A) \[\frac{G{{m}^{2}}}{{{l}^{2}}}\,\left[ \frac{5}{4}+\frac{1}{\sqrt{3}} \right]\]
B) \[\frac{G{{m}^{2}}}{{{l}^{2}}}\,\left[ \frac{3}{4}+\frac{1}{\sqrt{3}} \right]\]
C) \[\frac{G{{m}^{2}}}{{{l}^{2}}}\,\left[ \frac{5}{4}-\frac{1}{\sqrt{3}} \right]\]
D) \[\frac{G{{m}^{2}}}{{{l}^{2}}}\,\left[ \frac{3}{4}-\frac{1}{\sqrt{3}} \right]\]
Correct Answer: A
Solution :
From figure, |
\[AC=AM+MC=2AM=\,\,2l\,cos30{}^\circ =2l\frac{\sqrt{3}}{2}=\sqrt{3l}\] |
Similarly,\[AE=\sqrt{3}l\], |
\[\operatorname{AD} =AO+ON+ND= l sin\,30{}^\circ + l + l\,\,sin\,\,30{}^\circ \] |
\[=\,\,l\times \frac{1}{2}+l+l\times \frac{1}{2}=2\,l\] |
\[AB=AF=l\] |
Force on mass m at A due to mass m at B is |
\[{{F}_{AB}}=\frac{Gmm}{{{(AB)}^{2}}}=\frac{Gmm}{{{l}^{2}}}\,\,along\,\,AB\] |
Force on mass m at A due to mass m at C is |
\[{{F}_{AC}}=\frac{Gmm}{{{(AC)}^{2}}}=\frac{Gmm}{{{(\sqrt{3l})}^{2}}}\,=\frac{G{{m}^{2}}}{3{{l}^{2}}}\,\,along\,\,AC\] |
Force on mass m at A due to mass m at D is |
\[{{F}_{AD}}\,\,=\,\,\frac{Gmm}{{{(AD)}^{2}}}\,\,=\,\,\frac{Gmm}{(2\,{{l}^{2}})}\,\,=\,\,\frac{G{{m}^{2}}}{4\,{{l}^{2}}}\,\,along\,\,AD\] |
Force on mass m at A due to mass m at E is |
\[{{F}_{AE}}\,\,=\,\,\frac{Gmm}{{{(AE)}^{2}}}\,\,=\,\,\frac{Gmm}{{{(\sqrt{3\,l})}^{2}}}\,\,=\,\,\frac{G{{m}^{2}}}{3\,{{l}^{2}}}\,\,along\,\,AE\] |
Force on mass m at A due to mass m at F is |
\[{{F}_{AF}}\,\,=\,\,\frac{Gmm}{{{(AF)}^{2}}}\,\,=\,\,\frac{G{{m}^{2}}}{{{l}^{2}}}\,\,=\,\,along\,\,AF\] |
Resultant force due to \[{{F}_{AB}}\,\,and\,\,{{F}_{AF}}\] is |
\[{{F}_{{{R}_{1}}}}=\sqrt{F_{AB}^{3}+F_{AF}^{2}+2{{F}_{AB}}\,{{F}_{AF}}\,\cos \,\,120{}^\circ }\] |
\[=\,\,\,\sqrt{\left( \frac{G{{m}^{2}}}{{{l}^{2}}} \right)+\left( \frac{G{{m}^{2}}}{{{l}^{2}}} \right)+2\left( \frac{G{{m}^{2}}}{{{l}^{2}}} \right)\left( \frac{G{{m}^{2}}}{{{l}^{2}}} \right)\left( -\frac{1}{2} \right)}\] \[=\,\,\frac{G{{m}^{2}}}{{{l}^{2}}}\,\,along\,\,AD\] |
Resultant force due to \[{{\operatorname{F}}_{AC}}\,\,and\,\,{{F}_{AE}}\] is |
\[{{F}_{{{R}_{2}}}}=\sqrt{F_{AC}^{2}+F_{AE}^{2}+2{{F}_{AC}}\,{{F}_{AE}}\,\,\cos \,60{}^\circ }\] |
\[=\,\,\,\sqrt{\left( \frac{G{{m}^{2}}}{3{{l}^{2}}} \right)+\left( \frac{G{{m}^{2}}}{3{{l}^{2}}} \right)+2\left( \frac{G{{m}^{2}}}{3{{l}^{2}}} \right)\left( \frac{G{{m}^{2}}}{3{{l}^{2}}} \right)\left( \frac{1}{2} \right)}\]\[=\,\,\frac{\sqrt{3}\,G{{m}^{2}}}{3\,{{l}^{2}}}=\frac{G{{m}^{2}}}{\sqrt{3\,{{l}^{2}}}}\] along AD |
Net force on mass m along AD is |
\[{{F}_{R}}={{F}_{{{R}_{1}}}}+{{F}_{{{R}_{2}}}}+{{F}_{AD}}\,\,=\,\,\frac{G{{m}^{2}}}{{{l}^{2}}}+\frac{G{{m}^{2}}}{\sqrt{3\,{{l}^{2}}}}+\frac{G{{m}^{2}}}{4\,{{l}^{2}}}\] |
\[=\,\,\frac{G{{m}^{2}}}{{{l}^{2}}}\left[ 1+\frac{1}{\sqrt{3}}+\frac{1}{4} \right]=\frac{G{{m}^{2}}}{l{{\,}^{2}}}\,\left[ \frac{5}{4}+\frac{1}{\sqrt{3}} \right]\] |
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