Thank you for reporting, we will resolve it shortly
Q.
When the temperature of a metal wire is increased from $0^{\circ} C$ to $10^{\circ} C$, its length increases by $0.02 \% .$ The percentage change in its mass density will be closest to:
Given $\frac{\Delta L }{ L }=0.02 \%$
$\therefore \Delta L = L \alpha \Delta T \Rightarrow \frac{\Delta L }{ L }=\alpha \Delta T =0.02 \%$
$\therefore \beta=2 \alpha$ (Areal coefficient of expansion)
$\Rightarrow \beta \Delta T =2 \alpha \Delta T =0.04 \%$
Volume $=$ Area $\times$ Length
Density $(\rho)=\frac{\text { Mass }}{\text { Volume }}=\frac{\text { Mass }}{\text { Area } \times \text { Length }}=\frac{ M }{ AL }$
$\Rightarrow \frac{\Delta \rho}{\rho}=\frac{\Delta M}{M}-\frac{\Delta A }{ A }-\frac{\Delta L }{ L }$ (Mass remains constant)
$\Rightarrow \left(\frac{\Delta \rho}{\rho}\right)=\frac{\Delta A }{ A }+\frac{\Delta L }{ L }=\beta \Delta T +\alpha \Delta T$
$=0.04 \%+0.02 \%$
$=0.06 \%$