2025 IB DP Physics 模擬試題連答案詳解

In thermal equilibrium, the power absorbed by the planet must equal the power radiated. The planet absorbs radiation over its projected cross-sectional area \( \pi R^2 \), where \( R \) is the radius of the planet. Taking the albedo \( \alpha \) into account, the absorbed power is: \( P_{\text{abs}} = I_0 (1 - \alpha) \pi R^2 \). The planet radiates energy from its entire spherical surface area \( 4 \pi R^2 \). Thus, the average power radiated per unit area \( I_{\text{rad}} \) is: \( I_{\text{rad}} = \frac{P_{\text{abs}}}{4 \pi R^2} = \frac{I_0 (1 - \alpha) \pi R^2}{4 \pi R^2} = \frac{I_0 (1 - \alpha)}{4} \).

Award [1] for the correct answer of B.

The first polarizer polarizes the light vertically. According to Malus's Law, the intensity of light transmitted through a second polarizer is proportional to \( \cos^2\theta \), where \( \theta \) is the angle between the transmission axes of the two polarizers. For the transmitted intensity to be zero, the two axes must be perpendicular (\( \theta = 90^\circ \)). Since the second filter's axis is initially at \( 30^\circ \) to the vertical, the minimum rotation angle needed to reach \( 90^\circ \) is: \( 90^\circ - 30^\circ = 60^\circ \).

Award [1] for the correct answer of B.

Because the wires are connected in series, the electric current \( I \) flowing through both wires is identical: \( I_X = I_Y \). The relationship between current and drift speed is \( I = n A v q \), where \( A \) is the cross-sectional area of the wire. The cross-sectional area is related to the diameter \( d \) by \( A = \pi \frac{d^2}{4} \). Therefore: \( A_Y = \pi \frac{(2d)^2}{4} = 4 A_X \). Equating the currents: \( n A_X v_X q = n A_Y v_Y q \implies A_X v_X = (4 A_X) v_Y \implies v_Y = \frac{v_X}{4} \).

Award [1] for the correct answer of A.

The electric field due to a point charge \( q \) is given by \( E = \frac{k q}{r^2} \). The total electric field is zero where the fields due to the two charges are equal in magnitude but opposite in direction. This cannot happen between the charges because the fields point in the same direction. It must happen on the side of the smaller charge \( +Q \) to compensate for its smaller magnitude. Let the position be at a distance \( r \) to the left of \( +Q \). The distance to \( -4Q \) is \( d + r \). Equating the magnitudes: \( \frac{k Q}{r^2} = \frac{k (4Q)}{(d + r)^2} \implies \frac{1}{r} = \frac{2}{d + r} \implies d + r = 2r \implies r = d \). Thus, it is at distance \( d \) to the left of \( +Q \).

Award [1] for the correct answer of B.

From Stefan-Boltzmann law, \( L = 4\pi R^2 \sigma T^4 \). Since \( R_A = R_B \), the ratio of luminosities is: \( \frac{L_A}{L_B} = \left(\frac{T_A}{T_B}\right)^4 = 2^4 = 16 \). Using the relation \( L \propto M^{3.5} \), we have: \( \frac{L_A}{L_B} = \left(\frac{M_A}{M_B}\right)^{3.5} \implies 16 = \left(\frac{M_A}{M_B}\right)^{3.5} \implies \frac{M_A}{M_B} = 16^{1/3.5} = (2^4)^{1/3.5} = 2^{4/3.5} \approx 2^{1.14} \).

Award [1] for the correct answer of A.

The useful output power \( P_{\text{out}} \) of the motor is the rate at which gravitational potential energy is gained: \( P_{\text{out}} = F v = mgv \). The efficiency of the motor is defined as: \( \eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{mgv}{P} \). Rearranging this equation to solve for \( v \) yields: \( v = \frac{\eta P}{mg} \).

Award [1] for the correct answer of A.

The total energy \( E \) of a satellite of mass \( m \) in a circular orbit of radius \( r \) is given by: \( E = -\frac{GMm}{2r} \). The initial energy is \( E_{\text{initial}} = -\frac{GMm}{2R} \). The final energy is \( E_{\text{final}} = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R} \). The change in energy is: \( \Delta E = E_{\text{final}} - E_{\text{initial}} = -\frac{GMm}{6R} - \left(-\frac{GMm}{2R}\right) = \frac{GMm}{2R} - \frac{GMm}{6R} = \frac{3GMm - GMm}{6R} = \frac{2GMm}{6R} = +\frac{GMm}{3R} \). The total energy increases, which is consistent with work being done to move the satellite to a higher orbit.

Award [1] for the correct answer of D.

