Physics 0625 | Paper 4 Mock 01 | Theory (Extended)
Question 1 Report
Fig. 30.1 shows a simple pinhole camera made from a cardboard box. Light from a candle enters through a small pinhole at the front and forms an image on a translucent screen at the back. The candle is 20 cm tall and is placed 60 cm from the pinhole. The box is 15 cm long. A student uses this camera as part of a class project on the properties of light and image formation. She traces the image on the screen and measures its height to verify her calculation.
(a) State whether the image on the screen is upright or inverted. [1]
(b) Explain why the image is inverted. [1]
(c) Calculate the height of the image using similar triangles (image height / object height = image distance / object distance). [2]
(d) State and explain what would happen to the image if the pinhole were made larger. [2]
(e) State one similarity between the pinhole camera and the human eye. [1]
Marking Scheme
- (a) [1 mark]: The image is inverted.
- (b) [1 mark]: Light travels in straight lines through the pinhole, so rays from the top of the candle travel downward through the pinhole to the bottom of the screen, and rays from the bottom travel upward to the top of the screen.
- (c) [2 marks]: \(\dfrac{\text{image height}}{20} = \dfrac{15}{60}\) [1]. Image height = \(20 \times \dfrac{15}{60} = 5.0\) cm [1].
- (d) [2 marks]: The image becomes brighter because more light enters through the larger hole [1]. The image also becomes blurred (less sharp) because light from each point on the object can pass through a wider opening, spreading over a larger area on the screen [1].
- (e) [1 mark]: Any one valid similarity, e.g.: both form inverted images; both have a small aperture (pupil/pinhole); both detect light on a surface at the back (retina/screen).
Explanation
A pinhole camera works because light travels in straight lines. Each point on the object sends out light in all directions, but the tiny pinhole selects only a narrow cone of rays. A ray from the top of the candle passes through the pinhole and continues in a straight line to the bottom of the screen; a ray from the bottom of the candle ends up at the top. This crossing of rays produces an inverted image.
The image size is found using similar triangles. The triangle formed by the object and the pinhole is similar to the triangle formed by the image and the pinhole:
\(\dfrac{\text{image height}}{\text{object height}} = \dfrac{\text{image distance (box length)}}{\text{object distance}}\)
\(\dfrac{h_i}{20} = \dfrac{15}{60} = 0.25\)
\(h_i = 20 \times 0.25 = 5.0\) cm
Making the pinhole larger allows more light through (brighter image) but destroys sharpness. With a larger hole, multiple rays from the same point on the object can enter at slightly different angles, each hitting a different spot on the screen. The point becomes a disc of confusion, and the image blurs. The pinhole camera trades brightness for sharpness.
The human eye works on the same principle: light enters through a small aperture (the pupil) and forms an inverted image on the retina at the back. The eye also has a lens to focus the light more precisely, which a basic pinhole camera lacks.
Answer Details
Marking Scheme
- (a) [1 mark]: The image is inverted.
- (b) [1 mark]: Light travels in straight lines through the pinhole, so rays from the top of the candle travel downward through the pinhole to the bottom of the screen, and rays from the bottom travel upward to the top of the screen.
- (c) [2 marks]: \(\dfrac{\text{image height}}{20} = \dfrac{15}{60}\) [1]. Image height = \(20 \times \dfrac{15}{60} = 5.0\) cm [1].
- (d) [2 marks]: The image becomes brighter because more light enters through the larger hole [1]. The image also becomes blurred (less sharp) because light from each point on the object can pass through a wider opening, spreading over a larger area on the screen [1].
- (e) [1 mark]: Any one valid similarity, e.g.: both form inverted images; both have a small aperture (pupil/pinhole); both detect light on a surface at the back (retina/screen).
Explanation
A pinhole camera works because light travels in straight lines. Each point on the object sends out light in all directions, but the tiny pinhole selects only a narrow cone of rays. A ray from the top of the candle passes through the pinhole and continues in a straight line to the bottom of the screen; a ray from the bottom of the candle ends up at the top. This crossing of rays produces an inverted image.
The image size is found using similar triangles. The triangle formed by the object and the pinhole is similar to the triangle formed by the image and the pinhole:
\(\dfrac{\text{image height}}{\text{object height}} = \dfrac{\text{image distance (box length)}}{\text{object distance}}\)
\(\dfrac{h_i}{20} = \dfrac{15}{60} = 0.25\)
\(h_i = 20 \times 0.25 = 5.0\) cm
Making the pinhole larger allows more light through (brighter image) but destroys sharpness. With a larger hole, multiple rays from the same point on the object can enter at slightly different angles, each hitting a different spot on the screen. The point becomes a disc of confusion, and the image blurs. The pinhole camera trades brightness for sharpness.
The human eye works on the same principle: light enters through a small aperture (the pupil) and forms an inverted image on the retina at the back. The eye also has a lens to focus the light more precisely, which a basic pinhole camera lacks.
Question 2 Report
Radioactive materials are used in a number of everyday applications and in medicine. A hospital uses a radioactive isotope as a tracer to investigate the functioning of a patient's thyroid gland. A small quantity of iodine-131 is administered to the patient by injection. The isotope accumulates in the thyroid gland and its radiation can be detected externally. The doctor monitors the uptake of iodine-131 in the thyroid gland over a period of several days using a gamma camera positioned near the patient's neck. Iodine-131 has a half-life of 8.0 days and emits beta particles and gamma rays. The effective half-life inside the body is shorter than 8.0 days because the body also excretes the iodine naturally. The activity of the sample at the time of injection is 4.0 MBq.
(a) State what is meant by the term half-life of a radioactive isotope. [2]
(b) Explain why a gamma-emitting isotope is chosen for this application rather than one that emits only alpha particles. [2]
(c) The effective half-life inside the body is 5.0 days. Calculate the activity of the iodine-131 in the thyroid gland after 15 days. [3]
(a) Definition of half-life [2]
The half-life of a radioactive isotope is the time taken for the activity of the sample to decrease to half its original value. [1] Equivalently, it is the time taken for half the undecayed nuclei in a sample to decay. [1]
Both definitions are equivalent: as half the nuclei decay, the number of decays per second (activity) also halves.
