An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V1) Cambridge International A Level Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
卷二 (Extended MCQ)
Answer all 40 multiple-choice questions. For each question, choose the correct answer from the four options A, B, C or D.
40 題目 · 40 分
題目 1 · MCQ
1 分
A car travels along a straight road at a constant speed of \(15\text{ m/s}\) for \(20\text{ s}\). It then accelerates uniformly to a speed of \(25\text{ m/s}\) over the next \(10\text{ s}\). What is the total distance traveled by the car during this \(30\text{ s}\) journey?
A.400 m
B.500 m
C.550 m
D.750 m
查看答案詳解收起答案詳解
解題
In the first phase, the car travels at a constant speed of \(15\text{ m/s}\) for \(20\text{ s}\). The distance is \(d_1 = 15\text{ m/s} \times 20\text{ s} = 300\text{ m}\). In the second phase, the car accelerates uniformly from \(15\text{ m/s}\) to \(25\text{ m/s}\) over \(10\text{ s}\). The average speed is \(v_{\text{avg}} = \frac{15 + 25}{2} = 20\text{ m/s}\). The distance is \(d_2 = 20\text{ m/s} \times 10\text{ s} = 200\text{ m}\). The total distance is \(d_1 + d_2 = 300\text{ m} + 200\text{ m} = 500\text{ m}\).
評分準則
1 mark for the correct calculation of the total distance of 500 m.
題目 2 · MCQ
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 \(8.0\text{ s}\). The electrical power input to the motor is \(900\text{ W}\). What is the efficiency of the motor?
A.44%
B.53%
C.67%
D.75%
查看答案詳解收起答案詳解
解題
Useful work done in lifting the load is \(W = F \times d = 400\text{ N} \times 12\text{ m} = 4800\text{ J}\). The useful power output of the motor is \(P_{\text{out}} = \frac{W}{t} = \frac{4800\text{ J}}{8.0\text{ s}} = 600\text{ W}\). The efficiency is given by \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{600\text{ W}}{900\text{ W}} \times 100\% = 66.7\%\), which rounds to \(67\%\).
評分準則
1 mark for the correct calculation of efficiency to two significant figures.
題目 3 · MCQ
1 分
A uniform metal wire of length \(L\) and cross-sectional area \(A\) has a resistance \(R\). A second wire made of the same metal has a length of \(2L\) and twice the diameter of the first wire. What is the resistance of the second wire?
A.0.5 R
B.1.0 R
C.2.0 R
D.4.0 R
查看答案詳解收起答案詳解
解題
Resistance is defined by \(R = \rho \frac{L}{A}\). The second wire has length \(2L\) and double the diameter. Since cross-sectional area is proportional to the square of the diameter (\(A = \pi r^2 = \pi (d/2)^2\)), doubling the diameter increases the cross-sectional area by a factor of 4. Therefore, the new resistance is \(R' = \rho \frac{2L}{4A} = 0.5 R\).
評分準則
1 mark for correctly identifying that resistance is halved.
題目 4 · MCQ
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 for this boundary, and does total internal reflection occur if the angle of incidence is \(45^\circ\)?
A.Critical angle is \(42^\circ\); total internal reflection does not occur
B.Critical angle is \(42^\circ\); total internal reflection occurs
C.Critical angle is \(48^\circ\); total internal reflection does not occur
D.Critical angle is \(48^\circ\); total internal reflection occurs
查看答案詳解收起答案詳解
解題
The critical angle \(c\) is calculated using the formula \(\sin(c) = \frac{1}{n} = \frac{1}{1.50} \approx 0.667\), which gives \(c \approx 41.8^\circ\) (rounded to \(42^\circ\)). Since the angle of incidence (\(45^\circ\)) is greater than the critical angle (\(42^\circ\)) and the light is traveling from a more optically dense medium to a less optically dense medium, total internal reflection occurs.
評分準則
1 mark for identifying the correct critical angle and that total internal reflection occurs.
題目 5 · MCQ
1 分
A detector measures a total count rate of \(850\text{ counts/minute}\) near a radioactive source. The background count rate is constant at \(50\text{ counts/minute}\). The source has a half-life of \(6.0\text{ hours}\). What total count rate does the detector measure after \(18\text{ hours}\)?
A.100 counts/minute
B.106 counts/minute
C.150 counts/minute
D.250 counts/minute
查看答案詳解收起答案詳解
解題
First, subtract the background count rate to find the initial count rate from the source alone: \(850 - 50 = 800\text{ counts/minute}\). Since \(18\text{ hours}\) represents 3 half-lives (\(18 / 6.0 = 3\)), the source's activity halves 3 times: \(800 \rightarrow 400 \rightarrow 200 \rightarrow 100\text{ counts/minute}\). Finally, add the background count rate back to find the total measured count rate: \(100 + 50 = 150\text{ counts/minute}\).
評分準則
1 mark for the correct calculation of the final total count rate including background.
題目 6 · MCQ
1 分
Light from a distant galaxy is observed to have a longer wavelength than the light from a stationary source on Earth. What does this redshift indicate about the galaxy, and how does its recessional speed relate to its distance from Earth?
A.The galaxy is moving away from Earth; more distant galaxies move away slower
B.The galaxy is moving away from Earth; more distant galaxies move away faster
C.The galaxy is moving towards Earth; more distant galaxies move towards Earth slower
D.The galaxy is moving towards Earth; more distant galaxies move towards Earth faster
查看答案詳解收起答案詳解
解題
Redshift (an increase in the observed wavelength of light) indicates that the galaxy is moving away from Earth. According to Hubble's Law, the recessional speed of a galaxy is directly proportional to its distance from Earth, meaning more distant galaxies move away at a faster speed.
評分準則
1 mark for identifying both the direction of motion (away) and that more distant galaxies recede faster.
題目 7 · MCQ
1 分
A step-down transformer is used to reduce a mains voltage of \(240\text{ V}\) a.c. to \(12\text{ V}\) a.c. The primary coil has \(1200\text{ turns}\). Assuming the transformer is \(100\%\) efficient, what is the number of turns on the secondary coil and what is the current in the secondary coil if the primary current is \(0.15\text{ A}\)?
A.Number of turns = 60; secondary current = 0.0075 A
B.Number of turns = 60; secondary current = 3.0 A
C.Number of turns = 24000; secondary current = 0.0075 A
D.Number of turns = 24000; secondary current = 3.0 A
查看答案詳解收起答案詳解
解題
Using the transformer ratio equation: \(\frac{N_s}{N_p} = \frac{V_s}{V_p} \implies \frac{N_s}{1200} = \frac{12}{240} \implies N_s = 60\text{ turns}\). For a \(100\%\) efficient transformer, input power equals output power: \(I_p \times V_p = I_s \times V_s \implies 0.15\text{ A} \times 240\text{ V} = I_s \times 12\text{ V} \implies I_s = 3.0\text{ A}\).
評分準則
1 mark for the correct calculation of both the secondary turns and secondary current.
題目 8 · MCQ
1 分
A trolley of mass \(2.0\text{ kg}\) moving at a speed of \(6.0\text{ m/s}\) collides with a stationary trolley of mass \(4.0\text{ kg}\). After the collision, the two trolleys stick together. What is their common speed after the collision, and what is the loss in total kinetic energy?
A.Speed = 2.0 m/s; loss in kinetic energy = 12 J
B.Speed = 2.0 m/s; loss in kinetic energy = 24 J
C.Speed = 3.0 m/s; loss in kinetic energy = 18 J
D.Speed = 3.0 m/s; loss in kinetic energy = 27 J
查看答案詳解收起答案詳解
解題
Using conservation of momentum: \(m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \implies 2.0\text{ kg} \times 6.0\text{ m/s} + 0 = (2.0\text{ kg} + 4.0\text{ kg}) v_f \implies 12 = 6.0 v_f \implies v_f = 2.0\text{ m/s}\). Initial kinetic energy is \(E_{k,i} = \frac{1}{2} m_1 v_1^2 = 0.5 \times 2.0\text{ kg} \times (6.0\text{ m/s})^2 = 36\text{ J}\). Final kinetic energy is \(E_{k,f} = \frac{1}{2} (m_1 + m_2) v_f^2 = 0.5 \times 6.0\text{ kg} \times (2.0\text{ m/s})^2 = 12\text{ J}\). The loss in kinetic energy is \(36\text{ J} - 12\text{ J} = 24\text{ J}\).
評分準則
1 mark for both correct common velocity and kinetic energy loss calculations.
題目 9 · multiple_choice
1 分
A distant galaxy is located at a distance of \( 1.5 \times 10^{22} \text{ m} \) from Earth. The Hubble constant \( H_0 \) is estimated to be \( 2.2 \times 10^{-18} \text{ s}^{-1} \). What is the speed of this galaxy relative to Earth?
A.\( 1.5 \times 10^{4} \text{ m/s} \)
B.\( 3.3 \times 10^{4} \text{ m/s} \)
C.\( 6.8 \times 10^{39} \text{ m/s} \)
D.\( 1.5 \times 10^{22} \text{ m/s} \)
查看答案詳解收起答案詳解
解題
According to Hubble's Law, the recessional velocity \( v \) of a galaxy is directly proportional to its distance \( d \) from Earth: \( v = H_0 d \)
Substituting the given values: \( v = (2.2 \times 10^{-18} \text{ s}^{-1}) \times (1.5 \times 10^{22} \text{ m}) \) \( v = 3.3 \times 10^{4} \text{ m/s} \)
Therefore, the speed of the galaxy relative to Earth is \( 3.3 \times 10^{4} \text{ m/s} \).
評分準則
1 mark for using \( v = H_0 d \) with correct substitution and calculating the accurate velocity value.
題目 10 · multiple_choice
1 分
The count rate from a radioactive source is measured in a laboratory where the background count rate is constant at \( 20 \text{ counts/minute} \). Initially, the measured count rate is \( 340 \text{ counts/minute} \). After \( 12 \text{ hours} \), the measured count rate is \( 60 \text{ counts/minute} \). What is the half-life of the radioactive isotope?
