A) 50 cc
B) 100 cc
C) 150 cc
D) 200 cc
Correct Answer: C
Solution :
It is given that the volume of air in the flask remains the same. This means that the expansion in volume of the vessel is exactly equal to the volume expansion of mercury. i.e., \[\Delta {{V}_{g}}=\Delta {{V}_{L}}\] or \[{{V}_{g}}{{\gamma }_{g}}\Delta \theta ={{V}_{L}}{{\gamma }_{L}}\Delta \theta \] \ \[{{V}_{L}}=\frac{{{V}_{g}}{{\gamma }_{g}}}{{{\gamma }_{L}}}=\frac{1000\times (3\times 9\times {{10}^{-6}})}{1.8\times {{10}^{-4}}}=150\,cc\]
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