CBSE 2026 · Region 1 · Set 1 · Q22 · 2 marks
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 $\displaystyle 1 \mathrm{~mm}^{3} / \mathrm{min}$. Find the rate at which level of perfume is dropping at an instant when level of perfume in the bottle is $\displaystyle 10$ mm ,if the semi-vertical angle of conical bottle is $\displaystyle \frac{\pi}{6}$.
Official answer
From CBSE’s own marking scheme for this paper.
0.$\displaystyle 12$ mm/min or $\displaystyle 3$/($\displaystyle 25$π) mm/min
Marking-scheme solution
Let v, r, h, respectively, be the volume, radius and the height of perfume at any time t.
$\displaystyle \dfrac{dv}{dt} = -1\ \mathrm{mm^3/min}$
$\displaystyle \dfrac{r}{h} = \tan\dfrac{\pi}{6} \Rightarrow r = \dfrac{h}{\sqrt3}$
$\displaystyle v = \dfrac13\pi r^2 h = \dfrac{\pi}{9}h^3$
$\displaystyle \Rightarrow \dfrac{dv}{dt} = \dfrac{\pi}{3}h^2\dfrac{dh}{dt}$
$\displaystyle \dfrac{dh}{dt} = \dfrac{-3}{\pi h^2}$
At $\displaystyle h=10$: $\displaystyle \dfrac{dh}{dt} = \dfrac{-3}{100\pi}\ \mathrm{mm/min}$
The level of perfume is dropping at the rate of $\displaystyle \dfrac{3}{100\pi}\ \mathrm{mm/min}$
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.