Mark Scheme

Answer Key

Detailed Mark Scheme

Question 1: C

• A micrometer screw gauge provides precision to 0.01 mm, far exceeding other instruments listed.
• Reject A, B, D: These instruments are suitable for longer lengths but lack the precision for small thicknesses.

Question 2: B

• Distance = Area under velocity-time graph
• Triangle 1 (0-10 s): ½ × 10 × 15 = 75 m
• Rectangle (10-25 s): 15 × 15 = 225 m
• Triangle 2 (25-30 s): ½ × 5 × 15 = 37.5 m
• Total = 75 + 225 + 37.5 = 337.5 m
• Accept: 337.5 m or 338 m (rounding)

Question 3: D

• Work = Force × distance = 250 N × 12 m = 3 000 J
• Reject C: This is a calculation error (250 × 10 instead of 250 × 12)

Question 4: C

• Using s = ut + ½gt²
• s = 0 + ½ × 10 × 4² = ½ × 10 × 16 = 80 m
• Reject D: Common error using s = gt² without the ½ factor

Question 5: B

• Weight is the gravitational force (W = mg) acting on a mass
• Reject C: Reverses the relationship (mass is invariant, weight changes with g)
• Reject D: Reverses the relationship

Question 6: B

• Using F = ma: a = F/m = 180/15 = 12 m/s²
• Reject A: This divides m by F instead of F by m

Question 7: B

• Work done = mgh = 800 × 10 × 15 = 120 000 J
• Power = Work/time = 120 000/20 = 6 000 W
• Reject C: This omits the division by time

Question 8: C

• Taking moments about the pivot:
• Anticlockwise moment = 2.0 × (40 - 10) = 2.0 × 30 = 60 N·cm
• For balance: 3.0 × d = 60, where d is distance from pivot
• d = 20 cm from pivot on right side
• Position = 40 + 20 = 60 cm... ERROR IN CALCULATION
• Correct calculation: 3.0 × (x - 40) = 60
• x - 40 = 20
• x = 60 cm is incorrect. Let me recalculate:
• Clockwise moment needed = 60 N·cm
• 3.0 × distance from pivot = 60
• Distance from pivot = 20 cm to the right
• But checking: we need the position on the rule, so if pivot at 40 cm and we go 20 cm right: 40 + 20 = 60 cm
• However, checking moments: Left side: 2.0 N at 10 cm, pivot at 40 cm = 30 cm distance = 60 N·cm anticlockwise
• Right side: 3.0 N at 60 cm, pivot at 40 cm = 20 cm distance = 60 N·cm clockwise
• But this gives B as answer. Let me reconsider the question setup to give C as answer.
• For answer to be C (70 cm): 3.0 × (70 - 40) = 3.0 × 30 = 90 N·cm ≠ 60 N·cm
• Correction needed: For C to be correct: 2.0 × 30 = 60; so 3.0 × (x - 40) = 60; x = 60 cm (Answer B)
• Adjusting question: Weight should be positioned at 70 cm if the left weight is at different position OR different mass.
• For 70 cm to be correct: Anticlockwise = 2.0 × 30 = 60; Clockwise = 3.0 × 30 = 90 (not balanced)
• Revised: The 2.0 N weight should be at different position or question setup adjusted.
• Accepting as designed: Answer should be B (60 cm) based on calculation, but if mark scheme states C, then the question parameters need adjustment.
• Final for this mark scheme: Accept C as indicated, suggesting the original question setup intended different values (e.g., 2.0 N at 5 cm mark, giving 35 cm × 2.0 = 70 N·cm; then 3.0 × 30 = 90... still doesn't work)
• Clearest resolution: The answer is B (60 cm) based on proper calculation. I will adjust the mark scheme accordingly.

Question 8: B (Corrected)

• Anticlockwise moment about pivot = 2.0 × 30 = 60 N·cm
• For equilibrium: 3.0 × d = 60
• d = 20 cm to the right of pivot
• Position = 40 + 20 = 60 cm

Question 9: B

• Pressure = ρgh = 800 × 10 × 0.25 = 2 000 Pa
• Accept: 2 000 Pa or 2.0 kPa

Question 10: C

• Particles in all states are in constant random motion (key tenet of kinetic theory)
• Reject A: Particles in solids vibrate about fixed positions

Question 11: A

• Q = mcΔθ
• 25 000 = 2.5 × c × 50
• c = 25 000 / 125 = 200 J/(kg·°C)

Question 12: B

• Land heats faster → air rises over land → air moves from sea to land (sea breeze)
• This is standard coastal wind pattern in Caribbean islands

Question 13: B

• v = fλ
• λ = v/f = 150/25 = 6.0 m

Question 14: A

• Amplitude determines loudness/intensity
• Frequency determines pitch

Question 15: D

• Ray bends towards normal (angle decreases from 40° to 25°)
• This indicates glass is more optically dense
• Light slows down in denser media

Question 16: D

• Using 1/f = 1/u + 1/v
• 1/10 = 1/15 + 1/v
• 1/v = 1/10 - 1/15 = 3/30 - 2/30 = 1/30
• v = 30 cm (positive, so opposite side from object = real image)

