An original Thinka practice paper modelled on the structure and difficulty of the 2021 HKDSE Physics paper. Not affiliated with or reproduced from HKDSE.
卷一 甲部
回答全部33題選擇題。各題分值相同。
33 题目 · 33 分
题目 1 · 選擇題
1 分
A narrow beam of monochromatic light travels from medium \(X\) into medium \(Y\). The refractive indices of medium \(X\) and medium \(Y\) are \(n_X\) and \(n_Y\) respectively, where \(n_X > n_Y\). The angle of incidence in medium \(X\) is \(\theta\), and \(C\) is the critical angle at the interface. Which of the following statements is/are correct?
(1) The wavelength of the light in medium \(X\) is shorter than that in medium \(Y\). (2) As \(\theta\) increases from \(0^\circ\) to \(C\), the speed of the refracted light in medium \(Y\) increases. (3) When \(\theta > C\), no light enters medium \(Y\).
A.(1) only
B.(1) and (2) only
C.(1) and (3) only
D.(2) and (3) only
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解题
For (1): Since \(n = \frac{c}{v}\) and \(v = f\lambda\) (with frequency \(f\) remaining unchanged across media), \(\lambda = \frac{c}{n f} \propto \frac{1}{n}\). Since \(n_X > n_Y\), the wavelength in medium \(X\) is shorter than that in medium \(Y\). Thus, (1) is correct.
For (2): The speed of light in medium \(Y\) is given by \(v_Y = \frac{c}{n_Y}\), which depends solely on the medium's refractive index and is independent of the angle of incidence \(\theta\). Thus, (2) is incorrect.
For (3): When \(\theta > C\), total internal reflection occurs. All light is reflected back into medium \(X\), and no light is refracted into medium \(Y\). Thus, (3) is correct.
Hence, (1) and (3) only are correct.
评分标准
1A: Correctly identifies statements (1) and (3) as correct (Option C).
题目 2 · 選擇題
1 分
A ball \(P\) is projected horizontally with an initial speed \(u\) from the edge of a vertical cliff of height \(H\). At the same instant, a ball \(Q\) is projected vertically upwards from the ground directly below the cliff with an initial speed \(v\). The two balls collide in mid-air when ball \(P\) has descended through a vertical distance of \(\frac{1}{4}H\). Neglecting air resistance, what is the value of \(\frac{v}{\sqrt{gH}}\)? (Given: acceleration due to gravity is \(g\))
A.\(\frac{1}{\sqrt{2}}\)
B.\(1\)
C.\(\sqrt{2}\)
D.\(2\)
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解题
Let \(t\) be the time from launch to collision. For ball \(P\), vertical displacement downward is \(s_P = \frac{1}{4}H\): \(\frac{1}{4}H = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{H}{2g}}\).
At the moment of collision, ball \(Q\) is at a height of \(H - \frac{1}{4}H = \frac{3}{4}H\) above the ground. For ball \(Q\): \(s_Q = vt - \frac{1}{2}gt^2 = \frac{3}{4}H\).
Substitute \(t = \sqrt{\frac{H}{2g}}\): \(v \sqrt{\frac{H}{2g}} = H \implies v = \sqrt{2gH}\).
Therefore, \(\frac{v}{\sqrt{gH}} = \sqrt{2}\).
评分标准
1A: Correct algebraic deduction leading to \(\sqrt{2}\) (Option C).
题目 3 · 選擇題
1 分
A battery of constant e.m.f. and non-zero internal resistance \(r\) is connected to a circuit containing two fixed resistors, \(R_1\) and \(R_2\), an ideal ammeter, an ideal voltmeter, and a switch \(S\). A voltmeter is connected across the terminals of the battery. Resistor \(R_1\) is in series with the ammeter, and this entire branch is in parallel with resistor \(R_2\) and switch \(S\). Initially, switch \(S\) is open. When switch \(S\) is closed, what happen to the readings of the voltmeter and the ammeter?
Voltmeter reading | Ammeter reading
A.Decreases | Decreases
B.Decreases | Increases
C.Increases | Decreases
D.Increases | Increases
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解题
When switch \(S\) is closed, the branch containing \(R_2\) is connected in parallel with the branch containing \(R_1\) and the ammeter. 1. The total equivalent external resistance \(R_{ext}\) of the circuit decreases. 2. The total current supplied by the battery, \(I_{total} = \frac{\mathcal{E}}{R_{ext} + r}\), increases. 3. The terminal voltage \(V = \mathcal{E} - I_{total}r\) decreases. Thus, the voltmeter reading decreases. 4. The potential difference across the branch with \(R_1\) and the ammeter is equal to the terminal voltage \(V\). Since \(V\) decreases and the resistance \(R_1\) remains unchanged, the current through \(R_1\) (measured by the ammeter) \(I_1 = \frac{V}{R_1}\) decreases.
Therefore, both readings decrease.
评分标准
1A: Correct combination of decreases/decreases (Option A).
题目 4 · 選擇題
1 分
A fixed mass of an ideal gas is compressed to half of its original volume while its absolute temperature is doubled. Which of the following statements about the gas is/are correct?
(1) The root-mean-square speed of the gas molecules increases by a factor of \(\sqrt{2}\). (2) The average kinetic energy of the gas molecules is doubled. (3) The pressure of the gas increases by a factor of 4.
A.(1) and (2) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解题
Let initial state be \((p_1, V_1, T_1)\) and final state be \((p_2, V_2, T_2)\) with \(V_2 = \frac{1}{2}V_1\) and \(T_2 = 2T_1\).
For (1): The root-mean-square speed is \(c_{rms} = \sqrt{\frac{3RT}{M}} \propto \sqrt{T}\). Since \(T_2 = 2T_1\), \(c_{rms, 2} = \sqrt{2} c_{rms, 1}\). Thus, (1) is correct.
For (2): The average translational kinetic energy of a molecule is \(\overline{E_k} = \frac{3}{2}k_B T \propto T\). Since temperature is doubled, \(\overline{E_k}\) is doubled. Thus, (2) is correct.
For (3): From the ideal gas equation \(pV = nRT \implies p = \frac{nRT}{V}\), the new pressure is \(p_2 = \frac{nR(2T_1)}{\frac{1}{2}V_1} = 4\left(\frac{nRT_1}{V_1}\right) = 4p_1\). Thus, (3) is correct.
Hence, (1), (2) and (3) are all correct.
评分标准
1A: Correct selection of all three statements (Option D).
题目 5 · 選擇題
1 分
A square flat conducting loop of side length \(L\) and total resistance \(R\) lies in a plane perpendicular to a uniform magnetic field of flux density \(B\). The loop is pulled completely out of the magnetic field region at a constant velocity \(v\) perpendicular to one of its sides. What is the external mechanical power required to maintain this constant velocity while the loop is leaving the field?
A.\(\frac{BLv^2}{R}\)
B.\(\frac{B^2 L^2 v}{R}\)
C.\(\frac{B^2 L^2 v^2}{R}\)
D.\(\frac{B^2 L^2 v^2}{R^2}\)
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解题
As the loop leaves the magnetic field at constant speed \(v\): 1. The induced electromotive force generated across the moving side of length \(L\) is \(\mathcal{E} = BLv\). 2. The induced current in the loop is \(I = \frac{\mathcal{E}}{R} = \frac{BLv}{R}\). 3. The magnetic force opposing the motion is \(F_B = BIL = B\left(\frac{BLv}{R}\right)L = \frac{B^2 L^2 v}{R}\). 4. To keep the loop moving at a constant speed, an equal and opposite external pulling force is required: \(F_{ext} = F_B = \frac{B^2 L^2 v}{R}\). 5. The mechanical power provided by the external force is \(P = F_{ext} v = \frac{B^2 L^2 v^2}{R}\).
评分标准
1A: Correct power formula derivation (Option C).
题目 6 · 選擇題
1 分
A ray of monochromatic light in air enters one vertical face of a rectangular transparent block of refractive index 1.25 at an angle of incidence \(\theta\). The refracted ray travels inside the block and strikes the adjacent top horizontal surface. What is the condition on \(\theta\) such that total internal reflection occurs at the top surface?
A.\(\theta < 36.9^\circ\)
B.\(\theta < 48.6^\circ\)
C.\(\theta > 48.6^\circ\)
D.\(\theta > 53.1^\circ\)
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解题
Let \(r\) be the angle of refraction at the vertical surface. By Snell's law, \(1 \cdot \sin\theta = n \sin r\). The ray strikes the adjacent horizontal top surface at an angle of incidence \(i_2 = 90^\circ - r\). For total internal reflection at the top surface, \(i_2 > c\), where the critical angle \(c\) is given by \(\sin c = \frac{1}{n} = \frac{1}{1.25} = 0.80\). Thus, \(90^\circ - r > c \implies r < 90^\circ - c\), which gives \(\sin r < \cos c = \sqrt{1 - \sin^2 c} = \sqrt{1 - 0.80^2} = 0.60\). Since \(\sin\theta = n \sin r = 1.25 \sin r\), we have \(\sin\theta < 1.25 \times 0.60 = 0.75\). Hence, \(\theta < \arcsin(0.75) \approx 48.6^\circ\).
评分标准
1A for correct identification of option B.
题目 7 · 選擇題
1 分
A small ball is projected from horizontal ground with an initial kinetic energy \(E\). At the highest point of its parabolic trajectory, its kinetic energy is \(\frac{1}{4}E\). Neglecting air resistance, what is the angle of projection above the horizontal and the ratio of its maximum gravitational potential energy (taking the ground as the reference level) to \(E\)?
Angle of projection\(\frac{\text{Maximum potential energy}}{E}\)A.\(30^\circ\)0.75B.\(30^\circ\)0.50C.\(60^\circ\)0.75D.\(60^\circ\)0.50
A.Angle of projection: \(30^\circ\), Ratio: 0.75
B.Angle of projection: \(30^\circ\), Ratio: 0.50
C.Angle of projection: \(60^\circ\), Ratio: 0.75
D.Angle of projection: \(60^\circ\), Ratio: 0.50
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解题
Let \(u\) be the initial projection speed and \(\theta\) be the angle of projection. The initial kinetic energy is \(E = \frac{1}{2}m u^2\). At the highest point, the vertical component of velocity is zero, so the speed is \(u \cos\theta\). The kinetic energy at the highest point is \(E_{k,\text{top}} = \frac{1}{2}m (u \cos\theta)^2 = E \cos^2\theta\). Given \(E_{k,\text{top}} = \frac{1}{4}E\), we have \(\cos^2\theta = \frac{1}{4} \implies \cos\theta = 0.5 \implies \theta = 60^\circ\). By conservation of mechanical energy, \(E = E_{k,\text{top}} + E_p \implies E_p = E - \frac{1}{4}E = \frac{3}{4}E = 0.75E\). Thus, the ratio is 0.75.
评分标准
1A for correct identification of option C.
题目 8 · 選擇題
1 分
A cell of constant e.m.f. \(E\) and internal resistance \(r\) is connected in series with a variable resistor of resistance \(R\). As \(R\) is gradually increased from \(0.5r\) to \(2r\), which of the following statements is/are correct?
(1) The terminal voltage across the cell increases monotonically. (2) The total power supplied by the cell decreases monotonically. (3) The power dissipated in the variable resistor increases and then decreases.
A.(1) only
B.(1) and (2) only
C.(2) and (3) only
D.(1), (2) and (3)
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解题
(1) Terminal voltage \(V = E - Ir = E \frac{R}{R+r} = \frac{E}{1 + r/R}\). As \(R\) increases, \(r/R\) decreases, so \(V\) increases monotonically. Statement (1) is correct. (2) Total power supplied by the cell is \(P_{\text{total}} = E I = \frac{E^2}{R+r}\). As \(R\) increases, \(R+r\) increases, so \(P_{\text{total}}\) decreases monotonically. Statement (2) is correct. (3) Power dissipated in \(R\) is \(P_R = I^2 R = \frac{E^2 R}{(R+r)^2}\). According to the maximum power transfer theorem, \(P_R\) attains its maximum when \(R = r\). As \(R\) increases from \(0.5r\) to \(r\) and then to \(2r\), \(P_R\) increases to its maximum and then decreases. Statement (3) is correct.
评分标准
1A for correct identification of option D.
题目 9 · 選擇題
1 分
A flat circular coil with \(N\) turns and cross-sectional area \(A\) rotates at a constant angular frequency \(\omega\) in a uniform magnetic field of magnetic flux density \(B\) about an axis perpendicular to the field. Which of the following modifications will DOUBLE the peak induced electromotive force (e.m.f.) in the coil?
(1) Doubling the angular frequency \(\omega\) only. (2) Doubling both the number of turns \(N\) and the magnetic flux density \(B\). (3) Doubling the area \(A\) and halving the angular frequency \(\omega\).
A.(1) only
B.(3) only
C.(1) and (2) only
D.(2) and (3) only
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解题
The peak induced e.m.f. is given by \(\mathcal{E}_0 = N B A \omega\). (1) If \(\omega\) is doubled, \(\mathcal{E}'_0 = N B A (2\omega) = 2\mathcal{E}_0\) (doubled). (1) is correct. (2) If both \(N\) and \(B\) are doubled, \(\mathcal{E}'_0 = (2N)(2B)A\omega = 4\mathcal{E}_0\) (quadrupled). (2) is incorrect. (3) If \(A\) is doubled and \(\omega\) is halved, \(\mathcal{E}'_0 = N B (2A)(0.5\omega) = \mathcal{E}_0\) (unchanged). (3) is incorrect.