(b) Why a gamma-emitting isotope is chosen [2]
Gamma rays are highly penetrating and can pass through body tissue to be detected by the gamma camera positioned outside the patient's neck. [1] Alpha particles would be absorbed within millimetres of tissue and would never reach the external detector. Additionally, alpha particles are highly ionising and would cause excessive damage to the surrounding cells. [1]
(c) Activity after 15 days [3]
The effective half-life inside the body is 5.0 days (shorter than the physical half-life of 8.0 days because the body also excretes the iodine).
Number of half-lives in 15 days:
\(n = \frac{15}{5.0} = 3\) [1]
After 3 half-lives, the activity is:
\(A = 4.0 \times \left(\frac{1}{2}\right)^3 = \frac{4.0}{8}\) [1]
\(A = \mathbf{0.50 \text{ MBq}}\) [1]
Step by step: 4.0 \(\rightarrow\) 2.0 \(\rightarrow\) 1.0 \(\rightarrow\) 0.50 MBq.
Answer Details
(a) Definition of half-life [2]
The half-life of a radioactive isotope is the time taken for the activity of the sample to decrease to half its original value. [1] Equivalently, it is the time taken for half the undecayed nuclei in a sample to decay. [1]
Both definitions are equivalent: as half the nuclei decay, the number of decays per second (activity) also halves.
(b) Why a gamma-emitting isotope is chosen [2]
Gamma rays are highly penetrating and can pass through body tissue to be detected by the gamma camera positioned outside the patient's neck. [1] Alpha particles would be absorbed within millimetres of tissue and would never reach the external detector. Additionally, alpha particles are highly ionising and would cause excessive damage to the surrounding cells. [1]
(c) Activity after 15 days [3]
The effective half-life inside the body is 5.0 days (shorter than the physical half-life of 8.0 days because the body also excretes the iodine).
Number of half-lives in 15 days:
\(n = \frac{15}{5.0} = 3\) [1]
After 3 half-lives, the activity is:
\(A = 4.0 \times \left(\frac{1}{2}\right)^3 = \frac{4.0}{8}\) [1]
\(A = \mathbf{0.50 \text{ MBq}}\) [1]
Step by step: 4.0 \(\rightarrow\) 2.0 \(\rightarrow\) 1.0 \(\rightarrow\) 0.50 MBq.
Question 3 Report
A student stands 170 m from a vertical cliff face on a calm day with no wind. She claps her hands once sharply and hears the echo from the cliff after a measured time of 1.0 s. Fig. 3.1 shows the arrangement viewed from above. The student repeats the experiment ten times and calculates the average echo time. The air temperature during the experiment is 18 °C. A second cliff face is located 500 m behind the student, but no second echo is clearly heard.
(a) Calculate the total distance travelled by the sound from the clap to the echo returning. [1]
(b) Calculate the speed of sound in air from these results. [2]
(c) Explain why the echo is quieter than the original clap. [1]
(d) Suggest why no clear second echo is heard from the cliff 500 m behind the student. [2]
Marking Scheme
- (a) [1] Total distance = 2 x 170 = 340 m
- (b) [1+1] speed = distance/time = 340/1.0 = 340 m/s
- (c) [1] Some sound energy is absorbed by the cliff and the sound spreads out over a larger area, so less energy reaches the student on return
- (d) [1+1] The round trip to the second cliff is 2 x 500 = 1000 m, giving an echo time of about 2.9 s; the sound has travelled much further so it is much quieter and may be too faint to hear
Explanation
(a) The sound must travel from the student to the cliff and back again. The total distance is therefore twice the distance to the cliff: \(2 \times 170 = 340\) m. This is the key principle behind echo-based distance measurement.
(b) Using \(\text{speed} = \frac{\text{distance}}{\text{time}}\): \(v = \frac{340}{1.0} = 340\) m/s. This is consistent with the accepted value for the speed of sound in air at around 18-20 degrees C. The echo method is a practical way to measure the speed of sound outdoors.
(c) The echo is quieter than the original clap for two reasons. First, sound waves spread out spherically from the source, so the intensity (energy per unit area) decreases with distance according to the inverse square law. The sound must travel 340 m total, and the returning wavefront has spread over a much larger area than the outgoing one at the point of clapping. Second, the cliff face absorbs some of the sound energy rather than reflecting all of it - some energy is converted to heat in the rock. Together these effects mean the returning echo carries significantly less energy per unit area than the original clap.
(d) The second cliff is 500 m behind the student, so the round trip is \(2 \times 500 = 1000\) m. At 340 m/s, the echo return time would be \(\frac{1000}{340} \approx 2.94\) s. Two factors prevent a clear second echo: (1) the much greater distance means the sound has spread over a far larger area and lost much more energy, making the echo very faint; (2) the long delay (nearly 3 seconds) means the echo arrives well after the first echo and any residual reverberation, and background noise may mask it. Additionally, the cliff behind the student requires the sound to travel away from the first cliff, meaning neither cliff reinforces the other.
Answer Details
Marking Scheme
- (a) [1] Total distance = 2 x 170 = 340 m
- (b) [1+1] speed = distance/time = 340/1.0 = 340 m/s
- (c) [1] Some sound energy is absorbed by the cliff and the sound spreads out over a larger area, so less energy reaches the student on return
- (d) [1+1] The round trip to the second cliff is 2 x 500 = 1000 m, giving an echo time of about 2.9 s; the sound has travelled much further so it is much quieter and may be too faint to hear
Explanation
(a) The sound must travel from the student to the cliff and back again. The total distance is therefore twice the distance to the cliff: \(2 \times 170 = 340\) m. This is the key principle behind echo-based distance measurement.
