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By conservation of momentum, the initial momentum is equal to the final momentum: \(m_1 v_1 = (m_1 + m_2) v_f \implies 2.0 \times 4.0 = (2.0 + 3.0) v_f \implies v_f = 1.6 \text{ m s}^{-1}\). The initial kinetic energy of the system is \(E_{k,i} = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \times 2.0 \times 4.0^2 = 16.0 \text{ J}\). The final kinetic energy of the system is \(E_{k,f} = \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} \times 5.0 \times 1.6^2 = 6.4 \text{ J}\). The loss in kinetic energy is \(\Delta E_k = 16.0 \text{ J} - 6.4 \text{ J} = 9.6 \text{ J}\).

Award 1 mark for the correct answer C.

Initially, the potential difference across the thermistor is \(V = 6.0 \text{ V} \times \frac{5.0 \text{ k}\Omega}{5.0 \text{ k}\Omega + 10.0 \text{ k}\Omega} = 2.0 \text{ V}\). For an NTC thermistor, as the temperature increases, its resistance decreases. Since the resistance of the thermistor decreases relative to the fixed resistor, it takes a smaller fraction of the total supply voltage. Therefore, the voltmeter reading decreases from \(2.0 \text{ V}\).

Award 1 mark for the correct answer B.

Resolving the forces acting on the bob vertically, the vertical component of the tension must balance the weight of the bob because there is no vertical acceleration. This gives \(T \cos\theta = mg\). Rearranging this equation for tension gives \(T = \frac{mg}{\cos\theta}\).

Award 1 mark for the correct answer D.

The magnetic force provides the centripetal force: \(B e v = \frac{m v^2}{R} \implies v = \frac{B e R}{m}\). The kinetic energy is given by \(E_k = \frac{1}{2} m v^2\). Substituting the expression for \(v\): \(E_k = \frac{1}{2} m \left(\frac{B e R}{m}\right)^2 = \frac{B^2 e^2 R^2}{2m}\).

Award 1 mark for the correct answer C.

According to Einstein's photoelectric equation, \(h f = \phi + E_k \implies E_k = h f - \phi\), where \(\phi\) is the work function of the metal. When the frequency is doubled, the new maximum kinetic energy \(E_k'\) is given by \(E_k' = 2 h f - \phi\). Substituting \(h f = E_k + \phi\) gives \(E_k' = 2(E_k + \phi) - \phi = 2 E_k + \phi\). Since the work function \(\phi\) is a positive quantity, \(E_k'\) is greater than \(2 E_k\).

Award 1 mark for the correct answer B.

The number of undecayed nuclei at any time \(t\) is given by \(N(t) = N_0 \left(\frac{1}{2}\right)^{t/T}\). At \(t = T\), the number of undecayed nuclei remaining is \(N_1 = \frac{1}{2} N_0\). At \(t = 3T\), the number of undecayed nuclei remaining is \(N_2 = N_0 \left(\frac{1}{2}\right)^3 = \frac{1}{8} N_0\). The number of nuclei that decay in this interval is the difference: \(\Delta N_{\text{decayed}} = N_1 - N_2 = \frac{1}{2} N_0 - \frac{1}{8} N_0 = \frac{3}{8} N_0\).

Award 1 mark for the correct answer C.

According to the kinetic theory of gases, the average kinetic energy of gas molecules is directly proportional to the absolute temperature: \(\frac{1}{2} m \langle c^2 \rangle = \frac{3}{2} k_B T\). This shows that the mean square speed \(\langle c^2 \rangle\) is directly proportional to \(T\). If \(T\) is doubled, the mean square speed is also doubled. The root-mean-square speed \(c_{\text{rms}} = \sqrt{\langle c^2 \rangle}\) would increase by a factor of \(\sqrt{2}\). From \(pV = N k_B T\), since volume is fixed, pressure \(p\) must double when \(T\) doubles.

Award 1 mark for the correct answer A.

The elastic strain energy stored in a stretched wire is \(W = \frac{1}{2} F \Delta x\). The Young modulus is defined as \(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta x / L}\). Rearranging this gives \(F = \frac{E A \Delta x}{L}\). Substituting this expression for \(F\) into the energy equation gives \(W = \frac{1}{2} \left(\frac{E A \Delta x}{L}\right) \Delta x = \frac{E A (\Delta x)^2}{2L}\).

Award 1 mark for the correct answer C.

The terminal potential difference of the cell is given by \(V = V_0 - Ir\), where the current is \(I = \frac{V_0}{R + r}\). As \(R\) increases, the total resistance of the circuit increases, causing the current \(I\) to decrease. Because \(I\) decreases, the lost volts \(Ir\) decrease, and thus the terminal potential difference \(V\) increases.

The power dissipated in the variable resistor is given by \(P = I^2 R = \frac{V_0^2 R}{(R + r)^2}\). According to the maximum power transfer theorem, the power delivered to the load reaches a maximum when the load resistance equals the internal resistance (\(R = r\)). Since \(R\) starts at zero and increases, the power first increases until it reaches a maximum at \(R = r\), and then decreases as \(R\) increases further.

Therefore, the terminal potential difference increases, and the power dissipated increases then decreases.

Select Option A.
1 mark for the correct answer.

At the highest point of the bridge, the forces acting on the car are its weight \(mg\) acting downwards and the normal contact force \(N\) acting upwards.

Since the car is moving in a circular path, the resultant force must act towards the center of the circle (downwards):
\(F_{\text{net}} = mg - N = \frac{mv^2}{r}\)

Rearranging the equation to solve for the normal contact force \(N\):
\(N = mg - \frac{mv^2}{r} = m\left(g - \frac{v^2}{r}\right)\)

Select Option C.
1 mark for the correct answer.

The extension \(\Delta x\) of a wire is given by:
\(\Delta x = \frac{F L}{A E}\) where \(A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}\).

Thus, \(\Delta x = \frac{4 F L}{\pi d^2 E}\). Since both wires are made of the same material (same Young Modulus \(E\)) and are subjected to the same force \(F\), the extension is proportional to \(\frac{L}{d^2}\):
\(\Delta x \propto \frac{L}{d^2}\)

For wire X:
\(\Delta x_X \propto \frac{L}{d^2}\)

For wire Y:
\(\Delta x_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\)

Finding the ratio \(\frac{\Delta x_X}{\Delta x_Y}\):
\(\frac{\Delta x_X}{\Delta x_Y} = \frac{L/d^2}{L/(2d^2)} = 2\)

Select Option B.
1 mark for the correct answer.

Let the initial activity of isotope B be \(A_0\). Then the initial activity of isotope A is \(8A_0\).

After a time \(t\), the activity of isotope A is:
\(A_A(t) = 8A_0 \left(\frac{1}{2}\right)^{\frac{t}{2}}\)

The activity of isotope B is:
\(A_B(t) = A_0 \left(\frac{1}{2}\right)^{\frac{t}{4}}\)

We set the two activities equal:
\(8A_0 \left(\frac{1}{2}\right)^{\frac{t}{2}} = A_0 \left(\frac{1}{2}\right)^{\frac{t}{4}}\)

Dividing both sides by \(A_0\):
\(8 \left(\frac{1}{2}\right)^{\frac{t}{2}} = \left(\frac{1}{2}\right)^{\frac{t}{4}}\)

\(8 = \left(\frac{1}{2}\right)^{\frac{t}{4} - \frac{t}{2}} = \left(\frac{1}{2}\right)^{-\frac{t}{4}} = 2^{\frac{t}{4}}\)

Since \(8 = 2^3\), we have:
\(2^3 = 2^{\frac{t}{4}} \implies 3 = \frac{t}{4} \implies t = 12\text{ hours}\)

Select Option C.
1 mark for the correct answer.

The total energy in simple harmonic motion is given by:
\(E_{\text{total}} = \frac{1}{2} k A^2\)

The potential energy at displacement \(x\) is:
\(E_p = \frac{1}{2} k x^2\)

The kinetic energy is:
\(E_k = E_{\text{total}} - E_p = \frac{1}{2} k (A^2 - x^2)\)

We are given that the kinetic energy is three times the potential energy:
\(E_k = 3 E_p\)

\(\frac{1}{2} k (A^2 - x^2) = 3 \left(\frac{1}{2} k x^2\right)\)

Dividing both sides by \(\frac{1}{2} k\):
\(A^2 - x^2 = 3x^2\)

\(A^2 = 4x^2 \implies x^2 = \frac{A^2}{4} \implies x = \pm \frac{A}{2}\)

Select Option B.
1 mark for the correct answer.

The magnetic force provides the centripetal force:
\(q v B = \frac{m v^2}{r} \implies r = \frac{m v}{q B}\)

Since momentum \(p = m v = \sqrt{2 m E_k}\), where \(E_k\) is the kinetic energy, we can rewrite the radius as:
\(r = \frac{\sqrt{2 m E_k}}{q B}\)

Since both particles have the same kinetic energy \(E_k\) and are in the same magnetic field \(B\), the radius is proportional to \(\frac{\sqrt{m}}{q}\):
\(r \propto \frac{\sqrt{m}}{q}\)

For a proton (\(p\)), let the mass be \(m_p = m\) and charge be \(q_p = e\):
\(r_p \propto \frac{\sqrt{m}}{e}\)

For an alpha particle (\(\alpha\)), the mass is \(m_{\alpha} = 4m\) and the charge is \(q_{\alpha} = 2e\):
\(r_{\alpha} \propto \frac{\sqrt{4m}}{2e} = \frac{2\sqrt{m}}{2e} = \frac{\sqrt{m}}{e}\)

Thus, the ratio of the radii is:
\(\frac{r_p}{r_{\alpha}} = 1\) (or a ratio of 1:1).

Select Option B.
1 mark for the correct answer.

According to Einstein's photoelectric equation:
\(E_{max} = hf - \Phi\)

where \(\Phi\) is the work function of the metal. If the frequency is doubled to \(2f\), the new maximum kinetic energy \(E'_{max}\) is:
\(E'_{max} = h(2f) - \Phi = 2hf - \Phi\)

We can express \(hf\) as \(E_{max} + \Phi\):
\(E'_{max} = 2(E_{max} + \Phi) - \Phi = 2E_{max} + \Phi\)

Since the work function \(\Phi\) must be a positive quantity for photoelectrons to be emitted initially, the new maximum kinetic energy is \(2E_{max} + \Phi\), which is strictly greater than \(2E_{max}\).

Select Option C.
1 mark for the correct answer.

According to the kinetic theory of gases, the mean kinetic energy of the gas molecules is proportional to the absolute temperature \(T\):
\(\frac{1}{2} m \langle c^2 \rangle = \frac{3}{2} k T \implies \langle c^2 \rangle \propto T\)

First, convert the temperatures to Kelvin:
\(T_1 = 27 + 273.15 = 300\text{ K}\)
\(T_2 = 327 + 273.15 = 600\text{ K}\)

Now find the ratio of the mean square speeds:
\(\frac{\langle c^2 \rangle_2}{\langle c^2 \rangle_1} = \frac{T_2}{T_1} = \frac{600\text{ K}}{300\text{ K}} = 2\)

Select Option B.
1 mark for the correct answer.

During the upward motion, both gravity and the air resistance act downwards on the ball. Therefore, the net retarding force is \(F_{\text{net}} = mg + D\). According to the work-energy theorem, the initial kinetic energy is completely converted into work done against gravity and air resistance: \(E_k = mgh + Dh = h(mg + D)\). Substituting \(E_k = \frac{1}{2}mv^2\), we get \(\frac{1}{2}mv^2 = h(mg + D)\), which yields \(h = \frac{mv^2}{2(mg + D)}\).

1 mark for the correct answer (B). [1 AO1]

The Young modulus is given by \(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta x/L} = \frac{FL}{A\Delta x}\). Solving for extension: \(\Delta x = \frac{FL}{AE}\). For the second wire, the force is \(2F\), the length is \(2L\), and the area is \(2A\). The new extension is \(\Delta x' = \frac{(2F)(2L)}{(2A)E} = 2\left(\frac{FL}{AE}\right) = 2\Delta x\).

1 mark for the correct answer (C). [1 AO2]

Using the grating equation \(d \sin\theta = n\lambda\). For the first case, \(n = 3\) and wavelength is \(\lambda\), so \(d \sin\theta = 3\lambda\). For the second case, let the angle of diffraction be \(\phi\), \(n = 2\) and the wavelength is \(1.5\lambda\). This gives \(d \sin\phi = 2(1.5\lambda) = 3\lambda\). Since the right-hand sides are equal, \(d \sin\phi = d \sin\theta\), which means \(\sin\phi = \sin\theta\) and thus \(\phi = \theta\).

