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The percentage uncertainty in \(g\) is given by the sum of the percentage uncertainty in \(L\) and twice the percentage uncertainty in \(T\):

\[\frac{\Delta g}{g} \times 100\% = \left( \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \right) \times 100\%\]

Substituting the values:

\[\frac{\Delta L}{L} = \frac{0.005}{0.800} = 0.00625\text{ (or } 0.625\%\)\]
\[2 \frac{\Delta T}{T} = 2 \times \frac{0.02}{1.80} \approx 0.02222\text{ (or } 2.222\%\)\]

Total percentage uncertainty:

\[\text{Percentage uncertainty} = 0.625\% + 2.222\% = 2.847\% \approx 2.8\%\]

1 mark for calculating the correct sum of fractional uncertainties and obtaining 2.8%.

Using Newton's second law, the average force is the rate of change of momentum:

\[F = \frac{\Delta p}{\Delta t}\]

Taking the initial direction of motion as positive:

\[p_i = m u = 0.15 \times 20 = 3.0\text{ kg m s}^{-1}\]
\[p_f = m v = 0.15 \times (-15) = -2.25\text{ kg m s}^{-1}\]
\[\Delta p = p_f - p_i = -2.25 - 3.0 = -5.25\text{ kg m s}^{-1}\]

Thus, the magnitude of the change in momentum is \(5.25\text{ kg m s}^{-1}\). The average force is:

\[F = \frac{5.25}{0.050} = 105\text{ N}\]

1 mark for the correct calculation of momentum change and division by collision time.

First, calculate the volume \(V\) and weight \(W\) of the block:

\[V = A_1 \times h_1 = 0.20 \times 0.40 = 0.080\text{ m}^3\]
\[m = \rho V = 2500 \times 0.080 = 200\text{ kg}\]
\[W = mg = 200 \times 9.81 = 1962\text{ N}\]

In the first position, the pressure on the floor is:

\[P_1 = \frac{W}{A_1} = \frac{1962}{0.20} = 9810\text{ Pa} = 9.81\text{ kPa}\]

In the second position, the block rests on the smaller face of area:

\[A_2 = 0.20 \times 0.40 = 0.080\text{ m}^2\]

The new pressure is:

\[P_2 = \frac{W}{A_2} = \frac{1962}{0.080} = 24525\text{ Pa} = 24.525\text{ kPa}\]

The increase in pressure is:

\[\Delta P = P_2 - P_1 = 24.525 - 9.81 = 14.715\text{ kPa} \approx 14.7\text{ kPa}\]

1 mark for calculating both pressures correctly and subtracting them to obtain 14.7 kPa.

Comparing the given equation with the standard wave equation \(y = A \sin(\omega t - k x)\):

\[\omega = 15\pi\text{ rad s}^{-1}\]
\[k = 2.5\pi\text{ m}^{-1}\]

The wave speed \(v\) is given by:

\[v = \frac{\omega}{k} = \frac{15\pi}{2.5\pi} = 6.0\text{ m s}^{-1}\]

1 mark for calculating wave speed from the ratio of angular frequency to wave number.

The original resistance is \(R = \rho \frac{L}{A}\).

Since the volume \(V = L A\) remains constant when stretched to length \(L' = 3L\):

\[L' A' = L A \implies 3L A' = L A \implies A' = \frac{A}{3}\]

The new resistance is:

\[R' = \rho \frac{L'}{A'} = \rho \frac{3L}{A/3} = 9 \left( \rho \frac{L}{A} \right) = 9R\]

1 mark for correctly accounting for both the change in length and the corresponding change in cross-sectional area to get 9R.

The vertical component of the normal force \(N\) balances the weight of the car:

\[N \cos\theta = mg\]

The horizontal component of the normal force provides the centripetal force:

\[N \sin\theta = \frac{m v^2}{r}\]

Dividing the horizontal component equation by the vertical component equation:

\[\frac{N \sin\theta}{N \cos\theta} = \frac{m v^2 / r}{mg} \implies \tan\theta = \frac{v^2}{rg}\]

1 mark for setting up the equations for vertical equilibrium and horizontal centripetal acceleration, and dividing them to find the correct tangent relationship.

Since the cycle returns the gas to its initial state (returning to both the initial volume and initial temperature), the net change in internal energy for the complete cycle is \(\Delta U_{\text{net}} = 0\).

Let's apply the first law of thermodynamics, \(\Delta U = q + w\), to each process:

- **Process 1:** \(q_1 = +450\text{ J}\) (heat in), \(w_1 = -150\text{ J}\) (work done by the gas).

\[\Delta U_1 = +450 - 150 = +300\text{ J}\]

- **Process 2:** Isothermal compression: since the temperature does not change, \(\Delta U_2 = 0\).

- **Process 3:** Constant volume cooling to return to the initial temperature.

Since \(\Delta U_{\text{net}} = 0\):

\[\Delta U_1 + \Delta U_2 + \Delta U_3 = 0 \implies 300 + 0 + \Delta U_3 = 0 \implies \Delta U_3 = -300\text{ J}\]

At constant volume, the work done \(w_3 = 0\). Therefore:

\[\Delta U_3 = q_3 + w_3 \implies -300 = q_3 + 0 \implies q_3 = -300\text{ J}\]

This represents \(300\text{ J}\) of thermal energy transferred out of the gas.

1 mark for recognizing that net internal energy change in a closed cycle is zero, and applying the first law of thermodynamics to find that 300 J of heat is transferred out of the gas.

The centripetal force is provided by the magnetic force on the charge:

\[\frac{m v^2}{r} = B q v \implies r = \frac{m v}{B q}\]

Expressing the momentum \(p = m v\) in terms of kinetic energy \(E_k\):

\[p = \sqrt{2 m E_k} \implies r = \frac{\sqrt{2 m E_k}}{B q}\]

For the electron:

\[r_e = R = \frac{\sqrt{2 m E_k}}{B e}\]

For the alpha particle, with mass \(m_{\alpha} = 4m\), charge \(q_{\alpha} = 2e\), and the same kinetic energy \(E_k\):

\[r_{\alpha} = \frac{\sqrt{2 (4m) E_k}}{B (2e)} = \frac{2 \sqrt{2 m E_k}}{2 B e} = \frac{\sqrt{2 m E_k}}{B e} = R\]

1 mark for establishing the relationship between radius, mass, charge, and kinetic energy, and demonstrating that the ratio yields exactly R.

1. Express the base units of each quantity in the equation:
- Power \(P\): \(\text{W} = \text{J s}^{-1} = \text{kg m}^2 \text{s}^{-3}\\
- Area \)A\): \(\text{m}^2\)
- Temperature difference \(\Delta T\): \(\text{K}\\
- Length \)L\): \(\text{m}\\

2. Rearrange the equation for \)k\):
\(k = \frac{P L}{A \Delta T}\\

3. Substitute the base units:
\)\text{units of } k = \frac{(\text{kg m}^2 \text{s}^{-3}) \cdot \text{m}}{\text{m}^2 \cdot \text{K}} = \text{kg m s}^{-3} \text{K}^{-1}\\.

1 mark for the correct option A.

To find the percentage uncertainty of \(E\), we sum the percentage uncertainties of each variable, multiplying by their respective powers:

\(\frac{\Delta E}{E} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta e}{e}\\

Calculate each individual percentage uncertainty:
- For \)F\): \(\frac{2}{100} \times 100\% = 2\%\)
- For \(L\): \(\frac{0.05}{2.50} \times 100\% = 2\%\)
- For \(d\): \(\frac{0.01}{0.50} \times 100\% = 2\%\)
- For \(e\): \(\frac{0.1}{2.0} \times 100\% = 5\%\)

Combine these values:
\(\frac{\Delta E}{E} = 2\% + 2\% + 2(2\%) + 5\% = 13\%\\.

1 mark for the correct option C.

From conservation of momentum:
\(3m u = 3m v + m w \implies 3u = 3v + w \implies w = 3(u - v)\\t(Equation 1)

For a perfectly elastic collision, the relative velocity of approach equals the relative velocity of separation:
\)u - 0 = w - v \implies w = u + v\\ (Equation 2)

Equating Equation 1 and Equation 2:
\(3(u - v) = u + v\\
\)3u - 3v = u + v\\
\(2u = 4v \implies \frac{v}{u} = \frac{1}{2}\\.

1 mark for the correct option C.

Let \(A\) be the cross-sectional area of the cylinder. The volume of the cylinder in the upper liquid is \(A x\) and in the lower liquid is \(A (H - x)\).

For vertical equilibrium, the total upthrust equals the weight of the cylinder:

\(\rho_1 A x g + \rho_2 A (H - x) g = \rho_C A H g\\

Dividing both sides by \)A g\):
\(\rho_1 x + \rho_2 (H - x) = \rho_C H\\

Substitute the densities \)\rho_1 = 800\), \(\rho_2 = 1000\), and \(\rho_C = 850\):
\(800x + 1000(H - x) = 850H\\
\)800x + 1000H - 1000x = 850H\\
\(150H = 200x\\
\)\frac{x}{H} = \frac{150}{200} = 0.75\\.

1 mark for the correct option C.

Using the Doppler effect formulas:
For approach: \(f_1 = f_s \left( \frac{v}{v - v_s} \right)\\
For recession: \)f_2 = f_s \left( \frac{v}{v + v_s} \right)\\

Taking the ratio:
\(\frac{f_1}{f_2} = \frac{v + v_s}{v - v_s} = 1.20\\

Solve for \)v_s\):
\(v + v_s = 1.20(v - v_s)\\
\)v + v_s = 1.20v - 1.20v_s\\
\(2.20v_s = 0.20v\\
\)v_s = \frac{0.20}{2.20} v = \frac{1}{11} v \approx 0.091 v\\.

