2025 Cambridge IAS-Level Physics (9702) Practice Paper with Answers

To find the percentage uncertainty in the calculated density, we add the percentage uncertainties of each independent quantity, taking powers into account:

$$\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}$$

Calculate the percentage uncertainty of each individual measurement:
- Mass: \(\frac{0.07}{3.50} \times 100\% = 2.0\%\)
- Diameter: \(\frac{0.03}{1.50} \times 100\% = 2.0\%\)
- Length: \(\frac{0.5}{50.0} \times 100\% = 1.0\%\)

Now substitute these values into the uncertainty equation:
$$\frac{\Delta \rho}{\rho} = 2.0\% + 2(2.0\%) + 1.0\% = 7.0\%$$

Therefore, the percentage uncertainty is \(7.0\%\).

[1 mark] Correctly identifies that the percentage uncertainties of the terms must be added, with the diameter uncertainty doubled, and obtains the value of 7.0%.

Using the potential divider formula, we calculate the potential difference across the LDR in both states:

In the dark:
$$V_{\text{dark}} = V_{\text{supply}} \times \frac{R_{\text{LDR}}}{R_{\text{LDR}} + R}$$
$$V_{\text{dark}} = 12.0\text{ V} \times \frac{12.0\text{ k}\Omega}{12.0\text{ k}\Omega + 4.0\text{ k}\Omega} = 12.0\text{ V} \times \frac{12.0}{16.0} = 9.0\text{ V}$$

In bright light:
$$V_{\text{bright}} = V_{\text{supply}} \times \frac{R_{\text{LDR}}}{R_{\text{LDR}} + R}$$
$$V_{\text{bright}} = 12.0\text{ V} \times \frac{1.0\text{ k}\Omega}{1.0\text{ k}\Omega + 4.0\text{ k}\Omega} = 12.0\text{ V} \times \frac{1.0}{5.0} = 2.4\text{ V}$$

Calculating the change in potential difference:
$$\Delta V = 9.0\text{ V} - 2.4\text{ V} = 6.6\text{ V}$$

[1 mark] Calculates both voltmeter readings correctly and finds the absolute difference between them to be 6.6 V.

The potential energy of a simple harmonic oscillator at displacement \(x\) is given by:
$$E_p = \frac{1}{2} m \omega^2 x^2$$

The kinetic energy is given by:
$$E_k = \frac{1}{2} m \omega^2 (A^2 - x^2)$$

We are given that the kinetic energy is three times the potential energy:
$$E_k = 3 E_p$$
$$\frac{1}{2} m \omega^2 (A^2 - x^2) = 3 \left(\frac{1}{2} m \omega^2 x^2\right)$$
$$A^2 - x^2 = 3 x^2$$
$$A^2 = 4 x^2$$
$$x = \pm \frac{A}{2}$$

Given amplitude \(A = 4.0\text{ cm}\):
$$x = \pm \frac{4.0\text{ cm}}{2} = \pm 2.0\text{ cm}$$

[1 mark] Formulates the relationship between kinetic energy and potential energy in terms of amplitude and displacement, and correctly solves for x = 2.0 cm.

Since both wires are made of the same metal, they have the same Young modulus \(E\).
Young modulus is defined as:
$$E = \frac{\text{stress}}{\text{strain}} \implies \text{strain} = \frac{\text{stress}}{E}$$

Stress is given by force per unit cross-sectional area, \(\sigma = \frac{F}{A}\). The area of a wire of diameter \(D\) is \(A = \frac{\pi D^2}{4}\).
Therefore, stress is proportional to \(\frac{1}{D^2}\) because the tension \(F\) is identical in both wires.

Since strain \(\varepsilon = \frac{\sigma}{E}\) and \(E\) is constant:
$$\varepsilon \propto \sigma \propto \frac{1}{D^2}$$

This shows that strain is independent of the length of the wire!

Let the strain in wire \(X\) be \(\varepsilon_X\) and in wire \(Y\) be \(\varepsilon_Y\):
$$\frac{\varepsilon_X}{\varepsilon_Y} = \frac{d_Y^2}{d_X^2} = \frac{(2d)^2}{d^2} = 4$$

[1 mark] Recognises that strain is independent of length and depends inversely on the square of the diameter, resulting in a ratio of 4.

When unpolarised light of intensity \(I_0\) passes through the first polarizer, it becomes plane-polarised, and its intensity is reduced by half:
$$I_1 = \frac{1}{2} I_0$$

According to Malus's Law, when this polarised light passes through the second filter (analyser) at an angle of \(\theta = 60^\circ\), the transmitted intensity \(I_2\) is:
$$I_2 = I_1 \cos^2(\theta)$$
$$I_2 = \left(\frac{1}{2} I_0\right) \cos^2(60^\circ)$$

Since \(\cos(60^\circ) = 0.5\):
$$I_2 = \frac{1}{2} I_0 \times (0.5)^2 = \frac{1}{2} I_0 \times 0.25 = 0.125 I_0$$

[1 mark] Applies the unpolarised-to-polarised transition factor (0.5) and Malus's law with cos^2(60) correctly to yield 0.125 I_0.

Let \(v_1\) be the final velocity of the \(3m\) sphere and \(v_2\) be the final velocity of the \(m\) sphere.

By conservation of momentum:
$$3m u + m(0) = 3m v_1 + m v_2$$
$$3u = 3v_1 + v_2 \quad \text{--- (Equation 1)}$$

Since the collision is perfectly elastic, the relative speed of approach equals the relative speed of separation:
$$u - 0 = v_2 - v_1$$
$$v_2 = u + v_1 \quad \text{--- (Equation 2)}$$

Substitute Equation 2 into Equation 1:
$$3u = 3v_1 + (u + v_1)$$
$$2u = 4v_1 \implies v_1 = +0.5u$$

Now substitute \(v_1\) back to find \(v_2\):
$$v_2 = u + 0.5u = +1.5u$$

Thus, the velocity of the first sphere is \(+0.5u\) and the second is \(+1.5u\).

[1 mark] Sets up momentum conservation and relative velocity equations for an elastic collision and solves simultaneously for the correct velocities.

Let us model the system by taking the left-hand end (pivot \(A\)) as the reference point \(x=0\).
- Pivot \(A\) is located at \(x = 0\).
- Pivot \(B\) is located at \(x = 4.0\text{ m} - 1.0\text{ m} = 3.0\text{ m}\).
- The weight of the uniform plank acts at its center of mass, \(x = 2.0\text{ m}\). Its weight is \(W_p = m_p g = 20 \times 9.81\text{ N} = 196.2\text{ N}\).
- The child stands at \(x = 3.5\text{ m}\). The child's weight is \(W_c = m_c g = 30 \times 9.81\text{ N} = 294.3\text{ N}\).

To find the vertical force exerted by pivot \(B\) (\(F_B\)), we take moments about pivot \(A\) (which eliminates the force at pivot \(A\)):
$$\sum \text{Clockwise Moments} = \sum \text{Anticlockwise Moments}$$
$$(W_p \times 2.0\text{ m}) + (W_c \times 3.5\text{ m}) = F_B \times 3.0\text{ m}$$
$$(196.2 \times 2.0) + (294.3 \times 3.5) = 3.0 \times F_B$$
$$392.4 + 1030.05 = 3.0 \times F_B$$
$$1422.45 = 3.0 \times F_B$$
$$F_B = 474.15\text{ N} \approx 470\text{ N}$$

[1 mark] Formulates the moments equation about pivot A using the correct distances and weights, leading to a force of approximately 470 N.

