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Solution for If a Cone of Maximum Volume is Inscribed in a Given Sphere, Then the Ratio of the Height of the Cone to the Diameter of the Sphere is (A) 3 4 (B) 1 3 (C) 1 4 (D) 2 3 - CBSE (Commerce) Class 12 - Mathematics

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Question

If a cone of maximum volume is inscribed in a given sphere, then the ratio of the height of the cone to the diameter of the sphere is

(a) \[\frac{3}{4}\]

(b) \[\frac{1}{3}\]

(c) \[\frac{1}{4}\]

(d) \[\frac{2}{3}\]

Solution

(d) \[\frac{2}{3}\]

\[\text { Let h, r, V and R be the height, radius of the base, volume of the cone and the radius of the sphere, respectively } . \]

\[\text { Given:} h = R + \sqrt{R^2 - r^2}\]

\[ \Rightarrow h - R = \sqrt{R^2 - r^2}\]

\[\text { Squaring both side, we get }\]

\[ h^2 + R^2 - 2hR = R^2 - r^2 \]

\[ \Rightarrow r^2 = 2hr - h^2 . . . \left( 1 \right)\]

\[\text { Now,} \]

\[\text { Volume } = \frac{1}{3}\pi r^2 h\]

\[ \Rightarrow V = \frac{\pi}{3}\left( 2 h^2 R - h^3 \right) \left[\text {  From eq }. \left( 1 \right) \right]\]

\[ \Rightarrow \frac{dV}{dh} = \frac{\pi}{3}\left( 4hR - 3 h^2 \right)\]

\[\text { For maximum or minimum values of V, we must have }\]

\[\frac{dV}{dh} = 0\]

\[ \Rightarrow \frac{\pi}{3}\left( 4hR - 3 h^2 \right) = 0\]

\[ \Rightarrow 4hR - 3 h^2 = 0\]

\[ \Rightarrow 4hR = 3 h^2 \]

\[ \Rightarrow h = \frac{4R}{3}\]

\[\text { Now,} \]

\[\frac{d^2 V}{d h^2} = \frac{\pi}{3}\left( 4R - 6h \right) = \frac{\pi}{3}\left( 4R - 6 \times \frac{4R}{3} \right) = - \frac{4\piR}{3} < 0\]

\[\text { So, volume is maximum when h } = \frac{4R}{3} . \]

\[ \Rightarrow h = \frac{2\left( 2R \right)}{3}\]

\[ \Rightarrow \frac{h}{2R} = \frac{2}{3}\]

\[ \therefore \frac{\text { Height }}{\text { Diameter of sphere }} = \frac{2}{3}\]

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Solution If a Cone of Maximum Volume is Inscribed in a Given Sphere, Then the Ratio of the Height of the Cone to the Diameter of the Sphere is (A) 3 4 (B) 1 3 (C) 1 4 (D) 2 3 Concept: Graph of Maxima and Minima.
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