JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Gravitation

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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

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 '.

(A) $\frac{8}{9} \cdot \frac{GmM}{R}$
(B) $\frac{1}{3 8} \cdot \frac{GMm}{R ^{2}}$
(C) $\frac{8}{27} \cdot \frac{GmM}{R ^{2}}$ ✓ Correct
(D) $\frac{2 2}{3} \cdot \frac{GMm}{R ^{2}}$

Answer: (C)

Solution

A uniform sphere attracts, and is attracted, exactly as a point mass at its centre, so the force is

M

times the gravitational field of the ring at the sphere's centre. On the axis of a ring of mass

m

and radius

R

, at axial distance

d

from the centre, that field is

E_{g} = \frac{G m d}{( d ^{2} + R ^{2} ) ^{3/2}}

Here

d = 8 R

, so

d^{2} + R^{2} = 8 R^{2} + R^{2} = 9 R^{2}

and

(9 R^{2})^{3/2} = 27 R^{3}

. Therefore

F = M E_{g} = \frac{GM m 8 R}{27 R ^{3}} = \frac{8}{27} \cdot \frac{GM m}{R ^{2}}

so the ring and the sphere attract each other with force

\frac{8}{27} \cdot \frac{GM m}{R ^{2}}

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Gravitation

Solution

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