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Thinka Nov 2023 (V2) Cambridge International A Level-Style Mock — Physics (0625)

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An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V2) Cambridge International A Level Physics (0625) paper. Not affiliated with or reproduced from Cambridge.

Paper 22 Multiple Choice (Extended)

Answer all forty questions. Choose the correct option from A, B, C, or D.
40 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · multiple-choice
1 PastPaper.marks
An object accelerates from rest with a constant acceleration of \(3.0\text{ m/s}^2\) for \(4.0\text{ s}\). It then travels at a constant speed for \(5.0\text{ s}\). What is the total distance travelled by the object?
  1. A.24\text{ m}
  2. B.60\text{ m}
  3. C.84\text{ m}
  4. D.108\text{ m}
PastPaper.showAnswers

PastPaper.workedSolution

First, calculate the speed reached at the end of the first \(4.0\text{ s}\) of acceleration:
\(v = u + at = 0 + (3.0\text{ m/s}^2 \times 4.0\text{ s}) = 12\text{ m/s}\).

Next, calculate the distance travelled during each phase:
1. Acceleration phase: The distance is the area under the speed-time graph, which is a triangle.
\(\text{Distance}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 12\text{ m/s} = 24\text{ m}\).

2. Constant speed phase: The distance is the area of a rectangle.
\(\text{Distance}_2 = \text{speed} \times \text{time} = 12\text{ m/s} \times 5.0\text{ s} = 60\text{ m}\).

Total distance = \(24\text{ m} + 60\text{ m} = 84\text{ m}\).

PastPaper.markingScheme

1 mark for the correct calculation of the total distance of \(84\text{ m}\). [C]
PastPaper.question 2 · multiple-choice
1 PastPaper.marks
A student wants to determine the density of a collection of identical small metal spheres. She places a measuring cylinder containing \(50\text{ cm}^3\) of water on a balance, and the balance reads \(250\text{ g}\). She then gently drops 20 spheres into the water. The water level rises to \(74\text{ cm}^3\), and the balance reading increases to \(430\text{ g}\). What is the density of the metal of the spheres?
  1. A.1.8\text{ g/cm}^3
  2. B.5.8\text{ g/cm}^3
  3. C.7.5\text{ g/cm}^3
  4. D.9.0\text{ g/cm}^3
PastPaper.showAnswers

PastPaper.workedSolution

First, find the total mass of the 20 spheres by subtracting the initial reading from the final reading on the balance:
\(\text{Mass} = 430\text{ g} - 250\text{ g} = 180\text{ g}\).

Next, find the total volume of the 20 spheres by subtracting the initial water volume from the final volume in the cylinder:
\(\text{Volume} = 74\text{ cm}^3 - 50\text{ cm}^3 = 24\text{ cm}^3\).

Now, calculate the density of the metal:
\(\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{180\text{ g}}{24\text{ cm}^3} = 7.5\text{ g/cm}^3\).

PastPaper.markingScheme

1 mark for the correct calculation of density resulting in \(7.5\text{ g/cm}^3\). [C]
PastPaper.question 3 · multiple-choice
1 PastPaper.marks
An electric motor is used to lift a \(50\text{ kg}\) load vertically upwards through a height of \(12\text{ m}\) in a time of \(8.0\text{ s}\). The efficiency of the motor system is \(60\%\). What is the average electrical power input to the motor? (Take \(g = 10\text{ N/kg}\))
  1. A.450\text{ W}
  2. B.750\text{ W}
  3. C.1250\text{ W}
  4. D.7500\text{ W}
PastPaper.showAnswers

PastPaper.workedSolution

1. Find the useful work output (increase in gravitational potential energy of the load):
\(W_{\text{useful}} = m \times g \times h = 50\text{ kg} \times 10\text{ N/kg} \times 12\text{ m} = 6000\text{ J}\).

2. Find the useful power output:
\(P_{\text{useful}} = \frac{W_{\text{useful}}}{t} = \frac{6000\text{ J}}{8.0\text{ s}} = 750\text{ W}\).

3. Use the efficiency formula to find the electrical power input:
\(\text{Efficiency} = \frac{P_{\text{useful}}}{P_{\text{input}}} \times 100\%\)
\(0.60 = \frac{750\text{ W}}{P_{\text{input}}}\)
\(P_{\text{input}} = \frac{750\text{ W}}{0.60} = 1250\text{ W}\).

PastPaper.markingScheme

1 mark for the correct calculation of electrical power input of \(1250\text{ W}\). [C]
PastPaper.question 4 · multiple-choice
1 PastPaper.marks
A bubble of gas is released from the bottom of a lake. The temperature of the water is constant. At the bottom of the lake, the total pressure is \(3.0 \times 10^5\text{ Pa}\). At the surface of the lake, the atmospheric pressure is \(1.0 \times 10^5\text{ Pa}\). The initial volume of the bubble at the bottom is \(2.0\text{ cm}^3\). What is the volume of the bubble when it reaches the surface?
  1. A.0.67\text{ cm}^3
  2. B.2.0\text{ cm}^3
  3. C.4.0\text{ cm}^3
  4. D.6.0\text{ cm}^3
PastPaper.showAnswers

PastPaper.workedSolution

Since the temperature of the gas is constant, Boyle's law applies:
\(P_1 V_1 = P_2 V_2\)

Where:
\(P_1 = 3.0 \times 10^5\text{ Pa}\) (initial pressure at the bottom)
\(V_1 = 2.0\text{ cm}^3\) (initial volume at the bottom)
\(P_2 = 1.0 \times 10^5\text{ Pa}\) (final pressure at the surface)

Rearranging to solve for \(V_2\):
\(V_2 = \frac{P_1 V_1}{P_2} = \frac{3.0 \times 10^5\text{ Pa} \times 2.0\text{ cm}^3}{1.0 \times 10^5\text{ Pa}} = 6.0\text{ cm}^3\).

PastPaper.markingScheme

1 mark for the correct application of Boyle's law resulting in \(6.0\text{ cm}^3\). [D]
PastPaper.question 5 · multiple-choice
1 PastPaper.marks
A ray of light is incident on the boundary between a transparent liquid and air. The refractive index of the liquid is \(1.25\). What is the critical angle for this liquid-air boundary?
  1. A.37^\circ
  2. B.45^\circ
  3. C.53^\circ
  4. D.68^\circ
PastPaper.showAnswers

PastPaper.workedSolution

The critical angle \(c\) is related to the refractive index \(n\) of the denser medium (the liquid) by the formula:
\(\sin(c) = \frac{1}{n}\)

Substitute \(n = 1.25\) into the equation:
\(\sin(c) = \frac{1}{1.25} = 0.80\)

Now, calculate the angle \(c\):
\(c = \arcsin(0.80) \approx 53.1^\circ\).

To the nearest degree, the critical angle is \(53^\circ\).

PastPaper.markingScheme

1 mark for the correct calculation of the critical angle of \(53^\circ\). [C]
PastPaper.question 6 · multiple-choice
1 PastPaper.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 \(2L\) and a cross-sectional radius \(2r\). What is the resistance of the second wire in terms of \(R\)?
  1. A.0.5R
  2. B.R
  3. C.2R
  4. D.4R
PastPaper.showAnswers

PastPaper.workedSolution

The resistance of a wire is given by:
\(R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}\)

For the second wire:
- Length is \(2L\)
- Cross-sectional radius is \(2r\), so the cross-sectional area becomes:
\(A' = \pi (2r)^2 = 4 \pi r^2 = 4A\)

Now, substitute these into the resistance formula for the second wire \(R'\):
\(R' = \rho \frac{2L}{4A} = 0.5 \left(\rho \frac{L}{A}\right) = 0.5R\).

PastPaper.markingScheme

1 mark for showing that the resistance is halved to \(0.5R\). [A]
PastPaper.question 7 · multiple-choice
1 PastPaper.marks
A radioactive sample has a count rate of 640 counts per minute (including a background count rate of 40 counts per minute). After 12 hours, the count rate of the sample decreases to 115 counts per minute (including the same background count rate of 40 counts per minute). What is the half-life of the radioactive isotope?
  1. A.2.4\text{ hours}
  2. B.3.0\text{ hours}
  3. C.4.0\text{ hours}
  4. D.6.0\text{ hours}
PastPaper.showAnswers

PastPaper.workedSolution

First, subtract the background count rate to find the actual corrected count rate of the isotope at the start and at the end:
- Initial corrected count rate = \(640 - 40 = 600\text{ counts/minute}\)
- Final corrected count rate = \(115 - 40 = 75\text{ counts/minute}\)

Next, determine the fraction of the initial activity remaining:
\(\frac{75}{600} = \frac{1}{8}\)

Since \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\), the sample has undergone exactly 3 half-lives.

Now, find the duration of one half-life:
\(\text{Half-life} = \frac{\text{Total time}}{3} = \frac{12\text{ hours}}{3} = 4.0\text{ hours}\).

PastPaper.markingScheme

1 mark for the correct calculation of half-life as \(4.0\text{ hours}\). [C]
PastPaper.question 8 · multiple-choice
1 PastPaper.marks
An artificial satellite orbits the Earth in a circular orbit of radius \(8000\text{ km}\). It completes one orbit in \(2.0\text{ hours}\). What is the average orbital speed of the satellite?
  1. A.1.1\text{ km/s}
  2. B.3.5\text{ km/s}
  3. C.7.0\text{ km/s}
  4. D.44\text{ km/s}
PastPaper.showAnswers

PastPaper.workedSolution

The average orbital speed \(v\) is calculated using the formula:
\(v = \frac{2\pi r}{T}\)

Given:
- Orbital radius, \(r = 8000\text{ km}\)
- Time period, \(T = 2.0\text{ hours} = 2.0 \times 3600\text{ s} = 7200\text{ s}\)

Substitute these values into the formula:
\(v = \frac{2 \times \pi \times 8000\text{ km}}{7200\text{ s}} = \frac{50265\text{ km}}{7200\text{ s}} \approx 6.98\text{ km/s}\)

To two significant figures, this is \(7.0\text{ km/s}\).

PastPaper.markingScheme

1 mark for the correct calculation of orbital speed resulting in \(7.0\text{ km/s}\). [C]
PastPaper.question 9 · multiple-choice
1 PastPaper.marks
A student uses a micrometer screw gauge to measure the diameter of a steel ball. With the jaws closed, the reading is \(+0.04\text{ mm}\) (a positive zero error). When measuring the ball, the main scale reads \(4.5\text{ mm}\) and the thimble scale reads 32 divisions. Each division on the thimble represents \(0.01\text{ mm}\). What is the actual diameter of the steel ball?
  1. A.\(4.46\text{ mm}\)
  2. B.\(4.78\text{ mm}\)
  3. C.\(4.82\text{ mm}\)
  4. D.\(4.86\text{ mm}\)
PastPaper.showAnswers

PastPaper.workedSolution

First, calculate the uncorrected reading: \(4.5\text{ mm} + (32 \times 0.01\text{ mm}) = 4.82\text{ mm}\). Then, subtract the positive zero error to find the actual diameter: \(4.82\text{ mm} - 0.04\text{ mm} = 4.78\text{ mm}\).

