XMS 2025 Physics Prelim P2 MS

Marking Scheme Physics (6091) Preliminary Examination 2025 Accept 2 or 3 sf Paper 2 – Section A Qn Marking Scheme Marks Marker’s Report 1a F A = P = F A P = 5.0 2.0×10−4 P = 25 000 Pa F 0.080×10−4 = 25 000 Pa F = 0.20 N 1 1 1bi change of vol. at piston = A x d = 0.00020 × 0.0060 = 0.0000012 m3 1 Accept 1.2 cm3 and 1.2 ml 1bii Molecules in water are closely packed with little space between them …and cannot be pushed closer hence the volume remains the same /thus cannot be compressed. 1 1 1biii Vol = A x d 0.0000012 = 0.0000080 x d d = 0.15 m speed = 0.15/1 speed = 0.15 m/s 1 1 Accept alternative methods WD = WD WD = ½ mv2 (to explain not to use this method in the future) Accept 15 cm/s Mark scheme will use these abbreviations / + R A Ig ref ecf AW AVP ORA OWTTE underline ( ) alternatives statements on both sides of the + are needed for that mark reject accept (for answers correctly cued by the question) ignore as irrelevant with reference to error carried forward alternative wording (where responses vary more than usual) alternative valid point or reverse argument or words to that effect actual word given must be used by candidate (grammatical variants excepted) the word / phrase in brackets is not required but sets the context

7 Qn Marking Scheme Marks Marker’s Report 2a The point which the weight of the object appears to act on. 1 2b Pivot marked at C 1 2c moment = F x ⊥d moment = 500 x 0.40 moment = 200 Nm 1 1 2d When a system is in equilibrium, the sum of clockwise moment is equal to the sum of anticlockwise moment about the same pivot. 1 2e moment = F x ⊥d 200 = T x 2.0 m T = 100 N 1 1 2f Contact force acting at C, upwards Or Friction acting at C, leftwards 1 8 Qn Marking Scheme Marks Marker’s Report 3a The container is made from non- metals/cardboard/plastic which are poor conductors of heat. The layer of corrugated cardboard traps pockets of air which are poor conductors of heat. The lid prevents heated air from moving out of the cup restricting/preventing the formation of convection current. /the lid reduces the rate of evaporation 1 1 1 3b E to reach 100C = mc E = 1.6(4200)(75) E = 504 000 J E to vaporise 0.90 kg of water = ml E = 0.90 x 2.3 × 106 E = 2 070 000 J Total E = 504 000 + 2 070 000 = 2 574 000 = 2 600 000 J 1 1 1 6

Qn Marking Scheme Marks Marker’s Report 4a ‘ultra-sound’ should be replaced with ‘ultraviolet (waves/rays)’ ‘infra-red waves’ and ‘microwaves’ should swap positions ‘visible light’ is missing and should be after ‘ultraviolet (waves/rays)’ and before ‘infra-red waves’ 1 1 1 OWTTE 4b Any acceptable E.g. security screening, industrial defect detection, sterilisation. 1 4 Qn Marking Scheme Marks Marker’s Report 5a Potential difference across two points is the work done when a unit charge passes through two points on the circuit. Is the work done to drive a unit charge passes through two points on the circuit. Is the work done to drive a unit charge passes through/across a component. 1 OWTTE 5b I = Q/t I = 30/0.25 I = 120 A P = IV P = 120 x 7.5 × 108 P = 9.0 × 1010 W 1 1 Accept alternative method of: WD = V x Q P = WD/t 3 Qn Marking Scheme Marks Marker’s Report 6ai RT = V/I RT = 12/0.0015 RT = 8000  RT = RLDR + 5000 RLDR = 3000  1 1 6aii VLDR = 12 x 3000 8000 VLDR = 4.5 V 1 1

