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 14πcm/min. The surface area (in cm2) of the chocolate ball (without the ice-cream layer) is :

v=43πr3dvdt=4πr2drdt81=4πr2×14πr2=81r=9surface area of chocolate e=4π(r−1)2=256π

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 14πcm/min. The surface area (in cm2) of the chocolate ball (without the ice-cream layer) is :

v=43πr3dvdt=4πr2drdt81=4πr2×14πr2=81r=9surface area of chocolate e=4π(r−1)2=256π

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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 $\frac{1}{4 π} cm / \min .$ The surface area (in cm²) of the chocolate ball (without the ice-cream layer) is :

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225 π

196 π

128 π

256 π

answer is D.

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Detailed Solution

$\begin{array}{l} v = \frac{4}{3} {πr}^{3} \\ \frac{dv}{dt} = 4 {πr}^{2} \frac{dr}{dt} \\ 81 = 4 {πr}^{2} \times \frac{1}{4 π} \\ r^{2} = 81 \\ r = 9 \end{array}$
surface area of chocolate $e = 4 π (r - 1)^{2} = 256 π$

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