The grating equation is:
$$d \sin \theta = n \lambda$$

where:
- \(d\) is the grating spacing (distance between adjacent lines)
- \(\theta = 38.0^\circ\)
- \(n = 2\) (second-order maximum)
- \(\lambda = 633 \times 10^{-9}\text{ m}\)

Rearranging for \(d\):
$$d = \frac{n \lambda}{\sin \theta}$$
$$d = \frac{2 \times 633 \times 10^{-9}\text{ m}}{\sin(38.0^\circ)} = \frac{1.266 \times 10^{-6}\text{ m}}{0.61566} \approx 2.0563 \times 10^{-6}\text{ m}$$

The number of lines per metre \(N\) is given by:
$$N = \frac{1}{d} = \frac{1}{2.0563 \times 10^{-6}\text{ m}} \approx 4.863 \times 10^5 \text{ lines/m}$$

To convert this to lines per millimetre, we divide by 1000:
$$N_{\text{mm}} = \frac{4.863 \times 10^5}{1000} \approx 486 \text{ lines/mm}$$

This rounds to \(490\text{ lines/mm}\) to two significant figures.

[1 mark] Applies the diffraction grating equation to determine the spacing d, and then correctly calculates the number of lines per millimetre.

To find the percentage uncertainty in \(\rho\), we calculate the sum of the relative uncertainties for each variable, multiplying by the power of the variable where appropriate:

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

Calculating each term:

\(\frac{\Delta R}{R} = \frac{0.2}{4.0} = 0.050\) (or \(5.0\%\))

\(\frac{\Delta d}{d} = \frac{0.01}{0.50} = 0.020\) (or \(2.0\%\))

\(\frac{\Delta L}{L} = \frac{0.010}{1.000} = 0.010\) (or \(1.0\%\))

Therefore, the total percentage uncertainty is:

\(5.0\% + 2(2.0\%) + 1.0\% = 10.0\%\).

1 mark for the correct calculation of percentage uncertainty by summing the fractional uncertainties, ensuring the fractional uncertainty in diameter is multiplied by 2.

In bright light, the potential difference across the LDR is:

\(V_{\text{bright}} = 12 \times \frac{3.0}{6.0 + 3.0} = 4.0\text{ V}\)

In the dark, the potential difference across the LDR is:

\(V_{\text{dark}} = 12 \times \frac{18}{6.0 + 18} = 9.0\text{ V}\)

The change in the voltmeter reading is:

\(9.0\text{ V} - 4.0\text{ V} = 5.0\text{ V}\).

1 mark for calculating the two potential differences correctly and subtracting them to obtain 5.0 V.

The maximum kinetic energy \(E_{\text{max}}\) of an object in simple harmonic motion is equal to the total energy:

\(E_{\text{max}} = \frac{1}{2} m \omega^2 A^2\)

where \(\omega = 2\pi f\). Therefore:

\(E_{\text{max}} = 2 \pi^2 m f^2 A^2\)

Substituting the values:

\(E_{\text{max}} = 2 \times \pi^2 \times 0.25 \times 5.0^2 \times (0.040)^2\)

\(E_{\text{max}} = 2 \times \pi^2 \times 0.25 \times 25 \times 0.0016 = 0.02 \pi^2 \approx 0.197\text{ J}\)

To two significant figures, this is \(0.20\text{ J}\).

1 mark for the correct application of the maximum kinetic energy formula using angular frequency and amplitude in SI units.

The extension \(x\) of a wire is given by:

\(x = \frac{FL}{AE} = \frac{4FL}{\pi d^2 E}\)

where \(d\) is the diameter of the wire.

For the second wire:
- The length becomes \(2L\)
- The diameter becomes \(0.5d\)

Substituting these new values into the relationship:

\(x_{\text{new}} \propto \frac{2L}{(0.5d)^2} = \frac{2}{0.25} \frac{L}{d^2} = 8 \left(\frac{L}{d^2}\right)\)

Thus, the extension of the second wire is \(8x\).

1 mark for identifying how extension scales with length and diameter squared, and calculating the new extension factor correctly.

When unpolarised light passes through the first polarizing filter, its intensity is reduced to half of its initial value:

\(I_1 = \frac{1}{2}I_0\)

When this polarized light passes through the second filter, Malus's law applies:

\(I_2 = I_1 \cos^2(60^\circ)\)

Since \(\cos(60^\circ) = 0.5\), we have \(\cos^2(60^\circ) = 0.25\).

\(I_2 = \frac{1}{2}I_0 \times 0.25 = 0.125 I_0\).

1 mark for correctly applying both the factor of a half for the first unpolarized step and Malus's Law for the second step.

Let \(v_1'\) be the final velocity of glider \(m\), and \(v_2'\) be the final velocity of glider \(2m\).

From conservation of linear momentum:

\(m v = m v_1' + 2m v_2' \implies v = v_1' + 2v_2'\)

For a perfectly elastic collision, the relative velocity of approach equals the relative velocity of separation:

\(v = v_2' - v_1'\)

Subtracting the second equation from the first:

\(0 = 2v_1' + v_2' \implies v_2' = -2v_1'\)

Substituting this back into the first equation:

\(v = v_1' + 2(-2v_1') = -3v_1' \implies v_1' = -\frac{1}{3}v\)

Thus:

\(v_2' = \frac{2}{3}v\).

1 mark for setting up conservation of momentum and the elastic collision relative velocity equation, solving to find both velocities.

Using the equations of motion with upwards as the positive direction:

Initial velocity \(u = +15\text{ m s}^{-1}\)
Acceleration \(a = -9.81\text{ m s}^{-2}\)
Displacement \(s = -20\text{ m}\)

Using \(s = ut + \frac{1}{2}at^2\):

\(-20 = 15t - 4.905t^2\)

Rearranging into standard quadratic form:

\(4.905t^2 - 15t - 20 = 0\)

Using the quadratic formula:

\(t = \frac{15 \pm \sqrt{(-15)^2 - 4(4.905)(-20)}}{2(4.905)}\)

\(t = \frac{15 \pm \sqrt{225 + 392.4}}{9.81} = \frac{15 \pm 24.85}{9.81}\)

Since time must be positive, we take the positive root:

\(t = \frac{39.85}{9.81} \approx 4.1\text{ s}\).

1 mark for using equations of motion with consistent signs for displacement, velocity, and acceleration to find the total flight time.

We take moments about the hinge to eliminate the unknown reaction force at the hinge.

Clockwise moments:
- Due to the uniform shelf weight acting at its center of gravity (\(0.60\text{ m}\) from the hinge):
\(\tau_{\text{shelf}} = 40.0\text{ N} \times 0.60\text{ m} = 24.0\text{ N m}\)
- Due to the block (\(0.30\text{ m}\) from the hinge):
\(\tau_{\text{block}} = 80.0\text{ N} \times 0.30\text{ m} = 24.0\text{ N m}\)

Total clockwise moment = \(24.0 + 24.0 = 48.0\text{ N m}\).

Anticlockwise moments:
- Due to the tension \(T\) in the wire at the far end (\(1.20\text{ m}\)):
\(\tau_{\text{tension}} = T \sin(30.0^\circ) \times 1.20\text{ m} = T \times 0.5 \times 1.20 = 0.60 T\)

For rotational equilibrium:

\(0.60 T = 48.0 \implies T = 80.0\text{ N}\).

1 mark for taking moments about the hinge, resolving the perpendicular component of tension correctly, and solving for T.

The density \(\rho\) of a cylinder is given by \(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 l}\). The fractional uncertainty in density is \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta l}{l}\). First, calculate the individual percentage uncertainties: for mass, \(\frac{0.4}{40.0} \times 100\% = 1.0\%\); for diameter, \(\frac{0.1}{10.0} \times 100\% = 1.0\%\); for length, \(\frac{0.5}{50.0} \times 100\% = 1.0\%\). Substituting these values into the uncertainty equation gives: \(\frac{\Delta \rho}{\rho} \times 100\% = 1.0\% + 2(1.0\%) + 1.0\% = 4.0\%\).

