The resistance of a wire is 5 ohm at `50^@C` and 6 ohm at `100^@C`. The resistance of the wire at `0^@C` will be

Options

A. 3 ohm
B. 2 ohm
C. 1 ohm
D. 4 ohm

Explanation

To find the resistance of the wire at \(0^\circ C\), we can use the formula for the resistance of a wire as a function of temperature:

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

Where:
- \(R\) is the resistance at temperature \(T\),
- \(R_0\) is the resistance at \(0^\circ C\),
- \(\alpha\) is the temperature coefficient of resistance,
- \(\Delta T\) is the change in temperature from \(0^\circ C\).

### Step 1: Set up the equations for the two known resistances

We know:
- At \(50^\circ C\), \(R = 5 \, \Omega\)
- At \(100^\circ C\), \(R = 6 \, \Omega\)

Using the formula, we can write two equations:

1. For \(50^\circ C\):
\[ 5 = R_0 (1 + 50\alpha) \]

2. For \(100^\circ C\):
\[ 6 = R_0 (1 + 100\alpha) \]

### Step 2: Solve the equations for \(\alpha\)

We can express \(R_0\) from both equations:

From the first equation:
\[ R_0 = \frac{5}{1 + 50\alpha} \]

From the second equation:
\[ R_0 = \frac{6}{1 + 100\alpha} \]

Setting these two expressions for \(R_0\) equal to each other:
\[ \frac{5}{1 + 50\alpha} = \frac{6}{1 + 100\alpha} \]

### Step 3: Cross-multiply to eliminate the fractions

Cross-multiplying gives:
\[ 5(1 + 100\alpha) = 6(1 + 50\alpha) \]

Expanding both sides:
\[ 5 + 500\alpha = 6 + 300\alpha \]

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

Rearranging the equation:
\[ 500\alpha - 300\alpha = 6 - 5 \]
\[ 200\alpha = 1 \]
\[ \alpha = \frac{1}{200} \, \text{per } ^\circ C \]

### Step 5: Substitute \(\alpha\) back to find \(R_0\)

Now we can substitute \(\alpha\) back into one of the equations to find \(R_0\). Using the second equation:
\[ 6 = R_0 (1 + 100 \cdot \frac{1}{200}) \]
\[ 6 = R_0 (1 + 0.5) \]
\[ 6 = R_0 \cdot 1.5 \]
\[ R_0 = \frac{6}{1.5} = 4 \, \Omega \]

### Final Answer

The resistance of the wire at \(0^\circ C\) is \(4 \, \Omega\).

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