An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V3) Cambridge International A Level Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
卷二 (選擇題 - Extended)
Answer all forty multiple choice questions. For each question, choose the one correct option (A, B, C or D).
40 題目 · 40 分
題目 1 · 選擇題
1 分
An electric pump with an input power of \(80\text{ W}\) lifts \(12\text{ kg}\) of water per minute through a vertical height of \(25\text{ m}\). The gravitational field strength \(g\) is \(10\text{ N/kg}\). What is the efficiency of the pump?
1 mark for correct calculation of useful power output (50 W) and final efficiency of 62.5%.
題目 2 · 選擇題
1 分
A gas supply is connected to one side of a U-tube manometer. The other side is open to the atmosphere, where the atmospheric pressure is \(1.0 \times 10^5\text{ Pa}\). The liquid in the manometer has a density of \(1200\text{ kg/m}^3\). The liquid level on the side connected to the gas supply is \(25\text{ cm}\) lower than on the side open to the atmosphere. The gravitational field strength \(g\) is \(10\text{ N/kg}\). What is the pressure of the gas supply?
A.\(9.70 \times 10^4\text{ Pa}\)
B.\(1.00 \times 10^5\text{ Pa}\)
C.\(1.03 \times 10^5\text{ Pa}\)
D.\(1.30 \times 10^5\text{ Pa}\)
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解題
1. Calculate the pressure difference \(\Delta P\) due to the column of liquid: \(\Delta P = \rho g h = 1200\text{ kg/m}^3 \times 10\text{ N/kg} \times 0.25\text{ m} = 3000\text{ Pa} = 3.0 \times 10^3\text{ Pa}\). 2. Since the liquid level is lower on the side connected to the gas supply, the gas pressure is higher than atmospheric pressure: \(P_{\text{gas}} = P_{\text{atm}} + \Delta P = 1.0 \times 10^5\text{ Pa} + 3.0 \times 10^3\text{ Pa} = 1.03 \times 10^5\text{ Pa}\).
評分準則
1 mark for calculating the pressure difference of 3000 Pa and adding it to atmospheric pressure to get the correct absolute gas pressure.
題目 3 · 選擇題
1 分
An electrical heater of power \(60\text{ W}\) is used to heat \(0.50\text{ kg}\) of a liquid in a well-insulated container. The temperature of the liquid increases from \(20^\circ\text{C}\) to \(50^\circ\text{C}\) in \(5.0\text{ minutes}\). What is the specific heat capacity of the liquid?
A.\(120\text{ J}/(\text{kg}^\circ\text{C})\)
B.\(720\text{ J}/(\text{kg}^\circ\text{C})\)
C.\(1200\text{ J}/(\text{kg}^\circ\text{C})\)
D.\(7200\text{ J}/(\text{kg}^\circ\text{C})\)
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解題
1. Calculate the thermal energy \(E\) supplied by the heater: \(E = P \times t = 60\text{ W} \times (5.0 \times 60\text{ s}) = 18,000\text{ J}\). 2. Use the equation \(E = m c \Delta T\) where \(\Delta T = 50^\circ\text{C} - 20^\circ\text{C} = 30^\circ\text{C}\): \(18,000\text{ J} = 0.50\text{ kg} \times c \times 30^\circ\text{C}\) \(18,000 = 15 c\) \(c = 1200\text{ J}/(\text{kg}^\circ\text{C})\).
評分準則
1 mark for calculating total energy supplied (18 000 J) and resolving the formula to find the correct specific heat capacity.
題目 4 · 選擇題
1 分
A series of wavefronts in a ripple tank approach a gap in a barrier. Which change results in the waves diffracting the most as they pass through the gap?
A.Decrease the frequency of the wave source without changing the gap width.
B.Increase the frequency of the wave source without changing the gap width.
C.Increase the gap width without changing the wave source.
D.Decrease the amplitude of the waves without changing the gap width.
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解題
The extent of diffraction increases as the wavelength of the waves becomes comparable to or larger than the gap width. Since water wave speed is constant in constant depth, decreasing the frequency of the source increases the wavelength (from \(v = f\lambda\)), causing more significant diffraction. Changing the amplitude has no effect on the extent of diffraction.
評分準則
1 mark for identifying that decreasing the frequency increases the wavelength, which maximizes diffraction for a fixed gap width.
題目 5 · 選擇題
1 分
A cell of electromotive force (e.m.f.) \(1.5\text{ V}\) is connected to a bulb. The current in the circuit is \(0.20\text{ A}\). How much energy is transferred by the cell in \(5.0\text{ minutes}\)?
A.\(0.15\text{ J}\)
B.\(1.5\text{ J}\)
C.\(15\text{ J}\)
D.\(90\text{ J}\)
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解題
1. Calculate the total charge \(Q\) passing through the circuit: \(Q = I \times t = 0.20\text{ A} \times (5.0 \times 60\text{ s}) = 60\text{ C}\). 2. Calculate the energy transferred using \(E = V \times Q\): \(E = 1.5\text{ V} \times 60\text{ C} = 90\text{ J}\).
評分準則
1 mark for calculating charge (60 C) and finding the correct total energy transferred (90 J).
題目 6 · 選擇題
1 分
An LDR (light-dependent resistor) is connected in series with a fixed resistor of \(4.0\text{ k}\Omega\) and a \(12\text{ V}\) battery of negligible internal resistance. A voltmeter is connected across the \(4.0\text{ k}\Omega\) resistor. In bright light, the resistance of the LDR is \(2.0\text{ k}\Omega\). In darkness, the resistance of the LDR is \(20\text{ k}\Omega\). What are the voltmeter readings in bright light and in darkness?
1 mark for calculating both potential divider output voltages correctly (8.0 V and 2.0 V).
題目 7 · 選擇題
1 分
A radioactive source has a half-life of \(6.0\text{ hours}\). A detector is used to measure the activity of the source in a laboratory where the background radiation is constant at \(20\text{ counts/second}\). Initially, the detector registers a total count rate of \(820\text{ counts/second}\). What count rate does the detector register after \(18\text{ hours}\)?
A.\(100\text{ counts/second}\)
B.\(120\text{ counts/second}\)
C.\(205\text{ counts/second}\)
D.\(220\text{ counts/second}\)
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解題
1. Correct for background radiation to find the initial source activity: \(820 - 20 = 800\text{ counts/s}\). 2. Determine the number of half-lives in \(18\text{ hours}\): \(18 / 6.0 = 3\) half-lives. 3. Calculate the remaining source activity: \(800 \times (1/2)^3 = 100\text{ counts/s}\). 4. Add background radiation to find the total registered count rate: \(100 + 20 = 120\text{ counts/s}\).
評分準則
1 mark for subtracting background radiation, dividing by 8, and adding background back to obtain 120 counts/second.
題目 8 · 選擇題
1 分
A satellite is in a circular orbit around the Earth at a height of \(400\text{ km}\) above the Earth's surface. The radius of the Earth is \(6400\text{ km}\). The orbital period of the satellite is \(90\text{ minutes}\). What is the orbital speed of the satellite?
A.\(0.47\text{ km/s}\)
B.\(7.4\text{ km/s}\)
C.\(7.9\text{ km/s}\)
D.\(470\text{ km/s}\)
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解題
1. Calculate orbital radius \(r\) from the center of the Earth: \(r = 6400\text{ km} + 400\text{ km} = 6800\text{ km} = 6.8 \times 10^6\text{ m}\). 2. Convert the orbital period \(T\) into seconds: \(T = 90\text{ minutes} = 5400\text{ s}\). 3. Calculate orbital speed \(v\): \(v = \frac{2 \pi r}{T} = \frac{2 \pi \times 6.8 \times 10^6\text{ m}}{5400\text{ s}} \approx 7912\text{ m/s} \approx 7.9\text{ km/s}\).
評分準則
1 mark for finding orbital radius (6800 km), converting period to seconds, and calculating correct speed of 7.9 km/s.
題目 9 · 選擇題
1 分
A toy car accelerates from rest to a speed of \(v\) in time \(t\). It then travels at a constant speed \(v\) for a time \(2t\). Finally, it decelerates uniformly to rest in a time \(1.5t\). The total distance travelled is \(d\). What is the value of \(v\) in terms of \(d\) and \(t\)?
A.\(v = \frac{2d}{5t}\)
B.\(v = \frac{4d}{13t}\)
C.\(v = \frac{d}{3t}\)
D.\(v = \frac{4d}{15t}\)
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解題
First, find the distance travelled in each part of the motion. 1) Acceleration phase: average speed is \(v/2\) and time is \(t\), so distance \(d_1 = 0.5vt\). 2) Constant speed phase: speed is \(v\) and time is \(2t\), so distance \(d_2 = 2vt\). 3) Deceleration phase: average speed is \(v/2\) and time is \(1.5t\), so distance \(d_3 = 0.75vt\). The total distance is \(d = d_1 + d_2 + d_3 = 0.5vt + 2vt + 0.75vt = 3.25vt = \frac{13}{4}vt\). Solving for \(v\) gives \(v = \frac{4d}{13t}\).
評分準則
1 mark for the correct option B. Method: Calculate the distance for each segment of the motion in terms of v and t, sum them to get the total distance d, and rearrange to solve for v.
