
 

What is the life expectancy at birth in a country which has an infant mortality rate of 80?

  1. 20 years
  2. 21 years
  3. 61 years
  4. 62 years
Show Answers Only

\(D\)

Show Worked Solution

\(\text{When infant mortality rate is 80, life expectancy}\)

\(\text{at birth is 62 years (see below).}\)
 

\(\Rightarrow D\)

v1 Algebra, STD2 A2 2017 HSC 14 MC

Christopher is comparing two different models of 4WD cars. Car A uses fuel at the rate of 11.4 L/100 km. Car B uses 9.6 L/100 km.

Suppose Christopher plans on driving \(11\ 000\) km in the next year.

How much less fuel will he use driving car B instead of car A?

  1. 198 L
  2. 440 L
  3. 720 L
  4. 1000 L
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Difference in fuel usage}\)

\(=(11.4-9.6)\ \text{L/100 km}\)

\(=(1.8\ \text{L/100 km}\)
  

\(\therefore\ \text{Fuel saved using car}\ B\)

\(=\dfrac{11\ 000}{100}\times 1.8\)

\(=198\ \text{L}\)

\(\Rightarrow A\)

v1 Algebra, STD2 A4 EQ-Bank 8 MC

Water was poured into a container at a constant rate. The graph shows the depth of water in the container as it was being filled.
 


 

Which of the following containers could have been used to produce this result?

A. B.
C. D.
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Since the graph is a straight line, the container must have}\)
\(\text{vertical sides so the container will fill up at a constant rate.}\)
  

\(\Rightarrow C\)

v1 Algebra, STD2 A2 SM-Bank 4 MC

A car travels 480 km on 37 L of petrol.

What is its fuel consumption, correct to 1 decimal place?

  1. 5.2 L/100 km
  2. 7.7 L/100 km
  3. 13.0 L/100 km
  4. 15.4 L/100 km
Show Answers Only

\(B\)

Show Worked Solution

\(37\text{ litres are used to travel}\  4.8\times 100\ \text{km}\)

\(\text{Fuel consumption (L/100 km)}\)

\(=\dfrac{37}{4.8}\)

\(=7.7083\dots\)

\(=7.7\ \text{L/100 km}\)

\(\Rightarrow B\)

v1 Algebra, STD2 A2 2021 HSC 18

The fuel consumption for a medium SUV vehicle is 7.2 litres/100 km. On a road trip, the SUV travels a distance of 1325 km and the fuel cost is $2.15 per litre.

What is the total fuel cost for the trip?  (2 marks)

Show Answers Only

\($205.11\)

Show Worked Solution
\(\text{Total fuel used}\) \(=7.2\times \dfrac{1325}{100}\)  
  \(=95.4\ \text{litres}\)  

 

\(\text{Total fuel cost}\) \(=95.4\times 2.15\)  
  \(=$205.11\)  

v1 Algebra, STD2 A2 2022 HSC 16

Nicole is 38 years old, and likes to keep fit by doing cross-fit classes.

  1. Use this formula to find her maximum heart rate (bpm).
      
  2.      Maximum heart rate = 220 – age in years
      
  3. Nicole's maximum heart rate is ........................... bpm.   (1 mark)
  4. Nicole will get the most benefit from this exercise if her heart rate is between 50% and 85% of her maximum heart rate.
  5. Between what two heart rates should Nicole be aiming for to get the most benefit from her exercise?  (2 marks)

Show Answers Only

a.   \(182\ \text{bpm}\)

b.   \(91-155\ \text{bpm}\)

Show Worked Solution
a.     \(\text{Max heart rate}\) \(=220-38\)
    \(=182\ \text{bpm}\)

 

b.    \(\text{50% max heart rate}\ = 0.5\times 182 = 91\ \text{bpm}\)

\(\text{85% max heart rate}\ = 0.85\times 182 = 154.7\ \text{bpm}\)

\(\therefore\ \text{Nicole should aim for between 91 and 155 bpm during exercise.}\)

EXAMCOPY Functions, 2ADV F1 2009 HSC 1a

Sketch the graph of  `y-2x = 3`, showing the intercepts on both axes.   (2 marks)

Show Answers Only

 

Show Worked Solution

`y-2x=3\ \ =>\ \ y=2x+3`

`ytext{-intercept}\ = 3`

`text{Find}\ x\ text{when}\ y=0:`

`0-2x=3\ \ =>\ \ x=-3/2`
 

v1 Algebra, STD2 A1 SM-Bank 5

Fried's formula is used to calculate the medicine dosages for children aged 1-2 years.
 

