2025 Cambridge IAL Physics (9702) Practice Paper with Answers

To find the percentage uncertainty in the density \(\rho\), we use the rule for combining uncertainties in a product/quotient: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\).

Substituting the given percentage uncertainties:
\(\frac{\Delta \rho}{\rho} = 1.5\% + 2(2.0\%) + 1.0\%\)
\(\frac{\Delta \rho}{\rho} = 1.5\% + 4.0\% + 1.0\% = 6.5\%\).

1 mark for the correct calculation of the combined percentage uncertainty, showing the factor of 2 applied to the diameter's percentage uncertainty.

Rearranging the formula for \(\mu_0\) gives:
\(\mu_0 = \frac{2\pi d}{I_1 I_2} \cdot \frac{F}{L}\)

Now, substitute the SI base units for each quantity:
- Distance \([d] = \text{m}\)
- Current \([I_1 I_2] = \text{A}^2\)
- Force \([F] = \text{kg m s}^{-2}\)
- Length \([L] = \text{m}\)

This gives:
\([\mu_0] = \frac{\text{m}}{\text{A}^2} \cdot \frac{\text{kg m s}^{-2}}{\text{m}} = \text{kg m s}^{-2} \text{A}^{-2}\).

1 mark for correctly analyzing the base units of each quantity and simplifying to obtain the correct SI base unit for permeability.

First, make reasonable estimates of the mass and speed of the car:
- Mass of a family car, \(m \approx 1500 \text{ kg}\)
- Speed of the car, \(v = 100 \text{ km h}^{-1} = \frac{100 \times 10^3 \text{ m}}{3600 \text{ s}} \approx 28 \text{ m s}^{-1}\)

Now calculate the kinetic energy \(E_k\):
\(E_k = \frac{1}{2} m v^2 \approx \frac{1}{2} \times 1500 \times (28)^2 \approx 750 \times 784 \approx 5.9 \times 10^5 \text{ J}\)

This value is closest to \(6 \times 10^5 \text{ J}\).

1 mark for estimating the mass and speed of the car correctly and performing the kinetic energy calculation to identify the correct order of magnitude.

The delay in starting the stopwatch is due to human reaction time, which acts as a systematic error. Because the stopwatch is started slightly late, the measured time of fall is consistently shorter than the actual time of fall (\(t_{\text{measured}} < t_{\text{actual}}\)). Since \(g = \frac{2h}{t^2}\), a smaller recorded time \(t\) results in a calculated value of \(g\) that is systematically higher than the true value.

1 mark for identifying that the reaction time delay at the start represents a systematic error that makes the measured time too short and the calculated value of g too high.

First, calculate the redshift \(z\):
\(z = \frac{\Delta \lambda}{\lambda} = \frac{682.1 - 656.3}{656.3} = \frac{25.8}{656.3} \approx 0.03931\)

Next, determine the recession speed \(v\) of the galaxy:
\(v = z c = 0.03931 \times 3.00 \times 10^8 \text{ m s}^{-1} \approx 1.18 \times 10^7 \text{ m s}^{-1}\)

Finally, use Hubble's Law (\(v = H_0 d\)) to calculate the distance \(d\):
\(d = \frac{v}{H_0} = \frac{1.18 \times 10^7}{2.3 \times 10^{-18}} \approx 5.1 \times 10^{24} \text{ m}\).

1 mark for correctly applying the redshift equation, finding the recession velocity, and using Hubble's law to solve for distance.

Luminosity is given by the Stefan-Boltzmann law: \(L = 4\pi R^2 \sigma T^4\).

For the second star:
\(L_2 = 4\pi (2.0 R)^2 \sigma (1.5 T)^4\)
\(L_2 = 4\pi (4.0 R^2) \sigma (5.0625 T^4)\)
\(L_2 = 4.0 \times 5.0625 \times (4\pi R^2 \sigma T^4)\)
\(L_2 = 20.25 L \approx 20 L\).

1 mark for substituting the new radius and temperature into the luminosity equation and correctly working out the scaling factor.

Initial magnetic flux linkage, \(\Phi_1 = N B A \cos(0^\circ) = 150 \times 0.20 \times (4.0 \times 10^{-4}) = 0.012 \text{ Wb}\).

Final magnetic flux linkage, \(\Phi_2 = N B A \cos(90^\circ) = 0\).

The change in magnetic flux linkage, \(\Delta \Phi = |\Phi_2 - \Phi_1| = 0.012 \text{ Wb}\).

Using Faraday's law, the average induced e.m.f. \(E\) is:
\(E = \frac{\Delta \Phi}{\Delta t} = \frac{0.012 \text{ Wb}}{0.15 \text{ s}} = 0.080 \text{ V}\).

1 mark for calculating the change in flux linkage and dividing by the time interval to find the average induced e.m.f.

The Young modulus \(E\) is given by:
\(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}\)

Therefore, the extension \(x\) is:
\(x = \frac{FL}{AE} = \frac{FL}{\left(\frac{\pi d^2}{4}\right)E} = \frac{4FL}{\pi d^2 E}\)

For wire X:
\(x_X = \frac{4FL}{\pi d^2 E}\)

For wire Y:
\(x_Y = \frac{4F(2L)}{\pi (2d)^2 E} = \frac{8FL}{4\pi d^2 E} = \frac{2FL}{\pi d^2 E} = \frac{1}{2} x_X\)

Thus, the ratio is:
\(\frac{x_X}{x_Y} = 2\).

1 mark for setting up the ratio of extensions using the Young modulus expression and obtaining the correct value of 2.

The resistivity is given by the formula:

\(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\)

The fractional uncertainty in \(\rho\) is calculated by adding the fractional uncertainties of the independent terms, taking into account that the power of \(d\) is 2:

\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\)

First, calculate the percentage uncertainty of each individual measurement:

- For \(R\): \(\frac{0.09}{4.50} \times 100\\% = 2.0\\%\)
- For \(d\): \(\frac{0.01}{0.40} \times 100\\% = 2.5\\%\)
- For \(L\): \(\frac{0.01}{1.25} \times 100\\% = 0.8\\%\)

Now, sum these percentage uncertainties with the appropriate factor of 2 for \(d\):

\(\\% \Delta \rho = 2.0\\% + 2(2.5\\%) + 0.8\\% = 2.0\\% + 5.0\\% + 0.8\\% = 7.8\\%\)

- 1 mark for the correct calculation of individual percentage uncertainties (2.0%, 2.5%, 0.8%)
- 1 mark for correctly multiplying the diameter uncertainty by 2 and summing to obtain 7.8%

The temperature difference is \(\Delta \theta = \theta_2 - \theta_1 = 68.0 - 24.0 = 44.0\\ ^\circ\text{C}\).

When quantities are subtracted, their absolute uncertainties are added:

\(\Delta(\Delta \theta) = 0.5\\ ^\circ\text{C} + 0.5\\ ^\circ\text{C} = 1.0\\ ^\circ\text{C}\).

The percentage uncertainty in the temperature difference is:

\(\\% \Delta (\Delta \theta) = \frac{1.0}{44.0} \times 100\\% \approx 2.27\\%\).

The percentage uncertainty in the calculated specific heat capacity \(c\) is the sum of the percentage uncertainties of all the quantities in the formula:

\(\\% \Delta c = \\% \Delta Q + \\% \Delta m + \\% \Delta (\Delta \theta)\)

\(\\% \Delta c = 3\\% + 2\\% + 2.27\\% = 7.27\\% \approx 7.3\\%\).

- 1 mark for finding the absolute uncertainty of the temperature difference (1.0 °C)
- 1 mark for summing the individual percentage uncertainties to yield 7.3%

The total recorded time for 20 oscillations is \(t = 32.40 \pm 0.20\text{ s}\).

The period \(T\) is the time for a single oscillation:

\(T = \frac{t}{20} = \frac{32.40\text{ s}}{20} = 1.62\text{ s}\).

The absolute uncertainty in the period \(\Delta T\) is:

\(\Delta T = \frac{\Delta t}{20} = \frac{0.20\text{ s}}{20} = 0.01\text{ s}\).

Therefore, the period \(T\) with its absolute uncertainty is \((1.62 \pm 0.01)\text{ s}\).

- 1 mark for calculating the period (1.62 s) and dividing the absolute uncertainty of the total time by 20 to obtain 0.01 s.

Since the ammeter reading is consistently \(0.10\text{ A}\) greater than the true current under all conditions, it represents a constant offset, which is a systematic error (specifically, a zero error). Systematic errors cannot be reduced by averaging multiple readings, but they can be corrected or eliminated if the nature and magnitude of the error are known (by subtracting the offset of \(0.10\text{ A}\) from each reading).

- 1 mark for identifying that the constant offset is a systematic error that can be eliminated by subtraction.

First, find the redshift \(z\):

\(z = \frac{\Delta \lambda}{\lambda_0} = \frac{672.7 - 656.3}{656.3} = \frac{16.4}{656.3} \approx 0.02499\)

Next, determine the recessional velocity \(v\) using the relation \(v = z c\):

\(v = 0.02499 \times 3.00 \times 10^5\text{ km s}^{-1} \approx 7497\text{ km s}^{-1}\)

Using Hubble's law, \(v = H_0 d\), find the distance \(d\):

\(d = \frac{v}{H_0} = \frac{7497\text{ km s}^{-1}}{70\text{ km s}^{-1}\text{ Mpc}^{-1}} \approx 107\text{ Mpc}\)

To two significant figures, this is \(110\text{ Mpc}\).

- 1 mark for calculating redshift z and using v = z * c to find v.
- 1 mark for applying Hubble's Law to calculate the distance.

According to Wien's displacement law, the peak wavelength is inversely proportional to the absolute temperature:

\(T \propto \frac{1}{\lambda_{\text{max}}}\)

Therefore, the ratio of temperatures of Star Y and Star X is:

\(\frac{T_Y}{T_X} = \frac{\lambda_{\text{max}, X}}{\lambda_{\text{max}, Y}} = \frac{400\text{ nm}}{600\text{ nm}} = \frac{2}{3}\)

According to the Stefan-Boltzmann law, the luminosity \(L\) of a star of radius \(R\) and temperature \(T\) is:

\(L = 4 \pi R^2 \sigma T^4\)

Since both stars have the same radius \(R\), the luminosity is proportional to \(T^4\):

\(\frac{L_Y}{L_X} = \left(\frac{T_Y}{T_X}\right)^4 = \left(\frac{2}{3}\right)^4 = \frac{16}{81} \approx 0.198\)

Thus, \(L_Y \approx 0.20 L\).

- 1 mark for using Wien's law to relate the peak wavelength to temperature.
- 1 mark for using the Stefan-Boltzmann law to show that luminosity depends on the fourth power of temperature and finding the ratio 0.20.

According to Faraday's law of electromagnetic induction, the average induced e.m.f. \(E\) is given by the rate of change of magnetic flux linkage:

\(E = \frac{\Delta \Phi}{\Delta t} = \frac{N \Delta (B A)}{\Delta t}\)

Since the plane of the coil is perpendicular to the magnetic field, the flux is simply \(B A\). Substituting the given values:

\(E = \frac{120 \times (0.15 - 0) \times 8.0 \times 10^{-4}}{0.040}\)

\(E = \frac{0.0144}{0.040} = 0.36\text{ V}\).

- 1 mark for recalling Faraday's Law and correctly calculating the induced e.m.f. as 0.36 V.

The Young modulus \(E\) of a material is defined as:

\(E = \frac{\text{stress}}{\text{strain}} \implies \text{strain} = \frac{\text{stress}}{E}\)

Since both wires are made of the same metal, they have the exact same Young modulus \(E\). Thus, the ratio of their strains is equal to the ratio of their stresses:

\(\frac{\text{strain}_P}{\text{strain}_Q} = \frac{\text{stress}_P}{\text{stress}_Q}\)

Stress is force per unit cross-sectional area:

\(\text{stress} = \frac{F}{A} = \frac{F}{\pi d^2 / 4}\)

Since the applied force \(F\) is the same for both wires:

- For wire P: \(\text{stress}_P \propto \frac{1}{d^2}\)
- For wire Q: \(\text{stress}_Q \propto \frac{1}{(2d)^2} = \frac{1}{4d^2}\)

Therefore, the stress in wire P is 4 times the stress in wire Q:

\(\frac{\text{stress}_P}{\text{stress}_Q} = 4.0\)

This means the ratio of the strains is also 4.0. Note that the original length of the wires does not affect their strain under these conditions.

