Let the initial mass of the sphere is m′. Hence, mass of
a removed portion will be m′/8,F1=m.E.=9R2m.Gm′

F2=m[(3R)2G⋅m′−(5R/2)2G⋅m′/8]=9R2Gm′−8×25Gm′×4=(91−501)R2Gm′
F2=50×941⋅R2Gm′⇒F2F1=91×4150×9=4150
JEE Main 2021 — Physics Mechanics
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 (2R) is made in the sphere (as shown in figure) and the force becomes F2. The value of F1:F2 is:

Held on 25 Feb 2021 · Verified 6 Jul 2026.
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