A rod has variable coefficient of linear expansion `alpha=(x)/(5000)`. If the length of the rod is `1 m`, then determine the increase in length of the rod (in cm ) on increasing the temperature of the rod by `100^(circ) C`.
'(##CEN_KSR_PHY_JEE_C13_E01_035_Q11##)'

Answer

To determine the increase in length of a rod with a variable coefficient of linear expansion given by \(\alpha = \frac{x}{5000}\), where the length of the rod is \(1 \, \text{m}\) and the temperature increase is \(100^\circ C\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Linear Expansion**:
The change in length (\(dL\)) of a rod due to thermal expansion can be expressed as:
\[ dL = \alpha \cdot L_0 \cdot \Delta T \]
where:
- \(dL\) = change in length
- \(\alpha\) = coefficient of linear expansion
- \(L_0\) = original length of the rod
- \(\Delta T\) = change in temperature

2. **Set Up the Integral**:
Since \(\alpha\) is variable, we need to integrate over the length of the rod. The expression for the increase in length becomes:
\[ dL = \left(\frac{x}{5000}\right) \cdot L_0 \cdot dT \]
Here, \(L_0 = 1 \, \text{m}\) and \(\Delta T = 100^\circ C\).

3. **Integrate Over the Length of the Rod**:
We will integrate from \(x = 0\) to \(x = 1 \, \text{m}\):
\[ \Delta L = \int_0^1 \left(\frac{x}{5000}\right) \cdot 1 \cdot 100 \, dx \]
Simplifying this gives:
\[ \Delta L = \frac{100}{5000} \int_0^1 x \, dx \]

4. **Calculate the Integral**:
The integral \(\int_0^1 x \, dx\) can be calculated as:
\[ \int_0^1 x \, dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]

5. **Substitute Back into the Equation**:
Now substituting back:
\[ \Delta L = \frac{100}{5000} \cdot \frac{1}{2} = \frac{100}{10000} = 0.01 \, \text{m} \]

6. **Convert to Centimeters**:
Since we need the answer in centimeters:
\[ \Delta L = 0.01 \, \text{m} = 1 \, \text{cm} \]

### Final Answer:
The increase in length of the rod is \(1 \, \text{cm}\).