A.\( 1.7 \text{ hours} \)
B.\( 3.0 \text{ hours} \)
C.\( 4.0 \text{ hours} \)
D.\( 6.0 \text{ hours} \)
查看答案詳解收起答案詳解
解題
First, subtract the background count rate to find the corrected count rates from the source: - Initial corrected count rate = \( 340 - 20 = 320 \text{ counts/minute} \) - Final corrected count rate = \( 60 - 20 = 40 \text{ counts/minute} \)
Determine the fraction of the initial corrected count rate remaining: \( \frac{40}{320} = \frac{1}{8} \)
Since \( \left(\frac{1}{2}\right)^3 = \frac{1}{8} \), exactly \( 3 \) half-lives have elapsed in the \( 12 \text{ hours} \) period: \( 3 \times T_{1/2} = 12 \text{ hours} \) \( T_{1/2} = 4.0 \text{ hours} \)
評分準則
1 mark for subtracting background radiation from measured rates, finding that 3 half-lives elapsed, and dividing the total time of 12 hours by 3 to obtain 4.0 hours.
題目 11 · multiple_choice
1 分
A toy car starts from rest and accelerates uniformly for \( 4.0 \text{ s} \) to a speed of \( 8.0 \text{ m/s} \). It then travels at this constant speed for \( 6.0 \text{ s} \), before decelerating uniformly to rest in a further \( 2.0 \text{ s} \). What is the total distance traveled by the toy car?
Total distance = \( 16.0 \text{ m} + 48.0 \text{ m} + 8.0 \text{ m} = 72 \text{ m} \). Alternatively, using the area of a trapezium: \( \text{Area} = \frac{1}{2} \times (a + b) \times h = \frac{1}{2} \times (6.0 + 12.0) \times 8.0 = 9.0 \times 8.0 = 72 \text{ m} \).
評分準則
1 mark for calculating the total area under the velocity-time graph correctly to find 72 m.
題目 12 · multiple_choice
1 分
An electric motor is used to lift a load of mass \( 50 \text{ kg} \) vertically upwards through a height of \( 12 \text{ m} \) in a time of \( 15 \text{ s} \). The efficiency of the motor is \( 80\% \). Take the gravitational field strength \( g \) as \( 10 \text{ N/kg} \). What is the electrical power input to the motor?
Now, use the efficiency formula to find the electrical power input: \( \text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% \) \( 0.80 = \frac{400 \text{ W}}{P_{\text{in}}} \) \( P_{\text{in}} = \frac{400}{0.80} = 500 \text{ W} \)
評分準則
1 mark for calculating the useful output power (400 W) and dividing by the efficiency (0.80) to obtain the input power of 500 W.
題目 13 · multiple_choice
1 分
An unstretched spring has a length of \( 15.0 \text{ cm} \). When a load of \( 6.0 \text{ N} \) is suspended from the spring, its length becomes \( 19.0 \text{ cm} \). Assuming the limit of proportionality is not exceeded, what load is required to stretch the spring to a total length of \( 25.0 \text{ cm} \)?
A.\( 7.9 \text{ N} \)
B.\( 9.0 \text{ N} \)
C.\( 10.0 \text{ N} \)
D.\( 15.0 \text{ N} \)
查看答案詳解收起答案詳解
解題
Using Hooke's Law, \( F = kx \), where \( x \) is the extension of the spring.
First extension: \( x_1 = 19.0 \text{ cm} - 15.0 \text{ cm} = 4.0 \text{ cm} \) The spring constant is: \( k = \frac{F_1}{x_1} = \frac{6.0 \text{ N}}{4.0 \text{ cm}} = 1.5 \text{ N/cm} \)
For a total length of \( 25.0 \text{ cm} \), the required extension is: \( x_2 = 25.0 \text{ cm} - 15.0 \text{ cm} = 10.0 \text{ cm} \)
Using Hooke's Law again: \( F_2 = k \times x_2 = 1.5 \text{ N/cm} \times 10.0 \text{ cm} = 15.0 \text{ N} \)
評分準則
1 mark for calculating both extensions correctly (4.0 cm and 10.0 cm) and using Hooke's Law proportionality to find the load of 15.0 N.
題目 14 · multiple_choice
1 分
A uniform metal wire of length \( L \) and cross-sectional diameter \( d \) has a resistance of \( 16.0 \ \Omega \). A second wire is made of the same metal, but has a length of \( 2L \) and a cross-sectional diameter of \( 2d \). What is the resistance of the second wire?
A.\( 4.0 \ \Omega \)
B.\( 8.0 \ \Omega \)
C.\( 16.0 \ \Omega \)
D.\( 32.0 \ \Omega \)
查看答案詳解收起答案詳解
解題
The resistance of a wire is given by \( R = \rho \frac{L}{A} \), where \( A \) is the cross-sectional area. Since \( A = \pi \left(\frac{d}{2}\right)^2 \), area is proportional to the square of the diameter (\( A \propto d^2 \)).
Therefore, resistance is related to length and diameter by: \( R \propto \frac{L}{d^2} \)
For the second wire: \( R_2 \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \left( \frac{L}{d^2} \right) \)
This means the resistance of the second wire is half of the first wire's resistance: \( R_2 = \frac{1}{2} \times 16.0 \ \Omega = 8.0 \ \Omega \)
評分準則
1 mark for identifying that area is proportional to the square of the diameter and using the relationship R ∝ L/d² to calculate 8.0 Ω.
題目 15 · multiple_choice
1 分
An ideal step-down transformer is connected to a \( 240 \text{ V} \) a.c. mains supply. The primary coil has \( 1200 \text{ turns} \). The secondary coil is connected to a \( 12 \text{ V} \), \( 24 \text{ W} \) lamp which operates at its normal brightness. What is the current in the primary coil?
A.\( 0.10 \text{ A} \)
B.\( 2.0 \text{ A} \)
C.\( 10 \text{ A} \)
D.\( 20 \text{ A} \)
查看答案詳解收起答案詳解
解題
Since the transformer is ideal, there are no energy losses. Therefore, the power input to the primary coil equals the power output from the secondary coil: \( P_{\text{primary}} = P_{\text{secondary}} = 24 \text{ W} \)
Using the power formula for the primary coil: \( P_{\text{primary}} = V_{\text{primary}} \times I_{\text{primary}} \) \( 24 \text{ W} = 240 \text{ V} \times I_{\text{primary}} \) \( I_{\text{primary}} = \frac{24 \text{ W}}{240 \text{ V}} = 0.10 \text{ A} \)
評分準則
1 mark for using the principle of conservation of power for an ideal transformer (P_in = P_out) to calculate a primary current of 0.10 A.
題目 16 · multiple_choice
1 分
A ray of monochromatic light travels through a glass block and meets the glass-air boundary. The refractive index of the glass is \( 1.50 \). What is the critical angle for this glass-air boundary?
A.\( 30^\circ \)
B.\( 42^\circ \)
C.\( 49^\circ \)
D.\( 56^\circ \)
查看答案詳解收起答案詳解
解題
The relationship between the refractive index \( n \) and the critical angle \( c \) is given by: \( \sin(c) = \frac{1}{n} \)
Substituting \( n = 1.50 \): \( \sin(c) = \frac{1}{1.50} \approx 0.667 \) \( c = \arcsin(0.667) \approx 41.8^\circ \)
Rounding to two significant figures gives \( 42^\circ \).
評分準則
1 mark for using the formula sin(c) = 1/n and correctly calculating the critical angle as approximately 42 degrees.
題目 17 · MCQ
1 分
Light from a distant galaxy is observed to have a longer wavelength than light from a similar source on Earth. This is redshift. Which statement about the recession velocity of the galaxy and its distance from Earth is correct?
A.The recession velocity is directly proportional to the distance from Earth.
B.The recession velocity is inversely proportional to the distance from Earth.
C.The recession velocity is independent of the distance from Earth.
D.The recession velocity is proportional to the square of the distance from Earth..
查看答案詳解收起答案詳解
解題
According to Hubble's Law, the recession velocity \(v\) of a galaxy is directly proportional to its distance \(d\) from Earth. This is represented by the equation \(v = H_0 d\), where \(H_0\) is the Hubble constant.
評分準則
1 mark for identifying that the recession velocity of a galaxy is directly proportional to its distance from Earth.
題目 18 · MCQ
1 分
An object travels along a straight track. Its velocity-time graph shows an initial uniform acceleration from \(0\text{ m/s}\) to \(12\text{ m/s}\) in \(4\text{ s}\), followed by a period of constant velocity of \(12\text{ m/s}\) for \(6\text{ s}\), and finally a uniform deceleration to rest in \(2\text{ s}\). What is the total distance travelled by the object?
A.72\text{ m}
B.96\text{ m}
C.108\text{ m}
D.144\text{ m}
查看答案詳解收起答案詳解
解題
The total distance travelled is equal to the area under the velocity-time graph. The area can be calculated as a trapezium with parallel sides of length \(6\text{ s}\) (from \(t = 4\text{ s}\) to \(10\text{ s}\)) and \(12\text{ s}\) (total time), and a height of \(12\text{ m/s}\). Area = \(0.5 \times (6 + 12) \times 12 = 108\text{ m}\). Alternatively, the sum of the areas of the three sections is: First triangle: \(0.5 \times 4 \times 12 = 24\text{ m}\). Rectangle: \(6 \times 12 = 72\text{ m}\). Second triangle: \(0.5 \times 2 \times 12 = 12\text{ m}\). Total distance = \(24 + 72 + 12 = 108\text{ m}\).
評分準則
1 mark for calculating the area under the velocity-time graph to obtain the correct total distance of 108 m.
題目 19 · MCQ
1 分
The background radiation count rate in a laboratory is \(20\text{ counts/minute}\). A radioactive sample is placed near a detector, and the initial recorded count rate is \(340\text{ counts/minute}\). After \(24\text{ hours}\), the recorded count rate is \(60\text{ counts/minute}\). What is the half-life of the sample?
A.4.0\text{ hours}
B.6.0\text{ hours}
C.8.0\text{ hours}
D.12.0\text{ hours}
查看答案詳解收起答案詳解
解題
First, subtract the background radiation to find the corrected count rates of the sample. Initial corrected count rate = \(340 - 20 = 320\text{ counts/minute}\). Final corrected count rate after \(24\text{ hours}\) = \(60 - 20 = 40\text{ counts/minute}\). The corrected count rate halved from \(320\) to \(160\), then to \(80\), and finally to \(40\). This represents exactly 3 half-lives. Therefore, 3 half-lives equal \(24\text{ hours}\), so 1 half-life is \(24 / 3 = 8.0\text{ hours}\).
評分準則
1 mark for subtracting background radiation from both measurements and correctly calculating the half-life as 8.0 hours.
題目 20 · MCQ
1 分
A trolley of mass \(2.0\text{ kg}\) moving at a velocity of \(6.0\text{ m/s}\) collides with a stationary trolley of mass \(4.0\text{ kg}\). After the collision, the two trolleys stick together and move with a common velocity. What is their common velocity?
