A tank, as shown in the figure below, formed using a combination of a cylinder and a cone, offers better drainage as compared to a flat bottomed tank. A conical tank, whose conical part is filled with water, has a tap fitted and water is dripping from the tap at a rate of $2 \text{ cm}^3/\text{s}$. The semi-vertical angle of the conical tank is $45^\circ$. Based on the above information, answer the following questions: (i) Express the volume of water in the tank in terms of radius $r$. (ii) At what rate is the radius changing when $r = 2\sqrt{2} \text{ cm}$? (iii) (क) At what rate is the wet surface area of the conical tank decreasing when $r = 2\sqrt{2} \text{ cm}$? OR (iii) (ख) When the slant height is $4 \text{ cm}$, find the rate of change of height 'h'. CASE STUDY - 3

A tank, as shown in the figure below, formed using a combination of a cylinder and a cone, offers better drain

Question

2 \text{ cm}^3/\text{s}

(i) Express the volume of water in the tank in terms of radius

r

(ii) At what rate is the radius changing when

r = 2\sqrt{2} \text{ cm}

(iii) (क) At what rate is the wet surface area of the conical tank decreasing when

r = 2\sqrt{2} \text{ cm}

(OR)

(iii) (ख) When the slant height is

4 \text{ cm}

, find the rate of change of height 'h'. CASE STUDY - 3

CBSECBSE Class XII Board 2023Subjective· 4mImportance★★★★★

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Part (a)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 $r=5$ :

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

Part (b)Concept understanding — Related Rates

Related Rates

The idea: quantities that change together

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

\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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$r=h$ here. (i) $V=\tfrac13\pi r^3$ ; (ii) $dr/dt=-\tfrac1{4\pi}$ cm/s; (iii)(a) wet area falls at $2$ cm $^2$ /s; alt (b) $dh/dt=-\tfrac1{4\pi}$ cm/s.

Part (a)

A cone with semi-vertical angle $45^\circ$ has $\tan 45^\circ=\dfrac{r}{h}=1$ , so $r=h$ throughout. Water leaves at $2\text{ cm}^3/\text{s}$ , hence $\dfrac{dV}{dt}=-2$ .

(i) Volume of water: $V=\dfrac13\pi r^2 h$ . Substituting $h=r$ :

V=\frac13\pi r^3.

(ii) Differentiate w.r.t. $t$ : $\dfrac{dV}{dt}=\pi r^2\dfrac{dr}{dt}$ . At $r=2\sqrt2$ (so $r^2=8$ ):

-2=\pi(8)\frac{dr}{dt}\ \Rightarrow\ \frac{dr}{dt}=-\frac{2}{8\pi}=-\frac{1}{4\pi}\text{ cm/s}.

The radius is decreasing at $\dfrac{1}{4\pi}$ cm/s.

(iii)(a) Wet surface = curved surface of the cone, $S=\pi r l$ with slant $l=\sqrt{r^2+h^2}=\sqrt2\,r$ . So $S=\pi r(\sqrt2 r)=\sqrt2\,\pi r^2$ and

\frac{dS}{dt}=2\sqrt2\,\pi r\,\frac{dr}{dt}.

At $r=2\sqrt2$ , $\dfrac{dr}{dt}=-\dfrac{1}{4\pi}$ : …

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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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Part (a)

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