Question

A beaker of volume $140cc$. contains mercury so that the volume of the empty space above mercury is constant at all temperatures. What is the volume of mercury in cc in the beaker, if the linear coefficient of expansion of glass is $0.000009K^{-1}$and the volume coefficient of mercury is $0.000189K^{-1}$?.

Answer 1

Volume of beaker at $t_1^{\circ }C=V_1$
volume of mercury at $t_2^{\circ }C=V_2$
$\therefore$ volume of empty space at $t_1^{\circ }C=V_1-V_2$
On heating to $t_2^{\circ }C$ , volume of beaker $=V_1\{1+{\gamma }_1(t_2-t_1\left)\right\}$
volume of mercury $=V_2\{1+{\gamma }_2(t_2-t_1\left)\right\}$
$\therefore$ volume of empty space at $t_2^{\circ }C$
$=V_1\{1+{\gamma }_1(t_2-t_1\left)\right\}-V_2\{1+{\gamma }_2(t_2-t_1\left)\right\}=V_1-V_2+(V_1{\gamma }_1-V_2{\gamma }_2)(t_2-t_1)$
Since the volume of empty space must be the same at all temperatures, $V_1-V_2=V_1-V_2+(V_1{\gamma }_1-V_2{\gamma }_2)(t_2-t_1)$
$\therefore V_1{\gamma }_1-V_2{\gamma }_2=0$
$\therefore V_1{\gamma }_1=V_2{\gamma }_2\Rightarrow 140\times 3\times \alpha =V_2\times 0.000189$
$\therefore V_2=\frac{140\times 3\times 0.000009}{0.000189}=20cc$
Volume of mercury in beaker $=20cc$

Why this question? The volume of the beaker and liquid changes, inspite the difference in the volumes does not!