An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V3) Cambridge International A Level Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
Paper 13
Answer forty multiple-choice questions on Core Physics.
40 Question · 40 marks
Question 1 · multipleChoice
1 marks
A student wants to find the volume of a single small metal screw. She pours water into a measuring cylinder until the water level is at \(30\text{ cm}^{3}\). She then gently lowers 4 identical metal screws into the water, and the new water level is \(54\text{ cm}^{3}\). What is the volume of one screw?
A.\(6.0\text{ cm}^{3}\)
B.\(13.5\text{ cm}^{3}\)
C.\(24.0\text{ cm}^{3}\)
D.\(84\text{ cm}^{3}\)
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Worked solution
First, find the total volume of all 4 screws by calculating the change in water level: \(54\text{ cm}^{3} - 30\text{ cm}^{3} = 24\text{ cm}^{3}\). Next, find the volume of a single screw by dividing this total volume by the number of screws: \(24\text{ cm}^{3} / 4 = 6.0\text{ cm}^{3}\).
Marking scheme
1 mark for correct option A. (Calculation: 54 - 30 = 24, then 24 / 4 = 6.0)
Question 2 · multipleChoice
1 marks
A beaker containing water is heated from below at one side. A small piece of colored crystal is placed at the bottom to show the movement of the liquid. The colored water is observed to rise directly above the flame. What causes this upward movement?
A.The heated water contracts and its density increases.
B.The heated water contracts and its density decreases.
C.The heated water expands and its density increases.
D.The heated water expands and its density decreases.
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Worked solution
When water is heated, the thermal energy causes the molecules to move faster and push further apart, causing the water to expand. This expansion increases the volume for the same mass, which decreases its density. The less dense hot water then rises above the cooler, denser surrounding water.
Marking scheme
1 mark for correct option D. (Identifying that heating causes water to expand and its density to decrease)
Question 3 · multipleChoice
1 marks
An electric current of \(0.50\text{ A}\) flows through a filament lamp for a time of \(3.0\text{ minutes}\). What is the total charge that passes through the lamp during this time?
A.\(1.5\text{ C}\)
B.\(6.0\text{ C}\)
C.\(90\text{ C}\)
D.\(360\text{ C}\)
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Worked solution
First, convert the time from minutes to seconds: \(t = 3.0\text{ minutes} = 3.0 \times 60\text{ s} = 180\text{ s}\). Use the formula for charge: \(Q = I \times t\). Substitute the values: \(Q = 0.50\text{ A} \times 180\text{ s} = 90\text{ C}\).
Marking scheme
1 mark for correct option C. (Converting 3 minutes to 180 seconds and using Q = I * t to find 90 C)
Question 4 · multipleChoice
1 marks
A student wants to determine the density of an irregular metal object of mass 180 g. She pours 100 cm³ of water into a measuring cylinder and then lowers the object into the water. The object is completely submerged and the new reading on the measuring cylinder is 145 cm³. She notices that an air bubble of volume 5 cm³ is trapped underneath the object. What is the actual density of the metal of the object?
A.\(3.6\text{ g/cm}^3\)
B.\(4.0\text{ g/cm}^3\)
C.\(4.5\text{ g/cm}^3\)
D.\(5.0\text{ g/cm}^3\)
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Worked solution
1. Find the volume of the water displaced by both the object and the trapped air bubble: \(V_{\text{total}} = 145\text{ cm}^3 - 100\text{ cm}^3 = 45\text{ cm}^3\). 2. Subtract the volume of the trapped air bubble to find the volume of the object itself: \(V_{\text{object}} = 45\text{ cm}^3 - 5\text{ cm}^3 = 40\text{ cm}^3\). 3. Calculate the density of the metal: \(\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{180\text{ g}}{40\text{ cm}^3} = 4.5\text{ g/cm}^3\). Therefore, the correct option is C.
Marking scheme
[1 mark] for correct calculation of volume (40 cm³) and density (4.5 g/cm³). [0 marks] for incorrect options.
Question 5 · multipleChoice
1 marks
A ray of light in a semi-circular glass block is directed towards the centre of the flat face. The angle of incidence at the flat face is \(35^\circ\). The critical angle for this glass-air boundary is \(42^\circ\). What happens to the ray when it reaches the flat face?
A.It is totally internally reflected back into the glass, with an angle of reflection of \(35^\circ\).
B.It is totally internally reflected back into the glass, with an angle of reflection of \(55^\circ\).
C.It is partially reflected at \(35^\circ\) and partially refracted into the air, bending away from the normal.
D.It is partially reflected at \(35^\circ\) and partially refracted into the air, bending towards the normal.
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Worked solution
1. Compare the angle of incidence (\(i = 35^\circ\)) with the critical angle (\(c = 42^\circ\)). Since \(i < c\), total internal reflection does not occur. 2. Since light is traveling from a more optically dense medium (glass) to a less optically dense medium (air), it refracts (bends) away from the normal. 3. At the same time, some light is partially reflected back into the glass. The angle of reflection equals the angle of incidence, which is \(35^\circ\). Thus, the ray is partially reflected at \(35^\circ\) and partially refracted into the air, bending away from the normal.
Marking scheme
[1 mark] for identifying that partial reflection at 35° and refraction bending away from the normal occurs because the angle of incidence is less than the critical angle. [0 marks] for incorrect options.
Question 6 · multipleChoice
1 marks
A cylindrical metal wire of length \(L\) and cross-sectional area \(A\) has a resistance \(R\). A second wire is made of the same metal. It has a length of \(2L\) and a diameter that is twice the diameter of the first wire. What is the resistance of the second wire?
A.\(0.5 R\)
B.\(1.0 R\)
C.\(2.0 R\)
D.\(4.0 R\)
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Worked solution
1. The resistance of a wire is given by the formula: \(R = \rho \frac{\text{length}}{\text{area}}\). 2. For the first wire: \(R = \rho \frac{L}{A}\). Since the cross-section is circular, the area is related to the diameter \(d\) by: \(A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}\). 3. For the second wire: The length is \(2L\). The diameter is \(2d\), so its cross-sectional area \(A'\) is: \(A' = \pi \left(\frac{2d}{2}\right)^2 = \pi d^2 = 4A\). 4. Substituting these into the resistance formula for the second wire: \(R' = \rho \frac{2L}{4A} = 0.5 \left(\rho \frac{L}{A}\right) = 0.5 R\). Therefore, the resistance of the second wire is \(0.5 R\).
Marking scheme
[1 mark] for correctly calculating that doubling length and doubling diameter yields a resistance of 0.5R. [0 marks] for incorrect options.
Question 7 · multipleChoice
1 marks
A student wants to determine the density of steel. She uses 20 identical steel ball bearings. She measures the total mass of the 20 ball bearings on a digital balance as 160 g. Next, she pours 50 cm\(^{3}\) of water into a measuring cylinder, then gently lowers all 20 ball bearings into the water. The new water level in the measuring cylinder is 70 cm\(^{3}\). What is the density of the steel?
A.0.125 g/cm\(^{3}\)
B.2.29 g/cm\(^{3}\)
C.3.20 g/cm\(^{3}\)
D.8.00 g/cm\(^{3}\)
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Worked solution
First, calculate the volume of the 20 ball bearings by finding the change in the water level in the measuring cylinder: \(\text{Volume} = 70\text{ cm}^3 - 50\text{ cm}^3 = 20\text{ cm}^3\). Next, calculate the density using the density formula: \(\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{160\text{ g}}{20\text{ cm}^3} = 8.0\text{ g/cm}^3\). Therefore, the correct option is D.
Marking scheme
Award 1 mark for the correct answer D. (1 mark for calculating the volume as 20 cm\(^{3}\) and correctly applying the density formula \(\rho = m/V\) to get 8.0 g/cm\(^{3}\))
Question 8 · multipleChoice
1 marks
An electric motor lifts a crate of mass 45 kg vertically upwards through a height of 8.0 m. This process takes 6.0 s. The acceleration of free fall \(g\) is 10 m/s\(^{2}\). What is the average useful power output of the motor?
A.60 W
B.360 W
C.600 W
D.3600 W
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Worked solution
First, calculate the work done against gravity: \(\text{Work Done} = m \times g \times h = 45\text{ kg} \times 10\text{ m/s}^2 \times 8.0\text{ m} = 3600\text{ J}\). Next, calculate the power by dividing the work done by the time taken: \(\text{Power} = \frac{\text{Work Done}}{\text{time}} = \frac{3600\text{ J}}{6.0\text{ s}} = 600\text{ W}\). Therefore, the correct option is C.
Marking scheme
Award 1 mark for the correct answer C. (1 mark for calculating the work done as 3600 J and dividing it by 6.0 s to obtain 600 W)
Question 9 · multipleChoice
1 marks
Which statement correctly describes a difference between evaporation and boiling of a liquid?
A.Boiling occurs at any temperature, whereas evaporation only occurs at one specific temperature.
B.Evaporation occurs only at the surface of the liquid, whereas boiling occurs throughout the liquid.
C.During evaporation, bubbles of gas form inside the liquid, whereas no bubbles form during boiling.
D.During boiling, the temperature of the liquid increases, whereas during evaporation the temperature remains constant.
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Worked solution
Evaporation is a process that occurs only at the surface of the liquid and at any temperature below the boiling point. In contrast, boiling is a bulk process that occurs throughout the entire liquid and only at one specific temperature (the boiling point). Therefore, statement B is correct.
Marking scheme
Award 1 mark for the correct answer B. (1 mark for identifying that evaporation occurs only at the surface of the liquid whereas boiling occurs throughout the liquid)
Question 10 · multipleChoice
1 marks
A ray of light strikes a flat, horizontal mirror. The angle between the incident ray and the mirror surface is \(35^\circ\). What is the angle of reflection?
A.\(35^\circ\)
B.\(55^\circ\)
C.\(70^\circ\)
D.\(110^\circ\)
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Worked solution
The angle of incidence is the angle between the incident ray and the normal (an imaginary line perpendicular to the mirror surface). Therefore, the angle of incidence is \(90^\circ - 35^\circ = 55^\circ\). According to the law of reflection, the angle of reflection equals the angle of incidence. Thus, the angle of reflection is \(55^\circ\).
Marking scheme
Correct option B selected (1 mark). Any other option selected (0 marks).
Question 11 · multipleChoice
1 marks
A cyclist travels the first \(100\text{ m}\) of a journey in \(10\text{ s}\), and the remaining \(300\text{ m}\) in \(40\text{ s}\). What is the average speed of the cyclist for the entire journey?
A.\(7.5\text{ m/s}\)
B.\(8.0\text{ m/s}\)
C.\(8.75\text{ m/s}\)
D.\(10\text{ m/s}\)
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Worked solution
Average speed is calculated by dividing the total distance by the total time taken. Total distance = \(100\text{ m} + 300\text{ m} = 400\text{ m}\). Total time = \(10\text{ s} + 40\text{ s} = 50\text{ s}\). Average speed = \(\frac{400\text{ m}}{50\text{ s}} = 8.0\text{ m/s}\).
Marking scheme
Correct option B selected (1 mark). Any other option selected (0 marks).
Question 12 · multipleChoice
1 marks
A student places a metal spoon and a plastic spoon into a beaker of hot water. The end of the metal spoon outside the water becomes hot much faster than the end of the plastic spoon. Why does this happen?
A.Thermal energy is transferred through the metal spoon mainly by convection.
B.The metal spoon contains free electrons that rapidly transfer thermal energy.
C.The plastic spoon is a better conductor of thermal energy than the metal spoon.
D.The metal spoon loses less thermal energy to the surrounding air by radiation.
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Worked solution
Metals are excellent thermal conductors because they contain a large number of free (delocalised) electrons. These electrons can move rapidly through the metal lattice, colliding with metal ions and transferring kinetic (thermal) energy much faster than the vibration of atoms alone. Plastic is an insulator and lacks these free electrons.
Marking scheme
Correct option B selected (1 mark). Any other option selected (0 marks).
Question 13 · multipleChoice
1 marks
A student wants to find the volume of a single small metal sphere. She pours some water into a measuring cylinder and records the initial volume as \( 30\text{ cm}^3 \). She then lowers 50 identical metal spheres into the water so that they are completely submerged. The new volume reading is \( 48\text{ cm}^3 \). What is the average volume of one metal sphere?
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Worked solution
First, find the total volume of all 50 metal spheres by calculating the change in water level: \( 48\text{ cm}^3 - 30\text{ cm}^3 = 18\text{ cm}^3 \). Next, find the volume of a single sphere by dividing this total volume by the number of spheres: \( 18\text{ cm}^3 / 50 = 0.36\text{ cm}^3 \).
Marking scheme
1 mark for the correct calculation: 0.36 cm³.
Question 14 · multipleChoice
1 marks
Two identical metal containers are filled with equal volumes of hot water. One container has a shiny silver outer surface, and the other has a matt black outer surface. Which container cools down more quickly and why?
A.The container with the matt black surface, because it is a better emitter of radiation.
B.The container with the matt black surface, because it is a poorer conductor of heat.
C.The container with the shiny silver surface, because it is a better emitter of radiation.
D.The container with the shiny silver surface, because it is a poorer conductor of heat.
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Worked solution
Matt black surfaces are better emitters of thermal radiation than shiny silver surfaces. Therefore, the container with the matt black surface will lose heat energy to the surroundings at a faster rate, causing it to cool down more quickly.
Marking scheme
1 mark for identifying the matt black container as cooling faster due to being a better emitter of radiation.
Question 15 · multipleChoice
1 marks
An electric current of \( 2.5\text{ A} \) flows through a lamp for a time of \( 4.0\text{ minutes} \). What is the total charge that passes through the lamp during this time?
A.10\text{ C}
B.96\text{ C}
C.600\text{ C}
D.1500\text{ C}
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Worked solution
The relationship between charge, current, and time is given by the formula \( Q = I \times t \). First, convert the time from minutes to seconds: \( t = 4.0\text{ minutes} = 4.0 \times 60\text{ s} = 240\text{ s} \). Then, calculate the charge: \( Q = 2.5\text{ A} \times 240\text{ s} = 600\text{ C} \).
Marking scheme
1 mark for the correct calculation of charge (600 C) after converting time to seconds.
Question 16 · multipleChoice
1 marks
A student wants to determine the volume of a small irregular stone. She pours some water into a measuring cylinder and reads the volume as \(32\text{ cm}^3\). After gently lowering the stone into the cylinder so that it is completely submerged, the new water level is read as \(57\text{ cm}^3\). What is the volume of the stone?
A.\(25\text{ cm}^3\)
B.\(32\text{ cm}^3\)
C.\(57\text{ cm}^3\)
D.\(89\text{ cm}^3\)
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Worked solution
The volume of the irregular stone is calculated by subtracting the initial volume of water from the final volume of water when the stone is fully submerged: \(V = 57\text{ cm}^3 - 32\text{ cm}^3 = 25\text{ cm}^3\).
Marking scheme
Award 1 mark for the correct calculation and selection of option A.
Question 17 · multipleChoice
1 marks
Which statement about the Sun and the Solar System is correct?
A.The Sun is a planet that orbits the Earth.
B.The Sun is a star that releases energy by nuclear fusion of hydrogen.
C.The planets orbit the Sun in perfectly circular orbits.
D.The Moon is a star that shines by its own light.
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Worked solution
The Sun is a star. Its energy source is the nuclear fusion of hydrogen into helium. Planets have elliptical orbits, not perfectly circular ones, and the Moon is a non-luminous body that reflects the Sun's light.
Marking scheme
Award 1 mark for identifying that the Sun is a star which releases energy by nuclear fusion (Option B).
Question 18 · multipleChoice
1 marks
A student places a metal spoon in a cup of hot cocoa. The handle of the spoon becomes hot after a short time. By which process is thermal energy mainly transferred through the metal spoon to the handle?
A.conduction
B.convection
C.evaporation
D.radiation
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Worked solution
Metals are good thermal conductors. Thermal energy is transferred through the metal spoon from the hot liquid to the colder handle mainly by conduction, due to the vibrations of the particles and the movement of free electrons.
Marking scheme
Award 1 mark for correctly identifying conduction as the primary mechanism of thermal transfer through a solid metal (Option A).
Question 19 · multipleChoice
1 marks
A student measures the density of an irregularly shaped piece of rock. She first measures its mass as \(70\text{ g}\). Next, she pours water into a measuring cylinder until the volume is \(50\text{ cm}^3\). When the rock is completely submerged in the water, the new volume reading is \(78\text{ cm}^3\). What is the density of the rock?
A.\(0.40\text{ g/cm}^3\)
B.\(0.90\text{ g/cm}^3\)
C.\(1.4\text{ g/cm}^3\)
D.\(2.5\text{ g/cm}^3\)
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Worked solution
1. Find the volume of the rock by subtracting the initial water volume from the final volume: \(V = 78\text{ cm}^3 - 50\text{ cm}^3 = 28\text{ cm}^3\). 2. Use the density formula: \(\text{density} = \frac{\text{mass}}{\text{volume}}\). 3. Calculate the density: \(\text{density} = \frac{70\text{ g}}{28\text{ cm}^3} = 2.5\text{ g/cm}^3\).
Marking scheme
Award 1 mark for the correct answer D. Distractor A represents volume change divided by mass (28/70). Distractor B represents mass divided by final volume (70/78). Distractor C represents mass divided by initial volume (70/50).
Question 20 · multipleChoice
1 marks
A metal saucepan with a plastic handle contains water and is placed on a hotplate. Thermal energy is transferred through the metal base of the saucepan and then distributed throughout the water. What are the primary methods of thermal energy transfer through the metal base and through the water?
A.Through the metal base: conduction; through the water: conduction
B.Through the metal base: conduction; through the water: convection
C.Through the metal base: convection; through the water: conduction
D.Through the metal base: convection; through the water: convection
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Worked solution
1. Metals are excellent conductors of heat, so thermal energy passes through the solid metal base primarily by conduction. 2. Water is a fluid and is heated from below. As water near the bottom warms up, it expands, becomes less dense, and rises. Cooler, denser water sinks to take its place, setting up a convection current. Therefore, thermal energy is transferred through the water primarily by convection.
Marking scheme
Award 1 mark for the correct answer B. All other choices incorrectly pair the state of matter (solid metal base vs. liquid water) with the primary mode of thermal energy transfer.
