Secondary 3 Combined Science Practice Paper 3

TuitionGoWhere Practice Paper - Combined Science Secondary 3 - ANSWER KEY

Version: 3 of 5
Total Marks: 60 marks

Section A: Multiple Choice (15 marks)

Section B: Structured Questions (25 marks)

16. (a) GPE = $mgh$ (1) $= 400 \times 10 \times 25$ (0.5) $= 100\,000$ J or $1.0 \times 10^5$ J (0.5) [2 marks]

(b) Using conservation of energy: $mgh_A = \frac{1}{2}mv_B^2 + mgh_B$ (1) $400 \times 10 \times 25 = \frac{1}{2} \times 400 \times v_B^2 + 400 \times 10 \times 15$ (0.5) $100\,000 = 200v_B^2 + 60\,000$ (0.5) $40\,000 = 200v_B^2$ (0.5) $v_B^2 = 200$, so $v_B = \sqrt{200} = 14.1$ m/s (0.5) [3 marks; accept 14 m/s]

Alternative using energy from A to B: $mg\Delta h = \frac{1}{2}mv_B^2$ where $\Delta h = 10$ m. This gives $v_B = \sqrt{2 \times 10 \times 10} = 14.1$ m/s directly.

(c) Energy is lost to friction/air resistance (0.5); some GPE becomes thermal energy rather than kinetic energy (0.5). [1 mark]

(d) At the top of the loop, the car has downward centripetal acceleration requiring a centripetal force (0.5). The resultant of normal reaction R (downward) and weight mg (downward) provides this: $R + mg = \frac{mv^2}{r}$ (0.5). If speed is sufficient, $R > 0$ so track pushes down on car; the car pushes up on track (0.5). Minimum speed at top: $v_{min} = \sqrt{gr}$; if exceeded, car stays on track (0.5). [2 marks]

Expected visual from P2-Q16-fig1: Track profile with loop at C, height labels. Student must identify 10 m drop from A to B for part (b).

17. (a) Slip rings maintain continuous electrical contact with the rotating coil (0.5) while keeping the same coil connection to the external circuit (0.5). This produces alternating e.m.f./current because the coil connection doesn't reverse (0.5) — the current reverses naturally as the coil sides swap position in the field, but the external circuit sees the polarity flip each half-turn (0.5). Split-ring commutator would make it a d.c. motor. [2 marks]

(b) Area of coil: $A = 0.12 \times 0.08 = 9.6 \times 10^{-3}$ m² (0.5) Angular speed: $\omega = 2\pi f = 2\pi \times 50 = 314$ rad/s (0.5) $E_{max} = NBA\omega = 200 \times 0.15 \times 9.6 \times 10^{-3} \times 314$ (1) $= 200 \times 0.15 \times 3.014 = 90.4$ V ≈ 90 V (1) [3 marks]

Teaching note: Check: $200 \times 0.15 = 30$; $30 \times 0.0096 = 0.288$; $0.288 \times 314.16 = 90.5$ V. Accept 90 V or 91 V depending on rounding.

(c) Sinusoidal curve, alternating positive and negative peaks (1). Two complete cycles in 0.04 s (period T = 1/50 = 0.02 s) (0.5). Peaks at ±90 V (0.5). [2 marks]

Expected visual from P2-Q17-fig2: Sine wave with positive peak, negative peak, zero crossings at 0, 0.01, 0.02, 0.03, 0.04 s.

18. (a) Plotting marks [3]:

• Correct axes with labels and units (1)
• All 7 points correctly plotted (1) — allow ±1 small square tolerance
• Smooth curve (not line segments) through points, showing linear then kink (1)

(b) Linear region: first 5 points (0–8 N, 0–32 mm) (0.5). Gradient = $\frac{\Delta F}{\Delta x} = \frac{8}{32 \times 10^{-3}} = 250$ N/m (1). Spring constant $k = 250$ N/m (0.5). [2 marks]

Teaching note: Must convert mm to m for SI units. $k = F/x$ gives N/m.

(c) This conclusion is not correct (0.5). Hooke's law (F ∝ x) is only obeyed in the linear region where the graph is straight (0.5). Beyond ~32 mm / 8 N, the graph curves (non-linear), indicating the limit of proportionality has been exceeded (0.5). The spring may be permanently deformed if exceeded too far (0.5). [2 marks]

19. (a) Voltmeter in parallel: measures potential difference (voltage) across component without altering current through it — needs high resistance so minimal current diverts through it (1). Ammeter in series: measures current flowing through circuit — needs low resistance so it doesn't reduce current (1). [2 marks]

(b) Total resistance = $500 + 1500 = 2000$ Ω (0.5) Circuit current $I = V/R = 6.0/2000 = 0.003$ A = 3 mA (0.5) Voltmeter reading = $I \times R_{LDR} = 0.003 \times 500$ (0.5) $= 1.5$ V (0.5) [3 marks]

Alternative using potential divider: $V_{LDR} = \frac{500}{2000} \times 6.0 = 1.5$ V.

(c) As light intensity decreases, LDR resistance increases (0.5). In the potential divider, LDR gets larger share of total voltage (0.5). Therefore voltmeter reading increases (0.5). At very low light, $R_{LDR} \gg R$, so $V_{LDR} \approx 6$ V (0.5). [2 marks]

Expected visual from P2-Q19-fig1: Potential divider with LDR at bottom. As LDR R↑, its fraction of 6V increases.

