2023 AQA IAL Physics (9630) Practice Paper with Answers

First, calculate the mean time period: Mean = \( (4.0 + 4.2 + 3.8 + 4.1 + 3.9) / 5 = 4.0 \) s. Next, find the absolute uncertainty using half the range of the measurements: Range = \( 4.2 - 3.8 = 0.4 \) s. Absolute uncertainty = \( 0.4 / 2 = 0.2 \) s. Finally, calculate the percentage uncertainty: \( (0.2 / 4.0) \times 100 = 5.0\% \).

1 mark for calculating the absolute uncertainty as 0.2 s (or identifying the range as 0.4 s). 1 mark for the correct calculation of percentage uncertainty of 5.0%.

According to Faraday's law of electromagnetic induction, the induced emf is given by: \( \epsilon = N \frac{\Delta \Phi}{\Delta t} = N A \frac{\Delta B}{\Delta t} \). Substituting the given values: \( \epsilon = 150 \times (2.5 \times 10^{-3} \text{ m}^2) \times \frac{0.80 \text{ T} - 0.20 \text{ T}}{0.15 \text{ s}} = 150 \times 2.5 \times 10^{-3} \times 4.0 = 1.5 \text{ V} \).

1 mark for calculating the rate of change of magnetic flux density as 4.0 T s^-1 (or showing correct substitution into Faraday's law). 1 mark for the correct final emf of 1.5 V.

By conservation of energy, the elastic potential energy stored in the compressed spring is entirely converted into the kinetic energy of the car: \( \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \). This simplifies to: \( v = x \sqrt{\frac{k}{m}} \). Substituting the given values: \( v = 0.040 \times \sqrt{\frac{450}{0.12}} = 0.040 \times \sqrt{3750} \approx 2.45 \text{ m s}^{-1} \) (or \( 2.4 \text{ m s}^{-1} \) to 2 significant figures).

1 mark for equating kinetic energy and elastic potential energy, or calculating the stored energy as 0.36 J. 1 mark for the correct final velocity of 2.4 m s^-1 or 2.45 m s^-1.

The relationship between activity \( A \), decay constant \( \lambda \), and the number of nuclei \( N \) is given by: \( A = \lambda N \). Rearranging for \( \lambda \): \( \lambda = \frac{A}{N} = \frac{2.4 \times 10^{11} \text{ Bq}}{6.8 \times 10^{15}} \approx 3.53 \times 10^{-5} \text{ s}^{-1} \). to 2 significant figures, this is \( 3.5 \times 10^{-5} \text{ s}^{-1} \).

1 mark for rearranging the activity equation and substituting the values correctly. 1 mark for the correct numerical value of 3.5 * 10^-5 with the correct unit of s^-1.

The fringe spacing \( w \) is given by the double-slit formula: \( w = \frac{\lambda D}{s} \). Here, \( \lambda = 632 \times 10^{-9} \text{ m} \), \( D = 1.8 \text{ m} \), and \( s = 0.15 \times 10^{-3} \text{ m} \). Substituting the values: \( w = \frac{632 \times 10^{-9} \times 1.8}{0.15 \times 10^{-3}} = 7.58 \times 10^{-3} \text{ m} \) (or \( 7.6 \text{ mm} \)).

1 mark for correct substitution into the fringe spacing formula with all quantities converted to standard SI units. 1 mark for the correct answer of 7.6 * 10^-3 m or 7.6 mm.

First, convert the temperature from Celsius to Kelvin: \( T = 27 + 273.15 = 300.15 \text{ K} \) (using \( 300 \text{ K} \) is acceptable). The r.m.s. speed is given by: \( c_{\text{rms}} = \sqrt{\frac{3 R T}{M}} \). Substituting the values: \( c_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 300}{4.0 \times 10^{-3}}} = \sqrt{1.87 \times 10^6} \approx 1367 \text{ m s}^{-1} \). To 2 significant figures, this is \( 1.4 \times 10^3 \text{ m s}^{-1} \).

1 mark for converting temperature to kelvin and substituting values into the r.m.s. speed formula. 1 mark for the correct final answer of 1.4 * 10^3 m s^-1 (accept answers in range 1.36 * 10^3 to 1.4 * 10^3 m s^-1).

The capacitance of a parallel plate capacitor is given by: \( C = \frac{\varepsilon_0 A}{d} \). Here, \( A = 0.080 \text{ m}^2 \) and \( d = 0.45 \times 10^{-3} \text{ m} \). Substituting the values: \( C = \frac{8.85 \times 10^{-12} \times 0.080}{0.45 \times 10^{-3}} \approx 1.57 \times 10^{-9} \text{ F} \) (or \( 1.6 \text{ nF} \)).

1 mark for substituting values correctly into the parallel plate capacitance equation. 1 mark for the correct value of 1.6 * 10^-9 F (or 1.6 nF).

The maximum acceleration \( a_{\text{max}} \) in simple harmonic motion is given by: \( a_{\text{max}} = \omega^2 A \), where angular frequency \( \omega = \frac{2\pi}{T} \). Thus, \( \omega = \frac{2\pi}{0.80} \approx 7.854 \text{ rad s}^{-1} \). Now, calculate \( a_{\text{max}} = (7.854)^2 \times 0.050 \approx 3.08 \text{ m s}^{-2} \), which is \( 3.1 \text{ m s}^{-2} \) to 2 significant figures.

1 mark for calculating the angular frequency (7.9 rad s^-1) or showing a correct substituted expression for maximum acceleration. 1 mark for the correct maximum acceleration of 3.1 m s^-2 (accept 3.08 m s^-2).

