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

(b) The magnetic flux linkage through the coil is given by:
\(N\Phi = N B A = N B_0 A \sin(2\pi f t)\)
According to Faraday's law, the induced e.m.f. is:
\(\varepsilon = -\frac{d(N\Phi)}{dt} = -2\pi f N B_0 A \cos(2\pi f t)\)
The maximum e.m.f. occurs when \(\cos(2\pi f t) = \pm 1\), giving:
\(\varepsilon_{\text{max}} = 2\pi f N B_0 A\)
Substitute the given values:
\(\varepsilon_{\text{max}} = 2\pi \times 50 \times 250 \times (4.8 \times 10^{-3}\text{ T}) \times (1.5 \times 10^{-4}\text{ m}^2)\)
\(\varepsilon_{\text{max}} = 100\pi \times 250 \times 7.2 \times 10^{-7}\)
\(\varepsilon_{\text{max}} = 0.05655\text{ V} \approx 0.057\text{ V}\) (or \(5.7 \times 10^{-2}\text{ V}\)).

(c) A high input resistance ensures that negligible current is drawn from the coil circuit. This minimizes any potential drop across the internal resistance of the coil itself, ensuring the displayed voltage is equal to the true induced e.m.f.

(d) The maximum induced e.m.f. is proportional to the product of frequency and peak magnetic flux density: \(\varepsilon_{\text{max}} \propto f \cdot B_0\). Since the frequency is doubled (\(\times 2\)) and the flux density is halved (\(\times 0.5\)), the product remains unchanged. Thus, the maximum induced e.m.f. remains at \(0.057\text{ V}\).

(a)
- Correct statement: Induced e.m.f. is proportional to the rate of change of [1]
- magnetic flux linkage / flux through the coil [1]

(b)
- Recall or use of \(\Phi = B A\) and \(\text{flux linkage} = N \Phi\) [1]
- Recognition that \(\text{maximum rate of change of flux linkage} = 2\pi f N B_0 A\) [1]
- Correct substitution: \(2\pi \times 50 \times 250 \times 4.8 \times 10^{-3} \times 1.5 \times 10^{-4}\) [1]
- Calculates to \(0.05655\text{ V}\) (which rounds to \(0.057\text{ V}\)) [1]

(c)
- High resistance draws negligible / very low current from the coil [1]
- This prevents potential loss across the coil's own internal resistance / measures true terminal e.m.f. [1]

(d)
- Identifies \(\varepsilon_{\text{max}} \propto f \cdot B_0\) (or equivalent relationship) [1]
- Explains that doubling the frequency and halving the flux density keeps the product constant, so the maximum e.m.f. is unchanged [1]

(a) For a circular orbit, the gravitational force provides the centripetal acceleration:
\(\frac{G M m}{r^2} = \frac{m v^2}{r}\)
Rearranging for orbital speed \(v\):
\(v = \sqrt{\frac{G M}{r}}\)
Substitute the given values:
\(v = \sqrt{\frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{5.20 \times 10^6}}\)
\(v = \sqrt{\frac{4.282 \times 10^{13}}{5.20 \times 10^6}} = \sqrt{8.235 \times 10^6} = 2870\text{ m s}^{-1} \approx 2.9 \times 10^3\text{ m s}^{-1}\).

(b) The gravitational potential energy \(E_p\) of the satellite is:
\(E_p = -\frac{G M m}{r}\)
\(E_p = -\frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23} \times 1200}{5.20 \times 10^6}\)
\(E_p = -\frac{5.139 \times 10^{16}}{5.20 \times 10^6} = -9.88 \times 10^9\text{ J}\).

(c) The kinetic energy \(E_k\) of the satellite is:
\(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200 \times (2870)^2 = 4.94 \times 10^9\text{ J}\)
(or directly using \(E_k = \frac{G M m}{2r} = 4.94 \times 10^9\text{ J}\)).

The total energy of the orbiting satellite is:
\(E_{\text{total}} = E_p + E_k = -9.88 \times 10^9 + 4.94 \times 10^9 = -4.94 \times 10^9\text{ J}\).