題目 10 · 選擇題
1 分
An unstretched spring has a length of \(12.0\text{ cm}\). When a load of \(4.0\text{ N}\) is suspended from the spring, its length becomes \(15.0\text{ cm}\). Two of these identical springs are connected in parallel, and a total load of \(12.0\text{ N}\) is suspended from the combination. Assume the limit of proportionality is not exceeded. What is the new length of each spring?
A.\(13.5\text{ cm}\)
B.\(15.0\text{ cm}\)
C.\(16.5\text{ cm}\)
D.\(21.0\text{ cm}\)
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解題
The original length of the spring is \(12.0\text{ cm}\). Under a load of \(4.0\text{ N}\), the extension is \(15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\). This means a single spring extends by \(3.0\text{ cm}\) per \(4.0\text{ N}\) of load, giving a spring constant of \(k = 4.0/3.0 = 1.33\text{ N/cm}\). When two identical springs are connected in parallel, they share the total load equally. For a total load of \(12.0\text{ N}\), each spring experiences a load of \(6.0\text{ N}\). The extension of each spring is therefore \(6.0\text{ N} / 1.33\text{ N/cm} = 4.5\text{ cm}\). The new length of each spring is \(12.0\text{ cm} + 4.5\text{ cm} = 16.5\text{ cm}\).
評分準則
1 mark for the correct option C. Method: Determine the extension per unit force for a single spring, calculate the share of the load on each spring in parallel (half of the total load), find the corresponding extension, and add this to the unstretched length.
題目 11 · 選擇題
1 分
An electric motor is used to lift a crate of mass \(80\text{ kg}\) vertically upwards through a height of \(12\text{ m}\) in a time of \(6.0\text{ s}\). The efficiency of the motor system is \(64\\%\). The gravitational field strength \(g\) is \(10\text{ N/kg}\). What is the electrical power input to the motor?
A.\(1.0\text{ kW}\)
B.\(1.6\text{ kW}\)
C.\(2.5\text{ kW}\)
D.\(3.9\text{ kW}\)
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解題
First, calculate the useful work output: \(W_{\text{out}} = mgh = 80\text{ kg} \times 10\text{ N/kg} \times 12\text{ m} = 9600\text{ J}\). Next, calculate the useful power output: \(P_{\text{out}} = W_{\text{out}} / t = 9600\text{ J} / 6.0\text{ s} = 1600\text{ W}\). Use the efficiency formula to find the electrical power input: \(\text{Efficiency} = (P_{\text{out}} / P_{\text{in}}) \times 100\\%\). Therefore, \(0.64 = 1600\text{ W} / P_{\text{in}}\), which gives \(P_{\text{in}} = 1600 / 0.64 = 2500\text{ W} = 2.5\text{ kW}\).
評分準則
1 mark for the correct option C. Method: Calculate the change in gravitational potential energy to find the useful work done, divide by the time taken to find the useful power output, and then use the efficiency percentage to calculate the total electrical power input.
題目 12 · 選擇題
1 分
A container is filled with water of density \(1000\text{ kg/m}^3\). At a depth of \(h\), the hydrostatic pressure due to the water is \(P\). Another container is filled with a liquid of density \(800\text{ kg/m}^3\). At what depth in this liquid is the pressure equal to \(1.6P\)?
A.\(0.50 h\)
B.\(0.80 h\)
C.\(1.25 h\)
D.\(2.0 h\)
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解題
Hydrostatic pressure is given by \(P = \rho g d\), where \(d\) is the depth. For water, \(P = 1000 \times g \times h\). For the other liquid of density \(800\text{ kg/m}^3\) at depth \(d_2\), the pressure is \(P_2 = 800 \times g \times d_2\). We are looking for the depth \(d_2\) where \(P_2 = 1.6P\). Setting the equations equal: \(800 \times g \times d_2 = 1.6 \times (1000 \times g \times h)\) which simplifies to \(800 \times d_2 = 1600 \times h\), yielding \(d_2 = 2.0 h\).
評分準則
1 mark for the correct option D. Method: Relate hydrostatic pressure to density and depth using the formula P = rho * g * h, then set up a ratio or equality to solve for the unknown depth in terms of h.
題目 13 · 選擇題
1 分
A syringe contains a fixed mass of gas at a pressure of \(1.2 \times 10^5\text{ Pa}\) and a volume of \(30\text{ cm}^3\). The temperature of the gas remains constant. The piston of the syringe is slowly pushed in until the volume is reduced to \(18\text{ cm}^3\). What is the new pressure of the gas, and how does the average speed of the gas molecules compare to their initial speed?
A.New pressure = \(0.72 \times 10^5\text{ Pa}\); average speed decreases
B.New pressure = \(2.0 \times 10^5\text{ Pa}\); average speed increases
C.New pressure = \(2.0 \times 10^5\text{ Pa}\); average speed is the same
D.New pressure = \(3.6 \times 10^5\text{ Pa}\); average speed is the same
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解題
Since the temperature of the gas is constant, the average kinetic energy of the gas molecules, and therefore their average speed, remains unchanged (the same). To find the new pressure, apply Boyle's Law: \(P_1 V_1 = P_2 V_2\). Substituting the values: \(1.2 \times 10^5\text{ Pa} \times 30\text{ cm}^3 = P_2 \times 18\text{ cm}^3\). This gives \(P_2 = (3.6 \times 10^6) / 18 = 2.0 \times 10^5\text{ Pa}\).
評分準則
1 mark for the correct option C. Method: Recognize that constant temperature means unchanged average molecular speed, and use Boyle's Law (P1*V1 = P2*V2) to compute the new gas pressure.
題目 14 · 選擇題
1 分
A wire of length \(L\) and cross-sectional area \(A\) has a resistance \(R\). A second wire is made of the same material, has a length \(3L\), and has a uniform circular cross-section with twice the radius of the first wire. What is the resistance of the second wire?
A.\(0.375 R\)
B.\(0.75 R\)
C.\(1.5 R\)
D.\(3.0 R\)
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解題
The resistance of a wire is given by \(R = \rho \frac{L}{A}\). For the first wire, the cross-sectional area is \(A = \pi r^2\). For the second wire, the radius is doubled, so the new cross-sectional area is \(A_2 = \pi (2r)^2 = 4\pi r^2 = 4A\). The length is tripled, so \(L_2 = 3L\). Therefore, the new resistance is \(R_2 = \rho \frac{3L}{4A} = \frac{3}{4} \left(\rho \frac{L}{A}\right) = 0.75 R\).
評分準則
1 mark for the correct option B. Method: Identify how changing the radius of a wire affects its cross-sectional area (proportional to r^2), then use the resistance formula R = rho*L/A to find the proportional change in resistance.
題目 15 · 選擇題
1 分
A potential divider circuit consists of a \(12\text{ V}\) d.c. power supply connected in series with a fixed resistor of resistance \(300\\ \Omega\) and a light-dependent resistor (LDR). A voltmeter is connected across the LDR. In bright light, the resistance of the LDR is \(100\\ \Omega\). In the dark, the resistance of the LDR is \(900\\ \Omega\). What is the change in the voltmeter reading when the conditions change from bright light to dark?
A.It increases by \(3.0\text{ V}\)
B.It increases by \(6.0\text{ V}\)
C.It decreases by \(3.0\text{ V}\)
D.It decreases by \(6.0\text{ V}\)
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解題
First, calculate the voltmeter reading (voltage across the LDR) in bright light: \(V_{\text{bright}} = \frac{R_{\text{LDR}}}{R_{\text{fixed}} + R_{\text{LDR}}} \times V_{\text{supply}} = \frac{100}{300 + 100} \times 12\text{ V} = 3.0\text{ V}\). Next, calculate the voltmeter reading in the dark: \(V_{\text{dark}} = \frac{900}{300 + 900} \times 12\text{ V} = \frac{900}{1200} \times 12\text{ V} = 9.0\text{ V}\). The change in the voltmeter reading is \(9.0\text{ V} - 3.0\text{ V} = 6.0\text{ V}\) (an increase of \(6.0\text{ V}\)).
評分準則
1 mark for the correct option B. Method: Calculate the output voltage across the LDR in both bright light and dark conditions using the potential divider formula, then find the difference between these two values.
題目 16 · 選擇題
1 分
A dwarf planet travels in a circular orbit around the Sun. The radius of the orbit is \(6.0 \times 10^9\text{ km}\) and the orbital speed of the dwarf planet is \(4.7\text{ km/s}\). How many Earth years does it take for this dwarf planet to complete one full orbit around the Sun? (Assume \(1\text{ year} = 3.15 \times 10^7\text{ s}\).)
A.\(40\text{ years}\)
B.\(130\text{ years}\)
C.\(250\text{ years}\)
D.\(1600\text{ years}\)
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解題
The orbital speed is related to the orbital radius by \(v = \frac{2\pi r}{T}\), where \(T\) is the orbital period. First, ensure units are compatible: \(r = 6.0 \times 10^9\text{ km} = 6.0 \times 10^{12}\text{ m}\), and \(v = 4.7\text{ km/s} = 4700\text{ m/s}\). Solve for \(T\): \(T = \frac{2\pi r}{v} = \frac{2 \times \pi \times 6.0 \times 10^{12}\text{ m}}{4700\text{ m/s}} \approx 8.02 \times 10^9\text{ s}\) . Now, convert this period from seconds to Earth years: \(T_{\text{years}} = \frac{8.02 \times 10^9\text{ s}}{3.15 \times 10^7\text{ s/year}} \approx 254.6\text{ years}\), which is closest to \(250\text{ years}\).
