2025 Cambridge IAL Physics (9702) Practice Paper with Answers

The resistivity \(\rho\) is given by \(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\). The fractional uncertainty in \(\rho\) is given by \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\). Calculating the percentage uncertainties: for \(R\), \(\frac{0.5}{25.0} \times 100\% = 2.0\%\); for \(d\), \(\frac{0.02}{0.38} \times 100\% \approx 5.26\%\); for \(L\), \(\frac{0.01}{1.50} \times 100\% \approx 0.67\%\). Therefore, the percentage uncertainty in \(\rho\) is \(2.0\% + 2(5.26\%) + 0.67\% = 13.19\%\), which rounds to \(13\%\).

1 mark for the correct calculation of percentage uncertainty to 2 significant figures.

First, find the best value of \(g\): \(g_{\text{best}} = \frac{4\pi^2}{M_{\text{best}}} = \frac{4\pi^2}{4.02} \approx 9.820\text{ m s}^{-2}\). Next, find the value of \(g\) from the worst line: \(g_{\text{worst}} = \frac{4\pi^2}{M_{\text{worst}}} = \frac{4\pi^2}{4.14} \approx 9.535\text{ m s}^{-2}\). The absolute uncertainty in \(g\) is \(\Delta g = |g_{\text{best}} - g_{\text{worst}}| = |9.820 - 9.535| = 0.285\text{ m s}^{-2} \approx 0.3\text{ m s}^{-2}\). Thus, the acceleration of free fall is \((9.8 \pm 0.3)\text{ m s}^{-2}\).

1 mark for the correct value of g and its absolute uncertainty rounded to 1 significant figure.

The density is given by \(\rho = \frac{m}{V} = \frac{3 m}{4 \pi r^3}\). Substituting the values: \(\rho = \frac{3 \times 75.0}{4 \pi \times 1.50^3} \approx 5.305\text{ g cm}^{-3}\). The fractional uncertainty in density is \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta r}{r}\). Substituting the absolute uncertainties: \(\frac{\Delta \rho}{\rho} = \frac{0.5}{75.0} + 3 \times \frac{0.03}{1.50} = 0.00667 + 0.0600 = 0.06667\). The absolute uncertainty is \(\Delta \rho = 5.305 \times 0.06667 \approx 0.354\text{ g cm}^{-3}\). Rounding the uncertainty to 1 significant figure gives \(0.4\text{ g cm}^{-3}\), and the value to the same decimal place is \(5.3\text{ g cm}^{-3}\). Thus, \(\rho = (5.3 \pm 0.4)\text{ g cm}^{-3}\).

1 mark for the correct density and correctly propagated absolute uncertainty.

For static equilibrium, the vertical forces acting on the sphere must balance. The vertical component of the tension \(T\) must equal the weight of the sphere \(W\). Therefore, \(T \cos(35^\circ) = m g\). Substituting the given values: \(T \cos(35^\circ) = 0.20 \times 9.81 = 1.962\text{ N}\). Hence, \(T = \frac{1.962}{\cos(35^\circ)} \approx 2.40\text{ N}\).

1 mark for the correct application of vertical force equilibrium to find tension.

Let the hinge be at point A, and the other end of the shelf be at point B. The length of the shelf is \(1.6\text{ m}\). The weight \(W = 90\text{ N}\) acts downwards at the centre of gravity, which is \(0.8\text{ m}\) from A. The cable is connected from B to a point C on the wall, \(1.2\text{ m}\) above A. The length of the cable is \(\sqrt{1.6^2 + 1.2^2} = 2.0\text{ m}\). The angle \(\theta\) between the cable and the shelf satisfies \(\sin \theta = \frac{1.2}{2.0} = 0.60\). Taking moments about the hinge A: clockwise moment = \(W \times 0.8 = 90 \times 0.8 = 72\text{ N m}\). Anticlockwise moment = \(T \sin \theta \times 1.6 = T \times 0.60 \times 1.6 = 0.96 T\). For equilibrium, \(0.96 T = 72 \implies T = 75\text{ N}\).

1 mark for taking moments about the hinge and solving for tension correctly.

The fringe separation is given by the formula \(x = \frac{\lambda D}{a}\). With the new values \(\lambda' = 2\lambda\), \(D' = 2D\), and \(a' = \frac{a}{2}\), the new fringe separation is \(x' = \frac{\lambda' D'}{a'} = \frac{(2\lambda) (2D)}{\frac{a}{2}} = \frac{4 \lambda D}{\frac{a}{2}} = 8 \frac{\lambda D}{a} = 8x\).

1 mark for correctly determining the new fringe separation using the double-slit formula.

First, find the wavelength \(\lambda\) using the wave equation: \(\lambda = \frac{v}{f} = \frac{120}{180} = \frac{2}{3}\text{ m} \approx 0.667\text{ m}\). The distance between a node and the adjacent antinode is \(\frac{\lambda}{4}\). Thus, the distance is \(\frac{2/3}{4} = \frac{1}{6}\text{ m} \approx 0.167\text{ m} = 17\text{ cm}\).

1 mark for finding the wavelength and dividing by 4 to get the correct node-to-antinode distance.

Using the grating equation \(d \sin \theta = n \lambda\) with \(n = 1\), we find the grating spacing \(d = \frac{\lambda}{\sin(14.5^\circ)} = \frac{5.0 \times 10^{-7}}{0.2504} \approx 1.997 \times 10^{-6}\text{ m}\). For the maximum possible order, we set \(\theta = 90^\circ\): \(n_{\text{max}} = \frac{d}{\lambda} = \frac{1.997 \times 10^{-6}}{5.0 \times 10^{-7}} \approx 3.99\). Since the order \(n\) must be an integer, the highest observable order is \(n = 3\). The total number of maxima is \(2n_{\text{max}} + 1 = 2(3) + 1 = 7\).

1 mark for determining the maximum order as 3, and calculating the total number of maxima as 7.

First, calculate the volume \(V\) of the sphere: \(V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (2.00)^3 \approx 33.51\text{ cm}^3\). The calculated density \(\rho\) is: \(\rho = \frac{m}{V} = \frac{24.0}{33.51} \approx 0.7162\text{ g cm}^{-3}\). Now, determine the percentage uncertainties: - Percentage uncertainty in mass \(m\): \(\frac{\Delta m}{m} \times 100\% = \frac{0.6}{24.0} \times 100\% = 2.5\%\). - Percentage uncertainty in radius \(r\): \(\frac{\Delta r}{r} \times 100\% = \frac{0.04}{2.00} \times 100\% = 2.0\%\). - Since \(V \propto r^3\), the percentage uncertainty in volume \(V\) is: \(3 \times 2.0\% = 6.0\%\). - The percentage uncertainty in density \(\rho\) is: \(\frac{\Delta \rho}{\rho} \times 100\% = 2.5\% + 6.0\% = 8.5\%\). Now, find the absolute uncertainty in density \(\rho\): \(\Delta \rho = 8.5\% \times 0.7162\text{ g cm}^{-3} \approx 0.0609\text{ g cm}^{-3}\). An absolute uncertainty is usually expressed to 1 significant figure, which is \(0.06\text{ g cm}^{-3}\). The value of \(\rho\) must be rounded to match the decimal places of the absolute uncertainty (2 decimal places): \(\rho = 0.72\text{ g cm}^{-3}\). Therefore, the density is written as \((0.72 \pm 0.06)\text{ g cm}^{-3}\).

1 mark for the correct calculation of density and its absolute uncertainty, including correct rounding to appropriate significant figures.

A zero error (the reading when the jaws are closed is not zero) is a systematic error. Systematic errors cannot be reduced or eliminated by averaging multiple readings. To correct for a systematic zero error: \(\text{Correct Value} = \text{Measured Value} - \text{Zero Error}\). Here, the zero error is \(-0.02\text{ mm}\). Thus, \(\text{Correct Value} = \text{Measured Value} - (-0.02\text{ mm}) = \text{Measured Value} + 0.02\text{ mm}\). Therefore, \(0.02\text{ mm}\) must be added to each reading to obtain the correct diameter.

1 mark for identifying the error as systematic and correctly explaining how to apply the zero correction.

The relationship can be rewritten in the form \(y = mx\), where \(y = T^2\), \(x = L\), and the gradient is: \(m = \frac{4\pi^2}{g} \implies g = \frac{4\pi^2}{m}\). Therefore, the calculated value of \(g\) using the best-fit line is: \(g_{\text{best}} = \frac{4\pi^2}{m_{\text{best}}}\). Since \(4\pi^2\) is a constant, the fractional uncertainty in \(g\) is equal to the fractional uncertainty in the gradient \(m\): \(\frac{\Delta g}{g_{\text{best}}} = \frac{\Delta m}{m_{\text{best}}}\). Solving for \(\Delta g\): \(\Delta g = g_{\text{best}} \times \frac{\Delta m}{m_{\text{best}}} = \left(\frac{4\pi^2}{m_{\text{best}}}\right) \times \frac{\Delta m}{m_{\text{best}}} = \frac{4\pi^2 \Delta m}{m_{\text{best}}^2}\). Hence, option B is correct.

1 mark for relating the fractional uncertainty of the gradient to the fractional uncertainty of \(g\), and correctly substituting \(g_{\text{best}}\) to obtain the absolute uncertainty expression.

Three forces act on the sphere and meet at its centre: 1. Weight \(W\) acting vertically downwards. 2. Normal reaction force \(R\) from the wall acting horizontally. 3. Tension \(T\) acting along the line of the string. Let \(\theta\) be the angle the string makes with the vertical wall. The distance from the wall attachment point to the centre of the sphere is: \(\text{string length} + \text{radius} = r + r = 2r\). The horizontal distance from the wall to the centre of the sphere is the radius \(r\). Therefore, in the right-angled triangle formed: \(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{r}{2r} = 0.5 \implies \theta = 30^\circ\). For vertical equilibrium of the sphere: \(T \cos \theta = W \implies T \cos 30^\circ = W\). Since \(\cos 30^\circ = \frac{\sqrt{3}}{2}\): \(T = \frac{W}{\cos 30^\circ} = \frac{2W}{\sqrt{3}}\).

1 mark for determining the angle of the string to the vertical using geometry, and resolving forces vertically to find the tension.

First, notice that the angle between the two strings at the object is: \(180^\circ - (35^\circ + 55^\circ) = 90^\circ\). Since the two strings are perpendicular to each other, we can draw a right-angled triangle of forces representing the equilibrium of \(T_1\), \(T_2\), and the downward weight \(W\). In this right-angled triangle: - The hypotenuse represents the weight \(W\) (vertical). - The side representing \(T_1\) is at an angle of \(90^\circ - 35^\circ = 55^\circ\) to the vertical. - The side representing \(T_2\) is at an angle of \(90^\circ - 55^\circ = 35^\circ\) to the vertical. Using basic trigonometry on this force triangle: \(T_1 = W \cos(55^\circ) = W \sin(35^\circ)\) and \(T_2 = W \cos(35^\circ)\). Thus, \(T_1 = W \sin(35^\circ)\) and \(T_2 = W \cos(35^\circ)\), which matches option A.

