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(B) Let the radius of the metallic disc be $R$. When a sector is removed and the remaining part is bent to form a cone,the radius $R$ of the disc beco

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$6$

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(B) Let the radius of the metallic disc be $R$. When a sector is removed and the remaining part is bent to form a cone,the radius $R$ of the disc becomes the slant height $l$ of the cone.Let the radius of the base of the cone be $x$ and its height be $h$.We have the relation $R^2 = x^2 + h^2$,where $R = l$.The volume of the cone is given by $V = \frac{1}{3} \pi x^2 h = 2 \sqrt{3} \pi$.Thus,$x^2 h = 6 \sqrt{3}$,which implies $x^2 = \frac{6 \sqrt{3}}{h}$.Substituting this into the expression for $R^2$:$R^2 = \frac{6 \sqrt{3}}{h} + h^2$.To find the minimum diameter,we minimize $R^2$ with respect to $h$:$\frac{d(R^2)}{dh} = -\frac{6 \sqrt{3}}{h^2} + 2h$.Setting $\frac{d(R^2)}{dh} = 0$,we get $2h = \frac{6 \sqrt{3}}{h^2}$,so $h^3 = 3 \sqrt{3}$.This gives $h = \sqrt{3}$.Then $x^2 = \frac{6 \sqrt{3}}{\sqrt{3}} = 6$.Therefore,$R^2 = 6 + (\sqrt{3})^2 = 6 + 3 = 9$,so $R = 3$.The diameter of the disc is $2R = 2 	imes 3 = 6$.

$A$ sector is removed from a metallic disc and the remaining region is bent into the shape of a circular conical funnel with volume $2 \sqrt{3} \pi$. The least possible diameter of the disc is

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