2022 Edexcel AS Level Physics (8PH0) Practice Paper with Answers

Young modulus is defined as stress divided by strain. Stress has the unit of pressure, which is force per unit area, \(\text{N m}^{-2}\). In SI base units, force (Newtons) is \(\text{kg m s}^{-2}\), so stress is \(\text{kg m s}^{-2} \times \text{m}^{-2} = \text{kg m}^{-1} \text{s}^{-2}\). Since strain is a ratio of two lengths, it is dimensionless and has no units. Therefore, the SI base units of the Young modulus are \(\text{kg m}^{-1} \text{s}^{-2}\).

1 mark for selecting option B.

The horizontal component of the velocity remains constant at \(v_x = v\). The vertical component of the velocity \(v_y\) just before impact can be found using the equations of motion: \(v_y^2 = u_y^2 + 2as\). Since the stone is projected horizontally, \(u_y = 0\), which gives \(v_y^2 = 2gh\). The magnitude of the final velocity is given by \(V = \sqrt{v_x^2 + v_y^2} = \sqrt{v^2 + 2gh}\).

1 mark for selecting option B.

Since the block is stationary, it is in equilibrium. The component of the gravitational force (weight) acting down the slope is \(mg \sin\theta\). For the block to remain stationary, the static frictional force acting up the slope must balance this component. Therefore, the magnitude of the frictional force is \(f = mg \sin\theta\).

1 mark for selecting option B.

The elastic strain energy stored is given by \(E_{\text{el}} = \frac{1}{2} F \Delta x\). From the definition of Young modulus, \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta x/L}\), which simplifies to \(F = \frac{EA\Delta x}{L}\). Substituting this expression for force into the energy equation yields: \(E_{\text{el}} = \frac{1}{2} \left(\frac{EA\Delta x}{L}\right) \Delta x = \frac{EA(\Delta x)^2}{2L}\).

1 mark for selecting option A.

The initial direction of motion is positive, so the initial momentum is \(p_{\text{initial}} = mu\). Since the ball rebounds in the opposite direction, the final velocity is \(-v\), making the final momentum \(p_{\text{final}} = -mv\). The change in momentum is \(\Delta p = p_{\text{final}} - p_{\text{initial}} = -mv - mu = -m(u + v)\).

1 mark for selecting option B.

At terminal velocity, the downward force of gravity (weight) is equal to the sum of the upward forces (upthrust and viscous drag). Weight \(W = \frac{4}{3}\pi r^3 \rho g\). Upthrust \(U = \frac{4}{3}\pi r^3 \sigma g\). Viscous drag (Stokes' Law) \(F_{\text{drag}} = 6\pi \eta r v\). Equating the forces: \(W = U + F_{\text{drag}}\), which gives \(\frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \sigma g + 6\pi \eta r v\). Solving for \(v\) gives \(v = \frac{2r^2(\rho - \sigma)g}{9\eta}\).

1 mark for selecting option A.

When combining uncertainties for quantities that are multiplied or divided, the percentage uncertainties are added. For any quantity raised to a power, the percentage uncertainty is multiplied by that power. Thus, for \(g = 2s t^{-2}\), the percentage uncertainty in \(g\) is given by: \(\%\text{ uncertainty in } g = \%\text{ uncertainty in } s + 2 \times (\%\text{ uncertainty in } t) = 2.0\% + 2 \times 3.0\% = 8.0\%\).

1 mark for selecting option B.

Efficiency is the ratio of useful power output to total power input. The useful work done in lifting the crate is \(W = mgh\). The useful power output of the motor is the rate of doing this work: \(P_{\text{out}} = \frac{mgh}{t}\). Given that the power input is \(P\), the efficiency is: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{mgh/t}{P} = \frac{mgh}{Pt}\).

1 mark for selecting option C.

(a) The horizontal motion has a constant velocity:
\(x = u \cos(\theta) t \implies t = \frac{x}{u \cos(\theta)}\)

Substituting this into the vertical displacement equation:
\(y = u \sin(\theta) t - \frac{1}{2}gt^2\)
\(y = x \tan(\theta) - \frac{g x^2}{2 u^2 \cos^2(\theta)}\)

Substitute the given values:
\(3.5 = 42 \tan(35^\circ) - \frac{9.81 \times 42^2}{2 u^2 \cos^2(35^\circ)}\)
\(3.5 = 29.41 - \frac{17304.84}{1.342 u^2}\)
\(-25.91 = -\frac{12894.8}{u^2}\)
\(u^2 = 497.7 \implies u = 22.3\text{ m s}^{-1}\) (which is approximately \(22\text{ m s}^{-1}\)).

(b) Air resistance opposes motion, decreasing both horizontal and vertical components of velocity. This means the projectile takes longer to reach the wall, and its maximum height is lower, resulting in a lower height when it hits the wall.

(a) [Total: 4 marks]
- Use of horizontal motion equation to express time: \(t = \frac{42}{u \cos(35^\circ)}\) (1 mark)
- Substitution of \(t\) into vertical motion equation (1 mark)
- Correct algebraic rearrangement to solve for \(u^2\) or \(u\) (1 mark)
- Final calculated answer of \(22.3\text{ m s}^{-1}\) shown clearly (1 mark)

(b) [Total: 2 marks]
- Air resistance reduces both horizontal and vertical velocities / kinetic energy (1 mark)
- Hence, the projectile takes longer to travel the same distance or loses altitude faster, leading to a lower height on impact (1 mark)

(a) Calculate percentage uncertainty for each variable:
- For \(F\): \(\frac{0.5}{45.0} \times 100 = 1.11\%\)
- For \(L\): \(\frac{0.002}{2.150} \times 100 = 0.093\%\)
- For \(d\): \(\frac{0.01}{0.38} \times 100 = 2.63\%\). Since \(d\) is squared: \(2 \times 2.63\% = 5.26\%\)
- For \(x\): \(\frac{0.1}{1.4} \times 100 = 7.14\%\)

Total percentage uncertainty in \(E\) is the sum of these uncertainties:
\(\text{Total Uncertainty} = 1.11\% + 0.093\% + 5.26\% + 7.14\% = 13.603\% \approx 13.6\%\) (or \(14\%\) to 2 s.f.).

