BTYSS 2024 FYE 4E PHY MS

`1 Beatty Sec 4E Preliminary Examination 2024 Physics 6091 Beatty Sec 4E Physics (6091) Paper 1 and Paper 2 Preliminary Examination 2024 MARKING SCHEME Paper 1 [40 marks] 1 C 2 D 3 A 4 A 5 C 6 C 7 C 8 D 9 D 10 D 11 A 12 D 13 B 14 B 15 C 16 C 17 D 18 A 19 B 20 B 21 A 22 A 23 D 24 A 25 A 26 D 27 C 28 C 29 D 30 C 31 D 32 D 33 C 34 B 35 B 36 D 37 D 38 B 39 B 40 C Minus 1 for missing/wrong unit and minus 1 for wrong s.f [Max minus 2 for whole paper] Paper 2 Section A [70 marks] Qn Answer Mark 1a The force exerted by the mercury on the trapped air changes when the tube is rotated, causing the pressure exerted by the mercury on the trapped air to change. 1 1b Pliquid = ρgh = 14000 × 10 × (30 ÷ 100) = 42000 Pa Pgas = 42000 + 1.0 × 105 = 142000 Pa ≈ 140000 Pa (2 s.f) 1 1 1c 1.0 × 105 Pa 1 2a • When temperature increases, the pressure of the gas inside the elastic flask increases and the force exerted on the walls of the flask increases. • The elastic flask expands and the volume of the gas increases, leading to a decrease in the pressure of gas. 1 1 2b • When the temperature increases, the average kinetic energy of the particles increases. • The particles move faster (average velocity increases) and collide with the walls of the flask more frequently and vigorously. • The average force exerted on the walls increases and since P = F / A, the pressure of the gas increases. 1 1 1 3a Qheater – Qsurroundings = Qheating + Qboiling (1.0 × 103)t – 20t = (1.5)(4200)(100 – 25) + (0.050)(2.26 × 106) 980t = 585500 t = 597 s ≈ 600 s (2 s.f) 2 1 3b • The air above the hot water is heated and expands. • The heated air is less dense and rises while the cooler surrounding air which is denser sinks to replace it. • The cooler air is then heated by the hot water and the process repeats to remove energy from the hot water. 1 1 1 4a Converging / Convex 1 4b • When ray R enters, it moves from the optically less dense air to optically denser glass. • The speed of light decreases and the ray bends towards the normal. • When ray R exits, it moves from the optically denser glass to optically less dense air and the speed of light increases, causing the ray bends away from the normal. 1 1 1

`2 Beatty Sec 4E Preliminary Examination 2024 Physics 6091 4c Either emergent ray R or the ray through F extended to the left of the lens Both ray through F and R shown and image marked at intersection of extended rays 1 1 4d • The image forms further away from the lens. • The image becomes larger. 1 1 5a Direction of field lines Electric field pattern 1 1 5b I = Q / t = (4.8 × 10-9) ÷ (2.0 × 10-3) = 2.4 × 10-6 A 1 1 6a P = I2R = (0.0020)2(800) = 0.0032 W 1 1 6b VY = 6.0 – (0.0020)(800) = 4.4 V R = V / I = 4.4 ÷ 0.0020 = 2200 Ω 1 1 6c • When the thermistor is cooled, the resistance of the thermistor X increases and the potential difference across the thermistor X increases. • Given the total e.m.f remains unchanged, an increase in potential difference across thermistor X leads to the decreases in potential difference across resistor Y. OR The current through the resistor Y decrease and since V = IR, the potential difference across Y decreases. OR by potential divider principle, the ratio of the resistance of Y to the total resistance decreases and the potential difference across Y decreases. 1 1 7a To provide a return path for the current to flow back from the appliances. 1 7b • When the live wire touches the metal case, the earth wire provides a low resistance path for the current to flow from the metal case to the ground. • The large current flow exceeds the fuse rating and the fuse melts, disconnecting the appliance from the live terminal. 1 1

`3 Beatty Sec 4E Preliminary Examination 2024 Physics 6091 7c • The hairdryer is doubly insulated whereby two layers of insulation, one layer insulating the internal components from the external casing and another layer insulating the wires from the internal components. • This isolates the user from contact with any conducting metal parts. 1 1 8a P: North Q: South 1 8b • When current flows in the coil, the magnetic field generated around the coil interacts with the magnetic field of the magnets to produce a downward force to act on side AB and an upward force to act on side CD. • The equal forces acts in opposite direction to generate an anticlockwise moment. 1 1 8c Component X reverses the direction of current inside the coil and ensures the direction of the force at each side of the coil remains the same. 1 9ai T and Q or U and R 1 9aii Beta particle 1 9bi It means that it is not possible to predict which radioactive nucleus will emit radiation and the direction of the emitted radiation. 1 9bii 𝑃𝑜84 218 𝐻𝑒2 4 1 1 9biii To get 5.0 × 1013 atoms, (4.0 × 1014) × ½ × ½ × ½ = 5.0 × 1013 atoms No. of half-life = 3 Half-life = 11.4 ÷ 3 = 3.8 days 1 1 1 10ai • The power output increases at a constant rate from zero to a value of 0.40 MW between 5 to 15 m/s. • It then increases further at a higher constant rate to a value of 0.80 MW between 15 to 20 m/s. 1 1 10aii Rate of power output = 0.40 ÷ 10 = 0.040 MW / m/s 1 10aiii Total energy produced = [0 + 0 + 0 + 0 + 0.04 +0.62 + 0.36 + 0.40 + 0.80 + 0.80] × [2 × 60 × 106] = 3624000000 ≈ 3.6 × 109 J (2 s.f) 2 1 10aiv The wind speed may not be constant during the two-minute time interval. 1 10bi • The voltage produces an alternating current in the primary coil. • This produces an alternating (changing) magnetic field inside the iron core. • This alternating magnetic field cuts the secondary coil and induces an output voltage (e.m.f) in the secondary coil. 1 1 1 10bii • Increase the potential difference across the wire. • As Poutput = VI, delivering a large power through the wire using a higher potential difference leads to a lower current which in turn gives a lower energy loss as Ploss = I2R. OR • Increasing the cross-sectional area of the wire to decrease the resistance of the wire. • As Ploss = I2R, reducing the wire resistance leads to lower energy loss. 1 1 11a The centre of gravity is the point at which the whole weight of the object appears/seems to act on. 1 11b The force exerted by the slab on the bar. 1 11c Force × 0.90 = 200 × 0.60 Force = 133 N ≈ 130 N (2 s.f) 1 1 11d Force = 130 + 80 + 200 = 410 N OR 413 N (allow ECF from 11c) 1 11e Assuming length of slab = 1.0 m Weight × 0.50 = 133 × 1.0 Weight = 266 N (allow ECF from 11c) 1 11f • Decrease the distance between the stone and the slab. • The anticlockwise moments by the force of the slab decreases and there is a clockwise moment generated by the weight of the bar. Thus, the clockwise moments required by the force decreases. With an increase in perpendicular distance to the 1 1

`4 Beatty Sec 4E Preliminary Examination 2024 Physics 6091 stone, the force required will be less than 200 N. Paper 2 Section B [10 marks] Qn Answer Mark 12ai The graph shows the velocity of the man changing from positive to negative at t = 4.0 s. 1 12aii Acceleration = gradient of graph = (10 – 0) / (1.0 – 0) = 10 m/s2 (Accept use of any points between t = 0 to 1.4 s) 1 1 12aiii The air resistance is negligible / There is no air resistance. 1 12bi t = 4.0 s 12bii • At D, the weight of the man acts downward