2022 OCR AS Level Physics A - H156 Practice Paper with Answers

First, calculate the percentage uncertainty in each measured value: \%\Delta R = \frac{0.3}{12.0} \times 100\% = 2.5\% ; \%\Delta d = \frac{0.01}{0.50} \times 100\% = 2.0\% ; \%\Delta L = \frac{0.02}{2.00} \times 100\% = 1.0\% . Since resistivity is given by \(\rho = \frac{\pi R d^2}{4L}\), the percentage uncertainty in \(\rho\) is: \%\Delta \rho = \%\Delta R + 2(\%\Delta d) + \%\Delta L = 2.5\% + 2(2.0\%) + 1.0\% = 7.5\%.

1 mark for the correct calculation of individual percentage uncertainties and summing them appropriately, taking into account the square of the diameter, to arrive at 7.5%.

The energy of a single photon is given by: \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{633 \times 10^{-9}} = 3.14 \times 10^{-19}\text{ J}\). The number of photons emitted per second is: \(n = \frac{P}{E} = \frac{3.0 \times 10^{-3}}{3.14 \times 10^{-19}} = 9.55 \times 10^{15}\text{ s}^{-1} \approx 9.6 \times 10^{15}\text{ s}^{-1}\).

1 mark for calculating the photon energy and dividing the laser power by this energy to find the emission rate.

Since the beam is uniform, its center of gravity is at its midpoint, \(1.0\text{ m}\) from either end. The distance from the pivot to the center of gravity is \(1.0 - 0.60 = 0.40\text{ m}\) (acting to the right of the pivot). The load \(W\) acts at the left end, at a distance of \(0.60\text{ m}\) from the pivot. Taking moments about the pivot: Clockwise moment = Beam weight \(\times 0.40 = 120 \times 0.40 = 48\text{ N m}\). Anticlockwise moment = \(W \times 0.60\). For equilibrium, \(W \times 0.60 = 48\text{ N m} \implies W = 80\text{ N}\).

1 mark for identifying the correct distance from the pivot to the center of gravity and solving the equilibrium moments equation for W.

Using the kinematic equation \(s = ut + \frac{1}{2}at^2\) with upwards as positive: \(u = +15.0\text{ m s}^{-1}\), \(t = 4.00\text{ s}\), and \(a = -9.81\text{ m s}^{-2}\). Therefore, \(s = (15.0)(4.00) + 0.5(-9.81)(4.00)^2 = 60.0 - 78.48 = -18.48\text{ m}\). The height of the cliff is the magnitude of this displacement, which is \(18.5\text{ m}\) (to 3 s.f.).

1 mark for using the displacement equation with correct signs and values to find the magnitude of the displacement.

In bright light, \(R_{\text{LDR}} = 1.0\text{ k}\Omega\). Therefore, \(V_{\text{out, bright}} = 12 \times \frac{1.0}{4.0 + 1.0} = 2.4\text{ V}\). In darkness, \(R_{\text{LDR}} = 20\text{ k}\Omega\). Therefore, \(V_{\text{out, dark}} = 12 \times \frac{20}{4.0 + 20} = 10.0\text{ V}\). The change in output voltage is \(10.0\text{ V} - 2.4\text{ V} = 7.6\text{ V}\).

1 mark for calculating both output voltages correctly and subtracting them to find the difference.

The fringe width formula is \(x = \frac{\lambda D}{a}\). Initially, \(x = \frac{600 \times D}{a}\). For the new conditions, \(x_{\text{new}} = \frac{\lambda_{\text{new}} D_{\text{new}}}{a} = \frac{450 \times (2D)}{a} = \frac{900 \times D}{a}\). The ratio of the new fringe width to the original fringe width is \(x_{\text{new}}/x = 900 / 600 = 1.50\). Thus, the new fringe width is \(1.50x\).

1 mark for applying the double-slit equation and scaling the wavelength and screen distance factors correctly to get the ratio.

For the series combination, the effective spring constant is \(k_s = \frac{k}{2}\). The elastic potential energy is \(E_s = \frac{1}{2} \frac{W^2}{k_s} = \frac{W^2}{k}\). For the parallel combination, the effective spring constant is \(k_p = 2k\). The elastic potential energy is \(E_p = \frac{1}{2} \frac{W^2}{k_p} = \frac{W^2}{4k}\). The ratio of energy stored is \(E_s / E_p = \frac{W^2 / k}{W^2 / 4k} = 4.0\).

1 mark for correctly determining the effective spring constants and using them to calculate the ratio of the stored energy.

Resistance is given by \(R = \frac{\rho L}{A} = \frac{4\rho L}{\pi d^2}\). The second wire has length \(L_2 = 2L\) and diameter \(d_2 = 0.5d\). Therefore, \(R_2 = \frac{4\rho (2L)}{\pi (0.5d)^2} = \frac{8\rho L}{\pi (0.25d^2)} = 32 \frac{\rho L}{\pi d^2} = 8 R_1\). Thus, \(R_2 = 8 \times 1.6\,\Omega = 12.8\,\Omega\).

1 mark for relating the resistance of a wire to its length and diameter (inverse square relationship with diameter) and computing the final value of 12.8 ohms.

The volume \(V\) of a wire of length \(L\) and diameter \(d\) is given by:

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

The percentage uncertainty in each measurement is:

\%\text{ uncertainty in } d = \frac{0.02}{1.25} \times 100\% = 1.6\%

\%\text{ uncertainty in } L = \frac{0.01}{2.50} \times 100\% = 0.4\%

Using the rules for combining uncertainties:

\%\text{ uncertainty in } V = 2 \times (\%\text{ uncertainty in } d) + (\%\text{ uncertainty in } L)

\%\text{ uncertainty in } V = 2 \times 1.6\% + 0.4\% = 3.2\% + 0.4\% = 3.6\%

1 mark: Correctly identifies that percentage uncertainties are added, with the uncertainty of the diameter multiplied by 2 because it is squared, yielding 3.6%.

The phase difference \(\phi\) is related to path difference \(x\) and wavelength \(\lambda\) by:

\(\phi = \frac{2\pi x}{\lambda}\)

Substitute the given values:

\(\frac{\pi}{3} = \frac{2\pi \times 0.12}{\lambda}\)

\(\frac{1}{3} = \frac{0.24}{\lambda} \implies \lambda = 0.72\text{ m}\)

Now, calculate the wave speed \(v\) using \(v = f \lambda\):

\(v = 50 \times 0.72 = 36\text{ m s}^{-1}\)

1 mark: Calculate the wavelength as 0.72 m and then determine wave speed by multiplying by frequency to get 36 m s^-1.

