An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V2) Cambridge International A Level Physics (9702) paper. Not affiliated with or reproduced from Cambridge.
Paper 12 (Multiple Choice)
There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D.
40 Question · 40 marks
Question 1 · multiple_choice
1 marks
Two wires, X and Y, are made of the same metal. Wire X has length L and diameter d. Wire Y has length 2L and diameter 2d. Both wires are suspended vertically from a rigid support and each supports the same load F at its lower end. What is the ratio of the strain energy stored in wire X to the strain energy stored in wire Y?
A.0.5
B.1
C.2
D.4
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Worked solution
The strain energy in a wire is given by the formula U = 1/2 * F * x, where x is the extension. The extension is given by x = F * L / (A * E), where A is the cross-sectional area and E is the Young modulus. Since the cross-sectional area A is proportional to the square of the diameter d, the strain energy U is proportional to L / d^2. For wire X, the strain energy is proportional to L / d^2. For wire Y, the strain energy is proportional to 2L / (2d)^2 = L / (2d^2). Therefore, the ratio of the strain energy in X to that in Y is (L / d^2) / (L / 2d^2) = 2.
Marking scheme
1 mark for using the correct relationship between strain energy, wire length, and wire diameter to find the ratio of 2.
Question 2 · multiple_choice
1 marks
A uniform plank of length 3.0 m and weight 120 N rests horizontally on two supports. One support is at the left end of the plank, and the other support is 1.0 m from the right end of the plank. A metal block of weight 200 N is placed on the plank at a distance x from the left end. The upward force exerted on the plank by the left support is 100 N. What is the distance x?
A.0.80 m
B.1.1 m
C.1.3 m
D.1.5 m
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Worked solution
For vertical equilibrium of the plank, the total upward force must equal the total downward force. Thus, the right support force is R = (120 N + 200 N) - 100 N = 220 N. Since the plank is uniform, its center of gravity is at its midpoint (1.5 m from the left end). Taking moments about the left end (0 m): the clockwise moments are (120 N * 1.5 m) + (200 N * x), and the counter-clockwise moment is (220 N * 2.0 m). Setting these equal gives 180 + 200x = 440, which yields 200x = 260 and x = 1.3 m.
Marking scheme
1 mark for using vertical equilibrium to find the upward force on the right support and applying the principle of moments about the left end to solve for x.
Question 3 · multiple_choice
1 marks
A battery of electromotive force (e.m.f.) E and internal resistance r is connected across a variable resistor. The potential difference across the variable resistor is V and the current in the circuit is I. Which graph would be a straight line with a gradient equal to -r?
A.V on the y-axis against I on the x-axis
B.I on the y-axis against V on the x-axis
C.V on the y-axis against 1/I on the x-axis
D.I on the y-axis against 1/V on the x-axis
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Worked solution
The terminal potential difference V across the variable resistor is related to the e.m.f. E and internal resistance r by the equation V = E - Ir. This can be rewritten in the standard equation for a straight line, y = mx + c, as V = (-r)I + E. If V is plotted on the vertical axis (y-axis) and I is plotted on the horizontal axis (x-axis), the resulting graph is a straight line with a gradient equal to -r and a vertical intercept equal to E.
Marking scheme
1 mark for using the equation V = E - Ir to deduce that a plot of V against I gives a straight line with a gradient of -r.
Question 4 · multiple_choice
1 marks
A uniform metal wire of resistance 12.0 ohms is stretched uniformly to three times its original length. The volume and density of the wire do not change during the stretching. What is the resistance of the stretched wire?
A.12.0 ohms
B.36.0 ohms
C.108 ohms
D.324 ohms
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Worked solution
The resistance of a wire is given by R = rho * L / A, where rho is the resistivity, L is the length, and A is the cross-sectional area. Since the volume V = A * L remains constant, tripling the length (L' = 3L) means the cross-sectional area is reduced to one-third of its original value (A' = A/3). The new resistance is R' = rho * (3L) / (A/3) = 9 * (rho * L / A) = 9 * 12.0 ohms = 108 ohms.
Marking scheme
1 mark for identifying that the area decreases by a factor of 3 and correctly calculating the final resistance as 108 ohms.
Question 5 · multiple_choice
1 marks
A tube closed at one end has a fundamental frequency of 240 Hz. The speed of sound in air is 340 m s^-1. Which frequency is NOT a harmonic of this tube?
A.480 Hz
B.720 Hz
C.1200 Hz
D.1680 Hz
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Worked solution
For a tube closed at one end, the boundary conditions only allow odd harmonics to form. The frequencies of these harmonics are given by f_n = n * f_1, where f_1 = 240 Hz and n must be an odd integer (n = 1, 3, 5, 7, ...). The possible harmonic frequencies are 240 Hz, 720 Hz, 1200 Hz, 1680 Hz, etc. The frequency 480 Hz represents an even harmonic (n = 2), which cannot be formed in a tube closed at one end.
Marking scheme
1 mark for identifying that only odd harmonics are supported in a closed-end tube and selecting 480 Hz as the impossible frequency.
Question 6 · multiple_choice
1 marks
A toy car of mass 0.50 kg is released from rest at the top of a rough slope of height 2.0 m. The car reaches the bottom of the slope with a speed of 4.0 m s^-1. How much work is done against the frictional forces during the descent?
A.4.0 J
B.5.8 J
C.9.8 J
D.13.8 J
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Worked solution
The initial gravitational potential energy of the car is E_p = m * g * h = 0.50 kg * 9.81 m s^-2 * 2.0 m = 9.81 J. The final kinetic energy of the car is E_k = 1/2 * m * v^2 = 0.5 * 0.50 kg * (4.0 m s^-1)^2 = 4.0 J. By the conservation of energy, the work done against friction is the difference between the initial potential energy and the final kinetic energy: Work = E_p - E_k = 9.81 J - 4.0 J = 5.81 J, which rounds to 5.8 J.
Marking scheme
1 mark for calculating the loss in gravitational potential energy and gain in kinetic energy, and taking the difference to find the work done against friction.
Question 7 · multiple_choice
1 marks
Trolley P of mass 3m moves with a velocity v along a frictionless horizontal track. It collides with a stationary trolley Q of mass m. After the collision, the two trolleys stick together and move with a common velocity. What fraction of the initial kinetic energy is lost as thermal energy during the collision?
A.0.25
B.0.33
C.0.67
D.0.75
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Worked solution
Using conservation of linear momentum: (3m * v) + (m * 0) = (3m + m) * v_f, which gives the final velocity v_f = 3/4 * v. The initial kinetic energy is E_i = 1/2 * (3m) * v^2 = 1.5 * m * v^2. The final kinetic energy is E_f = 1/2 * (4m) * (3/4 * v)^2 = 2m * (9/16 * v^2) = 1.125 * m * v^2. The loss in kinetic energy is E_i - E_f = 1.5 * m * v^2 - 1.125 * m * v^2 = 0.375 * m * v^2. The fraction lost is (0.375 * m * v^2) / (1.5 * m * v^2) = 0.25.
Marking scheme
1 mark for calculating the final velocity of the combined mass and using it to find the ratio of final to initial kinetic energy, leading to a loss of 0.25.
Question 8 · multiple_choice
1 marks
The resistivity rho of a wire is determined using the formula rho = R * pi * d^2 / (4 * L). The measured values and their percentage uncertainties are shown: resistance R has an uncertainty of 3.0%, diameter d has an uncertainty of 2.0%, and length L has an uncertainty of 1.5%. What is the percentage uncertainty in the calculated value of the resistivity rho?
A.6.5%
B.8.5%
C.10.5%
D.11.5%
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Worked solution
The formula for combining independent uncertainties gives the percentage uncertainty in resistivity as: % uncertainty in rho = (% uncertainty in R) + 2 * (% uncertainty in d) + (% uncertainty in L). Note that the percentage uncertainty in the diameter d must be multiplied by 2 because it is squared in the formula. Substituting the given values: % uncertainty in rho = 3.0% + 2 * (2.0%) + 1.5% = 3.0% + 4.0% + 1.5% = 8.5%.
Marking scheme
1 mark for applying the rule for combining percentage uncertainties by doubling the uncertainty in d and adding all terms to obtain 8.5%.
Question 9 · MCQ
1 marks
The resistivity \(\rho\) of a metal wire is determined using the equation \(\rho = \frac{R \pi d^2}{4 L}\). The measurements obtained are: resistance \(R = (25.0 \pm 0.5)\ \Omega\), diameter \(d = (0.80 \pm 0.02)\text{ mm}\), and length \(L = (1.50 \pm 0.03)\text{ m}\). What is the percentage uncertainty in the calculated value of the resistivity?
A.4.5%
B.6.5%
C.9.0%
D.11.5%
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1 mark for calculating correct percentage uncertainties and combining them using the correct uncertainty equation to obtain 9.0%.
Question 10 · MCQ
1 marks
A uniform horizontal shelf of length \(1.2\text{ m}\) and weight \(36\text{ N}\) is attached to a wall by a frictionless hinge at point A. The shelf is held horizontal by a wire connected between the outer end of the shelf (point B) and a point C on the wall, which is vertically above A. The distance AC is \(0.9\text{ m}\). What is the tension in the wire?
