Practice city guilds model questions vol1 engineering 2019 Past Questions

Question 1 View Details

An inclined line AB is projected onto the top view and front view of an engineering drawing. In the top view its projected length is 30 mm, and in the front view its projected length is 40 mm. Assuming the line is inclined to both the horizontal and vertical planes, determine the true length of AB.

A. 60

Correct B. 50

C. 45

D. 70

Correct Answer: B

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Question 2 View Details

On a rectangular steel plate 200 mm long and 150 mm wide, a hole is nominally positioned 50 mm from the left edge and 30 mm from the bottom edge. The hole has a nominal diameter of 10 mm with a size tolerance of \(\pm0.05\) mm, and its positional tolerance is a circular tolerance zone of diameter 0.3 mm (at MMC). Determine the maximum possible distance between the outer edge of the hole and the nearest plate edge.

A. 45.0

Correct B. 44.8

C. 46.2

D. 44.5

Correct Answer: B

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Question 3 View Details

An initially stress‑free steel rod of length 2.00 m and diameter 20 mm is subjected to a tensile load of 50 kN at 20 °C. The rod's Young's modulus is 200 GPa and its coefficient of linear expansion is \(12\times10^{-6}\,^{\circ}\mathrm{C}^{-1}\). The rod is then uniformly heated to 70 °C while the load remains unchanged. Calculate the total elongation of the rod after heating.

A. 3.12

Correct B. 2.79

C. 1.98

D. 2.45

Correct Answer: B

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Question 4 View Details

During a turning operation on a mild‑steel workpiece, the recommended cutting speed for the material is given by \(V = \frac{C}{\sqrt{d}}\) where \(C = 180\) (m·mm\(^{0.5}\)/min) and \(d\) is the depth of cut in mm. The tool‑life relationship follows Taylor's equation \(V\,T^{n}=K\) with \(n = 0.25\) and \(K = 1200\) (m·min\(^{0.25}\))/min. The operation requires a depth of cut \(d = 2\) mm and a feed \(f = 0.2\) mm/rev. The workpiece diameter is 100 mm.\n\n(a) Determine the cutting speed \(V\) that will give a tool life of at least 30 min.\n(b) Using that cutting speed, compute the material‑removal rate (MRR) in mm\(^3\)/min.

A. V = 132.5 m/min, MRR ≈ 5.4×10⁴ mm³/min

Correct B. V = 127.3 m/min, MRR ≈ 5.1×10⁴ mm³/min

C. V = 115.0 m/min, MRR ≈ 4.8×10⁴ mm³/min

D. V = 120.0 m/min, MRR ≈ 5.0×10⁴ mm³/min

Correct Answer: B

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Question 5 View Details

An electric motor supplies a torque of 150 N·m at a speed of 1500 rpm. Determine the mechanical power output of the motor in kilowatts (kW), correct to one decimal place.

A. 24.8

B. 22.5

C. 21.0

Correct D. 23.6

Correct Answer: D

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Question 6 View Details

A motor drives a belt‑drive system as follows: the motor shaft has radius \(0.10\) m and supplies torque \(T_{m}\). A small pulley of radius \(0.05\) m is mounted on the same shaft and drives, via a belt of efficiency \(90\%\), a larger pulley of radius \(0.15\) m. The larger pulley is fixed to the same shaft as a drum of radius \(0.20\) m that lifts a load through a rope. The drum has an efficiency of \(85\%\). If the load to be lifted is \(500\) N, what is the minimum motor torque \(T_{m}\) (in N·m, to one decimal place) required to raise the load at constant speed?

A. 52.0 N·m

Correct B. 43.6 N·m

C. 49.1 N·m

D. 38.2 N·m

Correct Answer: B

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Question 7 View Details

Material A has tensile strength \(400\) MPa and density \(2.7\) g·cm\(^{-3}\). Material B has tensile strength \(150\) MPa and density \(7.8\) g·cm\(^{-3}\). A component is to be made from a mixture of the two materials, assuming linear mixing for both strength and density. The component must have a tensile strength of at least \(250\) MPa and a density not exceeding \(4.0\) g·cm\(^{-3}\). What is the minimum fraction (by weight) of Material A required? Express your answer as a decimal rounded to two places.