評分準則
1 mark for the correct option C. Method: Use the orbital speed formula v = 2*pi*r/T, convert units of radius and speed to a consistent system to find the period in seconds, and divide by the number of seconds in a year to get the answer in years.
題目 17 · 選擇題
1 分
A toy car accelerates from rest to a speed of (6.0\text{ m/s}) in (4.0\text{ s}). It then travels at a constant speed of (6.0\text{ m/s}) for another (6.0\text{ s}). Finally, it decelerates uniformly to rest in a further (2.0\text{ s}). What is the average speed of the toy car over the entire (12.0\text{ s}) journey?
A.\(4.5\text{ m/s}\)
B.\(4.8\text{ m/s}\)
C.\(5.0\text{ m/s}\)
D.\(6.0\text{ m/s}\)
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解題
To find the average speed, we calculate the total distance travelled and divide it by the total time. The distance travelled is the area under the speed-time graph: 1) During acceleration: \(\text{Area}_1 = \frac{1}{2} \times 4.0\text{ s} \times 6.0\text{ m/s} = 12.0\text{ m}\). 2) During constant speed: \(\text{Area}_2 = 6.0\text{ s} \times 6.0\text{ m/s} = 36.0\text{ m}\). 3) During deceleration: \(\text{Area}_3 = \frac{1}{2} \times 2.0\text{ s} \times 6.0\text{ m/s} = 6.0\text{ m}\). Total distance = \(12.0 + 36.0 + 6.0 = 54.0\text{ m}\). Total time = \(12.0\text{ s}\). Average speed = \(\frac{54.0\text{ m}}{12.0\text{ s}} = 4.5\text{ m/s}\).
評分準則
Correct option is A. 1 mark for calculating the correct total distance of 54.0 m and dividing by 12.0 s to find 4.5 m/s.
題目 18 · 選擇題
1 分
An electric motor is used to lift a load of weight (400\text{ N}) through a vertical height of (12\text{ m}) in a time of (8.0\text{ s}). The electrical power input to the motor is (800\text{ W}). What is the efficiency of the motor?
A.\(50\%\)
B.\(60\%\)
C.\(75\%\)
D.\(83\%\)
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解題
1) Calculate the useful work done in lifting the load: \(W = \text{force} \times \text{distance} = 400\text{ N} \times 12\text{ m} = 4800\text{ J}\). 2) Calculate the useful power output: \(P_{\text{out}} = \frac{W}{t} = \frac{4800\text{ J}}{8.0\text{ s}} = 600\text{ W}\). 3) Calculate the efficiency: \(\text{efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{600\text{ W}}{800\text{ W}} \times 100\% = 75\%\).
評分準則
Correct option is C. 1 mark for calculating the correct useful power output (600 W) and dividing by power input (800 W) to get 75%.
題目 19 · 選擇題
1 分
An alloy is made by mixing (200\text{ cm}^3) of metal X (density (8.0\text{ g/cm}^3)) with (300\text{ cm}^3) of metal Y (density (6.0\text{ g/cm}^3)). No change in total volume occurs during the mixing. What is the density of the alloy?
A.\(6.8\text{ g/cm}^3\)
B.\(7.0\text{ g/cm}^3\)
C.\(7.2\text{ g/cm}^3\)
D.\(14\text{ g/cm}^3\)
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解題
1) Find the mass of metal X: \(m_X = \text{volume} \times \text{density} = 200\text{ cm}^3 \times 8.0\text{ g/cm}^3 = 1600\text{ g}\). 2) Find the mass of metal Y: \(m_Y = 300\text{ cm}^3 \times 6.0\text{ g/cm}^3 = 1800\text{ g}\). 3) Total mass of alloy = \(1600\text{ g} + 1800\text{ g} = 3400\text{ g}\). 4) Total volume of alloy = \(200\text{ cm}^3 + 300\text{ cm}^3 = 500\text{ cm}^3\). 5) Density of alloy = \(\frac{\text{Total mass}}{\text{Total volume}} = \frac{3400\text{ g}}{500\text{ cm}^3} = 6.8\text{ g/cm}^3\).
評分準則
Correct option is A. 1 mark for determining the correct total mass (3400 g) and dividing it by the total volume (500 cm³).
題目 20 · 選擇題
1 分
A U-tube manometer containing liquid of density (800\text{ kg/m}^3) is connected to a gas cylinder. The level of liquid in the arm connected to the cylinder is (15\text{ cm}) lower than the level in the arm open to the atmosphere. The atmospheric pressure is (1.01 \times 10^5\text{ Pa}) and the gravitational field strength (g) is (9.8\text{ N/kg}). What is the gas pressure in the cylinder?
A.\(9.98 \times 10^4\text{ Pa}\)
B.\(1.01 \times 10^5\text{ Pa}\)
C.\(1.02 \times 10^5\text{ Pa}\)
D.\(1.13 \times 10^5\text{ Pa}\)
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解題
Since the liquid level is lower on the side connected to the cylinder, the gas pressure is higher than atmospheric pressure: \(P_{\text{gas}} = P_{\text{atm}} + \rho g h\). Convert height to metres: \(h = 15\text{ cm} = 0.15\text{ m}\). Calculate excess pressure: \(\Delta P = 800\text{ kg/m}^3 \times 9.8\text{ N/kg} \times 0.15\text{ m} = 1176\text{ Pa}\). Thus, \(P_{\text{gas}} = 101000\text{ Pa} + 1176\text{ Pa} = 102176\text{ Pa}\), which rounds to \(1.02 \times 10^5\text{ Pa}\) to three significant figures.
評分準則
Correct option is C. 1 mark for converting height to metres, calculating the pressure difference of 1176 Pa, and adding it to atmospheric pressure to get 1.02 x 10^5 Pa.
題目 21 · 選擇題
1 分
Two identical metal cans, one painted matte black and the other painted shiny silver, are filled with equal masses of hot water at (90\text{ }^\circ\text{C}). Both cans are placed in a cool room at (20\text{ }^\circ\text{C}) and their temperatures are recorded over time. Which statement correctly explains the difference in the rate of cooling of the water in the two cans?
A.The water in the matte black can cools slower because matte black surfaces are poor emitters of infrared radiation.
B.The water in the matte black can cools faster because matte black surfaces are good emitters of infrared radiation.
C.The water in the shiny silver can cools faster because shiny silver surfaces are good conductors of heat.
D.The water in both cans cools at the same rate because radiation is only affected by the temperature of the water, not the surface colour.
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解題
Matte black surfaces are excellent emitters of infrared radiation compared to shiny silver surfaces, which are poor emitters. Therefore, the can painted matte black will transfer thermal energy to the surroundings at a faster rate, causing the water inside it to cool down more rapidly.
評分準則
Correct option is B. 1 mark for correctly identifying that matte black is a good emitter of infrared radiation and thus the water in it cools faster.
題目 22 · 選擇題
1 分
A ray of light in glass is incident on a boundary with air. The refractive index of the glass is (1.50). What is the critical angle (c) for this glass-air boundary, and what happens to a ray incident at an angle of (45.0^\circ) to the normal?
A.\(c = 41.8^\circ\); the ray is refracted into the air.
B.\(c = 41.8^\circ\); the ray undergoes total internal reflection.
C.\(c = 48.2^\circ\); the ray is refracted into the air.
D.\(c = 48.2^\circ\); the ray undergoes total internal reflection.
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解題
The critical angle \(c\) is calculated using the formula: \(\sin c = \frac{1}{n}\). With \(n = 1.50\), we get \(\sin c = \frac{1}{1.50} \approx 0.667\), which gives \(c \approx 41.8^\circ\). Since the angle of incidence \(45.0^\circ\) is greater than the critical angle \(41.8^\circ\), the ray cannot refract into air and instead undergoes total internal reflection.
評分準則
Correct option is B. 1 mark for finding critical angle as 41.8 degrees and explaining that total internal reflection occurs because the angle of incidence exceeds the critical angle.
題目 23 · 選擇題
1 分
A battery of electromotive force (e.m.f.) (12\text{ V}) is connected to a circuit. During a certain time interval, a charge of (40\text{ C}) passes through the battery. How much electrical energy is transferred by the battery to the charge, and what is the current in the circuit if this charge transfers in (8.0\text{ s})?
A.Energy = \(96\text{ J}\), Current = \(0.20\text{ A}\)
B.Energy = \(96\text{ J}\), Current = \(5.0\text{ A}\)
C.Energy = \(480\text{ J}\), Current = \(0.20\text{ A}\)
D.Energy = \(480\text{ J}\), Current = \(5.0\text{ A}\)
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解題
1) Electrical energy transferred is calculated using \(E = Q \times V\). Here, \(E = 40\text{ C} \times 12\text{ V} = 480\text{ J}\). 2) Electric current is calculated using \(I = \frac{Q}{t}\). Here, \(I = \frac{40\text{ C}}{8.0\text{ s}} = 5.0\text{ A}\).
評分準則
Correct option is D. 1 mark for calculating correct energy (480 J) and correct current (5.0 A).
題目 24 · 選擇題
1 分
A satellite orbits the Earth in a circular orbit of radius (4.2 \times 10^7\text{ m}). The time taken for one complete orbit is (8.6 \times 10^4\text{ s}). What is the orbital speed of the satellite?
