Water is being filled at the rate of $1 c m^{3} / s e c$ in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm. When the height of the water level is 10 cm, the rate ( in cm² / sec ) at which the wet conical surface area of the vessel increases is

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$\frac{\sqrt{26}}{5}$

$\frac{\sqrt{26}}{10}$

$\frac{\sqrt{21}}{5}$

answer is C.

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Detailed Solution

From figure $\frac{r}{h} = \frac{7}{35} \Rightarrow h = 5 r$

Volume of the cone = $V = \frac{1}{3} π r^{2} h$

$\frac{d V}{d t} = 1 \Rightarrow \frac{d}{d t} (\frac{π r^{2} h}{3}) = 1$

$\Rightarrow \frac{d}{d t} (\frac{5 π}{3} r^{3}) = 1 \Rightarrow r^{2} \frac{d r}{d t} = \frac{1}{5 π}$

Let wet conical surface area =S

$= π r ℓ = π r \sqrt{h^{2} + r^{2}} = \sqrt{26} π r^{2} \Rightarrow \frac{d S}{d t} = 2 \sqrt{26} π r \frac{d r}{d t}$

When h =10 then r =2 $\Rightarrow \frac{d S}{d t} = \frac{2 \sqrt{26}}{10}$

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