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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

#### 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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#### Video TutorialsVIEW ALL [1]

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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