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 cm³/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:

• A. \( 225\pi \)
• B. \( 128\pi \)
• C. \( 196\pi \)
• D. \( 256\pi \)

Hint:

To solve problems involving the rate of change of volume, first express the volume in terms of the variables, then differentiate with respect to time. Don't forget to use the given rates of change and apply them correctly.

Answer 1

The Correct Option is D

To solve this problem, let us denote the radius of the spherical chocolate ball as \(R\) cm. The ice-cream layer on the chocolate ball has a uniform thickness of 1 cm, making the total radius of the chocolate ball with the ice-cream layer \((R+1)\) cm. 

We are given that the ice-cream melts at a rate of 81 cm³/min, which is the rate of change of its volume. Also, the thickness of the ice-cream layer decreases at the rate of \(\frac{1}{4\pi}\) cm/min. Using these values, we can find the surface area of the chocolate ball:

1. First, write the formula for the volume of the sphere (which represents the chocolate ball with ice-cream): \(V = \frac{4}{3}\pi (R+1)^3\).
2. The rate of change of the volume, when the thickness changes, is given by: \(\frac{dV}{dt} = 4\pi (R+1)^2 \cdot \frac{d(R+1)}{dt}\).
3. We are given that \(\frac{d(R+1)}{dt} = -\frac{1}{4\pi}\) cm/min. Substituting this into the formula, we have: \(\frac{dV}{dt} = 4\pi (R+1)^2 \cdot -\frac{1}{4\pi} = -(R+1)^2\).
4. Since the ice-cream layer alone melts (i.e., \(-\frac{dV}{dt} = 81\) cm³/min), we equate: \((R+1)^2 = 81\).
5. Solving \((R+1)^2 = 81\) gives \(R+1 = 9\), so \(R = 8\).
6. The surface area of the chocolate ball (without the ice-cream layer) is \(4\pi R^2\): \(= 4\pi (8)^2 = 256\pi\) cm².

Thus, the surface area of the chocolate ball without the ice-cream layer is \(256\pi\) cm².

Answer 2

Step 1: Define the Radii

Let the radius of the chocolate ball (without ice-cream) be \( r \) cm.

The thickness of the ice-cream layer is given as 1 cm.

Therefore, the radius of the full sphere (chocolate ball + ice-cream) is:

\( R = r + 1 \) cm

Step 2: Volume of the Ice-cream Layer

The volume of a sphere is given by the formula:

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

So, the volume of the ice-cream layer is the difference between the volume of the bigger sphere (with ice-cream) and the inner sphere (only chocolate ball):

\( V_{\text{ice-cream}} = \frac{4}{3} \pi (r + 1)^3 - \frac{4}{3} \pi r^3 \)

Step 3: Rate of Change Information

We are given the following:

• Rate of decrease of volume of the ice-cream layer: \( \frac{dV}{dt} = -81 \) cm³/min
• Rate of decrease of the outer radius \( R \): \( \frac{dR}{dt} = -\frac{1}{4\pi} \) cm/min

Step 4: Use Derivative of Volume with Respect to Time

We use the formula for the rate of change of volume of a sphere with respect to time:

\( \frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt} \)

Now plug in the values:

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

Step 5: Solve for R

Simplify the equation:

\( -81 = -R^2 \)

\( R^2 = 81 \Rightarrow R = \sqrt{81} = 9 \) cm

Step 6: Find the Radius of the Chocolate Ball

We know:

\( R = r + 1 \Rightarrow 9 = r + 1 \Rightarrow r = 8 \) cm

Step 7: Surface Area of the Chocolate Ball

The surface area \( S \) of a sphere is given by:

\( S = 4\pi r^2 \)

Substitute \( r = 8 \):

\( S = 4\pi (8)^2 = 4\pi \times 64 = 256\pi \) cm²

Final Answer:

The surface area of the chocolate ball is \( \boxed{256\pi \text{ cm}^2} \)