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PHYSICS, M7 EQ-Bank 26

The diagram shows a light source, slits and a translucent screen arranged for an experiment on light. Light and dark bands form on the screen. The light has a wavelength of 590 nm. The diagram is not to scale.
 

  1. Explain how any one of the dark bands forms on the screen.  (3 marks)
  1. The distance between the centres of the double slit is 0.15 mm, and the distance between the double slit and the screen is 0.75 m.
  2. Calculate the distance on the screen from the centre of the central maximum to the centre of a second-order bright band.  (3 marks)
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a.   Dark band formation:

  • When light arrives at the double slit, it undergoes diffraction.
  • The two slits act as sources of secondary wavefronts and light travels from both slits to the screen.
  • When light from one slit travels any odd multiple of `(lambda)/(2)` more than light from the other slit to the screen, it is out of phase and destructive interference occurs. Here waves superimpose and cancel each other out, creating dark bands.
  • For example, the dark bands either side of the central maximum form where light from one slit travels `(lambda)/(2)` further than light from the other slit.

b.   `0.0059  text{m}`

Show Worked Solution

a.  Dark band formation:

  • When light arrives at the double slit, it undergoes diffraction.
  • The two slits act as sources of secondary wavefronts and light travels from both slits to the screen.
  • When light from one slit travels any odd multiple of `(lambda)/(2)` more than light from the other slit to the screen, it is out of phase and destructive interference occurs. Here waves superimpose and cancel each other out, creating dark bands.
  • For example, the dark bands either side of the central maximum form where light from one slit travels `(lambda)/(2)` further than light from the other slit.

b.    Using  `d  sin theta=m lambda`

  `0.15 xx10^(-3)  sin theta` `=2xx590 xx10^(-9)`
  `sin theta` `=0.007867`

 

  • Using trigonometry, `tan theta=(d)/(0.75)`, where `theta` is the angular separation of the second order bright band from the central maximum and `d` is the distance between the centre of the central maximum and the centre of the second order bright band.
  • As the angle, `theta` is small, the approximation  `sin theta=tan theta`  is valid:
  •    \( \tan \theta \approx \dfrac{d}{0.75} \ \ \Rightarrow \ \ d=0.007867 \times 0.75 \approx 0.0059 \mathrm{~m} \)

PHYSICS, M7 EQ-Bank 28

In an experiment to investigate the photoelectric effect, a group of students used a piece of equipment containing a metal cathode inside a glass tube. The students were able to accurately measure both the current produced and the maximum energy of electrons in response to light hitting the cathode.

Explain how the choice of independent variable would give rise to different results. Sketch graphs to illustrate your answer.  (7 marks)

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Variables: the frequency of incident light (independent), and the maximum kinetic energy of ejected electrons (dependent).

  • Students would observe that below a certain frequency, no photoelectrons would be ejected. Photons with frequency less than the threshold frequency do not have enough energy to eject an electron.
  • Above this frequency, the students would observe that as the frequency increases, the kinetic energy of ejected electrons would increase linearly. This is because a specific amount of the photon’s energy is required to eject an electron, and any photon energy remaining is transferred to the electrons as kinetic energy, consistent with  `K_(max)=hf-Phi`.
     


 

Variables: intensity of incident light (independent) and the resultant photocurrent (dependent).

  • The frequency of light would be controlled and would be above the threshold frequency.
  • They would observe as the intensity of light increases the current produced would increase linearly.
  • This is because an increasing intensity of light increases the number of photons. This increases the rate at which photons strike the metal surface which increases the rate of photoelectron emission which in turn increases the photocurrent.
     

Show Worked Solution

Variables: the frequency of incident light (independent), and the maximum kinetic energy of ejected electrons (dependent).

  • Students would observe that below a certain frequency, no photoelectrons would be ejected. Photons with frequency less than the threshold frequency do not have enough energy to eject an electron.
  • Above this frequency, the students would observe that as the frequency increases, the kinetic energy of ejected electrons would increase linearly. This is because a specific amount of the photon’s energy is required to eject an electron, and any photon energy remaining is transferred to the electrons as kinetic energy, consistent with  `K_(max)=hf-Phi`.
     


 

Variables: intensity of incident light (independent) and the resultant photocurrent (dependent).

  • The frequency of light would be controlled and would be above the threshold frequency.
  • They would observe as the intensity of light increases the current produced would increase linearly.
  • This is because an increasing intensity of light increases the number of photons. This increases the rate at which photons strike the metal surface which increases the rate of photoelectron emission which in turn increases the photocurrent.
     

PHYSICS, M7 EQ-Bank 29

A proton travels along a particle accelerator at 10 m s ¯1 less than the speed of light.

Compare its speed and momentum with a proton travelling at 99% the speed of light. Support your answer with calculations.  (4 marks)

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  • First proton speed = `3 xx10^(8)-10  text{m s}^(-1)`, second proton speed = `0.99 xx 3 xx10^(8)  text{m s}^(-1)`
  • Comparing the two: `(v_(p1))/(v_(p2))=(3 xx10^(8)-10)/(0.99 xx 3 xx10^(8))=1.0101`.
  • The first proton is travelling 1% faster than the second proton.
  • Calculate the relativistic momentum of the first proton (`p_(1)`):
`p_(1)` `=(m_(0)v)/(sqrt(1-(v^(2))/(c^(2))))`  
  `=(1.673 xx10^(-27)xx(3.00 xx10^(8)-10))/(sqrt(1-((3.00 xx10^(8)-10)/(3.00 xx10^(8)))^(2)))`  
  `=1.94 xx10^(-15)\ text{kg m s}^(-1)`  
  •  Calculate the relativistic momentum of the second proton (`p_(2)`):
`p_(2)` `=(1.673 xx10^(-27)xx0.99xx3.00 xx10^(8))/(sqrt(1-0.99^(2)))`  
  `=3.52 xx10^(-18)  text{kg m s}^(-1)`  
  • Compare the two: `(p_(1))/(p_(2))=(1.94 xx10^(-15))/(3.52 xx10^(-18))=551`.
  • The momentum of the first proton is 551 times greater than the momentum of the second proton, while only going 1% faster.
Show Worked Solution
  • First proton speed = `3 xx10^(8)-10  text{m s}^(-1)`, second proton speed = `0.99 xx 3 xx10^(8)  text{m s}^(-1)`
  • Comparing the two: `(v_(p1))/(v_(p2))=(3 xx10^(8)-10)/(0.99 xx 3 xx10^(8))=1.0101`.
  • The first proton is travelling 1% faster than the second proton.
  • Calculate the relativistic momentum of the first proton (`p_(1)`):
`p_(1)` `=(m_(0)v)/(sqrt(1-(v^(2))/(c^(2))))`  
  `=(1.673 xx10^(-27)xx(3.00 xx10^(8)-10))/(sqrt(1-((3.00 xx10^(8)-10)/(3.00 xx10^(8)))^(2)))`  
  `=1.94 xx10^(-15)\ text{kg m s}^(-1)`  
  •  Calculate the relativistic momentum of the second proton (`p_(2)`):
`p_(2)` `=(1.673 xx10^(-27)xx0.99xx3.00 xx10^(8))/(sqrt(1-0.99^(2)))`  
  `=3.52 xx10^(-18)  text{kg m s}^(-1)`  
  •  Compare the two: `(p_(1))/(p_(2))=(1.94 xx10^(-15))/(3.52 xx10^(-18))=551`.
  • The momentum of the first proton is 551 times greater than the momentum of the second proton, while only going 1% faster.

PHYSICS, M8 EQ-Bank 28

Our understanding of matter is still incomplete and the Standard Model of matter is still being validated and tested. Technology plays a substantial role in this.

Explain the role of technology in developing both the Standard Model of matter and our understanding in ONE other area of physics.  (9 marks)

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Technology and the development of the Standard Model

  • Technology has played a significant role in developing the standard model of matter.
  • Scientists have used the technology of linear accelerators to accelerate a beam of electrons at stationary protons. Technology was then used to analyse the scattering patterns of the electrons which was inconsistent with protons being fundamental particles.
  • It was determined that protons were comprised of both positive and negative internal charges. This led to the discovery of quarks.
  • Further, the Large Hadron Collider (LHC) is technology which accelerates protons to speeds extremely close to the speed of light, and collides them with each other.
  • When these protons collide, their dilated kinetic energy is converted to mass in the form of new particles such as the Higgs Boson. This significantly develops our understanding of the standard model of matter. 

Technology and Special Relativity

  • Another area of physics in which technology has played a vital role is special relativity.
  • Einstein’s prediction of time dilation has been validated by the Hafele-Keating experiment. Technology such as atomic clocks and high speed aeroplanes were used to demonstrate time differences recorded when atomic clocks were flown around the world.
  • In this instance, technology made it possible to validate Einstein’s predictions, improving our understanding of special relativity.
Show Worked Solution

Technology and the development of the Standard Model

  • Technology has played a significant role in developing the standard model of matter.
  • Scientists have used the technology of linear accelerators to accelerate a beam of electrons at stationary protons. Technology was then used to analyse the scattering patterns of the electrons which was inconsistent with protons being fundamental particles.
  • It was determined that protons were comprised of both positive and negative internal charges. This led to the discovery of quarks.
  • Further, the Large Hadron Collider (LHC) is technology which accelerates protons to speeds extremely close to the speed of light, and collides them with each other.
  • When these protons collide, their dilated kinetic energy is converted to mass in the form of new particles such as the Higgs Boson. This significantly develops our understanding of the standard model of matter. 

Technology and Special Relativity

  • Another area of physics in which technology has played a vital role is special relativity.
  • Einstein’s prediction of time dilation has been validated by the Hafele-Keating experiment. Technology such as atomic clocks and high speed aeroplanes were used to demonstrate time differences recorded when atomic clocks were flown around the world.
  • In this instance, technology made it possible to validate Einstein’s predictions, improving our understanding of special relativity.

PHYSICS, M7 EQ-Bank 24

Parallel light rays of intensity `I_0` pass through two polarising filters `P_1` and `P_2` to a detector. The filters are initially aligned so that they produce the maximum amount of light, then filter `P_2` is slowly rotated through 180° as shown.
 

  1. On the axes provided sketch a graph showing how the intensity of light at the detector, `I`, changes as `P_2` rotates from zero to 180°.
     

  1. `P_2` is now rotated to a position such that no light reaches the detector. Without moving `P_1` or `P_2`, a third polarising filter is inserted between `P_1` and `P_2` and rotated at an angle of 30° from `P_1`.
  2. Explain, with the aid of calculations, why the light intensity at the detector is no longer zero.
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a.   

b.   Light intensity at detector:

  • `P_(2)` must be rotated 90° from `P_(1)` (no light reaches detector)
  • Calculate the light passing through the middle polariser:
  •    `I=I_(max)cos^2theta\ \ =>\ \ I_(m)=I_(1)cos^(2)30°=0.75 xxI_(1)`
  • `P_(2)` is a 60° rotation from the middle polariser:
  •    `I=I_(max)cos^2theta\ \ =>\ \ I_(2)=0.75xxI_(1)cos^(2)60°=0.1875xxI_(1)`
  • Light intensity at detector is therefore no longer zero.
Show Worked Solution

a.   

b.   Light intensity at detector:

  • `P_(2)` must be rotated 90° from `P_(1)` (no light reaches detector)
  • Calculate the light passing through the middle polariser:
  •    `I=I_(max)cos^2theta\ \ =>\ \ I_(m)=I_(1)cos^(2)30°=0.75 xxI_(1)`
  • `P_(2)` is a 60° rotation from the middle polariser:
  •    `I=I_(max)cos^2theta\ \ =>\ \ I_(2)=0.75xxI_(1)cos^(2)60°=0.1875xxI_(1)`
  • Light intensity at detector is therefore no longer zero.

PHYSICS, M8 EQ-Bank 25

An experiment was conducted to model Millikan's oil drop experiment. In the experiment, different numbers of dominoes were placed inside seven identical boxes. The boxes were then sealed and weighed. The table shows the mass of each sealed box and some preliminary analysis.
 

Analyse this experiment to assess its effectiveness in modelling Millikan's oil drop experiment.   (6 marks)

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Overview Statement

  • The domino experiment models key components of Millikan’s method. It demonstrates how a fundamental unit is discovered through mass differences and the experimental assumptions involved.

Mass Measurement and Fundamental Unit

  • The smallest mass difference (4.3 g) represents the fundamental unit, like Millikan’s elementary charge.
  • All other mass differences are multiples of this base value, which is similar to Millikan’s findings where the value of the charge of an oil drop was always an integer multiple of `1.6 xx10^(-19)  text{C}`
  • This pattern reveals that objects contain whole numbers of a basic unit.
  • In this way, both experiments found indivisible units through mathematical patterns.

Experimental Method and Limitations

  • Unknown domino numbers parallel Millikan’s unknown electron counts on oil drops.
  • Multiple measurements enable statistical confidence in the fundamental value.
  • The assumption that 4.3 g equals one domino reflects Millikan’s assumption about single electrons.
  • Uncertainty about measuring single units affects both experiments’ reliability.
  • This similarity shows how both experiments require repeated trials to verify results.

Implications and Synthesis

  • The model effectively demonstrates Millikan’s analytical method for finding fundamental quantities.
  • Complementary experimental components work together: mass differences reveal patterns while multiple trials confirm values.
  • The experiment illustrates both strengths and weaknesses of Millikan’s approach.
Show Worked Solution

Overview Statement

  • The domino experiment models key components of Millikan’s method. It demonstrates how a fundamental unit is discovered through mass differences and the experimental assumptions involved.

