Question 1 · multiple_choice
1 marksAn electric heater of power \(50\text{ W}\) is used to heat a solid block of mass \(0.2\text{ kg}\) initially at its melting point. During the heating process, heat is lost to the surroundings at a constant rate of \(10\text{ W}\). If the specific latent heat of fusion of the substance is \(2.0 \times 10^5\text{ J kg}^{-1}\), find the time \(t\) required to melt the block completely.
- A.800 s
- B.1000 s
- C.1250 s
- D.4000 s
Worked solution
The effective heating power is \(P_{\text{eff}} = 50\text{ W} - 10\text{ W} = 40\text{ W}\).
The energy required to melt the solid completely is:
\(Q = m L = 0.2\text{ kg} \times (2.0 \times 10^5\text{ J kg}^{-1}) = 40000\text{ J}\).
The time required is:
\(t = \frac{Q}{P_{\text{eff}}} = \frac{40000}{40} = 1000\text{ s}\).
Marking scheme
Correct Answer: B. Award 1 mark for selecting B.
Question 2 · multiple_choice
1 marksA fixed mass of an ideal gas is contained in a cylinder with a movable piston. The volume of the gas is slowly halved isothermally. Then, the gas is heated at constant pressure until its volume returns to the original volume. Which of the following statements about the final state of the gas (compared to its initial state) is correct?
- A.The average kinetic energy of the molecules is doubled, and the pressure is doubled.
- B.The average kinetic energy of the molecules is halved, and the pressure remains unchanged.
- C.The average kinetic energy of the molecules is doubled, and the pressure remains unchanged.
- D.The average kinetic energy of the molecules remains unchanged, and the pressure is doubled.
Worked solution
Let the initial state of the gas be \((P_0, V_0, T_0)\).
1. After isothermal compression to half its volume, the temperature remains \(T_0\) and pressure becomes \(2P_0\).
2. During isobaric heating back to \(V_0\), the pressure remains \(2P_0\). By the ideal gas law \(PV = nRT\), when volume doubles at constant pressure, the temperature must double to \(2T_0\).
Since the average kinetic energy of gas molecules is directly proportional to the absolute temperature \(T\), it is doubled. The final pressure is also doubled (\(2P_0\)).
Marking scheme
Correct Answer: A. Award 1 mark for selecting A.
Question 3 · multiple_choice
1 marksA block of mass \(4\text{ kg}\) is pushed up a rough inclined plane of inclination angle \(30^\circ\) by a constant horizontal force \(F = 50\text{ N}\). If the block moves up the incline at a constant speed, what is the magnitude of the frictional force acting on the block? (Take \(g = 9.8\text{ m s}^{-2}\))
- A.5.4 N
- B.19.6 N
- C.23.7 N
- D.43.3 N
Worked solution
Since the block moves up the incline at a constant speed, it is in equilibrium. Resolving forces parallel to the inclined plane:
\(F \cos 30^\circ = mg \sin 30^\circ + f\)
\(f = F \cos 30^\circ - mg \sin 30^\circ\)
\(f = 50 \times \cos 30^\circ - 4 \times 9.8 \times \sin 30^\circ\)
\(f \approx 43.30 - 19.6 = 23.7\text{ N}\).
Marking scheme
Correct Answer: C. Award 1 mark for selecting C.
Question 4 · multiple_choice
1 marksA toy car of mass \(0.5\text{ kg}\) starts from rest and moves along a horizontal track. The power \(P\) delivered to the car as a function of time \(t\) is given by \(P = 3t\text{ W}\) (for \(t \ge 0\)). Assuming that there is no energy loss, find the speed of the car at \(t = 4\text{ s}\).
- A.4.90 m/s
- B.6.93 m/s
- C.9.80 m/s
- D.13.9 m/s
Worked solution
The work done \(W\) on the car is the area under the \(P-t\) graph from \(t = 0\) to \(4\text{ s}\):
\(W = \int_0^4 3t \, dt = \left[ 1.5t^2 \right]_0^4 = 1.5 \times 16 = 24\text{ J}\).
By the work-energy theorem, \(W = \Delta KE = \frac{1}{2}mv^2 - 0\):
\(24 = \frac{1}{2} (0.5) v^2\)
\(v^2 = 96\)
\(v = \sqrt{96} \approx 9.80\text{ m s}^{-1}\).
Marking scheme
Correct Answer: C. Award 1 mark for selecting C.
Question 5 · multiple_choice
1 marksA small coin is placed on a horizontal turntable at a distance of \(0.2\text{ m}\) from the center. The turntable starts from rest and accelerates slowly. If the coefficient of static friction between the coin and the turntable is \(0.5\), find the maximum angular speed of the turntable before the coin starts to slip. (Take \(g = 9.8\text{ m s}^{-2}\))
- A.2.45 rad/s
- B.3.13 rad/s
- C.4.95 rad/s
- D.9.80 rad/s
Worked solution
The centripetal force is provided by the static friction. The maximum static friction is \(f_{\text{max}} = \mu_s m g\).
