2024 OCR AS Level Physics B (Advancing Physics) - H157 Practice Paper with Answers

The potential divider formula is given by: \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{thermistor}}}{R + R_{\text{thermistor}}}\).

Substituting the given values:
\(4.0 = 6.0 \times \frac{R_{\text{thermistor}}}{120 + R_{\text{thermistor}}}\)

Divide both sides by 2.0:
\(2.0 = 3.0 \times \frac{R_{\text{thermistor}}}{120 + R_{\text{thermistor}}}\)

Multiply out the denominator:
\(2.0(120 + R_{\text{thermistor}}) = 3.0 R_{\text{thermistor}}\)
\(240 + 2.0 R_{\text{thermistor}} = 3.0 R_{\text{thermistor}}\)
\(R_{\text{thermistor}} = 240\text{ }\Omega\).

[1 mark] - Correct choice C is selected.

First, calculate the total number of pixels in one image:
\(\text{Pixels} = 2048 \times 2048 = 4.194 \times 10^6\text{ pixels}\)

For 10 images, the total number of pixels is:
\(10 \times 4.194 \times 10^6 = 4.194 \times 10^7\text{ pixels}\)

Each pixel requires 12 bits of data:
\(\text{Total bits} = 4.194 \times 10^7 \times 12\text{ bits} = 5.033 \times 10^8\text{ bits}\)

Convert bits to bytes (8 bits = 1 byte):
\(\text{Total bytes} = \frac{5.033 \times 10^8}{8} \approx 6.29 \times 10^7\text{ bytes}\)

Convert bytes to megabytes (MB) using the standard metric prefix (\(1\text{ MB} = 10^6\text{ bytes}\)):
\(\text{Storage space} = \frac{6.29 \times 10^7}{10^6} \approx 63\text{ MB}\).
(Alternatively, using binary megabytes: \(6.29 \times 10^7 / (1024^2) \approx 60\text{ MiB}\), which still rounds closest to D).

[1 mark] - Correct choice D is selected.

Using the kinematic equation for displacement with vertical upward direction taken as positive:
\(s = u t + \frac{1}{2} a t^2\)

Where:
- \(u = +15\text{ m s}^{-1}\) (upwards)
- \(t = 4.0\text{ s}\)
- \(a = -9.8\text{ m s}^{-2}\) (downwards)

Calculate the displacement:
\(s = (15)(4.0) + \frac{1}{2}(-9.8)(4.0)^2\)
\(s = 60 - 4.9(16)\)
\(s = 60 - 78.4 = -18.4\text{ m}\)

The displacement is \(-18.4\text{ m}\), which means the final position is \(18.4\text{ m}\) below the starting point. Thus, the height of the cliff is approximately \(18\text{ m}\) (to 2 significant figures).

[1 mark] - Correct choice A is selected.

The formula for the fringe separation in double-slit interference is:
\(x = \frac{\lambda D}{d}\)

For the new setup, the parameters are modified as follows:
- New wavelength \(\lambda' = 2\lambda\)
- New distance to screen \(D' = 2D\)
- New slit separation \(d' = \frac{d}{2}\)

Substituting these into the formula for the new fringe separation \(x'\):
\(x' = \frac{\lambda' D'}{d'}\)
\(x' = \frac{(2\lambda)(2D)}{\frac{d}{2}} = \frac{4\lambda D}{\frac{d}{2}} = 8 \frac{\lambda D}{d} = 8x\)

Therefore, the new fringe separation is \(8x\).

[1 mark] - Correct choice C is selected.

The relationship between Young modulus \(E\), force \(F\), length \(L\), extension \(e\), and cross-sectional area \(A\) is:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{e/L} = \frac{F L}{A e}\)

Rearranging for extension \(e\):
\(e = \frac{F L}{A E}\)

Since the wire is circular, the area \(A\) is given by:
\(A = \pi \frac{d^2}{4}\)

Substituting this into the extension formula:
\(e = \frac{4 F L}{\pi d^2 E}\)

For the second wire:
- Length \(L' = 2L\)
- Diameter \(d' = 2d\)

Using the formula, the new extension \(e'\) is:
\(e' = \frac{4 F (2L)}{\pi (2d)^2 E} = \frac{8 F L}{\pi (4d^2) E} = \frac{1}{2} \left( \frac{4 F L}{\pi d^2 E} \right) = \frac{e}{2}\)

Therefore, the extension of the second wire is \(e/2\).

[1 mark] - Correct choice A is selected.

First, express \(g\) in terms of the measured quantities:
\(T^2 = 4\pi^2 \frac{L}{g} \implies g = \frac{4\pi^2 L}{T^2}\)

To find the percentage uncertainty in \(g\), sum the individual percentage uncertainties of the terms, multiplying by powers as necessary:
\(\frac{\Delta g}{g} \times 100\% = \left( \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \right) \times 100\%\)

Calculate the percentage uncertainty in \(L\):
\(\frac{\Delta L}{L} \times 100\% = \frac{0.008}{0.800} \times 100\% = 1.0\%\)

The percentage uncertainty in \(T\) is given as \(1.5\%\).

Substitute these values into the uncertainty equation:
\(\text{Percentage uncertainty in } g = 1.0\% + 2 \times 1.5\% = 1.0\% + 3.0\% = 4.0\%\).

[1 mark] - Correct choice C is selected.

Since the two horizontal forces act perpendicular to each other (East and North), we find the resultant horizontal force \(F_{\text{net}}\) using Pythagoras' theorem:
\(F_{\text{net}} = \sqrt{F_1^2 + F_2^2} = \sqrt{3.0^2 + 4.0^2} = \sqrt{9.0 + 16.0} = \sqrt{25.0} = 5.0\text{ N}\)

Using Newton's second law, \(F_{\text{net}} = m a\), the magnitude of the acceleration is:
\(a = \frac{F_{\text{net}}}{m} = \frac{5.0\text{ N}}{0.50\text{ kg}} = 10\text{ m s}^{-2}\).

[1 mark] - Correct choice C is selected.

From the photoelectric equation, the maximum kinetic energy of the photoelectrons initially is:
\(E_{k,\text{max}} = hf - \Phi \implies hf = E_{k,\text{max}} + \Phi\)

When the frequency of radiation is doubled to \(2f\), the new maximum kinetic energy \(E_{k,\text{max}}'\) is:
\(E_{k,\text{max}}' = h(2f) - \Phi = 2(hf) - \Phi\)

Substituting \(hf = E_{k,\text{max}} + \Phi\) into this equation:
\(E_{k,\text{max}}' = 2(E_{k,\text{max}} + \Phi) - \Phi = 2 E_{k,\text{max}} + 2\Phi - \Phi = 2 E_{k,\text{max}} + \Phi\).

[1 mark] - Correct choice B is selected.

Number of pixels = \(400 \times 600 = 2.4 \times 10^5\text{ pixels}\). Total number of bits = \(2.4 \times 10^5 \times 12\text{ bits} = 2.88 \times 10^6\text{ bits}\). Total number of bytes = \(\frac{2.88 \times 10^6}{8} = 3.6 \times 10^5\text{ bytes}\). Minimum storage size in kilobytes (kB) = \(\frac{3.6 \times 10^5}{1000} = 360\text{ kB}\).

1 mark: Correct calculation of total bits, division by 8 to obtain bytes, and conversion to kB to yield 360 kB (Option A).

