Question

Two forces of magnitudes $P$ and $Q$ are inclined at an angle $(\theta )$ the magnitude of their resultant is $3Q$. When the inclination is changed to $(180-\theta )$ the magnitude of the resultant force becomes $Q$. The ratio of the forces $\left(\frac{P}{Q}\right)$ is

• A. $\frac{4}{1}$
• B. $\frac{2}{1}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$(3Q{)}^2=P^2+Q^2+2PQ\cos \theta ...(i)$
$Q^2=P^2+Q^2+2PQ\cos \theta (180-\theta )...(ii)$
$9Q^2=P^2+Q^2+2PQ\cos \theta$
$Q^2=P^2+Q^2-2PQ\cos \theta$
Adding $10Q^2=2P^2+2Q^2$
or $8Q^2=2P^2$
or $\frac{P^2}{Q^2}=\frac{8}{2}=\frac{4}{1}$
or $\frac{P}{Q}=2$