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Initially, the total electromotive force (e.m.f.) of the power supply is \( V_{\text{total}} = 12\text{ V} \). The potential difference across the thermistor is \( V_{\text{th}} = 8.0\text{ V} \). Therefore, the potential difference across the fixed resistor is \( V_R = V_{\text{total}} - V_{\text{th}} = 12\text{ V} - 8.0\text{ V} = 4.0\text{ V} \).

Since the thermistor and the resistor are in series, the ratio of their potential differences is equal to the ratio of their resistances:
\( \frac{V_{\text{th}}}{V_R} = \frac{R_{\text{th}}}{R} \)
\( \frac{8.0\text{ V}}{4.0\text{ V}} = \frac{4.0\text{ k}\Omega}{R} \implies R = 2.0\text{ k}\Omega \).

At the higher temperature, the potential difference across the thermistor decreases to \( V_{\text{th}}' = 4.0\text{ V} \). The potential difference across the fixed resistor becomes:
\( V_R' = 12\text{ V} - 4.0\text{ V} = 8.0\text{ V} \).

Using the same ratio with the new values:
\( \frac{R_{\text{th}}'}{R} = \frac{V_{\text{th}}'}{V_R'} \)
\( \frac{R_{\text{th}}'}{2.0\text{ k}\Omega} = \frac{4.0\text{ V}}{8.0\text{ V}} \implies R_{\text{th}}' = 1.0\text{ k}\Omega \).

1 mark for the correct answer B.
- 1 mark for calculating the value of the fixed resistance R = 2.0 k\Omega.
- 1 mark for calculating the new resistance of the thermistor under the new potential split.

First, we use the conservation of linear momentum to find the final velocity \( v_f \) of the combined system. Let the direction to the right be positive.
Initial total momentum:
\( p_i = (3m)(v) + (m)(-2v) = 3mv - 2mv = mv \)

After the collision, the two trolleys stick together, so their combined mass is \( M = 3m + m = 4m \). Let \( v_f \) be their joint final velocity:
\( 4m v_f = mv \implies v_f = \frac{v}{4} \)

Now, calculate the initial kinetic energy \( E_{ki} \):
\( E_{ki} = \frac{1}{2}(3m)v^2 + \frac{1}{2}(m)(-2v)^2 = 1.5mv^2 + 2mv^2 = 3.5mv^2 = \frac{7}{2}mv^2 \)

Calculate the final kinetic energy \( E_{kf} \):
\( E_{kf} = \frac{1}{2}(4m)\left(\frac{v}{4}\right)^2 = 2m \left(\frac{v^2}{16}\right) = \frac{1}{8}mv^2 \)

Calculate the loss in kinetic energy \( \Delta E_k \):
\( \Delta E_k = E_{ki} - E_{kf} = \frac{28}{8}mv^2 - \frac{1}{8}mv^2 = \frac{27}{8}mv^2 \)

Find the fraction of the initial kinetic energy that is lost:
\( \text{Fraction lost} = \frac{\Delta E_k}{E_{ki}} = \frac{\frac{27}{8}mv^2}{\frac{7}{2}mv^2} = \frac{27}{8} \times \frac{2}{7} = \frac{27}{28} \).

1 mark for the correct answer D.
- 1 mark for finding the correct final velocity from conservation of momentum.
- 1 mark for finding the correct ratio of the kinetic energy loss to initial kinetic energy.

The Young modulus \( E \) is a property of the material and is defined as:
\( E = \frac{\text{stress}}{\text{strain}} \implies \text{strain} = \frac{\text{stress}}{E} \)

Since both wires are made of the same metal, they have the same Young modulus \( E \).
Stress is defined as \( \sigma = \frac{F}{A} \), where \( F \) is the applied load and \( A \) is the cross-sectional area of the wire. The cross-sectional area of a wire with diameter \( d \) is \( A = \frac{\pi d^2}{4} \).

Let the diameter of wire Q be \( d \). Thus, the diameter of wire P is \( 2d \).
The cross-sectional area of Q is \( A_Q \), and the cross-sectional area of P is \( A_P = 4A_Q \).

Since both wires support the same load \( F \):
\( \text{stress in P} = \frac{F}{A_P} = \frac{F}{4A_Q} \)
\( \text{stress in Q} = \frac{F}{A_Q} \)

Since strain is proportional to stress (as \( E \) is constant):
\( \frac{\text{strain in P}}{\text{strain in Q}} = \frac{\text{stress in P}}{\text{stress in Q}} = \frac{\frac{F}{4A_Q}}{\frac{F}{A_Q}} = \frac{1}{4} = 0.25 \).

Note that the original length of the wires does not affect their strain under a given stress.

1 mark for the correct answer B.
- 1 mark for showing that strain is independent of initial length for a given stress and material.
- 1 mark for identifying that the area ratio is 4, leading to a stress (and strain) ratio of 0.25.

The double-slit formula for fringe separation is given by:
\( x = \frac{\lambda D}{a} \)

Initially, for red light:
\( x = \frac{\lambda_1 D_1}{a} \implies D_1 = \frac{a x}{\lambda_1} \)

Finally, for green light:
\( x = \frac{\lambda_2 D_2}{a} \implies D_2 = \frac{a x}{\lambda_2} \)

Since \( a \) and the required fringe separation \( x \) remain constant:
\( \lambda_1 D_1 = \lambda_2 D_2 \)

Substitute the given wavelengths:
\( 640 \times D_1 = 480 \times D_2 \implies D_2 = \frac{640}{480} D_1 = \frac{4}{3} D_1 \approx 1.33 D_1 \)

To find the percentage change in \( D \):
\( \text{Percentage change} = \frac{D_2 - D_1}{D_1} \times 100\% = \frac{\frac{4}{3} D_1 - D_1}{D_1} \times 100\% = \frac{1}{3} \times 100\% \approx +33\% \)

Therefore, the distance \( D \) must be increased by \( 33\% \).

1 mark for the correct answer D.
- 1 mark for establishing the relationship between \( \lambda \) and \( D \) for a constant fringe separation.
- 1 mark for calculating the percentage increase of 33%.

The gravitational potential \( \phi \) at a distance \( r \) from a point mass \( M \) is given by:
\( \phi = -\frac{GM}{r} \)

Therefore, at radius \( R \):
\( \phi_1 = -\frac{GM}{R} \implies \frac{GM}{R} = -\phi_1 \)

At radius \( 3R \):
\( \phi_2 = -\frac{GM}{3R} = \frac{1}{3} \left( -\frac{GM}{R} \right) = \frac{1}{3} \phi_1 \)

Now, calculate the change in gravitational potential, \( \phi_2 - \phi_1 \):
\( \phi_2 - \phi_1 = \frac{1}{3} \phi_1 - \phi_1 = -\frac{2}{3} \phi_1 \).

1 mark for the correct answer A.
- 1 mark for identifying the correct potential formula and expressing the change in potential algebraically in terms of \( \phi_1 \).

According to the first law of thermodynamics:
\( \Delta U = q + w \)
where \( \Delta U \) is the change in internal energy, \( q \) is the heat energy transferred to the gas, and \( w \) is the work done on the gas.

For an ideal gas, the internal energy depends only on its absolute temperature. Since the process is isothermal, the temperature remains constant, and therefore:
\( \Delta U = 0 \)

Using the first law:
\( 0 = q + 150\text{ J} \implies q = -150\text{ J} \)

A negative value for \( q \) means that \( 150\text{ J} \) of heat energy is transferred *from* the gas to the surroundings.

1 mark for the correct answer A.
- 1 mark for realizing that \( \Delta U = 0 \) during an isothermal process for an ideal gas.
- 1 mark for applying the first law of thermodynamics to find that heat is transferred from the gas.

According to Wien's displacement law, the wavelength of peak intensity \( \lambda_{\text{max}} \) is inversely proportional to the absolute temperature \( T \):
\( \lambda_{\text{max}} \propto \frac{1}{T} \)

Since the surface temperature of Y is \( 2T \) (twice that of X), the wavelength of peak intensity is:
\( \lambda_Y = \frac{1}{2}\lambda_X \)

According to the Stefan-Boltzmann law, the luminosity \( L \) of a star modeled as a black-body radiator is given by:
\( L = 4\pi R^2 \sigma T^4 \implies L \propto R^2 T^4 \)

For star Y:
\( L_Y \propto (3R)^2 (2T)^4 = 9 R^2 \times 16 T^4 = 144 (R^2 T^4) \)

Therefore:
\( L_Y = 144 L_X \).

1 mark for the correct answer B.
- 1 mark for finding the peak wavelength using Wien's law.
- 1 mark for calculating the correct luminosity using the Stefan-Boltzmann law.

For a particle in simple harmonic motion, the relationship between acceleration \( a \) and displacement \( x \) is:
\( a = -\omega^2 x \)

The maximum acceleration magnitude is achieved at the maximum displacement (the amplitude \( x_0 \)):
\( a_{\text{max}} = \omega^2 x_0 \)

Given \( a_{\text{max}} = 20\text{ m s}^{-2} \) and \( x_0 = 5.0\text{ cm} = 0.050\text{ m} \):
\( 20 = \omega^2 \times 0.050 \)
\( \omega^2 = \frac{20}{0.050} = 400\text{ s}^{-2} \)
\( \omega = \sqrt{400} = 20\text{ rad s}^{-1} \)

The relationship between angular frequency \( \omega \) and frequency \( f \) is:
\( \omega = 2\pi f \implies f = \frac{\omega}{2\pi} \)

Therefore:
\( f = \frac{20}{2\pi} = \frac{10}{\pi} \approx 3.18\text{ Hz} \approx 3.2\text{ Hz} \).

1 mark for the correct answer B.
- 1 mark for finding the correct angular frequency \( \omega = 20\text{ rad s}^{-1} \).
- 1 mark for converting angular frequency to cyclic frequency \( f \approx 3.2\text{ Hz} \).

Let V be the potential difference across the parallel combination (and hence the 4.0 \(\Omega\) resistor). The current from the 12.0 V cell is \(I_1 = (12.0 - V) / 2.0\). The current from the 6.0 V cell is \(I_2 = (6.0 - V) / 4.0\). The total current through the 4.0 \(\Omega\) resistor is \(I = V / 4.0\). Using Kirchhoff's first law, \(I = I_1 + I_2\), so \(V / 4.0 = (12.0 - V) / 2.0 + (6.0 - V) / 4.0\). Multiplying the entire equation by 4.0 gives \(V = 2(12.0 - V) + (6.0 - V)\), which simplifies to \(V = 24.0 - 2V + 6.0 - V\), so \(4V = 30.0\) and \(V = 7.5\) V.

Award 1 mark for the correct calculation leading to 7.5 V (C).

Let the direction to the right be positive. The initial velocities are \(u_1 = +8.0\) m s\(^{-1}\) and \(u_2 = -2.0\) m s\(^{-1}\). Total initial momentum is \((2.0 \times 8.0) + (3.0 \times -2.0) = +10.0\) N s. For a perfectly elastic collision, the relative velocity of approach equals the relative velocity of separation: \(u_1 - u_2 = v_2 - v_1\), which gives \(8.0 - (-2.0) = v_2 - v_1\), so \(v_2 = v_1 + 10.0\). By conservation of momentum, \(2.0 v_1 + 3.0 v_2 = 10.0\). Substituting \(v_2\) gives \(2.0 v_1 + 3.0(v_1 + 10.0) = 10.0\), so \(5.0 v_1 = -20.0\) and \(v_1 = -4.0\) m s\(^{-1}\) (to the left). Thus, \(v_2 = -4.0 + 10.0 = +6.0\) m s\(^{-1}\) (to the right).

