Statement I: The expanded length l of a rod of original length `l_0` is not correctly given by (assuming `alpha` to be constant with T) `l=l_0(1+alphaDeltaT)` if `alphaDeltaT` is large.
Statement II: It is given by `l=l_0e^(alphaDeltaT)`, which cannot be treated as being approximately equal to `l=l_0(1+alphaDeltaT)` for large value a `DeltaT`.

Options

A. Statement I is true, Statement II is true and Statement II is the correct explanation for statement I.
B. Statement I is true, statement II is true and statement II NOT the correct explanation for Statement I
C. Statement I is true, Statement II is false.
D. Statement I is false, statement II is true.

Explanation

To solve the problem, we need to analyze the statements provided and understand the relationship between the length of a rod and its temperature change.

### Step-by-Step Solution:

1. **Understanding the Linear Expansion Formula**:
The linear expansion of a rod can be expressed as:
\[ l = l_0(1 + \alpha \Delta T) \]
where:
- \( l \) = expanded length of the rod
- \( l_0 \) = original length of the rod
- \( \alpha \) = coefficient of linear expansion
- \( \Delta T \) = change in temperature

2. **Condition for Validity**:
This formula is valid under the assumption that \( \alpha \Delta T \) is small. When \( \alpha \Delta T \) is large, the approximation \( 1 + \alpha \Delta T \) becomes inaccurate.

3. **Exponential Expansion Formula**:
For larger temperature changes, the correct expression for the length of the rod is given by:
\[ l = l_0 e^{\alpha \Delta T} \]
This formula arises from integrating the differential form of the expansion.

4. **Deriving the Exponential Formula**:
To derive this, we start from the differential relation:
\[ \frac{dl}{l} = \alpha dT \]
Integrating both sides gives:
\[ \int \frac{1}{l} dl = \int \alpha dT \]
This leads to:
\[ \ln(l) = \alpha T + C \]
where \( C \) is the integration constant. By applying limits and solving, we find:
\[ l = l_0 e^{\alpha \Delta T} \]

5. **Conclusion on Statements**:
- **Statement I**: The expanded length \( l \) is not correctly given by \( l = l_0(1 + \alpha \Delta T) \) if \( \alpha \Delta T \) is large. This statement is **True**.
- **Statement II**: The correct expression is \( l = l_0 e^{\alpha \Delta T} \), which cannot be approximated as \( l = l_0(1 + \alpha \Delta T) \) for large \( \Delta T \). This statement is also **True** and provides the correct explanation for Statement I.

### Final Answer:
Both Statement I and Statement II are true, and Statement II correctly explains Statement I.

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