JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Gravitation

Three identical particle A, B and C of mass 100 kg each are placed in a straight line with

AB = BC = 13 m

. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13 m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately :

(A)21 G
(B)100 G✓ Correct
(C)59 G
(D)42 G

Answer: (B)

Solution

Particle P is at distance 13 m from B (perpendicular bisector), so it forms a right triangle with A and C. The distance from P to A (or C) is

1 3^{2} + 1 3^{2} = 132

m.

The force from B on P:

F_{B} = \frac{G M ^{2}}{1 3 ^{2}}

The forces from A and C have equal magnitudes

F_{A} = F_{C} = \frac{G M ^{2}}{( 13 2 ) ^{2}} = \frac{G M ^{2}}{2 \times 1 3 ^{2}}

.

By symmetry, the horizontal components of

F_{A}

and

F_{C}

cancel, and their vertical components add:

F_{A + C} = 2 \times \frac{G M ^{2}}{2 \times 1 3 ^{2}} \times \frac{1}{2} = \frac{G M ^{2}}{1 3 ^{2} 2}

Total force:

F = F_{B} + F_{A + C} = \frac{G M ^{2}}{1 3 ^{2}} (1 + \frac{1}{2}) \approx \frac{G \times 1 0 ^{4}}{169} \times 1.707 \approx 100 G

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Gravitation

Solution

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