2021 HKDSE Physics Answers & Marking Scheme

The light ray is incident normally on the face opposite to the \( 30^\circ \) angle. This face is perpendicular to the adjacent face of the right angle. The ray passes undeflected and hits the hypotenuse. The angle of incidence \( i \) on the hypotenuse is \( 90^\circ - 30^\circ = 60^\circ \). For total internal reflection to occur, \( i \ge \theta_c \), where \( \sin \theta_c = \frac{n}{1.52} \). Therefore, \( \sin 60^\circ \ge \frac{n}{1.52} \Rightarrow n \le 1.52 \sin 60^\circ \approx 1.32 \).

Award 1 mark for the correct answer C. No marks for incorrect or multiple options selected.

The average kinetic energy of gas molecules is directly proportional to the absolute temperature \( T \). Since the average kinetic energy is doubled, the absolute temperature is also doubled: \( T_2 = 2 T_1 = 2 \times (27 + 273.15) \approx 600\text{ K} = 327^\circ\text{C} \). Since the container is rigid, the volume is constant. According to the pressure law, \( P \propto T \), so the pressure also doubles: \( P_2 = 2 \times 1.0 \times 10^5\text{ Pa} = 2.0 \times 10^5\text{ Pa} \).

Award 1 mark for the correct answer B. No marks for incorrect or multiple options selected.

The horizontal component of the velocity remains constant: \( v_x = u_x = 20 \cos 30^\circ = 10\sqrt{3}\text{ m s}^{-1} \). For the vertical motion, taking downwards as positive: \( u_y = -20 \sin 30^\circ = -10\text{ m s}^{-1} \). Using \( v_y^2 = u_y^2 + 2 g H \), we have \( v_y^2 = (-10)^2 + 2(10)(40) = 100 + 800 = 900 \Rightarrow v_y = 30\text{ m s}^{-1} \). Just before impact, the angle with the horizontal \( \theta \) satisfies \( \tan \theta = \frac{v_y}{v_x} = \frac{30}{10\sqrt{3}} = \sqrt{3} \Rightarrow \theta = 60^\circ \).

Award 1 mark for the correct answer C. No marks for incorrect or multiple options selected.

When S is open, R is disconnected, so P and Q are in series. The total resistance is \( 2R \), and the power of P is \( P_{\text{open}} = I^2 R = \left(\frac{V}{2R}\right)^2 R = \frac{V^2}{4R} \). When S is closed, Q and R are in parallel with an equivalent resistance of \( 0.5R \). The total resistance becomes \( 1.5R \). The current through P is \( I' = \frac{V}{1.5R} = \frac{2V}{3R} \). The new power of P is \( P_{\text{closed}} = (I')^2 R = \left(\frac{2V}{3R}\right)^2 R = \frac{4V^2}{9R} \). The ratio is \( \frac{P_{\text{closed}}}{P_{\text{open}}} = \frac{4/9}{1/4} = \frac{16}{9} \).

Award 1 mark for the correct answer B. No marks for incorrect or multiple options selected.

The induced electromotive force in the loop is \( \mathcal{E} = B L v = 0.5 \times 0.1 \times 4 = 0.2\text{ V} \). The induced current is \( I = \frac{\mathcal{E}}{R} = \frac{0.2}{0.2} = 1\text{ A} \). The magnetic force acting on the side of the loop inside the field is \( F_B = B I L = 0.5 \times 1 \times 0.1 = 0.05\text{ N} \). Since the speed is constant, the external force must balance the magnetic force, so \( F_{\text{ext}} = 0.05\text{ N} \).

Award 1 mark for the correct answer B. No marks for incorrect or multiple options selected.

Let \( a \) be the acceleration of the system and \( T \) be the tension. The kinetic friction force on A is \( f_k = \mu_k m_A g = 0.3 \times 3 \times 9.81 = 8.829\text{ N} \). The equations of motion for the blocks are: \( T - f_k = m_A a \) and \( m_B g - T = m_B a \). Adding these gives: \( m_B g - f_k = (m_A + m_B) a \Rightarrow 2 \times 9.81 - 8.829 = (3+2) a \Rightarrow 10.791 = 5a \Rightarrow a = 2.158\text{ m s}^{-2} \). The tension is \( T = m_B (g - a) = 2 \times (9.81 - 2.158) \approx 15.3\text{ N} \).

Award 1 mark for the correct answer C. No marks for incorrect or multiple options selected.

The total thermal energy supplied by the heater is \( E = P \times t = 200 \times 25 = 5000\text{ J} \). The useful heat absorbed by the metal block is \( Q = m c \Delta T = 0.5 \times 400 \times 20 = 4000\text{ J} \). The thermal energy lost is \( E_{\text{lost}} = 5000 - 4000 = 1000\text{ J} \). The percentage of energy lost is \( \frac{1000}{5000} \times 100\% = 20\% \).

Award 1 mark for the correct answer A. No marks for incorrect or multiple options selected.

The wavelength of the sound is \( \lambda = \frac{v}{f} = \frac{340}{850} = 0.4\text{ m} \). The path difference at \( P \) is \( \Delta s = |3.2 - 4.4| = 1.2\text{ m} \). Since \( \Delta s = 1.2\text{ m} = 3 \lambda \), constructive interference occurs, and a loud sound is heard at \( P \). If the frequency \( f \) increases, the wavelength \( \lambda = \frac{v}{f} \) decreases, meaning \( \Delta s / \lambda \) increases from \( 3.0 \) towards \( 3.5 \). At \( \Delta s / \lambda = 3.5 \), destructive interference occurs, so the sound becomes soft.

Award 1 mark for the correct answer A. No marks for incorrect or multiple options selected.

The electric field strength between the plates is \( E = \frac{V}{d} = \frac{400}{0.02} = 20000\text{ V m}^{-1} \). For the droplet to remain stationary, the upward electric force balances gravity: \( q E = m g \Rightarrow q = \frac{m g}{E} = \frac{1.6 \times 10^{-15} \times 10}{20000} = 8.0 \times 10^{-19}\text{ C} \). The number of excess elementary charges is \( N = \frac{q}{e} = \frac{8.0 \times 10^{-19}}{1.6 \times 10^{-19}} = 5 \).

Award 1 mark for the correct answer B. No marks for incorrect or multiple options selected.

