Water is being filled at the rate of $1 c m^{3} / \sec$ in a right circular conical vessel (vertex downwards) of height 35cm and diameter 14cm. When the height of the water level is $\sqrt{26} c m$ , the rate (in $c m^{3} / \sec$ ) at which the wet conical surface area of the vessel increases is

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Detailed Solution

$\frac{r}{h} = \frac{7}{35} = \frac{1}{5} \Rightarrow h = 5 r v = \frac{1}{3} π r^{2} h \Rightarrow v = \frac{π}{3} . \frac{h^{3}}{25}$

$\frac{d v}{d t} = \frac{π h^{2}}{25} \frac{d h}{d t} \Rightarrow l = \frac{π h^{2}}{25} \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{25}{π h^{2}}$

$s = π r l = π r \sqrt{r^{2} + h^{2}} = π \frac{h}{5} \sqrt{\frac{h^{2}}{25} + h^{2}}, s = \frac{π h}{5} \frac{\sqrt{26} h}{5} = \frac{π \sqrt{26}}{25} . h^{2}$

$\frac{d s}{d t} = \frac{π \sqrt{26}}{25} 2 h \frac{d h}{d t} = \frac{π \sqrt{26}}{25} .2 h \frac{25}{π h^{2}} = \frac{2 \sqrt{26}}{h}, \frac{d s}{d t} a t h = \sqrt{26} i s 2$