MI 2024 PU3 H2 PHYSICS PRELIM P3 ANS

1 2024 PU3 Preliminary Examinations H2 Physics Paper 3 Suggested Answers Section A 1 (a)(i) Period = 0.80 s Frequency = 1/T = 1/0.80 = 1.25 Hz C1 A1 (a)(ii) Magnitude of induced emf is proportional to the rate of change of flux linkage A1 (a)(iii) 1. As magnet oscillates in the coil, the flux linkage with the coil changes and emf is induced in the coil, from Faraday’s Law. 2. This causes an induced current to flow when the switch is closed. 3. The oscillating system loses energy as there is heat loss in the resistor. 4. Since energy of the system is proportional to the square of amplitude, amplitude decreases. OR 3. From Lenz Law, the flow of induced current produces a magnetic field which results in a force opposing the motion of the magnet. [B1] 4. The maximum resultant force on the magnet decreases and since amplitude is proportional to the maximum resultant force, amplitude decreases. [B1] B1 B1 B1 B1 OR B1 B1 (b)(i) 0.50 Hz A1 (b)(ii) Amplitude increases then decreases Amplitude is maximum at 1.25 Hz when resonance occurs. (Amplitude of oscillation increases as freq increases from 0.5 Hz to 1.25 Hz. At 1.25 Hz, resonance occurs. As frequency increases beyond 1.25 Hz to 5.0 Hz, the maximum amplitude decreases) B1 B1

2 2 (a)(i) Single slit: sin  = /b using small angle approximation sin   tan  𝑠𝑖𝑛𝜃 = 0.003 2.4 = 𝜆 𝑏 (Accept 0.0325) 𝑏 = 5.04 × 10−4 𝑚 (accept 4.65 x 10-4m) C1 A1 (a)(ii)1. (When B decreases↓ , then sin increases ↑, and hence  increases↑, therefore) this results in increase in width of central maximum and the maxima are further apart. The central maximum is also lower in intensity. B1 B1 (a)(ii)2. (When wavelength 𝝀 decreases↓, sin decreases↓, and hence  decreases↓) Therefore there is reduced width of central maximum and the maxima are closer. Intensity of the central maximum is unchanged (bright and dark regions) B1 B1 (b)(i) ( )1 188 160 142 = − =  ( ) ( ) 9sin sin 14 2 633 10d n d −= →  =  65.23 10d −= m 511 1.91 10dN Nd= → = =  C1 M1 C1 A1 2.4 0.003

3 3 (a) Electric potential at a point is the work done per unit positive charge in bringing a charge from infinity to that point. B1 (b)(i) A B D C F F’ Two arrows of equal length, correct direction with labels ( i.e. Force on C by B, Force on C by D) – 1 Mark B1 (b)(ii) F = 2 24 o Q r = (9 × 109) 62 22 (1.2 10 ) (2.0 10 ) − −   = 32.4 N Resultant force = √(F2 +F2) = F 2 = 32.4 2 = 45.8 N Direction: Along A to C (ecf (i)) C1 A1 A1 (b)(iii) Distance from corner to centre, r = 2 cm V = 4 o Q r = (9 × 109) 6 2 1.2 10 2 10 − − −  = −7.64 × 105 V Electric potential due to the three charges = 3V = 3 (−7.64 × 105) = −2.29 × 106 V C1 C1 A1 (b)(iv) At 100m away, potential due to the 3 charges is negligible W = q∆V = (1.2 × 10−6)(−2.29 × 106 - 0) = −2.75 J C1 A1

4 4 (a) Any of 2 of the following: Increase current flowing through the solenoid Increase the number of turns per unit length in the solenoid Insert a ferrous core into the solenoid B1 B1 B1 (Max 2) (b)(i) C to D B1 (b)(ii) By POM, F x 0.106 = 5.7x10-4 x 0.077 And F = BIL = B(4.9)(0.025) B = 3.38 x 10-3 T Tesla C1 C1 A1 B1 (c)(i) 7 3 4 () () 400 3.2(9)(0.010 0.018)(4 10 )( )( ) 0.50 10 10 5.21 10 V − − − =− = = = = =    = o o dE dt d NBA dt dBNA dt dnNA dt dNA n dt    I I C1 A1 (c)(ii) Horizontal lie of constant e.m.f. with correct values (allow for e.c.f.) Correct sign of V B1 B1 0 10 20 30 40 5.2110−4 time / ms e.m.f. / V - 5.2110−4 − 5.2110−4

