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ENGINEERING, PPT 2022 HSC 23c

Steel -beams have been used when large, open spans need to be created inside buildings.

Explain how microstructural changes take place in steel when an -beam is formed using the process of hot rolling. You may use a drawing to support your answer.   (4 marks)

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→ During hot rolling the steel is heated to above its recrystallisation temperature.

→ Therefore, due to the pressure of the rollers, the grains change from their original state to become elongated.

→ However, as the steel exits the rollers it is still above recrystallisation temperature, resulting in the recrystallisation of the elongated grains to create finer, equiaxed grains.

Show Worked Solution

→ During hot rolling the steel is heated to above its recrystallisation temperature.

→ Therefore, due to the pressure of the rollers, the grains change from their original state to become elongated.

→ However, as the steel exits the rollers it is still above recrystallisation temperature, resulting in the recrystallisation of the elongated grains to create finer, equiaxed grains.


♦ Mean mark 48%.

ENGINEERING, CS 2022 HSC 23b

The diagram represents a 120 kg beam that is being guided into place by a crane.
 

Use a scale drawing to graphically determine the tension in the two cables attached to the beam.   (3 marks)

Show Answers Only

Scale: 1 mm = 20 N


 

Tension in C1 = 1.3 kN

Tension in C2 = 1.06 kN

Show Worked Solution

Scale: 1 mm = 20 N


 

Tension in C1 = 1.3 kN

Tension in C2 = 1.06 kN

(Answers that are relatively similar in either direction will be correct as question asks for graphical solution, creating a margin of error)


♦ Mean mark 46%.

CHEMISTRY, M6 2016 HSC 29

A solution of hydrochloric acid was standardised by titration against a sodium carbonate solution using the following procedure.

  • All glassware was rinsed correctly to remove possible contaminants.
  • Hydrochloric acid was placed in the burette.
  • 25.0 mL of sodium carbonate solution was pipetted into the conical flask.

The titration was performed and the hydrochloric acid was found to be 0.200 mol L¯1.

  1. Identify the substance used to rinse the conical flask and justify your answer.   (2 marks)

  2. Seashells contain a mixture of carbonate compounds. The standardised hydrochloric acid was used to determine the percentage by mass of carbonate in a seashell using the following procedure.

• A 0.145 g sample of the seashell was placed in a conical flask.

• 50.0 mL of the standardised hydrochloric acid was added to the conical flask.

• At the completion of the reaction, the mixture in the conical flask was titrated with 0.250 mol L¯1 sodium hydroxide.

  1. The volume of sodium hydroxide used in the titration was 29.5 mL.
  2. Calculate the percentage by mass of carbonate in the sample of the seashell.   (4 marks)

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a.   → Distilled water.

→ This should be used to rinse the conical flask as this will not change the number of moles of placed in it.

b.   54.3%

Show Worked Solution

a.   → Distilled water.

→ This should be used to rinse the conical flask as this will not change the number of moles of placed in it.
 


♦ Mean mark (a) 42%.

b.   


 


♦♦ Mean mark (b) 40%.

ENGINEERING, TE 2022 HSC 22b

In the event of a fall or a medical emergency, smart watches are designed to alert emergency services when either of the following conditions is met.

Condition 1: the smart watch emergency alert is manually activated

Condition 2: the smart watch detects a sudden fall and no movement for 1 minute

An incomplete logic diagram showing the activation of the smart watch is given.
 

Complete the diagram by identifying the inputs and drawing the appropriate logic gates.   (3 marks)

Show Answers Only

Show Worked Solution


Mean mark 54%.

ENGINEERING, PPT 2022 HSC 21b

A pictorial drawing of an assembled chain master link is shown.   
 

Complete the orthogonal views below, to AS 1100 standards, by adding the following dimensions.   (3 marks)

  • • The overall height of the assembly – 55 mm
    • The thickness of one link plate – 5 mm
    • The radius of the link plate – 15 mm
    • The diameter of the link pin – 12 mm
    • The chamfer of the pin – 1 mm
    • Distance between centres of pins – 38 mm

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Notes on this drawing:

→ The correct radius (R) and diameter (∅) symbols must be used

→ The chamfer must be labelled with the width and angle

→ Dimensioning must be completed to AS1100 standards (including arrow heads and line thickness)

Show Worked Solution

Notes on this drawing:

→ The correct radius (R) and diameter (∅) symbols must be used

→ The chamfer must be labelled with the width and angle

→ Dimensioning must be completed to AS1100 standards (including arrow heads and line thickness)


Mean mark 52%.

ENGINEERING, PPT 2022 HSC 21a

The following images show an older bus and a contemporary bus.
 

  1. Older buses had flat, toughened glass windscreen panels.
  2. How is toughened glass manufactured?   (2 marks)
  1. Contemporary buses have windscreens which curve around the front of the bus.
  2. Describe the manufacturing process used to make these windscreens.   (3 marks)
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i.    Toughened glass is formed by:-

→ heating the glass to annealing temperature (approximately 600 degrees) and quenching/rapidly cooling the outside layers using a jet of water.

→ Alternatively toughened glass can be created by soaking the glass in potassium nitrate for an extended period.
 

ii.    Production of curved glass panels:

→ Windscreens are produced by placing a flat sheet of glass above a metal mould.

→ The glass is then evenly heated until it falls and takes the shape of the mould, and then allowed to cool.

→ Once cooled, the glass is then laminated.

→ Laminating involves placing a layer of polymer between two identically shaped sheets of glass, in this case curved windshields.

Show Worked Solution

i.    Toughened glass is formed by:-

→ heating the glass to annealing temperature (approximately 600 degrees) and quenching/rapidly cooling the outside layers using a jet of water.

→ Alternatively toughened glass can be created by soaking the glass in potassium nitrate for an extended period.


♦ Mean mark (i) 43%.

ii.    Production of curved glass panels:

→ Windscreens are produced by placing a flat sheet of glass above a metal mould.

→ The glass is then evenly heated until it falls and takes the shape of the mould, and then allowed to cool.

→ Once cooled, the glass is then laminated.

→ Laminating involves placing a layer of polymer between two identically shaped sheets of glass, in this case curved windshields.


♦ Mean mark (ii) 39%.

CHEMISTRY, M8 2016 HSC 27

The volume of gas formed at 25°C and 100 kPa as hydrochloric acid was added to a pure sample of aluminium is shown in the graph.
 

Calculate the original mass of the aluminium sample used in the reaction.   (4 marks)

Show Answers Only

0.109 grams

Show Worked Solution

→ The graph shows that all the aluminium has reacted when no more gas is being produced, at a volume of 0.150 L.


♦ Mean mark 48%.

CHEMISTRY, M7 2016 HSC 23

A spirit burner containing ethanol was used to heat water in a conical flask for three minutes to measure the molar heat of combustion of ethanol.

The results from the investigation are shown.
 

  1. On the grid, draw a line graph to represent the data contained in the table.   (3 marks)
     

  1. The following values were also recorded during the investigation:
  2.       Initial mass of spirit burner = 236.14 g
  3.       Final mass of spirit burner = 235.56 g
  4.       Calculated experimental molar heat of combustion of ethanol = – 827 kJ mol ¯1.
  5. Using information from the previous page and the above values, determine the mass of water that was in the conical flask.   (3 marks)

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a.   
     

b.  0.199 kg

Show Worked Solution

a.   