评分标准
1A for correct identification of option A.
题目 10 · 選擇題
1 分
A fixed mass of an ideal gas is enclosed in a container of variable volume. The absolute temperature of the gas is increased to \(1.44\) times its initial value, while its volume is simultaneously compressed by \(20\%\). What is the percentage change in the root-mean-square speed \(c_{\text{rms}}\) of the gas molecules, and what is the new gas pressure in terms of the initial pressure \(P\)?
Percentage change in \(c_{\text{rms}}\)New gas pressureA.\(+20\%\)\(1.80 P\)B.\(+20\%\)\(1.15 P\)C.\(+44\%\)\(1.80 P\)D.\(+44\%\)\(1.15 P\)
A.Percentage change in \(c_{\text{rms}}\) : \(+20\%\), New gas pressure: \(1.80 P\)
B.Percentage change in \(c_{\text{rms}}\) : \(+20\%\), New gas pressure: \(1.15 P\)
C.Percentage change in \(c_{\text{rms}}\) : \(+44\%\), New gas pressure: \(1.80 P\)
D.Percentage change in \(c_{\text{rms}}\) : \(+44\%\), New gas pressure: \(1.15 P\)
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解题
1. The root-mean-square speed is proportional to the square root of absolute temperature: \(c_{\text{rms}} = \sqrt{\frac{3RT}{M}} \propto \sqrt{T}\). When the temperature increases to \(1.44 T\), \(c_{\text{rms}}' = \sqrt{1.44} c_{\text{rms}} = 1.20 c_{\text{rms}}\). The percentage change is \((1.20 - 1) \times 100\% = +20\%\). 2. From the ideal gas law \(PV = nRT \implies P = \frac{nRT}{V}\). The new volume is \(V' = (1 - 0.20)V = 0.80V\). The new pressure is \(P' = \frac{nR(1.44T)}{0.80V} = \frac{1.44}{0.80} P = 1.80 P\).
评分标准
1A for correct identification of option A.
题目 11 · 選擇題
1 分
A monochromatic light ray enters normally through one face of an isosceles right-angled glass prism of refractive index 1.50. The ray is incident on the hypotenuse face at an angle of incidence of \(45^\circ\). The prism is immersed in a transparent liquid of refractive index \(n\). What is the maximum value of \(n\) such that total internal reflection still occurs at the hypotenuse face?
A.1.06
B.1.15
C.1.25
D.1.33
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解题
For total internal reflection to occur at the glass-liquid boundary, the angle of incidence must be greater than the critical angle \(c\): \(\sin 45^\circ > \sin c = \frac{n}{n_{\text{glass}}}\). Therefore, \(n < 1.50 \times \sin 45^\circ = 1.50 \times \frac{\sqrt{2}}{2} \approx 1.06\). Thus, the maximum refractive index of the liquid is 1.06.
评分标准
1A: Option A correctly identifies the maximum refractive index using the condition for total internal reflection \(\sin c = n_2 / n_1\).
题目 12 · 選擇題
1 分
Three identical light bulbs, \(X\), \(Y\), and \(Z\), and an open switch \(S\) are connected to an ideal battery of constant e.m.f. Bulb \(X\) is connected in series with the parallel combination of bulb \(Y\) and the branch containing bulb \(Z\) and switch \(S\). Switch \(S\) is initially open. When switch \(S\) is closed, which of the following statements is/are correct?
(1) The total power output of the battery increases. (2) Bulb \(X\) becomes brighter. (3) Bulb \(Y\) becomes dimmer.
A.(1) only
B.(1) and (2) only
C.(2) and (3) only
D.(1), (2) and (3)
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解题
Let the resistance of each bulb be \(R\). When \(S\) is open, the total resistance is \(R + R = 2R\), total current is \(I_1 = \frac{\mathcal{E}}{2R}\), and potential difference across \(Y\) is \(0.5\mathcal{E}\). When \(S\) is closed, the parallel combination of \(Y\) and \(Z\) has resistance \(0.5R\), so total resistance becomes \(1.5R\). The new total current is \(I_2 = \frac{\mathcal{E}}{1.5R} = \frac{2\mathcal{E}}{3R} > I_1\). (1) is correct: Total power \(P = \mathcal{E} I\) increases. (2) is correct: Current through bulb \(X\) increases, so it becomes brighter. (3) is correct: Potential difference across \(Y\) drops to \(I_2 \times 0.5R = \frac{\mathcal{E}}{3} < 0.5\mathcal{E}\), so bulb \(Y\) becomes dimmer. Hence, (1), (2), and (3) are all correct.
评分标准
1A: Option D correctly deduces the changes in total equivalent resistance, main current, and branch voltage after closing switch \(S\).
题目 13 · 選擇題
1 分
A small ball is projected horizontally with an initial speed \(u\) from the edge of a horizontal platform of height \(h\) above the ground. Just before the ball strikes the ground, its velocity makes an angle of \(45^\circ\) with the horizontal. If the ball is now projected horizontally from the same height with an initial speed of \(2u\), what is the tangent of the angle that its velocity makes with the horizontal just before striking the ground? (Air resistance is negligible.)
A.0.25
B.0.50
C.1.00
D.2.00
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解题
In horizontal projectile motion, the vertical velocity upon impact is \(v_y = \sqrt{2gh}\). Initially, \(\tan 45^\circ = \frac{v_y}{v_x} = \frac{\sqrt{2gh}}{u} = 1\), which implies \(\sqrt{2gh} = u\). When the initial speed is doubled to \(2u\), the vertical velocity remains unchanged at \(v_y' = \sqrt{2gh} = u\), while the horizontal velocity becomes \(v_x' = 2u\). Therefore, the new tangent of the angle is \(\tan \theta = \frac{v_y'}{v_x'} = \frac{u}{2u} = 0.50\).
评分标准
1A: Option B correctly applies kinematics to resolve horizontal and vertical velocity components and calculates the tangent of the angle.
题目 14 · 選擇題
1 分
A uniform magnetic field of flux density \(B\) is directed perpendicularly into the page. A conducting rod of length \(L\) and negligible resistance is pulled to the right at a constant velocity \(v\) along two frictionless, horizontal parallel conducting rails separated by distance \(L\). A resistor of resistance \(R\) is connected between the left ends of the rails. Which of the following statements is/are correct?
(1) An external force directed to the right is required to maintain the rod at constant velocity. (2) The induced current flows upwards through the moving rod. (3) If the resistance \(R\) is doubled while the speed \(v\) is kept constant, the mechanical power supplied by the external force is halved.
A.(1) only
B.(1) and (2) only
C.(2) and (3) only
D.(1), (2) and (3)
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解题
By Fleming's right-hand rule (or Lenz's law), moving the rod to the right increases magnetic flux into the page, inducing a counter-clockwise current. Thus, the induced current flows upwards through the rod (statement 2 is correct). By Fleming's left-hand rule, the magnetic force on the rod is directed to the left. To keep constant velocity, an external pulling force directed to the right is required (statement 1 is correct). The induced e.m.f. is \(\mathcal{E} = BLv\), the current is \(I = \frac{BLv}{R}\), and the required mechanical power is \(P = F_{\text{ext}} v = I L B v = \frac{B^2 L^2 v^2}{R}\). If \(R\) is doubled, \(P\) is halved (statement 3 is correct).
评分标准
1A: Option D correctly evaluates the direction of induced current, force balance for constant velocity, and the expression for mechanical power.
题目 15 · 選擇題
1 分
A sealed container of fixed volume contains a fixed mass of an ideal gas at an initial temperature of \(20^\circ\text{C}\). The gas is heated until its temperature reaches \(40^\circ\text{C}\). Which of the following statements is/are correct?
(1) The pressure of the gas is doubled. (2) The average kinetic energy of the gas molecules increases by about \(6.8\%\). (3) The root-mean-square speed of the gas molecules is doubled.
A.(2) only
B.(3) only
C.(1) and (2) only
D.(1) and (3) only
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解题
Absolute temperatures are \(T_1 = 20 + 273.15 = 293.15\text{ K}\) and \(T_2 = 40 + 273.15 = 313.15\text{ K}\). For a fixed volume, pressure \(P \propto T\), so \(P_2/P_1 = 313.15/293.15 \approx 1.068\), not doubled (statement 1 is incorrect). The average kinetic energy of molecules is proportional to absolute temperature \(T\), so the fractional increase is \(\frac{313.15 - 293.15}{293.15} \times 100\% \approx 6.8\%\) (statement 2 is correct). The root-mean-square speed is proportional to \(\sqrt{T}\), so \(c_{\text{rms},2}/c_{\text{rms},1} = \sqrt{313.15/293.15} \approx 1.034\), not doubled (statement 3 is incorrect).
评分标准
1A: Option A correctly relates ideal gas properties and molecular kinetic energy to thermodynamic temperature in kelvins.
题目 16 · 選擇題
1 分
A luminous object is placed on the principal axis of a thin convex lens of focal length \(f\). When the object is at a distance \(u_1\) from the lens, an upright image with linear magnification 3 is formed. When the object is moved to a distance \(u_2\) from the lens, an inverted image with linear magnification 3 is formed. Find the ratio \(u_1 : u_2\).
A.1 : 4
B.1 : 2
C.2 : 3
D.3 : 4 rock-paper-scissors pattern (or 3 : 4)
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解题
For a thin convex lens forming an upright (virtual) image of magnification \(m = 3\): \(v_1 = -3u_1\) Using the lens formula \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\): \(\frac{1}{u_1} - \frac{1}{3u_1} = \frac{1}{f} \implies \frac{2}{3u_1} = \frac{1}{f} \implies u_1 = \frac{2}{3}f\).
For an inverted (real) image of magnification \(m = 3\): \(v_2 = 3u_2\) \(\frac{1}{u_2} + \frac{1}{3u_2} = \frac{1}{f} \implies \frac{4}{3u_2} = \frac{1}{f} \implies u_2 = \frac{4}{3}f\).
A cell of constant e.m.f. \(E\) and non-negligible internal resistance \(r\) is connected in series with a variable resistor \(R\). An ideal voltmeter is connected across the terminals of the cell. As the resistance of \(R\) is gradually increased from a very small non-zero value, which of the following statements is/are correct?
(1) The reading of the voltmeter increases. (2) The product of the terminal voltage and the circuit current always decreases. (3) The power dissipated by \(R\) increases while \(R < r\).
A.(1) only
B.(3) only
C.(1) and (2) only
D.(1) and (3) only
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解题
(1) Terminal voltage \(V = E - Ir = \frac{E R}{R + r}\). As \(R\) increases, current \(I\) decreases, so the lost volts \(Ir\) decreases and \(V\) increases. Statement (1) is correct. (2) The product of terminal voltage and circuit current is the power delivered to the external resistor, \(P = I V = \frac{E^2 R}{(R+r)^2}\). This power increases as \(R\) increases from 0 up to \(R = r\) (maximum power transfer theorem), and only decreases for \(R > r\). Thus it does NOT always decrease. Statement (2) is incorrect. (3) By the maximum power transfer theorem, for \(R < r\), \(\frac{\mathrm{d}P}{\mathrm{d}R} > 0\), meaning power dissipated in \(R\) increases as \(R\) increases towards \(r\). Statement (3) is correct.
Hence, (1) and (3) only are correct.
评分标准
1A for selecting option D.
题目 18 · 選擇題
1 分
A small stone is projected horizontally at a speed of \(15\text{ m s}^{-1}\) from the top edge of a vertical cliff above horizontal ground. Neglecting air resistance, the stone strikes the ground at an angle of \(60^\circ\) to the horizontal. What is the height of the cliff? (Take \(g = 9.81\text{ m s}^{-2}\))
A.17.2 m
B.29.8 m
C.34.4 m
D.59.6 m
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解题
The horizontal component of velocity remains constant throughout the motion: \(v_x = u_x = 15\text{ m s}^{-1}\). When striking the ground at an angle of \(60^\circ\) below the horizontal: \(\tan 60^\circ = \frac{v_y}{v_x} \implies v_y = 15 \tan 60^\circ = 15\sqrt{3}\text{ m s}^{-1}\).
Using \(v_y^2 = u_y^2 + 2 g h\) with initial vertical velocity \(u_y = 0\): \((15\sqrt{3})^2 = 0 + 2(9.81)h\) \(675 = 19.62 h \implies h = \frac{675}{19.62} \approx 34.4\text{ m}\).
评分标准
1A for selecting option C.
题目 19 · 選擇題
1 分
A flat circular coil consisting of 200 tightly wound turns and of radius \(0.05\text{ m}\) is oriented perpendicular to a uniform magnetic field of \(0.40\text{ T}\). The coil is rotated by \(180^\circ\) about one of its diameters in a time interval of \(0.12\text{ s}\). What is the magnitude of the average e.m.f. induced in the coil during this rotation?
