The specific resistance and area of cross section of the potentiometer wire are `rho'` and `A` respectively. If a current `i` passes through the wire, its potential gradient will be

Options

A. `(I rho)/(A)`
B. `(I)/(A rho)`
C. `(I A)/(rho)`
D. `I A rho`

Explanation

To find the potential gradient of a potentiometer wire given its specific resistance (\(\rho\)), area of cross-section (\(A\)), and current (\(i\)), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between resistance, resistivity, and dimensions**:
The resistance \(R\) of a wire can be expressed using the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \(\rho\) = specific resistance (resistivity) of the material,
- \(L\) = length of the wire,
- \(A\) = area of cross-section of the wire.

2. **Relate potential drop to current and resistance**:
According to Ohm's Law, the potential drop \(V\) across a resistor is given by:
\[ V = I \cdot R \]
Substituting the expression for resistance from step 1, we have:
\[ V = I \cdot \left(\frac{\rho L}{A}\right) \]
Thus, the potential drop becomes:
\[ V = \frac{I \rho L}{A} \]

3. **Define potential gradient**:
The potential gradient \(k\) is defined as the potential drop per unit length of the wire:
\[ k = \frac{V}{L} \]

4. **Substitute the expression for potential drop**:
Now substituting the expression for \(V\) into the equation for potential gradient:
\[ k = \frac{\frac{I \rho L}{A}}{L} \]

5. **Simplify the equation**:
The \(L\) in the numerator and denominator cancels out:
\[ k = \frac{I \rho}{A} \]

6. **Final result**:
Therefore, the potential gradient of the potentiometer wire is given by:
\[ k = \frac{I \rho}{A} \]