Solve this following


When the temperature of a metal wire is increased from $0^{\circ} \mathrm{C}$ to $10^{\circ} \mathrm{C}$, its length increases by $0.02 \%$. The percentage change in its mass density will be closest to:


  1. $0.008$

  2. $0.06$

  3. $0.8$

  4. $2.3$

Correct Option: , 2


Given $\frac{\Delta \mathrm{L}}{\mathrm{L}}=0.02 \%$

$\therefore \Delta \mathrm{L}=\mathrm{L} \alpha \Delta \mathrm{T} \Rightarrow \frac{\Delta \mathrm{L}}{\mathrm{L}}=\alpha \Delta \mathrm{T}=0.02 \%$

$\therefore \beta=2 \alpha$ (Areal coefficient of expansion)

$\Rightarrow \beta \Delta \mathrm{T}=2 \alpha \Delta \mathrm{T}=0.04 \%$

Volume $=$ Area $\times$ Length

$\operatorname{Density}(\rho)=\frac{\text { Mass }}{\text { Volume }}=\frac{\text { Mass }}{\text { Area } \times \text { Length }}=\frac{\mathrm{M}}{\mathrm{AL}}$

$\Rightarrow\left(\frac{\Delta \rho}{\rho}\right)=\frac{\Delta \mathrm{A}}{\mathrm{A}}+\frac{\Delta \mathrm{L}}{\mathrm{L}}=\beta \Delta \mathrm{T}+\alpha \Delta \mathrm{T}$

$=0.04 \%+0.02 \%$

$=0.06 \%$


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