Question 21 · multipleChoice
1 marks
A steady current of \(0.50\text{ A}\) flows through a resistor in a simple electric circuit. How much electric charge passes through the resistor in a time interval of \(2.0\text{ minutes}\)?
A.\(1.0\text{ C}\)
B.\(4.0\text{ C}\)
C.\(60\text{ C}\)
D.\(240\text{ C}\)
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Worked solution
1. Convert the time interval from minutes to seconds: \(t = 2.0\text{ minutes} = 2.0 \times 60\text{ s} = 120\text{ s}\). 2. Use the equation relating charge, current, and time: \(Q = I \times t\). 3. Calculate the charge: \(Q = 0.50\text{ A} \times 120\text{ s} = 60\text{ C}\).
Marking scheme
Award 1 mark for the correct answer C. Distractor A is calculated without converting minutes to seconds (0.50 * 2). Distractor B is calculated as time in minutes divided by current (2.0 / 0.50). Distractor D is calculated as time in seconds divided by current (120 / 0.50).
Question 22 · multipleChoice
1 marks
Two identical metal cans, one painted matte black and the other shiny silver, are filled with equal amounts of hot water at \(80\\ ^\circ\text{C}\) and placed in a cool room. Which can cools down more rapidly, and what is the main process of thermal energy transfer involved?
A.The matte black can, because matte black surfaces are better emitters of infrared radiation than shiny silver surfaces.
B.The matte black can, because matte black surfaces are better conductors of heat than shiny silver surfaces.
C.The shiny silver can, because shiny silver surfaces are better absorbers of infrared radiation than matte black surfaces.
D.The shiny silver can, because shiny silver surfaces are poorer emitters of infrared radiation than matte black surfaces..
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Worked solution
Matte black surfaces are highly efficient emitters of infrared radiation compared to shiny silver surfaces, which are poor emitters. Since the water is hot and the room is cool, the primary method of heat loss from the surface of the cans to the surroundings is thermal radiation. Therefore, the matte black can loses thermal energy faster and cools down more rapidly.
Marking scheme
1 mark for identifying that the matte black can cools down more rapidly due to being a better emitter of infrared radiation.
Question 23 · multipleChoice
1 marks
A toy car of mass \(0.50\text{ kg}\) travelling at a velocity of \(4.0\text{ m/s}\) collides with a stationary toy car of mass \(1.50\text{ kg}\). The two cars couple (stick) together during the collision. What is the common velocity of the two cars immediately after the collision?
A.1.0 m/s
B.1.3 m/s
C.2.0 m/s
D.3.0 m/s
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Worked solution
Using the principle of conservation of momentum: Total initial momentum = Total final momentum. \(m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f\). Substituting the given values: \((0.50\text{ kg} \times 4.0\text{ m/s}) + (1.50\text{ kg} \times 0\text{ m/s}) = (0.50\text{ kg} + 1.50\text{ kg}) \times v_f\). This simplifies to \(2.0\text{ kg m/s} = 2.0\text{ kg} \times v_f\), which gives \(v_f = 1.0\text{ m/s}\).
Marking scheme
1 mark for calculating the correct final velocity of 1.0 m/s.
Question 24 · multipleChoice
1 marks
A \(12.0\text{ V}\) d.c. power supply is connected across a parallel combination of two resistors. One resistor has a resistance of \(4.0\\ \Omega\) and the other has a resistance of \(12.0\\ \Omega\). What is the total current drawn from the power supply?
A.0.75 A
B.3.0 A
C.4.0 A
D.16.0 A
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Worked solution
First, calculate the combined resistance \(R_p\) of the parallel network: \(1/R_p = 1/4.0 + 1/12.0 = 3/12.0 + 1/12.0 = 4/12.0\\ \Omega^{-1}\\, which gives \)R_p = 3.0\\ \\Omega\). Next, use Ohm\'s law to find the total current: \(I = V / R_p = 12.0\text{ V} / 3.0\\ \Omega = 4.0\text{ A}\\. Alternatively, calculate branch currents: \)I_1 = 12.0\\text{ V} / 4.0\\ \\Omega = 3.0\\text{ A}\) and \(I_2 = 12.0\text{ V} / 12.0\\ \Omega = 1.0\text{ A}\\. Total current \)I = I_1 + I_2 = 3.0\\text{ A} + 1.0\\text{ A} = 4.0\\text{ A}\\.
Marking scheme
1 mark for calculating the correct total current of 4.0 A.
Question 25 · multipleChoice
1 marks
Two identical metal cans contain equal volumes of hot water at \( 80\ ^\circ\text{C} \). Can X is painted matt black and Can Y is painted shiny silver. Both cans are left in a room at \( 20\ ^\circ\text{C} \). The temperature of the water in Can X decreases more rapidly than the temperature of the water in Can Y.
Which statement explains this observation?
A.Matt black surfaces are better conductors of thermal energy than shiny silver surfaces.
B.Matt black surfaces are better emitters of infrared radiation than shiny silver surfaces.
C.Shiny silver surfaces are better absorbers of infrared radiation than matt black surfaces.
D.Shiny silver surfaces reduce the rate of heat loss by convection.
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Worked solution
The main method of thermal energy transfer causing the rapid cooling of the cans to the cooler surrounding environment is infrared radiation.
- Matt black surfaces are excellent emitters of infrared radiation. - Shiny silver surfaces are poor emitters of infrared radiation.
Therefore, Can X (painted matt black) loses thermal energy to the room at a faster rate than Can Y (painted shiny silver), causing its temperature to decrease more rapidly.
Marking scheme
[1 mark] B - Correctly identifies that matt black surfaces are better emitters of infrared radiation than shiny silver surfaces. [0 marks] For any other selection.
Question 26 · multipleChoice
1 marks
A solid rectangular block has a weight of \( 24\text{ N} \) and dimensions of \( 2.0\text{ cm} \times 3.0\text{ cm} \times 4.0\text{ cm} \).
The block is placed on a flat horizontal table. What is the maximum pressure that the block can exert on the table?
A.\( 2.0\text{ Pa} \)
B.\( 3.0\text{ Pa} \)
C.\( 20\,000\text{ Pa} \)
D.\( 40\,000\text{ Pa} \)
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Worked solution
Pressure is defined by the formula:
\( P = \frac{F}{A} \)
To exert the maximum pressure, the block must be placed on its smallest face to minimize the surface area \( A \).
1. Find the smallest area of contact: \( A = 2.0\text{ cm} \times 3.0\text{ cm} = 6.0\text{ cm}^2 \)
2. Convert this area into square meters (\( \text{m}^2 \)): \( A = 6.0 \times 10^{-4}\text{ m}^2 = 0.0006\text{ m}^2 \)
3. Calculate the maximum pressure: \( P = \frac{24\text{ N}}{0.0006\text{ m}^2} = 40\,000\text{ Pa} \)
Marking scheme
[1 mark] D - Correctly calculates the maximum pressure as 40 000 Pa. [0 marks] For any other selection (e.g. C represents the minimum pressure when placed on the largest face).
Question 27 · multipleChoice
1 marks
The average distance from the Earth to the Sun is approximately \( 1.5 \times 10^8\text{ km} \). Light travels through space at a speed of \( 3.0 \times 10^8\text{ m/s} \).
How long does it take for light from the Sun to reach the Earth?
A.\( 0.5\text{ s} \)
B.\( 50\text{ s} \)
C.\( 500\text{ s} \)
D.\( 5000\text{ s} \)
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Worked solution
First, convert the distance from kilometers (\( \text{km} \)) to meters (\( \text{m} \)):
Thus, it takes \( 500\text{ s} \) (which is approximately 8.3 minutes) for light to reach the Earth.
Marking scheme
[1 mark] C - Correctly converts the unit of distance and calculates the time as 500 s. [0 marks] For any other selection (such as A, which results from failing to convert km to m).
Question 28 · multipleChoice
1 marks
A star is much more massive than the Sun. Which sequence correctly shows the stages in the life cycle of this star after it leaves the stable (main sequence) stage?
A.red giant -> white dwarf -> black dwarf
B.red giant -> supernova -> neutron star
C.red supergiant -> supernova -> neutron star
D.red supergiant -> planetary nebula -> white dwarf
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Worked solution
For a star much more massive than the Sun, after leaving the stable main sequence stage, it expands to form a red supergiant. This is followed by a supernova explosion, which leaves behind either a neutron star or a black hole. Therefore, the correct sequence is red supergiant -> supernova -> neutron star.
Marking scheme
C is the correct answer. 1 mark is awarded for identifying the correct evolutionary sequence for a high-mass star.
Question 29 · multipleChoice
1 marks
Two identical metal cans each contain the same mass of hot water at 80 degrees Celsius. Can X is painted matt black and Can Y is painted shiny silver. Both cans are left to cool in a laboratory at a room temperature of 20 degrees Celsius. Which can cools down faster and why?
A.Can X, because matt black surfaces are better emitters of infrared radiation than shiny silver surfaces.
B.Can X, because matt black surfaces are better absorbers of infrared radiation than shiny silver surfaces.
C.Can Y, because shiny silver surfaces are better emitters of infrared radiation than matt black surfaces.
D.Can Y, because shiny silver surfaces are better reflectors of infrared radiation than matt black surfaces.
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Worked solution
Matt black surfaces are excellent emitters of infrared radiation, whereas shiny silver surfaces are very poor emitters. Since both cans are hotter than their surroundings, they lose thermal energy to the surroundings primarily by radiation. Can X, with its matt black surface, will emit radiation at a higher rate and therefore cool down faster.
Marking scheme
A is the correct answer. 1 mark is awarded for identifying that Can X cools faster because matt black is a better emitter of radiation than shiny silver.
Question 30 · multipleChoice
1 marks
A student measures the density of an irregularly shaped rock. The mass of the rock is 135 g. She lowers the rock into a measuring cylinder containing 50 cm³ of water. The new water level is 80 cm³. What is the density of the rock?
A.1.7 g/cm³
B.2.7 g/cm³
C.4.5 g/cm³
D.5.4 g/cm³
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Worked solution
First, calculate the volume of the rock by finding the difference in water level in the measuring cylinder: Volume = 80 cm³ - 50 cm³ = 30 cm³. Next, use the density formula: Density = mass / volume. Substituting the values gives: Density = 135 g / 30 cm³ = 4.5 g/cm³.
Marking scheme
C is the correct answer. 1 mark is awarded for correctly calculating the volume (30 cm³) and using the formula density = mass / volume to get 4.5 g/cm³.
Question 31 · multipleChoice
1 marks
An object has a mass of 15 kg on the Earth, where the gravitational field strength \(g\) is 10 N/kg. The object is taken to the Moon, where the gravitational field strength \(g\) is 1.6 N/kg. What are the mass and the weight of the object on the Moon?
A.mass = 2.4 kg, weight = 24 N
B.mass = 15 kg, weight = 24 N
C.mass = 15 kg, weight = 150 N
D.mass = 150 kg, weight = 240 Nofff
Wait, let's keep the option clean: mass = 150 kg, weight = 240 N
Wait, I must ensure there is no typo in the options. Yes: mass = 150 kg, weight = 240 N is correct as a distractor.
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Worked solution
1. Mass is a measure of the amount of matter in an object and is constant regardless of the gravitational field strength. Therefore, the mass of the object on the Moon remains 15 kg. 2. Weight is calculated using the formula: \(W = m \times g\) Where \(m = 15\text{ kg}\) and \(g = 1.6\text{ N/kg}\) on the Moon. \(W = 15\text{ kg} \times 1.6\text{ N/kg} = 24\text{ N}\). 3. This corresponds to a mass of 15 kg and a weight of 24 N.
Marking scheme
[1 mark] Correctly identifies that mass remains 15 kg and weight is 24 N (Option B).
Question 32 · multipleChoice
1 marks
A sound wave is produced by a signal generator connected to a loudspeaker. The frequency of the sound wave is increased, and its amplitude is decreased. How do the pitch and the loudness of the sound change?
A.The pitch increases and the loudness decreases.
B.The pitch increases and the loudness increases.
C.The pitch decreases and the loudness decreases.
D.The pitch decreases and the loudness increases.
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Worked solution
1. Pitch is directly related to the frequency of a sound wave. An increase in frequency results in a higher pitch. 2. Loudness is directly related to the amplitude of a sound wave. A decrease in amplitude results in a quieter sound (decreased loudness). 3. Therefore, the pitch increases and the loudness decreases.
Marking scheme
[1 mark] Correctly identifies that increasing the frequency increases the pitch and decreasing the amplitude decreases the loudness (Option A).
Question 33 · multipleChoice
1 marks
A solid rectangular block of mass 4.0 kg has dimensions \(0.10\text{ m} \times 0.20\text{ m} \times 0.40\text{ m}\). The block is placed on a flat horizontal table. The gravitational field strength \(g\) is 10 N/kg. What is the minimum pressure that the block can exert on the table?
A.\(50\text{ Pa}\)
B.\(500\text{ Pa}\)
C.\(1000\text{ Pa}\)
D.\(2000\text{ Pa}\)
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Worked solution
1. Find the weight (downward force) of the block: \(W = m \times g = 4.0\text{ kg} \times 10\text{ N/kg} = 40\text{ N}\). 2. Pressure is given by: \(P = \frac{F}{A}\) To obtain the minimum pressure, the block must be placed on the face with the maximum surface area. 3. The three possible face areas are: - \(0.10\text{ m} \times 0.20\text{ m} = 0.02\text{ m}^2\) - \(0.10\text{ m} \times 0.40\text{ m} = 0.04\text{ m}^2\) - \(0.20\text{ m} \times 0.40\text{ m} = 0.08\text{ m}^2\) (maximum area) 4. Calculate the minimum pressure: \(P_{\text{min}} = \frac{40\text{ N}}{0.08\text{ m}^2} = 500\text{ Pa}\).
Marking scheme
[1 mark] Correctly identifies the maximum area and calculates the minimum pressure as 500 Pa (Option B).
Question 34 · multipleChoice
1 marks
A measuring cylinder contains \(40\text{ cm}^3\) of water. A metal object of mass \(120\text{ g}\) is lowered into the cylinder, and the water level rises to \(80\text{ cm}^3\). A second, identical metal object is lowered into the same cylinder. What is the final reading on the measuring cylinder, and what is the density of the metal?
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Worked solution
First, find the volume of one metal object by water displacement: \(V = 80\text{ cm}^3 - 40\text{ cm}^3 = 40\text{ cm}^3\). Next, calculate the density of the metal: \(\text{density} = \frac{\text{mass}}{\text{volume}} = \frac{120\text{ g}}{40\text{ cm}^3} = 3.0\text{ g/cm}^3\). When a second identical object is added, its volume is also \(40\text{ cm}^3\). The final volume reading will be \(80\text{ cm}^3 + 40\text{ cm}^3 = 120\text{ cm}^3\).
Marking scheme
Correct option B: 1 mark for identifying the correct final volume reading as \(120\text{ cm}^3\) and the correct density as \(3.0\text{ g/cm}^3\).
Question 35 · multipleChoice
1 marks
Two identical metal cans are filled with equal volumes of hot water at the same initial temperature. Can X is painted matt black and Can Y is painted shiny silver. Both cans are left in a cool room. Which statement correctly describes and explains which can cools down faster?
A.Can X cools faster because matt black surfaces are better emitters of thermal radiation than shiny silver surfaces.
B.Can X cools slower because matt black surfaces are better absorbers of thermal radiation than shiny silver surfaces.
C.Can Y cools faster because shiny silver surfaces are better emitters of thermal radiation than matt black surfaces.
D.Can Y cools slower because shiny silver surfaces are better absorbers of thermal radiation than matt black surfaces.
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Worked solution
The hot water in the cans transfers thermal energy to the surroundings primarily through thermal radiation. Matt black surfaces are excellent emitters of infrared radiation, whereas shiny silver surfaces are poor emitters. Consequently, Can X (matt black) will emit thermal radiation at a faster rate and cool down more quickly than Can Y.
Marking scheme
Correct option A: 1 mark for stating that Can X cools faster because matt black surfaces are better emitters of thermal radiation.
Question 36 · multipleChoice
1 marks
A bar magnet is pushed into a solenoid connected to a sensitive center-zero galvanometer. The pointer of the galvanometer deflects momentarily to the right. Which action will cause a larger momentary deflection of the pointer to the left?
A.Pushing the same pole of the magnet into the solenoid at a slower speed.
B.Pulling the same pole of the magnet out of the solenoid at a faster speed.
C.Holding the magnet stationary inside the solenoid.
D.Pulling the same pole of the magnet out of the solenoid at a slower speed.
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Worked solution
1. The direction of the induced electromotive force (EMF) and current depends on the direction of relative motion. Since pushing the magnet in causes a deflection to the right, moving the magnet in the opposite direction (pulling it out) will cause a deflection to the left. 2. The magnitude of the induced EMF depends on the rate at which the magnetic field lines are cut. Moving the magnet at a faster speed increases the rate of magnetic flux linkage change, which results in a larger deflection. Therefore, pulling the magnet out at a faster speed causes a larger deflection to the left.
Marking scheme
Correct option B: 1 mark for identifying that pulling the magnet out of the solenoid at a faster speed causes a larger deflection in the opposite direction (left).
Question 37 · multipleChoice
1 marks
A student determines the density of an irregularly shaped stone. The student measures the mass of the stone as \( 150\text{ g} \). The student then lowers the stone into a measuring cylinder containing \( 60\text{ cm}^3 \) of water. The water level rises to \( 85\text{ cm}^3 \). What is the density of the stone?
A.\( 1.8\text{ g/cm}^3 \)
B.\( 2.5\text{ g/cm}^3 \)
C.\( 6.0\text{ g/cm}^3 \)
D.\( 0.17\text{ g/cm}^3 \)
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Worked solution
The volume of the stone is calculated from the difference in the water levels in the measuring cylinder: \( V = 85\text{ cm}^3 - 60\text{ cm}^3 = 25\text{ cm}^3 \). The density is calculated using the formula: \( \rho = \frac{m}{V} = \frac{150\text{ g}}{25\text{ cm}^3} = 6.0\text{ g/cm}^3 \).
Marking scheme
Award 1 mark for the correct calculation of density using density = mass / volume, leading to option C.
Question 38 · multipleChoice
1 marks
A student heats one end of a copper rod. Thermal energy is transferred to the cooler end of the rod. Which process is mainly responsible for this transfer of thermal energy, and how does it occur?
