Cambridge IGCSE · Thinka-original Practice Paper

2024 Cambridge IGCSE Physics (0625) Practice Paper with Answers

Thinka Nov 2024 (V3) Cambridge International A Level-Style Mock — Physics (0625)

320 marks375 mins2024

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 toy car of mass 2.0 kg is pushed along a horizontal track. A forward force of 10 N acts on the car. A frictional force of 4 N opposes the motion. What is the acceleration of the toy car?

A.1.0 m/s²
B.3.0 m/s²
C.5.0 m/s²
D.7.0 m/s²

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Worked solution

First, calculate the resultant force: \( F = 10\text{ N} - 4\text{ N} = 6\text{ N} \). Next, use Newton's second law: \( a = \frac{F}{m} = \frac{6\text{ N}}{2.0\text{ kg}} = 3.0\text{ m/s}^2 \).

Marking scheme

1 mark: Correct calculation of acceleration leading to 3.0 m/s² (Option B).

Question 2 · multipleChoice

1 marks

A designer wants to increase the sensitivity of a liquid-in-glass thermometer. Which combination of changes to the bulb volume and the capillary tube diameter will achieve this?

A.larger bulb volume and wider capillary tube
B.larger bulb volume and narrower capillary tube
C.smaller bulb volume and wider capillary tube
D.smaller bulb volume and narrower capillary tube

Show answer & marking scheme

Worked solution

To increase sensitivity, there should be a larger expansion of liquid for a given temperature change, requiring a larger bulb volume. To make this expansion more visible, a narrower capillary tube is needed so that the liquid moves a greater distance along the tube.

Marking scheme

1 mark: Correct combination of larger bulb volume and narrower capillary tube (Option B).

Question 3 · multipleChoice

1 marks

An electric current of 0.50 A flows through a lamp for 4.0 minutes. How much charge passes through the lamp in this time?

A.2.0 C
B.8.0 C
C.120 C
D.480 C

Show answer & marking scheme

Worked solution

First, convert time into seconds: \( t = 4.0 \times 60 = 240\text{ s} \). Then use the formula for charge: \( Q = I \times t = 0.50\text{ A} \times 240\text{ s} = 120\text{ C} \).

Marking scheme

1 mark: Conversion of minutes to seconds and application of Q = I t to get 120 C (Option C).

Question 4 · multipleChoice

1 marks

Water waves in a ripple tank have a wavelength of 3.0 cm and a speed of 12 cm/s. What is the frequency of the waves?

A.0.25 Hz
B.4.0 Hz
C.9.0 Hz
D.36 Hz

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Worked solution

Use the wave equation: \( v = f \lambda \). Rearranging for frequency: \( f = \frac{v}{\lambda} = \frac{12\text{ cm/s}}{3.0\text{ cm}} = 4.0\text{ Hz} \).

Marking scheme

1 mark: Correct substitution and calculation to find frequency as 4.0 Hz (Option B).

Question 5 · multipleChoice

1 marks

A radioactive sample has a half-life of 20 minutes. Its initial count rate is 800 counts/s. What is the count rate after 1.0 hour?

A.100 counts/s
B.200 counts/s
C.267 counts/s
D.400 counts/s

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Worked solution

1.0 hour equals 60 minutes, which is exactly 3 half-lives (\( 60 / 20 = 3 \)). After 1 half-life: \( 800 / 2 = 400\text{ counts/s} \). After 2 half-lives: \( 400 / 2 = 200\text{ counts/s} \). After 3 half-lives: \( 200 / 2 = 100\text{ counts/s} \).

Marking scheme

1 mark: Determining that 1.0 hour is 3 half-lives and reducing 800 counts/s to 100 counts/s (Option A).

Question 6 · multipleChoice

1 marks

A ray of light strikes a plane mirror at an angle of incidence of 35°. What is the angle between the incident ray and the reflected ray?

A.35°
B.55°
C.70°
D.110°

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Worked solution

The law of reflection states that the angle of incidence equals the angle of reflection. Therefore, both angles are 35° relative to the normal. The total angle between the incident and reflected rays is \( 35^\circ + 35^\circ = 70^\circ \).

Marking scheme

1 mark: Using the law of reflection to calculate the total angle as 70° (Option C).

Question 7 · multipleChoice

1 marks

Which statement correctly describes the relationship between a planet's distance from the Sun and its orbital period?

A.Planets closer to the Sun have a shorter orbital period because they travel at a slower average speed.
B.Planets closer to the Sun have a shorter orbital period because they travel at a faster average speed and have a shorter path.
C.Planets further from the Sun have a shorter orbital period because they travel at a faster average speed.
D.All planets have the same orbital period but travel at different speeds depending on their mass.

Show answer & marking scheme

Worked solution

Planets closer to the Sun experience a stronger gravitational pull, meaning they travel at a faster average orbital speed. Combined with a shorter distance to travel (smaller orbital circumference), they have a significantly shorter orbital period.

Marking scheme

1 mark: Selection of statement B, which correctly accounts for both path length and orbital speed.

Question 8 · multipleChoice

1 marks

An electric motor lifts a weight of 50 N through a vertical height of 4.0 m in a time of 5.0 s. What is the useful power output of the motor?

A.10 W
B.40 W
C.250 W
D.1000 W

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Worked solution

First calculate the work done: \( W = F \times d = 50\text{ N} \times 4.0\text{ m} = 200\text{ J} \). Then calculate power: \( P = \frac{W}{t} = \frac{200\text{ J}}{5.0\text{ s}} = 40\text{ W} \).

Marking scheme

1 mark: Finding the work done (200 J) and dividing by time to find the power (40 W) (Option B).

Question 9 · multipleChoice

1 marks

A student wants to find the volume of a single small glass bead. She pours \( 35\text{ cm}^3 \) of water into a measuring cylinder. She then lowers 20 identical glass beads into the water, and the water level rises to \( 49\text{ cm}^3 \). What is the volume of a single glass bead?

A.0.70 cm^3
B.1.43 cm^3
C.1.75 cm^3
D.2.45 cm^3

Show answer & marking scheme

Worked solution

First, find the total volume of all 20 beads by subtracting the initial volume from the final volume: \( 49\text{ cm}^3 - 35\text{ cm}^3 = 14\text{ cm}^3 \). Then, divide this total volume by the number of beads to find the volume of one bead: \( 14\text{ cm}^3 / 20 = 0.70\text{ cm}^3 \).

Marking scheme

1 mark for the correct calculation leading to option A.

Question 10 · multipleChoice

1 marks

A car starts from rest and accelerates uniformly for \( 10\text{ s} \) to a speed of \( 20\text{ m/s} \). It then travels at this constant speed of \( 20\text{ m/s} \) for \( 15\text{ s} \), and finally decelerates uniformly to rest in a further \( 5\text{ s} \). What is the total distance travelled by the car?

A.350 m
B.400 m
C.450 m
D.600 m

Show answer & marking scheme

Worked solution

The total distance is the area under the speed-time graph. We can divide the motion into three parts: (1) Acceleration: \( \frac{1}{2} \times 10\text{ s} \times 20\text{ m/s} = 100\text{ m} \). (2) Constant speed: \( 15\text{ s} \times 20\text{ m/s} = 300\text{ m} \). (3) Deceleration: \( \frac{1}{2} \times 5\text{ s} \times 20\text{ m/s} = 50\text{ m} \). Total distance = \( 100\text{ m} + 300\text{ m} + 50\text{ m} = 450\text{ m} \).

Marking scheme

1 mark for the correct calculation leading to option C.

Question 11 · multipleChoice

1 marks

An astronaut has a mass of \( 75\text{ kg} \) on Earth, where the gravitational field strength \( g \) is \( 10\text{ N/kg} \). The astronaut travels to Mars, where the gravitational field strength \( g \) is \( 3.7\text{ N/kg} \). What are the mass and weight of the astronaut on Mars?

A.mass = 27.8 kg, weight = 278 N
B.mass = 75 kg, weight = 278 N
C.mass = 75 kg, weight = 750 N
D.mass = 203 kg, weight = 750 N

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Worked solution

Mass is a measure of the amount of matter and does not change with location, so the mass on Mars remains \( 75\text{ kg} \). Weight is the force of gravity on an object and is calculated using \( W = m \times g \). On Mars, \( W = 75\text{ kg} \times 3.7\text{ N/kg} = 277.5\text{ N} \), which rounds to \( 278\text{ N} \).

Marking scheme

1 mark for identifying that mass is constant and calculating the correct weight on Mars.

Question 12 · multipleChoice

1 marks

A uniform rectangular block of wood has a mass of \( 180\text{ g} \) and has dimensions of \( 5.0\text{ cm} \times 6.0\text{ cm} \times 8.0\text{ cm} \). What is the density of the wood?

A.0.75 g/cm^3
B.1.33 g/cm^3
C.2.40 g/cm^3
D.4.32 g/cm^3

Show answer & marking scheme

Worked solution

First, calculate the volume of the block: \( V = 5.0\text{ cm} \times 6.0\text{ cm} \times 8.0\text{ cm} = 240\text{ cm}^3 \). Next, calculate the density using the formula \( \text{density} = \text{mass} / \text{volume} \): \( \text{density} = 180\text{ g} / 240\text{ cm}^3 = 0.75\text{ g/cm}^3 \).

Marking scheme

1 mark for the correct calculation leading to option A.

Question 13 · multipleChoice

1 marks

A gas is sealed in a container of fixed volume. The gas is heated, causing its temperature to rise. Which statement describes what happens to the gas molecules?

A.The molecules expand, which increases the pressure on the container walls.
B.The molecules hit the container walls more frequently and with greater force.
C.The molecules move further apart, which increases their volume.
D.The molecules lose kinetic energy and collide less often.

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Worked solution

As the gas is heated, the average kinetic energy of the molecules increases, meaning they move faster. Since the volume of the container is fixed, the faster-moving molecules collide with the container walls more frequently and with greater force, which increases the pressure.

Marking scheme

1 mark for identifying the correct explanation of pressure increase due to kinetic theory.

