Question

Two forces of magnitudes $P$ and $Q$ are inclined at an angle $(\theta)$ the magnitude of their resultant is $3 Q$. 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)

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