2023 Cambridge IAS-Level Physics (9702) Practice Paper with Answers

The elastic potential energy stored in a wire under tensile force \(F\) is given by \(U = \frac{1}{2} F \Delta L\). Since Young Modulus \(E = \frac{F L}{A \Delta L}\), we have \ (\Delta L = \frac{F L}{A E}\). Substituting this gives \(U = \frac{F^2 L}{2 A E}\). For two wires of the same material, \(E\) is the same. Under the same force \(F\), the stored energy is proportional to \(\frac{L}{A}\), or \(\frac{L}{d^2}\) where \(d\) is the diameter. Therefore, the ratio is: \(\frac{E_X}{E_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2 = 2 \times 2^2 = 8\).

1 mark for the correct answer D. Show that energy stored is proportional to \(L/A\) and substitute the given ratios to find 8.

The work done in stretching a spring is equal to the increase in elastic potential energy: \(W = \frac{1}{2} k (x_1^2 - x_0^2)\). Converting extensions to meters: \(x_0 = 0.020\text{ m}\) and \(x_1 = 0.060\text{ m}\). Therefore, \(W = \frac{1}{2} \times 150 \times (0.060^2 - 0.020^2) = 75 \times (0.0036 - 0.0004) = 75 \times 0.0032 = 0.24\text{ J}\).

1 mark for the correct calculation of work done as change in elastic potential energy, yielding 0.24 J (B).

When a material is stretched beyond its elastic limit and then unloaded, it undergoes plastic deformation. The unloading line on a force-extension graph is parallel to the initial linear elastic portion of the loading curve, resulting in a permanent strain. The Young modulus of the material remains constant because it is a fundamental property determined by interatomic bonds.

1 mark for identifying that the unloading line is parallel to the elastic region and leaves a permanent plastic strain (A).

By Archimedes' principle, upthrust is equal to the weight of the displaced fluid. The volume of the cylinder is \(V = A h\). The mass of the displaced fluid is \(m_f = \rho_f V = \rho_f A h\). Therefore, the weight of the displaced fluid (and the upthrust) is \(\rho_f A h g\). This is independent of the depth \(d\) and the cylinder's density \(\rho_c\).

1 mark for identifying that upthrust depends only on the density of the fluid, the volume of the object, and \(g\), giving \(\rho_f A h g\) (A).

Since the plank is uniform, its center of gravity is at its geometric center, which is \(2.0\text{ m}\) from the left end. The pivot is \(1.5\text{ m}\) from the left end. Therefore, the distance from the pivot to the center of gravity is \(2.0 - 1.5 = 0.5\text{ m}\) to the right. The block of weight \(W\) is on the extreme left end, which is \(1.5\text{ m}\) to the left of the pivot. For rotational equilibrium, taking moments about the pivot: clockwise moments = anticlockwise moments. Thus, \(120\text{ N} \times 0.5\text{ m} = W \times 1.5\text{ m}\), which gives \(60 = 1.5 W\), so \(W = 40\text{ N}\).

1 mark for equating the clockwise and anticlockwise moments about the pivot to calculate \(W = 40\text{ N}\) (B).

The resistance of a wire is given by \(R = \frac{\rho L}{A}\). Since the volume \(V = A \times L\) remains constant, if the length becomes \(3L\), the cross-sectional area must become \(A/3\) to keep the product constant. The new resistance \(R'\) is \(R' = \frac{\rho (3L)}{(A/3)} = 9 \frac{\rho L}{A} = 9R\).

1 mark for using conservation of volume to determine the new area and calculating the final resistance as \(9R\) (D).

Kirchhoff's first law represents the conservation of charge. Kirchhoff's second law is a consequence of the conservation of energy, stating that around any closed loop in a circuit, the sum of the e.m.f.s is equal to the sum of the potential drops. For this single-loop circuit, this gives \(E = I R + I r = I(R + r)\).

1 mark for identifying that Kirchhoff's second law represents conservation of energy and leads to the loop equation \(E = I(R+r)\) (C).

Using the potential divider formula, the output voltage across the fixed resistor is given by: \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{fixed}}}{R_{\text{thermistor}} + R_{\text{fixed}}}\). Substituting the given values: \(V_{\text{out}} = 9.0\text{ V} \times \frac{1200\ \Omega}{600\ \Omega + 1200\ \Omega} = 9.0 \times \frac{1200}{1800} = 6.0\text{ V}\).

1 mark for using the potential divider formula correctly to obtain \(6.0\text{ V}\) (C).

Strain is defined as \(\epsilon = \frac{\Delta L}{L} = \frac{\sigma}{E} = \frac{F}{A E}\), where \(F\) is the load, \(A\) is the cross-sectional area, and \(E\) is the Young modulus. Since both wires are made of the same material, they have the same Young modulus \(E\). Both support the same load \(F\). Thus, strain is inversely proportional to the cross-sectional area: \(\epsilon \propto \frac{1}{A} \propto \frac{1}{d^2}\). The ratio of the strains is \(\frac{\epsilon_X}{\epsilon_Y} = \frac{A_Y}{A_X} = \frac{\pi (2d)^2 / 4}{\pi d^2 / 4} = 4\).

1 mark for the correct answer C. Awarded for correctly relating strain to stress and cross-sectional area, and determining the inverse square relation with diameter.

Since the wire is stretched within its limit of proportionality, extension is directly proportional to force (\(x \propto F\)). When the force is increased to \(3F\), the new extension is \(3x\). The initial elastic potential energy is \(E_s = \frac{1}{2}Fx\). The final elastic potential energy is \(E_{\text{final}} = \frac{1}{2}(3F)(3x) = 9 \left(\frac{1}{2}Fx\right) = 9 E_s\). The additional work done on the wire equals the increase in stored energy: \(\Delta E = E_{\text{final}} - E_s = 9E_s - E_s = 8E_s\).

1 mark for the correct answer C. Awarded for using Hooke's law to scale the extension and calculating the difference between final and initial strain energy.

The tension force in the wire is equal to the weight of the block, \(W = Mg\). From the definition of the Young modulus, \(E = \frac{W L}{A \Delta L}\), so the extension is \(\Delta L = \frac{M g L}{A E}\). The elastic potential energy stored is \(E_p = \frac{1}{2} W \Delta L = \frac{1}{2} M g \left(\frac{M g L}{A E}\right) = \frac{M^2 g^2 L}{2 A E}\).

