JEE Main 2020 (4 Sep, Shift 1) Physics Q5: Gravitation

Question

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

Accepted Answer

Option 2: $\frac{A}{(x^2+a^2)^{1/2}}$

Answer

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

Answer

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

Answer

$A(x^2+a^2)^{3/2}$

Answer

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

JEE Main 4 September 2020, Shift 1, Physics Q5

Same idea in other shifts