When the temperature of a rod increases from t to `t+Delta t`, its moment of inertia increases from I to `I+Delta I`. If `alpha` is coefficient of linear expansion, the value of `Delta I//I` is

Options

A. `2alpha Delta t`
B. `alpha Delta t`
C. `alpha Delta t//2`
D. `Delta t//alpha`

Explanation

To solve the problem, we need to determine the relationship between the change in moment of inertia (ΔI) and the change in temperature (Δt) of a rod, given the coefficient of linear expansion (α).

### Step-by-Step Solution:

1. **Understand the Moment of Inertia Formula**:
The moment of inertia (I) of a rod about an axis through one end is given by:
\[ I = \frac{1}{3} m L^2 \]
where \(m\) is the mass of the rod and \(L\) is its length.

2. **Change in Length with Temperature**:
When the temperature of the rod increases, its length changes due to thermal expansion. The change in length (ΔL) can be expressed as:
\[ \Delta L = \alpha L \Delta t \]
where \(α\) is the coefficient of linear expansion, and \(Δt\) is the change in temperature.

3. **New Length of the Rod**:
The new length of the rod after the temperature change becomes:
\[ L' = L + \Delta L = L + \alpha L \Delta t = L(1 + \alpha \Delta t) \]

4. **New Moment of Inertia**:
The new moment of inertia (I') after the temperature change is:
\[ I' = \frac{1}{3} m (L')^2 = \frac{1}{3} m (L(1 + \alpha \Delta t))^2 \]
Expanding this:
\[ I' = \frac{1}{3} m L^2 (1 + \alpha \Delta t)^2 = \frac{1}{3} m L^2 (1 + 2\alpha \Delta t + (\alpha \Delta t)^2) \]

5. **Change in Moment of Inertia**:
The change in moment of inertia (ΔI) is given by:
\[ \Delta I = I' - I = \frac{1}{3} m L^2 (2\alpha \Delta t + (\alpha \Delta t)^2) \]
Since we are interested in the ratio \( \frac{\Delta I}{I} \), we can simplify this.

6. **Calculating the Ratio**:
Substituting \(I = \frac{1}{3} m L^2\) into the ratio:
\[ \frac{\Delta I}{I} = \frac{\frac{1}{3} m L^2 (2\alpha \Delta t + (\alpha \Delta t)^2)}{\frac{1}{3} m L^2} = 2\alpha \Delta t + (\alpha \Delta t)^2 \]
For small changes, we can neglect the \((\alpha \Delta t)^2\) term, leading to:
\[ \frac{\Delta I}{I} \approx 2\alpha \Delta t \]

7. **Final Result**:
Thus, the value of \(\frac{\Delta I}{I}\) is:
\[ \frac{\Delta I}{I} = 2\alpha \Delta t \]