The question asks: **"Calculate the percentage cold work for a specimen having strain value as 0.15 in/in."** To solve for percentage cold work (%CW), we use the formula: \[ \% \text{Cold Work} = \left[ 1 - \frac{A_f}{A_0} \right] \times 100 \] Where: - \(A_0\) = original cross-sectional area - \(A_f\) = final cross-sectional area But, strain (\(\epsilon\)) is given as: \[ \epsilon = \ln \left(\frac{A_0}{A_f}\right) \] Rearranging for \(\frac{A_f}{A_0}\): \[ \epsilon = \ln \left(\frac{A_0}{A_f}\right) \implies e^\epsilon = \frac{A_0}{A_f} \implies \frac{A_f}{A_0} = \frac{1}{e^\epsilon} \] Now substitute in the %CW formula: \[ \% \text{Cold Work} = \left[1 - \frac{1}{e^{\epsilon}}\right] \times 100 \] Given \(\epsilon = 0.15\): \[ e^{0.15} \approx 1.1618 \] \[ \frac{1}{e^{0.15}} \approx 0.8607 \] \[ \% \text{Cold Work} = (1 - 0.8607) \times 100 \approx 13.04\% \] **Correct answer: 13.04%** (Option 5)