A room freshener bottle in the shape of an inverted cone sprays at regular intervals, due to which the volume of perfume in the bottle decreases at the rate of $1 \text{ mm}^3/\text{min}$. If the semi-vertical angle of the conical bottle is $\frac{\pi}{6}$, then find the rate at which the level of perfume in the bottle is decreasing, when the level of perfume in the bottle is $10 \text{ mm}$.

A room freshener bottle in the shape of an inverted cone sprays at regular intervals, due to which the volume

Question

1 \text{ mm}^3/\text{min}

CBSECBSE Class XII Board 2026Subjective· 2mImportance★★★★★

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Concept understanding — Related Rates

Related Rates

The idea: quantities that change together

Many real situations involve two or more quantities that all vary with time, linked by a fixed relationship. Inflate a balloon and its radius and volume both grow; slide a ladder down a wall and the top's height and the foot's distance both change. A related-rates problem gives you the rate at which one quantity is changing and asks for the rate of another, at some instant.

The key insight: if the quantities are tied together by an equation, then their rates are tied together too. We uncover that link by differentiating the equation with respect to time $t$ .

The core mechanism: differentiate with respect to time

Every variable is a function of $t$ , so differentiating brings in the chain rule — each variable's derivative picks up a factor of its own rate. For example, if the volume of a sphere is $V=\tfrac{4}{3}\pi r^{3}$ , then differentiating both sides with respect to $t$ gives

\frac{dV}{dt} = 4\pi r^{2}\,\frac{dr}{dt}.

This single equation connects the rate the volume grows, $\dfrac{dV}{dt}$ , to the rate the radius grows, $\dfrac{dr}{dt}$ . Knowing one (and the current $r$ ) gives the other.

The standard procedure

t

Worked example

Air is pumped into a spherical balloon at $\dfrac{dV}{dt}=100\ \text{cm}^3/\text{s}$ . How fast is the radius increasing when $r=5\ \text{cm}$ ?

From $\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}$ , substitute $\dfrac{dV}{dt}=100$ and $r=5$ :

$100 = 4\pi(5)^2\,\frac{dr}{dt} = 100\pi\,\frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{\pi}\ \text{cm/s}.$ …

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The problem is a classic related rates scenario: we know $dV/dt$ and want $dh/dt$ at a given $h$ . Using the cone’s geometry ( $r = h \tan(\pi/6)$ ) to relate $V$ and $h$ , differentiating gives $dh/dt = -1/(\pi h^2 \tan^2(\pi/6))$ . At $h = 10$ mm, the rate is $-\frac{3}{100\pi}$ mm/min.

Why related rates works here

When a quantity changes with time, any other quantity linked to it by a fixed geometric or physical relationship also changes. Here, the perfume volume $V$ and the height $h$ of the liquid are tied by the cone’s shape. If we know how fast $V$ drops, we can find how fast $h$ drops — provided we can write $V$ purely in terms of $h$ (eliminating the radius $r$ using the given semi-vertical angle).

The key is that the cone’s dimensions are proportional at every instant: the liquid surface is always a smaller, similar cone to the full bottle. That similarity gives a constant ratio between $r$ and $h$ , set by the semi-vertical angle.

Step-by-step solution

1. Relate radius and height using the semi-vertical angle

The semi-vertical angle is $\pi/6$ . In a right circular cone, if you take a vertical cross-section through the apex, you get an isosceles triangle. The semi-vertical angle is the angle between the axis and the slant edge. From the geometry:

\tan\left(\frac{\pi}{6}\right) = \frac{r}{h}

Since $\tan(\pi/6) = 1/\sqrt{3}$ , we have:

\frac{r}{h} = \frac{1}{\sqrt{3}} \quad\Rightarrow\quad r = \frac{h}{\sqrt{3}}

Tip

You never need the actual radius value — only the ratio $r/h$ . The semi-vertical angle directly gives that ratio as $\tan(\theta)$ .

