JEE Mains · Maths · STD 12 - 6. Application of derivatives
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 :
- A \(196 \pi\)
- B \(256 \pi\)
- C \(225 \pi\)
- D \(128 \pi\)
Answer & Solution
Correct Answer
(B) \(256 \pi\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{v}=\frac{4}{3} \pi \mathrm{r}^3 \\ & \frac{\mathrm{dv}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \\ & 81=4 \pi \mathrm{r}^2 \times \frac{1}{4 \pi} \\ & \mathrm{r}^2=81 \\ & \mathrm{r}=9 \end{aligned}\) surface area of chocolate…
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