2023 OCR AS Level Physics B (Advancing Physics) - H157 模擬試題連答案詳解

Total number of pixels = \(1200 \times 800 = 960,000\). Total bits = \(960,000 \times 24 = 23,040,000\) bits. Total bytes = \(23,040,000 / 8 = 2,880,000\) bytes. Converting to megabytes: \(2,880,000 / 10^6 = 2.88\text{ MB}\).

1 mark for the correct answer. Reject other options due to incorrect bit-to-byte conversion (b), power-of-10 errors (c), or failing to multiply by the color depth (d).

Using the potential divider equation for the voltage across the LDR: \(V_{out} = V_{in} \times \frac{R_{LDR}}{R_{fixed} + R_{LDR}}\). Substituting the values: \(V_{out} = 9.0\text{ V} \times \frac{1.2\text{ k}\Omega}{4.7\text{ k}\Omega + 1.2\text{ k}\Omega} = 9.0 \times \frac{1.2}{5.9} \approx 1.83\text{ V}\). To two significant figures, this is 1.8 V.

1 mark for the correct calculation. Option b is the voltage across the fixed resistor. Options c and d are incorrect calculations.

The formula for the volume of a cylinder is \(V = \frac{\pi d^2 L}{4}\). The percentage uncertainty in diameter \(d\) is \(\frac{0.02}{0.52} \times 100\% \approx 3.85\%\). The percentage uncertainty in length \(L\) is \(\frac{0.005}{1.240} \times 100\% \approx 0.40\%\). The total percentage uncertainty in \(V\) is \(2 \times (\%\text{ uncertainty in } d) + (\%\text{ uncertainty in } L) = 2(3.85\%) + 0.40\% = 8.1\%\).

1 mark for the correct percentage uncertainty. Option b is the sum of percentage uncertainties without doubling the diameter contribution. Option c is the diameter contribution only. Option d is double the total.

First, find the extension \(\Delta L\) using \(\Delta L = \frac{F L}{A E} = \frac{30 \times 2.0}{1.5 \times 10^{-7} \times 1.2 \times 10^{11}} = 3.33 \times 10^{-3}\text{ m}\). The strain energy \(E_s\) stored is \(E_s = \frac{1}{2} F \Delta L = 0.5 \times 30 \times 3.33 \times 10^{-3} = 0.050\text{ J}\).

1 mark for the correct calculation. Option b omits the factor of 1/2. Option c divides by 4. Option d is a calculation error.

Since the ball hits the ground at \(45^\circ\) to the horizontal, its final vertical velocity component \(v_y\) must equal its horizontal velocity component \(v_x\). Since horizontal velocity is constant, \(v_y = v_x = v\). Using \(v_y^2 = u_y^2 + 2gh\) with \(u_y = 0\), we get \(v^2 = 2gh\), which rearranges to \(h = \frac{v^2}{2g}\).

1 mark for the correct algebraic derivation. Other options represent incorrect kinematic formulas or mathematical errors.

The fringe spacing \(w\) is given by \(w = \frac{\lambda D}{s} = \frac{633 \times 10^{-9} \times 2.5}{0.15 \times 10^{-3}} = 10.55\text{ mm}\). The distance to the third-order bright fringe is \(3w = 3 \times 10.55\text{ mm} = 31.7\text{ mm}\).

1 mark for the correct distance. Option b is the distance to the first-order fringe. Option c is the distance to the second-order fringe. Option d is an incorrect scaling.

The resistance is \(R = \rho \frac{L}{A}\). Since volume \(V = A \times L\) is constant, stretching the wire to \(1.10 L\) causes the cross-sectional area to decrease to \(A / 1.10\). The new resistance is \(R_{new} = \rho \frac{1.10 L}{A / 1.10} = 1.21 \rho \frac{L}{A} = 1.21 R\).

1 mark for the correct answer. Option b ignores the area change. Option d is for a larger extension.

The momentum of the electron is \(p = m_e v = 9.11 \times 10^{-31} \times 1.5 \times 10^6 = 1.3665 \times 10^{-24}\text{ kg m s}^{-1}\). The wavelength of the photon with the same momentum is \(\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.3665 \times 10^{-24}} = 4.85 \times 10^{-10}\text{ m} = 0.49\text{ nm}\).

1 mark for the correct wavelength. Option b is half the correct wavelength. Option c is double. Option d has a power-of-10 error.

Using the wave front curvature (or thin lens) equation, the power \(P\) is given by \(P = \frac{1}{v} - \frac{1}{u}\). Under the standard Cartesian sign convention, the object is at \(u = -0.20\,\text{m}\) and the real image is at \(v = +0.30\,\text{m}\). Substituting these values: \(P = \frac{1}{0.30\,\text{m}} - \frac{1}{-0.20\,\text{m}} = 3.33\,\text{D} + 5.00\,\text{D} = 8.33\,\text{D}\).

1 mark: Correct calculation of the power of the lens leading to option C. Reject other options due to incorrect sign conventions or simple arithmetic errors.

First, calculate the initial output voltage at temperature \(T_1\): \(V_{\text{out1}} = V_{\text{in}} \times \frac{R_T}{R_T + R} = 6.0\,\text{V} \times \frac{800\,\Omega}{800\,\Omega + 1200\,\Omega} = 2.4\,\text{V}\). At temperature \(T_2\), the output voltage decreases by \(1.2\,\text{V}\), so the new output voltage is \(V_{\text{out2}} = 2.4\,\text{V} - 1.2\,\text{V} = 1.2\,\text{V}\). Now set up the potential divider equation for the new thermistor resistance \(R_{T2}\): \(1.2\,\text{V} = 6.0\,\text{V} \times \frac{R_{T2}}{R_{T2} + 1200\,\Omega}\). Simplifying gives \(0.2 = \frac{R_{T2}}{R_{T2} + 1200\,\Omega}\), which leads to \(0.2 R_{T2} + 240\,\Omega = R_{T2}\), or \(0.8 R_{T2} = 240\,\Omega\). Therefore, \(R_{T2} = 300\,\Omega\).

1 mark: Correct analysis of the potential divider circuit leading to the new resistance value of 300 ohms (option B).

The percentage uncertainty in a combined quantity is found by summing the fractional (or percentage) uncertainties of its factors, multiplied by their powers. For \(J = \frac{4I}{\pi d^2}\), the percentage uncertainty is given by \(\%\Delta J = \%\Delta I + 2 \times \%\Delta d\). Calculating the individual percentage uncertainties: \(\%\Delta I = \frac{0.05}{1.84} \times 100\% \approx 2.72\%\). \(\%\Delta d = \frac{0.02}{0.52} \times 100\% \approx 3.85\%\). Therefore, \(\%\Delta J = 2.72\% + 2 \times (3.85\%) = 2.72\% + 7.70\% = 10.42\% \approx 10.4\%\).

1 mark: Correct combination of independent percentage uncertainties (including doubling the uncertainty for the squared diameter term) to yield 10.4% (option C).

First, find the stiffness \(k\) of the wire using the formula \(k = \frac{E A}{L}\): \(k = \frac{1.2 \times 10^{11}\,\text{Pa} \times 1.5 \times 10^{-7}\,\text{m}^2}{2.0\,\text{m}} = 9.0 \times 10^3\,\text{N m}^{-1}\). Next, find the extension \(x\) produced by the \(45\,\text{N}\) force: \(x = \frac{F}{k} = \frac{45\,\text{N}}{9.0 \times 10^3\,\text{N m}^{-1}} = 5.0 \times 10^{-3}\,\text{m}\). Finally, calculate the elastic strain energy \(E_{\text{el}}\) stored: \(E_{\text{el}} = \frac{1}{2} F x = \frac{1}{2} \times 45\,\text{N} \times 5.0 \times 10^{-3}\,\text{m} = 0.1125\,\text{J} \approx 0.11\,\text{J}\).

