A) \[\frac{\sqrt{3}G{{M}^{2}}}{2{{d}^{2}}}\]in ST direction only
B) \[\frac{\sqrt{3}G{{M}^{2}}}{2{{d}^{2}}}\]in SQ direction and \[\frac{\sqrt{3}G{{M}^{2}}}{2{{d}^{2}}}\]in SU direction
C) \[\frac{\sqrt{3}G{{M}^{2}}}{2{{d}^{2}}}\] in SQ direction only
D) \[\frac{\sqrt{3}G{{M}^{2}}}{2{{d}^{2}}}\]in SQ direction and \[\frac{\sqrt{3}G{{M}^{2}}}{2{{d}^{2}}}\] in ST direction
Correct Answer: C
Solution :
[c] In horizontal direction \[Net\,force=\frac{G\sqrt{3}mm}{12{{d}^{2}}}\cos {{30}^{\operatorname{o}}}-\frac{G{{m}^{2}}}{4d}\cos {{60}^{\operatorname{o}}}\]\[=\frac{G{{m}^{2}}}{8{{d}^{2}}}-\frac{G{{m}^{2}}}{8{{d}^{2}}}=0\] In vertical direction, Net force \[=\frac{G\sqrt{3}{{m}^{2}}}{12{{d}^{2}}}\cos {{60}^{\operatorname{o}}}+\frac{G\sqrt{3}{{m}^{2}}}{3{{d}^{2}}}+\frac{G{{m}^{2}}}{4d}\cos {{30}^{\operatorname{o}}}\] \[=\frac{\sqrt{3}G{{m}^{2}}}{24{{d}^{2}}}+\frac{\sqrt{3}G{{m}^{2}}}{3{{d}^{2}}}+\frac{\sqrt{3}G{{m}^{2}}}{8d}\] \[=\frac{\sqrt{3}G{{m}^{2}}}{{{d}^{2}}}\left[ \frac{1+8+3}{24} \right]=\frac{\sqrt{3}G{{m}^{2}}}{2{{d}^{2}}}\]along SQ-
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