A.Conduction, because free electrons move through the metal and transfer energy by colliding with other particles.
B.Convection, because hot copper atoms expand and travel from the hot end to the cold end.
C.Radiation, because electromagnetic waves travel through the copper rod from atom to atom.
D.Conduction, because the air around the rod transfers energy to the colder end by movement of air currents.
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Worked solution
Copper is a metal containing free (delocalised) electrons. When heated, these electrons gain kinetic energy and move rapidly through the metal lattice, colliding with other electrons and metal ions to transfer thermal energy. This process is thermal conduction.
Marking scheme
Award 1 mark for identifying conduction as the correct process and free electron collision as the main mechanism (Option A).
Question 39 · multipleChoice
1 marks
An electric current of \( 1.5\text{ A} \) passes through a heating element for \( 4.0\text{ minutes} \). What is the total charge that passes through the heating element in this time?
A.\( 6.0\text{ C} \)
B.\( 160\text{ C} \)
C.\( 360\text{ C} \)
D.\( 0.375\text{ C} \)
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Worked solution
First, convert the time from minutes to seconds: \( t = 4.0\text{ minutes} = 4.0 \times 60\text{ s} = 240\text{ s} \). Next, use the formula for charge: \( Q = I \times t \). Substitute the values: \( Q = 1.5\text{ A} \times 240\text{ s} = 360\text{ C} \).
Marking scheme
Award 1 mark for converting time to seconds and applying the charge formula correctly to get 360 C (Option C).
Question 40 · multipleChoice
1 marks
A uniform metre rule is pivoted at its centre (the \(50\text{ cm}\) mark). A weight of \(3.0\text{ N}\) is hung from the rule at the \(20\text{ cm}\) mark.
At which mark on the rule must a \(2.0\text{ N}\) weight be hung to balance the rule?
A.\(65\text{ cm}\) mark
B.\(80\text{ cm}\) mark
C.\(90\text{ cm}\) mark
D.\(95\text{ cm}\) mark
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Worked solution
To find the correct position, we use the principle of moments about the pivot at the \(50\text{ cm}\) mark.
1. Calculate the perpendicular distance of the \(3.0\text{ N}\) weight from the pivot: $$\text{Distance} = 50\text{ cm} - 20\text{ cm} = 30\text{ cm}$$
3. For the rule to balance, the clockwise moment must equal the anticlockwise moment: $$\text{Clockwise moment} = 2.0\text{ N} \times d = 90\text{ N cm}$$ $$d = \frac{90}{2.0} = 45\text{ cm}$$
4. Therefore, the \(2.0\text{ N}\) weight must be placed at a distance of \(45\text{ cm}\) to the right of the \(50\text{ cm}\) pivot mark: $$\text{Position} = 50\text{ cm} + 45\text{ cm} = 95\text{ cm}$$
Hence, the correct option is D.
Marking scheme
- Correct calculation of distance from pivot to \(3.0\text{ N}\) load (\(30\text{ cm}\)) [1] - Correct application of the principle of moments to find the distance \(d = 45\text{ cm}\) [1] - Correct determination of the position on the rule (\(95\text{ cm}\)) [1]
Paper 23
Answer forty multiple-choice questions on Extended Physics.
42 Question · 42 marks
Question 1 · multipleChoice
1 marks
An object of mass \(180\text{ g}\) is made of a metal of density \(9.0\text{ g/cm}^3\). The object has an internal, completely sealed hollow cavity. When the object is fully submerged in a measuring cylinder containing water, the water level reading increases from \(40\text{ cm}^3\) to \(75\text{ cm}^3\).
What is the volume of the hollow cavity?
A.\(15\text{ cm}^3\)
B.\(20\text{ cm}^3\)
C.\(35\text{ cm}^3\)
D.\(55\text{ cm}^3\)
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Worked solution
1. Find the total volume of the object using the water displacement method: \(V_{\text{total}} = 75\text{ cm}^3 - 40\text{ cm}^3 = 35\text{ cm}^3\).
2. Calculate the actual volume occupied by the metal using the mass and density of the metal: \(V_{\text{metal}} = \frac{m}{\rho} = \frac{180\text{ g}}{9.0\text{ g/cm}^3} = 20\text{ cm}^3\).
3. The difference between the total volume and the metal volume is the volume of the internal hollow cavity: \(V_{\text{cavity}} = V_{\text{total}} - V_{\text{metal}} = 35\text{ cm}^3 - 20\text{ cm}^3 = 15\text{ cm}^3\).
Therefore, the correct option is A.
Marking scheme
Award 1 mark for the correct option A.
Award 1 mark for any of the following intermediate steps if evaluated individually: - Deducing the total volume of the object is \(35\text{ cm}^3\). - Calculating the metal volume to be \(20\text{ cm}^3\). - Correctly subtracting metal volume from total volume.
Question 2 · multipleChoice
1 marks
Light from a distant galaxy is redshifted.
Which row correctly describes the motion of the galaxy relative to Earth, the frequency of the observed light compared to the frequency emitted, and what this indicates about the Universe?
- Option A: Motion of galaxy: moving away from Earth; Frequency of light: lower; Indication: the Universe is expanding - Option B: Motion of galaxy: moving away from Earth; Frequency of light: higher; Indication: the Universe is contracting - Option C: Motion of galaxy: moving towards Earth; Frequency of light: lower; Indication: the Universe is expanding - Option D: Motion of galaxy: moving towards Earth; Frequency of light: higher; Indication: the Universe is contracting
A.Motion of galaxy: moving away from Earth; Frequency of light: lower; Indication: the Universe is expanding
B.Motion of galaxy: moving away from Earth; Frequency of light: higher; Indication: the Universe is contracting
C.Motion of galaxy: moving towards Earth; Frequency of light: lower; Indication: the Universe is expanding
D.Motion of galaxy: moving towards Earth; Frequency of light: higher; Indication: the Universe is contracting
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Worked solution
1. Redshift is the increase in the observed wavelength of light from galaxies moving away from us. Therefore, the galaxy is moving away from Earth. 2. Because the speed of light is constant, an increase in wavelength (\(\lambda\)) corresponds to a decrease in frequency (\(f\)), since \(v = f\lambda\). Thus, the frequency of the observed light is lower. 3. The redshift of light from distant galaxies indicates that they are moving away from Earth, which provides evidence that the Universe is expanding.
Therefore, option A is correct.
Marking scheme
Award 1 mark for selecting option A, which correctly identifies that redshifted light means the galaxy is moving away, the detected frequency is lower, and this indicates the expansion of the Universe.
Question 3 · multipleChoice
1 marks
Two cylindrical resistors, X and Y, are made of the same uniform metal. Resistor X has length \(l\) and radius \(r\). Resistor Y has length \(2l\) and radius \(2r\).
The resistors are connected in parallel across a power supply of constant potential difference \(V\).
What is the ratio of the current in X to the current in Y, \(\frac{I_{\text{X}}}{I_{\text{Y}}}\)?
A.\(0.25\)
B.\(0.5\)
C.\(1.0\)
D.\(2.0\)
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Worked solution
1. Express the resistance of each resistor using \(R = \rho \frac{\text{length}}{\text{area}}\). - The cross-sectional area of X is \(A_{\text{X}} = \pi r^2\). - The cross-sectional area of Y is \(A_{\text{Y}} = \pi (2r)^2 = 4\pi r^2 = 4A_{\text{X}}\).
2. Calculate the resistance of Y in terms of X's resistance (\(R_{\text{X}}\)): \(R_{\text{X}} = \rho \frac{l}{A_{\text{X}}}\) \(R_{\text{Y}} = \rho \frac{2l}{4A_{\text{X}}} = 0.5 R_{\text{X}}\)
3. Since the resistors are connected in parallel, they have the same potential difference \(V\) across them. - Current in X is \(I_{\text{X}} = \frac{V}{R_{\text{X}}}\). - Current in Y is \(I_{\text{Y}} = \frac{V}{R_{\text{Y}}} = \frac{V}{0.5 R_{\text{X}}} = 2 I_{\text{X}}\).
4. Find the ratio \(\frac{I_{\text{X}}}{I_{\text{Y}}}\): \(\frac{I_{\text{X}}}{I_{\text{Y}}} = \frac{I_{\text{X}}}{2 I_{\text{X}}} = 0.5\).
Therefore, the correct option is B.
Marking scheme
Award 1 mark for the correct option B.
Award 1 mark for any of the following intermediate steps if evaluated individually: - Finding that area is proportional to the square of radius, so \(A_{\text{Y}} = 4A_{\text{X}}\). - Deducing that \(R_{\text{Y}} = 0.5 R_{\text{X}}\). - Recognizing that voltage is constant in parallel, so current is inversely proportional to resistance.
Question 4 · multipleChoice
1 marks
A block of mass 2.0 kg is pulled up a rough slope by a constant force of 25 N parallel to the slope. The length of the slope is 5.0 m and its vertical height is 3.0 m. The block starts from rest at the bottom of the slope and reaches a speed of 5.0 m/s at the top. Take the acceleration of free fall, g, as 9.8 m/s^2. What is the work done against friction as the block moves up the slope?
A.41.2 J
B.66.2 J
C.90.0 J
D.125 J
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Worked solution
First, calculate the work done by the pulling force: \(W_{\text{input}} = F \times d = 25\text{ N} \times 5.0\text{ m} = 125\text{ J}\). Next, calculate the increase in gravitational potential energy: \(\Delta E_p = mgh = 2.0\text{ kg} \times 9.8\text{ m/s}^2 \times 3.0\text{ m} = 58.8\text{ J}\). Next, calculate the increase in kinetic energy: \(\Delta E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 2.0\text{ kg} \times (5.0\text{ m/s})^2 = 25\text{ J}\). The total mechanical energy gained is \(\Delta E_p + \Delta E_k = 58.8\text{ J} + 25\text{ J} = 83.8\text{ J}\). The work done against friction is the difference between the work input and the mechanical energy gained: \(W_{\text{friction}} = 125\text{ J} - 83.8\text{ J} = 41.2\text{ J}\).
Marking scheme
1 mark for the correct answer A. Method: Award 1 mark for calculating work input (125 J), GPE gain (58.8 J), KE gain (25 J) and subtracting total mechanical energy gain from work input to obtain 41.2 J.
Question 5 · multipleChoice
1 marks
A uniform beam of weight 30 N and length 4.0 m is pivoted at a point 1.0 m from its left end. A load of 50 N is hung from the left end of the beam. Where must a downward vertical force of 10 N be applied to keep the beam in horizontal equilibrium?
A.2.0 m to the right of the pivot
B.1.0 m to the right of the pivot
C.2.0 m to the left of the pivot
D.1.0 m to the left of the pivot
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Worked solution
For a uniform beam, its weight of 30 N acts at its centre of mass, which is at the midpoint (2.0 m from the left end). Since the pivot is 1.0 m from the left end, the centre of mass is \(2.0\text{ m} - 1.0\text{ m} = 1.0\text{ m}\) to the right of the pivot. This weight creates a clockwise moment: \(30\text{ N} \times 1.0\text{ m} = 30\text{ N m}\). The 50 N load hung at the left end is 1.0 m to the left of the pivot, creating an anticlockwise moment: \(50\text{ N} \times 1.0\text{ m} = 50\text{ N m}\). For horizontal equilibrium, the sum of clockwise moments must equal the sum of anticlockwise moments. Let the 10 N force be applied at a distance \(x\) to the right of the pivot to assist the clockwise moment: \(30\text{ N m} + (10\text{ N} \times x) = 50\text{ N m} \implies 10x = 20 \implies x = 2.0\text{ m}\). Therefore, the 10 N force must be applied 2.0 m to the right of the pivot.
Marking scheme
1 mark for the correct answer A. Method: Identify position of centre of mass (1.0 m to the right of pivot), calculate clockwise and anticlockwise moments, and apply the principle of moments to solve for the unknown distance.
Question 6 · multipleChoice
1 marks
Wire X and Wire Y are made of the same metal. Wire X has length \(L\) and cross-sectional radius \(r\). Wire Y has length \(3L\) and cross-sectional radius \(2r\). The resistance of Wire X is \(8.0\ \Omega\). What is the resistance of Wire Y?
A.6.0 \(\Omega\)
B.10.7 \(\Omega\)
C.12.0 \(\Omega\)
D.24.0 \(\Omega\)
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Worked solution
The resistance of a wire is given by \(R = \rho \frac{L}{A}\), where \(A = \pi r^2\). Thus, \(R \propto \frac{L}{r^2}\). For Wire X: \(R_X \propto \frac{L}{r^2} = 8.0\ \Omega\). For Wire Y: \(R_Y \propto \frac{3L}{(2r)^2} = \frac{3L}{4r^2} = \frac{3}{4} \left(\frac{L}{r^2}\right)\). Therefore, \(R_Y = \frac{3}{4} R_X = \frac{3}{4} \times 8.0\ \Omega = 6.0\ \Omega\).
Marking scheme
1 mark for the correct answer A. Method: Use the formula \(R = \rho \frac{L}{\pi r^2}\), substitute the relative values of length and radius for wire Y, and scale the original resistance of 8.0 ohms by a factor of 3/4.
Question 7 · multipleChoice
1 marks
A ray of light in a glass block of refractive index 1.50 is incident on the boundary with air at an angle of incidence of 45 degrees. What is the critical angle for the glass-air boundary, and what happens to this ray of light?
A.critical angle is 42 degrees; the ray undergoes total internal reflection
B.critical angle is 42 degrees; the ray refracts into the air
C.critical angle is 48 degrees; the ray undergoes total internal reflection
D.critical angle is 48 degrees; the ray refracts into the air
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Worked solution
The critical angle is calculated using the formula sin(c) = 1 / n. Substituting n = 1.50, we get sin(c) = 1 / 1.50 = 0.667, which gives c = 42 degrees. Since the angle of incidence (45 degrees) is greater than the critical angle (42 degrees), the ray undergoes total internal reflection.
Marking scheme
1 mark for the correct option A.
Question 8 · multipleChoice
1 marks
A uniform metal wire of resistance R is stretched so that its length doubles, while its volume remains constant. What is the new resistance of the stretched wire in terms of R?
A.0.5 R
B.R
C.2 R
D.4 R
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Worked solution
The resistance of a wire is given by R = p * L / A, where p is resistivity, L is length, and A is the cross-sectional area. Since the volume V = L * A remains constant, doubling the length (L' = 2L) means the area must be halved (A' = A / 2). The new resistance R' is therefore R' = p * 2L / (A / 2) = 4 * p * L / A = 4R.
Marking scheme
1 mark for the correct option D.
Question 9 · multipleChoice
1 marks
An electric motor is used to lift a load of 120 N through a vertical height of 5.0 m in a time of 4.0 s. The electrical power input to the motor is 200 W. What is the efficiency of the motor?
A.15 percent
B.60 percent
C.75 percent
D.80 percent
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Worked solution
The useful work done in lifting the load is W = force * distance = 120 N * 5.0 m = 600 J. The useful power output of the motor is P_out = W / t = 600 J / 4.0 s = 150 W. The efficiency is given by (P_out / P_in) * 100 = (150 W / 200 W) * 100 = 75 percent.
Marking scheme
1 mark for the correct option C.
Question 10 · multipleChoice
1 marks
An electric motor is used to lift a crate of mass \(50\text{ kg}\) vertically through a height of \(6.0\text{ m}\) in a time of \(5.0\text{ s}\). The electrical power supplied to the motor is \(800\text{ W}\). What is the efficiency of the motor? (Use \(g = 9.8\text{ m/s}^2\))
A.6.1%
B.37.5%
C.73.5%
D.75.0%
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Worked solution
1. Calculate the work done in lifting the crate (change in gravitational potential energy): \(W = mgh = 50 \times 9.8 \times 6.0 = 2940\text{ J}\)
2. Calculate the useful power output of the motor: \(P_{\text{out}} = \frac{W}{t} = \frac{2940}{5.0} = 588\text{ W}\)
3. Calculate the efficiency of the motor: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{588}{800} \times 100\% = 73.5\%\)
Marking scheme
1 mark for the correct option C.
Question 11 · multipleChoice
1 marks
A wire of length \(L\) and circular cross-sectional area \(A\) has a resistance \(R\). A second wire made of the exact same material has a length of \(3L\) and a diameter that is twice the diameter of the first wire. What is the resistance of the second wire?
A.0.75 R
B.1.5 R
C.3.0 R
D.6.0 R
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Worked solution
1. The resistance of a wire is given by \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity of the material.
2. The cross-sectional area \(A\) is related to the diameter \(d\) by \(A = \pi \left(\frac{d}{2}\right)^2\), which means \(A \propto d^2\).
3. Since the diameter of the second wire is twice that of the first wire, its cross-sectional area is \(4A\).
4. Substituting these dimensions into the resistance formula for the second wire: \(R_2 = \rho \frac{3L}{4A} = \frac{3}{4} \left(\rho \frac{L}{A}\right) = 0.75 R\)
Marking scheme
1 mark for the correct option A.
Question 12 · multipleChoice
1 marks
A ray of light in glass is incident on a boundary with air. The refractive index of the glass is \(1.50\). The angle of incidence is \(40^\circ\). What happens to the ray at the boundary?
A.It undergoes total internal reflection back into the glass at an angle of 40°.
B.It is refracted into the air at an angle of refraction of 25°.
C.It is refracted into the air at an angle of refraction of 75°.
D.It is refracted along the boundary at an angle of refraction of 90°.
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Worked solution
1. Calculate the critical angle \(c\) for the glass-air boundary: \(\sin(c) = \frac{1}{n} = \frac{1}{1.50} \approx 0.667\) \(c = \sin^{-1}(0.667) \approx 41.8^\circ\)
2. Compare the angle of incidence \(i\) with the critical angle \(c\): Since \(i = 40^\circ < 41.8^\circ\), the angle of incidence is less than the critical angle. Therefore, total internal reflection does not occur, and the ray refracts out into the air.
3. Use Snell's Law to find the angle of refraction \(r\): \(n = \frac{\sin(r)}{\sin(i)}\) \(1.50 = \frac{\sin(r)}{\sin(40^\circ)}\) \(\sin(r) = 1.50 \times \sin(40^\circ) = 1.50 \times 0.6428 = 0.9642\) \(r = \sin^{-1}(0.9642) \approx 74.6^\circ \approx 75^\circ\)
Thus, the ray is refracted into the air at an angle of refraction of \(75^\circ\).
