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When a wire is drawn until its radius decrease by 3%. Then percentage of increase in resistance is

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10%

12%

answer is D.

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Detailed Solution

$\begin{array}{l} R \propto \frac{1}{A^{2}} α \frac{1}{r^{4}} \\ \frac{ΔR}{R} \times 100 = 4 (\frac{Δr}{r} \times 100) \end{array}$
= 4(3) = 12%

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