A.18 N
B.30 N
C.36 N
D.45 N
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Worked solution
Let \(\theta\) be the angle between the wire and the horizontal shelf. The length of the wire is \(BC = \sqrt{1.2^2 + 0.9^2} = 1.5\text{ m}\). Therefore, \(\sin\theta = \frac{0.9}{1.5} = 0.6\). The weight of the shelf acts at its midpoint, \(0.6\text{ m}\) from A. Taking moments about the hinge A: \(T \sin\theta \times 1.2 = W \times 0.6\). Substituting the values: \(T \times 0.6 \times 1.2 = 36 \times 0.6 \implies 0.72 T = 21.6 \implies T = 30\text{ N}\).
Marking scheme
1 mark for taking moments about the hinge, resolving the vertical component of tension correctly, and obtaining 30 N.
Question 11 · MCQ
1 marks
A steel wire X has length \(L\) and diameter \(d\). Another steel wire Y made of the same material has length \(2L\) and diameter \(2d\). Both wires obey Hooke's law and support the same vertical load \(F\). What is the ratio \(\frac{\text{elastic potential energy in X}}{\text{elastic potential energy in Y}}\)?
A.0.5
B.1.0
C.2.0
D.4.0
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Worked solution
Elastic potential energy stored is \(E = \frac{1}{2} F x\) where \(x\) is the extension. Since the load \(F\) is the same for both wires, the ratio of energy is \(\frac{E_X}{E_Y} = \frac{x_X}{x_Y}\). The extension is given by \(x = \frac{F L}{A E} = \frac{4 F L}{\pi d^2 E}\). Since both wires are made of the same material, the Young's modulus \(E\) is identical. Thus, \(x \propto \frac{L}{d^2}\). For X, \(x_X \propto \frac{L}{d^2}\). For Y, \(x_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\). Thus, \(\frac{x_X}{x_Y} = 2\), which means the ratio of energy is 2.0.
Marking scheme
1 mark for expressing extension in terms of L and d, relating energy to extension, and calculating the ratio of 2.0.
Question 12 · MCQ
1 marks
A stationary wave is established on a stretched string of length \(1.20\text{ m}\) which is fixed at both ends. The string vibrates in its third harmonic with a frequency of \(150\text{ Hz}\). What is the speed of the transverse waves on the string?
A.60 m s⁻¹
B.90 m s⁻¹
C.120 m s⁻¹
D.240 m s⁻¹
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Worked solution
For the third harmonic of a string fixed at both ends, the length \(L\) consists of three half-wavelengths: \(L = \frac{3\lambda}{2}\). Therefore, the wavelength is \(\lambda = \frac{2 L}{3} = \frac{2 \times 1.20\text{ m}}{3} = 0.80\text{ m}\). The speed \(v\) of the wave is given by \(v = f \lambda = 150\text{ Hz} \times 0.80\text{ m} = 120\text{ m s}^{-1}\).
Marking scheme
1 mark for relating length to wavelength for the third harmonic, calculating the wavelength, and finding the wave speed of 120 m s⁻¹.
Question 13 · MCQ
1 marks
A cylindrical metal wire of length \(L\) and uniform cross-sectional area \(A\) has resistance \(R\). The wire is stretched at constant volume so that its length increases by \(10\%\). Assuming the resistivity of the metal remains unchanged, what is the new resistance of the wire?
A.1.10 R
B.1.21 R
C.1.33 R
D.1.44 R
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Worked solution
The volume of the wire is \(V = A L\), which remains constant. If the length increases by \(10\%\), the new length is \(L' = 1.10 L\). Since volume is conserved, the new area \(A'\) must satisfy \(A' L' = A L \implies A' = \frac{A}{1.10}\). The resistance is given by \(R = \rho \frac{L}{A}\). The new resistance is \(R' = \rho \frac{L'}{A'} = \rho \frac{1.10 L}{A / 1.10} = 1.21 \rho \frac{L}{A} = 1.21 R\).
Marking scheme
1 mark for using conservation of volume to find the new area and calculating the new resistance as 1.21 R.
Question 14 · MCQ
1 marks
A battery of electromotive force (e.m.f.) \(12\text{ V}\) and internal resistance \(r\) is connected to a variable resistor of resistance \(R\). When \(R = 4.0\ \Omega\), the terminal potential difference across the battery is \(8.0\text{ V}\). What is the terminal potential difference when \(R\) is increased to \(10.0\ \Omega\)?
A.9.0 V
B.9.6 V
C.10.0 V
D.10.8 V
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Worked solution
The terminal potential difference is given by \(V = \frac{E R}{R + r}\). When \(R = 4.0\ \Omega\), \(8.0 = \frac{12 \times 4.0}{4.0 + r} \implies 8.0(4.0 + r) = 48 \implies 32.0 + 8.0r = 48.0 \implies r = 2.0\ \Omega\). When \(R\) is increased to \(10.0\ \Omega\), the new terminal potential difference is \(V' = \frac{12 \times 10.0}{10.0 + 2.0} = \frac{120}{12} = 10.0\text{ V}\).
Marking scheme
1 mark for determining the internal resistance of the battery, and then using it to calculate the correct new terminal potential difference of 10.0 V.
Question 15 · MCQ
1 marks
A block of mass \(0.50\text{ kg}\) is launched up a rough inclined plane with an initial speed of \(6.0\text{ m s}^{-1}\). The block travels a distance of \(2.0\text{ m}\) along the slope, rising to a vertical height of \(1.2\text{ m}\) before coming to rest. What is the average frictional force acting on the block during its motion up the slope?
A.1.1 N
B.1.6 N
C.2.9 N
D.4.5 N
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Worked solution
The initial kinetic energy is \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} (0.50) (6.0)^2 = 9.0\text{ J}\). The gain in gravitational potential energy is \(E_p = m g h = 0.50 \times 9.81 \times 1.2 = 5.89\text{ J}\). The work done against friction is \(W_f = E_k - E_p = 9.0 - 5.89 = 3.11\text{ J}\). Since \(W_f = F_f \times d\), where \(d\) is the distance along the slope, the average frictional force is \(F_f = \frac{3.11}{2.0} = 1.56\text{ N} \approx 1.6\text{ N}\).
Marking scheme
1 mark for calculating kinetic energy and potential energy, using conservation of energy to find the work done against friction, and dividing by distance to get 1.6 N.
Question 16 · MCQ
1 marks
A nucleus of polonium-218 (\(^{218}_{84}\text{Po}\)) decays through a series of alpha (\(\alpha\)) and beta-minus (\(\beta^-\)) emissions to a stable isotope of lead-206 (\(^{206}_{82}\text{Pb}\)). How many \(\alpha\) and \(\beta^-\)-particles are emitted in this decay series?
A.3 ̑̑\alpha\text{ and } 2\ \beta^-
B.3 \alpha\text{ and } 4\ \beta^-
C.4 \alpha\text{ and } 2\ \beta^-
D.4 \alpha\text{ and } 4\ \beta^-
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Worked solution
Let \(x\) be the number of \(\alpha\)-particles (\(^{4}_{2}\text{He}\)) and \(y\) be the number of \(\beta^-\)-particles (\(^{0}_{-1}\text{e}\)) emitted. Conserving nucleon number: \(218 = 206 + 4x \implies 4x = 12 \implies x = 3\). Conserving proton number: \(84 = 82 + 2x - y \implies 84 = 82 + 2(3) - y \implies 84 = 88 - y \implies y = 4\). Thus, 3 alpha particles and 4 beta-minus particles are emitted.
Marking scheme
1 mark for using nucleon number conservation to find the number of alpha particles, and proton number conservation to find the number of beta-minus particles.
Question 17 · MCQ
1 marks
Two wires, X and Y, made of the same metal, are joined end-to-end and stretched by a constant tensile force. Wire X has length \(L\) and diameter \(d\). Wire Y has length \(2L\) and diameter \(2d\). What is the ratio \(\frac{\text{elastic potential energy stored in X}}{\text{elastic potential energy stored in Y}}\)?
A.\(\frac{1}{4}\)
B.\(\frac{1}{2}\)
C.\(2\)
D.\(4\)
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Worked solution
The wires are connected in series (end-to-end), so they experience the same tensile force \(F\). The elastic potential energy stored in a stretched wire is given by \(E_p = \frac{1}{2} F \Delta L\), where \(\Delta L\) is the extension. The extension is given by \(\Delta L = \frac{F L}{A E} = \frac{4 F L}{\pi d^2 E}\) (since cross-sectional area \(A = \frac{\pi d^2}{4}\)). For wire X: \(\Delta L_X = \frac{4 F L}{\pi d^2 E}\). For wire Y: \(\Delta L_Y = \frac{4 F (2L)}{\pi (2d)^2 E} = \frac{8 F L}{4 \pi d^2 E} = \frac{2 F L}{\pi d^2 E}\). Therefore, \(\Delta L_X = 2 \Delta L_Y\). Since the force \(F\) is the same for both, the ratio of the energy stored is \(\frac{E_X}{E_Y} = \frac{\Delta L_X}{\Delta L_Y} = 2\).
Marking scheme
1 mark for the correct option C. [1] for calculating the correct ratio of extensions using the stress-strain relationship and relating it to elastic potential energy.
