📚 Year 11 Eduqas Engineering: Unit Test Mock Paper Walkthrough | Year 11 Eduqas 工程：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for Year 11 Eduqas Engineering. Each question is analysed with explanations of the correct answers, key concepts, and common pitfalls. Use this to reinforce your understanding and exam technique.
本文详细解析了一套 Year 11 Eduqas 工程单元测试模拟卷。每道题都分析正确答案、关键概念和常见错误，帮助巩固学习与应试技巧。
1. Question 1: Material Selection for a Bicycle Frame | 第1题：自行车车架材料选择
A bicycle frame needs to be lightweight, stiff, and resistant to fatigue. Which of the following materials is most suitable? A. Mild steel B. Aluminium alloy 6061 C. Cast iron D. Copper
自行车车架需要重量轻、刚性好且抗疲劳。下列哪种材料最合适？ A. 低碳钢 B. 6061 铝合金 C. 铸铁 D. 铜
Answer: B. Aluminium alloy 6061. This alloy offers an excellent strength-to-weight ratio, good stiffness, and high fatigue resistance, making it ideal for bicycle frames. Mild steel is heavier and less corrosion-resistant; cast iron is brittle and too heavy; copper is too soft and dense.
答案：B. 6061 铝合金。该合金具有出色的强度重量比、良好的刚性以及高抗疲劳性，非常适合自行车车架。低碳钢较重且耐腐蚀性较差；铸铁易碎且过重；铜太软且密度大。
Aluminium alloy 6061 also responds well to heat treatment and can be easily extruded or welded, which suits modern frame manufacturing processes.
6061 铝合金还能很好地响应热处理，并且易于挤压或焊接，适合现代车架制造工艺。
2. Question 2: Disadvantages of Sand Casting | 第2题：砂型铸造的缺点
Which of the following is a significant disadvantage of sand casting compared to die casting? A. High tooling cost B. Poor surface finish C. Inability to produce large parts D. Limited material choice
与压铸相比，以下哪项是砂型铸造的显著缺点？ A. 模具成本高 B. 表面光洁度差 C. 无法生产大型零件 D. 材料选择有限
Answer: B. Poor surface finish. Sand casting uses a sand mould that leaves a rough texture on the component surface, requiring additional machining or finishing. Die casting uses permanent metal moulds, producing smoother finishes.
答案：B. 表面光洁度差。砂型铸造使用砂模，会在零件表面留下粗糙纹理，需要额外的机加工或精加工。压铸使用永久性金属模具，能产生更光滑的表面。
Sand casting is actually low in tooling cost and can produce very large parts with a wide range of metals, so options A, C and D are not correct disadvantages.
砂型铸造实际上模具成本低，能生产大型零件且可使用多种金属，因此选项 A、C 和 D 不是正确的缺点。
3. Question 3: LED Series Resistor Calculation | 第3题：LED 串联电阻计算
An LED has a forward voltage of 2 V and requires a forward current of 20 mA. If the supply voltage is 9 V, what value of series resistor is required? A. 35 Ω B. 100 Ω C. 350 Ω D. 450 Ω
某 LED 的正向电压为 2 V，所需正向电流为 20 mA。若电源电压为 9 V，需要多大阻值的串联电阻？ A. 35 Ω B. 100 Ω C. 350 Ω D. 450 Ω
Answer: C. 350 Ω. Use Ohm’s Law: the resistor must drop the excess voltage (9 V − 2 V = 7 V) while passing 20 mA (0.02 A).
答案：C. 350 Ω。使用欧姆定律：电阻需要降掉多余的电压（9 V − 2 V = 7 V），同时通过 20 mA（0.02 A）电流。
R = (Vsupply − VLED) ÷ I = 7 V ÷ 0.02 A = 350 Ω
Always convert milliamps to amps before calculating. A 350 Ω resistor will limit the current to the correct value and protect the LED from burning out. The nearest standard resistor value is 330 Ω or 360 Ω, but in theory 350 Ω is exact.
计算前务必将毫安转换为安培。350 Ω 电阻将电流限制在正确值，保护 LED 不致烧毁。最接近的标准电阻值是 330 Ω 或 360 Ω，但理论上 350 Ω 是准确的。
4. Question 4: Lever Mechanics and Moments | 第4题：杠杆力学与力矩
A first-class lever has an effort of 50 N applied at 0.4 m from the fulcrum. The load is placed 0.1 m from the fulcrum on the opposite side. Assuming no friction, what load can be lifted? A. 12.5 N B. 100 N C. 200 N D. 500 N
一第一类杠杆在距支点 0.4 m 处施加 50 N 的力。负载置于支点另一侧 0.1 m 处。假设无摩擦，能举起多大负载？ A. 12.5 N B. 100 N C. 200 N D. 500 N
Answer: C. 200 N. The principle of moments states that for equilibrium, the clockwise moment equals the anticlockwise moment: effort × its distance from fulcrum = load × its distance from fulcrum.
