Question

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

• A. $\frac{2}{3}$
• B. $\frac{3}{4}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A

$\frac{2}{3}$

Explanation for correct option:

Step 1: Calculate the volume of the cone.

Let, the sphere has a diameter of $AE=2r$.

The cone has a radius of $BD=x$ & height of $AD=y$.

Therefore, $DE=AE-AD=2r-y$.

Since,

$\begin{array}{rcl}{\left(BD\right)}^2& =& AD.DE\\ x^2& =& y(2r-y)\end{array}$

Therefore, the volume of the cone is

$$\begin{gathered} \mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathit{\pi }{\mathit{x}}^{\mathbf{2}}\mathbf{y} \\ =\frac{1}{3}\mathrm{\pi y}(2\mathrm{r}-\mathrm{y})\mathrm{y} \\ =\frac{1}{3}{\mathrm{\pi y}}^2(2\mathrm{r}-\mathrm{y}) \\ =\frac{1}{3}\mathrm{\pi }(2{\mathrm{ry}}^2-{\mathrm{y}}^3) \end{gathered}$$

Step: 2 Find out the value of $y$ for which the volume of the cone is maximum.

Now, $\frac{dv}{dy}=\frac{1}{3}\mathrm{\pi }(4ry-3{\mathrm{y}}^2)$

Here,

$\frac{dv}{dy}=0$

$\Rightarrow$$\frac{1}{3}\mathrm{\pi }(4\mathrm{ry}-3{\mathrm{y}}^2)=0$

$\Rightarrow$ $\mathrm{y}(4\mathrm{r}-3\mathrm{y})=0$

$\Rightarrow$ $\mathrm{y}=0,\frac{4}{3}\mathrm{r}$

Now,

$\frac{d^{2}y}{d{V}^2}=\frac{1}{3}\mathrm{\pi }(4\mathrm{r}-6\mathrm{y})$

Now, putting $y=\frac{4}{3}r$, then we get:

$$\begin{gathered} \frac{d^{2}y}{d{V}^2}=\frac{1}{3}\mathrm{\pi }[4\mathrm{r}-6(\frac{4}{3}r\left)\right] \\ =-ve \end{gathered}$$

Therefore, the volume of the cone is maximum at $y=\frac{4}{3}r$.

Step: 3 Calculate the ratio of the height and diameter of the cone.

Hence, the ratio of the height of the cone to the diameter of the sphere is

$\begin{array}{rcl}\frac{\mathbf{h}\mathbf{e}\mathbf{i}\mathbf{g}\mathbf{h}\mathbf{t}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{e}}{\mathbf{d}\mathbf{i}\mathbf{a}\mathbf{m}\mathbf{e}\mathbf{t}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{p}\mathbf{h}\mathbf{e}\mathbf{r}\mathbf{e}}& \mathbf{=}& \frac{\mathbf{y}}{\mathbf{2}\mathbf{r}}\\ & =& \frac{2}{3}\end{array}$

Therefore, Option(A) is the correct answer.