The resistance of the wire is 121 ohm. It is divided into 'n' equal parts and they are connected in parallel, then effective resistance is 1'ohm. The value of 'n' is

Options

A. 12
B. 13
C. 11
D. 3

Answer

To solve the problem, we need to determine the value of 'n' when a wire with a resistance of 121 ohms is divided into 'n' equal parts and connected in parallel, resulting in an effective resistance of 1 ohm.

### Step-by-Step Solution:

1. **Identify the Resistance of One Part**:
The total resistance of the wire is given as \( R = 121 \, \Omega \). When the wire is divided into 'n' equal parts, the resistance of each part, denoted as \( R' \), can be calculated using the formula:
\[ R' = \frac{R}{n} = \frac{121}{n} \, \Omega \]

2. **Calculate the Effective Resistance in Parallel**:
When 'n' resistances of \( R' \) are connected in parallel, the formula for the effective resistance \( R_{eff} \) is given by:
\[ \frac{1}{R_{eff}} = \frac{1}{R'} + \frac{1}{R'} + \ldots \text{(n times)} = \frac{n}{R'} \]
Thus, we can write:
\[ R_{eff} = \frac{R'}{n} = \frac{R}{n^2} \]

3. **Substitute the Known Values**:
We know that the effective resistance \( R_{eff} \) is given as 1 ohm. Therefore, we can set up the equation:
\[ 1 = \frac{121}{n^2} \]

4. **Solve for n**:
Rearranging the equation gives:
\[ n^2 = 121 \]
Taking the square root of both sides, we find:
\[ n = \sqrt{121} = 11 \]

5. **Conclusion**:
The value of 'n' is 11.