A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at an

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Q. A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.

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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}.

Watch out

Substitute the numerical values after differentiating, never before. If you plug $r=5$ into the volume formula first, $r$ becomes a constant and its rate $\dfrac{dr}{dt}$ vanishes from the equation.

Tip

The everyday cases are the sphere/circle ( $V,\,A$ vs. $r$ ), the sliding ladder ( $x^2+y^2=\ell^2$ ), and the filling cone. In each you differentiate the relation in $t$ and solve for the missing rate.

Related rates problems are a named application within the NCERT Class 12 Application of Derivatives chapter, and classic setups like the growing balloon or the sliding ladder are staples of CBSE board and JEE Main 'rate of change' questions. Students searching 'related rates problems class 12 examples' or 'rate of change of volume and radius' will find this differentiate-then-substitute method is exactly the five-step procedure boards expect to see written out.

The problem is a classic related rates application: we are told $dV/dt \propto S$ , and we need to show $dr/dt$ is constant. Using the formulas $V = \frac{4}{3}\pi r^3$ and $S = 4\pi r^2$ , differentiating, and substituting the proportionality gives $dr/dt = -k$ , a constant.

Why this works — the core idea

When a problem says "the rate of decrease of volume is proportional to the surface area," it gives you a direct link between $dV/dt$ and $S$ . But $V$ and $S$ are both functions of the radius $r$ . So if you differentiate $V$ with respect to time, you get an expression involving $dr/dt$ . Plugging in the proportionality then lets you solve for $dr/dt$ — and if that turns out to be independent of $r$ , you've proven it's constant.

The key is that the surface area $S$ itself depends on $r^2$ , and the volume depends on $r^3$ . When you differentiate, the $r^2$ terms cancel beautifully, leaving only a constant.

Step-by-step solution

1. Write down the known formulas

For a sphere of radius $r$ :

Volume: $V = \frac{4}{3}\pi r^3$
Surface area: $S = 4\pi r^2$

2. Translate the given condition into an equation

"The rate of decrease of the volume at any instant is proportional to the surface."

"Rate of decrease" means $dV/dt$ is negative. "Proportional to the surface" means $dV/dt \propto S$ . So we write:

\frac{dV}{dt} = -k S

where $k > 0$ is the constant of proportionality. The negative sign makes $dV/dt$ negative (decreasing volume) while $k$ is positive.

Watch out

A common mistake is to forget the negative sign. If you write $dV/dt = kS$ , then $dV/dt$ would be positive — meaning volume is increasing, which contradicts "dissolving". Always check the sign against the physical situation.

3. Differentiate the volume formula with respect to time

Since $V = \frac{4}{3}\pi r^3$ , differentiate both sides using the chain rule:

\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}

4. Substitute into the proportionality equation

We have two expressions for $dV/dt$ :

4\pi r^2 \frac{dr}{dt} = -k S

But $S = 4\pi r^2$ , so:

4\pi r^2 \frac{dr}{dt} = -k (4\pi r^2)

5. Cancel the common factor

Provided $r \neq 0$ (the ball hasn't fully dissolved yet), we can divide both sides by $4\pi r^2$ :

\frac{dr}{dt} = -k

Tip

Notice that $r^2$ cancels completely — the radius disappears from the equation. This is the mathematical reason why $dr/dt$ is constant: the dependence on $r$ from both sides cancels out.

6. Interpret the result

$\frac{dr}{dt} = -k$ means the radius decreases at a constant rate $k$ (the same $k$ from the proportionality). The negative sign just tells us the radius is shrinking, not growing.

✓Final answer

\frac{dr}{dt} = -k

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 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 $100\pi$ :

$\frac{dh}{dt} = 1 \text{ cm/s}.$
Watch out

A common mistake is to forget that the radius is constant and try to differentiate $V = \pi r^2 h$ using the product rule, treating $r$ as a variable. Here $r$ is fixed, so it’s just a constant factor. If the radius were also changing (e.g., a conical tank), you’d need a different approach.

