Question

A wire of resistance R and radius r is stretched till its radius became r/2. If new resistance of the stretched wire is x R, then value of x is _________.

Answer 1

We know that:

$R = \rho \frac{l}{A}, \quad R \propto \frac{l}{r^2}$

As we stretch the wire, its length increases, and its radius decreases, keeping the volume constant:

$V_i = V_f$

$\pi r^2 l = \pi r_f^2 l_f \implies l_f = 4l$

Now:

$\frac{R_\text{new}}{R_\text{old}} = \frac{l_f}{l} \cdot \frac{r^2}{(r/2)^2} = \frac{4l}{l} \cdot \frac{r^2}{r^2/4} = 4 \cdot 4 = 16$

$R_\text{new} = 16R$

Thus, $x = 16$.

Answer 2

We know that:

$R = \rho \frac{l}{A}, \quad R \propto \frac{l}{r^2}$

As we stretch the wire, its length increases, and its radius decreases, keeping the volume constant:

$V_i = V_f$

$\pi r^2 l = \pi r_f^2 l_f \implies l_f = 4l$

Now:

$\frac{R_\text{new}}{R_\text{old}} = \frac{l_f}{l} \cdot \frac{r^2}{(r/2)^2} = \frac{4l}{l} \cdot \frac{r^2}{r^2/4} = 4 \cdot 4 = 16$

$R_\text{new} = 16R$

Thus, $x = 16$.