The specific resistance of manganin is `50 xx 10^(-8)Omega` m.
The resistance of a cube of length 50 m will be:

Options

A. `10^(-6)Omega`
B. `2.5 xx 10^(-5)Omega`
C. `10^(-8)Omega`
D. `50 xx 10^(-8)Omega`

Explanation

To find the resistance of a cube of manganin with a specific resistance of \( 50 \times 10^{-8} \, \Omega \, \text{m} \) and a length of \( 50 \, \text{m} \), we can use the formula for resistance:

\[ R = \frac{\rho L}{A} \]

where:
- \( R \) is the resistance,
- \( \rho \) is the specific resistance (resistivity),
- \( L \) is the length of the conductor,
- \( A \) is the cross-sectional area.

### Step-by-Step Solution:

1. **Identify the given values:**
- Specific resistance \( \rho = 50 \times 10^{-8} \, \Omega \, \text{m} \)
- Length \( L = 50 \, \text{m} \)

2. **Determine the cross-sectional area \( A \):**
Since the shape is a cube, the cross-sectional area can be calculated as:
\[ A = \text{side}^2 = 50 \, \text{m} \times 50 \, \text{m} = 2500 \, \text{m}^2 \]

3. **Substitute the values into the resistance formula:**
\[ R = \frac{(50 \times 10^{-8} \, \Omega \, \text{m}) \times (50 \, \text{m})}{2500 \, \text{m}^2} \]

4. **Calculate the numerator:**
\[ 50 \times 10^{-8} \times 50 = 2500 \times 10^{-8} \, \Omega \, \text{m}^2 \]

5. **Now substitute back into the formula:**
\[ R = \frac{2500 \times 10^{-8}}{2500} \]

6. **Simplify the expression:**
The \( 2500 \) in the numerator and denominator cancels out:
\[ R = 10^{-8} \, \Omega \]

### Final Answer:
The resistance of the cube is \( 10^{-8} \, \Omega \).