A) 2
B) \[-1\]
C) 1
D) 0
Correct Answer: D
Solution :
[d] Let radius of the base of cone = r cm. Then, curved surface area (c) \[=\pi rl=\pi r\sqrt{{{r}^{2}}+{{h}^{2}}}\] Volume \[(v)=\frac{1}{3}\pi {{r}^{2}}h\] Now,\[3\pi v{{h}^{3}}-{{c}^{2}}{{h}^{2}}+9{{v}^{2}}\] \[=3\pi \frac{1}{3}\pi {{r}^{2}}h\,{{h}^{3}}-{{\pi }^{2}}{{r}^{2}}({{r}^{2}}+{{h}^{2}}){{h}^{2}}\] \[+\,9\cdot \frac{1}{9}{{\pi }^{2}}{{r}^{4}}{{h}^{2}}\] \[={{\pi }^{2}}{{r}^{2}}{{h}^{4}}-{{\pi }^{2}}{{r}^{2}}{{h}^{2}}-{{\pi }^{2}}{{r}^{2}}{{h}^{4}}+{{\pi }^{2}}{{r}^{4}}{{h}^{2}}=0\] |
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