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

Let the radius of the chocolate ball be \( r \) (in cm). The radius of the spherical ball with the ice-cream layer is \( r + 1 \) (since the thickness of the ice-cream layer is 1 cm). The volume of the ice-cream layer is: \[ V = \frac{4}{3}\pi \left[ (r+1)^3 - r^3 \right] \] We are told that the ice-cream melts at the rate of 81 cm³/min, i.e., \[ \frac{dV}{dt} = -81 \, \text{cm}^3/\text{min}. \] Now, differentiating the volume with respect to \( t \): \[ \frac{dV}{dt} = 4\pi (2r + 1) \frac{dr}{dt}. \] We are also told that the thickness decreases at the rate of \( \frac{1}{4\pi} \) cm/min: \[ \frac{dr}{dt} = \frac{1}{4\pi}. \] Substituting these values into the equation: \[ -81 = 4\pi (2r + 1) \times \frac{1}{4\pi}. \] Simplifying: \[ -81 = (2r + 1), \] which gives: \[ 2r + 1 = 81 \quad \Rightarrow \quad 2r = 80 \quad \Rightarrow \quad r = 40. \] The surface area of the chocolate ball is: \[ A = 4\pi r^2 = 4\pi (40)^2 = 4\pi \times 1600 = 6400\pi. \]