(b) The extension \(x\) contributes the most because it has the largest individual percentage uncertainty (\(7.14\%\)). This can be improved by using a longer wire to produce a larger extension for the same load, or using a more precise instrument such as a Vernier scale or travelling microscope instead of a ruler.

(a) [Total: 4 marks]
- Correct percentage uncertainty for \(F\) (1.11%) or \(L\) (0.093%) (1 mark)
- Correct percentage uncertainty for \(d\) and doubling it to get 5.26% (1 mark)
- Correct percentage uncertainty for \(x\) (7.14%) (1 mark)
- Sum of uncertainties giving \(13.6\%\) or \(14\%\) (1 mark)

(b) [Total: 2 marks]
- Identify extension \(x\) as the major source of uncertainty (1 mark)
- Suggestion of a valid improvement (e.g. use a longer wire, a more precise reading instrument, or a digital micrometer for diameter) (1 mark)

(a) At terminal velocity, the upward forces (Upthrust \(U\) and Drag \(F_D\)) equal the downward weight (\(W\)):
\(U + F_D = W \implies 6\pi \eta r v + \frac{4}{3}\pi r^3 \rho_{\text{fluid}} g = \frac{4}{3}\pi r^3 \rho_{\text{sphere}} g\)

Rearranging for terminal velocity \(v\):
\(v = \frac{2 r^2 (\rho_{\text{sphere}} - \rho_{\text{fluid}}) g}{9 \eta}\)

Substitute values:
\(v = \frac{2 \times (4.5 \times 10^{-5})^2 \times (1800 - 920) \times 9.81}{9 \times 0.085}\)
\(v = \frac{2 \times (2.025 \times 10^{-9}) \times 880 \times 9.81}{0.765}\)
\(v = \frac{3.496 \times 10^{-5}}{0.765} = 4.57 \times 10^{-5}\text{ m s}^{-1}\)

(b) Stokes' law is valid only if: 1. The particles are perfectly spherical. 2. The flow of fluid around the sphere is laminar. 3. The fluid is of infinite extent (no wall effects).

(a) [Total: 4 marks]
- Statement/use of equilibrium: \(U + F_D = W\) (1 mark)
- Substitution of formulas to arrive at \(v = \frac{2r^2(\Delta\rho)g}{9\eta}\) (1 mark)
- Correct substitution of values (1 mark)
- Value \(4.57 \times 10^{-5}\text{ m s}^{-1}\) with correct units (1 mark)

(b) [Total: 2 marks]
- Identify any two valid conditions: spherical particle, laminar flow/low speed, no container boundary effects (1 mark per condition, max 2)

(a) First, calculate the energy \(E\) of the incident photons:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{240 \times 10^{-9}} = 8.288 \times 10^{-19}\text{ J}\)

Convert the work function \(\Phi\) from eV to Joules:
\(\Phi = 2.36 \times 1.60 \times 10^{-19}\text{ J} = 3.776 \times 10^{-19}\text{ J}\)

Apply the photoelectric equation:
\(E_{k,\max} = E - \Phi = 8.288 \times 10^{-19} - 3.776 \times 10^{-19} = 4.512 \times 10^{-19}\text{ J} \approx 4.51 \times 10^{-19}\text{ J}\)

(b) The stopping potential \(V_s\) is related to the maximum kinetic energy by:
\(e V_s = E_{k,\max}\)
\(V_s = \frac{4.512 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ C}} = 2.82\text{ V}\)

(a) [Total: 4 marks]
- Correct calculation of photon energy using \(E = hc/\lambda\) (1 mark)
- Conversion of work function from eV to Joules (1 mark)
- Use of Einstein's photoelectric equation (1 mark)
- Correct value for \(E_{k,\max} = 4.51 \times 10^{-19}\text{ J}\) (1 mark)

(b) [Total: 2 marks]
- Relate stopping potential to maximum kinetic energy: \(E_{k} = eV_s\) (1 mark)
- Correct stopping potential calculated as \(2.82\text{ V}\) (1 mark)

(a) The total force opposing the motion along the incline is the sum of the component of the weight down the incline and the frictional force:
\(F_{\text{total}} = mg \sin(\theta) + F_{\text{fric}}\)
\(F_{\text{total}} = 450 \times 9.81 \times \sin(15^\circ) + 650 = 1142.5 + 650 = 1792.5\text{ N}\)

At a steady speed, the pulling force exerted by the winch equals the opposing force.

Using the power equation:
\(P = F \cdot v \implies v = \frac{P}{F_{\text{total}}} = \frac{1500\text{ W}}{1792.5\text{ N}} \approx 0.837\text{ m s}^{-1}\)

(b) The work done against friction is dissipated. This energy is transferred to the thermal energy store of the block, the ramp, and the surrounding air, raising their temperature.