The energy of the \( n \)-th level in hydrogen is \( E_n = -\frac{13.6}{n^2}\text{ eV} \). For \( n = 4 \), the energy is \( E_4 = -\frac{13.6}{4^2} = -0.85\text{ eV} \). The energy of the emitted photon is the difference: \( \Delta E = E_4 - E_1 = -0.85\text{ eV} - (-13.6\text{ eV}) = 12.75\text{ eV} \). Converting this energy to Joules: \( \Delta E = 12.75 \times 1.60 \times 10^{-19}\text{ J} = 2.04 \times 10^{-18}\text{ J} \). Using \( E = hf \), the frequency is: \( f = \frac{E}{h} = \frac{2.04 \times 10^{-18}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} \approx 3.08 \times 10^{15}\text{ Hz} \).

Award [1] for the correct answer of B.

1. The intensity of radiation from the star at the planet's distance is given by the inverse-square law: \(S_p = \frac{S_0}{2^2} = \frac{1360}{4} = 340\text{ W m}^{-2}\). 2. The power absorbed by the planet is \(P_{\text{abs}} = S_p(1 - \alpha_p)\pi R^2\), where \(\alpha_p = 0.20\) is the albedo and \(R\) is the radius of the planet. 3. The average absorbed intensity over the total surface area \(4\pi R^2\) of the spherical planet is: \(I_{\text{abs}} = \frac{P_{\text{abs}}}{4\pi R^2} = \frac{S_p(1 - \alpha_p)}{4} = \frac{340 \times (1 - 0.20)}{4} = 68\text{ W m}^{-2}\).

Award 1 mark for the correct calculation leading to choice A.

1. After passing through the first vertical polarizer, the light is vertically polarized with intensity \(I_1 = \frac{I_0}{2}\). 2. After passing through the second polarizer (axis at \(30^\circ\) to the vertical), the intensity is \(I_2 = I_1 \cos^2(30^\circ) = \frac{I_0}{2} \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{8}I_0\). 3. The light leaving the second polarizer is polarized at \(30^\circ\) to the vertical. The third polarizer is horizontal (which is at \(90^\circ - 30^\circ = 60^\circ\) to the light's polarization axis). 4. After passing through the third polarizer, the intensity is \(I_3 = I_2 \cos^2(60^\circ) = \frac{3}{8}I_0 \left(\frac{1}{2}\right)^2 = \frac{3}{32}I_0 \approx 0.094I_0\).

Award 1 mark for the correct mathematical determination of successive intensity reductions leading to choice A.

1. The parallel combination of two \(10\ \Omega\) resistors has an equivalent resistance \(R_p = \frac{10}{2} = 5\ \Omega\). 2. The potential difference across this combination is the terminal potential difference of the battery, \(V = 9.6\text{ V}\). 3. The current through the circuit is \(I = \frac{V}{R_p} = \frac{9.6}{5} = 1.92\text{ A}\). 4. Using the battery equation, \(E = V + Ir\), we get \(12 = 9.6 + 1.92r\), which yields \(2.4 = 1.92r\) and thus \(r = 1.25\ \Omega\).

Award 1 mark for finding the equivalent external resistance, the total current, and solving for the internal resistance to obtain choice B.

1. The total energy in simple harmonic motion is given by \(E_T = \frac{1}{2}kA^2\). 2. The potential energy is \(E_p = \frac{1}{2}kx^2\), and the kinetic energy is \(E_k = E_T - E_p = \frac{1}{2}k(A^2 - x^2)\). 3. Setting \(E_k = 3E_p\) gives \(k(A^2 - x^2) = 3 \left(k x^2\right)\), which simplifies to \(A^2 - x^2 = 3x^2\). 4. Thus, \(A^2 = 4x^2\), leading to \(x = \pm \frac{A}{2}\).

Award 1 mark for applying the conservation of energy in SHM and solving for x to obtain choice B.

1. For a source moving towards a stationary observer, the observed frequency is \(f_1 = f_0 \left(\frac{c}{c - v}\right)\). 2. For a source moving away from a stationary observer, the observed frequency is \(f_2 = f_0 \left(\frac{c}{c + v}\right)\). 3. The ratio of the two frequencies is: \(\frac{f_1}{f_2} = \frac{f_0 \left(\frac{c}{c - v}\right)}{f_0 \left(\frac{c}{c + v}\right)} = \frac{c + v}{c - v}\).

Award 1 mark for stating both Doppler equations correctly and dividing them to find the ratio as in choice B.