 

b.   Using the graph:

   

Mean mark (b) 53%.

CHEMISTRY, M7 2016 HSC 22

This apparatus was set up to produce methyl butanoate.
 

  1. Identify a safety issue in this experiment.   (1 mark)

  2. Using structural formulae, write the equation for the production of methyl butanoate.   (2 marks)

  3. Justify the use of apparatus in this experiment.   (2 marks)

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a.   Flame could ignite one of reagents which is flammable.

b.   
     

c.    → Esterification is a relatively slow reaction.

→ Heating the reaction makes it go faster. However, the low boiling points of the reactants make them volatile as they readily convert into gas.

→ The cooling condenser prevents the gas reactants from escaping the experiment by condensing them back into the reaction mixture. This process allows the reaction to proceed at higher temperatures.

Show Worked Solution

a.   Flame could ignite one of reagents which is flammable.
 

b.   
     


 


♦ Mean mark (b) 52%.

c.    → Esterification is a relatively slow reaction.

→ Heating the reaction makes it go faster. However, the low boiling points of the reactants make them volatile as they readily convert into gas.

→ The cooling condenser prevents the gas reactants from escaping the experiment by condensing them back into the reaction mixture. This process allows the reaction to proceed at higher temperatures.


♦♦♦ Mean mark (c) 14%.

ENGINEERING, CS 2022 HSC 20 MC

For a particular material, specimens of different sizes were tested. A load-extension diagram and a stress-strain diagram are to be drawn for each specimen.

The shape of which diagram(s) will be affected by the size of the specimen tested?

  1. Both diagrams
  2. Neither diagram
  3. The stress–strain diagram only
  4. The load–extension diagram only
Show Answers Only

Show Worked Solution

→ The stress–strain diagrams will be the same shapes as both stress and strain are calculated by dividing by constants (original area and original length).


♦♦ Mean mark 36%.

ENGINEERING, PPT 2022 HSC 19 MC

A gear system for a machine is shown.
 

What is the speed, in revolutions per minute (rpm), of the driven gear when the driving gear rotates at 1800 rpm?

  1. 150 rpm
  2. 450 rpm
  3. 600 rpm
  4. 1800 rpm
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Show Worked Solution
 
   
   

 


♦ Mean mark 44%.

ENGINEERING, PPT 2022 HSC 18 MC

A pictorial view of a machine component is shown.

When viewed from the direction of arrow A, which drawing correctly shows the half-sectional view?
 

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Show Worked Solution

By Elimination:

→ and are full sectional views of the actual body of the component.

→ Do not section webs in sectional drawings (eliminate )


♦ Mean mark 48%.

ENGINEERING, TE 2022 HSC 14 MC

What feature of fibre optic cables has made them a suitable replacement for traditional copper-based cables?

  1. Lower bandwidth
  2. Higher attenuation
  3. Do not need repeaters over long distances
  4. Not susceptible to electromagnetic interference
Show Answers Only

Show Worked Solution

By Elimination:

→ and are both negative impacts

→ may not be true over long enough distances


♦ Mean mark 49%.

ENGINEERING, AE 2022 HSC 10 MC

Which of the following identifies two causes of parasitic drag?

  1. Aircraft lift, angle of attack
  2. Aircraft lift, material of aircraft skin
  3. The movement of air over the wing, angle of attack
  4. The movement of air over the wing, material of aircraft skin
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Show Worked Solution

→ Parasitic drag is all drag that is caused by the shape, construction-type and material of an aircraft.


♦♦ Mean mark 37%.

ENGINEERING, CS 2022 HSC 7 MC

What is the thickness of a standard nut that fits a M20 × 2 bolt?

  1. 10 mm
  2. 16 mm
  3. 18 mm
  4. 20 mm
Show Answers Only

Show Worked Solution

→ AS1100 standards state the width of a nut is 0.8 x the diameter of the bolt.

→ 0.8 × 20 = 16 mm


Mean mark 52%.

BIOLOGY, M8 2015 HSC 31

'Renal dialysis and kidney transplants are very different treatments for the same medical condition. Each treatment was developed from a new application of biological knowledge.'

Justify these statements.  (8 marks)

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Consider each element of the statement separately:

Medical Conditions

→ Kidney failure is a recognised medical condition that can be treated using both kidney transplants or renal dialysis.
 

Different Treatments

→ Renal dialysis is the cleansing of wastes from the blood externally using a dialysis machine.

→ During a kidney transplant the diseased organ is removed and replaced with a healthy organ provided by a donor.
 

New Application of biological information: Renal dialysis

→ The movement of substances from regions of high concentration to regions of low concentration is known as diffusion.

→ Diffusion can occur across a semi-permeable membrane.

→The dialysis machine allows blood to flow through tubing which is permeable to urea.

→ The solution around the tubing is continually replaced to maintain a steep concentration gradient so the urea can be removed out of the blood through diffusion.
 

New Application of biological information: Kidney transplant

→ Increased knowledge of the immune system allowed for an improved understanding of organ rejection following organ transplantation.

→ B and T cells on donated organs are recognised by the immune system as foreign and it then attacks the transplanted organ.

→ Once an infection has been removed, suppressor T cells stop the immune cells and switch off the immune response.

→ This knowledge lead to the development of immunosuppressants or anti-rejection drugs designed to fight the immune symptoms’ response to reject donated organs.

→After transplantation, anti-rejection drugs are used by recipients for their lifetime.

Show Worked Solution

Medical Conditions

→ Kidney failure is a recognised medical condition that can be treated using both kidney transplants or renal dialysis.
 

Different Treatments

→ Renal dialysis is the cleansing of wastes from the blood externally using a dialysis machine.

→ During a kidney transplant the diseased organ is removed and replaced with a healthy organ provided by a donor.
 

New Application of biological information: Renal dialysis

→ The movement of substances from regions of high concentration to regions of low concentration is known as diffusion.

→ Diffusion can occur across a semi-permeable membrane.

→The dialysis machine allows blood to flow through tubing which is permeable to urea.

→ The solution around the tubing is continually replaced to maintain a steep concentration gradient so the urea can be removed out of the blood through diffusion.
 

New Application of biological information: Kidney transplant

→ Increased knowledge of the immune system allowed for an improved understanding of organ rejection following organ transplantation.

→ B and T cells on donated organs are recognised by the immune system as foreign and it then attacks the transplanted organ.

→ Once an infection has been removed, suppressor T cells stop the immune cells and switch off the immune response.

→ This knowledge lead to the development of immunosuppressants or anti-rejection drugs designed to fight the immune symptoms’ response to reject donated organs.

→After transplantation, anti-rejection drugs are used by recipients for their lifetime.


♦♦ Mean mark 41%.

CHEMISTRY, M8 2015 HSC 29

The procedure of a first-hand investigation conducted in a school laboratory to determine the percentage of sulfate in a lawn fertiliser is shown.
 

  1. Suggest modifications that could be made to the procedure to improve the results of this investigation. Justify your suggestions.   (4 marks)

  2. Calculate the percentage of sulfate in the original fertiliser sample.   (3 marks)

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a.   Modification 1:

→  The should be added slowly. This would help more of the sulfate ions come into contact with barium and form a precipitate.

Modification 2:

→ Barium sulfate is a very fine precipitate. By heating the mixture gently for some time, the precipitate is able to flocculate into larger particles, which would be trapped by the filter.

