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First, find the time period of one oscillation: T = t / 20 = 36.0 / 20 = 1.80 s. Next, use the formula for a simple pendulum: T = 2\pi \sqrt{L/g}. Rearranging for g gives g = 4\pi^2 L / T^2 = 4\pi^2 \times 0.800 / (1.80^2) = 9.7477 m s^-2. For the uncertainties: % uncertainty in L = (0.002 / 0.800) \times 100% = 0.25%. % uncertainty in T = (0.2 / 36.0) \times 100% = 0.556%. The percentage uncertainty in g is given by: % uncertainty in g = % uncertainty in L + 2 \times % uncertainty in T = 0.25% + 2 \times 0.556% = 1.362%. Therefore, the absolute uncertainty in g is: 9.7477 \times 0.01362 = 0.133 m s^-2. Expressing the final answer with consistent decimal places: g = 9.75 \pm 0.13 m s^-2.

Calculates period T = 1.80 s [1 mark]. Recalls or uses g = 4\pi^2 L / T^2 to calculate g = 9.75 m s^-2 [2 marks]. Calculates % uncertainty in L as 0.25% [1 mark]. Calculates % uncertainty in T as 0.556% [1 mark]. Adds twice the % uncertainty in T to the % uncertainty in L to find % uncertainty in g as 1.36% [2 marks]. Calculates absolute uncertainty as 0.13 m s^-2 [1.57 marks].

(a) The \Lambda^0 baryon has baryon number +1, meaning it contains three quarks. It has strangeness -1, indicating the presence of one strange (s) quark, which has a charge of -1/3. Since the overall charge is 0, the remaining two quarks must have a combined charge of +1/3. This is satisfied by one up (u) quark with charge +2/3 and one down (d) quark with charge -1/3. Hence, the quark composition is uds. (b) Let us check the conservation laws for the reaction: Charge: Initial: -1 (from e^-) + 1 (from p) = 0. Final: -1 (from e^-) + 0 (from \Lambda^0) + 0 (from \pi^0) + 1 (from K^+) = 0. Charge is conserved. Baryon number: Initial: 0 (e^-) + 1 (p) = 1. Final: 0 (e^-) + 1 (\Lambda^0) + 0 (\pi^0) + 0 (K^+) = 1. Baryon number is conserved. Strangeness: Initial: 0 + 0 = 0. Final: 0 (e^-) - 1 (\Lambda^0) + 0 (\pi^0) + 1 (K^+) = 0. Strangeness is conserved. Since strangeness is conserved, this interaction can proceed via the strong force.

Identifies s quark from strangeness [1 mark]. Deduces full uds composition [1.57 marks]. Demonstrates conservation of charge with values [2 marks]. Demonstrates conservation of baryon number with values [2 marks]. Demonstrates conservation of strangeness and states that this allows the strong force to govern the reaction [2 marks].

(a) Using the grating equation d \sin \theta = n \lambda, where n = 2, \lambda = 632.8 \times 10^-9 m, and \theta = 38.5 degrees. Rearranging for the slit spacing d gives d = (2 \times 632.8 \times 10^-9) / \sin(38.5) = 1.2656 \times 10^-6 / 0.62251 = 2.033 \times 10^-6 m. The number of lines per millimetre is given by N = 10^-3 / d = 10^-3 / (2.033 \times 10^-6) = 491.8 lines/mm. Rounding to three significant figures gives 492 lines/mm. (b) For the maximum number of orders, the maximum possible angle \theta is less than 90 degrees, so \sin \theta < 1. This means n < d / \lambda = 2.033 \times 10^-6 / 632.8 \times 10^-9 = 3.21. Since n must be an integer, the maximum observable order is n = 3.

States or uses d \sin \theta = n \lambda [1 mark]. Correctly calculates d = 2.03 \times 10^-6 m [2 marks]. Converts d to lines per millimetre to get 492 [2 marks]. Identifies that the limit of observation is when \sin \theta = 1 [1.57 marks]. Calculates n = 3.21 and correctly rounds down to deduce the maximum order is 3 [2 marks].

(a) Define the direction to the right as positive. Initial total momentum p_i = m_A u_A + m_B u_B = (0.45 \times 2.5) + (0.30 \times -1.8) = 1.125 - 0.54 = 0.585 kg m s^-1. After the collision, glider A moves to the left: v_A = -0.20 m s^-1. Final total momentum p_f = m_A v_A + m_B v_B = (0.45 \times -0.20) + (0.30 \times v_B) = -0.09 + 0.30 v_B. By conservation of momentum, p_i = p_f: 0.585 = -0.09 + 0.30 v_B, which gives 0.30 v_B = 0.675, so v_B = 2.25 m s^-1 (to the right). (b) Initial kinetic energy E_ki = 0.5 m_A u_A^2 + 0.5 m_B u_B^2 = 0.5 \times 0.45 \times 2.5^2 + 0.5 \times 0.30 \times 1.8^2 = 1.40625 + 0.486 = 1.89225 J. Final kinetic energy E_kf = 0.5 m_A v_A^2 + 0.5 m_B v_B^2 = 0.5 \times 0.45 \times 0.20^2 + 0.5 \times 0.30 \times 2.25^2 = 0.009 + 0.759375 = 0.768375 J. Loss in kinetic energy = 1.89225 - 0.768375 = 1.123875 J (approx 1.12 J). Since kinetic energy is not conserved (there is a loss), the collision is inelastic.

Applies conservation of momentum with correct signs for direction [2 marks]. Calculates final velocity of glider B as 2.25 m s^-1 to the right [2 marks]. Calculates initial kinetic energy as 1.89 J [1 mark]. Calculates final kinetic energy as 0.77 J [1 mark]. Subtracts kinetic energies to find loss of 1.12 J and states collision is inelastic [2.57 marks].

(a) The Young modulus is given by E = (F \times L) / (A \times \Delta L), where F = 120 N, L = 3.2 m, A = 1.5 \times 10^-6 m^2, and E = 2.0 \times 10^11 Pa. Rearranging for the extension \Delta L: \Delta L = (F \times L) / (A \times E) = (120 \times 3.2) / (1.5 \times 10^-6 \times 2.0 \times 10^11) = 384 / (3.0 \times 10^5) = 1.28 \times 10^-3 m (or 1.28 mm). (b) Assuming Hooke's law is obeyed, the elastic strain energy stored is given by E_el = 1/2 \times F \times \Delta L = 0.5 \times 120 \times 1.28 \times 10^-3 = 0.0768 J (or 7.68 \times 10^-2 J).

Recalls or writes formula for Young modulus [2 marks]. Rearranges and calculates extension as 1.28 \times 10^-3 m [2.57 marks]. Recalls or writes formula for elastic strain energy E_el = 1/2 F \Delta L [2 marks]. Calculates energy as 0.0768 J (accept 7.7 \times 10^-2 J) [2 marks].

(a) Since the battery voltage is 12.0 V and the potential difference across the thermistor is 4.0 V, the potential difference across the fixed resistor is V_R = 12.0 - 4.0 = 8.0 V. The current through the fixed resistor is I = V_R / R = 8.0 / 2400 = 3.333 \times 10^-3 A. Since it is a series circuit, this current flows through the thermistor. The resistance of the thermistor is R_th = V_th / I = 4.0 / (3.333 \times 10^-3) = 1200 \Omega (or 1.2 k\Omega). (b) When the thermistor resistance falls to R_th' = 800 \Omega, the total resistance of the circuit is R_total = 2400 + 800 = 3200 \Omega. The new reading on the voltmeter across the thermistor is V_th' = V_total \times (R_th' / R_total) = 12.0 \times (800 / 3200) = 12.0 \times 0.25 = 3.0 V.

Calculates voltage across fixed resistor as 8.0 V [1 mark]. Calculates current in the circuit as 3.33 mA [2 marks]. Calculates thermistor resistance as 1200 ohms [1.57 marks]. Applies potential divider formula with new resistance of 800 ohms [2 marks]. Calculates new voltmeter reading as 3.0 V [2 marks].

(a) First, calculate the energy of the incident photons: E = h c / \lambda = (6.63 \times 10^-34 \times 3.00 \times 10^8) / (240 \times 10^-9) = 8.2875 \times 10^-19 J. Convert the work function from eV to Joules: \Phi = 2.28 \times 1.60 \times 10^-19 = 3.648 \times 10^-19 J. Using Einstein's photoelectric equation, the maximum kinetic energy is: E_k(max) = E - \Phi = 8.2875 \times 10^-19 - 3.648 \times 10^-19 = 4.6395 \times 10^-19 J (or 4.64 \times 10^-19 J). (b) Doubling the intensity of the light has no effect on the maximum kinetic energy of the photoelectrons. This is because intensity represents the rate at which photons arrive, but does not alter the energy of individual photons. According to the photon model, each electron absorbs only a single photon; thus, the kinetic energy of individual emitted photoelectrons is determined solely by the frequency (or wavelength) of the light and the work function.

Calculates photon energy as 8.29 \times 10^-19 J [2 marks]. Converts work function to 3.65 \times 10^-19 J [1 mark]. Subtracts work function to get maximum kinetic energy as 4.64 \times 10^-19 J [1.57 marks]. States that there is no change to maximum kinetic energy [2 marks]. Explains that intensity increases photon delivery rate but not the energy of individual photons [2 marks].

Loss in gravitational potential energy of the block: \(\Delta E_p = m g d \sin\theta\).

Normal contact force exerted by the slope on the block: \(N = m g \cos\theta\).

Frictional force acting up the slope: \(f = \mu N = \mu m g \cos\theta\).

Work done against friction: \(W = f d = \mu m g d \cos\theta\).

By conservation of energy, the gain in kinetic energy \(E_k\) is:
\(E_k = \Delta E_p - W = m g d \sin\theta - \mu m g d \cos\theta = m g d (\sin\theta - \mu \cos\theta)\).

This matches option A.

1 mark for the correct answer A.

[1] Correctly identifies that kinetic energy is the difference between GPE lost and work done against friction, leading to the expression \(m g d (\sin\theta - \mu \cos\theta)\).

Let us analyze each conservation law for the hypothetical decay \( n \rightarrow p + e^- \):

1. Baryon number: Neutron has baryon number \( B = +1 \). Proton has baryon number \( B = +1 \). Electron has baryon number \( B = 0 \). Total baryon number is conserved (\( 1 \rightarrow 1 + 0 \)).

2. Lepton number: Neutron has lepton number \( L = 0 \). Proton has lepton number \( L = 0 \). Electron has lepton number \( L = +1 \). Total lepton number is not conserved (\( 0 \rightarrow 0 + 1 \)). Thus, a lepton with lepton number \( -1 \) (specifically, an electron antineutrino \( \bar{\nu}_e \)) must be emitted to conserve lepton number.

3. Charge: Neutron has charge \( Q = 0 \). Proton has charge \( Q = +1 \). Electron has charge \( Q = -1 \). Total charge is conserved (\( 0 \rightarrow +1 - 1 = 0 \)).

Therefore, only the conservation of lepton number prevents this decay from occurring. This matches option B.

1 mark for the correct answer B.

[1] Correctly identifies that only lepton number is violated in the decay \( n \rightarrow p + e^- \).