1. The electric potential at a point \(x\) is \(V = \frac{k(4Q)}{|x|} + \frac{k(-Q)}{|x - d|}\). 2. Setting \(V = 0\) yields \(\frac{4}{|x|} = \frac{1}{|x - d|}\), which simplifies to \(4|x - d| = |x|\). 3. Case 1 (between the charges, \(0 < x < d\)): \(4(d - x) = x \implies 4d = 5x \implies x = 0.8d\). 4. Case 2 (to the right of the negative charge, \(x > d\)): \(4(x - d) = x \implies 3x = 4d \implies x = \frac{4}{3}d \approx 1.33d\). 5. There are no solutions for \(x < 0\). Therefore, the potential is zero at both \(x = 0.8d\) and \(x = 1.33d\).

Award 1 mark for analyzing both regions along the axis and correctly solving the algebraic equations to find both points as in choice C.

1. The total mechanical energy \(E\) of a satellite in a circular orbit of radius \(r\) is \(E = -\frac{GMm}{2r}\). 2. In the new orbit of radius \(2r\), the total energy is \(E' = -\frac{GMm}{2(2r)} = -\frac{GMm}{4r}\). 3. The change in total mechanical energy is \(\Delta E = E' - E = -\frac{GMm}{4r} - \left(-\frac{GMm}{2r}\right) = +\frac{GMm}{4r}\).

Award 1 mark for using the correct expression for total energy in a circular orbit and subtracting to find the positive difference as in choice B.

1. The useful power output needed to lift the crate at constant speed \(v\) is \(P_{\text{out}} = F v = m g v\). 2. Efficiency \(\eta\) is defined as the ratio of useful power output to electrical power input: \(\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\). 3. Rearranging this relation for the electrical power input gives \(P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{mgv}{\eta}\).

Award 1 mark for defining useful output power and correctly applying the efficiency formula to find the input power as in choice B.

The electric potential \(V\) at a distance \(x\) from \(+Q\) (and thus \(d-x\) from \(-3Q\)) is given by: \(V = \frac{kQ}{x} - \frac{k(3Q)}{d - x} = 0\). Rearranging gives: \rac{1}{x} = \frac{3}{d - x}\), which simplifies to \(d - x = 3x\), so \(4x = d\), yielding \(\frac{x}{d} = 0.25\).

Award [1] for the correct calculation leading to option B.

After the first vertical polarizer, the intensity is halved: \(I_1 = \frac{1}{2} I_0\). After the second polarizer (at \(30^\circ\) to the first), the intensity is \(I_2 = I_1 \cos^2(30^\circ) = \frac{1}{2} I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{8} I_0\). The angle between the second polarizer and the third (horizontal, \(90^\circ\)) is \(90^\circ - 30^\circ = 60^\circ\). Thus, the final intensity is \(I_3 = I_2 \cos^2(60^\circ) = \frac{3}{8} I_0 \left(\frac{1}{2}\right)^2 = \frac{3}{32} I_0 \approx 0.094 I_0\).

Award [1] for the correct use of Malus's law across successive polarizers leading to option A.

The solar energy absorbed by the planet is given by \(P_{\text{absorbed}} = (1 - \alpha) S \pi R^2\), where \(\alpha = 0.30\) is the albedo, \(S = 1400\text{ W m}^{-2}\) is the solar constant, and \(R\) is the radius. For thermal equilibrium, this equals the power radiated, which is distributed over the entire surface area of \(4\pi R^2\). Thus, the average intensity radiated is \(I_{\text{radiated}} = \frac{(1 - \alpha) S \pi R^2}{4\pi R^2} = \frac{(1 - 0.30) \times 1400}{4} = 245\text{ W m}^{-2}\).

Award [1] for setting up the correct energy balance and calculating the radiated intensity leading to option A.

Luminosity is governed by the Stefan-Boltzmann law: \(L = 4\pi R^2 \sigma T^4\). Therefore, the ratio of their luminosities is \(\frac{L_{\text{Y}}}{L_{\text{X}}} = \left(\frac{R_{\text{Y}}}{R_{\text{X}}}\right)^2 \left(\frac{T_{\text{Y}}}{T_{\text{X}}}\right)^4 = 3^2 \times 2^4 = 9 \times 16 = 144\).

Award [1] for using the Stefan-Boltzmann law to scale radius and temperature correctly, yielding option D.

The total mechanical energy in a circular orbit is \(E = -\frac{GMm}{2r}\). The work required is the change in total energy: \(W = E_{\text{final}} - E_{\text{initial}} = -\frac{GMm}{2(3r)} - \left(-\frac{GMm}{2r}\right) = -\frac{GMm}{6r} + \frac{GMm}{2r} = \frac{GMm}{3r}\).

Award [1] for using the correct total energy equation and finding the difference, leading to option B.