At the verge of slipping:
\(m \omega^2 r = \mu_s m g\)
\(\omega = \sqrt{\frac{\mu_s g}{r}} = \sqrt{\frac{0.5 \times 9.8}{0.2}} = \sqrt{24.5} \approx 4.95\text{ rad s}^{-1}\).
Marking scheme
Correct Answer: C. Award 1 mark for selecting C.
Question 6 · multiple_choice
1 marksA light ray is incident at the boundary between medium X and medium Y. The refractive index of medium X is \(1.6\) and that of medium Y is \(1.2\). What is the critical angle for total internal reflection, and in which medium must the light be incident for total internal reflection to occur?
- A.\(48.6^\circ\), incident in medium X
- B.\(48.6^\circ\), incident in medium Y
- C.\(36.9^\circ\), incident in medium X
- D.\(36.9^\circ\), incident in medium Y
Worked solution
Total internal reflection can only occur when light travels from a medium with a higher refractive index to one with a lower refractive index. Since \(n_{\text{X}} = 1.6 > n_{\text{Y}} = 1.2\), the light must be incident in medium X.
The critical angle \(\theta_c\) is given by:
\(\sin \theta_c = \frac{n_{\text{Y}}}{n_{\text{X}}} = \frac{1.2}{1.6} = 0.75\)
\(\theta_c = \sin^{-1}(0.75) \approx 48.6^\circ\).
Marking scheme
Correct Answer: A. Award 1 mark for selecting A.
Question 7 · multiple_choice
1 marksThree identical resistors, each of resistance \(R\), are connected to an ideal battery of constant voltage \(V\). When they are connected in series, the total power dissipated in the circuit is \(P_s\). When they are connected in parallel, the total power dissipated is \(P_p\). What is the ratio \(P_p / P_s\)?
- A.1/9
- B.1/3
- C.3
- D.9
Worked solution
In series, the equivalent resistance is \(R_s = 3R\), so the total power is:
\(P_s = \frac{V^2}{3R}\).
In parallel, the equivalent resistance is \(R_p = \frac{R}{3}\), so the total power is:
\(P_p = \frac{V^2}{R/3} = \frac{3V^2}{R}\).
The ratio is:
\(\frac{P_p}{P_s} = \frac{3V^2/R}{V^2/(3R)} = 9\).
Marking scheme
Correct Answer: D. Award 1 mark for selecting D.
Question 8 · multiple_choice
1 marksA flat circular coil of \(200\) turns and area \(0.05\text{ m}^2\) is placed perpendicular to a uniform magnetic field of strength \(0.4\text{ T}\). The magnetic field is reduced to zero at a constant rate within a time interval of \(0.2\text{ s}\). What is the magnitude of the average induced electromotive force (e.m.f.) in the coil?
- A.0.1 V
- B.2.0 V
- C.20 V
- D.40 V
Worked solution
According to Faraday's law of electromagnetic induction, the magnitude of the average induced e.m.f. \(\mathcal{E}\) is:
\(\mathcal{E} = N \left| \frac{\Delta \Phi}{\Delta t} \right| = N A \left| \frac{\Delta B}{\Delta t} \right|\)
\(\mathcal{E} = 200 \times 0.05\text{ m}^2 \times \frac{0.4\text{ T}}{0.2\text{ s}} = 10 \times 2 = 20\text{ V}\).
Marking scheme
Correct Answer: C. Award 1 mark for selecting C.
Question 9 · multiple_choice
1 marksA sample contains a mixture of two radioactive isotopes, A and B. Initially, the activity of A is \(1600\text{ Bq}\) and the activity of B is \(100\text{ Bq}\). The half-life of A is \(3\text{ hours}\), and the half-life of B is \(6\text{ hours}\). After how many hours will the activities of the two isotopes in the sample become equal?
- A.12 hours
- B.18 hours
- C.24 hours
- D.30 hours
Worked solution
Let the activities of A and B at time \(t\) hours be \(A_A(t)\) and \(A_B(t)\) respectively.
\(A_A(t) = 1600 \times \left(\frac{1}{2}\right)^{t/3}\)
\(A_B(t) = 100 \times \left(\frac{1}{2}\right)^{t/6}\)
Set \(A_A(t) = A_B(t)\):
\(1600 \times \left(\frac{1}{2}\right)^{t/3} = 100 \times \left(\frac{1}{2}\right)^{t/6}\)
\(16 = \frac{\left(\frac{1}{2}\right)^{t/6}}{\left(\frac{1}{2}\right)^{t/3}} = \left(\frac{1}{2}\right)^{t/6 - t/3} = \left(\frac{1}{2}\right)^{-t/6} = 2^{t/6}\)
Since \(16 = 2^4\), we have:
\(\frac{t}{6} = 4 \implies t = 24\text{ hours}\).
