A) \[\frac{5}{6}\frac{GM}{{{x}^{2}}}\]
B) \[\frac{8}{9}\frac{GM}{{{x}^{2}}}\]
C) \[\frac{7}{8}\frac{GM}{{{x}^{2}}}\]
D) \[\frac{6}{7}\frac{GM}{{{x}^{2}}}\]
Correct Answer: C
Solution :
[c] Let mass of smaller sphere (which has to be removed) is m Radius \[=\frac{R}{2}\] (from figure) \[\frac{M}{\frac{4}{3}\pi {{R}^{3}}}=\frac{m}{\frac{4}{3}\pi {{\left( \frac{R}{2} \right)}^{3}}}\Rightarrow m=\frac{M}{8}\] Mass of the left over part of the sphere \[M'=M-\frac{M}{8}=\frac{7}{8}M\] Therefore gravitational field due to the left over part of the sphere \[=\frac{GM'}{{{x}^{2}}}=\frac{7}{8}\frac{GM}{{{x}^{2}}}\]
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