The grating spacing \( d \) is the distance between adjacent lines. Since there are \( N \) lines per millimetre, there are \( 10^3 N \) lines per metre. Therefore:

\( d = \frac{1}{10^3 N} \text{ m} \)

Using the grating equation for the third-order maximum (\( n = 3 \)):

\( d \sin\theta = n \lambda \)

\( \frac{1}{10^3 N} \sin(30^\circ) = 3 \lambda \)

Since \( \sin(30^\circ) = 0.5 = \frac{1}{2} \):

\( \frac{1}{2 \times 10^3 N} = 3 \lambda \implies \lambda = \frac{1}{6 \times 10^3 N} \text{ m} \)

This matches option A.

1 mark for the correct answer A.

[1] Correctly applies the grating equation \( d \sin\theta = n \lambda \) with \( d = \frac{10^{-3}}{N} \) and \( n = 3 \) to find the wavelength.

The net electric field strength is zero at a point where the electric field strengths due to both charges are equal in magnitude and opposite in direction.

Since the charges have opposite signs (\( +Q \) and \( -4Q \)), the field can only cancel outside the segment joining the two charges. Furthermore, it must be closer to the charge with the smaller magnitude (\( +Q \)).

Let this point be at a distance \( d \) to the left of \( +Q \) (away from \( -4Q \)). The distance from this point to \( -4Q \) is \( L + d \).

Setting the magnitudes of the electric fields equal:

\( \frac{1}{4\pi\varepsilon_0} \frac{Q}{d^2} = \frac{1}{4\pi\varepsilon_0} \frac{4Q}{(L + d)^2} \)

\( \frac{1}{d^2} = \frac{4}{(L + d)^2} \)

Taking the square root of both sides:

\( \frac{1}{d} = \frac{2}{L + d} \implies L + d = 2d \implies d = L \)

Therefore, the point is at distance \( L \) from \( +Q \), on the side away from \( -4Q \). This matches option C.

1 mark for the correct answer C.

[1] Equates the magnitudes of the electric fields for a point outside the charges and solves for the distance from \( +Q \) to find \( d = L \).

The energy stored in a capacitor of capacitance \( C \) at potential difference \( V \) is given by:

\( E_0 = \frac{1}{2} C V^2 \)

When the potential difference has decreased to \( V' = \frac{V}{e} \), the energy remaining in the capacitor is:

\( E = \frac{1}{2} C (V')^2 = \frac{1}{2} C \left(\frac{V}{e}\right)^2 = \frac{1}{2} C \frac{V^2}{e^2} = \frac{E_0}{e^2} \)

This matches option B.

1 mark for the correct answer B.

[1] Correctly uses the formula for capacitor energy \( E = \frac{1}{2} C V^2 \) and substitutes \( V' = V/e \) to get \( E = E_0 / e^2 \).

Using the equation for a cell connected to an external load:

\( \varepsilon = V + I r \)

Since \( I = \frac{V}{R} \), we have:

\( \varepsilon = V \left(1 + \frac{r}{R}\right) \)

For the two situations:

\( V_1 \left(1 + \frac{r}{R_1}\right) = V_2 \left(1 + \frac{r}{R_2}\right) \)

Expand the terms:

\( V_1 + \frac{V_1 r}{R_1} = V_2 + \frac{V_2 r}{R_2} \)

Rearranging to group the terms with \( r \):

\( V_2 - V_1 = r \left(\frac{V_1}{R_1} - \frac{V_2}{R_2}\right) = r \left(\frac{V_1 R_2 - V_2 R_1}{R_1 R_2}\right) \)

Solving for \( r \):

\( r = \frac{(V_2 - V_1) R_1 R_2}{V_1 R_2 - V_2 R_1} \)

This matches option A.

1 mark for the correct answer A.

[1] Correctly uses the terminal potential difference equation for two states and algebraically solves for the internal resistance \( r \).

The acceleration due to gravity on the surface of a planet is given by:

\( g = \frac{G M}{R^2} \)

For the new planet, \( M' = 2M \) and \( R' = 2R \). Thus, the new acceleration due to gravity is:

\( g' = \frac{G (2M)}{(2R)^2} = \frac{2}{4} \frac{G M}{R^2} = \frac{g}{2} \)

The period of a simple pendulum of length \( L \) is:

\( T = 2\pi \sqrt{\frac{L}{g}} \)

On the new planet, the period of oscillation \( T' \) is:

\( T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g/2}} = \sqrt{2} \times 2\pi \sqrt{\frac{L}{g}} = \sqrt{2} T \)

This matches option C.

1 mark for the correct answer C.

[1] Correctly deduces that gravity on the planet is half of Earth's gravity, and since \( T \propto 1/\sqrt{g} \), the new period is \( \sqrt{2} T \).

For a string of length \( L \) fixed at both ends vibrating in its third harmonic, there are 3 half-wavelengths along the length of the string:

\( L = 3 \left(\frac{\lambda}{2}\right) \implies \lambda = \frac{2L}{3} \)

Using the wave equation \( v = f \lambda \), where \( v \) is the speed of the progressive waves:

\( v = f \left(\frac{2L}{3}\right) = \frac{2 L f}{3} \)

This matches option B.

1 mark for the correct answer B.

[1] Correctly identifies that for the third harmonic, the wavelength is \( \lambda = \frac{2L}{3} \), and applies \( v = f \lambda \) to obtain the wave speed.

To determine which decay can only occur via the weak interaction, we examine the conservation of strangeness \(S\):

- In option A, \(\Delta^{++}\) (quark structure \(uuu\), strangeness \(S=0\)) decays to \(p\) (\(uud\), \(S=0\)) and \(\pi^+\) (\(u\bar{d}\), \(S=0\)). Strangeness is conserved (\(0 \rightarrow 0\)), and no quark flavours change. This occurs via the strong interaction.
- In option B, \(\Sigma^0\) (\(uds\), \(S=-1\)) decays to \(\Lambda^0\) (\(uds\), \(S=-1\)) and a photon \(\gamma\). This is an electromagnetic decay because strangeness is conserved and a photon is emitted.
- In option C, \(\Xi^-\space (dss, S = -2)\) decays to \(\Lambda^0\space (uds, S = -1)\) and \(\pi^-\space (d\bar{u}, S = 0)\). The total strangeness changes from \(-2\) to \(-1\) (so \(\Delta S = +1\)). A change in strangeness can only occur in weak interactions, so this decay must occur via the weak interaction.
- In option D, \(\rho^0\) is a vector meson of strangeness 0, which decays to two pions of strangeness 0. This is a strong interaction decay.

1 mark for correct option C.
- Correctly identifies that strangeness is violated by \(\Delta S = +1\) only in option C, signifying a weak interaction.
- Rejects other options as they are strong or electromagnetic interactions where strangeness is conserved.

Let the final velocities of the blocks of mass \(m\) and \(3m\) be \(v_1\) and \(v_2\) respectively.

From the conservation of linear momentum (since the initial momentum of the system is zero):
\(0 = m v_1 - 3m v_2 \implies v_1 = 3v_2\)

The total energy released is \(E\), which becomes the sum of the kinetic energies of the two blocks:
\(E = \frac{1}{2} m v_1^2 + \frac{1}{2} (3m) v_2^2\)

We can express \(v_2\) in terms of \(v_1\) as \(v_2 = \frac{v_1}{3}\):
\(E = \frac{1}{2} m v_1^2 + \frac{1}{2} (3m) \left(\frac{v_1}{3}\right)^2\)
\(E = \frac{1}{2} m v_1^2 + \frac{1}{6} m v_1^2 = \frac{2}{3} m v_1^2\)

Since the kinetic energy of the mass \(m\) is \(E_{k1} = \frac{1}{2} m v_1^2\):
\(E = \frac{4}{3} \left(\frac{1}{2} m v_1^2\right) = \frac{4}{3} E_{k1}\)

Solving for \(E_{k1}\):
\(E_{k1} = \frac{3}{4} E = 0.75 E\)

1 mark for correct option D.
- Correct conservation of momentum equation used to relate the velocities.
- Correct sum of kinetic energies set equal to energy \(E\).
- Accurate algebraic simplification.

At room temperature, the total resistance of the series circuit is:
\(R_{\text{total, 1}} = 3R + R = 4R\)

The current through the circuit is:
\(I_1 = \frac{V_0}{4R}\)

The power dissipated in the fixed resistor at room temperature is:
\(P_1 = I_1^2 R = \left(\frac{V_0}{4R}\right)^2 R = \frac{V_0^2}{16R}\)

At the higher temperature, the total resistance is:
\(R_{\text{total, 2}} = 0.5R + R = 1.5R\)

The current through the circuit is:
\(I_2 = \frac{V_0}{1.5R} = \frac{2V_0}{3R}\)

The power dissipated in the fixed resistor at the higher temperature is:
\(P_2 = I_2^2 R = \left(\frac{2V_0}{3R}\right)^2 R = \frac{4V_0^2}{9R}\)

The ratio of the powers is:
\(\frac{P_2}{P_1} = \frac{\frac{4V_0^2}{9R}}{\frac{V_0^2}{16R}} = \frac{4}{9} \times 16 = \frac{64}{9}\)

1 mark for correct option D.
- Correct expression of currents in terms of \(V_0\) and \(R\).
- Calculation of powers in terms of \(V_0^2/R\).
- Evaluation of the ratio of powers.

The fundamental frequency \(f\) of a stretched wire is given by:
\(f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\)
where \(\mu\) is the mass per unit length (constant).

Let the new fundamental frequency be \(f'\):
\(f' = \frac{1}{2(0.75L)} \sqrt{\frac{1.44T}{\mu}}\)

We can factor out the changes:
\(f' = \frac{\sqrt{1.44}}{0.75} \left( \frac{1}{2L} \sqrt{\frac{T}{\mu}} \right) = \frac{1.2}{0.75} f\)

Simplifying the fraction:
\(\frac{1.2}{0.75} = \frac{120}{75} = 1.60\)

So, \(f' = 1.60 f\).

1 mark for correct option D.
- Applies formula for the fundamental frequency of a vibrating string.
- Performs scale factor substitutions for length and tension correctly.
- Resolves the numerical coefficient to 1.60.

Using the grating equation:
\(d \sin\theta = n \lambda\)

First, calculate the grating spacing \(d\):
\(d = \frac{1\text{ mm}}{500} = \frac{1.0 \times 10^{-3}\text{ m}}{500} = 2.0 \times 10^{-6}\text{ m}\)

For the third-order maximum, \(n = 3\) and \(\sin\theta = \sin(53^\circ) = 0.80\):
\(\lambda = \frac{d \sin\theta}{n} = \frac{(2.0 \times 10^{-6}\text{ m}) \times 0.80}{3}\)
\(\lambda = \frac{1.6 \times 10^{-6}\text{ m}}{3} \approx 5.33 \times 10^{-7}\text{ m} = 533\text{ nm}\)

To two significant figures, this is \(530\text{ nm}\).

1 mark for correct option C.
- Correct determination of grating spacing \(d\).
- Correct rearrangement and calculation of \(\lambda\) using the diffraction formula.