Marking scheme
1 mark for the correct option C.
Question 13 · multipleChoice
1 marks
A ray of monochromatic light travels inside a semicircular glass block of refractive index (n = 1.60). It is incident on the flat boundary with air at an angle of incidence (\theta). What is the minimum angle of incidence (\theta) for which the light is totally internally reflected, and what happens if (\theta) is reduced to (30.0^\circ)?
A.Minimum angle for total internal reflection is (38.7^\circ). At (\theta = 30.0^\circ), light refracts into air at an angle of (53.1^\circ).
B.Minimum angle for total internal reflection is (38.7^\circ). At (\theta = 30.0^\circ), light refracts into air at an angle of (18.2^\circ).
C.Minimum angle for total internal reflection is (51.3^\circ). At (\theta = 30.0^\circ), light refracts into air at an angle of (53.1^\circ).
D.Minimum angle for total internal reflection is (51.3^\circ). At (\theta = 30.0^\circ), light is totally internally reflected.
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Worked solution
The critical angle (c) is found using (\sin(c) = 1/n = 1/1.60 = 0.625), which gives (c \approx 38.7^\circ). Total internal reflection occurs for any angle greater than or equal to this critical angle. When the angle of incidence is reduced to (30.0^\circ) (which is less than the critical angle), the light refracts into the air. According to Snell's law, (n_{\text{glass}} \sin(\theta) = n_{\text{air}} \sin(r)), so (1.60 \times \sin(30.0^\circ) = 1.00 \times \sin(r)), giving (\sin(r) = 0.80) and (r \approx 53.1^\circ).
Marking scheme
1 mark: Correctly identifies the minimum angle for TIR as (38.7^\circ) and the correct refraction angle of (53.1^\circ) using Snell's Law.
Question 14 · multipleChoice
1 marks
A toy car of mass (0.50\text{ kg}) travels in a straight line on a frictionless horizontal surface. It is initially moving at a speed of (2.0\text{ m/s}). A constant braking force of (3.0\text{ N}) is applied to the car in the direction opposite to its motion for a duration of (0.50\text{ s}). What is the final speed and direction of motion of the toy car?
A.Speed is (0.50\text{ m/s}) in the original direction of motion.
B.Speed is (1.0\text{ m/s}) in the opposite direction to its initial motion.
C.Speed is (1.0\text{ m/s}) in the original direction of motion.
D.Speed is (5.0\text{ m/s}) in the opposite direction to its initial motion.
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Worked solution
Using the impulse-momentum equation (F \Delta t = m(v - u)) and taking the original direction of motion as positive: (-3.0 \times 0.50 = 0.50(v - 2.0)). This simplifies to (-1.5 = 0.50v - 1.0), which gives (-0.5 = 0.50v), so (v = -1.0\text{ m/s}). The negative sign indicates the direction has reversed, so the final speed is (1.0\text{ m/s}) in the opposite direction.
Marking scheme
1 mark: Correctly calculates final speed as (1.0\text{ m/s}) and direction of motion as opposite to its initial motion.
Question 15 · multipleChoice
1 marks
A metal wire of length (L) and circular cross-section of radius (r) has a resistance (R). A second wire is made of the same metal but has a length of (3L) and a cross-sectional radius of (2r). What is the resistance of the second wire in terms of (R)?
A.(0.375 R)
B.(0.75 R)
C.(1.5 R)
D.(6.0 R)
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Worked solution
The resistance of a wire is given by (R = \rho \frac{L}{A}) where (A = \pi r^2), so (R \propto \frac{L}{r^2}). For the second wire, the length is multiplied by 3 and the radius is multiplied by 2. Thus, the new resistance is (R_2 = \rho \frac{3L}{\pi (2r)^2} = \frac{3}{4} \rho \frac{L}{\pi r^2} = 0.75 R).
Marking scheme
1 mark: Correctly applies the resistance formula showing (R \propto \frac{L}{r^2}) to calculate the new resistance as (0.75 R).
Question 16 · multipleChoice
1 marks
Liquid X has a density of \(0.80\text{ g/cm}^3\) and a volume of \(100\text{ cm}^3\). Liquid Y has a density of \(1.20\text{ g/cm}^3\) and a volume of \(300\text{ cm}^3\). The two liquids are mixed together thoroughly. Assuming no change in total volume occurs, what is the density of the mixture?
A.0.90 g/cm³
B.1.00 g/cm³
C.1.10 g/cm³
D.1.15 g/cm³
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Worked solution
To find the density of the mixture, we use the formula: \( \text{density} = \text{total mass} / \text{total volume} \). First, calculate the mass of each liquid: Mass of Liquid X = \( 0.80\text{ g/cm}^3 \times 100\text{ cm}^3 = 80\text{ g} \), Mass of Liquid Y = \( 1.20\text{ g/cm}^3 \times 300\text{ cm}^3 = 360\text{ g} \). Next, find the total mass and total volume: Total mass = \( 80\text{ g} + 360\text{ g} = 440\text{ g} \), Total volume = \( 100\text{ cm}^3 + 300\text{ cm}^3 = 400\text{ cm}^3 \). Finally, calculate the density of the mixture: \( \text{Density} = 440\text{ g} / 400\text{ cm}^3 = 1.10\text{ g/cm}^3 \).
Marking scheme
1 mark for the correct option C.
Question 17 · multipleChoice
1 marks
A step-up transformer has 200 turns on its primary coil and 1000 turns on its secondary coil. The primary coil is connected to a \(12\text{ V}\) d.c. power supply. What is the output voltage across the secondary coil?
A.0 V
B.2.4 V
C.60 V
D.120 V
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Worked solution
Transformers require an alternating current (a.c.) to operate because electromagnetic induction relies on a continuously changing magnetic field. A direct current (d.c.) supply provides a constant current, which produces a constant magnetic field. Since there is no change in magnetic flux linkage, no electromotive force (voltage) is induced in the secondary coil. Thus, the output voltage is \(0\text{ V}\).
Marking scheme
1 mark for the correct option A.
Question 18 · multipleChoice
1 marks
Which row correctly shows the sequence of stages in the life cycle of a star with a mass much larger than the Sun?
A.protostar \(\rightarrow\) stable star \(\rightarrow\) red giant \(\rightarrow\) white dwarf
B.protostar \(\rightarrow\) stable star \(\rightarrow\) red supergiant \(\rightarrow\) supernova \(\rightarrow\) neutron star
C.stable star \(\rightarrow\) protostar \(\rightarrow\) red supergiant \(\rightarrow\) white dwarf
D.protostar \(\rightarrow\) red giant \(\rightarrow\) supernova \(\rightarrow\) black dwarf
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Worked solution
A massive star (much larger than the Sun) begins as a protostar, develops into a stable main sequence star, and then expands into a red supergiant. When it runs out of fuel, it explodes in a supernova, leaving behind either a neutron star or a black hole. This path is correctly shown in option B.
Marking scheme
1 mark for the correct option B.
Question 19 · multipleChoice
1 marks
A tall cylindrical container contains a layer of liquid X of depth \(0.15\text{ m}\) floating on top of a layer of liquid Y of depth \(0.25\text{ m}\). Liquid X has a density of \(800\text{ kg/m}^3\) and liquid Y has a density of \(1200\text{ kg/m}^3\). The atmospheric pressure acting on the surface of liquid X is \(1.0 \times 10^5\text{ Pa}\). The acceleration of free fall \(g\) is \(9.8\text{ m/s}^2\). What is the total pressure at the bottom of the container?
A.\(4.1 \times 10^3\text{ Pa}\)
B.\(1.03 \times 10^5\text{ Pa}\)
C.\(1.04 \times 10^5\text{ Pa}\)
D.\(2.04 \times 10^5\text{ Pa}\)
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Worked solution
The pressure due to a column of liquid is given by \(p = \rho g h\). Pressure due to liquid X: \(p_X = 800\text{ kg/m}^3 \times 9.8\text{ m/s}^2 \times 0.15\text{ m} = 1176\text{ Pa}\). Pressure due to liquid Y: \(p_Y = 1200\text{ kg/m}^3 \times 9.8\text{ m/s}^2 \times 0.25\text{ m} = 2940\text{ Pa}\). The total pressure at the bottom is the sum of the atmospheric pressure and the pressure exerted by both liquid layers: \(p_{total} = p_{atm} + p_X + p_Y = 1.0 \times 10^5\text{ Pa} + 1176\text{ Pa} + 2940\text{ Pa} = 104\,116\text{ Pa} \approx 1.04 \times 10^5\text{ Pa}\).
Marking scheme
1 mark for the correct answer C. Correct application of hydrostatic pressure formula \(p = \rho g h\) for both layers and addition to atmospheric pressure.
Question 20 · multipleChoice
1 marks
A potential divider circuit consists of a fixed resistor of resistance \(6.0\text{ k}\Omega\) in series with a light-dependent resistor (LDR) connected across a \(12\text{ V}\) d.c. power supply of negligible internal resistance. A voltmeter is connected across the LDR. In bright light, the resistance of the LDR is \(2.0\text{ k}\Omega\). In darkness, the resistance of the LDR is \(18\text{ k}\Omega\). What is the change in the reading on the voltmeter when the light level changes from bright light to darkness?
A.a decrease of \(3.0\text{ V}\)
B.an increase of \(6.0\text{ V}\)
C.an increase of \(9.0\text{ V}\)
D.an increase of \(12\text{ V}\)
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Worked solution
In bright light, the total resistance of the circuit is \(R_{total} = 6.0\text{ k}\Omega + 2.0\text{ k}\Omega = 8.0\text{ k}\Omega\). The voltage across the LDR is \(V_{LDR} = 12\text{ V} \times \frac{2.0\text{ k}\Omega}{8.0\text{ k}\Omega} = 3.0\text{ V}\). In darkness, the total resistance of the circuit is \(R_{total} = 6.0\text{ k}\Omega + 18\text{ k}\Omega = 24\text{ k}\Omega\). The voltage across the LDR is \(V_{LDR} = 12\text{ V} \times \frac{18\text{ k}\Omega}{24\text{ k}\Omega} = 9.0\text{ V}\). The change in the voltmeter reading is \(9.0\text{ V} - 3.0\text{ V} = 6.0\text{ V}\) (an increase of \(6.0\text{ V}\)).
Marking scheme
1 mark for correct answer B. Correctly calculate the initial voltage across the LDR (3.0 V) and the final voltage across the LDR (9.0 V) to find an increase of 6.0 V.
Question 21 · multipleChoice
1 marks
A distant galaxy is observed to have a recession velocity of \(1.54 \times 10^6\text{ m/s}\). Using the Hubble constant value of \(H_0 = 2.2 \times 10^{-18}\text{ s}^{-1}\), what is the estimated distance from Earth to this galaxy?
A.\(3.4 \times 10^{-12}\text{ m}\)
B.\(7.0 \times 10^{23}\text{ m}\)
C.\(1.4 \times 10^{24}\text{ m}\)
D.\(7.0 \times 10^{24}\text{ m}\)
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Worked solution
The relationship between recession velocity \(v\), the Hubble constant \(H_0\), and distance \(d\) is given by Hubble's Law: \(v = H_0 d\). Rearranging for distance gives \(d = \frac{v}{H_0}\). Substituting the given values: \(d = \frac{1.54 \times 10^6\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} = 7.0 \times 10^{23}\text{ m}\).
Marking scheme
1 mark for the correct answer B. Correct rearrangement and substitution of Hubble's Law equation.
Question 22 · multipleChoice
1 marks
A distant galaxy is observed from Earth. The galaxy is at a distance of \(1.5 \times 10^{24}\text{ m}\) from Earth. The Hubble constant \(H_0\) is \(2.2 \times 10^{-18}\text{ s}^{-1}\). What is the speed at which this galaxy is moving away from Earth?
A.\(1.5 \times 10^{-42}\text{ m/s}\)
B.\(1.5 \times 10^{6}\text{ m/s}\)
C.\(3.3 \times 10^{6}\text{ m/s}\)
D.\(6.8 \times 10^{41}\text{ m/s}\)
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Worked solution
We use Hubble's Law, which relates the recessional velocity \(v\) of a galaxy to its distance \(d\) from Earth: \(v = H_0 d\). Substituting the given values: \(v = (2.2 \times 10^{-18}\text{ s}^{-1}) \times (1.5 \times 10^{24}\text{ m}) = 3.3 \times 10^{6}\text{ m/s}\). Therefore, the correct option is C.
Marking scheme
[1 mark] C is correct. Award 1 mark for the correct calculation of speed using Hubble's Law. Incorrect options: A represents incorrect exponential calculation, B is a distractor with incorrect magnitude, and D is the result of dividing distance by the Hubble constant instead of multiplying.
Question 23 · multipleChoice
1 marks
An ideal step-down transformer has a primary coil with \(1200\) turns and a secondary coil with \(300\) turns. The primary coil is connected to a \(240\text{ V}\) a.c. mains supply. A \(6.0\ \Omega\) resistor is connected across the secondary coil. What is the current in the primary coil?
A.\(0.63\text{ A}\)
B.\(2.5\text{ A}\)
C.\(10\text{ A}\)
D.\(40\text{ A}\)
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Worked solution
First, find the secondary voltage \(V_s\) using the transformer equation: \(V_s = V_p \times (N_s / N_p) = 240\text{ V} \times (300 / 1200) = 60\text{ V}\). Next, find the secondary current \(I_s\) using Ohm's law: \(I_s = V_s / R = 60\text{ V} / 6.0\ \Omega = 10\text{ A}\). For an ideal transformer, the input power equals the output power: \(I_p V_p = I_s V_s\). Therefore, \(I_p = (10\text{ A} \times 60\text{ V}) / 240\text{ V} = 2.5\text{ A}\). This corresponds to option B.
Marking scheme
[1 mark] B is correct. Award 1 mark for correctly determining the primary current. Incorrect options: A is a calculation error, C is the secondary current (10 A), and D is obtained by reversing the turns ratio to find secondary voltage.
Question 24 · multipleChoice
1 marks
Liquid P has a density of \(0.80\text{ g/cm}^3\) and a volume of \(150\text{ cm}^3\). Liquid Q has a density of \(1.20\text{ g/cm}^3\) and a volume of \(250\text{ cm}^3\). The two liquids are mixed together thoroughly. Assuming there is no change in total volume when they are mixed, what is the density of the mixture?
A.\(0.95\text{ g/cm}^3\)
B.\(1.00\text{ g/cm}^3\)
C.\(1.05\text{ g/cm}^3\)
D.\(2.00\text{ g/cm}^3\)
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Worked solution
To find the density of the mixture, we must find the total mass and divide it by the total volume. Mass of liquid P: \(m_P = \text{density} \times \text{volume} = 0.80\text{ g/cm}^3 \times 150\text{ cm}^3 = 120\text{ g}\). Mass of liquid Q: \(m_Q = 1.20\text{ g/cm}^3 \times 250\text{ cm}^3 = 300\text{ g}\). Total mass: \(m_{\text{total}} = 120\text{ g} + 300\text{ g} = 420\text{ g}\). Total volume: \(V_{\text{total}} = 150\text{ cm}^3 + 250\text{ cm}^3 = 400\text{ cm}^3\). Density of the mixture: \(\rho = 420\text{ g} / 400\text{ cm}^3 = 1.05\text{ g/cm}^3\). This corresponds to option C.
Marking scheme
[1 mark] C is correct. Award 1 mark for correctly calculating the mixture's density using total mass divided by total volume. Incorrect options: B is the simple arithmetic average of the two densities, which is incorrect because the volumes of the two liquids are not equal; D is the sum of the two densities; A is an incorrect calculation.
Question 25 · multipleChoice
1 marks
A metal wire of length \(L\) and cross-sectional area \(A\) has a resistance \(R\). A second wire is made of the same metal. It has length \(3L\) and a circular cross-section with twice the radius of the first wire. What is the resistance of the second wire?
A.0.75 R
B.1.5 R
C.3.0 R
D.6.0 R
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Worked solution
Resistance \(R\) of a wire is given by the formula \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area. The area of a circular cross-section of radius \(r\) is \(A = \pi r^2\). For the second wire, the length is \(3L\) and the radius is \(2r\), which means its area is \(A_{new} = \pi (2r)^2 = 4 \pi r^2 = 4A\). Substituting these into the formula gives \(R_{new} = \rho \frac{3L}{4A} = 0.75 R\).
Marking scheme
1 mark for identifying the correct option A.
Question 26 · multipleChoice
1 marks
A distant galaxy is located at a distance of \(2.0 \times 10^{22}\text{ m}\) from Earth. The Hubble constant is estimated to be \(2.2 \times 10^{-18}\text{ s}^{-1}\). What is the speed of recession of this galaxy?
A.1.1 \times 10^{-4}\text{ m/s}
B.4.4 \times 10^{4}\text{ m/s}
C.4.4 \times 10^{7}\text{ m/s}
D.1.1 \times 10^{40}\text{ m/s}
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Worked solution
The speed of recession \(v\) of a galaxy is related to its distance \(d\) by Hubble's Law: \(v = H_0 d\). Substituting the given values: \(v = (2.2 \times 10^{-18}\text{ s}^{-1}) \times (2.0 \times 10^{22}\text{ m}) = 4.4 \times 10^{4}\text{ m/s}\).
Marking scheme
1 mark for identifying the correct option B.
Question 27 · multipleChoice
1 marks
A trolley of mass \(2.0\text{ kg}\) travelling at \(6.0\text{ m/s}\) collides with a stationary trolley of mass \(4.0\text{ kg}\). The two trolleys couple together during the collision. What is their common speed after the collision?
A.1.5\text{ m/s}
B.3.0\text{ m/s}
C.2.0\text{ m/s}
D.4.0\text{ m/s}
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Worked solution
According to the principle of conservation of momentum, the total momentum before collision equals the total momentum after collision: \(m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f\). Substituting the given values: \((2.0\text{ kg} \times 6.0\text{ m/s}) + (4.0\text{ kg} \times 0) = (2.0\text{ kg} + 4.0\text{ kg}) v_f\), which simplifies to \(12 = 6.0 v_f\). Therefore, \(v_f = 2.0\text{ m/s}\).
Marking scheme
1 mark for identifying the correct option C.
