The length of a potentiometer wire is 10m and a potential difference of 2 volt is applied to its ends. If the length of its wire is increased by 1m, the value of potential gradient in volt/m will be (potential difference across potentiometer wire remain same)

Options

A. 0.18
B. 0.22
C. 1.3
D. 0.9

Explanation

To solve the problem, we need to calculate the potential gradient of the potentiometer wire before and after increasing its length.

### Step 1: Calculate the initial potential gradient
The potential gradient (V_g) is defined as the potential difference (V) per unit length (L) of the wire.

Given:
- Initial length of the wire, L = 10 m
- Potential difference, V = 2 V

Using the formula for potential gradient:
\[ V_g = \frac{V}{L} \]

Substituting the values:
\[ V_g = \frac{2 \, \text{V}}{10 \, \text{m}} = 0.2 \, \text{V/m} \]

### Step 2: Determine the new length of the wire
The length of the wire is increased by 1 m:
\[ \text{New length} = 10 \, \text{m} + 1 \, \text{m} = 11 \, \text{m} \]

### Step 3: Calculate the new potential gradient
Since the potential difference remains the same (2 V), we can calculate the new potential gradient using the new length.

Using the formula for potential gradient again:
\[ V_g' = \frac{V}{L'} \]

Where \( L' = 11 \, \text{m} \):
\[ V_g' = \frac{2 \, \text{V}}{11 \, \text{m}} \]

Calculating this:
\[ V_g' = \frac{2}{11} \approx 0.1818 \, \text{V/m} \]

### Final Answer
The value of the potential gradient after increasing the length of the wire by 1 m is approximately \( 0.1818 \, \text{V/m} \).

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