A.2.62 V
B.5.24 V
C.10.5 V
D.20.9 V
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解题
The cross-sectional area of the coil is \(A = \pi r^2 = \pi (0.05)^2 = 0.0025\pi\text{ m}^2\). Initial magnetic flux linkage: \(\Phi_1 = N B A = 200 \times 0.40 \times 0.0025\pi = 0.2\pi\text{ Wb}\). After a \(180^\circ\) rotation, the direction of the normal vector relative to the field reverses: \(\Phi_2 = -N B A = -0.2\pi\text{ Wb}\). Change in magnetic flux linkage: \(|\Delta \Phi| = |\Phi_2 - \Phi_1| = |-0.2\pi - 0.2\pi| = 0.4\pi\text{ Wb}\). Average induced e.m.f.: \(\mathcal{E}_{\text{avg}} = \frac{|\Delta \Phi|}{\Delta t} = \frac{0.4\pi}{0.12} \approx 10.47\text{ V} \approx 10.5\text{ V}\).
评分标准
1A for selecting option C.
题目 20 · 選擇題
1 分
A metal block of mass \(0.50\text{ kg}\) at \(100^\circ\text{C}\) is immersed in an insulated container of heat capacity \(120\text{ J K}^{-1}\) containing \(0.30\text{ kg}\) of water at \(20^\circ\text{C}\). The final steady temperature of the mixture is \(32^\circ\text{C}\). Assuming no heat is lost to the surroundings, find the specific heat capacity of the metal. (Given: specific heat capacity of water \(= 4200\text{ J kg}^{-1}\text{ K}^{-1}\))
Heat lost by the metal block: \(Q_{\text{lost}} = m_m c_m \Delta T_{\text{hot}} = 0.50 \times c_m \times (100 - 32) = 34 c_m\).
By conservation of energy (assuming no heat lost to surroundings): \(34 c_m = 16560 \implies c_m = \frac{16560}{34} \approx 487\text{ J kg}^{-1}\text{ K}^{-1}\).
评分标准
1A for selecting option C.
题目 21 · 選擇題
1 分
A fixed mass of an ideal gas is enclosed in a rigid container of fixed volume. The absolute temperature of the gas is increased from \(T\) to \(2T\). Which of the following statements is/are correct?
(1) The root-mean-square speed of the gas molecules increases by a factor of \(\sqrt{2}\). (2) The collision frequency of gas molecules with the container walls increases by a factor of 2. (3) The average kinetic energy of the gas molecules is doubled.
A.(1) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解题
For (1): The root-mean-square speed is given by \(c_{\text{rms}} = \sqrt{\frac{3RT}{M}}\). When absolute temperature doubles, \(c_{\text{rms}}\) increases by a factor of \(\sqrt{2}\). Thus, (1) is correct. For (2): The collision frequency with the container walls is proportional to the number density and the mean speed of molecules. Since volume is constant, number density is constant, so collision frequency is proportional to \(c_{\text{rms}}\), which increases by a factor of \(\sqrt{2}\), not 2. Thus, (2) is incorrect. For (3): The average translational kinetic energy of a molecule is \(\frac{3}{2}k_B T\), which is directly proportional to \(T\). Doubling \(T\) doubles the average kinetic energy. Thus, (3) is correct. Hence, (1) and (3) only are correct.
评分标准
1A: Correct option B selected.
题目 22 · 選擇題
1 分
A projectile is launched from horizontal ground with an initial speed \(u\) at an angle \(\theta\) above the horizontal. Air resistance is negligible. If the projectile attains a maximum height \(H\) and achieves a horizontal range \(R\), which of the following expressions represents \(\tan \theta\)?
A.\(\frac{H}{R}\)
B.\(\frac{2H}{R}\)
C.\(\frac{4H}{R}\)
D.\(\frac{R}{4H}\)
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解题
Maximum height is given by \(H = \frac{u^2 \sin^2 \theta}{2g}\). Horizontal range is given by \(R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}\). Taking the ratio: \(\frac{H}{R} = \frac{u^2 \sin^2 \theta / (2g)}{2u^2 \sin \theta \cos \theta / g} = \frac{\sin \theta}{4 \cos \theta} = \frac{\tan \theta}{4}\). Rearranging gives \(\tan \theta = \frac{4H}{R}\).
评分标准
1A: Correct option C selected.
题目 23 · 選擇題
1 分
A ray of monochromatic light travels in medium \(X\) with refractive index \(n_X\) and strikes the flat interface with medium \(Y\) with refractive index \(n_Y\). The angle of incidence in medium \(X\) is \(\theta\). Which of the following conditions is/are necessary for total internal reflection to occur at the interface?
(1) \(n_X > n_Y\) (2) \(\sin \theta > \frac{n_Y}{n_X}\) (3) The speed of light in medium \(X\) is greater than that in medium \(Y\).
A.(1) only
B.(1) and (2) only
C.(2) and (3) only
D.(1), (2) and (3)
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解题
For total internal reflection (TIR) to occur: (1) Light must travel from an optically denser medium to an optically less dense medium, so \(n_X > n_Y\). Thus, (1) is correct. (2) The critical angle \(C\) satisfies \(\sin C = \frac{n_Y}{n_X}\). The angle of incidence must exceed the critical angle (\(\theta > C\)), which implies \(\sin \theta > \sin C = \frac{n_Y}{n_X}\). Thus, (2) is correct. (3) Since \(n = \frac{c}{v}\) and \(n_X > n_Y\), the speed of light in medium \(X\) is less than that in medium \(Y\) (\(v_X < v_Y\)). Thus, (3) is incorrect. Therefore, only (1) and (2) are correct.
评分标准
1A: Correct option B selected.
题目 24 · 選擇題
1 分
In a circuit, a battery of constant electromotive force \(\mathcal{E}\) and negligible internal resistance is connected in series with a resistor \(R_1\) and a parallel combination of two identical resistors \(R_2\) and \(R_3\) (where \(R_1 = R_2 = R_3 = R\)). A switch \(S\) is connected in series with \(R_3\) in its branch. An ideal voltmeter is connected across resistor \(R_1\). What happens to the total power dissipated in the circuit and the reading of the voltmeter when switch \(S\) is closed?
Total power dissipatedVoltmeter reading
A.Increases | Increases
B.Increases | Decreases
C.Decreases | Increases
D.Decreases | Decreases
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解题
When switch \(S\) is open: The circuit consists of \(R_1\) and \(R_2\) in series. Equivalent resistance \(R_{\text{eq, open}} = R + R = 2R\). Total power dissipated \(P_{\text{open}} = \frac{\mathcal{E}^2}{2R}\). Voltmeter reading \(V_1 = \frac{R}{R + R}\mathcal{E} = 0.50\mathcal{E}\).
When switch \(S\) is closed: \(R_2\) and \(R_3\) are in parallel with equivalent resistance \(R_p = \frac{R}{2} = 0.5R\). Total equivalent resistance \(R_{\text{eq, closed}} = R + 0.5R = 1.5R\). Total power dissipated \(P_{\text{closed}} = \frac{\mathcal{E}^2}{1.5R} = \frac{2\mathcal{E}^2}{3R} > P_{\text{open}}\) (increases). Voltmeter reading \(V_1' = \frac{R}{1.5R}\mathcal{E} = \frac{2}{3}\mathcal{E} \approx 0.67\mathcal{E} > V_1\) (increases). Therefore, both the total power dissipated and the voltmeter reading increase.
评分标准
1A: Correct option A selected.
题目 25 · 選擇題
1 分
A uniform magnetic field is directed perpendicularly into the plane of the paper. A closed, rigid circular conducting loop lies completely within this magnetic field. Which of the following operations will NOT induce an electric current in the loop?
A.Rotating the loop about an axis lying along its diameter.
B.Translating the loop at a constant velocity wholly inside the uniform magnetic field.
C.Decreasing the magnitude of the magnetic field uniformly over time.
D.Deforming the circular loop into an elliptical shape of smaller area while keeping it in the plane.
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解题
According to Faraday's law of electromagnetic induction, an induced electromotive force (and thus induced current in a closed loop) is produced only when there is a change in the magnetic flux linkage through the loop (\(\Phi = B A \cos \theta\)). A. Rotating the loop about a diameter changes the angle \(\theta\), altering the magnetic flux and inducing a current. B. Translating the loop within a uniform magnetic field leaves \(B\), \(A\), and \(\theta\) unchanged, so \(\Delta \Phi = 0\) and no current is induced. C. Decreasing the magnetic field strength alters \(B\), changing flux and inducing a current. D. Deforming the loop into an ellipse decreases the enclosed area \(A\), changing flux and inducing a current. Therefore, operation B will NOT induce an electric current.
评分标准
1A: Correct option B selected.
题目 26 · 選擇題
1 分
A student sets up an illuminated object, a thin converging lens of focal length \(f = 15\text{ cm}\), and a screen. The distance between the object and the screen is fixed at \(80\text{ cm}\). Which of the following statements is/are correct?
(1) There are two distinct positions of the lens between the object and the screen where a sharp image is formed on the screen. (2) If the upper half of the lens is covered with opaque paper, only the bottom half of the image appears on the screen. (3) The minimum distance between an object and a screen to form a sharp real image with this lens is \(60\text{ cm}\).
A.(1) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解题
1. For a fixed distance \(D = 80\text{ cm}\) between object and screen, using the lens equation \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\) with \(v = D - u\) gives \(u^2 - Du + Df = 0\). The discriminant is \(\Delta = D^2 - 4Df = 80^2 - 4(80)(15) = 6400 - 4800 = 1600 > 0\). Since \(\Delta > 0\), there are two distinct real solutions for \(u\) (specifically \(u = 60\text{ cm}\) and \(u = 20\text{ cm}\)). Hence, statement (1) is correct.
2. Covering the upper half of the lens reduces the light collected, making the whole image dimmer, but every part of the uncovered lens still receives light from all parts of the object and forms a complete image. Thus, statement (2) is incorrect.
3. For a real image formed on a screen by a converging lens, the minimum separation between the real object and the screen is \(D_{\min} = 4f = 4 \times 15\text{ cm} = 60\text{ cm}\). Hence, statement (3) is correct.
Therefore, (1) and (3) only are correct.
评分标准
B (1A)
题目 27 · 選擇題
1 分
A battery of electromotive force \(\mathcal{E}\) and non-zero internal resistance \(r\) is connected to a circuit consisting of a fixed resistor \(R_1\) in parallel with a variable resistor \(R_2\). As the resistance of \(R_2\) is increased, which of the following quantities will INCREASE?
(1) The terminal voltage across the battery (2) The electric power dissipated by resistor \(R_1\) (3) The current passing through the battery
A.(1) only
B.(3) only
C.(1) and (2) only
D.(2) and (3) only
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解题
1. The equivalent external resistance is \(R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}\). As \(R_2\) increases, \(\frac{1}{R_2}\) decreases, so \(R_{\text{eq}}\) increases.
2. The total circuit current \(I = \frac{\mathcal{E}}{R_{\text{eq}} + r}\) decreases because \(R_{\text{eq}}\) increases. Hence, statement (3) is incorrect.
3. The terminal voltage of the battery is \(V = \mathcal{E} - Ir\). Since \(I\) decreases, \(Ir\) decreases and \(V\) increases. Hence, statement (1) is correct.
4. The voltage across \(R_1\) is the terminal voltage \(V\). The electric power dissipated by \(R_1\) is \(P_1 = \frac{V^2}{R_1}\). Since \(V\) increases and \(R_1\) is constant, \(P_1\) increases. Hence, statement (2) is correct.
Therefore, (1) and (2) only are correct.
评分标准
C (1A)
题目 28 · 選擇題
1 分
Ball \(P\) is projected horizontally from the edge of a cliff of height \(H\) with an initial speed \(u\) and lands on the ground at a horizontal distance \(D\) from the base of the cliff. Another ball \(Q\) of the same mass is projected horizontally from the same position with an initial speed \(2u\). Air resistance is negligible. Which of the following statements is/are correct?
(1) The time of flight of ball \(Q\) is equal to that of ball \(P\). (2) The horizontal distance travelled by ball \(Q\) is \(2D\). (3) The angle between the velocity vector and the horizontal just before landing for ball \(Q\) is half of that for ball \(P\).
A.(1) and (2) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解题
1. The vertical motion is governed by \(H = \frac{1}{2}gt^2\), so the time of flight is \(t = \sqrt{\frac{2H}{g}}\). Since \(H\) is identical, \(t_Q = t_P\). Statement (1) is correct.
2. The horizontal range is \(x = u_x t\). For \(Q\), \(x_Q = (2u)t = 2(ut) = 2D\). Statement (2) is correct.
3. The angle \(\theta\) with the horizontal just before landing satisfies \(\tan\theta = \frac{v_y}{v_x} = \frac{\sqrt{2gH}}{u_x}\). Thus \(\tan\theta_Q = \frac{1}{2}\tan\theta_P\). However, because \(\tan\theta\) is a non-linear function, \(\theta_Q \neq \frac{1}{2}\theta_P\). Statement (3) is incorrect.
Therefore, (1) and (2) only are correct.
评分标准
A (1A)
题目 29 · 選擇題
1 分
A conducting rod of length \(L\) and electrical resistance \(R\) moves at a constant speed \(v\) perpendicular to a uniform magnetic field \(B\) directed into the plane of the rails. The rod slides smoothly along two parallel, horizontal, frictionless rails of negligible resistance separated by a distance \(L\). A fixed resistor of resistance \(2R\) is connected between the ends of the rails. What is the mechanical power supplied by the external pulling force to maintain the constant speed \(v\)?