1 mark for the correct answer (A). [1 AO2]

The electrical power dissipated in the external resistor is given by \(P = I^2 R = \left(\frac{\varepsilon}{R + r}\right)^2 R\). For the two cases, the power is equal: \(\left(\frac{\varepsilon}{R + r}\right)^2 R = \left(\frac{\varepsilon}{4R + r}\right)^2 (4R)\). Taking the square root of both sides gives \(\frac{\sqrt{R}}{R + r} = \frac{2\sqrt{R}}{4R + r}\). Simplifying this, we get \(\frac{1}{R + r} = \frac{2}{4R + r}\). Cross-multiplying gives \(4R + r = 2(R + r) = 2R + 2r\). Rearranging this yields \(2R = r\), or \(r = 2R\).

1 mark for the correct answer (C). [1 AO2]

The frictional force providing the centripetal force is given by \(F = \frac{m v^2}{r}\). Since the maximum frictional force is proportional to the normal contact force, \(F_{\text{max}} = \mu m g\), we have \(\mu m g = \frac{m v^2}{r}\). The mass \(m\) cancels out, giving \(v = \sqrt{\mu g r}\). For the second car on the same road surface (same \(\mu\)), the radius is \(0.5r\). Thus, the new maximum speed is \(v' = \sqrt{\mu g (0.5r)} = \sqrt{0.5} v = \frac{v}{\sqrt{2}}\).

1 mark for the correct answer (B). [1 AO2]

The radius of the circular path of a charged particle in a magnetic field is given by \(r = \frac{mv}{qB}\). Since kinetic energy is \(E_k = \frac{p^2}{2m} = \frac{(mv)^2}{2m}\), the momentum is \(mv = \sqrt{2m E_k}\), which gives \(r = \frac{\sqrt{2m E_k}}{qB}\). Since \(r\) is proportional to the square root of the kinetic energy, doubling the kinetic energy increases the radius by a factor of \(\sqrt{2}\). Therefore, the new radius is \(\sqrt{2}r\).

1 mark for the correct answer (B). [1 AO2]

Let us check the conservation laws: 1. Charge (Q): \(Q(p) + Q(\pi^-) = +1 + (-1) = 0\). Therefore, \(Q(K^0) + Q(X) = 0 + Q(X) = 0\), so \(Q(X) = 0\). 2. Baryon number (B): \(B(p) + B(\pi^-) = +1 + 0 = +1\). Therefore, \(B(K^0) + B(X) = 0 + B(X) = +1\), so \(B(X) = +1\). Particle \(X\) is a baryon, consisting of three quarks. 3. Strangeness (S): \(S(p) + S(\pi^-) = 0 + 0 = 0\). Therefore, \(S(K^0) + S(X) = +1 + S(X) = 0\), so \(S(X) = -1\). Since \(X\) has strangeness \(-1\), it contains one strange (\(s\)) quark, which has charge \(-1/3\). To achieve a net charge of 0 and baryon number +1, the other two quarks must be an up (\(u\)) quark (charge \(+2/3\)) and a down (\(d\)) quark (charge \(-1/3\)). This gives a total charge of \(-1/3 + 2/3 - 1/3 = 0\). Thus, the quark structure of \(X\) is \(uds\).

1 mark for the correct answer (B). [1 AO2]

The temperatures must be converted to kelvin (K). \(T_1 = 27 + 273.15 = 300.15\text{ K}\), and \(T_2 = 327 + 273.15 = 600.15\text{ K}\). The absolute temperature doubles (\(T_2 / T_1 = 2\)). The mean kinetic energy of ideal gas molecules is directly proportional to the absolute temperature (\(E_k = \frac{3}{2}kT\)), so the mean kinetic energy doubles. Since the volume of the container is fixed, the pressure of an ideal gas is also directly proportional to the absolute temperature (\(p \propto T\)), so the pressure of the gas doubles.

1 mark for the correct answer (A). [1 AO1]

According to Newton's second law, the rate of change of momentum of an object is equal to the net resultant force acting on it: \(\frac{dp}{dt} = F_{\text{net}}\). For a projectile in flight where air resistance is negligible, the only force acting on it at any point in its trajectory (including at its maximum height) is the gravitational force, which is its weight \(mg\). Therefore, the magnitude of the rate of change of momentum is constantly \(mg\).

[1] C - correct application of Newton's second law where the net force is the constant gravitational force.

The extension of a wire is given by \(\Delta x = \frac{F L}{A E}\), where cross-sectional area \(A = \frac{\pi d^2}{4}\). Since the wires are connected in series, they experience the same tension force \(F\). Since they are made of the same metal, they have the same Young's modulus \(E\). Thus, extension is proportional to \(\frac{L}{d^2}\):

\(\frac{\Delta x_{\text{P}}}{\Delta x_{\text{Q}}} = \frac{L_{\text{P}}}{L_{\text{Q}}} \times \left(\frac{d_{\text{Q}}}{d_{\text{P}}}\right)^2\)

Given \(L_{\text{P}} = 2 L_{\text{Q}}\) and \(d_{\text{P}} = 0.5 d_{\text{Q}}\):

\(\frac{\Delta x_{\text{P}}}{\Delta x_{\text{Q}}} = 2 \times (2)^2 = 8\).

[1] C - correct deduction of proportionalities from Young's modulus formula and substitution of the ratio of length and diameter.

For a negative temperature coefficient (NTC) thermistor, an increase in temperature increases the number density of conduction electrons, which decreases its resistance \(R_{\text{T}}\). According to the potential divider formula, the output voltage across the thermistor is \(V_{\text{out}} = V \times \frac{R_{\text{T}}}{R + R_{\text{T}}}\). As \(R_{\text{T}}\) decreases, the fraction of the total voltage across the thermistor decreases, so \(V_{\text{out}}\) decreases.

[1] D - correct identification of thermistor behaviour with temperature and the resulting change in output potential difference.

The normal contact force \(N\) acts perpendicular to the banked track, and the weight \(mg\) acts vertically downwards. Resolving forces vertically (where there is no acceleration): \(N \cos\theta = mg\). Resolving forces horizontally towards the centre of the circle: \(N \sin\theta = \frac{m v^2}{r}\). Dividing the horizontal equation by the vertical equation yields: \(\tan\theta = \frac{v^2}{gr}\).

[1] C - correct derivation of the banking angle by resolving forces vertically and horizontally to find the centripetal force.

The magnetic force provides the centripetal force: \(B q v = \frac{m v^2}{r} \implies r = \frac{m v}{B q}\). The momentum is related to kinetic energy \(E_{\text{k}}\) by \(p = m v = \sqrt{2 m E_{\text{k}}}\). Thus, the radius is \(r = \frac{\sqrt{2 m E_{\text{k}}}}{B q}\). For the proton: \(R = \frac{\sqrt{2 m E_{\text{k}}}}{B e}\). For the alpha particle (mass \(4m\), charge \(2e\)) with the same kinetic energy: \(r_{\alpha} = \frac{\sqrt{2 (4m) E_{\text{k}}}}{B (2e)} = \frac{2\sqrt{2 m E_{\text{k}}}}{2 B e} = \frac{\sqrt{2 m E_{\text{k}}}}{B e} = R\).

[1] B - correct derivation of the orbital radius in terms of kinetic energy and comparison of the ratios.

A baryon consists of three quarks. A strangeness of \(-1\) means it contains exactly one strange quark \(s\) (charge \(-\frac{1}{3}\)). The other two quarks must be non-strange quarks (up \(u\) with charge \(+\frac{2}{3}\) or down \(d\) with charge \(-\frac{1}{3}\)). Since the total charge of the baryon is \(0\), the sum of the charges of the other two quarks must be \(+\frac{1}{3}\). This is satisfied by one up quark and one down quark: \(-\frac{1}{3} + \frac{2}{3} - \frac{1}{3} = 0\). This corresponds to the composition \(sud\).

[1] B - correct identification of quark charges and strangeness quantum numbers to yield a neutral baryon with strangeness \(-1\).

The total energy of the simple harmonic oscillator is given by \(E_{\text{total}} = \frac{1}{2} k A^2\). The potential energy at displacement \(x\) is \(E_{\text{p}} = \frac{1}{2} k x^2\). Since total energy is conserved, \(E_{\text{total}} = E_{\text{k}} + E_{\text{p}}\). We are given \(E_{\text{k}} = 3 E_{\text{p}}\), so \(E_{\text{total}} = 3 E_{\text{p}} + E_{\text{p}} = 4 E_{\text{p}}\). Substituting the equations: \(\frac{1}{2} k A^2 = 4 \left(\frac{1}{2} k x^2\right) \implies A^2 = 4 x^2 \implies x = \pm \frac{A}{2}\).

[1] B - correct substitution of energy relations and solving for the displacement \(x\) in terms of amplitude \(A\).

According to Wien's displacement law, \(\lambda_{\text{max}} T = \text{constant}\), so temperature is inversely proportional to peak wavelength: \(T \propto \frac{1}{\lambda_{\text{max}}}\). Thus, \(\frac{T_{\text{X}}}{T_{\text{Y}}} = \frac{\lambda_{\text{Y}}}{\lambda_{\text{X}}} = \frac{800\text{ nm}}{400\text{ nm}} = 2\). According to Stefan-Boltzmann's law, luminosity is given by \(L = 4\pi r^2 \sigma T^4\), which means \(L \propto r^2 T^4\). Comparing the stars: \(\frac{L_{\text{X}}}{L_{\text{Y}}} = \left(\frac{r_{\text{X}}}{r_{\text{Y}}}\right)^2 \left(\frac{T_{\text{X}}}{T_{\text{Y}}}\right)^4 \implies 16 = \left(\frac{r_{\text{X}}}{r_{\text{Y}}}\right)^2 (2)^4 \implies 16 = 16 \left(\frac{r_{\text{X}}}{r_{\text{Y}}}\right)^2 \implies \frac{r_{\text{X}}}{r_{\text{Y}}} = 1\).

[1] A - correct application of Wien's law and Stefan-Boltzmann's law to determine the radius ratio.

The initial kinetic energy is given by:
\( E_k = \frac{1}{2} m v^2 \)

At the highest point of the trajectory, the vertical component of the velocity is zero, and only the horizontal component remains:
\( v_x = v \cos\theta \)

Therefore, the kinetic energy at the highest point is:
\( E_{k,\text{max}} = \frac{1}{2} m (v \cos\theta)^2 = \left(\frac{1}{2} m v^2\right) \cos^2\theta = E_k \cos^2\theta \)

We are given that:
\( E_{k,\text{max}} = \frac{1}{4} E_k \)

So,
\( \cos^2\theta = \frac{1}{4} \implies \cos\theta = \frac{1}{2} \)

Thus, the angle of projection is:
\( \theta = 60^\circ \)

1 mark: Correct option selected (C).

First, calculate the equivalent resistance of the parallel combination:
\( \frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{12.0} = \frac{3}{12.0} \implies R_p = 4.0\ \Omega \)

Next, calculate the total external resistance of the circuit:
\( R_{\text{ext}} = 4.5\ \Omega + 4.0\ \Omega = 8.5\ \Omega \)

Now, calculate the total resistance of the entire circuit including the internal resistance \( r \):
\( R_{\text{total}} = R_{\text{ext}} + r = 8.5\ \Omega + 1.5\ \Omega = 10.0\ \Omega \)

Calculate the current in the circuit:
\( I = \frac{\varepsilon}{R_{\text{total}}} = \frac{6.0\text{ V}}{10.0\ \Omega} = 0.60\text{ A} \)

Finally, calculate the terminal potential difference \( V \):
\( V = \varepsilon - Ir = 6.0\text{ V} - (0.60\text{ A} \times 1.5\ \Omega) = 6.0\text{ V} - 0.9\text{ V} = 5.1\text{ V} \)

1 mark: Correct option selected (C).