Mass Measurement and Fundamental Unit

  • The smallest mass difference (4.3 g) represents the fundamental unit, like Millikan’s elementary charge.
  • All other mass differences are multiples of this base value, which is similar to Millikan’s findings where the value of the charge of an oil drop was always an integer multiple of `1.6 xx10^(-19)  text{C}`
  • This pattern reveals that objects contain whole numbers of a basic unit.
  • In this way, both experiments found indivisible units through mathematical patterns.

Experimental Method and Limitations

  • Unknown domino numbers parallel Millikan’s unknown electron counts on oil drops.
  • Multiple measurements enable statistical confidence in the fundamental value.
  • The assumption that 4.3 g equals one domino reflects Millikan’s assumption about single electrons.
  • Uncertainty about measuring single units affects both experiments’ reliability.
  • This similarity shows how both experiments require repeated trials to verify results.

Implications and Synthesis

  • The model effectively demonstrates Millikan’s analytical method for finding fundamental quantities.
  • Complementary experimental components work together: mass differences reveal patterns while multiple trials confirm values.
  • The experiment illustrates both strengths and weaknesses of Millikan’s approach.

PHYSICS, M7 EQ-Bank 16 MC

In 1972 , four caesium clocks were flown twice around the world on commercial jet flights, once eastward and once westward. The travelling clocks were compared with reference clocks at the US Naval Observatory and the results were compared with predictions from Einstein's theory of special relativity.
 

Which of the following is correct about the observed results in relation to Einstein's theory?

  1. Both of the results are inconclusive.
  2. Both of the results support the theory.
  3. One of the results supports the theory and the other is inconclusive.
  4. One of the results supports the theory and the other rejects the theory.
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`B`

Show Worked Solution
  • The tolerance range for the eastward journey is –63 to –17 nanoseconds. 
  • The tolerance range for the westward journey is 254 to 296 nanoseconds. 
  • Therefore, both of the results fall within the range predicted by Einstein’s theory of special relativity.

`=>B`

PHYSICS, M7 EQ-Bank 17 MC

The graph shows the maximum kinetic energy `(K)` with which photoelectrons are emitted as a function of frequency `(f)` for two different metals `X` and `Y`.
 

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 `X`, the number of photoelectrons emitted would not change but the maximum kinetic energy would increase.
  2. For metal `X`, the number of photoelectrons emitted would increase but the maximum kinetic energy remains unchanged.
  3. For metals `X` and `Y`, the number of photoelectrons emitted would not change but the maximum kinetic energy would increase.
  4. For metals `X` and `Y`, the number of photoelectrons emitted would increase but the maximum kinetic energy remains unchanged.
Show Answers Only

`B`

Show Worked Solution
  • Calculating the frequency of incident light:
  •    `v=f lambda`
  •   `f=(f)/(lambda)=(c)/(lambda)=(3 xx10^8)/(4.5 xx10^(-7))=6.67 xx10^(14)  text{Hz}` 
  • This is greater than the work function of metal `X`, but less than the work function of metal `Y`. So, when the light is incident upon both metals, photoelectron emission will occur for `X` but not `Y`.
  • Increasing the intensity of the light increases the rate at which photons strike the surface of metal `X`. This increases the rate of photoelectron emission.
  • The only way to increase the maximum kinetic energy is to increase the frequency of incident light. So, the maximum kinetic energy of photoelectrons remains unchanged. 

`=>B`

PHYSICS, M6 EQ-Bank 27

The diagram shows two rings `A` and `B`, connected to a balancing arm which swings freely on a pivot. Ring `A` has a split in it as shown.
 

When a bar magnet is pushed into one of the rings, the whole balancing arm begins to rotate on the pivot. When the magnet is pulled out, the balancing arm begins to rotate in the opposite direction. When the magnet is pushed in and out of the other ring, the apparatus does not move at all.

Account for these observations using Lenz's Law and conservation of energy.  (5 marks)

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  • When the magnet is pushed into ring `B`, a current is induced in the ring (Faraday’s Law). 
  • The induced current is in the direction such that the magnetic field it produces opposes the original change caused by the moving magnet (Lenz’s Law).
  • When the magnet is pushed into ring `B`, a like pole is produced which causes the magnet and the ring to repel. When the magnet is pulled out of ring `B`, an opposite pole is produced which causes the magnet and the ring to attract.
  • This is consistent with the law of conservation of energy as follows:
    • If the current were in the opposite direction then the field produced would cause a movement that increases the change in flux through the ring.
    • This would thereby produce an even greater induced current which in turn would accelerate the ring even more, in turn leading to an even greater change in flux through the ring.
    • This cycle would continue producing more kinetic and heat energy in the ring than was initially in the system thereby violating the law of conservation of energy. 
  • When the magnet is pushed into ring `A`, no movement of the ring is observed because the gap in the ring prevents a current from being induced.
  • In this instance, no magnetic field is created which means there is no attractive or repulsive force between the ring and the magnet.
Show Worked Solution
  • When the magnet is pushed into ring `B`, a current is induced in the ring (Faraday’s Law). 
  • The induced current is in the direction such that the magnetic field it produces opposes the original change caused by the moving magnet (Lenz’s Law).
  • When the magnet is pushed into ring `B`, a like pole is produced which causes the magnet and the ring to repel. When the magnet is pulled out of ring `B`, an opposite pole is produced which causes the magnet and the ring to attract.
  • This is consistent with the law of conservation of energy as follows:
    • If the current were in the opposite direction then the field produced would cause a movement that increases the change in flux through the ring.
    • This would thereby produce an even greater induced current which in turn would accelerate the ring even more, in turn leading to an even greater change in flux through the ring.
    • This cycle would continue producing more kinetic and heat energy in the ring than was initially in the system thereby violating the law of conservation of energy. 
  • When the magnet is pushed into ring `A`, no movement of the ring is observed because the gap in the ring prevents a current from being induced.
  • In this instance, no magnetic field is created which means there is no attractive or repulsive force between the ring and the magnet.

PHYSICS, M6 EQ-Bank 26

A solenoid was connected to a data logger to measure voltage. A magnet was dropped through the solenoid from above as shown.
 

On the axes provided, sketch a graph showing the change in voltage as the magnet falls completely through the solenoid.  (3 marks)
 

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Points to note on graph:

  • First peak negative, longer wavelength and shorter amplitude.
  • Second peak positive, shorter wavelength and higher amplitude.
Show Worked Solution

Points to note on graph:

  • First peak negative, longer wavelength and shorter amplitude.
  • Second peak positive, shorter wavelength and higher amplitude.

PHYSICS, M6 EQ-Bank 24

Negatively charged particles were accelerated from rest between a pair of parallel metal plates. The potential difference between the plates was varied, and the final velocity of the particles was measured for each variation.
 

The data in the table show the potential difference between the plates and the square of the corresponding final velocity of the particles.

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Potential difference}\ \text{(V)} \rule[-1ex]{0pt}{0pt}&\quad v^2\left(\times 10^9 \, \text{m}^2\, \text{s}^{-2}\right) \quad \\
\hline
\rule{0pt}{2.5ex}100\rule[-1ex]{0pt}{0pt}&0.8\\
\hline
\rule{0pt}{2.5ex}200\rule[-1ex]{0pt}{0pt}& 2.1\\
\hline
\rule{0pt}{2.5ex}300\rule[-1ex]{0pt}{0pt}& 3.1 \\
\hline
\rule{0pt}{2.5ex}400\rule[-1ex]{0pt}{0pt}& 4.1 \\
\hline
\rule{0pt}{2.5ex}500\rule[-1ex]{0pt}{0pt}& 5.2 \\
\hline
\end{array}

  1. Plot the data on the grid provided and draw a line of best fit.
     

   

  1. A student hypothesised that the charged particles are electrons. Justify whether the student's hypothesis is correct or not. Support your answer using the data provided and relevant calculations.
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a.   
         

       

b.    The gradient of the line = `(v^2)/(V)`

Calculate the gradient (choose values close to limits):

`text{gradient}` `=((5.2-0.9) xx10^(9))/(500-100)`  
  `=1.1 xx10^7  text{m}^2  text{s}^(-2)  text{V}^(-1)`  

 
The change in kinetic energy is equal to the work done by the electric field on the charged particles:

`W` `=Delta K`  
`qV` `=(1)/(2)mv^2`  
`(V^2)/(v)` `=(2q)/(m)`  

 
Charge to mass ratio of the particles:

`(q)/(m)` `=(V^2)/(2v)`  
  `=(text{gradient})/(2)`  
  `=5.4 xx10^6  text{C kg}^(-1)`  

 
 Charge to mass ratio of an electron:

`(q)/(m)` `=(1.602 xx10^(-19))/(9.109 xx10^(-31))`  
  `=1.8 xx10^(11)  text{C kg}^(-1)`  

  
Therefore, the charged particles are not electrons.

Show Worked Solution

a.   
         

      

b.    The gradient of the line = `(v^2)/(V)`

Calculate the gradient (choose values close to limits):

`text{gradient}` `=((5.2-0.9) xx10^(9))/(500-100)`  
  `=1.1 xx10^7  text{m}^2  text{s}^(-2)  text{V}^(-1)`  

 
The change in kinetic energy is equal to the work done by the electric field on the charged particles:

`W` `=Delta K`  
`qV` `=(1)/(2)mv^2`  
`(V^2)/(v)` `=(2q)/(m)`  

 
Charge to mass ratio of the particles:

`(q)/(m)` `=(V^2)/(2v)`  
  `=(text{gradient})/(2)`  
  `=5.4 xx10^6  text{C kg}^(-1)`  

 
 Charge to mass ratio of an electron:

`(q)/(m)` `=(1.602 xx10^(-19))/(9.109 xx10^(-31))`  
  `=1.8 xx10^(11)  text{C kg}^(-1)`  

  
Therefore, the charged particles are not electrons.

PHYSICS, M6 EQ-Bank 26

The diagram shows a stationary electron in a magnetic field. The magnetic field is surrounded by two parallel plates separated by a distance of `5.0 × 10^(-3) \ text {m}` and connected to a power supply and a switch.
 

The switch is initially open. At a later time the switch is closed.

Analyse the effects of the magnetic and electric fields on the acceleration of the electron both before and immediately after the switch is closed. In your answer, include calculation of the acceleration of the electron immediately after the switch is closed.   (5 marks)

Show Answers Only
  • Before the switch is closed, there is no electric field and so no electric force causing the electron to accelerate. As the electron is stationary, the magnetic field has no effect on it.
  • Once the switch is closed, the electric field causes the electron to accelerate down the page (towards the positive plate).
  • Calculating the acceleration of the electron immediately after the switch is closed:

`E=(V)/(d)=(100)/(0.005)=20\ 000\ text{V m}^(-1)`

`F=Eq=20\ 000 xx1.602 xx10^(-19)=3.2 xx10^(-15)  text{N}`

`a=(F)/(m)=(3.2 xx10^(-15))/(9.109 xx10^(-31))=3.5 xx10^(15)  text{m s}^(-2)  text{down page.}`

  • Now that the electron is moving, the magnetic field exerts a force on it towards the right (using the right hand palm rule).
  • The force and acceleration on the electron will increase due to its increasing velocity.
Show Worked Solution
  • Before the switch is closed, there is no electric field and so no electric force causing the electron to accelerate. As the electron is stationary, the magnetic field has no effect on it.
  • Once the switch is closed, the electric field causes the electron to accelerate down the page (towards the positive plate).
  • Calculating the acceleration of the electron immediately after the switch is closed:

`E=(V)/(d)=(100)/(0.005)=20\ 000\ text{V m}^(-1)`

`F=Eq=20\ 000 xx1.602 xx10^(-19)=3.2 xx10^(-15)  text{N}`

`a=(F)/(m)=(3.2 xx10^(-15))/(9.109 xx10^(-31))=3.5 xx10^(15)  text{m s}^(-2)  text{down page.}`

  • Now that the electron is moving, the magnetic field exerts a force on it towards the right (using the right hand palm rule).
  • The force and acceleration on the electron will increase due to its increasing velocity.

PHYSICS, M6 EQ-Bank 16 MC

An experiment was carried out to investigate the change in torque for a DC motor with a radial magnetic field. The data from start up to operating speed were graphed.

Which graph is most likely to represent this set of data?
 

Show Answers Only

`D`

Show Worked Solution
  • The DC motor has a radial magnetic field, so the coil is always perpendicular to the magnetic field. This means the torque for the motor does not depend on the angle of the coil.
  • The torque for this DC motor is therefore given by  `tau=NIAB`.
  • As the speed of the motor increases from start to operating speed, the rate of change of magnetic flux through its coil increases, increasing the induced emf in the motor (Faraday’s Law).
  • This induced back emf acts to oppose the rotation of the motor (Lenz’s Law). 
  • The back emf opposes the current supplied to the motor, decreasing the net current through the motor.
  • Therefore, the torque produced by the motor decreases as motor speed increases.

`=>D`

PHYSICS, M6 EQ-Bank 23

The diagram shows a circuit containing two ideal transformers connected with an ammeter. The current through the load is 5.0 A.
 

