2022 NCHS Phy Prelims Ans

1 NAN CHIAU HIGH SCHOOL Sec 4 Express Physics Papers 1, 2 and 3 Solutions 2022 Preliminary Examination Paper 1 (40 marks) 1 2 3 4 5 6 7 8 9 10 C C B C A A C B C C 11 12 13 14 15 16 17 18 19 20 A B B C A B C C D C 21 22 23 24 25 26 27 28 29 30 A B B B C C D A C B 31 32 33 34 35 36 37 38 39 40 B C C D B A B B D D Paper 2 Section A (50 marks) Question Answer Remarks 1 a > The ball is at equilibrium as > it is at rest, no resultant moment/ no resultant forces / stationary / not moving. B1 B1 b W = m.g = (2.0).(10) = 20 N (2sf) c The force must be from the centre of the ball and pointing downwards. d Deduct 1 mark each for: -No, wrong arrows; -Out of range; -Diagram too small,(minimum 1cm:2N -Wrong orientation *2 or 3sf is accepted 2 a D should be between 12 to 18 secs W 1 cm = 1.0 N T1 = 16.4 N T2 = 11.5 N 35 º 20 N 55 º

2 b (23, 8000), (31, 4000) acceleration = change of speed / time taken = (4000 – 8000) / (31- 23) = - 500 mm s-2 = -0.50 m s-2. Deceleration = 0.50 m s-2 allowance for error 0.48 to 0.52 m s-2 M1 A1 c > The rock is likely to have hit the bottom of the pond as > its speed changes from 2.0 m s-2 to 0 in an extremely short time, speed is zero, not moving (etc of the same meaning) B1 (position) B1 (motion) 3 a When an object is at equilibrium, the sum of clockwise moments about any point equal to the sum of anti-clockwise moments about the same point. B1 B1 b Sum of anti-clockwise moments = sum of clockwise moments 90.RB = (40).(25) + (90-80).(600) RB = 77.77 N = 78 N (2 sf) M1 A1 c The position of the centre of gravity of the man-ladder system will rise and the system to be less stable. *do not accept “CG increase” , as a position cannot increase. reasoning of system to be less stable must be correct also. M1 A1 4 a The force per unit mass acting on an object under gravity. bi and bii 1m for both shapes correct 1m for both labels P and K correct 1m for total energy is constant c GPE at 40 m = m.g.h = (5.0).(10).(40) = 2000 J GPE at 40 km = m.g.h = (5.0).(9.7).(40 000) = 1,940,000 J Difference in GPE = 1,940,000 – 2000 = 1,938,000 = 1,900,000 J (2 sf) (both gpe must be correct to get 1m) 1m 5 a (h.ρ.g)air + (h.ρ.g)H at position P= (h.ρ.g)Hg at foot of mountain hair.(1.23).(10) = (0.760 – 0.700).(13600).(10) hair = 663 m (No marks will be awarded if students only state pressure difference as 6.0 cm Hg) M1 M1 A1 b As point P has a lower atmospheric pressure than at sea level, the molecules do not need to overcome so high downward force exerted by the atmospheric pressure than when it is at sea level, so less energy is needed to boil water a point P, so water boils at a lower temperature. 1m 1m Speed / m s-1 D time t/s Energy/ J h /m0 P K