Question 28 · multipleChoice
1 marks
A ray of monochromatic light travels inside a glass block of refractive index \( 1.60 \). The ray strikes the flat boundary between the glass and air at an angle of incidence of \( 40.0^\circ \).
What is the behavior of the light ray at the boundary?
A.It is refracted into the air at an angle of refraction of \( 38.7^\circ \).
B.It is refracted into the air at an angle of refraction of \( 50.0^\circ \).
C.It is totally internally reflected back into the glass at an angle of reflection of \( 40.0^\circ \).
D.It is totally internally reflected back into the glass at an angle of reflection of \( 50.0^\circ \).
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Worked solution
First, calculate the critical angle \( c \) for the glass-air boundary:
Compare the angle of incidence \( i \) to the critical angle \( c \):
\( i = 40.0^\circ \)
Since \( i > c \) (\( 40.0^\circ > 38.7^\circ \)), the light ray undergoes total internal reflection.
According to the law of reflection, the angle of reflection equals the angle of incidence:
\( r = 40.0^\circ \)
Marking scheme
C is the correct answer. - 1 mark for identifying that \( i > c \) leads to total internal reflection and calculating the angle of reflection as equal to the angle of incidence (\( 40.0^\circ \)).
Question 29 · multipleChoice
1 marks
A toy car of mass \( 0.50 \text{ kg} \) is moving to the right at a speed of \( 4.0 \text{ m/s} \). It collides with a wall and rebounds to the left at a speed of \( 2.0 \text{ m/s} \). The collision lasts for \( 0.15 \text{ s} \).
What is the magnitude of the average force exerted by the wall on the car?
A.\( 6.7 \text{ N} \)
B.\( 13 \text{ N} \)
C.\( 20 \text{ N} \)
D.\( 3.0 \text{ N} \)
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Worked solution
Use the impulse formula:
\( F \Delta t = \Delta p = m(v - u) \)
Let the direction to the right be positive. Therefore: - Initial velocity, \( u = +4.0 \text{ m/s} \) - Final velocity, \( v = -2.0 \text{ m/s} \) (moving to the left)
\( F = \frac{\Delta p}{\Delta t} = \frac{-3.0 \text{ N s}}{0.15 \text{ s}} = -20 \text{ N} \)
The magnitude of the force is \( 20 \text{ N} \).
Marking scheme
C is the correct answer. - 1 mark for the correct calculation of impulse taking direction into account, leading to an average force of \( 20 \text{ N} \).
Question 30 · multipleChoice
1 marks
A cylindrical metal wire of length \( L \) and diameter \( d \) has a resistance \( R \).
A second wire is made of the same metal. It has a length of \( 2L \) and a diameter of \( 2d \).
What is the resistance of the second wire in terms of \( R \)?
A.\( 4.0 R \)
B.\( 2.0 R \)
C.\( 0.50 R \)
D.\( 0.25 R \)
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Worked solution
The resistance of a wire is given by the formula:
\( R = \rho \frac{L}{A} \)
where \( A \) is the cross-sectional area: \( A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \).
For the second wire: - Length is doubled: \( L' = 2L \) - Diameter is doubled: \( d' = 2d \), which means the area is quadrupled: \( A' = \frac{\pi (2d)^2}{4} = 4A \)
Substitute the new values into the resistance formula:
C is the correct answer. - 1 mark for correctly applying the resistance formula with a doubled length and quadrupled area, resulting in \( 0.50 R \).
Question 31 · multipleChoice
1 marks
An ideal step-down transformer is connected to a \(240\text{ V}\) a.c. supply. The primary coil has \(1000\) turns. The transformer has two separate secondary coils: Coil A has \(100\) turns and is connected to a \(12\ \Omega\) resistor. Coil B has \(50\) turns and is connected to a \(3.0\ \Omega\) resistor. What is the current in the primary coil?
A.0.10 A
B.0.20 A
C.0.40 A
D.0.80 A
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2. Calculate the power delivered to each resistor: - Power in A: \(P_A = \frac{V_A^2}{R_A} = \frac{24^2}{12} = 48\text{ W}\). - Power in B: \(P_B = \frac{V_B^2}{R_B} = \frac{12^2}{3.0} = 48\text{ W}\).
Award 1 mark for the correct answer C. Method: Deduce secondary voltages from turn ratios, find electrical power in each load, equate total power output to power input to find primary current.
Question 32 · multipleChoice
1 marks
A ray of light traveling inside a glass block of refractive index \(1.50\) is incident on the flat boundary with an unknown liquid. The critical angle for total internal reflection at this boundary is \(60^\circ\). What is the refractive index of the liquid?
A.0.87
B.1.15
C.1.30
D.1.73
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Worked solution
The formula for the critical angle \(\theta_c\) at the boundary between two media of refractive indices \(n_{\text{denser}}\) and \(n_{\text{rarer}}\) is given by:
Here, the denser medium is the glass (\(n_{\text{glass}} = 1.50\)) and the rarer medium is the liquid (\(n_{\text{liquid}} = n\)).
\(\sin(60^\circ) = \frac{n}{1.50}\)
\(n = 1.50 \times \sin(60^\circ)\)
\(n = 1.50 \times 0.866 = 1.30\)
Marking scheme
Award 1 mark for the correct answer C. Method: Recall and apply critical angle equation relation: sin(c) = n2 / n1.
Question 33 · multipleChoice
1 marks
A specific spectral line in the light from a distant galaxy has an observed wavelength of \(612\text{ nm}\). In a laboratory on Earth, this same line has a wavelength of \(600\text{ nm}\). The Hubble constant \(H_0\) is \(2.0 \times 10^{-18}\text{ s}^{-1}\) and the speed of light \(c\) is \(3.0 \times 10^8\text{ m/s}\). What is the approximate distance from Earth to this galaxy?
A.6.0 \times 10^{22}\text{ m}
B.1.5 \times 10^{23}\text{ m}
C.3.0 \times 10^{24}\text{ m}
D.1.5 \times 10^{26}\text{ m}
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Worked solution
1. Find the change in wavelength: \(\Delta \lambda = 612\text{ nm} - 600\text{ nm} = 12\text{ nm}\).
2. Use the redshift equation to find the recessional velocity \(v\): \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\)
Award 1 mark for the correct answer C. Method: Calculate recessional speed using redshift equation, then apply Hubble's law to solve for distance.
Question 34 · multipleChoice
1 marks
A trolley X of mass 2.0 kg is moving at a speed of 6.0 m/s. It collides with a stationary trolley Y of mass 4.0 kg. The two trolleys stick together and move off with a common velocity. What is the total kinetic energy of the two trolleys after the collision?
A.12 J
B.18 J
C.36 J
D.72 J
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Worked solution
First, find the common velocity after the collision using the conservation of momentum: Initial momentum = Final momentum. \( (m_X \times v_X) + (m_Y \times v_Y) = (m_X + m_Y) \times v_f \) which gives \( (2.0 \text{ kg} \times 6.0 \text{ m/s}) + 0 = (2.0 \text{ kg} + 4.0 \text{ kg}) \times v_f \). This simplifies to \( 12 \text{ kg m/s} = 6.0 \text{ kg} \times v_f \), so \( v_f = 2.0 \text{ m/s} \). Next, calculate the total kinetic energy after the collision: \( E_k = \frac{1}{2} M v_f^2 = \frac{1}{2} \times 6.0 \text{ kg} \times (2.0 \text{ m/s})^2 = 12 \text{ J} \). Therefore, the correct option is A.
Marking scheme
1 mark for the correct answer A. Award 1 mark for calculating the correct final velocity of 2.0 m/s and then using it to obtain 12 J.
Question 35 · multipleChoice
1 marks
An electric heater of power 50 W is used to heat 0.20 kg of a liquid in a well-insulated container. The temperature of the liquid increases by 15 °C in 2.0 minutes. What is the specific heat capacity of the liquid?
A.33 J/(kg °C)
B.120 J/(kg °C)
C.2000 J/(kg °C)
D.33000 J/(kg °C)
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Worked solution
First, calculate the energy supplied by the heater: \( E = P \times t \) where \( t = 2.0 \text{ minutes} = 120 \text{ s} \). Thus, \( E = 50 \text{ W} \times 120 \text{ s} = 6000 \text{ J} \). Next, use the thermal energy formula to find the specific heat capacity \( c \): \( E = m c \Delta\theta \). This gives \( 6000 \text{ J} = 0.20 \text{ kg} \times c \times 15 \, ^\circ\text{C} \), which simplifies to \( 6000 = 3.0 \times c \), so \( c = 2000 \text{ J/(kg }^\circ\text{C)} \). Therefore, the correct option is C.
Marking scheme
1 mark for the correct answer C. Award 1 mark for converting minutes to seconds (120 s) and calculating 2000 J/(kg °C).
Question 36 · multipleChoice
1 marks
A metal wire of length L and cross-sectional area A has a resistance of R. A second wire, made of the same metal, has a length of 2L and a diameter that is twice the diameter of the first wire. What is the resistance of the second wire?
A.0.5 R
B.R
C.2 R
D.4 R
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Worked solution
The resistance \( R \) of a wire is given by: \( R = \rho \frac{L}{A} \) where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. Since \( A = \frac{\pi d^2}{4} \), doubling the diameter \( d \) increases the cross-sectional area by a factor of 4 (so the new area is \( 4A \)). For the second wire: \( R_2 = \rho \frac{2L}{4A} = 0.5 \left(\rho \frac{L}{A}\right) = 0.5 R \). Therefore, the correct option is A.
Marking scheme
1 mark for the correct answer A. Award 1 mark for realizing that doubling the diameter quadruples the cross-sectional area, and using the formula to get 0.5 R.
Question 37 · multipleChoice
1 marks
Wire X has length \(l\) and diameter \(d\). Wire Y, made of the same material, has length \(3l\) and diameter \(2d\). The resistance of wire X is \(12\ \Omega\). What is the resistance of wire Y?
A.9.0 \(\Omega\)
B.16 \(\Omega\)
C.18 \(\Omega\)
D.48 \(\Omega\)
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Worked solution
The resistance \(R\) of a wire of length \(l\) and cross-sectional area \(A\) is given by \(R = \rho \frac{l}{A}\). Since the cross-section is circular, \(A = \frac{\pi d^2}{4}\), which means \(R \propto \frac{l}{d^2}\). For wire X: \(R_X = k \frac{l}{d^2} = 12\ \Omega\). For wire Y: \(R_Y = k \frac{3l}{(2d)^2} = k \frac{3l}{4d^2} = \frac{3}{4} R_X = \frac{3}{4} \times 12\ \Omega = 9.0\ \Omega\).
Marking scheme
1 mark for the correct option A. Award for correctly identifying the relationship between resistance, length, and diameter, and calculating the final resistance of 9.0 ohms.
Question 38 · multipleChoice
1 marks
An object is placed at a distance of 15 cm from a thin converging lens of focal length 10 cm. Which row describes the nature, orientation, and size of the image formed?
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Worked solution
The focal length \(f\) of the lens is 10 cm. The object distance is \(u = 15\text{ cm}\), which lies between \(F\) and \(2F\). When an object is placed between the focal point and twice the focal length of a converging lens, the image produced is real, inverted, and magnified.
Marking scheme
1 mark for the correct option A. Award for correctly identifying that the object position relative to the focal length results in a real, inverted, and magnified image.
Question 39 · multipleChoice
1 marks
Two identical metal cans, one painted matt black and the other painted shiny silver, are filled with the same mass of hot water at 90 °C. They are left to cool in a room at 20 °C. The water in the matt black can cools more rapidly. Which statement explains this result?
A.Matt black surfaces are better emitters of infrared radiation than shiny silver surfaces.
B.Matt black surfaces are better absorbers of infrared radiation than shiny silver surfaces.
C.Shiny silver surfaces are better conductors of heat than matt black surfaces.
D.Shiny silver surfaces are better emitters of infrared radiation than matt black surfaces.
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Worked solution
The cans are at a higher temperature than their surroundings, so they cool primarily by losing thermal energy via the emission of infrared radiation. Matt black surfaces are better emitters of infrared radiation than shiny silver surfaces, causing the matt black can to lose thermal energy faster and cool more rapidly. Although matt black surfaces are also better absorbers, absorption is not the primary cause of cooling in this scenario.
Marking scheme
1 mark for the correct option A. Award for identifying emission of infrared radiation as the primary mechanism of heat loss and matt black as the better emitter.
Question 40 · multipleChoice
1 marks
A uniform metal wire of length \(L\) and radius \(r\) has a resistance \(R\). A second wire is made of the same metal but has length \(2L\) and radius \(3r\). What is the resistance of the second wire?
A.\(\frac{2}{9}R\)
B.\(\frac{2}{3}R\)
C.\(\frac{4}{9}R\)
D.\(6R\)
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Worked solution
The resistance \(R\) of a wire is given by the formula:
\[R = \rho \frac{L}{A}\]
where \(\rho\) is the resistivity of the material, \(L\) is the length, and \(A\) is the cross-sectional area. Since the cross-section is circular, \(A = \pi r^2\), so:
\[R \propto \frac{L}{r^2}\]
For the second wire, let its resistance be \(R_{\text{new}}\). Its length is \(2L\) and its radius is \(3r\):
- Recall and use relationship \(R \propto \frac{L}{A}\) or \(R \propto \frac{L}{r^2}\) [1 mark] - Deduce that tripling the radius increases the area by a factor of 9, leading to the overall factor of \(\frac{2}{9}\) [1 mark]
Question 41 · multipleChoice
1 marks
Placeholder
A.A
B.B
C.C
D.D
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Worked solution
Placeholder
Marking scheme
Placeholder
Question 42 · multipleChoice
1 marks
A uniform metal wire of length \(L\) and radius \(r\) has a resistance \(R\). A second wire is made of the same metal but has length \(2L\) and radius \(3r\). What is the resistance of the second wire?
A.\(\frac{2}{9}R\)
B.\(\frac{2}{3}R\)
C.\(\frac{4}{9}R\)
D.\(6R\)
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Worked solution
The resistance \(R\) of a wire is given by the formula:
\[R = \rho \frac{L}{A}\]
where \(\rho\) is the resistivity of the material, \(L\) is the length, and \(A\) is the cross-sectional area. Since the cross-section is circular, \(A = \pi r^2\), so:
\[R \propto \frac{L}{r^2}\]
For the second wire, let its resistance be \(R_{\text{new}}\). Its length is \(2L\) and its radius is \(3r\):
- Recall and use relationship \(R \propto \frac{L}{A}\) or \(R \propto \frac{L}{r^2}\) [1 mark] - Deduce that tripling the radius increases the area by a factor of 9, leading to the overall factor of \(\frac{2}{9}\) [1 mark]
Paper 33
Answer core theory structured questions requiring short-answer, descriptive and numerical responses.
11 Question · 79.96999999999997 marks
Question 1 · structured
7.27 marks
A ray of light in air strikes the flat surface of a glass block.
(a) State the name of the imaginary line drawn at \(90^\circ\) to the glass surface at the point where the light ray enters.
(b) The angle of incidence of the light ray is \(45^\circ\). The light ray bends towards the normal as it enters the glass. State whether the angle of refraction is greater than, equal to, or less than \(45^\circ\).
(c) The speed of light in air is \(3.0 \times 10^8\text{ m/s}\) and its speed in the glass is \(2.0 \times 10^8\text{ m/s}\). Calculate the refractive index of the glass. Show your working.
(d) Explain why the light ray does not bend when it leaves the glass block through a curved boundary along the normal.
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Worked solution
(a) The line perpendicular to the surface is the normal.
(b) When light enters a denser medium (glass) from a less dense medium (air), it slows down and bends towards the normal. This means the angle of refraction is less than the angle of incidence, so it is less than \(45^\circ\).
(c) The formula for refractive index is \(n = \frac{v_{\text{air}}}{v_{\text{glass}}}\). Substituting the given speeds: \(n = \frac{3.0 \times 10^8\text{ m/s}}{2.0 \times 10^8\text{ m/s}} = 1.5\).
(d) When light travels along the normal, it hits the boundary at an angle of incidence of \(0^\circ\). Thus, the angle of refraction is also \(0^\circ\), meaning it does not change direction.
Marking scheme
(a) Normal [1 mark] (b) Less than \(45^\circ\) [1 mark] because light bends towards the normal as it slows down [1 mark] (c) Formula: \(n = \frac{v_1}{v_2}\) [1 mark]; substitution: \(\frac{3.0 \times 10^8}{2.0 \times 10^8}\) [1 mark]; final answer: 1.5 [1 mark] (d) Light ray is perpendicular to the boundary / angle of incidence is zero [1 mark]
Question 2 · structured
7.27 marks
A toy car is released from rest and moves down a ramp. Its motion is recorded.
(a) Describe how the speed of the toy car changes if its acceleration is constant and positive.
(b) A speed-time graph for the car shows: - Speed increases from \(0\text{ m/s}\) to \(4.0\text{ m/s}\) in \(5.0\text{ s}\). - Speed remains constant at \(4.0\text{ m/s}\) for another \(10.0\text{ s}\).
(i) Calculate the acceleration of the toy car during the first \(5.0\text{ s}\). Include the unit.
(ii) Calculate the total distance traveled by the car during the entire \(15.0\text{ s}\).
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Worked solution
(a) A constant positive acceleration means the speed of the toy car increases by the same amount each second (increases at a steady rate).
(b)(i) Acceleration is the gradient of a speed-time graph: \(a = \frac{v - u}{t} = \frac{4.0\text{ m/s} - 0\text{ m/s}}{5.0\text{ s}} = 0.80\text{ m/s}^2\).
(b)(ii) Total distance is the total area under the speed-time graph. - Area of the triangle (first \(5.0\text{ s}\)): \(\frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 5.0\text{ s} \times 4.0\text{ m/s} = 10\text{ m}\). - Area of the rectangle (next \(10.0\text{ s}\)): \(\text{width} \times \text{height} = 10.0\text{ s} \times 4.0\text{ m/s} = 40\text{ m}\). - Total distance = \(10\text{ m} + 40\text{ m} = 50\text{ m}\).
Marking scheme
(a) Speed increases at a constant/steady rate [1 mark] (b)(i) Formula: \(a = \frac{\Delta v}{t}\) [1 mark]; calculation: \(4.0 / 5.0 = 0.80\) [1 mark]; unit: \(\text{m/s}^2\) [1 mark] (b)(ii) Attempt to find area under graph [1 mark]; area of triangle = \(10\text{ m}\) AND area of rectangle = \(40\text{ m}\) [1 mark]; total distance = \(50\text{ m}\) [1 mark]
Question 3 · structured
7.27 marks
A student sets up a circuit to measure the resistance of a wire.