1 mark for the correct answer B. Awarded for combining the Young modulus formula and the elastic potential energy formula to express energy in terms of the given variables.

For a floating object, the upthrust equals the weight of the object. The weight of the cylinder is \(W = m g = (\rho_s V_{\text{total}}) g = \rho_s A h g\). The upthrust is the weight of the liquid displaced, which is \(U = \rho_l V_{\text{submerged}} g = \rho_l A d g\). Equating the two: \(\rho_l A d g = \rho_s A h g\). Solving for \(d\) gives \(d = h \frac{\rho_s}{\rho_l}\).

1 mark for the correct answer B. Awarded for applying Archimedes' principle and equating upthrust to the gravitational force.

Since the rod is uniform, its weight of \(30\text{ N}\) acts at its centre of gravity (midpoint), which is \(0.6\text{ m}\) from either end. The pivot is placed \(0.3\text{ m}\) from one end, so the distance from the pivot to the centre of gravity is \(0.6\text{ m} - 0.3\text{ m} = 0.3\text{ m}\). This weight creates a clockwise moment about the pivot of \(30\text{ N} \times 0.3\text{ m} = 9.0\text{ N m}\). The opposite end is at a distance of \(1.2\text{ m} - 0.3\text{ m} = 0.9\text{ m}\) from the pivot. For equilibrium, the sum of moments about the pivot must be zero: \(F \times 0.9\text{ m} = 30\text{ N} \times 0.3\text{ m}\), which gives \(F = 10\text{ N}\).

1 mark for the correct answer B. Awarded for identifying the point of action of the weight and applying the principle of moments.

Since the hook is at the midpoint of the string, each half has a length of \(0.5\text{ m}\). The hook and the two corners form an isosceles triangle with side lengths \(0.5\text{ m}\), \(0.5\text{ m}\), and a base of \(0.6\text{ m}\). Dividing this into two right-angled triangles, the horizontal side is \(0.3\text{ m}\) and the hypotenuse is \(0.5\text{ m}\). By Pythagoras, the vertical height is \(\sqrt{0.5^2 - 0.3^2} = 0.4\text{ m}\). The cosine of the angle \(\theta\) between each string segment and the vertical is \(\cos\theta = 0.4 / 0.5 = 0.8\). Resolving forces vertically for equilibrium: \(2 T \cos\theta = W \implies 2 T (0.8) = 24 \implies 1.6 T = 24 \implies T = 15\text{ N}\).

1 mark for the correct answer B. Awarded for using geometry to find the tension angle and applying the vertical equilibrium equation.

At temperature \(T_1\), the p.d. across the thermistor is \(9.0\text{ V}\), so the p.d. across the fixed resistor is \(12\text{ V} - 9.0\text{ V} = 3.0\text{ V}\). The ratio of resistances is equal to the ratio of their p.d.s: \(\frac{R_{th1}}{R} = \frac{9.0}{3.0} = 3\). Since \(R_{th1} = 3000\ \Omega\), we have \(R = 1000\ \Omega\). At temperature \(T_2\), the p.d. across the thermistor is \(4.0\text{ V}\), so the p.d. across the fixed resistor is \(12\text{ V} - 4.0\text{ V} = 8.0\text{ V}\). Using the ratio again: \(\frac{R_{th2}}{R} = \frac{4.0}{8.0} = 0.5\), which yields \(R_{th2} = 0.5 \times 1000\ \Omega = 500\ \Omega\).

1 mark for the correct answer A. Awarded for calculating the resistance of the fixed resistor first, then using the potential divider ratio at the second temperature.

Let the original length be \(L\) and area be \(A\). The original resistance is \(R = \rho \frac{L}{A}\). The new length is \(L' = 1.10 L\). Since the volume \(V = A L\) is constant, the new area \(A'\) must satisfy \(A' L' = A L \implies A' (1.10 L) = A L \implies A' = \frac{A}{1.10}\). The new resistance is \(R' = \rho \frac{L'}{A'} = \rho \frac{1.10 L}{A / 1.10} = 1.10^2 \left(\rho \frac{L}{A}\right) = 1.21 R\).

1 mark for the correct answer C. Awarded for relating the change in length to the change in cross-sectional area using conservation of volume, and calculating the new resistance.

The Young modulus \(E\) is given by \(E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{\text{strain}}\). Rearranging this gives \(\text{strain} = \frac{F}{A E}\). For wire \(X\), the area is \(A_X = \pi \left(\frac{d}{2}\right)^2 = A\) and the Young modulus is \(E_X = 2E_Y\). For wire \(Y\), the area is \(A_Y = \pi d^2 = 4A\) and the Young modulus is \(E_Y\). Thus, the ratio of the strains is: \(\frac{\text{strain}_X}{\text{strain}_Y} = \frac{F / (A_X E_X)}{F / (A_Y E_Y)} = \frac{A_Y E_Y}{A_X E_X} = \frac{4A \times E_Y}{A \times 2E_Y} = 2\).

1 mark for identifying the correct strain formula in terms of force, area, and Young modulus, and correctly determining the ratio of cross-sectional areas and Young moduli to obtain the ratio 2.

When a material is stretched beyond its elastic limit, plastic deformation occurs. Upon unloading, the material behaves elastically again, meaning the unloading path on the force-extension graph is a straight line that is parallel to the initial linear (elastic) portion of the loading curve. Because plastic deformation has occurred, the unloading curve does not return to the origin but instead meets the horizontal axis at a non-zero value, representing a permanent extension.

1 mark for identifying that the unloading line is straight and parallel to the initial elastic region, resulting in a permanent extension.

First, find the effective force constant \(k_s\) of the two springs in series: \(\frac{1}{k_s} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} \implies k_s = 0.5k\). Next, this combination is connected in parallel with a third spring of constant \(k\), so the total force constant \(k_{\text{total}}\) is: \(k_{\text{total}} = k_s + k = 0.5k + k = 1.5k\).

1 mark for calculating the series combination as \(0.5k\) and adding the parallel spring to get \(1.5k\).

By Archimedes' principle, the upthrust force is equal to the weight of the fluid displaced by the cylinder. The volume of the submerged part of the cylinder is \(V = A \times d\). The mass of the displaced liquid is \(m = \rho V = \rho A d\). Therefore, the upthrust (weight of displaced liquid) is \(U = m g = \rho A d g\).

