For turning $N i C r$ alloy steel at cutting speeds of $64 m / m i n$ and $100 m / m i n$ , the respective tool lives are $15 m i n$ and $12 m i n$ . The tool life for a cutting speed of $144 m / m i n$ is

8 m i n

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9 m i n

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10 m i n

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11.5 m i n

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Solution

The correct option is C $10 m i n$
$V_{1} = 64 m i n, V_{2} = 100 m / m i n$

$T_{1} = 15 m i n, T_{2} = 12 m i n$

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

$\Rightarrow V_{1} / V_{2} = [T_{2} / T_{1}]^{n}$

$\Rightarrow$ Taking $l o g_{e}$ on both side, we get

$n = \frac{l n \frac{V_{2}}{V_{1}}}{l n \frac{T_{1}}{T_{2}}} = \frac{l n \frac{100}{64}}{l n \frac{15}{12}} = \frac{0.4463}{0.2231} = 2$

$V_{1} T_{1}^{n} = V_{3} T_{3}^{n}$

$T_{3} = T_{1} {(\frac{V_{1}}{V_{3}})}^{1 / n} = 15 {(\frac{64}{144})}^{0.5} = 10 m i n$

$Note: Strictly speaking the value of n < 1$

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