A spherical chocolate ball has a layer of ice-cream of uniform th… — Solution | JEE Mathematics

MCQMediumJEE 2025Applications of Integrals (Area)

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is $1 \, \text{cm}$ , the ice-cream melts at the rate of $81 \, \text{cm}^3/\text{min}$ and the thickness of the ice-cream layer decreases at the rate of $\frac{1}{4\pi} \, \text{cm}/\text{min}$ . The surface area (in $\text{cm}^2$ ) of the chocolate ball (without the ice-cream layer) is:

A
$225\pi$
B
$128\pi$
C
$196\pi$
D
$256\pi$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Let the radius of the chocolate ball be $R$ cm. Since the thickness of the ice-cream layer is $1 \, \text{cm}$ , the total radius becomes $R+1$ cm.

Find: The surface area of the chocolate ball without the ice-cream layer.

The volume of the spherical body with outer radius $R+1$ is

V = \frac{4}{3}\pi (R+1)^3

Differentiating with respect to time,

\frac{dV}{dt} = 4\pi (R+1)^2 \cdot \frac{d(R+1)}{dt}

Step-by-step Evaluation

We are given that the thickness decreases at the rate

\frac{d(R+1)}{dt} = -\frac{1}{4\pi}

and the ice-cream melts at the rate $81 \, \text{cm}^3/\text{min}$ , so

\frac{dV}{dt} = -81

Substituting in the derivative formula,

-81 = 4\pi (R+1)^2 \left(-\frac{1}{4\pi}\right)

Therefore,

-81 = -(R+1)^2

(R+1)^2 = 81

Hence,

R+1 = 9

and therefore

R = 8

Direct Radius Method

Let the outer radius be $\rho$ . For a sphere,

\frac{dV}{dt} = 4\pi \rho^2 \frac{d\rho}{dt}

Using

\frac{dV}{dt} = -81, \qquad \frac{d\rho}{dt} = -\frac{1}{4\pi}

we get

-81 = 4\pi \rho^2 \left(-\frac{1}{4\pi}\right) = -\rho^2

Thus,

\rho^2 = 81 \Rightarrow \rho = 9

Since the ice-cream thickness is $1 \, \text{cm}$ , the chocolate ball radius is $9-1=8$ . Hence its surface area is

4\pi (8)^2 = 256\pi

Therefore, the correct option is D.

Common mistakes

Using the radius of the full sphere directly for the required surface area. This is wrong because $9 \, \text{cm}$ is the outer radius including ice-cream. Use the chocolate ball radius $8 \, \text{cm}$ instead.
Ignoring the negative sign in the rate of change. The volume and thickness are decreasing, so both rates are negative. Keep the signs consistent before solving for the radius.
Taking the volume of only the ice-cream layer and differentiating incorrectly with respect to time. Since the inner chocolate ball radius is constant, differentiating the outer sphere volume gives the same rate change for the melting layer.

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