\(\text{Child dosage}=\dfrac{\text{Age(in months)}\times \text{adult dosage}}{150}\)
 

Liam is 1.75 years old and receives a daily dosage of 350 mg of a medicine.

According to Fried's formula, what would the appropriate adult daily dosage of the medicine be?  (2 marks)

Show Answers Only

\(2500\ \text{mg}\)

Show Worked Solution

\(1.75\ \text{years}=21\ \text{months}\)

\(350\) \(=\dfrac{21\times \text{adult dosage}}{150}\)
\(\therefore\ \text{adult dosage}\) \(=\dfrac{350\times 150}{21}\)
   \(=2500\ \text{mg}\)

v1 Algebra, STD2 A1 2018 HSC 26b

Clark’s formula, given below, is used to determine the dosage of medicine for children.
 

\(\text{Dosage}=\dfrac{\text{weight in kg × adult dosage}}{70}\)

 
For a particular medicine, the adult dosage is 220 mg and the correct dosage for a specific child is 45 mg.

How much does the child weigh, to the nearest kg?  (2 marks)

Show Answers Only

\(14\ \text{kg}\)

Show Worked Solution

\(45 =\dfrac{\text{weight}\times 220}{70}\)

\(\therefore\ \text{weight}\) \(=\dfrac{70\times 45}{220}\)
  \(=14.318\dots\)
  \(\approx 14\ \text{kg  (nearest kg)}\)

v1 Algebra, STD2 A1 2015 HSC 26b

Clark’s formula is used to determine the dosage of medicine for children.
 

\(\text{Dosage}=\dfrac{\text{weight in kg × adult dosage}}{70}\)
 

The adult daily dosage of a medicine contains 1750 mg of a particular drug.

A child who weighs 30 kg is to be given tablets each containing 125 mg of this drug.

How many tablets should this child be given daily?  (2 marks)

Show Answers Only

\(6\)

Show Worked Solution
\(\text{Dosage}\) \(=\dfrac{30\times 1750}{70}\)
  \(=750\ \text{mg}\)

  
\(\text{Number of tablets per day}\)

\(=\dfrac{\text{Dosage}}{\text{mg per tablet}}\)

\(=\dfrac{750}{125}\)

\(=6\)
  

\(\therefore\ \text{The child should be given 6 tablets per day.}\)

v1 Algebra, STD2 A1 SM-Bank 15

Alonso drove 400 km in \(5\frac{1}{2}\) hours.

His average speed for the first 240 km was 80 km per hour.

How long did he take to travel the last 160 km?  (2 marks)

Show Answers Only

\(2\frac{1}{2}\ \text{hours}\)

Show Worked Solution

\(\text{Time for 1st 240 km}\)

\(=\dfrac{240}{80}\)

\(= 3\ \text{hours}\)

 

\(\therefore\ \text{Time for last 160 km}\)

\(=5\frac{1}{2}-3\)

\(=2\frac{1}{2}\ \text{hours}\)

v1 Algebra, STD2 A1 2020 HSC 3 MC

The distance between the Yarra Valley and Ballarat is 150 km. A person travels from the Yarra Valley to Ballarat at an average speed of 90 km/h.

How long does it take the person to complete the journey?

