Question

Three particles $P$ , $Q$ and $R$ are placed on a straight line as shown in the figure. The masses of $P$ , $Q$ and $R$ are $\sqrt{3}m$ , $\sqrt{3}m$ and $m$ respectively. The gravitational force on a fourth particle $S$ of mass $m$ is equal to

• A. $\frac{\sqrt{3}G{m}^2}{2{\text{d}}^2}$ in ST direction only
• B. $\frac{\sqrt{3}G{m}^2}{2{\text{d}}^2}$ in SQ direction and $\frac{\sqrt{3}G{m}^2}{2{\text{d}}^2}$ in SU direction
• C. $\frac{\sqrt{3}G{m}^2}{2{\text{d}}^2}$ in SQ direction only
• D. $\frac{\sqrt{3}G{m}^2}{2{\text{d}}^2}$ in SQ direction and $\frac{\sqrt{3}G{m}^2}{2{\text{d}}^2}$ in ST direction

Answer 1

Answer: (C)

Along the direction parallel to TU
Net force $=\frac{\text{G}\sqrt{3}mm}{12d^2}cos{\text{30}}^{{}^{\circ }}-\frac{{\text{Gm}}^2}{4d^2}cos{\text{60}}^{{}^{\circ }}$
$=\frac{{\text{Gm}}^2}{8{\text{d}}^2}-\frac{{\text{Gm}}^2}{8{\text{d}}^2}=0$
Along the direction perpendicular to TU
Net force $=\frac{\text{G}\sqrt{3}{m}^2}{12d^2}cos{\text{60}}^{{}^{\circ }}+\frac{\text{G}\sqrt{3}{m}^2}{3d^2}+\frac{{\text{Gm}}^2}{4d^2}cos{\text{30}}^{{}^{\circ }}$
$=\frac{\sqrt{3}G{m}^2}{24d^2}+\frac{\sqrt{3}G{m}^2}{3d^2}+\frac{\sqrt{3}G{m}^2}{8d^2}$
$=\frac{\sqrt{3}G{m}^2}{{\text{d}}^2}\left[\frac{1+8+3}{24}\right]=\frac{\sqrt{3}G{m}^2}{2d^2}\text{along SQ}$