The resistance of a uniform wire of length $L$ and area $A$ is $R$. If it is stretched to double its length (volume constant), the new resistance is:
Solution
Volume fixed: $L \to 2L$, $A \to A/2$. $R' = \rho L'/A' = 4R$.
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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