In a sphere of radius r , a right circular cone of height h having a…

The correct answer is Option (3) → \( \sqrt{2r}\,\pi\sqrt{(2h^2r-h^3)} \)

The expression asked for is the curved surface area S of the inscribed cone in terms of h, the function that is maximised.

Let the base radius of the cone be R. The vertex is on the sphere and the centre of the sphere is at distance h − r from the base, so by Pythagoras' theorem:

    \( \displaystyle R^2+(h-r)^2=r^2 \)

    \( \displaystyle R^2=2rh-h^2 \)

The slant height l:

    \( \displaystyle l^2=R^2+h^2=2rh \)

    \( \displaystyle l=\sqrt{2r}\sqrt{h} \)

Curved surface area \( S=\pi Rl \):

    \( \displaystyle S=\pi\sqrt{2rh-h^2}\cdot\sqrt{2r}\sqrt{h} \)

    \( \displaystyle =\sqrt{2r}\,\pi\sqrt{h(2rh-h^2)} \)

    \( \displaystyle =\sqrt{2r}\,\pi\sqrt{2h^2r-h^3} \)

Why the other options are incorrect

• Option (1) \( \sqrt{2}\pi\sqrt{(2h^2r-h^3)} \) — Incorrect. The factor \( \sqrt{2} \) should be \( \sqrt{2r} \), since \( l=\sqrt{2r}\sqrt{h} \).

• Option (2) \( \sqrt{2r}\,\pi\sqrt{(2hr-h^2)} \) — Incorrect. This is \( \pi Rl \) with \( l=\sqrt{2r} \); the factor \( \sqrt{h} \) of the slant height is missing.

• Option (4) \( \sqrt{2}\,\pi r\sqrt{(2h^2r-h^3)} \) — Incorrect. The factor outside is \( \sqrt{2r} \), not \( \sqrt{2}\,r \).