A) \[18\,mm\]
B) \[9\,mm\]
C) \[18\,cm\]
D) \[9\,cm\]
Correct Answer: B
Solution :
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, \[\Delta l=\frac{0.0018\times {{10}^{-6}}}{2\times {{10}^{-7}}}\] \[=0.009m-9\,m\,m\]
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