Since the volume \(V\) of the wire remains constant, we have:

\(V = A \times L\)

where \(A\) is the cross-sectional area and \(L\) is the length. If the length increases to \(L' = 1.10L\), then the cross-sectional area must decrease to:

\(A' = \frac{A}{1.10}\)

The resistance is given by \(R = \frac{\rho L}{A}\). The new resistance \(R'\) is:

\(R' = \frac{\rho L'}{A'} = \frac{\rho (1.10L)}{\frac{A}{1.10}} = 1.21 \left(\frac{\rho L}{A}\right) = 1.21R\)

1 mark: Correctly relates the change in length and cross-sectional area under constant volume to calculate the new resistance factor of 1.21.

Let the left end of the plank be at position \(x = 0\text{ m}\).
- The plank is uniform, so its weight of \(300\text{ N}\) acts at its centre of gravity, \(x = 3.0\text{ m}\).
- One pillar (Pillar A) is at \(x = 0\text{ m}\).
- The second pillar (Pillar B) is at \(x = 5.0\text{ m}\) (1.0 m from the right-hand end at \(x=6.0\text{ m}\)).
- The load of \(100\text{ N}\) is at \(x = 6.0\text{ m}\).

Taking moments about the first pillar (at \(x = 0\text{ m}\)):
- Clockwise moments = \((300\text{ N} \times 3.0\text{ m}) + (100\text{ N} \times 6.0\text{ m}) = 900 + 600 = 1500\text{ N m}\)
- Anticlockwise moments = \(F_B \times 5.0\text{ m}\)

For equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments:

\(5.0 F_B = 1500 \implies F_B = 300\text{ N}\)

1 mark: Correctly applies the principle of moments about the left pivot to find the force on the second pivot (300 N).

First, calculate the useful output energy, which is the gravitational potential energy gained by the mass:

\(E_{\text{out}} = m g h = 5.0 \times 9.81 \times 12 = 588.6\text{ J}\)

The motor's efficiency is \(60\%\), meaning:

\(\text{Efficiency} = \frac{E_{\text{out}}}{E_{\text{in}}} = 0.60\)

\(E_{\text{in}} = \frac{E_{\text{out}}}{0.60} = \frac{588.6}{0.60} = 981\text{ J}\)

1 mark: Calculate the output GPE (588.6 J) and divide by the efficiency of 0.60 to obtain the total electrical energy supplied (981 J).

The kinetic energy gained by the electron is \(E_k = e V = \frac{p^2}{2m}\).

Therefore, the momentum \(p\) of the electron is:

\(p = \sqrt{2m e V}\)

The de Broglie wavelength is given by:

\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m e V}}\)

This shows that \(\lambda\) is inversely proportional to the square root of the potential difference \(V\):

\(\lambda \propto \frac{1}{\sqrt{V}}\)

If the potential difference is increased to \(4V\), the new wavelength \(\lambda'\) is:

\(\lambda' = \frac{\lambda}{\sqrt{4}} = 0.5\lambda\)

1 mark: Realise that wavelength is inversely proportional to the square root of the potential difference, yielding 0.5λ.

First, calculate the output voltage in bright light:

\(V_{\text{out, bright}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R_{\text{fixed}} + R_{\text{LDR}}} = 12 \times \frac{1.0\text{ k}\Omega}{4.0\text{ k}\Omega + 1.0\text{ k}\Omega} = 12 \times 0.20 = 2.4\text{ V}\)

Next, calculate the output voltage in darkness:

\(V_{\text{out, dark}} = 12 \times \frac{20\text{ k}\Omega}{4.0\text{ k}\Omega + 20\text{ k}\Omega} = 12 \times \frac{20}{24} = 10.0\text{ V}\)

Calculate the change in \(V_{\text{out}}\):

\(\Delta V_{\text{out}} = 10.0\text{ V} - 2.4\text{ V} = 7.6\text{ V}\)

1 mark: Calculate both potential divider output voltages (2.4 V and 10.0 V) and correctly subtract them to get the change of 7.6 V.

We can use the equations of motion for uniform acceleration. Since the car starts from rest, the initial velocity \(u = 0\).

The formula relating displacement \(s\), initial velocity \(u\), final velocity \(v\), and time \(t\) is:

\(s = \frac{u+v}{2} t\)

Substitute the known values into the equation:

\(100 = \frac{0+v}{2} \times 5.0\)

\(100 = 2.5 v \implies v = 40\text{ m s}^{-1}\)

1 mark: Apply the correct SUVAT equation of motion to solve for the final velocity (40 m s^-1).

First, calculate the percentage uncertainty in mass \( m \): \( \frac{0.02}{0.50} \times 100\% = 4\% \). Next, calculate the percentage uncertainty in velocity \( v \): \( \frac{0.1}{2.0} \times 100\% = 5\% \). The kinetic energy is given by the formula \( E_k = \frac{1}{2} m v^2 \). Since \( E_k \) depends on the first power of \( m \) and the second power of \( v \), the percentage uncertainty in \( E_k \) is \( \%\Delta E_k = \%\Delta m + 2 \times (\%\Delta v) = 4\% + 2 \times 5\% = 14\% \).

1 mark for the correct choice C.

Resistance \( R \) is given by \( R = \rho \frac{L}{A} \), where \( A = \frac{\pi d^2}{4} \), so \( R \propto \frac{L}{d^2} \). For wire Y, the resistance is proportional to \( \frac{3L}{(2d)^2} = \frac{3}{4} \frac{L}{d^2} \). Because both wires are made of the same material, the resistivity \( \rho \) is identical. Therefore, the resistance of wire Y is \( 0.75 R \).

1 mark for the correct choice B.

The de Broglie wavelength is defined as \( \lambda = \frac{h}{p} \). Since kinetic energy \( E_k = \frac{p^2}{2m} \), the momentum is given by \( p = \sqrt{2mE_k} \). Substituting this into the wavelength formula yields \( \lambda = \frac{h}{\sqrt{2mE_k}} \). Since both particles have the same kinetic energy \( E_k \) and \( h \) is a constant, the wavelength is inversely proportional to the square root of the mass, \( \lambda \propto \frac{1}{\sqrt{m}} \). Therefore, the ratio of the wavelengths is \( \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} \).