Question 14 · multipleChoice

1 marks

A ray of light in air strikes the flat surface of a semi-circular glass block at an angle of incidence of \( 45^\circ \). The refractive index of the glass is \( 1.5 \). What is the angle of refraction inside the glass?

A.28 degrees
B.30 degrees
C.45 degrees
D.68 degrees

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Worked solution

Using Snell's Law: \( n = \frac{\sin i}{\sin r} \), where \( n = 1.5 \) and \( i = 45^\circ \). Therefore, \( \sin r = \frac{\sin 45^\circ}{1.5} = \frac{0.707}{1.5} \approx 0.471 \). Taking the inverse sine: \( r = \arcsin(0.471) \approx 28^\circ \).

Marking scheme

1 mark for the correct application of Snell's Law to find the angle of refraction.

Question 15 · multipleChoice

1 marks

A simple electric circuit has a current of \( 0.40\text{ A} \) flowing through a resistor. How much charge passes through the resistor in \( 5.0\text{ minutes} \)?

A.2.0 C
B.12 C
C.120 C
D.750 C

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Worked solution

The relationship between charge \( Q \), current \( I \), and time \( t \) is given by \( Q = I \times t \). First, convert the time to seconds: \( t = 5.0\text{ minutes} = 5.0 \times 60 = 300\text{ s} \). Then, calculate the charge: \( Q = 0.40\text{ A} \times 300\text{ s} = 120\text{ C} \).

Marking scheme

1 mark for converting time to seconds and correctly calculating the charge.

Question 16 · multipleChoice

1 marks

A radioactive source has a half-life of \( 4.0\text{ hours} \). Initially, the count rate from the source is \( 800\text{ counts per minute} \). After how long will the count rate from the source decrease to \( 100\text{ counts per minute} \)?

A.8.0 hours
B.12 hours
C.16 hours
D.32 hours

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Worked solution

Find the number of half-lives required: \( 800 \to 400 \) (1 half-life), \( 400 \to 200 \) (2 half-lives), \( 200 \to 100 \) (3 half-lives). Since each half-life is \( 4.0\text{ hours} \), the total time is \( 3 \times 4.0\text{ hours} = 12\text{ hours} \).

Marking scheme

1 mark for determining the number of half-lives and multiplying by the half-life duration.

Question 17 · multipleChoice

1 marks

A toy car moves along a straight path. The speed-time graph for the first 10 seconds of its journey is as follows: it starts from rest, accelerates uniformly to a speed of 6 m/s in 4 seconds, travels at this constant speed of 6 m/s for 4 seconds, and then decelerates uniformly to a stop in 2 seconds. What is the total distance travelled by the toy car?

A.36 m
B.42 m
C.48 m
D.60 m

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Worked solution

The distance travelled is represented by the area under a speed-time graph. The area can be calculated as the area of a trapezium: \(\text{Distance} = \frac{1}{2} \times (a + b) \times h\), where \(a = 4\text{ s}\) (the time spent at constant speed, from 4 s to 8 s), \(b = 10\text{ s}\) (the total time), and \(h = 6\text{ m/s}\) (the constant speed). Thus, \(\text{Distance} = \frac{1}{2} \times (4 + 10) \times 6 = 7 \times 6 = 42\text{ m}\). Alternatively, calculating the area of each phase: Phase 1 (acceleration): \(\frac{1}{2} \times 4\text{ s} \times 6\text{ m/s} = 12\text{ m}\). Phase 2 (constant speed): \(4\text{ s} \times 6\text{ m/s} = 24\text{ m}\). Phase 3 (deceleration): \(\frac{1}{2} \times 2\text{ s} \times 6\text{ m/s} = 6\text{ m}\). Total distance = \(12 + 24 + 6 = 42\text{ m}\).

Marking scheme

1 mark for the correct calculation of total distance under the graph (either using the trapezium formula or by summing the individual areas).

Question 18 · multipleChoice

1 marks

A student wants to find the density of an irregularly shaped piece of rock. First, she measures its mass using a digital balance and finds it is 135 g. Next, she pours 50 cm\(^3\) of water into a measuring cylinder. After lowering the rock carefully into the water, the water level rises to 80 cm\(^3\). What is the density of the rock?

A.1.69 g/cm\(^3\)
B.2.70 g/cm\(^3\)
C.4.50 g/cm\(^3\)
D.5.40 g/cm\(^3\)

Show answer & marking scheme

Worked solution

First, determine the volume of the rock by calculating the difference in the water levels: \(\text{Volume} = 80\text{ cm}^3 - 50\text{ cm}^3 = 30\text{ cm}^3\). Then, calculate the density using the formula: \(\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{135\text{ g}}{30\text{ cm}^3} = 4.50\text{ g/cm}^3\).

Marking scheme

1 mark for the correct volume calculation (30 cm\(^3\)) and correct application of the density formula to obtain 4.50 g/cm\(^3\).

Question 19 · multipleChoice

1 marks

An unstretched spring has a length of 12.0 cm. When a load of 4.0 N is suspended from the spring, its length becomes 15.0 cm. Assuming the limit of proportionality has not been exceeded, what is the length of the spring when a load of 12.0 N is suspended from it?

A.9.0 cm
B.21.0 cm
C.36.0 cm
D.45.0 cm

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Worked solution

First, find the extension produced by the 4.0 N load: \(\text{Extension} = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\). Since extension is directly proportional to the load (Hooke's Law), a load of 12.0 N (which is three times greater than 4.0 N) will produce three times the extension: \(\text{New Extension} = 3.0\text{ cm} \times 3 = 9.0\text{ cm}\). The new length of the spring is the original length plus the new extension: \(\text{Length} = 12.0\text{ cm} + 9.0\text{ cm} = 21.0\text{ cm}\).

Marking scheme

1 mark for identifying the extension of 3.0 cm, calculating the new extension of 9.0 cm, and adding it to the unstretched length to obtain 21.0 cm.

Question 20 · multipleChoice

1 marks

A sealed, rigid metal container holds a fixed mass of gas. The container is heated, causing the temperature of the gas to rise. Which statement describes what happens to the gas particles as the temperature increases?

A.The particles expand and become larger, increasing the pressure on the container walls.
B.The particles collide with each other less frequently but with greater force.
C.The particles move faster on average, leading to more frequent and more forceful collisions with the container walls.
D.The average distance between the particles increases, causing them to strike the walls harder.

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Worked solution

An increase in temperature increases the average kinetic energy of the gas particles, meaning they move faster on average. Because they move faster within a fixed volume, they collide with the walls of the container more frequently and with greater force, which increases the pressure.

Marking scheme

1 mark for identifying that higher temperature causes faster particle movement, resulting in more frequent and more forceful collisions with the walls.

Question 21 · multipleChoice

1 marks

A ray of light is travelling inside a glass block towards the boundary with air. The critical angle for this glass-air boundary is 42\(^\circ\). Which angle of incidence in the glass will result in the ray undergoing total internal reflection?

A.0\(^\circ\)
B.30\(^\circ\)
C.42\(^\circ\)
D.45\(^\circ\)

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Worked solution

Total internal reflection occurs when light is travelling from a more optically dense medium (glass) to a less optically dense medium (air), and the angle of incidence is greater than the critical angle. Since the critical angle is 42\(^\circ\), any angle of incidence greater than 42\(^\circ\) will result in total internal reflection. 45\(^\circ\) is the only option greater than 42\(^\circ\).

Marking scheme

1 mark for stating or applying the condition that the angle of incidence must be strictly greater than the critical angle for total internal reflection to occur.

Question 22 · multipleChoice

1 marks

A small electric motor is connected to a battery. A current of 0.25 A flows through the motor for exactly 3.0 minutes. How much electrical charge passes through the motor in this time?

A.0.75 C
B.12 C
C.45 C
D.720 C

Show answer & marking scheme

Worked solution

First, convert the time from minutes to seconds: \(t = 3.0\text{ minutes} = 3.0 \times 60 = 180\text{ s}\). Next, apply the formula for charge: \(Q = I \times t\). Substituting the values: \(Q = 0.25\text{ A} \times 180\text{ s} = 45\text{ C}\).

Marking scheme

1 mark for converting time to seconds (180 s) and correctly calculating the charge as 45 C.

Question 23 · multipleChoice

1 marks

A radioactive source emits three types of radiation: alpha (\(\alpha\)), beta (\(\beta\)), and gamma (\(\gamma\)). A sheet of thick paper is placed between the source and a radiation detector. Which types of radiation can pass through the sheet of paper and be detected on the other side?

A.alpha and beta only
B.beta and gamma only
C.alpha and gamma only
D.gamma only

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Worked solution

Alpha particles have very low penetrating power and are absorbed (stopped) by a thin sheet of paper. Beta particles and gamma rays are more penetrating and can easily pass through paper. Therefore, both beta and gamma radiation will pass through the sheet of paper and be detected on the other side.

Marking scheme

1 mark for identifying that alpha is stopped by paper, while beta and gamma pass through.

Question 24 · multipleChoice

1 marks

Which statement correctly describes the motion or characteristics of planets in our Solar System?

A.All planets orbit the Sun in perfectly circular orbits with the Sun at the exact centre.
B.Planets closer to the Sun have a shorter orbital period than planets further from the Sun.
C.The orbital speed of a planet remains completely constant at every point in its orbit.
D.Inner planets are made mostly of gas, while outer planets are made mostly of rock.

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Worked solution

Planets closer to the Sun experience a stronger gravitational attraction, travel a shorter orbital path, and move at faster average speeds, resulting in a much shorter orbital period. For example, Mercury's orbital period is 88 days, whereas outer planets like Neptune take many decades to orbit the Sun once.

Marking scheme

1 mark for identifying that orbital period increases with distance from the Sun.

Question 25 · multipleChoice

1 marks

A student wants to find the volume of an irregularly shaped small stone. The student pours some water into a measuring cylinder. The correct volume reading at the bottom of the water meniscus is \(35\text{ cm}^3\). However, the student misreads this initial level by looking at the top of the meniscus, recording it as \(36\text{ cm}^3\). The student then lowers the stone into the water. The correct final volume reading at the bottom of the meniscus is \(48\text{ cm}^3\). The student uses their recorded initial reading of \(36\text{ cm}^3\) and this correct final reading of \(48\text{ cm}^3\) to calculate the volume of the stone. What is the student's calculated volume, and how does it compare to the actual volume of the stone?

