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ENGINEERING, PPT 2020 HSC 6 MC

Which electronic component can be used to increase or decrease the speed of a train powered by an electric motor?

  1. Diode
  2. Capacitor
  3. Transistor
  4. Potentiometer
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Show Worked Solution

→ Potentiometers can act as variable resistors.

→ These can increase or decrease current and therefore adjust the speed of an electric motor.


♦♦♦ Mean mark 20%.

Vectors, EXT2 V1 2022 HSC 14a

  1. The two non-parallel vectors and satisfy    for some real numbers and .
  2. Show that  .  (2 marks)

  3. The two non-parallel vectors and satisfy    for some real numbers and .
  4. Using part (i), or otherwise, show that    and  . (1  mark)

The diagram below shows the tetrahedron with vertices and .

The point is defined by  , as shown in the diagram.

The point is the point of intersection of the straight lines and .
 
                       
 

  1. Using part (ii), or otherwise, determine the position of by showing that  .  (2 marks)

  2. The point is defined by  .
  3. Does lie on the line ? Justify your answer.  (2 marks)

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Show Worked Solution
i.   
 

 


 

ii. 

 
 

 


Mean mark (ii) 56%.
iii.  
   

 

 
   
   
   


♦♦♦ Mean mark (iii) 25%.

 
 
 
 

 


 

iv. 


♦♦♦ Mean mark (iv) 23%.
 
 
 
 

 

ENGINEERING, CS 2021 HSC 25b

A truss is fixed to a wall at and as shown. Ignore the mass of the truss.
 


 

  1. Determine the horizontal reaction at .   (2 marks)
  1. Determine, using method of joints, the internal reaction in member . Indicate the nature of the force in the member.   (3 marks)
  1. Determine, using method of sections, the internal reaction in member . Indicate the nature of the force.   (3 marks)
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i.   

ii.   

iii.   

Show Worked Solution

i.   Horizontal reaction at

 
 
 
   

 

ii.   Method of joints at A
 
     

→Σ  
 
 
   

♦♦ Mean mark (ii) 41%.

iii.  Method of Sections

 
 
 
   
   

♦♦♦ Mean mark (iii) 30%.

BIOLOGY, M7 2020 HSC 32c

Rabies is a disease that can affect all mammals and is caused by the rabies virus. It is transmitted by the bite of an infected animal. Without treatment it almost always results in death.
 
        

  1.  

The rabies virus is a single-stranded RNA virus. It contains and codes for only five proteins. The diagrams show the structure and reproduction of the virus.
 

 

Post exposure prophylaxis (PEP) is given to patients who have been bitten by a rabid animal.

PEP includes an injection of human rabies antibodies (HRIG) as well as injections of a rabies vaccine at 0, 3, 7 and 14 days after exposure to the virus.

The following graphs show a generalised response to rabies infection without and with PEP.
  

Explain how PEP prevents rabies developing after infection with the virus. Support your answer with reference to the information and data provided above.   (8 marks)

Show Answers Only

Once the rabies virus has entered the wound:

→ It will use the patient’s cells to replicate and the viral concentration will increase (as seen in the first five days).

→ Without PEP the virus will continue to replicate, migrate to the CNS (in the first graph this occurs by day 7), and eventually cause rabies and death.

→ Initially, the infected individual will not have the antibodies required to inactivate the virus.

→ A HRIG injection provides the required antibodies to inactivate the virus, through inhibiting replication or enhancing phagocytosis. 

→  The PEP graph shows that these antibodies will only last up to 21 days, but are essential in inactivating the initial virus, as seen by a reduction in viral concentration after 6-8 days.
 

The rabies vaccine works by:

→ containing an inactivated or weakened version of the rabies virus, which stimulates an immune response by the individual.

→ Initially, macrophages will display an MHC-antigen complex on its surface which helper T lymphocytes will bind to.

→ This then stimulates specific plasma B cells which can produce complementary antibodies, and memory B cells, which stay dormant and can rapidly differentiate into plasma B cells when exposed to the same virus.

→ The PEP graph shows the rapid production of antibodies on day 7, which coincides with rapid decrease in the virus concentration over the next few days. 

→ The antibodies then remain in the bloodstream and slowly decline over months, which allows quick diffusion if the virus is encountered within that timeframe.

Show Worked Solution

Once the rabies virus has entered the wound:

→ It will use the patient’s cells to replicate and the viral concentration will increase (as seen in the first five days).

→ Without PEP the virus will continue to replicate, migrate to the CNS (in the first graph this occurs by day 7), and eventually cause rabies and death.

→ Initially, the infected individual will not have the antibodies required to inactivate the virus.

→ A HRIG injection provides the required antibodies to inactivate the virus, through inhibiting replication or enhancing phagocytosis. 

→  The PEP graph shows that these antibodies will only last up to 21 days, but are essential in inactivating the initial virus, as seen by a reduction in viral concentration after 6-8 days.
 

The rabies vaccine works by:

→ containing an inactivated or weakened version of the rabies virus, which stimulates an immune response by the individual.

→ Initially, macrophages will display an MHC-antigen complex on its surface which helper T lymphocytes will bind to.

→ This then stimulates specific plasma B cells which can produce complementary antibodies, and memory B cells, which stay dormant and can rapidly differentiate into plasma B cells when exposed to the same virus.

→ The PEP graph shows the rapid production of antibodies on day 7, which coincides with rapid decrease in the virus concentration over the next few days. 

→ The antibodies then remain in the bloodstream and slowly decline over months, which allows quick diffusion if the virus is encountered within that timeframe.


♦♦ Mean mark 41%.

BIOLOGY, M8 2020 HSC 31

  1. The levels of glucose, insulin and glucagon were measured in the plasma of 24 healthy adults at intervals over a 5-hour period. After 1 hour at rest the patients ate a large carbohydrate meal. The results are shown.

  2. Use the data provided to explain how blood glucose is controlled in the body.   (6 marks)

  3. Outline how in humans, maintenance of temperature is different to the way that glucose is controlled.   (3 marks)

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a.   Glucose Levels

→ Plasma levels within the first 60 minutes represent the resting glucose levels of the individuals.

→ After the 1 hour mark, the rise in glucose is a result of the absorption of glucose into the bloodstream from the gut after eating the carb rich meal.
  

Insulin Levels

→ Insulin is a hormone secreted by beta cells in the pancreas. It causes excess glucose in the bloodstream to be taken by the liver and stored as glycogen.

→ This can be seen when the insulin levels directly correlate to the amount of glucose in the body; glucose absorbed in the meal stimulates release of insulin.

→ The reduction in glucose by the insulin then causes insulin to also fall.

  
Glucagon Levels

→ Glucagon is a hormone secreted by alpha cells in the pancreas. It can almost be seen as the opposite of insulin, and forms the negative feedback loop responsible for controlling blood glucose. 

→ It causes glycogen stores in the liver to decompose into glucose and be absorbed by the bloodstream when blood glucose levels drop too low.

→ When glucose levels rise between the 1-2 hour mark, glucagon levels drop significantly, as the build up of glucose from the meal has meant that glycogen stores are not needed.

→ Gradually, as glucose levels drop, glucagon levels will increase as the glucose from the meal is depleted.
 

b.  Differences in temperature vs glucose maintenance

→ Temperature changes are detected by the hypothalamus and sensory neurons, but changes in blood glucose levels (BGL) are detected by the pancreas.

→ Responses are carried out by the nervous system when temperature changes are detected, but carried out by hormones when changes in BGL are detected.

Show Worked Solution

a.   Glucose Levels

→ Plasma levels within the first 60 minutes represent the resting glucose levels of the individuals.

→ After the 1 hour mark, the rise in glucose is a result of the absorption of glucose into the bloodstream from the gut after eating the carb rich meal.
  

Insulin Levels

→ Insulin is a hormone secreted by beta cells in the pancreas. It causes excess glucose in the bloodstream to be taken by the liver and stored as glycogen.

→ This can be seen when the insulin levels directly correlate to the amount of glucose in the body; glucose absorbed in the meal stimulates release of insulin.

→ The reduction in glucose by the insulin then causes insulin to also fall.

  
Glucagon Levels

→ Glucagon is a hormone secreted by alpha cells in the pancreas. It can almost be seen as the opposite of insulin, and forms the negative feedback loop responsible for controlling blood glucose. 

→ It causes glycogen stores in the liver to decompose into glucose and be absorbed by the bloodstream when blood glucose levels drop too low.