- 1 mark for identifying that strain depends only on stress for the same material and calculating the correct ratio of 4.0 based on the cross-sectional area difference.

To find the percentage uncertainty in the volume \( V = \pi r^2 h \), we use the fractional uncertainty equation:

\( \frac{\Delta V}{V} = 2 \left( \frac{\Delta r}{r} \right) + \frac{\Delta h}{h} \)

First, calculate the individual fractional uncertainties:
- For the radius: \( \frac{\Delta r}{r} = \frac{0.05}{2.00} = 0.025 \) (or \( 2.5\% \))
- For the height: \( \frac{\Delta h}{h} = \frac{0.2}{10.0} = 0.020 \) (or \( 2.0\% \))

Now substitute these into the volume uncertainty relation:

\( \frac{\Delta V}{V} = 2(0.025) + 0.020 = 0.050 + 0.020 = 0.070 \)

Convert this to percentage uncertainty:

\( 0.070 \times 100\% = 7.0\% \)

Correct option is C.
- Award 1 mark for calculating the percentage uncertainty in the radius as 2.5% and height as 2.0%.
- Award 1 mark for adding twice the percentage uncertainty of the radius to that of the height to obtain 7.0%.

First, calculate the temperature rise \( \Delta \theta \):
\( \Delta \theta = 38.5 - 22.0 = 16.5\ ^\circ\text{C} \)

When subtracting two quantities, their absolute uncertainties are added together:
\( \Delta(\Delta \theta) = 0.5\ ^\circ\text{C} + 0.5\ ^\circ\text{C} = 1.0\ ^\circ\text{C} \)

Now, calculate the percentage uncertainty in \( \Delta \theta \):
\( \text{Percentage uncertainty} = \frac{1.0}{16.5} \times 100\% \approx 6.06\% \approx 6.1\% \)

Correct option is C.
- Award 1 mark for calculating absolute uncertainty in temperature change as 1.0 K / °C and dividing by the temperature change of 16.5 K / °C to get 6.1%.

We rearrange the formula to find the units of \( \sigma \):

\( \sigma = \frac{L}{4\pi r^2 T^4} \)

Now, substitute the SI base units for each quantity:
- Luminosity \( L \) is power, measured in watts (W):
\( 1\text{ W} = 1\text{ J s}^{-1} = 1\text{ kg m}^2 \text{s}^{-3} \)
- Radius \( r \) is measured in metres, so \( r^2 \) is in \( \text{m}^2 \).
- Temperature \( T \) is in kelvins, so \( T^4 \) is in \( \text{K}^4 \).

Substitute these base units into the expression for \( \sigma \):

\( [\sigma] = \frac{\text{kg m}^2 \text{s}^{-3}}{\text{m}^2 \text{K}^4} = \text{kg s}^{-3} \text{K}^{-4} \)

Correct option is B.
- Award 1 mark for expressing Watts in SI base units as kg m² s⁻³ and dividing by m² K⁴ to get kg s⁻³ K⁻⁴.

To reduce the effect of random errors, taking multiple measurements and calculating their mean (average) is the correct standard experimental procedure. This reduces the spread/scatter effect.

To correct for a systematic error where the instrument reads \( +0.02\text{ mm} \) higher than the true value, the student must subtract \( 0.02\text{ mm} \) from the final calculated average of the measurements. Therefore, option A is correct.

Correct option is A.
- Award 1 mark for identifying that averaging the readings reduces random error, and subtracting 0.02 mm corrects the systematic error.

The total power (luminosity \( L \)) radiated by a star is given by Stefan's law:

\( L = 4\pi r^2 \sigma T^4 \)

This means that \( L \propto r^2 T^4 \).

Let the first star have radius \( r_1 = r \) and temperature \( T_1 = 6000\text{ K} \). Its luminosity is:
\( L_0 \propto r^2 (6000)^4 \)

Let the second star have radius \( r_2 = 4r \) and temperature \( T_2 = 3000\text{ K} \). Its luminosity \( L_2 \) is:
\( L_2 \propto (4r)^2 (3000)^4 = 16 r^2 (3000)^4 \)

Taking the ratio of the two luminosities:

\( \frac{L_2}{L_0} = \frac{16 r^2 (3000)^4}{r^2 (6000)^4} = 16 \times \left(\frac{3000}{6000}\right)^4 = 16 \times \left(\frac{1}{2}\right)^4 = 16 \times \frac{1}{16} = 1 \)

Therefore, \( L_2 = L_0 \).

Correct option is B.
- Award 1 mark for showing that luminosity is proportional to R² and T⁴, and showing that the factors of 16 from the radius squared and 1/16 from the temperature to the fourth power cancel out.

First, find the recessional speed of the galaxy \( v \) using Hubble's law:
\( v = H_0 d = (2.3 \times 10^{-18}\text{ s}^{-1}) \times (1.2 \times 10^{24}\text{ m}) = 2.76 \times 10^6\text{ m s}^{-1} \)

Now, use the cosmological redshift formula for \( v \ll c \):
\( \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \)

Rearranging to find \( \Delta \lambda \):
\( \Delta \lambda = \lambda \frac{v}{c} = 656.3\text{ nm} \times \frac{2.76 \times 10^6\text{ m s}^{-1}}{3.00 \times 10^8\text{ m s}^{-1}} \)
\( \Delta \lambda = 656.3\text{ nm} \times 0.0092 = 6.038\text{ nm} \approx 6.0\text{ nm} \)

Correct option is C.
- Award 1 mark for calculating the recessional velocity of the galaxy (2.76 × 10⁶ m s⁻¹) and then correctly applying the redshift equation to find 6.0 nm.

According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. \( E \) is:

\( E = N \frac{\Delta \Phi}{\Delta t} = N A \frac{\Delta B}{\Delta t} \)

Where:
- \( N = 200 \)
- Area of circular coil \( A = \pi r^2 = \pi \times (0.050\text{ m})^2 = 7.854 \times 10^{-3}\text{ m}^2 \)
- Change in magnetic flux density \( \Delta B = 0.50\text{ T} - 0.10\text{ T} = 0.40\text{ T} \)
- Time interval \( \Delta t = 4.0\text{ s} \)

Now substitute the values:

\( E = 200 \times (7.854 \times 10^{-3}) \times \frac{0.40}{4.0} \)
\( E = 200 \times (7.854 \times 10^{-3}) \times 0.10 = 0.157\text{ V} \approx 0.16\text{ V} \)

Correct option is C.
- Award 1 mark for calculating the area of the circular coil (7.85 × 10⁻³ m²) and then substituting it along with N = 200 and ΔB/Δt = 0.10 T s⁻¹ to find 0.16 V.

First, find the extension \( x \) of the wire using the definition of Young modulus \( E \):

\( E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{F L}{A x} \)

Rearranging to solve for \( x \):

\( x = \frac{F L}{A E} = \frac{60 \times 2.0}{(1.5 \times 10^{-6}) \times (2.0 \times 10^{11})} = \frac{120}{3.0 \times 10^5} = 4.0 \times 10^{-4}\text{ m} \)

Next, the strain energy \( E_{\text{str}} \) stored in the wire is given by:

\( E_{\text{str}} = \frac{1}{2} F x = \frac{1}{2} \times 60\text{ N} \times 4.0 \times 10^{-4}\text{ m} = 1.2 \times 10^{-2}\text{ J} \)

Correct option is B.
- Award 1 mark for finding the elastic extension as 4.0 × 10⁻⁴ m and then applying the formula for strain energy (0.5 * F * x) to calculate 1.2 × 10⁻² J.

First, calculate the percentage uncertainty in each measured quantity: Percentage uncertainty in \(m\) is \(\frac{0.05}{5.00} \times 100\% = 1.0\%\). Percentage uncertainty in \(d\) is \(\frac{0.02}{0.80} \times 100\% = 2.5\%\). Percentage uncertainty in \(L\) is \(\frac{0.3}{15.0} \times 100\% = 2.0\%\). For the relationship \(\rho = \frac{4m}{\pi d^2 L}\), the fractional uncertainty equation is: \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}. Substituting the values: \frac{\Delta \rho}{\rho} = 1.0\% + 2(2.5\%) + 2.0\% = 8.0\%.

1 mark for calculating individual percentage uncertainties and using the correct compounding formula (adding the percentage uncertainties, with the diameter uncertainty doubled) to find 8.0%.

Systematic errors, like a zero error, introduce a constant offset to all measurements. Repeating measurements and averaging them does not reduce or eliminate systematic errors (it only reduces random errors). Precision, which relates to the spread of the results, is unaffected by this shift. Therefore, the systematic error can only be eliminated by subtracting \(0.3\text{ V}\) from each reading (calibration correction).

1 mark for identifying that a systematic error cannot be reduced by averaging, and that subtracting the zero error from each measurement corrects for it.

Using the relation for force on a current-carrying conductor, \(F = BIL\), we rearrange for magnetic flux density to get \(B = \frac{F}{IL}\). Substituting the SI units for each term: Force has units of \(\text{kg m s}^{-2}\), Current has units of \(\text{A}\), and Length has units of \(\text{m}\). Thus, the units of \(B\) are \(\frac{\text{kg m s}^{-2}}{\text{A m}} = \text{kg s}^{-2} \text{A}^{-1}\).

1 mark for deriving the SI base unit expression correctly using physical formulas like F = BIL.

First, convert the distance from light-years to metres. \(1\text{ year} = 365 \times 24 \times 3600 = 3.1536 \times 10^7\text{ s}\). \(1\text{ light-year} = c \times 1\text{ year} = 3.00 \times 10^8\text{ m s}^{-1} \times 3.1536 \times 10^7\text{ s} \approx 9.46 \times 10^{15}\text{ m}\). Hence, the distance is \(d = 4.2 \times 9.46 \times 10^{15}\text{ m} \approx 3.97 \times 10^{16}\text{ m}\). Now, compute the radiant flux intensity: \(F = \frac{L}{4\pi d^2} = \frac{3.8 \times 10^{26}}{4\pi (3.97 \times 10^{16})^2} \approx 1.9 \times 10^{-8}\text{ W m}^{-2}\).

1 mark for converting light-years to metres and correctly evaluating the radiant flux intensity formula.

Calculate the redshift parameter: \(z = \frac{\Delta \lambda}{\lambda_0} = \frac{682 - 656}{656} = \frac{26}{656} \approx 0.03963\). The recession speed \(v\) is given by: \(v = z c = 0.03963 \times 3.0 \times 10^8 \approx 1.189 \times 10^7\text{ m s}^{-1}\). Applying Hubble's law \(v = H_0 d\) gives: \(d = \frac{v}{H_0} = \frac{1.189 \times 10^7}{2.3 \times 10^{-18}} \approx 5.17 \times 10^{24}\text{ m}\). Rounding to 2 significant figures, we get \(5.2 \times 10^{24}\text{ m}\).

1 mark for computing the redshift, using it to calculate the recession velocity, and dividing by the Hubble constant to find the correct distance.

First, calculate the cross-sectional area: \(A = ̀́ r^2 = ̀́ (0.040\text{ m})^2 \approx 5.027 \times 10^{-3}\text{ m}^2\). The initial flux linkage is \(\Phi_i = N B A = 250 \times 0.12 \times 5.027 \times 10^{-3} \approx 0.1508\text{ Wb-turns}\). The final flux linkage is 0. According to Faraday's law: \(E = \frac{\Delta \Phi}{\Delta t} = \frac{0.1508}{0.15} \approx 1.0\text{ V}\).

