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Question

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


A

121+22G

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B

1+22G

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C

G222-1

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D

G21+22

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Solution

The correct option is A

121+22G


Step 1. Given Data,

Four identical particles of equal masses M=1kg,

Circle of radius R=1m

JEE Main 2021 24 Feb Physics Shift 1 Question 1

Step 2. Calculate the speed of each particle,

From the above diagram the net force on mass M,

Net Force Fc=F1+F2cos45°+F2cos45°

F1+F2cos45°+F2cos45°=Fc⇒F1+2F2cos45°=Fc

Since the four masses are moving in a circle, hence the net force acting on them will be centripetal force.

Fc=Centripetal force =mv2r (where, m is mass, v is velocity and r is radius).

We know that, Gravitational Force F=GMmR2

(where, G is universal gravitational constant.)

Here, m=M

⇒F1+2F2cos45°=Fc⇒GM22R2+2GM22R2cos45°=MV2R⇒GM24R2+GM222R2=MV2R⇒GM4R+GM22R=V2⇒GM4R+GM22R=V⇒GM4R1+22=V⇒12GMR1+22=V

Putting values of mass M=1kg and radius R=1m,

⇒V=12G1+22

Hence, the correct option is A.


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