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: Substitute the new dimensions directly.
Original resistance \(R = \rho\dfrac{L}{A} = 16\,\Omega\). New wire has length \(L' = L/2\) and, by volume conservation \(LA = L'A'\), area \(A' = 2A\).

Step 2: Plug into the resistance formula.
\(R' = \rho\dfrac{L'}{A'} = \rho\dfrac{L/2}{2A} = \rho\dfrac{L}{4A}\).

Step 3: Relate to the original resistance.
Notice \(\rho\dfrac{L}{4A} = \dfrac{1}{4}\left(\rho\dfrac{L}{A}\right) = \dfrac{R}{4}\).

Step 4: Insert the number.
\(R' = \dfrac{16}{4} = 4\,\Omega\).

Step 5: Sanity check.
Shortening the wire lowers \(L\) and thickening it raises \(A\); both changes reduce resistance, so a value below \(16\,\Omega\) is expected, and \(4\,\Omega\) fits.
\[\boxed{R' = 4\,\Omega}\]