2. Write volume in terms of height only

Volume of a cone: $V = \frac{1}{3}\pi r^2 h$ . Substitute $r = h/\sqrt{3}$ :

V = \frac{1}{3}\pi \left(\frac{h}{\sqrt{3}}\right)^2 h = \frac{1}{3}\pi \cdot \frac{h^2}{3} \cdot h = \frac{\pi}{9} h^3

So $V = \frac{\pi}{9} h^3$ . This is a clean cubic relation — no $r$ left.

V = \frac{\pi}{9} h^3

3. Differentiate with respect to time

Both $V$ and $h$ are functions of time $t$ . Differentiate implicitly:

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. The edge of a variable cube is increasing at the rate of 3 cm/s. The volume of the cube is increasing at the rate of __________ while the edge is 10 cm long.$
›Reveal solutionSolution

Use $V=e^3$ and the chain rule $dV/dt = 3e^2\,de/dt$ .

Let $e$ be the edge; $V=e^3$ , so $\dfrac{dV}{dt}=3e^2\dfrac{de}{dt}$ .

Given $\dfrac{de}{dt}=3$ cm/s and $e=10$ cm:

…

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CBSE 2026Set ANNUAL1 mark
$1/2$
›Reveal solutionSolution

Use $V=\dfrac43\pi r^3$ and $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ .

Given $\dfrac{dr}{dt}=\dfrac12$ cm/s, at $r=1$ cm:

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Radius of a circle is increasing at the rate of 1/π m/s. Rate of change of its circumference is:$
(a) 4π m/s
(b) 2 m/s
(c) 2π m/s
(d) 4 m/s
›Reveal solutionSolution

Since $C = 2\pi r$ , differentiating both sides w.r.t. time gives $\frac{dC}{dt} = 2\pi \frac{dr}{dt}$ directly.

The circumference of a circle of radius $r$ is $C = 2\pi r$ .

Differentiating with respect to time $t$ :

$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$

Given $\frac{dr}{dt} = \frac{1}{\pi}$ m/s, substitute: …

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CBSE 2025Set 65/2/11 markMCQ
$100\pi$
›Reveal solutionSolution
The volume of a cylinder is $V = \pi r^2 h$ . Since the radius is constant, the rate of change of volume with respect to time is $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$ . Given $\frac{dV}{dt} = 100\pi$ cm³/s and $r = 10$ cm, solving gives $\frac{dh}{dt} = 1$ cm/s. The correct option is (C).

This is a classic Related Rates problem. The core idea is that when two quantities are linked by a geometric formula (here, volume and height of a cylinder), their rates of change with respect to time are also linked. You differentiate the relationship with respect to time, plug in what you know, and solve for the unknown rate.

The key insight: the tank’s radius is fixed at 10 cm. So as sugar pours in, the height increases, but the cross-sectional area stays the same. That means the volume increases at a constant rate per unit height — specifically, each 1 cm rise in height adds $\pi (10)^2 = 100\pi$ cm³ of volume. Since sugar is being added at exactly $100\pi$ cm³/s, the height must be rising at 1 cm/s.

Let’s work it out formally.

Write the relationship between volume and height. For a cylinder, $V = \pi r^2 h$ . Here $r = 10$ cm, so

$V = \pi (10)^2 h = 100\pi h.$

Differentiate both sides with respect to time $t$ .

Since $r$ is constant, $\frac{dV}{dt} = 100\pi \frac{dh}{dt}$ .

This is the related rates equation — it tells us how fast the volume changes in terms of how fast the height changes.

Substitute the given rate.

We know $\frac{dV}{dt} = 100\pi$ cm³/s. So:

$100\pi = 100\pi \frac{dh}{dt}.$

Solve for $\frac{dh}{dt}$ . Divide both sides by $100\pi$ : …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Radius of a circle is increasing at the rate of 2 m/s. Rate of change of its circumference is:$
(a) 4π m/s
(b) 2 m/s
(c) 2π m/s
(d) 4 m/s
›Reveal solutionSolution

Differentiate the circumference formula $C=2\pi r$ with respect to time and plug in $\dfrac{dr}{dt}$ .

