2023 Cambridge IAS-Level Physics (9702) 模擬試題連答案詳解

The work done during loading represents the total energy stored in the string, which is \(0.075\text{ J}\). The work done by the string during unloading represents the elastic potential energy recovered, which is \(0.045\text{ J}\). The difference between these two values is the energy dissipated as heat (thermal energy) due to plastic deformation: \(\text{Energy dissipated} = 0.075\text{ J} - 0.045\text{ J} = 0.030\text{ J}\). The permanent extension is the extension remaining when the force is reduced to zero, which is explicitly given as \(2.0\text{ mm}\).

1 mark for the correct calculation of thermal energy dissipated (0.030 J) and identifying the permanent extension as 2.0 mm.

The total energy in simple harmonic motion is given by \(E = E_k + E_p = \frac{1}{2} m \omega^2 x_0^2\). The potential energy at displacement \(x\) is \(E_p = \frac{1}{2} m \omega^2 x^2\), and the kinetic energy is \(E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2)\). Since we are given that \(E_k = 3 E_p\), we can substitute these expressions: \(\frac{1}{2} m \omega^2 (x_0^2 - x^2) = 3 \left(\frac{1}{2} m \omega^2 x^2\right) \implies x_0^2 - x^2 = 3 x^2 \implies x_0^2 = 4 x^2 \implies x^2 = 0.25 x_0^2 \implies x = \pm 0.50 x_0\).

1 mark for correctly setting up the ratio of kinetic to potential energy in terms of amplitude and displacement, and solving for displacement x.

Using the potential divider formula, the output voltage in dark conditions is \(V_{\text{out, dark}} = E \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}} = 12 \times \frac{16\text{ k}\Omega}{4.0\text{ k}\Omega + 16\text{ k}\Omega} = 12 \times 0.8 = 9.6\text{ V}\). In bright light, the output voltage is \(V_{\text{out, bright}} = 12 \times \frac{1.0\text{ k}\Omega}{4.0\text{ k}\Omega + 1.0\text{ k}\Omega} = 12 \times 0.2 = 2.4\text{ V}\). The change in output voltage is the difference between these two states: \(\Delta V_{\text{out}} = 9.6\text{ V} - 2.4\text{ V} = 7.2\text{ V}\).

1 mark for correctly calculating the potential divider outputs for both conditions and finding the correct difference of 7.2 V.

The double-slit fringe spacing is given by \(x = \frac{\lambda D}{a}\). When the wavelength increases by \(20\%\), \(\lambda' = 1.2\lambda\). When the slit separation is halved, \(a' = 0.5a\). When the screen distance is doubled, \(D' = 2D\). Substituting these values into the formula: \(x' = \frac{\lambda' D'}{a'} = \frac{1.2\lambda \times 2D}{0.5a} = \frac{2.4}{0.5} \frac{\lambda D}{a} = 4.8 x\).

1 mark for correctly applying the double-slit fringe width equation to scale the parameters and determine the new fringe width.

According to the principle of conservation of linear momentum, the total momentum before the explosion equals the total momentum after. Taking the positive x-direction as positive: \(p_{\text{initial}} = (3m)v = 3mv\). After the explosion: \(p_{\text{final}} = m(-v) + (2m)v_2 = -mv + 2mv_2\). Setting these equal: \(3mv = -mv + 2mv_2 \implies 4mv = 2mv_2 \implies v_2 = 2v\). Since \(v_2\) is positive, the fragment moves in the positive x-direction with a speed of \(2v\).

1 mark for correctly applying the conservation of linear momentum, accounting for the direction of velocity, and solving for the unknown velocity.

For a floating cylinder in equilibrium, the upward buoyant force (upthrust) must equal the downward gravitational force (weight). Upthrust is the sum of the buoyant forces from the oil and the water: \(U = U_{\text{oil}} + U_{\text{water}} = \rho_{\text{oil}} V_{\text{sub, oil}} g + \rho_{\text{water}} V_{\text{sub, water}} g\). Since the cylinder is uniform, the submerged volumes are \(V_{\text{sub, oil}} = A \left(\frac{1}{3}h\right)\) and \(V_{\text{sub, water}} = A \left(\frac{2}{3}h\right)\). The weight of the cylinder is \(W = \rho_{\text{cyl}} A h g\). Setting \(W = U\): \(\rho_{\text{cyl}} A h g = 800 \times A \left(\frac{1}{3}h\right) g + 1000 \times A \left(\frac{2}{3}h\right) g\). Dividing both sides by \(A h g\): \(\rho_{\text{cyl}} = \frac{1}{3}(800) + \frac{2}{3}(1000) = 266.7 + 666.7 = 933.3\text{ kg m}^{-3}\), which rounds to \(933\text{ kg m}^{-3}\).

1 mark for correctly setting up the buoyant force balance equation and solving for the density of the floating object.

The extension of a wire under load is given by \(\Delta L = \frac{F L}{A E}\), where \(A = \pi \frac{d^2}{4}\). Thus, \(\Delta L \propto \frac{L}{d^2}\) for wires of the same material under the same force. For wire \(X\): \(\Delta L_X \propto \frac{L}{d^2}\). For wire \(Y\): \(\Delta L_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = 0.5 \frac{L}{d^2}\). Taking the ratio: \(\frac{\Delta L_X}{\Delta L_Y} = \frac{1}{0.5} = 2.0\).

1 mark for using the Young modulus formula to relate extension to length and diameter, and evaluating the ratio of the extensions correctly.

The relationship between the terminal potential difference \(V\) and the current \(I\) for a cell is given by \(V = E - Ir\), which can be rewritten as \(V = -rI + E\). Comparing this to the equation of a straight line, \(y = mx + c\), where \(y = V\) and \(x = I\), the vertical intercept \(c\) is equal to the e.m.f. \(E\), and the gradient \(m\) is equal to \(-r\).

1 mark for correctly expressing the relationship between terminal potential difference, e.m.f., and internal resistance, and identifying the vertical intercept and gradient.

The strain energy \(E\) stored in a wire within its elastic limit is given by:
\(E = \frac{1}{2} F x\)
where extension \(x\) is given by \(x = \frac{FL}{AY}\) (with \(Y\) being the Young modulus).
Substituting \(x\) gives:
\(E = \frac{F^2 L}{2AY}\)

For the second wire, the parameters are length \(2L\), area \(3A\), and force \(2F\). The stored energy \(E_2\) is:
\(E_2 = \frac{(2F)^2 (2L)}{2 (3A) Y} = \frac{8 F^2 L}{6 AY} = \frac{4}{3} \frac{F^2 L}{AY}\)

Since \(\frac{F^2 L}{2AY} = E\), then \(\frac{F^2 L}{AY} = 2E\). Substituting this into the equation for \(E_2\):
\(E_2 = \frac{4}{3} (2E) = \frac{8}{3} E\)

1 mark for identifying the correct relationship between strain energy, force, and dimensions, and calculating the ratio of the energy in the second wire to the first wire as 8/3.

