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The semi-vertical angle of the cone of the maximum volume and of given slant height is.

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\tan^{- 1} \sqrt{5}

\tan^{- 1} \sqrt{8}

\tan^{- 1} \sqrt{3}

\tan^{- 1} \sqrt{2}

answer is D.

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Detailed Solution

Let, height of cone

= h

Radius of cone

= r

Slant height

= l

We know slant height is

l^{2} = r^{2} + h^{2}

Volume

(V)

of the cone

= \frac{π}{3} r^{2} h

= \frac{π}{3} (l^{2} - h^{2}) h

= \frac{π}{3} (l^{2} h - h^{3})

Now

V = \frac{π}{3} (l^{2} h - h^{3})

\frac{dV}{dh} = \frac{π}{3} (l^{2} - 3 h^{2}) = 0

\Rightarrow l^{2} = 3 h^{2}

\Rightarrow l = \sqrt{3} h

Suppose,

θ

be the semi vertical angle. So,

\Rightarrow \cos θ = \frac{height}{slant height} = \frac{h}{l}

\cos θ = \frac{h}{h \sqrt{3}}

Now we know that,

∴ \tan^{2} θ = \sec^{2} θ - 1,

{\Rightarrow \tan}^{2} θ = \frac{1}{\cos^{2} θ} - 1

(since,

\sec θ = \frac{1}{\cos θ}

)

\Rightarrow \tan^{2} θ = \frac{1 - \cos^{2} θ}{\cos^{2} θ}

\Rightarrow \tan θ = \frac{\sqrt{1 - \cos^{2} θ}}{\cos θ}

\Rightarrow \tan θ = \frac{\sqrt{1 - \frac{1}{3}}}{\frac{1}{\sqrt{3}}}

\Rightarrow \tan θ = \frac{\frac{\sqrt{2}}{\sqrt{3}}}{\sqrt{3}}

\Rightarrow \tan θ = \sqrt{2}

\Rightarrow θ = \tan^{- 1} \sqrt{2}

Correct option is 4.

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