Using the Taylor's tool life equation with exponent n = 0.5. If the cutting speed is reduced by 50%, the ratio of new tool life to original tool life is

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0.5

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Solution

The correct option is A 4
If the cutting speed is reduced by 50%
$V_{2} = \frac{V_{1}}{2}$

Taylor's tool life equation: $V_{1} T_{1}^{0.5} = V_{2} T_{2}^{0.5}$

$V_{1} T_{1}^{0.5} = \frac{V_{1}}{2} . T_{2}^{0.5}$

${(\frac{T_{2}}{T_{1}})}^{0.5} = 2$

$\frac{T_{2}}{T_{1}} = 2^{\frac{1}{0.5}} = 4$

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