JEE Mains · Physics · STD 11 - 7. 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\(....G\)
- A \(21\)
- B \(100\)
- C \(59\)
- D \(42\)
Answer & Solution
Correct Answer
(B) \(100\)
Step-by-step Solution
Detailed explanation
\(F =\frac{ GMM }{ r ^{2}}+\sqrt{2} \frac{ GMM }{(\sqrt{2 r })^{2}}\) \(=\frac{ GMM }{ r ^{2}}\left(1+\frac{1}{\sqrt{2}}\right)\) \(=\frac{ G \times 10^{4}}{13^{2}}\left(1+\frac{1}{\sqrt{2}}\right)\) \(F \simeq 100 G\)
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