O Level Combined Science Practice Paper 2

TuitionGoWhere Exam Practice (AI) - Combined Science O-Level

Physical Sciences Practice Paper (Version 2 of 5) - Answer Key

Total Marks: 65

Section A: Energy, Forces and Motion (26 marks)

Q1 [2 marks]
Answer: Energy cannot be created or destroyed, only converted from one form to another (or total energy in a closed system is conserved).
Teaching note: This is the principle of conservation of energy. Students must mention both "not created/destroyed" and "converted/transformed" for full marks. Common mistake: stating only "energy cannot be created or destroyed" without conversion → 1 mark only.

Q2 [3 marks]
Step 1: Total height = 25 × 15 cm = 375 cm = 3.75 m. [1]
Step 2: Work done against gravity = Weight × height = 500 N × 3.75 m = 1875 J. [1]
Step 3: Power = Work / time = 1875 J / 12 s = 156.25 W ≈ 156 W. [1]
Answer: 156 W (or 156.25 W).
Teaching note: Convert cm to m before calculating. Using mass instead of weight is a trap; weight given directly so no need for mg.

Q3 [2 marks]
Answer: Diagram must show:

• Weight (W) downward from centre of block. [1]
• Normal reaction (N) upward from base, equal length to W. [1]
  Teaching note: Block at rest on horizontal table → forces balanced. No friction if no horizontal push. From image placeholder: rectangle on table line, downward arrow W, upward arrow N equal size.

Q4 [2 marks]
Answer: Weight (downward), Tension in string (upward). [1+1]
Teaching note: At rest → two forces only, equal and opposite. Do not include air resistance unless specified.

Q5 [1 mark]
Answer: Gravitational potential energy (G.P.E.) is maximum at A.
Teaching note: Highest point = max height = max G.P.E., min kinetic energy.

Q6 [2 marks]
Answer: G.P.E. transforms to kinetic energy (K.E.) as it descends. [1] At B, G.P.E. is min, K.E. is max. [1]
Teaching note: Energy conservation in pendulum; from A to B height decreases.

Q7 [3 marks]
Step 1: Acceleration a = (v - u)/t = (20 - 0)/10 = 2 m/s². [1]
Step 2: Resultant force F = ma = 1200 kg × 2 m/s² = 2400 N. [1]
Step 3: Direction: forward (in direction of motion). [1]
Answer: 2400 N forward.
Teaching note: Use Newton's 2nd law. Mass given, not weight.

Q8 [2 marks]
Answer: Due to inertia (Newton's first law), the passenger's body tends to continue in its state of motion (forward) when the bus stops. [1+1]
Teaching note: Inertia = resistance to change in motion. Bus stops, passenger not acted on by same stopping force immediately.

Section B: Thermal Physics and Waves (19 marks)

Q9 [1 mark]
Answer: B. Free electrons and vibrating particles transfer kinetic energy.
Teaching note: Conduction in metals via free electrons; A is wrong (no bulk movement), C is convection, D is radiation.

Q10 [1 mark]
Answer: Conduction transfers heat without bulk movement of material; convection involves bulk movement of fluid. (Any one valid difference.)
Teaching note: e.g., conduction in solids, convection needs fluid circulation.

Q11 [2 marks]
Step 1: From diagram, one full wave = 4 cm. [1]
Step 2: Wavelength λ = 4 cm. [1]
Answer: 4 cm.
Teaching note: Distance between two successive crests/troughs. Image shows 4 cm per wave.

Q12 [2 marks]
Answer: Speed decreases; direction bends towards normal. [1+1]
Teaching note: Light enters denser medium (glass) → slows, refracts toward normal.

Q13 [2 marks]
Answer: Travel at speed of light in vacuum (3×10⁸ m/s); transverse waves; can travel vacuum. (Any two correct.) [1+1]
Teaching note: All EM waves share speed in vacuum and transverse nature.

Q14 [3 marks]
Answer: Black surface is a better emitter and absorber of infrared radiation. [1] Cloth radiates heat away faster. [1] Shiny metal reflects radiation, reducing loss. [1]
Teaching note: Radiation depends on surface colour/texture; black > shiny for emission.

Section C: Electricity and Magnetism (20 marks)

Q15 [1 mark]
Answer: Current through a metallic conductor is directly proportional to potential difference across it, at constant temperature.
Teaching note: Must mention constant temp for full definition.

Q16 [2 marks]
Step 1: V = IR → I = V/R = 6 / 10 = 0.6 A. [1 for formula, 1 for answer]
Answer: 0.6 A.

Q17 [3 marks]
Step 1: R_total = 4 + 6 = 10 Ω. [1]
Step 2: I = V / R = 10 / 10 = 1.0 A. [1]
Step 3: State units: R = 10 Ω, I = 1 A. [1]
Answer: 10 Ω, 1 A.

Q18 [3 marks]
Step 1: 1/R_eff = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2. [1]
Step 2: R_eff = 2 Ω. [1]
Step 3: Units stated. [1]
Answer: 2 Ω.
Teaching note: Parallel formula required; from image R1=3, R2=6.

Q19 [2 marks]
Answer: Concentric circles around wire, arrow anticlockwise (viewed from top) for upward current. [1 for circles, 1 for direction]
Teaching note: Right-hand grip rule; from image placeholder wire with I up.

Q20 [4 marks]
Step 1: V_s / V_p = N_s / N_p → V_s = 12 × (200/100) = 24 V. [2]
Step 2: Since N_s > N_p, step-up transformer. [1]
Step 3: Units and conclusion. [1]
Answer: 24 V, step-up.
Teaching note: Transformer equation; turns ratio >1 → step-up.

End of Answer Key – Total 65 marks