2026 Cambridge IAS-Level Physics (9702) Practice Paper with Answers

Resistivity \(\rho\) is defined by \(R = \frac{\rho L}{A} \implies \rho = \frac{R A}{L}\).
Resistance \(R = \frac{V}{I}\), where potential difference \(V = \frac{W}{Q} = \frac{W}{I t}\).
In SI base units:
- Work \(W = F s = (m a) s \implies \text{kg}\cdot(\text{m s}^{-2})\cdot\text{m} = \text{kg m}^2 \text{s}^{-2}\)
- Potential difference \(V = \frac{\text{kg m}^2 \text{s}^{-2}}{\text{A s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-1}\)
- Resistance \(R = \frac{\text{kg m}^2 \text{s}^{-3} \text{A}^{-1}}{\text{A}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\)
- Resistivity \(\rho = \frac{R A}{L} = \frac{(\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}) \cdot \text{m}^2}{\text{m}} = \text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\).

A is correct [1 mark].
B is incorrect (unit of potential difference / e.m.f.).
C has incorrect power of seconds (\(\text{s}^{-2}\) instead of \(\text{s}^{-3}\)).
D is the unit of resistance (\(\Omega\)), forgetting to multiply by area and divide by length.

Let the initial speed be \(u\) and the launch angle to the horizontal be \(\theta\).
The initial kinetic energy is \(E_{\text{k}} = \frac{1}{2}mu^2\).
At the highest point, the vertical component of velocity is zero, so the speed is equal to the horizontal component \(v_x = u\cos\theta\).
The kinetic energy at the highest point is:
\(E_{\text{k, top}} = \frac{1}{2}m(u\cos\theta)^2 = \left(\frac{1}{2}mu^2\right)\cos^2\theta = E_{\text{k}}\cos^2\theta\).
Given \(E_{\text{k, top}} = 0.25 E_{\text{k}}\):
\(\cos^2\theta = 0.25 \implies \cos\theta = 0.50 \implies \theta = 60^\circ\).

C is correct [1 mark].
A corresponds to \(\sin\theta = 0.50\) (measuring from the vertical).
B corresponds to \(\cos^2\theta = 0.50\).
D is incorrect.

The forces acting on the submerged block in equilibrium are:
1. Upthrust \(U = \rho_{\text{liquid}} g V = 1.2 \times 10^3 \times 9.81 \times 1.0 \times 10^{-3} = 11.77\text{ N}\) acting upwards.
2. Weight \(W = mg = 0.72 \times 9.81 = 7.06\text{ N}\) acting downwards.
3. Tension \(T\) acting downwards.

For vertical equilibrium:
\(U = W + T \implies T = U - W = 11.77 - 7.06 = 4.71\text{ N} \approx 4.7\text{ N}\).

B is correct [1 mark].
A is calculated using incorrect density difference or arithmetic slip.
C is the weight of the block (\(W = 7.1\text{ N}\)).
D is the total upthrust without subtracting the weight (\(U = 11.8\text{ N}\)).

For a wire obeying Hooke's law, the extension is \(e = \frac{F L}{A E_Y}\) where \(A = \pi r^2\).
The strain energy stored is \(E = \frac{1}{2} F e = \frac{F^2 L}{2 A E_Y} = \frac{F^2 L}{2 \pi r^2 E_Y}\).

For the second wire:
- Force \(F_2 = 2F\)
- Length \(L_2 = 2L\)
- Cross-sectional area \(A_2 = \pi (2r)^2 = 4 \pi r^2 = 4A\)

Strain energy in the second wire:
\(E_2 = \frac{F_2^2 L_2}{2 A_2 E_Y} = \frac{(2F)^2 (2L)}{2 (4A) E_Y} = \frac{4 F^2 \cdot 2L}{8 A E_Y} = \frac{F^2 L}{A E_Y} = 2 \left(\frac{F^2 L}{2 A E_Y}\right) = 2E\).

C is correct [1 mark].
A corresponds to assuming \(E_2 \propto \frac{F L}{r^2}\) linearly without squaring force.
B assumes no change due to cancelation of factors.
D corresponds to neglecting the factor of 4 from the area \(r^2\).

The source moves towards the reflecting wall at speed \(v_{\text{s}} = 30\text{ m s}^{-1}\).
The frequency of the sound waves incident on the wall is given by the Doppler equation for an approaching source:
\(f_{\text{wall}} = f_{\text{s}} \left(\frac{v}{v - v_{\text{s}}}\right) = 720 \times \left(\frac{330}{330 - 30}\right) = 720 \times \frac{330}{300} = 792\text{ Hz}\).

The stationary wall reflects these waves at this same frequency \(792\text{ Hz}\). Since the observer is also stationary relative to the wall, the observer detects the reflected waves at \(792\text{ Hz}\).

D is correct [1 mark].
A is the frequency heard directly from the receding source (\(720 \times \frac{330}{360} = 660\text{ Hz}\)).
B is the unshifted source frequency (\(720\text{ Hz}\)).
C is an incorrect arithmetic calculation.

The grating spacing \(d\) is:
\(d = \frac{1}{N} = \frac{1}{4.0 \times 10^5\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m} = 2500\text{ nm}\).

The condition for principal maxima is \(d \sin\theta = n \lambda\).
The maximum order \(n_{\text{max}}\) occurs when \(\sin\theta \le 1\):
\(n_{\text{max}} \le \frac{d}{\lambda} = \frac{2500\text{ nm}}{600\text{ nm}} \approx 4.17\).

Since \(n\) must be an integer, \(n_{\text{max}} = 4\).
The possible orders are \(n = 0, \pm 1, \pm 2, \pm 3, \pm 4\).
Total number of maxima \(= 2 n_{\text{max}} + 1 = 2(4) + 1 = 9\).

C is correct [1 mark].
A is only the maximum positive order \(n = 4\).
B considers \(n = \pm 1, \pm 2, \pm 3, \pm 4\) but omits the central zero-order maximum (\(n = 0\)).
D incorrectly rounds 4.17 up to 5.

At \(20^\circ\text{C}\), the potential difference across the thermistor is:
\(V_{\text{th},1} = 12.0 \times \frac{18.0}{6.0 + 18.0} = 12.0 \times \frac{18.0}{24.0} = 9.0\text{ V}\).

At \(60^\circ\text{C}\), the potential difference across the thermistor decreases by \(3.0\text{ V}\):
\(V_{\text{th},2} = 9.0 - 3.0 = 6.0\text{ V}\).

Since the total supply voltage is \(12.0\text{ V}\), the potential difference across the \(6.0\text{ k}\Omega\) fixed resistor is:
\(V_R = 12.0 - 6.0 = 6.0\text{ V}\).

Because the potential difference is shared equally (\(6.0\text{ V}\) across each component), the resistance of the thermistor at \(60^\circ\text{C}\) must equal that of the fixed resistor:
\(R_{\text{th},2} = 6.0\text{ k}\Omega\).

C is correct [1 mark].
A is calculated assuming thermistor p.d. drops to 3.0 V (\(12 \times \frac{R}{R+6} = 3 \implies R = 2.0\text{ k}\Omega\)).
B and D are incorrect values from faulty potential divider ratios.

