Question

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

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

Answer 1

Answer: (C)

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

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 $r=2$
$\Rightarrow \frac{d S }{ dt }=\frac{2 \sqrt{26}}{10}$