答案：C. 200 N。力矩原理指出，平衡时顺时针力矩等于逆时针力矩：力 × 力到支点距离 = 负载 × 负载到支点距离。
50 N × 0.4 m = Load × 0.1 m → Load = (50 N × 0.4 m) ÷ 0.1 m = 20 N·m ÷ 0.1 m = 200 N
This demonstrates mechanical advantage: a small effort further from the fulcrum can lift a larger load closer to the fulcrum. Incorrect options typically arise from swapping distances or misapplying the formula.
这展示了机械优势：距支点较远的小力可以举起距支点较近的大负载。错误选项通常源于距离混淆或公式误用。
5. Question 5: Orthographic Projection – Identifying the Front View | 第5题：正交投影——识别正视图
An engineering drawing of a stepped block is shown in orthographic projection. Which of the following views represents the front elevation? (A) A rectangle with two hidden lines (B) An L-shaped outline (C) A rectangle with a smaller rectangle on top (D) A rectangle with a central hole
一张阶梯状块体的工程图以正交投影展示。下列哪个视图表示正视图？ (A) 带两条隐藏线的矩形 (B) L 形轮廓 (C) 下方大矩形上方小矩形 (D) 带中心孔的矩形
Answer: C. A rectangle with a smaller rectangle on top. The front view shows the object as seen from the front, indicating the overall height and width plus any steps. The stepped block typically appears as two stacked rectangles when viewed from the front.
答案：C. 下方大矩形上方小矩形。正视图显示从正面观察物体时的形状，指示总高度和宽度以及阶梯特征。阶梯块从正面看通常呈现为两个叠放的矩形。
Option A with hidden lines suggests a hole or internal feature viewed from the top or side. Option B (L-shape) might be a side view. Option D (hole) could be a top view. Understanding the alignment of views in first-angle or third-angle projection is essential for the exam.
选项 A 中的隐藏线表示从顶部或侧面观察到的孔或内部特征。选项 B（L 形）可能是侧视图。选项 D（孔）可能是俯视图。理解第一角或第三角投影中视图的对齐方式对考试至关重要。
6. Question 6: Tolerance in Engineering Drawings | 第6题：工程图中的公差
What is meant by ‘tolerance’ in engineering drawings? A. The allowed variation from a specified dimension B. The surface roughness of a finished part C. The maximum load a component can withstand D. The colour coding of materials
工程图中的“公差”指什么？ A. 规定尺寸的允许变动量 B. 成品零件的表面粗糙度 C. 部件能承受的最大载荷 D. 材料的颜色编码
Answer: A. The allowed variation from a specified dimension. Tolerance defines the upper and lower limits between which a part dimension can vary and still be considered acceptable. It accounts for inherent manufacturing inaccuracies.
答案：A. 规定尺寸的允许变动量。公差定义了零件尺寸可变化的上下限，在该范围内仍被视为合格。它考虑了固有的制造误差。
For example, a shaft diameter of 20 ±0.1 mm means the actual diameter can be anywhere between 19.9 mm and 20.1 mm. Tolerances are critical for interchangeable parts and quality assurance.
例如，轴径标注 20 ±0.1 mm 意味着实际直径可在 19.9 mm 至 20.1 mm 之间。公差对于互换性零件和质量保证至关重要。
7. Question 7: Sustainable Engineering Strategies | 第7题：可持续工程策略
Which of the following is NOT a strategy for sustainable engineering? A. Using recycled materials B. Designing for disassembly C. Increasing product weight D. Minimising packaging
以下哪项不是可持续工程策略？ A. 使用回收材料 B. 可拆解设计 C. 增加产品重量 D. 最小化包装
Answer: C. Increasing product weight. Sustainability aims to reduce material usage, energy consumption, and environmental impact. Increasing weight typically requires more material and energy, contradicting those goals.
