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1. Calculate the work output (gravitational potential energy gained by the mass):
\(W = mgh = 4.0\text{ kg} \times 9.8\text{ m/s}^2 \times 5.0\text{ m} = 196\text{ J}\).

2. Calculate the useful power output:
\(P_{\text{out}} = \frac{W}{t} = \frac{196\text{ J}}{8.0\text{ s}} = 24.5\text{ W}\).

3. Use the efficiency formula to find the electrical power input:
\(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies 0.70 = \frac{24.5\text{ W}}{P_{\text{in}}}\)
\(P_{\text{in}} = \frac{24.5}{0.70} = 35\text{ W}\).

1 mark for the correct calculation of useful power output (24.5 W) and dividing by the efficiency (0.70) to obtain the correct input power of 35 W (Option C).

1. Calculate the refractive index \(n\) of the glass using the critical angle \(c = 42^\circ\):
\(n = \frac{1}{\sin c} = \frac{1}{\sin(42^\circ)} \approx 1.49\).

2. Use Snell's Law to find the angle of refraction \(r\) for an angle of incidence \(i = 30^\circ\):
\(n \sin i = 1 \cdot \sin r\)
\(1.49 \times \sin(30^\circ) = \sin r\)
\(\sin r = 1.49 \times 0.5 = 0.745\)
\(r = \arcsin(0.745) \approx 48^\circ\).

1 mark for correctly determining the refractive index (1.49) and applying Snell's Law to find the correct angle of refraction of 48 degrees (Option B).

1. Calculate the total charge \(Q\) passing through the cross-section:
\(Q = I \times t\)
Convert minutes to seconds: \(t = 5.0 \times 60 = 300\text{ s}\).
\(Q = 2.0\text{ A} \times 300\text{ s} = 600\text{ C}\).

2. Calculate the number of electrons \(N\):
\(N = \frac{Q}{e} = \frac{600\text{ C}}{1.6 \times 10^{-19}\text{ C}} = 3.75 \times 10^{21}\).

1 mark for converting time to seconds, calculating total charge (600 C), and dividing by the electron charge to get 3.75 x 10^21 (Option B).

1. Find the maximum velocity reached during the acceleration phase:
\(v = u + at = 0 + (2.0 \times 6.0) = 12\text{ m/s}\).

2. Calculate the distance during each phase (equivalent to the area under the velocity-time graph):
- Phase 1 (acceleration): \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6.0 \times 12 = 36\text{ m}\).
- Phase 2 (constant speed): \(d_2 = \text{base} \times \text{height} = 4.0 \times 12 = 48\text{ m}\).
- Phase 3 (deceleration): \(d_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3.0 \times 12 = 18\text{ m}\).

3. Sum the distances:
\(D_{\text{total}} = 36 + 48 + 18 = 102\text{ m}\).

1 mark for calculating the maximum speed of 12 m/s, finding the areas under the three parts of the velocity-time graph, and summing them to 102 m (Option C).

1. Determine the voltage across the fixed resistor \(V_R\):
\(V_R = V_{\text{total}} - V_{\text{thermistor}} = 6.0\text{ V} - 2.4\text{ V} = 3.6\text{ V}\).

2. Use the potential divider relation:
\(\frac{R_{\text{thermistor}}}{R_{\text{fixed}}} = \frac{V_{\text{thermistor}}}{V_R}\)
\(\frac{R_{\text{thermistor}}}{4.0\text{ k}\Omega} = \frac{2.4\text{ V}}{3.6\text{ V}}\)
\(R_{\text{thermistor}} = 4.0 \times \frac{2}{3} \approx 2.67\text{ k}\Omega \approx 2.7\text{ k}\Omega\).

1 mark for calculating the voltage drop across the fixed resistor (3.6 V) and applying the potential divider ratio to find the thermistor resistance of approximately 2.7 k\Omega (Option B).

1. Calculate the change in wavelength:
\(\Delta \lambda = 672.7\text{ nm} - 656.3\text{ nm} = 16.4\text{ nm}\).

2. Calculate the redshift \(z\):
\(z = \frac{\Delta \lambda}{\lambda_0} = \frac{16.4\text{ nm}}{656.3\text{ nm}} \approx 0.025\).

3. Since the observed wavelength is longer (shifted toward the red end of the spectrum), the light is redshifted, which indicates that the galaxy is moving away from the Earth.

1 mark for calculating redshift z = 0.025 and stating that a positive change in wavelength (redshift) indicates the galaxy is moving away from the Earth (Option A).

1. Identify the forces and their points of action:
- The rule is uniform, so its weight of \(1.5\text{ N}\) acts at its center of gravity, which is the \(50\text{ cm}\) mark.
- The pivot is at \(40\text{ cm}\).
- The load \(W\) acts at \(10\text{ cm}\).

2. Calculate distances from the pivot:
- Distance to the weight: \(50\text{ cm} - 40\text{ cm} = 10\text{ cm}\).
- Distance to the load \(W\): \(40\text{ cm} - 10\text{ cm} = 30\text{ cm}\).

3. Apply the principle of moments about the pivot:
\(\text{Clockwise Moments} = \text{Counter-Clockwise Moments}\)
\(1.5\text{ N} \times 10\text{ cm} = W \times 30\text{ cm}\)
\(15 = 30 W \implies W = \frac{15}{30} = 0.5\text{ N}\).