→ When glucose levels rise between the 1-2 hour mark, glucagon levels drop significantly, as the build up of glucose from the meal has meant that glycogen stores are not needed.

→ Gradually, as glucose levels drop, glucagon levels will increase as the glucose from the meal is depleted.


♦ Mean mark (a) 48%.

b.  Differences in temperature vs glucose maintenance:

→ Temperature changes are detected by the hypothalamus and sensory neurons, but changes in blood glucose levels (BGL) are detected by the pancreas.

→ Responses are carried out by the nervous system when temperature changes are detected, but carried out by hormones when changes in BGL are detected.


♦♦ Mean mark (b) 36%.

BIOLOGY, M5 2020 HSC 28

  1. A student drew a diagram to model part of the process of meiosis.
     

  1. Explain the misunderstanding of meiosis shown in this model.   (3 marks)

  2. Explain the effect of meiosis on genetic variation.   (3 marks)

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a.
 
       

 
→ In the diagram above, the paired homologous chromosomes are incorrectly drawn.

→ Prior to crossing over, each chromosome duplicates forming the sister chromatids with one being maternal and one being paternal.

→ In the model above, they are shown as a mix of both, when in actuality the sister chromatids should be identical.
 

b.   The Effect of Meiosis on Genetic Variation

Independent Assortment

→ when chromosomes are lined up along the cells equator in independent order and orientation to all other chromosomes.

Random Segregation  

→ different combinations of maternal and paternal chromosomes end up in resulting gametes, increasing variability amongst them.

Crossing Over

→ the process of exchanging genetic material between chromatids of homologous chromosomes during Meiosis I, leading to unique combinations of alleles on each chromatid.

Show Worked Solution

a.
 
       

 
→ In the diagram above, the paired homologous chromosomes are incorrectly drawn.

→ Prior to crossing over, each chromosome duplicates forming the sister chromatids with one being maternal and one being paternal.

→ In the model above, they are shown as a mix of both, when in actuality the sister chromatids should be identical.


♦♦♦ Mean mark (a) 25%.

b.   The Effect of Meiosis on Genetic Variation

Independent Assortment

→ when chromosomes are lined up along the cells equator in independent order and orientation to all other chromosomes.

Random Segregation  

→ different combinations of maternal and paternal chromosomes end up in resulting gametes, increasing variability amongst them.

Crossing Over

→ the process of exchanging genetic material between chromatids of homologous chromosomes during Meiosis I, leading to unique combinations of alleles on each chromatid.


♦ Mean mark (b) 43%.

CHEMISTRY, M6 2020 HSC 33

Excess solid calcium hydroxide is added to a beaker containing 0.100 L of 2.00 mol L¯1 hydrochloric acid and the mixture is allowed to come to equilibrium.

  1. Show that the amount (in mol) of calcium hydroxide that reacts with the hydrochloric acid is 0.100 mol.   (2 marks)

  2. It is valid in this instance to make the simplifying assumption that the amount of calcium ions present at equilibrium is equal to the amount generated in the reaction in part (a).
  3. Calculate the pH of the resulting solution.  (4 marks)

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a.   

 

b.  

Show Worked Solution

a.   

 

b.   


 

 
 
 
   
 
   

 


♦♦♦ Mean mark (b) 20%.

PHYSICS, M6 2018 HSC 18 MC

An experiment is set up as shown.
 

When the switch is closed, the reading on the spring balance changes immediately, then returns to the initial reading.

Which row of the table correctly shows the direction of the current through the straight conductor and the direction in which the pointer on the spring balance initially moves?

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Show Worked Solution

→ Using the right hand grip rule, when the switch is closed the current through the lower solenoid produces a south pole at its top.

→ So, a south pole is induced at the bottom of the solenoid connected to the spring balance (Lenz’s Law).

→ Using the right hand grip rule a second time. The current through the straight conductor goes from to .

→ As the two solenoids repel each other, the force on the higher solenoid counteracts its weight and the pointer moves up.


♦♦♦ Mean mark 34%.

PHYSICS, M7 2018 HSC 16 MC

When a train is at rest in a tunnel, the train is slightly longer than the tunnel.
 

In a thought experiment, the train is travelling from left to right fast enough relative to the tunnel that its length contracts and it fits inside the tunnel.

An observer on the ground sets up two cameras, at and , to take photos at exactly the same time. The photos show that both ends of the train are inside the tunnel.
 

A passenger travelling on the train at its centre can see both ends of the tunnel and is later shown the photos.

From the point of view of the passenger, what is observed and what can be deduced about the photos?

  1. The tunnel's length contracts so the train does not fit, and photo 2 is taken before photo 1.
  2. The tunnel's length contracts so the train does not fit, and photos 1 and 2 are taken at the same time.
  3. The tunnel appears to expand due to length contraction of the train, allowing it to fit in the tunnel, and photo 1 is taken before photo 2 .
  4. The tunnel appears to expand due to length contraction of the train, allowing it to fit in the tunnel, and photos 1 and 2 are taken at the same time.
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Show Worked Solution

→ From the frame of reference of an observer on the train, the tunnel’s length will contract (all objects outside the train will appear length contracted).

→ Due to the relativity of simultaneity, the photo that the observer on the train sees as taken first is the photo taken at the end of the tunnel that the train was moving towards at that moment.


♦♦♦ Mean mark 36%.

ENGINEERING, PPT 2021 HSC 24d

A drive mechanism used to control the satellite dish of a mobile TV transmission van is shown.

Key dimensions of components are given.
 


 

The dimensions of each of the five holes in the gear are  on one side.

A partially sectioned drawing of the gear and shaft is provided below.

Complete the sectioned assembly of this drive mechanism to AS 1100 from the direction indicated by the arrow. Include break lines where appropriate.   (6 marks)
 

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Show Worked Solution

Critical aspects of this drawing:

→ Do not section wedges

→ Do not section counterbores or holes

→ Break line must be included

→ Section lines of adjacent areas should go in different directions

→ Section lines should be at 45° and 3mm apart

→ Answer may be drawn half sectioned as it is symmetrical


♦♦ Mean mark 41%.

CHEMISTRY, M5 2020 HSC 19 MC

Nitrogen dioxide reacts to form dinitrogen tetroxide in a sealed flask according to the following equation.

Which graph best represents the rates of both the forward and reverse reactions when an equilibrium system containing these gases is cooled at time ?
 

 

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Show Worked Solution

→ When the temperature decreases, both the forward and reverse reaction rates will decrease (eliminate A).

→ Exothermic reactions have a lower activation energy threshold (i.e. their reaction rates are less affected by cooling).

→ Therefore, the exothermic reaction rate will decrease to a smaller extent (eliminate B and C).


♦♦♦ Mean mark 22%.

CHEMISTRY, M6 2020 HSC 18 MC

An aqueous solution of sodium hydrogen carbonate has a pH greater than 7 .

Which statement best explains this observation?

  1. is a stronger acid than .
  2. is a weaker acid than .
  3. reacts with water to produce the strong base .
  4. The conjugate acid of is a stronger acid than .
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Show Worked Solution

→ can act as an acid or a base (amphiprotic).

→ is a stronger acid than , so donates a proton to .

→ As a result, this produces ions in solution, causing the pH to be greater than 7.


♦♦♦ Mean mark 23%.

CHEMISTRY, M8 2022 HSC 33

Analyse how a student could design a chemical synthesis process to be undertaken in the school laboratory. In your response, use a specific process relating to the synthesis of an organic compound, including a chemical equation, and refer to:

  • selection of reagent(s)
  • reaction conditions
  • any potential hazards and any safety precautions to minimise the risk
  • yield and purity of the product(s).   (8 marks)
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Selecting reagents

→ The student could synthesise ethyl ethanoate through esterification between acetic acid and ethanol.

→ Both readily available in the school laboratory and are relatively safe.

Acetic acid + ethanol  ⇌  Ethyl ethanoate + water

→ Concentrated sulfuric acid should be used as the acid catalyst as it is a strong acid that is also readily available.
 

Reaction conditions

→ Increasing the temperature of the system increases the reaction rate because it increases the average kinetic energy of the reactant molecules, and thus increases the likelihood of successful collisions, producing more product.

→ Additionally, the addition of reactants would increase the likelihood of successful collisions, thus increasing the reaction rate.

→ The reaction should also be undertaken under reflux allowing vaporised molecules to condense and return back to the reaction vessel, increasing the amount of reactants, and thus increasing the rate of reaction. 

→ Concentrated sulfuric acid should also be utilised as a catalyst in order to speed up the reaction and lower the activation energy.
 

