Year 12 OCR Engineering: Unit Test Mock Exam Analysis | Year 12 OCR 工程：单元测试模拟卷解析

📚 Year 12 OCR Engineering: Unit Test Mock Exam Analysis | Year 12 OCR 工程：单元测试模拟卷解析

This article provides a detailed walkthrough of a typical Year 12 OCR Engineering unit test mock paper. By examining common question types and their solutions, students can reinforce key principles in mechanics, materials, electronics and design. Each section below breaks down a representative question, offers a step-by-step solution and highlights the underlying engineering concepts relevant to the OCR specification.

本文详细解析了一份典型的 Year 12 OCR 工程单元测试模拟卷。通过分析常见题型及其解答，学生可以巩固力学、材料、电子学和设计中的关键原理。以下每个部分分解一道代表性试题，提供逐步求解过程，并强调与 OCR 考试大纲相关的工程概念。

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1. Engineering Stress and Strain | 工程应力与应变

Question: A cylindrical steel specimen of original gauge length 50 mm and diameter 8 mm is subjected to an axial tensile load of 15 kN. The gauge length extends by 0.12 mm. Calculate the engineering stress, engineering strain, and the Young’s modulus of the steel.

题目：一根原始标距长度为 50 毫米、直径为 8 毫米的圆柱形钢试样承受 15 kN 的轴向拉伸载荷。标距长度延伸了 0.12 毫米。计算工程应力、工程应变以及钢的杨氏模量。

Step 1: Compute the original cross-sectional area A₀. For a circular section, A₀ = π × (d/2)² = π × (4 × 10⁻³ m)² = 5.027 × 10⁻⁵ m².

步骤1：计算原始横截面积 A₀。对于圆形截面，A₀ = π × (d/2)² = π × (4 × 10⁻³ m)² = 5.027 × 10⁻⁵ m²。

Step 2: Engineering stress σ = F / A₀. Using F = 15 × 10³ N, σ = (15 × 10³) / (5.027 × 10⁻⁵) ≈ 2.98 × 10⁸ Pa = 298 MPa.

步骤2：工程应力 σ = F / A₀。F = 15 × 10³ N，σ = (15 × 10³) / (5.027 × 10⁻⁵) ≈ 2.98 × 10⁸ Pa = 298 MPa。

Step 3: Engineering strain ε = ΔL / L₀ = 0.12 mm / 50 mm = 0.0024 (dimensionless, or 0.24%).

步骤3：工程应变 ε = ΔL / L₀ = 0.12 毫米 / 50 毫米 = 0.0024（无量纲，或者说 0.24%）。

Step 4: Young’s modulus E = σ / ε = (298 × 10⁶ Pa) / 0.0024 ≈ 1.24 × 10¹¹ Pa = 124 GPa. This value is typical for steel.

步骤4：杨氏模量 E = σ / ε = (298 × 10⁶ Pa) / 0.0024 ≈ 1.24 × 10¹¹ Pa = 124 GPa。该数值是钢的典型值。

Key insight: Engineering stress uses the original area, and strain is based on original length; E is the stiffness measure in the linear elastic region.

关键点：工程应力使用原始面积，应变基于原始长度；E 是线弹性区域的刚度度量。

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2. Static Equilibrium and Reaction Forces | 静力平衡与支反力

Question: A uniform beam of length 4 m and weight 200 N rests on two supports at its ends. A concentrated load of 500 N is applied 1.5 m from the left end. Determine the reaction forces at the left support (R_A) and right support (R_B).

题目：一根长 4 米、重 200 N 的均匀梁两端简支。在距离左端 1.5 米处施加一个 500 N 的集中载荷。求左支座 (R_A) 和右支座 (R_B) 的反力。

Step 1: Draw the free-body diagram. Forces: downward weight 200 N at midpoint 2 m from left, 500 N at 1.5 m from left; upward reactions R_A and R_B.

步骤1：绘制受力图。力：向下的自重 200 N 作用在距左端 2 米处的中点；500 N 作用在距左端 1.5 米处；向上的支反力 R_A 和 R_B。

Step 2: Apply vertical equilibrium: ΣF_y = 0 → R_A + R_B = 200 N + 500 N = 700 N.

