If h, c, v are respectively the height,curved | vedclass

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(D) Let the radius of the base of the cone be $r$ units.The volume $v$ is given by $v = \frac{1}{3} \pi r^{2} h$.The curved surface area $c$ is given

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(D) Let the radius of the base of the cone be $r$ units.The volume $v$ is given by $v = \frac{1}{3} \pi r^{2} h$.The curved surface area $c$ is given by $c = \pi r \sqrt{h^{2} + r^{2}}$.Now,substitute these into the expression $3 \pi v h^{3} - c^{2} h^{2} + 9 v^{2}$:$3 \pi v h^{3} = 3 \pi (\frac{1}{3} \pi r^{2} h) h^{3} = \pi^{2} r^{2} h^{4}$.$c^{2} h^{2} = (\pi r \sqrt{h^{2} + r^{2}})^{2} h^{2} = \pi^{2} r^{2} (h^{2} + r^{2}) h^{2} = \pi^{2} r^{2} h^{4} + \pi^{2} r^{4} h^{2}$.$9 v^{2} = 9 (\frac{1}{3} \pi r^{2} h)^{2} = 9 (\frac{1}{9} \pi^{2} r^{4} h^{2}) = \pi^{2} r^{4} h^{2}$.Substituting these back into the expression:$(\pi^{2} r^{2} h^{4}) - (\pi^{2} r^{2} h^{4} + \pi^{2} r^{4} h^{2}) + (\pi^{2} r^{4} h^{2}) = \pi^{2} r^{2} h^{4} - \pi^{2} r^{2} h^{4} - \pi^{2} r^{4} h^{2} + \pi^{2} r^{4} h^{2} = 0$.

If $h, c, v$ are respectively the height,curved surface area,and volume of a right circular cone,then the value of $3 \pi v h^{3} - c^{2} h^{2} + 9 v^{2}$ is

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