To escape Mars' gravitational field completely, the final total energy must be at least \(0\). Therefore, the minimum additional energy that must be supplied is:
\(\Delta E = 0 - E_{\text{total}} = 4.94 \times 10^9\text{ J}\).

(a)
- Equating gravitational force to centripetal force: \(\frac{G M m}{r^2} = \frac{m v^2}{r}\) [1]
- Rearranging to make \(v\) the subject: \(v = \sqrt{\frac{G M}{r}}\) [1]
- Correct substitution to show \(v = 2870\text{ m s}^{-1}\) (rounds to \(2.9 \times 10^3\text{ m s}^{-1}\)) [1]

(b)
- Recall of formula \(E_p = -\frac{G M m}{r}\) [1]
- Substitution: \(E_p = -\frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23} \times 1200}{5.20 \times 10^6}\) [1]
- Correct final calculation of potential energy: \(-9.88 \times 10^9\text{ J}\) (allow negative sign omitted if context of magnitude is clear, but award full marks for negative value) [1]

(c)
- Recognizes total energy of satellite is \(E_{\text{total}} = E_k + E_p\) or \(E_{\text{total}} = -\frac{G M m}{2 r}\) [1]
- Calculation of Kinetic Energy: \(E_k = \frac{1}{2} m v^2 = 4.94 \times 10^9\text{ J}\) [1]
- Determination of total energy: \(E_{\text{total}} = -4.94 \times 10^9\text{ J}\) [1]
- Calculates escape energy required: \(+4.94 \times 10^9\text{ J}\) (accept \(4.9 \times 10^9\text{ J}\)) [1]

(a) The cell, the fixed resistor, and the thermistor must be connected in series to form a complete circuit loop. The output voltage \(V_{\text{out}}\) is measured across the terminals of the fixed resistor. When the temperature increases, the resistance of the thermistor decreases, reducing its share of the total input voltage, which in turn increases the potential difference across the fixed resistor.

(b) At \(20\text{ }^\circ\text{C}\), the thermistor resistance is \(R_T = 3.8\text{ k}\Omega = 3800\text{ }\Omega\). The potential divider equation gives:
\(V_{\text{out}} = V_{\text{in}} \times \left( \frac{R}{R + R_T} \right)\)
\(V_{\text{out}} = 9.0 \times \left( \frac{1200}{1200 + 3800} \right)\)
\(V_{\text{out}} = 9.0 \times \left( \frac{1200}{5000} \right) = 9.0 \times 0.24 = 2.16\text{ V}\).

(c) At \(50\text{ }^\circ\text{C}\), the thermistor resistance is \(R_T = 850\text{ }\Omega\). Under these conditions:
\(V_{\text{out}} = 9.0 \times \left( \frac{1200}{1200 + 850} \right)\)
\(V_{\text{out}} = 9.0 \times \left( \frac{1200}{2050} \right) \approx 5.27\text{ V}\).

(d) Around \(20\text{ }^\circ\text{C}\), the thermistor's resistance is \(3.8\text{ k}\Omega\). A potential divider circuit is most sensitive (shows the largest rate of change of output voltage per unit change in resistance) when the resistance of the fixed resistor is comparable to the resistance of the sensor. Since \(3.9\text{ k}\Omega\) is much closer to \(3.8\text{ k}\Omega\) than \(1.2\text{ k}\Omega\) is, changing the fixed resistor to \(3.9\text{ k}\Omega\) will significantly increase the sensitivity of the sensor circuit around \(20\text{ }^\circ\text{C}\).

(a)
- States that the fixed resistor and thermistor are connected in series across the cell [1]
- States that \(V_{\text{out}}\) is measured in parallel across the fixed resistor [1]

(b)
- Recall of potential divider formula: \(V_{\text{out}} = V_{\text{in}} \times \frac{R}{R + R_T}\) [1]
- Substitution: \(V_{\text{out}} = 9.0 \times \frac{1200}{1200 + 3800}\) [1]
- Calculation to give: \(2.16\text{ V}\) (accept \(2.2\text{ V}\)) [1]

(c)
- Substitution with new values: \(V_{\text{out}} = 9.0 \times \frac{1200}{1200 + 850}\) [1]
- Correct calculation of \(5.27\text{ V}\) (or \(5.3\text{ V}\)) [1]