In bright light: \(V_{\text{out, bright}} = 6.0\text{ V} \times \frac{0.4\text{ k}\Omega}{0.4\text{ k}\Omega + 2.2\text{ k}\Omega} \approx 0.92\text{ V}\). In the dark: \(V_{\text{out, dark}} = 6.0\text{ V} \times \frac{10\text{ k}\Omega}{10\text{ k}\Omega + 2.2\text{ k}\Omega} \approx 4.92\text{ V}\). Change in output voltage is \(\Delta V_{\text{out}} = 4.92\text{ V} - 0.92\text{ V} = 4.0\text{ V}\).

1 mark: Correct calculation of both output voltages and finding the difference to be 4.0 V (Option C).

The percentage uncertainty in a combined quantity is the sum of the percentage uncertainties of the independent terms, taking power factors into account: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta h}{h}\). This gives \(\% \text{ uncertainty} = 1.5\% + 2(2.0\%) + 1.0\% = 6.5\%\).

1 mark: Correctly multiplying the uncertainty of the diameter by 2 and summing all percentage uncertainties to obtain 6.5% (Option C).

Using the formula for elastic strain energy \(E = \frac{1}{2} F \Delta x\) and Hooke's law for extension \(\Delta x = \frac{F L}{A E}\), we get \(E = \frac{F^2 L}{2 A E}\). Substituting the given values: \(E = \frac{120^2 \times 2.0}{2 \times (1.5 \times 10^{-6}) \times (2.0 \times 10^{11})} = 4.8 \times 10^{-2}\text{ J}\).

1 mark: Correct calculation of extension (8.0 x 10^-4 m) or direct substitution into the combined energy formula to obtain 4.8 x 10^-2 J (Option B).

The vertical component of the initial velocity is \(u_y = 20 \sin(30^\circ) = 10\text{ m s}^{-1}\). At maximum height, the vertical velocity \(v_y = 0\). Using \(v_y^2 = u_y^2 - 2gy\), we get \(0 = 10^2 - 2 \times 9.8 \times y\), which yields \(y = \frac{100}{19.6} \approx 5.1\text{ m}\).

1 mark: Correct calculation of the vertical component of velocity (10 m/s) and application of SUVAT to find the height of 5.1 m (Option B).

The kinetic energy gained by the electron is \(E_k = q V = 1.60 \times 10^{-19}\text{ C} \times 150\text{ V} = 2.40 \times 10^{-17}\text{ J}\). The momentum \(p\) of the electron is \(p = \sqrt{2 m_e E_k} = \sqrt{2 \times (9.11 \times 10^{-31}\text{ kg}) \times (2.40 \times 10^{-17}\text{ J})} \approx 6.61 \times 10^{-24}\text{ kg m s}^{-1}\). The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{6.61 \times 10^{-24}\text{ kg m s}^{-1}} \approx 1.0 \times 10^{-10}\text{ m}\).

1 mark: Correct calculation of kinetic energy, momentum, and de Broglie wavelength of 1.0 x 10^-10 m (Option B).

In equilibrium, the vertical forces balance: \(2 T \cos(35^\circ) = m g\). Therefore, \(2 T \cos(35^\circ) = 4.0 \times 9.8 = 39.2\text{ N}\). Rearranging for tension gives \(T = \frac{39.2}{2 \cos(35^\circ)} \approx 23.9\text{ N} \approx 24\text{ N}\).

1 mark: Correct setting up of the equilibrium equation resolving forces vertically and solving for tension to get 24 N (Option C).

The wavelength \(\lambda\) of the sound wave is \(\lambda = \frac{v}{f} = \frac{340\text{ m s}^{-1}}{680\text{ Hz}} = 0.50\text{ m}\). In a stationary wave, the distance between adjacent nodes is half a wavelength, \(\frac{\lambda}{2}\). Thus, the distance is \(\frac{0.50\text{ m}}{2} = 0.25\text{ m}\).

1 mark: Correct calculation of wavelength (0.50 m) and division by 2 to find the inter-nodal distance of 0.25 m (Option B).

The potential divider equation is given by: \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{th}}}{R_{\text{th}} + R_{\text{fixed}}}\). Substituting the given values: \(V_{\text{out}} = 9.0\text{ V} \times \frac{1.8\text{ k}\Omega}{1.8\text{ k}\Omega + 1.2\text{ k}\Omega} = 9.0\text{ V} \times \frac{1.8}{3.0} = 5.4\text{ V}\).

[1] for correct calculation of potential difference resulting in 5.4 V (B).

Calculate the vertical component of the final velocity: \(v_y^2 = u_y^2 + 2 g s_y = 0 + 2 \times 9.8\text{ m s}^{-2} \times 45\text{ m} = 882\text{ m}^2\text{ s}^{-2}\). The horizontal component remains constant: \(v_x = 20\text{ m s}^{-1}\). The final velocity magnitude is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 882} = \sqrt{400 + 882} = \sqrt{1282} \approx 35.8\text{ m s}^{-1}\).

[1] for correct vertical velocity calculation and finding the vector sum resultant of 35.8 m/s (B).

First, calculate the energy of an incident photon: \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{250 \times 10^{-9}\text{ m}} = 7.956 \times 10^{-19}\text{ J}\). Convert the work function into Joules: \(\phi = 2.3\text{ eV} \times 1.60 \times 10^{-19}\text{ J/eV} = 3.68 \times 10^{-19}\text{ J}\). The maximum kinetic energy is \(E_{\text{k, max}} = E - \phi = 7.956 \times 10^{-19}\text{ J} - 3.68 \times 10^{-19}\text{ J} = 4.276 \times 10^{-19}\text{ J} \approx 4.3 \times 10^{-19}\text{ J}\).

[1] for correct calculation of photon energy and subtracting work function in consistent units to obtain 4.3 x 10^-19 J (A).

The tension force in the wire is \(F = mg = 6.0\text{ kg} \times 9.8\text{ m s}^{-2} = 58.8\text{ N}\). Using the formula for Young modulus: \(E = \frac{F L}{A \Delta L}\), rearrange for extension: \(\Delta L = \frac{F L}{A E} = \frac{58.8 \times 2.5}{1.2 \times 10^{-7} \times 2.0 \times 10^{11}} = \frac{147}{2.4 \times 10^4} = 6.125 \times 10^{-3}\text{ m} \approx 6.1\text{ mm}\).

[1] for correct calculation of extension resulting in 6.1 mm (C).

(a) At \(20^\circ\text{C}\), the output voltage across the thermistor is \(V_{\text{out}} = 4.0\text{ V}\). Using the potential divider equation:
\(V_{\text{out}} = V_{\text{supply}} \times \frac{R_{\text{thermistor}}}{R + R_{\text{thermistor}}}\)
\(4.0 = 6.0 \times \frac{2400}{R + 2400}\)
\(4.0(R + 2400) = 14400\)
\(4.0R + 9600 = 14400\)
\(4.0R = 4800 \implies R = 1200\ \Omega = 1.2\text{ k}\Omega\).

(b) At \(60^\circ\text{C}\), the thermistor resistance drops to \(650\ \Omega\). Using the potential divider equation with \(R = 1200\ \Omega\):
\(V_{\text{out}} = 6.0 \times \frac{650}{1200 + 650} = 6.0 \times \frac{650}{1850} \approx 2.11\text{ V}\).

(c) No, the circuit is not suitable because as the temperature increases, the resistance of the thermistor decreases, which reduces the fraction of the total voltage dropped across it. Consequently, \(V_{\text{out}}\) decreases at higher temperatures, so it cannot be used directly to trigger a device that requires a high voltage at high temperatures.