1 mark for the correct option A.

First, find the initial resistance of the thermistor \(R_T\) at \(20^\circ\text{C}\):
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_T}{R + R_T}\\
\)4.0 = 6.0 \times \frac{R_T}{4.0 + R_T}\\
\(\frac{2}{3} = \frac{R_T}{4.0 + R_T} \implies 8.0 + 2 R_T = 3 R_T \implies R_T = 8.0\text{ k}\Omega\\

At \)80^\circ\text{C}\), the new resistance \(R'_T\) is:
\(R'_T = \frac{1}{4} \times 8.0\text{ k}\Omega = 2.0\text{ k}\Omega\\

Calculate the new \)V_{\text{out}}\):
\(V_{\text{out}}' = 6.0 \times \frac{2.0}{4.0 + 2.0} = 6.0 \times \frac{2.0}{6.0} = 2.0\text{ V}\\.

1 mark for the correct option C.

1. Identify the forces acting on the bob:
- Tension \(T\) along the string
- Weight \(mg\) vertically downwards

2. Resolve forces:
- Vertically (no acceleration): \(T \cos \theta = mg\\
- Horizontally (centripetal force): \)T \sin \theta = m \omega^2 r\\

3. Relate the radius \(r\) to string length \(L\):
\(r = L \sin \theta\\

4. Substitute \)r\) into the horizontal equation:
\(T \sin \theta = m \omega^2 L \sin \theta \implies T = m \omega^2 L\\

5. Substitute \)T\) back into the vertical equation:
\((m \omega^2 L) \cos \theta = mg \implies \cos \theta = \frac{mg}{m \omega^2 L} = \frac{g}{\omega^2 L}\\.

1 mark for the correct option B.

The first law of thermodynamics is written as:
\(\Delta U = q + w\\
where \)q\) is the heat supplied to the system, and \(w\) is the work done on the system.

1. For path \(A \rightarrow B\):
- Work is done on the gas, so \(w = +400\text{ J}\\.
- Heat is released, so \)q = -150\text{ J}\\.
\(\Delta U_{AB} = -150 + 400 = +250\text{ J}\\.

2. For path \)B \rightarrow C\):
- Gas does work on surroundings, so work done on the gas is \(w = -200\text{ J}\\.
- Heat is absorbed, so \)q = +600\text{ J}\\.
\(\Delta U_{BC} = +600 - 200 = +400\text{ J}\\.

3. Since the process is cyclic, the total change in internal energy over a complete loop is zero:
\)\Delta U_{AB} + \Delta U_{BC} + \Delta U_{CA} = 0\\
\(250 + 400 + \Delta U_{CA} = 0\\
\)\Delta U_{CA} = -650\text{ J}\\.

1 mark for the correct option A.

First, calculate the percentage uncertainties for each measured quantity: \% uncertainty in \(R = \frac{0.5}{25.0} \times 100\% = 2.0\%\); \% uncertainty in \(d = \frac{0.02}{0.40} \times 100\% = 5.0\%\); \% uncertainty in \(L = \frac{0.003}{1.500} \times 100\% = 0.2\%\). According to the formula for resistivity, the total percentage uncertainty is given by: \frac{\Delta \rho}{\rho} \times 100\% = \left(\frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\right) \times 100\% = 2.0\% + 2(5.0\%) + 0.2\% = 12.2\%.

1 mark for the correct calculation of percentage uncertainty by summing individual percentage uncertainties, doubling the uncertainty of the diameter.

The rate of mass of water striking the wall is \(\frac{\Delta m}{\Delta t} = \rho A v\). Since the water does not rebound, its final horizontal velocity is zero. The change in momentum per second is the force: \(F = \frac{\Delta p}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right) v = \rho A v^2\). Substituting the values: \(F = 1000 \times (4.0 \times 10^{-4}) \times 15^2 = 0.40 \times 225 = 90\text{ N}\).

1 mark for the correct application of Newton's second law in terms of momentum change of a fluid stream, yielding 90 N.

The forces acting on the sphere in equilibrium are the upward tension \(T\), the upward upthrust \(U\), and the downward weight \(W\). Thus, \(T + U = W \implies T = W - U\). The weight is \(W = \rho_{\text{glass}} V g = (2.5 \times 10^3) \times (8.0 \times 10^{-5}) \times 9.81 = 1.962\text{ N}\). The upthrust is \(U = \rho_{\text{water}} V g = (1.0 \times 10^3) \times (8.0 \times 10^{-5}) \times 9.81 = 0.785\text{ N}\). Therefore, the tension in the string is \(T = 1.962 - 0.785 = 1.177\text{ N} \approx 1.18\text{ N}\).

1 mark for correctly setting up the equilibrium equation with upthrust and weight, calculating the correct tension of 1.18 N.

The phase difference \(\phi\) is related to the path difference \(x\) by \(\phi = \frac{2\pi}{\lambda} x\). Substituting the given values: \(\frac{2\pi}{3} = \frac{2\pi}{\lambda} (0.40)\), which simplifies to \(\lambda = 3 \times 0.40 = 1.20\text{ m}\). The wave speed is given by \(v = f \lambda = 50 \times 1.20 = 60\text{ m s}^{-1}\).

1 mark for calculating the wavelength as 1.20 m and then using the wave equation to find the speed of 60 m/s.

Let the total current from the cell be \(I\). The two parallel resistors are identical, so the current splits equally, meaning the current through each parallel resistor is \(\frac{I}{2}\). The equivalent resistance of the parallel pair is \(\frac{R}{2}\). The total resistance of the circuit is \(R_{\text{total}} = \frac{R}{2} + R = 1.5R\). The total power supplied by the cell is \(P_{\text{total}} = I^2 R_{\text{total}} = 1.5 I^2 R\). The power dissipated in one of the parallel resistors is \(P_1 = \left(\frac{I}{2}\right)^2 R = 0.25 I^2 R\). The ratio is \(\frac{P_1}{P_{\text{total}}} = \frac{0.25}{1.5} = \frac{1}{6}\).

1 mark for finding the correct expression for both individual power and total power, and dividing them to get 1/6.

At the highest point of the vertical circle, both the normal contact force \(N\) and the weight \(mg\) act downwards towards the center of the circular path. This combined force provides the centripetal acceleration: \(N + mg = \frac{m v^2}{r}\). Rearranging for the normal force gives \(N = m\left(\frac{v^2}{r} - g\right)\). Substituting the values: \(N = 0.20 \times \left(\frac{5.0^2}{0.80} - 9.81\right) = 0.20 \times (31.25 - 9.81) = 0.20 \times 21.44 = 4.288\text{ N} \approx 4.3\text{ N}\).

1 mark for correctly formulating the forces at the top of a vertical circle and calculating the normal force of 4.3 N.

According to the first law of thermodynamics, \(\Delta U = q + w\), where \(\Delta U\) is the change in internal energy, \(q\) is the heat energy transferred to the gas, and \(w\) is the work done on the gas. Since work is done on the gas, \(w = +180\text{ J}\). Since heat is lost by the gas, \(q = -120\text{ J}\). Substituting these values: \(\Delta U = -120 + 180 = +60\text{ J}\). Thus, the internal energy increases by \(60\text{ J}\).

1 mark for correctly assigning the positive and negative signs according to the first law of thermodynamics and getting +60 J.

For a charged particle in a magnetic field, the centripetal force is provided by the magnetic force: \(qvB = \frac{mv^2}{r}\), which gives \(r = \frac{mv}{qB}\). The kinetic energy of the particle is \(K = \frac{p^2}{2m} = \frac{(mv)^2}{2m}\), so the momentum is \(mv = \sqrt{2mK}\). Substituting this into the radius formula yields \(r = \frac{\sqrt{2mK}}{qB}\). Since both particles have the same kinetic energy \(K\) and move in the same magnetic field \(B\), the radius is proportional to \(\frac{\sqrt{m}}{q}\). Comparing the alpha particle to the proton: \(\frac{r_{\alpha}}{R} = \frac{\sqrt{4m_{\text{p}}}}{2e} \times \frac{e}{\sqrt{m_{\text{p}}}} = \frac{2\sqrt{m_{\text{p}}}}{2e} \times \frac{e}{\sqrt{m_{\text{p}}}} = 1\). Thus, the radius of the alpha particle's path is also \(R\).

1 mark for relating the radius to kinetic energy and finding that the ratio is exactly 1.

The density \(\rho\) is calculated using the formula:
\rho = \frac{m}{V} = \frac{4m}{\pi d^2 l}

The fractional uncertainty in density is given by:
\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}

Calculate each fractional uncertainty:
\frac{\Delta m}{m} = \frac{0.02}{2.45} \approx 0.00816

\frac{\Delta l}{l} = \frac{0.1}{12.4} \approx 0.00806

2\frac{\Delta d}{d} = 2 \times \frac{0.04}{1.62} \approx 0.04938

Summing these up:
\frac{\Delta \rho}{\rho} = 0.00816 + 0.00806 + 0.04938 = 0.06560

To find the percentage uncertainty, multiply by 100%:
0.06560 \times 100\% \approx 6.6\%

1 mark for the correct calculation of the percentage uncertainty by summing the percentage uncertainty of mass, length, and twice the percentage uncertainty of the diameter.

The force exerted on the ball is equal to its rate of change of momentum. We consider the direction perpendicular to the wall (the normal direction):

Initial perpendicular momentum component = \(mv \cos 30^\circ\)
Final perpendicular momentum component = \(-mv \cos 30^\circ\)

Therefore, the magnitude of the change in momentum is:
\Delta p = 2mv \cos 30^\circ
\Delta p = 2 \times 0.15 \times 12 \times \cos 30^\circ \approx 3.118 \text{ kg m s}^{-1}

Using Newton's second law, the average force is:
F = \frac{\Delta p}{\Delta t} = \frac{3.118}{0.045} \approx 69.3 \text{ N}

1 mark for correctly determining the change in perpendicular momentum component and dividing by the contact time to get approximately 69 N.

