Question

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:

• A. $0.008$
• B. $0.06$
• C. $0.8$
• D. $2.3$

Answer 1

Answer: (B)

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