Let's start by understanding the basic relationships in play when a force is applied to rods within the elastic limit. The extension ($$\Delta L$$) seen in the rod when a tensile force is applied can be described by Hooke's law for elastic deformation, which is given by:

$$\Delta L = \frac{FL}{AE}$$

where:

$$F$$ is the force applied,
$$L$$ is the original length of the rod,
$$A$$ is the cross-sectional area of the rod,
$$E$$ is the Young's modulus of the material (a property of the material).

Given that both rods are of the same material, the value of $$E$$ is constant for both. Additionally, since both rods are subjected to the same tensile force ($$T$$), $$F$$ is constant as well. The question tells us that the volume of both rods is the same, and the diameter of the first rod is half that of the second rod. Let's denote the diameter of the first rod as $$D$$ and that of the second rod as $$2D$$.

The cross-sectional area of a rod with a circular cross-section is given by $$A = \pi \left(\frac{D}{2}\right)^2$$. So, for the first rod, the area $$A_1$$ is:

$$A_1 = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4}$$

And for the second rod, the area $$A_2$$ is:

$$A_2 = \pi \left(\frac{2D}{2}\right)^2 = \pi D^2$$

Since the volume of both rods is the same and they are made up of the same material, their volume is $$V = AL$$, indicating that their length and area are inversely proportional. If the diameter of the second rod is twice that of the first, its cross-sectional area is quadruple the first rod's area, which implies that the length of the first rod is four times that of the second to maintain the same volume for both rods.

Therefore, the length of the first rod $$L_1 = 4L_2$$, where $$L_2$$ is the length of the second rod.

Substituting these values back into the extension formula and comparing the extensions of the first and second rods, we get:

$$\Delta L_1 = \frac{T \times 4L_2}{\frac{\pi D^2}{4}E} = \frac{16TL_2}{\pi D^2E}$$

$$\Delta L_2 = \frac{T \times L_2}{\pi D^2E}$$

Comparing $$\Delta L_1$$ to $$\Delta L_2$$, we find:

$$\frac{\Delta L_1}{\Delta L_2} = \frac{\frac{16TL_2}{\pi D^2E}}{\frac{TL_2}{\pi D^2E}} = 16$$

Therefore, the correct ratio of extension of the first rod to the second rod is $$16:1$$, which corresponds to Option C.

Elasticity

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