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To find the percentage uncertainty in \(\rho\), we add the percentage uncertainties of the components. For a power term like \(d^2\), the percentage uncertainty is multiplied by its exponent (2).

1. Percentage uncertainty in \(R\):
\[\frac{0.05}{2.40} \times 100\% \approx 2.083\%\]

2. Percentage uncertainty in \(d\):
\[\frac{0.01}{0.38} \times 100\% \approx 2.632\%\]

3. Percentage uncertainty in \(L\):
\[\frac{0.002}{1.250} \times 100\% \approx 0.160\%\]

4. Combining these terms:
\[\frac{\Delta \rho}{\rho}\% = \frac{\Delta R}{R}\% + 2 \left(\frac{\Delta d}{d}\%\right) + \frac{\Delta L}{L}\% \]
\[\frac{\Delta \rho}{\rho}\% = 2.083\% + 2(2.632\%) + 0.160\% = 2.083\% + 5.264\% + 0.160\% = 7.507\% \approx 7.5\%\]

Award 1 mark for the correct answer C.
- Reject answers that do not double the uncertainty of the diameter d (which yields 4.9%, option A).

1. Let rightward motion be positive.
Initial momentum \(p_i = (3.0 \times 4.0) + (2.0 \times (-3.0)) = 12.0 - 6.0 = 6.0\text{ kg m s}^{-1}\).

2. After the collision, the total mass is \(M = 3.0 + 2.0 = 5.0\text{ kg}\).
By conservation of momentum, \(M v_f = p_i \implies 5.0 \times v_f = 6.0 \implies v_f = 1.2\text{ m s}^{-1}\).

3. Initial kinetic energy:
\[E_{k,i} = \frac{1}{2}(3.0)(4.0)^2 + \frac{1}{2}(2.0)(-3.0)^2 = 24.0 + 9.0 = 33.0\text{ J}\]

4. Final kinetic energy:
\[E_{k,f} = \frac{1}{2}(5.0)(1.2)^2 = 3.6\text{ J}\]

5. Loss in kinetic energy:
\[\Delta E_k = 33.0 - 3.6 = 29.4\text{ J} \approx 29\text{ J}\]

Award 1 mark for the correct answer C.
- Option A is the final kinetic energy.
- Option D is the initial kinetic energy.

The strain energy stored in an elastically deformed wire is given by \(U = \frac{1}{2} F x\), where \(F = W\) is the applied load and \(x\) is the extension.

Using Young's Modulus, \(E = \frac{\text{stress}}{\text{strain}} = \frac{F L}{A x}\), where \(L\) is the original length and \(A\) is the cross-sectional area. Since both wires are made of the same metal and have the same initial length, \(E\) and \(L\) are constants.

Thus, extension is given by:
\[x = \frac{W L}{A E} \propto \frac{1}{A} \propto \frac{1}{d^2}\]

Since the diameter of wire X is twice that of Y (\(d_X = 2d_Y\)), the area of X is four times that of Y (\(A_X = 4A_Y\)).

Therefore, the extension of X is one-quarter of the extension of Y:
\[x_X = \frac{1}{4} x_Y\]

Since both wires support the same load \(W\), the strain energy ratio is:
\[\frac{U_X}{U_Y} = \frac{\frac{1}{2} W x_X}{\frac{1}{2} W x_Y} = \frac{x_X}{x_Y} = 0.25\]

Award 1 mark for the correct answer B.

For a single slit diffraction pattern, the first minimum occurs at an angle \(\theta\) given by the single slit formula:
\[\sin\theta \approx \theta = \frac{\lambda}{b}\]

To double \(\theta\), the ratio \(\frac{\lambda}{b}\) must double.
- Option A: Doubling both \(\lambda\) and \(b\) keeps \(\frac{\lambda}{b}\) constant.
- Option B: Doubling \(\lambda\) and halving \(b\) increases \(\frac{\lambda}{b}\) by a factor of 4.
- Option C: Doubling \(\lambda\) and keeping \(b\) constant doubles \(\frac{\lambda}{b}\), which doubles \(\theta\).
- Option D: Halving \(\lambda\) and keeping \(b\) constant halves \(\theta\).

Award 1 mark for the correct answer C.

1. Find initial reading across the LDR (\(R_1 = 24.0\ \text{k}\Omega\)):
\[V_{\text{initial}} = E \times \frac{R_1}{R_{\text{fixed}} + R_1} = 9.0 \times \frac{24.0}{12.0 + 24.0} = 9.0 \times \frac{24.0}{36.0} = 6.00\text{ V}\]

2. Find final reading across the LDR (\(R_2 = 4.0\ \text{k}\Omega\)):
\[V_{\text{final}} = E \times \frac{R_2}{R_{\text{fixed}} + R_2} = 9.0 \times \frac{4.0}{12.0 + 4.0} = 9.0 \times \frac{4.0}{16.0} = 2.25\text{ V}\]

3. Calculate the change in the voltmeter reading:
\[\Delta V = 6.00\text{ V} - 2.25\text{ V} = 3.75\text{ V} \text{ (decrease)}\]

Award 1 mark for the correct answer B.

1. Find the combined capacitance of the series branch (\(C_{\text{series}}\)):
\[\frac{1}{C_{\text{series}}} = \frac{1}{6.0\ \mu\text{F}} + \frac{1}{12.0\ \mu\text{F}} = \frac{2 + 1}{12.0\ \mu\text{F}} = \frac{3}{12.0\ \mu\text{F}}\]
\[C_{\text{series}} = 4.0\ \mu\text{F}\]

2. This branch is in parallel with the third capacitor (\(4.0\ \mu\text{F}\)):
\[C_{\text{total}} = C_{\text{series}} + C_3 = 4.0\ \mu\text{F} + 4.0\ \mu\text{F} = 8.0\ \mu\text{F}\]

Award 1 mark for the correct answer C.
- Option A corresponds to all three capacitors in series.
- Option D corresponds to all three capacitors in parallel.

Using the redshift formula for non-relativistic speeds:
\[z = \frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\]

where:
- \(\lambda_0 = 656\text{ nm}\) (the source/laboratory wavelength)
- \(\Delta \lambda = 682\text{ nm} - 656\text{ nm} = 26\text{ nm}\)
- \(c = 3.00 \times 10^8\text{ m s}^{-1}\)

Calculate speed \(v\):
\[v = c \times \frac{\Delta \lambda}{\lambda_0} = 3.00 \times 10^8 \times \frac{26}{656} \approx 1.19 \times 10^7\text{ m s}^{-1}\]

Award 1 mark for the correct answer B.
- Option A is obtained by incorrectly using 682 nm as the denominator \(\lambda_0\).

Let \(C\) be the heat capacity of the system (liquid + container).
In both cases, the temperature rise is \(\Delta T = 10^\circ\text{C}\), so the net thermal energy required is the same: \(Q = C \Delta T\).

1. In the first case:
- Heating time \(t_1 = 5.0\text{ minutes} = 300\text{ s}\).
- Energy supplied = \(P_1 t_1 = 40 \times 300 = 12000\text{ J}\).
- Heat lost to surroundings = \(h t_1 = 300h\).
- Net heat transferred = \(12000 - 300h\).

2. In the second case:
- Heating time \(t_2 = 2.5\text{ minutes} = 150\text{ s}\).
- Energy supplied = \(P_2 t_2 = 70 \times 150 = 10500\text{ J}\).
- Heat lost to surroundings = \(h t_2 = 150h\).
- Net heat transferred = \(10500 - 150h\).

Since \(Q\) is the same in both cases:
\[12000 - 300h = 10500 - 150h\]
\[12000 - 10500 = 300h - 150h\]
\[1500 = 150h \implies h = 10\text{ W}\]

Award 1 mark for the correct answer B.

First, calculate the percentage uncertainty for each measured value:
1. For resistance \(R\):
\[ \frac{\Delta R}{R} \times 100\% = \frac{0.6}{24.0} \times 100\% = 2.5\% \]

2. For diameter \(d\):
\[ \frac{\Delta d}{d} \times 100\% = \frac{0.01}{0.50} \times 100\% = 2.0\% \]

3. For length \(L\):
\[ \frac{\Delta L}{L} \times 100\% = \frac{0.005}{1.250} \times 100\% = 0.4\% \]

Now, find the total percentage uncertainty in \(\rho\). Since \(\rho \propto \frac{R d^2}{L}\), we sum the fractional uncertainties, doubling the fractional uncertainty of \(d\) due to the power of 2:
\[ \frac{\Delta \rho}{\rho} \times 100\% = \left(\frac{\Delta R}{R} \times 100\%\right) + 2\left(\frac{\Delta d}{d} \times 100\%\right) + \left(\frac{\Delta L}{L} \times 100\%\right) \]
\[ \frac{\Delta \rho}{\rho} \times 100\% = 2.5\% + 2(2.0\%) + 0.4\% = 2.5\% + 4.0\% + 0.4\% = 6.9\% \]

Award 1 mark for the correct answer C. Method breakdown: calculating the individual percentage uncertainties (2.5%, 2.0%, 0.4%) and combining them correctly by doubling the uncertainty of the diameter term.

Let \(E_{k,i}\) be the initial kinetic energy and \(E_{k,f}\) be the final kinetic energy.

Before the collision:
- Only the mass \(3M\) is moving.
- Total momentum: \(p_i = 3M v\).
- Total kinetic energy: \(E_{k,i} = \frac{1}{2} (3M) v^2 = 1.5 M v^2\).

After the collision:
- The two blocks stick together, so the combined mass is \(3M + M = 4M\).
- By conservation of linear momentum:
\[ p_f = p_i \implies 4M v_f = 3M v \implies v_f = \frac{3}{4} v = 0.75 v \]
- Total kinetic energy after collision:
\[ E_{k,f} = \frac{1}{2} (4M) v_f^2 = 2 M (0.75 v)^2 = 2 M (0.5625 v^2) = 1.125 M v^2 \]

Now calculate the ratio:
\[ \frac{E_{k,f}}{E_{k,i}} = \frac{1.125 M v^2}{1.5 M v^2} = 0.75 \]

Award 1 mark for the correct option C. Correct application of conservation of momentum to find the final velocity (0.75v) and ratio of kinetic energies.

Since the cyclist is moving at a constant speed, her kinetic energy is constant.
Therefore, the rate of loss of gravitational potential energy must equal the rate at which thermal energy is generated due to resistive forces.

Let \(s\) be the distance travelled along the slope. In a time interval of \(\Delta t = 1\text{ s}\), the distance travelled along the slope is:
\[ s = v \Delta t = 8.0\text{ m} \]

The vertical height drop \(\Delta h\) corresponding to this distance is:
\[ \Delta h = 8.0 \times \frac{5.0}{100} = 0.40\text{ m} \]

The rate of loss of gravitational potential energy is:
\[ \frac{\Delta E_p}{\Delta t} = \frac{m g \Delta h}{\Delta t} = \frac{80\text{ kg} \times 9.81\text{ m s}^{-2} \times 0.40\text{ m}}{1\text{ s}} = 313.92\text{ W} \approx 310\text{ W} \]

This is the rate at which thermal energy is generated.

Award 1 mark for the correct option B. Method: Determine the vertical rate of descent (0.40 m/s) and use P = mg * (dh/dt) to find the rate of energy dissipation.

The Young modulus \(E\) is given by:
\[ E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{F L}{A x} \]

Rearranging for extension \(x\):
\[ x = \frac{F L}{A E} \]

Since both wires are made of the same metal, \(E\) is identical. The load \(F\) is also identical. Thus, the extension is:
\[ x \propto \frac{L}{A} \propto \frac{L}{d^2} \]
where \(d\) is the diameter because cross-sectional area \(A = \frac{\pi d^2}{4}\).

