Temperature coefficient of resistance of a wire at `0^(@)C` is `0.00125^(@)C^(-1)` . At `25^(@)C` its resistance is `1 Omega ` . The resitance of the wire will be `1.2 Omega ` at

Options

A. 225 K
B. `190^(@) C`
C. `260^(@)C`
D. ` 185 K `

Answer

To solve the problem, we will use the formula for the resistance of a wire as a function of temperature:

\[ R = R_0 (1 + \alpha T) \]

Where:
- \( R \) is the resistance at temperature \( T \)
- \( R_0 \) is the resistance at 0°C
- \( \alpha \) is the temperature coefficient of resistance
- \( T \) is the temperature in °C

### Step 1: Find \( R_0 \)

We know that at \( T = 25°C \), the resistance \( R = 1 \, \Omega \). We can substitute the values into the equation:

\[ 1 = R_0 (1 + 0.00125 \times 25) \]

Calculating \( 0.00125 \times 25 \):

\[ 0.00125 \times 25 = 0.03125 \]

Now substituting back:

\[ 1 = R_0 (1 + 0.03125) \]
\[ 1 = R_0 \times 1.03125 \]

Now, solving for \( R_0 \):

\[ R_0 = \frac{1}{1.03125} \approx 0.968 \, \Omega \]

### Step 2: Set up the equation for \( R = 1.2 \, \Omega \)

Now we want to find the temperature \( T' \) at which the resistance \( R = 1.2 \, \Omega \):

\[ 1.2 = R_0 (1 + \alpha T') \]

Substituting \( R_0 \) and \( \alpha \):

\[ 1.2 = 0.968 (1 + 0.00125 T') \]

### Step 3: Solve for \( T' \)

First, divide both sides by \( 0.968 \):

\[ \frac{1.2}{0.968} = 1 + 0.00125 T' \]

Calculating \( \frac{1.2}{0.968} \):

\[ \frac{1.2}{0.968} \approx 1.237 \]

Now we have:

\[ 1.237 = 1 + 0.00125 T' \]

Subtracting 1 from both sides:

\[ 0.237 = 0.00125 T' \]

Now, solving for \( T' \):

\[ T' = \frac{0.237}{0.00125} \approx 189.6 \, °C \]

### Step 4: Round the result

Rounding \( 189.6 \, °C \) gives us approximately \( 190 \, °C \).

### Final Answer

The resistance of the wire will be \( 1.2 \, \Omega \) at approximately \( 190 \, °C \).

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