The density of a substance at $20^{\circ} C$ and $120^{\circ} C$ are $1.1 g c . c^{- 1}$ and $1 g c . c^{- 1}$ respecttively. Find the coefficient of cubical expansion of the solid.

$0.001 /^{\circ} C$

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$0.002 /^{\circ} C$

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0.003 /^{\circ} C

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$0.004 /^{\circ} C$

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Solution

The correct option is A
$0.001 /^{\circ} C$

Given: Initial density, $d_{1} = 1.1 g c . c^{- 1}$
Final density, $d_{2} = 1 g c . c^{- 1}$
Initial tempereature, $t_{1} = 20^{\circ} C$
Final temperature, $t_{2} = 120^{\circ} C$

Volumetric expansion of solids can be expressed as,
$V_{2} = V_{1} (1 + γ (t_{2} - t_{1}))$
where, $V_{1}$ is the initial volume, $V_{2}$ is the final volume, $t_{1}$ is the initial temperature, $t_{2}$ is the final temperature and $γ$ is the coefficient of cubical expansion of solid.
We know $v o l u m e = \frac{m a s s}{d e n s i t y}$
Let $d_{1}$ be the density at volume $V_{1}$ and $d_{2}$ be the density at volume $V_{2}$
Since mass of the substance, $m$ remains same, we have,
or, $\frac{m}{d_{2}} = \frac{m}{d_{1}} (1 + γ (t_{2} - t_{1}))$
or, $γ = \frac{d_{1} - d_{2}}{d_{2} (t_{2} - t_{1})}$
or, $γ = \frac{1.1 - 1}{1 (120 - 20)} = \frac{0.1}{100} = 0.001$
$∴ γ = {0.001}^{\circ} C^{- 1}$

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The density of a substance at 20∘C and 120∘C are 1.1g c.c−1 and 1 g c.c−1 respecttively. Find the coefficient of cubical expansion of the solid.

The density of a substance at $20^{\circ} C$ and $120^{\circ} C$ are $1.1 g c . c^{- 1}$ and $1 g c . c^{- 1}$ respecttively. Find the coefficient of cubical expansion of the solid.