Question

$\text{Using Taylor equation}V{T}^n=C,\text{what is teh percentage increase in tool life when the cutting speed is reduced by 75\%. Tool life exponent is}\frac{2}{3}.$

• A. 8000%
• B. 700%
• C. 600%
• D. 500%

Answer 1

The correct option is B 700%

$V_1{T}_1^n=C...........\left(1\right)$
$\left(0.25V_1\right)T_2^n=C...........\left(2\right)$
$\Rightarrow \frac{\left(0.25V_1\right)T_2^n}{V_1{T}_1^n}=1$
$\Rightarrow 0.25{\left(\frac{T_2}{T_1}\right)}^{\frac{2}{3}}=1$
$\Rightarrow \left(\frac{T_2}{T_1}\right)=4$
$\Rightarrow \frac{T_2}{T_1}=8$
$\Rightarrow \frac{T_2-T_1}{T_1}=7\text{or}700\%$