(a) [Total: 4 marks]
- Component of weight down incline: \(mg \sin(\theta) = 1142.5\text{ N}\) (1 mark)
- Total force calculation: \(1792.5\text{ N}\) (1 mark)
- Correct recall of \(P = Fv\) (1 mark)
- Final velocity of \(0.837\text{ m s}^{-1}\) (1 mark)

(b) [Total: 2 marks]
- Heat/thermal energy transfer identified (1 mark)
- Dissipation to surroundings, block, or ramp (1 mark)

(a) First, calculate the cross-sectional area \(A\):
\(A = \pi r^2 = \pi \left(\frac{3.2 \times 10^{-3}}{2}\right)^2 = 8.042 \times 10^{-6}\text{ m}^2\)

Now, calculate stress \(\sigma\) and strain \(\varepsilon\):
\(\sigma = \frac{F}{A} = \frac{120}{8.042 \times 10^{-6}} = 1.492 \times 10^7\text{ Pa}\)

\(\varepsilon = \frac{\Delta L}{L} = \frac{1.4 \times 10^{-3}}{8.5 \times 10^{-2}} = 0.01647\)

Calculate Young modulus \(E\):
\(E = \frac{\sigma}{\varepsilon} = \frac{1.492 \times 10^7}{0.01647} = 9.06 \times 10^8\text{ Pa}\) (or \(0.91\text{ GPa}\))

(b) The elastic strain energy \(E_{\text{el}}\) stored (assuming Hooke's Law is obeyed) is:
\(E_{\text{el}} = \frac{1}{2} F \Delta L = 0.5 \times 120 \times 1.4 \times 10^{-3} = 0.084\text{ J}\)

(a) [Total: 4 marks]
- Correct calculation of cross-sectional area \(A\) (1 mark)
- Correct calculation of stress or strain (1 mark)
- Correct use of \(E = \frac{\text{Stress}}{\text{Strain}}\) (1 mark)
- Correct Young modulus of \(9.06 \times 10^8\text{ Pa}\) or \(0.91\text{ GPa}\) (1 mark)

(b) [Total: 2 marks]
- Use of \(E_{\text{el}} = \frac{1}{2}F\Delta L\) (1 mark)
- Correct value of \(0.084\text{ J}\) (1 mark)

(a) First, calculate the grating spacing \(d\):
\(d = \frac{1}{400 \times 10^3\text{ lines m}^{-1}} = 2.50 \times 10^{-6}\text{ m}\)

Determine the angle \(\theta\) of the second-order maximum:
\(\sin(\theta) = \frac{x}{\sqrt{D^2 + x^2}} = \frac{0.95}{\sqrt{1.80^2 + 0.95^2}} = \frac{0.95}{2.0353} = 0.4668\)

Apply the diffraction grating equation \(d \sin(\theta) = n \lambda\) for \(n=2\):
\(2.50 \times 10^{-6} \times 0.4668 = 2 \lambda\)

\(\lambda = 5.835 \times 10^{-7}\text{ m} \approx 5.83 \times 10^{-7}\text{ m}\) (or \(583\text{ nm}\))

(b) If a grating with \(600\text{ lines per mm}\) is used, the grating spacing \(d\) decreases. From \(\sin(\theta) = \frac{n\lambda}{d}\), a smaller \(d\) leads to a larger value for \(\sin(\theta)\), meaning the angle \(\theta\) increases. Consequently, the second-order maximum moves further away from the central zero-order maximum on the screen.

(a) [Total: 4 marks]
- Correct calculation of grating spacing \(d = 2.50 \times 10^{-6}\text{ m}\) (1 mark)
- Correct determination of \(\theta\) or \(\sin(\theta) = 0.467\) (1 mark)
- Correct substitution into \(d \sin(\theta) = n \lambda\) (1 mark)
- Wavelength calculated as \(5.83 \times 10^{-7}\text{ m}\) or \(583\text{ nm}\) (1 mark)

(b) [Total: 2 marks]
- Recognize that more lines per mm decreases the spacing \(d\) (1 mark)
- Conclude that angle \(\theta\) (and distance from central maximum) increases (1 mark)

(a) Use conservation of linear momentum:
\(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\)
\((12000 \times 4.2) + (8500 \times 0) = (12000 + 8500) v\)
\(50400 = 20500 v \implies v = 2.4585 \approx 2.46\text{ m s}^{-1}\)

(b) The impulse \(I\) is the change in momentum of either carriage. For the second carriage:
\(I = \Delta p = m_2 (v - u_2) = 8500 \times (2.4585 - 0) = 2.09 \times 10^4\text{ N s}\)

Alternatively, checking with the first carriage:
\(I = |12000 \times (2.4585 - 4.2)| = |12000 \times (-1.7415)| = 2.09 \times 10^4\text{ N s}\).

(a) [Total: 3 marks]
- Correct expression for conservation of momentum (1 mark)
- Correct substitution of values (1 mark)
- Correct calculation of velocity as \(2.46\text{ m s}^{-1}\) (1 mark)

(b) [Total: 3 marks]
- Correct definition of impulse as change in momentum (1 mark)
- Calculation showing change in momentum of one carriage (1 mark)
- Correct magnitude of \(2.09 \times 10^4\text{ N s}\) (or \(\text{kg m s}^{-1}\)) (1 mark)

**Part (a)**
The resistance of a uniform conductor of length \(l\) and cross-sectional area \(A\) is given by:
\(R = \frac{\rho l}{A}\)
Since resistivity \(\rho\) and cross-sectional area \(A\) are constants, \(R \propto l\).
Comparing this to the equation of a straight line, \(y = mx + c\), a plot of \(R\) against \(l\) yields a straight line with gradient \(m = \frac{\rho}{A}​\) and a y-intercept \(c = 0\), meaning it passes through the origin.

**Part (b)**
To obtain a reliable and accurate value for the diameter:
1. Check the micrometer for zero error before starting and record any offset to correct future readings.
2. Measure the diameter at several different positions along the wire, and at perpendicular orientations at each position to account for any non-circularity.
3. Calculate the mean diameter from these readings, subtracting any zero error.