Einstein's photoelectric equation states \( e V_s = h f - \phi \). If the frequency of light is doubled, the new stopping potential \( V_s' \) satisfies \( e V_s' = h (2f) - \phi = 2(h f - \phi) + \phi = 2 e V_s + \phi \). Since the work function \( \phi > 0 \), we have \( e V_s' > 2 e V_s \Rightarrow V_s' > 2 V_s \).

Award 1 mark for the correct answer C. No marks for incorrect or multiple options selected.

By the principle of conservation of energy: Heat lost by water = Heat gained by metal block. \(m_w c_w (T_w - T_f) = m_m c_m (T_f - T)\). Substituting the given values: \(1.0 \times 4200 \times (20 - 15) = 0.5 \times 400 \times (15 - T)\). \(21000 = 200 \times (15 - T)\). \(105 = 15 - T\), which gives \(T = -90^\circ\text{C}\).

Correct option is A (1 mark).

Since the stone strikes the ground at an angle of \(45^\circ\) to the horizontal, the vertical component of velocity \(v_y\) is equal in magnitude to the horizontal component \(v_x\). Since the horizontal velocity remains constant, \(v_y = v_x = u\). Using the equations of motion for vertical motion: \(v_y^2 = 2gH\), we have \(u^2 = 2gH \implies H = \frac{u^2}{2g}\).

Correct option is B (1 mark).

In series, the equivalent resistance is \(R_S = 3R\), so the power is \(P_S = \frac{V^2}{3R}\). In parallel, the equivalent resistance is \(R_P = \frac{R}{3}\), so the power is \(P_P = \frac{V^2}{R/3} = \frac{3V^2}{R}\). The ratio is \(\frac{P_P}{P_S} = \frac{3V^2/R}{V^2/(3R)} = 9\).

Correct option is C (1 mark).

By Snell's Law at the flat boundary: \(n_1 \sin \theta_1 = n_2 \sin \theta_2 \implies 1.50 \sin(30.0^\circ) = 1.00 \sin \theta_2 \implies \sin \theta_2 = 0.75 \implies \theta_2 \approx 48.6^\circ\). The angle of deviation is the difference between the angle of refraction and the angle of incidence: \(d = \theta_2 - \theta_1 = 48.6^\circ - 30.0^\circ = 18.6^\circ\).

Correct option is A (1 mark).

As the loop is pulled out, the induced electromotive force (EMF) is \(\mathcal{E} = BLv\). The induced current is \(I = \frac{\mathcal{E}}{R} = \frac{BLv}{R}\). The magnetic force acting on the wire inside the field opposes the motion and has magnitude \(F_M = BIL = B \left(\frac{BLv}{R}\right) L = \frac{B^2 L^2 v}{R}\). To maintain a constant speed, the external pulling force must balance this magnetic force, so \(F = \frac{B^2 L^2 v}{R}\).

Correct option is C (1 mark).

For the combined system of mass \(M + m\), the equation of motion is: \(T_1 - (M + m)g = (M + m)a \implies T_1 = (M + m)(g + a)\). For the bottom block of mass \(m\), the equation of motion is: \(T_2 - mg = ma \implies T_2 = m(g + a)\). Thus, the ratio is: \(\frac{T_1}{T_2} = \frac{(M + m)(g + a)}{m(g + a)} = \frac{M + m}{m}\). This is independent of the acceleration \(a\).

Correct option is A (1 mark).

The melting process occurs from \(t = 4.0\text{ minutes}\) to \(t = 12.0\text{ minutes}\), so the duration of melting is \(\Delta t = 8.0\text{ minutes} = 480\text{ s}\). The energy supplied during this period is \(Q = P \Delta t = 100 \times 480 = 48000\text{ J}\). Since \(Q = m l_f\), the specific latent heat of fusion is \(l_f = \frac{Q}{m} = \frac{48000}{0.20} = 2.4 \times 10^5\text{ J kg}^{-1}\).

Correct option is B (1 mark).

Comparing the wave equation with the standard equation \(y = A \sin(\omega t - k x)\), we have: Angular frequency \(\omega = 4\pi\text{ rad s}^{-1}\), Wave number \(k = 2\pi\text{ m}^{-1}\). The wave speed is given by \(v = \frac{\omega}{k} = \frac{4\pi}{2\pi} = 2\text{ m s}^{-1}\). Since the signs of the \(t\) and \(x\) terms are opposite, the wave is travelling in the positive \(x\)-direction.

Correct option is A (1 mark).

Let \(v_{\max}\) be the maximum velocity reached by the object. Since the object accelerates uniformly from rest to \(v_{\max}\) and then decelerates uniformly back to rest, the velocity-time graph of the motion is a triangle. The total displacement \(s\) is represented by the area under the triangle: \(s = \frac{1}{2} \times \text{total time } T \times v_{\max}\). The average velocity \(v_{\text{avg}}\) of the journey is \(v_{\text{avg}} = \frac{s}{T} = \frac{1}{2} v_{\max}\). Thus, the ratio of the average velocity to the maximum velocity is \(1 : 2\). This ratio is independent of the values of the accelerations \(a\) and \(2a\).

Correct option is B (1 mark).

In the dark: \(V_{\text{LDR}} = 12 \times \frac{8}{4 + 8} = 12 \times \frac{8}{12} = 8\text{ V}\). Under bright light: \(V_{\text{LDR}}' = 12 \times \frac{2}{4 + 2} = 12 \times \frac{2}{6} = 4\text{ V}\). Therefore, the potential difference across the LDR decreases by: \(8\text{ V} - 4\text{ V} = 4\text{ V}\).

Correct option is A (1 mark).

Using \(Q = mc\Delta T\) and \(Q = Pt\), we have \(Pt = mc\Delta T\), which gives \(c = \frac{Pt}{m\Delta T}\). Since the power \(P\) is identical for both heaters:
For liquid A: \(c_A \propto \frac{120}{0.1 \times 10} = 120\)
For liquid B: \(c_B \propto \frac{180}{0.2 \times 15} = 60\)
Therefore, \(c_A : c_B = 120 : 60 = 2 : 1\).

Award 1 mark for the correct option (C). No marks are awarded if more than one option is selected.

Let the projection speed be \(u\) and angle of projection be \(\theta\). The kinetic energy at the highest point is \(K_{\text{top}} = \frac{1}{2} m (u\cos\theta)^2\). The initial kinetic energy is \(K_0 = \frac{1}{2} mu^2\). Since \(K_{\text{top}} = \frac{1}{2} K_0\), we have \(\cos^2\theta = \frac{1}{2}\), which gives \(\theta = 45^\circ\). The ratio of maximum height \(H\) to horizontal range \(R\) is given by \(\frac{H}{R} = \frac{u^2 \sin^2\theta / 2g}{u^2 \sin(2\theta) / g} = \frac{\tan\theta}{4}\). Since \(\theta = 45^\circ\), \(\tan 45^\circ = 1\), so \(\frac{H}{R} = \frac{1}{4}\).