1 mark for correctly determining the area, using the flux linkage formula with the number of turns, and applying Faraday's Law to calculate the induced e.m.f.

The extension \(e\) is related to Young's modulus by \(e = \frac{FL}{AE}\). Since both wires are steel, \(E\) is constant. The cross-sectional area \(A\) is proportional to the square of the diameter \(d^2\). The second wire has half the diameter of the first, so its cross-sectional area is \(0.25 A\). The second wire has length \(2L\). Thus, the new extension is \(e_2 = \frac{F (2L)}{(0.25A)E} = 8 \frac{FL}{AE} = 8e\).

1 mark for analyzing the scaling factors of length and cross-sectional area (which scales as diameter squared) and finding the extension increases by a factor of 8.

Let's check each option: A: Displacement (vector), force (vector), kinetic energy (scalar) - two vectors and one scalar. B: Velocity (vector), acceleration (vector), momentum (vector) - three vectors. C: Electric current (scalar), temperature (scalar), weight (vector) - one vector and two scalars. D: Work done (scalar), power (scalar), gravitational potential (scalar) - three scalars. Thus, option C is the correct answer.

1 mark for identifying the vector and scalar nature of all listed quantities and selecting the option with exactly one vector and two scalars.

To find the percentage uncertainty in \(g\), we use the rules for combining fractional/percentage uncertainties.

From the equation \(g = \frac{4\pi^2 L}{T^2}\), the percentage uncertainty in \(g\) is given by:

\[\frac{\Delta g}{g} \times 100\% = \left( \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \right) \times 100\%\]

1. Calculate the percentage uncertainty in \(L\):
\[\frac{\Delta L}{L} \times 100\% = \frac{0.4}{80.0} \times 100\% = 0.5\%\]

2. Calculate the percentage uncertainty in \(T\):
\[\frac{\Delta T}{T} \times 100\% = \frac{0.03}{1.80} \times 100\% \approx 1.67\%\]

3. Calculate the total percentage uncertainty in \(g\):
\[\text{Percentage uncertainty in } g = 0.5\% + 2 \times 1.67\% = 0.5\% + 3.33\% = 3.83\% \approx 3.8\%\]

Therefore, the correct percentage uncertainty is \(3.8\%\).

[1 mark] B is correct.
- Award 1 mark for calculating the percentage uncertainty of length as 0.5%, the percentage uncertainty of period as 1.67%, doubling the percentage uncertainty of the period, and adding them to get 3.8%.

To find the corrected diameter of the wire:

\[\text{Corrected diameter} = \text{Measurement reading} - \text{Zero reading}\]
\[\text{Corrected diameter} = 1.42\text{ mm} - (-0.04\text{ mm}) = 1.46\text{ mm}\]

When subtracting measurements, the absolute uncertainties are added together:

\[\text{Absolute uncertainty} = 0.01\text{ mm} + 0.02\text{ mm} = 0.03\text{ mm}\]

Therefore, the corrected measurement with its uncertainty is \((1.46 \pm 0.03)\text{ mm}\).

[1 mark] D is correct.
- Award 1 mark for adding the negative zero error to the reading (1.46 mm) and adding the absolute uncertainties (0.03 mm).

Precision is a measure of the repeatability of the results; it is determined by the spread of the data (random errors). The readings are closely clustered between \(35.3\text{ s}\) and \(35.5\text{ s}\) (a range of only \(0.2\text{ s}\)), indicating high precision.

Accuracy is a measure of how close the experimental average is to the true value. The average of the readings is approximately \(35.4\text{ s}\), which is significantly different from the true value of \(32.1\text{ s}\). This indicates low accuracy, likely caused by a systematic error (such as the student's reaction time or a faulty stopwatch).

Therefore, the measurements have high precision but low accuracy.

[1 mark] B is correct.
- Award 1 mark for recognizing that the small spread in values indicates high precision and the large deviation from the true value indicates low accuracy.

First, find the recession speed \(v\) of the galaxy using Hubble's law:

\[v = H_0 d = (2.3 \times 10^{-18}\text{ s}^{-1}) \times (1.2 \times 10^{24}\text{ m}) = 2.76 \times 10^6\text{ m s}^{-1}\]

Next, use the redshift formula to find the change in wavelength \(\Delta \lambda\):

\[\frac{\Delta \lambda}{\lambda} = \frac{v}{c}\]
\[\Delta \lambda = \lambda \times \frac{v}{c} = 656\text{ nm} \times \frac{2.76 \times 10^6\text{ m s}^{-1}}{3.0 \times 10^8\text{ m s}^{-1}} = 656 \times 0.0092 = 6.04\text{ nm}\]

Since the galaxy is receding from Earth, its light is redshifted, meaning the observed wavelength is longer:

\[\lambda_{\text{obs}} = \lambda + \Delta \lambda = 656\text{ nm} + 6.04\text{ nm} \approx 662\text{ nm}\]

[1 mark] C is correct.
- Award 1 mark for finding the recession velocity, calculating the change in wavelength as 6.0 nm, and adding it to the source wavelength to find the redshifted wavelength.

According to Wien's displacement law, the peak wavelength of radiation \(\lambda_{\text{max}}\) is inversely proportional to the thermodynamic temperature \(T\):

\[T \propto \frac{1}{\lambda_{\text{max}}}\]

Therefore, the ratio of the temperatures of Star X and Star Y is:

\[\frac{T_X}{T_Y} = \frac{\lambda_{\text{max}, Y}}{\lambda_{\text{max}, X}} = \frac{960\text{ nm}}{480\text{ nm}} = 2\]

According to the Stefan-Boltzmann law, the luminosity \(L\) of a star is given by:

\[L = 4\pi R^2 \sigma T^4 \implies L \propto R^2 T^4\]

Thus, the ratio of their luminosities is:

\[\frac{L_X}{L_Y} = \left( \frac{R_X}{R_Y} \right)^2 \times \left( \frac{T_X}{T_Y} \right)^4\]
\[\frac{L_X}{L_Y} = \left( \frac{R}{2R} \right)^2 \times (2)^4 = \frac{1}{4} \times 16 = 4\]

[1 mark] C is correct.
- Award 1 mark for using Wien's law to find the temperature ratio (TX/TY = 2) and combining this with Stefan-Boltzmann's law to obtain the luminosity ratio of 4.

According to Faraday's law of electromagnetic induction, the average induced e.m.f. \(E\) is given by:

\[E = \frac{\Delta \Phi}{\Delta t}\]

where \(\Delta \Phi\) is the change in magnetic flux linkage.

Initially, the plane of the coil is perpendicular to the magnetic field, so the flux linkage \(\Phi_1\) is maximum:

\[\Phi_1 = N B A = 150 \times 0.40\text{ T} \times (8.0 \times 10^{-4}\text{ m}^2) = 0.048\text{ Wb-turns}\]

After rotating by \(90^\circ\), the plane of the coil is parallel to the magnetic field lines, so no magnetic flux passes through the coil:

\[\Phi_2 = 0\text{ Wb-turns}\]

The change in magnetic flux linkage is:

\[\Delta \Phi = 0.048 - 0 = 0.048\text{ Wb-turns}\]

Now, calculate the average induced e.m.f.:

\[E = \frac{0.048\text{ Wb-turns}}{0.12\text{ s}} = 0.40\text{ V}\]

[1 mark] C is correct.
- Award 1 mark for calculating the initial flux linkage (0.048 Wb-turns), recognizing the final flux linkage is zero, and dividing by the time interval to get 0.40 V.

According to Lenz's law, the direction of the induced e.m.f. (and current) is such that it opposes the change in magnetic flux that produces it.

As the North pole of the magnet approaches the copper ring, the change in flux is an increase in magnetic flux pointing towards the ring. To oppose this approach, the ring must generate its own magnetic field with a North pole on the side facing the magnet to repel it.

By the right-hand grip rule, to produce a North pole on the side facing the observer (the approaching magnet), the current must flow in an anticlockwise direction.

Because the induced current opposes the motion of the magnet, the magnetic force acting on the magnet must oppose its motion (it acts to repel and slow down the magnet).

[1 mark] A is correct.
- Award 1 mark for identifying that the induced current must flow anticlockwise to produce a repulsive North pole, and the magnetic force must oppose the motion of the magnet.

The strain \(\epsilon\) in a wire is given by:

\[\epsilon = \frac{\sigma}{E} = \frac{F}{A E}\]

where:
- \(\sigma\) is the tensile stress,
- \(E\) is the Young modulus of the metal,
- \(F\) is the applied tensile force,
- \(A\) is the cross-sectional area of the wire.

Since both wires are made of the same metal, they have the same Young modulus \(E\). Both are subjected to the same tensile force \(F\).

Thus, the strain is inversely proportional to the cross-sectional area:

\[\epsilon \propto \frac{1}{A} \propto \frac{1}{d^2}\]

Notice that the strain is independent of the original length \(L\) of the wire.

Therefore, the ratio of the strain in wire P to the strain in wire Q is:

\[\frac{\epsilon_P}{\epsilon_Q} = \left( \frac{d_Q}{d_P} \right)^2 = \left( \frac{2d}{d} \right)^2 = 4\]

[1 mark] C is correct.
- Award 1 mark for expressing strain as strain = F/(AE) and recognizing that because E and F are constant, strain is inversely proportional to diameter squared, giving a ratio of 4.

(a) Convert all values to SI base units:
\(m = 5.12 \times 10^{-3}\text{ kg}\)
\(L = 0.125\text{ m}\)
\(d = 2.42 \times 10^{-3}\text{ m}\)

The cross-sectional area \(A\) is:
\(A = \frac{\pi d^2}{4} = \frac{\pi (2.42 \times 10^{-3})^2}{4} = 4.5996 \times 10^{-6}\text{ m}^2\)

The volume \(V\) is:
\(V = A L = (4.5996 \times 10^{-6}\text{ m}^2) \times 0.125\text{ m} = 5.7495 \times 10^{-7}\text{ m}^3\)

The density \(\rho\) is:
\(\rho = \frac{m}{V} = \frac{5.12 \times 10^{-3}}{5.7495 \times 10^{-7}} = 8905\text{ kg m}^{-3} \approx 8.9 \times 10^3\text{ kg m}^{-3}\).

(b)
(i) The area \(A\) is given by \(A = \frac{\pi d^2}{4}\).
The percentage uncertainty in \(A\) is:
\(\frac{\Delta A}{A} \times 100\% = 2 \times \frac{\Delta d}{d} \times 100\% = 2 \times \frac{0.02}{2.42} \times 100\% = 1.65\% \approx 1.7\%\).

(ii) The density \(\rho\) is given by \(\rho = \frac{m}{A L}\).
The percentage uncertainty in \(\rho\) is:
\(\frac{\Delta \rho}{\rho} \times 100\% = \left(\frac{\Delta m}{m} + \frac{\Delta L}{L} + \frac{\Delta A}{A}\right) \times 100\%\)
Calculate individual percentage uncertainties:
\(\frac{\Delta m}{m} \times 100\% = \frac{0.05}{5.12} \times 100\% = 0.98\%\)
\(\frac{\Delta L}{L} \times 100\% = \frac{0.1}{12.5} \times 100\% = 0.80\%\)
Sum of percentage uncertainties:
\(\frac{\Delta \rho}{\rho} \times 100\% = 0.98\% + 0.80\% + 1.65\% = 3.43\% \approx 3.4\%\).

(c) The measurement of the diameter \(d\) contributes the most. Even though its individual percentage uncertainty (\(0.83\%\)) is slightly smaller than that of mass (\(0.98\%\)), the diameter is squared in the volume calculation, meaning its fractional uncertainty is multiplied by 2, contributing \(1.65\%\) to the final uncertainty.

(a)
M1: Correct substitution of values into density formula \(\rho = \frac{4m}{\pi d^2 L}\).
A1: Clear unit conversions shown (g to kg, cm and mm to m).
A1: Final calculated value shown as \(8900\text{ kg m}^{-3}\) or \(8.9 \times 10^3\text{ kg m}^{-3}\).

