A) \[60{}^\circ \]
B) \[120{}^\circ \]
C) \[70{}^\circ \]
D) \[180{}^\circ \]
Correct Answer: B
Solution :
The resultant force is given by \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\cos \alpha \] In first case \[{{R}^{2}}={{(3P)}^{2}}+{{(2P)}^{2}}+2\times 3P\times 2P\cos \alpha \] \[{{R}^{2}}=9{{P}^{2}}+4{{P}^{2}}+12{{P}^{2}}\cos \alpha \] \[{{R}^{2}}=13{{P}^{2}}+12{{P}^{2}}\cos \alpha \] ?(1) In second case \[{{(2R)}^{2}}={{(6P)}^{2}}+{{(2P)}^{2}}+2\times 6P\times 2P\cos \alpha \] \[{{R}^{2}}=10{{P}^{2}}+6{{P}^{2}}\cos \alpha \] From equation (1) and equation (2) we get \[\cos \alpha =-\frac{1}{2}\] thus, \[\alpha =120{}^\circ \]
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