A.1.5\text{ m/s}
B.2.0\text{ m/s}
C.3.0\text{ m/s}
D.4.0\text{ m/s}
查看答案詳解收起答案詳解
解題
Using the principle of conservation of momentum: total momentum before collision = total momentum after collision. Before collision: \(p = m_1 v_1 + m_2 v_2 = (2.0 \times 6.0) + (4.0 \times 0) = 12.0\text{ kg m/s}\). After collision: both trolleys move together with a combined mass of \(2.0 + 4.0 = 6.0\text{ kg}\) and common velocity \(v\). Thus, \(12.0 = 6.0 \times v\), which gives \(v = 2.0\text{ m/s}\).
評分準則
1 mark for applying conservation of momentum to find the common velocity of 2.0 m/s.
題目 21 · MCQ
1 分
A ray of light travels from a glass block into air. The critical angle for the glass-air boundary is \(42^\circ\). What is the refractive index of the glass?
A.0.67
B.1.2
C.1.5
D.2.2
查看答案詳解收起答案詳解
解題
The relationship between the refractive index \(n\) of a material and its critical angle \(c\) when light passes into air is given by \(n = 1 / \sin c\). Substituting \(c = 42^\circ\): \(n = 1 / \sin(42^\circ) \approx 1 / 0.6691 = 1.494\), which is approximately \(1.5\).
評分準則
1 mark for using the formula n = 1 / sin(c) and arriving at the correct refractive index of 1.5.
題目 22 · MCQ
1 分
A total charge of \(180\text{ C}\) flows through a resistor in a time of \(1.5\text{ minutes}\). What is the current in the resistor?
A.2.0\text{ A}
B.120\text{ A}
C.270\text{ A}
D.12000\text{ A}
查看答案詳解收起答案詳解
解題
Current \(I\) is calculated using the formula \(I = Q / t\), where \(Q\) is the charge and \(t\) is the time in seconds. First, convert time to seconds: \(t = 1.5 \times 60 = 90\text{ s}\). Then, calculate current: \(I = 180\text{ C} / 90\text{ s} = 2.0\text{ A}\).
評分準則
1 mark for converting minutes to seconds and calculating the current to be 2.0 A.
題目 23 · MCQ
1 分
A diver is swimming at a depth of \(15\text{ m}\) below the surface of a freshwater lake. The density of the water is \(1000\text{ kg/m}^3\) and the gravitational field strength \(g\) is \(9.8\text{ N/kg}\). If the atmospheric pressure at the surface is \(1.0 \times 10^5\text{ Pa}\), what is the total pressure acting on the diver?
A.1.5 \times 10^5\text{ Pa}
B.2.5 \times 10^5\text{ Pa}
C.3.5 \times 10^5\text{ Pa}
D.1.5 \times 10^6\text{ Pa}
查看答案詳解收起答案詳解
解題
The total pressure acting on the diver is the sum of the atmospheric pressure at the surface and the pressure due to the water. The hydrostatic pressure of the water is calculated using \(p = \rho g h = 1000\text{ kg/m}^3 \times 9.8\text{ N/kg} \times 15\text{ m} = 1.47 \times 10^5\text{ Pa}\). Total pressure = \(1.0 \times 10^5\text{ Pa} + 1.47 \times 10^5\text{ Pa} = 2.47 \times 10^5\text{ Pa}\), which to two significant figures is \(2.5 \times 10^5\text{ Pa}\).
評分準則
1 mark for calculating the pressure from the water column and adding the atmospheric pressure to get 2.5 x 10^5 Pa.
題目 24 · MCQ
1 分
A student places four metal canisters of identical size and shape, but with different surface finishes, filled with hot water at \(90^\circ\text{C}\) on a laboratory bench. Which surface finish will cause the water inside the canister to cool down the fastest by emitting thermal radiation?
A.Dull black
B.Shiny silver
C.Dull white
D.Shiny black
查看答案詳解收起答案詳解
解題
Dull black surfaces are the most effective emitters of thermal infrared radiation. Since a dull black surface radiates energy more quickly than shiny or lighter-colored surfaces, the hot water inside the dull black canister will lose thermal energy at the fastest rate and cool down the quickest.
評分準則
1 mark for identifying dull black as the most efficient emitter of thermal radiation.
題目 25 · MCQ
1 分
A toy car travels along a straight track. Between \(t = 0\) and \(t = 4.0\text{ s}\), its speed increases at a constant rate from \(0\) to \(8.0\text{ m/s}\). From \(t = 4.0\text{ s}\) to \(t = 10.0\text{ s}\), its speed remains constant at \(8.0\text{ m/s}\). What is the average speed of the toy car for the entire \(10.0\text{ s}\) journey?
A.\(4.0\text{ m/s}\)
B.\(4.8\text{ m/s}\)
C.\(6.4\text{ m/s}\)
D.\(8.0\text{ m/s}\)
查看答案詳解收起答案詳解
解題
To find the average speed, we first determine the total distance travelled by calculating the area under the speed-time graph.
- Award 1 mark for calculating total distance as the area under the speed-time graph (64 m) and dividing by total time (10 s) to get 6.4 m/s.
題目 26 · MCQ
1 分
An electric water pump with an efficiency of \(60\%\) raises \(5.0\text{ kg}\) of water through a vertical height of \(12\text{ m}\) in a time of \(8.0\text{ s}\). The acceleration of free fall \(g\) is \(9.8\text{ m/s}^2\). What is the minimum electrical power input to the pump?
A.\(44\text{ W}\)
B.\(74\text{ W}\)
C.\(120\text{ W}\)
D.\(200\text{ W}\)
查看答案詳解收起答案詳解
解題
1. Calculate the useful work output (increase in gravitational potential energy): \(W_{\text{out}} = mgh = 5.0\text{ kg} \times 9.8\text{ m/s}^2 \times 12\text{ m} = 588\text{ J}\).
2. Calculate the useful power output: \(P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{588\text{ J}}{8.0\text{ s}} = 73.5\text{ W}\).
3. Calculate the electrical power input using efficiency: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{73.5\text{ W}}{0.60} = 122.5\text{ W}\).
Rounding to two significant figures gives \(120\text{ W}\).
評分準則
- Award 1 mark for finding useful power output (73.5 W) and using the efficiency formula correctly to get 120 W.
題目 27 · MCQ
1 分
A spring has an unstretched length of \(15.0\text{ cm}\). When a load of \(6.0\text{ N}\) is hung from it, the length of the spring becomes \(21.0\text{ cm}\). Two of these identical springs are connected in parallel to support a total load of \(18.0\text{ N}\). What is the new length of each spring?
A.\(18.0\text{ cm}\)
B.\(24.0\text{ cm}\)
C.\(33.0\text{ cm}\)
D.\(48.0\text{ cm}\)
查看答案詳解收起答案詳解
解題
1. Find the extension of a single spring with a \(6.0\text{ N}\) load: \(\text{Extension } x_1 = 21.0\text{ cm} - 15.0\text{ cm} = 6.0\text{ cm}\).
2. Find the spring constant \(k\) of a single spring: \(k = \frac{F}{x_1} = \frac{6.0\text{ N}}{6.0\text{ cm}} = 1.0\text{ N/cm}\).
3. For two identical springs in parallel sharing an \(18.0\text{ N}\) load: Each spring carries half of the total load: \(F_{\text{each}} = \frac{18.0\text{ N}}{2} = 9.0\text{ N}\).
4. Calculate the extension \(x_2\) of each spring under this shared load: \(x_2 = \frac{F_{\text{each}}}{k} = \frac{9.0\text{ N}}{1.0\text{ N/cm}} = 9.0\text{ cm}\).
5. Calculate the new length of each spring: \(\text{New Length} = \text{Unstretched Length} + x_2 = 15.0\text{ cm} + 9.0\text{ cm} = 24.0\text{ cm}\).
評分準則
- Award 1 mark for calculating the correct load per spring (9.0 N), finding the new extension (9.0 cm), and adding it to the unstretched length to get 24.0 cm.
題目 28 · MCQ
1 分
A tennis ball of mass \(0.20\text{ kg}\) hits a wall horizontally at a speed of \(6.0\text{ m/s}\). It rebounds horizontally in the opposite direction with a speed of \(4.0\text{ m/s}\). The ball is in contact with the wall for \(0.050\text{ s}\). What is the magnitude of the average force exerted on the ball by the wall?
A.\(8.0\text{ N}\)
B.\(24\text{ N}\)
C.\(40\text{ N}\)
D.\(200\text{ N}\)
查看答案詳解收起答案詳解
解題
1. Determine the velocity vectors, taking the initial direction toward the wall as positive: Initial velocity, \(u = +6.0\text{ m/s}\) Final velocity, \(v = -4.0\text{ m/s}\)
2. Calculate the change in momentum (impulse): \(\Delta p = m(v - u) = 0.20\text{ kg} \times (-4.0\text{ m/s} - 6.0\text{ m/s}) = 0.20 \times (-10.0) = -2.0\text{ kg m/s}\). So the magnitude of the change in momentum is \(2.0\text{ N s}\).
3. Calculate the average force using Newton's second law in terms of momentum: \(F = \frac{\Delta p}{\Delta t} = \frac{2.0\text{ N s}}{0.050\text{ s}} = 40\text{ N}\).
評分準則
- Award 1 mark for correctly determining the magnitude of the change in momentum (2.0 kg m/s) and dividing by the contact time to get 40 N.
題目 29 · MCQ
1 分
The constant background radiation count rate in a laboratory is \(24\text{ counts/minute}\). A radioactive source is placed near a detector, and the total measured count rate at time \(t = 0\) is \(360\text{ counts/minute}\). After \(12.0\text{ hours}\), the total measured count rate is \(66\text{ counts/minute}\). What is the half-life of the radioactive isotope?
A.\(2.0\text{ hours}\)
B.\(3.0\text{ hours}\)
C.\(4.0\text{ hours}\)
D.\(6.0\text{ hours}\)
查看答案詳解收起答案詳解
解題
1. Subtract background radiation to find the count rate due only to the source: - At \(t = 0\): \(R_0 = 360 - 24 = 336\text{ counts/minute}\). - At \(t = 12.0\text{ hours}\): \(R_t = 66 - 24 = 42\text{ counts/minute}\).