PastPaper.markingScheme

Award 1 mark for selecting the correct option B. 0 marks for other options.
PastPaper.question 10 · multiple-choice
1 PastPaper.marks
A ball of mass \(0.15\text{ kg}\) is dropped onto a concrete floor. It strikes the floor vertically downwards at a speed of \(8.0\text{ m/s}\) and rebounds vertically upwards at a speed of \(6.0\text{ m/s}\). The ball is in contact with the floor for \(0.070\text{ s}\). What is the average force exerted by the floor on the ball during the collision?
  1. A.\(4.3\text{ N}\)
  2. B.\(12\text{ N}\)
  3. C.\(17\text{ N}\)
  4. D.\(30\text{ N}\)
PastPaper.showAnswers

PastPaper.workedSolution

Taking upwards as the positive direction, the initial velocity is \(u = -8.0\text{ m/s}\) and the final velocity is \(v = +6.0\text{ m/s}\). The change in momentum is \(\Delta p = m(v - u) = 0.15\text{ kg} \times (6.0 - (-8.0))\text{ m/s} = 0.15 \times 14.0 = 2.1\text{ N s}\). The average force is \(F = \frac{\Delta p}{\Delta t} = \frac{2.1\text{ N s}}{0.070\text{ s}} = 30\text{ N}\).

PastPaper.markingScheme

Award 1 mark for selecting the correct option D. 0 marks for other options.
PastPaper.question 11 · multiple-choice
1 PastPaper.marks
An electrical heater of power \(150\text{ W}\) is used to heat a \(0.50\text{ kg}\) block of ice at \(0^\circ\text{C}\). The heater is switched on for \(5.0\text{ minutes}\). During this time, \(120\text{ g}\) of the ice melts to form water at \(0^\circ\text{C}\). How much thermal energy from the heater is lost to the surroundings during this process? (Specific latent heat of fusion of ice = \(3.3 \times 10^5\text{ J/kg}\))
  1. A.\(5.4\text{ kJ}\)
  2. B.\(39.6\text{ kJ}\)
  3. C.\(45.0\text{ kJ}\)
  4. D.\(120\text{ kJ}\)
PastPaper.showAnswers

PastPaper.workedSolution

Total energy supplied by the heater: \(E = P \times t = 150\text{ W} \times (5.0 \times 60\text{ s}) = 45,000\text{ J} = 45.0\text{ kJ}\). Energy used to melt the ice: \(Q = m \times L_f = 0.12\text{ kg} \times 3.3 \times 10^5\text{ J/kg} = 39,600\text{ J} = 39.6\text{ kJ}\). Thermal energy lost to the surroundings: \(E_{\text{lost}} = E - Q = 45.0\text{ kJ} - 39.6\text{ kJ} = 5.4\text{ kJ}\).

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Award 1 mark for selecting the correct option A. 0 marks for other options.
PastPaper.question 12 · multiple-choice
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A potential divider circuit consists of a light-dependent resistor (LDR) and a fixed resistor of resistance \(4.0\text{ k}\Omega\) connected in series across a \(6.0\text{ V}\) d.c. power supply. A voltmeter is connected in parallel with the fixed resistor. The resistance of the LDR is \(2.0\text{ k}\Omega\) in bright light and \(8.0\text{ k}\Omega\) in the dark. What are the voltmeter readings in bright light and in the dark?
  1. A.Bright light: \(4.0\text{ V}\), Dark: \(2.0\text{ V}\)
  2. B.Bright light: \(2.0\text{ V}\), Dark: \(4.0\text{ V}\)
  3. C.Bright light: \(4.5\text{ V}\), Dark: \(1.5\text{ V}\)
  4. D.Bright light: \(1.5\text{ V}\), Dark: \(4.5\text{ V}\)
PastPaper.showAnswers

PastPaper.workedSolution

Using the potential divider formula, the voltmeter reading across the fixed resistor is given by \(V = V_{\text{in}} \times \frac{R_{\text{fixed}}}{R_{\text{LDR}} + R_{\text{fixed}}}\). In bright light: \(V = 6.0\text{ V} \times \frac{4.0\text{ k}\Omega}{2.0\text{ k}\Omega + 4.0\text{ k}\Omega} = 4.0\text{ V}\). In the dark: \(V = 6.0\text{ V} \times \frac{4.0\text{ k}\Omega}{8.0\text{ k}\Omega + 4.0\text{ k}\Omega} = 2.0\text{ V}\).

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Award 1 mark for selecting the correct option A. 0 marks for other options.
PastPaper.question 13 · multiple-choice
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Which statement about redshift and the expansion of the Universe is correct?
  1. A.Light from more distant galaxies has a smaller redshift, indicating that they are moving away from us at a slower speed.
  2. B.Light from more distant galaxies has a greater redshift, indicating that they are moving away from us at a faster speed.
  3. C.Light from more distant galaxies has a greater redshift, indicating that they are moving away from us at a slower speed.
  4. D.Light from more distant galaxies has a smaller redshift, indicating that they are moving away from us at a faster speed.
PastPaper.showAnswers

PastPaper.workedSolution

Based on observations of distant galaxies, the light from more distant galaxies shows a greater redshift. This indicates that they are moving away from us at a faster speed (Hubble's Law).

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Award 1 mark for selecting the correct option B. 0 marks for other options.
PastPaper.question 14 · multiple-choice
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An ideal step-down transformer is connected to a \(240\text{ V}\) a.c. mains supply. The primary coil has 600 turns and the secondary coil has 30 turns. The transformer supplies power to a lamp rated \(12\text{ V}, 24\text{ W}\). What is the current in the primary coil?
  1. A.\(0.10\text{ A}\)
  2. B.\(2.0\text{ A}\)
  3. C.\(4.0\text{ A}\)
  4. D.\(5.0\text{ A}\)
PastPaper.showAnswers

PastPaper.workedSolution

The voltage across the secondary is \(V_s = V_p \times \frac{N_s}{N_p} = 240\text{ V} \times \frac{30}{600} = 12\text{ V}\), which is the operating voltage of the lamp. Since the transformer is ideal, power in the primary coil equals power in the secondary coil: \(P_p = P_s = 24\text{ W}\). The current in the primary coil is \(I_p = \frac{P_p}{V_p} = \frac{24\text{ W}}{240\text{ V}} = 0.10\text{ A}\).

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Award 1 mark for selecting the correct option A. 0 marks for other options.
PastPaper.question 15 · multiple-choice
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A radioactive source has a half-life of \(4.0\text{ hours}\). A detector is placed near the source and measures a count rate of 340 counts per minute. The background count rate is constant at 20 counts per minute. What is the measured count rate after \(12\text{ hours}\)?
  1. A.\(42.5\text{ counts per minute}\)
  2. B.\(45\text{ counts per minute}\)
  3. C.\(60\text{ counts per minute}\)
  4. D.\(85\text{ counts per minute}\)
PastPaper.showAnswers

PastPaper.workedSolution

First, find the initial count rate due to the source alone by subtracting the background count rate: \(340 - 20 = 320\text{ counts per minute}\). Since \(12\text{ hours}\) represents 3 half-lives (\(12 / 4.0 = 3\)), the source activity decreases to \(320 \times (1/2)^3 = 40\text{ counts per minute}\). The final measured count rate includes background: \(40 + 20 = 60\text{ counts per minute}\).

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Award 1 mark for selecting the correct option C. 0 marks for other options.
PastPaper.question 16 · multiple-choice
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A ray of light travels through a liquid and strikes the boundary with air. The refractive index of the liquid is 1.40. What is the critical angle for light in this liquid, and what happens to the ray of light if its angle of incidence at the boundary is \(50^\circ\)?
  1. A.Critical angle is \(45.6^\circ\); the ray undergoes total internal reflection.
  2. B.Critical angle is \(45.6^\circ\); the ray refracts into the air.
  3. C.Critical angle is \(54.4^\circ\); the ray undergoes total internal reflection.
  4. D.Critical angle is \(54.4^\circ\); the ray refracts into the air.
PastPaper.showAnswers

PastPaper.workedSolution

The critical angle \(c\) is calculated using the formula \(\sin(c) = \frac{1}{n} = \frac{1}{1.40} \approx 0.714\), which gives \(c \approx 45.6^\circ\). Since the angle of incidence \(50^\circ\) is greater than the critical angle \(45.6^\circ\), the ray of light undergoes total internal reflection.

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Award 1 mark for selecting the correct option A. 0 marks for other options.
PastPaper.question 17 · multiple-choice
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A toy car of mass \(0.50\text{ kg}\) moving at a speed of \(4.0\text{ m/s}\) collides with a stationary toy car of mass \(1.5\text{ kg}\). After the collision, the two cars stick together and move forward. What is the total kinetic energy lost during the collision?
  1. A.\(1.0\text{ J}\)
  2. B.\(2.0\text{ J}\)
  3. C.\(3.0\text{ J}\)
  4. D.\(4.0\text{ J}\)
PastPaper.showAnswers

PastPaper.workedSolution

First, use the principle of conservation of momentum to find the final velocity \(v\) of the combined cars: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\). Substituting the values: \(0.50 \times 4.0 + 1.5 \times 0 = (0.50 + 1.5) v\), which gives \(2.0 = 2.0 v\), so \(v = 1.0\text{ m/s}\). Next, calculate the total initial kinetic energy: \(E_{ki} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} (0.50) (4.0)^2 = 4.0\text{ J}\). Calculate the total final kinetic energy: \(E_{kf} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} (2.0) (1.0)^2 = 1.0\text{ J}\). The kinetic energy lost is: \(\Delta E_k = 4.0\text{ J} - 1.0\text{ J} = 3.0\text{ J}\).

PastPaper.markingScheme

1 mark: Correct calculation of the final velocity using conservation of momentum and finding the difference in kinetic energy to arrive at 3.0 J.
PastPaper.question 18 · multiple-choice
1 PastPaper.marks
A motorcycle accelerates from rest at a constant rate of \(3.0\text{ m/s}^2\) for \(6.0\text{ s}\). It then travels at a constant speed for \(10\text{ s}\) before decelerating uniformly to rest in a further \(4.0\text{ s}\). What is the total distance travelled by the motorcycle?
  1. A.\(180\text{ m}\)
  2. B.\(234\text{ m}\)
  3. C.\(270\text{ m}\)
  4. D.\(360\text{ m}\)
PastPaper.showAnswers

PastPaper.workedSolution

First, find the maximum speed reached: \(v = u + at = 0 + (3.0 \times 6.0) = 18\text{ m/s}\). Then, calculate the distance in each stage by finding the area under the speed-time graph: Distance during acceleration (first stage): \(d_1 = \frac{1}{2} \times 6.0 \times 18 = 54\text{ m}\). Distance during constant speed (second stage): \(d_2 = 10 \times 18 = 180\text{ m}\). Distance during deceleration (third stage): \(d_3 = \frac{1}{2} \times 4.0 \times 18 = 36\text{ m}\). Summing these distances: \(D = 54 + 180 + 36 = 270\text{ m}\).