1 mark: Correct calculation of the elastic strain energy to 2 significant figures, giving 0.11 J (option B).

The horizontal velocity \(v_x\) remains constant at \(15\,\text{m s}^{-1}\). Since the ball strikes the ground at \(45^\circ\) to the horizontal, the vertical component of the velocity \(v_y\) must have the same magnitude as the horizontal component: \(v_y = 15\,\text{m s}^{-1}\). Using the equation of motion for the vertical direction: \(v_y^2 = u_y^2 + 2 g h\). Since the initial vertical velocity is zero (\(u_y = 0\)), we get \(15^2 = 2 \times 9.8 \times h\), which simplifies to \(225 = 19.6 h\). Therefore, \(h = \frac{225}{19.6} \approx 11.5\,\text{m}\).

1 mark: Correct identification of the vertical speed component and subsequent application of the SUVAT equation to solve for the height, resulting in 11.5 m (option B).

According to Einstein\'s photoelectric equation, the maximum kinetic energy of the photoelectrons is \(E_{k,\text{max}} = h f - \phi\). The energy of the incoming photon is: \(E = h f = 6.63 \times 10^{-34}\,\text{J s} \times 7.5 \times 10^{14}\,\text{Hz} = 4.9725 \times 10^{-19}\,\text{J}\). The work function is \(\phi = 2.1\,\text{eV} = 2.1 \times 1.60 \times 10^{-19}\,\text{J} = 3.36 \times 10^{-19}\,\text{J}\). Thus, the maximum kinetic energy is: \(E_{k,\text{max}} = 4.9725 \times 10^{-19}\,\text{J} - 3.36 \times 10^{-19}\,\text{J} = 1.6125 \times 10^{-19}\,\text{J}\). Since \(E_{k,\text{max}} = \frac{1}{2} m_e v_{\text{max}}^2\), we find the maximum speed: \(v_{\text{max}} = \sqrt{\frac{2 \times 1.6125 \times 10^{-19}\,\text{J}}{9.11 \times 10^{-31}\,\text{kg}}} \approx 5.95 \times 10^5\,\text{m s}^{-1} \approx 6.0 \times 10^5\,\text{m s}^{-1}\).

1 mark: Correct conversion of eV to Joules, correct calculation of photon energy, and correct application of kinetic energy to find the maximum electron speed of 6.0 * 10^5 m s^-1 (option C).

For an inverting operational amplifier, the voltage gain \(G\) is given by \(G = -\frac{R_f}{R_{\text{in}}}\). Here, \(R_f = 120\,\text{k}\Omega\) and \(R_{\text{in}} = 15\,\text{k}\Omega\), giving a gain of \(G = -\frac{120}{15} = -8.0\). The output voltage is \(V_{\text{out}} = G \times V_{\text{in}} = -8.0 \times (-0.50\,\text{V}) = +4.0\,\text{V}\).

1 mark: Correct determination of the inverting amplifier\'s gain and subsequent output voltage, resulting in +4.0 V (option D).

According to the Nyquist theorem, to prevent aliasing, the minimum sampling frequency \(f_s\) must be at least twice the maximum frequency present in the signal: \(f_s = 2 \times 3.4\,\text{kHz} = 6.8\,\text{kHz} = 6800\,\text{Hz}\). The bit rate is the number of samples per second multiplied by the number of bits per sample: \(\text{Bit Rate} = 6800\,\text{samples s}^{-1} \times 8\,\text{bits sample}^{-1} = 54400\,\text{bit s}^{-1} = 54.4\,\text{kbit s}^{-1}\).

1 mark: Correct calculation of the minimum sampling rate (6.8 kHz) and subsequent calculation of the bit rate, leading to 54.4 kbit s^-1 (option B).

We use the lens equation in standard Cartesian convention: \(P = \frac{1}{v} - \frac{1}{u}\). Here, the power \(P = \frac{1}{f} = \frac{1}{0.050\text{ m}} = +20\text{ D}\). The object is placed to the left of the lens, so the object distance is \(u = -2.0\text{ m}\). Substituting these into the equation: \(20 = \frac{1}{v} - \frac{1}{-2.0}\) which simplifies to \(20 = \frac{1}{v} + 0.5\). Therefore, \(\frac{1}{v} = 19.5\text{ D}\), giving the image distance \(v = \frac{1}{19.5} \approx 0.0513\text{ m}\), which is \(51.3\text{ mm}\).

1 mark: Correctly applies the lens equation with correct signs to find \(v = 51.3\text{ mm}\), selecting option C.

First, calculate the initial output voltage at \(20^\circ\text{C}\): \(V_{\text{out, 1}} = V_{\text{supply}} \times \frac{R_{\text{th}}}{R + R_{\text{th}}} = 5.0\text{ V} \times \frac{3.6}{1.2 + 3.6} = 5.0\text{ V} \times \frac{3.6}{4.8} = 3.75\text{ V}\). Next, calculate the final output voltage at \(60^\circ\text{C}\): \(V_{\text{out, 2}} = 5.0\text{ V} \times \frac{0.80}{1.2 + 0.80} = 5.0\text{ V} \times \frac{0.80}{2.0} = 2.00\text{ V}\). The magnitude of the change in output voltage is \(|3.75\text{ V} - 2.00\text{ V}| = 1.75\text{ V}\).

1 mark: Correctly calculates the two output voltages and finds the difference of \(1.75\text{ V}\), selecting option B.

The Young modulus is defined as \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta L}\), which gives extension \(\Delta L = \frac{F L}{A E}\). Since both wires are made of the same metal, \(E\) is the same. The force \(F\) is also the same. Therefore, \(\Delta L \propto \frac{L}{A}\). The cross-sectional area \(A\) is proportional to the square of the diameter \(d^2\), so \(\Delta L \propto \frac{L}{d^2}\). For wire X: \(L_X = 2 L_Y\) and \(d_X = 0.5 d_Y\). Thus, \(\frac{\Delta L_X}{\Delta L_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2 = 2 \times \left(\frac{1}{0.5}\right)^2 = 2 \times 4 = 8\).

1 mark: Correctly identifies the relation between extension, length, and diameter, and calculates the ratio as 8, selecting option D.

A path difference of \(0.25\lambda\) (or \(\frac{\lambda}{4}\)) corresponds to a phase difference of \(90^\circ\) (or \(\frac{\pi}{2}\) radians). Therefore, the two rotating phasors representing the waves are perpendicular to each other. We can find the resultant amplitude using Pythagoras' theorem: \(A_{\text{resultant}} = \sqrt{(4.0\ \mu\text{V})^2 + (3.0\ \mu\text{V})^2} = \sqrt{16 + 9} = 5.0\ \mu\text{V}\).

1 mark: Correctly identifies the \(90^\circ\) phase difference from the path difference of \(0.25\lambda\) and uses Pythagoras' theorem to find \(5.0\ \mu\text{V}\), selecting option B.

