To determine the new resistance of a wire after it has been drawn out to twice its original length and its cross-sectional area is halved, we can use the formula for resistance in terms of resistivity, length, and cross-sectional area: ### Formula for Resistance The resistance \( R \) of a wire is given by the formula: \[ R = \rho \frac{L}{A} \] 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 Explanation 1. **Original Resistance**: The original resistance of the wire is given as \( R_1 = 5 \, \Omega \). 2. **Original Length and Area**: Let's denote the original length of the wire as \( L \) and the original cross-sectional area as \( A \). Therefore, we can express the original resistance as: \[ R_1 = \rho \frac{L}{A} = 5 \, \Omega \] 3. **New Length**: The wire is drawn out to a new length that is twice the original length: \[ L' = 2L \] 4. **New Cross-Sectional Area**: The cross-sectional area is halved: \[ A' = \frac{A}{2} \] 5. **Calculating New Resistance**: We can now calculate the new resistance \( R_2 \) using the new length and new area: \[ R_2 = \rho \frac{L'}{A'} = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{2L \cdot 2}{A} = \rho \frac{4L}{A} \] 6. **Relating New Resistance to Original Resistance**: We know from the original resistance that: \[ R_1 = \rho \frac{L}{A} = 5 \, \Omega \] Therefore, we can express \( R_2 \) in terms of \( R_1 \): \[ R_2 = 4 \cdot R_1 = 4 \cdot 5 \, \Omega = 20 \, \Omega \] ### Conclusion Thus, the new resistance of the wire after it has been drawn out and its cross-sectional area halved is: \[ \boxed{20 \, \Omega} \] ### Explanation of Other Options - **Option A (40Ω)**: This option is incorrect because it assumes an incorrect factor of change in resistance. The resistance increases by a factor of 4, not 8. - **Option C (10Ω)**: This option is incorrect as it underestimates the increase in resistance due to the doubling of length and halving of area. - **Option D (5Ω)**: This option is incorrect because it suggests that the resistance remains unchanged, which contradicts the effects of changing both length and area. ### Revision Summary - Resistance \( R \) is directly proportional to length \( L \) and inversely proportional to cross-sectional area \( A \). - Doubling the length of the wire and halving the area results in a resistance increase by a factor of 4. - The new resistance can be calculated using the original resistance and the changes in dimensions. - The final answer for the new resistance is \( 20 \, \Omega \).