The elastic strain energy stored in a wire under tension is given by:
\(U = \frac{1}{2} F \Delta L\)

From the definition of the Young Modulus \(E = \frac{F L}{A \Delta L}\), we can express the extension as:
\(\Delta L = \frac{F L}{A E}\)

Substituting this back into the energy formula gives:
\(U = \frac{F^2 L}{2 A E}\)

Since both wires are made of the same metal, \(E\) is constant. The force \(F\) is also the same for both. Therefore, \(U \propto \frac{L}{A}\).

Since the cross-sectional area \(A = \frac{\pi d^2}{4}\), we have \(A \propto d^2\), which means:
\(U \propto \frac{L}{d^2}\)

Comparing wire \(Y\) to wire \(X\):
- \(L_Y = 2 L_X\)
- \(d_Y = 0.5 d_X\)

So,
\(\frac{U_Y}{U_X} = \frac{L_Y / d_Y^2}{L_X / d_X^2} = \frac{2 L_X / (0.5 d_X)^2}{L_X / d_X^2} = \frac{2}{0.25} = 8\)

1 mark for correct option D.
- Derives expression relating strain energy to length and cross-sectional area.
- Incorporates diameter relationship correctly (inverse-square dependence).
- Computes ratio accurately.

The formula for the resistivity \(\rho\) is:
\(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\)

The percentage uncertainty in \(\rho\) is calculated by adding the individual percentage uncertainties, multiplying by powers where applicable:
\(\%\Delta\rho = \%\Delta R + 2(\%\Delta d) + \%\Delta L\)

Calculate the percentage uncertainty for each term:
- \(\%\Delta R = \frac{0.08}{4.00} \times 100\% = 2.0\%\)
- \(\%\Delta d = \frac{0.02}{0.40} \times 100\% = 5.0\%\)
- \(\%\Delta L = \frac{0.005}{1.250} \times 100\% = 0.4\%\)

Combine these values:
\(\%\Delta\rho = 2.0\% + 2(5.0\%) + 0.4\%\)
\(\%\Delta\rho = 2.0\% + 10.0\% + 0.4\% = 12.4\%\)

1 mark for correct option C.
- Correctly calculates individual percentage uncertainties.
- Correctly applies the power rule to the uncertainty of diameter (multiplies by 2).
- Adds uncertainties to find the final percentage uncertainty.

According to Einstein's photoelectric equation, the maximum kinetic energy of photoelectrons emitted by light of frequency \(f\) is:
\(E_k = hf - \Phi\)
This can be rewritten as:
\(hf = E_k + \Phi\)

When the incident frequency is doubled to \(2f\), the new maximum kinetic energy \(E_k'\) is:
\(E_k' = h(2f) - \Phi = 2(hf) - \Phi\)

Substitute the expression for \(hf\):
\(E_k' = 2(E_k + \Phi) - \Phi\)
\(E_k' = 2E_k + 2\Phi - \Phi = 2E_k + \Phi\)

1 mark for correct option C.
- Formulates the initial and final photoelectric equations.
- Uses substitution to express the new kinetic energy in terms of \(E_k\) and \(\Phi\).

Let us test the conservation of baryon number \( B \) for each decay:

* **For option A:** \( p \rightarrow e^+ + \pi^0 \)
On the left-hand side: proton is a baryon, so \( B = +1 \).
On the right-hand side: positron is a lepton (\( B = 0 \)) and pion is a meson (\( B = 0 \)).
Total baryon number on the right-hand side is \( 0 + 0 = 0 \).
Since \( 1 \neq 0 \), baryon number is **not conserved**.

* **For option B:** \( \mu^- \rightarrow e^- + \bar{\nu}_e + \nu_\mu \)
All particles involved are leptons (or anti-leptons), so baryon number on both sides is \( 0 \). Baryon number is conserved.

* **For option C:** \( n \rightarrow p + e^- + \bar{\nu}_e \)
The neutron and proton each have \( B = +1 \). The electron and anti-neutrino have \( B = 0 \). Thus \( B \) is conserved (\( 1 = 1 + 0 + 0 \)).

* **For option D:** \( \pi^+ \rightarrow \mu^+ + \nu_\mu \)
All particles are mesons or leptons, so baryon number on both sides is \( 0 \). Baryon number is conserved.

1 mark for identifying that option A violates the conservation of baryon number because the proton has a baryon number of +1, while the decay products are leptons and mesons which all have a baryon number of 0.

We can determine the impulse from the change in momentum of the ball.

1. Find the speed \( v_1 \) of the ball immediately before impact using \( v^2 = u^2 + 2as \):
\( v_1 = \sqrt{2gh} \)
Taking the downwards direction as negative, the initial velocity is \( \vec{v}_1 = -\sqrt{2gh} \).

2. Find the speed \( v_2 \) of the ball immediately after impact, knowing it rebounds to height \( 0.64h \):
\( v_2 = \sqrt{2g(0.64h)} = 0.8\sqrt{2gh} \)
Taking the upwards direction as positive, the final velocity is \( \vec{v}_2 = +0.8\sqrt{2gh} \).

3. Calculate the impulse \( J = \Delta p = m\vec{v}_2 - m\vec{v}_1 \):
\( J = m(0.8\sqrt{2gh}) - m(-\sqrt{2gh}) = 1.8m\sqrt{2gh} \).

1 mark for calculating the magnitude of momentum change by properly accounting for the change in direction of the velocity vectors, yielding \( 1.8m\sqrt{2gh} \).

We use the conservation of mechanical energy. The total energy remains constant, so the gain in kinetic energy \( E_k \) equals the loss in electric potential energy \( E_p \).

1. The formula for the electric potential energy between two point charges separated by distance \( x \) is:
\( E_p = \frac{Qq}{4\pi\varepsilon_0 x} \)

2. The initial potential energy at distance \( r \) is:
\( E_{p,i} = \frac{Qq}{4\pi\varepsilon_0 r} \)

3. The final potential energy at distance \( 3r \) is:
\( E_{p,f} = \frac{Qq}{4\pi\varepsilon_0 (3r)} = \frac{Qq}{12\pi\varepsilon_0 r} \)

4. The gain in kinetic energy is:
\( E_k = E_{p,i} - E_{p,f} = \frac{Qq}{4\pi\varepsilon_0 r} - \frac{Qq}{12\pi\varepsilon_0 r} = \frac{Qq}{4\pi\varepsilon_0 r} \left(1 - \frac{1}{3}\right) = \frac{2Qq}{12\pi\varepsilon_0 r} = \frac{Qq}{6\pi\varepsilon_0 r} \).

1 mark for applying the conservation of energy principle and correctly subtracting the final potential energy from the initial potential energy to find \( \frac{Qq}{6\pi\varepsilon_0 r} \).

From the diffraction grating equation:
\( d \sin\theta_n = n\lambda \)

For the first case (third-order, \( n=3 \), wavelength \( \lambda_1 = \lambda \)):
\( d \sin\theta = 3\lambda \implies \sin\theta = \frac{3\lambda}{d} \)

For the second case (second-order, \( n=2 \), wavelength \( \lambda_2 = 1.5\lambda \)):
\( d \sin\theta_2 = 2 \times (1.5\lambda) = 3\lambda \implies \sin\theta_2 = \frac{3\lambda}{d} \)

Since \( \sin\theta_2 = \sin\theta \), the angle of diffraction for the second-order maximum is also \( \theta \).

1 mark for establishing that the product of order and wavelength \( n\lambda \) is identical in both scenarios, meaning the path difference remains the same, hence \( \theta_2 = \theta \).

The fundamental frequency of a stretched wire is given by:
\( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \)

Let \( f' \) be the new fundamental frequency when the length becomes \( L' = 2L \) and the tension becomes \( T' = 4T \):
\( f' = \frac{1}{2(2L)} \sqrt{\frac{4T}{\mu}} \)
\( f' = \frac{1}{4L} \times 2 \sqrt{\frac{T}{\mu}} = \frac{1}{2L} \sqrt{\frac{T}{\mu}} = f \)

Therefore, the new fundamental frequency remains unchanged at \( f \).

1 mark for demonstrating that doubling the length and quadrupling the tension results in a net multiplier of 1 for the fundamental frequency.

1. At \( 60^\circ\text{C} \), the potential difference across the fixed resistor is \( V_{\text{fixed}} = 8.0\text{ V} \).
2. Since the supply is \( 12.0\text{ V} \), the potential difference across the thermistor is:
\( V_{\text{th}} = 12.0\text{ V} - 8.0\text{ V} = 4.0\text{ V} \)
3. For components connected in series, the potential difference across each component is proportional to its resistance:
\( \frac{R_{\text{th}}}{R_{\text{fixed}}} = \frac{V_{\text{th}}}{V_{\text{fixed}}} \)
\( R_{\text{th}} = R_{\text{fixed}} \times \frac{V_{\text{th}}}{V_{\text{fixed}}} = 4.0\ \text{k}\Omega \times \frac{4.0\text{ V}}{8.0\text{ V}} = 2.0\ \text{k}\Omega \).

1 mark for calculating the potential difference across the thermistor (4.0 V) and applying the potential divider ratio to yield a thermistor resistance of 2.0 kΩ.

1. When two identical springs of spring constant \( k \) are connected in parallel, the effective spring constant is:
\( k_{\text{eff}} = k + k = 2k \)
2. The formula for the time period of a mass-spring system is:
\( T = 2\pi \sqrt{\frac{m}{k}} \)
3. The new time period \( T' \) is:
\( T' = 2\pi \sqrt{\frac{m}{2k}} = \frac{1}{\sqrt{2}} \left( 2\pi \sqrt{\frac{m}{k}} \right) = \frac{T}{\sqrt{2}} \).

1 mark for determining that parallel springs double the stiffness, leading to a new period of \( \frac{T}{\sqrt{2}} \).

The elastic strain energy \( E_s \) stored in a wire under tension \( F \) is:
\( E_s = \frac{1}{2} F \Delta L \)

Since the extension is \( \Delta L = \frac{F L}{A E} \), we can rewrite the strain energy as:
\( E_s = \frac{F^2 L}{2 A E} \)

Both wires are made of the same material, so the Young modulus \( E \) is identical. The applied force \( F \) is also identical. Thus:
\( E_s \propto \frac{L}{A} \)

Since \( A \propto d^2 \), where \( d \) is the diameter:
\( E_s \propto \frac{L}{d^2} \)

For wire P:
\( E_{s,P} \propto \frac{L}{d^2} \)

For wire Q:
\( E_{s,Q} \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \frac{L}{d^2} \)

Therefore, the ratio of the energy stored in wire P to wire Q is:
\( \frac{E_{s,P}}{E_{s,Q}} = \frac{1}{\left(\frac{1}{2}\right)} = 2 \).

1 mark for correctly determining the relationship between strain energy, length, and diameter, and calculating the ratio of 2.

The horizontal velocity \(v\) of each particle remains constant as it passes through the plates. The time \(t\) spent between the plates of length \(L\) is:

\(t = \frac{L}{v}\)

The vertical acceleration \(a\) due to the uniform electric field \(E\) is given by Newton's second law:

\(a = \frac{F}{m} = \frac{qE}{m}\)

where \(q\) is the charge and \(m\) is the mass of the particle.