PastPaper.markingScheme

1 mark: Correct determination of maximum speed (18 m/s) and summing of the area under the speed-time graph to get 270 m.
PastPaper.question 19 · multiple-choice
1 PastPaper.marks
A ray of light traveling in a glass block of refractive index \(1.52\) is incident on the boundary with an unknown liquid. The critical angle for total internal reflection at this boundary is \(58.0^\circ\). What is the refractive index of the liquid? (Given: \(\sin(58.0^\circ) = 0.848\))
  1. A.\(0.81\)
  2. B.\(1.21\)
  3. C.\(1.29\)
  4. D.\(1.79\)
PastPaper.showAnswers

PastPaper.workedSolution

The relationship between the critical angle \(c\), the refractive index of the denser medium \(n_1\), and the less dense medium \(n_2\) is given by \(\sin(c) = \frac{n_2}{n_1}\). Here, \(n_1 = 1.52\) and \(c = 58.0^\circ\). Solving for \(n_2\): \(n_2 = n_1 \sin(c) = 1.52 \times \sin(58.0^\circ) = 1.52 \times 0.848 \approx 1.29\).

PastPaper.markingScheme

1 mark: Correct application of the critical angle formula relating the two media to find the refractive index of the liquid.
PastPaper.question 20 · multiple-choice
1 PastPaper.marks
An artificial satellite orbits the Earth in a circular orbit of radius \(8.0 \times 10^6\text{ m}\). The orbital speed of the satellite is \(7.0\text{ km/s}\). How long does it take for the satellite to complete one full orbit?
  1. A.\(1.1 \times 10^3\text{ s}\)
  2. B.\(3.6 \times 10^3\text{ s}\)
  3. C.\(7.2 \times 10^3\text{ s}\)
  4. D.\(5.0 \times 10^4\text{ s}\)
PastPaper.showAnswers

PastPaper.workedSolution

The orbital speed \(v\) is given by \(v = \frac{2\pi r}{T}\). Converting speed to meters per second: \(v = 7000\text{ m/s}\). Solving for the period \(T\): \(T = \frac{2\pi r}{v} = \frac{2 \pi \times 8.0 \times 10^6}{7000} \approx 7181\text{ s}\). Rounding to two significant figures yields \(7.2 \times 10^3\text{ s}\).

PastPaper.markingScheme

1 mark: Correct conversion of speed to m/s, application of the circular orbit formula, and calculation of the orbital period to 2 significant figures.
PastPaper.question 21 · multiple-choice
1 PastPaper.marks
A uniform metal wire of length \(L\) and cross-sectional area \(A\) has a resistance of \(16.0\ \Omega\). Another wire made of the same metal has a length of \(2L\) and a diameter that is twice that of the first wire. What is the resistance of the second wire?
  1. A.\(4.0\ \Omega\)
  2. B.\(8.0\ \Omega\)
  3. C.\(16.0\ \Omega\)
  4. D.\(32.0\ \Omega\)
PastPaper.showAnswers

PastPaper.workedSolution

Resistance is \(R = \rho \frac{L}{A}\), where \(A = \pi \left(\frac{d}{2}\right)^2\). Since the diameter is doubled, the cross-sectional area becomes \(4A\). The new resistance is \(R_2 = \rho \frac{2L}{4A} = \frac{1}{2} R_1 = \frac{1}{2} \times 16.0\ \Omega = 8.0\ \Omega\).

PastPaper.markingScheme

1 mark: Correct deduction of the area scaling factor (4x) and applying it with the length factor (2x) to find the final resistance of 8.0 ohms.
PastPaper.question 22 · multiple-choice
1 PastPaper.marks
A cylinder with a movable piston contains a fixed mass of gas at a constant temperature. The volume of the gas is initially \(120\text{ cm}^3\) and its pressure is \(1.5 \times 10^5\text{ Pa}\). The piston is slowly pushed in, decreasing the volume of the gas to \(40\text{ cm}^3\). What is the new pressure of the gas?
  1. A.\(0.50 \times 10^5\text{ Pa}\)
  2. B.\(1.5 \times 10^5\text{ Pa}\)
  3. C.\(3.0 \times 10^5\text{ Pa}\)
  4. D.\(4.5 \times 10^5\text{ Pa}\)
PastPaper.showAnswers

PastPaper.workedSolution

Applying Boyle's Law: \(p_1 V_1 = p_2 V_2\). Substituting the given values: \(1.5 \times 10^5\text{ Pa} \times 120\text{ cm}^3 = p_2 \times 40\text{ cm}^3\), which yields \(p_2 = 4.5 \times 10^5\text{ Pa}\).

PastPaper.markingScheme

1 mark: Correct application of Boyle's Law equation to solve for the final pressure.
PastPaper.question 23 · multiple-choice
1 PastPaper.marks
An electric heater of power \(600\text{ W}\) is used to heat a \(2.0\text{ kg}\) block of metal. The temperature of the block rises 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?
  1. A.\(300\text{ J}/(\text{kg}^\circ\text{C})\)
  2. B.\(900\text{ J}/(\text{kg}^\circ\text{C})\)
  3. C.\(1800\text{ J}/(\text{kg}^\circ\text{C})\)
  4. D.\(3600\text{ J}/(\text{kg}^\circ\text{C})\)
PastPaper.showAnswers

PastPaper.workedSolution

Calculate energy supplied: \(E = P \times t = 600\text{ W} \times (3.0 \times 60)\text{ s} = 108,000\text{ J}\). The temperature change is \(\Delta \theta = 50^\circ\text{C} - 20^\circ\text{C} = 30^\circ\text{C}\). Using \(E = m c \Delta \theta\), we get \(108,000 = 2.0 \times c \times 30\), which leads to \(c = 1800\text{ J}/(\text{kg}^\circ\text{C})\).

PastPaper.markingScheme

1 mark: Correct calculation of energy supplied (108,000 J) and application of the heat equation to find 1800 J/(kg °C).
PastPaper.question 24 · multiple-choice
1 PastPaper.marks
A radioactive source has a half-life of \(20\text{ minutes}\). The measured count rate near the source, which includes a constant background radiation of \(15\text{ counts/minute}\), is initially \(255\text{ counts/minute}\). What will the measured count rate be after \(1.0\text{ hour}\)?
  1. A.\(30\text{ counts/minute}\)
  2. B.\(45\text{ counts/minute}\)
  3. C.\(47\text{ counts/minute}\)
  4. D.\(95\text{ counts/minute}\)
PastPaper.showAnswers

PastPaper.workedSolution

First, find the initial corrected count rate: \(255 - 15 = 240\text{ counts/minute}\). The elapsed time of \(1.0\text{ hour}\) (\(60\text{ minutes}\)) corresponds to \(3\) half-lives. The corrected count rate after \(3\) half-lives is \(240 / 2^3 = 30\text{ counts/minute}\). Adding back the background radiation gives \(30 + 15 = 45\text{ counts/minute}\).

PastPaper.markingScheme

1 mark: Correct subtraction of background, division by 8, and addition of background radiation to find the final measured count rate of 45 counts/minute.
PastPaper.question 25 · multiple-choice
1 PastPaper.marks
A car travels along a straight road. It accelerates from rest at a constant rate of \(2.0\text{ m/s}^2\) for \(6.0\text{ s}\), then travels at a constant speed for \(10\text{ s}\), and finally decelerates to rest at a constant rate in \(4.0\text{ s}\). What is the total distance traveled by the car?
  1. A.144 m
  2. B.168 m
  3. C.180 m
  4. D.240 m
PastPaper.showAnswers

PastPaper.workedSolution

First, find the maximum speed reached during the acceleration phase: \(v = u + at = 0 + (2.0 \times 6.0) = 12\text{ m/s}\). The distance traveled during this first phase is \(d_1 = \frac{1}{2} \times v \times t = \frac{1}{2} \times 12 \times 6.0 = 36\text{ m}\). During the constant speed phase, the distance is \(d_2 = v \times t = 12 \times 10 = 120\text{ m}\). During the deceleration phase, the distance is \(d_3 = \frac{1}{2} \times v \times t = \frac{1}{2} \times 12 \times 4.0 = 24\text{ m}\). The total distance is the sum of these three distances: \(d_{\text{total}} = 36 + 120 + 24 = 180\text{ m}\).

PastPaper.markingScheme

1 mark for the correct calculation of the total distance (180 m). Correct option is C.
PastPaper.question 26 · multiple-choice
1 PastPaper.marks
Liquid X has a density of \(0.80\text{ g/cm}^3\) and liquid Y has a density of \(1.20\text{ g/cm}^3\). A mixture is made by mixing \(150\text{ cm}^3\) of liquid X with \(250\text{ cm}^3\) of liquid Y. There is no change in total volume when the liquids are mixed. What is the density of the mixture?
  1. A.0.95 g/cm³
  2. B.1.00 g/cm³
  3. C.1.05 g/cm³
  4. D.1.10 g/cm³
PastPaper.showAnswers

PastPaper.workedSolution

First, calculate the mass of liquid X: \(m_X = \text{density} \times \text{volume} = 0.80\text{ g/cm}^3 \times 150\text{ cm}^3 = 120\text{ g}\). Next, calculate the mass of liquid Y: \(m_Y = 1.20\text{ g/cm}^3 \times 250\text{ cm}^3 = 300\text{ g}\). The total mass of the mixture is \(m_{\text{total}} = 120\text{ g} + 300\text{ g} = 420\text{ g}\). The total volume is \(V_{\text{total}} = 150\text{ cm}^3 + 250\text{ cm}^3 = 400\text{ cm}^3\). The density of the mixture is \(\rho = \frac{m_{\text{total}}}{V_{\text{total}}} = \frac{420\text{ g}}{400\text{ cm}^3} = 1.05\text{ g/cm}^3\).

PastPaper.markingScheme

1 mark for the correct calculation of the mixture density (1.05 g/cm³). Correct option is C.
PastPaper.question 27 · multiple-choice
1 PastPaper.marks
An electric pump is used to lift water from a well. The pump raises \(40\text{ kg}\) of water through a vertical height of \(12\text{ m}\) in a time of \(8.0\text{ s}\). The electrical power input to the pump is \(800\text{ W}\). The acceleration of free fall \(g\) is \(10\text{ m/s}^2\). What is the efficiency of the pump?
  1. A.6.0%
  2. B.48%
  3. C.60%
  4. D.75%
PastPaper.showAnswers

PastPaper.workedSolution

The work done in lifting the water is \(W = mgh = 40 \times 10 \times 12 = 4800\text{ J}\). The useful power output of the pump is \(P_{\text{out}} = \frac{W}{t} = \frac{4800}{8.0} = 600\text{ W}\). The efficiency is given by \(\text{efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{600}{800} \times 100\% = 75\%\).

