A) \[\frac{10R}{9}\]
B) \[\frac{9\,R}{7}\]
C) \[\frac{9\,R}{8}\]
D) \[\frac{10\,R}{3}\]
Correct Answer: B
Solution :
Decrease in KE = Increase in PE \[\Rightarrow \,\,\,\frac{1}{2}\,m\,{{\left( \frac{3}{4}{{v}_{e}} \right)}^{2}}\,\,-0=\frac{mgh}{1+\frac{h}{R}}\] \[\Rightarrow \,\,\frac{9}{12}\,\,m{{v}^{2}}_{e}\,=\,\,\frac{mgh}{1+\frac{h}{R}}\] \[\Rightarrow \,\,\,\frac{9}{32}\,\,\frac{2GM}{R}\,=\,\,\frac{gh}{1+\frac{h}{R}}\] \[\Rightarrow \,\,\,\frac{9}{16}\,\frac{g{{R}^{2}}}{R}\,\,=\,\,\frac{gh}{1+\frac{h}{R}}\] \[\Rightarrow \,\,\,\frac{9}{16}\,R\,\,=\,\,\frac{h}{1+\frac{h}{R}}\] \[\Rightarrow \,\,\,\frac{9}{16}\,R\,\,+\,\,\frac{9}{16}\,h=h\] \[9R\,\,+\,\,9h\,\,=\,\,16h\] \[7h=9R\] \[h=\frac{9}{7}R\]
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