In \(\beta^+\) decay, a proton converts into a neutron:
\(p \to n + e^+ + \nu_e\)

The quark composition of a proton is \(uud\) and of a neutron is \(udd\).
Therefore, an up quark (\(u\)) changes into a down quark (\(d\)):
\(u \to d + e^+ + \nu_e\)

The lepton emitted alongside the positron (\(e^+\)) is an electron neutrino (\(\nu_e\)) to conserve lepton number (\(L = 0 \to -1 + 1 = 0\)).

A is correct [1 mark].
B incorrectly pairs \(\beta^+\) with an antineutrino (violating lepton number conservation).
C and D represent \(\beta^-\) decay where \(d \to u\).

Resistivity is given by \(\rho = \frac{RA}{L}\).
Resistance is \(R = \frac{V}{I} = \frac{W}{Q I} = \frac{W}{I^2 t}\).
In SI base units:
Work \(W\) has units \(\text{kg m}^2 \text{s}^{-2}\).
Current \(I\) has units \(\text{A}\), and time \(t\) has units \(\text{s}\).
Therefore, the unit of resistance is:
\([R] = \frac{\text{kg m}^2 \text{s}^{-2}}{\text{A}^2 \text{s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\).
Substituting into the expression for resistivity:
\([\rho] = (\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}) \times \frac{\text{m}^2}{\text{m}} = \text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\).

B [1 mark]: Correctly derives the base units of resistivity as \(\text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\).

The horizontal velocity remains constant: \(v_x = 15.0\text{ m s}^{-1}\).
The vertical velocity \(v_y\) just before impact is found using \(v_y^2 = u_y^2 + 2gs\):
\(v_y^2 = 0 + 2 \times 9.81 \times 19.6 = 384.55\text{ m}^2\text{ s}^{-2}\).

The final speed \(v\) is the magnitude of the resultant velocity vector:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{15.0^2 + 384.55} = \sqrt{225 + 384.55} = \sqrt{609.55} \approx 24.7\text{ m s}^{-1}\).

C [1 mark]: Calculates vertical velocity component and combines vectorially with horizontal velocity to give \(24.7\text{ m s}^{-1}\).

For a floating object, Upthrust = Weight of object.
Let \(V\) be the total volume of the block and \(\rho_b\) be its density.
In liquid X: \(\text{Upthrust} = \rho_X (0.40 V) g = \rho_b V g \implies \rho_X \times 0.40 = \rho_b\).
In liquid Y: \(\text{Upthrust} = \rho_Y (0.60 V) g = \rho_b V g \implies \rho_Y \times 0.60 = \rho_b\).

Equating both expressions:
\(\rho_X \times 0.40 = \rho_Y \times 0.60\)
\(\frac{\rho_X}{\rho_Y} = \frac{0.60}{0.40} = 1.5\).

D [1 mark]: Correct application of the principle of flotation giving \(1.5\).

Total initial momentum taking right as positive:
\(p_i = (0.60 \times 2.0) + (0.40 \times (-3.0)) = 1.20 - 1.20 = 0\text{ kg m s}^{-1}\).

Since the total momentum is zero, after sticking together the combined mass is stationary (final kinetic energy = \(0\text{ J}\)).

Total initial kinetic energy:
\(E_{k,i} = \frac{1}{2}(0.60)(2.0)^2 + \frac{1}{2}(0.40)(3.0)^2 = 1.2\text{ J} + 1.8\text{ J} = 3.0\text{ J}\).

Therefore, the loss in kinetic energy is \(3.0\text{ J} - 0 = 3.0\text{ J}\).

D [1 mark]: Correct calculation of total initial kinetic energy and recognition that final kinetic energy is zero, giving \(3.0\text{ J}\).

The extension is given by \(x = \frac{F L}{A E}\).
For the first wire: \(A_1 = \frac{\pi d^2}{4}\), so \(x = \frac{F L}{A_1 E}\).
For the second wire: Length is \(2L\) and diameter is \(2d\), so cross-sectional area \(A_2 = 4 A_1\).
Let \(F_2\) be the required force:
\(x = \frac{F_2 (2L)}{A_2 E} = \frac{F_2 (2L)}{(4 A_1) E} = \frac{F_2 L}{2 A_1 E}\).

Equating the expressions for \(x\):
\(\frac{F L}{A_1 E} = \frac{F_2 L}{2 A_1 E} \implies F_2 = 2F\).

C [1 mark]: Uses \(A \propto d^2\) and \(x = \frac{FL}{AE}\) to deduce \(F_2 = 2F\).

For a source moving towards a stationary observer, the observed frequency \(f_o\) is given by:
\(f_o = f_s \left( \frac{v}{v - v_s} \right)\)
where:
\(f_s = 600\text{ Hz}\)
\(v = 330\text{ m s}^{-1}\)
\(v_s = 30\text{ m s}^{-1}\)

\(f_o = 600 \times \left( \frac{330}{330 - 30} \right) = 600 \times \frac{330}{300} = 660\text{ Hz}\).

D [1 mark]: Correct substitution into Doppler effect formula for an approaching source yielding \(660\text{ Hz}\).

Grating spacing \(d = \frac{1}{N} = \frac{1}{4.00 \times 10^5\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m}\).
The condition for diffraction maxima is \(d \sin \theta = n \lambda\).
Since \(\sin \theta \le 1\):
\(n \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}\text{ m}}{532 \times 10^{-9}\text{ m}} = 4.699\).

Thus, the highest observable order is \(n_{\text{max}} = 4\).

The total number of maxima observed is \(2 n_{\text{max}} + 1 = 2(4) + 1 = 9\) (the central zero-order maximum plus 4 on each side).

D [1 mark]: Correctly calculates maximum order \(n=4\) and includes negative orders and zero order to give \(9\).

Using \(E = I(R + r)\):
For \(R = 4.0\,\Omega\): \(E = 2.0(4.0 + r) = 8.0 + 2.0r\) --- (1)
For \(R = 10.0\,\Omega\): \(E = 1.0(10.0 + r) = 10.0 + 1.0r\) --- (2)

Equating (1) and (2):
\(8.0 + 2.0r = 10.0 + 1.0r\)
\(1.0r = 2.0 \implies r = 2.0\,\Omega\).

Substituting \(r\) back into equation (1):
\(E = 8.0 + 2.0(2.0) = 12\text{ V}\).

D [1 mark]: Forms and solves simultaneous equations for internal resistance and e.m.f. to obtain \(12\text{ V}\).

Rearranging the equation for viscosity gives:
\[\eta = \frac{F}{6\pi r v}\]

Substituting the base units for each quantity:
- Force \(F = m a \rightarrow \text{kg m s}^{-2}\)
- Radius \(r \rightarrow \text{m}\)
- Speed \(v \rightarrow \text{m s}^{-1}\)
- \(6\pi\) is a dimensionless constant.

\[[\eta] = \frac{\text{kg m s}^{-2}}{\text{m} \times \text{m s}^{-1}} = \frac{\text{kg m s}^{-2}}{\text{m}^2 \text{s}^{-1}} = \text{kg m}^{-1} \text{s}^{-1}\]

A is correct [1 mark].
B is incorrect (incorrect power of metres).
C is incorrect (base units of force).
D is incorrect (incorrect powers).

