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 (R/2) is made in the sphere (as shown in figure) and the force becomes F2. The value of F1: F2 is (a) 36: 25 (b) 41: 50 (c) 50: 41 (d) 25: 36

Answer: (c). F₁ = (GMm)/((3R)²) = (GMm)/(9R²) Cavity of radius R/2 has mass M/8; its centre B is R/2 from O, so 3R - R/2 = 5R/2 from m. Pull of the cavity part: (G(M/8)m)/((5R/2)²) = (GMm)/(50R²) F₂ = GMm(1/9 - 1/50)1/R² = (41GMm)/(450R²) F₁: F₂ = 50/450: 41/450 = 50: 41, so the answer is (c).

A solid sphere of radius R gravitationally attracts a particle placed… | JEE Main 2021 (25 Feb, Shift 1)

Q15JEE Main 2021 · 25 FEB · SHIFT 1Difficult

Newton's Law of Gravitation

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 $\left(\frac{R}{2}\right)$ is made in the sphere (as shown in figure) and the force becomes F2. The value of F1: F2 is

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Answer: (c)

$F_1 = \frac{GMm}{(3R)^2} = \frac{GMm}{9R^2}$
Cavity of radius $\frac{R}{2}$ has mass $\frac{M}{8}$ ; its centre B is $\frac{R}{2}$ from O, so $3R - \frac{R}{2} = \frac{5R}{2}$ from $m$ .
Pull of the cavity part: $\frac{G(M/8)m}{(5R/2)^2} = \frac{GMm}{50R^2}$
$F_2 = GMm\left(\frac{1}{9} - \frac{1}{50}\right)\frac{1}{R^2} = \frac{41GMm}{450R^2}$
$F_1 : F_2 = \frac{50}{450} : \frac{41}{450} = 50 : 41$ , so the answer is (c).

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