Question

Two forces simultaneously act on a particle making an angle ${120}^o$ with each other, if one of them is reversed the acceleration of the particle becomes $\sqrt{3}$ times its initial value. The ratio of the magnitude of the forces is:

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

Answer 1

The correct option is B $1:1$

Let $P$ and $Q$ be the forces,

$\begin{array}{c}Infirstcase:\\ R^2=P^2+Q^2+2PQ\cos 120\\ =P^2+Q^2-PQ\left(\sin ce\cos 60=\frac{1}{2}and\cos 120=-\cos 60\right)\\ In\sec ondcase;\\ S^2=P^2+Q^2+2PQ\cos 60\\ =P^2+Q^2+PQ\\ Given,\\ S=\sqrt{3}\times R\\ or{S}^2=3R^2\end{array}$

Plug and we get $P=Q$

Hence the ratio of forces $1:1$