2024 IB DP Physics 模擬試題連答案詳解

First, apply the conservation of linear momentum to find the velocity \(v'\) of the combined mass immediately after the collision: \(m v = (m + 3m)v' \implies v' = \frac{v}{4}\). Next, apply the conservation of mechanical energy as the combined mass compresses the spring to its maximum compression \(x\): \(\frac{1}{2} (4m) (v')^2 = \frac{1}{2} k x^2 \implies 4m \left(\frac{v}{4}\right)^2 = k x^2 \implies 4m \frac{v^2}{16} = k x^2 \implies \frac{m v^2}{4} = k x^2 \implies x = \frac{v}{2} \sqrt{\frac{m}{k}}\).

Award 1 mark for the correct option B. (Method: Identify momentum conservation to find speed after collision, then apply conservation of energy for the spring compression.)

According to the Stefan-Boltzmann law, the total power radiated by a blackbody is \(P = \sigma A T^4 = \sigma (4\pi R^2) T^4 \propto R^2 T^4\). Comparing the power of star X to star Y: \(\frac{P_X}{P_Y} = \frac{R_X^2 T_X^4}{R_Y^2 T_Y^4} = \frac{R^2 T^4}{(3R)^2 (\frac{T}{2})^4} = \frac{R^2 T^4}{9R^2 \frac{T^4}{16}} = \frac{1}{\frac{9}{16}} = \frac{16}{9}\).

Award 1 mark for the correct option B. (Method: Relate radiated power to radius squared and temperature to the fourth power, then compute the ratio.)

For a tube open at both ends, the fundamental standing wave has a wavelength \(\lambda_{\text{open}} = 2L\), so the fundamental frequency is \(f = \frac{v}{2L}\). For a tube closed at one end, the fundamental standing wave has a wavelength \(\lambda_{\text{closed}} = 4L\), so the new fundamental frequency is \(f' = \frac{v}{4L} = \frac{f}{2}\).

Award 1 mark for the correct option B. (Method: Identify fundamental wavelengths for both tube configurations and relate their frequencies.)

Using the potential divider formula, the voltage across the LDR is \(V_{\text{out}} = V \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\). In high intensity light, \(R_{\text{LDR}} \approx 0\), so \(V_{\text{out}} \approx 0\). In complete darkness, \(R_{\text{LDR}} \gg R\), so \(V_{\text{out}} \approx V \times \frac{R_{\text{LDR}}}{R_{\text{LDR}}} = V\).

Award 1 mark for the correct option C. (Method: Use potential divider relationship under extreme resistance limits for the LDR.)

The energy of the emitted photon is equal to the difference between the energy levels: \(\Delta E = \frac{hc}{\lambda}\). For the first transition: \(\Delta E_{3 \to 2} = E_3 - E_2 = E_0 \left(\frac{1}{4} - \frac{1}{9}\right) = E_0 \left(\frac{5}{36}\right)\). For the second transition: \(\Delta E_{4 \to 2} = E_4 - E_2 = E_0 \left(\frac{1}{4} - \frac{1}{16}\right) = E_0 \left(\frac{3}{16}\right)\). Taking the ratio of wavelengths: \(\frac{\lambda'}{\lambda} = \frac{\Delta E_{3 \to 2}}{\Delta E_{4 \to 2}} = \frac{5/36}{3/16} = \frac{5}{36} \times \frac{16}{3} = \frac{20}{27}\). Therefore, \(\lambda' = \frac{20}{27}\lambda\).

Award 1 mark for the correct option A. (Method: Calculate energy level differences for both transitions and relate them inversely to wavelength.)

For a charged particle with charge \(q\) to pass through perpendicular electric and magnetic fields undeflected, the electric and magnetic forces must balance: \(q E = q v B \implies v = \frac{E}{B}\). Since this condition for zero deflection is independent of both the mass and the charge of the particle, any charged particle with velocity \(v = \frac{E}{B}\) (and with positive charge) will also pass through undeflected.

Award 1 mark for the correct option D. (Method: Recognize that the velocity selector balance condition is independent of mass and charge.)

Using the radioactive decay law \(N = N_0 \left(\frac{1}{2}\right)^{t/T}\), we substitute \(t = 2.5 T\) to get: \(N = N_0 \left(\frac{1}{2}\right)^{2.5} = N_0 \left(\frac{1}{2}\right)^{5/2} = N_0 \frac{1}{2^2 \cdot 2^{1/2}} = \frac{N_0}{4\sqrt{2}}\).

Award 1 mark for the correct option B. (Method: Use radioactive decay formula with exponential base 2 or 1/2 and simplify the non-integer index.)

When unpolarized light passes through the first polarizer, its intensity is reduced by half: \(I_1 = \frac{I_0}{2}\). It then becomes vertically polarized. When it passes through the second polarizer, Malus's Law applies: \(I_2 = I_1 \cos^2(60^\circ) = \frac{I_0}{2} \left(\frac{1}{2}\right)^2 = \frac{I_0}{8}\).

Award 1 mark for the correct option B. (Method: Identify initial intensity reduction to 50% for unpolarized light, then apply Malus's law with the given angle.)

The impulse given to the body is equal to the area under the force-time graph.

- From \(t = 0 \text{ s}\) to \(t = 4.0 \text{ s}\), the shape is a triangle with area:
\(\text{Area}_1 = \frac{1}{2} \times 4.0 \text{ s} \times 10 \text{ N} = 20 \text{ N s}\)

- From \(t = 4.0 \text{ s}\) to \(t = 6.0 \text{ s}\), the shape is a rectangle with area:
\(\text{Area}_2 = 2.0 \text{ s} \times 10 \text{ N} = 20 \text{ N s}\)

Total impulse \(\Delta p = 20 \text{ N s} + 20 \text{ N s} = 40 \text{ N s}\).

