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 |
| \[\Delta V={{V}_{0}}\left[ \gamma L-\gamma \right]\Delta \text{ }\!\!\theta\!\!\text{ = V}\left[ {{\gamma }_{m}}-3{{\alpha }_{g}} \right]\Delta \text{ }\!\!\theta\!\!\text{ }\] |
| \[=50\left[ 180\times {{10}^{-6}}-3\times 9\times {{10}^{-6}} \right]\left( 38-18 \right)\] |
| \[=0.153\,cc\] |
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