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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:
$0.008$
$0.06$
$0.8$
$2.3$
Solution
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)
$\Rightarrow\left(\frac{\Delta \rho}{\rho}\right)=\frac{\Delta A }{ A }+\frac{\Delta L }{ L }=\beta \Delta T +\alpha \Delta T$
$=0.04 \%+0.02 \%$
$=0.06 \%$