6b V across LDR when bright = 12 x 1000 6000 = 2.0 V V across resistor when bright = 12 x 5000 6000 = 10 V The electronic needs a pd > 8.0 V to trigger, therefore it needs to be placed across QR. 1 1 1 7 Qn Marking Scheme Marks Marker’s Report 7ai South pole 1 7aii When current passes through the coil and it becomes an electromagnet. An unlike pole is induced closer to R in the iron armature causing it to be attracted to R. The armature rotates about the pivot and pushes the contacts together completing the circuit of the motor and turning it on. 1 1 7b Steel is a hard magnetic material and loses its magnetism slowly. Thus, when switch S is opened, the steel core still attracts the iron armature and the motor remains on. 1 1 5 Qn Marking Scheme Marks Marker’s Report 8a 8bi 1 1 ecf 8bii Using RHFR at X, with the thumb as the force pointing upwards and index finger pointing along the direction of magnetic field, the middle finger is the direction of the current. 1 8c To allow for continuous sliding contact between the coil and the external circuit and preventing the entangling of wires. 1 4

Qn Marking Scheme Marks Marker’s Report 9a X64 149 → Y65 149 + 𝑒−1 0 + γ0 0 nucleus Y correct Beta particle and gamma ray correct 1 1 Accept ‘’ in place of ‘e’ 9bi 20 counts per sec 1 9bii Allow acceptable concrete, lights, sunlight 1 9biii Shows indication of removal of background radiation Removing background: 0h – 60 1h – 30 2h – 15 Half-life = 1.0 h 1 1 6 Qn Marking Scheme Marks Marker’s Report 10a w = mg w = 65 (10) w = 650 N 1 10bi FR = ma FR = 65 (0.50) FR = 33 N 1 1 10bii weight acting vertically downwards normal contact force vertically upwards 1 Accept correct values 10biii Force on the women by floor = 33 + 650 = 683 N = 680 N (2sf) 1 10ci a = VF-VI t 0.50 = VF-0 4.0 VF= 2.0 m/s 1 10cii t = 0s to t = 4.0s: straight line with positive gradient. t = 4.0s to t = 8.0s: horizontal straight line 1 1

10ciii t = 0s to t = 4.0s: curve with increasing positive gradient t = 4.0s to t = 8.0s: straight line with constant positive gradient 1 1 10 Qn Marking Scheme Marks Marker’s Report 11a Energy is transferred to the solar panel by the propagation of waves (light). Some energy is transferred into the internal store of the solar panel and some energy is transferred out of the solar panel electrically. Alternatively: Light energy is… …converted to thermal energy and electrical energy. 1 1 1 1 11b Its output is dependent on the amount of sunlight and cloud cover. /Takes up lots of space and space is limited in Singapore. 1 Do not accept: Its costly. 11ci 200 W 1 11cii Power is the rate of work done. Power is the work done per unit time. 1 11ciii E = 200(60) + 200(60) + 0 + 0 + 0 + 0 + 800(60) + 800(60) + 600(60) + 600(60) E = 192 000 J E = 190 000 J (2sf) 1 1 11civ Part ii is calculated with the assumption that the intensity of sunlight remains as stated by the table for the entire 1 min interval without changing. 1 11cv 1. E per second falling on panel = 600 J  0.3 = 2000 J 2. Surface area of cell = 2000  200 = 10 m2 1 1 Total 10

Paper 2 – Section B Qn Marking Scheme Marks Marker’s Report 12a sin c = 1/n sin c = 1/1.3 c = 50 1 12b i = 28 and labelled on diagram 1.3 sin 28 = 1 sin r r = 38 and labelled on diagram light ray bends away from normal 1 1 1 12c Ray Q is travelling from optical denser medium to an optical less dense medium. Its angle of incidence is 54 (if not stated can be show on diagram) and greater than the critical angle, hence it would undergo total internal reflection. 1 1 12di Red light can only exit the plastic panel at angles of incidence lesser or equal to 50 and would not exit the plastic panel at incident angles of greater than 50 Thus, light only emerges from a circular area centred on the LED. 1 1 12dii Blue light is slower in the plastic panel. This means the refractive index of blue light in plastic is greater than 1.3. This me