• # question_answer The coefficient of volumetric expansion of mercury is$1.8\times {{10}^{-5}}^{o}C$. The thermometer bulb has a velocity of ${{10}^{-6}}{{m}^{3}}$ and cross-section of stern$0.002\,c{{m}^{2}}$. Assuming that bulb is filled with mercury at ${{0}^{o}}C,$ the length of mercury column at ${{100}^{o}}C$ will be: A)  $18\,mm$                  B)  $9\,mm$C)  $18\,cm$                   D)  $9\,cm$

Coefficient of volumetric expansion $(\gamma )$ $=1.8\times {{10}^{-5}}{{/}^{o}}C$ Initial volume $V={{10}^{-6}}{{m}^{3}}$ Area of cross section                         $A=0.02c{{m}^{2}}=2\times {{10}^{-7}}\,c{{m}^{2}}$ Initial temperature ${{T}_{1}}={{0}^{o}}C$ Final temperature ${{T}_{2}}={{100}^{o}}C$ The final volume is $V'=V\left[ 1+\gamma ({{T}_{2}}-{{T}_{1}}) \right]$                 $={{10}^{-6}}\,[1+1.8\times {{10}^{-5}}(100-0)]$                 $=1.0018\times {{10}^{-6}}$ Change in volume $\Delta V=A\times \Delta l=V'-V$ (Where $\Delta l$ is the length of mercury column) $\Rightarrow$ $(2\times {{10}^{-7}})\times \Delta l=1.0018\times {{10}^{-6}}-{{10}^{-6}}$ Hence,         $\Delta l=\frac{0.0018\times {{10}^{-6}}}{2\times {{10}^{-7}}}$                 $=0.009m-9\,m\,m$