A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of…
- $196 \pi$
- $256 \pi$
- $225 \pi$
- $128 \pi$
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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