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 find the increase in the moment of inertia of a rod about its perpendicular bisector when the temperature is increased by ΔT, we can follow these steps:

### Step-by-Step Solution:

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

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

3. **New Length of the Rod**:
The new length \( l' \) of the rod after the temperature increase is:
\[ l' = l + \Delta l = l + \alpha l \Delta T = l(1 + \alpha \Delta T) \]

4. **Calculate the New Moment of Inertia**:
The new moment of inertia \( I' \) about the same axis (the perpendicular bisector) can be calculated using the new length:
\[ I' = \frac{1}{12} m (l')^2 = \frac{1}{12} m (l(1 + \alpha \Delta T))^2 \]
Expanding this:
\[ I' = \frac{1}{12} m \left( l^2(1 + 2\alpha \Delta T + (\alpha \Delta T)^2) \right) \]
Since \( (\alpha \Delta T)^2 \) is very small compared to \( 2\alpha \Delta T \), we can neglect it:
\[ I' \approx \frac{1}{12} m l^2 (1 + 2\alpha \Delta T) = I (1 + 2\alpha \Delta T) \]

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

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