For (i): The potential divider equation is given by \(V_{\text{out}} = V_{\text{in}} \times \frac{R_1}{R_1 + R_T}\). Substituting the values at \(20^\circ\text{C}\): \(V_{\text{out}} = 6.0\text{ V} \times \frac{2.2\text{ k}\Omega}{2.2\text{ k}\Omega + 4.7\text{ k}\Omega} = 6.0 \times \frac{2.2}{6.9} = 1.91\text{ V}\), which rounds to \(1.9\text{ V}\). For (ii): At \(50^\circ\text{C}\), \(V_{\text{out}} = 3.8\text{ V}\). Rearranging the divider equation: \(3.8 = 6.0 \times \frac{2.2\text{ k}\Omega}{2.2\text{ k}\Omega + R_T}\). Thus, \(2.2\text{ k}\Omega + R_T = 2.2\text{ k}\Omega \times \frac{6.0}{3.8} = 3.47\text{ k}\Omega\). This gives \(R_T = 3.47\text{ k}\Omega - 2.2\text{ k}\Omega = 1.27\text{ k}\Omega\), which rounds to \(1.3\text{ k}\Omega\) (or \(1300\ \Omega\)).

(i) [1] Recall and use potential divider formula: \(V_{\text{out}} = V_{\text{in}} \times R_1 / (R_1 + R_T)\). [1] Obtain \(V_{\text{out}} = 1.9\text{ V}\) (allow 1.91 V). (ii) [1] Substitute values for \(50^\circ\text{C}\): \(3.8 = 6.0 \times 2.2 / (2.2 + R_T)\) or equivalent ratio method. [1] Rearrange to find total resistance of the circuit (\(3.5\text{ k}\Omega\) or \(3474\ \Omega\)). [1] Calculate final thermistor resistance: \(1.3\text{ k}\Omega\) (allow \(1.27\text{ k}\Omega\) to \(1.3\text{ k}\Omega\)).

For (i): The cross-sectional area is \(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.48 \times 10^{-3}\text{ m})^2}{4} = 1.81 \times 10^{-7}\text{ m}^2\), which is approximately \(1.8 \times 10^{-7}\text{ m}^2\). For (ii): Tensile stress \(\sigma = \frac{F}{A} = \frac{35\text{ N}}{1.81 \times 10^{-7}\text{ m}^2} = 1.93 \times 10^8\text{ Pa}\) (or \(\text{N m}^{-2}\)). Tensile strain \(\varepsilon = \frac{\Delta L}{L} = \frac{4.8 \times 10^{-3}\text{ m}}{2.50\text{ m}} = 1.92 \times 10^{-3}\). For (iii): The Young modulus \(E = \frac{\sigma}{\varepsilon} = \frac{1.93 \times 10^8\text{ Pa}}{1.92 \times 10^{-3}} = 1.01 \times 10^{11}\text{ Pa} \approx 1.0 \times 10^{11}\text{ Pa}\).

(i) [1] Correctly apply \(A = \pi r^2\) or \(A = \pi d^2 / 4\) to show \(1.81 \times 10^{-7}\text{ m}^2\). (ii) [1] Calculate tensile stress = \(1.93 \times 10^8\text{ Pa}\) (accept \(1.9 \times 10^8\text{ Pa}\)). [1] Calculate tensile strain = \(1.92 \times 10^{-3}\) (accept \(1.9 \times 10^{-3}\)). (iii) [1] Use \(E = \text{stress} / \text{strain}\) or equivalent formula. [1] Calculate Young modulus = \(1.0 \times 10^{11}\text{ Pa}\) (accept \(1.01 \times 10^{11}\text{ Pa}\) or answers using the show-that value of \(1.8 \times 10^{-7}\text{ m}^2\) which yield \(1.02 \times 10^{11}\text{ Pa}\)).

For (i): The energy of a photon is given by \(E = \frac{h c}{\lambda}\). Assuming the electrical work done on each electron crossing the junction, \(e V_0\), is converted entirely into the energy of one photon, we have \(e V_0 = \frac{h c}{\lambda}\). For (ii): Rearranging the equation for the Planck constant gives \(h = \frac{e V_0 \lambda}{c}\). Substituting the values: \(h = \frac{1.60 \times 10^{-19}\text{ C} \times 1.90\text{ V} \times 635 \times 10^{-9}\text{ m}}{3.00 \times 10^8\text{ m s}^{-1}} = \frac{1.9304 \times 10^{-25}}{3.00 \times 10^8} = 6.43 \times 10^{-34}\text{ J s}\). This is approximately \(6.4 \times 10^{-34}\text{ J s}\). For (iii): A key systematic uncertainty is the difficulty in determining the exact threshold voltage by eye. The point at which the LED first glows is subjective and influenced by the background lighting of the room, which usually leads to an overestimation of the threshold voltage.

(i) [1] Correct equation: \(e V_0 = h c / \lambda\) or \(h = e V_0 \lambda / c\). (ii) [1] Substitute correct values with matching powers of ten (e.g., \(635 \times 10^{-9}\text{ m}\) and \(1.60 \times 10^{-19}\text{ C}\)). [1] Correct final calculated value of \(h\) to 2 or 3 sig figs: \(6.4 \times 10^{-34}\text{ J s}\) (or \(6.43 \times 10^{-34}\text{ J s}\)). (iii) [1] Any valid physical suggestion: difficulty in seeing the first light emission due to ambient room light / different visual thresholds of observers / voltage drop across internal resistance of the LED.

For (i): Use the equation of motion \(v^2 = u^2 + 2 a s\). Here, the initial velocity is \(u = 45\text{ m s}^{-1}\) upwards, the final velocity at maximum height is \(v = 0\), and the acceleration is \(a = -9.81\text{ m s}^{-2}\) (downwards under gravity). Substituting these values: \(0 = 45^2 + 2 (-9.81) s \Rightarrow 19.62 s = 2025 \Rightarrow s = 103.2\text{ m}\). The maximum height above the ground is the sum of the height at engine shutdown and the subsequent displacement: \(30\text{ m} + 103.2\text{ m} = 133.2\text{ m}\), which rounds to \(133\text{ m}\). For (ii): For the return motion from maximum height to the ground, the initial velocity is \(u = 0\), the displacement is \(s = -133.2\text{ m}\), and the acceleration is \(a = -9.81\text{ m s}^{-2}\). Using \(s = u t + \frac{1}{2} a t^2\): \(-133.2 = 0 + \frac{1}{2} (-9.81) t^2 \Rightarrow t^2 = \frac{133.2}{4.905} = 27.16\). Taking the square root gives \(t = 5.21\text{ s}\), which rounds to \(5.2\text{ s}\).

(i) [1] Use \(v^2 = u^2 + 2 a s\) with appropriate sign convention. [1] Calculate additional displacement \(s = 103\text{ m}\) (or \(103.2\text{ m}\)). [1] Add original \(30\text{ m}\) height to get maximum height = \(133\text{ m}\) (accept \(130\text{ m}\) if using 2 sig figs). (ii) [1] Use \(s = \frac{1}{2} g t^2\) (or another valid kinematic approach) with the value of maximum height from (i). [1] Obtain correct time \(t = 5.2\text{ s}\) (allow ecf from part (i); e.g., if height was 130m, \(t = 5.1\text{ s}\)).

(a) A correct circuit diagram must show:
- A series loop containing the DC power supply, the fixed resistor, and the thermistor.
- A voltmeter connected in parallel across the thermistor.
- Correct standard symbols: a rectangle for the fixed resistor, and a rectangle with a diagonal line (with a flat base) for the thermistor.

(b) Experimental details:
- Set up: Place the thermistor and a thermometer in a beaker of water with ice to get \(0\,^{\circ}\text{C}\). Heat the beaker using a Bunsen burner or electric hotplate to reach temperatures up to \(100\,^{\circ}\text{C}\).
- Techniques for reliability and accuracy:
1. Stir the water continuously before taking any readings to ensure a uniform temperature throughout the beaker.
2. Ensure the thermometer bulb is close to, but not touching, the thermistor or the beaker walls to measure the temperature actually experienced by the thermistor.