3 Note: energy required to overcome intermolecular forces of attractions at P and at the foot of the mountain is constant thus the assumption by most students that less energy is required to overcome the intermolecular forces of attraction at P is incorrect. 6 ⮚The marker is too thick, thinner markers should be used. ⮚Should divide the length into 100 equal divisions of 1 oC, not 8 divisions. ⮚Should use boi ling point of water instead of 80 oC as it is not easy to reproduce exactly 80 oC. ⮚Ice point was marked after only 10 s which is too fast, should mark after at least 2 min for the reading to be stabilised. ⮚Placing the thermometer above the boiling water/in the steam and not inside the hot water. Do not accept : > use pure ice as 0 deg C is stated in the question > use equal mass/volume for 0 deg C ice/water and for 80 deg C water as it does not affect the temperature > repeating the experiment is not an improvement as there is no change to the setup > Any methods to reduce heat loss such as insulation is irrelevant. 1 m for each point, max 3 marks 7 a It is a process of charging a conductor without contact between the charging body and the conductor. b The electrons (BOD: negative charges) inside insulators cannot move even in an electric field, so they cannot be charged by induction. OR Insulators have no mobile electrons. c ⮚Bring Plate X near to (BOD: close together) but not touching Plate Y, connect the earth w ire to Plate Y. (Accept “connect earth wire first, then bring X near”) ⮚Disconnect the earth wire and bring Plate X far far way from Plate Y. 1 m 1 m 8 a ⮚The resistor is ohmic as ⮚its I-V graph is a straight line passing through the origin or Current is directly proportional to voltage or Resistance is constant. ⮚The lamp is not ohmic as ⮚its I-V graph is not a straight line or Current is not directly proportional to voltage or Resistance is increasing. Minus 1 m for each wrong/ missing part Max minus 3 m b From Fig. 8.1, the resistances of the lamp and of the resistor are equal (where the I-V graphs intercept) when the lamp has pd of 7.0 V (2 sf) and current of 4.6 A C ⮚From Fig. 8.1, when the total pd is 7.6 V and when current is the same for lamp and resistor, current of lamp is 3.0 A and pd across the lamp is 3.0 V. ⮚Resistance of lamp = 3.0 V / 3.0 A = 1.0 Ω. 9 a ⮚It can be attracted by magnets. ⮚It can be made to be magnets (or be magnetised). 1 m 1 m bi Steel/ Cobalt/ Lodestone/ Nickel bii As the iron bar is placed near the permanent magnet, its right-hand side is induced a South pole and left-hand side a North pole. As unlike poles attract and like poles repel, the attractive forces between the North pole of the permanent magnet and the South pole of the induced magnet is stronger than the repulsive force between the North-pole of the permanent magnet and North-pole of the induced magnet as the distance between the unlike poles are smaller, so there is a net attractive force.

4 c ⮚ Place a magnet inside a solenoid connected to an alternating current (a.c.) supply ⮚ Without switching off the current , withdraw the magnet slowly in the East- West direction. 1 m for correct diagram 1 m 1 m Paper 2 Section B (30 marks) 10 ai When the heater is placed at the bottom, it heats up the liquid at the bottom. The liquid expands and becomes less dense, it then rises to the top. The colder, denser liquid sinks to the bottom. This creates a convection current in the liquid and will ensure the heat is distributed evenly. aii I.V.t = C.∆θ C = (I.V.t) / ∆θ = (5.0).(20).(12).(60) / (38 – 30) = 9000 J / oC M1 A1 Deduct 1 mark if uses min or hour for time aiii ⮚Yes, the student is right. ⮚From 26 oC to 30 oC (room temperature), the liquid gain heat from the surrounding and from 30 oC to 34 oC, the liquid lose heat to the surrounding. ⮚The heat gained and loss from and to the surrounding can potentially cancelled off and the heat capacity calculate will be more accurate. Max 2 marks No mark if stated yes but with wrong reasons. bi ⮚Real ⮚Diminished ⮚Inverted 1 m for 2 correct, 2 m for 3 correct. bii ⮚f = 10 mm. ⮚When object distance is 20 mm, image distance is also 20 mm, this means that the object size is equal to the image size, which means the object is placed at 2f from lens. So since 2f = 20 mm, f = 10 mm. ⮚1/f = 1/u + 1/v 1/f = 1/30 + 1/15 f = 10 mm 11 ai ⮚ Type I : ultraviolet radiation ⮚ Type II : visible light ⮚ Type III : infrared radiation 1 m for 2 correct, 2 m for 3 correct Aii All the brands can be used in vacuum as all electromagnetic waves can travel in vacuum.

5 bi ICE (through LDR and 20 Ω resistor) = V / R = 12 / (20 + 180) = 0.060 A ICE (through lamp and 60 Ω resistor) = 0.66 – 0.060 = 0.60 A I60Ω = V / R = 12 / 60 = 0.20 A Therefore Ilamp =0.60 – 0.20 = 0.40 A Rlamp = 12 / Ilamp = 12 / 0.40 = 30 Ω Alternatively, V/I = [1/(20 + RLDR) + 1/60 + 1/Rlamp]-1 12/0.66 = [1/(20+180] + 1/60 + 1/Rlamp]-1 18.18 = [0.00500 + 0.0167 + 1/Rlamp]-1 0.055 = 0.0217 + 1/Rlamp 0.0333 = 1/Rlamp Rlamp = 30 Ω B1 B2 A1 Deduct 1 m for each mistake bii1 No change in the brightness of the lamp as both potential difference