Question

A metal wire of resistance 8 Ω is stretched so that its length is increased 4 times the original length. What is the new resistance of the wire?

Answer 1

Given: Original resistance R = 8 Ω

Let original length [l] and original area of cross-section [A].

When the wire is stretched, its length becomes \[l' = 4l\].

Since the volume of the wire remains constant, \[l'A' = lA \Rightarrow A' = \frac{A}{4}\]

Formula: For a wire

\[R = \rho \frac{l}{A}\]

Therefore, new resistance \[R' = \rho \frac{l'}{A'}\]

\[= \rho \frac{4l}{A/4}\]

\[= 16 \rho \frac{l}{A}\]

= 16 R

= 16 × 8 Ω

= 128 Ω

The new resistance of the wire is 128 Ω.