The vertical deflection \(y\) upon exiting the plates is given by:

\(y = \frac{1}{2} a t^2 = \frac{1}{2} \left(\frac{qE}{m}\right) \left(\frac{L}{v}\right)^2 = \frac{q E L^2}{2 m v^2}\)

Since the initial kinetic energy is \(E_k = \frac{1}{2} m v^2\), we can substitute \(m v^2 = 2 E_k\) into the equation for \(y\):

\(y = \frac{q E L^2}{4 E_k}\)

For a given electric field, plate length, and initial kinetic energy, the vertical deflection \(y\) is directly proportional to the charge \(q\) of the particle:

\(y \propto q\)

An alpha particle has a charge of \(q_{\alpha} = 2e\) and a proton has a charge of \(q_p = e\). Therefore, the ratio of their deflections is:

\(\frac{y_{\alpha}}{y_p} = \frac{q_{\alpha}}{q_p} = \frac{2e}{e} = 2\)

C is the correct answer.

Award 1 mark for selecting the correct option.
No partial marks are available for this multiple-choice question.

(a) Using Kepler's Third Law: \(T^2 \propto r^3\).
\(\left(\frac{T_B}{T_A}\right)^2 = \left(\frac{r_B}{r_A}\right)^3 = \left(\frac{4.8 \times 10^5}{2.4 \times 10^5}\right)^3 = 2^3 = 8\).
\(T_B = T_A \times \sqrt{8} = 4.2 \times 2.83 = 11.9\text{ days}\).

(b) Converting period to seconds: \(T_A = 4.2 \times 24 \times 3600 = 3.63 \times 10^5\text{ s}\).
Converting radius to meters: \(r_A = 2.4 \times 10^8\text{ m}\).
Using the orbital equation: \(T^2 = \frac{4\pi^2 r^3}{G M_p}\).
\(M_p = \frac{4\pi^2 r_A^3}{G T_A^2} = \frac{4 \pi^2 (2.4 \times 10^8)^3}{(6.67 \times 10^{-11}) (3.63 \times 10^5)^2} = 6.21 \times 10^{25}\text{ kg}\), which rounds to \(6.2 \times 10^{25}\text{ kg}\).

(c) Escape velocity formula: \(v_{\text{esc}} = \sqrt{\frac{2 G M_{\text{moon}}}{R_{\text{moon}}}}\).
\(v_{\text{esc}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 1.5 \times 10^{20}}{3.5 \times 10^5}} = \sqrt{5.72 \times 10^4} = 239\text{ m s}^{-1}\) (or \(240\text{ m s}^{-1}\) to 2 s.f.).

(a)
- Use of Kepler's Third Law relationship: \(T^2 \propto r^3\) [1 mark]
- Correct ratio substituted: \(T_B = 4.2 \times \sqrt{8}\) [1 mark]
- Correct final answer: 12 days (or 11.9 days) [1 mark]

(b)
- Conversion of period to seconds AND radius to meters [1 mark]
- Rearrangement of orbital formula to make \(M_p\) the subject [1 mark]
- Calculation showing exoplanet mass \(M_p = 6.2 \times 10^{25}\text{ kg}\) [1 mark]

(c)
- Recall of escape velocity formula: \(v_{\text{esc}} = \sqrt{\frac{2GM}{R}}\) [1 mark]
- Correct substitution of values [1 mark]
- Correct final answer: 240 or 239 \(\text{m s}^{-1}\) [1 mark]

(a) The direction of an electric field is the direction of the force on a positive test charge. Since the upper plate is positive, a positive charge would be repelled downwards. Thus, the electric field is directed vertically downwards.

(b) Electric field strength \(E = \frac{V}{d} = \frac{120}{15 \times 10^{-3}} = 8000\text{ V m}^{-1}\).
Electrostatic force \(F = E e = 8000 \times 1.60 \times 10^{-19} = 1.28 \times 10^{-15}\text{ N}\).

(c) Vertical acceleration \(a_y = \frac{F}{m_e} = \frac{1.28 \times 10^{-15}}{9.11 \times 10^{-31}} = 1.405 \times 10^{15}\text{ m s}^{-2}\).
Vertical distance to upper plate \(y = 7.5\text{ mm} = 7.5 \times 10^{-3}\text{ m}\).
Using \(y = \frac{1}{2} a_y t^2\), time taken \(t = \sqrt{\frac{2 y}{a_y}} = \sqrt{\frac{2 \times 7.5 \times 10^{-3}}{1.405 \times 10^{15}}} = 3.27 \times 10^{-9}\text{ s}\).
Horizontal distance \(x = v_x t = 4.5 \times 10^6 \times 3.27 \times 10^{-9} = 0.0147\text{ m} = 1.47 \times 10^{-2}\text{ m}\) (or \(15\text{ mm}\)).

(a)
- Field direction is downwards [1 mark]
- Explanation: direction is from positive to negative plate (or force direction on a positive charge) [1 mark]

(b)
- Calculation of electric field strength \(E = 8000\text{ V m}^{-1}\) [1 mark]
- Recall and substitution into \(F = E e\) [1 mark]
- Correct evaluation: \(1.3 \times 10^{-15}\text{ N}\) (accept \(1.28 \times 10^{-15}\text{ N}\)) [1 mark]

(c)
- Calculation of vertical acceleration: \(a_y = 1.4 \times 10^{15}\text{ m s}^{-2}\) [1 mark]
- Use of vertical displacement \(y = 7.5\text{ mm}\) to calculate flight time \(t = 3.3 \times 10^{-9}\text{ s}\) [1 mark]
- Calculation of horizontal distance using \(x = v_x t\) [1 mark]
- Correct final answer: \(0.015\text{ m}\) (or \(14.7\text{ mm}\) / \(15\text{ mm}\)) [1 mark]

(a) The time constant \(\tau = RC\) is the time taken for the charge, voltage, or current of a discharging capacitor to decrease to \(\frac{1}{e}\) (or approximately \(37\%\)) of its initial value.

(b) The discharge equation is \(V = V_0 e^{-t / RC}\).
Substituting the values: \(3.0 = 9.0 e^{-15 / (R \times 470 \times 10^{-6})}\).
\(\frac{1}{3} = e^{-15 / (R \times 470 \times 10^{-6})}\).
Taking natural logarithms on both sides:
\(\ln(3) = \frac{15}{R \times 470 \times 10^{-6}}\).
\(R = \frac{15}{470 \times 10^{-6} \times \ln(3)} = \frac{15}{5.163 \times 10^{-4}} = 2.905 \times 10^4\ \Omega\), which is approximately \(2.9 \times 10^4\ \Omega\).

(c) Energy stored in a capacitor is \(E = \frac{1}{2} C V^2\).
Substituting \(C = 470 \times 10^{-6}\text{ F}\) and \(V = 3.0\text{ V}\):
\(E = \frac{1}{2} \times (470 \times 10^{-6}) \times (3.0)^2 = 2.115 \times 10^{-3}\text{ J} = 2.1\text{ mJ}\).

(a)
- Defines time constant as time taken for charge/voltage/current to fall to \(1/e\) (or \(37\%\)) [1 mark]
- Of its initial value [1 mark]

(b)
- Uses \(V = V_0 e^{-t/RC}\) [1 mark]
- Rearranges and takes natural logarithms correctly: \(\ln(3) = \frac{t}{RC}\) or equivalent [1 mark]
- Substitutes \(C = 470 \times 10^{-6}\text{ F}\) and \(t = 15\text{ s}\) [1 mark]
- Shows calculation leading to \(2.9 \times 10^4\ \Omega\) [1 mark]

(c)
- Recalls formula \(E = \frac{1}{2} C V^2\) [1 mark]
- Substitutes \(C = 470 \times 10^{-6}\text{ F}\) and \(V = 3.0\text{ V}\) [1 mark]
- Calculates correct value: \(2.1 \times 10^{-3}\text{ J}\) (or \(2.1\text{ mJ}\)) [1 mark]

(a) Kinetic energy gained \(E_k = q V = (1.6 \times 10^{-19}\text{ C}) \times (2500\text{ V}) = 4.0 \times 10^{-16}\text{ J}\).

(b) The magnetic force on a moving charge is given by \(F = Bqv\) and is always perpendicular to both the velocity of the ion and the magnetic field. Since the force is perpendicular to velocity, it performs no work and causes a change in direction only, providing a constant centripetal acceleration, which results in a circular path.

(c) First, find the velocity \(v\):
\(E_k = \frac{1}{2} m v^2 \implies v = \sqrt{\frac{2 E_k}{m}} = \sqrt{\frac{2 \times 4.0 \times 10^{-16}}{2.0 \times 10^{-26}}} = 2.0 \times 10^5\text{ m s}^{-1}\).
Equating centripetal and magnetic forces: \(\frac{m v^2}{r} = B q v \implies r = \frac{m v}{B q}\).
\(r = \frac{2.0 \times 10^{-26} \times 2.0 \times 10^5}{0.18 \times 1.6 \times 10^{-19}} = \frac{4.0 \times 10^{-21}}{2.88 \times 10^{-20}} = 0.139\text{ m}\) (or \(0.14\text{ m}\)).

(d) Substituting \(v = \sqrt{\frac{2 E_k}{m}}\), the radius expression is \(r = \frac{\sqrt{2 m E_k}}{B q}\).
Since both ions have the same charge and are accelerated through the same potential difference, they have the same kinetic energy \(E_k\). Since \(r \propto \sqrt{m}\), the carbon-14 ion, having a greater mass, will travel in a circular path of larger radius.

(a)
- States \(E_k = q V\) [1 mark]
- Substitutes values to show \(4.0 \times 10^{-16}\text{ J}\) [1 mark]

(b)
- Identifies that magnetic force is perpendicular to velocity [1 mark]
- Explains that this perpendicular force acts as a centripetal force causing circular motion [1 mark]

(c)
- Calculates velocity \(v = 2.0 \times 10^5\text{ m s}^{-1}\) [1 mark]
- Uses circular motion equation \(r = \frac{mv}{Bq}\) [1 mark]
- Obtains correct radius of \(0.14\text{ m}\) (or \(0.139\text{ m}\)) [1 mark]

(d)
- Identifies that both ions have the same kinetic energy [1 mark]
- Concludes that radius is larger because mass is greater (using \(r \propto \sqrt{m}\) or by calculating new velocity and radius) [1 mark]

(a) Convert temperature to Kelvin: \(T = 20 + 273.15 = 293.15\text{ K}\).
Using the ideal gas equation: \(p V = N k T\).
\(N = \frac{p V}{k T} = \frac{1.2 \times 10^5 \times 0.045}{1.38 \times 10^{-23} \times 293.15} = 1.335 \times 10^{24}\text{ atoms}\) (accept \(1.3 \times 10^{24}\)).

(b) At constant volume: \(\frac{p_1}{T_1} = \frac{p_2}{T_2}\).
\(T_2 = T_1 \times \frac{p_2}{p_1} = 293.15 \times \frac{1.8 \times 10^5}{1.2 \times 10^5} = 439.7\text{ K}\).
Converting to Celsius: \(t_2 = 439.7 - 273.15 = 166.6\ ^\circ\text{C}\) (or \(167\ ^\circ\text{C}\)).

(c) Average kinetic energy of an ideal gas molecule: \(E_k = \frac{3}{2} k T\).
Using \(T = 439.7\text{ K}\):
\(E_k = 1.5 \times 1.38 \times 10^{-23} \times 439.7 = 9.10 \times 10^{-21}\text{ J}\) (accept \(9.1 \times 10^{-21}\text{ J}\)).