A.\(\frac{B^2 L^2 v^2}{9R}\)
B.\(\frac{B^2 L^2 v^2}{3R}\)
C.\(\frac{2B^2 L^2 v^2}{3R}\)
D.\(\frac{B^2 L^2 v^2}{R}\)
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解题
1. The induced electromotive force in the moving rod is \(\mathcal{E} = BLv\).
2. The total resistance of the closed circuit loop is \(R_{\text{total}} = R + 2R = 3R\).
3. The induced current circulating in the loop is \(I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{BLv}{3R}\).
4. The magnetic force acting on the rod is \(F_B = BIL = B\left(\frac{BLv}{3R}\right)L = \frac{B^2 L^2 v}{3R}\).
5. To maintain constant speed, the external pulling force must balance the magnetic force: \(F_{\text{ext}} = F_B\).
6. The mechanical power supplied is \(P = F_{\text{ext}} v = \left(\frac{B^2 L^2 v}{3R}\right)v = \frac{B^2 L^2 v^2}{3R}\).
评分标准
B (1A)
题目 30 · 選擇題
1 分
A fixed mass of an ideal gas is sealed in a rigid container of fixed volume. Initially, the gas is at a pressure of \(p_0\) and a temperature of \(27\ ^\circ\text{C}\). The gas is heated until the root-mean-square (r.m.s.) speed of its molecules is doubled. Find the new pressure and the new temperature (in \(^\circ\text{C}\)) of the gas.
\begin{tabular}{lcc} & \textbf{Pressure} & \textbf{Temperature} \\ A. & \(2p_0\) & \(54\ ^\circ\text{C}\) \\ B. & \(2p_0\) & \(327\ ^\circ\text{C}\) \\ C. & \(4p_0\) & \(108\ ^\circ\text{C}\) \\ D. & \(4p_0\) & \(927\ ^\circ\text{C}\) \end{tabular}
A.Pressure = \(2p_0\), Temperature = \(54\ ^\circ\text{C}\)
B.Pressure = \(2p_0\), Temperature = \(327\ ^\circ\text{C}\)
C.Pressure = \(4p_0\), Temperature = \(108\ ^\circ\text{C}\)
D.Pressure = \(4p_0\), Temperature = \(927\ ^\circ\text{C}\)
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解题
1. The r.m.s. speed of ideal gas molecules is given by \(c_{\text{rms}} = \sqrt{\frac{3kT}{m}}\), where \(T\) is absolute temperature in kelvins. Hence, \(c_{\text{rms}} \propto \sqrt{T}\), or \(T \propto c_{\text{rms}}^2\).
3. When the r.m.s. speed is doubled (\(c_{\text{rms},2} = 2 c_{\text{rms},1}\)), the new absolute temperature becomes \(T_2 = 2^2 \times T_1 = 4 \times 300\text{ K} = 1200\text{ K}\).
4. Converting back to Celsius: \(\theta_2 = 1200 - 273 = 927\ ^\circ\text{C}\).
5. For a gas in a rigid container of constant volume, the pressure law states \(\frac{p}{T} = \text{constant}\). Therefore, \(p_2 = p_1 \left(\frac{T_2}{T_1}\right) = p_0 \left(\frac{1200}{300}\right) = 4p_0\).
Thus, the new pressure is \(4p_0\) and the new temperature is \(927\ ^\circ\text{C}\).
评分标准
D (1A)
题目 31 · 選擇題
1 分
A ray of monochromatic light travels through three transparent media with flat, parallel boundaries, denoted as Layer 1, Layer 2, and Layer 3, with refractive indices \(n_1\), \(n_2\), and \(n_3\) respectively. The light enters Layer 2 from Layer 1 at an angle of incidence \(\theta\), and undergoes total internal reflection at the boundary between Layer 2 and Layer 3.
Which of the following statements MUST be correct?
(1) \(n_2 < n_1\) (2) \(n_2 > n_3\) (3) If the angle of incidence \(\theta\) in Layer 1 is increased, total internal reflection will still occur at the boundary between Layer 2 and Layer 3.
A.(1) only
B.(2) only
C.(1) and (3) only
D.(2) and (3) only
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解题
Let \(r\) be the angle of refraction at the boundary between Layer 1 and Layer 2. Since the boundaries are parallel, the normal to the 1–2 boundary is parallel to the normal to the 2–3 boundary. Thus, the angle of incidence at the 2–3 boundary is also \(r\).
1. For total internal reflection (TIR) to occur at the boundary between Layer 2 and Layer 3, the light must travel from a medium of higher refractive index to a medium of lower refractive index, which requires \(n_2 > n_3\). Therefore, statement (2) is correct.
2. From Snell's law at the 1–2 interface, \(n_1 \sin\theta = n_2 \sin r\). Total internal reflection at the 2–3 interface requires \(\sin r > \frac{n_3}{n_2}\), which gives \(n_1 \sin\theta > n_3\). This condition can be satisfied whether \(n_2\) is greater than, equal to, or less than \(n_1\). Thus, statement (1) is not necessarily correct.
3. If the angle of incidence \(\theta\) is increased, \(\sin\theta\) increases. Consequently, \(\sin r = \frac{n_1 \sin\theta}{n_2}\) increases, which means the angle of incidence \(r\) at the 2–3 interface increases. Since \(r\) was already greater than the critical angle \(c = \arcsin\left(\frac{n_3}{n_2}\right)\), it remains greater than \(c\). Hence, total internal reflection must still occur. Therefore, statement (3) is correct.
Hence, (2) and (3) only are correct.
评分标准
D (1A) - Statement (1) is incorrect because TIR depends on \(n_1\sin\theta > n_3\), placing no requirement on the relative magnitude of \(n_1\) and \(n_2\). - Statement (2) is correct as TIR only occurs when propagating from an optically denser to a less dense medium (\(n_2 > n_3\)). - Statement (3) is correct as increasing \(\theta\) increases the angle of incidence at the second boundary, keeping it above the critical angle.
题目 32 · 選擇題
1 分
In the circuit shown below, a battery of constant e.m.f. \(\mathcal{E}\) and non-zero internal resistance \(r\) is connected to three identical light bulbs \(X\), \(Y\), and \(Z\), and an open switch \(S\).
Bulb \(X\) is connected in series with the parallel combination of bulb \(Y\) and the branch containing bulb \(Z\) and switch \(S\).
What happens to the brightness of bulb \(X\) and bulb \(Y\) after switch \(S\) is closed?
$$\begin{array}{ccc} & \text{\textbf{Brightness of bulb }} X & \text{\textbf{Brightness of bulb }} Y \\ \text{A.} & \text{Increases} & \text{Increases} \\ \text{B.} & \text{Increases} & \text{Decreases} \\ \text{C.} & \text{Decreases} & \text{Increases} \\ \text{D.} & \text{Decreases} & \text{Decreases} \end{array}$$
A.Increases | Increases
B.Increases | Decreases
C.Decreases | Increases
D.Decreases | Decreases
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解题
Let the resistance of each identical bulb be \(R\).
1. Before switch \(S\) is closed, the branch with bulb \(Z\) is an open circuit. The total equivalent resistance of the external circuit is: \[ R_{\text{ext, initial}} = R_X + R_Y = 2R \] The total circuit current is \(I_{\text{initial}} = \frac{\mathcal{E}}{2R + r}\).
2. When switch \(S\) is closed, bulbs \(Y\) and \(Z\) are connected in parallel, with equivalent resistance \(R_{YZ} = \frac{R}{2}\). The new total equivalent external resistance is: \[ R_{\text{ext, final}} = R_X + R_{YZ} = R + \frac{R}{2} = 1.5R \] Since the total resistance of the circuit decreases, the total current \(I_{\text{final}} = \frac{\mathcal{E}}{1.5R + r}\) increases.
3. Brightness of bulb \(X\): Bulb \(X\) carries the total current \(I\). Since the total current increases, the power dissipated by bulb \(X\) (\(P_X = I^2 R\)) increases, so its brightness increases.
4. Brightness of bulb \(Y\): The terminal potential difference is \(V_{\text{term}} = \mathcal{E} - Ir\). The voltage across bulb \(Y\) is: \[ V_Y = V_{\text{term}} - V_X = \mathcal{E} - I(R + r) \] Since \(I\) increases, \(V_Y\) decreases. The power dissipated by bulb \(Y\) (\(P_Y = \frac{V_Y^2}{R}\)) decreases, so its brightness decreases.
Therefore, option B is correct.
评分标准
B (1A) - Brightness of \(X\) increases because total circuit resistance decreases, leading to an increased main current through \(X\). - Brightness of \(Y\) decreases because the increased current causes larger potential drops across \(X\) and internal resistance \(r\), reducing the p.d. across parallel branch \(Y\).
题目 33 · 選擇題
1 分
Two identical particles, \(P\) and \(Q\), are projected simultaneously from flat horizontal ground into the air. Air resistance is negligible. Both particles reach the same maximum vertical height \(H\), but particle \(P\) achieves a greater horizontal range than particle \(Q\).
Which of the following statements MUST be correct?
(1) Both particles take the same total time to return to the ground. (2) The initial kinetic energy of particle \(P\) is greater than that of particle \(Q\). (3) The angle of projection of particle \(P\) above the horizontal is greater than that of particle \(Q\).
A.(1) and (2) only
B.(1) and (3) only
C.(2) and (3) only
D.(1), (2) and (3)
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解题
Let \(u_x\) and \(u_y\) be the initial horizontal and vertical velocity components of a particle, and let \(m\) be the mass of each particle.
1. Maximum height reached is given by: \[ H = \frac{u_y^2}{2g} \] Since both particles reach the same maximum height \(H\), their initial vertical velocity components must be identical: \(u_{P,y} = u_{Q,y}\).
2. Time of flight is given by: \[ T = \frac{2u_y}{g} \] Since \(u_{P,y} = u_{Q,y}\), both particles have the same time of flight \(T_P = T_Q\). Thus, statement (1) is correct.
3. Horizontal range is given by: \[ R = u_x T \] Given \(R_P > R_Q\) and \(T_P = T_Q\), it follows that \(u_{P,x} > u_{Q,x}\).
4. The initial kinetic energy is: \[ E_k = \frac{1}{2}m(u_x^2 + u_y^2) \] Since \(u_{P,x} > u_{Q,x}\) and \(u_{P,y} = u_{Q,y}\), the initial kinetic energy of \(P\) is greater than that of \(Q\). Thus, statement (2) is correct.
5. The angle of projection \(\theta\) satisfies: \[ \tan\theta = \frac{u_y}{u_x} \] Since \(u_{P,x} > u_{Q,x}\) and \(u_{P,y} = u_{Q,y}\), we have \(\tan\theta_P < \tan\theta_Q\), which implies \(\theta_P < \theta_Q\). Thus, the angle of projection of \(P\) is smaller than that of \(Q\). Statement (3) is incorrect.
Therefore, statements (1) and (2) only are correct.
评分标准
A (1A) - Statement (1) is correct: same maximum height implies equal initial vertical velocity and thus equal flight time. - Statement (2) is correct: larger range with equal flight time requires larger horizontal velocity component, leading to higher initial speed and kinetic energy. - Statement (3) is incorrect: larger horizontal velocity with equal vertical velocity gives a smaller launch angle relative to the horizontal.
A stunt performer rides a motorcycle horizontally off a cliff of height \( h = 45.0\text{ m} \) at a speed of \( u = 24.0\text{ m s}^{-1} \). At the same instant, a drone takes off vertically upwards from the flat ground below at a horizontal distance of \( d = 72.0\text{ m} \) from the cliff base, moving with a constant upward acceleration \( a \). Assume air resistance is negligible. (Take \( g = 9.81\text{ m s}^{-2} \))
(a) Calculate the time taken for the motorcycle to reach the horizontal position of the drone. (2 marks)
(b) Find the vertical distance fallen by the motorcycle when it reaches the horizontal position of the drone. (2 marks)
(c) The drone is intended to be at the exact same height as the motorcycle at the moment the motorcycle passes over its horizontal position. Determine the required acceleration \( a \) of the drone. (3 marks)
(d) State and explain whether the motorcycle hits the ground before or after passing the drone's position. (2 marks)
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解题
(a) Horizontal motion of the motorcycle is at constant velocity \( u = 24.0\text{ m s}^{-1} \). \( t = \frac{d}{u} = \frac{72.0}{24.0} = 3.00\text{ s} \)
(b) Vertical displacement fallen by the motorcycle: \( s_y = \frac{1}{2}gt^2 = \frac{1}{2}(9.81)(3.00)^2 = 44.145\text{ m} \approx 44.1\text{ m} \)
(c) The height of the motorcycle above the ground at \( t = 3.00\text{ s} \) is: \( H_{\text{motorcycle}} = h - s_y = 45.0 - 44.145 = 0.855\text{ m} \) For the drone starting from rest \( (u_0 = 0) \): \( s_{\text{drone}} = \frac{1}{2}at^2 \) \( 0.855 = \frac{1}{2}a(3.00)^2 \) \( a = \frac{2 \times 0.855}{9.00} = 0.190\text{ m s}^{-2} \) (or \( 0.200\text{ m s}^{-2} \) if taking \( g = 9.80\text{ m s}^{-2} \)) Using \( g = 9.81\text{ m s}^{-2} \): \( a = 0.190\text{ m s}^{-2} \).