A. 0.55

B. 0.80

C. 0.65

Correct D. 0.75

Correct Answer: D

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Question 8 View Details

A workshop operates 8 hours per day, 6 days a week. Three machines require preventive maintenance (PM):\n- Machine M1: 2 h per PM, interval 30 days;\n- Machine M2: 3 h per PM, interval 45 days;\n- Machine M3: 4 h per PM, interval 60 days.\nOnly one PM may be performed on any given operating day. What is the minimum number of whole weeks required to schedule all PM tasks while respecting the interval requirements?

A. 12 weeks

B. 8 weeks

Correct C. 9 weeks

D. 10 weeks

Correct Answer: C

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Question 9 View Details

In a workshop two hazards are identified. Hazard H1 occurs with probability \(0.02\) per year and costs an average of \(\text{₦}150{,}000\) per incident. Hazard H2 occurs with probability \(0.05\) per year and costs an average of \(\text{₦}80{,}000\) per incident. A safety measure S costs \(\text{₦}20{,}000\) per year and reduces the probability of H1 by 60 % and of H2 by 30 %. Should the workshop adopt S based on expected monetary loss? Show the net monetary effect per year.

A. Yes; adopting S increases yearly cost by ₦17,000

Correct B. No; adopting S increases yearly cost by ₦17,000

C. Yes; adopting S reduces yearly cost by ₦5,000

D. No; adopting S increases yearly cost by ₦5,000

Correct Answer: B

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Question 10 View Details

A turning operation follows the tool‑life equation \(T = \dfrac{C}{V^{n} f^{m}}\) where \(T\) is tool life in minutes, \(V\) is cutting speed (m·min\(^{-1}\)), \(f\) is feed (mm·rev\(^{-1}\)), \(C = 1.2\times10^{5}\), \(n = 0.25\) and \(m = 0.5\). The workpiece requires removal of \(0.12\) m\(^3\). The workpiece diameter is \(D = 80\) mm. Material removal rate is \(\text{MRR}=\pi D f N\) where \(N\) is spindle speed (rev·min\(^{-1}\)). Cutting speed relates to spindle speed by \(V = \dfrac{\pi D N}{1000}\). The machine can operate at most \(V_{max}=120\) m·min\(^{-1}\) and \(f_{max}=0.30\) mm·rev\(^{-1}\). Machining cost is \(\text{₦}0.5\) per minute and each tool costs \(\text{₦}150\). Total cost = machining time × 0.5 + (number of tools used) × 150. Assuming a tool is replaced only when its life expires, determine the values of \(V\) and \(f\) (to the nearest integer for \(V\) and two decimal places for \(f\)) that minimise the total cost, and give that minimum cost (rounded to the nearest naira).

Correct A. V = 120 m·min⁻¹, f = 0.30 mm·rev⁻¹; minimum cost = ₦1,817

B. V = 120 m·min⁻¹, f = 0.25 mm·rev⁻¹; minimum cost = ₦1,880

C. V = 115 m·min⁻¹, f = 0.28 mm·rev⁻¹; minimum cost = ₦1,950

D. V = 110 m·min⁻¹, f = 0.30 mm·rev⁻¹; minimum cost = ₦1,900

Correct Answer: A

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Question 11 View Details

A composite material is made from two constituents, A and B. The Young's modulus of A is (200\,\text{GPa}\) and that of B is unknown. The densities are (\rho_A=8000\,\text{kg/m}^3\) and (\rho_B=2700\,\text{kg/m}^3\). The overall modulus of the composite is (150\,\text{GPa}\) and the volume fraction of A is twice that of B. Determine (i) the modulus of B, (ii) the volume fractions of A and B, and (iii) the mass fractions of A and B in the composite.

Correct A. Modulus of B = 50 GPa; Volume fractions - A = 2/3, B = 1/3; Mass fractions - A = 160/187 ≈ 0.856, B = 27/187 ≈ 0.144