  1.  60 minutes
  2.  66 minutes
  3.  1 hour 30 minutes
  4.  1 hour 40 minutes
Show Answers Only

\(D\)

Show Worked Solution
\(\text{Time}\) \(=\dfrac{\text{Distance}}{\text{Speed}}\)
  \(=\dfrac{150}{90}\)
  \(=1.\dot{6}\ \text{hours}\)
  \(=1\ \text{hour}\ 40\ \text{minutes}\)

 
\(\Rightarrow D\)

v1 Algebra, STD2 A1 2007 HSC 24b

The distance in kilometres (\(D\)) of an observer from the centre of a thunderstorm can be estimated by counting the number of seconds (\(t\)) between seeing the lightning and first hearing the thunder.

Use the formula  \(D=\dfrac{t}{3}\)  to estimate the number of seconds between seeing the lightning and hearing the thunder if the storm is 2.1 km away.   (1 mark)

Show Answers Only

\(6.3\ \text{seconds}\)

Show Worked Solution

\(D=\dfrac{t}{3}\)

\(\text{When}\ \ D = 2.1,\)

\(\dfrac{t}{3}\) \(=2.1\)
\(t\) \(=6.3\ \text{seconds}\)

v1 Algebra, STD2 A1 SM-Bank 7

If  \(S = V_0(1 - r)^n\), find \(S\) given  \(V_0 = $57\ 000\), \(r = 0.12\) and \(n=5\). (give your answer to the nearest cent).  (2 marks)

Show Answers Only

\($30\ 080.72\ \text{(to nearest cent)}\)

Show Worked Solution
\(S\) \(=V_0(1 – r)^n\)
  \(=57\ 000 (1-0.12)^5\)
  \(=57\ 000 (0.88)^5\)
  \(=$30\ 080.719\dots\)
  \(=$30\ 080.72\ \text{(to nearest cent)}\)

EXAMCOPY Functions, MET2 2022 VCAA 4

Consider the function `f`, where `f:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, f(x)=\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right).`

Part of the graph of `y=f(x)` is shown below.
 

  1. State the range of `f(x)`.   (1 mark)

  2.  i. Find `f^{\prime}(0)`.   (2 marks)

  3. ii. State the maximal domain over which `f` is strictly increasing.   (1 mark)

  4. Show that `f(x)+f(-x)=0`.   (1 mark)

  5. Find the domain and the rule of `f^{-1}`, the inverse of `f`.   (3 marks)

  6. Let `h` be the function `h:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, h(x)=\frac{1}{k}\left(\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)\right)`, where `k \in R` and `k>0`.
  7. The inverse function of `h` is defined by `h^{-1}: R \rightarrow R, h^{-1}(x)=\frac{e^{k x}-1}{2\left(e^{k x}+1\right)}`.
  8. The area of the regions bound by the functions `h` and `h^{-1}` can be expressed as a function, `A(k)`.
  9. The graph below shows the relevant area shaded.
     

  1. You are not required to find or define `A(k)`.
  1. Determine the range of values of `k` such that `A(k)>0`.   (1 mark)

  2. Explain why the domain of `A(k)` does not include all values of `k`.   (1 mark)

Show Answers Only
a.     `R`
b.i `f^{\prime}(0)=4`
b.ii `\left(-\frac{1}{2}, \frac{1}{2}\right)`
c. `0`
d. `x \in \mathbb{R}`
e.i  ` k > 4`
e.ii No bounded area for `0<k \leq 4`
Show Worked Solution

a.   `R` is the range.

b.i    `f(x)`
 `= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)`  
  `f^{\prime}(x)` `= \frac{1}{x+\frac{1}{2}}+\frac{1}{\frac{1}{2}-x}`  
    `= \frac{2}{2 x+1}-\frac{2}{2 x-1}`  
  `f^{\prime}(0)` `= \frac{2}{2 xx 0+1}-\frac{2}{2 xx 0-1}`  
    `= 4`  

 
b.ii  `\left(-\frac{1}{2}, \frac{1}{2}\right)`

c.   `f(x)+f(-x)` `= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)+\log _e\left(-x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}+x\right)`  
  `= 0`  