PastPaper.markingScheme

1 mark for the correct calculation of efficiency (75%). Correct option is D.
PastPaper.question 28 · multiple-choice
1 PastPaper.marks
A uniform block of mass \(15\text{ kg}\) has dimensions \(0.20\text{ m} \times 0.30\text{ m} \times 0.50\text{ m}\). It lies on a flat, horizontal table. What is the difference between the maximum pressure and the minimum pressure that the block can exert on the table? (Take \(g = 10\text{ m/s}^2\))
  1. A.1000 Pa
  2. B.1500 Pa
  3. C.2500 Pa
  4. D.3500 Pa
PastPaper.showAnswers

PastPaper.workedSolution

The weight of the block is \(W = mg = 15 \times 10 = 150\text{ N}\). The minimum surface area is \(A_{\text{min}} = 0.20 \times 0.30 = 0.06\text{ m}^2\), which produces the maximum pressure: \(P_{\text{max}} = \frac{150}{0.06} = 2500\text{ Pa}\). The maximum surface area is \(A_{\text{max}} = 0.30 \times 0.50 = 0.15\text{ m}^2\), which produces the minimum pressure: \(P_{\text{min}} = \frac{150}{0.15} = 1000\text{ Pa}\). The difference between the maximum and minimum pressure is \(P_{\text{max}} - P_{\text{min}} = 2500 - 1000 = 1500\text{ Pa}\).

PastPaper.markingScheme

1 mark for calculating the difference in pressure correctly (1500 Pa). Correct option is B.
PastPaper.question 29 · multiple-choice
1 PastPaper.marks
A bubble of gas has a volume of \(2.0\text{ cm}^3\) at the bottom of a lake where the depth is \(20\text{ m}\). The temperature of the water is constant. The atmospheric pressure at the surface of the lake is \(1.0 \times 10^5\text{ Pa}\). The density of water is \(1000\text{ kg/m}^3\) and \(g = 10\text{ m/s}^2\). What is the volume of the bubble of gas when it reaches the surface?
  1. A.4.0 cm³
  2. B.6.0 cm³
  3. C.8.0 cm³
  4. D.12.0 cm³
PastPaper.showAnswers

PastPaper.workedSolution

The pressure at the bottom of the lake is \(P_1 = P_{\text{atm}} + \rho g h = 1.0 \times 10^5 + (1000 \times 10 \times 20) = 1.0 \times 10^5 + 2.0 \times 10^5 = 3.0 \times 10^5\text{ Pa}\). The volume at the bottom is \(V_1 = 2.0\text{ cm}^3\). At the surface, the pressure is \(P_2 = 1.0 \times 10^5\text{ Pa}\). Since the temperature is constant, Boyle's law applies: \(P_1 V_1 = P_2 V_2\). Therefore, \(V_2 = \frac{P_1 V_1}{P_2} = \frac{3.0 \times 10^5 \times 2.0}{1.0 \times 10^5} = 6.0\text{ cm}^3\).

PastPaper.markingScheme

1 mark for the correct final volume of the bubble (6.0 cm³). Correct option is B.
PastPaper.question 30 · multiple-choice
1 PastPaper.marks
A ray of light travels from glass into air. The refractive index of the glass is \(1.50\). The angle of incidence in the glass is \(45.0^\circ\). Which statement correctly describes what happens to the ray?
  1. A.The ray is refracted into the air with an angle of refraction of 45.0°.
  2. B.The ray is refracted into the air with an angle of refraction of 70.5°.
  3. C.The ray undergoes total internal reflection with an angle of reflection of 45.0°.
  4. D.The ray undergoes total internal reflection with an angle of reflection of 41.8°.
PastPaper.showAnswers

PastPaper.workedSolution

First, determine the critical angle \(c\) for the glass-air boundary: \(\sin(c) = \frac{1}{n} = \frac{1}{1.50}\), so \(c = \arcsin(0.667) \approx 41.8^\circ\). Since the angle of incidence in the glass is \(45.0^\circ\), which is greater than the critical angle, total internal reflection occurs. By the law of reflection, the angle of reflection must equal the angle of incidence, which is \(45.0^\circ\).

PastPaper.markingScheme

1 mark for identifying total internal reflection and the correct angle of reflection (45.0°). Correct option is C.
PastPaper.question 31 · multiple-choice
1 PastPaper.marks
Two copper wires, P and Q, have different dimensions. Wire P has length \(L\) and diameter \(d\). Wire Q has length \(2L\) and diameter \(2d\). What is the ratio \(\frac{\text{resistance of P}}{\text{resistance of Q}}\)?
  1. A.1/2
  2. B.1
  3. C.2
  4. D.4
PastPaper.showAnswers

PastPaper.workedSolution

The resistance of a wire is given by \(R = \rho \frac{L}{A}\), where \(A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}\). Thus, \(R \propto \frac{L}{d^2}\). For wire P, \(R_P \propto \frac{L}{d^2}\). For wire Q, \(R_Q \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \frac{L}{d^2}\). Comparing the two resistances, we have \(R_P = 2 R_Q\), so the ratio \(\frac{R_P}{R_Q} = 2\).

PastPaper.markingScheme

1 mark for correctly determining the ratio of the resistances (2). Correct option is C.
PastPaper.question 32 · multiple-choice
1 PastPaper.marks
A student measures the count rate from a radioactive source near a Geiger-Müller (GM) tube. Before bringing the source near, the background count rate is measured as \(24\text{ counts/minute}\). With the source in place, the measured count rate is \(360\text{ counts/minute}\). After exactly \(6.0\text{ hours}\), the measured count rate decreases to \(66\text{ counts/minute}\). What is the half-life of the radioactive source?
  1. A.1.2 hours
  2. B.1.5 hours
  3. C.2.0 hours
  4. D.3.0 hours
PastPaper.showAnswers

PastPaper.workedSolution

First, subtract the background count rate to find the initial and final corrected count rates of the source. Initial corrected count rate = \(360 - 24 = 336\text{ counts/minute}\). Final corrected count rate = \(66 - 24 = 42\text{ counts/minute}\). The fraction of activity remaining is \(\frac{42}{336} = \frac{1}{8}\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly 3 half-lives have elapsed in \(6.0\text{ hours}\). Therefore, the half-life of the source is \(\frac{6.0\text{ hours}}{3} = 2.0\text{ hours}\).

PastPaper.markingScheme

1 mark for the correct determination of the half-life (2.0 hours). Correct option is C.
PastPaper.question 33 · multiple_choice
1 PastPaper.marks
A model rocket is launched vertically upwards. Its velocity increases at a constant rate of \(15\text{ m/s}^2\) for the first \(4.0\text{ s}\). The engine then shuts off, and the rocket is in free fall with an acceleration of \(10\text{ m/s}^2\) downwards until it reaches its maximum height. Air resistance is negligible. What is the total time from launch until the rocket reaches its maximum height?
  1. A.4.0 s
  2. B.6.0 s
  3. C.10 s
  4. D.15 s
PastPaper.showAnswers

PastPaper.workedSolution

During the first phase of motion, the rocket accelerates from rest (\(u = 0\text{ m/s}\)) at \(a_1 = 15\text{ m/s}^2\) for \(t_1 = 4.0\text{ s}\). The velocity at the end of this phase is: \(v_1 = u + a_1 t_1 = 0 + 15 \times 4.0 = 60\text{ m/s}\). During the second phase of motion, the engine is off and the rocket is in free fall with deceleration due to gravity (\(a_2 = -10\text{ m/s}^2\)) until it reaches its maximum height where its velocity is \(v_2 = 0\text{ m/s}\). The time taken for this phase is: \(t_2 = \frac{v_2 - v_1}{a_2} = \frac{0 - 60}{-10} = 6.0\text{ s}\). The total time from launch to maximum height is: \(t_{\text{total}} = t_1 + t_2 = 4.0 + 6.0 = 10\text{ s}\).

PastPaper.markingScheme

1 mark for calculating the velocity at the end of the first phase (\(60\text{ m/s}\)) and the time for the second phase (\(6.0\text{ s}\)), leading to the correct total time of \(10\text{ s}\).
PastPaper.question 34 · multiple_choice
1 PastPaper.marks
An alloy is made by mixing \(200\text{ g}\) of metal X (density \(8.0\text{ g/cm}^3\)) with \(300\text{ g}\) of metal Y (density \(6.0\text{ g/cm}^3\)). No change in volume occurs during mixing. What is the density of the alloy?
  1. A.6.7 g/cm³
  2. B.7.0 g/cm³
  3. C.7.2 g/cm³
  4. D.14 g/cm³
PastPaper.showAnswers

PastPaper.workedSolution

First, find the volume of each metal. For metal X: \(V_{\text{X}} = \frac{\text{mass}}{\text{density}} = \frac{200\text{ g}}{8.0\text{ g/cm}^3} = 25\text{ cm}^3\). For metal Y: \(V_{\text{Y}} = \frac{\text{mass}}{\text{density}} = \frac{300\text{ g}}{6.0\text{ g/cm}^3} = 50\text{ cm}^3\). The total mass of the alloy is: \(M = 200\text{ g} + 300\text{ g} = 500\text{ g}\). The total volume of the alloy is: \(V = 25\text{ cm}^3 + 50\text{ cm}^3 = 75\text{ cm}^3\). The density of the alloy is: \(\rho = \frac{M}{V} = \frac{500\text{ g}}{75\text{ cm}^3} \approx 6.7\text{ g/cm}^3\).

PastPaper.markingScheme

1 mark for calculating the total mass (\(500\text{ g}\)) and total volume (\(75\text{ cm}^3\)) of the mixture, and dividing to find the correct density of \(6.7\text{ g/cm}^3\).
PastPaper.question 35 · multiple_choice
1 PastPaper.marks
A trolley of mass \(2.0\text{ kg}\) travels at \(6.0\text{ m/s}\) to the right. It collides with a second trolley of mass \(3.0\text{ kg}\) travelling at \(2.0\text{ m/s}\) to the left. After the collision, the two trolleys stick together and move with a common velocity. What is their common velocity?
  1. A.1.2 m/s to the left
  2. B.1.2 m/s to the right
  3. C.3.6 m/s to the left
  4. D.3.6 m/s to the right
PastPaper.showAnswers

PastPaper.workedSolution

Let the direction to the right be positive. Total momentum before collision is: \(p_{\text{initial}} = (m_1 \times u_1) + (m_2 \times u_2) = (2.0 \times 6.0) + (3.0 \times -2.0) = 12.0 - 6.0 = 6.0\text{ kg m/s}\). Total mass after collision is: \(M = m_1 + m_2 = 2.0 + 3.0 = 5.0\text{ kg}\). Using conservation of momentum: \(p_{\text{final}} = M \times v \implies 6.0 = 5.0 \times v \implies v = 1.2\text{ m/s}\). Since the value is positive, the velocity is directed to the right.

