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

Since the conical vessel has height $$H = 35$$ cm and radius $$R = 7$$ cm, by similar triangles at water level height $$h$$ one obtains $$\frac{r}{h} = \frac{7}{35} = \frac{1}{5} \Rightarrow r = \frac{h}{5}$$.

The volume of water is given by $$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{5}\right)^2 h = \frac{\pi h^3}{75}$$, so differentiating with respect to time yields $$\frac{dV}{dt} = \frac{\pi h^2}{25} \cdot \frac{dh}{dt} = 1$$. Substituting $$h = 10$$ gives $$\frac{\pi (100)}{25} \cdot \frac{dh}{dt} = 1 \Rightarrow 4\pi \cdot \frac{dh}{dt} = 1 \Rightarrow \frac{dh}{dt} = \frac{1}{4\pi}$$.

Next, the slant height satisfies $$l = \sqrt{r^2 + h^2} = \sqrt{\frac{h^2}{25} + h^2} = h\sqrt{\frac{26}{25}} = \frac{h\sqrt{26}}{5}$$, and hence the wet conical surface area is $$S = \pi r l = \pi \cdot \frac{h}{5} \cdot \frac{h\sqrt{26}}{5} = \frac{\pi h^2 \sqrt{26}}{25}$$.

Therefore, differentiating gives $$\frac{dS}{dt} = \frac{2\pi h \sqrt{26}}{25} \cdot \frac{dh}{dt}$$, and at $$h = 10$$ one finds $$\frac{dS}{dt} = \frac{2\pi \cdot 10 \cdot \sqrt{26}}{25} \cdot \frac{1}{4\pi} = \frac{20\pi\sqrt{26}}{25} \cdot \frac{1}{4\pi} = \frac{20\sqrt{26}}{100} = \frac{\sqrt{26}}{5}$$.

Therefore, the rate of increase of wet conical surface area is $$\frac{\sqrt{26}}{5}$$ cm$$^2$$/sec, which is Option C.