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 ________.

<p>The resistance $R$ of a wire is given by the formula:</p> $R = \rho \frac{l}{A},$ <p>where:</p> <ul> <li>$\rho$ is the resistivity of the material,</li> <li>$l$ is the length of the wire,</li> <li>$A$ is the cross-sectional area of the wire.</li> </ul> <p>If we have a cylindrical wire, the cross-sectional area can be expressed as $A = \pi r^2$, where $r$ is the radius of the cylinder. Therefore, the resistance of the original wire can be written as:</p> $R = \rho \frac{l}{\pi r^2}.$ <p>When the wire is stretched such that its radius becomes $r / 2$, its volume would remain constant, given that the volume of a cylinder is $V = A \cdot l = \pi r^2 \cdot l$. Assuming the volume before and after the stretching is the same, and since the area is now a quarter of the original (because when the radius is halved, the area, which is proportional to the square of the radius, is reduced to a quarter), the length must have increased to four times the original to preserve the volume. That is,</p> $l' = 4l,$ <p>and the new area,</p> $A' = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4}.$ <p>Therefore, the new resistance, $R'$, of the wire can be calculated using the original formula for resistance:</p> $$ R' = \rho \frac{l'}{A'} = \rho \frac{4l}{\frac{\pi r^2}{4}} = \rho \frac{4l}{\pi r^2} \cdot 4 = 16 \rho \frac{l}{\pi r^2} = 16R. $$ <p>Hence, the new resistance of the wire is $16R$, which means $x = 16$.</p>