Award 1 mark for the correct option (B). No marks are awarded if more than one option is selected.

When the block is on the verge of sliding down, the static friction \(f\) reaches its maximum value \(f_{\text{max}} = \mu_s N\) acting up the incline. Resolving forces perpendicular to the incline: \(N = mg \cos 45^\circ + F \sin 45^\circ\). Resolving forces parallel to the incline: \(mg \sin 45^\circ = F \cos 45^\circ + f_{\text{max}} = F \cos 45^\circ + \mu_s (mg \cos 45^\circ + F \sin 45^\circ)\). Since \(\sin 45^\circ = \cos 45^\circ\), we have \(mg = F + \mu_s(mg + F)\), which simplifies to \(F = mg \frac{1 - \mu_s}{1 + \mu_s}\). Substituting \(mg = 20\text{ N}\) and \(\mu_s = 0.25\), we get \(F = 20 \times \frac{0.75}{1.25} = 12\text{ N}\).

Award 1 mark for the correct option (B). No marks are awarded if more than one option is selected.

Due to the symmetry of the path inside the prism, the angle of refraction \(r\) at the first surface is \(r = \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ\). Applying Snell's Law at the first boundary: \(n_{\text{air}} \sin \theta = n_{\text{prism}} \sin r \Rightarrow 1 \times \sin \theta = \sqrt{3} \times \sin 30^\circ = \frac{\sqrt{3}}{2}\). Thus, the angle of incidence is \(\theta = 60^\circ\).

Award 1 mark for the correct option (C). No marks are awarded if more than one option is selected.

The light travels from the bird (in air, \(n_{\text{object}} = 1\)) to the fish (in water, \(n_{\text{observer}} = \frac{4}{3}\)). The apparent height of the bird above the water surface is \(h' = h \times \frac{n_{\text{observer}}}{n_{\text{object}}} = 1.8 \times \frac{4/3}{1} = 2.4\text{ m}\). Since the fish is \(1.2\text{ m}\) below the surface, the total apparent distance of the bird from the fish is \(2.4\text{ m} + 1.2\text{ m} = 3.6\text{ m}\).

Award 1 mark for the correct option (D). No marks are awarded if more than one option is selected.

There are four possible configurations for three identical resistors:
1. All in series: \(R_{\text{eq}} = 3R\)
2. All in parallel: \(R_{\text{eq}} = \frac{1}{3}R\)
3. Two in series in parallel with the third: \(R_{\text{eq}} = \frac{2R \times R}{2R + R} = \frac{2}{3}R\)
4. Two in parallel in series with the third: \(R_{\text{eq}} = \frac{R}{2} + R = \frac{3}{2}R\).
Thus, \(\frac{5}{4}R\) is not a possible equivalent resistance.

Award 1 mark for the correct option (C). No marks are awarded if more than one option is selected.

In the dark: \(V_{\text{out, dark}} = 12 \times \frac{40\text{ k}\Omega}{40\text{ k}\Omega + 10\text{ k}\Omega} = 9.6\text{ V}\).
In bright light: \(V_{\text{out, bright}} = 12 \times \frac{2\text{ k}\Omega}{2\text{ k}\Omega + 10\text{ k}\Omega} = 2.0\text{ V}\).
The change in output voltage is \(\Delta V_{\text{out}} = 9.6\text{ V} - 2.0\text{ V} = 7.6\text{ V}\).

Award 1 mark for the correct option (C). No marks are awarded if more than one option is selected.

The angular speed of rotation is \(\omega = 300 \times \frac{2\pi}{60} = 10\pi\text{ rad s}^{-1}\). The induced e.m.f. between the center of rotation and the outer end of the rod is \(E = \frac{1}{2} B \omega L^2 = \frac{1}{2} \times 0.5 \times 10\pi \times 0.4^2 = 0.4\pi \approx 1.26\text{ V}\).

Award 1 mark for the correct option (C). No marks are awarded if more than one option is selected.

The magnetic force per unit length between two parallel wires is \(f = \frac{\mu_0 I_X I_Y}{2\pi d}\). Since the currents are in opposite directions, the force is repulsive. Under the new conditions, the force per unit length is \(f' = \frac{\mu_0 (2I_X) (0.5I_Y)}{2\pi (3d)} = \frac{1}{3} f\). Since the current directions are still opposite, the force remains repulsive.

Award 1 mark for the correct option (B). No marks are awarded if more than one option is selected.

The absolute initial temperature is \(T_1 = 27 + 273 = 300\text{ K}\), and the final temperature is \(T_2 = 327 + 273 = 600\text{ K}\). Since the absolute temperature is doubled:
(1) For fixed volume, \(P \propto T\), so pressure is doubled. (Correct)
(2) Average kinetic energy of gas molecules is \(E_k = \frac{3}{2} k_B T \propto T\), so it is doubled. (Correct)
(3) Root-mean-square speed is \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \propto \sqrt{T}\), so it increases by a factor of \(\sqrt{2}\) instead of 2. (Incorrect)

Award 1 mark for the correct option (B). No marks are awarded if more than one option is selected.

Let the angle of refraction at the air-core interface be \(\theta_r\), and the angle of incidence at the core-cladding interface be \(\phi\). For total internal reflection at the core-cladding interface: \(\sin \phi \ge \frac{n_2}{n_1} = \frac{1.40}{1.50}\), which gives \(\phi \ge 68.96^\circ\). From geometry, \(\theta_r + \phi = 90^\circ\), so the maximum refraction angle is \(\theta_r = 90^\circ - 68.96^\circ = 21.04^\circ\). Applying Snell's Law at the air-core interface: \(1.00 \times \sin \theta = 1.50 \times \sin \theta_r\). Thus, \(\sin \theta \le 1.50 \times \sin 21.04^\circ = 0.5385\), which gives \(\theta \le 32.6^\circ\). Hence, the maximum value of \(\theta\) is \(32.6^\circ\).

Award 1 mark for selecting the correct option B. Award 0 marks for any other option.

