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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$8 \sqrt{3} π$

$6 \sqrt{2} π$

$\sqrt{3} π$

$8 \sqrt{2} π$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Sphere of radius $r = 3 c m$
Let $b, h$ be base radius and height of conc respectively.
So, volume of conc= $\frac{1}{2} π b^{2} h$
In right angles $Δ A B C$ by Pythagoras theorem
$(h^{2} - r^{2}) + b^{2} = r^{2}$
$\Rightarrow b^{2} = r^{2} - {(h - r)}^{2} = r^{2} - (h^{2} - 2 h r + r^{2}) = 2 h r - h^{2} .... (i)$
Volume $(v) = \frac{1}{3} π h [2 h r - h^{2}] = \frac{1}{3} [2 h^{2} r - h^{3}]$
$\frac{d v}{d h} = \frac{1}{3} [4 h r - 3 h^{2}] = 0 \Rightarrow h (4 r - 3 h) = 0$
$∴ h \neq 0 \Rightarrow \frac{4 r}{3} = 4; \frac{d^{2} v}{d h^{2}} = \frac{1}{3} [4 r - 6 h] (∵ r = 3)$
At $h = 4 \frac{d^{2} v}{d h^{2}} = - 4 < 0$
Maximum volume occurs at $h = 4 c m$
As from(i) $b^{2} = 2 h r - h^{2} = 8 \Rightarrow b = 2 \sqrt{2}$
Therefore curved surface area $= π b l$ $= π b \sqrt{h^{2} + b^{2}} = π 2 \sqrt{2} \sqrt{4^{2} + 8} = 8 \sqrt{3} π c m^{2}$