Statement-1: The temperature dependence of resistance is usually given as `R=R_0(1+alphaDeltaT)`. The resistance of a wire changes from `100Omega to 150Omega` when its temperature is increased from `27^@C to 227^@C`. This implies that `alpha=2.5xx10^-3//^@C`.
Statement 2: `R=R_0(1+alphaDeltaT)` is valid only when the change in the temperature `DeltaT` is small and `DeltaR=(R-R_0) lt ltR_0`.

Options

A. Statement-1 is true, Statement-2 is false
B. Statement-1 is true, Statement-2 is true, Statement-2 is the correct explaination of Statement-1
C. Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1
D. Statement-1 is false, Statement-2 is true

Answer

To solve the problem, we will analyze both statements and derive the necessary calculations step by step.

### Step 1: Understand the given information
We know that the resistance of a wire changes from \( R_0 = 100 \, \Omega \) to \( R = 150 \, \Omega \) when the temperature increases from \( T_0 = 27^\circ C \) to \( T = 227^\circ C \).

### Step 2: Calculate the change in resistance
The change in resistance \( \Delta R \) can be calculated as:
\[ \Delta R = R - R_0 = 150 \, \Omega - 100 \, \Omega = 50 \, \Omega \]

### Step 3: Calculate the change in temperature
The change in temperature \( \Delta T \) is:
\[ \Delta T = T - T_0 = 227^\circ C - 27^\circ C = 200^\circ C \]

### Step 4: Use the resistance-temperature relationship
The relationship between resistance and temperature is given by:
\[ R = R_0(1 + \alpha \Delta T) \]
Substituting the known values:
\[ 150 = 100(1 + \alpha \cdot 200) \]

### Step 5: Solve for \( \alpha \)
Rearranging the equation:
\[ 1 + \alpha \cdot 200 = \frac{150}{100} = 1.5 \]
\[ \alpha \cdot 200 = 1.5 - 1 = 0.5 \]
\[ \alpha = \frac{0.5}{200} = 2.5 \times 10^{-3} \, ^\circ C^{-1} \]

### Step 6: Analyze Statement 1
Statement 1 claims that the calculated value of \( \alpha = 2.5 \times 10^{-3} \, ^\circ C^{-1} \) is correct. However, we must check if the assumptions of the equation \( R = R_0(1 + \alpha \Delta T) \) hold true.

### Step 7: Check the validity of the equation
The equation is valid only when:
1. The change in temperature \( \Delta T \) is small.
2. The change in resistance \( \Delta R \) is much less than \( R_0 \).

In this case:
- \( \Delta T = 200^\circ C \) is not small.
- \( \Delta R = 50 \, \Omega \) is not much less than \( R_0 = 100 \, \Omega \).

Thus, Statement 1 is **false**.

### Step 8: Analyze Statement 2
Statement 2 states that the equation \( R = R_0(1 + \alpha \Delta T) \) is valid only when \( \Delta T \) is small and \( \Delta R \ll R_0 \). Since we have established that these conditions are not met in this case, Statement 2 is **true**.

### Conclusion
- Statement 1 is false.
- Statement 2 is true.

### Final Answer
- Statement 1: False
- Statement 2: True

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