The volume of a uniform wire is given by \(V = \frac{\pi d^2 L}{4}\). The fractional uncertainty in the volume is given by \(\frac{\Delta V}{V} = 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). The percentage uncertainty in the diameter is \(\frac{0.01}{1.42} \times 100\% = 0.704\%\). The percentage uncertainty in the length is \(\frac{0.1}{84.0} \times 100\% = 0.119\%\). Therefore, the total percentage uncertainty is \(2 \times 0.704\% + 0.119\% = 1.408\% + 0.119\% = 1.527\%\), which is \(1.5\%\) to two significant figures.

1 mark for correctly identifying that the percentage uncertainty in diameter is doubled: \(2 \times \frac{0.01}{1.42} \times 100 = 1.41\%\). 1 mark for adding the percentage uncertainty in length (\(0.12\%\)) to obtain the correct final answer of \(1.5\%\) (or \(1.53\%\)).

The open circuit voltage is the electromotive force (emf) of the battery, \(E = 12.0\text{ V}\). When the resistor \(R = 6.0\ \Omega\) is connected, the terminal potential difference is \(V = 9.0\text{ V}\). The current in the circuit is \(I = \frac{V}{R} = \frac{9.0\text{ V}}{6.0\ \Omega} = 1.5\text{ A}\). Using the equation \(E = V + Ir\), we find \(12.0 = 9.0 + 1.5r\), which gives \(1.5r = 3.0\), hence \(r = 2.0\ \Omega\).

1 mark for calculating the current in the circuit: \(I = 1.5\text{ A}\) (or for setting up the potential divider equation \(\frac{9.0}{12.0} = \frac{6.0}{6.0 + r}\)). 1 mark for calculating the correct internal resistance of \(2.0\ \Omega\).

Using Faraday's law of induction, the magnitude of the induced emf is given by \(\epsilon = N \frac{\Delta \Phi}{\Delta t} = N A \frac{\Delta B}{\Delta t}\). Substituting the given values: \(\epsilon = 150 \times (2.4 \times 10^{-3}\text{ m}^2) \times \frac{0.35\text{ T} - 0.05\text{ T}}{0.12\text{ s}} = 0.36 \times \frac{0.30}{0.12} = 0.90\text{ V}\).

1 mark for calculating the rate of change of magnetic flux density (\(2.5\text{ T s}^{-1}\)) or the change in flux linkage (\(0.108\text{ Wb-turns}\)). 1 mark for obtaining the correct magnitude of the induced emf of \(0.90\text{ V}\) (or \(0.9\text{ V}\)).

The initial gravitational potential energy is \(E_p = mgh = 0.25 \times 9.81 \times 1.2 = 2.943\text{ J}\). The final kinetic energy is \(E_k = \frac{1}{2}mv^2 = 0.5 \times 0.25 \times 3.8^2 = 1.805\text{ J}\). The energy transferred to the surroundings due to friction is the loss of mechanical energy: \(W = E_p - E_k = 2.943 - 1.805 = 1.138\text{ J}\), which rounds to \(1.1\text{ J}\) (to 2 significant figures).

1 mark for calculating both the initial potential energy (\(2.94\text{ J}\)) and the final kinetic energy (\(1.81\text{ J}\)). 1 mark for calculating the difference to find the energy transferred to the surroundings: \(1.1\text{ J}\) (accept \(1.14\text{ J}\)).

First, convert the elapsed time into seconds: \(t = 8.0\text{ hours} \times 3600\text{ s hour}^{-1} = 28800\text{ s}\). Using the radioactive decay equation: \(A = A_0 e^{-\lambda t} = (5.60 \times 10^5) \times e^{-(2.10 \times 10^{-5} \times 28800)} = (5.60 \times 10^5) \times e^{-0.6048} = (5.60 \times 10^5) \times 0.54618 = 3.0586 \times 10^5\text{ Bq}\). This is \(3.1 \times 10^5\text{ Bq}\) to 2 significant figures.

1 mark for converting time to seconds (\(28800\text{ s}\)) and substituting into the exponential decay formula. 1 mark for obtaining the correct final activity of \(3.1 \times 10^5\text{ Bq}\) (or \(3.06 \times 10^5\text{ Bq}\)).

Using the double-slit equation \(w = \frac{\lambda D}{d}\), where \(\lambda = 589 \times 10^{-9}\text{ m}\), \(D = 1.5\text{ m}\), and \(d = 0.30 \times 10^{-3}\text{ m}\). Substituting these values gives \(w = \frac{589 \times 10^{-9} \times 1.5}{0.30 \times 10^{-3}} = 2.945 \times 10^{-3}\text{ m} = 2.9\text{ mm}\) (to 2 significant figures).

1 mark for substituting the values into the correct formula with consistent units. 1 mark for the correct fringe spacing of \(2.9\text{ mm}\) (or \(2.95\text{ mm}\), or \(2.9 \times 10^{-3}\text{ m}\)).

The absolute temperature is \(T = 27 + 273 = 300\text{ K}\). The mean kinetic energy of a gas atom is given by \(E_k = \frac{3}{2}kT\). Substituting the values: \(E_k = 1.5 \times (1.38 \times 10^{-23}\text{ J K}^{-1}) \times 300\text{ K} = 6.21 \times 10^{-21}\text{ J}\). To 2 significant figures, this is \(6.2 \times 10^{-21}\text{ J}\).

1 mark for converting the temperature to Kelvin (\(300\text{ K}\)) and identifying the correct formula. 1 mark for calculating the correct kinetic energy of \(6.2 \times 10^{-21}\text{ J}\) (or \(6.21 \times 10^{-21}\text{ J}\)).

