1.

The gravitational field due to a mass distribution is E=(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

Answer»

`(K)/(x)`
`(K)/(x^(2))`
`(K)/(2x^(2))`
`(K)/(3x^(2))`

SOLUTION :Gravitational Potential
`V=overset(oo)underset(x)INT E.DX=overset(oo)underset(x)int(K)/(x^(3))dx=(K)/(2x^(2))`


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