1 mark for calculating individual component uncertainties as 1.0% each, and applying the correct power relation to double the uncertainty of the diameter, leading to a total percentage uncertainty of 4.0%.

By Malus's law, the intensity after the first filter is \(I_1 = I_0 \cos^2(60^\circ) = I_0 (0.5)^2 = \frac{1}{4}I_0\). The light leaving the first filter is polarised at \(60^\circ\) to the vertical. The second filter's transmission axis is horizontal (at \(90^\circ\) to the vertical). The angle between the polarisation of the light and the second filter is \(90^\circ - 60^\circ = 30^\circ\). Applying Malus's law again: \(I_2 = I_1 \cos^2(30^\circ) = \frac{1}{4}I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4}I_0 \times \frac{3}{4} = \frac{3}{16}I_0\).

1 mark for correctly applying Malus's law for both stages, resolving the angle between the polarization axis of the first filter and the second filter as 30 degrees, and showing that the final intensity is 3/16 of the initial intensity.

Young modulus is given by \(E = \frac{F L}{A \Delta L}\), so extension is \(\Delta L = \frac{F L}{A E}\). Since \(A = \frac{\pi d^2}{4}\), we have \(\Delta L \propto \frac{L}{d^2}\) for the same material and force. Thus, the ratio of extensions is \(\frac{\Delta L_X}{\Delta L_Y} = \left(\frac{L_X}{L_Y}\right) \times \left(\frac{d_Y}{d_X}\right)^2 = 2 \times \left(\frac{1}{0.5}\right)^2 = 2 \times 4 = 8\).

1 mark for establishing the relationship between extension, length, and diameter, and substituting the ratios correctly to obtain a value of 8.

The total resistance of the primary circuit is \(3.0\ \Omega + 5.0\ \Omega = 8.0\ \Omega\). The current through the uniform wire is \(I = \frac{E_1}{8.0}\). The potential difference across the entire \(1.00\text{ m}\) slide-wire is \(V_w = I R_w = \frac{5}{8}E_1\). The balance point is at \(60.0\text{ cm} = 0.60\text{ m}\), so the potential difference across this part of the wire (which equals \(E_2\)) is \(E_2 = \frac{5}{8}E_1 \times 0.60 = \frac{5}{8}E_1 \times \frac{3}{5} = \frac{3}{8}E_1\). Hence, \(\frac{E_2}{E_1} = \frac{3}{8}\).

1 mark for calculating the potential drop across the wire in terms of E1, and scaling it to the balanced fraction of the wire to find the correct ratio of 3/8.

Let the initial direction of motion be positive. The initial momentum is \(p_i = mv\) and the final momentum is \(p_f = -\frac{2}{3}mv\). The change in momentum is \(\Delta p = p_f - p_i = -\frac{2}{3}mv - mv = -\frac{5}{3}mv\). The magnitude of the change in momentum is \(|\Delta p| = \frac{5}{3}mv\). From Newton's second law, the average force is \(F = \frac{|\Delta p|}{\Delta t} = \frac{5mv}{3\Delta t}\).

1 mark for calculating the magnitude of the change in momentum as 5mv/3, accounting for the change in direction of velocity, and dividing by the collision time to obtain the average force.

Consider the downward direction as positive. Let the launch point at the top of the cliff be the origin. The initial velocity of the stone is \(u_i = -u\), its final velocity is \(v_f = +3u\), the acceleration is \(a = +g\), and the displacement is \(s = +h\). Using the equation \(v_f^2 = u_i^2 + 2as\), we substitute these values: \((3u)^2 = (-u)^2 + 2gh \implies 9u^2 = u^2 + 2gh \implies 8u^2 = 2gh \implies h = \frac{4u^2}{g}\).

1 mark for using the correct equations of motion with consistent signs for displacement and velocities, and correctly solving for h.

Take moments about the pivot. The weight of the uniform beam acts at its midpoint, \(1.5\text{ m}\) from the pivot. The clockwise moment due to the weight is \(200\text{ N} \times 1.5\text{ m} = 300\text{ N m}\). The anticlockwise moment is produced by the vertical component of the tension, which is \(T \sin(30^\circ)\), acting at a distance of \(2.0\text{ m}\): \(\text{Moment} = T \sin(30^\circ) \times 2.0 = T \times 0.5 \times 2.0 = 1.0 T\). Equating the clockwise and anticlockwise moments gives: \(1.0 T = 300 \implies T = 300\text{ N}\).

1 mark for establishing the moment balance about the pivot, noting that the weight acts at the midpoint of the beam, and resolving the component of the tension to solve for T.

Using the grating equation \(d \sin\theta = n\lambda\) for \(n = 2\) and \(\theta = 30^\circ\): \(d \sin(30^\circ) = 2\lambda \implies 0.5 d = 2\lambda \implies d = 4\lambda\). To find the maximum possible order \(n\) of diffraction visible on a screen, \(\sin\theta < 1\) (at \(90^\circ\) the beam is parallel to the grating and cannot be projected on a screen): \(\frac{n\lambda}{d} < 1 \implies n < \frac{d}{\lambda} \implies n < 4\). Since \(n\) must be an integer, the highest observable order is \(n = 3\). The observable orders are therefore \(0, \pm 1, \pm 2, \pm 3\). This gives a total of \(2 \times 3 + 1 = 7\) maxima.

1 mark for calculating the grating spacing in terms of wavelength, finding the highest observable order (n = 3), and correctly calculating the total number of maxima as 7.

Initially, the total resistance of the circuit is \(R_{\text{total1}} = r + R_{\text{fixed}} + R_{\text{thermistor}} = 2.0 + 4.0 + 6.0 = 12.0\ \Omega\). The initial current is \(I_1 = \frac{E}{R_{\text{total1}}} = \frac{12\text{ V}}{12.0\ \Omega} = 1.0\text{ A}\). The initial potential difference across the thermistor is \(V_{\text{thermistor1}} = I_1 \times R_{\text{thermistor}} = 1.0\text{ A} \times 6.0\ \Omega = 6.0\text{ V}\). Finally, when the thermistor resistance decreases to \(2.0\ \Omega\), the new total resistance is \(R_{\text{total2}} = 2.0 + 4.0 + 2.0 = 8.0\ \Omega\). The final current is \(I_2 = \frac{12\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\). The final potential difference across the thermistor is \(V_{\text{thermistor2}} = I_2 \times R_{\text{thermistor}} = 1.5\text{ A} \times 2.0\ \Omega = 3.0\text{ V}\). The change in potential difference across the thermistor is \(6.0\text{ V} - 3.0\text{ V} = 3.0\text{ V}\) (a decrease of \(3.0\text{ V}\)).

1 mark for calculating both the initial and final potential differences across the thermistor correctly, and finding their difference to show a decrease of 3.0 V.

The angular frequency of the oscillation is given by \(\omega = \frac{2\pi}{T} = \frac{2\pi}{4.0} = 0.5\pi\text{ rad s}^{-1}\). The magnitude of the acceleration is related to displacement by \(|a| = \omega^2 |x|\). Substituting the values, we get \(|a| = (0.5\pi)^2 \times 3.0 = 0.25 \pi^2 \times 3.0 \approx 7.4\text{ cm s}^{-2}\).

1 mark for calculating the angular frequency and using the SHM acceleration relation to find the correct magnitude.

First, calculate the percentage uncertainty for each measured value: \(\%\Delta R = \frac{0.05}{2.50} \times 100\% = 2.0\%\), \(\%\Delta d = \frac{0.01}{0.40} \times 100\% = 2.5\%\), and \(\%\Delta L = \frac{0.4}{80.0} \times 100\% = 0.5\%\). For the resistivity formula \(\rho = \frac{R \pi d^2}{4 L}\), the combined fractional uncertainty is \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). Therefore, \(\%\Delta \rho = 2.0\% + 2(2.5\%) + 0.5\% = 7.5\%\).

