A) \[(2+\sqrt{2})N\]and\[(2-\sqrt{2})N\]
B) \[(2+\sqrt{3})N\]and\[(2-\sqrt{3})N\]
C) \[\left( 2+\frac{1}{2}\sqrt{2} \right)N\]and \[\left( 2-\frac{1}{2}\sqrt{2} \right)N\]
D) \[\left( 2+\frac{1}{2}\sqrt{3} \right)N\]and\[\left( 2-\frac{1}{2}\sqrt{3} \right)N\]
Correct Answer: C
Solution :
Let P and Q be the two forces. We know that \[R=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta }\] When\[\theta ={{0}^{o}},R=4N\] \[R=4N\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ}\] \[P+Q=4\] ??.(i) When\[\theta =90{}^\circ ,R=3N\] \[{{P}^{2}}+{{Q}^{2}}=9\] ...(ii) From Eq.(i),\[{{(P+Q)}^{2}}=16\] \[\Rightarrow \]\[{{P}^{2}}+{{Q}^{2}}+2PQ=16\] \[\Rightarrow \] \[9+2PQ=16\] [using Eq. (iii)] \[\Rightarrow \] \[2PQ=7\] Now, \[{{(P-Q)}^{2}}={{P}^{2}}+{{Q}^{2}}-2PQ\] \[\Rightarrow \] \[P-Q=\sqrt{2}\] ...(iii) On solving Eqs. (i) and (iii), \[P=\left( 2+\frac{1}{2}\sqrt{2} \right)N\]and\[Q=\left( 2-\frac{1}{2}\sqrt{2} \right)N\]
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