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

Options

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

Answer

To solve the problem, we need to find the ratio \( \alpha = \frac{\Delta I}{I} \) when the temperature of a rod increases from \( t \) to \( t + \Delta t \). Here's a step-by-step solution:

### Step 1: Understand the Initial Moment of Inertia
The moment of inertia \( I \) of a rod can be expressed as:
\[ I = M \cdot r^2 \]
where \( M \) is the mass of the rod and \( r \) is the radius.

### Step 2: Define the Final Moment of Inertia
When the temperature increases, the radius of the rod changes. The new radius \( r_2 \) can be expressed as:
\[ r_2 = r_1 \cdot (1 + \alpha \cdot \Delta t) \]
where \( \alpha \) is the coefficient of linear expansion.

### Step 3: Write the Final Moment of Inertia
The final moment of inertia \( I_2 \) can then be expressed as:
\[ I_2 = M \cdot r_2^2 = M \cdot (r_1 \cdot (1 + \alpha \cdot \Delta t))^2 \]
Expanding this, we get:
\[ I_2 = M \cdot r_1^2 \cdot (1 + \alpha \cdot \Delta t)^2 \]

### Step 4: Expand the Square
Using the binomial expansion, we can write:
\[ (1 + \alpha \cdot \Delta t)^2 = 1 + 2\alpha \cdot \Delta t + (\alpha \cdot \Delta t)^2 \]
Neglecting the higher-order term \( (\alpha \cdot \Delta t)^2 \) for small \( \Delta t \), we have:
\[ I_2 \approx M \cdot r_1^2 \cdot (1 + 2\alpha \cdot \Delta t) \]
Thus, we can express \( I_2 \) as:
\[ I_2 \approx I \cdot (1 + 2\alpha \cdot \Delta t) \]

### Step 5: Calculate the Change in Moment of Inertia
The change in moment of inertia \( \Delta I \) is given by:
\[ \Delta I = I_2 - I = I \cdot (1 + 2\alpha \cdot \Delta t) - I = I \cdot (2\alpha \cdot \Delta t) \]

### Step 6: Find the Ratio \( \frac{\Delta I}{I} \)
Now, we can find the ratio:
\[ \frac{\Delta I}{I} = \frac{I \cdot (2\alpha \cdot \Delta t)}{I} = 2\alpha \cdot \Delta t \]

### Conclusion
Thus, the value of \( \alpha \) is:
\[ \alpha = 2\alpha \cdot \Delta t \]

### Final Answer
The final expression for the ratio \( \frac{\Delta I}{I} \) is:
\[ \frac{\Delta I}{I} = 2\alpha \cdot \Delta t \]

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