**Part (c)**
The gradient of the graph is given by:
\(m = \frac{\rho}{A} \implies \rho = m \times A\)

The cross-sectional area \(A\) is:
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3} \text{ m})^2}{4} = 1.134 \times 10^{-7} \text{ m}^2\)

Substitute the gradient and area to calculate \(\rho\):
\(\rho = 4.52 \text{ }\Omega\text{ m}^{-1} \times 1.134 \times 10^{-7} \text{ m}^2 = 5.126 \times 10^{-7} \text{ }\Omega\text{ m}\)

Rounding to 2 significant figures gives \(5.1 \times 10^{-7} \text{ }\Omega\text{ m}\).

**Part (d)**
As the temperature of the constantan wire rises, its resistance increases. This causes the measured resistances at larger lengths to be higher than they would be at constant temperature, leading to a steeper slope (larger gradient) and an overestimate of the resistivity \(\rho\).
To minimize this effect, the student should use a switch to turn off the current between measurements so that the wire does not heat up continually, or use a variable resistor to keep the current as small as possible.

**Part (a) [2 Marks]**
- State \(R = \frac{\rho l}{A}\) and identify that resistivity \(\rho\) and area \(A\) are constant. (1 mark)
- Explain that because \(R \propto l\), the graph is a straight line through the origin. (1 mark)

**Part (b) [3 Marks]**
- Check for and correct any zero error on the micrometer. (1 mark)
- Measure diameter at multiple positions and orientations along the wire. (1 mark)
- Calculate the mean of these measurements. (1 mark)

**Part (c) [3.5 Marks]**
- Correct equation for cross-sectional area: \(A = \frac{\pi d^2}{4}\). (1 mark)
- Correct calculation of area: \(A = 1.13 \times 10^{-7} \text{ m}^2\). (0.5 marks)
- Correct algebraic rearrangement: \(\rho = \text{gradient} \times A\). (1 mark)
- Correct final calculation of resistivity in the range \(5.1 \times 10^{-7}\) to \(5.13 \times 10^{-7} \text{ }\Omega\text{ m}\) with correct unit. (1 mark)

**Part (d) [2 Marks]**
- Explain that higher temperature increases resistance/resistivity, leading to an overestimated gradient/resistivity value. (1 mark)
- Suggest a suitable modification, such as opening the switch between readings to minimize heating. (1 mark)

**Part (a)**
Safety precautions include:
1. Wear safety goggles to protect eyes in case the wire snaps under tension.
2. Place a padded box, cushion, or sand tray directly beneath the suspended masses to catch falling weights and protect feet.

**Part (b)**
The Young modulus is defined as:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{x/L} = \frac{F \cdot L}{A \cdot x}\)

Since the gradient of the force-extension graph is \(m = \frac{F}{x}\):
\(E = m \times \frac{L}{A}\)

The cross-sectional area \(A\) is calculated as:
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.28 \times 10^{-3} \text{ m})^2}{4} = 6.158 \times 10^{-8} \text{ m}^2\)

Substitute the values:
\(E = (2.52 \times 10^3 \text{ N m}^{-1}) \times \frac{2.45 \text{ m}}{6.158 \times 10^{-8} \text{ m}^2} = 1.0026 \times 10^{11} \text{ Pa}\)

Rounding to 2 significant figures (due to the 0.28 mm diameter) gives \(1.0 \times 10^{11} \text{ Pa}\).

**Part (c)**
Slippage of the wire causes the recorded extension \(x\) to be larger than the actual extension of the wire for a given load \(F\).
- This results in an increased recorded value of \(x\) for each load, which decreases the gradient \(\frac{F}{x}\) of the graph.
- Since the calculated Young modulus \(E\) is directly proportional to the gradient, the calculated value of \(E\) will be lower than the true value (an underestimate).

**Part (d)**
The gradient \(m\) is given by:
\(m = \frac{E \cdot A}{L}\)
Since \(A \propto d^2\), doubling the diameter of the wire increases the cross-sectional area by a factor of \(2^2 = 4\).
As the material (Young modulus \(E\)) and the length \(L\) remain unchanged, the gradient of the force-extension graph will increase by a factor of 4.

**Part (a) [2 Marks]**
- Wear safety goggles to protect eyes from a snapping wire. (1 mark)
- Place a sand tray/soft padding under the weights to protect feet/floor. (1 mark)

**Part (b) [3.5 Marks]**
- Identify the relationship between the Young modulus and the gradient: \(E = \text{gradient} \times \frac{L}{A}\). (1 mark)
- Correct calculation of cross-sectional area: \(A = 6.16 \times 10^{-8} \text{ m}^2\). (1 mark)
- Correct substitution of values into the Young modulus equation. (0.5 marks)
- Correct calculation of the Young modulus: \(1.0 \times 10^{11} \text{ Pa}\) (allow \(1.00 \times 10^{11} \text{ Pa}\) or values in the range \(1.00 \times 10^{11}\) to \(1.01 \times 10^{11} \text{ Pa}\)). (1 mark)

**Part (c) [3 Marks]**
- Identify that slippage increases the recorded extension \(x\) for a given load. (1 mark)
- State that this decreases the gradient of the graph. (1 mark)
- Conclude that the calculated Young modulus is underestimated/lower than the true value. (1 mark)

**Part (d) [2 Marks]**
- State that the gradient is proportional to the cross-sectional area (or \(d^2\)). (1 mark)
- Conclude that the gradient increases by a factor of 4. (1 mark)

Unpolarised light passing through the first polariser has its intensity halved, so \(I_1 = 0.5 I_0\). According to Malus's Law, the intensity after the second polariser is \(I_2 = I_1 \cos^2(\theta) = 0.5 I_0 \cos^2(30^\circ) = 0.5 I_0 \times 0.75 = 0.375 I_0\).