(c) Using the potential divider equation:
\(V_{\text{out}} = V_s \times \frac{R_T}{R_T + R}\)
\(2.40 = 6.00 \times \frac{R_T}{R_T + 1200}\)
\(0.40(R_T + 1200) = R_T\)
\(0.40 R_T + 480 = R_T\)
\(0.60 R_T = 480 \implies R_T = 800\,\Omega\) (or \(0.800\text{ k}\Omega\)).

(d) In an NTC thermistor, as temperature increases, more charge carriers are released into the conduction band, causing a rapid decrease in resistance. However, the rate of change of resistance with temperature (\(\frac{\Delta R}{\Delta T}\)) decreases at higher temperatures. Consequently, the output voltage change per unit change in temperature becomes smaller, resulting in reduced sensitivity of the sensor circuit at higher temperatures.

**(a) Circuit Diagram [3 marks]**
- **M1:** Series loop with power supply, fixed resistor, and thermistor. (1)
- **M2:** Voltmeter connected in parallel across the thermistor. (1)
- **M3:** Correct circuit symbols used throughout. (1)

**(b) Experimental Procedure [3 marks]**
- **M1:** Method to vary temperature: thermistor in water, heated gradually with Bunsen burner/heater. (1)
- **M2:** Continuous stirring of water to ensure uniform temperature. (1)
- **M3:** Thermometer placed close to thermistor, ensuring readings are taken only after a stable state is reached. (1)

**(c) Calculation [2 marks]**
- **M1:** Employs the potential divider equation or calculates current first (\(I = \frac{6.00 - 2.40}{1200} = 3.00\text{ mA}\)). (1)
- **A1:** Correctly evaluates \(R_T = 800\,\Omega\) (or \(0.800\text{ k}\Omega\)). (1)

**(d) Sensitivity Explanation [2 marks]**
- **M1:** Mentions that the resistance of the NTC thermistor decreases at a decreasing rate as temperature rises. (1)
- **A1:** Concludes that the change in output voltage per unit temperature (sensitivity) decreases at high temperatures. (1)

(a) The four measurements are:
1. Original length of the wire, \(L\), using a metre ruler or tape measure.
2. Diameter of the wire, \(d\), using a micrometer screw gauge.
3. Added load/mass, \(m\), using standard masses (or mass balance to weigh them).
4. Extension, \(\Delta L\), using a vernier scale arrangement, traveling microscope, or high-resolution ruler.

(b) To minimize uncertainty in the diameter:
- Take diameter measurements at several different positions along the wire's length.
- Take measurements at different perpendicular orientations at each position to account for any non-circularity.
- Check and record any zero error on the micrometer screw gauge before measuring and correct the final mean value.

(c) Young modulus calculation:
- Cross-sectional area:
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.1341 \times 10^{-7}\text{ m}^2\)
- Tension Force:
\(F = m g = 2.50 \times 9.81 = 24.525\text{ N}\)
- Stress:
\(\sigma = \frac{F}{A} = \frac{24.525}{1.1341 \times 10^{-7}} = 2.1625 \times 10^8\text{ Pa}\)
- Strain:
\(\epsilon = \frac{\Delta L}{L} = \frac{3.60 \times 10^{-3}}{2.20} = 1.6364 \times 10^{-3}\)
- Young Modulus:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{2.1625 \times 10^8}{1.6364 \times 10^{-3}} = 1.3215 \times 10^{11}\text{ Pa}\).
To 2 significant figures, \(E = 1.3 \times 10^{11}\text{ Pa}\).

(d) Since extension \(\Delta L = \frac{F L}{A E}\), a larger initial length \(L\) and a smaller cross-sectional area \(A\) (thin wire) will significantly increase the extension \(\Delta L\) for a given tension force. This larger extension is much easier to measure with standard laboratory instruments, thereby greatly reducing the percentage uncertainty in \(\Delta L\).

**(a) Measurements and Instruments [3 marks]**
- **M1:** Original length with metre ruler AND diameter with micrometer. (1)
- **M2:** Mass/Force with mass hanger/balance AND extension with vernier scale/traveling microscope. (1)
- **A1:** All four physical measurements matched to correct instruments. (1)

**(b) Minimizing Uncertainty [2 marks]**
- **M1:** Take multiple readings along the length and at perpendicular angles. (1)
- **A1:** Calculate a mean diameter AND correct for the micrometer's zero error. (1)

**(c) Calculation [3 marks]**
- **M1:** Correct calculation of area \(A = 1.13 \times 10^{-7}\text{ m}^2\). (1)
- **M2:** Correct calculation of force \(F = 24.5\text{ N}\) and correct substitution of values into \(E = \frac{F L}{A \Delta L}\). (1)
- **A1:** Calculates \(E = 1.3 \times 10^{11}\text{ Pa}\) (or \(1.32 \times 10^{11}\text{ Pa}\)). (1)

**(d) Explanation [2 marks]**
- **M1:** Identifies that a larger \(L\) and smaller \(A\) maximizes the extension \(\Delta L\). (1)
- **A1:** Explains that a larger extension decreases the percentage uncertainty of the measurement. (1)

(a) Experimental setup and alignment:
- Direct the laser pointer normal to the plane of the diffraction grating, with a screen placed a distance \(D\) away.
- Ensure the grating is perpendicular to the laser beam by checking that the diffracted pattern is symmetrical on both sides of the central maximum on the screen, and that the spots lie along a single horizontal line.
- Measure the grating-to-screen distance \(D\) using a metre ruler.

(b) Slit spacing calculation:
- The number of lines per metre \(N = 400\text{ lines/mm} = 400,000\text{ lines/m}\).
- Slit spacing \(d = \frac{1}{N} = \frac{1}{400,000} = 2.50 \times 10^{-6}\text{ m}\).

(c) Angle and Wavelength:
- Angle \(\theta\):
\(\tan\theta = \frac{y}{D} = \frac{0.825\text{ m}}{1.50\text{ m}} = 0.55\)
\(\theta = \arctan(0.55) = 28.81^{\circ}\) (or \(28.8^{\circ}\))
- Wavelength \(\lambda\):
Using \(d \sin\theta = n \lambda\) with \(n = 2\):
\(\lambda = \frac{d \sin\theta}{2} = \frac{2.50 \times 10^{-6} \times \sin(28.81^{\circ})}{2} = 6.024 \times 10^{-7}\text{ m}\)
\(\lambda \approx 6.02 \times 10^{-7}\text{ m}\) (or \(602\text{ nm}\)).

(d) Percentage uncertainty:
Method 1: Max/Min Method (preferred as \(\theta\) is large):
- Max wavelength \(\lambda_{\text{max}}\) using \(y_{\text{max}} = 0.830\text{ m}\) and \(D_{\text{min}} = 1.49\text{ m}\):
\(\tan\theta_{\text{max}} = \frac{0.830}{1.49} = 0.55705 \implies \theta_{\text{max}} = 29.117^{\circ}\)
\(\lambda_{\text{max}} = \frac{2.50 \times 10^{-6} \times \sin(29.117^{\circ})}{2} = 6.082 \times 10^{-7}\text{ m}\)
- Min wavelength \(\lambda_{\text{min}}\) using \(y_{\text{min}} = 0.820\text{ m}\) and \(D_{\text{max}} = 1.51\text{ m}\):
\(\tan\theta_{\text{min}} = \frac{0.820}{1.51} = 0.54305 \implies \theta_{\text{min}} = 28.504^{\circ}\)
\(\lambda_{\text{min}} = \frac{2.50 \times 10^{-6} \times \sin(28.504^{\circ})}{2} = 5.965 \times 10^{-7}\text{ m}\)
- Absolute uncertainty in \(\lambda\):
\(\Delta \lambda = \frac{\lambda_{\text{max}} - \lambda_{\text{min}}}{2} = \frac{(6.082 - 5.965) \times 10^{-7}}{2} \approx 5.85 \times 10^{-9}\text{ m}\)
- Percentage uncertainty:
\(\% \text{ uncertainty} = \frac{5.85 \times 10^{-9}}{6.02 \times 10^{-7}} \times 100\% \approx 1.0\%\).