Question 18 · MCQ
1 marks
A uniform beam of weight \(W\) and length \(L\) is supported by a pivot located at a distance of \(\frac{L}{3}\) from its left end. A block of weight \(W\) is hung from the extreme left end. A vertical downward force \(F\) is applied at the extreme right end to keep the beam in horizontal equilibrium. What is the magnitude of the force \(F\) in terms of \(W\)?
A.\(0.13 W\)
B.\(0.25 W\)
C.\(0.50 W\)
D.\(0.75 W\)
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Worked solution
To find \(F\), take moments about the pivot. The pivot is at \(\frac{L}{3}\) from the left end. The block of weight \(W\) is at the left end, at a distance of \(\frac{L}{3}\) from the pivot, causing an anticlockwise moment: \(\text{Moment}_{\text{anticlockwise}} = W \times \frac{L}{3}\). The uniform beam's weight \(W\) acts at its center of gravity, which is at \(\frac{L}{2}\) from the left end. The distance from the pivot to the center of gravity is \(\frac{L}{2} - \frac{L}{3} = \frac{L}{6}\) to the right. This causes a clockwise moment: \(\text{Moment}_{\text{beam}} = W \times \frac{L}{6}\). The force \(F\) acts downward at the right end, at a distance of \(L - \frac{L}{3} = \frac{2L}{3}\) to the right of the pivot, causing a clockwise moment: \(\text{Moment}_{F} = F \times \frac{2L}{3}\). For rotational equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments: \(W \left(\frac{L}{3}\right) = W \left(\frac{L}{6}\right) + F \left(\frac{2L}{3}\right)\). Multiplying by \(\frac{6}{L}\) gives: \(2W = W + 4F\), which simplifies to \(W = 4F \implies F = 0.25W\).
Marking scheme
1 mark for the correct option B. [1] for correctly setting up the principle of moments equation about the pivot and solving for F.
Question 19 · MCQ
1 marks
A cell of electromotive force (e.m.f.) \(6.0\text{ V}\) and internal resistance \(1.0\ \Omega\) is connected in series with a \(2.0\ \Omega\) fixed resistor and a thermistor. As the temperature of the thermistor increases, its resistance decreases from \(3.0\ \Omega\) to \(1.0\ \Omega\). What are the initial and final values of the potential difference across the thermistor?
A.Initial = \(3.0\text{ V}\), Final = \(1.5\text{ V}\)
B.Initial = \(3.6\text{ V}\), Final = \(2.0\text{ V}\)
C.Initial = \(3.0\text{ V}\), Final = \(2.0\text{ V}\)
D.Initial = \(4.5\text{ V}\), Final = \(1.5\text{ V}\)
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Worked solution
Initial state: Thermistor resistance \(R_{\text{th}} = 3.0\ \Omega\). Total resistance of the circuit \(R_{\text{total}} = R_{\text{th}} + R_{\text{fixed}} + r = 3.0 + 2.0 + 1.0 = 6.0\ \Omega\). Circuit current \(I = \frac{E}{R_{\text{total}}} = \frac{6.0\text{ V}}{6.0\ \Omega} = 1.0\text{ A}\). Potential difference across the thermistor \(V_{\text{th}} = I \times R_{\text{th}} = 1.0\text{ A} \times 3.0\ \Omega = 3.0\text{ V}\). Final state: Thermistor resistance \(R_{\text{th}}' = 1.0\ \Omega\). Total resistance \(R_{\text{total}}' = 1.0 + 2.0 + 1.0 = 4.0\ \Omega\). New circuit current \(I' = \frac{6.0\text{ V}}{4.0\ \Omega} = 1.5\text{ A}\). New potential difference across the thermistor \(V_{\text{th}}' = I' \times R_{\text{th}}' = 1.5\text{ A} \times 1.0\ \Omega = 1.5\text{ V}\).
Marking scheme
1 mark for the correct option A. [1] for calculating both initial and final potential differences across the thermistor including the internal resistance of the cell.
Question 20 · MCQ
1 marks
Two cylindrical conductors, P and Q, are made of the same metal and have the same mass. Conductor Q has twice the length of conductor P. What is the ratio \(\frac{\text{resistance of Q}}{\text{resistance of P}}\)?
A.\(1\)
B.\(2\)
C.\(4\)
D.\(8\)
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Worked solution
Since conductors P and Q are made of the same metal and have the same mass, they must have the same volume \(V\). The volume of a cylindrical conductor is \(V = A \times L\), where \(A\) is the cross-sectional area and \(L\) is the length. Given \(L_Q = 2L_P\) and \(V_Q = V_P\), we have \(A_Q \times (2L_P) = A_P \times L_P \implies A_Q = \frac{A_P}{2}\). The resistance is given by \(R = \rho \frac{L}{A}\). For conductor P: \(R_P = \rho \frac{L_P}{A_P}\). For conductor Q: \(R_Q = \rho \frac{L_Q}{A_Q} = \rho \frac{2L_P}{A_P / 2} = 4 \rho \frac{L_P}{A_P} = 4 R_P\). Thus, the ratio \(\frac{R_Q}{R_P} = 4\).
Marking scheme
1 mark for the correct option C. [1] for using conservation of volume to find the new area and substituting into the resistivity formula.
Question 21 · MCQ
1 marks
A stationary wave is set up in an air column inside a pipe of length \(0.85\text{ m}\) that is closed at one end and open at the other. The speed of sound in air is \(340\text{ m s}^{-1}\). Which of the following frequencies is NOT a possible resonant frequency of this air column?
A.\(100\text{ Hz}\)
B.\(200\text{ Hz}\)
C.\(300\text{ Hz}\)
D.\(500\text{ Hz}\)
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Worked solution
For a pipe closed at one end, stationary waves can only form at odd harmonics. The resonant frequencies are given by \(f_n = \frac{n v}{4L}\) where \(n = 1, 3, 5, \dots\) is an odd integer, \(v = 340\text{ m s}^{-1}\), and \(L = 0.85\text{ m}\). The fundamental frequency (\(n = 1\)) is \(f_1 = \frac{1 \times 340}{4 \times 0.85} = \frac{340}{3.4} = 100\text{ Hz}\). The subsequent possible resonant frequencies (harmonics) are odd multiples of the fundamental frequency: \(300\text{ Hz}\), \(500\text{ Hz}\), \(700\text{ Hz}\), etc. Therefore, \(200\text{ Hz}\) is not a possible resonant frequency.
Marking scheme
1 mark for the correct option B. [1] for calculating the fundamental frequency and identifying that only odd harmonics are possible for a pipe closed at one end.
Question 22 · MCQ
1 marks
The density \(\rho\) of a solid cylinder of mass \(m\), diameter \(d\), and height \(h\) is determined using the equation \(\rho = \frac{4m}{\pi d^2 h}\). The measurements obtained are: \(m = (120.0 \pm 1.2)\text{ g}\), \(d = (2.00 \pm 0.04)\text{ mm}\), and \(h = (5.0 \pm 0.1)\text{ cm}\). What is the percentage uncertainty in the calculated value of \(\rho\)?
A.\(5.0\%\)
B.\(6.0\%\)
C.\(7.0\%\)
D.\(9.0\%\)
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Worked solution
The fractional uncertainty in \(\rho\) is given by: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta h}{h}\). First, calculate the percentage uncertainty for each measured quantity: Percentage uncertainty in \(m = \frac{1.2}{120.0} \times 100 = 1.0\%\). Percentage uncertainty in \(d = \frac{0.04}{2.00} \times 100 = 2.0\%\). Percentage uncertainty in \(h = \frac{0.1}{5.0} \times 100 = 2.0\%\). Now combine them: Percentage uncertainty in \(\rho = 1.0\% + 2 \times (2.0\%) + 2.0\% = 7.0\%\).
Marking scheme
1 mark for the correct option C. [1] for summing individual percentage uncertainties with correct weighting for the squared term.
Question 23 · MCQ
1 marks
Unpolarised light of intensity \(I_0\) is incident on a pair of polarising filters. The angle between the transmission axes of the two filters is \(60^\circ\). What is the intensity of the light emerging from the second filter?
A.\(\frac{1}{8} I_0\)
B.\(\frac{1}{4} I_0\)
C.\(\frac{3}{8} I_0\)
D.\(\frac{1}{2} I_0\)
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Worked solution
When unpolarised light of intensity \(I_0\) passes through the first polarising filter, its intensity is reduced by half, so the intensity of the transmitted light is \(I_1 = \frac{1}{2} I_0\). This light is now plane-polarised. When it passes through the second polarising filter, we apply Malus's Law: \(I_2 = I_1 \cos^2 \theta\), where \(\theta = 60^\circ\). Substituting the values: \(I_2 = \left(\frac{1}{2} I_0\right) \cos^2(60^\circ) = \left(\frac{1}{2} I_0\right) \times (0.5)^2 = \frac{1}{8} I_0\).
Marking scheme
1 mark for the correct option A. [1] for applying the half-intensity rule for unpolarised light followed by Malus's Law.
Question 24 · MCQ
1 marks
A free proton decays via \(\beta^+\) decay to form a neutron, a positron, and an electron neutrino. Which row correctly identifies the change in quark flavour and the leptons produced during this process?
