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$A wire of resistance R = 10 Ω is stretched to twice its original length. What will be the new resistance of the wire?$

A. $20 Ω$
B. $40 Ω$
C. $10 Ω$
D. $5 Ω$

Show correct answer & step-by-step solution

Correct answer: B — $40 Ω$

Solution

Since volume V = A \cdot L is constant, A^{'} = A / n when L^{'} = n L . Resistance R = ρ L / A becomes R^{'} = ρ (n L) / (A / n) = n^{2} R . For n = 2, R^{'} = 4 \times 10 = 40 Ω .

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