A small point of mass m is placed at a distance 2R from the centre ' O ' of a big uniform solid sphere of mass M and radius R. The gravitational force on ' m ' due to M is F₁. A spherical part of radius R/3 is removed from the big sphere as shown in the figure and the gravitational force on m due to remaining part of M is found to be F₂. The value of ratio F₁: F₂ is (a) 11: 10 (b) 12: 11 (c) 16: 9 (d) 12: 9

Answer: (b). F₁ = (GMm)/((2R)²) = (GMm)/(4R²) Removed sphere: mass M/27, centre at R - R/3 = 2R/3 from O, so 4R/3 from m. Its pull: (G(M/27)m)/((4R/3)²) = (GMm)/(48R²) F₂ = (GMm)/(4R²) - (GMm)/(48R²) = (11GMm)/(48R²) F₁: F₂ = 12: 11, so the answer is (b).

A small point of mass m is placed at a distance 2R from the centre '… | JEE Main 2025 (22 Jan, Shift 1)

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Newton's Law of Gravitation

A small point of mass $m$ is placed at a distance $2R$ from the centre ' $O$ ' of a big uniform solid sphere of mass M and radius R. The gravitational force on ' m ' due to M is $F_1$ . A spherical part of radius $R/3$ is removed from the big sphere as shown in the figure and the gravitational force on m due to remaining part of $M$ is found to be $F_2$ . The value of ratio $F_1 : F_2$ is

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Answer: (b)

$F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$
Removed sphere: mass $\frac{M}{27}$ , centre at $R - \frac{R}{3} = \frac{2R}{3}$ from $O$ , so $\frac{4R}{3}$ from $m$ .
Its pull: $\frac{G(M/27)m}{(4R/3)^2} = \frac{GMm}{48R^2}$
$F_2 = \frac{GMm}{4R^2} - \frac{GMm}{48R^2} = \frac{11GMm}{48R^2}$
$F_1 : F_2 = 12 : 11$ , so the answer is (b).

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