First, calculate the energy \(E\) of a single photon using:
E = \frac{hc}{\lambda}

E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{633 \times 10^{-9}} \approx 3.14 \times 10^{-19} \text{ J}

The total number of photons emitted per second \(n\) is given by dividing the power \(P\) by the energy of a single photon:
n = \frac{P}{E} = \frac{3.0 \times 10^{-3}}{3.14 \times 10^{-19}} \approx 9.6 \times 10^{15} \text{ s}^{-1}

1 mark for the correct calculation of the photon energy and the total number of photons emitted per second.

Since the car moves at a constant speed, the net force along the slope is zero. Therefore, the driving force \(F\) of the engine must balance the parallel component of weight and the resistive force:
F = mg \sin\theta + F_{\text{resistive}}
F = (1200 \times 9.81 \times 0.080) + 450 = 941.8 + 450 = 1391.8 \text{ N}

Now, calculate the useful power output \(P\):
P = F v = 1391.8 \times 15 \approx 20877 \text{ W} \approx 21 \text{ kW}

1 mark for calculating the total driving force required to overcome both gravity and resistance, and then multiplying by the constant speed to obtain the correct power of 21 kW.

The resistance \(R\) of a wire of resistivity \(\rho\), length \(L\), and diameter \(d\) is given by:
R = \rho \frac{L}{A} = \rho \frac{L}{\pi (d/2)^2} = \frac{4 \rho L}{\pi d^2}

This shows that \(R \propto \frac{L}{d^2}\).

Therefore, the ratio of resistances is:
\frac{R_X}{R_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2

Given \(L_X = 2 L_Y\) and \(d_X = 0.5 d_Y\), we substitute these into the ratio:
\frac{R_X}{R_Y} = 2 \times (2)^2 = 8

1 mark for correctly applying the resistivity equation to relate the resistance ratio to length and the square of diameter, obtaining 8.

Using the Doppler effect formula for a moving source approaching a stationary observer:
f_o = f_s \left( \frac{v}{v - v_s} \right)

Substitute the given values into the equation:
f_o = 800 \times \left( \frac{340}{340 - 30} \right) = 800 \times \left( \frac{340}{310} \right) \approx 877 \text{ Hz}

1 mark for selecting and calculating with the correct Doppler formula for an approaching source, giving approximately 877 Hz.

First, calculate the fringe width \(w\) using:
w = \frac{\lambda D}{a}

w = \frac{600 \times 10^{-9} \times 2.0}{0.40 \times 10^{-3}} = 3.0 \times 10^{-3} \text{ m} = 3.0 \text{ mm}

The central bright fringe is at the center (\(0\)). The dark fringes occur at half-integer multiples of the fringe width from the center:
- First-order dark fringe: \(0.5 w\)
- Second-order dark fringe: \(1.5 w\)
- Third-order dark fringe: \(2.5 w\)

Therefore, the distance is:
distance = 2.5 \times 3.0 \text{ mm} = 7.5 \text{ mm}

1 mark for calculating the fringe width and multiplying by 2.5 to find the distance to the third-order dark fringe.

First, calculate the total thermal energy \(E\) supplied by the heater in \(5.0 \text{ minutes}\):
t = 5.0 \times 60 = 300 \text{ s}
E = P \times t = 80 \times 300 = 24000 \text{ J}

Since \(20\%\) is lost, \(80\%\) of the total energy is useful for heating the metal:
Q = 0.80 \times 24000 = 19200 \text{ J}

Now, use the specific heat capacity equation:
Q = m c \Delta\theta
19200 = 0.50 \times c \times (60 - 20)
19200 = 20 c
c = \frac{19200}{20} = 960 \text{ J kg}^{-1} \text{K}^{-1}

1 mark for converting time to seconds, applying the efficiency factor to find the useful heat energy, and correctly solving for the specific heat capacity.

The percentage uncertainty of a product or quotient is found by summing the individual percentage uncertainties, each multiplied by the absolute value of its power. For \(\rho = \frac{\pi R d^2}{4 L}\), we have: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). Substituting the given values: \(\frac{\Delta \rho}{\rho} = 2\% + 2(1.5\%) + 1\% = 2\% + 3\% + 1\% = 6.0\%\).

1 mark for correctly adding the percentage uncertainties using the power rule to obtain 6.0%.

First, calculate the useful power output: \(P_{\text{out}} = F v = m g v = 120\text{ kg} \times 9.81\text{ m s}^{-2} \times 0.50\text{ m s}^{-1} = 588.6\text{ W}\). Next, calculate the electrical power input: \(P_{\text{in}} = V I = 240\text{ V} \times 3.5\text{ A} = 840\text{ W}\). Finally, find the efficiency: \(\text{efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{588.6}{840} \times 100\% \approx 70.1\%\), which is approximately \(70\%\).

1 mark for calculating the power output and power input and finding the correct efficiency of 70%.

The relationship between phase difference \(\Delta\phi\) and path difference \(\Delta x\) is given by: \(\Delta\phi = \frac{2\pi \Delta x}{\lambda}\). Given \(\Delta\phi = \frac{\pi}{3}\text{ rad}\) and \(\Delta x = 0.15\text{ m}\), we have: \(\frac{\pi}{3} = \frac{2\pi \times 0.15}{\lambda}\). Solving for the wavelength \(\lambda\) gives \(\lambda = 6 \times 0.15 = 0.90\text{ m}\). The wave speed \(v\) is then: \(v = f \lambda = 50.0\text{ Hz} \times 0.90\text{ m} = 45\text{ m s}^{-1}\).

1 mark for calculating the wavelength as 0.90 m and using v = f * lambda to get 45 m/s.

Using the potential divider formula, \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R_{\text{fixed}} + R_{\text{LDR}}}\). In bright light: \(V_{\text{out, bright}} = 12 \times \frac{1.0}{4.0 + 1.0} = 2.4\text{ V}\). In the dark: \(V_{\text{out, dark}} = 12 \times \frac{16.0}{4.0 + 16.0} = 12 \times 0.80 = 9.6\text{ V}\). The change in output voltage is: \(\Delta V_{\text{out}} = 9.6\text{ V} - 2.4\text{ V} = 7.2\text{ V}\).

1 mark for calculating both output voltages and finding the difference to be 7.2 V.

The centripetal force is: \(F = \frac{m v^2}{r} = \frac{0.20 \times 3.0^2}{0.80} = 2.25\text{ N}\). The kinetic energy is: \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.20 \times 3.0^2 = 0.90\text{ J}\).

1 mark for the correct calculations of both the centripetal force (2.25 N) and kinetic energy (0.90 J).

According to the first law of thermodynamics: \(\Delta U = q + w\), where \(q\) is the heat supplied to the system and \(w\) is the work done on the system. Here, \(q = +450\text{ J}\) (heat is added to the gas) and \(w = -180\text{ J}\) (since work is done by the gas on the surroundings). Therefore: \(\Delta U = +450 - 180 = +270\text{ J}\).

1 mark for using the first law of thermodynamics with correct signs for heat input and work output to find +270 J.

The magnetic force provides the centripetal force: \(B e v = \frac{m v^2}{R}\), which simplifies to \(R = \frac{m v}{B e}\). The period \(T\) of the circular orbit is the distance divided by the speed: \(T = \frac{2̇\pi R}{v} = \frac{2\pi m}{B e}\).

1 mark for equating the magnetic force to the centripetal force, substituting for radius in terms of period, and obtaining the correct algebraic expression.

Let the number of \(\alpha\) particles be \(x\) and \(\beta^-\PD\) particles be \(y\). The change in nucleon number is: \(212 - 4x = 208 \implies 4x = 4 \implies x = 1\). Thus, there is 1 \(\alpha\) decay. The change in proton number is: \(83 - 2x + y = 82 \implies 83 - 2(1) + y = 82 \implies 81 + y = 82 \implies y = 1\). Thus, 1 \(\beta^-\D\) decay occurs. The correct combination is 1 \(\alpha\) particle and 1 \(\beta^-\PD\) particle.

1 mark for using conservation of nucleon and proton numbers to calculate the correct number of alpha and beta-minus particles.

(a) Convert the diameter: \(d = 0.38 \times 10^{-3}\text{ m}\).
Using the formula:
$$\rho = \frac{\pi \times (0.38 \times 10^{-3})^2 \times 12.4}{4 \times 1.500}$$
$$\rho = \frac{3.14159 \times 1.444 \times 10^{-7} \times 12.4}{6.000}$$
$$\rho = 9.375 \times 10^{-7}\ \Omega\text{ m} \approx 9.4 \times 10^{-7}\ \Omega\text{ m}$$.

(b) The fractional uncertainty formula is:
$$\frac{\Delta \rho}{\rho} = 2 \frac{\Delta d}{d} + \frac{\Delta R}{R} + \frac{\Delta L}{L}$$
- \(\frac{\Delta d}{d} = \frac{0.02}{0.38} = 0.05263\) (or \(5.26\%\))
- \(\frac{\Delta R}{R} = \frac{0.2}{12.4} = 0.01613\) (or \(1.61\%\))
- \(\frac{\Delta L}{L} = \frac{0.005}{1.500} = 0.00333\) (or \(0.33\%\))

Percentage uncertainty in \(\rho\):
$$\% \Delta \rho = (2 \times 5.263\%) + 1.613\% + 0.333\% = 10.526\% + 1.613\% + 0.333\% = 12.47\% \approx 12.5\%\text{ (or } 12\%\text{)}$$

(c) The diameter \(d\) contributes most to the uncertainty because its fractional uncertainty is the largest and is doubled due to the squared term. To reduce this uncertainty, use a micrometer screw gauge to measure the diameter at multiple points along the wire and average the results.

