A container is the shape of an inverted cone. Its height is $6 \mathrm{~m}$ and radius is $4 \mathrm{~m}$ at…

1. $\frac{3}{4 \pi}$
2. $\frac{2}{9 \pi}$
3. $16 \pi$
4. $2 \pi$

Solution

Let $V$ be the volume, $r$ be the radius and $h$ be the height of an inverted cone at any time $t$. Then
Given, $\frac{d v}{d t}=3 \mathrm{~m}^3 / \mathrm{min}$ We know that,
From similarity of triangle $\begin{aligned} & \frac{B O}{B A}=\frac{O D}{A C} \\ & \Rightarrow \frac{6}{h} \quad \frac{4}{r} \Rightarrow r=\frac{2}{3} h \end{aligned}$ From Eq. (i), $V=\frac{1}{3} \pi\left(\frac{2}{3} h\right)^2 h=\frac{4}{27} \pi h^3$ On differentiating w.r.t. $t$, we get $\begin{aligned} & \Rightarrow \quad \frac{d v}{d t}=\frac{4}{27} \pi 3 h^2 \frac{d h}{d t} \\ & \Rightarrow \quad 3=\frac{4}{27} \pi 3 h^2 \frac{d h}{d t} \\ & \Rightarrow \quad \frac{d h}{d t}=\frac{3 \times 27}{4 \pi 3 h^2} \\ & \Rightarrow \quad\left(\frac{d h}{d t}\right)_{h=3}=\frac{3 \times 27}{4 \pi 27}=\frac{3}{4 \pi} \mathrm{mt} / \mathrm{min} \end{aligned}$

Asked in: MHT CET Full Test 7

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