(a)
- M1: Uses potential divider equation or equivalent ratio of voltage and resistance: \(V_{\text{out}}/V_{\text{supply}} = R_{\text{thermistor}} / (R + R_{\text{thermistor}})\) [1 mark]
- A1: Correct substitution of values leading to \(R = 1200\ \Omega\) (must see working showing how \(1200\ \Omega\) is calculated). [1 mark]

(b)
- M1: Substitution of \(650\ \Omega\) and \(1200\ \Omega\) into potential divider equation. [1 mark]
- A1: Correct calculation of voltage: \(2.1\text{ V}\) (allow \(2.11\text{ V}\)). [1 mark]

(c)
- A1: States 'No' with valid explanation that resistance decreases as temperature increases, resulting in a lower output voltage \(V_{\text{out}}\) (or vice versa). [1 mark]

(a) First, find the energy of the photon in Joules:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{2.50 \times 10^{-7}} = 7.956 \times 10^{-19}\text{ J}\)
Now convert to electron-volts by dividing by the elementary charge:
\(E_{\text{eV}} = \frac{7.956 \times 10^{-19}}{1.60 \times 10^{-19}} \approx 4.97\text{ eV}\) (or \(4.9725\text{ eV}\)).

(b) Use Einstein's photoelectric equation:
\(E_{\text{k,max}} = hf - \Phi = E_{\text{photon}} - \Phi\)
First calculate the work function in Joules:
\(\Phi = 2.28 \times 1.60 \times 10^{-19}\text{ C} = 3.648 \times 10^{-19}\text{ J}\)
Now calculate the maximum kinetic energy in Joules:
\(E_{\text{k,max}} = 7.956 \times 10^{-19}\text{ J} - 3.648 \times 10^{-19}\text{ J} = 4.308 \times 10^{-19}\text{ J} \approx 4.31 \times 10^{-19}\text{ J}\).
Alternatively, calculating in eV first:
\(E_{\text{k,max}} = 4.9725\text{ eV} - 2.28\text{ eV} = 2.6925\text{ eV}\)
Converting back to Joules:
\(2.6925 \times 1.60 \times 10^{-19} = 4.31 \times 10^{-19}\text{ J}\).

(c) There is no change to the maximum kinetic energy. Doubling the intensity of the radiation increases the rate at which photons arrive (and thus the rate of photoelectron emission), but the energy of individual photons remains unchanged.

(a)
- M1: Correct use of \(E = hc/\lambda\) to find photon energy in Joules: \(7.96 \times 10^{-19}\text{ J}\). [1 mark]
- A1: Correct conversion to eV to yield \(4.97\text{ eV}\) (accept range \(4.95\text{ eV}\) to \(5.00\text{ eV}\)). [1 mark]

(b)
- M1: Recall/use of Einstein's photoelectric equation with consistent units (e.g., subtracting work function in Joules from photon energy in Joules). [1 mark]
- A1: Correct calculation of maximum kinetic energy: \(4.31 \times 10^{-19}\text{ J}\) (accept range \(4.3 \times 10^{-19}\text{ J}\) to \(4.35 \times 10^{-19}\text{ J}\)). [1 mark]

(c)
- A1: States there is no change and explains that intensity increases the rate of emission/photon arrival but does not change the energy of individual photons. [1 mark]

(a) The diameter of the wire is \(d = 0.64\text{ mm} = 0.64 \times 10^{-3}\text{ m}\). The radius is \(r = 0.32 \times 10^{-3}\text{ m}\).
\(Area = \pi r^2 = \pi \times (0.32 \times 10^{-3})^2 \approx 3.22 \times 10^{-7}\text{ m}^2\) (or \(3.2 \times 10^{-7}\text{ m}^2\)).

(b) Stress is defined as force per unit cross-sectional area:
\(\sigma = \frac{F}{A} = \frac{35}{3.217 \times 10^{-7}} \approx 1.09 \times 10^8\text{ Pa}\) (or \(1.1 \times 10^8\text{ Pa}\)).

(c) Young modulus is \(E = \frac{\text{stress}}{\text{strain}}\), where \(\text{strain} = \frac{\text{extension } e}{\text{original length } L}\).
Rearranging this gives:
\(e = \frac{\text{stress} \times L}{E} = \frac{1.088 \times 10^8 \times 2.2}{1.1 \times 10^{11}} = 2.18 \times 10^{-3}\text{ m}\) (or \(2.2 \times 10^{-3}\text{ m}\)).

(a)
- A1: Correct calculation of cross-sectional area: \(3.2 \times 10^{-7}\text{ m}^2\) (accept range \(3.20 \times 10^{-7}\) to \(3.22 \times 10^{-7}\)). [1 mark]

(b)
- M1: Recall/use of formula for tensile stress: \(\text{stress} = \frac{\text{Force}}{\text{Area}}\). [1 mark]
- A1: Correct calculation of stress: \(1.1 \times 10^8\text{ Pa}\) (allow \(1.09 \times 10^8\text{ Pa}\), accept error carried forward from area in (a)). [1 mark]

(c)
- M1: Recall/use of Young modulus relationship to calculate extension: \(e = \frac{F L}{A E}\) or \(e = \frac{\text{stress} \times L}{E}\). [1 mark]
- A1: Correct calculation of extension: \(2.2 \times 10^{-3}\text{ m}\) (or \(2.2\text{ mm}\), accept \(2.18 \times 10^{-3}\text{ m}\), allow error carried forward from previous steps). [1 mark]

(a) The vertical component of the initial velocity is:
\(u_y = u \sin \theta = 18 \sin(40^\circ) \approx 11.57\text{ m s}^{-1} \approx 11.6\text{ m s}^{-1}\).

(b) At maximum vertical height, the vertical velocity \(v_y = 0\).
Using the equations of motion:
\(v_y^2 = u_y^2 + 2 a s_y\)
\(0 = (11.57)^2 - 2 \times 9.81 \times s_y\)
\(19.62 s_y = 133.86\)
\(s_y = \frac{133.86}{19.62} \approx 6.82\text{ m}\).
This is approximately \(6.8\text{ m}\).

(c) First, determine the total time of flight \(t\) by setting the displacement \(s_y = 0\):
\(s_y = u_y t + \frac{1}{2} a t^2\)
\(0 = 11.57 t - 4.905 t^2\)
\(t = \frac{11.57}{4.905} \approx 2.359\text{ s}\) (or find time to peak \(= 11.57 / 9.81 = 1.179\text{ s}\) and double it).
Now, calculate the horizontal distance using the constant horizontal component of velocity:
\(u_x = 18 \cos(40^\circ) \approx 13.79\text{ m s}^{-1}\)
\(\text{Horizontal distance } s_x = u_x \times t = 13.79 \times 2.359 \approx 32.5\text{ m}\) (or \(33\text{ m}\) to two significant figures).

