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VITEEE 2009 Paper

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Question : 13 of 120

Marks: +1, -0

A piece of metal weighs 45g in air and 25g in a liquid of density

1.5\times10^{3}\,\mathrm{kg}\,\mathrm{m}^{-3}

kept at 30°C. When the temperature of the liquid is raised to 40°C, the metal piece weights 27g. The density of liquid at 40°C is

1.25\times10^{3}\,\mathrm{kg}\,\mathrm{m}^{-3}.

The coefficient of linear expansion of metal is

$1.3\times10^{-3}$ /°C
$5.2\times10^{-3}$ /°C
$2.6\times10^{-3}$ /°C
$0.26\times10^{-3}$ /°C

Solution:

Volume of the metal at 30°C

V_{30}=\frac{\text{loss of weight}}{\text{specific gravity} \times g}=\frac{45-25}{1.5 \times g}=13.33\,\mathrm{cm}^{3}

Similarly, volume of metal at 40°C

V_{40}=\frac{(45-27)g}{1.25 \times g}=14.40\,\mathrm{cm}^{3}

Now,

\;V_{40}=V_{30}[1+\gamma(t_{2}-t_{1})]

or

\;\gamma=\frac{V_{40}-V_{30}}{V_{30}(t_{2}-t_{1})}=\frac{14.40-13.33}{13.33(40-30)}=8.03\times10^{-3}/^{\circ}\mathrm{C}

∴ Coefficient of linear expansion of the metal

\alpha = \frac{\gamma}{3}=\frac{8.03\times10^{-3}}{3}\approx 2.6\times10^{-3}/^{\circ}\mathrm{C}

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