Physics - NLM Question with Solution | TestHub

By symmetry the particle $P$ hangs vertically below the centroid $O$ of the equilateral triangle $ABC$ of side $2b$. The distance of each attachment point from the centroid is the circumradius:

$$\text{cos}30^o=\frac{b}{BO}=\frac{\sqrt{3}}{2}\Rightarrow BO=\frac{2b}{\sqrt{3}}$$

If $\theta$ is the angle each string makes with the vertical, then from the string of length $a$,

$$\text{cos}\theta =\frac{\sqrt{a^2-{\left(\frac{2b}{\sqrt{3}}\right)}^2}}{a}=\frac{\sqrt{3a^2-4b^2}}{\sqrt{3}a}$$

Vertical equilibrium of the particle gives

$$3T_i\text{cos}\theta =mg\Rightarrow T_i=\frac{mg}{3}\text{sec}\theta =\frac{mga}{\sqrt{3(3a^2-4b^2)}}$$

which is option (B).