The horizontal component of velocity remains constant:
\[v_x = 15\text{ m s}^{-1}\]

For the vertical component of velocity, using \(v_y^2 = u_y^2 + 2gs\) where \(u_y = 0\):
\[v_y^2 = 0 + 2 \times 9.81 \times 45 = 882.9\text{ m}^2\text{ s}^{-2}\]

The resultant speed \(v\) is given by:
\[v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 882.9} = \sqrt{225 + 882.9} = \sqrt{1107.9} \approx 33\text{ m s}^{-1}\]

C is correct [1 mark].
A is incorrect (only considers horizontal speed \(v_x\)).
B is incorrect (only considers vertical speed \(v_y \approx 29.7\text{ m s}^{-1}\)).
D is incorrect (adds speeds directly \(15 + 30 = 45\)).

For an elastic collision, relative speed of approach equals relative speed of separation:
\[u_1 - u_2 = v_2 - v_1\]
\[3.0 - 0 = v_2 - v_1 \implies v_2 = v_1 + 3.0\]

Applying conservation of linear momentum:
\[m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\]
\[(0.40)(3.0) + (0.20)(0) = 0.40 v_1 + 0.20(v_1 + 3.0)\]
\[1.20 = 0.60 v_1 + 0.60\]
\[0.60 = 0.60 v_1 \implies v_1 = +1.0\text{ m s}^{-1}\]

The positive sign indicates it continues moving in its original direction of motion.

A is correct [1 mark].
B is incorrect (reverses direction incorrectly).
C is incorrect (corresponds to an inelastic collision).
D is incorrect (this is the velocity of the \(0.20\text{ kg}\) trolley, \(v_2 = 4.0\text{ m s}^{-1}\)).

Take moments about the hinge \(P\) in equilibrium:
- The weight of the uniform beam acts at its centre of gravity, \(1.2\text{ m}\) from \(P\).
- Clockwise moments: \((180\text{ N} \times 1.2\text{ m}) + (240\text{ N} \times 2.4\text{ m}) = 216 + 576 = 792\text{ N m}\).
- Anticlockwise moment from wire tension: \(T \sin(30^\circ) \times 1.8\text{ m} = T \times 0.50 \times 1.8 = 0.90 T\).

Equating clockwise and anticlockwise moments:
\[0.90 T = 792 \implies T = \frac{792}{0.90} = 880\text{ N}\]

C is correct [1 mark].
A is incorrect (omits \(\sin(30^\circ)\) or factor of 2).
B is incorrect (takes weight of beam at end \(Q\)).
D is incorrect (uses \(\cos(30^\circ)\) instead of \(\sin(30^\circ)\)).

The Young modulus is given by:
\[E = \frac{\sigma}{\varepsilon} = \frac{F/A}{x/L} = \frac{4FL}{\pi d^2 x} \implies x = \frac{4FL}{\pi d^2 E}\]

For the second wire:
\[x_2 = \frac{4 F_2 (2L)}{\pi (2d)^2 E} = \frac{8 F_2 L}{4 \pi d^2 E} = \frac{2 F_2 L}{\pi d^2 E}\]

We require \(x_2 = x\):
\[\frac{2 F_2 L}{\pi d^2 E} = \frac{4 F L}{\pi d^2 E} \implies 2 F_2 = 4 F \implies F_2 = 2F\]

C is correct [1 mark].
A is incorrect (inverts the diameter squared ratio).
B is incorrect (assumes force is unchanged).
D is incorrect (scales directly with area without accounting for length doubling).

When the source approaches the stationary observer:
\[f_{\text{approach}} = f_s \left(\frac{v}{v - v_s}\right) = 750 \left(\frac{340}{340 - 28.0}\right) = 750 \times \frac{340}{312} \approx 817.3\text{ Hz}\]

When the source moves away from the stationary observer:
\[f_{\text{receding}} = f_s \left(\frac{v}{v + v_s}\right) = 750 \left(\frac{340}{340 + 28.0}\right) = 750 \times \frac{340}{368} \approx 692.9\text{ Hz}\]

The total change in observed frequency is:
\[\Delta f = 817.3 - 692.9 = 124.4\text{ Hz} \approx 124\text{ Hz}\]

C is correct [1 mark].
A is incorrect (calculates difference from source frequency to receding frequency \(\approx 57\text{ Hz}\) or half the change).
B is incorrect (calculates only the shift on approach \(817.3 - 750 = 67.3\text{ Hz}\)).
D is incorrect (uses \(f_s \times \frac{2v_s}{v}\) approximation improperly).

1. For an NTC thermistor, increasing temperature causes its resistance \(R_{\text{th}}\) to decrease.
2. The total resistance of the circuit \(R_{\text{total}} = R + R_{\text{th}} + r\) decreases.
3. The current in the circuit \(I = \frac{E}{R_{\text{total}}}\) increases.
4. The potential difference across the fixed resistor is \(V_R = I R\); since \(I\) increases, \(V_R\) increases.
5. The terminal potential difference across the battery is \(V = E - I r\); since \(I\) increases and \(r > 0\), the lost volts \(Ir\) increase, so the terminal p.d. \(V\) decreases.

B is correct [1 mark].
A is incorrect (incorrectly deduces current decreases for \(V_R\)).
C is incorrect (inverts both relationships).
D is incorrect (confuses terminal p.d. with fixed resistor p.d.).

In \(\beta^+\) decay, a proton in the nucleus transforms into a neutron:
\[\text{p} \rightarrow \text{n} + \text{e}^+ + \nu_e\]

1. By lepton number conservation, the emission of a positron (antilepton, \(L = -1\)) must be accompanied by an electron neutrino (lepton, \(L = +1\)), so particle \(X\) is a neutrino (\(\nu_e\)).
2. In terms of quark structure:
- Proton = \(\text{uud}\)
- Neutron = \(\text{udd}\)
Therefore, an up quark changes into a down quark (\(\text{u} \rightarrow \text{d}\)).

D is correct [1 mark].
A is incorrect (describes \(\beta^-\text{ decay}\)).
B is incorrect (identifies \(X\) as an antineutrino).
C is incorrect (quark change is for \(\beta^-\text{ decay}\)).

Rearranging the equation for \(k\):
$$k = \frac{Q \Delta x}{t A \Delta T}$$
Substituting the SI base units for each quantity:
- Energy \(Q\) has base units of \(\text{kg m}^2 \text{s}^{-2}\)
- Thickness \(\Delta x\) has base units of \(\text{m}\)
- Time \(t\) has base units of \(\text{s}\)
- Area \(A\) has base units of \(\text{m}^2\)
- Temperature difference \(\Delta T\) has base units of \(\text{K}\)

Combining these:
$$[k] = \frac{(\text{kg m}^2 \text{s}^{-2})(\text{m})}{(\text{s})(\text{m}^2)(\text{K})} = \frac{\text{kg m}^3 \text{s}^{-2}}{\text{m}^2 \text{s K}} = \text{kg m s}^{-3} \text{K}^{-1}$$

A is correct [1].
B is incorrect (incorrect power of metre and second).
C is incorrect (incorrect power of metre and second).
D is incorrect (power of metre omitted).