The photoelectric equation is:
\( E_{k,\text{max}} = \frac{hc}{\lambda} - \phi \)

When the wavelength is halved, the new wavelength is \( \lambda' = \frac{\lambda}{2} \). The new photon energy is:
\( E_{\text{photon}}' = \frac{hc}{\lambda'} = \frac{hc}{\lambda/2} = 2 \left(\frac{hc}{\lambda}\right) \)

Substituting this into the photoelectric equation for the new maximum kinetic energy \( E' \):
\( E' = 2 \left(\frac{hc}{\lambda}\right) - \phi \)

Since \( \frac{hc}{\lambda} = E_{k,\text{max}} + \phi \), we have:
\( E' = 2(E_{k,\text{max}} + \phi) - \phi = 2 E_{k,\text{max}} + \phi \)

Since the work function \( \phi \) is a positive value, \( 2 E_{k,\text{max}} + \phi > 2 E_{k,\text{max}} \). Therefore, the new maximum kinetic energy is increased to a value greater than twice \( E_{k,\text{max}} \).

1 mark: Correct option selected (D).

The radius of the circular path of a charged particle in a magnetic field is given by:
\( r = \frac{mv}{Bq} = \frac{p}{Bq} \)

Since the momentum \( p \) is related to kinetic energy \( E_k \) by \( p = \sqrt{2mE_k} \), the radius is:
\( r = \frac{\sqrt{2mE_k}}{Bq} \)

For a proton (mass \( m_p = m \), charge \( q_p = q \)), the radius is:
\( R = \frac{\sqrt{2mE_k}}{Bq} \)

An alpha particle consists of two protons and two neutrons, so its mass is approximately \( m_\alpha = 4m \) and its charge is \( q_\alpha = 2q \).

Substituting these values for the alpha particle with the same kinetic energy:
\( r_\alpha = \frac{\sqrt{2(4m)E_k}}{B(2q)} = \frac{2\sqrt{2mE_k}}{2Bq} = \frac{\sqrt{2mE_k}}{Bq} = R \)

1 mark: Correct option selected (B).

At the highest point of the vertical circle, both the weight of the car \( mg \) and the normal reaction force \( N \) of the track act vertically downwards towards the centre of the circular path.

The net centripetal force is:
\( F_c = N + mg = \frac{mv^2}{R} \)

We are given that the normal reaction force is twice the weight:
\( N = 2mg \)

Substituting this in the equation:
\( 2mg + mg = \frac{mv^2}{R} \implies 3mg = \frac{mv^2}{R} \implies v^2 = 3gR \implies v = \sqrt{3gR} \)

1 mark: Correct option selected (C).

Applying conservation laws for a strong interaction:
1. Charge conservation:
\( (-1) + (+1) = 0 + Q_X \implies Q_X = 0 \)

2. Baryon number conservation:
\( 0 + 1 = 0 + B_X \implies B_X = 1 \) (which indicates \( X \) is a baryon made of three quarks).

3. Strangeness conservation (since it is a strong interaction):
\( 0 + 0 = S_{K^0} + S_X \)
Since \( K^0 \) has quark structure \( d\bar{s} \), its strangeness is \( +1 \).
\( 0 = +1 + S_X \implies S_X = -1 \)

Baryon \( X \) must contain one strange quark (\( s \)), which has a charge of \( -1/3 \). To achieve a net charge of 0 and baryon number 1, the other two quarks must be one up quark (\( u \)) and one down quark (\( d \)), because:
\( Q(u) + Q(d) + Q(s) = \frac{2}{3} - \frac{1}{3} - \frac{1}{3} = 0 \)

Thus, the quark composition of \( X \) is \( uds \).

1 mark: Correct option selected (C).

The root-mean-square speed of ideal gas molecules is given by:
\( v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \)

This means the r.m.s. speed is directly proportional to the square root of the absolute temperature in Kelvin:
\( v_{\text{rms}} \propto \sqrt{T} \)

Convert the temperatures from Celsius to Kelvin:
\( T_1 = 27^\circ\text{C} = 27 + 273 = 300\text{ K} \)
\( T_2 = 327^\circ\text{C} = 327 + 273 = 600\text{ K} \)

Calculate the ratio of the final r.m.s. speed to the initial r.m.s. speed:
\( \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{600}{300}} = \sqrt{2} \approx 1.41 \)

1 mark: Correct option selected (B).

The fraction of active nuclei remaining undecayed after \( n \) half-lives is given by:
\( N_{\text{remaining}} = \left(\frac{1}{2}\right)^n \)

For \( n = 3 \) half-lives:
\( N_{\text{remaining}} = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \)

The fraction of the original active nuclei that have decayed is:
\( N_{\text{decayed}} = 1 - N_{\text{remaining}} = 1 - \frac{1}{8} = \frac{7}{8} \)

1 mark: Correct option selected (D).

1. Air resistance (or drag) acts on the football in the direction opposite to its motion. 2. The horizontal deceleration caused by air resistance reduces the horizontal velocity throughout the flight, which directly decreases the horizontal distance travelled. 3. The upward vertical motion is also opposed by air resistance, reducing the maximum height reached and the total time of flight, further reducing the range.

MP1: State that air resistance/drag acts on the football opposing its direction of motion. (1) MP2: Explain that the horizontal component of air resistance causes a horizontal deceleration (or reduces horizontal velocity). (1) MP3: Explain that air resistance reduces the time of flight (or reduces maximum height / causes a steeper angle of descent), leading to a smaller range. (1)

1. Rubber consists of long, tangled polymer chains. When stretched, these chains untangle and align. 2. During stretching and releasing, work is done to slide the chains past one another against intermolecular forces. Some of this energy is dissipated as thermal energy, raising the temperature of the rubber band. 3. The unloading curve of the stress-strain graph does not coincide with the loading curve (hysteresis). The delayed recovery to original length is due to the time required for chains to return to their random, tangled configurations, or some permanent deformation has occurred.

MP1: Reference to polymer chains untangling/sliding past each other when stretched. (1) MP2: Work is done against internal friction/intermolecular forces between chains, which is dissipated as thermal energy (making it warm). (1) MP3: Explain that energy is lost in a hysteresis cycle, meaning not all elastic potential energy is recovered, causing delayed or incomplete return to original length. (1)

1. An increase in light intensity causes the resistance of the LDR to decrease. 2. Since the LDR and the fixed resistor are in series, the total resistance of the circuit decreases, which increases the circuit current \( I \) (or the ratio of fixed resistance to total resistance increases). 3. Since \( V_{\text{out}} = I \times R \), and \( R \) is constant, the increased current results in an increased potential difference across the fixed resistor (or by potential divider ratio \( V_{\text{out}} = V \times \frac{R}{R + R_{\text{LDR}}} \), as \( R_{\text{LDR}} \) decreases, \( V_{\text{out}} \) increases).

MP1: State that the resistance of the LDR decreases as light intensity increases. (1) MP2: Explain that this decreases the total circuit resistance, thereby increasing the circuit current. (1) MP3: Conclude that the voltage across the fixed resistor increases, supported by \( V = IR \) or the potential divider formula. (1)

1. According to the photon model, electromagnetic radiation consists of discrete packets of energy called photons, where the energy of a single photon is \( E = hf \). 2. The photoelectric effect is a one-to-one interaction between a single photon and a single electron. If the frequency \( f \) is below the threshold frequency \( f_0 \), the energy of a photon \( hf \) is less than the work function \( \phi \) of the metal, so no electron can gain enough energy to escape, regardless of the number of photons (intensity). 3. Increasing the frequency increases the energy of each individual photon. Once \( hf > \phi \), each photon has sufficient energy to liberate an electron, resulting in immediate emission.

MP1: State that energy of a photon is proportional to frequency (\( E = hf \)) and light-matter interaction is a one-to-one collision between a photon and an electron. (1) MP2: Explain that if \( f < f_0 \), photon energy \( hf \) is less than the work function \( \phi \), so increasing intensity (more photons per second) still cannot provide enough energy for a single electron to escape. (1) MP3: Explain that increasing the frequency increases the photon energy above the work function, enabling electrons to escape immediately. (1)

1. The protons accelerate in the gaps between the drift tubes due to the electric field, so their speed increases as they progress along the accelerator. 2. Inside the drift tubes, there is no electric field (shielded), so the protons travel at a constant speed. 3. Since the frequency of the alternating voltage supply is constant, the time spent inside each drift tube must be constant (equal to half the period of the AC supply) to ensure that the electric field in the gaps is always in the correct accelerating direction when the protons emerge. Since speed \( v \) increases, the length \( d = v \times t \) must increase.

MP1: State that protons accelerate in the gaps and therefore their speed increases as they move along the LINAC. (1) MP2: State that the frequency/period of the alternating voltage is constant, so the time spent inside each drift tube must remain constant. (1) MP3: Relate speed and length (\( d = v \times t \)) to explain that since speed increases and time is constant, the tube length must increase. (1)

1. As the copper ring falls towards the magnet, there is a change in magnetic flux linkage through the ring. 2. According to Faraday's law, this change in flux induces an electromotive force (emf) in the ring, and since copper is a conductor, an induced current flows. 3. According to Lenz's law, the direction of the induced current is such that it opposes the change that created it. It produces a magnetic field with a North pole facing downwards to repel the approaching North pole of the magnet, resulting in an upward magnetic force on the ring.

MP1: Explain that as the ring falls, there is a change in magnetic flux linkage, inducing an emf (Faraday's Law) and causing a current to flow. (1) MP2: State Lenz's law (the direction of induced emf/current opposes the change in magnetic flux producing it). (1) MP3: Explain that the induced current creates a magnetic field that repels the magnet (e.g. creating a North pole at the bottom of the ring), which results in an upward force on the falling ring. (1)

1. An increase in temperature increases the average kinetic energy of the gas molecules, meaning their average speed increases. 2. Since the molecules are moving faster, they collide with the walls of the container more frequently (more collisions per unit time). 3. Each collision involves a larger change in momentum because the molecules strike the walls at higher speeds. Since force is the rate of change of momentum, the average force exerted on the walls increases, leading to an increase in pressure (\( P = F/A \)).

MP1: Link temperature to the average kinetic energy (or average speed) of the gas molecules. (1) MP2: State that the frequency of collisions between molecules and the container walls increases. (1) MP3: State that the change in momentum per collision is greater, so the total average force (and hence pressure) increases. (1)

1. A very heavy nucleus (like Uranium-235) has a lower binding energy per nucleon than the lighter, more stable nuclei produced during fission (which are closer to the peak of the binding energy per nucleon curve). 2. The total binding energy of the product nuclei is therefore greater than the binding energy of the original heavy reactant nucleus. 3. This increase in binding energy means that mass is converted into energy (according to \( \Delta E = \Delta m c^2 \)), which is released during the fission reaction.

MP1: State that the products of fission (medium-sized nuclei) have a higher binding energy per nucleon than the original heavy nucleus. (1) MP2: Explain that the total binding energy of the system increases after fission (or the product nuclei are more tightly bound). (1) MP3: Explain that this increase in binding energy results in a decrease in total mass, which is released as energy (according to \( E=mc^2 \) / mass defect). (1)

The magnetic force acting on a moving charged particle in a magnetic field is given by \(F = Bqv \sin\theta\). According to Fleming's Left Hand Rule, the direction of this force is always perpendicular to the velocity \(v\) of the particle. The work done by a force is given by \(W = Fd \cos\theta\), where \(\theta\) is the angle between the force and displacement. Since the force is perpendicular to the displacement at all times, \(\theta = 90^\circ\) and the work done on the particle by the magnetic field is zero. By the work-energy theorem, since no work is done, the kinetic energy of the particle remains constant, even though its direction of motion (velocity vector) is continually changed by the centripetal force.

MP1: State that the magnetic force is perpendicular to the velocity / direction of motion of the charged particle. (1)
MP2: State that since force is perpendicular to motion, the work done on the particle by the magnetic field is zero. (1)
MP3: Link zero work done to constant kinetic energy (as speed remains constant). (1)

When a skier moves over a circular hump, the forces acting on the skier are the downward gravitational force (weight, \(mg\)) and the upward normal contact force (\(R\)). The resultant downward force provides the necessary centripetal force for circular motion: \(mg - R = \frac{mv^2}{r}\). Rearranging this equation for the normal contact force gives \(R = m\left(g - \frac{v^2}{r}\right)\). As the speed \(v\) increases, the required centripetal force \(\frac{mv^2}{r}\) increases. This causes the normal contact force \(R\) to decrease. The skier loses contact with the snow when \(R = 0\), which is more likely to occur at higher speeds.