Marking scheme
Correct Answer: C. Award 1 mark for selecting C.
Question 10 · multiple_choice
1 marksA ball of mass \(0.2\text{ kg}\) is dropped from a height of \(2.50\text{ m}\) onto a hard horizontal floor. It rebounds to a maximum height of \(0.90\text{ m}\). If the collision between the ball and the floor lasts for \(0.05\text{ s}\), what is the magnitude of the average force exerted by the floor on the ball? (Take \(g = 9.8\text{ m s}^{-2}\))
- A.11.2 N
- B.44.8 N
- C.46.8 N
- D.56.0 N
Worked solution
Taking the upward direction as positive:
Velocity just before impact: \(v_1 = -\sqrt{2 g h_1} = -\sqrt{2 \times 9.8 \times 2.50} = -7\text{ m s}^{-1}\).
Velocity just after impact: \(v_2 = +\sqrt{2 g h_2} = +\sqrt{2 \times 9.8 \times 0.90} = +4.2\text{ m s}^{-1}\).
The change in momentum of the ball is:
\(\Delta p = m(v_2 - v_1) = 0.2 \times (4.2 - (-7)) = 2.24\text{ kg m s}^{-1}\).
The average net force is:
\(F_{\text{net}} = \frac{\Delta p}{\Delta t} = \frac{2.24}{0.05} = 44.8\text{ N}\) (directed upwards).
Since \(F_{\text{net}} = F_{\text{floor}} - m g\):
\(F_{\text{floor}} = F_{\text{net}} + m g = 44.8 + 0.2 \times 9.8 = 44.8 + 1.96 = 46.76\text{ N} \approx 46.8\text{ N}\).
Marking scheme
Correct Answer: C. Award 1 mark for selecting C.
Question 11 · Multiple-choice
1 marksAn insulated container of negligible heat capacity contains 0.2 kg of water at 20°C. A 0.05 kg block of ice at 0°C is added to the container. Assume no heat loss to the surroundings. Which of the following statements about the mixture when it reaches thermal equilibrium is/are correct? (Given: specific heat capacity of water = 4200 J kg^{-1} °C^{-1}; specific latent heat of fusion of ice = 3.34 \times 10^5 J kg^{-1}) (1) All ice has melted. (2) The final temperature of the mixture is 0°C. (3) If the initial temperature of the water were 15°C, some ice would remain unmelted.
- A.(1) only
- B.(3) only
- C.(1) and (3) only
- D.(1), (2) and (3)
Worked solution
First, calculate the maximum heat released by water cooling to 0°C: Q_{\text{release}} = m_w c_w \Delta T = 0.2 \times 4200 \times 20 = 16800 J. The heat required to completely melt the ice: Q_{\text{melt}} = m_{\text{ice}} \ell_f = 0.05 \times 3.34 \times 10^5 = 16700 J. Since Q_{\text{release}} > Q_{\text{melt}}, all ice melts. Thus, statement (1) is correct. Since Q_{\text{release}} is greater than Q_{\text{melt}} by 100 J, this excess heat will raise the temperature of the 0.25 kg mixture of water slightly above 0°C. Therefore, statement (2) is incorrect. If the initial temperature of the water were 15°C, the maximum heat released would be 0.2 \times 4200 \times 15 = 12600 J. Since this is less than 16700 J, the ice cannot melt completely, and some ice will remain unmelted at 0°C. Thus, statement (3) is correct.
Marking scheme
1 mark for the correct option (C). 0 marks for incorrect options.
Question 12 · Multiple-choice
1 marksAn ideal gas is enclosed in a rigid container of fixed volume. When the absolute temperature of the gas is doubled, which of the following statements is/are correct? (1) The root-mean-square speed of the gas molecules is doubled. (2) The collision frequency between gas molecules and the container walls is doubled. (3) The pressure of the gas is doubled.
- A.(3) only
- B.(1) and (2) only
- C.(1) and (3) only
- D.(1), (2) and (3)
Worked solution
(1) The root-mean-square speed of molecules is v_{\text{rms}} = \sqrt{\frac{3RT}{M}}. Since v_{\text{rms}} \propto \sqrt{T}, doubling T increases v_{\text{rms}} by a factor of \sqrt{2}, not 2. Thus (1) is incorrect. (2) At constant volume, the number density of gas molecules is constant. The rate of collisions with the walls is proportional to the average speed of molecules, which is proportional to \sqrt{T}. Thus, collision frequency increases by a factor of \sqrt{2}, not 2. Thus (2) is incorrect. (3) According to the ideal gas law PV = nRT, at constant volume, pressure P \propto T. Doubling T doubles P. Thus (3) is correct.
