2024 Cambridge IAS-Level Physics (9702) Practice Paper with Answers

(a) From \(\rho = \frac{R A}{L}\), the unit of \(R\) is \(\Omega\) (which is \(\text{V A}^{-1} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\)). Since \(A\) has unit \(\text{m}^2\) and \(L\) has unit \(\text{m}\), the SI base units of \(\rho\) are \(\text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\). (b) Area \(A = \frac{\pi d^2}{4} = \frac{\pi (0.38 \times 10^{-3})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\). Resistivity \(\rho = \frac{R A}{L} = \frac{8.4 \times 1.134 \times 10^{-7}}{1.250} = 7.62 \times 10^{-7}\ \Omega\text{ m}\). (c) The fractional uncertainty relation is \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). Percentage uncertainty in \(R = \frac{0.2}{8.4} \times 100\% = 2.38\%\). Percentage uncertainty in \(d = 2 \times \frac{0.02}{0.38} \times 100\% = 10.53\%\). Percentage uncertainty in \(L = \frac{0.002}{1.250} \times 100\% = 0.16\%\). Total percentage uncertainty = \(2.38\% + 10.53\% + 0.16\% = 13.07\% \approx 13\%\). (d) Absolute uncertainty \(\Delta \rho = 0.1307 \times 7.62 \times 10^{-7} = 1.0 \times 10^{-7}\ \Omega\text{ m}\).

(a) [2 marks] 1 mark for showing \(\Omega = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\), 1 mark for multiplying by \(\text{m}\) to get \(\text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\). (b) [2.5 marks] 1 mark for correct calculation of cross-sectional area \(1.13 \times 10^{-7}\text{ m}^2\), 1 mark for formula substitution, 0.5 mark for final answer of \(7.6 \times 10^{-7}\ \Omega\text{ m}\). (c) [3 marks] 1 mark for adding fractional uncertainties with 2 times for diameter, 1 mark for calculating individual percentages, 1 mark for total of \(13\%\) (accept 13.1%). (d) [1 mark] 1 mark for absolute uncertainty \(\pm 1.0 \times 10^{-7}\ \Omega\text{ m}\) (must match the decimal precision of the value).

(a) There are no horizontal forces acting on the ball because air resistance is neglected, so from Newton's first/second law, the horizontal acceleration is zero and horizontal velocity remains constant. (b)(i) Horizontal motion: \(x = v_x \times t \implies 24 = 12 \times t \implies t = 2.0\text{ s}\). (b)(ii) Vertical motion: \(h = u_y t + \frac{1}{2} g t^2\). Since initial vertical velocity \(u_y = 0\), \(h = 0.5 \times 9.81 \times 2.0^2 = 19.62\text{ m} \approx 20\text{ m}\) (or \(19.6\text{ m}\)). (b)(iii) Vertical velocity component before impact: \(v_y = u_y + g t = 0 + 9.81 \times 2.0 = 19.62\text{ m s}^{-1}\). Horizontal velocity \(v_x = 12\text{ m s}^{-1}\). Magnitude of velocity \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 19.62^2} = 23.0\text{ m s}^{-1}\). Direction \(\theta = \tan^{-1}(\frac{19.62}{12}) = 58.6^\circ \approx 59^\circ\) below the horizontal.

(a) [1.5 marks] 1 mark for stating no horizontal force / negligible air resistance, 0.5 mark for concluding zero horizontal acceleration. (b)(i) [2 marks] 1 mark for formula \(x = v_x t\), 1 mark for showing \(t = 2.0\text{ s}\). (b)(ii) [2 marks] 1 mark for using \(s = \frac{1}{2} g t^2\) with \(g = 9.81\text{ m s}^{-2}\), 1 mark for correct value \(19.6\text{ m}\) (or \(20\text{ m}\)). (b)(iii) [3 marks] 1 mark for finding \(v_y = 19.6\text{ m s}^{-1}\), 1 mark for calculating final speed \(23\text{ m s}^{-1}\), 1 mark for angle \(59^\circ\) below horizontal.

