Water is being filled at the rate of $1 \mathrm{~cm}^{3} / \mathrm{sec}$ in a right circular conical

Question

Water is being filled at the rate of $1 \mathrm{~cm}^{3} / \mathrm{sec}$ in a right circular conical vessel (vertex downwards) of height $35 \mathrm{~cm}$ and diameter $14 \mathrm{~cm}$. When the height of the water levels is $10 \mathrm{~cm}$, the rate (in $\mathrm{cm}^{2} / \mathrm{sec}$ ) at which the wet conical surface area of the vessel increases is\

Professor PC Thomas Classes · Answer

(c) : Let $r$ be the radius and $h$ be the height of cone.\ $	herefore \frac{r}{h}=\frac{7}{35}=\frac{1}{5} \Rightarrow r=\frac{h}{5}$\ Now, volume of cone ( $v$ )\ $=\frac{1}{3} \pi r^{2} h=\frac{1}{75} \pi h^{3}$\ $\Rightarrow \frac{d v}{d t}=\frac{1}{75} \pi \cdot 3 h^{2} \frac{d h}{d t}$\ $\Rightarrow \frac{d h}{d t}=\frac{1 	imes 75}{\pi 	imes 3(h)^{2}}$\ $\left(\because \frac{d v}{d t}=1 \mathrm{~cm}^{3} / \mathrm{sec}ight)$\ Now, $A=\pi r l=\pi r \sqrt{r^{2}+h^{2}}$\ $\Rightarrow A=\pi r^{2} \sqrt{1+\frac{h^{2}}{r^{2}}}=\frac{\pi \cdot h^{2}}{25} \sqrt{1+25}$\ $\Rightarrow A=\frac{\sqrt{26}}{25} \pi h^{2}$\ $	herefore \frac{d A}{d t}=\frac{\sqrt{26}}{25} \pi \cdot 2 h \frac{d h}{d t}$\ $=\frac{\sqrt{26}}{25} \pi \cdot 2 \cdot 10 \cdot \frac{75}{\pi 	imes 3 	imes 10^{2}}=\frac{\sqrt{26}}{5}$