The horizontal component of velocity remains constant:
$$v_x = 15\text{ m s}^{-1}$$
For the vertical motion, using \(v_y^2 = u_y^2 + 2gs\) where \(u_y = 0\):
$$v_y^2 = 0 + 2(9.81)(45) = 882.9\text{ m}^2\text{ s}^{-2}$$
$$v_y = \sqrt{882.9} \approx 29.71\text{ m s}^{-1}$$
The resultant speed \(v\) is:
$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 882.9} = \sqrt{225 + 882.9} = \sqrt{1107.9} \approx 33.3\text{ m s}^{-1}$$

B is correct [1].
A is incorrect (calculates only the vertical velocity \(v_y = 29.7\text{ m s}^{-1}\)).
C is incorrect (adds speeds directly without vector addition, \(15 + 29.7 = 44.7\text{ m s}^{-1}\)).
D is incorrect (uses \(2gh\) incorrectly without square root).

Let \(F_1\) be the upward force at the left support (at \(x = 0.50\text{ m}\)) and \(F_2\) be the upward force at the right support (at \(x = 3.5\text{ m}\)).

The distance between the supports is \(3.5 - 0.50 = 3.0\text{ m}\).

Weights of the components:
- Weight of block: \(W_b = mg = 8.0 \times 9.81 = 78.48\text{ N}\), located \(3.5 - 1.0 = 2.5\text{ m}\) from the right support.
- Weight of uniform plank: \(W_p = Mg = 12 \times 9.81 = 117.72\text{ N}\), located at the centre of gravity (\(x = 2.0\text{ m}\)), which is \(3.5 - 2.0 = 1.5\text{ m}\) from the right support.

Taking moments about the right support for rotational equilibrium:
$$F_1 \times 3.0 = (W_b \times 2.5) + (W_p \times 1.5)$$
$$F_1 \times 3.0 = (78.48 \times 2.5) + (117.72 \times 1.5) = 196.2 + 176.58 = 372.78\text{ N m}$$
$$F_1 = \frac{372.78}{3.0} \approx 124\text{ N}$$

C is correct [1].
A is incorrect (calculates the force on the right support \(F_2 \approx 72\text{ N}\)).
B is incorrect (assumes the total weight is shared equally between the two supports).
D is incorrect (calculates total weight \(196\text{ N}\)).

The extension \(e\) of a wire of Young modulus \(E_Y\) is given by:
$$e = \frac{F L}{A E_Y} = \frac{F L}{\pi r^2 E_Y}$$
For the second wire:
$$e_2 = \frac{(2F)(2L)}{\pi (2r)^2 E_Y} = \frac{4 F L}{4 \pi r^2 E_Y} = \frac{F L}{\pi r^2 E_Y} = e$$
So both wires have the same extension \(e\).

The elastic potential energy stored in the first wire is:
$$E = \frac{1}{2} F e$$
The elastic potential energy stored in the second wire is:
$$E_2 = \frac{1}{2} (2F) e_2 = \frac{1}{2} (2F) e = 2 \left(\frac{1}{2} F e\right) = 2E$$

C is correct [1].
A is incorrect (inverts the factor of 2).
B is incorrect (assumes energy stored is unchanged).
D is incorrect (multiplies by 4 assuming area is not squared or force is squared).

Using the Doppler effect formula for a moving source approaching a stationary observer:
$$f_o = f_s \left(\frac{v}{v - v_s}\right)$$
$$f_o = 680 \times \left(\frac{340}{340 - 20.0}\right) = 680 \times \left(\frac{340}{320}\right) = 722.5\text{ Hz} \approx 723\text{ Hz}$$

The wavelength \(\lambda'\) of the compressed waves in front of the source is:
$$\lambda' = \frac{v - v_s}{f_s} = \frac{340 - 20.0}{680} = \frac{320}{680} \approx 0.471\text{ m}$$
(Alternatively, \(\lambda' = \frac{v}{f_o} = \frac{340}{722.5} = 0.471\text{ m}\)).

A is correct [1].
B is incorrect (uses unshifted wavelength \(\lambda = 340/680 = 0.500\text{ m}\)).
C is incorrect (uses Doppler formula for a receding source \(f_o = 642\text{ Hz}\)).
D is incorrect (uses receding source frequency and receding source wavelength).

The grating spacing \(d\) is:
$$d = \frac{1}{4.00 \times 10^5\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m}$$
For diffraction grating maxima:
$$d \sin \theta = n \lambda$$
Since the maximum possible value of \(\sin \theta\) is 1:
$$n_{\text{max}} \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}}{532 \times 10^{-9}} \approx 4.699$$
Therefore, the highest integer order observed is \(n = 4\).

The total number of maxima includes the central maximum (\(n = 0\)) and the \(n = 1, 2, 3, 4\) orders on both sides:
$$\text{Total} = 2n_{\text{max}} + 1 = 2(4) + 1 = 9$$

D is correct [1].
A is incorrect (considers only \(n_{\text{max}} = 4\)).
B is incorrect (considers only one side plus the central maximum, \(4 + 1 = 5\)).
C is incorrect (omits the zero-order central maximum, \(2 \times 4 = 8\)).

The terminal potential difference \(V\) is given by:
$$V = I R = \left(\frac{E}{R + r}\right) R$$
For the first case:
$$8.0 = \frac{4.0 E}{4.0 + r} \implies 8.0(4.0 + r) = 4.0 E \implies 32.0 + 8.0r = 4.0E \implies E = 8.0 + 2.0r$$
For the second case:
$$9.0 = \frac{9.0 E}{9.0 + r} \implies 9.0(9.0 + r) = 9.0 E \implies E = 9.0 + r$$
Equating the two expressions for \(E\):
$$8.0 + 2.0r = 9.0 + r \implies r = 1.0\ \Omega$$
Substituting \(r = 1.0\ \Omega\) into either equation:
$$E = 9.0 + 1.0 = 10.0\text{ V}$$

A is correct [1].
B is incorrect (inconsistent algebraic solution).
C is incorrect (incorrect internal resistance value).
D is incorrect (incorrect e.m.f. value).

In \(\beta^+\) decay, a proton changes into a neutron:
$$p \to n + e^+ + \nu_e$$
Since a proton has quark composition \(uud\) and a neutron has quark composition \(udd\), one up quark transforms into a down quark:
$$u \to d$$
The leptons emitted are a positron (\(e^+\)) and an electron neutrino (\(\nu_e\)) to conserve lepton number (\(L = 0 \to -1 + 1 = 0\)).

A is correct [1].
B is incorrect (an electron antineutrino is emitted in \(\beta^-\), not \(\beta^+\)).
C is incorrect (describes \(\beta^-\) decay where \(d \to u\)).
D is incorrect (describes wrong quark change for \(\beta^+\) decay).

Resistance \(R\) is defined by \(R = \frac{V}{I}\).
Potential difference \(V\) is work done per unit charge, \(V = \frac{W}{Q} = \frac{W}{I t}\).
Therefore, \(R = \frac{W}{I^2 t}\).
In SI base units:
Work \(W = F \times s = (\text{kg}\text{ m}\text{ s}^{-2}) \times \text{m} = \text{kg}\text{ m}^2\text{ s}^{-2}\).
Current is in \(\text{A}\) and time in \(\text{s}\).
Thus, base units of \(R = \frac{\text{kg}\text{ m}^2\text{ s}^{-2}}{\text{A}^2\text{ s}} = \text{kg}\text{ m}^2\text{ s}^{-3}\text{ A}^{-2}\).

A is correct [1 mark].
B: Incorrect power of time and current (represents unit of magnetic flux or volt-second).
C: Missing square on metre (represents electric field per current).
D: Incorrect current power (has \(\text{A}^{-1}\) instead of \(\text{A}^{-2}\)).