B. Modulus of B = 50 GPa; Volume fractions - A = 1/2, B = 1/2; Mass fractions - A = 160/187 ≈ 0.856, B = 27/187 ≈ 0.144

C. Modulus of B = 50 GPa; Volume fractions - A = 2/3, B = 1/3; Mass fractions - A = 150/187 ≈ 0.802, B = 37/187 ≈ 0.198

D. Modulus of B = 75 GPa; Volume fractions - A = 2/3, B = 1/3; Mass fractions - A = 160/187 ≈ 0.856, B = 27/187 ≈ 0.144

Correct Answer: A

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Question 12 View Details

A critical component has a constant exponential failure rate of \(\lambda = 0.0002\ \text{h}^{-1}\). A preventive maintenance action can be performed at any chosen interval \(T\) (hours) and costs \(\text{₦}500\) each time it is carried out. If the component fails before the scheduled preventive maintenance, a corrective maintenance is required costing \(\text{₦}5000\) plus a downtime cost of \(\text{₦}200\) per hour of lost production; the downtime for a failure is assumed to be 1 hour. Determine the preventive‑maintenance interval \(T\) that minimises the expected cost per operating hour. State the optimal policy and the corresponding minimum expected cost per hour.

A. Perform preventive maintenance every 500 h (T = 500 h); minimum expected cost ≈ ₦0.95 per hour

B. Perform preventive maintenance every 200 h (T = 200 h); minimum expected cost ≈ ₦0.80 per hour

C. Perform preventive maintenance every 1000 h (T = 1000 h); minimum expected cost ≈ ₦1.20 per hour

Correct D. Do not perform preventive maintenance (T → ∞); minimum expected cost ≈ ₦1.14 per hour

Correct Answer: D

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Question 13 View Details

In a workshop three hazards have been identified: (i) Electric shock - probability of occurrence per month = 0.02, severity = 8; (ii) Fire - probability per month = 0.01, severity = 9; (iii) Chemical spill - probability per month = 0.03, severity = 6. Risk rating for each hazard is defined as \(R = \text{probability}\times\text{severity}\).\n(a) Compute the individual risk ratings, the total risk score for the workshop, and identify which hazard requires the highest‑priority control.\n(b) If a control measure reduces the probability of fire by 60 %, recalculate the total risk score.

Correct A. Individual risks - electric: 0.16, fire: 0.09, chemical: 0.18; total risk = 0.43; highest‑priority hazard = chemical spill. After reducing fire probability by 60 %, new total risk = 0.376.

B. Individual risks - electric: 0.16, fire: 0.09, chemical: 0.18; total risk = 0.43; highest‑priority hazard = electric shock. After reducing fire probability by 60 %, new total risk = 0.376.

C. Individual risks - electric: 0.04, fire: 0.09, chemical: 0.18; total risk = 0.31; highest‑priority hazard = chemical spill. After reducing fire probability by 60 %, new total risk = 0.304.

D. Individual risks - electric: 0.16, fire: 0.09, chemical: 0.18; total risk = 0.43; highest‑priority hazard = chemical spill. After reducing fire probability by 60 %, new total risk = 0.34.

Correct Answer: A

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Question 14 View Details

A car of mass \(1500\,\text{kg}\) accelerates uniformly from rest to a speed of \(20\,\text{m s}^{-1}\) in \(10\,\text{s}\). Ignoring air resistance and rolling friction, calculate the average power output of the engine during this interval. Express your answer in kilowatts (kW).

A. 20 kW

B. 15 kW

C. 45 kW

Correct D. 30 kW

Correct Answer: D

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Question 15 View Details

Water flows steadily through a horizontal pipe that narrows from a diameter of \(0.20\,\text{m}\) to \(0.10\,\text{m}\). The volumetric flow rate is \(0.05\,\text{m}^3\!\,\text{s}^{-1}\). The pressure at the wider section is \(300\,\text{kPa}\). Assuming negligible elevation change and no friction losses, determine the pressure at the narrower section using the continuity equation and Bernoulli's principle. Give your answer in kilopascals (kPa) to the nearest kilopascal.

Correct A. 281 kPa

B. 260 kPa

C. 300 kPa

D. 290 kPa

Correct Answer: A

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Question 16 View Details

An AC source supplies (230\text{ V rms}\) at (50\text{ Hz}\) to a series circuit containing a resistor \(R\) and an ideal inductor \(L\). The magnitude of the total impedance is \(250\ \Omega\) and the real power consumed is \(150\text{ W}\). Determine the resistance \(R\) (in ohms, to the nearest whole ohm) and the inductance \(L\) (in henries, to two decimal places).