Award 1 mark for the correct combination of velocities (A).

The Young modulus is defined as \(E = \frac{F L}{A \Delta L}\), so the extension is \(\Delta L = \frac{F L}{A E} = \frac{4 F L}{\pi d^2 E}\) where d is the diameter. Since the force F and material (Young modulus E) are the same, the extension is proportional to \(L/d^2\). For wire X, \(\Delta L_X\) is proportional to \(2 L / (0.5 d)^2 = 8 L / d^2\). Thus, the ratio of the extension of X to Y is 8.0.

Award 1 mark for determining the correct ratio of 8.0 (D).

The fringe width is given by the formula \(x = \frac{\lambda D}{a}\). The new fringe width is \(x' = \frac{(1.5\lambda) (2D)}{(0.5a)} = \frac{3.0 \lambda D}{0.5 a} = 6.0 \frac{\lambda D}{a} = 6.0x\).

Award 1 mark for the correct calculation showing the new fringe width is 6.0x (C).

For a closed cycle, the net change in internal energy is zero. In Process 1, work done by the gas is \(+150\) J. In Process 2, 250 J of work is done on the gas, so work done by the gas is \(-250\) J. In Process 3, since the volume is constant, work done is 0. Net work done by the gas in the cycle is \(150 - 250 = -100\) J. Thus, net thermal energy transfer to the gas must also be \(-100\) J. Since 400 J was transferred to the gas in Process 1, and 0 in Process 2, the heat transfer in Process 3 is \(Q_3 = -100 - 400 = -500\) J. A negative sign means 500 J of thermal energy is transferred from the gas.

Award 1 mark for identifying that 500 J is transferred from the gas (B).

The redshift is defined as \(z = \frac{\Delta \lambda}{\lambda_0} = \frac{682 - 656}{656} = \frac{26}{656} \approx 0.0396\). Rounding to two significant figures, \(z = 0.040\). The recession speed is \(v = z c = 0.0396 \times 3.00 \times 10^8 \approx 1.2 \times 10^7\) m s\(^{-1}\).

Award 1 mark for the correct calculation of redshift and speed (B).

In simple harmonic motion, the maximum acceleration \(a_{\text{max}}\) is related to the amplitude \(x_0\) and angular frequency \(\omega\) by \(a_{\text{max}} = \omega^2 x_0\). Given \(a_{\text{max}} = 2.0\) m s\(^{-2}\) and \(x_0 = 5.0\) cm \(= 0.050\) m, we find \(\omega^2 = \frac{2.0}{0.050} = 40\) s\(^{-2}\), which gives \(\omega = \sqrt{40} \approx 6.32\) rad s\(^{-1}\). The frequency is \(f = \frac{\omega}{2\pi} = \frac{6.32}{2\pi} \approx 1.0\) Hz.

Award 1 mark for the correct frequency of 1.0 Hz (B).

Let the charge \(+Q\) be at \(x = 0\) and the charge \(-3Q\) be at \(x = d\). We seek a point \(x\) between them (\(0 < x < d\)) where the total electric potential \(V = 0\). The potential due to a point charge is given by \(V = \frac{q}{4 \pi \varepsilon_0 r}\). Thus, \(V = \frac{Q}{4 \pi \varepsilon_0 x} - \frac{3Q}{4 \pi \varepsilon_0 (d - x)} = 0\). This simplifies to \(\frac{1}{x} = \frac{3}{d - x}\). Solving for \(x\) gives \(d - x = 3x \Rightarrow 4x = d \Rightarrow x = 0.25d\).

Award 1 mark for calculating the correct distance of 0.25 d (B).

Let the potential at junction Y be \( 0\text{ V} \) and the potential at junction X be \( V \). Applying Kirchhoff's first law (current law) at junction X: \( I_1 + I_2 = I_R \) where \( I_1 \) is the current flowing from the \( 9.0\text{ V} \) cell into junction X, \( I_2 \) is the current flowing from the \( 3.0\text{ V} \) cell into junction X, and \( I_R \) is the current flowing from junction X through the \( 1.0\,\Omega \) resistor to junction Y. Using Ohm's law and the e.m.f. values: \( I_1 = \frac{9.0 - V}{2.0} \), \( I_2 = \frac{3.0 - V}{2.0} \), \( I_R = \frac{V}{1.0} \). Substitute these expressions into Kirchhoff's current law: \( \frac{9.0 - V}{2.0} + \frac{3.0 - V}{2.0} = \frac{V}{1.0} \). Multiply the entire equation by \( 2.0 \): \( (9.0 - V) + (3.0 - V) = 2V \) which simplifies to \( 12.0 - 2V = 2V \), leading to \( 4V = 12.0 \implies V = 3.0\text{ V} \). Now, calculate the current \( I_2 \) in the \( 3.0\text{ V} \) cell: \( I_2 = \frac{3.0 - 3.0}{2.0} = 0\text{ A} \).

Award 1 mark for the correct answer A. No partial marks.

Let the direction to the right be positive. Initial velocity of block 1 (\( m_1 = 3m \)): \( u_1 = +v \). Initial velocity of block 2 (\( m_2 = m \)): \( u_2 = -v \). For a perfectly elastic collision, the relative velocity of approach equals the relative velocity of separation: \( u_1 - u_2 = v_2 - v_1 \implies v - (-v) = v_2 - v_1 \implies 2v = v_2 - v_1 \implies v_2 = v_1 + 2v \). By conservation of linear momentum: \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \implies 3m(v) + m(-v) = 3m v_1 + m v_2 \implies 2v = 3v_1 + v_2 \). Substitute \( v_2 = v_1 + 2v \) into the momentum equation: \( 2v = 3v_1 + (v_1 + 2v) \implies 2v = 4v_1 + 2v \implies 4v_1 = 0 \implies v_1 = 0 \). Substitute \( v_1 = 0 \) back to find \( v_2 \): \( v_2 = 0 + 2v = 2v \). Thus, the \( 3m \) block is stationary and the \( m \) block moves at speed \( 2v \) to the right.

Award 1 mark for the correct answer A. No partial marks.

The Young modulus \( E \) of a material is given by \( E = \frac{\text{stress}}{\text{strain}} = \frac{FL}{A \Delta L} \), which gives the extension as \( \Delta L = \frac{FL}{AE} \). The elastic potential energy \( U \) stored in a wire is \( U = \frac{1}{2} F \Delta L = \frac{F^2 L}{2AE} \). Since both wires are made of the same metal, they have the same Young modulus \( E \). Since they support equal weights, the tension \( F \) is the same in both wires. Thus, the elastic potential energy stored is proportional to \( \frac{L}{A} \): \( U \propto \frac{L}{A} \propto \frac{L}{d^2} \) where \( d \) is the diameter of the wire. We are given \( L_X = 2 L_Y \) and \( d_X = 0.5 d_Y \). Therefore, the ratio is \( \frac{U_X}{U_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2 = 2 \times (2)^2 = 8 \).

Award 1 mark for the correct answer C. No partial marks.

The fringe separation \( x \) in a double-slit experiment is given by \( x = \frac{\lambda D}{d} \) where \( \lambda \) is the wavelength of the light. The wavelength \( \lambda \) and frequency \( f \) of electromagnetic waves are related by \( c = f \lambda \implies \lambda = \frac{c}{f} \). When the frequency is increased to \( 2f \), the new wavelength \( \lambda' \) is \( \lambda' = \frac{c}{2f} = 0.5\lambda \). Using the modified values: New wavelength \( \lambda' = 0.5\lambda \), New slit separation \( d' = 0.5d \), New screen distance \( D' = 2D \). The new fringe separation \( x' \) is \( x' = \frac{\lambda' D'}{d'} = \frac{(0.5\lambda) (2D)}{0.5d} = 2 \frac{\lambda D}{d} = 2x \).

Award 1 mark for the correct answer C. No partial marks.

The gravitational potential energy \( U \) of a mass \( m \) at a distance \( r \) from the centre of a planet of mass \( M \) is \( U = -\frac{GMm}{r} \). Initially, at the surface of the planet, the distance from the centre is \( r_1 = R \), so \( U_1 = -\frac{GMm}{R} \). At an altitude equal to \( R \), the distance from the centre is \( r_2 = R + R = 2R \), so \( U_2 = -\frac{GMm}{2R} \). The acceleration of free fall at the surface of the planet is \( g = \frac{GM}{R^2} \implies GM = g R^2 \). Substitute \( GM \) in terms of \( g \): \( U_1 = -mgR \) and \( U_2 = -0.5mgR \). The increase in gravitational potential energy is \( \Delta U = U_2 - U_1 = -0.5mgR - (-mgR) = 0.5mgR \).

Award 1 mark for the correct answer B. No partial marks.

The first law of thermodynamics is \( \Delta U = q + w \), where \( \Delta U \) is the change in internal energy, \( q \) is thermal energy absorbed, and \( w \) is work done on the gas. For process X: Work done by the gas is \( 500\text{ J} \) so \( w_X = -500\text{ J} \); thermal energy absorbed is \( q_X = +800\text{ J} \); thus \( \Delta U_X = 800 - 500 = +300\text{ J} \). For process Y: volume is constant, so \( w_Y = 0 \); thermal energy released is \( 450\text{ J} \), so \( q_Y = -450\text{ J} \); thus \( \Delta U_Y = -450\text{ J} \). For the complete cycle: \( \Delta U_{\text{total}} = \Delta U_X + \Delta U_Y + \Delta U_Z = 0 \implies +300 - 450 + \Delta U_Z = 0 \implies \Delta U_Z = +150\text{ J} \). For process Z: Work done on the gas is \( w_Z = +200\text{ J} \). Using the first law: \( +150 = q_Z + 200 \implies q_Z = -50\text{ J} \). Since \( q_Z \) is negative, \( 50\text{ J} \) of thermal energy is released by the gas.

Award 1 mark for the correct answer B. No partial marks.

In simple harmonic motion, the speed \( v \) at displacement \( x \) is given by: \( v = \omega \sqrt{A^2 - x^2} \). The maximum speed occurs at \( x = 0 \), which is \( v_0 = \omega A \). When the displacement is \( x = 0.6A \), the speed is: \( v = \omega \sqrt{A^2 - (0.6A)^2} = \omega \sqrt{A^2 - 0.36A^2} = \omega \sqrt{0.64A^2} = 0.8 \omega A \). Since \( v_0 = \omega A \), we have \( v = 0.8 v_0 \).

Award 1 mark for the correct answer C. No partial marks.

The magnetic force provides the centripetal force: \( q v B = \frac{m v^2}{r} \implies r = \frac{mv}{qB} \). The momentum \( p = mv \) is related to kinetic energy \( K \) by \( p = \sqrt{2mK} \). Therefore, the radius is: \( r = \frac{\sqrt{2mK}}{qB} \). Since both particles have the same kinetic energy \( K \) and move in the same magnetic field \( B \), the radius of the path is proportional to \( \frac{\sqrt{m}}{q} \): \( r \propto \frac{\sqrt{m}}{q} \). The ratio of the radii is: \( \frac{r_{\alpha}}{r_p} = \frac{\sqrt{m_{\alpha}/m_p}}{q_{\alpha}/q_p} \). Since \( m_{\alpha}/m_p = 4 \) and \( q_{\alpha}/q_p = 2 \), the ratio is \( \frac{r_{\alpha}}{r_p} = \frac{\sqrt{4}}{2} = \frac{2}{2} = 1 \).

