A) \[4.5\times {{10}^{-3}}/K\]
B) \[5.4\times {{10}^{-5}}/K\]
C) \[4.5\times {{10}^{-4}}/K\]
D) \[5.4\times {{10}^{-6}}/K\]
Correct Answer: C
Solution :
[c] \[V={{V}_{0}}(1+ \alpha \,\Delta \theta )\]where \[\Delta \theta \] is the difference in temperature, \[\alpha \] is the coefficient of volume expansion. i.e., \[\frac{V-{{V}_{0}}}{{{V}_{0}}}\frac{1}{\Delta \theta }=\alpha \] \[\alpha =\frac{\frac{m}{{{d}_{2}}}-\frac{m}{{{d}_{1}}}}{m/{{d}_{1}}}.\frac{1}{\Delta \theta }=\frac{\left( {{d}_{1}}-{{d}_{2}} \right)}{\left( {{d}_{1}}{{d}_{2}} \right)}\times \frac{{{d}_{1}}}{\Delta \theta }\] \[=\frac{\left( {{d}_{1}}-{{d}_{2}} \right)}{\left( {{d}_{1}}{{d}_{2}} \right)}.\frac{{{d}_{1}}}{\Delta \theta }=\frac{1000-958.4}{958.4\times 96}\] \[=4.5\times l{{0}^{-4}}\]per degree K.
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