1 mark for the correct calculation leading to 0.375 \(I_0\) (C).

Young Modulus \(E = \frac{FL}{A x}\), which rearranges to \(x = \frac{FL}{AE}\). For the second wire, the length is \(2L\) and the area is \(2A\). Substituting these values gives \(x_2 = \frac{F(2L)}{(2A)E} = \frac{FL}{AE} = x\).

1 mark for identifying that the factors of 2 cancel out, yielding the same extension (B).

The critical angle \(\theta_c\) is calculated using \(\sin\theta_c = \frac{n_2}{n_1}\) where \(n_1 > n_2\). This gives \(\sin\theta_c = \frac{1.33}{1.52} \approx 0.875\). Taking the inverse sine yields \(\theta_c \approx 61.0^\circ\).

1 mark for the correct application of the critical angle formula leading to 61.0 degrees (C).

The photoelectric equation is \(hf = \phi + E_k\), which means \(E_k = hf - \phi\). When the frequency is doubled, the new maximum kinetic energy is \(E_k' = h(2f) - \phi = 2hf - \phi\). Since \(hf = E_k + \phi\), substituting this gives \(E_k' = 2(E_k + \phi) - \phi = 2E_k + \phi\).

1 mark for using the photoelectric equation to show the new kinetic energy is \(2E_k + \phi\) (B).

The work done to extend the spring by \(x\) is \(W = \frac{1}{2} k x^2\). The total work done to extend it by \(2x\) is \(W_{\text{total}} = \frac{1}{2} k (2x)^2 = 4 \left(\frac{1}{2} k x^2\right) = 4W\). Therefore, the additional work done for the second extension is \(W_{\text{total}} - W = 4W - W = 3W\).

1 mark for calculating the difference between total work at 2x and work at x to get 3W (C).

Using \(d \sin \theta = n \lambda\), we find the grating spacing \(d = \frac{n \lambda}{\sin \theta} = \frac{3 \times 6.0 \times 10^{-7}\text{ m}}{\sin(30^\circ)} = 3.6 \times 10^{-6}\text{ m}\). The number of lines per millimetre is \(\frac{1 \times 10^{-3}\text{ m}}{d} = \frac{10^{-3}\text{ m}}{3.6 \times 10^{-6}\text{ m}} \approx 278\text{ lines/mm}\).

1 mark for calculating d and converting to lines per millimetre to get 278 (B).

At terminal velocity, the net downward force is zero: \(\text{Weight} - \text{Upthrust} = \text{Stokes' Drag}\). This gives \(\frac{4}{3}\pi r^3 (\rho_s - \rho_f)g = 6\pi \eta r v\). Solving for terminal velocity yields \(v = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}\), showing \(v \propto r^2\). Doubling the radius \(r\) increases the terminal velocity by a factor of \(2^2 = 4\).

1 mark for identifying the proportional relationship \(v \propto r^2\) and finding the new velocity as 4v (B).

The fundamental frequency of a string is given by \(f_0 = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\). If the tension \(T\) is doubled to \(2T\), the new frequency is \(f' = \frac{1}{2L}\sqrt{\frac{2T}{\mu}} = \sqrt{2} \times f_0 \approx 1.41 f_0\).

1 mark for identifying that frequency is proportional to the square root of tension, giving \(1.41 f_0\) (B).

(a) First, calculate the grating spacing \(d\):
\(d = \frac{1}{N} = \frac{1}{600 \times 10^3\text{ m}^{-1}} = 1.67 \times 10^{-6}\text{ m}\).

Using the grating equation \(d \sin\theta = n\lambda\) for \(n = 2\):
\(\lambda = \frac{d \sin\theta}{n} = \frac{1.667 \times 10^{-6}\text{ m} \times \sin(38.5^\circ)}{2}\)
\(\lambda = \frac{1.667 \times 10^{-6} \times 0.6225}{2} = 5.19 \times 10^{-7}\text{ m}\) (which is approximately \(5.2 \times 10^{-7}\text{ m}\)).

(b) For the maximum possible order, the maximum angle of diffraction is \(\theta = 90^\circ\), so \(\sin\theta = 1\):
\(n_{\text{max}} = \frac{d}{\lambda} = \frac{1.667 \times 10^{-6}\text{ m}}{5.19 \times 10^{-7}\text{ m}} = 3.21\).

Since the order \(n\) must be an integer, the highest observable order is \(n = 3\).

The total number of visible bright fringes on the screen includes the central maximum (zero-order) and three orders on either side:
\(N_{\text{fringes}} = 2(3) + 1 = 7\).

Part (a) [4 Marks]:
- Convert lines per mm to lines per m to find \(d\) (1)
- Correct calculation of \(d = 1.67 \times 10^{-6}\text{ m}\) (1)
- Use of \(d \sin\theta = n\lambda\) rearranged for \(\lambda\) (1)
- Substitution leading to correct wavelength of \(5.19 \times 10^{-7}\text{ m}\) (1)

Part (b) [3 Marks]:
- Use of \(\theta = 90^\circ\) or \(\sin\theta = 1\) (1)
- Calculation of maximum order \(n = 3\) as integer (1)
- Final calculation of 7 total fringes (1)

(a) The critical angle \(\theta_c\) is given by:
\(\sin\theta_c = \frac{n_{\text{cladding}}}{n_{\text{core}}} = \frac{1.42}{1.48} = 0.9595\)
\(\theta_c = \sin^{-1}(0.9595) = 73.64^\circ \approx 73.6^\circ\).

(b) The speed of light in the core is given by:
\(v = \frac{c}{n_{\text{core}}} = \frac{3.00 \times 10^8\text{ m s}^{-1}}{1.48} = 2.027 \times 10^8\text{ m s}^{-1} \approx 2.03 \times 10^8\text{ m s}^{-1}\).

