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3-2.Motion in Plane
hard
A single wire $ACB$ passes through a smooth ring at $C$ which revolves at a constant speed in the horizontal circle of radius $r$ as shown in the figure. The speed of revolution is
A$\sqrt{rg}$
B$\sqrt{2rg}$
C$2\sqrt{2rg}$
D$2\sqrt{rg}$
Solution

$\mathrm{T} \cos 30^{\circ}+\mathrm{T} \cos 60^{\circ}=\mathrm{mg}$
$\mathrm{T} \times\left(\frac{\sqrt{3}+1}{2}\right)=\mathrm{mg}$ $…(i)$
along $x-axis$
$\mathrm{T} \sin 30^{\circ}+\mathrm{T} \sin 60^{\circ}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$
$\mathrm{T}\left(\frac{\sqrt{3}+1}{2}\right)=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$ $…(ii)$
$(ii)$ $/( i)$ we get
$\mathrm{v}=\sqrt{\mathrm{rg}}$
Standard 11
Physics