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

Solution

1. Given the initial resistance $R=10\Omega$ and the new length $L^{\prime }=2L$.
2. Since the volume $V=A\cdot L$ is constant, stretching the wire to twice its length reduces the cross-sectional area to $A^{\prime }=A/2$.
3. Using the resistance formula $R=\rho \frac{L}{A}$, 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 initial value, $$\begin{array}{rl}{R}^{\prime }& =4\times 10\Omega \\ & =40\Omega \end{array}$$.
5. Hence the answer is (B).