question_answer

The resultant of forces P and Q is R. If Q is doubled, then R is doubled. If the direction of Q is reversed, then R is again doubled, then? \[{{P}^{2}}:{{Q}^{2}}:{{R}^{2}}\] is     AIEEE  Solved  Paper-2003

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),                 \[4{{R}^{2}}={{R}^{2}}+2{{Q}^{2}}\] \[\Rightarrow \]               \[3{{R}^{2}}=2{{Q}^{2}}\] \[\therefore \]  \[\frac{{{P}^{2}}}{2}=\frac{{{Q}^{2}}}{3}=\frac{{{R}^{2}}}{2}\] \[\Rightarrow \]               \[{{P}^{2}}:{{Q}^{2}}:{{R}^{2}}=2:3:2\]