A.Quark change: \(d \to u\), Leptons produced: positron and electron neutrino
B.Quark change: \(u \to d\), Leptons produced: positron and electron neutrino
C.Quark change: \(d \to u\), Leptons produced: electron and electron antineutrino
D.Quark change: \(u \to d\), Leptons produced: electron and electron antineutrino
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Worked solution
In \(\beta^+\) decay, a proton (\(uud\)) turns into a neutron (\(udd\)). This means one of the up (\(u\)) quarks changes flavour into a down (\(d\)) quark (\(u \to d\)). To conserve charge and lepton number, the decay also produces a positron (\(e^+\)) and an electron neutrino (\(\nu_e\)). Thus, row B is correct.
Marking scheme
1 mark for the correct option B. [1] for identifying both the correct quark transformation (u to d) and the correct leptons emitted in beta-plus decay.
Question 25 · MCQ
1 marks
The resistivity \(\rho\) of a uniform metal wire is calculated using the formula \(\rho = \frac{R \pi d^2}{4 L}\). The percentage uncertainties in the measurements are: resistance \(R\) is \(2\%\), diameter \(d\) is \(2\%\), and length \(L\) is \(1\%\). What is the percentage uncertainty in the calculated value of the resistivity?
A.3%
B.5%
C.7%
D.9%
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Worked solution
The fractional uncertainty equation for the resistivity is given by: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). Substituting the given percentage uncertainties into this relation: percentage uncertainty in \(\rho = 2\% + 2(2\%) + 1\% = 7\%\).
Marking scheme
1 mark for correct application of absolute/percentage uncertainty rules with power relations to obtain 7%.
Question 26 · MCQ
1 marks
Two trolleys, X and Y, are placed on a horizontal frictionless track. Trolley X has mass \(2m\) and moves with velocity \(v\) towards trolley Y, of mass \(3m\), which is initially at rest. The trolleys collide and stick together. What percentage of the initial kinetic energy is lost as thermal energy and sound during the collision?
A.40%
B.50%
C.60%
D.80%
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Worked solution
By conservation of linear momentum: \(2mv + 0 = (2m + 3m)v_f\) which simplifies to \(v_f = 0.4v\). The initial kinetic energy of the system is \(E_i = \frac{1}{2}(2m)v^2 = mv^2\). The final kinetic energy is \(E_f = \frac{1}{2}(5m)(0.4v)^2 = 0.4mv^2\). The loss in kinetic energy is therefore \(E_i - E_f = mv^2 - 0.4mv^2 = 0.6mv^2\). The percentage of energy lost is \(\frac{0.6mv^2}{mv^2} \times 100\% = 60\%\).
Marking scheme
1 mark for calculating the final velocity from conservation of momentum and finding the percentage of kinetic energy lost to be 60%.
Question 27 · MCQ
1 marks
A block of mass \(0.50\text{ kg}\) is launched up a constant slope of angle \(30^\circ\) to the horizontal with an initial kinetic energy of \(20\text{ J}\). The block travels a distance of \(2.0\text{ m}\) along the slope before coming to rest. What is the average resistive force acting on the block during its ascent?
A.5.1 N
B.7.5 N
C.10 N
D.15 N
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Worked solution
The vertical height gained by the block is \(h = 2.0 \sin(30^\circ) = 1.0\text{ m}\). The gain in gravitational potential energy is \(E_p = mgh = 0.50 \times 9.81 \times 1.0 = 4.9\text{ J}\). The total mechanical energy lost due to work done against friction is \(W = E_{k,\text{initial}} - E_p = 20 - 4.9 = 15.1\text{ J}\). Since the work done against friction is given by \(W = F \times d\), where \(d = 2.0\text{ m}\) is the distance along the slope, the average resistive force is \(F = \frac{15.1}{2.0} = 7.55\text{ N} \approx 7.5\text{ N}\).
Marking scheme
1 mark for calculating the vertical height and using conservation of energy to find the average resistive force of 7.5 N.
Question 28 · MCQ
1 marks
Two wires, X and Y, are made of the same material and are suspended vertically from a ceiling. Wire X has length \(L\) and diameter \(d\). Wire Y has length \(2L\) and diameter \(2d\). If both wires support equal loads, what is the ratio of the extension of wire X to the extension of wire Y?
A.0.5
B.1.0
C.2.0
D.4.0
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Worked solution
The Young modulus \(E\) is given by \(E = \frac{F L}{A \Delta L}\), so the extension is \(\Delta L = \frac{F L}{A E}\). Given that the cross-sectional area is \(A = \frac{\pi d^2}{4}\), we have \(\Delta L \propto \frac{L}{d^2}\) since the load \(F\) and material \(E\) are the same for both. For wire X, \(\Delta L_X \propto \frac{L}{d^2}\). For wire Y, \(\Delta L_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = 0.5 \frac{L}{d^2}\). Thus, the ratio of their extensions is \(\frac{\Delta L_X}{\Delta L_Y} = \frac{1}{0.5} = 2.0\).
Marking scheme
1 mark for using the definition of Young modulus to find the relationship between extension, length, and diameter, and computing the correct ratio of 2.0.
Question 29 · MCQ
1 marks
A uniform beam of weight \(120\text{ N}\) and length \(3.0\text{ m}\) is pivoted at one end. The beam is held horizontally in equilibrium by a cable attached to the other end. The cable makes an angle of \(30^\circ\) with the horizontal beam. What is the tension in the cable?
A.60 N
B.120 N
C.240 N
D.360 N
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Worked solution
Taking moments about the pivot: The clockwise moment due to the weight acting at the center of gravity (at the midpoint, \(1.5\text{ m}\) from the pivot) is \(120\text{ N} \times 1.5\text{ m} = 180\text{ N m}\). The anticlockwise moment is due to the vertical component of the tension, \(T \sin(30^\circ)\), acting at the end of the beam (\(3.0\text{ m}\) from the pivot): \((T \sin(30^\circ)) \times 3.0 = 1.5 T\). For equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments: \(1.5 T = 180\text{ N m} \implies T = 120\text{ N}\).
Marking scheme
1 mark for setting up the moments equation about the pivot and solving for a tension of 120 N.
Question 30 · MCQ
1 marks
A stationary wave is set up on a stretched string of length \(1.2\text{ m}\) fixed at both ends. The string vibrates in its third harmonic. What is the distance between a node and an adjacent antinode?
A.0.10 m
B.0.20 m
C.0.40 m
D.0.60 m
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Worked solution
For the third harmonic of a fixed string, there are three half-wavelength loops along the length \(L\) of the string. Hence, \(L = 3 \times \frac{\lambda}{2}\). Substituting \(L = 1.2\text{ m}\), we find the wavelength \(1.2 = 1.5\lambda \implies \lambda = 0.80\text{ m}\). The distance between a node and its adjacent antinode is one-quarter of a wavelength: \(\frac{\lambda}{4} = \frac{0.80}{4} = 0.20\text{ m}\).
Marking scheme
1 mark for calculating the wavelength from the harmonic description and dividing by 4 to find the correct node-to-antinode distance.
Question 31 · MCQ
1 marks
A uniform metal wire has resistance \(R\). The wire is stretched until its length is doubled. Assuming that the volume and resistivity of the metal do not change, what is the new resistance of the wire?
A.R
B.2R
C.4R
D.8R
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Worked solution
Since the volume \(V = A \times L\) is constant, doubling the length (\(L \to 2L\)) requires the cross-sectional area to be halved (\(A \to A/2\)). The resistance of the wire is given by \(R = \rho \frac{L}{A}\). The new resistance of the stretched wire is \(R_{\text{new}} = \rho \frac{2L}{A/2} = 4 \rho \frac{L}{A} = 4R\).
Marking scheme
1 mark for identifying that resistance is proportional to \(L^2\) when volume is conserved, resulting in a value of 4R.
Question 32 · MCQ
1 marks
A battery of electromotive force (e.m.f.) \(E\) and internal resistance \(r\) is connected in series with a variable resistor. When the resistance of the variable resistor is varied, the current in the circuit is \(I\) and the potential difference across the battery terminals is \(V\). A graph is plotted with \(V\) on the vertical axis and \(I\) on the horizontal axis. What is the gradient of this graph?
A.\(E\)
B.\(-r\)
C.\(r\)
D.\(-\frac{E}{r}\)
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Worked solution
The terminal potential difference \(V\) is related to the e.m.f. \(E\) and current \(I\) by the equation \(V = E - Ir\). Rearranging this in the form \(y = mx + c\) gives \(V = -rI + E\). Comparing terms, a plot of \(V\) against \(I\) has a gradient equal to \(-r\).
Marking scheme
1 mark for matching the terminal potential difference formula with the equation of a straight line to find the gradient \(-r\).
Question 33 · MCQ
1 marks
Two wires, \(X\) and \(Y\), are made of the same material. Wire \(X\) has length \(L\) and diameter \(d\). Wire \(Y\) has length \(2L\) and diameter \(2d\). Both wires obey Hooke's law. They are extended by equal tensile forces. What is the ratio of the elastic potential energy stored in wire \(X\) to that stored in wire \(Y\)?