Let the properties of wire \(X\) be \(L_X\) and \(d_X\). For wire \(Y\):
- \(L_Y = 2 L_X\)
- \(d_Y = 0.5 d_X\)

Therefore, the ratio of extensions is:
\[ \frac{x_Y}{x_X} = \frac{L_Y}{L_X} \times \left(\frac{d_X}{d_Y}\right)^2 = (2) \times \left(\frac{1}{0.5}\right)^2 = 2 \times 2^2 = 8 \]

Award 1 mark for the correct option C. This relies on knowing the relationship between extension, length, and diameter (x is proportional to L/d^2), showing that halving the diameter increases extension by a factor of 4, and doubling the length increases it by a factor of 2, yielding 8 times the extension.

The grating spacing \(d\) is given by:
\[ d = \frac{1}{N} = \frac{1}{5.00 \times 10^5 \text{ m}^{-1}} = 2.00 \times 10^{-6}\text{ m} = 2000\text{ nm} \]

The diffraction grating formula is:
\[ d \sin\theta = n \lambda \]

For a bright spot to be observed, \(\sin\theta\) must be less than or equal to 1. The maximum possible order of diffraction \(n_{\text{max}}\) is:
\[ n_{\text{max}} \le \frac{d}{\lambda} = \frac{2000\text{ nm}}{633\text{ nm}} \approx 3.16 \]

Since order \(n\) must be an integer, the maximum observed order is \(n = 3\).

The total number of bright spots includes:
- The central zero-order maximum (\(n = 0\))
- Three orders of maxima on each side of the central maximum (\(n = 1, 2, 3\) on both sides)

Total number of spots = \(2n_{\text{max}} + 1 = 2(3) + 1 = 7\).

Award 1 mark for the correct option D. Method: Calculate line spacing d, use d*sin(theta) = n*lambda with sin(theta) <= 1 to find maximum order n = 3, and calculate total spots as 2n + 1 = 7.

For a series potential divider, the output voltage across the thermistor is given by:
\[ V_{\text{out}} = V_{\text{supply}} \times \frac{R_{\text{th}}}{R + R_{\text{th}}} \]

At temperature \(10^\circ\text{C}\):
- \(R_{\text{th}} = 8.0\text{ k}\Omega\)
- \(V_{\text{out, 10}} = 9.0 \times \frac{8.0}{2.0 + 8.0} = 9.0 \times \frac{8.0}{10.0} = 7.2\text{ V}\)

At temperature \(40^\circ\text{C}\):
- \(R_{\text{th}} = 1.0\text{ k}\Omega\)
- \(V_{\text{out, 40}} = 9.0 \times \frac{1.0}{2.0 + 1.0} = 9.0 \times \frac{1.0}{3.0} = 3.0\text{ V}\)

The change in output voltage is:
\[ \Delta V_{\text{out}} = |7.2\text{ V} - 3.0\text{ V}| = 4.2\text{ V} \]

Award 1 mark for the correct option C. Calculate V_out at both temperatures and subtract to find the difference of 4.2 V.

The formula for gravitational potential \(\phi\) at a distance \(r\) from the center of a uniform spherical mass of mass \(M\) is:
\[ \phi = -\frac{GM}{r} \]

At the surface of the planet, the distance from the center is \(r = R\). The potential is:
\[ \phi_{\text{surface}} = -\frac{GM}{R} = -64\text{ MJ kg}^{-1} \]

At a height of \(3R\) above the surface of the planet, the distance from the center is:
\[ r' = R + 3R = 4R \]

The potential at this distance is:
\[ \phi' = -\frac{GM}{4R} = \frac{1}{4} \left(-\frac{GM}{R}\right) = \frac{1}{4} (-64\text{ MJ kg}^{-1}) = -16\text{ MJ kg}^{-1} \]

Award 1 mark for the correct option A. Method: Recognize that height above surface must be added to radius to find the total distance from the center (4R), and scale potential inversely with distance.

By Einstein's photoelectric equation:
\[ E_{\text{max}} = hf - \Phi = hf - hf_0 \]
where \(f_0\) is the threshold frequency.

From the first scenario:
\[ E_{\text{max}} = hf - hf_0 \quad \text{--- (Equation 1)} \]

From the second scenario:
\[ 2.0 E_{\text{max}} = h(1.5f) - hf_0 = 1.5hf - hf_0 \quad \text{--- (Equation 2)} \]

Substitute Equation 1 into Equation 2:
\[ 2(hf - hf_0) = 1.5hf - hf_0 \]
\[ 2hf - 2hf_0 = 1.5hf - hf_0 \]

Rearranging terms to solve for \(f_0\):
\[ 2hf - 1.5hf = 2hf_0 - hf_0 \]
\[ 0.5hf = hf_0 \]
\[ f_0 = 0.50f \]

Award 1 mark for the correct option B. Method: Formulate the two photoelectric equations, eliminate the maximum kinetic energy, and solve for the threshold frequency as 0.50f.

First, rearrange the equation for \(\eta\) to get \(\eta = \frac{F}{6\pi r v}\). Next, substitute the SI base units for each quantity in the expression: Force \(F\) has units of \(\text{kg m s}^{-2}\), radius \(r\) has units of \(\text{m}\), and speed \(v\) has units of \(\text{m s}^{-1}\). The constant \(6\pi\) is dimensionless. Substituting these units into the expression gives \([\eta] = \frac{\text{kg m s}^{-2}}{\text{m} \cdot \text{m s}^{-1}} = \text{kg m}^{-1}\text{s}^{-1}\).

1 mark for the correct option B.

By the principle of conservation of linear momentum, the total momentum before the explosion equals the total momentum after the explosion. Initial momentum is \(M v\). Final momentum is \(0.6M(1.5v) + 0.4M v_B = 0.9M v + 0.4M v_B\). Equating the two expressions gives \(M v = 0.9 M v + 0.4 M v_B\), which simplifies to \(0.1 M v = 0.4 M v_B\). Solving for the velocity of fragment B gives \(v_B = 0.25v\) in the original direction.

1 mark for the correct option B.

Using the principle of conservation of energy: Work done by the pulling force = Gain in GPE + Gain in KE + Work done against friction. First, the work done by the pulling force is \(W = F \cdot d = 35 \times 6.0 = 210\text{ J}\). Second, the gain in gravitational potential energy is \(\Delta E_p = m g d \sin 30^\circ = 4.0 \times 9.81 \times 6.0 \times 0.5 = 117.72\text{ J}\). Third, the gain in kinetic energy is \(\Delta E_k = 0.5 m v^2 = 0.5 \times 4.0 \times 4.5^2 = 40.5\text{ J}\). Finally, the work done against friction is \(W_{\text{friction}} = 210 - 117.72 - 40.5 = 51.78\text{ J}\), which rounds to \(52\text{ J}\).

1 mark for the correct option B.

The relationship between force \(F\), extension \(x\), original length \(L\), cross-sectional area \(A\), and Young modulus \(Y\) is given by \(x = \frac{F L}{A Y}\). Since the wires are in series, the tension \(F\) is the same. For the steel wire, the extension is \(x_s = \frac{F L}{(\pi d^2 / 4)(2E)} = \frac{2 F L}{\pi d^2 E}\). For the brass wire, the extension is \(x_b = \frac{F (2L)}{(\pi (2d)^2 / 4)(E)} = \frac{2 F L}{\pi d^2 E}\). The ratio of the extensions is therefore \(\frac{x_s}{x_b} = 1.0\).

1 mark for the correct option C.

The grating spacing is \(d = \frac{1}{5.0 \times 10^5} = 2.0 \times 10^{-6}\text{ m} = 2000\text{ nm}\). Using the grating equation \(d \sin \theta = n \lambda\), the maximum order \(n\) is found when \(\sin \theta \le 1\), which gives \(n \le \frac{2000}{630} \approx 3.17\). Thus, the maximum integer order is \(n = 3\). The total number of bright fringes is \(2n + 1 = 2(3) + 1 = 7\).

1 mark for the correct option D.

At room temperature, the potential difference across the fixed resistor is \(V_{\text{out}} = 6.0\text{ V} \times \frac{4.0\text{ k}\Omega}{8.0\text{ k}\Omega + 4.0\text{ k}\Omega} = 2.0\text{ V}\). As temperature increases, the resistance of the thermistor decreases, lowering the total resistance of the circuit. This increases the circuit current, thereby increasing the potential difference across the fixed resistor. The voltmeter reading therefore increases.

1 mark for the correct option A.

The acceleration of free fall at the surface is given by \(g = \frac{G M}{R^2}\). For the planet with mass \(2M\) and radius \(2R\), the new acceleration of free fall is \(g' = \frac{G (2M)}{(2R)^2} = \frac{2}{4} \frac{G M}{R^2} = 0.50 g\).

1 mark for the correct option B.

The maximum kinetic energy of the emitted photoelectrons depends only on the frequency of the incident light, which remains constant, so the maximum kinetic energy is unchanged. However, doubling the intensity at constant frequency doubles the rate at which photons arrive, which doubles the rate of emission of photoelectrons.

1 mark for the correct option C.

The percentage uncertainty in the volume \( V \) is determined by the sum of the percentage uncertainties of the terms in the formula. For \( V = \frac{\pi d^2 L}{4} \), we have: \( \frac{\Delta V}{V} = 2 \frac{\Delta d}{d} + \frac{\Delta L}{L} \). Calculating the individual fractional uncertainties: \( \frac{\Delta d}{d} = \frac{0.02}{1.20} \approx 0.0167 \) (or \( 1.67\% \)), and \( \frac{\Delta L}{L} = \frac{0.2}{50.0} = 0.0040 \) (or \( 0.40\% \)). Substituting these values: \( \frac{\Delta V}{V} = 2(0.0167) + 0.0040 = 0.0333 + 0.0040 = 0.0373 \). Expressed as a percentage, this is \( 3.7\% \).

Award 1 mark for the correct option B. Method: Identify that percentage uncertainty in \( V \) is \( 2 \times \% \text{ uncertainty in } d + \% \text{ uncertainty in } L \). Calculate \( 2 \times 1.67\% + 0.40\% = 3.7\% \).

According to Newton's second law, force is the rate of change of momentum: \( F = \frac{\Delta p}{\Delta t} \). For a continuous mass flow, this is given by \( F = \frac{\Delta m}{\Delta t} \Delta v \). Taking the initial direction of the sand as positive: \( u = +15 \text{ m s}^{-1} \) and \( v = -3.0 \text{ m s}^{-1} \). The change in velocity is \( \Delta v = v - u = -3.0 - 15 = -18 \text{ m s}^{-1} \). The magnitude of the force is: \( F = 0.40 \text{ kg s}^{-1} \times 18 \text{ m s}^{-1} = 7.2 \text{ N} \).

Award 1 mark for the correct option C. Method: Calculate change in velocity including direction (\( 15 - (-3) = 18 \text{ m s}^{-1} \)), then multiply by mass flow rate (\( 0.40 \times 18 = 7.2 \text{ N} \)).

Using the principle of conservation of energy: \( E_{\text{initial}} = E_{\text{final}} + W_{\text{lost}} \). The initial energy is entirely gravitational potential energy relative to the ground: \( E_{\text{initial}} = m g h_1 = 0.20 \times 9.81 \times 0.80 = 1.57 \text{ J} \). The final energy is the sum of kinetic energy and gravitational potential energy at the exit point: \( E_{\text{final}} = m g h_2 + \frac{1}{2} m v^2 = 0.20 \times 9.81 \times 0.30 + \frac{1}{2} \times 0.20 \times 2.5^2 = 0.59 \text{ J} + 0.63 \text{ J} = 1.22 \text{ J} \). The energy lost is: \( W_{\text{lost}} = E_{\text{initial}} - E_{\text{final}} = 1.57 - 1.22 = 0.35 \text{ J} \) (which rounds to \( 0.36 \text{ J} \) using unrounded values: \( 1.5696 - 1.2136 = 0.356 \text{ J} \)).

