Question

If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm²) of this cone is :

• A. \(6\sqrt{2}\,\pi\)
• B. \(6\sqrt{3}\,\pi\)
• C. \(8\sqrt{2}\,\pi\)
• D. \(8\sqrt{3}\,\pi\)

Hint:

Express the cone volume in terms of its height \(h\) using \(r^2=h(2R-h)\), maximise to get \(h=4\), then use \(\pi r l\).

Answer 1

The Correct Option is D

Let the cone have height \(h\) and base radius \(r\) in a sphere of radius \(R=3\). Then \(r^2=h(2R-h)=h(6-h)\).
\(V=\frac{1}{3}\pi r^2h=\frac{\pi}{3}(6h^2-h^3)\).
\(\frac{dV}{dh}=\frac{\pi}{3}(12h-3h^2)=0\Rightarrow h=4\), and \(\frac{d^2V}{dh^2}=\frac{\pi}{3}(12-6h)<0\) at \(h=4\), so V is maximum.
\(r^2=4\times2=8\Rightarrow r=2\sqrt{2}\). Slant height \(l=\sqrt{r^2+h^2}=\sqrt{8+16}=2\sqrt{6}\).
Curved surface area \(=\pi r l=\pi\cdot2\sqrt{2}\cdot2\sqrt{6}=4\pi\sqrt{12}=8\sqrt{3}\,\pi\) cm².
Answer: (4) \(8\sqrt{3}\,\pi\)