When 16 wires which are identical are connected in series, the effective resistance exceeds that when they are in parallel by `31 (7)/(8) Omega `Then the resistance of each wire is

Options

A. `0.5 Omega `
B. `1 Omega `
C. `2 Omega `
D. `8 Omega `

Answer

To solve the problem, we need to find the resistance of each wire when 16 identical wires are connected in series and in parallel, and the difference in effective resistance is given.

### Step-by-Step Solution:

1. **Define Variables**:
Let the resistance of each wire be \( r \).

2. **Calculate Series Resistance**:
When 16 wires are connected in series, the total resistance \( R_s \) is given by:
\[ R_s = 16r \]

3. **Calculate Parallel Resistance**:
When the same 16 wires are connected in parallel, the total resistance \( R_p \) is given by:
\[ R_p = \frac{r}{16} \]

4. **Set Up the Equation**:
According to the problem, the effective resistance in series exceeds that in parallel by \( 31 \frac{7}{8} \) ohms. We can express this as:
\[ R_s = R_p + 31 \frac{7}{8} \]
Converting \( 31 \frac{7}{8} \) to an improper fraction:
\[ 31 \frac{7}{8} = \frac{31 \times 8 + 7}{8} = \frac{248 + 7}{8} = \frac{255}{8} \]
Thus, we can rewrite the equation as:
\[ 16r = \frac{r}{16} + \frac{255}{8} \]

5. **Clear the Fractions**:
To eliminate the fractions, multiply the entire equation by 16:
\[ 16 \times 16r = 16 \times \frac{r}{16} + 16 \times \frac{255}{8} \]
This simplifies to:
\[ 256r = r + 510 \]

6. **Rearrange the Equation**:
Now, rearranging gives:
\[ 256r - r = 510 \]
\[ 255r = 510 \]

7. **Solve for \( r \)**:
Divide both sides by 255:
\[ r = \frac{510}{255} = 2 \text{ ohms} \]

### Conclusion:
The resistance of each wire is \( 2 \) ohms.