1 mark for applying Archimedes' principle to find that the submerged volume is \(A d\) and the resulting upthrust is \(\rho A d g\).

The weight of the block is \(W = m g = 12 \times 9.81 = 117.72\text{ N}\). The pressure is given by \(P = \frac{F}{A}\). The three possible contact areas are: \(A_1 = 0.10 \times 0.20 = 0.02\text{ m}^2\) (smallest area), \(A_2 = 0.10 \times 0.50 = 0.05\text{ m}^2\), and \(A_3 = 0.20 \times 0.50 = 0.10\text{ m}^2\) (largest area). Maximum pressure occurs with the smallest area: \(P_{\text{max}} = \frac{117.72}{0.02} = 5886\text{ Pa}\). Minimum pressure occurs with the largest area: \(P_{\text{min}} = \frac{117.72}{0.10} = 1177\text{ Pa}\). The difference is \(\Delta P = P_{\text{max}} - P_{\text{min}} = 5886 - 1177 = 4709\text{ Pa} \approx 4.7 \times 10^3\text{ Pa}\).

1 mark for calculating the weight of the block, finding the maximum and minimum surface areas, calculating both pressures, and determining the correct difference.

The plank is on the verge of tipping about the pivot at \(3.5\text{ m}\) when the normal contact force at the other pivot (at \(1.0\text{ m}\)) becomes zero. Let the left end of the plank be at position \(x = 0\). The center of gravity of the uniform plank is at \(x = 2.0\text{ m}\), where its weight of \(120\text{ N}\) acts. Taking moments about the pivot at \(x = 3.5\text{ m}\): Counter-clockwise moment = \(120 \times (3.5 - 2.0) = 120 \times 1.5 = 180\text{ N m}\). Clockwise moment = \(W \times (4.0 - 3.5) = 0.5 W\). Equating moments for rotational equilibrium: \(0.5 W = 180 \implies W = 360\text{ N}\).

1 mark for identifying that the pivot point for tipping is at \(3.5\text{ m}\), setting up the moment equation about this pivot, and solving for \(W = 360\text{ N}\).

The resistance is given by \(R = \rho \frac{L}{A}\). Since the volume \(V = A \times L\) is kept constant, when the length is tripled (\(L_2 = 3L\)), the cross-sectional area must be reduced to one-third of its original value (\(A_2 = A/3\)) to maintain \(A_2 L_2 = A L\). Substituting these values into the resistance equation: \(R_{\text{new}} = \rho \frac{3L}{A/3} = 9 \left(\rho \frac{L}{A}\right) = 9R\).

1 mark for recognizing that volume conservation means area is reduced to \(A/3\) when length is tripled, and applying this to get \(9R\).

At high temperature, the thermistor resistance is \(1.0\text{ k}\Omega\). Using the potential divider formula: \(V_{\text{out, high}} = 12 \times \frac{1.0}{1.0 + 4.0} = 2.4\text{ V}\). At low temperature, the thermistor resistance is \(8.0\text{ k}\Omega\). Using the potential divider formula: \(V_{\text{out, low}} = 12 \times \frac{8.0}{8.0 + 4.0} = 8.0\text{ V}\). The change in output voltage is \(\Delta V_{\text{out}} = 8.0\text{ V} - 2.4\text{ V} = 5.6\text{ V}\).

1 mark for calculating both output voltages using the potential divider relation and finding their difference to be \(5.6\text{ V}\).

The Young modulus \(E\) of the material is given by:
\(E = \frac{\sigma}{\epsilon} = \frac{F / A}{x / L} = \frac{FL}{Ax}\)

Rearranging for extension \(x\):
\(x = \frac{FL}{AE}\)

Let the length of wire Y be \(L\) and its cross-sectional area be \(A\). Since wire Y has diameter \(d\), \(A = \frac{\pi d^2}{4}\).
For wire X:
- Length \(L_X = 2L\)
- Diameter \(d_X = 0.5d\), so its cross-sectional area is \(A_X = \frac{\pi (0.5d)^2}{4} = 0.25 A\)

Both wires are of the same material (same \(E\)) and subjected to the same tensile force \(F\). Comparing the extensions:
\(\frac{x_X}{x_Y} = \frac{L_X}{L_Y} \times \frac{A_Y}{A_X} = 2 \times \frac{A}{0.25A} = 2 \times 4 = 8\)

1 mark for the correct calculation of the ratio of extensions, showing that extension is proportional to length and inversely proportional to cross-sectional area (which depends on diameter squared).

For an elastic material obeying Hooke's law, the force \(F\) is directly proportional to the extension \(x\) (i.e., \(F = kx\)).

The work done \(W\) in stretching the wire to extension \(x\) is the area under the force-extension graph:
\(W = \frac{1}{2} F x\)

When the force is increased to \(2F\), the new extension is \(2x\).
The total work done in stretching the wire from zero to \(2x\) is:
\(W_{\text{total}} = \frac{1}{2} (2F) (2x) = 2 F x = 4 W\)

The additional work done in stretching the wire from extension \(x\) to \(2x\) is:
\(\Delta W = W_{\text{total}} - W = 4W - W = 3W\)

1 mark for identifying that doubling the force doubles the extension, calculating total work done as 4W, and subtracting the initial work to find the additional work of 3W.

The picture frame is in equilibrium, so the vertical forces must balance. The forces acting on the frame are its weight \(W = mg\) acting downwards, and the tensions \(T\) in the two sections of the string acting upwards and inwards.

Let the string form an isosceles triangle with the top edge of the frame.
- The base of this triangle is the width of the frame, \(w = 0.80\text{ m}\). The half-base is \(0.40\text{ m}\).
- The total length of the string is \(1.00\text{ m}\), so each side of the string has a length of \(0.50\text{ m}\).

Let \(\alpha\) be the angle between the string and the vertical. From the right-angled triangle formed by the half-base, the vertical height, and the string:
\(\sin\alpha = \frac{0.40}{0.50} = 0.8\)
\(\cos\alpha = \frac{\sqrt{0.50^2 - 0.40^2}}{0.50} = \frac{0.30}{0.50} = 0.6\)

Resolving vertically for equilibrium:
\(2T\cos\alpha = mg\)
\(2T(0.6) = 2.4 \times 9.81\)
\(1.2T = 23.54\)
\(T \approx 20\text{ N}\)

1 mark for using the geometry of the string to find the angle, resolving forces vertically for equilibrium, and calculating the tension correctly.