What is the reading on the ammeter?   (3 marks)

Show Answers Only

`1.0  text{A}`

Show Worked Solution
  • The total power `(P=VI)` of the circuit is constant. This can be calculated from the data given regarding the load.
  •    `P=VI=24 xx5.0=120\ text{W}`
  • Calculating the voltage in the secondary coil of transformer 1:
`(V_(p))/(V_(s))` `=(N_(p))/(N_(s))`  
`V_(s)` `=(V_(p)N_(s))/(N_(p))=(240 xx500)/(1000)120\ text{V}`  

 

  • Calculating the reading on the ammeter:
`P` `=VI`  
`120` `=120 xxI`  
`I` `=1.0  text{A}`  

PHYSICS, M6 EQ-Bank 8 MC

A positively-charged ion travelling at 250 ms ¯1 is fired between two parallel charged plates, \(M\) and \(N\). There is also a magnetic field present in the region between the two plates. The direction of the magnetic field is into the page as shown. The ion is travelling perpendicular to both the electric and the magnetic fields.
 

The electric field between the plates has a magnitude of 200 V m ¯1. The magnetic field is adjusted so that the ion passes through undeflected.

What is the magnitude of the adjusted magnetic field, and the polarity of the \(M\) terminal relative to the \(N\) terminal?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ & \\
\ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Magnitude of magnetic field }& \textit{Polarity of M relative to N} \\
\text{(teslas)}\rule[-1ex]{0pt}{0pt}& \text{} \\
\hline
\rule{0pt}{2.5ex}0.8\rule[-1ex]{0pt}{0pt}&\text{positive}\\
\hline
\rule{0pt}{2.5ex}0.8\rule[-1ex]{0pt}{0pt}& \text{negative}\\
\hline
\rule{0pt}{2.5ex}1.25\rule[-1ex]{0pt}{0pt}& \text{positive} \\
\hline
\rule{0pt}{2.5ex}1.25\rule[-1ex]{0pt}{0pt}& \text{negative} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • The ion passes through undeflected, so the magnitude of the force it experiences due to the magnetic field is equal to the magnitude of the force it experiences due to the electric field. 
  • As the ion travels perpendicular to the magnetic field:
\(F\) \(=qvB\)  
\(E q\) \(=qvB\)  
\(B\) \(=\dfrac{E}{v}\)  
  \(=\dfrac{V}{d} \times \dfrac{1}{v}=\dfrac{200}{250}=0.8 \ \text{T}\)  
     
  •  Using the right hand palm rule, the magnetic field exerts a force up the page on the positive ion. The electric field must therefore exert a force down the page on the ion.
  • \(M\) must be positive relative to \(N\).

\(\Rightarrow A\)

BIOLOGY, M8 EQ-Bank 14

Glucose is a chemical that must be maintained at concentrations between 70 to 130 mg/dL in the blood in order for the body to function normally.

  1. Draw a diagram that illustrates how the body maintains blood glucose within this range.  (3 marks)

  2. Sketch a graph on the axes provided showing the expected blood glucose levels of both a healthy person and a diabetic person after consuming a fruit juice. On the same graph, show what would happen when the diabetic person injects themself with insulin 20 minutes after consuming the fruit juice.  (3 marks)
     
     

Show Answers Only

a.  

 
b.

   

Show Worked Solution

a.  

 
b.

   

BIOLOGY, M8 EQ-Bank 15

The diagram shows a rural coastal area and the towns, rivers and associated industry for each of the townships.
 

An epidemic of a disease has broken out in Nanavale. The symptoms are stomach ache, vomiting and tiredness. Many families in Nanavale have only one member with the disease, therefore it appears to be non-infectious. The symptoms are worse in infants than in adults.

Isolated cases of this disease have occurred in the nearby towns of Dairyville and Beefville. No cases have been reported on Gull Island.

Design an epidemiological study to investigate the origin of the disease. Refer to features of validity and reliability in your answer.   (7 marks)

Show Answers Only
  • When planning an epidemiological study it is important to first analyse all the initial evidence to better construct an effective study.
  • The disease is most likely infectious as an outbreak that affects many people is very unlikely to be due to a previously masked non-infectious disease.
  • The fact that the disease is not present on gull island also supports this and may indicate that the disease is not waterborne and may spread through physical touch, close proximity, food or radioactive toxic elements.
  • The disease also affects children more severely. This fact must be addressed in the study and measures taken to protect and monitor this vulnerable group.
  • The study should survey affected families to try to pinpoint the transmission of the disease.
  • Appropriate fact finding questions should include
    • Where have you travelled to?
    • What have you eaten/drunk and where did you get it from?
    • Who else have you been in contact with and where are they from?
  • These results should be analysed for common factors and then compared to results from the same set of questions asked of unaffected families, thus increasing the study’s validity.
  • The more people that can be reached and questioned, the more accurate the findings of the study will be.
  • Geiger readings, screening and soil extraction may pinpoint whether the disease is caused by a carcinogen on the coastal area.
  • Common factors found in the study may also reveal an antidote or treatment for affected individuals.
Show Worked Solution
  • When planning an epidemiological study it is important to first analyse all the initial evidence to better construct an effective study.
  • The disease is most likely infectious as an outbreak that affects many people is very unlikely to be due to a previously masked non-infectious disease.
  • The fact that the disease is not present on gull island also supports this and may indicate that the disease is not waterborne and may spread through physical touch, close proximity, food or radioactive toxic elements.
  • The disease also affects children more severely. This fact must be addressed in the study and measures taken to protect and monitor this vulnerable group.
  • The study should survey affected families to try to pinpoint the transmission of the disease.
  • Appropriate fact finding questions should include
    • Where have you travelled to?
    • What have you eaten/drunk and where did you get it from?
    • Who else have you been in contact with and where are they from?
  • These results should be analysed for common factors and then compared to results from the same set of questions asked of unaffected families, thus increasing the study’s validity.
  • The more people that can be reached and questioned, the more accurate the findings of the study will be.
  • Geiger readings, screening and soil extraction may pinpoint whether the disease is caused by a carcinogen on the coastal area.
  • Common factors found in the study may also reveal an antidote or treatment for affected individuals.

BIOLOGY, M8 EQ-Bank 16

How effective is renal dialysis in compensating for the loss of kidney function?   (7 marks)

Show Answers Only

Normal kidney function

  • The kidneys are the main components of the mammalian urinary system. They are organs which filter blood and maintain water, pH, ion and salt concentration in the body through varying concentrations of each in excreted urine dependent on the body’s needs.
  • Each kidney contains 1 million nephrons, the main unit responsible for filtration.
  • Each one also contains a Bowman’s capsule, proximal and distal tubules as well as the Loop of Henle which acts as a site for selective re-absorption of certain components of the blood. This is controlled by both passive diffusion of unwanted substances through a concentration gradient (e.g. urea) or by hormonal control.
  • Aldosterone and ADH are hormones secreted by the hypothalamus which increase the permeability of the distal convoluted tubule to salt and water respectively.

Kidney disfunction and dialysis

  • Kidney function can however be impaired by diseases or disorders (such as polycystic kidney disease), many of which can kill affected individuals in a number of months if left untreated.
  • When kidney function drops below 80%, haemodialysis is an effective treatment to replace kidney function.
  • Haemodialysis involves the removal of blood from the body into the dialysis machine, which will clean the blood before returning it to the body. This is achieved by running a fluid known as dialysate, countercurrent to the blood.
  • The dialysate contains a similar composition to the blood with low urea and toxins to allow the passive diffusion of the substances via the concentration gradient into the dialysate. This is then removed and constantly replenished during a session. The countercurrent direction also improves effectiveness of this process. The dialysate can also be altered to have varying amounts of salt and ions depending on the concentration in the patients body.
  • Haemodialysis can provide an effective treatment for individuals until death or an effective transplant can be found, however the process often requires 3-4 sessions per week each of which is 4 hours long.
  • Without haemodialysis loss of kidney function is often fatal, but this life-saving technology is extremely effective in preventing many deaths despite its inconvenience.
Show Worked Solution

Normal kidney function

  • The kidneys are the main components of the mammalian urinary system. They are organs which filter blood and maintain water, pH, ion and salt concentration in the body through varying concentrations of each in excreted urine dependent on the body’s needs.
  • Each kidney contains 1 million nephrons, the main unit responsible for filtration.
  • Each one also contains a Bowman’s capsule, proximal and distal tubules as well as the Loop of Henle which acts as a site for selective re-absorption of certain components of the blood. This is controlled by both passive diffusion of unwanted substances through a concentration gradient (e.g. urea) or by hormonal control.
  • Aldosterone and ADH are hormones secreted by the hypothalamus which increase the permeability of the distal convoluted tubule to salt and water respectively.

Kidney disfunction and dialysis

  • Kidney function can however be impaired by diseases or disorders (such as polycystic kidney disease), many of which can kill affected individuals in a number of months if left untreated.
  • When kidney function drops below 80%, haemodialysis is an effective treatment to replace kidney function.
  • Haemodialysis involves the removal of blood from the body into the dialysis machine, which will clean the blood before returning it to the body. This is achieved by running a fluid known as dialysate, countercurrent to the blood.
  • The dialysate contains a similar composition to the blood with low urea and toxins to allow the passive diffusion of the substances via the concentration gradient into the dialysate. This is then removed and constantly replenished during a session. The countercurrent direction also improves effectiveness of this process. The dialysate can also be altered to have varying amounts of salt and ions depending on the concentration in the patients body.
  • Haemodialysis can provide an effective treatment for individuals until death or an effective transplant can be found, however the process often requires 3-4 sessions per week each of which is 4 hours long.
  • Without haemodialysis loss of kidney function is often fatal, but this life-saving technology is extremely effective in preventing many deaths despite its inconvenience.

BIOLOGY, M8 EQ-Bank 14

  1. Identify a disorder or disease, and describe how it affects the normal function of an organ.   (2 marks)

  2. Evaluate the effectiveness of a technology in managing the disorder or disease described in part (a).   (4 marks)

Show Answers Only

a.   Polycystic kidney disease

  • A disease in which cysts grow in the kidney.
  • This interferes with filtration and overtime, adversely affects the overall function of the kidney.

Other disorders/disease can include

  • Astigmatism, myopia or hyperopia.
  • Single sided deafness, hearing loss.

b.   Use of technology in treating polycystic kidney disease

  • Polycystic kidney disease will often progress in patients until kidney failure has occurred.
  • When 80% of kidney function is lost and a transplant cannot be found, dialysis is often the only viable option.
  • There are two types of dialysis, both of which replace the function of the kidney and clean the blood.
    • Haemodialysis is where blood is removed from the body and passed through a dialysis machine. The main component of the machine runs a special fluid known as dialysate countercurrent to the blood next to a permeable membrane. This allows the blood to effectively remove the toxins into the dialyse by diffusion. While this is extremely effective, it involves multiple visits per week and is very time consuming for the patients.
    • Peritoneal dialysis involves the flushing of dialysate directly into the abdominal cavity through a catheter. The peritoneal membrane then acts as the filter and blood is cleaned via diffusion. The dialysate containing the waste products then exits the body into another section of the catheter. In contrast to haemodialysis, this method requires a more involved initial surgery but provides the benefit of patients being able to do this by themselves at night.
  • Both forms of dialysis are extremely effective in replacing kidney function. Despite being time consuming and uncomfortable, they allow patients suffering from polycystic kidney disease to survive for extended periods, during which a transplant may become available.

Other technologies can include

  • Glasses.
  • Any form of hearing aid dependent on the ear disease/disorder, including standard hearing aids, bone conduction implants or cochlear implant.
Show Worked Solution

a.   Polycystic kidney disease

  • A disease in which cysts grow in the kidney.
  • This interferes with filtration and overtime, adversely affects the overall function of the kidney.

Other disorders/disease can include

  • Astigmatism, myopia or hyperopia.
  • Single sided deafness, hearing loss.

b.   Use of technology in treating polycystic kidney disease

  • Polycystic kidney disease will often progress in patients until kidney failure has occurred.
  • When 80% of kidney function is lost and a transplant cannot be found, dialysis is often the only viable option.
  • There are two types of dialysis, both of which replace the function of the kidney and clean the blood.
    • Haemodialysis is where blood is removed from the body and passed through a dialysis machine. The main component of the machine runs a special fluid known as dialysate countercurrent to the blood next to a permeable membrane. This allows the blood to effectively remove the toxins into the dialyse by diffusion. While this is extremely effective, it involves multiple visits per week and is very time consuming for the patients.
    • Peritoneal dialysis involves the flushing of dialysate directly into the abdominal cavity through a catheter. The peritoneal membrane then acts as the filter and blood is cleaned via diffusion. The dialysate containing the waste products then exits the body into another section of the catheter. In contrast to haemodialysis, this method requires a more involved initial surgery but provides the benefit of patients being able to do this by themselves at night.
  • Both forms of dialysis are extremely effective in replacing kidney function. Despite being time consuming and uncomfortable, they allow patients suffering from polycystic kidney disease to survive for extended periods, during which a transplant may become available.

Other technologies can include

  • Glasses.
  • Any form of hearing aid dependent on the ear disease/disorder, including standard hearing aids, bone conduction implants or cochlear implant.

BIOLOGY, M7 EQ-Bank 23

The diagram shows the immune response after primary exposure to a pathogen.
 

  1. On the diagram, continue the graph to show the immune response upon secondary exposure to the same pathogen.   (2 marks)
  2. Using annotations on the diagram, explain the shape of the entire graph.   (4 marks)

Show Answers Only

a. & b.

Show Worked Solution

a. & b.

