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A piece of metal weighs 46g in air. When immersed in a liquid of specific gravity 1.24 at $27^{o} C$ it weighs 30g. When the temperature of liquid is raised to $42^{o} C$ the metal piece weighs 30.5g. Specific gravity of liquid at $42^{o} C$ is 1.20. The coefficient of linear expansion of metal is $_____\times 10^{- 5} / ° C$ .

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Detailed Solution

Low of weight of metal piece in liquid = $46 - 30 = 16 g$
Weight of liquid displaced = volume of liquid displaced density
$16 = V_{1} \times 1.24$
$V_{1} = \frac{16}{1.24} c m^{3}$
Similarly
$V_{2} = V_{1} (1 + γ + θ)$

$= V_{1} (1 + γ + 15)$

$1 + 15 γ = \frac{V_{2}}{V_{1}} = \frac{\frac{15.5}{1.20}}{\frac{16}{1.24}} = 1.0010$

$γ = 6.7 \times 10^{- 5} /^{o} C$

$α = 2.23 \times 10^{- 5} /^{o} C$

$2.23 \times 10^{- 5} /^{o} C$

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