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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2024 · Paper 2 · Mathematics · Application of Derivatives

The maximum volume (in cu. units) of the cylinder which can be...

The maximum volume (in cu. units) of the cylinder which can be inscribed in a sphere of radius 12 units isThe maximum volume (in cu. units) of the cylinder which can be inscribed in a sphere of radius 12 units is

Options

  1. A3843π384 \sqrt{3} \pi
  2. B7683π768 \sqrt{3} \pi
  3. C768π3\frac{768 \pi}{\sqrt{3}}
  4. D1152π3\frac{1152 \pi}{\sqrt{3}}

Correct answer

B. 7683π768 \sqrt{3} \pi

Step-by-step solution

122=r2+(h2)2⇒V=πr2h⇒V=π(144−h24)h12^2=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 The maximum volume (in cu. units) of the cylinder which can be inscribed in a sphere of radius 12 units is ⇒V=144πh−π4h3⇒dVdh=144π−3π4h2⇒dVdh=0⇒144π=3π4h2⇒h2=48×4⇒h=83∴122=r2+48⇒r2=96\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 =πr2h=π×96×83=7683πcm3=\pi r^2 h=\pi \times 96 \times 8 \sqrt{3}=768 \sqrt{3} \pi \mathrm{ cm}^3.
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