Concept:Changing the physical dimensions of a wire alters its resistance, which inversely affects current for a constant voltage.
Formula:$$ R = \rho \frac{l}{\pi r^2} \quad \text{and} \quad I = \frac{V}{R} $$
Solution:- Original Resistance: \( R = \rho \frac{l}{\pi r^2} \).
- New length \( l' = 2l \). New radius \( r' = 2r \), meaning new area \( A' = \pi (2r)^2 = 4\pi r^2 \).
- New Resistance: \( R' = \rho \frac{2l}{4\pi r^2} = \frac{1}{2} \left( \rho \frac{l}{\pi r^2} \right) = \frac{R}{2} \).
- Since voltage is constant, \( I' = \frac{V}{R'} = \frac{V}{R/2} = 2 \left( \frac{V}{R} \right) = 2I \).
Why other options are incorrect:- Opt A, B, D: Arise from failing to square the radius when calculating the new area, or directly applying the resistance ratio to current without taking the inverse.
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