According to Archimedes' principle, the total upthrust \(U\) on the cylinder is equal to the total weight of the fluids displaced by it.

- Volume of upper liquid displaced: \(V_1 = A \times \frac{2}{3}h\)
- Weight of upper liquid displaced: \(W_1 = \rho_1 V_1 g = \frac{2}{3}\rho_1 Ahg\)

- Volume of lower liquid displaced: \(V_2 = A \times \frac{1}{3}h\)
- Weight of lower liquid displaced: \(W_2 = \rho_2 V_2 g = \frac{1}{3}\rho_2 Ahg\)

Adding these two weights together gives the total upthrust:
\(U = W_1 + W_2 = \frac{2}{3}\rho_1 Ahg + \frac{1}{3}\rho_2 Ahg = \frac{1}{3}Ahg(2\rho_1 + \rho_2)\)

1 mark for correctly applying Archimedes' principle to find the sum of the weights of the two displaced fluids.

Since the bar is uniform, its weight of 30.0 N acts at its centre of gravity, which is at its midpoint (0.60 m from either end).

Let's find the perpendicular distances from the pivot (located 0.30 m from the left end):
- Distance to weight: \(0.60\text{ m} - 0.30\text{ m} = 0.30\text{ m}\) (to the right of pivot)
- Distance to force \(F\) (applied at the right-hand end): \(1.20\text{ m} - 0.30\text{ m} = 0.90\text{ m}\) (to the right of pivot)

Taking moments about the pivot for equilibrium:
\(\text{Clockwise Moment} = \text{Anticlockwise Moment}\)
\(30.0\text{ N} \times 0.30\text{ m} = F \times 0.90\text{ m}\)
\(9.00 = 0.90 F\)
\(F = 10.0\text{ N}\)

1 mark for identifying the correct point of action of the weight, calculating the distances from the pivot, and applying the principle of moments.

The resistance of a wire is given by \(R = \frac{\rho L}{A}\).

Since the volume \(V = AL\) is constant, if the length increases to \(L' = 1.20 L\), the cross-sectional area must decrease to:
\(A' = \frac{A}{1.20}\)

The new resistance \(R'\) is:
\(R' = \frac{\rho L'}{A'} = \frac{\rho (1.20 L)}{A / 1.20} = 1.44 \frac{\rho L}{A} = 1.44 R\)

1 mark for determining the change in cross-sectional area due to volume conservation and substituting the new parameters into the resistivity formula to find the correct resistance factor.

Since the two cells are connected to oppose each other, the net electromotive force (e.m.f.) in the circuit is:
\(E_{\text{net}} = 12.0\text{ V} - 4.0\text{ V} = 8.0\text{ V}\)

The total resistance of the series circuit is the sum of the external resistance and the internal resistances of both cells:
\(R_{\text{total}} = 6.0\ \Omega + 1.5\ \Omega + 0.5\ \Omega = 8.0\ \Omega\)

The current \(I\) in the circuit is:
\(I = \frac{E_{\text{net}}}{R_{\text{total}}} = \frac{8.0\text{ V}}{8.0\ \Omega} = 1.0\text{ A}\)

The direction of the current is determined by the 12.0 V cell, so it flows out of its positive terminal and enters the positive terminal of the 4.0 V cell. This means the 4.0 V cell is being charged.

The terminal potential difference \(V\) across a cell being charged is:
\(V = E + Ir = 4.0\text{ V} + (1.0\text{ A} \times 0.5\ \Omega) = 4.5\text{ V}\)

1 mark for determining the net e.m.f. and total resistance, calculating the correct current, and using the formula for the terminal potential difference of a charging cell.

For a negative temperature coefficient (NTC) thermistor, as its temperature increases, its resistance decreases.

The output voltage across the thermistor in a potential divider is given by:
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{thermistor}}}{R_{\text{thermistor}} + R_{\text{fixed}}}\)

As the resistance of the thermistor decreases, it takes a smaller fraction of the total resistance, so the voltage dropped across it, \(V_{\text{out}}\), decreases.

1 mark for identifying that the resistance of an NTC thermistor decreases with temperature, and using the potential divider principle to conclude that the output voltage across it also decreases.

The extension of a wire is given by the formula \(x = \frac{FL}{AE}\), where \(E\) is the Young modulus of the material. For the second wire, the length is \(2L\), the area is \(A/2\), and the force is \(2F\). Substituting these values gives the new extension \(x' = \frac{(2F)(2L)}{(A/2)E} = 8\frac{FL}{AE} = 8x\).

1 mark for selecting the correct option D. 0 marks for incorrect options.

The extension \(x\) of the rubber band is the difference between its stretched and unstretched lengths: \(x = 20\text{ cm} - 15\text{ cm} = 5.0\text{ cm} = 0.050\text{ m}\). The elastic potential energy \(E_p\) stored is given by \(E_p = \frac{1}{2} F x = \frac{1}{2} \times 6.0\text{ N} \times 0.050\text{ m} = 0.15\text{ J}\).

1 mark for selecting the correct option A. 0 marks for incorrect options.

For a floating object, the upthrust equals the weight of the object: \(U = mg\). According to Archimedes' principle, upthrust is also equal to the weight of the fluid displaced: \(U = \rho_{water} V_{sub} g = \rho_{water} A h g\), where \(h\) is the submerged depth. Equating the two expressions gives \(\rho_{water} A h g = mg\), which simplifies to \(h = \frac{m}{\rho_{water} A}\). Substituting the values: \(h = \frac{0.48}{1.0 \times 10^3 \times 8.0 \times 10^{-4}} = 0.60\text{ m}\).

1 mark for selecting the correct option D. 0 marks for incorrect options.

First, use vertical equilibrium to find the tension in the right-hand rope: \(T_{left} + T_{right} = W \Rightarrow 80\text{ N} + T_{right} = 200\text{ N} \Rightarrow T_{right} = 120\text{ N}\). Next, take moments about the left end (pivot) to find the distance \(d\) to the centre of gravity: clockwise moment = anticlockwise moment \
\(W \times d = T_{right} \times L \Rightarrow 200 \times d = 120 \times 3.0 \Rightarrow 200d = 360 \Rightarrow d = 1.8\text{ m}\).

1 mark for selecting the correct option D. 0 marks for incorrect options.

