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 answer
    B
    $768 \sqrt{3} \pi$
  • C
    $\frac{768 \pi}{\sqrt{3}}$
  • D
    $\frac{1152 \pi}{\sqrt{3}}$

Solution

$12^2=r^2+\left(\frac{h}{2}\right)^2 \Rightarrow V=\pi r^2 h$ $\Rightarrow \quad V=\pi\left(144-\frac{h^2}{4}\right) h$ Figure $\Rightarrow \quad V=144 \pi h-\frac{\pi}{4} h^3 \Rightarrow \frac{d V}{d h}=144 \pi-\frac{3 \pi}{4} h^2$ $\Rightarrow \frac{d V}{d h}=0 \Rightarrow 144 \pi=\frac{3 \pi}{4} h^2$ $\Rightarrow \quad h^2=48 \times 4 \Rightarrow h=8 \sqrt{3}$ $\therefore \quad 12^2=r^2+48 \Rightarrow r^2=96$ 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'.

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