(b) Using \(\text{speed} = \frac{\text{distance}}{\text{time}}\): \(v = \frac{340}{1.0} = 340\) m/s. This is consistent with the accepted value for the speed of sound in air at around 18-20 degrees C. The echo method is a practical way to measure the speed of sound outdoors.
(c) The echo is quieter than the original clap for two reasons. First, sound waves spread out spherically from the source, so the intensity (energy per unit area) decreases with distance according to the inverse square law. The sound must travel 340 m total, and the returning wavefront has spread over a much larger area than the outgoing one at the point of clapping. Second, the cliff face absorbs some of the sound energy rather than reflecting all of it - some energy is converted to heat in the rock. Together these effects mean the returning echo carries significantly less energy per unit area than the original clap.
(d) The second cliff is 500 m behind the student, so the round trip is \(2 \times 500 = 1000\) m. At 340 m/s, the echo return time would be \(\frac{1000}{340} \approx 2.94\) s. Two factors prevent a clear second echo: (1) the much greater distance means the sound has spread over a far larger area and lost much more energy, making the echo very faint; (2) the long delay (nearly 3 seconds) means the echo arrives well after the first echo and any residual reverberation, and background noise may mask it. Additionally, the cliff behind the student requires the sound to travel away from the first cliff, meaning neither cliff reinforces the other.
Question 4 Report
A student performs a comprehensive investigation of gas behaviour. She uses the apparatus shown in Fig. 67.1 to study both Boyle's law and the pressure law. The apparatus consists of a sealed round-bottom flask of fixed volume connected to both a Bourdon pressure gauge and a gas syringe via a three-way tap. Parts labelled W and X must be identified. For the Boyle's law part, she uses the syringe alone (tap closed to flask) starting with 80 cm³ of air at 100 kPa and 20 °C. For the pressure law part, she uses the flask alone (tap closed to syringe) and heats it in a water bath from 20 °C to 80 °C while recording the pressure. The initial flask pressure is 100 kPa. She must describe both procedures, perform calculations, discuss sources of error, and suggest improvements. No gas leaks during either experiment.
(a) State Boyle's law. [1]
(b) Label the parts W and X of the apparatus. [2]
(c) Describe the procedure for the pressure law investigation using the flask. [2]
(d) Calculate the following. Show all working. (i) The syringe volume needed to give a pressure of 200 kPa (Boyle's law). (ii) The flask pressure at 80 °C (pressure law). [4]
(e) Explain why the experimental results might differ from the calculated theoretical values. [2]
(f) Suggest two improvements to the experimental procedures. [2]
Part (a) [1 mark]
For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume [1].
Mathematically: \(pV = \text{constant}\), or equivalently \(p \propto \frac{1}{V}\). Doubling the volume halves the pressure, because the same number of molecules now has twice the space and strikes the walls half as often per unit area.
Part (b) [2 marks]
- W is the Bourdon gauge (pressure gauge) [1]
- X is the plunger (piston) of the gas syringe [1]
The Bourdon gauge contains a curved metal tube that straightens under increased pressure, moving a pointer across a scale. The plunger slides within the syringe barrel to change the volume of the trapped gas.
Part (c) [2 marks]
Close the three-way tap to the syringe so that only the flask is connected to the Bourdon gauge [1]. Heat the water bath slowly, stir to ensure uniform temperature, and record the pressure reading at regular temperature intervals after allowing each reading to stabilise [1].
Stirring ensures all the water (and therefore the flask) reaches a uniform temperature. Waiting for the reading to stabilise ensures the gas inside the flask has reached thermal equilibrium with the water bath.
Part (d) [4 marks]
(i) Boyle's law calculation:
\(p_1 V_1 = p_2 V_2\) [1]
\(V_2 = \frac{p_1 V_1}{p_2} = \frac{100 \times 80}{200} = 40\) cm³ [1]
To double the pressure, the volume must be halved (from 80 cm³ to 40 cm³).
(ii) Pressure law calculation:
Convert to kelvin: \(T_1 = 20 + 273 = 293\) K, \(T_2 = 80 + 273 = 353\) K [1]
\(\frac{p_1}{T_1} = \frac{p_2}{T_2}\)
\(p_2 = \frac{p_1 \times T_2}{T_1} = \frac{100 \times 353}{293} = 120.5\) kPa (accept 120 kPa) [1]
Heating the gas by 60 °C increases the absolute temperature by about 20%, so the pressure also rises by about 20%. The pressure law requires absolute temperatures because pressure is proportional to the average kinetic energy of the particles, which is measured from absolute zero.
Part (e) [2 marks]
Compressing the gas may cause a slight temperature rise even if done slowly, affecting the Boyle's law results [1]. The gas may not behave perfectly as an ideal gas, there may be small gas leaks at the joints, or the gauge may have a zero error [1].
Part (f) [2 marks]
Any two valid improvements, for example: use a data logger for more precise and frequent readings [1]; repeat each measurement three times and calculate an average to reduce random errors [1].
Other acceptable improvements include insulating the syringe to maintain constant temperature, using a more precise digital pressure sensor, or allowing longer equilibration time between readings.
Answer Details
Part (a) [1 mark]
For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume [1].
Mathematically: \(pV = \text{constant}\), or equivalently \(p \propto \frac{1}{V}\). Doubling the volume halves the pressure, because the same number of molecules now has twice the space and strikes the walls half as often per unit area.
Part (b) [2 marks]
- W is the Bourdon gauge (pressure gauge) [1]
- X is the plunger (piston) of the gas syringe [1]
The Bourdon gauge contains a curved metal tube that straightens under increased pressure, moving a pointer across a scale. The plunger slides within the syringe barrel to change the volume of the trapped gas.
Part (c) [2 marks]
Close the three-way tap to the syringe so that only the flask is connected to the Bourdon gauge [1]. Heat the water bath slowly, stir to ensure uniform temperature, and record the pressure reading at regular temperature intervals after allowing each reading to stabilise [1].