2. Find the fraction of source activity remaining: \(\frac{R_t}{R_0} = \frac{42}{336} = \frac{1}{8}\).
3. Express this fraction in terms of half-lives: \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), so exactly \(3\) half-lives have elapsed.
- Award 1 mark for subtracting background radiation from both measurements, identifying that 3 half-lives have elapsed, and calculating the correct half-life of 4.0 hours.
題目 30 · MCQ
1 分
A ray of light is travelling inside a glass block toward a boundary with air. The refractive index of the glass is \(1.50\). The angle of incidence at the boundary is \(45^\circ\). What is the critical angle \(\theta_c\) for the glass-air boundary, and what happens to the ray of light?
A.\(\theta_c = 42^\circ\); the ray undergoes total internal reflection
B.\(\theta_c = 49^\circ\); the ray undergoes total internal reflection
C.\(\theta_c = 42^\circ\); the ray refracts into the air
D.\(\theta_c = 49^\circ\); the ray refracts into the air
查看答案詳解收起答案詳解
解題
1. Use the critical angle formula: \(\sin(\theta_c) = \frac{1}{n} = \frac{1}{1.50} \approx 0.667\). \(\theta_c = \sin^{-1}(0.667) \approx 41.8^\circ\), which rounds to \(42^\circ\).
2. Compare the angle of incidence \(i = 45^\circ\) with the critical angle \(\theta_c = 42^\circ\): Since the angle of incidence is greater than the critical angle (\(45^\circ > 42^\circ\)), the light cannot refract into the air. Instead, all the light is reflected back into the glass, meaning it undergoes total internal reflection.
評分準則
- Award 1 mark for calculating the critical angle as 42° and correctly identifying that total internal reflection occurs because the angle of incidence is greater than the critical angle.
題目 31 · MCQ
1 分
A cylindrical copper wire has length \(L\), diameter \(d\), and resistance \(R\). A second copper wire at the same temperature has length \(3L\) and diameter \(2d\). What is the resistance of the second copper wire in terms of \(R\)?
A.\(0.38 R\)
B.\(0.75 R\)
C.\(1.5 R\)
D.\(6.0 R\)
查看答案詳解收起答案詳解
解題
1. Use the formula for electrical resistance of a uniform wire: \(R = \rho \frac{\text{length}}{\text{area}}\)
2. The cross-sectional area of a cylindrical wire of diameter \(d\) is: \(A = \pi \left(\frac{d}{2}\right)^2 \propto d^2\)
3. This gives the proportional relationship: \(R \propto \frac{L}{d^2}\)
4. For the second wire, let the resistance be \(R'\): \(R' \propto \frac{3L}{(2d)^2} = \frac{3L}{4d^2} = 0.75 \left(\frac{L}{d^2}\right)\)
5. Therefore, \(R' = 0.75 R\).
評分準則
- Award 1 mark for applying the resistivity relation to show that resistance is proportional to length and inversely proportional to the square of diameter, resulting in 0.75 R.
題目 32 · MCQ
1 分
Astronomical observations show that light from distant galaxies is redshifted. Which statement correctly describes the relationship between redshift and the movement of galaxies?
A.Light from more distant galaxies shows a smaller redshift, indicating that they are moving away from us more slowly.
B.Light from more distant galaxies shows a greater redshift, indicating that they are moving away from us more quickly.
C.Light from more distant galaxies shows a smaller redshift, indicating that they are moving away from us more quickly.
D.Light from more distant galaxies shows a greater redshift, indicating that they are moving away from us more slowly.
查看答案詳解收起答案詳解
解題
According to Hubble's Law, the recessional velocity \(v\) of a galaxy is directly proportional to its distance \(d\) from Earth (\(v = H_0 d\)).
This means: - More distant galaxies are moving away from us more quickly. - This faster movement causes a greater shift in spectral lines toward longer, redder wavelengths, resulting in a larger (greater) redshift.
Therefore, light from more distant galaxies shows a greater redshift, indicating that they are moving away from us more quickly.
評分準則
- Award 1 mark for identifying that more distant galaxies have a greater redshift and that this indicates they are moving away at a faster speed.
題目 33 · MCQ
1 分
A distant galaxy is observed to have a speed of recession of \( 1.4 \times 10^4 \text{ km/s} \). Assuming the Hubble constant is \( 70 \text{ km/(s Mpc)} \), what is the approximate distance to this galaxy from Earth?
A.\( 2.0 \times 10^{-3} \text{ Mpc} \)
B.\( 5.0 \times 10^{-3} \text{ Mpc} \)
C.\( 200 \text{ Mpc} \)
D.\( 9.8 \times 10^5 \text{ Mpc} \)
查看答案詳解收起答案詳解
解題
Using Hubble's Law: \( v = H_0 d \), where \( v \) is the speed of recession, \( H_0 \) is the Hubble constant, and \( d \) is the distance. Rearranging the formula to solve for distance gives: \( d = \frac{v}{H_0} \). Substituting the given values: \( d = \frac{1.4 \times 10^4 \text{ km/s}}{70 \text{ km/(s Mpc)}} = 200 \text{ Mpc} \).
評分準則
1 mark for the correct substitution into Hubble's Law formula and correct calculation of distance.
題目 34 · MCQ
1 分
An object starts from rest and accelerates uniformly at \( 2.5 \text{ m/s}^2 \) for \( 6.0 \text{ s} \). It then travels at a constant velocity for \( 4.0 \text{ s} \), before decelerating uniformly to rest in a further \( 2.0 \text{ s} \). What is the total distance traveled by the object?
A.\( 15 \text{ m} \)
B.\( 90 \text{ m} \)
C.\( 120 \text{ m} \)
D.\( 180 \text{ m} \)
查看答案詳解收起答案詳解
解題
The motion can be split into three phases: 1. Acceleration: The final velocity reached is \( v = a t = 2.5 \times 6.0 = 15 \text{ m/s} \). The distance covered is represented by the area under this phase on a velocity-time graph: \( s_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6.0 \times 15 = 45 \text{ m} \). 2. Constant velocity: The velocity remains \( 15 \text{ m/s} \) for \( 4.0 \text{ s} \). Distance covered: \( s_2 = 15 \times 4.0 = 60 \text{ m} \). 3. Deceleration: The object decelerates from \( 15 \text{ m/s} \) to rest in \( 2.0 \text{ s} \). Distance covered: \( s_3 = \frac{1}{2} \times 2.0 \times 15 = 15 \text{ m} \). Total distance \( s = s_1 + s_2 + s_3 = 45 + 60 + 15 = 120 \text{ m} \).
評分準則
1 mark for the correct calculation of maximum velocity and summing the correct distance values for all three phases of motion.
題目 35 · MCQ
1 分
The background radiation count rate in a laboratory is steady at \( 15 \text{ counts/minute} \). A radioactive sample is placed near a detector, and the total measured count rate is \( 135 \text{ counts/minute} \). After a time period of \( 6.0 \text{ hours} \), the total measured count rate is \( 30 \text{ counts/minute} \). What is the half-life of the radioactive sample?
A.\( 1.5 \text{ hours} \)
B.\( 2.0 \text{ hours} \)
C.\( 3.0 \text{ hours} \)
D.\( 4.5 \text{ hours} \)
查看答案詳解收起答案詳解
解題
First, subtract the background radiation to find the corrected count rates from the source. Initial corrected count rate = \( 135 - 15 = 120 \text{ counts/minute} \). Final corrected count rate = \( 30 - 15 = 15 \text{ counts/minute} \). The ratio of final to initial corrected activity is \( \frac{15}{120} = \frac{1}{8} \). Since \( \frac{1}{8} = \left(\frac{1}{2}\right)^3 \), exactly 3 half-lives have elapsed. Therefore, the half-life is \( \frac{6.0 \text{ hours}}{3} = 2.0 \text{ hours} \).
評分準則
1 mark for correcting the count rates for background radiation, identifying that 3 half-lives have elapsed, and calculating the half-life correctly.
題目 36 · MCQ
1 分
An electric motor with an efficiency of \( 75\% \) is used to lift a load of mass \( 120 \text{ kg} \) vertically upwards through a height of \( 5.0 \text{ m} \) in a time of \( 4.0 \text{ s} \). What is the electrical power input to the motor? (Take \( g = 9.8 \text{ N/kg} \))
A.\( 1.10 \text{ kW} \)
B.\( 1.47 \text{ kW} \)
C.\( 1.96 \text{ kW} \)
D.\( 2.61 \text{ kW} \)
查看答案詳解收起答案詳解
解題
The useful work done in lifting the load is equal to the gain in gravitational potential energy: \( E_p = mgh = 120 \text{ kg} \times 9.8 \text{ N/kg} \times 5.0 \text{ m} = 5880 \text{ J} \). The useful power output of the motor is: \( P_{\text{out}} = \frac{\text{Work Done}}{t} = \frac{5880 \text{ J}}{4.0 \text{ s}} = 1470 \text{ W} \). Since efficiency is \( 75\% \) (or 0.75), the required electrical power input is: \( P_{\text{in}} = \frac{P_{\text{out}}}{0.75} = \frac{1470 \text{ W}}{0.75} = 1960 \text{ W} = 1.96 \text{ kW} \).
評分準則
1 mark for calculating work done and power output, and correctly dividing by efficiency to find the power input.
題目 37 · MCQ
1 分
A ray of light travels inside a glass block towards a boundary with air. The refractive index of the glass is \( 1.62 \). What is the critical angle for light in this glass block, and what happens to the ray if its angle of incidence at the boundary is \( 40^\circ \)?
A.Critical angle is \( 38.1^\circ \); the ray is totally internally reflected.
B.Critical angle is \( 38.1^\circ \); the ray is refracted into the air.
C.Critical angle is \( 51.9^\circ \); the ray is totally internally reflected.
D.Critical angle is \( 51.9^\circ \); the ray is refracted into the air.
查看答案詳解收起答案詳解
解題
The critical angle \( c \) is related to the refractive index by: \( \sin(c) = \frac{1}{n} = \frac{1}{1.62} \approx 0.6173 \). Calculating the inverse sine gives: \( c \approx 38.1^\circ \). Since the angle of incidence \( i = 40^\circ \) is greater than the critical angle \( 38.1^\circ \), total internal reflection occurs.
評分準則
1 mark for the correct calculation of the critical angle and correctly identifying that total internal reflection occurs since the angle of incidence is greater than the critical angle.