The volume of a cylinder is \(V = A \times L\). Since volume is constant, doubling the length (\(L' = 2L\)) means the cross-sectional area must be halved (\(A' = A/2\)). The resistance is given by \(R = \frac{\rho L}{A}\). The new resistance is \(R' = \frac{\rho L'}{A'} = \frac{\rho (2L)}{A/2} = 4 \frac{\rho L}{A} = 4R\).

1 mark for selecting the correct option D. 0 marks for incorrect options.

Using the potential divider equation, the output potential difference across the thermistor (\(V_{out}\)) is: \(V_{out} = V_{in} \times \frac{R_{th}}{R_{fixed} + R_{th}} = 12\text{ V} \times \frac{1000\,\Omega}{3000\,\Omega + 1000\,\Omega} = 12 \times \frac{1000}{4000} = 3.0\text{ V}\).

1 mark for selecting the correct option A. 0 marks for incorrect options.

The weight of the sphere is \(W = mg = 0.50 \times 9.81 = 4.905\text{ N}\). Resolving the forces on the sphere in equilibrium: horizontally, \(T \sin(30^\circ) = F\), and vertically, \(T \cos(30^\circ) = W\). Dividing these equations gives \(F = W \tan(30^\circ)\). Substituting the values, we find \(F = 4.905 \times \tan(30^\circ) \approx 2.83\text{ N}\). To two significant figures, this is 2.8 N.

1 mark for selecting the correct option B. 0 marks for incorrect options.

The current in the circuit is given by \(I = \frac{E}{R + r}\). When \(R\) is decreased, the current \(I\) increases. The terminal potential difference is given by \(V = E - Ir\). Since \(I\) increases, \(Ir\) increases, and therefore \(V\) decreases. The power lost inside the cell is given by \(P = I^2 r\). Since \(I\) increases, the power \(P\) increases.

1 mark for selecting the correct option B. 0 marks for incorrect options.

(a) The extension of the brass wire is calculated using the formula: \(x_b = \frac{F L_b}{A_b E_b}\). Substituting the given values: \(x_b = \frac{180 \times 1.2}{(2.0 \times 10^{-6}) \times (1.0 \times 10^{11})} = 1.08 \times 10^{-3}\text{ m}\). (b)(i) The extension of the steel wire is: \(x_s = \text{total extension} - x_b = 1.62 \times 10^{-3} - 1.08 \times 10^{-3} = 0.54 \times 10^{-3}\text{ m}\). (b)(ii) The cross-sectional area of the steel wire is: \(A_s = \frac{F L_s}{x_s E_s}\). Substituting the calculated extension and values: \(A_s = \frac{180 \times 1.8}{(0.54 \times 10^{-3}) \times (2.0 \times 10^{11})} = \frac{324}{1.08 \times 10^8} = 3.0 \times 10^{-6}\text{ m}^2\).

Part (a): [1 mark] for correct formula shown or used for the extension of the brass wire: \(x_b = \frac{FL}{AE}\). [1 mark] for correct substitution of values leading to \(1.08 \times 10^{-3}\text{ m}\). Part (b)(i): [1 mark] for subtracting the brass wire extension from the total extension: \(1.62 \times 10^{-3} - 1.08 \times 10^{-3}\). [1 mark] for obtaining \(0.54 \times 10^{-3}\text{ m}\). Part (b)(ii): [1 mark] for rearrangement of the Young modulus formula for area: \(A_s = \frac{FL_s}{x_s E_s}\). [1 mark] for correct substitution of values. [1 mark] for calculating the intermediate step (e.g., denominator \(1.08 \times 10^8\)). [1 mark] for the final correct answer with units: \(3.0 \times 10^{-6}\text{ m}^2\).

(a) The volume of the cylinder is: \(V = A \times h = 1.5 \times 10^{-3} \times 0.12 = 1.8 \times 10^{-4}\text{ m}^3\). The mass of the cylinder is: \(m = \rho_c \times V = 2.7 \times 10^3 \times 1.8 \times 10^{-4} = 0.486\text{ kg}\). The weight is: \(W = m \times g = 0.486 \times 9.81 = 4.77\text{ N}\). (b) The upthrust \(U\) equals the weight of the oil displaced by the fully submerged cylinder: \(U = \rho_{\text{oil}} \times V \times g = 850 \times 1.8 \times 10^{-4} \times 9.81 = 1.50\text{ N}\). (c) For vertical equilibrium: \(T + U = W\). Therefore, \(T = W - U = 4.77 - 1.50 = 3.27\text{ N}\).

Part (a): [1 mark] for calculation of the cylinder volume \(1.8 \times 10^{-4}\text{ m}^3\) and mass \(0.486\text{ kg}\). [1 mark] for weight calculation of \(4.77\text{ N}\) (accept \(4.8\text{ N}\) if 2 s.f. is used consistently). Part (b): [1 mark] for stating or using Archimedes' principle formula: \(U = \rho_{\text{fluid}} V g\). [1 mark] for correct substitution of values: \(850\text{ kg m}^{-3}\), \(1.8 \times 10^{-4}\text{ m}^3\), and \(9.81\text{ m s}^{-2}\). [1 mark] for calculating upthrust as \(1.50\text{ N}\) (accept \(1.5\text{ N}\)). Part (c): [1 mark] for writing the equilibrium equation: \(T = W - U\). [1 mark] for substitution of previous values. [1 mark] for the final answer of \(3.27\text{ N}\) (accept \(3.3\text{ N}\)).

(a) Using \(E = I(R + r)\): For the first state: \(E = 1.2(4.5 + r)\). For the second state: \(E = 0.60(10.5 + r)\). Equating the two expressions for \(E\): \(1.2(4.5 + r) = 0.60(10.5 + r)\). Dividing both sides by 0.60 gives: \(2(4.5 + r) = 10.5 + r \implies 9.0 + 2r = 10.5 + r \implies r = 1.5\ \Omega\). (b) Substituting \(r = 1.5\ \Omega\) into the first equation: \(E = 1.2(4.5 + 1.5) = 1.2 \times 6.0 = 7.2\text{ V}\). (c) The power dissipated in the internal resistance \(r\) when \(R = 4.5\ \Omega\) (so \(I = 1.2\text{ A}\)) is: \(P = I^2 r = (1.2)^2 \times 1.5 = 1.44 \times 1.5 = 2.16\text{ W}\) (or \(2.2\text{ W}\)).