Award 1 mark for the correct answer B. No partial marks.

Using the conservation of linear momentum: the initial momentum of the space probe is \( M v \). Since fragment A is projected backward, its velocity is \( -0.50 v \). Let the velocity of fragment B be \( v_B \). The mass of fragment B is \( M - 0.40 M = 0.60 M \). The equation for conservation of momentum is: \( M v = 0.40 M (-0.50 v) + 0.60 M v_B \). Solving this gives: \( M v = -0.20 M v + 0.60 M v_B \), which simplifies to \( 1.20 M v = 0.60 M v_B \). Hence, \( v_B = 2.0 v \).

1 mark for the correct choice B. Method: setting up the conservation of momentum equation with correct mass fractions and direction sign, solving for the velocity of fragment B.

The extension \( x \) of a wire of Young modulus \( E \), length \( L \), and area \( A \) under a force \( F \) is given by \( x = \frac{F L}{A E} \). For the second wire, the length is \( 2L \), the force is \( 2F \), and the diameter is halved, which means its cross-sectional area is \( \frac{1}{4} A \) since \( A \propto d^2 \). Substituting these into the formula for the second extension \( x_2 \): \( x_2 = \frac{(2F)(2L)}{(\frac{1}{4}A)E} = 16 \left( \frac{F L}{A E} \right) = 16x \).

1 mark for the correct choice D. Method: applying the formula for extension in terms of Young modulus, substituting the modified parameters, and calculating the factor of 16.

The fringe width \( w \) is given by the formula \( w = \frac{\lambda D}{d} \). Under the new conditions, the new fringe width \( w' \) is: \( w' = \frac{(1.5\lambda)(2D)}{0.5d} = \frac{3.0 \lambda D}{0.5 d} = 6 \left( \frac{\lambda D}{d} \right) = 6w \).

1 mark for the correct choice C. Method: using the double-slit fringe spacing equation, replacing variables with their scaled values, and simplifying the ratio.

The gravitational field strength at a distance \( r \) from the center of a planet of mass \( M \) is \( g = \frac{GM}{r^2} \). At the surface of the planet, the distance from the center is \( r = R \), so \( g = \frac{GM}{R^2} \). At a height of \( 3R \) above the surface, the distance from the center is \( r = R + 3R = 4R \). Therefore, the field strength is \( g' = \frac{GM}{(4R)^2} = \frac{GM}{16 R^2} = \frac{g}{16} \).

1 mark for the correct choice D. Method: identifying the total distance from the center as \( 4R \) and applying the inverse-square law for gravitational field strength.

According to the first law of thermodynamics: \( \Delta U = q + w \). Here, \( q = -150\text{ J} \) because thermal energy is transferred out of the gas. The gas does work on the surroundings, which means the work done on the gas is negative: \( w = -450\text{ J} \). Therefore, the change in internal energy is: \( \Delta U = -150\text{ J} + (-450\text{ J}) = -600\text{ J} \).

1 mark for the correct choice A. Method: using the first law of thermodynamics with correct sign conventions for energy transfer and work done during expansion.

Radiant flux intensity is related to luminosity and distance by \( F = \frac{L}{4 \pi d^2} \). Since both stars have the same luminosity \( L \), we have \( F \propto \frac{1}{d^2} \). Therefore, \( \frac{F_2}{F_1} = \left( \frac{d_1}{d_2} \right)^2 \). Substituting \( F_2 = 0.040 F_1 \) gives \( 0.040 = \left( \frac{d_1}{d_2} \right)^2 \). Taking the square root of both sides: \( 0.20 = \frac{d_1}{d_2} \), which gives \( d_2 = \frac{d_1}{0.20} = 5.0 d_1 \).

1 mark for the correct choice B. Method: applying the inverse-square law for radiant flux intensity, setting up the ratio, and solving for the new distance.

Let the distance from the \( +Q \) charge be \( d \), where \( 0 < d < x \). The distance from the \( -2Q \) charge is then \( x - d \). The total electric potential \( V \) at this point is the sum of the potentials due to each charge: \( V = \frac{Q}{4\pi\varepsilon_0 d} + \frac{-2Q}{4\pi\varepsilon_0 (x-d)} \). For the potential to be zero: \( \frac{Q}{d} = \frac{2Q}{x-d} \implies x - d = 2d \implies 3d = x \implies d = \frac{x}{3} \).

1 mark for the correct choice B. Method: expressing the total electric potential as the sum of potentials from both charges, setting the sum to zero, and solving for \( d \).

The terminal potential difference \( V \) across a cell with e.m.f. \( E \) and internal resistance \( r \) delivering a current \( I \) is given by the equation \( V = E - Ir \). Comparing this to the equation of a straight line, \( y = mx + c \), where \( y = V \) and \( x = I \), the gradient \( m \) is \( -r \) (a negative constant value) and the y-intercept is \( E \) (a positive constant). Thus, the graph of \( V \) against \( I \) is a straight line of negative gradient that does not pass through the origin.

1 mark for the correct choice B. Method: relating terminal potential difference, e.m.f., and internal resistance to a linear equation and identifying its gradient and intercept.

Initially, at \(20^\circ\text{C}\), the potential difference across the fixed resistor is given by the potential divider formula: \(V_{\text{out1}} = V \times \frac{R}{R + R_{\text{th}}} = 9.0 \times \frac{3.0\text{ k}\Omega}{3.0\text{ k}\Omega + 6.0\text{ k}\Omega} = 3.0\text{ V}\). At \(80^\circ\text{C}\), the potential difference is: \(V_{\text{out2}} = 9.0 \times \frac{3.0\text{ k}\Omega}{3.0\text{ k}\Omega + 1.5\text{ k}\Omega} = 9.0 \times \frac{3.0}{4.5} = 6.0\text{ V}\). The change in the output voltage is \(V_{\text{out2}} - V_{\text{out1}} = 6.0\text{ V} - 3.0\text{ V} = +3.0\text{ V}\), which represents an increase of \(3.0\text{ V}\).

1 mark for the correct calculation showing an increase of \(3.0\text{ V}\).

By conservation of linear momentum, taking the direction to the right as positive: \(p_i = m_A u_A + m_B u_B = (2.0 \times 6.0) + (3.0 \times (-4.0)) = 12.0 - 12.0 = 0\text{ kg m s}^{-1}\). Since the initial total momentum is zero and the blocks stick together, their combined velocity after the collision must be zero. The initial kinetic energy of the system is \(E_{\text{ki}} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2 = \frac{1}{2}(2.0)(6.0)^2 + \frac{1}{2}(3.0)(-4.0)^2 = 36.0 + 24.0 = 60.0\text{ J}\). The final kinetic energy is \(0\text{ J}\). Therefore, the loss in kinetic energy is \(60.0\text{ J}\).

1 mark for calculating the correct kinetic energy loss of \(60\text{ J}\).

The Young modulus \(E\) of the material is given by \(E = \frac{F L}{A x}\). Rearranging for extension gives \(x = \frac{F L}{A E} \propto \frac{L}{d^2}\) where \(d\) is the diameter of the wire, since the force \(F\) and material Young modulus \(E\) are identical for both wires. The ratio of extensions is \(\frac{x_X}{x_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2\). Substituting the given values \(L_X = 2 L_Y\) and \(d_X = 0.5 d_Y\), we obtain \(\frac{x_X}{x_Y} = 2 \times \left(\frac{1}{0.5}\right)^2 = 2 \times 4 = 8\).

1 mark for identifying the correct ratio relationship and calculating \(8\).

Let the charge \(+Q\) be at position \(x = 0\) and the charge \(-3Q\) be at position \(x = d\). Let the point where the electric potential is zero be at a distance \(r\) from \(+Q\) between the charges (where \(0 < r < d\)). The potential due to a point charge is given by \(V = \frac{q}{4\pi\varepsilon_0 r}\). The total potential is \(V = \frac{Q}{4\pi\varepsilon_0 r} - \frac{3Q}{4\pi\varepsilon_0 (d - r)} = 0\). Simplifying this gives \(\frac{1}{r} = \frac{3}{d - r}\). Cross-multiplying, we find \(d - r = 3r\), which leads to \(4r = d\) and \(r = 0.25d\).

1 mark for calculating the correct distance fraction of \(0.25d\).

The redshift \(z\) is related to the recession velocity \(v\) by \(z = \frac{v}{c}\). Thus, the velocity of the galaxy is \(v = z c = 0.050 \times 3.00 \times 10^8\text{ m s}^{-1} = 1.50 \times 10^7\text{ m s}^{-1} = 1.50 \times 10^4\text{ km s}^{-1}\). According to Hubble's law, \(v = H_0 d\). Re-arranging for distance \(d\) gives \(d = \frac{v}{H_0} = \frac{1.50 \times 10^4\text{ km s}^{-1}}{70\text{ km s}^{-1}\text{ Mpc}^{-1}} \approx 214\text{ Mpc}\), which is closest to \(210\text{ Mpc}\).

1 mark for the correct calculation of distance using redshift and Hubble's law.

According to Faraday's law of electromagnetic induction, the average induced e.m.f. \(E\) is given by \(E = \frac{\Delta \Phi}{\Delta t}\), where \(\Delta \Phi\) is the change in magnetic flux linkage. The initial flux linkage is \(\Phi_i = N B A = 400 \times 0.30\text{ T} \times 2.5 \times 10^{-3}\text{ m}^2 = 0.30\text{ Wb}\). The final magnetic flux linkage is \(0\text{ Wb}\). The average induced e.m.f. is therefore \(E = \frac{0.30\text{ Wb}}{0.15\text{ s}} = 2.0\text{ V}\).

1 mark for correctly calculating the average induced e.m.f. of \(2.0\text{ V}\).

The first law of thermodynamics is given by \(\Delta U = q + w\), where \(q\) is the thermal energy transferred to the gas and \(w\) is the work done on the gas. The work done by the expanding gas is \(W_{\text{by}} = p \Delta V = 1.2 \times 10^5\text{ Pa} \times (5.0 \times 10^{-3} - 2.0 \times 10^{-3})\text{ m}^3 = 360\text{ J}\). The work done on the gas is therefore \(w = -360\text{ J}\). The thermal energy transferred to the gas is \(q = +450\text{ J}\). Hence, the change in internal energy is \(\Delta U = +450\text{ J} - 360\text{ J} = +90\text{ J}\).

1 mark for applying the first law of thermodynamics with correct signs to get \(+90\text{ J}\).

For a charged particle in a uniform magnetic field, the magnetic force provides the centripetal force: \(B q v = \frac{m v^2}{R}\), which gives \(R = \frac{m v}{B q}\). The momentum of the particle is \(p = m v = \sqrt{2 m E_k}\), where \(E_k\) is the kinetic energy. Therefore, the radius of the path is \(R = \frac{\sqrt{2 m E_k}}{B q}\). For the electron, \(r = \frac{\sqrt{2 m E_k}}{B e}\). For the second particle, the mass is \(2m\), the charge is \(e\), and the kinetic energy is \(2E_k\). The radius is \(R' = \frac{\sqrt{2 (2m) (2E_k)}}{B e} = \frac{\sqrt{8 m E_k}}{B e} = 2 \frac{\sqrt{2 m E_k}}{B e} = 2r\).

1 mark for identifying that the radius scales with the square root of the product of mass and kinetic energy, resulting in a radius of \(2r\).