(a)
- [1] Substitution of values into formula with consistent SI units.
- [1] Final calculated value shown to at least 3 s.f. (\(9.38 \times 10^{-7}\ \Omega\text{ m}\)).

(b)
- [1] Correct equation for combining fractional uncertainties (with factor of 2 for \(d\)).
- [1] Correct individual percentage uncertainties (\(5.26\%\), \(1.61\%\), \(0.33\%\)).
- [1] Final percentage uncertainty correct (\(12\%\) or \(12.5\%\)).

(c)
- [1] Identifies diameter \(d\) as the main contributor with valid reasoning (largest contribution/doubled).
- [1.5] Suggests using a micrometer screw gauge / taking measurements at different positions and orientations.

(a) By conservation of momentum:
$$m_A u_A + m_B u_B = (m_A + m_B) v$$
$$0.80 \times 2.5 + 0 = (0.80 + 1.2) v$$
$$2.0 = 2.0 v \implies v = 1.0\text{ m s}^{-1}$$.

(b) Initial kinetic energy:
$$E_{k,i} = \frac{1}{2} m_A u_A^2 = 0.5 \times 0.80 \times 2.5^2 = 2.5\text{ J}$$
Final kinetic energy:
$$E_{k,f} = \frac{1}{2} (m_A + m_B) v^2 = 0.5 \times 2.0 \times 1.0^2 = 1.0\text{ J}$$
Loss in total kinetic energy:
$$\Delta E_k = 2.5 - 1.0 = 1.5\text{ J}$$.

(c) The average force exerted on B is equal to the rate of change of momentum of B:
$$\Delta p_B = m_B v - 0 = 1.2 \times 1.0 = 1.2\text{ N s}$$
$$F = \frac{\Delta p_B}{\Delta t} = \frac{1.2\text{ N s}}{0.15\text{ s}} = 8.0\text{ N}$$
Direction: In the direction of initial velocity of trolley A (to the right).

(a)
- [1] States or uses conservation of linear momentum equation.
- [1] Correctly substitutes values and shows \(v = 1.0\text{ m s}^{-1}\).

(b)
- [1] Correct calculation of initial kinetic energy (\(2.5\text{ J}\)).
- [1] Correct calculation of final kinetic energy (\(1.0\text{ J}\)).
- [0.5] Correct final answer for loss in kinetic energy (\(1.5\text{ J}\)).

(c)
- [1] Formula \(F = \frac{\Delta p}{\Delta t}\) stated or implied.
- [1] Correct calculation of average force (\(8.0\text{ N}\)).
- [1] Correct direction specified (forward / direction of initial motion of A).

(a) The center of gravity is the point at which the entire weight of an object is considered to act.

(b) For vertical equilibrium:
$$\text{Total upward forces} = \text{Total downward forces}$$
$$T_A + T_B = W$$
$$60 + T_B = 150 \implies T_B = 90\text{ N}$$.

(c) Taking moments about end A:
$$\text{Clockwise moment} = \text{Counter-clockwise moment}$$
$$W \times x = T_B \times L$$
$$150 \times x = 90 \times 2.4$$
$$150 x = 216 \implies x = 1.44\text{ m}$$.

(d) When the rope at A is moved inwards, the pivot/support A is closer to the center of gravity. Consequently, rope A supports a greater share of the weight, and the tension in rope B decreases (from \(90\text{ N}\) to \(78\text{ N}\)).

(a)
- [1] Defines center of gravity as the point where weight appears to act.
- [0.5] Mentions 'entire' weight / 'single point'.

(b)
- [1] Mentions vertical forces must balance for equilibrium (\(T_A + T_B = W\)).
- [0.5] Calculates \(T_B = 90\text{ N}\).

(c)
- [1] States principle of moments.
- [1] Correctly sets up equation taking moments about A: \(150 \times x = 90 \times 2.4\).
- [1] Calculates \(x = 1.44\text{ m}\).

(d)
- [0.5] States that tension in rope B decreases.
- [1] Explains that support A is closer to the center of gravity, taking a greater portion of the load (or shows calculation yielding \(78\text{ N}\)).

(a) Sound waves travel down the tube from the tuning fork and reflect off the water surface at the closed end. The incident and reflected waves, which have the same frequency and amplitude, travel in opposite directions and superpose to form a stationary wave.

(b) The sketch should show a node (zero amplitude) at the water surface (closed end) and an antinode (maximum amplitude) near the open end of the tube.

(c) For a tube closed at one end, the difference in length between two successive resonances is equal to half a wavelength:
$$\Delta L = L_2 - L_1 = 74.0\text{ cm} - 24.0\text{ cm} = 50.0\text{ cm} = 0.500\text{ m}$$
$$\frac{\lambda}{2} = 0.500\text{ m} \implies \lambda = 1.00\text{ m}$$
Using the wave equation:
$$v = f \lambda = 340\text{ Hz} \times 1.00\text{ m} = 340\text{ m s}^{-1}$$
This cancels the effect of the end correction \(c\) because both lengths are affected by the same end correction \(c\): \(L_1 + c = \lambda/4\) and \(L_2 + c = 3\lambda/4\).

(a)
- [1] Mention of reflection of sound waves at the boundary / water surface.
- [1] Mention of superposition of incident and reflected waves.
- [0.5] Mentions waves have same frequency/amplitude and travel in opposite directions.

(b)
- [1] Correct shape showing node at the bottom and antinode near the top.
- [0.5] Labels 'N' at the node and 'A' at the antinode.

(c)
- [1] States relationship \(L_2 - L_1 = \frac{\lambda}{2}\).
- [1] Calculates wavelength \(\lambda = 1.00\text{ m}\).
- [1] Uses \(v = f\lambda\) to find the speed of sound.
- [0.5] Correct final value with unit (\(340\text{ m s}^{-1}\)).

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

(b) For the first wire:
$$R_1 = \rho \frac{L}{A} = 4.5\ \Omega$$
For the second wire:
- Length \(L_2 = 2 L\)
- Diameter \(d_2 = 0.5 d \implies A_2 = \frac{\pi d_2^2}{4} = 0.25 A\)
$$R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2 L}{0.25 A} = 8 R_1$$
$$R_2 = 8 \times 4.5 = 36\ \Omega$$.

(c) The rate of thermal energy dissipation is the electrical power:
$$P = \frac{V^2}{R_2} = \frac{12^2}{36} = \frac{144}{36} = 4.0\text{ W}$$
This corresponds to \(4.0\text{ J s}^{-1}\).

(a)
- [1] Correctly states \(\rho = R A / L\) or equivalent verbal definition.
- [0.5] Clearly defines all symbols used.

(b)
- [1] Relates change in diameter to change in cross-sectional area (\(A_2 = 0.25 A_1\)).
- [1.5] Sets up ratio of resistance \(R_2 / R_1 = (L_2 / L_1) \times (A_1 / A_2) = 2 / 0.25 = 8\).
- [1] Obtains final value of \(36\ \Omega\).

(c)
- [1] Uses correct formula for power (\(P = V^2 / R\) or \(I^2 R\) with calculated \(I\)).
- [1] Correct substitution of \(12\text{ V}\) and \(36\ \Omega\).
- [0.5] Obtains final power value of \(4.0\text{ W}\).

(a) The diagram must feature a series loop containing a DC power source (battery), a fixed resistor, and an LDR. Output terminals for \(V_{\text{out}}\) must be drawn across the LDR.

(b) The potential divider equation gives:
$$V_{\text{out}} = \frac{R_{\text{LDR}}}{R_{\text{LDR}} + R_{\text{fixed}}} \times V_{\text{in}}$$
$$V_{\text{out}} = \frac{300}{300 + 1200} \times 9.0 = \frac{300}{1500} \times 9.0 = 0.20 \times 9.0 = 1.8\text{ V}$$.

(c) Using the potential divider equation:
$$6.0 = \frac{R_{\text{LDR}}}{R_{\text{LDR}} + 1200} \times 9.0$$
$$\frac{6.0}{9.0} = \frac{2}{3} = \frac{R_{\text{LDR}}}{R_{\text{LDR}} + 1200}$$
$$2(R_{\text{LDR}} + 1200) = 3 R_{\text{LDR}}$$
$$2 R_{\text{LDR}} + 2400 = 3 R_{\text{LDR}} \implies R_{\text{LDR}} = 2400\ \Omega\text{ (or } 2.4\text{ k}\Omega\text{)}$$.

(a)
- [1] Correct circuit symbols for LDR and fixed resistor in series with battery.
- [1] Correct connections showing \(V_{\text{out}}\) measured across the LDR.

(b)
- [1] Recall and use of potential divider formula.
- [1] Correct substitution of \(300\ \Omega\) and \(1200\ \Omega\).
- [0.5] Correct final voltage of \(1.8\text{ V\text{ (or } 1.80\text{ V}\)}).

(c)
- [1] Formulates equation with \(V_{\text{out}} = 6.0\text{ V}\).
- [1] Rearranges equation correctly to solve for \(R_{\text{LDR}}\).
- [1] Correct final resistance \(2400\ \Omega\) (or \(2.4\text{ k}\Omega\)).

(a) Acceleration is the rate of change of velocity. Velocity is a vector quantity with magnitude (speed) and direction. Even though speed is constant, the direction of motion is continuously changing, which means velocity is changing, resulting in centripetal acceleration.

(b) Resolving vertically for equilibrium:
$$T \cos(30^\circ) = mg$$
$$T \cos(30^\circ) = 0.15 \times 9.81$$
$$T \times 0.8660 = 1.4715$$
$$T = \frac{1.4715}{0.8660} = 1.699\text{ N} \approx 1.7\text{ N}$$.