1 mark for identifying that the weight acts at the 50 cm mark, calculating the distances from the pivot, and applying the principle of moments to find W = 0.5 N (Option B).

1. Calculate the secondary voltage \(V_s\):
\(V_s = V_p \times \frac{N_s}{N_p} = 240\text{ V} \times \frac{60}{1200} = 12\text{ V}\).

2. Calculate the secondary current \(I_s\):
\(I_s = \frac{V_s}{R} = \frac{12\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\).

3. For an ideal transformer, input electrical power equals output electrical power:
\(V_p I_p = V_s I_s\)
\(240 \times I_p = 12 \times 1.5\)
\(240 I_p = 18 \implies I_p = \frac{18}{240} = 0.075\text{ A}\).

1 mark for calculating secondary voltage (12 V) and current (1.5 A), then using the power equivalence or turn ratio equation for currents to find the primary current of 0.075 A (Option A).

1. Find the distance traveled in the three stages using the area under a speed-time graph or SUVAT equations:
- Stage 1 (acceleration): \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 8.0\text{ m/s} = 16.0\text{ m}\).
- Stage 2 (constant speed): \(d_2 = \text{speed} \times \text{time} = 8.0\text{ m/s} \times 6.0\text{ s} = 48.0\text{ m}\).
- Stage 3 (deceleration): \(d_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.0\text{ s} \times 8.0\text{ m/s} = 8.0\text{ m}\).

2. Calculate the total distance:
\(d_{\text{total}} = 16.0\text{ m} + 48.0\text{ m} + 8.0\text{ m} = 72.0\text{ m}\).

3. Calculate the average speed:
\(\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{72.0\text{ m}}{12.0\text{ s}} = 6.0\text{ m/s}\).

Award 1 mark for the correct answer B.

Method:
- Find total distance by calculating area under speed-time graph: 16 m + 48 m + 8 m = 72 m.
- Apply average speed = total distance / total time = 72 m / 12 s = 6.0 m/s.

1. Calculate the useful work done in lifting the mass:
\(W = m \cdot g \cdot h = 120\text{ kg} \times 9.8\text{ N/kg} \times 5.0\text{ m} = 5880\text{ J}\).

2. Calculate the useful power output:
\(P_{\text{out}} = \frac{W}{t} = \frac{5880\text{ J}}{4.0\text{ s}} = 1470\text{ W}\).

3. Use the efficiency formula to find the power input:
\(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\)
\(0.75 = \frac{1470\text{ W}}{P_{\text{in}}}\)
\(P_{\text{in}} = \frac{1470}{0.75} = 1960\text{ W}\).

Award 1 mark for the correct answer C.

Method:
- Calculate useful work = 5880 J.
- Calculate useful power output = 1470 W.
- Rearrange efficiency equation to find input power = 1960 W.

1. Use Newton's second law to find the net (resultant) force acting on the crate:
\(F_{\text{net}} = m \cdot a = 40\text{ kg} \times 2.5\text{ m/s}^2 = 100\text{ N}\).

2. The net force is the difference between the forward tension force (\(T\)) and the opposing frictional force (\(f\)):
\(F_{\text{net}} = T - f\)
\(100\text{ N} = 180\text{ N} - f\)
\(f = 180\text{ N} - 100\text{ N} = 80\text{ N}\).

Award 1 mark for the correct answer A.

Method:
- Calculate net force = 40 x 2.5 = 100 N.
- Deduct net force from tension to find friction = 180 N - 100 N = 80 N.

1. Determine the speed of light in glass using the refractive index formula:
\(v = \frac{c}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.5} = 2.0 \times 10^8\text{ m/s}\).

2. Determine the wavelength inside the glass. Since the frequency (\(f\)) remains unchanged when light enters a new medium, we use the wave equation:
\(v = f \cdot \lambda\)
\(\lambda = \frac{v}{f} = \frac{2.0 \times 10^8\text{ m/s}}{5.0 \times 10^{14}\text{ Hz}} = 4.0 \times 10^{-7}\text{ m}\).

Award 1 mark for the correct answer A.

Method:
- Find the speed in glass using v = c/n = 2.0 x 10^8 m/s.
- Calculate the wavelength in glass using lambda = v/f = 4.0 x 10^-7 m.

1. Use the definition of electromotive force (e.m.f.), which is the energy converted per unit charge:
\(E = \frac{W}{Q}\)

2. Rearrange the formula to solve for the work done (energy converted):
\(W = E \cdot Q\)
\(W = 12\text{ V} \times 180\text{ C} = 2160\text{ J}\).

Award 1 mark for the correct answer C.

Method:
- Apply the formula W = E x Q to find energy = 12 V x 180 C = 2160 J.

1. Since the transformer is ideal, there are no energy losses, meaning:
\(\text{Power Input (primary)} = \text{Power Output (secondary)}\)
\(P_{\text{primary}} = P_{\text{secondary}} = 24\text{ W}\).

2. Calculate the primary current using the power formula:
\(P_{\text{primary}} = V_{\text{primary}} \cdot I_{\text{primary}}\)
\(24\text{ W} = 240\text{ V} \times I_{\text{primary}}\)
\(I_{\text{primary}} = \frac{24\text{ W}}{240\text{ V}} = 0.10\text{ A}\).

Award 1 mark for the correct answer A.

Method:
- Identify that for an ideal transformer, input power equals output power (24 W).
- Apply P = V x I to solve for primary current: Ip = 24 W / 240 V = 0.10 A.