題目 38 · MCQ
1 分
Two identical springs, each having a spring constant of \( 120 \text{ N/m} \), are connected in parallel to support a load of mass \( 6.0 \text{ kg} \). What is the extension of this spring system? (Take \( g = 9.8 \text{ N/kg} \))
A.\( 12.3 \text{ cm} \)
B.\( 24.5 \text{ cm} \)
C.\( 49.0 \text{ cm} \)
D.\( 98.0 \text{ cm} \)
查看答案詳解收起答案詳解
解題
The total downward force exerted by the load is: \( F = mg = 6.0 \text{ kg} \times 9.8 \text{ N/kg} = 58.8 \text{ N} \). When two identical springs are arranged in parallel, the effective spring constant \( k_{\text{eff}} \) of the combination is: \( k_{\text{eff}} = k_1 + k_2 = 120 + 120 = 240 \text{ N/m} \). Using Hooke's Law \( F = k_{\text{eff}} x \), the extension is: \( x = \frac{F}{k_{\text{eff}}} = \frac{58.8 \text{ N}}{240 \text{ N/m}} = 0.245 \text{ m} = 24.5 \text{ cm} \).
評分準則
1 mark for calculating the force, finding the effective spring constant for a parallel arrangement, and correctly solving for the extension in centimeters.
題目 39 · MCQ
1 分
An ideal step-down transformer has 400 turns on its primary coil and 100 turns on its secondary coil. The primary coil is connected to a \( 240 \text{ V} \) alternating current (a.c.) power supply. A resistor of resistance \( 15 \ \Omega \) is connected across the secondary coil. What is the current in the primary coil?
A.\( 0.25 \text{ A} \)
B.\( 1.0 \text{ A} \)
C.\( 4.0 \text{ A} \)
D.\( 16 \text{ A} \)
查看答案詳解收起答案詳解
解題
First, find the secondary voltage \( V_s \) using the transformer equation: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = 240 \times \frac{100}{400} = 60 \text{ V} \). Second, find the secondary current \( I_s \) using Ohm's Law on the secondary circuit: \( I_s = \frac{V_s}{R} = \frac{60 \text{ V}}{15 \ \Omega} = 4.0 \text{ A} \). Finally, for an ideal transformer, the input power equals the output power: \( V_p I_p = V_s I_s \implies I_p = I_s \times \frac{V_s}{V_p} = 4.0 \times \frac{60}{240} = 1.0 \text{ A} \).
評分準則
1 mark for calculating the secondary voltage, secondary current, and applying power conservation to find the correct primary current.
題目 40 · MCQ
1 分
A charge of \( 360 \text{ C} \) passes through a lamp in a time of \( 5.0 \text{ minutes} \). What is the current in the lamp, and how many electrons pass through the lamp during this time? (The charge on an electron is \( 1.6 \times 10^{-19} \text{ C} \))
A.Current = \( 1.2 \text{ A} \); Number of electrons = \( 2.25 \times 10^{21} \)
B.Current = \( 1.2 \text{ A} \); Number of electrons = \( 5.76 \times 10^{-17} \)
C.Current = \( 72 \text{ A} \); Number of electrons = \( 2.25 \times 10^{21} \)
D.Current = \( 72 \text{ A} \); Number of electrons = \( 5.76 \times 10^{-17} \)
查看答案詳解收起答案詳解
解題
First, convert time into seconds: \( t = 5.0 \times 60 = 300 \text{ s} \). The electric current is calculated as: \( I = \frac{Q}{t} = \frac{360 \text{ C}}{300 \text{ s}} = 1.2 \text{ A} \). The number of electrons \( N \) is the total charge divided by the charge on a single electron: \( N = \frac{Q}{e} = \frac{360 \text{ C}}{1.6 \times 10^{-19} \text{ C}} = 2.25 \times 10^{21} \).
評分準則
1 mark for converting minutes to seconds, calculating the correct current, and dividing the charge by the elementary charge to find the correct number of electrons.
Paper 4 (Extended Theory)
Answer all structured questions in the spaces provided on the question paper. Show all your working and use appropriate units.
10 題目 · 80 分
題目 1 · Structured
8 分
Answer the following questions about stars and cosmology. (a) Describe how a protostar becomes a stable main sequence star, and explain the physical factors that keep it in a state of stable equilibrium. (b) State and describe what happens to a star with a mass much greater than the Sun after it leaves the main sequence stage, up to the end of its lifecycle. (c) Light observed from distant galaxies exhibits redshift. Explain how this redshift observation supports the Big Bang theory.
查看答案詳解收起答案詳解
解題
(a) A protostar forms when gravitational forces pull interstellar dust and gas together, raising the internal temperature and pressure. Once the core is hot enough, nuclear fusion of hydrogen into helium begins, releasing energy. The star reaches stable equilibrium (main sequence) when the outward thermal pressure and radiation pressure from fusion perfectly balance the inward gravitational attraction. (b) When a high-mass star runs out of hydrogen fuel in its core, it expands and cools to become a red supergiant. Eventually, its core collapses catastrophically, triggering a massive explosion called a supernova. The remnants of the core collapse to form either an extremely dense neutron star or, if the mass is large enough, a black hole. (c) Light from distant galaxies is shifted to longer wavelengths (redshift). This indicates that these galaxies are moving away from us. Observations show that more distant galaxies are moving away faster (recessional velocity is proportional to distance). This expansion of space implies that in the past, all matter in the Universe must have started from a single, extremely dense and hot point, supporting the Big Bang theory.
評分準則
(a) [3 marks total] - Gravitational attraction pulls dust/gas together and increases temperature/pressure (1) - Nuclear fusion of hydrogen begins in the core (1) - Inward gravitational collapse is balanced by outward thermal/radiation pressure (1). (b) [3 marks total] - Star expands and cools to become a red supergiant (1) - Star explodes as a supernova (1) - Core collapses to form a neutron star or a black hole (1). (c) [2 marks total] - Redshift indicates distant galaxies are moving away / space is expanding (1) - Shows that all matter in the Universe was once concentrated at a single point (1).
題目 2 · Structured
8 分
A toy car of mass \(1.2\text{ kg}\) starts from rest and accelerates uniformly to a velocity of \(15\text{ m/s}\) in a time of \(6.0\text{ s}\). (a) Calculate the acceleration of the car during these first \(6.0\text{ s}\). (b) Calculate the resultant force acting on the car during this acceleration. (c) After the first \(6.0\text{ s}\), the car travels at a constant velocity of \(15\text{ m/s}\) for a further \(10\text{ s}\), and then decelerates uniformly to rest in \(4.0\text{ s}\). (i) Calculate the total distance traveled by the car during the entire \(20\text{ s}\) motion. (ii) State the value of the acceleration when the car is traveling at the constant velocity.
查看答案詳解收起答案詳解
解題
(a) Using the acceleration formula: \(a = \frac{v - u}{t}\). Here, \(u = 0\text{ m/s}\), \(v = 15\text{ m/s}\), and \(t = 6.0\text{ s}\). So, \(a = \frac{15 - 0}{6.0} = 2.5\text{ m/s}^2\). (b) Using Newton's second law: \(F = m a\). Here, \(m = 1.2\text{ kg}\) and \(a = 2.5\text{ m/s}^2\). So, \(F = 1.2 \times 2.5 = 3.0\text{ N}\). (c) (i) The motion consists of three parts. The total distance is the total area under the velocity-time graph: - Part 1 (acceleration, \(0\) to \(6\text{ s}\)): \(\text{Distance}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6.0 \times 15 = 45\text{ m}\). - Part 2 (constant velocity, \(6\) to \(16\text{ s}\)): \(\text{Distance}_2 = \text{base} \times \text{height} = 10 \times 15 = 150\text{ m}\). - Part 3 (deceleration, \(16\) to \(20\text{ s}\)): \(\text{Distance}_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0 \times 15 = 30\text{ m}\). - Total Distance = \(45 + 150 + 30 = 225\text{ m}\). (c) (ii) Since the velocity is constant during this phase, there is no change in velocity, so the acceleration is \(0\text{ m/s}^2\).
評分準則
(a) [2 marks total] - Correct formula or working: \(a = \frac{15}{6}\) (1) - Correct value and unit: \(2.5\text{ m/s}^2\) (1). (b) [2 marks total] - Correct formula or working: \(F = 1.2 \times 2.5\) (1) - Correct value and unit: \(3.0\text{ N}\) (1). (c) (i) [3 marks total] - Calculation of distance during acceleration (45 m) or deceleration (30 m) (1) - Calculation of distance during constant speed (150 m) (1) - Correct total distance: \(225\text{ m}\) (1) [Alternative: correct use of trapezoid area formula: \(\frac{1}{2} \times (10 + 20) \times 15 = 225\text{ m}\) gets 3 marks]. (c) (ii) [1 mark] - Correct answer: \(0\) / \(0\text{ m/s}^2\) (1).
題目 3 · Structured
8 分
(a) Carbon-14 (\(^{14}_{6}\text{C}\)) is a radioactive isotope that undergoes beta-minus (\(\beta^-\)) decay to form nitrogen (\(\text{N}\)). Write down the complete balanced nuclear equation for this decay. (b) A physicist measures the activity of a radioactive isotope in a laboratory. The constant background radiation count rate is measured to be \(25\text{ counts/minute}\). The initial count rate measured from the isotope sample (including background) is \(425\text{ counts/minute}\). After a time of \(15\text{ hours}\), the measured count rate (including background) drops to \(75\text{ counts/minute}\). (i) State the corrected initial count rate of the sample. (ii) Calculate the half-life of this radioactive isotope.
查看答案詳解收起答案詳解
解題
(a) In beta-minus decay, a neutron decays into a proton and an electron (beta particle). The nucleon number remains unchanged at \(14\), and the proton number increases by \(1\) from \(6\) to \(7\). The beta particle is represented as \(^{0}_{-1}\text{e}\) (or \(^{0}_{-1}\beta\)). The balanced nuclear equation is: \(^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}\text{e}\) (b) (i) To find the corrected initial count rate, subtract the background radiation: \(425\text{ counts/minute} - 25\text{ counts/minute} = 400\text{ counts/minute}\). (b) (ii) Find the corrected final count rate after \(15\text{ hours}\): \(75\text{ counts/minute} - 25\text{ counts/minute} = 50\text{ counts/minute}\). Now, determine how many half-lives have passed as the corrected activity drops from \(400\) to \(50\): \(400 \xrightarrow{1\text{ half-life}} 200 \xrightarrow{2\text{ half-lives}} 100 \xrightarrow{3\text{ half-lives}} 50\). Therefore, exactly \(3\) half-lives have elapsed in \(15\text{ hours}\). \(\text{Half-life} = \frac{15\text{ hours}}{3} = 5.0\text{ hours}\).
