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 option is 4.

Taylor's tool life equation is given by:
$$V{T}^n=\text{Constant}$$
where:
• $V$ is the cutting speed,
• $T$ is the tool life, and
• $n$ is the tool life exponent.

From the problem description and the details visible in the formula steps in the image, we have:
• Exponent, $n=0.5$
• The cutting speed is reduced by 50%, meaning the new cutting speed ($V_2$) is half of the original cutting speed ($V_1$):
$$V_2=\frac{V_1}{2}$$

Applying the relation to the initial and final states:
$$V_1{T_1}^{0.5}=V_2{T_2}^{0.5}$$

Substitute $V_2=\frac{V_1}{2}$ into the equation:
$$V_1{T_1}^{0.5}=\frac{V_1}{2}{T_2}^{0.5}$$

Divide both sides by $V_1$ to simplify:
$${T_1}^{0.5}=\frac{{T_2}^{0.5}}{2}$$

Rearrange to group the tool life terms together:
$${\left(\frac{T_2}{T_1}\right)}^{0.5}=2$$

To solve for the ratio of the new tool life to the original tool life ($\frac{T_2}{T_1}$), we raise both sides to the power of $\frac{1}{0.5}$:
$$\frac{T_2}{T_1}=2^{\frac{1}{0.5}}=2^2=4$$

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