MP1: Identify the forces acting and write the correct circular motion equation: \(mg - R = \frac{mv^2}{r}\) (or equivalent). (1)
MP2: State that as speed \(v\) increases, the term \(\frac{mv^2}{r}\) increases, causing the normal contact force \(R\) to decrease. (1)
MP3: Explain that losing contact corresponds to \(R = 0\), which occurs when \(v = \sqrt{gr}\). (1)

An increase in temperature increases the average kinetic energy of the gas molecules, meaning they move with a higher average speed. Because the molecules are moving faster, they collide with the walls of the container more frequently. Additionally, each collision involves a larger change in momentum because the molecules strike the walls with greater velocity. Since force is defined as the rate of change of momentum, the average force exerted on the walls increases. Since pressure is force per unit area (\(P = F/A\)) and the volume (and therefore wall area) is constant, the overall pressure of the gas increases.

MP1: State that an increase in temperature increases the average kinetic energy and speed of the gas molecules. (1)
MP2: State that molecules collide with the container walls more frequently AND experience a greater change in momentum per collision. (1)
MP3: Relate this to an increased rate of change of momentum (force), and conclude that pressure increases as area is constant. (1)

According to the photon model of light, electromagnetic radiation consists of discrete packets of energy called photons, where the energy of a single photon is \(E = hf\). The interaction between a photon and an electron in the metal is a one-to-one process. If the frequency \(f\) of the light is below the threshold frequency \(f_0\), the energy of a single photon is less than the work function \(\Phi\) of the metal (\(hf < \Phi\)), meaning no single photon has enough energy to liberate an electron. Increasing the intensity of the light increases the number of photons arriving per second, but does not increase the energy of individual photons. Therefore, no electrons can escape, regardless of the intensity.

MP1: State that the interaction between photons and electrons is a one-to-one interaction, where photon energy is \(E = hf\). (1)
MP2: Explain that below the threshold frequency, the energy of an individual photon is less than the work function of the metal (\(hf < \Phi\)), so emission cannot occur. (1)
MP3: Explain that increasing the intensity only increases the number of photons per second, but not the energy of individual photons, so no emission occurs. (1)

When a rubber band is stretched, work is done to straighten and align its polymer chains. During unloading, the chains return to a disordered state, but some of the energy supplied during loading is dissipated as thermal energy due to internal friction between the sliding polymer chains. Consequently, less tension is exerted at any given extension during unloading, causing the unloading curve to lie below the loading curve. The area under the loading curve represents the work done in stretching the rubber, while the area under the unloading curve is the work done by the rubber as it relaxes. The difference—represented by the area between the curves—is the net energy converted into thermal energy (heat) during the cycle.

MP1: Explain that internal friction between polymer chains occurs during stretching/unstretching, causing energy dissipation. (1)
MP2: State that the area under the loading curve is work done on the rubber, and the area under the unloading curve is work done by the rubber. (1)
MP3: State that the area between the curves represents the net energy transferred to thermal energy (dissipated as heat) during the loading-unloading cycle. (1)

Alpha decay is a two-body decay process in which a parent nucleus decays into a daughter nucleus and an alpha particle. To satisfy the laws of conservation of energy and conservation of momentum simultaneously, the available decay energy must be shared in a fixed, unique ratio between the alpha particle and the recoiling daughter nucleus. Thus, the alpha particles always have discrete energies. In contrast, beta decay is a three-body decay process, yielding a daughter nucleus, a beta particle, and an antineutrino (or neutrino). Because there are three particles sharing the released momentum and kinetic energy, the kinetic energy can be shared among them in an infinite number of ways. Consequently, the beta particles exhibit a continuous range of energies up to a maximum limit.

MP1: Identify that alpha decay is a two-body decay (daughter nucleus and alpha particle), so conservation of energy and momentum restricts the alpha particle to discrete energy levels. (1)
MP2: Identify that beta decay is a three-body decay, consisting of the daughter nucleus, beta particle, and neutrino/antineutrino. (1)
MP3: Explain that the energy and momentum can be shared in varying proportions among the three products, resulting in a continuous range of beta energies. (1)

When the unbalanced drum spins, its asymmetric mass distribution generates a periodic driving force. The frequency of this force depends on the spin speed. When the spin speed reaches a value where the driving frequency matches the natural frequency of the washing machine's suspension system, resonance occurs. At resonance, there is maximum energy transfer from the drum to the frame, causing a dramatic increase in the amplitude of the vibrations. To minimize this effect, washing machines are designed with high-damping materials, such as heavy concrete weights (to lower the natural frequency by increasing mass) and shock absorbers (to dissipate kinetic energy as thermal energy, reducing the resonance amplitude).

MP1: Identify that the spinning unbalanced drum acts as a periodic driving force. (1)
MP2: Explain that resonance occurs when the driving frequency equals the natural frequency of the suspension system, resulting in maximum amplitude of vibration. (1)
MP3: State that heavy masses (to lower natural frequency) or shock absorbers/dampers (to dissipate energy) are used to minimize the resonance amplitude. (1)

A real battery possesses internal resistance, denoted by \(r\). The electromotive force (e.m.f.), \(\mathcal{E}\), is the total energy supplied per unit charge by the battery. When a current \(I\) flows through the battery, some of this potential difference is dropped across the internal resistance. This internal voltage drop is referred to as 'lost volts' and is given by \(Ir\). The terminal potential difference, \(V\), is the potential difference available to the external circuit, represented by the equation \(V = \mathcal{E} - Ir\). Since the e.m.f. \(\mathcal{E}\) and internal resistance \(r\) are constant, an increase in current \(I\) leads to an increase in the lost volts (\(Ir\)), which in turn decreases the terminal potential difference \(V\).

MP1: Identify that a real battery has an internal resistance, \(r\). (1)
MP2: State or use the relationship \(V = \mathcal{E} - Ir\) (or explain that 'lost volts' occur across the internal resistance). (1)
MP3: Explain that as current \(I\) increases, the lost volts (\(Ir\)) increase, causing the terminal potential difference \(V\) to decrease. (1)

When the temperature of the gas is raised, the average kinetic energy of the gas molecules increases, which means their mean square speed increases. Because the molecules are moving faster, they collide with the walls of the container more frequently. Additionally, each collision involves a larger change in momentum because the molecules are moving faster. This results in a larger rate of change of momentum, and hence a larger average force exerted on the walls. Since pressure is force per unit area, the pressure increases.

MP1: States that an increase in temperature increases the average kinetic energy or speed of the molecules. (1) MP2: States that this leads to a higher frequency of collisions with the walls of the container. (1) MP3: States that there is a larger change in momentum per collision, leading to an increased average force on the walls and hence greater pressure. (1)

As the ring falls, the magnetic flux passing through the ring increases, which induces an emf and hence a current in the ring. According to Lenz's law, the direction of the induced current will oppose the change in flux, creating a magnetic field that repels the magnet. This results in an upward magnetic force on the falling ring. According to Newton's third law, if the magnet exerts an upward force on the ring, the ring must exert an equal and opposite downward force on the magnet.

MP1: Identifies that the falling ring experiences a change in magnetic flux linkage, which induces an emf or current in the ring. (1) MP2: Uses Lenz's law to explain that the induced current opposes the motion, resulting in an upward repulsive force on the ring. (1) MP3: Uses Newton's third law to state that the ring exerts an equal and opposite downward force on the magnet. (1)

(a) The thrust force \( F \) is given by:
\( F = \frac{\Delta m}{\Delta t} v = 0.12\text{ kg s}^{-1} \times 140\text{ m s}^{-1} = 16.8\text{ N} \)

The initial weight of the rocket is:
\( W = mg = 0.65\text{ kg} \times 9.81\text{ m s}^{-2} = 6.38\text{ N} \)

The resultant force \( F_{\text{net}} \) is:
\( F_{\text{net}} = F - W = 16.8\text{ N} - 6.38\text{ N} = 10.42\text{ N} \)

The initial acceleration \( a_0 \) is:
\( a_0 = \frac{F_{\text{net}}}{m} = \frac{10.42\text{ N}}{0.65\text{ kg}} \approx 16.0\text{ m s}^{-2} \)

(b) After \( 1.5\text{ s} \), the mass of the rocket has decreased due to fuel consumption:
\( m = 0.65\text{ kg} - (0.12\text{ kg s}^{-1} \times 1.5\text{ s}) = 0.47\text{ kg} \)

The new weight of the rocket is:
\( W' = 0.47\text{ kg} \times 9.81\text{ m s}^{-2} = 4.61\text{ N} \)

The new resultant force is:
\( F_{\text{net}}' = 16.8\text{ N} - 4.61\text{ N} = 12.19\text{ N} \)

The acceleration is:
\( a = \frac{12.19\text{ N}}{0.47\text{ kg}} \approx 25.9\text{ m s}^{-2} \)

For part (a):
- Use of \( F = v \frac{\Delta m}{\Delta t} \) to calculate thrust of \( 16.8\text{ N} \) (1)
- Use of \( F - mg = ma \) with correct initial mass (1)
- Value of \( 16.0\text{ m s}^{-2} \) (or \( 16\text{ m s}^{-2} \)) (1)

For part (b):
- Correct calculation of new mass \( 0.47\text{ kg} \) (1)
- Correct calculation of acceleration \( 25.9\text{ m s}^{-2} \) (accept \( 26\text{ m s}^{-2} \)) (1)

(a) The current in the circuit is given by:
\( I = \frac{E}{R + r} \)

The power dissipated in the variable resistor is:
\( P = I^2 R = \frac{E^2 R}{(R + r)^2} \)

To find the maximum power, we differentiate \( P \) with respect to \( R \) and set it to zero:
\( \frac{dP}{dR} = E^2 \left[ \frac{(R + r)^2 \cdot 1 - R \cdot 2(R + r)}{(R + r)^4} \right] = 0 \)
\( (R + r)^2 - 2R(R + r) = 0 \)
\( (R + r)(R + r - 2R) = 0 \)
Since \( R + r \neq 0 \), we have:
\( r - R = 0 \implies R = r \)

(b) Substituting \( R = r \) into the power equation:
\( P_{\text{max}} = \frac{E^2 (r)}{(r + r)^2} = \frac{E^2}{4r} \)

Using the given values:
\( P_{\text{max}} = \frac{(6.0\text{ V})^2}{4 \times 1.5\ \Omega} = \frac{36}{6.0} = 6.0\text{ W} \)

For part (a):
- Expression for \( I \) or \( P \) in terms of \( E, R, r \) (1)
- Differentiation or algebraic manipulation to set up maximum condition (e.g., setting derivative to 0 or completing the square) (1)
- Clear algebraic steps leading to \( R = r \) (1)

For part (b):
- Identification of \( P_{\text{max}} = \frac{E^2}{4r} \) or calculation of current \( I = 2.0\text{ A} \) (1)
- Correct value of \( 6.0\text{ W} \) with unit (1)

(a) The decay constant \( \lambda \) is calculated using:
\( \lambda = \frac{\ln 2}{T_{1/2}} \)

Convert half-life to seconds:
\( T_{1/2} = 4.2\text{ hours} \times 3600\text{ s hour}^{-1} = 15120\text{ s} \)

\( \lambda = \frac{0.6931}{15120\text{ s}} \approx 4.58 \times 10^{-5}\text{ s}^{-1} \)

(b) The activity \( A \) after a time \( t \) is given by:
\( A = A_0 e^{-\lambda t} \)
where the initial activity is:
\( A_0 = \lambda N_0 = (4.584 \times 10^{-5}\text{ s}^{-1}) \times (2.5 \times 10^{18}) = 1.146 \times 10^{14}\text{ Bq} \)

For \( t = 10\text{ hours} = 36000\text{ s} \):
\( A = (1.146 \times 10^{14}\text{ Bq}) \times e^{-(4.584 \times 10^{-5} \times 36000)} = (1.146 \times 10^{14}\text{ Bq}) \times e^{-1.65} \approx 2.2 \times 10^{13}\text{ Bq} \)

Alternatively, using the half-life directly:
\( N = N_0 \times (0.5)^{\frac{t}{T_{1/2}}} = 2.5 \times 10^{18} \times (0.5)^{\frac{10}{4.2}} = 4.8 \times 10^{17}\text{ nuclei} \)
\( A = \lambda N = 4.584 \times 10^{-5}\text{ s}^{-1} \times 4.8 \times 10^{17} \approx 2.2 \times 10^{13}\text{ Bq} \)