Taking upwards as the positive direction:
Initial velocity \(u = +15\text{ m s}^{-1}\)
Acceleration \(a = -9.81\text{ m s}^{-2}\)
Displacement \(s = -45\text{ m}\)

Using the equation of motion:
\(s = ut + \frac{1}{2}at^2\)
\(-45 = 15t - 4.905t^2\)
\(4.905t^2 - 15t - 45 = 0\)

Using the quadratic formula:
\(t = \frac{15 \pm \sqrt{(-15)^2 - 4(4.905)(-45)}}{2(4.905)} = \frac{15 + \sqrt{225 + 882.9}}{9.81} = \frac{15 + 33.285}{9.81} \approx 4.92\text{ s}\).

C is correct [1 mark].
A (2.1 s): Miscalculates time to reach maximum height and adds it incorrectly.
B (3.1 s): Time to fall 45 m from rest.
D (6.2 s): Arithmetic error in quadratic solution.

Let \(A\) be the cross-sectional area of the block.
Initially, the block is floating in water:
Weight of block \(W = \rho_w g A h_1 = 1000 \times g \times A \times 0.100\).

When fully submerged in two layers, let \(x\) be the depth in water. The depth in oil is \(0.120 - x\).
Total upthrust \(U = \rho_w g A x + \rho_{\text{oil}} g A (0.120 - x)\).
At equilibrium, \(U = W\):
\(1000 g A x + 750 g A (0.120 - x) = 1000 g A (0.100)\)
Dividing through by \(g A\):
\(1000x + 750(0.120 - x) = 100\)
\(1000x + 90 - 750x = 100\)
\(250x = 10\)
\(x = \frac{10}{250} = 0.040\text{ m} = 4.0\text{ cm}\).

B is correct [1 mark].
A (2.5 cm): Calculated using ratio of densities without considering total height constraint.
C (6.0 cm): Assumes half the block is in water.
D (8.0 cm): Subtracts oil layer incorrectly.

Strain energy stored in an elastic wire is given by:
\(U = \frac{1}{2} F e\)
Extension \(e = \frac{F L}{A E}\)
Substituting \(e\) into the strain energy formula:
\(U = \frac{F^2 L}{2 A E}\)

For the second wire with length \(2L\), area \(2A\), and force \(F_2\):
\(U_2 = \frac{F_2^2 (2L)}{2 (2A) E} = \frac{F_2^2 L}{2 A E}\)

Since \(U_2 = U\):
\(\frac{F_2^2 L}{2 A E} = \frac{F^2 L}{2 A E} \implies F_2 = F\).

B is correct [1 mark].
A: Assumes energy is directly proportional to force.
C: Arises from forgetting that both length and area are doubled.
D: Assumes doubling dimension doubles the required force.

Grating spacing \(d = \frac{1 \times 10^{-3}\text{ m}}{400} = 2.50 \times 10^{-6}\text{ m}\).
Using the grating equation \(d \sin \theta = n \lambda\):
\(\lambda = \frac{d \sin \theta}{n} = \frac{2.50 \times 10^{-6} \times \sin(42.0^\circ)}{3}\)
\(\sin(42.0^\circ) = 0.66913\)
\(\lambda = \frac{2.50 \times 10^{-6} \times 0.66913}{3} = 5.58 \times 10^{-7}\text{ m} = 558\text{ nm}\).

C is correct [1 mark].
A (186 nm): Divided by \(n^2 = 9\) instead of \(n = 3\).
B (372 nm): Used order \(n = 2\) instead of \(n = 3\).
D (836 nm): Multiplied by 1.5 due to inverted order substitution.

Electric current is given by \(I = A n v q\), where \(A = \frac{\pi d^2}{4}\).
Rearranging for drift speed \(v\):
\(v = \frac{I}{n q A} = \frac{4 I}{n q \pi d^2} \propto \frac{I}{d^2}\)

For the second wire:
\(v_2 = v_1 \times \left(\frac{I_2}{I_1}\right) \times \left(\frac{d_1}{d_2}\right)^2 = v \times 3 \times \left(\frac{1}{2}\right)^2 = \frac{3}{4} v = 0.75 v\).

B is correct [1 mark].
A (0.38 v): Used \(2^3 = 8\) in denominator.
C (1.33 v): Inverted ratio \(4/3\).
D (1.50 v): Failed to square the diameter ratio (used \(3/2\)).

Current in first case: \(I_1 = \frac{V_1}{R_1} = \frac{8.0}{4.0} = 2.0\text{ A}\).
Using \(E = V + I r\):
\(E = 8.0 + 2.0 r\) --- (equation 1)

Current in second case: \(I_2 = \frac{V_2}{R_2} = \frac{10.8}{9.0} = 1.2\text{ A}\).
\(E = 10.8 + 1.2 r\) --- (equation 2)

Equating (1) and (2):
\(8.0 + 2.0 r = 10.8 + 1.2 r\)
\(0.8 r = 2.8 \implies r = 3.5\ \Omega\)

Substitute \(r\) back into equation 1:
\(E = 8.0 + 2.0(3.5) = 8.0 + 7.0 = 15.0\text{ V}\).

C is correct [1 mark].
A (12.0 V): Calculated e.m.f. assuming internal resistance is negligible.
B (13.5 V): Arithmetic error in setting up current difference.
D (18.0 V): Added terminal voltages directly.

In \(\beta^+\) decay, a proton is converted into a neutron inside the nucleus:
\(p \rightarrow n + \beta^+ + \nu_e\)
Proton composition is \(uud\) and neutron composition is \(udd\).
Thus, an up quark (\(u\)) changes into a down quark (\(d\)).
The emitted particles are a positron (\(\beta^+\)) and an electron neutrino (\(\nu_e\)) to conserve lepton number.

A is correct [1 mark].
B: Incorrect neutrino (\(\overline{\nu}_e\) is emitted in \(\beta^-\), not \(\beta^+\)).
C: Describes \(\beta^-\) decay.
D: Incorrect quark change for \(\beta^+\) decay.

Precision refers to the closeness of agreement between independent measurements of the same quantity obtained under stipulated conditions (indicating small random errors / small spread). Accuracy refers to how close a measured value (or the mean of repeated measurements) is to the true value of the quantity being measured (indicating small systematic errors).

B1: Precision: closeness of repeated readings to each other / size of random errors.
B0.5: Accuracy: closeness of the measured value(s) / average value to the true value (or value with small systematic error).
[Total: 1.5 marks]

A couple consists of two equal and opposite parallel forces whose lines of action do not coincide. The torque of a couple is defined as the product of the magnitude of one of the forces and the perpendicular distance between the lines of action of the two forces.

B1: Product of (one of) the force(s) and the perpendicular distance.
B0.5: between the lines of action of the forces.
[Total: 1.5 marks]

The Young modulus \(E\) of a material is defined as the ratio of tensile stress \(\sigma\) to tensile strain \(\varepsilon\), provided the limit of proportionality is not exceeded:
\[ E = \frac{\text{stress}}{\text{strain}} \]

B1: Ratio of (tensile/compressive) stress to (tensile/compressive) strain.
B0.5: Explicit mention of stress defined as force per unit cross-sectional area and strain as extension per unit original length (or symbols clearly defined).
[Total: 1.5 marks]

In a transverse wave, the particles of the medium vibrate or oscillate at right angles (perpendicular) to the direction of energy transfer (or wave propagation).