1 mark for recognizing that the force system forms a right-angled triangle of forces, and applying trigonometry to find the relationships.

The formula for double-slit fringe spacing is: \(x = \frac{\lambda D}{a}\). Initially, the fringe spacing is \(x = \frac{\lambda D}{a}\). For the new arrangement, let \(a'\) be the new slit separation. The new fringe spacing \(x'\) is given by: \(x' = \frac{\lambda' D'}{a'} = \frac{(1.5\lambda) (0.5D)}{a'} = \frac{0.75 \lambda D}{a'}\). Since we require the fringe spacing to remain unchanged (\(x' = x\)): \(\frac{0.75 \lambda D}{a'} = \frac{\lambda D}{a} \implies a' = 0.75a\).

1 mark for using the double-slit formula and setting up the ratio to solve for the new slit separation.

A vertical tube closed at the bottom behaves as an air column closed at one end and open at the other. The first resonance occurs at the fundamental mode, where the length of the column is: \(L_1 + c = \frac{\lambda}{4}\) where \(c\) is the end correction and \(\lambda\) is the wavelength of the sound. The next resonance (the third harmonic) occurs when: \(L_2 + c = \frac{3\lambda}{4}\). Subtracting the two equations eliminates the end correction \(c\): \(L_2 - L_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{\lambda}{2} \implies \lambda = 2(L_2 - L_1)\). Using the wave equation \(v = f\lambda\), the speed of sound \(v\) is: \(v = f \times 2(L_2 - L_1) = 2f(L_2 - L_1)\).

1 mark for identifying the relation between successive resonance lengths and wavelength, and substituting into the wave equation.

For the maxima of the two wavelengths to coincide at an angle \(\theta\), the condition is: \(d \sin\theta = n_1 \lambda_1 = n_2 \lambda_2\) where \(n_1\) and \(n_2\) are integers representing the orders of the maxima, \(\lambda_1 = 480\text{ nm}\), and \(\lambda_2 = 640\text{ nm}\). Thus: \(\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{640\text{ nm}}{480\text{ nm}} = \frac{4}{3}\). The first coincidence occurs at the smallest integer values for \(n_1\) and \(n_2\), which are \(n_1 = 4\) and \(n_2 = 3\). The slit spacing \(d\) of the grating is: \(d = \frac{1}{5.0 \times 10^5\text{ m}^{-1}} = 2.0 \times 10^{-6}\text{ m}\). Now, calculate the angle \(\theta\): \(\sin\theta = \frac{n_1 \lambda_1}{d} = \frac{4 \times 480 \times 10^{-9}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 0.96\). \(\theta = \sin^{-1}(0.96) \approx 73.7^\circ \approx 74^\circ\).

1 mark for setting up the coincidence condition, finding the correct order numbers, calculating the grating spacing, and finding the angle \(\theta\).

The percentage uncertainty in a quantity calculated by multiplication and division with powers is the sum of the percentage uncertainties of each individual measurement multiplied by their respective powers: \(\frac{\Delta \rho}{\rho} \times 100\% = \left(\frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\right) \times 100\%\). Calculating each: \(\frac{\Delta R}{R} \times 100\% = \frac{0.05}{2.50} \times 100\% = 2.0\%\); \(\frac{\Delta d}{d} \times 100\% = \frac{0.01}{0.40} \times 100\% = 2.5\%\); \(\frac{\Delta L}{L} \times 100\% = \frac{0.006}{1.200} \times 100\% = 0.5\%\). Therefore, the percentage uncertainty is: \(2.0\% + 2(2.5\%) + 0.5\% = 7.5\%\).

1 mark for the correct calculation of percentage uncertainty by summing 2.0% + 2*(2.5%) + 0.5% = 7.5%.

When subtracting two quantities, their absolute uncertainties are added: \(\text{Temperature rise} = \theta_2 - \theta_1 = 65.8 - 24.3 = 41.5\ ^\circ\text{C}\). The absolute uncertainty in this change is: \(\Delta(\Delta \theta) = 0.2\ ^\circ\text{C} + 0.3\ ^\circ\text{C} = 0.5\ ^\circ\text{C}\). So the final value with its uncertainty is \((41.5 \pm 0.5)\ ^\circ\text{C}\).

1 mark for the correct calculation of both the difference in temperature (41.5) and the sum of absolute uncertainties (0.5).

For the picture frame to be in equilibrium, the upward vertical components of the tension in the two strings must balance the downward weight of the frame. Let \(T\) be the tension in each string. The vertical component of each tension is \(T \sin(35^\circ)\). Therefore: \(2T \sin(35^\circ) = W\) which gives \(2T \sin(35^\circ) = 12\), and solving for \(T\) yields \(T = \frac{6}{\sin(35^\circ)} \approx 10.46\text{ N} \approx 10.5\text{ N}\).

1 mark for resolving forces vertically and setting up the correct equation 2 T sin(35) = 12, yielding T = 10.5 N.

The beam is uniform, so its weight of \(80\text{ N}\) acts at its centre of gravity, which is at the midpoint of the beam, \(1.0\text{ m}\) from either end. The pivot is located \(0.60\text{ m}\) from the left end. The distance from the pivot to the block at the left end is \(0.60\text{ m}\). The distance from the pivot to the centre of gravity of the beam is \(1.0\text{ m} - 0.60\text{ m} = 0.40\text{ m}\). Taking moments about the pivot: \(\text{Sum of anticlockwise moments} = \text{Sum of clockwise moments}\), so \(W \times 0.60\text{ m} = 80\text{ N} \times 0.40\text{ m}\). This gives \(0.60 W = 32\) which leads to \(W = \frac{32}{0.60} \approx 53.3\text{ N} \approx 53\text{ N}\).

1 mark for applying the principle of moments about the pivot and correctly solving for W = 53 N.

First, calculate the fringe width \(x\) using the formula: \(x = \frac{\lambda D}{a}\). Given: \(\lambda = 600 \times 10^{-9}\text{ m}\), \(D = 1.5\text{ m}\), and \(a = 0.30 \times 10^{-3}\text{ m}\). Therefore: \(x = \frac{(600 \times 10^{-9}) \times 1.5}{0.30 \times 10^{-3}} = 3.0 \times 10^{-3}\text{ m} = 3.0\text{ mm}\). The distance from the central maximum to the third-order dark fringe is given by: \(y_3 = 2.5 x = 2.5 \times 3.0\text{ mm} = 7.5\text{ mm}\).

1 mark for calculating the fringe width (3.0 mm) and multiplying by 2.5 to find the distance to the 3rd dark fringe (7.5 mm).

The wavelength is given by \(\lambda = \frac{v}{f} = \frac{340}{340} = 1.0\text{ m}\). For a closed tube, consecutive resonances occur when the change in length of the air column is equal to half a wavelength: \(\Delta L = \frac{\lambda}{2}\). Therefore, \(\Delta L = \frac{1.0\text{ m}}{2} = 0.50\text{ m}\).

1 mark for calculating wavelength = 1.0 m and determining that the difference in length between consecutive resonances is lambda / 2 = 0.50 m.

By conservation of energy, the initial kinetic energy of the block equals the maximum elastic potential energy of the spring: \(\frac{1}{2} m v^2 = \frac{1}{2} k x^2 \implies m v^2 = k x^2\). Substituting the values: \(2.0 \times 8.0^2 = 800 \times x^2 \implies 128 = 800 x^2 \implies x^2 = 0.16\). Taking the square root gives \(x = 0.40\text{ m}\).

1 mark for equating kinetic energy to elastic potential energy and solving for the maximum compression to get 0.40 m.

The resistance is given by \(R = \rho \frac{L}{A}\), where area \(A = \frac{\pi d^2}{4}\). Thus \(R \propto \frac{L}{d^2}\). The ratio of resistances is \(\frac{R_X}{R_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2\). Substituting the given relationships: \(\frac{R_X}{R_Y} = \left(\frac{L}{2L}\right) \times \left(\frac{2d}{d}\right)^2 = \frac{1}{2} \times 4 = 2\).

1 mark for writing the resistance formula in terms of length and diameter, and correctly setting up and evaluating the ratio of resistances to be 2.

The density of the cylindrical wire is calculated using the formula \(\rho = \frac{4m}{\pi d^2 l}\). The fractional uncertainty is given by \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}\). Calculating the percentage uncertainties: for mass, \(\frac{0.005}{0.640} \times 100\% \approx 0.78\%\); for diameter, \(\frac{0.02}{0.80} \times 100\% = 2.5\%\); and for length, \(\frac{0.1}{15.0} \times 100\% \approx 0.67\%\). Summing these terms, with the diameter uncertainty doubled, we get: percentage uncertainty = \(0.78\% + 2(2.5\%) + 0.67\% = 6.45\% \approx 6.4\%\).

1 mark for the correct calculation of individual percentage uncertainties, doubling the uncertainty of the diameter, and summing them to get 6.4%.

The kinetic energy is calculated as \(E_k = \frac{1}{2}mv^2 = 0.5 \times 2.00 \times 6.0^2 = 36.0\text{ J}\). The fractional uncertainty in kinetic energy is given by \(\frac{\Delta E_k}{E_k} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v} = \frac{0.05}{2.00} + 2\left(\frac{0.3}{6.0}\right) = 0.025 + 0.100 = 0.125\). Therefore, the absolute uncertainty in kinetic energy is \(\Delta E_k = 0.125 \times 36.0\text{ J} = 4.5\text{ J}\).

1 mark for calculating kinetic energy as 36 J, determining the combined fractional uncertainty as 0.125 (including the factor of 2 for speed), and obtaining the absolute uncertainty of 4.5 J.

For the rod to be in rotational equilibrium, taking moments about end \(\text{A}\) gives: Clockwise moment = Anticlockwise moment, which is \(W \times x = T_B \times 1.20\), where \(x\) is the distance of the centre of gravity from end \(\text{A}\), \(W = 40.0\text{ N}\), and \(T_B = 25.0\text{ N}\). This leads to \(40.0 \times x = 25.0 \times 1.20 = 30.0\text{ N m}\). Solving for \(x\) gives \(x = 0.75\text{ m}\).

1 mark for taking moments about end A (or B) correctly and solving for the distance of the centre of gravity to obtain 0.75 m.

