A) \[K/x\]
B) \[K/2x\]
C) \[K/{{x}^{2}}\]
D) \[K/2{{x}^{2}}\]
Correct Answer: D
Solution :
Gravitational potential \[=\int_{{}}^{{}}{I\ dx=\int_{x}^{\infty }{{}}\frac{K}{{{x}^{3}}}dx}\] \[=K\,\left( \frac{{{x}^{-3+1}}}{-3+1} \right)_{x}^{\infty }=\left| \frac{-K}{2{{x}^{2}}} \right|_{x}^{\infty }=\frac{K}{2{{x}^{2}}}\]
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