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JEE Main 202126 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
26 February
Shift
Morning Shift
Subject
Physics
Chapter
Gravitation

JEE Main 2021 · 26 February · Morning Shift · Physics · Gravitation

Find the gravitational force of attraction between the ring and...

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 8R\sqrt{8}R is the distance between the centres of a ring (of mass mm) and a sphere (mass MM) where both have equal radius RR

Find the gravitational force of attraction between the ring and sphere as shown in the diagram, where the plane of the

Options

  1. A223GMmR2\dfrac{2\sqrt{2}}{3}\dfrac{GMm}{{R}^{2}}
  2. B89GmMR\dfrac{\sqrt{8}}{9}\dfrac{GmM}{R}
  3. C138GMmR2\dfrac{1}{3\sqrt{8}}\dfrac{GMm}{{R}^{2}}
  4. D827GmMR2\dfrac{\sqrt{8}}{27}\dfrac{GmM}{{R}^{2}}

Correct answer

D. 827GmMR2\dfrac{\sqrt{8}}{27}\dfrac{GmM}{{R}^{2}}

Step-by-step solution

Gravitational field of ring

=−Gmx(R2+x2)3/2= - \dfrac{Gmx}{{\left({R}^{2} + {x}^{2}\right)}^{3 / 2}}

Force between sphere & ring

=GmM(8R)(R2+8R2)3/2= \dfrac{GmM\left(\sqrt{8}R\right)}{{\left({R}^{2} + 8{R}^{2}\right)}^{3 / 2}}

=GmMR2×827= \dfrac{GmM}{{R}^{2}} \times \dfrac{\sqrt{8}}{27}

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