📚 Year 11 SQA Engineering: Unit Test Mock Paper Analysis | Year 11 SQA 工程：单元测试模拟卷解析
Welcome to this in-depth analysis of a Year 11 SQA Engineering unit test mock paper. This article will guide you through typical exam-style questions, offering step-by-step worked solutions and revision notes. Whether you are preparing for National 5 Engineering Science or consolidating your knowledge, these explanations will sharpen your problem-solving skills and deepen your understanding of core concepts.
欢迎阅读这篇针对 Year 11 SQA 工程单元测试模拟试卷的深度解析。本文将带领你剖析典型的考试题型，提供逐步的解题过程和复习要点。无论你是在备战 National 5 工程科学，还是想巩固知识，这些解释都能提升你的解题能力，加深你对核心概念的理解。
1. Direct Stress Calculation | 直接应力计算
A common question asks: A steel rod with a diameter of 10 mm is subjected to a tensile force of 5 kN. Determine the direct tensile stress in the rod.
一个常见问题是：一根直径为 10 mm 的钢杆承受 5 kN 的拉力。计算杆中的直接拉伸应力。
A = 5000 N / 7.854 × 10⁻⁵ m² = 63.66 × 10⁶ Pa, or 63.7 MPa. Always ensure consistent units — convert millimetres to metres and kilonewtons to newtons before substituting.
解答：首先求横截面积 A = πd²/4。已知 d = 10 mm = 0.01 m，A = π × (0.01)² / 4 = 7.854 × 10⁻⁵ m²。然后应力 σ = F / A = 5000 N / 7.854 × 10⁻⁵ m² = 63.66 × 10⁶ Pa，即 63.7 MPa。务必保持单位一致——代入公式前将毫米转换为米，千牛转换为牛顿。
σ = F / A
2. Strain and Young's Modulus | 应变与杨氏模量
In a tensile test, the original gauge length L₀ of a steel specimen is 200 mm. Under a load it extends by 0.3 mm. Calculate the direct strain ε.
在拉伸试验中，钢试样的原始标距 L₀ 为 200 mm。在负载下它伸长了 0.3 mm。计算直接应变 ε。
Strain is dimensionless: ε = ΔL / L₀ = 0.3 mm / 200 mm = 0.0015 (or 0.15%). If Young' Pa / 0.0015 = 200 × 10⁹ Pa = 200 GPa. This matches the typical value for structural steel.
应变是无量纲的：ε = ΔL / L₀ = 0.3 mm / 200 mm = 0.0015（或 0.15%）。如果钢的杨氏模量 E 定义为应力/应变，且已知应力为 300 MPa，则 E = σ / ε = 300 × 10⁶ Pa / 0.0015 = 200 × 10⁹ Pa = 200 GPa。这与结构钢的典型值相符。
ε = ΔL / L₀ E = σ / ε
3. Ohm's Law and Resistivity | 欧姆定律与电阻率
A wire carries a current of 3 A when connected to a 12 V supply. Calculate its resistance, and then determine the resistivity of the wire material if its length is 2.5 m and cross-sectional area is 0.5 mm².
一根导线连接到 12 V 电源时承载 3 A 电流。计算其电阻，然后如果导线长度为 2.5 m、截面积为 0.5 mm²，求导线材料的电阻率。
Using Ohm' n ρ = 4 Ω × 0.5 × 10⁻⁶ m² / 2.5 m = 8 × 10⁻⁷ Ω·m. This is close to the value for nichrome.
利用欧姆定律，R = V / I = 12 V / 3 A = 4 Ω。电阻率 ρ 由 R = ρL / A 求得，故 ρ = R × A / L。将面积换算为 m²：0.5 mm² = 0.5 × 10⁻⁶ m²。则 ρ = 4 Ω × 0.5 × 10⁻⁶ m² / 2.5 m = 8 × 10⁻⁷ Ω·m。这接近于镍铬合金的数值。
R = V / I ρ = R A / L
4. Series and Parallel Resistors | 串联与并联电阻
Two resistors, 10 Ω and 20 Ω, are connected first in series and then in parallel across a 6 V battery. For each arrangement, calculate the total resistance and the current drawn from the battery.
两个电阻器，10 Ω 和 20 Ω，先串联然后并联在 6 V 电池两端。针对每种接法，计算总电阻和从电池吸取的电流。
Series: Rtotal = 10 Ω + 20 Ω = 30 Ω. Current I = V / R = 6 V / 30 Ω = 0.2 A. Parallel: 1/Rtotal = 1/10 + 1/20 = 3/20, so Rtotal .9 A. Always remember that parallel branches reduce total resistance and increase current.
串联：Rtotal = 10 Ω + 20 Ω = 30 Ω。电流 I = V / R = 6 V / 30 Ω = 0.2 A。并联：1/Rtotal = 1/10 + 1/20 = 3/20，故 Rtotal = 20/3 ≈ 6.67 Ω。电流 I = 6 V / 6.67 Ω ≈ 0.9 A。始终记住并联支路会降低总电阻并增大电流。
5. Pneumatics: Force and Pressure | 气动系统：力与压强
A pneumatic cylinder has a piston diameter of 50 mm. If the supplied air pressure is 0.6 MPa, calculate the theoretical force output during extension. Ignore friction.
一个气动缸的活塞直径为 50 mm。如果供气压为 0.6 MPa，计算伸出时的理论输出力。忽略摩擦。
l applications, the return stroke force is slightly lower due to the rod area on one side.