For part (a):
- Conversion of half-life to seconds (1)
- Correct value of \( \lambda = 4.6 \times 10^{-5}\text{ s}^{-1} \) (accept \( 4.58 \times 10^{-5}\text{ s}^{-1} \)) (1)

For part (b):
- Use of \( A = \lambda N \) to find initial activity OR use of \( N = N_0 (0.5)^{t/T_{1/2}} \) (1)
- Substitution of \( t = 36000\text{ s} \) (or \( 10\text{ hours} \)) (1)
- Correct final activity \( 2.2 \times 10^{13}\text{ Bq} \) (accept range \( 2.15 \times 10^{13}\text{ Bq} \) to \( 2.25 \times 10^{13}\text{ Bq} \)) (1)

(a) At the highest point, the forces providing centripetal acceleration are Tension \( T_{\text{top}} \) and weight \( mg \), both acting downwards:
\( T_{\text{top}} + mg = \frac{m v_{\text{top}}^2}{r} \)

Substitute the values:
\( 1.2\text{ N} + (0.20\text{ kg} \times 9.81\text{ m s}^{-2}) = \frac{0.20\text{ kg} \times v_{\text{top}}^2}{0.80\text{ m}} \)
\( 1.2 + 1.962 = 0.25 v_{\text{top}}^2 \)
\( 3.162 = 0.25 v_{\text{top}}^2 \implies v_{\text{top}}^2 = 12.65\text{ m}^2\text{ s}^{-2} \)
\( v_{\text{top}} = \sqrt{12.65} \approx 3.56\text{ m s}^{-1} \)

(b) By conservation of energy, the gain in kinetic energy equals the loss in gravitational potential energy from top to bottom:
\( \frac{1}{2} m v_{\text{bottom}}^2 - \frac{1}{2} m v_{\text{top}}^2 = m g (2r) \)
\( v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4 g r \)
\( v_{\text{bottom}}^2 = 12.65 + (4 \times 9.81 \times 0.80) = 12.65 + 31.39 = 44.04\text{ m}^2\text{ s}^{-2} \)

At the lowest point, the tension \( T_{\text{bottom}} \) acts upwards and weight \( mg \) acts downwards:
\( T_{\text{bottom}} - mg = \frac{m v_{\text{bottom}}^2}{r} \)
\( T_{\text{bottom}} = mg + \frac{m v_{\text{bottom}}^2}{r} = 1.962\text{ N} + \frac{0.20\text{ kg} \times 44.04\text{ m}^2\text{ s}^{-2}}{0.80\text{ m}} = 1.962 + 11.01 = 12.97\text{ N} \approx 13.0\text{ N} \)

Alternatively, using the standard derivation \( T_{\text{bottom}} = T_{\text{top}} + 6mg \):
\( T_{\text{bottom}} = 1.2\text{ N} + 6(0.20\text{ kg} \times 9.81\text{ m s}^{-2}) = 1.2 + 11.77 = 12.97\text{ N} \approx 13.0\text{ N} \)

For part (a):
- Use of \( T + mg = \frac{mv^2}{r} \) at the top (1)
- Correct substitution of values (1)
- Value of \( 3.56\text{ m s}^{-1} \) (1)

For part (b):
- Use of conservation of energy to find \( v_{\text{bottom}}^2 \) OR use of the relation \( T_{\text{bottom}} = T_{\text{top}} + 6mg \) (1)
- Correct value of \( T_{\text{bottom}} = 13.0\text{ N} \) (accept \( 12.97\text{ N} \) or \( 13\text{ N} \)) (1)

(a) The electrical work done on the alpha particle by the potential difference \( V \) is:
\( W = q V = (2e) V = 2eV \)

This work is entirely converted into kinetic energy:
\( E_k = \frac{1}{2} m v^2 \)

Equating the two:
\( \frac{1}{2} m v^2 = 2eV \implies v^2 = \frac{4eV}{m} \implies v = \sqrt{\frac{4eV}{m}} \)

(b) For circular motion in a magnetic field, the magnetic force provides the centripetal force:
\( B q v = \frac{m v^2}{R} \implies v = \frac{q B R}{m} \)

Substitute \( q = 2e \):
\( v = \frac{2e B R}{m} \)

Equating both expressions for \( v^2 \):
\( \frac{4eV}{m} = \frac{4 e^2 B^2 R^2}{m^2} \implies V = \frac{e B^2 R^2}{m} \)

Substitute the values:
\( V = \frac{(1.60 \times 10^{-19}\text{ C}) \times (0.35\text{ T})^2 \times (0.12\text{ m})^2}{6.64 \times 10^{-27}\text{ kg}} \)
\( V = \frac{1.60 \times 10^{-19} \times 0.1225 \times 0.0144}{6.64 \times 10^{-27}} = \frac{2.8224 \times 10^{-22}}{6.64 \times 10^{-27}} \approx 4.25 \times 10^4\text{ V} \approx 43\text{ kV} \)

For part (a):
- Equating work done \( qV \) to kinetic energy \( \frac{1}{2} m v^2 \) with \( q = 2e \) (1)
- Correct algebraic rearrangement to show \( v = \sqrt{\frac{4eV}{m}} \) (1)

For part (b):
- Equating magnetic force to centripetal force to obtain \( r = \frac{mv}{Bq} \) or \( v = \frac{qBR}{m} \) (1)
- Correct algebraic combination to find \( V = \frac{e B^2 R^2}{m} \) or separate step-by-step calculation of velocity and potential difference (1)
- Final value of \( 4.3 \times 10^4\text{ V} \) (accept range \( 4.2 \times 10^4\text{ V} \) to \( 4.3 \times 10^4\text{ V} \)) (1)

(a) The angular frequency \( \omega \) is given by:
\( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{45\text{ N m}^{-1}}{0.35\text{ kg}}} = \sqrt{128.57} \approx 11.34\text{ rad s}^{-1} \)

The maximum velocity \( v_{\text{max}} \) is:
\( v_{\text{max}} = \omega A = 11.34\text{ rad s}^{-1} \times 0.050\text{ m} = 0.567\text{ m s}^{-1} \approx 0.57\text{ m s}^{-1} \)

(b) Since the mass is released from rest at \( t = 0 \), the displacement \( x \) from equilibrium is described by:
\( x = A \cos(\omega t) \)

When the mass has traveled \( 2.5\text{ cm} \) from its release point (which was at \( x = 5.0\text{ cm} \)), its new displacement from equilibrium is:
\( x = 5.0\text{ cm} - 2.5\text{ cm} = 2.5\text{ cm} \)

Therefore:
\( 2.5 = 5.0 \cos(\omega t) \)
\( \cos(\omega t) = 0.5 \)
\( \omega t = \arccos(0.5) = \frac{\pi}{3}\text{ rad} \approx 1.047\text{ rad} \)

Using \( \omega = 11.34\text{ rad s}^{-1} \):
\( t = \frac{1.047}{11.34} \approx 0.0924\text{ s} \)

For part (a):
- Calculation of \( \omega = 11.3\text{ rad s}^{-1} \) (1)
- Calculation of \( v_{\text{max}} = \omega A \) to show \( 0.57\text{ m s}^{-1} \) (1)

For part (b):
- Correct displacement \( x = 2.5\text{ cm} \) or use of cosine function (1)
- Solving \( \cos(\omega t) = 0.5 \) to find phase angle \( \frac{\pi}{3} \) or \( 1.05\text{ rad} \) (1)
- Correct time of \( 0.092\text{ s} \) (1)

(a) Using the formula for Young Modulus:
\( E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \implies \Delta L = \frac{F L}{A E} \)

The force \( F \) is the weight of the mass:
\( F = m g = 8.0\text{ kg} \times 9.81\text{ m s}^{-2} = 78.48\text{ N} \)

\( \Delta L = \frac{78.48\text{ N} \times 2.2\text{ m}}{(1.5 \times 10^{-6}\text{ m}^2) \times (2.0 \times 10^{11}\text{ Pa})} \)
\( \Delta L = \frac{172.656}{3.0 \times 10^5} \approx 5.76 \times 10^{-4}\text{ m} = 0.58\text{ mm} \)

(b) The elastic strain energy \( E_{\text{el}} \) stored in the wire is:
\( E_{\text{el}} = \frac{1}{2} F \Delta L = \frac{1}{2} \times 78.48\text{ N} \times 5.76 \times 10^{-4}\text{ m} \approx 0.0226\text{ J} = 2.26 \times 10^{-2}\text{ J} \)

The volume \( V \) of the wire is:
\( V = A \times L = 1.5 \times 10^{-6}\text{ m}^2 \times 2.2\text{ m} = 3.3 \times 10^{-6}\text{ m}^3 \)

The energy density \( u \) is:
\( u = \frac{E_{\text{el}}}{V} = \frac{0.0226\text{ J}}{3.3 \times 10^{-6}\text{ m}^3} \approx 6848\text{ J m}^{-3} \approx 6.8 \times 10^3\text{ J m}^{-3} \)

For part (a):
- Use of \( \Delta L = \frac{FL}{AE} \) with \( F = mg \) (1)
- Correct extension value \( 5.8 \times 10^{-4}\text{ m} \) (accept \( 5.76 \times 10^{-4}\text{ m} \)) (1)

For part (b):
- Correct calculation of elastic strain energy \( 2.3 \times 10^{-2}\text{ J} \) (accept range \( 2.2 \times 10^{-2}\text{ J} \) to \( 2.3 \times 10^{-2}\text{ J} \)) (1)
- Use of volume \( V = AL \) to set up energy density calculation (1)
- Correct value of energy density \( 6.8 \times 10^3\text{ J m}^{-3} \) (accept \( 6800\text{ J m}^{-3} \) to \( 6900\text{ J m}^{-3} \)) (1)

(a) The energy of each incident photon is:
\( E = \frac{h c}{\lambda} = \frac{(6.63 \times 10^{-34}\text{ J s}) \times (3.00 \times 10^8\text{ m s}^{-1})}{410 \times 10^{-9}\text{ m}} = 4.85 \times 10^{-19}\text{ J} \)

Convert the work function from eV to Joules:
\( \Phi = 2.28\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.65 \times 10^{-19}\text{ J} \)

According to Einstein's photoelectric equation:
\( E_k = E - \Phi = 4.85 \times 10^{-19}\text{ J} - 3.65 \times 10^{-19}\text{ J} = 1.20 \times 10^{-19}\text{ J} \)

(b) The momentum \( p \) of the photoelectrons is related to kinetic energy by:
\( p = \sqrt{2 m E_k} \)

Using the mass of an electron \( m = 9.11 \times 10^{-31}\text{ kg} \):
\( p = \sqrt{2 \times (9.11 \times 10^{-31}\text{ kg}) \times (1.20 \times 10^{-19}\text{ J})} = \sqrt{2.186 \times 10^{-49}} \approx 4.68 \times 10^{-25}\text{ kg m s}^{-1} \)

The de Broglie wavelength is:
\( \lambda_{\text{dB}} = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{4.68 \times 10^{-25}\text{ kg m s}^{-1}} \approx 1.42 \times 10^{-9}\text{ m} \approx 1.4\text{ nm} \)

For part (a):
- Correct calculation of photon energy in Joules (1)
- Conversion of work function into Joules and subtraction to find \( E_k \) (1)
- Correct maximum kinetic energy of \( 1.2 \times 10^{-19}\text{ J} \) (accept range \( 1.15 \times 10^{-19}\text{ J} \) to \( 1.25 \times 10^{-19}\text{ J} \)) (1)

For part (b):
- Correct use of \( p = \sqrt{2mE_k} \) or calculating velocity first (1)
- Correct de Broglie wavelength \( 1.4 \times 10^{-9}\text{ m} \) (accept range \( 1.38 \times 10^{-9}\text{ m} \) to \( 1.44 \times 10^{-9}\text{ m} \)) (1)

1. Express the horizontal and vertical displacements in terms of time \(t\):
\(x = u \cos\theta \cdot t\) and \(y = u \sin\theta \cdot t - \frac{1}{2}gt^2\).

2. Substitute \(t = \frac{x}{u \cos\theta}\) into the vertical equation:
\(y = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta}\).

3. Rearrange the equation to solve for \(u^2\):
\(u^2 = \frac{g x^2}{2 \cos^2\theta (x \tan\theta - y)}\).

4. Substitute the given values (\(x = 45.0\text{ m}\), \(y = 12.0\text{ m}\), \(\theta = 35.0^\circ\), \(g = 9.81\text{ m s}^{-2}\)):
\(u^2 = \frac{9.81 \times 45.0^2}{2 \cos^2(35.0^\circ) (45.0 \tan(35.0^\circ) - 12.0)}\).

