JEE Main 2025 Jan 23 Shift 2, Mathematics Q9: Some Basic Applications Of Derivatives

JEE Main2025Jan 23, Shift 2Mathematics

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
$256 π$

Solution

Let $R$ denote the outer radius of the (chocolate + ice-cream) sphere. The chocolate ball has fixed radius $R - 1$ at the instant being considered (when the layer is $1$ cm thick).

Outer volume: $V = \frac{4}{3} π R^{3}$ , so $\frac{d V}{d t} = 4 π R^{2} \frac{d R}{d t}$ .

The melting rate is $- \frac{d V}{d t} = 81$ and the layer thickness decreases at $- \frac{d R}{d t} = \frac{1}{4 π}$ (since the chocolate radius is fixed, the layer thickness equals $R -$ const, and decreases at the same rate as $R$ ).

Substitute: $81 = 4 π R^{2} \cdot \frac{1}{4 π} = R^{2}$ , so $R = 9$ . The chocolate radius is $R - 1 = 8$ .

Surface area of chocolate ball $= 4 π (8)^{2} = 256 π$ cm $^{2}$ .

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JEE Main 2025 Jan 23 Shift 2, Mathematics Q9: Some Basic Applications Of Derivatives

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