Since the body starts from rest, the final momentum is equal to the impulse:
\(p = m v \implies 40 \text{ kg m s}^{-1} = 2.0 \text{ kg} \times v \implies v = 20 \text{ m s}^{-1}\).

Award [1] for the correct identification of the total impulse as 40 N s and computing the velocity to be 20 m s\(^{-1}\).

Using the thermal energy relation \(Q = m c \Delta T\), where \(Q = P t\):

For the first liquid:
\(P t = m c_1 \Delta T \implies c_1 = \frac{P t}{m \Delta T}\)

For the second liquid:
\((2P) (2t) = (2m) c_2 (3\Delta T) \implies 4 P t = 6 m c_2 \Delta T \implies c_2 = \frac{4 P t}{6 m \Delta T} = \frac{2 P t}{3 m \Delta T}\)

Taking the ratio:
\(\frac{c_1}{c_2} = \frac{\frac{P t}{m \Delta T}}{\frac{2 P t}{3 m \Delta T}} = \frac{3}{2} = 1.5\).

Award [1] for obtaining the expressions for \(c_1\) and \(c_2\) and calculating the ratio to be 1.5.

For the first pipe of length \(L\) (closed at one end, open at the other):
The fundamental wavelength is \(\lambda_1 = 4L\), so the fundamental frequency is:
\(f_0 = \frac{v}{4L}\)

For the second pipe of length \(2L\) (open at both ends):
The harmonics of an open-open pipe of length \(L'\) are given by:
\(f'_n = n \frac{v}{2L'}\)

Here, \(L' = 2L\), so the frequency of the \(n\)-th harmonic is:
\(f'_n = n \frac{v}{2(2L)} = n \frac{v}{4L} = n f_0\)

For the first overtone (which corresponds to the second harmonic, \(n = 2\)):
\(f'_2 = 2 f_0\).

Award [1] for the correct analysis of standing wave conditions for both pipes leading to \(2f_0\).

The current in the circuit is given by \(I = \frac{E}{R + r}\), and the power dissipated in the external resistor is \(P = I^2 R\).

Case 1: \(R = r\)
\(I_1 = \frac{E}{r + r} = \frac{E}{2r}\)
\(P = I_1^2 R = \left(\frac{E}{2r}\right)^2 r = \frac{E^2}{4r}\)

Case 2: \(R = 3r\)
\(I_2 = \frac{E}{3r + r} = \frac{E}{4r}\)
\(P_2 = I_2^2 (3r) = \left(\frac{E}{4r}\right)^2 (3r) = \frac{3E^2}{16r}\)

Expressing \(P_2\) in terms of \(P\):
\(P_2 = \frac{3}{4} \left(\frac{E^2}{4r}\right) = 0.75 P\).

Award [1] for writing correct expressions for power in both cases and taking their ratio to find 0.75P.

The energy levels of a hydrogen-like atom are given by \(E_n = -\frac{E_0}{n^2}\).

For the transition from \(n=4\) to \(n=2\), the energy change is:
\(\Delta E_{4 \to 2} = E_4 - E_2 = -\frac{E_0}{16} - \left(-\frac{E_0}{4}\right) = E_0 \left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3}{16} E_0\)
Since \(h f = \Delta E_{4 \to 2}\), we have \(\frac{3}{16} E_0 = h f\).

For the transition from \(n=4\) to \(n=1\), the energy change is:
\(\Delta E_{4 \to 1} = E_4 - E_1 = -\frac{E_0}{16} - (-E_0) = E_0 \left(1 - \frac{1}{16}\right) = \frac{15}{16} E_0\)

Comparing the two energy changes:
\(\frac{f'}{f} = \frac{\Delta E_{4 \to 1}}{\Delta E_{4 \to 2}} = \frac{\frac{15}{16} E_0}{\frac{3}{16} E_0} = 5\)
Therefore, the new frequency is \(f' = 5f\).

Award [1] for calculating the energy differences of both transitions and showing that the ratio of frequencies is 5.

The electric field \(E\) between parallel plates is uniform and its magnitude is given by:
\(E = \frac{V}{d}\)

Since the upper plate is positive (\(+V\)) and the lower plate is at \(0\text{ V}\), the electric field points vertically downwards.

For a positive charge \(+q\), the electric force \(F_e\) acts in the direction of the field, which is downwards:
\(F_e = qE = \frac{qV}{d}\)

The gravitational force \(F_g = mg\) also acts vertically downwards.

For the sphere to be in equilibrium, the upward tension \(T\) in the string must balance both downward forces:
\(T = F_g + F_e = mg + \frac{qV}{d}\).

Award [1] for identifying the downward direction of both gravitational and electric forces and correctly expressing the sum.

Let the initial number of nuclei for both isotopes be \(N_0\).

At \(t = 4T\):
- For isotope X (half-life \(T\)), the number of half-lives is \(4\).
The remaining number of nuclei is: \(N_X = N_0 \left(\frac{1}{2}\right)^4 = \frac{N_0}{16}\)
- For isotope Y (half-life \(2T\)), the number of half-lives is \(\frac{4T}{2T} = 2\).
The remaining number of nuclei is: \(N_Y = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4}\)

The decay constant \(\lambda\) is inversely proportional to the half-life \(T_{1/2}\):
\(\frac{\lambda_X}{\lambda_Y} = \frac{T_{1/2, Y}}{T_{1/2, X}} = \frac{2T}{T} = 2\)

The activity of a sample is \(A = \lambda N\).
Thus, the ratio of activities is:
\(\frac{A_X}{A_Y} = \frac{\lambda_X N_X}{\lambda_Y N_Y} = \left(\frac{\lambda_X}{\lambda_Y}\right) \left(\frac{N_X}{N_Y}\right) = 2 \times \frac{\frac{N_0}{16}}{\frac{N_0}{4}} = 2 \times \frac{1}{4} = 0.5\).