Modification 3:

→ The pores of most filter paper are still too large to effectively capture the precipitate. The experiment should use a small pore filter, like a sintered glass crucible.

Modification 4:

→ Multiple samples of the fertiliser should be analysed and the average value calculated to achieve more valid results. This would mitigate inaccuracies caused by unevenly distributed sulfate in the mixture.

b.   45.9%

Show Worked Solution

a.   Modification 1:

→  The should be added slowly. This would help more of the sulfate ions come into contact with barium and form a precipitate.

Modification 2:

→ Barium sulfate is a very fine precipitate. By heating the mixture gently for some time, the precipitate is able to flocculate into larger particles, which would be trapped by the filter.

Modification 3:

→ The pores of most filter paper are still too large to effectively capture the precipitate. The experiment should use a small pore filter, like a sintered glass crucible.

Modification 4:

→ Multiple samples of the fertiliser should be analysed and the average value calculated to achieve more valid results. This would mitigate inaccuracies caused by unevenly distributed sulfate in the mixture.
 


♦ Mean mark (a) 49%.

b.   


Mean mark (b) 54%.

BIOLOGY, M6 2015 HSC 29

'The application of modern reproductive techniques in plant and animal breeding limits genetic diversity.'

Discuss this statement.   (6 marks)

Show Answers Only

→ Artificial pollination occurs when humans perform the natural pollination process and transfer the pollen manually to the stigma of one plant from the anther of another.

→ Artificial insemination is the manual transfer of collected semen into the female reproductive system of animals and humans.

→ Both methods can result in an increase in the number of offspring that can be produced by one parent.

→ Whilst the increase in offspring produced may be beneficial, there is also the added risk of reducing the genetic diversity in future populations as one parent can produce many more offspring.

→ On the other hand where species are endangered, or natural processes are interrupted (for example declining bee populations), the techniques provide an opportunity for numbers to stabilise or even increase in subsequent generations.

→ Another advantage is that the process allows for banks of sperm and pollen to be established that can be accessed by other countries, thus creating genetic diversity, by enabling endangered species to survive across continents.

Show Worked Solution

→ Artificial pollination occurs when humans perform the natural pollination process and transfer the pollen manually to the stigma of one plant from the anther of another.

→ Artificial insemination is the manual transfer of collected semen into the female reproductive system of animals and humans.

→ Both methods can result in an increase in the number of offspring that can be produced by one parent.

→ Whilst the increase in offspring produced may be beneficial, there is also the added risk of reducing the genetic diversity in future populations as one parent can produce many more offspring.

→ On the other hand where species are endangered, or natural processes are interrupted (for example declining bee populations), the techniques provide an opportunity for numbers to stabilise or even increase in subsequent generations.

→ Another advantage is that the process allows for banks of sperm and pollen to be established that can be accessed by other countries, thus creating genetic diversity, by enabling endangered species to survive across continents.


♦ Mean mark 49%.

CHEMISTRY, M6 2015 HSC 28

The equipment shown is set up. After some time a ring of white powder is seen to form on the inside of the glass tube.

  1. Why would this NOT be an acid-base reaction according to Arrhenius?   (1 mark)

  2. Explain why this would be considered a Bronsted-Lowry acid-base reaction. Include an equation in your answer.   (2 marks)

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a.   According to Arrhenius:

→ An acid is a solution that produces hydrogen ions when in a solution.

→ A base is a solution that produces hydroxide ions when in a solution.

→ This reaction does not occur in an aqueous solution and would not be an acid-base reaction according to Arrhenius.
 

b.   

→ A Bronsted-Lowry acid donates a proton while a base accepts a proton.

→ This reaction involves proton transfer ( donates, receives) and would therefore be considered a Bronsted-Lowry acid-base reaction.

Show Worked Solution

a.   According to Arrhenius:

→ An acid is a solution that produces hydrogen ions when in a solution.

→ A base is a solution that produces hydroxide ions when in a solution.

→ This reaction does not occur in an aqueous solution and would not be an acid-base reaction according to Arrhenius.
 


♦♦ Mean mark (a) 37%.

b.   

→ A Bronsted-Lowry acid donates a proton while a base accepts a proton.

→ This reaction involves proton transfer ( donates, receives) and would therefore be considered a Bronsted-Lowry acid-base reaction.


Mean mark (b) 53%.

CHEMISTRY, M6 2015 HSC 26

A sodium hydroxide solution was titrated against citric acid which is triprotic.

  1. Draw the structural formula of citric acid (2-hydroxypropane-1,2,3-tricarboxylic acid).   (1 mark)
     

  1. How could a computer-based technology be used to identify the equivalence point of this titration?   (2 marks)

  2. The sodium hydroxide solution was titrated against 25.0 mL samples of 0.100 mol L ¯1 citric acid. The average volume of sodium hydroxide used was 41.50 mL.
  3. Calculate the concentration of the sodium hydroxide solution.   (4 marks)

Show Answers Only

a.   
         

b.   Technology solution

→ A digital pH probe could be placed in the flask and used to collect data that plots the pH of the solution against the volume of sodium hydroxide added.

→ The equivalence point would be identified by a steep rise in the pH on the graph.
 

c.   

Show Worked Solution
a.   
         

♦ Mean mark (a) 46%.

b.   Technology solution

→ A digital pH probe could be placed in the flask and used to collect data that plots the pH of the solution against the volume of sodium hydroxide added.

→ The equivalence point would be identified by a steep rise in the pH on the graph.
 


♦ Mean mark (b) 42%.

c.   


Mean mark (c) 56%.

CHEMISTRY, M7 2015 HSC 25a

Describe the steps involved in the process of addition polymerisation.   (3 marks)

Show Answers Only

Step 1:

→ An organic peroxide (initiator – ) attacks the ethylene molecule, breaking its double bond and forming a free radical.

→ An unpaired reactive electron can be seen on the end of the growing chain (initiation).

Step 2:

→ The free radical will then attack another ethylene molecule, increasing the length of the growing polymer chain (propagation).

Step 3:

→ Chain length increases in this fashion until two growing chains combine (termination).

Show Worked Solution

Step 1:

→ An organic peroxide (initiator – ) attacks the ethylene molecule, breaking its double bond and forming a free radical.

→ An unpaired reactive electron can be seen on the end of the growing chain (initiation).

Step 2:

→ The free radical will then attack another ethylene molecule, increasing the length of the growing polymer chain (propagation).

Step 3:

→ Chain length increases in this fashion until two growing chains combine (termination).


♦ Mean mark 50%.

CHEMISTRY, M6 2015 HSC 24

  1. Explain why the salt, sodium acetate, forms a basic solution when dissolved in water. Include an equation in your answer.   (2 marks)

  2. A solution is prepared by using equal volumes and concentrations of acetic acid and sodium acetate.
  3. Explain how the pH of this solution would be affected by the addition of a small amount of sodium hydroxide solution. Include an equation in your answer.   (3 marks)

Show Answers Only

a.   

→ Sodium acetate is a basic salt.

→ Acetate is a strong base that accepts a proton, producing hydroxide.

→ The presence of ions produced by the hydrolysis of increases the pH, producing a basic solution.
 

b. 

→ The ions introduced into the solution will react with the ions, reducing their concentration in the equilibrium mixture.