The useful power output is the rate of increase of gravitational potential energy: \(P_{\text{out}} = \frac{mgh}{t} = \frac{50 \times 9.8 \times 12}{8.0} = 735\text{ W}\). The electrical power input is \(P_{\text{in}} = VI = 120 \times 8.5 = 1020\text{ W}\). The efficiency is \(\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{735}{1020} \approx 0.72\), which is \(72\%\).

Award [1] for calculating output and input power and finding the efficiency, leading to option C.

The power dissipated in the external resistor is \(P = I^2 R = \left(\frac{\varepsilon}{R+r}\right)^2 R\). By the maximum power transfer theorem, maximum power dissipation occurs when the external resistance equals the internal resistance (\(R = r\)). Substituting \(R = r\) into the expression gives: \(P_{\text{max}} = \left(\frac{\varepsilon}{2r}\right)^2 r = \frac{\varepsilon^2}{4r^2} r = \frac{\varepsilon^2}{4r}\).

Award [1] for using the condition \(R = r\) and correctly finding the maximum power, leading to option C.

The energy change for the transition from level 3 to level 1 is \(E_{31} = E_{32} + E_{21}\). Since energy is related to wavelength by \(E = \frac{hc}{\lambda}\), we can write: \(\frac{hc}{\lambda_{31}} = \frac{hc}{\lambda_{32}} + \frac{hc}{\lambda_{21}}\), which simplifies to \(\frac{1}{\lambda_{31}} = \frac{1}{\lambda_{32}} + \frac{1}{\lambda_{21}}\). Substituting the values: \(\frac{1}{\lambda_{31}} = \frac{1}{656} + \frac{1}{122} \approx 0.001524 + 0.008197 = 0.009721\text{ nm}^{-1}\). Thus, \(\lambda_{31} = \frac{1}{0.009721} \approx 103\text{ nm}\).

Award [1] for adding reciprocal wavelengths to calculate the combined wavelength, leading to option A.

**(a)**
1. Calculate the cross-sectional area \(A\):
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\)

2. Calculate the fractional uncertainty in \(d\):
\(\frac{\Delta d}{d} = \frac{0.01}{0.38} \approx 0.0263\) (or \(2.63\%\))

3. Calculate the percentage uncertainty in \(A\):
Since \(A \propto d^2\), \(\frac{\Delta A}{A} = 2 \times \frac{\Delta d}{d} = 2 \times 2.63\% = 5.26\%\)

4. Calculate the absolute uncertainty in \(A\):
\(\Delta A = 1.134 \times 10^{-7} \times 0.0526 = 5.97 \times 10^{-9}\text{ m}^2 \approx 0.06 \times 10^{-7}\text{ m}^2\)

Thus, \(A = (1.13 \pm 0.06) \times 10^{-7}\text{ m}^2\).

**(b)**
1. Calculate the resistivity \(\rho\):
\(\rho = R \frac{A}{L} = m A\) (since \(m = R/L\))
\(\rho = 4.15 \times 1.1341 \times 10^{-7} = 4.7065 \times 10^{-7}\ \Omega\text{ m}\)

2. Calculate the total percentage uncertainty in \(\rho\):
\(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta A}{A} = \frac{0.12}{4.15} + 5.26\% = 2.89\% + 5.26\% = 8.15\%\)

3. Calculate the absolute uncertainty in \(\rho\):
\(\Delta \rho = 4.7065 \times 10^{-7} \times 0.0815 = 0.384 \times 10^{-7}\ \Omega\text{ m} \approx 0.4 \times 10^{-7}\ \Omega\text{ m}\) (rounded to 1 s.f.)

Thus, \(\rho = (4.7 \pm 0.4) \times 10^{-7}\ \Omega\text{ m}\).

**(a) [3 marks maximum]**
* \(A = 1.13 \times 10^{-7}\text{ m}^2\) [1]
* Percentage uncertainty in \(A\) is \(5.3\%\) (or fractional uncertainty is \(0.053\)) [1]
* Final answer with correct absolute uncertainty: \((1.13 \pm 0.06) \times 10^{-7}\text{ m}^2\) [1]

**(b) [3 marks maximum]**
* \(\rho = m \times A = 4.7 \times 10^{-7}\ \Omega\text{ m}\) (allow ECF from (a)) [1]
* Sums fractional uncertainties: \(\frac{0.12}{4.15} + 0.053 = 0.082\) (or \(8.2\%\)) [1]
* Final answer with absolute uncertainty to 1 s.f. and matching decimal places: \((4.7 \pm 0.4) \times 10^{-7}\ \Omega\text{ m}\) (allow \((4.71 \pm 0.38) \times 10^{-7}\ \Omega\text{ m}\)) [1]

**(a)**
1. Calculate period \(T\):
\(T = \frac{t_{20}}{20} = \frac{35.8}{20} = 1.79\text{ s}\)

2. Calculate the absolute uncertainty in \(T\):
\(\Delta T = \frac{\Delta t_{20}}{20} = \frac{0.2}{20} = 0.01\text{ s}\)

So, \(T = (1.79 \pm 0.01)\text{ s}\).