(a)
- A1: Correct calculation: \(11.6\text{ m s}^{-1}\) (accept \(11.57\text{ m s}^{-1}\)). [1 mark]

(b)
- M1: Recall/use of \(v^2 = u^2 + 2as\) with vertical components (setting \(v_y = 0\) and \(a = -9.81\text{ m s}^{-2}\)). [1 mark]
- A1: Correct calculation of maximum height: \(6.82\text{ m}\) (which rounds to \(6.8\text{ m}\)). [1 mark]

(c)
- M1: Correct method to find the total time of flight \(t \approx 2.36\text{ s}\) (or \(2.4\text{ s}\)). [1 mark]
- A1: Correct calculation of horizontal range: \(33\text{ m}\) (accept range \(32.5\text{ m}\) to \(33\text{ m}\), allow error carried forward from (a)). [1 mark]

(a) The cross-sectional area is given by \(A = \frac{\pi d^2}{4}\).
The percentage uncertainty in the diameter \(d\) is:
\%\Delta d = \frac{0.02\text{ mm}}{0.38\text{ mm}} \times 100\% \approx 5.263\%
Since \(A\) depends on \(d^2\), the percentage uncertainty in the area is:
\%\Delta A = 2 \times \%\Delta d = 2 \times 5.263\% = 10.53\% \approx 10.5\%.

(b) The Young modulus is defined as:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\)
Since the gradient of the graph of \(x\) against \(F\) is \(m = \frac{x}{F}\), we can write:
\(E = \frac{L}{A m}\)
First, calculate the cross-sectional area \(A\):
\(A = \frac{\pi (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\)
Now calculate \(E\):
\(E = \frac{2.05\text{ m}}{(1.134 \times 10^{-7}\text{ m}^2) \times (1.64 \times 10^{-4}\text{ m N}^{-1})} = 1.102 \times 10^{11}\text{ Pa}\)
To 3 significant figures, \(E = 1.10 \times 10^{11}\text{ Pa}\).

(c) The formula for \(E\) is \(E = \frac{L}{A m}\).
The percentage uncertainty in \(E\) is the sum of the percentage uncertainties of the constituent quantities:
\%\Delta E = \%\Delta L + \%\Delta A + \%\Delta m
\%\Delta L = \frac{0.01\text{ m}}{2.05\text{ m}} \times 100\% \approx 0.49\%
\%\Delta A = 10.53\%
\%\Delta m = 3.0\%
\%\Delta E = 0.49\% + 10.53\% + 3.0\% = 14.02\%
Therefore, the absolute uncertainty in \(E\) is:
\(\Delta E = 14.02\% \times (1.102 \times 10^{11}\text{ Pa}) = 0.1545 \times 10^{11}\text{ Pa} \approx 0.15 \times 10^{11}\text{ Pa}\) (or \(1.5 \times 10^{10}\text{ Pa}\)).

(d) One potential source of systematic error is 'slack' in the wire at the start of the experiment, which causes the initial extensions to appear larger than they are, or a zero error on the micrometer screw gauge.
To minimize the impact of slack, a small initial 'pre-tension' load can be applied to the wire before recording the baseline length and starting measurements. To minimize micrometer zero error, the gauge should be closed fully before the experiment to check for and record any zero offset, which is then subtracted from the diameter measurements.

(a)
- [1] Calculates percentage uncertainty in \(d\): \(\frac{0.02}{0.38} \times 100\% = 5.26\%\).
- [1] Doubles the percentage uncertainty in \(d\) to find \(\%\Delta A = 10.53\%\) and rounds to \(10.5\%\).

(b)
- [1] Correctly recalls or derives \(E = \frac{L}{A m}\).
- [1] Calculates area \(A = 1.13 \times 10^{-7}\text{ m}^2\) (allow rounding errors here).
- [1] Obtains \(E = 1.10 \times 10^{11}\text{ Pa}\) (allow \(1.1 \times 10^{11}\text{ Pa}\)).

(c)
- [1] Calculates percentage uncertainty in \(L\): \(\frac{0.01}{2.05} \times 100\% = 0.49\%\).
- [1] Adds percentage uncertainties: \(0.49\% + 10.53\% + 3.0\% = 14.0\%\) (allow ecf from (a)).
- [1] Calculates absolute uncertainty as \(0.15 \times 10^{11}\text{ Pa}\) (accept \(0.15\) to \(0.16 \times 10^{11}\text{ Pa}\)).

(d)
- [1] Identifies a valid systematic error (e.g., slack in wire, zero error in micrometer, temperature fluctuations altering length).
- [1] Provides a valid method of mitigation (e.g., pre-tension weight, zero correction calibration, controlling room temperature).

(a) A variable resistor allows for smooth, continuous adjustments of current without breaking the circuit, which prevents unwanted self-induction transients and keeps the circuit conditions stable. It is also quicker and more convenient than manually swapping out individual fixed resistors, minimizing the time current flows and reducing cell discharge.

(b) The equation is \(V = -r I + \varepsilon\). A graph of \(V\) against \(I\) is a straight line with gradient \(-r\).
Using extreme data points to calculate the gradient:
\(\text{Gradient } m = \frac{V_5 - V_1}{I_5 - I_1} = \frac{1.05\text{ V} - 1.41\text{ V}}{0.75\text{ A} - 0.15\text{ A}} = \frac{-0.36\text{ V}}{0.60\text{ A}} = -0.60\text{ V A}^{-1}\).
Since \(\text{gradient} = -r\), the internal resistance \(r = 0.60\ \Omega\) (or \(\text{V A}^{-1}\)).

(c) Using the equation \(V = \varepsilon - I r\) at any data point, e.g., \(I = 0.15\text{ A}\), \(V = 1.41\text{ V}\):
\(\varepsilon = V + I r = 1.41 + (0.15 \times 0.60) = 1.50\text{ V}\).
Checking with another point, e.g., \(I = 0.75\text{ A}\), \(V = 1.05\text{ V}\):
\(\varepsilon = 1.05 + (0.75 \times 0.60) = 1.50\text{ V}\).
So the intercept (EMF) is \(\varepsilon = 1.50\text{ V}\).
The percentage uncertainty in \(\varepsilon\) is:
\(\%\Delta \varepsilon = \frac{0.04\text{ V}}{1.50\text{ V}} \times 100\% = 2.67\% \approx 2.7\%\).

(d) As the cell heats up, its internal resistance \(r\) increases. This means the term \(I r\) increases for a given current, causing the measured terminal potential difference \(V\) to decrease more than expected (systematic downward shift of values at later/higher current stages).
To prevent this, the student should open the switch between readings to allow the cell to cool back to room temperature, only closing it long enough to take a stable reading.

(a)
- [1] States that a rheostat allows continuous adjustment of current without disconnecting/breaking the circuit.
- [1] Explains that this saves time or prevents cell depletion/discharge during changes.

(b)
- [1] Shows correct gradient calculation using data points, e.g., \(\Delta V / \Delta I = -0.36 / 0.60 = -0.60\).
- [1] Equates negative gradient to internal resistance: \(r = 0.60\).
- [1] States correct unit: \(\Omega\) or \(\text{V A}^{-1}\).

(c)
- [1] Correctly uses \(V = \varepsilon - I r\) to calculate \(\varepsilon = 1.50\text{ V}\) (allow range \(1.49 - 1.51\text{ V}\)).
- [1] Uses the correct division: \(\frac{0.04}{1.50}\).
- [1] Obtains percentage uncertainty of \(2.7\%\) (accept \(2.67\%\) or ecf from calculated \(\varepsilon\)).

(d)
- [1] Identifies that heating increases internal resistance, which decreases the measured terminal PD (or shifts points downwards).
- [1] Suggests opening the switch between readings to minimize current flow time and keep temperature constant.

(a) Setup a circuit with the variable DC power supply, a protective resistor, and the LED in series with a microammeter. Connect a voltmeter in parallel across the LED. Gradually increase the voltage from zero. Record the voltage at which the microammeter first shows a non-zero current reading (e.g., \(5\ \mu\text{A}\)), or use a viewing tube to identify the exact voltage at which light emission first becomes visible in a dark room.