A.\(12\text{ cm}^3\), which is \(1\text{ cm}^3\) less than the actual volume
B.\(12\text{ cm}^3\), which is \(1\text{ cm}^3\) more than the actual volume
C.\(13\text{ cm}^3\), which is equal to the actual volume
D.\(14\text{ cm}^3\), which is \(1\text{ cm}^3\) more than the actual volume

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Worked solution

The actual volume of the stone is the difference between the actual final volume and the actual initial volume: \(48\text{ cm}^3 - 35\text{ cm}^3 = 13\text{ cm}^3\). The student's calculated volume is based on the misread initial value: \(48\text{ cm}^3 - 36\text{ cm}^3 = 12\text{ cm}^3\). Therefore, the calculated volume is \(12\text{ cm}^3\), which is \(1\text{ cm}^3\) less than the actual volume.

Marking scheme

1 mark: Correct calculation of calculated volume (12 cm³) and comparison (1 cm³ less).

Question 26 · multipleChoice

1 marks

A skydiver falls from an aircraft. After accelerating, they reach terminal velocity. They then open their parachute. Which statement describes the motion of the skydiver from the moment the parachute is opened until they reach their new, lower terminal velocity?

A.The speed decreases and the deceleration increases.
B.The speed decreases and the deceleration decreases.
C.The speed increases and the acceleration decreases.
D.The speed decreases and the deceleration remains constant.

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Worked solution

When the parachute is opened, the upward drag force suddenly becomes much larger than the downward weight, resulting in a large upward net force. This causes a high initial deceleration. As the skydiver's downward speed decreases, the air resistance decreases as well, which reduces the upward net force. Consequently, the deceleration decreases until it becomes zero, at which point the new, slower terminal velocity is reached.

Marking scheme

1 mark: Correctly identifying that both speed and deceleration decrease.

Question 27 · multipleChoice

1 marks

A student mixes \(100\text{ cm}^3\) of liquid X (density \(0.80\text{ g/cm}^3\)) with \(200\text{ cm}^3\) of liquid Y (density \(1.10\text{ g/cm}^3\)). The liquids mix thoroughly without any change in the total volume. What is the density of the mixture?

A.\(0.90\text{ g/cm}^3\)
B.\(0.95\text{ g/cm}^3\)
C.\(1.00\text{ g/cm}^3\)
D.\(1.05\text{ g/cm}^3\)

Show answer & marking scheme

Worked solution

First, calculate the mass of each liquid: \\ \\ \(\text{Mass of X} = \text{volume} \times \text{density} = 100\text{ cm}^3 \times 0.80\text{ g/cm}^3 = 80\text{ g}\)\
\(\text{Mass of Y} = 200\text{ cm}^3 \times 1.10\text{ g/cm}^3 = 220\text{ g}\)\
\
Next, find the total mass and total volume: \\ \\ \(\text{Total mass} = 80\text{ g} + 220\text{ g} = 300\text{ g}\)\
\(\text{Total volume} = 100\text{ cm}^3 + 200\text{ cm}^3 = 300\text{ cm}^3}\)\
\
Finally, calculate the density of the mixture: \\ \\ \(\text{Density} = \frac{\text{Total mass}}{\text{Total volume}} = \frac{300\text{ g}}{300\text{ cm}^3} = 1.00\text{ g/cm}^3\)

Marking scheme

1 mark: Correct application of density formula to find total mass and then mixture density.

Question 28 · multipleChoice

1 marks

An unstretched spring has a length of \(12.0\text{ cm}\). When a \(4.0\text{ N}\) load is hung from it, the length becomes \(15.0\text{ cm}\). Assuming the limit of proportionality is not exceeded, what load is needed to produce a total spring length of \(21.0\text{ cm}\)?

A.\(7.0\text{ N}\)
B.\(8.0\text{ N}\)
C.\(12.0\text{ N}\)
D.\(28.0\text{ N}\)

Show answer & marking scheme

Worked solution

Calculate the initial extension: \(15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\). The load per unit extension (spring constant) is: \(\frac{4.0\text{ N}}{3.0\text{ cm}} = \frac{4}{3}\text{ N/cm}\). The new desired length is \(21.0\text{ cm}\), which corresponds to an extension of: \(21.0\text{ cm} - 12.0\text{ cm} = 9.0\text{ cm}\). Using Hooke's law, the load needed is: \(F = kx = \left(\frac{4.0\text{ N}}{3.0\text{ cm}}\right) \times 9.0\text{ cm} = 12.0\text{ N}\).

Marking scheme

1 mark: Correct calculation of extension and proportional load to get 12.0 N.

Question 29 · multipleChoice

1 marks

A beaker of water is placed on a tripod and heated from below with a Bunsen burner. Which statement describes how a convection current is formed in the water?

A.Water at the bottom is heated, expands, becomes less dense, and rises.
B.Water at the bottom is heated, contracts, becomes more dense, and rises.
C.Water at the bottom is heated, expands, becomes more dense, and sinks.
D.Water at the bottom is heated, contracts, becomes less dense, and sinks.

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Worked solution

When the water at the bottom of the beaker is heated, the thermal energy causes the water molecules to move faster and spread further apart (expand). This expansion decreases the density of the heated water. The less dense hot water rises to the top, while the cooler, denser water at the top sinks to replace it, forming a convection current.

Marking scheme

1 mark: Correct explanation of convection current formation (heated water expands, density decreases, and rises).

Question 30 · multipleChoice

1 marks

A ray of light in air is incident on the surface of a rectangular glass block at an angle of incidence of \(40^\circ\). How does the path and speed of the light change as it enters the glass block?

A.It bends away from the normal because its speed increases.
B.It bends away from the normal because its speed decreases.
C.It bends towards the normal because its speed increases.
D.It bends towards the normal because its speed decreases.

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Worked solution

Glass is optically denser than air. When a ray of light enters an optically denser medium, its speed decreases. Because it slows down, the ray bends towards the normal.

Marking scheme

1 mark: Correctly identifying that the light slows down and bends towards the normal.

Question 31 · multipleChoice

1 marks

A current of \(0.25\text{ A}\) flows through a filament lamp for a time of \(4.0\text{ minutes}\). What is the total charge that passes through the lamp during this time?

A.\(1.0\text{ C}\)
B.\(16\text{ C}\)
C.\(60\text{ C}\)
D.\(960\text{ C}\)

Show answer & marking scheme

Worked solution

Charge \(Q\) is given by the formula: \\ \\ \(Q = I \times t\) \\ \\ where \(I = 0.25\text{ A}\) and \(t = 4.0\text{ minutes} = 4.0 \times 60\text{ seconds} = 240\text{ s}\). Therefore: \\ \\ \(Q = 0.25\text{ A} \times 240\text{ s} = 60\text{ C}\).

Marking scheme

1 mark: Correct conversion of time to seconds and calculation of charge.

Question 32 · multipleChoice

1 marks

A radioactive sample has a half-life of \(6.0\text{ hours}\). The initial emission rate from the sample is measured as \(800\text{ counts per minute}\). What is the expected emission rate from the sample after \(18\text{ hours}\) have elapsed, assuming background radiation is negligible?

A.\(100\text{ counts per minute}\)
B.\(200\text{ counts per minute}\)
C.\(267\text{ counts per minute}\)
D.\(400\text{ counts per minute}\)

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Worked solution

Determine the number of half-lives that have elapsed in \(18\text{ hours}\): \\ \\ \(\text{Number of half-lives} = \frac{18\text{ hours}}{6.0\text{ hours}} = 3\) \\ \\ After each half-life, the activity is halved: \\ \\ - After 1 half-life (6 hours): \(800 / 2 = 400\text{ counts per minute}\) \\ - After 2 half-lives (12 hours): \(400 / 2 = 200\text{ counts per minute}\) \\ - After 3 half-lives (18 hours): \(200 / 2 = 100\text{ counts per minute}\)

Marking scheme

1 mark: Correct calculation of the number of half-lives and the resulting count rate.

Question 33 · multipleChoice

1 marks

A ray of light travelling in air strikes the flat surface of a glass block at an angle of incidence of \(35^\circ\). What is a possible angle of refraction inside the glass, and in which direction does the ray bend?

A.\(22^\circ\), bending towards the normal
B.\(22^\circ\), bending away from the normal
C.\(50^\circ\), bending towards the normal
D.\(50^\circ\), bending away from the normal

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Worked solution

When light travels from an optically less dense medium (air) to an optically denser medium (glass), it refracts (bends) towards the normal. This means the angle of refraction must be smaller than the angle of incidence. Since the angle of incidence is \(35^\circ\), the angle of refraction must be less than \(35^\circ\) (i.e. \(22^\circ\)) and it bends towards the normal.

Marking scheme

1 mark for identifying that the angle of refraction must be smaller than the angle of incidence because the ray bends towards the normal.

Question 34 · multipleChoice

1 marks

A student heats a beaker of water from below. How is thermal energy mainly transferred through the water, and how does the density of the heated water change?

A.convection, density decreases
B.convection, density increases
C.conduction, density decreases
D.conduction, density increases

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Worked solution

In liquids, thermal energy is transferred mainly by convection. When water at the bottom of the beaker is heated, it expands, becomes less dense, and rises, setting up a convection current.

Marking scheme

1 mark for correctly identifying convection and the decrease in density of the heated water.

Question 35 · multipleChoice

1 marks

An unstretched spring has a length of \(12.0\text{ cm}\). When a load of \(4.0\text{ N}\) is suspended from the spring, its length becomes \(15.0\text{ cm}\). What is the total length of the spring when a load of \(8.0\text{ N}\) is suspended from it, assuming the limit of proportionality is not exceeded?

