MSHS 2025 PHY PRELIM P1+P2 MS

1 2025 MSHS 6091Physics Prelim Exam Solutions PAPER 1 1 2 3 4 5 6 7 8 9 10 C B D A C A A D A B 11 12 13 14 15 16 17 18 19 20 C C A A B B D A C A 21 22 23 24 25 26 27 28 29 30 A A D B D A A A A D 31 32 33 34 35 36 37 38 39 40 D C D B D D B A D B

2 Deductions U (units) – 1M for each occurrence capped at 1 M per question SF (significant figures) & P (presentation) – 1M for each occurrence capped at 1M per paper PAPER 2 SECTION A 1(ai) Distance = Area under v-t graph 5 = 1 2 (∆𝑣 10) (8.4 + 8.4 + ∆𝑣) ∆𝑣2 + 16.8∆𝑣 − 100 = 0 v = 4.66 m/s v = 8.4 + 4.66 = 13 m/s C1 A1 1(aii) Acceleration = gradient of v-t graph 10 = 13 − 8.4 𝑡 t = 0.46 s C1 A1 1(b) reasonable shape suitable scale correctly plotted 1st and last points at (0,8.4) and (0.90,0) with non-vertical line at 0.46 s M1 A1 A1 0.20 0.40 0.60 0.80 10 -10

3 2(a) all correct any 3 correct A2 A1 2(bi) Taking moments about point A, sum of anti-clockwise moments = sum of clockwise moments TB x 4.0 = 4.0 x 10 x 1.0 +25.0 x 10 x 2.0 + 60.0 x 10 x 3.6 TB = 675N C1 A1 2(bii) Sum of forces acting on scaffold = 0 N Upward forces = downward forces TA + TB = 4.0g + 25.0g + 60.0g TA = 89 x 10 - 675 = 215 N C1 A1 2(c) As the painter walks towards point A, the tension in rope A will increase and the tension in rope B will decrease. By considering the moment of the forces about point A, as x decreases, the sum of clockwise moment decreases. (Since the beam is in equilibrium), tension in rope B must also decrease. Hence the tension in rope A increases, as total downward force remains the constant. B1 B1 B1 3(a) height difference = 32 - 15 = 17 cm Pgas = 76 + 17 = 93 cmHg Pgas = gh = 13600 x 0.93 x 10 = 126000 Pa M1 A1

4 3(b) The number of gas molecules per unit volume is equal both at the bung and at the walls. Hence, the frequency of collision of the gas molecules per unit area against the walls and the bung are equal, exerting equal force per unit area, which is the same gas pressure. B1 B1 B1 3(c) Use a liquid of smaller density. For any gas pressure increase, the height increase will be more because the density is lesser, according to the formula P = gh. B1 B1 4(a) During the change in state, energy is being transferred to the internal potential store of the molecules. Since there is no increase in energy in the internal kinetic store, temperature remains constant. B1 4(bi) Energy transferred from flame = Energy transferred to soup (Pt) x 30% = mlv (300) (3.0) (3600) (0.30) = m (2.26 x 106) m = 0.430 kg C1 A1 4(bii) When high flame is used, large amount of soup will be vaporized in 3 hours. Hence a large amount of energy will be wasted in preparing the soup. B1 4(ci) • Meat is a poor conductor of heat. If it is too thick, its interior might not be fully cooked. • To increase the size of contact area between the meat and the soup. Either one of the above. B1 4(cii) The layer of oil helps to reduce the rate of evaporation from the soup. This reduces the energy transferred from the soup to the surroundings. B1 4(ciii) Let n be no. of slices of meat Energy transferred from soup = Energy transferred to meat slices C1

5 (1.0) (4200) (97 – 82) = (0.020) (3500) (82 – 27) n n = 16.4 Maximum no. of slices is 16. A1 4(d) Energy will be transferred from the soup to mixian which will reduce temperature of the soup significantly. This might result in the meat not able to reach 82 oC subsequently. B1 5(a) F F 1 cm o P (a) (a) (c) Ray through optical centre drawn correctly and ray through focal point drawn correctly. Arrows included for light rays. Dotted lines used for construction. B1 B1 5(b) Magnification = 2 / 1 = 2 A1 5(c) Ray P extends backwards from lens to base of image. Light ray from lens forms a straight line to the base of image. B1

