A) 8
B) \[4\sqrt{2}\]
C) 4
D) 2
Correct Answer: C
Solution :
Total energy at the earth?s surface =Total energy at the maximum height. \[\frac{1}{2}m{{v}^{2}}+\left( -\frac{GMm}{R} \right)=-\frac{GMm}{R+h}\] \[\therefore \] \[{{v}^{2}}=\frac{2Mm}{R}\left( 1-\frac{R}{R+h} \right)\] \[=2gR\left( \frac{h}{R+h} \right)\]\[\left[ \because \,GM=g{{R}^{2}} \right]\] Now, for \[A,\,v_{A}^{2}=2gR\left( \frac{{{h}_{A}}}{R+{{h}_{A}}} \right)=\frac{4gR}{3}\] or, \[\frac{{{h}_{A}}}{R+{{h}_{A}}}=\frac{2}{3}\Rightarrow {{h}_{A}}=2R\] and for \[B,\,v_{B}^{2}=2gR\left( \frac{{{h}_{B}}}{R+{{h}_{B}}} \right)=\frac{2gR}{3}\] or, \[\frac{{{h}_{B}}}{R+{{h}_{B}}}=\frac{1}{3}\Rightarrow {{h}_{B}}=\frac{R}{2}\] \[\therefore \] \[\frac{{{h}_{A}}}{{{h}_{B}}}=4\] Hence, the correction option is [c].You need to login to perform this action.
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