(d) Total time for the motorcycle to reach the ground: \( 45.0 = \frac{1}{2}(9.81)t_{\text{total}}^2 \implies t_{\text{total}} = \sqrt{\frac{2 \times 45.0}{9.81}} = 3.03\text{ s} \). Since \( t_{\text{total}} = 3.03\text{ s} > 3.00\text{ s} \), the motorcycle passes the drone before hitting the ground.
评分标准
(a) \( t = \frac{d}{u} = \frac{72.0}{24.0} \) (1M) \( t = 3.00\text{ s} \) (1A)
(c) Height of motorcycle \( = 45.0 - 44.145 = 0.855\text{ m} \) (1M) \( 0.855 = \frac{1}{2}a(3.00)^2 \) (1M) \( a = 0.190\text{ m s}^{-2} \) (Accept \( 0.19\text{ m s}^{-2} \) to \( 0.20\text{ m s}^{-2} \)) (1A)
(d) Calculation of total flight time: \( t_{\text{total}} = \sqrt{\frac{2 \times 45.0}{9.81}} = 3.03\text{ s} \) (1A) Comparison showing \( t = 3.00\text{ s} < 3.03\text{ s} \), so it passes the drone before hitting the ground. (1A)
题目 2 · 結構題
9 分
A glass block of semicircular cross-section with radius \( R = 8.0\text{ cm} \) and refractive index \( n = 1.60 \) is surrounded by air. A narrow beam of monochromatic light is directed towards the flat surface at an angle of incidence \( \theta = 30.0^\circ \) at the center \( O \) of the flat face.
(a) Calculate the angle of refraction \( r \) inside the glass block. (2 marks)
(b) State the angle of incidence when this refracted ray reaches the curved surface, and explain why the ray emerges into the air without changing direction at the curved surface. (2 marks)
(c) Calculate the critical angle \( C \) for the glass-air boundary. (2 marks)
(d) The beam's angle of incidence \( \theta \) at the flat face is now increased. Determine the maximum value of \( \theta \) such that total internal reflection will NOT occur at the flat surface, or explain why total internal reflection can never occur at this first boundary. (3 marks)
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解题
(a) By Snell's law: \( 1.00 \sin 30.0^\circ = 1.60 \sin r \) \( \sin r = \frac{0.500}{1.60} = 0.3125 \) \( r = 18.21^\circ \approx 18.2^\circ \)
(b) The ray originates from the center of the semicircle \( O \), so it travels along the radius of the circular boundary. Thus, it strikes the curved surface perpendicularly (angle of incidence \( = 0^\circ \)). Since the angle of incidence is \( 0^\circ \), the angle of refraction is also \( 0^\circ \), so the ray passes straight through without bending.
(c) Critical angle \( C \): \( \sin C = \frac{1}{n} = \frac{1}{1.60} = 0.625 \) \( C = 38.68^\circ \approx 38.7^\circ \)
(d) Total internal reflection can only occur when light travels from an optically denser medium to an optically less dense medium (i.e., from a medium of higher refractive index to one of lower refractive index). Since light is traveling from air (\( n = 1.00 \)) to glass (\( n = 1.60 \)), \( n_1 < n_2 \), the angle of refraction is always less than the angle of incidence. Therefore, total internal reflection can NEVER occur at the flat entrance surface for any angle of incidence \( 0^\circ \le \theta < 90^\circ \).
评分标准
(a) \( 1.00 \sin 30.0^\circ = 1.60 \sin r \) (1M) \( r = 18.2^\circ \) (1A)
(b) Angle of incidence is \( 0^\circ \) (1A) Explanation: The ray travels along a normal/radial line of the circular arc, so no refraction/deviation occurs. (1A)
(c) \( \sin C = \frac{1}{1.60} \) (1M) \( C = 38.7^\circ \) (1A)
(d) Statement: Total internal reflection cannot occur. (1A) Reason: Light travels from an optically less dense medium (air) to a denser medium (glass) / \( n_{\text{air}} < n_{\text{glass}} \). (1A) The refracted ray always bends towards the normal, so \( r < 90^\circ \) for all \( \theta < 90^\circ \). (1A)
题目 3 · 結構題
9 分
A circuit contains a real battery of electromotive force (e.m.f.) \( \mathcal{E} \) and internal resistance \( r \), connected in series with a variable resistor \( R \) and an ideal ammeter. A voltmeter of infinite resistance is connected across the terminals of the battery.
When the resistance \( R \) is set to \( 5.0\ \Omega \), the ammeter reads \( 1.50\text{ A} \). When \( R \) is changed to \( 11.0\ \Omega \), the ammeter reads \( 0.75\text{ A} \).
(a) Write an equation relating \( \mathcal{E} \), terminal voltage \( V \), current \( I \), and internal resistance \( r \). (1 mark)
(b) Determine: (i) the e.m.f. \( \mathcal{E} \) and internal resistance \( r \) of the battery; (3 marks) (ii) the reading of the voltmeter when \( R = 11.0\ \Omega \). (2 marks)
(c) A student claims: 'To maximize the power delivered to \( R \), the value of \( R \) should be as close to zero as possible.' Without deriving the maximum power theorem mathematically, explain whether this claim is correct. (3 marks)
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解题
(a) \( V = \mathcal{E} - Ir \) or \( \mathcal{E} = I(R + r) \).
(b)(ii) Terminal voltage \( V = IR = 0.75 \times 11.0 = 8.25\text{ V} \) (or \( V = \mathcal{E} - Ir = 9.0 - 0.75 \times 1.0 = 8.25\text{ V} \)).
(c) The student's claim is incorrect. Power delivered to \( R \) is \( P = I^2 R = \frac{\mathcal{E}^2 R}{(R+r)^2} \). When \( R \to 0 \), the potential difference across \( R \) tends to 0, so the power delivered to \( R \) tends to zero (almost all power is dissipated inside the internal resistance \( r \)). Maximum power transfer occurs when the external resistance equals the internal resistance (\( R = r \)).
评分标准
(a) \( V = \mathcal{E} - Ir \) or \( \mathcal{E} = I(R + r) \) (1A)
(b)(i) Formulating two equations: \( \mathcal{E} = 1.50(5.0 + r) \) and \( \mathcal{E} = 0.75(11.0 + r) \) (1M) Solving for \( r = 1.0\ \Omega \) (1A) Solving for \( \mathcal{E} = 9.0\text{ V} \) (1A)
(b)(ii) \( V = IR = 0.75 \times 11.0 \) or \( V = 9.0 - 0.75(1.0) \) (1M) \( V = 8.25\text{ V} \) (1A)
(c) State that the claim is incorrect. (1A) Reasoning: As \( R \to 0 \), although current is large, the potential difference across \( R \) approaches zero (or power dissipated in \( R \) tends to zero / most power is wasted in \( r \)). (1A) State that maximum power occurs when \( R = r \) (or power increases then decreases as \( R \) increases). (1A)
题目 4 · 結構題
10 分
A rigid rectangular metal frame of width \( L = 0.20\text{ m} \), length \( 0.40\text{ m} \), and total resistance \( R = 0.50\ \Omega \) enters a region of uniform magnetic field \( B = 0.80\text{ T} \) at a constant velocity \( v = 2.5\text{ m s}^{-1} \) perpendicular to the field boundary. The magnetic field is directed perpendicularly into the page.
(a) Explain why an induced e.m.f. is generated in the frame as it enters the magnetic field. (2 marks)
(b) Calculate the magnitude of the induced electromotive force (e.m.f.) and the induced current in the frame while it is entering the magnetic field. (3 marks)
(c) State the direction of the induced current (clockwise or counter-clockwise) and deduce the direction of the magnetic force acting on the frame. (2 marks)
(d) Calculate the external mechanical power required to maintain the frame moving at the constant velocity of \( 2.5\text{ m s}^{-1} \) during entry. (3 marks)
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解题
(a) As the frame enters the field, the magnetic flux linkage \( \Phi = B A \) through the loop increases with time. According to Faraday's Law of Electromagnetic Induction, an induced e.m.f. is produced which is equal to the rate of change of magnetic flux linkage (\( \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} \)).
(b) Magnitude of induced e.m.f.: \( \mathcal{E} = B L v = (0.80)(0.20)(2.5) = 0.40\text{ V} \) Induced current: \( I = \frac{\mathcal{E}}{R} = \frac{0.40}{0.50} = 0.80\text{ A} \)
(c) By Lenz's law, the induced current creates an outward magnetic field to oppose the increase of inward flux. Thus, by the right-hand grip rule, the induced current flows in the counter-clockwise direction. Using Fleming's left-hand rule / \( \mathbf{F} = I\mathbf{L} \times \mathbf{B} \), the magnetic force on the leading edge acts to the left (opposite to the velocity), opposing the motion.
(d) Magnetic force opposing motion: \( F_{\text{mag}} = B I L = (0.80)(0.80)(0.20) = 0.128\text{ N} \) To keep velocity constant, external force \( F_{\text{ext}} = F_{\text{mag}} = 0.128\text{ N} \). Mechanical power required: \( P = F_{\text{ext}} v = 0.128 \times 2.5 = 0.32\text{ W} \) (Alternative method: \( P = I^2 R = (0.80)^2 \times 0.50 = 0.32\text{ W} \)).
评分标准
(a) Recognition of magnetic flux / area inside field changing with time (1A) Reference to Faraday's law: \( \mathcal{E} \propto \frac{\Delta \Phi}{\Delta t} \) (1A)
(b) \( \mathcal{E} = B L v = (0.80)(0.20)(2.5) = 0.40\text{ V} \) (1M + 1A) \( I = \frac{\mathcal{E}}{R} = \frac{0.40}{0.50} = 0.80\text{ A} \) (1A)
(c) Direction of induced current: counter-clockwise (1A) Direction of magnetic force: to the left / opposite to the direction of motion (1A)
(d) \( F_{\text{ext}} = B I L = (0.80)(0.80)(0.20) = 0.128\text{ N} \) (1M) \( P = F v = 0.128 \times 2.5 \) OR \( P = I^2 R = (0.80)^2(0.50) \) (1M) \( P = 0.32\text{ W} \) (1A)
题目 5 · 結構題
10 分
An electric kettle containing \( 0.80\text{ kg} \) of water at \( 25.0^\circ\text{C} \) is heated by an immersion heater rated at \( 1500\text{ W} \). The water reaches its boiling point of \( 100.0^\circ\text{C} \) in \( 180\text{ s} \).
(Given: specific heat capacity of water \( c_w = 4200\text{ J kg}^{-1}\text{ }^\circ\text{C}^{-1} \); specific latent heat of vaporization of water \( l_v = 2.26 \times 10^6\text{ J kg}^{-1} \))
(a) Calculate the useful energy absorbed by the water to heat it from \( 25.0^\circ\text{C} \) to \( 100.0^\circ\text{C} \). (2 marks)
(b) Calculate the total electrical energy supplied by the heater in \( 180\text{ s} \), and hence determine the efficiency of the heating process. (3 marks)
(c) After boiling begins, the kettle continues to operate at \( 1500\text{ W} \) for another \( 120\text{ s} \). Assuming that heat loss to the surroundings occurs at a constant rate of \( 240\text{ W} \) during boiling, calculate the mass of water converted into steam during this \( 120\text{ s} \). (3 marks)
(d) Explain in molecular terms why the temperature of the water remains constant at \( 100.0^\circ\text{C} \) during boiling even though heat is continuously supplied. (2 marks)
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解题
(a) Useful energy absorbed: \( Q = m c_w \Delta T = (0.80)(4200)(100.0 - 25.0) = (0.80)(4200)(75.0) = 2.52 \times 10^5\text{ J} \)
(b) Total electrical energy supplied: \( E = P t = 1500 \times 180 = 2.70 \times 10^5\text{ J} \) Efficiency: \( \eta = \frac{Q}{E} \times 100\% = \frac{2.52 \times 10^5}{2.70 \times 10^5} \times 100\% = 93.33\% \approx 93.3\% \)
(c) Net rate of heat supplied to water during boiling: \( P_{\text{net}} = P_{\text{heater}} - P_{\text{loss}} = 1500 - 240 = 1260\text{ W} \) Total net energy received in \( 120\text{ s} \): \( Q_{\text{boil}} = P_{\text{net}} \times t = 1260 \times 120 = 1.512 \times 10^5\text{ J} \) Mass of steam formed \( m_s \): \( Q_{\text{boil}} = m_s l_v \implies m_s = \frac{1.512 \times 10^5}{2.26 \times 10^6} \approx 0.0669\text{ kg} = 66.9\text{ g} \)
(d) During boiling, the supplied thermal energy is absorbed to overcome/break the intermolecular forces of attraction between water molecules and do work against atmospheric pressure as volume increases. This increases the molecular potential energy, while the average random kinetic energy of the molecules remains unchanged, so the temperature remains constant.
