Q. A uniform ring of mass $M$ and radius $r$ is placed directly above a uniform sphere of mass $8M$ and of same radius $R$. The centre of the ring is at a distance of $d = \sqrt{3}R$ from the centre of the sphere. The gravitational attraction beween the sphere and the ring is
A) $\frac{GM^{2}}{R^{2}}$
B) $\frac{3GM^{2}}{{2R}^{2}}$
C) $\frac{2GM^{2}}{\sqrt{2}R^{2}}$
D) $\frac{\sqrt{3}GM^{2}}{R^{2}}$
Explanation
(d) From the figure the gravitational intensity due to the ring at a distance $d = \sqrt{3}R$ on its axis is
[image:8_image70.emf]
$I = \frac{GM}{\left( d^{2} + R^{2} \right)^{3/2}} = \frac{GM \times \sqrt{3}R}{\left( 3R^{2} + R^{2} \right)^{3/2}} = \frac{\sqrt{3}GM}{8R^{2}}$
Force on sphere $= \ (8M)I = (8M) \times \frac{\sqrt{3}GM}{8R^{2}}$
$= \frac{\sqrt{3}GM^{2}}{R^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.