Question

A wire of resistance $4\Omega$ is stretched to twice its original length. In the process of stretching, its area of cross section gets halved. Now, the resistance of the wire is:

• A. $8\Omega$
• B. $16\Omega$
• C. $1\Omega$
• D. $4\Omega$

Answer 1

The correct option is B $16\Omega$

Let length and cross-section area of the wire be $l$ and $A$ respectively.

$\therefore$ Resistance $R=\rho \frac{l}{A}$ where $\rho$ is the resistivity of the material

Given : $R=4\Omega$

For $l^{\prime }=2l$ and $A^{\prime }=\frac{A}{2}$, resistance of the wire be $R^{\prime }$

$\therefore$ $R^{\prime }=\rho \frac{l^{\prime }}{A^{\prime }}=\rho \frac{2l}{A/2}=4\rho \frac{l}{A}=4R$

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