**(b)**
Rearrange the formula for \(g\):
\(T = 2\pi\sqrt{\frac{L}{g}} \implies T^2 = 4\pi^2 \frac{L}{g} \implies g = \frac{4\pi^2 L}{T^2}\)

Calculate \(g\):
\(g = \frac{4\pi^2 \times 0.800}{(1.79)^2} \approx 9.857\text{ m s}^{-2}\)

**(c)**
1. Calculate the percentage uncertainty in \(L\):
\(\frac{\Delta L}{L} = \frac{0.002}{0.800} \times 100\% = 0.25\%\)

2. Calculate the percentage uncertainty in \(T\):
\(\frac{\Delta T}{T} = \frac{0.01}{1.79} \times 100\% \approx 0.559\%\)

3. Calculate the total percentage uncertainty in \(g\):
\(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} = 0.25\% + 2(0.559\%) = 1.368\% \approx 1.4\%\)

4. Calculate the absolute uncertainty in \(g\):
\(\Delta g = 9.857 \times 0.01368 = 0.135\text{ m s}^{-2} \approx 0.1\text{ m s}^{-2}\) (to 1 s.f.)

Thus, \(g = (9.9 \pm 0.1)\text{ m s}^{-2}\).

**(a) [2 marks maximum]**
* \(T = 1.79\text{ s}\) [1]
* \(\Delta T = 0.01\text{ s}\) [1]

**(b) [2 marks maximum]**
* Rearranges formula to make \(g\) the subject: \(g = \frac{4\pi^2 L}{T^2}\) [1]
* Correct calculation of \(g\): \(9.86\text{ m s}^{-2}\) or \(9.9\text{ m s}^{-2}\) [1]

**(c) [3 marks maximum]**
* Adds percentage uncertainties correctly: \(\frac{\Delta g}{g}\% = \frac{\Delta L}{L}\% + 2 \frac{\Delta T}{T}\%\) [1]
* Obtains percentage uncertainty in \(g\) as \(1.4\%\) (or \(1.37\%\)) [1]
* Final formatted answer with consistent decimal places: \((9.9 \pm 0.1)\text{ m s}^{-2}\) (allow ECF from (b); accept \(9.86 \pm 0.14\text{ m s}^{-2}\)) [1]

**(a)**
According to Boyle's Law, \(p \propto \frac{1}{V}\). A plot of \(p\) against \(\frac{1}{V}\) yields a straight line passing through the origin. A straight line is much easier to identify, fit, and verify than a curved hyperbola (which is obtained from a plot of \(p\) against \(V\)).

**(b)**
1. Calculate the value of \(\frac{1}{V}\):
\(\frac{1}{V} = \frac{1}{12.5} = 0.0800\text{ cm}^{-3}\)

2. Calculate the percentage uncertainty in \(V\):
\(\frac{\Delta V}{V} = \frac{0.5}{12.5} \times 100\% = 4.0\%\)

3. Calculate the absolute uncertainty in \(\frac{1}{V}\):
Since the percentage uncertainty in \(\frac{1}{V}\) is the same as that of \(V\) (which is \(4\%\)):
\(\Delta \left(\frac{1}{V}\right) = 0.0800 \times 0.040 = 0.0032\text{ cm}^{-3} \approx 0.003\text{ cm}^{-3}\)

Thus, \(\frac{1}{V} = (0.080 \pm 0.003)\text{ cm}^{-3}\).

**(c)**
1. Calculate the percentage uncertainty in \(p\):
\(\frac{\Delta p}{p} = \frac{4}{200} \times 100\% = 2.0\%\)

2. Calculate the percentage uncertainty in the product \(pV\):
\(\frac{\Delta (pV)}{pV}\% = \frac{\Delta p}{p}\% + \frac{\Delta V}{V}\% = 2.0\% + 4.0\% = 6.0\%\).