BIOLOGY, M8 EQ-Bank 13

A scientist performed an epidemiological study to investigate the cause and effect relationship of smoking and lung cancer as follows:

    1. Handed out a scientifically valid questionnaire to all employees at her daughter's high school (n=96)
    2. Checked that there were an equal number of male and female respondents
    3. Discovered that there were more non-smoking respondents than smoking respondents. Removed some of the non-smokers until both groups had equal numbers
    4. Required all respondents to have a general check up from their doctor before participating 
    5. Analysed data, wrote the paper and published it in a scientific blog

From the information provided, assess the suitability of the methodology for this investigation.   (5 marks)

Show Answers Only
  • This study cannot be referred to as a suitable comparison between smoking and lung cancer due to several flaws in its methodology.
  • The sample size of the study is too small and confined to one workplace.
  • The participants should represent a number of equal categories such as age and ethnicity, not just equal representation of males and females.
  • The fact that participants were removed in order to create equal smoking/non-smoking categories reduces the validity of the findings.
  • The medical check ups should not be general and instead be in relation to the study such as cancer and lung checks.
  • The study will also benefit from peer review.
Show Worked Solution
  • This study cannot be referred to as a suitable comparison between smoking and lung cancer due to several flaws in its methodology.
  • The sample size of the study is too small and confined to one workplace.
  • The participants should represent a number of equal categories such as age and ethnicity, not just equal representation of males and females.
  • The fact that participants were removed in order to create equal smoking/non-smoking categories reduces the validity of the findings.
  • The medical check ups should not be general and instead be in relation to the study such as cancer and lung checks.
  • The study will also benefit from peer review.

BIOLOGY, M8 EQ-Bank 3 MC

A student wanted to test the claim, made by an agricultural seed company, that their variety of wheat was more salt tolerant than other wheat varieties on the market. The company explained that their variety can better maintain water balance by increasing the organic salt concentration in the roots of the plants which increases osmotic pressure.

Which row of the table shows the independent and dependent variables that the student should use to test the company's claim?
 

Show Answers Only

`D`

Show Worked Solution

By Elimination

  • When testing this aim, the variety of wheat plant should vary and the salt concentration given to each must remain the same. (Eliminate A and B).
  • Each variety’s turgor will give an accurate representation of ability to maintain water balance, the initial claim by the company as to its wheat’s benefit. (Eliminate C).

`=>D`

BIOLOGY, M7 EQ-Bank 26

The immune system's primary role is to defend against pathogens. For this to be effective the immune system must be able to recognise cells that belong to the body and cells that do not.

  1. Describe the mechanism that the immune system uses to distinguish between body cells and potential pathogens. Support your answer with an example.   (3 marks)

  2. Explain why this mechanism means that patients who receive an organ donation require immune suppression drugs.   (3 marks)

Show Answers Only

a.    Immune system blood type mechanisms

  • Every cell has proteins on its surface which the body can use to distinguish them from self and non-self.
  • The body’s own cells produce a specific composition of surface proteins referred to as the major histocompatibility complex (MHC), as opposed to other proteins which are not recognised by the immune system and are called antigens.
  • This provides the basis of blood types, where A and B are specific surface proteins while O is blood with no surface proteins.
  • AB blood type individuals are therefore universal recipients, as their immune system has recognised both A and B proteins as self. This also explains why type O individuals are universal donors as their blood has no surface proteins, hence type O, A, B and AB individuals will not recognise it as foreign.
     

b.    Immune systems detect new donated organs as foreign bodies.

  • When foreign cells/material are detected by the body, an immune response will attempt to kill or remove it from the body.
  • The same scenario occurs during an organ transplant, as the donor’s MHC complex is different to that of the recipient.
  • To counteract this, specialists run genetic testing on donors as well as the recipient to find people with a similar chromosome 6 genetic sequence. MHC is coded by genes found on this chromosome and compatible organ donors are often family members.
  • As a precaution, specialists administer immune suppression drugs which will slow the patient’s immune response, reducing the chance of an attack on the donated organ.
Show Worked Solution

a.    Immune system blood type mechanisms

  • Every cell has proteins on its surface which the body can use to distinguish them from self and non-self.
  • The body’s own cells produce a specific composition of surface proteins referred to as the major histocompatibility complex (MHC), as opposed to other proteins which are not recognised by the immune system and are called antigens.
  • This provides the basis of blood types, where A and B are specific surface proteins while O is blood with no surface proteins.
  • AB blood type individuals are therefore universal recipients, as their immune system has recognised both A and B proteins as self. This also explains why type O individuals are universal donors as their blood has no surface proteins, hence type O, A, B and AB individuals will not recognise it as foreign. 

b.    Immune systems detect new donated organs as foreign bodies.

  • When foreign cells/material are detected by the body, an immune response will attempt to kill or remove it from the body.
  • The same scenario occurs during an organ transplant, as the donor’s MHC complex is different to that of the recipient.
  • To counteract this, specialists run genetic testing on donors as well as the recipient to find people with a similar chromosome 6 genetic sequence. MHC is coded by genes found on this chromosome and compatible organ donors are often family members.
  • As a precaution, specialists administer immune suppression drugs which will slow the patient’s immune response, reducing the chance of an attack on the donated organ.

BIOLOGY, M7 EQ-Bank 25

A silver birch tree, Butula pendula, has been attacked by a fungal stem canker. Fungal cankers are opportunistic plant pathogens that gain access to the inner layers of the stem as a result of damage to the protective outer layer of bark. The inner layers provide suitable conditions for the fungus to grow. Once established it destroys the bark cells that protect the tree. The tree responds to the presence of the canker under its bark by producing an excess of resinous sap at the wound site.
 

  1. Using the information provided, suggest a hypothesis to explain how the tree is responding to the presence of the canker.   (2 marks)

  2. Consider another plant with a different response to a specific pathogen.
  3. Compare the necessity and limitations of this plant's response with the response of the birch tree described above.   (4 marks)

Show Answers Only

a.    Hypothesis of birch tree response:

  • Excess sap production alters the growth conditions of the canker.
  • The sap will change the chemical environment around the wound site and inflame the site to reduce the spread of the pathogen, much similar to inflammation in humans. 

b.   Red river gum comparison

  • The river red gum responds to halo leaf spot cankers by dropping its leaves.
  • While this means the tree will have to expend energy and resources regrowing leaves, it simultaneously counteracts the canker as it blocks the ability of the leaves to photosynthesise.
  • By dropping the affected leaves, the tree can grow them quicker than if the leaves were left to die.
  • The birch tree will counteract a stem canker by producing excess sap. While this may not completely get rid of any pathogens, the tree cannot simply remove its own stems like the river red gum.
  • This method isolates the infected area, similar to inflammation, retarding its growth and potentially killing the fungus.
  • The isolation reduces the canker’s ability to adversely effect the birch tree’s supply of water and nutrients via the xylem and phloem of the tree.
  • This process occurs with the canker still attached to the tree, making the fungus relatively more difficult to kill.
Show Worked Solution

a.    Hypothesis of birch tree response:

  • Excess sap production alters the growth conditions of the canker.
  • The sap will change the chemical environment around the wound site and inflame the site to reduce the spread of the pathogen, much similar to inflammation in humans. 

b.   Red river gum comparison

  • The river red gum responds to halo leaf spot cankers by dropping its leaves.
  • While this means the tree will have to expend energy and resources regrowing leaves, it simultaneously counteracts the canker as it blocks the ability of the leaves to photosynthesise.
  • By dropping the affected leaves, the tree can grow them quicker than if the leaves were left to die.
  • The birch tree will counteract a stem canker by producing excess sap. While this may not completely get rid of any pathogens, the tree cannot simply remove its own stems like the river red gum.
  • This method isolates the infected area, similar to inflammation, retarding its growth and potentially killing the fungus.
  • The isolation reduces the canker’s ability to adversely effect the birch tree’s supply of water and nutrients via the xylem and phloem of the tree.
  • This process occurs with the canker still attached to the tree, making the fungus relatively more difficult to kill.

BIOLOGY, M7 EQ-Bank 8 MC

Eight sick animals had the same symptoms. Blood tests showed that they were infected with the same type of bacterium.

Which of the following would be the best course of action to determine if this particular type of bacterium is the cause of the symptoms?

  1. Treat all eight animals with the antibiotic known to kill this type of bacterium. Check if they recover.
  2. Find other animals with the same symptoms. Attempt to isolate the same type of bacterium from their blood.
  3. Inject blood from animals with the symptoms into suitable host individuals. Check if they develop the same symptoms.
  4. Use bacteria from the blood of affected animals to inoculate healthy animals. If these healthy animals develop the symptoms, attempt to isolate the same bacterium from their blood.
Show Answers Only

`D`

Show Worked Solution
  • The blood from an affected animal will contain bacteria that is presumed to be responsible for the disease.
  • If it is injected into healthy organisms that subsequently develop the same symptoms, and their blood is isolated and observed to have the same bacteria, then it is certain that the bacteria is responsible for the disease.
  • This follows Koch’s postulates.

`=>D`

BIOLOGY, M7 EQ-Bank 9 MC

Melanomas are characterised by uncontrolled cell division caused by mutations that continue to occur once the tumour has developed. Scientists have discovered that vaccines produced using antigens extracted from the patient's own melanoma cells can be useful in treating melanoma. When injected, the vaccines stimulate an immune response.

The effect of the melanoma vaccine is to stimulate

  1. T cells which produce antibodies.
  2. cytotoxic T cells which activate B cells.
  3. cell division to produce more lymphocytes.
  4. production of B cells which destroy melanoma cells.
Show Answers Only

`C`

Show Worked Solution
  • The vaccine will encourage the division of lymphocytes (B and T cells) as a response to melanoma antigens.

`=>C`

BIOLOGY, M7 EQ-Bank 7 MC

Melanomas are characterised by uncontrolled cell division caused by mutations that continue to occur once the tumour has developed. Scientists have discovered that vaccines produced using antigens extracted from the patient's own melanoma cells can be useful in treating melanoma. When injected, the vaccines stimulate an immune response.

What can be inferred from the scientists' discovery?

  1. Cancer cells carry unique antigens.
  2. Self-antigens are not present on cancer cells.
  3. The melanoma patient has a dysfunctional immune system.
  4. The body cannot mount an immune response against cancer cells.
Show Answers Only

`A`

Show Worked Solution
  • To provide a vaccine, there must be antigens present on cancer cells in order to trigger an immune response.

`=>A`

BIOLOGY, M7 EQ-Bank 7 MC

The diagram shows a model of disease transmission.
 

An epidemiologist suspected that bats were acting as a reservoir for an infectious disease in humans.

Which condition would need to be met to confirm the epidemiologist's suspicion?

  1. The infectious agent would need to have a mode of entry into humans.
  2. The infectious agent would need a mode of transmission from bats to humans.
  3. The bats would have to be able to transmit the infectious agent between each other.
  4. The susceptible human host must be able to transmit the infectious agent to the reservoir of bats.
Show Answers Only

`B`

Show Worked Solution
  • For the assumption to be true, the pathogen would need to have developed an adaptation allowing it to transfer from bats to humans.

`=>B`

BIOLOGY, M6 EQ-Bank 28

Describe how technological developments led to the advancement of our knowledge and understanding of inheritance. Support your answer with examples.   (7 marks)

Show Answers Only
  • Artificial pollination is the process by which pollen with known characteristics is manually applied to another plant with known genotypes anther.
  • This is used in agriculture to preserve favourable characteristics in plants. However, it can also be used to study inheritance patterns and genotype/phenotype relationships.
  • Mendel used this process to investigate ‘pure-breeding’ individuals/plants (ones which always produce the same offspring) to develop the idea of recessive and dominant alleles and the offspring they produced depending on the genotype. This laid the basis for the concept of ‘Mendelian ratios’.
  • Mendel used artificial pollination on purple and white pea plants which were known pure-breeding plants and found that all the offspring (F1 Generation) were purple, determining that the purple allele was dominant.
  • When he again used artificial pollination between F1 Generation plants, he found that it always produced 3/4 of purple plants and 1/4 white. This showed that F1 Generation purple plants were not pure-breed and instead ‘heterozygous’ and contained a masked allele for the white colour. This 3:1 ratio is what became known as a Mendelian ratio. 

Other suggested examples include:

  • Gene editing technology CRISPR
  • Microscopy and staining – behaviour of chromosomes (Sutton and Boveri)
  • X‐ray crystallography and DNA structure (Rosalind Franklin)
  • Radiation – one gene, one polypeptide (Beadle and Tatum)
Show Worked Solution
  • Artificial pollination is the process by which pollen with known characteristics is manually applied to another plant with known genotypes anther.
  • This is used in agriculture to preserve favourable characteristics in plants. However, it can also be used to study inheritance patterns and genotype/phenotype relationships.
  • Mendel used this process to investigate ‘pure-breeding’ individuals/plants (ones which always produce the same offspring) to develop the idea of recessive and dominant alleles and the offspring they produced depending on the genotype. This laid the basis for the concept of ‘Mendelian ratios’.
  • Mendel used artificial pollination on purple and white pea plants which were known pure-breeding plants and found that all the offspring (F1 Generation) were purple, determining that the purple allele was dominant.
  • When he again used artificial pollination between F1 Generation plants, he found that it always produced 3/4 of purple plants and 1/4 white. This showed that F1 Generation purple plants were not pure-breed and instead ‘heterozygous’ and contained a masked allele for the white colour. This 3:1 ratio is what became known as a Mendelian ratio. 

Other suggested examples include:

  • Gene editing technology CRISPR
  • Microscopy and staining – behaviour of chromosomes (Sutton and Boveri)
  • X‐ray crystallography and DNA structure (Rosalind Franklin)

BIOLOGY, M6 EQ-Bank 25

A woman recently conceived a baby guaranteed to be free from hereditary breast cancer. Doctors screened for an embryo that was free from a gene that can cause breast cancer.

