A) 2
B) 1
C) \[\sqrt{2}\]
D) \[\sqrt{3}\]
Correct Answer: C
Solution :
Volume of cone, \[V=\frac{\pi }{3}{{r}^{2}}h\] \[\Rightarrow \] \[V=\frac{\pi }{3}{{r}^{2}}\sqrt{{{l}^{2}}-{{r}^{2}}}\] On differentiating w.r.t., r we get \[\frac{dV}{dr}=\frac{\pi }{3}\left[ 2r\sqrt{{{l}^{2}}-{{r}^{2}}}+\frac{{{r}^{2}}}{2\sqrt{{{l}^{2}}-{{r}^{2}}}}(-2r) \right]\] Put \[\frac{dV}{dr}=0\] \[\Rightarrow \]\[2r(\sqrt{{{l}^{2}}-{{r}^{2}}})-\frac{{{r}^{3}}}{\sqrt{{{l}^{2}}-{{r}^{2}}}}=0\] \[\Rightarrow \]\[r[2({{l}^{2}}-{{r}^{2}})-{{r}^{3}}]=0\] \[\Rightarrow \] \[2{{l}^{2}}-3{{r}^{2}}=0\] \[\Rightarrow \] \[r=\pm \,\,l\sqrt{\frac{2}{3}}\] \[\therefore \]At \[r=l\sqrt{\frac{2}{3}},\frac{{{d}^{2}}V}{d{{r}^{2}}}<0,\]maxima \[\therefore \] \[h=\sqrt{{{l}^{2}}-\frac{2}{3}{{l}^{2}}}=\frac{l}{\sqrt{3}}\] In \[\Delta \Alpha BC,\tan \theta =\frac{r}{h}\] \[=\frac{l\sqrt{\frac{2}{3}}}{\frac{l}{\sqrt{3}}}=\sqrt{2}\]
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