A.\(4.9 \times 10^2\text{ m/s}\)
B.\(3.1 \times 10^3\text{ m/s}\)
C.\(1.9 \times 10^4\text{ m/s}\)
D.\(4.2 \times 10^5\text{ m/s}\)
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解題
The orbital speed \(v\) in a circular orbit is given by: \(v = \frac{2 \pi r}{T}\). Substituting the given values: \(v = \frac{2 \times \pi \times 4.2 \times 10^7\text{ m}}{8.6 \times 10^4\text{ s}} \approx 3.07 \times 10^3\text{ m/s}\). To two significant figures, this is \(3.1 \times 10^3\text{ m/s}\).
評分準則
Correct option is B. 1 mark for using the circular orbital speed formula and calculating the speed as 3.1 x 10^3 m/s.
題目 25 · 選擇題
1 分
A car travels along a straight road. It accelerates from rest with a constant acceleration of \(2.0\text{ m/s}^2\) for \(5.0\text{ s}\). It then travels at a constant speed for \(10\text{ s}\), and finally decelerates uniformly to rest in a further \(4.0\text{ s}\). What is the total distance travelled by the car?
A.135 m
B.145 m
C.170 m
D.190 m
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解題
The motion has three phases. Phase 1 (acceleration): \(u = 0\text{ m/s}\), \(a = 2.0\text{ m/s}^2\), \(t = 5.0\text{ s}\). The final speed reached is \(v = a \times t = 2.0 \times 5.0 = 10\text{ m/s}\). The distance travelled is \(d_1 = \frac{1}{2} \times 5.0 \times 10 = 25\text{ m}\). Phase 2 (constant speed): \(v = 10\text{ m/s}\), \(t = 10\text{ s}\). The distance travelled is \(d_2 = 10 \times 10 = 100\text{ m}\). Phase 3 (deceleration): The car decelerates from \(10\text{ m/s}\) to rest in \(t = 4.0\text{ s}\). The distance travelled is \(d_3 = \frac{1}{2} \times 4.0 \times 10 = 20\text{ m}\). Total distance = \(25 + 100 + 20 = 145\text{ m}\).
評分準則
1 mark for calculating the correct distance in each of the three stages (25 m, 100 m, and 20 m) and adding them to obtain 145 m. Reject calculation errors that ignore the acceleration/deceleration shapes (resulting in 190 m).
題目 26 · 選擇題
1 分
An electric motor is used to lift a load of weight \(300\text{ N}\) through a vertical height of \(8.0\text{ m}\) in a time of \(5.0\text{ s}\). The electrical power input to the motor is \(800\text{ W}\). What is the efficiency of the motor?
1 mark for calculating the useful power output (480 W) and dividing by the input power (800 W) to find the correct efficiency of 60%.
題目 27 · 選擇題
1 分
A submarine is at a depth of \(150\text{ m}\) below the surface of the sea. The density of seawater is \(1020\text{ kg/m}^3\). The atmospheric pressure at the surface is \(1.0 \times 10^5\text{ Pa}\) and the acceleration of free fall \(g\) is \(9.8\text{ m/s}^2\). What is the total pressure exerted on the outer surface of the submarine?
A.\(1.5 \times 10^5\text{ Pa}\)
B.\(1.5 \times 10^6\text{ Pa}\)
C.\(1.6 \times 10^6\text{ Pa}\)
D.\(1.6 \times 10^7\text{ Pa}\)
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解題
The pressure due to the liquid column is \(p_{\text{liquid}} = \rho g h = 1020\text{ kg/m}^3 \times 9.8\text{ m/s}^2 \times 150\text{ m} = 1,499,400\text{ Pa} = 1.5 \times 10^6\text{ Pa}\). The total pressure is the sum of hydrostatic pressure and atmospheric pressure: \(p_{\text{total}} = 1.5 \times 10^6\text{ Pa} + 1.0 \times 10^5\text{ Pa} = 1.6 \times 10^6\text{ Pa}\).
評分準則
1 mark for calculating the pressure from the liquid column and correctly adding the atmospheric pressure to get the total pressure of 1.6 x 10^6 Pa. Reject if atmospheric pressure is omitted (yielding 1.5 x 10^6 Pa).
題目 28 · 選擇題
1 分
A \(2.0\text{ kg}\) block of metal is heated by an electric heater rated at \(150\text{ W}\) for a time of \(4.0\text{ minutes}\). The temperature of the block rises from \(20^\circ\text{C}\) to \(65^\circ\text{C}\). Assuming no thermal energy is lost to the surroundings, what is the specific heat capacity of the metal?
A.\(6.7\text{ J}/(\text{kg}^\circ\text{C})\)
B.\(400\text{ J}/(\text{kg}^\circ\text{C})\)
C.\(800\text{ J}/(\text{kg}^\circ\text{C})\)
D.\(24000\text{ J}/(\text{kg}^\circ\text{C})\)
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解題
The electrical energy supplied by the heater is \(E = P \times t = 150\text{ W} \times (4.0 \times 60)\text{ s} = 36,000\text{ J}\). The change in temperature is \(\Delta T = 65^\circ\text{C} - 20^\circ\text{C} = 45^\circ\text{C}\). Since \(E = m c \Delta T\), we have \(36,000 = 2.0 \times c \times 45\), which gives \(c = \frac{36,000}{90} = 400\text{ J}/(\text{kg}^\circ\text{C})\).
評分準則
1 mark for converting time to seconds (240 s) and calculating energy (36,000 J), and correctly applying the specific heat capacity formula to get 400 J/(kg °C). Reject if minutes are not converted to seconds.
題目 29 · 選擇題
1 分
A ray of light travels from a transparent plastic block into air. The refractive index of the plastic is \(1.50\). What is the critical angle for light in this plastic, and does total internal reflection occur if the angle of incidence is \(45^\circ\)?
A.critical angle is \(34^\circ\), total internal reflection does not occur
B.critical angle is \(42^\circ\), total internal reflection does not occur
C.critical angle is \(42^\circ\), total internal reflection occurs
D.critical angle is \(49^\circ\), total internal reflection occurs
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解題
The critical angle \(c\) is calculated using \(\sin c = \frac{1}{n} = \frac{1}{1.50}\), which gives \(c \approx 42^\circ\). Total internal reflection occurs when the light travels from a optically denser medium to a less dense medium, and the angle of incidence is greater than the critical angle. Since the angle of incidence is \(45^\circ\), which is greater than \(42^\circ\), total internal reflection occurs.
評分準則
1 mark for calculating the critical angle as 42 degrees and correctly identifying that total internal reflection occurs because the angle of incidence is greater than the critical angle.
題目 30 · 選擇題
1 分
Two copper wires, X and Y, have different dimensions. Wire X has length \(L\) and radius \(r\). Wire Y has length \(2L\) and radius \(2r\). The resistance of wire X is \(R\). What is the resistance of wire Y?
A.0.5 R
B.R
C.2 R
D.4 R
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解題
The resistance of a wire is given by \(R = \rho \frac{\text{length}}{\text{Area}}\). The cross-sectional area of wire X is \(A_{\text{X}} = \pi r^2\), so \(R_{\text{X}} = \rho \frac{L}{\pi r^2} = R\). The area of wire Y is \(A_{\text{Y}} = \pi (2r)^2 = 4 \pi r^2\). Thus, the resistance of wire Y is \(R_{\text{Y}} = \rho \frac{2L}{4 \pi r^2} = 0.5 \left(\rho \frac{L}{\pi r^2}\right) = 0.5 R\).
評分準則
1 mark for correctly showing that doubling the radius quadruples the area, which results in a net factor of 0.5 on the resistance value, leading to 0.5 R.
題目 31 · 選擇題
1 分
A student measures the count rate from a radioactive source. The background count rate is constant at \(24\text{ counts/minute}\). At time \(t = 0\), the measured count rate is \(344\text{ counts/minute}\). After \(6.0\text{ hours}\), the measured count rate is \(64\text{ counts/minute}\). What is the half-life of the radioactive source?
A.1.5 hours
B.2.0 hours
C.3.0 hours
D.6.0 hours
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解題
First, find the corrected initial and final count rates by subtracting the background: Initial corrected count rate = \(344 - 24 = 320\text{ counts/minute}\). Final corrected count rate = \(64 - 24 = 40\text{ counts/minute}\). The ratio of final to initial corrected count rate is \(\frac{40}{320} = \frac{1}{8} = (\frac{1}{2})^3\). This corresponds to exactly 3 half-lives. Therefore, 3 half-lives = \(6.0\text{ hours}\), so 1 half-life = \(2.0\text{ hours}\).
評分準則
1 mark for subtracting the background count rate from both measurements to find corrected counts, identifying that 3 half-lives occurred, and dividing the total time by 3 to find a half-life of 2.0 hours. Reject if background is not subtracted.
題目 32 · 選擇題
1 分
Light from a distant galaxy is observed to be redshifted. What does this redshift indicate about the galaxy, and how does the redshift of more distant galaxies compare to closer ones?
A.The galaxy is moving away from Earth, and more distant galaxies have a smaller redshift.
B.The galaxy is moving away from Earth, and more distant galaxies have a larger redshift.
C.The galaxy is moving towards Earth, and more distant galaxies have a smaller redshift.
D.The galaxy is moving towards Earth, and more distant galaxies have a larger redshift.
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解題
Redshift of light indicates that the wavelength has increased because the galaxy is moving away from the Earth. According to Hubble's law, the recessional speed of a galaxy is proportional to its distance from Earth. Therefore, more distant galaxies are moving away at higher speeds and their light shows a larger redshift.