(b)(i)
M1: Realises that \(\% \text{ uncertainty in } A = 2 \times \% \text{ uncertainty in } d\).
A1: Correct calculation to give \(1.7\%\) (accept \(1.65\%\)).

(b)(ii)
C1: Calculation of percentage uncertainty in \(m\) (\(0.98\%\)) AND \(L\) (\(0.80\%\)).
M1: Summing the percentage uncertainties of \(m\), \(L\), and \(A\).
A1: Correct final answer of \(3.4\%\) (or \(3.43\%\)).

(c)
B1: Identifies the diameter \(d\) as the largest contributor.
B1: Explains that because \(d\) is squared in the equation, its percentage uncertainty is doubled, making its contribution (\(1.65\%\)) larger than mass (\(0.98\%\)) or length (\(0.80\%\)).

(a) (i) Force: \(\text{kg m s}^{-2}\)
Electric charge: \(\text{A s}\)

(ii) Resistivity \(\rho\) is defined by \(\rho = \frac{R A}{l}\).
The resistance \(R = \frac{V}{I}\), and potential difference \(V = \frac{\text{Work done}}{\text{Charge}}\).
In SI base units:
\([V] = \frac{\text{kg m}^2\text{ s}^{-2}}{\text{A s}} = \text{kg m}^2\text{ s}^{-3}\text{ A}^{-1}\).
Thus, \([R] = \frac{\text{kg m}^2\text{ s}^{-3}\text{ A}^{-1}}{\text{A}} = \text{kg m}^2\text{ s}^{-3}\text{ A}^{-2}\).
Now substitute \(R\) into the resistivity formula:
\([\rho] = [R] \times \frac{[A]}{[l]} = (\text{kg m}^2\text{ s}^{-3}\text{ A}^{-2}) \times \frac{\text{m}^2}{\text{m}} = \text{kg m}^3\text{ s}^{-3}\text{ A}^{-2}\).

(b) LHS unit of speed \(v = \text{m s}^{-1}\).
RHS expression: \(\sqrt{\frac{T}{\mu}}\).
SI base unit of tension \(T\) (force) \(= \text{kg m s}^{-2}\).
SI base unit of mass per unit length \(\mu = \text{kg m}^{-1}\).
Substitute units into the RHS:
\(\left[\sqrt{\frac{T}{\mu}}\right] = \left( \frac{\text{kg m s}^{-2}}{\text{kg m}^{-1}} \right)^{0.5} = (\text{m}^2\text{ s}^{-2})^{0.5} = \text{m s}^{-1}\).
Since the units of the LHS match the units of the RHS, the equation is homogeneous.

(c) Electric field strength \(E = \frac{F}{q} \implies [E] = \frac{\text{kg m s}^{-2}}{\text{A s}} = \text{kg m s}^{-3}\text{ A}^{-1}\).
Magnetic flux density \(B = \frac{F}{I L} \implies [B] = \frac{\text{kg m s}^{-2}}{\text{A m}} = \text{kg s}^{-2}\text{ A}^{-1}\).
Therefore, \([X] = \frac{[E]}{[B]} = \frac{\text{kg m s}^{-3}\text{ A}^{-1}}{\text{kg s}^{-2}\text{ A}^{-1}} = \text{m s}^{-1}\).
The physical quantity represented by \(X\) is speed (or velocity).

(a)(i)
B1: \(\text{kg m s}^{-2}\) for force.
B1: \(\text{A s}\) for charge.

(a)(ii)
C1: Expresses \(V\) or \(R\) in base units, e.g., \([V] = \text{kg m}^2\text{ s}^{-3}\text{ A}^{-1}\).
M1: Correct base units for resistance, \([R] = \text{kg m}^2\text{ s}^{-3}\text{ A}^{-2}\).
A1: Correctly multiplies by \(\text{m}\) to obtain \(\text{kg m}^3\text{ s}^{-3}\text{ A}^{-2}\).

(b)
B1: States LHS unit is \(\text{m s}^{-1}\).
B1: States unit of \(\mu\) is \(\text{kg m}^{-1}\).
B1: Correct simplification of RHS to \(\text{m s}^{-1}\) with a clear conclusion of homogeneity.

(c)
B1: Derives base units of \(X\) as \(\text{m s}^{-1}\).
B1: Identifies the quantity as speed / velocity.

(a) Elastic deformation: the material returns to its original length/shape when the load is removed.
Plastic deformation: the material is permanently deformed and does not return to its original length/shape when the load is removed.

(b) (i) Stress \(\sigma = \frac{F}{A}\)
\(\sigma = \frac{90}{4.5 \times 10^{-7}} = 2.0 \times 10^8\text{ Pa}\).

(ii) Young modulus \(E = \frac{\text{Stress}}{\text{Strain}}\)
Strain \(\epsilon = \frac{\text{Stress}}{E} = \frac{2.0 \times 10^8}{2.0 \times 10^{11}} = 1.0 \times 10^{-3}\).
Extension \(x = \epsilon L = (1.0 \times 10^{-3}) \times 2.2 = 2.2 \times 10^{-3}\text{ m}\) (or \(2.2\text{ mm}\)).

(iii) Elastic potential energy \(E_p = \frac{1}{2} F x\)
\(E_p = 0.5 \times 90 \times (2.2 \times 10^{-3}) = 0.099\text{ J}\) (or \(0.10\text{ J}\)).

(c) Ultimate tensile stress is the maximum stress a material can withstand before breaking or fracturing.

(a)
B1: Defines elastic deformation as returning to its original shape/size when load is removed.
B1: Defines plastic deformation as permanent extension / not returning to original shape when load is removed.

(b)(i)
M1: Use of \(\sigma = \frac{F}{A}\).
A1: \(2.0 \times 10^8\text{ Pa}\) (or \(\text{N m}^{-2}\)).

(b)(ii)
M1: Use of \(E = \frac{\sigma}{\epsilon}\) or equivalent.
A1: \(2.2 \times 10^{-3}\text{ m}\) (or \(2.2\text{ mm}\)).

(b)(iii)
M1: Use of \(E_p = \frac{1}{2} F x\) or \(\frac{1}{2} k x^2\).
A1: \(0.099\text{ J}\) or \(0.10\text{ J}\).

(c)
B1: Mentions 'maximum stress' or 'maximum force per unit cross-sectional area'.
B1: Mentions 'before breaking' or 'before fracture'.

(a) The magnitude of the induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage.

(b) (i) Area of the coil \(A = \pi r^2 = \pi (0.028\text{ m})^2 = 2.463 \times 10^{-3}\text{ m}^2\).
Magnetic flux linkage \(\Phi = N B A\).
\(\Phi = 450 \times 0.12\text{ T} \times 2.463 \times 10^{-3}\text{ m}^2 = 0.133\text{ Wb}\) (or \\text{Wb-turns}).
To 2 significant figures, this is \(0.13\text{ Wb}\).

(ii) Induced e.m.f. \(E = \frac{\Delta \Phi}{\Delta t}\).
\(E = \frac{0.133\text{ Wb} - 0}{2.0\text{ s}} = 0.0665\text{ V}\).
To 2 significant figures, this is \(0.067\text{ V}\) (or \(67\text{ mV}\)).

(c) According to Lenz's law, the direction of the induced e.m.f. is such that it opposes the change in magnetic flux that produces it.
During the first \(2.0\text{ s}\), the magnetic flux through the coil is increasing.
Therefore, the induced current in the coil will flow in a direction that creates its own magnetic field opposing the increasing external magnetic field.

(a)
B1: Induced e.m.f. is proportional to the rate of change of flux linkage.
B1: Explicitly mentions 'magnitude of e.m.f.' and 'magnetic flux linkage'.

(b)(i)
M1: Calculate area \(A = \pi (0.028)^2 = 2.46 \times 10^{-3}\text{ m}^2\).
M1: Correct use of \(\Phi = N B A\).
A1: \(0.13\text{ Wb}\) (accept \(0.133\text{ Wb}\)).

(b)(ii)
M1: Use of \(E = \frac{\Delta \Phi}{\Delta t}\).
A1: \(0.067\text{ V}\) or \(67\text{ mV}\) (allow ecf from (b)(i)).

(c)
B1: State Lenz's law: induced e.m.f./current opposes the change in magnetic flux.
B1: Identify that the external magnetic flux is increasing.
B1: State that the induced current creates a magnetic field in the opposite direction to the external field to oppose this increase.

(a) Redshift is the increase in the observed wavelength (or decrease in frequency) of electromagnetic radiation from an astronomical source moving away from the observer.

(b) (i) Redshift \(z = \frac{\Delta \lambda}{\lambda_0}\).
\(\Delta \lambda = 671.5\text{ nm} - 656.3\text{ nm} = 15.2\text{ nm}\).
\(z = \frac{15.2}{656.3} = 0.02316 \approx 0.023\).

(ii) Since \(z = \frac{v}{c}\):
\(v = z c = 0.02316 \times 3.00 \times 10^8\text{ m s}^{-1} = 6.95 \times 10^6\text{ m s}^{-1} \approx 6.9 \times 10^6\text{ m s}^{-1}\).

(c) (i) Using Hubble's law: \(v = H_0 d\)
\(d = \frac{v}{H_0} = \frac{6.95 \times 10^6\text{ m s}^{-1}}{2.2 \times 10^{-18}\text{ s}^{-1}} = 3.16 \times 10^{24}\text{ m} \approx 3.2 \times 10^{24}\text{ m}\).

(ii) Age of the Universe \(T \approx \frac{1}{H_0}\)
\(T = \frac{1}{2.2 \times 10^{-18}\text{ s}^{-1}} = 4.545 \times 10^{17}\text{ s}\).
Convert to years:
\(T = \frac{4.545 \times 10^{17}}{365.25 \times 24 \times 3600} = 1.44 \times 10^{10}\text{ years} \approx 1.4 \times 10^{10}\text{ years}\) (or 14 billion years).

(a)
B1: Increase in wavelength / decrease in frequency.
B1: Caused by the source moving away from the observer.

(b)(i)
M1: Use of \(z = \frac{\Delta \lambda}{\lambda}\).
A1: \(0.023\) (or \(0.0232\)).

(b)(ii)
M1: Use of \(v = z c\).
A1: \(6.9 \times 10^6\text{ m s}^{-1}\) or \(7.0 \times 10^6\text{ m s}^{-1}\).

(c)(i)
M1: Use of \(v = H_0 d\).
A1: \(3.2 \times 10^{24}\text{ m}\) (allow ecf from (b)(ii)).

(c)(ii)
M1: Realises that age \(T \approx \frac{1}{H_0}\).
A1: \(1.4 \times 10^{10}\text{ years}\) (accept \(1.4 \times 10^{10}\) to \(1.5 \times 10^{10}\text{ years}\)).

(a) Systematic errors cause measurements to be consistently larger or consistently smaller than the true value (cannot be reduced by repeating and averaging).
Random errors cause measurements to fluctuate randomly around the true value (can be reduced by repeating and averaging).

(b) (i) Period \(T = \frac{t}{20} = \frac{37.0}{20} = 1.85\text{ s}\).
The absolute uncertainty in the time \(t\) is \(\Delta t = 0.2\text{ s}\).
The absolute uncertainty in the period \(T\) is \(\Delta T = \frac{\Delta t}{20} = \frac{0.2}{20} = 0.01\text{ s}\).
Therefore, \(T = (1.85 \pm 0.01)\text{ s}\).

(ii) \(g = \frac{4\pi^2 L}{T^2} = \frac{4\pi^2 \times 0.850}{(1.85)^2} = \frac{33.558}{3.4225} = 9.805\text{ m s}^{-2}\).
Since both \(L\) and \(t\) (and hence \(T\)) are measured to 3 significant figures, \(g\) should be expressed to 3 significant figures:
\(g = 9.81\text{ m s}^{-2}\).

