Question

A wire of length $l$ has a resistance $R$ . If half of the length is stretched to make the radius half of its original value, then the final resistance of the wire is

• A. $9R$
• B. $5R$
• C. $\frac{17R}{2}$
• D. $3R$

Answer 1

Answer: (C)

$R=\frac{\rho l}{A}=\frac{\rho l}{\pi r^2}$
$\pi r^2\frac{l}{2}=\pi {\left(\frac{r}{2}\right)}^2{l}^{\prime }$
$l^{\prime }=2l$
$R^{\prime }=\frac{\rho l^{\prime }}{A^{\prime }}=\frac{\rho 2l}{\pi {\frac{r}{2}}^2}=8R$
$R^{\prime \prime }=\frac{\rho \frac{l}{2}}{\pi r^2}=\frac{R}{2}$
$R_{eq}=R^{\prime }+R^{\prime \prime }=\frac{17R}{2}$