(a) For equilibrium, the net force on the body in any direction must be zero, and the net moment about any point must be zero. (b)(i) Since the plank is in translational equilibrium, the total upward force equals the total downward force: \(T_A + T_C = W \implies 120 + 180 = 300\text{ N}\). (b)(ii) Let the center of gravity be at a distance \(x\) from \(A\). Taking moments about \(A\): Clockwise moment = \(W \times x = 300 \times x\). Anticlockwise moment = \(T_C \times \text{distance } AC = 180 \times (3.0 - 0.50) = 180 \times 2.50 = 450\text{ N m}\). For rotational equilibrium: \(300 x = 450 \implies x = 1.50\text{ m}\). (b)(iii) Taking moments about \(A\) with the added weight at \(B\): New anticlockwise moment \(T_C \times 2.5 = (300 \times 1.5) + (60 \times 3.0) = 450 + 180 = 630\text{ N m}\). New tension \(T_C = \frac{630}{2.5} = 252\text{ N}\). Thus, the tension increases because the added block creates an additional clockwise moment about \(A\), requiring a larger anticlockwise moment to maintain balance.

(a) [2 marks] 1 mark for sum of forces is zero, 1 mark for sum of moments is zero. (b)(i) [1.5 marks] 1 mark for stating \(T_A + T_C = W\), 0.5 mark for answer \(300\text{ N}\). (b)(ii) [3 marks] 1 mark for taking moments about a point (e.g., A), 1 mark for equating clockwise and anticlockwise moments, 1 mark for finding \(1.5\text{ m}\). (b)(iii) [2 marks] 1 mark for stating that tension increases, 1 mark for explanation involving the increase in clockwise moment about A.

(a) Resistance is defined as the ratio of potential difference across a conductor to the current through it (\(R = V / I\)). (b) \(R = \frac{\rho L}{A} = \frac{3.5 \times 10^{-5} \times 0.15}{2.4 \times 10^{-6}} = 2.1875\ \Omega \approx 2.19\ \Omega\). (c)(i) \(I = \frac{V}{R} = \frac{6.0}{2.1875} = 2.74\text{ A}\) (or \(2.7\text{ A}\)). (c)(ii) Number of electrons per second \(n = \frac{I}{e} = \frac{2.743}{1.60 \times 10^{-19}} = 1.71 \times 10^{19}\text{ electrons s}^{-1}\). (c)(iii) Power \(P = V I = 6.0 \times 2.743 = 16.46\text{ W}\), which is approximately \(16\text{ W}\) to 2 significant figures (or \(P = \frac{V^2}{R} = \frac{36}{2.1875} = 16.46\text{ W}\)).

(a) [1 mark] 1 mark for \(R = V/I\) with terms defined as potential difference and current. (b) [2 marks] 1 mark for converting length to meters and substituting into formula, 1 mark for \(2.2\ \Omega\) (or \(2.19\ \Omega\)). (c)(i) [2 marks] 1 mark for formula \(I = V/R\), 1 mark for \(2.7\text{ A}\). (c)(ii) [2 marks] 1 mark for using \(Q = I t\) and dividing by electron charge, 1 mark for \(1.7 \times 10^{19}\text{ s}^{-1}\). (c)(iii) [1.5 marks] 1 mark for formula \(P = V I\) or \(P = V^2/R\) and showing substitution, 0.5 mark for showing final value rounds to \(16\text{ W}\).

(a) Potential difference is the electrical energy converted into other forms of energy per unit charge passing through a component. (b) When the temperature of the thermistor increases, its resistance decreases. Since the total circuit resistance decreases, the current in the circuit increases. Because the resistance of the fixed resistor is constant, the potential difference across it (\(V = I R\)) increases. (c)(i) Using the potential divider formula, \(V_{\text{out}} = V_{\text{in}} \left( \frac{R_{\text{fixed}}}{R_{\text{fixed}} + R_{\text{thermistor}}} \right) = 9.0 \left( \frac{1200}{1200 + 800} \right) = 9.0 \times 0.60 = 5.4\text{ V}\). (c)(ii) The potential difference across the thermistor is \(V_{\text{thermistor}} = 9.0 - 5.4 = 3.6\text{ V}\). Power dissipated in the thermistor is \(P = \frac{V^2}{R} = \frac{3.6^2}{800} = \frac{12.96}{800} = 0.0162\text{ W}\) (or \(16\text{ mW}\)).