(b) Rearranging the given equation for \(V_0\):
\(e V_0 = \frac{h c}{l} \implies V_0 = \left(\frac{h c}{e}\right) \frac{1}{l}\)
Comparing this to the equation of a straight line, \(y = m x + c\):
- \(y = V_0\)
- \(x = \frac{1}{l}\)
- \(m = \frac{h c}{e}\)
- intercept \(c = 0\)
Thus, the gradient of the line is indeed \(m = \frac{h c}{e}\).

(c) First, calculate the values of \(\frac{1}{l}\) in \(\text{m}^{-1}\):
- For Red: \(x_1 = \frac{1}{635 \times 10^{-9}\text{ m}} = 1.575 \times 10^6\text{ m}^{-1}\)
- For Blue: \(x_2 = \frac{1}{470 \times 10^{-9}\text{ m}} = 2.128 \times 10^6\text{ m}^{-1}\)

Calculate the change in \(x\) and \(V_0\):
\(\Delta x = 2.128 \times 10^6 - 1.575 \times 10^6 = 0.553 \times 10^6\text{ m}^{-1}\)
\(\Delta V_0 = 2.65 - 1.95 = 0.70\text{ V}\)

Calculate the gradient \(m\):
\(m = \frac{\Delta V_0}{\Delta x} = \frac{0.70\text{ V}}{0.553 \times 10^6\text{ m}^{-1}} = 1.266 \times 10^{-6}\text{ V m}\)

Since \(m = \frac{h c}{e}\):
\(h = \frac{m e}{c} = \frac{(1.266 \times 10^{-6}\text{ V m}) \times (1.60 \times 10^{-19}\text{ C})}{3.00 \times 10^8\text{ m s}^{-1}} = 6.752 \times 10^{-34}\text{ J s}\)
To 3 significant figures, \(h = 6.75 \times 10^{-34}\text{ J s}\).

(d) If the measured threshold voltage \(V_0\) is always slightly higher than the true thermodynamic value by a constant offset, this constant systematic error (zero error equivalent) shifts the entire line vertically upwards. Since the gradient represents the rate of change of \(V_0\) with respect to \(1/l\), a uniform vertical shift does not alter the gradient of the line. Therefore, using the gradient of a multi-point graph instead of a single-point calculation successfully eliminates the effect of this constant systematic offset.

(a)
- [1] Describes a correct series circuit containing the LED, variable power supply, and microammeter, with the voltmeter across the LED.
- [1] Explains the action of varying the voltage and monitoring current/light.
- [1] Defines threshold voltage as when current first starts to rise significantly/continuously (or when light is first visible in a dark room).

(b)
- [1] Rearranges equation to make \(V_0\) the subject: \(V_0 = \left(\frac{h c}{e}\right) \frac{1}{l}\).
- [1] Compares directly to \(y = m x\) to identify gradient \(m = \frac{h c}{e}\).

(c)
- [1] Correctly converts wavelengths to \(1/l\) in \(\text{m}^{-1}\): \(1.58 \times 10^6\text{ m}^{-1}\) and \(2.13 \times 10^6\text{ m}^{-1}\).
- [1] Calculates gradient \(m \approx 1.27 \times 10^{-6}\text{ V m}\) (allow range \(1.25 \times 10^{-6}\) to \(1.28 \times 10^{-6}\)).
- [1] Correctly calculates \(h = 6.75 \times 10^{-34}\text{ J s}\) (allow range \(6.65 \times 10^{-34}\) to \(6.85 \times 10^{-34}\)).

(d)
- [1] Explains that a constant offset shifts the y-intercept but leaves the gradient unchanged.
- [1] Concludes that using the gradient of the line mitigates/eliminates this constant offset error.

(a) Place the thermistor in a water bath with a thermometer. Heat the water bath slowly, recording the output voltage \(V_{\text{out}}\) at regular temperature intervals (e.g., every \(5^{\circ}\text{C}\)). Plot a calibration curve of \(V_{\text{out}}\) against temperature \(T\) to read off the temperature for any given voltage.

(b) Using the potential divider equation: \(V_{\text{out}} = V_{\text{in}} \times \frac{R}{R + R_T} = 6.0\text{ V} \times \frac{1200\ \Omega}{1200\ \Omega + 800\ \Omega} = 6.0 \times 0.60 = 3.6\text{ V}\).

(c) At \(26.0^{\circ}\text{C}\), \(R_T = 775\ \Omega\). The new output voltage is: \(V_{\text{out}} = 6.0\text{ V} \times \frac{1200}{1200 + 775} = 3.6456\text{ V}\).
Change in output voltage \(\Delta V_{\text{out}} = 3.6456\text{ V} - 3.6000\text{ V} = 0.0456\text{ V}\).
Sensitivity \(= \frac{\Delta V_{\text{out}}}{\Delta T} = \frac{0.0456\text{ V}}{1.0^{\circ}\text{C}} = 0.046\text{ V }^{\circ}\text{C}^{-1}\).

(a) 1 mark: Describe placing thermistor in water bath and measuring voltage and temperature simultaneously. 1 mark: Describe plotting a calibration curve of voltage against temperature.
(b) 1 mark: Correct substitution into the potential divider equation showing \(3.6\text{ V}\).
(c) 1 mark: Calculates new voltage of \(3.65\text{ V}\) (or \(3.646\text{ V}\)). 1 mark: Calculates correct sensitivity of \(0.046\text{ V }^{\circ}\text{C}^{-1}\) (allow \(0.045\text{ to } 0.046\)).

(a) For vertical motion (downwards positive): \(u_y = 0\), \(a_y = g\), \(s_y = h\). From \(s = ut + \frac{1}{2}at^2\), we get \(h = \frac{1}{2}gt^2\), which rearranges to time of flight \(t = \sqrt{\frac{2h}{g}}\). For horizontal motion, velocity is constant: \(x = v \times t = v \sqrt{\frac{2h}{g}}\).

(b) Central value calculation: \(x = 3.40 \times \sqrt{\frac{2 \times 1.25}{9.81}} = 3.40 \times 0.5048\text{ s} = 1.716\text{ m} \approx 1.72\text{ m}\).
Uncertainties: Fractional/percentage uncertainties are added. \(\%\Delta x = \%\Delta v + \frac{1}{2}\%\Delta h\) (due to the square root of \(h\)).
\%\Delta v = \frac{0.10}{3.40} \times 100 = 2.94\%\).
\%\Delta h = \frac{0.02}{1.25} \times 100 = 1.60\%\).
\%\Delta x = 2.94\% + \frac{1}{2}(1.60\%) = 3.74\%\).
Absolute uncertainty \(\Delta x = 1.716 \times 0.0374 = 0.064\text{ m} \approx 0.06\text{ m}\).
Combining these gives: \(x = 1.72 \pm 0.06\text{ m}\).

(a) 1 mark: Derives time of flight formula \(t = \sqrt{\frac{2h}{g}}\) using vertical SUVAT. 1 mark: Links this to horizontal distance via \(x = vt\).
(b) 1 mark: Calculates central value of \(x = 1.72\text{ m}\). 1 mark: Correctly sums fractional uncertainties, including halving the fractional uncertainty of \(h\) (totaling \(3.7\%\) or \(0.037\) fractional). 1 mark: Correct absolute uncertainty of \(0.06\text{ m}\) and presents final value and uncertainty to consistent decimal places with units.