Award [1] for calculating the remaining fraction of nuclei and the ratio of decay constants, leading to a ratio of 0.5.

The total power absorbed by a planet of radius \(R\) is:
\(P_{\text{absorbed}} = I \times \pi R^2 \times (1 - \alpha)\)
where \(\pi R^2\) is the projected cross-sectional area of the planet intercepting the star's radiation, and \((1 - \alpha)\) is the fraction of radiation absorbed.

In thermal equilibrium, the total power radiated by the planet's surface must equal the power absorbed. The surface area of the spherical planet is \(4\pi R^2\).

Therefore, the average intensity radiated by the surface of the planet is:
\(I_{\text{radiated}} = \frac{P_{\text{absorbed}}}{4\pi R^2} = \frac{I \pi R^2 (1 - \alpha)}{4\pi R^2} = \frac{I(1 - \alpha)}{4}\).

Award [1] for calculating the power absorbed and dividing by the total surface area \(4\pi R^2\) to get the final result.

By conservation of linear momentum: \(3m u = (3m + m)v\), where \(v\) is the common velocity after collision. This gives \(v = \frac{3}{4}u\). The initial kinetic energy is \(E_i = \frac{1}{2}(3m)u^2 = \frac{3}{2}mu^2\). The final kinetic energy is \(E_f = \frac{1}{2}(4m)v^2 = 2m\left(\frac{3}{4}u\right)^2 = \frac{9}{8}mu^2\). The kinetic energy lost is \(\Delta E = E_i - E_f = \frac{12}{8}mu^2 - \frac{9}{8}mu^2 = \frac{3}{8}mu^2\). The fraction of energy lost is \(\frac{\Delta E}{E_i} = \frac{3/8}{3/2} = \frac{1}{4} = 25\%\).

Award 1 mark for the correct calculation showing that the lost fraction is 25%.

For a pipe closed at one end, the boundary conditions require a node at the closed end and an antinode at the open end. The possible standing wave wavelengths are given by \(\lambda_n = \frac{4L}{n}\) for odd integers \(n = 1, 3, 5, \dots\). The fundamental mode (first harmonic) has \(n=1\). The second resonant mode is the third harmonic with \(n=3\). The frequency is \(f = \frac{v}{\lambda_3} = \frac{3v}{4L}\).

Award 1 mark for identifying the formula for the third harmonic wavelength and calculating the correct frequency.

According to Wien's displacement law, \(\lambda_{\text{max}} T = \text{constant}\). Therefore, \(\lambda_X T_X = \lambda_Y T_Y \Rightarrow \frac{T_X}{T_Y} = \frac{\lambda_Y}{\lambda_X} = \frac{1}{2}\). According to the Stefan-Boltzmann law, the total power radiated by a blackbody of surface area \(A\) at temperature \(T\) is \(P = \sigma A T^4\). Since the surface areas are identical, the ratio of power is \(\frac{P_X}{P_Y} = \left(\frac{T_X}{T_Y}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}\).

Award 1 mark for using Wien's law to relate temperature to wavelength, and the Stefan-Boltzmann law to find the power ratio.

The drift speed \(v\) of electrons is given by \(I = n A v e\), where \(I = \frac{V}{R}\) and \(R = \frac{\rho L}{A}\). Substituting these into the drift speed formula gives \(v = \frac{V / R}{n A e} = \frac{V A}{\rho L n A e} = \frac{V}{n e \rho L}\). This shows that the drift speed is independent of the cross-sectional area \(A\) (and thus the radius \(r\)) of the wire, and is inversely proportional to the length \(L\) of the wire under a constant potential difference \(V\). Therefore, \(\frac{v_2}{v_1} = \frac{L_1}{L_2} = \frac{L}{2L} = \frac{1}{2}\).

Award 1 mark for realizing that drift speed in this parallel configuration depends only on length, yielding a ratio of \(\frac{1}{2}\).

The energy levels of a hydrogen atom are given by \(E_n = -\frac{E_0}{n^2}\), where \(E_0 = 13.6\text{ eV}\). The energy of the emitted photon for the transition \(3 \to 1\) is \(h f = E_3 - E_1 = -\frac{E_0}{9} - (-E_0) = \frac{8}{9} E_0\). The energy of the emitted photon for the transition \(2 \to 1\) is \(h f' = E_2 - E_1 = -\frac{E_0}{4} - (-E_0) = \frac{3}{4} E_0\). Taking the ratio of the two frequencies: \(\frac{f'}{f} = \frac{\frac{3}{4} E_0}{\frac{8}{9} E_0} = \frac{3}{4} \times \frac{9}{8} = \frac{27}{32}\). Therefore, \(f' = \frac{27}{32} f\).

Award 1 mark for expressing the energy of both transitions correctly and finding their ratio to obtain the correct frequency.

Because the charges have opposite signs, the point where the net electric field is zero must lie outside the region between them. Since the magnitude of \(+4q\) is larger than \(-q\), the field can only cancel on the side of the smaller charge, so \(x > d\). Let this point be at coordinate \(x\). The electric field due to \(+4q\) is \(E_1 = \frac{k(4q)}{x^2}\) pointing in the positive x-direction, and the field due to \(-q\) is \(E_2 = \frac{kq}{(x-d)^2}\) pointing in the negative x-direction. Setting the magnitudes equal: \(\frac{4kq}{x^2} = \frac{kq}{(x-d)^2} \Rightarrow \frac{4}{x^2} = \frac{1}{(x-d)^2}\). Taking the square root of both sides gives \(\frac{2}{x} = \frac{1}{x-d} \Rightarrow 2x - 2d = x \Rightarrow x = 2d\).

