The gravitational field due to a mass distribution is $$E = \frac{K}{{{x^3}}}$$ in the $$x$$-direction. ($$K$$ is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance $$x$$ is
A.
$$\frac{K}{x}$$
B.
$$\frac{K}{{2x}}$$
C.
$$\frac{K}{{{x^2}}}$$
D.
$$\frac{K}{{2{x^2}}}$$
Answer :
$$\frac{K}{{2{x^2}}}$$
Solution :
$$V = - \int_\infty ^x E dx = \int_\infty ^x K {x^{ - 3}}dx = \frac{K}{{2{x^2}}}.$$
Releted MCQ Question on Basic Physics >> Gravitation
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