A.\(18.0\text{ cm}\)
B.\(20.0\text{ cm}\)
C.\(24.0\text{ cm}\)
D.\(30.0\text{ cm}\)

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Worked solution

First, calculate the extension produced by the \(4.0\text{ N}\) load: \(15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\). Since extension is directly proportional to load, doubling the load to \(8.0\text{ N}\) doubles the extension: \(3.0\text{ cm} \times 2 = 6.0\text{ cm}\). The total length is the original length plus this new extension: \(12.0\text{ cm} + 6.0\text{ cm} = 18.0\text{ cm}\).

Marking scheme

1 mark for calculating the new extension as \(6.0\text{ cm}\) and adding it to the unstretched length to obtain \(18.0\text{ cm}\).

Question 36 · multipleChoice

1 marks

A car travels at a constant speed of \(15\text{ m/s}\) for \(20\text{ s}\). It then accelerates uniformly to a speed of \(25\text{ m/s}\) over the next \(10\text{ s}\). What is the total distance travelled by the car during the entire \(30\text{ s}\) journey?

A.\(400\text{ m}\)
B.\(450\text{ m}\)
C.\(500\text{ m}\)
D.\(750\text{ m}\)

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Worked solution

The motion is divided into two parts. Distance in the first part: \(d_1 = 15\text{ m/s} \times 20\text{ s} = 300\text{ m}\). Distance in the second part (uniform acceleration): \(d_2 = \text{average speed} \times \text{time} = \frac{15 + 25}{2} \times 10 = 20 \times 10 = 200\text{ m}\). Total distance = \(300\text{ m} + 200\text{ m} = 500\text{ m}\).

Marking scheme

1 mark for calculating both distances correctly and summing them to get \(500\text{ m}\).

Question 37 · multipleChoice

1 marks

A small lamp is connected to a battery. A current of \(0.40\text{ A}\) flows through the lamp. How much charge passes through the lamp in \(5.0\text{ minutes}\)?

A.\(2.0\text{ C}\)
B.\(12\text{ C}\)
C.\(120\text{ C}\)
D.\(200\text{ C}\)

Show answer & marking scheme

Worked solution

Using the equation \(Q = I \times t\). First, convert minutes to seconds: \(t = 5.0\text{ minutes} = 5.0 \times 60 = 300\text{ s}\). Then, calculate charge: \(Q = 0.40\text{ A} \times 300\text{ s} = 120\text{ C}\).

Marking scheme

1 mark for converting time to seconds and correctly calculating the total charge as \(120\text{ C}\).

Question 38 · multipleChoice

1 marks

Which row correctly identifies the ionizing power and the penetrating power of alpha (\(\alpha\)) radiation compared to beta (\(\beta\)) and gamma (\(\gamma\)) radiation?

A.Ionizing power: High | Penetrating power: Low
B.Ionizing power: High | Penetrating power: High
C.Ionizing power: Low | Penetrating power: Low
D.Ionizing power: Low | Penetrating power: High

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Worked solution

Alpha particles have the largest mass and charge (+2), which makes them highly ionizing. Because they interact strongly with other atoms, they lose energy quickly and therefore have very low penetrating power, being stopped by a sheet of paper or a few centimetres of air.

Marking scheme

1 mark for correctly matching high ionizing power with low penetrating power for alpha radiation.

Question 39 · multipleChoice

1 marks

A student measures the mass of an irregular piece of rock as \(160\text{ g}\). She pours \(50\text{ cm}^3\) of water into a measuring cylinder and lowers the rock into it, causing the water level to rise to \(90\text{ cm}^3\). What is the density of the rock?

A.\(1.8\text{ g/cm}^3\)
B.\(3.2\text{ g/cm}^3\)
C.\(4.0\text{ g/cm}^3\)
D.\(8.0\text{ g/cm}^3\)

Show answer & marking scheme

Worked solution

The volume of the rock is found by displacement: \(V = 90\text{ cm}^3 - 50\text{ cm}^3 = 40\text{ cm}^3\). Using the density formula: \(\rho = \frac{m}{V} = \frac{160\text{ g}}{40\text{ cm}^3} = 4.0\text{ g/cm}^3\).

Marking scheme

1 mark for calculating the correct volume and using the density formula to find \(4.0\text{ g/cm}^3\).

Question 40 · multipleChoice

1 marks

Two musical notes, X and Y, are played. Note X is louder and has a lower pitch than note Y. How do the amplitude and the frequency of the sound wave for note X compare with those for note Y?

A.Note X has a larger amplitude and a higher frequency than note Y.
B.Note X has a larger amplitude and a lower frequency than note Y.
C.Note X has a smaller amplitude and a higher frequency than note Y.
D.Note X has a smaller amplitude and a lower frequency than note Y.

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Worked solution

The loudness of a sound depends on the amplitude of the wave: a louder sound has a larger amplitude. The pitch of a sound depends on the frequency: a lower pitch corresponds to a lower frequency. Therefore, Note X must have a larger amplitude and a lower frequency than Note Y.

Marking scheme

1 mark for correctly matching amplitude to loudness and frequency to pitch.

Paper 23

Answer forty multiple-choice questions on Extended Physics.

40 Question · 40 marks

Question 1 · multipleChoice

1 marks

A student uses a micrometer screw gauge to measure the diameter of a thin copper wire. Before taking the measurement, the student closes the jaws fully and notes a zero error reading of \(+0.03\text{ mm}\). When measuring the wire, the micrometer reading is \(2.44\text{ mm}\). What is the actual diameter of the wire?

A.2.41 mm
B.2.44 mm
C.2.47 mm
D.2.04 mm

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Worked solution

To find the actual diameter, subtract the positive zero error from the measured value: \(\text{Actual diameter} = \text{Measured reading} - \text{Zero error} = 2.44\text{ mm} - 0.03\text{ mm} = 2.41\text{ mm}\).

Marking scheme

1 mark for selecting the correct option A (2.41 mm).

Question 2 · multipleChoice

1 marks

A toy car starts from rest and accelerates uniformly to a speed of \(6.0\text{ m/s}\) in \(4.0\text{ s}\). It then travels at this constant speed for \(5.0\text{ s}\), before decelerating uniformly to rest in a further \(3.0\text{ s}\). What is the total distance travelled by the toy car?

A.36 m
B.48 m
C.51 m
D.72 m

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Worked solution

The total distance is the area under the speed-time graph. This can be calculated as the area of a trapezium: \(\text{Distance} = \frac{1}{2} \times (a + b) \times h\), where \(a = 5.0\text{ s}\) (time spent at constant speed) and \(b = 12.0\text{ s}\) (total time of travel: \(4.0 + 5.0 + 3.0\)). Height \(h = 6.0\text{ m/s}\). Therefore, \(\text{Distance} = \frac{1}{2} \times (5.0 + 12.0) \times 6.0 = 51\text{ m}\).

Marking scheme

1 mark for calculating the total area under the graph to get 51 m (Option C).

Question 3 · multipleChoice

1 marks

A metal cylinder has a mass of \(160\text{ g}\) and a density of \(8.0\text{ g/cm}^3\). It is completely submerged in a measuring cylinder containing \(50\text{ cm}^3\) of water. What is the new reading of the water level in the measuring cylinder?

A.30 cm³
B.50 cm³
C.70 cm³
D.210 cm³

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Worked solution

First, find the volume of the metal cylinder: \(\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{160\text{ g}}{8.0\text{ g/cm}^3} = 20\text{ cm}^3\). The new water level is the sum of the initial water volume and the volume of the cylinder: \(50\text{ cm}^3 + 20\text{ cm}^3 = 70\text{ cm}^3\).

Marking scheme

1 mark for correctly calculating the cylinder volume as 20 cm³ and finding the final water level of 70 cm³ (Option C).

Question 4 · multipleChoice

1 marks

A trolley of mass \(2.0\text{ kg}\) moving at a speed of \(4.0\text{ m/s}\) collides with a stationary trolley of mass \(3.0\text{ kg}\). After the collision, the two trolleys stick together and move off with a common velocity \(v\). What is the value of \(v\)?

A.1.6 m/s
B.2.0 m/s
C.2.7 m/s
D.4.0 m/s

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Worked solution

Using the principle of conservation of momentum: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\). Substituting the known values: \((2.0 \times 4.0) + (3.0 \times 0) = (2.0 + 3.0) v\), which gives \(8.0 = 5.0 v\). Solving for \(v\) yields \(v = 1.6\text{ m/s}\).

Marking scheme

1 mark for applying conservation of momentum to obtain 1.6 m/s (Option A).

Question 5 · multipleChoice

1 marks

A fixed mass of gas is kept in a sealed container of constant volume. The temperature of the gas is increased. How does this affect the frequency of collisions of the gas molecules with the walls of the container, and how does it affect the average force of each collision?

A.Frequency of collisions: Increases | Average force of each collision: Increases
B.Frequency of collisions: Increases | Average force of each collision: Remains constant
C.Frequency of collisions: Remains constant | Average force of each collision: Increases
D.Frequency of collisions: Remains constant | Average force of each collision: Remains constant

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Worked solution

Increasing the temperature increases the average kinetic energy and speed of the gas molecules. Because the molecules move faster, they collide with the walls of the container more frequently. In addition, each collision involves a larger change in momentum, meaning the average force exerted during each collision also increases.

Marking scheme

1 mark for identifying that both the collision frequency and the average force of each collision increase (Option A).

Question 6 · multipleChoice

1 marks

A ray of monochromatic light travels through glass and meets the glass-air boundary. The refractive index of the glass is \(1.50\). What is the critical angle for this boundary?

A.30°
B.42°
C.49°
D.56°

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Worked solution

The critical angle \(c\) is calculated using \(\sin(c) = \frac{1}{n}\), where \(n\) is the refractive index. Substituting \(n = 1.50\) gives \(\sin(c) = \frac{1}{1.50} \approx 0.667\). Therefore, \(c = \arcsin(0.667) \approx 41.8^\circ\), which is approximately \(42^\circ\).

Marking scheme

1 mark for calculating the critical angle as approximately 42° (Option B).