PastPaper.markingScheme

1 mark for applying the law of conservation of momentum with appropriate signs for directions, obtaining the correct magnitude and direction of \(1.2\text{ m/s}\) to the right.
PastPaper.question 36 · multiple_choice
1 PastPaper.marks
An electric heater rated at \(50\text{ W}\) is used to heat \(200\text{ g}\) of a liquid in a well-insulated container. The liquid is already at its boiling point. It takes \(400\text{ s}\) of heating to completely vaporise all of the liquid. What is the specific latent heat of vaporisation of the liquid?
  1. A.1.0 × 10⁴ J/kg
  2. B.2.5 × 10⁴ J/kg
  3. C.1.0 × 10⁵ J/kg
  4. D.4.0 × 10⁵ J/kg
PastPaper.showAnswers

PastPaper.workedSolution

The energy supplied by the heater is: \(E = P \times t = 50\text{ W} \times 400\text{ s} = 20,000\text{ J}\). The mass of the liquid is: \(m = 200\text{ g} = 0.20\text{ kg}\). Using the formula for latent heat: \(E = m \times L_v \implies L_v = \frac{E}{m} = \frac{20,000}{0.20} = 100,000\text{ J/kg} = 1.0 \times 10^5\text{ J/kg}\).

PastPaper.markingScheme

1 mark for finding the total energy transferred (\(20,000\text{ J}\)) and converting the mass to kilograms (\(0.20\text{ kg}\)), leading to the correct specific latent heat value of \(1.0 \times 10^5\text{ J/kg}\).
PastPaper.question 37 · multiple_choice
1 PastPaper.marks
A ray of light in air strikes the flat surface of a transparent 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 block?
  1. A.28°
  2. B.30°
  3. C.45°
  4. D.68°
PastPaper.showAnswers

PastPaper.workedSolution

According to Snell's Law: \(n_{\text{air}} \sin(\theta_{\text{air}}) = n_{\text{glass}} \sin(\theta_{\text{glass}})\). Since \(n_{\text{air}} = 1.0\), we have: \(1.0 \times \sin(45^\circ) = 1.5 \times \sin(\theta_{\text{glass}})\). This gives: \(\sin(\theta_{\text{glass}}) = \frac{\sin(45^\circ)}{1.5} \approx \frac{0.707}{1.5} \approx 0.471\). Finding the inverse sine: \(\theta_{\text{glass}} = \arcsin(0.471) \approx 28.1^\circ\), which rounds to \(28^\circ\).

PastPaper.markingScheme

1 mark for applying Snell's law correctly to calculate the angle of refraction as \(28^\circ\).
PastPaper.question 38 · multiple_choice
1 PastPaper.marks
A wire of length \(L\) and cross-sectional area \(A\) has a resistance of \(8.0\ \Omega\). A second wire made of the same metal has a length of \(2L\) and a cross-sectional area of \(4A\). What is the resistance of the second wire?
  1. A.2.0 Ω
  2. B.4.0 Ω
  3. C.16 Ω
  4. D.32 Ω
PastPaper.showAnswers

PastPaper.workedSolution

The resistance of a wire is given by \(R = \rho \frac{L}{A}\). For the first wire, \(R_1 = \rho \frac{L}{A} = 8.0\ \Omega\). For the second wire, \(R_2 = \rho \frac{2L}{4A} = \frac{1}{2} \left(\rho \frac{L}{A}\right) = \frac{1}{2} R_1 = \frac{1}{2} \times 8.0\ \Omega = 4.0\ \Omega\).

PastPaper.markingScheme

1 mark for using the relationship of resistance to length and cross-sectional area to find the correct resistance of \(4.0\ \Omega\).
PastPaper.question 39 · multiple_choice
1 PastPaper.marks
A radioactive sample has a measured count rate of \(360\text{ counts/minute}\) in a laboratory where the background radiation level is constant at \(40\text{ counts/minute}\). After \(12\text{ hours}\), the measured count rate of the sample (including background) has fallen to \(80\text{ counts/minute}\). What is the half-life of the radioactive isotope?
  1. A.2.0 hours
  2. B.3.0 hours
  3. C.4.0 hours
  4. D.6.0 hours
PastPaper.showAnswers

PastPaper.workedSolution

First, subtract the background count rate to find the actual count rate of the sample. Initial source count rate = \(360 - 40 = 320\text{ counts/minute}\). Final source count rate = \(80 - 40 = 40\text{ counts/minute}\). The ratio of the final to initial count rate of the source is: \(\frac{40}{320} = \frac{1}{8}\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly 3 half-lives have elapsed. The total time is \(12\text{ hours}\), so: \(3 \times T_{1/2} = 12\text{ hours} \implies T_{1/2} = 4.0\text{ hours}\).

PastPaper.markingScheme

1 mark for correctly subtracting the background count rate, identifying that 3 half-lives have passed, and calculating the half-life as \(4.0\text{ hours}\).
PastPaper.question 40 · multiple_choice
1 PastPaper.marks
An artificial satellite orbits the Earth in a circular path of radius \(r\) with a constant speed \(v\). The time taken for one complete orbit is \(T\). If the orbital radius is increased to \(4r\) and the orbital speed decreases to \(0.5v\), what is the new orbital period in terms of \(T\)?
  1. A.2T
  2. B.4T
  3. C.8T
  4. D.16T
PastPaper.showAnswers

PastPaper.workedSolution

The orbital period is given by the formula \(T = \frac{2\pi r}{v}\). For the new orbit, the radius is \(4r\) and the speed is \(0.5v\). Therefore, the new period \(T'\) is: \(T' = \frac{2\pi (4r)}{0.5v} = \frac{4}{0.5} \times \frac{2\pi r}{v} = 8 \times T = 8T\).

PastPaper.markingScheme

1 mark for using the relationship \(T \propto \frac{r}{v}\) and substituting the new parameters to obtain \(8T\).

Paper 42 Theory (Extended)

Answer all questions. Show all working and write answers in the spaces provided.
10 PastPaper.question · 80 PastPaper.marks
PastPaper.question 1 · Structured
8 PastPaper.marks
A cyclist starts from rest and accelerates.

(a) Define acceleration. [1]

(b) The cyclist accelerates uniformly at \(1.5\text{ m/s}^2\) for \(6.0\text{ s}\).

(i) Calculate the velocity of the cyclist at \(t = 6.0\text{ s}\). [2]

(ii) The cyclist then travels at this constant velocity for another \(12.0\text{ s}\). Finally, she decelerates uniformly to rest in a further \(4.0\text{ s}\). Describe the features of a velocity-time graph representing this entire \(22.0\text{ s}\) journey, specifying the key coordinates \((t, v)\) at each phase. [2]

(iii) Calculate the total distance traveled by the cyclist during the entire \(22.0\text{ s}\) journey. [3]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Acceleration is defined as the rate of change of velocity per unit time (or change in velocity divided by time taken). [1 mark]

(b) (i) Initial velocity \(u = 0\), \(a = 1.5\text{ m/s}^2\), \(t = 6.0\text{ s}\).
\(v = u + at = 0 + (1.5 \times 6.0) = 9.0\text{ m/s}\). [2 marks]

(ii) The graph starts at \((0, 0)\), goes up as a straight line to \((6.0, 9.0)\).
It continues as a horizontal line from \((6.0, 9.0)\) to \((18.0, 9.0)\).
Finally, it goes down as a straight line from \((18.0, 9.0)\) to \((22.0, 0)\). [2 marks]

(iii) Total distance is the area under the velocity-time graph.
The area is a trapezium:
\(\text{Area} = \frac{1}{2}(a + b)h\) where \(a = 12.0\text{ s}\) (from \(t=6.0\) to \(t=18.0\)) and \(b = 22.0\text{ s}\), and \(h = 9.0\text{ m/s}\).
\(\text{Area} = \frac{1}{2}(12.0 + 22.0) \times 9.0 = 17.0 \times 9.0 = 153\text{ m}\).

Alternatively, sum the areas of the individual phases:
- Phase 1 (triangle): \(\frac{1}{2} \times 6.0 \times 9.0 = 27\text{ m}\)
- Phase 2 (rectangle): \(12.0 \times 9.0 = 108\text{ m}\)
- Phase 3 (triangle): \(\frac{1}{2} \times 4.0 \times 9.0 = 18\text{ m}\)
Total distance = \(27 + 108 + 18 = 153\text{ m}\). [3 marks]

PastPaper.markingScheme

**Part (a)**
- Rate of change of velocity OR change of velocity per unit time. [1 mark]

**Part (b)(i)**
- \(v = a \times t\) or \(1.5 \times 6.0\) [1 mark]
- \(9.0\text{ m/s}\) [1 mark]

**Part (b)(ii)**
- Straight line from \((0,0)\) to \((6.0, 9.0)\) AND horizontal line to \(t = 18.0\text{ s}\). [1 mark]
- Straight line from \((18.0, 9.0)\) to \((22.0, 0)\). [1 mark]

**Part (b)(iii)**
- Method to find area under graph: \(\text{Area} = \text{sum of areas of shapes}\) OR \(\frac{1}{2}(a+b)h\) [1 mark]
- Correct substitution of values: \(\frac{1}{2} \times (12 + 22) \times 9.0\) OR \(27 + 108 + 18\) [1 mark]
- \(153\text{ m}\) (accept \(150\text{ m}\) to 2 s.f.) [1 mark]
PastPaper.question 2 · Structured
8 PastPaper.marks
A small trolley A of mass \(2.5\text{ kg}\) travels at a velocity of \(4.0\text{ m/s}\) on a friction-free horizontal track. It collides with a stationary trolley B of mass \(1.5\text{ kg}\). After the collision, the two trolleys stick together and move with a common velocity \(v\).

(a) State the principle of conservation of momentum. [2]

(b) Calculate the common velocity \(v\) of the trolleys after the collision. [3]

(c) Calculate the magnitude of the impulse exerted on trolley B during the collision. [3]
PastPaper.showAnswers

PastPaper.workedSolution

(a) The total momentum of an isolated or closed system remains constant, provided no external forces act on it. [2 marks]

(b) Initial total momentum \(p_{\text{initial}} = m_A u_A + m_B u_B = (2.5 \times 4.0) + (1.5 \times 0) = 10\text{ kg m/s}\).
Final total momentum \(p_{\text{final}} = (m_A + m_B)v = (2.5 + 1.5)v = 4.0 v\).
By conservation of momentum:
\(4.0 v = 10\text{ kg m/s} \implies v = 2.5\text{ m/s}\). [3 marks]

(c) Impulse on trolley B is equal to its change in momentum:
\(\text{Impulse} = \Delta p_B = m_B (v_B - u_B)\)
\(\text{Impulse} = 1.5 \times (2.5 - 0) = 3.75\text{ N s}\) (or \(\text{kg m/s}\)). [3 marks]

PastPaper.markingScheme

**Part (a)**
- Total momentum before collision is equal to total momentum after collision [1 mark]
- In a closed system / provided no external forces act [1 mark]

**Part (b)**
- Use of \(m_1 u_1 + m_2 u_2 = (m_1 + m_2)v\) or calculates initial momentum as \(10\text{ kg m/s}\) [1 mark]
- Substitution: \(10 = 4.0 \times v\) [1 mark]
- \(2.5\text{ m/s}\) [1 mark]

**Part (c)**
- Recall of \(\text{Impulse} = \text{change in momentum} = m\Delta v\) [1 mark]
- Correct substitution: \(1.5 \times 2.5\) [1 mark]
- \(3.75\text{ N s}\) (or \(\text{kg m/s}\), accept \(3.8\text{ N s}\)) [1 mark]
PastPaper.question 3 · Structured
8 PastPaper.marks
An electric heater rated at \(150\text{ W}\) is placed inside a solid block of lead of mass \(0.80\text{ kg}\).