Stirring ensures all the water (and therefore the flask) reaches a uniform temperature. Waiting for the reading to stabilise ensures the gas inside the flask has reached thermal equilibrium with the water bath.
Part (d) [4 marks]
(i) Boyle's law calculation:
\(p_1 V_1 = p_2 V_2\) [1]
\(V_2 = \frac{p_1 V_1}{p_2} = \frac{100 \times 80}{200} = 40\) cm³ [1]
To double the pressure, the volume must be halved (from 80 cm³ to 40 cm³).
(ii) Pressure law calculation:
Convert to kelvin: \(T_1 = 20 + 273 = 293\) K, \(T_2 = 80 + 273 = 353\) K [1]
\(\frac{p_1}{T_1} = \frac{p_2}{T_2}\)
\(p_2 = \frac{p_1 \times T_2}{T_1} = \frac{100 \times 353}{293} = 120.5\) kPa (accept 120 kPa) [1]
Heating the gas by 60 °C increases the absolute temperature by about 20%, so the pressure also rises by about 20%. The pressure law requires absolute temperatures because pressure is proportional to the average kinetic energy of the particles, which is measured from absolute zero.
Part (e) [2 marks]
Compressing the gas may cause a slight temperature rise even if done slowly, affecting the Boyle's law results [1]. The gas may not behave perfectly as an ideal gas, there may be small gas leaks at the joints, or the gauge may have a zero error [1].
Part (f) [2 marks]
Any two valid improvements, for example: use a data logger for more precise and frequent readings [1]; repeat each measurement three times and calculate an average to reduce random errors [1].
Other acceptable improvements include insulating the syringe to maintain constant temperature, using a more precise digital pressure sensor, or allowing longer equilibration time between readings.
Question 5 Report
Two students, P and Q, run along a straight track. Fig. 10.1 shows the distance-time graphs for both students. Student P starts at time t = 0 and runs at a constant speed. Student Q starts 5.0 s later and also runs at a constant speed, but faster than P. Both students run in the same direction from the same starting line. The track is 150 m long. A teacher uses two stopwatches to time each student independently and records the results on a whiteboard for the class to analyse.
(a) Calculate the speed of student P. [2]
(b) Calculate the speed of student Q. [2]
(c) Determine the time at which Q overtakes P. [2]
(d) State the distance from the start at which Q overtakes P. [1]
(a) Speed of student P
From the graph, student P starts at t = 0 and reaches 150 m at t = 30 s. Since P travels at constant speed, the distance-time graph is a straight line.
\(\text{speed of P} = \frac{\text{distance}}{\text{time}} = \frac{150}{30}\) [1]
= 5.0 m/s [1]
(b) Speed of student Q
Student Q starts at t = 5.0 s and reaches 150 m at t = 25 s. The time Q is running = 25 - 5 = 20 s.
\(\text{speed of Q} = \frac{150}{20}\) [1]
= 7.5 m/s [1]
(c) Time at which Q overtakes P
At the moment of overtaking, both students have covered the same distance from the start line.
- Distance covered by P at time t: \(d_P = 5.0t\)
- Distance covered by Q at time t (Q started at t = 5.0 s): \(d_Q = 7.5(t - 5)\)
Setting \(d_P = d_Q\):
\(5.0t = 7.5(t - 5)\)
\(5.0t = 7.5t - 37.5\)
\(37.5 = 2.5t\) [1]
\(t = 15\) s [1]
(d) Distance from the start at which Q overtakes P
\(d = 5.0 \times 15 = \) 75 m [1]
This can be confirmed from Q's equation: \(7.5 \times (15 - 5) = 7.5 \times 10 = 75\) m. Both answers agree, confirming the overtake point.
Answer Details
(a) Speed of student P
From the graph, student P starts at t = 0 and reaches 150 m at t = 30 s. Since P travels at constant speed, the distance-time graph is a straight line.
\(\text{speed of P} = \frac{\text{distance}}{\text{time}} = \frac{150}{30}\) [1]
= 5.0 m/s [1]
(b) Speed of student Q
Student Q starts at t = 5.0 s and reaches 150 m at t = 25 s. The time Q is running = 25 - 5 = 20 s.
\(\text{speed of Q} = \frac{150}{20}\) [1]
= 7.5 m/s [1]
(c) Time at which Q overtakes P
At the moment of overtaking, both students have covered the same distance from the start line.
- Distance covered by P at time t: \(d_P = 5.0t\)
- Distance covered by Q at time t (Q started at t = 5.0 s): \(d_Q = 7.5(t - 5)\)
Setting \(d_P = d_Q\):
\(5.0t = 7.5(t - 5)\)
\(5.0t = 7.5t - 37.5\)
\(37.5 = 2.5t\) [1]
\(t = 15\) s [1]
(d) Distance from the start at which Q overtakes P
\(d = 5.0 \times 15 = \) 75 m [1]
This can be confirmed from Q's equation: \(7.5 \times (15 - 5) = 7.5 \times 10 = 75\) m. Both answers agree, confirming the overtake point.
Question 6 Report
Fig. 27.1 shows three sealed containers, A, B and C, each holding the same substance (water) in a different state. In container A the molecules are arranged in a regular open pattern and vibrate in fixed positions. In container B the molecules are close together but move freely past each other. In container C the molecules are far apart and move rapidly in all directions. The mass of substance in container A is 460 g and its volume is 500 cm³. Container B holds the same mass of 460 g. Container C holds 0.46 g of the substance as a vapour in a volume of 760 cm³. The density of liquid water is 1.0 g/cm³. The student compares densities across the three states.