1 mark for the correct choice D.

For rotational equilibrium, the sum of clockwise moments about the pivot must equal the sum of counter-clockwise moments about the pivot. Taking the pivot at the left end: the clockwise moments are produced by the weight of the uniform beam (acting at its midpoint, \( 1.5 \text{ m} \) from the pivot) and the weight of the block (at \( 0.6 \text{ m} \) from the pivot). \( \text{Total clockwise moment} = (120 \text{ N} \times 1.5 \text{ m}) + (50 \text{ N} \times 0.6 \text{ m}) = 180 \text{ N m} + 30 \text{ N m} = 210 \text{ N m} \). The counter-clockwise moment is produced by the tension \( T \) in the vertical wire acting at \( 2.0 \text{ m} \): \( \text{Counter-clockwise moment} = T \times 2.0 \text{ m} \). Setting them equal: \( 2.0 T = 210 \implies T = 105 \text{ N} \).

1 mark for the correct choice B.

a) Phase difference measures the difference in position or timing of the cycles of two points on a wave. Specifically, it is the fraction of a wave cycle by which one point lags behind or leads another, expressed as an angle in degrees or radians.

b) i) The relationship between phase difference \(\phi\), path difference \(d\), and wavelength \(\lambda\) is given by:
\(\phi = \frac{2\pi d}{\lambda}\)

Substituting the values:
\(\frac{2\pi}{5} = \frac{2\pi \times 0.24}{\lambda}\)

Rearranging to solve for \(\lambda\):
\(\frac{1}{5} = \frac{0.24}{\lambda} \implies \lambda = 5 \times 0.24 = 1.20\text{ m}\)

ii) The wave speed is given by:
\(v = f \lambda\)

Substituting the values:
\(v = 15.0\text{ Hz} \times 1.20\text{ m} = 18.0\text{ m s}^{-1}\)

iii) To convert from radians to degrees:
\(\text{Phase difference in degrees} = \frac{2\pi}{5} \times \frac{180^\circ}{\pi} = 72^\circ\)

a)
- 1 mark: Mention of the fraction of a cycle / angle between the oscillations of two points.
- 1 mark: Reference to angular measurement in radians/degrees or time lag/fraction of a period.

b) i)
- 1 mark: Correct formula \(\phi = \frac{2\pi d}{\lambda}\) or equivalent ratio.
- 1 mark: Correct substitution: \(\frac{2\pi}{5} = \frac{2\pi \times 0.24}{\lambda}\).
- 1 mark: Final correct value of \(1.20\text{ m}\) (with units).

b) ii)
- 1 mark: Correct formula \(v = f\lambda\) used with their value from b) i).
- 1 mark: Final correct value of \(18.0\text{ m s}^{-1}\) (with units).

b) iii)
- 1.33 marks: Correct calculation of \(72^\circ\).

a) i) Mean diameter \(d\):
\(d = \frac{0.42 + 0.44 + 0.43 + 0.43}{4} = 0.43\text{ mm}\)

ii) Absolute uncertainty in \(d\) using half the range:
\(\text{Range} = 0.44 - 0.42 = 0.02\text{ mm}\)
\(\text{Uncertainty} = \frac{0.02}{2} = 0.01\text{ mm}\)
So, \(d = 0.43 \pm 0.01\text{ mm}\).

b) i) The formula for the Young modulus is:
\(E = \frac{4FL}{\pi d^2 e}\)

The percentage uncertainty in \(E\) is given by adding the individual percentage uncertainties, multiplying the percentage uncertainty of \(d\) by 2 because it is squared:
\(\% \Delta E = \% \Delta F + \% \Delta L + 2 \times \% \Delta d + \% \Delta e\)

First, find the percentage uncertainty in \(d\):
\(\% \Delta d = \frac{0.01}{0.43} \times 100\% \approx 2.33\%\)

Now, substitute the values:
\(\% \Delta E = 2.0\% + 1.5\% + 2 \times (2.33\%) + 3.0\%\)
\(\% \Delta E = 6.5\% + 4.66\% = 11.16\% \approx 11\%\)

ii) The absolute uncertainty in \(E\) is:
\(\Delta E = 1.25 \times 10^{11}\text{ Pa} \times 0.1116 \approx 0.14 \times 10^{11}\text{ Pa}\) (or \(1.4 \times 10^{10}\text{ Pa}\))

Quoting the absolute uncertainty to 1 significant figure gives \(\pm 0.1 \times 10^{11}\text{ Pa}\).
Therefore, the value of the Young modulus \(E\) should be rounded to the same decimal place:
\(E = (1.3 \pm 0.1) \times 10^{11}\text{ Pa}\)
(Alternatively, quoting to 2 significant figures for uncertainty: \(E = (1.25 \pm 0.14) \times 10^{11}\text{ Pa}\))

a) i)
- 1 mark: Correct mean value of \(0.43\text{ mm}\).

a) ii)
- 1.33 marks: Correct absolute uncertainty of \(\pm 0.01\text{ mm}\).

b) i)
- 1 mark: Expression for total percentage uncertainty showing that \(\%\Delta d\) is doubled: \(\%\Delta E = \%\Delta F + \%\Delta L + 2\%\Delta d + \%\Delta e\).
- 1 mark: Correct calculation of \(\%\Delta d = 2.3\%\).
- 1 mark: Summing to get \(11.16\%\) or \(11.2\%\), which is approximately \(11\%\).

b) ii)
- 1 mark: Calculation of absolute uncertainty in \(E\) (approx. \(0.14 \times 10^{11}\text{ Pa}\) or \(1.4 \times 10^{10}\text{ Pa}\)).
- 1 mark: Absolute uncertainty rounded appropriately to 1 or 2 sig figs.
- 1 mark: Young modulus \(E\) stated to a matching decimal place: \((1.3 \pm 0.1) \times 10^{11}\text{ Pa}\) or \((1.25 \pm 0.14) \times 10^{11}\text{ Pa}\).

a) For a rigid body to be in equilibrium:
1. The net (resultant) force acting on the body must be zero (translational equilibrium).
2. The net (resultant) moment (or torque) about any point must be zero (rotational equilibrium).