(a) Define specific heat capacity. [2]

(b) The block of lead is initially at a temperature of \(20^\circ\text{C}\). The specific heat capacity of lead is \(130\text{ J/(kg }^\circ\text{C)}\).

(i) Calculate the energy required to heat the lead block to its melting point of \(327^\circ\text{C}\). [2]

(ii) Assuming there is no loss of thermal energy to the surroundings, calculate the time taken by the heater to raise the temperature of the lead block to \(327^\circ\text{C}\). [2]

(c) Once the block reaches \(327^\circ\text{C}\), it begins to melt. Explain, in terms of intermolecular forces and particle arrangement, why the temperature of the lead remains constant during melting even though the heater is still supplying thermal energy. [2]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Specific heat capacity is the energy required per unit mass to raise the temperature of a substance by one degree Celsius (or one Kelvin). [2 marks]

(b) (i) First, find the change in temperature: \(\Delta T = 327 - 20 = 307^\circ\text{C}\).
\(Q = mc\Delta T = 0.80 \times 130 \times 307 = 31920\text{ J}\) (or \(3.2 \times 10^4\text{ J}\)). [2 marks]

(ii) Since \(Q = P \times t\):
\(t = Q / P = 31920 / 150 = 212.8\text{ s}\) (or \(210\text{ s}\) to 2 s.f.). [2 marks]

(c) The energy supplied by the heater is used to break or weaken the intermolecular bonds between lead atoms, changing their regular solid lattice structure into an irregular liquid arrangement. It does not increase the average kinetic energy of the particles, which means the temperature remains constant during the phase change. [2 marks]

PastPaper.markingScheme

**Part (a)**
- Energy required per unit mass (or per kg) [1 mark]
- To raise the temperature by one degree (Celsius or Kelvin) [1 mark]

**Part (b)(i)**
- Calculates temperature change \(\Delta T = 307^\circ\text{C}\) and uses \(Q = mc\Delta T\) [1 mark]
- \(31920\text{ J}\) (accept \(3.2 \times 10^4\text{ J}\) or \(31900\text{ J}\)) [1 mark]

**Part (b)(ii)**
- Uses \(t = Q / P\) with their value from (b)(i) [1 mark]
- \(212.8\text{ s}\) (accept \(213\text{ s}\) or \(210\text{ s}\)) [1 mark]

**Part (c)**
- Energy is used to overcome/break intermolecular bonds / change particle arrangement from regular solid to irregular liquid [1 mark]
- Kinetic energy of particles does not increase (so temperature remains constant) [1 mark]
PastPaper.question 4 · Structured
8 PastPaper.marks
A ray of monochromatic light is incident on the boundary between a glass prism and the air. The refractive index of the glass is \(1.52\).

(a) State what is meant by the critical angle. [2]

(b) Calculate the critical angle \(c\) for the boundary between the glass and air. [2]

(c) The light ray is traveling inside the glass prism and strikes the inner surface of the glass at an angle of incidence of \(42.0^\circ\).

(i) Determine whether the light ray will emerge into the air or undergo total internal reflection. Explain your answer with reference to the critical angle calculated in (b). [2]

(ii) State the name of one practical application of total internal reflection in telecommunications. [2]
PastPaper.showAnswers

PastPaper.workedSolution

(a) The critical angle is the angle of incidence in the more dense medium for which the angle of refraction in the less dense medium is \(90^\circ\). [2 marks]

(b) Use the relation: \(\sin(c) = 1/n\) where \(n = 1.52\).
\(\sin(c) = 1/1.52 = 0.6579 \implies c = \sin^{-1}(0.6579) = 41.1^\circ\) (or \(41^\circ\)). [2 marks]

(c) (i) The angle of incidence is \(i = 42.0^\circ\), which is greater than the critical angle \(c = 41.1^\circ\). Since \(i > c\) and the light is traveling in the optically denser medium (glass) towards the less dense medium (air), the ray undergoes total internal reflection and does not emerge into the air. [2 marks]

(ii) Optical fibers (used for high-speed broadband communication and telecommunications). [2 marks]

PastPaper.markingScheme

**Part (a)**
- Angle of incidence in the optically denser medium [1 mark]
- That results in an angle of refraction of \(90^\circ\) / light traveling along the boundary [1 mark]

**Part (b)**
- Use of \(\sin(c) = 1 / n\) [1 mark]
- \(41.1^\circ\) (accept \(41^\circ\)) [1 mark]

**Part (c)(i)**
- Total internal reflection (TIR) occurs [1 mark]
- Because the angle of incidence (\(42.0^\circ\)) is greater than the critical angle (\(41.1^\circ\)) [1 mark]

**Part (c)(ii)**
- Optical fibers / fiber optic cables [2 marks]
PastPaper.question 5 · Structured
8 PastPaper.marks
A cell of electromotive force (e.m.f.) \(9.0\text{ V}\) and negligible internal resistance is connected in series with a fixed resistor of resistance \(12\\ \Omega\) and a thermistor.

(a) Define electromotive force (e.m.f.). [2]

(b) At a certain room temperature, the resistance of the thermistor is \(18\\ \Omega\).

(i) Calculate the total resistance of the series circuit. [1]

(ii) Calculate the current in the circuit. [2]

(iii) Calculate the electrical energy delivered to the \(12\\ \Omega\) fixed resistor in a time of \(5.0\text{ minutes}\). [3]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Electromotive force (e.m.f.) is the energy supplied by a source in driving charge around a complete circuit per unit charge. [2 marks]

(b) (i) Total resistance is the sum of the series resistances:
\(R_{\text{total}} = R_1 + R_2 = 12 + 18 = 30\\ \Omega\). [1 mark]

(ii) Current \(I = V / R_{\text{total}} = 9.0 / 30 = 0.30\text{ A}\). [2 marks]

(iii) First convert minutes to seconds: \(t = 5.0\text{ minutes} = 5.0 \times 60 = 300\text{ s}\).
Power dissipated in the \(12\\ \Omega\) resistor is:
\(P = I^2 R = (0.30)^2 \times 12 = 0.09 \times 12 = 1.08\text{ W}\).
Energy delivered \(E = P \times t = 1.08 \times 300 = 324\text{ J}\). [3 marks]

PastPaper.markingScheme

**Part (a)**
- Energy supplied by a source per unit charge / work done per unit charge [1 mark]
- In driving charge around a complete circuit [1 mark]

**Part (b)(i)**
- \(30\\ \Omega\) [1 mark]

**Part (b)(ii)**
- Uses \(I = V/R\) [1 mark]
- \(0.30\text{ A}\) [1 mark]

**Part (b)(iii)**
- Converts \(5.0\text{ minutes}\) to \(300\text{ s}\) [1 mark]
- Use of \(E = I^2 R t\) OR \(E = V I t\) where \(V_{\text{resistor}} = 0.30 \times 12 = 3.6\text{ V}\) [1 mark]
- \(324\text{ J}\) [1 mark]
PastPaper.question 6 · Structured
8 PastPaper.marks
An experiment is conducted to demonstrate electromagnetic induction.

(a) State Faraday's law of electromagnetic induction. [2]

(b) A student drops a bar magnet vertically, north pole pointing downwards, through a long solenoid connected to a sensitive center-zero millivoltmeter.

(i) Describe and explain the movement of the meter's pointer as the north pole enters the top of the solenoid. [3]

(ii) Describe the movement of the pointer as the south pole of the magnet leaves the bottom of the solenoid. Explain why the maximum deflection in this stage is larger than the maximum deflection observed in (b)(i). [3]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Faraday's law states that the induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage (or the rate of cutting of magnetic field lines). [2 marks]

(b) (i) The pointer deflects to one side (e.g., to the right) and then returns to zero. This happens because the falling magnet causes a changing magnetic field inside the solenoid, which induces an e.m.f. and hence a current. Lenz's law indicates a magnetic field is created to oppose the entry of the N pole. [3 marks]

(ii) The pointer deflects in the opposite direction (e.g., to the left) and then returns to zero. The maximum deflection is greater because the magnet has accelerated under gravity as it fell through the length of the solenoid. Therefore, it is moving faster as it leaves than when it entered, causing a greater rate of change of magnetic flux linkage and inducing a larger e.m.f. [3 marks]

PastPaper.markingScheme

**Part (a)**
- Induced e.m.f. is proportional to / is equal to [1 mark]
- Rate of change of magnetic flux linkage / rate at which magnetic field lines are cut [1 mark]

**Part (b)(i)**
- Pointer deflects to one side (and returns to zero) [1 mark]
- Due to changing magnetic flux / field lines being cut by coil [1 mark]
- Inducing an e.m.f. / current in the circuit [1 mark]

**Part (b)(ii)**
- Pointer deflects in the opposite direction [1 mark]
- Larger maximum deflection because the magnet is moving faster (accelerated by gravity) [1 mark]
- Resulting in a higher rate of change of magnetic flux linkage (or rate of cutting lines of force) [1 mark]
PastPaper.question 7 · Structured
8 PastPaper.marks
An artificial satellite orbits the Earth in a circular path.

(a) State the name of the force that keeps the satellite in its circular orbit, and state the direction in which this force acts. [2]

(b) The satellite orbits at a height of \(600\text{ km}\) above the Earth's surface. The radius of the Earth is \(6400\text{ km}\).

(i) State the radius of the satellite's circular orbit. [1]

(ii) The satellite travels with a constant orbital speed of \(7.5\text{ km/s}\). Calculate the time taken, in minutes, for the satellite to complete one full orbit. [3]

(c) Explain why the satellite is constantly accelerating even though it is traveling at a constant speed. [2]
PastPaper.showAnswers

PastPaper.workedSolution

(a) The force is the gravitational force (or gravity). It acts radially inwards towards the center of the Earth. [2 marks]

(b) (i) Radius of the orbit \(r = R_{\text{Earth}} + h = 6400 + 600 = 7000\text{ km}\) (or \(7.0 \times 10^6\text{ m}\)). [1 mark]

(ii) Distance for one orbit \(d = 2\pi r = 2 \times \pi \times 7000\text{ km} = 43982\text{ km}\).
Orbital speed \(v = 7.5\text{ km/s}\).
Time taken \(t = d / v = 43982 / 7.5 = 5864\text{ seconds}\).
In minutes: \(t = 5864 / 60 = 97.7\text{ minutes}\) (or \(98\text{ minutes}\)). [3 marks]

(c) Acceleration is defined as the rate of change of velocity. Velocity is a vector quantity that has both magnitude (speed) and direction. As the satellite moves in a circular orbit, its direction of travel is continuously changing, so its velocity is continuously changing, which means it is accelerating. [2 marks]

PastPaper.markingScheme

**Part (a)**
- Gravitational force / gravity [1 mark]
- Towards the center of the Earth [1 mark]

**Part (b)(i)**
- \(7000\text{ km}\) (or \(7.0 \times 10^6\text{ m}\) - must include correct unit) [1 mark]

**Part (b)(ii)**
- Uses \(v = 2\pi r / T\) or finds circumference \(d = 2\pi r = 43982\text{ km}\) [1 mark]
- Calculates time in seconds: \(t = 5864\text{ s}\) [1 mark]
- Converts to minutes correctly: \(97.7\text{ minutes}\) (accept \(98\text{ minutes}\) to 2 s.f.) [1 mark]

**Part (c)**
- Velocity has both magnitude and direction / is a vector [1 mark]
- Direction is constantly changing, so velocity is changing, meaning there is acceleration [1 mark]
PastPaper.question 8 · Structured
8 PastPaper.marks
A sample containing a radioactive isotope decays by emitting beta-minus (\(\beta^-\)) particles.

