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Thermodynamics Part 2

Section: Physics

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Question : 52 of 100

Marks: +1, -0

When the temperature of a metal wire isincreased from

0^{\circ}\mathrm{C}

10^{\circ}\mathrm{C},

its length increasesby

0.02\ \%.

The percentage change in its massdensity will be closest to:

[2 Sep 2020 Shift 2]

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\ \%

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