AI-generated practice question — model-generated for extra practice, not a previous-year CBSE board question.
(D) 4R
When length is doubled, area of cross-section is halved (volume is constant). Using $R = \rho \frac{l}{A}$, new resistance $= \rho \frac{2l}{A/2} = 4\rho\frac{l}{A} = 4R$.
Source: Chapter 11, Section 11.5
The key idea is volume conservation: stretching doubles the length but halves the cross-sectional area. Since $R \propto \frac{l}{A}$, both changes multiply the resistance — doubling $l$ gives ×2 and halving $A$ gives another ×2, so the net effect is ×4. Always remember to account for the change in both $l$ and $A$ when a wire is stretched.