→ By Le Chatelier’s principle, this will subsequently move the reaction to the left to increase the ions, thus minimising any change in pH.

Show Worked Solution

a.   

→ Sodium acetate is a basic salt.

→ Acetate is a strong base that accepts a proton, producing hydroxide.

→ The presence of ions produced by the hydrolysis of increases the pH, producing a basic solution.
 


♦♦ Mean mark (a) 37%.

b. 

→ The ions introduced into the solution will react with the ions, reducing their concentration in the equilibrium mixture.

→ By Le Chatelier’s principle, this will subsequently move the reaction to the left to increase the ions, thus minimising any change in pH.


♦♦♦ Mean mark (b) 25%.

CHEMISTRY, M7 2015 HSC 20 MC

The table shows the heat of combustion of four straight chain alkanols.
 

What is the mass of water that could be heated from 20°C to 45°C by the complete combustion of 1.0 g of heptan-1-ol?

  1. 0.032 kg
  2. 0.044 kg
  3. 0.36 kg
  4. 0.38 kg
Show Answers Only

Show Worked Solution

 
   
   
   

 


♦ Mean mark 51%.

CHEMISTRY, M8 2015 HSC 18-19 MC

Question 18

How could the reliability of the analysis of the pond water be improved?

  1. Analyse more samples from the same pond
  2. Use 50 mL of distilled water as a control sample
  3. Analyse samples from different ponds on the site
  4. Remove other contaminants from the sample before the analysis

 
Question 19

What was the concentration of lead ions in the sample?

Show Answers Only

Show Worked Solution

Question 18


 

Question 19


♦ Mean mark (Q19) 52%.

BIOLOGY, M8 2015 HSC 28

A group of students hypothesised that the height of plants decreases with increased elevation.

The students planted ten plant cuttings from the same plant at each of five locations. The locations were at varying elevations in the same mountain range. All the cuttings were provided with the same volume of water on planting, and no fertiliser was applied. The students returned after the same growth period and measured the height of the plants.

The cross-section shown indicates the average height of the plants in metres after the growth period at each location in the mountain range.
 

  1. Evaluate the validity of the experiment.   (3 marks)

  2. Complete both columns of the table to best present the data for the analysis of any trend.   (2 marks)
      


Show Answers Only

a.    Factors affecting experiment validity:

→ There are many variables in this experiment that are not controlled, hence the experiment is invalid.

→ There is no real evidence that the differences in plant heights observed in the different locations are due to elevation alone.

→ Variables with adequate controls in the experiment include planting methods and genotype.

→ Factors other than these can affect plant growth and should have been considered.

→ Variables not controlled well include exposure to the prevailing wind, available sunlight at the various locations and size and health of cuttings at time of planting.
  

b.   Mountain Range cross-section data

²

Show Worked Solution

a.    Factors affecting experiment validity:

→ There are many variables in this experiment that are not controlled, hence the experiment is invalid.

→ There is no real evidence that the differences in plant heights observed in the different locations are due to elevation alone.

→ Variables with adequate controls in the experiment include planting methods and genotype.

→ Factors other than these can affect plant growth and should have been considered.

→ Variables not controlled well include exposure to the prevailing wind, available sunlight at the various locations and size and health of cuttings at time of planting.
  


♦ Mean mark (a) 47%.

b.   Mountain Range cross-section data

²


Mean mark (b) 55%.

BIOLOGY, M5 2015 HSC 23

The diagram shows a model involving DNA.
 

  1. What process is being modelled?   (1 mark)

  2. Identify TWO structural features of the DNA molecule which are NOT shown in this model.   (2 marks)

Show Answers Only

a.   DNA replication

b.   Successful answers should include two of the following:

→ the double helix.

→ the sugar-phosphate backbone.

→ hydrogen bonds (connecting base pairs)

Show Worked Solution

a.   DNA replication

b.   Successful answers should include two of the following:

→ the double helix.

→ the sugar-phosphate backbone.

→ hydrogen bonds (connecting base pairs)


Mean mark (a) 53%.

BIOLOGY, M7 2015 HSC 19-20 MC

Refer to the following information to answer Questions 19 and 20 .

The intestinal tract of a human foetus is sterile.
After birth, microflora from the mother are transferred to the baby's mouth through close contact. After a year, the microflora of the baby is similar to the mother's, with the baby's immune system ignoring these microbes.

Also during the first year of life, breast milk from the mother provides antibodies to the baby for any disease the mother has already experienced. When breastfeeding ceases, these antibody levels in the baby start to fall.

After the first year, any new species of invading bacteria is treated as a pathogen by the baby's immune system.

 

Question 19

A medical consequence for six-month-old babies that have only been bottle-fed with formula milk and not breastfed is that

  1. they will not develop microflora.
  2. their immune system will be damaged.
  3. their consumption of milk cannot be quantified.
  4. they will be at increased risk of infectious disease.

 
Question 20

Strict hygiene practices are followed in the care of newborns, whereas hygiene practices in the care of older babies are less emphasised.

Which of the following is the best reason for this difference?

  1. Vaccinations render personal hygiene unnecessary for older babies.
  2. Procaryotic cells are not identified as antigens in early development.
  3. Antibiotic treatments kill bacterial populations in the digestive system.
  4. Early exposure to pathogens helps to build a strong immune system.

 

Show Answers Only

Q19. 

Q20. 

Show Worked Solution

Q 19.

→ Bottle fed babies will not have the added immunity that is provided by the mother’s breast milk.

→ They will therefore have immune systems that are more susceptible to infectious disease.


  

Q20.

→ The immune systems of newborns will not work as efficiently as that of older babies as their immune systems are underdeveloped.


♦ Mean mark Q20 45%.

BIOLOGY, M7 2015 HSC 12-13 MC

Refer to the following information to answer Questions 12 and 13.

The larvae of fruit flies damage fruit in Western Australia. To control the problem, growers are advised to spray fruit trees with pesticides. Any already damaged fruit is boiled and disposed of as chicken food or landfill. Another control measure is the release of genetically engineered infertile flies of this species.

 
Question 12

Which reason best explains why the corresponding control measure reduces this problem?

Question 13

The following measures could be used to prevent the spread of this fruit fly across Australia.

    1. Australia-wide release of infertile fruit flies
    2. Aerial spraying of orchards throughout the country
    3. Spot spraying of newly affected orchards in Western Australia
    4. Stopping the transport of fruit from Western Australia to other states

To prevent the spread of this fruit fly across Australia, which combination of measures would be most practical to use?

  1. 1 and 2
  2. 1 and 4
  3. 2 and 3
  4. 3 and 4
Show Answers Only

Q12. 

Q13. 

Show Worked Solution

Q12.

→ Future generations of flies will eventually be reduced in numbers and may even be wiped out altogether.


 

Q13.

→ It may be possible to contain the outbreak to local areas by restricting the movement of affected fruit to other states and countries, thus making it easier to manage.


♦ Mean mark (Q13) 49%.

BIOLOGY, M8 2015 HSC 7-8 MC

Refer to the following information to answer Questions 7 and 8 .

The diagram shows a homeostatic mechanism in a mammal.
 

Question 7

What does represent in the diagram?

  1. The heart
  2. The brain
  3. A thermoreceptor in the skin
  4. A pressure receptor in a blood vessel

 
Question 8

Which of the following describes what happens to the muscles and the arteriole walls in the skin when the core body temperature is below normal?
 

