A) \[\sqrt{{{F}^{2}}\,{{\cos }^{2}}\,\alpha \,+{{G}^{2}}\,{{\sin }^{2}}\alpha }\]
B) \[\sqrt{{{F}^{2}}\,{{\sin }^{2}}\alpha +{{G}^{2}}\,{{\cos }^{2}}\alpha }\]
C) \[\sqrt{{{F}^{2}}+{{G}^{2}}}\]
D) \[\sqrt{{{F}^{2}}-{{G}^{2}}}\]
Correct Answer: A
Solution :
Let the two forces are \[{{P}_{1}}\] and \[{{P}_{2}}\]. Then, greatest resultant \[F={{P}_{1}}+{{P}_{2}}\] and least resultant, \[G={{P}_{1}}-{{P}_{2}}\] Thus, \[{{P}_{1}}=\frac{F+G}{2},{{P}_{2}}=\frac{F-G}{2}\] When forces act at an angle \[2\alpha ,\] then \[R=\sqrt{\begin{align} & {{\left( \frac{F+G}{2} \right)}^{2}}+{{\left( \frac{F-G}{2} \right)}^{2}}+\left( \frac{F+G}{2} \right) \\ & \left( \frac{F-G}{2} \right)\,\cos \,2\alpha \\ \end{align}}\] \[=\frac{1}{2}\sqrt{\begin{align} & {{F}^{2}}+{{G}^{2}}+2FG+{{F}^{2}}+{{G}^{2}}-2FG \\ & +2({{F}^{2}}-{{G}^{2}})\,\,cos\,2\alpha \\ \end{align}}\] \[=\frac{1}{2}\sqrt{2{{F}^{2}}+2{{G}^{2}}+2({{F}^{2}}-{{G}^{2}})\,\cos \,2\alpha }\] \[=\frac{1}{2}\sqrt{2{{F}^{2}}(1+\cos \,2\alpha )+2{{G}^{2}}\,(1-\cos \,2\alpha )}\] \[=\frac{1}{2}\sqrt{2{{F}^{2}}.\,2\,{{\cos }^{2}}\,\alpha +2{{G}^{2}}.\,2{{\sin }^{2}}\,\alpha }\] \[=\sqrt{{{F}^{2}}\,{{\cos }^{2}}\alpha +{{G}^{2}}{{\sin }^{2}}\alpha }\]
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