A) \[\frac{41\,G{{M}^{2}}}{3600\,{{R}^{2}}}\]
B) \[\frac{41\,G{{M}^{2}}}{450\,{{R}^{2}}}\]
C) \[\frac{59\,G{{M}^{2}}}{450\,{{R}^{2}}}\]
D) \[\frac{G{{M}^{2}}}{225\,{{R}^{2}}}\]
Correct Answer: A
Solution :
[a] Volume of removed sphere \[{{V}_{remo}}=\frac{4}{3}\pi {{\left( \frac{R}{2} \right)}^{3}}=\frac{4}{3}\pi {{R}^{3}}\left( \frac{1}{8} \right)\] Volume of the sphere (remaining) \[{{V}_{remain}}=\frac{4}{3}\pi {{R}^{3}}-\frac{4}{3}\pi {{R}^{3}}\left( \frac{1}{8} \right)=\frac{4}{3}\pi {{R}^{3}}\left( \frac{7}{8} \right)\] Therefore mass of sphere carved and remaining sphere are at respectively \[\frac{1}{8}M\]and\[\frac{7}{8}M\] Therefore, gravitational force between these two sphere, \[F=\frac{GMm}{{{r}^{2}}}=\frac{G\frac{7M}{8}\times \frac{1}{8}M}{{{\left( 3R \right)}^{2}}}=\frac{7}{64\times 9}\frac{GM}{{{R}^{2}}}\] \[\simeq \frac{41}{3600}\frac{G{{M}^{2}}}{{{R}^{2}}}\]You need to login to perform this action.
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