Question

A container is the shape of an inverted cone. Its height is $6m$ and radius is $4m$ at the top. If it is filled with water at the rate of $3m^3/min$ then the rate of change of height of water (in $mt/min$ ) when the water level is $3m$ is

• A. $\frac{3}{4\pi }$
• B. $\frac{2}{9\pi }$
• C. $16\pi$
• D. $2\pi$

Answer 1

Answer: (A)

Let $V$ be the volume, $r$ be the radius and $h$ be the height of cone at any time $t$. Then

We have, $\frac{dv}{dt}=3m^3/\min$
We have to find,: $\frac{dh}{dt}$ when $h=3m$
Clearly, $V=\frac{1}{3}\pi r^{2}h$ ...(i)
Since, $\Delta BOD-\Delta BAC$ [By AA-similarity criterion]
$\therefore \frac{BO}{BA}=\frac{OD}{AC}$
$\Rightarrow \frac{6}{h}=\frac{4}{r}$
$\Rightarrow r=\frac{2}{3}h$
Now, from Eq. (i), we get
$V=\frac{1}{3}\pi {\left(\frac{2}{3}h\right)}^{2}h=\frac{4}{27}\pi h^3$
$\Rightarrow \frac{dV}{dt}=\frac{4}{27}\pi 3h^2\frac{dh}{dt}$
$\Rightarrow 3=\frac{4}{27}\pi 3h^2\frac{dh}{dt}$
$\Rightarrow \frac{db}{dt}=\frac{3\times 27}{4\pi 3h^2}$
${\Rightarrow \frac{dh}{dt}|}_{h=3}$
$=\frac{3\times 27}{4\pi 27}=\frac{3}{4\pi }m/\min$