Question

If a wire of resistance 1 $\Omega$ is stretched to double its length, then the resistance will become

• A. $\frac{1}{2}\Omega$
• B. $2\Omega$
• C. $\frac{1}{4}\Omega$
• D. $4\Omega$

Answer 1

The correct option is D $4\Omega$

Given : $L_2=2L_1$ $R=1\Omega$

As volume of the wire is constant, thus $A_1{L}_1=A_2{L}_2$

$\therefore$ $A_1{L}_2=A_2\left(2L_1\right)$ $⟹A_1=2A_2$

Initial resistance of the wire $R=\rho \frac{L_1}{A_1}$

Final resistance of the wire $R^{\prime }=\rho \frac{L_2}{A_2}=\rho \frac{2L_1}{A_1/2}=4R=4\Omega$

Thus resistance will become $4\Omega$.