Award 1 mark for the correct option A. Method: Calculate initial GPE (1.57 J), final GPE (0.59 J), and final KE (0.63 J). Compute loss as \( 1.57 - (0.59 + 0.63) = 0.36 \text{ J} \).

The Young modulus is given by \( E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{e/L} = \frac{F L}{A e} \). Rearranging for extension \( e \): \( e = \frac{F L}{A E} = \frac{4 F L}{\rm \pi d^2 E} \). Since both wires are made of the same metal (same \( E \)) and are subjected to the same force \( F \), the extension is proportional to \( \frac{L}{d^2} \). Therefore, the ratio of the extensions is: \( \frac{e_X}{e_Y} = \frac{L_X}{L_Y} \times \left( \frac{d_Y}{d_X} \right)^2 \). Given \( L_X = 2 L_Y \) and \( d_X = 0.5 d_Y \): \( \frac{e_X}{e_Y} = 2 \times \left( \frac{1}{0.5} \right)^2 = 2 \times 2^2 = 8 \).

Award 1 mark for the correct option D. Method: Relate extension to length and diameter: \( e \propto L/d^2 \). Determine the ratio as \( 2 \times (2)^2 = 8 \).

For a tube closed at one end and open at the other, a node forms at the closed end and an antinode forms at the open end. The possible wavelengths \( \lambda \) of stationary waves are given by \( L = n \frac{\lambda}{4} \) where \( n \) is an odd integer (\( n = 1, 3, 5, \dots \)). The possible resonant frequencies are \( f = \frac{v}{\lambda} = \frac{n v}{4 L} \). The fundamental frequency (\( n=1 \)) is: \( f_1 = \frac{340}{4 \times 0.85} = 100 \text{ Hz} \). Since only odd harmonics are possible, the allowed frequencies are \( 100 \text{ Hz} \), \( 300 \text{ Hz} \), \( 500 \text{ Hz} \), etc. Therefore, \( 200 \text{ Hz} \) (which is an even harmonic) cannot produce a stationary wave.

Award 1 mark for the correct option B. Method: Determine the allowed resonant frequencies of a closed-open tube using \( f_n = \frac{n v}{4L} \) for odd \( n \). Identify that \( 200 \text{ Hz} \) is not an allowed harmonic.

Using the potential divider equation, \( V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R_{\text{fixed}} + R_{\text{LDR}}} \). In the dark: \( V_{\text{out, dark}} = 12 \times \frac{12}{4.0 + 12} = 9.0 \text{ V} \). In bright light: \( V_{\text{out, bright}} = 12 \times \frac{2.0}{4.0 + 2.0} = 4.0 \text{ V} \). The change in the output voltage is \( \Delta V_{\text{out}} = 9.0 \text{ V} - 4.0 \text{ V} = 5.0 \text{ V} \).

Award 1 mark for the correct option C. Method: Calculate \( V_{\text{out}} \) in the dark (9.0 V) and in bright light (4.0 V). Calculate the difference (5.0 V).

From Einstein's photoelectric equation, \( E_{\text{photon}} = \Phi + E_{K,\text{max}} \), where \( \Phi \) is the work function. For wavelength \( \lambda \): \( \frac{hc}{\lambda} = \Phi + E_K \) (Equation 1). For wavelength \( \frac{\lambda}{2} \), the photon energy is doubled: \( \frac{2hc}{\lambda} = \Phi + 3E_K \) (Equation 2). Substitute Equation 1 into Equation 2: \( 2(\Phi + E_K) = \Phi + 3E_K \). Expanding and solving for \( \Phi \): \( 2\Phi + 2E_K = \Phi + 3E_K \implies \Phi = E_K \).

Award 1 mark for the correct option B. Method: Formulate the two photoelectric equations for \( \lambda \) and \( \lambda/2 \). Substitute and solve for work function \( \Phi \) in terms of \( E_K \).

A meson must consist of a quark-antiquark pair. The charge of an up quark \( u \) is \( +\frac{2}{3}e \), and the charge of a down quark \( d \) is \( -\frac{1}{3}e \). The corresponding antiquarks have opposite charges: \( \bar{u} \) has charge \( -\frac{2}{3}e \), and \( \bar{d} \) has charge \( +\frac{1}{3}e \). Checking the options: A: \( u\bar{d} \) has charge \( +\frac{2}{3}e + \frac{1}{3}e = +1e \). B: \( d\bar{u} \) has charge \( -\frac{1}{3}e - \frac{2}{3}e = -1e \). C: \( u\bar{u} \) has charge \( +\frac{2}{3}e - \frac{2}{3}e = 0 \). D: \( d\bar{d}\bar{d} \) consists of three particles and is not a meson. Thus, \( u\bar{u} \) is the correct combination.

Award 1 mark for the correct option C. Method: Recognize that a meson is a quark-antiquark pair and calculate the net charge of each combination to find the one that equals 0.

To find the systematic error, we look at the relation for the calculated value: \(g_{\text{calc}} = \frac{4\pi^2 L_{\text{calc}}}{T_{\text{calc}}^2}\).
Letting \(L_{\text{calc}} = 1.01 L_{\text{true}}\) and \(T_{\text{calc}} = 0.98 T_{\text{true}}\), the ratio of the calculated to the true value of \(g\) is:
\[\frac{g_{\text{calc}}}{g_{\text{true}}} = \frac{1.01}{0.98^2} \approx 1.051\]
Thus, the systematic error in \(g\) is approximately \(+5\%\). Alternatively, using the linear approximation for small percentage changes:
\[\% \text{ error in } g \approx \% \text{ error in } L - 2(\% \text{ error in } T) = +1\% - 2(-2\%) = +5\%\].

For random uncertainty, the fractional uncertainties of the independent quantities add together (weighted by their powers):
\[\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} = 2\% + 2(3\%) = 8\%\]
Therefore, the random uncertainty is \(\pm 8\%\).

1 mark: Correctly calculate both the systematic error and random uncertainty to select D.

The initial momentum of the block is:
\[p_i = m v_i = 2.0\text{ kg} \times 8.0\text{ m s}^{-1} = 16.0\text{ kg m s}^{-1}\]

The impulse \(I\) of the constant resistive force \(F\) over time \(\Delta t\) is given by:
\[I = F \Delta t = -4.0\text{ N} \times 5.0\text{ s} = -20.0\text{ N s} = -20.0\text{ kg m s}^{-1}\]
(the negative sign denotes that the force acts in the opposite direction to the initial velocity).

Using the impulse-momentum theorem, the final momentum \(p_f\) is:
\[p_f = p_i + I = 16.0 - 20.0 = -4.0\text{ kg m s}^{-1}\]

The magnitude of the final momentum is \(4.0\text{ kg m s}^{-1}\), and the negative sign indicates it is directed in the opposite direction to the initial velocity.

1 mark: Calculate initial momentum and impulse, then determine the final momentum magnitude and direction to select B.

The grating equation is:
\[d \sin\theta = n \lambda\]
where \(d\) is the grating spacing, \(n = 3\), and \(\lambda = 600\text{ nm} = 6.0 \times 10^{-7}\text{ m}\).

Using \(\sin(53.1^\circ) = 0.80\):
\[d \times 0.80 = 3 \times (6.0 \times 10^{-7}\text{ m}) = 1.8 \times 10^{-6}\text{ m}\]
\[d = \frac{1.8 \times 10^{-6}\text{ m}}{0.80} = 2.25 \times 10^{-6}\text{ m}\]

The number of lines per millimetre, \(N\), is:
\[N = \frac{10^{-3}\text{ m}}{d} = \frac{10^{-3}\text{ m}}{2.25 \times 10^{-6}\text{ m}} \approx 4.44 \times 10^2\text{ lines mm}^{-1}\]

Thus, \(N \approx 4.4 \times 10^2\text{ lines mm}^{-1}\).

1 mark: Correctly use the grating equation to find grating spacing \(d\) and convert to lines per millimetre to select B.

The potential divider equation for the voltage across the LDR is:
\[V_{\text{out}} = V_{\text{supply}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\]

1. In bright light, with \(R_{\text{LDR}} = 2.0\text{ k}\Omega\):
\[V_{\text{out, light}} = 12\text{ V} \times \frac{2.0\text{ k}\Omega}{4.0\text{ k}\Omega + 2.0\text{ k}\Omega} = 12 \times \frac{2.0}{6.0} = 4.0\text{ V}\]

2. In darkness, with \(R_{\text{LDR}} = 20\text{ k}\Omega\):
\[V_{\text{out, dark}} = 12\text{ V} \times \frac{20\text{ k}\Omega}{4.0\text{ k}\Omega + 20\text{ k}\Omega} = 12 \times \frac{20}{24} = 10.0\text{ V}\]

The increase in output voltage is:
\[\Delta V_{\text{out}} = 10.0\text{ V} - 4.0\text{ V} = 6.0\text{ V}\]

1 mark: Calculate \(V_{\text{out}}\) for both states and find their difference to select C.

The total energy \(E\) of a satellite in a circular orbit of radius \(r\) is the sum of its kinetic energy \(E_k\) and gravitational potential energy \(E_p\).

Here, \(r = 3R\).
\[E_p = -\frac{GMm}{r} = -\frac{GMm}{3R}\]

For a stable circular orbit, the centripetal force is provided by the gravitational force:
\[\frac{m v^2}{r} = \frac{GMm}{r^2} \implies m v^2 = \frac{GMm}{3R}\]

Therefore, the kinetic energy is:
\[E_k = \frac{1}{2} m v^2 = \frac{GMm}{6R}\]

The total energy in the orbit is:
\[E_{\text{orbit}} = E_k + E_p = \frac{GMm}{6R} - \frac{GMm}{3R} = -\frac{GMm}{6R}\]

At infinity and at rest, the total energy is \(E_{\infty} = 0\).

The minimum work \(W\) that must be done is the difference in total energy:
\[W = E_{\infty} - E_{\text{orbit}} = 0 - \left(-\frac{GMm}{6R}\right) = \frac{GMm}{6R}\]

1 mark: Express the total orbital energy of the satellite and calculate the change in energy to escape to infinity to select A.

The initial energy stored in the first capacitor of capacitance \(C\) charged to potential difference \(V\) is:
\[E_i = \frac{1}{2} C V^2 = \frac{Q_0^2}{2C}\]
where \(Q_0 = C V\) is the initial charge on the capacitor.

After the battery is disconnected and the capacitor is connected in parallel with an identical uncharged capacitor, the total charge \(Q_0\) is shared between the two capacitors. The combined capacitance of two identical capacitors in parallel is:
\[C_{\text{total}} = C + C = 2C\]

The total energy stored in the system after connection is:
\[E_f = \frac{Q_0^2}{2C_{\text{total}}} = \frac{Q_0^2}{2(2C)} = \frac{Q_0^2}{4C} = \frac{1}{2} E_i\]

Therefore, the ratio of final energy to initial energy is:
\[\frac{E_f}{E_i} = 0.50\]

1 mark: Use charge conservation and the formula for energy stored in a capacitor to find the ratio to select B.

Let \(P\) be the constant power of the heater and \(m\) be the mass of the sample.

1. For the solid heating phase (from \(t = 0\) to \(t = 120\text{ s}\)):
- Time interval: \(\Delta t_1 = 120\text{ s}\)
- Temperature increase: \(\Delta T = 80^\circ\text{C} - 20^\circ\text{C} = 60\text{ K}\)
- Specific heat capacity: \(c = 1.2\text{ kJ kg}^{-1}\text{ K}^{-1}\)

Using \(Q_1 = P \Delta t_1 = m c \Delta T\):
\[P \times 120 = m \times 1.2 \times 60\]
\[120 P = 72 m \implies P = 0.60 m\text{ (with } P \text{ in kW and } m \text{ in kg)}\]

2. For the melting phase (from \(t = 120\text{ s}\) to \(t = 300\text{ s}\)):
- Time interval: \(\Delta t_2 = 300 - 120 = 180\text{ s}\)
- Heat supplied: \(Q_2 = P \Delta t_2 = m L\), where \(L\) is the specific latent heat of fusion.

