Answer: (1)

Solution

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, $\Delta l = 0.02%$, $\Delta T = 10°$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, $\gamma = 3\alpha = 6 \times 10^{-5}$ $\therefore \rho = \frac{M}{V}$ $\frac{\Delta V}{V} \times 100 = \gamma\Delta T = (6 \times 10^{-5} \times 10 \times 100) = 6 \times 10^{-2}$ Volume increase by 0.06% therefore density decrease by 0.06%.