In simple harmonic motion, the total energy \(E_{\text{total}}\) of the system is constant and is equal to the sum of the kinetic energy \(E_k\) and potential energy \(E_p\):
\(E_{\text{total}} = E_k + E_p = \frac{1}{2} k x_0^2\)

We are given that \(E_k = 3 E_p\). Substituting this in:
\(3 E_p + E_p = E_{\text{total}} \implies 4 E_p = E_{\text{total}}\)

The potential energy at displacement \(x\) is given by \(E_p = \frac{1}{2} k x^2\). Therefore:
\(4 \left(\frac{1}{2} k x^2\right) = \frac{1}{2} k x_0^2\)
\(4 x^2 = x_0^2\)
\(x = \pm 0.50 x_0\)

1 mark for setting up the energy conservation equation with given kinetic/potential ratio and solving for the displacement.

The potential difference across the variable resistor is equal to the terminal potential difference of the battery, which is given by:
\(V_R = V - I r\)

Comparing this with the equation of a straight line, \(y = mx + c\), where \(y = V_R\) and \(x = I\):
- The gradient \(m\) is equal to \(-r\).
- The vertical intercept \(c\) is equal to \(V\).

Therefore, the graph is a straight line with a negative gradient equal to \(-r\) and a vertical intercept equal to \(V\).

1 mark for using the terminal potential difference equation to deduce that the graph is a straight line with negative gradient -r and intercept V.

The formula for the double-slit fringe spacing is:
\(w = \frac{\lambda D}{a}\)

Since the slit separation \(a\) is unchanged, the fringe spacing \(w\) is directly proportional to the product \(\lambda D\).
Therefore, the ratio of the new fringe spacing \(w_2\) to the original fringe spacing \(w_1\) is:
\(\frac{w_2}{w_1} = \frac{\lambda_2 D_2}{\lambda_1 D_1}\)

Substitute the given values:
\(\frac{w_2}{w_1} = \frac{400\text{ nm} \times 2.4\text{ m}}{600\text{ nm} \times 1.5\text{ m}} = \frac{960}{900} \approx 1.07\)

1 mark for applying the double-slit equation and calculating the ratio of the new to the original fringe spacing.

According to Newton's second law, average force is equal to the rate of change of momentum:
\(F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}\)

Let the initial direction of travel be positive.
Initial velocity, \(u = +24\text{ m s}^{-1}\)
Final velocity, \(v = -16\text{ m s}^{-1}\) (since it rebounds in the opposite direction)

Change in momentum \(\Delta p\) is:
\(\Delta p = 0.15\text{ kg} \times (-16\text{ m s}^{-1} - 24\text{ m s}^{-1}) = 0.15 \times (-40) = -6.0\text{ kg m s}^{-1}\)

The magnitude of the average force is:
\(F = \frac{6.0\text{ kg m s}^{-1}}{0.050\text{ s}} = 120\text{ N}\)

1 mark for using the correct sign convention to calculate change in momentum, and dividing by time to obtain 120 N.

The difference between the hydrostatic forces on the bottom and top faces of the cylinder is equal to the upthrust \(U\) acting on the cylinder:
\(\Delta F = F_{\text{bottom}} - F_{\text{top}} = U = \rho g V\)

where:
- \(\rho = 1200\text{ kg m}^{-3}\)
- \(g = 9.81\text{ m s}^{-2}\)
- \(V = A \times h = 2.5 \times 10^{-3}\text{ m}^2 \times 0.80\text{ m} = 2.0 \times 10^{-3}\text{ m}^3\)

Substitute these values:
\(\Delta F = 1200 \times 9.81 \times 2.0 \times 10^{-3} = 2.4 \times 9.81 = 23.54\text{ N} \approx 24\text{ N}\)

1 mark for recognizing that the difference in force equals upthrust, calculating the volume of the cylinder, and correctly applying Archimedes' principle to find 24 N.

- Option A is incorrect because stretching past the elastic limit causes permanent plastic deformation, meaning the wire will not return to its original length (origin).
- Option B is incorrect because the area under the unloading curve represents the elastic strain energy recovered.
- Option C is correct because the area of the hysteresis loop (the area between the loading and unloading curves) represents the net work done on the wire that cannot be recovered and is dissipated as thermal energy.
- Option D is incorrect because the gradient of the load-extension graph is \(\frac{F}{x} = \frac{AY}{L}\), which is not equal to the Young modulus \(Y\) alone.

1 mark for identifying that the area between the loading and unloading curves represents the energy dissipated as thermal energy.

The general equation for the displacement in simple harmonic motion is:
\(x = x_0 \cos(\omega t)\)

By comparing this with the given equation \(x = 0.080 \cos(5.0 t)\):
- The amplitude \(x_0 = 0.080\text{ m}\)
- The angular frequency \(\omega = 5.0\text{ rad s}^{-1}\)

The maximum speed \(v_{\text{max}}\) is:
\(v_{\text{max}} = \omega x_0 = 5.0\text{ rad s}^{-1} \times 0.080\text{ m} = 0.40\text{ m s}^{-1}\)

1 mark for identifying the amplitude and angular frequency from the equation and using them to calculate the maximum speed.

The Young modulus \(E\) is defined as \(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}\). Rearranging this gives the force applied to the wire: \(F = \frac{EAx}{L}\). The elastic potential energy \(E_p\) stored in a wire obeying Hooke's law is given by \(E_p = \frac{1}{2}Fx\). Substituting the expression for \(F\) yields: \(E_p = \frac{1}{2} \left(\frac{EAx}{L}\right) x = \frac{EAx^2}{2L}\).

1 mark for the correct option B. Award 1 mark for correctly substituting the force from the Young modulus formula into the elastic potential energy equation.

The total energy of the simple harmonic oscillator is the sum of its kinetic energy \(E_k\) and potential energy \(E_p\): \(E_{\text{total}} = E_k + E_p\). We are given \(E_k = 3E_p\). Substituting this into the energy conservation equation gives \(E_{\text{total}} = 3E_p + E_p = 4E_p\). The potential energy at displacement \(x\) is \(E_p = \frac{1}{2}m\omega^2 x^2\), and the total energy is \(E_{\text{total}} = \frac{1}{2}m\omega^2 x_0^2\). Therefore, \(\frac{1}{2}m\omega^2 x_0^2 = 4 \left(\frac{1}{2}m\omega^2 x^2\right)\), which simplifies to \(x_0^2 = 4x^2\). Taking the square root of both sides gives \(x = \pm 0.50x_0\).

