A)
B)
C)
D)
Correct Answer: C
Solution :
\[{{V}_{in}}=\frac{-Gm}{2R}\left[ 3-{{\left( \frac{r}{R} \right)}^{2}} \right]\] \[{{V}_{surface}}=\frac{-Gm}{R}\] \[{{V}_{out}}=-\frac{Gm}{r}\]
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