(a)
- Converts temperature to Kelvin (\(293\text{ K}\)) [1 mark]
- Recalls and rearranges \(p V = N k T\) [1 mark]
- Correct calculation of \(1.3 \times 10^{24}\) atoms [1 mark]

(b)
- Uses pressure law (constant volume) relation \(T_2 = T_1 \times \frac{p_2}{p_1}\) with temperatures in Kelvin [1 mark]
- Converts result back to Celsius to get \(167\ ^\circ\text{C}\) (accept \(166\ ^\circ\text{C}\) to \(167\ ^\circ\text{C}\)) [1 mark]

(c)
- Recalls kinetic energy formula \(E_k = \frac{3}{2} k T\) [1 mark]
- Substitutes temperature in Kelvin (\(440\text{ K}\)) [1 mark]
- Correct final kinetic energy \(9.1 \times 10^{-21}\text{ J}\) [1 mark]

(a) Converting hours to seconds: \(t = 6.0 \times 3600 = 21600\text{ s}\).
Using \(A = A_0 e^{-\lambda t}\):
\(1.5 \times 10^5 = 8.4 \times 10^5 e^{-\lambda \times 21600}\).
\(\frac{1.5}{8.4} = e^{-\lambda \times 21600} \implies \ln\left(\frac{1.5}{8.4}\right) = -21600 \lambda\).
\(-1.723 = -21600 \lambda \implies \lambda = 7.98 \times 10^{-5}\text{ s}^{-1}\) (or \(8.0 \times 10^{-5}\text{ s}^{-1}\)).
If calculated in hours: \(\lambda = \frac{\ln(8.4/1.5)}{6.0} = 0.287\text{ hr}^{-1}\).

(b) Half-life \(T_{1/2} = \frac{\ln(2)}{\lambda}\).
Using \(\lambda\) in \(\text{hr}^{-1}\): \(T_{1/2} = \frac{0.693}{0.287} = 2.41\text{ hours}\) (or \(2.4\text{ hours}\)).

(c) Activity \(A = \lambda N\), so \(N = \frac{A}{\lambda}\).
After \(6.0\text{ hours}\), \(A = 1.5 \times 10^5\text{ Bq}\).
Using \(\lambda\) in \(\text{s}^{-1}\):
\(N = \frac{1.5 \times 10^5}{7.98 \times 10^{-5}} = 1.88 \times 10^9\) nuclei (or \(1.9 \times 10^9\) nuclei).

(a)
- Correct formula used: \(A = A_0 e^{-\lambda t}\) [1 mark]
- Correct rearrangement using natural logarithms [1 mark]
- Correct value of \(\lambda\): \(8.0 \times 10^{-5}\) or \(0.29\) [1 mark]
- Correct unit matching value: \(\text{s}^{-1}\) or \(\text{hr}^{-1}\) [1 mark]

(b)
- Recalls \(T_{1/2} = \frac{\ln(2)}{\lambda}\) [1 mark]
- Calculates correct half-life: \(2.4\text{ hours}\) (or \(8700\text{ s}\)) [1 mark]

(c)
- Recalls and rearranges \(A = \lambda N\) [1 mark]
- Correctly calculates \(1.9 \times 10^9\) nuclei remaining [1 mark]

(a) Power \(P = \frac{\Delta Q}{\Delta t}\).
Since \(\Delta Q = m c \Delta T\), we have \(P = m c \frac{\Delta T}{\Delta t}\).
Substituting the values:
\(150 = 0.45 \times c \times 0.68\).
\(c = \frac{150}{0.45 \times 0.68} = \frac{150}{0.306} = 490\text{ J kg}^{-1}\text{ K}^{-1}\) (or \(490\text{ J kg}^{-1}\ ^\circ\text{C}^{-1}\)).

(b) If heat is lost to the surroundings, the actual thermal energy absorbed by the block is less than the electrical energy delivered by the heater (\(P \Delta t\)). Using \(P \Delta t\) in the calculations overestimates the energy entering the block, leading to an overestimation (larger calculated value) of the specific heat capacity.

(c) Thermal energy needed to melt the ice: \(Q = m L = 0.20 \times 3.34 \times 10^5 = 6.68 \times 10^4\text{ J}\).
Time required: \(t = \frac{Q}{P} = \frac{6.68 \times 10^4}{150} = 445\text{ s}\) (or \(450\text{ s}\) to 2 s.f.).

(a)
- Recalls thermal energy power relation \(P = m c \frac{\Delta T}{\Delta t}\) [1 mark]
- Substitutes rate of temperature change (\(0.68\)) and mass (\(0.45\)) correctly [1 mark]
- Obtains correct specific heat capacity: \(490\text{ J kg}^{-1}\text{ K}^{-1}\) [1 mark]

(b)
- States that actual energy transferred to the metal is less than calculated from heater power [1 mark]
- Concludes that the calculated specific heat capacity is larger than the true value [1 mark]

(c)
- Uses \(Q = m L\) to find heat required [1 mark]
- Equates energy to \(P \times t\) and rearranges [1 mark]
- Calculates correct time: \(450\text{ s}\) (or \(445\text{ s}\)) [1 mark]

The pressure \( p \) of an ideal gas is given by the kinetic theory equation: \( p = \frac{1}{3} \frac{N m c^2_{rms}}{V} \), where \( c^2_{rms} \) is the mean square speed of the molecules.

Let the initial pressure be \( p_1 = \frac{1}{3} \frac{N m c^2_{rms}}{V} \).

When the number of molecules is doubled to \( 2N \) and the mean square speed is halved to \( 0.5 c^2_{rms} \), the new pressure \( p_2 \) is:
\( p_2 = \frac{1}{3} \frac{(2N) m (0.5 c^2_{rms})}{V} = \frac{1}{3} \frac{N m c^2_{rms}}{V} = p_1 \).

Thus, the pressure remains unchanged.

1 mark for selecting the correct option (B).
- Award 1 mark for the correct reasoning that the pressure depends on the product of the number of molecules and their mean square speed, resulting in no change.

The work done on the satellite is equal to the increase in its gravitational potential energy (\( \Delta E_p \)).

The formula for gravitational potential energy is:
\( E_p = -\frac{GMm}{r} \)

Therefore, the change in potential energy is:
\( \Delta E_p = E_{p, \text{final}} - E_{p, \text{initial}} \)
\( \Delta E_p = -\frac{GMm}{3R} - \left(-\frac{GMm}{R}\right) \)
\( \Delta E_p = \frac{GMm}{R} - \frac{GMm}{3R} = \frac{2GMm}{3R} \)

1 mark for selecting the correct option (B).
- Award 1 mark for calculating the difference between the final and initial gravitational potential energy values.

Let the charge \( +2Q \) be located at \( x = 0 \) and the charge \( -Q \) be at \( x = d \).

The electric potential \( V \) at a point \( x \) along the line is given by:
\( V = \frac{1}{4\pi\varepsilon_0} \left( \frac{2Q}{|x|} + \frac{-Q}{|x-d|} \right) = 0 \)

This simplifies to:
\( \frac{2}{|x|} = \frac{1}{|x-d|} \)

We consider two regions where the potential can be zero:

1) Between the charges (\( 0 < x < d \)):
\( \frac{2}{x} = \frac{1}{d-x} \implies 2(d-x) = x \implies 2d = 3x \implies x = \frac{2}{3}d \)
The distance from \( -Q \) (at \( x=d \)) is \( d - x = d - \frac{2}{3}d = \frac{1}{3}d \).

2) Outside the charges, to the right of \( -Q \) (\( x > d \)):
\( \frac{2}{x} = \frac{1}{x-d} \implies 2(x-d) = x \implies 2x - 2d = x \implies x = 2d \)
The distance from \( -Q \) (at \( x=d \)) is \( x - d = 2d - d = d \).

Therefore, the potential is zero at both a distance of \( \frac{d}{3} \) and \( d \) from the charge \( -Q \).

1 mark for selecting the correct option (C).
- Award 1 mark for analyzing both possible positions (between and outside the charges) along the line where the potentials sum to zero.

The energy \( E \) stored in a capacitor at any instant is given by:
\( E = \frac{1}{2} C V^2 \)

During discharge, the potential difference \( V \) decays exponentially with time \( t \):
\( V = V_0 e^{-\frac{t}{RC}} \)

Substituting this into the energy formula:
\( E = \frac{1}{2} C \left( V_0 e^{-\frac{t}{RC}} \right)^2 = E_0 e^{-\frac{2t}{RC}} \)

We want to find the time \( t \) when the energy \( E \) is half of the initial energy \( E_0 \):
\( 0.5 E_0 = E_0 e^{-\frac{2t}{RC}} \)
\( 0.5 = e^{-\frac{2t}{RC}} \)

Taking the natural logarithm of both sides:
\( \ln(0.5) = -\frac{2t}{RC} \)
\( -\ln(2) = -\frac{2t}{RC} \)
\( t = \frac{RC \ln(2)}{2} \)

1 mark for selecting the correct option (C).
- Award 1 mark for using the relationship \( E \propto V^2 \) and solving the exponential decay equation for energy.

For a charged particle of mass \( m \) and charge \( q \) moving at speed \( v \) perpendicular to a magnetic field \( B \), the magnetic force provides the centripetal force:
\( B q v = \frac{m v^2}{r} \implies r = \frac{m v}{B q} \)

The momentum \( p = m v \) can be related to the kinetic energy \( E_k \) by:
\( E_k = \frac{p^2}{2m} \implies p = \sqrt{2 m E_k} \)

Substituting this into the radius formula:
\( r = \frac{\sqrt{2 m E_k}}{B q} \)

Let the proton mass be \( m_p = m \) and its charge be \( q_p = e \):
\( r_p = \frac{\sqrt{2 m E_k}}{B e} \)

An alpha particle consists of 2 protons and 2 neutrons, so its mass is approximately \( m_\alpha = 4m \) and its charge is \( q_\alpha = 2e \):
\( r_\alpha = \frac{\sqrt{2 (4m) E_k}}{B (2e)} = \frac{2 \sqrt{2 m E_k}}{2 B e} = \frac{\sqrt{2 m E_k}}{B e} \)

Thus, the ratio \( \frac{r_p}{r_\alpha} = 1 \), which is \( 1 : 1 \).

1 mark for selecting the correct option (B).
- Award 1 mark for relating circular radius to kinetic energy and substituting the relative mass and charge values for a proton and an alpha particle.

Let the initial number of nuclei of Y be \( N_{Y0} = N_0 \). Since there are twice as many nuclei of X initially, \( N_{X0} = 2N_0 \).

The number of nuclei remaining after time \( t \) is given by:
\( N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \)

For X:
\( N_X = 2N_0 \left(\frac{1}{2}\right)^{\frac{t}{4}} \)

For Y:
\( N_Y = N_0 \left(\frac{1}{2}\right)^{\frac{t}{8}} \)

We set the two numbers equal:
\( 2N_0 \left(\frac{1}{2}\right)^{\frac{t}{4}} = N_0 \left(\frac{1}{2}\right)^{\frac{t}{8}} \)

Dividing both sides by \( N_0 \):
\( 2 \cdot 2^{-\frac{t}{4}} = 2^{-\frac{t}{8}} \)
\( 2^{1 - \frac{t}{4}} = 2^{-\frac{t}{8}} \)

Equating the exponents:
\( 1 - \frac{t}{4} = -\frac{t}{8} \)
\( 1 = \frac{t}{4} - \frac{t}{8} \)
\( 1 = \frac{t}{8} \implies t = 8 \text{ hours} \)

1 mark for selecting the correct option (B).
- Award 1 mark for setting up the equations for decay of both isotopes and solving for \( t \).