Substituting \(P = 0.60 m\):
\[(0.60 m) \times 180 = m L\]
\[L = 0.60 \times 180 = 108\text{ kJ kg}^{-1}\]

1 mark: Use the heating phase parameters to find the heater power in terms of mass, and substitute into the melting phase equation to select B.

First, find the redshift \(z\) of the galaxy:
\[z = \frac{\Delta \lambda}{\lambda_0} = \frac{672.4\text{ nm} - 656.3\text{ nm}}{656.3\text{ nm}} = \frac{16.1}{656.3} \approx 0.02453\]

Next, calculate the recession velocity \(v\) of the galaxy using the redshift equation \(v = z c\) (where \(c = 3.00 \times 10^8\text{ m s}^{-1} = 3.00 \times 10^5\text{ km s}^{-1}\)):
\[v = 0.02453 \times 3.00 \times 10^5\text{ km s}^{-1} \approx 7360\text{ km s}^{-1}\]

Finally, apply Hubble’s law \(v = H_0 d\) to find the distance \(d\):
\[d = \frac{v}{H_0} = \frac{7360\text{ km s}^{-1}}{70\text{ km s}^{-1}\text{ Mpc}^{-1}} \approx 105\text{ Mpc}\]

1 mark: Calculate the redshift, find the recession velocity, and use Hubble's law to determine the distance to select B.

(a) A systematic error causes all measurements to deviate from the true value in the same direction and by a constant bias (cannot be reduced by taking averages). A random error causes measurements to scatter randomly on both sides of the mean value (can be reduced by repeating and averaging).

(b)(i) The cross-sectional area \(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\).
The resistivity \(\rho = \frac{R A}{L} = \frac{4.20 \times 1.134 \times 10^{-7}}{1.250} = 3.81 \times 10^{-7}\ \Omega\text{ m}\).

(b)(ii) The fractional uncertainty formula is:
\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + \frac{\Delta L}{L} + 2 \frac{\Delta d}{d}\)
\(\frac{\Delta R}{R} = \frac{0.05}{4.20} \approx 0.0119\)
\(\frac{\Delta L}{L} = \frac{0.002}{1.250} = 0.0016\)
\(\frac{\Delta d}{d} = \frac{0.01}{0.38} \approx 0.0263\)
\(\frac{\Delta \rho}{\rho} = 0.0119 + 0.0016 + 2(0.0263) = 0.0661\)
Percentage uncertainty = \(0.0661 \times 100\% \approx 6.6\%\) (or \(7\%\) to 1 significant figure).

(a) Systematic error: constant bias, same direction deviation [1 mark]; Random error: scatter of readings about a mean, reduced by averaging [1 mark].
(b)(i) Correct area formula or substitution [1 mark]; substitution of values into resistivity formula [1 mark]; final answer \(3.8 \times 10^{-7}\ \Omega\text{ m}\) (accept \(3.81 \times 10^{-7}\ \Omega\text{ m}\)) [1 mark].
(b)(ii) Correct sum of fractional uncertainties with factor of 2 for diameter [1.57 marks]; substitution of individual values [1 mark]; final percentage uncertainty to 1 or 2 s.f. \(7\%\) or \(6.6\%\) [1 mark].

(a) The total momentum of a system remains constant, provided no external forces act on the system.

(b)(i) Let the direction of trolley A's initial motion be the positive direction.
Initial momentum: \(p_i = m_A u_A + m_B u_B = 0.75(2.4) + 1.25(-1.2) = 1.8 - 1.5 = 0.30\text{ kg m s}^{-1}\).
Final momentum: \(p_f = m_A v_A + m_B v_B = 0.75(-0.60) + 1.25 v_B = -0.45 + 1.25 v_B\).
By conservation of momentum: \(0.30 = -0.45 + 1.25 v_B \implies 1.25 v_B = 0.75 \implies v_B = +0.60\text{ m s}^{-1}\) (in the direction of A's initial velocity).

(b)(ii) Initial Kinetic Energy: \(E_{k,i} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2 = 0.5(0.75)(2.4)^2 + 0.5(1.25)(1.2)^2 = 2.16 + 0.90 = 3.06\text{ J}\).
Final Kinetic Energy: \(E_{k,f} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = 0.5(0.75)(-0.6)^2 + 0.5(1.25)(0.6)^2 = 0.135 + 0.225 = 0.36\text{ J}\).
Since the final kinetic energy is less than the initial kinetic energy, the collision is inelastic.

(a) Total momentum remains constant [1 mark], in an isolated system / when no external force acts [1 mark].
(b)(i) Conservation of momentum equation set up with signs correct [1 mark]; calculation of initial momentum \(0.30\text{ kg m s}^{-1}\) [1 mark]; correct final velocity \(0.60\text{ m s}^{-1}\) with positive sign/correct direction [1 mark].
(b)(ii) Calculation of initial KE as \(3.1\text{ J}\) or \(3.06\text{ J}\) [1 mark]; calculation of final KE as \(0.36\text{ J}\) [1.57 marks]; correct comparison showing kinetic energy is not conserved and concluding the collision is inelastic [1 mark].

(a) The torque of a couple is the product of one of the forces and the perpendicular distance between the lines of action of the two forces.

(b)(i) Let us take moments about the fulcrum at \(x = 1.2\text{ m}\) from the left end.
- Weight of uniform board: \(W_b = 24 \times 9.81 = 235.4\text{ N}\), acting at the center of gravity \(x = 1.8\text{ m}\) (distance from fulcrum = \(0.6\text{ m}\) to the right).
- Weight of diver: \(W_d = 65 \times 9.81 = 637.7\text{ N}\), acting at the right end \(x = 3.6\text{ m}\) (distance from fulcrum = \(2.4\text{ m}\) to the right).
- Anchoring force \(F\) acts downwards at the left end \(x = 0\text{ m}\) (distance from fulcrum = \(1.2\text{ m}\) to the left).
Sum of clockwise moments = Sum of anticlockwise moments:
\((W_b \times 0.6) + (W_d \times 2.4) = F \times 1.2\)
\((235.4 \times 0.6) + (637.7 \times 2.4) = 1.2 F\)
\(141.2 + 1530.5 = 1.2 F\)
\(1671.7 = 1.2 F \implies F = 1393\text{ N} \approx 1400\text{ N}\) (or \(1.4\text{ kN}\)).

(b)(ii) For vertical equilibrium, total upward force equals total downward force.
\(R = F + W_b + W_d\)
\(R = 1393 + 235.4 + 637.7 = 2266\text{ N} \approx 2270\text{ N}\) (or \(2.3\text{ kN}\)).

(a) Product of one force and perpendicular distance between the forces [2 marks].
(b)(i) Moment equation with correct weights and distances from the pivot [1.57 marks]; calculation of clockwise moments [1 mark]; final answer for \(F = 1400\text{ N}\) (accept \(1390\text{ N}\)) [1 mark].
(b)(ii) Equating sum of downward forces to upward reaction force [1.5 marks]; correct calculation of \(R = 2270\text{ N}\) (accept \(2300\text{ N}\) or \(2260\text{ N}\)) [1.5 marks].

(a) Work done is the product of the force and the displacement in the direction of the force.

(b)(i) The vertical height gained is \(h = 4.5 \sin(25^\circ) = 1.902\text{ m}\).
Gain in GPE: \(\Delta E_p = mgh = 8.0 \times 9.81 \times 1.902 = 149.3\text{ J} \approx 150\text{ J}\).

(b)(ii) Work done by pulling force: \(W_{\text{pull}} = F \times d = 55 \times 4.5 = 247.5\text{ J} \approx 248\text{ J}\).

(b)(iii) The final kinetic energy of the box is: \(E_k = \frac{1}{2} m v^2 = 0.5 \times 8.0 \times (2.2)^2 = 19.36\text{ J}\).
By the conservation of energy:
\(W_{\text{pull}} = \Delta E_p + E_k + W_{\text{friction}}\)
\(247.5 = 149.3 + 19.36 + W_{\text{friction}}\)
\(247.5 = 168.66 + W_{\text{friction}} \implies W_{\text{friction}} = 78.8\text{ J} \approx 79\text{ J}\).

(a) Force multiplied by displacement in the direction of the force [2 marks].
(b)(i) Calculation of vertical height \(h\) [1 mark]; calculation of GPE to get \(150\text{ J}\) (or \(149\text{ J}\)) [1 mark].
(b)(ii) Correct calculation of work done \(W = Fd = 248\text{ J}\) (or \(250\text{ J}\)) [1.57 marks].
(b)(iii) Correct formula and calculation of final KE (\(19.4\text{ J}\)) [1 mark]; setting up the energy conservation equation [1 mark]; final value for work against friction \(79\text{ J}\) [1 mark].

(a)(i) Stress is force per unit cross-sectional area.
(a)(ii) Strain is extension per unit original length.
(a)(iii) Young modulus is the ratio of tensile stress to tensile strain.

(b)(i) The load force is \(F = m g = 15 \times 9.81 = 147.15\text{ N}\).
Cross-sectional area \(A = 0.45 \times 10^{-6}\text{ m}^2\).
From Young modulus \(E = \frac{F L}{A e}\):
\(e = \frac{F L}{A E} = \frac{147.15 \times 2.2}{0.45 \times 10^{-6} \times 2.0 \times 10^{11}} = \frac{323.73}{90000} = 3.60 \times 10^{-3}\text{ m} = 3.6\text{ mm}\).

(b)(ii) The elastic potential energy stored is:
\(E_p = \frac{1}{2} F e = 0.5 \times 147.15 \times 3.60 \times 10^{-3} = 0.265\text{ J} \approx 0.27\text{ J}\).

(a)(i) force / cross-sectional area [1 mark].
(a)(ii) extension / original length [1 mark].
(a)(iii) stress / strain [1 mark].
(b)(i) Calculation of force \(F = 147\text{ N}\) [1 mark]; correct substitution of values including area conversion [1.57 marks]; final answer \(3.6\text{ mm}\) [1 mark].
(b)(ii) Elastic energy formula used [1 mark]; correct calculation \(0.26\text{ J}\) or \(0.27\text{ J}\) [1 mark].

(a) Sound waves from the source travel down the tube and reflect at the closed boundary. The incident and reflected waves, which have the same frequency and speed, travel in opposite directions and superpose. This produces a stationary wave with nodes where destructive interference occurs and antinodes where constructive interference occurs.

(b)(i) For a tube closed at one end, the fundamental mode occurs when the length of the tube is a quarter wavelength: \(L = \frac{\lambda}{4} \implies \lambda = 4L\).
Using wave equation \(v = f \lambda\):
\(v = f (4L) \implies 340 = 140 \times 4L\)
\(560 L = 340 \implies L = \frac{340}{560} \approx 0.607\text{ m} \approx 0.61\text{ m}\).

(b)(ii) The boundary conditions dictate that only odd harmonics can exist: \(f_n = n \times f_1\) for \(n = 1, 3, 5, \dots\).
The fundamental is \(140\text{ Hz}\) (first resonance), the next is \(3 \times 140 = 420\text{ Hz}\) (second resonance), and the third resonance corresponds to the fifth harmonic:
\(f_5 = 5 \times f_1 = 5 \times 140 = 700\text{ Hz}\).

