Tool life equations for two tools under consideration are as follows
$H S S : V T^{0.2} = 150$
$C a r b i d e : V T^{0.3} = 250$
Where $V$ is the cutting speed in $m / m i n$ and $T$ is the tool life in $m i n$ . The breakeven cutting speed above which the carbide tool will be beneficial is

54 m / m i n

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150 m / m i n

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194 m / m i n

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250 m / m i n

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Solution

The correct option is A $54 m / m i n$
At $B E P V_{H S S} = V_{C a r b i d e}$
$T_{H S S} = T_{C a r b i d e}$
${(\frac{150}{V})}^{1 / 0.2} = {(\frac{150}{V})}^{1 / 0 / 3}$
$\frac{150^{5}}{250^{3.33}} = \frac{V^{5}}{V^{3.33}} = V^{1.67}$
$V = 54 m / m i n$

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