Jurong Prelim P2 Ans

1 JSS 2024 Graduation Examination Secondary 4E Pure Physics (6091) MARKING SCHEME Paper 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 B D D D A B C B B C A C A B A C D B C C 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 C D C B B B A A D A C B D A C B A D D A Paper 2 Section A Question Answer Mark s Markers’ Comments 1 (a) random error B1 (b) Digital vernier calliper. Degree of accuracy is 0.01 mm B1 B1 (c) volume =mass ÷ density = 1380 ÷ 7.65 = 180 cm3 A1 2 (ai) B1 B1 Correct labeling/Directio n of forces (aii) 30 – T = 3a T – 20 = 2a a = 2 m/s2 M1 A1 (b) Applying Principle of Energy Conservation, magha = ½ mav2 + mbghb hb = 10.8 m M1 A1 3 (a) moment of weight = weight ×perpendicular distance of weight from pivot = 150 ×1.8 = 270 Nm A1 (bi) The Principle of Moments states that when a body is balanced, the total clockwise moment about a point equals the total anticlockwise moment about the same point. B1

2 Question Answer Mark s Markers’ Comments (bii)  By M = Fd  Higher perpendicular distance  increase ACW moment  By POM  for the same distance of the weight of the barrier  an additional force is needed to increase the CW moment to balance thr barrier. B1 B1 4 (a) Choose two factors: density of water/ gravitational field strength/depth of water Accept temperate and humidity of air (b) P = hρg + Patm 3x105 = h(1030)(10) + 1x105 h = 19.4 m M1 A1 (ci) Correct arrangement of LDR and light bulb (Parallel) B1 0 mark if no labelling (cii) Deeper  darker  brightness decreases  resistance of LDR increases  Voltage of LDR increase due to potential dividing Light bulb parallel to LDR  voltage increases  P increases  brighter B1 B1 5 (a) from elastic potential store to kinetic store and gravitational potential store B1 (b) kinetic store and gravitational potential store to gravitational potential store and elastic potential store B1 6 (a) Cooling Evaporated ethanol gain latent heat in the form of PE to change state  remove thermal energy from skin B1 B1 (b) Rate increases. Moving air remove evaporated ethanol particles so that other ethanol particles can leave the body easily B1 B1 7 (a) ultraviolet rays, X-rays B1 (b) detection and treatment of cancer OR imaging/gamma photography OR telescopes OR sterilisation of food/medical equipment B1 (c)  radiations arrive) at same time.  because they have same speed in a vacuum B1 B1 8 (a) sin c = 1 / refractive index c = 48.8° A1 (b) B1 Correct path of light / angles labelled correctly

3 Question Answer Mark s Markers’ Comments (c) i > c and light travels from optically denser to less dense medium  TIR occurs B1 9 (a)  Correct ray A and B drawn with arrows  Image drawn correct at 2F B1 B1 (b) Correct ray C drawn with arrows B1 10 (a) two points labelled C and R at the centre of the one compression and rarefaction B1 (b) Measured wavelength = 5 cm = 0.05 m f = V/λ = 330/ 0.05 = 6600 Hz M1 A1 11 (a) B1 (b)  The negative charges on the cloud repels the negative electrons to the lower part of the aircraft due to electrostatic force of repulsion.  The upper part of the aircraft now has an induced positive charge while the lower part an induced negative charge. B1 B1 12 (a) work done by the source in driving a unit charge around a complete circuit is 12 J B1 (bi) current in R = 4 A, V in R = 9 V R = V/I = 9/4 = 2.25 Ω M1 A1 (bii) resistance proportional to length (so twice length twice resistance) + resistance inversely proportional to area (so twice diameter decreases resistance by factor of 4) R = 2.25x2/4= 1.13 Ω M1 A1 13 (a)  Current in the coil creates a magnetic field, this magnetises the iron core into an electromagnet.  The iron bar  magnetic material  attracted to the core  movable pivot to move to the right and pulls the steel bolt out of the door B1 B1 O L F F Princip A B C __ __ __ __ __ __ __ __ __ __ + + + + _ _ _ _

4 Question Answer Mark s Markers’ Comments (b) Current upwards Core left – North, Core Right –South, Bar left – North B1 14 (a) B1 B1 Transformer/a. suppy/load Ammeter/voltmet er -1 mark if no labelling (b) Setup 3. No of secondary turns are more than no of primary turns B1 B1 (c) Using Ns/Np = Vs/Vp Vs1 = 2 V, Vs = 6 V, Vs3 = 12 V M1 A1 (di) Setup 1 efficiency = 4.61/7.2 x 100 = 64% B1 A1 (dii) Internal resistance of the wire Since highest current  Power loss = I2R (highest) B1 B1 15 (a) α-emissions (helium nucleus, +2) β- emissions (electron, -1)  γ-emissions (EM radiation, 0) B1 B1 (b)  226 86 Ra = 222 84 Rn + 42 He A2 Correct mass no, proton no (ci) B1 B1 Plot Best fit line (cii) half life = 15 hour (+/- 1 hour) A1 (di)  Gamma. Passes through paper and alumnium, absorbed by concrete B1 B1 (dii) Cause cancer/ stunt children growth B1 0204060801001201400 12 24 36 48 60

5 Section B Question Answer Marks Markers’ Comments 16 (a) B1 B1 B1 (bi) the change of velocity per second is contant B1 (bii) Stage 2. Gradient = acceleration = steeper B1 B1 (ci) t = Q/ I = (3400/1000)(3600) / 10 = 1224 s M1 A1 (cii) W = VQ = 12 (3400/1000)(3600) = 146880 J M1 A1 17 (a)  Air which is a bad conductor is trapped in woolly fibre  Prevent heat gain in the pipe through conduction  Aluminium is shiny, a bad absorber of infra-red radiation  Prevent heat gain in the pipe through radiation. B1 B1 (bi)  The coolant changes its state from solid to liquid.  Heat (PE) absorbed is used to break the intermolecular bonds B1 B1 (bii) Q = mlf = (0.3)(2x105) = 60000J M1 A1 (biii) Q = mcθ = (0.3)(4000)(9) = 10 800 J M1 A1 biv) Energy lost by food = energy gained by coolant 70800 / (20 – 4) 4425 J °C-1 M1 A1

6 Subtract from the total score if incorrect [up to 4 marks]: s.f. [up to 2 marks] units [up to 2 marks]