Marking scheme
1 mark for the correct option (A). 0 marks for incorrect options.
Question 13 · Multiple-choice
1 marksTwo blocks of masses m_1 = 3 kg and m_2 = 1 kg are connected by a light inextensible string passing over a smooth light pulley. Block m_1 is placed on a rough horizontal table with coefficient of kinetic friction \mu = 0.2. Block m_2 hangs vertically. When the system is released from rest, what is the magnitude of the acceleration of the blocks? (Take the acceleration due to gravity as g.)
- A.0.10 g
- B.0.15 g
- C.0.25 g
- D.0.40 g
Worked solution
The normal force on m_1 is R = m_1 g = 3g. The kinetic frictional force on m_1 is f = \mu R = 0.2 \times 3g = 0.6g. The downward pulling force on the hanging mass m_2 is m_2 g = 1g. Since m_2 g > f (1g > 0.6g), the system will accelerate. Applying Newton's second law to the whole system along the direction of motion: F_{\text{net}} = m_2 g - f = (m_1 + m_2) a \implies 1g - 0.6g = (3 + 1) a \implies 0.4g = 4a \implies a = 0.1g.
Marking scheme
1 mark for the correct option (A). 0 marks for incorrect options.
Question 14 · Multiple-choice
1 marksTrolley A of mass 2 kg moving at 3 m s^{-1} to the right collides head-on with trolley B of mass 1 kg moving at 1 m s^{-1} to the left. After the collision, trolley A continues to move to the right at 1 m s^{-1}. Which of the following describes the change in momentum of trolley B and the nature of the collision?
- A.Change in momentum of B is 4 kg m s^{-1} to the right; the collision is inelastic.
- B.Change in momentum of B is 4 kg m s^{-1} to the right; the collision is elastic.
- C.Change in momentum of B is 2 kg m s^{-1} to the right; the collision is inelastic.
- D.Change in momentum of B is 2 kg m s^{-1} to the right; the collision is elastic.
Worked solution
Let the direction to the right be positive. Initial momentum of A: p_{A,i} = 2 \times 3 = 6 kg m s^{-1}. Initial momentum of B: p_{B,i} = 1 \times (-1) = -1 kg m s^{-1}. Total initial momentum P_i = 6 - 1 = 5 kg m s^{-1}. Final momentum of A: p_{A,f} = 2 \times 1 = 2 kg m s^{-1}. By conservation of linear momentum, total final momentum P_f = P_i = 5 kg m s^{-1}. Final momentum of B: p_{B,f} = 5 - 2 = 3 kg m s^{-1}. Change in momentum of B: \Delta p_B = p_{B,f} - p_{B,i} = 3 - (-1) = 4 kg m s^{-1} to the right. Initial kinetic energy: K_i = 0.5 \times 2 \times 3^2 + 0.5 \times 1 \times (-1)^2 = 9.5 J. Final kinetic energy: K_f = 0.5 \times 2 \times 1^2 + 0.5 \times 1 \times 3^2 = 5.5 J. Since K_f < K_i, the collision is inelastic.
Marking scheme
1 mark for the correct option (A). 0 marks for incorrect options.
Question 15 · Multiple-choice
1 marksA satellite orbits the Earth in a circular orbit of radius r with a speed v and an orbital period T. If the satellite is moved to a new circular orbit of radius 4r, what will be its new speed and orbital period?
- A.Speed is \frac{1}{2} v; period is 8T.
- B.Speed is \frac{1}{4} v; period is 8T.
- C.Speed is \frac{1}{2} v; period is 4T.
- D.Speed is \frac{1}{4} v; period is 16T.
Worked solution
For a circular orbit: GMm/r^2 = mv^2/r, so v = \sqrt{GM/r}. Thus, v \propto 1/\sqrt{r}. When r increases to 4r, the speed becomes v' = \sqrt{GM/(4r)} = 0.5 v. By Kepler's Third Law, T^2 \propto r^3, meaning T \propto r^{3/2}. When r increases to 4r, the period becomes T' = 4^{3/2} T = 8T.
Marking scheme
1 mark for the correct option (A). 0 marks for incorrect options.
Question 16 · Multiple-choice
1 marksAn object is placed at a distance u from a thin converging lens of focal length 10 cm. A real, inverted image is formed. If the magnification of the image is 2, find the value of u.
- A.5 cm
- B.15 cm
- C.20 cm
- D.30 cm
Worked solution
The magnification m is given by m = v/u = 2, which gives v = 2u. Since a real image is formed, v is positive. Applying the lens formula: 1/f = 1/u + 1/v. Substitute f = 10 cm and v = 2u: 1/10 = 1/u + 1/(2u) = 3/(2u). Thus, 2u = 30, which gives u = 15 cm.
Marking scheme
1 mark for the correct option (B). 0 marks for incorrect options.