Award 1 mark for setting up the electric field equality and solving for the coordinate \(x = 2d\).

Let \(N_0\) be the initial number of nuclei of X. At time \(t\), the number of X nuclei is \(N_X\) and the number of Y nuclei is \(N_Y\), such that \(N_X + N_Y = N_0\). We are given \(\frac{N_Y}{N_X} = 7\), which means \(N_Y = 7 N_X\). Substituting this gives \(N_X + 7 N_X = 8 N_X = N_0\), so \(N_X = \frac{1}{8} N_0\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly 3 half-lives have passed during time \(t\). Thus, \(3 T_{1/2} = t \Rightarrow T_{1/2} = \frac{t}{3}\).

Award 1 mark for finding that the fraction of parent isotope remaining is \(\frac{1}{8}\) and deducing that the half-life is \(\frac{t}{3}\).

The intensity of radiation from the star reaching the planet's orbit is \(I_0 = \frac{L}{4 \pi d^2}\). The total power intercepted by the planet of radius \(R\) is \(P_{\text{intercepted}} = I_0 \times \pi R^2\). The fraction of this power absorbed is \((1 - \alpha)\), so the absorbed power is \(P_{\text{absorbed}} = I_0 \pi R^2 (1 - \alpha) = \frac{L \pi R^2 (1 - \alpha)}{4 \pi d^2} = \frac{L R^2 (1 - \alpha)}{4 d^2}\). The average intensity of this absorbed radiation over the entire surface area of the planet (\(4 \pi R^2\)) is \(I_{\text{absorbed}} = \frac{P_{\text{absorbed}}}{4 \pi R^2} = \frac{L R^2 (1 - \alpha)}{16 \pi d^2 R^2} = \frac{L(1 - \alpha)}{16 \pi d^2}\).

Award 1 mark for expressing the total power absorbed and dividing by the total surface area of the sphere to obtain the correct average intensity.

Using the impulse-momentum theorem: \( I = \Delta p = F \Delta t \). Let the initial direction of motion be positive. The force acts in the opposite direction, so its value is \( -F \). The impulse is \( I = -F \times \frac{3mu}{2F} = -1.5mu \). The final momentum is \( p_f = p_i + I = mu - 1.5mu = -0.5mu \). Since the final momentum is negative, the particle moves in the opposite direction with a speed of \( 0.5u \).

Award [1] for the correct answer. No partial marks are available for multiple-choice questions.

The total energy supplied by the heater in time \( t \) is \( P t \). This energy is used to increase the thermal energy of the liquid and to supply the energy lost to the surroundings. Therefore, \( P t = mc(T_2 - T_1) + L t \). Dividing both sides by \( t \) gives \( P = \frac{mc(T_2 - T_1)}{t} + L \). Solving for \( L \) gives \( L = P - \frac{mc(T_2 - T_1)}{t} \).

Award [1] for the correct answer. No partial marks are available for multiple-choice questions.

For a pipe closed at one end, the boundary conditions require a displacement node at the closed end (\( x = 0 \)) and a displacement antinode at the open end (\( x = L \)). The allowed wavelengths for harmonic \( n \) (where \( n \) is an odd integer) are given by \( L = \frac{n\lambda}{4} \). For the third harmonic (\( n = 3 \)), we have \( L = \frac{3\lambda}{4} \), which means the wavelength is \( \lambda = \frac{4L}{3} \). The standing wave has displacement nodes at \( x = 0 \) and \( x = \frac{\lambda}{2} = \frac{2L}{3} \), and displacement antinodes at \( x = \frac{\lambda}{4} = \frac{L}{3} \) and \( x = \frac{3\lambda}{4} = L \). Therefore, the distance from the closed end to the first displacement antinode is \( \frac{L}{3} \).

Award [1] for the correct answer. No partial marks are available for multiple-choice questions.

Let the emf of the cell be \( V \). When connected in series, the total equivalent resistance is \( R_{\text{eq1}} = 3R \). The total power dissipated is \( P_1 = \frac{V^2}{3R} \). When connected in parallel, the total equivalent resistance is \( R_{\text{eq2}} = \frac{R}{3} \). The total power dissipated is \( P_2 = \frac{V^2}{R/3} = \frac{3V^2}{R} \). The ratio of the powers is \( \frac{P_2}{P_1} = \frac{3V^2/R}{V^2/3R} = 9 \).

Award [1] for the correct answer. No partial marks are available for multiple-choice questions.

The energy of the emitted photon is equal to the difference in energy between the two states. For the transition from \( n = 3 \) to \( n = 2 \): \( hf = E_3 - E_2 = -\frac{E_0}{9} - \left(-\frac{E_0}{4}\right) = E_0 \left(\frac{1}{4} - \frac{1}{9}\right) = E_0 \left(\frac{5}{36}\right) \). For the transition from \( n = 4 \) to \( n = 2 \): \( hf' = E_4 - E_2 = -\frac{E_0}{16} - \left(-\frac{E_0}{4}\right) = E_0 \left(\frac{1}{4} - \frac{1}{16}\right) = E_0 \left(\frac{3}{16}\right) \). Dividing the second equation by the first: \( \frac{f'}{f} = \frac{3/16}{5/36} = \frac{3}{16} \times \frac{36}{5} = \frac{27}{20} \). Therefore, \( f' = \frac{27}{20}f \).

Award [1] for the correct answer. No partial marks are available for multiple-choice questions.

The formula for the double-slit fringe width is \( s = \frac{\lambda D}{d} \). Under the new conditions, the slit separation is \( d' = 2d \) and the screen distance is \( D' = \frac{D}{2} \). The new fringe width \( s' \) is given by: \( s' = \frac{\lambda D'}{d'} = \frac{\lambda (D/2)}{2d} = \frac{1}{4} \frac{\lambda D}{d} = \frac{s}{4} \).

