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 the ratio of their specific resistances and diameters are 1:2.

### Step-by-Step Solution:

1. **Understand the Formula for Resistance**:
The resistance \( R \) of a cylindrical wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the specific resistance (resistivity), \( L \) is the length of the wire, and \( A \) is the cross-sectional area of the wire.

2. **Express the Cross-Sectional Area**:
For a cylindrical wire, the cross-sectional area \( A \) can be expressed in terms of the radius \( r \):
\[ A = \pi r^2 \]

3. **Set Up the Resistance Equations for Wires A and B**:
Let the specific resistances of wires A and B be \( \rho_A \) and \( \rho_B \), and their lengths be \( L_A \) and \( L_B \). The resistances can be expressed as:
\[ R_A = \frac{\rho_A L_A}{\pi r_A^2} \]
\[ R_B = \frac{\rho_B L_B}{\pi r_B^2} \]

4. **Use Given Ratios**:
We know that:
- The ratio of specific resistances is \( \frac{\rho_A}{\rho_B} = \frac{1}{2} \) which implies \( \rho_B = 2\rho_A \).
- The ratio of diameters is \( \frac{d_A}{d_B} = \frac{1}{2} \), which means \( d_B = 2d_A \). Since radius is half of the diameter, we have \( r_B = 2r_A \).

5. **Substitute the Values into the Resistance Equations**:
Substitute \( \rho_B \) and \( r_B \) into the equation for \( R_B \):
\[ R_B = \frac{(2\rho_A) L_B}{\pi (2r_A)^2} = \frac{(2\rho_A) L_B}{\pi (4r_A^2)} = \frac{\rho_A L_B}{2\pi r_A^2} \]

6. **Set the Resistances Equal**:
Since \( R_A = R_B \):
\[ \frac{\rho_A L_A}{\pi r_A^2} = \frac{\rho_A L_B}{2\pi r_A^2} \]

7. **Cancel Common Terms**:
Cancel \( \rho_A \) and \( \pi r_A^2 \) from both sides:
\[ L_A = \frac{L_B}{2} \]

8. **Find the Ratio of Lengths**:
Rearranging gives:
\[ L_B = 2L_A \]
Therefore, the ratio of lengths \( \frac{L_B}{L_A} = 2 \).

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