A.\(\frac{1}{4}\)
B.\(\frac{1}{2}\)
C.2
D.4
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Worked solution
The extension \(x\) of a wire under force \(F\) is given by \(x = \frac{F L}{A E}\), where \(A = \frac{\pi d^2}{4}\) is the cross-sectional area and \(E\) is the Young modulus. Thus, \(x = \frac{4 F L}{\pi d^2 E}\).
For wire \(X\): \(x_X = \frac{4 F L}{\pi d^2 E}\)
For wire \(Y\): \(x_Y = \frac{4 F (2L)}{\pi (2d)^2 E} = \frac{8 F L}{4 \pi d^2 E} = \frac{2 F L}{\pi d^2 E} = \frac{1}{2} x_X\)
The elastic potential energy stored is given by \(E_p = \frac{1}{2} F x\). Since the force \(F\) is the same for both wires: \(\frac{E_X}{E_Y} = \frac{x_X}{x_Y} = 2\)
Marking scheme
1 mark for the correct answer C.
Question 34 · MCQ
1 marks
A uniform bar of length \(2.0\text{ m}\) and weight \(40\text{ N}\) is supported horizontally by a pivot at a distance of \(0.40\text{ m}\) from its left end. A vertical string attached to the right end of the bar exerts an upward force of \(10\text{ N}\). A load of weight \(W\) is suspended from the extreme left end of the bar so that the bar remains in horizontal equilibrium. What is the weight \(W\)?
A.10\text{ N}
B.20\text{ N}
C.30\text{ N}
D.40\text{ N}
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Worked solution
To find the weight \(W\), take moments about the pivot. The bar is uniform, so its weight of \(40\text{ N}\) acts at its midpoint, which is \(1.0\text{ m}\) from either end. The distance from the pivot (at \(0.40\text{ m}\) from the left) to the center of gravity is: \(1.0\text{ m} - 0.40\text{ m} = 0.60\text{ m}\) (to the right of the pivot).
The vertical string is at the right end, which is at a distance of: \(2.0\text{ m} - 0.40\text{ m} = 1.6\text{ m}\) (to the right of the pivot).
The load \(W\) is at the left end, which is at a distance of \(0.40\text{ m}\) (to the left of the pivot).
Taking moments about the pivot: Sum of clockwise moments = Sum of counter-clockwise moments \((40\text{ N} \times 0.60\text{ m}) = (W \times 0.40\text{ m}) + (10\text{ N} \times 1.6\text{ m})\) \(24\text{ N m} = 0.40 W + 16\text{ N m}\) \(0.40 W = 8\text{ N m}\) \(W = 20\text{ N}\)
Marking scheme
1 mark for the correct answer B.
Question 35 · MCQ
1 marks
A battery of electromotive force (e.m.f.) \(E\) and internal resistance \(r\) is connected in series with a variable resistor. A voltmeter of infinite resistance is connected across the terminals of the battery.
The resistance of the variable resistor is varied, and the terminal potential difference \(V\) across the battery is recorded for different currents \(I\). A graph is plotted of \(V\) on the vertical axis against \(I\) on the horizontal axis.
Which statement about this graph is correct?
A.The gradient of the graph is positive.
B.The intercept on the horizontal axis is equal to the e.m.f. \(E\).
C.The magnitude of the gradient is equal to the internal resistance \(r\).
D.The area under the graph is equal to the total power dissipated in the circuit.
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Worked solution
The terminal potential difference \(V\) is related to the current \(I\) by the equation: \(V = E - Ir = -rI + E\) This represents a straight line with a negative gradient of \(-r\) and a vertical intercept of \(E\). Therefore, the magnitude of the gradient of the graph is equal to the internal resistance \(r\).
Marking scheme
1 mark for the correct answer C.
Question 36 · MCQ
1 marks
A cylindrical wire of length \(L\) and cross-sectional area \(A\) has a resistance \(R\). The wire is stretched uniformly so that its length increases by \(10\%\) while its volume remains constant. What is the new resistance of the wire in terms of \(R\)?
A.1.00 R
B.1.10 R
C.1.21 R
D.1.44 R
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Worked solution
The initial resistance is \(R = \rho \frac{L}{A}\). Since the volume \(V = A L\) remains constant, if the length increases by \(10\%\), the new length is \(L' = 1.10 L\). Because \(A' L' = A L\), the new cross-sectional area is: \(A' = \frac{A}{1.10}\)
Now, the new resistance \(R'\) is: \(R' = \rho \frac{L'}{A'} = \rho \frac{1.10 L}{A / 1.10} = 1.21 \rho \frac{L}{A} = 1.21 R\)
Marking scheme
1 mark for the correct answer C.
Question 37 · MCQ
1 marks
A stationary wave of wavelength \(\lambda\) is set up on a stretched string. Two points, \(X\) and \(Y\), on the string are oscillating. The distance between \(X\) and \(Y\) is exactly \(\frac{1}{3}\lambda\), and there is no node between them. What is the phase difference between the oscillations of \(X\) and \(Y\)?
A.0
B.\(\frac{\pi}{3}\text{ rad}\)
C.\(\frac{2\pi}{3}\text{ rad}\)
D.\(\pi\text{ rad}\)
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Worked solution
In a stationary wave, all points within a single loop (i.e., between two adjacent nodes) oscillate in phase with each other. Since there is no node between \(X\) and \(Y\), they must be in the same loop. Therefore, they oscillate in phase, and their phase difference is 0.
Marking scheme
1 mark for the correct answer A.
Question 38 · MCQ
1 marks
A block of mass \(2.0\text{ kg}\) is pushed up a rough slope inclined at an angle of \(30^\circ\) to the horizontal. An average frictional force of \(5.0\text{ N}\) acts on the block parallel to the slope. The block is pushed with a constant force of \(20\text{ N}\) parallel to the slope for a distance of \(4.0\text{ m}\). What is the increase in the kinetic energy of the block over this distance?
A.21\text{ J}
B.41\text{ J}
C.60\text{ J}
D.80\text{ J}
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Worked solution
By the principle of conservation of energy: Work done by applied force = Increase in G.P.E. + Work done against friction + Increase in K.E.
1. Work done by applied force: \(W = F \times d = 20\text{ N} \times 4.0\text{ m} = 80\text{ J}\) 2. Increase in G.P.E.: \(\Delta E_p = m g \Delta h = m g (d \sin 30^\circ) = 2.0 \times 9.81 \times (4.0 \times 0.5) = 39.24\text{ J}\) 3. Work done against friction: \(W_f = f \times d = 5.0\text{ N} \times 4.0\text{ m} = 20\text{ J}\)
Now, substitute these values into the energy equation: \(80 = 39.24 + 20 + \Delta E_k\) \(\Delta E_k = 80 - 59.24 = 20.76\text{ J} \approx 21\text{ J}\)
Marking scheme
1 mark for the correct answer A.
Question 39 · MCQ
1 marks
Unpolarised light of intensity \(I_0\) is incident on a polarizing filter. The light transmitted by this first filter then passes through a second polarizing filter (the analyser). The transmission axis of the analyser is at an angle of \(60^\circ\) to the transmission axis of the first filter. What is the intensity of the light emerging from the analyser?
A.\(\frac{1}{8} I_0\)
B.\(\frac{1}{4} I_0\)
C.\(\frac{3}{8} I_0\)
D.\(\frac{1}{2} I_0\)
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Worked solution
When unpolarised light of intensity \(I_0\) passes through the first polarizer, it becomes plane-polarised, and its intensity is halved: \(I_1 = \frac{1}{2} I_0\)
According to Malus's Law, when this polarised light passes through the analyser with its transmission axis at \(\theta = 60^\circ\), the intensity of the emerging light \(I_2\) is: \(I_2 = I_1 \cos^2(60^\circ) = \left(\frac{1}{2} I_0\right) \times (0.5)^2 = \frac{1}{2} I_0 \times \frac{1}{4} = \frac{1}{8} I_0\)
Marking scheme
1 mark for the correct answer A.
Question 40 · MCQ
1 marks
An experimenter determines the acceleration of free fall \(g\) by measuring the period \(T\) of a simple pendulum of length \(L\). The equation used is:
\(g = \frac{4\pi^2 L}{T^2}\)
The percentage uncertainty in the measurement of \(L\) is \(1.5\%\). The percentage uncertainty in the measurement of \(T\) is \(2.0\%\).
What is the percentage uncertainty in the calculated value of \(g\)?
A.3.5%
B.5.5%
C.7.5%
D.9.5%
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Worked solution
To find the percentage uncertainty in \(g\), we add the percentage uncertainty in \(L\) to twice the percentage uncertainty in \(T\):
Answer all questions. Write your answers in the spaces provided. Show all working and use appropriate units.