The screening was performed due to the long history of this form of cancer in the family and the fact that any daughter born with the gene would have a 50-80% chance of developing breast cancer.

  1. Explain the possible impact of this reproductive technology on the genetic composition of the population.   (2 marks)

  2. Discuss the use of this genetic technology in the treatment of medical conditions.   (3 marks)

Show Answers Only

a.    Impact on genetic composition of population

  • This use of this technology will reduce the frequency of the gene that is likely to produce breast cancer in the population.
  • In the longer term, proceeding generations of offspring will also not have the gene, resulting in fewer people suffering from breast cancer. 

b.    Genetic technology treatments

  • Genetic technologies are a great tool against preventing life-threatening genetic diseases, as well as being used as a treatment for other serious conditions such as cancer.
  • These genetic technologies can treat and potentially cure many genetic conditions, improving life expectancy, quality of life and reducing incidence. 
  • Although many of the treatments are expensive upfront, they can reduce future ongoing medical costs.
  • Genetic technologies are relatively new and experimental on humans. This poses a risk of unknown side-effects later on in the patient’s life.
  • Scepticism surrounding many genetic technologies has brought up several ethical considerations including the unnatural attributes of the procedures and concerns of personal liberty and choice.
Show Worked Solution

a.    Impact on genetic composition of population

  • This use of this technology will reduce the frequency of the gene that is likely to produce breast cancer in the population.
  • In the longer term, proceeding generations of offspring will also not have the gene, resulting in fewer people suffering from breast cancer. 

b.    Genetic technology treatments

  • Genetic technologies are a great tool against preventing life-threatening genetic diseases, as well as being used as a treatment for other serious conditions such as cancer.
  • These genetic technologies can treat and potentially cure many genetic conditions, improving life expectancy, quality of life and reducing incidence. 
  • Although many of the treatments are expensive upfront, they can reduce future ongoing medical costs.
  • Genetic technologies are relatively new and experimental on humans. This poses a risk of unknown side-effects later on in the patient’s life.
  • Scepticism surrounding many genetic technologies has brought up several ethical considerations including the unnatural attributes of the procedures and concerns of personal liberty and choice.

BIOLOGY, M6 EQ-Bank 24

The yeast Saccharomyces cerevisiae cannot naturally ferment the sugar xylose. Low value biomass, such as straw and wood fibres, contains up to 20% xylose. S. cerevisiae was modified to enable it to produce ethanol from xylose. Information on the two species involved in making the modified S. cerevisiae is shown in the table.
 

  1. Explain why biotechnology was needed to modify S. cerevisiae.   (2 marks)

  2. Two strains of genetically modified S.cerevisiae were produced. The two strains were compared under the same conditions. The results are shown.
     
          

  3. Justify which of these two strains would be better to use to produce commercial quantities of ethanol using low value biomass. In your answer, refer to information from the graph.   (3 marks)

Show Answers Only

a.    The table shows that the organisms are from different genera.

  • Genetic material generally can’t be transferred between organisms of different genera.
  • Genetic technology was therefore needed to cut out and insert the required genes from B. cenocepacia to create the genetically modified S. cerevisiae as this process could not occur naturally. 

b.    Best strain for commercial production

  •  Strain B would be the more effective strain to use as it consistently produces double the ethanol of Strain A.
  • This can be seen even at the plateau of both strains, where at 30 hrs Strain B produced 20g/L of ethanol whilst Strain A only produced 10g/L.
  • These data in the graph indicates that Strain B will be more efficient at producing commercial ethanol in any given time-frame.
Show Worked Solution

a.    The table shows that the organisms are from different genera.

  • Genetic material generally can’t be transferred between organisms of different genera.
  • Genetic technology was therefore needed to cut out and insert the required genes from B. cenocepacia to create the genetically modified S. cerevisiae as this process could not occur naturally. 

b.    Best strain for commercial production

  •  Strain B would be the more effective strain to use as it consistently produces double the ethanol of Strain A.
  • This can be seen even at the plateau of both strains, where at 30 hrs Strain B produced 20g/L of ethanol whilst Strain A only produced 10g/L.
  • These data in the graph indicates that Strain B will be more efficient at producing commercial ethanol in any given time-frame.

BIOLOGY, M6 EQ-Bank 23

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

To what extent is this statement correct?   (6 marks)

Show Answers Only
  • Reproductive technologies intervene in natural reproduction and allow offspring with a higher chance of favourable characteristics.
  • Examples includes artificial insemination in animals, artificial pollination in plants, as well as IVF and selective breeding.
  • In all the above processes, male gametes with favourable traits are isolated and delivered into the reproductive system of selected females using technology.
  • The processes greatly increase the number of offspring generated by one parent, thus decreasing the genetic variability amongst them.
  • The repetitive use of the same male gametes across generations will eventually reduce the genetic diversity amongst the entire species, as the favoured characteristics will increase and less favourable alleles will be lost.
  • It should be noted that these technologies can also increase genetic diversity of a species through the collection and transportation of male gametes. This process helps genes to overcome geographical and time barriers, allowing for genotypes and hybrids which would otherwise not be possible by natural processes.
Show Worked Solution
  • Reproductive technologies intervene in natural reproduction and allow offspring with a higher chance of favourable characteristics.
  • Examples includes artificial insemination in animals, artificial pollination in plants, as well as IVF and selective breeding.
  • In all the above processes, male gametes with favourable traits are isolated and delivered into the reproductive system of selected females using technology.
  • The processes greatly increase the number of offspring generated by one parent, thus decreasing the genetic variability amongst them.
  • The repetitive use of the same male gametes across generations will eventually reduce the genetic diversity amongst the entire species, as the favoured characteristics will increase and less favourable alleles will be lost.
  • It should be noted that these technologies can also increase genetic diversity of a species through the collection and transportation of male gametes. This process helps genes to overcome geographical and time barriers, allowing for genotypes and hybrids which would otherwise not be possible by natural processes.

BIOLOGY, M6 EQ-Bank 27

Compare the processes and effects of point mutations and chromosomal mutations. Include examples in your answer.   (8 marks)

Show Answers Only
  • Mutations are the change in an individuals genome. Their effects are dependent on
    • the location of the mutation. Mutations in coding DNA can affect proteins and have adverse physiological affects while mutations in non-coding DNA often have no affect, however may slightly impact cell activity.
    • the type of cell it occurs in. Somatic cell mutations only affect the individual, while germ-line mutations occur in germ-line cells which go on to produce gametes, thus potentially affecting offspring. 
  • Point mutations are mutations which change only a single base in a DNA sequence. They can be a single base substitution or the deletion or insertion of an extra base. Deletion or insertions are referred to as ‘frameshift’ mutations as they impact all codons proceeding it.
  • Sometimes substitution mutations are neutral and occur in non-coding DNA, code for an amino acid of very similar chemical composition or even code for the same amino acid. When a mutation does code for a dissimilar amino acid or a stop codon causing premature termination, this interferes with the shape of the polypeptide chain and the protein it becomes part of, rendering it faulty or completely unusable (e.g. sickle cell anaemia, NRAS mutation). The nature of frameshift mutations affecting every codon proceeding it are often more severe, as there is more potential to cause a faulty polypeptide chain (e.g. cystic fibrosis).
  • Chromosomal mutations can refer to the removal, insertion, translocation or inversion of large sections of a chromosome. These often affect genes by either splitting them up or relocating them to a new chromosome.
  • Chromosomal mutations can also refer to the relocation of entire chromosomes during meiosis of germ-line cells where they fail to separate. This is called non-disjunction and results in aneuploidy of two gametes, one which will result in having an extra chromosome (trisomy) and one will lack a chromosome (monosomy). Conditions such as Down’s syndrome (trisomy-21) or Turner syndrome (XO) are aneuploidy conditions but most often cause a miscarriage or unsuccessful zygote development.
  • Polyploidy is the complete development of an entirely new set of chromosomes. This is known to occur in strawberries which have 8 sets of chromosomes but cannot occur in humans.
  • While both can cause radical health defects, it is important to note that these mutations are the basis of evolution. Beneficial mutations are more likely to be carried through a species through generations according to natural selection.
Show Worked Solution
  • Mutations are the change in an individuals genome. Their effects are dependent on
    • the location of the mutation. Mutations in coding DNA can affect proteins and have adverse physiological affects while mutations in non-coding DNA often have no affect, however may slightly impact cell activity.
    • the type of cell it occurs in. Somatic cell mutations only affect the individual, while germ-line mutations occur in germ-line cells which go on to produce gametes, thus potentially affecting offspring. 
  • Point mutations are mutations which change only a single base in a DNA sequence. They can be a single base substitution or the deletion or insertion of an extra base. Deletion or insertions are referred to as ‘frameshift’ mutations as they impact all codons proceeding it.
  • Sometimes substitution mutations are neutral and occur in non-coding DNA, code for an amino acid of very similar chemical composition or even code for the same amino acid. When a mutation does code for a dissimilar amino acid or a stop codon causing premature termination, this interferes with the shape of the polypeptide chain and the protein it becomes part of, rendering it faulty or completely unusable (e.g. sickle cell anaemia, NRAS mutation). The nature of frameshift mutations affecting every codon proceeding it are often more severe, as there is more potential to cause a faulty polypeptide chain (e.g. cystic fibrosis).
  • Chromosomal mutations can refer to the removal, insertion, translocation or inversion of large sections of a chromosome. These often affect genes by either splitting them up or relocating them to a new chromosome.
  • Chromosomal mutations can also refer to the relocation of entire chromosomes during meiosis of germ-line cells where they fail to separate. This is called non-disjunction and results in aneuploidy of two gametes, one which will result in having an extra chromosome (trisomy) and one will lack a chromosome (monosomy). Conditions such as Down’s syndrome (trisomy-21) or Turner syndrome (XO) are aneuploidy conditions but most often cause a miscarriage or unsuccessful zygote development.
  • Polyploidy is the complete development of an entirely new set of chromosomes. This is known to occur in strawberries which have 8 sets of chromosomes but cannot occur in humans.
  • While both can cause radical health defects, it is important to note that these mutations are the basis of evolution. Beneficial mutations are more likely to be carried through a species through generations according to natural selection.

BIOLOGY, M6 EQ-Bank 22

The flow chart illustrates the effect of a point mutation on an organism.
 

  1. Outline the series of events from stages 2 to 4 that resulted in the faulty protein.   (3 marks)

  2. Describe how a type of mutagen may have caused the changes observed in stage 2.   (2 marks)

  3. Given the information in the chart shown, describe the effect caused by the mutation in stage 4 and the effect this would have on the organism.   (3 marks)
     
     

Show Answers Only

a.    Events causing faulty protein:

  • Stage 2: A mutagen causes a point mutation in the DNA.
  • Stage 3: The mutated DNA transcription into mRNA.
  • Stage 4: The translation of the mRNA into a polypeptide chain which is not functional due to the mutation.

b.    Possible causes of Stage 2 observed changes:

  • High frequency photons including gamma and X-rays are regarded as ionising radiation due to their extremely high energy.
  • They can interfere with DNA undergoing replication causing misalignment and a possible substation change.
  • High energy gamma radiation may also cause mutation when directly striking a nitrogenous base.  

Answers could also include

  • Chemical mutagens (alkylating, delaminating and structurally similar chemicals).
  • Carcinogenic metals/elements (radon, cadmium)

c.   Effects of Stage 4 mutation:

  • When considering the above sequence as 2 codons, the chart shows us that this mutation results in an arginine amino becoming a stop codon.
  • This mutation will hence cause a premature termination of the polypeptide and result in a non-functional polypeptide/protein it forms a part of.
  • Due to the severity of this type of mutation it usually results in extensive health complications for affected individuals.
Show Worked Solution

a.    Events causing faulty protein:

  • Stage 2: A mutagen causes a point mutation in the DNA.
  • Stage 3: The mutated DNA transcription into mRNA.
  • Stage 4: The translation of the mRNA into a polypeptide chain which is not functional due to the mutation.

b.    Possible causes of Stage 2 observed changes:

  • High frequency photons including gamma and X-rays are regarded as ionising radiation due to their extremely high energy.
  • They can interfere with DNA undergoing replication causing misalignment and a possible substation change.
  • High energy gamma radiation may also cause mutation when directly striking a nitrogenous base.  

Answers could also include

  • Chemical mutagens (alkylating, delaminating and structurally similar chemicals).
  • Carcinogenic metals/elements (radon, cadmium)

c.   Effects of Stage 4 mutation:

  • When considering the above sequence as 2 codons, the chart shows us that this mutation results in an arginine amino becoming a stop codon.
  • This mutation will hence cause a premature termination of the polypeptide and result in a non-functional polypeptide/protein it forms a part of.
  • Due to the severity of this type of mutation it usually results in extensive health complications for affected individuals.

BIOLOGY, M6 EQ-Bank 14 MC

Which of the following is true of a mutation that produces an allele that is dominant?

  1. It would be expected to cause death.
  2. It could give an observable phenotype in a heterozygous genotype.
  3. It could give an observable phenotype only in a homozygous genotype.
  4. It would be expected to spread more quickly through a population than a recessive mutation.
Show Answers Only

`B`

Show Worked Solution

By Elimination

  • Mutations are not expected to cause death as death is dependant on the location and type of mutation (Eliminate A).
  • A dominant allele mutation vs recessive has no impact on spread in a population (Eliminate D).
  • The change in phenotype is not only possible in homozygous recessive phenotypes but also in heterozygous individuals (Eliminate C).