Given $\dfrac{dr}{dt} = 2$ m/s. Circumference $C = 2\pi r$ .

Differentiating both sides with respect to time $t$ : …

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CBSE 2024Set ANNUAL1 mark
$Q. The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.$
›Reveal solutionSolution

Use related rates: differentiate $A=\pi r^2$ w.r.t. time and substitute the given $\dfrac{dr}{dt}$ and $r$ .

Given $\dfrac{dr}{dt} = 3$ cm/s, find $\dfrac{dA}{dt}$ at $r = 10$ cm.

$A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$

…

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CBSE 2024Set ANNUAL1 mark
$0.7$
›Reveal solutionSolution

Differentiate the circumference formula $C=2\pi r$ with respect to time.

The circumference of a circle of radius $r$ is

$C = 2\pi r.$

Given $\dfrac{dr}{dt} = 0.7$ cm/s. Differentiating with respect to $t$ : …

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CBSE 2023Set ANNUAL1 markMCQ
$0.3$
(a) $0.4\pi$ cm/sec
(b) $0.6\pi$ cm/sec
(c) $0.8\pi$ cm/sec
(d) none of these
›Reveal solutionSolution

This is a related-rates problem: differentiate the perimeter formula w.r.t. time.

Perimeter (circumference) $P = 2\pi r$ . Differentiating w.r.t. time $t$ : $\dfrac{dP}{dt} = 2\pi\dfrac{dr}{dt}$ .

…

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CBSE 2023Set ANNUAL1 mark
$3$
›Reveal solutionSolution

Differentiate $A=\pi r^2$ with respect to time and substitute $r=10$ , $dr/dt=3$ .

Area of circle: $A=\pi r^2$ . Differentiating both sides with respect to $t$ :

$\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. The radius of a circle is increasing at the rate of 0.7 cm/s. The rate of increase of its circumference is -$
(a) 7.1 cm/s
(b) 4.0 cm/s
(c) 3.9 cm/s
(d) 4.4 cm/s
›Reveal solutionSolution

Differentiate $C = 2\pi r$ with respect to time and substitute the given rate of change of the radius.

Circumference $C = 2\pi r$ . Differentiating with respect to time $t$ :

$\frac{dC}{dt} = 2\pi\frac{dr}{dt}.$

Given $\dfrac{dr}{dt} = 0.7$ cm/s: …

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CBSE 2018Set ANNUAL1 markMCQ
$cm^3$
(a) 5 $cm^2$ /sec.
(b) 10 $cm^2$ /sec.
(c) 15 $cm^2$ /sec.
(d) 20 $cm^2$ /sec.
›Reveal solutionSolution

related rates: relate dV/dt to dr/dt, then dS/dt to dr/dt

Volume $V=\dfrac43\pi r^3\Rightarrow \dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ .

At $r=5$ : $25=4\pi(25)\dfrac{dr}{dt}\implies \dfrac{dr}{dt}=\dfrac{25}{100\pi}=\dfrac{1}{4\pi}$ .

Surface area $S=4\pi r^2\Rightarrow \dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$ .

…

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CBSE 2018Set ANNUAL1 markMCQ
$x$
(a) $\dfrac{\pi}{3}$
(b) $\dfrac{\pi}{2}$
(c) $\pi$
(d) $\dfrac{3\pi}{2}$
›Reveal solutionSolution

Translate "x increases twice as fast as sin x" into $dx/dt = 2\,d(\sin x)/dt$ and solve for x.

"x increases twice as fast as its sine" means $\dfrac{dx}{dt} = 2\dfrac{d(\sin x)}{dt} = 2\cos x\dfrac{dx}{dt}$ …

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Why related rates works here

Step-by-step solution

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