(a) Calculate resistivity:
\[\rho = \frac{2.40 \times \pi \times (0.38 \times 10^{-3})^2}{4 \times 1.25 \times 0.750}\]
\[\rho = \frac{2.40 \times 4.536 \times 10^{-7}}{3.75} = 2.903 \times 10^{-7}\ \Omega\text{ m} \approx 2.9 \times 10^{-7}\ \Omega\text{ m}\]

(b) Percentage uncertainties:
- \(\frac{\Delta V}{V} = \frac{0.05}{2.40} = 0.0208\) (or \(2.08\%\))
- \(\frac{\Delta I}{I} = \frac{0.02}{1.25} = 0.0160\) (or \(1.60\%\))
- \(\frac{\Delta d}{d} = \frac{0.01}{0.38} = 0.0263\) (or \(2.63\%\))
- \(\frac{\Delta L}{L} = \frac{0.003}{0.750} = 0.0040\) (or \(0.40\%\))

Using the uncertainty equation:
\[\frac{\Delta \rho}{\rho} = \frac{\Delta V}{V} + \frac{\Delta I}{I} + 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta L}{L}\]
\[\frac{\Delta \rho}{\rho} = 0.0208 + 0.0160 + 2(0.0263) + 0.0040 = 0.0934\text{ (or } 9.34\%\)\]

Absolute uncertainty:
\[\Delta \rho = 0.0934 \times 2.903 \times 10^{-7} = 0.271 \times 10^{-7}\ \Omega\text{ m} \approx 0.3 \times 10^{-7}\ \Omega\text{ m}\]

(c) The measurement contributing the most is the diameter \(d\). This is because its fractional uncertainty is doubled in the calculation of \(\rho\) as it is squared (\(2 \times 2.63\% = 5.26\%\)), which is larger than the percentage uncertainties of all other single quantities.

(a)
- C1: Recall of formula or substitution of values into formula with correct conversion of units (diameter in m).
- A1: Value of \(2.9 \times 10^{-7}\ \Omega\text{ m}\) (accept \(2.90 \times 10^{-7}\ \Omega\text{ m}\)).

(b)
- C1: Correctly calculate the individual fractional/percentage uncertainties (e.g., any two correct).
- C1: Use of uncertainty equation with \(2 \frac{\Delta d}{d}\) correctly included.
- A1: Final absolute uncertainty calculated as \(0.3 \times 10^{-7}\ \Omega\text{ m}\) (accept \(0.27 \times 10^{-7}\ \Omega\text{ m}\)).

(c)
- B1: Identifies diameter \(d\) as the main contributor.
- B1: Explains that its percentage uncertainty is doubled because the diameter is squared in the resistivity formula, resulting in a contribution of \(5.3\%\).

(a) The total momentum of an isolated system (or closed system) remains constant, provided no external forces act on it.

(b) Let the direction to the right be positive.
Initial momentum:
\[p_{\text{initial}} = m_A u_A + m_B u_B\]
\[p_{\text{initial}} = (0.40 \times 2.5) + (0.60 \times [-1.2])\]
\[p_{\text{initial}} = 1.0 - 0.72 = +0.28\text{ kg m s}^{-1}\]

Final momentum:
\[p_{\text{final}} = m_A v_A + m_B v_B\]
Since trolley A rebounds to the left, \(v_A = -0.80\text{ m s}^{-1}\).
\[p_{\text{final}} = (0.40 \times [-0.80]) + (0.60 \times v_B)\]
\[p_{\text{final}} = -0.32 + 0.60 v_B\]

By conservation of momentum:
\[+0.28 = -0.32 + 0.60 v_B\]
\[0.60 v_B = 0.60\]
\[v_B = +1.0\text{ m s}^{-1}\]
Since the sign is positive, the direction of trolley B is to the right.

(c) Total initial kinetic energy:
\[E_{k,\text{initial}} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2\]
\[E_{k,\text{initial}} = \frac{1}{2}(0.40)(2.5)^2 + \frac{1}{2}(0.60)(-1.2)^2\]
\[E_{k,\text{initial}} = 1.25 + 0.432 = 1.682\text{ J}\]

Total final kinetic energy:
\[E_{k,\text{final}} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2\]
\[E_{k,\text{final}} = \frac{1}{2}(0.40)(-0.80)^2 + \frac{1}{2}(0.60)(1.0)^2\]
\[E_{k,\text{final}} = 0.128 + 0.300 = 0.428\text{ J}\]

Since \(E_{k,\text{final}} < E_{k,\text{initial}}\), kinetic energy is not conserved, hence the collision is inelastic.

(a)
- B1: Momentum remains constant / is conserved.
- B1: For a closed/isolated system OR where no external force acts.

(b)
- C1: Correct calculation of initial momentum (taking signs into account, \(0.28\text{ N s}\)).
- C1: Correct formulation of conservation equation: \(0.28 = -0.32 + 0.60 v_B\).
- A1: Final velocity magnitude \(1.0\text{ m s}^{-1}\) and direction specified as to the right.

(c)
- C1: Correct calculation of initial kinetic energy (\(1.68\text{ J}\) or \(1.7\text{ J}\)).
- C1: Correct calculation of final kinetic energy (\(0.43\text{ J}\)).
- A1: Compares values and concludes that collision is inelastic because kinetic energy is not conserved.

(a) Volume of the cylinder:
\[V = A \times h = (4.50 \times 10^{-3}\text{ m}^2) \times 0.150\text{ m} = 6.75 \times 10^{-4}\text{ m}^3\]

Upthrust is equal to the weight of the displaced oil:
\[U = \rho_{\text{oil}} V g\]
\[U = 860 \times 6.75 \times 10^{-4} \times 9.81 = 5.6947\text{ N} \approx 5.7\text{ N}\]

(b) Weight of the cylinder:
\[W = m g = \rho_{\text{brass}} V g\]
\[W = 8400 \times 6.75 \times 10^{-4} \times 9.81 = 55.62\text{ N} \approx 55.6\text{ N}\text{ (or } 56\text{ N with 2 s.f.)}\]

(c) For the submerged cylinder in equilibrium:
\[T + U = W\]
where \(T\) is the tension in the wire (the reading on the newton meter).
\[T = W - U = 55.62 - 5.695 = 49.925\text{ N} \approx 50\text{ N}\text{ (or } 49.9\text{ N)}\]

(a)
- C1: Calculates the volume of the cylinder as \(6.75 \times 10^{-4}\text{ m}^3\).
- A1: Uses \(U = \rho V g\) with density of oil and \(g = 9.81\text{ m s}^{-2}\) to show \(5.69\text{ N} \approx 5.7\text{ N}\).

(b)
- C1: Recalls and uses \(m = \rho_{\text{brass}} V\).
- A1: Obtains a value of \(55.6\text{ N}\) or \(56\text{ N}\).

(c)
- C1: States or applies condition for equilibrium: \(T = W - U\).
- C1: Substitutes values for \(W\) and \(U\) consistently.
- A1: Obtains \(49.9\text{ N}\) or \(50\text{ N}\).

(a) Tensile stress is the force applied per unit cross-sectional area perpendicular to the force: \(\text{stress} = \frac{F}{A}\).

(b)
- (i) Cross-sectional area:
\[A = \frac{\pi d^2}{4} = \frac{\pi \times (0.68 \times 10^{-3})^2}{4} = 3.632 \times 10^{-7}\text{ m}^2 \approx 3.6 \times 10^{-7}\text{ m}^2\]

- (ii) Tensile stress:
\[\sigma = \frac{F}{A} = \frac{45}{3.632 \times 10^{-7}} = 1.239 \times 10^8\text{ Pa} \approx 1.2 \times 10^8\text{ Pa}\]

- (iii) Extension:
\[E = \frac{\text{stress}}{\text{strain}} \implies \text{strain} = \frac{\sigma}{E} = \frac{1.239 \times 10^8}{2.0 \times 10^{11}} = 6.195 \times 10^{-4}\]
\[\text{extension } x = \text{strain} \times L = 6.195 \times 10^{-4} \times 2.20 = 1.36 \times 10^{-3}\text{ m} = 1.36\text{ mm} \approx 1.4\text{ mm}\]

(c) The elastic limit is the maximum force or stress that can be applied to the wire such that it will return to its original length when the load is removed (i.e., beyond this point, plastic/permanent deformation occurs).

(a)
- B1: Force per unit cross-sectional area.

(b)
- (i) A1: Correctly calculates \(A = 3.6 \times 10^{-7}\text{ m}^2\) (accept \(3.63 \times 10^{-7}\text{ m}^2\)).
- (ii) A1: Correctly calculates \(\text{stress} = 1.2 \times 10^8\text{ Pa}\) (accept \(1.24 \times 10^8\text{ Pa}\)).
- (iii) C1: Recalls Young modulus formula or relates extension to Young modulus and stress/strain.
- A1: Correctly calculates extension \(x = 1.4\text{ mm}\) (accept \(1.36\text{ mm}\)).

(c)
- B1: Explains that it is the maximum load/stress the wire can experience and still return to its original length.
- B1: Reference to no plastic/permanent deformation occurring below this limit.

(a) Coherent light sources have a constant phase difference (and the same frequency/wavelength). If they were not coherent, the phase difference at any point on the screen would change rapidly and randomly, causing the bright and dark fringes to continuously shift, resulting in a uniform average illumination instead of a stable fringe pattern.

(b) Double-slit interference formula:
\[x = \frac{\lambda D}{a}\]
where:
- \(\lambda = 633 \times 10^{-9}\text{ m}\)
- \(D = 1.80\text{ m}\)
- \(a = 0.240 \times 10^{-3}\text{ m}\)

\[x = \frac{(633 \times 10^{-9}) \times 1.80}{0.240 \times 10^{-3}} = 4.7475 \times 10^{-3}\text{ m} \approx 4.75\text{ mm}\]

(c)
- The width of the individual slits determines the degree of diffraction of the light passing through them.
- When the slits are made narrower, the diffraction envelope of light from each slit becomes wider (it spreads through a larger angle).
- Consequently, the region of overlap on the screen is greater, so more interference fringes are visible.
- Also, because less light passes through the narrower slits, the peak intensity of the fringes decreases (they become dimmer).

(a)
- B1: Defines coherence as constant phase difference.
- B1: Explains that without coherence, the phase difference shifts rapidly/randomly, washing out the fringe pattern / creating uniform illumination.

(b)
- C1: Identifies and uses the formula \(x = \frac{\lambda D}{a}\).
- C1: Converts units correctly (e.g., \(633\text{ nm}\) to \(633 \times 10^{-9}\text{ m}\) and \(0.240\text{ mm}\) to \(0.240 \times 10^{-3}\text{ m}\)).
- A1: Final answer \(4.75\text{ mm}\) (or \(4.7 \times 10^{-3}\text{ m}\)).

(c)
- B1: Mentions that narrower slits cause more diffraction/spreading of light.
- B1: Explains that the region of overlap increases, so more fringes are visible.
- B1: Mentions that the maximum/overall intensity of the fringes decreases (due to less light entering).

(a) Electromotive force (e.m.f.) is the energy converted from other forms of energy (e.g., chemical energy in the battery) into electrical energy per unit charge.