The energy stored in a capacitor is given by \(E = \frac{1}{2} C V^2\). The initial energy is \(E_i = 0.5 \times (47 \times 10^{-6}\text{ F}) \times (15\text{ V})^2 = 5.29 \times 10^{-3}\text{ J}\). The final energy is \(E_f = 0.5 \times (47 \times 10^{-6}\text{ F}) \times (5.0\text{ V})^2 = 0.59 \times 10^{-3}\text{ J}\). The energy discharged is the difference: \(\Delta E = E_i - E_f = 5.29 \times 10^{-3} - 0.59 \times 10^{-3} = 4.70 \times 10^{-3}\text{ J}\), which is \(4.7\text{ mJ}\).

1 mark for calculating the difference in stored energy using the formula \(\Delta E = \frac{1}{2}C(V_i^2 - V_f^2)\) (award 0 marks if the formula is incorrectly written as \(\Delta E = \frac{1}{2}C(V_i - V_f)^2\)). 1 mark for the correct energy value of \(4.7 \times 10^{-3}\text{ J}\) or \(4.7\text{ mJ}\).

1. Find the mean diameter:
\(d_{\text{mean}} = \frac{1.22 + 1.25 + 1.21 + 1.24}{4} = 1.23\text{ mm}\)

2. Find the absolute uncertainty using half the range:
\(\text{Range} = 1.25 - 1.21 = 0.04\text{ mm}\)
\(\text{Uncertainty} = \frac{0.04}{2} = 0.02\text{ mm}\)

3. Calculate percentage uncertainty:
\(\text{Percentage uncertainty} = \left( \frac{0.02}{1.23} \right) \times 100 = 1.626\% \approx 1.6\%\)

• [1 mark] Correct calculation of the mean (\(1.23\text{ mm}\)) and the absolute uncertainty (\(0.02\text{ mm}\)).
• [1 mark] Correct calculation of the percentage uncertainty to 2 significant figures: \(1.6\%\) (accept \(1.63\%\)).

Using the equation for terminal potential difference:
\(V = \mathcal{E} - Ir\)

For the first scenario:
\(1.40 = \mathcal{E} - 0.40r\) (Equation 1)

For the second scenario:
\(1.20 = \mathcal{E} - 0.80r\) (Equation 2)

Subtract Equation 2 from Equation 1:
\(1.40 - 1.20 = (\mathcal{E} - 0.40r) - (\mathcal{E} - 0.80r)\)
\(0.20 = 0.40r\)
\(r = 0.50\ \Omega\)

• [1 mark] Setting up simultaneous equations or expressing the relation \(r = -\frac{\Delta V}{\Delta I}\).
• [1 mark] Correct calculation of the internal resistance: \(0.50\ \Omega\) (with unit \(\Omega\)).

Using Faraday's Law of Electromagnetic Induction:
\(\text{emf} = N \frac{\Delta \Phi}{\Delta t} = N A \frac{\Delta B}{\Delta t}\)

Where:
\(N = 50\)
\(A = 4.0 \times 10^{-4}\text{ m}^2\)
\(\Delta B = 0.80 - 0.20 = 0.60\text{ T}\)
\(\Delta t = 0.15\text{ s}\)

Substitute the values:
\(\text{emf} = 50 \times (4.0 \times 10^{-4}) \times \frac{0.60}{0.15}\)
\(\text{emf} = 50 \times (4.0 \times 10^{-4}) \times 4.0 = 0.080\text{ V}\) (or \(80\text{ mV}\))

• [1 mark] Correct substitution into the Faraday's Law formula, demonstrating conversion of quantities.
• [1 mark] Correct final value: \(0.080\text{ V}\) (or \(80\text{ mV}\)) with appropriate units.

By conservation of energy, the loss in gravitational potential energy (GPE) is equal to the gain in kinetic energy (KE):
\(mg\Delta h = \frac{1}{2}mv^2\)

Since mass \(m\) cancels out on both sides:
\(g(h - 8.0) = \frac{1}{2}v^2\)
\(9.81 \times (h - 8.0) = 0.5 \times (15)^2\)
\(9.81 \times (h - 8.0) = 112.5\)
\(h - 8.0 = \frac{112.5}{9.81} \approx 11.47\text{ m}\)
\(h = 11.47 + 8.0 = 19.47\text{ m} \approx 19.5\text{ m}\)

• [1 mark] Correct application of conservation of energy principle to find the change in height \(\Delta h \approx 11.5\text{ m}\).
• [1 mark] Correct calculation of the initial height \(h = 19.5\text{ m}\) (accept \(19\text{ m}\) to \(20\text{ m}\) depending on rounding).

The fraction of activity remaining is:
\(\frac{A}{A_0} = \frac{1.0 \times 10^5}{8.0 \times 10^5} = \frac{1}{8}\)

Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), the sample has undergone exactly 3 half-lives.

Let \(T_{1/2}\) be the half-life:
\(3 \times T_{1/2} = 12\text{ hours}\)
\(T_{1/2} = 4.0\text{ hours}\)

• [1 mark] Deducing that the sample has decayed for 3 half-lives (or equivalent calculation using decay constant \(\lambda = 0.173\text{ h}^{-1}\)).
• [1 mark] Correct calculation of the half-life: \(4.0\text{ hours}\) (accept \(1.4 \times 10^4\text{ s}\) if converted correctly).