(a) State two differences between an alpha particle and a beta-minus particle. [2]

(b) The activity of the sample is monitored. The constant background radiation in the laboratory is measured to be \(24\text{ counts/minute}\). The initial raw count rate of the sample, including background, is \(412\text{ counts/minute}\). After \(15.0\text{ hours}\), the raw count rate of the sample is \(72.5\text{ counts/minute}\).

(i) Calculate the corrected initial count rate and the corrected count rate after \(15.0\text{ hours}\). [1]

(ii) Calculate the half-life of the radioactive isotope. [3]

(c) State one risk of exposure to beta radiation and describe a suitable safety precaution when handling this source. [2]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Differences include: Alpha is a helium nucleus (2 protons, 2 neutrons) while beta is a high-speed electron; alpha is positively charged (+2e) while beta is negatively charged (-1e); alpha is much more massive than beta; alpha is highly ionizing but poorly penetrating, whereas beta is moderately ionizing and moderately penetrating. (Any two) [2 marks]

(b) (i) Corrected initial count rate = \(412 - 24 = 388\text{ counts/minute}\).
Corrected count rate after 15.0 hours = \(72.5 - 24 = 48.5\text{ counts/minute}\). [1 mark]

(ii) Find the number of half-lives that have passed by calculating the ratio of final to initial corrected activity:
\(48.5 / 388 = 0.125 = 1/8 = (1/2)^3\).
So, exactly \(3\) half-lives have passed in \(15.0\text{ hours}\).
\(3 \times t_{1/2} = 15.0\text{ hours} \implies t_{1/2} = 5.0\text{ hours}\). [3 marks]

(c) Risk: Beta radiation is ionizing and can penetrate living tissue to cause gene mutations or cancer.
Precaution: Handle the source using long tongs; keep a safe distance; store in a lead-lined container when not in use. [2 marks]

PastPaper.markingScheme

**Part (a)**
Any two from:
- Alpha has a positive charge, beta has a negative charge
- Alpha is a helium nucleus (2p + 2n), beta is a fast electron
- Alpha is much heavier / has mass 4u, beta has negligible mass / 1/1840 u
- Alpha is more ionizing than beta
- Alpha is less penetrating than beta [2 marks]

**Part (b)(i)**
- Correctly subtracts background: \(388\text{ counts/minute}\) AND \(48.5\text{ counts/minute}\) [1 mark]

**Part (b)(ii)**
- Finds ratio \(48.5 / 388 = 0.125\) or \(1/8\) [1 mark]
- Identifies this represents 3 half-lives [1 mark]
- Calculates half-life: \(15.0 / 3 = 5.0\text{ hours}\) [1 mark]

**Part (c)**
- Risk: Can cause mutation of cells / cancer / cell damage [1 mark]
- Precaution: Use of tongs / increase distance from source / store in lead container when not in use [1 mark]
PastPaper.question 9 · Structured Theory
8 PastPaper.marks
A toy truck of mass \(1.2\text{ kg}\) travelling at a speed of \(2.5\text{ m/s}\) collides with a stationary toy car of mass \(0.80\text{ kg}\). After the collision, the truck and the car couple together and move with a common velocity. (a) Define momentum. [1] (b) State the condition necessary for the total momentum of a system of interacting objects to remain constant. [1] (c) Calculate the common speed of the coupled truck and car immediately after the collision. [3] (d) Calculate the magnitude of the impulse exerted by the truck on the car during the collision. State the unit. [3]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Momentum is defined mathematically as the product of an object's mass and its velocity (\(p = mv\)). (b) For the total momentum of a system to remain constant, there must be no external resultant force acting on the system (an isolated system). (c) Apply the law of conservation of momentum: Total momentum before collision = Total momentum after collision. \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\). Substituting the given values: \((1.2 \times 2.5) + (0.80 \times 0) = (1.2 + 0.80) v\), which simplifies to \(3.0 = 2.0 v\). Solving for \(v\): \(v = 1.5\text{ m/s}\). (d) The impulse on the car is equal to the change in momentum of the car: \(\text{Impulse} = \Delta p = m_2 (v - u_2) = 0.80 \times (1.5 - 0) = 1.2\text{ N s}\) (or \(1.2\text{ kg m/s}\)).

PastPaper.markingScheme

(a) product of mass and velocity [1 mark]. (b) no external forces acting / closed system / isolated system [1 mark]. (c) conservation of momentum equation or substitution: \(1.2 \times 2.5 = (1.2 + 0.80) v\) [1 mark], correct calculation of total mass of 2.0 kg or initial momentum of 3.0 kg m/s [1 mark], final speed of 1.5 m/s [1 mark]. (d) recognition that impulse equals change in momentum or formula \(m \Delta v\) [1 mark], value of 1.2 [1 mark], correct unit: N s or kg m/s [1 mark].
PastPaper.question 10 · Structured Theory
8 PastPaper.marks
An electric heater is used to heat a block of metal. The heater is rated at \(60\text{ W}\). The metal block has a mass of \(1.5\text{ kg}\). The heater is switched on for \(5.0\text{ minutes}\) and the temperature of the block rises from \(20\text{ }^\circ\text{C}\) to \(32\text{ }^\circ\text{C}\). (a) Calculate the total thermal energy supplied by the heater in this time. [2] (b) Calculate the specific heat capacity of the metal from these results. State the unit. [4] (c) In practice, some thermal energy is lost to the surroundings during heating. (i) State the effect of this thermal energy loss on the calculated value of the specific heat capacity. [1] (ii) Suggest one improvement to the apparatus or procedure to reduce this heat loss. [1]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Thermal energy supplied is calculated using energy = power \(\times\) time. Converting time to seconds: \(t = 5.0\text{ minutes} = 5.0 \times 60 = 300\text{ s}\). Therefore, \(E = 60\text{ W} \times 300\text{ s} = 18\,000\text{ J}\) (or \(18\text{ kJ}\)). (b) The specific heat capacity formula is \(Q = m c \Delta \theta\). Rearranging for \(c\) gives \(c = Q / (m \Delta \theta)\). Here, \(Q = 18\,000\text{ J}\), \(m = 1.5\text{ kg}\), and \(\Delta \theta = 32 - 20 = 12\text{ }^\circ\text{C}\). Substituting these values: \(c = 18\,000 / (1.5 \times 12) = 18\,000 / 18 = 1000\text{ J/(kg }^\circ\text{C)}\). (c)(i) Since some energy escapes to the surroundings, the actual thermal energy absorbed by the metal block is less than the calculated 18000 J. This leads to an overestimation of the specific heat capacity, making the calculated value larger than the true value. (c)(ii) Wrap the metal block with an insulator (lagging) or place it on an insulating stand to minimize heat transfer to the environment.

PastPaper.markingScheme

(a) conversion of time to 300 s [1 mark], correct thermal energy calculation of 18000 J or 18 kJ [1 mark]. (b) temperature difference \(\Delta \theta = 12\text{ }^\circ\text{C}\) calculated or used [1 mark], formula rearrangement or substitution seen: \(18000 = 1.5 \times c \times 12\) [1 mark], value 1000 [1 mark], unit: J / (kg °C) or J / (kg K) [1 mark]. (c)(i) value is higher / larger [1 mark]. (c)(ii) wrap in lagging / insulate the metal block [1 mark].

Paper 62 Alternative to Practical

Answer all questions. Use a ruler, sharp pencil, and show experimental calculations.
4 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · Practical
10 PastPaper.marks
A student investigates the focal length of a converging lens. Fig 1.1 shows the experimental setup. An illuminated object (a cross-wire) is placed at the zero end of a metre rule. A lens is placed on the rule at some distance \(u_1\) from the object. A screen is moved along the rule until a sharp, inverted image of the object is formed on the screen. The distance from the lens to the screen is the image distance, \(v_1\). (a) For the first trial, the lens is positioned at the \(35.0\text{ cm}\) mark, and the screen is at the \(68.0\text{ cm}\) mark. (i) Calculate the image distance \(v_1\). (ii) Calculate the focal length \(f_1\) of the lens using the equation: \(f_1 = \frac{u_1 v_1}{u_1 + v_1}\) where the object distance \(u_1 = 22.0\text{ cm}\). Give your answer to a suitable number of significant figures and include the unit. (b) The student repeats the procedure for a second trial, obtaining \(u_2 = 30.0\text{ cm}\) and \(v_2 = 24.5\text{ cm}\). Calculate the focal length \(f_2\) for this trial. (c) Calculate the average focal length \(f_{\text{avg}}\) of the lens using your values of \(f_1\) and \(f_2\). (d) Suggest two reasons why it is difficult to obtain an exact, precise position of the screen for the sharpest image. (e) State one precaution that the student should take to ensure that the measurements of \(u\) and \(v\) are accurate. (f) The student wants to check if the focal length is constant. State whether the two values \(f_1\) and \(f_2\) are within the limits of experimental accuracy, and justify your answer.
PastPaper.showAnswers

PastPaper.workedSolution

To find the solutions: (a)(i) \(v_1 = 68.0\text{ cm} - 35.0\text{ cm} = 33.0\text{ cm}\). (a)(ii) Using \(u_1 = 22.0\text{ cm}\) and \(v_1 = 33.0\text{ cm}\), \(f_1 = \frac{22.0 \times 33.0}{22.0 + 33.0} = \frac{726}{55.0} = 13.2\text{ cm}\). (b) For \(u_2 = 30.0\text{ cm}\) and \(v_2 = 24.5\text{ cm}\), \(f_2 = \frac{30.0 \times 24.5}{30.0 + 24.5} = \frac{735}{54.5} = 13.5\text{ cm}\) (to 3 significant figures). (c) The average is \(f_{\text{avg}} = \frac{13.2 + 13.5}{2} = 13.35\text{ cm}\). (d) It is difficult to find the exact point because the image may look sharp over a small range of screen movements, or because of a low-contrast image. (e) Precaution: Avoid parallax error when measuring distances by looking perpendicularly at the ruler, or ensure the optical components are strictly parallel. (f) The difference between \(f_1\) and \(f_2\) is \(13.5 - 13.2 = 0.3\text{ cm}\). Percentage difference is \(\frac{0.3}{13.2} \times 100\% \approx 2.3\%\). Since this is less than \(10\%\), the two values are within the limits of experimental accuracy.