1 mark for calculating individual percentage uncertainties and combining them correctly, taking into account the squared dependency of diameter.

Since the wires are in series, they experience the same tensile force \(F\). Extension is given by \(\Delta L = \frac{F L}{A E} = \frac{4 F L}{\pi d^2 E}\). Thus, the ratio of extensions is \(\frac{\Delta L_s}{\Delta L_b} = \left(\frac{L_s}{L_b}\right) \times \left(\frac{d_b}{d_s}\right)^2 \times \left(\frac{E_b}{E_s}\right)\). Given that \(\frac{L_s}{L_b} = 2\), \(\frac{d_b}{d_s} = 2\), and \(\frac{E_b}{E_s} = 0.5\), we obtain \(\frac{\Delta L_s}{\Delta L_b} = 2 \times 2^2 \times 0.5 = 4\).

1 mark for establishing the correct dependency of extension on the parameters and substituting the ratios correctly.

When unpolarised light of intensity \(I_0\) passes through the first polariser, its intensity is reduced by half, so \(I_1 = 0.50 I_0\). By Malus's Law, when this polarised light passes through the second polariser rotated by \(30^\circ\), the emergent intensity is \(I_2 = I_1 \cos^2(30^\circ) = 0.50 I_0 \times \left(\frac{\sqrt{3}}{2}\right)^2 = 0.50 I_0 \times 0.75 = 0.375 I_0 \approx 0.38 I_0\).

1 mark for applying both the factor of 0.5 for unpolarised light passing through a polariser and Malus's law for the second filter.

Initial kinetic energy: \(E_{\text{k,initial}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} (2.0) (6.0)^2 = 36.0\text{ J}\). Using conservation of momentum: \(m_1 u_1 = (m_1 + m_2) v \implies 2.0 \times 6.0 = (2.0 + 4.0) v \implies v = 2.0\text{ m s}^{-1}\). Final kinetic energy: \(E_{\text{k,final}} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} (6.0) (2.0)^2 = 12.0\text{ J}\). Loss in kinetic energy: \(\Delta E_k = 36.0\text{ J} - 12.0\text{ J} = 24.0\text{ J}\).

1 mark for determining the final common velocity using momentum conservation and finding the difference between the initial and final kinetic energies.

Taking upwards as the positive direction, the initial velocity is \(u = +15.0\text{ m s}^{-1}\), the acceleration is \(a = -9.81\text{ m s}^{-2}\), and the time of flight is \(t = 4.00\text{ s}\). Using the equation of motion: \(s = ut + \frac{1}{2}at^2 = (15.0)(4.00) + \frac{1}{2}(-9.81)(4.00)^2 = 60.0 - 78.48 = -18.48\text{ m}\). The negative sign indicates that the final position is below the starting point. Thus, the height of the cliff is \(18.5\text{ m}\).

1 mark for using the correct kinematic equation with consistent sign conventions to calculate the vertical displacement.

Because the beam is uniform, its weight of \(120\text{ N}\) acts at its centre of gravity, which is at the midpoint: \(1.5\text{ m}\) from \(P\). Taking moments about the pivot \(P\) for rotational equilibrium: Clockwise moments = Counter-clockwise moments, which gives: \(120\text{ N} \times 1.5\text{ m} = T \times 2.0\text{ m} \implies 180\text{ N m} = 2.0 T \implies T = 90\text{ N}\).

1 mark for correctly placing the weight at the midpoint of the uniform beam and taking moments about the pivot to solve for the tension.

The formula for the density of a cylindrical wire is:

\(\rho = \frac{m}{V} = \frac{m}{\pi (d/2)^2 L} = \frac{4m}{\pi d^2 L}\)

The fractional uncertainty in density is given by:

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

Calculate each percentage uncertainty:

- \(\frac{\Delta m}{m} \times 100\% = \frac{0.05}{2.50} \times 100\% = 2.0\%\)
- \(\frac{\Delta d}{d} \times 100\% = \frac{0.01}{0.50} \times 100\% = 2.0\%\)
- \(\frac{\Delta L}{L} \times 100\% = \frac{0.2}{20.0} \times 100\% = 1.0\%\)

Substitute these into the equation:

Percentage uncertainty in \(\rho = 2.0\% + 2(2.0\%) + 1.0\% = 7.0\%\).

C is the correct answer. 1 mark for calculating the percentage uncertainties of each quantity and combining them correctly with the factor of 2 for diameter.

The Young modulus \(E\) of a material is given by:

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

where \(A = \frac{\pi d^2}{4}\). Thus,

\(\Delta L = \frac{4 F L}{\pi d^2 E}\)

For the first wire:

\(\Delta L_1 = \frac{4 W L}{\pi d^2 E} = \Delta L\)

For the second wire with \(F_2 = 2W\), \(L_2 = 2L\), and \(d_2 = 2d\):

\(\Delta L_2 = \frac{4 (2W) (2L)}{\pi (2d)^2 E} = \frac{16 W L}{4 \pi d^2 E} = \frac{4 W L}{\pi d^2 E} = \Delta L_1 = \Delta L\).

B is the correct answer. 1 mark for expressing the extension in terms of force, length, and diameter, and showing that the factors of 2 in the numerator and denominator cancel out.

When unpolarised light of intensity \(I_0\) passes through the first polarising filter, its intensity becomes halved:

\(I_1 = \frac{1}{2} I_0\)

When this polarised light passes through the second polarising filter, the transmitted intensity \(I_2\) is given by Malus's Law:

\(I_2 = I_1 \cos^2\theta = \frac{1}{2} I_0 \cos^2\theta\)

Given that \(I_2 = \frac{3}{8} I_0\):

\(\frac{1}{2} I_0 \cos^2\theta = \frac{3}{8} I_0 \implies \cos^2\theta = \frac{3}{4} \implies \cos\theta = \frac{\sqrt{3}}{2}\)

Thus, \( \theta = 30^\circ \).

A is the correct answer. 1 mark for identifying that the first filter halves the intensity, applying Malus's law for the second filter, and solving for the angle of 30 degrees.

By the conservation of linear momentum, the total initial momentum of the system is zero. After release, the total momentum must still be zero:

\(P_{\text{total}} = 3m(-v) + m(v_Y) = 0 \implies v_Y = 3v\)

So, the speed of trolley Y is \(3v\).

The kinetic energy of trolley X is:

\(E_K = \frac{1}{2} (3m) v^2 = 1.5 m v^2\)

The kinetic energy of trolley Y is:

\(E_Y = \frac{1}{2} m (v_Y)^2 = \frac{1}{2} m (3v)^2 = 4.5 m v^2\)

Comparing \(E_Y\) to \(E_K\):

\(E_Y = 3 \times (1.5 m v^2) = 3 E_K\).

C is the correct answer. 1 mark for using the conservation of momentum to find the speed of Y as 3v and the kinetic energy as 3 times that of X.

Since the beam is uniform, its weight of \(120\text{ N}\) acts vertically downwards at its center of gravity, which is at the midpoint of the beam (\(1.0\text{ m}\) from the pivot).

Let the tension in the string be \(T\). The string is attached at the far end (\(2.0\text{ m}\) from the pivot) at an angle of \(30^\circ\) to the horizontal.

Taking moments about the pivot at equilibrium:

\(\text{Clockwise moment} = \text{Counter-clockwise moment}\)

\(120\text{ N} \times 1.0\text{ m} = (T \sin 30^\circ) \times 2.0\text{ m}\)

\(120 = T \times 0.5 \times 2.0\)

\(120 = T\)

Therefore, the tension in the string is \(120\text{ N}\).

B is the correct answer. 1 mark for applying the principle of moments about the pivot, identifying the correct distance for the weight, and solving for the tension.