The three coplanar forces acting on the block are the weight \(W\) acting vertically downwards, the normal reaction \(R\) perpendicular to the incline, and the force \(F\) parallel to the incline. Because \(R\) and \(F\) are perpendicular to each other, the closed vector triangle representing the equilibrium must be a right-angled triangle where \(R\) and \(F\) form the two legs and \(W\) forms the hypotenuse. Applying Pythagoras' theorem, we get \(W^2 = F^2 + R^2\).

1 mark for identifying that the force parallel to the incline and the normal contact force are perpendicular, making weight the hypotenuse in the closed vector triangle, and applying Pythagoras' theorem correctly.

The formula for the fringe width is \(x = \frac{\lambda D}{a}\). For the new configuration, the wavelength becomes \(\lambda' = 0.75\lambda\), the slit separation is \(a' = 0.5a\), and the screen distance is \(D' = 2D\). The new fringe width is \(x' = \frac{\lambda' D'}{a'} = \frac{(0.75\lambda)(2D)}{0.5a} = \frac{1.5}{0.5} \frac{\lambda D}{a} = 3x\). Thus, the new fringe width is \(3.00x\).

1 mark for using the double-slit equation and scaling each variable correctly to find that the fringe width increases by a factor of 3.

For a tube closed at one end, only odd harmonics are possible. The first harmonic (fundamental) has frequency \(f_1 = f\) and wavelength \(\lambda_1 = 4L\). The next possible mode of vibration is the third harmonic, which has a frequency of \(f_3 = 3f\). Its wavelength \(\lambda_3\) satisfies the boundary condition \(L = \frac{3}{4}\lambda_3\), which gives \(\lambda_3 = \frac{4}{3}L\).

1 mark for recognizing that only odd harmonics exist in a closed tube, so the next mode has a frequency of 3f, and correctly calculating its wavelength as 4L/3.

Repeating measurements 20 times and calculating the mean reduces the effect of random errors (timing reaction time), which increases the precision of the result. However, the calibration error is a systematic error that shifts all measurements in the same direction, which means the mean value remains inaccurate (not close to the true value). Therefore, the precision is high, but the accuracy is low.

1 mark for identifying that repeating measurements improves precision by reducing random uncertainties, but systematic calibration errors limit the accuracy of the result.

The grating spacing \(d\) is calculated as \(d = \frac{1}{6.00 \times 10^5\text{ m}^{-1}} \approx 1.667 \times 10^{-6}\text{ m}\). For diffraction maxima, we use the formula \(d \sin\theta = n\lambda\). The maximum value of \(\sin\theta\) is 1, which means the maximum order \(n\) is given by \(n \le \frac{d}{\lambda} = \frac{1.667 \times 10^{-6}\text{ m}}{589 \times 10^{-9}\text{ m}} \approx 2.83\). Since \(n\) must be an integer, the highest observable order is \(n = 2\). This means we can see the central maximum (\(n = 0\)), and the first and second orders on both sides of the center (\(n = \pm 1, \pm 2\)). Thus, the total number of bright fringes is \(2(2) + 1 = 5\).

1 mark for calculating the grating spacing, finding the maximum integer order of 2, and concluding that there are 5 fringes in total (including the central maximum).

The formula for density is \(\rho = \frac{4 m}{\pi d^2 L}\). The fractional uncertainty in \(\rho\) is given by \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\). Substituting the given percentage uncertainties: \(\frac{\Delta \rho}{\rho} = 1.1\% + 2(2.5\%) + 0.7\% = 1.1\% + 5.0\% + 0.7\% = 6.8\%\).

1 mark for the correct calculation of percentage uncertainty by summing the individual percentage uncertainties, doubling the uncertainty for the squared term.

The actual potential difference is \(V_{\text{true}} = V - 0.1\), where \(V\) is the measured potential difference. Since \(I = \frac{V_{\text{true}}}{R} = \frac{V - 0.1}{R} = \frac{1}{R}V - \frac{0.1}{R}\), the graph of \(I\) against \(V\) is a straight line with gradient \(\frac{1}{R}\). The zero error only shifts the line horizontally (changing the intercept) but does not affect the gradient. Therefore, the gradient is unchanged, and the value of \(R = \frac{1}{\text{gradient}}\) calculated from the gradient remains correct.

1 mark for identifying that the zero error shifts the line horizontally without changing its gradient, meaning both the gradient and the calculated resistance remain correct.

First, find the mean time for 20 oscillations: \(t_{\text{mean}} = \frac{38.2 + 38.5 + 38.1 + 38.6 + 38.1}{5} = 38.30\text{ s}\). The range of these measurements is \(38.6 - 38.1 = 0.5\text{ s}\). The uncertainty in the total time is half the range: \(\Delta t = \frac{0.5}{2} = 0.25\text{ s}\). The period of one oscillation is \(T = \frac{t_{\text{mean}}}{20} = \frac{38.30}{20} = 1.915\text{ s}\). The uncertainty in the period is \(\Delta T = \frac{\Delta t}{20} = \frac{0.25}{20} = 0.0125\text{ s}\). Rounding the uncertainty to one significant figure gives \(\Delta T = 0.01\text{ s}\), which means the period should be quoted to two decimal places: \(T = 1.92 \pm 0.01\text{ s}\).

1 mark for calculating the correct mean and half-range uncertainty, dividing both by 20, and formatting to appropriate significant figures.

The weight of the frame is \(W = mg = 1.2 \times 9.81 = 11.77\text{ N}\). Let the tension in each string be \(T\). Since the strings have equal length and support the weight symmetrically, the angle each string makes with the vertical is \(35^\circ\). Resolving the forces vertically: \(2 T \cos(35^\circ) = W \implies 2 T \cos(35^\circ) = 11.77\text{ N}\). Solving for \(T\): \(T = \frac{11.77}{2 \cos(35^\circ)} \approx 7.2\text{ N}\).

1 mark for correctly resolving forces vertically with the correct angle of 35 degrees and finding the tension.

The three forces acting on the sphere are: the downward weight \(W\), the horizontal normal reaction \(R\) pushing away from the wall, and the tension \(T\) pulling along the string. Since the wall is vertical, the angle between the string and the vertical is \(\theta\). Resolving forces: Vertically, \(T \cos\theta = W\). Horizontally, \(T \sin\theta = R\). Dividing the two equations gives \(\frac{R}{W} = \frac{T \sin\theta}{T \cos\theta} = \tan\theta\), so \(R = W \tan\theta\).

1 mark for correctly setting up the equations of equilibrium for horizontal and vertical components and deriving the relation for normal reaction.

The fringe spacing \(x\) (the distance between adjacent bright fringes) is given by the formula \(x = \frac{\lambda D}{a}\), where \(\lambda = 650 \times 10^{-9}\text{ m}\), \(D = 1.8\text{ m}\), and \(a = 0.45 \times 10^{-3}\text{ m}\). Calculating \(x\): \(x = \frac{650 \times 10^{-9} \times 1.8}{0.45 \times 10^{-3}} = 2.6 \times 10^{-3}\text{ m} = 2.6\text{ mm}\). The distance from the central bright fringe (zeroth order) to the third-order bright fringe is \(3x\). \(3x = 3 \times 2.6\text{ mm} = 7.8\text{ mm}\).

1 mark for correctly calculating the fringe spacing and multiplying by 3 to find the distance to the third-order fringe.

First, find the wavelength \(\lambda\) of the sound wave: \(\lambda = \frac{v}{f} = \frac{340}{340} = 1.0\text{ m}\). For a tube closed at one end, resonance occurs when the length of the tube \(L\) is an odd number of quarter-wavelengths: \(L = (2n - 1)\frac{\lambda}{4}\) for \(n = 1, 2, 3, \dots\). Thus, the allowed resonance lengths are: 0.25 m, 0.75 m, 1.25 m, 1.75 m, etc. A length of 0.50 m (which is \(\frac{\lambda}{2}\)) represents a half-wavelength, which cannot support a stationary wave with a node at one end and an antinode at the other.

1 mark for calculating the wavelength and identifying that even multiples of quarter-wavelengths (like 0.50 m) do not support resonance in a tube closed at one end.

The phase difference \(\phi\) is related to the path difference \(x\) by the formula \(\phi = \frac{2\pi x}{\lambda}\). Substituting the given values: \(\frac{\pi}{3} = \frac{2\pi \times 0.15}{\lambda} \implies \frac{1}{3} = \frac{0.30}{\lambda} \implies \lambda = 0.90\text{ m}\). The wave speed \(v\) is given by \(v = f \lambda = 50 \times 0.90 = 45\text{ m s}^{-1}\).

1 mark for calculating the wavelength using the phase difference formula and then calculating the wave speed correctly.

(a) Convert units: \(d = 0.38 \times 10^{-3} \text{ m}\), \(L = 0.750 \text{ m}\). Calculate \(\rho = \frac{4.20 \times \pi \times (0.38 \times 10^{-3})^2}{4 \times 0.750} = 6.35 \times 10^{-7} \ \Omega \text{ m}\). (b) Percentage uncertainties: For \(R\): \(\frac{0.05}{4.20} \times 100\% = 1.19\%\). For \(d\): \(\frac{0.02}{0.38} \times 100\% = 5.26\%\). For \(L\): \(\frac{0.5}{75.0} \times 100\% = 0.67\%\). Since \(d\) is squared, the total percentage uncertainty is: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L} = 1.19\% + 2(5.26\%) + 0.67\% = 12.38\% \approx 12.4\%\). (c) Absolute uncertainty in \(\rho = 12.38\% \times 6.35 \times 10^{-7} \ \Omega \text{ m} = 0.786 \times 10^{-7} \ \Omega \text{ m} \approx 0.8 \times 10^{-7} \ \Omega \text{ m}\). Rounding resistivity to match: \(\rho = (6.4 \pm 0.8) \times 10^{-7} \ \Omega \text{ m}\). (d) The diameter \(d\) contributes the most (approx \(10.5\%\)). This can be made more precise by using a digital micrometer screw gauge and taking multiple readings at different points along the wire to calculate an average.

(a) [2 marks] C1 for correct substitution of values with unit conversion. A1 for correct value with unit: \(6.4 \times 10^{-7} \ \Omega \text{ m}\) (accept \(6.3 \times 10^{-7}\) to \(6.4 \times 10^{-7}\)). (b) [4 marks] C1 for \(\% \Delta R = 1.2\%\). C1 for \(\% \Delta d = 5.3\%\). C1 for doubling the percentage uncertainty in diameter (\(10.5\%\)). A1 for correct total: \(12\%\) (accept \(12.3\%\) to \(12.4\%\)). (c) [2 marks] C1 for calculating absolute uncertainty \(\Delta \rho \approx 0.8 \times 10^{-7} \ \Omega \text{ m}\). A1 for expressing the final answer correctly: \((6.4 \pm 0.8) \times 10^{-7} \ \Omega \text{ m}\) (accept \((6.3 \pm 0.8) \times 10^{-7} \ \Omega \text{ m}\)). (d) [2 marks] B1 for identifying the diameter \(d\). B1 for suggesting a micrometer screw gauge / taking multiple readings.

