Sample 2025-26 Class 12 Maths Q27 Solution: A melting ball of ice: the radius decreases at a constant rate

The question

A spherical ball of ice melts in such a way that the rate at which its volume decreases at any instant is directly proportional to its surface area. Prove that the radius of the ice ball decreases at a constant rate.

Exam answer

Let $r$ be the radius of the ice ball at time $t$ ⁠, $V$ its volume and $S$ its surface area.

V = \frac43\pi r^3\quad\ldots(\text{i})

S = 4\pi r^2\quad\ldots(\text{ii})

Why?

These are the standard formulas for a sphere of radius $r$ ⁠. The ball keeps its spherical shape as it melts, so only $r$ changes with time.

Given $\frac{dV}{dt} \propto S$ ⁠, and the volume decreases, so

\frac{dV}{dt} = -kS\quad\ldots(\text{iii})

where $k$ is some positive constant.

Why?

"Directly proportional" means equal to a constant times: the rate is $k$ times $S$ ⁠.

The volume is decreasing, so $\frac{dV}{dt}$ is negative. With $k > 0$ and $S > 0$ ⁠, the minus sign makes $-kS$ negative, as it must be.

Differentiating (i) with respect to $t$ ⁠,

\begin{aligned} \frac{dV}{dt} &= \frac43\pi\cdot 3r^2\frac{dr}{dt} \\ &= 4\pi r^2\frac{dr}{dt}\quad\ldots(\text{iv}) \end{aligned}

Why?

$r$ depends on $t$ ⁠, so the chain rule gives

\frac{d}{dt}(r^3) = 3r^2\frac{dr}{dt}

Then $\frac43 \times 3 = 4$ ⁠.

\begin{aligned} \frac43\pi \cdot 3r^2\frac{dr}{dt} &= \frac{4 \times 3}{3}\pi r^2\frac{dr}{dt} \\ &= 4\pi r^2\frac{dr}{dt} \end{aligned}

From (iii) and (iv), and using (ii),

\begin{aligned} -kS &= 4\pi r^2\frac{dr}{dt} = S\frac{dr}{dt} \\ \Rightarrow \frac{dr}{dt} &= -k \end{aligned}

Why?

Both (iii) and (iv) give $\frac{dV}{dt}$ ⁠, so their right-hand sides are equal.

By (ii), $4\pi r^2$ is exactly $S$ ⁠. The ball has a positive radius, so $S \ne 0$ and we can divide both sides by $S$ ⁠.

\begin{aligned} -kS &= S\frac{dr}{dt} \\ \frac{-kS}{S} &= \frac{dr}{dt} \\ -k &= \frac{dr}{dt} \end{aligned}

The result does not involve $r$ or $t$ ⁠: the rate is the same at every instant. The minus sign shows the radius is decreasing.

$\therefore\ \dfrac{dr}{dt} = -k$ ⁠, a negative constant, so the radius of the ice ball decreases at a constant rate. Hence proved.

Where this comes from

Chapter 6, Application of Derivatives: Rate of change: a melting ball of ice.

Question type: Rates in time

How to spot it: click to see

Type 2. Rates in time: how fast is it changing now?

Spot it: words like "per second", "how fast", "at the rate of", and one moment, such as "when the radius is 10 cm".

Here: The volume decreases at a rate proportional to the surface area, so $\frac{dV}{dt} = -kS$ ⁠. Differentiate $V = \frac43\pi r^3$ with respect to $t$ to get $S\frac{dr}{dt}$ ⁠, and $\frac{dr}{dt} = -k$ comes out constant.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Q27: A melting ball of ice: the radius decreases at a constant rate