When switch \(S\) is closed, the second resistor \(R_2\) is connected in parallel with \(R_1\). This decreases the equivalent resistance of the external circuit. As a result, the total current delivered by the battery increases, which increases the internal potential drop (lost volts, \(I_{total}r\)). Consequently, the terminal voltage \(V = E - I_{total}r\) decreases, so the voltmeter reading decreases. Since the ammeter is in series with \(R_1\) which is directly across the terminals, the current through \(R_1\) is \(I = V/R_1\). Since the terminal voltage \(V\) decreases and \(R_1\) remains constant, the ammeter reading \(I\) must decrease.

Award 1 mark for selecting the correct option D. Award 0 marks for any other option.

Let \(a\) be the upward acceleration of the blocks. Consider the bottom block of mass \(3m\) as a system: \(T_2 - 3mg = 3ma \implies T_2 = 3m(g+a)\). Now consider the system consisting of the middle and bottom blocks (total mass \(2m + 3m = 5m\)): \(T_1 - 5mg = 5ma \implies T_1 = 5m(g+a)\). Therefore, the ratio of the tensions is \(\frac{T_1}{T_2} = \frac{5m(g+a)}{3m(g+a)} = \frac{5}{3}\).

Award 1 mark for selecting the correct option C. Award 0 marks for any other option.

(a) The critical angle \(\theta_c\) is given by:
\(\sin \theta_c = \frac{1}{n} = \frac{1}{1.60} = 0.625\)
\(\theta_c = \sin^{-1}(0.625) \approx 38.68^\circ \approx 38.7^\circ\)

(b) Since the light ray enters \(AB\) normally, it passes straight without deviation.
It hits the face \(BC\). Let's find the angle of incidence \(\theta_i\) on \(BC\).
Since the ray is initially horizontal (parallel to \(AC\)), the angle it makes with \(BC\) is \(30^\circ\).
Thus, the normal to \(BC\) makes an angle of \(90^\circ - 30^\circ = 60^\circ\) with the ray.
Therefore, the angle of incidence on \(BC\) is \(\theta_i = 60^\circ\).
Since \(\theta_i = 60^\circ > \theta_c = 38.7^\circ\), total internal reflection occurs at \(BC\).

(c) After total internal reflection at \(BC\), the reflected ray makes an angle of \(30^\circ\) with the face \(BC\).
The ray travels towards face \(AC\).
In the triangle formed by the reflection point on \(BC\), vertex \(C\), and the incidence point on \(AC\):
The angle at \(C\) is \(30^\circ\).
The angle between the reflected ray and \(BC\) is \(30^\circ\).
Thus, the angle between the ray and \(AC\) inside the prism is \(180^\circ - (30^\circ + 30^\circ) = 120^\circ\).
The normal to \(AC\) makes an angle of \(\theta_2 = 120^\circ - 90^\circ = 30^\circ\) with the ray.
Since \(\theta_2 = 30^\circ < \theta_c = 38.7^\circ\), the ray refracts out into the air.
Using Snell's Law:
\(n \sin \theta_2 = 1 \cdot \sin \theta_3\)
\(1.60 \sin 30^\circ = \sin \theta_3 \Rightarrow \sin \theta_3 = 0.80\)
\(\theta_3 \approx 53.13^\circ \approx 53.1^\circ\)
The emergent ray is at \(53.1^\circ\) to the normal of \(AC\).
Since the normal is perpendicular to \(AC\) (vertical), the angle of the emergent ray with the horizontal is \(90^\circ - 53.1^\circ = 36.9^\circ\).
Since the original ray was horizontal, the angle of deviation is \(36.9^\circ\).

(a) 1M for critical angle formula sin(c)=1/n, 1A for 38.7 degrees.
(b) 1M for identifying angle of incidence is 60 degrees, 1M for comparing with critical angle, 1A for stating TIR occurs.
(c) 1M for determining the angle of incidence at AC is 30 degrees, 1M for applying Snell's law at AC, 1A for finding the angle of refraction in air is 53.1 degrees, 1A for finding deviation is 36.9 degrees.

(a) The current in the circuit is \(I = \frac{E}{R + r}\).
The power dissipated in \(R\) is \(P = I^2 R = \left(\frac{E}{R+r}\right)^2 R = \frac{E^2 R}{(R+r)^2}\).

(b) (i) Expanding the denominator of the given form:
\((\sqrt{R} - \frac{r}{\sqrt{R}})^2 + 4r = R - 2r + \frac{r^2}{R} + 4r = R + 2r + \frac{r^2}{R} = \frac{R^2 + 2rR + r^2}{R} = \frac{(R+r)^2}{R}\).
Therefore, \(P = \frac{E^2}{\frac{(R+r)^2}{R}} = \frac{E^2 R}{(R+r)^2}\). (Shown)

(ii) To maximize \(P\), the denominator must be minimized. The term \((\sqrt{R} - \frac{r}{\sqrt{R}})^2\) is a perfect square and is minimum (equal to 0) when \(\sqrt{R} = \frac{r}{\sqrt{R}}\), which gives \(R = r = 2.0\ \Omega\).
The maximum power is:
\(P_{max} = \frac{E^2}{4r} = \frac{12.0^2}{4 \times 2.0} = 18.0\text{ W}\).

(c) The terminal voltage is given by \(V = E - Ir\).
When \(R\) decreases, the total resistance of the circuit decreases, so the current \(I\) increases.
Since \(V = E - Ir\), and both \(E\) and \(r\) are constant, an increase in \(I\) leads to an increase in lost volts \(Ir\).
Therefore, the terminal voltage \(V\) decreases.
Specifically:
At \(R = 8.0\ \Omega\), \(V = 12.0 \times \frac{8.0}{8.0+2.0} = 9.6\text{ V}\).
At \(R = 4.0\ \Omega\), \(V = 12.0 \times \frac{4.0}{4.0+2.0} = 8.0\text{ V}\).
Thus, the terminal voltage decreases from \(9.6\text{ V}\) to \(8.0\text{ V}\).

(a) 1M for current expression, 1A for P expression.
(b) (i) 1M for algebraic expansion of the denominator, 1A for simplification.
(ii) 1M for setting R = r, 1A for calculating 18.0 W.
(c) 1M for stating total resistance decreases so current increases, 1M for linking increased current to increased lost volts, 1A for concluding terminal voltage decreases.