(a) (i) Systematic error is a constant bias causing readings to deviate from the true value in one direction only. (ii) Random error causes readings to scatter about a mean value due to unpredictable fluctuations. (b) (i) Mean time \(t_{\text{mean}} = (35.8 + 36.2 + 35.9) / 3 = 35.97 \text{ s}\). Period \(T = t_{\text{mean}} / 20 = 1.798 \text{ s} \approx 1.80 \text{ s}\). (ii) Absolute uncertainty in \(t\) from half-range: \(\Delta t = (36.2 - 35.8) / 2 = 0.2 \text{ s}\). Uncertainty in period: \(\Delta T = 0.2 / 20 = 0.01 \text{ s}\). (c) Value of \(g = \frac{4 \pi^2 \times 0.800}{1.798^2} = 9.77 \text{ m s}^{-2}\) (or \(9.75 \text{ m s}^{-2}\) using \(1.80 \text{ s}\)). Percentage uncertainty in \(L = (0.002 / 0.800) \times 100\% = 0.25\%\). Percentage uncertainty in \(T = (0.01 / 1.80) \times 100\% = 0.56\%\). Total percentage uncertainty in \(g = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} = 0.25\% + 2(0.56\%) = 1.37\% \approx 1.4\%\).

(a) [2 marks] (i) B1 for constant deviation in one direction. (ii) B1 for random scatter of readings about a mean. (b) [4 marks] (i) C1 for mean time of \(35.97 \text{ s}\) or \(36.0 \text{ s}\). A1 for dividing by 20 to get \(T = 1.80 \text{ s}\) (or \(1.798 \text{ s}\)). (ii) C1 for calculating half-range of time \(= 0.2 \text{ s}\). A1 for \(\Delta T = 0.01 \text{ s}\). (c) [4 marks] C1 for calculating \(g = 9.75\) or \(9.77 \text{ m s}^{-2}\). C1 for \(\% \Delta L = 0.25\%\). C1 for \(\% \Delta T = 0.56\%\). A1 for correct sum \(1.4\%\) (accept \(1.37\%\) to \(1.4\%\)).

(a) (i) Accuracy is the closeness of a measurement to the true or accepted value. (ii) Precision is the closeness of agreement between independent, repeated measurements. (b) (i) Volume of sphere \(V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.25)^3 = 8.181 \text{ cm}^3\). Density \(\rho = m / V = 45.2 / 8.181 = 5.525 \text{ g cm}^{-3} \approx 5.5 \text{ g cm}^{-3}\). (ii) \(\% \Delta m = (0.5 / 45.2) \times 100\% = 1.11\%\). \(\% \Delta r = (0.02 / 1.25) \times 100\% = 1.60\%\). Since \(V \propto r^3\), \(\% \Delta \rho = \% \Delta m + 3 \% \Delta r = 1.11\% + 3(1.60\%) = 5.91\% \approx 5.9\%\). (iii) Absolute uncertainty \(\Delta \rho = 5.91\% \times 5.525 \text{ g cm}^{-3} = 0.327 \text{ g cm}^{-3} \approx 0.3 \text{ g cm}^{-3}\). Thus, density is written as \(\rho = (5.5 \pm 0.3) \text{ g cm}^{-3}\).

(a) [2 marks] (i) B1 for closeness to true value. (ii) B1 for closeness of repeated readings to each other. (b) [8 marks] (i) C1 for calculating \(V = 8.18 \text{ cm}^3\). C1 for using \(\rho = m / V\). A1 for obtaining \(5.525 \text{ g cm}^{-3}\) showing it rounds to \(5.5\). (ii) C1 for calculating \(\% \Delta m = 1.11\%\) and \(\% \Delta r = 1.60\%\). C1 for applying fractional uncertainty formula with factor of 3 for radius. A1 for \(5.9\%\) (accept \(6.0\%\) or \(5.91\%\)). (iii) C1 for calculating \(\Delta \rho \approx 0.3 \text{ g cm}^{-3}\). A1 for expressing density correctly as \((5.5 \pm 0.3) \text{ g cm}^{-3}\).

(a) The four forces are: 1. Weight of beam \(120 \text{ N}\) acting downwards at its midpoint (\(1.2 \text{ m}\) from \(P\)). 2. Suspended load \(250 \text{ N}\) acting downwards at end \(Q\) (\(2.4 \text{ m}\) from \(P\)). 3. Tension \(T\) in the cable acting at end \(Q\), upwards and leftwards at \(30^\circ\) to the horizontal. 4. Hinge force \(F\) acting at \(P\) upwards and rightwards. (b) Taking moments about hinge \(P\): Clockwise moments \(= (120 \times 1.2) + (250 \times 2.4) = 144 + 600 = 744 \text{ N m}\). Anticlockwise moments \(= T \sin(30^\circ) \times 2.4 = 1.2 T\). Equating moments: \(1.2 T = 744 \implies T = 620 \text{ N}\). (c) Resolve forces for equilibrium: Horizontal: \(F_x = T \cos(30^\circ) = 620 \cos(30^\circ) = 536.9 \text{ N}\). Vertical: \(F_y + T \sin(30^\circ) = 120 + 250 \implies F_y = 370 - 620 \sin(30^\circ) = 60 \text{ N}\). Hinge force magnitude \(F = \sqrt{F_x^2 + F_y^2} = \sqrt{536.9^2 + 60^2} = 540.2 \text{ N} \approx 540 \text{ N}\).

(a) [2 marks] B1 for correct representation of downwards beam weight and load. B1 for correct representation of tension at \(30^\circ\) and hinge force. (b) [4 marks] C1 for correct calculation of clockwise moments (\(744 \text{ N m}\)). C1 for correct algebraic expression of anticlockwise moment (\(1.2 T\)). C1 for equating moments. A1 for calculating \(T = 620 \text{ N}\). (c) [4 marks] C1 for horizontal resolution: \(F_x = 537 \text{ N}\). C1 for vertical resolution: \(F_y = 60 \text{ N}\). C1 for using Pythagoras formula. A1 for final answer of \(540 \text{ N}\) (accept \(539 \text{ N}\) to \(541 \text{ N}\)).

(a) Weight \(W = m g = 15.0 \times 9.81 = 147.15 \text{ N} \approx 147 \text{ N}\). (b) (i) In the closed vector triangle: The downward force is \(W = 147 \text{ N}\). The angle between \(W\) and \(T_B\) is \(90.0^\circ - 50.0^\circ = 40.0^\circ\). The angle between \(W\) and \(T_A\) is \(90.0^\circ - 35.0^\circ = 55.0^\circ\). The angle between \(T_A\) and \(T_B\) is \(180.0^\circ - (40.0^\circ + 55.0^\circ) = 85.0^\circ\). (ii) Applying the Sine Rule: \(\frac{W}{\sin(85.0^\circ)} = \frac{T_A}{\sin(40.0^\circ)} = \frac{T_B}{\sin(55.0^\circ)}\). 1. \(T_A = 147.15 \times \frac{\sin(40.0^\circ)}{\sin(85.0^\circ)} = 147.15 \times \frac{0.6428}{0.9962} = 94.8 \text{ N} \approx 95 \text{ N}\). 2. \(T_B = 147.15 \times \frac{\sin(55.0^\circ)}{\sin(85.0^\circ)} = 147.15 \times \frac{0.8192}{0.9962} = 121 \text{ N}\).

(a) [1 mark] B1 for showing \(15.0 \times 9.81 = 147 \text{ N}\). (b) (i) [3 marks] B1 for three vectors forming a closed triangle with consistent arrows. B1 for correct labels of \(T_A\), \(T_B\), and \(W\). B1 for correct interior angles shown: \(40^\circ\), \(55^\circ\), and \(85^\circ\). (b) (ii) Part 1 [3 marks] C1 for formulating equations (resolving horizontally: \(T_A \cos(35^\circ) = T_B \cos(50^\circ)\)). C1 for vertical resolution substitution. A1 for \(T_A = 95 \text{ N}\) (accept \(94.8 \text{ N}\)). (b) (ii) Part 2 [3 marks] C1 for setting up formula for \(T_B\). C1 for substitution of values. A1 for \(T_B = 121 \text{ N}\) (accept \(120.9 \text{ N}\)).

(a) Coherent light from the two slits overlaps and superposes. Constructive interference occurs at points where the waves arrive in phase (path difference is an integer number of wavelengths, \(n\lambda\)), adding to give maximum intensity (a bright fringe). Destructive interference occurs where waves arrive in anti-phase (path difference is an odd number of half-wavelengths, \((n+0.5)\lambda\)), cancelling to give minimum intensity (a dark fringe). (b) The distance between the central maximum (0th order) and the 5th order bright maximum contains exactly 5 fringe separations: \(5x = 2.45 \text{ cm} = 24.5 \text{ mm} \implies x = 24.5 / 5 = 4.90 \text{ mm}\). (c) Using \(\lambda = \frac{xd}{D}\): \(d = \frac{\lambda D}{x} = \frac{632.8 \times 10^{-9} \times 1.85}{4.90 \times 10^{-3}} = 2.389 \times 10^{-4} \text{ m} = 0.239 \text{ mm}\). (d) (i) Moving the screen closer reduces \(D\). Since \(x \propto D\), the fringe separation \(x\) decreases. (ii) Since \(514.5 \text{ nm} < 632.8 \text{ nm}\), the wavelength \(\lambda\) decreases. Since \(x \propto \lambda\), the fringe separation \(x\) decreases.

(a) [3 marks] B1 for stating that waves overlap and superpose. B1 for constructive interference / in phase / path difference \(n\lambda\) for bright fringe. B1 for destructive interference / anti-phase / path difference \((n+0.5)\lambda\) for dark fringe. (b) [2 marks] C1 for identifying 5 fringe widths in \(2.45 \text{ cm}\). A1 for showing division resulting in \(4.90 \text{ mm}\). (c) [3 marks] C1 for rearranging to \(d = \frac{\lambda D}{x}\). C1 for correct substitutions with unit consistency. A1 for obtaining \(d = 2.39 \times 10^{-4} \text{ m}\) (or \(0.239 \text{ mm}\)). (d) [2 marks] (i) B1 for stating \(x\) decreases because \(D\) is reduced. (ii) B1 for stating \(x\) decreases because \(\lambda\) is reduced.

