Question

A wire of resistance $100\Omega$ is stretched so that its radius becomes half the initial value. What will be the new resistance of the wire?

• A. $400\Omega$
• B. $800\Omega$
• C. $1200\Omega$
• D. $1600\Omega$

Answer 1

The correct option is D $1600\Omega$

Resistance of the wire is,

$R=\frac{\rho l}{A}=\frac{\rho lA}{A^2}=\frac{\rho V}{(\pi r^2{)}^2}$

$\because \rho \text{and}V\text{(volume) are constant}$

$\Rightarrow R\propto \frac{1}{r^4}$

$\Rightarrow \frac{R^{{}^{\prime }}}{R}={\left(\frac{r}{r^{{}^{\prime }}}\right)}^4[\because r^{{}^{\prime }}=\frac{r}{2}]$

$\Rightarrow \frac{R^{{}^{\prime }}}{R}={\left[\frac{r}{\left(\frac{r}{2}\right)}\right]}^4=2^4$

$R^{{}^{\prime }}=16\times R$

$R^{{}^{\prime }}=16\times 100=1600\Omega$

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