Method 2: Summing percentage uncertainties (small-angle approximation):
- \(\% \Delta y = \frac{0.5}{82.5} \times 100\% = 0.61\%\)
- \(\% \Delta D = \frac{0.01}{1.50} \times 100\% = 0.67\%\)
- Total percentage uncertainty \(\approx 0.61\% + 0.67\% = 1.28\% \approx 1.3\%\).

**(a) Alignment and Setup [3 marks]**
- **M1:** Laser aligned through grating to screen, with measured grating-screen distance \(D\). (1)
- **M2:** Grating set perpendicular to beam. (1)
- **A1:** Verification of perpendicularity by ensuring symmetrical pattern on either side of central maximum. (1)

**(b) Slit Spacing [2 marks]**
- **M1:** Uses \(d = \frac{1}{N}\) with consistent units (e.g., converting lines/mm to lines/m). (1)
- **A1:** Correct calculation showing \(d = 2.50 \times 10^{-6}\text{ m}\). (1)

**(c) Angle and Wavelength [3 marks]**
- **M1:** Uses trigonometry to find \(\theta = 28.8^{\circ}\) (or \(28.81^{\circ}\)). (1)
- **M2:** Substitutes values correctly into the grating equation \(n\lambda = d\sin\theta\). (1)
- **A1:** Obtains \ \lambda = 6.02 \times 10^{-7}\text{ m}\) (or \(602\text{ nm}\)). (1)

**(d) Percentage Uncertainty [2 marks]**
- **M1:** Attempt to calculate uncertainty using max/min bounds on \(\lambda\) OR summing the percentage uncertainties of \(y\) and \(D\). (1)
- **A1:** Arrives at a consistent percentage uncertainty: \(1.0\%\) (for max/min method) or \(1.3\%\) (for sum of percentages). (1)

(a) The sampling frequency is:
\(f_{\text{sample}} = 2.5 \times 8.0\text{ kHz} = 20\text{ kHz}\)

(b) The bit rate is the sampling frequency multiplied by the number of bits per sample:
\(\text{Bit rate} = 20,000\text{ samples s}^{-1} \times 12\text{ bits sample}^{-1} = 240,000\text{ bits s}^{-1} = 240\text{ kbps}\)

(c) According to the Nyquist criterion, the minimum sampling rate required to avoid signal distortion is twice the highest frequency component present in the analogue signal:
\(f_{\text{Nyquist}} = 2 \times 8.0\text{ kHz} = 16\text{ kHz}\)
If the sampling rate is reduced to \(12\text{ kHz}\), which is below this minimum threshold, high-frequency components will be folded back into lower frequencies. This causes aliasing, leading to irreversible distortion of the reconstructed audio signal.

(a)
- 1 mark for correct calculation: \(2.5 \times 8.0\text{ kHz} = 20\text{ kHz}\).
- 0.2 marks for stating correct unit (kHz or Hz).

(b)
- 1 mark for method: multiplying the calculated sampling rate by 12.
- 1 mark for correct value with unit: \(240\text{ kbps}\) (accept \(2.4 \times 10^5\text{ bps}\)).

(c)
- 1 mark for identifying that the Nyquist frequency is \(16\text{ kHz}\) and that \(12\text{ kHz}\) violates this limit.
- 1 mark for explaining that this leads to aliasing / overlap of frequency spectra / high-frequency distortion.

(a) At \(10^\circ\text{C}\), the thermistor resistance \(R_{\text{th}} = 3.0\text{ k}\Omega\).
Using the potential divider equation:
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{th}}}{R_{\text{th}} + R_{\text{fixed}}}\)
\(V_{\text{out}} = 9.0\text{ V} \times \frac{3.0\text{ k}\Omega}{3.0\text{ k}\Omega + 1.5\text{ k}\Omega} = 9.0 \times \frac{3.0}{4.5} = 6.0\text{ V}\)

(b) At \(40^\circ\text{C}\), the thermistor resistance \(R_{\text{th}} = 0.60\text{ k}\Omega\).
\(V_{\text{out}} = 9.0\text{ V} \times \frac{0.60\text{ k}\Omega}{0.60\text{ k}\Omega + 1.5\text{ k}\Omega} = 9.0 \times \frac{0.60}{2.10} \approx 2.57\text{ V}\) (or \(2.6\text{ V}\))

(c) As temperature increases, the resistance of the NTC thermistor decreases non-linearly, with the change in resistance per unit temperature becoming smaller. Consequently, the change in output voltage per degree change in temperature decreases, meaning the circuit sensitivity decreases at higher temperatures.

(a)
- 1 mark for correct substitution into potential divider equation.
- 1 mark for correct output voltage: \(6.0\text{ V}\).

(b)
- 1 mark for correct substitution with new resistance values.
- 1 mark for correct calculation: \(2.57\text{ V}\) (accept \(2.6\text{ V}\)).

(c)
- 1 mark for stating that the sensitivity decreases as temperature rises.
- 0.2 marks for explanation relating to the non-linear decrease of NTC thermistor resistance (less change in resistance per unit temperature rise).

(a) The force acting on the wire is the weight of the sign:
\(F = mg = 45\text{ kg} \times 9.8\text{ m s}^{-2} = 441\text{ N}\)
Tensile stress \(\sigma\) is:
\(\sigma = \frac{F}{A} = \frac{441\text{ N}}{1.5 \times 10^{-6}\text{ m}^2} = 2.94 \times 10^8\text{ Pa}\) (or \(2.9 \times 10^8\text{ Pa}\) to 2 s.f.)

(b) Using the definition of Young modulus \(E = \frac{\text{stress}}{\text{strain}}\):
\(\text{Strain } \varepsilon = \frac{\sigma}{E} = \frac{2.94 \times 10^8\text{ Pa}}{2.0 \times 10^{11}\text{ Pa}} = 1.47 \times 10^{-3}\) (or \(1.5 \times 10^{-3}\) to 2 s.f.)

(c) Using the definition of strain \(\varepsilon = \frac{\Delta L}{L}\):
\(\Delta L = \varepsilon \times L = 1.47 \times 10^{-3} \times 3.2\text{ m} = 4.7 \times 10^{-3}\text{ m}\) (or \(4.7\text{ mm}\))

(a)
- 1 mark for correct determination of tension: \(F = 441\text{ N}\) (or \(441.5\text{ N}\) if using \(9.81\)).
- 1 mark for correct stress calculation: \(2.9 \times 10^8\text{ Pa}\) (accept \(2.94 \times 10^8\text{ Pa}\)).

(b)
- 1 mark for correct method using Young modulus: \(\varepsilon = \frac{\text{stress}}{E}\).
- 0.2 marks for calculating strain correctly as \(1.5 \times 10^{-3}\) (or \(1.47 \times 10^{-3}\)) without any units.

(c)
- 1 mark for correct calculation method: multiplying strain by original length.
- 1 mark for correct final value with appropriate unit: \(4.7 \times 10^{-3}\text{ m}\) (accept \(4.7\text{ mm}\)).