8 Question · 60 marks
Question 1 · Structured
7.5 marks
A copper wire of original length \(2.40\text{ m}\) and cross-sectional area \(3.2 \times 10^{-7}\text{ m}^2\) is suspended vertically. A load of \(48\text{ N}\) is applied to the free end, causing an elastic extension. The Young modulus of copper is \(1.2 \times 10^{11}\text{ Pa}\). (a) Define the term tensile strain. [1.5 marks] (b) Calculate: (i) the stress in the wire due to the load, [2 marks] (ii) the extension produced by this load. [2 marks] (c) State and explain how the extension would differ if a steel wire of the same initial length and diameter, but with a Young modulus of \(2.0 \times 10^{11}\text{ Pa}\), were used under the same load. [2 marks]
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Worked solution
(a) Tensile strain is defined as the extension per unit original length: \(\text{strain} = \frac{\Delta L}{L_0}\). It is a dimensionless ratio. (b)(i) \(\text{Stress} = \frac{\text{Force}}{\text{Area}} = \frac{48}{3.2 \times 10^{-7}} = 1.5 \times 10^8\text{ Pa}\). (b)(ii) \(\text{Young modulus } E = \frac{\text{Stress}}{\text{Strain}} \Rightarrow \text{Strain} = \frac{\text{Stress}}{E} = \frac{1.5 \times 10^8}{1.2 \times 10^{11}} = 1.25 \times 10^{-3}\). Since \(\text{Strain} = \frac{x}{L_0}\), extension \(x = \text{Strain} \times L_0 = 1.25 \times 10^{-3} \times 2.40 = 3.0 \times 10^{-3}\text{ m}\) (or \(3.0\text{ mm}\)). (c) Since \(x = \frac{FL}{AE}\), extension is inversely proportional to the Young modulus \(E\) for a constant force, length, and area. Because steel has a higher Young modulus than copper, the extension of the steel wire will be smaller: \(x_{\text{steel}} = x_{\text{copper}} \times \frac{1.2 \times 10^{11}}{2.0 \times 10^{11}} = 1.8\text{ mm}\).
Marking scheme
(a) [1.5 marks] For stating 'extension divided by original length' (1 mark) and noting it is ratio of similar quantities / has no units (0.5 marks). (b)(i) [2 marks] Formula stress = F/A (1 mark), final answer 1.5 * 10^8 Pa with correct unit (1 mark). (b)(ii) [2 marks] Strain formula or E = FL/Ax (1 mark), final answer 3.0 * 10^-3 m or 3.0 mm with unit (1 mark). (c) [2 marks] Stating extension is smaller (1 mark), explaining that extension is inversely proportional to Young modulus / steel is stiffer (1 mark).
Question 2 · Structured
7.5 marks
A non-uniform wooden plank AB of length \(3.0\text{ m}\) and weight \(120\text{ N}\) is supported horizontally by two vertical ropes. One rope is attached at end A and the other rope is attached at a distance of \(0.60\text{ m}\) from end B. The tension in the rope at end A is \(40\text{ N}\). (a) State the two conditions required for a rigid body to be in equilibrium. [2 marks] (b) Calculate: (i) the tension in the second rope, [1.5 marks] (ii) the distance of the centre of gravity of the plank from end A. [4 marks]
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Worked solution
(a) The two conditions for equilibrium are: 1. The resultant force acting on the body in any direction must be zero. 2. The resultant torque (or sum of moments) about any point must be zero. (b)(i) Let the tension in the second rope be \(T_B\). For vertical equilibrium, the sum of upward forces equals the sum of downward forces: \(T_A + T_B = W \Rightarrow 40 + T_B = 120 \Rightarrow T_B = 80\text{ N}\). (b)(ii) Let the centre of gravity be at a distance \(d\) from end A. The second rope is located at a distance of \(3.0 - 0.60 = 2.4\text{ m}\) from end A. Taking moments about end A: \(\text{Clockwise moment} = W \times d = 120 \times d\). \(\text{Counter-clockwise moment} = T_B \times 2.4 = 80 \times 2.4 = 192\text{ N m}\). For rotational equilibrium: \(120 d = 192 \Rightarrow d = 1.6\text{ m}\).
Marking scheme
(a) [2 marks] 1 mark for stating zero resultant force, 1 mark for stating zero resultant torque/moment about any point. (b)(i) [1.5 marks] Sum of upward forces = downward forces equation (0.5 marks), final answer 80 N (1 mark). (b)(ii) [4 marks] Identifying distance of second rope from A is 2.4 m (1 mark), writing correct moment equation about a chosen pivot (e.g. A) (1.5 marks), calculation of distance d (1 mark), final answer 1.6 m with unit (0.5 marks).
Question 3 · Structured
7.5 marks
A cell of electromotive force (e.m.f.) \(E = 6.0\text{ V}\) and internal resistance \(r = 1.5\ \Omega\) is connected in series with a resistor R and a switch. A high-resistance voltmeter is connected across the terminals of the cell. (a) Explain what is meant by the internal resistance of a cell. [1.5 marks] (b) When the switch is open, the reading on the voltmeter is \(V_1\). When the switch is closed, the reading is \(V_2 = 4.8\text{ V}\). (i) State the value of \(V_1\). [1 mark] (ii) Calculate the current in the circuit when the switch is closed. [2 marks] (iii) Calculate the resistance of resistor R. [3 marks]
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Worked solution
(a) Internal resistance is the resistance to current flow within the cell or battery itself, which results in some electrical energy being transformed into thermal energy ('lost volts') inside the cell when there is a current. (b)(i) When the switch is open, no current flows through the circuit. Thus, there is no potential drop across the internal resistance. The terminal potential difference equals the e.m.f. of the cell, so \(V_1 = 6.0\text{ V}\). (b)(ii) When the switch is closed, the 'lost volts' across the internal resistance is \(v = E - V_2 = 6.0 - 4.8 = 1.2\text{ V}\). Using Ohm's law for the internal resistance: \(I = \frac{v}{r} = \frac{1.2}{1.5} = 0.80\text{ A}\). (b)(iii) The terminal potential difference \(V_2\) is the voltage across the external resistor R. Thus, \(V_2 = I R \Rightarrow 4.8 = 0.80 R \Rightarrow R = 6.0\ \Omega\).
Marking scheme
(a) [1.5 marks] 1 mark for stating resistance inside the cell / of the cell material, 0.5 marks for mentioning that it causes lost volts / heat energy loss. (b)(i) [1 mark] Correctly stating 6.0 V. (b)(ii) [2 marks] Calculation of lost volts (1.2 V) or formula E = V + Ir (1 mark), final current 0.80 A (1 mark). (b)(iii) [3 marks] Formula V = IR or E = I(R+r) (1 mark), substitution of values (1 mark), final answer 6.0 ohms (1 mark).
Question 4 · Structured
7.5 marks
A uniform wire X of length \(1.5\text{ m}\) and diameter \(0.40\text{ mm}\) has a resistance of \(6.4\ \Omega\). (a) Show that the resistivity of the material of wire X is approximately \(5.4 \times 10^{-7}\ \Omega\text{ m}\). [3 marks] (b) A second wire Y is made of the same material as wire X. Wire Y has twice the length and half the diameter of wire X. Calculate: (i) the cross-sectional area of wire Y, [2.5 marks] (ii) the resistance of wire Y. [2 marks]
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Worked solution
(a) First, find the cross-sectional area of wire X: \(A_X = \pi r^2 = \pi \left(\frac{0.40 \times 10^{-3}}{2}\right)^2 = \pi \left(2.0 \times 10^{-4}\right)^2 \approx 1.257 \times 10^{-7}\text{ m}^2\). Using the resistivity formula \(R = \frac{\rho L}{A}\): \(\rho = \frac{R A_X}{L} = \frac{6.4 \times 1.257 \times 10^{-7}}{1.5} \approx 5.36 \times 10^{-7}\ \Omega\text{ m}\). This rounds to the shown value of \(5.4 \times 10^{-7}\ \Omega\text{ m}\). (b)(i) The diameter of wire Y is half of wire X, so \(d_Y = 0.20\text{ mm}\). Radius \(r_Y = 0.10\text{ mm} = 1.0 \times 10^{-4}\text{ m}\). Area \(A_Y = ̅\pi r_Y^2 = \pi \left(1.0 \times 10^{-4}\right)^2 = 3.14 \times 10^{-8}\text{ m}^2\). (b)(ii) Since Y is made of the same material, its resistivity is also \(\rho = 5.36 \times 10^{-7}\ \Omega\text{ m}\). Its length is \(L_Y = 2 \times 1.5 = 3.0\text{ m}\). Its resistance is \(R_Y = \frac{\rho L_Y}{A_Y} = \frac{5.36 \times 10^{-7} \times 3.0}{3.14 \times 10^{-8}} = 51.2\ \Omega\) (or exactly \(8 \times 6.4 = 51.2\ \Omega\)).
Marking scheme
(a) [3 marks] Area calculation (1 mark), resistivity formula rearranged (1 mark), substitution showing value of 5.36 * 10^-7 or 5.4 * 10^-7 ohm m (1 mark). (b)(i) [2.5 marks] Radius is 0.10 mm (0.5 marks), area formula (1 mark), calculation of 3.14 * 10^-8 m^2 (1 mark). (b)(ii) [2 marks] Using R proportional to L/A or using new parameters in resistivity formula (1 mark), final answer 51.2 ohms (or 51 ohms) (1 mark).
