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On the  x -axis and at a distance x  from the origin, the gravitational field due to a mass distribution is given by A x x 2 + a 2 3 / 2  in the x  -direction. The magnitude of gravitational potential on the x  -axis at a distance x  , taking its value to be zero at infinity is:

Option 1 -

A x 2 + a 2 1 / 2

Option 2 -

A x 2 + a 2 3 / 2

Option 3 -

A x 2 + a 2 3 / 2

Option 4 -

A x 2 + a 2 1 / 2

0 4 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    a month ago
    Correct Option - 4


    Detailed Solution:

    g = A x x 2 + a 2 3 / 2  

    ⇒ ∫ v 0 ? d V = - ∫ x ∞ ? g d x

    ⇒ O - V = - ∫ A x a 2 + x 2 3 / 2

    Let, a 2 + x 2 = t 2

    ⇒ 2 x d x = 2 t d t

    ⇒ x d x = t d t

    ⇒ V = ∫ A t d t t 3 ⇒ - A t ⇒ - A a 2 + x 2 x ∞

    ⇒ V = A a 2 + x 2

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