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 $81 \mathrm{~cm}^{3} / \mathrm{min}$ and the thickness of the ice-cream layer decreases at the rate of $\frac{1}{4 \pi} \mathrm{~cm} / \mathrm{min}$. The surface area (in $\mathrm{cm}^{2}$) of the chocolate ball (without the ice-cream layer) is :

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Application of Derivatives

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 π} cm / min

. The surface area (in

cm^{2}

) of the chocolate ball (without the ice-cream layer) is :

(A) $225 π$
(B) $128 π$
(C) $196 π$
(D) $256 π$ ✓ Correct

Answer: (D)

Solution

Let

r

be the outer radius of the sphere formed by the chocolate ball together with the ice-cream layer, and let

V = \frac{4}{3} π r^{3}

be its volume.

\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}

The melting rate is

\frac{d V}{d t} = 81 cm^{3} / min

, and the layer thickness — hence

r

, since the chocolate ball's own radius is fixed — decreases at

\frac{d r}{d t} = \frac{1}{4 π}

cm/min. At the instant the layer thickness is 1 cm:

81 = 4 π r^{2} \cdot \frac{1}{4 π} = r^{2} \Rightarrow r = 9 cm

The chocolate ball's own radius is

9 - 1 = 8

cm, so its surface area is

4 π (8)^{2} = 256 π cm^{2}

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Application of Derivatives

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