A) 3 : 1 : 1
B) 2 : 3 : 2
C) 1 : 2 : 3
D) 2 : 3 : 1
Correct Answer: B
Solution :
Resultant force of two forces P and Q acting at an angle , is given by \[R=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ\,\cos \,\alpha }\] If is the angle between P and Q, then \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\,\cos \alpha \] ... (i) If Q is doubled, then R is doubled. \[4{{R}^{2}}={{P}^{2}}+4{{Q}^{2}}+4PQ\cos \alpha \] ... (ii) Again, if the direction of Q is reversed, then R is again doubled. \[\Rightarrow \] \[4{{R}^{2}}={{P}^{2}}+{{(-Q)}^{2}}+2P(-Q)\cos \alpha \] \[\Rightarrow \] \[4{{R}^{2}}={{P}^{2}}+{{Q}^{2}}-2PQ\,\,\cos \alpha \] ... (iii) On adding Eqs. (i) and (iii), we get \[5{{R}^{2}}=2{{P}^{2}}+2{{Q}^{2}}\] ?. (iv) On multiplying by 2 in Eq. (iii) and adding in Eq. (ii). we get \[12{{R}^{2}}=3{{P}^{2}}+6{{Q}^{2}}\] \[\Rightarrow \] \[4{{R}^{2}}={{P}^{2}}+2{{Q}^{2}}\] ... (v) On subtracting Eq. (v) from Eq. (iv), we get \[{{R}^{2}}={{P}^{2}}\] On putting in Eq. (v), \[4{{R}^{2}}={{R}^{2}}+2{{Q}^{2}}\] \[\Rightarrow \] \[3{{R}^{2}}=2{{Q}^{2}}\] \[\therefore \] \[\frac{{{P}^{2}}}{2}=\frac{{{Q}^{2}}}{3}=\frac{{{R}^{2}}}{2}\] \[\Rightarrow \] \[{{P}^{2}}:{{Q}^{2}}:{{R}^{2}}=2:3:2\]You need to login to perform this action.
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