Q57 / 100 · BITSAT 2024 (Memory Based Paper 2) · Mathematics
The maximum volume (in cu. units) of the cylinder which can be inscribed in a sphere of radius 12 units is
- A$384 \sqrt{3} \pi$
- Correct answerB$768 \sqrt{3} \pi$
- C$\frac{768 \pi}{\sqrt{3}}$
- D$\frac{1152 \pi}{\sqrt{3}}$
Solution
$12^2=r^2+\left(\frac{h}{2}\right)^2 \Rightarrow V=\pi r^2 h$
$\Rightarrow \quad V=\pi\left(144-\frac{h^2}{4}\right) h$
$\Rightarrow \quad V=144 \pi h-\frac{\pi}{4} h^3 \Rightarrow \frac{d V}{d h}=144 \pi-\frac{3 \pi}{4} h^2$
$\Rightarrow \frac{d V}{d h}=0 \Rightarrow 144 \pi=\frac{3 \pi}{4} h^2$
$\Rightarrow \quad h^2=48 \times 4 \Rightarrow h=8 \sqrt{3}$
$\therefore \quad 12^2=r^2+48 \Rightarrow r^2=96$
Volume $=\pi r^2 h=\pi \times 96 \times 8 \sqrt{3}=768 \sqrt{3} \pi \mathrm{~cm}^3$.
$\Rightarrow \quad V=144 \pi h-\frac{\pi}{4} h^3 \Rightarrow \frac{d V}{d h}=144 \pi-\frac{3 \pi}{4} h^2$
$\Rightarrow \frac{d V}{d h}=0 \Rightarrow 144 \pi=\frac{3 \pi}{4} h^2$
$\Rightarrow \quad h^2=48 \times 4 \Rightarrow h=8 \sqrt{3}$
$\therefore \quad 12^2=r^2+48 \Rightarrow r^2=96$
Volume $=\pi r^2 h=\pi \times 96 \times 8 \sqrt{3}=768 \sqrt{3} \pi \mathrm{~cm}^3$.Hence, the correct answer is option: 'B'.