📚 Year 11 CAIE Engineering: Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析
Welcome to this detailed walkthrough of a Year 11 CAIE Engineering unit test mock paper. This mock covers essential topics including materials, mechanics, electronics and control systems. Each question is followed by a clear explanation to help you understand the concepts and avoid common mistakes.
欢迎阅读这份 Year 11 CAIE 工程单元测试模拟卷解析。这份模拟卷涵盖了材料、力学、电子学和控制系统等重要主题。每道题都附有清晰的解释，帮助你理解概念并避免常见错误。
1. Question 1: Ferrous and Non-Ferrous Metals | 第1题：铁金属与非铁金属
Question: Which of the following is a ferrous metal?
题目： 以下哪种材料是铁金属？
A) Copper B) Aluminium C) Mild steel D) Brass
A) 铜 B) 铝 C) 低碳钢 D) 黄铜
Answer and Explanation: Ferrous metals contain iron. Mild steel is an alloy of iron and carbon, therefore it is a ferrous metal. Copper, aluminium and brass are non-ferrous.
答案与解析：铁金属含有铁。低碳钢是铁和碳的合金，因此属于铁金属。铜、铝和黄铜均为非铁金属。
2. Question 2: Stress-Strain Calculation | 第2题：应力-应变计算
Question: A steel rod with a cross-sectional area of 50 mm² experiences a tensile force of 10 kN. Calculate the stress in the rod.
题目： 一根横截面积为 50 mm² 的钢杆受到 10 kN 的拉力。计算杆中的应力。
Solution:
解答：
Stress = Force / Area
应力 = 力 / 面积
orce = 10 kN = 10000 N. Area = 50 mm². So Stress = 10000 N / 50 mm² = 200 N/mm² = 200 MPa.
力 = 10 kN = 10000 N。面积 = 50 mm²。所以 应力 = 10000 N ÷ 50 mm² = 200 N/mm² = 200 MPa。
Key point: Always check units. Using N and mm² gives MPa directly.
要点： 请注意单位换算。使用牛顿和平方毫米可以直接得到兆帕。
3. Question 3: Casting Process | 第3题：铸造工艺
Question: Describe the basic steps of sand casting.
题目： 描述砂型铸造的基本步骤。
Model Answer: 1) A pattern is placed in a moulding box and sand is packed around it. 2) The pattern is removed, leaving a cavity. 3) A runner and riser are cut to allow metal flow and gas escape. 4) Molten metal is poured into the cavity. 5) After solidification, the sand mould is broken and the casting is removed for cleaning and finishing.
参考答案： 1) 将模型放入砂箱，填充型砂。 2) 取出模型，形成型腔。 3) 开设浇道和冒口，便于金属流动和排气。 4) 将熔融金属浇入型腔。 5) 冷却凝固后，打破砂型，取出铸件进行清理和修整。
4. Question 4: Ohm’s Law Application | 第4题：欧姆定律应用
Question: A resistor has a resistance of 220 Ω and a current of 0.05 A flowing through it. Calculate the voltage across the resistor.
题目： 一个电阻的阻值为 220 Ω，通过它的电流为 0.05 A。计算电阻两端的电压。
Solution:
解答：
V = I × R
V = I × R
V = 0.05 A × 220 Ω = 11 V.
V = 0.05 A × 220 Ω = 11 V。
Therefore, the voltage is 11 volts.
因此，电压为 11 伏特。
5. Question 5: Lever Mechanical Advantage | 第5题：杠杆的机械利益
Question: A first-class lever has an effort arm of 0.8 m and a load arm of 0.2 m. Calculate its mechanical advantage (MA).
题目： 一个第一类杠杆，动力臂长 0.8 m，阻力臂长 0.2 m。计算其机械利益（MA）。
Solution: MA = effort arm / load arm = 0.8 / 0.2 = 4.
解答： 机械利益 = 动力臂 / 阻力臂 = 0.8 / 0.2 = 4。
This means the lever multiplies the effort force by 4.
这意味着该杠杆将输入力放大 4 倍。
6. Question 6: Gear Ratio Calculation | 第6题：齿轮比计算
Question: A driver gear has 20 teeth and the driven gear has 60 teeth. Find the gear ratio and the output speed if the input speed is 300 rpm.
题目： 主动轮有 20 齿，从动轮有 60 齿。求齿轮比，以及输入转速为 300 rpm 时的输出转速。
Solution:
解答：
Gear ratio = driven teeth / driver teeth = 60 / 20 = 3 : 1 (a speed reduction).
齿轮比 = 从动轮齿数 / 主动轮齿数 = 60 / 20 = 3:1（减速）。
Output speed = input speed / gear ratio = 300 / 3 = 100 rpm.
输出转速 = 输入转速 / 齿轮比 = 300 / 3 = 100 rpm。
7. Question 7: Welding vs. Soldering | 第7题：焊接与钎焊的区别
Question: State two key differences between welding and soldering.
题目： 说出焊接与钎焊的两个主要区别。
Answer:
答案：
1) Welding melts the base metals to fuse them, while soldering uses a filler metal with a lower melting point and does not melt the base metals.
1) 焊接会熔化母材使其熔合，而钎焊使用熔点较低的填充金属，不熔化母材。
2) Welded joints are generally stronger and used for structural applications; soldered joints are weaker and used for electrical connections or sealing.
2) 焊接接头通常更坚固，用于结构件；钎焊接头强度较低，用于电气连接或密封。
8. Question 8: Basic CNC G-Code | 第8题：基础数控G代码
Question: dly to position X=50, Y=30 before starting a cut. Write the G-code command and explain it.
题目： CNC铣床在开始切削前需要将刀具快速移动到 X=50, Y=30 的位置。写出G代码指令并解释。
Answer: G00 X50 Y30
答案： G00 X50 Y30
G00 is the rapid positioning command. The tool moves at maximum speed to the specified coordinates without cutting.
G00 是快速定位指令。刀具以最快速度移动到指定坐标而不进行切削。
9. Question 9: Material Hardness Testing | 第9题：材料硬度测试
Question: Explain how the Brinell hardness test is performed and what the hardness number indicates.
题目： 解释布氏硬度测试的方法以及硬度值的含义。
Answer: A hardened steel ball (typically 10 mm) is pressed into the material surface under a specified load (e.g., 3000 kg for steel). The diameter of the indentation is measured. The Brinell Hardness Number (BHN) is calculated as the load divided by the surface area of the indentation. A higher BHN indicates a harder material.
答案： 使用一个硬化钢球（通常直径 10 mm），在规定的载荷（例如钢件为 3000 kg）下压入材料表面。测量压痕直径。布氏硬度值（BHN）为载荷除以压痕表面积。BHN 越高，材料越硬。
10. Question 10: Open vs. Closed Loop Control | 第10题：开环与闭环控制
Question: Describe the difference between an open-loop and a closed-loop control system, giving an example of each.
题目： 描述开环控制系统与闭环控制系统的区别，并分别举例。
Answer:
答案：
An open-loop system has no feedback; the output does not affect the control action. Example: a simple electric heater without a thermostat – it runs for a set time regardless of room temperature. A closed-loop system uses feedback to adjust the output. Example: a room thermostat that switches the heater on/off based on the measured temperature to maintain a set point.
开环系统没有反馈，输出不影响控制动作。例如：一个没有恒温器的简单电暖器——它按照设定时间运行，不管室温如何。闭环系统利用反馈调节输出。例如：室内恒温器根据测得的温度开关加热器，以保持设定值。
Published by TutorHao | Engineering Revision Series | aleveler.com
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