(a) Reflection of wave at boundary [1 mark]; superposition of waves traveling in opposite directions [1 mark]; waves have same frequency and speed, producing nodes and antinodes [1 mark].
(b)(i) Identification that \(\lambda = 4L\) for fundamental [1 mark]; substitution of \(f\) and \(v\) [1.57 marks]; correct answer \(0.61\text{ m}\) (or \(0.607\text{ m}\)) [1 mark].
(b)(ii) Recognition that third resonance is 5th harmonic [1 mark]; final answer \(700\text{ Hz}\) [1 mark].

(a) e.m.f. is the energy converted from chemical (or other) forms to electrical energy per unit charge. Potential difference is the electrical energy converted to other forms of energy per unit charge.

(b)(i) Using the potential divider formula:
\(V_{\text{out}} = E \times \frac{R_{\text{thermistor}}}{R_{\text{fixed}} + R_{\text{thermistor}}}\)
\(V_{\text{out}} = 9.0 \times \frac{1800}{1200 + 1800} = 9.0 \times \frac{1800}{3000} = 5.4\text{ V}\).

(b)(ii) As the temperature of the thermistor increases, its resistance decreases (negative temperature coefficient thermistor). Since its resistance is a smaller fraction of the total circuit resistance, the share of the potential difference dropped across it decreases. Thus, the voltmeter reading decreases.

(a) e.m.f. is non-electrical energy converted to electrical per unit charge AND p.d. is electrical energy converted to other forms per unit charge [2 marks].
(b)(i) Correct potential divider expression [1 mark]; substitution of values [1 mark]; final answer \(5.4\text{ V}\) [1 mark].
(b)(ii) Statement that thermistor resistance decreases with temperature [1 mark]; explanation based on proportion of resistance or current increase [1.57 marks]; conclusion that reading decreases [1 mark].

### Step-by-Step Worked Solution

**Part (a)**
- For \(L = 0.800\text{ m}\):
- \(h_0 = 42.5\text{ cm}\)
- \(h_1 = 38.4\text{ cm}\)
- \(y = 42.5 - 38.4 = 4.1\text{ cm} = 0.041\text{ m}\)

**Part (b)**
- The absolute uncertainty in measuring the deflection \(y\) using a metre rule is typically \(\Delta y = \pm 0.1\text{ cm}\) (or \(\pm 1\text{ mm}\)) due to reading two positions (the zero and the loaded position) and parallax.
- Percentage uncertainty:
\[ \frac{0.1\text{ cm}}{4.1\text{ cm}} \times 100\% \approx 2.4\% \]

**Part (c) - Table of Results**
Let's construct a representative data set:

| \(L \ / \text{ m}\) | \(L^3 \ / \text{ m}^3\) | \(h_0 \ / \text{ m}\) | \(h_1 \ / \text{ m}\) | \(y \ / \text{ m}\) |
|---|---|---|---|---|
| 0.500 | 0.125 | 0.425 | 0.414 | 0.011 |
| 0.600 | 0.216 | 0.425 | 0.407 | 0.018 |
| 0.700 | 0.343 | 0.425 | 0.397 | 0.028 |
| 0.800 | 0.512 | 0.425 | 0.384 | 0.041 |
| 0.850 | 0.614 | 0.425 | 0.376 | 0.049 |
| 0.900 | 0.729 | 0.425 | 0.367 | 0.058 |

**Part (d) - Graph Plotting**
- Graph plotted with \(y\) on the vertical axis (scale: \(0.01\text{ m}\) per large square) and \(L^3\) on the horizontal axis (scale: \(0.1\text{ m}^3\) per large square).
- Points are plotted accurately within half a small square, and a straight line of best fit is drawn.

**Part (e) - Gradient and y-Intercept**
- Using two points on the line of best fit: \((0.150, 0.013)\) and \((0.700, 0.056)\):
\[ \text{Gradient } a = \frac{0.056 - 0.013}{0.700 - 0.150} = \frac{0.043}{0.550} \approx 0.0782\text{ m}^{-2} \]
- Calculating the y-intercept \(b\):
\[ y = a(L^3) + b \implies 0.013 = 0.0782(0.150) + b \implies b = 0.013 - 0.0117 = 0.0013\text{ m} \]

**Part (f) - Values of Constants**
- \(a = 0.078\text{ m}^{-2}\) (to 2 significant figures)
- \(b = 0.001\text{ m}\) (to 1 significant figure)

**[1 mark]** Measurement of a single value of \(y\) to nearest mm with unit.
**[1 mark]** Correct calculation of first deflection \(y\).
**[1 mark]** Estimate of absolute uncertainty of \(y\) is \(1\text{ mm}\) to \(2\text{ mm}\) with correct calculation of percentage uncertainty.
**[1 mark]** Range of \(L\) used: minimum \(L \le 0.550\text{ m}\) and maximum \(L \ge 0.850\text{ m}\).
**[1 mark]** Column headings: table must have headings and correct units for each column (e.g., \(L / \text{m}\), \(L^3 / \text{m}^3\), \(y / \text{m}\)).
**[1 mark]** Consistency of raw readings: all raw values of length and vertical position must be recorded to the nearest mm.
**[1 mark]** Significant figures: \(L^3\) must be given to the same number of significant figures as, or one more than, the raw \(L\) value.
**[2 marks]** 6 sets of data recorded in the table with no arithmetic errors in calculations.
**[1 mark]** Linear axes on graph: scale chosen so that points occupy at least half the graph grid in both directions.
**[1 mark]** Plotting of points: points plotted correctly within half a small grid square.
**[2 marks]** Line of best fit drawn with balanced distribution of points above and below the line.
**[1 mark]** Gradient calculation: uses coordinates from a line interval spanning at least half the drawn line length.
**[1 mark]** y-intercept: read directly from the y-axis (if x=0 is on the grid) or calculated using a point on the line.
**[2 marks]** Value of \(a\) determined from gradient with correct unit (\(\text{m}^{-2}\) or equivalent depending on units used).
**[2 marks]** Value of \(b\) determined from intercept with correct unit (\(\text{m}\) or equivalent).

### Step-by-Step Worked Solution

**Part (a)**
- Unstretched length: \(l_0 = 12.5\text{ cm} = 0.125\text{ m}\)
- Loaded (empty cup): \(l_1 = 15.2\text{ cm} = 0.152\text{ m}\)
- Extension: \(x_1 = 15.2 - 12.5 = 2.7\text{ cm} = 0.027\text{ m}\)

**Part (b)**
- Time for 10 oscillations: \(t_1 = 3.3\text{ s}\)
- Period: \(T_1 = 3.3 / 10 = 0.33\text{ s}\)
- Estimate percentage uncertainty: human reaction time error is approx. \(\pm 0.2\text{ s}\) for timing a sequence.
\[ \% \text{ uncertainty} = \frac{0.2\text{ s}}{3.3\text{ s}} \times 100\% \approx 6.1\% \]

**Part (c)**
- Length with 100g water: \(l_2 = 24.5\text{ cm} = 0.245\text{ m}\)
- Extension: \(x_2 = 24.5 - 12.5 = 12.0\text{ cm} = 0.120\text{ m}\)
- Time for 10 oscillations: \(t_2 = 6.9\text{ s}\)
- Period: \(T_2 = 6.9 / 10 = 0.69\text{ s}\)

**Part (d)**
- Calculate \(k = T^2 / x\):
- \(k_1 = \frac{0.33^2}{0.027} = \frac{0.1089}{0.027} \approx 4.03\text{ s}^2\text{ m}^{-1}\)
- \(k_2 = \frac{0.69^2}{0.120} = \frac{0.4761}{0.120} \approx 3.97\text{ s}^2\text{ m}^{-1}\)
- Comparison:
- Percentage difference between \(k_1\) and \(k_2\):
\[ \% \text{ diff} = \frac{|4.03 - 3.97|}{4.00} \times 100\% = 1.5\% \]
- Since \(1.5\%\) is less than a typical experimental tolerance limit of \(10\%\), the results strongly support the suggested relationship.

**Part (e) - Limitations and Improvements**
1. *Limitation*: Two sets of readings are insufficient to validate a mathematical relationship.
*Improvement*: Take multiple readings with different volumes of water (e.g., 20 ml, 40 ml, 60 ml, 80 ml, 100 ml) and plot a graph of \(T^2\) against \(x\).
2. *Limitation*: Water inside the cup sloshes during vertical movement, disturbing the simple harmonic motion.
*Improvement*: Use solid slotted masses or dry sand of equivalent weight instead of water to avoid liquid sloshing.
3. *Limitation*: The cup exhibits lateral (side-to-side) oscillations in addition to vertical oscillations, making timing difficult.
*Improvement*: Use a vertical guide tube or guide wire to constrain the motion of the cup to a single vertical axis.
4. *Limitation*: Human reaction time is significant relative to the short total time of 10 oscillations.
*Improvement*: Use a video camera with a slow-motion playback timer, or a motion sensor with a data logger to record position against time and extract \(T\).

**[1 mark]** Value of \(l_0\) and \(l_1\) recorded to the nearest mm with units.
**[1 mark]** Correct calculation of extension \(x_1\).
**[1 mark]** Value of \(t_1\) recorded to at least 0.1 s.
**[1 mark]** Correct calculation of period \(T_1\).
**[1 mark]** Percentage uncertainty calculation for \(T_1\) showing correct working (based on absolute timing uncertainty of 0.1 s to 0.2 s).
**[1 mark]** Value of \(l_2\) and calculated \(x_2\) recorded correctly.
**[1 mark]** Value of \(t_2\) and calculated \(T_2\) recorded correctly.
**[1 mark]** Raw measurements of \(l_0, l_1, l_2\) must be consistently formatted.
**[1 mark]** Correct calculation of both constant values \(k_1\) and \(k_2\).
**[1 mark]** Correct units for \(k\) (\(\text{s}^2\text{ m}^{-1}\) or equivalent depending on units used).
**[1 mark]** Quantitative comparison showing percentage difference calculation between \(k_1\) and \(k_2\).
**[1 mark]** Clear conclusion stating whether relationship is supported based on a stated criterion (e.g., "percentage difference is less than 10%").
**[8 marks]** Identification of 4 distinct limitations of the experiment and 4 corresponding realistic improvements (1 mark per matched pair of limitation and improvement).

(a) Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point.

(b) Using the formula for gravitational potential:
\(\phi = -\frac{GM}{R}\)
\(\phi = -\frac{(6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}) \times (4.80 \times 10^{24}\text{ kg})}{5.40 \times 10^6\text{ m}}\)
\(\phi = -5.9288 \times 10^7\text{ J kg}^{-1} \approx -5.93 \times 10^7\text{ J kg}^{-1}\)

(c) (i) For escape velocity, the total energy of the probe at infinity is zero:
\(\frac{1}{2} m v_{\text{esc}}^2 - \frac{GMm}{R} = 0\)
\(v_{\text{esc}} = \sqrt{\frac{2GM}{R}} = \sqrt{2 \times |\phi|}\)
\(v_{\text{esc}} = \sqrt{2 \times 5.9288 \times 10^7}\)
\(v_{\text{esc}} = \sqrt{1.1858 \times 10^8} \approx 1.0889 \times 10^4\text{ m s}^{-1} \approx 1.09 \times 10^4\text{ m s}^{-1}\)

(ii) The minimum kinetic energy required to escape is:
\(E_{k,\text{min}} = \frac{1}{2} m v_{\text{esc}}^2 = -m\phi\)
\(E_{k,\text{min}} = 1500\text{ kg} \times 5.9288 \times 10^7\text{ J kg}^{-1} = 8.89 \times 10^{10}\text{ J}\)
Since the initial kinetic energy of the probe is \(6.00 \times 10^{10}\text{ J}\), which is less than the required escape energy of \(8.89 \times 10^{10}\text{ J}\), the probe will not escape.

