Linked Question:
In a machining experiment, tool life was found to vary with the cutting speed in the following manner:

Cutting speed $(m / m i n)$ Tool life $(m i n u t e s)$

$60$ $81$

$90$ $36$

What is the percentage increase in tool life when the cutting speed is havled?

50

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200

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300

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400

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Solution

The correct option is C $300$ %
Taylor's tool life equation

$V T^{0.5} = K = 540$

If cutting speed is havled.

$V_{1} = V; T_{1} = T$

$V_{2} = V / 2; T_{2} = ?$

$V_{1} T_{1}^{0.5} = V_{2} T_{2}^{0.5}$

$T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{2} = T_{1} {(\frac{V}{V / 2})}^{2}$

$T_{2} = 4 T_{1} = 4 T$

Percentage increase in tool life

$= \frac{T_{2} - T_{1}}{T_{1}} \times 100 = \frac{4 T - T}{T} \times 100 = 300 %$

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