On the $x$-axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{\left(x^{2}+a^{2}\right)^{3 / 2}}$ 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:

(D) $\frac{A}{\left(x^{2}+a^{2}\right)^{1 / 2}}$

(A) $A\left(x^{2}+a^{2}\right)^{1 / 2}$

(B) $\frac{A}{\left(x^{2}+a^{2}\right)^{3 / 2}}$

(C) $A\left(x^{2}+a^{2}\right)^{3 / 2}$

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Gravitation

On the

x

-axis and at a distance

x

from the origin, the gravitational field due to a mass distribution is given by

\frac{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:

Answer: (D)

The field is related to the potential by

g = - \frac{d V}{d x}

, so taking

V = 0

at infinity,

V (x) = - \int_{\infty}^{x} g d x^{'} = - \int_{\infty}^{x} \frac{A x ^{'}}{( x ^{'2} + a ^{2} ) ^{3/2}} d x^{'}

. Substituting

t = x^{'2} + a^{2}

d t = 2 x^{'} d x^{'}

, the antiderivative of

\frac{A x ^{'}}{( x ^{'2} + a ^{2} ) ^{3/2}}

- \frac{A}{x ^{'2} + a ^{2}}

. Evaluating between the limits gives

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

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option