Question 17 · Multiple-choice
1 marksTwo resistors R_1 and R_2, each of resistance 6 \Omega, are connected in parallel. This combination is connected in series with a third resistor R_3, also of resistance 6 \Omega. The entire circuit is connected across an ideal battery of electromotive force (emf) 18 V. What are the potential difference across R_1 and the total power dissipated in the circuit?
- A.Potential difference across R_1 is 6 V; total power is 36 W.
- B.Potential difference across R_1 is 12 V; total power is 36 W.
- C.Potential difference across R_1 is 6 V; total power is 54 W.
- D.Potential difference across R_1 is 12 V; total power is 54 W.
Worked solution
The equivalent resistance of R_1 and R_2 in parallel is R_{12} = (6 \times 6)/(6 + 6) = 3 \Omega. The total resistance is R_{\text{total}} = R_{12} + R_3 = 3 + 6 = 9 \Omega. The total current is I = V/R_{\text{total}} = 18/9 = 2 A. The potential difference across the parallel branch (and R_1) is V_1 = I \times R_{12} = 2 \times 3 = 6 V. The total power is P = I^2 R_{\text{total}} = 2^2 \times 9 = 36 W.
Marking scheme
1 mark for the correct option (A). 0 marks for incorrect options.
Question 18 · Multiple-choice
1 marksA bar magnet is dropped vertically, north pole pointing downwards, through a horizontal copper ring. Which of the following statements is/are correct? (1) As the north pole approaches the ring from above, the induced current in the ring (viewed from above) is anticlockwise. (2) As the south pole of the magnet leaves the ring from below, the induced current in the ring (viewed from above) is anticlockwise. (3) The acceleration of the falling magnet is less than the acceleration due to gravity g both when it is approaching and when it is leaving the ring.
- A.(1) and (2) only
- B.(1) and (3) only
- C.(2) and (3) only
- D.(1), (2) and (3)
Worked solution
(1) As the N-pole approaches from above, downward magnetic flux increases. Lenz's law dictates an upward induced field, which corresponds to an anticlockwise current (viewed from above). Thus (1) is correct. (2) The S-pole is at the top. As the S-pole leaves downwards, the downward flux decreases. The ring induces a downward magnetic field to oppose this decrease, which is a clockwise current (viewed from above). Thus (2) is incorrect. (3) Lenz's law always opposes the relative motion, creating an upward retarding force. Hence, acceleration is always less than g. Thus (3) is correct.
Marking scheme
1 mark for the correct option (B). 0 marks for incorrect options.
Question 19 · Multiple-choice
1 marksAn alternating voltage v = 220\sqrt{2} \sin(100\pi t) V is connected across a resistor of resistance 44 \Omega. What are the root-mean-square (rms) current in the resistor and the average power dissipated?
- A.rms current is 5 A; average power is 1100 W.
- B.rms current is 5\sqrt{2} A; average power is 1100 W.
- C.rms current is 5 A; average power is 2200 W.
- D.rms current is 5\sqrt{2} A; average power is 2200 W.
Worked solution
The voltage equation has peak value V_0 = 220\sqrt{2} V. The rms voltage is V_{\text{rms}} = V_0/\sqrt{2} = 220 V. The rms current is I_{\text{rms}} = V_{\text{rms}}/R = 220/44 = 5 A. The average power dissipated is P_{\text{avg}} = I_{\text{rms}} V_{\text{rms}} = 5 \times 220 = 1100 W.
Marking scheme
1 mark for the correct option (A). 0 marks for incorrect options.
Question 20 · Multiple-choice
1 marksA radioactive sample contains a mixture of two radioisotopes, X and Y. Initially, the activity of X is 1600 Bq and the activity of Y is 200 Bq. The half-life of X is 3 hours, and the half-life of Y is 6 hours. After how many hours will the activities of X and Y be equal?
- A.6 hours
- B.9 hours
- C.12 hours
- D.18 hours
Worked solution
Let t be the elapsed time in hours. The activity of X is A_X(t) = 1600 \times (1/2)^{t/3}. The activity of Y is A_Y(t) = 200 \times (1/2)^{t/6}. Equating them: 1600 \times (1/2)^{t/3} = 200 \times (1/2)^{t/6} \implies 8 \times (1/2)^{t/3} = (1/2)^{t/6} \implies 8 = (1/2)^{t/6 - t/3} = 2^{t/6}. Since 8 = 2^3, we get 3 = t/6 \implies t = 18 hours.
Marking scheme
1 mark for the correct option (D). 0 marks for incorrect options.
