6 . The coefficient of linear expansion of lead is $29 \times 10^{-6} \mathrm{~K}^{-1}$. What change in temperature will cause a $10-\mathrm{m}$ long lead bar to change in length by $3.0 \mathrm{~mm}$ ? a) $5.0 \mathrm{~K}$ b) $10 \mathrm{~K}$ c) $15 \mathrm{~K}$ d) $20 \mathrm{~K}$

Step 1 ：The coefficient of linear expansion is a measure of how much a material expands per degree change in temperature. It is given by the formula: \(\Delta L = \alpha \cdot L \cdot \Delta T\) where \(\Delta L\) is the change in length, \(\alpha\) is the coefficient of linear expansion, \(L\) is the original length, and \(\Delta T\) is the change in temperature.

Step 2 ：We are given that \(\Delta L = 3.0 \, mm = 0.003 \, m\), \(\alpha = 29 \times 10^{-6} \, K^{-1}\), and \(L = 10 \, m\). We are asked to find \(\Delta T\).

Step 3 ：Rearranging the formula for \(\Delta T\) gives: \(\Delta T = \Delta L / (\alpha \cdot L)\)

Step 4 ：We can substitute the given values into this formula to find the change in temperature.

Step 5 ：\(\alpha = 2.9e-05\), \(L = 10\), \(\Delta_L = 0.003\), \(\Delta_T = 10.344827586206897\)

Step 6 ：The calculated change in temperature is approximately 10.34 K. However, the options provided in the question are all integers. Therefore, we should round the calculated value to the nearest integer.

Step 7 ：\(\alpha = 2.9e-05\), \(L = 10\), \(\Delta_L = 0.003\), \(\Delta_T = 10.344827586206897\), \(\Delta_T_{rounded} = 10\)

Step 8 ：Final Answer: The change in temperature that will cause a 10-m long lead bar to change in length by 3.0 mm is approximately \(\boxed{10 \, K}\).