(c) Resolving horizontally, the centripetal force is provided by the horizontal component of the tension:
$$T \sin(30^\circ) = \frac{m v^2}{r}$$
The radius of the circular path is:
$$r = L \sin(30^\circ) = 0.80 \times 0.50 = 0.40\text{ m}$$
Substitute values:
$$1.699 \times \sin(30^\circ) = \frac{0.15 \times v^2}{0.40}$$
$$0.8495 = 0.375 \times v^2$$
$$v^2 = 2.265 \implies v = 1.505\text{ m s}^{-1} \approx 1.5\text{ m s}^{-1}$$.

(a)
- [1] States that velocity is a vector / has direction, and direction is changing.
- [0.5] Concludes that change in velocity implies acceleration.

(b)
- [1] Resolves forces vertically (\(T \cos \theta = mg\)).
- [1] Substitutes \(m = 0.15\text{ kg}\), \(g = 9.81\text{ m s}^{-2}\), and \(\theta = 30^\circ\).
- [0.5] Shows calculated tension of \(1.7\text{ N}\).

(c)
- [1] Identifies radius \(r = L \sin \theta = 0.40\text{ m}\).
- [1] Sets horizontal component of tension equal to centripetal force (\(T \sin \theta = m v^2 / r\)).
- [1] Solves for \(v^2\) or \(v\).
- [0.5] Correct final answer of \(1.5\text{ m s}^{-1}\) (or \(1.50\text{ m s}^{-1}\)).

(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 heat supplied to the system.
- \(w\) is the work done on the system.
(Note: \(\Delta Q = \Delta U + \Delta W\) is also acceptable if symbols are correctly defined with work done *by* the system).

(b) Work done by the gas during expansion at constant pressure:
$$W_{\text{by}} = p \Delta V$$
$$W_{\text{by}} = 3.0 \times 10^5\text{ Pa} \times (5.0 \times 10^{-3}\text{ m}^3 - 2.0 \times 10^{-3}\text{ m}^3)$$
$$W_{\text{by}} = 3.0 \times 10^5 \times 3.0 \times 10^{-3} = 900\text{ J}$$.

(c) Using \(\Delta U = q + w\), where \(q = +1200\text{ J}\) (heat supplied to gas) and \(w = -900\text{ J}\) (work done *on* gas, which is negative because the gas expands):
$$\Delta U = 1200 + (-900) = +300\text{ J}$$
Since \(\Delta U\) is positive, the internal energy of the gas increases by \(300\text{ J}\).

(a)
- [1] States algebraic form of the first law (e.g., \(\Delta U = q + w\)).
- [1] Correctly defines all three terms, including correct sign convention (e.g., work done *on* system).

(b)
- [1] Recalls and uses formula \(W = p \Delta V\).
- [1] Correct substitution of pressure and change in volume values.
- [0.5] Correct final work value of \(900\text{ J}\).

(c)
- [1] Correctly identifies that work done *on* the gas is \(-900\text{ J}\) or uses equivalent formula correctly.
- [1] Computes absolute change in internal energy as \(300\text{ J}\).
- [1] Explicitly states that internal energy increases.

Typical experimental data: For x = 0.550 m, h_0 = 0.853 m, and h = 0.811 m, giving deflection y = 0.042 m. A set of six measurements is taken for x in the range 0.400 m to 0.750 m. The values are tabulated with correct column headings and units: x/m, h_0/m, h/m, y/m, and x^3/m^3. Plotted points of y against x^3 show a linear trend. Best-fit line is drawn. Using two distant points on the line, say (0.064, 0.010) and (0.343, 0.080), the gradient is calculated as (0.080 - 0.010)/(0.343 - 0.064) = 0.251 m^-2. The y-intercept is calculated as 0.010 - 0.251 * 0.064 = -0.006 m. Therefore, constant a = 0.251 m^-2 and constant b = -0.006 m.

Table of Results (6 marks): - 1 mark: At least 6 sets of readings of x and heights recorded with a range of x of at least 30 cm. - 1 mark: Correct column headings with units. - 1 mark: All raw values of x, h_0, and h recorded to the nearest millimeter (0.001 m). - 1 mark: Values of x^3 calculated to the same or one more significant figure than the raw x values. - 1 mark: Deflection y calculated correctly. - 1 mark: Quality of results (points lie very close to a straight line). Graph (4 marks): - 1 mark: Linear, simple scales chosen so that points occupy at least half the grid. Axes labeled with units. - 1 mark: Points plotted accurately to within half a small square. - 1 mark: Single, sharp, balanced best-fit straight line drawn. - 1 mark: Quality of line matching coordinates. Gradient and Intercept (4 marks): - 1 mark: Large triangle used for gradient calculation (hypotenuse > 50% of drawn line length). - 1 mark: Coordinates read correctly to half a small square. - 1 mark: Correct gradient calculation. - 1 mark: Correct y-intercept calculation using y - m*x^3. Constants and Units (6 marks): - 1 mark: Value of a equated to gradient. - 1 mark: Value of b equated to y-intercept. - 1 mark: Unit of a is m^-2 (or cm^-2). - 1 mark: Unit of b is m (or cm). - 1 mark: Values of a and b given to 2 or 3 significant figures. - 1 mark: All calculations and working shown clearly.

Typical measurements: First height h_1 = 6.2 cm. Times recorded: 2.45 s, 2.39 s; average t_1 = 2.42 s. Absolute uncertainty in t is estimated using human reaction time as 0.15 s. Percentage uncertainty = (0.15 / 2.42) * 100% = 6.2%. First constant k_1 = 2.42 * sqrt(6.2) = 6.03 s cm^0.5. Second measurements: h_2 = 12.1 cm. Times recorded: 1.70 s, 1.76 s; average t_2 = 1.73 s. Second constant k_2 = 1.73 * sqrt(12.1) = 6.02 s cm^0.5. Percentage difference between the two k values is |6.03 - 6.02| / 6.025 * 100% = 0.17%. Since the percentage difference (0.17%) is less than the percentage uncertainty in t (6.2%), the results strongly support the suggested relationship that k is constant.

Measurements and Calculations (6 marks): - 1 mark: First height h recorded to nearest mm and at least two raw times t recorded. - 1 mark: Correct calculation of average t. - 1 mark: Correct calculation of percentage uncertainty in t using half the range or human reaction time (0.1 s to 0.2 s). - 1 mark: Second values of h and t recorded correctly, with shorter times for larger h. - 1 mark: Correct calculation of both k values. - 1 mark: Correct units of k (s cm^0.5 or s m^0.5). Analysis (2 marks): - 1 mark: Calculation of percentage difference between the two k values. - 1 mark: Conclusion clearly stated comparing the percentage difference with a reasonable threshold (e.g. 10%) or the percentage uncertainty of t. Limitations and Improvements (12 marks): - 8 marks: Four pairs of limitations and corresponding improvements (1 mark for each limitation, 1 mark for each corresponding improvement): (L1) Difficulty in releasing cylinder consistently / (I1) Use mechanical release mechanism (hinged block/card). (L2) Cylinder veers off a straight line / (I2) Use guide rails or straight track on the board. (L3) Human reaction time error in timing / (I3) Use light gates connected to timer or high-speed video. (L4) Parallax error in measuring height h / (I4) Use set square or spirit level to align ruler. - 4 marks for specific quality parameters: 1 mark for keeping distance L constant to within 2 mm, 1 mark for ensuring wooden blocks are stable, 1 mark for giving k to 2 or 3 significant figures, 1 mark for taking at least three repeat readings of t to get a reliable average.

(a) The gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point. (b)(i) Potential \(\phi = -GM/r\). Rearranging gives \(M = -\phi r / G\). Substituting the values: \(M = -(-1.5 \times 10^6 \times 3.2 \times 10^6) / (6.67 \times 10^{-11}) = 7.2 \times 10^{22}\text{ kg}\). (b)(ii) For a stable circular orbit, centripetal force equals gravitational force: \(m v^2 / r = G M m / r^2\), which simplifies to \(m v^2 = G M m / r\). The kinetic energy \(E_k = (1/2) m v^2 = G M m / (2r)\). Since \(\phi = -GM/r\), we can write \(E_k = -1/2 \times m \times \phi = -0.5 \times 450 \times (-1.5 \times 10^6) = 3.4 \times 10^8\text{ J}\). (c) The total energy of the probe in orbit is \(E_t = E_k + E_p = 3.375 \times 10^8\text{ J} + (-1.5 \times 10^6 \times 450)\text{ J} = 3.375 \times 10^8 - 6.75 \times 10^8 = -3.375 \times 10^8\text{ J}\). To escape the gravitational field completely, the total energy of the probe must be at least zero at infinity. Therefore, the minimum work done by the thrusters is equal to the increase in total energy required, which is \(+3.4 \times 10^8\text{ J}\).

(a) Work done per unit mass [1], bringing a test mass from infinity to the point [1]. (b)(i) Formula \(\phi = -GM/r\) used [1], substitution shown [1], final value \(7.2 \times 10^{22}\text{ kg}\) [1]. (b)(ii) Relationship \(E_k = -0.5 m \phi\) or centripetal force equivalence [1], substitution shown [1], final value \(3.4 \times 10^8\text{ J}\) (accept \(3.38 \times 10^8\text{ J}\)) [1]. (c) Mention of total energy in orbit being negative or \(-3.4 \times 10^8\text{ J}\) [1], explanation that work done must equal the binding energy to reach zero total energy at infinity, leading to \(+3.4 \times 10^8\text{ J}\) [1].

