Question

A wire of radius $r$ has resistance $R$. If it is stretched to a wire of radius $\frac{r}{2},$ then the resistance becomes

• A. $2R$
• B. $4R$
• C. $16R$
• D. zero

Answer 1

The correct option is C $16R$

Resistance of a conducting wire:
$R=\frac{\rho l}{A}...\left(i\right)$

As the volume of the wire remains constant.
$V_i=V_f$
$\pi r_1^2{l}_1=\pi r_2^2{l}_2$
$\left[\frac{l_2}{l_1}\right]={\left[\frac{r_1}{r_2}\right]}^2={\left[\frac{r}{\frac{r}{2}}\right]}^2=4$
$\frac{A_1}{A_2}=\frac{\pi r_1^2}{\pi r_2^2}=4$

Now, from equatuion (i)
$\frac{R_2}{R_1}=\left[\frac{l_2}{l_1}\right]\left[\frac{A_1}{A_2}\right]$

$\frac{R_2}{R_1}=\left[\frac{4}{1}\right]\left[4\right]=16$

$R_2=16R$