The energy released in fission is the difference between the total binding energy of the final product nuclei and the total binding energy of the initial parent nucleus.

Total initial binding energy of parent nucleus:
\( E_{\text{initial}} = 240 \times 7.6 \text{ MeV} = 1824 \text{ MeV} \)

Total final binding energy of the two identical product fragments:
\( E_{\text{final}} = 2 \times (120 \times 8.5 \text{ MeV}) = 240 \times 8.5 \text{ MeV} = 2040 \text{ MeV} \)

Energy released:
\( \Delta E = E_{\text{final}} - E_{\text{initial}} = 2040 \text{ MeV} - 1824 \text{ MeV} = 216 \text{ MeV} \)

1 mark for selecting the correct option (B).
- Award 1 mark for calculating the total initial and final binding energies and subtracting them to find the energy released.

The heat energy \( Q \) supplied by the heater is given by:
\( Q = P \Delta t \)
where:
\( P = 50 \text{ W} \)
\( \Delta t = 4.0 \text{ minutes} = 4.0 \times 60 \text{ s} = 240 \text{ s} \)

\( Q = 50 \times 240 = 12,000 \text{ J} \)

During melting, the thermal energy is used solely to change state without changing temperature:
\( Q = m L_f \)
where:
\( m = 0.20 \text{ kg} \)
\( L_f \) is the specific latent heat of fusion.

\( 12,000 = 0.20 \times L_f \implies L_f = \frac{12,000}{0.20} = 60,000 \text{ J kg}^{-1} = 6.0 \times 10^4 \text{ J kg}^{-1} \)

1 mark for selecting the correct option (B).
- Award 1 mark for converting minutes to seconds, calculating total energy supplied, and correctly applying \( Q = m L_f \).

Let \(m\) be the mass of ice added.
The total heat energy supplied \(Q = 150\text{ kJ} = 150 \times 10^3\text{ J}\).

This energy is consumed in three parts:
1. Melting the ice of mass \(m\) at \(0^\circ\text{C}\):
\(Q_1 = m L_{\text{fusion}} = m \times 3.3 \times 10^5\text{ J}\)

2. Raising the temperature of the melted ice of mass \(m\) from \(0^\circ\text{C}\) to \(20^\circ\text{C}\):
\(Q_2 = m c \Delta T = m \times 4200 \times 20 = m \times 84\,000\text{ J}\)

3. Raising the temperature of the existing \(0.20\text{ kg}\) of water from \(0^\circ\text{C}\) to \(20^\circ\text{C}\):
\(Q_3 = 0.20 \times 4200 \times 20 = 16\,800\text{ J}\)

Using conservation of energy:
\(Q = Q_1 + Q_2 + Q_3\)

\(150\,000 = m(3.3 \times 10^5 + 84\,000) + 16\,800\)

\(133\,200 = 414\,000 m\)

\(m = \frac{133\,200}{414\,000} \approx 0.32\text{ kg}\)

1 mark for the correct answer (C).

The total energy \(E\) of a satellite of mass \(m\) in a stable circular orbit of radius \(r\) around a planet of mass \(M\) is given by:
\(E = E_{\text{kinetic}} + E_{\text{potential}} = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}\)

Initial total energy at \(r_1 = 3R\):
\(E_1 = -\frac{GMm}{6R}\)

Final total energy at \(r_2 = 4R\):
\(E_2 = -\frac{GMm}{8R}\)

Change in total energy \(\Delta E\):
\(\Delta E = E_2 - E_1 = -\frac{GMm}{8R} - \left(-\frac{GMm}{6R}\right) = \frac{GMm}{R} \left( \frac{1}{6} - \frac{1}{8} \right) = +\frac{GMm}{24R}\)

1 mark for the correct answer (B).

Let the charge \(+4q\) be at position \(x = 0\) and the charge \(-q\) be at position \(x = d\).
The net electric field can only be zero in the region where the individual field vectors from the two charges point in opposite directions and have equal magnitude.
- Between the charges (\(0 < x < d\)), both fields point to the right, so they cannot cancel.
- To the left of \(+4q\) (\(x < 0\)), the field from \(+4q\) is stronger than that from \(-q\) because \(4q > q\) and the distance is smaller, so they cannot cancel.
- To the right of \(-q\) (\(x > d\)), the fields point in opposite directions. Let this point be at a distance \(x\) to the right of \(-q\). The distance from \(+4q\) is \(d + x\).

Equating the magnitudes of the electric field strengths:
\(\frac{1}{4\pi\varepsilon_0} \frac{4q}{(d+x)^2} = \frac{1}{4\pi\varepsilon_0} \frac{q}{x^2}\)

Taking the square root of both sides:
\(\frac{2}{d+x} = \frac{1}{x}\)

Solving for \(x\):
\(2x = d + x \implies x = d\)

Thus, the net electric field is zero at a distance \(d\) to the right of the charge \(-q\).

1 mark for the correct answer (C).

The original energy stored in the first capacitor is:
\(U_0 = \frac{1}{2} C V^2\)

The original charge on the first capacitor is:
\(Q = C V\)

When connected in parallel with the uncharged \(2C\) capacitor, the total capacitance becomes:
\(C_{\text{total}} = C + 2C = 3C\)

By conservation of charge, the new potential difference \(V'\) across the parallel combination is:
\(V' = \frac{Q}{C_{\text{total}}} = \frac{CV}{3C} = \frac{V}{3}\)

The final electrostatic energy stored in both capacitors is:
\(U_{\text{final}} = \frac{1}{2} C_{\text{total}} (V')^2 = \frac{1}{2} (3C) \left(\frac{V}{3}\right)^2 = \frac{1}{6} C V^2\)

The energy dissipated as heat and radiation is:
\(\Delta U = U_0 - U_{\text{final}} = \frac{1}{2} C V^2 - \frac{1}{6} C V^2 = \frac{1}{3} C V^2\)

The fraction of the original energy dissipated is:
\(\frac{\Delta U}{U_0} = \frac{\frac{1}{3} C V^2}{\frac{1}{2} C V^2} = \frac{2}{3}\)

1 mark for the correct answer (D).

As the coil is pulled out of the magnetic field, the rate of change of magnetic flux linkage is:
\(\frac{d\Phi}{dt} = N B \frac{dA}{dt} = N B w v\)

According to Faraday's law, the magnitude of the induced electromotive force (emf) is:
\(\varepsilon = N B w v\)

The induced current in the coil is:
\(I = \frac{\varepsilon}{R} = \frac{N B w v}{R}\)

This current experiences a magnetic force that opposes the motion of the coil. The force acts on the side of width \(w\) that is still inside the field. For \(N\) turns, this opposing magnetic force is:
\(F_{\text{magnetic}} = N B I w = N B \left(\frac{N B w v}{R}\right) w = \frac{N^2 B^2 w^2 v}{R}\)

Since the coil is pulled at a constant speed, the pulling force must be equal in magnitude to this magnetic force:
\(F_{\text{pull}} = \frac{N^2 B^2 w^2 v}{R}\)

1 mark for the correct answer (C).

The number of active parent nuclei remaining at time \(t = t_1\) is given by:
\(N(t_1) = N_0 e^{-\lambda t_1}\)

The number of active parent nuclei remaining at time \(t = t_2\) is:
\(N(t_2) = N_0 e^{-\lambda t_2}\)

Each decay of a parent nucleus produces exactly one stable daughter nucleus. Therefore, the number of daughter nuclei produced between \(t_1\) and \(t_2\) is equal to the number of parent nuclei that decayed in this time interval:
\(\Delta N_{\text{decayed}} = N(t_1) - N(t_2) = N_0 e^{-\lambda t_1} - N_0 e^{-\lambda t_2} = N_0 (e^{-\lambda t_1} - e^{-\lambda t_2})\)

1 mark for the correct answer (B).

From the kinetic theory of gases, the root-mean-square speed of molecules is given by:
\(v = \sqrt{\frac{3kT}{m}}\)
which means \(v \propto \sqrt{T}\), where \(T\) is the absolute temperature.

From the ideal gas equation \(PV = NkT\), we have:
\(T = \frac{PV}{Nk}\)

Let the initial state of the gas be \(P_1\), \(V_1\), \(T_1\). The final state has \(P_2 = 4P_1\) and \(V_2 = \frac{1}{2}V_1\).
The new temperature \(T_2\) is:
\(T_2 = \frac{P_2 V_2}{Nk} = \frac{(4P_1)(\frac{1}{2}V_1)}{Nk} = 2 \frac{P_1 V_1}{Nk} = 2T_1\)

Since the absolute temperature is doubled, the new r.m.s. speed \(v_2\) is:
\(v_2 \propto \sqrt{T_2} = \sqrt{2T_1} \implies v_2 = \sqrt{2} v\)

1 mark for the correct answer (B).

The energy required to separate the nucleus into these specific components is the difference in total binding energies between the initial nucleus and the final products.

Initial state: one \(\text{}_3^7\text{Li}\) nucleus
Total binding energy \(E_{\text{Li}} = 7 \times 5.60\text{ MeV} = 39.20\text{ MeV}\)

Final state products:
- One \(\text{}_2^4\text{He}\) nucleus: total binding energy \(E_{\text{He}} = 4 \times 7.07\text{ MeV} = 28.28\text{ MeV}\)
- One \(\text{}_1^2\text{H}\) nucleus: total binding energy \(E_{\text{H}} = 2 \times 1.11\text{ MeV} = 2.22\text{ MeV}\)
- One free neutron \(\text{}_0^1\text{n}\): binding energy = \(0\text{ MeV}\)

Total binding energy of the products:
\(E_{\text{products}} = 28.28 + 2.22 + 0 = 30.50\text{ MeV}\)

Minimum energy required:
\(\Delta E = E_{\text{Li}} - E_{\text{products}} = 39.20 - 30.50 = 8.70\text{ MeV}\)

1 mark for the correct answer (B).

The root-mean-square (r.m.s.) speed \(c_{\text{rms}}\) of molecules in an ideal gas is given by the formula \(c_{\text{rms}} = \sqrt{\frac{3 k_{\text{B}} T}{m}}\), where \(T\) is the absolute temperature and \(m\) is the mass of a single molecule.

For container X: \(c_{\text{X}} = \sqrt{\frac{3 k_{\text{B}} T}{m}}\)

For container Y: \(c_{\text{Y}} = \sqrt{\frac{3 k_{\text{B}} (4T)}{2m}} = \sqrt{2} \sqrt{\frac{3 k_{\text{B}} T}{m}}\)

Therefore, the ratio of the r.m.s. speed in Y to that in X is:

\frac{c_{\text{Y}}}{c_{\text{X}}} = \frac{\sqrt{2} c_{\text{X}}}{c_{\text{X}}} = \sqrt{2}.

1 mark for the correct answer A.