 
d.   To find the inverse swap `x` and `y` in `y=f(x)`

`x` `= \log _e\left(y+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-y\right)`  
`x` `= \log _e\left(\frac{y+\frac{1}{2}}{\frac{1}{2}-y}\right)`  
`e^x` `=\frac{y+\frac{1}{2}}{\frac{1}{2}-y}`  
`y+\frac{1}{2}` `= e^x\left(-y+\frac{1}{2}\right)`  
`y+\frac{1}{2}` `= -e^x y+\frac{e^x}{2}`  
`y\left(e^x+1\right)` `= \frac{e^x-1}{2}`  
`:.\ f^(-1)(x)` `= \frac{e^x-1}{2(e^x + 1)}`  

 
  `:.`  Domain: `x \in \mathbb{R}`
  

e.i   The vertical dilation factor of  `f(x)` is  `1/k`

For `A(k)>=0` , `h^{\prime}(0)<1`

`\frac{1}{k}(4)<1`   [Using CAS]

`:.\  k > 4`


♦♦♦♦ Mean mark (e.i) 10%.
MARKER’S COMMENT: Incorrect responses included `k>0` and `4<k<33`.

e.ii  When `h \geq h^{-1}` for  `x>0` (or `h \leq h^{-1}` for  `x<0`) there is no bounded area.

`:.`  There will be no bounded area for `0<k \leq 4`.


♦♦♦♦ Mean mark (e.ii) 10%.

CHEMISTRY, M8 2013 VCE 8* MC

A forensic chemist tests mud from a crime scene to determine whether the mud contains zinc. Which one of the following analytical techniques would be best suited to this task?

  1. infrared spectroscopy
  2. mass spectroscopy
  3. atomic absorption spectroscopy
  4. nuclear magnetic resonance spectroscopy
Show Answers Only

\(C\)

Show Worked Solution

→ AAS is the only analytical technique that is specific to identifying metals within a solution.

\(\Rightarrow C\)

CHEMISTRY, M7 2018 VCE 1a

Organic compounds are numerous and diverse due to the nature of the carbon atom. There are international conventions for the naming and representation of organic compounds.

  1. Draw the structural formula of 2-methyl-propan-2-ol.   (1 mark)

  2. Give the molecular formula of but-2-yne.   (1 mark)

  1. Give the IUPAC name of the compound that has the structural formula shown above.   (1 mark)

Show Answers Only

i.    
         

ii.    \(\ce{C4H6}\)

→ Structural or semi-structural formulas are not appropriate.
 

iii.  → Longest carbon chain is 6, hence hexane

→ Number carbons to give functional group lowest possible numbers: 2,3-dibromo, 4-methyl

→ Compound name: 2,3-dibromo-4-methylhexane

Show Worked Solution

i.    
         

ii.    \(\ce{C4H6}\)

→ Structural or semi-structural formulas are not appropriate.
 

iii.  → Longest carbon chain is 6, hence hexane

→ Number carbons to give functional group lowest possible numbers: 2,3-dibromo, 4-methyl

→ Compound name: 2,3-dibromo-4-methylhexane

♦♦♦ Mean mark (c) 27%.

CHEMISTRY, M4 2012 VCE 5 MC

Nitrogen dioxide decomposes as follows.

\(\ce{2NO2(g) \rightarrow N2(g) + 2O2(g)}\ \quad \quad \Delta H = -66 \text{ kJ mol}^{-1}\)

The enthalpy change for the reaction represented by the equation  \(\ce{\frac{1}{2}N2(g) + O2(g) \rightarrow NO2(g)}\)  is

  1. \(-66 \text{ kJ mol} ^{-1}\)
  2. \(-33 \text{ kJ mol} ^{-1}\)
  3. \(+33 \text{ kJ mol} ^{-1}\)
  4. \(+66 \text{ kJ mol} ^{-1}\)
Show Answers Only

\(C\)

Show Worked Solution

\(\ce{2NO2(g) \rightarrow N2(g) + 2O2(g),}\ \quad \Delta H = -66 \text{ kJ mol}^{-1}\)

