Question

The apparatus shown in figure consists of four glass columns connected by horizontal sections. The heights of two central columns $B$ and $C$ are $49cm$ each. The two outer columns $A$ and $D$ are open to atmosphere. The columns $A$ and $C$ are maintained at a temperature of ${95}^{\circ }C$, while the column $B$ and $D$ are maintained at $5^{\circ }C$. The heights of liquid in $A$ and $D$ measured from base line are $52.8cm$ and $51cm$ respectively. Determine the coefficient of thermal expansion of the liquid.

Answer 1

Applying Pascal's law from limb $A$ to limb $D$
$P_0+{\rho }_{A}g{h}_A-{\rho }_{B}g{h}_B+{\rho }_{C}g{h}_C-{\rho }_{D}g{h}_D=P_0$
and ${\rho }_A={\rho }_C+{\rho }_2\left(say\right)$ and ${\rho }_B={\rho }_D={\rho }_1\left(say\right)$
$\therefore {\rho }_{2}g\left(52.8\right)-{\rho }_{1}g\left(49\right)+{\rho }_2\left(49\right)-{\rho }_{1}g\left(51\right)=0$
or, $\frac{{\rho }_2}{{\rho }_1}=\frac{51+49}{52.8+49}=\frac{100}{101.8}$
As density decreases with rise in temperature, ${\rho }_2={\rho }_1[1-\gamma △t)$
$\therefore \frac{{\rho }_1[1-\gamma (t_2-t_1\left)\right]}{{\rho }_1}=\frac{100}{101.8}$
or, $1-\gamma (95-5)=\frac{100}{101.8}$
$\Rightarrow \gamma =1.96\times {10}^{-4}{/}^{\circ }C$.