A glass flask is filled up to a mark with 50 cc of mercury at 18°C. If the flask and contents are heated to 38°C, how much mercury will be above the mark ? ($\mathrm{\alpha }$ for glass is  9 x 10^-6/°C and coefficient of real expansion of mercury is 0.18 x 10^-3/°C)

Volume of mercury at 18°C = V₀ = 50 cc
Volume of mercury at 38°C = (V₃₈)_r
                   $={\mathrm{V}}_0\left(1+{\mathrm{\gamma }}_{\mathrm{m}}\cdot \mathrm{\Delta t}\right)$
Volume of flask at ${38}^{\circ }\mathrm{C}={\left({\mathrm{V}}_{38}\right)}_{\mathrm{f}}$
             $={\mathrm{V}}_0\left(1+{\mathrm{\gamma }}_{\mathrm{f}}\mathrm{\Delta t}\right)$
$\therefore$Volume of mercury at 38°C above the mark 
           $\begin{array}{l}={\mathrm{V}}_0\left(1+{\mathrm{\gamma }}_{\mathrm{m}}\mathrm{\Delta t}\right)-{\mathrm{V}}_0\left(1+{\mathrm{\gamma }}_{\mathrm{f}}\mathrm{\Delta t}\right)\\ ={\mathrm{V}}_0\left({\mathrm{\gamma }}_{\mathrm{m}}-{\mathrm{\gamma }}_{\mathrm{f}}\right)\mathrm{\Delta t}\\ =50\left[0.18\times {10}^{-3}-3\times 9\times {10}^{-6}\right](38-18)\\ =0\cdot 153 \mathrm{cc}\end{array}$