The resistivity of a wire at `20^circ C` and `100^circ C` is `3 Omega-m` and ` 4 Omega-m` respectively. The resistivity of the wire at `0^circ C` is

Options

A. `11/4 Omega-m`
B. `4/11 Omega-m`
C. `3/4 Omega-m`
D. `11/3 Omega-m`

Answer

To find the resistivity of the wire at 0°C, we can use the formula that relates resistivity at different temperatures:

\[ \rho(T) = \rho_0 (1 + \alpha (T - T_0)) \]

Where:
- \(\rho(T)\) is the resistivity at temperature \(T\),
- \(\rho_0\) is the resistivity at a reference temperature \(T_0\),
- \(\alpha\) is the temperature coefficient of resistivity,
- \(T\) is the temperature in degrees Celsius,
- \(T_0\) is the reference temperature in degrees Celsius.

### Step 1: Set up the equations for the two given temperatures

1. For \(T = 20°C\) and \(\rho(20°C) = 3 \, \Omega \cdot m\):
\[ 3 = \rho_0 (1 + \alpha (20 - 0)) \]
This simplifies to:
\[ 3 = \rho_0 (1 + 20\alpha) \quad \text{(Equation 1)} \]

2. For \(T = 100°C\) and \(\rho(100°C) = 4 \, \Omega \cdot m\):
\[ 4 = \rho_0 (1 + \alpha (100 - 0)) \]
This simplifies to:
\[ 4 = \rho_0 (1 + 100\alpha) \quad \text{(Equation 2)} \]

### Step 2: Solve the equations

Now we have two equations:
1. \(3 = \rho_0 (1 + 20\alpha)\)
2. \(4 = \rho_0 (1 + 100\alpha)\)

From Equation 1, we can express \(\rho_0\):
\[ \rho_0 = \frac{3}{1 + 20\alpha} \quad \text{(Substituting this into Equation 2)} \]

Substituting \(\rho_0\) into Equation 2:
\[ 4 = \frac{3}{1 + 20\alpha} (1 + 100\alpha) \]

### Step 3: Cross-multiply and simplify

Cross-multiplying gives:
\[ 4(1 + 20\alpha) = 3(1 + 100\alpha) \]
Expanding both sides:
\[ 4 + 80\alpha = 3 + 300\alpha \]
Rearranging gives:
\[ 4 - 3 = 300\alpha - 80\alpha \]
\[ 1 = 220\alpha \]
\[ \alpha = \frac{1}{220} \]

### Step 4: Substitute \(\alpha\) back to find \(\rho_0\)

Now substitute \(\alpha\) back into Equation 1 to find \(\rho_0\):
\[ 3 = \rho_0 \left(1 + 20 \cdot \frac{1}{220}\right) \]
Calculating:
\[ 3 = \rho_0 \left(1 + \frac{20}{220}\right) \]
\[ 3 = \rho_0 \left(1 + \frac{1}{11}\right) \]
\[ 3 = \rho_0 \left(\frac{12}{11}\right) \]
\[ \rho_0 = 3 \cdot \frac{11}{12} \]
\[ \rho_0 = \frac{33}{12} = 2.75 \, \Omega \cdot m \]

### Final Answer

The resistivity of the wire at \(0°C\) is:
\[ \rho(0°C) = 2.75 \, \Omega \cdot m \]