The specific resistance of a wire is `rho`, its volume is `3 m^(3)` and its resistance is `3 ohms`, then its length will be

Options

A. `sqrt((1)/(rho))`
B. `(3)/(sqrtrho)`
C. `(1)/(rho) sqrt3`
D. `rho sqrt((1)/(3))`

Explanation

To find the length of the wire given its specific resistance (ρ), volume (V), and resistance (R), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between resistance, resistivity, length, and area**:
The formula for resistance (R) in terms of resistivity (ρ), length (L), and cross-sectional area (A) is given by:
\[ R = \frac{\rho L}{A} \]

2. **Substitute the known values**:
We know that the resistance \( R = 3 \, \Omega \) and we will keep \( \rho \) as it is for now. Thus, we can rewrite the equation as:
\[ 3 = \frac{\rho L}{A} \]

3. **Relate volume to area and length**:
The volume (V) of the wire is given by the product of its cross-sectional area (A) and its length (L):
\[ V = A \cdot L \]
Given that \( V = 3 \, m^3 \), we can express the area as:
\[ A = \frac{V}{L} = \frac{3}{L} \]

4. **Substitute area back into the resistance equation**:
Now substitute \( A = \frac{3}{L} \) into the resistance equation:
\[ 3 = \frac{\rho L}{\frac{3}{L}} \]
This simplifies to:
\[ 3 = \frac{\rho L^2}{3} \]

5. **Solve for \( L^2 \)**:
Multiply both sides by 3 to eliminate the fraction:
\[ 9 = \rho L^2 \]
Rearranging gives:
\[ L^2 = \frac{9}{\rho} \]

6. **Find the length \( L \)**:
Taking the square root of both sides:
\[ L = \sqrt{\frac{9}{\rho}} = \frac{3}{\sqrt{\rho}} \]

### Final Answer:
Thus, the length \( L \) of the wire is:
\[ L = \frac{3}{\sqrt{\rho}} \]

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