`=>B`

BIOLOGY, M6 EQ-Bank 6 MC

An evolutionary biologist was investigating the timeframes of genetic divergence between different species of Acacia. She hypothesised that she would get a better indication of the time at which Acacia species diverged by using non-coding DNA segments rather than coding DNA segments.

Why is this hypothesis most likely to be supported by the evidence?

  1. Coding DNA segments never undergo mutation.
  2. Coding DNA segments are less stable over time due to the selection pressures of the environment.
  3. Non-coding DNA segments do not undergo mutation as they are not subject to selection pressures.
  4. Non-coding DNA segments will show greater diversity after divergence as they are not exposed to selection pressures.
Show Answers Only

`D`

Show Worked Solution
  • Coding DNA cannot vary as much as even point mutations can be harmful or fatal.
  • Non-coding DNA does not impact the individual or their offspring and will therefore show greater variance which can be used as a tool to track species divergence.

`=>D`

BIOLOGY, M5 EQ-Bank 25

A student plans to investigate whether the development of insulin has affected the prevalence of Type 1 diabetes in the human population and subsequently influenced human evolution. She has access to data on Australians with diabetes extending back to 1973 .

  1. Propose a suitable hypothesis for this investigation.   (2 marks)

  2. Identify TWO variables that need to be controlled for this investigation and explain their importance.   (4 marks)

Show Answers Only

a.   Two possible hypotheses include:

  • The development of insulin has led to a decrease in the prevalence of Type 1 diabetics.
  • The use of insulin by people who have Type 1 diabetes has increased the prevalence of Type 1 diabetes in the human population.
     

b.  Variables to be controlled:

  • The growth of the Australian population in general will statistically lead to a bias in more recent years, where higher a number of individuals with Type 1 diabetes is to be expected.
  • To control this variable, the investigation should be measured using a fixed rate such as ‘per 100 000’.
  • The time of diagnosis and length of time an individual has used insulin can impact the trend of the obtained data.
  • This data must be obtained and accounted for to be able to increase the validity of findings.
Show Worked Solution

a.   Two possible hypotheses include:

  • The development of insulin has led to a decrease in the prevalence of Type 1 diabetics.
  • The use of insulin by people who have Type 1 diabetes has increased the prevalence of Type 1 diabetes in the human population.

b.  Variables to be controlled

  • The growth of the Australian population in general will statistically lead to a bias in more recent years, where higher a number of individuals with Type 1 diabetes is to be expected.
  • To control this variable, the investigation should be measured using a fixed rate such as ‘per 100 000’.
  • The time of diagnosis and length of time an individual has used insulin can impact the trend of the obtained data.
  • This data must be obtained and accounted for to be able to increase the validity of findings.

BIOLOGY, M5 EQ-Bank 26

The diagram shows a model developed in the early 20th century of crossing over of homologous chromosomes.
 

Explain how the difference between this model and our current model of crossing over reflects an increased understanding of the way in which new combinations of genotypes are produced. Support your answer with a diagram.  (4 marks)

Show Answers Only
  • The model shows that crossing over occurs between homologous chromosomes before DNA replication. This means that when sister chromatids are formed, they will contain the post-crossing over sequence.
  • In the new model, DNA replication occurs prior to crossing over, where crossing over only occurs when replicated homologous chromosomes line up in tetrads, and only 2 homologous pairs exchange information. 
  • In this way, the new model will produce 4 unique chromatids while the old model will produce only 2. Therefore new model produces a wider range of genetic variability in offspring.
  • Possible unique chromatids:

Show Worked Solution
  • The model shows that crossing over occurs between homologous chromosomes before DNA replication. This means that when sister chromatids are formed, they will contain the post-crossing over sequence.
  • In the new model, DNA replication occurs prior to crossing over, where crossing over only occurs when replicated homologous chromosomes line up in tetrads, and only 2 homologous pairs exchange information. 
  • In this way, the new model will produce 4 unique chromatids while the old model will produce only 2. Therefore new model produces a wider range of genetic variability in offspring.
  • Possible unique chromatids:

BIOLOGY, M5 EQ-Bank 28

  1. Complete the following diagram to show the process by which gametes are formed.   (3 marks)
     

      
  2. How does the segregation of chromosomes during meiosis lead to a wide variety of gametes being produced?   (2 marks)

Show Answers Only

a.   


 

b.    Process creating wide variety of gametes

  • Independent assortment is the process by which homologous pairs are separated during meiosis into daughter cells.
  • During this process, daughter cell orientation and the cell they are separated into is random and not dependent on any factors.
  • This leads to a great variety in gametes due to the numerous combinations of chromosomes.
Show Worked Solution

a.   


 

b.    Process creating wide variety of gametes

  • Independent assortment is the process by which homologous pairs are separated during meiosis into daughter cells.
  • During this process, daughter cell orientation and the cell they are separated into is random and not dependent on any factors.
  • This leads to a great variety in gametes due to the numerous combinations of chromosomes.

BIOLOGY, M5 EQ-Bank 23

Students conducted preliminary experiments to analyse the DNA base composition of five different individuals.

The table shows the experimental data collected.
 

  1. On the grid provided, plot the % Guanine against % Adenine of the individuals analysed and draw a suitable line of best fit.   (3 marks)
     
     

  2. Identify the relationship shown by the data.   (1 mark)

  3. Explain the relationship shown by the data.   (3 marks)

Show Answers Only

a.   
       

b.  Relationship: As the % of Adenine increases, the % of Guanine decreases.

c.   Adenine/Guanine relationship

→ Because Adenine pairs with Thymine and Cytosine pairs with Guanine occurring to DNA base pairing rules, each should directly correlate to each other and account for the same % within the DNA composition of any species.

→ There are the only 4 components that make up all DNA composition. This also means that when comparing % of bases that do not pair, such as Adenine and Guanine, they must combine to form 50% of all DNA. This is evident when looking at individuals A and B, as well as C, D and E which are only off by 1-2%, most likely due to human error.

→ This always explains the relationship where an increase in Adenine (and Thymine) % will result in a decrease in Guanine (and Cytosine) % as the A and T pairs now occupy more space in the genome.

Show Worked Solution

a.   
       

b.  Relationship: As the % of Adenine increases, the % of Guanine decreases.

c.   Adenine/Guanine relationship

→ Because Adenine pairs with Thymine and Cytosine pairs with Guanine occurring to DNA base pairing rules, each should directly correlate to each other and account for the same % within the DNA composition of any species.

→ There are the only 4 components that make up all DNA composition. This also means that when comparing % of bases that do not pair, such as Adenine and Guanine, they must combine to form 50% of all DNA. This is evident when looking at individuals A and B, as well as C, D and E which are only off by 1-2%, most likely due to human error.

→ This always explains the relationship where an increase in Adenine (and Thymine) % will result in a decrease in Guanine (and Cytosine) % as the A and T pairs now occupy more space in the genome.

BIOLOGY, M5 EQ-Bank 27

Explain why internal fertilisation is a more robust process than external fertilisation in maintaining a species generation after generation.   (5 marks)

Show Answers Only
  • In order for a species to successfully continue and avoid extinction, a sufficient number of offspring must be able to survive until gestational maturity and produce offspring themselves, hence continuing the cycle.
  • External fertilisation is used mainly in sea animals such as fish where the survival rate of offspring is low due to the plentiful numbers of predators in such a vast environment.
  • In response to this environment, external fertilisation methods must produce large numbers of sperm and eggs to increase the number of fertilised embryos. This process requires of a significant amount of energy.
  • Fertilised embryos typically experience minimal or no parental care, further lowering the survival rate of offspring. Therefore, external fertilisation has numerous disadvantages to overcome in achieving the continuity of the species.
  • Internal fertilisation is generally more common amongst varying species (such as humans). Despite often only producing one child at a time, internal fertilisation as a process protects the developing embryo inside the female uterus.
  • Although the number of fertilised embryos is lower, greater parental care throughout most of early life and a significant decrease in predation creates a higher survival rate.
  • In this way, internal fertilisation is more advantageous to ensure the continuity of a species.
Show Worked Solution
  • In order for a species to successfully continue and avoid extinction, a sufficient number of offspring must be able to survive until gestational maturity and produce offspring themselves, hence continuing the cycle.
  • External fertilisation is used mainly in sea animals such as fish where the survival rate of offspring is low due to the plentiful numbers of predators in such a vast environment.
  • In response to this environment, external fertilisation methods must produce large numbers of sperm and eggs to increase the number of fertilised embryos. This process requires of a significant amount of energy.
  • Fertilised embryos typically experience minimal or no parental care, further lowering the survival rate of offspring. Therefore, external fertilisation has numerous disadvantages to overcome in achieving the continuity of the species.
  • Internal fertilisation is generally more common amongst varying species (such as humans). Despite often only producing one child at a time, internal fertilisation as a process protects the developing embryo inside the female uterus.
  • Although the number of fertilised embryos is lower, greater parental care throughout most of early life and a significant decrease in predation creates a higher survival rate.
  • In this way, internal fertilisation is more advantageous to ensure the continuity of a species.

BIOLOGY, M5 EQ-Bank 16 MC

A student conducted a survey to determine the phenotype prevalence of cats that had long hair in comparison to the number that had short hair in a population of cats.

She asked her classmates to describe the coat length of their cats and tallied the results. Out of 26 cats that were counted, she found that 42% of the cats had long hair and 58% had short hair, and that the trait did not follow a Mendelian ratio.

Which of the following best explains why the results did not follow a Mendelian ratio?

  1. The student tallied the numbers incorrectly.
  2. The length of cat hair may be determined by more than one gene.
  3. The student cannot determine the genotype from the phenotype alone.
  4. The students were unclear about whether their cat had long or short hair.
Show Answers Only

`B`

Show Worked Solution

By Elimination

  • A and D relate to human error and cannot be deemed a valid reason for an experiment not appealing to a hypothesis (Eliminate A and D).
  • Many characteristics (phenotypes) are controlled by a variety of genes, including standard Mendelian alleles, sex-linked genes, multiple alleles and incomplete dominance/co-dominace. A combination of these may relate to the length of hair in cats, and is the best explanation of the results.

`=>B`

BIOLOGY, M5 EQ-Bank 15 MC

It is suspected that a child has a recessive, sex-linked condition. An initial pedigree was developed.

Which of the following is most likely to depict this initial pedigree?
 

Show Answers Only

`B`

Show Worked Solution

By Elimination

  • Pedigree C displays no information related to the question as no children are affected (Eliminate C).
  • Both pedigree A and D display crosses which are inaccurate and impossible to occur as daughters cannot be affected if the mother does not have recessive sex-linked traits (Eliminate A and D).

`=>B`

BIOLOGY, M5 EQ-Bank 13 MC

Haemophilia A is a blood clotting disorder that arises from a defect in the gene which is carried on the `text{X}` chromosome.

A couple is considering starting a family. However, the father suffers from Haemophilia A. The mother is healthy with no family history of the disease.

What is the probability that a potential grandson will have Haemophilia A if they have a daughter who partners with an unaffected man?

  1. 0%
  2. 50%
  3. 75%
  4. 100%
Show Answers Only

`B`

Show Worked Solution
  • The probability can be explained by the cross below, where the daughter has a heterozygous phenotype (mother was homozygous recessive and father had haemophilia).
  • Daughter : `text{X}^text{A}text{X}^text{a}`   Husband : `text{X}^text{a}text{Y}`

\begin{array} {|l|c|c|}
\hline  & \text{X}^\text{a} & \text{Y} \\
\hline \text{X}^\text{A} & \text{X}^\text{A}\text{X}^\text{a} & \text{X}^\text{A}\text{Y}\\
\hline \text{X}^\text{a} & \text{X}^\text{a}\text{X}^\text{a} & \text{X}^\text{a}\text{Y} \\
\hline \end{array}

  • This shows that 1 in 2 grandsons will have haemophilia and 1 in 2 granddaughters will have haemophilia, therefore there is a 50% likelihood of grandchildren being affected.

`=>B`

BIOLOGY, M5 EQ-Bank 12 MC

The table shows the base triplets in mRNA for amino acids.

From the table, the amino acid Tryptophan (Trp) can be coded for by the base triplet `text{UGG}`.
 

Which base triplet could code for the amino acid Arginine (Arg)?

  1. `text{AAU}`
  2. `text{UGC}`
  3. `text{GCC}`
  4. `text{CGG}`
Show Answers Only

`D`

Show Worked Solution
  • `text{CGG}` is the only correct combination for Arginine from options given.

`=>D`

BIOLOGY, M5 EQ-Bank 14 MC

Haemophilia A is a blood clotting disorder that arises from a defect in the gene F8 which is carried on the X chromosome. The disorder affects the production of a glycoprotein that is one of many components needed to form the platelets which form blood clots when a bleed occurs. It is typically treated with infusions of FVIII product, an inactive single chain polypeptide of 2332 amino acids, which is manufactured using DNA technology on human endothelial cells.

Why is the inactive FVIII polypeptide chain used in the treatment of Haemophilia A?

  1. It will prevent bleeds from occurring.
  2. It can take the place of platelets in clotting blood.
  3. It can be used to manufacture the glycoprotein that is affected by the defective F8 gene.
  4. It is used as a gene therapy to help the patient manufacture FVIII in their own endothelial cells.
Show Answers Only

`C`

Show Worked Solution
  • An inactive single chain polypeptide that has been manufactured by DNA technology from endothelial cells (the single cell layer of blood vessels) would imply that it is the mRNA strand which codes for the glycoprotein.
  • Haemophilia A is a mutation of the F8 which codes for the glycoprotein (or part of it), and therefore injections of this mRNA strand will ‘bypass’ the mutation and allow the body to manufacture the protein itself.