Using the diffraction grating equation:

\(d \sin\theta = n \lambda\)

where:
- \(n = 2\) (second-order)
- \(\lambda = 500\text{ nm} = 5.0 \times 10^{-7}\text{ m}\)
- \(\theta = 30^\circ\)

Substitute the values to find the grating spacing \(d\):

\(d \sin 30^\circ = 2 \times (5.0 \times 10^{-7}\text{ m})\)

\(d \times 0.5 = 1.0 \times 10^{-6}\text{ m}\)

\(d = 2.0 \times 10^{-6}\text{ m}\)

The number of lines per metre \(N\) is:

\(N = \frac{1}{d} = \frac{1}{2.0 \times 10^{-6}\text{ m}} = 5.0 \times 10^5\text{ lines m}^{-1}\)

To find the number of lines per millimetre:

\(N_{\text{mm}} = \frac{5.0 \times 10^5}{1000} = 500\text{ lines mm}^{-1}\).

C is the correct answer. 1 mark for calculating the slit spacing d, finding the lines per metre, and converting to lines per millimetre.

Let the resistance of the wire be \(R_W\).

Initially, the full e.m.f. of the driver cell is dropped across the \(1.00\text{ m}\) wire. The potential gradient is:

\(k_1 = \frac{2.0\text{ V}}{1.00\text{ m}} = 2.0\text{ V m}^{-1}\)

The e.m.f. of the test cell \(E_2\) corresponds to a balance length of \(40.0\text{ cm} = 0.400\text{ m}\):

\(E_2 = k_1 \times 0.400 = 2.0 \times 0.400 = 0.80\text{ V}\)

When a resistor of resistance \(R = R_W\) is connected in series with the wire, the total resistance of the driver circuit becomes \(2 R_W\).

The potential difference across the wire is now halved by the potential divider effect:

\(V_W' = 2.0\text{ V} \times \frac{R_W}{2 R_W} = 1.0\text{ V}\)

The new potential gradient along the wire is:

\(k_2 = \frac{1.0\text{ V}}{1.00\text{ m}} = 1.0\text{ V m}^{-1}\)

The new balance length \(L_2\) for cell \(E_2\) is:

\(L_2 = \frac{E_2}{k_2} = \frac{0.80\text{ V}}{1.0\text{ V m}^{-1}} = 0.800\text{ m} = 80.0\text{ cm}\).

D is the correct answer. 1 mark for finding the value of the test cell e.m.f., calculating the new potential gradient across the wire, and determining the new balance length.

We can apply the conservation of energy principle:

\(W_{\text{input}} = \Delta E_p + \Delta E_k + W_{\text{friction}}\)

1. Calculate the work input by the pushing force:
\(W_{\text{input}} = F \times d = 25\text{ N} \times 4.0\text{ m} = 100\text{ J}\)

2. Calculate the gain in gravitational potential energy:
The vertical height gained is \(h = d \sin 30^\circ = 4.0 \times 0.5 = 2.0\text{ m}\).
\(\Delta E_p = m g h = 2.0\text{ kg} \times 9.81\text{ m s}^{-2} \times 2.0\text{ m} = 39.24\text{ J}\)

3. Calculate the gain in kinetic energy:
\(\Delta E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 2.0\text{ kg} \times (5.0\text{ m s}^{-1})^2 = 25.0\text{ J}\)

4. Calculate the work done against friction:
\(100\text{ J} = 39.24\text{ J} + 25.0\text{ J} + W_{\text{friction}}\)
\(W_{\text{friction}} = 100 - 64.24 = 35.76\text{ J} \approx 36\text{ J}\).

B is the correct answer. 1 mark for setting up the energy conservation equation, calculating the three individual energy values, and solving for work done against friction.

(a) The volume \(V\) of the cylinder is given by:
\(V = \frac{\pi d^2 L}{4}\\
Converting measurements to cm:
\)d = 1.24\\text{ cm}\\
\(L = 5.02\text{ cm}\\
\)V = \\frac{\\pi \\times 1.24^2 \\times 5.02}{4} \\approx 6.0624\\text{ cm}^3\\n
Density:
\(\rho = \frac{M}{V} = \frac{48.3}{6.0624} = 7.967\text{ g cm}^{-3} \approx 7.97\text{ g cm}^{-3}\\

(b) The fractional uncertainty in density is:
\)\\frac{\\Delta \\rho}{\\rho} = \\frac{\\Delta M}{M} + 2\\frac{\\Delta d}{d} + \\frac{\\Delta L}{L}\\
\(\frac{\Delta M}{M} = \frac{0.1}{48.3} \approx 0.00207\n\)\\frac{\\Delta d}{d} = \\frac{0.1}{12.4} \\approx 0.00806\\n\(\frac{\Delta L}{L} = \frac{0.2}{50.2} \approx 0.00398\n
Sum of fractional uncertainties:
\)\\frac{\\Delta \\rho}{\\rho} = 0.00207 + 2(0.00806) + 0.00398 = 0.02217\\n
Absolute uncertainty:
\(\Delta \rho = 0.02217 \times 7.967 \approx 0.177\text{ g cm}^{-3}\\

(c) The absolute uncertainty is expressed to 1 significant figure as \pm 0.2\text{ g cm}^{-3}\\. Therefore, the density value must be quoted to 1 decimal place:
\)\rho = 8.0 \\pm 0.2\\text{ g cm}^{-3}\)

(a)
- Use of \(V = \pi d^2 L / 4\) to find \(V = 6.06\text{ cm}^3\) [1 mark]
- Use of \(\rho = M/V\) [1 mark]
- Correct value of \(\rho = 7.97\text{ g cm}^{-3}\) (accept \(7.97 \times 10^3\text{ kg m}^{-3}\)) [1 mark]

(b)
- Expression for fractional uncertainty: \(\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\) [1 mark]
- Correct substitution of values into the fractional uncertainty equation [1 mark]
- Correct absolute uncertainty \(\Delta \rho = 0.18\text{ g cm}^{-3}\) or \(0.2\text{ g cm}^{-3}\) [1 mark]

(c)
- Final answer written as \(8.0 \pm 0.2\text{ g cm}^{-3}\) (matching the decimal places of the density to the decimal place of the 1 s.f. uncertainty) [1 mark]

(a) The moment of a force is defined as the product of the force and the perpendicular distance from the pivot to the line of action of the force.

(b) Taking moments about point \(A\):
- Clockwise moment due to the weight of the shelf (acting at its midpoint, \(0.40\text{ m}\) from \(A\)):
\(M_1 = 15\text{ N} \times 0.40\text{ m} = 6.0\text{ N m}\\
- Clockwise moment due to the weight of the book (acting at \)0.20\\text{ m}\) from \(A\)):
\(M_2 = 25\text{ N} \times 0.20\text{ m} = 5.0\text{ N m}\\
- Total clockwise moment = \)6.0 + 5.0 = 11.0\\text{ N m}\\
- Anticlockwise moment due to the tension \(T\) in the wire:
\(M_{\text{anti}} = T \sin(30^\circ) \times 0.80\text{ m} = T \times 0.50 \times 0.80 = 0.40 T\n
Setting clockwise moments equal to anticlockwise moments:
\)0.40 T = 11.0 \\implies T = 27.5\\text{ N} \\approx 28\\text{ N}\\

(c) For vertical equilibrium, the sum of upward forces must equal the sum of downward forces.
Let \(V\) be the vertical force exerted by the hinge on the shelf (assume upward direction):
\(V + T \sin(30^\circ) = W_{\text{shelf}} + W_{\text{book}}\\
\)V + 27.5 \\sin(30^\\circ) = 15 + 25\\n\(V + 13.75 = 40 \implies V = 26.25\text{ N} \approx 26\text{ N}\)

(a)
- Force multiplied by perpendicular distance from pivot to the line of action of the force [1 mark]

(b)
- Calculation of clockwise moments: \(15 \times 0.40\) and \(25 \times 0.20\) [1 mark]
- Expression for anticlockwise moment: \(T \sin(30^\circ) \times 0.80\) [1 mark]
- Equating clockwise and anticlockwise moments: \(0.40 T = 11.0\) [1 mark]
- Correct calculation of \(T = 28\text{ N}\) (or \(27.5\text{ N}\)) [1 mark]

(c)
- Realisation that vertical forces must balance: \(\Sigma F_y = 0\) [1 mark]
- Correct substitution of values: \(V + 27.5 \sin(30^\circ) = 40\) [1 mark]
- Correct final vertical force \(V = 26\text{ N}\) (or \(26.3\text{ N}\)) [1 mark]

(a) The principle of conservation of momentum states that the total momentum of a closed system remains constant, provided no external forces act on it.

