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Three identical particles A, B and C of mass $100 \mathrm{~kg}$ each are placed in a straight line with $\mathrm{A B=B C=13 \mathrm{~m}}$ . The gravitational force on a fourth particle $\mathrm{P}$ of the same mass is $\mathrm{F}$ , when placed at a distance $13 \mathrm{~m}$ from the particle $\mathrm{B}$ on the perpendicular bisector of the line $\mathrm{AC}$ . The value of F will be approximately :
Option: 1
$21\: \mathrm{G}$

Option: 2
$100\: \mathrm{G}$

Option: 3
$59\: \mathrm{G}$

Option: 4
$42\: \mathrm{G}$

Answers (1)

best_answer

Let, $\mathrm{F_{A},F_{B}\: and\:F_{c} }$ be the gravitational force on particle $\mathrm{P}$ by particle $\mathrm{A,B\, and\: C}$ respectively .

$\mathrm{\begin{aligned} \bar{F} &=\bar{F}_A+\bar{F}_B+\bar{F}_C \quad \text { (Given) } \\ \end{aligned}}$

$\mathrm{F =F_A \cos \theta+F_B+F_{C \cdot} \cos \theta }$

$\mathrm{=\frac{2 G(100)(100)}{r^2} \times \frac{13}{r}+\frac{G(100)^2}{13^2}}$

$\mathrm{r=\sqrt{\left ( 13 \right )^{2}+\left ( 13 \right )^{2}}}$

$\mathrm{F=\frac{2G\left ( 100 \right )^{2}\times 13}{\left ( 13 \right )^{2}\times 2\times 13\sqrt{2}}}+\mathrm{\frac{G\left ( 100 \right )^{2}}{13^{2}}}$

$=\mathrm{\frac{G\left ( 10 \right )^{4}}{\left ( 13 \right )^{2}}}\left [ \frac{1}{\sqrt{2}} +1\right ]$

$\mathrm{F=100\: G}$

Hence (2) is correct option

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