評分準則
1 mark for identifying that redshift means the galaxy is moving away from Earth and that more distant galaxies exhibit a larger redshift due to higher recessional speeds.
題目 33 · multiple_choice
1 分
A car accelerates uniformly from rest to a speed of \(12.0\text{ m/s}\) in \(4.0\text{ s}\). It then travels at this constant speed of \(12.0\text{ m/s}\) for \(6.0\text{ s}\), and finally decelerates uniformly to rest in a further \(2.0\text{ s}\). What is the average speed of the car for the entire journey?
A.\(6.0\text{ m/s}\)
B.\(9.0\text{ m/s}\)
C.\(10.0\text{ m/s}\)
D.\(12.0\text{ m/s}\)
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解題
To find the average speed, we first calculate the total distance traveled during the three stages: 1. Acceleration stage: \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 12.0\text{ m/s} = 24.0\text{ m}\). 2. Constant speed stage: \(d_2 = \text{speed} \times \text{time} = 12.0\text{ m/s} \times 6.0\text{ s} = 72.0\text{ m}\). 3. Deceleration stage: \(d_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.0\text{ s} \times 12.0\text{ m/s} = 12.0\text{ m}\).
Total distance = \(24.0 + 72.0 + 12.0 = 108.0\text{ m}\). Total time = \(4.0 + 6.0 + 2.0 = 12.0\text{ s}\).
Award 1 mark for the correct option B. - Award 0 marks for incorrect options. - Method: Calculate total distance as area under the speed-time graph, divide by total time. - Accuracy: \(9.0\text{ m/s}\).
題目 34 · multiple_choice
1 分
Two identical springs are connected in parallel to support a load of \(6.0\text{ N}\). Each spring has an unstretched length of \(10.0\text{ cm}\). The spring constant of each spring is \(150\text{ N/m}\). What is the total length of the parallel combination of springs when supporting this load?
A.\(2.0\text{ cm}\)
B.\(12.0\text{ cm}\)
C.\(14.0\text{ cm}\)
D.\(18.0\text{ cm}\)
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解題
Since the two identical springs are in parallel, they share the \(6.0\text{ N}\) load equally. Therefore, the load on each spring is: \(F = \frac{6.0\text{ N}}{2} = 3.0\text{ N}\).
Using Hooke's Law (\(F = kx\)), the extension \(x\) of each spring is: \(x = \frac{F}{k} = \frac{3.0\text{ N}}{150\text{ N/m}} = 0.02\text{ m} = 2.0\text{ cm}\).
The total length of the combination is the unstretched length plus the extension: \(L = 10.0\text{ cm} + 2.0\text{ cm} = 12.0\text{ cm}\).
評分準則
Award 1 mark for the correct option B. - Option A incorrectly gives only the extension. - Option C incorrectly assumes only one spring carries the entire load. - Option D incorrectly treats the arrangement as series.
題目 35 · multiple_choice
1 分
An electric motor is used to lift a block of mass \(25\text{ kg}\) vertically upwards through a height of \(12\text{ m}\) in a time of \(5.0\text{ s}\). The electrical power input to the motor is \(980\text{ W}\). What is the efficiency of the motor? (Use \(g = 9.8\text{ m/s}^2\))
A.\(30\%\)
B.\(50\%\)
C.\(60\%\)
D.\(82\%\)
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解題
First, calculate the useful work output (gravitational potential energy gained by the block): \(W_{\text{out}} = mgh = 25\text{ kg} \times 9.8\text{ m/s}^2 \times 12\text{ m} = 2940\text{ J}\).
Next, calculate the useful power output: \(P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{2940\text{ J}}{5.0\text{ s}} = 588\text{ W}\).
Award 1 mark for the correct option C. - Method: Work done = \(mgh\), Power output = \(W/t\), Efficiency = \(P_{\text{out}}/P_{\text{in}}\). - Accuracy: 60%.
題目 36 · multiple_choice
1 分
A U-tube manometer containing water (density \(1000\text{ kg/m}^3\)) is connected to a gas supply. The water level in the arm connected to the gas supply is \(25\text{ cm}\) lower than the water level in the open arm. The atmospheric pressure is \(1.01 \times 10^5\text{ Pa}\). What is the absolute pressure of the gas supply? (Use \(g = 9.8\text{ m/s}^2\))
A.\(2.45 \times 10^3\text{ Pa}\)
B.\(0.99 \times 10^5\text{ Pa}\)
C.\(1.01 \times 10^5\text{ Pa}\)
D.\(1.03 \times 10^5\text{ Pa}\)
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解題
The height difference of water is \(h = 25\text{ cm} = 0.25\text{ m}\). The pressure difference between the gas supply and atmospheric pressure is: \(\Delta p = \rho g h = 1000\text{ kg/m}^3 \times 9.8\text{ m/s}^2 \times 0.25\text{ m} = 2450\text{ Pa}\).
Since the water level in the gas-connected arm is lower, the pressure of the gas is greater than atmospheric pressure: \(p_{\text{gas}} = p_{\text{atm}} + \Delta p = 1.01 \times 10^5\text{ Pa} + 2450\text{ Pa} = 101,000\text{ Pa} + 2450\text{ Pa} = 103,450\text{ Pa} \approx 1.03 \times 10^5\text{ Pa}\).
評分準則
Award 1 mark for correct option D. - Option A is only the gauge pressure. - Option B results from incorrectly subtracting the gauge pressure. - Option C is just the atmospheric pressure.
題目 37 · multiple_choice
1 分
A \(2.0\text{ kg}\) block of metal is heated by a \(150\text{ W}\) electric heater. The heater is switched on for \(4.0\text{ minutes}\). The temperature of the block rises from \(20^\circ\text{C}\) to \(65^\circ\text{C}\). Assuming no thermal energy is lost to the surroundings, what is the specific heat capacity of the metal?
A.\(6.7\text{ J/(kg}\cdot^\circ\text{C)}\)
B.\(280\text{ J/(kg}\cdot^\circ\text{C)}\)
C.\(400\text{ J/(kg}\cdot^\circ\text{C)}\)
D.\(900\text{ J/(kg}\cdot^\circ\text{C)}\)
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解題
The thermal energy \(E\) supplied by the heater is: \(E = P \times t = 150\text{ W} \times (4.0 \times 60)\text{ s} = 150 \times 240 = 36,000\text{ J}\).
The temperature rise is \(\Delta \theta = 65^\circ\text{C} - 20^\circ\text{C} = 45^\circ\text{C}\).
Using the formula for specific heat capacity: \(E = m c \Delta \theta\) \(36,000\text{ J} = 2.0\text{ kg} \times c \times 45^\circ\text{C}\) \(36,000 = 90 c\) \(c = 400\text{ J/(kg}\cdot^\circ\text{C)}\).
評分準則
Award 1 mark for correct option C. - Option A fails to convert time from minutes to seconds. - Option B uses \(T = 65^\circ\text{C}\) instead of the temperature change. - Option D uses \(T = 20^\circ\text{C}\) instead of the temperature change.
題目 38 · multiple_choice
1 分
A charge of \(180\text{ C}\) flows through a lamp in a circuit during a time interval of \(1.5\text{ minutes}\). The potential difference across the lamp is \(6.0\text{ V}\). Which row in the table shows the current in the lamp and the thermal energy transferred?
A.Current = \(2.0\text{ A}\), Energy = \(12\text{ J}\)
B.Current = \(2.0\text{ A}\), Energy = \(1080\text{ J}\)
C.Current = \(120\text{ A}\), Energy = \(18\text{ J}\)
D.Current = \(120\text{ A}\), Energy = \(1080\text{ J}\)
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解題
First, convert time into seconds: \(t = 1.5\text{ minutes} = 1.5 \times 60\text{ s} = 90\text{ s}\).
The current \(I\) is calculated as: \(I = \frac{Q}{t} = \frac{180\text{ C}}{90\text{ s}} = 2.0\text{ A}\).
The energy transferred \(E\) is given by: \(E = V Q = 6.0\text{ V} \times 180\text{ C} = 1080\text{ J}\).
評分準則
Award 1 mark for correct option B. - Option A uses the incorrect energy calculation. - Options C and D fail to convert minutes to seconds for the current calculation.
題目 39 · multiple_choice
1 分
A potential divider circuit consists of a stable \(12\text{ V}\) d.c. power supply, a fixed resistor of resistance \(8.0\ \Omega\), and a thermistor connected in series. At room temperature, the resistance of the thermistor is \(4.0\ \Omega\). What is the potential difference across the thermistor, and how does this potential difference change when the temperature of the thermistor increases?
A.Potential difference is \(4.0\text{ V}\); it decreases when temperature increases.
B.Potential difference is \(4.0\text{ V}\); it increases when temperature increases.
C.Potential difference is \(8.0\text{ V}\); it decreases when temperature increases.
D.Potential difference is \(8.0\text{ V}\); it increases when temperature increases.
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解題
The total resistance is \(R_{\text{total}} = 8.0\ \Omega + 4.0\ \Omega = 12.0\ \Omega\). The current is \(I = \frac{V}{R} = \frac{12\text{ V}}{12.0\ \Omega} = 1.0\text{ A}\).
The initial potential difference across the thermistor is: \(V_{\text{th}} = I \times R_{\text{thermistor}} = 1.0\text{ A} \times 4.0\ \Omega = 4.0\text{ V}\).
