Question

A given right circular cone has a volume $p$ and the longest right circular cylinder that can be inscribed in the cone has a volume $q$. Then $p:q$ is :

• A. $9:4$
• B. $4:5$
• C. $7:3$
• D. None of these

Answer 1

The correct option is A

$9:4$

Explanation for the correct option:

Step 1- From the given information we get.

Let $H$ and $R$ be the height and radius of the cylinder.

Let $h$ and $r$ be the height and radius of the cone.

Let the angle be $\alpha$

The volume of the cone be $p$ and the volume of the cylinder be $q$.

The formula for volume of cylinder is, $\mathit{V}\mathbf{=}\mathit{\pi }{\mathit{R}}^{\mathbf{2}}\mathbf{H}$

Differentiate the volume twice to get the relation between $p$ and $q$.

Step 2: Finding the volume of the cylinder.

From the figure, $\tan \left(\alpha \right)=\frac{r}{h}$

⇒ $r=h\tan \left(\alpha \right)$ ------------ (1)

Also, $h=r\cot \left(\alpha \right)$

Now, $H=h-r\cot \left(\alpha \right)$

Volume of the cylinder,

$$\begin{gathered} V={\mathrm{\pi R}}^2\mathrm{H} \\ ={\mathrm{\pi R}}^2\left(\mathrm{h}-\mathrm{Rcot}\left(\mathrm{\alpha }\right)\right) \\ ={\mathrm{\pi R}}^2\mathrm{h}-{\mathrm{\pi R}}^3\cot \left(\mathrm{\alpha }\right) \end{gathered}$$

Step 3: Taking first derivative.

Differentiate $V$ with respect to $R$,

$\Rightarrow \frac{dV}{dR}=2\mathrm{\pi Rh}-3{\mathrm{\pi R}}^2\cot \left(\mathrm{\alpha }\right)$

Put, $\frac{dV}{dR}=0,$

$$\begin{gathered} 0=\mathrm{\pi }\left(2\mathrm{Rh}-3{\mathrm{R}}^2\cot \left(\mathrm{\alpha }\right)\right) \\ 2\mathrm{Rh}=3{\mathrm{R}}^2\cot \left(\mathrm{\alpha }\right) \\ \mathrm{R}=\frac{2}{3}\mathrm{htan}\left(\mathrm{\alpha }\right) \end{gathered}$$

Step 4: Taking the second derivative.

Differentiate $\frac{dV}{dR}$again with respect to $R$,

$$\begin{gathered} \frac{d^{2}V}{d{R}^2}=2\mathrm{\pi h}-6\mathrm{\pi Rcot}\left(\mathrm{\alpha }\right) \\ =2\mathrm{\pi h}-6\mathrm{\pi }\left(\frac{2}{3}\mathrm{htan}\left(\mathrm{\alpha }\right)\right)\cot \left(\mathrm{\alpha }\right)\left[\because R=\frac{2}{3}htan\left(\alpha \right)\right] \\ =2\mathrm{\pi h}-4\mathrm{\pi h} \\ =-2\mathrm{\pi h}<0 \end{gathered}$$

Thus the volume of the cylinder is maximum, when $R=\frac{2}{3}h\tan \left(\alpha \right)$.

Step 5: Finding the height H.

From diagram,

$$\begin{gathered} H=\frac{h}{r}\left(r-R\right) \\ =\frac{h}{h\tan \left(\alpha \right)}\left(h\tan \left(\alpha \right)-R\right)\mathbf{}\mathbf{}\left[\because r=htan\left(\alpha \right)\right] \\ =\frac{h\tan \left(\alpha \right)-R}{\tan \left(\alpha \right)} \\ =h-\frac{1}{\tan \left(\alpha \right)}\frac{2}{3}h\tan \left(\alpha \right)\mathbf{}\mathbf{}\left[\because R=\frac{2}{3}htan\left(\alpha \right)\right] \\ =\frac{h}{3} \end{gathered}$$

Step 6: Finding $p$and $q$.

The volume of a cone,

$$\begin{gathered} p=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} \\ =\frac{1}{3}\mathrm{\pi }{\left(\mathrm{htan}\left(\mathrm{\alpha }\right)\right)}^2\mathrm{h} \\ =\frac{1}{3}{\mathrm{\pi h}}^3{\tan }^2\left(\mathrm{\alpha }\right) \end{gathered}$$

The volume of the cylinder,

Substitute the value of R and H

$$\begin{gathered} q={\mathrm{\pi R}}^2\mathrm{H} \\ =\mathrm{\pi }{\left(\frac{2}{3}\mathrm{htan}\left(\mathrm{\alpha }\right)\right)}^2\left(\frac{\mathrm{h}}{3}\right) \\ =\frac{4}{27}{\mathrm{h}}^3{\tan }^2\left(\mathrm{\alpha }\right) \end{gathered}$$

Step 7: Finding the ratio $p:q$

$$\begin{gathered} \frac{p}{q}=\frac{{\displaystyle \frac{1}{3}}{\mathrm{\pi h}}^3{\tan }^2\left(\mathrm{\alpha }\right)}{{\displaystyle \frac{4}{27}}{\mathrm{\pi h}}^3{\tan }^2\left(\mathrm{\alpha }\right)} \\ =\frac{9}{4} \end{gathered}$$

Hence, option (A) is the correct answer.