(iii) The percentage uncertainty in \(g\) is given by:
\(\frac{\Delta g}{g} \times 100\% = \left(\frac{\Delta L}{L} + 2\frac{\Delta T}{T}\right) \times 100\%\)
\(\frac{\Delta L}{L} \times 100\% = \frac{0.005}{0.850} \times 100\% = 0.588\%\).
\(\frac{\Delta T}{T} \times 100\% = \frac{0.01}{1.85} \times 100\% = 0.541\%\).
\(\frac{\Delta g}{g} \times 100\% = 0.588\% + 2(0.541\%) = 0.588\% + 1.082\% = 1.67\% \approx 1.7\%\).

(a)
B1: Systematic error is a constant bias in one direction / cannot be eliminated by averaging.
B1: Random error causes spread about the mean / can be reduced by averaging.

(b)(i)
M1: Calculates period \(T = 1.85\text{ s}\).
M1: Divides uncertainty in time by 20 to find \(\Delta T\).
A1: Correct absolute uncertainty \(\Delta T = 0.01\text{ s}\) (to 1 s.f.).

(b)(ii)
M1: Substitute values into formula for \(g\).
A1: \(g = 9.81\text{ m s}^{-2}\) (must be given to 3 s.f. to match raw data).

(b)(iii)
C1: Calculate percentage uncertainty of \(L\) (\(0.59\%\)) and \(T\) (\(0.54\%\)).
M1: Use of \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}\) (doubling uncertainty of period is essential).
A1: \(1.7\%\) (or \(1.67\%\)).

Representative data: (a)(i) \(L = 45.0\text{ cm} = 0.450\text{ m}\). (a)(ii) Time for 10 oscillations: \(t_1 = 12.1\text{ s}\), \(t_2 = 12.0\text{ s}\). Mean \(t = 12.05\text{ s}\). Period \(T = \frac{t}{10} = 1.21\text{ s}\). (b) Table of results: For \(L\) values of 0.450, 0.400, 0.350, 0.300, 0.250, and 0.200 m, the calculated values of \(L^3\) and \(T^2\) are recorded. Example data: at \(L = 0.450\text{ m}\), \(L^3 = 0.0911\text{ m}^3\), \(T^2 = 1.46\text{ s}^2\); at \(L = 0.200\text{ m}\), \(L^3 = 0.0080\text{ m}^3\), \(T^2 = 0.24\text{ s}^2\). (c) Graph of \(T^2\) against \(L^3\) is plotted with a linear scale covering more than half the grid. Best-fit and worst acceptable lines are drawn. (d) Gradient of best-fit line: \(m = \frac{1.44 - 0.27}{0.090 - 0.010} = 14.6\text{ s}^2\text{ m}^{-3}\). y-intercept: \(c = 1.44 - (14.6 \times 0.090) = 0.13\text{ s}^2\). (e) Comparing \(T^2 = p L^3 + q\) with \(y = mx + c\) yields \(p = \text{gradient} = 14.6\text{ s}^2\text{ m}^{-3}\) and \(q = \text{y-intercept} = 0.13\text{ s}^2\).

(a)(i) [1 mark] Value of L recorded to nearest millimetre, with unit, in range 0.440 m to 0.460 m. (a)(ii) [2 marks] First mark for measurement of t with at least two trials and t > 5 s. Second mark for calculation of T = t/10 correct to 3 s.f. (b) [6 marks] Table of results: [1] Successful collection of 6 sets of readings of L and t. [1] Wide range of L used (max L >= 0.44 m and min L <= 0.25 m). [1] Column headings with correct units, e.g., L/m, t/s, T/s, L^3/m^3, T^2/s^2. [1] Consistency of raw L values recorded to nearest mm. [1] Significant figures for L^3 and T^2 consistent with raw data. [1] Correct calculations of L^3 and T^2. (c) [5 marks] Graph plotting: [1] Linear scales with labels, points occupy at least half the grid. [1] All points plotted to within half a small square. [1] Line of best fit drawn symmetrically. [1] Worst acceptable line clearly labelled. [1] Scatter quality: points lie close to the line. (d) [2 marks] [1] Gradient calculation: triangle vertices on line and separated by at least half the line length. [1] y-intercept calculation using a point from the line, or direct read-off if x = 0 is plotted. (e) [4 marks] [1] p equated to gradient with unit s^2 m^-3. [1] q equated to y-intercept with unit s^2. [1] Values given to 2 or 3 s.f.

(a)(ii) With \(W_1 = 3.0\text{ cm}\), initial amplitude \(A_0 = 10.0\text{ cm}\). Measured amplitude after 5 oscillations: \(A_{1,\text{trial 1}} = 7.5\text{ cm}\), \(A_{1,\text{trial 2}} = 7.3\text{ cm}\). Mean \(A_1 = 7.4\text{ cm}\). (b) Choosing absolute uncertainty in measuring oscillating amplitude as \(\Delta A_1 = \pm 0.3\text{ cm}\) (due to motion of bob). Percentage uncertainty = \(\frac{0.3}{7.4} \times 100\% = 4.1\%\). (c) With \(W_2 = 6.0\text{ cm}\), measured amplitude after 5 oscillations: \(A_{2,\text{trial 1}} = 4.9\text{ cm}\), \(A_{2,\text{trial 2}} = 5.1\text{ cm}\). Mean \(A_2 = 5.0\text{ cm}\). (d)(i) Calculate constants: \(k_1 = \frac{10.0 - 7.4}{3.0} = 0.87\text{ cm}\ \text{cm}^{-1}\). \(k_2 = \frac{10.0 - 5.0}{6.0} = 0.83\text{ cm}\ \text{cm}^{-1}\). (d)(ii) Percentage difference = \(\frac{|0.87 - 0.83|}{0.85} \times 100\% = 4.7\%\). Criterion: experimental relationship is supported if the percentage difference is within 10%. Since 4.7% < 10%, the relationship is supported. (e) See marking scheme for limitations and improvements.

(a) [3 marks] [1] A_0 recorded as 10.0 cm with unit. [1] Raw values of A_1 recorded to nearest mm with unit. [1] Repeated measurements for A_1 taken and mean calculated correctly. (b) [2 marks] [1] Sensible absolute uncertainty selected (e.g. 0.2 to 0.5 cm) with justification (moving bob). [1] Correct calculation of percentage uncertainty. (c) [2 marks] [1] A_2 recorded to nearest mm with unit. [1] Repeated measurements of A_2 show A_2 < A_1. (d)(i) [2 marks] [1] Calculation of k_1 and k_2 correct. [1] Correct units for k (dimensionless or cm/cm) and 2 or 3 s.f. (d)(ii) [3 marks] [1] Correct percentage difference between k_1 and k_2. [1] Sensible criterion stated (e.g., 10% or 15%). [1] Valid conclusion drawn. (e) [8 marks] 4 marks for 4 limitations, 4 marks for 4 improvements. Pairs: [1] L1: Two values of W are not enough to draw a conclusion. I1: Take readings with at least 5 different card widths and plot a graph. [1] L2: Difficulty in determining exact turning point of moving bob. I2: Use video recording with a scale behind the pendulum, playback frame-by-frame. [1] L3: Card twists during oscillations. I3: Use a bifilar suspension to prevent rotational motion. [1] L4: Hand release introduces force/parallax error. I4: Use an electromagnet or mechanical release mechanism.

(a) Systematic errors cause measurements to deviate from the true value by a constant amount in the same direction each time, whereas random errors cause measurements to vary unpredictably about the mean value.

(b) First, calculate the value of \(n\):
\(n = \frac{0.250 \times (12.0 \times 10^{-3})}{1.60 \times 10^{-19} \times (0.12 \times 10^{-3}) \times (5.4 \times 10^{-3})} = 2.8935 \times 10^{22}\text{ m}^{-3}\).

Next, calculate individual percentage uncertainties:
- For \(B\): \(\frac{0.005}{0.250} \times 100\% = 2.0\%\)
- For \(I\): \(\frac{0.2}{12.0} \times 100\% = 1.67\%\)
- For \(t\): \(\frac{0.01}{0.12} \times 100\% = 8.33\%\)
- For \(V_H\): \(\frac{0.1}{5.4} \times 100\% = 1.85\%\)

Sum of percentage uncertainties in \(n\):
\(\% \Delta n = 2.0\% + 1.67\% + 8.33\% + 1.85\% = 13.85\%\)

Absolute uncertainty in \(n\):
\(\Delta n = 2.8935 \times 10^{22} \times 0.1385 = 4.01 \times 10^{21}\text{ m}^{-3} \approx 0.4 \times 10^{22}\text{ m}^{-3}\).

Expressing the final result to the correct number of significant figures matching the uncertainty:
\(n = (2.9 \pm 0.4) \times 10^{22}\text{ m}^{-3}\).

(c) The thickness \(t\) contributes most to the uncertainty of \(n\) because it has the largest percentage uncertainty (\(8.33\%\)) among all measurements.

Part (a):
- 1 mark for explaining systematic error: fixed deviation in the same direction from the true value.
- 1 mark for explaining random error: unpredictable scatter about the mean value.

Part (b):
- 1 mark for calculating \(n = 2.9 \times 10^{22}\text{ m}^{-3}\).
- 1 mark for calculating individual percentage uncertainties of \(B\), \(I\), and \(V_H\).
- 1 mark for calculating percentage uncertainty of \(t\) (\(8.3\%\)).
- 1 mark for summing percentage uncertainties to get \(13.9\%\).
- 1 mark for absolute uncertainty \(\Delta n = 4 \times 10^{21}\text{ m}^{-3}\) (or \(0.4 \times 10^{22}\text{ m}^{-3}\)).
- 1 mark for stating final answer as \(n = (2.9 \pm 0.4) \times 10^{22}\text{ m}^{-3}\).

Part (c):
- 1 mark for identifying the thickness \(t\).
- 1 mark for stating it has the largest fractional / percentage uncertainty.

(a) Taking the natural logarithm of both sides of the equation:
\(\ln(V) = \ln(V_0) - \frac{t}{RC}\)
\(\frac{t}{RC} = \ln(V_0) - \ln(V) = \ln(V_0 / V)\)
Rearranging for \(C\) gives:
\(C = \frac{t}{R \ln(V_0/V)}\) (Shown).

(b) \(V_0 / V = 9.00 / 3.30 = 2.7273\)
\(\ln(V_0/V) = \ln(2.7273) = 1.0033\)
\(C = \frac{15.0}{220 \times 10^3 \times 1.0033} = 6.796 \times 10^{-5}\text{ F} = 68.0\ \mu\text{F}\).

(c) (i) Let \(y = V_0 / V\).
\(\% \Delta y = \% \Delta V_0 + \% \Delta V = \frac{0.15}{9.00} \times 100\% + \frac{0.10}{3.30} \times 100\% = 1.67\% + 3.03\% = 4.70\%\).
Absolute uncertainty in \(y\):
\(\Delta y = 2.7273 \times 0.0470 = 0.128\).
Let \(x = \ln(y)\).
Absolute uncertainty in \(x\):
\(\Delta x = \frac{\Delta y}{y} = 0.0470\).
Percentage uncertainty in \(x = \ln(V_0/V)\):
\(\% \Delta x = \frac{\Delta x}{x} \times 100\% = \frac{0.0470}{1.0033} \times 100\% = 4.68\% \approx 4.7\%\).

(ii) Percentage uncertainty in \(C\):
\(\% \Delta C = \% \Delta t + \% \Delta R + \% \Delta [\ln(V_0/V)]\)
\(\% \Delta t = \frac{0.5}{15.0} \times 100\% = 3.33\%\)
\(\% \Delta R = \frac{5}{220} \times 100\% = 2.27\%\)
\(\% \Delta C = 3.33\% + 2.27\% + 4.68\% = 10.28\% \approx 10.3\%\).

(d) Absolute uncertainty in \(C\):
\(\Delta C = 68.0\ \mu\text{F} \times 0.1028 = 6.99\ \mu\text{F} \approx 7\ \mu\text{F}\).
Thus, \(C = (68 \pm 7)\ \mu\text{F}\) or \((6.8 \pm 0.7) \times 10^{-5}\text{ F}\).