A \(0.5\text{ kg}\) block of ice at \(-10 ^\circ\text{C}\) is heated by a \(100\text{ W}\) heater. Find the time taken to melt all the ice into water at \(0 ^\circ\text{C}\). (Given: specific heat capacity of ice \(= 2100\text{ J kg}^{-1}{^\circ}\text{C}^{-1}\), specific latent heat of fusion of ice \(= 3.34 \times 10^5\text{ J kg}^{-1}\))
- A.\(105\text{ s}\)
- B.\(1670\text{ s}\)
- C.\(1775\text{ s}\)
- D.\(1880\text{ s}\)
Worked solution
Energy required to raise the temperature of ice from \(-10 ^\circ\text{C}\) to \(0 ^\circ\text{C}\) is \(E_1 = m c_{\text{ice}} \Delta T = 0.5 \times 2100 \times 10 = 10500\text{ J}\). Energy required to melt the ice completely at \(0 ^\circ\text{C}\) is \(E_2 = m \ell_f = 0.5 \times 3.34 \times 10^5 = 167000\text{ J}\). Total energy needed is \(E = E_1 + E_2 = 177500\text{ J}\). The heating time is \(t = \frac{E}{P} = \frac{177500}{100} = 1775\text{ s}\).
Marking scheme
Correct answer is C. Award 1 mark for selecting C. No mark for other options.
A rigid gas cylinder contains an ideal gas at a pressure of \(P\) and a temperature of \(27 ^\circ\text{C}\). After some gas leaks out, the pressure becomes \(0.6 P\) and the temperature decreases to \(7 ^\circ\text{C}\). Find the fraction of the original mass of gas that remains in the cylinder.
- A.0.56
- B.0.60
- C.0.64
- D.0.69
Worked solution
From the ideal gas law \(PV = nRT\), the mass of gas \(m \propto n = \frac{PV}{RT}\) since the molar mass and volume \(V\) are constant. The initial mass \(m_1 \propto \frac{P_1}{T_1} = \frac{P}{300}\). The final mass \(m_2 \propto \frac{P_2}{T_2} = \frac{0.6P}{280}\). The remaining fraction is \(\frac{m_2}{m_1} = \frac{0.6 P / 280}{P / 300} = 0.6 \times \frac{300}{280} \approx 0.643\).
Marking scheme
Correct answer is C. Award 1 mark for selecting C.
A car starts from rest and accelerates uniformly to a speed of \(10\text{ m s}^{-1}\) in \(4\text{ s}\). It then travels at this constant speed for \(4\text{ s}\), and finally decelerates uniformly to rest in \(2\text{ s}\). What is the average speed of the car for the entire \(10\text{ s}\) journey?
- A.\(5.0\text{ m s}^{-1}\)
- B.\(6.0\text{ m s}^{-1}\)
- C.\(7.0\text{ m s}^{-1}\)
- D.\(8.0\text{ m s}^{-1}\)
Worked solution
The total displacement is the area under the \(v-t\) graph, which is a trapezoid. The bottom base is \(10\text{ s}\), top base is \(4\text{ s}\) (from \(t=4\) to \(t=8\)), and height is \(10\text{ m s}^{-1}\). Area \(= \frac{1}{2} \times (10 + 4) \times 10 = 70\text{ m}\). Average speed \(= \frac{\text{Total displacement}}{\text{Total time}} = \frac{70\text{ m}}{10\text{ s}} = 7.0\text{ m s}^{-1}\).
Marking scheme
Correct answer is C. Award 1 mark for selecting C.
A block of mass \(2\text{ kg}\) is placed on a rough inclined plane making an angle of \(30^\circ\) with the horizontal. The coefficient of static friction between the block and the plane is \(0.6\). What is the magnitude of the friction force acting on the block? (Take \(g = 9.8\text{ m s}^{-2}\))
- A.\(4.9\text{ N}\)
- B.\(9.8\text{ N}\)
- C.\(10.2\text{ N}\)
- D.\(19.6\text{ N}\)
Worked solution
First, check if the block slides down. The component of weight parallel to the incline is \(F_{\text{pull}} = mg \sin 30^\circ = 2 \times 9.8 \times 0.5 = 9.8\text{ N}\). The maximum possible static friction force is \(f_{\text{max}} = \mu_s N = \mu_s mg \cos 30^\circ = 0.6 \times 2 \times 9.8 \times \cos 30^\circ \approx 10.18\text{ N}\). Since \(F_{\text{pull}} < f_{\text{max}}\), the block does not slide and remains in static equilibrium. The static friction force must balance the downward gravitational component, so \(f = F_{\text{pull}} = 9.8\text{ N}\).
Marking scheme
Correct answer is B. Award 1 mark for selecting B.
A light spring of force constant \(100\text{ N m}^{-1}\) hangs vertically with a \(0.5\text{ kg}\) block attached to its lower end. The block is initially at rest at its equilibrium position. It is then pulled vertically downwards by \(0.1\text{ m}\) and released from rest. What is the maximum kinetic energy of the block during the subsequent motion?
