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HKDSE Design and Applied Technology Paper 1 Mock Paper | 10 Questions, 100 Marks
HKDSE Design and Applied Technology
HKDSE Design and Applied Technology mock: 10 questions, 100 marks, 105 minutes, covering cantilever bending moment and stress, gear ratio and belt speed, series and parallel circuits with an LED resistor, material density and cost, batch yield and process capability, solar and battery backup, drawing scale and CAD file size, pneumatic thrust and air use, ergonomic percentiles, and a smart lock battery and FMEA discussion, with marking notes.
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Paper details Level: HKDSE | Subject: Design and Applied Technology | Time: 1 hour 45 minutes | Marks: 100 | Questions: 10
Section A (Questions 1 to 5, 50 marks)
Answer ALL questions. In calculations show the formula, the substitution and the unit, and give answers to the accuracy asked; in discussion questions state a position and support it with reasons.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Question 1 (Structures and mechanics) A cantilever beam is 2.4 m long and carries a vertical load of 350 N at its free end. Its cross-section is rectangular, 60 mm wide and 120 mm deep (the load acts along the depth). (a) Calculate the bending moment at the fixed end (N m). (3 marks) (b) Calculate the second moment of area I (mm^4), I = b h^3 / 12. (3 marks) (c) Calculate the maximum bending stress at the fixed end, sigma = M y / I with y = h / 2. (2 marks) (d) If the cantilever is shortened to 1.6 m, by what percentage does the bending moment fall? (2 marks) |
(a) M = 350 x 2.4 = 840 N m. (2 marks working, 1 mark answer) (b) I = 60 x 120^3 / 12 = 8,640,000 mm^4. (2 marks + 1 mark) (c) sigma = 5.83 N/mm^2 (MPa). (1 mark + 1 mark) (d) New moment = 350 x 1.6 = 560 N m, a fall of 280 N m = 33.33%. (1 mark + 1 mark) |
| 2 | Question 2 (Mechanical transmission) Data: a spur gear pair has an 18-tooth driver at 1,450 rpm and a 54-tooth driven gear; the drive efficiency is 96% and the input torque is 2.5 N m. A 120 mm diameter pulley is mounted on the same input shaft. (a) Find the gear ratio and the driven speed (rpm, 2 decimal places). (4 marks) (b) Find the output torque (N m, 2 decimal places). (3 marks) (c) Find the belt linear speed (m/s, 2 decimal places). (3 marks) |
(a) Ratio = 54 / 18 = 3:1; driven speed = 1,450 / 3 = 483.33 rpm. (2 marks + 2 marks) (b) Output torque = 2.5 x 3 x 0.96 = 7.20 N m. (2 marks working, 1 mark answer) (c) v = pi D n / 60 = pi x 0.12 x 1,450 / 60 = 9.11 m/s. (2 marks + 1 mark) |
| 3 | Question 3 (Electronics and circuits) Data: a 12 V dc supply feeds R1 = 220 ohm in series with the parallel combination of R2 = 330 ohm and R3 = 680 ohm. (a) Find the parallel resistance and the total resistance (ohm, 2 decimal places). (3 marks) (b) Find the total current (mA, 2 decimal places) and the voltage across R1 (V). (3 marks) (c) Find the currents in R2 and R3 (mA, 2 decimal places each). (2 marks) (d) A 5 V supply drives an LED with a 2.0 V forward drop at 15 mA. Find the series resistor and its power dissipation. (2 marks) |
(a) Rp = 330 x 680 / (330 + 680) = 222.18 ohm; Rt = 220 + 222.18 = 442.18 ohm. (2 marks + 1 mark) (b) I = 12 / 442.18 = 27.14 mA; V_R1 = 5.97 V, so 6.03 V across the parallel pair. (1.5 marks each) (c) I2 = 18.27 mA, I3 = 8.87 mA. (1 mark each) (d) R = (5 – 2) / 0.015 = 200 ohm; P = 3 x 0.015 = 0.045 W, so a quarter-watt resistor or larger is needed. (1 mark each) |
| 4 | Question 4 (Materials and cost) Data: a sheet 1,000 mm x 2,000 mm x 2 mm. Aluminium alloy: density 2,700 kg/m^3 at 38 dollars per kg; mild steel: 7,850 kg/m^3 at 12 dollars per kg; PLA for 3D printing: 1,240 kg/m^3. (a) Find the sheet volume (m^3) and the masses of the aluminium and steel sheets (kg, 2 decimal places). (3 marks) (c) How many times heavier is the steel sheet than the aluminium one (2 decimal places)? (2 marks) (d) Find the material cost of each sheet and the percentage by which aluminium is dearer. (3 marks) (e) If the same part is 3D printed in PLA at 20% infill, find its mass (kg, 3 decimal places). (2 marks) |
