When the temperature of a metal wire is increased from 0oC to 10oC, its length increases by 0.02%. The percentage change in its mass density will be closest to:
Option: 1 3
Option: 2 0.06
Option: 3 0.36
Option: 4 300

Answers (1)

\begin{array}{l} \Delta \ell=\ell \alpha \Delta t \\ \\ \alpha=\frac{\Delta \ell}{\ell \Delta T}=\frac{0.02}{100 \times 10} ; \quad \alpha=2 \times 10^{-5} \end{array}

Now

\rho=\frac{\mathrm{M}}{\mathrm{V}} \Rightarrow\left|\frac{\Delta \rho}{\rho} \times 100\right|= \left|\frac{\Delta V}{V} \times 100\right|

\begin{array}{l} As \ \frac{\Delta V}{V}=\gamma \times \Delta T ; \\ \Rightarrow \ Using \ \gamma=3 \alpha=6 \times 10^{-5} \\ \Rightarrow \quad \frac{\Delta V}{V} \times 100=\left(6 \times 10^{-5} \times 10 \times 100\right)=6 \times 10^{-2} \end{array}

So density decrease by 0.06 \%

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