Question : If h, c, v are respectively the height, curved surface area, and volume of a right circular cone, then the value of $3\pi v h^{3}-c^{2}h^{2}+9v^{2}$ is:
Option 1: 2
Option 2: –1
Option 3: 1
Option 4: 0
Correct Answer: 0
Solution :
Let $r$ be the radius, $h$ be the height, $l$ be the slant height and $c$ be the curved surface area of the cone.
We know, $l=\sqrt{r^2+h^2},v=\frac{1}{3}\pi r^2 h,$ and $c=\pi r l$
According to the question:
$3\pi v h^{3}-c^{2}h^{2}+9v^{2}$
= $3\pi ×\frac{1}{3}\pi r^2 h× h^{3}-(\pi r l)^{2}h^{2}+9(\frac{1}{3}\pi r^2 h)^{2}$
= $\pi^2 r^2 h^4-\pi^2 r^2 l^2h^{2}+\pi^2 r^4 h^{2}$
Putting the value of $l$, we get:
= $\pi^2 r^2 h^4-\pi^2 r^2 h^{2}(r^2+h^2)+\pi^2 r^4 h^{2}$
= $\pi^2 r^2 h^4-\pi^2 r^4 h^{2}-\pi^2 r^2 h^4+\pi^2 r^4 h^{2}$
= $0$
Hence, the correct answer is 0.
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