(a) The components of the initial velocity are:
- Vertical component: \(u_y = 12 \sin(38^\circ) = 7.39\text{ m s}^{-1} \approx 7.4\text{ m s}^{-1}\)
- Horizontal component: \(u_x = 12 \cos(38^\circ) = 9.46\text{ m s}^{-1} \approx 9.5\text{ m s}^{-1}\)

(b) At the maximum height, the vertical component of velocity is \(v_y = 0\).
Using the equation of motion:
\(v_y^2 = u_y^2 + 2as_y\)
\(0 = (7.39)^2 - 2 \times 9.8 \times s_y\)
\(s_y = \frac{54.6}{19.6} = 2.79\text{ m}\)
The maximum height above the ground is the initial launch height plus this vertical displacement:
\(h_{\text{max}} = 1.8\text{ m} + 2.79\text{ m} = 4.59\text{ m} \approx 4.6\text{ m}\)

(c) Air resistance always opposes the velocity vector. On the ascent, air resistance has a downward vertical component, which increases the total downward acceleration (gravity + drag). This causes the shot to lose vertical speed more rapidly, thereby reducing the maximum height reached.

(a)
- 1 mark for correct vertical component calculation: \(7.4\text{ m s}^{-1}\).
- 1 mark for correct horizontal component calculation: \(9.5\text{ m s}^{-1}\).

(b)
- 1 mark for calculation of vertical height gain from the launch point: \(2.79\text{ m}\) (accept \(2.8\text{ m}\)).
- 1 mark for adding the initial launch height \(1.8\text{ m}\) to show \(4.6\text{ m}\) (or \(4.59\text{ m}\)).

(c)
- 1 mark for stating that air resistance opposes the motion and increases the downward acceleration on the way up.
- 0.2 marks for stating that the maximum height reached is reduced.

(a) Using the double-slit equation:
\(x = \frac{\lambda D}{a}\)
where:
- \(\lambda = 633 \times 10^{-9}\text{ m}\)
- \(D = 2.4\text{ m}\)
- \(a = 0.25 \times 10^{-3}\text{ m}\)

\(x = \frac{633 \times 10^{-9}\text{ m} \times 2.4\text{ m}}{0.25 \times 10^{-3}\text{ m}} = 6.08 \times 10^{-3}\text{ m} \approx 6.1\text{ mm}\)

(b) Since \(x \propto \lambda\) (fringe separation is directly proportional to wavelength for a constant slit-to-screen distance and slit separation), a shorter wavelength will result in smaller fringe spacing. Since green light (\(532\text{ nm}\)) has a shorter wavelength than red light (\(633\text{ nm}\)), the fringe separation will decrease.

(c) If one slit is covered, light can no longer interfere from two coherent sources. The fine, closely spaced double-slit interference fringes vanish. Instead, you observe the single-slit diffraction pattern of the remaining open slit, which features a wide central maximum that is much broader and has less peak intensity than the original pattern.

(a)
- 1 mark for correct substitution into \(x = \frac{\lambda D}{a}\) (with consistent metric units).
- 1 mark for correct calculation: \(6.1\text{ mm}\) (or \(6.1 \times 10^{-3}\text{ m}\); accept \(6.08\text{ mm}\)).

(b)
- 1 mark for noting that green light has a shorter wavelength than red light.
- 1 mark for linking shorter wavelength to smaller fringe spacing using \(x \propto \lambda\).

(c)
- 1 mark for stating that double-slit interference fringes disappear.
- 0.2 marks for describing the emergence of a single-slit diffraction pattern with a broader central maximum.

Part (a): Thermistors are made of semiconducting materials. As the temperature of the semiconductor increases, the atoms/lattice ions vibrate with more thermal energy. This thermal energy is sufficient to liberate extra charge carriers (conduction electrons) from the valence band to the conduction band, increasing the charge carrier density, \(n\). Although the increased lattice vibrations cause more collisions/scattering of these carriers, the massive increase in \(n\) dominates. From the relation \(I = nAve\), a higher \(n\) increases the current for a given voltage, meaning the electrical resistance \(R\) decreases significantly.

Part (b)(i): The total resistance of the series circuit is \(R_{\text{total}} = R + R_T = 1.2\text{ k}\Omega + 2.8\text{ k}\Omega = 4.0\text{ k}\Omega\). The output voltage across the fixed resistor is given by the potential divider formula: \(V_{\text{out}} = V_{\text{in}} \times \frac{R}{R_{\text{total}}} = 6.0\text{ V} \times \frac{1.2\text{ k}\Omega}{4.0\text{ k}\Omega} = 1.80\text{ V}\).

Part (b)(ii): At \(35^\circ\text{C}\), the output voltage is \(3.6\text{ V}\). Using the potential divider formula again: \(V_{\text{out}} = V_{\text{in}} \times \frac{R}{R + R_T}\). Rearranging this gives: \(3.6 = 6.0 \times \frac{1.2}{1.2 + R_T} \implies 0.60 = \frac{1.2}{1.2 + R_T}\). Therefore, \(1.2 + R_T = \frac{1.2}{0.60} = 2.0\text{ k}\Omega\). This gives \(R_T = 2.0 - 1.2 = 0.8\text{ k}\Omega\) (or \(800\ \Omega\)).

Part (c): Sensitivity is defined as the change in output voltage per unit change in temperature, \(\frac{\Delta V_{\text{out}}}{\Delta \theta}\). Because the resistance of an NTC thermistor decreases exponentially with increasing temperature, the rate of change of resistance with respect to temperature, \(\frac{\Delta R_T}{\Delta \theta}\), is much greater at lower temperatures (around \(15^\circ\text{C}\)) than at higher temperatures (around \(35^\circ\text{C}\)). Consequently, the circuit's sensitivity decreases as the temperature rises. To optimize sensitivity at lower temperatures (such as \(5^\circ\text{C}\)), where the thermistor's resistance \(R_T\) is very high, the value of the fixed resistor \(R\) should be increased. This ensures that the ratio \(R / (R + R_T)\) remains in a region where small changes in \(R_T\) produce a significant change in \(V_{\text{out}}\), ideally when \(R \approx R_T\).

Part (a) [4 marks]:
1. Mention that the thermistor is a semiconductor. [1]
2. State that increasing temperature provides thermal energy to release charge carriers (electrons). [1]
3. State that this increases the charge carrier density \(n\). [1]
4. Explain that this increase in \(n\) far outweighs any increase in carrier scattering due to lattice vibrations, so resistance decreases. [1]

Part (b)(i) [3 marks]:
1. Calculate total resistance \(R_{\text{total}} = 4.0\text{ k}\Omega\). [1]
2. Correct substitution into potential divider equation. [1]
3. Correct final answer of \(1.80\text{ V}\) (must have units and be to a consistent number of significant figures, 2 or 3 s.f.). [1]

Part (b)(ii) [3.5 marks]:
1. Correctly set up the formula with \(V_{\text{out}} = 3.6\text{ V}\). [1]
2. Correct algebraic rearrangement to isolate \(R_T\). [1.5]
3. Correct calculation of \(R_T = 0.8\text{ k}\Omega\) or \(800\ \Omega\) with units. [1]

Part (c) [4 marks]:
1. Define sensitivity in this context as change in voltage per unit change in temperature. [1]
2. Explain that thermistor resistance changes more rapidly at lower temperatures, so sensitivity is higher at \(15^\circ\text{C}\) than at \(35^\circ\text{C}\). [1]
3. Suggest increasing the value of the fixed resistor \(R\) for low temperature operation. [1]
4. Explain that this matches the higher thermistor resistance at low temperatures to keep the potential divider output in its most responsive range. [1]

Part (a): To measure the diameter accurately, the student should use a micrometer screw gauge or digital caliper. Readings of the diameter should be taken at multiple different positions along the length of the polymer line to check for uniformity. At each position, measurements should be made in at least two perpendicular directions to check for any non-circularity. The student must check for and record any zero error on the micrometer before starting, and subtract this from all measurements. Finally, the mean of these measurements should be calculated to minimize random uncertainty.

