JEE Main 2018 — Mathematics Calculus

Sphere of radius $r=3 \mathrm{~cm}$r=3 cm Let $b, h$b,h be base radius and height of cone respectively. So, volume of cone $=\frac{1}{2} \pi b^2 h$=21​πb2h

In right angled $\triangle A B C$△ABC by Pythagoras theorem $$\begin{aligned} &(h-r)^2+b^2=r^2 \\ \Rightarrow & b^2=r^2-(h-r)^2=r^2-\left(h^2-2 h r+\mathrm{r}^2\right)=2 h r-h^2 \\ \therefore & \text { Volume }(v) \frac{1}{3} \pi h\left[2 h r-h^2\right]=\frac{1}{3}\left[2 h r^2-h^3\right] \\ & \frac{d v}{d h}=\frac{1}{3}\left[4 h r-3 h^2\right]=0 \Rightarrow h(4 r-3 h)=0 \\ & \frac{d^2 v}{d h^2}=\frac{1}{3}[4 r-6 h] \end{aligned}$$⇒∴​(h−r)2+b2=r2b2=r2−(h−r)2=r2−(h2−2hr+r2)=2hr−h2 Volume (v)31​πh[2hr−h2]=31​[2hr2−h3]dhdv​=31​[4hr−3h2]=0⇒h(4r−3h)=0dh2d2v​=31​[4r−6h]​ At $h=\frac{4 r}{3}, \frac{d^2 v}{d h^2}=\frac{1}{3}\left[4 r-\frac{4 r}{3} \times 6\right]=\frac{1}{3}[4 r-8 r < ] 0$h=34r​,dh2d2v​=31​[4r−34r​×6]=31​[4r−8r<]0 $\Rightarrow$⇒ maximum volume ocurs at $h$h $$=\frac{4 r}{3}=\frac{4}{3} \times 3=4 \mathrm{~cm}$$=34r​=34​×3=4 cm As from (1), $$(h-r)^2+b^2=r^2$$(h−r)2+b2=r2 $$\Rightarrow b^2=2 h r-h^2=2 \cdot \frac{4 r}{3} r-\frac{16 r^2}{9}$$⇒b2=2hr−h2=2⋅34r​r−916r2​ $$\begin{aligned} &=\frac{8 r^2}{3}-\frac{16 r^2}{9} \\ &=\frac{(24-16) r^2}{9}=\frac{8 r^2}{9} \\ &\Rightarrow b=\frac{2 \sqrt{2}}{3} r=2 \sqrt{2} \mathrm{~cm} \end{aligned}$$​=38r2​−916r2​=9(24−16)r2​=98r2​⇒b=322​​r=22​ cm​ Therefore curved surface area $$\begin{aligned} &=\pi b l=\pi b \sqrt{h^2+r^2} \\ &=\pi 2 \sqrt{2} \sqrt{4^2+8}=8 \sqrt{3} \pi \mathrm{cm}^2 \end{aligned}$$​=πbl=πbh2+r2​=π22​42+8​=83​πcm2​