A) a year
B) nearly 4 years
C) nearly 1/4 year
D) 2.5 years
Correct Answer: C
Solution :
\[\frac{{{T}_{mercury}}}{{{T}_{earth}}}\,\,=\,\,{{\left( \frac{{{r}_{mercury}}}{{{r}_{earth}}} \right)}^{3/2}}\] \[=\,\,{{\left( \frac{6\,\,\times \,\,{{10}^{10}}}{1.5\times {{10}^{11}}} \right)}^{3/2}}\,\,=\,\,\frac{1}{4}\,\,\,\,\,\,\,\,\,\,\,\,(approx.)\] \[\therefore \,\,\,{{T}_{mercury}}\,=\,\,\frac{1}{4}\,\,year\]
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