The height of the maximum cone that can be obtained by revolving a right angled triangle of hypotenuse 'l' units about a side is

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$\frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{3}}$

$\frac{1}{3}$

$2 l$

answer is C.

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Detailed Solution

$l^{2} = h^{2} + r^{2} volume of cone v = \frac{π}{3} r^{2} h = \frac{π}{3} (l^{2} - h^{2}) h v = \frac{π}{3} (l^{2} h - h^{3}) v^{1} = \frac{π}{3} (l^{2} - 3 h^{2}) vismin or \max \Rightarrow v^{1} = 0 l^{2} = 3 h^{2} \Rightarrow h = \frac{l}{\sqrt{3}}$

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