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Question: Answered & Verified by Expert
Three solid spheres each of mass $1 \mathrm{~kg}$ and radius $2 \mathrm{~m}$ are arranged at the three corners of an equilateral triangle of side $10 \mathrm{~m}$, such that centres of the spheres coincide with the comers of the triangle. When they are released from that position, the speed of any one sphere at the time of collision would be ($\mathrm{G}$ is universal gravitational constant).
PhysicsGravitationAP EAMCETAP EAMCET 2017 (25 Apr Shift 2)
Options:
  • A $\sqrt{\frac{3 G}{10}}$
  • B $\sqrt{\frac{10 G}{3}}$
  • C $\sqrt{30 G}$
  • D $\sqrt{3 G}$
Solution:
1037 Upvotes Verified Answer
The correct answer is: $\sqrt{\frac{3 G}{10}}$
No solution. Refer to answer key.

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