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 (π)/(6).

Level drops at (3)/(100π) mm/min

MathsCBSE Class 12 · Board exam 2026, set 65/1/1 · 2 marks · Very short answer

CBSE Class 12 Mathematics 2026: question 22

Set in the CBSE Class 12 board exam of 2026, paper 65/1/1, section B, question 22, for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q22 · 2 marks · Very short answerCompetency-basedApplication of Derivatives

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 mm $^{3}$ /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 $\frac{π}{6}$ .

Figure in the paper: Left: a room freshener bottle shaped as an inverted cone spraying perfume. Right: an inverted cone with shaded perfume; the semi-vertical angle is marked 30° and the perfume level height is marked 10 mm. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 65/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Relate the radius and height of the liquid using the semi-vertical angle.

Hint 2

Write the volume of the cone in terms of $h$ only.

Hint 3

Differentiate with respect to $t$ and put $\frac{d V}{d t} = - 1$ , $h = 10$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Let $V$ , $r$ , $h$ be the volume, radius and height of perfume at time $t$ . Given $\frac{d V}{d t} = - 1$ mm³/min.
From the semi-vertical angle: $\frac{r}{h} = tan \frac{π}{6} = \frac{1}{3} \Rightarrow r = \frac{h}{3}$
½ mark
$V = \frac{1}{3} π r^{2} h = \frac{1}{3} π \cdot \frac{h ^{2}}{3} \cdot h = \frac{π h ^{3}}{9}$
½ mark
Differentiating w.r.t. $t$ : $\frac{d V}{d t} = \frac{π}{9} \cdot 3 h^{2} \frac{d h}{d t} = \frac{π h ^{2}}{3} \frac{d h}{d t}$
½ mark
So $\frac{d h}{d t} = - \frac{3}{π h ^{2}}$ . At $h = 10$ mm: $\frac{d h}{d t} = - \frac{3}{100 π}$ .
The level of perfume is dropping at $\frac{3}{100 π}$ mm/min.
½ mark

Answer: Level drops at $\frac{3}{100 π}$ mm/min

Where the marks are

Uses $tan \frac{π}{6}$ to get $r = \frac{h}{3}$ (½ mark)
Writes $V = \frac{π h ^{3}}{9}$ (½ mark)
Differentiates: $\frac{d V}{d t} = \frac{π h ^{2}}{3} \frac{d h}{d t}$ (½ mark)
Gets rate $\frac{3}{100 π}$ mm/min at $h = 10$ (½ mark)

Revise the chapter

Chapter: Application of Derivatives (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Application of Derivatives, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 65/1/1: every question, by section.

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CBSE Class 12 Mathematics 2026: question 22

Hints

Step-by-step solution

Revise the chapter

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