JEE Main 2021 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 24 February 2021, Shift 1, Physics Q3

Question 3 of 90 in this shift, Physics question 3 of 30, Section A.

Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be :

(1) $\frac{\sqrt{(1+2\sqrt{2})G}}{2}$ Official answer
(2) $\sqrt{\frac{G}{2}(1+2\sqrt{2})}$
(3) $\sqrt{G(1+2\sqrt{2})}$
(4) $\sqrt{\frac{G}{2}(2\sqrt{2}-1)}$

Official answer

Option 1

NTA final key.

Chapter: Gravitation
Topic: Orbital Mechanics
Idea tested: Multi-Star Polygonal Orbits

Same topic in other shifts

All Orbital Mechanics questions

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.