Using the double-slit formula for fringe spacing:
\(w = \frac{\lambda D}{s}\)

Where:
\(\lambda = 633 \times 10^{-9}\text{ m}\)
\(D = 1.80\text{ m}\)
\(s = 0.250 \times 10^{-3}\text{ m}\)

Substitute the values:
\(w = \frac{633 \times 10^{-9} \times 1.80}{0.250 \times 10^{-3}}\)
\(w = 4.558 \times 10^{-3}\text{ m} = 4.56\text{ mm}\)

• [1 mark] Correct substitution of all quantities converted into meters into the formula \(w = \frac{\lambda D}{s}\).
• [1 mark] Correct final answer: \(4.56 \times 10^{-3}\text{ m}\) (accept \(4.6 \times 10^{-3}\text{ m}\) or \(4.56\text{ mm}\)).

1. Convert the temperature to kelvins:
\(T = 27.0 + 273.15 = 300.15\text{ K}\) (accept \(300\text{ K}\))

2. Use the kinetic theory formula for rms speed:
\(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\)

Substitute the values:
\(v_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 300.15}{4.00 \times 10^{-3}}}\)
\(v_{\text{rms}} = \sqrt{\frac{7482.7}{4.00 \times 10^{-3}}} = \sqrt{1.871 \times 10^6} \approx 1370\text{ m s}^{-1}\) (or \(1.37 \times 10^3\text{ m s}^{-1}\))

• [1 mark] Correct temperature conversion to Kelvin (\(300\text{ K}\)) and correct selection of the formula.
• [1 mark] Correct final answer of \(1.37 \times 10^3\text{ m s}^{-1}\) (accept range \(1360\text{ m s}^{-1}\) to \(1370\text{ m s}^{-1}\)).

Using the formula for energy stored in a capacitor:
\(E = \frac{1}{2} C V^2\)

Where:
\(C = 470 \times 10^{-6}\text{ F}\)
\(V = 12.0\text{ V}\)

Substitute the values:
\(E = 0.5 \times (470 \times 10^{-6}) \times (12.0)^2\)
\(E = 0.5 \times 4.70 \times 10^{-4} \times 144 = 0.03384\text{ J}\) (or \(3.38 \times 10^{-2}\text{ J}\))

• [1 mark] Correct substitution into the energy formula, with capacitance properly converted into Farads.
• [1 mark] Correct final energy value: \(3.38 \times 10^{-2}\text{ J}\) (accept \(0.034\text{ J}\) or \(33.8\text{ mJ}\)).

Using Faraday's law of electromagnetic induction: \(\varepsilon = N \frac{\Delta \Phi}{\Delta t}\). Since the field is perpendicular, \(\Phi = BA\). The change in magnetic flux is \(\Delta \Phi = (0.25 \times 0.040) - 0 = 0.010\text{ Wb}\). The induced emf is therefore \(\varepsilon = 150 \times \frac{0.010}{0.12} = 12.5\text{ V}\).

[1 mark] for substituting correctly into Faraday's law: \(\varepsilon = \frac{150 \times 0.25 \times 0.040}{0.12}\) (or showing a change in flux linkage of \(1.5\text{ Wb-turns}\)). [1 mark] for the correct final answer of \(12.5\text{ V}\) (accept \(13\text{ V}\)).

Initial gravitational potential energy: \(E_p = mgh = 1.5 \times 9.81 \times 2.0 = 29.43\text{ J}\). Final kinetic energy: \(E_k = \frac{1}{2}mv^2 = 0.5 \times 1.5 \times 5.0^2 = 18.75\text{ J}\). By conservation of energy, the work done against friction is the loss in mechanical energy: \(W = E_p - E_k = 29.43 - 18.75 = 10.68\text{ J}\). Rounding to two significant figures gives \(11\text{ J}\) (or \(10.7\text{ J}\) to 3 s.f.).

[1 mark] for calculating both \(E_p = 29.4\text{ J}\) and \(E_k = 18.8\text{ J}\) (or setting up the energy conservation equation \(mgh = \frac{1}{2}mv^2 + W\)). [1 mark] for the correct final answer of \(11\text{ J}\) or \(10.7\text{ J}\).

First, convert the half-life to seconds: \(T_{1/2} = 5730 \times 3.16 \times 10^7\text{ s} = 1.81 \times 10^{11}\text{ s}\). Next, calculate the decay constant: \(\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{1.81 \times 10^{11}} = 3.83 \times 10^{-12}\text{ s}^{-1}\). Finally, calculate the activity: \(A = \lambda N = 3.83 \times 10^{-12} \times 2.5 \times 10^{18} = 9.57 \times 10^6\text{ Bq}\). Rounding to two significant figures gives \(9.6 \times 10^6\text{ Bq}\).

[1 mark] for calculating the decay constant \(\lambda \approx 3.83 \times 10^{-12}\text{ s}^{-1}\) (or showing the correct conversion of half-life to seconds and formula \(\lambda = \ln 2 / T_{1/2}\)). [1 mark] for the correct final answer of \(9.6 \times 10^6\text{ Bq}\) (accept \(9.5 \times 10^6\text{ Bq}\) to \(9.6 \times 10^6\text{ Bq}\)).

The volume of a cylinder is \(V = \frac{\pi d^2 L}{4}\). The percentage uncertainty in \(V\) is given by \(\frac{\Delta V}{V} \times 100 = 2\frac{\Delta d}{d}\times 100 + \frac{\Delta L}{L}\times 100\). Calculating individual percentage uncertainties: \(\%\text{ uncertainty in } d = \frac{0.01}{0.35} \times 100 = 2.86\%\) and \(\%\text{ uncertainty in } L = \frac{0.02}{1.24} \times 100 = 1.61\%\). Total percentage uncertainty = \(2 \times 2.86\% + 1.61\% = 7.33\%\). Rounded to two significant figures, this is \(7.3\%\).