B1: Oscillations / vibrations are perpendicular to the direction of wave travel / energy transfer.
B0.5: Clear reference to transfer of energy / progressive nature.
[Total: 1.5 marks]

Coherent waves have a constant phase relationship (constant phase difference) over time. To maintain a constant phase difference, they must also have the same frequency (or wavelength).

B1: Constant phase difference.
B0.5: Same frequency / wavelength.
[Total: 1.5 marks]

Electromotive force (e.m.f.) is defined as the electrical energy produced per unit charge transferred from non-electrical forms into electrical form:
\[ E = \frac{W}{Q} \]

B1: Energy converted from other forms into electrical energy per unit charge.
B0.5: Reference to charge passing through the source / complete circuit (or \(E = W/Q\) with \(W\) and \(Q\) fully defined).
[Total: 1.5 marks]

A neutron consists of one up quark (charge \(+\frac{2}{3}e\)) and two down quarks (each of charge \(-\frac{1}{3}e\)), giving total charge 0. Because it is a hadron made of three quarks, it is classified as a baryon.

B1: (quark composition) udd / up, down, down.
B0.5: (classification) baryon.
[Total: 1.5 marks]

1. Calculate the vertical component of velocity \(v_y\) just before impact:
\[ v_y^2 = u_y^2 + 2gs = 0 + 2 \times 9.81 \times 45.0 = 882.9 \implies v_y = \sqrt{882.9} \approx 29.71\text{ m s}^{-1} \]

2. The horizontal component of velocity remains constant:
\[ v_x = 18.0\text{ m s}^{-1} \]

3. Calculate the magnitude of the resultant velocity \(v\):
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{18.0^2 + 882.9} = \sqrt{324 + 882.9} = \sqrt{1206.9} \approx 34.74\text{ m s}^{-1} \approx 34.7\text{ m s}^{-1} \]

[C1] for calculating vertical velocity component \(v_y = 29.7\text{ m s}^{-1}\) (or \(v_y^2 = 883\text{ m}^2\text{ s}^{-2}\))
[C1] for combining perpendicular components using Pythagoras' theorem: \(v = \sqrt{v_x^2 + v_y^2}\)
[A0.5] for final answer \(34.7\text{ m s}^{-1}\) (accept 2 or 3 s.f., \(35\text{ m s}^{-1}\) or \(34.7\text{ m s}^{-1}\))

1. Calculate the volume \(V\) of the cylinder:
\[ V = A \times L = (4.50 \times 10^{-3}) \times 0.120 = 5.40 \times 10^{-4}\text{ m}^3 \]

2. Calculate the upthrust \(U\) acting on the cylinder:
\[ U = \rho_{\text{liquid}} g V = 1.15 \times 10^3 \times 9.81 \times 5.40 \times 10^{-4} = 6.092\text{ N} \]

3. Determine the weight \(W\) and mass \(m\) of the cylinder:
\[ W = T + U = 3.80 + 6.092 = 9.892\text{ N} \]
\[ m = \frac{W}{g} = \frac{9.892}{9.81} = 1.008\text{ kg} \]

4. Calculate the density of the cylinder \(\rho\):
\[ \rho = \frac{m}{V} = \frac{1.008}{5.40 \times 10^{-4}} \approx 1.867 \times 10^3\text{ kg m}^{-3} \approx 1.87 \times 10^3\text{ kg m}^{-3} \]

[C1] for calculating volume \(V = 5.40 \times 10^{-4}\text{ m}^3\) and upthrust \(U = 6.09\text{ N}\)
[C1] for equating forces \(W = T + U\) to find mass \(m = 1.01\text{ kg}\) (or \(W = 9.89\text{ N}\))
[A0.5] for final density \(1.87 \times 10^3\text{ kg m}^{-3}\) (accept \(1870\text{ kg m}^{-3}\))

1. Calculate cross-sectional area \(A\):
\[ A = \pi r^2 = \pi \times (0.350 \times 10^{-3})^2 = 3.8485 \times 10^{-7}\text{ m}^2 \]

2. Calculate the extension \(x\) using \(E = \frac{FL}{Ax}\):
\[ x = \frac{FL}{EA} = \frac{85.0 \times 2.40}{(2.00 \times 10^{11}) \times (3.8485 \times 10^{-7})} = \frac{204}{7.697 \times 10^4} = 2.650 \times 10^{-3}\text{ m} \]

3. Calculate the elastic potential energy \(E_p\):
\[ E_p = \frac{1}{2} F x = 0.5 \times 85.0 \times (2.650 \times 10^{-3}) = 0.1126\text{ J} \approx 0.113\text{ J} \]

[C1] for calculating cross-sectional area \(A = 3.85 \times 10^{-7}\text{ m}^2\)
[C1] for calculating extension \(x = 2.65 \times 10^{-3}\text{ m}\) using \(x = FL / (EA)\)
[A0.5] for final energy \(0.113\text{ J}\)

1. Use the Doppler effect formula for an approaching source:
\[ f_o = f_s \left( \frac{v}{v - v_s} \right) \]
\[ f_o = 750 \times \left( \frac{340}{340 - 28.0} \right) = 750 \times \left( \frac{340}{312} \right) \approx 817.31\text{ Hz} \]

2. Calculate the difference in frequency \(\Delta f\):
\[ \Delta f = f_o - f_s = 817.31 - 750 = 67.31\text{ Hz} \approx 67.3\text{ Hz} \]

[C1] for using \(f_o = f_s v / (v - v_s)\) with correct substitution of signs
[C1] for observed frequency \(f_o = 817\text{ Hz}\) (or \(817.3\text{ Hz}\))
[A0.5] for frequency difference \(67.3\text{ Hz}\) (accept \(67\text{ Hz}\))

1. Calculate the fringe separation \(x\) between adjacent maxima:
\[ x = \frac{\lambda D}{a} = \frac{(590 \times 10^{-9}) \times 1.80}{0.450 \times 10^{-3}} = 2.36 \times 10^{-3}\text{ m} \]

2. The distance \(y_n\) from the central maximum to the \(n\)-th dark fringe is given by \(y_n = (n - 0.5) x\).
For the 3rd dark fringe (\(n = 3\)):
\[ y_3 = 2.5 x = 2.5 \times 2.36 \times 10^{-3}\text{ m} = 5.90 \times 10^{-3}\text{ m} \text{ (or } 5.90\text{ mm)} \]

[C1] for calculating fringe separation \(x = 2.36 \times 10^{-3}\text{ m}\)
[C1] for identifying the third dark fringe at \(2.5 x\) from the central maximum
[A0.5] for final distance \(5.90 \times 10^{-3}\text{ m}\) (or \(5.90\text{ mm}\))

1. Calculate the resistance \(R\) of the wire:
\[ R = \frac{\rho L}{A} = \frac{(1.70 \times 10^{-8}) \times 15.0}{1.20 \times 10^{-6}} = 0.2125\,\Omega \]