b) The three forces acting on the beam are:
1. The weight of the beam, \(W = 120\text{ N}\), acting vertically downwards at the beam's geometric centre (its centre of gravity, \(1.2\text{ m}\) from the pivot).
2. The tension \(T\) in the wire, acting at the far end (\(2.4\text{ m}\) from the pivot) at an angle of \(30^\circ\) upwards and towards the wall.
3. The reaction force \(R\) exerted by the pivot at the pivoted end of the beam, acting upwards and outwards.

c) Take moments about the pivot to eliminate the unknown pivot force \(R\):
\(\Sigma M_{\text{pivot}} = 0\)

Clockwise moment due to weight:
\(M_{\text{clockwise}} = W \times 1.2\text{ m} = 120\text{ N} \times 1.2\text{ m} = 144\text{ N m}\)

Anticlockwise moment due to tension:
\(M_{\text{anticlockwise}} = (T \sin 30^\circ) \times 2.4\text{ m}\)

Equating moments:
\(144 = T \times 0.50 \times 2.4\)
\(144 = 1.2 T \implies T = 120\text{ N}\)

d) To find the force \(R\) exerted by the pivot, we resolve forces horizontally and vertically:

Horizontal equilibrium:
\(R_x = T \cos 30^\circ = 120 \cos 30^\circ = 103.92\text{ N}\)

Vertical equilibrium:
\(R_y + T \sin 30^\circ = W\)
\(R_y + 120 \sin 30^\circ = 120\)
\(R_y + 60 = 120 \implies R_y = 60\text{ N}\)

The magnitude of the pivot force \(R\) is:
\(R = \sqrt{R_x^2 + R_y^2} = \sqrt{103.92^2 + 60^2} = \sqrt{10800 + 3600} = \sqrt{14400} = 120\text{ N}\)

a)
- 1 mark: Resultant / net force on the object is zero.
- 1 mark: Resultant / net moment about any point is zero.

b)
- 1.33 marks: Correct identification of the three forces (weight acting downwards at center, tension acting at \(30^\circ\) at the end, pivot reaction force acting at the pivot).

c)
- 1 mark: Principle of moments stated or applied clearly.
- 1 mark: Correct equation: \(120 \times 1.2 = T \sin 30^\circ \times 2.4\).
- 1 mark: Correct calculation of \(T = 120\text{ N}\).

d)
- 1 mark: Finding horizontal and vertical components of pivot force: \(R_x = 104\text{ N}\) and \(R_y = 60\text{ N}\).
- 1 mark: Correctly calculating the overall magnitude to get \(120\text{ N}\).

a) The de Broglie wavelength is the wavelength associated with a moving particle, representing its wave-like nature, given by \(\lambda = \frac{h}{p}\) where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

b) i) The electrical work done on the electron equals its gain in kinetic energy \(E_k\):
\(E_k = e V\)

Since kinetic energy is also related to momentum \(p\) by:
\(E_k = \frac{p^2}{2m_e}\)

We can write:
\(\frac{p^2}{2m_e} = e V \implies p^2 = 2m_e e V \implies p = \sqrt{2m_e e V}\)

Using de Broglie's relation:
\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_e e V}}\)

ii) Substitute the known physical constants and given values into the equation:
\(h = 6.63 \times 10^{-34}\text{ J s}\)
\(m_e = 9.11 \times 10^{-31}\text{ kg}\)
\(e = 1.60 \times 10^{-19}\text{ C}\)
\(V = 150\text{ V}\)

\(p = \sqrt{2 \times 9.11 \times 10^{-31}\text{ kg} \times 1.60 \times 10^{-19}\text{ C} \times 150\text{ V}\)
\(p = \sqrt{4.3728 \times 10^{-47}} = 6.613 \times 10^{-24}\text{ kg m s}^{-1}\)

\(\lambda = \frac{6.63 \times 10^{-34}}{6.613 \times 10^{-24}} = 1.0026 \times 10^{-10}\text{ m} \approx 1.0 \times 10^{-10}\text{ m}\) (or \(0.10\text{ nm}\))

a)
- 1.33 marks: Statement identifying \(\lambda\) as the wavelength associated with a particle related to its momentum by \(\lambda = h/p\).

b) i)
- 1 mark: Stating kinetic energy \(E_k = eV\).
- 1 mark: Linking kinetic energy to momentum, e.g., \(E_k = \frac{p^2}{2m}\).
- 1 mark: Substituting \(p\) into the de Broglie equation \(\lambda = \frac{h}{p}\) to arrive at the shown formula.

b) ii)
- 1 mark: Correct substitution of values into the formula.
- 1 mark: Correct calculation of kinetic energy \(2.40 \times 10^{-17}\text{ J}\) or momentum \(6.6 \times 10^{-24}\text{ kg m s}^{-1}\).
- 1 mark: Correct calculation of wavelength as \(1.0 \times 10^{-10}\text{ m}\) (accept \(1.00 \times 10^{-10}\text{ m}\)).
- 1 mark: Correct unit given (\(\text{m}\) or \(\text{nm}\)).

a) Resistivity \(\rho\) is defined as the resistance of a sample of the material of unit length and unit cross-sectional area. It is defined by the equation \(\rho = \frac{R A}{L}\), where \(R\) is resistance, \(A\) is cross-sectional area, and \(L\) is length.

b) i) Using the resistivity equation:
\(R = \frac{\rho L}{A}\)

Substitute the values:
\(R = \frac{1.1 \times 10^{-6}\ \Omega\text{ m} \times 3.5\text{ m}}{1.2 \times 10^{-7}\text{ m}^2}\)
\(R = 32.08\ \Omega \approx 32\ \Omega\)

ii) At the operating temperature, using Ohm's law:
\(R_{\text{operating}} = \frac{V}{I} = \frac{230\text{ V}}{5.8\text{ A}} \approx 39.7\ \Omega\) (or \(40\ \Omega\))

The operating resistance is higher than the room-temperature resistance. This is because when current flows through the wire, electrical energy is converted to thermal energy (Joule heating), which raises the temperature of the wire. As temperature increases, the positive lattice ions in the metal vibrate with larger amplitude. This increases the rate of collisions between the conduction electrons and the lattice ions, resulting in a higher resistance.

a)
- 1 mark: Define resistivity using the equation \(\rho = R A / L\) with all terms defined.
- 1 mark: State that it is the property of a material representing resistance per unit length and unit cross-sectional area.

b) i)
- 1 mark: Use of correct formula \(R = \frac{\rho L}{A}\).
- 1.33 marks: Correct substitution and calculation to show \(32.08\ \Omega\) (must show calculation to at least 3 sig figs to justify "approximately 32").

