JEE Main 2024 — Current Electricity Question with Solution

The resistance $$R$$ of a wire is given by the formula:

$$R=\rho \frac{l}{A},$$

where:

• $$\rho$$ is the resistivity of the material,
• $$l$$ is the length of the wire,
• $$A$$ is the cross-sectional area of the wire.

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:

$$R=\rho \frac{l}{\pi r^2}.$$

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,

$$l^{\prime }=4l,$$

and the new area,

$$A^{\prime }=\pi {\left(\frac{r}{2}\right)}^2=\frac{\pi r^2}{4}.$$

Therefore, the new resistance, $$R^{\prime }$$, of the wire can be calculated using the original formula for resistance:

$$R^{\prime }=\rho \frac{l^{\prime }}{A^{\prime }}=\rho \frac{4l}{\frac{\pi r^2}{4}}=\rho \frac{4l}{\pi r^2}\cdot 4=16\rho \frac{l}{\pi r^2}=16R.$$

Hence, the new resistance of the wire is $$16R$$, which means $$x=16$$.