1. Redshift represents a shift towards longer wavelengths (red end of the spectrum), which indicates that the source of light is moving away from the observer (Earth).
2. The observation of redshift in distant galaxies shows that almost all distant galaxies are moving away from us, which supports the Big Bang theory and indicates that the Universe is expanding.

Award 1 mark for the correct answer D.

Method:
- Associate longer wavelength shift (redshift) with moving away from the observer.
- Connect the redshift of distant galaxies to the expanding Universe model.

1. Calculate the energy needed to melt the ice at \(0^\circ\text{C}\):
\(Q_1 = m \cdot L_f = 0.50\text{ kg} \times 3.3 \times 10^5\text{ J/kg} = 165\,000\text{ J}\).

2. Calculate the energy needed to heat the resulting water from \(0^\circ\text{C}\) to \(20^\circ\text{C}\):
\(Q_2 = m \cdot c \cdot \Delta \theta = 0.50\text{ kg} \times 4200\text{ J/(kg}^\circ\text{C)} \times (20 - 0)^\circ\text{C} = 42\,000\text{ J}\).

3. Calculate the total energy required:
\(Q_{\text{total}} = Q_1 + Q_2 = 165\,000\text{ J} + 42\,000\text{ J} = 207\,000\text{ J}\).

4. Calculate the time taken using power:
\(t = \frac{Q_{\text{total}}}{P} = \frac{207\,000\text{ J}}{100\text{ W}} = 2070\text{ s}\).

Award 1 mark for the correct answer C.

Method:
- Calculate latent heat required to melt the ice (165,000 J).
- Calculate specific heat required to raise temperature (42,000 J).
- Sum the energy (207,000 J) and divide by power (100 W) to find time (2070 s).

First, calculate the weight of the crate: \(W = m \times g = 4.0\text{ kg} \times 9.8\text{ N/kg} = 39.2\text{ N}\). Next, calculate the useful work output (gravitational potential energy gained): \(E_p = W \times h = 39.2\text{ N} \times 2.5\text{ m} = 98\text{ J}\). Since this happens in \(1.0\text{ s}\), the useful power output is \(P_{\text{out}} = \frac{98\text{ J}}{1.0\text{ s}} = 98\text{ W}\). Now, calculate the efficiency: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{98\text{ W}}{120\text{ W}} \times 100\% \approx 81.7\%\), which rounds to \(82\%\).

1 mark for the correct calculation of efficiency, leading to option C.

The resistance of a wire is given by \(R = \rho \frac{L}{A}\), where \(A = \pi \frac{d^2}{4}\). Thus, resistance is proportional to length \(L\) and inversely proportional to the square of the diameter \(d\), so \(R \propto \frac{L}{d^2}\). For wire Q, the length is \(2L\) and the diameter is \(0.5d\). Therefore, \(R_Q \propto \frac{2L}{(0.5d)^2} = \frac{2L}{0.25d^2} = 8 \frac{L}{d^2}\). This means wire Q has a resistance of \(8R\).

1 mark for identifying the correct dependency on length and diameter, leading to option D.

First, find the refractive index \(n\) of the glass using Snell's law: \(n = \frac{\sin i}{\sin r} = \frac{\sin(45^\circ)}{\sin(28^\circ)} \approx \frac{0.7071}{0.4695} \approx 1.506\). Next, use the relationship between refractive index and wave speed: \(n = \frac{c}{v}\), where \(c\) is the speed of light in air and \(v\) is the speed in glass. Rearranging gives \(v = \frac{c}{n} = \frac{3.0 \times 10^8\text{ m/s}}{1.506} \approx 2.0 \times 10^8\text{ m/s}\).

1 mark for correct application of Snell's law and the refractive index speed formula, leading to option C.

The total resistance in the series circuit is \(R_{\text{total}} = R_{\text{thermistor}} + R_{\text{fixed}} = 8.0\text{ k}\Omega + 4.0\text{ k}\Omega = 12.0\text{ k}\Omega\). The potential difference across the thermistor \(V_{\text{thermistor}}\) can be found using the potential divider formula: \(V_{\text{thermistor}} = \left(\frac{R_{\text{thermistor}}}{R_{\text{total}}}\right) \times V_{\text{supply}} = \left(\frac{8.0\text{ k}\Omega}{12.0\text{ k}\Omega}\right) \times 12\text{ V} = 8.0\text{ V}\).

1 mark for the correct potential divider calculation, leading to option C.

The time interval between consecutive claps is \(t = \frac{1.0\text{ s}}{4.0} = 0.25\text{ s}\). During this time interval, the sound travels to the wall and back, which is a distance of \(2d\). Therefore, \(2d = v \times t = 340\text{ m/s} \times 0.25\text{ s} = 85\text{ m}\). Solving for \(d\) gives \(d = \frac{85\text{ m}}{2} = 42.5\text{ m}\), which is approximately \(43\text{ m}\).

1 mark for the calculation of the echo time interval and distance, leading to option A.

According to Hubble's Law, the recessional speed \(v\) of a galaxy is proportional to its distance \(d\) from Earth: \(v = H_0 \times d\). Substituting the given values: \(v = (2.2 \times 10^{-18}\text{ s}^{-1}) \times (3.0 \times 10^{22}\text{ m}) = 6.6 \times 10^4\text{ m/s}\).

1 mark for correctly applying Hubble's Law equation, leading to option B.

