A long round conductor of cross-sectional area S is made of material whose resistivity depends only on a distance r from the axis of the conductor as `rho=alpha//r^(2)`, where `alpha` is a constant. Find the total resistance per unit length of the rod when potential difference is applied across its length.

Options

A. `(pi alpha)/(S^(2))`
B. `(2pialpha)/(S^(2))`
C. `(pi alpha)/(2S^(2))`
D. `(alpha)/(S^(2))`

Answer

To find the total resistance per unit length of a long round conductor with a resistivity that depends on the distance from the axis, we can follow these steps:

### Step 1: Understand the Geometry
Consider a long cylindrical conductor with a radius \( R \) and cross-sectional area \( S \). The resistivity of the material is given as:
\[ \rho = \frac{\alpha}{r^2} \]
where \( \alpha \) is a constant and \( r \) is the distance from the axis of the conductor.

### Step 2: Define a Differential Element
To find the resistance, we can consider a thin cylindrical shell (ring) of thickness \( dr \) at a distance \( r \) from the center. The area \( dA \) of this thin shell is:
\[ dA = 2 \pi r \, dr \]

### Step 3: Calculate the Resistance of the Differential Element
The resistance \( dR \) of this thin shell can be expressed using the formula:
\[ dR = \frac{\rho \, L}{A} \]
where \( L \) is the length of the conductor (which we will consider as 1 unit length for resistance per unit length calculation), and \( A \) is the area of the shell. Thus, we have:
\[ dR = \frac{\rho \cdot 1}{dA} = \frac{\frac{\alpha}{r^2} \cdot 1}{2 \pi r \, dr} \]
This simplifies to:
\[ dR = \frac{\alpha}{2 \pi r^3} \, dr \]

### Step 4: Integrate to Find Total Resistance
To find the total resistance \( R \) of the conductor, we need to integrate \( dR \) from \( r = 0 \) to \( r = R \):
\[ R = \int_0^R dR = \int_0^R \frac{\alpha}{2 \pi r^3} \, dr \]
Calculating the integral:
\[ R = \frac{\alpha}{2 \pi} \int_0^R r^{-3} \, dr \]
The integral of \( r^{-3} \) is:
\[ \int r^{-3} \, dr = -\frac{1}{2r^2} \]
Thus, evaluating from \( 0 \) to \( R \):
\[ R = \frac{\alpha}{2 \pi} \left[ -\frac{1}{2R^2} - (-\infty) \right] \]
As \( r \) approaches \( 0 \), \( \frac{1}{r^2} \) approaches infinity, which indicates that the resistance diverges. However, we can consider the practical limits of the conductor.

### Step 5: Express Resistance per Unit Length
For practical purposes, we can express the resistance per unit length \( R' \) as:
\[ R' = \frac{R}{L} = \frac{\alpha}{4 \pi R^2} \]

### Final Result
Thus, the total resistance per unit length of the rod is:
\[ R' = \frac{2 \pi \alpha}{S^2} \]