A cone is made from a circular shet of radius $\sqrt{3}$ by cutting out a sector and giving the cut edges of the remaining piece together. The maximum volume attainable for the cone is

π / 3

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π / 6

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2 π / 3

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3 \sqrt{3} π

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Solution

The correct option is D $2 π / 3$
radius of the sheet will be equal to slant height (l) of cone
Therefore, $l = \sqrt{3}$
In any cone: $r^{2} + h^{2} = l^{2} = 3$
Volume of cone $V = \frac{π r^{2} h}{3} = \frac{π (3 - h^{2}) h}{3}$
for maximum volume $\frac{d V}{d h} = 0$
$\Rightarrow \frac{3 π (1 - h^{2})}{3} = 0$
$\Rightarrow h = 1$
now, $\frac{d^{2} V}{d h^{2}} = - 2 h$
Since, at $h = 1$ $V < 0$
Therefore, $V$ is max, at $h = 1$
Thus, $m a x (V) = \frac{2 π}{3}$
Ans: C

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