A) \[\frac{k}{x}\]
B) \[\frac{k}{{{x}^{3}}}\]
C) \[\frac{k}{2{{x}^{2}}}\]
D) \[\frac{k}{3{{x}^{3}}}\]
Correct Answer: C
Solution :
Gravitational potential \[=\int{/\,dx}\] Given, \[l=E=\frac{k}{{{x}^{3}}}\] \[\therefore \] \[\int_{x}^{\infty }{\frac{k}{{{x}^{3}}}dk}\] \[=k\left[ \frac{{{x}^{-3+1}}}{-3+1} \right]_{x}^{\infty }=k\left[ \frac{-1}{2{{x}^{2}}} \right]_{x}^{\infty }=\frac{k}{2{{x}^{2}}}\]
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