(a) State the name of the instrument used to measure: (i) electrical current (ii) potential difference (voltage)
(b) The current in the wire is \(0.40\text{ A}\). Calculate the charge that passes through a point in the wire in \(2.0\text{ minutes}\). Show your working.
(c) The potential difference across the wire is \(6.0\text{ V}\). Calculate the resistance of the wire. State the unit of resistance.
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Worked solution
(a)(i) Current is measured using an ammeter. (a)(ii) Potential difference is measured using a voltmeter.
(b) Charge is calculated using \(Q = I \times t\). First, convert the time from minutes to seconds: \(t = 2.0 \times 60 = 120\text{ s}\). Now, calculate charge: \(Q = 0.40\text{ A} \times 120\text{ s} = 48\text{ C}\).
(c) Resistance is calculated using Ohm's Law: \(R = \frac{V}{I}\). \(R = \frac{6.0\text{ V}}{0.40\text{ A}} = 15\ \Omega\) (ohms).
Marking scheme
(a)(i) Ammeter [1 mark] (a)(ii) Voltmeter [1 mark] (b) Convert time: \(2.0\text{ min} = 120\text{ s}\) [1 mark]; use of \(Q = I \times t\) [1 mark]; charge = \(48\text{ C}\) (or A s) [1 mark] (c) Use of \(R = V/I = 6.0 / 0.40 = 15\) [1 mark]; unit of ohms (or \(\Omega\)) [1 mark]
Question 4 · structured
7.27 marks
An experiment is carried out to find the density of an irregularly shaped piece of mineral.
(a) State the name of the measuring instrument used to find the mass of the mineral.
(b) The mass of the mineral is found to be \(108\text{ g}\). The mineral is then placed into a measuring cylinder containing water. The initial volume of water is \(55\text{ cm}^3\). When the mineral is fully submerged, the water level rises to \(95\text{ cm}^3\). (i) Calculate the volume of the mineral. (ii) Calculate the density of the mineral. State the formula used and show your working.
(c) Describe how the student could adapt this experiment to determine the density of a block of wood that floats in water.
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Worked solution
(a) The standard laboratory instrument to measure mass is a balance (electronic balance or beam balance).
(b)(i) The volume of the mineral is found by subtraction: \(V = 95\text{ cm}^3 - 55\text{ cm}^3 = 40\text{ cm}^3\).
(b)(ii) The formula for density is \(\rho = \frac{m}{V}\). Substituting the given values: \(\rho = \frac{108\text{ g}}{40\text{ cm}^3} = 2.7\text{ g/cm}^3\).
(c) To find the volume of a floating object, it must be fully submerged. This can be achieved by tying a heavy sinker to the block of wood. The volume of the sinker is measured first, then the combined volume of the wood and sinker is measured. The difference gives the volume of the wood. Alternatively, a thin rod/pin can be used to push the wood under water.
Marking scheme
(a) [1 mark] balance / electronic balance / scales. (b)(i) [1 mark] \(40\text{ cm}^3\). (b)(ii) [3 marks] - 1 mark for formula: \(\text{density} = \frac{\text{mass}}{\text{volume}}\) - 1 mark for substitution: \(\frac{108}{40}\) - 1 mark for correct calculation with unit: \(2.7\text{ g/cm}^3\) (accept \(2700\text{ kg/m}^3\) if converted correctly). (c) [2 marks] - 1 mark for mentioning using a sinker / heavy object to fully submerge the wood (or pushing it down with a thin rod). - 1 mark for detailing how to calculate the volume (e.g., subtracting the volume of the sinker/rod from the combined volume).
Question 5 · structured
7.27 marks
A student stands \(150\text{ m}\) away from a tall brick wall. She claps two wooden blocks together and hears an echo.
(a) Explain what is meant by an echo.
(b) She claps the blocks regularly so that each clap coincides with the echo of the previous clap. The total time for 20 complete round trips of sound is \(18.0\text{ s}\). (i) Calculate the time interval between one clap and its echo. (ii) Calculate the speed of sound in air from these measurements. Show your working.
(c) State how the sound heard by the student changes if the frequency of the sound is increased.
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Worked solution
(a) An echo is a sound heard after reflection from a solid surface.
(b)(i) The total time for 20 round trips is \(18.0\text{ s}\). Therefore, the time for 1 round trip (the time interval between a clap and its echo) is: \(t = \frac{18.0\text{ s}}{20} = 0.90\text{ s}\).
(b)(ii) The distance traveled by the sound in one round trip is twice the distance to the wall: \(d = 2 \times 150\text{ m} = 300\text{ m}\). Using the formula for speed: \(v = \frac{d}{t} = \frac{300\text{ m}}{0.90\text{ s}} = 333.3\text{ m/s}\) (or \(333\text{ m/s}\) rounded to 3 significant figures).
(c) An increase in the frequency of sound waves corresponds to an increase in pitch, making the sound sound higher.
Marking scheme
(a) [1 mark] Reflection of a sound wave (from a surface/barrier). (b)(i) [2 marks] - 1 mark for recognizing 20 intervals: \(\frac{18.0}{20}\) - 1 mark for correct calculation: \(0.90\text{ s}\) (allow 0.9). (b)(ii) [3 marks] - 1 mark for calculating total distance for one round trip: \(300\text{ m}\) (or showing \(2 \times 150\)) - 1 mark for formula/working: \(v = \frac{d}{t} = \frac{300}{0.90}\) - 1 mark for correct value: \(333\text{ m/s}\) (accept range \(330\text{ m/s}\) to \(333.3\text{ m/s}\) based on rounding). (c) [1 mark] The pitch becomes higher / frequency determines pitch.
Question 6 · structured
7.27 marks
A student sets up a circuit consisting of a \(6.0\text{ V}\) battery connected in series with a switch, an ammeter, and a resistor.
(a) State the quantity measured by the ammeter.
(b) When the switch is closed, the reading on the ammeter is \(0.25\text{ A}\). (i) Calculate the charge that passes through the resistor in \(2.0\text{ minutes}\). State the unit. (ii) Calculate the resistance of the resistor.
(c) The temperature of the resistor increases. State what happens to the resistance of a typical metal wire resistor as its temperature rises.
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Worked solution
(a) An ammeter measures electric current.
(b)(i) The relation between charge \(Q\), current \(I\), and time \(t\) is given by: \(Q = I \times t\) Convert the time to seconds: \(t = 2.0 \times 60 = 120\text{ s}\). \(Q = 0.25\text{ A} \times 120\text{ s} = 30\text{ C}\). The unit is Coulombs (\(\text{C}\)).
(b)(ii) Ohm's law gives the resistance \(R\) as: \(R = \frac{V}{I} = \frac{6.0\text{ V}}{0.25\text{ A}} = 24\ \Omega\).
(c) For a typical metal wire resistor, as temperature rises, lattice vibrations increase, which increases resistance.
Marking scheme
(a) [1 mark] Electric current (accept current, reject charge/voltage). (b)(i) [3 marks] - 1 mark for time conversion: \(120\text{ s}\) - 1 mark for formula and calculation: \(Q = I \times t = 0.25 \times 120 = 30\) - 1 mark for unit: Coulombs / \(\text{C}\). (b)(ii) [2 marks] - 1 mark for formula: \(R = \frac{V}{I}\) (or substitution \(\frac{6.0}{0.25}\)) - 1 mark for correct value: \(24\) (unit \(\Omega\) not strictly penalized if omitted here, but correct value required). (c) [1 mark] (Resistance) increases.
Question 7 · structured
7.27 marks
A student wants to determine the density of an irregularly shaped stone. (a) Describe how the student can find the volume of the stone using a measuring cylinder containing some water. [3] (b) The mass of the stone is measured using a balance and is found to be \( 54.0\text{ g} \). When the stone is lowered into the measuring cylinder, the water level rises from \( 30.0\text{ cm}^3 \) to \( 50.0\text{ cm}^3 \). Calculate the density of the stone. [3] (c) State the safety precaution the student should take when lowering the stone into the glass measuring cylinder. [1]
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Worked solution
(a) Record the initial volume of water in the measuring cylinder. Gently lower the stone into the cylinder so it is fully submerged. Record the new water level. The volume of the stone is calculated as the difference between the final and initial water level readings. (b) Volume of the stone \( V = 50.0\text{ cm}^3 - 30.0\text{ cm}^3 = 20.0\text{ cm}^3 \). Using the density formula: \( \rho = \frac{m}{V} = \frac{54.0\text{ g}}{20.0\text{ cm}^3} = 2.7\text{ g/cm}^3 \). (c) The stone should be lowered gently using a piece of thread to avoid cracking or breaking the bottom of the glass measuring cylinder.
Marking scheme
(a) [3 marks] - Record initial volume of water [1 mark] - Submerge stone fully and record final volume [1 mark] - Subtract initial volume from final volume [1 mark]
(b) [3 marks] - Calculate volume of stone: \( 20.0\text{ cm}^3 \) [1 mark] - Use of formula \( \rho = m/V \) [1 mark] - Correct density value with unit: \( 2.7\text{ g/cm}^3 \) [1 mark]
(c) [1 mark] - Lower gently / use string to prevent breaking glass cylinder [1 mark]
Question 8 · structured
7.27 marks
A solar panel is used to heat water for a house. The water flows through copper pipes painted matte black behind a glass cover. (a) Explain why the copper pipes are painted matte black. [2] (b) Explain why copper is a suitable material for the pipes. [2] (c) Explain how the glass cover helps to reduce thermal energy loss by convection. [3]
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Worked solution
(a) Matte black surfaces are excellent absorbers of infrared radiation from the Sun, maximizing the transfer of thermal energy to the pipes. (b) Copper is a very good thermal conductor, which allows heat to be transferred quickly through the pipe walls to the water flowing inside. (c) The glass cover traps a layer of air between itself and the pipes. This prevents the warm air from rising and being carried away by external winds (convection currents), thereby reducing convection heat loss.
Marking scheme
(a) [2 marks] - Black is a good absorber of (infrared) radiation [1 mark] - Matte surface absorbs more / reflects less than a shiny surface [1 mark]
(b) [2 marks] - Copper is a good (thermal) conductor [1 mark] - Allows rapid transfer of heat to the water [1 mark]
(c) [3 marks] - Traps a layer of air [1 mark] - Prevents warm air from rising / escaping [1 mark] - Minimizes convection currents to the outside atmosphere [1 mark]
Question 9 · structured
7.27 marks
A circuit contains a \( 9.0\text{ V} \) battery connected in series with two resistors of resistance \( 6.0\\ \Omega \) and \( 12.0\\ \Omega \). (a) Calculate the total combined resistance of the two resistors in series. [2] (b) Calculate the current in the circuit. [2] (c) Describe how the total resistance of the circuit changes if the \( 12.0\\ \Omega \) resistor is replaced by one with a lower resistance. [1] (d) State the function of a fuse in such a circuit. [2]
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Worked solution
(a) For series resistors, the combined resistance is the sum of the individual resistances: \( R_{\text{total}} = R_1 + R_2 = 6.0\\ \Omega + 12.0\\ \Omega = 18.0\\ \Omega \). (b) Using Ohm's law: \( I = \frac{V}{R_{\text{total}}} = \frac{9.0\text{ V}}{18.0\\ \Omega} = 0.50\text{ A} \). (c) If a smaller resistor is used, the total resistance of the series combination will decrease. (d) A fuse contains a thin wire that melts and breaks the circuit when the current exceeds a safe value, protecting the components from damage and preventing fires.
(b) [2 marks] - Formula: \( I = V/R \) [1 mark] - Correct value and unit: \( 0.50\text{ A} \) [1 mark]
(c) [1 mark] - Total resistance decreases [1 mark]
(d) [2 marks] - Fuse wire melts when current is too high [1 mark] - This breaks the circuit/stops current flow [1 mark]
Question 10 · structured
7.27 marks
A student heats water in a metal saucepan on a hotplate.
(a) State the main process by which thermal energy is transferred through: (i) the metal bottom of the saucepan, (ii) the water inside the saucepan.
(b) Describe, in terms of particles (atoms/electrons), how thermal energy is transferred through the metal bottom of the saucepan.
(c) The outside of the saucepan is highly polished and shiny. Explain how this helps to keep the food inside hot for a longer time after the saucepan is removed from the heat source.
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Worked solution
(a)(i) Conduction (a)(ii) Convection
(b) When the saucepan is heated, metal atoms at the bottom gain kinetic energy and vibrate more. These vibrating atoms collide with neighboring atoms, transferring kinetic energy. In addition, metals contain free (delocalised) electrons. These electrons gain kinetic energy, move rapidly through the metal lattice, and transfer energy quickly by colliding with atoms and other electrons.
(c) Highly polished, shiny surfaces are poor emitters of infrared radiation (thermal radiation). This reduces the rate of thermal energy transfer from the outer surface of the saucepan to the cooler surroundings, keeping the food inside hot for longer.
(b) Max [3 marks] from: - Atoms/particles gain kinetic energy and vibrate more [1 mark] - Energy is passed on via collisions with neighboring atoms/particles [1 mark] - Metals contain free (delocalised) electrons [1 mark] - Free electrons move/diffuse through the metal to transfer energy [1 mark]
(c) Max [2 marks] from: - Shiny/polished surfaces are poor/weak emitters of radiation [1 mark] - Reduces the rate of heat loss / thermal energy transfer to the surroundings [1 mark]
Question 11 · structured
7.27 marks
A ray of light is incident on the flat face of a rectangular glass block.
(a) Define the term *refraction*.
(b) The angle of incidence in air is \(45^\circ\) and the angle of refraction in the glass is \(28^\circ\). (i) Calculate the refractive index of the glass. Show your working. (ii) State what happens to the speed of light as it enters the glass block.
(c) Under certain conditions, total internal reflection occurs at the boundary. State the two conditions required for total internal reflection to happen.
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Worked solution
(a) Refraction is the change in direction of a wave (or light) when it passes from one medium to another of different optical density (due to a change in speed).
(b)(ii) The speed of light decreases (slows down).
(c) The two conditions required for total internal reflection are: 1. Light must travel from an optically denser medium to an optically less dense medium (e.g., from glass to air). 2. The angle of incidence must be greater than the critical angle.
Marking scheme
(a) Bending / change in direction of light as it passes into a different medium [1 mark]
(c) - Light travels from a more dense to a less dense medium [1 mark] - Angle of incidence is greater than the critical angle [1 mark]
Paper 43
Answer extended theory structured questions demanding deep explanation and calculation.
10 Question · 80 marks
Question 1 · structured
8 marks
A small toy car A of mass 0.40 kg is moving at a velocity of 3.0 m/s along a frictionless horizontal track. It collides with a stationary toy car B of mass 0.60 kg. After the collision, car A rebounds in the opposite direction at a speed of 0.60 m/s. (a) Calculate the momentum of car A before the collision. (b) Calculate the velocity of car B after the collision. (c) The collision lasts for 0.15 s. Calculate the average force exerted on car A during the collision and state its direction.
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Worked solution
(a) Momentum is given by \(p = mv\). For car A before the collision: \(p = 0.40 \text{ kg} \times 3.0 \text{ m/s} = 1.2 \text{ kg m/s}\). (b) According to the conservation of momentum, total momentum before collision equals total momentum after collision. Let the initial direction of car A be positive. Initial momentum = \(1.2 \text{ kg m/s}\). After collision, car A moves in the opposite direction, so its velocity is \(-0.60 \text{ m/s}\). Final momentum of A = \(0.40 \text{ kg} \times (-0.60 \text{ m/s}) = -0.24 \text{ kg m/s}\). Therefore, \(1.2 = -0.24 + m_B v_B\), which gives \(1.2 + 0.24 = 0.60 \times v_B\). Solving for \(v_B\), we get \(v_B = \frac{1.44}{0.60} = 2.4 \text{ m/s}\). Since the value is positive, car B moves in the original direction of car A. (c) The average force is given by \(F = \frac{\Delta p}{\Delta t}\). The change in momentum of car A is \(\Delta p = p_{\text{final}} - p_{\text{initial}} = -0.24 - 1.2 = -1.44 \text{ kg m/s}\). Thus, \(F = \frac{-1.44}{0.15} = -9.6 \text{ N}\). The magnitude of the average force is \(9.6 \text{ N}\) and the negative sign indicates the direction is opposite to car A's initial motion.
Marking scheme
(a) [2 marks] \(p = mv\) formula or substitution shown: 1 mark. Correct answer with unit \(1.2 \text{ kg m/s}\): 1 mark. (b) [3 marks] Statement or application of conservation of momentum: 1 mark. Substitution of correct momentum values including correct sign for rebounded car A: 1 mark. Correct calculation of velocity with direction \(2.4 \text{ m/s}\) in the original direction of car A: 1 mark. (c) [3 marks] Impulse equation \(F = \frac{\Delta p}{\Delta t}\) or substitution shown: 1 mark. Correct calculation of force magnitude \(9.6 \text{ N}\): 1 mark. Statement that direction is opposite to initial motion of car A: 1 mark.
Question 2 · structured
8 marks
A metal block is heated to a high temperature. One half of its outer surface is painted matte black, and the other half is painted shiny silver. The block is suspended inside a vacuum chamber. (a) State and explain the method of thermal energy transfer from the block to the surroundings, explaining why other methods cannot occur in a vacuum. (b) Two identical temperature sensors are placed equal distances from the block: Sensor X faces the black side and Sensor Y faces the silver side. State and explain which sensor will detect a faster rate of temperature rise. (c) In a separate setup, the same hot block is suspended in air. Describe how convection currents are set up in the air surrounding the block.
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Worked solution
(a) The method of thermal energy transfer is thermal radiation (or infrared radiation). This is because radiation consists of electromagnetic waves which can travel through a vacuum. Conduction and convection both require a physical medium (particles) to transfer energy. Since a vacuum has no particles, these two methods cannot occur. (b) Sensor X will detect a faster rate of temperature rise. This is because matte black surfaces are much better emitters of thermal radiation than shiny silver surfaces, resulting in a higher rate of energy transfer to Sensor X. (c) When suspended in air, the hot block heats the air in contact with it. This air expands, making it less dense than the surrounding cooler air. The warmer, less dense air rises, and cooler, denser air sinks to take its place. This continuous cycle creates convection currents.