5. Perform the calculation:
\(45.0 \tan(35.0^\circ) \approx 31.51\text{ m}\).

\(31.51 - 12.0 = 19.51\text{ m}\).

\(2 \cos^2(35.0^\circ) \approx 2 \times 0.6710 = 1.342\).

\(u^2 = \frac{19865.25}{1.342 \times 19.51} = \frac{19865.25}{26.18} \approx 758.75\text{ m}^2\text{ s}^{-2}\).

\(u = \sqrt{758.75} \approx 27.5\text{ m s}^{-1}\).

• Use of horizontal and vertical displacement equations to eliminate time \(t\) to find projectile path equation (1 mark)
• Algebraic rearrangement to express \(u\) or \(u^2\) as the subject (1 mark)
• Substitution of correct values into the rearranged equation (1 mark)
• Correct calculation of intermediate terms, such as \(45.0 \tan(35.0^\circ) = 31.5\) (1 mark)
• Correct final value of \(u = 27.5\text{ m s}^{-1}\) (or \(27.5\text{ m s}^{-1}\) to 3 s.f.) (1 mark)

(a) Derivation:
1. The total resistance of the circuit is \(R + r\).
2. The current \(I\) in the circuit is \(I = \frac{E}{R+r}\).
3. The power \(P\) dissipated in the external resistor is \(P = I^2 R\).
4. Substituting \(I\) gives: \(P = \left(\frac{E}{R+r}\right)^2 R = \frac{E^2 R}{(R+r)^2}\).

(b) Calculation:
1. Substitute the known values into the derived equation:
\(4.0 = \frac{6.0^2 R}{(R + 2.0)^2}\)

2. Expand and simplify the equation:
\(4.0(R^2 + 4.0R + 4.0) = 36.0R\)

\(4.0R^2 + 16.0R + 16.0 = 36.0R\)

\(4.0R^2 - 20.0R + 16.0 = 0\)

3. Divide by 4.0 to form a simple quadratic equation:
\(R^2 - 5.0R + 4.0 = 0\)

4. Factorise the quadratic equation:
\((R - 4.0)(R - 1.0) = 0\)

5. Thus, the two possible values of \(R\) are \(R = 1.0\,\Omega\) and \(R = 4.0\,\Omega\).

• Part (a): Expresses current in terms of \(E\), \(r\) and \(R\) AND relates power to \(I^2 R\) (1 mark)
• Part (a): Combines equations correctly to show \(P = \frac{E^2 R}{(R+r)^2}\) (1 mark)
• Part (b): Substitutes values and forms a correct quadratic equation, e.g., \(4R^2 - 20R + 16 = 0\) (1 mark)
• Part (b): Factorises or solves the quadratic equation correctly (1 mark)
• Part (b): States both solutions with correct units: \(1.0\,\Omega\) and \(4.0\,\Omega\) (1 mark)

(a) Derivation:
1. The maximum force of static friction providing the centripetal force is \(F_{\text{f}} = \mu R = \mu mg\).
2. The centripetal force required to keep the block in circular motion is \(F_{\text{c}} = \frac{mv^2}{r}\).
3. For no slipping, \(F_{\text{c}} \le F_{\text{f}}\), so \(\frac{mv^2}{r} \le \mu mg\).
4. Thus, \(v^2 \le \mu g r \implies v = \sqrt{\mu g r}\) at the threshold of sliding.

(b) Calculation:
1. The relationship between linear speed \(v\) and angular velocity \(\omega\) is \(v = \omega r\).
2. Substituting \(v\) into the derivation: \(\omega r = \sqrt{\mu g r} \implies \omega = \sqrt{\frac{\mu g}{r}}\).
3. Substitute the values: \(\omega = \sqrt{\frac{0.35 \times 9.81}{0.40}}\).
4. \(\omega = \sqrt{8.58375} \approx 2.93\text{ rad s}^{-1}\).

• Part (a): Identifies that maximum frictional force is \(\mu mg\) (1 mark)
• Part (a): Equates centripetal force \(\frac{mv^2}{r}\) to \(\mu mg\) and simplifies to find \(v = \sqrt{\mu g r}\) (1 mark)
• Part (b): Uses \(v = \omega r\) to express \(\omega = \sqrt{\frac{\mu g}{r}}\) or calculates \(v = 1.17\text{ m s}^{-1}\) first (1 mark)
• Part (b): Substitutes numerical values correctly (1 mark)
• Part (b): Correct value for \(\omega = 2.9\text{ rad s}^{-1}\) (accept \(2.93\text{ rad s}^{-1}\)) (1 mark)

(a) Derivation and Calculation:
1. The work done on the proton by the electric field is \(W = eV\).
2. This is equal to the gain in kinetic energy: \(eV = \frac{1}{2} m_{\text{p}} v^2\).
3. Rearranging for \(v\): \(v = \sqrt{\frac{2eV}{m_{\text{p}}}}\).
4. Substitute values: \(v = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 2500}{1.67 \times 10^{-27}}} = \sqrt{4.79 \times 10^{11}} \approx 6.92 \times 10^5\text{ m s}^{-1}\).

(b) Calculation:
1. The magnetic force provides the centripetal force: \(B e v = \frac{m_{\text{p}} v^2}{R}\).
2. Solve for \(B\): \(B = \frac{m_{\text{p}} v}{e R}\).
3. Substitute values: \(B = \frac{1.67 \times 10^{-27} \times 6.92 \times 10^5}{1.60 \times 10^{-19} \times 0.12}\).
4. \(B = 6.02 \times 10^{-2}\text{ T}\) (or \(6.0 \times 10^{-2}\text{ T}\)).

• Part (a): Equates electrical work done \(eV\) to kinetic energy \(\frac{1}{2}mv^2\) to show the formula (1 mark)
• Part (a): Correct substitution of values into the speed formula (1 mark)
• Part (a): Correct value for \(v = 6.9 \times 10^5\text{ m s}^{-1}\) (or \(6.92 \times 10^5\text{ m s}^{-1}\)) (1 mark)
• Part (b): Equates magnetic force to centripetal force to obtain \(B = \frac{mv}{eR}\) (1 mark)
• Part (b): Correct value for \(B = 6.0 \times 10^{-2}\text{ T}\) (or \(6.02 \times 10^{-2}\text{ T}\)) (1 mark)

(a) Derivation:
1. By definition, when the time \(t = T_{1/2}\), the number of undecayed nuclei is \(N = \frac{N_0}{2}\).
2. Substitute these into the decay law: \(\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}}\).
3. Divide both sides by \(N_0\): \(\frac{1}{2} = e^{-\lambda T_{1/2}}\).
4. Take the natural logarithm of both sides: \(\ln(0.5) = -\lambda T_{1/2} \implies -\ln(2) = -\lambda T_{1/2} \implies \lambda = \frac{\ln(2)}{T_{1/2}}\).

(b) Calculation:
1. Calculate the decay constant \(\lambda\) in terms of days: \(\lambda = \frac{\ln(2)}{5.0\text{ days}} = 0.1386\text{ day}^{-1}\).
2. Use the activity decay law \(A = A_0 e^{-\lambda t}\) with \(t = 12.0\text{ days}\):
\(A = 8.0 \times 10^6 \times e^{-0.1386 \times 12.0}\).
3. Evaluate the exponent: \(A = 8.0 \times 10^6 \times e^{-1.663} = 8.0 \times 10^6 \times 0.1895 \approx 1.52 \times 10^6\text{ Bq}\) (or \(1.5 \times 10^6\text{ Bq}\)).

• Part (a): Sets \(N = \frac{N_0}{2}\) at \(t = T_{1/2}\) (1 mark)
• Part (a): Takes natural logarithms correctly to show \(\lambda = \frac{\ln(2)}{T_{1/2}}\) (1 mark)
• Part (b): Calculates decay constant \(\lambda = 0.139\text{ day}^{-1}\) or converts to seconds correctly (1 mark)
• Part (b): Uses decay equation with activity \(A = A_0 e^{-\lambda t}\) or fractional half-lives \(A = A_0 (0.5)^{t/T_{1/2}}\) (1 mark)
• Part (b): Correct value for \(A = 1.5 \times 10^6\text{ Bq}\) (accept \(1.52 \times 10^6\text{ Bq}\)) (1 mark)

(a) Derivation:
1. The density of the gas is \(\rho = \frac{N m}{V}\), where \(N\) is the number of molecules and \(m\) is the mass of one molecule.
2. Substitute \(\rho\) into the kinetic theory equation: \(P = \frac{1}{3} \frac{N m}{V} \langle c^2 \rangle \implies P V = \frac{1}{3} N m \langle c^2 \rangle\).
3. The average translational kinetic energy of a single molecule is \(E_{\text{mean}} = \frac{1}{2} m \langle c^2 \rangle\).
4. The total kinetic energy of \(N\) molecules is \(E_{\text{k}} = N \times \left(\frac{1}{2} m \langle c^2 \rangle\right) = \frac{1}{2} N m \langle c^2 \rangle\).
5. Substitute \(N m \langle c^2 \rangle = 2 E_{\text{k}}\) into the equation for \(P V\):
\(P V = \frac{1}{3} (2 E_{\text{k}}) \implies E_{\text{k}} = \frac{3}{2} P V\).

(b) Calculation:
1. Substitute the given values into the derived equation:
\(E_{\text{k}} = \frac{3}{2} \times (1.5 \times 10^5\text{ Pa}) \times (2.4 \times 10^{-3}\text{ m}^3)\).
2. \(E_{\text{k}} = 1.5 \times 360 = 540\text{ J}\).

• Part (a): Expresses density as \(\rho = \frac{Nm}{V}\) and writes \(PV = \frac{1}{3}Nm\langle c^2 \rangle\) (1 mark)
• Part (a): Identifies total kinetic energy as \(E_{\text{k}} = \frac{1}{2}Nm\langle c^2 \rangle\) (1 mark)
• Part (a): Substitutes correctly to show \(E_{\text{k}} = \frac{3}{2}PV\) (1 mark)
• Part (b): Substitutes values of \(P\) and \(V\) into the energy expression (1 mark)
• Part (b): Correct value for \(E_{\text{k}} = 540\text{ J}\) (1 mark)

(a) Derivation and Calculation:
1. For SHM, the restoring force is \(F = -kx\). By Newton's second law, \(ma = -kx \implies a = -\frac{k}{m} x\).
2. Comparing with the defining equation of SHM \(a = -\omega^2 x\), we have \(\omega^2 = \frac{k}{m}\).
3. Maximum acceleration occurs at maximum displacement \(x = A\), so \(a_{\text{max}} = \omega^2 A\).
4. Calculate \(\omega^2 = \frac{40}{0.25} = 160\text{ s}^{-2}\).
5. Calculate \(a_{\text{max}} = 160 \times 0.080 = 12.8\text{ m s}^{-2}\).

(b) Calculation:
1. The velocity is given by \(v = \pm \omega \sqrt{A^2 - x^2}\).
2. Find \(\omega = \sqrt{160} \approx 12.65\text{ rad s}^{-1}\).
3. Substitute values: \(v = \pm 12.65 \times \sqrt{0.080^2 - 0.050^2} = \pm 12.65 \times \sqrt{0.0039}\).
4. \(v = \pm 12.65 \times 0.06245 = \pm 0.79\text{ m s}^{-1}\).

• Part (a): Relates restoring force and acceleration to show \(a_{\text{max}} = \frac{k}{m}A = \omega^2 A\) (1 mark)
• Part (a): Correct calculation of \(a_{\text{max}} = 13\text{ m s}^{-2}\) (or \(12.8\text{ m s}^{-2}\)) (1 mark)
• Part (b): Identifies and uses the SHM velocity-displacement formula (1 mark)
• Part (b): Correct substitution of \(\omega\), \(A\), and \(x\) (1 mark)
• Part (b): Correct value for velocity \(v = 0.79\text{ m s}^{-1}\) (or \(\pm 0.79\text{ m s}^{-1}\)) (1 mark)

(a) Derivation:
1. From Einstein's photoelectric equation: \(hf = \Phi + E_{\text{k,max}}\).
2. Since the wave speed is \(c = f\lambda\), the frequency is \(f = \frac{c}{\lambda}\).
3. Substituting this gives: \(\frac{hc}{\lambda} = \Phi + E_{\text{k,max}} \implies E_{\text{k,max}} = \frac{hc}{\lambda} - \Phi\).

