Q 11-07-153JEE MainJEE Main 2021 (26 Feb, Shift 1)Medium
Find the gravitational force of attraction between the ring and sphere as shown in the diagram, where the plane of the ring is perpendicular to the line joining the centres. If $\sqrt{8}R$ is the distance between the centres of a ring (of mass $m$) and a sphere (mass $M$) where both have equal radius $R$
Answer: (D) $\frac{\sqrt{8}}{27}\frac{GmM}{R^2}$
The sphere acts as a point mass $M$ at its centre. The field of a ring of mass $m$ at distance $x$ on its axis is $\dfrac{Gm x}{(R^2+x^2)^{3/2}}$, so the force is
$$F = \frac{GMm\,x}{(R^2+x^2)^{3/2}}$$
With $x = \sqrt{8}R$: $R^2 + x^2 = 9R^2$, $(9R^2)^{3/2} = 27R^3$.
$$F = \frac{GMm\sqrt{8}R}{27R^3} = \frac{\sqrt{8}}{27}\frac{GmM}{R^2}$$
Solution by Sreeraj P, M.Sc Physics