(b) Taking the direction of glider \(A\)'s initial movement (to the right) as positive:
- Initial momentum of glider \(A\): \(p_A = 0.40\text{ kg} \times 2.5\text{ m s}^{-1} = 1.0\text{ kg m s}^{-1}\\
- Initial momentum of glider \)B\): \(p_B = 0.20\text{ kg} \times (-1.0\text{ m s}^{-1}) = -0.20\text{ kg m s}^{-1}\\
- Total initial momentum: \)p_{\\text{total}} = 1.0 - 0.20 = 0.80\\text{ kg m s}^{-1}\\

After the collision, the combined mass is \(0.40 + 0.20 = 0.60\text{ kg}\\
\)0.60 \\times v = 0.80 \\implies v = \\frac{0.80}{0.60} \\approx 1.33\\text{ m s}^{-1}\) to the right (since positive).

(c) Initial Kinetic Energy \(E_{k,i}\):
\(E_{k,i} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2\n\)E_{k,i} = 0.5 \\times 0.40 \\times 2.5^2 + 0.5 \\times 0.20 \\times (-1.0)^2 = 1.25 + 0.10 = 1.35\\text{ J}\\

Final Kinetic Energy \(E_{k,f}\):
\(E_{k,f} = \frac{1}{2} (m_A + m_B) v^2 = 0.5 \times 0.60 \times 1.333^2 \approx 0.533\text{ J}\\

Change in kinetic energy:
\)\\Delta E_k = 0.533 - 1.35 = -0.817\\text{ J}\\
There is a loss of kinetic energy of approximately \(0.82\\text{ J}\\, meaning the collision is inelastic.

(a)
- Total momentum of a closed system is constant / conserved [1 mark]
- Specified 'closed system' / 'no external forces act' [1 mark]

(b)
- Calculation of total initial momentum as \(0.80\text{ kg m s}^{-1}\) (accepting sign convention) [1 mark]
- Equating to final momentum: \(0.60 \times v = 0.80\) [1 mark]
- Correct calculation of magnitude \(1.3\text{ m s}^{-1}\) (or \(1.33\text{ m s}^{-1}\)) and direction 'to the right' [1 mark]

(c)
- Correct calculation of initial kinetic energy (\(1.35\text{ J}\)) [1 mark]
- Correct calculation of final kinetic energy (\(0.53\text{ J}\)) [1 mark]
- Correct final conclusion that the collision is inelastic because there is a loss of kinetic energy (\(0.82\text{ J}\)) [1 mark]

(a) First, find the cross-sectional area \(A\) of the wire:
\(d = 0.80\text{ mm} = 8.0 \times 10^{-4}\text{ m}\\
\)A = \\frac{\\pi d^2}{4} = \\frac{\\pi (8.0 \\times 10^{-4})^2}{4} \\approx 5.027 \\times 10^{-7}\\text{ m}^2\\n
Tensile stress \(\sigma = \frac{F}{A}\):
\(\sigma = \frac{45}{5.027 \times 10^{-7}} \approx 8.95 \times 10^7\text{ Pa}\) (or \(\text{N m}^{-2}\)).

(b) Tensile strain \(\varepsilon = \frac{e}{L}\):
\(e = 1.2\text{ mm} = 1.2 \times 10^{-3}\text{ m}\\
\)\varepsilon = \\frac{1.2 \\times 10^{-3}}{2.2} \\approx 5.45 \\times 10^{-4}\) (or \(5.5 \times 10^{-4}\).

(c) Young modulus \(E = \frac{\text{stress}}{\text{strain}}\\
\)E = \\frac{8.95 \\times 10^7}{5.45 \\times 10^{-4}} \\approx 1.64 \\times 10^{11}\\text{ Pa}\) (or \(1.6 \times 10^{11}\text{ Pa}\).

(a)
- Correct equation for cross-sectional area \(A = \pi d^2/4\) [1 mark]
- Calculation of area as \(5.03 \times 10^{-7}\text{ m}^2\) [1 mark]
- Correct stress calculation to give \(8.9 \times 10^7\text{ Pa}\) or \(9.0 \times 10^7\text{ Pa}\) [1 mark]

(b)
- Formula for strain \(\varepsilon = e/L\) with correct unit conversion [1 mark]
- Correct strain calculation as \(5.5 \times 10^{-4}\) (or \(5.45 \times 10^{-4}\)) [1 mark]

(c)
- Formula for Young modulus \(E = \text{stress}/\text{strain}\) [1 mark]
- Correct calculation to give \(1.6 \times 10^{11}\text{ Pa}\) (or \(1.64 \times 10^{11}\text{ Pa}\)) [1 mark]

(a) Polarised light refers to transverse waves in which the oscillations of the electric field vector occur in only a single plane containing the direction of wave propagation.

(b) When unpolarised light of intensity \(I_0\) is incident on a linear polariser, the intensity of the transmitted light is halved:
\(I_A = 0.5 I_0\n
(c) Using Malus's law, the intensity \)I_f\) of the light transmitted through filter \(B\) is:
\(I_f = I_A \cos^2(\theta)\\
Where \)\theta = 35^\\circ\) is the angle of rotation between the transmission axes of \(A\) and \(B\).
\(I_f = 0.5 I_0 \cos^2(35^\circ)\\
\)\\cos(35^\\circ) \\approx 0.8192\\n\(\cos^2(35^\circ) \approx 0.6710\n\)I_f = 0.5 \\times 0.6710 \\times I_0 \\approx 0.336 I_0\\n
Therefore, the ratio \(\frac{I_f}{I_0} \approx 0.34\).

(a)
- Reference to oscillations in a single plane [1 mark]
- Reference to plane containing the direction of wave travel / energy propagation [1 mark]

(b)
- \(0.5 I_0\) or \(\frac{1}{2} I_0\) [1 mark]

(c)
- Statement of Malus's Law \(I = I_{\text{max}} \cos^2\theta\) [1 mark]
- Identifying that \(I_{\text{max}}\) after filter A is \(0.5 I_0\) [1 mark]
- Substitution of \(\theta = 35^\circ\) to calculate \(\cos^2(35^\circ) = 0.671\) [1 mark]
- Correct calculation of the ratio \(\frac{I_f}{I_0} = 0.34\) [1 mark]

(a) Using the grating equation:
\(d \sin\theta = n\lambda\nwhere \)n = 2\), \(\lambda = 589\text{ nm} = 5.89 \times 10^{-7}\text{ m}\), and \(\theta = 38.4^\circ\).
\(d = \frac{2 \times 5.89 \times 10^{-7}}{\sin(38.4^\circ)}\\
\)\\sin(38.4^\\circ) \\approx 0.6211\\n\(d = \frac{1.178 \times 10^{-6}}{0.6211} \approx 1.90 \times 10^{-6}\text{ m}\) (or \(1.9\\ \mu\text{m}\)).

