Concept:When a wire is mechanically compressed or stretched, its volume remains constant. Therefore, a change in radius forces a corresponding change in length.
Formula:$$ \text{Volume} = A \cdot L = \text{constant} $$
$$ R = \rho \frac{L}{A} $$
Solution:- Original Area \( A = \pi r^2 \).
- New radius \( r' = 2r \). New Area \( A' = \pi (2r)^2 = 4\pi r^2 = 4A \).
- Since Volume is constant: \( A \cdot L = A' \cdot L' \implies L' = \frac{A \cdot L}{4A} = \frac{L}{4} \). The wire is 4 times thicker and 4 times shorter.
- New Resistance \( R' = \rho \frac{L'}{A'} = \rho \frac{L/4}{4A} = \frac{1}{16} \rho \frac{L}{A} = \frac{1}{16} R \).
Why other options are incorrect:- Opt A: Occurs if you stretch the wire (making it thinner) rather than compress it.
- Opt D: Occurs if you forget that length also shrinks by a factor of 4.
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