(a) For vertical motion, the initial vertical velocity is \(u_y = 0\).
Using \(s_y = u_y t + \frac{1}{2} g t^2\):
\(125 = 0 + \frac{1}{2} (9.81) t^2\)
\(t^2 = \frac{250}{9.81} \approx 25.484\)
\(t = \sqrt{25.484} \approx 5.048\text{ s} \approx 5.05\text{ s}\). (Shown)

(b) For horizontal motion, the velocity remains constant: \(u_x = 40.0\text{ m s}^{-1}\).
The horizontal distance \(s_x = u_x t = 40.0 \times 5.048 \approx 202\text{ m}\).

(c) The horizontal component of the final velocity is \(v_x = 40.0\text{ m s}^{-1}\).
The vertical component of the final velocity is \(v_y = u_y + g t = 0 + 9.81 \times 5.048 \approx 49.52\text{ m s}^{-1}\) (downwards).
The magnitude of the final velocity (speed) is:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{40.0^2 + 49.52^2} = \sqrt{1600 + 2452.2} = \sqrt{4052.2} \approx 63.7\text{ m s}^{-1}\).
The angle \(\theta\) with the horizontal is given by:
\(\tan \theta = \frac{v_y}{v_x} = \frac{49.52}{40.0} = 1.238\)
\(\theta = \tan^{-1}(1.238) \approx 51.1^circ\) (below the horizontal).

(a) 1M for using s_y = 0.5 g t^2, 1A for 5.05 s.
(b) 1M for s_x = u_x t, 1A for 202 m.
(c) 1M for v_y = g t calculation, 1M for speed formula, 1A for 63.7 m s^-1, 1M for tangent of the angle, 1A for 51.1 degrees below horizontal.

(a) Faraday's law states that the magnitude of the induced electromotive force (e.m.f.) in a conductor is directly proportional to the rate of change of magnetic flux linkage through it.

(b) The area of the circular coil is \(A =
\pi r^2 = \pi (0.050)^2 = 7.854 \times 10^{-3}\text{ m}^2\).
The magnetic flux linkage is \(\Phi_N = N B A\).
The rate of change of magnetic flux linkage is:
\(\frac{\Delta \Phi_N}{\Delta t} = N A \frac{\Delta B}{\Delta t}\).
From \(B(t) = 0.50 - 1.20t\), we have \(\frac{\Delta B}{\Delta t} = -1.20\text{ T s}^{-1}\).
Thus, \(\frac{\Delta \Phi_N}{\Delta t} = 200 \times (7.854 \times 10^{-3}) \times (-1.20) = -1.885\text{ Wb s}^{-1} \approx -1.89\text{ Wb s}^{-1}\).
The magnitude of the rate of change is \(1.89\text{ Wb s}^{-1}\).

(c) According to Faraday's law, the magnitude of the induced e.m.f. \(\mathcal{E}\) is:
\(\mathcal{E} = \left| \frac{\Delta \Phi_N}{\Delta t} \right| = 1.89\text{ V}\).

(d) The induced current is \(I = \frac{\mathcal{E}}{R} = \frac{1.885}{4.0} \approx 0.471\text{ A}\).
Since the magnetic field points vertically upwards and its strength is decreasing, by Lenz's law, the induced current must create an induced magnetic field pointing vertically upwards to oppose the decrease in magnetic flux. According to the right-hand grip rule, this corresponds to a counter-clockwise induced current when viewed from above.

(a) 2M for clear statement of Faraday's Law (1M for linking e.m.f. to rate of change, 1M for magnetic flux linkage).
(b) 1M for area calculation, 1M for rate of change of B, 1A for -1.89 Wb s^-1 (accept 1.89 Wb s^-1 if magnitude is stated).
(c) 1M for linking e.m.f. to flux linkage change rate, 1A for 1.89 V.
(d) 1M for calculation of current (0.471 A), 1A for counter-clockwise direction with justification.

(a) For block \(A\):
- Tension \(T\) pointing to the right.
- Friction force \(f_k\) pointing to the left.
- Normal reaction \(N\) pointing upwards.
- Weight \(m_A g\) pointing downwards.
For block \(B\):
- Tension \(T\) pointing upwards.
- Weight \(m_B g\) pointing downwards.

(b) For Block \(A\):
\(T - f_k = m_A a\) where \(f_k = \mu_k N = \mu_k m_A g\).
Thus: \(T - \mu_k m_A g = m_A a\). (Equation 1)
For Block \(B\):
\(m_B g - T = m_B a\). (Equation 2)

(c) Adding Equation 1 and Equation 2:
\(m_B g - \mu_k m_A g = (m_A + m_B) a\)
\(a = \frac{m_B - \mu_k m_A}{m_A + m_B} g\).
Substituting the given values:
\(a = \frac{6.0 - 0.25 \times 4.0}{4.0 + 6.0} \times 9.81 = \frac{5.0}{10.0} \times 9.81 = 4.905\text{ m s}^{-2} \approx 4.91\text{ m s}^{-2}\).

(d) Using Equation 2 to find tension \(T\):
\(T = m_B g - m_B a = m_B (g - a) = 6.0 \times (9.81 - 4.905) = 29.43\text{ N} \approx 29.4\text{ N}\).

(a) 1A for block A FBD (all 4 forces correct), 1A for block B FBD (both forces correct).
(b) 1M for block A equation of motion, 1A for block B equation of motion.
(c) 1M for combining equations, 1M for substituting values, 1A for 4.91 m s^-2.
(d) 1M for substituting a back into equation 1 or 2, 1A for 29.4 N.

(a) Heat energy needed to raise the temperature of water to \(100^\circ\text{C}\):
\(Q_1 = m c_w \Delta T = 0.80 \times 4200 \times (100.0 - 20.0) = 268800\text{ J}\).
Electrical energy input needed:
\(E_{in} = \frac{Q_1}{\eta} = \frac{268800}{0.85} \approx 316235\text{ J}\).
Time taken \(t\):
\(t = \frac{E_{in}}{P} = \frac{316235}{2200} \approx 143.7\text{ s} \approx 144\text{ s}\).

(b) During the \(2.0\text{ minutes}\) (\(120\text{ s}\)) of boiling, the electrical energy input is:
\(E_{input} = P \times t_{boil} = 2200 \times 120 = 264000\text{ J}\).
Useful heat energy transferred to the water:
\(Q_2 = \eta \times E_{input} = 0.85 \times 264000 = 224400\text{ J}\).
Let \(m_{evap}\) be the mass of water evaporated:
\(Q_2 = m_{evap} l_v\)
\(224400 = m_{evap} \times 2.26 \times 10^6\)
\(m_{evap} = \frac{224400}{2.26 \times 10^6} \approx 0.0993\text{ kg} \approx 0.099\text{ kg}\) (or \(99\text{ g}\)).

