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