JEE Main 2021 · 24 Feb · Shift 1 · Physics Q3

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PhysicsSingle correct+4 / -1Question 3 of 90

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 :

\dfrac{\sqrt{(1+2\sqrt2)G}}2

\sqrt{\dfrac G2(1+2\sqrt2)}

\sqrt{G(1+2\sqrt2)}

\sqrt{\dfrac G2(2\sqrt2-1)}

Answer and worked solution

Official answer: A

Net pull towards the centre on one particle (square of side $\sqrt2$ , diagonal 2): $G\left(2\cdot\frac12\cdot\frac1{\sqrt2}+\frac14\right)=\frac G4(2\sqrt2+1)$ .

Circular motion $v^2=\frac G4(1+2\sqrt2)$ , so $v=\dfrac{\sqrt{(1+2\sqrt2)G}}2$ .

(A)

Topics:Newton's law of gravitation Circular motion dynamics

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JEE Main 2021 · 24 Feb · Shift 1, Physics question 3