Marking scheme
(a) [3 marks] Radiation/infrared radiation identified: 1 mark. Conduction and convection require a medium/particles: 1 mark. Vacuum contains no medium/particles, preventing conduction and convection: 1 mark. (b) [2 marks] Sensor X identified: 1 mark. Matte black is a better/more effective emitter of radiation (or shiny silver is a poor emitter): 1 mark. (c) [3 marks] Air near block expands and density decreases when heated: 1 mark. Warm air rises: 1 mark. Cool air sinks to replace the rising warm air: 1 mark.
Question 3 · structured
8 marks
A student conducts an experiment with a solenoidal coil connected to a sensitive center-zero galvanometer. The student pushes the North pole of a bar magnet into the left-hand end of the coil. (a) Explain why a deflection is observed on the galvanometer as the magnet enters the coil. (b) State Lenz's law and use it to identify the magnetic pole induced at the left-hand end of the coil as the North pole enters. (c) The student now pulls the magnet out of the coil at a much higher speed than it was inserted. Compare the deflection on the galvanometer during removal with the deflection during insertion, explaining the differences.
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Worked solution
(a) When the magnet is pushed into the coil, the magnetic field lines cut the turns of the coil, creating a changing magnetic flux linkage. This induces an electromotive force (e.m.f.) and therefore an electrical current in the closed circuit, which causes the galvanometer to deflect. (b) Lenz's law states that the direction of the induced current is such that it opposes the change that produces it. To oppose the entry of the North pole of the magnet, a North pole is induced at the left-hand end of the coil to create a repulsive force. (c) When the magnet is pulled out, the direction of the induced e.m.f. reverses because the magnetic flux is now decreasing instead of increasing, causing a deflection in the opposite direction. Additionally, because the magnet is pulled out at a higher speed, the rate of change of magnetic flux linkage is greater, which induces a larger e.m.f. and results in a larger maximum deflection.
Marking scheme
(a) [2 marks] Moving magnet causes magnetic field lines to cut the coil / changing magnetic flux: 1 mark. Induction of e.m.f. or current: 1 mark. (b) [3 marks] Lenz's law defined (opposition to change): 1 mark. North pole identified at the left-hand end: 1 mark. Explanation that like poles repel to oppose entry: 1 mark. (c) [3 marks] Deflection is in the opposite direction: 1 mark. Magnitude of deflection is larger: 1 mark. Explanation that higher speed leads to a greater rate of change of flux linkage / rate of cutting field lines: 1 mark.
Question 4 · structured
8 marks
A student investigates electromagnetic induction using a coil of wire connected to a sensitive center-zero galvanometer and a bar magnet.
(a) Explain what is meant by electromagnetic induction. [2]
(b) The student pushes the north pole of the bar magnet into the coil. (i) Describe and explain the deflection observed on the galvanometer during this action. [3] (ii) State two changes to the procedure that would increase the magnitude of the deflection on the galvanometer. [2] (iii) The magnet is then held completely stationary inside the coil. State the reading on the galvanometer. [1]
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Worked solution
(a) Electromagnetic induction is the process where an electromotive force (e.m.f.) or voltage is induced across a conductor when it experiences a changing magnetic field (or cuts across magnetic field lines).
(b)(i) The galvanometer needle deflects to one side (e.g., to the right) and then returns to zero. This is because the moving magnet causes a changing magnetic field through the coil, which induces an e.m.f. and hence a current. Once the magnet stops moving, there is no change in magnetic field, so the current drops back to zero.
(ii) Any two of: - Push the magnet into the coil faster. - Use a stronger magnet. - Use a coil with more turns of wire.
(iii) Zero / no deflection.
Marking scheme
(a) - e.m.f. / voltage is induced in a conductor [1 mark] - when there is a changing magnetic field / conductor cuts magnetic field lines [1 mark]
(b)(i) - needle deflects to one side and returns to zero [1 mark] - moving magnet creates a changing magnetic field / cuts field lines which induces an e.m.f. and current [1 mark] - when motion stops, there is no change in magnetic field so the current returns to zero [1 mark]
(b)(ii) - Any two from: move magnet faster, use a stronger magnet, use a coil with more turns of wire. [2 marks]
(b)(iii) - Zero / no deflection / center position [1 mark]
Question 5 · structured
8 marks
Our understanding of the Universe is built on observations of stars and distant galaxies.
(a) State what is meant by redshift and explain how it supports the Big Bang theory of the origin of the Universe. [3]
(b) Explain the difference between a stable main sequence star and a supernova, in terms of the balance of forces acting within them. [2]
(c) A distant galaxy is observed to be moving away from the Earth at a speed of \(4.4 \times 10^6\text{ m/s}\). (i) Calculate the distance of this galaxy from Earth in meters. Use the Hubble constant \(H_0 = 2.2 \times 10^{-18}\text{ s}^{-1}\). [2] (ii) Convert this distance into light-years. Use the speed of light \(c = 3.0 \times 10^8\text{ m/s}\) and \(1\text{ year} = 3.15 \times 10^7\text{ s}\). [1]
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Worked solution
(a) Redshift is the increase in the observed wavelength (or decrease in frequency) of light from distant galaxies because they are moving away from us. It supports the Big Bang theory because light from almost all distant galaxies is redshifted, demonstrating that the entire Universe is expanding from a single point of origin.
(b) In a stable main sequence star, the inward gravitational force is balanced by the outward thermal and radiation pressure from nuclear fusion. In a supernova, the outward pressure drops due to fuel exhaustion, and gravity causes the core to collapse rapidly, leading to a massive explosion.
(ii) Distance in one light-year = \(c \times 1\text{ year} = (3.0 \times 10^8\text{ m/s}) \times (3.15 \times 10^7\text{ s}) = 9.45 \times 10^{15}\text{ m}\) Distance in light-years = \(\frac{2.0 \times 10^{24}\text{ m}}{9.45 \times 10^{15}\text{ m}} \approx 2.12 \times 10^8\text{ light-years}\) (accept \(2.1 \times 10^8\) to \(2.12 \times 10^8\))
Marking scheme
(a) - Redshift: increase in wavelength / decrease in frequency of light from distant galaxies [1 mark] - Shows galaxies are moving away / receding [1 mark] - Supports Big Bang because it shows the entire Universe is expanding [1 mark]
(b) - In main sequence: inward gravitational force balances outward radiation and thermal pressure [1 mark] - In supernova: outward pressure decreases, gravity wins, causing rapid collapse and explosion [1 mark]
(c)(i) - State or use formula: \(d = v / H_0\) [1 mark] - \(d = 2.0 \times 10^{24}\text{ m}\) [1 mark]
(c)(ii) - Calculates \(1\text{ ly} = 9.45 \times 10^{15}\text{ m}\) and divides \(d\) by this value to get \(2.1 \times 10^8\text{ light-years}\) (accept \(2.1 \times 10^8\) to \(2.12 \times 10^8\)) [1 mark]
Question 6 · structured
8 marks
An experiment is set up to determine the specific heat capacity of aluminum. An aluminum block of mass \(0.80\text{ kg}\) is heated using a \(40\text{ W}\) electrical heater for \(5.0\text{ minutes}\). The temperature of the block rises from \(20^\circ\text{C}\) to \(35^\circ\text{C}\).
(a) Define specific heat capacity. [2]
(b)(i) Calculate the electrical energy supplied to the heater during the \(5.0\text{ minutes}\). [2] (ii) Calculate the experimental value for the specific heat capacity of aluminum. [2] (iii) Explain why the experimental value calculated in (b)(ii) is likely to be greater than the actual accepted value for aluminum. Suggest one practical modification to the setup to reduce this error. [2]
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Worked solution
(a) Specific heat capacity is the energy required per unit mass per unit temperature increase (or the energy needed to raise the temperature of 1 kg of a substance by 1 \(^\circ\)C).
(ii) \(E = m c \Delta \theta\) \(\Delta \theta = 35^\circ\text{C} - 20^\circ\text{C} = 15^\circ\text{C}\) \(12\\,000 = 0.80 \times c \times 15\) \(12\\,000 = 12 \times c\) \(c = \frac{12\\,000}{12} = 1000\text{ J/(kg}\cdot^\circ\text{C)}\)
(iii) Explanation: Some thermal energy is lost to the surroundings (air and thermometer), so not all of the 12,000 J of electrical energy is used to heat the aluminum block. This means the temperature rise (\(\Delta \theta\)) is smaller than it would be without heat loss, leading to a calculated value of \(c\) that is too high.
Modification: Wrap the block in an insulating material (such as cotton wool) or put petroleum jelly in the thermometer hole to improve thermal contact.
Marking scheme
(a) - energy required per unit mass / per kg [1 mark] - per unit temperature increase / per \(^\circ\)C / per K [1 mark]
(b)(i) - Use of \(E = P \times t\) with time converted to seconds (300 s) [1 mark] - \(E = 12\\,000\text{ J}\) [1 mark]
(b)(ii) - Use of \(c = \frac{E}{m \Delta \theta}\) with \(\Delta \theta = 15^\circ\text{C}\) [1 mark] - \(c = 1000\text{ J/(kg}\cdot^\circ\text{C)}\) [1 mark]
(b)(iii) - Heat is lost to surroundings, making \(\Delta \theta\) smaller than expected for the energy supplied [1 mark] - Insulate the block / add oil or petroleum jelly to the thermometer hole [1 mark]
Question 7 · structured
8 marks
An alternating current (a.c.) generator consists of a rectangular coil rotating in a uniform magnetic field.
(a) Explain why an electromotive force (e.m.f.) is induced in the coil as it rotates. [3]
(b) State two ways to increase the maximum value of the induced e.m.f. [2]
(c) The output of the generator is connected to a step-up transformer. The primary coil of the transformer has 200 turns and is connected to an input voltage of \(12\text{ V}\) a.c. The secondary coil has 1500 turns.
(i) Calculate the output voltage of the transformer. [2]
(ii) Assuming the transformer is \(100\%\) efficient and the current in the secondary coil is \(0.080\text{ A}\), calculate the current in the primary coil. [1]
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Worked solution
**(a)** - As the coil rotates, its sides cut the magnetic field lines, causing a change in the magnetic flux linkage. (1) - This changing magnetic flux linkage induces an e.m.f. across the coil. (1) - According to Faraday's law, the magnitude of the induced e.m.f. is directly proportional to the rate of change of flux linkage (or rate of cutting lines). (1)
**(b)** Any two from: - Rotate the coil faster (increase the frequency of rotation). (1) - Use a stronger magnet (increase the magnetic field strength). (1) - Increase the number of turns on the coil. (1) - Increase the cross-sectional area of the coil. (1) - Use a soft iron core inside the coil. (1)
**(c) (i)** - Using the transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) - Substituting the given values: \(\frac{12}{V_s} = \frac{200}{1500}\) (1) - \(V_s = \frac{12 \times 1500}{200} = 90\text{ V}\) (1)
**(a)** - 1 mark: mention of cutting magnetic field lines / change in magnetic flux linkage. - 1 mark: statement that the change in flux induces an e.m.f. - 1 mark: reference to Faraday's law or rate of change of flux linkage determining the magnitude.
**(b)** - 2 marks: any two valid methods to increase the induced e.m.f. (1 mark per correct suggestion).
**(c) (i)** - 1 mark: correct formula or substitution of values. - 1 mark: correct final answer with unit \(90\text{ V}\).
**(c) (ii)** - 1 mark: correct calculation of primary current \(0.60\text{ A}\) (allow error carried forward from (c)(i)).
Question 8 · structured
8 marks
(a) Describe the life cycle of a star that is much more massive than our Sun, starting from its stable main sequence stage until its final stage. [4]
(b) Light emitted from a distant galaxy is observed to be redshifted.
(i) Explain what is meant by the term redshift. [2]
(ii) State and explain how redshift provides evidence for the Big Bang theory. [2]
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Worked solution
**(a)** - A massive star in the main sequence runs out of hydrogen fuel in its core and expands to become a red supergiant. (1) - The core collapses rapidly and the outer layers of the star are ejected in a massive explosion called a supernova. (1) - The remaining core collapses under its own gravity. (1) - Depending on the mass of the remnant, it forms either a highly dense neutron star or a black hole. (1)
**(b) (i)** - Redshift is the increase in the observed wavelength (or decrease in frequency) of light from a source. (1) - This occurs because the light source (galaxy) is moving away from the observer. (1)
**(b) (ii)** - Redshift of light from almost all distant galaxies shows that they are moving away from us, which means the Universe is expanding. (1) - If the Universe is currently expanding, tracing this expansion backward in time suggests that it must have started from a single hot, dense point in the past. (1)
Marking scheme
**(a)** - 1 mark: expansion into a red supergiant stage when hydrogen runs out. - 1 mark: description of the supernova explosion. - 1 mark: gravitational collapse of the remaining core. - 1 mark: identification of the final remnant as either a neutron star or a black hole.
**(b) (i)** - 1 mark: definition that wavelength of light increases / frequency decreases. - 1 mark: link to the source moving away from the observer.
**(b) (ii)** - 1 mark: connection that redshift shows the expansion of the universe / galaxies moving apart. - 1 mark: explanation that tracing expansion backwards leads to a single origin / hot dense starting point.
Question 9 · structured
8 marks
A rectangular block of concrete has dimensions \(0.50\text{ m} \times 0.40\text{ m} \times 1.2\text{ m}\). The density of concrete is \(2400\text{ kg/m}^3\).
(a) The block is placed on flat, horizontal ground.
(i) Calculate the mass of the block. [2]
(ii) Calculate the minimum pressure that the block can exert on the ground. (Take the acceleration of free fall \(g = 9.8\text{ m/s}^2\)) [3]
(b) A U-tube manometer containing water of density \(1000\text{ kg/m}^3\) is connected to a gas supply. The difference in the water levels between the two columns is \(25\text{ cm}\).
Calculate the difference in pressure between the gas supply and the atmosphere. [3]
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Worked solution
**(a) (i)** - Volume of the block \(V = 0.50 \times 0.40 \times 1.2 = 0.24\text{ m}^3\) (1) - Mass \(m = \rho \times V = 2400 \times 0.24 = 576\text{ kg}\) (1)
**(a) (ii)** - Weight \(W = m \times g = 576 \times 9.8 = 5644.8\text{ N}\) (or \(5760\text{ N}\) if using \(g=10\text{ m/s}^2\)) (1) - For minimum pressure, the block must be placed on its largest face to maximize contact area. - Largest surface area \(A = 1.2 \times 0.50 = 0.60\text{ m}^2\) (1) - Pressure \(P = \frac{F}{A} = \frac{5644.8}{0.60} = 9408\text{ Pa} \approx 9400\text{ Pa}\) (or \(9600\text{ Pa}\) if using \(g=10\text{ m/s}^2\)) (1)
**(b)** - Pressure difference formula: \(\Delta P = \rho g h\) - Height \(h = 25\text{ cm} = 0.25\text{ m}\) (1) - \(\Delta P = 1000 \times 9.8 \times 0.25 = 2450\text{ Pa}\) (or \(2500\text{ Pa}\) if using \(g = 10\text{ m/s}^2\)) (2) - (Award 1 mark for substitution, 1 mark for final calculated answer).
**(a) (ii)** - 1 mark: calculation of weight: \(5644.8\text{ N}\) (or \(5760\text{ N}\)). - 1 mark: identifying and calculating the largest surface area: \(0.60\text{ m}^2\). - 1 mark: correct calculation of minimum pressure: \(9400\text{ Pa}\) or \(9408\text{ Pa}\) (allow \(9600\text{ Pa}\) if using \(g = 10\text{ m/s}^2\)).
**(b)** - 1 mark: conversion of height to metres: \(0.25\text{ m}\). - 1 mark: correct substitution into formula: \(1000 \times 9.8 \times 0.25\). - 1 mark: correct pressure value: \(2450\text{ Pa}\) or \(2450\text{ N/m}^2\) (allow \(2500\text{ Pa}\) if using \(g = 10\text{ m/s}^2\)).
Question 10 · structured
8 marks
(a) Define specific heat capacity. [2]
(b) An electric heater of power 45 W is used to heat a metal block of mass 1.2 kg. The heater is switched on for 6.0 minutes. The temperature of the block rises from 20 °C to 45 °C.
(i) Calculate the specific heat capacity of the metal, assuming no thermal energy is transferred to the surroundings.
Specific heat capacity = ................................... J/(kg °C) [3]
(ii) In practice, some thermal energy is lost to the surroundings during the heating process. State and explain the effect of this thermal energy loss on the calculated value of the specific heat capacity. [3]
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Worked solution
(a) Specific heat capacity is defined as the thermal energy required to raise the temperature of a unit mass of a substance by 1 °C (or 1 K).
(b)(i) 1. Calculate the energy supplied by the heater: \(\Delta E = P \times t = 45\text{ W} \times (6.0 \times 60\text{ s}) = 45 \times 360\text{ s} = 16200\text{ J}\)
2. Calculate the temperature change of the block: \(\Delta \theta = 45\text{ °C} - 20\text{ °C} = 25\text{ °C}\)
3. Use the specific heat capacity formula: \(c = \frac{\Delta E}{m \Delta \theta} = \frac{16200}{1.2 \times 25} = \frac{16200}{30} = 540\text{ J/(kg °C)}\)
(b)(ii) The calculated value of the specific heat capacity will be larger than the true value. Some energy supplied by the heater is lost to the surroundings rather than warming the block. Consequently, the energy actually absorbed by the block is less than the calculated 16200 J. Because the calculation uses the higher value of energy supplied (16200 J) to find \(c\) for the measured temperature change, the resulting calculated value is too high.
Marking scheme
(a) - (thermal) energy required per unit mass / per kg [1] - per unit temperature change / per °C / per K [1] (Accept: formula \(c = \frac{\Delta E}{m \Delta \theta}\) with all terms correctly defined for [2], or with one omission/error for [1])
(b)(i) - conversion of time to seconds: \(6.0 \times 60 = 360\text{ s}\) OR calculation of energy \(E = 16200\text{ J}\) [1] - temperature change \(\Delta \theta = 25\text{ °C}\) OR substitution of values into the formula \(c = \frac{E}{m \Delta \theta}\) [1] - correct calculation: \(540\text{ J/(kg °C)}\) (accept 540 without unit if unit is printed on answer line) [1]
(b)(ii) - (calculated value is) larger / higher / too high [1] - energy absorbed by the block is less than the energy supplied by the heater (or some energy is lost to the surroundings) [1] - therefore, a larger energy input is assumed for the measured temperature rise / the calculated value of \(c\) is based on an overestimate of energy absorbed [1]
Paper 53
Perform four practical experiments in laboratory conditions.