1 mark for the correct option B. Award 1 mark for expressing total energy in terms of potential energy and solving for the displacement.

Initially, at \(20^\circ\text{C}\), the resistance of the thermistor is \(R_T = 2400\ \Omega\). The voltmeter reading across the thermistor is given by the potential divider formula: \(V_{\text{initial}} = E \times \frac{R_T}{R_T + R} = 6.0 \times \frac{2400}{2400 + 1200} = 4.0\text{ V}\). When the temperature increases, the thermistor resistance decreases to \(R_T = 800\ \Omega\). The new voltmeter reading is \(V_{\text{final}} = 6.0 \times \frac{800}{800 + 1200} = 2.4\text{ V}\). The change in the voltmeter reading is \(V_{\text{final}} - V_{\text{initial}} = 2.4\text{ V} - 4.0\text{ V} = -1.6\text{ V}\), which represents a decrease of \(1.6\text{ V}\).

1 mark for the correct option A. Award 1 mark for calculating both initial and final voltmeter readings and finding the difference.

The fringe spacing \(x\) is given by the formula \(x = \frac{\lambda D}{a}\). Under the modified conditions, the new slit separation is \(a' = 2a\) and the new distance to the screen is \(D' = D/2\). We want the new fringe spacing \(x'\) to be equal to \(x\): \(x' = \frac{\lambda' D'}{a'} = \frac{\lambda' (D/2)}{2a} = \frac{\lambda' D}{4a}\). Setting \(x' = x\) gives \(\frac{\lambda' D}{4a} = \frac{\lambda D}{a}\), which simplifies to \(\lambda' = 4\lambda\).

1 mark for the correct option D. Award 1 mark for setting up the double-slit equation and solving for the new wavelength.

The mass of water striking the wall per unit time \(\Delta t\) is given by \(\frac{\Delta m}{\Delta t} = \rho A v\). Since the water does not rebound, its final horizontal velocity is zero, so the change in horizontal velocity is \(\Delta v = v\). The magnitude of the rate of change of momentum of the water is \(F = \frac{\Delta p}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right) \Delta v = (\rho A v) v = \rho A v^2\). According to Newton's third law, the force exerted by the water on the wall is equal in magnitude to the force exerted by the wall on the water, which is \(\rho A v^2\).

1 mark for the correct option C. Award 1 mark for expressing the mass flow rate and calculating the rate of change of momentum.

At the horizontal level of the oil-water interface, the pressure in both limbs of the U-tube must be equal. In the limb containing oil, the pressure at this level is due to the oil column: \(P_{\text{oil}} = \rho_o g h_o\), where \(h_o = 15.0\text{ cm}\). In the other limb, the pressure at this level is due to the water column above it: \(P_{\text{water}} = \rho_w g h_w\), where \(h_w = 12.0\text{ cm}\). Equating the pressures, we get \(\rho_o g h_o = \rho_w g h_w\), which simplifies to \(\rho_o h_o = \rho_w h_w\). Substituting the known values: \(\rho_o \times 15.0\text{ cm} = 1000\text{ kg m}^{-3} \times 12.0\text{ cm}\), giving \(\rho_o = \frac{12000}{15.0} = 800\text{ kg m}^{-3}\).

1 mark for the correct option B. Award 1 mark for equating the hydrostatic pressures at the interface level and solving for oil density.

The total work done on the wire during the loading process is given by the area under the loading force-extension curve, which is \(W_1\). When the wire is unloaded, some elastic strain energy is recovered, which is represented by the area under the unloading curve, \(W_2\). The difference between the work done during loading and the energy recovered during unloading, \(W_1 - W_2\), represents the work done in permanently deforming the wire. This energy cannot be recovered and is dissipated as thermal energy (heat) within the material.

1 mark for the correct option D. Award 1 mark for identifying that the area between the loading and unloading curves represents dissipated thermal energy.

The velocity \(v\) of a particle in simple harmonic motion as a function of displacement \(x\) is given by \(v = \pm \omega \sqrt{x_0^2 - x^2}\). The maximum speed \(v_0\) occurs at the equilibrium position (where \(x = 0\)), so \(v_0 = \omega x_0\). When the displacement is \(x = 0.50 x_0\), the speed of the particle is: \(v = \omega \sqrt{x_0^2 - (0.50 x_0)^2} = \omega \sqrt{x_0^2 - 0.25 x_0^2} = \omega \sqrt{0.75 x_0^2} = \sqrt{0.75} \omega x_0 = \sqrt{0.75} v_0 \approx 0.866 v_0\). Rounding to two decimal places, this is \(0.87 v_0\).

1 mark for the correct option C. Award 1 mark for using the velocity-displacement relation in SHM and expressing the result in terms of maximum speed.

The work done \(W\) in stretching a wire within its limit of proportionality is equal to the elastic potential energy stored, which is given by:

\(W = \frac{1}{2} F x\)

From the definition of the Young modulus \(E\):

\(E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\)

Rearranging this expression for the force \(F\):

\(F = \frac{E A x}{L}\)

Substituting this back into the work done formula:

\(W = \frac{1}{2} \left( \frac{E A x}{L} \right) x = \frac{E A x^2}{2 L}\)

A: Correct. Uses the relation between Young modulus and force, and substitutes into the work done formula.
B: Incorrect. Missing the factor of 1/2.
C: Incorrect. This represents 1/2 of the force, not the work done.
D: Incorrect. Incorrect power of the area.

The total energy of the oscillating system is constant and is equal to the maximum kinetic energy:

\(E_{\text{total}} = E_{k} + E_{p} = \frac{1}{2} m \omega^2 x_0^2\)

We are looking for the time when the kinetic energy \(E_k\) is three-quarters of the maximum kinetic energy:

\(E_k = \frac{3}{4} E_{\text{total}}\)

This means the potential energy \(E_p\) must be one-quarter of the total energy:

\(E_p = \frac{1}{4} E_{\text{total}}\)

Since \(E_p = \frac{1}{2} m \omega^2 x^2\), we can write:

\(\frac{1}{2} m \omega^2 x^2 = \frac{1}{4} \left( \frac{1}{2} m \omega^2 x_0^2 \right)\)

\(x^2 = \frac{1}{4} x_0^2 \implies x = \pm \frac{1}{2} x_0\)

Starting from \(t = 0\), the displacement is \(x = x_0\). The first time it reaches \(x = \frac{1}{2} x_0\) is when:

\(x_0 \cos(\omega t) = \frac{1}{2} x_0 \implies \cos(\omega t) = \frac{1}{2}\)

This first occurs when \(\omega t = \frac{\pi}{3}\).