(a) The energy of a photon emitted by an LED is related to the threshold voltage by \(e V_0 = \frac{hc}{\lambda}\). Rearranging this in the form \(y = mx + c\) gives \(V_0 = \left(\frac{hc}{e}\right) \frac{1}{\lambda}\). A plot of \(V_0\) (y-axis) against \(1/\lambda\) (x-axis) will be a straight line through the origin. The gradient \(m\) of this line is \(m = \frac{hc}{e}\). Therefore, \(h = \frac{m e}{c}\).

(b) Using the relation \(h = \frac{m e}{c}\) where \(m = 1.24 \times 10^{-6}\text{ V m}\), \(e = 1.60 \times 10^{-19}\text{ C}\), and \(c = 3.00 \times 10^8\text{ m s}^{-1}\):
\(h = \frac{1.24 \times 10^{-6} \times 1.60 \times 10^{-19}}{3.00 \times 10^8} = 6.613 \times 10^{-34}\text{ J s}\). To 3 significant figures, \(h = 6.61 \times 10^{-34}\text{ J s}\).

(a) 1 mark: States relation \(e V_0 = \frac{hc}{\lambda}\) or equivalent. 1 mark: Explains that plotting \(V_0\) against \(1/\lambda\) yields a straight line. 1 mark: Clearly states that the gradient of this line equals \(\frac{hc}{e}\) and so \(h = \frac{\text{gradient} \times e}{c}\).
(b) 1 mark: Correct substitution of values into rearranged equation. 1 mark: Correct value calculated to 2 or 3 sig figs: \(6.6 \times 10^{-34}\text{ J s}\) or \(6.61 \times 10^{-34}\text{ J s}\) with correct unit.

(a) Use a digital micrometer screw gauge (or digital callipers). To ensure reliability, take measurements at multiple different positions along the active length of the strip and calculate the mean thickness. Additionally, check for and correct any zero error before taking measurements.

(b) Cross-sectional area \(A = w \times t = (5.0 \times 10^{-3}\text{ m}) \times (0.40 \times 10^{-3}\text{ m}) = 2.0 \times 10^{-6}\text{ m}^2\).
Tensile stress \(\sigma = \frac{F}{A} = \frac{24.0\text{ N}}{2.0 \times 10^{-6}\text{ m}^2} = 1.20 \times 10^7\text{ Pa} = 12.0\text{ MPa}\).
Uncertainties:
\%\Delta F = \frac{0.2}{24.0} \times 100\% = 0.83\%\).
\%\Delta w = \frac{0.1}{5.0} \times 100\% = 2.00\%\).
\%\Delta t = \frac{0.02}{0.40} \times 100\% = 5.00\%\).
Total percentage uncertainty in \(\sigma = \%\Delta F + \%\Delta w + \%\Delta t = 0.83\% + 2.00\% + 5.00\% = 7.83\%\).
Absolute uncertainty \(\Delta \sigma = 12.0\text{ MPa} \times 0.0783 = 0.94\text{ MPa} \approx 1.0\text{ MPa}\).
Therefore, \(\sigma = 12.0 \pm 1.0\text{ MPa}\) (or \(1.20 \pm 0.10 \times 10^7\text{ Pa}\)).

(a) 1 mark: Identifies micrometer screw gauge or digital callipers. 1 mark: Explains repeating measurements at different positions to average, or checking for zero error.
(b) 1 mark: Correct calculation of stress \(1.2 \times 10^7\text{ Pa}\) or \(12\text{ MPa}\). 1 mark: Adds fractional/percentage uncertainties of force, width, and thickness to find total percentage uncertainty of \(7.8\%\) (accept \(8.0\%\) if rounded early). 1 mark: Correctly computes absolute uncertainty of \(1.0\text{ MPa}\) (or \(0.10 \times 10^7\text{ Pa}\)) and quotes final answer consistently.

(a) Number of bits per frame \(= 1024 \times 1024 \times 12\text{ bits} = 12,582,912\text{ bits}\).
Bit rate \(= 12,582,912\text{ bits/frame} \times 30.0\text{ frames/s} = 377,487,360\text{ bits s}^{-1} \approx 377\text{ Mbit s}^{-1}\).

(b) Digital signals consist of discrete states (high/low or 1/0). Although noise is added and the signal is attenuated (loses amplitude) along the fiber, the signal can be completely restored. As long as the noise level is not large enough to cause high states to be confused with low states, regenerator amplifiers can detect the discrete pulses and reconstruct a perfect copy of the original sequence of 1s and 0s, eliminating the noise. This is not possible with analog signals, where noise and distortion are amplified alongside the signal.

(a) 1 mark: Calculates total bits per frame correctly (\(1.26 \times 10^7\text{ bits}\)). 1 mark: Calculates correct bit rate of \(377\text{ Mbit s}^{-1}\) (or \(377.5\text{ Mbit s}^{-1}\)).
(b) 1 mark: Mentions that digital signals consist of only two discrete states (1s and 0s). 1 mark: Explains that regenerators can completely reconstruct/re-shape pulses to eliminate noise and attenuation. 1 mark: Contrasts with analog signals where any added noise degrades the signal quality permanently.

(a) In an NTC thermistor, conduction occurs via charge carriers (electrons) in a semiconductor. As temperature increases, thermal energy is supplied to the lattice, which liberates electrons from the valence band to the conduction band (increasing the number density \(n\) of charge carriers). Although lattice vibrations increase, leading to more frequent collisions, the exponential increase in carrier concentration \(n\) dominates, resulting in a significant reduction in resistance.

(b) At \(60^\circ\text{C}\), the resistance of the thermistor is \(R_T = 1.2\text{ k}\Omega = 1200\ \Omega\).
Total resistance of the series loop is:
\(R_{\text{total}} = R + R_T + r = 2200\ \Omega + 1200\ \Omega + 150\ \Omega = 3550\ \Omega\).
The current in the circuit is:
\(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{6.0\text{ V}}{3550\ \Omega} \approx 1.690 \times 10^{-3}\text{ A}\).
The potential difference across the thermistor is:
\(V_{\text{out}} = I \times R_T = 1.690 \times 10^{-3}\text{ A} \times 1200\ \Omega \approx 2.03\text{ V}\).
The given quantities are stated to 2 or 3 significant figures, so the final value is appropriately expressed to 3 s.f. as \(2.03\text{ V}\).

(c) The sensitivity of the output voltage is given by the derivative \(\frac{dV_{\text{out}}}{dT} = \frac{dV_{\text{out}}}{dR_T} \times \frac{dR_T}{dT}\).
To maximise the sensitivity at a given temperature, we must maximise the magnitude of \(\frac{dV_{\text{out}}}{dR_T}\).
Using the potential divider equation with \(R' = R + r\):
\(V_{\text{out}} = \varepsilon \frac{R_T}{R_T + R'}\).
Differentiating \(V_{\text{out}}\) with respect to \(R_T\):
\(\frac{dV_{\text{out}}}{dR_T} = \varepsilon \frac{R'}{(R_T + R')^2}\).
To find the value of \(R'\) that maximises this derivative for a fixed \(R_T\), we differentiate with respect to \(R'\) and set to zero:
\(\frac{d}{dR'}\left( \frac{R'}{(R_T + R')^2} \right) = 0 \implies R_T + R' - 2R' = 0 \implies R' = R_T\).
Thus, maximum sensitivity is achieved when the external series resistance (including internal resistance) equals the thermistor resistance: \(R + r \approx R_T\).
At \(60^\circ\text{C}\), \(R_T = 1.2\text{ k}\Omega\).
Currently, \(R + r = 2.2\text{ k}\Omega + 0.15\text{ k}\Omega = 2.35\text{ k}\Omega\), which is greater than \(1.2\text{ k}\Omega\).
To make \(R + r\) closer to \(1.2\text{ k}\Omega\), we must decrease the fixed resistance \(R\). Therefore, a smaller value of \(R\) (ideally around \(1.05\text{ k}\Omega\)) should be used.