(b)
- (i) Total resistance of the circuit:
\[R_{\text{total}} = 1200 + 2400 = 3600\ \Omega\]
Using the potential divider equation for the thermistor:
\[V_{\text{thermistor}} = \frac{R_{\text{thermistor}}}{R_{\text{total}}} \times E = \frac{2400}{3600} \times 6.0 = 4.0\text{ V}\]

- (ii) The potential difference across the fixed resistor is:
\[V_{\text{fixed}} = E - V_{\text{thermistor}} = 6.0 - 4.0 = 2.0\text{ V}\]
Power dissipated in the fixed resistor:
\[P = \frac{V_{\text{fixed}}^2}{R_{\text{fixed}}} = \frac{2.0^2}{1200} = 3.33 \times 10^{-3}\text{ W} = 3.33\text{ mW} \approx 3.3\text{ mW}\]
Alternatively, current \(I = \frac{6.0}{3600} = 1.67 \times 10^{-3}\text{ A}\), and \(P = I^2 R = (1.67 \times 10^{-3})^2 \times 1200 = 3.33\text{ mW}\).

(c)
- As temperature increases, the resistance of the negative temperature coefficient (NTC) thermistor decreases.
- This reduces the total resistance of the circuit, which increases the current \(I\) flowing through the circuit.
- Since the potential difference across the fixed resistor is given by \(V_{\text{fixed}} = I R_{\text{fixed}}\), and \(R_{\text{fixed}}\) is constant, the potential difference across the fixed resistor increases.

(a)
- B1: Work done / energy converted per unit charge.
- B1: Energy converted from chemical/other forms to electrical.

(b)
- (i) C1: Calculates total resistance \(3600\ \Omega\) or current \(1.67\text{ mA}\).
- A1: Correct potential difference of \(4.0\text{ V}\).
- (ii) C1: Recalls power formula \(P = I^2 R\) or \(P = \frac{V^2}{R}\).
- A1: Correct calculation of power as \(3.3\text{ mW}\) (accept \(3.33\text{ mW}\)).

(c)
- B1: States that thermistor resistance decreases with temperature.
- B1: Explains that total resistance decreases, so current increases, resulting in a higher potential difference across the fixed resistor (or because the fixed resistor takes a greater share of the total resistance).

(a) Nuclear decay equation:
\[^1_0\text{n} \rightarrow ^1_1\text{p} + \text{e}^- + \bar{\nu}_e\]
(Accept \(\text{n} \rightarrow \text{p} + \text{e}^- + \bar{\nu}_e\) or equivalent nucleon notation).

(b) A neutron is composed of one up quark and two down quarks (\(udd\)). A proton is composed of two up quarks and one down quark (\(uud\)). During \(\beta^-\)\-decay, a down quark (\(d\)) decays into an up quark (\(u\)):
\[d \rightarrow u + \text{e}^- + \bar{\nu}_e\]

(c)
- Proton: Hadron, and specifically a baryon (as it consists of three quarks).
- Electron: Lepton (it is a fundamental particle and does not experience the strong nuclear force).

(d)
- The fundamental force is the weak force (or weak nuclear interaction).
- The exchange particles (\(\text{W}^-\), \(\text{W}^+\), \(\text{Z}^0\) bosons) belong to the class of particles called gauge bosons (or simply bosons).

(a)
- B1: Correct symbols for neutron, proton, and electron.
- B1: Inclusion of the electron antineutrino \(\bar{\nu}_e\) with correct symbol (must not be neutrino \(\nu\)).

(b)
- B1: States neutron is \(udd\) and proton is \(uud\).
- B1: States a down quark (\(d\)) changes to an up quark (\(u\)).

(c)
- B1: Correctly classifies the proton as a hadron / baryon.
- B1: Correctly classifies the electron as a lepton.

(d)
- B1: Identifies the weak nuclear force / weak interaction.
- B1: Identifies the exchange particles as gauge bosons (accept 'bosons').

**Sample Results:**
* Resistance wire of length 1.00 m with \(k = 0.044\ \Omega\ \text{cm}^{-1}\).
* Cell of e.m.f. \(E = 1.5\text{ V}\).

**Table of Results:**
* \(x = 20.0\text{ cm}\): \(V = 0.40\text{ V}\), \(1/x = 0.050\text{ cm}^{-1}\), \(1/V = 2.50\text{ V}^{-1}\)
* \(x = 30.0\text{ cm}\): \(V = 0.50\text{ V}\), \(1/x = 0.033\text{ cm}^{-1}\), \(1/V = 2.00\text{ V}^{-1}\)
* \(x = 40.0\text{ cm}\): \(V = 0.59\text{ V}\), \(1/x = 0.025\text{ cm}^{-1}\), \(1/V = 1.70\text{ V}^{-1}\)
* \(x = 50.0\text{ cm}\): \(V = 0.67\text{ V}\), \(1/x = 0.020\text{ cm}^{-1}\), \(1/V = 1.50\text{ V}^{-1}\)
* \(x = 60.0\text{ cm}\): \(V = 0.73\text{ V}\), \(1/x = 0.017\text{ cm}^{-1}\), \(1/V = 1.37\text{ V}^{-1}\)
* \(x = 80.0\text{ cm}\): \(V = 0.81\text{ V}\), \(1/x = 0.013\text{ cm}^{-1}\), \(1/V = 1.23\text{ V}^{-1}\)

**Graph:**
* Plot \(1/V\) vs \(1/x\).
* Gradient \(m = \frac{2.50 - 1.23}{0.050 - 0.0125} = \frac{1.27}{0.0375} \approx 33.9\text{ V}^{-1}\text{ cm}\).
* Intercept \(c \approx 0.81\text{ V}^{-1}\).

**Calculations:**
* From \(c = \frac{1}{E} (1 + R_1/R_2)\):
\(0.81 = \frac{1}{E} (1 + 2.2 / 10) = \frac{1.22}{E} \implies E \approx 1.5\text{ V}\).
* From \(m = \frac{R_1}{E k}\):
\(33.9 = \frac{2.2}{1.5 \times k} \implies k = \frac{2.2}{1.5 \times 33.9} \approx 0.043\ \Omega\ \text{cm}^{-1}\) (or \(4.3\ \Omega\ \text{m}^{-1}\)).

**1. Table of Results (6 Marks)**
* **Successful Collection of Data (1 mark):** Six sets of values of \(x\) and \(V\) recorded.
* **Range and spacing (1 mark):** \(x\) values must cover a range of at least 50.0 cm.
* **Column headings (1 mark):** Each column must have a heading and unit: \(x/\text{cm}\), \(V/\text{V}\), \(1/x\ /\text{cm}^{-1}\), \(1/V\ /\text{V}^{-1}\).
* **Consistency of raw readings (1 mark):** All \(x\) values to the nearest mm (e.g. 20.0 cm) and \(V\) to a consistent decimal place (usually 0.01 V).
* **Significant figures (1 mark):** The number of significant figures for \(1/x\) and \(1/V\) must be the same as, or one more than, the significant figures of the raw values of \(x\) and \(V\).
* **Calculations (1 mark):** \(1/V\) values are correctly calculated.

**2. Graph and Plotting (5 Marks)**
* **Axes (1 mark):** Sensible, linear scales with \(1/V\) on the vertical axis and \(1/x\) on the horizontal axis. Labels must include units.
* **Plotting (1 mark):** All points plotted to within half a small square. No thick dots.
* **Line of Best Fit (1 mark):** Straight line drawn with balanced points on either side.
* **Size (1 mark):** Plotted points must occupy at least half the vertical and horizontal range of the grid.
* **Quality of Graph (1 mark):** No anomalous points or large scatter; points lie very close to the line.

**3. Gradient and Intercept (2 Marks)**
* **Gradient (1 mark):** Calculated using a triangle whose hypotenuse is at least half the length of the drawn line.
* **y-intercept (1 mark):** Read directly from the y-axis if \(1/x = 0\), or calculated using \(y = mx + c\) from a point on the line.

**4. Analysis and Calculations (7 Marks)**
* **Calculation of \(E\) (2 marks):** Correct calculation of \(E\) using the intercept: \(E = \frac{1.22}{c}\). Expect \(E\) in the range \(1.3\text{ V} - 1.7\text{ V}\).
* **Calculation of \(k\) (2 marks):** Correct calculation of \(k\) using the gradient: \(k = \frac{R_1}{E m}\). Expect \(k\) in the range \(0.03\ - 0.06\ \Omega\ \text{cm}^{-1}\).
* **Units (1 mark):** Correct units given for \(E\) (\text{V}) and \(k\) (\Omega\ \text{cm}^{-1} or \Omega\ \text{m}^{-1}).
* **Working (2 marks):** Clear, logical steps shown for both calculations.

**Sample Results:**
* Diameter of cylinder \(D = 4.0\text{ cm} = 0.040\text{ m}\).
* Absolute uncertainty in \(D\) is \(\pm 0.1\text{ cm}\) (using a ruler) or \(\pm 0.01\text{ cm}\) (using vernier calipers).
* Percentage uncertainty in \(D = \frac{0.1}{4.0} \times 100\% = 2.5\%\).
* For \(h_1 = 0.4\text{ cm}\):
* \(F_{1,1} = 1.4\text{ N}\), \(F_{1,2} = 1.6\text{ N}\), Average \(F_1 = 1.5\text{ N}\).
* \(k_1 = F_1 \times \frac{D - 2h_1}{\sqrt{D h_1 - h_1^2}} = 1.5 \times \frac{4.0 - 0.8}{\sqrt{4.0 \times 0.4 - 0.16}} = 1.5 \times \frac{3.2}{1.2} = 4.0\text{ N}\).
* For \(h_2 = 1.0\text{ cm}\):
* \(F_{2,1} = 3.4\text{ N}\), \(F_{2,2} = 3.6\text{ N}\), Average \(F_2 = 3.5\text{ N}\).
* \(k_2 = 3.5 \times \frac{4.0 - 2.0}{\sqrt{4.0 \times 1.0 - 1.0}} = 3.5 \times \frac{2.0}{1.73} \approx 4.04\text{ N}\).

**Evaluation:**
* Percentage difference between \(k_1\) and \(k_2\) is \(\frac{|4.04 - 4.0|}{4.0} \times 100\% = 1.0\%\).
* Since this percentage difference is less than 10%, the suggested relationship is valid.

**1. Measurements and Uncertainty (4 Marks)**
* **Measurement of \(D\) (1 mark):** Measured to the nearest 0.1 mm (calipers) or 1 mm (ruler) with correct unit.
* **Percentage uncertainty in \(D\) (1 mark):** Correctly calculated with absolute uncertainty of 0.1 mm (or 1 mm) and shown with working.
* **Measurement of \(h_1\) and \(h_2\) (1 mark):** Recorded to the nearest 1 mm with correct units.
* **Repeat readings of \(F\) (1 mark):** Raw values of \(F\) repeated for each height to find the average.

**2. Results and Calculations (4 Marks)**
* **First and second force values (2 marks):** Appropriate values of \(F_1\) and \(F_2\) recorded with correct precision and units.
* **Calculation of \(k_1\) and \(k_2\) (2 marks):** Correct substitution and calculation of \(k\) values.

**3. Evaluation and Conclusion (4 Marks)**
* **Percentage difference (2 marks):** Correctly calculated percentage difference between \(k_1\) and \(k_2\).
* **Conclusion (2 marks):** Valid conclusion based on a comparison of the percentage difference with a stated criterion (e.g., 10%).