(a) Identify the state of matter in each container. [3]
(b) Describe the arrangement and movement of particles in container B. [2]
(c) Calculate the density of the substance in container A. [2]
(d) Calculate the density of the substance in container C. [2]
(e) Explain why the density in container B is greater than the density in container A, even though both hold the same mass. [2]
(a) Identify the state of matter in each container. [3]
- A is solid (ice): particles in a regular pattern, vibrating in fixed positions. [1]
- B is liquid (water): particles close together but free to move past each other. [1]
- C is gas (steam / water vapour): particles far apart, moving rapidly in all directions. [1]
(b) Describe the arrangement and movement of particles in container B. [2]
The particles are close together but not arranged in a regular pattern (they are disordered). [1]
They move randomly and can slide past each other, which is why liquids flow and take the shape of their container. [1]
(c) Calculate the density of the substance in container A. [2]
\(\rho = \frac{m}{V} = \frac{460}{500}\) [1]
\(\rho = 0.92 \text{ g/cm}^3\) [1]
This is the density of ice, which is less than that of liquid water (1.0 g/cm3). Ice is one of the rare substances whose solid form is less dense than its liquid form.
(d) Calculate the density of the substance in container C. [2]
\(\rho = \frac{m}{V} = \frac{0.46}{760}\) [1]
\(\rho = 6.1 \times 10^{-4} \text{ g/cm}^3\) (or 0.00061 g/cm3) [1]
This is about 1600 times less dense than liquid water, reflecting the very large spacing between gas particles.
(e) Explain why the density in container B is greater than the density in container A, even though both hold the same mass. [2]
In ice (A), the water molecules form an open crystalline structure with relatively large gaps between molecules. In liquid water (B), the molecules are closer together because the regular open structure has collapsed. [1]
Since both containers hold the same mass (460 g) but the liquid occupies a smaller volume (460 cm3 at 1.0 g/cm3) than the ice (500 cm3), the liquid has a higher density: the same mass in a smaller volume gives a higher density. [1]
Answer Details
(a) Identify the state of matter in each container. [3]
- A is solid (ice): particles in a regular pattern, vibrating in fixed positions. [1]
- B is liquid (water): particles close together but free to move past each other. [1]
- C is gas (steam / water vapour): particles far apart, moving rapidly in all directions. [1]
(b) Describe the arrangement and movement of particles in container B. [2]
The particles are close together but not arranged in a regular pattern (they are disordered). [1]
They move randomly and can slide past each other, which is why liquids flow and take the shape of their container. [1]
(c) Calculate the density of the substance in container A. [2]
\(\rho = \frac{m}{V} = \frac{460}{500}\) [1]
\(\rho = 0.92 \text{ g/cm}^3\) [1]
This is the density of ice, which is less than that of liquid water (1.0 g/cm3). Ice is one of the rare substances whose solid form is less dense than its liquid form.
(d) Calculate the density of the substance in container C. [2]
\(\rho = \frac{m}{V} = \frac{0.46}{760}\) [1]
\(\rho = 6.1 \times 10^{-4} \text{ g/cm}^3\) (or 0.00061 g/cm3) [1]
This is about 1600 times less dense than liquid water, reflecting the very large spacing between gas particles.
(e) Explain why the density in container B is greater than the density in container A, even though both hold the same mass. [2]
In ice (A), the water molecules form an open crystalline structure with relatively large gaps between molecules. In liquid water (B), the molecules are closer together because the regular open structure has collapsed. [1]
Since both containers hold the same mass (460 g) but the liquid occupies a smaller volume (460 cm3 at 1.0 g/cm3) than the ice (500 cm3), the liquid has a higher density: the same mass in a smaller volume gives a higher density. [1]
Question 7 Report
A student heats 0.30 kg of olive oil in a metal pan using a gas burner. The initial temperature is 22 °C and the desired cooking temperature is 180 °C. The specific heat capacity of olive oil is 1970 J/(kg °C). Fig. 57.1 shows the pan on the gas burner. The gas burner supplies energy at an effective rate of 800 W to the oil. Some energy is lost from the pan to the surroundings.
(a) Calculate the thermal energy required to heat the oil from 22 °C to 180 °C. [2]
(b) Calculate the minimum time needed to reach 180 °C at 800 W. [2]
(c) Explain why the actual time is longer than the calculated value. [1]
(d) Explain why cooking oil is chosen for frying instead of water for cooking at temperatures above 100 °C. [2]
(a) Thermal energy required to heat the oil [2]
Using the thermal energy equation with the specific heat capacity of olive oil:
\( E = mc\Delta\theta = 0.30 \times 1970 \times (180 - 22) \) [1]
\( E = 0.30 \times 1970 \times 158 = \mathbf{93\,378 \text{ J}} \) (accept 93 400 J or 93.4 kJ) [1]
Note that the specific heat capacity of oil (1970 J/(kg°C)) is less than half that of water (4200 J/(kg°C)), meaning oil requires less energy per kilogram per degree to heat up. However, the very large temperature rise (158°C) still requires a substantial amount of energy.
(b) Minimum time to reach 180°C [2]
The gas burner delivers energy at an effective rate of 800 W. The minimum time assumes all this energy goes into heating the oil:
\( t = \frac{E}{P} = \frac{93\,378}{800} \) [1]
\( t = \mathbf{116.7 \text{ s}} \) (accept 117 s or approximately 1.9 minutes) [1]
(c) Why the actual time is longer [1]
Some energy is lost to the surroundings from the pan by convection (hot air rising from the pan) and radiation (infrared emission from the hot surfaces), so the effective heating rate reaching the oil is less than 800 W. [1] The pan itself also absorbs some energy as its temperature rises.
(d) Why oil is used instead of water for frying above 100°C [2]
Cooking oil has a much higher boiling point than water (well above 180°C for olive oil), so it remains in the liquid state at typical frying temperatures. [1] Water, on the other hand, boils at 100°C at atmospheric pressure and would simply evaporate away long before reaching 180°C, making it impossible to use as a frying medium at high temperatures. [1] The high boiling point of oil allows it to transfer heat efficiently to food at the elevated temperatures needed for frying.