5 5 (a)(i) <V2> = [(32 x 3) + (62 x 1)] / 4 = 15.75 V2 Vrms = (15.75)0.5 = 4.0 V C1 A1 (a)(ii) Shape (not rectified) with correct period – 1m Correct values either (3.0 V, -2.0 V) Or (-6.0 V, 1.0 V) – 1m B1 B1 (b)(i) Turns ratio = 4.50 :500 = 9:1000 = 0.009 (allow for 500:4.50 = 111) A1 (b)(ii) transmission line resistance = 6.44 x 105 x 4.50 x 10-4 = 289.8 Ω Given the voltage is stepped up to 500 kV at the output of the transformer, The current flowing in the tx line = 5 x 106 / 500x103 = 10 A Power loss = 102(289.8) = 2.9x104 W C1 C1 A1 (b)(iii) By stepping up the voltage, the current in the transmission line will be lowered and hence the amount of power loss in the transmission line will be lowered. A1 V / V 3.0 −2.0 1 2 3 4 5 6 7 8 t / s

6 6 (a)(i) The spread of the electron beam to form bright rings due to diffraction, The presence of dark regions between bright rings indicate that destructive interference is taking place B1 B1 (a)(ii)1. KE = eV = p2/2m 𝜆 = ℎ 𝑝 = ℎ √2𝑚𝑒𝑉 M1 M1 A0 (a)(ii)2. 𝜆 = ℎ √2𝑚𝑒𝑉 = 6.63×10−34 √2(9.11×10−31)(1.60×10−19)(250) = 7.77 x 10-12 m C1 A1 (a)(iii) As potential difference V increases, the de Broglie wavelength decreases since 𝜆 𝛼 1 √𝑉. Since dsin  = n, the angles at which the maxima are produced decrease. The rings become smaller/closer together. B1 B1 (b) ∆p ∆x ≳ h ∆p (1.2) ≳ 6.63 x 10-34 ∆p ≳ 5.53 x 10-34 kg m s–1 p = 80 x 2.0 = 160 kg m s-1 Hence, the uncertainty in his momentum is negligible as compared to his linear momentum. OR p p 34 365.53 10tan 3.46 10 0 160 − −= = =   Hence, the angle of deflection is negligible. M1 B1

7 2024 PU3 Preliminary Examinations H2 Physics Paper 3 Suggested Answers Section B 7 (a) The moment of a force about a pivot is the product of that force and the perpendicular distance between the line of action of the force and the pivot. The torque of a couple between two forces is the product of one of the forces and the perpendicular distance between their lines of action. One force for moment & two forces for torque OR about a pivot for moment & no need for pivot for torque B1 B1 B1 (b)(i) The anti-clockwise moment due to the weight of the hook is balanced by the total clockwise moments due to the weight of the sliding weights. OR c.g of the weighing machine is at the pivot B1 (b)(ii) For the same load on the hook, a smaller perpendicular distance from the hook to the pivot will result in smaller anti-clockwise moment to be balanced by the sliding weights. A longer rigid rod will allow for smaller sliding weights to be used. OR A longer rigid rod will give a better precision to the reading. OR With the same sliding weights, a bigger range of loads can be measured. B1 B1 (b)(iii) Weight of sack of flour = mg mg × 4.8 = (12 × 84) + (2.5 × 72) m = 25.2 kg C1 A1 (b)(iv) The movement of the sliding weights will be much smaller for such a small load hence the corresponding fractional uncertainty of the weight of the object will be higher. B1 B1 (c)(i) A force F acts on a mass m to move it vertically upwards at constant speed (so that no change in Ek) by a displacement h in the direction of the force. Since the object moves at constant speed, the upward force, F, must be equal to the weight of the object, mg. (no resultant force). Using Work done on object = Fh = (mg)h Since Fh is the work done on the object and is equal to the increase in potential energy, Ep = mgh B1 B1 B1 (c)(