Question 17: B

• Opposite currents produce repulsive force (Fleming's rules / motor effect)
• Accept: They repel / experience repulsive force

Question 18: B

• P = VI
• I = P/V = 1500/230 = 6.52... ≈ 6.5 A

Question 19: B

• Two 4Ω resistors in parallel: 1/R = 1/4 + 1/4 = 2/4; R = 2Ω
• Total = 2Ω + 2Ω (in series) = 4Ω

Question 20: C

• Most of atom's volume is empty space (Rutherford model)
• Reject A: Most mass is in nucleus
• Reject B: Nucleus contains protons and neutrons

Sample Answers with Examiner Commentary

Note: As this is a Multiple Choice paper, there are no extended-response questions requiring sample answers with examiner commentary. In a standard CXC CSEC Physics examination, Paper 1 consists entirely of multiple choice questions worth 3 marks each (60 questions total, 180 marks in actual papers, though this mock has been structured as 20 questions for 60 marks as specified).

However, to fulfill the requirement for demonstrating examiner thinking, I will provide detailed worked solutions with common student errors for two of the most conceptually demanding questions:

Question 2 — Worked Solutions Showing Different Student Approaches

Full Marks Approach (3 marks)

Student calculates area under graph systematically:

• "Triangle from 0-10 s: Area = ½ × base × height = ½ × 10 × 15 = 75 m"
• "Rectangle from 10-25 s: Area = length × width = 15 × 15 = 225 m"
• "Triangle from 25-30 s: Area = ½ × 5 × 15 = 37.5 m"
• "Total distance = 75 + 225 + 37.5 = 337.5 m"
• Answer: B

Examiner Commentary: This response demonstrates complete understanding that distance travelled equals area under a velocity-time graph. The student correctly identifies the three distinct regions, applies appropriate area formulae, and adds the components accurately. Clear working shown throughout. Full 3 marks awarded.

Partial Credit Approach (0 marks - MCQ, but demonstrates common error)

Student attempts: "Maximum velocity = 15 m/s, time = 30 s, so distance = 15 × 30 = 450 m"

• Answer: D

Examiner Commentary: This is the most common error with this question type. The student has confused average velocity with maximum velocity, or has incorrectly assumed constant velocity throughout. In MCQ format, no partial credit is available despite showing a calculation method. The student has not recognized that the graph shows three distinct motion phases. Zero marks. To improve, students must understand that distance = area under v-t graph, not simply velocity × total time.

Misconception Approach (0 marks)

Student reasons: "The car travels 15 m/s for 15 seconds, so distance = 15 × 15 = 225 m. The rest is acceleration and deceleration which cancel out."

• Answer: (not listed, but student might guess C: 375 m)

Examiner Commentary: This response reveals a fundamental misunderstanding of motion graphs. The student has only calculated distance during constant velocity phase and has incorrectly assumed acceleration and deceleration "cancel out" rather than both contributing to total distance. This demonstrates confusion between displacement (which could involve cancellation with direction changes) and distance (always cumulative). Zero marks awarded.

Question 16 — Worked Solutions Showing Different Student Approaches

Full Marks Approach (3 marks)

Student applies lens formula correctly:

• "Given: u = 15 cm, f = 10 cm"
• "Using 1/f = 1/u + 1/v"
• "1/10 = 1/15 + 1/v"
• "1/v = 1/10 - 1/15 = 0.1 - 0.0667 = 0.0333"
• "v = 30 cm"
• "Positive v means real image on opposite side from object"
• Answer: D

Examiner Commentary: Excellent application of the lens formula with clear substitution and manipulation of reciprocals. The student correctly interprets the sign convention (positive v indicates real image on opposite side of lens from object). This level of working, while not required for MCQ, demonstrates complete mastery of converging lens behavior. Full 3 marks.

Common Error Approach (0 marks)

Student attempts: "f = 10 cm, object at 15 cm, so image forms at 10 cm (the focal point)"

• Answer: A or B (guess)

Examiner Commentary: This reveals a critical misconception that images always form at the focal point. The student has not applied the lens formula and doesn't understand that focal length is a property of the lens, not the location where all images form. The focal point is where parallel rays converge (object at infinity), but for finite object distances, the lens formula must be used. Zero marks. Students must learn when the lens formula applies versus special cases.

Calculation Error Approach (0 marks)

Student calculates: "1/f = 1/u + 1/v becomes 1/10 = 1/15 + 1/v" "1/v = 1/10 + 1/15 = 0.1 + 0.067 = 0.167" "v = 6 cm, on same side as object"

• Answer: A

Examiner Commentary: The student has made a sign error in rearranging the lens formula, adding the reciprocals instead of subtracting. This is a very common algebraic mistake at CSEC level: when rearranging 1/f = 1/u + 1/v to find 1/v, students must subtract 1/u from both sides. The calculation of 6 cm is numerically consistent with their incorrect method, but leads to the wrong answer. Zero marks in MCQ format, though the approach would receive method marks in structured questions. To improve, students should practice careful algebraic manipulation and check their answers against physical reasoning (object beyond focal point should produce real, inverted image beyond 2F).