When the temperature of a metal wire is increased from

(1) Change in length of the metal wire $(\Delta l)$ when its temperature is changed by $\Delta T$ is given by

$\Delta l=l \alpha \Delta T$

Here, $\alpha=$ Coefficient of linear expansion

Here, $\Delta l=0.02 \%, \Delta T=10^{\circ} \mathrm{C}$

$\therefore \alpha=\frac{\Delta l}{l \Delta T}=\frac{0.02}{100 \times 10} \Rightarrow \alpha=2 \times 10^{-5}$

Volume coefficient of expansion, $\gamma=3 \alpha=6 \times 10^{-5}$

$\because \rho=\frac{M}{V}$

$\frac{\Delta V}{V} \times 100=\gamma \Delta T=\left(6 \times 10^{-5} \times 10 \times 100\right)=6 \times 10^{-2}$

Volume increase by $0.06 \%$ therefore density decrease by $0.06 \%$