Question 7 · multipleChoice

1 marks

A uniform metal wire of length \(L\) and cross-sectional area \(A\) has a resistance \(R\). A second wire made of the same metal has a length of \(2L\) and a cross-sectional area of \(2A\). 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 of a wire is given by \(R = \rho \frac{L}{A}\). For the second wire, its resistance is \(R' = \rho \frac{2L}{2A} = \rho \frac{L}{A} = R\). Therefore, its resistance remains exactly \(R\).

Marking scheme

1 mark for correctly showing that doubling both length and area leaves the resistance unchanged (Option B).

Question 8 · multipleChoice

1 marks

Light from a distant galaxy is observed to have a longer wavelength than the light emitted by a stationary source of the same elements on Earth. Which statement is correct about this observation?

A.The galaxy is moving away from Earth, and the light is redshifted.
B.The galaxy is moving towards Earth, and the light is redshifted.
C.The galaxy is moving away from Earth, and the light is blueshifted.
D.The galaxy is moving towards Earth, and the light is blueshifted.

Show answer & marking scheme

Worked solution

An increase in observed wavelength of light (shifting towards the red end of the visible spectrum) is known as redshift. This shift occurs when the light source (the galaxy) is moving away from the observer on Earth.

Marking scheme

1 mark for correctly identifying that a longer wavelength represents redshift and implies movement away from Earth (Option A).

Question 9 · multipleChoice

1 marks

A toy car starts from rest and accelerates uniformly to a maximum speed of \(15.0\text{ m/s}\) in \(4.0\text{ s}\). It then travels at this constant speed for \(10.0\text{ s}\), before decelerating uniformly to rest in a final \(6.0\text{ s}\). What is the average speed of the car for the entire journey?

A.7.50 m/s
B.11.3 m/s
C.12.5 m/s
D.15.0 m/s

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Worked solution

1. Calculate the distance covered in each phase:
- Acceleration phase: \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 15.0\text{ m/s} = 30.0\text{ m}\).
- Constant speed phase: \(d_2 = \text{speed} \times \text{time} = 15.0\text{ m/s} \times 10.0\text{ s} = 150.0\text{ m}\).
- Deceleration phase: \(d_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6.0\text{ s} \times 15.0\text{ m/s} = 45.0\text{ m}\).

2. Calculate the total distance:
\(D = 30.0 + 150.0 + 45.0 = 225.0\text{ m}\).

3. Calculate the total time:
\(T = 4.0 + 10.0 + 6.0 = 20.0\text{ s}\).

4. Calculate the average speed:
\(\text{Average speed} = \frac{D}{T} = \frac{225.0\text{ m}}{20.0\text{ s}} = 11.25\text{ m/s}\), which rounds to \(11.3\text{ m/s}\) to 3 significant figures.

Marking scheme

1 mark for the correct calculation of total distance and total time, leading to the correct average speed of 11.3 m/s (Option B).

Question 10 · multipleChoice

1 marks

An electric motor is used to lift a crate of mass \(40\text{ kg}\) vertically upwards through a height of \(12\text{ m}\). The motor has an efficiency of \(60\%\). The acceleration of free fall \(g\) is \(10\text{ m/s}^2\). What is the total electrical energy input to the motor?

A.2.9 kJ
B.4.8 kJ
C.8.0 kJ
D.29 kJ

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Worked solution

1. Calculate the useful work output (gravitational potential energy gained):
\(\Delta E_p = mgh = 40\text{ kg} \times 10\text{ m/s}^2 \times 12\text{ m} = 4800\text{ J}\).

2. Use the efficiency formula to find the electrical energy input:
\(\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\%\)
\(0.60 = \frac{4800\text{ J}}{\text{Total energy input}}\)
\(\text{Total energy input} = \frac{4800\text{ J}}{0.60} = 8000\text{ J} = 8.0\text{ kJ}\).

Marking scheme

1 mark for calculating the potential energy (4.8 kJ) and dividing by the efficiency (0.60) to obtain the correct input energy of 8.0 kJ (Option C).

Question 11 · multipleChoice

1 marks

An electric heater with a power rating of \(800\text{ W}\) is used to heat a \(2.0\text{ kg}\) block of metal. The temperature of the block increases from \(20^\circ\text{C}\) to \(50^\circ\text{C}\) in \(3.0\text{ minutes}\). Assuming there is no thermal energy loss to the surroundings, what is the specific heat capacity of the metal?

A.40 J / (kg °C)
B.1200 J / (kg °C)
C.2400 J / (kg °C)
D.4320 J / (kg °C)

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Worked solution

1. Calculate the total energy supplied by the heater:
\(Q = P \times t = 800\text{ W} \times (3.0 \times 60)\text{ s} = 800 \times 180 = 144,000\text{ J}\).

2. Calculate the temperature rise:
\(\Delta \theta = 50^\circ\text{C} - 20^\circ\text{C} = 30^\circ\text{C}\).

3. Use the thermal energy equation \(Q = mc\Delta \theta\) to solve for \(c\):
\(144,000\text{ J} = 2.0\text{ kg} \times c \times 30^\circ\text{C}\)
\(144,000 = 60 \times c\)
\(c = \frac{144,000}{60} = 2400\text{ J}/(\text{kg}^\circ\text{C})\).

Marking scheme

1 mark for correctly converting time to seconds, calculating total energy, and solving for the correct specific heat capacity of 2400 J/(kg °C) (Option C).

Question 12 · multipleChoice

1 marks

A ray of light traveling inside a glass block is incident on the boundary with air. The refractive index of the glass is \(1.50\). What is the critical angle for this boundary, and what happens to the ray if the angle of incidence in the glass is \(45^\circ\)?

A.Critical angle is \(41.8^\circ\); the ray undergoes total internal reflection.
B.Critical angle is \(41.8^\circ\); the ray refracts into the air.
C.Critical angle is \(48.2^\circ\); the ray undergoes total internal reflection.
D.Critical angle is \(48.2^\circ\); the ray refracts into the air.

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Worked solution

1. Calculate the critical angle \(c\) using the formula:
\(\sin(c) = \frac{1}{n} = \frac{1}{1.50} \approx 0.6667\)
\(c = \arcsin(0.6667) \approx 41.8^\circ\).

2. Compare the angle of incidence \(i\) with the critical angle \(c\):
Here, \(i = 45^\circ\). Since \(i > c\) (\(45^\circ > 41.8^\circ\)) and the light is traveling from a more optically dense medium (glass) to a less optically dense medium (air), the ray undergoes total internal reflection.

Marking scheme

1 mark for calculating the critical angle as 41.8° and identifying that total internal reflection occurs because the angle of incidence exceeds the critical angle (Option A).

Question 13 · multipleChoice

1 marks

A wire of length \(L\) and uniform cross-sectional area \(A\) has a resistance \(R\). 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 in terms of \(R\)?

A.0.5 R
B.1.0 R
C.2.0 R
D.4.0 R

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Worked solution

1. Express the resistance of the first wire: \(R = \rho \frac{L}{A}\), where \(A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}\).

2. Determine the cross-sectional area of the second wire, \(A'\):
Since the diameter of the second wire is twice that of the first (\(2d\)), the area scales with the square of the diameter: \(A' = \pi \left(\frac{2d}{2}\right)^2 = \pi d^2 = 4A\).

3. Calculate the resistance of the second wire:
\(R' = \rho \frac{2L}{A'} = \rho \frac{2L}{4A} = 0.5 \left(\rho \frac{L}{A}\right) = 0.5 R\).

Marking scheme

1 mark for identifying that area is proportional to the square of diameter, setting up the new resistance equation, and correctly simplifying to 0.5 R (Option A).

Question 14 · multipleChoice

1 marks

An ideal transformer has \(800\) turns on its primary coil and \(40\) turns on its secondary coil. The primary coil is connected to a \(230\text{ V}\) AC mains supply. A \(5.75\ \Omega\) resistor is connected across the secondary coil. What is the current in the primary coil?

A.0.10 A
B.2.0 A
C.4.0 A
D.40 A

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Worked solution

1. Calculate the secondary voltage \(V_s\):
\(\frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = 230\text{ V} \times \frac{40}{800} = 11.5\text{ V}\).

2. Calculate the secondary current \(I_s\):
\(I_s = \frac{V_s}{R} = \frac{11.5\text{ V}}{5.75\ \Omega} = 2.0\text{ A}\).

3. Use the power conservation for an ideal (100% efficient) transformer:
\(P_p = P_s \Rightarrow V_p I_p = V_s I_s\)
\(230\text{ V} \times I_p = 11.5\text{ V} \times 2.0\text{ A} = 23.0\text{ W}\)
\(I_p = \frac{23.0\text{ W}}{230\text{ V}} = 0.10\text{ A}\).

Marking scheme

1 mark for calculating secondary voltage (11.5 V), secondary current (2.0 A), and using primary-secondary power balance to find the correct primary current of 0.10 A (Option A).

Question 15 · multipleChoice

1 marks

A radioactive source has a half-life of \(6.0\text{ hours}\). The background radiation in the laboratory is constant at \(20\text{ counts/minute}\). Initially, a detector placed near the source registers a total count rate of \(340\text{ counts/minute}\). What total count rate does the detector register after \(18\text{ hours}\)?

A.40 counts/minute
B.42.5 counts/minute
C.60 counts/minute
D.113 counts/minute

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Worked solution

1. Determine the initial count rate due to the source alone:
\(\text{Source count rate} = \text{Total count rate} - \text{Background rate} = 340 - 20 = 320\text{ counts/minute}\).

2. Determine the number of half-lives that have elapsed in 18 hours:
\(n = \frac{18\text{ hours}}{6.0\text{ hours}} = 3.0\text{ half-lives}\).

3. Calculate the source activity after 3 half-lives:
\(\text{Final source count rate} = \frac{320}{2^3} = \frac{320}{8} = 40\text{ counts/minute}\).

4. Calculate the total count rate registered by the detector:
\(\text{Total count rate} = \text{Final source count rate} + \text{Background rate} = 40 + 20 = 60\text{ counts/minute}\).

