A wire of resistance $4\Omega$ is stretched to twice its original length. The resistance of stretched wire would be

Resistance of a wire,

$R=\rho \frac{l}{A}=4\Omega ---\left(i\right)$

When wire is stretched twice, its new length be ‘l’. Then $l^{\text{'}}=2l$

On stretching volume of the wire remains constant.

$$\begin{gathered} \therefore lA=l^{\text{'}}{A}^{\text{'}}where A^{\text{'}} is the new cross-sectional area \\ Or A^{\text{'}}=\frac{l}{l^{\text{'}}}A=\frac{l}{2l}A=\frac{A}{2} \end{gathered}$$

$\therefore Resis\tan ce of a stretched wire is R^{\text{'}}=\rho \frac{l^{\text{'}}}{A^{\text{'}}}=\rho \frac{2l}{\left(A/2\right)}=4\rho \frac{l}{A}$

$=4\left(4\Omega \right)=16\Omega \left(U\sin g\left(i\right)\right)$