Since \(\omega = \frac{2\pi}{T}\), we have:

\(\left(\frac{2\pi}{T}\right) t = \frac{\pi}{3} \implies t = \frac{T}{6}\)

A: Incorrect. This corresponds to when potential energy is 3/4 of the total energy.
B: Incorrect. This is a common miscalculation if assuming a linear relationship.
C: Correct. Derived from the energy relations and cosine of the angle.
D: Incorrect. This is when displacement is zero (maximum kinetic energy).

1. As the temperature of the NTC thermistor increases, its resistance \(R_T\) decreases.
2. The total resistance of the series circuit, \(R_{\text{total}} = R + R_T + r\), therefore decreases.
3. Since the e.m.f. \(V_s\) is constant, the circuit current \(I = \frac{V_s}{R_{\text{total}}}\) increases.
4. The potential difference across the internal resistance is \(V_r = I r\). Since \(I\) increases and \(r\) is constant, \(V_r\) must increase.
5. The potential difference measured by the voltmeter across the thermistor is \(V_T = V_s - I(R + r)\). Since \(I\) increases, the term \(I(R + r)\) increases, and therefore \(V_T\) must decrease.

A: Incorrect. The increase in current means the internal p.d. must increase.
B: Correct. Decreased resistance of the thermistor reduces its share of the potential, and the increased overall current increases the p.d. across the internal resistance.
C: Incorrect. This would be the case if the temperature decreased.
D: Incorrect. The thermistor's share of the voltage decreases due to its lower resistance relative to the other series components.

The relationship for the fringe spacing \(x\) in double-slit interference is given by:

\(x = \frac{\lambda D}{a}\)

For the modified experiment, we substitute the new values into the formula:

\(x' = \frac{\lambda' D'}{a'}\)

Given:
- \(\lambda' = 2\lambda\)
- \(D' = 2D\)
- \(a' = \frac{a}{2}\)

Substituting these values:

\(x' = \frac{(2\lambda) (2D)}{\frac{a}{2}} = \frac{4 \lambda D}{\frac{a}{2}} = 8 \left( \frac{\lambda D}{a} \right) = 8x\)

A: Incorrect. This would be the result if only wavelength or distance were changed.
B: Incorrect. This ignores the halving of the slit separation in the denominator.
C: Correct. Incorporating all factor changes leads to a factor of 8.
D: Incorrect. This is a result of squaring some factor incorrectly.

First, we apply the conservation of linear momentum to find the common velocity \(v\) after the collision:

\(p_{\text{initial}} = p_{\text{final}}\)

\(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\)

\((2.0\text{ kg} \times 6.0\text{ m s}^{-1}) + (4.0\text{ kg} \times 0) = (2.0\text{ kg} + 4.0\text{ kg}) v\)

\(12\text{ kg m s}^{-1} = 6.0\text{ kg} \times v \implies v = 2.0\text{ m s}^{-1}\)

Next, calculate the initial kinetic energy \(E_{ki}\):

\(E_{ki} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} (2.0) (6.0)^2 = 36\text{ J}\)

Now, calculate the final kinetic energy \(E_{kf}\):

\(E_{kf} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} (6.0) (2.0)^2 = 12\text{ J}\)

Finally, find the loss in kinetic energy:

\(\Delta E_k = E_{ki} - E_{kf} = 36\text{ J} - 12\text{ J} = 24\text{ J}\)

A: Incorrect. This is the final kinetic energy of the system.
B: Incorrect. This results from a calculation error in momentum or velocities.
C: Correct. Calculated by subtracting the final kinetic energy (12 J) from the initial kinetic energy (36 J).
D: Incorrect. This is the total initial kinetic energy of the system.

The total pressure \(P\) at the bottom of the beaker is the sum of the atmospheric pressure and the hydrostatic pressures of both liquids:

\(P = P_{\text{atm}} + P_X + P_Y\)

Where:
- \(P_X = \rho_X g h_X = 800\text{ kg m}^{-3} \times 9.81\text{ m s}^{-2} \times 0.15\text{ m} = 1177.2\text{ Pa}\)
- \(P_Y = \rho_Y g h_Y = 1200\text{ kg m}^{-3} \times 9.81\text{ m s}^{-2} \times 0.25\text{ m} = 2943.0\text{ Pa}\)

Sum of hydrostatic pressures from the liquids:

\(P_{\text{liquids}} = 1177.2\text{ Pa} + 2943.0\text{ Pa} = 4120.2\text{ Pa} \approx 4.1 \times 10^3\text{ Pa}\)

Total pressure:

\(P = 1.01 \times 10^5\text{ Pa} + 4120.2\text{ Pa} = 101000\text{ Pa} + 4120.2\text{ Pa} = 105120.2\text{ Pa} \approx 1.05 \times 10^5\text{ Pa}\)

A: Incorrect. This is only the hydrostatic pressure of the liquids, neglecting atmospheric pressure.
B: Incorrect. This is just the atmospheric pressure, neglecting the liquids.
C: Correct. The total pressure correctly adds atmospheric pressure and both hydrostatic pressures.
D: Incorrect. This is the result if the densities are reversed or added incorrectly.

By definition, the elastic limit (Point Q) is the maximum stress or strain that a material can withstand without permanent deformation. Beyond this point, any further stress leads to plastic deformation, meaning the material will not return to its original shape when the load is removed.

Point P is the limit of proportionality, which is the end of the linear region, but the material can still behave elastically between P and Q (returning to its original shape, even if the graph is not strictly linear). Point R is the ultimate tensile strength, and Point S is the breaking point.

A: Incorrect. Point P is the limit of proportionality, not the elastic limit.
B: Correct. Point Q is the definition of the elastic limit.
C: Incorrect. Point R is the ultimate tensile strength.
D: Incorrect. Point S is the breaking/fracture point.

