QJEEM24Pha8ab: JEE Main 2024 Physics - Class 12 JEEM24Pha8ab Solution: A wire of resistance and radius is stretched till its radius…

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

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:

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,

and the new area,

Therefore, the new resistance, $R'$ , of the wire can be calculated using the original formula for resistance:

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