The cross-section area and length of cylindrical conductor are A and l respectively. The specific conductivity varies as `sigma(x)=sigma_(0)(l)/(sqrt(x))`, where x is the distance along the axis of the cylinder from one of its ends. Compute the resistance of the system along the cylindrical axis.

Options

A. `(3sqrt(l))/(2A sigma_(0))`
B. `(2l)/(3A sigma_(0))`
C. `(2sqrt(l))/(3A sigma_(0))`
D. `(sqrt(l))/(A sigma_(0))`

Answer

To compute the resistance of a cylindrical conductor with varying specific conductivity, we will follow these steps:

### Step 1: Understand the problem
We are given a cylindrical conductor with:
- Cross-sectional area \( A \)
- Length \( l \)
- Specific conductivity \( \sigma(x) = \frac{\sigma_0 l}{\sqrt{x}} \), where \( x \) is the distance from one end of the cylinder.

### Step 2: Relate resistance to resistivity
The resistance \( R \) of a conductor can be expressed as:
\[ R = \frac{\rho l}{A} \]
where \( \rho \) is the resistivity. Since resistivity is the inverse of conductivity, we have:
\[ \rho(x) = \frac{1}{\sigma(x)} = \frac{\sqrt{x}}{\sigma_0 l} \]

### Step 3: Consider a small element of the conductor
We will consider a small element of the conductor of length \( dx \) at a distance \( x \) from one end. The resistance \( dR \) of this small element is given by:
\[ dR = \frac{\rho(x) \, dx}{A} = \frac{\sqrt{x}}{\sigma_0 l} \cdot \frac{dx}{A} \]

### Step 4: Integrate to find total resistance
To find the total resistance \( R \), we need to integrate \( dR \) from \( x = 0 \) to \( x = l \):
\[ R = \int_0^l dR = \int_0^l \frac{\sqrt{x}}{\sigma_0 l A} \, dx \]

### Step 5: Solve the integral
The integral can be solved as follows:
\[ R = \frac{1}{\sigma_0 l A} \int_0^l \sqrt{x} \, dx \]
The integral \( \int_0^l \sqrt{x} \, dx \) can be computed:
\[ \int \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \]
Evaluating from \( 0 \) to \( l \):
\[ \int_0^l \sqrt{x} \, dx = \frac{2}{3} l^{3/2} \]

### Step 6: Substitute back into the equation for resistance
Substituting the result of the integral back into the equation for \( R \):
\[ R = \frac{1}{\sigma_0 l A} \cdot \frac{2}{3} l^{3/2} \]
This simplifies to:
\[ R = \frac{2}{3} \cdot \frac{l^{3/2}}{\sigma_0 A} \]

### Final Result
Thus, the resistance of the cylindrical conductor is:
\[ R = \frac{2}{3} \cdot \frac{l^{3/2}}{\sigma_0 A} \]