步骤2：应用竖直方向平衡：ΣF_y = 0 → R_A + R_B = 200 N + 500 N = 700 N。

Step 3: Take moments about the left support to eliminate R_A. ΣM_A = 0: (R_B × 4 m) – (200 N × 2 m) – (500 N × 1.5 m) = 0.

步骤3：对左支座取矩消除 R_A。ΣM_A = 0：(R_B × 4 m) – (200 N × 2 m) – (500 N × 1.5 m) = 0。

Step 4: Solve for R_B: 4 R_B = 400 + 750 = 1150 → R_B = 287.5 N. Then R_A = 700 – 287.5 = 412.5 N.

步骤4：求解 R_B：4 R_B = 400 + 750 = 1150 → R_B = 287.5 N。然后 R_A = 700 – 287.5 = 412.5 N。

Check: Sum of moments about the right support yields consistent results – an essential verification habit.

检查：对右支座取矩得出了一致的结果——这是必不可少的验证习惯。

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3. Series and Parallel Resistor Networks | 串并联电阻网络

Question: Three resistors are arranged: R1 = 100 Ω in series with a parallel combination of R2 = 200 Ω and R3 = 300 Ω. A 12 V battery is connected across the whole network. Calculate the total equivalent resistance and the current supplied by the battery.

题目：三个电阻的连接方式为：R1 = 100 Ω 与 R2 = 200 Ω 和 R3 = 300 Ω 的并联组合串联。总电路连接着一个 12 V 的电池。计算总等效电阻和电池提供的电流。

Step 1: Find the equivalent resistance of the parallel pair R2 and R3: 1/R_para = 1/200 + 1/300 = 3/600 + 2/600 = 5/600 → R_para = 120 Ω.

步骤1：计算 R2 和 R3 并联的等效电阻：1/R_para = 1/200 + 1/300 = 3/600 + 2/600 = 5/600 → R_para = 120 Ω。

Step 2: Total resistance R_total = R1 + R_para = 100 Ω + 120 Ω = 220 Ω.

步骤2：总电阻 R_total = R1 + R_para = 100 Ω + 120 Ω = 220 Ω。

Step 3: Apply Ohm’s law: I = V / R_total = 12 V / 220 Ω ≈ 0.0545 A = 54.5 mA.

步骤3：应用欧姆定律：I = V / R_total = 12 V / 220 Ω ≈ 0.0545 A = 54.5 mA。

Step 4: Verify voltage division: Voltage across R1 is I × R1 = 5.45 V; across parallel block 6.55 V, consistent with total 12 V.

步骤4：验证电压分配：R1 两端的电压为 I × R1 = 5.45 V；并联块两端的电压为 6.55 V，与总电压 12 V 一致。

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4. Fluid Statics and Gauge Pressure | 流体静力学与表压

Question: An open water tank has a water column of height 8 m. Water density ρ = 1000 kg m⁻³. Determine the gauge pressure at the base of the tank and the absolute pressure if atmospheric pressure is 101.3 kPa.

题目：一个敞口水箱的水柱高度为 8 m。水的密度 ρ = 1000 kg m⁻³。求水箱底部的表压，以及当大气压为 101.3 kPa 时的绝对压力。

Step 1: Gauge pressure p_gauge = ρ × g × h = 1000 kg m⁻³ × 9.81 m s⁻² × 8 m = 78,480 Pa = 78.48 kPa.

步骤1：表压 p_gauge = ρ × g × h = 1000 kg m⁻³ × 9.81 m s⁻² × 8 m = 78,480 Pa = 78.48 kPa。

Step 2: Absolute pressure p_abs = p_atm + p_gauge = 101.3 kPa + 78.48 kPa = 179.78 kPa.

步骤2：绝对压力 p_abs = p_atm + p_gauge = 101.3 kPa + 78.48 kPa = 179.78 kPa。

Step 3: In engineering, gauge pressure is often specified to exclude ambient pressure; ensure correct conversion when using fluid systems.

步骤3：在工程中，通常指定表压以排除环境压力；在使用流体系统时确保正确换算。

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5. Linear Thermal Expansion | 线性热膨胀

Question: An aluminium rod is exactly 2.000 m long at 20 °C. Its coefficient of linear expansion α = 23 × 10⁻⁶ °C⁻¹. What is its length when heated to 120 °C?

