2023 TPSS Phy Prelims Ans

Marking Scheme for 4E 2023 Physics Prelim P1 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 C B D B A B D D D B Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 D D A B C C D D C B Q21 Q22 Q3 Q24 Q25 Q26 Q27 Q28 Q29 Q30 B A D D C C A A D C Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40 D B C C A C B D C A Marking Scheme for 4E 2023 Physics Prelim P2 Section A Qn Answers Mark 1ai  Gradient of the graph [1] 1 1aii 4 min = 240 s [1] (as long as student use this in Any part of the working) **1m only if a is positive (correct value) + t is correct value 2 1b  Displacement of the car [1] *do not accept “distance” 1 1c  At 120 s or 2 min [1] 1 1d Direction = right/opposite direct (B.O.D) [1] **Displacement no penalty if student use t in min **Direction  no mark for “negative direction or backwards” 3 2a 2 2b 2

3a  A point where all the weight appears/considered to act on. [1] 1 3b 2 3c 1 4ai 2 4aii (Prove 01 angle)  Since Angle iAB < angle C [1]  TIR will not occur. [1] 3 4bi  Principal Focus F correctly positioned & Labelled correctly [1]  using one/two OTHER light rays with correct dotted and solid portions with arrow [1]  Completing the existing one ray with correct dotted and solid portions and direction. [1] 3 4bii  Image will be dimmer/less bright OR intensity will be reduced [1] 1 5a 1 5b OR 2 5c  4.0s [1] 1 5d  speed @ Q is lower/decreased or the wave is slower [1]  frequency remains the same/constant [1]  depth of sea @ Q is shallower/decreases [1] 3

6a  Friction between car and air causes electrons to be lost, [1]  Making the car positively charged. **student’s answer must state/display knowledge that the surfaces mentioned is between the car & air. 1 6b Any sensible answer:  Tyres are non-conductors or Tyres act as insulating material between car and ground. [1] 1 6c  Metal strap conducts electrons from the ground to the car [1]  to neutralise the positive charges on the car [1] 2 7a  Reading of Ammeter A1: increases  Reading of Ammeter A2: increases  Reading of Voltmeter V1: no change  Reading of Voltmeter V2: increases (2 correct =1 mark, 3 correct = 2 marks, 4 correct = 3 marks) 3 7b **1 mark for value of I if Rtotal is calculated wrongly 1 8a  The index finger (magnetic field) is perpendicular to middle finger (current) [1]  The direction of the force is perpendicular to the current and the magnetic field lines + The force induced (thumb) on the U frame will be upwards/out of paper. [1]  Causing the U frame to rotate Anti-Clockwise/upwards. [1] 3 8b (OWTTE)  Parallel wires carrying current in the same direction.  Has a stronger magnetic field at its exterior as compared to its interior. OR N-pole on right side & S-pole of left side of each turn OR opposite poles between each turn [1]  This results in a net attractive force [1]  Pushing the coil of wire together/closer. 2 9a 2 9b 2 9c  The transformer has an efficiency of 100%.  No heat loss to the surroundings (any one) 1

Section B Qn Answers Mark 10a  An electronic device that convert non-electrical energy to electrical energy. [1] 1 10bi As temperature increases, the resistance of the thermistor decreases  [1]  at a decreasing rate. OR  From 14 oC to 18 oC, the decrease is 760 per 4 degree Celsius.  From 40 oC to 44 oC, the decrease is 200 per 4 degree Celsius. (Compare any range) 1 10bi & bii points all plotted correctly = 1 mark good curve = 1 mark 2 10ci  As temperature increases, resistance of transducer decreases. + As p.d is proportionate to resistance, the p.d across transducer decreases [1]  Since total pd remains constant, pd of 4800Ω/Vout increases. [1] 2 10cii  Final voltage across transducer = 12 – 7.2 = 4.8 V  Final resistance of transducer = 4.8 × 4800=3200 Ω [1]  When resistance is 3200, temperature = 24℃. (refer to from student’s graph) [1] 2 resistance / Ω temperature / oC 10 15 4530 352520 5040 1000 1500 0 2000 2500 5000 3000 3500 4000 4500 5000

10ciii (OWTTE) When the lamp is connected in parallel,  the effective/total resistance in parallel (R//) decreased.  This cause pd across R// or lamp will reduce/drop OR pd across transducer will be higher [1]  Thus power output of lamp will be reduced/lowered [1]  Hence the lamp will be much dimmer. 2 11a  Gases are highly compressible. [1]  Gas molecules are very far apart / large intermolecular distance / very weak intermolecular force/attractive force between them. [1]. *Do not accept “force” 2 11b  The molecules collide with the wall of the container and exert an average force on the wall. [1]  The (average) force per unit area exerted by the gas molecules on the wall of the container is the gas pressure. [1] 2 11c (OWTTE)  Gas molecules are in continuous random motion / no preferred direction [1]  At each instant, there will be as many molecules colliding with one part of  the wall, as there will be for other parts. [1] 2 11d As volume decreases,  the number of gas molecules per unit volume increases .  The rate of collision of the gas molecules with the wall of the container increases [1] and  The force exerted per unit area will increase.  And thus the pressure exerted will double [1]. 2 11e  The number of molecules remain the same [1] 1 11f  During heating, the kinetic energy of the particles increases + The particles move faster /speed increases [1]  The particles are further apart from each other. 1 12 Either a 1 B  10J (pond surface as reference pt) OR 25J (bottom of pond as reference pt) [1] 1 c  Air resistance is negligible/No energy is lost to the surroundings [1] 1 d 2

e 1 fi  The velocity of the bob is increasing at decreasing rate OR Resistive force (in pond/water) is increasing OR acceleration of the bob is decreasing. [1]  Thus the resultant force acting on the bob is decreasing [1] 2 fii  Object is moving at constant speed / terminal velocity OR acceleration is zero [1]  The resultant force acting on the bob is zero [1] 2 12 OR A X = 2160 or 2200 g [1] Y = 1440 or 1400 g [1] Z = 6400 g [1] 3 b  Heat capacity is the amount of thermal energy required to raise the temperature of a cookware by 1 o C [1] 1 ci  Water in pot/metal Y will be the first to boil. [1]  because the heat capacity of metal Y (216 J/oC) is the lowest. OR For the same amount of thermal energy supplied, temperature of metal Y will increase the most. OR Thermal energy required for temperature of metal Y to increase by 1oC is the smallest. [1]  For the same amount of thermal energy supplied, water in the pot Y will be heated up faster. 2 cii  Water in pot/metal Z will be the at the highest temperature [1]  because it has the highest heat capacity OR For the same amount of thermal energy lost, temperature of metal Z will dropped by the least. OR thermal energy needed for the temperature of metal Z to drop 1 o C is the highest. [1]  And thus the water (inside Z) will be still be highest 2 d  Heat loss to the surrounding for all three pots is the same /negligible. [1]  Power outpu t (of heater) is consistent/constant during the heating process. [1] **no mark if student writes “heat absorbed (by pots) is the same” 2