Show Answers Only

Q7. 
Q8. 

Show Worked Solution

Q7.    = the brain

→ Messages are sent by the brain to the effectors.

→ This promotes a stimulus response.


 


♦♦ Mean mark (Q7) 38%.

Q8.   When body temperature is below normal:

→ Hairs stand on end and arteriole walls of skin contract to prevent the loss of heat.

CHEMISTRY, M7 2017 HSC 28b

The molar heat of combustion for ethanol is 1360 kJ mol ¯1.

Calculate the energy generated per kg of released by the combustion of ethanol.   (3 marks)

Show Answers Only

Show Worked Solution

 
   

♦ Mean mark 48%.

CHEMISTRY, M7 2017 HSC 28a

Outline TWO advantages and TWO disadvantages of using ethanol as an alternative fuel for motor vehicles.   (4 marks)

Show Answers Only

Advantages:

→ Ethanol can be produced from biomass. These renewable sources include crops such as sugarcane as opposed to other fuels such as petrol which come from non-renewable fossil fuels, which are finite resources.

→ Ethanol undergoes complete combustion more easily than octane, producing less soot which can adversely affect the efficiency and running of motors, and less which is poisonous.
 

Disadvantages:

→ Ethanol releases less energy, on a mole or per kilogram basis, than octane. This results in a greater mass of fuel being required to supply an equivalent amount of energy.

→ Producing ethanol from renewable crops requires a huge amount of arable land. This reduces the availability of land for other crops.

Show Worked Solution

Advantages:

→ Ethanol can be produced from biomass. These renewable sources include crops such as sugarcane as opposed to other fuels such as petrol which come from non-renewable fossil fuels, which are finite resources.

→ Ethanol undergoes complete combustion more easily than octane, producing less soot which can adversely affect the efficiency and running of motors, and less which is poisonous.
 

Disadvantages:

→ Ethanol releases less energy, on a mole or per kilogram basis, than octane. This results in a greater mass of fuel being required to supply an equivalent amount of energy.

→ Producing ethanol from renewable crops requires a huge amount of arable land. This reduces the availability of land for other crops.


♦ Mean mark 52%.

CHEMISTRY, M7 2017 HSC 27

The boiling points and molar masses of three compounds are shown in the table.
 

Acetic acid, butan-1-ol and butyl acetate have very different molar masses but similar boiling points. Explain why in terms of the structure and bonding of the three compounds.   (5 marks)

Show Answers Only

→ Although the three listed compounds different molar masses, they have similar boiling points due to their different structures and resulting intermolecular forces.

→ Butyl acetate has the largest molar mass and therefore greatest dispersion forces but it is only slightly polar and has no hydrogen bonding.

→ Butan-1-ol has lower molar mass than butyl acetate and therefore smaller dispersion forces but it is polar and contains a hydrogen bound to an oxygen. Therefore, it exhibits hydrogen bonding resulting in strong intermolecular forces and a boiling point in the middle of the three compounds.

→ Acetic acid has the lowest molar mass and hence the weakest dispersion forces. It is however highly polar due to the presence of the carboxyl group and contains a hydrogen bound to an oxygen allowing the formation of hydrogen bonds between molecules.

→ The presence of a second oxygen in acetic acid increases the hydrogen bonding compared with butan-1-ol. 

→ These factors lead to acetic acid possessing the highest boiling point despite its molar mass being the lowest.

→ In summary, the totality of the intermolecular forces of all three molecules is similar and therefore similar boiling points.

Show Worked Solution

→ Although the three listed compounds different molar masses, they have similar boiling points due to their different structures and resulting intermolecular forces.

→ Butyl acetate has the largest molar mass and therefore greatest dispersion forces but it is only slightly polar and has no hydrogen bonding.

→ Butan-1-ol has lower molar mass than butyl acetate and therefore smaller dispersion forces but it is polar and contains a hydrogen bound to an oxygen. Therefore, it exhibits hydrogen bonding resulting in strong intermolecular forces and a boiling point in the middle of the three compounds.

→ Acetic acid has the lowest molar mass and hence the weakest dispersion forces. It is however highly polar due to the presence of the carboxyl group and contains a hydrogen bound to an oxygen allowing the formation of hydrogen bonds between molecules.

→ The presence of a second oxygen in acetic acid increases the hydrogen bonding compared with butan-1-ol. 

→ These factors lead to acetic acid possessing the highest boiling point despite its molar mass being the lowest.

→ In summary, the totality of the intermolecular forces of all three molecules is similar and therefore similar boiling points.


♦♦ Mean mark 41%.

CHEMISTRY, M6 2017 HSC 24

A solution of sodium hydroxide was titrated against a standardised solution of acetic acid which had a concentration of 0.5020 mol L¯1.

  1. The end point was reached when 19.30 mL of sodium hydroxide solution had been added to 25.00 mL of the acetic acid solution.
  2. Calculate the concentration of the sodium hydroxide solution.   (3 marks)

  3. Explain why the pH of the resulting salt solution was not 7. Include a relevant chemical equation in your answer.   (2 marks)

Show Answers Only

a.   
 

b.   

→ The acetate ion is a weak base.

→ As a result, it has accepted a proton from the water resulting in production of hydroxide ions.

→ Therefore the solution has a pH > 7.

Show Worked Solution

a.   

 

b.   

→ The acetate ion is a weak base.

→ As a result, it has accepted a proton from the water resulting in production of hydroxide ions.

→ Therefore the solution has a pH > 7.


♦♦ Mean mark (b) 34%.

CHEMISTRY, M7 2019 HSC 34

The following reaction scheme can be used to synthesise ethyl ethanoate.
 


 

Outline the reagents and conditions required for each step and how the product of each step could be identified.   (7 marks) 

Show Answers Only

Step 1:

→ To synthesise chloroethane (A) into ethanol (B), is added and heated. is then added and heated.

→ The mixture is then treated with concentrated sulfuric acid and refluxed.

→ Ethanol (B) can be identified using infrared spectroscopy by looking for a broad absorption between 3230 cm ¯1 and 3550 cm ¯1, which indicates the presence of an bond. This absorption would not be present in chloroethane (A).

→ Alternative ways to identify ethanol include: mass spectrum analysis (single ion peak at m/z = 46), reactivity tests, and spectrum analysis (3 signals vs 2 for chloroethane).
 

Step 2:

→ Ethanol (B) can be converted into ethanoic acid (C) by combining it with a strong oxidant like sodium carbonate, which produces carbon dioxide bubbles, confirming the presence of a carboxylic acid.

→ Ethanol will not react as above and the compounds can be distinguished.

→ Alternative ways to identify ethanoic acid include: IR or spectrum analysis, litmus indicators, mass spectrum analysis (ion peak at m/z = 60 vs m/z = 46)
 

Step 3

→ Ethyl ethanoate (D) can be synthesised by heating a mixture of ethanol, ethanoic acid and concentrated sulfuric acid under reflux.

→ A spectrum can be used to identify ethyl ethanoate as it will have 3 signals versus ethanol and ethanoic acid that will only have 2 each.

→ Alternative ways to identify ethyl ethanoate include: a distinct smell, no peaks in the IR spectrum or mass spectrum analysis (ion peak at m/z = 102).