6 6(a) 6(b) At least 4 charges At least 4 lines (originate from centre) B1 B1 6(c) Negative charges nearer to sphere Attractive force stronger than repulsive force B1 B1 6(d) Aluminum drawn displaced to the left Negative charges move from aluminum to sphere, resulting in net positive charge of both objects. B1 B1 6(d) Copper is a non-magnetic material whereas steel is a magnetic material. Steel will magnetise and interfere with the function of the solenoid.is this excess? Can delete? B1 7(a) total resistance = 20 k current = 12 / 20 mA or potential divider formula p.d. = [12 / 20] × 12 = 7.2 V C1 A1 7(b) parallel resistance = 3 k total resistance 8 + 3 = 11 k current = 12 / 11 × 103 = 1.09 × 10–3 or 1.1 × 10–3 A C1 A1 7(ci) LDR resistance decreases total resistance (of circuit) is less hence current increases M1 A1 7(cii) resistance across XY is less less proportion of 12 V across XY hence p.d. is less M1 A1 insulating stands conducting sphere + + + + + insulating stands conducting sphere (d)

7 8(a) Inserting the iron core gives rise to rate of change of flux within the solenoid and e.m.f is induced in solenoid induced e.m.f. opposes applied d.c. emf so current smaller/acts to reduce current M1 A1 8(b) 𝑉𝑆 𝑉𝑃 = 𝑁𝑆 𝑁𝑃 𝑉𝑆 240 = 350 5600 VS = 15 V IS = P / VS = 90 / 15 = 6.0 A C1 C1 A1 9(a) Number of half-lives elapsed = 45/15 = 3 A1 9(b) Initial activity of 6.0 cm3 blood = ((5 x 2) x 2) x 2 = 40 Let V be volume of blood 𝑉 6 = 32 × 103 40 V = 4800 cm3 C1 A1 9(c) The half-life of 15 hours is short and hence the source will not remain inside the body for a long time. B1 10(ai) Resistance of the bulb is the ratio of the potential across the bulb to the current flowing through it. B1 10(aii) 𝑅 = 𝑉2 𝑃 𝑅 = 2402 60 R = 960  C1 A1 10(b) 𝑅 = 𝜌𝑙 𝐴

8 𝑅 = 𝜌𝑙 (𝜋 4⁄ )𝑑2 𝑑 = √ 𝜌𝑙 (𝜋 4⁄ )(𝑅) 𝑑 = √ (7.9 × 10−7)(0.14) (𝜋 4⁄ )(960) d = 0.0121 mm C1 A1 10(c) By varying the length and cross-sectional area of the wires so that the ratio 𝑙 𝐴⁄ remains constant. B1 10(d) Filament A. It has the lowest resistance per unit length. Hence power dissipated per unit length is the lowest and it will take a longer time to break. B1 B1

9 10(e) A B C resistance/  960 960 960 diameter/ mm 0.0129 0.0121 0.0113 length/ m 0.158 0.14 0.122 All values correct Three to five values correct A2 A1 10(f) Running cost = cost of electrical consumption + cost to make bulbs = ((60/1000) x 2000) x 0.172 + (2000/500) x 0.50 = $22.6 C1 A1

10 SECTION B 11(a) The position with greatest acceleration is Q. At point Q, there is no net change in magnetic flux of the coil, hence there will not be e.m.f. or opposing magnetic field produced by the induced current to slow down the magnet. When the magnet is just falling into (P) or out of the coil (R), it produces a change of magnetic flux within the coil. This induces a current in the coil which produces an opposing magnetic field to slow down the magnet according to Lenz’s Law. B1 B1 11(bi) As the magnet falls through the coil, it speeds up due to gravitational force, this results in a greater rate of change of magnetic flux. Based on Faraday’s Law, greater rate of change o