(d)
- Identifies that \(3.9\text{ k}\Omega\) is closer in value to the thermistor's resistance at \(20\text{ }^\circ\text{C}\) (which is \(3.8\text{ k}\Omega\)) [1]
- Explains that potential divider sensitivity is maximum when the two series resistances are approximately equal [1]
- Concludes that the sensitivity of the output voltage to temperature changes around \(20\text{ }^\circ\text{C}\) will increase [1]

(a) As the coil rotates at angular speed \(\omega = 2\pi f\), the angle \(\theta\) between the normal to the plane of the coil and the magnetic field lines changes continuously according to \(\theta = \omega t\). The magnetic flux linkage is given by \(N\Phi = NAB \cos(\omega t)\), which varies sinusoidally over time. According to Faraday's law of electromagnetic induction, the induced EMF \(\varepsilon\) is proportional to the rate of change of magnetic flux linkage: \(\varepsilon = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t} = NAB\omega \sin(\omega t)\). Since the rate of change of flux linkage alternates in sign as the coil rotates through each half-turn, the induced EMF alternates in direction, producing an alternating voltage.

(b) The maximum EMF occurs when \(\sin(\omega t) = 1\), so:
\(\varepsilon_0 = N B A \omega\)
First, calculate the cross-sectional area of the square coil:
\(A = L^2 = (0.050\text{ m})^2 = 2.5 \times 10^{-3}\text{ m}^2\)
Next, calculate the angular frequency:
\(\omega = 2\pi f = 2\pi \times 25 = 50\pi \approx 157.08\text{ rad s}^{-1}\)
Now, substitute the values into the equation for maximum EMF:
\(\varepsilon_0 = 200 \times 0.15\text{ T} \times (2.5 \times 10^{-3}\text{ m}^2) \times 157.08\text{ rad s}^{-1}\)
\(\varepsilon_0 = 0.075 \times 157.08 = 11.78\text{ V}\)
This is approximately \(12\text{ V}\).
Exact value to 3 s.f. = \(11.8\text{ V}\).

(c)(i) The maximum current in the circuit \(I_0\) is given by Ohm's Law:
\(I_0 = \frac{\varepsilon_0}{R_{\text{total}}} = \frac{\varepsilon_0}{R + r} = \frac{11.78\text{ V}}{8.0\ \Omega + 2.0\ \Omega} = 1.178\text{ A}\)
The maximum power dissipated in the external resistor \(R\) is:
\(P_{\text{ext, max}} = I_0^2 R = (1.178\text{ A})^2 \times 8.0\ \Omega \approx 11.1\text{ W}\) (using \(11.78\text{ V}\) gives \(11.1\text{ W}\); using \(11.8\text{ V}\) gives \(11.14\text{ W}\)).

(c)(ii) The mean power dissipated in the entire circuit is given by:
\(P_{\text{mean}} = I_{\text{rms}}^2 R_{\text{total}} = \left(\frac{I_0}{\sqrt{2}}\right)^2 (R + r) = \frac{I_0^2 (R + r)}{2}\)
\(P_{\text{mean}} = \frac{(1.178\text{ A})^2 \times 10.0\ \Omega}{2} = 6.94\text{ W}\) (using \(11.8\text{ V}\) gives \(6.96\text{ W}\)).

(d) The student's statement is incorrect.
From \(\varepsilon_0 = NAB\omega\), the induced peak voltage is directly proportional to frequency (\(\varepsilon_0 \propto f\)). Since peak current \(I_0 = \frac{\varepsilon_0}{R_{\text{total}}}\), current is also directly proportional to frequency (\(I_0 \propto f\)).
The mean power is given by \(P_{\text{mean}} = I_{\text{rms}}^2 R_{\text{total}} = \frac{I_0^2 R_{\text{total}}}{2}\). Therefore, \(P_{\text{mean}} \propto f^2\).
If the frequency is doubled, the mean power will increase by a factor of \(2^2 = 4\), not \(2\).