(d) Self-heating occurs when the current passing through the thermistor dissipates electrical energy as heat inside the thermistor. This raises its temperature above the ambient temperature of the system being monitored, leading to an artificially lower thermistor resistance, which is a systematic error. This can be minimised by reducing the current in the thermistor, e.g., by choosing a larger fixed resistor \(R\), reducing the supply voltage \(\varepsilon\), or using a pulsed circuit to only apply potential difference when taking a reading.

(a) [3 marks total]:
- [1 mark] State that conduction is due to charge carriers (electrons/holes) in a semiconductor.
- [1 mark] Explain that rising temperature releases/liberates more charge carriers (increases number density \(n\)).
- [1 mark]* Quality of Written Communication: Clearly connect the dominating effect of increased carrier density over lattice collisions to the resulting drop in resistance.

(b) [4 marks total]:
- [1 mark] Sums resistances correctly: \(R_{\text{total}} = 3550\ \Omega\).
- [1 mark] Calculates circuit current correctly: \(I = 1.69 \times 10^{-3}\text{ A}\).
- [1 mark] Calculates final potential difference across the thermistor: \(V_{\text{out}} = 2.03\text{ V}\) (allow \(2.0\text{ V}\)).
- [1 mark] Expresses final answer to an appropriate number of significant figures (2 or 3 s.f.) consistent with the raw data.

(c) [5 marks total]:
- [1 mark] States that sensitivity depends on \(\frac{dV_{\text{out}}}{dT}\) or \(\frac{dV_{\text{out}}}{dR_T}\).
- [1 mark] Shows or states that the sensitivity factor \(\frac{R'}{(R_T+R')^2}\) is maximised when \(R + r \approx R_T\).
- [1 mark] Calculates current total external resistance: \(R + r = 2.35\text{ k}\Omega\).
- [1 mark] Compares this with \(R_T = 1.2\text{ k}\Omega\) and identifies that \(R + r > R_T\).
- [1 mark] Concludes that a smaller value of \(R\) is needed to increase sensitivity around \(60^\circ\text{C}\).

(d) [3 marks total]:
- [1 mark] Identifies self-heating due to current flow (dissipation of power \(I^2 R_T\)).
- [1 mark] Explains that this raises thermistor temperature above ambient, leading to a systematic underestimate of resistance / overestimate of temperature.
- [1 mark] Suggests a valid remedy (e.g., lower supply emf, pulsed measurement, or higher overall circuit resistances).

(a) The limit of proportionality is the point beyond which stress is no longer directly proportional to strain (Hooke's law is no longer obeyed). The elastic limit is the maximum stress/force that can be applied to a material without causing permanent (plastic) deformation when the load is removed.

(b) Cross-sectional area:
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.44 \times 10^{-3}\text{ m})^2}{4} = 1.5205 \times 10^{-7}\text{ m}^2 \approx 1.52 \times 10^{-7}\text{ m}^2\), which is approximately \(1.5 \times 10^{-7}\text{ m}^2\).
Percentage uncertainty in \(d\):
\(\%\Delta d = \frac{0.02}{0.44} \times 100\% \approx 4.545\%\).
Since \(A \propto d^2\), the percentage uncertainty in \(A\) is twice the percentage uncertainty in \(d\):
\(\%\Delta A = 2 \times 4.545\% \approx 9.09\% \approx 9.1\%\).

(c) Young Modulus:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\).
\(E = \frac{8.0\text{ N} \times 1.85\text{ m}}{(1.5205 \times 10^{-7}\text{ m}^2) \times (10.4 \times 10^{-3}\text{ m})} = \frac{14.8}{1.5813 \times 10^{-9}} \approx 9.359 \times 10^9\text{ Pa} = 9.359\text{ GPa}\).

Now calculate the individual percentage uncertainties:
- \(\%\Delta L = \frac{0.01}{1.85} \times 100\% \approx 0.541\%\)
- \(\%\Delta x = \frac{0.1}{10.4} \times 100\% \approx 0.962\%\)
- \(\%\Delta A \approx 9.091\%\)

Total percentage uncertainty in \(E\):
\(\%\Delta E = \%\Delta L + \%\Delta x + \%\Delta A = 0.541\% + 0.962\% + 9.091\% \approx 10.594\%\).

Absolute uncertainty in \(E\):
\(\Delta E = 10.594\% \times 9.359\text{ GPa} \approx 0.991\text{ GPa} \approx 1.0\text{ GPa}\).

Rounding both values to 2 s.f. (since \(d\) and the uncertainty in \(d\) limit the precision to 2 s.f.):
\(E = (9.4 \pm 1.0)\text{ GPa}\) (or \((9.4 \pm 1.0) \times 10^9\text{ Pa}\)).

(d) A suitable setup consists of a vertical polymer wire suspended from a rigid clamp, with a second parallel reference wire suspended from the same support to eliminate support yield and temperature expansion errors. A vernier scale or a travel microscope is attached between the reference wire and the test wire. Loads are applied to the hanger at the bottom of the test wire.

To minimise uncertainty in extension:
- Use a very long original length \(L\) of filament so that the absolute extension \(x\) is larger for any given force, reducing its percentage uncertainty.
- Apply a small initial load ('pre-tension') to ensure the filament is completely straight before taking the reference/zero reading.
- Use a travel microscope or vernier scale to achieve a precision of \(0.01\text{ mm}\) or better instead of a standard ruler.
- Use the reference wire configuration to automatically compensate for thermal expansion during the experiment.

(a) [2 marks total]:
- [1 mark] Limit of proportionality: the point beyond which stress is no longer proportional to strain / force is no longer proportional to extension.
- [1 mark] Elastic limit: the maximum stress/load up to which the material behaves elastically / beyond which permanent plastic deformation remains.

(b) [3 marks total]:
- [1 mark] Calculates area correctly showing working: \(A = 1.52 \times 10^{-7}\text{ m}^2\).
- [1 mark] Calculates percentage uncertainty in diameter: \(\%\Delta d = 4.5\%\).
- [1 mark] Doubles percentage uncertainty of diameter to get percentage uncertainty in area: \(\%\Delta A = 9.1\%\) (allow \(9\%\)).

(c) [6 marks total]:
- [1 mark] Correctly recalls/uses \(E = \frac{FL}{Ax}\).
- [1 mark] Correct value calculated for Young modulus: \(9.4\text{ GPa}\) (or \(9.36 \times 10^9\text{ Pa}\)).
- [1 mark] Correct percentage uncertainty in length: \(0.54\%\).
- [1 mark] Correct percentage uncertainty in extension: \(0.96\%\).
- [1 mark] Sums percentage uncertainties to find \(\%\Delta E \approx 10.6\%\).
- [1 mark] Expresses absolute uncertainty and final value to consistent, appropriate significant figures: \(9.4 \pm 1.0\text{ GPa}\) (allow \(9.36 \pm 0.99\text{ GPa}\)).

