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 \(0^\circ C\), we can use the relationship between resistivity and temperature. The resistivity \(\rho\) of a material changes with temperature according to the formula:

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

where:
- \(\rho\) is the resistivity at temperature \(T\),
- \(\rho_0\) is the resistivity at a reference temperature (in this case, \(0^\circ C\)),
- \(\alpha\) is the temperature coefficient of resistivity,
- \(\Delta T\) is the change in temperature from the reference temperature.

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

1. At \(20^\circ C\) (which is \(20^\circ C - 0^\circ C\)):
\[ \rho_{20} = \rho_0 (1 + 20\alpha) = 3 \, \Omega \cdot m \]
This is our first equation.

2. At \(100^\circ C\) (which is \(100^\circ C - 0^\circ C\)):
\[ \rho_{100} = \rho_0 (1 + 100\alpha) = 4 \, \Omega \cdot m \]
This is our second equation.

### Step 2: Write the equations explicitly

From the first equation:
\[ \rho_0 (1 + 20\alpha) = 3 \]
From the second equation:
\[ \rho_0 (1 + 100\alpha) = 4 \]

### Step 3: Solve for \(\rho_0\) and \(\alpha\)

Now, we can express \(\rho_0\) from both equations:

1. From the first equation:
\[ \rho_0 = \frac{3}{1 + 20\alpha} \tag{1} \]

2. From the second equation:
\[ \rho_0 = \frac{4}{1 + 100\alpha} \tag{2} \]

### Step 4: Set the two expressions for \(\rho_0\) equal to each other

Equating (1) and (2):
\[ \frac{3}{1 + 20\alpha} = \frac{4}{1 + 100\alpha} \]

### Step 5: Cross-multiply and simplify

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

### Step 6: Rearrange to find \(\alpha\)

Rearranging gives:
\[ 300\alpha - 80\alpha = 4 - 3 \]
\[ 220\alpha = 1 \]
\[ \alpha = \frac{1}{220} \]

### Step 7: Substitute \(\alpha\) back to find \(\rho_0\)

Now substitute \(\alpha\) back into either equation for \(\rho_0\). Using equation (1):
\[ \rho_0 = \frac{3}{1 + 20 \cdot \frac{1}{220}} = \frac{3}{1 + \frac{20}{220}} = \frac{3}{1 + \frac{1}{11}} = \frac{3}{\frac{12}{11}} = \frac{3 \cdot 11}{12} = \frac{33}{12} = \frac{11}{4} \, \Omega \cdot m \]

### Final Answer

The resistivity of the wire at \(0^\circ C\) is:
\[ \rho_0 = \frac{11}{4} \, \Omega \cdot m \]