Award [1] for the correct answer. No partial marks are available for multiple-choice questions.

(a) Linear momentum is defined as the product of mass and velocity of an object.

(b) Using conservation of momentum: \(m_1 v_1 + m_2 v_2 = m_1 v_{1f} + m_2 v_{2f}\).
Let the initial direction of motion be positive.
\(0.40 \times 3.0 + 0 = 0.40 \times (-0.60) + 0.80 \times v_{2f}\)
\(1.2 = -0.24 + 0.80 v_{2f}\)
\(0.80 v_{2f} = 1.44 \implies v_{2f} = 1.8\text{ m s}^{-1}\).

(c) Initial kinetic energy: \(E_{k,i} = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} (0.40)(3.0)^2 = 1.80\text{ J}\).
Final kinetic energy: \(E_{k,f} = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 = \frac{1}{2} (0.40)(-0.60)^2 + \frac{1}{2} (0.80)(1.8)^2 = 0.072 + 1.296 = 1.368\text{ J} \approx 1.37\text{ J}\).
Change in kinetic energy: \(\Delta E_k = E_{k,f} - E_{k,i} = 1.368 - 1.80 = -0.432\text{ J}\).
Since kinetic energy is not conserved (the total kinetic energy decreased), the collision is inelastic.

(a) 1 mark for: product of mass and velocity.

(b) 1 mark for correct statement of momentum conservation; 1 mark for correct substitution (including negative sign for rebounding velocity); 1 mark for correct final answer with appropriate units.

(c) 1 mark for correct initial kinetic energy; 1 mark for correct final kinetic energy; 1 mark for showing that kinetic energy is not conserved; 1 mark for final conclusion clearly supported by the calculations.

(a) Specific heat capacity is the thermal energy required per unit mass to change the temperature of a substance by \(1\text{ K}\) without a change of phase. Specific latent heat is the thermal energy required per unit mass to change the phase of a substance at a constant temperature.

(b) Total energy supplied by the heater: \(Q = P \times t = 45\text{ W} \times (5.0 \times 60\text{ s}) = 13500\text{ J}\).
Using \(Q = m c \Delta T\):
\(13500 = 0.25 \times c \times (50.0 - 20.0)\)
\(13500 = 7.5 \times c \implies c = 1800\text{ J kg}^{-1}\text{ K}^{-1}\).

(c) If heat is lost to the surroundings, the temperature increase \(\Delta T\) of the liquid will be smaller than it would be if all energy was retained. Since specific heat capacity is calculated using \(c = \frac{P \times t}{m \Delta T}\), a smaller \(\Delta T\) for a given energy input results in a larger calculated value of \(c\). Thus, the experimental value is an overestimation of the true specific heat capacity.

(a) 1 mark for: defining specific heat capacity with reference to temperature change per unit mass; 1 mark for: defining specific latent heat with reference to phase change at constant temperature.

(b) 1 mark for: calculating energy supplied \(Q = 13500\text{ J}\); 1 mark for: correct substitution into \(Q = m c \Delta T\); 1 mark for: final correct answer with correct units \(\text{J kg}^{-1}\text{ K}^{-1}\) or \(\text{J kg}^{-1}\ ^\circ\text{C}^{-1}\).

(c) 1 mark for: stating that heat loss leads to a smaller temperature rise for the same energy input; 1 mark for: explaining that \(c \propto \frac{1}{\Delta T}\); 1 mark for: concluding that the calculated value of specific heat capacity is higher than the true value.

(a) Sound waves from a source travel down the tube and reflect at the closed boundary. The incident and reflected waves, which have the same frequency and amplitude but travel in opposite directions, superpose. This creates a standing wave with nodes (points of zero displacement at the closed end) and antinodes (points of maximum displacement at the open end).

(b) The sketch should show a node (zero amplitude) at the closed end and an antinode (maximum amplitude) at the open end. The distance from the closed end to the open end is a quarter wavelength \(\frac{\lambda}{4} = L\).

(c) \(\lambda = 4L = 4 \times 0.60\text{ m} = 2.40\text{ m}\).
Using \(v = f \lambda\):
\(f = \frac{v}{\lambda} = \frac{340}{2.40} \approx 141.67\text{ Hz} \approx 142\text{ Hz}\).

(d) If both ends are closed, there must be displacement nodes at both ends. The fundamental standing wave now has a wavelength of \(\lambda' = 2L = 1.20\text{ m}\). Since the wavelength is halved and the speed of sound remains constant, the new fundamental frequency is doubled: \(f' = \frac{340}{1.20} \approx 283\text{ Hz}\).

(a) 1 mark for: reference to reflection at the boundary; 1 mark for: superposition/interference of two waves of equal frequency and amplitude travelling in opposite directions.

(b) 1 mark for: correct drawing showing a node at the closed end and an antinode at the open end; 1 mark for: labeling the closed end as a Node (N) and the open end as an Antinode (A).

(c) 1 mark for: identifying that \(\lambda = 4L = 2.4\text{ m}\); 1 mark for: correct calculation of frequency (141.7 Hz or 142 Hz) with correct unit.

(d) 1 mark for: identifying that both ends must now be nodes, so \(\lambda = 2L\); 1 mark for: stating that the fundamental frequency doubles or calculating \(283\text{ Hz}\).

(a) Electromotive force (emf) is the total energy transferred per unit charge from other forms to electrical energy (or the work done per unit charge in moving charge completely around a circuit).

(b) Terminal potential difference: \(V = \varepsilon - I r = 9.0 - (0.40 \times 1.5) = 8.4\text{ V}\).

