If specific resistance of a potentiometer wire is `10^(–7 ) Omega m`, the current flow through it is `0.1 A` and the cross-sectional area of wire is `10^(–6) m^(2)` then potential gradient will be

Options

A. `10^(-2)` V/m
B. `10^(-4)` V/m
C. `10^(-6)` V/m
D. `10^(-8)` V/m

Explanation

To find the potential gradient \( K \) of the potentiometer wire, we can follow these steps:

### Step 1: Identify the given values
- Specific resistance \( \rho = 10^{-7} \, \Omega \, m \)
- Current \( I = 0.1 \, A \)
- Cross-sectional area \( A = 10^{-6} \, m^2 \)

### Step 2: Calculate the resistance \( R \) of the wire
The resistance \( R \) of the wire can be calculated using the formula:
\[ R = \frac{\rho L}{A} \]
Where:
- \( L \) is the length of the wire (which we will keep as \( L \) for now).

Substituting the values we have:
\[ R = \frac{10^{-7} \cdot L}{10^{-6}} = 10^{-1} L = 0.1 L \, \Omega \]

### Step 3: Calculate the potential difference \( V \)
Using Ohm's law, the potential difference \( V \) across the wire can be calculated as:
\[ V = I \cdot R \]
Substituting the values:
\[ V = 0.1 \cdot (0.1 L) = 0.01 L \, V \]

### Step 4: Calculate the potential gradient \( K \)
The potential gradient \( K \) is defined as the potential difference per unit length:
\[ K = \frac{V}{L} \]
Substituting the expression for \( V \):
\[ K = \frac{0.01 L}{L} = 0.01 \, V/m \]

### Step 5: Final result
Thus, the potential gradient \( K \) is:
\[ K = 10^{-2} \, V/m \]

### Summary
The potential gradient of the potentiometer wire is \( 10^{-2} \, V/m \).

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