A sector of circle of radius 3 cm has an angle of 120°. If it is modulated into a cone, find then volume of the cone is

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\frac{π}{\sqrt{3}} c m^{3}

\frac{2 \sqrt{3}}{π} c m^{3}

\frac{2 \sqrt{2} π}{3} c m^{3}

\frac{\sqrt{3}}{π} c m^{3}

answer is B.

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Detailed Solution

When a sector modulated into cone

Length of the sector $=$ circumference of the base of the cone

$\frac{120 °}{360 °} \times 2 π (3) = 2 π r$

$r = 1 c m$ (radius of the base of the cone)

Slant height of the cone $=$ radius of the sector

$∴ l = 3 c m$

Then $h = \sqrt{l^{2} - r^{2}} = \sqrt{3^{2} - 1^{2}} = \sqrt{8} = 2 \sqrt{2}$

Volume of the cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times π {(1)}^{2} 2 \sqrt{2} = \frac{2 \sqrt{2} π}{3}$

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