Question 11 of 223

Three identical particle $A,B$ 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

A$21G$B$100G$C$59G$D$42G$

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Answer: (B) $100G$

Given here, $m=100kg$

Gravitational force, ${F}_{AP}=\dfrac{G{m}^{2}}{{\left(13\sqrt{2}\right)}^{2}}$,
${F}_{BP}=\dfrac{G{m}^{2}}{{\left(13\right)}^{2}}$ and ${F}_{CP}=\dfrac{G{m}^{2}}{{\left(13\sqrt{2}\right)}^{2}}$
Net gravitational force,
${F}_{net}={F}_{BP}+{F}_{AP}\cos 45^{\circ}+{F}_{CP}\cos 45^{\circ}$
$=\dfrac{G\times {10}^{4}}{{13}^{2}}\left(1+\dfrac{1}{\sqrt{2}}\right)$
$F≃100G$