b) ii)
- 1 mark: Correct calculation of operating resistance: \(R = \frac{230}{5.8} = 39.7\ \Omega\) (or \(40\ \Omega\)).
- 1 mark: Explanation that current causes temperature rise in the wire (heating effect of current).
- 1 mark: Stating that higher temperature leads to increased lattice vibrations of metal ions.
- 1 mark: Stating that this increases the frequency of collisions with conduction electrons, increasing resistance.

a) i) During the reaction time, the car travels at constant velocity:
\(s_1 = v \times t = 24\text{ m s}^{-1} \times 0.60\text{ s} = 14.4\text{ m}\)

ii) During braking, the car decelerates to rest:
\(u = 24\text{ m s}^{-1}\), \(v = 0\), \(a = -4.5\text{ m s}^{-2}\)
Using: \(v^2 = u^2 + 2 a s_2\)
\(0 = 24^2 + 2(-4.5)s_2\)
\(0 = 576 - 9 s_2 \implies s_2 = \frac{576}{9} = 64\text{ m}\)

iii) The total stopping distance is the sum of the reaction distance and the braking distance:
\(s_{\text{total}} = s_1 + s_2 = 14.4\text{ m} + 64\text{ m} = 78.4\text{ m}\) (rounds to \(78\text{ m}\))

b) The velocity-time graph has two distinct phases:
1. A horizontal line at \(v = 24\text{ m s}^{-1}\) from time \(t = 0\) to \(t = 0.60\text{ s}\) (representing the reaction phase).
2. A straight line with a constant negative slope from \(t = 0.60\text{ s}\) to the stopping time \(t_{\text{total}}\).

To find the stopping time:
\(v = u + a t_{\text{braking}} \implies 0 = 24 - 4.5 t_{\text{braking}}\)
\(t_{\text{braking}} = \frac{24}{4.5} = 5.33\text{ s}\)
\(t_{\text{total}} = 0.60 + 5.33 = 5.93\text{ s}\)

Critical points for the graph axes:
- Vertical axis (Velocity): starts at \(24\text{ m s}^{-1}\), ends at \(0\text{ m s}^{-1}\).
- Horizontal axis (Time): critical time coordinates at \(t = 0.60\text{ s}\) (start of deceleration) and \(t = 5.93\text{ s}\) (car at rest).

a) i)
- 1.33 marks: Correct calculation: \(24 \times 0.60 = 14.4\text{ m}\).

a) ii)
- 1 mark: Selection of correct equation of motion: \(v^2 = u^2 + 2as\).
- 1 mark: Correct substitution: \(0 = 24^2 + 2(-4.5)s_2\).
- 1 mark: Correct calculation of braking distance: \(64\text{ m}\).

a) iii)
- 1 mark: Addition of both distances to get \(78.4\text{ m}\) (or \(78\text{ m}\)).

b)
- 1 mark: Description of a horizontal line at \(24\text{ m s}^{-1}\) from \(t = 0\) to \(t = 0.60\text{ s}\).
- 1 mark: Description of a straight downward sloping line ending at the time axis.
- 1 mark: Correct calculated total stopping time of \(5.9\text{ s}\) (or \(5.93\text{ s}\)).

(a) Convert values to SI units:
\( m = 3.54 \times 10^{-3} \text{ kg} \)
\( L = 0.624 \text{ m} \)
\( d = 0.76 \times 10^{-3} \text{ m} \)

Using the formula for density:
\( \rho = \frac{m}{V} = \frac{4m}{\pi d^2 L} \)
\( \rho = \frac{4 \times 3.54 \times 10^{-3}}{\pi \times (0.76 \times 10^{-3})^2 \times 0.624} \approx 1.251 \times 10^4 \text{ kg m}^{-3} \)

(b) Percentage uncertainties for each variable:
\( \% \Delta m = \frac{0.02}{3.54} \times 100\% \approx 0.565\% \)
\( \% \Delta L = \frac{0.2}{62.4} \times 100\% \approx 0.321\% \)
\( \% \Delta d = \frac{0.01}{0.76} \times 100\% \approx 1.316\% \)

Total percentage uncertainty in \( \rho \):
\( \% \Delta \rho = \% \Delta m + \% \Delta L + 2(\% \Delta d) \)
\( \% \Delta \rho = 0.565\% + 0.321\% + 2(1.316\%) = 3.518\% \approx 3.5\% \)

(c) Absolute uncertainty:
\( \Delta \rho = 1.251 \times 10^4 \times 0.03518 \approx 4.4 \times 10^2 \text{ kg m}^{-3} = 0.04 \times 10^4 \text{ kg m}^{-3} \)

Thus, the density with absolute uncertainty is:
\( \rho = (1.25 \pm 0.04) \times 10^4 \text{ kg m}^{-3} \)

(a) [3 marks]
- 1 mark for converting all values to SI units (kg, m).
- 1 mark for correct formula of volume of a cylinder \( V = \pi r^2 L \) or \( \frac{\pi d^2 L}{4} \).
- 1 mark for correct substitution leading to \( \approx 1.25 \times 10^4 \text{ kg m}^{-3} \).

(b) [3 marks]
- 1 mark for calculating correct percentage uncertainties for \( m \) and \( L \).
- 1 mark for doubling the percentage uncertainty of diameter \( d \) because of \( d^2 \).
- 1 mark for adding them to get \( 3.5\% \) (accept \( 3.5\% \) to \( 3.6\% \)).

(c) [3 marks]
- 1 mark for using percentage uncertainty to calculate absolute uncertainty (\( 4.4 \times 10^2 \) or \( 0.04 \times 10^4 \)).
- 1 mark for quoting absolute uncertainty to 1 significant figure (or 2 if starting with 1).
- 1 mark for final value and uncertainty to matching decimal places: \( (1.25 \pm 0.04) \times 10^4 \text{ kg m}^{-3} \).