When temperature increases, the resistance of an NTC thermistor decreases. This decreases its share of the total resistance, so the potential difference across it decreases.
評分準則
Award 1 mark for correct option A. - Options C and D calculate the voltage across the fixed resistor (8.0 V) instead of the thermistor. - Option B incorrectly suggests the potential difference increases with temperature.
題目 40 · multiple_choice
1 分
An ideal transformer steps down a mains voltage of \(240\text{ V}\) a.c. to \(12\text{ V}\) a.c. The primary coil has \(1200\text{ turns}\). A lamp rated \(12\text{ V}, 36\text{ W}\) is connected to the secondary coil and operates normally. What is the number of turns on the secondary coil, and what is the current in the primary coil?
A.Secondary turns = \(60\); primary current = \(0.15\text{ A}\)
B.Secondary turns = \(60\); primary current = \(3.0\text{ A}\)
C.Secondary turns = \(24,000\); primary current = \(0.15\text{ A}\)
D.Secondary turns = \(24,000\); primary current = \(3.0\text{ A}\)
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解題
First, find the number of secondary turns \(N_{\text{s}}\): \(\frac{V_{\text{p}}}{V_{\text{s}}} = \frac{N_{\text{p}}}{N_{\text{s}}}\) \(\frac{240}{12} = \frac{1200}{N_{\text{s}}} \implies N_{\text{s}} = 60\text{ turns}\).
Since the transformer is ideal (100% efficient), the electrical power input equals the power output: \(P_{\text{p}} = P_{\text{s}} = 36\text{ W}\).
Thus, the current in the primary coil \(I_{\text{p}}\) is: \(I_{\text{p}} = \frac{P_{\text{p}}}{V_{\text{p}}} = \frac{36\text{ W}}{240\text{ V}} = 0.15\text{ A}\).
評分準則
Award 1 mark for correct option A. - Options B and D confuse primary and secondary currents (secondary current is \(3.0\text{ A}\)). - Options C and D calculate the secondary turns incorrectly by multiplying instead of dividing.
Paper 4 (Theory - Extended)
Answer all structured and quantitative questions. Show all your working and state units where appropriate.
10 題目 · 80 分
題目 1 · structured
8 分
An electric test car starts from rest and accelerates uniformly along a straight test track. It reaches a speed of \(30\text{ m/s}\) in a time of \(5.0\text{ s}\). It then travels at this constant speed of \(30\text{ m/s}\) for a further \(10\text{ s}\). Finally, the car decelerates uniformly to rest in a time of \(3.0\text{ s}\).
(a) Calculate the acceleration of the car during the first \(5.0\text{ s}\). [2 marks] (b) Show that the total distance travelled by the car during the entire \(18\text{ s}\) motion is \(420\text{ m}\). [4 marks] (c) Calculate the average speed of the car for the entire journey. [2 marks]
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解題
(a) Acceleration \(a = \frac{v - u}{t} = \frac{30 - 0}{5.0} = 6.0\text{ m/s}^2\). (b) The total distance is the area under the speed-time graph: - Phase 1 (0 to 5 s): Area of triangle = \(\frac{1}{2} \times 5.0 \times 30 = 75\text{ m}\) - Phase 2 (5 to 15 s): Area of rectangle = \(10 \times 30 = 300\text{ m}\) - Phase 3 (15 to 18 s): Area of triangle = \(\frac{1}{2} \times 3.0 \times 30 = 45\text{ m}\) - Total distance = \(75 + 300 + 45 = 420\text{ m}\). (c) Average speed = \(\frac{\text{total distance}}{\text{total time}} = \frac{420\text{ m}}{18.0\text{ s}} \approx 23.3\text{ m/s}\).
評分準則
(a) - Formula used: \(a = \frac{\Delta v}{\Delta t}\) [1 mark] - Correct calculation with units: \(6.0\text{ m/s}^2\) [1 mark]
(b) - Calculates area of first triangle (75 m) or last triangle (45 m) [1 mark] - Calculates area of rectangle (300 m) [1 mark] - Adds all three areas together: \(75 + 300 + 45\) [1 mark] - Obtains the final value of \(420\text{ m}\) clearly [1 mark]
An electric pump is used to lift water from a reservoir to a storage tank located at a vertical height of \(15\text{ m}\) above the reservoir. The pump lifts \(1200\text{ kg}\) of water every minute. Take the acceleration of free fall \(g = 9.8\text{ m/s}^2\).
(a) Calculate the gravitational potential energy gained by the water lifted in one minute. [3 marks] (b) Calculate the minimum useful power output of the pump. [2 marks] (c) The electrical power input to the motor of the pump is \(4.0\text{ kW}\). Calculate the efficiency of the pump system. [3 marks]
An electric heater of power \(200\text{ W}\) is placed inside a well-insulated container containing \(0.40\text{ kg}\) of ice at its melting point of \(0^\circ\text{C}\). The specific latent heat of fusion of ice is \(3.3 \times 10^5\text{ J/kg}\), and the specific heat capacity of water is \(4200\text{ J/(kg}^\circ\text{C)}\).
(a) Calculate the energy required to melt all the ice to water at \(0^\circ\text{C}\). [2 marks] (b) Calculate the time taken for the heater to melt all the ice. [2 marks] (c) After all the ice has melted, the heater continues to heat the resulting water. Calculate the thermal energy needed to raise the temperature of the water from \(0^\circ\text{C}\) to \(25^\circ\text{C}\). [2 marks] (d) State and explain how the presence of thermal energy losses to the surroundings would affect the time taken to melt the ice. [2 marks]
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解題
(a) \(E_f = m L_f = 0.40 \times 3.3 \times 10^5 = 132000\text{ J}\). (b) \(t = \frac{E}{P} = \frac{132000}{200} = 660\text{ s}\). (c) \(\Delta E = m c \Delta \theta = 0.40 \times 4200 \times 25 = 42000\text{ J}\). (d) If thermal energy is lost to the surroundings, less energy from the heater is transferred to the ice per second. Therefore, it will take more time to melt the ice.
(d) - Time taken increases [1 mark] - Explanation: some thermal energy escapes to the surroundings instead of melting the ice [1 mark]
題目 4 · structured
8 分
A ray of light is incident on the boundary between a glass block and air. The refractive index of the glass is \(1.52\).
(a) Calculate the critical angle \(c\) for the boundary between this glass and air. [3 marks] (b) The angle of incidence in the glass is set to \(30^\circ\). Calculate the angle of refraction in the air. [3 marks] (c) State and explain what happens to the ray of light if the angle of incidence in the glass is increased to \(45^\circ\). [2 marks]
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解題
(a) \(\sin(c) = \frac{1}{n} = \frac{1}{1.52} \approx 0.6579\), which gives \(c = \arcsin(0.6579) \approx 41.1^\circ\). (b) Using Snell's law: \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \implies 1.52 \times \sin(30^\circ) = 1.0 \times \sin(r)\). Thus, \(\sin(r) = 1.52 \times 0.5 = 0.76\\, leading to \)r = \\arcsin(0.76) \\approx 49.5^\\circ\). (c) Since the angle of incidence (\(45^\circ\)) is greater than the critical angle (\(41.1^\circ\)), total internal reflection occurs, and all light is reflected back into the glass.
(c) - Total internal reflection occurs [1 mark] - Reason: the angle of incidence exceeds the critical angle [1 mark]
題目 5 · structured
8 分
A potential divider circuit is set up using a \(12.0\text{ V}\) d.c. power supply, a fixed resistor \(R\) of resistance \(400\\ \Omega\), and a thermistor.
(a) At a certain temperature, the reading on a voltmeter connected across the thermistor is \(8.0\text{ V}\). (i) Determine the voltage across the fixed resistor \(R\). [1 mark] (ii) Calculate the current in the circuit. [2 marks] (iii) Calculate the resistance of the thermistor at this temperature. [2 marks] (b) The temperature of the thermistor is now increased. State and explain how this affects: (i) the resistance of the thermistor. [1 mark] (ii) the reading on the voltmeter connected across the thermistor. [2 marks]
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解題
(a) (i) The voltage across \(R\) is \(V_R = V_{\text{total}} - V_{\text{thermistor}} = 12.0 - 8.0 = 4.0\text{ V}\). (ii) The current is \(I = \frac{V_R}{R} = \frac{4.0}{400} = 0.010\text{ A}\) (or \(10\text{ mA}\)). (iii) The resistance of the thermistor is \(R_{\text{thermistor}} = \frac{V_{\text{thermistor}}}{I} = \frac{8.0}{0.010} = 800\\ \Omega\). (b) (i) The resistance of the thermistor decreases as the temperature increases. (ii) As the resistance of the thermistor decreases, it takes a smaller fraction of the total supply voltage, so the voltmeter reading across it decreases.
(ii) - Voltmeter reading decreases [1 mark] - Reason: thermistor makes up a smaller fraction of the total circuit resistance [1 mark]
題目 6 · structured
8 分
An ideal step-down transformer is connected to a \(230\text{ V}\) a.c. mains supply. The primary coil has \(1500\) turns, and the secondary coil has \(60\) turns. The secondary coil is connected to a \(12\text{ W}\) lamp which operates at its normal brightness.