(a)
- M1: Explains that flux linkage varies sinusoidally because \(\theta = \omega t\) and \(N\Phi = NAB\cos(\omega t)\).
- A1: States Faraday's Law (induced EMF is rate of change of flux linkage) or \(\varepsilon = -\mathrm{d}(N\Phi)/\mathrm{d}t\).
- A1: Explains that the direction of EMF reverses as the rate of change of flux linkage reverses sign, producing AC.

(b)
- C1: Area calculation \(A = 2.5 \times 10^{-3}\text{ m}^2\) and angular speed \(\omega = 157.1\text{ rad s}^{-1}\).
- C1: Correct substitution into \(\varepsilon_0 = BAN\omega\).
- A1: Show value is approximately \(12\text{ V}\) (must show unrounded value e.g. \(11.78\text{ V}\)).
- A1: Correct final answer to 3 s.f.: \(11.8\text{ V}\).

(c)(i)
- C1: Total resistance \(R + r = 10.0\ \Omega\) used to find peak current \(I_0 = 1.18\text{ A}\).
- C1: Use of \(P = I^2 R\) with external resistor \(R = 8.0\ \Omega\).
- A1: Correct final answer: \(11.1\text{ W}\) (accept \(11.1 - 11.2\text{ W}\)).

(c)(ii)
- C1: Use of \(I_{\text{rms}} = I_0 / \sqrt{2}\) or \(P_{\text{mean}} = P_{\text{max}} / 2\) with total resistance.
- A1: Correct final answer: \(6.94\text{ W}\) (accept \(6.9\text{ W}\) to \(7.0\text{ W}\)).

(d)
- R1: Statement is incorrect.
- R1: Clear physical step showing that peak EMF (or peak current) is directly proportional to frequency (\(V_0 \propto f\)).
- R1: Concludes mathematically that power is proportional to the square of frequency (\(P \propto f^2\)), hence doubling frequency quadruples the mean power.

(a) The orbital radius \(r_1\) is the sum of the planet's radius and the height of the orbit:
\(r_1 = R_p + h = 3.4 \times 10^6\text{ m} + 4.0 \times 10^5\text{ m} = 3.8 \times 10^6\text{ m}\)
The gravitational force provides the centripetal force:
\(\frac{G M m}{r_1^2} = \frac{m v^2}{r_1} \implies v = \sqrt{\frac{G M}{r_1}}\)
Substitute the values:
\(v = \sqrt{\frac{6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2} \times 6.4 \times 10^{23}\text{ kg}}{3.8 \times 10^6\text{ m}}}\)
\(v = \sqrt{\frac{4.2688 \times 10^{13}}{3.8 \times 10^6}} = \sqrt{1.123 \times 10^7} \approx 3352\text{ m s}^{-1} \approx 3.35\text{ km s}^{-1}\)
This is approximately \(3.4\text{ km s}^{-1}\).

(b) The period \(T_1\) of the circular orbit is given by:
\(T_1 = \frac{2\pi r_1}{v} = \frac{2\pi \times 3.8 \times 10^6\text{ m}}{3352\text{ m s}^{-1}} \approx 7123\text{ s}\)
Converting seconds to minutes:
\(T_1 = \frac{7123}{60} \approx 119\text{ minutes}\) (or \(1.99\text{ hours}\)).

(c)(i) According to Kepler's Third Law, the square of the orbital period is proportional to the cube of the orbital radius:
\(T^2 \propto r^3 \implies \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3\)
We are given that \(T_2 = 2 T_1\), so:
\((2)^2 = \left(\frac{r_2}{r_1}\right)^3 \implies 4 = \left(\frac{r_2}{r_1}\right)^3\)
\(r_2 = 4^{1/3} \times r_1\)
\(r_2 \approx 1.5874 \times 3.8 \times 10^6\text{ m} \approx 6.03 \times 10^6\text{ m}\) (or \(6.0 \times 10^6\text{ m}\)).