The acceleration \(a\) of an object in simple harmonic motion is related to its displacement \(x\) by the equation:

\(a = -\omega^2 x\)

where \(\omega\) is the angular frequency. The magnitude of the acceleration is:

\(|a| = \omega^2 |x|\)

The angular frequency \(\omega\) is calculated from the frequency \(f\):

\(\omega = 2\pi f = 2\pi \times 2.0\text{ Hz} = 4\pi\text{ rad s}^{-1}\)

Converting the displacement to meters:

\(|x| = 5.0\text{ cm} = 0.050\text{ m}\)

Now, substitute these values into the magnitude of acceleration formula:

\(|a| = (4\pi)^2 \times 0.050 = 16\pi^2 \times 0.050 = 0.80\pi^2 \approx 7.9\text{ m s}^{-2}\)

A: Incorrect. This is the product of \(\omega\) and \(x\) (the maximum velocity instead of acceleration magnitude, if amplitude was 5 cm).
B: Incorrect. This corresponds to omitting \(\pi\) or other terms from the calculation.
C: Correct. Applying \(|a| = \omega^2 x\) with proper conversions yields \(7.9\text{ m s}^{-2}\).
D: Incorrect. This uses the frequency directly without multiplying by \(2\pi\) or similar arithmetic error.

The extension of a wire is given by \(x = \frac{FL}{AE}\), where \(A = \frac{\pi d^2}{4}\). This means \(x = \frac{4FL}{\pi d^2 E}\). The elastic strain energy is \(U = \frac{1}{2} F x = \frac{2 F^2 L}{\pi d^2 E}\). Since \(F\) and \(E\) are the same for both wires, the strain energy is proportional to \(\frac{L}{d^2}\). For wire X: \(U_X \propto \frac{L}{d^2}\). For wire Y: \(U_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\). Therefore, \(\frac{U_X}{U_Y} = \frac{L/d^2}{L/(2d^2)} = 2\).

C1: Recall of extension and strain energy equations.
A1: Correct determination of the ratio as 2.0 (Option D).

The work done during loading is the area under the loading force-extension graph: \(W_{\text{loading}} = \frac{1}{2} \times 120\text{ N} \times 4.0 \times 10^{-3}\text{ m} = 0.24\text{ J}\). The energy recovered during unloading is the area under the unloading graph, which is linear from 120 N at 4.0 mm to 0 N at 1.0 mm extension: \(W_{\text{unloading}} = \frac{1}{2} \times 120\text{ N} \times (4.0 - 1.0) \times 10^{-3}\text{ m} = 0.18\text{ J}\). The work done in plastic deformation is the difference between these two areas: \(\Delta W = 0.24\text{ J} - 0.18\text{ J} = 0.06\text{ J}\).

C1: Correct calculation of loading and unloading work done.
A1: Subtracting unloading energy from loading energy to obtain 0.06 J (Option A).

Using the displacement equation \(x(t) = A \sin(\omega t)\), the time \(t_1\) at which the displacement is \(+\frac{1}{2}A\) satisfies \(\sin(\omega t_1) = 0.5\), so \(\omega t_1 = \frac{\pi}{6}\). The time \(t_2\) at which the displacement is \(-\frac{1}{2}A\) satisfies \(\sin(\omega t_2) = -0.5\), so \(\omega t_2 = -\frac{\pi}{6}\). The phase difference is \(\Delta \theta = \frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{\pi}{3}\). Since a phase of \(2\pi\) corresponds to a time period \(T\), the minimum time interval is \(\Delta t = \frac{\pi/3}{2\pi} T = \frac{T}{6}\).

C1: Determination of the phase angle difference between +A/2 and -A/2.
A1: Calculation of the minimum time interval as T/6 (Option B).

The total energy is \(E_T = \frac{1}{2} m \omega^2 A^2\) and the potential energy is \(E_p = \frac{1}{2} m \omega^2 x^2\). The kinetic energy is \(E_k = E_T - E_p = \frac{1}{2} m \omega^2 (A^2 - x^2)\). We are given \(E_k = 3 E_p\), which means: \(\frac{1}{2} m \omega^2 (A^2 - x^2) = 3 \left(\frac{1}{2} m \omega^2 x^2\right) \implies A^2 - x^2 = 3x^2 \implies A^2 = 4x^2 \implies x = \pm \frac{1}{2}A\).

C1: Setting up the equation relating potential and kinetic energy in terms of displacement.
A1: Solving for displacement to obtain x = +/- 0.5A (Option B).

The potential difference across the fixed resistor is given by the potential divider formula: \(V_{\text{out}} = V_{\text{supply}} \times \frac{R_{\text{fixed}}}{R_{\text{fixed}} + R_{\text{thermistor}}}\). Substituting the known values: \(4.0 = 12 \times \frac{2.0}{2.0 + R_{\text{thermistor}}} \implies \frac{4.0}{12.0} = \frac{1}{3} = \frac{2.0}{2.0 + R_{\text{thermistor}}} \implies 2.0 + R_{\text{thermistor}} = 6.0 \implies R_{\text{thermistor}} = 4.0\text{ k}\Omega\).

C1: Correct application of the potential divider formula.
A1: Calculation of the thermistor resistance to be 4.0 kOhm (Option C).

The fringe separation is given by the formula \(x = \frac{\lambda D}{d}\). For the modified setup: \(x_{\text{new}} = \frac{(1.2\lambda) D}{0.5d} = \frac{1.2}{0.5} \frac{\lambda D}{d} = 2.4 x\).

C1: Recall and manipulation of the double-slit fringe spacing formula.
A1: Evaluating the numerical factor to find the new fringe separation as 2.4x (Option C).

Let the initial direction of the ball of mass \(m\) be positive. By conservation of linear momentum: \(m u = m v_1 + 2m v_2 \implies u = v_1 + 2v_2\). For a perfectly elastic collision, the relative speed of approach equals the relative speed of separation: \(u - 0 = v_2 - v_1 \implies v_2 = v_1 + u\). Substituting this into the momentum equation: \(u = v_1 + 2(v_1 + u) \implies u = 3v_1 + 2u \implies 3v_1 = -u \implies v_1 = -\frac{1}{3}u\). The negative sign indicates that the velocity is in the opposite direction to its initial velocity.

C1: Simultaneous use of conservation of momentum and relative speed of approach/separation equations.
A1: Solving for v_1 to obtain a speed of 1/3 u in the opposite direction (Option A).

The pressure difference is the sum of the hydrostatic pressures exerted by each of the two liquid layers: \(\Delta P = \Delta P_{\text{upper}} + \Delta P_{\text{lower}} = \rho g \left(\frac{1}{3}h\right) + (3\rho) g \left(\frac{2}{3}h\right)\). Simplifying this expression: \(\Delta P = \frac{1}{3}\rho g h + 2\rho g h = \frac{7}{3}\rho g h\).

C1: Calculation of individual pressure contributions from each liquid layer.
A1: Addition of pressure components to yield 7/3 rho g h (Option D).

(a) Tensile strain is defined as the change in length (extension) per unit original length: \(\text{strain} = e / L\). Since it is the ratio of two lengths, it is dimensionless and has no unit.