**Sample Table of Results:**
* \(L_0 = 3.5\text{ cm}\)

| \(d / \text{cm}\) | \(L / \text{cm}\) | \(e / \text{cm}\) |
|---|---|---|
| 20.0 | 7.8 | 4.3 |
| 30.0 | 8.8 | 5.3 |
| 40.0 | 9.8 | 6.3 |
| 50.0 | 10.8 | 7.3 |
| 60.0 | 11.8 | 8.3 |
| 70.0 | 12.8 | 9.3 |

**Graph Plotting:**
* The points \((20.0, 4.3)\), \((30.0, 5.3)\), \((40.0, 6.3)\), \((50.0, 7.3)\), \((60.0, 8.3)\), and \((70.0, 9.3)\) are plotted on a grid.
* \(x\)-axis scale: 2 cm = 10 cm. \(y\)-axis scale: 2 cm = 1.0 cm.
* A straight line of best fit is drawn directly through all points.

**Gradient and Intercept Calculations:**
* Using coordinates on the line of best fit: \((20.0, 4.3)\) and \((70.0, 9.3)\):
\(\text{gradient } a = \frac{9.3 - 4.3}{70.0 - 20.0} = \frac{5.0}{50.0} = 0.100\)
* y-intercept \(b = e - a d = 4.3 - (0.100 \times 20.0) = 2.3\text{ cm} = 0.023\text{ m}\).

**Calculating Physical Constants:**
1. Spring Constant \(k\):
\(a = \frac{M g}{k D} \implies 0.100 = \frac{0.200 \times 9.81}{k \times 0.800} \implies k = \frac{1.962}{0.080} = 24.5\text{ N m}^{-1}\)
2. Mass of Rule \(m_{\text{rule}}\):
\(b = \frac{m_{\text{rule}} g d_{\text{cm}}}{k D} \implies 0.023 = \frac{m_{\text{rule}} \times 9.81 \times 0.400}{24.5 \times 0.800} \implies m_{\text{rule}} = \frac{0.023 \times 19.6}{3.924} = 0.115\text{ kg} = 115\text{ g}\).

**Total Marks: 20**

* **(a) Setup and Initial Measurement [1 mark]**
* [1] \(L_0\) recorded to the nearest 0.1 cm (1 mm) with correct unit.

* **(b) Table of Results [6 marks]**
* [1] At least 6 sets of readings of \(d\) and \(e\) showing the correct trend (extension increases as mass distance increases).
* [1] Column headings: each column must have a heading and a separating solidus with unit (e.g., \(d/\text{cm}\), \(e/\text{cm}\)).
* [1] Consistency: all raw values of \(d\) and \(L\) must be recorded to the nearest millimetre.
* [1] Significant figures: values of \(e\) must be calculated to the same number of decimal places as the raw values of \(L\).
* [1] Range: \(d\) values must span at least 50.0 cm.
* [1] Quality: points must lie close to a linear path with minimal scatter.

* **(c) Graph [4 marks]**
* [1] Axes: scales must be simple (multiples of 1, 2, or 5). Plotted points must occupy at least half the grid height and half the grid width. Correct labels with units.
* [1] Plotting: all points plotted within \(\pm 0.5\text{ mm}\). Point diameters must be \(\le 1\text{ mm}\).
* [1] Line of best fit: straight line with balanced distribution of points on either side.
* [1] Quality: scatter is small (all points within \(0.2\text{ cm}\) of the line of best fit).

* **(d) Analysis [4 marks]**
* [2] Gradient: calculated using a triangle with hypotenuse length greater than half the length of the drawn line.
* [2] y-intercept: correct calculation using a point from the line of best fit, or read directly from the y-axis if \(d=0\) is on the grid.

* **(e) Calculations and Constants [5 marks]**
* [1] Constant \(a\) calculated with unit (dimensionless or \(\text{cm/cm}\)).
* [1] Constant \(b\) calculated with unit (e.g., \(\text{cm}\) or \(\text{m}\)).
* [1] Correct value and unit of \(k\) (e.g., \(\text{N m}^{-1}\)).
* [1] Correct value and unit of \(m_{\text{rule}}\) (in range \(0.08\text{ kg}\) to \(0.15\text{ kg}\)).
* [1] Precision: \(k\) and \(m_{\text{rule}}\) given to 2 or 3 significant figures.

**Step-by-Step Calculations:**

1. **Uncertainty Analysis:**
* Distance \(w = 30.0\text{ cm}\).
* Absolute uncertainty \(\Delta w = 0.3\text{ cm}\) (to account for parallax, alignment of the two vertical strings, and thickness of the string).
* Percentage uncertainty \(= \frac{0.3}{30.0} \times 100\% = 1.0\%\).

2. **First Measurement (\(h_1 = 40.0\text{ cm} = 0.400\text{ m}\)):**
* \(t_1 = 15.12\text{ s}\), \(t_2 = 15.04\text{ s}\)
* \(t_{\text{avg}} = 15.08\text{ s}\)
* \(T_1 = 15.08 / 10 = 1.508\text{ s}\)
* \(k_1 = \frac{T_1^2}{h_1} = \frac{1.508^2}{0.400} = 5.69\text{ s}^2\text{ m}^{-1}\)

3. **Second Measurement (\(h_2 = 25.0\text{ cm} = 0.250\text{ m}\)):**
* \(t_1 = 11.98\text{ s}\), \(t_2 = 11.92\text{ s}\)
* \(t_{\text{avg}} = 11.95\text{ s}\)
* \(T_2 = 11.95 / 10 = 1.195\text{ s}\)
* \(k_2 = \frac{T_2^2}{h_2} = \frac{1.195^2}{0.250} = 5.71\text{ s}^2\text{ m}^{-1}\)

4. **Conclusion:**
* Percentage difference in \(k\) values:
\(\text{Difference} = \frac{|5.71 - 5.69|}{5.69} \times 100\% = 0.35\%\)
* Since this percentage difference is well below the 10% limit, the experimental results support the suggested relationship: \(T^2 = k h\).

**Total Marks: 20**

* **(a) Distance \(w\) and Uncertainty [2 marks]**
* [1] Measurement of \(w\) with appropriate unit in range \(25.0\text{ cm} \le w \le 35.0\text{ cm}\), recorded to the nearest 1 mm.
* [1] Percentage uncertainty in \(w\) calculated using an absolute uncertainty in range \(2\text{ mm} \le \Delta w \le 5\text{ mm}\) with clear working shown.

* **(b) First Oscillation Measurement (\(h = 40.0\text{ cm}\)) [3 marks]**
* [1] At least two measurements of \(t\) recorded to the nearest 0.1 s or 0.01 s.
* [1] Correct calculation of the average time \(t_{\text{avg}}\) and period \(T\).
* [1] Repeat measurements show reasonable agreement (within 0.3 s).

* **(c) Second Oscillation Measurement (\(h = 25.0\text{ cm}\)) [3 marks]**
* [1] Sets of raw times \(t\) recorded for the second configuration.
* [1] Correct calculation of second period \(T\).
* [1] Correct trend shown (period \(T\) is smaller for smaller height \(h\)).

* **(d) Evaluation of Constant \(k\) [2 marks]**
* [1] Correct calculation of both values of \(k = T^2 / h\) with appropriate unit (e.g., \(\text{s}^2\text{ m}^{-1}\) or \(\text{s}^2\text{ cm}^{-1}\)).
* [1] Both values of \(k\) given to 2 or 3 significant figures.

* **(e) Conclusion [2 marks]**
* [1] Calculation of percentage difference between \(k_1\) and \(k_2\).
* [1] Correct concluding statement comparing the calculated percentage difference with a specified limit (e.g., 10%).

* **(f) Table of Limitations and Improvements [8 marks]**
* **Limitations (any 4, 1 mark each):**
1. Measuring \(h\) or \(w\) is difficult due to flexible threads / knot sizes / parallax.
2. Rule does not oscillate in a pure torsional mode (wobbles or swings sideways).
3. Human reaction time error in starting/stopping the stopwatch.
4. Difficult to determine the exact end of an oscillation due to rapid motion.
* **Improvements (any 4, 1 mark each):**
1. Use a rigid metal wire instead of thin string / clamp threads between wooden blocks to define precise suspension boundaries.
2. Attach a vertical guide pin through the rule's center of mass to restrict motion to torsional rotation only.
3. Use a photogate connected to an electronic timer / use a slow-motion video recorder with a frame counter.
4. Use a clear vertical pointer (fiducial marker) at the equilibrium position to assist in timing.

(a)(i) A systematic error is an error that causes all readings to be consistently shifted from the true value in one direction.
(a)(ii) A random error is an error that causes readings to scatter about the mean value with no constant pattern.

(b) The formula for resistivity is:
\(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\)
Substitute the measured values:
\(d = 0.38 \times 10^{-3}\ \text{m}\)
\(A = \frac{\pi}{4} (0.38 \times 10^{-3})^2 = 1.1341 \times 10^{-7}\ \text{m}^2\)
\(\rho = \frac{4.5 \times 1.1341 \times 10^{-7}}{1.240} = 4.116 \times 10^{-7}\ \Omega\,\text{m} \approx 4.1 \times 10^{-7}\ \Omega\,\text{m}\)

(c) The percentage uncertainties of the measured quantities are:
- For \(R\): \(\frac{\Delta R}{R} = \frac{0.2}{4.5} \times 100\% = 4.44\%\)
- For \(L\): \(\frac{\Delta L}{L} = \frac{0.002}{1.240} \times 100\% = 0.16\%\)
- For \(d\): \(\frac{\Delta d}{d} = \frac{0.02}{0.38} \times 100\% = 5.26\%\)
Since \(\rho \propto \frac{R d^2}{L}\):
\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\)
\(\frac{\Delta \rho}{\rho} = 4.44\% + 2(5.26\%) + 0.16\% = 15.12\% \approx 15\%\)

(d) First calculate the absolute uncertainty \(\Delta \rho\):
\(\Delta \rho = 15.12\% \times 4.116 \times 10^{-7}\ \Omega\,\text{m} = 0.622 \times 10^{-7}\ \Omega\,\text{m}\)
Rounding \(\Delta \rho\) to 1 significant figure gives \(0.6 \times 10^{-7}\ \Omega\,\text{m}\).
Therefore, the final value of resistivity is written as:
\(\rho = (4.1 \pm 0.6) \times 10^{-7}\ \Omega\,\text{m}\)

(a)(i) Constant error in all measurements in one direction (e.g., zero error) [1 mark]
(a)(ii) Error that leads to spread of values about the mean value [1 mark]

(b) Correct formula for cross-sectional area or direct substitution of \(A = \pi d^2 / 4\) [1 mark]
Correct calculation of cross-sectional area \(A = 1.13 \times 10^{-7}\ \text{m}^2\) [1 mark]
Correct calculation of \(\rho = 4.1 \times 10^{-7}\ \Omega\,\text{m}\) [1 mark]

(c) Individual percentage uncertainties calculated correctly (at least two) [1 mark]
Summing of fractional uncertainties including doubling the uncertainty of diameter: \(\frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\) [1 mark]
Final percentage uncertainty of \(15\%\) (or \(15.1\%\)) [1 mark]

(d) Absolute uncertainty calculated as \(0.6 \times 10^{-7}\ \Omega\,\text{m}\) (rounded to 1 s.f.) [1 mark]
Final value matching the precision of uncertainty: \((4.1 \pm 0.6) \times 10^{-7}\ \Omega\,\text{m}\) [1 mark]

(a) Condition 1: The net external force acting on the body must be zero (translational equilibrium).
Condition 2: The net external torque (sum of moments) about any axis must be zero (rotational equilibrium).