`=>C`

PHYSICS, M8 EQ-Bank 26

Observations and mathematical ideas are critical to the improvement of scientific models.

Discuss this statement with reference to scientific discoveries that have contributed to our understanding of the atom.   (8 marks)

Show Answers Only
  • [P] Observations have been critical in advancing atomic models.
  • [E] When experiments reveal unexpected results, scientists must revise their models to match reality.
  • [Ev] The Geiger-Marsden experiment showed alpha particles bouncing back from gold foil, contradicting the plum pudding model. This led Rutherford to propose a dense nucleus.
  • [L] This demonstrates how observations drive scientific progress in understanding atoms.
     
  • [P] Mathematical ideas provide essential frameworks for atomic models.
  • [E] Mathematics allows scientists to make precise predictions and test theories quantitatively.
  • [Ev] Bohr developed a mathematical model explaining the specific wavelengths emitted which relies on Rydberg’s equation: `(1)/(lambda)=R((1)/(n_(f)^2)-(1)/(n_(i)^2))`.
  • [Ev] This model was limited as it was only able to explain the hydrogen atom. De Broglie built upon it by postulating that electrons behaved as a wave, eventually describing the equation `\lambda=h/{mv}` which helped explain electron wave behaviour.
  • [L] These mathematical models transformed vague ideas into testable predictions about atomic structure.
     
  • [P] However, observations and mathematics alone have limitations.
  • [E] Models based purely on observations can miss underlying principles without theoretical insight.
  • [Ev] Bohr’s model perfectly matched hydrogen spectra but failed for other atoms because it lacked deeper quantum understanding.
  • [L] This shows that observation and mathematics need theoretical frameworks to truly advance atomic understanding.
     
  • [P] Nevertheless, observations and mathematical ideas remain fundamental to atomic theory development.
  • [E] Despite limitations, these tools work together to progressively refine scientific understanding.
  • [Ev] The progression from Rutherford to Bohr to de Broglie shows each model building on previous observations and mathematical frameworks, creating increasingly accurate atomic models.
  • [L] Therefore, the statement is valid as both elements are critical for advancing our understanding of the atom.

Answers could also reference:

  • Thomson’s discovery of the electron.
  • Further contributions from Davisson-Germer as well as Schrodinger and Heisenberg.
Show Worked Solution
  • [P] Observations have been critical in advancing atomic models.
  • [E] When experiments reveal unexpected results, scientists must revise their models to match reality.
  • [Ev] The Geiger-Marsden experiment showed alpha particles bouncing back from gold foil, contradicting the plum pudding model. This led Rutherford to propose a dense nucleus.
  • [L] This demonstrates how observations drive scientific progress in understanding atoms.
     
  • [P] Mathematical ideas provide essential frameworks for atomic models.
  • [E] Mathematics allows scientists to make precise predictions and test theories quantitatively.
  • [Ev] Bohr developed a mathematical model explaining the specific wavelengths emitted which relies on Rydberg’s equation: `(1)/(lambda)=R((1)/(n_(f)^2)-(1)/(n_(i)^2))`.
  • [Ev] This model was limited as it was only able to explain the hydrogen atom. De Broglie built upon it by postulating that electrons behaved as a wave, eventually describing the equation `\lambda=h/{mv}` which helped explain electron wave behaviour.
  • [L] These mathematical models transformed vague ideas into testable predictions about atomic structure.
     
  • [P] However, observations and mathematics alone have limitations.
  • [E] Models based purely on observations can miss underlying principles without theoretical insight.
  • [Ev] Bohr’s model perfectly matched hydrogen spectra but failed for other atoms because it lacked deeper quantum understanding.
  • [L] This shows that observation and mathematics need theoretical frameworks to truly advance atomic understanding.
     
  • [P] Nevertheless, observations and mathematical ideas remain fundamental to atomic theory development.
  • [E] Despite limitations, these tools work together to progressively refine scientific understanding.
  • [Ev] The progression from Rutherford to Bohr to de Broglie shows each model building on previous observations and mathematical frameworks, creating increasingly accurate atomic models.
  • [L] Therefore, the statement is valid as both elements are critical for advancing our understanding of the atom.

Answers could also reference:

  • Thomson’s discovery of the electron.
  • Further contributions from Davisson-Germer as well as Schrodinger and Heisenberg.

PHYSICS, M8 EQ-Bank 27

Explain how the analysis of quantitative observations contributed to the development of the concept that certain matter and energy are quantised.   (9 marks)

Show Answers Only

Experiments such as Millikan’s oil drop experiment and others testing the photoelectric effect have demonstrated that certain quantities of matter and energy are quantised which means they are multiples of some fundamental value.

Millikan’s Oil Drop Experiment

  • Millikan’s oil drop experiment was able to show that charge is quantised. 
  • Millikan levitated oil drops in an electric field by balancing the electric and gravitational forces on them. This allowed him to find the electric force acting on each oil drop, and using the mass of the oil drop he found its charge.
  • Analysing his results, he found that the charge on every oil drop was an integer multiple of `1.602 xx10^(-19) C`. This was determined to be the fundamental charge on an electron.
  • Further, with Thompson’s later discovery of the charge to mass ratio of an electron, its mass could be determined. 

Photoelectric Effect

  • Photoelectric effect experiments showed the quantum properties of light which seemingly contradicted the view of light as a wave.
  • It was found that there was a minimum frequency (energy) of light that would cause photoemission when it was incident upon a metal plate, and no photoemission occurred with light lower than this frequency, regardless of intensity.
  • As one photon would strike one electron on the metal surface, the electron would receive a discrete amount of energy from that photon determined by its frequency `E=hf`. If a photon didn’t have enough energy, an electron couldn’t be removed.
  • This experimental evidence changed the conceptual understanding of energy within physics and provided a basis for the quantisation of the energy of light. 

Other quantitative experiments that could be explored include:

  • Bohr’s analysis of emission spectra to demonstrate the existence of quantised energy levels in atoms.
  • Cathode ray experiments showing the particle nature of electrons.
  • Blackbody radiation experiments. 
Show Worked Solution

Experiments such as Millikan’s oil drop experiment and others testing the photoelectric effect have demonstrated that certain quantities of matter and energy are quantised which means they are multiples of some fundamental value.

Millikan’s Oil Drop Experiment

  • Millikan’s oil drop experiment was able to show that charge is quantised. 
  • Millikan levitated oil drops in an electric field by balancing the electric and gravitational forces on them. This allowed him to find the electric force acting on each oil drop, and using the mass of the oil drop he found its charge.
  • Analysing his results, he found that the charge on every oil drop was an integer multiple of `1.602 xx10^(-19) C`. This was determined to be the fundamental charge on an electron.
  • Further, with Thompson’s later discovery of the charge to mass ratio of an electron, its mass could be determined. 

Photoelectric Effect

  • Photoelectric effect experiments showed the quantum properties of light which seemingly contradicted the view of light as a wave.
  • It was found that there was a minimum frequency (energy) of light that would cause photoemission when it was incident upon a metal plate, and no photoemission occurred with light lower than this frequency, regardless of intensity.
  • As one photon would strike one electron on the metal surface, the electron would receive a discrete amount of energy from that photon determined by its frequency `E=hf`. If a photon didn’t have enough energy, an electron couldn’t be removed.
  • This experimental evidence changed the conceptual understanding of energy within physics and provided a basis for the quantisation of the energy of light. 

Other quantitative experiments that could be explored include:

  • Bohr’s analysis of emission spectra to demonstrate the existence of quantised energy levels in atoms.
  • Cathode ray experiments showing the particle nature of electrons.
  • Blackbody radiation experiments. 

PHYSICS, M8 EQ-Bank 22

Einstein's equation `E = mc^(2)`  is one of the most important equations in the history of physics.

Justify this statement.   (7 marks)

Show Answers Only

Position Statement

  • Einstein’s equation `E=mc^2` is among physics’ most important equations because it reveals mass-energy equivalence and explains fundamental processes from atomic to cosmic scales.

Nuclear Applications

  • The equation explains energy from nuclear fission and fusion processes. It shows how tiny amounts of matter convert to enormous energy in nuclear reactions.
  • The mass defect in atoms directly relates to binding energy through `E=mc^2` and this principle drives nuclear reactors that power cities and nuclear weapons.
  • These applications are responsible for revolutionising both energy production and global politics.

Cosmic and Particle Physics

  • Stars produce energy by converting mass to energy through fusion reactions. The Sun converts 4 million tonnes of mass to energy every second using this principle
  • Particle accelerators create new particles by converting kinetic energy into mass. This allows scientists to study fundamental matter structure and discover new particles.
  • Furthermore, mass dilation near light speed also follows from this mass-energy relationship.

Reinforcement

  • While other equations like Newton’s gravity law are important, they lack the broad applicability of `E=mc^2`.
  • No other single equation explains phenomena from subatomic particles to stellar processes.
  • The equation unified our understanding of matter and energy as different forms of the same thing.
  • This justifies calling `E=mc^2` one of history’s most important physics equations.
Show Worked Solution

Position Statement

  • Einstein’s equation `E=mc^2` is among physics’ most important equations because it reveals mass-energy equivalence and explains fundamental processes from atomic to cosmic scales.

Nuclear Applications

  • The equation explains energy from nuclear fission and fusion processes. It shows how tiny amounts of matter convert to enormous energy in nuclear reactions.
  • The mass defect in atoms directly relates to binding energy through `E=mc^2` and this principle drives nuclear reactors that power cities and nuclear weapons.
  • These applications are responsible for revolutionising both energy production and global politics.

Cosmic and Particle Physics

  • Stars produce energy by converting mass to energy through fusion reactions. The Sun converts 4 million tonnes of mass to energy every second using this principle
  • Particle accelerators create new particles by converting kinetic energy into mass. This allows scientists to study fundamental matter structure and discover new particles.
  • Furthermore, mass dilation near light speed also follows from this mass-energy relationship.

Reinforcement

  • While other equations like Newton’s gravity law are important, they lack the broad applicability of `E=mc^2`.
  • No other single equation explains phenomena from subatomic particles to stellar processes.
  • The equation unified our understanding of matter and energy as different forms of the same thing.
  • This justifies calling `E=mc^2` one of history’s most important physics equations.

PHYSICS, M8 EQ-Bank 23

A patient is given an injection containing `6.0 × 10^(-18)` kg of radioactive technetium-99m which has a half-life of 6 hours.

How much remains undecayed when a scan is taken 3 hours later?   (3 marks)

Show Answers Only

`4.2 xx10^(-18)\ text{kg}`

Show Worked Solution

Calculating the decay constant:

   `lambda=(ln2)/(t_((1)/(2)))=(ln2)/(6)=0.1155\ text{hour}^(-1)`

Amount remaining undecayed:

`N` `=N_(0)e^(-lambdat)`  
  `=6.0 xx10^(-18) xxe^(-0.1155 xx3)`  
  `=4.2429… xx10^(-18)`  
  `=4.2 xx10^(-18)\ text{kg}`  

PHYSICS, M8 EQ-Bank 17 MC

The graph shown plots isotopes in terms of their numbers of protons and neutrons. When an isotope undergoes nuclear decay, it will move to a different location on the graph. The movement can be represented with an arrow.
 

Which arrow would correctly describe beta negative `(beta^(-))` decay on the graph?
 

Show Answers Only

`A`

Show Worked Solution
  • In `beta^(-)` decay, a neutron is converted into a proton and an electron.
  • The number of neutrons decreases by one and the number of protons increase by one.
  • This corresponds to a movement downwards and to the right on the graph.

`=>A`

PHYSICS, M8 EQ-Bank 24

The table shows the quantum numbers of the four lowest states of the hydrogen atom, together with the energies of those states.

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex}\textit{Quantum number,} \ n \rule[-1ex]{0pt}{0pt}& \quad \textit{Energy}\ \text{(joules)} \quad  \\
\hline \rule{0pt}{2.5ex}1 \text { (ground state) } \rule[-1ex]{0pt}{0pt}& 0 \\
\hline \rule{0pt}{2.5ex}2 \rule[-1ex]{0pt}{0pt}& 1.63 \times 10^{-18} \\
\hline \rule{0pt}{2.5ex}3 \rule[-1ex]{0pt}{0pt}& 1.94 \times 10^{-18} \\
\hline \rule{0pt}{2.5ex}4 \rule[-1ex]{0pt}{0pt}& 2.04 \times 10^{-18} \\
\hline
\end{array}

Using appropriate calculations, explain a quantum transition that will absorb a photon of wavelength 102 nm?   (3 marks)

Show Answers Only

See Worked Solutions

Show Worked Solution

Calculating the energy of the photon:

\(E\) \(=h f\)  
  \(=\dfrac{h c}{\lambda}\)  
  \(=\dfrac{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^8\right)}{1.02 \times 10^{-7}}\)  
  \(=1.949 \times 10^{-18} \  \text{J}\)  
     
  •  When a photon with wavelength 102 nm strikes an electron in the ground state of the hydrogen atom, it will transfer its energy to the electron causing it to occupy a shell with greater energy. 
  • The electron absorbing the photon will therefore transition between shells one (ground state) and three, which is approximately the energy difference between these levels as shown in the table.