(a) Calculate the output voltage across the secondary coil. [3 marks] (b) Calculate the current in the secondary coil. [2 marks] (c) Assuming the transformer is \(100\\%\) efficient, calculate the current in the primary coil. [3 marks]
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解題
(a) Use the transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). Thus, \(V_s = V_p \times \frac{N_s}{N_p} = 230 \times \frac{60}{1500} = 9.2\text{ V}\). (b) Power is given by \(P = V \times I\). Thus, secondary current \(I_s = \frac{P_s}{V_s} = \frac{12}{9.2} \approx 1.3\text{ A}\). (c) For an ideal transformer, input power equals output power: \(P_p = P_s = 12\text{ W}\). Thus, primary current \(I_p = \frac{P_p}{V_p} = \frac{12}{230} \approx 0.052\text{ A}\) (or \(52\text{ mA}\)).
(c) - State \(P_{\text{in}} = P_{\text{out}} = 12\text{ W}\) OR use formula \(\frac{I_p}{I_s} = \frac{N_s}{N_p}\) [1 mark] - Substitution: \(I_p = \frac{12}{230}\) [1 mark] - Correct current: \(0.052\text{ A}\) (or \(52\text{ mA}\)) [1 mark]
題目 7 · structured
8 分
A radioactive isotope decays by alpha (\(\alpha\)) emission.
(a) State the composition of an alpha particle. [2 marks] (b) A detector is used to measure the background radiation in a laboratory. Over a period of 10 minutes, the total count recorded is 240. Calculate the average background count rate in counts per second. [2 marks] (c) A sample of the alpha-emitting isotope is placed near the detector. The initial total count rate measured by the detector is \(180\text{ counts/s}\) (which includes background). The half-life of the isotope is \(20\text{ minutes}\). (i) Determine the initial count rate due to the source alone. [1 mark] (ii) Calculate the expected total count rate measured by the detector after \(40\text{ minutes}\). [3 marks]
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解題
(a) An alpha particle consists of 2 protons and 2 neutrons (equivalent to a helium-4 nucleus). (b) Background count rate = \(\frac{240\text{ counts}}{10 \times 60\text{ s}} = 0.4\text{ counts/s}\). (c) (i) Initial count rate due to source alone = \(180 - 0.4 = 179.6\text{ counts/s}\). (ii) In 40 minutes, the number of half-lives is \(\frac{40}{20} = 2\). After 2 half-lives, the source activity decreases to \(\frac{179.6}{2^2} = \frac{179.6}{4} = 44.9\text{ counts/s}\). Total count rate = source count rate + background = \(44.9 + 0.4 = 45.3\text{ counts/s}\).
(b) - Converts time to seconds: \(600\text{ s}\) [1 mark] - Correct calculation: \(0.4\text{ counts/s}\) [1 mark]
(c) (i) - Correct calculation: \(179.6\text{ counts/s}\) (accept \(180\) as a common approximation) [1 mark]
(ii) - Identifies that \(40\text{ min}\) is 2 half-lives [1 mark] - Divides source activity by 4: \(44.9\text{ counts/s}\) [1 mark] - Adds background rate back to find total rate: \(45.3\text{ counts/s}\) (accept \(45\text{ counts/s}\) if background subtraction was omitted) [1 mark]
題目 8 · structured
8 分
(a) Describe the life cycle of a star with a mass much greater than the Sun, starting from its stable main-sequence phase until its final state. [4 marks] (b) Light from a distant galaxy is observed to have a redshift. (i) Explain what is meant by the term 'redshift'. [2 marks] (ii) The redshift of a galaxy is used to find its recessional speed, which is \(3.3 \times 10^6\text{ m/s}\). Using the Hubble constant \(H_0 = 2.2 \times 10^{-18}\text{ s}^{-1}\), calculate the distance of this galaxy from Earth. [2 marks]
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解題
(a) A massive star leaves the main sequence and expands to become a red supergiant. It then undergoes a supernova explosion. The remaining core collapses to form either a neutron star or, if the mass is sufficiently high, a black hole. (b) (i) Redshift is the increase in the observed wavelength (or decrease in frequency) of light from a source moving away from the observer. (ii) Using Hubble's law: \(v = H_0 d \implies d = \frac{v}{H_0} = \frac{3.3 \times 10^6}{2.2 \times 10^{-18}} = 1.5 \times 10^{24}\text{ m}\).
評分準則
(a) - Expands into a Red Supergiant [1 mark] - Explodes as a Supernova [1 mark] - Core collapses [1 mark] - Forms a Neutron Star OR Black Hole [1 mark]
(b) (i) - Wavelength of light increases / moves towards red end of spectrum [1 mark] - Occurs because the galaxy/source is moving away [1 mark]
An electric motor-driven pump is used to raise water from a well to a storage tank. The pump lifts 150 kg of water through a vertical height of 12 m in a time of 25 s. Take the acceleration of free fall g to be 9.8 m/s^{2}. (a) Calculate the useful work done by the pump in lifting this water. (b) Calculate the useful power output of the pump. (c) The electrical power input to the motor is 1.2 kW. (i) Calculate the efficiency of the pump system. (ii) State what happens to the energy that is not transferred usefully, and describe how this affects the temperature of the pump motor.
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解題
**(a) Work done calculation:** \(W = mgh\) \(W = 150 \times 9.8 \times 12 = 17640\text{ J}\) (or \(17600\text{ J}\) to 3 s.f., or \(18000\text{ J}\) if using \(g = 10\text{ m/s}^2\))
**(c)(i) Efficiency calculation:** Input Power = \(1.2\text{ kW} = 1200\text{ W}\) \(\text{Efficiency} = \frac{\text{Useful Power Output}}{\text{Total Power Input}} \times 100\\%\) \(\text{Efficiency} = \frac{705.6}{1200} \times 100 = 58.8\\%\) (or \(59\\%\) to 2 s.f.)
**(c)(ii) Energy dissipation:** The wasted energy is transferred to the surroundings as thermal energy (and sound) due to friction in the pump and electrical resistance in the windings. This causes the temperature of the motor to rise.
評分準則
**(a)** - Formula used: \(W = mgh\) [1 mark] - Correct calculation: \(17600\text{ J}\) or \(17640\text{ J}\) (accept \(18000\text{ J}\) if \(g=10\text{ m/s}^2\) is used) [1 mark]
**(b)** - Formula used: \(P = \frac{W}{t}\) [1 mark] - Correct calculation using candidate's value from (a): \(706\text{ W}\) or \(710\text{ W}\) (accept \(720\text{ W}\) if using \(18000\text{ J}\)) [1 mark]
**(c)(i)** - Correct conversion: \(1.2\text{ kW} = 1200\text{ W}\) OR correct formula for efficiency [1 mark] - Correct calculation: \(58.8\\%\) or \(59\\%\) (accept \(60\\%\) if using \(720\text{ W}\)) [1 mark]
**(c)(ii)** - State that energy is dissipated/wasted/transferred to the surroundings as thermal energy/heat (or sound) [1 mark] - State that the motor temperature increases/heats up [1 mark]
題目 10 · structured
8 分
A ray of monochromatic light traveling in air is incident on the flat surface of a glass block. The angle of incidence is 42^{\circ}. The refractive index of the glass is 1.52. (a) Calculate the angle of refraction of the light as it enters the glass. Show your working. (b) Calculate the critical angle for the glass-air boundary. Show your working. (c) Inside the glass block, the ray of light approaches a different glass-air boundary at an angle of incidence of 45^{\circ}. State and explain what happens to the ray of light at this boundary.
**(b) Critical angle:** Using the formula: \(\sin c = \frac{1}{n}\) \(\sin c = \frac{1}{1.52} = 0.6579\) \(c = \sin^{-1}(0.6579) = 41.1^\circ\) (or \(41^\circ\))
**(c) Behavior at the second boundary:** The angle of incidence inside the glass is \(45^\circ\). Since \(45^\circ > 41.1^\circ\) (the angle of incidence is greater than the critical angle), the light ray cannot refract out into the air. Instead, it undergoes total internal reflection (TIR) back into the glass block.
評分準則
**(a)** - Use of \(n = \frac{\sin i}{\sin r}\) [1 mark] - Correct rearrangement: \(\sin r = \frac{\sin 42^\circ}{1.52}\) [1 mark] - Final answer: \(26^\circ\) or \(26.1^\circ\) [1 mark]
**(b)** - Use of \(\sin c = \frac{1}{n}\) [1 mark] - Correct substitution: \(\sin c = \frac{1}{1.52}\) [1 mark] - Final answer: \(41^\circ\) or \(41.1^\circ\) [1 mark]
**(c)** - State: Total internal reflection (or TIR) occurs [1 mark] - Explain: The angle of incidence (\(45^\circ\)) is greater than the critical angle (\(41^\circ\) / \(41.1^\circ\)) [1 mark]
Paper 6 (Alternative to Practical)
Answer all questions. Use your knowledge of practical physics to plot graphs, identify sources of error, and design experiments.
A student is investigating how the material of a beaker affects the rate of cooling of hot water. Two containers, Container A (made of glass) and Container B (made of copper), are filled with equal volumes of hot water.
(a) At the start of the experiment, a thermometer placed in Container A shows the temperature. The scale of the thermometer at \(t = 30\text{ s}\) is shown in Fig 1.1. [Fig 1.1 shows a thermometer scale with a meniscus pointing exactly to \(78.5^\circ\text{C}\)] State the temperature shown in Fig 1.1.
(b) Table 1.1 shows the temperature readings recorded by the student.
(i) Complete the column headings in Table 1.1 by writing the appropriate units. (ii) Enter the temperature value from part (a) into the table.
