Calculate the percentage increase in the moment of inertia of a ring of iron. Given that `alpha_("iron")=11xx10^(-6).^(@)C^(-1)` and rise in temperature is `20^(@)C`

Answer

To calculate the percentage increase in the moment of inertia of a ring of iron due to a rise in temperature, we can follow these steps:

### Step 1: Understand the Moment of Inertia
The moment of inertia \( I \) of a ring is given by the formula:
\[ I = M R^2 \]
where \( M \) is the mass of the ring and \( R \) is the radius of the ring.

### Step 2: Identify the Change in Radius
When the temperature of the ring increases, the radius will also change due to thermal expansion. The change in radius \( \Delta R \) can be expressed in terms of the coefficient of linear expansion \( \alpha \) and the change in temperature \( \Delta T \):
\[ \frac{\Delta R}{R} = \alpha \Delta T \]

### Step 3: Substitute for Percentage Change in Moment of Inertia
The percentage increase in the moment of inertia can be expressed as:
\[ \frac{\Delta I}{I} = 2 \frac{\Delta R}{R} \]
Substituting the expression for \( \frac{\Delta R}{R} \):
\[ \frac{\Delta I}{I} = 2 \alpha \Delta T \]

### Step 4: Plug in the Values
Given:
- \( \alpha = 11 \times 10^{-6} \, \text{°C}^{-1} \)
- \( \Delta T = 20 \, \text{°C} \)

Substituting these values into the equation:
\[ \frac{\Delta I}{I} = 2 \times (11 \times 10^{-6}) \times 20 \]

### Step 5: Calculate the Result
Calculating the above expression:
\[ \frac{\Delta I}{I} = 2 \times 11 \times 20 \times 10^{-6} = 440 \times 10^{-6} \]

### Step 6: Convert to Percentage
To convert the fraction to a percentage, multiply by 100:
\[ \text{Percentage Increase} = 440 \times 10^{-6} \times 100 = 0.044\% \]

### Final Answer
The percentage increase in the moment of inertia of the ring of iron is:
\[ \text{Percentage Increase} = 0.044\% \]

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