Part (b)(i): The radius \(r\) of the polymer line is \(r = d / 2 = 0.80\text{ mm} / 2 = 0.40\text{ mm} = 4.0 \times 10^{-4}\text{ m}\). The cross-sectional area \(A\) is calculated using the formula: \(A = \pi r^2 = \pi \times (4.0 \times 10^{-4}\text{ m})^2 = 5.027 \times 10^{-7}\text{ m}^2\). This is approximately \(5.0 \times 10^{-7}\text{ m}^2\) (as shown).

Part (b)(ii): The Young modulus \(E\) is defined as the ratio of tensile stress to tensile strain in the linear elastic region: \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{x/L} = \frac{F L}{A x}\). Given: \(F = 45\text{ N}\), \(L = 2.50\text{ m}\), \(x = 15\text{ mm} = 1.5 \times 10^{-2}\text{ m}\), and the calculated area \(A = 5.027 \times 10^{-7}\text{ m}^2\). Substituting these values into the expression: \(E = \frac{45 \times 2.50}{5.027 \times 10^{-7} \times 1.5 \times 10^{-2}} = \frac{112.5}{7.541 \times 10^{-9}} = 1.492 \times 10^{10}\text{ Pa}\). To 2 significant figures, \(E = 1.5 \times 10^{10}\text{ Pa}\) (or \(15\text{ GPa}\)). (Using \(A = 5.0 \times 10^{-7}\text{ m}^2\) yields exactly \(1.50 \times 10^{10}\text{ Pa}\)).

Part (c)(i): In the elastic region, the stretching forces cause the coiled polymer chains to slightly uncoil or align, and the intermolecular bonds (such as weak Van der Waals forces) between chains are stretched but not broken. When the load is removed, these intermolecular forces pull the chains back to their initial configuration, restoring original length. In the plastic region, the stress is high enough to break the intermolecular bonds between chains, allowing the polymer chains to slide past one another into new, permanent relative positions. When the stress is removed, the chains remain in these new slipped configurations, causing permanent extension.

Part (c)(ii): Cross-linking involves the formation of strong, covalent chemical bonds between adjacent polymer chains. If the polymer is highly cross-linked, these covalent bonds act as anchors that prevent the polymer chains from sliding past each other. This drastically increases the stiffness (higher Young modulus) and hardness of the material, making it more resistant to deformation. However, it also eliminates the possibility of plastic deformation; instead of stretching plastically, the material becomes brittle and will fracture suddenly once the stress exceeds the strength of the covalent cross-links.

Part (a) [3 marks]:
1. State the use of a micrometer screw gauge or digital caliper. [1]
2. Explain the need to take measurements at multiple positions and orientations along the line. [1]
3. Describe checking for zero error and taking a mean value to reduce uncertainty. [1]

Part (b)(i) [1.5 marks]:
1. Calculate radius as \(0.40\text{ mm}\) or \(4.0 \times 10^{-4}\text{ m}\). [0.5]
2. Correct substitution into \(A = \pi r^2\) to yield a value of \(5.03 \times 10^{-7}\text{ m}^2\) (showing the transition to the approximation). [1]

Part (b)(ii) [4 marks]:
1. State formula for Young modulus \(E = \frac{FL}{Ax}\) or define Stress and Strain. [1]
2. Convert extension unit to meters (\(0.015\text{ m}\)). [1]
3. Correct substitution of all values including a valid cross-sectional area. [1]
4. Correct final calculated value of \(1.5 \times 10^{10}\text{ Pa}\) (or \(1.49 \times 10^{10}\text{ Pa}\) or \(15\text{ GPa}\)) with appropriate units (\(\text{Pa}\) or \(\text{N m}^{-2}\)). [1]

Part (c)(i) [3 marks]:
1. Explain elastic deformation as involving the stretching of intermolecular bonds/temporary uncoiling of chains that return to original positions when unloaded. [1.5]
2. Explain plastic deformation as involving the breaking of intermolecular bonds and the permanent sliding/slipping of polymer chains past one another. [1.5]

Part (c)(ii) [3 marks]:
1. Define cross-linking as covalent bonds between polymer chains. [1]
2. Explain that cross-links prevent chains from sliding past each other, which increases stiffness / Young modulus / strength. [1]
3. State that this reduces or eliminates plastic deformation, making the material more brittle. [1]

### 1. Experimental Setup and Diagram
* **Setup:** Clamp one end of a long length (typically 1.5 m to 2.0 m) of nylon line firmly between two blocks of wood held in a G-clamp on a laboratory bench. Pass the other end over a low-friction pulley clamped at the edge of the bench, and attach a mass hanger to this end.
* **Reference Marker:** Attach a small tape marker to the nylon line near the pulley. Place a metre rule flat on the bench directly underneath the marker to measure its displacement.

### 2. Measurements and Minimising Uncertainty
* **Original Length \(L\):** Measure the distance from the inside edge of the wooden blocks (fixed end) to the marker using a metre rule. To minimise percentage uncertainty, use a long original length of line (e.g., \(> 1.5\text{ m}\)).
* **Diameter \(d\):** Use a digital micrometer screw gauge to measure the diameter of the line. Measure at at least 5 different positions along the line and at different orientations (perpendicular angles) at each position to account for non-circular cross-section. Calculate the mean diameter \(d\).
* **Extension \(x\):** Add slotted masses in 100 g increments up to a maximum (e.g., 1.0 kg). For each load, read the position of the marker on the metre rule. To avoid parallax error, look vertically down over the marker onto the scale. Subtract the initial position (under a small initial load to keep the wire taut) from each subsequent reading to find the extension \(x\).
* **Systematic Error Control:** Measuring the extension of the marker rather than the movement of the mass hanger avoids systematic errors caused by the line slipping in the clamp or the knot stretching.

### 3. Data Analysis
* Calculate the cross-sectional area of the nylon line: \(A = \frac{\pi d^2}{4}\).
* Plot a graph of Force \(F = mg\) (on the y-axis) against Extension \(x\) (on the x-axis).
* Identify the linear elastic region of the graph and draw a straight line of best fit through the origin for this section. Measure the gradient \(m = \frac{\Delta F}{\Delta x}\) of this linear region.
* Since the Young modulus is defined as \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\), the Young modulus is calculated using:
\(E = \text{gradient} \times \frac{L}{A}\).
* Alternatively, calculate stress \(\sigma = \frac{F}{A}\) and strain \(\varepsilon = \frac{x}{L}\), plot stress vs strain, and the gradient of the linear region directly equals \(E\).

### 4. Safety Precautions
* **Hazard:** The nylon line stores elastic strain energy under tension. If it snaps, it can whip back rapidly and cause serious eye injury.
* **Precaution:** Everyone in the vicinity must wear safety goggles/eyeshields. Place a soft padded box or foam directly under the hanging masses to cushion their fall and protect feet/flooring when the line snaps.

**Level 3 (11-15 marks):**
- A comprehensive description with a detailed, feasible experimental setup (diagram included).
- Clearly explains how to measure original length, diameter (with multiple readings and mean), and extension (using a marker to avoid slip/knot errors).
- Provides a complete graphical analysis method to find \(E\) (plotting \(F\) vs \(x\) or stress vs strain, identifying the gradient, and giving the correct formula relating the gradient to \(E\)).
- Identifies the major hazard of the snapping wire and specifies wearing safety goggles and using a catch box for falling masses.

