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

Answer 1

Answer: (D)

$V=\frac{1}{3}\pi r^{2}h$
where r is radius and h is height of coin
$\Rightarrow V=\frac{1}{3}\pi {\left(3sin2\theta \right)}^2\left(3+3cos2\theta \right)$
$=72\pi si{n}^2\theta co{s}^4\theta$
$\frac{dv}{d\theta }=72\pi \left[2sin\theta co{s}^5\theta -4si{n}^3\theta co{s}^2\theta \right]=0\Rightarrow 0ta{n}^2\theta =\frac{1}{2}$
$V_{mx}iftan=\frac{1}{\sqrt{2}}$
Hence curved surface area S $=\pi r\ell$
$=\pi r\sqrt{{\left(3+3cos20\right)}^2+{\left(3sin2\theta \right)}^2}$
$=\pi \left(3sin2\theta \right)\sqrt{36si{n}^2\theta }=18\pi \left(2sin\theta co{s}^2\theta \right)=36\pi .$
$\frac{1}{\sqrt{3}}.\frac{2}{3}=\frac{24\pi }{3}=8\sqrt{3}\pi$