Part (a): [1 mark] for setting up the two simultaneous equations: \(E = 1.2(4.5 + r)\) and \(E = 0.60(10.5 + r)\). [1 mark] for equating and expanding to solve for \(r\). [1 mark] for correct internal resistance \(r = 1.5\ \Omega\). Part (b): [1 mark] for substituting \(r\) into either of the starting equations. [1 mark] for correct e.m.f. \(E = 7.2\text{ V}\). Part (c): [1 mark] for using the correct power formula: \(P = I^2 r\). [1 mark] for correct substitution: \(1.2^2 \times 1.5\). [1 mark] for final answer \(2.16\text{ W}\) or \(2.2\text{ W}\).

(a) The force acting on the spring is: \(F = mg = 0.45 \times 9.81 = 4.41\text{ N}\). The extension is: \(x = 18.6\text{ cm} - 15.0\text{ cm} = 3.6\text{ cm} = 0.036\text{ m}\). The spring constant is: \(k = \frac{F}{x} = \frac{4.41}{0.036} = 123\text{ N m}^{-1}\) (or \(122.6\text{ N m}^{-1}\)). (b)(i) Stored elastic potential energy: \(E_p = \frac{1}{2} k x^2 \implies 0.40 = 0.5 \times 122.6 \times x^2 \implies x^2 \approx 0.006525 \implies x = 0.0808\text{ m}\) (or \(8.08\text{ cm}\)). (b)(ii) The new length is: \(L = L_0 + x = 15.0\text{ cm} + 8.08\text{ cm} = 23.1\text{ cm}\) (or \(0.231\text{ m}\)).

Part (a): [1 mark] for calculating the tension force \(F = 4.41\text{ N}\). [1 mark] for obtaining the extension \(x = 0.036\text{ m}\). [1 mark] for calculating the spring constant \(k = 123\text{ N m}^{-1}\) (or \(122.6\text{ N m}^{-1}\)). Part (b)(i): [1 mark] for using the elastic potential energy equation: \(E_p = \frac{1}{2} k x^2\). [1 mark] for substituting \(E_p = 0.40\text{ J}\) and the value of \(k\). [1 mark] for obtaining the extension \(x = 0.081\text{ m}\) or \(8.1\text{ cm}\) (allow \(0.0808\text{ m}\) or \(8.08\text{ cm}\)). Part (b)(ii): [1 mark] for adding the calculated extension to the unstretched length \(15.0\text{ cm}\). [1 mark] for the correct final length of \(23.1\text{ cm}\) or \(0.231\text{ m}\).

(a) Weight of the plank: \(W_p = mg = 12 \times 9.81 = 117.7\text{ N}\) (or \(118\text{ N}\)). (b) Taking moments about Rope A (left-hand end, where \(x = 0\)): The plank is uniform, so its weight acts at its center of gravity: \(x = 1.2\text{ m}\). Rope B is at a distance: \(x = 2.4 - 0.60 = 1.8\text{ m}\). The load is at \(x = 0.40\text{ m}\). Setting clockwise moments equal to anticlockwise moments: \((M \times g \times 0.40) + (W_p \times 1.2) = T_B \times 1.8\). Substituting the knowns: \((M \times 9.81 \times 0.40) + (117.72 \times 1.2) = 110 \times 1.8 \implies 3.924M + 141.26 = 198 \implies 3.924M = 56.74 \implies M = 14.5\text{ kg}\) (or \(14\text{ kg}\) to 2 s.f.). (c) For vertical equilibrium, total upward force equals total downward force: \(T_A + T_B = (M \times g) + W_p \implies T_A + 110 = (14.46 \times 9.81) + 117.72 = 141.85 + 117.72 = 259.6\text{ N}\). Therefore, \(T_A = 259.6 - 110 = 150\text{ N}\) (or \(149.6\text{ N}\)).

Part (a): [1 mark] for weight formula: \(W = mg\). [1 mark] for correct calculation of \(118\text{ N}\) (or \(117.7\text{ N}\)). Part (b): [1 mark] for identifying distance of Rope B from Rope A as \(1.8\text{ m}\). [1 mark] for a correct moment equation about Rope A: \(M g (0.4) + W_p (1.2) = T_B (1.8)\). [1 mark] for correct substitution of values. [1 mark] for the final mass of the load: \(14.5\text{ kg}\) (allow \(14\text{ kg}\) for 2 s.f.). Part (c): [1 mark] for equating the upward and downward forces: \(T_A + T_B = M g + W_p\). [1 mark] for calculating the tension in Rope A: \(150\text{ N}\) (allow \(149.6\text{ N}\) or \(150\text{ N}\) depending on rounding).

In the elastic region, the tensile force causes the distance between adjacent metal atoms to increase slightly. The atoms remain in their original relative positions, and the attractive electrostatic forces between them act as restoring forces. When the force is removed, these interatomic forces pull the atoms back to their initial equilibrium positions, and the wire returns to its original length. In the plastic region, the stress is sufficient to cause entire planes of atoms to slide past one another (dislocation movement). This sliding breaks old metallic bonds and forms new ones in new positions. When the force is removed, there are no restoring forces to return the planes of atoms to their original relative positions, so the atoms remain in their new positions, leaving a permanent extension.

1. Elastic deformation: Atoms are pulled apart from their equilibrium positions (or interatomic separation increases). [1 mark] 2. Elastic recovery: When force is removed, interatomic attractive forces pull atoms back to their original equilibrium positions (so original shape/length is restored). [1 mark] 3. Plastic deformation: Layers/planes of atoms slide over/past each other (breaking and reforming of bonds). [1 mark] 4. Plastic behavior: When force is removed, atoms do not return to their original positions (they remain in their new positions, resulting in permanent extension). [1 mark]

Upthrust is the net upward force experienced by an object when it is partially or fully submerged in a fluid. The pressure \(p\) in a fluid at rest of uniform density \(\rho\) increases with depth \(h\) according to the relation \(p = h\rho g\). Therefore, the bottom surface of the submerged object is at a greater depth than the top surface, meaning the fluid pressure acting on the bottom surface is greater than the fluid pressure acting on the top surface. Since force is the product of pressure and area (\(F = pA\)), the upward force exerted by the fluid on the bottom surface is greater than the downward force exerted on the top surface. This difference in forces results in a net upward force on the object, which is the upthrust.