题目：一根铝棒在 20 °C 时恰好长 2.000 m。其线性热膨胀系数 α = 23 × 10⁻⁶ °C⁻¹。加热到 120 °C 时，它的长度是多少？

Step 1: Temperature change ΔT = 120 °C – 20 °C = 100 °C.

步骤1：温度变化 ΔT = 120 °C – 20 °C = 100 °C。

Step 2: Length change ΔL = α × L₀ × ΔT = (23 × 10⁻⁶) × 2.000 m × 100 = 4.6 × 10⁻³ m = 4.6 mm.

步骤2：长度变化 ΔL = α × L₀ × ΔT = (23 × 10⁻⁶) × 2.000 m × 100 = 4.6 × 10⁻³ m = 4.6 mm。

Step 3: Final length L = L₀ + ΔL = 2.000 m + 0.0046 m = 2.0046 m (or 2004.6 mm).

步骤3：最终长度 L = L₀ + ΔL = 2.000 m + 0.0046 m = 2.0046 m（即 2004.6 mm）。

Note: Thermal expansion must be accommodated in design to avoid stress build-up, often using expansion joints.

注意：设计中必须考虑热膨胀以避免应力积累，通常使用伸缩缝。

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6. Engineering Drawings and Dimensional Tolerances | 工程制图与尺寸公差

Question: A shaft is specified as 25.00 ± 0.05 mm in diameter. Calculate the upper and lower limits of size, and the bilateral tolerance. Explain why a hole designated as 25.10 ± 0.02 mm would create a clearance fit with this shaft.

题目：一根轴的直径标注为 25.00 ± 0.05 mm。计算尺寸的上下极限以及双边公差。说明为何一个标注为 25.10 ± 0.02 mm 的光孔会与该轴形成间隙配合。

Step 1: Shaft upper limit = 25.00 + 0.05 = 25.05 mm; lower limit = 24.95 mm. The bilateral tolerance is ± 0.05 mm.

步骤1：轴的上限 = 25.00 + 0.05 = 25.05 mm；下限 = 24.95 mm。双边公差为 ± 0.05 mm。

Step 2: Hole upper limit = 25.10 + 0.02 = 25.12 mm; lower limit = 25.08 mm.

步骤2：光孔上限 = 25.10 + 0.02 = 25.12 mm；下限 = 25.08 mm。

Step 3: For a clearance fit, the minimum hole diameter (25.08 mm) must be larger than the maximum shaft diameter (25.05 mm). Here, 25.08 > 25.05, guaranteeing clearance.

步骤3：对于间隙配合，光孔的最小直径（25.08 mm）必须大于轴的最大直径（25.05 mm）。此处 25.08 > 25.05，确保存在间隙。

Step 4: Interpret tolerance symbols: Uppercase for holes, lowercase for shafts, and the fundamental deviation designator – a core skill in reading engineering drawings.

步骤4：解释公差符号：大写字母用于孔，小写字母用于轴，以及基本偏差代号——这是阅读工程图纸的核心技能。

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7. Power, Work and Efficiency in Mechanical Systems | 机械系统中的功率、功与效率

Question: An electric motor lifts a 250 kg mass vertically at a constant speed of 1.2 m s⁻¹. The motor input electrical power is 3.5 kW. Calculate the useful mechanical output power, the work done in 10 seconds, and the overall efficiency of the lifting system.

题目：一台电动机以 1.2 m s⁻¹ 的恒定速度垂直提升 250 kg 的重物。电动机输入电功率为 3.5 kW。计算有用机械输出功率、10 秒内所做的功以及提升系统的总效率。

Step 1: The tension in the cable equals the weight: F = m × g = 250 kg × 9.81 m s⁻² = 2452.5 N.

步骤1：钢丝绳中的张力等于重量：F = m × g = 250 kg × 9.81 m s⁻² = 2452.5 N。

Step 2: Mechanical output power P_out = F × v = 2452.5 N × 1.2 m s⁻¹ = 2943 W = 2.943 kW.

步骤2：机械输出功率 P_out = F × v = 2452.5 N × 1.2 m s⁻¹ = 2943 W = 2.943 kW。

Step 3: Work done in 10 s = P_out × t = 2943 W × 10 s = 29,430 J (or 29.43 kJ).

