Question Bank › General Aptitude › Volume and Surface Area › ABC26GN4962ABC26GN4962 · Volume and Surface Area Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: · Marks: · Difficulty:
If the height, curved surface area and the volume of a cone are $h, c$ and $v$ respectively, then $3 \pi v h^{3}-c^{2} h^{2}$ $+9 v^{2}$ will be equal to
(a) 0
(b) 1
(c) chv
(d) $v^{2} h$
Answer Explanation Volume of the cone, $v=\frac{1}{3} \pi r^{2} h$. Curved surface area, $c=\pi r l=\pi r \sqrt{r^{2}+h^{2}}$ $$\begin{aligned} & \Rightarrow c^{2}=\pi^{2} r^{2}\left(r^{2}+h^{2}\right) . \\ & \therefore 3 \pi v h^{3}-c^{2} h^{2}+9 v^{2} \\ & =3 \pi \times \frac{1}{3} \pi r^{2} h \times h^{3}-\pi^{2} r^{2}\left(r^{2}+h^{2}\right) h^{2}+9 \times \frac{1}{9} \pi^{2} r^{4} h^{2} \\ & \quad=\pi^{2} r^{2} h^{4}-\pi^{2} r^{4} h^{2}-\pi^{2} r^{2} h^{4}+\pi^{2} r^{4} h^{2}=0 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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