Part (a):
- 1 mark for correctly taking logs on both sides: \(\ln(V) = \ln(V_0) - \frac{t}{RC}\).
- 1 mark for clear steps showing rearrangement to get \(C = \frac{t}{R \ln(V_0/V)}\).

Part (b):
- 1 mark for \(\ln(V_0/V) \approx 1.0\).
- 1 mark for \(C = 6.8 \times 10^{-5}\text{ F}\) (or \(68\ \mu\text{F}\)).

Part (c) (i):
- 1 mark for calculating percentage uncertainty in \(V_0/V\) as \(4.7\%\).
- 1 mark for calculating absolute uncertainty in \(\ln(V_0/V)\) as \(0.047\).
- 1 mark for calculating percentage uncertainty in \(\ln(V_0/V)\) as \(4.7\%\) (accept range \(4.6\%\) to \(4.8\%\)).

Part (c) (ii):
- 1 mark for \(\% \Delta t = 3.3\%\) and \(\% \Delta R = 2.3\%\).
- 1 mark for summing the three percentage uncertainties to get \(10.3\%\) (or \(10\%\)).

Part (d):
- 1 mark for \(C = (68 \pm 7)\ \mu\text{F}\) or \((6.8 \pm 0.7) \times 10^{-5}\text{ F}\).

(a) \(\theta = 46.8^\circ - 10.0^\circ = 36.8^\circ\).
Since \(\theta\) is determined by subtracting two measured values, their absolute uncertainties are added:
\(\Delta \theta = 0.1^\circ + 0.1^\circ = 0.2^\circ\).
Thus, \(\theta = (36.8 \pm 0.2)^\circ\).

(b) \(\lambda = \frac{\sin(36.8^\circ)}{5.00 \times 10^5 \times 2} = \frac{0.59902}{1.00 \times 10^6} = 5.99 \times 10^{-7}\text{ m} = 599\text{ nm}\).

(c) Calculate \(\sin \theta\) at the upper and lower limits of the angle:
\(\sin(36.8^\circ + 0.2^\circ) = \sin(37.0^\circ) = 0.60182\)
\(\sin(36.8^\circ - 0.2^\circ) = \sin(36.6^\circ) = 0.59622\)
The absolute uncertainty in \(\sin \theta\) is:
\(\Delta(\sin \theta) = 0.60182 - 0.59902 = 0.00280\).
Percentage uncertainty in \(\sin \theta\):
\(\% \Delta(\sin \theta) = \frac{0.00280}{0.59902} \times 100\% = 0.467\% \approx 0.47\%\).

(d) Percentage uncertainty in \(\lambda\):
\(\% \Delta \lambda = \% \Delta N + \% \Delta(\sin \theta)\)
\(\% \Delta N = \frac{0.05}{5.00} \times 100\% = 1.00\%\)
\(\% \Delta \lambda = 1.00\% + 0.47\% = 1.47\%\).

Absolute uncertainty in \(\lambda\):
\(\Delta \lambda = 599\text{ nm} \times 0.0147 = 8.81\text{ nm} \approx 9\text{ nm}\).
Thus, \(\lambda = (599 \pm 9)\text{ nm}\).

Part (a):
- 1 mark for \(\theta = 36.8^\circ\).
- 1 mark for adding absolute uncertainties to get \(\Delta \theta = 0.2^\circ\).

Part (b):
- 1 mark for substituting values correctly into \(\lambda = \frac{\sin \theta}{2 N}\).
- 1 mark for calculating \(\lambda = 599\text{ nm}\) (or \(5.99 \times 10^{-7}\text{ m}\)).

Part (c):
- 1 mark for calculating \(\sin(37.0^\circ)\) and \(\sin(36.6^\circ)\).
- 1 mark for finding the absolute uncertainty in \(\sin \theta\) is \(0.0028\).
- 1 mark for showing \(\% \Delta(\sin \theta) \approx 0.47\%\).

Part (d):
- 1 mark for calculating \(\% \Delta N = 1.0\%\).
- 1 mark for summing percentage uncertainties to get \(1.47\%\) (or \(1.5\%\)).
- 1 mark for stating final result as \(\lambda = (599 \pm 9)\text{ nm}\) (or \((5.99 \pm 0.09) \times 10^{-7}\text{ m}\)).

(a) Luminosity is defined as the total radiant power emitted by a star.

(b) \(L = 4 \pi (7.2 \times 10^8)^2 \times (5.67 \times 10^{-8}) \times (5800)^4\)
\(L = 4 \pi \times (5.184 \times 10^{17}) \times (5.67 \times 10^{-8}) \times (1.1316 \times 10^{15}) = 4.180 \times 10^{26}\text{ W}\).

(c) (i) \(\% \Delta (R^2) = 2 \times \% \Delta R = 2 \times \left(\frac{0.3}{7.2}\right) \times 100\% = 2 \times 4.167\% = 8.33\% \approx 8.3\%\).

(ii) \(\% \Delta (T^4) = 4 \times \% \Delta T = 4 \times \left(\frac{120}{5800}\right) \times 100\% = 4 \times 2.069\% = 8.28\% \approx 8.3\%\).

(iii) Since \(L \propto R^2 T^4\), the percentage uncertainty in \(L\) is:
\(\% \Delta L = \% \Delta(R^2) + \% \Delta(T^4) = 8.33\% + 8.28\% = 16.61\% \approx 17\%\).

(d) Absolute uncertainty in \(L\):
\(\Delta L = 4.180 \times 10^{26}\text{ W} \times 0.1661 = 6.94 \times 10^{25}\text{ W} \approx 0.7 \times 10^{26}\text{ W}\).
Thus, \(L = (4.2 \pm 0.7) \times 10^{26}\text{ W}\).

Part (a):
- 1 mark for stating 'total power emitted/radiated by the star'.

Part (b):
- 1 mark for substituting values into the formula correctly.
- 1 mark for calculating \(L = 4.18 \times 10^{26}\text{ W}\) (or \(4.2 \times 10^{26}\text{ W}\)).

Part (c) (i):
- 1 mark for finding percentage uncertainty of \(R\) is \(4.2\%\).
- 1 mark for multiplying by 2 to get \(8.3\%\) (accept range \(8.3\%\) to \(8.4\%\)).

Part (c) (ii):
- 1 mark for finding percentage uncertainty of \(T\) is \(2.1\%\).
- 1 mark for multiplying by 4 to get \(8.3\%\) (accept range \(8.2\%\) to \(8.4\%\)).

Part (c) (iii):
- 1 mark for adding the percentage uncertainties of \(R^2\) and \(T^4\) to get \(16.6\%\) (or \(17\%\)).

Part (d):
- 1 mark for calculating absolute uncertainty as \(0.7 \times 10^{26}\text{ W}\) (or \(0.69 \times 10^{26}\text{ W}\)).
- 1 mark for final result \(L = (4.2 \pm 0.7) \times 10^{26}\text{ W}\) (or \(L = (4.18 \pm 0.69) \times 10^{26}\text{ W}\)).

(a) The base unit equation is:
\([E] = \frac{[F][L]}{[d]^2[e]}\).
- Force \(F\) has SI base units of \(\text{kg m s}^{-2}\).
- Length \(L\) and extension \(e\) have units of \(\text{m}\).
- Diameter squared \(d^2\) has units of \(\text{m}^2\).
Substituting these units:
\([E] = \frac{\text{kg m s}^{-2} \cdot \text{m}}{\text{m}^2 \cdot \text{m}} = \text{kg m}^{-1}\text{ s}^{-2}\) (Shown).

(b) Substitute values into the formula:
\(E = \frac{4 \times 65.0 \times 2.035}{\pi \times (0.42 \times 10^{-3})^2 \times 1.8 \times 10^{-3}} = \frac{529.1}{9.9752 \times 10^{-10}} = 5.304 \times 10^{11}\text{ Pa} = 530\text{ GPa}\).

(c) Calculate individual percentage uncertainties:
- For \(F\): \(\frac{0.5}{65.0} \times 100\% = 0.77\%\)
- For \(L\): \(\frac{0.002}{2.035} \times 100\% = 0.10\%\)
- For \(d\): \(\frac{0.01}{0.42} \times 100\% = 2.38\%\) (which is doubled for \(d^2\): \(2 \times 2.38\% = 4.76\%\))
- For \(e\): \(\frac{0.1}{1.8} \times 100\% = 5.56\%\)
Total percentage uncertainty in \(E\):
\(\% \Delta E = 0.77\% + 0.10\% + 4.76\% + 5.56\% = 11.19\% \approx 11.2\%\).

(d) Calculate absolute uncertainty in \(E\):
\(\Delta E = 530.4\text{ GPa} \times 0.1119 = 59.4\text{ GPa} \approx 60\text{ GPa}\).
Thus, \(E = (530 \pm 60)\text{ GPa}\).

Part (a):
- 1 mark for expressing unit of force as \(\text{kg m s}^{-2}\).
- 1 mark for clear steps of cancelling and simplifying units to get \(\text{kg m}^{-1}\text{ s}^{-2}\).

Part (b):
- 1 mark for substitution of values with correct powers of ten.
- 1 mark for calculating \(E = 5.3 \times 10^{11}\text{ Pa}\) (or \(530\text{ GPa}\)).

Part (c):
- 1 mark for percentage uncertainties of \(F\) (\(0.77\%\)) and \(L\) (\(0.10\%\)).
- 1 mark for percentage uncertainty of \(d\) (\(2.38\%\)).
- 1 mark for multiplying percentage uncertainty of \(d\) by 2 (\(4.76\%\)).
- 1 mark for summing to get \(11\%\) or \(11.2\%\).

Part (d):
- 1 mark for absolute uncertainty \(\Delta E = 60\text{ GPa}\) (or \(59\text{ GPa}\)).
- 1 mark for final result \(E = (530 \pm 60)\text{ GPa}\) with matching decimal precision.

(a) Hubble's law states that the speed \(v\) at which a galaxy is moving away from an observer is directly proportional to its distance \(d\) from the observer.
Formula: \(v = H_0 d\) where \(H_0\) is the Hubble constant.

(b) Redshift \(z = \frac{\Delta \lambda}{\lambda_0} = \frac{682.1 - 656.3}{656.3} = \frac{25.8}{656.3} = 0.0393 \approx 0.039\) (Shown).

(c) For small redshift \(z \ll 1\), the recessional speed \(v\) is given by:
\(v = z c = 0.0393 \times 3.00 \times 10^8\text{ m s}^{-1} = 1.18 \times 10^7\text{ m s}^{-1}\).

(d) (i) Convert \(v\) to \(\text{km s}^{-1}\):
\(v = 1.18 \times 10^4\text{ km s}^{-1}\).
Rearranging Hubble's law:
\(d = \frac{v}{H_0} = \frac{1.18 \times 10^4}{70} = 168.6\text{ Mpc} \approx 170\text{ Mpc}\).

(ii) The age of the Universe is estimated as \(T \approx \frac{1}{H_0}\).
Convert \(H_0\) to SI units (\(\text{s}^{-1}\)):
\(H_0 = 70\text{ km s}^{-1}\text{ Mpc}^{-1} = \frac{70 \times 10^3\text{ m s}^{-1}}{10^6 \times 3.09 \times 10^{16}\text{ m}} = 2.265 \times 10^{-18}\text{ s}^{-1}\).
\(T = \frac{1}{H_0} = \frac{1}{2.265 \times 10^{-18}} = 4.415 \times 10^{17}\text{ s}\).
Converting to years:
\(T = \frac{4.415 \times 10^{17}}{365.25 \times 24 \times 3600} = 1.40 \times 10^{10}\text{ years}\) (or 14 billion years).

Part (a):
- 1 mark for stating that recessional speed is directly proportional to distance.
- 1 mark for defining all three symbols \(v\), \(d\), and \(H_0\) correctly.