Potential hazards and safety precautions

→ The acetic acid and sulfuric acid used is corrosive and may cause skin and eye burns, therefore, appropriate lab coat and safety glasses should be utilised.

→ The organic reactants are highly flammable and may cause fires. The reaction mixture should be heated on a hot plate or heating mantle instead of a bunsen burner.

→ Refluxing may cause pressure build-up, therefore, ensure the reflux condenser is open.

→ Superheating and bumping may occur in apparatus. Boiling chips should be utilised to provide nucleation sites allowing liquids to boil smoothly.
 

Yield and purity

→ Concentrated sulfuric acid, used as a catalyst, also acts as a dehydrating agent that removes water from the system and improves yield.

→ When the reaction reaches equilibrium, the ester can be separated from the mixture by adding excess sodium carbonate solution in order to neutralise the acid.

→ Transfer to a separation funnel to separate the organic layer (containing the ester) from the aqueous layer.

→ Then use fractional distillation to separate the ester from the organic layer.

Show Worked Solution

Selecting reagents

→ The student could synthesise ethyl ethanoate through esterification between acetic acid and ethanol.

→ Both readily available in the school laboratory and are relatively safe.

Acetic acid + ethanol  ⇌  Ethyl ethanoate + water

→ Concentrated sulfuric acid should be used as the acid catalyst as it is a strong acid that is also readily available.
 

Reaction conditions

→ Increasing the temperature of the system increases the reaction rate because it increases the average kinetic energy of the reactant molecules, and thus increases the likelihood of successful collisions, producing more product.

→ Additionally, the addition of reactants would increase the likelihood of successful collisions, thus increasing the reaction rate.

→ The reaction should also be undertaken under reflux allowing vaporised molecules to condense and return back to the reaction vessel, increasing the amount of reactants, and thus increasing the rate of reaction. 

→ Concentrated sulfuric acid should also be utilised as a catalyst in order to speed up the reaction and lower the activation energy.
 

Potential hazards and safety precautions

→ The acetic acid and sulfuric acid used is corrosive and may cause skin and eye burns, therefore, appropriate lab coat and safety glasses should be utilised.

→ The organic reactants are highly flammable and may cause fires. The reaction mixture should be heated on a hot plate or heating mantle instead of a bunsen burner.

→ Refluxing may cause pressure build-up, therefore, ensure the reflux condenser is open.

→ Superheating and bumping may occur in apparatus. Boiling chips should be utilised to provide nucleation sites allowing liquids to boil smoothly.
 

Yield and purity

→ Concentrated sulfuric acid, used as a catalyst, also acts as a dehydrating agent that removes water from the system and improves yield.

→ When the reaction reaches equilibrium, the ester can be separated from the mixture by adding excess sodium carbonate solution in order to neutralise the acid.

→ Transfer to a separation funnel to separate the organic layer (containing the ester) from the aqueous layer.

→ Then use fractional distillation to separate the ester from the organic layer.


♦♦ Mean mark 52%.

CHEMISTRY, M6 2022 HSC 32

The concentration of citric acid, a triprotic acid, in a carbonated soft drink was to be determined.

Step 1: A solution of was standardised by titrating it against 25.00 mL aliquots of a solution of the monoprotic acid potassium hydrogen phthalate . The solution was produced by dissolving 4.989 g in enough water to make 100.0 mL of solution. The molar mass of is 204.22 g mol ¯1.

The results of the standardisation titration are given in the table.
 

Step 2: A 75.00 mL bottle of the drink was opened and the contents quantitatively transferred to a beaker. The soft drink was gently heated to remove .

Step 3: The cooled drink was quantitatively transferred to a 250.0 mL volumetric flask and distilled water was added up to the mark.

Step 4: 25.00 mL samples of the solution were titrated with the solution. The average volume of used was 13.10 mL.

  1. Calculate the concentration of the triprotic citric acid in the soft drink.  (6 marks)
  1. Explain how your answer to part (a) would be different if the carbon dioxide was not removed from the soft drink.  (2 marks)
Show Answers Only

a. 

b.  can dissolve in water to produce :

→ This would enable to react with

→ Therefore, if was not removed, more would be required to reach the endpoint.

→ This would result in a higher citric acid concentration calculation.

Show Worked Solution

a.   

 
   
   

 

 
   
   

 

 

Eliminate the first trial because it is an outlier.

 
   
   

 

 

 
Therefore, the concentration of citric acid in the soft drink is 0.1298 mol L¯1.


♦ Mean mark (a) 48%.

b.    can dissolve in water to produce :

This would enable to react with

→ Therefore, if was not removed, more would be required to reach the endpoint.

→ This would result in a higher citric acid concentration calculation.


♦♦ Mean mark (b) 28%.

CHEMISTRY, M7 2022 HSC 18 MC

A low molecular weight biopolymer is being investigated for its suitability for medical use. In one trial a molecular weight of proved to be optimum.

A section of this biopolymer is shown.
 

Which will produce the suitable biopolymer?
 

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Show Worked Solution

This polymer is a condensation polymer, meaning that it is formed through the reaction between monomers that consist of a carboxylic acid and/or an alcohol functional group, with the elimination of water.

Thus, if monomers react to form this polyester, molecules of water would be eliminated.

Calculating their molar masses:

 


♦♦ Mean mark 23%.

CHEMISTRY, M5 2022 HSC 17 MC

A 2.0 g sample of silver carbonate (MM = 275.81 g mol ¯1) was added to 100.0 mL of water in a beaker. The solubility of silver carbonate at this temperature is mol L ¯1. It was then diluted by adding another 100.0 mL of water.

What is the ratio of the concentration of silver ions in solution before and after dilution?

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Show Worked Solution

The maximum moles of that can be dissolved in 100.0 mL is:

 

 
The number of moles of silver carbonate added to the water is:

 
 

 
→ Thus, the moles of   added is more than the maximum moles of able to be dissolved. i.e. the solution would be saturated before and after dilution.

→ As a result, the solution would have the same concentration before and after, thus, the ratio is 1:1.


♦♦♦ Mean mark 15%.

CHEMISTRY, M6 2021 HSC 35

A manufacturer requires that its product contains at least 85% v/v ethanol.

The concentration of ethanol in water can be determined by a back titration. Ethanol is first oxidised to ethanoic acid using an excess of acidified potassium dichromate solution.

   

The remaining dichromate ions are reacted with excess iodide ions to produce iodine

The iodine produced is then titrated with sodium thiosulfate .

   

A 25.0 mL sample of the manufacturer's product was diluted with distilled water to 1.00 L. A 25.0 mL aliquot of the diluted solution was added to 20.0 mL of 0.500 mol L¯1 acidified potassium dichromate solution in a conical flask. Potassium iodide (5.0 g) was added and the solution titrated with 0.900 mol L¯1 sodium thiosulfate. This was repeated three times.

The following results were obtained.
 


 

The density of ethanol is 0.789 g mL¯1.

Does the sample meet the manufacturer's requirements? Support your answer with calculations.   (7 marks)

Show Answers Only
 
 

 
→ The first titration is an outlier and so is excluded from the average.


 


 


 


 

→ Thus, 0.3949 g of ethanol is in a diluted 25 mL solution.
 

Find the mass of ethanol in the original solution:


 

→ Therefore, the product doesn’t meet the manufacturer’s requirement as the concentration is less than 85%.

Show Worked Solution
 
 

 
→ The first titration is an outlier and so is excluded from the average.


 


 


 


 

→ Thus, 0.3949 g of ethanol is in a diluted 25 mL solution.
 

Find the mass of ethanol in the original solution:


 

→ Therefore, the product doesn’t meet the manufacturer’s requirement as the concentration is less than 85%.


♦♦♦ Mean mark 39%.

CHEMISTRY, M6 2021 HSC 32

The molar enthalpies of neutralisation of three reactions are given.

Reaction 1:

               

Reaction 2:

       

Reaction 3:

           

 
Explain why the first two reactions have the same enthalpy value but the third reaction has a different value.   (4 marks)

Show Answers Only

→ Reaction 1 and reaction 2 are both neutralisation reactions between strong acids and strong bases. These reactions completely ionise in solution when added to water.

→ Both reactions have the same net ionic equation:

→ Therefore, the enthalpy values obtained are the same for both reactions.

→ In reaction 3, is a weak acid that only partially ionises in an equilibrium reaction with water.

→ As the reaction continues, will further ionise as the equilibrium shifts to the right.

