Question

A container is the shape of an inverted cone. Its height is $6\, m$ and radius is $4\, m$ at the top. If it is filled with water at the rate of $3\, m ^{3} / min$ then the rate of change of height of water (in $mt / min$ ) when the water level is $3\, m$ 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, $\frac{d v}{d t}=3\, m ^{3} / \min$
We have to find,: $\frac{d h}{d t}$ when $h=3\, m$
Clearly, $V=\frac{1}{3} \pi r^{2} h$ ...(i)
Since, $\Delta B O D-\Delta B A C$ [By AA-similarity criterion]
$\therefore \frac{B O}{B A}=\frac{O D}{A C}$
$\Rightarrow \frac{6}{h} =\frac{4}{r}$
$\Rightarrow r =\frac{2}{3} h$
Now, from Eq. (i), we get
$V=\frac{1}{3} \pi\left(\frac{2}{3} h\right)^{2} h=\frac{4}{27} \pi h^{3}$
$\Rightarrow \frac{d V}{d t}=\frac{4}{27} \pi 3 h^{2} \frac{d h}{d t}$
$\Rightarrow 3=\frac{4}{27} \pi 3 h^{2} \frac{d h}{d t}$
$\Rightarrow \frac{d b}{d t}=\frac{3 \times 27}{4 \pi 3 h^{2}}$
$\left.\Rightarrow \frac{d h}{d t}\right|_{h=3}$
$=\frac{3 \times 27}{4 \pi 27}=\frac{3}{4 \pi} m / \min$