Board 2026 Class 12 Maths Q22 Solution: Rate at which the perfume level drops in an inverted cone

The question

A room freshner bottle in the shape of an inverted cone sprays the perfume at regular intervals such that volume of the perfume in the bottle decreases at the steady rate of 1 mm³/min. Find the rate at which level of perfume is dropping at an instant when level of perfume in the bottle is 10 mm, if the semi-vertical angle of conical bottle is $\dfrac{\pi}{6}$ ⁠.

Exam answer

Let $V$ ⁠, $r$ and $h$ be the volume, radius and height (level) of the perfume at time $t$ ⁠.

Given $\frac{dV}{dt} = -1$ mm³/min.

Semi-vertical angle $= \frac{\pi}{6}$ ⁠, so $\frac{r}{h} = \tan\frac{\pi}{6} = \frac{1}{\sqrt3}$ ⁠, i.e. $r = \frac{h}{\sqrt3}$ ⁠.

Why?

The semi-vertical angle is the angle between the axis of the cone and its slant side. The perfume fills a smaller cone with the same angle. Its height $h$ runs along the axis and its radius $r$ is opposite the angle, so in that right triangle

\tan\frac{\pi}{6} = \frac{r}{h} = \frac{1}{\sqrt3}

This links $r$ to $h$ ⁠, so the volume can be written with $h$ alone.

\begin{aligned} V &= \frac13\pi r^2h = \frac13\pi\cdot\frac{h^2}{3}\cdot h \\ &= \frac{\pi h^3}{9} \end{aligned}

Why?

Volume of a cone:

V = \frac13\pi r^2h

From $r = \frac{h}{\sqrt3}$ we get $r^2 = \frac{h^2}{3}$ ⁠. With only $h$ left, we can differentiate.

\begin{aligned} r^2 &= \left(\frac{h}{\sqrt3}\right)^2 \\ &= \frac{h^2}{(\sqrt3)^2} = \frac{h^2}{3} \\[4pt] V &= \frac13\pi\cdot\frac{h^2}{3}\cdot h \\ &= \frac{\pi}{3 \times 3}\cdot h^2 \cdot h \\ &= \frac{\pi h^3}{9} \end{aligned}

Differentiating w.r.t. $t$ ⁠,

\begin{aligned} \frac{dV}{dt} &= \frac{\pi}{9}\cdot 3h^2\frac{dh}{dt} = \frac{\pi h^2}{3}\cdot\frac{dh}{dt} \\ \Rightarrow \frac{dh}{dt} &= \frac{3}{\pi h^2}\cdot\frac{dV}{dt} = -\frac{3}{\pi h^2} \end{aligned}

Why?

$h$ changes with time, so by the chain rule

\frac{d}{dt}(h^3) = 3h^2\frac{dh}{dt}

\begin{aligned} \frac{dV}{dt} &= \frac{\pi}{9}\cdot 3h^2\frac{dh}{dt} \\ &= \frac{3\pi h^2}{9}\cdot\frac{dh}{dt} \\ &= \frac{\pi h^2}{3}\cdot\frac{dh}{dt} \end{aligned}

( $\frac{3}{9} = \frac{3 \div 3}{9 \div 3} = \frac13$ ⁠.)

Then divide by $\frac{\pi h^2}{3}$ to get $\frac{dh}{dt}$ on its own, and put $\frac{dV}{dt} = -1$ ⁠.

\begin{aligned} \frac{dh}{dt} &= \frac{dV}{dt} \div \frac{\pi h^2}{3} \\ &= \frac{3}{\pi h^2}\cdot\frac{dV}{dt} \\ &= \frac{3}{\pi h^2}(-1) = -\frac{3}{\pi h^2} \end{aligned}

The rate is negative because the volume is going down.

At $h = 10$ mm,

\left(\frac{dh}{dt}\right)_{h = 10} = -\frac{3}{100\pi}

Why?

With $h = 10$ ⁠, $h^2 = 100$ ⁠.

\begin{aligned} \frac{dh}{dt} &= -\frac{3}{\pi(10)^2} \\ &= -\frac{3}{\pi \times 100} = -\frac{3}{100\pi} \end{aligned}

The minus sign says the level is falling. The question asks for the rate at which the level is dropping, so the answer is the positive number $\frac{3}{100\pi}$ ⁠.

Watch out: put $h = 10$ only after differentiating. If you put it in earlier, $h$ becomes a constant and its rate of change is lost.

$\therefore$ The level of perfume is dropping at the rate of $\dfrac{3}{100\pi}$ mm/min.

Where this comes from

Chapter 6, Application of Derivatives: Rate of change: liquid level in an inverted cone.

Question type: Rates in time

How to spot it: click to see

Type 2. Rates in time: how fast is it changing now?

Spot it: words like "per second", "how fast", "at the rate of", and one moment, such as "when the radius is 10 cm".

Here: The volume falls at 1 mm³/min, so $\frac{dV}{dt} = -1$ ⁠, and the moment is a level of 10 mm. Use $r = \frac{h}{\sqrt3}$ to write $V = \frac{\pi h^3}{9}$ ⁠, differentiate with respect to $t$ ⁠, then put in $h = 10$ ⁠.

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Q22: Rate at which the perfume level drops in an inverted cone