GATE ME 2026 Set 1 — Question 64
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Manufacturing Processes → Machining & Machine Tools → Tool Life (Taylor's Equation)
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Question
A drill bit during its lifetime can produce 150 through holes in a plate at a drill-speed of 200 RPM. If the drill-speed increases to 300 RPM, then it can produce 60 through holes in the same plate before the drill bit fails. Assume all other parameters remain constant. The value of the exponent in Taylor’s tool life equation is ________ (rounded off to 2 decimal places).
Correct answer
0.29 to 0.33
Solution
Taylor’s tool life equation is given by , where is the cutting speed, is the tool life, and is the tool life exponent.
Total tool life , where is the number of holes produced.
Case 2: RPM, holes
Rounding off to 2 decimal places, the value of the exponent is 0.31.
1.Relationship between Speed and RPM:
Cutting speed , where is the drill diameter and is the rotational speed in RPM. Since the diameter is constant, .2.Relationship between Tool Life and Number of Holes:
Tool life is the total cutting time. If is the depth of the hole and is the feed rate (mm/rev), the time taken to drill one hole is .Total tool life , where is the number of holes produced.
3.Substituting into Taylor's Equation:
Since and are constant:4.Applying for two cases:
Case 1: RPM, holesCase 2: RPM, holes
5.Solving for :
Taking natural logarithm on both sides:Rounding off to 2 decimal places, the value of the exponent is 0.31.
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