(a) The acceleration of the object is directly proportional to its displacement from the equilibrium position, and is always directed towards the equilibrium position. (b)(i) The angular frequency \(\omega = \sqrt{k/m} = \sqrt{140 / 0.35} = \sqrt{400} = 20\text{ rad s}^{-1}\). (ii) The maximum acceleration \(a_{max} = \omega^2 x_0 = 20^2 \times 0.080 = 400 \times 0.080 = 32\text{ m s}^{-2}\). (c)(i) The amplitude decreases exponentially over time, and the total energy decreases continuously as work is done against the resistive forces. (ii) The period of the oscillation remains constant (or increases very slightly, but to a first approximation, it is unaffected by light damping).

(a) Acceleration proportional to displacement [1], directed towards equilibrium point [1]. (b)(i) Formula \(\omega = \sqrt{k/m}\) [1], calculation leading to \(20\text{ rad s}^{-1}\) [1]. (b)(ii) Formula \(a_{max} = \omega^2 x_0\) [1], substitution leading to \(32\text{ m s}^{-2}\) [1]. (c)(i) Amplitude decreases (exponentially) [1], total energy decreases [1]. (c)(ii) Period remains constant [1], explanation that light damping does not significantly alter the natural frequency of oscillation [1].

(a) 1. The volume of the gas molecules is negligible compared to the volume of the container. 2. There are no intermolecular forces of attraction or repulsion except during collisions. 3. All collisions between molecules and with the container walls are perfectly elastic. (b)(i) Using the equation of state \(pV = nRT\), we have \(p = nRT / V = 1.8 \times 8.31 \times (27 + 273.15) / 0.024 = 1.87 \times 10^5\text{ Pa}\). (b)(ii) Mean kinetic energy of an atom is \(E_k = (3/2) k_B T\). Here, \(T = 300\text{ K}\). Thus, \(E_k = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21}\text{ J}\). (b)(iii) For a monatomic ideal gas, the total internal energy \(U\) is equal to the sum of the kinetic energies of all atoms: \(U = (3/2) N k_B T = (3/2) n R T\). At \(T = 127^\circ\text{C} = 400\text{ K}\), \(U = 1.5 \times 1.8 \times 8.31 \times 400 = 8975\text{ J} \approx 9.0 \times 10^3\text{ J}\).

(a) Any three assumptions: negligible molecular volume, no intermolecular forces (except during collisions), perfectly elastic collisions, motion is continuous and random, duration of collision is negligible [3, 1 mark for each]. (b)(i) \(pV = nRT\) stated [1], \(T = 300.15\text{ K}\) or \(300\text{ K}\) used [1], answer \(1.9 \times 10^5\text{ Pa}\) (or \(1.87 \times 10^5\text{ Pa}\)) [1]. (b)(ii) Formula \(E_k = (3/2) k_B T\) used [1], correct answer \(6.2 \times 10^{-21}\text{ J}\) [1]. (b)(iii) Total internal energy formula \(U = 1.5 n R T\) or equivalent used [1], substitution of \(T = 400\text{ K}\) [1], correct answer \(9.0 \times 10^3\text{ J}\) (or \(8970\text{ J}\)) [1].

(a) Coulomb's law states that the electrostatic force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of their separation. (b)(i) Potential \(V = V_A + V_B = Q_A / (4̀̂̐\varepsilon_0 r_A) + Q_B / (4̀̂̐\varepsilon_0 r_B)\). Here, \(r_A = 0.080\text{ m}\) and \(r_B = 0.120 - 0.080 = 0.040\text{ m}\). \(V = (8.99 \times 10^9) \times (6.0 \times 10^{-9} / 0.080 - 3.0 \times 10^{-9} / 0.040) = 8.99 \times (75 - 75) \times 10^{-9} = 0\text{ V}\). (b)(ii) The electric field strength is a vector. Since \(A\) is positive, its field points away from \(A\) (towards \(B\)). Since \(B\) is negative, its field points towards \(B\) (away from \(A\)). Thus, the fields add: \(E = E_A + E_B = Q_A / (4̀̂̐\varepsilon_0 r_A^2) + |Q_B| / (4̀̂̐\varepsilon_0 r_B^2) = (8.99 \times 10^9) \times (6.0 \times 10^{-9} / 0.080^2 + 3.0 \times 10^{-9} / 0.040^2) = 8.99 \times (937.5 + 1875) = 2.53 \times 10^4\text{ V m}^{-1}\). (c) At this point, although the potential is zero, the electric field is not zero. It points towards \(B\). Thus, a positive test charge experiences a force towards \(B\) and will accelerate towards the right (towards \(B\)).

(a) Force is proportional to product of charges [1], inversely proportional to square of separation [1]. (b)(i) Clear statement that \(V = V_A + V_B\) [1], correct substitution of distances \(0.080\text{ m}\) and \(0.040\text{ m}\) [1], calculation showing the result is zero [1]. (b)(ii) Recognition that both field components have the same direction and must be added [1], correct calculation of individual fields \(E_A = 8.43 \times 10^3\text{ V m}^{-1}\) and \(E_B = 1.69 \times 10^4\text{ V m}^{-1}\) [1], correct total field \(2.5 \times 10^4\text{ V m}^{-1}\) [1]. (c) State that electric field is non-zero (or directed towards \(B\)) [1], positive charge experiences force in the direction of the field and accelerates towards \(B\) [1].

(a) Capacitance is the charge stored per unit potential difference. (b) When connected to a supply, electrons are transferred from one plate to another. Work is done to separate these positive and negative charges, and this work is stored as electric potential energy in the electric field between the plates. (c)(i) Energy stored \(E = (1/2) C V^2 = 0.5 \times 470 \times 10^{-6} \times 9.0^2 = 0.019\text{ J}\). (c)(ii) 1. Time constant \(\tau = R C = 15 \times 10^3 \times 470 \times 10^{-6} = 7.05\text{ s} \approx 7\text{ s}\). 2. Potential difference during discharge is given by \(V = V_0 e^{-t / \tau}\). Substituting \(t = 10\text{ s}\), we get \(V = 9.0 \times e^{-10 / 7.05} = 9.0 \times e^{-1.418} = 2.2\text{ V}\).

(a) Charge / potential difference [1]. (b) Work is done to separate charges / move electrons from one plate to the other [1], energy is stored in the electric field between plates [1]. (c)(i) Formula \(E = 0.5 C V^2\) used [1], correct answer \(1.9 \times 10^{-2}\text{ J}\) (or \(0.019\text{ J}\)) [1]. (c)(ii) 1. Formula \(\tau = RC\) used [1], substitution showing \(7.05\text{ s}\) [1]. 2. Formula \(V = V_0 e^{-t/\tau}\) used [1], correct substitution of numbers [1], final value \(2.2\text{ V}\) [1].

(a) Faraday's law states that the magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage. (b)(i) Magnetic flux linkage \(\Phi_{link} = N B A = 450 \times 0.080 \times 1.6 \times 10^{-4} = 5.76 \times 10^{-3}\text{ Wb-turns} \approx 5.8 \times 10^{-3}\text{ Wb-turns}\). (b)(ii) Induced e.m.f. \(E = \Delta \Phi_{link} / \Delta t = (5.76 \times 10^{-3}) / 0.15 = 0.0384\text{ V} \approx 0.038\text{ V}\) (or \(3.8 \times 10^{-2}\text{ V}\)). (c) Lenz's law states that the direction of the induced e.m.f. is such that it opposes the change in magnetic flux that causes it. If it did not oppose, the induced current would create a magnetic field that reinforces the original change, leading to an ever-increasing current and field without doing any work. This would violate the conservation of energy. Hence, work must be done against the opposing force to generate electrical energy.

(a) Induced e.m.f. is proportional to [1], rate of change of magnetic flux linkage [1]. (b)(i) Formula \(\Phi = N B A\) used [1], correct calculation giving \(5.8 \times 10^{-3}\text{ Wb-turns}\) (or \(5.76 \times 10^{-3}\)) [1]. (b)(ii) Formula \(E = \Delta \Phi / \Delta t\) used [1], substitution shown [1], correct answer \(3.8 \times 10^{-2}\text{ V}\) [1]. (c) Statement that induced field opposes the change [1], if it did not oppose, electrical energy would be created from nothing [1], work must be done against the opposing force to conserve energy [1].

(a) Work function energy is the minimum energy required to remove a single electron from the surface of a metal. (b)(i) Photon energy \(E = h c / \lambda = (6.63 \times 10^{-34} \times 3.00 \times 10^8) / (240 \times 10^{-9}) = 8.2875 \times 10^{-19}\text{ J}\). In eV: \(E = 8.2875 \times 10^{-19} / (1.60 \times 10^{-19}) = 5.18\text{ eV}\). (b)(ii) Using Einstein's photoelectric equation: \(E_{k,max} = E - \Phi = 5.18 - 4.3 = 0.88\text{ eV}\). Converting to Joules: \(E_{k,max} = 0.88 \times 1.60 \times 10^{-19} = 1.41 \times 10^{-19}\text{ J}\). (c) Photons interact with electrons at various depths below the surface of the metal. Electrons deeper inside lose kinetic energy through collisions with lattice ions as they travel to the surface. Electrons at the surface experience minimal collisions and are emitted with the maximum kinetic energy.

(a) Minimum energy [1], to release an electron from metal surface [1]. (b)(i) Formula \(E = hc/\lambda\) used [1], conversion factor \(1.6 \times 10^{-19}\text{ J eV}^{-1}\) used [1], correct answer \(5.2\text{ eV}\) or \(5.18\text{ eV}\) [1]. (b)(ii) Photoelectric equation \(E_{k,max} = hf - \Phi\) [1], subtraction of \(4.3\text{ eV}\) to get \(0.88\text{ eV}\) [1], conversion to \(1.4 \times 10^{-19}\text{ J}\) [1]. (c) Description of collisions with metal lattice for deeper electrons [1], surface electrons escape with no such losses, leading to a range of energies up to the maximum [1].