Marking scheme

1 mark for subtracting background first, halving the source activity three times to get 40 counts/minute, and adding background back to find the correct final total of 60 counts/minute (Option C).

Question 16 · multipleChoice

1 marks

Light from a distant galaxy is observed to be redshifted. What does this redshift indicate about the movement of the galaxy, and how does the speed of recession of the galaxy relate to its distance from Earth?

A.The galaxy is moving away from Earth, and its speed of recession is directly proportional to its distance from Earth.
B.The galaxy is moving away from Earth, and its speed of recession is inversely proportional to its distance from Earth.
C.The galaxy is moving towards Earth, and its speed of recession is directly proportional to its distance from Earth.
D.The galaxy is moving towards Earth, and its speed of recession is inversely proportional to its distance from Earth.

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Worked solution

1. Redshift refers to an increase in the observed wavelength of light from distant galaxies, which shows that the source is moving away from the observer (Earth).
2. According to Hubble's Law, the speed of recession \(v\) of a galaxy is directly proportional to its distance \(d\) from us (\(v = H_0 d\)).

Marking scheme

1 mark for identifying that redshift shows the galaxy is moving away from Earth, and that the speed of recession is directly proportional to its distance (Option A).

Question 17 · multipleChoice

1 marks

Wire X has length \(L\) and diameter \(d\). Wire Y of the same material has length \(2L\) and diameter \(2d\). If the resistance of wire X is \(R\), what is the resistance of wire Y?

A.\(R/2\)
B.\(R\)
C.\(2R\)
D.\(4R\)

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Worked solution

The resistance \(R\) of a wire of length \(L\) and cross-sectional area \(A\) is given by \(R = \rho \frac{L}{A}\). The cross-sectional area is proportional to the square of the diameter, \(A \propto d^2\). Therefore, wire Y has 4 times the area of wire X. Substituting these relationships into the formula for wire Y gives: \(R_Y = \rho \frac{2L}{4A_X} = \frac{1}{2} R\). Hence, the resistance of wire Y is \(R/2\).

Marking scheme

1 mark: Correctly identifies that resistance is proportional to length and inversely proportional to the square of the diameter, leading to option A.

Question 18 · multipleChoice

1 marks

Two identical metal canisters are filled with equal volumes of hot water at \(80^\circ\text{C}\). One canister has a polished silver outer surface, and the other has a matte black outer surface. They are left in a cool room at \(20^\circ\text{C}\). After 10 minutes, their temperatures are measured. Which canister has cooled more, and why?

A.The matte black canister, because it is a better emitter of infrared radiation.
B.The matte black canister, because it is a better absorber of infrared radiation.
C.The polished silver canister, because it is a better emitter of infrared radiation.
D.The polished silver canister, because it is a better reflector of infrared radiation.

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Worked solution

Matte black surfaces are highly efficient emitters of thermal infrared radiation compared to polished silver surfaces, which are poor emitters (and good reflectors). Since emission is the primary radiative mechanism for a hot object cooling down in cooler surroundings, the matte black canister loses heat more rapidly and cools down more.

Marking scheme

1 mark: Correctly identifies the matte black canister as the better emitter, leading to greater cooling.

Question 19 · multipleChoice

1 marks

A toy car starts from rest and accelerates uniformly to a speed of \(6.0\text{ m/s}\) in \(4.0\text{ s}\). It then travels at this constant speed for \(5.0\text{ s}\), and finally decelerates uniformly to rest in \(3.0\text{ s}\). What is the total distance travelled by the toy car?

A.\(42.0\text{ m}\)
B.\(48.0\text{ m}\)
C.\(51.0\text{ m}\)
D.\(72.0\text{ m}\)

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Worked solution

The total distance is the area under the velocity-time graph, which is a trapezoid of height \(h = 6.0\text{ m/s}\), bottom parallel side (total time) \(b = 4.0 + 5.0 + 3.0 = 12.0\text{ s}\), and top parallel side (constant speed duration) \(a = 5.0\text{ s}\). Using the trapezoid area formula: \(\text{Distance} = \frac{1}{2} \times (a + b) \times h = \frac{1}{2} \times (5.0 + 12.0) \times 6.0 = 17.0 \times 3.0 = 51.0\text{ m}\).

Marking scheme

1 mark: Correctly calculates the total distance from the area under the graph to get 51.0 m.

Question 20 · multipleChoice

1 marks

An unstretched spring has a length of \(12.0\text{ cm}\). When a load of \(4.0\text{ N}\) is suspended from the spring, its length becomes \(15.0\text{ cm}\). Assuming the limit of proportionality is not exceeded, what is the total length of the spring when a load of \(10.0\text{ N}\) is suspended from it?

A.\(7.5\text{ cm}\)
B.\(19.5\text{ cm}\)
C.\(22.0\text{ cm}\)
D.\(37.5\text{ cm}\)

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Worked solution

First, find the initial extension: \(\Delta x_1 = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\) for a load of \(4.0\text{ N}\). By Hooke's Law, extension is directly proportional to load: \(\Delta x \propto F\). For a \(10.0\text{ N}\) load, the new extension \(\Delta x_2\) is given by: \(\Delta x_2 = 3.0\text{ cm} \times \frac{10.0\text{ N}}{4.0\text{ N}} = 7.5\text{ cm}\). The total length of the spring is the original length plus this new extension: \(12.0\text{ cm} + 7.5\text{ cm} = 19.5\text{ cm}\).

Marking scheme

1 mark: Finds the correct extension of 7.5 cm for 10.0 N and adds it to the original length of 12.0 cm to get 19.5 cm.

Question 21 · multipleChoice

1 marks

Light from a distant galaxy is observed to have a longer wavelength than light from a similar source on Earth. What is the explanation for this observation and what does it tell us about the galaxy?

A.The galaxy is moving towards Earth, showing blueshift.
B.The galaxy is moving towards Earth, showing redshift.
C.The galaxy is moving away from Earth, showing blueshift.
D.The galaxy is moving away from Earth, showing redshift.

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Worked solution

An increase in the observed wavelength of light from distant galaxies is called redshift. Redshift indicates that the source of light is moving away from the observer on Earth, which supports the expansion of the Universe.

Marking scheme

1 mark: Identifies that an increase in wavelength corresponds to redshift and indicates motion away from the Earth.

Question 22 · multipleChoice

1 marks

A radiation detector is used to measure the activity of a radioactive sample. The background radiation count rate is constantly \(20\text{ counts/minute}\). At time \(t = 0\), the recorded count rate is \(340\text{ counts/minute}\). After \(6.0\text{ hours}\), the recorded count rate is \(100\text{ counts/minute}\). What is the half-life of the sample?

A.\(1.5\text{ hours}\)
B.\(2.0\text{ hours}\)
C.\(3.0\text{ hours}\)
D.\(4.5\text{ hours}\)

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Worked solution

Subtract the background count rate of \(20\text{ counts/minute}\) from the initial and final rates to find the active count rates of the sample alone. Initial sample rate: \(340 - 20 = 320\text{ counts/minute}\). Final sample rate: \(100 - 20 = 80\text{ counts/minute}\). The count rate decreased from \(320\) to \(80\) which is a factor of \(1/4\), or two half-lives (\(320 \rightarrow 160 \rightarrow 80\)). Therefore, \(2 \times \text{half-life} = 6.0\text{ hours}\), meaning the half-life is \(3.0\text{ hours}\).

Marking scheme

1 mark: Correctly subtracts background radiation, determines that the sample underwent two half-lives in 6.0 hours, and calculates the half-life as 3.0 hours.

Question 23 · multipleChoice

1 marks

A ray of monochromatic light passes from air into a glass block. The angle of incidence in air is \(45^\circ\) and the angle of refraction in glass is \(28^\circ\). What is the speed of light in this glass block? (Speed of light in air = \(3.0 \times 10^8\text{ m/s}\))

A.\(1.5 \times 10^8\text{ m/s}\)
B.\(2.0 \times 10^8\text{ m/s}\)
C.\(3.0 \times 10^8\text{ m/s}\)
D.\(4.5 \times 10^8\text{ m/s}\)

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Worked solution

The refractive index \(n\) of the glass is given by Snell's Law: \(n = \frac{\sin(i)}{\sin(r)} = \frac{\sin(45^\circ)}{\sin(28^\circ)} \approx \frac{0.7071}{0.4695} \approx 1.51\). The speed of light in glass \(v\) is given by: \(v = \frac{c}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.51} \approx 1.99 \times 10^8\text{ m/s}\), which rounds to \(2.0 \times 10^8\text{ m/s}\).

Marking scheme

1 mark: Correctly calculates refractive index as ~1.51 and uses it to find the speed of light in glass as 2.0 * 10^8 m/s.

Question 24 · multipleChoice

1 marks

An ideal step-down transformer has a primary coil with \(1200\text{ turns}\) and a secondary coil with \(300\text{ turns}\). The primary coil is connected to a \(240\text{ V}\) a.c. supply. A resistor of resistance \(12\ \Omega\) is connected across the secondary coil. What is the current in the primary coil?

A.\(0.31\text{ A}\)
B.\(1.25\text{ A}\)
C.\(5.0\text{ A}\)
D.\(20\text{ A}\)

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Worked solution

First, calculate the secondary voltage: \(V_s = V_p \times \frac{N_s}{N_p} = 240\text{ V} \times \frac{300}{1200} = 60\text{ V}\). Next, find the current in the secondary circuit using Ohm's Law: \(I_s = \frac{V_s}{R} = \frac{60\text{ V}}{12\ \Omega} = 5.0\text{ A}\). For an ideal transformer, the electrical power is fully transferred: \(P = V_p I_p = V_s I_s\). Therefore, \(240 \times I_p = 60 \times 5.0\), leading to \(I_p = \frac{300}{240} = 1.25\text{ A}\).

Marking scheme

1 mark: Correctly determines the secondary voltage (60 V), the secondary current (5.0 A), and uses power conservation to find the primary current of 1.25 A.

