To solve the problem, we need to understand the concept of specific resistance (or resistivity) and how it applies to the combination of two cubes of manganin in series.
### Step-by-Step Solution:
1. **Understanding Specific Resistance**:
- The specific resistance (or resistivity) of a material is a property that indicates how strongly the material opposes the flow of electric current. It is denoted by the symbol ρ (rho) and is measured in ohm-meters (Ω·m).
- In this case, the specific resistance of manganin is given as \( 50 \times 10^{-8} \, \Omega \, m \).
2. **Identifying the Configuration**:
- The question states that we have two cubes of manganin, each with a length of 50 m, connected in series.
- When resistors are connected in series, the total resistance is the sum of the individual resistances.
3. **Calculating the Resistance of Each Cube**:
- The resistance \( R \) of a cube can be calculated using the formula:
\[
R = \rho \frac{L}{A}
\]
where:
- \( R \) is the resistance,
- \( \rho \) is the specific resistance,
- \( L \) is the length of the cube,
- \( A \) is the cross-sectional area.
- Since we are dealing with cubes, the length \( L \) is given as 50 m. However, we need to know the cross-sectional area \( A \) to calculate the resistance.
4. **Assuming the Cross-Sectional Area**:
- For the sake of this problem, let's assume the cross-sectional area \( A \) is constant and does not affect the specific resistance given (as it is a property of the material).
- Therefore, the resistance of each cube can be expressed as:
\[
R_1 = R_2 = \rho \frac{50}{A}
\]
- Since both cubes are identical, they will have the same resistance.
5. **Calculating Total Resistance in Series**:
- The total resistance \( R_{total} \) for two resistors in series is given by:
\[
R_{total} = R_1 + R_2 = 2R
\]
- Substituting the expression for \( R \):
\[
R_{total} = 2 \left( \rho \frac{50}{A} \right)
\]
6. **Conclusion**:
- However, since the specific resistance of the material does not change regardless of the configuration, the specific resistance of the combination remains the same as that of a single cube:
\[
\text{Specific Resistance of the combination} = 50 \times 10^{-8} \, \Omega \, m
\]
### Final Answer:
The specific resistance of the combination of two cubes of manganin in series is \( 50 \times 10^{-8} \, \Omega \, m \).
