A) \[\gamma =\alpha \]
B) \[\gamma =2\alpha \]
C) \[\gamma =3\alpha \]
D) \[\gamma =\frac{\alpha }{3}\]
Correct Answer: B
Solution :
When length of the liquid column remains constant, then the level of liquid moves down with respect to the container, thus must be less than \[3\alpha \] Now we can write \[V={{V}_{0}}(1+\gamma \Delta T)\] Since \[V=A{{l}_{0}}\,\left[ {{A}_{0}}(1+2\alpha \Delta T) \right]\,{{l}_{0}}\] \[=\,\,\,{{V}_{0}}\,\,(1+2\alpha \Delta T)\] Hence \[{{V}_{0}}\,(1+\gamma \Delta T)\,\,=\,\,{{V}_{0}}(1+2\alpha \Delta T)\,\,\Rightarrow \,\,\,\gamma =2\alpha \]
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