(c) The potential difference across the resistor \(R\) is: \(V_R = I R = 0.40 \times 12.0 = 4.8\text{ V}\).
The terminal potential difference is the sum of the potential differences across the external components: \(V = V_R + V_{\text{LED}}\).
Therefore, \(V_{\text{LED}} = V - V_R = 8.4 - 4.8 = 3.6\text{ V}\).

(d) When the temperature of an NTC thermistor increases, its resistance decreases. Since the other components and the emf remain constant, the overall resistance of the circuit decreases. This causes the current in the circuit to increase. Since the power dissipated by the LED is related to the current, the brightness of the LED increases.

(a) 1 mark for: definition as work done per unit charge around the complete circuit (or energy converted from chemical to electrical energy per unit charge).

(b) 1 mark for: using \(V = \varepsilon - Ir\); 1 mark for: correct calculation of \(8.4\text{ V}\).

(c) 1 mark for: calculating \(V_R = 4.8\text{ V}\); 1 mark for: subtracting \(V_R\) from terminal PD to find \(3.6\text{ V}\).

(d) 1 mark for: stating that the resistance of the NTC thermistor decreases with temperature; 1 mark for: stating that this increases the current in the circuit; 1 mark for: relating the increased current to an increase in LED brightness.

(a) When an electron falls from a higher energy level to a lower one, a photon of energy \(\Delta E = hf\) is emitted. Since the emission spectrum consists of discrete wavelengths/frequencies rather than a continuous spectrum, the energy differences \(\Delta E\) must have specific, discrete values. This implies that the energy levels within the atom are restricted to specific, discrete states.

(b) The energy difference is: \(\Delta E = E_3 - E_2 = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}\).
In Joules: \(\Delta E = 1.89 \times 1.60 \times 10^{-19}\text{ J} = 3.024 \times 10^{-19}\text{ J}\).
Using \(\Delta E = h f\):
\(f = \frac{3.024 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} \approx 4.561 \times 10^{14}\text{ Hz}\).
Accept answers between \(4.56 \times 10^{14}\text{ Hz}\) and \(4.57 \times 10^{14}\text{ Hz}\).

(c) For a hydrogen atom to absorb a photon, the photon energy must precisely match the difference between its current energy level (ground state \(E_1\)) and a higher energy level.
Energy to transition to \(E_2\): \(-3.40 - (-13.6) = 10.20\text{ eV}\).
Energy to transition to \(E_3\): \(-1.51 - (-13.6) = 12.09\text{ eV}\).
Since \(11.5\text{ eV}\) does not correspond to any possible transition from the ground state, these photons cannot be absorbed.

(a) 1 mark for: photon energy is determined by frequency \(E = hf\); 1 mark for: transitions between specific levels emit specific photon energies; 1 mark for: discrete line spectrum demonstrates that only specific energy transitions are allowed, proving discrete energy levels.

(b) 1 mark for: calculating \(\Delta E = 1.89\text{ eV}\); 1 mark for: converting \(\Delta E\) to Joules (\(3.02 \times 10^{-19}\text{ J}\)); 1 mark for: calculating frequency with correct unit (\(4.56 \times 10^{14}\text{ Hz}\) to \(4.57 \times 10^{14}\text{ Hz}\)).

(c) 1 mark for: stating that the photon energy must match the difference between two energy levels; 1 mark for: calculating the transition energies from ground state (\(10.2\text{ eV}\) and \(12.1\text{ eV}\)) and concluding that \(11.5\text{ eV}\) cannot be absorbed.

(a) An electric field is a region of space where a charged particle experiences a force (or force per unit positive test charge).

(b) The distance from each charge to the midpoint P is \(r = 0.060\text{ m}\).
The electric field strength \(E\) due to a point charge is: \(E = \frac{k |q|}{r^2}\).
For the positive charge \(q_1\):
\(E_1 = \frac{8.99 \times 10^9 \times 5.0 \times 10^{-6}}{0.060^2} \approx 1.25 \times 10^7\text{ N C}^{-1}\) (directed away from \(q_1\), i.e., towards \(q_2\)).
For the negative charge \(q_2\):
\(E_2 = \frac{8.99 \times 10^9 \times 5.0 \times 10^{-6}}{0.060^2} \approx 1.25 \times 10^7\text{ N C}^{-1}\) (directed towards \(q_2\)).
Since both fields point in the same direction, the total electric field strength is:
\(E_{\text{total}} = E_1 + E_2 = 2.5 \times 10^7\text{ N C}^{-1}\) directed towards \(q_2\).

(c) Electric potential \(V\) is a scalar quantity. The total potential at P is the algebraic sum of the potentials from both charges:
\(V_{\text{total}} = V_1 + V_2 = \frac{k q_1}{r} + \frac{k q_2}{r}\).
Since \(q_1 = +5.0\ \mu\text{C}\) and \(q_2 = -5.0\ \mu\text{C}\), and both are at the same distance \(r = 0.060\text{ m}\) from P:
\(V_1 = -V_2\).
Therefore, \(V_{\text{total}} = 0\text{ V}\).

(a) 1 mark for: force per unit positive charge on a small test charge placed at that point.

(b) 1 mark for: calculating distance \(r = 0.060\text{ m}\); 1 mark for: correct formula and substitution for one charge; 1 mark for: recognizing that both fields point in the same direction and adding them; 1 mark for: correct final answer (\(2.5 \times 10^7\text{ N C}^{-1}\)) and direction (towards \(q_2\) / the negative charge).

(c) 1 mark for: identifying that electric potential is a scalar quantity; 1 mark for: showing that the potential of \(q_1\) is equal in magnitude and opposite in sign to that of \(q_2\); 1 mark for: concluding that \(V = 0\text{ V}\).

(a)
* **Systematic Error**: Heat loss to the environment OR thermal lag (the thermometer does not instantly reach the same temperature as the liquid).
* **Minimization**: To minimize heat loss, start the experiment with the liquid temperature as far below room temperature as it will end up above room temperature (so that net heat exchange is zero), or use better insulation. To minimize thermal lag, stir the liquid continuously.