(a)
- Work done per unit mass [1]
- in bringing a small test mass from infinity to that point [1]

(b)
- \(\phi = -GM/R\) substitution [1]
- Correct value and negative sign: \(-5.93 \times 10^7\text{ J kg}^{-1}\) [1]

(c)(i)
- Equating gain in gravitational potential energy to loss in kinetic energy (or stating total energy = 0 at infinity) [1]
- Substitution: \(v_{\text{esc}} = \sqrt{2 \times 6.67 \times 10^{-11} \times 4.80 \times 10^{24} / 5.40 \times 10^6}\) [1]
- Calculation leading to \(1.09 \times 10^4\text{ m s}^{-1}\) (must show at least 3 sig figs: e.g., \(1.089 \times 10^4\)) [1]

(c)(ii)
- Calculation of minimum escape kinetic energy: \(E_k = 1.50 \times 10^9 \times 59.3 = 8.89 \times 10^{10}\text{ J}\) (or alternative comparison using velocity, where launch velocity \(v = \sqrt{2E_k/m} = 8.94 \times 10^3\text{ m s}^{-1}\)) [1]
- Comparison of values: \(6.00 \times 10^{10}\text{ J} < 8.89 \times 10^{10}\text{ J}\) (or \(8.94 \times 10^3\text{ m s}^{-1} < 1.09 \times 10^4\text{ m s}^{-1}\)) [1]
- Explicit conclusion that the probe will not escape [1]

(a) The first law of thermodynamics is written as:
\(\Delta U = q + w\)
where \(\Delta U\) is the increase in internal energy of the system,
\(q\) is the heating of the system (thermal energy supplied to it),
and \(w\) is the work done on the system.

(b) Using the ideal gas equation at state B:
\(P_B V_B = n R T_B\)
Given \(P_B = 3.00 \times 10^5\text{ Pa}\), \(V_B = 5.00 \times 10^{-3}\text{ m}^3\), \(n = 0.240\text{ mol}\), and \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\):
\(T_B = \frac{P_B V_B}{n R} = \frac{(3.00 \times 10^5) \times (5.00 \times 10^{-3})}{0.240 \times 8.31}\)
\(T_B = \frac{1500}{1.9944} \approx 752\text{ K}\)

(c) (i) For an isobaric process, the work done on the gas is given by:
\(w = -P \Delta V\)
Here, compression is at constant pressure \(P_A = 1.20 \times 10^5\text{ Pa}\).
\(\Delta V = V_A - V_C = 5.00 \times 10^{-3}\text{ m}^3 - 1.25 \times 10^{-2}\text{ m}^3 = -7.50 \times 10^{-3}\text{ m}^3\)
\(w = -(1.20 \times 10^5\text{ Pa}) \times (-7.50 \times 10^{-3}\text{ m}^3) = +900\text{ J}\)

(ii) Using the first law of thermodynamics:
\(\Delta U = q + w\)
Given \(\Delta U = -1350\text{ J}\) (decrease in internal energy) and \(w = +900\text{ J}\):
\(-1350 = q + 900\)
\(q = -1350 - 900 = -2250\text{ J}\)
Since \(q\) is negative, the thermal energy transferred out of the gas is \(2250\text{ J}\).

(a)
- \(\Delta U = q + w\) or worded equivalent [1]
- Correct identification of all terms: \(\Delta U\) is increase in internal energy, \(q\) is heat added to the system, and \(w\) is work done on the system [1]

(b)
- Substitution into \(PV = nRT\) using \(3.00 \times 10^5\) and \(5.00 \times 10^{-3}\) [1]
- Correct calculation leading to \(752\text{ K}\) (accept \(750\text{ K}\) if 8.3 is used) [1]

(c)(i)
- Recognition that work done \(W = P \Delta V\) [1]
- Substitution: \(1.20 \times 10^5 \times (1.25 \times 10^{-2} - 5.00 \times 10^{-3})\) [1]
- Correct calculation to give \(900\text{ J}\) (sign must represent work done on the gas, i.e. positive) [1]

(c)(ii)
- Recall of \(\Delta U = q + w\) with correct signs [1]
- Substitution: \(-1350 = q + 900\) [1]
- Answer: \(2250\text{ J}\) (must specify energy transferred out or show negative sign) [1]

(a) The relationship between angular frequency and period is:
\(\omega = \frac{2\pi}{T}\)
\(\omega = \frac{2\pi}{0.800\text{ s}} = 2.50\pi \approx 7.854 \approx 7.85\text{ rad s}^{-1}\)

(b) The maximum kinetic energy is equal to the total energy of the oscillator:
\(E_{k,\text{max}} = \frac{1}{2} m \omega^2 x_0^2\)
where \(x_0 = 6.00\text{ cm} = 0.0600\text{ m}\) and \(m = 0.450\text{ kg}\).
\(E_{k,\text{max}} = 0.5 \times 0.450 \times (7.854)^2 \times (0.0600)^2\)
\(E_{k,\text{max}} = 0.5 \times 0.450 \times 61.685 \times 0.0036 = 0.04996\text{ J} \approx 0.0500\text{ J}\)

(c) When kinetic energy equals potential energy:
\(E_k = E_p\)
Since total energy \(E_T = E_k + E_p\), we have:
\(E_T = 2 E_p\)
\(\frac{1}{2} m \omega^2 x_0^2 = 2 \left( \frac{1}{2} m \omega^2 x^2 \right)\)
\(x_0^2 = 2 x^2 \implies x = \frac{x_0}{\sqrt{2}}\)
\(x = \frac{6.00\text{ cm}}{\sqrt{2}} \approx 4.24\text{ cm}\) (or \(0.0424\text{ m}\))

(d) In simple harmonic motion, acceleration is always directed opposite to displacement (represented by the defining equation \(a = -\omega^2 x\)). Therefore, the phase difference is exactly \(\pi\text{ rad}\) (or \(180^\circ\)).

(a)
- Use of \(\omega = 2\pi / T\) [1]
- Calculation leading to \(7.85\text{ rad s}^{-1}\) (must show 3 sig figs, e.g., 7.854) [1]

(b)
- State or use \(E_k = \frac{1}{2} m \omega^2 x_0^2\) (or equivalent) [1]
- Convert amplitude to meters: \(0.0600\text{ m}\) [1]
- Final answer: \(0.0500\text{ J}\) (accept \(0.0499\text{ J}\) to \(0.0501\text{ J}\)) [1]

(c)
- State that total energy \(E_T = 2 E_p\) (or equate potential energy to half of total energy) [1]
- Write equation \(x^2 = \frac{1}{2} x_0^2\) or \(x = x_0 / \sqrt{2}\) [1]
- Calculate correct displacement: \(4.24\text{ cm}\) (or \(0.0424\text{ m}\)) [1]

(d)
- Identify that acceleration and displacement are in antiphase [1]
- Answer: \(\pi\text{ rad}\) (accept \(180^\circ\)) [1]

(a) Capacitance is defined as the charge stored per unit potential difference:
\(C = \frac{Q}{V}\)

(b) (i) The discharge of a capacitor is modeled by:
\(V = V_0 e^{-t/RC}\)
Given \(V = 4.00\text{ V}\), \(V_0 = 12.0\text{ V}\), \(t = 8.50\text{ s}\), and \(C = 220 \times 10^{-6}\text{ F}\):
\(4.00 = 12.0 e^{-8.50 / (R \times 220 \times 10^{-6})}\)
\(\frac{4.00}{12.0} = \frac{1}{3} = e^{-8.50 / (2.20 \times 10^{-4} R)}\)
Taking natural logarithms on both sides:
\(\ln(3) = \frac{8.50}{2.20 \times 10^{-4} R}\)
\(R = \frac{8.50}{2.20 \times 10^{-4} \times \ln(3)}\)
\(R = \frac{8.50}{2.417 \times 10^{-4}} \approx 3.5168 \times 10^4\text{ }\Omega \approx 3.5 \times 10^4\text{ }\Omega\)

(ii) The energy stored in the capacitor when \(V = 4.00\text{ V}\) is:
\(E = \frac{1}{2} C V^2\)
\(E = 0.5 \times (220 \times 10^{-6}\text{ F}) \times (4.00\text{ V})^2\)
\(E = 1.10 \times 10^{-4} \times 16 = 1.76 \times 10^{-3}\text{ J}\)

(iii) The initial charge is:
\(Q_0 = C V_0 = 220 \times 10^{-6} \times 12.0 = 2.64 \times 10^{-3}\text{ C}\)
The charge at \(t = 8.50\text{ s}\) is:
\(Q = C V = 220 \times 10^{-6} \times 4.00 = 0.88 \times 10^{-3}\text{ C}\)
The charge lost during this time is:
\(\Delta Q = Q_0 - Q = 1.76 \times 10^{-3}\text{ C}\)
The average current is:
\(I_{\text{avg}} = \frac{\Delta Q}{\Delta t} = \frac{1.76 \times 10^{-3}\text{ C}}{8.50\text{ s}} \approx 2.07 \times 10^{-4}\text{ A}\) (or \(0.207\text{ mA}\))

(a)
- Charge / potential difference [1]
- Charge is on one plate and potential difference is between the plates [1]

(b)(i)
- Recall of formula: \(V = V_0 e^{-t/RC}\) [1]
- Substitution: \(4.00 = 12.0 e^{-8.50 / (R \times 220 \times 10^{-6})}\) [1]
- Rearrangement using natural logarithms: \(\ln(3) = 8.50 / (2.20 \times 10^{-4} R)\) [1]
- Correct calculation leading to \(3.52 \times 10^4\text{ }\Omega\) (must see at least 3 sig figs to justify the "show that") [1]

(b)(ii)
- Recall of \(E = \frac{1}{2} C V^2\) with correct substitution [1]
- Answer: \(1.76 \times 10^{-3}\text{ J}\) (or \(1.8 \times 10^{-3}\text{ J}\)) [1]

(b)(iii)
- Use of \(\Delta Q = C \Delta V\) to find charge loss of \(1.76 \times 10^{-3}\text{ C}\) [1]
- Answer: \(2.07 \times 10^{-4}\text{ A}\) (or \(2.1 \times 10^{-4}\text{ A}\)) [1]

(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) An e.m.f. is induced because the magnetic flux density \(B\) varies sinusoidal with time, which causes a continuous change in the magnetic flux passing through the area of each turn of the coil. Since there is a change in magnetic flux linkage, an e.m.f. is induced according to Faraday's law.

(c) The magnetic flux linkage \(\Phi\) is given by:
\(\Phi = N B A\)
The maximum magnetic flux linkage occurs when \(B\) is maximum (\(B = B_0\)):
\(\Phi_{\text{max}} = N B_0 A = 150 \times 0.24\text{ T} \times 4.5 \times 10^{-3}\text{ m}^2\)
\(\Phi_{\text{max}} = 150 \times 1.08 \times 10^{-3} = 0.162\text{ Wb} \approx 0.16\text{ Wb}\)

(d) The induced e.m.f. is given by the derivative of the flux linkage:
\(E = -\frac{d(N B A)}{dt} = -N A \frac{dB}{dt}\)
Given \(B = B_0 \sin(\omega t)\):
\(\frac{dB}{dt} = B_0 \omega \cos(\omega t)\)
Therefore, \(E = -N A B_0 \omega \cos(\omega t)\).
The maximum magnitude of the induced e.m.f. is:
\(E_{\text{max}} = N A B_0 \omega\)
\(E_{\text{max}} = \Phi_{\text{max}} \omega\)
\(E_{\text{max}} = 0.162\text{ Wb} \times 50\pi\text{ rad s}^{-1}\)
\(E_{\text{max}} = 8.10\pi \approx 25.4\text{ V}\)

(a)
- Induced e.m.f. is proportional to the rate of change of magnetic flux linkage [1]
- Correct formulation showing magnitude, e.m.f. \(\propto d\Phi/dt\) [1]

(b)
- Magnetic flux density varies with time [1]
- Changing magnetic flux linkage induces an e.m.f. [1]

(c)
- Formula used: \(N B_0 A\) [1]
- Substitution: \(150 \times 0.24 \times 4.5 \times 10^{-3} = 0.162\text{ Wb}\) (which rounds to \(0.16\text{ Wb}\)) [1]

(d)
- Recognize that \(E = -d(N\Phi)/dt\) leads to \(E_{\text{max}} = N A B_0 \omega\) [1]
- Identify \(\omega = 50\pi\text{ rad s}^{-1}\) [1]
- Correct substitution of values: \(0.162 \times 50\pi\) [1]
- Answer: \(25.4\text{ V}\) (accept \(25\text{ V}\) if \(0.16\) is used) [1]

(a) The work function energy is the minimum energy required to liberate an electron from the surface of a metal.