**4. Limitations and Improvements (8 Marks)**
Four pairs of limitations and corresponding improvements (1 mark for each limitation, 1 mark for each matching improvement):
* **Limitation 1:** Only two sets of readings are not enough to draw a valid conclusion.
* **Improvement 1:** Take more readings of \(F\) for different heights \(h\) and plot a graph of \(F\) against \(\frac{\sqrt{Dh - h^2}}{D - 2h}\).
* **Limitation 2:** The cylinder slips on the step instead of rolling over it.
* **Improvement 2:** Use a high-friction material (e.g., sandpaper) on the step or cylinder to prevent slipping.
* **Limitation 3:** Difficult to keep the thread perfectly horizontal while pulling.
* **Improvement 3:** Use a guide, level, or a low-friction pulley at a fixed horizontal height to align the thread.
* **Limitation 4:** Difficult to read the maximum force on the spring balance at the exact instant of release.
* **Improvement 4:** Use a digital force sensor connected to a data logger to record the peak force.

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

(b) (i) \(\phi_1 = -\frac{GM}{r_1} = -\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{24}}{1.2 \times 10^7} = -3.56 \times 10^7\text{ J kg}^{-1}\).
(ii) Centripetal force is provided by the gravitational force:
\(\frac{m v_1^2}{r_1} = \frac{G M m}{r_1^2} \implies v_1 = \sqrt{\frac{GM}{r_1}} = \sqrt{-\phi_1} = \sqrt{3.5573 \times 10^7} = 5.96 \times 10^3\text{ m s}^{-1}\).

(c) The total energy of a satellite in a circular orbit of radius \(r\) is:
\(E = E_k + E_p = \frac{1}{2} m v^2 - \frac{G M m}{r} = \frac{G M m}{2 r} - \frac{G M m}{r} = -\frac{G M m}{2 r}\).
Initial total energy: \(E_1 = -\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{24} \times 450}{2 \times 1.2 \times 10^7} = -8.00 \times 10^9\text{ J}\).
Final total energy: \(E_2 = -\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{24} \times 450}{2 \times 1.8 \times 10^7} = -5.34 \times 10^9\text{ J}\).
Change in total energy: \(\Delta E = E_2 - E_1 = -5.34 \times 10^9 - (-8.00 \times 10^9) = +2.67 \times 10^9\text{ J}\).
Since \(\Delta E\) is positive, the total energy of the satellite increases.

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

(b)
(i)
- \(\phi = -\frac{GM}{r}\) formula used [1]
- Correct value: \(-3.56 \times 10^7\text{ J kg}^{-1}\) (ignore minus sign if explained, accept 2 or 3 s.f.) [1]
(ii)
- \(v = \sqrt{\frac{GM}{r}}\) or equivalent substitution [1]
- Correct calculation: \(5.96 \times 10^3\text{ m s}^{-1}\) (or \(5.96 \times 10^3\)) [1]

(c)
- Correct formula for total energy: \(E = -\frac{GMm}{2r}\) (or calculates kinetic and potential energies separately) [1]
- Correct calculation of \(E_1\) and \(E_2\) [1]
- Final value: \(\Delta E = 2.67 \times 10^9\text{ J}\) (accept range \(2.66 \times 10^9\) to \(2.68 \times 10^9\)) [1]
- Statement: Total energy increases [1]

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

(b) (i) Work done by the gas is \(P \Delta V = 1.5 \times 10^5 \times (3.6 \times 10^{-3} - 2.0 \times 10^{-3}) = 1.5 \times 10^5 \times 1.6 \times 10^{-3} = 240\text{ J}\).
(ii) Using the ideal gas equation, \(P \Delta V = n R \Delta T \implies \Delta T = \frac{P \Delta V}{n R} = \frac{240}{0.12 \times 8.31} = 240.7\text{ K} \approx 241\text{ K}\).

(c) Let us calculate \(W\), \(Q\), and \(\Delta U\) for each path:
1. Path \(A \to B\):
- Work done *on* the gas: \(W_{AB} = -240\text{ J}\).
- Change in internal energy: \(\Delta U_{AB} = \frac{3}{2} n R \Delta T = 1.5 \times 240 = 360\text{ J}\).
- Heating supplied to the gas: \(Q_{AB} = \Delta U_{AB} - W_{AB} = 360 - (-240) = 600\text{ J}\).

2. Path \(B \to C\):
- Constant volume, so work done on the gas: \(W_{BC} = 0\text{ J}\).
- Change in pressure: \(\Delta P = P_C - P_B = 0.90 \times 10^5 - 1.50 \times 10^5 = -0.60 \times 10^5\text{ Pa}\).
- Change in internal energy: \(\Delta U_{BC} = \frac{3}{2} V_B \Delta P = 1.5 \times 3.6 \times 10^{-3} \times (-0.60 \times 10^5) = -324\text{ J}\).
- Heating: \(Q_{BC} = \Delta U_{BC} - W_{BC} = -324\text{ J}\).

3. Path \(C \to A\):
- Total change in internal energy over a complete cycle is zero: \(\Delta U_{AB} + \Delta U_{BC} + \Delta U_{CA} = 0 \implies 360 - 324 + \Delta U_{CA} = 0 \implies \Delta U_{CA} = -36\text{ J}\).
- Work done on the gas: \(W_{CA} = +210\text{ J}\).
- Heating supplied: \(Q_{CA} = \Delta U_{CA} - W_{CA} = -36 - 210 = -246\text{ J}\).

(a)
- \(\Delta U = Q + W\) with all terms correctly defined [1]
- Consistent signs: increase in internal energy, heat added to, and work done on [1]

(b)
(i)
- Use of \(W = P \Delta V\) [1]
- \(240\text{ J}\) (accept \(-240\text{ J}\) if they state work done *by* the gas) [1]
(ii)
- Use of \(P\Delta V = nR\Delta T\) [1]
- Correct calculation: \(241\text{ K}\) or \(240\text{ K}\) [1]

(c)
- Correct values for \(A \to B\): \(W = -240\text{ J}\), \(\Delta U = +360\text{ J}\), \(Q = +600\text{ J}\) [1]
- Correct values for \(B \to C\): \(W = 0\text{ J}\), \(\Delta U = -324\text{ J}\), \(Q = -324\text{ J}\) [1]
- Correct value for \(\Delta U_{CA} = -36\text{ J}\) using cycle rule [1]
- Correct value for \(Q_{CA} = -246\text{ J}\) using \(Q = \Delta U - W\) [1]

(a) The angular frequency is:
\(\omega =
\sqrt{\frac{k}{M}} = \sqrt{\frac{140}{0.35}} = \sqrt{400} = 20\text{ rad s}^{-1}\).

(b) Since the block is released from rest at maximum displacement at \(t = 0\), the displacement as a function of time is given by a cosine function:
\(x = x_0 \cos(\omega t) = 0.050 \cos(20 t)\).

(c) (i) The maximum kinetic energy is equal to the total elastic potential energy of the spring at maximum displacement:
\(E_{k,\text{max}} = \frac{1}{2} k x_0^2 = \frac{1}{2} \times 140 \times (0.050)^2 = 70 \times 0.0025 = 0.175\text{ J}\).
(ii) The magnitude of the acceleration is:
\(a = \omega^2 x = (20)^2 \times 0.030 = 400 \times 0.030 = 12\text{ m s}^{-2}\).

(d) The graph of kinetic energy versus displacement \(x\) is a symmetrical parabola opening downwards, with its peak at \(x = 0\) with value \(0.175\text{ J}\), and touching the horizontal axis at \(x = \pm 0.050\text{ m}\).

(a)
- Correct formula \(\omega = \sqrt{\frac{k}{m}}\) stated [1]
- Correct calculation shown to yield \(20\text{ rad s}^{-1}\) [1]

(b)
- Cosine form chosen (as displacement is max at \(t = 0\)) [1]
- Correct equation: \(x = 0.050 \cos(20 t)\) (with correct numbers) [1]

(c)
(i)
- Use of \(E_k = \frac{1}{2} k x_0^2\) or \(\frac{1}{2} m \omega^2 x_0^2\) [1]
- Correct answer: \(0.175\text{ J}\) (or \(0.18\text{ J}\)) [1]
(ii)
- Use of \(a = \omega^2 x\) [1]
- Correct answer: \(12\text{ m s}^{-2}\) [1]

(d)
- Symmetrical downward parabola centered on the y-axis [1]
- Correctly labeled coordinates: peak at \(0.175\text{ J}\) and x-intercepts at \(\pm 0.05\text{ m}\) [1]

(a) Electric potential at a point is defined as the work done per unit positive charge in bringing a small positive test charge from infinity to that point.

(b) Let the potential be zero at distance \(x\) from \(q_1\):
\(V = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1}{x} + \frac{q_2}{0.10 - x} \right) = 0\).
Since the constant factor is non-zero, we must have:
\(\frac{4.0 \times 10^{-9}}{x} - \frac{9.0 \times 10^{-9}}{0.10 - x} = 0 \implies \frac{4}{x} = \frac{9}{0.10 - x}\).
\(4(0.10 - x) = 9x \implies 0.40 - 4x = 9x \implies 13x = 0.40 \implies x = 0.031\text{ m}\).

(c) The net electric field strength is zero where the electric fields due to \(q_1\) and \(q_2\) are equal in magnitude and opposite in direction.
- Between \(q_1\) and \(q_2\), both fields point in the same direction (from positive \(q_1\) to negative \(q_2\)), so they cannot cancel.
- To the right of \(q_2\), the field of \(q_2\) is always greater than the field of \(q_1\) because \(|q_2| > |q_1|\) and the distance to \(q_2\) is smaller.
- Therefore, the point must be to the left of \(q_1\) (where \(x < 0\)).

Let \(d\) be the distance to the left of \(q_1\):
\(E_1 = E_2 \implies \frac{q_1}{4\pi\varepsilon_0 d^2} = \frac{|q_2|}{4\pi\varepsilon_0 (0.10 + d)^2}\).
\(\frac{4.0}{d^2} = \frac{9.0}{(0.10 + d)^2}\).
Taking the square root of both sides:
\(\frac{2}{d} = \frac{3}{0.10 + d}\).
\(2(0.10 + d) = 3d \implies 0.20 + 2d = 3d \implies d = 0.20\text{ m}\).
So the point is \(0.20\text{ m}\) to the left of \(q_1\) (or \(x = -0.20\text{ m}\)).

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

(b)
- Correct formula for sum of potentials: \(\frac{q_1}{x} + \frac{q_2}{0.10-x} = 0\) [1]
- Correct algebraic equation: \(4(0.10 - x) = 9x\) [1]
- Correct value: \(0.031\text{ m}\) (or \(3.1\text{ cm}\)) [1]

(c)
- Correct reasoning that the point must lie to the left of \(q_1\) [1]
- Correct formula setup: \(\frac{q_1}{d^2} = \frac{q_2}{(0.10+d)^2}\) [1]
- Square root simplification or quadratic equation expansion shown [1]
- Correct value for distance: \(0.20\text{ m}\) [1]
- Clearly states position is to the left of \(q_1\) [1]

(a) Faraday’s law of electromagnetic induction states that the magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage.

(b) The minus sign indicates that the direction of the induced e.m.f. (and hence any induced current) is such that it opposes the change in magnetic flux that produced it. This is a consequence of the conservation of energy.

(c) (i) The cross-sectional area of the coil is:
\(A = \pi r^2 = \pi (0.040)^2 = 5.027 \times 10^{-3}\text{ m}^2\).
The change in magnetic flux linkage is:
\(\Delta \Phi = N \Delta B A = 250 \times (0.20 - 0.80) \times 5.027 \times 10^{-3} = 250 \times (-0.60) \times 5.027 \times 10^{-3} = -0.754\text{ Wb-turns}\).
Magnitude of change in flux linkage is \(0.75\text{ Wb}\).

(ii) The magnitude of the induced e.m.f. is given by:
\(E = \left| \frac{\Delta \Phi}{\Delta t} \right| = \frac{0.754}{0.15} = 5.03\text{ V} \approx 5.0\text{ V}\).

