Question

If a right circular cone, having maximum volume, is inscribed in a sphere of radius $3cm$, then the curved surface area (in $c{m}^2)$ of this cone is

• A. $8\sqrt{3}\pi$
• B. $6\sqrt{2}\pi$
• C. $6\sqrt{3}\pi$
• D. $8\sqrt{2}\pi$

Answer 1

The correct option is A $8\sqrt{3}\pi$

Sphere of radius : $3cm(r=3)$

Let $b,h$ be radius and height of sphere, respectively.

$\therefore$ volume of cone $=\frac{1}{3}\pi b^{2}h$

In $△ABC$, using Pythagoras theorem

$(h–r{)}^2+b^2=r^2\dots \dots (i)$

$b^2=r^2–(h–r{)}^2=r^2–(h^2–2hr+r^2)=2hr–h^2$

$\therefore$ Volume $v=\frac{1}{3}h\pi [r^2-(h-r{)}^2]$

$=\frac{1}{3}\pi h[2hr–h^2]=\frac{1}{3}[2h^{2}r–h^3]$

$\frac{dv}{dh}=\frac{1}{3}[4hr–3h^2]=0\Rightarrow h(4r–3h)=0$

$\frac{d^{2}v}{d{h}^2}=\frac{1}{3}[4r–6h]$

At $h=\frac{4r}{3},\frac{d^{2}v}{d{h}^2}=\frac{1}{3}[4r-\frac{4r}{3}\times 6]$

$=\frac{1}{3}[4r–8r]<0\Rightarrow$ maximum volume at $h=\frac{4r}{3}$

$h=\frac{4r}{3}=4$

$\therefore$ From $\left(1\right)$

$(h–r{)}^2+b^2=r^2$

$\Rightarrow b^2=2hr–h^2$

$=2\cdot \frac{4r}{3}r-\frac{16r^2}{9}$

$=\frac{8r^2}{3}-\frac{16r^2}{9}$

$=\frac{(24–16{)}^{r^2}}{9}=\frac{8r^2}{9}$

$\Rightarrow b=\frac{2\sqrt{2}}{3}r\Rightarrow 2\sqrt{2}$

Curved surface area $=\pi bl$

$=\pi b\sqrt{h^2+r^2}$

$=\pi 2\sqrt{2}\sqrt{4^2+8}$

$=\pi 2\sqrt{2}\sqrt{24}$

$=\pi 2\sqrt{2}2\sqrt{3}\sqrt{2}$

$=8\sqrt{3}\pi$.