(c)(ii) Gravitational potential energy \(E_p = -\frac{G M m}{r}\).
The change in gravitational potential energy \(\Delta E_p\) is:
\(\Delta E_p = E_{p2} - E_{p1} = -G M m \left( \frac{1}{r_2} - \frac{1}{r_1} \right) = G M m \left( \frac{1}{r_1} - \frac{1}{r_2} \right)\)
Substitute the values:
\(G M m = 6.67 \times 10^{-11} \times 6.4 \times 10^{23} \times 450 = 1.921 \times 10^{16}\text{ J m}\)
\(\Delta E_p = 1.921 \times 10^{16} \times \left( \frac{1}{3.8 \times 10^6} - \frac{1}{6.03 \times 10^6} \right)\)
\(\Delta E_p = 1.921 \times 10^{16} \times (2.632 \times 10^{-7} - 1.658 \times 10^{-7})\)
\(\Delta E_p = 1.921 \times 10^{16} \times 9.74 \times 10^{-8} \approx +1.87 \times 10^9\text{ J}\) (accept \(1.9 \times 10^9\text{ J}\)).

(d) Redshift measurements of characteristic spectral lines from distant galaxies show they are shifted to longer wavelengths, indicating that the galaxies are moving away from us (receding). Hubble observed that the recessional velocity \(v\) is directly proportional to the distance \(d\) to the galaxy (\(v = H_0 d\)).
Assuming the expansion rate of the Universe has been constant since the Big Bang, the time taken for galaxies to reach their current distance is \(t = \frac{d}{v} = \frac{1}{H_0}\). Thus, the reciprocal of the Hubble constant, \(1/H_0\), gives an upper limit estimate for the age of the Universe.

(a)
- C1: Identifies orbital radius \(r = 3.8 \times 10^6\text{ m}\) (adding planet radius and altitude).
- C1: Uses circular motion equating gravitational and centripetal forces: \(v = \sqrt{GM/r}\).
- A1: Shows calculated speed is \(3.35\text{ km s}^{-1}\) or \(3352\text{ m s}^{-1}\) (which rounds to \(3.4\text{ km s}^{-1}\)).

(b)
- C1: Uses \(T = 2\pi r / v\) or Kepler's relation with appropriate orbital radius.
- C1: Obtains value of period in seconds (\(\approx 7120\text{ s}\)).
- A1: Correctly converts to minutes: \(119\text{ minutes}\) (accept range \(118 - 120\text{ minutes}\)).

(c)(i)
- C1: Applies Kepler's Third Law \(T^2 \propto r^3\) or equivalent formula.
- C1: Finds the scale factor \(4^{1/3} \approx 1.59\).
- A1: Calculates the correct new radius: \(6.03 \times 10^6\text{ m}\) (accept \(6.0 \times 10^6\text{ m}\)).

(c)(ii)
- C1: Uses formula for gravitational potential energy: \(E_p = -GMm/r\).
- C1: Sets up the correct subtraction structure: \(\Delta E_p = GMm(1/r_1 - 1/r_2)\).
- A1: Correct final value: \(+1.87 \times 10^9\text{ J}\) (accept \(1.8 \times 10^9\text{ J}\) to \(1.9 \times 10^9\text{ J}\); must be positive as it moves to a higher orbit).

(d)
- R1: Links redshift to recessional velocity (wavelength shift indicates expansion/movement away).
- R1: States Hubble's Law as \(v = H_0 d\) (velocity proportional to distance).
- R1: Shows that the age of the universe is estimated by \(t = 1/H_0\) (by setting time \(= \text{distance}/\text{velocity}\)).

(a) Step-by-step calculations:
For \(t = 0.0\text{ s}\):
\(V = 9.00\text{ V}\)
\(I = \frac{9.00\text{ V}}{1500\ \Omega} = 6.00 \times 10^{-3}\text{ A} = 6.00\text{ mA}\)
\(\Delta Q = -I \Delta t = -(6.00 \times 10^{-3}\text{ A}) \times 2.0\text{ s} = -1.20 \times 10^{-2}\text{ C} = -12.00\text{ mC}\)
\(\Delta V = \frac{\Delta Q}{C} = \frac{-1.20 \times 10^{-2}\text{ C}}{4.7 \times 10^{-3}\text{ F}} \approx -2.553\text{ V} \approx -2.55\text{ V}\)