(b) Three forces act on the beam:
1. Weight \(W = m g = 15 \times 9.81 = 147.15\ \text{N}\) acting vertically downwards at the midpoint (\(1.2\ \text{m}\) from the hinge).
2. Tension \(T\) acting at point B (right end) upwards and to the left, at an angle of \(35^\circ\) to the horizontal.
3. Hinge force \(F\) acting at point A (left end) upwards and to the right.

(c) Take moments about point A (hinge) to eliminate the hinge force:
\(\text{Clockwise moment} = \text{Anticlockwise moment}\)
\(W \times 1.2 = T \sin(35^\circ) \times 2.4\)
\(147.15 \times 1.2 = T \sin(35^\circ) \times 2.4\)
\(176.58 = T \times 1.3766\)
\(T = 128.3\ \text{N} \approx 128\ \text{N}\) (or \(130\ \text{N}\) to 2 s.f.).

(d) For translational equilibrium:
Horizontal component of hinge force, \(H = T \cos(35^\circ) = 128.3 \cos(35^\circ) = 105.1\ \text{N}\).
Vertical component of hinge force, \(V = W - T \sin(35^\circ) = 147.15 - 128.3 \sin(35^\circ) = 147.15 - 73.58 = 73.57\ \text{N}\).
Magnitude of hinge force:
\(F_H = \sqrt{H^2 + V^2} = \sqrt{105.1^2 + 73.57^2} = 128.3\ \text{N} \approx 128\ \text{N}\).

(a) Net/resultant force is zero [1 mark]
Net/resultant moment/torque is zero [1 mark]

(b) Weight \(W\) shown acting vertically downwards at center of gravity [1 mark]
Tension shown along cable and hinge force shown with both vertical and horizontal components (or angled upwards and outwards) [1 mark]

(c) Correct moment equation about hinge: \(15 \times 9.81 \times 1.2 = T \sin(35^\circ) \times 2.4\) [1 mark]
Correct substitution and rearrangement [1 mark]
Calculation of \(T = 128\ \text{N}\) (accept range \(128 - 130\ \text{N}\)) [1 mark]

(d) Correct horizontal component: \(H = 105\ \text{N}\) [1 mark]
Correct vertical component: \(V = 73.6\ \text{N}\) [1 mark]
Correct magnitude \(128\ \text{N}\) (accept range \(128 - 130\ \text{N}\)) [1 mark]

(a) Coherent light sources have a constant phase difference between them (which also implies they have the same frequency/wavelength).

(b) The distance between the 1st bright fringe and the 9th bright fringe corresponds to exactly 8 fringe widths (\(8w\)).
\(8w = 4.4\ \text{cm} = 0.044\ \text{m}\)
\(w = \frac{0.044}{8} = 5.5 \times 10^{-3}\ \text{m}\)
Using the double-slit formula:
\(w = \frac{\lambda D}{a}\)
\(\lambda = \frac{w a}{D} = \frac{5.5 \times 10^{-3} \times 0.18 \times 10^{-3}}{1.80} = 5.5 \times 10^{-7}\ \text{m} = 550\ \text{nm}\).

(c) Since the polarization axes of the light emerging from the two slits are perpendicular to each other, the two waves cannot interfere with one another (as perpendicular electric field vectors do not cancel or reinforce to change the total average intensity). Consequently, the interference fringes disappear, and a uniform illumination (or single-slit diffraction pattern) is observed on the screen.

(d) When the apparatus is submerged in fluid, the speed of light decreases, which decreases the wavelength of the light (\(\lambda_f = \lambda / n\)). Since the fringe spacing is given by \(w = \lambda D / a\), the decrease in wavelength causes the fringe spacing to decrease.

(a) Constant phase difference between the sources [1 mark]
Same frequency or wavelength [1 mark]

(b) Correct deduction of 8 fringe widths: \(w = 5.5\ \text{mm}\) [1 mark]
Correct rearrangement of formula \(\lambda = w a / D\) [1 mark]
Calculation of wavelength \(\lambda = 5.5 \times 10^{-7}\ \text{m}\) (or \(550\ \text{nm}\)) [1 mark]

(c) No interference pattern/fringes are observed (or uniform intensity) [1 mark]
Because the light waves from the two slits are polarized at right angles / perpendicular to each other [1 mark]
Mutually perpendicular waves cannot interfere to form areas of constructive and destructive interference [1 mark]

(d) Wavelength decreases when entering a medium with higher refractive index (\(\lambda \propto 1/n\)) [1 mark]
Since \(w \propto \lambda\), the fringe spacing decreases [1 mark]

(a) Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point.

(b) The orbit radius is:
\(r = R_E + h = 6.4 \times 10^6\ \text{m} + 8.0 \times 10^5\ \text{m} = 7.2 \times 10^6\ \text{m}\)
The gravitational potential energy is:
\(E_p = -\frac{G M m}{r}\)
\(E_p = -\frac{(6.67 \times 10^{-11}) \times (6.0 \times 10^{24}) \times 450}{7.2 \times 10^6} = -2.501 \times 10^{10}\ \text{J} \approx -2.50 \times 10^{10}\ \text{J}\).

(c) For a circular orbit, the centripetal force is provided by the gravitational force:
\(\frac{m v^2}{r} = \frac{G M m}{r^2} \implies m v^2 = \frac{G M m}{r}\)
The kinetic energy is:
\(E_k = \frac{1}{2} m v^2 = \frac{G M m}{2r} = -\frac{1}{2} E_p\)
\(E_k = \frac{2.501 \times 10^{10}}{2} = 1.25 \times 10^{10}\ \text{J}\).

(d) Total energy in orbit:
\(E_{\text{orbit}} = E_p + E_k = -2.501 \times 10^{10} + 1.250 \times 10^{10} = -1.251 \times 10^{10}\ \text{J}\)
Total energy on Earth's surface (at rest):
\(E_{\text{surface}} = -\frac{G M m}{R_E} = -\frac{(6.67 \times 10^{-11}) \times (6.0 \times 10^{24}) \times 450}{6.4 \times 10^6} = -2.814 \times 10^{10}\ \text{J}\)
Energy required for launch:
\(\Delta E = E_{\text{orbit}} - E_{\text{surface}} = -1.251 \times 10^{10} - (-2.814 \times 10^{10}) = 1.56 \times 10^{10}\ \text{J}\).

(a) Work done per unit mass [1 mark]
Bringing a mass from infinity to the point [1 mark]

(b) Identification of orbit radius as \(r = 7.2 \times 10^6\ \text{m}\) [1 mark]
Formula for gravitational potential energy \(E_p = -GMm/r\) stated or used [1 mark]
Calculation showing \(E_p = -2.50 \times 10^{10}\ \text{J}\) [1 mark]

(c) Statement linking centripetal force to gravitational force to find \(v^2\) or directly stating \(E_k = \frac{GMm}{2r}\) [1 mark]
Calculation of \(E_k\) using correct numbers [1 mark]
Final kinetic energy value of \(1.25 \times 10^{10}\ \text{J}\) [1 mark]

(d) Calculation of potential energy on surface \(= -2.81 \times 10^{10}\ \text{J}\) [1 mark]
Correct subtraction of surface energy from total orbital energy to give \(1.56 \times 10^{10}\ \text{J}\) [1 mark]

(a) Capacitance is defined as charge stored per unit potential difference (\(C = Q/V\)).

(b) The equation is:
\(V = V_0 e^{-\frac{t}{R C}}\)

(c) Substitute the given values into the discharge equation:
\(4.4 = 12.0 e^{-\frac{8.5}{R \times 220 \times 10^{-6}}}\)
\(\frac{4.4}{12.0} = 0.3667 = e^{-\frac{8.5}{R \times 220 \times 10^{-6}}}\)
Take the natural logarithm of both sides:
\(\ln(0.3667) = -1.003 = -\frac{8.5}{R \times 220 \times 10^{-6}}\)
\(R \times 220 \times 10^{-6} = \frac{8.5}{1.003} = 8.475\ \text{s}\)
\(R = \frac{8.475}{220 \times 10^{-6}} = 3.85 \times 10^4\ \Omega \approx 3.9 \times 10^4\ \Omega\).

(d) The initial energy stored in the capacitor is:
\(E_i = \frac{1}{2} C V_0^2 = 0.5 \times 220 \times 10^{-6} \times 12.0^2 = 0.01584\ \text{J}\)
The final energy stored in the capacitor is:
\(E_f = \frac{1}{2} C V^2 = 0.5 \times 220 \times 10^{-6} \times 4.4^2 = 0.00213\ \text{J}\)
Energy lost:
\(\Delta E = E_i - E_f = 0.01584 - 0.00213 = 0.01371\ \text{J} \approx 1.37 \times 10^{-2}\ \text{J}\).

(e) Since charge is proportional to potential difference, \(Q = Q_0 e^{-t/\tau}\).
After a time \(t = 2\tau\):
\(\frac{Q}{Q_0} = e^{-2} \approx 0.135\) (or \(13.5\%\)).

(a) Ratio of charge to potential difference [1 mark]

(b) \(V = V_0 e^{-t/RC}\) [1 mark]

(c) Correct substitution of numbers: \(4.4 / 12 = e^{-8.5 / (R \times 220 \times 10^{-6})}\) [1 mark]
Taking natural logarithms correctly to get \(-1.00\) [1 mark]
Calculation of \(R = 3.9 \times 10^4\ \Omega\) (accept range \(3.8 - 3.9 \times 10^4\ \Omega\)) [1 mark]

(d) Formula for energy stored \(E = \frac{1}{2} C V^2\) stated or implied [1 mark]
Correct substitutions of initial and final potential differences [1 mark]
Energy lost calculated as \(1.37 \times 10^{-2}\ \text{J}\) (accept range \(1.35 - 1.40 \times 10^{-2}\ \text{J}\)) [1 mark]

(e) Identification that fraction remaining is \(e^{-2}\) [1 mark]
Correct calculation of \(0.135\) or \(13.5\%\) [1 mark]

(a) Defining equation:
\(a = -\omega^2 x\)
where \(a\) is the acceleration of the object, \(\omega\) is the angular frequency of the oscillation, and \(x\) is the displacement from the equilibrium position.