(a) The decay constant is the probability of decay of a nucleus per unit time. (b)(i) Half-life in seconds: \(t_{1/2} = 8.0 \times 24 \times 3600 = 691200\text{ s}\). The decay constant \(\lambda = \ln 2 / t_{1/2} = 0.693 / 691200 = 1.00 \times 10^{-6}\text{ s}^{-1}\). (b)(ii) Using \(A = \lambda N\), we get \(N = A / \lambda = 3.5 \times 10^7 / (1.00 \times 10^{-6}) = 3.5 \times 10^{13}\) nuclei. (b)(iii) Since \(24\text{ days}\) is exactly \(3\) half-lives (\(24 / 8.0 = 3\)), the activity will halve three times: \(A = A_0 / 2^3 = 3.5 \times 10^7 / 8 = 4.38 \times 10^6\text{ Bq} \approx 4.4 \times 10^6\text{ Bq}\).

(a) Probability of decay of a nucleus [1], per unit time [1]. (b)(i) Correct conversion of half-life to seconds (\(6.91 \times 10^5\text{ s}\)) [1], calculation of \(\lambda\) showing \(1.0 \times 10^{-6}\text{ s}^{-1}\) [1]. (b)(ii) Formula \(A = \lambda N\) [1], substitution shown [1], correct answer \(3.5 \times 10^{13}\) [1]. (b)(iii) Recognition of 3 half-lives or use of \(A = A_0 e^{-\lambda t}\) [1], calculation shown [1], correct final answer \(4.4 \times 10^6\text{ Bq}\) (or \(4.38 \times 10^6\text{ Bq}\)) [1].

(a) Internal energy is the sum of a random distribution of kinetic and potential energies associated with the molecules (or atoms) of the system.

(b) (i) The work done \(W\) during an isobaric compression is given by:
\(W = P \Delta V\)
\(W = 1.2 \times 10^5\text{ Pa} \times (5.0 \times 10^{-3}\text{ m}^3 - 2.5 \times 10^{-3}\text{ m}^3) = 300\text{ J}\)
Since the volume decreases, work is done on the gas, so \(W = +300\text{ J}\).

(ii) 1. During process \(A \rightarrow B\), the volume is constant (\(\Delta V = 0\)), so the work done on the gas is \(0\text{ J}\).
2. For any complete cycle that returns to its initial state, the total change in internal energy \(\Delta U_{\text{cycle}} = 0\text{ J}\).

(iii) Using the first law of thermodynamics, \(\Delta U = Q + W\), where \(Q\) is heat supplied and \(W\) is work done on the system:
1. For \(A \rightarrow B\):
\(\Delta U_{A \rightarrow B} = 675\text{ J} + 0\text{ J} = +675\text{ J}\).
2. For \(B \rightarrow C\):
Since the process is adiabatic, \(Q = 0\). The work done by the gas is \(430\text{ J}\), which means the work done ON the gas is \(W = -430\text{ J}\).
\(\Delta U_{B \rightarrow C} = 0 + (-430\text{ J}) = -430\text{ J}\).
3. For the complete cycle:
\(\Delta U_{\text{cycle}} = \Delta U_{A \rightarrow B} + \Delta U_{B \rightarrow C} + \Delta U_{C \rightarrow A} = 0\)
\(675 - 430 + \Delta U_{C \rightarrow A} = 0\)
\(245 + \Delta U_{C \rightarrow A} = 0 \Rightarrow \Delta U_{C \rightarrow A} = -245\text{ J}\).

(iv) For the process \(C \rightarrow A\):
\(\Delta U_{C \rightarrow A} = Q_{C \rightarrow A} + W_{C \rightarrow A}\)
\(-245\text{ J} = Q_{C \rightarrow A} + 300\text{ J}\)
\(Q_{C \rightarrow A} = -245 - 300 = -545\text{ J}\)
Since \(Q_{C \rightarrow A}\) is negative, \(545\text{ J}\) of thermal energy is released from the gas.

(a)
- sum of a random distribution of kinetic and potential energies [1]
- associated with the molecules/atoms/particles of the system [1]

(b)(i)
- use of \(W = P \Delta V\) [1]
- substitution: \(1.2 \times 10^5 \times 2.5 \times 10^{-3} = 300\text{ J}\) and states that since volume decreases, work is done on the gas [1]

(ii)
- 1. \(0\text{ J}\) [1]
- 2. \(0\text{ J}\) [1]

(iii)
- 1. \(\Delta U = +675\text{ J}\) [1]
- 2. \(\Delta U = -430\text{ J}\) [1]
- 3. \(\Delta U = -245\text{ J}\) (allow ecf from previous parts, i.e., \(0 - \text{part 1} - \text{part 2}\)) [1]

(iv)
- use of \(\Delta U = Q + W\) to obtain numerical value of \(545\text{ J}\) [1]
- states clearly that the thermal energy is "released" or "lost" from the gas (due to the negative sign) [1]

(a) Faraday's law of electromagnetic induction 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) (i) Magnetic flux linkage is the product of the magnetic flux through a coil and the number of turns on the coil (or \(\Phi N = B A N \cos \theta\), where \(\theta\) is the angle between the field and the normal to the plane of the coil).

(ii) From Faraday's law, the average induced e.m.f. \(E\) is given by:
\(E = \frac{\Delta (\Phi N)}{\Delta t} = \frac{N A \Delta B}{\Delta t}\)
Since the coil is pulled completely out of the field, \(\Delta B = B\).
Substitute the given values:
\(1.8 \times 10^{-3}\text{ V} = \frac{150 \times (4.5 \times 10^{-4}\text{ m}^2) \times B}{0.12\text{ s}}\)
\(1.8 \times 10^{-3} \times 0.12 = 0.0675 \times B\)
\(2.16 \times 10^{-4} = 0.0675 \times B\)
\(B = \frac{2.16 \times 10^{-4}}{0.0675} = 3.2 \times 10^{-3}\text{ T}\) (or \(3.2\text{ mT}\)).

(c) (i) The magnetic flux linkage \(\Phi N\) is:
\(\Phi N = B A N = [B_0 \sin(100\pi t)] \times A \times N\)
\(\Phi N = N A B_0 \sin(100\pi t)\)
Calculate the constant term:
\(N A B_0 = 150 \times (4.5 \times 10^{-4}) \times (3.2 \times 10^{-3}) = 2.16 \times 10^{-4}\text{ Wb}\) (or \(\text{Wb-turn}\)).
Therefore, the expression is:
\(\Phi N = 2.16 \times 10^{-4} \sin(100\pi t)\).

(ii) The induced e.m.f. \(E\) is given by \(-\frac{d(\Phi N)}{dt}\):
\(E = -\frac{d}{dt} [2.16 \times 10^{-4} \sin(100\pi t)] = - (2.16 \times 10^{-4} \times 100\pi) \cos(100\pi t)\)
\(E = -2.16 \times 10^{-2}\pi \cos(100\pi t)\)
The maximum value of the induced e.m.f., \(E_{\text{max}}\), occurs when \(\cos(100\pi t) = \pm 1\):
\(E_{\text{max}} = 2.16 \times 10^{-2} \pi \approx 0.0679\text{ V}\) (or \(6.8 \times 10^{-2}\text{ V}\) / \(68\text{ mV}\)).

(a)
- induced e.m.f. is proportional to / equal to [1]
- the rate of change of magnetic flux linkage [1]

(b)(i)
- product of magnetic flux and the number of turns of the coil (or \(N\Phi\) where \(\Phi = BA\)) [1]

(ii)
- recall of \(E = \frac{N A \Delta B}{\Delta t}\) [1]
- substitution: \(1.8 \times 10^{-3} = \frac{150 \times (4.5 \times 10^{-4}) \times B}{0.12}\) [1]
- answer: \(B = 3.2 \times 10^{-3}\text{ T}\) [1]

(c)(i)
- calculation of peak flux linkage amplitude: \(150 \times (4.5 \times 10^{-4}) \times (3.2 \times 10^{-3}) = 2.16 \times 10^{-4}\text{ Wb}\) [1]
- expression: \(\Phi N = 2.16 \times 10^{-4} \sin(100\pi t)\) (accept ecf from (b)(ii)) [1]

(ii)
- differentiates expression or uses \(E_0 = \omega N A B_0\) [1]
- answer: \(0.068\text{ V}\) (or \(6.8 \times 10^{-2}\text{ V}\) or \(68\text{ mV}\)) [1]

### 1. Diagram of Experimental Setup
Draw a labeled diagram showing:
- A uniform metal cantilever firmly clamped between two wooden blocks using a G-clamp, which is secured to a heavy bench or stand.
- A small cylindrical magnet attached to the free end of the cantilever.
- An electromagnet (or coil) positioned close to, but not touching, the cylindrical magnet. The electromagnet is connected to a variable AC signal generator.
- A laser pointer directed at the end of the cantilever, with a screen or photo-detector behind it to clearly observe the amplitude of oscillations, or a travel microscope to monitor amplitude.

### 2. Experimental Procedures
- Secure the metal strip so that a known length \(L\) protrudes from the clamp. Measure \(L\) using a meter rule from the clamp edge to the center of the magnet.
- Attach the magnet of mass \(m\) to the free end.
- Turn on the AC signal generator to power the electromagnet, creating an alternating magnetic field that exerts an oscillating force on the magnet, driving the cantilever.
- Gradually vary the frequency of the signal generator while watching the amplitude of oscillation. Identify the frequency at which the amplitude of oscillation is maximum. This is the resonant frequency \(f\).
- Record \(f\) directly from the signal generator's digital display or use a calibrated oscilloscope connected across the signal generator to measure the period \(T\) of the driving signal and calculate \(f = 1/T\).
- Repeat the procedure for at least six different lengths \(L\).

