Question

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

• A. $16\Omega$
• B. $2\Omega$
• C. $4\Omega$
• D. $8\Omega$

Answer 1

The correct option is A $16\Omega$

$R=\frac{\rho l}{A}=\frac{\rho l^2}{Al}=\frac{\rho l^2}{V}(V-\text{volume of the wire})$

Due to elongation, the length of the wire will increase, its area of cross-section will decrease, but its volume will remain constant.

$\Rightarrow R\propto l^2$

$\Rightarrow \frac{R}{R^{\prime }}=\frac{l^2}{(2l{)}^2}$

$R^{\prime }=4R$

$\therefore R^{\prime }=4\times 4=16\Omega$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.