For a satellite of mass \(m\) in a stable circular orbit of radius \(r\) around a planet of mass \(M\), the orbital speed \(v\) is given by:

\frac{m v^2}{r} = \frac{G M m}{r^2} \implies E_{\text{k}} = \frac{1}{2} m v^2 = \frac{G M m}{2r}

The initial kinetic energy is \(E_{\text{k}} = \frac{G M m}{2R}\), which means \frac{G M m}{R} = 2 E_{\text{k}}.

The gravitational potential energy of the satellite in orbit is \(E_{\text{p}} = -\frac{G M m}{r}\).

- Initial potential energy (\(r = R\)): \(E_{\text{p, initial}} = -\frac{G M m}{R} = -2 E_{\text{k}}\)
- Final potential energy (\(r = 3R\)): \(E_{\text{p, final}} = -\frac{G M m}{3R} = -\frac{2}{3} E_{\text{k}}\)

The change in gravitational potential energy is:

\Delta E_{\text{p}} = E_{\text{p, final}} - E_{\text{p, initial}} = -\frac{2}{3} E_{\text{k}} - (-2 E_{\text{k}}) = +\frac{4}{3} E_{\text{k}}.

1 mark for the correct answer C.

Let the \(+2Q\) charge be at position \(x = 0\) and the \(-Q\) charge be at \(x = d\).

For a point between the charges at a distance \(x\) from \(+2Q\) (where \(0 < x < d\)), the total electric potential is:

V = \frac{1}{4\pi\varepsilon_0} \left( \frac{2Q}{x} + \frac{-Q}{d - x} \right)

Setting \(V = 0\) gives:

\frac{2}{x} = \frac{1}{d - x} \implies 2(d - x) = x \implies 2d - 2x = x \implies 3x = 2d \implies x = \frac{2}{3}d

This position is at a distance of \(\frac{2}{3}d\) from the \(+2Q\) charge, which corresponds to a distance of \(d - \frac{2}{3}d = \frac{1}{3}d\) from the \(-Q\) charge.

1 mark for the correct answer A.

The initial charge on the first capacitor is \(Q_0 = C V\).

The initial energy stored is:

E_{\text{initial}} = \frac{1}{2} \frac{Q_0^2}{C}

When connected in parallel with an uncharged capacitor of capacitance \(2C\), the total capacitance becomes \(C_{\text{total}} = C + 2C = 3C\). The total charge \(Q_0\) remains conserved and is shared between the two capacitors.

The final total energy stored is:

E_{\text{final}} = \frac{1}{2} \frac{Q_0^2}{C_{\text{total}}} = \frac{1}{2} \frac{Q_0^2}{3C} = \frac{1}{3} E_{\text{initial}}

The energy lost is:

\Delta E = E_{\text{initial}} - E_{\text{final}} = E_{\text{initial}} - \frac{1}{3} E_{\text{initial}} = \frac{2}{3} E_{\text{initial}}

Therefore, the fraction of the initial energy lost is \(\frac{2}{3}\).

1 mark for the correct answer C.

For a charged particle in a magnetic field, the centripetal force is provided by the magnetic force:

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

Since kinetic energy is \(E_{\text{k}} = \frac{1}{2} m v^2\), the momentum is \(p = m v = \sqrt{2 m E_{\text{k}}}\).

Substituting this into the radius formula gives:

r = \frac{\sqrt{2 m E_{\text{k}}}}{q B}

For the second particle:
- mass \(m_2 = 2m\)
- charge \(q_2 = 2q\)
- kinetic energy \(E_{\text{k2}} = \frac{1}{2} E_{\text{k}}\)

The radius of its path \(r_2\) is:

r_2 = \frac{\sqrt{2 (2m) (\frac{1}{2} E_{\text{k}})}}{2q B} = \frac{\sqrt{2 m E_{\text{k}}}}{2 q B} = \frac{1}{2} r.

1 mark for the correct answer B.

Let \(N_0\) be the initial number of radioactive nuclei. After three half-lives, the number of remaining radioactive nuclei \(N\) is:

N = N_0 \left(\frac{1}{2}\right)^3 = \frac{1}{8} N_0

The number of decayed nuclei \(N_{\text{decayed}}\) is:

N_{\text{decayed}} = N_0 - N = N_0 - \frac{1}{8} N_0 = \frac{7}{8} N_0

The ratio of the number of decayed nuclei to the number of remaining nuclei is:

\frac{N_{\text{decayed}}}{N} = \frac{\frac{7}{8} N_0}{\frac{1}{8} N_0} = 7.

1 mark for the correct answer C.

The time period of a simple pendulum is given by:

T = 2\pi \sqrt{\frac{L}{g}}

For the pendulum on the planet:
- length \(L_{\text{P}} = 2 L\)
- gravitational field strength \(g_{\text{P}} = \frac{1}{4} g\)

The new period \(T_{\text{P}}\) is:

T_{\text{P}} = 2\pi \sqrt{\frac{2 L}{\frac{1}{4} g}} = 2\pi \sqrt{8 \frac{L}{g}} = \sqrt{8} \times 2\pi \sqrt{\frac{L}{g}} = 2\sqrt{2} T.

1 mark for the correct answer D.

An electron (negative charge) and a proton (positive charge) have opposite charges, so the electrostatic force between them is attractive. Since both have mass, the gravitational force between them is also attractive.

Comparing the magnitudes of these forces, the electrostatic force is many orders of magnitude stronger than the gravitational force (\(F_{\text{E}} \gg F_{\text{G}}\)). Therefore, \(F_{\text{E}} > F_{\text{G}}\) and both forces are attractive.

1 mark for the correct answer A.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

To find the point where the net gravitational field strength is zero, we equate the magnitudes of the gravitational fields due to both masses. Let this point be at a distance \(r\) from mass \(M\), which means it is at a distance \(d - r\) from mass \(4M\). The condition is: \(\frac{GM}{r^2} = \frac{G(4M)}{(d-r)^2}\). Taking the square root of both sides gives \(\frac{1}{r} = \frac{2}{d-r}\). Solving for \(r\) gives \(d - r = 2r\), which simplifies to \(3r = d\), or \(r = \frac{d}{3}\). The distance from mass \(4M\) is therefore \(d - r = \frac{2d}{3}\). Now, we calculate the total gravitational potential \(V\) at this point: \(V = -\frac{GM}{r} - \frac{G(4M)}{d-r}\). Substituting the values of \(r\) and \(d-r\) gives: \(V = -\frac{GM}{\frac{d}{3}} - \frac{4GM}{\frac{2d}{3}} = -\frac{3GM}{d} - \frac{6GM}{d} = -\frac{9GM}{d}\). Therefore, the correct answer is B.

Award 1 mark for the correct answer B. Method mark: Find that the field is zero at \(r = d/3\) from mass \(M\). Accuracy mark: Correctly sum the two negative potentials to find the total potential is \(-9GM/d\). Reject other options.

**Part a**
- Place the thermistor in a water bath, ensuring it is completely submerged.
- Stir the water continuously before taking any reading to prevent local temperature gradients.
- Wait for the temperature on the digital thermometer to stabilise at each point before recording the output voltage to ensure the thermistor is in thermal equilibrium with the water.

**Part b**
- The formula for the output voltage of a potential divider is:
\( V_{\text{out}} = V_0 \frac{R}{R_T + R} \)
- Rearranging this for \( R_T \) gives:
\( R_T = R \left( \frac{V_0}{V_{\text{out}}} - 1 \right) \)
- Substituting the nominal values:
\( R_T = 1000 \left( \frac{5.00}{2.15} - 1 \right) = 1000 \times (2.3256 - 1) = 1325.6 \ \Omega \approx 1330 \ \Omega \)
- To find the uncertainty in \( R_T \), let \( y = \frac{V_0}{V_{\text{out}}} \):
\( \%\Delta y = \%\Delta V_0 + \%\Delta V_{\text{out}} = \left(\frac{0.05}{5.00} \times 100\%\right) + \left(\frac{0.02}{2.15} \times 100\%\right) = 1.0\% + 0.93\% = 1.93\% \)
\( \Delta y = 2.3256 \times 0.0193 = 0.0449 \)
- Since \( R_T = Ry - R \) and \( R \) has no uncertainty:
\( \Delta R_T = R \Delta y = 1000 \times 0.0449 = 44.9 \ \Omega \approx 45 \ \Omega \)
So, \( R_T = 1330 \pm 45 \ \Omega \) (or \( 1326 \pm 45 \ \Omega \)).

**Part c**
- Taking natural logarithms on both sides:
\( \ln(R_T) = \ln(R_{\infty} e^{B/T}) \Rightarrow \ln(R_T) = \ln(R_{\infty}) + \frac{B}{T} \)
- This is in the form \( y = mx + c \) where \( y = \ln(R_T/\Omega) \), \( x = 1/T \), \( m = B \), and \( c = \ln(R_{\infty}/\Omega) \).
- Therefore:
- The constant \( B \) is equal to the gradient of the graph. Its unit is Kelvin (\( \text{K} \)), as the exponent \( B/T \) must be dimensionless.
- The constant \( R_{\infty} \) is equal to \( e^{\text{intercept}} \). Its unit is the Ohm (\( \Omega \)), as it is a pre-exponential resistance factor.

**Part d**
- The absolute uncertainty in the gradient \( B \) is:
\( \Delta B = |m_{\text{worst}} - m_{\text{best}}| = |3350 - 3200| = 150 \text{ K} \)
- The percentage uncertainty in \( B \) is:
\( \%\Delta B = \frac{150}{3200} \times 100\% = 4.69\% \approx 4.7\% \)
- Expressing \( B \) with its absolute uncertainty to appropriate significant figures:
\( B = 3200 \pm 150 \text{ K} \) (or \( 3.2 \times 10^3 \pm 0.2 \times 10^3 \text{ K} \)).

**Part a (3 marks)**
- **[1 mark]** Submerges thermistor in water bath/oil bath.
- **[1 mark]** Stirs the bath continuously before each reading.
- **[1 mark]** Waits for temperature reading to stabilise before recording voltage.

**Part b (4 marks)**
- **[1 mark]** Uses correct rearranged formula \( R_T = R \left( \frac{V_0}{V_{\text{out}}} - 1 \right) \) and finds nominal value of \( R_T \approx 1330 \ \Omega \).
- **[1 mark]** Adds percentage uncertainties of \( V_0 \) and \( V_{\text{out}} \) to get \( 1.93\% \) (or uses max/min values: \( R_{T,\text{max}} \approx 1371 \ \Omega \), \( R_{T,\text{min}} \approx 1281 \ \Omega \)).
- **[1 mark]** Calculates absolute uncertainty in \( R_T \) as \( 45 \ \Omega \) (or \( 44.9 \ \Omega \)).
- **[1 mark]** Expresses final answer as \( 1330 \pm 45 \ \Omega \) (or \( 1326 \pm 45 \ \Omega \)), with correct significant figures (uncertainty to 1 or 2 s.f., value matching decimal place precision).

**Part c (4 marks)**
- **[1 mark]** Rearranges to logarithmic form: \( \ln(R_T) = \ln(R_{\infty}) + \frac{B}{T} \).
- **[1 mark]** Identifies \( B = \text{gradient} \) and \( R_{\infty} = e^{\text{intercept}} \).
- **[1 mark]** Correctly identifies unit of \( B \) as \( \text{K} \).
- **[1 mark]** Correctly identifies unit of \( R_{\infty} \) as \( \Omega \).

