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