A uniform wire (Young's modulus $$2 \times 10^{11}$$ Nm$$^{-2}$$) is subjected to longitudinal tensile stress of $$5 \times 10^7$$ Nm$$^{-2}$$. If the overall volume change in the wire is 0.02%, the fractional decrease in the radius of the wire is close to:

A uniform wire is subjected to a longitudinal tensile stress, and we are given Young's modulus $$Y = 2 \times 10^{11}$$ N/m² and stress $$\sigma = 5 \times 10^7$$ N/m². The overall volume change is 0.02%, which means $$\frac{\Delta V}{V} = \frac{0.02}{100} = 0.0002$$. We need to find the fractional decrease in the radius, which is $$\left| \frac{\Delta r}{r} \right|$$.

When a wire is stretched, the volume change relates to Poisson's ratio ($$\nu$$) and the longitudinal strain. The formula for volume change is:

$$\frac{\Delta V}{V} = (1 - 2\nu) \cdot \epsilon_{\text{long}}$$

where $$\epsilon_{\text{long}}$$ is the longitudinal strain. Using Hooke's law, $$\sigma = Y \cdot \epsilon_{\text{long}}$$, so $$\epsilon_{\text{long}} = \frac{\sigma}{Y}$$.

First, calculate $$\epsilon_{\text{long}}$$:

$$\epsilon_{\text{long}} = \frac{\sigma}{Y} = \frac{5 \times 10^7}{2 \times 10^{11}} = \frac{5}{2} \times 10^{7-11} = 2.5 \times 10^{-4}$$

Now substitute into the volume change formula:

$$\frac{\Delta V}{V} = (1 - 2\nu) \cdot \epsilon_{\text{long}}$$

$$0.0002 = (1 - 2\nu) \cdot (2.5 \times 10^{-4})$$

Solve for $$1 - 2\nu$$:

$$1 - 2\nu = \frac{0.0002}{2.5 \times 10^{-4}} = \frac{2 \times 10^{-4}}{2.5 \times 10^{-4}} = \frac{2}{2.5} = 0.8$$

Now solve for $$\nu$$:

$$1 - 2\nu = 0.8$$

$$-2\nu = 0.8 - 1 = -0.2$$

$$\nu = \frac{-0.2}{-2} = 0.1$$

Poisson's ratio $$\nu$$ is 0.1. The fractional decrease in radius is given by the magnitude of the lateral strain. Poisson's ratio relates lateral strain ($$\epsilon_{\text{lat}}$$) to longitudinal strain:

$$\nu = -\frac{\epsilon_{\text{lat}}}{\epsilon_{\text{long}}}$$

So,

$$\epsilon_{\text{lat}} = -\nu \cdot \epsilon_{\text{long}}$$

The fractional change in radius is $$\frac{\Delta r}{r} = \epsilon_{\text{lat}}$$, so the fractional decrease is $$\left| \frac{\Delta r}{r} \right| = \nu \cdot \epsilon_{\text{long}}$$.

Substitute the values:

$$\left| \frac{\Delta r}{r} \right| = \nu \cdot \epsilon_{\text{long}} = 0.1 \cdot (2.5 \times 10^{-4}) = 0.25 \times 10^{-4}$$

This matches option C. Therefore, the fractional decrease in the radius is $$0.25 \times 10^{-4}$$.

Hence, the correct answer is Option C.