**Part d (4 marks)**
- **[1 mark]** Identifies absolute uncertainty in gradient as \( 150 \text{ K} \).
- **[1 mark]** Calculates percentage uncertainty in \( B \) as \( 4.7\% \) (accept \( 4.69\% \)).
- **[1 mark]** Rounds absolute uncertainty to 1 or 2 significant figures (e.g. \( 150 \text{ K} \) or \( 200 \text{ K} \)).
- **[1 mark]** Formats final value consistently with uncertainty, including correct unit (e.g. \( 3200 \pm 150 \text{ K} \)).

**Part a**
- A small resistance \( R \) results in a very small time constant \( \tau = RC \), meaning the discharge happens extremely quickly.
- Human reaction time when using a manual stopwatch (typically \( \sim 0.2 \text{ s} \)) would introduce a massive percentage uncertainty into the time measurements.
- A digital data logger can capture voltage readings at high sampling rates (e.g. hundreds of readings per second), allowing a precise, automated trace of the rapid decay curve without human delay errors.

**Part b**
- Taking natural logarithms of both sides of \( V = V_0 e^{-t/RC} \):
\( \ln(V) = \ln(V_0) - \frac{1}{RC}t \)
- Comparing this to \( y = mx + c \), the gradient of the graph of \( \ln(V) \) against \( t \) is:
\( \text{gradient} = -\frac{1}{RC} = -\frac{1}{\tau} \)
- Therefore, the time constant \( \tau \) is equal to \( -\frac{1}{\text{gradient}} \).
- Since the relationship involves a simple reciprocal (which acts like a power of \( -1 \)), the percentage uncertainty in \( \tau \) is exactly the same as the percentage uncertainty in the gradient, which is \( 4.5\% \).

**Part c**
- The nominal value of \( \tau \) is:
\( \tau = RC = (10 \times 10^3 \ \Omega) \times (470 \times 10^{-6} \text{ F}) = 4.70 \text{ s} \)
- The percentage uncertainty in a product is the sum of the individual percentage uncertainties:
\( \%\Delta \tau = \%\Delta R + \%\Delta C = 5\% + 20\% = 25\% \)
- The maximum possible value of \( \tau \) occurs when both \( R \) and \( C \) are at their maximum values:
\( R_{\text{max}} = 10 \times 10^3 \times 1.05 = 10.5 \text{ k}\Omega \)
\( C_{\text{max}} = 470 \times 10^{-6} \times 1.20 = 564 \ \mu\text{F} \)
\( \tau_{\text{max}} = 10.5 \times 10^3 \times 564 \times 10^{-6} = 5.922 \text{ s} \approx 5.92 \text{ s} \)
(Alternatively, using the exact combined percentage factor: \( 4.70 \times (1 + 0.25) \) is accepted as \( 5.88 \text{ s} \)).

**Part d**
- At \( t = 3\tau \), the voltage is:
\( V = V_0 e^{-3} = 9.00 \times e^{-3} = 9.00 \times 0.049787 = 0.448 \text{ V} \)
- The absolute uncertainty is the resolution of the sensor: \( \Delta V = 0.01 \text{ V} \).
- The percentage uncertainty in this reading is:
\( \%\Delta V = \frac{0.01}{0.448} \times 100\% = 2.23\% \approx 2.2\% \)
- Comment: For even larger discharge times (e.g. \( t > 5\tau \)), the voltage drops to extremely low values (e.g. \( V \approx 0.06 \text{ V} \) at \( 5\tau \)). Because the absolute resolution uncertainty remains fixed at \( 0.01 \text{ V} \), the percentage uncertainty rises rapidly (exceeding \( 16\% \)). This makes the data point readings highly inaccurate and creates significant scatter on the logarithmic plot at large discharge times.

**Part a (3 marks)**
- **[1 mark]** Identifies that low resistance means a small time constant / very fast discharge.
- **[1 mark]** Explains that human reaction time introduces significant/unacceptable percentage uncertainty in manual timing.
- **[1 mark]** Explains that a data logger can take readings at a high sampling rate to capture the rapid curve accurately.

**Part b (4 marks)**
- **[1 mark]** Show algebraic derivation of \( \ln(V) = \ln(V_0) - \frac{t}{RC} \).
- **[1 mark]** States that the gradient is equal to \( -1/RC \) (or \( -1/\tau \)).
- **[1 mark]** Clearly states \( \tau = -1/\text{gradient} \).
- **[1 mark]** States that the percentage uncertainty is \( 4.5\% \).

**Part c (4 marks)**
- **[1 mark]** Calculates nominal \( \tau = 4.70 \text{ s} \).
- **[1 mark]** States that percentage uncertainty in \( \tau \) is \( 25\% \) (sum of \( 5\% \) and \( 20\% \)).
- **[1 mark]** Calculates \( R_{\text{max}} = 10.5 \text{ k}\Omega \) AND \( C_{\text{max}} = 564 \ \mu\text{F} \).
- **[1 mark]** Finds max \( \tau = 5.92 \text{ s} \) (accept \( 5.88 \text{ s} \) if using combined \( 25\% \) uncertainty limit).

**Part d (4 marks)**
- **[1 mark]** Calculates voltage at \( 3\tau \) as \( 0.448 \text{ V} \) (or \( 0.45 \text{ V} \)).
- **[1 mark]** Calculates percentage uncertainty in this voltage as \( 2.2\% \) (accept range \( 2.2\% \) to \( 2.3\% \)).
- **[1 mark]** Explains that at larger times, the voltage becomes extremely small.
- **[1 mark]** Concludes that the fixed resolution of the sensor results in high percentage uncertainty, making the data noisy/unreliable at large times.

**Part a**
- The student can observe the retro-reflected laser light that rebounds off the flat glass surface of the diffraction grating.
- The student should adjust the angle of the diffraction grating until this reflected spot is aligned precisely with the exit aperture of the laser source.
- Additionally, they can verify normal incidence by checking that the distance from the central maximum to the \( +n \) and \( -n \) diffraction orders on either side is symmetrical on the screen.

**Part b**
- Convert the line density to lines per metre:
\( N = 300 \text{ lines/mm} = 3.00 \times 10^5 \text{ lines/m} \)
- The slit spacing \( d \) is the reciprocal of the line density:
\( d = \frac{1}{N} = \frac{1}{3.00 \times 10^5} = 3.33 \times 10^{-6} \text{ m} \)
- The percentage uncertainty in \( d \) is equal to the percentage uncertainty in \( N \):
\( \%\Delta d = \frac{\Delta N}{N} \times 100\% = \frac{2}{300} \times 100\% = 0.67\% \)

**Part c**
- Calculate the diffraction angle \( \theta_2 \):
\( \tan \theta_2 = \frac{y_2}{D} = \frac{0.428}{1.000} = 0.428 \Rightarrow \theta_2 = \arctan(0.428) = 23.17^\circ \)
- To find the absolute uncertainty in \( \theta_2 \), calculate the maximum and minimum possible values:
\( \tan\theta_{2,\text{max}} = \frac{0.428 + 0.002}{1.000 - 0.005} = \frac{0.430}{0.995} = 0.43216 \Rightarrow \theta_{2,\text{max}} = 23.37^\circ \)
\( \tan\theta_{2,\text{min}} = \frac{0.428 - 0.002}{1.000 + 0.005} = \frac{0.426}{1.005} = 0.42388 \Rightarrow \theta_{2,\text{min}} = 22.97^\circ \)
- The uncertainty is:
\( \Delta\theta_2 = \frac{\theta_{2,\text{max}} - \theta_{2,\text{min}}}{2} = \frac{23.37^\circ - 22.97^\circ}{2} = 0.20^\circ \) (or \( 0.0035 \text{ rad} \))

**Part d**
- From \( n\lambda = d \sin \theta \), for \( n = 2 \):
\( \lambda = \frac{d \sin \theta_2}{2} = \frac{3.333 \times 10^{-6} \times \sin(23.17^\circ)}{2} = \frac{3.333 \times 10^{-6} \times 0.39345}{2} = 6.557 \times 10^{-7} \text{ m} = 656 \text{ nm} \)
- Find the absolute uncertainty in \( \sin \theta_2 \):
\( \sin(23.37^\circ) = 0.3967 \) and \( \sin(22.97^\circ) = 0.3902 \)
\( \Delta(\sin\theta_2) = \frac{0.3967 - 0.3902}{2} = 0.00325 \)
\( \%\Delta(\sin\theta_2) = \frac{0.00325}{0.39345} \times 100\% = 0.83\% \)
- Sum the percentage uncertainties to find the percentage uncertainty of \( \lambda \):
\( \%\Delta\lambda = \%\Delta d + \%\Delta(\sin\theta_2) = 0.67\% + 0.83\% = 1.50\% \)
- Convert this to absolute uncertainty:
\( \Delta\lambda = 656 \text{ nm} \times 0.0150 = 9.84 \text{ nm} \approx 10 \text{ nm} \)
- Thus, \( \lambda = 656 \pm 10 \text{ nm} \).

**Part a (3 marks)**
- **[1 mark]** Explains that laser light reflects back from the surface of the grating.
- **[1 mark]** Identifies that the reflected spot must overlap with the exit aperture of the laser.
- **[1 mark]** Mentions verifying symmetry of diffraction orders on either side of the zero-order beam on the screen.

**Part b (4 marks)**
- **[1 mark]** Converts line density correctly to lines per metre: \( N = 3.00 \times 10^5 \text{ m}^{-1} \).
- **[1 mark]** Calculates slit spacing \( d = 3.33 \times 10^{-6} \text{ m} \) (accept \( 3.3 \times 10^{-6} \text{ m} \)).
- **[1 mark]** Shows calculation of percentage uncertainty: \( \frac{2}{300} \times 100\% \).
- **[1 mark]** States percentage uncertainty as \( 0.67\% \) (accept \( 0.7\% \)).

**Part c (4 marks)**
- **[1 mark]** Calculates nominal angle \( \theta_2 = 23.2^\circ \) (or \( 23.17^\circ \) or \( 0.404 \text{ rad} \)).
- **[1 mark]** Calculates maximum and minimum values of \( \tan\theta_2 \) or calculates uncertainty of ratio \( y_2/D \) as \( 0.97\% \).
- **[1 mark]** Correctly obtains maximum angle \( 23.37^\circ \) and minimum angle \( 22.97^\circ \).
- **[1 mark]** States absolute uncertainty \( \Delta\theta_2 = 0.20^\circ \) (or \( 0.0035 \text{ rad} \)).

**Part d (4 marks)**
- **[1 mark]** Calculates nominal wavelength \( \lambda = 656 \text{ nm} \) (or \( 6.56 \times 10^{-7} \text{ m} \)).
- **[1 mark]** Determines uncertainty in \( \sin \theta_2 \) as \( 0.83\% \) (or absolute value of \( 0.00325 \)).
- **[1 mark]** Correctly combines the percentage uncertainties to find \( \%\Delta\lambda = 1.5\% \).
- **[1 mark]** Expresses the final result with appropriate absolute uncertainty as \( 656 \pm 10 \text{ nm} \) (or equivalent form in metres with consistent significant figures).