JEE Main 2019 — Gravitation Question with Solution

Net force on mass M at position B towards centre of circle is

$$F_{BOnet}=F_{BD}+F_{BA}\sin 45^{\circ }+F_{BC}\sin 45^{\circ }$$

$$=\frac{G{M}^2}{{(\sqrt{2}a)}^2}+\frac{G{M}^2}{a^2}\left(\frac{1}{\sqrt{2}}\right)+\frac{G{M}^2}{a^2}\left(\frac{1}{\sqrt{2}}\right)$$

[where, diagonal length BD is $$\sqrt{2}$$a]

$$=\frac{G{M}^2}{2a^2}+\frac{G{M}^2}{a^2}\left(\frac{2}{\sqrt{2}}\right)=\frac{G{M}^2}{a^2}\left(\frac{1}{2}+\sqrt{2}\right)$$

This force will act as centripetal force.

Distance of particle from centre of circle is $$\frac{a}{\sqrt{2}}$$.

Here, $$F_{centripetal}=\frac{M{v}^2}{r}=\frac{M{v}^2}{\frac{a}{\sqrt{2}}}=\frac{\sqrt{2}M{v}^2}{a}$$ ($$\because$$ $$r=\frac{a}{\sqrt{2}}$$)

So, for rotation about the centre,

$$F_{centripetal}=F_{BO(net)}$$

$$\Rightarrow \sqrt{2}\frac{M{v}^2}{a}=\frac{G{M}^2}{a^2}\left(\frac{1}{2}+\sqrt{2}\right)$$

$$\Rightarrow v^2=\frac{GM}{a}\left(1+\frac{1}{2\sqrt{2}}\right)=\frac{GM}{a}(1.35)$$

$$\Rightarrow v=1.16\sqrt{\frac{GM}{a}}$$