\(\ce{N2(g) + 2O2(g) \rightarrow 2NO2(g),}\ \quad \Delta H = +66 \text{ kJ mol}^{-1}\)

\(\ce{\frac{1}{2}N2(g) + O2(g) \rightarrow NO2(g),}\ \quad \Delta H = +33 \text{ kJ mol}^{-1}\)

\(\Rightarrow C\)

CHEMISTRY, M4 2013 VCE 16*

\(\ce{C(s) + O2(g)\rightarrow CO2(g)}\) \(\quad \quad \Delta H = -393.5 \text{ kJ mol}^{-1}\)
\(\ce{2H(g) + O2(g)\rightarrow 2H2O(l)}\) \(\quad \quad \Delta H = -571.6 \text{ kJ mol}^{-1}\)

Given the information above, what is the enthalpy change for the following reaction?   (2 marks)

\(\ce{C(s) + 2H2O(l)\rightarrow CO2(g) + 2H2(g)}\)

Show Answers Only

\(\Delta H = 178.1\ \text{kJ mol}^{-1}\)

Show Worked Solution

1st reaction is unchanged:

\(\ce{C(s) + O2(g)\rightarrow CO2(g)}\quad \Delta H = -393.5 \text{ kJ mol}^{-1}\)
 

Reverse 2nd reaction:

\(\ce{2H2O(l) \rightarrow 2H(g) + O2(g) }\quad \Delta H = +571.6 \text{ kJ mol}^{-1}\)
 

Combine both reactions:

\(\ce{C(s) + 2H2O(l)\rightarrow CO2(g) + 2H2(g)}\)

\(\Delta H = -393.5+571.6 = 178.1\ \text{kJ mol}^{-1}\)

CHEMISTRY, M4 2016 VCE 17*

The combustion of hexane takes place according to the equation

\(\ce{C6H14(g) + \dfrac{19}{2}O2(g)\rightarrow 6CO2(g) + 7H2O(g)}\) \(\quad \quad \Delta H = -4158\ \text{kJ mol}^{-1}\)

Consider the following reaction.

\(\ce{ 12CO2(g) + 14H2O(g)\rightarrow 2C6H14(g) + 19O2(g)}\)

  1. Calculate the value of \(\Delta H\), in kJ mol\(^{-1}\), for this reaction   (2 marks)

  2. Is the reaction exothermic or endothermic?   (1 mark)

Show Answers Only

a.    \(\Delta H = +8316\ \text{kJ mol}^{-1}\)

b.    Endothermic

Show Worked Solution

a.    \(\ce{C6H14(g) + \dfrac{19}{2}O2(g)\rightarrow 6CO2(g) + 7H2O(g)}\) \(\quad \Delta H = -4158\ \text{kJ mol}^{-1}\)

→ Reverse equation:

\(\ce{6CO2(g) + 7H2O(g)\rightarrow C6H14(g) + \dfrac{19}{2}O2(g)}\) \(\quad \Delta H = +4158\ \text{kJ mol}^{-1}\)

→ Double equation:

\(\ce{12CO2(g) + 14H2O(g)\rightarrow 2C6H14(g) + 19O2(g)}\) \(\quad \Delta H = 2 \times +4158\ \text{kJ mol}^{-1}\)

→ \(\Delta H = +8316\ \text{kJ mol}^{-1}\)

b.    Endothermic

CHEMISTRY, M4 2015 VCE 17 MC

The oxidation of sulfur dioxide is an exothermic reaction. The reaction is catalysed by vanadium\(\text{(V)}\) oxide.

\(\ce{2SO2(g) + O2(g)\rightleftharpoons 2SO3(g)}\)

Which one of the following energy profile diagrams correctly represents both the catalysed and the uncatalysed reaction?
 


 

Show Answers Only

\(B\)

Show Worked Solution

→ Exothermic reaction releases energy (eliminate C and D).

→ Catalysts only effect the enthalpy during the chemical reaction by reducing the activation energy required.