(a) [1 mark] 1 mark for energy transferred per unit charge. (b) [3 marks] 1 mark for stating thermistor resistance decreases, 1 mark for stating circuit current increases, 1 mark for concluding the potential difference across the fixed resistor increases. (c)(i) [2.5 marks] 1 mark for potential divider equation or finding total current, 1 mark for substituting values, 0.5 mark for final answer of \(5.4\text{ V}\). (c)(ii) [2 marks] 1 mark for finding current (\(4.5\text{ mA}\)) or p.d. across thermistor (\(3.6\text{ V}\)), 1 mark for calculating power \(1.6 \times 10^{-2}\text{ W}\).

(a) Progressive waves transfer energy through the medium from the source to other regions, whereas stationary waves store/confine energy in the wave pattern, with no net transfer of energy. (b)(i) For the fundamental mode in a closed-open pipe, there is a node (N) at the closed end and an antinode (A) at the open end. (b)(ii) For the fundamental frequency, the length \(L\) of the tube is equal to a quarter of a wavelength, so \(L = \frac{\lambda}{4}\). Using \(v = f \lambda\), we have \(\lambda = \frac{v}{f} = \frac{340}{250} = 1.36\text{ m}\). Therefore, \(L = \frac{1.36}{4} = 0.34\text{ m}\). (b)(iii) For a tube closed at one end, only odd harmonics can form. The next resonant mode has \(L = \frac{3\lambda}{4}\), corresponding to the 3rd harmonic. The frequency is \(f_3 = 3 \times f_1 = 3 \times 250\text{ Hz} = 750\text{ Hz}\).

(a) [2 marks] 1 mark for stating progressive waves transfer energy, 1 mark for stating stationary waves do not transfer net energy (energy is localized). (b)(i) [2 marks] 1 mark for Node at closed end, 1 mark for Antinode at open end. (b)(ii) [2.5 marks] 1 mark for relating \(L\) to \(\lambda / 4\), 1 mark for finding \(\lambda = 1.36\text{ m}\), 0.5 mark for answer \(0.34\text{ m}\). (b)(iii) [2 marks] 1 mark for identifying the next mode as the 3rd harmonic (or showing \(L = 3\lambda/4\)), 1 mark for final frequency of \(750\text{ Hz}\).

(a) The equation is: \(^{14}_{6}\text{C} \rightarrow\ ^{14}_{7}\text{N} + \text{e}^- + \bar{\nu}_e\) (or \(^0_{-1}\beta + \bar{\nu}\)). (b) The interaction is the weak nuclear interaction (weak force). The emitted electron and electron antineutrino belong to the class of particles called leptons. (c) In \(\beta^-\)-decay, a neutron decays into a proton. The quark composition of a neutron is \(udd\) and a proton is \(uud\). Thus, a down quark (\(d\)) changes into an up quark (\(u\)). (d) Initial charge: Charge of 6 protons = \(+6e\). Final charge: Charge of nitrogen nucleus (\(+7e\)) + charge of electron (\(-1e\)) + charge of antineutrino (\(0\)) = \(+6e\). Since the initial total charge is equal to the final total charge, charge is conserved.

(a) [2.5 marks] 1 mark for Nitrogen-14 with correct numbers, 1 mark for electron/beta particle with correct numbers, 0.5 mark for electron antineutrino. (b) [2 marks] 1 mark for identifying the weak interaction, 1 mark for stating 'leptons'. (c) [2 marks] 1 mark for stating that a neutron changes to a proton, 1 mark for stating that a down quark changes to an up quark (\(d \rightarrow u\)). (d) [2 marks] 1 mark for calculating initial charge \(+6e\), 1 mark for showing total final charge is \(7e - 1e = +6e\) and stating they are equal.