Wire Diameter Changes Explained

When a metal wire is subjected to external forces or temperature changes, its physical dimensions, including length and diameter, can change. The question describes two identical metal wires, one stretched along its length and the other heated. Both wires end up having the same final length. We need to compare their final diameters.

Stretched Wire

When a wire is stretched along its length, its length increases. This deformation is related to the material's elastic properties. A key concept here is Poisson's ratio ($\nu$), which describes the ratio of transverse strain to axial strain.

• Axial strain ($\epsilon_{axial}$) is the change in length divided by the original length. Stretching means $\epsilon_{axial}$ is positive.
• Transverse strain ($\epsilon_{transverse}$) is the change in diameter divided by the original diameter.
• Poisson's ratio is given by $\nu = -\frac{\epsilon_{transverse}}{\epsilon_{axial}}$.

For most common materials like metals, Poisson's ratio is positive ($\nu > 0$). Since the axial strain is positive (stretching), the transverse strain must be negative. A negative transverse strain means the diameter decreases. This is often visualized as the material getting thinner as it gets longer, conserving volume (or nearly so for elastic deformation).

So, the stretched wire increases in length and decreases in diameter compared to its original dimensions.

Heated Wire

When a wire is heated, it undergoes thermal expansion. Thermal expansion causes materials to expand in all dimensions if they are free to do so. The change in length ($\Delta L$) and the change in diameter ($\Delta D$) due to a temperature change ($\Delta T$) are related to the original length ($L_0$), original diameter ($D_0$), and the coefficient of linear expansion ($\alpha$):

• $\Delta L = \alpha L_0 \Delta T$
• $\Delta D = \alpha D_0 \Delta T$

For a positive temperature change ($\Delta T > 0$), both the length and the diameter increase, assuming $\alpha$ is positive (which it is for most metals). The wire becomes longer and thicker.

Comparing Final Diameters

Both wires started identical and reached the same final length. Let's consider the original diameter $D_{original}$ and final length $L_{final}$.

• The stretched wire increased its length from $L_{original}$ to $L_{final}$ ($L_{final} > L_{original}$) by decreasing its diameter from $D_{original}$ to $D_{stretched}$ ($D_{stretched} < D_{original}$).
• The heated wire increased its length from $L_{original}$ to $L_{final}$ ($L_{final} > L_{original}$) by increasing its diameter from $D_{original}$ to $D_{heated}$ ($D_{heated} > D_{original}$).

Since the stretched wire got thinner to achieve the final length, and the heated wire got thicker while expanding to achieve the same final length, the final diameter of the heated wire must be larger than the final diameter of the stretched wire.

The heated wire expanded in all directions, making it longer and thicker. The stretched wire elongated by sacrificing its thickness. Therefore, the wire that was heated will have a larger diameter than the wire that was stretched, even though their final lengths are the same.

Based on this analysis:

• Option 1 is incorrect because diameter changes are not necessarily equal, and one decreases while the other increases.
• Option 2 is incorrect because one diameter decreases and the other increases.
• Option 4 is incorrect because the hot wire expands in diameter, making it larger.
• Option 3 correctly states that the hot wire has a larger diameter than the stretched wire.