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 are forces. We know that \[R=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta }\] when \[\theta =0{}^\circ ,\text{ }R=4\text{ }N\] \[R=4N=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ}\] \[P+Q=4\] ...(i) When \[\theta \,\text{= }90{}^\circ ,\text{ }R=3\text{ }N\] \[{{P}^{2}}+{{Q}^{2}}=9\] ?(ii) From Eq. (i) \[{{(P+Q)}^{2}}=16\] \[\Rightarrow \] \[{{P}^{2}}+{{Q}^{2}}+2PQ=16\] \[\Rightarrow \] \[9+2\text{ }PQ=16\] [using Eq. (ii)] \[\Rightarrow \] \[2PQ=7\] Now, \[{{(P-Q)}^{2}}={{P}^{2}}+{{Q}^{2}}-2PQ\] \[\Rightarrow \] \[{{(P-Q)}^{2}}=9-7\] \[P-Q=\sqrt{2}\] ...(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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