PastPaper.markingScheme

1 mark: v1 = 33.0 cm. 1 mark: f1 = 13.2. 1 mark: Correct unit (cm) and 2 or 3 sig figs for f1. 1 mark: f2 = 13.5 cm. 1 mark: favg = 13.35 cm (or 13.4 cm). 2 marks: Two valid reasons for screen positioning difficulty (e.g., range of focus, faint image). 1 mark: One valid precaution (e.g., darkened room, aligned heights). 2 marks: Yes + justification showing that the percentage difference is within 10% (or similar reasonable limit).
PastPaper.question 2 · Practical
10 PastPaper.marks
A student is investigating how the resistance of a wire depends on its length. The student connects a circuit containing a power supply, a switch, an ammeter, a voltmeter, and a length of resistance wire \(XY\) of total length \(100.0\text{ cm}\) fixed to a metre rule. (a) Draw a circuit diagram to show how this apparatus is connected to measure the current in the wire \(XY\) and the potential difference across a length \(L\) of the wire. (b) The student measures the current \(I_1\) in the wire and the potential difference \(V_1\) across a length \(L_1 = 20.0\text{ cm}\) of the wire. The pointer of the voltmeter has major markings at 0 V, 1 V, 2 V, and 3 V, with 10 small subdivisions between each. The pointer is at the 6th subdivision after the 1 V mark. The ammeter scale has major markings at 0 A, 0.5 A, and 1.0 A, with 10 small subdivisions between 0 and 0.5 A. The pointer is at the 8th subdivision after 0 A. (i) State the voltmeter reading \(V_1\). (ii) State the ammeter reading \(I_1\). (c) Calculate the resistance \(R_1\) of the \(20.0\text{ cm}\) length of wire using \(R_1 = \frac{V_1}{I_1}\). Include the unit. (d) The student repeats the measurements for a length \(L_2 = 60.0\text{ cm}\) and obtains \(V_2 = 2.4\text{ V}\) and \(I_2 = 0.20\text{ A}\). Calculate the resistance \(R_2\) for \(L_2 = 60.0\text{ cm}\). (e) Suggest why the current in the second measurement is lower than the current in the first measurement. (f) A student suggests that the resistance of the wire is directly proportional to its length. State whether the results support this suggestion. Justify your answer with a calculation.
PastPaper.showAnswers

PastPaper.workedSolution

(a) The diagram must show a standard series circuit of cell/power supply, switch, ammeter, and resistance wire. A voltmeter must be connected in parallel specifically across the length \(L\) of the wire. (b)(i) \(V_1 = 1.0\text{ V} + 6 \times 0.1\text{ V} = 1.6\text{ V}\). (ii) \(I_1 = 0\text{ A} + 8 \times 0.05\text{ A} = 0.40\text{ A}\). (c) \(R_1 = \frac{1.6}{0.40} = 4.0\ \Omega\). (d) \(R_2 = \frac{2.4}{0.20} = 12.0\ \Omega\). (e) When the length of the wire in the circuit is tripled, the resistance increases, which reduces the total current since current is inversely proportional to resistance for a given voltage. (f) Ratio of resistance: \(\frac{R_2}{R_1} = \frac{12.0}{4.0} = 3.0\). Ratio of length: \(\frac{L_2}{L_1} = \frac{60.0}{20.0} = 3.0\). Since the ratios are identical, the resistance is directly proportional to the length of the wire.

PastPaper.markingScheme

1 mark: Correct circuit symbols with power supply, ammeter, switch, and wire in series. 1 mark: Voltmeter connected in parallel with the wire segment of length L. 1 mark: V1 = 1.6 V. 1 mark: I1 = 0.40 A. 1 mark: R1 = 4.0. 1 mark: Correct unit (ohms or Omega) for R1. 1 mark: R2 = 12.0 ohms. 1 mark: Explaining that increasing length increases resistance, lowering current. 1 mark: Statement 'Yes' supporting the proportional relationship. 1 mark: Showing equal ratios (R/L is constant or ratios are equal) to justify.
PastPaper.question 3 · Practical
10 PastPaper.marks
A student is investigating the rate of cooling of hot water in two beakers under different conditions. Beaker \(A\) is uninsulated. Beaker \(B\) is insulated with a single layer of bubble wrap. Both beakers are filled with hot water. The temperature \(\theta\) of the water in each beaker is recorded every \(30\text{ s\) for \(180\text{ s}\). (a) The thermometer for beaker \(A\) at the start of the experiment (\(t = 0\)) has major markings at 70 \(^\circ\text{C}\), 80 \(^\circ\text{C}\), 90 \(^\circ\text{C}\) with 10 small divisions between 80 \(^\circ\text{C}\) and 90 \(^\circ\text{C}\). The liquid column ends exactly 4 divisions above 80 \(^\circ\text{C}\). (i) Record the initial temperature \(\theta_0\) of the water in beaker \(A\). (ii) State one precaution the student must take when reading the thermometer to ensure the temperature is recorded accurately. (b) Table 3.1 shows the recorded temperature readings: Time \(t\) / \(\text{s}\) | Temperature of \(A\), \(\theta_A\) / \(^\circ\text{C}\) | Temperature of \(B\), \(\theta_B\) / \(^\circ\text{C}\) [0, 84.0, 84.0], [30, 79.5, 81.5], [60, 76.0, 79.5], [90, 73.0, 77.5], [120, 70.5, 76.0], [150, 68.5, 74.5], [180, 67.0, 73.5]. (i) Calculate the average rate of cooling \(R_A\) of the water in beaker \(A\) during the first \(90\text{ s}\). Use the equation: \(R_A = \frac{\theta_0 - \theta_{90}}{t}\) where \(t = 90\text{ s}\). Include the unit. (ii) Calculate the average rate of cooling \(R_B\) of the water in beaker \(B\) during the first \(90\text{ s}\). (c) Compare the rates of cooling and state whether the bubble wrap insulation is effective. Justify your answer by reference to the calculated rates of cooling. (d) A student suggests that using a lid on beaker \(B\) would make the experiment a more fair comparison of the insulation properties of the wrap itself. Suggest why using lids on both beakers is a good idea to focus on the insulation on the sides. (e) Another student wants to repeat the experiment to see if the thickness of the insulation affects the cooling rate. Describe how the apparatus can be modified and state what variables should be kept constant.
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(a)(i) The thermometer reading is \(80 + 4 = 84.0\ ^\circ\text{C}\). (ii) Precaution: view the scale perpendicularly to avoid parallax error, or stir the water before reading. (b)(i) \(R_A = \frac{84.0 - 73.0}{90} = \frac{11.0}{90} = 0.122\ ^\circ\text{C/s}\). (ii) \(R_B = \frac{84.0 - 77.5}{90} = \frac{6.5}{90} = 0.072\ ^\circ\text{C/s}\). (c) Comparing rates, \(R_B < R_A\). This shows that heat is lost more slowly from beaker \(B\), meaning the bubble wrap is effective. (d) Lids block evaporation and upward convection, so the majority of heat transfer occurs through the cylinder sides, which are being tested. (e) Modification: Wrap different numbers of bubble wrap layers (e.g., 1, 2, 3, 4 layers) around similar beakers. Control variables: volume of water, initial temperature of water, and room temperature.

PastPaper.markingScheme

1 mark: theta_0 = 84.0 °C. 1 mark: Valid precaution (parallax, stirring). 1 mark: RA = 0.122 (or 0.12). 1 mark: Correct unit for rate (°C/s or °C s^-1). 1 mark: RB = 0.072 (or 0.07). 2 marks: State 'Yes' and justify with comparison of cooling rates (RB is lower than RA). 1 mark: Lid minimizes evaporation/top heat loss. 1 mark: Modification (different numbers of layers of bubble wrap). 1 mark: Identify at least one key control variable (volume of water, initial temperature).
PastPaper.question 4 · Practical
10 PastPaper.marks
A student is investigating the balancing of a metre rule to determine its mass \(M\). The student places a pivot under the metre rule at a distance \(d\) from the zero end. A known load \(P = 1.0\text{ N}\) is suspended from the rule at a distance \(x = 10.0\text{ cm}\) from the zero end. The position of the pivot \(d\) is adjusted until the metre rule balances horizontally. For \(x = 10.0\text{ cm}\), the rule balances when the pivot is at the \(d = 34.0\text{ cm}\) mark. (a) (i) Explain what is meant by the 'centre of gravity' of the metre rule, and state its likely position on a uniform metre rule. (ii) Describe the forces acting on the rule and their respective positions when balanced. (b) Write down: (i) the distance \(a\) from the pivot to the load \(P\). (ii) the distance \(b\) from the pivot to the centre of gravity of the rule. (c) Calculate the weight \(W\) of the metre rule using the principle of moments: \(P \times a = W \times b\). (d) Calculate the mass \(M\) of the metre rule in grams, using \(g = 10\text{ N/kg}\). (e) Suggest one practical reason why it is difficult to find the exact balance point \(d\) of the metre rule. (f) Describe a method to find the centre of gravity of an irregularly-shaped piece of cardboard.
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(a)(i) The centre of gravity is the point through which the entire weight of an object acts. For a uniform metre rule, it lies at its geometric centre, which is the \(50.0\text{ cm}\) mark. (ii) The downward weight of load \(P\) acts at \(10.0\text{ cm}\), the downward weight of the rule \(W\) acts at the centre of gravity (\(50.0\text{ cm}\)), and the upward normal contact force from the pivot acts at the \(34.0\text{ cm}\) mark. (b)(i) \(a = d - x = 34.0 - 10.0 = 24.0\text{ cm}\). (ii) \(b = 50.0 - d = 50.0 - 34.0 = 16.0\text{ cm}\). (c) Principle of moments: \(1.0 \times 24.0 = W \times 16.0 \implies W = \frac{24.0}{16.0} = 1.5\text{ N}\). (d) \(M = \frac{W}{g} = \frac{1.5}{10} = 0.15\text{ kg} = 150\text{ g}\). (e) Practical difficulty: the rule has a finite thickness and may slide off, or the pivot is broad and not a perfect knife-edge, making the exact balancing center hard to specify. (f) Hang the cardboard from a pin/hole, attach a plumb line to the same pin, trace the plumb line's path on the cardboard. Repeat this for another hole. The intersection of the two drawn lines is the centre of gravity.

PastPaper.markingScheme

1 mark: Definition of centre of gravity. 1 mark: Center of gravity of uniform rule is at 50.0 cm mark. 2 marks: Correct description of the forces (direction and position). 1 mark: a = 24.0 cm. 1 mark: b = 16.0 cm. 1 mark: W = 1.5 N. 1 mark: M = 150 g. 1 mark: Practical reason for difficulty in balancing (e.g. flat pivot, rule slides). 1 mark: Method for irregular shape (plumb line method, tracing lines from pinpoints).

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