(b)
Using the conservation of energy:
\[ Q = V I \Delta t = m c \Delta T \]
\[ c = \frac{V I \Delta t}{m \Delta T} \]
\[ c = \frac{12.0 \times 2.00 \times 300}{0.450 \times 15.0} \]
\[ c = \frac{7200}{6.75} = 1067 \approx 1070 \text{ J kg}^{-1}\text{ K}^{-1} \] (or \(1.1 \times 10^3 \text{ J kg}^{-1}\text{ K}^{-1}\) to 2 significant figures).

(c) (i)
The fractional uncertainty in \(c\) is:
\[ \frac{\Delta c}{c} = \frac{\Delta V}{V} + \frac{\Delta I}{I} + \frac{\Delta(\Delta t)}{\Delta t} + \frac{\Delta m}{m} + \frac{\Delta(\Delta T)}{\Delta T} \]
\[ \frac{\Delta c}{c} = \frac{0.2}{12.0} + \frac{0.05}{2.00} + \frac{1}{300} + \frac{0.005}{0.450} + \frac{0.5}{15.0} \]
\[ \frac{\Delta c}{c} = 0.0167 + 0.0250 + 0.0033 + 0.0111 + 0.0333 = 0.0894 \]
Percentage uncertainty \(\approx 8.94\% \approx 9\%\).

(c) (ii)
It leads to an **overestimate**.
Because some of the electrical work supplied by the heater is absorbed by the calorimeter itself, the actual heat energy absorbed by the liquid alone is less than the calculated electrical energy \(V I \Delta t\). Since we assume all of the energy goes to the liquid, \(c\) is calculated using a value of energy that is too high, leading to an overestimate.

(a)
* Award [1] for identifying a correct systematic error (e.g., heat loss to surroundings / thermal lag / thermometer calibration error).
* Award [1] for a valid corresponding minimization method (e.g., insulate container / stir liquid / calibrate thermometer).

(b)
* Award [1] for correct substitution of terms into the heat energy equation: \(c = \frac{12.0 \times 2.00 \times 300}{0.450 \times 15.0}\).
* Award [1] for the final answer \(1070\) or \(1100\) with correct unit \(\text{J kg}^{-1}\text{ K}^{-1}\) (or \(\text{J kg}^{-1}\text{ }^\circ\text{C}^{-1}\)).

(c) (i)
* Award [1] for summing the fractional uncertainties correctly: \(1.67\% + 2.50\% + 0.33\% + 1.11\% + 3.33\%\).
* Award [1] for the correct final percentage uncertainty of \(9\%\) (accept values between \(8.9\%\) and \(9.0\%\)).

(c) (ii)
* Award [1] for identifying 'overestimate' with a valid explanation showing that the electrical energy supplied is larger than the energy actually absorbed by the liquid.

(a)
* Standing waves are formed in the air column by the superposition/interference of the incident wave and the wave reflected from the water surface.
* Resonance occurs when the frequency of the tuning fork matches a natural frequency of the air column. This corresponds to a displacement node at the water surface and a displacement antinode near the open end, creating a maximum in amplitude/loudness.

(b)
* The antinode at the open end does not form exactly at the top of the tube, but slightly outside it at a distance \(e\) known as the end correction.
* For the first resonance, \(L_1 + e = \frac{\lambda}{4}\), and for the second resonance, \(L_2 + e = \frac{3\lambda}{4}\).
* Subtracting these equations gives \(L_2 - L_1 = \frac{\lambda}{2}\). Using the difference eliminates the systematic error introduced by the unknown end correction \(e\).

(c) (i)
\[ \frac{\lambda}{2} = L_2 - L_1 = 0.498 - 0.165 = 0.333 \text{ m} \]
\[ \lambda = 2 \times 0.333 = 0.666 \text{ m} \]
\[ v = f \lambda = 512 \times 0.666 = 340.99 \approx 341 \text{ m s}^{-1} \] (accept 340 to 341)

(c) (ii)
* Uncertainty in difference \(\Delta L = L_2 - L_1\) is \(\Delta (\Delta L) = 0.2 + 0.2 = 0.4 \text{ cm} = 0.004 \text{ m}\).
* Fractional uncertainty in \(\Delta L\) (and thus in \(\lambda\)) is:
\[ \frac{\Delta \lambda}{\lambda} = \frac{0.004}{0.333} \approx 0.01201 \ (1.20\%) \]
* Fractional uncertainty in \(f\) is:
\[ \frac{\Delta f}{f} = \frac{1}{512} \approx 0.00195 \ (0.20\%) \]
* Fractional uncertainty in \(v\) is:
\[ \frac{\Delta v}{v} = \frac{\Delta f}{f} + \frac{\Delta \lambda}{\lambda} = 0.00195 + 0.01201 = 0.01396 \ (1.40\%) \]
* Absolute uncertainty in \(v\) is:
\[ \Delta v = 340.99 \times 0.01396 = 4.76 \approx 5 \text{ m s}^{-1} \]

(a)
* Award [1] for explaining that a standing wave is formed by the superposition of the incident and reflected waves.
* Award [1] for stating that resonance occurs when a node forms at the water boundary and an antinode forms at/near the open end.

(b)
* Award [1] for referencing the concept of 'end correction' (that the antinode is not exactly at the boundary of the tube).
* Award [1] for demonstrating/explaining mathematically that taking the difference \(L_2 - L_1\) cancels out the end correction.

(c) (i)
* Award [1] for finding wavelength \(\lambda = 0.666 \text{ m}\).
* Award [1] for finding speed of sound \(v = 341 \text{ m s}^{-1}\) (or \(340 \text{ m s}^{-1}\)).

