Let's solve the question step by step. ### Given: - Taylor tool life equation: \( VT^n = C \) - Initial cutting speed \( V_1 \) - Tool life exponent \( n = 0.3 \) - Cutting speed is reduced by 20%: \( V_2 = 0.8V_1 \) - Initial tool life \( T_1 = 7 \) min ### To Find: - The percentage change in tool life (\( T_2 \)) when cutting speed is reduced by 20%. --- #### Step 1: Taylor's Equation Setup \[ V_1 T_1^n = V_2 T_2^n \] #### Step 2: Substitute \( V_2 = 0.8V_1 \) \[ V_1 T_1^n = 0.8V_1 T_2^n \] \[ T_1^n = 0.8 T_2^n \] \[ \frac{T_2^n}{T_1^n} = \frac{1}{0.8} \] \[ \left(\frac{T_2}{T_1}\right)^n = \frac{1}{0.8} \] \[ \frac{T_2}{T_1} = \left(\frac{1}{0.8}\right)^{1/n} \] #### Step 3: Substitute \( n = 0.3 \) \[ \frac{T_2}{T_1} = (1.25)^{1/0.3} \] \[ \frac{T_2}{T_1} = 1.25^{3.33} \] Calculate \( 1.25^{3.33} \): \[ 1.25^{3.33} \approx 2.07 \] So, \[ T_2 \approx 2.07 \times T_1 \] That is, the tool life increases by about **107%**. --- ### Correct Answer: **The tool life increased by 107%.** This matches the last option in your list.