For \(t = 2.0\text{ s}\):
New voltage \(V = 9.00\text{ V} + \Delta V = 9.00 - 2.553 = 6.447\text{ V}\) (accept \(6.45\text{ V}\))
New current \(I = \frac{6.447\text{ V}}{1500\ \Omega} = 4.298 \times 10^{-3}\text{ A} \approx 4.30\text{ mA}\)
New charge transferred \(\Delta Q = -I \Delta t = -(4.298 \times 10^{-3}\text{ A}) \times 2.0\text{ s} = -8.596 \times 10^{-3}\text{ C} \approx -8.60\text{ mC}\)
New change in voltage \(\Delta V = \frac{\Delta Q}{C} = \frac{-8.596 \times 10^{-3}\text{ C}}{4.7 \times 10^{-3}\text{ F}} \approx -1.829\text{ V} \approx -1.83\text{ V}\)

For \(t = 4.0\text{ s}\):
New voltage \(V = 6.447\text{ V} + \Delta V = 6.447 - 1.829 = 4.618\text{ V}\) (accept \(4.62\text{ V}\))

Completed row values:
- At \(t = 2.0\text{ s}\): \(V = 6.45\text{ V}\), \(I = 4.30\text{ mA}\), \(\Delta Q = -8.60\text{ mC}\), \(\Delta V = -1.83\text{ V}\)
- At \(t = 4.0\text{ s}\): \(V = 4.62\text{ V}\)

(b) First, calculate the time constant \(\tau = RC\):
\(\tau = 1.5 \times 10^3\ \Omega \times 4.7 \times 10^{-3}\text{ F} = 7.05\text{ s}\)
Theoretical voltage at \(t = 4.0\text{ s}\):
\(V = V_0 e^{-t/RC} = 9.00 \times e^{-4.0 / 7.05} = 9.00 \times e^{-0.5674} = 9.00 \times 0.5670 = 5.10\text{ V}\)

Comparison:
The modeled value at \(t = 4.0\text{ s}\) (\(4.62\text{ V}\)) is lower than the analytical value (\(5.10\text{ V}\)).

Explanation:
The numerical model assumes that the current remains constant at its initial (higher) value during each step interval of \(\Delta t = 2.0\text{ s}\). In reality, the voltage and current decrease continuously during the interval, so the actual rate of discharge is slower than modeled, causing the actual remaining voltage to be higher.

(c)(i) Reducing the step size \(\Delta t\) increases the accuracy of the model, as the assumption of constant current over the interval becomes a closer approximation to continuous decay.
(ii) The computational effort increases because more steps (iterations) are required to model the same total duration of discharge.

(d) When the bright light is turned on, the resistance of the LDR drops significantly compared to the original \(1.5 \times 10^3\ \Omega\) fixed resistor.
Because \(R\) is much smaller, the time constant \(\tau = RC\) is greatly reduced.
At \(t = 0\), the initial current \(I_0 = \frac{V_0}{R}\) is much larger, meaning charge flows off the capacitor at a much faster initial rate.
On a graph of \(V\) against \(t\), the discharge curve will start with a much steeper negative gradient compared to the original.
The curve will still be exponential (assuming the light and thus resistance remain constant), but it will decay to near zero much more quickly, squeezed tightly towards the vertical axis.

(a)
- C1: Calculates \(V = 6.45\text{ V}\) (or \(6.44\text{ V}\)) for \(t = 2.0\text{ s}\).
- C1: Calculates \(I = 4.30\text{ mA}\) and \(\Delta Q = -8.60\text{ mC}\) for the second step.
- C1: Calculates \(\Delta V = -1.83\text{ V}\) for the second step.
- A1: Correct final voltage of \(4.62\text{ V}\) (accept range \(4.61 - 4.63\text{ V}\)) at \(t = 4.0\text{ s}\).

(b)
- C1: Calculates correct time constant \(\tau = 7.05\text{ s}\).
- A1: Obtains theoretical potential difference \(V = 5.10\text{ V}\).
- A1: States that the modeled value is lower than the theoretical value.
- A1: Explains that the numerical method assumes current is constant over each interval at the start-of-interval maximum, leading to an overestimation of charge lost in each step.

