On the x-axis and at a distance x from the origin, the gravitational field due a mass distribution is given by Ax \over ( 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:

Gravitation PYQ — JEE Main 2020 · 4 Sep · Shift 1 · Q45 with Solution

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Correct answer: C

On the x-axis, the gravitational field is

\vec g(x)=\frac{Ax}{(x^2+a^2)^{3/2}}\,\hat i

We need the magnitude of gravitational potential $V(x)$ , with

V(\infty)=0.

For one-dimensional motion along x,

g_x=-\frac{dV}{dx}

So,

-\frac{dV}{dx}=\frac{Ax}{(x^2+a^2)^{3/2}}

Hence,

\frac{dV}{dx}=-\frac{Ax}{(x^2+a^2)^{3/2}}.

Integrate with respect to $x$ :

V(x)=\int -\frac{Ax}{(x^2+a^2)^{3/2}}\,dx

Let

u=x^2+a^2 \quad \Rightarrow \quad d\nu=2x\,dx

Then,

V(x)=-A\int \frac{x\,dx}{(x^2+a^2)^{3/2}}

Using the standard result,

\int \frac{x\,dx}{(x^2+a^2)^{3/2}}=-\frac{1}{\sqrt{x^2+a^2}}+C

Therefore,

V(x)= -A\left(-\frac{1}{\sqrt{x^2+a^2}}\right)+C

V(x)=\frac{A}{\sqrt{x^2+a^2}}+C

Given $V(\infty)=0$ .

As $x\to\infty$ ,

\frac{A}{\sqrt{x^2+a^2}}\to 0

So,

C=0.

Thus,

V(x)=\frac{A}{\sqrt{x^2+a^2}}.

The required magnitude is

\boxed{\frac{A}{\sqrt{x^2+a^2}}}

This corresponds to:

Option C

\boxed{\frac{A}{(x^2+a^2)^{1/2}}}