A wire has a resistance of `3.1 Omega` at `30°C` and a resistance `4.5 Omega` at 100°C. The temperature coefficient of resistance of the wire

Options

A. `0.008^(@) C^(-1)`
B. `0.0034^(@) C^(-1)`
C. `0.0025^(@) C^(-1)`
D. `0.0012^(@) C^(-1)`

Explanation

To find the temperature coefficient of resistance (α) of the wire, we can use the formula for resistance at different temperatures:

\[ R_t = R_0 (1 + \alpha t) \]

Where:
- \( R_t \) 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: Set up the equations for the two given temperatures

We are given:
- At \( 30°C \), \( R_{30} = 3.1 \, \Omega \)
- At \( 100°C \), \( R_{100} = 4.5 \, \Omega \)

Using the formula for resistance, we can write two equations:

1. For \( 30°C \):
\[ 3.1 = R_0 (1 + 30\alpha) \] (Equation 1)

2. For \( 100°C \):
\[ 4.5 = R_0 (1 + 100\alpha) \] (Equation 2)

### Step 2: Solve the equations to eliminate \( R_0 \)

We can divide Equation 1 by Equation 2 to eliminate \( R_0 \):

\[ \frac{3.1}{4.5} = \frac{R_0(1 + 30\alpha)}{R_0(1 + 100\alpha)} \]

This simplifies to:

\[ \frac{3.1}{4.5} = \frac{1 + 30\alpha}{1 + 100\alpha} \]

### Step 3: Cross-multiply and simplify

Cross-multiplying gives us:

\[ 3.1(1 + 100\alpha) = 4.5(1 + 30\alpha) \]

Expanding both sides:

\[ 3.1 + 310\alpha = 4.5 + 135\alpha \]

### Step 4: Rearrange the equation to solve for \( \alpha \)

Now, rearranging the equation to isolate \( \alpha \):

\[ 310\alpha - 135\alpha = 4.5 - 3.1 \]

This simplifies to:

\[ 175\alpha = 1.4 \]

### Step 5: Solve for \( \alpha \)

Now, divide both sides by 175:

\[ \alpha = \frac{1.4}{175} \]

Calculating this gives:

\[ \alpha = 0.008 \, \text{°C}^{-1} \]

### Final Answer

The temperature coefficient of resistance of the wire is:

\[ \alpha = 0.008 \, \text{°C}^{-1} \]

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