1. Given data

• Temperature increases from $0^\circ C$ to $10^\circ C$, so $\Delta T = 10^\circ C$
• Increase in length = $0.02\%$

Thus, $\frac{\Delta L}{L} = 0.02\% = 0.0002$

2. Find coefficient of linear expansion

Using $\frac{\Delta L}{L} = \alpha \Delta T$ we get $\alpha = \frac{0.0002}{10} = 2 \times 10^{-5}\, ^\circ C^{-1}$

3. Find volume expansion

For an isotropic metal, $\beta = 3\alpha = 6 \times 10^{-5}\, ^\circ C^{-1}$

Hence fractional change in volume for $10^\circ C$ rise is $\frac{\Delta V}{V} = \beta \Delta T = 6 \times 10^{-5} \times 10 = 6 \times 10^{-4}$

In percentage, $\frac{\Delta V}{V} = 0.06\%$

4. Relate density with volume

Density is $\rho = \frac{m}{V}$ Since mass of the wire does not change, density varies inversely with volume.

So, $\frac{\Delta \rho}{\rho} \approx -\frac{\Delta V}{V}$

Therefore, $\frac{\Delta \rho}{\rho} = -6 \times 10^{-4}$

So the percentage decrease in density is $0.06\%$

5. Match with options

The closest value is:

• Option B: $0.06$

Therefore, the density decreases by about $0.06\%$.