[1 mark] for doubling the percentage uncertainty of the diameter, i.e., \(2 \times 2.86\% = 5.71\%\) (or showing correct sum of percentage uncertainties formula). [1 mark] for the correct final answer of \(7.3\%\) (accept range \(7.3\%\) to \(7.4\%\)).

The percentage uncertainty in the length \(L\) is:
\[ \frac{0.004}{0.800} \times 100\% = 0.50\% \]

The period \(T\) is calculated by dividing the total time \(t\) by the number of oscillations, \(N = 20\). Therefore, the percentage uncertainty in \(T\) is equal to the percentage uncertainty in the total time \(t\):
\[ \frac{0.2}{36.0} \times 100\% \approx 0.556\% \]

Since \(g = \frac{4\pi^2 L}{T^2}\), the percentage uncertainty in \(g\) is:
\[ \%\Delta g = \%\Delta L + 2 \times \%\Delta T = 0.50\% + 2 \times 0.556\% = 1.61\% \]

1 mark for the correct answer D. Award 1 mark for correct calculation of fractional uncertainties and combining them using the correct uncertainty propagation rules.

Using the terminal potential difference formula \(V = E - Ir\):

1. For \(R = 3.0\ \Omega\):
\[ I_1 = \frac{V_1}{R_1} = \frac{6.0\text{ V}}{3.0\ \Omega} = 2.0\text{ A} \]
\[ E = V_1 + I_1 r \implies E = 6.0 + 2.0r \]

2. For \(R = 8.0\ \Omega\):
\[ I_2 = \frac{V_2}{R_2} = \frac{8.0\text{ V}}{8.0\ \Omega} = 1.0\text{ A} \]
\[ E = V_2 + I_2 r \implies E = 8.0 + 1.0r \]

Equating the two expressions for \(E\):
\[ 6.0 + 2.0r = 8.0 + 1.0r \implies r = 2.0\ \Omega \]

Substitute \(r = 2.0\ \Omega\) back to find \(E\):
\[ E = 6.0 + 2.0(2.0) = 10.0\text{ V} \]

1 mark for the correct answer B. Award 1 mark for setting up and solving the simultaneous equations from the emf relationship.

According to Faraday's law of electromagnetic induction, the magnitude of the induced emf is given by:
\[ \varepsilon = N \frac{\Delta \Phi}{\Delta t} = N A \frac{\Delta B}{\Delta t} \]

First, calculate the area \(A\) of the square coil:
\[ A = (0.10\text{ m})^2 = 0.010\text{ m}^2 \]

The change in magnetic flux density is:
\[ \Delta B = 0.80\text{ T} - 0.20\text{ T} = 0.60\text{ T} \]

Substitute the values into the formula:
\[ \varepsilon = 50 \times 0.010\text{ m}^2 \times \frac{0.60\text{ T}}{0.15\text{ s}} = 2.0\text{ V} \]

1 mark for the correct answer C. Award 1 mark for calculating the area of the coil and successfully applying Faraday's law.

By the principle of conservation of energy:
\[ E_{\text{ki}} = E_{\text{pf}} + W_f \]

Calculate the vertical height \(h\) gained by the block:
\[ h = d \sin(30^\circ) = 4.0\text{ m} \times \sin(30^\circ) = 2.0\text{ m} \]

Calculate the gravitational potential energy gained:
\[ E_{\text{pf}} = mgh = 2.5\text{ kg} \times 9.81\text{ m s}^{-2} \times 2.0\text{ m} = 49.05\text{ J} \]

The work done against friction is:
\[ W_f = E_{\text{ki}} - E_{\text{pf}} = 120\text{ J} - 49.05\text{ J} = 70.95\text{ J} \]

Since \(W_f = f \times d\), where \(f\) is the average frictional force:
\[ f = \frac{W_f}{d} = \frac{70.95\text{ J}}{4.0\text{ m}} \approx 17.7\text{ N} \]

This rounds to \(18\text{ N}\).

1 mark for the correct answer C. Award 1 mark for applying the work-energy theorem to include gravitational potential energy and the work done against friction.

The activity \(A\) of a radioisotope as a function of time \(t\) is given by:
\[ A(t) = A_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]

Equating the activities of X and Y:
\[ 800 \left(\frac{1}{2}\right)^{\frac{t}{4.0}} = 100 \left(\frac{1}{2}\right)^{\frac{t}{8.0}} \]
\[ 8 \times 2^{-\frac{t}{4.0}} = 2^{-\frac{t}{8.0}} \]

Since \(8 = 2^3\):
\[ 2^{3 - \frac{t}{4.0}} = 2^{-\frac{t}{8.0}} \]

Equating the powers:
\[ 3 - \frac{t}{4.0} = -\frac{t}{8.0} \implies 3 = \frac{t}{8.0} \implies t = 24\text{ hours} \]

1 mark for the correct answer D. Award 1 mark for setting up the equations of exponential decay and resolving for the time \(t\).

The formula for the fringe width is:
\[ w = \frac{\lambda D}{s} \]

With the modifications:
- \(\lambda' = 1.20 \lambda\)
- \(s' = 0.5 s\)
- \(D' = 2 D\)

Substituting these values into the expression:
\[ w' = \frac{\lambda' D'}{s'} = \frac{(1.20 \lambda) (2 D)}{0.5 s} = 4.8 \left(\frac{\lambda D}{s}\right) = 4.8 w \]

1 mark for the correct answer C. Award 1 mark for expressing the modified variables correctly and evaluating the new proportional fringe width.

The root-mean-square speed of molecules is related to absolute temperature by \(c_{\text{rms}} \propto \sqrt{T}\).