Question 5 · Structured
7.5 marks
A tuning fork of frequency \(340\text{ Hz}\) is held above the open end of a vertical tube that is closed at the bottom by water. The length of the air column in the tube is gradually increased from zero by lowering the water level. A loud sound (resonance) is first heard when the length of the air column is \(24.0\text{ cm}\). (a) Explain how a stationary wave is formed in the air column. [2.5 marks] (b) Calculate: (i) the speed of sound in the air, [3 marks] (ii) the length of the air column when the next resonance is heard. [2 marks]
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Worked solution
(a) Sound waves are generated by the tuning fork and travel down the tube. They are reflected at the water surface, which acts as a closed, fixed boundary. The incident and reflected waves, which have the same frequency and amplitude, travel in opposite directions and superpose (interfere). This produces a stationary wave with a node at the water boundary and an antinode near the open end. (b)(i) The first resonance occurs when the length of the air column is equal to a quarter of a wavelength (assuming no end correction): \(L_1 = \frac{\lambda}{4} \Rightarrow \lambda = 4 \times 24.0\text{ cm} = 96.0\text{ cm} = 0.960\text{ m}\). The speed of sound \(v\) is given by: \(v = f \lambda = 340 \times 0.960 = 326.4\text{ m s}^{-1}\) (or \(330\text{ m s}^{-1}\) to 2 significant figures). (b)(ii) The next resonance occurs at the next odd harmonic, which is three-quarters of a wavelength: \(L_2 = \frac{3\lambda}{4} = 3 \times L_1 = 3 \times 24.0 = 72.0\text{ cm}\) (or \(0.720\text{ m}\)).
Marking scheme
(a) [2.5 marks] Sound waves reflect at the water boundary (1 mark), incident and reflected waves travel in opposite directions and superpose/interfere (1 mark), nodes and antinodes are formed / node at water surface (0.5 marks). (b)(i) [3 marks] Identifying first resonance is lambda/4 (1 mark), calculation of wavelength as 0.96 m (1 mark), calculation of speed as 326 m/s (or 330 m/s) with correct unit (1 mark). (b)(ii) [2 marks] Identifying next resonance is at 3*lambda/4 (1 mark), final answer 72.0 cm or 0.72 m (1 mark).
Question 6 · Structured
7.5 marks
A block of mass \(0.50\text{ kg}\) is released from rest at the top of a rough ramp inclined at \(30^\circ\) to the horizontal. The length of the ramp is \(2.0\text{ m}\). The block slides down the ramp and reaches the bottom with a speed of \(3.6\text{ m s}^{-1}\). (a) Calculate, for the block's motion from the top to the bottom of the ramp: (i) the loss in gravitational potential energy, [2 marks] (ii) the gain in kinetic energy. [2 marks] (b) Calculate the work done against friction as the block slides down the ramp. [1.5 marks] (c) Determine the average frictional force acting on the block. [2 marks]
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Worked solution
(a)(i) The vertical height through which the block falls is \(h = 2.0 \times \sin(30^\circ) = 1.0\text{ m}\). The loss in gravitational potential energy is: \(\Delta E_p = m g h = 0.50 \times 9.81 \times 1.0 = 4.905\text{ J}\) (or \(4.9\text{ J}\)). (a)(ii) The gain in kinetic energy is: \(\Delta E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.50 \times (3.6)^2 = 3.24\text{ J}\) (or \(3.2\text{ J}\)). (b) The work done against friction is equal to the difference between the loss of potential energy and the gain of kinetic energy: \(W_f = \Delta E_p - \Delta E_k = 4.905 - 3.24 = 1.665\text{ J}\) (which rounds to \(1.7\text{ J}\)). (c) The work done against friction is given by \(W_f = F_f \times d\), where \(d\) is the distance along the ramp: \(1.665 = F_f \times 2.0 \Rightarrow F_f = \frac{1.665}{2.0} = 0.8325\text{ N} \approx 0.83\text{ N}\).
Marking scheme
(a)(i) [2 marks] Formula h = L sin(theta) or equivalent used (1 mark), final value 4.9 J (or 4.91 J) (1 mark). (a)(ii) [2 marks] Formula KE = 1/2 m v^2 used (1 mark), final value 3.2 J (or 3.24 J) (1 mark). (b) [1.5 marks] Stating work done against friction is the energy difference (0.5 marks), calculation of 1.7 J (or 1.67 J) (1 mark). (c) [2 marks] Formula W_f = F_f * d (1 mark), final answer 0.83 N (or 0.833 N) with unit (1 mark).
Question 7 · Structured
7.5 marks
A nucleus of bismuth-212 (\(_{83}^{212}\text{Bi}\)) is unstable and can decay by emitting either an \(\alpha\)-particle to form a nucleus of thallium (\(\text{Tl}\)) or a \(\beta^-\)-particle to form a nucleus of polonium (\(\text{Po}\)). (a) Write a nuclear equation to represent: (i) the \(\alpha\)-decay of bismuth-212, [2 marks] (ii) the \(\beta^-\)-decay of bismuth-212. [2.5 marks] (b) During \(\beta^-\)-decay, a neutron in the nucleus decays into a proton. State the change in quark composition of this nucleon. [1 mark] (c) A detector measures the activity of a sample of bismuth-212. Explain why the measured count rate is always less than the actual activity of the sample. [2 marks]
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Worked solution
(a)(i) In alpha decay, the nucleon number decreases by 4 and the proton number decreases by 2: \(_{83}^{212}\text{Bi} \rightarrow _{81}^{208}\text{Tl} + _{2}^{4}\text{He}\) (or \(_{2}^{4}\alpha\)). (a)(ii) In beta-minus decay, a neutron converts to a proton, emitting an electron and an electron antineutrino: \(_{83}^{212}\text{Bi} \rightarrow _{84}^{212}\text{Po} + _{-1}^{0}\text{e} + \overline{\nu}_e\). (b) A neutron consists of quarks \(udd\) and a proton consists of \(uud\). Therefore, during beta-minus decay, a down quark (\(d\)) decays into an up quark (\(u\)): \(d \rightarrow u + \text{e}^- + \overline{\nu}_e\). (c) The reasons why the measured count rate is less than the true activity include: 1. Radiation is emitted in all directions, so only a fraction of the emitted particles enter the detector (solid angle limit). 2. Some radiation is absorbed by the air or the detector's window before triggering a count. 3. Some radiation is absorbed within the source itself (self-absorption).
Marking scheme
(a)(i) [2 marks] Correct numbers for Tl (nucleon number 208, proton number 81) (1 mark), correct alpha particle symbol and numbers (1 mark). (a)(ii) [2.5 marks] Correct numbers for Po (nucleon number 212, proton number 84) (1 mark), correct beta particle (1 mark), correct antineutrino symbol (0.5 marks). (b) [1 mark] Stating down quark changes to up quark / d -> u. (c) [2 marks] Any two valid reasons: 1. Radiation emitted in all directions / doesn't all enter tube (1 mark), 2. Absorption of particles by window/air/source (1 mark), 3. Dead-time of G-M tube (1 mark).
Question 8 · Structured
7.5 marks
A student determines the density \(\rho\) of a metal cylinder. The following measurements are made: mass \(m = 48.2 \pm 0.1\text{ g}\), diameter \(d = 1.24 \pm 0.02\text{ cm}\), length \(L = 5.80 \pm 0.05\text{ cm}\). (a) Distinguish between systematic errors and random errors. [2 marks] (b) Calculate: (i) the percentage uncertainty in the volume of the cylinder, [3.5 marks] (ii) the density \(\rho\), with its absolute uncertainty, in \(\text{g cm}^{-3}\). [2 marks]
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Worked solution
(a) Systematic errors cause measurements to deviate from the true value by a constant offset in the same direction each time, whereas random errors cause measurements to fluctuate randomly around a mean value with different directions and magnitudes. (b)(i) The volume of a cylinder is given by \(V = \frac{\pi d^2 L}{4}\). The percentage uncertainty in \(V\) is: \(\frac{\Delta V}{V} = 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\). Percentage uncertainty in \(d\): \(\frac{0.02}{1.24} \times 100\% \approx 1.613\%\). Percentage uncertainty in \(L\): \(\frac{0.05}{5.80} \times 100\% \approx 0.862\%\). Therefore: \(\frac{\Delta V}{V} = 2(1.613\%) + 0.862\% = 3.226\% + 0.862\% = 4.088\% \approx 4.1\%\). (b)(ii) The calculated volume is \(V = \frac{\pi \times (1.24)^2 \times 5.80}{4} = 7.004\text{ cm}^3\). The density is \(\rho = \frac{m}{V} = \frac{48.2}{7.004} = 6.882\text{ g cm}^{-3}\). The percentage uncertainty in density \(\rho\) is: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V} = \left(\frac{0.1}{48.2}\right) \times 100\% + 4.088\% = 0.207\% + 4.088\% = 4.295\%\). The absolute uncertainty in \(\rho\) is: \(\Delta \rho = 6.882 \times 0.04295 \approx 0.296\text{ g cm}^{-3}\) (which rounds to \(0.3\text{ g cm}^{-3}\)). Thus, \(\rho = 6.9 \pm 0.3\text{ g cm}^{-3}\).
