For tool $A$ , Talor's tool life exponent $(n)$ is $0.45$ and constant $(K)$ is $90$ . Similarly for tools $B$ , $n = 0.3$ and $K = 60$ . The cutting speed (in $m / m i n$ ) above which tool $A$ will have a higher tool life than tool $B$ is

26.7

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42.5

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80.7

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142.9

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Solution

The correct option is A $26.7$
By Taylor's tool life equation
$V T^{n} = K$

Let $V$ is the speed above which tool $A$ will have a higher tool life than tool $B$ , but at $V, T_{A} = T_{B}$ .
Then
$V T_{A}^{0.45} = 90$
$V T_{B}^{0.3} = 60$
Taking $l o g_{e}$ both sides
$l o g_{e} V + 0.45 l o g_{e} T_{A} = l o g_{e} 90$

$l o g_{e} V + 0.3 l o g_{e} T_{B} = l o g_{e} 60$

$l o g T_{A} - l o g_{e} T_{B} = \frac{l o g_{e} 90 - l o e_{e} V}{0.45} - \frac{l o g_{e} 60 - l o e_{e} V}{0.3}$

as $T_{A} = T_{B}$
$0 = \frac{0.3 l o g_{e} 90 - 0.3 l o g_{e} V - 0.45 l o g_{e} 60 + 0.45 l o g_{e} V}{0.45 \times 0.3}$

$0.4925 = 0.15 l o g V$

$V = 26.67 m / m i n$

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Q. 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 $(m i n u t e s)$
$60$	$81$
$90$	$36$

The exponent

(n)

and constant

(k)

of the Taylor's tool life equation are

Taylor's tool life equation is given by $V T^{n} = C$ where V is in m/min and T is in min. In a turning operation, two tools X and Y are used. For tool X, n = 03 and C = 60 and for tool Y, n = 0.6 and C = 90. Both the tools will have the same tool life for the cutting speed (in m/min, round off to one decimal place) of