A. R ≈ 150 Ω, L ≈ 0.70 H

B. R ≈ 165 Ω, L ≈ 0.60 H

C. R ≈ 200 Ω, L ≈ 0.45 H

Correct D. R ≈ 177 Ω, L ≈ 0.56 H

Correct Answer: D

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Question 17 View Details

A metal rod is measured with a steel rule that has a reading uncertainty of \(\pm0.5\text{ mm}\). The measured length at \(30^{\circ}\text{C}\) is \(250.0\text{ mm}\). The coefficient of linear expansion of the metal is \(12\times10^{-6}\ \text{°C}^{-1}\). Determine the length of the rod at \(20^{\circ}\text{C}\) and the combined uncertainty, taking into account the temperature uncertainty of \(\pm0.5^{\circ}\text{C}\). Give the corrected length to three decimal places and the uncertainty to one decimal place (in mm).

A. 250.030 mm ± 0.6 mm

B. 249.950 mm ± 0.4 mm

Correct C. 249.970 mm ± 0.5 mm

D. 250.010 mm ± 0.5 mm

Correct Answer: C

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Question 18 View Details

A motor supplies \(5\text{ kW}\) at \(1500\text{ rpm}\) to a conveyor system through a gear train. The driver gear attached to the motor shaft has a diameter of \(0.20\text{ m}\) and drives a larger gear of diameter \(0.50\text{ m}\) on the conveyor shaft. The overall mechanical efficiency of the drive is \(85\%\). The belt that drives the conveyor wraps a pulley of radius \(0.10\text{ m}\) on the conveyor shaft. Assuming the belt tension is uniform, calculate the tension in the belt (in newtons, to the nearest whole newton).

A. 735 N

B. 540 N

C. 620 N

Correct D. 677 N

Correct Answer: D

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Question 19 View Details

In a DC circuit a 12 V ideal source is in series with a 4 Ω resistor \(R_{1}\). From the node after \(R_{1}\) a branch contains a 6 Ω resistor \(R_{2}\) in series with an ideal inductor (which behaves as a short circuit at steady state). A load resistor \(R_{L}\) is to be connected across the terminals of the branch containing \(R_{2}\). Determine the Thevenin equivalent voltage \(V_{\text{th}}\) and resistance \(R_{\text{th}}\) seen by \(R_{L}\). Then find the value of \(R_{L}\) that will draw maximum power and compute that maximum power (in watts, to one decimal place).

A. Vth = 8.0 V, Rth = 2.0 Ω, RL = 2.0 Ω, Pmax = 6.0 W

Correct B. Vth = 7.2 V, Rth = 2.4 Ω, RL = 2.4 Ω, Pmax = 5.4 W

C. Vth = 7.2 V, Rth = 3.6 Ω, RL = 3.6 Ω, Pmax = 3.6 W

D. Vth = 6.5 V, Rth = 3.0 Ω, RL = 3.0 Ω, Pmax = 4.5 W

Correct Answer: B

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Question 20 View Details

Water flows steadily through a horizontal pipe that narrows from a diameter of \(0.30\text{ m}\) to \(0.15\text{ m}\). The upstream pressure is \(250\text{ kPa}\) and the downstream pressure is \(200\text{ kPa}\). The volumetric flow rate is \(0.020\text{ m}^{3}\!\!/\text{s}\). Assuming water density \(\rho = 1000\text{ kg}\!\!/\text{m}^{3}\) and neglecting elevation changes, calculate (a) the head loss due to the contraction (in metres, to two decimal places) and (b) the power that must be supplied by a pump that adds \(30\text{ kPa}\) to the upstream pressure to maintain the flow (in kilowatts, to two decimal places).

A. Head loss = 5.10 m, Pump power = 0.60 kW

B. Head loss = 4.95 m, Pump power = 0.45 kW

Correct C. Head loss = 5.04 m, Pump power = 0.60 kW

D. Head loss = 5.04 m, Pump power = 0.90 kW

Correct Answer: C

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Question 21 View Details

A motor delivers 1500 W at 1500 rpm and drives a belt‑drive system. The driver pulley has a diameter of \(0.20\) m and the driven pulley a diameter of \(0.40\) m. The belt efficiency is 85 %. The load on the driven shaft requires a torque of \(120\) N·m. Determine (a) whether the motor can meet the load requirement, and (b) if it cannot, the maximum torque that can be transmitted at the driven shaft.

