JEE MAIN - Mathematics (2025 - 23rd January Evening Shift - No. 18)
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 :
$128 \pi$
$196 \pi$
$225 \pi$
$256 \pi$
Explanation
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$$\begin{aligned} &\begin{aligned} & \mathrm{v}=\frac{4}{3} \pi \mathrm{r}^3 \\ & \frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \\ & 81=4 \pi \mathrm{r}^2 \times \frac{1}{4 \pi} \\ & \mathrm{r}^2=81 \\ & \mathrm{r}=9 \end{aligned}\\ &\text { surface area of chocolate }=4 \pi(r-1)^2=256 \pi \end{aligned}$$
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