(b)(i) The force applied by the hanging mass is \(F = m g = 15.0 \times 9.81 = 147.15\text{ N}\).
Using the Young modulus formula:
\(E = \frac{\text{stress}}{\text{strain}} = \frac{F L}{A e}\)
\(e = \frac{F L}{A E} = \frac{147.15 \times 2.50}{1.20 \times 10^{-6} \times 2.00 \times 10^{11}} = 1.533 \times 10^{-3}\text{ m} = 1.53\text{ mm}\).

(b)(ii) The elastic strain energy \(W\) stored is given by:
\(W = \frac{1}{2} F e = 0.5 \times 147.15 \times 1.533 \times 10^{-3} = 0.113\text{ J}\).

(a) Correct definition as extension/original length [1.0 mark]; state that it has no unit [0.5 mark].
(b)(i) Calculation of tension \(F = 147\text{ N}\) [1.0 mark]; correct rearrangement and substitution of Young modulus [1.0 mark]; final answer \(1.53\text{ mm}\) (or \(1.53 \times 10^{-3}\text{ m}\)) [1.0 mark].
(b)(ii) Correct formula for strain energy \(W = \frac{1}{2} F e\) [1.0 mark]; substitution of values [1.0 mark]; final answer \(0.113\text{ J}\) (accept \(0.11\text{ J}\)) [1.0 mark].

(a) Under elastic deformation, the material returns to its original shape and size when the deforming force is removed. Under plastic deformation, the material remains permanently deformed and does not return to its original shape and size when the force is removed.

(b)(i) The limit of proportionality is the point on the force-extension graph beyond which the graph is no longer a straight line (force is no longer directly proportional to extension).

(b)(ii) The elastic limit is the point beyond which any further deformation is plastic (the wire will not return to its original length when unloaded).

(c) The cross-sectional area of the fiber is:
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (5.0 \times 10^{-5})^2}{4} = 1.9635 \times 10^{-9}\text{ m}^2\).
The breaking stress is:
\(\sigma = \frac{F}{A} = \frac{1.8}{1.9635 \times 10^{-9}} = 9.17 \times 10^8\text{ Pa} = 9.2 \times 10^8\text{ Pa}\).

(a) Explaining elastic behavior (returns to original shape) [1.0 mark]; explaining plastic behavior (permanent deformation) [1.0 mark].
(b)(i) Identifying the end of the linear portion/straight line [1.5 marks].
(b)(ii) Identifying the onset of permanent/plastic extension [1.5 marks].
(c) Correct area calculation [1.0 mark]; correct calculation of stress with appropriate unit (Pa or \(\text{N m}^{-2}\)) [1.5 marks].

(a) For two springs connected in parallel, both springs undergo the same extension \(x\) when a load \(F\) is applied.
The total restoring force \(F\) is the sum of the individual restoring forces exerted by the two springs:
\(F = F_1 + F_2\)
Applying Hooke's Law (\(F = k x\)) to each spring:
\(F_1 = k_1 x\) and \(F_2 = k_2 x\)
Substitute these into the force equation:
\(F = k_1 x + k_2 x = (k_1 + k_2) x\)
By comparison with the definition of the effective spring constant \(F = k_p x\), we obtain:
\(k_p = k_1 + k_2\).

(b)(i) For identical springs in series:
\(\frac{1}{k_s} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} \implies k_s = \frac{k}{2} = \frac{450}{2} = 225\text{ N m}^{-1}\).

(b)(ii) The total extension \(x\) is given by:
\(x = \frac{F}{k_s} = \frac{15}{225} = 0.0667\text{ m}\).
Work done:
\(W = \frac{1}{2} F x = 0.5 \times 15 \times 0.0667 = 0.50\text{ J}\).

(a) Stating that extension \(x\) is the same for both springs [1.0 mark]; expressing total force as sum of forces (\(F = F_1 + F_2\)) [1.0 mark]; substituting Hooke's Law [1.0 mark]; final algebraic comparison to show \(k_p = k_1 + k_2\) [0.5 mark].
(b)(i) Correct value of \(225\text{ N m}^{-1}\) [1.0 mark].
(b)(ii) Correct formula for work done \(W = \frac{1}{2} F x\) or \(W = \frac{1}{2} \frac{F^2}{k_s}\) [1.0 mark]; calculation of extension or correct substitution [1.0 mark]; final answer \(0.50\text{ J}\) [1.0 mark].

(a) Simple harmonic motion is defined as periodic motion where the acceleration of the body is directly proportional to its displacement from its equilibrium position, and is always directed towards that equilibrium position (opposite to the displacement: \(a = -\omega^2 x\)).

(b)(i) The angular frequency \(\omega\) is:
\(\omega = 2\pi f = 2\pi \times 2.0 = 4\pi \approx 12.57\text{ rad s}^{-1}\).
The maximum velocity is:
\(v_{\max} = \omega x_0 = 12.57 \times 0.045 = 0.565\text{ m s}^{-1} \approx 0.57\text{ m s}^{-1}\).

(b)(ii) The maximum acceleration magnitude is:
\(a_{\max} = \omega^2 x_0 = (12.57)^2 \times 0.045 = 7.11\text{ m s}^{-2} \approx 7.1\text{ m s}^{-2}\).

(a) Acceleration proportional to displacement [1.0 mark]; acceleration directed towards equilibrium position [1.5 marks].
(b)(i) Formula for maximum velocity \(v_{\max} = \omega x_0\) [1.0 mark]; calculation of \(\omega\) [0.5 mark]; final answer \(0.57\text{ m s}^{-1}\) [1.0 mark].
(b)(ii) Formula for maximum acceleration \(a_{\max} = \omega^2 x_0\) [1.0 mark]; substitution [0.5 mark]; final answer \(7.1\text{ m s}^{-2}\) [1.0 mark].

(a)(i) Comparing the equation to \(x = x_0 \cos(\omega t)\), the amplitude is \(x_0 = 0.080\text{ m}\).

(a)(ii) The angular frequency is \(\omega = 5.0\text{ rad s}^{-1}\).
Period \(T = \frac{2\pi}{\omega} = \frac{2\pi}{5.0} = 1.26\text{ s} \approx 1.3\text{ s}\).

(b) The speed \(v\) of an object in simple harmonic motion is:
\(v = \omega \sqrt{x_0^2 - x^2}\)
The maximum speed is \(v_{\max} = \omega x_0\).
We are given \(v = 0.5 v_{\max} = 0.5 \omega x_0\).
Thus:
\(0.5 \omega x_0 = \omega \sqrt{x_0^2 - x^2}\)
\(0.25 x_0^2 = x_0^2 - x^2\)
\(x^2 = 0.75 x_0^2\)
\(x = \sqrt{0.75} x_0 = 0.866 \times 0.080 = 0.0693\text{ m} \approx 0.069\text{ m}\) (or \(6.9\text{ cm}\)).

