The moment of inertia of a rod about its perpendicular bisector is I . When the temperature of the rod is increased by `Delta T` , the increase in the moment of inertia of the rod about the same axis is (Here , `alpha` is the coefficient of linear expansion of the rod )

Options

A. `alpha I Delta T `
B. `2 alpha I Delta T`
C. `(alphaI Delta T)/(2)`
D. `(2 I Delta T )/(alpha)`

Explanation

To solve the problem, we need to determine the increase in the moment of inertia of a rod when its temperature is increased by \( \Delta T \). The moment of inertia of a rod about its perpendicular bisector is given as \( I \).

### Step-by-Step Solution:

1. **Understand the Moment of Inertia**:
The moment of inertia \( I \) of a rod of mass \( m \) and length \( L \) about its perpendicular bisector is given by the formula:
\[ I = \frac{1}{12} m L^2 \]

2. **Determine the Change in Length**:
When the temperature of the rod is increased by \( \Delta T \), the length of the rod changes due to thermal expansion. The change in length \( \Delta L \) can be expressed as:
\[ \Delta L = L \alpha \Delta T \]
where \( \alpha \) is the coefficient of linear expansion of the material of the rod.

3. **Calculate the New Length**:
The new length \( L' \) of the rod after the temperature increase will be:
\[ L' = L + \Delta L = L + L \alpha \Delta T = L(1 + \alpha \Delta T) \]

4. **Calculate the New Moment of Inertia**:
The new moment of inertia \( I' \) about the same axis (perpendicular bisector) can be calculated using the new length \( L' \):
\[ I' = \frac{1}{12} m (L')^2 = \frac{1}{12} m \left(L(1 + \alpha \Delta T)\right)^2 \]
Expanding this:
\[ I' = \frac{1}{12} m \left(L^2(1 + 2\alpha \Delta T + (\alpha \Delta T)^2)\right) \]

5. **Neglect Higher Order Terms**:
Since \( \Delta T \) is small, we can neglect the term \( (\alpha \Delta T)^2 \):
\[ I' \approx \frac{1}{12} m L^2 (1 + 2\alpha \Delta T) = I(1 + 2\alpha \Delta T) \]

6. **Find the Increase in Moment of Inertia**:
The increase in moment of inertia \( \Delta I \) is given by:
\[ \Delta I = I' - I = I(1 + 2\alpha \Delta T) - I = 2\alpha I \Delta T \]

### Final Answer:
The increase in the moment of inertia of the rod about the same axis when the temperature is increased by \( \Delta T \) is:
\[ \Delta I = 2\alpha I \Delta T \]