→ The bond-breaking is an endothermic process and thus will consume energy to break the bonds. As a result, the overall reaction is less exothermic than reaction 1 and reaction 2.

Show Worked Solution

→ Reaction 1 and reaction 2 are both neutralisation reactions between strong acids and strong bases. These reactions completely ionise in solution when added to water.

→ Both reactions have the same net ionic equation:

→ Therefore, the enthalpy values obtained are the same for both reactions.

→ In reaction 3, is a weak acid that only partially ionises in an equilibrium reaction with water.

→ As the reaction continues, will further ionise as the equilibrium shifts to the right.

→ The bond-breaking is an endothermic process and thus will consume energy to break the bonds. As a result, the overall reaction is less exothermic than reaction 1 and reaction 2.


♦ Mean mark 44%.

ENGINEERING, CS 2021 HSC 19 MC

The diagram shows a simply supported beam in equilibrium. It is loaded with a single force () as shown.
 

Which of the following angles is closest to the angle of the reaction force to the horizontal at the fixed bearing?

  1. 7°
  2. 14°
  3. 31°
  4. 90°
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Show Worked Solution

→ The fixed joint takes half of the vertical force and all of the horizontal force, therefore the angle will be half that of the original force .

♦♦♦ Mean mark 18%.

ENGINEERING, CS 2021 HSC 17 MC

Which of the following is a specialised test used to determine the compressive strength of cured concrete?

  1. Slump test
  2. Janka hardness test
  3. Shore hardness test
  4. Rebound hammer test
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Show Worked Solution

Considering each option:

→ Slump test – tests the workability and consistency of wet concrete.

→ Janka hardness test – tests hardwood flooring.

→ Shore hardness test – tests soft elastomers and other soft polymers.

→ Rebound hammer test – tests the compressive strength of cured concrete.

♦♦♦ Mean mark 19%.

ENGINEERING, TE 2021 HSC 15 MC

Which of the following would be the most suitable material to use for a spring in a telecommunication component?

  1. Annealed copper zinc alloy
  2. Electrolytic tough pitch copper
  3. Age hardened aluminium silicon alloy
  4. Age hardened copper–beryllium alloy
Show Answers Only

Show Worked Solution

→ Age hardened copper-beryllium alloy is used due to its conductivity, low sparking capabilities and physical capabilities of hardness and strength in high temperature conditions.

♦♦♦ Mean mark 29%.

BIOLOGY, M5 2020 HSC 19 MC

Which diagram correctly models one phase of meiosis in an organism that has six chromosomes in its somatic cells?

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Show Worked Solution

By Elimination:

→ Option D would end up with 12 chromosomes in its somatic cells (Eliminate D).

→ Option B and C show meiosis 1, as they are arranged in tetrads. Option C will end up with 12 chromosomes in its somatic cells, while option B is an incorrect model as it does not correctly display homologous pairs arranging in tetrads. (Eliminate B and C).

→ The diagram in option shows metaphase II of meiosis in a cell with 6 chromosomes.

♦ Mean mark 19%.

BIOLOGY, M6 2020 HSC 18 MC

SNP databases have been used in forensic investigations. One is outlined below.

  1. DNA was collected at a crime scene 30 years ago.
  2. Recently the crime scene DNA was analysed at 700 000 SNP locations.
  3. An SNP profile was created and uploaded to a genealogy database.
  4. The SNP profile from the crime scene indicated some shared SNPs with two individuals (who did not have SNPs in common).
  5. The pedigrees were constructed for the two individuals.
     

Which person is most likely to be the suspect who should be investigated?

Show Answers Only

Show Worked Solution

Individual is the only individual that shares DNA with both I and II and is the most likely suspect.

♦♦♦ Mean mark 17%.

BIOLOGY, M5 2020 HSC 16 MC

Analysis of DNA shows that adenine and guanine always make up 50% of the total amount of nitrogenous bases in DNA.

Which structural feature of DNA does this provide evidence for?

  1. DNA is helical in structure.
  2. DNA is always a double-stranded molecule.
  3. DNA always has adenine paired with guanine.
  4. DNA is made up of equal amounts of nitrogenous bases.
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Show Worked Solution

→ In DNA, adenine always pairs with thymine and guanine always pairs with cytosine.

→ The fact that adenine and guanine make up 50% of the total nitrogenous bases in DNA is therefore evidence that DNA is double stranded (held together by paired bases).

♦ Mean mark 27%.

BIOLOGY, M5 2020 HSC 14 MC

A normal allele results in liver cells with sufficient cholesterol receptors. A different allele results in liver cells without cholesterol receptors. Individuals who are heterozygous have liver cells with insufficient cholesterol receptors.

What type of inheritance is the most likely explanation for this?

  1. Sex-linked
  2. Autosomal dominant
  3. Autosomal recessive
  4. Incomplete dominance
Show Answers Only

Show Worked Solution

→ A heterozygous individual exhibiting faulty cholesterol receptors suggests that neither allele is completely expressed.

→ There is also no evidence of the condition being sex-linked.

♦♦♦ Mean mark 31%.

BIOLOGY, M6 2021 HSC 33c

Genetically engineered Atlantic salmon have been produced and approved for aquaculture in the US. These salmon have a transgene that includes a protein-coding sequence from a Chinook salmon's growth hormone gene and the promoter region of an Ocean Pout's antifreeze protein gene. The following diagram provides an overview of the production of the transgenic salmon.
 

Transgenic fish can reproduce and pass on the dominant transgene (T).

Reproduction for aquaculture is strictly controlled using a variety of techniques in order to protect and preserve biodiversity.

Some of these techniques are outlined below. 

1. Homozygous (TT) female (XX) breeding stock are kept in quarantine.

2. The female fish undergo hormone treatment that results in sex reversal and the development of male sex organs and sperm.

3. The sperm produced is collected and used to fertilise eggs obtained from wild-type, non-transgenic salmon.

4. The eggs are treated with pressure shock to prevent the completion of meiosis II. As a result, offspring are triploid (three copies of each chromosome).

All offspring are transgenic female fish and have XXX (XXX fish cannot develop sex organs).

5. Offspring are transported to inland aquaculture tanks to be grown to market size.

Analyse how these techniques protect and preserve biodiversity.   (9 marks)

Show Answers Only

Protecting and Preserving Biodiversity:

→ Biodiversity refers to the variety of genes within the gene pool of a species as well as the variety of species within an ecosystem.

→ The nature of the transgenic salmon carries the risk of reducing biodiversity and has the potential to do great harm.

→ If transgenic salmon escape captivity, they may have a survival advantage over natural salmon and other species. In this way, they could outcompete other species and reduce species diversity, as well as outcompete natural salmon.

→ As a result of the transgene being dominant, it will be passed onto offspring, and in time reduce the gene pool.

→ The physical separation techniques 1 and 5 prevent the transgenic salmon from being released into the environment, preserving biodiversity.

→ Reproductive techniques such as the hormone treatment in technique 2 allow sperm to be produced that only carries the X chromosome as the parents are genetically female.

→ This sperm is then used to fertilise from wild-type salmon, and all offspring will be female as sperm does not contain a Y chromosome, and contain the transgene.

→ Technique 4 uses pressure shock to prevent meiosis 2, resulting in infertile, triploid offspring. This produces transgenic salmon incapable of breeding, meaning the transgene cannot be passed on, preserving biodiversity within the gene pool.

→ The use of wild-type salmon eggs to produce transgenic fish is also crucial in preserving biodiversity as it reduces the likelihood of the accumulation of mutations via interbreeding of transgenic salmon.

→ The physical separation techniques of quarantine and inland aquaculture tanks coupled with reproductive techniques are essential biotechnologies to protect and preserve biodiversity from the dangers of transgenic salmon to the ecosystem and gene pool.

Show Worked Solution

Protecting and Preserving Biodiversity:

→ Biodiversity refers to the variety of genes within the gene pool of a species as well as the variety of species within an ecosystem.

→ The nature of the transgenic salmon carries the risk of reducing biodiversity and has the potential to do great harm.

→ If transgenic salmon escape captivity, they may have a survival advantage over natural salmon and other species. In this way, they could outcompete other species and reduce species diversity, as well as outcompete natural salmon.

→ As a result of the transgene being dominant, it will be passed onto offspring, and in time reduce the gene pool.

→ The physical separation techniques 1 and 5 prevent the transgenic salmon from being released into the environment, preserving biodiversity.

→ Reproductive techniques such as the hormone treatment in technique 2 allow sperm to be produced that only carries the X chromosome as the parents are genetically female.

