A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of…

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 :
  1. $196 \pi$
  2. $256 \pi$
  3. $225 \pi$
  4. $128 \pi$

Solution


$\begin{aligned}
& \mathrm{v}=\frac{4}{3} \pi \mathrm{r}^3 \\ & \frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \\ & 81=4 \pi \mathrm{r}^2 \times \frac{1}{4 \pi} \\ & \mathrm{r}^2=81 \\ & \mathrm{r}=9
\end{aligned}$
surface area of chocolate $=4 \pi(r-1)^2=256 \pi$

Asked in: JEE Main 2025 (23 Jan Shift 2)

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