2. Calculate the current \(I\) in the wire:
\[ I = \frac{V}{R} = \frac{0.425}{0.2125} = 2.00\text{ A} \]

3. Calculate the drift velocity \(v\) using \(I = A n v q\):
\[ v = \frac{I}{n A e} = \frac{2.00}{(8.50 \times 10^{28}) \times (1.20 \times 10^{-6}) \times (1.60 \times 10^{-19})} \]
\[ v = \frac{2.00}{16320} \approx 1.2255 \times 10^{-4}\text{ m s}^{-1} \approx 1.23 \times 10^{-4}\text{ m s}^{-1} \]

[C1] for calculating resistance \(R = 0.2125\,\Omega\) and current \(I = 2.00\text{ A}\)
[C1] for rearrangement and substitution into \(I = Anvq\)
[A0.5] for final drift speed \(1.23 \times 10^{-4}\text{ m s}^{-1}\)

1. Calculate the circuit current \(I\) from the fixed resistor:
\[ I = \frac{V_1}{R_1} = \frac{4.80}{18.0} = 0.2667\text{ A} \]

2. Calculate the total circuit resistance \(R_{\text{total}}\):
\[ R_{\text{total}} = \frac{E}{I} = \frac{9.00}{0.2667} = 33.75\,\Omega \]

3. Calculate the resistance of the variable resistor \(R_v\):
\[ R_{\text{total}} = R_1 + R_v + r \implies 33.75 = 18.0 + R_v + 1.50 \]
\[ R_v = 33.75 - 19.50 = 14.25\,\Omega \approx 14.3\,\Omega \]

[C1] for calculating current \(I = 0.267\text{ A}\) (or \(4/15\text{ A}\))
[C1] for applying \(E = I(R_1 + R_v + r)\) or calculating total resistance \(R_{\text{total}} = 33.75\,\Omega\)
[A0.5] for final answer \(14.3\,\Omega\) (accept \(14.2\,\Omega\) to \(14.3\,\Omega\))

1. Calculate the acceleration \(a\) and final velocity \(v\) of the crate:
\[ s = ut + \frac{1}{2}at^2 \implies 12.0 = 0 + \frac{1}{2} a (4.00)^2 \implies a = 1.50\text{ m s}^{-2} \]
\[ v = u + at = 0 + (1.50)(4.00) = 6.00\text{ m s}^{-1} \]

2. Calculate the total useful work done \(W\) (gain in \(E_p\) + gain in \(E_k\)):
\[ \Delta E_p = mgh = 45.0 \times 9.81 \times 12.0 = 5297.4\text{ J} \]
\[ \Delta E_k = \frac{1}{2}mv^2 = 0.5 \times 45.0 \times (6.00)^2 = 810.0\text{ J} \]
\[ W = 5297.4 + 810.0 = 6107.4\text{ J} \]

3. Calculate the useful power output \(P_{\text{out}}\) and electrical input power \(P_{\text{in}}\):
\[ P_{\text{out}} = \frac{W}{t} = \frac{6107.4}{4.00} = 1526.85\text{ W} \]
\[ P_{\text{in}} = \frac{P_{\text{out}}}{0.680} = \frac{1526.85}{0.680} \approx 2245.4\text{ W} \approx 2.25 \times 10^3\text{ W} \]

[C1] for finding acceleration \(a = 1.50\text{ m s}^{-2}\) and final velocity \(v = 6.00\text{ m s}^{-1}\) (or total tension \(F = 509\text{ N}\))
[C1] for total useful mechanical energy \(6110\text{ J}\) or useful power output \(1530\text{ W}\)
[A0.5] for final electrical input power \(2.25 \times 10^3\text{ W}\) (or \(2250\text{ W}\))

1. Calculate the volume of the wooden block:
\[V = A \times h = 1.60 \times 10^{-2} \times 0.250 = 4.00 \times 10^{-3}\text{ m}^3\]

2. Calculate the upward buoyant force (upthrust) \(U\) exerted by the water:
\[U = \rho_{\text{water}} \times V \times g = 1.00 \times 10^3 \times 4.00 \times 10^{-3} \times 9.81 = 39.24\text{ N}\]

3. Calculate the weight \(W\) of the block:
\[W = m g = \rho_{\text{wood}} \times V \times g = 620 \times 4.00 \times 10^{-3} \times 9.81 = 24.33\text{ N}\]

4. Set up the equilibrium condition for the submerged block:
\[U = W + T \implies T = U - W\]
\[T = 39.24 - 24.33 = 14.91\text{ N} \approx 14.9\text{ N}\]

C1: Calculation of volume \(V = 4.00 \times 10^{-3}\text{ m}^3\) and determination of upthrust \(U = 39.2\text{ N}\) or weight \(W = 24.3\text{ N}\)
C1: Correct formulation of equilibrium condition \(T = U - W\) or \(T = (\rho_{\text{water}} - \rho_{\text{wood}}) V g\)
A0.5: Correct final value of tension given to 2 or 3 significant figures: \(14.9\text{ N}\) (allow \(15\text{ N}\))

1. At release, velocity is zero, so drag force is zero; the resultant downward force is maximum, giving maximum acceleration.
2. As speed increases, the upward viscous drag force increases, which reduces the resultant downward force (\[F_{\text{net}} = W - U - F_{\text{drag}}\]).
3. According to Newton's second law (\[a = F_{\text{net}} / m\]), the acceleration progressively decreases until viscous drag plus upthrust equals weight, at which point resultant force and acceleration become zero (terminal velocity).

B1: Acceleration decreases (from maximum) to zero / becomes zero.
B1: Viscous drag increases (with speed / velocity), so resultant force decreases (until drag + upthrust = weight / resultant force is zero).

1. The unloading graph is a straight line parallel to the initial Hooke's law (elastic) region of the loading graph.
2. It meets the extension axis at a non-zero value because the wire has been plastically deformed; atomic planes have slipped past each other permanently, leaving a permanent extension.

B1: (Unloading line is a) straight line parallel to the initial linear / elastic line (intercepting the extension axis at a value greater than zero).
B1: Wire has undergone plastic deformation / permanent extension / atomic planes have slid (over each other).

1. The \(I\)–\(V\) characteristic is a curve passing through the origin with a decreasing gradient as \(V\) (or \(I\)) increases in magnitude.
2. As current increases, electrical energy is dissipated as thermal energy, causing the temperature of the metal filament to increase.
3. Increased lattice vibrations cause greater rate of collisions with charge carriers, raising resistance (\(R = V/I\)), which corresponds to a decreasing gradient (\(I/V\)).

B1: Curve through the origin with decreasing gradient / rate of increase of current decreases (symmetrical in both quadrants).
B1: Higher current increases temperature, which increases resistance (due to increased lattice vibrations / electron-ion collisions).

1. Particles situated in adjacent loops separated by a single node oscillate in antiphase, so the phase difference is \(\pi\text{ rad}\) (or \(180^\circ\)).
2. A displacement node occurs where the two constituent progressive waves traveling in opposite directions superpose with a phase difference of \(180^\circ\) (in antiphase) at all times, causing complete destructive superposition (cancellation).

B1: Phase difference is \(180^\circ\) or \(\pi\text{ rad}\) (allow: in antiphase).
B1: Two waves meet / superpose in antiphase / have a phase difference of \(180^\circ\) producing destructive interference / cancellation.

