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) =