Question

If a cone of maximum volume is inscribed in a given sphere, then the ratio of the height of the cone to the diameter of the sphere is

(a) $\frac{3}{4}$

(b) $\frac{1}{3}$

(c) $\frac{1}{4}$

(d) $\frac{2}{3}$

Answer 1

(d) $\frac{2}{3}$


$$\begin{gathered} \mathrm{Let}h,r,V\mathrm{and}R\mathrm{be}\mathrm{the}\mathrm{height},\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{base},\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cone}\mathrm{and}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{sphere},\mathrm{respectively}. \\ \mathrm{Given}:h=R+\sqrt{R^2-r^2} \\ \Rightarrow h-R=\sqrt{R^2-r^2} \\ \text{Squaring both side, we get} \\ {h}^2+R^2-2hR=R^2-r^2 \\ \Rightarrow r^2=2hr-h^2...\left(1\right) \\ \mathrm{Now}, \\ \mathrm{Volume}=\frac{1}{3}\pi r^{2}h \\ \Rightarrow V=\frac{\pi }{3}\left(2h^{2}R-h^3\right)\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow \frac{dV}{dh}=\frac{\pi }{3}\left(4hR-3h^2\right) \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}V,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{h}}=0 \\ \Rightarrow \frac{\pi }{3}\left(4hR-3h^2\right)=0 \\ \Rightarrow 4hR-3h^2=0 \\ \Rightarrow 4hR=3h^2 \\ \Rightarrow h=\frac{4R}{3} \\ \mathrm{Now}, \\ \frac{d^{2}V}{d{h}^2}=\frac{\pi }{3}\left(4R-6h\right)=\frac{\pi }{3}\left(4R-6\times \frac{4R}{3}\right)=-\frac{4\pi R}{3}<0 \\ \text{So, volume is maximum when}\mathit{\text{h}}=\frac{4R}{3}. \\ \Rightarrow h=\frac{2\left(2R\right)}{3} \\ \Rightarrow \frac{h}{2R}=\frac{2}{3} \\ \therefore \frac{\mathrm{Height}}{\mathrm{Diameter}\mathrm{of}\mathrm{sphere}}=\frac{2}{3} \end{gathered}$$