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

(b)
- Induced e.m.f. / current opposes the change in flux [1]
- In accordance with conservation of energy [1]

(c)
(i)
- Correct calculation of area \(A = 5.03 \times 10^{-3}\text{ m}^2\) [1]
- Correct formula for change in flux linkage \(N A \Delta B\) [1]
- Correct value: \(0.75\text{ Wb}\) (or \(0.754\text{ Wb}\); allow negative sign) [1]
(ii)
- Use of \(E = \frac{\Delta \Phi}{\Delta t}\) [1]
- Substitution of their flux linkage change and \(0.15\text{ s}\) [1]
- Correct final answer: \(5.0\text{ V}\) (accept \(5.03\text{ V}\)) [1]

(a) (i) By comparing \(v = 15 \sin(120\pi t)\) with \(v = V_0 \sin(\omega t)\), the peak voltage is \(V_0 = 15\text{ V}\).
(ii) The angular frequency is \(\omega = 120\pi\text{ rad s}^{-1}\).
Since \(\omega = 2\pi f\), then \(f = \frac{120\pi}{2\pi} = 60\text{ Hz}\).
(iii) The r.m.s. voltage is:
\(V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{15}{\sqrt{2}} \approx 10.61\text{ V}\).
The r.m.s. current is:
\(I_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{10.61}{30} = 0.354\text{ A}\).

(b) The r.m.s. value of an alternating current is that value of direct current which produces the same average heating power in a given resistor as the alternating current.

(c) The mean power \(P\) is:
\(P = I_{\text{rms}}^2 R = (0.3536)^2 \times 30 = 3.75\text{ W}\)
(Alternatively, \(P = \frac{V_{\text{rms}}^2}{R} = \frac{V_0^2}{2R} = \frac{15^2}{2 \times 30} = 3.75\text{ W}\)).

(a)
(i)
- \(15\text{ V}\) [1]
(ii)
- Use of \(\omega = 2\pi f\) [1]
- \(60\text{ Hz}\) [1]
(iii)
- Use of \(V_{\text{rms}} = \frac{V_0}{\sqrt{2}}\) [1]
- Use of \(I_{\text{rms}} = \frac{V_{\text{rms}}}{R}\) [1]
- Correct answer: \(0.35\text{ A}\) or \(0.354\text{ A}\) [1]

(b)
- Value of steady / direct current [1]
- that produces same heating power in a given resistor (as the AC) [1]

(c)
- Correct formula used, e.g., \(P = \frac{V_0^2}{2R}\) or \(P = I_{\text{rms}}^2 R\) [1]
- Correct answer: \(3.75\text{ W}\) [1]

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

(b) (i) Photon energy \(E = \frac{hc}{\lambda}\).
\(E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{310 \times 10^{-9}} = 6.416 \times 10^{-19}\text{ J}\).
To convert to \(\text{eV}\):
\(E = \frac{6.416 \times 10^{-19}}{1.60 \times 10^{-19}} = 4.01\text{ eV}\).

(ii) According to Einstein’s photoelectric equation:
\(E_k = hf - \Phi = 4.01\text{ eV} - 2.28\text{ eV} = 1.73\text{ eV}\).

(iii) Maximum kinetic energy in Joules:
\(E_k = 1.73 \times 1.60 \times 10^{-19} = 2.768 \times 10^{-19}\text{ J}\).
Since \(E_k = \frac{1}{2} m_e v^2\):
\(v = \sqrt{\frac{2 E_k}{m_e}} = \sqrt{\frac{2 \times 2.768 \times 10^{-19}}{9.11 \times 10^{-31}}} = \sqrt{6.077 \times 10^{11}} = 7.79
\times 10^5\text{ m s}^{-1}\).

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

(b)
(i)
- Use of \(E = \frac{hc}{\lambda}\) [1]
- Division by \(1.60 \times 10^{-19}\) [1]
- Correct answer: \(4.01\text{ eV}\) (accept \(4.0\text{ eV}\)) [1]
(ii)
- Use of \(E_{k,\text{max}} = E - \Phi\) [1]
- Correct answer: \(1.73\text{ eV}\) (accept range \(1.72\) to \(1.73\)) [1]
(iii)
- Conversion of kinetic energy to Joules [1]
- Use of \(E_k = \frac{1}{2} m_e v^2\) [1]
- Correct answer: \(7.79 \times 10^5\text{ m s}^{-1}\) (accept \(7.8 \times 10^5\text{ m s}^{-1}\)) [1]

(a) Linear attenuation coefficient \(\mu\) is defined by the equation \(I = I_0 e^{-\mu x}\), representing the fractional decrease in X-ray intensity per unit thickness of the attenuating medium.

(b) (i) Using \(I = I_0 e^{-\mu d}\):
\(0.15 I_0 = I_0 e^{-2.3 d} \implies 0.15 = e^{-2.3 d}\).
Taking the natural logarithm of both sides:
\(\ln(0.15) = -2.3 d \implies -1.897 = -2.3 d \implies d = \frac{1.897}{2.3} = 0.825\text{ cm} \approx 0.83\text{ cm}\).
(ii) The half-value thickness is:
\(x_{1/2} = \frac{\ln(2)}{\mu} = \frac{0.693}{2.3} = 0.301\text{ cm} \approx 0.30\text{ cm}\).

(c) Contrast is achieved because different tissues (bone and muscle) have significantly different linear attenuation coefficients. Bone has a much higher density and higher average atomic number than muscle tissue, so it absorbs much more X-ray energy. Consequently, fewer X-rays reach the detector behind the bone, creating a clear difference in exposure (contrast) between bone and surrounding tissue.

(a)
- Defined via \(I = I_0 e^{-\mu x}\) or fractional decrease in intensity per unit thickness [1]
- All terms identified (intensity, initial intensity, thickness) [1]

(b)
(i)
- Formula \(I/I_0 = e^{-\mu d}\) used [1]
- Substitution: \(0.15 = e^{-2.3 d}\) [1]
- Correct answer: \(0.83\text{ cm}\) (or \(8.3\text{ mm}\)) [1]
(ii)
- Formula \(x_{1/2} = \frac{\ln(2)}{\mu}\) used [1]
- Substitution: \(x_{1/2} = \frac{0.693}{2.3}\) [1]
- Correct answer: \(0.30\text{ cm}\) (or \(3.0\text{ mm}\)) [1]

(c)
- Mentions that bone and muscle have different absorption (linear attenuation) coefficients [1]
- Explains that high atomic number/density of bone leads to greater absorption, producing a dark/light contrast on the image [1]

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

(b)(i) The relation for gravitational potential at the surface of a sphere is:
\(\phi = -\frac{GM}{R}\)
Substituting the values:
\(-3.8 \times 10^7 = -\frac{(6.67 \times 10^{-11}) M}{4.2 \times 10^6}\)
\(M = \frac{3.8 \times 10^7 \times 4.2 \times 10^6}{6.67 \times 10^{-11}} \approx 2.39 \times 10^{24}\text{ kg} \approx 2.4 \times 10^{24}\text{ kg}\).

(ii) The escape energy required for the probe to reach infinity is equal to its change in gravitational potential energy:
\(\Delta E_p = 0 - U_{\text{surface}} = -m\phi\)
\(\Delta E_p = -180 \times (-3.8 \times 10^7) = 6.84 \times 10^9\text{ J}\).
Since the initial kinetic energy of the probe (\(3.2 \times 10^9\text{ J}\)) is less than the required energy to escape (\(6.84 \times 10^9\text{ J}\)), the probe does not escape.
Alternatively, the total energy \(E = E_k + U = 3.2 \times 10^9 - 6.84 \times 10^9 = -3.64 \times 10^9\text{ J}\). Since the total energy is negative, the probe remains gravitationally bound and cannot escape.

(iii) By conservation of energy, at maximum height \(h\), the kinetic energy is zero, and the total energy is equal to the gravitational potential energy:
\(E_k + U_i = U_f\)
\(3.2 \times 10^9 + (-6.84 \times 10^9) = -\frac{GMm}{R + h}\)
\(-3.64 \times 10^9 = -\frac{6.67 \times 10^{-11} \times 2.39 \times 10^{24} \times 180}{4.2 \times 10^6 + h}\)
\(-3.64 \times 10^9 = -\frac{2.87 \times 10^{16}}{4.2 \times 10^6 + h}\)
\(4.2 \times 10^6 + h = 7.89 \times 10^6\)
\(h = 3.69 \times 10^6\text{ m}\) (or \(3.7 \times 10^6\text{ m}\)).

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

(b)(i)
- Formula \(\phi = -GM/R\) or equivalent used [1]
- Correct substitution shown resulting in \(2.39 \times 10^{24}\text{ kg}\) [1]

(b)(ii)
- Work done / GPE change required to escape calculated as \(6.84 \times 10^9\text{ J}\) (or \(6.8 \times 10^9\text{ J}\)) [1]
- States total energy at infinity is zero or escape energy is the energy needed [1]
- Compares initial KE of \(3.2 \times 10^9\text{ J}\) to the escape energy and concludes that since KE < escape energy, it cannot escape [1]

(b)(iii)
- Applies conservation of energy: \(E_k + U_i = U_f\) (or gain in GPE = loss in KE) [1]
- Obtains final separation distance \(r = 7.9 \times 10^6\text{ m}\) from center of planet [1]
- Subtracts radius \(R\) to find \(h = 3.7 \times 10^6\text{ m}\) (accept \(3.68 \times 10^6\text{ m}\) to \(3.70 \times 10^6\text{ m}\)) [1]

(a) The root-mean-square (r.m.s.) value of an alternating current is that value of direct current which produces the same average power (or heating effect) in a given resistor as the alternating current.

(b)(i) Comparing the equation to the standard form \(V = V_0 \sin(\omega t)\), we have:
\(\omega = 120 \pi\)
Since \(\omega = 2 \pi f\):
\(2 \pi f = 120 \pi\)
\(f = 60\text{ Hz}\).

(b)(ii) The peak voltage is \(V_0 = 15\text{ V}\).
The peak current is:
\(I_0 = \frac{V_0}{R} = \frac{15}{45} = 0.333\text{ A}\).
The r.m.s. current is:
\(I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{0.333}{\sqrt{2}} \approx 0.236\text{ A}\) (or \(0.24\text{ A}\)).

(b)(iii) The mean power dissipated in the resistor is:
\(P = I_{\text{rms}}^2 R = 0.2357^2 \times 45 = 2.5\text{ W}\).
Alternatively, \(P = \frac{V_{\text{rms}}^2}{R} = \frac{(15 / \sqrt{2})^2}{45} = \frac{112.5}{45} = 2.5\text{ W}\).

(c)(i) The peak current is unchanged (remains \(0.33\text{ A}\)) because the diode has zero resistance in the forward direction and still allows the maximum positive voltage of \(15\text{ V}\) across the resistor.

(c)(ii) The mean power is halved (becomes \(1.25\text{ W}\)) because power is only dissipated during the half-cycle when the diode is forward-biased, and no current flows (zero power) during the other half of the cycle.

