The area of cross-section of a metal wire is doubled, keeping its length same, then its resistance is:

Options

A. No change
B. doubled
C. four times
D. halved

Answer

To solve the problem, we need to understand how the resistance of a wire changes when its cross-sectional area is modified while keeping its length constant.

### Step-by-Step Solution:

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

2. **Identify the initial conditions**:
Let the initial resistance of the wire be \( R_1 \) with length \( L \) and area \( A \):
\[ R_1 = \frac{\rho L}{A} \]

3. **Change the area of cross-section**:
According to the problem, the area of cross-section is doubled. Therefore, the new area \( A_2 \) is:
\[ A_2 = 2A \]

4. **Calculate the new resistance**:
The new resistance \( R_2 \) with the same length \( L \) but with the new area \( A_2 \) is:
\[ R_2 = \frac{\rho L}{A_2} = \frac{\rho L}{2A} \]

5. **Relate the two resistances**:
Now, we can express the ratio of the initial resistance \( R_1 \) to the new resistance \( R_2 \):
\[ \frac{R_1}{R_2} = \frac{\frac{\rho L}{A}}{\frac{\rho L}{2A}} = \frac{2A}{A} = 2 \]
This implies:
\[ R_2 = \frac{R_1}{2} \]

6. **Conclusion**:
Therefore, when the area of cross-section of the wire is doubled while keeping the length constant, the new resistance \( R_2 \) is half of the original resistance \( R_1 \).

### Final Answer:
The resistance is halved.