Two cylindrical wire A and B have the same resistance . The ratio of their specific resistances and diameters are 1 : 2 each, then what is the ratio of the length of B to the length of A ?

Answer

To solve the problem, we need to find the ratio of the lengths of two cylindrical wires A and B, given that they have the same resistance and specific resistances and diameters in the ratio of 1:2. Let's denote the specific resistances of A and B as \( \rho_A \) and \( \rho_B \), and their diameters as \( d_A \) and \( d_B \).

### Step-by-Step Solution:

1. **Understanding the Given Ratios**:
- The ratio of specific resistances: \( \frac{\rho_B}{\rho_A} = 2 \) (since \( \rho_B = 2 \rho_A \)).
- The ratio of diameters: \( \frac{d_B}{d_A} = 2 \).

2. **Relating Diameter to Radius**:
- The diameter is related to the radius by \( d = 2r \). Therefore, we can express the radii as:
\[ r_B = \frac{d_B}{2} = \frac{2d_A}{2} = d_A \quad \text{(since } d_B = 2d_A\text{)} \]
- Thus, the ratio of the radii is:
\[ \frac{r_B}{r_A} = \frac{d_A}{d_A} = 1 \]

3. **Using the Formula for Resistance**:
- The resistance \( R \) of a cylindrical wire is given by:
\[ R = \frac{\rho L}{A} \]
- The cross-sectional area \( A \) of a cylinder is given by \( A = \pi r^2 \). Therefore, we can write the resistances of wires A and B as:
\[ R_A = \frac{\rho_A L_A}{\pi r_A^2} \]
\[ R_B = \frac{\rho_B L_B}{\pi r_B^2} \]

4. **Setting the Resistances Equal**:
- Since the resistances are equal, we have:
\[ R_A = R_B \implies \frac{\rho_A L_A}{\pi r_A^2} = \frac{\rho_B L_B}{\pi r_B^2} \]
- Canceling \( \pi \) from both sides gives:
\[ \frac{\rho_A L_A}{r_A^2} = \frac{\rho_B L_B}{r_B^2} \]

5. **Substituting the Known Ratios**:
- Substitute \( \rho_B = 2\rho_A \) and \( r_B = r_A \):
\[ \frac{\rho_A L_A}{r_A^2} = \frac{2\rho_A L_B}{r_A^2} \]
- Cancel \( \rho_A \) and \( r_A^2 \) from both sides:
\[ L_A = 2L_B \]

6. **Finding the Ratio of Lengths**:
- Rearranging gives:
\[ \frac{L_B}{L_A} = \frac{1}{2} \]

### Final Answer:
The ratio of the length of wire B to the length of wire A is:
\[ \frac{L_B}{L_A} = \frac{1}{2} \]