→ This sperm is then used to fertilise from wild-type salmon, and all offspring will be female as sperm does not contain a Y chromosome, and contain the transgene.

→ Technique 4 uses pressure shock to prevent meiosis 2, resulting in infertile, triploid offspring. This produces transgenic salmon incapable of breeding, meaning the transgene cannot be passed on, preserving biodiversity within the gene pool.

→ The use of wild-type salmon eggs to produce transgenic fish is also crucial in preserving biodiversity as it reduces the likelihood of the accumulation of mutations via interbreeding of transgenic salmon.

→ The physical separation techniques of quarantine and inland aquaculture tanks coupled with reproductive techniques are essential biotechnologies to protect and preserve biodiversity from the dangers of transgenic salmon to the ecosystem and gene pool.

♦ Mean mark 44%.

Mechanics, EXT2 M1 2022 HSC 8 MC

As a projectile of mass kilograms travels through air, it experiences a frictional force. The magnitude of this force is proportional to the square of the speed of the projectile. The constant of proportionality is the positive number . The position of the particle at time is denoted by . The acceleration due to gravity is .

Based on Newton's laws of motion, which equation models the motion of this projectile?

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Show Worked Solution


 


♦♦♦ Mean mark 26%.

BIOLOGY, M8 2021 HSC 31

Millions of people around the world take drugs known as statins, which have been shown to reduce the incidence of heart attacks and strokes in vulnerable patients. However, up to 20% of people stop taking statins due to side-effects such as muscle aches, fatigue, feeling sick and joint pain.

A recent study at a public hospital focused on 60 patients who had all stopped taking statins in the past due to severe side-effects. Patients took statin tablets for four months, placebo tablets for four months and no tablets for four months.

Every day for the year the patients scored, from zero to 100, how bad their symptoms were. The results are shown.
 

Evaluate this study and its results.   (6 marks)

Show Answers Only

→ The data shows that when no tablets were taken, patients still displayed symptoms of the side effects (score 8.0). Taking the statin tablets doubled the severity of the side effects (score 16.3), however, placebo tablets with no active ingredients still produced similar results (score 15.4).

→ This seems to indicate that the side effects may arise as a result of other factors.

→ However, this study cannot be regarded as valid. Only 60 patients were involved in comparison to the millions that take statin tablets.

→ The results were also taken in a qualitative way and cannot be regarded as reliable.

→ The study should also have included a control group of people who have never taken statins to be used for a comparison to improve validity.

→ While the raw data provides data which is valuable and should be followed up with a large, randomised control trial, the study itself cannot be regarded as valid.

Show Worked Solution

→ The data shows that when no tablets were taken, patients still displayed symptoms of the side effects (score 8.0). Taking the statin tablets doubled the severity of the side effects (score 16.3), however, placebo tablets with no active ingredients still produced similar results (score 15.4).

→ This seems to indicate that the side effects may arise as a result of other factors.

→ However, this study cannot be regarded as valid. Only 60 patients were involved in comparison to the millions that take statin tablets.

→ The results were also taken in a qualitative way and cannot be regarded as reliable.

→ The study should also have included a control group of people who have never taken statins to be used for a comparison to improve validity.

→ While the raw data provides data which is valuable and should be followed up with a large, randomised control trial, the study itself cannot be regarded as valid.

♦♦ Mean mark 38%.

PHYSICS, M8 2019 HSC 36

A radon-198 atom, initially at rest, undergoes alpha decay. The masses of the atoms involved are shown in atomic mass units .  
 

The kinetic energy of the polonium atom produced is  J.

By considering mass defect, calculate the kinetic energy of the alpha particle, and explain why it is significantly greater than that of the polonium atom.   (7 marks)

Show Answers Only
 
  u  
  (Converting to kg)
  kg  

 

 
   
  J  

 
Applying the law of conservation of energy:

 
   
  J  

 
→ As the radon atom is initially at rest, the initial momentum of this reaction system is zero. So, by applying the principle of conservation of momentum, the decay products must move away from each other with equal and opposite momenta.

→ The alpha particle has a significantly lower mass compared to the polonium atom and therefore has a significantly higher velocity.

→   and the momenta,   are equal for both the alpha particle and polonium atom. The alpha particle therefore has a significantly greater kinetic energy due to its greater velocity.

Show Worked Solution
 
  u   
  (Converting to kg)
  kg  

 

 
   
  J  

 
Applying the law of conservation of energy:

 
   
  J  

 
→ As the radon atom is initially at rest, the initial momentum of this reaction system is zero. So, by applying the principle of conservation of momentum, the decay products must move away from each other with equal and opposite momenta.

→ The alpha particle has a significantly lower mass compared to the polonium atom and therefore has a significantly higher velocity.

→   and the momenta,   are equal for both the alpha particle and polonium atom. The alpha particle therefore has a significantly greater kinetic energy due to its greater velocity.


♦♦♦ Mean mark 30%.

PHYSICS, M8 2019 HSC 34

Use the following information to answer this question.
 

Describe both the production and radiation of energy by the sun. In your answer, include a quantitative analysis of both the power output and the surface temperature of the sun.   (9 marks)

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Production of the sun’s energy is predominantly done through the proton-proton chain. Four protons react in the sun’s core to produce one helium atom. The mass of one helium atom is less than the mass of four protons so mass is converted into energy through Einstein’s mass-energy equivalence,

The sun acts as a black body and so radiates its energy in the form of black body radiation. It radiates energy as light with a variety of wavelengths, as shown in the black body curve above. It peaks at a specific wavelength which can be used to calculate its temperature:

m  (see graph above)

 
 
  K  
  K  

 
Energy radiated from the sun spreads out over a sphere as it travels away from the sun. The intensity of radiated energy decreases at distances further from the sun, consistent with the inverse square law.

Using the intensity of the sun’s radiation at earth, its power output can be calculated:

 
   
   
  W  
Show Worked Solution

Production of the sun’s energy is predominantly done through the proton-proton chain. Four protons react in the sun’s core to produce one helium atom. The mass of one helium atom is less than the mass of four protons, so mass is converted into energy through Einstein’s mass-energy equivalence,

The sun acts as a black body and so radiates its energy in the form of black body radiation. It radiates energy as light with a variety of wavelengths, as shown in the black body curve above. It peaks at a specific wavelength which can be used to calculate its temperature:

m  (see graph above)

 
 
  K  
  K  

 
Energy radiated from the sun spreads out over a sphere as it travels away from the sun. The intensity of radiated energy decreases at distances further from the sun, consistent with the inverse square law.

Using the intensity of the sun’s radiation at earth, its power output can be calculated:

 
   
   
  W    

♦♦♦ Mean mark 38%.

PHYSICS, M6 2020 HSC 33

A strong magnet of mass 0.04 kg falls 0.78 m under the action of gravity from position above a hollow copper cylinder. It then travels a distance of 0.20 m through the cylinder from to before falling freely again.
 

The magnet takes 0.5 seconds to pass through the cylinder. The displacement-time graph of the magnet is shown.
 

Analyse the motion of the magnet by applying the law of conservation of energy.

Your analysis should refer to gravity and the copper cylinder, and include both qualitative and quantitative information.   (9 marks)

Show Answers Only

During the first 0.4 seconds:

→  The magnet is accelerating downwards at 9.8 m s–2 due to gravity.

→  The magnet’s gravitational potential energy is being converted into kinetic energy consistent with the law of conservation of energy.

→  Quantitatively:

 
   
   
  J  

 

→  Hence 0.30576 J of the magnet’s gravitational potential energy is converted into kinetic energy as it falls under gravity.

 
As magnet reaches the copper cylinder:

→  Its downwards motion causes the formation of induced currents in the cylinder which produce a magnetic field that opposes the magnet’s motion (Lenz’s law).

→  This causes the magnet to decelerate to 0.4 m s–1 and lose kinetic energy. 

→  Finding the magnets speed before entering the copper cylinder:

 
 
  m s–1  

 
→  Quantifying the kinetic energy loss:

 
   
  J  

 
→  Applying the law of conservation of energy shows that 0.30256 J of the magnet’s kinetic energy is being converted to heat energy within the cylinder as the magnet decelerates.

→  Finally, as the magnet passes through the copper cylinder, its gravitational potential energy decreases while its velocity remains constant.

→  Quantifying the decrease in gravitational potential energy:

 
   
   

 

→  Hence, 0.0784 J of the magnet’s gravitational potential energy is converted into heat energy in the cylinder, consistent with the law of conservation of energy.