First, find the spring constant \(k\) using Hooke's Law: \(F = kx \implies k = \frac{F}{x} = \frac{6.0\text{ N}}{1.5\text{ cm}} = 4.0\text{ N/cm}\). Next, find the load required for a \(4.0\text{ cm}\) extension: \(F = kx = 4.0\text{ N/cm} \times 4.0\text{ cm} = 16\text{ N}\).

1 mark for calculating the correct spring constant and corresponding load, leading to option B.

The dominant process of thermal energy transfer to the surroundings from the outer surfaces of the cans at this temperature is infrared radiation. Matt black surfaces are much better emitters of infrared radiation than shiny silver surfaces. Therefore, the matt black can will emit radiation at a faster rate and cool down faster.

1 mark for identifying that matt black is a better emitter than shiny silver, causing it to cool faster, leading to option C.

First, calculate the useful work output of the pump: W = m * g * h = 150 kg * 9.8 N/kg * 10 m = 14700 J. Next, calculate the useful power output: P_out = W / t = 14700 J / 15 s = 980 W. The input power is 1.4 kW = 1400 W. Finally, calculate the efficiency: Efficiency = (P_out / P_in) * 100% = (980 / 1400) * 100% = 70%.

1 mark for the correct calculation of efficiency (70%).

The critical angle \(\theta_c\) at the boundary between two media is given by the formula sin(\(\theta_c\)) = n_2 / n_1, where n_1 is the refractive index of the denser medium (glass, 1.52) and n_2 is the refractive index of the less dense medium (plastic, 1.35). Therefore, sin(\(\theta_c\)) = 1.35 / 1.52 = 0.8882. Taking the inverse sine: \(\theta_c\) = arcsin(0.8882) = 62.6 degrees.

1 mark for the correct critical angle (62.6 degrees).

Resistance is given by R = \(\rho\) * L / A. The area of a cylindrical wire is proportional to the square of its diameter d, so A = \(\pi\) * (d/2)^2. For the second wire, the length is multiplied by 3 and the diameter is multiplied by 2, which increases the cross-sectional area by a factor of 2^2 = 4. Therefore, the new resistance R' = R * 3 / 4 = 8.0 ohms * 0.75 = 6.0 ohms.

1 mark for the correct resistance value (6.0 ohms).

The voltmeter measures the potential difference across the 4.0 k\(\Omega\) resistor. Initial: Total resistance = 20 k\(\Omega\) + 4.0 k\(\Omega\) = 24 k\(\Omega\). Voltmeter reading = (4.0 / 24) * 12 V = 2.0 V. Final: Total resistance = 1.0 k\(\Omega\) + 4.0 k\(\Omega\) = 5.0 k\(\Omega\). Voltmeter reading = (4.0 / 5.0) * 12 V = 9.6 V.

1 mark for the correct initial (2.0 V) and final (9.6 V) readings.

The time interval between successive claps is T = 1 / f = 1 / 2.5 Hz = 0.40 s. For each clap to coincide with the echo of the previous clap, the sound must travel to the wall and back in this time interval. Therefore, the distance traveled by the sound is 2d. Using the wave speed equation: 2d = v * T, which gives d = (v * T) / 2 = (340 m/s * 0.40 s) / 2 = 68 m.

1 mark for the correct distance of 68 m.

First, find the secondary voltage V_s using the turns ratio: V_s / V_p = N_s / N_p, so V_s = 240 V * (30 / 600) = 12 V. Next, find the secondary current I_s using Ohm's law: I_s = V_s / R = 12 V / 4.0 \(\Omega\) = 3.0 A. Since the transformer is ideal, the input power equals the output power: V_p * I_p = V_s * I_s, which gives I_p = (V_s * I_s) / V_p = (12 V * 3.0 A) / 240 V = 0.15 A.

1 mark for the correct primary current of 0.15 A.

The observed wavelength (671.1 nm) is longer than the laboratory wavelength (656.3 nm). This increase in wavelength is known as redshift. Redshift indicates that the light source is moving away from the observer (receding). According to cosmological observations, the recession of distant galaxies shows that the Universe is expanding.

1 mark for identifying that the galaxy is moving away due to redshift, indicating the Universe is expanding.

First, calculate the energy required to melt the ice: Q_1 = m * L_f = 0.20 kg * 330000 J/kg = 66000 J. Second, calculate the energy required to raise the temperature of the water: Q_2 = m * c * \(\Delta\theta\) = 0.20 kg * 4200 J/(kg degrees Celsius) * 20 degrees Celsius = 16800 J. The total thermal energy required is Q_total = Q_1 + Q_2 = 66000 J + 16800 J = 82800 J. Since power P = Q / t, the time taken is t = Q_total / P = 82800 J / 80 W = 1035 s.

1 mark for the correct total time of 1035 s.

First, calculate the refractive index \(n\) of the plastic using Snell's Law: \(n = \frac{\sin i}{\sin r} = \frac{\sin 35.0^\circ}{\sin 21.0^\circ} \approx \frac{0.5736}{0.3584} \approx 1.60\). Next, use the relationship between the refractive index and the speed of light in the medium: \(n = \frac{c}{v}\) which gives \(v = \frac{c}{n} = \frac{3.00 \times 10^8\text{ m/s}}{1.60} \approx 1.87 \times 10^8\text{ m/s}\).

1 mark for the correct calculation of the speed of light in the plastic block, matching option B.

