A) \[\sqrt{3}\]
B) \[\frac{5}{\sqrt{3}}\]
C) \[\sqrt{\frac{3}{2}}\]
D) \[\frac{2}{\sqrt{3}}\]
Correct Answer: D
Solution :
\[{{F}_{A}}=\frac{\mu mg}{\sin \theta -\mu \cos \theta }\]Similarly, \[{{F}_{B}}=\frac{\mu mg}{\sin \theta +\mu \cos \theta }\] \[\therefore \] \[\frac{{{F}_{A}}}{{{F}_{B}}}=\frac{\frac{\mu mg}{\sin \theta -\mu \cos \theta }}{\frac{\mu mg}{\sin \theta +\mu \cos \theta }}\] \[\frac{\frac{\mu mg}{\sin {{60}^{o}}-\frac{\sqrt{3}}{5}\cos {{60}^{o}}}}{\frac{\mu mg}{\sin {{30}^{o}}+\frac{\sqrt{3}}{5}\cos {{30}^{o}}}}\] \[\left[ \mu =\frac{\sqrt{3}}{5}\text{given} \right]\] \[=\frac{\sin {{30}^{o}}+\frac{\sqrt{3}}{5}\cos {{30}^{o}}}{\sin {{60}^{o}}-\frac{\sqrt{3}}{5}\cos {{60}^{o}}}\] \[=\frac{\frac{1}{2}+\frac{\sqrt{3}}{5}\times \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{5}\times \frac{1}{2}}\] \[=\frac{\frac{1}{2}\left( 1+\frac{3}{5} \right)}{\frac{\sqrt{3}}{5}\left( 1-\frac{1}{5} \right)}=\frac{\frac{1}{2}\times \frac{8}{5}}{\frac{\sqrt{3}\times 4}{10}}\] \[=\frac{\frac{8}{10}}{\frac{\sqrt{3}\times 4}{10}}=\frac{8}{\sqrt{3}\times 4}=\frac{2}{\sqrt{3}}\]You need to login to perform this action.
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