(a)(i) State amplitude as \(0.080\text{ m}\) [1.0 mark].
(a)(ii) Correct formula \(T = 2\pi / \omega\) [1.0 mark]; correct period \(1.3\text{ s}\) (or \(1.26\text{ s}\)) [1.0 mark].
(b) Correct formula relating velocity and displacement [1.0 mark]; setting up the condition \(0.5 v_{\max} = v\) [1.0 mark]; simplifying algebra to get \(x^2 = 0.75 x_0^2\) or equivalent [1.5 marks]; final answer \(0.069\text{ m}\) (or \(6.9\text{ cm}\)) [1.0 mark].

(a) At the balance point, no current is drawn from the cell under test. Consequently, there is no potential difference lost across the cell's internal resistance, so the terminal potential difference equals the e.m.f. of the cell. A voltmeter always draws some current, measuring a value slightly lower than the actual e.m.f.

(b)(i) The potential difference across the entire wire \(AB\) is \(2.0\text{ V}\).
The potential gradient along the wire is:
\(k = \frac{2.0\text{ V}}{1.00\text{ m}} = 2.0\text{ V m}^{-1}\).
At the balance point, the potential difference across the balanced section of length \(l = 0.650\text{ m}\) equals \(E_Y\):
\(E_Y = k \times l = 2.0\text{ V m}^{-1} \times 0.650\text{ m} = 1.30\text{ V}\).
Note: The internal resistance of cell \(Y\) does not affect the balance point as no current flows through it at balance.

(b)(ii) Adding a series resistor reduces the current flowing in the primary circuit (containing cell \(X\) and the wire \(AB\)). This decreases the potential difference across the slide-wire \(AB\), reducing the potential gradient \(k\). Since \(E_Y\) is constant and \(E_Y = k \times l\), a smaller potential gradient requires a longer balance length \(l\). Therefore, the balance point moves further towards \(B\) (the balance length increases).

(a) Stating that zero current is drawn from the secondary circuit at balance [1.0 mark]; stating that this means no potential drop occurs across the internal resistance [1.0 mark]; concluding that the true e.m.f. is measured [0.5 mark].
(b)(i) Correct potential gradient calculation [1.0 mark]; formula for e.m.f. \(E = k l\) [1.0 mark]; final answer \(1.30\text{ V}\) [1.0 mark].
(b)(ii) Mentioning that the current in wire \(AB\) (or potential gradient) decreases [1.0 mark]; stating that the balance length must increase to balance the same e.m.f. [1.0 mark].

(a) Two waves are coherent if they have a constant phase difference between them.

(b)(i) Using the double-slit interference equation:
\(x = \frac{\lambda D}{a}\)
where \(\lambda = 633 \times 10^{-9}\text{ m}\), \(D = 1.80\text{ m}\), and \(a = 0.240 \times 10^{-3}\text{ m}\).
\(x = \frac{633 \times 10^{-9} \times 1.80}{0.240 \times 10^{-3}} = 4.75 \times 10^{-3}\text{ m} = 4.75\text{ mm}\).

(b)(ii)
- Fringe separation is determined by \(x = \frac{\lambda D}{a}\). Since \(\lambda\), \(D\), and \(a\) are unchanged, the fringe separation remains unchanged.
- The brightness (or maximum intensity) of the bright fringes increases because of the greater intensity of the incident wave.

(a) Stating 'constant phase difference' [1.5 marks].
(b)(i) Correct formula \(x = \frac{\lambda D}{a}\) [1.0 mark]; correct substitutions with powers of ten [1.0 mark]; final answer \(4.75\text{ mm}\) [1.0 mark].
(b)(ii) Stating that fringe separation is unchanged [1.5 marks]; stating that bright fringes become brighter/more intense [1.5 marks].

(a) The principle of conservation of momentum states that the total momentum of a system remains constant, provided no external forces act on it.

(b)(i) Taking the direction to the right as positive:
Initial momentum \(p_i = m_P u_P + m_Q u_Q\)
\(p_i = (0.80 \times 3.0) + (1.20 \times (-1.5)) = 2.4 - 1.8 = 0.60\text{ kg m s}^{-1}\).
After collision, the two stick together, so their combined mass is \(M = 0.80 + 1.20 = 2.0\text{ kg}\).
Final momentum \(p_f = M v = 2.0 v\).
By conservation of momentum: \(2.0 v = 0.60 \implies v = 0.30\text{ m s}^{-1}\).
Since \(v\) is positive, the direction is to the right.

(b)(ii) Initial Kinetic Energy:
\(E_{k,i} = \frac{1}{2} m_P u_P^2 + \frac{1}{2} m_Q u_Q^2\)
\(E_{k,i} = (0.5 \times 0.80 \times 3.0^2) + (0.5 \times 1.20 \times 1.5^2) = 3.6 + 1.35 = 4.95\text{ J}\).
Final Kinetic Energy:
\(E_{k,f} = \frac{1}{2} M v^2 = 0.5 \times 2.0 \times (0.30)^2 = 0.090\text{ J}\).
Since \(E_{k,f} < E_{k,i}\), kinetic energy is not conserved, hence the collision is inelastic.

(a) Stating total momentum is constant [1.0 mark]; stating condition 'in an isolated system' or 'no external forces' [1.0 mark].
(b)(i) Expressing momentum conservation equation with negative sign for opposite velocity [1.0 mark]; correct calculation of common speed \(0.30\text{ m s}^{-1}\) [1.0 mark]; stating direction is to the right [1.0 mark].
(b)(ii) Calculating total initial KE as \(4.95\text{ J}\) [1.0 mark]; calculating final KE as \(0.09\text{ J}\) [1.0 mark]; stating that since KE is not conserved, the collision is inelastic [0.5 mark].

Sample Results:
When \(L = 15.0\text{ cm}\), raw time for 20 oscillations \(t = 2.40\text{ s}\), giving \(T = 0.120\text{ s}\).

