MCQ
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
  • $\sqrt{rg}$
  • B
    $\sqrt{2rg}$
  • C
    $2\sqrt{2rg}$
  • D
    $2\sqrt{rg}$

Answer

Correct option: A.
$\sqrt{rg}$
a
Force equation from ring frame along $y-axis$

$\mathrm{T} \cos 30^{\circ}+\mathrm{T} \cos 60^{\circ}=\mathrm{mg}$

$\mathrm{T} \times\left(\frac{\sqrt{3}+1}{2}\right)=\mathrm{mg}$          $...(i)$

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

$\mathrm{v}=\sqrt{\mathrm{rg}}$

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