**(a) [1 mark maximum]**
* Mentions that a straight line (through the origin) is easier to identify/verify than a curve/hyperbola [1]

**(b) [3 marks maximum]**
* \(\frac{1}{V} = 0.080\text{ cm}^{-3}\) (or \(0.0800\)) [1]
* Determines percentage uncertainty of \(V\) is \(4\%\) [1]
* Correctly calculates absolute uncertainty as \(0.003\text{ cm}^{-3}\) and formats final answer with matching decimal places: \((0.080 \pm 0.003)\text{ cm}^{-3}\) [1]

**(c) [2 marks maximum]**
* Determines percentage uncertainty of \(p\) is \(2\%\) [1]
* Sums percentage uncertainties to find total percentage uncertainty of \(6\%\) (or \(6.0\%\)) [1]

a) Since the upper plate is positive and the lower plate must be negative, the electric field lines point from positive to negative, which is vertically downwards.

b) The electric field between parallel plates is given by:
\(E = \frac{V}{d}\)
\(E = \frac{1500\text{ V}}{0.025\text{ m}} = 6.0 \times 10^4\text{ V m}^{-1}\) (or \(\text{N C}^{-1}\)).

c) For the droplet to remain stationary, the upwards electric force must balance the downwards gravitational force:
\(F_e = F_g \Rightarrow qE = mg\)
\(q = \frac{mg}{E}\)
\(q = \frac{2.94 \times 10^{-15}\text{ kg} \times 9.81\text{ m s}^{-2}}{6.0 \times 10^4\text{ V m}^{-1}}\)
\(q = 4.807 \times 10^{-19}\text{ C} \approx 4.8 \times 10^{-19}\text{ C}\).

d) The number of excess electrons is:
\(n = \frac{q}{e} = \frac{4.81 \times 10^{-19}\text{ C}}{1.60 \times 10^{-19}\text{ C}} = 3.0\)
So there are 3 excess electrons on the droplet. (The charge must be negative to experience an upward force in a downward electric field).

a) [1 mark] Downwards (accept from top plate to bottom plate).

b) [2 marks]
- [1 mark] for correct formula and substitution: \(E = 1500 / 0.025\).
- [1 mark] for final answer \(6.0 \times 10^4\text{ V m}^{-1}\) (or \(\text{N C}^{-1}\)).

c) [3.33 marks]
- [1 mark] for stating the condition for equilibrium: \(qE = mg\) or equating electric force to weight.
- [1 mark] for calculating weight: \(2.884 \times 10^{-14}\text{ N}\).
- [1.33 marks] for calculating charge: \(q = 4.81 \times 10^{-19}\text{ C}\) (must show at least 3 significant figures to justify 'approximately').

d) [2 marks]
- [1 mark] for dividing the charge by the elementary charge: \(4.8 \times 10^{-19} / 1.6 \times 10^{-19}\).
- [1 mark] for final integer answer of 3.

a) The total solar power intercepted by the planet is \(S \times \pi R^2\), where \(R\) is the planet's radius. A fraction \(\alpha\) (albedo) is reflected, so the absorbed power is \(S(1 - \alpha)\pi R^2\). Since this power is distributed over the entire surface area of the spherical planet (\(4\pi R^2\)), the average absorbed intensity is:
\(I_{\text{absorbed}} = \frac{S(1 - \alpha)}{4} = \frac{800 \times (1 - 0.30)}{4} = 140\text{ W m}^{-2}\).

b) At thermal equilibrium, the power emitted by the planet's surface per unit area equals the average power absorbed per unit area:
\(\sigma T^4 = I_{\text{absorbed}}\)
\(T = \left(\frac{140}{5.67 \times 10^{-8}}\right)^{1/4} = (2.469 \times 10^9)^{1/4} \approx 223\text{ K}\).

c) The solar radiation incident on the planet consists of short-wavelength radiation (primarily visible and near-UV), which mostly passes through the atmosphere with minimal absorption. The warmed planet surface then emits long-wavelength (infrared) radiation. Greenhouse gases in the atmosphere (such as carbon dioxide and water vapor) have vibrational modes that allow them to absorb this infrared radiation. These excited molecules then re-emit infrared radiation in all directions, including back towards the planet's surface, resulting in an additional warming effect.

a) [2 marks]
- [1 mark] for recognizing the factor of \(1/4\) to account for the spherical surface area: \(I = S(1-\alpha)/4\).
- [1 mark] for correct substitution: \(800 \times 0.70 / 4 = 140\text{ W m}^{-2}\).

b) [3 marks]
- [1 mark] for equating emitted power to absorbed power: \(\sigma T^4 = 140\).
- [1 mark] for correct substitution of Stefan-Boltzmann constant: \(5.67 \times 10^{-8}\).
- [1 mark] for correct final answer \(223\text{ K}\) (or \(-50^\circ\text{C}\)).

c) [3.33 marks]
- [1 mark] for explaining that solar radiation (short wavelength) is largely transmitted through the atmosphere.
- [1 mark] for explaining that the Earth's surface re-radiates infrared (long wavelength) radiation.
- [1.33 marks] for explaining that greenhouse gases absorb this infrared radiation and re-emit it in all directions (including back towards the surface).

a) Coherent sources are sources of waves that have a constant phase difference between them and have the same frequency (or wavelength).

