GATE 2022 AE – Question 36

Correct answer: (C) $4R/3$

Explanation

**Setup.** A right circular cone of height $h$ is inscribed in a sphere of radius $R$, with its apex at the top of the sphere. The base circle is a chord section of the sphere at depth $h$ below the apex. By the intersecting-chords relation (or Pythagoras), the base radius $r$ satisfies
$$r^2=h(2R-h).$$

**Volume of the cone:**
$$V=\tfrac13\pi r^2h=\tfrac13\pi h^2(2R-h)=\tfrac{\pi}{3}\left(2Rh^2-h^3\right).$$

**Maximise:**
$$\frac{dV}{dh}=\tfrac\pi3\left(4Rh-3h^2\right)=0\;\Rightarrow\;h(4R-3h)=0\;\Rightarrow\;h=\frac{4R}{3}.$$

**Check:** $\dfrac{d^2V}{dh^2}=\tfrac\pi3(4R-6h)$, which at $h=4R/3$ is $\tfrac\pi3(4R-8R)<0$, so it is a maximum.

Answer $h=\dfrac{4R}{3}$ (option C).