4 Question · 40 marks
Question 1 · practical
10 marks
A student investigates the rate of cooling of water in a beaker under different conditions. They set up a beaker containing hot water with no lid, and record its temperature every 60 seconds for 3 minutes. They then repeat the procedure using a beaker with a cardboard lid. (a) Figure 1.1 shows a diagram of the thermometer at the start of the first experiment (no lid). The meniscus of the thermometer liquid is aligned with the mark exactly halfway between 84 and 85 °C. Record this initial temperature \(\theta_0\). (b) The student's recorded temperatures are shown: With no lid: \(t = 0\text{ s}\), \(\theta = \theta_0\); \(t = 60\text{ s}\), \(\theta = 76.5^\circ\text{C}\); \(t = 120\text{ s}\), \(\theta = 70.0^\circ\text{C}\); \(t = 180\text{ s}\), \(\theta = 65.0^\circ\text{C}\). With cardboard lid: \(t = 0\text{ s}\), \(\theta = \theta_0\); \(t = 60\text{ s}\), \(\theta = 79.5^\circ\text{C}\); \(t = 120\text{ s}\), \(\theta = 75.0^\circ\text{C}\); \(t = 180\text{ s}\), \(\theta = 71.5^\circ\text{C}\). (i) Calculate the temperature drop \(\Delta\theta_1\) during the first 180 s for the beaker with no lid, including the correct unit. (ii) Calculate the temperature drop \(\Delta\theta_2\) during the first 180 s for the beaker with the cardboard lid, including the correct unit. (c) Suggest two precautions that should be taken in this experiment to ensure the temperature readings are accurate. (d) State whether the cardboard lid significantly reduces the cooling rate. Justify your answer by reference to the results. (e) The student wants to extend the experiment to investigate how the material of the lid affects the rate of cooling. Identify two variables that must be kept constant.
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Worked solution
(a) \(\theta_0 = 84.5^\circ\text{C}\). (b)(i) \(\Delta\theta_1 = 84.5 - 65.0 = 19.5^\circ\text{C}\). (ii) \(\Delta\theta_2 = 84.5 - 71.5 = 13.0^\circ\text{C}\). (c) Two precautions: 1. Stir the water before reading the temperature. 2. Read the scale with line of sight perpendicular to prevent parallax error. (d) Yes, the cardboard lid significantly reduces the cooling rate because the temperature drop with the lid (\(13.0^\circ\text{C}\)) is significantly smaller than the temperature drop without the lid (\(19.5^\circ\text{C}\)). (e) Two variables to keep constant: 1. Initial volume/mass of hot water. 2. Initial temperature of the water. 3. Room temperature / presence of draughts.
Marking scheme
Total: 10 marks. (a) \(\theta_0 = 84.5^\circ\text{C}\) [1 mark]. (b)(i) Correct subtraction (\(19.5\)) and unit (\(^\circ\text{C}\)) [2 marks]. (ii) Correct subtraction (\(13.0\)) with unit [1 mark]. (c) Any two correct precautions: e.g. stir before reading, thermometer not touching sides/bottom, avoid parallax by reading perpendicularly [2 marks, 1 mark each]. (d) Statement 'yes' / 'lid reduces rate' [1 mark] and comparison justification referring to \(19.5^\circ\text{C}\) vs \(13.0^\circ\text{C}\) [1 mark]. (e) Any two correct constant variables: volume of water, initial temperature of water, same beaker type, ambient room temperature [2 marks, 1 mark each].
Question 2 · practical
10 marks
A student investigates the focal length \(f\) of a converging lens by obtaining a sharp image of an illuminated object on a screen. (a) Figure 2.1 shows a diagram of the experimental setup, drawn to a scale of 1:5. On the diagram, the distance representing the object-to-lens distance is measured as \(x = 4.0\text{ cm}\), and the distance representing the lens-to-screen distance is measured as \(y = 8.0\text{ cm}\). (i) State the actual distance \(u\) from the object to the lens, in cm, using the scale. (ii) State the actual distance \(v\) from the lens to the screen, in cm, using the scale. (b) Calculate the focal length \(f_0\) of the lens using the equation: \(f = \frac{uv}{u+v}\). (c) The student repeats the procedure for several positions, yielding the values in Table 2.1. Table 2.1: Run 1: \(u = 25.0\text{ cm}\), \(v = 50.0\text{ cm}\); Run 2: \(u = 30.0\text{ cm}\), \(v = 37.5\text{ cm}\); Run 3: \(u = 40.0\text{ cm}\), \(v = 26.7\text{ cm}\). (i) For Run 1, Run 2, and Run 3, calculate the values of the focal length \(f_1\), \(f_2\), and \(f_3\). (ii) Calculate the average focal length \(f_{\text{avg}}\) of the lens from these three runs. (d) State one difficulty in obtaining an accurate, sharp image on the screen in this experiment, and suggest how the student can overcome it. (e) Why is it important to perform this experiment in a darkened room?
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Worked solution
(a)(i) \(u = 4.0 \times 5 = 20.0\text{ cm}\). (ii) \(v = 8.0 \times 5 = 40.0\text{ cm}\). (b) \(f_0 = \frac{20.0 \times 40.0}{20.0 + 40.0} = \frac{800}{60} = 13.3\text{ cm}\). (c)(i) \(f_1 = \frac{25.0 \times 50.0}{25.0 + 50.0} = 16.7\text{ cm}\); \(f_2 = \frac{30.0 \times 37.5}{30.0 + 37.5} = 16.7\text{ cm}\); \(f_3 = \frac{40.0 \times 26.7}{40.0 + 26.7} = 16.0\text{ cm}\). (ii) \(f_{\text{avg}} = \frac{16.7 + 16.7 + 16.0}{3} = 16.5\text{ cm}\). (d) Difficulty: It is hard to judge the exact point of sharpest focus because the image remains in focus over a small range of screen movement. Solution: Move the screen back and forth until the image just begins to blur in both directions, then place the screen halfway between these two points. (e) It makes the image on the screen brighter and clearly visible, allowing the sharpest focus to be identified more easily.
Marking scheme
Total: 10 marks. (a)(i) \(u = 20.0\text{ cm}\) [1 mark]. (ii) \(v = 40.0\text{ cm}\) [1 mark]. (b) \(f_0 = 13.3\text{ cm}\) (or 13 cm) [1 mark]. (c)(i) Calculated values of \(f_1\) (\(16.7\)), \(f_2\) (\(16.7\)), and \(f_3\) (\(16.0\)) all correct [2 marks, deduct 1 mark for each incorrect value]. (ii) Calculated average \(f_{\text{avg}} = 16.5\text{ cm}\) [1 mark]. (d) Difficulty identified: image faint, or difficulty finding exact sharp point [1 mark]. Suggestion: move screen back and forth, or use grid, or align components [1 mark]. (e) Brighter image / easier to see sharp focus [1 mark].
Question 3 · practical
10 marks
A student investigates the resistance of a constantan wire. They set up a circuit to measure the current in the wire and the potential difference across various lengths \(L\) of the wire. (a) Draw a circuit diagram containing a cell, a switch, an ammeter, a voltmeter, and a length of resistance wire \(XY\) connected in series, with the voltmeter connected in parallel across the wire \(XY\. (b) The student connects the voltmeter across a length \)L = 40.0\text{ cm}\) of the wire. The ammeter shows a reading of 0.32 A, and the voltmeter shows a reading of 1.28 V. Record the current \(I\) and potential difference \(V\). (c) Calculate the resistance \(R\) of this \(40.0\text{ cm}\) length of wire. (d) The student records the resistance for other lengths as shown: \(L = 20.0\text{ cm}\): \(R = 1.98\ \Omega\); \(L = 60.0\text{ cm}\): \(R = 6.02\ \Omega\); \(L = 80.0\text{ cm}\): \(R = 8.01\ \Omega\). (i) Calculate the ratio \(\frac{R}{L}\) for each length (including the \(40.0\text{ cm}\) length). State the unit for this ratio. (ii) State whether your values for \(\frac{R}{L}\) suggest that the resistance \(R\) is directly proportional to the length \(L\) of the wire. Justify your answer by reference to your calculated ratios. (e) Explain why the switch should be opened between readings.
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Worked solution
(a) A standard circuit containing a cell connected to a switch, an ammeter, and a resistance wire \(XY\) in series, with a voltmeter connected in parallel across the wire \(XY\). (b) \(I = 0.32\text{ A}\) and \(V = 1.28\text{ V}\). (c) \(R = \frac{1.28\text{ V}}{0.32\text{ A}} = 4.00\ \Omega\). (d)(i) Ratios \(\frac{R}{L}\): For \(L = 20.0\text{ cm}\): \(\frac{R}{L} = 0.099\ \Omega/\text{cm}\). For \(L = 40.0\text{ cm}\): \(\frac{R}{L} = 0.100\ \Omega/\text{cm}\). For \(L = 60.0\text{ cm}\): \(\frac{R}{L} = 0.100\ \Omega/\text{cm}\). For \(L = 80.0\text{ cm}\): \(\frac{R}{L} = 0.100\ \Omega/\text{cm}\). Unit is \(\Omega/\text{cm}\) (or \(\Omega/\text{m}\) if lengths converted, yielding 9.9, 10.0, 10.0, 10.0 \(\Omega/\text{m}\)). (ii) Yes, the resistance \(R\) is directly proportional to the length \(L\) because the ratio \(\frac{R}{L}\) is constant within the limits of experimental accuracy. (e) To prevent the wire from heating up, which would increase its resistance and affect the accuracy of the readings.
Marking scheme
Total: 10 marks. (a) Correct circuit symbols for cell, switch, ammeter in series, voltmeter in parallel across wire \(XY\) [1 mark]; completely correct series loop with voltmeter in parallel [1 mark]. (b) Correct reading for \(I = 0.32\text{ A}\) [1 mark] and \(V = 1.28\text{ V}\) [1 mark]. (c) Correct calculation of \(R = 4.0\ \Omega\) (or \(4.00\ \Omega\)) with unit [1 mark]. (d)(i) Correctly calculated ratios [1 mark] and correct unit (\(\Omega/\text{cm}\) or \(\Omega/\text{m}\)) [1 mark]. (ii) Statement 'yes' [1 mark] and justification that ratios are very close / constant within experimental limits [1 mark]. (e) Correct reason: to prevent heating of the wire (which alters its resistance) [1 mark].
Question 4 · practical
10 marks
A student is determining the focal length \(f\) of a convex lens in a practical experiment. (a)(i) The student sets up the apparatus on an optical bench. The illuminated object is placed at the 0.0 cm mark. The lens is placed at the 20.0 cm mark, so the object distance \(u = 20.0\text{ cm}\). The screen is moved until a sharp image of the object is formed on it. The screen is at the 60.2 cm mark. Calculate the image distance \(v\) from the lens. (ii) Calculate the focal length \(f_1\) using the formula: \(f_1 = \frac{u \times v}{u + v}\). (b)(i) The lens is then moved to the 30.0 cm mark, so \(u = 30.0\text{ cm}\). A sharp image is formed when the screen is at the 60.1 cm mark, giving \(v = 30.1\text{ cm}\). Calculate the focal length \(f_2\) using the same formula. (ii) Calculate the average focal length \(f_{\text{avg}}\) of the lens. (c) State two practical precautions that should be taken in this experiment to obtain accurate position readings. (d) Describe the shape of the graph of \(v\) against \(u\) and explain how the value of \(f\) can be determined from the point on the graph where \(u = v\). (e) State one characteristic of the image other than it being inverted when \(u = 30.0\text{ cm}\). (f) State why it is difficult to determine the exact position of the screen for the sharpest image.
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Worked solution
Part (a)(i): The image distance \(v\) is the distance from the lens to the screen. Since the lens is at 20.0 cm and the screen is at 60.2 cm, \(v = 60.2 - 20.0 = 40.2\text{ cm}\). Part (a)(ii): \(f_1 = \frac{20.0 \times 40.2}{20.0 + 40.2} = \frac{804}{60.2} \approx 13.4\text{ cm}\). Part (b)(i): \(f_2 = \frac{30.0 \times 30.1}{30.0 + 30.1} = \frac{903}{60.1} \approx 15.0\text{ cm}\). Part (b)(ii): The average focal length is \(f_{\text{avg}} = \frac{13.36 + 15.02}{2} = 14.19 \approx 14.2\text{ cm}\) (or \\frac{13.4 + 15.0}{2} = 14.2\text{ cm}\)). Part (c): Useful precautions include: 1. Carrying out the experiment in a dark room to see the image clearly. 2. Moving the screen slowly back and forth to locate the position of maximum sharpness. 3. Ensuring the object, lens, and screen are aligned along the same horizontal axis. 4. Avoiding parallax error when reading the ruler by looking perpendicularly at the markings. Part (d): The graph of \(v\) against \(u\) is a downward-sloping curve (hyperbolic-type curve). Since \(1/u + 1/v = 1/f\), when \(u = v\), we have \(2/u = 1/f\), which simplifies to \(u = 2f\). Therefore, at this point, \(f = u/2\) (or \(f = v/2\)). Part (e): When \(u = 30.0\text{ cm}\), which is approximately equal to \(2f\) (since \(f \approx 15.0\text{ cm}\)), the image is real and of the same size as the object. Part (f): The image remains in relatively sharp focus over a small range of screen movement (known as the depth of focus), making the exact center of sharpness hard to judge by eye.
Marking scheme
(a)(i) 1 Mark: Correct calculation of v = 40.2 cm. (a)(ii) 1 Mark: Correct calculation of f1 = 13.4 cm (accept 13 or 13.36). (b)(i) 1 Mark: Correct calculation of f2 = 15.0 cm (accept 15 or 15.02). (b)(ii) 1 Mark: Correct average calculation favg = 14.2 cm (accept 14 or 14.19). (c) 2 Marks: Any two valid precautions (e.g., use a dark room, move screen back and forth, align heights, view ruler perpendicularly to avoid parallax). (d) 2 Marks: 1 mark for describing the curve (decreasing curve/slope), 1 mark for explaining that at u = v, f = u/2 (or f = v/2). (e) 1 Mark: Real or 'same size' (or unmagnified). (f) 1 Mark: Difficulty in identifying the exact point of sharpest focus / image remains sharp over a small range of movement (depth of focus).
Paper 63
Answer alternative-to-practical questions covering experimental design and data analysis.
1 Question · 10 marks
Question 1 · alternativeToPractical
10 marks
A student investigates how the resistance of a constantan wire depends on its length. The student sets up a circuit with a power source, an ammeter, a voltmeter connected across a variable length L of the wire, and a sliding contact. (a) (i) For a length of wire L = 20.0 cm, the student records a potential difference V = 1.20 V and a current I = 0.80 A. Calculate the resistance R of this length of wire, and state its unit. (ii) State one practical precaution the student should take when measuring the length L of the wire to ensure that the measurement is accurate. (b) The incomplete table shows the student's measurements and calculated resistance values. Table: Length L / cm | Potential Difference V / V | Current I / A | Resistance R / \(\Omega\). Row 1: 20.0 | 1.20 | 0.80 | [to be calculated]. Row 2: 40.0 | 1.80 | 0.60 | 3.00. Row 3: 60.0 | 2.16 | 0.48 | 4.50. Row 4: 80.0 | 2.40 | 0.40 | 6.00. Row 5: 100.0 | 2.55 | 0.34 | 7.50. (i) Complete the table by writing in the value of the resistance for L = 20.0 cm. (ii) State why it is important to record the values of resistance to the same number of significant figures or decimal places as the other calculated resistance values in the table. (c) Another student suggests that the resistance R of the wire is directly proportional to its length L. (i) State how the data can be used to test this suggestion by calculating a constant ratio. (ii) Calculate the value of the ratio R/L for L = 40.0 cm and for L = 80.0 cm. Include the unit for this ratio. (iii) State whether your results from (c)(ii) support the suggestion. Justify your answer by referring to your calculations. (d) Current passing through the wire causes it to warm up during the experiment. (i) State the effect that a temperature rise has on the resistance of a metal wire. (ii) Suggest one practical method used in this experiment to minimize the heating effect of the current.
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Worked solution
(a) (i) Resistance R = V / I = 1.20 / 0.80 = 1.50 \(\Omega\). (ii) Precaution: Align the ruler parallel and close to the wire, or view the alignment perpendicularly to prevent parallax error. (b) (i) The value 1.50 must be written in the table (using 2 decimal places to match the consistency of other values like 3.00, 4.50, etc.). (ii) Recording values to the same number of decimal places maintains consistent precision across experimental data. (c) (i) If R is directly proportional to L, then the ratio R/L should be constant. (ii) For L = 40.0 cm: R/L = 3.00 / 40.0 = 0.075 \(\Omega\)/cm. For L = 80.0 cm: R/L = 6.00 / 80.0 = 0.075 \(\Omega\)/cm. (iii) Yes, the results support the suggestion because the ratio R/L is constant (both are exactly 0.075 \(\Omega\)/cm, well within the limits of experimental accuracy). (d) (i) An increase in temperature increases the resistance of the metal wire. (ii) Open the switch (turn off the current) immediately after taking each reading to prevent continuous heating.
Marking scheme
(a)(i) [1 mark] Correct calculation of R = 1.50 with correct unit \(\Omega\). (a)(ii) [1 mark] Valid precaution (e.g. view perpendicular to scale, keep ruler close to wire, ensure wire is straight). (b)(i) [1 mark] Value 1.50 entered in the table (reject 1.5). (b)(ii) [1 mark] Statement referring to maintaining consistent precision/decimal places. (c)(i) [1 mark] Statement that the ratio R/L should be constant. (c)(ii) [1 mark] Correct calculations showing both values as 0.075 with unit \(\Omega\)/cm (or \(\Omega\) cm^{-1}). (c)(iii) [1 mark] Yes, with justification that the values are equal/constant. (d)(i) [1 mark] Resistance increases. (d)(ii) [1 mark] Turn off current/switch off circuit between readings.
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