(c) Calculate the average cooling rate \(R\) during the first \(150\text{ s}\) of the experiment for both containers. Use the formula: \[R = \frac{\theta_{\text{initial}} - \theta_{\text{final}}}{\Delta t}\] where \(\Delta t = 150\text{ s}\).
(d) State two variables that must be kept constant to ensure a fair comparison between the cooling rates of the water in Container A and Container B.
(e) State one precaution that the student should take when reading the thermometer to ensure that the temperature values obtained are accurate.
(f) State one safety precaution required when handling hot water in this experiment.
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解題
(a) Looking at the thermometer meniscus, it lies halfway between 78 and 79, which reads \(78.5^\circ\text{C}\).
(b) (i) Time is measured in seconds (\(\text{s}\)) and temperature is measured in degrees Celsius (\(^\circ\text{C}\)). (ii) The value for Container A at \(t = 30\text{ s}\) is \(78.5\).
(d) To make it a fair test, factors that affect heat loss other than the material itself must be controlled. These include the volume of water, the initial temperature, the surrounding room temperature, and the surface area of the water exposed to the air.
(e) To avoid parallax error, the thermometer must be read with the line of sight perpendicular to the scale.
(f) Safety when dealing with boiling/hot liquids requires securing them on a stable, heat-resistant surface (heatproof mat) and avoiding direct contact with the hot container skin.
評分準則
(a) 1 Mark: Correct temperature value of \(78.5^\circ\text{C}\) (allow \(78.5\) without unit since unit is tested in b). (b) (i) 1 Mark: Correct units for both time (\(\text{s}\)) and temperature (\(^\circ\text{C}\)). (ii) 1 Mark: Correctly transferred value from (a) into the table. (c) 2 Marks: 1 mark for correct calculation of \(R_A\) (accept \(0.15\) or \(0.153\)) and 1 mark for correct calculation of \(R_B\) (accept \(0.26\) or \(0.260\)). Units must be stated (e.g. \(^\circ\text{C/s}\) or \(^\circ\text{C/min}\)). (d) 2 Marks: 1 mark for each valid control variable (up to 2). (e) 1 Mark: For mentioning eye level perpendicular to scale / avoiding parallax. (f) 1 Mark: For a valid safety action (e.g. use of heatproof mat, carrying with care, tongs).
A student determines the focal length \(f\) of a thin converging lens by placing an illuminated object on an optical bench and moving a screen until a sharp image is formed.
(a) The student sets up the apparatus. Fig 2.1 shows a portion of the metric scale on the optical bench. - The object is at the \(10.0\text{ cm}\) mark. - The lens is at the \(35.0\text{ cm}\) mark. Determine the object distance \(u\).
(b) The student obtains three sets of sharp focus measurements of object distance \(u\) and image distance \(v\). The readings are recorded in Table 2.1.
Calculate the focal length \(f\) for each set using the equation: \[f = \frac{uv}{u+v}\] Record your values in the table, and state the unit of \(f\).
(c) Calculate the average focal length, \(f_{\text{avg}}\), of the lens from your calculated values.
(d) State two difficulties encountered in this experiment when attempting to obtain an accurate position for the screen to show a sharp image.
(e) Describe one practical precaution the student should take to ensure that the measurements of \(u\) and \(v\) are as accurate as possible on the optical bench.
(d) Finding the exact point of focus is subjective because the image remains reasonably sharp over a small physical range. Ambient light can also make the image less distinct.
(e) To ensure proper geometry, the optical centers must be aligned. This avoids distortion and ensures the measured distances along the bench represent the true axial path.
評分準則
(a) 1 Mark: Correct object distance \(u = 25.0\text{ cm}\) (must include decimal place or unit if not given elsewhere). (b) 3 Marks: 1 mark for each correct calculation of \(f\) (all \(15.0\)), with 1 mark for the correct unit (\(\text{cm}\)). (c) 1 Mark: Correct average calculation matching the values in (b). (d) 2 Marks: 1 mark for each valid difficulty (e.g., subjective 'sharpness' definition, low brightness of image, screen movement increment resolution). (e) 3 Marks: 1 mark for identifying alignment (same height/coaxial), 1 mark for holding the ruler/lens perpendicular, and 1 mark for conducting the experiment in a darkened room to improve image contrast.
A student is investigating the relationship between the length of a resistance wire and its resistance.
(a) Draw a standard circuit diagram showing a cell, an ammeter, a switch, and a length of resistance wire connected in series. Include a voltmeter connected to measure the potential difference across a length \(L\) of the resistance wire.
(b) The student sets up the circuit. The ammeter reading \(I\) is shown in Fig 3.1. [Fig 3.1 shows an analogue ammeter scale with a needle pointing exactly halfway between \(0.4\text{ A}\) and \(0.5\text{ A}\)] State the value of the current \(I\).
(c) The voltmeter readings for different lengths \(L\) of the wire are shown in Table 3.1. The current is kept constant at the value determined in part (b).
Calculate and record the resistance \(R\) for each length \(L\) using the equation: \[R = \frac{V}{I}\] Record the values of \(R\) to 2 decimal places.
(d) Use your results to describe the relationship between length \(L\) and resistance \(R\).
(e) Explain why it is important to open the switch between taking readings of current and voltage.
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解題
(a) A standard series circuit contains the power source (cell), the ammeter, the test wire, and the switch in a single loop. The voltmeter is connected in parallel (shunted) across the sliding jockey connections on the wire.
(b) The needle on the ammeter is exactly halfway between 0.4 and 0.5, giving a reading of \(0.45\text{ A}\).
(c) Using \(R = \frac{V}{I}\) where \(I = 0.45\text{ A}\): - For \(20.0\text{ cm}\): \(R = \frac{0.60}{0.45} = 1.33\ \Omega\) - For \(40.0\text{ cm}\): \(R = \frac{1.20}{0.45} = 2.67\ \Omega\) - For \(60.0\text{ cm}\): \(R = \frac{1.80}{0.45} = 4.00\ \Omega\) - For \(80.0\text{ cm}\): \(R = \frac{2.40}{0.45} = 5.33\ \Omega\) - For \(100.0\text{ cm}\): \(R = \frac{3.00}{0.45} = 6.67\ \Omega\)
(d) Looking at the values: \(1.33 \times 2 = 2.66 \approx 2.67\), \(1.33 \times 3 = 3.99 \approx 4.00\), etc. Since the ratio \(R/L\) remains constant, \(R\) is directly proportional to \(L\).
(e) Keeping the switch closed causes a continuous current to flow, heating the wire. Resistance increases with temperature, which introduces systematic error.
評分準則
(a) 2 Marks: 1 mark for correct circuit symbols (cell, ammeter, voltmeter, switch, resistor/wire). 1 mark for correct connections (voltmeter in parallel across wire; others in series). (b) 1 Mark: Correctly read current \(0.45\text{ A}\). (c) 2 Marks: All 5 resistance values calculated correctly to 2 decimal places (1.33, 2.67, 4.00, 5.33, 6.67). Deduct 1 mark if not rounded consistently to 2 d.p. (d) 3 Marks: 1 mark for stating "directly proportional", 2 marks for justifying with data (e.g. "doubling the length from 20 cm to 40 cm doubles the resistance from 1.33 to 2.67 ohms"). (e) 2 Marks: 1 mark for mentioning preventing heating/temperature rise. 1 mark for linking temperature rise to a change in resistance.
A student wants to investigate how the mass of a falling object affects its terminal velocity.
Design an experiment to investigate this relationship using paper cones. You are provided with several identical paper templates (which can be folded and taped to make paper cones), adhesive tape, a electronic balance, a stopwatch, and a meter rule.
Write a plan for the experiment. You should: - Describe how you will vary the mass of the falling cones while keeping their size, shape, and surface area exposed to air resistance the same. - Describe how you will measure the terminal velocity of the cones, including the measurements you need to take. - Explain how you will ensure that the cones have reached terminal velocity before you begin timing. - Draw a table with column headings (and units) to show how you would record your results (no actual data is required). - Explain how you will use your results to reach a conclusion.
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解題
To conduct a fair test on terminal velocity versus mass, the shape and surface area facing the air flow must be kept completely constant. Stacking/nesting identical paper cones is the standard method for this.
When a cone is dropped, it initially accelerates due to gravity. After falling a certain distance, the upward air resistance balances the downward weight, and it reaches constant terminal velocity. Timing should only begin after this point. By placing two markers a known distance \(d\) apart near the floor, we can measure the time \(t\) taken to cover this interval and use \(v = \frac{d}{t}\).
A table format with units is required to record the mass, transit time, and calculated terminal velocities.
評分準則
- 2 Marks: For describing the method of nesting/stacking identical cones to vary mass while maintaining constant outer shape and surface area. - 1 Mark: For measuring mass of the cones using the electronic balance. - 2 Marks: For detailing how terminal velocity is measured (using a meter rule to mark a fixed distance \(d\), timing the fall \(t\) through this interval, and calculating \(v = d/t\)). - 1 Mark: For ensuring terminal velocity has been reached (dropping the cone from a height above the first marker so it has time to accelerate to terminal velocity before timing begins). - 2 Marks: For a clearly laid out table with correct columns, including units (e.g. Mass \(m / \text{g}\), Time \(t / \text{s}\), Velocity \(v / \text{m/s}\)). - 2 Marks: For explaining how to analyze the results (plotting a graph of terminal velocity vs mass or looking for constant ratio/trends to determine the pattern).
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