On the x-axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by Ax/(x²+a²)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(x² + a²)¹/²

Option 2 -

A/(x²+a²)³/²

Option 3 -

A(x² + a²)³/²

Option 4 -

A/(x²+a²)¹/²

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

g = Ax / (x² + a²)³?²
⇒ ∫? dV = -∫?^∞ gdx
⇒ 0 - V = -∫?^∞ [Ax / (a² + x²)³?²] dx
Let, a² + x² = t²
⇒ 2xdx = 2tdt
⇒ xdx = tdt
⇒ V = ∫?^∞ (Atdt / t³) ⇒ [-A / t]?^∞ ⇒ [-A / √(a² + x²)]?^∞
⇒ V = A / √(a² + x²)

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