b) Using the double-slit formula:
\(s = \frac{\lambda D}{d}\)
where:
\(\lambda = 633 \times 10^{-9}\text{ m}\)
\(D = 2.4\text{ m}\)
\(d = 0.15 \times 10^{-3}\text{ m}\)
\(s = \frac{633 \times 10^{-9} \times 2.4}{0.15 \times 10^{-3}} = 1.01 \times 10^{-2}\text{ m} = 10.1\text{ mm}\) (or \(1.0 \times 10^{-2}\text{ m}\)).

c) At a bright fringe maximum, the waves from the two slits interfere constructively, meaning they are in phase. The resultant amplitude of the combined wave is the sum of the individual amplitudes:
\(A_{\text{total}} = A + A = 2A\)
Since the intensity of a wave is proportional to the square of its amplitude (\(I \propto A^2\)):
\(I_{\text{total}} \propto (2A)^2 = 4A^2\)
Therefore, \(I_{\text{total}} = 4I_0\). It is not \(2I_0\) because intensity depends non-linearly on amplitude; we are adding the wave fields (amplitudes) constructively before squaring, rather than simply adding the intensities.

a) [2 marks]
- [1 mark] for mentioning 'constant phase difference'.
- [1 mark] for mentioning 'same frequency' or 'constant wavelength'.

b) [3 marks]
- [1 mark] for correct rearrangement/use of \(s = \lambda D / d\).
- [1 mark] for correct substitution of values in SI units.
- [1 mark] for correct final answer: \(1.0 \times 10^{-2}\text{ m}\) (or \(1.01 \times 10^{-2}\text{ m}\) or \(1.0\text{ cm}\)).

c) [3.33 marks]
- [1 mark] for stating \(4I_0\) as the correct maximum intensity.
- [1 mark] for stating that the amplitudes add constructively to double the individual amplitude (\(A_{\text{total}} = 2A\)).
- [1.33 marks] for explaining that intensity is proportional to the square of amplitude (\(I \propto A^2\)), hence \((2)^2 = 4\).

a) The mass of the star is \(3.0\text{ }M_{\odot}\).
Using the mass-luminosity relationship:
\(\frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^{3.5} = (3.0)^{3.5} = 46.77 \approx 47\)
Therefore, the luminosity is \(47\text{ }L_{\odot}\).

b) The lifetime \(T\) of a star on the main sequence is proportional to the ratio of its available fuel (which is proportional to its mass \(M\)) to the rate at which it consumes fuel (its luminosity \(L\)):
\(T \propto \frac{M}{L} \propto \frac{M}{M^{3.5}} = M^{-2.5}\)
Therefore, the ratio of lifetimes is:
\(\frac{T}{T_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^{-2.5} = (3.0)^{-2.5} \approx 0.06415\)
\(T = 0.06415 \times 10 \times 10^9\text{ years} = 6.4 \times 10^8\text{ years}\) (or \(640\text{ million years}\)).

c) Since the progenitor mass is \(3.0\text{ }M_{\odot}\) (which is below the threshold of roughly \(8\text{ }M_{\odot}\) for supernova progenitors):
- The star will expand and cool to become a red giant as hydrogen in the core is depleted and helium fusion begins.
- Eventually, its outer layers will be gently ejected into space, forming a planetary nebula.
- The remaining carbon-oxygen core will collapse until stopped by electron degeneracy pressure, leaving a stable white dwarf remnant.

a) [2 marks]
- [1 mark] for utilizing the mass-luminosity ratio formula: \((3.0)^{3.5}\).
- [1 mark] for the final correct answer: \(47\text{ }L_{\odot}\) (accept 46.8).

b) [3 marks]
- [1 mark] for recognizing that lifetime is proportional to \(M / L\) or \(M^{-2.5}\).
- [1 mark] for setting up the calculation: \(T = 10 \times 10^9 \times (3.0)^{-2.5}\).
- [1 mark] for the final correct lifetime: \(6.4 \times 10^8\text{ years}\) (accept \(6.42 \times 10^8\text{ years}\)).

c) [3.33 marks]
- [1 mark] for identifying the expansion into a red giant.
- [1 mark] for describing the ejection of outer layers as a planetary nebula.
- [1.33 marks] for identifying the remnant core as a white dwarf, and stating it is stable due to electron degeneracy pressure.

a) By conservation of energy on a frictionless track, all gravitational potential energy at point A is converted to kinetic energy at point B:
\(m g h_A = \frac{1}{2} m v_B^2\)
\(v_B = \sqrt{2 g h_A} = \sqrt{2 \times 9.81 \times 24} = 21.7\text{ m s}^{-1} \approx 22\text{ m s}^{-1}\).