面积 A = πd²/4 = π × (0.05 m)² / 4 = 1.9635 × 10⁻³ m²。力 F = P × A = 0.6 × 10⁶ Pa × 1.9635 × 10⁻³ m² = 1178.1 N。在实际应用中，由于一侧存在活塞杆面积，回程力会略小。
F = P × A
6. Mechanical Advantage of Lever Systems | 杠杆系统的机械效益
In a first-class lever, the distance from the fulcrum to the effort is 0.4 m, and the distance from the fulcrum to the load is 0.1 m. Calculate the ideal mechanical advantage (MA) and the velocity ratio (VR).
在一类杠杆中，支点到施力点的距离为 0.4 m，支点到负载点的距离为 0.1 m。计算理想机械效益 (MA) 和速比 (VR)。
means the effort moves four times the distance of the load, and the force required is one-quarter of the load weight. In practice, friction reduces MA slightly below the VR.
对于无摩擦杠杆，MA = VR = 力臂 / 重臂 = 0.4 m / 0.1 m = 4。这意味着施力点移动距离是负载移动距离的四倍，所需力是负载重量的四分之一。实际中，摩擦会使 MA 略低于 VR。
7. Material Selection Using a Property Table | 利用属性表选择材料
Engineers often choose materials by comparing density, tensile strength, and cost. Study the table below and answer: which material offers the best strength-to-weight ratio for a lightweight beam?
工程师常通过比较密度、抗拉强度和成本来选择材料。研究下表并回答：哪种材料对轻量化梁具有最佳比强度？
| Material | Density (kg/m³) | Tensile Strength (MPa) | Strength/Density (kN·m/kg) |
|---|---|---|---|
| Mild Steel | 7850 | 400 | 51 |
| Aluminium Alloy | 2700 | 310 | 115 |
| Titanium Alloy | 4500 | 900 | 200 |
The specific strength (strength/density) indicates how much load a material can carry per unit mass. Titanium alloy gives 200 kN·m/kg, making it the best choice for weight-critical applications. Aluminium alloy also performs well and is cheaper, but titanium leads in strength-to-weight ratio.
比强度（强度/密度）表示材料单位质量所能承受的载荷。钛合金的比强度为 200 kN·m/kg，是重量敏感场合的最佳选择。铝合金也表现良好且更便宜，但钛合金在比强度上处于领先地位。
8. Energy and Power in Electrical Circuits | 电路中的能量与功率
An electric heater rated at 230 V draws 2.5 A. Find the power rating and the energy consumed if it runs for 4 hours. Express the energy in joules and kilowatt-hours.
一台额定 230 V 的电热器取用电流 2.5 A。求额定功率及它运行 4 小时所消耗的能量。能量用焦耳和千瓦时表示。
kilowatt-hours, E = 0.575 kW × 4 h = 2.3 kWh. This is the unit used on electricity bills.
功率 P = V × I = 230 V × 2.5 A = 575 W（0.575 kW）。能量 E = P × t = 575 W × (4 × 3600 s) = 8.28 × 10⁶ J。以千瓦时计，E = 0.575 kW × 4 h = 2.3 kWh。这正是电费账单上所用的单位。
P = V I E = P t
9. Interpreting Orthographic Drawings | 识读正投影图
You are given a front view and a side view of a simple bracket. The front view shows a rectangle of height 60 mm and width 40 mm with a circular hole of diameter 10 mm centred 20 mm from the top edge. Identify the dimension that must appear in the top view.
给出了一个简单支架的主视图和侧视图。主视图显示一个高 60 mm、宽 40 mm 的矩形，中心距顶边 20 mm 处有一个直径 10 mm 的圆孔。确定必须在俯视图中出现的尺寸。
The top view would show the depth of the bracket and the hole's position along the width. If the side view indicates a depth of 25 mm, the top view must include this 25 mm dimension and the 40 mm width. Always project views systematically to ensure full dimensioning.
俯视图将显示支架的深度以及孔沿宽度方向的位置。如果侧视图指示深度为 25 mm，则俯视图必须包含此 25 mm 尺寸和 40 mm 宽度。始终系统地投影视图以确保完整标注尺寸。
10. Cost Estimation and Bill of Materials | 成本估算与材料清单
ct requires two steel plates (each 0.5 m × 0.3 m × 5 mm) and four bolts. Steel plate costs £12 per kilogram (density 7850 kg/m³), and each bolt costs £0.50. Prepare a simple bill of materials and calculate the total material cost.
一个项目需要两块钢板（每块 0.5 m × 0.3 m × 5 mm）和四个螺栓。钢板每公斤成本 £12（密度 7850 kg/m³），每个螺栓 £0.50。编制一份简单的材料清单并计算总材料成本。
× 7.5 × 10⁻⁴ = 5.8875 kg. Total plate mass = 2 × 5.8875 = 11.775 kg. Plate cost = 11.775 × £12 = £141.30. Bolts cost = 4 × £0.50 = £2.00. Total = £143.30. Always add a contingency of 5-10% for waste and cuts.
每块钢板体积 = 0.5 × 0.3 × 0.005 = 7.5 × 10⁻⁴ m³。每块质量 = 7850 × 7.5 × 10⁻⁴ = 5.8875 kg。钢板总质量 = 2 × 5.8875 = 11.775 kg。钢板成本 = 11.775 × £12 = £141.30。螺栓成本 = 4 × £0.50 = £2.00。总计 = £143.30。始终为废料和切割增加 5-10% 的余量。
Published by TutorHao | Engineering Revision Series | aleveler.com
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