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

The correct answer is 4.

Step-by-Step Explanation:

Taylor's tool life equation relates cutting speed and tool life as follows:
$V{T}^n=C$
where:
- $V$ is the cutting speed,
- $T$ is the tool life,
- $n$ is the tool life exponent, and
- $C$ is a constant depending on the tool and work material combination.

Given:
- The tool life exponent, $n=0.5$
- The cutting speed is reduced by 50%, which means the new cutting speed ($V_2$) is half of the original cutting speed ($V_1$):
$V_2=0.5V_1$

Let $T_1$ be the original tool life and $T_2$ be the new tool life. Since the product remains constant under both conditions, we can write:
$V_1{T_1}^n=V_2{T_2}^n$

Substitute $n=0.5$ and $V_2=0.5V_1$ into the equation:
$V_1{T_1}^{0.5}=\left(0.5V_1\right){T_2}^{0.5}$

Divide both sides by $V_1$:
${T_1}^{0.5}=0.5{T_2}^{0.5}$

Rearrange the terms to find the ratio of the new tool life to the original tool life:
$\frac{{T_2}^{0.5}}{{T_1}^{0.5}}=\frac{1}{0.5}$

Simplify the right side of the equation:
${\left(\frac{T_2}{T_1}\right)}^{0.5}=2$

Square both sides to find the ratio of $T_2$ to $T_1$:
$\frac{T_2}{T_1}=2^2=4$

Thus, the ratio of the new tool life to the original tool life is 4.