The length of a metallic rod is 5 m at `0^(@)C` and becomes `5.01 m` , on heating upto `100^(@)C` . The linear expansion of the metal will be

Options

A. `2.33xx10^(-5)//^(@)C`
B. `6.0xx10^(-5)//^(@)C`
C. `4.0xx10^(-5)//^(@)C`
D. `2.0xx10^(-5)//^(@)C`

Answer

To find the linear expansion of the metallic rod, we can use the formula for linear expansion:

\[ L_f = L_i (1 + \alpha \Delta T) \]

Where:
- \( L_f \) = final length of the rod
- \( L_i \) = initial length of the rod
- \( \alpha \) = coefficient of linear expansion
- \( \Delta T \) = change in temperature

### Step-by-step Solution:

1. **Identify the given values:**
- Initial length \( L_i = 5 \, m \)
- Final length \( L_f = 5.01 \, m \)
- Initial temperature \( T_i = 0^\circ C \)
- Final temperature \( T_f = 100^\circ C \)

2. **Calculate the change in temperature (\( \Delta T \)):**
\[ \Delta T = T_f - T_i = 100^\circ C - 0^\circ C = 100^\circ C \]

3. **Substitute the known values into the linear expansion formula:**
\[ 5.01 = 5 (1 + \alpha \cdot 100) \]

4. **Rearrange the equation to solve for \( \alpha \):**
\[ 5.01 = 5 + 5 \alpha \cdot 100 \]
\[ 5.01 - 5 = 5 \alpha \cdot 100 \]
\[ 0.01 = 5 \alpha \cdot 100 \]

5. **Solve for \( \alpha \):**
\[ \alpha = \frac{0.01}{5 \cdot 100} \]
\[ \alpha = \frac{0.01}{500} = 2 \times 10^{-5} \, \text{per degree centigrade} \]

### Conclusion:
The coefficient of linear expansion \( \alpha \) of the metal is \( 2 \times 10^{-5} \, \text{per degree centigrade} \).

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