(c)
- A1: States that reducing step size increases accuracy because it approximates continuous change better.
- A1: States that computational effort increases because more calculation steps are needed.

(d) Level of Response / Structured explanation:
- Clear description of LDR behavior (resistance drops when illuminated) [1].
- Links lower resistance to a much smaller time constant \(\tau = RC\) [1].
- Connects smaller \(\tau\) to a much higher initial current / faster initial rate of discharge [1].
- Graph shape: Describes curve starting with a much steeper initial gradient [1].
- Comparison: Identifies that the curve decays to zero much faster, representing a faster rate of energy/charge depletion [1].

1. Since the tether is vertical and its motion is horizontal (eastward), the wire cuts across the horizontal component of the magnetic field (which points northward). The vertical component of the magnetic field is parallel to the vertical wire and does not result in any magnetic force on charges along the length of the wire (the cross product \(\vec{v} \times \vec{B}_v\) is horizontal and perpendicular to the wire length, so it does not drive current along the wire). Therefore, only the northward horizontal component \(B_h\) induces the EMF.
2. The formula for the induced EMF in a moving conductor of length \(L\) is \(\mathcal{E} = B_h L v\) when the velocity, field, and length are mutually perpendicular. Here, the tether is vertical (along the z-axis), motion is East (along the x-axis), and \(B_h\) is North (along the y-axis). They are mutually perpendicular.
\(\mathcal{E} = (1.8 \times 10^{-5}\text{ T}) \times (500\text{ m}) \times (25\text{ m s}^{-1})\)
\(\mathcal{E} = 0.225\text{ V}\).

1. Identifies horizontal component \(B_h\) [1 mark]; explains that only this component is perpendicular to both the conductor length and the direction of horizontal motion, creating a force along the wire [1 mark].
2. Recalls or uses \(\mathcal{E} = B L v\) [1 mark]; substitutes correct values: \(1.8 \times 10^{-5} \times 500 \times 25\) [1 mark]; calculates correct numerical value: \(0.225\text{ V}\) (or \(0.23\text{ V}\)) [1 mark]; states correct unit: \(\text{V}\) [1 mark].

1. From the ideal gas equation, \(PV = nRT\). Since the mass of helium is constant, \(n\) is constant.
Therefore: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\) where state 1 is launch and state 2 is peak altitude.
Rearranging for the volume ratio: \(\frac{V_2}{V_1} = \frac{P_1}{P_2} \times \frac{T_2}{T_1}\)
\(\frac{V_2}{V_1} = \frac{1.01 \times 10^5\text{ Pa}}{1.20 \times 10^3\text{ Pa}} \times \frac{225\text{ K}}{293\text{ K}}\)
\(\frac{V_2}{V_1} = 84.17 \times 0.7679 \approx 64.6\) (or \(65\) to 2 s.f.).

2. The number of moles of helium is \(n = \frac{\text{mass}}{\text{molar mass}} = \frac{12.0\text{ kg}}{4.00 \times 10^{-3}\text{ kg mol}^{-1}} = 3000\text{ mol}\).
Using the ideal gas equation at launch: \(P_1 V_1 = n R T_1\)
\(V_1 = \frac{n R T_1}{P_1} = \frac{3000\text{ mol} \times 8.31\text{ J mol}^{-1}\text{ K}^{-1} \times 293\text{ K}}{1.01 \times 10^5\text{ Pa}}\)
\(V_1 = \frac{7.304 \times 10^6}{1.01 \times 10^5} \approx 72.3\text{ m}^3\) (accept \(72.5\text{ m}^3\) if using \(R = 8.314\)).

1. Recognizes \(PV/T = \text{constant}\) [1 mark]; substitutes values correctly into ratio formula [1 mark]; calculates ratio of \(65\) or \(64.6\) [1 mark].
2. Calculates the number of moles as \(3000\text{ mol}\) [1 mark]; rearranges \(PV = nRT\) for \(V\) [1 mark]; calculates correct volume of \(72.3\text{ m}^3\) or \(72.5\text{ m}^3\) with correct unit [1 mark].