PHYSICS, M5 EQ-Bank 27

A toy car was placed facing outwards on a rotating turntable. The car was held in place by a force sensor connected to the centre of the turntable. The centre of mass of the car was 0.25 metres from the centre of the turntable. The reading from the force sensor was recorded at varying speeds of rotation. A stopwatch was used to time the rotation of the turntable. The linear velocity was calculated from the period of rotation. The graph shows the force on the car versus the square of the linear velocity of the car.
 


 

  1. Use the graph to determine the mass of the car.   (3 marks)
  1. Identify possible errors in the data and outline how to reduce their effects on the estimation of the mass of the car.   (4 marks)
Show Answers Only

a.   0.034 kg

b.   Errors and reduction strategies:

  • If the sensor does not produce accurate readings then systemic errors will result. 
  • Inaccurate readings can be minimised by calibrating the sensor. This can be done against a known force such as the force of gravity on a 1 kg mass.
  • If the stopwatch used to time rotations is manually operated, then error due to human inaccuracy and varying reaction times is introduced. Inaccurate calculations of the motion’s period (i.e. timing of one rotation) makes linear velocity estimations (`v=romega`) less accurate.
  • A strategy for mitigating this inaccuracy is by using the stopwatch to measure several rotations at a time and then dividing the result by the total number of rotations.
Show Worked Solution

a.   Graph passes through `(2,0)` and `(25, 3.1)`:

`text{Gradient}` `=(3.1-0)/(25-2)=0.135`  
`text{Gradient}` `=(F)/(v^2)=0.135`  

 
Since the car is undergoing uniform circular motion:

`F_(c)` `=(mv^2)/(r)`  
`(F_(c))/(v^2)` `=(m)/(r)`  
`0.135` `=(m)/(r)`  
`:. m` `=0.135 xx 0.25=0.03375=0.034\ text{kg}`  

 

b.   Errors and reduction strategies:

  • If the sensor does not produce accurate readings then systemic errors will result. 
  • Inaccurate readings can be minimised by calibrating the sensor. This can be done against a known force such as the force of gravity on a 1 kg mass.
  • If the stopwatch used to time rotations is manually operated, then error due to human inaccuracy and varying reaction times is introduced. Inaccurate calculations of the motion’s period (i.e. timing of one rotation) makes linear velocity estimations (`v=romega`) less accurate.
  • A strategy for mitigating this inaccuracy is by using the stopwatch to measure several rotations at a time and then dividing the result by the total number of rotations.

PHYSICS, M5 EQ-Bank 26

A baseball is hit with a velocity of 28 m s ¯1 at an angle of 30° to the horizontal at an initial height of 1.0 m above the plate. Ignore air resistance in your calculations.
 

  1. How long does it take the ball to return to the initial height above the ground?   (3 marks)
  1. The ball is hit directly towards a stationary outfielder who is 85 m from the plate. At the instant the ball is hit, the outfielder begins to run towards the plate with constant acceleration.
  2. What is the magnitude of her acceleration if she catches the ball when it is 0.50 m above the ground?   (4 marks)
Show Answers Only

a.   `t=2.9\ text{seconds}`

b.   `a=3.6  text{m s}^(-2)`

Show Worked Solution

a.   At initial height: `s_y=0`

Initial vertical velocity:  `u_y=28sin30°=14\ text{m s}^(-1)`

Vertical acceleration: `-9.8\ text{m s}^(-2)`

`s_(y)` `=u_(y)t+(1)/(2)a_(y)t^2`  
`0` `=14 t+(1)/(2)xx(-9.8)xxt^(2)`  
`0` `=14 t-4.9 xxt^(2)`  
`0` `=t(14-4.9 t)`  
`t` `=(14)/(4.9)=2.9\ text{seconds},\ \ (t>0)`  

 

b.   At  `s_y=0.50\ text{m}`:

`v_(y)^2` `=u_(y)^2+2a_(y)s_(y)`  
`v_(y)^2` `=14^(2)+2(-9.8)(-0.50)`  
`v_(y)` `=sqrt(205.8)~~14.3457\ text{m s}^(-1)`  

 
Using  `v_y=u_(y)+a_(y)t:`

`-14.3457` `=14-9.8t`  
`t` `=(14+14.3457)/(9.8)=2.892\ text{s}`  

 
Horizontal range travelled (`s_x`):

`s_(x)=u_(x)t=28 cos 30^{\circ} xx 2.892=70.127\ text{m}`
 

COMMENT: Note `v_y` and `u_y` are in opposite directions and have opposite signs.

  Distance fielder travelled `=85-70.127 = 14.873\ text{m}`

`s` `=ut+(1)/(2)at^(2)`  
`14.873` `=0+(1)/(2)xx a xx2.892^(2)`  
`:.a` `=(2 xx 14.873)/2.892^(2)`  
  `=3.6\ text{m s}^(-2)\ \ text{(to 1 d.p.)}`  

PHYSICS, M5 EQ-Bank 17 MC

Two identical masses are placed at points `P` and `Q`. The escape velocity and circular orbital velocity of the mass at point `P` are `v_{P_(esc)}` and `v_{P_{o rb}}`. The escape velocity and circular orbital velocity of the mass at point `Q` are `v_{Q_(e s c)}` and `v_{Q_{o rb}}`. The diagram is drawn to scale and `X` denotes the centre of Earth.
 

The velocity for a body in circular orbit is given by  `v_{o rb} = sqrt((GM)/r`.

What is the value of `(v_{Q_(e s c)})/v_{P_{o rb}}`?

  1. `0.5`
  2. `1`
  3. `sqrt(2)`
  4. `2`
Show Answers Only

`B`

Show Worked Solution
  • The escape velocity of an object is given by `v_(esc)=sqrt((2GM)/(r))`
  • As the diagram is to scale, it can be measured that the distance from `Q` to `X` is twice that from `P` to `X`.
  • Let  `r` = distance from `P` to `X`  and  `2r` = distance from `Q` to `X`:
  • `v_(Q_(esc))=sqrt((2GM)/(2r))=sqrt((GM)/(r))`
  • `v_(p_(text{orb}))=sqrt((GM)/(r))` 
  • Hence, `v_(Q_(esc))=v_(P_(text{orb}))` → `(v_(Q_(esc)))/(v_(p_(text{orb})))=1`

`=>B`

PHYSICS, M5 EQ-Bank 15 MC

The table shows data about the solar system.

\begin{array} {|l|c|c|}
\hline  \rule{0pt}{2.5ex} \quad\textit{Planet} \quad  & \textit{Average distance} & \textit{Period (days)} \\
\quad\textit{} \quad  \rule[-1ex]{0pt}{0pt}& \textit{from the sun (AU)} & \textit{} \\
\hline \rule{0pt}{2.5ex} \text{Venus} \rule[-1ex]{0pt}{0pt}& 0.72 & 224.7 \\
\hline \rule{0pt}{2.5ex} \text{Earth} \rule[-1ex]{0pt}{0pt}& 1 & 365 \\
\hline \rule{0pt}{2.5ex} \text{Mars} \rule[-1ex]{0pt}{0pt}& 1.54 & 686.2 \\
\hline \end{array}

What would be the period of another planet if it orbited the Sun at an average distance of 4.5 AU ?

  1. \(7.9 \times 10^2\) days
  2. \(1.5 \times 10^3\) days
  3. \(1.6 \times 10^3\) days
  4. \(6.6 \times 10^3\) days
Show Answers Only

\(B\)

Show Worked Solution
  • \(  \dfrac{r^3}{T^2}=\dfrac{G M}{4 \pi^2}\)
  • \( \text{Since all planets are orbiting the same central body with mass, \(M, G\) and \(4 \pi^2\) are constants.}\)
  • \( \text{The ratio \(\dfrac{r^3}{T^2}\) will be the same for all planets in the solar system.}\)

\(\begin{aligned}
\rule{0pt}{2.5ex} \dfrac{r^3}{T^2} \rule[-1ex]{0pt}{0pt}& =\dfrac{r_e^3}{T_e^2} \\
\rule{0pt}{2.5ex} \dfrac{4.5^3}{T^2} \rule[-1ex]{0pt}{0pt}& =\dfrac{1^3}{365^2} \\
\rule{0pt}{2.5ex} T^2 \rule[-1ex]{0pt}{0pt}& =\dfrac{365^2}{4.5^3} \\
\rule{0pt}{2.5ex} \therefore T \rule[-1ex]{0pt}{0pt}& =\sqrt{\frac{365^2}{4.5^3}} \\
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt}& =1462 \ \text{days} \\
\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt}& \approx 1.5 \times 10^3 \text {days}
\end{aligned}\)

\(\Rightarrow B\)

PHYSICS, M5 EQ-Bank 9 MC

A satellite is orbiting a planet at a fixed altitude.

Which row of the table correctly identifies the magnitude of the work done by the forces on the satellite and the reason for this being the case?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex}\textit{}& \textit{} \\
\textit{}\rule[-2ex]{0pt}{0pt}& \textit{} \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|l|}
\hline
\rule{0pt}{2.5ex}\textit{Magnitude of}&  \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\textit{Reason} \\
\textit{work done}\rule[-1ex]{0pt}{0pt}& \textit{} \\
\hline
\rule{0pt}{2.5ex}\text{Zero}\rule[-1ex]{0pt}{0pt}&\text{The net force on the satellite is zero.}\\
\hline
\rule{0pt}{2.5ex}\text{Zero}& \text{Gravity acts at 90 degrees to the direction of motion of the}\\
\text{}\rule[-1ex]{0pt}{0pt}& \text{satellite} \\
\hline
\rule{0pt}{2.5ex}\text{Greater than}& \text{The work done equals the kinetic energy of the satellite.} \\
\text{zero}\rule[-1ex]{0pt}{0pt}& \text{} \\
\hline
\rule{0pt}{2.5ex}\text{Greater than}& \text{The work done equals the gravitational force multiplied by} \\
\text{zero}\rule[-1ex]{0pt}{0pt}& \text{the length of the orbital path of the satellite.} \\
\hline
\end{array}
\end{align*}

Show Answers Only

`B`

Show Worked Solution
  • The work done by a force on an object is the product of the displacement of the object and the magnitude of the force in the direction of the displacement.
  • The satellite is orbiting the planet in uniform circular motion, so the force of gravity acts perpendicular to the direction of motion of the satellite.
  • So, zero work is done on the satellite.

`=>B`

Calculus, EXT1 C3 EQ-Bank 14

Find the particular solution to the differential equation  `(dy)/(dx)=(2y+1)(x-3)`  that passes through the point `(2,-1)`.  (4 marks)

Show Answers Only

`y=-1/2(e^((x-2)(x-4))+1)`

Show Worked Solution
`(dy)/(dx)` `=(2y+1)(x-3)`  
`dy/(2y+1)` `=x-3\ dx`  
`int 1/(2y+1)\ dy` `=int x-3\ dx`  
`1/2ln|2y+1|` `=x^2/2-3x+c`  

 
`text{Passes through}\ (2,-1):`

`1/2ln|-1|` `=2-6+c`  
`c` `=4`  

 

`1/2ln|2y+1|` `=x^2/2-3x+4`  
`ln|2y+1|` `=x^2-6x+8`  
`ln|2y+1|` `=(x-4)(x-2)`  
`2y+1` `=+-e^((x-2)(x-4))`  
`2y` `=-e^((x-2)(x-4))-1,\ \ (text{passes through}\ ( 2,-1))`  
`y` `=-1/2(e^((x-2)(x-4))+1)`  

Vectors, EXT2 V1 EQ-Bank 11

Point `O` is the circumcentre of triangle `ABC` which is the centre of the circle that passes through each of its vertices.

Point `P` is the centroid of triangle `ABC` where the bisectors of each angle within the triangle intersect.

Point `Q` is such that `vec(OQ)=3vec(OP)`.
 

Prove that  `vec(CQ) ⊥ vec(AB)`   (5 marks)

Show Answers Only

`text{See Worked Solution}` 

Show Worked Solution

`vec(OP)=(vec(OA)+vec(OB)+vec(OC))/3`

`vec(OQ)` `=3\ vec(OP)`  
  `=3((vec(OA)+vec(OB)+vec(OC))/3)`  
  `=vec(OA)+vec(OB)+vec(OC)`  
  `=((x_1+x_2+x_3),(y_1+y_2+y_3))`  

 

`vec(CQ)` `=vec(OQ)-vec(OC)`  
  `=((x_1+x_2+x_3),(y_1+y_2+y_3))-((x_3),(y_3))`  
  `=((x_1+x_2),(y_1+y_2))`  

 

`vec(AB)` `=vec(OB)-vec(OA)`  
  `=((x_2),(y_2))-((x_1),(y_1))`  
  `=((x_2-x_1),(y_2-y_1))`  

 

`text{Test}\ \ vec(CQ)*vec(AB)=0:`

`text{LHS}` `=((x_1+x_2),(y_1+y_2))((x_2-x_1),(y_2-y_1))`  
  `=(((x_1+x_2)(x_2-x_1)),((y_1+y_2)(y_2-y_1)))`  
  `=(x_2)^2-(x_1)^2+(y_2)^2-(y_1)^2`  
  `=(x_2)^2+(y_2)^2-((x_1)^2+(y_1)^2)`  
  `=abs(vec(OB))^2-abs(vec(OA))^2`  
  `=OB^2-OA^2\ \ \ text{(radii)}`  
  `=0`  
 
`:.vec(CQ) ⊥ vec(AB)\ \ text{… as required}`