1. Find the weight of the water: \(W = mg = 300\text{ kg} \times 9.8\text{ N/kg} = 2940\text{ N}\). 2. Find the useful work done in lifting the water: \(E_{\text{out}} = W \times h = 2940\text{ N} \times 15\text{ m} = 44100\text{ J}\). 3. Calculate the useful power output: \(P_{\text{out}} = \frac{E_{\text{out}}}{t} = \frac{44100\text{ J}}{50\text{ s}} = 882\text{ W}\). 4. Determine the efficiency using the input power of \(1.2\text{ kW} = 1200\text{ W}\): \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{882}{1200} \times 100\% = 73.5\%\), which rounds to \(74\%\).

1 mark for the correct calculation of the pump efficiency, matching option C.

The average speed is the total distance divided by the total time. The distance is the area under the velocity-time graph. 1. Distance during acceleration: \(d_1 = \frac{1}{2} \times 4.0\text{ s} \times 12\text{ m/s} = 24\text{ m}\). 2. Distance during constant velocity: \(d_2 = 6.0\text{ s} \times 12\text{ m/s} = 72\text{ m}\). 3. Distance during deceleration: \(d_3 = \frac{1}{2} \times 2.0\text{ s} \times 12\text{ m/s} = 12\text{ m}\). Total distance = \(24 + 72 + 12 = 108\text{ m}\). Total time = \(12.0\text{ s}\). \(\text{Average speed} = \frac{108\text{ m}}{12.0\text{ s}} = 9.0\text{ m/s}\).

1 mark for the correct calculation of average speed, matching option B.

Since the beam is uniform and pivoted at its center, its own weight of \(60\text{ N}\) acts through the pivot, producing no moment. For horizontal equilibrium, the sum of the clockwise moments must equal the sum of the anticlockwise moments about the pivot: \(\text{Anticlockwise moment} = 40\text{ N} \times 0.60\text{ m} = 24\text{ N m}\). To balance this, the \(30\text{ N}\) load must be placed on the right side of the pivot at a distance \(x\): \(\text{Clockwise moment} = 30\text{ N} \times x\). Setting them equal: \(30 x = 24 \implies x = 0.80\text{ m}\) to the right of the pivot.

1 mark for the correct distance and direction of the load, matching option B.

1. Convert the current to amperes and time to seconds: \(I = 8.0 \times 10^{-3}\text{ A}\), \(t = 5.0 \times 60\text{ s} = 300\text{ s}\). 2. Calculate the total charge \(Q\) that flows: \(Q = I \times t = (8.0 \times 10^{-3}\text{ A}) \times 300\text{ s} = 2.4\text{ C}\). 3. Calculate the number of electrons \(N\): \(N = \frac{Q}{e} = \frac{2.4\text{ C}}{1.6 \times 10^{-19}\text{ C}} = 1.5 \times 10^{19}\).

1 mark for calculating the correct number of electrons, matching option C.

1. Calculate the equivalent resistance \(R_p\) of the parallel combination of the \(6.0\ \Omega\) and \(3.0\ \Omega\) resistors: \(R_p = \frac{6.0 \times 3.0}{6.0 + 3.0} = 2.0\ \Omega\). 2. Calculate the total resistance \(R_{\text{total}}\) of the circuit: \(R_{\text{total}} = 4.0\ \Omega + 2.0\ \Omega = 6.0\ \Omega\). 3. Calculate the total circuit current \(I\): \(I = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\). 4. Calculate the potential difference across the \(4.0\ \Omega\) resistor: \(V = I \times R = 2.0\text{ A} \times 4.0\ \Omega = 8.0\text{ V}\).

1 mark for calculating the correct potential difference, matching option C.

The sound travels from the student to the wall and back, so the total distance traveled is \(2d\). Using the relation \(\text{distance} = \text{speed} \times \text{time}\), we get \(2d = 340\text{ m/s} \times 0.60\text{ s} = 204\text{ m}\). Therefore, \(d = 102\text{ m}\).

1 mark for the correct calculation of distance, matching option B.

Redshift is the increase in the observed wavelength of electromagnetic radiation from objects that are moving away from the observer. When a galaxy is moving away from Earth, the light waves are stretched, making the observed wavelength longer than the emitted wavelength. This provides strong evidence that the universe is expanding.

1 mark for identifying the correct movement and wavelength change, matching option D.

(a) The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the less dense medium is 90°.

(b) Use the formula:
\(\sin(c) = \frac{1}{n}\)
\(\sin(c) = \frac{1}{1.52} \approx 0.6579\)
\(c = \arcsin(0.6579) = 41.1^\circ\)

(c) Light enters the fiber at an angle greater than the critical angle at the core-cladding boundary. It undergoes continuous total internal reflection inside the core, bouncing along the length of the fiber. An advantage over copper cables is that optical fibers have higher bandwidth (less signal attenuation / lower interference).

(a)
- Mention of angle of incidence in the denser medium [1]
- Refraction angle is 90° / light travels along the boundary [1]

(b)
- State formula: \(\sin(c) = 1/n\) [1]
- Substitution: \(\sin(c) = 1/1.52\) [1]
- Correct calculation of critical angle: \(41.1^\circ\) (or \(41^\circ\)) [1]

(c)
- Core concept of Total Internal Reflection (TIR) [1]
- Angle of incidence is greater than the critical angle [1]
- Correct advantage (e.g., lower signal loss, higher speed/bandwidth, no electromagnetic interference) [1]

(a) In the main sequence state, a star is stable because the inward gravitational force is balanced by the outward radiation pressure. This pressure is generated by the thermal energy from nuclear fusion in the star's core.