(a) The distance between a peak and an adjacent trough is half of a wavelength:
\( \frac{\lambda}{2} = 4.5 - 1.5 = 3.0 \text{ cm} \implies \lambda = 6.0 \text{ cm} = 0.060 \text{ m} \)

Amplitude \( A \) is the maximum displacement from equilibrium:
\( A = 3.0 \text{ mm} = 3.0 \times 10^{-3} \text{ m} \)

Wave speed \( v \):
\( v = f \lambda = 4.0 \times 0.060 = 0.24 \text{ m s}^{-1} \)

(b) Distance between the two points:
\( \Delta x = 5.0 - 2.0 = 3.0 \text{ cm} \)

Since \( \lambda = 6.0 \text{ cm} \), this distance corresponds to half a wavelength:
\( \Delta x = 0.5 \lambda \)

Therefore, the phase difference is:
\( \phi = \pi \text{ radians} \) (or \( 3.14 \text{ rad} \))

(c) The time period of the wave is:
\( T = \frac{1}{f} = \frac{1}{4.0} = 0.25 \text{ s} \)

At \( t = 0 \), the particle at \( x = 1.5 \text{ cm} \) is at a peak (displacement \( y = +3.0 \text{ mm} \), velocity is zero).
During one complete period (\( t = 0 \) to \( 0.25 \text{ s} \)), the particle completes one full oscillation:
- It moves downwards, passing through equilibrium (\( y = 0 \)) with maximum speed.
- It reaches maximum negative displacement (\( y = -3.0 \text{ mm} \)) at \( t = 0.125 \text{ s} \) (momentarily at rest).
- It then moves upwards through equilibrium back to its initial position of \( y = +3.0 \text{ mm} \) at \( t = 0.25 \text{ s} \).

(a) [3 marks]
- 1 mark for identifying \( \lambda = 6.0 \text{ cm} \) or \( 0.060 \text{ m} \).
- 1 mark for amplitude \( A = 3.0 \text{ mm} \) or \( 3.0 \times 10^{-3} \text{ m} \).
- 1 mark for calculating speed \( v = f \lambda = 0.24 \text{ m s}^{-1} \).

(b) [3 marks]
- 1 mark for identifying path difference as \( 3.0 \text{ cm} \).
- 1 mark for using the ratio \( \frac{\Delta x}{\lambda} \) or the formula \( \phi = \frac{2\pi \Delta x}{\lambda} \).
- 1 mark for getting \( \pi \text{ rad} \) (or \( 180^{\circ} \), but accept \( 3.14 \text{ rad} \) since radians are requested).

(c) [3 marks]
- 1 mark for stating that \( t = 0.25 \text{ s} \) is exactly one period \( T \).
- 1 mark for stating the starting position is the maximum positive displacement (peak) and ending position is the same.
- 1 mark for a complete description of the downward movement to \( -3.0 \text{ mm} \) at \( t = 0.125 \text{ s} \) and return upward.

(a) Since the plank is uniform, its weight \( W_p = 25 \times 9.81 = 245.25 \text{ N} \) acts at its midpoint, \( 2.0 \text{ m} \) from each end.

Rope A is at \( x = 0 \).

Rope B is at \( 4.0 - 1.2 = 2.8 \text{ m} \) from Rope A.

Take moments about the point of attachment of Rope A (\( x = 0 \)):
\( \Sigma M_A = 0 \implies (W_p \times 2.0) = (T_B \times 2.8) \)
\( 245.25 \times 2.0 = 2.8 T_B \)
\( 490.5 = 2.8 T_B \implies T_B = 175.2 \text{ N} \approx 180 \text{ N} \)

Using vertical equilibrium:
\( T_A + T_B = W_p \)
\( T_A + 175.2 = 245.25 \implies T_A = 70.15 \text{ N} \approx 70 \text{ N} \)

(b) When the plank is about to tilt, it will pivot about Rope B. At this exact threshold, the plank lifts off Rope A, meaning the tension in Rope A becomes zero:
\( T_A = 0 \)

Let the builder walk a distance \( d \) past Rope B.

Take moments about the pivot at Rope B:
- The weight of the plank acts at \( x = 2.0 \text{ m} \), which is \( 2.8 - 2.0 = 0.8 \text{ m} \) to the left of Rope B.
- The builder's weight acts at distance \( d \) to the right of Rope B.

For rotational equilibrium at the tipping point:
\( W_p \times 0.8 = W_b \times d \)
\( (25 \times g) \times 0.8 = (80 \times g) \times d \)
\( 20 = 80 d \implies d = 0.25 \text{ m} \)

(a) [5 marks]
- 1 mark for locating the center of gravity of the plank at \( 2.0 \text{ m} \).
- 1 mark for stating Rope B is at \( 2.8 \text{ m} \) from Rope A.
- 1 mark for setting up a correct moment equation about Rope A (or Rope B).
- 1 mark for calculating \( T_B = 175 \text{ N} \) (shows \( \approx 180 \text{ N} \)).
- 1 mark for calculating \( T_A = 70 \text{ N} \) (or \( 70.2 \text{ N} \)).

(b) [4.67 marks]
- 1 mark for recognizing that \( T_A = 0 \) when the plank is about to tilt.
- 1 mark for identifying the pivot point is at Rope B.
- 1 mark for identifying the distance of the plank's weight to Rope B is \( 0.8 \text{ m} \).
- 1.67 marks for setting up the balance of moments about Rope B and calculating \( d = 0.25 \text{ m} \).

(a) The graph should have velocity \( v \) in \( \text{ m s}^{-1} \) on the y-axis and time \( t \) in \( \text{ s} \) on the x-axis.
- From \( t = 0 \) to \( t = 0.60 \text{ s} \), the graph is a flat horizontal line at \( v = 24 \text{ m s}^{-1} \).
- From \( t = 0.60 \text{ s} \) to \( t = t_{\text{stop}} \), the graph is a straight line sloping down to \( v = 0 \).

(b) First, calculate the thinking distance (distance traveled during reaction time):
\( s_{\text{thinking}} = v \times t_{\text{reaction}} = 24 \times 0.60 = 14.4 \text{ m} \)

Next, find the braking distance:
\( s_{\text{braking}} = s_{\text{total}} - s_{\text{thinking}} = 52 - 14.4 = 37.6 \text{ m} \)

Using the equations of motion for uniform deceleration:
\( v^2 = u^2 + 2as \)
\( 0^2 = 24^2 + 2a(37.6) \)
\( 0 = 576 + 75.2a \implies a = -\frac{576}{75.2} \approx -7.66 \text{ m s}^{-2} \)

Therefore, the deceleration is \( 7.7 \text{ m s}^{-2} \).