### 3. Control of Variables
- Keep the mass \(m\) of the magnet constant throughout the experiment.
- Use the same metal strip throughout (or strips of identical width \(w\) and thickness \(t\)) to ensure \(w\) and \(t\) remain constant.

### 4. Data Analysis
- The relationship can be rearranged as:
\[ f^2 = \left( \frac{3 E I}{4\pi^2 m} \right) \frac{1}{L^3} \]
- Substituting \(I = \frac{wt^3}{12}\):
\[ f^2 = \left( \frac{E w t^3}{16\pi^2 m} \right) \frac{1}{L^3} \]
- Plot a graph of \(f^2\) (y-axis) against \(1/L^3\) (x-axis).
- If the relationship is valid, the graph should be a straight line passing through the origin.
- Determine the gradient \(M\) of the straight line.
- Calculate the Young modulus \(E\) using:
\[ E = \frac{16\pi^2 m M}{w t^3} \]

### 5. Safety Precautions
- Wear safety goggles to protect eyes in case the cantilever snaps or the magnet becomes detached during high-amplitude resonant oscillations.
- Place a soft cushion or tray below the cantilever to safely catch the magnet if it falls.

### 6. Additional Details for Accuracy and Reliability
- Use a micrometer screw gauge to measure the thickness \(t\) of the strip at several points along its length and average the results.
- Use vernier calipers to measure the width \(w\) of the strip at several points along its length and average the results.
- Use an electronic balance to measure the mass \(m\) of the magnet before attaching it.
- To locate the peak amplitude precisely, approach the resonance frequency from both above and below and take the average value of the frequency.
- Ensure the G-clamp is tightened extremely firmly to prevent energy losses at the support, which would damp the oscillations and shift the resonant frequency.
- Keep the distance between the electromagnet and the cantilever's magnet constant during setup so that the driving force amplitude is consistent.

**Defining the Problem (2 marks):**
- [1] identify \(L\) as the independent variable and \(f\) as the dependent variable.
- [1] State that mass \(m\), width \(w\), and thickness \(t\) are kept constant.

**Methods of Data Collection (4 marks):**
- [1] Labeled diagram showing cantilever firmly clamped between blocks and a driving system (electromagnet near the end or vibration generator).
- [1] Method to drive cantilever: electromagnet driven by AC signal generator to interact with magnet, or direct mechanical drive from a vibration generator.
- [1] Method to measure \(L\): use a meter rule from clamp edge to center of magnet.
- [1] Method to determine resonant frequency \(f\): adjust frequency until amplitude of oscillation is maximum; record frequency from signal generator dial, digital counter, or oscilloscope.

**Analysis of Data (3 marks):**
- [1] Plot a graph of \(f^2\) against \(1/L^3\) (or \(\log f\) against \(\log L\)).
- [1] State that a straight line passing through the origin confirms the relationship (or a straight line with gradient \(-1.5\) for log-log plot).
- [1] Correct expression for \(E\) from the gradient \(M\): \(E = \frac{16\pi^2 m M}{w t^3}\) (or correct expression if log-log graph is used).

**Additional Details and Safety (6 marks):**
- [1] Safety precaution: wear goggles to protect eyes from snapping metal or thrown magnet.
- [1] Use micrometer screw gauge to measure thickness \(t\) at several positions and calculate average.
- [1] Use vernier calipers to measure width \(w\) at several positions and calculate average.
- [1] Use an electronic balance to measure the mass \(m\) of the magnet.
- [1] Rigid mounting / clamping to a heavy table to prevent energy loss to supports.
- [1] Approach resonant frequency from above and below to locate maximum amplitude accurately.
- [1] Mark the magnet's position on the strip to ensure the same position is maintained for all lengths.

### Part (a)
Taking natural logarithms of both sides:
\[ \ln C = \ln C_0 - \mu x \]
This equation is of the form \(y = mx + c\), where:
- \(y = \ln C\)
- \(x = x\)
- \(m = -\mu\) (gradient)
- \(c = \ln C_0\) (y-intercept)
A graph of \(\ln C\) against \(x\) is suitable because it yields a straight line with a negative gradient of constant value \(-\mu\) if the relationship holds.

### Part (b)
We calculate \(\ln C\) and the absolute uncertainty \(\Delta(\ln C)\) using \(\Delta(\ln C) \approx \frac{\Delta C}{C}\) or \(\frac{1}{2} [\ln(C + \Delta C) - \ln(C - \Delta C)]\):

1. For \(C = 370 \pm 15\):
\(\ln(370) = 5.91\)
\(\Delta(\ln C) = \frac{15}{370} = 0.04\)
2. For \(C = 275 \pm 12\):
\(\ln(275) = 5.62\)
\(\Delta(\ln C) = \frac{12}{275} = 0.04\)
3. For \(C = 205 \pm 10\):
\(\ln(205) = 5.32\)
\(\Delta(\ln C) = \frac{10}{205} = 0.05\)
4. For \(C = 150 \pm 8\):
\(\ln(150) = 5.01\)
\(\Delta(\ln C) = \frac{8}{150} = 0.05\)
5. For \(C = 112 \pm 6\):
\(\ln(112) = 4.72\)
\(\Delta(\ln C) = \frac{6}{112} = 0.05\)
6. For \(C = 82 \pm 5\):
\(\ln(82) = 4.41\)
\(\Delta(\ln C) = \frac{5}{82} = 0.06\)

Completed column values:
- \(\ln C\): \(5.91\), \(5.62\), \(5.32\), \(5.01\), \(4.72\), \(4.41\)
- \(\Delta(\ln C)\): \(0.04\), \(0.04\), \(0.05\), \(0.05\), \(0.05\), \(0.06\)

### Part (c) & (d)
- Plotting points with y-error bars of length \(\pm \Delta(\ln C)\) and x-error bars of length \(\pm 0.1\text{ mm}\).
- The line of best fit goes through the center of the points.
- The worst acceptable line is the steepest or shallowest possible line that passes through all error bars.

### Part (e)
- Gradient of line of best fit:
Using two distant points on the line, e.g., \((2.0, 5.91)\) and \((12.0, 4.41)\):
\[ \text{gradient} = \frac{4.41 - 5.91}{12.0 - 2.0} = -0.150 \text{ mm}^{-1} \]
- Steepest worst acceptable line:
Using \((1.9, 5.95)\) and \((12.1, 4.35)\):
\[ \text{gradient}_{\text{worst}} = \frac{4.35 - 5.95}{12.1 - 1.9} = -0.157 \text{ mm}^{-1} \]
- Uncertainty in gradient:
\[ \Delta m = |-0.150 - (-0.157)| = 0.007 \text{ mm}^{-1} \]
So, \(\text{gradient} = -0.150 \pm 0.007 \text{ mm}^{-1}\).

### Part (f)
- Linear absorption coefficient:
\(\mu = -\text{gradient} = 0.150 \pm 0.007 \text{ mm}^{-1}\)
- Y-intercept:
\[ c = y - mx = 5.91 - (-0.150)(2.0) = 6.21 \]
Uncertainty in intercept \(\Delta c = c_{\text{worst}} - c \approx 0.04\).
So \(c = 6.21 \pm 0.04\).
- Calculating \(C_0\):
\[ C_0 = e^{c} = e^{6.21} = 498 \text{ s}^{-1} \approx 500 \text{ s}^{-1} \]
\[ \Delta C_0 = C_0 \times \Delta c = 498 \times 0.04 \approx 20 \text{ s}^{-1} \]
So, \(C_0 = 500 \pm 20 \text{ s}^{-1}\).

**Part (a) [2 marks]**
- [1] State \(\ln C = -\mu x + \ln C_0\).
- [1] Explain that comparing with \(y = mx+c\), the graph is a straight line where gradient \(= -\mu\) is constant.

**Part (b) [3 marks]**
- [1] All \(\ln(C / \text{s}^{-1})\) values calculated correctly (5.91, 5.62, 5.32, 5.01, 4.72, 4.41).
- [1] All absolute uncertainties in \(\ln C\) calculated correctly (0.04, 0.04, 0.05, 0.05, 0.05, 0.06).
- [1] All values recorded consistently to 2 decimal places.

**Part (c) [3 marks]**
- [1] Plotted points must be within half a small square of the correct coordinates.
- [1] Correct vertical error bars representing \(\pm \Delta(\ln C)\).
- [1] Correct horizontal error bars representing \(\pm 0.1\text{ mm}\).

**Part (d) [2 marks]**
- [1] Line of best fit drawn cleanly through all error bars.
- [1] Worst acceptable line (steepest or shallowest) drawn passing through all error boxes, clearly labeled.

**Part (e) [3 marks]**
- [1] Gradient of best-fit line calculated with a large triangle (hypotenuse > 50% of the drawn line).
- [1] Gradient of worst acceptable line calculated.
- [1] Uncertainty in gradient calculated as difference in gradients, with correct negative or positive sign and appropriate units (\(\text{mm}^{-1}\)). Accept values in range \(0.140\text{ to }0.160\text{ mm}^{-1}\) for magnitude.

**Part (f) [2 marks]**
- [1] Value of \(\mu\) stated correctly with uncertainty (\(0.150 \pm 0.007 \text{ mm}^{-1}\)).
- [1] Value of \(C_0\) calculated correctly from \(e^{\text{intercept}}\), in range \(480 - 520 \text{ s}^{-1}\), with correct absolute uncertainty (\(\pm 20 \text{ s}^{-1}\)) and unit (\(\text{s}^{-1}\)).