Question

The resistance of a wire is $R \, \Omega.$ The wire is stretched to double its length keeping volume constant. Now the resistance of the wire will become

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

Answer 1

Answer: (A)

When wire is stretched to doubled its length, its resistance becomes four times
$R=\frac{\rho l}{A}\times \frac{l}{l}=\frac{\rho l^{2}}{V}$
$R \propto l^{2}so\frac{R_{1}}{R_{2}}=\frac{l_{1}^{2}}{l_{2}^{2}}$
$\frac{R}{R_{2}}=\frac{l^{2}}{\left(2 l\right)^{2}}$
$R_{2}=4R$