(b)(i) Angular frequency:
\(\omega = 2 \pi f = 2 \pi \times 2.5 = 5.0\pi \approx 15.7\ \text{rad}\,\text{s}^{-1}\)

(b)(ii) The magnitude of the maximum acceleration is:
\(a_{\text{max}} = \omega^2 x_0\)
where \(x_0 = 6.0\ \text{cm} = 0.060\ \text{m}\)
\(a_{\text{max}} = (15.71)^2 \times 0.060 = 246.7 \times 0.060 = 14.8\ \text{m}\,\text{s}^{-2}\)

(b)(iii) The maximum kinetic energy is equal to the total energy of the oscillation:
\(E_k = \frac{1}{2} m v_{\text{max}}^2 = \frac{1}{2} m (\omega x_0)^2\)
\(E_k = 0.5 \times 0.35 \times (15.71 \times 0.060)^2 = 0.175 \times 0.888 = 0.155\ \text{J}\)

(c) The graph of \(a\) against \(x\) is a straight line passing through the origin with a negative gradient.
- The line starts at \(x = -6.0\ \text{cm}\), where \(a = 14.8\ \text{m}\,\text{s}^{-2}\).
- The line terminates at \(x = 6.0\ \text{cm}\), where \(a = -14.8\ \text{m}\,\text{s}^{-2}\).

(a) Equation: \(a = -\omega^2 x\) [1 mark]
Symbols: \(a\) is acceleration, \(x\) is displacement from equilibrium, \(\omega\) is angular frequency (or constant) [1 mark]

(b)(i) Correct equation: \(\omega = 2 \pi f\) [1 mark]
Value: \(\omega = 15.7\ \text{rad}\,\text{s}^{-1}\) (or \(16\ \text{rad}\,\text{s}^{-1}\)) [1 mark]

(b)(ii) Correct equation: \(a_{\text{max}} = \omega^2 x_0\) [1 mark]
Value: \(14.8\ \text{m}\,\text{s}^{-2}\) (accept range \(14.5 - 15.0\ \text{m}\,\text{s}^{-2}\)) [1 mark]

(b)(iii) Correct equation: \(E_k = \frac{1}{2} m \omega^2 x_0^2\) or \(E_k = \frac{1}{2} m v^2\) [1 mark]
Value: \(0.155\ \text{J}\) (accept range \(0.15 - 0.16\ \text{J}\)) [1 mark]

(c) Straight line with a negative gradient passing through the origin [1 mark]
Endpoints of the line labeled correctly with values \(x = \pm 6.0\ \text{cm}\) and \(a = \mp 14.8\ \text{m}\,\text{s}^{-2}\) [1 mark]

(a) The decay constant \(\lambda\) is the probability of decay of a nucleus per unit time interval.

(b) First, convert half-life into seconds:
\(t_{1/2} = 6.0\ \text{hours} = 6.0 \times 3600\ \text{s} = 2.16 \times 10^4\ \text{s}\)
The decay constant is:
\(\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.6931}{2.16 \times 10^4\ \text{s}} = 3.21 \times 10^{-5}\ \text{s}^{-1}\).

(c) The initial activity is:
\(A_0 = \lambda N_0\)
\(A_0 = (3.21 \times 10^{-5}\ \text{s}^{-1}) \times (3.2 \times 10^{18}) = 1.03 \times 10^{14}\ \text{Bq}\) (or \(\text{s}^{-1}\)).

(d) The activity after \(t = 15\ \text{hours}\) is:
\(A = A_0 e^{-\lambda t}\) or \(A = A_0 \times (0.5)^{\frac{t}{t_{1/2}}}\)
Since \(\frac{t}{t_{1/2}} = \frac{15}{6.0} = 2.5\):
\(A = 1.03 \times 10^{14} \times (0.5)^{2.5} = 1.03 \times 10^{14} \times 0.1768 = 1.82 \times 10^{13}\ \text{Bq}\).

(e) Radioactive decay is a random process. The calculated activity is only the average rate of decay of a large number of nuclei, meaning there are statistical fluctuations from second to second.

(a) Probability of decay of a nucleus [1 mark]
Per unit time [1 mark]

(b) Conversion of hours to seconds: \(21600\ \text{s}\) [1 mark]
Calculation of decay constant: \(3.21 \times 10^{-5}\ \text{s}^{-1}\) [1 mark]

(c) Formula \(A_0 = \lambda N_0\) used [1 mark]
Correct value: \(1.03 \times 10^{14}\ \text{Bq}\) (accept \(1.0 \times 10^{14}\ \text{Bq}\)) [1 mark]

(d) Identification of number of half-lives as \(2.5\) or correct use of \(e^{-\lambda t}\) [1 mark]
Correct value: \(1.8 \times 10^{13}\ \text{Bq}\) (accept \(1.82 \times 10^{13}\ \text{Bq}\)) [1 mark]

(e) Radioactive decay is a random process [1 mark]
Activity is a statistical average / actual decays fluctuate about the mean [1 mark]

(a) Faraday's law states that the magnitude of the induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage.

(b) Magnetic flux linkage is the product of the magnetic flux through a single turn of a coil and the number of turns in the coil (\(\Psi = N \Phi = N B A\)).

(c)(i) Calculate the cross-sectional area of the circular coil:
\(A = \pi r^2 = \pi \times (0.025)^2 = 1.963 \times 10^{-3}\ \text{m}^2\)
The initial magnetic flux linkage is:
\(\Psi = N B A = 150 \times 0.080 \times 1.963 \times 10^{-3}\)
\(\Psi = 0.02356\ \text{Wb-turns} \approx 2.4 \times 10^{-2}\ \text{Wb-turns}\) (or \(2.36 \times 10^{-2}\ \text{Wb-turns}\)).

(c)(ii) Since the field is reduced to zero at a constant rate, the rate of change of flux linkage is constant.
Using Faraday's law:
\(E = \frac{\Delta \Psi}{\Delta t} = \frac{\Psi_{\text{final}} - \Psi_{\text{initial}}}{\Delta t}\)
\(E = \frac{0 - 0.02356}{0.12} = -0.196\ \text{V}\)
The magnitude of the induced e.m.f. is \(0.20\ \text{V}\) (or \(0.196\ \text{V}\)).

(a) Induced e.m.f. is proportional to the rate of change of magnetic flux linkage [2 marks]

(b) Product of magnetic flux and the number of turns [1 mark]
Formula \(\Psi = N B A \sin\theta\) (where \(\theta\) is angle to the plane) or \(\Psi = N \Phi\) with symbols explained [1 mark]

(c)(i) Calculation of area \(A = \pi (0.025)^2 = 1.96 \times 10^{-3}\ \text{m}^2\) [1 mark]
Use of formula \(\Psi = N B A\) [1 mark]
Calculation of flux linkage \(= 2.36 \times 10^{-2}\ \text{Wb}\) (or \(2.4 \times 10^{-2}\ \text{Wb}\)) [1 mark]

(c)(ii) Formula \(E = \Delta(N\Phi) / \Delta t\) stated or implied [1 mark]
Correct substitution of \(\Delta t = 0.12\ \text{s}\) and initial flux linkage [1 mark]
Induced e.m.f. \(= 0.20\ \text{V}\) (or \(0.196\ \text{V}\)) [1 mark]

(a) The electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point.

(b) Using the formula for the potential of an isolated sphere:
\(V = \frac{Q}{4\pi\varepsilon_0 R}\)

Rearranging for \(Q\):
\(Q = V \cdot 4\pi\varepsilon_0 R\)
\(Q = 3.0 \times 10^3 \times 4\pi \times 8.85 \times 10^{-12} \times 0.15\)
\(Q \approx 5.01 \times 10^{-8}\text{ C}\) (or \(5.0 \times 10^{-8}\text{ C}\) to 2 significant figures).

(c) (i) There is initially a potential difference between the spheres (since sphere \(A\) is at \(3.0\text{ kV}\) and sphere \(B\) is at \(0\text{ V}\)). Charge flows from the region of higher potential (sphere \(A\)) to the region of lower potential (sphere \(B\)) until both spheres reach the same electric potential.

(ii) Let \(Q_A\) and \(Q_B\) be the final charges on spheres \(A\) and \(B\). Since they are connected by a wire, their final potential \(V_f\) is equal:
\(V_f = \frac{Q_A}{4\pi\varepsilon_0 R} = \frac{Q_B}{4\pi\varepsilon_0 r}\)

Therefore:
\(\frac{Q_A}{R} = \frac{Q_B}{r} \implies \frac{Q_A}{15} = \frac{Q_B}{5.0} \implies Q_A = 3 Q_B\)

By conservation of total charge:
\(Q_A + Q_B = Q_{\text{total}} = 5.01 \times 10^{-8}\text{ C}\)
\(3 Q_B + Q_B = 5.01 \times 10^{-8}\text{ C} \implies 4 Q_B = 5.01 \times 10^{-8}\text{ C}\)
\(Q_B = 1.25 \times 10^{-8}\text{ C}\)
\(Q_A = 3.76 \times 10^{-8}\text{ C}\)

Now, calculate the final potential \(V_f\):
\(V_f = \frac{3.76 \times 10^{-8}}{4\pi \times 8.85 \times 10^{-12} \times 0.15} \approx 2.25 \times 10^3\text{ V} = 2.3\text{ kV}\).

(iii) The final charge on sphere \(B\) is:
\(q_B = Q_B = 1.25 \times 10^{-8}\text{ C} \approx 1.3 \times 10^{-8}\text{ C}\).

(a)
- M1: Work done per unit positive charge
- A1: in bringing a small test charge from infinity to that point

(b)
- C1: \(V = \frac{Q}{4\pi\varepsilon_0 R}\) or substitution of values into this formula
- A1: \(Q = 5.0 \times 10^{-8}\text{ C}\) (accept \(5.01 \times 10^{-8}\text{ C}\))

(c) (i)
- B1: Reference to an initial difference in potential between sphere A and sphere B
- B1: Charge flows until there is no potential difference / they are at the same potential

(ii)
- C1: States or uses \(V_f = V_A = V_B\) to show \(\frac{Q_A}{Q_B} = \frac{R}{r} = 3.0\)
- C1: Solves for final charge on sphere A, e.g., \(Q_A = 0.75 \times 5.01 \times 10^{-8}\text{ C} = 3.76 \times 10^{-8}\text{ C}\)
- A1: \(V_f = 2.3\text{ kV}\) (or \(2.25\text{ kV}\), allow \(2250\text{ V}\))

(iii)
- A1: \(q_B = 1.3 \times 10^{-8}\text{ C}\) (or \(1.25 \times 10^{-8}\text{ C}\))

(a) Magnetic flux linkage is defined as the product of the magnetic flux and the number of turns of the coil, where magnetic flux is the product of the magnetic flux density and the area normal to the magnetic field lines.

