A) 0.025
B) 0.010
C) 0.015
D) 0.020
Correct Answer: D
Solution :
\[\operatorname{V}={{V}_{o}}\left[ 1+\gamma \Delta T \right]\therefore \rho =\frac{M}{V}\] \[\frac{M}{{{d}_{r}}}=\frac{M}{{{d}_{\operatorname{o}}}}\left[ 1+\gamma \Delta T \right]\] \[{{d}_{r}}=\frac{{{d}_{\operatorname{o}}}}{1+\gamma \Delta T}\] \[{{d}_{r}}={{d}_{\operatorname{o}}}-{{d}_{\operatorname{o}}}\left( 1-\gamma \Delta T \right)\] \[{{d}_{r}}={{d}_{\operatorname{o}}}-{{d}_{\operatorname{o}}}\gamma \Delta T\] \[{{d}_{\operatorname{o}}}\gamma \Delta T={{d}_{\operatorname{o}}}-{{d}_{r}}\] \[\left[ \frac{{{d}_{\operatorname{o}}}-{{d}_{r}}}{{{d}_{\operatorname{o}}}} \right]=\gamma \Delta T\] Fraction change in density =\[\gamma \Delta T\] \[=5\times 10-1\times 40K=0.020\]
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