(c) Heat is lost to the surroundings (air) by convection/radiation, and some heat is used to raise the temperature of the kettle's heating element and container wall itself.

(a) 1M for Q = mc*dT, 1M for introducing efficiency (E = Q/0.85), 1A for 144 s.
(b) 1M for total electrical energy (264000 J), 1M for multiplying by 0.85 to find useful energy (224400 J), 1M for applying Q = m*lv, 1A for 0.099 kg.
(c) 2M for any correct physical explanation (e.g. heat loss to air, heat absorbed by kettle wall/element).

(a) First, convert temperature to Kelvin:
\(T_1 = 27.0 + 273.15 = 300.15\text{ K}\) (accept \(300\text{ K}\) for calculation).
Using the ideal gas law \(P V = N k T\):
\(N = \frac{P_1 V_1}{k T_1} = \frac{(1.5 \times 10^5) \times (3.0 \times 10^{-3})}{(1.38 \times 10^{-23}) \times 300} = \frac{450}{4.14 \times 10^{-21}} \approx 1.087 \times 10^{23} \approx 1.09 \times 10^{23}\).

(b) Since the compression is isothermal, temperature \(T\) is constant. Boyle's law applies:
\(P_1 V_1 = P_2 V_2\)
\(P_2 = \frac{P_1 V_1}{V_2} = \frac{(1.5 \times 10^5) \times (3.0 \times 10^{-3})}{1.2 \times 10^{-3}} = 3.75 \times 10^5\text{ Pa}\).

(c) Since volume is constant, the Pressure Law applies between state 2 and state 3:
\(\frac{P_2}{T_2} = \frac{P_3}{T_3}\)
where \(T_2 = 300\text{ K}\) (since process 1-2 was isothermal).
\(T_3 = T_2 \times \frac{P_3}{P_2} = 300 \times \frac{4.5 \times 10^5}{3.75 \times 10^5} = 300 \times 1.20 = 360\text{ K}\).
To find the temperature in Celsius:
\(t_3 = 360 - 273 = 87^\circ\text{C}\).

(a) 1M for converting temperature to K, 1M for applying PV = NkT, 1A for 1.09 x 10^23.
(b) 1M for P_1*V_1 = P_2*V_2, 1A for 3.75 x 10^5 Pa.
(c) 1M for using Pressure Law P/T = constant, 1M for finding T_3 = 360 K, 1M for converting to Celsius, 1A for 87 degrees Celsius.

(a) Period \(T = 2.0\text{ hours} = 2.0 \times 3600\text{ s} = 7200\text{ s}\).
Angular velocity \(\omega = \frac{2\pi}{T} = \frac{2\pi}{7200} \approx 8.727 \times 10^{-4}\text{ rad s}^{-1} \approx 8.73 \times 10^{-4}\text{ rad s}^{-1}\).

(b) The gravitational force provides the necessary centripetal force for circular motion:
\(\frac{G M m}{r^2} = m \omega^2 r = m \left(\frac{2\pi}{T}\right)^2 r\).
Dividing both sides by \(m\):
\(\frac{G M}{r^2} = \frac{4\pi^2 r}{T^2}\).
Rearranging for \(r^3\):
\(r^3 = \frac{G M T^2}{4\pi^2}\).
Calculate \(r\):
\(r^3 = \frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times (7200)^2}{4\pi^2} \approx 5.228 \times 10^{20}\text{ m}^3\).
\(r = (5.228 \times 10^{20})^{1/3} \approx 8.056 \times 10^6\text{ m} \approx 8.06 \times 10^6\text{ m}\).

(c) The gravitational force is \(F_g = \frac{G M m}{r^2}\) (or \(F_g = m \omega^2 r\)).
Using \(F_g = m \omega^2 r\):
\(F_g = 800 \times (8.727 \times 10^{-4})^2 \times 8.056 \times 10^6 \approx 4908\text{ N} \approx 4910\text{ N}\).

(a) 1M for converting period to seconds (7200 s), 1A for 8.73 x 10^-4 rad s^-1.
(b) 1M for equating gravitational force to centripetal force, 1M for Kepler's 3rd law derivation formula, 1M for substitution, 1A for 8.06 x 10^6 m.
(c) 1M for using gravitational force formula or centripetal force formula, 1M for substitution, 1A for 4910 N.

(a) Mass defect is the difference between the sum of the individual rest masses of the reactants (parent nuclei) and the sum of the rest masses of the products (daughter products) in a nuclear reaction.

(b) Sum of reactant masses:
\(m_{reactants} = m({}^2_1\text{H}) + m({}^3_1\text{H}) = 2.014102 + 3.016049 = 5.030151\text{ u}\).
Sum of product masses:
\(m_{products} = m({}^4_2\text{He}) + m({}^1_0\text{n}) = 4.002603 + 1.008665 = 5.011268\text{ u}\).
Mass defect:
\(\Delta m = m_{reactants} - m_{products} = 5.030151 - 5.011268 = 0.018883\text{ u}\).

(c) Energy released \(\Delta E\):
\(\Delta E = \Delta m \times 931.5\text{ MeV} = 0.018883 \times 931.5 \approx 17.5895\text{ MeV} \approx 17.6\text{ MeV}\).

(d) The binding energy per nucleon of the product Helium-4 (\({}^4_2\text{He}\)) is significantly higher than those of the reactants Deuterium and Tritium. This means the products are more tightly bound and have less total mass-energy, so the difference in mass-energy is released as kinetic energy.

(a) 2M for defining mass defect (1M for mentioning the reactants and products, 1M for mentioning difference in rest masses).
(b) 1M for calculating mass of reactants, 1M for calculating mass of products, 1A for 0.018883 u.
(c) 1M for using E = delta_m * 931.5, 1A for 17.6 MeV.
(d) 1M for mentioning higher binding energy per nucleon of products, 1M for explaining that this leads to lesser mass-energy and the release of energy.

For a circular orbit under gravity, the orbital speed is given by \(v = \sqrt{\frac{GM}{r}}\). Thus, the speed is inversely proportional to the square root of the orbital radius, \(v \propto \frac{1}{\sqrt{r}}\). Therefore, \(\frac{v_2}{v_1} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{1}{4}} = 0.5\), which gives \(v_2 = 0.5 v_1\).

Award 1 mark for identifying the relationship between orbital speed and radius, and calculating the correct speed ratio.