評分準則
(a) [3 marks total] - Correct reactant side: \(^{14}_{6}\text{C}\) (1) - Correct product nitrogen: \(^{14}_{7}\text{N}\) (1) - Correct beta particle: \(^{0}_{-1}\text{e}\) or \(^{0}_{-1}\beta\) (1). (b) (i) [1 mark] - Correct calculation: \(400\text{ counts/minute}\) (1). (b) (ii) [4 marks total] - Calculate corrected final count rate: \(75 - 25 = 50\text{ counts/minute}\) (1) - Deduce that the activity has halved 3 times (or fraction is \(\frac{1}{8}\)) (1) - Set up equation: \(3 \times t_{1/2} = 15\text{ hours}\) (1) - Correct final value with unit: \(5.0\text{ hours}\) (1).
題目 4 · Structured
8 分
(a) Define the term *power*. (b) An electric motor is used to lift a metal block of mass \(45\text{ kg}\) vertically through a height of \(12\text{ m}\). (i) Calculate the increase in gravitational potential energy (\(\Delta E_p\)) of the block. (Take \(g = 9.8\text{ m/s}^2\)). (ii) The motor takes \(8.0\text{ s}\) to lift the block. During this time, the electrical power input to the motor is \(900\text{ W}\). Calculate the percentage efficiency of the motor system. (c) Suggest two reasons why the efficiency of the motor system is less than \(100\%\).
查看答案詳解收起答案詳解
解題
(a) Power is defined as the work done per unit time, or the rate at which energy is transferred: \(P = \frac{W}{t}\). (b) (i) The change in gravitational potential energy is given by: \(\Delta E_p = m g h\). Here, \(m = 45\text{ kg}\), \(g = 9.8\text{ m/s}^2\), and \(h = 12\text{ m}\). So, \(\Delta E_p = 45 \times 9.8 \times 12 = 5292\text{ J}\) (or \(5300\text{ J}\) to 2 s.f.). (b) (ii) The total electrical energy input is: \(E_{\text{input}} = P_{\text{input}} \times t = 900\text{ W} \times 8.0\text{ s} = 7200\text{ J}\). Useful work done (output energy) is the change in gravitational potential energy, \(5292\text{ J}\). \(\text{Efficiency} = \left( \frac{\text{Useful output energy}}{\text{Total input energy}} \right) \times 100\% = \left( \frac{5292}{7200} \right) \times 100\% = 73.5\%\) (or \(74\%\) to 2 s.f.). (c) The efficiency is less than \(100\%\) because energy is dissipated as waste thermal energy. This is due to: 1. Electrical resistance in the coils/wires of the motor (Joule heating). 2. Friction in the moving parts (motor bearings or pulley system) which transfers kinetic energy into thermal energy.
評分準則
(a) [1 mark] - Correct definition: rate of doing work OR rate of energy transfer OR work done per unit time (1). (b) (i) [2 marks total] - Correct formula or working: \(45 \times 9.8 \times 12\) (1) - Correct value with unit: \(5292\text{ J}\) or \(5300\text{ J}\) (1). (b) (ii) [3 marks total] - Calculate input energy: \(900 \times 8 = 7200\text{ J}\) OR useful output power: \(\frac{5292}{8} = 661.5\text{ W}\) (1) - Correct substitution in efficiency formula: \(\frac{5292}{7200}\) or \(\frac{661.5}{900}\) (1) - Correct final percentage: \(73.5\%\) or \(74\%\) (1). (c) [2 marks total] - Thermal energy lost in coils due to electrical resistance / heating (1) - Thermal energy lost due to friction in bearings / moving parts (1).
題目 5 · Structured
8 分
A ray of light traveling in air enters a semi-circular glass block of refractive index \(1.52\) at an angle of incidence of \(40^\circ\). (a) (i) Calculate the angle of refraction inside the glass block. (ii) Calculate the speed of light inside the glass block. (Speed of light in air = \(3.0 \times 10^8\text{ m/s}\)). (b) The ray of light travels through the glass block and meets the straight boundary back into air. (i) Define the term *critical angle*. (ii) Calculate the critical angle for this glass-air boundary.
查看答案詳解收起答案詳解
解題
(a) (i) Using Snell's law at the air-to-glass boundary: \(n = \frac{\sin i}{\sin r}\). Heres, \(n = 1.52\), \(i = 40^\circ\). \(1.52 = \frac{\sin 40^\circ}{\sin r} \Rightarrow \sin r = \frac{\sin 40^\circ}{1.52} = \frac{0.6428}{1.52} \approx 0.4229\). \(r = \sin^{-1}(0.4229) \approx 25.0^\circ\). (a) (ii) Using the relationship between refractive index and the speed of light: \(n = \frac{c}{v}\). \(1.52 = \frac{3.0 \times 10^8}{v} \Rightarrow v = \frac{3.0 \times 10^8}{1.52} \approx 1.97 \times 10^8\text{ m/s}\) (or \(2.0 \times 10^8\text{ m/s}\)). (b) (i) The critical angle is defined as the angle of incidence in the optically denser medium for which the angle of refraction is \(90^\circ\) in the less dense medium. (b) (ii) Using the critical angle formula: \(\sin c = \frac{1}{n}\). \(\sin c = \frac{1}{1.52} \approx 0.6579\). \(c = \sin^{-1}(0.6579) \approx 41.1^\circ\) (or \(41^\circ\)).
評分準則
(a) (i) [2 marks total] - Correct formula or substitution: \(1.52 = \frac{\sin 40^\circ}{\sin r}\) (1) - Correct angle: \(25^\circ\) or \(25.0^\circ\) (1). (a) (ii) [2 marks total] - Correct formula or substitution: \(v = \frac{3.0 \times 10^8}{1.52}\) (1) - Correct value with unit: \(1.97 \times 10^8\text{ m/s}\) or \(2.0 \times 10^8\text{ m/s}\) (1). (b) (i) [2 marks total] - Angle of incidence in denser medium (1) - Angle of refraction is \(90^\circ\) in less dense medium (1). (b) (ii) [2 marks total] - Correct formula or substitution: \(\sin c = \frac{1}{1.52}\) (1) - Correct angle: \(41.1^\circ\) or \(41^\circ\) (1).
題目 6 · Structured
8 分
A student conducts an experiment to investigate Hooke's Law using a spring. (a) State Hooke's Law. (b) The original length of the spring is \(15.0\text{ cm}\). When a load of \(6.0\text{ N}\) is hung from the spring, its length becomes \(21.0\text{ cm}\). (i) Show that the spring constant \(k\) is \(100\text{ N/m}\). (ii) When a total load of \(14.0\text{ N}\) is applied, the length of the spring becomes \(31.0\text{ cm}\). Determine whether the spring has exceeded its limit of proportionality. Show your calculations clearly. (c) Suggest two precautions that the student should take during this experiment to obtain accurate length measurements.
查看答案詳解收起答案詳解
解題
(a) Hooke's Law states that the extension of a spring is directly proportional to the force (load) applied to it, provided the limit of proportionality is not exceeded. (b) (i) First, calculate the extension \(x\) for a \(6.0\text{ N}\) load: \(x = \text{New length} - \text{Original length} = 21.0\text{ cm} - 15.0\text{ cm} = 6.0\text{ cm} = 0.060\text{ m}\). Using Hooke's Law \(F = k x\): \(k = \frac{F}{x} = \frac{6.0\text{ N}}{0.060\text{ m}} = 100\text{ N/m}\) (as shown). (b) (ii) If the spring is still proportional, the expected extension \(x_{\text{expected}}\) for a load of \(14.0\text{ N}\) is: \(x_{\text{expected}} = \frac{F}{k} = \frac{14.0\text{ N}}{100\text{ N/m}} = 0.140\text{ m} = 14.0\text{ cm}\). This would give an expected length of: \(15.0\text{ cm} + 14.0\text{ cm} = 29.0\text{ cm}\). However, the actual measured length is \(31.0\text{ cm}\), which means the actual extension is \(31.0\text{ cm} - 15.0\text{ cm} = 16.0\text{ cm}\). Since the actual extension (\(16.0\text{ cm}\)) is greater than the proportional extension (\(14.0\text{ cm}\)), the spring has exceeded its limit of proportionality. (c) Two precautions to take: 1. View the ruler scale at eye level (perpendicularly) to avoid parallax errors. 2. Attach a horizontal pointer (fiducial marker) to the bottom of the spring to align accurately with the ruler scale.
評分準則
(a) [1 mark] - Extension is directly proportional to load / force (1) [do not accept unless 'limit of proportionality' is mentioned or implied, e.g., 'until it reaches a limit']. (b) (i) [2 marks total] - Calculate extension: \(21.0 - 15.0 = 6.0\text{ cm} = 0.06\text{ m}\) (1) - Correct calculation: \(k = \frac{6}{0.06} = 100\text{ N/m}\) (1). (b) (ii) [3 marks total] - Calculate actual extension for 14 N: \(31.0 - 15.0 = 16.0\text{ cm}\) (1) - Calculate expected proportional extension: \(x = \frac{14}{100} = 14.0\text{ cm}\) (1) - Draw correct conclusion based on comparing \(16.0\text{ cm} > 14.0\text{ cm}\) (1). (c) [2 marks total] - Any two of: Read scale at eye level to avoid parallax (1) / Use horizontal pointer (fiducial marker) (1) / Ensure the ruler is vertical (1) / Wait for the spring to stop moving before reading (1).
題目 7 · Structured
8 分
(a) Explain the principle of operation of a step-down transformer. In your answer, refer to magnetic fields and electromagnetic induction. (b) A transformer is designed to step down a mains supply of \(240\text{ V}\) a.c. to \(12\text{ V}\) a.c. in order to power a lamp. (i) The primary coil of the transformer has \(1200\text{ turns}\). Calculate the number of turns in the secondary coil. (ii) The lamp has a power rating of \(36\text{ W}\). Assuming the transformer is \(100\%\) efficient, calculate the current in the primary coil when the lamp is operating at normal brightness. (iii) State one structural feature of a real transformer that reduces thermal energy losses.