b) By conservation of mechanical energy between point A and point C:
\(E_{k,A} + E_{p,A} = E_{k,C} + E_{p,C}\)
Since the cart starts from rest at A, \(E_{k,A} = 0\):
\(E_{k,C} = m g h_A - m g h_C = m g (h_A - h_C)\)
\(E_{k,C} = 450 \times 9.81 \times (24 - 15) = 450 \times 9.81 \times 9 = 3.97 \times 10^4\text{ J}\) (or \(4.0 \times 10^4\text{ J}\)).

c) Taking the resistive force into account, mechanical energy is lost as work done against friction:
\(W_f = f \times d = 120\text{ N} \times 85\text{ m} = 10200\text{ J}\)
Using the work-energy theorem:
\(E_{k,\text{actual}} = E_{k,C} - W_f = 39730.5\text{ J} - 10200\text{ J} = 29530.5\text{ J}\)
Now find the actual speed:
\(\frac{1}{2} m v_C^2 = 29530.5\)
\(v_C = \sqrt{\frac{2 \times 29530.5}{450}} = 11.45\text{ m s}^{-1} \approx 11\text{ m s}^{-1}\).

a) [2 marks]
- [1 mark] for setting up correct energy conservation: \(v = \sqrt{2gh}\).
- [1 mark] for final answer of \(22\text{ m s}^{-1}\) (or \(21.7\text{ m s}^{-1}\)).

b) [3 marks]
- [1 mark] for calculating initial potential energy or setting up conservation statement: \(E_p = m g \Delta h\).
- [1 mark] for correct substitution of numbers: \(450 \times 9.81 \times 9\).
- [1 mark] for correct final answer: \(4.0 \times 10^4\text{ J}\) (or \(3.97 \times 10^4\text{ J}\)).

c) [3.33 marks]
- [1 mark] for calculating work done against friction: \(120 \times 85 = 10200\text{ J}\).
- [1 mark] for subtracting work done from the ideal kinetic energy: \(E_k = 29530.5\text{ J}\).
- [1.33 marks] for calculating the final speed: \(11\text{ m s}^{-1}\) (or \(11.5\text{ m s}^{-1}\)).

a) The orbital radius \(r\) is the sum of the planet's radius \(R\) and the altitude \(h\):
\(r = R + h = 6.37 \times 10^6\text{ m} + 1.20 \times 10^6\text{ m} = 7.57 \times 10^6\text{ m}\).

b) The centripetal force is provided entirely by the gravitational force:
\(\frac{G M m}{r^2} = \frac{m v^2}{r} \Rightarrow v = \sqrt{\frac{G M}{r}}\)
\(v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{7.57 \times 10^6}} = \sqrt{5.260 \times 10^7} \approx 7.25 \times 10^3\text{ m s}^{-1}\).

c) The initial energy at the surface of the planet is purely potential energy (neglecting rotation of the planet):
\(E_{\text{surface}} = -\frac{G M m}{R}\)
\(E_{\text{surface}} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 850}{6.37 \times 10^6} = -5.3135 \times 10^{10}\text{ J}\).

The total energy of the satellite in its stable circular orbit is:
\(E_{\text{orbit}} = -\frac{G M m}{2r}\)
\(E_{\text{orbit}} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 850}{2 \times 7.57 \times 10^6} = -2.2356 \times 10^{10}\text{ J}\).

The minimum launch energy required is the difference in energy between these two states:
\(\Delta E = E_{\text{orbit}} - E_{\text{surface}}\)
\(\Delta E = -2.2356 \times 10^{10} - (-5.3135 \times 10^{10}) = 3.0779 \times 10^{10}\text{ J} \approx 3.08 \times 10^{10}\text{ J}\).

a) [1 mark]
- [1 mark] for adding \(R + h\) with correct values: \(6.37 \times 10^6 + 1.20 \times 10^6 = 7.57 \times 10^6\text{ m}\).

b) [3 marks]
- [1 mark] for equating centripetal force to gravitational force to find \(v = \sqrt{GM/r}\).
- [1 mark] for correct substitution of \(G\), \(M\), and \(r\).
- [1 mark] for correct final speed: \(7.25 \times 10^3\text{ m s}^{-1}\) (or \(7250\text{ m s}^{-1}\)).

c) [4.33 marks]
- [1 mark] for formula/calculation of surface energy: \(E_p = -5.31 \times 10^{10}\text{ J}\).
- [1 mark] for formula/calculation of orbital total energy: \(E_{\text{total}} = -2.24 \times 10^{10}\text{ J}\).
- [1 mark] for establishing that \(\Delta E = E_{\text{orbit}} - E_{\text{surface}}\).
- [1.33 marks] for the final correct positive energy value: \(3.08 \times 10^{10}\text{ J}\).