Water is being filled at the rate of 1 cm 3 sec - 1 in a right circular conical vessel (vertex downwards) of…

Let the volume of the cone be $V \mathrm{c}.\mathrm{c}.$

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

i.e. $\frac{h}{r}=5$

We know for a cone $l^2=r^2+h^2$

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

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

$\Rightarrow \frac{dV}{dt}=\frac{\pi }{25}{h}^2\frac{dh}{dt}\Rightarrow \frac{\pi }{25}{h}^2\frac{dh}{dt}=1$

$\Rightarrow \frac{dh}{dt}=\frac{25}{\pi h^2} \dots \left(i\right)$

Now $\frac{dS}{dt}=\frac{\pi \sqrt{26}}{25}\times 2h\frac{dh}{dt}=\frac{2\sqrt{26}}{h}$    (from $\left(i\right)$)

${\left(\frac{dS}{dt}\right)}_{h=10}=\frac{\sqrt{26}}{5}$