Converting the temperatures to kelvin:
- \(T_1 = 27 + 273.15 = 300.15\text{ K} \approx 300\text{ K}\)
- \(T_2 = 327 + 273.15 = 600.15\text{ K} \approx 600\text{ K}\)

The ratio of the speeds is:
\[ \frac{c_{\text{rms}, 2}}{c_{\text{rms}, 1}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{600}{300}} = \sqrt{2} \approx 1.41 \]

1 mark for the correct answer B. Award 1 mark for converting temperatures into Kelvin scale and utilizing the relation \(c_{\text{rms}} \propto \sqrt{T}\).

First, find the energy \(E_1\) stored in the parallel combination:
- Equivalent capacitance: \(C_{\text{parallel}} = 2C\)
- Energy: \(E_1 = \frac{1}{2} C_{\text{parallel}} V^2 = C V^2\)

Next, find the potential difference \(V_{\text{series}}\) across the series combination for the same energy:
- Equivalent capacitance: \(C_{\text{series}} = \frac{C}{2}\)
- Energy: \(E_1 = \frac{1}{2} C_{\text{series}} V_{\text{series}}^2 = \frac{C}{4} V_{\text{series}}^2\)

Equating the two energy equations:
\[ C V^2 = \frac{C}{4} V_{\text{series}}^2 \implies V_{\text{series}}^2 = 4 V^2 \implies V_{\text{series}} = 2V \]

1 mark for the correct answer C. Award 1 mark for calculating the equivalent capacitance of both configurations and resolving the potential difference relationship.

1. Calculate the mean of the raw measurements: \(\text{Mean raw value} = \frac{1.22 + 1.25 + 1.21 + 1.24 + 1.23}{5} = 1.23\text{ mm}\). 2. Correct for the zero error of \(+0.03\text{ mm}\): \(\text{Corrected mean value} = 1.23 - 0.03 = 1.20\text{ mm}\). 3. Determine the absolute uncertainty in the readings using half the range of the raw measurements: \(\text{Uncertainty} = \frac{1.25 - 1.21}{2} = 0.02\text{ mm}\). 4. Calculate the percentage uncertainty in the corrected mean: \(\text{Percentage uncertainty} = \frac{0.02}{1.20} \times 100\% \approx 1.67\% \approx 1.7\%\).

1 mark for the correct option B. Award 1 mark for calculating the corrected mean as \(1.20\text{ mm}\), the absolute uncertainty as \(0.02\text{ mm}\), and finding the percentage uncertainty to be \(1.7\%\). Reject A (calculated using the uncorrected mean \(1.23\text{ mm}\)). Reject C (calculated using the full range \(0.04\text{ mm}\) as the uncertainty).

Since the three resistors are connected in parallel, the potential difference across each resistor is equal to the emf \(\varepsilon\) of the cell. The power dissipated in a resistor of resistance \(r\) is given by \(P = \frac{\varepsilon^2}{r}\). 1. Power dissipated in the resistor of resistance \(R\): \(P_R = \frac{\varepsilon^2}{R}\). 2. Total power dissipated in the circuit: \(P_{\text{total}} = \frac{\varepsilon^2}{R} + \frac{\varepsilon^2}{2R} + \frac{\varepsilon^2}{3R} = \frac{\varepsilon^2}{R} \left(1 + \frac{1}{2} + \frac{1}{3}\right) = \frac{11\varepsilon^2}{6R}\). 3. Ratio of \(P_R\) to \(P_{\text{total}}\): \(\frac{P_R}{P_{\text{total}}} = \frac{\frac{\varepsilon^2}{R}}{\frac{11\varepsilon^2}{6R}} = \frac{6}{11}\).

1 mark for the correct option C. Award 1 mark for correctly identifying that power in parallel is given by \(V^2/R\), writing expressions for individual and total powers, and finding the ratio \(\frac{6}{11}\). Reject A (which would be correct for a series circuit).

According to Faraday's law of electromagnetic induction, the magnitude of the average induced emf is given by: \(E = \frac{\Delta \Phi}{\Delta t}\) where \(\Delta \Phi\) is the change in magnetic flux linkage. Initially, the plane of the coil is perpendicular to the field, so the initial magnetic flux linkage is: \(\Phi_{\text{initial}} = B A N\). When the coil is rotated through \(180^\circ\), its orientation relative to the field is reversed, so the final magnetic flux linkage is: \(\Phi_{\text{final}} = -B A N\). The change in magnetic flux linkage is: \(\Delta \Phi = |\Phi_{\text{final}} - \Phi_{\text{initial}}| = |-B A N - B A N| = 2 B A N\). Thus, the average induced emf is: \(E = \frac{2 B A N}{\Delta t}\).

1 mark for the correct option C. Award 1 mark for demonstrating that the change in flux linkage is \(2 B A N\) and applying Faraday's law to get \(E = \frac{2 B A N}{\Delta t}\). Reject B (which corresponds to a \(90^\circ\) rotation).

1. Calculate the total electrical energy input in \(3.0\text{ s}\): \(E_{\text{in}} = P \times t = 50\text{ W} \times 3.0\text{ s} = 150\text{ J}\). 2. Calculate the useful energy output, which consists of the gain in gravitational potential energy (\(\Delta E_p\)) and the gain in kinetic energy (\(\Delta E_k\)): \(\Delta E_p = m g h = 2.0\text{ kg} \times 9.81\text{ m s}^{-2} \times 4.5\text{ m} = 88.29\text{ J}\), \(\Delta E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 2.0\text{ kg} \times (3.0\text{ m s}^{-1})^2 = 9.0\text{ J}\), and \(E_{\text{out}} = \Delta E_p + \Delta E_k = 88.29 + 9.0 = 97.29\text{ J}\). 3. Calculate the average efficiency: \(\text{Efficiency} = \frac{E_{\text{out}}}{E_{\text{in}}} \times 100\% = \frac{97.29}{150} \times 100\% \approx 65\%\).

