Question

The resultant of forces $\vec{ p }$ and $\vec{ Q }$ is $\vec{ R }$. If $\vec{ Q }$ is doubled, then $\vec{ R }$ is doubled. If the direction of $\vec{ Q }$ is reversed, then $\vec{ R }$ is again doubled, then $p^{2}: Q^{2}: R^{2}$ is :

• A. 3: 1: 1
• B. 2: 3: 2
• C. 1: 2: 3
• D. 2: 3: 1

Answer 1

Answer: (B)

Key Idea : Resultant force of two forces $P$ and $Q$ acting at an angle $\alpha,$ is given by
$R=\sqrt{P^{2}+Q^{2}+2 P Q \cos \alpha}$
If $\alpha$ is the angle between $\vec{ P }$ and $\vec{ Q }$, then
$R^{2}=P^{2}+Q^{2}+2 P Q \cos \alpha \dots$(i)
If $\vec{ Q }$ is doubled, then $\vec{ R }$ is doubled
$\Rightarrow 4 R^{2}=P^{2}+4 Q^{2}+4 P Q \cos \alpha \dots$(ii)
Again if the direction of $\vec{ Q }$ is reversed, then $\vec{ R }$ is again doubled
$\Rightarrow 4 R^{2}=P^{2}+(-Q)^{2}+2 P(-Q) \cos \alpha $
$ 4 R^{2}=P^{2}+Q^{2}-2 P Q \cos \alpha\dots$ (iii)
Adding (i) and (iii)
$5 R^{2}=2 P^{2}+2 Q^{2} \dots$(iv)
Multiplying by 2 in (iii) and adding (ii)
$12 R^{2} =3 P^{2}+6 Q^{2} $
$\Rightarrow 4 R^{2} =P^{2}+2 Q^{2} \dots$(v)
Subtracting (v) from (iv), we get
$R^{2}=P^{2}$
Putting in (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$