Part (b):
- 1 mark for the correct formula \(z = \frac{\Delta \lambda}{\lambda}\).
- 1 mark for showing substitution and obtaining \(0.0393 \approx 0.039\).

Part (c):
- 1 mark for using \(v = z c\).
- 1 mark for calculating \(v = 1.18 \times 10^7\text{ m s}^{-1}\).

Part (d) (i):
- 1 mark for converting recessional speed to \(\text{km s}^{-1}\).
- 1 mark for calculating \(d = 170\text{ Mpc}\) (accept \(168 - 170\text{ Mpc}\)).

Part (d) (ii):
- 1 mark for converting \(H_0\) to \(\text{s}^{-1}\).
- 1 mark for estimating the age as \(1.4 \times 10^{10}\text{ years}\) (accept range \(1.38 \times 10^{10}\) to \(1.42 \times 10^{10}\text{ years}\)).

(a) A distance of 1 light-year is defined as:
\(1\text{ ly} = (3.00 \times 10^8\text{ m s}^{-1}) \times (365.25 \times 24 \times 3600\text{ s}) = 9.467 \times 10^{15}\text{ m}\).
For \(8.61\text{ light-years}\):
\(d = 8.61 \times 9.467 \times 10^{15}\text{ m} = 8.151 \times 10^{16}\text{ m} \approx 8.15 \times 10^{16}\text{ m}\) (Shown).

(b) Use the radiant flux density formula:
\(F = \frac{L}{4 \pi d^2} \implies L = 4 \pi d^2 F\)
\(L = 4 \pi \times (8.151 \times 10^{16}\text{ m})^2 \times (1.17 \times 10^{-7}\text{ W m}^{-2})\)
\(L = 4 \pi \times (6.644 \times 10^{33}) \times (1.17 \times 10^{-7}) = 9.77 \times 10^{27}\text{ W} \approx 9.8 \times 10^{27}\text{ W}\).

(c) Wien's displacement law states that the wavelength of maximum intensity of emission \(\lambda_{\text{max}}\) from a black body is inversely proportional to its thermodynamic temperature \(T\):
\(\lambda_{\text{max}} T = 2.898 \times 10^{-3}\text{ m K}\).
Solving for \(\lambda_{\text{max}}\):
\(\lambda_{\text{max}} = \frac{2.898 \times 10^{-3}\text{ m K}}{9940\text{ K}} = 2.916 \times 10^{-7}\text{ m} \approx 2.9 \times 10^{-7}\text{ m}\) (or \(292\text{ nm}\)).

(d) First, obtain the continuous black-body emission spectrum of the star. Determine the wavelength at peak intensity (\(\lambda_{\text{max}}\)), and then apply Wien's displacement law (\(\lambda_{\text{max}} \propto 1/T\)) to compute the surface temperature.

Part (a):
- 1 mark for converting \(8.61\text{ years}\) into seconds correctly (\(2.72 \times 10^8\text{ s}\)).
- 1 mark for multiplying by the speed of light to get \(8.15 \times 10^{16}\text{ m}\).

Part (b):
- 1 mark for recalling \(F = \frac{L}{4 \pi d^2}\).
- 1 mark for correct substitution of values.
- 1 mark for \(L = 9.8 \times 10^{27}\text{ W}\) (accept \(9.77 \times 10^{27}\text{ W}\)).

Part (c):
- 1 mark for stating that peak wavelength is inversely proportional to thermodynamic temperature.
- 1 mark for stating the equation \(\lambda_{\text{max}} T = \text{constant}\).
- 1 mark for calculating \(\lambda_{\text{max}} = 2.9 \times 10^{-7}\text{ m}\).

Part (d):
- 1 mark for explaining that peak wavelength is found from the black-body spectrum.
- 1 mark for explaining that Wien's displacement law is used to calculate temperature.

(a) Magnetic flux linkage is defined as the product of the magnetic flux through a coil and the number of turns of the coil.

(b) First, calculate the cross-sectional area \(A\) of the circular coil:
\(A = \pi r^2 = \pi \times (0.035\text{ m})^2 = 3.848 \times 10^{-3}\text{ m}^2\).

At \(t = 2.0\text{ s}\), the magnetic flux density is \(B = 0.18\text{ T}\).
Since the field is perpendicular to the plane of the coil, the magnetic flux linkage is:
\(N \Phi = N B A = 250 \times 0.18\text{ T} \times 3.848 \times 10^{-3}\text{ m}^2 = 0.173\text{ Wb} \approx 0.17\text{ Wb}\) (or \(\text{V s}\)).

(c) (i) By Faraday's law, the induced e.m.f. \(E\) is given by the rate of change of magnetic flux linkage:
\(E = \frac{\Delta(N\Phi)}{\Delta t}\).
During the first 2.0 s:
\(E = \frac{0.173 - 0}{2.0} = 0.0866\text{ V} \approx 0.087\text{ V}\) (or \(87\text{ mV}\)).

(ii) During the interval from \(t = 5.0\text{ s}\) to \(t = 6.0\text{ s}\):
\(E = \frac{0 - 0.173}{1.0} = -0.173\text{ V}\).
The magnitude of the induced e.m.f. is \(0.173\text{ V} \approx 0.17\text{ V}\) (or \(170\text{ mV}\)).

(d) Lenz's law states that the direction of the induced e.m.f. is such that it opposes the change in magnetic flux that produces it.

Part (a):
- 1 mark for stating product of magnetic flux and number of turns.
- 1 mark for expressing this as \(N \Phi\) where \(\Phi = B A\) (or \(B A \cos\theta\)).

Part (b):
- 1 mark for calculating cross-sectional area \(A = 3.85 \times 10^{-3}\text{ m}^2\).
- 1 mark for multiplying \(N\), \(B\), and \(A\).
- 1 mark for correct value and unit: \(0.17\text{ Wb}\) (or \(\text{V s}\)).

Part (c) (i):
- 1 mark for stating or using \(E = \frac{\Delta(N\Phi)}{\Delta t}\).
- 1 mark for calculating \(E = 0.087\text{ V}\) (or \(87\text{ mV}\)).

Part (c) (ii):
- 1 mark for using \(\Delta t = 1.0\text{ s}\).
- 1 mark for calculating magnitude \(E = 0.17\text{ V}\) (or \(170\text{ mV}\)).

Part (d):
- 1 mark for stating that the induced e.m.f. opposes the change in magnetic flux.

\textbf{(a)} Wien's displacement law states that the wavelength \(\lambda_{\text{max}}\) of the peak of the black-body radiation spectrum is inversely proportional to the thermodynamic temperature \(T\) of the black body. Mathematically: \(\lambda_{\text{max}} T = b\), where \(\lambda_{\text{max}}\) is the wavelength at which maximum intensity is emitted, \(T\) is the absolute (thermodynamic) temperature, and \(b\) is Wien's displacement constant.

\textbf{(b)(i)} Using Wien's displacement law:
\(T = \frac{b}{\lambda_{\text{max}}} = \frac{2.90 \times 10^{-3}}{410 \times 10^{-9}} = 7073\text{ K} \approx 7070\text{ K}\) (or \(7100\text{ K}\) to 2 significant figures).

\textbf{(b)(ii)} Using the Stefan-Boltzmann law:
\(L = 4 \pi R^2 \sigma T^4\)

Rearranging for \(R\):
\(R^2 = \frac{L}{4 \pi \sigma T^4} = \frac{9.2 \times 10^{28}}{4 \pi \times 5.67 \times 10^{-8} \times (7073)^4}\)
\(R^2 = \frac{9.2 \times 10^{28}}{4 \pi \times 5.67 \times 10^{-8} \times 2.502 \times 10^{15}} = \frac{9.2 \times 10^{28}}{1.783 \times 10^9} \approx 5.16 \times 10^{19}\text{ m}^2\)
\(R = \sqrt{5.16 \times 10^{19}} = 7.18 \times 10^9\text{ m} \approx 7.2 \times 10^9\text{ m}\).

\textbf{(c)(i)} The redshift is given by:
\(\frac{\Delta \lambda}{\lambda} = \frac{v}{c}\)
where \(\Delta \lambda = 682.1\text{ nm} - 656.3\text{ nm} = 25.8\text{ nm}\).
\(v = c \times \frac{\Delta \lambda}{\lambda} = 3.00 \times 10^8 \times \frac{25.8}{656.3} = 1.18 \times 10^7\text{ m s}^{-1}\).

\textbf{(c)(ii)} The observed wavelength is larger than the laboratory wavelength (redshifted), which indicates that the galaxy is moving away from the Earth (receding).

(a)
- Wien's displacement law stated (e.g., peak wavelength is inversely proportional to thermodynamic temperature) [1]
- Symbols defined: \(\lambda_{\text{max}}\) is peak wavelength and \(T\) is thermodynamic temperature [1]

(b)(i)
- Formula used \(T = b / \lambda_{\text{max}}\) [1]
- Correct calculation to give \(7070\text{ K}\) (or \(7100\text{ K}\)) [1]

(b)(ii)
- Formula used \(L = 4\pi R^2 \sigma T^4\) [1]
- Correct substitution of values including candidate's \(T\) [1]
- Correct calculation leading to \(R = 7.18 \times 10^9\text{ m}\) (or \(7.2 \times 10^9\text{ m}\)) [1]

(c)(i)
- Correct calculation of change in wavelength \(\Delta \lambda = 25.8\text{ nm}\) [1]
- Correct calculation of speed \(v = 1.18 \times 10^7\text{ m s}^{-1}\) [1]

(c)(ii)
- Galaxy is moving away from Earth (or receding) [1]

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

\textbf{(b)(i)} The area \(A\) of the circular coil is given by:
\(A = \pi r^2 = \pi (0.025\text{ m})^2 = 1.963 \times 10^{-3}\text{ m}^2\)

At \(t = 0.40\text{ s}\), the magnetic flux density is \(B = 0.12\text{ T}\).
Since the plane of the coil is perpendicular to the magnetic field, the magnetic flux linkage is:
\(N \Phi = N B A = 150 \times 0.12 \times 1.963 \times 10^{-3} = 0.353\text{ Wb}\) (or \(0.35\text{ Wb}\)).

\textbf{(b)(ii) 1.} During the interval \(t = 0\) to \(t = 0.40\text{ s}\), the rate of change of magnetic flux linkage is constant. The magnitude of the induced e.m.f. is:
\(E = \frac{\Delta (N \Phi)}{\Delta t} = \frac{0.353 - 0}{0.40} = 0.883\text{ V} \approx 0.88\text{ V}\).

\textbf{2.} During the interval \(t = 1.00\text{ s}\) to \(t = 1.20\text{ s}\), the magnetic flux density decreases to zero in a time interval of \(0.20\text{ s}\). The magnitude of the induced e.m.f. is:
\(E = \frac{\Delta (N \Phi)}{\Delta t} = \frac{0.353}{0.20} = 1.77\text{ V} \approx 1.8\text{ V}\).

\textbf{(b)(iii)} The induced e.m.f. is zero when there is no change in magnetic flux linkage, which occurs when the magnetic flux density is constant. This is during the interval from \(t = 0.40\text{ s}\) to \(t = 1.00\text{ s}\).

(a)
- induced e.m.f. is proportional to rate of change of [1]
- magnetic flux linkage [1]

(b)(i)
- Correct calculation of area (\(1.96 \times 10^{-3}\text{ m}^2\)) [1]
- Formula \(N B A\) used [1]
- Correct calculation to give \(0.35\text{ Wb}\) or \(0.353\text{ Wb}\) [1]

(b)(ii) 1.
- Correct use of \(E = \Delta (N \Phi) / \Delta t\) [1]
- Value \(0.88\text{ V}\) (or \(0.883\text{ V}\)) [1]

(b)(ii) 2.
- Correct time interval of \(0.20\text{ s}\) used [1]
- Value \(1.8\text{ V}\) (or \(1.77\text{ V}\)) [1]

(b)(iii)
- Interval specified as from \(t = 0.40\text{ s}\) to \(t = 1.00\text{ s}\) [1]

### Apparatus Design and Diagram
Draw a diagram showing a horizontal air track connected to a constant air supply. A plastic glider of mass \(m\) with a card of width \(d\) and a small neodymium magnet is placed on the track. At one end of the air track, place the launcher coil. The launcher coil is connected to a circuit containing a double-throw switch (two-way switch), a DC power supply, a digital voltmeter (to measure the charging voltage \(V\)), and a capacitor bank of variable capacitance \(C\).

