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

Options

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

Answer

To find the length of the wire given its resistivity, volume, and resistance, we can follow these steps:

### Step 1: Understand the relationship between resistance, resistivity, length, and area.
The resistance \( R \) of a wire can be expressed using the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) = resistance (in ohms)
- \( \rho \) = resistivity of the material (in ohm-meters)
- \( L \) = length of the wire (in meters)
- \( A \) = cross-sectional area of the wire (in square meters)

### Step 2: Express the volume of the wire in terms of area and length.
The volume \( V \) of the wire can be expressed as:
\[ V = A \times L \]
Given that the volume of the wire is \( 3 \, m^3 \), we can write:
\[ 3 = A \times L \]
From this, we can express the area \( A \) in terms of length \( L \):
\[ A = \frac{3}{L} \]

### Step 3: Substitute the expression for area into the resistance formula.
Now, substitute \( A \) into the resistance formula:
\[ R = \frac{\rho L}{\frac{3}{L}} = \frac{\rho L^2}{3} \]
Given that the resistance \( R \) is \( 3 \, \Omega \), we can set up the equation:
\[ 3 = \frac{\rho L^2}{3} \]

### Step 4: Solve for \( L^2 \).
To isolate \( L^2 \), multiply both sides by \( 3 \):
\[ 9 = \rho L^2 \]
Now, rearranging gives:
\[ L^2 = \frac{9}{\rho} \]

### Step 5: Solve for \( L \).
Taking the square root of both sides, we find:
\[ L = \sqrt{\frac{9}{\rho}} = \frac{3}{\sqrt{\rho}} \]

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

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