(d) [4 marks total]:
- [1 mark] Provides a labelled diagram illustrating a valid tension setup (clamped wire, weights, marker/scale).
- [1 mark] Shows/mentions a reference wire/scale system or travel microscope to measure small extension.
- [1 mark]* Explains a valid method to increase extension size (longer wire) or improve measurement resolution (travel microscope/vernier).
- [1 mark] Explains use of pre-tensioning or reference wire to eliminate systematic alignment/support issues.

### Part (a) Solution
1. **Calculate Cross-Sectional Area (\(A\)):**
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.46 \times 10^{-3}\text{ m})^2}{4} = 1.662 \times 10^{-7}\text{ m}^2\) (accept \(1.66 \times 10^{-7}\text{ m}^2\)).

2. **Percentage Uncertainty in \(d\):**
\(\%\Delta d = \frac{0.01\text{ mm}}{0.46\text{ mm}} \times 100\% \approx 2.17\%\)

3. **Percentage Uncertainty in \(A\):**
Since \(A \propto d^2\), the percentage uncertainty in \(A\) is:
\(\%\Delta A = 2 \times \%\Delta d = 2 \times 2.17\% \approx 4.35\%\) (accept \(4.3\%\) to \(4.4\%\)).

---

### Part (b) Solution
1. **Derive the Equation:**
By definition:
\(\text{Stress} = \frac{F}{A}\) and \(\text{Strain} = \frac{x}{L}\)
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\)
Since the gradient \(k = \frac{F}{x}\), substituting this into the equation yields:
\(E = \frac{k L}{A}\) *(Shown)*

2. **Calculate \(E\):**
\(E = \frac{(2.10 \times 10^4\text{ N m}^{-1}) \times 1.50\text{ m}}{1.662 \times 10^{-7}\text{ m}^2} = 1.895 \times 10^{11}\text{ Pa}\) (or \(\text{N m}^{-2}\)).

---

### Part (c) Solution
1. **Find Percentage Uncertainty in \(L\):**
\(\%\Delta L = \frac{0.01\text{ m}}{1.50\text{ m}} \times 100\% = 0.67\%\)

2. **Find Percentage Uncertainty in \(k\):**
\(\%\Delta k = \frac{0.08 \times 10^4}{2.10 \times 10^4} \times 100\% = 3.81\%\)

3. **Calculate Total Percentage Uncertainty in \(E\):**
\(\%\Delta E = \%\Delta k + \%\Delta L + \%\Delta A = 3.81\% + 0.67\% + 4.35\% = 8.83\%\) (accept \(8.8\%\) to \(9.0\%\) depending on intermediate rounding).

4. **Calculate Absolute Uncertainty in \(E\):**
\(\Delta E = 8.83\% \times 1.895 \times 10^{11}\text{ Pa} = 0.167 \times 10^{11}\text{ Pa} \approx 0.17 \times 10^{11}\text{ Pa}\) (or \(0.2 \times 10^{11}\text{ Pa}\)).

5. **State Final Value with Uncertainty:**
\(E = (1.90 \pm 0.17) \times 10^{11}\text{ Pa}\) or \(E = (1.9 \pm 0.2) \times 10^{11}\text{ Pa}\) (or in GPa: \(190 \pm 17\text{ GPa}\)).

---

### Part (d) Level of Response Evaluation
- **Evaluation of the suggestion:**
- *Measurement of \(L\)*: The student is correct that using digital calipers on a shorter wire length reduces the percentage uncertainty in the original length \(L\) to near zero (\(\%\Delta L = \frac{0.01\text{ mm}}{150\text{ mm}} \times 100\% \approx 0.007\%\)).
- *Effect on Extension*: However, because extension \(x = \frac{F L}{A E}\), a wire that is 10 times shorter will yield an extension that is 10 times smaller for the same range of loads.
- *Effect on Percentage Uncertainty of \(x\)*: Measuring a 10-fold smaller extension with the same apparatus will drastically increase the percentage uncertainty in \(x\), compounding the error in the gradient \(k\).
- *Systematic Errors*: Any slippage of the wire within the clamps/grips acts as a constant absolute systematic error (e.g., \(\approx 0.05\text{ mm}\)). For a short wire with tiny extensions, this slippage becomes a much larger percentage of the measured extension, causing a severe underestimation of the calculated Young Modulus.
- **Recommended Improvements:**
- *Improvement 1*: Attach reference markers directly onto the wire away from the clamps/grips, and measure the displacement of these markers using a travel microscope or high-resolution optical sensor. This eliminates clamp slippage systematic error entirely.
- *Improvement 2*: Take multiple diameter measurements along different parts of the wire and in orthogonal directions using a digital micrometer to calculate an accurate average cross-sectional area, reducing random and systematic variation in \(A\).

### Part (a) [3 Marks]
- **1 Mark:** Correct calculation of \(A = 1.66 \times 10^{-7}\text{ m}^2\) (allow \(1.7 \times 10^{-7}\text{ m}^2\)).
- **1 Mark:** Correct percentage uncertainty in \(d\) (\(2.17\%\) or \(2.2\%\)).
- **1 Mark:** Recognises that the percentage uncertainty in \(A\) is twice that of \(d\), giving \(4.3\%\) to \(4.4\%\).

### Part (b) [3 Marks]
- **1 Mark:** Correct derivation showing \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\) and links gradient \(k = \frac{F}{x}\) to obtain \(E = \frac{k L}{A}\).
- **1 Mark:** Correct substitution of values: \(E = \frac{2.10 \times 10^4 \times 1.50}{1.66 \times 10^{-7}}\).
- **1 Mark:** Calculated value of \(E = 1.9 \times 10^{11}\text{ Pa}\) (allow range \(1.89 \times 10^{11}\) to \(1.90 \times 10^{11}\)).

### Part (c) [3 Marks]
- **1 Mark:** Sums percentage uncertainties correctly: \(\%\Delta E = \%\Delta k + \%\Delta L + \%\Delta A = 3.81\% + 0.67\% + 4.35\% \approx 8.8\%\) (allow ecf from (a)).
- **1 Mark:** Correct absolute uncertainty \(\Delta E\) calculated (\(0.17 \times 10^{11}\text{ Pa}\) or \(0.2 \times 10^{11}\text{ Pa}\)).
- **1 Mark:** Final answer written with consistent significant figures and unit: \(E = (1.9 \pm 0.2) \times 10^{11}\text{ Pa}\) or \((1.90 \pm 0.17) \times 10^{11}\text{ Pa}\) (accept \(\text{N m}^{-2}\) or \(\text{GPa}\)).

### Part (d) [6 Marks] - Level of Response
- **Level 3 (5-6 marks):** Detailed evaluation of both the advantages (reduced \(\%\Delta L\)) and major disadvantages (reduced extension leading to increased \(\%\Delta x\), and severe impact of clamp slippage). Two valid, well-explained improvements are suggested. The discussion is logically structured and uses precise physical terminology.
- **Level 2 (3-4 marks):** Reasonable evaluation of the proposal. Identifies that a shorter wire leads to a smaller extension and thus a higher percentage uncertainty in extension. Suggests at least one valid improvement. The structure is mostly clear but may lack detail on systematic slippage.
- **Level 1 (1-2 marks):** Fragmented discussion. Identifies either the reduction in \(\%\Delta L\) or the reduction in extension size. Minimal or no practical improvements suggested.
- **0 marks:** No response worthy of credit.