(a)
- The value of direct current [1]
- that produces the same power/heating effect in a resistor [1]

(b)(i)
- Equates \(2 \pi f = 120 \pi\) [1]
- \(f = 60\text{ Hz}\) [1]

(b)(ii)
- Calculates \(V_{\text{rms}} = 10.6\text{ V}\) or \(I_0 = 0.333\text{ A}\) [1]
- Calculates \(I_{\text{rms}} = 0.24\text{ A}\) (or \(0.236\text{ A}\)) [1]

(b)(iii)
- Uses suitable power equation (e.g., \(P = I_{\text{rms}}^2 R\) or \(P = V_{\text{rms}}^2 / R\)) [1]
- \(P = 2.5\text{ W}\) [1]

(c)(i)
- Unchanged/no change because the ideal diode has zero resistance when conducting [1]

(c)(ii)
- Halved (or is \(1.25\text{ W}\)) because current only flows during one half-cycle [1]

### 1. Diagram & Experimental Arrangement
- The electric motor is clamped firmly to a bench. A small cylindrical neodymium magnet is attached securely to its shaft.
- The flat search coil is clamped using a boss and stand such that its face is parallel to the plane of the spinning magnet and its central axis aligns coaxially with the motor shaft.
- The search coil is connected to a Cathode-Ray Oscilloscope (CRO) using coaxial cables to minimize external noise.
- A metre rule or Vernier caliper is positioned to measure the distance \(x\) between the search coil and the magnet.

### 2. Variables
- **Independent Variable:** Distance \(x\) between the magnet and the search coil.
- **Dependent Variable:** Peak induced e.m.f. \(V_0\) in the search coil.
- **Controlled Variables:**
- Rotational frequency (speed) of the motor/magnet \(f\).
- Number of turns and cross-sectional area of the search coil.
- Orientation of the search coil (always coaxial to the shaft).

### 3. Data Collection
- Set the motor into motion by connecting it to a variable DC power supply.
- Measure \(x\) using a metre rule (or calipers for smaller distances). To avoid parallax errors, align the rule parallel to the axis of rotation and use a set square.
- The AC voltage signal induced in the search coil is displayed on the CRO. Adjust the time base and voltage sensitivity dial to display a clear sinusoidal wave.
- Measure the peak-to-peak voltage \(V_{\text{p-p}}\) from the grid of the screen and calculate the peak e.m.f. using \(V_0 = \frac{1}{2} V_{\text{p-p}}\).
- Monitor the rotational frequency \(f\) of the magnet. This can be done by using a digital non-contact tachometer pointing at a reflective strip on the motor shaft, or by measuring the period \(T\) of the sinusoidal wave on the CRO screen, where \(f = 1/T\). Keep this frequency constant throughout the experiment by adjusting the variable power supply if necessary.

### 4. Analysis of Data
- The relationship is given by:
\[V_0 = C e^{-kx}\]
- Taking the natural logarithm of both sides:
\[\ln(V_0) = -kx + \ln(C)\]
- Plot a graph of \(\ln(V_0)\) against \(x\).
- If the relationship is valid, the graph of \(\ln(V_0)\) against \(x\) will be a straight line with a negative gradient.
- The constants are determined as:
- \(k = -\text{gradient}\)
- \(C = e^{y\text{-intercept}}\)

### 5. Safety and Detail
- **Safety:** The rotating magnet spinning at high speed presents a risk of flying off or causing injury. Mount a transparent safety screen between the motor assembly and the observer, and wear safety goggles.
- **Detail:** Use a search coil with a large number of turns to ensure a measurable induced e.m.f. even at greater distances \(x\).

### Mark Scheme (Total 15 marks)

#### Defining the Problem (max 3 marks)
- **[1]** Identify \(x\) as the independent variable and \(V_0\) as the dependent variable.
- **[1]** State that the rotational speed (or frequency) of the motor is kept constant.
- **[1]** State that the orientation/alignment of the search coil is kept constant.

#### Methods of Data Collection (max 4 marks)
- **[1]** Draw a clear, labelled diagram showing the motor, spinning magnet, search coil, and a metre rule / calipers positioned to measure distance \(x\).
- **[1]** Describe how \(x\) is measured from the center of the magnet to the center of the coil face.
- **[1]** Describe the use of an oscilloscope (or a peak voltmeter / data logger) to measure the peak voltage \(V_0\).
- **[1]** Describe a method to monitor and keep the rotational speed constant (e.g. using a digital tachometer or checking the constant time-period of the wave on the CRO screen).

#### Method of Analysis (max 3 marks)
- **[1]** Plot a graph of \(\ln(V_0)\) against \(x\).
- **[1]** State that a straight-line graph confirms the suggested relationship.
- **[1]** State that \(k = -\text{gradient}\) and \(C = e^{y\text{-intercept}}\).

#### Safety Considerations (max 1 mark)
- **[1]** Identify risk of high-speed rotating parts and mitigate by using a safety screen / wearing safety goggles / clamping components securely.

#### Additional Details (max 4 marks)
- **[1]** Use of a variable power supply or rheostat to adjust and control the motor speed.
- **[1]** Describe alignment technique to ensure the coil is perfectly coaxial with the rotating shaft (e.g. using a guide rail).
- **[1]** Measure \(x\) several times at different orientations and average to reduce systematic/random error.
- **[1]** Use coaxial/shielded cables to connect the search coil to the CRO to eliminate stray electromagnetic noise.
- **[1]** Detail on finding \(V_0\) from the CRO (e.g., peak-to-peak height multiplied by Y-gain setting, divided by 2).

### Part (a)
Calculate \(1/d\), \(T = t / 20\), and \(\Delta T = \Delta t / 20 = 0.2 / 20 = 0.01\text{ s}\):

- For \(d = 0.100\text{ m}\):
- \(1/d = 10.0\text{ m}^{-1}\)
- \(T = 4.01\text{ s}\), \(\Delta T = \pm 0.01\text{ s}\)
- For \(d = 0.150\text{ m}\):
- \(1/d = 6.67\text{ m}^{-1}\)
- \(T = 2.68\text{ s}\), \(\Delta T = \pm 0.01\text{ s}\)
- For \(d = 0.200\text{ m}\):
- \(1/d = 5.00\text{ m}^{-1}\)
- \(T = 2.01\text{ s}\), \(\Delta T = \pm 0.01\text{ s}\)
- For \(d = 0.250\text{ m}\):
- \(1/d = 4.00\text{ m}^{-1}\)
- \(T = 1.61\text{ s}\), \(\Delta T = \pm 0.01\text{ s}\)
- For \(d = 0.300\text{ m}\):
- \(1/d = 3.33\text{ m}^{-1}\)
- \(T = 1.34\text{ s}\), \(\Delta T = \pm 0.01\text{ s}\)
- For \(d = 0.350\text{ m}\):
- \(1/d = 2.86\text{ m}^{-1}\)
- \(T = 1.15\text{ s}\), \(\Delta T = \pm 0.01\text{ s}\)

### Part (b) & (c)
- **Plotting:** Points plotted on grid. Vertical error bars of size \(\pm 0.01\text{ s}\) are added to each point.
- **Gradient of Best-Fit Line:**
Using two distant points on the best-fit line:
- \((10.0, 4.01)\) and \((2.86, 1.15)\)
- \(m = \frac{4.01 - 1.15}{10.0 - 2.86} = \frac{2.86}{7.14} = 0.401\text{ s m}\) (accept range \(0.395\) to \(0.405\text{ s m}\)).
- **Worst Acceptable Line Gradient:**
A line passing through the top of the first point's error bar and the bottom of the last point's error bar has a gradient:
- \(m_{\text{worst}} = \frac{4.02 - 1.14}{10.0 - 2.86} = \frac{2.88}{7.14} = 0.403\text{ s m}\).
- The absolute uncertainty in gradient \(\Delta m = m_{\text{worst}} - m = 0.403 - 0.401 = 0.002\text{ s m}\) (or up to \(0.004\text{ s m}\) depending on choice of worst line).
- Thus, \(m = (0.401 \pm 0.003)\text{ s m}\).

### Part (d)
- **Equation for \(I\):**
The equation of the straight line is \(T = m \left(\frac{1}{d}\right)\), where:
\[m = 4\pi \sqrt{\frac{I L}{M g}} \implies m^2 = 16\pi^2 \frac{I L}{M g}\]
\[I = \frac{M g m^2}{16\pi^2 L}\]
- **Calculation:**
\[I = \frac{0.350 \times 9.81 \times 0.401^2}{16\pi^2 \times 0.850} = \frac{0.350 \times 9.81 \times 0.1608}{16 \times 9.8696 \times 0.850} \approx 4.11 \times 10^{-3}\text{ kg m}^2\]
- **Uncertainty Analysis:**
Using fractional uncertainties:
\[\frac{\Delta I}{I} = \frac{\Delta M}{M} + \frac{\Delta L}{L} + 2 \frac{\Delta m}{m}\]
\[\frac{\Delta I}{I} = \frac{0.002}{0.350} + \frac{0.005}{0.850} + 2 \left(\frac{0.003}{0.401}\right) = 0.00571 + 0.00588 + 0.01496 = 0.02655\]
\[\Delta I = I \times 0.02655 = 4.11 \times 10^{-3} \times 0.02655 \approx 0.11 \times 10^{-3}\text{ kg m}^2\]
- **Final Answer:**
\[I = (4.11 \pm 0.11) \times 10^{-3}\text{ kg m}^2\]
(or \((4.1 \pm 0.1) \times 10^{-3}\text{ kg m}^2\))

### Mark Scheme (Total 15 marks)

#### Part (a) (max 2 marks)
- **[1]** Correct values of \(1/d\) to 3 significant figures (\(10.0\), \(6.67\), \(5.00\), \(4.00\), \(3.33\), \(2.86\)).
- **[1]** All \(T\) values calculated correctly (\(4.01\), \(2.68\), \(2.01\), \(1.61\), \(1.34\), \(1.15\)) with an absolute uncertainty of \(\pm 0.01\text{ s}\) for all entries.

#### Part (b) (max 4 marks)
- **[1]** Graph axes labelled with quantities and correct units: \(T / \text{s}\) on y-axis, and \((1/d) / \text{m}^{-1}\) on x-axis.
- **[1]** Error bars plotted accurately to within half a small grid square.
- **[1]** Best-fit line drawn carefully with points evenly balanced on either side.
- **[1]** Worst acceptable line drawn (steeper or shallower) passing through all error bars.

#### Part (c) (max 3 marks)
- **[1]** Gradient of best-fit line calculated from points separated by at least half the length of the line drawn.
- **[1]** Correct value of best-fit gradient in the range \(0.395\) to \(0.405\text{ s m}\).
- **[1]** Uncertainty in gradient determined from difference between best-fit and worst acceptable gradients (range \(0.002\) to \(0.005\text{ s m}\)).

#### Part (d) (max 6 marks)
- **[1]** Correct algebraic step relating the gradient \(m\) to \(I\): \(I = \frac{M g m^2}{16\pi^2 L}\).
- **[1]** Correct numerical calculation of \(I\) (approx. \(4.11 \times 10^{-3}\)).
- **[1]** Correct unit of moment of inertia: \(\text{kg m}^2\).
- **[1]** Show formula for fractional uncertainty: \(\frac{\Delta I}{I} = \frac{\Delta M}{M} + \frac{\Delta L}{L} + 2\frac{\Delta m}{m}\).
- **[1]** Calculate the absolute uncertainty in \(I\) to be approximately \(0.11 \times 10^{-3}\text{ kg m}^2\) (or consistent with \(\Delta m\)).
- **[1]** Final value of \(I\) with absolute uncertainty presented to a consistent number of significant figures/decimal places (e.g. \((4.1 \pm 0.1) \times 10^{-3}\text{ kg m}^2\) or \((4.11 \pm 0.11) \times 10^{-3}\text{ kg m}^2\)).