(a) V = 0.002 x 1.0 x 2.0 = 0.0040 m^3; aluminium = 10.80 kg, steel = 31.40 kg. (1 mark each) (c) 31.40 / 10.80 = 2.91 times. (2 marks) (d) Aluminium 410.4 dollars, steel 376.8 dollars, so aluminium costs 33.6 dollars more, i.e. 8.9%. (1 mark each) (e) 1,240 x 0.0040 x 0.20 = 0.992 kg. (2 marks) |
| 5 | Question 5 (Manufacturing and quality) Data: a batch of 480 parts with a 92.5% yield. Material costs 18.5 dollars per part; machining takes 12 minutes at 180 dollars per machine hour. A bore is specified as 20.00 +/- 0.05 mm and the process standard deviation is 0.015 mm. (a) Find the number of good and defective parts, the unit cost and the total cost. (4 marks) (b) Find the cost per good part (dollars, 2 decimal places). (2 marks) (c) State the upper and lower limits and the tolerance band; if a bore measures 20.07 mm, by how much does it exceed the upper limit? (2 marks) (d) Find the process capability Cp = tolerance band / (6 sigma) and judge whether it meets the usual 1.33 requirement. (2 marks) |
(a) Good = 480 x 92.5% = 444; defective = 36; unit cost = 18.5 + 12/60 x 180 = 54.50 dollars; total = 26,160.0 dollars. (1 mark each) (b) 26,160.0 / 444 = 58.92 dollars. (1 mark + 1 mark) (c) Upper 20.05 mm, lower 19.95 mm, band 0.10 mm; 20.07 mm is 0.02 mm over the limit and must be rejected or reworked. (1 mark each) (d) Cp = 0.10 / (6 x 0.015) = 1.111, below 1.33, so the process does not meet the usual requirement and variation must be reduced or control tightened. (1 mark + 1 mark) |
Section B (Questions 6 to 10, 50 marks)
Answer ALL questions. In calculations show the formula, the substitution and the unit, and give answers to the accuracy asked; in discussion questions state a position and support it with reasons.
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | Question 6 (Energy and sustainable design) Data: an off-grid monitoring station has a 12 V, 2.5 A load running 6 hours a day; overall system efficiency is 75%. The solar panel is rated 100 W peak and the site gets 3.2 equivalent sun hours a day. The battery is 12 V, 100 Ah with a 50% depth of discharge. (a) Find the daily load energy and the generation needed after efficiency (Wh). (3 marks) (b) Find the actual daily output of the panel and judge whether it is sufficient. (3 marks) (c) Find the usable battery energy and the number of days it can support without sun (2 decimal places). (3 marks) (d) If the load rises to 3.0 A, what is the new backup duration (2 decimal places)? Suggest one design measure to cut night-time consumption. (3 marks) |
(a) Daily load = 12 x 2.5 x 6 = 180 Wh; with efficiency, 180 / 0.75 = 240 Wh is needed. (1.5 marks each) (b) Output = 100 x 3.2 = 320 Wh, above 240 Wh, so it is sufficient with 80 Wh to spare. (2 marks + 1 mark) (c) Usable = 12 x 100 x 0.5 = 600 Wh; duration = 600 / 180 = 3.33 days. (1.5 marks each) (d) New load = 216 Wh, so 600 / 216 = 2.78 days; measures include switching sensors and the communication module off with a presence sensor or timer, or using a low-power sleep mode. (2 marks + 1 mark) |
| 7 | Question 7 (Graphical communication and CAD) Data: an engineering drawing at 1:5 shows a length of 48 mm; the sheet is A3 (420 mm x 297 mm) with a 10 mm border on every side. The CAD STL file has 12,480 triangular facets, each recorded in 50 bytes. (a) Find the real length (mm). (2 marks) (b) Find the drawing area and the largest real extent it can represent at 1:5 (mm). (4 marks) (c) Find the STL file size (KB, 1 decimal place). (2 marks) (d) A line 0.25 mm thick on the drawing represents what real edge thickness? Why must an engineering drawing state its scale? (2 marks) |
(a) 48 x 5 = 240 mm. (2 marks) (b) Drawing area = 400 x 277 mm; real extent = 2000 mm by 1385 mm. (1 mark each) (c) 12,480 x 50 = 624,000 bytes = 609.4 KB. (1 mark each) (d) 0.25 x 5 = 1.25 mm; the scale fixes the linear relation between drawing and product so dimensions can be measured and checked, and the same part can be made accurately from sheets of different sizes. (1 mark + 1 mark) |
