City Guilds Model Questions Vol1 2019 Engineering Question 10

Practice objective / multiple choice question 10 from the 2019 City Guilds Model Questions Vol1 Engineering examination.

City Guilds Model Questions Vol1 2019 Engineering Objective / Multiple Choice Hard Difficulty
Question 10 CITY_GUILDS_MODEL_QUESTIONS_VOL1 • 2019 • ENGINEERING • objective

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).

Answer Options

Correct Answer 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
Correct Option:
V = 120 m·min⁻¹, f = 0.30 mm·rev⁻¹; minimum cost = ₦1,817

About This Question

This is City Guilds Model Questions Vol1 2019 Engineering Question 10. It is one of the objective questions from the 2019 City Guilds Model Questions Vol1 Engineering examination.

Difficulty level: Hard .

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