A. Yes; maximum transmissible torque ≈ 18.5 N·m

B. No; maximum transmissible torque ≈ 20.0 N·m

C. Yes; maximum transmissible torque ≈ 15.0 N·m

Correct D. No; maximum transmissible torque ≈ 16.2 N·m

Correct Answer: D

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Question 22 View Details

In a workshop a flammable‑liquid spill creates a vapor cloud that initially raises the concentration to \(2\)% (by volume). The room volume is \(120\) m³ and the ventilation provides \(12\) air changes per hour. (a) Assuming perfect mixing and first‑order decay, calculate the minimum time required for the concentration to fall to \(10\)% of the lower explosive limit (LEL), where the LEL for the liquid is \(1.5\)% by volume. (b) A Class B fire extinguisher contains \(2\) kg of CO₂; each kilogram can suppress \(0.8\) m³ of vapor. Determine whether this extinguisher is sufficient to extinguish the fire once the concentration reaches the safe level.

A. ≈15 minutes; the 2 kg CO₂ extinguisher is sufficient

B. ≈13 minutes; the 2 kg CO₂ extinguisher is insufficient

C. ≈9 minutes; the 2 kg CO₂ extinguisher is insufficient

Correct D. ≈13 minutes; the 2 kg CO₂ extinguisher is sufficient

Correct Answer: D

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Question 23 View Details

A cylindrical steel rod is to be machined to a nominal length of \(500\) mm (tolerance \(\pm0.2\) mm) and a nominal diameter of \(50\) mm (tolerance \(\pm t\) mm, where \(t\) is to be determined). The material's coefficient of linear thermal expansion is \(12\times10^{-6}\ \text{°C}^{-1}\). The rod will operate between \(20\)°C and \(80\)°C. It must fit into a clearance hole whose nominal diameter is \(50.10\) mm with a tolerance of \(\pm0.03\) mm. Assuming uniform expansion of both parts, find the maximum allowable machining tolerance \(t\) for the rod's diameter at \(20\)°C so that the fit is guaranteed at \(80\)°C.

A. ±0.050 mm

B. ±0.028 mm

C. ±0.040 mm

Correct D. ±0.034 mm

Correct Answer: D

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Question 24 View Details

A digital caliper has a systematic error of \(+0.02\) mm and a random error with standard deviation \(0.01\) mm. Nine successive measurements of a metal block give the following readings (mm): 100.12, 100.15, 100.09, 100.13, 100.11, 100.14, 100.10, 100.16, 100.08. (a) Determine the best estimate of the true length of the block and the 95 % confidence interval, taking both error sources into account. (b) The specification requires the length to be \(100.10\) mm \(\pm0.05\) mm. State whether the block meets the specification.

A. True length 100.10 mm; 95 % CI (100.060 mm, 100.140 mm); meets specification

B. True length 100.10 mm; 95 % CI (100.079 mm, 100.121 mm); does not meet specification

Correct C. True length 100.10 mm; 95 % CI (100.079 mm, 100.121 mm); meets specification

D. True length 100.12 mm; 95 % CI (100.085 mm, 100.135 mm); meets specification

Correct Answer: C

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Question 25 View Details

A CNC milling operation must remove material from a steel workpiece that is \(200\) mm long, \(50\) mm wide and \(10\) mm thick. The cutter is a carbide end‑mill of diameter \(10\) mm with four teeth. The recommended cutting speed for the steel is \(120\) m/min and the feed per tooth is \(0.08\) mm. (a) Calculate the required spindle speed in rpm. (b) Assuming a single‑pass depth of cut equal to the cutter diameter and 100 % material removal per pass, determine the total number of cutter revolutions needed to clear the whole workpiece. (c) Using the Taylor tool‑life equation \(V T^{n}=C\) with \(V=120\) m/min, \(n=0.25\) and \(C=150\), estimate the tool life \(T\) in minutes and state whether a single tool will survive the operation.

A. Spindle speed ≈3820 rpm; revolutions ≈3500; machining time ≈0.95 min; tool life ≈1.80 min - tool is insufficient

B. Spindle speed ≈3500 rpm; revolutions ≈3400; machining time ≈0.90 min; tool life ≈2.30 min - tool is sufficient

C. Spindle speed ≈4000 rpm; revolutions ≈3000; machining time ≈0.75 min; tool life ≈2.00 min - tool is insufficient

Correct D. Spindle speed ≈3820 rpm; revolutions ≈3125; machining time ≈0.82 min; tool life ≈2.44 min - tool is sufficient

Correct Answer: D

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