Question

Resistance of a wire is 16 ohm. By melting it, its length is stretched to half of its original length. What will be the resistance of the new wire?

Hint:

Volume stays constant, so \(R \propto L^2\). Halving the length makes \(R' = R/4\).

Answer 1

Step 1: Write the resistance formula.
For a wire of length \(L\), cross-sectional area \(A\) and resistivity \(\rho\),
\[ R = \rho\frac{L}{A} \]
Given \(R = 16\,\Omega\).

Step 2: Use conservation of volume.
When the wire is melted and reshaped, its material (and hence its volume) does not change:
\[ V = L\,A = L'\,A' \]
The new length is \(L' = \dfrac{L}{2}\).

Step 3: Find the new area.
\[ A' = \frac{L\,A}{L'} = \frac{L\,A}{L/2} = 2A \]
So the area doubles when the length is halved.

Step 4: Express resistance in terms of length only.
Since \(A = \dfrac{V}{L}\), we can write \(R = \rho\dfrac{L}{V/L} = \dfrac{\rho L^2}{V}\). For fixed volume, \(R \propto L^2\). Therefore
\[ \frac{R'}{R} = \left(\frac{L'}{L}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

Step 5: Compute the new resistance.
\[ R' = \frac{R}{4} = \frac{16}{4} = 4\,\Omega \]
\[\boxed{R' = 4\,\Omega}\]