GATE 2018 ME (ME2) – Question 59
Identical HSS twist drills make through holes at constant feed in 20 mm mild-steel plates. In test 1 a drill lasts 300 holes at 150 rpm; in test 2 another lasts 200 holes at 300 rpm. Using Taylor’s tool-life equation, find the maximum number of holes another drill can make at 200 rpm, to two decimal places.
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 252.00 to 254.00
Explanation
At constant feed per revolution and hole depth, tool-life time scales as $T\propto H/N$, where H is hole count and N rpm. With cutting speed proportional to N, Taylor’s equation gives $H\propto N^b$, $b=1-1/n$. From the two tests, $b=\ln(200/300)/\ln(300/150)=-0.584963$. Hence $H=300(200/150)^b=$ **253.53 holes**. If only completed holes count, floor to 253; both lie within the official range.