(b) (i) The energy \(E\) of a photon is given by:
\(E = \frac{hc}{\lambda}\)
\(E = \frac{(6.63 \times 10^{-34}\text{ J s}) \times (3.00 \times 10^8\text{ m s}^{-1})}{240 \times 10^{-9}\text{ m}}\)
\(E = \frac{1.989 \times 10^{-25}}{2.40 \times 10^{-7}} = 8.2875 \times 10^{-19}\text{ J} \approx 8.29 \times 10^{-19}\text{ J}\)

(ii) In electron-volts:
\(E = \frac{8.2875 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} \approx 5.18\text{ eV}\)

(c) From the photoelectric equation:
\(E_{k,\text{max}} = hf - \Phi = E - \Phi\)
\(E_{k,\text{max}} = 5.18\text{ eV} - 2.36\text{ eV} = 2.82\text{ eV}\)

(d) (i) There is no change in the maximum kinetic energy, because the maximum kinetic energy is determined only by the frequency (or wavelength) of the incident photons and the work function of the metal.

(ii) The rate of emission of photoelectrons increases. Higher intensity means more photons are incident on the surface per unit time. Since each photoelectron is released by a one-to-one interaction with a single photon, more photoelectrons are emitted per unit time.

(a)
- Minimum energy [1]
- to remove/liberate an electron from the surface of a metal [1]

(b)(i)
- Use of \(E = hc/\lambda\) [1]
- Answer: \(8.29 \times 10^{-19}\text{ J}\) [1]

(b)(ii)
- Divide by \(1.60 \times 10^{-19}\) to get \(5.18\text{ eV}\) (accept \(5.2\text{ eV}\)) [1]

(c)
- Photoelectric equation used: \(E_{k,\text{max}} = E - \Phi\) [1]
- Answer: \(2.82\text{ eV}\) (or \(2.8\text{ eV}\)) [1]

(d)(i)
- States no change [1]

(d)(ii)
- Rate of emission increases [1]
- Because higher intensity means more photons per second and there is a 1-to-1 interaction [1]

(a) The decay constant is the probability of decay per unit time of a single nucleus.

(b) The relationship for activity is:
\(A = A_0 e^{-\lambda t}\)
Given \(A_0 = 3.60 \times 10^5\text{ Bq}\), \(A = 4.50 \times 10^4\text{ Bq}\), and \(t = 18.0\text{ hours} = 18.0 \times 3600\text{ s} = 64800\text{ s}\):
\(\frac{4.50 \times 10^4}{3.60 \times 10^5} = e^{-\lambda (64800)}\)
\(0.125 = \frac{1}{8} = e^{-64800 \lambda}\)
Taking natural logarithms on both sides:
\(\ln(8) = 64800 \lambda\)
\(\lambda = \frac{\ln(8)}{64800}\)
\(\lambda = \frac{2.07944}{64800} \approx 3.209 \times 10^{-5}\text{ s}^{-1} \approx 3.21 \times 10^{-5}\text{ s}^{-1}\)

(c) The relationship between activity and number of nuclei is:
\(A_0 = \lambda N_0\)
\(N_0 = \frac{A_0}{\lambda} = \frac{3.60 \times 10^5}{3.209 \times 10^{-5}} \approx 1.12 \times 10^{10}\) nuclei.

(d) The half-life \(t_{1/2}\) is given by:
\(t_{1/2} = \frac{\ln(2)}{\lambda}\)
\(t_{1/2} = \frac{\ln(2)}{3.209 \times 10^{-5}\text{ s}^{-1}} \approx 21600\text{ s}\)
Converting to hours:
\(t_{1/2} = \frac{21600}{3600} = 6.00\text{ hours}\)

(a)
- Probability of decay of a nucleus [1]
- per unit time [1]

(b)
- Use of \(A = A_0 e^{-\lambda t}\) [1]
- Correct conversion of time: \(18.0 \times 3600 = 64800\text{ s}\) [1]
- Ratio substitution: \(\ln(8) = 64800 \lambda\) or equivalent [1]
- Correct calculation to at least 3 sig figs: \(3.21 \times 10^{-5}\text{ s}^{-1}\) [1]

(c)
- Use of \(A = \lambda N\) [1]
- Correct calculation: \(N_0 = 1.12 \times 10^{10}\) (accept \(1.1 \times 10^{10}\)) [1]

(d)
- Use of \(t_{1/2} = \ln(2)/\lambda\) [1]
- Correct calculation leading to \(6.00\text{ hours}\) [1]

(a) From Wien's displacement law:
\(\lambda_{\text{max}} T = b\)
\(\lambda_{\text{max}} = \frac{2.90 \times 10^{-3}\text{ m K}}{4800\text{ K}} \approx 6.04 \times 10^{-7}\text{ m}\) (or \(604\text{ nm}\))

(b) From Stefan-Boltzmann law:
\(L = 4\pi R^2 \sigma T^4\)
\(L = 4 \pi \times (1.20 \times 10^9\text{ m})^2 \times (5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}) \times (4800\text{ K})^4\)
\(L = 4\pi \times 1.44 \times 10^{18} \times 5.67 \times 10^{-8} \times 5.3084 \times 10^{14}\)
\(L \approx 5.44 \times 10^{26}\text{ W}\)

(c) (i) Redshift \(z\) is defined as:
\(z = \frac{\Delta \lambda}{\lambda_0} = \frac{678.5\text{ nm} - 656.3\text{ nm}}{656.3\text{ nm}}\)
\(z = \frac{22.2\text{ nm}}{656.3\text{ nm}} \approx 0.0338\)

(ii) The recessional velocity \(v\) of the galaxy is:
\(v = z c = 0.03382 \times (3.00 \times 10^5\text{ km s}^{-1}) \approx 1.015 \times 10^4\text{ km s}^{-1}\)
Using Hubble's law, \(v = H_0 d\):
\(d = \frac{v}{H_0} = \frac{1.015 \times 10^4\text{ km s}^{-1}}{70.0\text{ km s}^{-1}\text{ Mpc}^{-1}} \approx 145\text{ Mpc}\)

(a)
- Recall and substitute into \(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\) [1]
- Answer: \(6.04 \times 10^{-7}\text{ m}\) (or \(6.0 \times 10^{-7}\text{ m}\)) [1]

(b)
- Recall \(L = 4\pi R^2 \sigma T^4\) [1]
- Substitution: \(4 \times \pi \times (1.20 \times 10^9)^2 \times 5.67 \times 10^{-8} \times (4800)^4\) [1]
- Answer: \(5.44 \times 10^{26}\text{ W}\) (accept \(5.4 \times 10^{26}\text{ W}\)) [1]

(c)(i)
- Formula: \(z = \Delta\lambda / \lambda\) [1]
- Answer: \(0.0338\) (or \(0.034\)) [1]

(c)(ii)
- Calculation of velocity \(v = z c\): \(0.0338 \times 3.00 \times 10^8 = 1.01 \times 10^7\text{ m s}^{-1} = 1.01 \times 10^4\text{ km s}^{-1}\) [1]
- Recall \(v = H_0 d\) and substitute [1]
- Answer: \(145\text{ Mpc}\) (accept range \(143 - 147\text{ Mpc}\)) [1]

(a)
• M1: Work done per unit mass.
• A1: In bringing a small / test mass from infinity to that point.

(b)(i)
• C1: Formula \(\phi = -GM/R\) written or clearly implied by substitution.
• A1: Correct substitution and calculation showing at least 3 sig figs (\(-1.259 \times 10^7\text{ J kg}^{-1}\)) before rounding to \(-1.26 \times 10^7\text{ J kg}^{-1}\).

(b)(ii)
• C1: Calculates total radial distance from center \(r = 5.90 \times 10^6\text{ m}\).
• C1: Uses correct difference formula, e.g., \(\Delta E_p = GMm (1/R - 1/r)\) or computes potential difference \(\Delta \phi \approx 5.34 \times 10^6\text{ J kg}^{-1}\).
• A1: Correct final answer: \(6.40 \times 10^9\text{ J}\) (accept \(6.4 \times 10^9\text{ J}\)).

(b)(iii)
• C1: Equates centripetal force to gravitational force to get \(v^2 = GM/r\) or \(v = \sqrt{GM/r}\).
• C1: Substitutes correct values including \(r = 5.90 \times 10^6\text{ m}\).
• A1: Correct final answer: \(2.69 \times 10^3\text{ m s}^{-1}\) (or \(2.7 \times 10^3\text{ m s}^{-1}\)).

(a)
• M1: Time taken for the charge / current / p.d. to fall to \(1/e\) (or \(37\%\)) of its initial value.
• A1: State expression \(\tau = RC\).

(b)(i)
• C1: Recalls or uses the exponential discharge relation: \(V = V_0 e^{-t/RC}\).
• C1: Substitutes the correct values: \(3.00 = 12.0 e^{-8.50 / (R \times 470 \times 10^{-6})}\).
• C1: Applies logarithms correctly: \(\ln(0.25) = -8.50 / (R \times 470 \times 10^{-6})\).
• A1: Correct final answer: \(1.30 \times 10^4\ \Omega\) (or \(13.0\text{ k}\Omega\); accept \(1.3 \times 10^4\ \Omega\)).

(b)(ii)
• C1: Recalls and uses \(Q = CV\) with \(V = 3.00\text{ V}\).
• A1: Correct calculation: \(1.41 \times 10^{-3}\text{ C}\) (or \(1.41\text{ mC}\)).

(b)(iii)
• C1: Recognizes energy lost is \(\Delta E = \frac{1}{2} C (V_0^2 - V^2)\) or calculates individual energy values correctly.
• A1: Correct calculation: \(0.0317\text{ J}\) (or \(3.17 \times 10^{-2}\text{ J}\) or \(31.7\text{ mJ}\)).

### 1. Experimental Setup and Diagram
As shown in the diagram, a flat circular coil of known turns \(N\) is mounted vertically. It is connected in a series circuit containing a variable DC power supply, a switch, an ammeter, and a rheostat to control and monitor the current \(I\).

A Hall probe is mounted on a non-magnetic linear track (or carriage with a ruler) aligned precisely along the longitudinal axis passing through the center of the coil. The probe is connected to a calibrated Tesla meter to measure the axial magnetic flux density \(B\).

### 2. Variables
- **Independent Variable**: Distance \(x\) from the center of the coil.
- **Dependent Variable**: Magnetic flux density \(B\).
- **Control Variables**:
- Current \(I\) through the coil (monitored via the ammeter and adjusted using the rheostat).
- Number of turns \(N\) of the coil.
- Radius \(R\) of the coil.

### 3. Procedure and Measurements
1. Measure the radius \(R\) of the coil using a vernier caliper or a ruler (take the average of internal and external diameters if the coil has thickness, then divide by 2).
2. Set up the circuit and position the Hall probe exactly at the center of the coil (where \(B\) is maximum) to define the reference position \(x = 0\).
3. Switch on the current and adjust the rheostat so that a constant current \(I\) (e.g., 2.0 A) flows through the coil.
4. Move the Hall probe along the axis of the coil to a distance \(x\) (measured using the scale/ruler).
5. Record the distance \(x\) and the corresponding magnetic flux density \(B\) from the Tesla meter.
6. Repeat the measurements for at least six different values of \(x\) in regular increments.
7. Periodically check and adjust the rheostat to ensure the current \(I\) remains strictly constant throughout the experiment.