Answer Details
(a) Thermal energy required to heat the oil [2]
Using the thermal energy equation with the specific heat capacity of olive oil:
\( E = mc\Delta\theta = 0.30 \times 1970 \times (180 - 22) \) [1]
\( E = 0.30 \times 1970 \times 158 = \mathbf{93\,378 \text{ J}} \) (accept 93 400 J or 93.4 kJ) [1]
Note that the specific heat capacity of oil (1970 J/(kg°C)) is less than half that of water (4200 J/(kg°C)), meaning oil requires less energy per kilogram per degree to heat up. However, the very large temperature rise (158°C) still requires a substantial amount of energy.
(b) Minimum time to reach 180°C [2]
The gas burner delivers energy at an effective rate of 800 W. The minimum time assumes all this energy goes into heating the oil:
\( t = \frac{E}{P} = \frac{93\,378}{800} \) [1]
\( t = \mathbf{116.7 \text{ s}} \) (accept 117 s or approximately 1.9 minutes) [1]
(c) Why the actual time is longer [1]
Some energy is lost to the surroundings from the pan by convection (hot air rising from the pan) and radiation (infrared emission from the hot surfaces), so the effective heating rate reaching the oil is less than 800 W. [1] The pan itself also absorbs some energy as its temperature rises.
(d) Why oil is used instead of water for frying above 100°C [2]
Cooking oil has a much higher boiling point than water (well above 180°C for olive oil), so it remains in the liquid state at typical frying temperatures. [1] Water, on the other hand, boils at 100°C at atmospheric pressure and would simply evaporate away long before reaching 180°C, making it impossible to use as a frying medium at high temperatures. [1] The high boiling point of oil allows it to transfer heat efficiently to food at the elevated temperatures needed for frying.
Question 8 Report
A student tests how well four different surfaces absorb infrared radiation. She uses four identical aluminium plates, each coated with a different surface finish. Each plate has the same area and thickness, and a temperature sensor is taped to the back. An identical radiant heater is placed 20 cm from each plate in turn. The starting temperature is 20 °C. The heater is switched on for exactly 4 minutes and the final temperature is recorded. The results are shown in the table below.
| Surface finish | Starting temp / °C | Final temp / °C | Temp rise / °C |
|---|---|---|---|
| Matt black | 20 | 39 | 19 |
| Shiny black | 20 | 33 | 13 |
| Matt white | 20 | 29 | 9 |
| Polished silver | 20 | 23 | 3 |
(a) State which surface is the best absorber of infrared radiation. [1]
(b) Explain why the matt black surface shows the largest temperature rise. [2]
(c) State which surface is the best reflector of infrared radiation. Explain your reasoning. [2]
(d) Suggest why the student uses plates that are all the same size, thickness and metal. [1]
(e) State the relationship between a surface that is a good absorber of infrared radiation and a good emitter. [1]
Marking Scheme
- (a) [1] Matt black.
- (b) [1] Matt black surfaces absorb the largest proportion of the infrared radiation that falls on them. [1] More energy is absorbed per second, so the temperature rises faster and reaches a higher value in the same time.
- (c) [1] Polished silver. [1] It has the smallest temperature rise, meaning it absorbed the least radiation; the radiation not absorbed was reflected.
- (d) [1] To ensure a fair test: the only variable that changes is the surface finish, so any difference in temperature rise must be due to the absorbing properties of the surface.
- (e) [1] A good absorber of infrared radiation is also a good emitter of infrared radiation.
Explanation
(a) The matt black plate shows the greatest temperature rise (19 °C), which means it absorbed the most infrared radiation in the 4-minute heating period.
(b) When infrared radiation hits a matt black surface, almost all of it is absorbed and converted to internal energy of the plate. A larger fraction of incident energy absorbed per second means the plate's temperature climbs more quickly. The matt texture scatters any reflection rather than bouncing it away cleanly, and the black colour absorbs across all visible and infrared wavelengths.
(c) The polished silver plate rose only 3 °C. Since every plate received the same amount of radiation, the one that heated least must have reflected most of the radiation away. A polished (smooth, shiny) silver surface is an excellent reflector of infrared radiation.
(d) This is a controlled experiment. Variables such as plate material, area, thickness, starting temperature, heater distance, and heating time are all kept the same. The only factor that changes is the surface finish (the independent variable). This ensures the temperature rise (dependent variable) can be attributed solely to differences in absorption properties rather than any other factor.
(e) This is a fundamental principle of thermal radiation. A surface that is good at absorbing a particular wavelength of radiation is equally good at emitting that wavelength when hot. Matt black is the best absorber and the best emitter. Polished silver is the worst absorber and the worst emitter. This symmetry arises because absorption and emission involve the same physical interaction between radiation and the surface material.
Answer Details
Marking Scheme
- (a) [1] Matt black.
- (b) [1] Matt black surfaces absorb the largest proportion of the infrared radiation that falls on them. [1] More energy is absorbed per second, so the temperature rises faster and reaches a higher value in the same time.
- (c) [1] Polished silver. [1] It has the smallest temperature rise, meaning it absorbed the least radiation; the radiation not absorbed was reflected.
- (d) [1] To ensure a fair test: the only variable that changes is the surface finish, so any difference in temperature rise must be due to the absorbing properties of the surface.
- (e) [1] A good absorber of infrared radiation is also a good emitter of infrared radiation.
Explanation
(a) The matt black plate shows the greatest temperature rise (19 °C), which means it absorbed the most infrared radiation in the 4-minute heating period.
(b) When infrared radiation hits a matt black surface, almost all of it is absorbed and converted to internal energy of the plate. A larger fraction of incident energy absorbed per second means the plate's temperature climbs more quickly. The matt texture scatters any reflection rather than bouncing it away cleanly, and the black colour absorbs across all visible and infrared wavelengths.
(c) The polished silver plate rose only 3 °C. Since every plate received the same amount of radiation, the one that heated least must have reflected most of the radiation away. A polished (smooth, shiny) silver surface is an excellent reflector of infrared radiation.
(d) This is a controlled experiment. Variables such as plate material, area, thickness, starting temperature, heater distance, and heating time are all kept the same. The only factor that changes is the surface finish (the independent variable). This ensures the temperature rise (dependent variable) can be attributed solely to differences in absorption properties rather than any other factor.