1. Definition: Upthrust is the net upward force exerted on a submerged/immersed body by a fluid. [1 mark] 2. Pressure variation: Pressure in a fluid increases with depth (or states/uses \(p = h\rho g\)). [1 mark] 3. Pressure difference: The pressure acting on the bottom surface of the body is greater than the pressure on the top surface. [1 mark] 4. Force difference: Since \(F = pA\), the upward force on the bottom is greater than the downward force on the top, resulting in a net upward force. [1 mark]

When the light intensity incident on the LDR increases, the number of charge carriers in the semiconductor material of the LDR increases, which causes the resistance of the LDR (\(R_{\text{LDR}}\)) to decrease. The LDR is in series with a fixed resistor of resistance \(R\), so the total resistance of the circuit (\(R + R_{\text{LDR}}\)) also decreases. This leads to an increase in the circuit current \(I\). The potential difference across the fixed resistor (\(V_{\text{fixed}} = I R\)) must increase because \(I\) has increased while \(R\) remains constant. Because the e.m.f. of the cell is constant and is shared between the fixed resistor and the LDR, the potential difference across the LDR (which is the voltmeter reading) must decrease (\(V_{\text{LDR}} = \text{e.m.f.} - V_{\text{fixed}}\)). Alternatively, by the potential divider formula, \(V_{\text{out}} = \text{e.m.f.} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\). As \(R_{\text{LDR}}\) decreases, the ratio decreases, leading to a smaller output voltage.

1. LDR Resistance: States that resistance of the LDR decreases as light intensity increases. [1 mark] 2. Total resistance/current: States that total circuit resistance decreases, so current in the circuit increases. [1 mark] 3. Fixed resistor p.d.: Explains that the potential difference (p.d.) across the fixed resistor increases because the current increases (while its resistance is constant). [1 mark] 4. Voltmeter reading: Concludes that the voltmeter reading (p.d. across the LDR) must decrease because the total e.m.f. is constant (or correctly uses the potential divider equation to show the fraction of voltage decreases). [1 mark]

### **Sample Data and Table**

| \(d / \text{cm}\) | \(h / \text{cm}\) |
|---|---|
| 15.0 | 42.1 |
| 30.0 | 35.2 |
| 45.0 | 28.1 |
| 60.0 | 21.0 |
| 70.0 | 16.2 |
| 75.0 | 13.9 |

### **Graph Analysis**

1. **Plotting the Points:**
- \(d\) is plotted on the horizontal axis (scale: \(10\text{ cm}\) per \(2\text{ cm}\) grid division).
- \(h\) is plotted on the vertical axis (scale: \(10\text{ cm}\) per \(2\text{ cm}\) grid division).

2. **Determining the Gradient:**
- Choose two points on the line of best fit: \((15.0, 42.1)\) and \((75.0, 13.9)\).
- \(\text{Gradient} = \frac{13.9 - 42.1}{75.0 - 15.0} = \frac{-28.2}{60.0} = -0.470\).

3. **Determining the y-intercept:**
- Using the point \((15.0, 42.1)\):
\(h = (\text{gradient} \times d) + y\text{-intercept}\)
\(42.1 = (-0.470 \times 15.0) + y\text{-intercept}\)
\(y\text{-intercept} = 42.1 + 7.05 = 49.15\text{ cm}\).

4. **Finding Constants \(P\) and \(Q\):**
- Comparing the equation \(h = -P d + Q\) to the linear format \(y = m x + c\):
- \(P = -\text{gradient} = 0.470\)
- \(Q = y\text{-intercept} = 49.2\text{ cm}\) (or \(0.492\text{ m}\))

### **Units:**
- \(P\) is the ratio of two heights/distances, so it is dimensionless (no units or \(\text{cm cm}^{-1}\)).
- \(Q\) has the same units as \(h\), which is \(\text{cm}\) (or \(\text{m}\)).

### **Marking Scheme (Total: 20 Marks)**

#### **1. Data Collection (6 Marks)**
- **[1]** Successful collection of 6 sets of readings of \(d\) and \(h\) with correct negative trend (as \(d\) increases, \(h\) decreases).
- **[1]** Range of \(d\) values is wide, extending from at least \(15.0\text{ cm}\) to \(75.0\text{ cm}\) (span of at least \(55.0\text{ cm}\)).
- **[1]** Raw values of \(h\) and \(d\) are recorded to the nearest \(0.1\text{ cm}\) (1 millimetre), representing the precision of the rules used.
- **[1]** All columns in the table have correct headings and units formatted as \(d / \text{cm}\) and \(h / \text{cm}\).
- **[1]** Consistency: All values of \(d\) and \(h\) in the table are written to the same number of decimal places (e.g., \(15.0\), not \(15\)).
- **[1]** Quality: Scatter of points is small. All plotted points must lie within \(\pm 1.0\text{ cm}\) on the vertical scale of the line of best fit.

#### **2. Graph Plotting (4 Marks)**
- **[1]** **Axes:** Sensible linear scales where the plotted points cover more than \(50\%\) of the grid in both directions. Scales are labelled with quantity and unit.
- **[1]** **Plotting:** Points plotted accurately to within half a small square. No thick 'blobs' (points must be fine crosses or small dots inside circles).
- **[1]** **Line of best fit:** Drawn with a ruler, representing a fair balance of points on either side with no systematic bias.
- **[1]** **Anomalies:** Any clearly anomalous points are identified (or a statement is made if none are present).

#### **3. Gradient and Intercept (4 Marks)**
- **[1]** **Gradient calculation:** Uses a triangle with a hypotenuse of at least half the length of the drawn line. Coordinates must be read to within half a small square.
- **[1]** **Sign check:** Gradient is calculated correctly as negative.
- **[1]** **y-intercept calculation:** Calculated correctly using a point on the line of best fit, or read directly from the y-axis if \(d=0\) is on the grid.
- **[1]** **Significant figures:** Gradient and intercept are quoted to 2 or 3 significant figures, consistent with the raw data.

#### **4. Evaluation of Constants (6 Marks)**
- **[1]** Identifies that \(P = -\text{gradient}\).
- **[1]** Identifies that \(Q = y\text{-intercept}\).
- **[1]** Correct numerical value calculated for \(P\) (must be positive, typically between \(0.35\) and \(0.65\)).
- **[1]** Correct numerical value calculated for \(Q\) (must be positive, typically between \(40\text{ cm}\) and \(60\text{ cm}\)).
- **[1]** Unit of \(P\) is stated as dimensionless (no unit, or \(\text{cm/cm}\)), and unit of \(Q\) is given as \(\text{cm}\) or \(\text{m}\).
- **[1]** Final answers for \(P\) and \(Q\) are given to an appropriate number of significant figures (2 or 3).

