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A copper wire of length $L$ and area $A$ has resistance $R$. If length is doubled and area halved, the new resistance is:
A$4R$, since $R = \rho L/A$ scales by 2 and by 2
B$R$, since the two changes cancel each other
C$R/4$, the inverse of the correct factor here
D$2R$, only the length doubling matters here
Answer & Solution
Correct answer: A. $4R$, since $R = \rho L/A$ scales by 2 and by 2
$R' = \rho(2L)/(A/2) = 4\rho L/A = 4R$.
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