A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickne

Q: Wait, why do we treat the inner radius $r$ as a constant?

The problem specifies a 'chocolate ball' with a layer of ice-cream. The chocolate ball is the core, and it does not melt. In calculus, constants have a derivative of zero, which simplifies our life significantly!

Q: Why does the rate of change of the ice-cream volume equal the rate of change of the total volume?

Great question! The total volume $V_{total}$ is $V_{chocolate} + V_{ice}$. Since $V_{chocolate}$ is constant, $\frac{d}{d\tau}(V_{total}) = \frac{d}{d\tau}(V_{chocolate}) + \frac{d}{d\tau}(V_{ice}) = 0 + \frac{d}{d\tau}(V_{ice})$. Thus, the rate of change of the ice-cream volume is identical to the rate of change of the total volume.

Q: Do I really need to worry about the negative signs for the rates?

Yes! In physics and calculus, signs represent direction. If the volume is 'decreasing', the rate is negative. If the thickness is 'decreasing', that rate is also negative. If you ignore them, you might end up with a negative radius, which is physically impossible. Always respect the signs!

Question

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 cm, the ice-cream melts at the rate of 81 $cm^{3}/min$ and the thickness of the ice-cream layer decreases at the rate of $\frac{1}{4\pi}$ cm/min. The surface area (in cm²) of the chocolate ball (without the ice-cream layer) is:

Options:

(A) 196π

(B) 256π

(C) 225π

(D) 128π

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a kitchen, holding a perfectly spherical chocolate ball. It is a solid, unchanging core. Now, imagine coating it with a uniform layer of ice cream.

This creates a larger sphere. Let the radius of the chocolate ball be $r$ and the thickness of the ice cream layer be $t$. The total radius of this delicious system is simply $r+t$.

This is the geometric foundation of our problem. We are dealing with a sphere within a sphere, and as the ice cream melts, the outer boundary shrinks while the inner core remains steadfast.

### The Calculus of Change

Since the entire object forms a sphere, we can describe its total volume $V$ using the standard formula:

$$V = \frac{4}{3}\pi(r+t)^3$$

Now, the ice cream is melting. This means the volume $V$ and the thickness $t$ are both functions of time $	au$. To understand how they change, we must differentiate with respect to $	au$.

Using the chain rule, we find:

$$\frac{dV}{d	au} = 4\pi(r+t)^2 \frac{dt}{d	au}$$

Notice that because $r$ is constant, its derivative is zero, which is why it disappears from the rate equation. This equation is the bridge between the physical change we observe and the mathematical reality of the system.

### The Moment of Truth

We are given specific data for a single instant: the thickness $t$ is $1$ cm, the volume is decreasing at $81$ cm$^3$/min, and the thickness is decreasing at $\frac{1}{4\pi}$ cm/min.

Because these quantities are decreasing, we must assign them negative values:

$$\frac{dV}{d	au} = -81 \quad 	ext{and} \quad \frac{dt}{d	au} = -\frac{1}{4\pi}$$

Substituting these into our derivative equation, we get:

$$-81 = 4\pi(r+1)^2 \left( -\frac{1}{4\pi} ight)$$

### The Elegant Cancellation

This is where the magic happens. Look at the right side of the equation. The $4\pi$ in the numerator and the $4\pi$ in the denominator cancel out perfectly, and the negative signs on both sides also cancel.

We are left with:

$$81 = (r+1)^2$$

This is a beautiful, clean result. Taking the square root of both sides, we get $9 = r+1$, which means the radius of our chocolate ball is **$r = 8$ cm**. We have successfully worked backward from the rate of melting to the physical dimension of the object.

### Final Calculation

Finally, the problem asks for the surface area of the chocolate ball itself, without the ice cream. The formula for the surface area of a sphere is $A = 4\pi r^2$.

Substituting our value $r=8$, we calculate:

$$A = 4\pi(8)^2 = 4\pi(64) = 256\pi 	ext{ cm}^2$$

We have navigated the calculus, respected the geometry, and arrived at the solution. Remember, in JEE problems, always set up your geometry first, and the calculus will follow. The final answer is **$256\pi$ cm$^2$**.

Select Answer:

Visualized Solution

Visualizing the Layered Sphere

Total Volume of the System

Differentiating with Respect to Time

Substituting Given Values

Simplifying the Equation

Solving for the Radius

Formula for Surface Area

Calculating Final Surface Area

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Calculus of Change

The Moment of Truth

The Elegant Cancellation

Final Calculation

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