评分标准
(a) \( Q = m c \Delta T = (0.80)(4200)(75.0) \) (1M) \( Q = 2.52 \times 10^5\text{ J} \) (1A)
(b) \( E = P t = 1500 \times 180 = 2.70 \times 10^5\text{ J} \) (1A) \( \text{Efficiency} = \frac{2.52 \times 10^5}{2.70 \times 10^5} \times 100\% \) (1M) \( \eta = 93.3\% \) (Accept 93% or 0.933) (1A)
(d) Heat energy is used to break/overcome intermolecular bonds (or do work against atmospheric expansion) / increases molecular potential energy (1A) Average kinetic energy of molecules does not change, hence temperature stays constant (1A)
题目 6 · 結構題
9 分
A student investigates the optical properties of a transparent liquid contained in a thin-walled semi-circular glass container of radius \( R = 8.0\text{ cm} \). A narrow ray of monochromatic light is directed towards the curved surface at an angle of incidence \( \theta_1 \) to the normal and enters the liquid at point \( P \), directed towards the centre \( O \) on the flat surface, as shown below.
[Diagram: A semi-circular container filled with liquid. The flat surface lies horizontally at the bottom with centre \( O \). A light ray enters normally through the curved surface at point \( P \) and strikes point \( O \) on the flat boundary between the liquid and air. The angle between the ray in liquid and the normal to the flat surface is \( \phi \).]
(a) State why the ray does NOT bend at point \( P \) when entering the liquid from air. (1 mark)
(b) The angle \( \phi \) inside the liquid at point \( O \) is varied. When \( \phi = 38.0^\circ \), the refracted ray in air emerges into air along the flat surface. (i) State the name of the phenomenon occurring when \( \phi \) is slightly increased beyond \( 38.0^\circ \). (1 mark) (ii) Calculate the refractive index \( n \) of the liquid. (2 marks)
(c) The container is now replaced with another identical one containing a different oil of refractive index \( n_{\text{oil}} = 1.60 \). A thin layer of water (refractive index \( n_{\text{water}} = 1.33 \)) is placed on top of the flat surface in contact with the oil. (i) Calculate the critical angle for a ray travelling from the oil into the water layer. (2 marks) (ii) A ray in the oil strikes the oil-water interface at an angle of incidence of \( 60.0^\circ \). Determine the angle of refraction of this ray when it eventually emerges into the air above the water, or state with justification if it cannot emerge into air. (3 marks)
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解题
(a) Since the ray is directed towards the centre \( O \) of the semi-circle, it travels along the radius and is perpendicular (normal) to the curved surface at point \( P \). The angle of incidence at \( P \) is \( 0^\circ \), so no refraction (bending) occurs.
(b) (i) Total internal reflection.
(ii) At the critical angle \( \phi_c = 38.0^\circ \): \[ n \sin(\phi_c) = 1 \] \[ n = \frac{1}{\sin(38.0^\circ)} = \frac{1}{0.61566} \approx 1.624 \approx 1.62 \]
(ii) Since the angle of incidence at the oil-water interface is \( 60.0^\circ \), which is greater than the critical angle \( \theta_c = 56.2^\circ \), total internal reflection occurs at the oil-water interface. Therefore, the light does not enter the water layer at all and cannot emerge into the air.
评分标准
(a) Ray is incident along the normal / radius of the curved surface (angle of incidence is 0°). (1A)
(b)(i) Total internal reflection. (1A)
(b)(ii) \( n = \frac{1}{\sin 38.0^\circ} \) (1M) \( n = 1.62 \) (accept 1.624) (1A)
(c)(ii) Angle of incidence \( 60.0^\circ > 56.2^\circ \) (critical angle) (1M) Total internal reflection occurs at the oil-water boundary (1A) The ray does not enter water and thus cannot emerge into air. (1A)
题目 7 · 結構題
9 分
A student sets up a circuit to investigate the electrical characteristics of a DC power supply with an electromotive force (emf) \( \mathcal{E} \) and internal resistance \( r \). The power supply is connected in series with an ammeter of negligible resistance, a switch \( S \), and a variable resistor (rheostat) \( R \). A high-resistance voltmeter is connected directly across the terminals of the power supply.
(a) Draw a circuit diagram representing the experimental setup. (2 marks)
(b) By adjusting the resistance of the rheostat, the student records several pairs of terminal voltage \( V \) and current \( I \). The linear relationship is plotted on a graph of \( V \) against \( I \). (i) Write down the equation relating \( V \), \( I \), \( \mathcal{E} \), and \( r \). (1 mark) (ii) State how \( \mathcal{E} \) and \( r \) can be found from the graph of \( V \) against \( I \). (2 marks)
(c) The graph yields \( \mathcal{E} = 6.0\text{ V} \) and \( r = 2.5\ \Omega \). (i) Calculate the maximum power that the power supply can deliver to the external load \( R \). (2 marks) (ii) Find the efficiency of the power supply when it delivers maximum power to the external load. (2 marks)
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解题
(a) A series loop containing the cell/battery (with internal resistance \( r \)), ammeter \( A \), switch \( S \), and variable resistor \( R \), with a voltmeter \( V \) connected in parallel across the terminals of the battery.
(b) (i) The equation is: \[ V = \mathcal{E} - I r \]
(ii) The emf \( \mathcal{E} \) is given by the vertical intercept (\( y \)-intercept) of the graph. The internal resistance \( r \) is equal to the negative of the slope (gradient) of the graph, i.e., \( r = -\text{slope} \).
(c) (i) Maximum power transfer occurs when the external load equals the internal resistance, i.e., \( R = r = 2.5\ \Omega \). The current at maximum power is: \[ I = \frac{\mathcal{E}}{R + r} = \frac{6.0}{2.5 + 2.5} = 1.2\text{ A} \] The maximum output power is: \[ P_{\text{max}} = I^2 R = (1.2)^2 \times 2.5 = 3.6\text{ W} \] Or using \( P_{\text{max}} = \frac{\mathcal{E}^2}{4r} = \frac{6.0^2}{4 \times 2.5} = 3.6\text{ W} \).
(ii) Efficiency \( \eta \) is: \[ \eta = \frac{P_{\text{out}}}{P_{\text{total}}} \times 100\% = \frac{I^2 R}{I^2(R + r)} \times 100\% = \frac{R}{R+r} \times 100\% \] Since \( R = r \): \[ \eta = \frac{2.5}{2.5 + 2.5} \times 100\% = 50\% \]
评分标准
(a) - Correct symbols for battery, ammeter, rheostat, switch in series. (1A) - Voltmeter connected in parallel across the power supply. (1A)
(b)(i) \( V = \mathcal{E} - Ir \) (1A)
(b)(ii) - \( \mathcal{E} \) is the vertical / y-intercept. (1A) - \( r \) is the negative of the gradient / slope (or magnitude of slope). (1A)
(c)(i) - Recognition that maximum power occurs when \( R = r = 2.5\ \Omega \) OR use of \( P = \frac{\mathcal{E}^2}{4r} \). (1M) - \( P_{\text{max}} = 3.6\text{ W} \) (1A)
A small ball of mass \( m = 0.20\text{ kg} \) is projected horizontally with an initial speed \( u = 12.0\text{ m s}^{-1} \) from the edge of a horizontal platform at height \( h = 19.6\text{ m} \) above level ground. Air resistance is negligible. (Take \( g = 9.81\text{ m s}^{-2} \))
(a) Calculate the time taken for the ball to reach the ground. (2 marks)
(b) Find the horizontal distance \( d \) from the base of the platform to the landing point. (1 mark)
(c) Determine the magnitude and direction of the velocity of the ball just before it hits the ground. (4 marks)
(d) A second identical ball is thrown from the same point at the same instant with the same speed \( u = 12.0\text{ m s}^{-1} \), but at an angle of \( 30.0^\circ \) above the horizontal. State and explain whether the speed of this second ball just before hitting the ground is greater than, less than, or equal to that of the first ball. (2 marks)
(b) Horizontal distance: \[ d = u_x t = (12.0)(2.00) = 24.0\text{ m} \]
(c) Horizontal component of final velocity: \[ v_x = u = 12.0\text{ m s}^{-1} \] Vertical component of final velocity: \[ v_y = u_y + g t = 0 + (9.81)(2.00) = 19.62\text{ m s}^{-1} \quad (\text{downwards}) \] Magnitude of velocity: \[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(12.0)^2 + (19.62)^2} = \sqrt{144 + 384.94} = \sqrt{528.94} \approx 23.0\text{ m s}^{-1} \] Direction \( \theta \) below the horizontal: \[ \tan \theta = \frac{v_y}{v_x} = \frac{19.62}{12.0} = 1.635 \] \[ \theta = \tan^{-1}(1.635) \approx 58.6^\circ \]
(d) The speeds are equal. By conservation of mechanical energy: \[ \frac{1}{2} m u^2 + mgh = \frac{1}{2} m v^2 \implies v = \sqrt{u^2 + 2gh} \] Since both balls have the same initial speed \( u \) and fall through the same vertical height \( h \), their final kinetic energies and hence their final speeds must be identical (ignoring air resistance).
(c) - \( v_y = gt = 19.62\text{ m s}^{-1} \) (or via \( v_y^2 = 2gh \)) (1M) - \( v = \sqrt{12.0^2 + 19.62^2} \) (1M) - \( v = 23.0\text{ m s}^{-1} \) (accept 22.9 to 23.0) (1A) - \( \theta = 58.6^\circ \) below horizontal (accept 58.5° to 58.6°) (1A)
(d) - Stating that the speeds are equal. (1A) - Explanation based on conservation of mechanical energy (or \( v = \sqrt{u^2 + 2gh} \) depends only on initial speed and vertical drop \( h \)). (1A)
题目 9 · 結構題
9 分
A conducting rod \( PQ \) of length \( L = 0.40\text{ m} \) and resistance \( R = 1.5\ \Omega \) rests on two parallel, horizontal, frictionless metal rails separated by distance \( L \). The rails have negligible resistance and are connected to a resistor of resistance \( R_L = 3.5\ \Omega \) at one end. A uniform magnetic field of flux density \( B = 0.75\text{ T} \) points vertically downwards into the plane of the rails.
A constant horizontal external force \( F_{\text{ext}} = 0.18\text{ N} \) pulls the rod towards the right from rest at time \( t = 0 \).
(a) Explain why an induced current flows through the rod when it moves, and state the direction of the induced current in rod \( PQ \) (from \( P \) to \( Q \) or from \( Q \) to \( P \), where \( P \) is the top end and \( Q \) is the bottom end). (2 marks)
(b) When the rod moves at speed \( v \), express the induced electromotive force \( \mathcal{E} \) and the magnetic braking force \( F_B \) acting on the rod in terms of \( B \), \( L \), \( R_{\text{total}} \), and \( v \). (2 marks)
(c) The rod eventually reaches a constant terminal speed \( v_t \). (i) Calculate the terminal speed \( v_t \). (3 marks) (ii) Show that at terminal speed, the rate of mechanical work done by \( F_{\text{ext}} \) is equal to the total rate of electrical energy dissipation in the circuit. (2 marks)
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解题
(a) As the rod moves to the right, the area enclosed by the circuit increases, leading to an increasing magnetic flux linkage through the circuit. By Faraday's law, an emf is induced, producing a current. By Lenz's law / Fleming's right-hand rule, to oppose the increase in flux (or produce an opposing force to the left), the induced current must flow from \( Q \) to \( P \) along the rod (anticlockwise in the circuit).
(b) Induced emf: \[ \mathcal{E} = B L v \] Total resistance of the circuit: \[ R_{\text{total}} = R + R_L \] Induced current: \[ I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{B L v}{R_{\text{total}}} \] Magnetic braking force (opposing force): \[ F_B = B I L = B \left(\frac{B L v}{R_{\text{total}}}\right) L = \frac{B^2 L^2 v}{R_{\text{total}}} \]
(c) (i) At terminal speed \( v_t \), the net force on the rod is zero, so \( F_{\text{ext}} = F_B \): \[ F_{\text{ext}} = \frac{B^2 L^2 v_t}{R + R_L} \] \[ 0.18 = \frac{(0.75)^2 (0.40)^2 v_t}{1.5 + 3.5} \] \[ 0.18 = \frac{(0.5625)(0.16) v_t}{5.0} = \frac{0.090 v_t}{5.0} = 0.018 v_t \] \[ v_t = \frac{0.18}{0.018} = 10.0\text{ m s}^{-1} \]
(ii) Mechanical power input: \[ P_{\text{mech}} = F_{\text{ext}} \times v_t = 0.18 \times 10.0 = 1.80\text{ W} \] Total electrical power dissipated: \[ \mathcal{E} = B L v_t = 0.75 \times 0.40 \times 10.0 = 3.0\text{ V} \] \[ P_{\text{elec}} = \frac{\mathcal{E}^2}{R_{\text{total}}} = \frac{(3.0)^2}{5.0} = \frac{9.0}{5.0} = 1.80\text{ W} \] Since \( P_{\text{mech}} = P_{\text{elec}} = 1.80\text{ W} \), the rate of work done by \( F_{\text{ext}} \) equals the rate of electrical energy dissipation.
