A) \[\sqrt{\frac{2Gm}{r}\left( \sqrt{2}-1 \right)}\]
B) \[\sqrt{\frac{Gm}{r}}\]
C) \[\sqrt{\frac{2Gm}{r}\left( 1-\frac{1}{\sqrt{2}} \right)}\]
D) \[\sqrt{\frac{2Gm}{r}}\]
Correct Answer: C
Solution :
Let 'M' be the mass of the particle Now,\[{{E}_{initial}}={{E}_{final}}\] i.e.\[\frac{GMm}{\sqrt{2}r}+0=\frac{GMm}{r}+\frac{1}{2}M{{V}^{2}}\]or\[\frac{1}{2}M{{V}^{2}}=\frac{GMm}{r}\left[ 1-\frac{1}{\sqrt{2}} \right]\] \[\Rightarrow \]\[\frac{1}{2}{{V}^{2}}=\frac{Gm}{r}\left[ 1-\frac{1}{\sqrt{2}} \right]\] or\[V=\sqrt{\frac{2Gm}{r}\left[ 1-\frac{1}{\sqrt{2}} \right]}\]
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