Section C: Data Analysis and Evaluation (20 marks)

20. (a) Plotting marks [3]:

• Correct axes: Time / min and Temperature / °C (1)
• All 9 points correctly plotted (±1 mm tolerance) (1)
• Smooth curve of best fit, showing rapid initial drop then flattening (1)

(b) The curve is steep initially then gradually flattens to approach 25 °C asymptotically (1). This is because rate of heat loss depends on temperature difference between water and surroundings (Newton's law of cooling) (0.5). Initially large ΔT → rapid heat loss → steep gradient (0.5). As water cools, ΔT decreases → slower heat loss → gentler gradient (0.5). Finally water reaches thermal equilibrium at room temperature (ΔT = 0) so no net heat flow (0.5). [3 marks]

(c) After 5 min: reading from graph should be approximately 48 °C (accept 47–49 °C) (1). Working shown on graph: vertical line at t = 5 min to curve, horizontal to temperature axis (must be indicated). [1 mark]

(d) Sketch: Shallower curve starting at same point, same asymptote at 25 °C, but cooling more slowly throughout (1). Below original curve at all points t > 0 (0.5). Explanation: Insulation reduces heat loss by conduction/convection/radiation (0.5). Same mechanism (Newton's law) but lower rate constant (0.5). Same final temperature (room temperature) reached more slowly (0.5). [3 marks]

Expected visual from P3-Q20-fig1: Original curve descending from (0, 80) to asymptote at 25°C. Insulated curve should be below original curve, same start and end.

21. (a) Plotting marks [3]:

• Correct axes with labels and units (1)
• All 6 points correctly plotted (1)
• Best-fit straight line through origin (or near origin) (1)

(b) From graph: gradient = $R/L$ ≈ 7.5 Ω/m (accept 7.3–7.7) (1) Cross-sectional area: $d = 0.38$ mm = $0.38 \times 10^{-3}$ m (0.5) $A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3})^2}{4}$ (0.5) $= \frac{\pi \times 1.444 \times 10^{-7}}{4} = 1.13 \times 10^{-7}$ m² (0.5) Resistivity $\rho = \frac{RA}{L} = (\text{gradient}) \times A = 7.5 \times 1.13 \times 10^{-7}$ (0.5) $= 8.5 \times 10^{-7}$ Ωm (or nm scale: ≈ $1.1 \times 10^{-6}$ Ωm with gradient ≈ 7.5) (0.5) [4 marks]

Note: Using $R = 7.5$ Ω at $L = 1.0$ m: $\rho = 7.5 \times 1.13 \times 10^{-7} = 8.5 \times 10^{-7}$ Ωm. This is close to nichrome ($1.1 \times 10^{-6}$ Ωm) or constantan ($4.9 \times 10^{-7}$ Ωm). Accept reasonable values.

(c) Source 1: Contact resistance at clips/clamps — reduce by soldering connections or using clean, tight clips (1.5) Source 2: Temperature change during measurement heating wire — reduce by using brief measurement pulses or waiting for thermal equilibrium (1.5) Source 3: Measuring length with ruler (parallax, limited precision) — use tape measure with finer divisions or measure multiple times (1.5) [Any two sources with improvements, 3 marks]

Other valid answers: Wire thickness non-uniform → measure diameter at multiple points; instrument zero errors → calibrate ammeter/voltmeter.

(d) Method [4]:

1. Use same material and length of wire, but different cross-sectional areas (different gauges/swaging) (0.5)
2. Control variables: Keep temperature constant, use same current/voltage measurement method (0.5)
3. Measure diameter with micrometer screw gauge at multiple points along each wire; calculate $A = \pi d^2/4$ (0.5)
4. Measure resistance using voltmeter-ammeter method or ohmmeter (0.5)
5. Tabulate $A$ and $R$; plot $R$ vs $1/A$ or verify $R \times A$ = constant for same $L$ (0.5)
6. Repeat for multiple wire thicknesses to establish pattern; check if $R \propto 1/A$ (straight line through origin on R vs 1/A graph) (0.5)
7. Conclusion: If straight line through origin obtained, $R \propto 1/A$ is verified (0.5); if curved, relationship is more complex (0.5)

Key validity: Must control material, length, temperature. Must measure A directly (not assume from manufacturer). Multiple data points needed.

22. (a) Using Fleming's left-hand rule: First finger (Field): N to S (horizontal, say left to right) (0.5). Second finger (Current): direction in rod (given by circuit) (0.5). Thumb (Motion/Force): perpendicular to both, vertically upward or downward depending on current direction (0.5). Based on standard diagram with conventional current and typical N-S arrangement, force is vertically upwards (or into/out of page if field vertical) — must be consistent with diagram (0.5). [2 marks]

Teaching note: Need clear diagram interpretation. With rod horizontal between N (above) and S (below), current into page, field down, force is horizontal along rails.

(b) $F = BIL$ (maximum, when perpendicular) (0.5) $= 0.25 \times 5.0 \times 0.08$ (0.5) $= 0.10$ N (0.5) [Accept 0.1 N]

Direction: Perpendicular to both B and I by Fleming's left-hand rule. [2 marks]

(c) Constant velocity means zero net force (Newton's first law) (0.5). The motor effect force is balanced by friction/drag/resistive force from rails/air/at contacts (0.5). This opposing force increases with speed until it equals the driving force; once balanced, no acceleration occurs (0.5). Similar to terminal velocity in fluids (0.5). [2 marks]

Expected visual from P3-Q22-fig1: Rod on rails between magnets, complete circuit. Student should identify B field direction (N to S) and current direction to apply FLHR.

TOTAL MARKS: 60

Section A: 15 marks
Section B: 25 marks
Section C: 20 marks
Total: 60 marks ✓