(e) This is a fundamental principle of thermal radiation. A surface that is good at absorbing a particular wavelength of radiation is equally good at emitting that wavelength when hot. Matt black is the best absorber and the best emitter. Polished silver is the worst absorber and the worst emitter. This symmetry arises because absorption and emission involve the same physical interaction between radiation and the surface material.
Question 9 Report
A student is asked to determine the half-life of a radioactive source in a school laboratory. The apparatus available includes a sealed radioactive source stored in a lead-lined box, a Geiger-Müller tube, a counter (scaler), a timer, and tongs. Fig. 50.1 shows the experimental setup. The student must first measure the background count rate before bringing the source near the detector. She must then place the source at a fixed distance from the tube and record the count rate at regular time intervals. The background count rate must be subtracted from each reading to give the corrected count rate. The student then uses the corrected readings to determine the half-life. The teacher provides the student with a checklist of safety precautions to follow throughout.
(a) Describe how the student should measure the background count rate. [2]
(b) Describe the procedure the student should follow to collect data for determining the half-life. [3]
(c) Explain how the student uses the corrected count rate data to determine the half-life. [2]
(d) State two safety precautions the student should follow during this experiment. [2]
(a) Measuring the background count rate [2]
First, remove all radioactive sources from the area (or ensure the source remains sealed in its lead-lined box) so that only natural background radiation reaches the detector [1]. Then record the total count on the counter over a measured time period (for example, 5 minutes), and calculate the count rate in counts per minute. Repeat this measurement several times and take an average to reduce the effect of random variation in the background count [1].
(b) Procedure for collecting half-life data [3]
- Using tongs, place the sealed radioactive source at a fixed distance from the GM tube [1].
- Start the timer and record the count rate at regular time intervals (for example, every 30 seconds or every minute) [1].
- Continue recording readings for at least four or five half-lives to obtain enough data points for an accurate determination. This gives a clear exponential decay curve [1].
Keep the distance between source and detector constant throughout, and do not move the apparatus during the experiment.
(c) Using the data to find the half-life [2]
Subtract the background count rate from each measured count rate to obtain the corrected count rate for each time interval [1]. Then find the time taken for the corrected count rate to fall to exactly half its initial value. This time interval is the half-life. Alternatively, plot the corrected count rate against time and read off the time for the value to halve [1]. Repeating this for several halving intervals (e.g., from the initial value to half, then from half to a quarter) and averaging gives a more reliable result.
(d) Two safety precautions [2]
- Handle the source with tongs (never bare hands) to keep it at a distance from the body and reduce the radiation dose to the hands [1].
- Return the source to its lead-lined container immediately after the experiment is finished, and minimise the total time spent near the source [1].
Answer Details
(a) Measuring the background count rate [2]
First, remove all radioactive sources from the area (or ensure the source remains sealed in its lead-lined box) so that only natural background radiation reaches the detector [1]. Then record the total count on the counter over a measured time period (for example, 5 minutes), and calculate the count rate in counts per minute. Repeat this measurement several times and take an average to reduce the effect of random variation in the background count [1].
(b) Procedure for collecting half-life data [3]
- Using tongs, place the sealed radioactive source at a fixed distance from the GM tube [1].
- Start the timer and record the count rate at regular time intervals (for example, every 30 seconds or every minute) [1].
- Continue recording readings for at least four or five half-lives to obtain enough data points for an accurate determination. This gives a clear exponential decay curve [1].
Keep the distance between source and detector constant throughout, and do not move the apparatus during the experiment.
(c) Using the data to find the half-life [2]
Subtract the background count rate from each measured count rate to obtain the corrected count rate for each time interval [1]. Then find the time taken for the corrected count rate to fall to exactly half its initial value. This time interval is the half-life. Alternatively, plot the corrected count rate against time and read off the time for the value to halve [1]. Repeating this for several halving intervals (e.g., from the initial value to half, then from half to a quarter) and averaging gives a more reliable result.
(d) Two safety precautions [2]
- Handle the source with tongs (never bare hands) to keep it at a distance from the body and reduce the radiation dose to the hands [1].
- Return the source to its lead-lined container immediately after the experiment is finished, and minimise the total time spent near the source [1].
Question 10 Report
A rechargeable lithium-ion cell does 540 J of electrical work in moving 100 C of charge around a complete circuit. A student wants to determine the electromotive force of the cell using the definition that relates e.m.f. to work done and charge. The cell is connected to a fixed resistor of 5.4 Ω in a simple series circuit. Fig. 5.1 shows the circuit. The cell has negligible internal resistance and the connecting wires have negligible resistance.
(a) State the equation that defines electromotive force in terms of work done and charge. [1]
(b) Calculate the e.m.f. of the cell. [2]
(c) Calculate the current flowing through the resistor. [2]
(d) State the unit of e.m.f. [1]
Marking Scheme
- (a) e.m.f. = W / Q (or: e.m.f. = work done per unit charge) [1]
- (b) e.m.f. = 540 / 100 [1]; e.m.f. = 5.4 V [1]
- (c) \(I = V/R = 5.4 / 5.4\) [1]; \(I = 1.0\) A [1]
- (d) volt (V) [1]
Explanation
(a) Electromotive force (e.m.f.) is defined as the work done by a source of electrical energy per unit charge that passes through it:
\(\text{e.m.f.} = \frac{W}{Q}\)
where \(W\) is the work done (energy transferred) in joules and \(Q\) is the charge in coulombs. Although it has "force" in its name, e.m.f. is actually measured in volts and represents an energy-per-charge quantity, not a force.
(b) Substituting the given values:
\(\text{e.m.f.} = \frac{540 \text{ J}}{100 \text{ C}} = 5.4\) V
This means the cell does 5.4 J of work on every coulomb of charge that passes through it.