- A.\(0.1\text{ J}\)
- B.\(0.2\text{ J}\)
- C.\(0.5\text{ J}\)
- D.\(1.0\text{ J}\)
Worked solution
In vertical spring-mass systems, the total mechanical energy relative to the equilibrium position can be treated as purely elastic potential energy of a simple harmonic oscillator: \(E = \frac{1}{2} k A^2\), where \(A = 0.1\text{ m}\) is the amplitude of oscillation. The maximum kinetic energy occurs as the block passes through the equilibrium position, which equals the total energy: \(K_{\text{max}} = \frac{1}{2} k A^2 = \frac{1}{2} \times 100 \times 0.1^2 = 0.5\text{ J}\).
Marking scheme
Correct answer is C. Award 1 mark for selecting C.
Two satellites, \(P\) and \(Q\), orbit the Earth in circular orbits. The mass of \(P\) is half that of \(Q\), and the kinetic energy of \(P\) is twice that of \(Q\). What is the ratio of the orbital radius of \(P\) to that of \(Q\)?
- A.\(1/4\)
- B.\(1/2\)
- C.\(2\)
- D.\(4\)
Worked solution
The gravitational force provides the centripetal force: \(\frac{GMm}{R^2} = \frac{mv^2}{R} \implies v^2 = \frac{GM}{R}\). The kinetic energy is \(K = \frac{1}{2}mv^2 = \frac{GMm}{2R}\). Therefore, \(R = \frac{GMm}{2K}\). This gives \(\frac{R_P}{R_Q} = \frac{m_P}{m_Q} \times \frac{K_Q}{K_P}\). Given \(m_P = 0.5 m_Q \implies \frac{m_P}{m_Q} = \frac{1}{2}\), and \(K_P = 2 K_Q \implies \frac{K_Q}{K_P} = \frac{1}{2}\). Thus, \(\frac{R_P}{R_Q} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\).
Marking scheme
Correct answer is A. Award 1 mark for selecting A.
A ray of light travels from medium \(X\) to medium \(Y\). The angle of incidence in \(X\) is \(30^\circ\) and the angle of refraction in \(Y\) is \(45^\circ\). What is the critical angle for total internal reflection when light travels from one medium to the other?
- A.\(30^\circ\)
- B.\(45^\circ\)
- C.\(60^\circ\)
- D.Total internal reflection cannot occur.
Worked solution
By Snell's Law, \(n_X \sin \theta_X = n_Y \sin \theta_Y \implies n_X \sin 30^\circ = n_Y \sin 45^\circ \implies \frac{n_Y}{n_X} = \frac{\sin 30^\circ}{\sin 45^\circ} = \frac{0.5}{1/\sqrt{2}} = \frac{1}{\sqrt{2}}\). Total internal reflection can only happen when light travels from a denser medium (\(X\)) to a rarer medium (\(Y\)). The critical angle \(\theta_c\) satisfies \(\sin \theta_c = \frac{n_Y}{n_X} = \frac{1}{\sqrt{2}}\), which gives \(\theta_c = 45^\circ\).
Marking scheme
Correct answer is B. Award 1 mark for selecting B.
A string of fixed length is clamped at both ends. When it vibrates in its third harmonic, the frequency is \(120\text{ Hz}\). What is the frequency of the vibration when it is excited to its fifth harmonic?
- A.\(72\text{ Hz}\)
- B.\(120\text{ Hz}\)
- C.\(200\text{ Hz}\)
- D.\(300\text{ Hz}\)
Worked solution
For a string fixed at both ends, the frequency of the \(n\)-th harmonic is given by \(f_n = n f_1\), where \(f_1\) is the fundamental frequency. We have \(f_3 = 3 f_1 = 120\text{ Hz}\), which means \(f_1 = 40\text{ Hz}\). Thus, the fifth harmonic frequency is \(f_5 = 5 f_1 = 5 \times 40 = 200\text{ Hz}\).
Marking scheme
Correct answer is C. Award 1 mark for selecting C.
Three identical resistors are connected to a DC source of constant voltage \(V\). When they are connected in series, the total power dissipated is \(P_1\). When they are connected in parallel to the same source, the total power dissipated is \(P_2\). What is the ratio \(\frac{P_1}{P_2}\)?
- A.\(1/9\)
- B.\(1/3\)
- C.\(3\)
- D.\(9\)
Worked solution
In series, the equivalent resistance is \(R_1 = 3R\). The power dissipated is \(P_1 = \frac{V^2}{3R}\). In parallel, the equivalent resistance is \(R_2 = \frac{R}{3}\). The power dissipated is \(P_2 = \frac{V^2}{R/3} = \frac{3V^2}{R}\). Therefore, \(\frac{P_1}{P_2} = \frac{V^2 / 3R}{3V^2 / R} = \frac{1}{9}\).
Marking scheme
Correct answer is A. Award 1 mark for selecting A.