Show Worked Solution

During the first 0.4 seconds:

→  The magnet is accelerating downwards at 9.8 m s–2 due to gravity.

→  The magnet’s gravitational potential energy is being converted into kinetic energy consistent with the law of conservation of energy.

→  Quantitatively:

 
   
   
  J  

 

→  Hence 0.30576 J of the magnet’s gravitational potential energy is converted into kinetic energy as it falls under gravity.

 
As magnet reaches the copper cylinder:

→  Its downwards motion causes the formation of induced currents in the cylinder which produce a magnetic field that opposes the magnet’s motion (Lenz’s law).

→  This causes the magnet to decelerate to 0.4 m s–1 and lose kinetic energy. 

→  Finding the magnets speed before entering the copper cylinder:

 
 
m s–1  

 
→  Quantifying the kinetic energy loss:

 
   
  J  

 
→  Applying the law of conservation of energy shows that 0.30256 J of the magnet’s kinetic energy is being converted to heat energy within the cylinder as the magnet decelerates.

→  Finally, as the magnet passes through the copper cylinder, its gravitational potential energy decreases while its velocity remains constant.

→  Quantifying the decrease in gravitational potential energy:

 
   
   

 

→  Hence, 0.0784 J of the magnet’s gravitational potential energy is converted into heat energy in the cylinder, consistent with the law of conservation of energy.

♦ Mean mark 51%.

PHYSICS M6 2022 HSC 34

Three charged particles, and , travelling along straight, parallel trajectories at the same speed, enter a region in which there is a uniform magnetic field which causes them to follow the paths shown. Assume that the particles do not exert any significant force on each other.
 


 

Explain the different paths that the particles follow through the magnetic field.   (7 marks)

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→ The charged particles each experience a force equal to    as they travel perpendicular to the magnetic field lines.

→ This force acts perpendicular to the direction of their velocity, so the charged particles undergo uniform circular motion with the force due to the magnetic field acting as a centripetal force.

→ The radii of their paths can be described by the equation:

 
 
 

 
→ As the strength of the magnetic field and the speed of the particles is the same, their trajectory is only affected by their mass to charge ratio   which affects their radius of motion and their sign, which determines the direction which they deflect.

→ Both and initially curve upwards, using the right hand palm rule, these must both be positive charges.

→  has a greater radius than , so ’s charge to mass ratio is greater than ’s.

→ initially curves downwards, using the right hand palm rule, it must have a negative charge.

→ has the same radius of motion as , so the charge to mass ratios of and are the same.

→ has a smaller radius than , so ’s charge to mass ratio is less than that of .

Show Worked Solution

→ The charged particles each experience a force equal to    as they travel perpendicular to the magnetic field lines.

→ This force acts perpendicular to the direction of their velocity, so the charged particles undergo uniform circular motion with the force due to the magnetic field acting as a centripetal force.

→ The radii of their paths can be described by the equation:

 
 
 

 
→ As the strength of the magnetic field and the speed of the particles is the same, their trajectory is only affected by their mass to charge ratio   which affects their radius of motion and their sign, which determines the direction which they deflect.

→ Both and initially curve upwards, using the right hand palm rule, these must both be positive charges.

→  has a greater radius than , so ’s charge to mass ratio is greater than ’s.

→ initially curves downwards, using the right hand palm rule, it must have a negative charge.

→ has the same radius of motion as , so the charge to mass ratios of and are the same.

→ has a smaller radius than , so ’s charge to mass ratio is less than that of .


Mean mark 60%.

PHYSICS M8 2022 HSC 31

Following the Geiger-Marsden experiment, Rutherford proposed a model of the atom.
 


 

Bohr modified this model to explain the spectrum of hydrogen observed in experiments.
 


 

The Bohr-Rutherford model of the atom consists of electrons in energy levels around a positive nucleus.

How do features of this model account for all the experimental evidence above? Support your answer with a sample calculation and a diagram, and refer to energy, forces and photons.  (9 marks)

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The Geiger-Marsden experiment, which involved firing alpha particles at a thin sheet of gold foil produced results which can be explained by the Bohr-Rutherford model:

→ The majority of fired alpha particles passed through the gold foil undeflected. Rutherford concluded from this that the atom had a small, central nucleus.

→ Some alpha particles were deflected and some of these were deflected at very large angles. Rutherford concluded from this that the nucleus was dense and positively charged exerting a repulsive electromagnetic force on the fired alpha particles.

→ The model accounts for Rutherford’s conclusions, placing electrons in orbits around a small positive nucleus.
 

Rutherford’s model alone could not explain the emission spectra of elements such as hydrogen. Bohr’s contribution to the Bohr-Rutherford model amended this:
 


 

→ Bohr proposed that electrons orbited the atomic nucleus in quantised orbits at fixed energies. He proposed that electrons could move from a higher energy orbit (eg. n=1) to a lower energy orbit (n=3) by emitting a photon with energy    equal to the energy difference between the two orbits.

→ Additionally, he proposed that electrons could move from a lower energy orbit to a higher energy orbit by absorbing a photon with energy    equal to the energy difference between the two orbits. 

→ This is able to account for the given emission spectra of hydrogen, where emission lines correspond to electron transitions from higher energy orbits to the second energy orbit which produce photons within the spectrum of visible light.
 

Using Rydberg’s equation it is possible to predict the emission lines of hydrogen, using an electron moving from the sixth to the second Bohr energy orbit as an example:
 

 
   
   
   
 

 
→ This value corresponds to the leftmost line on the given spectrum, reflecting how Bohr’s model can account for the emission spectra of hydrogen.

 

Show Worked Solution

The Geiger-Marsden experiment, which involved firing alpha particles at a thin sheet of gold foil produced results which can be explained by the Bohr-Rutherford model:

→ The majority of fired alpha particles passed through the gold foil undeflected. Rutherford concluded from this that the atom had a small, central nucleus.

→ Some alpha particles were deflected and some of these were deflected at very large angles. Rutherford concluded from this that the nucleus was dense and positively charged exerting a repulsive electromagnetic force on the fired alpha particles.

→ The model accounts for Rutherford’s conclusions, placing electrons in orbits around a small positive nucleus.
 

Rutherford’s model alone could not explain the emission spectra of elements such as hydrogen. Bohr’s contribution to the Bohr-Rutherford model amended this:
 


 

→ Bohr proposed that electrons orbited the atomic nucleus in quantised orbits at fixed energies. He proposed that electrons could move from a higher energy orbit (eg. n=1) to a lower energy orbit (n=3) by emitting a photon with energy    equal to the energy difference between the two orbits.

→ Additionally, he proposed that electrons could move from a lower energy orbit to a higher energy orbit by absorbing a photon with energy    equal to the energy difference between the two orbits. 

→ This is able to account for the given emission spectra of hydrogen, where emission lines correspond to electron transitions from higher energy orbits to the second energy orbit which produce photons within the spectrum of visible light.
 

Using Rydberg’s equation it is possible to predict the emission lines of hydrogen, using an electron moving from the sixth to the second Bohr energy orbit as an example:
 

 
   
   
   
 

 
→ This value corresponds to the leftmost line on the given spectrum, reflecting how Bohr’s model can account for the emission spectra of hydrogen.


♦ Mean mark 51%.

PHYSICS, M7 2013 HSC 20 MC

The graph shows the maximum kinetic energy with which photoelectrons are emitted as a function of frequency for two different metals and .
 


 

The metals are illuminated with light of wavelength 450 nm.

What would be the effect of doubling the intensity of this light without changing the wavelength?

  1. For metal , the number of photoelectrons emitted would not change but the maximum kinetic energy would increase.
  2. For metal , the number of photoelectrons emitted would increase but the maximum kinetic energy would remain unchanged.
  3. For both metals and , the number of photoelectrons emitted would not change but the maximum kinetic energy would increase.
  4. For both metals and , the number of photoelectrons emitted would increase but the maximum kinetic energy would remain unchanged.
Show Answers Only

 

Show Worked Solution

PHYSICS M6 2022 HSC 19 MC

An AC generator is operated by turning a handle, which rotates a coil in a magnetic field.

The handle is turned at a constant speed and the AC voltage output of the generator causes a light globe connected to it to light up, as shown in Circuit 1.
 


 

A second identical light globe is then connected in series to the generator output, as shown in Circuit 2 . The handle is turned at the same constant speed.
 

Which statement describes and explains the effort required to turn the handle in Circuit 2, compared to Circuit 1 ?

