To find the temperature at which the steel plug will fit into the hole, we use the formula for linear thermal expansion: $$L = L_0 (1 + \alpha \Delta T)$$ where: $L$ is the final length (diameter) $L_0$ is the initial length (diameter) $\alpha$ is the coefficient of linear expansion $\Delta T$ is the change in temperature ($\Delta T = T - T_0$) Step 1: Identify the given values. Initial diameter, $L_0 = 5 \, \text{cm}$ Initial temperature, $T_0 = 30^\circ \text{C}$ Final diameter, $L = 4.997 \, \text{cm}$ Coefficient of linear expansion, $\alpha = 11 \times 10^{-6} \, \text{C}^{-1}$ Step 2: Calculate the change in length ($\Delta L$). $$\Delta L = L - L_0 = 4.997 \, \text{cm} - 5 \, \text{cm} = -0.003 \, \text{cm}$$ Step 3: Rearrange the thermal expansion formula to solve for $\Delta T$. $$L = L_0 (1 + \alpha \Delta T)$$ $$\frac{L}{L_0} = 1 + \alpha \Delta T$$ $$\frac{L}{L_0} - 1 = \alpha \Delta T$$ $$\frac{L - L_0}{L_0} = \alpha \Delta T$$ $$\Delta T = \frac{L - L_0}{L_0 \alpha}$$ Step 4: Substitute the values into the formula to find $\Delta T$. $$\Delta T = \frac{-0.003 \, \text{cm}}{(5 \, \text{cm}) \times (11 \times 10^{-6} \, \text{C}^{-1})}$$ $$\Delta T = \frac{-0.003}{55 \times 10^{-6}} \, \text{C}$$ $$\Delta T = \frac{-0.003}{0.000055} \, \text{C}$$ $$\Delta T \approx -54.545 \, \text{C}$$ Step 5: Calculate the final temperature ($T$). $$\Delta T = T - T_0$$ $$T = T_0 + \Delta T$$ $$T = 30^\circ \text{C} + (-54.545^\circ \text{C})$$ $$T = 30^\circ \text{C} - 54.545^\circ \text{C}$$ $$T \approx -24.545^\circ \text{C}$$ Rounding to one decimal place, $T \approx -24.5^\circ \text{C}$. Step 6: Compare with the given options. The calculated temperature matches option B. The final answer is $\boxed{\text{B}}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.