(b) Calculation:
1. Convert work function to Joules: \(\Phi = 2.10 \times 1.60 \times 10^{-19} = 3.36 \times 10^{-19}\text{ J}\).
2. Calculate photon energy: \(E_{\text{photon}} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{450 \times 10^{-9}} = 4.42 \times 10^{-19}\text{ J}\).
3. Calculate maximum kinetic energy: \(E_{\text{k,max}} = 4.42 \times 10^{-19} - 3.36 \times 10^{-19} = 1.06 \times 10^{-19}\text{ J}\).
4. Relate kinetic energy to momentum: \(p = \sqrt{2 m_{\text{e}} E_{\text{k,max}}}\).
\(p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.06 \times 10^{-19}} = \sqrt{1.931 \times 10^{-49}} = 4.39 \times 10^{-25}\text{ kg m s}^{-1}\).
5. Calculate de Broglie wavelength: \(\lambda_{\text{dB}} = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{4.39 \times 10^{-25}} = 1.51 \times 10^{-9}\text{ m}\) (or \(1.51\text{ nm}\)).

• Part (a): Identifies photoelectric equation and wave speed equation correctly to derive expression (1 mark)
• Part (b): Calculates \(E_{\text{k,max}}\) in Joules correctly as \(1.06 \times 10^{-19}\text{ J}\) (1 mark)
• Part (b): Relates momentum to kinetic energy, \(p = \sqrt{2mE_k}\) (1 mark)
• Part (b): Substitutes values into the de Broglie equation \(\lambda = \frac{h}{p}\) (1 mark)
• Part (b): Correct value for de Broglie wavelength \(\lambda_{\text{dB}} = 1.5 \times 10^{-9}\text{ m}\) (accept \(1.51 \times 10^{-9}\text{ m}\)) (1 mark)

Upward forces are upthrust and drag (viscous drag), while the downward force is weight. At the instant of release (\( t = 0 \)), velocity is zero, so drag is zero. The resultant force is downwards (\( W - U \)), producing an initial downward acceleration. As the sphere's velocity increases, the upward drag force increases (\( D \propto v \) or \( D \propto v^2 \)). Since weight and upthrust remain constant, the resultant downward force (\( W - U - D \)) decreases. According to Newton's second law (\( F = ma \)), the acceleration of the sphere decreases, although it is still speeding up. Eventually, the drag force increases to a value where the sum of drag and upthrust equals the weight (\( U + D = W \)). The resultant force becomes zero, so the acceleration becomes zero. The sphere has reached a constant maximum velocity, known as the terminal velocity.

Indicative content:
1. Identifies the three forces acting on the sphere: weight acting downwards, upthrust acting upwards, and drag (or viscous resistance) acting upwards.
2. At the moment of release, velocity is zero, so drag is zero; the resultant force is downwards (\( W - U \)), causing initial downward acceleration.
3. As the sphere accelerates, its velocity increases, which causes the drag force to increase.
4. The weight and upthrust remain constant, so the increased drag force reduces the resultant force (\( F_{\text{res}} = W - U - D \)).
5. Since the resultant force decreases, the acceleration of the sphere decreases (but it is still speeding up).
6. Acceleration becomes zero when the upward forces equal the downward force (\( U + D = W \)). The sphere then travels at a constant terminal velocity.

Marking points:
- 1-2 marks: 1 or 2 of the indicative content points are clearly explained.
- 3-4 marks: 3 or 4 of the indicative content points are clearly explained.
- 5-6 marks: 5 or 6 of the indicative content points are clearly explained with logical structure and appropriate scientific terminology.

A progressive wave is sent along the string by the vibration generator. This wave travels to the fixed end and is reflected, reversing its direction of travel. The incident and reflected waves travel in opposite directions along the string. These two waves have the same frequency (or wavelength) and amplitude. The waves superpose (or interfere) with each other. At points where the waves meet in phase, constructive interference occurs, creating points of maximum amplitude called antinodes. At points where the waves meet in anti-phase (out of phase by \( 180^{\circ} \) or \( \pi \) radians), destructive interference occurs, creating points of zero amplitude called nodes. Because both ends of the string are fixed, they must be nodes. This boundary condition means that standing waves can only be established when the length \( L \) of the string is equal to an integer number of half-wavelengths: \( L = \frac{n\lambda}{2} \) where \( n = 1, 2, 3, \dots \). Since \( v = f\lambda \), the allowed frequencies are \( f = \frac{nv}{2L} \). Therefore, only specific discrete frequencies can produce standing waves.

Indicative content:
1. Waves travel along the string and are reflected at the fixed boundaries.
2. The incident and reflected waves travel in opposite directions and superpose/interfere.
3. The waves must have the same frequency/wavelength and similar amplitude.
4. Nodes are formed by destructive interference (waves in anti-phase) where amplitude is zero, and antinodes are formed by constructive interference (waves in phase) where amplitude is maximum.
5. The fixed ends must act as nodes (boundary condition).
6. This limits the wavelengths to \( \lambda = \frac{2L}{n} \) (where \( n \) is an integer), and since \( v = f\lambda \), standing waves can only occur at discrete frequencies \( f = \frac{nv}{2L} \).

Marking points:
- 1-2 marks: 1 or 2 points clearly explained.
- 3-4 marks: 3 or 4 points clearly explained with some physical reasoning.
- 5-6 marks: 5 or 6 points clearly explained with logical structure and clear link between boundary conditions and discrete frequencies.

The electric field exists only in the gap between the dees. This electric field accelerates the protons across the gap, increasing their kinetic energy. Inside the dees, the metal shields the protons from the electric field (Faraday cage effect), so they travel at constant speed. However, a uniform magnetic field acts perpendicular to the plane of the dees. The magnetic force (\( F = Bqv \)) acts perpendicular to the velocity of the protons, providing the centripetal force (\( Bqv = \frac{mv^2}{r} \)), which causes them to follow a circular path of radius \( r = \frac{mv}{Bq} \). As the protons are accelerated in the gaps, their speed \( v \) increases, which increases the radius \( r \) of their path inside the dees. The time spent in one dee is \( t = \frac{\pi r}{v} \). Substituting \( r = \frac{mv}{Bq} \) gives \( t = \frac{\pi m}{Bq} \). This time is independent of the speed \( v \) and radius \( r \). Therefore, the total period of one orbit is \( T = \frac{2\pi m}{Bq} \), and the required frequency of the alternating potential difference is \( f = \frac{Bq}{2\pi m} \). Since \( B \), \( q \), and \( m \) are constant (non-relativistic limit), the frequency must be constant to ensure that the electric field changes direction at the exact moment the protons arrive at the gap, ensuring they are always accelerated.

Indicative content:
1. Electric field across the gap accelerates the protons / increases their kinetic energy.
2. Magnetic field is perpendicular to the motion and exerts a force perpendicular to velocity (\( F = Bqv \)).
3. This magnetic force acts as a centripetal force (\( Bqv = \frac{mv^2}{r} \)), causing circular motion inside the dees where there is no electric field.
4. As speed \( v \) increases, the radius \( r \) increases (\( r = \frac{mv}{Bq} \)).
5. The time spent in each dee is \( t = \frac{\pi r}{v} = \frac{\pi m}{Bq} \), which is independent of the speed/radius.
6. Thus, the period of the alternating voltage is constant (\( T = \frac{2\pi m}{Bq} \)), so a constant frequency alternating potential difference is required to ensure acceleration occurs in the correct direction each time protons cross the gap.

Marking points:
- 1-2 marks: 1 or 2 points clearly explained.
- 3-4 marks: 3 or 4 points clearly explained.
- 5-6 marks: 5 or 6 points clearly explained, including the mathematical derivation or detailed reasoning showing why the time/frequency is independent of radius/speed.

When the hydrogen fuel in the core of a massive star is depleted, nuclear fusion in the core slows down, and the core contracts under gravity. This contraction raises the temperature and pressure, allowing hydrogen fusion to occur in a shell around the core, and causing the outer layers of the star to expand and cool, forming a red supergiant. Because of the massive star's immense gravitational pull, its core becomes hot enough to fuse helium into carbon and oxygen, and then successively fuse heavier elements (such as neon, magnesium, silicon) in layers. This creates an 'onion-skin' structure of concentric fusing shells. This nucleosynthesis process continues until iron is produced in the core. The fusion of iron is endothermic (requires energy rather than releasing it), so fusion reactions cease in the core. Without the outward radiation and thermal pressure of fusion to oppose gravity, the core collapses catastrophically. The outer layers of the star collapse onto the core and rebound violently, resulting in a supernova explosion, which ejects most of the star's outer envelope and synthesizes elements heavier than iron. The remnant core is compressed. If its mass is below the Tolman-Oppenheimer-Volkoff limit (approx. 3 solar masses), it collapses into a highly dense neutron star. If the remnant core mass exceeds this limit, gravity overcomes neutron degeneracy pressure, and it collapses completely to form a black hole.

Indicative content:
1. Star leaves main sequence when hydrogen in the core is depleted, leading to core contraction and outer layer expansion to form a red supergiant.
2. Due to high temperature/pressure, fusion of heavier elements occurs in the core/shells (carbon, oxygen, up to iron).
3. Fusion stops at iron because fusing iron does not release energy (is endothermic).
4. Loss of outward radiation pressure causes the core to collapse catastrophically under gravity.
5. The sudden collapse and subsequent rebound of outer layers results in a supernova explosion.
6. The remnant core collapses to form either a neutron star (supported by neutron degeneracy pressure) or, if extremely massive, a black hole.

Marking points:
- 1-2 marks: 1 or 2 points clearly described.
- 3-4 marks: 3 or 4 points clearly described.
- 5-6 marks: 5 or 6 points clearly described with logical progression from red supergiant, shell fusion, iron core limit, supernova, to final remnants.

(a) Taking the natural logarithm of both sides of the decay equation: \(\ln(V) = \ln(V_0 e^{-t/RC}) = \ln(V_0) + \ln(e^{-t/RC}) = -\frac{1}{RC}t + \ln(V_0)\). Comparing this to the equation of a straight line, \(y = mx + c\), where \(y = \ln(V)\), \(x = t\), the gradient is \(m = -\frac{1}{RC}\), and the y-intercept is \(c = \ln(V_0)\). Since \(R\) and \(C\) are constants, the graph is a straight line.

(b) The calculated values of \(\ln(V/\text{V})\) to 3 significant figures are:
At \(t = 0\text{ s}\), \(\ln(V/\text{V}) = 2.20\)
At \(t = 10\text{ s}\), \(\ln(V/\text{V}) = 1.90\)
At \(t = 20\text{ s}\), \(\ln(V/\text{V}) = 1.61\)
At \(t = 30\text{ s}\), \(\ln(V/\text{V}) = 1.31\)
At \(t = 40\text{ s}\), \(\ln(V/\text{V}) = 1.03\)
At \(t = 50\text{ s}\), \(\ln(V/\text{V}) = 0.742\)
At \(t = 60\text{ s}\), \(\ln(V/\text{V}) = 0.470\).

(c) The graph of \(\ln(V/\text{V})\) against \(t/\text{s}\) should be plotted with suitable linear scales, points plotted within half a small square, and a straight line of best fit drawn.

(d) Using a large triangle from the line of best fit, the gradient \(m\) is calculated as: \(m = \frac{0.470 - 2.197}{60.0 - 0.0} = -0.0288\text{ s}^{-1}\). Since \(m = -\frac{1}{RC}\), the capacitance \(C\) is: \(C = -\frac{1}{mR} = \frac{1}{0.0288 \times 47.0 \times 10^3\ \Omega} \approx 7.39 \times 10^{-4}\text{ F} = 740\ \mu\text{F}\).

(e) The percentage uncertainty in \(C\) is the sum of the percentage uncertainties of the gradient and resistance: \(\%\text{ uncertainty in } C = \%\text{ uncertainty in } R + \%\text{ uncertainty in } m = 5\% + 4\% = 9\%\).

(a) Attempt to take natural logarithms of both sides to obtain the linear form [1 mark]. Compare to \(y = mx + c\) showing that gradient is \(-1/RC\) (constant) and intercept is \(\ln(V_0)\) [1 mark].