Complete Table of Results:
\begin{array}{|c|c|c|c|c|} \hline L / \text{cm} & t / \text{s} & T / \text{s} & \lg(L/\text{cm}) & \lg(T/\text{s}) \\ \hline 15.0 & 2.40 & 0.120 & 1.176 & -0.921 \\ 17.5 & 3.02 & 0.151 & 1.243 & -0.821 \\ 20.0 & 3.68 & 0.184 & 1.301 & -0.735 \\ 22.5 & 4.38 & 0.219 & 1.352 & -0.660 \\ 25.0 & 5.14 & 0.257 & 1.398 & -0.590 \\ 28.0 & 6.08 & 0.304 & 1.447 & -0.517 \\ \hline \end{array}

Graph Analysis:
- Plotting \(\lg(T/\text{s})\) against \(\lg(L/\text{cm})\) gives a straight line.
- Gradient \(m = \frac{-0.517 - (-0.921)}{1.447 - 1.176} = \frac{0.404}{0.271} \approx 1.49\).
- y-intercept \(c\): Using the point \((1.301, -0.735)\):
\(-0.735 = 1.49(1.301) + c \implies c = -2.673\).

Constants:
- \(n = m = 1.5\) (no units).
- \(k = 10^c = 10^{-2.673} = 2.12 \times 10^{-3}\text{ s cm}^{-1.5}\).

(a) Raw Observations [2 marks]
- 1 mark: \(L\) recorded to the nearest millimetre with correct unit in the range \(14.5\text{ cm} \le L \le 15.5\text{ cm}\).
- 1 mark: Raw time \(t\) recorded to 0.01 s or 0.1 s with unit.

(b) Table of Results [6 marks]
- 1 mark: Six sets of readings with no empty cells.
- 1 mark: Column headings include quantity and unit in standard form, e.g., \(L/\text{cm}\), \(t/\text{s}\), \(T/\text{s}\), \(\lg(L/\text{cm})\). Note: \(\lg(T/\text{s})\) has no unit.
- 1 mark: All raw values of \(L\) recorded to 1 mm consistency.
- 1 mark: All logarithmic values calculated to the correct number of decimal places (matching the significant figures of the raw data).
- 1 mark: Arithmetic calculation of \(T = t/20\) and logarithms is correct.
- 1 mark: Quality of data: points show very little scatter about a linear trend line.

(c) Graph [5 marks]
- 1 mark: Sensible scales chosen so that points occupy at least half the grid. No awkward scales (no 3, 7, etc.).
- 1 mark: All points plotted to within half a small square. Points must be fine crosses or circled dots.
- 1 mark: Best-fit line drawn with a balanced scatter of points above and below the line.
- 1 mark: No thick lines, all lines must be fine and drawn with a ruler.
- 1 mark: Any anomalous point clearly identified (if none, then this is awarded for general plot quality).

(d) Gradient and Intercept [3 marks]
- 2 marks: Gradient calculation using a large triangle where the hypotenuse is at least half the length of the line. Coordinates of vertices read correctly to half a small square.
- 1 mark: y-intercept calculated using a point on the line and \(y = mx + c\), or read directly from the y-axis if the scale starts at \(x = 0\).

(e) Constants [4 marks]
- 1 mark: \(n\) equated to the gradient and given to 2 or 3 s.f. with no units.
- 1 mark: \(k\) calculated correctly from \(10^{\text{intercept}}\).
- 1 mark: Correct working shown for both constants.
- 1 mark: Correct unit for \(k\) (e.g., \(\text{s cm}^{-n}\) or equivalent).

Sample Observations:
- First run: \(D_1 = 40.0\text{ cm}\), \(y_1 = 22.4\text{ cm}\)
- Second run: \(D_2 = 50.0\text{ cm}\), \(y_2 = 16.6\text{ cm}\)

Uncertainty in \(y_1\):
- Absolute uncertainty \(\Delta y \approx 0.3\text{ cm}\) (due to the difficulty in locating the horizontal reference line and midpoint).
- Percentage uncertainty in \(y_1 = \frac{0.3}{22.4} \times 100\% \approx 1.3\%\).

Calculations of \(k\):
- For run 1: \(k_1 = (40.0)^2 + 4(22.4)^2 = 1600 + 4(501.76) = 3607\text{ cm}^2 \approx 3610\text{ cm}^2\).
- For run 2: \(k_2 = (50.0)^2 + 4(16.6)^2 = 2500 + 4(275.56) = 3602\text{ cm}^2 \approx 3600\text{ cm}^2\).

Comparison:
- Percentage difference: \(\frac{|3610 - 3600|}{3605} \times 100\% \approx 0.3\%\).
- Since the percentage difference is much less than the typical 10% experimental limit of accuracy, the relationship is strongly supported.

(a) Measurements [2 marks]
- 1 mark: First value of \(D\) and \(y\) recorded to the nearest millimetre with units.
- 1 mark: First value of \(y\) is in the expected physical range (\(20.0\text{ cm}\) to \(24.0\text{ cm}\)).

(b) Percentage Uncertainty [1 mark]
- 1 mark: Correct calculation of percentage uncertainty in \(y_1\) with an absolute uncertainty \(\Delta y\) in the range \(2\text{ mm}\) to \(5\text{ mm}\).

(c) Second Run [2 marks]
- 1 mark: Second values of \(D\) and \(y\) recorded with appropriate units.
- 1 mark: Quality of data: \(y_2\) is clearly smaller than \(y_1\) for larger distance \(D_2 = 50.0\text{ cm}\).

(d) Calculations of \(k\) [3 marks]
- 1 mark: Correct substitution of values to calculate \(k_1\) and \(k_2\).
- 1 mark: Values of \(k\) calculated to the same number of significant figures as the raw measurements (3 s.f.).
- 1 mark: Correct unit of \(k\) (\(\text{cm}^2\) or \(\text{m}^2\)).

(e) Evaluation [1 mark]
- 1 mark: Logical conclusion comparing percentage difference of \(k\) values to a specified limit (e.g., 10% or the calculated percentage uncertainty in \(y\)).

(f) Quality of Observations [3 marks]
- 1 mark: Repeated readings taken and averaged for \(y\).
- 1 mark: Effort made to ensure symmetry (mass hung at exact center).
- 1 mark: Parallax error addressed (e.g., set-squares used to keep ruler vertical).

(g) Limitations and Improvements [8 marks - 4 pairs]
- Limitation 1: Difficult to keep the two clamps at exactly the same vertical height.
- Improvement 1: Use a spirit level to level the clamps, or measure height from a flat desk surface to both clamp jaws.
- Limitation 2: Difficult to align the vertical ruler perpendicularly to measure \(y\) accurately.
- Improvement 2: Use a plumb line or a set-square aligned with the bench surface to ensure vertical alignment of the rule.
- Limitation 3: The thread stretches slightly under the load, changing the overall length \(S\).
- Improvement 3: Use a non-stretchy material like thin steel wire or braided fishing line.
- Limitation 4: Two-point measurement has parallax error.
- Improvement 4: Use a digital camera to capture a photograph of the setup in front of a grid scale, then analyze the coordinates digitally.