(c) Let \(L = 500\text{ m}\). The straight path travel time is:
\(t_1 = \frac{L}{v} = \frac{500}{2.027 \times 10^8} = 2.467 \times 10^{-6}\text{ s}\).

The maximum path length taken by a ray at the critical angle is:
\(s = \frac{L}{\sin\theta_c} = L \times \frac{n_{\text{core}}}{n_{\text{cladding}}} = 500 \times \frac{1.48}{1.42} = 521.13\text{ m}\).

The time taken along this path is:
\(t_2 = \frac{s}{v} = \frac{521.13}{2.027 \times 10^8} = 2.571 \times 10^{-6}\text{ s}\).

The maximum time delay is:
\(\Delta t = t_2 - t_1 = 2.571 \times 10^{-6}\text{ s} - 2.467 \times 10^{-6}\text{ s} = 1.04 \times 10^{-7}\text{ s}\).

Part (a) [2 Marks]:
- Use of \(\sin\theta_c = n_2 / n_1\) (1)
- Correct critical angle of \(73.6^\circ\) (1)

Part (b) [2 Marks]:
- Use of \(v = c / n\) (1)
- Correct speed of \(2.03 \times 10^8\text{ m s}^{-1}\) (1)

Part (c) [3 Marks]:
- Correct expression or value for total path length at critical angle (1)
- Correctly calculating times \(t_1\) and \(t_2\) (1)
- Final time delay calculated as \(1.04 \times 10^{-7}\text{ s}\) (1)

(a) Calculate energy using Planck's relationship:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{225 \times 10^{-9}\text{ m}} = 8.84 \times 10^{-19}\text{ J}\).

(b) First, convert the work function from eV to J:
\(\Phi = 2.36\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.776 \times 10^{-19}\text{ J}\).

According to Einstein's photoelectric equation:
\(hf = \Phi + E_{k,\text{max}}\)
\(E_{k,\text{max}} = hf - \Phi = 8.84 \times 10^{-19}\text{ J} - 3.776 \times 10^{-19}\text{ J} = 5.064 \times 10^{-19}\text{ J} \approx 5.06 \times 10^{-19}\text{ J}\).

(c) The maximum speed of the photoelectrons is calculated using kinetic energy:
\(E_{k,\text{max}} = \frac{1}{2} m_e v^2 \implies v = \sqrt{\frac{2 E_{k,\text{max}}}{m_e}}\)
\(v = \sqrt{\frac{2 \times 5.064 \times 10^{-19}\text{ J}}{9.11 \times 10^{-31}\text{ kg}}} = 1.054 \times 10^6\text{ m s}^{-1} \approx 1.05 \times 10^6\text{ m s}^{-1}\).

Part (a) [2 Marks]:
- Correct use of \(E = hf\) or \(E = hc/\lambda\) (1)
- Correct calculation to get \(8.84 \times 10^{-19}\text{ J}\) (1)

Part (b) [3 Marks]:
- Correct conversion of work function from eV to J (1)
- Recall/use of photoelectric equation \(E = \Phi + E_{k,\text{max}}\) (1)
- Correct value of \(5.06 \times 10^{-19}\text{ J}\) (1)

Part (c) [2 Marks]:
- Use of \(E_k = \frac{1}{2} m v^2\) rearranged for speed (1)
- Correct speed value of \(1.05 \times 10^6\text{ m s}^{-1}\) (1)

(a) A plane polarised wave is a transverse wave in which the oscillations are restricted to a single plane that is perpendicular to the direction of energy propagation.

(b) Since unpolarised light contains waves oscillating in all random orientations perpendicular to propagation, passing through an ideal linear polariser absorbs all component oscillations except those parallel to its axis. This reduces the overall intensity of the beam by exactly half:
\(I_1 = \frac{I_0}{2} = \frac{360\text{ W m}^{-2}}{2} = 180\text{ W m}^{-2}\).

(c) Applying Malus's Law to the polarised light entering the second filter:
\(I_2 = I_1 \cos^2\theta\)
\(I_2 = 180 \times \cos^2(40^\circ) = 180 \times (0.7660)^2 = 180 \times 0.5868 = 105.6\text{ W m}^{-2} \approx 106\text{ W m}^{-2}\).

Part (a) [2 Marks]:
- Reference to oscillations being restricted to a single direction/plane (1)
- Stating that this plane is perpendicular to the direction of propagation (1)

Part (b) [2 Marks]:
- Explanation that components perpendicular to the axis are absorbed, yielding half intensity (1)
- Correct calculation of \(I_1 = 180\text{ W m}^{-2}\) (1)

Part (c) [3 Marks]:
- Recall of Malus's Law \(I = I_0 \cos^2\theta\) (1)
- Correct substitution of \(180\) and \(40^\circ\) (1)
- Correct calculation yielding \(106\text{ W m}^{-2}\) (accept \(105.6\text{ W m}^{-2}\)) (1)

(a) Mass per unit length is defined as mass divided by length:
\(\mu = \frac{m}{L} = \frac{4.1 \times 10^{-3}\text{ kg}}{0.82\text{ m}} = 5.0 \times 10^{-3}\text{ kg m}^{-1}\).

(b) Wave speed on a stretched string is given by:
\(v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{120\text{ N}}{5.0 \times 10^{-3}\text{ kg m}^{-1}}} = \sqrt{24000} = 154.9\text{ m s}^{-1} \approx 155\text{ m s}^{-1}\).

(c) For a wire fixed at both ends vibrating in its fundamental mode, the length of the wire is equal to half a wavelength:
\(L = \frac{\lambda}{2} \implies \lambda = 2 L = 2 \times 0.82\text{ m} = 1.64\text{ m}\).