(b) Correct column heading with no unit for \(\ln(V/\text{V})\) [1 mark]. Calculated values of \(\ln(V/\text{V})\) correct to 3 s.f. [1 mark]. All values tabulated with consistent decimal places (2 d.p. or 3 s.f.) [1 mark].

(c) Axes labeled with quantity and unit on both axes [1 mark]. Scales chosen so points occupy more than half the grid [1 mark]. All points plotted to within half a small square [1 mark]. Straight line of best fit drawn with a balanced distribution of points [1 mark].

(d) Large gradient triangle used with \(\Delta t \ge 40\text{ s}\) [1 mark]. Value of gradient in the range \(-0.028\text{ s}^{-1}\) to \(-0.030\text{ s}^{-1}\) [1 mark]. Calculation of \(C\) using \(C = -1/(mR)\) giving \(7.1 \times 10^{-4}\text{ F}\) to \(7.7 \times 10^{-4}\text{ F}\) [1 mark].

(e) Statement that percentage uncertainties add: \(\%\Delta C = \%\Delta R + \%\Delta m\) [1 mark]. Correct final calculation of \(9\%\) [1 mark].

(a) Background radiation from environmental sources adds an extra count rate to the measured value. This must be subtracted so that only the radiation from the source is analyzed, ensuring that the inverse-square law can be verified.

(b) The corrected count rates are calculated by \(C = R - 0.4\text{ s}^{-1}\). Then, the values of \(C^{-1/2}\) are calculated. The results are:
For \(d = 5.0\text{ cm}\): \(C = 36.0\text{ s}^{-1}\), \(C^{-1/2} = 0.167\text{ s}^{1/2}\)
For \(d = 10.0\text{ cm}\): \(C = 12.4\text{ s}^{-1}\), \(C^{-1/2} = 0.284\text{ s}^{1/2}\)
For \(d = 15.0\text{ cm}\): \(C = 6.4\text{ s}^{-1}\), \(C^{-1/2} = 0.395\text{ s}^{1/2}\)
For \(d = 20.0\text{ cm}\): \(C = 3.9\text{ s}^{-1}\), \(C^{-1/2} = 0.506\text{ s}^{1/2}\)
For \(d = 25.0\text{ cm}\): \(C = 2.7\text{ s}^{-1}\), \(C^{-1/2} = 0.609\text{ s}^{1/2}\)
For \(d = 30.0\text{ cm}\): \(C = 1.9\text{ s}^{-1}\), \(C^{-1/2} = 0.725\text{ s}^{1/2}\).

(c) The graph is plotted with \(C^{-1/2} / \text{s}^{1/2}\) on the y-axis and \(d / \text{cm}\) on the x-axis. A line of best fit is drawn and extrapolated to meet the negative x-axis.

(d)(i) Since \(C^{-1/2} = \frac{1}{\sqrt{k}}(d + c)\), when \(C^{-1/2} = 0\), \(d = -c\). Thus, \(c\) is determined by finding the negative of the intercept on the x-axis. From the graph, the x-intercept is \(-2.5\text{ cm}\), so \(c = 2.5\text{ cm}\) (acceptable range: \(2.2\text{ cm}\) to \(2.8\text{ cm}\)).

(ii) The gradient of the graph is \(m = \frac{1}{\sqrt{k}}\). Using the line of best fit, the gradient \(m = \frac{0.725 - 0.167}{30.0 - 5.0} = 0.0223\text{ s}^{1/2}\text{ cm}^{-1}\). Therefore, \(k = \frac{1}{m^2} = \frac{1}{(0.0223)^2} \approx 2010\text{ s}^{-1}\text{ cm}^2\) (acceptable range: \(1800\text{ s}^{-1}\text{ cm}^2\) to \(2200\text{ s}^{-1}\text{ cm}^2\)).

(a) Background radiation must be subtracted to isolate the radiation emitted solely by the source, which is necessary to test the inverse-square relationship [1 mark].

(b) Correct calculation of all corrected count rates \(C\) [1 mark]. Correct calculation of all values of \(C^{-1/2}\) [1 mark]. Headers for both columns including units \(C / \text{s}^{-1}\) and \(C^{-1/2} / \text{s}^{1/2}\) [1 mark]. Consistent use of significant figures (3 s.f.) [1 mark].

(c) Axes labeled with quantity and unit [1 mark]. Linear scale that allows extrapolation to the negative x-axis [1 mark]. Correct plotting of all points to within half a small square [1 mark]. Straight line of best fit drawn [1 mark].

(d)(i) Formulate relationship \(C^{-1/2} = \frac{d}{\sqrt{k}} + \frac{c}{\sqrt{k}}\) [1 mark]. State that \(c\) is the negative of the x-intercept [1 mark]. Value of \(c\) in the range \(2.2\text{ cm}\) to \(2.8\text{ cm}\) from correct extrapolation [1 mark].

(ii) Calculation of gradient from a large triangle of at least half the length of the line [1 mark]. Calculation of \(k = 1/m^2\) in range \(1800\text{ s}^{-1}\text{ cm}^2\) to \(2200\text{ s}^{-1}\text{ cm}^2\) [1 mark].

(a) Mean time for 20 oscillations: \(\text{Mean} = \frac{37.12 + 37.28 + 37.19}{3} = 37.197 \text{ s} \approx 37.20 \text{ s}\). Uncertainty (half-range) = \(\frac{\text{max} - \text{min}}{2} = \frac{37.28 - 37.12}{2} = 0.08 \text{ s}\). So, time = \(37.20 \pm 0.08 \text{ s}\).
(b) Period \(T\) is the time for 1 oscillation: \(T = \frac{37.20}{20} = 1.860 \text{ s}\). Percentage uncertainty in \(T\) is the same as the percentage uncertainty in the total time: \(\% \text{ uncertainty} = \frac{0.08}{37.20} \times 100\% = 0.215\% \approx 0.22\%\).
(c) Percentage uncertainty in \(L\) = \(\frac{0.002}{0.850} \times 100\% = 0.235\% \approx 0.24\%\).
(d) Using \(g = \frac{4\pi^2 L}{T^2}\): \(g = \frac{4\pi^2 \times 0.850}{1.860^2} = 9.7005 \text{ m s}^{-2} \approx 9.70 \text{ m s}^{-2}\). Percentage uncertainty in \(g\) = \(\% \text{ uncertainty in } L + 2 \times (\% \text{ uncertainty in } T) = 0.235\% + 2 \times 0.215\% = 0.665\% \approx 0.67\%\). Absolute uncertainty in \(g\) = \(9.70 \times 0.00665 = 0.065 \text{ m s}^{-2} \approx 0.07 \text{ m s}^{-2}\).
(e) Combining results gives: \(g = (9.70 \pm 0.07) \text{ m s}^{-2}\) (or \((9.70 \pm 0.06) \text{ m s}^{-2}\) if intermediate rounding was not done).
(f) The experimental range for \(g\) is \(9.63 \text{ m s}^{-2}\) to \(9.77 \text{ m s}^{-2}\). The accepted value of \(9.81 \text{ m s}^{-2}\) lies outside this range, indicating the experiment has systematic inaccuracies. This could be due to air resistance slowing the pendulum (increasing \(T\)), or the length \(L\) being measured to the top of the pendulum bob instead of its center of mass.

Part (a):
- Mean calculated as \(37.20 \text{ s}\) (1 mark)
- Uncertainty calculated as \(0.08 \text{ s}\) using half-range (1 mark)

Part (b):
- Period \(T = 1.86 \text{ s}\) or \(1.860 \text{ s}\) (1 mark)
- Percentage uncertainty in \(T = 0.22\%\) (allow \(0.2\%\)) (1 mark)

Part (c):
- Percentage uncertainty in \(L = 0.24\%\) (allow \(0.2\%\)) (1 mark)

Part (d):
- Value of \(g = 9.70 \text{ m s}^{-2}\) (1 mark)
- Percentage uncertainty in \(g = 0.67\%\) or \(0.68\%\) (1 mark)
- Absolute uncertainty in \(g = 0.07 \text{ m s}^{-2}\) (allow \(0.06 \text{ m s}^{-2}\)) (1 mark)

Part (e):
- Expresses \(g = (9.70 \pm 0.07) \text{ m s}^{-2}\) with matching decimal places and units (1 mark)

Part (f):
- States experimental range is \(9.63\) to \(9.77 \text{ m s}^{-2}\) and notes \(9.81 \text{ m s}^{-2}\) is outside, so it is inaccurate (1 mark)
- Identifies a valid systematic error (e.g. length measured to top of bob, reaction time offset, air resistance) (1 mark)

(a) Mean diameter \(d = \frac{0.38 + 0.40 + 0.39 + 0.37 + 0.39}{5} = 0.386 \text{ mm}\). Uncertainty in \(d\) (half-range) = \(\frac{0.40 - 0.37}{2} = 0.015 \text{ mm}\). Percentage uncertainty in \(d\) = \(\frac{0.015}{0.386} \times 100\% = 3.89\% \approx 3.9\%\) (if rounded to 1 sig fig: \(0.39 \pm 0.02 \text{ mm}\), leading to \(5.1\%\)).
(b) Cross-sectional area \(A = \frac{\pi d^2}{4}\). The percentage uncertainty in \(A\) is twice the percentage uncertainty in \(d\): \(\% \Delta A = 2 \times 3.89\% = 7.78\% \approx 7.8\%\), which is approximately \(8\%\). (If using \(5.1\%\), \(\% \Delta A = 10.2\%\)).
(c) Area \(A = \frac{\pi \times (0.386 \times 10^{-3} \text{ m})^2}{4} = 1.170 \times 10^{-7} \text{ m}^2\). Resistivity \(\rho = \frac{R A}{L} = \frac{4.56 \times 1.170 \times 10^{-7}}{0.724} = 7.37 \times 10^{-7} \ \Omega \text{ m} \approx 7.4 \times 10^{-7} \ \Omega \text{ m}\).
(d) Percentage uncertainty in \(R\) = \(\frac{0.05}{4.56} \times 100\% = 1.10\%\). Percentage uncertainty in \(L\) = \(\frac{0.002}{0.724} \times 100\% = 0.28\%\). Percentage uncertainty in \(\rho\) = \(\% \Delta R + \% \Delta L + \% \Delta A = 1.10\% + 0.28\% + 7.78\% = 9.16\% \approx 9.2\%\) (or \(11.6\%\) if using \(10.2\%\) for area). Absolute uncertainty in \(\rho\) = \(7.37 \times 10^{-7} \times 0.0916 = 0.68 \times 10^{-7} \ \Omega \text{ m} \approx 0.7 \times 10^{-7} \ \Omega \text{ m}\) (or \(0.9 \times 10^{-7} \ \Omega \text{ m}\) if using \(11.6\%\)).
(e) Measuring at different positions checks for variation in diameter/thickness along the length of the wire. Measuring in different orientations checks for non-circularity (elliptical cross-section) of the wire.

Part (a):
- Mean diameter = \(0.386 \text{ mm}\) (allow \(0.39 \text{ mm}\)) (1 mark)
- Absolute uncertainty (half-range) = \(0.015 \text{ mm}\) (allow \(0.02 \text{ mm}\)) (1 mark)
- Percentage uncertainty = \(3.9\%\) (allow range \(3.8\%\) to \(5.2\%\) depending on rounding) (1 mark)

Part (b):
- Recalls that \(\% \Delta A = 2 \times \% \Delta d\) (1 mark)
- Calculates percentage uncertainty in area as \(7.8\%\) (or \(10\%\)), demonstrating that it is approximately \(8\%\) (1 mark)

Part (c):
- Correctly calculates area \(A = 1.17 \times 10^{-7} \text{ m}^2\) (1 mark)
- Correctly calculates resistivity \(\rho = 7.37 \times 10^{-7} \ \Omega \text{ m}\) (allow \(7.4 \times 10^{-7} \ \Omega \text{ m}\)) (1 mark)

Part (d):
- Calculates compounded percentage uncertainty in \(\rho\) = \(9.2\%\) (allow range \(9\%\) to \(12\%\)) (1 mark)
- Calculates absolute uncertainty in \(\rho\) = \(0.7 \times 10^{-7} \ \Omega \text{ m}\) (allow range \(0.6 \times 10^{-7}\) to \(0.9 \times 10^{-7} \ \Omega \text{ m}\)) (1 mark)

Part (e):
- Explains that multiple positions account for diameter variations along the length (1 mark)
- Explains that multiple orientations check if the wire's cross-section is non-circular (1 mark)