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. $\frac{17 R}{2}$
• D. $3R$

Answer 1

Answer: (C)

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