Tip

You can often avoid calculus entirely for constant-cross-section tanks: the rate of height increase is simply (volume flow rate) ÷ (cross-sectional area). Here, area = $\pi (10)^2 = 100\pi$ cm², so $\frac{dh}{dt} = \frac{100\pi}{100\pi} = 1$ cm/s. This shortcut works because the shape is a right cylinder.

✓Final answer
$The height increases at 1 cm/s, so the correct option is (C) .$
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:

$\dfrac{dV}{dt}=3(10)^2(3)=900$ cm³/s.

✓Final answer
$900\text{ cm}^3/\text{s}$
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:

$\dfrac{dV}{dt}=4\pi(1)^2\left(\dfrac12\right)=2\pi$ cm³/s.

✓Final answer
$2\pi\text{ cm}^3/\text{s}$
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:

$\frac{dC}{dt} = 2\pi \times \frac{1}{\pi} = 2 \text{ m/s}$

✓Final answer
$The rate of change of the circumference is 2 m/s (option b).$
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 $t$ :

$\frac{dC}{dt} = 2\pi \frac{dr}{dt} = 2\pi(2) = 4\pi \text{ m/s}.$

Note the rate is constant — it does not depend on the actual value of $r$ .

✓Final answer
$4\pi$
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.

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

$\dfrac{dA}{dt} = 2\pi (10)(3) = 60\pi$ cm² per second

✓Final answer
$60\pi$
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 $t$ :

$\dfrac{dC}{dt} = 2\pi\,\dfrac{dr}{dt} = 2\pi(0.7) = 1.4\pi\ \text{cm/s}.$

✓Final answer
$\boldsymbol{1.4\pi\ \text{cm/s}\ (\approx 4.4\ \text{cm/s})}$
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 $t$ : $\dfrac{dP}{dt} = 2\pi\dfrac{dr}{dt}$ .

Given $\dfrac{dr}{dt}=0.3$ cm/sec: $\dfrac{dP}{dt} = 2\pi(0.3) = 0.6\pi$ cm/sec.

✓Final answer
$0.6\pi$
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 $t$ :

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

Given $\dfrac{dr}{dt}=3$ cm/sec and $r=10$ cm:

$\dfrac{dA}{dt}=2\pi(10)(3)=60\pi$ cm²/sec.

✓Final answer
$\dfrac{dA}{dt}=60\pi$
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:

$\frac{dC}{dt} = 2\pi(0.7) = 1.4\pi \approx 4.4 \text{ cm/s}.$

✓Final answer
$Rate of increase of circumference \approx 4.4 cm/s — option (d).$
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 $S=4\pi r^2\Rightarrow \dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$ .

At $r=5$ : $\dfrac{dS}{dt}=8\pi(5)\left(\dfrac{1}{4\pi}\right)=\dfrac{40\pi}{4\pi}=10\ \text{cm}^2/\text{sec}$

✓Final answer
$\dfrac{dS}{dt}=10\ \text{cm}^2/\text{sec}$
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 $\dfrac{dx}{dt} = 2\dfrac{d(\sin x)}{dt} = 2\cos x\dfrac{dx}{dt}$

Since $\dfrac{dx}{dt}\ne0$ , divide both sides by it: $1=2\cos x \Rightarrow \cos x=\dfrac12$

$x=\dfrac\pi3$ (the standard angle with $\cos x=1/2$ ).

✓Final answer
$x=\dfrac{\pi}{3}$

←Previous section · A.2.3Principles of Mathematical Modelling →Next questionQ2 — If the area of a circle increases at a uniform rate, then prove that t…

Q.A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.

Concept understanding — Related Rates

Related Rates

The idea: quantities that change together

The core mechanism: differentiate with respect to time

The standard procedure

Worked example

Why this works — the core idea

Step-by-step solution

Method: Proving a Rate of Change Is Constant Using Related Rates

Steps

Common Mistakes

Q.A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.

Why this works — the core idea

Step-by-step solution

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