Question 25 · multipleChoice

1 marks

A student uses a micrometer screw gauge to measure the diameter of a uniform thin copper wire. Before making the measurement, they close the jaws of the micrometer fully and note that the scale reads \(-0.04\text{ mm}\). When they measure the wire, the reading on the main scale and thimble scale combined is \(1.32\text{ mm}\). What is the actual diameter of the wire?

A.\(1.28\text{ mm}\)
B.\(1.32\text{ mm}\)
C.\(1.36\text{ mm}\)
D.\(1.40\text{ mm}\)

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Worked solution

The micrometer has a negative zero error of \(-0.04\text{ mm}\). The corrected reading is calculated as: \(\text{Corrected reading} = \text{Measured reading} - \text{Zero error} = 1.32\text{ mm} - (-0.04\text{ mm}) = 1.36\text{ mm}\).

Marking scheme

Correct option is C. 1 mark for calculating the actual diameter by subtracting the negative zero error.

Question 26 · multipleChoice

1 marks

An irregular stone of mass \(120\text{ g}\) is lowered into a measuring cylinder containing \(50\text{ cm}^3\) of water. The water level rises to \(98\text{ cm}^3\). The stone is then removed, and a wooden block of mass \(40\text{ g}\) is tied to the stone. When both are submerged together in the cylinder, the water level rises to \(148\text{ cm}^3\). What is the density of the wooden block?

A.\(0.40\text{ g/cm}^3\)
B.\(0.80\text{ g/cm}^3\)
C.\(1.25\text{ g/cm}^3\)
D.\(2.50\text{ g/cm}^3\)

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Worked solution

First, find the volume of the stone: \(V_{\text{stone}} = 98\text{ cm}^3 - 50\text{ cm}^3 = 48\text{ cm}^3\). Next, find the volume of the wooden block: \(V_{\text{wood}} = 148\text{ cm}^3 - 98\text{ cm}^3 = 50\text{ cm}^3\). The density of the wooden block is: \(\rho = \frac{\text{mass}}{\text{volume}} = \frac{40\text{ g}}{50\text{ cm}^3} = 0.80\text{ g/cm}^3\).

Marking scheme

Correct option is B. 1 mark for using the difference in water levels to find the volume of the wooden block and correctly calculating its density.

Question 27 · multipleChoice

1 marks

A toy car of mass \(0.50\text{ kg}\) travels at a velocity of \(6.0\text{ m/s}\) to the right. It collides with a stationary toy truck of mass \(1.5\text{ kg}\). After the collision, the toy car rebounds to the left at a speed of \(1.5\text{ m/s}\). What is the velocity of the toy truck immediately after the collision?

A.\(1.5\text{ m/s}\) to the right
B.\(1.5\text{ m/s}\) to the left
C.\(2.5\text{ m/s}\) to the right
D.\(2.5\text{ m/s}\) to the left

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Worked solution

Using the principle of conservation of momentum: \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\). Let the direction to the right be positive. Thus: \(u_1 = +6.0\text{ m/s}\), \(u_2 = 0\text{ m/s}\), and \(v_1 = -1.5\text{ m/s}\). Substituting the values: \((0.50 \times 6.0) + (1.5 \times 0) = (0.50 \times -1.5) + (1.5 \times v_2) \Rightarrow 3.0 = -0.75 + 1.5 v_2 \Rightarrow 3.75 = 1.5 v_2 \Rightarrow v_2 = +2.5\text{ m/s}\). Since \(v_2\) is positive, the velocity is \(2.5\text{ m/s}\) to the right.

Marking scheme

Correct option is C. 1 mark for correctly applying conservation of momentum with appropriate signs for directions.

Question 28 · multipleChoice

1 marks

A sealed gas syringe contains a fixed mass of gas at a pressure of \(1.0 \times 10^5\text{ Pa}\). The piston is slowly pushed in, reducing the volume of the gas to one-third of its original volume, while the temperature of the gas is kept constant. According to the kinetic particle model, why does the pressure of the gas increase?

A.The average kinetic energy of the gas particles increases, so they strike the walls with greater force.
B.The gas particles collide with each other more frequently, causing them to speed up and exert more force on the walls.
C.The frequency of collisions between the gas particles and the cylinder walls increases, whilst the average force of each collision remains constant.
D.The attractive forces between the gas particles increase as they are closer together, causing them to pull harder on the cylinder walls.

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Worked solution

Since the temperature is constant, the average kinetic energy and speed of the gas particles remain unchanged, meaning the average force exerted during each collision remains the same. However, because the volume is reduced, the particles are closer together, which increases the frequency of their collisions with the container walls, thus increasing the pressure.

Marking scheme

Correct option is C. 1 mark for identifying that constant temperature means constant average collision force, and smaller volume increases collision frequency with the walls.

Question 29 · multipleChoice

1 marks

Four identical copper cans are filled with equal masses of water at \(90\text{ }^\circ\text{C}\). The outer surfaces of the cans are finished differently. Can 1: Dull, black surface, with a lid. Can 2: Dull, black surface, without a lid. Can 3: Shiny, silver surface, with a lid. Can 4: Shiny, silver surface, without a lid. The cans are left in a draught-free room at \(20\text{ }^\circ\text{C}\). Which list shows the order of the cans from the one that cools down the fastest to the one that cools down the slowest?

A.Can 2 \(\rightarrow\) Can 1 \(\rightarrow\) Can 4 \(\rightarrow\) Can 3
B.Can 2 \(\rightarrow\) Can 4 \(\rightarrow\) Can 1 \(\rightarrow\) Can 3
C.Can 3 \(\rightarrow\) Can 1 \(\rightarrow\) Can 4 \(\rightarrow\) Can 2
D.Can 3 \(\rightarrow\) Can 4 \(\rightarrow\) Can 1 \(\rightarrow\) Can 2

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Worked solution

Evaporation is the fastest mechanism of heat loss here; hence, the cans without lids (Can 2 and Can 4) will cool down faster than the cans with lids (Can 1 and Can 3). Between Can 2 and Can 4, Can 2 has a dull black surface which is a better emitter of radiation than the shiny silver surface of Can 4, so Can 2 cools faster than Can 4. Similarly, between the lidded cans, Can 1 (dull black) is a better emitter than Can 3 (shiny silver), so Can 1 cools faster than Can 3. Thus, the order of cooling rate from fastest to slowest is Can 2, Can 4, Can 1, Can 3.

Marking scheme

Correct option is B. 1 mark for correctly identifying that evaporation (no lid) is faster than radiation alone, and dull black is a better emitter than shiny silver.

Question 30 · multipleChoice

1 marks

An object is placed at a distance \(d\) in front of a thin converging lens of focal length \(f\). Which of the following conditions will produce a virtual, magnified, and upright image?

A.\(d < f\)
B.\(f < d < 2f\)
C.\(d = 2f\)
D.\(d > 2f\)

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Worked solution

A converging lens acts as a magnifying glass when the object is placed between the optical center and the principal focus (i.e., at an object distance \(d < f\)). The image produced under this condition is virtual, upright, and magnified.

Marking scheme

Correct option is A. 1 mark for recognizing that a virtual, magnified, and upright image is formed when the object distance is less than the focal length.

Question 31 · multipleChoice

1 marks

A high-resistance voltmeter is connected across a cell of electromotive force (e.m.f.) \(E\) and internal resistance \(r\). When a resistor of resistance \(R\) is connected in parallel with the voltmeter, the reading on the voltmeter drops from \(E\) to \(V\). Which expression gives the current \(I\) in the circuit?

A.\(I = \frac{E}{r}\)
B.\(I = \frac{E + V}{r}\)
C.\(I = \frac{E - V}{r}\)
D.\(I = \frac{V}{R + r}\)

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Worked solution

The difference between the electromotive force \(E\) and the terminal potential difference \(V\) is equal to the 'lost volts' across the internal resistance \(r\). That is, \(\text{lost volts} = E - V\). Since this potential drop occurs across the internal resistance \(r\), the current \(I\) is given by Ohm's law as: \(I = \frac{E - V}{r}\).

Marking scheme

Correct option is C. 1 mark for relating terminal potential difference, e.m.f., and internal resistance to determine the expression for current.

Question 32 · multipleChoice

1 marks

Light from a distant galaxy is observed to have a longer wavelength than light from a similar source in a laboratory on Earth. What does this redshift indicate, and how is it used to support the Big Bang theory?

A.The galaxy is moving towards Earth, indicating that the Universe is contracting.
B.The galaxy is moving away from Earth, indicating that the Universe is expanding.
C.The galaxy is rotating rapidly, indicating the presence of a supermassive black hole.
D.The galaxy is cooling down, indicating that the Universe is reaching thermal equilibrium.

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Worked solution

The increase in the observed wavelength of light from distant galaxies is known as redshift, which indicates that these galaxies are moving away from us. Since almost all distant galaxies show redshift, and the speed of recession is proportional to their distance, this provides strong evidence that the Universe is expanding, which supports the Big Bang theory.

Marking scheme

Correct option is B. 1 mark for identifying redshift as evidence of galaxies moving away and the expansion of the Universe.

Question 33 · multipleChoice

1 marks

A car accelerates from rest at a constant rate of \( 2.0\text{ m/s}^2 \) for \( 6.0\text{ s} \). It then travels at a constant speed for \( 10\text{ s} \), and finally decelerates uniformly to rest in \( 4.0\text{ s} \). What is the average speed of the car for the entire journey?

A.\( 6.0\text{ m/s} \)
B.\( 9.0\text{ m/s} \)
C.\( 10\text{ m/s} \)
D.\( 12\text{ m/s} \)

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Worked solution

First, calculate the distance in each phase. Phase 1: \( s_1 = \frac{1}{2} a t^2 = 0.5 \times 2.0 \times 6.0^2 = 36\text{ m} \). The speed reached is \( v = a t = 2.0 \times 6.0 = 12\text{ m/s} \). Phase 2: \( s_2 = v t = 12 \times 10 = 120\text{ m} \). Phase 3: \( s_3 = \frac{1}{2}(u + v)t = 0.5 \times 12 \times 4.0 = 24\text{ m} \). Total distance \( S = 36 + 120 + 24 = 180\text{ m} \). Total time \( T = 6.0 + 10 + 4.0 = 20\text{ s} \). Average speed \( = S / T = 180 / 20 = 9.0\text{ m/s} \).