(b) After leaving the main sequence, a massive star expands to become a red supergiant. When nuclear fuel runs out, the core collapses rapidly, resulting in a giant explosion called a supernova. The remaining core becomes either a neutron star or a black hole.

(c) The phenomenon is called redshift. It indicates that the galaxy is moving away from the Earth (receding).

(a)
- Outward force from radiation pressure / nuclear fusion [1]
- Inward force due to gravity [1]
- Forces are balanced / in equilibrium [1]

(b)
- Becomes a red supergiant [1]
- Explodes as a supernova [1]
- Leaves behind a neutron star or a black hole [1]

(c)
- Identifies phenomenon as redshift [1]
- States that the galaxy is moving away from Earth / receding [1]

(a) Acceleration is given by:
\(a = \frac{v - u}{t} = \frac{12 - 0}{4.0} = 3.0\text{ m/s}^2\)

(b) The total distance is the area under the speed-time graph:
- Distance during acceleration (0 to 4 s):
\(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0 \times 12 = 24\text{ m}\)
- Distance during constant speed (4 to 10 s):
\(d_2 = \text{base} \times \text{height} = 6.0 \times 12 = 72\text{ m}\)
- Distance during deceleration (10 to 13 s):
\(d_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3.0 \times 12 = 18\text{ m}\)

Total distance = \(24 + 72 + 18 = 114\text{ m}\)

(c) Magnitude of deceleration in the last 3.0 s:
\(|a| = \frac{12}{3.0} = 4.0\text{ m/s}^2\)
Since \(4.0\text{ m/s}^2 > 3.0\text{ m/s}^2\), the magnitude of the deceleration is greater than the initial acceleration.

(a)
- Formula used: \(a = \Delta v / t\) [1]
- Correct answer: \(3.0\text{ m/s}^2\) [1]

(b)
- Method of finding area under graph (or kinematics equations) [1]
- Area of central rectangle: \(72\text{ m}\) [1]
- Area of two triangles: \(24\text{ m}\) and \(18\text{ m}\) [1]
- Correct sum: \(114\text{ m}\) [1]

(c)
- Deceleration calculation: \(4.0\text{ m/s}^2\) [1]
- Comparison statement: Deceleration is greater than acceleration [1]

(a) An increase in current leads to more collisions between flowing electrons and metal ions, heating up the filament. As temperature increases, the metal ions vibrate more rapidly and with larger amplitude, which further obstructs the path of the drifting electrons, increasing resistance.

(b) Using \(P = I \times V\):
\(I = \frac{P}{V} = \frac{36}{12} = 3.0\text{ A}\)

(c) Convert time to seconds: \(t = 5.0 \times 60 = 300\text{ s}\)
Using \(Q = I \times t\):
\(Q = 3.0 \times 300 = 900\text{ C}\) (or Coulombs)

(a)
- Increased current increases temperature / heating effect [1]
- Ions vibrate with larger amplitude / more violently [1]
- More frequent collisions between electrons and ions [1]

(b)
- Formula used: \(P = IV\) or \(I = P/V\) [1]
- Correct current: \(3.0\text{ A}\) (accept with unit) [1]

(c)
- Correct time conversion: \(300\text{ s}\) [1]
- Formula used: \(Q = It\) [1]
- Correct answer with unit: \(900\text{ C}\) (or Coulombs) [1]

(a) Matt black surfaces are excellent absorbers of infrared radiation (thermal energy) compared to shiny silver surfaces, which are poor absorbers (good reflectors).

(b) Air is a very poor thermal conductor, so a trapped layer of air significantly reduces heat loss by conduction. Because the air is trapped in a narrow space, it cannot circulate easily, which prevents/minimizes heat loss by convection currents.

(c) Hot water entering from the bottom has a lower density than the colder water above it. This hot water rises, and the colder, denser water sinks to replace it, setting up a convection current.

(a)
- Matt black is a good absorber of radiation [1]
- Shiny silver is a poor absorber / reflector of radiation [1]

(b)
- Air is a poor conductor / good insulator [1]
- Reduces heat loss by conduction [1]
- Trapped air cannot circulate [1]
- Reduces heat loss by convection [1]

(c)
- Hot water is less dense (so it rises) [1]
- Cold/denser water sinks to replace it [1]

(a) Hooke's Law states that the extension of an object is directly proportional to the applied load/force, provided the limit of proportionality is not exceeded. The limit of proportionality is the point beyond which load and extension are no longer directly proportional.

(b) Extension, \(x = 18.6 - 15.0 = 3.6\text{ cm} = 0.036\text{ m}\).
Using Hooke's Law: \(F = k \times x\)
\(6.0 = k \times 0.036\)
\(k = \frac{6.0}{0.036} \approx 167\text{ N/m}\) (or \(1.67\text{ N/cm}\))

(c) Using \(F = k \times x\) where \(F = 10.0\text{ N}\):
\(x = \frac{10.0\text{ N}}{1.67\text{ N/cm}} = 6.0\text{ cm}\)
Total length = original length + extension = \(15.0\text{ cm} + 6.0\text{ cm} = 21.0\text{ cm}\)

(a)
- Hooke's Law: Force is proportional to extension [1]
- Reference to limit of proportionality not being exceeded [1]
- Limit of proportionality definition: point where linear relationship ceases [1]

(b)
- Calculation of extension: \(3.6\text{ cm}\) or \(0.036\text{ m}\) [1]
- Use of \(F = kx\) [1]
- Calculation of \(k = 167\text{ N/m}\) (or \(170\text{ N/m}\)) [1]

(c)
- Correct extension calculated: \(6.0\text{ cm}\) (or \(0.06\text{ m}\)) [1]
- Correct total length: \(21.0\text{ cm}\) (or \(0.21\text{ m}\)) [1]

(a) Faraday's law states that the size of an induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage (or rate of cutting magnetic field lines).