步骤3：10 秒内所做的功 = P_out × t = 2943 W × 10 s = 29,430 J（即 29.43 kJ）。

Step 4: Efficiency η = (P_out / P_in) × 100% = (2.943 kW / 3.5 kW) × 100% ≈ 84.1%. Losses are due to friction, heat, and motor losses.

步骤4：效率 η = (P_out / P_in) × 100% = (2.943 kW / 3.5 kW) × 100% ≈ 84.1%。损失是由摩擦、发热和电动机损失造成的。

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8. Material Indices and Lightweight Beam Design | 材料指数与轻量化梁设计

Question: A cantilever beam must be stiff and as light as possible. The performance index to minimise mass for a given stiffness is E^(1/2)/ρ, where E is Young’s modulus and ρ is density. Using the table below, identify the best material among aluminium alloy (E=70 GPa, ρ=2.7 Mg m⁻³), titanium alloy (E=110 GPa, ρ=4.4 Mg m⁻³) and CFRP (E=130 GPa, ρ=1.6 Mg m⁻³).

题目：一根悬臂梁要求刚度大且重量尽可能轻。对于给定刚度，最小化质量的性能指数为 E^(1/2)/ρ，其中 E 是杨氏模量，ρ 是密度。根据下表，在铝合金（E=70 GPa，ρ=2.7 Mg m⁻³）、钛合金（E=110 GPa，ρ=4.4 Mg m⁻³）和 CFRP（E=130 GPa，ρ=1.6 Mg m⁻³）中找出最佳材料。

Material 材料	E (GPa)	ρ (Mg m⁻³)
Aluminium alloy 铝合金	70	2.7
Titanium alloy 钛合金	110	4.4
CFRP 碳纤维聚合物	130	1.6

Step 1: Convert units consistently: E in GPa → Pa? For the index E^(1/2)/ρ, using GPa and Mg m⁻³ is permissible as long as consistent. We compute (√E)/ρ.

步骤1：单位统一换算：E 由 GPa 转换为 Pa？对于指数 E^(1/2)/ρ，只要保持一致，使用 GPa 和 Mg m⁻³ 是允许的。我们计算 (√E)/ρ。

Step 2: Aluminium: √70 ≈ 8.367; index = 8.367 / 2.7 ≈ 3.10.

步骤2：铝合金：√70 ≈ 8.367；指数 = 8.367 / 2.7 ≈ 3.10。

Step 3: Titanium: √110 ≈ 10.488; index = 10.488 / 4.4 ≈ 2.38.

步骤3：钛合金：√110 ≈ 10.488；指数 = 10.488 / 4.4 ≈ 2.38。

Step 4: CFRP: √130 ≈ 11.402; index = 11.402 / 1.6 ≈ 7.13. CFRP offers the highest index, making it the best choice for a light, stiff beam.

步骤4：CFRP：√130 ≈ 11.402；指数 = 11.402 / 1.6 ≈ 7.13。CFRP 提供了最高的指数，使其成为轻质高刚度梁的最佳选择。

Note: Consider composite material anisotropy and cost; the index is a starting point for material selection in engineering design.

注意：要考虑复合材料的各向异性和成本；该指数是工程设计中材料选择的起点。

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9. Truss Analysis – Method of Joints | 桁架分析 – 节点法

Question: A simple triangular truss has a vertical load of 2 kN applied downward at the apex joint C. Joints A and B are pin supports 4 m apart horizontally, and the apex C is 3 m above the midpoint of AB. Determine the axial forces in members AC and BC, stating whether tension or compression.

题目：一个简单的三角形桁架在顶点 C 处承受竖直向下的 2 kN 载荷。支座 A 和 B 为铰支座，水平间距 4 m，顶点 C 位于 AB 中点正上方 3 m 处。求杆件 AC 和 BC 的轴力，并说明是拉力还是压力。

Step 1: Geometry – AC and BC are identical. Length L_AC = √(2² + 3²) = √13 ≈ 3.606 m. The angle between member AC and horizontal is θ where tan θ = 3/2 → θ ≈ 56.31°.

步骤1：几何尺寸——AC 和 BC 相同。长度 L_AC = √(2² + 3²) = √13 ≈ 3.606 m。杆 AC 与水平线的夹角为 θ，其中 tan θ = 3/2 → θ ≈ 56.31°。

Step 2: Apply joint equilibrium at C. Two unknown member forces F_AC and F_BC. Assume both in tension (pulling away from C). Vertical equilibrium: -2 kN + F_AC sin θ + F_BC sin θ = 0.