\(\Rightarrow B\)

CHEMISTRY, M4 2018 VCE 14*

An equation for the complete combustion of methanol is

\(\ce{2CH3OH(l) + 3O2(g)\rightarrow 2CO2(g) + 4H2O(g)}\ \ \ \ \ \ \ \Delta H=-726\ \text{kJ mol}^{-1}\)

  1. State whether this reaction exothermic or endothermic.   (1 mark)

  2. Calculate the total enthalpy change of the equation, in kilojoules.   (1 mark)

Show Answers Only

a.    → Combustion is an exothermic reaction.

→ \(\Delta H=-726\ \text{kJ mol}^{-1}\)  also indicates it is exothermic.
 

b.   → \(\Delta H\ \ce{(CH3OH) = 726\ \text{kJ mol}^{-1}}\)

→ 2 moles of \(\ce{CH3OH}\) are involved.

→ \(\text{Total}\ \Delta H = 2 \times -726 = -1452\ \text{kJ}\)

Show Worked Solution

a.    → Combustion is an exothermic reaction.

→ \(\Delta H=-726\ \text{kJ mol}^{-1}\)  also indicates it is exothermic.
 

b.   → \(\Delta H\ \ce{(CH3OH) = 726\ \text{kJ mol}^{-1}}\)

→ 2 moles of \(\ce{CH3OH}\) are involved.

→ \(\text{Total}\ \Delta H = 2 \times -726 = -1452\ \text{kJ}\)

PHYSICS, M5 2019 VCE 10

A projectile is launched from the ground at an angle of 39° and at a speed of 25 m s\(^{-1}\), as shown in Figure 10. The maximum height that the projectile reaches above the ground is labelled \(h\).
 

  1. Ignoring air resistance, show that the projectile's time of flight from the launch to the highest point is equal to 1.6 seconds. Give your answer to two significant figures. Show your working and indicate your reasoning.  (2 marks)

  2. Calculate the range, \(R\), of the projectile. Show your working.  (2 marks)

Show Answers Only

a.    \(t=1.6\ \text{s}\)

b.    \(62\ \text{m}\)

Show Worked Solution

a.    
       

\(v_v=25\,\sin39^{\circ}=15.733\ \text{ms}^{-1}\)

\(\text{At max height:}\ v_v=0\)

\(v\) \(=u+at_1\)  
\(0\) \(=15.733 -9.8t_1\)  
\(9.8t_1\) \(=15.733\)  
\(t_1\) \(=1.6\ \text{s  (2 sig.fig)}\)  

 

b.    \(\text{Since path is symmetrical:}\)

\(\text{Time of flight}\ (t_2) =1.6 \times 2=3.2\ \text{s}\).

\(v_h=25\,\cos39^{\circ}=19.43\ \text{ms}^{-1}\ \ \text{(see part (a) diagram)}\) 

\(\therefore R=v_ht_2=19.43 \times 3.2= 62\ \text{m}\)

PHYSICS, M6 2019 VCE 7*

Students in a Physics practical class investigate the piece of electrical equipment shown in Figure 5. It consists of a single rectangular loop of wire that can be rotated within a uniform magnetic field. The loop has dimensions 0.50 m × 0.25 m and is connected to the output terminals with slip rings. The loop is in a uniform magnetic field of strength 0.40 T.
 

  1. What name best describes the piece of electrical equipment shown in Figure 5.   (1 mark)

  2. What is the magnitude of the flux through the loop when it is in the position shown in Figure 5? Explain your answer.   (2 marks)

The students connect the output terminals of the piece of electrical equipment to an oscilloscope. One student rotates the loop at a constant rate of 20 revolutions per second.

