A copper wire of resistance $R=10\Omega$ is stretched to double its original length. What is the new resistance of the wire?

Solution

1. The resistance of a wire is given by $R=\rho \frac{L}{A}$.
2. When length $L$ becomes $2L$, the area $A$ becomes $A/2$ to keep volume constant.
3. The new resistance $$\begin{array}{rl}{R}^{\prime }& =\rho \frac{2L}{A/2}\\ & =4\rho \frac{L}{A}\\ & =4R\end{array}$$.
4. Substituting $R=10\Omega$, we get $$\begin{array}{rl}{R}^{\prime }& =4\times 10\\ & =40\Omega \end{array}$$.
5. Hence the answer is (B).