**Level 2 (6-10 marks):**
- A functional experimental setup described (may lack some detail in diagram or use a less reliable method for extension).
- Explains measurements of length, diameter, and extension, but may omit details on how to reduce uncertainties (e.g., repeating diameter measurements, avoiding parallax, or clamp slippage).
- Suggests a graphical analysis method but may have minor algebraic errors in calculating \(E\) from the gradient.
- Mentions safety but with less specific detail on hazards/precautions.

**Level 1 (1-5 marks):**
- Identifies basic apparatus and some relevant measurements (e.g., length, extension).
- Mentions a basic formula for Young modulus but fails to describe a valid graphical analysis method.
- Safety precautions are weak or generic (e.g., 'be careful').

**Specific Mark Allocation Points (indicative of what examiners look for):**
- **Setup (Max 3 marks):**
- Labeled diagram of horizontal/vertical setup with fixed clamp and tension applied by loads [1]
- Marker on wire with fixed scale to measure extension directly [1]
- Use of G-clamp/wooden blocks to prevent slippage [1]
- **Measurements (Max 5 marks):**
- Measure original length \(L\) from clamp to marker with a metre rule [1]
- Measure diameter \(d\) with micrometer at multiple positions and orientations, calculate mean [1]
- Explain how extension is calculated: position under load minus initial position under preload [1]
- Reference to reducing parallax error (viewing marker perpendicularly) or using a travelling microscope [1]
- Explain that using a marker eliminates errors from clamp slippage/knot tightening [1]
- **Analysis (Max 4 marks):**
- Calculate cross-sectional area \(A = \pi d^2 / 4\) [1]
- Plot graph of Force vs Extension or Stress vs Strain [1]
- Measure the gradient of the linear region [1]
- Correctly relate gradient to Young modulus (e.g., \(E = \text{gradient} \times L / A\) for \(F\) vs \(x\) graph) [1]
- **Safety and Reliability (Max 3 marks):**
- Wear safety goggles due to risk of wire snapping under tension [1]
- Place a box/cushion under the masses to protect feet/floor [1]
- Explain why a long wire is used (to increase absolute extension and reduce percentage uncertainty) [1]

### 1. Experimental Setup and Diagram
* **Setup:** Clamp one end of a long length (typically 1.5 m to 2.0 m) of nylon line firmly between two blocks of wood held in a G-clamp on a laboratory bench. Pass the other end over a low-friction pulley clamped at the edge of the bench, and attach a mass hanger to this end.
* **Reference Marker:** Attach a small tape marker to the nylon line near the pulley. Place a metre rule flat on the bench directly underneath the marker to measure its displacement.

### 2. Measurements and Minimising Uncertainty
* **Original Length \(L\):** Measure the distance from the inside edge of the wooden blocks (fixed end) to the marker using a metre rule. To minimise percentage uncertainty, use a long original length of line (e.g., \(> 1.5\text{ m}\)).
* **Diameter \(d\):** Use a digital micrometer screw gauge to measure the diameter of the line. Measure at at least 5 different positions along the line and at different orientations (perpendicular angles) at each position to account for non-circular cross-section. Calculate the mean diameter \(d\).
* **Extension \(x\):** Add slotted masses in 100 g increments up to a maximum (e.g., 1.0 kg). For each load, read the position of the marker on the metre rule. To avoid parallax error, look vertically down over the marker onto the scale. Subtract the initial position (under a small initial load to keep the wire taut) from each subsequent reading to find the extension \(x\).
* **Systematic Error Control:** Measuring the extension of the marker rather than the movement of the mass hanger avoids systematic errors caused by the line slipping in the clamp or the knot stretching.

### 3. Data Analysis
* Calculate the cross-sectional area of the nylon line: \(A = \frac{\pi d^2}{4}\).
* Plot a graph of Force \(F = mg\) (on the y-axis) against Extension \(x\) (on the x-axis).
* Identify the linear elastic region of the graph and draw a straight line of best fit through the origin for this section. Measure the gradient \(m = \frac{\Delta F}{\Delta x}\) of this linear region.
* Since the Young modulus is defined as \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\), the Young modulus is calculated using:
\(E = \text{gradient} \times \frac{L}{A}\).
* Alternatively, calculate stress \(\sigma = \frac{F}{A}\) and strain \(\varepsilon = \frac{x}{L}\), plot stress vs strain, and the gradient of the linear region directly equals \(E\).

### 4. Safety Precautions
* **Hazard:** The nylon line stores elastic strain energy under tension. If it snaps, it can whip back rapidly and cause serious eye injury.
* **Precaution:** Everyone in the vicinity must wear safety goggles/eyeshields. Place a soft padded box or foam directly under the hanging masses to cushion their fall and protect feet/flooring when the line snaps.

**Level 3 (11-15 marks):**
- A comprehensive description with a detailed, feasible experimental setup (diagram included).
- Clearly explains how to measure original length, diameter (with multiple readings and mean), and extension (using a marker to avoid slip/knot errors).
- Provides a complete graphical analysis method to find \(E\) (plotting \(F\) vs \(x\) or stress vs strain, identifying the gradient, and giving the correct formula relating the gradient to \(E\)).
- Identifies the major hazard of the snapping wire and specifies wearing safety goggles and using a catch box for falling masses.

**Level 2 (6-10 marks):**
- A functional experimental setup described (may lack some detail in diagram or use a less reliable method for extension).
- Explains measurements of length, diameter, and extension, but may omit details on how to reduce uncertainties (e.g., repeating diameter measurements, avoiding parallax, or clamp slippage).
- Suggests a graphical analysis method but may have minor algebraic errors in calculating \(E\) from the gradient.
- Mentions safety but with less specific detail on hazards/precautions.

**Level 1 (1-5 marks):**
- Identifies basic apparatus and some relevant measurements (e.g., length, extension).
- Mentions a basic formula for Young modulus but fails to describe a valid graphical analysis method.
- Safety precautions are weak or generic (e.g., 'be careful').

**Specific Mark Allocation Points (indicative of what examiners look for):**
- **Setup (Max 3 marks):**
- Labeled diagram of horizontal/vertical setup with fixed clamp and tension applied by loads [1]
- Marker on wire with fixed scale to measure extension directly [1]
- Use of G-clamp/wooden blocks to prevent slippage [1]
- **Measurements (Max 5 marks):**
- Measure original length \(L\) from clamp to marker with a metre rule [1]
- Measure diameter \(d\) with micrometer at multiple positions and orientations, calculate mean [1]
- Explain how extension is calculated: position under load minus initial position under preload [1]
- Reference to reducing parallax error (viewing marker perpendicularly) or using a travelling microscope [1]
- Explain that using a marker eliminates errors from clamp slippage/knot tightening [1]
- **Analysis (Max 4 marks):**
- Calculate cross-sectional area \(A = \pi d^2 / 4\) [1]
- Plot graph of Force vs Extension or Stress vs Strain [1]
- Measure the gradient of the linear region [1]
- Correctly relate gradient to Young modulus (e.g., \(E = \text{gradient} \times L / A\) for \(F\) vs \(x\) graph) [1]
- **Safety and Reliability (Max 3 marks):**
- Wear safety goggles due to risk of wire snapping under tension [1]
- Place a box/cushion under the masses to protect feet/floor [1]
- Explain why a long wire is used (to increase absolute extension and reduce percentage uncertainty) [1]