Question

What fraction of the volume of a glass flask must be filled with mercury so that the volume of the empty space may be the same at all temperatures? $\left(\alpha_{\text {glass }}=9 \times 10^{-6} /{ }^{\circ} C , \gamma_{ Hg }=18.9 \times 10^{-5} /{ }^{\circ} C \right)$

• A. $\frac{1}{2}$
• B. $\frac{1}{7}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

Answer: (B)

$\frac{1}{7}$

Answer 2

If $V$ is the volume of glass flask, $V_{L} o f$ mercury and $V_{A}$ of air in it
$V=V_{L}+V_{A}$
Now as with change in temperature volume of air remains constant, the expansion of mercury will be equal to that of the glass flask,
ie, $\Delta V=\Delta V_{L}$
or $V_{\gamma G} \Delta \theta=V H g \gamma_{H g} \Delta \theta$
( as $\Delta V=V \gamma \Delta \theta$)
or $\frac{V_{H g}}{V}=\frac{\gamma_{G}}{\gamma_{H g}}=\frac{3 \times 9 \times 10^{-6}}{18.9 \times 106-5}=\frac{1}{7}$