A) 0.85 cc
B) 0.46 cc
C) 0.153 cc
D) 0.05 cc
Correct Answer: C
Solution :
Due to volume expansion of both mercury and flask, the change in volume of mercury relative to flask is given by \[\Delta V={{V}_{0}}[{{\gamma }_{L}}-{{\gamma }_{g}}]\Delta \theta =V[{{\gamma }_{m}}-3{{\alpha }_{g}}]\Delta \theta \] \[=50\,[180\times {{10}^{-6}}-3\times 9\times {{10}^{-6}}]\,(38-18)=0.153\,cc\]
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