(b) The formula for magnetic flux linkage is:
\(\text{Flux Linkage} = N \Phi = N B A\)

Substitute the given values:
\(N \Phi = 180 \times 0.40\text{ T} \times 3.5 \times 10^{-4}\text{ m}^2\)
\(N \Phi = 0.0252\text{ Wb-turns} \approx 2.5 \times 10^{-2}\text{ Wb}\).

(c) (i) Faraday's law of electromagnetic induction states that the magnitude of the induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage.

(ii) According to Faraday's law, the induced e.m.f. \(E\) is given by:
\(E = \frac{\Delta (N \Phi)}{\Delta t}\)

Substitute the change in flux linkage and time interval:
\(E = \frac{0.0252\text{ Wb}}{0.15\text{ s}} = 0.168\text{ V} \approx 0.17\text{ V}\).

(iii) The total resistance of the circuit is:
\(R_{\text{total}} = R_{\text{coil}} + R_{\text{external}} = 3.0\ \Omega + 12.0\ \Omega = 15.0\ \Omega\)

The electrical energy \(W\) dissipated as thermal energy in the entire circuit is:
\(W = \frac{E^2}{R_{\text{total}}} \Delta t\)
\(W = \frac{(0.168)^2}{15.0} \times 0.15 = 2.82 \times 10^{-4}\text{ J} \approx 2.8 \times 10^{-4}\text{ J}\).

(a)
- M1: Product of magnetic flux and the number of turns
- A1: where flux is the product of magnetic flux density and area perpendicular to the field

(b)
- C1: Uses \(\text{Flux Linkage} = N B A\) with correct substitution
- A1: \(2.5 \times 10^{-2}\text{ Wb}\) (or \(2.52 \times 10^{-2}\text{ Wb}\))

(c) (i)
- B1: Magnitude of induced e.m.f. is directly proportional to...
- B1: ...the rate of change of magnetic flux linkage

(ii)
- C1: Use of \(E = \frac{\Delta (N\Phi)}{\Delta t}\)
- A1: \(0.17\text{ V}\) (or \(0.168\text{ V}\))

(iii)
- C1: Total resistance \(R_{\text{total}} = 15.0\ \Omega\) and substitution into \(W = I^2 R t\) or \(W = \frac{E^2}{R} t\)
- A1: \(2.8 \times 10^{-4}\text{ J}\) (or \(2.82 \times 10^{-4}\text{ J}\))

### Experimental Design

1. **Apparatus Setup**:
- Set up two parallel horizontal conductors (rigid copper rods). Fix the upper conductor to a heavy non-magnetic wooden stand.
- Suspend the lower conductor directly below and parallel to the first using two identical, light steel springs of known spring constant \(k\).
- Connect the two conductors in series with a DC power supply, a rheostat (variable resistor), a switch, and a high-current ammeter. Connect them such that the current flows in opposite directions in the two conductors (causing a repulsive force).

2. **Measurement of Variables**:
- **Independent Variable**: Current \(I\), measured using the ammeter.
- **Dependent Variable**: Vertical deflection \(y\). Focus a traveling microscope (or cathetometer) on a reference mark on the suspended conductor. Record the position before the switch is closed (zero current) and after the current is established. The deflection \(y\) is the difference between these two readings.
- **Control Variables**:
- The initial separation distance \(d\) between the conductors when \(I = 0\) must be kept constant. Measure this at both ends using a vernier caliper to ensure the wires are parallel.
- The length \(L\) of the parallel section of the conductors.
- The spring constant \(k\) of the supporting springs.

3. **Analysis of Data**:
- Take log of both sides of the suggested equation:
\[ \lg y = \beta \lg I + \lg \alpha \]
- Plot a graph of \(\lg y\) against \(\lg I\).
- A straight-line graph confirms the suggested relationship.
- The gradient of the line gives the value of \(\beta\).
- The y-intercept of the line gives \(\lg \alpha\), so \(\alpha = 10^{\text{intercept}}\).

4. **Safety and Practical Details**:
- High currents (several amperes) are needed to produce measurable deflections. Switch off the current immediately after taking each reading to prevent the conductors from overheating, which causes thermal expansion and alters the physical dimensions.
- Conduct the experiment in a transparent draft-shield box to prevent draft/air currents from displacing the light suspended conductor.

**Defining the Problem (3 marks):**
- **DP1**: Identify current \(I\) as the independent variable and deflection \(y\) as the dependent variable.
- **DP2**: State that the initial separation \(d\) between the conductors must be kept constant.
- **DP3**: State that the length \(L\) of the conductors (or the spring constant \(k\)) must be kept constant.

**Methods of Data Collection (5 marks):**
- **DC1**: Draw a workable diagram showing both parallel conductors, springs supporting the lower conductor, and a series circuit containing a DC power supply, ammeter, and rheostat.
- **DC2**: Explicitly state how the connections are made so that the current runs in opposite directions (e.g., from positive terminal to left of upper wire, right of upper wire to right of lower wire, left of lower wire to negative terminal).
- **DC3**: Describe how to measure the vertical deflection \(y\) using a traveling microscope focused on a reference mark on the wire.
- **DC4**: Describe how to measure the initial separation \(d\) using vernier calipers at multiple points to ensure parallel alignment.
- **DC5**: Use of a rheostat/variable resistor to vary the current \(I\) systematically.

**Method of Analysis (2 marks):**
- **MA1**: State that a graph of \(\lg y\) against \(\lg I\) (or \(\ln y\) against \(\ln I\)) should be plotted.
- **MA2**: State that \(\beta = \text{gradient}\) and \(\alpha = 10^{\text{intercept}}\) (or \(e^{\text{intercept}}\)).

**Additional Details / Safety (5 marks):**
- **AD1**: Safety precaution: Switch off the circuit between readings to prevent overheating of the wires.
- **AD2**: Detail on using non-magnetic materials (e.g., wood, brass clamps) for the supporting frame to prevent interference with the magnetic force.
- **AD3**: Method to calibrate or determine the spring constant \(k\) of the springs by suspending known masses and measuring extensions.
- **AD4**: Detail on using a draft shield to eliminate air current disturbances.
- **AD5**: Repeat measurements of \(y\) for each value of \(I\) and calculate the average.

### Experimental Design

1. **Apparatus Setup**:
- Connect a signal generator in series with a large flat circular coil and a small standard non-inductive resistor \(R\) (e.g., \(1.0\, \Omega\)).
- Place a small flat search coil at the exact center of the large circular coil. Ensure they are coplanar and coaxial.
- Connect a cathode-ray oscilloscope (CRO) or digital storage oscilloscope across the search coil to measure the induced e.m.f. \(V_0\).
- Connect a second channel of the CRO across the standard resistor \(R\) to monitor the alternating current amplitude \(I_0\) in the circular coil.

2. **Measurement of Variables**:
- **Independent Variable**: Frequency \(f\) of the alternating current, varied via the signal generator.
- **Dependent Variable**: Peak induced e.m.f. \(V_0\), measured from the peak-to-peak amplitude on the CRO screen (where \(V_0 = \frac{V_{\text{p-p}}}{2}\)).
- **Control Variables**:
- The peak current \(I_0\) in the circular coil must be kept constant. Since the inductive reactance of the coil increases with frequency (\(X_L = 2\pi f L\)), the current will naturally drop as frequency increases. At each frequency step, adjust the output amplitude of the signal generator to keep the voltage amplitude across \(R\) (and thus the current \(I_0 = V_R/R\)) constant.
- The geometric arrangement (coaxial and coplanar) of the search coil relative to the primary coil.

3. **Analysis of Data**:
- Take log of both sides of the suggested equation:
\[ \lg V_0 = n \lg f + \lg K \]
- Plot a graph of \(\lg V_0\) against \(\lg f\).
- A straight line confirms the validity of the relationship.
- Determine the gradient of this line to find \(n\).
- Determine the y-intercept of the line to find \(\lg K\), which gives \(K = 10^{\text{intercept}}\).

4. **Safety and Practical Details**:
- Keep the apparatus away from mains wiring, power supplies, or large metal bodies to avoid stray magnetic fields inducing noise in the search coil.
- Use a search coil with a high number of turns to maximize the induced e.m.f. so it can be accurately resolved above background noise.

**Defining the Problem (3 marks):**
- **DP1**: Identify frequency \(f\) as the independent variable and peak induced e.m.f. \(V_0\) as the dependent variable.
- **DP2**: State that the current amplitude \(I_0\) in the large circular coil must be kept constant.
- **DP3**: State that the geometric alignment (axial position and orientation) of the search coil relative to the circular coil must be kept constant.

**Methods of Data Collection (5 marks):**
- **DC1**: Draw a workable diagram showing a signal generator, circular coil, and series resistor connected in a loop, with a search coil positioned centrally and a CRO connected to measure both coils' voltages.
- **DC2**: Describe how to measure the peak induced e.m.f. \(V_0\) using the vertical scale (Y-gain) on the CRO screen.
- **DC3**: Describe how to determine frequency \(f\) using the CRO time-base setting to find the period \(T\) (\(f = 1/T\)) or from a digital display on the signal generator.
- **DC4**: Explain how to monitor the current \(I_0\) using the CRO to measure the peak voltage across a small standard resistor in series with the large coil.
- **DC5**: Detail on how to align the search coil to be coplanar and coaxial with the circular coil using non-magnetic templates or clamps.

**Method of Analysis (2 marks):**
- **MA1**: State that a graph of \(\lg V_0\) against \(\lg f\) should be plotted.
- **MA2**: State that \(n = \text{gradient}\) and \(K = 10^{\text{intercept}}\).

**Additional Details / Safety (5 marks):**
- **AD1**: Explicit explanation that the output voltage of the signal generator must be increased at higher frequencies to overcome the increased inductive reactance of the coil and maintain a constant current.
- **AD2**: Use of a search coil with a very large number of turns to increase the magnitude of the signal.
- **AD3**: Use of shielded/coaxial cables for connecting the search coil to the CRO to reduce high-frequency noise pickup.
- **AD4**: Perform the experiment away from other electrical appliances (mains transformers) to minimize electromagnetic interference.
- **AD5**: Safety precaution: Use low voltage from the signal generator to avoid electric shocks.