Two copper wires A and B of equal masses are taken. The length of A is double the length of B. If the resistance of wire A is `160 Omega `, then calculate the resistance of the wire B.

Answer

To solve the problem, we need to use the relationship between resistance, length, and cross-sectional area of the wires. The resistance \( R \) of a wire is given by the formula:

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

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

### Step 1: Understand the relationship between the wires
Given:
- Length of wire A, \( L_A = 2L_B \) (Length of A is double that of B)
- Resistance of wire A, \( R_A = 160 \, \Omega \)
- Mass of wire A = Mass of wire B

Since both wires have the same mass and are made of the same material (copper), we can say that their volumes are equal.

### Step 2: Express the volume in terms of mass and density
The volume \( V \) of a wire can be expressed as:

\[ V = A \cdot L \]

Thus, for both wires A and B, we have:

\[ V_A = A_A \cdot L_A \]
\[ V_B = A_B \cdot L_B \]

Since \( V_A = V_B \) and both wires have the same mass and density, we can write:

\[ A_A \cdot L_A = A_B \cdot L_B \]

### Step 3: Substitute the lengths
Substituting \( L_A = 2L_B \) into the volume equation gives:

\[ A_A \cdot (2L_B) = A_B \cdot L_B \]

Dividing both sides by \( L_B \) (assuming \( L_B \neq 0 \)):

\[ 2A_A = A_B \]

Thus, we can express the area of wire B in terms of the area of wire A:

\[ A_B = 2A_A \]

### Step 4: Write the resistance equations
Now we can express the resistances of both wires:

For wire A:

\[ R_A = \frac{\rho L_A}{A_A} \]

For wire B:

\[ R_B = \frac{\rho L_B}{A_B} \]

### Step 5: Substitute for \( A_B \)
Substituting \( A_B = 2A_A \) into the equation for \( R_B \):

\[ R_B = \frac{\rho L_B}{2A_A} \]

### Step 6: Relate \( R_A \) and \( R_B \)
Now, we can relate \( R_A \) and \( R_B \):

From \( R_A \):

\[ R_A = \frac{\rho (2L_B)}{A_A} \]

Now, we can find the ratio of the resistances:

\[ \frac{R_A}{R_B} = \frac{\frac{\rho (2L_B)}{A_A}}{\frac{\rho L_B}{2A_A}} = \frac{2L_B}{\frac{L_B}{2}} = 4 \]

### Step 7: Calculate \( R_B \)
Given \( R_A = 160 \, \Omega \):

\[ \frac{R_A}{R_B} = 4 \implies R_B = \frac{R_A}{4} = \frac{160}{4} = 40 \, \Omega \]

### Final Answer
Thus, the resistance of wire B is:

\[ R_B = 40 \, \Omega \]