Three identical particles A, B and C of mass 100kg each are placed in a straight line with AB = BC = 13m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately:

Option 1 -

21 G

Option 2 -

100 G

Option 3 -

59 G

Option 4 -

42 G

3 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

F on P = $\frac{}{^{}} \frac{}{^{}}$ $\frac{G M M}{r^{2}} + \sqrt[]{2} \frac{G M M}{{(\sqrt[]{2} r)}^{2}}$

$F o n P = \frac{G \times 1 0^{4}}{1 3^{2}} (1 + \frac{1}{\sqrt[]{2}}) \approx 100 G$

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