(b) For the maximum order \(n\), we set \(\theta \le 90^\circ\), which implies \(\sin\theta \le 1\).
\(n \le \frac{d}{\lambda} = \frac{1.897 \times 10^{-6}}{5.89 \times 10^{-7}} \approx 3.22\nSince \)n\) must be an integer, the maximum observable order is \(n_{\text{max}} = 3\).

The total number of bright maxima observed is given by:
\(N = 2 n_{\text{max}} + 1 = 2(3) + 1 = 7\).

(c) From \(d \sin\theta = n\lambda\), \(\sin\theta = \frac{n\lambda}{d}\). If a shorter wavelength \(\lambda\) is used, \(\sin\theta\) decreases for the second-order maximum. Thus, the diffraction angle \(\theta\) decreases, causing the maxima to be closer to the central axis.

(a)
- State the equation \(d \sin\theta = n\lambda\) [1 mark]
- Correct substitution of \(n=2\) and \(\lambda = 5.89 \times 10^{-7}\text{ m}\) [1 mark]
- Calculation of \(d = 1.90 \times 10^{-6}\text{ m}\) [1 mark]

(b)
- Condition for maximum order: \(\sin\theta \le 1\) [1 mark]
- Finding \(n_{\text{max}} = 3\) [1 mark]
- Total number of maxima = 7 (using \(2n + 1\)) [1 mark]

(c)
- Angle of diffraction \(\theta\) decreases [1 mark]
- Explanation: Since \(\sin\theta \propto \lambda\), a shorter wavelength requires a smaller angle for the same order [1 mark]

(a) Electromotive force (e.m.f.) is the energy transferred per unit charge from other forms of energy (e.g., chemical) into electrical energy. Potential difference (p.d.) is the electrical energy converted per unit charge to other forms of energy (e.g., thermal, light) in a component.

(b)
(i) Using the equation \(E = I(R + r)\):
For the first case: \(E = 1.2(4.5 + r) = 5.4 + 1.2r\) --- (Eq. 1)
For the second case: \(E = 0.70(8.5 + r) = 5.95 + 0.70r\) --- (Eq. 2)

Equating both equations:
\(5.4 + 1.2r = 5.95 + 0.70r\n\)0.50r = 0.55 \\implies r = 1.1\\ \\Omega\).

(ii) Substitute \(r = 1.1\\ \Omega\) into Eq. 1:
\(E = 1.2(4.5 + 1.1) = 1.2(5.6) = 6.72\text{ V} \approx 6.7\text{ V}\\

(c) When \)R = 4.5\\ \\Omega\), the current is \(I = 1.2\text{ A}\).
Power dissipated in the internal resistance \(r\) is:
\(P = I^2 r = (1.2)^2 \times 1.1 = 1.44 \times 1.1 = 1.584\text{ W} \approx 1.6\text{ W}\)

(a)
- e.m.f.: energy transferred from other forms to electrical per unit charge [1 mark]
- p.d.: energy transferred from electrical to other forms per unit charge [1 mark]

(b)
- (i) Two correct simultaneous equations formed, e.g., \(E = 1.2(4.5 + r)\) and \(E = 0.7(8.5 + r)\) [1 mark]
- Showing step of equating or substitution to eliminate \(E\) [1 mark]
- Finding \(r = 1.1\\ \Omega\) [1 mark]
- (ii) Correct calculation of \(E = 6.7\text{ V}\) (or \(6.72\text{ V}\)) [1 mark]

(c)
- Correct formula used: \(P = I^2 r\) [1 mark]
- Correct calculation to give \(1.6\text{ W}\) [1 mark]

(a) Using the resistance formula:
\(R = \frac{\rho L}{A}\\
Rearranging to find the cross-sectional area \)A\):
\(A = \frac{\rho L}{R} = \frac{1.1 \times 10^{-6} \times 3.5}{15}\\
\)A = \\frac{3.85 \\times 10^{-6}}{15} \\approx 2.57 \\times 10^{-7}\\text{ m}^2 \\approx 2.6 \\times 10^{-7}\\text{ m}^2\\.

(b) The cross-sectional area is related to diameter \(d\) by:
\(A = \frac{\pi d^2}{4} \implies d = \sqrt{\frac{4A}{\pi}}\\
\)d = \\sqrt{\\frac{4 \\times 2.567 \\times 10^{-7}}{\\pi}} = \\sqrt{3.268 \\times 10^{-7}} \\approx 5.72 \\times 10^{-4}\\text{ m} = 0.57\\text{ mm}\\

(c) The rate of thermal energy production is the electrical power \(P\) dissipated:
\(P = \\frac{V^2}{R} = \\frac{24^2}{15} = \\frac{576}{15} = 38.4\\text{ W} \\approx 38\\text{ W}\\.

(a)
- State resistance formula \(R = \rho L / A\) [1 mark]
- Correct rearrangement and substitution of given values [1 mark]
- Correct calculation of \(A = 2.6 \times 10^{-7}\text{ m}^2\) (or \(2.57 \times 10^{-7}\text{ m}^2\)) [1 mark]

(b)
- Equating \(A\) to \(\pi d^2/4\) or \(\pi r^2\) [1 mark]
- Correct calculation of \(d = 0.57\text{ mm}\) (or \(5.7 \times 10^{-4}\text{ m}\)) [1 mark]

(c)
- Correct formula for power: \(P = V^2 / R\) or \(P = VI\) with calculated \(I\) [1 mark]
- Correct calculation of \(38\text{ W}\) (or \(38.4\text{ W}\)) [1 mark]

### Sample Measurements and Table
Let the true values of the parameters be \(E = 1.50\text{ V}\), \(\rho = 5.00\ \Omega\text{ m}^{-1}\), and \(R = 12.0\ \Omega\). The theoretical relationship is:
\[ \frac{1}{I} = \frac{\rho}{E} x + \frac{R}{E} = 3.33 x + 8.00 \]

| \(x\ /\text{ m}\) | \(I\ /\text{ A}\) | \(1/I\ /\text{ A}^{-1}\) |
|---|---|---|
| 0.200 | 0.115 | 8.70 |
| 0.350 | 0.109 | 9.17 |
| 0.500 | 0.103 | 9.71 |
| 0.650 | 0.098 | 10.20 |
| 0.800 | 0.094 | 10.64 |
| 0.950 | 0.090 | 11.11 |

### Graph Analysis
- Plot \(1/I\) (vertical axis) against \(x\) (horizontal axis).
- Draw the line of best fit through the points.
- **Gradient \(m\)**:
Using points on the line of best fit, e.g., \((0.200, 8.67)\) and \((0.950, 11.17)\):
\[ m = \frac{11.17 - 8.67}{0.950 - 0.200} = \frac{2.50}{0.750} \approx 3.33\text{ A}^{-1}\text{ m}^{-1} \]
- **y-intercept \(c\)**:
Using \(y = mx + c\):
\[ 8.67 = 3.33(0.200) + c \implies c = 8.67 - 0.67 = 8.00\text{ A}^{-1} \]

### Calculations
- Since \(c = \frac{R}{E}\):
\[ E = \frac{R}{c} = \frac{12.0}{8.00} = 1.50\text{ V} \]
- Since \(m = \frac{\rho}{E}\):
\[ \rho = m \times E = 3.33 \times 1.50 = 5.00\ \Omega\text{ m}^{-1} \]

### Mark Breakdown

**Data Collection (6 marks)**
- **(a) [1 mark]** Correct measurement of \(I\) for \(x = 20.0\text{ cm}\) with appropriate unit. Value must be realistic for the apparatus setup.
- **(b) [1 mark]** At least 6 sets of readings of \(x\) and \(I\) recorded.
- **(b) [1 mark]** Range of \(x\) values must span at least \(50.0\text{ cm}\).
- **(b) [1 mark]** Column headings in the table must have correct quantities and units, e.g., \(x\ /\text{ m}\) (or \(\text{cm}\)), \(I\ /\text{ A}\) (or \(\text{mA}\)), and \(1/I\ /\text{ A}^{-1}\).
- **(b) [1 mark]** Raw values of \(x\) recorded to the nearest millimeter (e.g. \(20.0\text{ cm}\) or \(0.200\text{ m}\)).
- **(b) [1 mark]** Raw values of \(I\) recorded to a consistent decimal place (e.g., to the nearest 1 mA or 0.1 mA depending on meter precision).