  1. Calculate the period of the rotation of the loop.  (1 mark)

  2. Calculate the maximum flux through the loop. Show your working.  (1 mark)

  3. The loop starts in the position shown in Figure 5.
  4. What is the average voltage measured across the output terminals for the first quarter turn? Show your working.  (2 marks)

Show Answers Only

a.    Alternator.

b.   → \(0\ \text{Wb}\)

→ This is because the plane of the area of the coil is parallel to the direction of the magnetic field, not perpendicular.

c.    \(0.05\ \text{s}\)

d.    \(0.05\ \text{Wb}\)

e.    \(4\ \text{V}\)

Show Worked Solution

a.    → The device is an alternator.

→ This is due to the input being mechanical motion and the output being AC current, whereas an AC motor is the opposite.

♦ Mean mark (a) 44%.
COMMENT: Many students incorrectly answered AC motor.

b.   → \(0\ \text{Wb}\)

→ This is because the plane of the area of the coil is parallel to the direction of the magnetic field, not perpendicular.
 

c.    \(T=\dfrac{1}{f}=\dfrac{1}{20}=0.05\ \text{s}\)
 

d.    \(\Phi=BA=0.4 \times (0.5 \times 0.25)=0.05\ \text{Wb}\)
 

e.     \(\varepsilon\) \(=\dfrac{\Delta \Phi}{\Delta t_{\text{1/4 rotation}}}\)
    \(=\dfrac{0.05}{0.0125}\)
    \(=4\ \text{V}\)
♦ Mean mark (e) 51%.

PHYSICS, M6 2019 VCE 6

A home owner on a large property creates a backyard entertainment area. The entertainment area has a low-voltage lighting system. To operate correctly, the lighting system requires a voltage of 12 V. The lighting system has a resistance of 12 \(\Omega\).

  1. Calculate the power drawn by the lighting system.   (1 mark)

To operate the lighting system, the home owner installs an ideal transformer at the house to reduce the voltage from 240 V to 12 V. The home owner then runs a 200 m long heavy-duty outdoor extension lead, which has a total resistance of 3 \( \Omega\), from the transformer to the entertainment area.

  1. The lights are a little dimmer than expected in the entertainment area.
  2. Give one possible reason for this and support your answer with calculations.   (4 marks)

  3. Using the same equipment, what changes could the home owner make to improve the brightness of the lights? Explain your answer.   (2 marks)

Show Answers Only

a.    \(12\ \text{W}\)

b.   Potential cause of dimmer lights: 

→ Power loss in the extension lead.

→ The total resistance (\(R_T\)) in the circuit \(=12 + 3=15\ \Omega\)

→ The new current in the circuit \(=\dfrac{V}{R_T}=\dfrac{12}{15}=0.8\ \text{A}\)

→ The power loss across the extension lead, \(P_{\text{loss}}=I^2R_l=0.8^2 \times 3=1.92\ \text{W}\)

→ Therefore, power supplied to lights \(=12.0-1.92=10.08\ \text{W}\).

c.    Changes using the same equipment:

→ The homeowner needs to move the transformer from inside the home (start of the lead) to the outside entertainment area (end of the lead).

→ This would mean a higher voltage/lower current is running through the lead which would decrease the power loss through the lead and improve the brightness of the lights.

Show Worked Solution

a.    \(I=\dfrac{V}{R}=\dfrac{12}{12}=1\ \text{A}\)

\(P=IV=1 \times 12 = 12\ \text{W}\)
 

b.   Potential cause of dimmer lights: 

→ Power loss in the extension lead.

→ The total resistance (\(R_T\)) in the circuit \(=12 + 3=15\ \Omega\)

→ The new current in the circuit \(=\dfrac{V}{R_T}=\dfrac{12}{15}=0.8\ \text{A}\)

→ The power loss across the extension lead, \(P_{\text{loss}}=I^2R_l=0.8^2 \times 3=1.92\ \text{W}\)

→ Therefore, power supplied to lights \(=12.0-1.92=10.08\ \text{W}\).

♦♦♦ Mean mark (b) 23%.
COMMENT: Many students used 1 Amp in their calculations instead of finding the total current in the system.

c.    Changes using the same equipment:

→ The homeowner needs to move the transformer from inside the home (start of the lead) to the outside entertainment area (end of the lead).

