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 $$(in cm^{2})$$ of the chocolate ball (without the ice-cream layer) is :

A spherical chocolate ball has an ice-cream layer of uniform thickness. When the thickness is 1 cm, the ice-cream melts at 81 cm³/min and the thickness decreases at $$\frac{1}{4\pi}$$ cm/min.

Set up variables

Let r = radius of the chocolate ball and t = thickness of ice-cream layer.

The outer radius is R = r + t.

Volume of ice-cream: $$V = \frac{4}{3}\pi(r+t)^3 - \frac{4}{3}\pi r^3$$

Find the rate of change of volume

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

(since r is constant, $$\frac{d(r+t)}{dt_{time}} = \frac{dt}{dt_{time}}$$)

Note: The volume is decreasing, so $$\frac{dV}{dt_{time}} = -81$$ cm³/min

And the thickness decreases, so $$\frac{dt}{dt_{time}} = -\frac{1}{4\pi}$$ cm/min

Substitute values when t = 1 cm

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

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

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

$$r+1 = 9$$

$$r = 8$$ cm

Find the surface area of the chocolate ball

$$S = 4\pi r^2 = 4\pi(8)^2 = 256\pi \text{ cm}^2$$

The correct answer is Option 2: $$256\pi$$.