Water is being filled at the rate of $1 {cm}^{3} / \sec$ 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 ${cm}^{2} / \sec)$ at which the wet conical surface area of the vessel increases is

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

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

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

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{dv}{dt} = 1 {cm}^{3} / \sec, h = 35 cm, r = 7 cm, \frac{r}{h} = \frac{1}{5} \\ \begin{array}{l} ∵ \frac{dv}{dt} = 1 \Rightarrow \frac{d}{dt} (\frac{1}{3} {πr}^{2} . 5 r) = 1 \\ \Rightarrow \frac{5}{3} π (3 r^{2} \frac{dr}{dt}) = 1 \\ \Rightarrow \frac{dr}{dt} = \frac{1}{5 π (4)} = \frac{1}{20 π} \end{array} \\ Area, A = πrl = πr \sqrt{h^{2} + r^{2}} = \sqrt{26} {πr}^{2} \\ \frac{dA}{dt} = 2 π \sqrt{26} r \frac{dr}{dt} \\ (∴ at h = 10, r = 2) \\ = \frac{2 π \sqrt{26}}{20 π} (2) = \frac{\sqrt{26}}{5} \end{array}$