→ This would mean a higher voltage/lower current is running through the lead which would decrease the power loss through the lead and improve the brightness of the lights.

♦♦♦ Mean mark (c) 22%.
COMMENT: Students incorrectly added or changed the equipment used in the question.

PHYSICS, M6 2019 VCE 3

Figure 3 shows a schematic diagram of a DC motor. The motor has a coil, \(JKLM\), consisting of 100 turns. The permanent magnets provide a uniform magnetic field of 0.45 T.

The commutator connectors, \(X\) and \(Y\), provide a constant DC current, \(I\), to the coil. The length of the side \(JK\) is 5.0 cm.

The current \(I\) flows in the direction shown in the diagram.
 

  1. Which terminal of the commutator is connected to the positive terminal of the current supply?   (1 mark)

  2. Draw an arrow on Figure 3 to indicate the direction of the magnetic force acting on the side \(JK\).   (1 mark)

  3. Explain the role of the commutator in the operation of the DC motor.   (2 marks)

  4. A current of 6.0 A flows through the 100 turns of the coil \(JKLM\).
  5. The side \(JK\) is 5.0 cm in length.
  6. Calculate the size of the magnetic force on the side \(JK\) in the orientation shown in Figure 3. Show your working.   (2 marks)

Show Answers Only

a.    → Current runs from the positive terminal to the negative terminal.

→ \(X\) is connected to the positive terminal of the current supply.
 

b.    Using the right-hand rule: force on \(JK\) is down the page.

c.    Role of the split-ring commutator:

→ Changes the direction of the current through the arms of the coil every 180\(^{\circ}\).

→ This ensures that the torque through the motor will always be in the same direction so the motor will rotate in the same direction.
 

d.    \(F=nlIB=100 \times 0.05 \times 6 \times 0.45 = 13.5\ \text{N}\)

Show Worked Solution

a.    → Current runs from the positive terminal to the negative terminal.

→ \(X\) is connected to the positive terminal of the current supply.
 

b.    Using the right-hand rule: force on \(JK\) is down the page.

c.    Role of the split-ring commutator:

→ Changes the direction of the current through the arms of the coil every 180\(^{\circ}\).

→ This ensures that the torque through the motor will always be in the same direction so the motor will rotate in the same direction.
 

d.    \(F=nlIB=100 \times 0.05 \times 6 \times 0.45 = 13.5\ \text{N}\)

PHYSICS, M7 2019 VCE 16 MC

Students are conducting a photoelectric effect experiment. They shine light of known frequency onto a metal and measure the maximum kinetic energy of the emitted photoelectron.

The students increase the intensity of the incident light.

The effect of this increase would most likely be

  1. lower maximum kinetic energy of the emitted photoelectrons.
  2. higher maximum kinetic energy of the emitted photoelectrons.
  3. fewer emitted photoelectrons but of higher maximum kinetic energy.
  4. more emitted photoelectrons but of the same maximum kinetic energy.
Show Answers Only

\(D\)

Show Worked Solution

→ The energy of a photon (which is transferred to the photoelectron) is independent of the intensity of the light.

→ Increasing the intensity of light will not change the maximum kinetic energy of the photoelectrons.

→ However, a higher intensity of light corresponds to more photons and thus more emitted photoelectrons.

\(\Rightarrow D\)

PHYSICS, M8 2019 VCE 14* MC

Electrons are accelerated in an electron gun to a speed of 1.0 × 10\(^7\) m s\(^{-1}\).

The best estimate of the de Broglie wavelength of these electrons is

  1. 4.5 × 10\(^{-6}\) m
  2. 7.3 × 10\(^{-8}\) m
  3. 7.3 × 10\(^{-11}\) m
  4. 4.5 × 10\(^{-12}\) m
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\(C\)

Show Worked Solution

\(\lambda=\dfrac{h}{mv}=\dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 1 \times 10^7}=7.3 \times 10^{-11}\ \text{m}\)

\(\Rightarrow C\)