Marking scheme
(a) [2 marks] Systematic error definition / constant offset (1 mark), random error definition / scatter about mean (1 mark). (b)(i) [3.5 marks] Expressing volume uncertainty equation (1 mark), calculating fractional/percentage uncertainty of diameter (1.61%) and length (0.86%) (1 mark), doubling the diameter percentage uncertainty (0.5 marks), final answer 4.1% (or 4.09%) (1 mark). (b)(ii) [2 marks] Calculation of density as 6.9 g cm^-3 (1 mark), calculation of absolute uncertainty as 0.3 g cm^-3 with correct representation (1 mark).
Paper 32 (Advanced Practical Skills)
Answer all questions. Record all your observations in the spaces provided as soon as these observations are made.
2 Question · 40 marks
Question 1 · Practical
20 marks
Answer all questions. Record all your observations in the spaces provided as soon as these observations are made. In this experiment, you will investigate how the deflection of a clamped metre rule (acting as a cantilever) depends on the length of the overhanging section. Apparatus: Metre rule clamped to a bench (to act as the cantilever), G-clamp and wooden blocks, Mass hanger of mass \(M = 500\text{ g}\), Second vertical metre rule to measure heights above the floor, Set square. (a)(i) Set up the apparatus where a horizontal metre rule is clamped to the edge of a bench. Adjust the position of the rule in the clamp so that the projecting length \(L\) is \(60.0\text{ cm}\). Record \(L\). (ii) With no mass suspended from the free end of the rule, measure the vertical height \(h_0\) of the bottom edge of the free end of the rule above the floor. Use the set square to ensure that the vertical rule is perpendicular to the floor. Record \(h_0\). (iii) Suspend the mass \(M = 500\text{ g}\) from the free end of the rule. Measure the new vertical height \(h\) of the bottom edge of the free end of the rule above the floor. Record \(h\). (iv) Calculate the deflection \(y\), where \(y = h_0 - h\). (b) Change \(L\) and repeat (a)(ii), (a)(iii) and (a)(iv) to obtain six sets of values of \(L\), \(h_0\), \(h\) and \(y\). The values of \(L\) should be in the range \(50.0\text{ cm} \le L \le 80.0\text{ cm}\). In your table of results, include calculated values of \(L^2\) and \(y/L\). (c)(i) Plot a graph of \(y/L\) on the y-axis against \(L^2\) on the x-axis. (ii) Draw the straight line of best fit. (d) Determine the gradient and y-intercept of this line. (e) The quantities \(y\) and \(L\) are related by the equation: \(\frac{y}{L} = p L^2 + q\) where \(p\) and \(q\) are constants. Using your answers from (d), determine values for \(p\) and \(q\). Include appropriate units for both constants.
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Worked solution
1. Measurement: At L = 60.0 cm, the heights are h0 = 85.0 cm and h = 81.2 cm, yielding deflection y = 3.8 cm. 2. Multi-reading range: Six sets of measurements are taken for L = 50.0, 55.0, 60.0, 65.0, 70.0, 75.0, and 80.0 cm. As L increases, the deflection y increases. 3. Compilation: A complete table of results is compiled with columns L (cm), h0 (cm), h (cm), y (cm), L^2 (cm^2), and y/L. All raw heights are recorded to 1 mm precision. 4. Graphing: A graph of y/L against L^2 is plotted. It displays a strong linear trend. 5. Analysis: The gradient p of the best-fit line is calculated using the change in y/L over the change in L^2 for two points far apart on the line. The y-intercept q is read directly or calculated using y/L = p * L^2 + q. 6. Final Values: Expected values are p = 1.95 * 10^-5 cm^-2 and q = -0.007.
Marking scheme
a(i)-(iv) [2 marks]: 1 mark for raw h0 and h to nearest millimetre with units. 1 mark for correct calculation of y. b [6 marks]: 1 mark for 6 sets of data with correct trend (y increases as L increases). 1 mark for range of L (must include L <= 55.0 cm and L >= 75.0 cm). 1 mark for column headings with correct units (e.g., L / cm, L^2 / cm^2, y/L). 1 mark for raw data to 1 mm precision. 1 mark for significant figures of calculated columns (L^2 and y/L) matching raw data. 1 mark for correct calculation of L^2 and y/L. c [5 marks]: 2 marks for axes (linear, sensible scale, labelled with units). 2 marks for accurate plotting (within half a small square). 1 mark for best-fit straight line. d [2 marks]: 1 mark for gradient calculated using a large triangle (hypotenuse > 50% of line length). 1 mark for y-intercept. e [5 marks]: 1 mark for equating p to gradient. 1 mark for equating q to intercept. 1 mark for correct unit of p (e.g., cm^-2 or m^-2). 1 mark for correct unit of q (dimensionless or none). 1 mark for sensible values of p and q.
Question 2 · Practical
20 marks
Answer all questions. Record all your observations in the spaces provided as soon as these observations are made. In this experiment, you will investigate the tipping point of an overhanging metre rule to determine its mass. Apparatus: Metre rule (wooden, of mass \(M\)), Flat bench or table with a straight, horizontal edge, A mass \(m = 200\text{ g}\), Small piece of tape. (a)(i) Place the metre rule on the flat table surface perpendicular to the edge of the table. Let a length \(x\) of the rule project beyond the edge of the table. Set \(x = 30.0\text{ cm}\). Securely tape a mass \(m = 200\text{ g}\) to the rule. The mass should be on the overhanging section, at a distance \(d\) from the table edge. Adjust the position of the mass along the overhanging section until the rule is just on the point of tipping (the end on the table is just about to lift). Measure the distance \(d\) from the table edge to the center of the mass \(m\). Record \(d\). (ii) Repeat your measurements to find an average value of \(d\). Record the average \(d\). (iii) Estimate the percentage uncertainty in your average value of \(d\). Show your working. (b) Describe the technique you used to determine the tipping point as precisely as possible. (c)(i) Change \(x\) to \(40.0\text{ cm}\). Repeat (a)(i) and (a)(ii) to find the new average value of \(d\). Record \(x\) and average \(d\). (ii) Check if the trend is correct (\(d\) is smaller for \(x = 40.0\text{ cm}\) than for \(x = 30.0\text{ cm}\)). (d) The relationship between \(d\), \(x\), and the mass \(M\) of the rule is given by: \(M = \frac{m d}{50.0 - x}\) where \(m = 200\text{ g}\). (i) Calculate the value of \(M\) for \(x = 30.0\text{ cm}\) and \(x = 40.0\text{ cm}\). Include appropriate units. (ii) State the number of significant figures you have chosen for your values of \(M\) and explain your choice. (e) It is suggested that the percentage difference between your two values of \(M\) is within the limits of experimental accuracy. State whether your results support this suggestion. Justify your conclusion with a calculation. (f) Describe four sources of uncertainty or limitations of the procedure in this experiment. (g) Describe four improvements that could be made to this experiment to address the sources of uncertainty or limitations described in (f).
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Worked solution
1. Measurement: At x = 30.0 cm, the average tipping distance d1 is 12.0 cm (derived from trials of 12.1 cm and 11.9 cm). 2. Uncertainty calculation: Absolute uncertainty in d is 0.5 cm due to difficulties in defining the exact tipping pivot and mass center. Percentage uncertainty = (0.5 / 12.0) * 100% = 4.2%. 3. Technique: The mass is moved outward in increments of 1 mm until the far end of the rule just separates from the table. 4. Second set: At x = 40.0 cm, average d2 is 6.0 cm. This is smaller than d1, confirming the correct trend. 5. Calculations of M: M1 = (200 * 12.0) / (50.0 - 30.0) = 120 g. M2 = (200 * 6.0) / (50.0 - 40.0) = 120 g. 6. Significant figures: M is given to 3 s.f. because raw values of x and d are measured to 3 s.f. 7. Comparison: Percentage difference = (|120 - 120| / 120) * 100% = 0%. Since 0% is less than the standard 10% limit of experimental accuracy, the relationship is supported.
Marking scheme
a(i)-(iii) [3 marks]: 1 mark for first value of d1 to 1 mm precision with unit. 1 mark for repeated trials of d1 and correct average. 1 mark for percentage uncertainty in d1 with working, absolute uncertainty 0.2 cm to 0.5 cm. b [1 mark]: 1 mark for describing a precise method to find the tipping point (e.g., sliding mass in 1 mm steps, or tapping the end on the table). c [3 marks]: 2 marks for x2 and average d2 with units. 1 mark for correct trend (d2 < d1). d [3 marks]: 1 mark for correct calculations of M1 and M2. 1 mark for correct unit (g or kg). 1 mark for significant figures of M being 2 or 3 (matching raw measurements x and d). e [2 marks]: 1 mark for calculating the percentage difference between M1 and M2. 1 mark for comparing this to a criterion (usually 10% or 20%) and drawing a valid conclusion. f-g [8 marks]: 1 mark for each of 4 limitations and 1 mark for each of 4 corresponding improvements. Limitations: 1. Hard to find center of mass of taped mass. 2. Metre rule may not have a uniform mass distribution (center of gravity not exactly at 50.0 cm). 3. Rule slips on the table edge during tipping. 4. Only two values of x are investigated, which is not enough to confirm a relationship. Improvements: 1. Use a mass with a hook or a knife-edge indicator to align with scale. 2. Find the actual center of gravity of the rule by balancing it on a knife edge first. 3. Use tape or a rubber strip on the table edge to prevent slipping. 4. Take more readings of x and d and plot a graph.
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