First calculate the redshift: \(z = \frac{\Delta \lambda}{\lambda_0} = \frac{672.7 - 656.3}{656.3} \approx 0.024987\). The recessional velocity is \(v = z c = 0.024987 \times 3 \times 10^5 \text{ km s}^{-1} \approx 7496 \text{ km s}^{-1}\). From Hubble's Law, \(d = \frac{v}{H_0} = \frac{7496}{70} \approx 107 \text{ Mpc}\).

Award 1 mark for calculating the redshift, recessional velocity, and then using Hubble's Law to find the correct distance.

A massive star (typically with mass between 8 and 20 solar masses) evolves from a main-sequence star into a red supergiant. It eventually undergoes a core-collapse supernova explosion, leaving behind a neutron star as its final remnant.

Award 1 mark for selecting the correct evolutionary path and remnant according to the star's mass.

From the photoelectric equation, \(K_1 = \frac{hc}{\lambda} - \phi\). When the wavelength is halved, the photon energy doubles: \(K_2 = \frac{2hc}{\lambda} - \phi = 2(K_1 + \phi) - \phi = 2K_1 + \phi\). Since the work function \(\phi\) of the metal is positive, \(K_2 > 2K_1\).

Award 1 mark for analyzing Einstein's photoelectric equation to find the relation between kinetic energies.

According to Bohr's quantization condition, the circumference of the orbit must contain an integer number of de Broglie wavelengths: \(2\pi r_n = n \lambda_n\). Since the radius satisfies \(r_n \propto n^2\), we have \(\lambda_n = \frac{2\pi r_n}{n} \propto \frac{n^2}{n} = n\). Thus, \(\lambda_3 = 3\lambda_1\).

Award 1 mark for relating de Broglie wavelength, orbital radius, and the quantum number \(n\) to get the correct wavelength.

The resolution of any microscope is limited by the diffraction of the probe wave, which depends on its wavelength. The de Broglie wavelength of accelerated electrons in a TEM is orders of magnitude smaller than the wavelength of visible light, significantly reducing diffraction and enabling much higher resolution.

Award 1 mark for identifying the relationship between resolution, diffraction, and de Broglie wavelength of electrons.

The required electrical output power is \(P_{\text{out}} = 3.6\text{ kW} = 3600\text{ W}\). Since the system efficiency is \(15\%\), the input solar power required is \(P_{\text{in}} = \frac{P_{\text{out}}}{0.15} = \frac{3600}{0.15} = 24000\text{ W}\). Given the solar intensity is \(I = 800\text{ W m}^{-2}\), the area is \(A = \frac{P_{\text{in}}}{I} = \frac{24000}{800} = 30.0\text{ m}^2\).

Award 1 mark for correctly calculating the required solar panel area from output power, efficiency, and solar intensity.

The maximum theoretical wind power is given by \(P = \frac{1}{2}\rho A v^3\), where \(A = \pi L^2\) is the swept area of the blades. Therefore, \(P \propto L^2 v^3\). Let \(P'\) be the new power: \(\frac{P'}{P} = \left(\frac{1.5L}{L}\right)^2 \left(\frac{2v}{v}\right)^3 = 1.5^2 \times 2^3 = 2.25 \times 8 = 18\). Hence, \(P' = 18 P\).

Award 1 mark for analyzing the proportional relation of wind power with respect to blade length and wind speed, and obtaining the correct factor.

The reflection coefficient \(R\), which is the ratio of the reflected intensity to the incident intensity, is given by: \(R = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2 = \left(\frac{7.80 \times 10^6 - 1.70 \times 10^6}{7.80 \times 10^6 + 1.70 \times 10^6}\right)^2 = \left(\frac{6.10}{9.50}\right)^2 \approx 0.412 = 41.2\%\).

Award 1 mark for using the acoustic impedance reflection formula and calculating the correct percentage.

The object is placed at \(u = +0.25\text{ m}\). The lens must form a virtual image of the object at the person's near point, so the image distance is \(v = -0.80\text{ m}\). The power \(P\) of the lens is: \(P = \frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{0.25} + \frac{1}{-0.80} = 4.00 - 1.25 = +2.75\text{ D}\).

Award 1 mark for applying the lens equation with correct signs for virtual image, and calculating the refractive power in dioptres.

Using Stefan-Boltzmann law, the luminosity of a star is given by \(L = 4\pi R^2 \sigma T^4\). Comparing the two stars:

\(\frac{L_A}{L_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4\)

Substitute the given values:

\(\frac{100}{16} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{3000}{6000}\right)^4\)

\(\frac{100}{16} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{1}{2}\right)^4\)

\(\frac{100}{16} = \left(\frac{R_A}{R_B}\right)^2 \frac{1}{16}\)

\(\left(\frac{R_A}{R_B}\right)^2 = 100\)

\(\frac{R_A}{R_B} = 10\)

C (1 mark)

According to the conservation of angular momentum, \(L = m r v \sin\theta\) remains constant throughout the orbit. At both the perihelion and aphelion, the velocity vector is perpendicular to the position vector, so \(\theta = 90^\circ\) and \(\sin\theta = 1\).

Therefore, \(r_p v_p = r_a v_a\), where the subscripts \(p\) and \(a\) represent perihelion and aphelion respectively.

Since \(\frac{r_a}{r_p} = 3\), we have:

\(v_a = v_p \frac{r_p}{r_a} = v \times \frac{1}{3} = \frac{v}{3}\).

B (1 mark)

According to Einstein's photoelectric equation:

\(e V_s = h f - \Phi\) where \(\Phi\) is the work function.

For the first case:
\(e V_0 = h f - \Phi\) --- (Equation 1)

For the second case:
\(2 e V_0 = 1.5 h f - \Phi\) --- (Equation 2)

Substitute Equation 1 into Equation 2:
\(2 (h f - \Phi) = 1.5 h f - \Phi\)

\(2 h f - 2 \Phi = 1.5 h f - \Phi\)

\(0.5 h f = \Phi\)

Therefore, the work function is \(\Phi = 0.5 h f\).

B (1 mark)

The orbital period \(T\) of the electron is given by the distance of one orbit divided by the speed:

\(T = \frac{2 \pi r}{v}\)

Since \(r \propto n^2\) and \(v \propto \frac{1}{n}\), we have:

\(T \propto \frac{n^2}{1/n} = n^3\)

Therefore, the ratio of the period in \(n = 3\) to that in \(n = 1\) is:

\(\frac{T_3}{T_1} = \left(\frac{3}{1}\right)^3 = 27\).