Show Worked Solution

Step 1:

→ To synthesise chloroethane (A) into ethanol (B), is added and heated. is then added and heated.

→ The mixture is then treated with concentrated sulfuric acid and refluxed.

→ Ethanol (B) can be identified using infrared spectroscopy by looking for a broad absorption between 3230 cm ¯1 and 3550 cm ¯1, which indicates the presence of an bond. This absorption would not be present in chloroethane (A).

→ Alternative ways to identify ethanol include: mass spectrum analysis (single ion peak at m/z = 46), reactivity tests, and spectrum analysis (3 signals vs 2 for chloroethane).
 

Step 2:

→ Ethanol (B) can be converted into ethanoic acid (C) by combining it with a strong oxidant like sodium carbonate, which produces carbon dioxide bubbles, confirming the presence of a carboxylic acid.

→ Ethanol will not react as above and the compounds can be distinguished.

→ Alternative ways to identify ethanoic acid include: IR or spectrum analysis, litmus indicators, mass spectrum analysis (ion peak at m/z = 60 vs m/z = 46)
 

Step 3

→ Ethyl ethanoate (D) can be synthesised by heating a mixture of ethanol, ethanoic acid and concentrated sulfuric acid under reflux.

→ A spectrum can be used to identify ethyl ethanoate as it will have 3 signals versus ethanol and ethanoic acid that will only have 2 each.

→ Alternative ways to identify ethyl ethanoate include: a distinct smell, no peaks in the IR spectrum or mass spectrum analysis (ion peak at m/z = 102).


♦♦ Mean mark 38%.

CHEMISTRY, M8 2019 HSC 29

Stormwater from a mine site has been found to be contaminated with copper and lead ions. The required discharge limit is 1.0 mg L¯1 for each metal ion. Treatment of the stormwater with solid to remove the metal ions is recommended.

  1. Explain the recommended treatment with reference to solubility. Include a relevant chemical equation.   (2 marks)

  2. Explain why atomic absorption spectroscopy can be used to determine the concentrations of and ions in a solution containing both species.   (2 marks)

  3. The data below were obtained after treatment of the stormwater.
     
         
     
    To what extent is the treatment effective in meeting the required discharge limit of 1.0 mg L¯1 for each metal ion? Support your conclusion with calibration curves and calculations.  (7 marks)
     
         

Show Answers Only

a.   Recommended Treatment:

→ Calcium hydroxide is a slightly soluble compound, while copper hydroxide and lead hydroxide are very insoluble in water.

→ When these compounds are added to water, the metal ions tend to precipitate out of solution.

→ For example, the addition of solid calcium hydroxide to water produces calcium ions and hydroxide ions , which can then react with lead ions ( and copper ions to form precipitates of lead hydroxide and copper hydroxide, respectively.

→ These reactions are represented by the equations:


 

b.   Atomic absorption spectroscopy (AAS):

→ Can be used for determining the concentration of metal ions in a sample by measuring the absorbance of light at specific wavelengths that are characteristic of each metal.

→ AAS uses light wavelengths that correspond to atomic absorption by the element of interest, and since each element has unique wavelengths that are absorbed, the concentration of that element can be selectively measured in the presence of other species.

→ As a result, AAS can be used to independently measure the concentrations of different metal ions, such as lead ions and copper ions in a sample containing both types.
 

c.   Concentrations of ions:

→ Concentrations of copper and lead have been significantly reduced.

→ Convert concentrations to compare with standard:

 

   

Conclusion:

→ The concentration of copper ions has been reduced to a level that is lower than the discharge limit (0.16 < 1.0) but the lead ion concentration has not (1.76 > 1.0).

→ The treatment has only been partially successful.

Show Worked Solution

a.   Recommended Treatment:

→ Calcium hydroxide is a slightly soluble compound, while copper hydroxide and lead hydroxide are very insoluble in water.

→ When these compounds are added to water, the metal ions tend to precipitate out of solution.

→ For example, the addition of solid calcium hydroxide to water produces calcium ions and hydroxide ions , which can then react with lead ions ( and copper ions to form precipitates of lead hydroxide and copper hydroxide, respectively.

→ These reactions are represented by the equations:


 


♦ Mean mark (a) 46%.

b.   Atomic absorption spectroscopy (AAS):

→ Can be used for determining the concentration of metal ions in a sample by measuring the absorbance of light at specific wavelengths that are characteristic of each metal.

→ AAS uses light wavelengths that correspond to atomic absorption by the element of interest, and since each element has unique wavelengths that are absorbed, the concentration of that element can be selectively measured in the presence of other species.

→ As a result, AAS can be used to independently measure the concentrations of different metal ions, such as lead ions and copper ions in a sample containing both types.
 

c.   Concentrations of ions:


♦♦ Mean mark (b) 32%.

→ Concentrations of copper and lead have been significantly reduced.

→ Convert concentrations to compare with standard:

 

   

Conclusion:

→ The concentration of copper ions has been reduced to a level that is lower than the discharge limit (0.16 < 1.0) but the lead ion concentration has not (1.76 > 1.0).

→ The treatment has only been partially successful.


♦ Mean mark (c) 53%.

CHEMISTRY, M6 2019 HSC 28

Assess the usefulness of the Brønsted-Lowry model in classifying acids and bases. Support your answer with at least TWO chemical equations.   (5 marks)

Show Answers Only

→ The Bronsted-Lowry model is a way of classifying acids and bases based on their ability to donate or accept protons.

→ This model is more comprehensive than the Arrhenius model, as it can explain the acid-base behaviour of more species, including those that do not contain ions, and non-aqueous acid-base reactions.

→ Consider the reaction where a proton is transferred from hydrogen chloride to ammonia (according to the Bronsted-Lowry model). Ammonia is not an Arrhenius base as it doesn’t dissociate to produce ions and the reaction cannot be described using the Arrhenius model.

→ However, the Bronsted-Lowry model does have some limitation, such as its inability to explain the acidity of certain acidic oxides and their reactions with basic oxides.

→ e.g. is an acid-base reaction but since there is no proton transfer, it cannot be described using the Bonsted-Lowry model.

Show Worked Solution

→ The Bronsted-Lowry model is a way of classifying acids and bases based on their ability to donate or accept protons.

→ This model is more comprehensive than the Arrhenius model, as it can explain the acid-base behaviour of more species, including those that do not contain ions, and non-aqueous acid-base reactions.

→ Consider the reaction where a proton is transferred from hydrogen chloride to ammonia (according to the Bronsted-Lowry model). Ammonia is not an Arrhenius base as it doesn’t dissociate to produce ions and the reaction cannot be described using the Arrhenius model.

→ However, the Bronsted-Lowry model does have some limitation, such as its inability to explain the acidity of certain acidic oxides and their reactions with basic oxides.

→ e.g. is an acid-base reaction but since there is no proton transfer, it cannot be described using the Bonsted-Lowry model.


♦♦♦ Mean mark 35%.
MARKER’S COMMENT: Two relevant equations required to achieve full marks.

CHEMISTRY, M6 2019 HSC 27

The relationship between the acid dissociation constant, , and the corresponding conjugate base dissociation constant, , is given by:

Assume that the temperature for part (a) and part (b) is 25°C.