(c) Use the equation of motion:
\( v = u + at \) during braking
\( 0 = 24 - 7.66 t_{\text{braking}} \implies t_{\text{braking}} = \frac{24}{7.66} \approx 3.13 \text{ s} \)

Alternatively, \( s_{\text{braking}} = \frac{u+v}{2} t_{\text{braking}} \implies 37.6 = 12 t_{\text{braking}} \implies t_{\text{braking}} = 3.13 \text{ s} \)

Rounding to 2 significant figures gives \( 3.1 \text{ s} \).

(a) [3 marks]
- 1 mark for axes labeled correctly with quantities and units (Velocity / \( \text{m s}^{-1} \) and Time / \( \text{s} \)).
- 1 mark for a horizontal line from \( 0 \) to \( 0.6 \text{ s} \) at \( 24 \text{ m s}^{-1} \).
- 1 mark for a straight line with a constant negative gradient from \( 0.6 \text{ s} \) to the final time.

(b) [4 marks]
- 1 mark for calculating thinking distance \( 14.4 \text{ m} \).
- 1 mark for calculating braking distance \( 37.6 \text{ m} \).
- 1 mark for using \( v^2 = u^2 + 2as \) with appropriate values.
- 1 mark for final deceleration value of \( 7.7 \text{ m s}^{-2} \) (accept \( 7.66 \text{ m s}^{-2} \)).

(c) [2.67 marks]
- 1 mark for a correct kinematic equation used to solve for braking time.
- 1.67 marks for the correct answer of \( 3.1 \text{ s} \) (or \( 3.13 \text{ s} \)).

(a) The de Broglie wavelength is the wavelength associated with a particle that is moving with momentum \( p \), given by \( \lambda = \frac{h}{p} \) where \( h \) is the Planck constant.

(b)(i) Diffraction is a wave property. The concentric rings observed on the screen result from constructive and destructive interference of the electron waves passing through the atomic gaps in the graphite lattice (which acts as a diffraction grating). Particle-only theories cannot explain interference and diffraction.

(b)(ii) The kinetic energy \( E_k \) gained by an electron accelerated through potential difference \( V \) is:
\( E_k = eV \)

Also, kinetic energy is related to momentum \( p \) by:
\( E_k = \frac{p^2}{2m_e} \implies p = \sqrt{2m_e E_k} = \sqrt{2m_e eV} \)

Substituting this into the de Broglie relation:
\( \lambda = \frac{h}{p} \implies \lambda = \frac{h}{\sqrt{2m_e eV}} \)

(b)(iii) Given:
\( V = 4.5 \text{ kV} = 4500 \text{ V} \)
\( m_e = 9.11 \times 10^{-31} \text{ kg} \)
\( e = 1.60 \times 10^{-19} \text{ C} \)
\( h = 6.63 \times 10^{-34} \text{ J s} \)

Calculate momentum:
\( p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 4500} \)
\( p = \sqrt{1.3118 \times 10^{-45}} \approx 3.622 \times 10^{-23} \text{ kg m s}^{-1} \)

Calculate wavelength:
\( \lambda = \frac{6.63 \times 10^{-34}}{3.622 \times 10^{-23}} \approx 1.83 \times 10^{-11} \text{ m} \) (or \( 0.0183 \text{ nm} \))

(a) [2 marks]
- 1 mark for stating that it is the wavelength of a moving particle.
- 1 mark for referencing momentum, e.g., \( \lambda = h/p \) where symbols are defined.

(b) [7.67 marks]
- (i) [2 marks]: 1 mark for noting that diffraction/interference is uniquely a wave behavior; 1 mark for stating that the atomic spacing of graphite acts as a diffraction grating to produce the observed pattern.
- (ii) [3 marks]: 1 mark for equating kinetic energy to electrical work done: \( E_k = eV \); 1 mark for relating momentum to kinetic energy: \( p = \sqrt{2m_e E_k} \); 1 mark for combining with \( \lambda = h/p \) to obtain the final target formula.
- (iii) [2.67 marks]: 1 mark for converting \( 4.5 \text{ kV} \) to \( 4500 \text{ V} \); 1.67 marks for correct substitution of values leading to \( 1.8 \times 10^{-11} \text{ m} \) (allow \( 1.8 \times 10^{-11} \text{ m} \) to \( 1.84 \times 10^{-11} \text{ m} \)).

(a) First, calculate the cross-sectional area \( A \) of the wire:
\( A = \pi r^2 = \pi \times (0.18 \times 10^{-3})^2 \approx 1.018 \times 10^{-7} \text{ m}^2 \)

Now, calculate the resistance \( R \):
\( R = \rho \frac{L}{A} = 1.1 \times 10^{-6} \times \frac{2.4}{1.018 \times 10^{-7}} \approx 25.93 \ \Omega \approx 26 \ \Omega \)

(b)(i) The rate of electrical energy dissipated is the power \( P \):
\( P = \frac{V^2}{R} = \frac{12^2}{25.93} = \frac{144}{25.93} \approx 5.55 \text{ W} \approx 5.6 \text{ W} \)

(b)(ii) Let the original wire have resistance \( R \), length \( L \), and area \( A = \pi r^2 \).

The new wire has length \( L' = 2L \) and radius \( r' = 0.5r \).
Therefore, its cross-sectional area is:
\( A' = \pi (r')^2 = \pi (0.5r)^2 = 0.25 A \)

The resistance of the new wire \( R' \) is:
\( R' = \rho \frac{L'}{A'} = \rho \frac{2L}{0.25A} = 8 \left( \rho \frac{L}{A} \right) = 8R \)

Since both wires are connected to the same voltage supply \( V \):
\( P_{\text{original}} = \frac{V^2}{R} \)
\( P_{\text{new}} = \frac{V^2}{R'} = \frac{V^2}{8R} = \frac{1}{8} P_{\text{original}} \)

The ratio of the power dissipated in the original wire to that in the new wire is:
\( \frac{P_{\text{original}}}{P_{\text{new}}} = 8 \)

(a) [4 marks]
- 1 mark for converting radius to meters: \( 0.18 \times 10^{-3} \text{ m} \).
- 1 mark for calculating area correctly: \( A = 1.02 \times 10^{-7} \text{ m}^2 \).
- 1 mark for using the resistivity equation: \( R = \rho L / A \).
- 1 mark for calculating \( R = 25.9 \ \Omega \) and showing it rounds to \( 26 \ \Omega \).

