A 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 $R=\rho \frac{L}{A}$, where $L$ is length and $A$ is cross-sectional area.
2. Since the volume $V=A\cdot L$ remains constant, doubling the length to $L^{\prime }=2L$ forces the area to become $A^{\prime }=\frac{A}{2}$.
3. The new resistance is $$\begin{array}{rl}{R}^{\prime }& =\rho \frac{2L}{A/2}\\ & =4\left(\rho \frac{L}{A}\right)\\ & =4R\end{array}$$.
4. Substituting the given value $R=10\Omega$, we find $$\begin{array}{rl}{R}^{\prime }& =4\times 10\Omega \\ & =40\Omega \end{array}$$.
5. Hence the answer is (C).