Marking scheme

1 mark for calculating the total distance of 180 m and dividing by the total time of 20 s to obtain 9.0 m/s.

Question 34 · multipleChoice

1 marks

An electric motor with an efficiency of \( 60\% \) is used to lift \( 120\text{ kg} \) of water through a vertical height of \( 5.0\text{ m} \) in \( 10\text{ s} \). What is the electrical power input to the motor? (Take \( g = 9.8\text{ N/kg} \))

A.\( 350\text{ W} \)
B.\( 590\text{ W} \)
C.\( 980\text{ W} \)
D.\( 1600\text{ W} \)

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Worked solution

The useful work output is \( W = mgh = 120 \times 9.8 \times 5.0 = 5880\text{ J} \). The useful power output is \( P_{out} = W / t = 5880 / 10 = 588\text{ W} \). Since efficiency is \( 60\% \), the input power is \( P_{in} = P_{out} / 0.60 = 588 / 0.60 = 980\text{ W} \).

Marking scheme

1 mark for the correct calculation of input power using the gravitational potential energy formula and efficiency relation.

Question 35 · multipleChoice

1 marks

A ray of light in a transparent plastic block strikes the boundary with air at an angle of incidence of \( 40^\circ \). The refractive index of the plastic is \( 1.6 \). What happens to the ray of light at the boundary?

A.It is completely refracted into the air at an angle of refraction of \( 24^\circ \).
B.It is completely refracted into the air at an angle of refraction of \( 74^\circ \).
C.It undergoes total internal reflection back into the plastic block.
D.It travels along the boundary between the plastic and the air.

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Worked solution

The critical angle \( c \) is given by \( \sin(c) = 1/n = 1/1.6 = 0.625 \), which gives \( c \approx 38.7^\circ \). Since the angle of incidence \( 40^\circ \) is greater than the critical angle, the light undergoes total internal reflection.

Marking scheme

1 mark for calculating the critical angle and correctly identifying that total internal reflection occurs because the angle of incidence exceeds it.

Question 36 · multipleChoice

1 marks

A wire of length \( L \) and cross-sectional area \( A \) has a resistance \( R \). Another wire of the same material has a length of \( 2L \) and a circular cross-section with 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.\( 8 R \)

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Worked solution

Resistance is \( R = \rho L / A \). The diameter is doubled, so the cross-sectional area increases by a factor of \( 2^2 = 4 \), giving a new area of \( 4A \). The new resistance is \( R' = \rho (2L) / (4A) = 0.5 R \).

Marking scheme

1 mark for applying the resistivity formula and correctly accounting for the effect of doubling the diameter on the cross-sectional area to get 0.5R.

Question 37 · multipleChoice

1 marks

Four identical metal cans are filled with the same volume of hot water at \( 80^\circ\text{C} \) and left in a room at \( 20^\circ\text{C} \). The outer surfaces of the cans are painted with different finishes: Can 1 is matt black, Can 2 is shiny black, Can 3 is matt white, and Can 4 is shiny silver. Which can cools down the fastest, and which can cools down the slowest?

A.Fastest: Can 1; Slowest: Can 4
B.Fastest: Can 4; Slowest: Can 1
C.Fastest: Can 3; Slowest: Can 2
D.Fastest: Can 1; Slowest: Can 3

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Worked solution

Matt black surfaces are the best emitters of infrared radiation, so Can 1 cools down the fastest. Shiny silver surfaces are the worst emitters of infrared radiation, so Can 4 cools down the slowest.

Marking scheme

1 mark for identifying that Can 1 (matt black) cools fastest and Can 4 (shiny silver) cools slowest based on emission properties.

Question 38 · multipleChoice

1 marks

The background radiation count rate in a laboratory is \( 25\text{ counts/minute} \). A detector placed near a radioactive sample records a count rate of \( 225\text{ counts/minute} \). After \( 6.0\text{ hours} \), the recorded count rate is \( 50\text{ counts/minute} \). What is the half-life of the radioactive sample?

A.\( 1.5\text{ hours} \)
B.\( 2.0\text{ hours} \)
C.\( 3.0\text{ hours} \)
D.\( 4.5\text{ hours} \)

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Worked solution

First, subtract the background radiation: initial corrected count rate \( = 225 - 25 = 200\text{ counts/minute} \), and final corrected count rate \( = 50 - 25 = 25\text{ counts/minute} \). The corrected rate decreases from 200 to 25, which represents three half-lives (\( 200 \rightarrow 100 \rightarrow 50 \rightarrow 25 \)). Thus, \( 3 \times t_{1/2} = 6.0\text{ hours} \), which gives \( t_{1/2} = 2.0\text{ hours} \).

Marking scheme

1 mark for subtracting background radiation, identifying that three half-lives have elapsed, and calculating the half-life to be 2.0 hours.

Question 39 · multipleChoice

1 marks

Light from a distant galaxy is observed to have a longer wavelength than light from a stationary source on Earth. What does this observation indicate about the galaxy, and how does the change in wavelength depend on the galaxy's distance from Earth?

A.The galaxy is moving towards Earth; the change in wavelength is greater for more distant galaxies.
B.The galaxy is moving away from Earth; the change in wavelength is greater for more distant galaxies.
C.The galaxy is moving towards Earth; the change in wavelength is smaller for more distant galaxies.
D.The galaxy is moving away from Earth; the change in wavelength is smaller for more distant galaxies.

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Worked solution

An increased wavelength (redshift) indicates that the galaxy is moving away from Earth. According to Hubble's Law, the recessional velocity of a galaxy is proportional to its distance, meaning the change in wavelength is greater for more distant galaxies.

Marking scheme

1 mark for identifying that the galaxy is moving away and that the shift in wavelength is greater for more distant galaxies.

Question 40 · multipleChoice

1 marks

A transformer has \( 100 \) turns on its primary coil and \( 500 \) turns on its secondary coil. An alternating voltage of \( 12\text{ V} \) is applied across the primary coil. Assuming the transformer is \( 100\% \) efficient and the primary current is \( 2.5\text{ A} \), what are the secondary voltage and secondary current?

A.Secondary voltage = \( 2.4\text{ V} \), Secondary current = \( 12.5\text{ A} \)
B.Secondary voltage = \( 60\text{ V} \), Secondary current = \( 0.50\text{ A} \)
C.Secondary voltage = \( 60\text{ V} \), Secondary current = \( 12.5\text{ A} \)
D.Secondary voltage = \( 2.4\text{ V} \), Secondary current = \( 0.50\text{ A} \)

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Worked solution

Using the transformer equation: \( V_s = V_p \times (N_s / N_p) = 12 \times (500 / 100) = 60\text{ V} \). Since efficiency is \( 100\% \), the input power equals the output power: \( I_p V_p = I_s V_s \), which gives \( I_s = (2.5 \times 12) / 60 = 0.50\text{ A} \).

Marking scheme

1 mark for correctly calculating the secondary voltage as 60 V and the secondary current as 0.50 A.

Paper 33

Answer core theory structured questions requiring short-answer, descriptive and numerical responses.

(a) 1 mark for: stating copper is a good conductor / better conductor than plastic.
1 mark for: so thermal energy is transferred to the water quickly/efficiently.
(b) 1 mark for: dull black is a good / excellent absorber of radiation.
1 mark for: reference to infrared / thermal radiation / heat from the Sun.
(c) 1 mark for: glass traps air / creates a sealed space.
1 mark for: prevents warm air from escaping / rising out of the box.
1 mark for: thereby preventing/reducing convection currents to the outside.

Paper 43

Answer extended theory structured questions demanding deep explanation and calculation.

(a) [2 marks]
- use of \(\sin(c) = 1/n\) [1]
- \(41.1^\circ\) (or \(41^\circ\)) [1]

(b)(i) [1 mark]
- states that the angle of incidence (\(35^\circ\)) is less than the critical angle (\(41.1^\circ\)) [1]

(b)(ii) [3 marks]
- use of \(n = \frac{\sin(r)}{\sin(i)}\) [1]
- substitution: \(\sin(r) = 1.52 \times \sin(35^\circ)\) [1]
- \(60.7^\circ\) (or \(61^\circ\)) [1]

(c) [2 marks]
- states total internal reflection occurs [1]
- explains that angle of incidence is greater than critical angle (or \(50^\circ > 41.1^\circ\)) [1]

Paper 53

Perform four practical experiments in laboratory conditions.

- (a) [1 mark]: v = 60.0 cm (must include unit). - (b) [2 marks]: 1 mark for correct substitution: (20.0 * 60.0) / (20.0 + 60.0); 1 mark for correct value f1 = 15.0 cm with unit. - (c)(i) [1 mark]: v2 = 41.0 cm. - (c)(ii) [1 mark]: f2 = 15.1 cm (accept 15.1 or 15.14). - (c)(iii) [1 mark]: favg = 15.1 cm (accept 15.05 or 15.07). - (d) [2 marks]: 1 mark for each valid precaution (e.g., dark room, object/lens/screen aligned, slow adjustment of screen). - (e) [1 mark]: light rays do not actually meet / diverge from the virtual image point. - (f) [2 marks]: 1 mark for stating 'enlarged'; 1 mark for showing magnification calculation: m = v/u = 60.0/20.0 = 3.0, and explaining that since m > 1 the image is larger.

Paper 63

Answer alternative-to-practical questions covering experimental design and data analysis.

(a) \(72.4\ \text{g}\) [1]
(b) (i) \(54\ \text{cm}^3\) [1]
(ii) \(81\ \text{cm}^3\) [1]
(iii) Eye line perpendicular to scale / read bottom of meniscus [1]
(c) (i) \(27\ \text{cm}^3\) [1]
(ii) \(2.68\) or \(2.7\) [1], correct unit \(\text{g/cm}^3\) [1]
(d) Air bubbles increase the volume reading [1], which decreases the calculated density [1]
(e) Use of a sinker [1], subtract volume of sinker alone from combined volume [1]

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