1. Hydrostatic pressure in a fluid increases with depth according to \(\Delta p = \rho g \Delta h\), so the pressure on the lower face is greater than the pressure on the upper face.
2. Because force is \(F = p \times A\) and both faces have equal surface area, the upward force exerted by the water on the lower face exceeds the downward force on the upper face, creating a net upward force (upthrust).

B1: Pressure on bottom/lower surface is greater than pressure on top/upper surface (because pressure increases with depth).
B1: Upward force on lower face is greater than downward force on upper face, producing a net upward force.

1. In \(\beta^-\)-decay, a fixed total amount of energy (equal to the mass defect between parent and daughter nuclei) is released.
2. Three particles share the kinetic energy and momentum: the recoil daughter nucleus, the \(\beta^-\)-particle (electron), and an electron antineutrino (\(\bar{\nu}_e\)).
3. Because the emitted energy is shared between the electron and the antineutrino in variable proportions, the \(\beta^-\)-particle can have any kinetic energy up to a maximum value.

B1: An (electron) antineutrino is also emitted (along with the beta particle and daughter nucleus).
B1: The (fixed total) decay energy is shared between the beta particle and the (anti)neutrino in variable proportions.

### (a) Initial Measurement
- Mass $m = 100\text{ g} = 0.100\text{ kg}$, weight $W = mg = 0.100 \times 9.81 = 0.981\text{ N}$.
- For a uniform metre rule of mass $M \approx 0.122\text{ kg}$ (weight $W_{\text{rule}} \approx 1.20\text{ N}$ acting at $50.0\text{ cm}$):
Taking moments about the pivot string at $d_1 = 10.0\text{ cm}$:
$$T(d_2 - d_1) = W(x - d_1) + W_{\text{rule}}(50.0 - d_1)$$
$$T(80.0) = (0.981)(20.0 - 10.0) + (1.20)(50.0 - 10.0)$$
$$T(80.0) = 9.81 + 48.0 = 57.81 \implies T \approx 0.72\text{ N}$$

### (b) Table of Results
Six sets of readings across a range spanning from $20.0\text{ cm}$ to $80.0\text{ cm}$:

| $x / \text{cm}$ | $(x - d_1) / \text{cm}$ | $T_1 / \text{N}$ | $T_2 / \text{N}$ | $\langle T \rangle / \text{N}$ |
| :---: | :---: | :---: | :---: | :---: |
| 20.0 | 10.0 | 0.72 | 0.74 | 0.73 |
| 30.0 | 20.0 | 0.85 | 0.85 | 0.85 |
| 40.0 | 30.0 | 0.96 | 0.98 | 0.97 |
| 50.0 | 40.0 | 1.10 | 1.08 | 1.09 |
| 65.0 | 55.0 | 1.28 | 1.26 | 1.27 |
| 80.0 | 70.0 | 1.46 | 1.46 | 1.46 |

### (c) Graph Plotting
- Axes: Scaled so points occupy $> 50\%$ of grid along both dimensions. Scales clearly labeled with quantity and unit.
- Plotting: Points plotted precisely to within $\pm 0.5$ small square.
- Line of Best Fit: Balanced number of points above and below the drawn straight line, no kinks/thick lines.

### (d) Gradient and $y$-intercept
- Points chosen on the line: $(x_1, y_1) = (10.0\text{ cm}, 0.725\text{ N})$ and $(x_2, y_2) = (70.0\text{ cm}, 1.460\text{ N})$ (where hypotenuse is $> 50\%$ of the plotted line length):
$$\text{gradient} = \frac{1.460 - 0.725}{70.0 - 10.0} = \frac{0.735}{60.0} = 0.01225\text{ N cm}^{-1} = 1.225\text{ N m}^{-1}$$
- $y$-intercept:
$$y = mx + c \implies c = 0.725 - (0.01225 \times 10.0) = 0.6025\text{ N} \approx 0.60\text{ N}$$

### (e) Constants and SI Units
(i) Comparing $T = \frac{W}{L}(x - d_1) + T_0$ with $y = mx + c$:
- Gradient $m = \frac{W}{L} \implies W = m \times L = 1.225\text{ N m}^{-1} \times 0.800\text{ m} = 0.98\text{ N}$
- Intercept $c = T_0 = 0.60\text{ N}$

(ii) SI base units of gradient:
$$\text{gradient} = \frac{\text{Force}}{\text{length}} = \frac{\text{N}}{\text{m}} = \frac{\text{kg m s}^{-2}}{\text{m}} = \text{kg s}^{-2}$$

### (a) Initial measurement [1 mark]
- A1: Value of $T$ recorded with appropriate unit ($\text{N}$) to the precision of the newton meter (typically to $0.02\text{ N}$ or $0.05\text{ N}$).

### (b) Table of results [6 marks]
- B1 [Range]: Range of values of $x$ covers at least $x_{\text{max}} - x_{\text{min}} \geqslant 50.0\text{ cm}$.
- B1 [Six sets]: At least six sets of values of $(x, T)$ recorded without assistance.
- B1 [Column headings]: Each column heading contains quantity and valid unit in standard slash or bracket notation (e.g., $x / \text{cm}$, $\langle T \rangle / \text{N}$, $(x - d_1) / \text{cm}$).
- B1 [Consistency of raw data]: All raw values of $x$ recorded to the same precision (nearest $1\text{ mm}$ / $0.1\text{ cm}$) and raw $T$ values recorded to consistent precision of the instrument.
- B1 [Repetition]: Evidence of repeat trials for $T$ at each position and correctly calculated mean $\langle T \rangle$.
- B1 [Significant figures in calculated values]: Number of significant figures in $(x - d_1)$ is identical to or one more than the significant figures in raw $x$.

### (c) Graph plotting [4 marks]
- B1 [Axes & scales]: Sensible linear scales without awkward multipliers ($1:3$, $1:7$, etc.). Plotted data points span more than half of the grid in both $x$ and $y$ directions.
- B1 [Plotting]: Points plotted accurately within $\pm 0.5$ of a small square.
- B1 [Line of best fit]: Straight line of best fit drawn with even distribution of points on either side; line is thin, continuous, and drawn with a ruler.
- B1 [Quality of data]: All plotted points lie close to the trend line (scatter within $\pm 0.05\text{ N}$).

### (d) Gradient and Intercept [4 marks]
- M1 [Gradient method]: Clear triangle shown on graph where the length of the hypotenuse is greater than half the length of the drawn line.
- A1 [Gradient value]: Accurate gradient calculation consistent with candidate's graph points.
- M1 [Intercept method]: Value determined either by direct readout from a true $(0,0)$ axis or by correct algebraic substitution $c = y_1 - m x_1$ using a point on the line.
- A1 [Intercept value]: Accurate value of $y$-intercept determined with correct sign and appropriate significant figures.

### (e) Determination of constants and units [5 marks]
- A1: Correct evaluation of $W$ using $W = \text{gradient} \times 0.800\text{ m}$ (with ECF from (d)).
- A1: Correct evaluation of $T_0$ equating directly to the $y$-intercept value from (d).
- A1: Both values of $W$ and $T_0$ given with valid units ($\text{N}$) and given to 2 or 3 significant figures.
- M1: Working shown expressing $\text{N m}^{-1}$ in terms of base units ($\text{N} = \text{kg m s}^{-2}$).
- A1: Correct SI base units of gradient: $\text{kg s}^{-2}$ (or $\text{kg}\cdot\text{s}^{-2}$).