| 8 | Question 8 (Pneumatics and automation) Data: a double-acting cylinder with a 32 mm bore runs at a supply pressure of 6 bar gauge, stroke 150 mm, mechanical efficiency 85%, completing 12 cycles per minute. (a) Find the piston area (mm^2, 2 decimal places) and the theoretical thrust (N, 1 decimal place). (4 marks) (b) Find the actual thrust (N, 1 decimal place). (2 marks) (c) Find the volume per stroke (cm^3, 1 decimal place) and, taking free air as about 7 times the gauge volume, the air consumption per minute (L/min, 1 decimal place). (4 marks) |
(a) A = pi x 16^2 = 804.25 mm^2; 6 bar = 0.6 N/mm^2, so F = 0.6 x 804.25 = 482.5 N. (2 marks each) (b) Actual thrust = 482.5 x 0.85 = 410.2 N. (1 mark + 1 mark) (c) Volume per stroke = 804.25 x 150 = 120,637 mm^3, about 120.6 cm^3; free air about 0.844 L per stroke, and with 24 strokes a minute the consumption is 20.3 L/min. (2 marks each) |
| 9 | Question 9 (Ergonomics and design data) Data: an office desk is 720 mm high and the seat is 450 mm high. User stature is normally distributed with a mean of 1,700 mm and a standard deviation of 60 mm (z = -1.645 for P5, +1.645 for P95). (a) Find the difference between desk and seat height and explain its effect on writing posture. (3 marks) (b) Find the 5th and 95th percentile statures (mm, 1 decimal place) and their difference. (3 marks) (c) To cover the desk-height needs of users from P5 to P95, estimated as seat height plus that difference, explain how the desk and chair should be made adjustable. (3 marks) |
(a) 720 – 450 = 270 mm. (1 mark) At about 270 mm the forearms are close to horizontal and the shoulders relax, reducing neck and shoulder fatigue; too high a desk makes users shrug and bend the wrist, too low a desk forces a stooped back. (2 marks) (b) P5 = 1,700 – 1.645 x 60 = 1601.3 mm; P95 = 1798.7 mm; difference 197.4 mm. (1 mark each) (c) Both chair and desk should adjust: seat about 380-520 mm, desk about 650-800 mm, plus a footrest; set seat height so the feet rest flat, then set the desk about 270 mm below elbow height so users from P5 (1601.3 mm) to P95 (1798.7 mm) can keep a neutral posture. (1 mark each) |
| 10 | Question 10 (Product review and risk analysis, discussion) Data: a smart door lock runs on 4 AA cells (2,900 mAh each) with a 45 uA standby current; each unlock drives the motor at 180 mA for 1.2 s and the lock is used about 8 times a day. In the FMEA, motor seizure has severity 8, occurrence 4 and detection 3; adding a Hall sensor cuts occurrence to 2. (a) Find the theoretical standby life in days (whole days) and years (2 decimal places). (3 marks) (b) Find the daily energy used for unlocking (mAh, 2 decimal places) and comment on whether it is significant against standby drain. (2 marks) (c) Find the risk priority number before and after the sensor is added and state the improvement. (2 marks) (d) Take a stand on whether a smart lock should keep a mechanical key as backup and give two reasons. (2 marks) |
(a) 2,900 mAh = 2,900,000 uAh; 2,900,000 / 45 = 64,444 h = 2,685 days, about 7.36 years (self-discharge not counted). (1.5 marks each) (b) 8 x 180 x 1.2 / 3,600 = 0.48 mAh a day, about 0.4 times the 1.08 mAh standby drain, but negligible against 2,900 mAh, so standby current dominates battery life. (1 mark + 1 mark) (c) Before: 8 x 4 x 3 = 96; after: 8 x 2 x 3 = 48, a fall of 50% because faults are detected earlier. (1 mark each) (d) Sample stand: keep it. First, during a power cut or electronic fault the household can still get in, which is safety redundancy. Second, a mechanical cylinder gives a backup for escape and for elderly or child users and lets technicians intervene when the system is offline. The counter-argument is that a keyway adds picking risk, which must be offset by a high-security cylinder and drill plate. (1 mark stand, 0.5 mark each reason) |
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