查看答案詳解收起答案詳解
解題
(a) An alternating current (a.c.) in the primary coil creates a continuously changing magnetic field around it. This magnetic field is guided by the soft iron core to pass through the secondary coil. Because the magnetic field is constantly changing, it cuts through the secondary coil and induces an alternating electromotive force (e.m.f. / voltage) across its ends by electromagnetic induction. (b) (i) Using the transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). \(\frac{240}{12} = \frac{1200}{N_s} \Rightarrow 20 = \frac{1200}{N_s} \Rightarrow N_s = \frac{1200}{20} = 60\text{ turns}\). (b) (ii) Since the transformer is \(100\%\) efficient, \(\text{Power input} = \text{Power output} = 36\text{ W}\). Using \(P = I_p V_p\) for the primary coil: \(36 = I_p \times 240 \Rightarrow I_p = \frac{36}{240} = 0.15\text{ A}\). (b) (iii) Energy losses can be reduced by: using a laminated iron core (which reduces eddy currents), or using thick copper wires with low electrical resistance for the coils.
評分準則
(a) [3 marks total] - Alternating current in primary produces changing magnetic field (1) - Core links/conducts magnetic field to secondary coil (1) - Changing magnetic field induces alternating e.m.f. / voltage in secondary (1). (b) (i) [2 marks total] - Correct substitution in formula: \(\frac{240}{12} = \frac{1200}{N_s}\) (1) - Correct answer: \(60\) (1). (b) (ii) [2 marks total] - Use of \(P = I V\) or substitution: \(36 = I_p \times 240\) (1) - Correct current: \(0.15\text{ A}\) (1). (b) (iii) [1 mark] - Correct structural feature: Laminated core / thick copper wires / soft iron core (1).
題目 8 · Structured
8 分
(a) Define *momentum* and state whether it is a scalar or vector quantity. (b) Trolley A of mass \(2.0\text{ kg}\) is traveling to the right at a velocity of \(4.5\text{ m/s}\). It collides head-on with Trolley B of mass \(1.5\text{ kg}\), which is traveling to the left at a velocity of \(2.0\text{ m/s}\). During the collision, the two trolleys couple together. (i) Calculate the velocity of the combined trolleys after the collision, and state the direction of motion. (ii) Calculate the change in momentum of Trolley A as a result of the collision.
查看答案詳解收起答案詳解
解題
(a) Momentum is defined as the product of mass and velocity: \(p = m v\). Because it has both magnitude and direction, it is a vector quantity. (b) (i) Let the direction to the right be positive. - Mass of A, \(m_A = 2.0\text{ kg}\), initial velocity, \(u_A = +4.5\text{ m/s}\). - Mass of B, \(m_B = 1.5\text{ kg}\), initial velocity, \(u_B = -2.0\text{ m/s}\) (since it moves to the left). Total initial momentum, \(p_i\): \(p_i = m_A u_A + m_B u_B = (2.0 \times 4.5) + (1.5 \times (-2.0)) = 9.0 - 3.0 = +6.0\text{ kg}\cdot\text{m/s}\). By the law of conservation of momentum, total final momentum must be equal to \(p_i\): \(p_f = (m_A + m_B) v_f\) \(6.0 = (2.0 + 1.5) v_f = 3.5 v_f\) \(v_f = \frac{6.0}{3.5} \approx +1.71\text{ m/s}\) (or \(1.7\text{ m/s}\)). Since the sign of the velocity is positive, the combined trolleys move to the right. (b) (ii) The change in momentum (\(\Delta p_A\)) of Trolley A is given by: \(\Delta p_A = m_A (v_f - u_A)\) \(\Delta p_A = 2.0 \times (1.71 - 4.5) = 2.0 \times (-2.79) = -5.58\text{ kg}\cdot\text{m/s}\). This means Trolley A has a change of momentum of \(5.58\text{ kg}\cdot\text{m/s}\) (or \(5.6\text{ kg}\cdot\text{m/s}\)) directed to the left.
評分準則
(a) [2 marks total] - Correct definition: mass \(\times\) velocity (1) - Correct category: vector (1). (b) (i) [4 marks total] - Correct calculation of initial momentum with correct sign: \(9.0 - 3.0 = 6.0\text{ kg}\cdot\text{m/s}\) (2) [deduct 1 mark if direction of B is not taken as negative] - Correct equation and calculation: \(v_f = \frac{6.0}{3.5} = 1.71\text{ m/s}\) (1) - Correct direction: to the right (1). (b) (ii) [2 marks total] - Correct calculation working: \(2.0 \times (1.71 - 4.5)\) (1) - Correct answer: \(-5.6\text{ kg}\cdot\text{m/s}\) or magnitude of \(5.6\text{ kg}\cdot\text{m/s}\) to the left (1).
題目 9 · Structured
8 分
(a) Describe the life cycle of a star with a mass much greater than that of the Sun. Your answer should start from its formation in a nebula and end with its final stages. [4]
(b) Explain how observations of redshift in the light from distant galaxies provide evidence for the expanding Universe and support the Big Bang theory. [4]
查看答案詳解收起答案詳解
解題
### Part (a) 1. Interstellar clouds of gas and dust (nebula) collapse under gravitational attraction to form a protostar. As temperature increases, nuclear fusion of hydrogen begins, and the star becomes a stable main sequence star. 2. When hydrogen in the core runs out, the outer layers expand and cool, and the star becomes a red supergiant. 3. The core eventually collapses rapidly, resulting in a gigantic explosion called a supernova. 4. The remaining core collapses further to form either a highly dense neutron star or, if the mass is large enough, a black hole.
### Part (b) 1. Light observed from distant galaxies has a longer wavelength than expected, which is shifted towards the red end of the spectrum (redshift). 2. This redshift indicates that distant galaxies are moving away from the Earth (and each other). 3. The further away a galaxy is, the greater its redshift, which means it is moving away faster. 4. This expansion implies that at some point in the distant past, all matter in the Universe must have been concentrated at a single, extremely hot and dense point, supporting the Big Bang theory.
評分準則
### Part (a) [Maximum 4 marks] - Nebula collapses under gravity to form a protostar / nuclear fusion starts [1] - Star becomes a stable / massive main sequence star [1] - Star expands to become a red supergiant [1] - Supernova explosion occurs [1] - Remnant forms a neutron star OR a black hole [1]
### Part (b) [Maximum 4 marks] - Light from distant galaxies has its wavelength increased / shifted to red [1] - This indicates galaxies are moving away from us [1] - More distant galaxies have a greater redshift / are moving faster [1] - Suggests they started from a single point in the past (Big Bang theory) [1]
題目 10 · Structured
8 分
An ideal transformer is used to step down an alternating voltage of \(240\text{ V}\) from the mains supply to \(12\text{ V}\) to operate a low-voltage lamp. The primary coil of the transformer has \(1200\) turns.
(a) (i) Calculate the number of turns on the secondary coil. [2] (ii) State the principle of electromagnetic induction that explains how a voltage is induced in the secondary coil of the transformer. [2]
(b) In a real transformer, there are energy losses. (i) State one cause of thermal energy loss in a real transformer and describe how this loss is minimized. [2] (ii) The real transformer operates with an efficiency of \(85\%\). When the lamp is switched on, the current in the secondary coil is \(2.5\text{ A}\). Calculate the current in the primary coil. [2]
查看答案詳解收起答案詳解
解題
### Part (a)(i) Using the transformer equation: $$\frac{V_p}{V_s} = \frac{N_p}{N_s}$$ Substitute the given values: $$\frac{240}{12} = \frac{1200}{N_s}$$ $$N_s = 1200 \times \frac{12}{240} = 60\text{ turns}$$
### Part (a)(ii) An alternating current in the primary coil creates a continuously changing magnetic field in the iron core. This changing magnetic field passes through the secondary coil, cutting across its turns and inducing an alternating electromotive force (e.m.f.) in the secondary coil.
### Part (b)(i) One common cause of thermal energy loss is the resistance of the copper wires in the coils, which causes heating (Joule heating). This is minimized by using low-resistance, thick copper wires. Another cause is eddy currents induced in the iron core by the changing magnetic field. This is minimized by using a laminated iron core (thin sheets of iron separated by an insulating layer).
### Part (a)(ii) [2 marks] - Alternating current in primary creates a changing magnetic field / flux in core [1] - Changing magnetic field cuts secondary coil, inducing an e.m.f. / voltage [1]
### Part (b)(i) [2 marks] - Identify cause: Resistance of coils / heating of wires OR Eddy currents in iron core [1] - Identify mitigation: Use thick copper wires / low-resistance wires OR Use a laminated iron core [1] (Note: mitigation must match the identified cause)
### Part (b)(ii) [2 marks] - Correct formula for efficiency with power, e.g., \(\eta = \frac{V_s I_s}{V_p I_p}\) OR \(P_{\text{in}} = \frac{30}{0.85} = 35.3\text{ W}\) [1] - Correct calculation of primary current: \(0.15\text{ A}\) or \(0.147\text{ A}\) (accept \(0.147\text{ A}\) to \(0.15\text{ A}\)) [1]
Paper 6 (Alternative to Practical)
Answer all questions. You will need to show understanding of experimental techniques, graph plotting, gradient calculations, and practical design planning.
4 題目 · 40 分
題目 1 · Practical Structured
10 分
A student investigates the rate of cooling of water in two identical beakers under different conditions.
Beaker A is uninsulated. Beaker B is insulated with a layer of bubble wrap.
(a) Figure 1.1 shows the thermometer scale at the start of the experiment. Record the initial temperature \(\theta_0\) of the hot water.
[Assume the liquid level is exactly halfway between the \(84^\circ\text{C}\) and \(85^\circ\text{C}\) marks.]
(b) Table 1.1 shows the temperatures recorded for both beakers over a period of 180 seconds. Complete the column headings in Table 1.1 by inserting the correct units.
(a) The liquid level is halfway between 84 and 85, so \(\theta_0 = 84.5\ ^\circ\text{C}\).
(b) The unit for time \(t\) is seconds (\(\text{s}\)), and for temperatures \(\theta_A\) and \(\theta_B\) is degrees Celsius (\(^\circ\text{C}\)).
(c) Using the formula: \(R_1 = \frac{84.5 - 71.0}{90} = \frac{13.5}{90} = 0.15\ ^\circ\text{C/s}\).
(d) Using the formula: \(R_2 = \frac{71.0 - 63.5}{90} = \frac{7.5}{90} \approx 0.083\ ^\circ\text{C/s}\).
(e) Since \(R_2 < R_1\), the rate of cooling decreases as the temperature of the water decreases.
(f) Variables to keep constant include: volume of water in both beakers, starting temperature, and environmental conditions (room temperature, drafts).
