A) \[mgR\]
B) \[\frac{mgR}{2}\]
C) \[\frac{mgR}{4}\]
D) None of these
Correct Answer: C
Solution :
[c] \[{{E}_{i}}=-\frac{GMm}{2{{r}_{0}}}=-\frac{mg{{R}^{2}}}{2\left( 2R \right)}=-\frac{mgR}{4}\] Just after explosion, half part is at rest \[\Rightarrow \] other half will move at \[v=2{{v}_{0}}=2\sqrt{\frac{GM}{{{r}_{0}}}}\,\left( \,\because \,\,{{{\vec{p}}}_{i}}={{{\vec{p}}}_{r}} \right)\] \[\therefore \,\,\,\,{{E}_{f}}=\frac{1}{2}\left( \frac{m}{2} \right){{v}^{2}}-\frac{GMm}{{{r}_{0}}}=0,\] Increase in Mechanical energy \[={{E}_{f}}-{{E}_{i}}=\frac{mgR}{4}\]
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