If I is the moment of inertia of a solid body having `alpha`-coefficient of linear expansion then the change in I corresponding to a small change in temperature `DeltaT` is

Options

A. `alpha I DeltaT`
B. `(1)/(2)alpha I DeltaT`
C. `2 alpha I Delta T`
D. `3 alpha I DeltaT`

Answer

To solve the problem of finding the change in the moment of inertia (I) of a solid body due to a small change in temperature (ΔT), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Moment of Inertia**:
The moment of inertia (I) of a solid body is proportional to the square of its radius (r). We can express this relationship as:
\[ I = k r^2 \]
where \( k \) is a proportionality constant.

2. **Differentiating the Moment of Inertia**:
To find the change in moment of inertia, we differentiate I with respect to r:
\[ \frac{dI}{dr} = 2kr \]
From this, we can express the change in moment of inertia (dI) in terms of the change in radius (dR):
\[ dI = 2kr \, dR \]

3. **Expressing Change in Radius**:
According to the linear expansion formula, the change in any length (in this case, the radius) due to a change in temperature is given by:
\[ \Delta R = \alpha \Delta T \]
where \( \alpha \) is the coefficient of linear expansion.

4. **Substituting Change in Radius**:
We can substitute ΔR into the equation for dI:
\[ dI = 2kr \, \Delta R = 2kr \, (\alpha \Delta T) \]

5. **Relating to Moment of Inertia**:
Since we know that \( I = kr^2 \), we can replace \( kr \) in the equation:
\[ dI = 2 \left(\frac{I}{r}\right) \alpha \Delta T \]
Simplifying this gives:
\[ dI = 2I \alpha \Delta T \]

6. **Final Result**:
Therefore, the change in moment of inertia corresponding to a small change in temperature ΔT is:
\[ \Delta I = 2I \alpha \Delta T \]

### Conclusion:
The answer to the question is:
\[ \Delta I = 2I \alpha \Delta T \]