A flat coil of \(50\) turns has an area of \(0.2\text{ m}^2\). It is placed in a uniform magnetic field perpendicular to the plane of the coil. If the magnetic field strength decreases from \(0.8\text{ T}\) to \(0.2\text{ T}\) at a constant rate in \(0.3\text{ s}\), what is the magnitude of the average induced electromotive force (emf) in the coil?
- A.\(2.0\text{ V}\)
- B.\(10\text{ V}\)
- C.\(20\text{ V}\)
- D.\(40\text{ V}\)
Worked solution
According to Faraday's law of electromagnetic induction, the average induced emf is \(\varepsilon = N \frac{\Delta \Phi}{\Delta t} = N A \frac{\Delta B}{\Delta t}\). Substituting the given values: \(\varepsilon = 50 \times 0.2 \times \frac{0.8 - 0.2}{0.3} = 10 \times \frac{0.6}{0.3} = 20\text{ V}\).
Marking scheme
Correct answer is C. Award 1 mark for selecting C.
Question 31 · multiple-choice
1 marksA small block of mass \(m\) is placed on a rough horizontal turntable rotating at a constant angular speed \(\omega\). The distance from the block to the center of rotation is \(r\). The coefficient of static friction between the block and the turntable is \(\mu\). If the angular speed is doubled and the distance from the center is halved, what is the new minimum coefficient of static friction required to prevent the block from slipping?
- A.Half of the original value
- B.The same as the original value
- C.Double the original value
- D.Four times the original value
Worked solution
To prevent slipping, the static friction must provide the necessary centripetal force: \(\mu m g \ge m \omega^2 r \implies \mu \ge \frac{\omega^2 r}{g}\). The minimum coefficient is \(\mu_{min} = \frac{\omega^2 r}{g}\). When \(\omega' = 2\omega\) and \(r' = r/2\), we have \(\mu'_{min} = \frac{(2\omega)^2 (r/2)}{g} = 2 \frac{\omega^2 r}{g} = 2 \mu_{min}\). Thus, the minimum coefficient is doubled.
Marking scheme
1 mark for the correct option C. Method: set up the friction inequality \(\mu g \ge \omega^2 r\) and substitute the changes.
Question 32 · multiple-choice
1 marksA copper ring is dropped vertically through the center of a strong bar magnet fixed vertically on a table (with its North pole facing upwards). Which of the following statements about the magnetic force acting on the ring is/are correct as it falls? (1) Before the ring passes the magnet, the magnetic force on it is upwards. (2) When the ring is exactly at the midpoint of the magnet, the magnetic force on it is zero. (3) After the ring passes the magnet, the magnetic force on it is downwards.
- A.(1) only
- B.(3) only
- C.(1) and (2) only
- D.(2) and (3) only
Worked solution
According to Lenz's law, the induced current always opposes the relative motion. As the ring approaches the magnet, the upward magnetic flux increases, inducing a downward magnetic field, resulting in an upward repulsive force. Statement (1) is correct. At the exact midpoint of a symmetric bar magnet, the magnetic flux through the ring reaches its maximum, meaning the rate of change of flux is zero (\(\frac{d\Phi}{dt} = 0\)). Hence, the induced EMF and current are zero, so the magnetic force is zero. Statement (2) is correct. As the ring leaves the magnet, the upward magnetic flux decreases, inducing an upward magnetic field, resulting in an upward attractive force to oppose it moving away. Statement (3) is incorrect.
Marking scheme
1 mark for the correct option C. Analyze the rate of change of magnetic flux and Lenz's law for each stage of the motion.
Question 33 · multiple-choice
1 marksA ray of monochromatic light travels from medium X to medium Y. The angle of incidence in medium X is \(30^\circ\), and the angle of refraction in medium Y is \(45^\circ\). What is the ratio of the speed of light in medium X to that in medium Y?
- A.1 : \(\sqrt{2}\)
- B.\(\sqrt{2}\) : 1
- C.1 : 2
- D.\(\sqrt{3}\) : 2
Worked solution
By Snell's Law, \(n_X \sin \theta_X = n_Y \sin \theta_Y\). Since the refractive index \(n = c/v\), we can write \(\frac{\sin \theta_X}{v_X} = \frac{\sin \theta_Y}{v_Y}\). Rearranging gives the ratio of speeds: \(\frac{v_X}{v_Y} = \frac{\sin \theta_X}{\sin \theta_Y} = \frac{\sin 30^\circ}{\sin 45^\circ} = \frac{0.5}{1/\sqrt{2}} = \frac{1}{\sqrt{2}}\). Thus, the ratio of the speed in X to Y is \(1 : \sqrt{2}\).
Marking scheme
1 mark for the correct option A. Apply Snell's Law in terms of wave speeds: \(\frac{v_X}{v_Y} = \frac{\sin \theta_X}{\sin \theta_Y}\).