1 mark for the correct option C. Award 1 mark for calculating total input energy (150 J), sum of potential and kinetic energy changes (97.3 J), and obtaining an efficiency of approximately 65%. Reject B (if only GPE is used, giving 59%). Reject A (if only KE is used, giving 6.0%).

The activity of a radioactive sample is given by \(A = \lambda N\), and the number of remaining nuclei at time \(t\) is given by \(N = N_0 e^{-\lambda t}\). Let \(N_0\) be the initial number of nuclei for both X and Y. At any time \(t\), the activities of X and Y are: \(A_X(t) = \lambda N_0 e^{-\lambda t}\) and \(A_Y(t) = 3\lambda N_0 e^{-3\lambda t}\). At \(t = \frac{1}{\lambda}\): \(A_X = \lambda N_0 e^{-1}\) and \(A_Y = 3\lambda N_0 e^{-3}\). Taking the ratio of \(A_X\) to \(A_Y\): \(\frac{A_X}{A_Y} = \frac{\lambda N_0 e^{-1}}{3\lambda N_0 e^{-3}} = \frac{e^{-1}}{3 e^{-3}} = \frac{e^2}{3}\).

1 mark for the correct option C. Award 1 mark for writing the expressions for activities, substituting \(t = \frac{1}{\lambda}\), and simplifying the ratio of the exponentials to obtain \(\frac{e^2}{3}\). Reject A (inverse ratio) and D (missing the factor of 3 in denominator).

The fringe spacing \(w\) in a double-slit experiment is given by: \(w = \frac{\lambda D}{s}\). The distance \(y\) from the central maximum to the third-order bright fringe is: \(y = 3w = \frac{3 \lambda D}{s}\). When the slit separation becomes \(s' = \frac{s}{2}\) and the screen distance becomes \(D' = 2D\), the new distance \(y'\) becomes: \(y' = \frac{3 \lambda D'}{s'} = \frac{3 \lambda (2D)}{\frac{s}{2}} = 4 \times \left(\frac{3 \lambda D}{s}\right) = 4y\).

1 mark for the correct option C. Award 1 mark for identifying the proportionality \(y \propto \frac{D}{s}\), finding that doubling \(D\) and halving \(s\) increases \(y\) by a factor of 4, yielding \(4y\). Reject B (only one parameter changed) and D (incorrect multiplication).

According to the kinetic theory of gases, the mean kinetic energy of a gas molecule is directly proportional to its absolute temperature: \(E_k = \frac{3}{2} k T\). Therefore, when the absolute temperature is doubled from \(T\) to \(2T\), the mean kinetic energy of the molecules also doubles. Option A is incorrect because the mean square speed is doubled, not halved. Option B is incorrect because the root-mean-square speed increases by a factor of \(\sqrt{2}\), not 2. Option D is incorrect because collision frequency is proportional to the average molecular speed, which increases by a factor of \(\sqrt{2}\), not 2.

1 mark for the correct option C. Award 1 mark for explaining that mean kinetic energy is proportional to \(T\), and identifying why the other statements are incorrect. Reject B (confusing root-mean-square speed proportionality with temperature instead of its square root).

The total mechanical energy \(E\) of the satellite is the sum of its kinetic energy \(E_k\) and gravitational potential energy \(E_p\). 1. Gravitational potential energy: \(E_p = -\frac{G M m}{r}\). 2. For a circular orbit, the gravitational force provides the centripetal force: \(\frac{G M m}{r^2} = \frac{m v^2}{r} \implies m v^2 = \frac{G M m}{r}\). 3. Kinetic energy: \(E_k = \frac{1}{2} m v^2 = \frac{G M m}{2r}\). 4. Total mechanical energy: \(E = E_k + E_p = \frac{G M m}{2r} - \frac{G M m}{r} = -\frac{G M m}{2r}\).

1 mark for the correct option B. Award 1 mark for expressing both kinetic energy and gravitational potential energy in terms of gravitational parameters and adding them to find \(-\frac{G M m}{2r}\). Reject A (just potential energy) and C (just kinetic energy).

The density \( \rho \) of the wire is given by the formula:
\( \rho = \frac{m}{V} = \frac{m}{\pi (d/2)^2 L} = \frac{4m}{\pi d^2 L} \)

To find the percentage uncertainty in the density, we sum the percentage uncertainties of each variable, taking into account the power of 2 for the diameter:
\( \frac{\Delta \rho}{\rho} \times 100\% = \left( \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L} \right) \times 100\% \)

Let's calculate each individual percentage uncertainty:
- Mass: \( \frac{0.01}{0.55} \times 100\% \approx 1.82\% \)
- Diameter: \( \frac{0.01}{0.40} \times 100\% = 2.5\% \). Doubling this for \( d^2 \) gives \( 5.0\% \).
- Length: \( \frac{0.1}{50.0} \times 100\% = 0.2\% \)

Total percentage uncertainty:
\( 1.82\% + 5.0\% + 0.2\% = 7.02\% \approx 7.0\% \)

1 mark for the correct calculation of percentage uncertainty to 2 significant figures, showing clearly that the diameter uncertainty is doubled.