步骤2：在节点 C 处应用平衡条件。两个未知杆力 F_AC 和 F_BC。假设两者均为拉力（从 C 向外拉）。竖直方向平衡：-2 kN + F_AC sin θ + F_BC sin θ = 0。

Step 3: By symmetry, F_AC = F_BC = F. Then 2 F sin θ = 2 kN → F = 1 kN / sin 56.31° ≈ 1 kN / 0.832 = 1.202 kN. The positive sign indicates tension. Thus AC and BC are in tension at 1.20 kN each.

步骤3：由对称性，F_AC = F_BC = F。那么 2 F sin θ = 2 kN → F = 1 kN / sin 56.31° ≈ 1 kN / 0.832 = 1.202 kN。正号表示拉力。因此 AC 和 BC 均受拉，各为 1.20 kN。

Step 4: Check horizontal equilibrium at C: F_AC cos θ equals F_BC cos θ but in opposite directions → net zero, confirming solution.

步骤4：检查节点 C 的水平平衡：F_AC cos θ 等于 F_BC cos θ 但方向相反 → 合力为零，验证了解答。

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10. Kirchhoff’s Voltage Law in a Two-Loop Circuit | 双回路电路中的基尔霍夫电压定律

Question: A circuit has two voltage sources and three resistors. Loop 1 contains V1 = 9 V, R1 = 10 Ω, R2 = 20 Ω; Loop 2 contains V2 = 6 V, R2 (shared), R3 = 30 Ω. Using Kirchhoff’s Voltage Law, write the loop equations and solve for the mesh currents I₁ and I₂.

题目：一个电路包含两个电压源和三个电阻。回路 1 包含 V1 = 9 V，R1 = 10 Ω，R2 = 20 Ω；回路 2 包含 V2 = 6 V，R2（共用），R3 = 30 Ω。使用基尔霍夫电压定律，写出回路方程并求解网孔电流 I₁ 和 I₂。

Step 1: Assign mesh currents I₁ clockwise in left loop, I₂ clockwise in right loop. Through R2, current = I₁ – I₂ (if both clock, the net is difference).

步骤1：指定网孔电流 I₁ 在左回路顺时针，I₂ 在右回路顺时针。通过 R2 的电流 = I₁ – I₂（如果都为顺时针，净电流为差值）。

Step 2: KVL for Loop 1: -V1 + R1·I₁ + R2·(I₁ – I₂) = 0 → 10 I₁ + 20(I₁ – I₂) = 9 → 30 I₁ – 20 I₂ = 9. (Equation 1)

步骤2：回路 1 的 KVL：-V1 + R1·I₁ + R2·(I₁ – I₂) = 0 → 10 I₁ + 20(I₁ – I₂) = 9 → 30 I₁ – 20 I₂ = 9。（方程 1）

Step 3: KVL for Loop 2: -V2 + R3·I₂ + R2·(I₂ – I₁) = 0 → 30 I₂ + 20(I₂ – I₁) = 6 → -20 I₁ + 50 I₂ = 6. (Equation 2)

步骤3：回路 2 的 KVL：-V2 + R3·I₂ + R2·(I₂ – I₁) = 0 → 30 I₂ + 20(I₂ – I₁) = 6 → -20 I₁ + 50 I₂ = 6。（方程 2）

Step 4: Solve simultaneous equations. Multiply Eq1 by 5 and Eq2 by 2: 150 I₁ – 100 I₂ = 45; -40 I₁ + 100 I₂ = 12. Add: 110 I₁ = 57 → I₁ ≈ 0.518 A. Substitute: 30×0.518 – 20 I₂ = 9 → 15.54 – 20 I₂ = 9 → I₂ = 0.327 A. Positive values mean assumed clockwise directions are correct.

步骤4：解联立方程。将方程 1 乘以 5，方程 2 乘以 2：150 I₁ – 100 I₂ = 45；-40 I₁ + 100 I₂ = 12。相加：110 I₁ = 57 → I₁ ≈ 0.518 A。代入：30×0.518 – 20 I₂ = 9

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