Question

A given wire of residence $1\Omega$ is stretched to double its length. What will be its new resistance.

Answer 1

Given resistance$=1\Omega$ and initial length =L and final length =2 L.

Let Initial area$A$ and final area$A_f=\frac{A}{2}$

Though the length and cross sectional area of the wire changes it volume will remain constant.

$V_i=V_f$

Thus initial resistance$R_i=\rho \frac{L}{A}$--------(1)

Final resistance $R_f=\rho \frac{2L}{\frac{A}{2}}$--------(2)

Computing 1 and 2 we get

$R_f=4R_i$

$R_f=4\Omega$