(b) [5.67 marks]
- (i) [2 marks]: 1 mark for utilizing \( P = V^2/R \); 1 mark for calculating power as \( 5.6 \text{ W} \) (or \( 5.5 \text{ W} \)).
- (ii) [3.67 marks]: 1 mark for determining that halving the radius quarters the area; 1 mark for showing the resistance of the new wire is \( 8 \) times greater than the original (\( R' = 8R \)); 1.67 marks for establishing that power is inversely proportional to resistance at constant voltage, giving the ratio of \( 8 \).

Experimental Setup:
- A potential divider circuit is used to vary the voltage across the LED from 0 up to its threshold voltage. The circuit contains a d.c. power supply, a variable resistor or potentiometer connected as a potential divider, a voltmeter in parallel with the LED, and a sensitive milliammeter or microammeter in series with the LED.

Procedure:
1. Place the first LED (of known wavelength \(\lambda\)) in the circuit.
2. Cover the LED with a dark viewing tube to exclude ambient light.
3. Slowly adjust the potential divider to increase the potential difference across the LED from 0 V until the LED just begins to emit light (or when a small threshold current begins to flow, e.g., 10 \(\mu\text{A}\)). Record this voltage as the threshold voltage \(V_0\).
4. Repeat the procedure for several other LEDs of different known wavelengths.

Analysis:
- The energy of a photon is given by \(E = hf = \frac{hc}{\lambda}\).
- At the threshold voltage, the work done on each electron crossing the junction is equal to the photon energy: \(e V_0 = \frac{hc}{\lambda}\), where \(e\) is the elementary charge (\(1.60 \times 10^{-19}\text{ C}\)) and \(c\) is the speed of light in a vacuum (\(3.00 \times 10^8\text{ m s}^{-1}\)).
- Rearranging this gives \(V_0 = \left(\frac{hc}{e}\right) \frac{1}{\lambda}\).
- Plot a graph of \(V_0\) (y-axis) against \(1/\lambda\) (x-axis).
- The graph is a straight line through the origin with gradient \(m = \frac{hc}{e}\).
- Determine the gradient \(m\) of the line of best fit and calculate the Planck constant using \(h = \frac{m e}{c}\).

Level 3 (5-6 marks):
- Draws/describes a complete and workable potential divider circuit with voltmeter across the LED and ammeter in series.
- Clear, detailed procedure explaining how \(V_0\) is determined (visual onset of light and current reading).
- Complete graphical analysis identifying \(eV_0 = \frac{hc}{\lambda}\), plotting \(V_0\) against \(1/\lambda\), and calculating \(h\) from the gradient \(m = \frac{hc}{e}\).
- Information is organised clearly and logically, using specialist terminology.

Level 2 (3-4 marks):
- Draws/describes a circuit with minor omissions (e.g., standard series variable resistor instead of potential divider, or missing meter).
- Explains the procedure for determining \(V_0\) and mentions multiple LEDs.
- Mentions \(eV = hf\) or \(eV_0 = \frac{hc}{\lambda}\) and attempts to describe a graph or averages, but analysis may lack detail (e.g., does not show how to extract \(h\) from the gradient of a plot of \(V_0\) against \(1/\lambda\)).
- Mostly structured and coherent.

Level 1 (1-2 marks):
- Identifies basic circuit components or basic equation.
- Brief description of the procedure without clear definition of how threshold voltage is found.
- Analysis is incomplete or incorrect.
- Unstructured or lacks scientific terminology.

0 marks:
- No response or no relevant physics.

Experimental Setup:
- A long plastic or glass resonance tube is placed vertically inside a tall measuring cylinder filled with water. The water level inside the tube creates a closed boundary at that end.
- A tuning fork of known frequency \(f\) is held horizontally just above the open end of the tube.
- A metre rule is used to measure the length of the air column inside the tube.

Procedure:
1. Strike the tuning fork on a rubber pad to make it vibrate and hold it just above the top of the resonance tube.
2. Slowly raise the tube out of the water. Listen for a distinct increase in loudness (resonance).
3. Mark this first resonance position and use the metre rule to measure the length \(L_1\) from the top of the tube to the water surface.
4. Continue raising the tube further until a second loud resonance sound is heard. Measure this second length \(L_2\).
5. Repeat the measurements for several tuning forks of different known frequencies \(f\).

Analysis:
- For a closed tube, the first resonance occurs at: \(L_1 + c = \frac{\lambda}{4}\), where \(c\) is the end correction of the tube.
- The second resonance occurs at: \(L_2 + c = \frac{3\lambda}{4}\).
- Subtracting these two equations eliminates the end correction \(c\), giving: \(L_2 - L_1 = \frac{\lambda}{2}\), which simplifies to \(\lambda = 2(L_2 - L_1)\).
- The wave equation is \(v = f \lambda\). Substituting for \(\lambda\) gives \(v = 2f(L_2 - L_1)\).
- Rearranging this formula yields: \(L_2 - L_1 = \frac{v}{2} \left(\frac{1}{f}\right)\).
- Plot a graph of \((L_2 - L_1)\) (y-axis) against \(1/f\) (x-axis).
- The gradient \(m\) of the line of best fit is equal to \(\frac{v}{2}\).
- Determine \(v\) using \(v = 2m\). This graphical method averages all data points and eliminates systematic errors due to end correction.

Level 3 (5-6 marks):
- Clear, detailed description of experimental setup (tube, water cylinder, tuning fork, ruler).
- Explains how to find both the first and second resonance lengths \(L_1\) and \(L_2\) for each frequency.
- Rigorous graphical analysis: derives \(\lambda = 2(L_2 - L_1)\) to eliminate the end correction, relates to \(v = f \lambda\), and explains how to plot \((L_2 - L_1)\) against \(1/f\) to find \(v\) from the gradient (\(v = 2m\)).
- Structure is logical, clear, and uses accurate physics terms.

Level 2 (3-4 marks):
- Describes setup and procedure to find resonance, but may only focus on the first resonance length \(L_1\) or lacks clarity on finding both resonances.
- Mentions \(v = f \lambda\) and attempts to relate length to wavelength, but treatment of end correction is incomplete or omitted.
- Mentions plotting a graph (e.g., \(L_1\) vs \(1/f\)) but doesn't correctly explain how to find \(v\) from it without end correction error.
- Mostly structured and coherent.

Level 1 (1-2 marks):
- Basic description of a resonance tube experiment.
- Identifies that frequency and length are measured.
- Shows some understanding of \(v = f \lambda\) but no graphical analysis or end correction discussion.
- Unstructured or vague.

0 marks:
- No response or no relevant physics.