(b) When the coil is perpendicular to the magnetic field lines, its sides are moving parallel to the magnetic field lines at that instantaneous moment. Therefore, no magnetic field lines are being cut by the sides of the coil, resulting in zero rate of change of flux linkage and thus zero induced e.m.f.

(c) The maximum induced e.m.f. can be increased by:
1. Rotating the coil faster (increasing frequency of rotation).
2. Increasing the number of turns on the coil.
3. Using a stronger magnet (increasing magnetic field strength).
4. Using a soft-iron core inside the coil.

(a)
- e.m.f. is induced when magnetic flux changes [1]
- magnitude is proportional to the rate of change of flux linkage [1]

(b)
- Movement of coil sides is parallel to field lines [1]
- No cutting of magnetic flux lines [1]
- Rate of change of magnetic flux linkage is zero [1]

(c)
- Any three of: increase speed of rotation, increase number of turns, use stronger magnets, use soft-iron core [3]

(a) Sound travels as a longitudinal wave. Compressions are regions of high pressure where air molecules are pushed close together, and rarefactions are regions of low pressure where air molecules are spread far apart. The wave propagates as these vibrations are passed from molecule to molecule.

(b) (i) There are 15 intervals in \(15.0\text{ s}\).
\(\text{Time for 1 interval} = \frac{15.0\text{ s}}{15} = 1.0\text{ s}\).

(ii) The distance traveled by the sound to the wall and back is \(2 \times d = 2 \times 165\text{ m} = 330\text{ m}\).
\(\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{330\text{ m}}{1.0\text{ s}} = 330\text{ m/s}\).

(c) The speed of sound in water is greater than in air. Water is a liquid where molecules are much closer together than gas molecules in air, allowing the vibrations to transfer faster from particle to particle.

(a)
- Longitudinal wave / vibration parallel to energy travel [1]
- Compressions (molecules close) and rarefactions (molecules far apart) [1]

(b) (i)
- Interval calculation: \(15.0 / 15\) [1]
- Correctly showing \(1.0\text{ s}\) [1]

(b) (ii)
- Distance \(= 2d = 330\text{ m}\) [1]
- Speed \(= 330\text{ m/s}\) [1]

(c)
- States: Speed is greater in water [1]
- Reason: Particles in water are closer together/more dense, transferring vibrations faster [1]

(a) Since the number of turns on the secondary coil (100) is less than the number of turns on the primary coil (400), this is a step-down transformer.
(b) Using the transformer relationship:
\(\frac{V_p}{V_s} = \frac{N_p}{N_s}\)
\(\frac{240}{V_s} = \frac{400}{100} = 4\)
\(V_s = \frac{240}{4} = 60\text{ V}\)
(c) (i) Using Ohm's law for the secondary circuit:
\(I_s = \frac{V_s}{R} = \frac{60}{15} = 4.0\text{ A}\)
(ii) For a 100% efficient transformer, the input power equals the output power:
\(P_{\text{in}} = P_{\text{out}}\)
\(V_p \times I_p = V_s \times I_s\)
\(240 \times I_p = 60 \times 4.0 = 240\)
\(I_p = 1.0\text{ A}\)

(a) Step-down [1]
(b) \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) or correct substitution seen: \(\frac{240}{V_s} = \frac{400}{100}\) [1] (C1)
\(V_s = 60\text{ V}\) (unit required) [1] (A1)
(c) (i) \(I = \frac{V}{R}\) or \(\frac{60}{15}\) [1] (C1)
\(I_s = 4.0\text{ A}\) (unit required, accept 4 A) [1] (A1)
(ii) \(V_p I_p = V_s I_s\) or \(\frac{I_p}{I_s} = \frac{N_s}{N_p}\) [1] (C1)
Correct substitution: \(240 \times I_p = 60 \times 4.0\) or \(\frac{I_p}{4.0} = \frac{100}{400}\) [1] (C1)
\(I_p = 1.0\text{ A}\) (unit required, accept 1 A) [1] (A1)

(a) Redshift refers to the increase in the observed wavelength (or decrease in the observed frequency) of light emitted by distant galaxies because they are moving away from Earth.
(b) (i) The Hubble equation is: \(v = H_0 d\)
(ii) First, convert the speed of recession from km/s to m/s:
\(v = 1.8 \times 10^4\text{ km/s} = 1.8 \times 10^7\text{ m/s}\)
Now, rearrange the Hubble equation to solve for distance \(d\):
\(d = \frac{v}{H_0}\)
\(d = \frac{1.8 \times 10^7\text{ m/s}}{2.2 \times 10^{-18}\text{ s}^{-1}} = 8.18 \times 10^{24}\text{ m}\)
This is approximately \(8.2 \times 10^{24}\text{ m}\).
(c) The observation of redshift shows that distant galaxies are moving away from us, meaning the Universe is expanding. If we run this expansion backward in time, it implies that all matter and galaxies must have originated from a single, extremely dense and hot starting point at some definite time in the past.