答案：C. 增加产品重量。可持续性旨在减少材料使用、能源消耗和环境影响。增加重量通常需要更多材料和能源，与这些目标相悖。
Other common sustainable practices include life-cycle assessment, using renewable energy in manufacturing, and designing for recyclability. Eduqas often tests knowledge of the 6Rs (Reduce, Reuse, Recycle, Repair, Refuse, Rethink).
其他常见的可持续实践包括生命周期评估、在制造中使用可再生能源，以及为可回收性而设计。Eduqas 经常考察 6R 原则（减量、重用、回收、修复、拒绝、再思考）。
8. Question 8: Tensile Test and Young’s Modulus | 第8题：拉伸试验与杨氏模量
In a tensile test, which property is indicated by the gradient of the stress-strain curve in the elastic region? A. Toughness B. Hardness C. Stiffness (Young’s modulus) D. Ductility
在拉伸试验中，应力-应变曲线弹性区的斜率表示哪种特性？ A. 韧性 B. 硬度 C. 刚度（杨氏模量） D. 延展性
Answer: C. Stiffness (Young’s modulus). The elastic region follows Hooke’s Law, where stress is proportional to strain. The slope (stress ÷ strain) equals the Young’s modulus, a measure of material stiffness. A steeper slope means a stiffer material.
答案：C. 刚度（杨氏模量）。弹性区遵循胡克定律，应力与应变成正比。斜率（应力 ÷ 应变）等于杨氏模量，衡量材料的刚度。斜率越陡，材料越刚硬。
Toughness relates to the total area under the stress-strain curve, hardness to resistance to indentation, and ductility to the plastic deformation before fracture. It is important not to confuse these mechanical properties in the exam.
韧性涉及应力-应变曲线下的总面积，硬度指抗压痕能力，延展性指断裂前的塑性变形。考试中切勿混淆这些力学性能。
9. Question 9: PPE for Welding | 第9题：焊接的个人防护装备
When performing arc welding, which piece of personal protective equipment is essential to protect the eyes from ultraviolet and infrared radiation? A. Safety glasses with clear lenses B. A face shield with a grinding visor C. A welding helmet with an appropriate shade filter D. Cotton gloves
进行电弧焊接时，哪件个人防护装备对保护眼睛免受紫外线和红外线辐射至关重要？ A. 透明镜片安全眼镜 B. 带打磨面罩的头盔 C. 具有合适遮光度滤光片的焊接面罩 D. 棉手套
Answer: C. A welding helmet with an appropriate shade filter. Arc welding generates intense light, UV and IR radiation that can cause ‘arc eye’ or permanent retinal damage. A proper auto-darkening or fixed-shade helmet is required.
答案：C. 具有合适遮光度滤光片的焊接面罩。电弧焊产生强光、紫外线和红外线辐射，可导致“电光性眼炎”或永久性视网膜损伤。需要合适的自动变光或固定遮光度面罩。
Additional PPE includes flame-resistant overalls, welding gauntlets, and safety boots. Safety glasses alone are insufficient; they do not provide the required shade against arc brightness.
额外的个人防护装备包括阻燃工作服、焊接手套和安全靴。仅使用安全眼镜是不够的；它们无法提供抵御电弧亮度的必要遮光度。
10. Question 10: Open-Loop vs Closed-Loop Systems | 第10题：开环与闭环系统
A domestic central heating system uses a thermostat to maintain the room temperature at 21 °C. Is this system open-loop or closed-loop? A. Open-loop, because it uses a timer B. Closed-loop, because it has feedback from the thermostat C. Open-loop, because it does not use a microcontroller D. Closed-loop, because it operates on electricity
家用中央供暖系统使用恒温器将室温维持在 21 °C。该系统是开环还是闭环？ A. 开环，因为使用了定时器 B. 闭环，因为有恒温器的反馈 C. 开环，因为没有使用微控制器 D. 闭环，因为依靠电力运行
Answer: B. Closed-loop, because it has feedback from the thermostat. A closed-loop control system continuously monitors the output (room temperature) and compares it with the desired set point. The difference (error) is used to adjust the input to the boiler.
答案：B. 闭环，因为有恒温器的反馈。闭环控制系统持续监测输出（室温），并将其与期望的设定点比较。差值（误差）用于调节锅炉的输入。
An open-loop system, such as a simple timer-operated boiler, does not have feedback and cannot automatically correct deviations. The presence of a sensor feeding back information makes this a closed-loop system.
开环系统，例如简单的定时器控制锅炉，没有反馈，无法自动纠正偏差。传感器反馈信息的存在使该系统成为闭环系统。
Published by TutorHao | Engineering Revision Series | aleveler.com
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