Using the wave equation \(v = f\lambda\):
\(f = \frac{v}{\lambda} = \frac{154.9\text{ m s}^{-1}}{1.64\text{ m}} = 94.45\text{ Hz} \approx 94.5\text{ Hz}\).

Part (a) [1 Mark]:
- Correct division of mass by length yielding \(5.0 \times 10^{-3}\text{ kg m}^{-1}\) (1)

Part (b) [3 Marks]:
- Recall/write formula \(v = \sqrt{T/\mu}\) (1)
- Substitute correct values of tension and mass per unit length (1)
- Correct calculation of speed as \(155\text{ m s}^{-1}\) (accept \(154.9\text{ m s}^{-1}\)) (1)

Part (c) [3 Marks]:
- Connect fundamental wavelength to wire length \(\lambda = 2L\) (1)
- Use of \(v = f\lambda\) (1)
- Correct calculation of \(94.5\text{ Hz}\) (accept range \(94.4\text{ Hz} - 95.0\text{ Hz}\) depending on rounding) (1)

(a) Using the wave equation:
\(v = f\lambda \implies \lambda = \frac{v}{f} = \frac{340\text{ m s}^{-1}}{850\text{ Hz}} = 0.40\text{ m}\).

(b) The relationship between phase difference \(\Delta\phi\) and path difference (distance) \(\Delta x\) is given by:
\(\Delta\phi = \frac{2\pi \Delta x}{\lambda}\)

Rearranging to solve for \(\Delta x\):
\ \Delta x = \frac{\Delta\phi \times \lambda}{2\pi} = \frac{1.5\text{ rad} \times 0.40\text{ m}}{2\pi} = \frac{0.60}{6.283} = 0.0955\text{ m} \approx 0.095\text{ m}\) (or \(9.5\text{ cm}\)).

(c) Since sound is a longitudinal wave, an air molecule at point \(A\) oscillates with simple harmonic motion back and forth along a line parallel to the direction of energy propagation. The molecule does not experience any net displacement.

Part (a) [2 Marks]:
- Use of \(v = f\lambda\) (1)
- Correct calculation of \(0.40\text{ m}\) (1)

Part (b) [3 Marks]:
- State or use relation \(\Delta\phi / 2\pi = \Delta x / \lambda\) (1)
- Substitute \(\lambda = 0.40\text{ m}\) and \(\Delta\phi = 1.5\text{ rad}\) (1)
- Correctly solve for distance as \(0.095\text{ m}\) (accept \(0.095\text{ m}\) to \(0.096\text{ m}\)) (1)

Part (c) [2 Marks]:
- State that the molecule oscillates / moves back and forth (1)
- State that this motion is parallel to the direction of wave travel / energy transfer (1)

(a) Tensile Stress (\(\sigma\)) is force per unit area:
\(\sigma = \frac{F}{A} = \frac{45\text{ N}}{1.5 \times 10^{-6}\text{ m}^2} = 3.0 \times 10^7\text{ Pa}\).

Tensile Strain (\(\epsilon\)) is extension per unit original length:
\(\epsilon = \frac{\Delta L}{L} = \frac{0.60 \times 10^{-3}\text{ m}}{2.2\text{ m}} = 2.73 \times 10^{-4}\).

(b) Young Modulus (\(E\)) is the ratio of stress to strain:
\(E = \frac{\sigma}{\epsilon} = \frac{3.0 \times 10^7\text{ Pa}}{2.73 \times 10^{-4}} = 1.10 \times 10^{11}\text{ Pa} \approx 1.1 \times 10^{11}\text{ Pa}\).

(c) The expression for extension is:
\(\Delta L = \frac{F L}{A E} = \frac{F L}{\pi (d/2)^2 E} = \frac{4 F L}{\pi d^2 E}\)

Thus, extension is directly proportional to length \(L\) and inversely proportional to diameter squared \(d^2\):
\(\Delta L \propto \frac{L}{d^2}\)

If diameter is doubled, the area becomes four times larger, which quarters the extension. Since length is also doubled, this doubles the extension. Combining these effects:
\(\text{New extension} = 2 \times \frac{1}{4} \times \Delta L_0 = \frac{1}{2} \Delta L_0 = 0.30\text{ mm}\).
The extension is halved.

Part (a) [3 Marks]:
- Calculate stress as \(3.0 \times 10^7\text{ Pa}\) (with unit) (1)
- Calculate strain as \(2.7 \times 10^{-4}\) (no unit) (1)
- Conversion of mm to m correct in strain formula (1)

Part (b) [2 Marks]:
- Recall and use of \(E = \sigma / \epsilon\) (1)
- Correct calculation yielding \(1.1 \times 10^{11}\text{ Pa}\) (1)

Part (c) [2 Marks]:
- State that extension is proportional to \(L/d^2\) or analyze individual effects (1)
- Conclude that the extension is halved / becomes 0.30 mm (1)

(a)
1. Calculate the mass of the bronze statue:
m = \rho_b \times V = 8600\text{ kg m}^{-3} \times 0.0450\text{ m}^3 = 387\text{ kg}

2. Calculate the weight of the statue:
W = mg = 387\text{ kg} \times 9.81\text{ m s}^{-2} = 3796.5\text{ N}

3. Calculate the upthrust acting on the statue:
U = \rho_w \times V \times g = 1025\text{ kg m}^{-3} \times 0.0450\text{ m}^3 \times 9.81\text{ m s}^{-2} = 452.5\text{ N}

4. Since the velocity is constant, the forces are in equilibrium:
T + U = W \implies T = W - U = 3796.5\text{ N} - 452.5\text{ N} = 3344\text{ N} \approx 3340\text{ N}

(b)
1. Viscous drag force increases as the velocity of the statue increases (either proportionally or with velocity squared).
2. When accelerating upwards, a net upward resultant force is required, so
T + U - W - F_D = ma \implies T = W - U + F_D + ma
3. The tension must now overcome both the downward drag force and provide the mass with upward acceleration, hence the tension is significantly higher than in the constant-velocity case.

