A solid sphere of radius R gravitationally attracts a particle placed at 3 R from its centre with a force F…

A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F1. Now a spherical cavity of radius R2 is made in the sphere (as shown in figure) and the force becomes F2. The value of F1:F2 is:

  1. 41:50
  2. 50:41
  3. 25:36
  4. 36:25

Solution

Let the initial mass of the sphere is m'. Hence, mass of
a removed portion will be m'/8,F1=m.E.=m.Gm'9R2

F2=mG·m'3R2-G·m'/85R/22=Gm'9R2-Gm'×48×25=19-150Gm'R2

F2=4150×9·Gm'R2⇒F1F2=19×50×941=5041

Asked in: JEE Main 2021 (25 Feb Shift 1)

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