1. Distance to travel: \(d = 15.0\text{ km} = 15000\text{ m}\).
Speed of muons: \(v = 0.998 \times 3.00 \times 10^8 = 2.994 \times 10^8\text{ m s}^{-1}\).
Time taken to reach sea level in Earth's frame: \(t = \frac{d}{v} = \frac{15000}{2.994 \times 10^8} = 5.01 \times 10^{-5}\text{ s}\).
Number of half-lives elapsed in classical physics: \(N = \frac{t}{t_{1/2}} = \frac{5.01 \times 10^{-5}}{1.52 \times 10^{-6}} \approx 33.0\).
Fraction of muons remaining: \((1/2)^{33} \approx 1.2 \times 10^{-10}\). This is extremely close to zero, so virtually no muons would reach the surface.

2. The Lorentz factor is calculated by: \(\gamma = \frac{1}{\sqrt{1 - (v/c)^2}} = \frac{1}{\sqrt{1 - 0.998^2}} = \frac{1}{\sqrt{1 - 0.996004}} = \frac{1}{\sqrt{0.003996}} \approx 15.8\).
Due to relativistic time dilation, the half-life of the muons in the Earth's reference frame is prolonged to:
\(t'_{1/2} = \gamma \times t_{1/2} = 15.8 \times 1.52 \times 10^{-6}\text{ s} = 2.40 \times 10^{-5}\text{ s}\).
Number of half-lives elapsed in Earth's frame: \(N' = \frac{t}{t'_{1/2}} = \frac{5.01 \times 10^{-5}\text{ s}}{2.40 \times 10^{-5}\text{ s}} = 2.09\).
Fraction of muons remaining: \((1/2)^{2.09} \approx 0.235\) or \(23.5\%\) (approx. \(24\%\)). This is a significant, easily measurable fraction.

1. Calculates travel time as \(5.01 \times 10^{-5}\text{ s}\) [1 mark]; calculates number of elapsed half-lives (~33) [1 mark]; computes extremely small fraction (~\(10^{-10}\)) to conclude virtually none survive [1 mark].
2. Calculates \(\gamma = 15.8\) (accept 16) [1 mark]; calculates dilated half-life in Earth frame as \(2.40 \times 10^{-5}\text{ s}\) (or shows equivalent length contraction argument) [1 mark]; calculates new number of half-lives elapsed as \(2.09\) [1 mark]; calculates remaining fraction as \(24\%\) (or \(23.5\%\)) to show they are detectable [1 mark].

1. An NTC thermistor's resistance increases as the temperature decreases. Therefore, as temperature drops, the resistance \(R_{\text{th}}\) of the thermistor increases.
To get a higher output voltage \(V_{\text{out}}\) when \(R_{\text{th}}\) increases, the output voltage must be measured across the thermistor. The circuit should consist of a fixed resistor \(R\) connected in series with the thermistor across the \(5.0\text{ V}\) supply, with \(V_{\text{out}}\) taken across the thermistor.

2. Using the potential divider equation:
\(V_{\text{out}} = V_s \times \frac{R_{\text{th}}}{R + R_{\text{th}}}\)
At \(-50^\circ\text{C}\), \(V_{\text{out}} = 4.0\text{ V}\), \(V_s = 5.0\text{ V}\), and \(R_{\text{th}} = 120\text{ k}\Omega\).
\(4.0 = 5.0 \times \frac{120}{R + 120}\)
\(0.8 = \frac{120}{R + 120}\)
\(0.8(R + 120) = 120\)
\(0.8R + 96 = 120\)
\(0.8R = 24\)
\(R = 30\text{ k}\Omega\) (or \(3.0 \times 10^4\ \Omega\)).

1. States that thermistor resistance increases as temperature decreases [1 mark]; states that \(V_{\text{out}}\) must be taken across the thermistor [1 mark]; describes the circuit configuration correctly (fixed resistor in series with the thermistor across the power supply) [1 mark].
2. Recalls or uses the potential divider equation [1 mark]; substitutes the correct values: \(4.0 = 5.0 \times \frac{120}{R + 120}\) [1 mark]; solves to obtain \(R = 30\text{ k}\Omega\) (or \(30000\ \Omega\)) [1 mark].