The temperature coefficient of resistance of the material of a wire is `0.00125^(@)C^(-1)`. Its resistance at `27^(@)C` is `1 Omega`. At what temperature will its resistance be `2 Omega` ?

Options

A. 1154 K
B. 1100 K
C. 1400 K
D. 1127 K

Answer

To find the temperature at which the resistance of the wire will be 2 Ohms, we can use the formula that relates resistance and temperature:

\[ R = R_0 (1 + \alpha (T - T_0)) \]

Where:
- \( R \) is the resistance at temperature \( T \)
- \( R_0 \) is the resistance at a reference temperature \( T_0 \)
- \( \alpha \) is the temperature coefficient of resistance
- \( T \) is the temperature in degrees Celsius

### Given:
- Temperature coefficient of resistance, \( \alpha = 0.00125 \, ^\circ C^{-1} \)
- Resistance at \( 27^\circ C \) (which we will take as \( T_0 \)), \( R_0 = 1 \, \Omega \)
- Resistance at unknown temperature \( T \), \( R = 2 \, \Omega \)

### Step 1: Write the equation for the initial condition
Using the formula for resistance at \( T_0 = 27^\circ C \):

\[ R_0 = R (1 + \alpha (T_0 - T_0)) \]

This simplifies to:

\[ 1 = R_0 (1 + \alpha (27 - 27)) \]

This confirms that at \( 27^\circ C \), the resistance is indeed \( 1 \, \Omega \).

### Step 2: Write the equation for the final condition
Now, we want to find the temperature \( T \) when the resistance \( R = 2 \, \Omega \):

\[ 2 = 1 (1 + \alpha (T - 27)) \]

### Step 3: Rearrange the equation
We can rearrange this equation to solve for \( T \):

\[ 2 = 1 + \alpha (T - 27) \]

Subtract \( 1 \) from both sides:

\[ 1 = \alpha (T - 27) \]

### Step 4: Substitute the value of \( \alpha \)
Now substitute \( \alpha = 0.00125 \):

\[ 1 = 0.00125 (T - 27) \]

### Step 5: Solve for \( T \)
Now, divide both sides by \( 0.00125 \):

\[ T - 27 = \frac{1}{0.00125} \]

Calculating the right side:

\[ T - 27 = 800 \]

Now, add \( 27 \) to both sides:

\[ T = 800 + 27 = 827 \, ^\circ C \]

### Final Answer
Thus, the temperature at which the resistance will be \( 2 \, \Omega \) is:

\[ T = 827 \, ^\circ C \]