Question

Water is being filled at the rate of $1c{m}^3/sec$ in a right circular conical vessel (vertex downwards) of height $35cm$ and diameter $14cm$. When the height of the water level is $10cm$, the rate (in $c{m}^2/\sec$ ) at which the wet conical surface area of the vessel increases is

• A. $5$
• B. $\frac{\sqrt{21}}{5}$
• C. $\frac{\sqrt{26}}{5}$
• D. $\frac{\sqrt{26}}{10}$

Answer 1

Answer: (C)

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

Given $\frac{dV}{dt}=1$
$\Rightarrow \frac{d}{dt}\left(\frac{\pi r^{2}h}{3}\right)=1$
$\Rightarrow \frac{d}{dt}\left(\frac{5\pi }{3}{r}^3\right)=1$
$\Rightarrow r^2\frac{dr}{dt}=\frac{1}{5\pi }$
Let wet conical surface area $=S$
$=\pi r\ell =\pi r\sqrt{h^2+r^2}$
$=\sqrt{26}\pi r^2$
$\Rightarrow \frac{dS}{dt}=2\sqrt{26}\pi r\frac{dr}{dt}$
When $h=10$ then $r=2$
$\Rightarrow \frac{dS}{dt}=\frac{2\sqrt{26}}{10}$