C (1 mark)

1. Calculate the useful energy absorbed by the water:
\(Q = m c \Delta T = 50 \times 4200 \times (60 - 20) = 8.4 \times 10^6\text{ J}\)

2. Calculate the total solar energy incident on the collector:
\(E_{\text{in}} = P_{\text{in}} \times A \times t = 700\text{ W m}^{-2} \times 2.5\text{ m}^2 \times (2 \times 3600\text{ s}) = 1.26 \times 10^7\text{ J}\)

3. Calculate the efficiency \(\eta\):
\(\eta = \frac{Q}{E_{\text{in}}} \times 100\% = \frac{8.4 \times 10^6}{1.26 \times 10^7} \times 100\% \approx 66.7\%\).

C (1 mark)

1. To correct distance vision, the lens must form a virtual image of an object at infinity (\(u = \infty\)) at the patient's far point (\(v = -300\text{ cm} = -3.0\text{ m}\)).
Using the lens formula:
\(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\)
\(\frac{1}{\infty} + \frac{1}{-3.0} = P\)
\(P = -0.333\text{ D} \approx -0.33\text{ D}\)

2. To find the new near point \(u_{\text{new}}\), the lens must form a virtual image of an object at \(u_{\text{new}}\) at the patient's original near point (\(v = -50\text{ cm} = -0.5\text{ m}\)).
\(\frac{1}{u_{\text{new}}} + \frac{1}{-0.5} = \frac{1}{-3.0}\)
\(\frac{1}{u_{\text{new}}} - 2 = -0.333\)
\(\frac{1}{u_{\text{new}}} = 1.667 = \frac{5}{3}\)
\(u_{\text{new}} = 0.6\text{ m} = 60\text{ cm}\).

B (1 mark)

(a)(i)\n\(\Phi = 2.28\text{ eV} = 2.28 \times 1.60 \times 10^{-19}\text{ J} = 3.65 \times 10^{-19}\text{ J}\)\nThreshold frequency \(f_0 = \frac{\Phi}{h} = \frac{3.65 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.50 \times 10^{14}\text{ Hz}\)\n\n(a)(ii)\nEnergy of each photon \(E = hf = (6.63 \times 10^{-34}) \times (7.0 \times 10^{14}) = 4.64 \times 10^{-19}\text{ J} = 2.90\text{ eV}\)\nMaximum kinetic energy \(K_{\max} = E - \Phi = 2.90 - 2.28 = 0.62\text{ eV}\)\n\n(b)(i)\nThe stopping potential remains unchanged.\nThe stopping potential depends only on the maximum kinetic energy of the photoelectrons, which is determined by the frequency of light (which is unchanged).\n\n(b)(ii)\nThe saturation photocurrent doubles.\nDoubling the light intensity doubles the rate of photons arriving at the cathode. As each photon can emit at most one photoelectron, the rate of emission of photoelectrons doubles, leading to a doubled saturation current.\n\n(c)\nGraph: Both curves \(I_1\) and \(I_2\) start at the same negative voltage (stopping potential) \(V = -0.62\text{ V}\). Both curves rise with voltage and level off to saturation current. The saturation current for \(I_2\) is twice that of \(I_1\).

(a)(i)\n- Convert eV to J or use \(f_0 = \Phi/h\) [1M]\n- Correct threshold frequency with unit: \(5.50 \times 10^{14}\text{ Hz}\) [1A]\n\n(a)(ii)\n- Calculate photon energy [1M]\n- Correct maximum kinetic energy in eV: \(0.62\text{ eV}\) (or \(9.9 \times 10^{-20}\text{ J}\) if they correctly write the unit in J) [1A]\n\n(b)(i)\n- State 'unchanged' [1A]\n- Explain that \(K_{\max}\) depends only on frequency, which is unchanged [1A]\n\n(b)(ii)\n- State 'doubles' [1A]\n- Explain that photon rate doubles, leading to doubled photoelectron emission rate [1A]\n\n(c)\n- Both curves starting from the same negative potential [1A]\n- Saturation current of \(I_2\) is twice that of \(I_1\) [1A]

(a)\nArea swept by blades \(A = \pi L^2\).\nIn one second, the volume of air passing through the blades is \(V_{\text{vol}} = A v = \pi L^2 v\).\nMass of air passing through per second \(\frac{dm}{dt} = \rho V_{\text{vol}} = \pi \rho L^2 v\).\nKinetic energy of this mass of air per second is:\n\(P_w = \frac{1}{2} \left(\frac{dm}{dt}\right) v^2 = \frac{1}{2} (\pi \rho L^2 v) v^2 = \frac{1}{2} \pi \rho L^2 v^3\).\n\nSubstitute values:\n\(P_w = \frac{1}{2} \times \pi \times 1.2 \times (25)^2 \times (8.0)^3 = 603185\text{ W} \approx 603\text{ kW}\).\n\n(b)\n\(P_e = 35\% \times P_w = 0.35 \times 6.03185 \times 10^5\text{ W} = 2.11 \times 10^5\text{ W}\) (or \(211\text{ kW}\)).\n\n(c)(i)\nTo reduce transmission current, thereby minimizing the power loss (Joule heating, \(I^2R\)) in the transmission cables.\n\n(c)(ii)\nTransmission current \(I = \frac{P_e}{V} = \frac{2.111 \times 10^5\text{ W}}{33000\text{ V}} = 6.40\text{ A}\).\nPower loss \(P_{\text{loss}} = I^2 R = (6.40)^2 \times 4.0 = 164\text{ W}\) (or \(163.7\text{ W}\)).\n\n(d)\nAdvantage: It is a renewable energy source / does not release greenhouse gases during operation.\nLimitation: Wind is unstable and intermittent / occupies a large area of land / noise pollution.

(a)\n- Express mass flow rate as \(\pi \rho L^2 v\) [1M]\n- Derivation of formula for \(P_w\) [1M]\n- Correct calculation of \(P_w = 603\text{ kW}\) [1A]\n\n(b)\n- Multiply by 0.35 [1M]\n- Correct electrical power output: \(211\text{ kW}\) [1A]\n\n(c)(i)\n- State that higher voltage reduces current to minimize power loss [1A]\n\n(c)(ii)\n- Calculate the transmission current: \(I = 6.40\text{ A}\) [1M]\n- Correct power loss: \(164\text{ W}\) (or \(163\text{ W}\)) [1A]\n\n(d)\n- Correct environmental advantage [1A]\n- Correct limitation [1A]