Place a light gate connected to a digital timer/data logger along the air track, just in front of the glider's initial position, so that the card interrupts the beam immediately after the glider leaves the launcher coil.

### Procedure
1. Set up the circuit as described. Connect the capacitor bank in parallel combinations to vary the capacitance \(C\) (e.g., using five identical \(100\text{ \mu F}\) capacitors to get \(100, 200, 300, 400, 500\text{ \mu F}\)).
2. With the switch in the 'charge' position, adjust the power supply until the voltmeter reads a predetermined constant voltage \(V\) (e.g., \(12.0\text{ V}\)).
3. Position the glider at the same starting mark at the end of the air track (e.g., touching a mechanical stop right next to the launcher coil).
4. Throw the switch to the 'discharge' position to pass the current pulse through the coil and launch the glider.
5. The card on the glider passes through the light gate. Record the time of interruption \(\Delta t\).
6. Calculate the maximum speed \(v\) using \(v = \frac{d}{\Delta t}\).
7. Repeat the measurement at least three times to find an average value of \(v\) for each capacitance value.
8. Repeat steps 2-7 for different values of capacitance \(C\).

### Control of Variables
- **Mass of the glider**: Use the same glider throughout the experiment, ensuring no attachments are changed.
- **Charging Voltage**: Ensure the voltmeter reads exactly the same voltage \(V\) before each launch.
- **Initial Position**: Use a fixed spacer or ruler to ensure the distance between the neodymium magnet and the coil is identical before every launch.
- **Friction**: Maintain a constant pressure from the air supply to the track to minimize frictional retardation.

### Analysis of Data
Take natural logarithms of both sides of the relationship:
\[\ln(v) = n \ln(C) + \ln(k)\]
Plot a graph of \(\ln(v)\) on the y-axis against \(\ln(C)\) on the x-axis.
If the relationship is valid, the graph will be a straight line.
- Gradient \(= n\)
- y-intercept \(= \ln(k)\), so \(k = e^{\text{y-intercept}}\).

### Safety Precautions
- Use a soft foam buffer or net at the far end of the air track to prevent the glider from flying off and causing injury or damage.
- Do not exceed the maximum voltage rating of the electrolytic capacitors to prevent overheating or explosion.

**Defining the problem (2 marks)**
- **DP1**: Identify \(C\) as the independent variable and \(v\) as the dependent variable.
- **DP2**: State that the charging voltage \(V\) (or mass \(m\) of the glider) must be kept constant.

**Methods of data collection (4 marks)**
- **MC1**: Draw a clear, labelled diagram showing the glider on an air track, launcher coil, capacitor circuit, and a light gate.
- **MC2**: Describe a workable circuit showing a double-throw switch to charge the capacitor from a DC supply and discharge it through the coil, with a voltmeter across the capacitor bank.
- **MC3**: Describe how the maximum velocity is measured using \(v = d/t\) where \(d\) is the width of a card and \(t\) is the interruption time measured by a light gate.
- **MC4**: Explain how the capacitance \(C\) is varied (e.g., by adding identical capacitors in parallel and using \(C = C_1 + C_2 + ...\)).

**Method of analysis (3 marks)**
- **MA1**: State that a graph of \(\ln(v)\) against \(\ln(C)\) should be plotted.
- **MA2**: State that the relationship is valid if a straight line is obtained.
- **MA3**: Identify that the gradient is equal to \(n\) and the y-intercept is equal to \(\ln(k)\).

**Safety considerations (1 mark)**
- **SF1**: Use a padded barrier or catch-box at the end of the air track to safely stop the glider; OR ensure the charging voltage does not exceed the capacitors' working voltage.

**Additional detail (5 marks)**
- **AD1**: Ensure the glider always starts from the same starting position using a mechanical stop or alignment mark.
- **AD2**: Keep the air-supply pressure constant to ensure consistent, negligible friction.
- **AD3**: Wait for the voltmeter reading to stabilize at the set charging voltage \(V\) before each discharge.
- **AD4**: Repeat the time measurement for each capacitance and take the average.
- **AD5**: Use a high-resolution digital voltmeter and a light-gate timer with millisecond resolution.

### Part (a)
The given formula is:
\[y = \frac{F L^3}{4 E w d^3}\]
When \(y\) in mm is plotted on the y-axis and \(L^3\) in \(\text{m}^3\) is plotted on the x-axis, we can write the equation as:
\[y / \text{ mm} = 10^3 \left( \frac{F}{4 E w d^3} \right) L^3\]
Therefore, the gradient \(G\) of the graph is:
\[G = \frac{1000 F}{4 E w d^3}\]

### Part (b)
Calculating the values of \(L^3 / \text{ m}^3\):
- For \(L = 0.50\text{ m}\): \(L^3 = 0.50^3 = 0.125\text{ m}^3\)
- For \(L = 0.60\text{ m}\): \(L^3 = 0.60^3 = 0.216\text{ m}^3\)
- For \(L = 0.70\text{ m}\): \(L^3 = 0.70^3 = 0.343\text{ m}^3\)
- For \(L = 0.80\text{ m}\): \(L^3 = 0.80^3 = 0.512\text{ m}^3\)
- For \(L = 0.90\text{ m}\): \(L^3 = 0.90^3 = 0.729\text{ m}^3\)
- For \(L = 1.00\text{ m}\): \(L^3 = 1.00^3 = 1.000\text{ m}^3\)

### Part (c)
Plot the six points on a graph of \(y / \text{ mm}\) against \(L^3 / \text{ m}^3\). Each point must have a vertical error bar of length \(\pm 0.5\text{ mm}\).
- **Line of Best Fit (LOBF)**: Draw a single straight line that represents the best average of all the data points. This line passes approximately through \((0.125, 3.2)\) and \((1.000, 25.4)\).
- **Worst Acceptable Straight Line (WASL)**: Draw a line that passes through the top of the error bar of the first data point \((0.125, 3.7)\) and the bottom of the error bar of the last data point \((1.000, 25.0)\) (or vice versa).

### Part (d)
Using coordinates from the LOBF:
- Point 1: \((0.125, 3.2)\)
- Point 2: \((1.000, 25.4)\)
\[G_{\text{best}} = \frac{25.4 - 3.2}{1.000 - 0.125} = \frac{22.2}{0.875} = 25.37 \approx 25.4\text{ mm m}^{-3}\]
Using coordinates from the WASL:
- Point 1: \((0.125, 3.7)\)
- Point 2: \((1.000, 25.0)\)
\[G_{\text{worst}} = \frac{25.0 - 3.7}{1.000 - 0.125} = \frac{21.3}{0.875} = 24.34\text{ mm m}^{-3}\]

Uncertainty in gradient:
\[\Delta G = G_{\text{best}} - G_{\text{worst}} = 25.37 - 24.34 = 1.03 \approx 1.1\text{ mm m}^{-3}\]
Therefore, the gradient is \(25.4 \pm 1.1\text{ mm m}^{-3}\) (accept \(25.4 \pm 1.2\text{ mm m}^{-3}\)).

y-intercept of the LOBF:
\[y = G x + c \implies 25.4 = 25.37(1.000) + c \implies c = 0.03 \approx 0.0\text{ mm}\]
y-intercept of the WASL:
\[c_{\text{worst}} = 25.0 - 24.34(1.000) = 0.66 \approx 0.7\text{ mm}\]

Uncertainty in y-intercept:
\[\Delta c = 0.66 - 0.03 = 0.63 \approx 0.7\text{ mm}\]
Therefore, the y-intercept is \(0.0 \pm 0.7\text{ mm}\).

### Part (e)
(i) Calculating the Young modulus \(E\):
From \(G = \frac{1000 F}{4 E w d^3}\), we have:
\[E = \frac{1000 F}{4 w d^3 G}\]
Substitute the given values and the best-fit gradient:
\[E = \frac{1000 \times 15.0}{4 \times 0.0240 \times (0.0080)^3 \times 25.37}\]
\[E = \frac{15000}{0.0960 \times 5.12 \times 10^{-7} \times 25.37} = \frac{15000}{1.247 \times 10^{-6}} = 1.20 \times 10^{10}\]

(ii) Calculating the uncertainty \(\Delta E\):
Using the fractional uncertainty relation:
\[\frac{\Delta E}{E} = \frac{\Delta F}{F} + \frac{\Delta w}{w} + 3\frac{\Delta d}{d} + \frac{\Delta G}{G}\]
\[\frac{\Delta E}{E} = \frac{0.2}{15.0} + \frac{0.0010}{0.0240} + 3 \left( \frac{0.0002}{0.0080} \right) + \frac{1.1}{25.4}\]
\[\frac{\Delta E}{E} = 0.0133 + 0.0417 + 0.0750 + 0.0433 = 0.1733\]
Therefore, the absolute uncertainty in \(E\) is:
\[\Delta E = 1.20 \times 10^{10} \times 0.1733 = 2.08 \times 10^9 \approx 2.1 \times 10^9\]
So, \(E = (1.2 \pm 0.2) \times 10^{10}\).

**(a) Expression for Gradient (1 mark)**
- **A1**: Identifies the gradient as \(G = \frac{1000 F}{4 E w d^3}\) or \(G = \frac{F}{4 E w d^3}\) (if vertical axis is in m, but accept either with clear formulation of units).

**(b) Data Processing (1 mark)**
- **A1**: Correct values of \(L^3\) in \(\text{m}^3\) to 3 significant figures: 0.125, 0.216, 0.343, 0.512, 0.729, 1.000.

**(c) Graph Plotting (4 marks)**
- **GP1**: Linear axes with appropriate scales. Points should occupy more than half of the grid in both directions.
- **GP2**: All six points plotted accurately within half a small square.
- **GP3**: Vertical error bars plotted accurately at \(\pm 0.5\text{ mm}\) for all six points.
- **GP4**: Both the best-fit line and the worst acceptable straight line drawn, clearly labelled. The worst acceptable line must pass through the top of the first error bar and the bottom of the last error bar (or vice versa).

**(d) Gradient and Intercept determination (4 marks)**
- **GI1**: Large triangle used (at least half of the drawn line) to find the gradient of the best-fit line. Gradient value in range \(24.5\) to \(26.5\).
- **GI2**: Uncertainty in gradient calculated as \(G_{\text{best}} - G_{\text{worst}}\). Value is in range \(1.0\) to \(1.3\).
- **GI3**: y-intercept determined by direct reading or using \(y = mx + c\) for LOBF. Intercept is in range \(-0.2\) to \(0.2\).
- **GI4**: Uncertainty in y-intercept calculated as \(c_{\text{best}} - c_{\text{worst}}\). Value is in range \(0.5\) to \(0.8\).

**(e) Calculations and Uncertainties (5 marks)**
- **EU1**: Correct calculation of \(E\) using the formula \(E = \frac{1000 F}{4 w d^3 G}\). Expect a value in range \(1.1 \times 10^{10}\) to \(1.3 \times 10^{10}\).
- **EU2**: Correct calculation of fractional or percentage uncertainties of all constituents:
- \(\Delta F/F = 0.0133\) or \(1.3\%\)
- \(\Delta w/w = 0.0417\) or \(4.2\%\)
- \(3\Delta d/d = 0.0750\) or \(7.5\%\)
- \(\Delta G/G \approx 0.0433\) or \(4.3\%\)
- **EU3**: Sums all fractional uncertainties correctly to obtain overall fractional uncertainty (expect \(\approx 0.17\) or \(17\%\)) and multiplies by \(E\) to obtain absolute uncertainty \(\Delta E\) (expect \(\approx 2 \times 10^9\)).