Taylor’s tool life equation is given by VTⁿ = 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 = 0.3 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 ________.

The correct answer is 40.

Step-by-Step Derivation:

Taylor's tool life equation is given by:
$V{T}^n=C$
where:
$V$ is the cutting speed in m/min,
$T$ is the tool life in minutes,
$n$ is the tool life exponent, and
$C$ is a constant.

We are given the parameters for two tools, X and Y:
For tool X: $n_X=0.3$ and $C_X=60$
For tool Y: $n_Y=0.6$ and $C_Y=90$

We want to find the cutting speed $V$ at which both tools have the same tool life, i.e., $T_X=T_Y=T$.

From Taylor's equation, we can express tool life $T$ in terms of $V$, $n$, and $C$:
$T={\left(\frac{C}{V}\right)}^{\frac{1}{n}}$

Equating the tool lives for both tools:
$T_X=T_Y$
${\left(\frac{60}{V}\right)}^{\frac{1}{0.3}}={\left(\frac{90}{V}\right)}^{\frac{1}{0.6}}$

To simplify the equation, we can raise both sides to the power of 0.6:
${\left[{\left(\frac{60}{V}\right)}^{\frac{1}{0.3}}\right]}^{0.6}={\left[{\left(\frac{90}{V}\right)}^{\frac{1}{0.6}}\right]}^{0.6}$
${\left(\frac{60}{V}\right)}^{\frac{0.6}{0.3}}=\frac{90}{V}$
${\left(\frac{60}{V}\right)}^2=\frac{90}{V}$

Expanding the left side:
$\frac{3600}{V^2}=\frac{90}{V}$

Assuming $V\ne 0$, we can multiply both sides by $V^2$:
$3600=90V$

Solving for $V$:
$V=\frac{3600}{90}=40$

Thus, the cutting speed at which both tools have the same tool life is 40 m/min.