### (a) Preliminary Measurements
(i) Width of the rule, \(w\):
Typical width \(w = 2.64\text{ cm}\) (or \(26.4\text{ mm}\)).
(ii) Percentage uncertainty:
The smallest division of a standard vernier caliper is \(0.1\text{ mm}\) (or \(0.01\text{ cm}\)).
\[ \text{Percentage uncertainty} = \frac{0.1\text{ mm}}{26.4\text{ mm}} \times 100\% \approx 0.38\% \]

### (b) First Run of the Experiment
- Mass \(m_1 = 100\text{ g}\)
- Measured displacement \(y_1 = 4.2\text{ cm}\) (or \(42\text{ mm}\)). Note: Measured with a meter rule, so recorded to the nearest \(1\text{ mm}\).

### (c) Percentage Uncertainty in \(y\)
- Absolute uncertainty chosen: \(\Delta y = 2\text{ mm}\) (or \(0.2\text{ cm}\)).
- Justification: Parallax error due to the gap between the rule scale and the tilted meter rule, and slight oscillation of the suspended ruler.
- Percentage uncertainty:
\[ \frac{2\text{ mm}}{42\text{ mm}} \times 100\% \approx 4.8\% \]

### (d) Second Run of the Experiment
- Mass \(m_2 = 200\text{ g}\)
- Measured displacement \(y_2 = 8.3\text{ cm}\) (or \(83\text{ mm}\)).

### (e) Calculation of Constants
(i) First run constant \(k_1\):
\[ k_1 = \frac{y_1}{m_1} = \frac{42\text{ mm}}{100\text{ g}} = 0.42\text{ mm g}^{-1} \]
(ii) Second run constant \(k_2\):
\[ k_2 = \frac{y_2}{m_2} = \frac{83\text{ mm}}{200\text{ g}} = 0.415\text{ mm g}^{-1} \]
Both values are written to 2 or 3 significant figures, matching the precision of the raw data.

### (f) Analysis and Evaluation
- Calculate the percentage difference between \(k_1\) and \(k_2\):
\[ \text{Percentage difference} = \frac{|k_1 - k_2|}{\text{mean}} \times 100\% = \frac{|0.42 - 0.415|}{0.4175} \times 100\% \approx 1.2\% \]
- Compare with a typical criterion of \(10\%\) (or \(20\%\)):
Since \(1.2\% < 10\%\), the difference is within the limits of experimental accuracy. Therefore, the experimental results support the suggested relationship.

### (g) Limitations and Improvements
| Limitation / Source of Uncertainty | Improvement |
| :--- | :--- |
| 1. Two sets of readings are not enough to draw a valid conclusion. | Take multiple sets of readings for different masses and plot a graph of \(y\) against \(m\). |
| 2. Difficult to align ruler vertically to measure \(y\) / parallax error when reading the vertical displacement scale. | Use a set square or plumb line to ensure the vertical ruler is perpendicular, or use a fiducial marker / mirror scale to eliminate parallax. |
| 3. The rule slips on the knife-edge pivot during tilt, altering the pivot position. | Secure the pivot point by drilling a small hole through the center of the rule and using a needle or pin axle, or file a shallow groove on the underside. |
| 4. The spring is no longer vertical when the rule tilts, introducing non-vertical forces. | Use a longer string to attach the spring to the retort stand clamp to keep the spring near-vertical, or adjust the horizontal position of the clamp for each run. |

### Part (a) [2 Marks]
- **[1]** Value of \(w\) measured to \(0.1\text{ mm}\) (or \(0.01\text{ cm}\)) with appropriate unit.
- **[1]** Correct calculation of percentage uncertainty using an absolute uncertainty of \(0.1\text{ mm}\) (or \(0.2\text{ mm}\) if justified by difficult access).

### Part (b) [3 Marks]
- **[1]** Measurement of \(y_1\) recorded to the nearest \(1\text{ mm}\).
- **[1]** Appropriate unit for \(y_1\) (e.g., \(\text{cm}\) or \(\text{mm}\)).
- **[1]** Setup done correctly and \(y_1\) has a realistic value (typically between \(2.0\text{ cm}\) and \(6.0\text{ cm}\)).

### Part (c) [1 Mark]
- **[1]** Correct calculation of percentage uncertainty with absolute uncertainty of \(1\text{ mm}\) to \(3\text{ mm}\) justified by parallax/oscillation.

### Part (d) [3 Marks]
- **[1]** Measurement of \(y_2\) recorded to the nearest \(1\text{ mm}\).
- **[1]** Quality mark: \(y_2 > y_1\) for the larger mass \(m_2 = 200\text{ g}\).
- **[1]** Raw values of both \(y_1\) and \(y_2\) are recorded cleanly without scratch-outs.

### Part (e) [2 Marks]
- **[1]** Correct calculation of \(k_1\) and \(k_2\) with consistent arithmetic.
- **[1]** Units of \(k\) correctly stated (e.g., \(\text{mm g}^{-1}\), \(\text{cm g}^{-1}\), or \(\text{m kg}^{-1}\)) and significant figures of \(k\) match those of the raw values (2 or 3 SF).

### Part (f) [1 Mark]
- **[1]** Correct calculation of percentage difference between \(k_1\) and \(k_2\), leading to a clear conclusion stating whether the relationship is supported, referencing a chosen criterion (usually \(10\%\) or \(20\%\)).

### Part (g) [8 Marks]
- **[4]** One mark for each valid limitation described (up to 4):
- L1: Two sets of readings are not enough.
- L2: Parallax error / difficult to position vertical rule perpendicular to table.
- L3: Knife-edge pivot slips / moves from 50.0 cm mark.
- L4: Spring does not remain vertical when the rule tilts.
- **[4]** One mark for each corresponding, specific improvement suggested (up to 4):
- I1: Take more readings and plot a graph of \(y\) against \(m\).
- I2: Use set square / plumb line to align scale, or use a fiducial marker / mirror scale.
- I3: Locate pivot using a pre-drilled pin hole or physical groove.
- I4: Use a long string to support the spring / adjust clamp horizontally so spring remains vertical.