评分标准
(a) - Movement causes change/increase in magnetic flux linkage through the closed loop. (1A) - Direction of current: from \( Q \) to \( P \). (1A)
The \(H_\beta\) spectral line in the absorption spectrum of a distant galaxy is observed at a wavelength of \(510.3\text{ nm}\). The rest wavelength of this line measured in a laboratory is \(486.0\text{ nm}\). If the Hubble constant is taken as \(70\text{ km s}^{-1}\text{ Mpc}^{-1}\), estimate the distance of this galaxy from the Earth. (Given: speed of light in vacuum \(c = 3.00 \times 10^8\text{ m s}^{-1}\))
Next, find the recessional speed \(v\): \(v = z c = 0.0500 \times 3.00 \times 10^5\text{ km s}^{-1} = 1.50 \times 10^4\text{ km s}^{-1}\)
According to Hubble's law, \(v = H_0 d\): \(d = \frac{v}{H_0} = \frac{1.50 \times 10^4\text{ km s}^{-1}}{70\text{ km s}^{-1}\text{ Mpc}^{-1}} \approx 214\text{ Mpc}\)
评分标准
C (1 mark) - Correct application of redshift relation \(z = \Delta \lambda / \lambda_0 = v / c\) - Correct use of Hubble's Law \(v = H_0 d\)
题目 2 · MC
1 分
Monochromatic light of frequency \(f\) and intensity \(I\) is incident on a clean metal surface with work function \(\Phi\), resulting in the emission of photoelectrons. Which of the following modifications, when applied alone, will INCREASE the stopping potential of the emitted photoelectrons?
(1) Increasing the intensity \(I\) of the incident light while keeping \(f\) unchanged (2) Increasing the frequency \(f\) of the incident light while keeping \(I\) unchanged (3) Replacing the metal with another metal having a smaller work function \(\Phi\)
A.(1) only
B.(3) only
C.(1) and (2) only
D.(2) and (3) only
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解题
By Einstein's photoelectric equation: \(e V_s = h f - \Phi \implies V_s = \frac{h f - \Phi}{e}\)
(1) Incorrect: Increasing intensity \(I\) increases the rate of photon arrival and saturation current, but does not alter photon energy \(h f\); thus \(V_s\) remains unchanged. (2) Correct: Increasing frequency \(f\) increases the kinetic energy of emitted photoelectrons \(h f - \Phi\), thereby increasing \(V_s\). (3) Correct: Decreasing work function \(\Phi\) increases the maximum kinetic energy for the same photon frequency, thus increasing \(V_s\).
Therefore, (2) and (3) only are correct.
评分标准
D (1 mark) - Identifies that stopping potential depends only on photon energy \(hf\) and work function \(\Phi\), independent of light intensity.
题目 3 · MC
1 分
A wind turbine with blades of length \(20\text{ m}\) operates in a region where the average wind speed is \(12\text{ m s}^{-1}\). The density of air is \(1.2\text{ kg m}^{-3}\). If the turbine has an overall conversion efficiency of \(35\%\) in generating electricity from the total kinetic energy of the incoming air swept by the blades, what is the electrical power output of the turbine? (Take \(\pi \approx 3.14\))
A.\(76\text{ kW}\)
B.\(228\text{ kW}\)
C.\(456\text{ kW}\)
D.\(1303\text{ kW}\)
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解题
Area swept by blades \(A = \pi r^2 = \pi (20)^2 = 400\pi \approx 1256.6\text{ m}^2\).
Total kinetic power of the wind entering the swept area: \(P_{\text{wind}} = \frac{1}{2} \rho A v^3 = \frac{1}{2} (1.2)(1256.6)(12)^3 = 0.6 \times 1256.6 \times 1728 \approx 1.303 \times 10^6\text{ W} = 1303\text{ kW}\)
C (1 mark) - Correct formula for wind power \(P = \frac{1}{2}\rho A v^3\) - Correct application of efficiency factor \(\eta = 0.35\)
题目 4 · MC
1 分
An ultrasound pulse travels through muscle tissue (acoustic impedance \(Z_1 = 1.70 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}\)) and hits the boundary with bone (acoustic impedance \(Z_2 = 6.80 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}\)) at normal incidence. What percentage of the incident ultrasound intensity is reflected back into the muscle tissue?
A.\(36\%\)
B.\(40\%\)
C.\(60\%\)
D.\(64\%\)
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解题
The intensity reflection coefficient \(\alpha\) at normal incidence is given by: \(\alpha = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2\)
A (1 mark) - Correct substitution into reflection coefficient formula \(\alpha = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2\)
题目 5 · MC
1 分
A narrow beam of monoenergetic X-rays passes through a lead shield. A lead sheet of thickness \(4.0\text{ mm}\) is found to reduce the transmitted intensity to \(25\%\) of its incident value. What total thickness of lead is required to reduce the incident beam intensity by \(87.5\%\)?
A.\(6.0\text{ mm}\)
B.\(8.0\text{ mm}\)
C.\(12.0\text{ mm}\)
D.\(14.0\text{ mm}\)
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解题
1. Transmitted intensity is \(25\% = \left(\frac{1}{2}\right)^2\) of the incident intensity. This means \(4.0\text{ mm}\) corresponds to \(2\) half-value thicknesses (HVT). Therefore, \(\text{HVT} = \frac{4.0\text{ mm}}{2} = 2.0\text{ mm}\).
2. Reducing the intensity by \(87.5\%\) means the remaining transmitted intensity is: \(100\% - 87.5\% = 12.5\% = \left(\frac{1}{2}\right)^3\) of the initial intensity.
A (1 mark) - Deduces \(\text{HVT} = 2.0\text{ mm}\) from the first condition. - Recognises that reduction by \(87.5\%\) leaves \(12.5\% = (1/2)^3\), requiring \(3 \times \text{HVT} = 6.0\text{ mm}\).
题目 6 · MCQ
1 分
A star is at a distance of \(40\text{ pc}\) from Earth and has an apparent magnitude of \(+2.5\). What is the absolute magnitude of the star?
A.\(-3.01\)
B.\(-0.51\)
C.\(+0.51\)
D.\(+5.51\)
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解题
Using the distance modulus formula: \[ m - M = 5 \log_{10}\left(\frac{d}{10}\right) \] where \(m = +2.5\) and \(d = 40\text{ pc}\): \[ 2.5 - M = 5 \log_{10}\left(\frac{40}{10}\right) = 5 \log_{10}(4) \approx 5 \times 0.6021 = 3.01 \] \[ M = 2.5 - 3.01 = -0.51 \]
评分标准
Correct application of distance modulus formula \(m - M = 5 \log_{10}(d/10)\) leading to \(M = -0.51\). Option B.
题目 7 · MCQ
1 分
In a photoelectric effect experiment, monochromatic light of frequency \(f\) is shone onto a clean metal surface with work function \(\Phi\), causing photoelectrons to be emitted. Which of the following statements is/are correct?
(1) If the intensity of the incident light is doubled while keeping the frequency unchanged, the stopping potential remains unchanged. (2) If the frequency of the incident light is doubled, the maximum kinetic energy of the emitted photoelectrons is doubled. (3) The threshold frequency of the metal increases when the intensity of the incident light increases.
A.(1) only
B.(3) only
C.(1) and (2) only
D.(2) and (3) only
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解题
(1) is correct: Stopping potential \(V_s = \frac{hf - \Phi}{e}\), which depends on frequency \(f\) and work function \(\Phi\), but is independent of light intensity. (2) is incorrect: \(K_{\max} = hf - \Phi\). When frequency is doubled to \(2f\), \(K_{\max}' = 2hf - \Phi = 2(hf - \Phi) + \Phi = 2K_{\max} + \Phi > 2K_{\max}\). It is more than doubled. (3) is incorrect: Threshold frequency \(f_0 = \frac{\Phi}{h}\) is a property of the metal surface only and is independent of light intensity. Therefore, (1) only is correct.
评分标准
Correct identification that only statement (1) is valid. Option A.
题目 8 · MCQ
1 分
An air-conditioning system operating in cooling mode has a coefficient of performance (COP) of \(3.2\). It extracts thermal energy from an indoor room at a rate of \(4.8\text{ kW}\). At what rate is thermal energy rejected to the outdoor surroundings?
A.\(1.5\text{ kW}\)
B.\(3.3\text{ kW}\)
C.\(4.8\text{ kW}\)
D.\(6.3\text{ kW}\)
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解题
The coefficient of performance (COP) in cooling mode is given by: \[ \text{COP} = \frac{Q_c}{W} \] where \(Q_c = 4.8\text{ kW}\) is the rate of heat removed from the room, and \(W\) is the rate of electrical work input: \[ W = \frac{Q_c}{\text{COP}} = \frac{4.8\text{ kW}}{3.2} = 1.5\text{ kW} \] By conservation of energy, the rate of heat rejected to the outdoors \(Q_h\) is: \[ Q_h = Q_c + W = 4.8\text{ kW} + 1.5\text{ kW} = 6.3\text{ kW} \]
评分标准
Calculation of input power \(W = 1.5\text{ kW}\) and summation \(Q_h = Q_c + W = 6.3\text{ kW}\). Option D.
题目 9 · MCQ
1 分
An ultrasound pulse is directed perpendicularly across a boundary from muscle into bone.
Given: Acoustic impedance of muscle \(= 1.70 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}\) Acoustic impedance of bone \(= 7.80 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}\)
What percentage of the incident ultrasound intensity is reflected at the boundary?
A.\(17.7\%\)
B.\(41.2\%\)
C.\(58.8\%\)
D.\(64.2\%\)
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解题
The intensity reflection coefficient \(\alpha\) is given by: \[ \alpha = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2 \] Substituting the given values: \[ \alpha = \left(\frac{7.80 \times 10^6 - 1.70 \times 10^6}{7.80 \times 10^6 + 1.70 \times 10^6}\right)^2 = \left(\frac{6.10}{9.50}\right)^2 \approx (0.6421)^2 \approx 0.412 = 41.2\% \]
评分标准
Correct use of acoustic impedance reflection formula \(\alpha = ((Z_2 - Z_1)/(Z_2 + Z_1))^2\) yielding \(41.2\%\). Option B.
题目 10 · MCQ
1 分
A spectral line emitted by hydrogen in a distant galaxy is observed at a wavelength of \(520\text{ nm}\). In a laboratory, the rest wavelength of this line is \(480\text{ nm}\).
Given: Hubble constant \(H_0 = 70\text{ km s}^{-1}\text{ Mpc}^{-1}\) Speed of light \(c = 3.00 \times 10^8\text{ m s}^{-1}\)
Estimate the distance of the galaxy from Earth.
A.\(180\text{ Mpc}\)
B.\(290\text{ Mpc}\)
C.\(357\text{ Mpc}\)
D.\(429\text{ Mpc}\)
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解题
The redshift \(z\) of the galaxy is: \[ z = \frac{\Delta \lambda}{\lambda_0} = \frac{520\text{ nm} - 480\text{ nm}}{480\text{ nm}} = \frac{40}{480} = \frac{1}{12} \approx 0.08333 \] Using the Doppler formula for recession velocity \(v\): \[ v = c z = (3.00 \times 10^5\text{ km s}^{-1}) \times \frac{1}{12} = 25000\text{ km s}^{-1} \] Using Hubble's Law \(v = H_0 d\): \[ d = \frac{v}{H_0} = \frac{25000\text{ km s}^{-1}}{70\text{ km s}^{-1}\text{ Mpc}^{-1}} \approx 357\text{ Mpc} \]
评分标准
Calculation of redshift \(z = 1/12\), recession velocity \(v = 25000\text{ km s}^{-1}\), and distance \(d \approx 357\text{ Mpc}\). Option C.
题目 11 · Elective 選擇題
1 分
Star \(X\) has a peak radiation wavelength of \(500\text{ nm}\) and a luminosity of \(L_X\). Star \(Y\) has a peak radiation wavelength of \(1000\text{ nm}\) and its radius is \(3\) times that of Star \(X\). Assuming both stars radiate as ideal black bodies, what is the luminosity of Star \(Y\) in terms of \(L_X\)?
A.\(\frac{9}{4} L_X\)
B.\(\frac{9}{16} L_X\)
C.\(\frac{3}{16} L_X\)
D.\(\frac{3}{4} L_X\)
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解题
According to Wien's displacement law, \(\lambda_{\max} T = \text{constant}\).
Since \(\lambda_{\max, Y} = 2 \lambda_{\max, X}\), the surface temperature of Star \(Y\) is: \[ T_Y = \frac{1}{2} T_X \]
According to Stefan-Boltzmann law, the luminosity of a spherical black body is given by \(L = 4\pi R^2 \sigma T^4\).
Therefore, the ratio of luminosities is: \[ \frac{L_Y}{L_X} = \left(\frac{R_Y}{R_X}\right)^2 \left(\frac{T_Y}{T_X}\right)^4 = (3)^2 \left(\frac{1}{2}\right)^4 = 9 \times \frac{1}{16} = \frac{9}{16} L_X = 0.563 L_X \]
评分标准
1A: Correct option B.
题目 12 · Elective 選擇題
1 分
According to the Bohr model of the hydrogen atom, the orbital radius of an electron at the \(n\)-th energy level is proportional to \(n^2\). What is the ratio of the de Broglie wavelength of the orbital electron in the \(n = 4\) state to that in the \(n = 2\) state?
