Resistivity of a material of wire is `3 * 10^(-6) Omega-m` and resistance of a particular thickness and length of wire is `2 Omega`. If the diameter of the wire gets doubled then the resistivity will be

Options

A. `1.1 * 10^(-2) Omega-m`
B. `1.5 * 10^(-3) Omega-m`
C. `2 * 10^(-6) Omega-m`
D. `3 * 10^(-6) Omega-m`

Answer

To solve the problem, we need to determine how the resistivity of a wire changes when its diameter is doubled. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the given values
- The resistivity of the material, \( \rho = 3 \times 10^{-6} \, \Omega \cdot m \)
- The resistance of the wire, \( R = 2 \, \Omega \)

### Step 2: Recall the formula for resistance
The resistance \( R \) of a wire can be expressed as:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) = resistance
- \( \rho \) = resistivity
- \( L \) = length of the wire
- \( A \) = cross-sectional area of the wire

### Step 3: Determine the cross-sectional area
The cross-sectional area \( A \) of a wire with diameter \( D \) is given by:
\[ A = \frac{\pi D^2}{4} \]
Thus, substituting for \( A \) in the resistance formula gives:
\[ R = \frac{\rho L}{\frac{\pi D^2}{4}} = \frac{4\rho L}{\pi D^2} \]

### Step 4: Set up the equation with the given values
Substituting the known values into the resistance equation:
\[ 2 = \frac{4(3 \times 10^{-6})L}{\pi D^2} \]
From this, we can rearrange to find:
\[ \frac{L}{\pi D^2} = \frac{2}{4(3 \times 10^{-6})} = \frac{1}{6 \times 10^{-6}} \]
Let’s denote this equation as (1).

### Step 5: Analyze the effect of doubling the diameter
When the diameter is doubled, the new diameter \( D' = 2D \). The new cross-sectional area \( A' \) becomes:
\[ A' = \frac{\pi (D')^2}{4} = \frac{\pi (2D)^2}{4} = \frac{\pi \cdot 4D^2}{4} = \pi D^2 \]

### Step 6: Write the new resistance equation
The new resistance \( R' \) when the diameter is doubled is:
\[ R' = \frac{\rho' L}{A'} = \frac{\rho' L}{\pi D^2} \]
Substituting \( A' \) into the resistance formula gives:
\[ R' = \frac{\rho' L}{\pi D^2} \]

### Step 7: Substitute equation (1) into the new resistance equation
Since we know from equation (1) that \( \frac{L}{\pi D^2} = \frac{1}{6 \times 10^{-6}} \), we can substitute this into the new resistance equation:
\[ 2 = \rho' \cdot \frac{1}{6 \times 10^{-6}} \]

### Step 8: Solve for the new resistivity \( \rho' \)
Rearranging gives:
\[ \rho' = 2 \cdot 6 \times 10^{-6} = 12 \times 10^{-6} \, \Omega \cdot m \]

### Final Answer
The new resistivity when the diameter of the wire is doubled is:
\[ \rho' = 12 \times 10^{-6} \, \Omega \cdot m \]

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