(a)
- Increase in wavelength / decrease in frequency of light (from distant galaxies) [1]
- Due to the galaxy moving away from Earth [1]
(b)
(i) \(v = H_0 d\) [1]
(ii)
- Conversion of speed: \(1.8 \times 10^4\text{ km/s} = 1.8 \times 10^7\text{ m/s}\) [1] (C1)
- Rearrangement and substitution: \(d = \frac{1.8 \times 10^7}{2.2 \times 10^{-18}}\) [1] (C1)
- Calculation showing \(8.18 \times 10^{24}\text{ m}\) (accept 8.18 to 8.2) [1] (A1)
(c)
- Redshift shows galaxies are moving away / Universe is expanding [1]
- If traced back in time, all matter/galaxies must have originated from a single extremely hot and dense point / single start point [1]

(a) The thermometer reading \(\theta_0 = 74.0^\circ\text{C}\). (b) The temperature decrease is \(\Delta\theta = 74.0^\circ\text{C} - 60.5^\circ\text{C} = 13.5^\circ\text{C}\). (c) Two precautions: 1. Keep eye level perpendicular to the scale to avoid parallax error. 2. Ensure the thermometer bulb is fully immersed in the liquid. (d) Use a clamp and stand to suspend the thermometer in the center of the beaker. (e) Stirring ensures that the temperature is uniform throughout the volume of the water.

Total 10 marks: (a) 2 marks: correct temperature reading of 74.0 with unit. (b) 2 marks: correct subtraction of 60.5 from student's value in (a), with unit. (c) 2 marks: any two valid precautions (1 mark each). (d) 2 marks: suggestion of clamp stand / clamp. (e) 2 marks: explanation of uniform temperature / distribution of heat.

(a) Instrument: Protractor. Step to use accurately: Align the central origin of the protractor precisely with the point of incidence and its baseline with the normal. (b) If the pins are too close, any slight error in alignment creates a much larger angular deviation when drawing the ray line. (c) \(n = \frac{\sin 45^\circ}{\sin 28^\circ} = \frac{0.7071}{0.4695} \approx 1.51\). (d) Precautions: 1. Use thin, sharp pencil lines. 2. Look at the bases of the pins when aligning them.

Total 10 marks: (a) 2 marks: Identifying protractor (1 mark) and correct alignment description (1 mark). (b) 2 marks: Explanation of reducing angular error/improving alignment accuracy. (c) 3 marks: Correct substitution (1 mark), calculation to 2 or 3 sig figs (1 mark), and no units given (1 mark). (d) 3 marks: Any two correct precautions (1.5 marks each or 2 marks for one, 1 mark for second).

(a) The voltmeter reading is \(1.2\text{ V}\). (b) \(R = \frac{1.2\text{ V}}{0.35\text{ A}} = 3.4\ \Omega\) (or \(3.43\ \Omega\)). (c) Yes, because doubling the length from \(40.0\text{ cm}\) to \(80.0\text{ cm}\) should double the resistance. \(2 \times 3.4\ \Omega = 6.8\ \Omega\), which is very close to the measured \(6.9\ \Omega\) (within typical experimental uncertainty). (d) The switch is used to control the current. Opening it between readings prevents the wire from heating up, which would otherwise increase its resistance and cause an unfair test.

Total 10 marks: (a) 2 marks: Voltmeter reading 1.2 V. (b) 2 marks: Correct calculation with ohm (\(\Omega\)) unit. (c) 4 marks: Clear decision 'yes' or 'supported' (1 mark), calculation showing predicted value of 6.8 or 6.86 (2 marks), and statement that they are within experimental limits (1 mark). (d) 2 marks: Mentioning prevention of temperature rise/heating effect.

1. Independent variable: Surface area of the water container. Dependent variable: Mass of water evaporated (or rate of evaporation). 2. Procedure: Pour a fixed volume of hot water into a container. Place it on the balance and record the initial mass. Start the stopwatch. After 15 minutes, record the final mass of the container. Repeat this process for containers of at least 4 different surface areas. 3. Control variables: Initial water temperature, duration of evaporation, environment temperature, volume of water. 4. Surface area calculation: Measure the internal diameter \(d\) of the circular container opening using a ruler. Calculate area using \(A = \pi \left(\frac{d}{2}\right)^2\). 5. Rate of evaporation: Calculate mass lost \(\Delta m = \text{initial mass} - \text{final mass}\). The rate of evaporation is \(\frac{\Delta m}{\Delta t}\). 6. Table: Columns for 'Container ID', 'Diameter \(d\) / cm', 'Surface Area \(A\) / cm\(^2\)', 'Initial Mass \(m_1\) / g', 'Final Mass \(m_2\) / g', 'Mass Loss \(\Delta m\) / g'.

Total 10 marks: 1. Variables: 2 marks (1 mark for independent, 1 mark for dependent). 2. Procedure steps: 3 marks (step-by-step, repeating with different containers, logical flow). 3. Control variables: 2 marks (any two correct variables). 4. Calculating area: 1 mark (formula using diameter/radius). 5. Rate determination: 1 mark (mass change divided by time). 6. Table: 1 mark (appropriate headings and units shown).