(c)
1. The weight of the statue in air is
W_{air} = 387\text{ kg} \times 9.81\text{ m s}^{-2} = 3796.5\text{ N}. This equals the static tension in the cable.
2. Calculate the extension \(\Delta x\) of the cable:
\Delta x = \frac{FL}{AE} = \frac{3796.5\text{ N} \times 25.0\text{ m}}{(1.20 \times 10^{-4}\text{ m}^2) \times (2.00 \times 10^{11}\text{ Pa})} = 3.955 \times 10^{-3}\text{ m}
3. Calculate the elastic strain energy stored:
E_{el} = \frac{1}{2} F \Delta x = 0.5 \times 3796.5\text{ N} \times 3.955 \times 10^{-3}\text{ m} = 7.51\text{ J}

(a) [4 marks]
* M1: Use of \(m = \rho V\) to calculate mass of the statue (387 kg).
* M1: Use of \(W = mg\) and \(U = \rho_w V g\) to calculate weight (3796.5 N) and upthrust (452.5 N).
* M1: Setting up correct equilibrium equation: \(T = W - U\).
* A1: Correct final answer for tension to 3 s.f. (3340 N) [Accept range: 3340 N - 3350 N].

(b) [3 marks]
* B1: States that drag force increases as velocity increases.
* B1: Explains that acceleration requires a net upward force (\(F = ma\)).
* B1: Links both factors to explain that the cable must support the weight, overcome drag, and provide accelerating force, leading to a higher tension (\(T = W - U + F_D + ma\)).

(c) [3 marks]
* M1: Substitution of \(F = 3800\text{ N}\) (or \(3796.5\text{ N}\)) into \(\Delta x = \frac{FL}{AE}\) or equivalent energy formula \(E_{el} = \frac{F^2 L}{2AE}\).
* M1: Correct calculation of extension \(\Delta x = 3.96 \times 10^{-3}\text{ m}\) (or correct substitution in energy equation).
* A1: Correct elastic strain energy: \(7.51\text{ J}\) (accept range 7.50 J - 7.53 J).

(a)
1. Calculate the grating spacing \(d\):
d = \frac{1 \times 10^{-3}\text{ m}}{600} = 1.667 \times 10^{-6}\text{ m}
2. Use the grating equation \(d \sin\theta = n \lambda\) for \(n = 1\):
\lambda = (1.667 \times 10^{-6}\text{ m}) \times \sin(20.7^\circ)
3. Calculate the value of \(\lambda\):
\lambda = 1.667 \times 10^{-6} \times 0.35347 = 5.895 \times 10^{-7}\text{ m} \approx 5.90 \times 10^{-7}\text{ m} (shown to be approximately \(5.9 \times 10^{-7}\text{ m}\)).

(b)
1. Calculate the photon energy in Joules using \(E = \frac{hc}{\lambda}\):
E = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{5.90 \times 10^{-7}\text{ m}} = 3.371 \times 10^{-19}\text{ J}
2. Convert this photon energy to electronvolts:
E_{\text{eV}} = \frac{3.371 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J/eV}} = 2.11\text{ eV}
3. Set up the transition energy equation:
\Delta E = E_{\text{upper}} - E_{\text{lower}} \implies 2.11\text{ eV} = -1.93\text{ eV} - E_{\text{lower}}
4. Solve for \(E_{\text{lower}}\):
E_{\text{lower}} = -1.93\text{ eV} - 2.11\text{ eV} = -4.04\text{ eV}

(c)
1. The maximum energy of a photon in the incident light beam is \(E_{\text{photon}} = 2.11\text{ eV}\).
2. The minimum energy required to liberate an electron from the zinc plate is its work function, \(\Phi = 4.3\text{ eV}\).
3. Since \(E_{\text{photon}} < \Phi\) (\(2.11\text{ eV} < 4.3\text{ eV}\)), no photoelectrons will be emitted because the photons do not have sufficient energy to overcome the electrostatic attraction holding the conduction electrons to the metal surface.

(a) [3 marks]
* M1: Correct calculation of grating spacing \(d = 1.67 \times 10^{-6}\text{ m}\) (or substitution of \(1/600\text{ mm}\)).
* M1: Rearrangement of \(d \sin\theta = n \lambda\) to make \(\lambda\) the subject and substitution of \(n = 1\) and \(\theta = 20.7^\circ\).
* A1: Show a calculated value of at least 3 s.f. (e.g., \(5.90 \times 10^{-7}\text{ m}\) or \(5.89 \times 10^{-7}\text{ m}\)) to justify the approximation.

(b) [4 marks]
* M1: Use of \(E = hc/\lambda\) or \(E = hf\).
* A1: Calculation of energy in Joules as \(3.37 \times 10^{-19}\text{ J}\).
* M1: Dividing the energy in Joules by \(1.60 \times 10^{-19}\text{ J/eV}\) to obtain energy in eV (\(2.11\text{ eV}\)).
* A1: Correct calculation of lower energy level as \(-4.04\text{ eV}\) (accept \(-4.0\text{ eV}\) to \(-4.1\text{ eV}\)).

(c) [3 marks]
* B1: Identification of photon energy as \(2.11\text{ eV}\) and work function as \(4.3\text{ eV}\).
* B1: Quantitative comparison showing \(E_{\text{photon}} < \Phi\).
* B1: Clear conclusion that no photoelectrons will be emitted because individual photon energy is insufficient.