(c) (ii)
* Award [1] for finding the absolute uncertainty in length difference \(\pm 0.4 \text{ cm}\) (or \(0.004 \text{ m}\)) and calculating its fractional contribution.
* Award [1] for correctly combining the fractional uncertainties to find the final absolute uncertainty of \(5 \text{ m s}^{-1}\) (accept range \(4.7\) to \(5\)).

Part a: The stellar parallax angle is half of the maximum angular shift of a star's apparent position relative to distant background stars, when observed from opposite sides of the Earth's orbit around the Sun. Part b: First, calculate the distance in parsecs: \(d = \frac{1}{p} = \frac{1}{0.080} = 12.5\text{ pc}\). Convert parsecs to meters: \(1\text{ pc} \approx 3.09 \times 10^{16}\text{ m}\), so \(d = 12.5 \times 3.09 \times 10^{16}\text{ m} = 3.86 \times 10^{17}\text{ m}\), which rounds to \(3.9 \times 10^{17}\text{ m}\). Part c: The relationship between apparent brightness \(b\), luminosity \(L\), and distance \(d\) is: \(b = \frac{L}{4\pi d^2}\). Rearranging for \(L\): \(L = 4\pi d^2 b\). Substituting the values: \(L = 4\pi (3.86 \times 10^{17}\text{ m})^2 (2.5 \times 10^{-9}\text{ W m}^{-2}) = 4.68 \times 10^{27}\text{ W}\). Using the show-that value of \(3.9 \times 10^{17}\text{ m}\) gives \(L = 4.78 \times 10^{27}\text{ W}\). Either value rounds to \(4.7 \times 10^{27}\text{ W}\) or \(4.8 \times 10^{27}\text{ W}\) respectively.

Part a: Award [1 mark] for defining the angle subtended at the star by the radius of the Earth's orbit (1 AU). Part b: Award [1 mark] for calculating the distance in parsecs as 12.5 pc. Award [1 mark] for converting parsecs to meters to show the value of approximately \(3.9 \times 10^{17}\text{ m}\). Part c: Award [1 mark] for correctly identifying and rearranging the brightness-luminosity formula. Award [1 mark] for the final answer of \(4.7 \times 10^{27}\text{ W}\) (or \(4.8 \times 10^{27}\text{ W}\)).

Part a: According to Wien's displacement law, \(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\text{ m K}\). Substituting the peak wavelength: \(T = \frac{2.90 \times 10^{-3}}{410 \times 10^{-9}} = 7073\text{ K}\), which is approximately \(7100\text{ K}\). Part b: The luminosity is given by the Stefan-Boltzmann law \(L = 4\pi R^2 \sigma T^4\). Therefore, the ratio of the luminosities is \(\frac{L_Y}{L_{\odot}} = (\frac{R_Y}{R_{\odot}})^2 \times (\frac{T_Y}{T_{\odot}})^4\). Substituting the known values: \(\frac{L_Y}{L_{\odot}} = (3.5)^2 \times (\frac{7073}{5800})^4 = 12.25 \times (1.2195)^4 = 12.25 \times 2.209 = 27.06\), which rounds to 27.

Part a: Award [1 mark] for stating Wien's law or substituting values. Award [1 mark] for the final temperature of \(7100\text{ K}\) or \(7073\text{ K}\). Part b: Award [1 mark] for writing the ratio formula in terms of radius and temperature. Award [1 mark] for correct substitution. Award [1 mark] for the final ratio of 27 (accept 27 to 28).

Part a: The Jeans criterion states that a gas cloud will collapse to form a star if its gravitational potential energy is greater than its thermal kinetic energy, meaning its mass exceeds the Jeans mass for its given temperature and density. Part b: In a main sequence star, the inward force of gravity is perfectly balanced by the outward radiation pressure and gas pressure produced by nuclear fusion reactions in the core, maintaining a stable state called hydrostatic equilibrium. Part c: Since its mass is 1.2 solar masses, which is below the Chandrasekhar limit of 1.4 solar masses, its core will end its life as a white dwarf, supported by electron degeneracy pressure.

Part a: Award [1 mark] for stating that gravitational potential energy must exceed the random kinetic energy of the particles. Award [1 mark] for stating that the mass must exceed the Jeans mass for the given temperature and density. Part b: Award [1 mark] for identifying the inward gravitational force. Award [1 mark] for identifying the outward radiation or gas pressure from fusion. Part c: Award [1 mark] for stating white dwarf.

Part a: The CMB is isotropic and homogeneous radiation that matches a blackbody spectrum at 2.7 K. This represents the highly redshifted (stretched) remnant of radiation from the hot, dense early universe, providing direct evidence for the Big Bang expansion. Part b (i): Using the redshift formula, \(v = z c = 0.045 \times 3.00 \times 10^8\text{ m s}^{-1} = 1.35 \times 10^7\text{ m s}^{-1}\) (or \(13500\text{ km s}^{-1}\)). Part b (ii): From Hubble's Law, \(d = \frac{v}{H_0}\). Substituting the values: \(d = \frac{13500\text{ km s}^{-1}}{70\text{ km s}^{-1}\text{ Mpc}^{-1}} = 192.9\text{ Mpc}\), which rounds to \(190\text{ Mpc}\).

Part a: Award [1 mark] for describing CMB as isotropic and matching a blackbody spectrum at 2.7 K. Award [1 mark] for explaining that this is the redshifted remnant of radiation from the hot, dense early universe. Part b (i): Award [1 mark] for the recession speed of \(1.35 \times 10^7\text{ m s}^{-1}\) or \(13500\text{ km s}^{-1}\). Part b (ii): Award [1 mark] for using Hubble's Law. Award [1 mark] for the final distance of \(190\text{ Mpc}\) (or \(193\text{ Mpc}\)).