### 4. Analysis of Results
The relationship is given by:
\[B = \frac{\mu_0 N I R^2}{2} (R^2 + x^2)^{-1.5}\]

- Plot a graph of \(B\) on the y-axis against \((R^2 + x^2)^{-1.5}\) on the x-axis.
- If the relationship is valid, the plot will yield a straight line passing through the origin.
- Determine the gradient \(m\) of the straight line.
- Since the gradient \(m = \frac{\mu_0 N I R^2}{2}\), the value of \(\mu_0\) is calculated using:
\[\mu_0 = \frac{2 m}{N I R^2}\]

### 5. Safety and Quality Control
- **Safety**: The coil can become hot when carrying current for long periods. Switch off the current between readings to prevent overheating and potential burns.
- **Quality/Accuracy**:
- To eliminate the effect of the Earth's magnetic field or background stray fields, measure the magnetic flux density with the power switched off at each position and subtract this zero-offset from the active reading.
- Ensure the Hall probe is kept perpendicular to the axis of the coil to measure the maximum axial magnetic flux density.

**Defining the Problem (max 3 marks):**
- [1] State that \(x\) is the independent variable and \(B\) is the dependent variable.
- [1] Identify that the current \(I\) must be kept constant.
- [1] Identify that the number of turns \(N\) or radius \(R\) of the coil must be kept constant.

**Methods of Data Collection (max 5 marks):**
- [1] Draw a clear, labeled diagram of the apparatus showing the coil connected in series with a power supply, ammeter, and a Hall probe placed along the central axis of the coil connected to a Tesla meter.
- [1] Describe the use of a ruler/scale to measure the distance \(x\) from the center of the coil.
- [1] Explain how the center of the coil \(x = 0\) is located (e.g., finding the position where \(B\) is maximum or using geometric symmetry).
- [1] Explain the use of a caliper/ruler to measure the radius \(R\) of the coil.
- [1] Describe how the current \(I\) is kept constant using a variable resistor/rheostat and monitored with an ammeter.

**Method of Analysis (max 3 marks):**
- [1] State that a graph of \(B\) against \((R^2 + x^2)^{-1.5}\) should be plotted.
- [1] State that a straight line passing through the origin confirms the suggested relationship.
- [1] State that \(\mu_0 = \frac{2m}{N I R^2}\), where \(m\) is the gradient of the graph.

**Safety Precautions (max 1 mark):**
- [1] Identify the hazard of hot coils and suggest a precaution (e.g., switch off current when not taking readings / use insulated gloves).

**Additional Details / Good Practice (max 3 marks):**
- [1] Describe how to eliminate stray/background magnetic fields (e.g., zeroing the Tesla meter or subtracting background readings with the current switched off).
- [1] Explain how to ensure the Hall probe is correctly oriented (e.g., rotating the probe until a maximum reading is obtained).
- [1] Suggest repeating measurements of \(B\) for each \(x\) on both sides of the coil and calculating the average.

### Part (a)
Squaring both sides of the relation:
\[f^2 = \frac{k}{4\pi^2 (M + \alpha)}\]
Taking the reciprocal:
\[\frac{1}{f^2} = \frac{4\pi^2}{k} (M + \alpha) = \left(\frac{4\pi^2}{k}\right) M + \frac{4\pi^2 \alpha}{k}\]
Comparing this to the equation of a straight line, \(y = mx + c\), where \(y = 1/f^2\) and \(x = M\):
- **Gradient** \(m = \frac{4\pi^2}{k}\)
- **y-intercept** \(c = \frac{4\pi^2 \alpha}{k} = m \alpha\)

### Part (b)
The absolute uncertainty in \(y = f^{-2}\) is given by:
\[\Delta y = 2 \left(\frac{\Delta f}{f}\right) y\]

Calculating for each row:
1. \(M = 50\text{ g}\):
- \(1/f^2 = 1 / 2.25^2 = 0.198\text{ s}^2\)
- \( \Delta(1/f^2) = 2 \times (0.08 / 2.25) \times 0.198 = 0.014\text{ s}^2\)
- Value: \(0.198 \pm 0.014\text{ s}^2\)

2. \(M = 100\text{ g}\):
- \(1/f^2 = 1 / 1.74^2 = 0.330\text{ s}^2\)
- \( \Delta(1/f^2) = 2 \times (0.06 / 1.74) \times 0.330 = 0.023\text{ s}^2\)
- Value: \(0.330 \pm 0.023\text{ s}^2\)

3. \(M = 150\text{ g}\):
- \(1/f^2 = 1 / 1.47^2 = 0.463\text{ s}^2\)
- \( \Delta(1/f^2) = 2 \times (0.05 / 1.47) \times 0.463 = 0.031\text{ s}^2\)
- Value: \(0.463 \pm 0.031\text{ s}^2\)

4. \(M = 200\text{ g}\):
- \(1/f^2 = 1 / 1.30^2 = 0.592\text{ s}^2\)
- \( \Delta(1/f^2) = 2 \times (0.04 / 1.30) \times 0.592 = 0.036\text{ s}^2\)
- Value: \(0.592 \pm 0.036\text{ s}^2\)

5. \(M = 250\text{ g}\):
- \(1/f^2 = 1 / 1.18^2 = 0.718\text{ s}^2\)
- \( \Delta(1/f^2) = 2 \times (0.04 / 1.18) \times 0.718 = 0.049\text{ s}^2\)
- Value: \(0.718 \pm 0.049\text{ s}^2\)

6. \(M = 300\text{ g}\):
- \(1/f^2 = 1 / 1.08^2 = 0.857\text{ s}^2\)
- \( \Delta(1/f^2) = 2 \times (0.03 / 1.08) \times 0.857 = 0.048\text{ s}^2\)
- Value: \(0.857 \pm 0.048\text{ s}^2\)

### Part (c)
- Plot a graph of \(1/f^2\) on the y-axis against \(M\) on the x-axis.
- Plot the data points along with vertical error bars of length \(2 \Delta(1/f^2)\).
- Draw a line of best fit (the straight line passing closest to all points with balanced scatter).
- Draw a worst acceptable line (the steepest or shallowest possible line that passes through all error bars, usually from the bottom of the first error bar to the top of the last error bar, or vice versa).

### Part (d)
From linear regression / graph analysis:
- **Best-fit line**:
- Convert \(M\) to SI units (kg) to find \(k\) in \(\text{N m}^{-1}\):
- At \(M = 0.050\text{ kg}\), \(y = 0.198\text{ s}^2\)
- At \(M = 0.300\text{ kg}\), \(y = 0.857\text{ s}^2\)
- Gradient \(m_{\text{best}} = \frac{0.857 - 0.198}{0.300 - 0.050} = \frac{0.659}{0.250} = 2.636\text{ s}^2\text{ kg}^{-1}\) (or more precisely \(2.622\text{ s}^2\text{ kg}^{-1}\))
- y-intercept \(c_{\text{best}} = 0.067\text{ s}^2\)
- **Steepest Worst-Acceptable Line**:
- Passes through \((0.050, 0.198 - 0.014 = 0.184)\) and \((0.300, 0.857 + 0.048 = 0.905)\)
- Gradient \(m_{\text{steep}} = \frac{0.905 - 0.184}{0.250} = 2.884\text{ s}^2\text{ kg}^{-1}\)
- Intercept \(c_{\text{steep}} = 0.905 - 2.884 \times 0.30 = 0.040\text{ s}^2\)
- **Shallowest Worst-Acceptable Line**:
- Passes through \((0.050, 0.198 + 0.014 = 0.212)\) and \((0.300, 0.857 - 0.048 = 0.809)\)
- Gradient \(m_{\text{shallow}} = \frac{0.809 - 0.212}{0.250} = 2.388\text{ s}^2\text{ kg}^{-1}\)
- Intercept \(c_{\text{shallow}} = 0.809 - 2.388 \times 0.30 = 0.093\text{ s}^2\)

#### (i) Calculating \(k\):
\[k_{\text{best}} = \frac{4\pi^2}{m_{\text{best}}} = \frac{39.478}{2.62} \approx 15.1\text{ N m}^{-1}\]
\[\Delta m = \frac{2.884 - 2.388}{2} = 0.248\text{ s}^2\text{ kg}^{-1}\]
\[\Delta k = k_{\text{best}} \times \frac{\Delta m}{m_{\text{best}}} = 15.1 \times \frac{0.25}{2.62} \approx 1.4\text{ N m}^{-1}\]
Thus, \(k = 15.1 \pm 1.4\text{ N m}^{-1}\) (accept \(15 \pm 1\)).

#### (ii) Calculating \(\alpha\):
\[\alpha_{\text{best}} = \frac{c_{\text{best}}}{m_{\text{best}}} = \frac{0.067\text{ s}^2}{2.62\text{ s}^2\text{ kg}^{-1}} = 0.0256\text{ kg} = 25.6\text{ g}\]
Using the extreme lines:
\[\alpha_{\text{steep}} = \frac{0.040}{2.884} = 0.0139\text{ kg} = 13.9\text{ g}\]
\[\alpha_{\text{shallow}} = \frac{0.093}{2.388} = 0.0389\text{ kg} = 38.9\text{ g}\]
\[\Delta \alpha = \frac{38.9 - 13.9}{2} = 12.5\text{ g}\]
Thus, \(\alpha = 26 \pm 13\text{ g}\).

**Part (a):**
- [1] Clear algebraic derivation leading to \(\frac{1}{f^2} = \left(\frac{4\pi^2}{k}\right) M + \frac{4\pi^2 \alpha}{k}\).
- [1] Correctly identifies \(\text{gradient} = \frac{4\pi^2}{k}\) and \(\text{y-intercept} = \frac{4\pi^2 \alpha}{k}\).

**Part (b):**
- [1] All values of \(1/f^2\) calculated correctly (0.198, 0.330, 0.463, 0.592, 0.718, 0.857).
- [1] Absolute uncertainties calculated correctly using \(\Delta(1/f^2) = 2 \times \frac{\Delta f}{f} \times \frac{1}{f^2}\) (values: 0.014, 0.023, 0.031, 0.036, 0.049, 0.048).
- [1] Values and uncertainties are written to consistent decimal places / significant figures (usually 3 decimal places for \(1/f^2\)).

**Part (c):**
- [1] Mentions plotting \(1/f^2\) against \(M\) with appropriate scales, labeled axes, and error bars.
- [1] Explains drawing the line of best fit through the center of gravity of data points.
- [1] Explains drawing the worst acceptable line passing through the top of the first error bar and the bottom of the last error bar (or vice versa).

**Part (d)(i):**
- [1] Correct determination of the gradient \(m\) of the best-fit line (approx. \(2.62 \pm 0.05\text{ s}^2\text{ kg}^{-1}\) or \(0.00262\text{ s}^2\text{ g}^{-1}\)).
- [1] Correct determination of the worst acceptable gradient (from steepest/shallowest line).
- [1] Calculated value of \(k\) in range \(14.8 - 15.4\text{ N m}^{-1}\).
- [1] Value of \(\Delta k\) correctly computed from the difference between best-fit and worst acceptable gradients (approx. \(1.4\text{ N m}^{-1}\)).

**Part (d)(ii):**
- [1] Correct determination of the y-intercept \(c\) of the best-fit line (approx. \(0.067 \pm 0.005\text{ s}^2\)).
- [1] Calculated value of \(\alpha\) in range \(24\text{ g} - 27\text{ g}\).
- [1] Value of \(\Delta \alpha\) calculated correctly using extreme values or fractional uncertainties (approx. \(11\text{ g} - 13\text{ g}\)).