Question

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

• A. ${0.001}^{\circ }{C}^{-1}$
• B. ${0.002}^{\circ }{C}^{-1}$
• C. ${0.003}^{\circ }{C}^{-1}$
• D. ${0.004}^{\circ }{C}^{-1}$

Answer 1

The correct option is A

${0.001}^{\circ }{C}^{-1}$

Given: Initial density, $d_1=1.1gc.c^{-1}$
Final density, $d_2=1gc.c^{-1}$
Initial tempereature, $t_1={20}^{\circ }C$
Final temperature, $t_2={120}^{\circ }C$

Volumetric expansion of solids can be expressed as,
$V_2=V_1(1+\gamma (t_2-t_1\left)\right)$
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 $\gamma$ is the coefficient of cubical expansion of solid.
We know $volume=\frac{mass}{density}$
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, $\frac{m}{d_2}=\frac{m}{d_1}(1+\gamma (t_2-t_1\left)\right)$
or, $\gamma =\frac{d_1-d_2}{d_2(t_2-t_1)}$
or, $\gamma =\frac{1.1-1}{1(120-20)}=\frac{0.1}{100}=0.001$
$\therefore \gamma ={0.001}^{\circ }{C}^{-1}$