A) 2/3
B) 3/4
C) 1/3
D) ¼
Correct Answer: A
Solution :
Let diameter of sphere \[AE=2r\] Let radius of cone is x and height is y \ \[AD=y\], since \[B{{D}^{2}}=AD.DE\] or \[{{x}^{2}}=y(2r-y)\] ??(i) Volume of cone \[V=\frac{1}{3}\pi {{x}^{2}}y=\frac{1}{3}\pi y(2r-y)y\] \[=\frac{1}{3}\pi (2r{{y}^{2}}-{{y}^{3}})\] Þ \[\frac{dV}{dy}=\frac{1}{3}\pi (4ry-3{{y}^{2}})\] Þ \[\frac{dV}{dy}=0\] Þ \[\frac{1}{3}\pi (4ry-3{{y}^{2}})=0\] Þ \[y(4r-3y)=0\]Þ \[y=\frac{4}{3}r,\,0\] Now \[\frac{{{d}^{2}}V}{d{{y}^{2}}}=\frac{1}{3}\pi (4r-6y)\], put \[y=\frac{4}{3}r\] Þ \[\frac{{{d}^{2}}V}{d{{y}^{2}}}=\frac{1}{3}\pi \,\left( 4r-6\times \frac{4}{3}r \right)\] = negative value So, volume of cone is maximum at \[y=\frac{4}{3}r\] Þ \[\frac{\text{Height}}{\text{Diametre}}\]=\[\frac{y}{2r}=\frac{2}{3}\].You need to login to perform this action.
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