The power dissipated in the external resistor is given by:
\( P = I^2 R = \left( \frac{E}{R+r} \right)^2 R \)

According to the maximum power transfer theorem, maximum power is dissipated when the external resistance equals the internal resistance (\( R = r \)):
\( P_{\text{max}} = \left( \frac{E}{2r} \right)^2 r = \frac{E^2}{4r} \implies \frac{E^2}{r} = 4P_{\text{max}} \)

When \( R = 3r \), the power dissipated is:
\( P = \left( \frac{E}{3r+r} \right)^2 (3r) = \frac{3E^2 r}{16r^2} = \frac{3}{16} \frac{E^2}{r} \)

Substituting \( \frac{E^2}{r} = 4P_{\text{max}} \):
\( P = \frac{3}{16} (4 P_{\text{max}}) = \frac{12}{16} P_{\text{max}} = 0.75 P_{\text{max}} \)

1 mark for obtaining the correct fraction of maximum power (0.75).

The magnetic flux linkage through the coil as a function of time \( t \) is given by:
\( \Phi = BAN \cos(\omega t) \)

According to Faraday's law of electromagnetic induction, the induced emf \( \epsilon \) is:
\( \epsilon = -\frac{d\Phi}{dt} = BAN\omega \sin(\omega t) \)

- The maximum emf occurs when \( \sin(\omega t) = 1 \), which gives \( \epsilon_{\text{max}} = BAN\omega \).
- When the plane of the coil is perpendicular to the magnetic field lines, the angle between the normal to the coil and the magnetic field is \( 0 \) (i.e., \( \omega t = 0 \)). At this instant, \( \sin(\omega t) = 0 \), so the induced emf is zero.

1 mark for identifying both the correct maximum emf expression and the zero value when the coil plane is perpendicular to the magnetic field.

By the principle of conservation of energy, the initial gravitational potential energy at the top of the ramp is converted into kinetic energy at the bottom plus work done against friction:
\( E_p = E_k + W_{\text{friction}} \)

The height of the ramp is \( h = L \sin\theta \). Therefore:
\( mg(L \sin\theta) = \frac{1}{2}mv^2 + F L \)

Divide the entire equation by \( L \) to solve for \( F \):
\( mg \sin\theta = \frac{mv^2}{2L} + F \implies F = mg \sin\theta - \frac{mv^2}{2L} \)

1 mark for using conservation of energy and rearranging correctly to find the expression for the average frictional force.

After each half-life \( T \), the number of remaining radioactive nuclei is halved:
- After \( 1T \): \( N = \frac{1}{2}N_0 \)
- After \( 2T \): \( N = \frac{1}{4}N_0 \)
- After \( 3T \): \( N = \left(\frac{1}{2}\right)^3 N_0 = \frac{1}{8}N_0 = 0.125 \, N_0 \)

The number of nuclei that have decayed is the difference between the initial number of nuclei and the remaining nuclei:
\( N_{\text{decayed}} = N_0 - \frac{1}{8}N_0 = \frac{7}{8}N_0 = 0.875 \, N_0 \)

1 mark for calculating the decayed fraction as 7/8 (or 0.875) of the initial sample.

The formula for the fringe width \( w \) is:
\( w = \frac{\lambda D}{d} \)

In the second setup:
- New slit separation \( d' = 2d \)
- New screen distance \( D' = \frac{D}{2} \)
- Desired fringe width \( w' = 2w \)

Let \( \lambda' \) be the new wavelength. Using the formula:
\( w' = \frac{\lambda' D'}{d'} \implies 2w = \frac{\lambda' \left( \frac{D}{2} \right)}{2d} = \frac{\lambda' D}{4d} \)

Substitute \( w = \frac{\lambda D}{d} \) into the equation:
\( 2 \left( \frac{\lambda D}{d} \right) = \frac{\lambda' D}{4d} \implies 2\lambda = \frac{\lambda'}{4} \implies \lambda' = 8\lambda \)

1 mark for using the double-slit formula and correctly scaling the variables to find the new wavelength.

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

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

For the speed to double, the absolute temperature must increase by a factor of \( 2^2 = 4 \).

First, convert the initial temperature to Kelvin:
\( T_1 = 27^\circ\text{C} + 273 = 300\text{ K} \)

Calculate the new absolute temperature:
\( T_2 = 4 \times T_1 = 4 \times 300\text{ K} = 1200\text{ K} \)

Convert back to degrees Celsius:
\( t_2 = 1200 - 273 = 927^\circ\text{C} \)

1 mark for converting to Kelvin, applying the proportional relationship \( T \propto v_{\text{rms}}^2 \), and converting back to Celsius correctly.

The total energy stored in a network of capacitors connected to a constant potential difference \( V \) is given by:
\( E = \frac{1}{2} C_{\text{eq}} V^2 \)

**Case 1: Parallel combination**
The equivalent capacitance is:
\( C_{\text{parallel}} = C + 2C = 3C \)

The energy stored is:
\( E_1 = \frac{1}{2} (3C) V^2 = 1.5 \, C V^2 \)

**Case 2: Series combination**
The equivalent capacitance is:
\( C_{\text{series}} = \frac{C \times 2C}{C + 2C} = \frac{2}{3}C \)

The energy stored is:
\( E_2 = \frac{1}{2} \left(\frac{2}{3}C\right) V^2 = \frac{1}{3} \, C V^2 \)

**Ratio \( \frac{E_1}{E_2} \):**
\( \frac{E_1}{E_2} = \frac{1.5 \, C V^2}{\frac{1}{3} \, C V^2} = \frac{1.5}{1/3} = 4.5 \)

1 mark for calculating the equivalent capacitances of both combinations and finding their ratio to be 4.5.