**Data Quality & Graphical Presentation (7 marks)**
- **(c) [1 mark]** \(1/I\) values calculated correctly to a consistent number of significant figures (must match or be one more than the S.F. of raw \(I\)).
- **(c) [1 mark]** Axes on the graph must be labelled with quantity and units. Scales must be chosen so that the plotted points occupy at least half the grid in both directions, and must not be awkward (e.g., multiples of 3, 7, etc.).
- **(c) [1 mark]** Points must be plotted within half a small grid square. No thick or excessively large dots.
- **(c) [2 marks]** Line of best fit drawn correctly, showing a balanced distribution of points about the line. No 'point-to-point' lines. Deduct 1 mark if more than one outlier is ignored without explanation.
- **(c) [2 marks]** Quality of results: all points must lie very close to the line of best fit (low scatter).

**Analysis of Graph (3 marks)**
- **(d) [2 marks]** Gradient determined from a triangle whose hypotenuse is at least half the length of the drawn line of best fit. Check calculation.
- **(d) [1 mark]** y-intercept determined by reading directly from the y-axis (if scale starts at \(x = 0\)) or by calculation using \(y = mx + c\) with coordinates from a point on the line.

**Calculations and Constants (4 marks)**
- **(e) [1 mark]** Value of \(E\) calculated correctly with unit (V).
- **(e) [1 mark]** Value of \(\rho\) calculated correctly with unit (\(\Omega\text{ m}^{-1}\) or \(\Omega\text{ cm}^{-1}\)).
- **(e) [1 mark]** Final answers given to 2 or 3 significant figures.
- **(e) [1 mark]** Correct use of algebraic substitution: \(E = R/c\) and \(\rho = m \times E\).

### Sample Calculations and Analysis

- **Part (a):**
- Measured overhang length \(L_1 = 40.0\text{ cm} = 0.400\text{ m}\).
- Uncertainty in \(L_1\): Since the edge of the bench has a small radius, the uncertainty is estimated as \(\Delta L = 0.1\text{ cm}\) (or \(1\text{ mm}\)).
- Percentage uncertainty:
\[ \%\text{ uncertainty} = \frac{0.1\text{ cm}}{40.0\text{ cm}} \times 100\% = 0.25\% \]

- **Part (b):**
- For \(L_1 = 40.0\text{ cm}\):
- Trial 1: \(t = 15.02\text{ s}\)
- Trial 2: \(t = 14.98\text{ s}\)
- Average time \(t_{\text{avg}} = 15.00\text{ s}\)
- Period \(T_1 = \frac{15.00}{10} = 1.50\text{ s}\)

- **Part (c):**
- For \(L_2 = 30.0\text{ cm}\):
- Trial 1: \(t = 9.78\text{ s}\)
- Trial 2: \(t = 9.82\text{ s}\)
- Average time \(t_{\text{avg}} = 9.80\text{ s}\)
- Period \(T_2 = \frac{9.80}{10} = 0.98\text{ s}\)

- **Part (d):**
- Calculate the constant \(k = T^2 / L^3\):
- For \(L_1 = 40.0\text{ cm}\):
\[ k_1 = \frac{1.50^2}{40.0^3} = \frac{2.25}{64000} \approx 3.52 \times 10^{-5}\text{ s}^2\text{ cm}^{-3} \]
- For \(L_2 = 30.0\text{ cm}\):
\[ k_2 = \frac{0.98^2}{30.0^3} = \frac{0.9604}{27000} \approx 3.56 \times 10^{-5}\text{ s}^2\text{ cm}^{-3} \]
- Comparing the values:
- Percentage difference between \(k_1\) and \(k_2\):
\[ \%\text{ difference} = \frac{|3.56 - 3.52|}{3.52} \times 100\% \approx 1.1\% \]
- Since the percentage difference (1.1%) is well within a reasonable experimental tolerance criterion of 10%, the relationship \(T^2 = k L^3\) is strongly supported by the experimental results.

### Mark Breakdown

**Measurements and Uncertainty (6 marks)**
- **(a)(i) [1 mark]** Overhang length \(L_1\) recorded with correct unit (cm or m) to the nearest mm (e.g., \(40.0\text{ cm}\)).
- **(a)(ii) [2 marks]** Percentage uncertainty calculation shown correctly: \(\frac{\Delta L}{L} \times 100\%\). Uncertainty \(\Delta L\) must be between \(1\text{ mm}\) and \(2\text{ mm}\). Value calculated correctly to 2 S.F.
- **(b)(i) [1 mark]** At least two trials of \(t\) recorded to at least 0.01 s. Average \(t_{\text{avg}}\) calculated correctly.
- **(b)(ii) [1 mark]** Period \(T_1\) calculated correctly with unit (s).
- **(b)(ii) [1 mark]** Value of \(T_1\) lies in a realistic range (0.5 s to 2.0 s).

**Second Set of Readings (2 marks)**
- **(c) [1 mark]** Measurement of second overhang length \(L_2 = 30.0\text{ cm}\) and time for 10 oscillations recorded.
- **(c) [1 mark]** Period \(T_2\) calculated correctly, demonstrating that period decreases as length decreases.

**Analysis and Evaluation of Formula (4 marks)**
- **(d)(i) [2 marks]** Calculation of \(k_1\) and \(k_2\) correct, including correct units (e.g., \(\text{s}^2\text{ cm}^{-3}\) or \(\text{s}^2\text{ m}^{-3}\)). Deduct 1 mark if unit is missing or incorrect.
- **(d)(ii) [2 marks]** Quantitative comparison of \(k_1\) and \(k_2\). Must calculate the percentage difference, compare it to a stated threshold (e.g., 10%), and state a clear, consistent conclusion (e.g. "Since 1.1% is less than 10%, the relationship is supported").

**Limitations and Improvements (8 marks)**
- Award 1 mark for each valid limitation (up to 4) and 1 mark for each corresponding improvement (up to 4).

*Limitations (max 4):*
- **L1:** Two readings of \(L\) and \(T\) are not enough to draw a valid conclusion.
- **L2:** Difficult to judge the precise beginning or end of an oscillation due to high speed / human reaction time.
- **L3:** The oscillations damp out too quickly, making it hard to count 10 complete cycles.
- **L4:** The ruler slips or shifts in the G-clamp during oscillation, causing changing boundary conditions.
- **L5:** Ruler oscillates in more than one direction (e.g., twisting or lateral motion).
- **L6:** The tape/Blu-tack securing the mass shifts or detaches.

*Improvements (max 4):*
- **I1:** Take more readings of \(L\) and \(T\) and plot a graph of \(T^2\) against \(L^3\).
- **I2:** Use a fiducial marker placed at the equilibrium position to improve timing precision, or record the oscillation with a high-speed video camera with a timer and analyze frame-by-frame.
- **I3:** Use a slightly larger mass (within elastic limit) to increase the period and reduce damping rate.
- **I4:** Place high-friction rubber pads or sandpaper between the wooden blocks and the ruler to prevent slipping.
- **I5:** Ensure the ruler is pulled vertically downward without twist and released cleanly, or use a vertical guide-rail to restrict lateral movement.
- **I6:** Use a slotted mass hanger secured firmly with a clamp-screw to prevent movement.