A.\(0.25\)
B.\(0.50\)
C.\(2.0\)
D.\(4.0\)
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解题
According to Bohr's quantization condition, the circumference of the \(n\)-th orbit is an integral multiple of the de Broglie wavelength: \[ 2\pi r_n = n \lambda_n \implies \lambda_n = \frac{2\pi r_n}{n} \]
Since \(r_n \propto n^2\), we have: \[ \lambda_n \propto \frac{n^2}{n} = n \]
Therefore, the ratio of the de Broglie wavelengths is: \[ \frac{\lambda_4}{\lambda_2} = \frac{4}{2} = 2 \]
评分标准
1A: Correct option C.
题目 13 · Elective 選擇題
1 分
A vertical external wall of an air-conditioned room has a surface area of \(15\text{ m}^2\) and a thickness of \(0.20\text{ m}\). The thermal conductivity of the wall material is \(0.80\text{ W m}^{-1}\text{ K}^{-1}\). The outdoor temperature is \(34^\circ\text{C}\) and the indoor temperature is maintained at \(22^\circ\text{C}\).
Find the U-value of the wall and the rate of heat conduction through the wall.
A.U-value = \(4.0\text{ W m}^{-2}\text{ K}^{-1}\), Rate = \(720\text{ W}\)
B.U-value = \(4.0\text{ W m}^{-2}\text{ K}^{-1}\), Rate = \(144\text{ W}\)
C.U-value = \(0.16\text{ W m}^{-2}\text{ K}^{-1}\), Rate = \(28.8\text{ W}\)
D.U-value = \(0.16\text{ W m}^{-2}\text{ K}^{-1}\), Rate = \(720\text{ W}\)
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解题
The U-value (thermal transmittance) of a single uniform layer is: \[ U = \frac{k}{d} = \frac{0.80}{0.20} = 4.0\text{ W m}^{-2}\text{ K}^{-1} \]
The rate of heat conduction \(Q/t\) through the wall is: \[ \frac{Q}{t} = U A \Delta T = 4.0 \times 15 \times (34 - 22) = 4.0 \times 15 \times 12 = 720\text{ W} \]
Hence, option A is correct.
评分标准
1A: Correct option A.
题目 14 · Elective 選擇題
1 分
A parallel beam of ultrasound travels through muscle and strikes the boundary with bone at normal incidence.
Given: - Acoustic impedance of muscle = \(1.70 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}\) - Acoustic impedance of bone = \(7.80 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}\)
What percentage of the incident ultrasound intensity is transmitted into the bone?
A.\(35.8\%\)
B.\(41.2\%\)
C.\(50.0\%\)
D.\(58.8\%\)
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解题
The intensity reflection coefficient \(\alpha\) at normal incidence is: \[ \alpha = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2 = \left(\frac{7.80 \times 10^6 - 1.70 \times 10^6}{7.80 \times 10^6 + 1.70 \times 10^6}\right)^2 = \left(\frac{6.10}{9.50}\right)^2 \approx (0.6421)^2 \approx 0.4123 = 41.2\% \]
Assuming no absorption at the boundary, the percentage of intensity transmitted is: \[ 100\% - \alpha = 100\% - 41.2\% = 58.8\% \]
评分标准
1A: Correct option D.
题目 15 · Elective 選擇題
1 分
In a transmission electron microscope (TEM), electrons are accelerated from rest through an anode potential difference \(V\). If the accelerating potential difference is increased from \(25\text{ kV}\) to \(100\text{ kV}\), assuming non-relativistic kinematics, how do the de Broglie wavelength of the electrons and the minimum resolvable detail (limit of resolution) change?
A.de Broglie wavelength is halved; Minimum resolvable detail is halved
B.de Broglie wavelength is halved; Minimum resolvable detail is doubled
C.de Broglie wavelength is quartered; Minimum resolvable detail is halved
D.de Broglie wavelength is quartered; Minimum resolvable detail is quartered
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解题
The kinetic energy of the accelerated electrons is \(E_k = e V = \frac{p^2}{2m}\), so \(p = \sqrt{2m e V}\).
The de Broglie wavelength is: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2m e V}} \propto \frac{1}{\sqrt{V}} \]
When \(V\) increases by a factor of \(4\) (from \(25\text{ kV}\) to \(100\text{ kV}\)), \(\lambda\) decreases by a factor of \(\sqrt{4} = 2\) (i.e. halved).
The theoretical limit of resolution (minimum resolvable detail, \(d_{\min}\)) is directly proportional to the wavelength (\(d_{\min} \propto \lambda\)). Therefore, the minimum resolvable detail is also halved (meaning the resolving power is doubled).
评分标准
1A: Correct option A.
题目 16 · MC
1 分
Two stars, \(X\) and \(Y\), have the same apparent brightness when observed from the Earth. Star \(X\) is at a distance of \(15\text{ pc}\) from the Earth and has a surface temperature of \(6000\text{ K}\). Star \(Y\) is at a distance of \(45\text{ pc}\) from the Earth and has a surface temperature of \(12\,000\text{ K}\). Find the ratio of the radius of star \(X\) to the radius of star \(Y\), \(R_X : R_Y\).
A.\(3 : 4\)
B.\(4 : 3\)
C.\(9 : 16\)
D.\(16 : 9\)
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解题
Apparent brightness is given by \(b = \frac{L}{4\pi d^2}\).
Since both stars have the same apparent brightness (\(b_X = b_Y\)): \[\frac{L_X}{4\pi d_X^2} = \frac{L_Y}{4\pi d_Y^2} \implies \frac{L_X}{L_Y} = \left(\frac{d_X}{d_Y}\right)^2 = \left(\frac{15}{45}\right)^2 = \frac{1}{9}\]
A horizontal-axis wind turbine is installed in a coastal wind farm. The turbine has rotor blades of length \( R = 38\text{ m} \).
(a) (i) Derive an expression for the theoretical kinetic energy per second (wind power \( P \)) of air of density \( \rho \) passing perpendicularly through the swept area of the turbine blades at speed \( v \). (2 marks)
(ii) Given that the air density is \( 1.22\text{ kg m}^{-3} \) and the wind speed is \( 12.0\text{ m s}^{-1} \), show that the maximum theoretical wind power entering the turbine is approximately \( 4.8\text{ MW} \). (2 marks)
(b) In actual operation under the conditions in (a)(ii), the turbine delivers an electrical power output of \( 1.40\text{ MW} \).
(i) Calculate the overall efficiency of the wind turbine system. (2 marks)
(ii) State TWO physical reasons why the electrical power generated is significantly less than the theoretical wind power available. (2 marks)
(c) The operator considers replacing this turbine with a newer model with blade radius \( 1.25R \) placed at an inland site where the average wind speed is \( 10.0\text{ m s}^{-1} \). Without calculating the exact numerical power, deduce whether the maximum theoretical wind power available to the new turbine will increase or decrease compared to the original turbine. (2 marks)
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解题
(a) (i) Swept area \( A = \pi R^2 \). In time \( \Delta t \), the volume of air passing through the swept area is \( V = A v \Delta t = \pi R^2 v \Delta t \). Mass of air passing through in \( \Delta t \) is \( m = \rho V = \rho \pi R^2 v \Delta t \). The kinetic energy of this air is \( E_k = \frac{1}{2} m v^2 = \frac{1}{2} (\rho \pi R^2 v \Delta t) v^2 = \frac{1}{2} \pi \rho R^2 v^3 \Delta t \). Therefore, kinetic energy per second (wind power) is \( P = \frac{E_k}{\Delta t} = \frac{1}{2} \pi \rho R^2 v^3 \).
(ii) Substitute \( \rho = 1.22\text{ kg m}^{-3} \), \( R = 38\text{ m} \), and \( v = 12.0\text{ m s}^{-1} \): \( P = \frac{1}{2} \pi (1.22)(38)^2(12.0)^3 \approx 4.781 \times 10^6\text{ W} = 4.78\text{ MW} \approx 4.8\text{ MW} \).
(ii) 1. The air behind the turbine cannot be brought to a complete standstill, as air must flow away to allow incoming air to pass (Betz's limit). 2. Energy losses due to aerodynamic drag on the blades, mechanical friction in bearings/gearbox, and electrical/resistive losses in the generator.
(c) Maximum available wind power is proportional to \( R^2 v^3 \). Ratio of new power to original power: \( \frac{P_{\text{new}}}{P_{\text{orig}}} = \left(\frac{R_{\text{new}}}{R}\right)^2 \left(\frac{v_{\text{new}}}{v}\right)^3 = (1.25)^2 \times \left(\frac{10.0}{12.0}\right)^3 = 1.5625 \times 0.5787 \approx 0.904 < 1 \). Since the ratio is less than 1, the maximum available wind power will decrease (the effect of the reduction in wind speed outweighs the increase in blade radius).
评分标准
(a)(i) - Correct mass of air passing per second \( \frac{\Delta m}{\Delta t} = \rho \pi R^2 v \) (1M) - Correct derivation showing \( P = \frac{1}{2}\pi\rho R^2 v^3 \) (1A)
(a)(ii) - Correct substitution of all numerical values: \( P = \frac{1}{2} \pi (1.22)(38)^2(12.0)^3 \) (1M) - Obtains \( 4.78\text{ MW} \) (or \( 4.781 \times 10^6\text{ W} \)) and shows it is approximately \( 4.8\text{ MW} \) (1A)
(b)(i) - Correct ratio of powers: \( \frac{1.40}{4.78} \times 100\% \) (1M) - Correct efficiency: \( 29.3\% \) (accept \( 29.2\% \) to \( 29.5\% \)) (1A)
(b)(ii) - Stating that air cannot be brought to rest / must retain kinetic energy to move away (Betz limit) (1A) - Stating frictional/thermal losses in mechanical components (gearbox/bearings) OR resistive/electrical losses in generator OR aerodynamic drag/turbulence (1A)
(c) - Deducing that \( P \propto R^2 v^3 \) and setting up the ratio comparison \( (1.25)^2 \times (10/12)^3 \approx 0.904 \) (1M) - Concluding that the available power will DECREASE with correct justification (1A)
题目 18 · Elective 結構題
10 分
An ultrasound transducer emitting pulses of frequency \( 3.5\text{ MHz} \) is used to locate a kidney stone in a patient's abdomen.
Given: - Speed of ultrasound in soft tissue \( c = 1540\text{ m s}^{-1} \) - Acoustic impedance of soft tissue \( Z_1 = 1.63 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1} \) - Acoustic impedance of kidney stone \( Z_2 = 6.45 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1} \)
(a) (i) Explain why ultrasound is emitted in short pulses rather than as a continuous wave in diagnostic A-scan imaging. (2 marks)
(ii) In an A-scan trace, the reflected echo from the anterior surface of the kidney stone is received \( 52.0\text{ }\mu\text{s} \) after the pulse is emitted. Calculate the depth of the kidney stone beneath the skin surface. (2 marks)
(b) (i) Calculate the intensity reflection coefficient \( \alpha \) at the boundary between soft tissue and the kidney stone. (2 marks)
(ii) Explain why an acoustic coupling gel must be applied between the ultrasound probe and the patient's skin. (2 marks)
(c) The physician considers changing to a transducer with an ultrasound frequency of \( 7.0\text{ MHz} \). State and explain ONE advantage and ONE disadvantage of using this higher frequency for detecting this kidney stone. (2 marks)
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解题
(a) (i) Pulses allow the transducer to switch between transmitting and receiving modes so that the time delay between emission and echo return can be measured to determine depth/distance. With continuous waves, returning echoes would overlap with emitted waves, making time-of-flight measurement impossible.
(ii) Time taken to reach the kidney stone is \( t = \frac{52.0 \times 10^{-6}\text{ s}}{2} = 26.0\text{ }\mu\text{s} \). Depth \( d = c t = 1540\text{ m s}^{-1} \times 26.0 \times 10^{-6}\text{ s} = 0.04004\text{ m} = 4.00\text{ cm} \) (or \( 4.00 \times 10^{-2}\text{ m} \)).
(ii) Air has a very low acoustic impedance compared to skin/soft tissue, which creates a huge impedance mismatch that causes almost \( 100\% \) of ultrasound energy to be reflected at the probe-air-skin boundary. Coupling gel has an acoustic impedance close to skin, eliminating air pockets and allowing ultrasound energy to transmit into the body.
(c) Advantage: Higher frequency corresponds to a shorter wavelength (\( \lambda = c/f \)), which provides higher spatial/axial resolution and allows finer detail of the stone to be resolved. Disadvantage: Ultrasound attenuation/absorption increases with frequency, resulting in poorer penetration depth into deep tissue.
评分标准
(a)(i) - Transducer needs to alternate between emitting and detecting echoes / prevent interference between emitted and reflected waves (1A) - To measure the time delay / time-of-flight to determine the depth of the organ/boundary (1A)
(a)(ii) - Correct formula for two-way travel: \( d = \frac{c \Delta t}{2} \) (1M) - Correct depth: \( 0.0400\text{ m} \) or \( 4.00\text{ cm} \) (1A)
(b)(ii) - Explaining the large acoustic impedance mismatch between air and skin causes almost complete reflection of ultrasound (1A) - Stating that gel matches the acoustic impedance of skin / excludes air pockets to maximize sound transmission into the body (1A)