The temperature coefficient of resistance for a wire is `0.00125^(@)C^(-1)`. At 300 K its resistance is `1Omega`. The temperature at which the resistance becomes `2Omega` is

Options

A. 450 K
B. 1127 K
C. 454 K
D. 900 K

Explanation

To find the temperature at which the resistance of the wire becomes 2 ohms, we can use the formula for resistance as a function of 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,
- \( T_0 \) is the reference temperature in degrees Celsius.

Given:
- \( \alpha = 0.00125 \, ^\circ C^{-1} \)
- \( R_0 = 1 \, \Omega \) at \( T_0 = 300 \, K \) (which is \( 27^\circ C \))

We need to find the temperature \( T \) at which the resistance \( R = 2 \, \Omega \).

### Step 1: Set up the equation for the initial resistance
Using the reference resistance:
\[ R_0 = 1 \, \Omega \text{ at } T_0 = 27^\circ C \]
So, we can write:
\[ 1 = 1 (1 + 0.00125 (27 - 27)) \]
This confirms our reference point.

### Step 2: Set up the equation for the final resistance
Now, we set up the equation for the resistance when it becomes \( 2 \, \Omega \):
\[ 2 = 1 (1 + 0.00125 (T - 27)) \]

### Step 3: Simplify the equation
This simplifies to:
\[ 2 = 1 + 0.00125 (T - 27) \]

### Step 4: Rearranging the equation
Subtract 1 from both sides:
\[ 1 = 0.00125 (T - 27) \]

### Step 5: Solve for \( T \)
Now, divide both sides by \( 0.00125 \):
\[ T - 27 = \frac{1}{0.00125} \]
\[ T - 27 = 800 \]

Now, add 27 to both sides:
\[ T = 800 + 27 \]
\[ T = 827 \, ^\circ C \]

### Step 6: Convert to Kelvin
To convert Celsius to Kelvin:
\[ T_{K} = T + 273 \]
\[ T_{K} = 827 + 273 = 1100 \, K \]

Thus, the temperature at which the resistance becomes \( 2 \, \Omega \) is \( 1100 \, K \).

### Final Answer
The temperature at which the resistance becomes \( 2 \, \Omega \) is \( 1100 \, K \).

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