The resistivity of cylindrical conductor carrying steady current along its length varies linearly with the distance from the current carrying end as given by `rho=rho_(0)(1+(x)/(l))` where `l` is the length of the conductor and x is the distance from the current entry and. `rho_(o)` is a positive constant.

Options

A. Electric field varies linearly with x
B. Electric potential difference across the length varies linearly with x
C. Volume charge density in the conductor is zero
D. volume charge density in the conductor is non zero.

Answer

To solve the problem, we need to analyze the given information about the cylindrical conductor and its resistivity, and then determine the behavior of the electric field, electric potential difference, and volume charge density.

### Step 1: Understand the resistivity variation
The resistivity of the cylindrical conductor is given by:
\[ \rho(x) = \rho_0 \left(1 + \frac{x}{l}\right) \]
where:
- \(\rho_0\) is a positive constant,
- \(x\) is the distance from the current entry point,
- \(l\) is the total length of the conductor.

### Step 2: Determine the electric field
Using the relationship between electric field \(E\), current density \(J\), and resistivity \(\rho\):
\[ E = \rho \cdot J \]
Since the resistivity varies with \(x\), we can substitute \(\rho(x)\):
\[ E(x) = \rho_0 \left(1 + \frac{x}{l}\right) J \]
This shows that the electric field \(E\) also varies linearly with \(x\) because it is directly proportional to \(\left(1 + \frac{x}{l}\right)\).

### Step 3: Determine the electric potential difference
The electric potential difference \(V\) across a length \(dx\) can be expressed as:
\[ dV = -E \, dx \]
Substituting for \(E\):
\[ dV = -\rho_0 \left(1 + \frac{x}{l}\right) J \, dx \]
To find the total potential difference \(V\) from \(0\) to \(x\), we integrate:
\[ V = -\int_0^x \rho_0 \left(1 + \frac{x'}{l}\right) J \, dx' \]
Calculating the integral:
\[ V = -J \rho_0 \left[ x' + \frac{x'^2}{2l} \right]_0^x = -J \rho_0 \left( x + \frac{x^2}{2l} \right) \]
This shows that the potential difference \(V\) varies as a quadratic function of \(x\) (i.e., \(V \propto x + \frac{x^2}{2l}\)), indicating it does not vary linearly with \(x\).

### Step 4: Determine the volume charge density
Using the relationship between electric field and volume charge density \(\rho_v\):
\[ \frac{dE}{dx} = \frac{\rho_v}{\epsilon} \]
Substituting for \(E\):
\[ \frac{d}{dx} \left(J \rho_0 \left(1 + \frac{x}{l}\right)\right) = \frac{\rho_v}{\epsilon} \]
Calculating the derivative:
\[ \frac{dE}{dx} = J \frac{\rho_0}{l} \]
Thus,
\[ \rho_v = \epsilon J \frac{\rho_0}{l} \]
Since \(J\), \(\rho_0\), and \(l\) are constants, \(\rho_v\) is non-zero and constant.

### Conclusion
1. **Electric field \(E\) varies linearly with \(x\)**.
2. **Electric potential difference \(V\) does not vary linearly with \(x\)** (it varies quadratically).
3. **Volume charge density \(\rho_v\) is non-zero and constant**.