  1. The handle in Circuit 2 is easier to turn because the smaller current in Circuit 2 produces less opposing torque.
  2. The handle in Circuit 2 is easier to turn because the voltage output is shared equally across the two identical light globes.
  3. The handle in Circuit 2 is more difficult to turn because the larger current in Circuit 2 produces more opposing torque.
  4. The handle in Circuit 2 is more difficult to turn because it takes more power to operate the two identical globes than it does to operate the single globe.
Show Answers Only

Show Worked Solution

→ Circuit 2 has a greater resistance due to the second light globe in series.

→ The current in circuit 2 is smaller.

→ As this current is the induced current which opposes the rotation of the handle, the handle in circuit 2 is easier to turn.


♦♦♦ Mean mark 17%.

PHYSICS M6 2022 HSC 18 MC

A charged oil droplet was observed between metal plates, as shown.
 


 

While the switch was open, the oil droplet moved downwards at a constant speed. After the switch was closed, the oil droplet moved upwards at the same constant speed.

Assume that the only three forces that may act on the oil droplet are the force of gravity, the force due to the electric field and the frictional force between the air and the oil droplet. The magnitudes of these forces are (due to gravity), (due to the electric field) and (due to the frictional force).

Which row of the table shows all the forces affecting the motion of the oil droplet in the direction indicated, and the relationship between these forces?

Show Answers Only

Show Worked Solution

→ For the droplet to move at a constant speed, the net force acting on it must be zero.

→ For the downwards motion, this means the downwards gravitational force is equal in magnitude to the upwards frictional force (i.e. ).

→ For the upwards motion, this means the upwards electric force is equal in magnitude to the sum of the downwards frictional and gravitational forces (i.e. ).


♦♦♦ Mean mark 26%.

PHYSICS, M8 2019 HSC 32

Describe how specific experiments have contributed to our understanding of the electron and ONE other fundamental particle.   (5 marks)

Show Answers Only

Millikan’s Oil Drop Experiment:

→ Millikan’s oil drop experiment involved first measuring the terminal velocity of charged oil droplets in a gravitational field and calculating their mass.

→ An electric field was applied to balance the gravitational field, allowing Millikan to find the electric force and hence, charge on an oil droplet.

→ This allowed him to find the charge on an electron as the smallest difference in charges between two oil drops.

Linear accelerator experiment discovering quarks:

→ An experiment involved using a linear accelerator to speed up and fire a beam of electrons at protons. The scattering pattern of the electrons was analysed and was consistent with protons having an internal structure with both positive and negative charges.

→ This contributed to our understanding of the existence of quarks.
 

Other possible answers could include:

  • Thomson’s experiment showing the mass to charge ratio of an electron.
  • Experiments involving synchrotrons discovering particles predicted by the standard model of matter, such as the Higgs-Boson.
Show Worked Solution

Millikan’s Oil Drop Experiment:

→ Millikan’s oil drop experiment involved first measuring the terminal velocity of charged oil droplets in a gravitational field and calculating their mass.

→ An electric field was applied to balance the gravitational field, allowing Millikan to find the electric force and hence, charge on an oil droplet.

→ This allowed him to find the charge on an electron as the smallest difference in charges between two oil drops.

Linear accelerator experiment discovering quarks:

→ An experiment involved using a linear accelerator to speed up and fire a beam of electrons at protons. The scattering pattern of the electrons was analysed and was consistent with protons having an internal structure with both positive and negative charges.

→ This contributed to our understanding of the existence of quarks.

Other possible answers could include:

  • Thomson’s experiment showing the mass to charge ratio of an electron.
  • Experiments involving synchrotrons discovering particles predicted by the standard model of matter, such as the Higgs-Boson.

♦♦ Mean mark 43%.

Statistics, EXT1 S1 2022 HSC 14d

An airline company that has empty seats on a flight is not maximising its profit.

An airline company has found that there is a probability of 5% that a passenger books a flight but misses it. The management of the airline company decides to allow for overbooking, which means selling more tickets than the number of seats available on each flight.

To protect their reputation, management makes the decision that no more than 1% of their flights should have more passengers showing up for the flight than available seats.

Given management's decision and using the attached normal distribution probability table to find a suitable approximation, find the maximum number of tickets that can be sold for a flight which has 350 seats.  (4 marks)

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Show Worked Solution


 

 
 

 

 
   
   
   

  


♦♦♦ Mean mark 22%.

Vectors, EXT1 V1 2022 HSC 14b

The vectors and are not parallel. The vector is the projection of onto the vector .

The vector is parallel to so it can be written for some real number . (Do NOT prove this.)

Prove that    is smallest when by showing that, for all real numbers .  (3 marks)

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Show Worked Solution
 
 
 

 


 


♦♦♦ Mean mark 22%.

Calculus, EXT1 C2 2022 HSC 9 MC

A given function    has an inverse  .

The derivatives of    and    exist for all real numbers .

The graphs    and    have at least one point of intersection.

Which statement is true for all points of intersection of these graphs?

  1. All points of intersection lie on the line  .
  2. None of the points of intersection lie on the line  .
  3. At no point of intersection are the tangents to the graphs parallel.
  4. At no point of intersection are the tangents to the graphs perpendicular.
Show Answers Only

Show Worked Solution


 


♦♦♦ Mean mark 13%.

Financial Maths, 2ADV M1 2022 HSC 32

In a reducing-balance loan, an amount is borrowed for a period of months at an interest rate of 0.25% per month, compounded monthly. At the end of each month, a repayment of is made. After the th repayment has been made, the amount owing, , is given by

(Do NOT prove this.)

  1. Jane borrows $200 000 in a reducing-balance loan as described.

  2. The loan is to be repaid in 180 monthly repayments.

  3. Show that  = 1381.16, when rounded to the nearest cent.  (2 marks)

  4. After 100 repayments of $1381.16 have been made, the interest rate changes to 0.35% per month.

  5. At this stage, the amount owing to the nearest dollar is $100 032. (Do NOT prove this.)

  6. Jane continues to make the same monthly repayments.

  7. For how many more months will Jane need to make full monthly payments of $1381.16?  (3 marks)

  8. The final payment will be less than $1381.16.
  9. How much will Jane need to pay in the final payment in order to pay off the loan?   (2 marks)

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Show Worked Solution

a.   

 
 
 

   
   
   

 

b.  

 
 
 
   
 
 
 
   
   

 


♦♦ Mean mark part (b) 38%.

c.   

 
   

 

 
   

♦♦♦ Mean mark part (c) 14%.

Calculus, 2ADV C3 2022 HSC 31

A line passes through the point    and meets the axes at    and  , where .
 

  1. Show that  .  (2 marks)

  2. Find the minimum value of the area of triangle .  (4 marks)

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Show Worked Solution

a.  

 
 
 
   
   
   

 


♦♦♦ Mean mark part (a) 17%.
COMMENT: is the expression of a relationship between the intercepts and not the equation of the line.
b.   
   
   

 

 
   
   

 


 

 
   

♦♦ Mean mark part (b) 29%.

Calculus, 2ADV C4 2022 HSC 29

  1. The diagram shows the graph of  . Also shown on the diagram are the first 5 of an infinite number of rectangular strips of width 1 unit and height    for non-negative integer values of . For example, the second rectangle shown has width 1 and height . 
     


 

  1. The sum of the areas of the rectangles forms a geometric series.
  2. Show that the limiting sum of this series is 2. (1 mark)

  3. Show that .  (2 marks)

  4. Use parts (a) and (b) to show that  .  (2 marks)

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Show Worked Solution

a.   


 

b.   

 
   
   
   
   

 


Mean mark (b) 56%.

c.   

 
 
 
 
 
 

♦♦♦ Mean mark (c) 9%.

Probability, 2ADV S1 2022 HSC 9 MC

Liam is playing two games. He is equally likely to win each game. The probability that Liam will win at least one of the games is 80%.

Which of the following is closest to the probability that Liam will win both games?

  1.  31%
  2.  40%
  3.  55%
  4.  64%
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… Mean marks/comments here
♦♦♦ Mean mark 27%.

Financial Maths, STD2 F4 2022 HSC 36

Frankie borrows $200 000 from a bank. The loan is to be repaid over 23 years at a rate of 7.2% per annum, compounded monthly. The repayments have been set at $1485 per month.

The interest charged and the balance owing for the first three months of the loan are shown in the spreadsheet below.
 

