When the temperature of a metal wire is increased from 0$$°$$C to 10$$°$$C, its length increases by 0.02%. The percentage change in its mass density will be closed to:

We are told that when the temperature of a metal wire rises from $$0^{\circ}\text C$$ to $$10^{\circ}\text C$$ its length increases by $$0.02\%$$. First, let us convert this percentage increase in length into a fractional form because all expansion formulae use fractional changes.

$$0.02\% = \frac{0.02}{100}=0.0002$$

So the fractional change in length is

$$\frac{\Delta L}{L}=0.0002$$

The formula for linear (one-dimensional) thermal expansion is first stated:

$$\frac{\Delta L}{L}=\alpha\,\Delta T$$

where $$\alpha$$ is the coefficient of linear expansion and $$\Delta T$$ is the rise in temperature. We substitute the known values:

$$0.0002=\alpha\,(10^{\circ}\text C)$$

Solving for $$\alpha$$, we divide both sides by $$10$$:

$$\alpha=\frac{0.0002}{10}=2\times10^{-5}\;{}^{\circ}\text C^{-1}$$

Now we turn to the volume change. For an isotropic solid, the relation between the coefficient of volume expansion $$\beta$$ and the linear coefficient $$\alpha$$ is stated:

$$\beta = 3\alpha$$

Therefore, substituting $$\alpha = 2\times10^{-5}\;{}^{\circ}\text C^{-1}$$, we get

$$\beta = 3 \times 2\times10^{-5}=6\times10^{-5}\;{}^{\circ}\text C^{-1}$$

The fractional change in volume is then

$$\frac{\Delta V}{V}=\beta\,\Delta T = (6\times10^{-5})(10)=6\times10^{-4}$$

To convert this fractional change to a percentage, we multiply by $$100$$:

$$6\times10^{-4}\times100 = 0.06\%$$

Thus, the volume increases by $$0.06\%$$ when the temperature rises from $$0^{\circ}\text C$$ to $$10^{\circ}\text C$$.

Because the mass of the wire does not change, density $$\rho$$ varies inversely with volume. The relation is written as

$$\rho = \frac{M}{V}\quad\Longrightarrow\quad \frac{\Delta\rho}{\rho} = -\frac{\Delta V}{V}$$

So the fractional change in density is the negative of the fractional change in volume. Hence

$$\frac{\Delta\rho}{\rho}= -6\times10^{-4}$$

In percentage terms this is

$$-6\times10^{-4}\times100 = -0.06\%$$

The negative sign merely tells us the density decreases; the magnitude of the percentage change is $$0.06\%$$.

Hence, the correct answer is Option A.