(g) Precautions for accuracy: 1. Read the thermometer scale perpendicularly (eye level with meniscus) to avoid parallax error. 2. Ensure the thermometer bulb is fully immersed but not touching the bottom or sides of the beaker.
(h) Stirring ensures that the temperature is uniform throughout the volume of water, preventing localized hot or cold spots.
評分準則
(a) 1 Mark: 84.5 (accept 84.5) (b) 1 Mark: Units for time (s) and temperature (\(^\circ\text{C}\) or C) correct. (c) 2 Marks: 1 mark for correct working / substitution \(\frac{84.5 - 71.0}{90}\), 1 mark for correct value (0.15) with unit (\(^\circ\text{C/s}\) or \(^\circ\text{C}/\text{s}\)). (d) 1 Mark: Correct calculation (0.083 or 0.08) with or without unit. (e) 1 Mark: Statement that matches calculated values (e.g., rate of cooling decreases / gets slower). (f) 1 Mark: Any one correct variable (e.g., volume of water, initial temperature, material/thickness of beaker, ambient room temperature). (g) 2 Marks: 1 mark for each valid precaution (up to 2): - Avoid parallax error / look at meniscus perpendicularly. - Do not let bulb touch beaker sides/bottom. - Keep thermometer at same depth. (h) 1 Mark: To ensure the temperature is uniform/equal throughout.
題目 2 · Practical Structured
10 分
A student investigates the resistance of a constantan wire.
(a) Draw a circuit diagram showing a power source, a switch, an ammeter, a voltmeter, and a length \(L\) of resistance wire connected in series, such that the voltmeter measures the potential difference across the resistance wire, and the ammeter measures the total circuit current. [2]
(b) Table 2.1 shows the measurements of potential difference \(V\) and current \(I\) for five different lengths \(L\) of the wire.
Plot a graph of Resistance \(R\) / \(\Omega\) (y-axis) against Length \(L\) / \(\text{cm}\) (x-axis) on a grid. Draw the best-fit straight line. [4]
(c) Determine the gradient \(G\) of the graph. Show clearly on your graph how you obtained the necessary coordinates, choosing points that are at least half the length of your drawn line apart. [2]
\(G = \) ........................................
(d) State and explain whether the results show that resistance \(R\) is directly proportional to length \(L\). [2]
(a) The diagram must have the power supply, switch, ammeter, and resistance wire in a single loop (series). The voltmeter must be connected in parallel (across) the length of resistance wire.
(b) Graph details: - Axis labels: y-axis: \(R\ /\ \Omega\), x-axis: \(L\ /\ \text{cm}\). - Scales: suitable linear scales where plotted points occupy more than half the grid. For instance, x-axis: 0 to 100 cm, y-axis: 0 to 7.0 \(\Omega\). - Plotting: all points plotted within half a small square. - Line: best-fit straight line drawn with a ruler, passing through the origin \((0,0)\).
(c) Choosing two points far apart on the best-fit line, e.g., \((20, 1.37)\) and \((100, 6.84)\): \(G = \frac{R_2 - R_1}{L_2 - L_1} = \frac{6.84 - 1.37}{100 - 20} = \frac{5.47}{80} = 0.0684\ \Omega/\text{cm}\). (Accept values between \(0.067\) and \(0.070\)).
(d) Statement: Yes. Explanation: The graph is a straight line that passes directly through the origin \((0,0)\).
評分準則
(a) 2 Marks: - 1 mark for correct symbols of cell/battery, switch, ammeter, and resistance wire in series. - 1 mark for voltmeter connected in parallel with the resistance wire.
(b) 4 Marks: - 1 mark: Axes labeled with quantity and unit (e.g., \(R\ /\ \Omega\) and \(L\ /\ \text{cm}\)). - 1 mark: Suitable scales chosen so that points occupy at least half the grid in both directions. - 1 mark: All five points plotted accurately (within half a small square). - 1 mark: A single, thin, best-fit straight line drawn.
(c) 2 Marks: - 1 mark: Clear construction lines on the graph (or a large triangle) showing the coordinates used, with points at least half the length of the line apart. - 1 mark: Correct calculation of gradient \(G\) to 2 or 3 significant figures, within the range \(0.067\) to \(0.070\).
(d) 2 Marks: - 1 mark: Correct statement ('Yes'). - 1 mark: Justification referring to a straight line through the origin.
題目 3 · Practical Structured
10 分
A student investigates the focal length of a converging lens by locating a sharp image of an illuminated object on a screen.
(a) The student sets the distance from the illuminated object to the lens to be \(u = 20.0\text{ cm}\). They move the screen until a sharp image is formed at a distance \(v = 60.0\text{ cm}\) from the lens.
Calculate the focal length \(f_1\) using the equation:
(f) Suggest one characteristic of the image formed when the object distance \(u = 20.0\text{ cm}\) and the image distance \(v = 60.0\text{ cm}\), besides the fact that it is inverted.
(c) The average is: \(f_{\text{avg}} = \frac{15.0 + 15.0 + 15.0}{3} = 15.0\text{ cm}\).
(d) Suitable precautions: 1. Perform the experiment in a dark room so that the image on the screen is clearly visible. 2. Ensure that the object, lens, and screen are all aligned along the same horizontal axis, and that the lens is perpendicular to the bench.
(e) Suitable difficulties: 1. The "range of sharpness" effect, where the image looks almost equally sharp over a small range of screen positions, making the exact point hard to locate. 2. The image can have low brightness/contrast, making details blurry and hard to focus.
(f) Since \(v > u\) (\(60.0\text{ cm} > 20.0\text{ cm}\)), the image is magnified (larger than the object). Alternatively, because it is projected onto a screen, it must be a real image.
評分準則
(a) 2 Marks: 1 mark for correct substitution into the formula, 1 mark for correct value (15.0) with unit (\(\text{cm}\)). (b) 2 Marks: - 1 mark for \(f_2 = 15.0\text{ cm}\) (or 15). - 1 mark for \(f_3 = 15.0\text{ cm}\) (or 15). (c) 1 Mark: Correct calculation of average based on the student's values in (a) and (b). (d) 2 Marks: 1 mark for each valid precaution (up to 2): - Darkened room. - Object, lens, and screen must be perpendicular to the bench / centered at the same height. - Avoid parallax when measuring with ruler / use a set square. - Move screen slowly back and forth to find mid-point of sharp image. (e) 2 Marks: 1 mark for each valid difficulty (up to 2): - Hard to judge the exact point of sharpest focus / image is sharp over a range of positions. - Low intensity / dim light source makes image faint. - Hand blocks light / shadows. (f) 1 Mark: Magnified / larger than object / real.
題目 4 · Practical Structured
10 分
A student wants to investigate how the density of saltwater varies with the concentration of salt dissolved in it.
Plan an experiment to investigate how the density of a salt solution depends on the concentration of dissolved salt (expressed as the mass of salt dissolved in a fixed volume of water).
You are provided with the following apparatus: - A supply of distilled water - A container of table salt - A beaker - A spatula / stirring rod - A measuring cylinder - An electronic balance
In your plan, you should: - State any additional apparatus needed. - Describe a method for carrying out the investigation, including how to prepare different concentrations. - State the key variables to control. - Present a sample table with column headings (and units) to show how you would record the results. - Explain how you would use your results to draw a conclusion.
查看答案詳解收起答案詳解
解題
Plan details:
1. **Apparatus**: No major additional apparatus is strictly needed, though a weighing boat/paper to transfer salt to the balance is helpful.
2. **Method**: - Use the measuring cylinder to measure a fixed, known volume of distilled water (e.g., \(100\text{ cm}^3\)) and transfer it to a beaker. - Place a weighing boat on the electronic balance, tare (zero) the balance, and weigh a specific mass of salt (e.g., \(5.0\text{ g}\)) using the spatula. - Add this salt to the beaker of water and stir with the stirring rod until all the salt is completely dissolved. - Measure the mass of the empty measuring cylinder, then pour the entire solution into it to measure its total mass \(m_s\) and record the exact final volume \(V_s\). - Calculate the density \(\rho\) of the solution using \(\rho = \frac{m_s}{V_s}\). - Repeat the entire process for at least four other different masses of salt (e.g., \(10.0\text{ g}\), \(15.0\text{ g}\), \(20.0\text{ g}\), \(25.0\text{ g}\)), maintaining the same initial volume of distilled water.
3. **Control Variables**: - Temperature of the water (as density varies with temperature). - The initial volume of distilled water used to dissolve the salt.
4. **Table of Results**:
| Mass of salt dissolved \(m_d\) / \(\text{g}\) | Total Mass of solution \(m_s\) / \(\text{g}\) | Total Volume of solution \(V_s\) / \(\text{cm}^3\) | Density of solution \(\rho\) / \(\text{g/cm}^3\) | | :--- | :--- | :--- | :--- |
5. **Analysis and Conclusion**: - Plot a graph of the density of the solution \(\rho\) (y-axis) against the mass of dissolved salt \(m_d\) (x-axis). - If the graph is a straight line with a positive gradient, it shows that the density increases linearly with salt concentration.
評分準則
Max 10 Marks breakdown: - **Apparatus** [1 Mark]: - Identify that the electronic balance is used to weigh the salt and the measuring cylinder is used to measure the volume of liquid. - **Method** [4 Marks]: - 1 mark: Measure a fixed initial volume of water using the measuring cylinder. - 1 mark: Weigh a specific mass of salt on the balance. - 1 mark: Dissolve the salt fully in the water (by stirring). - 1 mark: Repeat the procedure with at least 5 different masses of salt. - **Control Variables** [2 Marks]: - 1 mark: Temperature of the water / solution. - 1 mark: Initial volume of distilled water (or same grade of water). - **Results Table** [2 Marks]: - 1 mark: Table presented with clear column headings. - 1 mark: Correct units included for all quantities: Mass of salt (\(\text{g}\)), mass of solution (\(\text{g}\)), volume (\(\text{cm}^3\) or \(\text{ml}\)), density (\(\text{g/cm}^3\) or \(\text{g/ml}\)). - **Analysis / Conclusion** [1 Mark]: - Explain that plotting a graph of density against mass of salt added (or salt concentration) allows the relationship to be analyzed (e.g., checking if it is a straight line / directly proportional).
想知道自己有幾分把握？
Thinka 是 DSE 學生用的 AI 練習應用程式，有無限量練習題、即時自動批改和詳細解題步驟。逾 100,000 名學生用它確認自己真的識，而不只是「以為識」。