When length of a metal wire is doubled and area of cross-section is reduced to half, then its resistance becomes

Options

A. half
B. double
C. four times
D. No changes

Answer

To solve the problem, we need to understand how the resistance of a wire changes when its length and cross-sectional area are altered. The formula for the resistance \( R \) of a wire is given by:

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

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-by-Step Solution:

1. **Identify Initial Conditions**:
Let the initial length of the wire be \( L_1 = L \) and the initial area of cross-section be \( A_1 = A \). The initial resistance \( R_1 \) can be expressed as:
\[ R_1 = \frac{\rho L_1}{A_1} = \frac{\rho L}{A} \]

2. **Change in Length and Area**:
According to the problem, the length of the wire is doubled:
\[ L_2 = 2L \]
and the area of cross-section is reduced to half:
\[ A_2 = \frac{A}{2} \]

3. **Calculate New Resistance**:
The new resistance \( R_2 \) can be calculated using the new length and area:
\[ R_2 = \frac{\rho L_2}{A_2} = \frac{\rho (2L)}{\frac{A}{2}} = \frac{\rho (2L) \cdot 2}{A} = \frac{4\rho L}{A} \]

4. **Relate New Resistance to Initial Resistance**:
Now, we can express \( R_2 \) in terms of \( R_1 \):
\[ R_2 = 4 \left(\frac{\rho L}{A}\right) = 4R_1 \]

5. **Conclusion**:
Therefore, the resistance of the wire after the changes is four times the initial resistance:
\[ R_2 = 4R_1 \]

### Final Answer:
The resistance becomes \( 4 \) times the original resistance.

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