(c) Since the cell has negligible internal resistance, the terminal voltage equals the e.m.f. (5.4 V). All this voltage appears across the external resistor. Applying Ohm's law:
\(I = \frac{V}{R} = \frac{5.4}{5.4} = 1.0\) A
The numerical coincidence that \(V = R\) gives \(I = 1.0\) A is simply because the resistance was chosen to equal the e.m.f. value.
(d) The unit of e.m.f. is the volt (V), the same as for potential difference. One volt equals one joule per coulomb (1 V = 1 J/C).
Answer Details
Marking Scheme
- (a) e.m.f. = W / Q (or: e.m.f. = work done per unit charge) [1]
- (b) e.m.f. = 540 / 100 [1]; e.m.f. = 5.4 V [1]
- (c) \(I = V/R = 5.4 / 5.4\) [1]; \(I = 1.0\) A [1]
- (d) volt (V) [1]
Explanation
(a) Electromotive force (e.m.f.) is defined as the work done by a source of electrical energy per unit charge that passes through it:
\(\text{e.m.f.} = \frac{W}{Q}\)
where \(W\) is the work done (energy transferred) in joules and \(Q\) is the charge in coulombs. Although it has "force" in its name, e.m.f. is actually measured in volts and represents an energy-per-charge quantity, not a force.
(b) Substituting the given values:
\(\text{e.m.f.} = \frac{540 \text{ J}}{100 \text{ C}} = 5.4\) V
This means the cell does 5.4 J of work on every coulomb of charge that passes through it.
(c) Since the cell has negligible internal resistance, the terminal voltage equals the e.m.f. (5.4 V). All this voltage appears across the external resistor. Applying Ohm's law:
\(I = \frac{V}{R} = \frac{5.4}{5.4} = 1.0\) A
The numerical coincidence that \(V = R\) gives \(I = 1.0\) A is simply because the resistance was chosen to equal the e.m.f. value.
(d) The unit of e.m.f. is the volt (V), the same as for potential difference. One volt equals one joule per coulomb (1 V = 1 J/C).
Question 11 Report
Fig. 44.1 shows an aluminium ring placed loosely over the top of a vertical solenoid. The solenoid is connected to a d.c. power supply through a switch. When the switch is closed rapidly, the ring jumps upward off the solenoid. A student repeats the experiment with different ring materials and observes that a copper ring also jumps but a plastic ring does not move. The solenoid has 600 turns and the power supply provides 12 V. When the switch is closed, the current rises rapidly from zero to its maximum value. The aluminium ring has a mass of 5.0 g. The solenoid is clamped vertically to a heavy stand.
(a) Explain why the aluminium ring jumps upward when the switch is closed. Use Lenz's law in your explanation. [4]
(b) Explain why the plastic ring does not move. [1]
(c) State the effect on the height the ring jumps if the power supply voltage is increased. [1]
(d) State what happens to the ring if the switch is held closed for a long time and then opened suddenly. [2]
(a) This phenomenon is explained in four steps using electromagnetic induction and Lenz's law:
- When the switch is closed, the current in the solenoid rises rapidly from zero to its maximum value, creating a rapidly increasing magnetic field through and around the solenoid
- The changing magnetic field passes through the aluminium ring, which is a conductor. By Faraday's law, this changing flux induces an e.m.f. and hence a current in the ring
- By Lenz's law, the induced current must flow in a direction that opposes the change causing it. The change is an increasing upward magnetic field (assuming north pole at top), so the ring's induced current creates its own magnetic field that opposes the increase
- The ring's magnetic field is directed opposite to the solenoid's field, making the ring behave like a magnet with the same pole facing the solenoid's top. Like poles repel, so the ring experiences an upward repulsive force that launches it off the solenoid
(b) Plastic is an electrical insulator. No current can be induced in an insulator because it has no free electrons to carry a current. Without an induced current, there is no magnetic field produced by the ring, so there is no magnetic force and the ring does not move.
(c) If the power supply voltage is increased, the current in the solenoid rises to a larger value and does so more rapidly. This produces a greater rate of change of magnetic field, which induces a larger current in the ring and a stronger repulsive force. The ring therefore jumps higher.
(d) When the switch is opened suddenly after being held closed for a long time:
- The current in the solenoid drops rapidly to zero, causing the magnetic field to decrease rapidly
- This decreasing field again induces a current in the ring (by Faraday's law), but now Lenz's law dictates that the ring's field opposes the decrease, so the ring is attracted downward towards the solenoid rather than repelled upward
Answer Details
(a) This phenomenon is explained in four steps using electromagnetic induction and Lenz's law:
- When the switch is closed, the current in the solenoid rises rapidly from zero to its maximum value, creating a rapidly increasing magnetic field through and around the solenoid
- The changing magnetic field passes through the aluminium ring, which is a conductor. By Faraday's law, this changing flux induces an e.m.f. and hence a current in the ring
- By Lenz's law, the induced current must flow in a direction that opposes the change causing it. The change is an increasing upward magnetic field (assuming north pole at top), so the ring's induced current creates its own magnetic field that opposes the increase
- The ring's magnetic field is directed opposite to the solenoid's field, making the ring behave like a magnet with the same pole facing the solenoid's top. Like poles repel, so the ring experiences an upward repulsive force that launches it off the solenoid
(b) Plastic is an electrical insulator. No current can be induced in an insulator because it has no free electrons to carry a current. Without an induced current, there is no magnetic field produced by the ring, so there is no magnetic force and the ring does not move.
(c) If the power supply voltage is increased, the current in the solenoid rises to a larger value and does so more rapidly. This produces a greater rate of change of magnetic field, which induces a larger current in the ring and a stronger repulsive force. The ring therefore jumps higher.
(d) When the switch is opened suddenly after being held closed for a long time:
- The current in the solenoid drops rapidly to zero, causing the magnetic field to decrease rapidly
- This decreasing field again induces a current in the ring (by Faraday's law), but now Lenz's law dictates that the ring's field opposes the decrease, so the ring is attracted downward towards the solenoid rather than repelled upward
