A) \[\gamma =\alpha \]
B) \[\gamma =2\alpha \]
C) \[\gamma =3\alpha \]
D) \[\gamma =0\]
Correct Answer: B
Solution :
[b] Let \[{{A}_{0}}\] and At be the areas of cross-section of the tube at temperature 0 and t respectively, 1 = length of the liquid column (constant) \[{{V}_{0}}\] and \[{{V}_{t}}\] be the volumes of the liquid at temperature 0 and t respectively, \[{{V}_{0}}=\ell {{A}_{0}}{{V}_{t}}=\ell {{A}_{t}}\] \[{{\operatorname{V}}_{t}}={{V}_{0}}\left( l+\gamma t \right){{A}_{t}}={{A}_{0}}\left( l+2\alpha t \right)\] \[\therefore {{V}_{t}}=\ell {{A}_{0}}\left( l+2\alpha t \right)={{V}_{0}}\left( l+\gamma t \right)\] \[=\ell {{A}_{0}}\left( l+\gamma t \right)or\gamma =2a.\]
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