  1. The of hypochlorous acid is  .
  2. Show that the of the hypochlorite ion, , is  .   (1 mark)

  3. The conjugate base dissociation constant, , is the equilibrium constant for the following equation:
  4.      
  5. Calculate the pH of a 0.20 mol L¯1 solution of sodium hypochlorite .   (4 mark)

Show Answers Only
Show Worked Solution

a.   

 
   

 

b.   
 

 

Assume    because is negligible:

 
 
   

 


♦ Mean mark (b) 45%.

CHEMISTRY, M7 2015 HSC 10 MC

Which of the equations correctly describes incomplete combustion?

  1. +   +
  2. + +
  3. + +
  4. + + +
Show Answers Only

Show Worked Solution

→ Incomplete combustion produces carbon and/or carbon monoxide (eliminate B and C).

→ Option D is not balanced (oxygen atoms do not equate)


♦ Mean mark 44%.

CHEMISTRY, M5 2017 HSC 18 MC

Three gases , and were mixed in a closed container and allowed to reach equilibrium. A change was imposed at time and the equilibrium was re-established. The concentration of each gas is plotted against time.
 

Which reaction is represented by the graph?

Show Answers Only

Show Worked Solution

Concentration changes at T:

→ Concentrations of X ↓ 25%, Y ↓ 25%, Z ↓ 25%
 

New equilibrium after T:

→ Concentrations of X ↓ 0.005 mol L ¯1, Y ↑ 0.005 mol L ¯1, Z ↑ 0.005 mol L ¯1

→ Shift is equimolar (1:1:1) and Y and Z are on the same side of the equation (both increase)


♦ Mean mark 42%.

CHEMISTRY, M5 2017 HSC 16 MC

The following equilibrium is established in a closed system.

How can the gas pressure in the system be decreased?

  1. Add more
  2. Add hydroxide ions to the solution
  3. Decrease the volume of the container
  4. Increase the temperature of the system
Show Answers Only

Show Worked Solution

→ Gas pressure can be decreased if the equilibrium shifts to the right.

→ If hydroxide ions are added to the solution by adding NaOH, this will neutralise the carbonic acid and cause an equilibrium shift to the right.


♦♦ Mean mark 32%.

CHEMISTRY, M6 2017 HSC 14 MC

One litre of an aqueous solution is formed from mixing equal volumes of 0.2 mol L hydrochloric acid and 0.2 mol L sodium chloride .

How effective as a buffer is the aqueous solution formed?

  1. Ineffective, because is a strong acid
  2. Effective, because is the conjugate base of
  3. Ineffective, because forms a neutral salt solution
  4. Effective, because the pH would change when a solution of is added
Show Answers Only

Show Worked Solution

→ is a strong acid

→ Aqueous solution formed will be ineffective as a buffer (no equilibrium in solution will be formed).


♦ Mean mark 51%.

CHEMISTRY, M7 2017 HSC 12 MC

What is the product when propene undergoes addition polymerisation?
 


 
 

Show Answers Only

Show Worked Solution

Polymerisation of propene:

→ No double bonds (eliminate A and C)

→ Methyl functional group appear on every second carbon atom


♦ Mean mark 43%.

CHEMISTRY, M8 2017 HSC 8 MC

There are two unlabelled solutions. One is barium nitrate and the other lead nitrate.

Which of the following could be added to the two unlabelled solutions to distinguish between them?

  1. Sodium sulfate
  2. Sodium nitrate
  3. Sodium chloride
  4. Sodium carbonate
Show Answers Only

Show Worked Solution

→ If sodium chloride is added to barium nitrate, barium chloride (soluble) is formed but no precipitate results.

→ If sodium chloride is added to lead nitrate, lead chloride (insoluble) is formed and a precipitate results.

→ All other options will produce either a precipitate or no precipitate in both solutions and not distinguish between them.


♦ Mean mark 49%.

CHEMISTRY, M7 2017 HSC 7 MC

Three test tubes were set up as shown.


 

Bromine water was added to and in the absence of UV light.

Which of the following best represents the changes in test tubes and ?
 

Show Answers Only

Show Worked Solution

By Elimination:

→ Liquids are immiscible (eliminate A and D).

→ Bromine water does not decolourise with with alkanes, but does with alkenes (eliminate B).


♦♦ Mean mark 33%.

BIOLOGY, M7 2018 HSC 3 MC

Which defence adaptation in the table is correctly matched with one of its features?

 

Show Answers Only

Show Worked Solution

→ In some cases, such as in response to leprosy or tuberculosis, the body seals off a pathogen by forming a granuloma, a collection of cells containing macrophages, lymphocytes then fibroblasts (a structural cell involved in connective tissue), which forms a tough outer wall.


♦ Mean mark 45%.

PHYSICS, M7 2015 HSC 29

In the Large Hadron Collider (LHC), protons travel in a circular path at a speed greater than 0.9999 .

  1. What are the advantages of using superconductors to produce the magnetic fields used to guide protons around the LHC?   (2 marks)
  1. Discuss the application of special relativity to the protons in the LHC.   (3 marks)
Show Answers Only

a.    Advantages:

→ Strong magnetic fields are necessary to deflect protons travelling around the LHC due to their high velocities and the effect of mass dilation.

→ Superconductors can achieve very high current flow. This produces the very strong magnetic fields that are required.
 

b.    Applications of special relativity:

→ Since the protons are travelling so close to the speed of light, special relativity states that mass dilation will be significant.

→ This effect means that an increasing amount of energy is needed to accelerate the proton.
 

Other answers could include:

→ Effects of length contraction/time dilation.

→ Identification of the non-inertial frame of reference.

Show Worked Solution

a.   Advantages:

→ Strong magnetic fields are necessary to deflect protons travelling around the LHC due to their high velocities and the effect of mass dilation.

→ Superconductors can achieve very high current flow. This produces the very strong magnetic fields that are required.


♦ Mean mark (a) 45%.

b.   Applications of special relativity:

→ Since the protons are travelling so close to the speed of light, special relativity states that mass dilation will be significant.

→ This effect means that an increasing amount of energy is needed to accelerate the proton.

Other answers could include:

→ Effects of length contraction/time dilation.

→ Identification of the non-inertial frame of reference.


♦ Mean mark (b) 53%.

PHYSICS, M5 2015 HSC 26

Consider the following two models used to calculate the work done when a 300 kg satellite is taken from Earth's surface to an altitude of 200 km.

You may assume that the calculations are correct.
 

  1. What assumptions are made about Earth's gravitational field in models and that lead to the different results shown?   (2 marks)
  1. Why do models and produce results that, although different, are close in value?   (1 mark)
  1. Calculate the orbital velocity of the satellite in a circular orbit at the altitude of 200 km.   (3 marks)
Show Answers Only

a.  Model :

→ Assumes Earth’s gravitational field strength remains constant moving upwards from the surface.

Model :

→ Assumes Earth’s gravitational field strength changes with altitude.

b.   Similarity of results due to:

→ The variation in gravitational field strength from Earth’s surface to an altitude of 200 km is minimal, so both models and produce similar results.

c.  

Show Worked Solution

a.   Model :

→ Assumes Earth’s gravitational field strength remains constant moving upwards from the surface.

Model :

→ Assumes Earth’s gravitational field strength changes with altitude.


♦ Mean mark (a) 54%.

b.   Similarity of results due to:

→ The variation in gravitational field strength from Earth’s surface to an altitude of 200 km is minimal, so both models and produce similar results.


