At `20^@C` , a rod is exactly 20 .05 cm long on a steel ruler. Both are placed in an over at `250^@C` , where the rod now measure 20.11 cm on the same ruler. What is the coefficient of linear expansion for the material of which the rod is made ?

Answer

To find the coefficient of linear expansion for the material of the rod, we will follow these steps:

### Step 1: Identify the initial conditions
- Initial length of the rod, \( L_0 = 20.05 \, \text{cm} \)
- Initial temperature, \( T_0 = 20^\circ C \)
- Final temperature, \( T_f = 250^\circ C \)

### Step 2: Calculate the change in temperature
\[ \Delta T = T_f - T_0 = 250^\circ C - 20^\circ C = 230^\circ C \]

### Step 3: Measure the final length of the rod
- Final length of the rod after heating, \( L_f = 20.11 \, \text{cm} \)

### Step 4: Calculate the change in length of the rod
\[ \Delta L = L_f - L_0 = 20.11 \, \text{cm} - 20.05 \, \text{cm} = 0.06 \, \text{cm} \]

### Step 5: Calculate the change in length of the steel ruler
- The coefficient of linear expansion for steel is approximately \( \alpha_{\text{steel}} = 11 \times 10^{-6} \, \text{°C}^{-1} \).
- Change in length of the steel ruler can be calculated as:
\[ \Delta L_{\text{ruler}} = L_0 \cdot \alpha_{\text{steel}} \cdot \Delta T \]
\[ \Delta L_{\text{ruler}} = 20.05 \, \text{cm} \cdot (11 \times 10^{-6} \, \text{°C}^{-1}) \cdot 230 \, \text{°C} \]
\[ \Delta L_{\text{ruler}} = 20.05 \cdot 11 \cdot 10^{-6} \cdot 230 \]
\[ \Delta L_{\text{ruler}} \approx 0.051 \, \text{cm} \]

### Step 6: Calculate the actual change in length of the rod
The actual change in length of the rod, considering the expansion of the ruler, is:
\[ \Delta L_{\text{actual}} = \Delta L + \Delta L_{\text{ruler}} = 0.06 \, \text{cm} + 0.051 \, \text{cm} = 0.111 \, \text{cm} \]

### Step 7: Use the formula for the coefficient of linear expansion
The formula for the coefficient of linear expansion \( \alpha \) is given by:
\[ \alpha = \frac{\Delta L_{\text{actual}}}{L_0 \cdot \Delta T} \]
Substituting the values:
\[ \alpha = \frac{0.111 \, \text{cm}}{20.05 \, \text{cm} \cdot 230 \, \text{°C}} \]
Calculating:
\[ \alpha = \frac{0.111}{20.05 \cdot 230} \approx \frac{0.111}{4611.5} \approx 2.41 \times 10^{-5} \, \text{°C}^{-1} \]

### Final Answer
The coefficient of linear expansion for the material of which the rod is made is approximately \( 2.41 \times 10^{-5} \, \text{°C}^{-1} \).

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