A) \[\frac{1}{2}\]
B) \[\frac{1}{7}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{5}\]
Correct Answer: B
Solution :
If \[V\] is the volume of glass flask, \[{{V}_{L}}\] of mercury and \[{{V}_{A}}\] of air in it \[V={{V}_{L}}+{{V}_{A}}\] Now as with change in temperature volume of air remains constant, the expansion of mercury will be equal to that of the glass flask, ie, \[\Delta V=\Delta {{V}_{L}}\] or \[{{V}_{{{\gamma }_{G}}}}\Delta \theta ={{V}_{Hg}}{{\gamma }_{Hg}}\Delta \theta \] \[(as\Delta V={{V}_{\gamma }}\Delta \theta )\] or \[\frac{{{V}_{Hg}}}{V}=\frac{{{\gamma }_{G}}}{{{\gamma }_{Hg}}}=\frac{3\times 9\times {{10}^{-6}}}{18.9\times {{10}^{-5}}}=\frac{1}{7}\]
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