  1. What are the values of and ?  (2 marks)
  2. After 50 months of repaying the loan, Frankie decides to make a lump sum payment of $ 40 000 and to continue making the monthly repayments of $1485. The loan will then be fully repaid after a further 146 monthly repayments.
  3. How much less will Frankie pay overall by making the lump sum payment?  (3 marks)
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a.   

 
   

 

 
   
   

 

b.   


 


 


♦♦♦ Mean mark (b) 17%.
 

Statistics, STD2 S1 2022 HSC 15 MC

The cumulative frequency graph shows the distribution of the number of movie downloads made by 100 people in one month.
 

Which box-plot best represents the same data as displayed in the cumulative frequency graph?
 

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♦♦♦ Mean mark 30%.

Statistics, STD2 S5 2022 HSC 13 MC

A random variable is normally distributed with mean 0 and standard deviation 1 . The table gives the probability that this random variable lies below for some positive values of .
 

The probability values given in the table are represented by the shaded area in the following diagram.

What is the probability that a normally distributed random variable with mean 0 and standard deviation 1 lies between 0 and 1.94 ?

  1. 0.0262
  2. 0.4738
  3. 0.5262
  4. 0.9738
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♦♦♦ Mean mark 22%.

PHYSICS, M6 2019 HSC 18 MC

A circular loop of wire is connected to a battery and a lamp. The apparatus is moved from to along the path shown at a constant velocity through a region containing a uniform magnetic field.
 

Which graph shows the brightness of the lamp as the apparatus moves between and ?
 

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→ Initially, the current travels clockwise through the loop of wire. As it enters the magnetic field, an anticlockwise current is induced in the loop in order to induce a magnetic field out of page, opposing the external magnetic field (Lenz’s Law).

→ This decreases the net current through the loop, causing a decrease in brightness.

→ As the loop exits the magnetic field, a clockwise current is induced to create a magnetic field into the page, opposing the decrease in magnetic flux passing through it (Lenz’s Law).

→ This increases the net current in the loop, causing an increase in brightness.


♦♦♦ Mean mark 25%.

PHYSICS, M8 2019 HSC 8 MC

A typical galaxy has a diameter of 100 000 light years (∼30 000 pc).

Which graph is consistent with Hubble's measurements of the recessional velocity of galaxies?
 

 

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By elimination:

Hubble found that the recessional velocity of a galaxy was directly proportional to its distance (linear graph).

→ Eliminate C and D

Hubble measured distance using parsecs.

→ Eliminate B


♦ Mean mark 25%.

CHEMISTRY, M8 2021 HSC 9 MC

The amount of paracetamol in a sample needs to be determined.

The UV absorption spectrum of paracetamol is shown.
 

Based on the absorption spectra provided, which solvent should be used to determine the amount of paracetamol?
 

 

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→ The solvent used shouldn’t have an absorption spectrum with a maximum that corresponds to that of the paracetamol.

→ This is to ensure that the absorption of solvent will have little to no impact on the measured absorption of the paracetamol sample.


♦♦♦ Mean mark 24%.

PHYSICS, M6 2020 HSC 19 MC

A conductor is in a uniform magnetic field. The conductor rotates around the end at a constant angular velocity.
 

Which graph shows the induced emf between and as the conductor completes one revolution from the position shown?
 

 

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At the starting position shown, electrons in the rod are moving to the right, parallel to the magnetic field lines. So, there is no force acting on the.

→ EMF of zero.

After a quarter of a rotation, electrons in the rod are moving up the page. Using the right hand palm rule, they experience a force out of the page. This will not induce an EMF between and

→ the graph will show an EMF of zero at both and after a quarter of a rotation.


♦♦ Mean mark 10%.

PHYSICS, M7 2020 HSC 18 MC

An observer sees Io complete one orbit of Jupiter as Earth moves from to , and records the observed orbital period as . Similarly, the time for one orbit of Io around Jupiter was measured as Earth moved between the pairs of points at , and , with the corresponding measured periods of Io being , and .
 

Which measurement of the orbital period would be the longest?

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When the Earth is travelling between the pairs of points at , it is moving away from Jupiter:-

→ light must travel further to signal the end of an orbit than it does to signal the start of an orbit.

→   would be the longest measured orbital period.


♦ Mean mark 21%.

PHYSICS, M8 2020 HSC 16 MC

A model of the core of a nuclear fission reactor is shown.
 

When the reactor is operating normally, the moderator, control rods and coolant work in combination to maintain a controlled nuclear reaction in the fuel rods.

The moderator is a liquid which slows down neutrons to increase the rate of fission. The control rods absorb free neutrons. The coolant reduces the core temperature.

A fault causes some of the moderator to leak out of the core.

Which action would compensate for the effect of the loss of moderator?

  1. Withdraw the control rods from the core.
  2. Lower the control rods further into the core.
  3. Pump the coolant through the core at a faster rate.
  4. Reduce the temperature of the coolant before pumping it into the core.
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Loss of moderator leads to a reduction in the rate of fission. Withdrawing the control rods decreases the amount of neutrons absorbed, increasing the rate of fission.

♦ Mean mark 25%.

PHYSICS, M7 2021 HSC 33

Two experiments are performed with identical light sources having a wavelength of 400 nm.

In experiment , the light is incident on a pair of narrow slits  m  apart, producing a pattern on a screen located 3.0 m behind the slits.
 

In experiment , the light is incident on different metal samples inside an evacuated tube as shown. The kinetic energy of any emitted photoelectrons can be measured.
 

Some results from experiment are shown.
 

How do the results from Experiment and Experiment support TWO different models of light? In your answer, include a quantitative analysis of each experiment.   (9 marks)

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→ Experiment A supports the wave model of light as it demonstrates light undergoing diffraction as well as constructive and destructive interference, which are wave properties.

→ When light is incident upon the slits, it diffracts and causes the slit to act as a source of wavefronts. When light from the slits arrives at the screen, bright bands are produced when light waves arrive in phase and undergo constructive interference.

→ Dark bands are produced when light waves arrive at the screen out of phase and undergo destructive interference.

→ The spacing between adjacent bright bands can be calculated:

             
 
 
 

 
→ Experiment B supports Einstein’s particle, or photon model of light. This model can calculate the photon energy of incident light and explain why photons are emitted from calcium but not nickel:


 

→ This energy is greater than the work function of calcium, explaining why one photon has enough energy to liberate a photoelectron from the calcium sample. However, this energy is less than the work function of nickel, explaining why no photoelectrons were observed from the nickel sample.

→ These observations support the particle model of light. Applying the particle model, the kinetic energy of photoelectrons emitted from calcium can be calculated:

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→ Experiment A supports the wave model of light as it demonstrates light undergoing diffraction as well as constructive and destructive interference, which are wave properties.

→ When light is incident upon the slits, it diffracts and causes the slit to act as a source of wavefronts. When light from the slits arrives at the screen, bright bands are produced when light waves arrive in phase and undergo constructive interference.

→ Dark bands are produced when light waves arrive at the screen out of phase and undergo destructive interference.

→ The spacing between adjacent bright bands can be calculated:

             
 
 
 

 
→ Experiment B supports Einstein’s particle, or photon model of light. This model can calculate the photon energy of incident light and explain why photons are emitted from calcium but not nickel:


 

→ This energy is greater than the work function of calcium, explaining why one photon has enough energy to liberate a photoelectron from the calcium sample. However, this energy is less than the work function of nickel, explaining why no photoelectrons were observed from the nickel sample.

→ These observations support the particle model of light. Applying the particle model, the kinetic energy of photoelectrons emitted from calcium can be calculated:


♦ Mean mark 52%.

PHYSICS, M7 2021 HSC 20 MC

A metal cylinder is located in a uniform magnetic field. The work function of the metal is .

Photons having an energy of 2 strike the side of the cylinder, liberating photoelectrons which travel perpendicular to the magnetic field in a circular path. The maximum radius of the path is .
 

If the photon energy is doubled, what will the maximum radius of the path become?

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… (1)  

 

Substitute into (1):

When the photon energy is

When the photon energy is doubled to

∴ As the radius increases by a factor of


♦ Mean mark 24%.

PHYSICS, M5 2021 HSC 14 MC

Which of the following statements correctly describes the gravitational interaction between the Earth and the Moon?

  1. The Earth accelerates towards the Moon.
  2. The net force acting on the Earth is zero.
  3. The Moon and Earth experience equal and opposite accelerations.
  4. The force acting on the Moon is smaller than the force acting on the Earth.
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→ According to Newton’s Third Law, the earth and moon experience equal forces in opposite directions. This causes them to accelerate towards each other.


♦ Mean mark 19%.