Water is being filled at the rate of 1 cm ^ 3 ^ - 1 in a right...

Let the volume of the cone be $Vc . c .$

Given $\dfrac{dv}{dt} = 1cc / sec, h = 35cm, r = 7cm$

i.e. $\dfrac{h}{r} = 5$

We know for a cone ${l}^{2} = {r}^{2} + {h}^{2}$

Lateral surface area, $S = \pir\sqrt{{r}^{2} + {h}^{2}}$
$S = \pi\dfrac{h}{5}\sqrt{\dfrac{{h}^{2}}{25} + {h}^{2}} = \pi\dfrac{\sqrt{26}}{25}{h}^{2}$

Also $V = \dfrac{1}{3}\pi{r}^{2}h = \dfrac{1}{3}\pi{\left(\dfrac{h}{5}\right)}^{2}h = \dfrac{\pi}{75}{h}^{3}$

$⇒ \dfrac{dV}{dt} = \dfrac{\pi}{25}{h}^{2}\dfrac{dh}{dt} ⇒ \dfrac{\pi}{25}{h}^{2}\dfrac{dh}{dt} = 1$

$⇒ \dfrac{dh}{dt} = \dfrac{25}{\pi{h}^{2}} \ldots \left(i\right)$

Now $\dfrac{dS}{dt} = \dfrac{\pi\sqrt{26}}{25} \times 2h\dfrac{dh}{dt} = \dfrac{2\sqrt{26}}{h}$ (from \left(i\right)$)</p><p>${\left(\dfrac{dS}{dt}\right)}_{h = 10} = \dfrac{\sqrt{26}}{5}$