$\mathrm{R}=\frac{2}{3} \mathrm{AD}=\frac{2}{3} \times \mathrm{L} \sin 60^{\circ}=\frac{2 \mathrm{L}}{3} \times \frac{\sqrt{3}}{2}$
$=\frac{\mathrm{L}}{\sqrt{3}} \quad$ or $\quad \mathrm{L}=\sqrt{3} \mathrm{R}$
Centripetal force on any one mass $\mathrm{M}$ is
$=\frac{2 \mathrm{GM}^{2}}{\mathrm{L}^{2}} \cos \left(\frac{60^{\circ}}{2}\right)$
$\therefore \frac{\mathrm{Mv}^{2}}{\mathrm{R}}=\frac{2 \mathrm{GM}^{2}}{\mathrm{L}^{2}} \times \frac{\sqrt{3}}{2}$
or $\frac{\mathrm{GM}^{2}}{(\sqrt{3} \mathrm{R})^{2}} \sqrt{3}=\frac{\mathrm{Mv}^{2}}{\mathrm{R}}$
$\therefore v=\sqrt{\frac{\mathrm{GM}}{\sqrt{3} \mathrm{R}}}$
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Simultaneously at $t=0$, a small pebble is projected with speed $v$ from point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \ cm$ from $O$. If the pebble hits the block at $t=1 \ s$, the value of $v$ is (take $g =10 \ m / s ^2$ )