♦♦ Mean mark (b) 38%.

c.    Centripetal force = force due to gravity:

 
 
 
   
   

PHYSICS, M6 2015 HSC 25b

The diagram shows a label on a transformer used in an appliance.

     
Explain why the information provided on the label is not correct. Support your answer with calculations.   (3 marks)

Show Answers Only

Calculating the input power:

 
   
   

 
Calculating the output power:

 
   
   

 
→ The output power is greater than the input power.

→ The label is incorrect as this is inconsistent with the law of conservation of energy.

Show Worked Solution

Calculating the input power:

 
   
   

 
Calculating the output power:

 
   
   

 
→ The output power is greater than the input power.

→ The label is incorrect as this is inconsistent with the law of conservation of energy.


♦ Mean mark 45%.

PHYSICS, M5 2015 HSC 20 MC

A projectile was launched from the ground. It had a range of 70 metres and was in the air for 3.5 seconds.

At what angle to the horizontal was it launched?

  1. 30°
  2. 40°
  3. 50°
  4. 60°
Show Answers Only

Show Worked Solution

Find

 
 
   

 
Find (projectile has a vertical velocity of zero at its maximum height):

 
 
 

  
Find launch angle:

 
 
   

 


♦ Mean mark 55%.

PHYSICS, M6 2015 HSC 18 MC

The diagram shows an ideal transformer.
 

When the switch is closed, the pointer on the galvanometer deflects.

How could the size of the deflection be increased?

  1. Decrease the number of primary coils
  2. Decrease the number of secondary coils
  3. Replace the iron core with a copper core
  4. Place a resistor in series with the galvanometer
Show Answers Only

Show Worked Solution

In order to increase the deflection of the galvanometer, the current through the secondary coil must increase. 

→ Decreasing the number of secondary coils will decrease
 

→ Decreasing the number of secondary coils will increase


♦ Mean mark 45%.

PHYSICS, M6 2015 HSC 15 MC

A circular loop of wire is stationary in a magnetic field. The sides are then pushed together to change the shape, as shown in the diagram.
 

As the loop is compressed, a current is induced.

Which row of the table shows the direction of the current and explains why it is induced?

Show Answers Only

Show Worked Solution

→ Compressing the loop decreases the area of the loop.

→ There is a change (reduction) in magnetic flux passing through the loop.

→ A current is generated in the loop in order to mitigate this change by producing a magnetic field directed into the page.

→ Using the right hand grip rule, the current direction is clockwise.


♦♦ Mean mark 38%.

PHYSICS, M6 2015 HSC 12 MC

A simple AC generator was connected to a cathode ray oscilloscope and the coil was rotated at a constant rate. The output is shown on this graph.
 

Which of the following graphs best represents the output if the rate of rotation is decreased to half of the original value?
 

 

Show Answers Only

Show Worked Solution

Halving the rate of rotation of the bar magnet:

→ Doubles the period of the output graph (eliminate A and B).

→ Halves the rate of change of flux through the coil of the generator.

→ Halves of the maximum output voltage.


♦ Mean mark 52%.

PHYSICS, M6 2015 HSC 9 MC

, and are straight, current-carrying conductors. They all carry currents of the same magnitude . Conductors and are fixed in place. The magnitude of the force between conductors and is newtons.
 

What is the net force on conductor when it is in the position shown?

  1. newtons to the left
  2. newtons to the right
  3. newtons to the left
  4. newtons to the right
Show Answers Only

Show Worked Solution

→ As the currents in and are in opposite directions, they repel. The force on due to is to the right.

→ As the currents in and are in the same direction, they attract. The force on due to is to the left.

→ The force on due to is half the magnitude of the force on due to .

→ The force on due to is newtons to the left.

→ experiences a force of newtons to the right due to and a force of newtons to the left due to .

→ The net force on is newtons to the right.


Mean mark 54%.

PHYSICS, M7 2015 HSC 6 MC

Which of the following is a true statement about scientific theories, such as Einstein's theory of special relativity?

  1. They are valid but unreliable ideas.
  2. They are useful in making predictions.
  3. They are concepts that lack an experimental basis.
  4. They are ideas that can't be accepted until they have been tested.
Show Answers Only

Show Worked Solution

By Elimination:

→ Scientific theories are supported by experiments that when performed yield similar results (reliability). Examples include observations of muons and atomic clock experiments for Einstein’s theory of special relativity (eliminate A and C).

→ Scientific theories have large amounts of supporting evidence and are accepted (eliminate D).


♦♦ Mean mark 37%.

PHYSICS, M6 2016 HSC 30

The following makeshift device was made to provide lighting for a stranded astronaut on Mars.

The mass of Mars is .
 

The 2 kg mass falls, turning the DC generator, which supplies energy to the light bulb. The mass falls from a point that is 3 376 204 m from the centre of Mars.

  1. Calculate the maximum possible energy released by the light bulb as the mass falls through a distance of one metre.   (3 marks)
  1. Explain the difference in the behaviour of the falling mass when the switch is open.   (3 marks)
Show Answers Only

a.   7.48 J

b.   When switch is opened:

→ When the switch is opened, there is no induced current opposing the downwards motion of the mass (Lenz’s Law).

→ Hence, the mass will fall more quickly.

Show Worked Solution

a.   


 

→ 7.48 J is lost by the falling mass.

→ The light bulb released 7.48 J of energy.


♦ Mean mark (a) 52%.

b.   When switch is opened:

→ When the switch is opened, there is no induced current opposing the downwards motion of the mass (Lenz’s Law).

→ Hence, the mass will fall more quickly.


♦♦♦ Mean mark (b) 27%.

PHYSICS, M5 2016 HSC 28

The following diagram shows the acceleration of a rocket during the first stage of its launch.
 

Explain the acceleration of the rocket with reference to the law of conservation of momentum.   (5 marks)

Show Answers Only

→ The combustion of fuel and expulsion of gases causes propulsion of the rocket. The rocket exerts a force on these gases in order to expel them backwards.

→ An equal and opposite force is exerted back on the rocket by the gases (Newton’s Third Law).

→ As the rocket and gases form a closed system, the law of conservation of momentum applies, meaning the momentum of the rocket is equal in magnitude and opposite in direction to the momentum of the fuel.

→ The momentum of the rocket is given by  . Since the force (and hence momentum) of the fuel is constant, the velocity of the rocket must increase as its mass decreases due to the expulsion of its fuel.

→ The rocket will accelerate. Also, using  , as the force acting on the rocket due to the gases is constant, the acceleration of the rocket must increase as its mass decreases.

Show Worked Solution

→ The combustion of fuel and expulsion of gases causes propulsion of the rocket. The rocket exerts a force on these gases in order to expel them backwards.

→ An equal and opposite force is exerted back on the rocket by the gases (Newton’s Third Law).

→ As the rocket and gases form a closed system, the law of conservation of momentum applies, meaning the momentum of the rocket is equal in magnitude and opposite in direction to the momentum of the fuel.

→ The momentum of the rocket is given by  . Since the force (and hence momentum) of the fuel is constant, the velocity of the rocket must increase as its mass decreases due to the expulsion of its fuel.

→ The rocket will accelerate. Also, using  , as the force acting on the rocket due to the gases is constant, the acceleration of the rocket must increase as its mass decreases.


♦ Mean mark 52%.